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Simplification of Numerical Expressions - Learn and Solve Questions
Overview of Numerical Expressions
Simplifying expressions means rewriting the same algebraic expression with no like terms and in a compact manner. To simplify expressions, we combine all the like terms and solve all the given
brackets, if any, and then in the simplified expression, we will be only left with unlike terms that cannot be reduced further.
In this article, we will learn a bunch of things. We will understand what numerical means in Maths and get to know more through numerical expression examples. We will learn the popular BODMAS rule
that is used to simplify numerical expressions.
Numerical Expression Examples
A numerical expression consists of only numbers or integers. It includes basic mathematical operations such as addition, subtraction, multiplication, or division. Therefore, a numerical expression
can be defined as a combination of numbers or integers and mathematical operators.
Example: k (16 - k) + 8 (16 - k), 9 (15 - k) = 145, etc.
What Does It Mean to Simplify Numerical Expressions?
Simplifying a numerical expression means solving it in the simplest way possible. Simplification refers to reducing complex expressions into simple Maths numerical that can be solved easily with
basic arithmetics.
There are a bunch of things that you need to take care of while simplifying an expression such as the BODMAS rule, removing brackets, ordering from left to right, etc. Let us learn all of them, one
by one.
Rule of BODMAS
The rule of BODMAS is one of the popular concepts in the field of algebra. The term BODMAS is an acronym. It stands for Brackets, Order, Division, Multiplication, Addition, and Subtraction. The
acronym tells you the order in which you need to simplify the expression.
Full Form of BODMAS
Firstly, you need to resolve the brackets. The second preference is for orders or exponents. Then you have to do the division or multiplication in order from left to right, whichever comes first.
Similarly, in the end, you must solve addition or subtraction, from left to right, whichever operation comes first.
Example: 5 x 8 - 12 ÷ 6 + 4\[^{2}\]
Ans: First of all we will solve the square of 4. i.e. 16.
= 5 x 8 - 12 ÷ 6 + 16
Now as per the BODMAS rule, first of all, the division will be done, then multiplication. So, the equation will become
= 40 - 2 + 16
Now the addition and subtraction will be done in the next step, which will give the result as:
= 56 - 2
= 54
Types of Brackets Used in Expressions
Generally, three types of brackets are used in a numerical expression and are very important while knowing how to simplify expressions.
1. The First Bracket is Parentheses, which is denoted by ( ).
2. The Second Bracket is the Curly Bracket, which is denoted by { }.
3. The third Bracket is the Square Bracket, which is denoted by $\left[ \right]$.
The rule in the bracket classification is that the operation in the first bracket needs to be performed first. Then the second bracket and finally, the third.
If you are confused with all these rules, you need not worry. We have got some solved examples for you to understand the concepts better.
Solved Examples
Go through these two solved examples to understand how the BODMAS rule is applied while simplifying the numerical expression calculator. Also, notice how the brackets are removed with priority. Below
are some examples of numerical expressions with answers:
Example 1: $[8+\{6-(6 \div 2)\}] \times 4$
Ans: First of all-round brackets will be removed
$=[8+\{6-3\}] \times 4$
Then, curly brackets will be removed
$=[8+3] \times 4$
In the end, Square brackets will be removed
$=11 \times 4$ $=44$
Example 2: $12+[16-\{6+(4 \div 2)\}]$
Ans: First of all-round brackets will be removed
Then, curly brackets will be removed
In the end, Square brackets will be removed
Practice Questions
Simplify numerical expressions given below:
Q1. 15 × 3 - 15 ÷ 3
Ans: 40
Q2. $64-[\{48 \div 6\} \times 4]+8$
Ans: 40
Q3. $(9 \div 3) \times 7-5 \times 4$
Ans: 1
Q4. $22-\{8+(6 \div 2)\}$
Ans: 11
Firstly, we learned what numerical expressions are, with the help of some examples. The major focus of the article was on the important topic of simplification of numerical expressions for Class 5.
We understood the types of brackets used in these expressions.
We also got to know about the BODMAS rule and learned how to use it to simplify complex expressions by going through some solved examples. If an expression is simplified without using the BODMAS
rule, you might still get an answer but it will be incorrect. Therefore, it is necessary to follow the BODMAS rule carefully.
FAQs on Simplification of Numerical Expressions
Q1. What is the PEMDAS rule?
Ans: The PEMDAS rule is similar to the BODMAS rule. In India and the UK, what we know as the BODMAS rule is called the PEMDAS rule in the US. It stands for Parentheses, Exponents, Multiplication,
Division (from left to right), Addition, and Subtraction (from left to right).
Q2. Can power be expressed as a numerical expression?
Ans: Power has two parts: an exponent and a base. Yes, it can be expressed as a numerical expression. For example, 2 raised to power 3 is nothing but 3 times the number 2. Therefore, it can be
expressed as 2 × 2 × 2.
Q3. How do simplifying expressions and solving equations differ?
Ans: Equations refer to those statements that have an equal to "=" sign between the term(s) written on the left side and the term(s) written on the right side. Solving equations means finding the
value of the unknown variable given. On the other hand, simplifying expressions means only reducing the expression to its lowest form. It does not intend to find the value of an unknown quantity. | {"url":"https://www.vedantu.com/maths/simplification-of-numerical-expressions","timestamp":"2024-11-02T20:30:06Z","content_type":"text/html","content_length":"205831","record_id":"<urn:uuid:b321b2b2-97d0-4450-9642-9f867b0391e0>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00198.warc.gz"} |
An introduction to GARMA models
Richard Hunt
Introduction to long memory models and GARMA models.
GARMA models are a type of time series models with special properties. These models are related to the “arima” models which can be fit in R by the standard “arima” function or the “Arima” function in
the forecast package.
If you are not very familiar with ARIMA models, it might be advisable to first have a look at the excellent online book by Rob Hyndman and George Athanasopoulos:- Forecasting: Principals and Practice
GARMA models are a type of long memory model, also known as fractal models or fractionally differenced models. The reason for this is that they can model processes which have a high correlation
between observations which are far apart in time.
For instance, consider the data on the minimum level of the Nile (Tousson 1925), as measured from 622 to 1281 AD (from the longmemo package):
> library(tidyverse)
> library(forecast)
> data(NileMin,package='longmemo')
> # we'll just set the correct start year, for display purposes.
> NileMin<-ts(as.numeric(NileMin),start=622,frequency=1)
> ggtsdisplay(NileMin,lag.max=350, main='NileMin time series.', theme=theme_bw())
The first thing to note here is that the ACF (the autocorrelation function) is showing highly significant correlations between Nile measurements, even at 350 years apart!
We won’t attempt to address the question of how it is that a flood or minima level of a river flow today could be significantly related to what was happening 350 years ago, but scientists believe
this is a real affect, not an artifact of measurement. This time series has been investigated many times by researchers.
The phenomena of long memory has been identified in a number of different types of time series from sunspots (Gray, Zhang, and Woodward 1989), the timing of xrays from Cygnus X1 (Greenhough et al.
2002), inflation rates, exchange rates (Caporale and Gil-Alana 2014) and even in non-coding DNA (Lopes and Nunes 2006).
Generically, long memory models (sometimes known as ‘fractionally differenced’ models) can also have ‘short memory’ components, and a number of these have been explored. One popular type of short
memory model is the ARMA model. There are a number of R packages which can fit these models, including fracdiff, longmemo, rugarch and even forecast, among others.
These models are considered useful because they produce long range forecasts which tend to be more accurate than other short-memory models, particularly if the model assumptions are correct.
At time of writing there are few R packages on CRAN which can fit GARMA models.
Package tswge provides a function “est.garma.wge” which can fit a GARMA model, but unfortunately can only do so via a grid search technique (searching a grid of values of the likelihood function),
which tends to be slow and can result in rather inaccurate results. It also does not allow any MA terms to be fitted.
Package waveslim has functions “spp.mle” and “spp2.mle” which can identify the parameters of a white noise process only (ie a process without any short memory component), using a wavelet method.
Package VGAM to have a function called “garma” but this actually fits very different models - it fits short-memory models which can include count data. Package gsarima also fits these short memory
count models.
The models fit by the garma package are models which are not just long memory models, but where the long memory phenomena cycles in strength. An example of this is the Southern Oscillation Index.
> soi <-
+ read_fwf('soi.long.data',fwf_widths(c(5, rep(7,12)), c("year", 1:12)),col_types='nnnnnnnnnnnnn') %>%
+ gather(mth_no,soi,-year) %>%
+ mutate(mth = lubridate::make_date(year,as.numeric(mth_no),1)) %>%
+ select(mth,soi) %>%
+ arrange(mth)
> soi_ts <- ts(soi$soi,start=c(1951,2),frequency=12)
> ggtsdisplay(soi_ts, lag.max=400, main='Southern Oscillation Index', theme=theme_bw())
The first hing to note is that there are significant correlations in the ACF even up to lag 400, so this qualifies as a long memory process.
But also the ACF shows the cyclical nature which is very typical of Gegenbauer models.
When a long memory Gegenbauer model is combined with a short memory ARMA model, the result is known as a GARMA model.
GARMA stands for Gegenbauer AR-MA, and the AR and MA letters have their usual meaning in time series of “auto-regressive” and “moving average”. ‘Gegenbauer’ refers to a mathematician Leopold
Gegenbauer who lived in the 19th century and developed a series of orthogonal polynomials now called Gegenbauer polynomials, and sometimes called ‘ultraspherical’ polynomials (Szegò 1959). These
polynomials feature strongly in the mathematics of GARMA models (Gray, Zhang, and Woodward 1989), (Gray, Zhang, and Woodward 1994).
All long memory models, including GARMA models, share one thing in common. When you examine them on the frequency domain, they will show evidence of an unbounded (read: very large) peak in the
spectrum. The more traditional models - known as “short memory” models do not have such a peak - whilst the spectrum may vary up and down, a short memory model has a spectrum which has absolute
bounds beyond which it will not go.
In the chart below we illustrate two long memory processes, one which would be well modelled by an ARFIMA model and the other which would be modelled by a Gegenbauer model.
> spectrum_nilemin <- spectrum(NileMin, plot=FALSE)
> spectrum_soi <- spectrum(soi_ts, plot=FALSE)
> # now munge these lists together into a single dataframe.
> spec_df <- rbind(data.frame(freq=spectrum_nilemin$freq,
+ spec=spectrum_nilemin$spec,
+ process='NileMin'),
+ data.frame(freq=spectrum_soi$freq,
+ spec=spectrum_soi$spec,
+ process='SOI'))
> # and plot
> ggplot(filter(spec_df,freq<pi), aes(x=freq,y=spec)) +
+ geom_line() +
+ facet_wrap(.~process,scales='free_y') +
+ ggtitle('Spectrum of NileMin and SOI') +
+ ylab('Intensity') +
+ xlab(bquote('Frequency (0 -' ~ pi ~')' )) + xlim(0,pi) +
+ theme_bw()
From the above you can see that the NileMin spectrum shows an (essentially) unbounded peak at 0 which is the marker of a traditional long memory process.
However the SOI has 3 main peaks, all separated away from 0. Although these are not as large as the NileMin peaks, compared with the rest of the spectrum they are quite large, so it is not
inappropriate to at least try to model the SOI by a 3-factor (or \(k=3\)) Gegenbauer model. This was first pointed out by (Lustig, Charlot, and Marimoutou 2017).
Technical details of the model.
The GARMA model as fit by the garma package is specified as \[ \displaystyle{\phi(B)\prod_{i=1}^{k}(1-2u_{i}B+B^{2})^{d_{i}}(1-B)^{id}(X_{t}-\mu)= \theta(B) \epsilon _{t}} \] where
1. \(\phi(B)\) represents the short-memory Autoregressive component of order p,
2. \(\theta(B)\) represents the short-memory Moving Average component of order q,
3. \((1-2u_{i}B+B^{2})^{d_{i}}\) represents the long-memory Gegenbauer component (there may in general be k of these),
4. \(id\) represents integer differencing (currently only \(id\)=0 or 1 is supported).
5. \(X_{t}\) represents the observed process,
6. \(\epsilon_{t}\) represents the random component of the model - these are assumed to be uncorrelated but identically distributed variates. Generally the routines in this package will work best if
these have an approximate Gaussian distribution.
7. \(B\) represents the Backshift operator, defined by \(B X_{t}=X_{t-1}\).
When \(k=0\), then this is just a short memory model as fit by the stats “arima” function.
Fitting a short memory model
We have deliberately kept the fitting process close to that of the “arima” and “forecast::Arima” functions.
To illustrate basic usage of the routine, we will first look at fitting a simple ARIMA model to the “AirPassengers” data supplied with R. To achieve stationarity with this data, we’ll need to
seasonally difference it. The “arima” function can do this but unfortunately “garma” as yet does not fit a seasonal model - seasonality is essentially modelled by the Gegenbauer components, but we
won’t initially use that.
To ensure the results are as comparable as possible we will work with the seasonally differenced series, and specify the estimation method as ‘CSS’ for both.
> # GARMA model
> # Note in the below we specify k=0.
> # This tells the routine is not to fit a Gegenbauer/GARMA model.
> garma_mdl <- garma_fit(ap,order=c(9,1,0),k=0,method='CSS')
> summary(garma_mdl)
garma_fit(x = ap, order = c(9, 1, 0), k = 0, method = "CSS")
No Mean term was fitted.
Drift (trend) term was fitted.
Convergence Code: 0
Optimal Value found: 16150.37572243
ar1 ar2 ar3 ar4 ar5 ar6 ar7 ar8 ar9 drift
-0.32003 -0.01625 -0.13821 -0.23457 -0.01306 0.03086 -0.1504 -0.09221 0.2211 0.19353
s.e. 0.07194 0.08475 0.08014 0.06856 0.06669 0.07748 0.0744 0.07860 0.0794 0.03675
sigma^2 estimated as 123.2853: part log likelihood = -501.230781
AR Factor Table.
Coefficients of Original polynomial:
-0.3200 -0.0163 -0.1382 -0.2346 -0.0131 0.0309 -0.1504 -0.0922 0.2211
Factor Roots Abs Recip System Freq
1+1.0357B+0.8090B^2 -0.6401+-0.9090i 0.8995 0.3476
1+1.6600B+0.7690B^2 -1.0793+-0.3681i 0.8769 0.4477
1-0.4062B+0.7464B^2 0.2721+-1.1250i 0.8639 0.2122
1-1.2885B+0.6991B^2 0.9215+-0.7623i 0.8361 0.1100
1-0.6809B 1.4686 0.6809 0.0000
We note here that (Box et al. 2015) advise taking logarithms as the more correct way to model this data, however we are illustrating the similarity of methods here and feel additional code may
obscure the intention. The interested reader may care to repeat this using logs themselves.
As can be seen above, the coefficients produced are similar but not identical - the log-likelihood from the “garma” run is a larger value than that produced by the “arima” run, indicating that the
routine has in fact found a (slightly) better solution (estimating parameters like these involves non-linear optimisation - “arima” uses R’s built-in optimiser called “optim”; garma however by
default uses “solnp” from the Rsolnp package).
Fitting a GARMA model.
In this section we’ll look at fitting a GARMA model to the Sunspots data, which is supplied with R. This data has been analysed many times in the literature; the first time with a GARMA model was by
(Gray, Zhang, and Woodward 1989). Generally, authors have used a standard subset of this data from 1749 to 1924.
The Sunspot data consists of counts of sunspots as observed over a considerable period. In terms of ARIMA models, (Box et al. 2015) suggest an AR(9) model may be best.
First we fit an ARIMA model to this data:
> data(sunspot.year)
> # Next we subset the data to ensure we are using the years used in the literature.
> sunspots <- ts(sunspot.year[49:224],start=1749,end=1924)
> # Now as in Gray et al 1989 we fit a garma_fit(1,0) model:
> sunspots_arima_mdl <- arima(sunspots, order=c(9,0,0),method='CSS')
> summary(sunspots_arima_mdl)
arima(x = sunspots, order = c(9, 0, 0), method = "CSS")
ar1 ar2 ar3 ar4 ar5 ar6 ar7 ar8 ar9 intercept
1.2419 -0.4976 -0.1406 0.1943 -0.1767 0.0979 -0.1139 0.1435 0.0570 45.1427
s.e. 0.0753 0.1195 0.1245 0.1247 0.1245 0.1234 0.1266 0.1228 0.0766 5.6748
sigma^2 estimated as 213.5: part log likelihood = -721.72
Training set error measures:
ME RMSE MAE MPE MAPE MASE ACF1
Training set -0.0002709438 14.23214 10.49252 -Inf Inf 0.6405913 -0.001348032
Next we look at a GARMA model, but instead of 9 AR parameters, we will only fit 1.
> library(garma)
> data(sunspot.year)
> # Next we subset the data to ensure we are using the years used in the literature.
> sunspots <- ts(sunspot.year[49:224],start=1749,end=1924)
> # Now as in Gray et al 1989 we fit a garma_fit(1,0) model:
> sunspots_garma_mdl <- garma_fit(sunspots, order=c(1,0,0),k=1,method='CSS')
> summary(sunspots_garma_mdl)
garma_fit(x = sunspots, order = c(1, 0, 0), k = 1, method = "CSS")
Mean term was fitted.
No Drift (trend) term was fitted.
Convergence Code: 0
Optimal Value found: 40145.33980554
intercept u1 fd1 ar1
45.13719 0.847533 0.4228 0.4937
s.e. 0.02675 0.007103 0.1260 0.1119
Gegenbauer frequency: 0.0890
Gegenbauer Period: 11.2305
Gegenbauer Exponent: 0.4228
sigma^2 estimated as 228.0985: part log likelihood = -727.553615
AR Factor Table.
Coefficients of Original polynomial:
Factor Roots Abs Recip System Freq
1-0.4937B 2.0257 0.4937 0.0000
For the GARMA model, we have specified method=‘CSS’ to ensure we are using a method as close as possible to that used by (Gray, Zhang, and Woodward 1989). The “garma” function uses a frequency-domain
method known as the “Whittle” method by default, since this method not only produces very accurate results very quickly, but also has a lot of theoretical results available to support its use - for
example (Giraitis, Hidalgo, and Robinson 2001).
The following table compares the values found by Gray et al and by the “garma” function:
intercept 44.78 45.1350
u 0.85 0.8475
d 0.42 0.4228
ar1 0.49 0.4937
As you can see, the results are quite close (Gray et al only published their results to 2 decimal places).
Notice also that the routine displays the Gegenbauer Period - in this case 11.2 years - which corresponds nicely with the “known” 11 year sunspot cycle.
Also shown is the degree of fractional differencing and the Fractional Dimension of the original series.
Whilst this model does fit the data quite well, is quite simple and does not produce very effective forecasts. A better model is the garma_fit(8,0) model, which was also examined by (Gray, Zhang, and
Woodward 1989), however to compare with the ARIMA model, we can fit the same number of AR terms - 9 - and then forecast below, for the next sunspot cycle.
> library(yardstick)
> (sunspots_garma_mdl <- garma_fit(sunspots, order = c(9, 0, 0), k = 1))
garma_fit(x = sunspots, order = c(9, 0, 0), k = 1)
Mean term was fitted.
No Drift (trend) term was fitted.
intercept u1 fd1 ar1 ar2 ar3 ar4 ar5 ar6 ar7 ar8 ar9
45.03 0.8413 0.1213 0.9994 -0.2925 -0.16877 0.18380 -0.1445 0.03879 0.01112 0.04964 0.09302
s.e. 136.53 0.0539 0.2833 0.4598 0.3662 0.07609 0.09225 0.1093 0.08272 0.07614 0.07520 0.07293
Gegenbauer frequency: 0.0909
Gegenbauer Period: 11.0000
Gegenbauer Exponent: 0.1213
sigma^2 estimated as 34.3504: approx. log likelihood = -342.309594, aic = 710.619187
AR Factor Table.
Coefficients of Original polynomial:
0.9994 -0.2925 -0.1688 0.1838 -0.1445 0.0388 0.0111 0.0496 0.0930
Factor Roots Abs Recip System Freq
1-1.4790B+0.8427B^2 0.8776+-0.6454i 0.9180 0.1009
1-0.9014B 1.1094 0.9014 0.0000
1-0.4277B+0.6241B^2 0.3427+-1.2186i 0.7900 0.2064
1+0.5167B+0.4431B^2 -0.5830+-1.3844i 0.6657 0.3134
1+1.2920B+0.4428B^2 -1.4588+-0.3606i 0.6655 0.4614
> compare_df <- data.frame(yr=1925:1935,
+ Actuals=as.numeric(sunspot.year[225:235]),
+ ARIMA=forecast(sunspots_arima_mdl,h=11)$mean,
+ GARMA=forecast(sunspots_garma_mdl,h=11)$mean)
> cat(sprintf('\n\nEvaluating Test set data from 1925 to 1936.\n\nARIMA RMSE: %.2f\nGARMA RMSE: %.2f\n',
+ yardstick::rmse(compare_df,Actuals,ARIMA)$.estimate,
+ yardstick::rmse(compare_df,Actuals,GARMA)$.estimate))
Evaluating Test set data from 1925 to 1936.
ARIMA RMSE: 13.59
GARMA RMSE: 11.73
The RMSE for the ARIMA model is 13.59 and the RMSE for the GARMA model is 11.73 - which shows in terms of future predictions the GARMA model has a smaller error.
But visually there appears to be little difference between the models.
> arima_df <- data.frame(yr=1925:1935, sunspots=forecast(sunspots_arima_mdl,h=11)$mean,grp='AR(11) forecasts')
> garma_df <- data.frame(yr=1925:1935, sunspots=forecast(sunspots_garma_mdl,h=11)$mean,grp='garma_fit(1),k=1 forecasts')
> actuals_df <- data.frame(yr=1749:1935,sunspots=as.numeric(sunspot.year[49:235]),grp='Actuals')
> df <- rbind(actuals_df,garma_df,arima_df)
> ggplot(df,aes(x=yr,y=sunspots,color=grp)) + geom_line() +
+ scale_color_manual(name='',values=c('Actuals'='darkgray','AR(11) forecasts'='darkgreen','garma_fit(1),k=1 forecasts'='blue')) +
+ xlab('') + ylab('Sunspot Counts') +
+ geom_vline(xintercept=1924,color='red',linetype='dashed',size=0.3) +
+ theme_bw() +
+ ggtitle('Sunspot Forecasts: Comparing ARIMA(9,0,0) and garma_fit(9,0),k=1')
Box, George EP, Gwilym M Jenkins, Gregory C Reinsel, and Greta M Ljung. 2015. Time Series Analysis : Forecasting and Control. John Wiley & Sons Inc., Hoboken New Jersey.
Caporale, G, and L Gil-Alana. 2014. “Long-Run and Cyclical Dynamics in the US Stock Market.” Journal of Forecasting 33: 147–61.
Giraitis, L, J Hidalgo, and P Robinson. 2001. “Gaussian Estimation of Parametric Spectral Density with Unknown Pole.” The Annals of Statistics 29 (4): 987–1023.
Gray, H, N Zhang, and W Woodward. 1989. “On Generalized Fractional Processes.” Journal of Time Series Analysis 10 (3): 233–57.
———. 1994. “On Generalized Fractional Processes - a Correction.” Journal of Time Series Analysis 15 (5): 561–62.
Greenhough, J, S Chapman, S Chaty, R Dendy, and G Rowlands. 2002. “Characterising Anomalous Transport in Accretion Disksfrom x-Ray Observations.” Astronomy and Astrophysics 385: 693–700.
Lopes, S, and M Nunes. 2006. “Long Memory Analysis in DNA Sequences.” Physica A 361: 569–88.
Lustig, A, P Charlot, and V Marimoutou. 2017. “The Memory of ENSO Revisited by a 2-Factor Gegenbauer Process.” International Journal of Climatology 37: 2295–2303.
Szegò, G. 1959. Orthogonal Polynomials. AMS, New York.
Tousson, O. 1925. Mémoire Sur l’histoire Du Nil; Mémoire de l’institut d’egypte. | {"url":"https://www.stats.bris.ac.uk/R/web/packages/garma/vignettes/introduction.html","timestamp":"2024-11-05T02:35:34Z","content_type":"text/html","content_length":"280387","record_id":"<urn:uuid:361ef603-769c-438c-ae37-a2bcc8374d27>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00643.warc.gz"} |
greek mathematician crossword clue
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Time evolution of effective central cha
SciPost Submission Page
Time evolution of effective central charge and signatures of RG irreversibility after a quantum quench
by Axel Cortes Cubero
This Submission thread is now published as
Submission summary
Authors (as registered SciPost users): Axel Cortes Cubero
Submission information
Preprint Link: https://arxiv.org/abs/1707.05671v3 (pdf)
Date accepted: 2018-03-19
Date submitted: 2018-02-26 01:00
Submitted by: Cortes Cubero, Axel
Submitted to: SciPost Physics
Ontological classification
Academic field: Physics
• Condensed Matter Physics - Theory
Specialties: • High-Energy Physics - Theory
• Statistical and Soft Matter Physics
Approach: Theoretical
At thermal equilibrium, the concept of effective central charge for massive deformations of two-dimensional conformal field theories (CFT) is well understood, and can be defined by comparing the
partition function of the massive model to that of a CFT. This temperature-dependent effective charge interpolates monotonically between the central charge values corresponding to the IR and UV fixed
points at low and high temperatures, respectively. We propose a non-equilibrium, time-dependent generalization of the effective central charge for integrable models after a quantum quench, $c_{\rm
eff}(t)$, obtained by comparing the return amplitude to that of a CFT quench. We study this proposal for a large mass quench of a free boson, where the charge is seen to interpolate between $c_{\rm
eff}=0$ at $t=0$, and $c_{\rm eff}\sim 1$ at $t\to\infty$, as is expected. We use our effective charge to define an "Ising to Tricritical Ising" quench protocol, where the charge evolves from $c_{\rm
eff}=1/2$ at $t=0$, to $c_{\rm eff}=7/10$ at $t\to\infty$, the corresponding values of the first two unitary minimal CFT models. We then argue that the inverse "Tricritical Ising to Ising" quench is
impossible with our methods. These conclusions can be generalized for quenches between any two adjacent unitary minimal CFT models. We finally study a large mass quench into the "staircase model"
(sinh-Gordon with a particular complex coupling). At short times after the quench, the effective central charge increases in a discrete "staircase" structure, where the values of the charge at the
steps can be computed in terms of the central charges of unitary minimal CFT models. When the initial state is a pure state, one always finds that $c_{\rm eff}(t\to\infty)\geq c_{\rm eff}(t=0)$,
though $c_{\rm eff}(t)$, generally oscillates at finite times. We explore how this constraint may be related to RG flow irreversibility.
Author comments upon resubmission
I thank the referees for their time and work reviewing this manuscript. Attached is the newest version of this manuscript, which I hope addresses the issues that have been raised.
First, it seems Referee 1 was satisfied with the previous version and no changes were suggested. I am happy to see that in his/her opinion the quality of the manuscript has already improved, and I
thank again the referee for their work.
In this version I have incorporated several changes, in attempt to answer for some of the issues raised by Referee 2. Here I will review the modifications step by step, corresponding to the items
enumerated in the referee's report (see the list of changes below).
List of changes
1) I have made many changes throughout the paper regarding the discussion of "RG irreversibility", including several mentions in the introduction and the conclusion sections. The main theme has been
to soften the tone of some claims. When the value of the effective central charge exhibits some apparent irreversibility properties, c_{\rm eff}(\infty)\geq c_{\rm eff}(0), it is commented that this
is suggestive and may point to a connection with the concept of RG flow and irreversibility, by comparing to the known properties of the equilibrium c-function. This is in contrast with previous
versions where it may seem like I claim "this IS RG irreversibility", and made it sound like a claim of a rigorous proof.
Accordingly, I have modified the title with the phrase "signatures of RG irreversibility", suggesting that this is not a direct proof of RG irreversibility, but that arguments are presented to
suggest what we see may be connected with it. I have opted not to entirely remove "RG irreversibility" from the title, since I feel that would be slightly dishonest to the paper, since this is a
subject that is widely discussed, and a central theme in the paper. I hope that this softening the tone of the claims, and the title, will be enough to reflect the nature of what is actually
discussed in the paper.
2)The discussion regarding the Calabrese-Cardy initial states has been enhanced. I attempt to make it more clear that this is indeed just a simplified model that is only useful as a first step. A new
paragraph augmenting this discussion was added after Eq. 15, just after introducing these states. There is also more discussion added on this at the end of the same section. I have also added in the
conclusions section a brief discussion of how one could a attempt a generalization to other kinds of initial states, which do not lead to effective thermalization. This should make it clear that it
should be conceptually possible to define similarly an effective central charge for wider classes of quantum quenches, the only limitation is a practical computational one.
3)The discussion of the relation between the return amplitude and the partition function on the strip is clarified. Indeed I agree, that the roles of the space and time coordinates are exchanged in
this analogy, and this was not very clear in my previous discussion. This discussion has been modified, and hopefully the relation is now clear.
4) Together with the modifications in the discussion of CC initial states, the perhaps misleading claim has been corrected, along with a more careful discussion of when thermalization is expected to
5) The issue of the limits and conditions which lead effective thermalization has been more carefully addressed, including the limit \lim_{(m_0/m)\to\infty}, throughout the paper.
6)The labels in the plots have been modified.
7) There have been several modifications in the conclusions. Any claims regarding RG irreversibility have been softened. It is explained that the results are suggestive at some connection with this
concept, given how the effective central charge is related to the known equilibrium c-function, but any hard claims that make it seem like we have found some rigorous proof are dropped. The
discussion regarding the finite-time values of the central charge is also enhanced, making it clear that it is difficult to see if and how it may be connected with RG flow, and making it clear that
for now one can only speculate about this. As mentioned before, the discussion of initial states was also enhanced in the conclusions.
Published as SciPost Phys. 4, 016 (2018) | {"url":"https://scipost.org/submissions/1707.05671v3/","timestamp":"2024-11-04T05:21:58Z","content_type":"text/html","content_length":"35738","record_id":"<urn:uuid:e15166c8-9f1b-4df2-817d-e4b0704db785>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00532.warc.gz"} |
Complex Analysis
Subject MAST30021 (2015)
Note: This is an archived Handbook entry from 2015.
Credit Points: 12.5
Level: 3 (Undergraduate)
This subject has the following teaching availabilities in 2015:
Semester 1, Parkville - Taught on campus.
Pre-teaching Period Start not applicable
Teaching Period 02-Mar-2015 to 31-May-2015
Assessment Period End 26-Jun-2015
Last date to Self-Enrol 13-Mar-2015
Census Date 31-Mar-2015
Last date to Withdraw without fail 08-May-2015
Dates &
Locations: Semester 2, Parkville - Taught on campus.
Pre-teaching Period Start not applicable
Teaching Period 27-Jul-2015 to 25-Oct-2015
Assessment Period End 20-Nov-2015
Last date to Self-Enrol 07-Aug-2015
Census Date 31-Aug-2015
Last date to Withdraw without fail 25-Sep-2015
Timetable can be viewed here. For information about these dates, click here.
Contact Hours: 3 x one hour lectures per week, 1 x one hour practice class per week
Total Time Commitment:
Time Commitment:
Estimated total time commitment of 170 hours
One of
Study Period Commencement:
Prerequisites: Credit Points:
Semester 2
And any other second year level subject from the Department of Mathematics and Statistics.
Corequisites: None
Background None
Non Allowed
For the purposes of considering request for Reasonable Adjustments under the Disability Standards for Education (Cwth 2005), and Student Support and Engagement Policy, academic
Core requirements for this subject are articulated in the Subject Overview, Learning Outcomes, Assessment and Generic Skills sections of this entry.
Requirements: It is University policy to take all reasonable steps to minimise the impact of disability upon academic study, and reasonable adjustments will be made to enhance a student's
participation in the University's programs. Students who feel their disability may impact on meeting the requirements of this subject are encouraged to discuss this matter with a
Faculty Student Adviser and Student Equity and Disability Support: http://services.unimelb.edu.au/disability
Dr Anita Ponsaing, Prof Peter Forrester
Complex analysis is a core subject in pure and applied mathematics, as well as the physical and engineering sciences. While it is true that physical phenomena are given in terms
of real numbers and real variables, it is often too difficult and sometimes not possible, to solve the algebraic and differential equations used to model these phenomena without
introducing complex numbers and complex variables and applying the powerful techniques of complex analysis.
Subject Overview:
Topics include:the topology of the complex plane; convergence of complex sequences and series; analytic functions, the Cauchy-Riemann equations, harmonic functions and
applications; contour integrals and the Cauchy Integral Theorem; singularities, Laurent series, the Residue Theorem, evaluation of integrals using contour integration, conformal
mapping; and aspects of the gamma function.
At the completion of this subject, students should understand the concepts of analytic function and contour integral and should be able to:
• apply the Cauchy-Riemann equations
Learning Outcomes: • use the complex exponential and logarithm
• apply Cauchy’s theorems concerning contour integrals
• apply the residue theorem in a variety of contexts
• understand theoretical implications of Cauchy’s theorems such as the maximum modulus principle, Liouville’s Theorem and the fundamental theorem of algebra
Three or four written assignments due at regular intervals during semester amounting to a total of up to 50 pages (20%), and a 3-hour written examination in the examination period
Assessment: (80%).
Prescribed Texts: None
Recommended Texts: Spiegel, Lipschutz, Schiller and Spellman, Schaum's Outline of Complex Variables, 2nd edition, McGraw-Hill, 2009
This subject potentially can be taken as a breadth subject component for the following courses:
Breadth Options: You should visit learn more about breadth subjects and read the breadth requirements for your degree, and should discuss your choice with your student adviser, before deciding on
your subjects.
Fees Information: Subject EFTSL, Level, Discipline & Census Date
In addition to learning specific skills that will assist students in their future careers in science, they will have the opportunity to develop generic skills that will assist
them in any future career path. These include:
Generic Skills: • problem-solving skills: the ability to engage with unfamiliar problems and identify relevant solution strategies;
• analytical skills: the ability to construct and express logical arguments and to work in abstract or general terms to increase the clarity and efficiency of analysis;
• collaborative skills: the ability to work in a team;
• time-management skills: the ability to meet regular deadlines while balancing competing commitments.
Notes: This subject is available for science credit to students enrolled in the BSc (both pre-2008 and new degrees), BASc or a combined BSc course.
Applied Mathematics
Applied Mathematics
Applied Mathematics
Applied Mathematics
Applied Mathematics (specialisation of Mathematics and Statistics major)
Discrete Mathematics / Operations Research
Discrete Mathematics / Operations Research
Related Majors/ Discrete Mathematics / Operations Research
Minors/ Discrete Mathematics / Operations Research
Specialisations: Discrete Mathematics and Operations Research (specialisation of Mathematics and Statistics major)
Mathematical Physics
Pure Mathematics
Pure Mathematics
Pure Mathematics
Pure Mathematics
Pure Mathematics (specialisation of Mathematics and Statistics major)
Science-credited subjects - new generation B-SCI and B-ENG.
Selective subjects for B-BMED
Download PDF version. | {"url":"https://archive.handbook.unimelb.edu.au/view/2015/mast30021/","timestamp":"2024-11-07T00:19:27Z","content_type":"text/html","content_length":"10708","record_id":"<urn:uuid:665f5f6d-f5e7-41ca-a2b6-0ab8793bc153>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00746.warc.gz"} |
Declarative Programming
This course explores the use of "declarative languages" for programming. Such programming languages enable specifying the characteristics of a solution to a problem, rather than specifying the steps
that have to be followed to come to this solution. The system itself will find the solution using a general-purpose problem solving procedure. In most declarative programming languages, the problem
is specified by means of assertions in a logic. The problem solving strategy is a proof procedure which can be queried for answers that follow from the assertions. We will explore the theoretical
underpinnings of the declarative programming paradigm, the diveristy among the programming languages that different logics and proof procedures gives rise to and their applications in the domain of
artificial intelligence for which they are particulary well-suited.
The official course description can be found here and here.
The exam consists of an oral test with written preparation about the entire course (theory and exercises) and an oral defense of an individually completed programming project. The end result is
calculated as the average of the results on both parts. If one of both results is 7 or less, however, the end result cannot exceed 7.
When and Where
The theory is lectured every Wednesday from 13:00 till 15:00 in room D.2.19 (in room F.4.104 on the 1st of December).
The exercise sessions are organized after the lectures (from 15:00 till 17:00) of October 6th, 13th, 20th, 27th and November 3rd in room E.1.2.
Course Material
Peter Flach, "Simply Logical: Intelligent Reasoning by Example", J. Wiley & Sons, 1994 , ISBN 0471 94152 2
This book is currently out of print, but available for free on-line.
The material for the individual lectures can be found below (or here as one large [PDF] or one with 4 slides per page [PDF]).
Theoretical Backgrounds
Logic programming and Prolog
Blind and informed search through state spaces and proving as a search process
Natural language processing using definite clause grammars
Completed on 17/11/2010
□ Slides: [PDF]
□ Further reading: Chapter 7 of Simply Logical, for those interested in natural language processing: Sections 2.7 and 3.7 of the digital edition of the book Prolog and Natural Language Analysis
, for those interested in parsers and compilers for programming languages: the 1987 paper Parsing and Compiling Using Prolog
Efficiency issues, best practices and coding guidelines
Group discussion on 1/12/2010
Reasoning with incomplete information
Completed on 8/12/2010
□ Slides: default reasoning and abduction [PDF], program completion [PDF], inductive reasoning [PDF]
□ Further reading: Chapter 8 and 9 of Simply Logical
Interesting loose ends (will not be asked on the exam)
Completed on 15/12/2010
□ Slides: meta-interpreters for programming with quantified and qualified truth, programming with constraints on integer domains, implicit parallel evaluation, software engineering applications
Additional Material
□ I maintain a list of logic programming links on delicious.
□ The Prolog Dictionary is valuable resource for starting logic programmers.
□ Your programming project is likely to benefit from these Coding Guidelines for Prolog. Refactoring for Prolog Programs lists some interesting refactorings for Prolog (e.g., renaming a
predicate, reordering its arguments, replacing particular uses of cut by if-then-else) that can improve the readibility, structure and even the performance of your code. ViPress automates
these refactorings as a plugin for the VIM editor. | {"url":"http://prog.vub.ac.be/~cderoove/declarative_programming/","timestamp":"2024-11-11T17:09:07Z","content_type":"application/xhtml+xml","content_length":"13003","record_id":"<urn:uuid:ead05894-ef3c-4c2f-a2a7-486a2e592918>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00689.warc.gz"} |
Section 2.1 and 2.2 in Matter and Interactions (4th edition)
The Momentum Principle
The motion of a system is governed by the Momentum Principle. This principle describes how a system changes its motion when it experiences a net force. We observe that when objects move in a straight
line at constant speed they experience no net force. This observation is critical to our understanding of motion (This observation is often called "Newton's First Law of Motion").
In these notes, you will be introduced to the idea of a system, net force, and how a system's momentum and the net force it experiences are related (i.e., through "Newton's Second Law of Motion"). In
another set of notes, you find a few useful formula for when the net force acting on a system is a constant vector (fixed magnitude and direction).
Lecture Video
System and Surroundings
You can consider a single object or a collection of objects to be a “system.” Anything that you choose to not be in your system exists in the “surroundings.” In mechanics, you will choose systems by
considering what objects you want to predict or explain the motion of. That is, the choice of system is arbitrary to the extent that you only care about predicting or explaining the motion of objects
in your system.
Through interactions with the surroundings, systems can change their momentum, energy, angular momentum, and entropy. For the time being, you will work with systems that consist of a single-particle,
and you will consider only how single particles change their momentum.
An Example: A fan cart changing its momentum
Consider the video below where a fan cart is released from rest and moves to the right.
If we choose the system to be the fan cart, you can observe that the momentum of that system changes (the fan cart accelerates to the right). This is due to the interaction between the air
(surroundings) and the fan blades (part of the system), namely, that the air exerts a force on the fan blades. As a result, the fan cart increases its momentum.
The Momentum Principle
The Momentum Principle is one of three fundamental principles of mechanics and is also known as "Newton's Second Law of Motion". No matter what system you choose the Momentum Principle will correctly
predict the motion of that system. It is the quantitative form of Newton's First Law; it tells you precisely how the momentum (and thus the velocity) of an system will evolve when it experiences a
net force.
If a system experiences a net force, it can experience either:
• a change in the magnitude of its momentum,
• a change in the direction of its momentum, or
• a simultaneous change in the magnitude and the direction of its momentum
Mathematically, the Momentum Principle states:
$$\Delta \vec{p} = \vec{p}_f - \vec{p}_i = \vec{F}_{net,avg} \Delta t$$
which you can think of as the change in a system's momentum = the average net force acting on the system multiplied by the time interval over which the net force acts. In this formulation of the
Momentum Principle, it must be that the time interval over which the net force acts is sufficiently small that the net force is can be approximated as constant. This is why this net force in this
principle is the average net force. Notice that this is a vector principle.
If you divide both sides of this version of the Momentum Principle by this time interval ($\Delta t$) and take the limit as the time interval goes to zero, ($\Delta t \rightarrow 0$), you obtain the
exact definition of the net force acting on the system at any instant,
$$\lim_{\Delta t \rightarrow 0} \dfrac{\Delta \vec{p}}{\Delta t} = \vec{F}_{net,avg} \longrightarrow \vec{F}_{net} = \dfrac{d\vec{p}}{dt}$$
Net Force
A force is a vector that quantifies the interactions between two objects. The units of force in SI are Newtons (N). 1 Newton is equal to 1 kilogram-meter-per-second squared (1 N = 1 $\dfrac{kg\:m}{s^
There are two types of forces that you will work with in mechanics: gravitational forces and electrostatic forces. As you will learn, all interactions that you will consider in mechanics are a result
of objects either having mass and, thus, attracting gravitationally, or being charged, and thus, interacting through electrical repulsion or attraction.
Systems might interact with several objects in their surroundings, and thus, experience a variety of forces. Fortunately to make predictions of the motion, the Momentum Principle only requires that
you know the net force.
The Net Force is the vector sum of all forces acting (at an instant) on a system as due to the systems' surroundings.
Mathematically, we can represent this sum using vector notation:
$$\vec{F}_{net} = \vec{F}_{1} + \vec{F}_{2} + \vec{F}_{3} + \dots$$
where each interaction/force (at an instant) is counted as a specific $\vec{F}_{i}$. These interactions may be “field interactions” (e.g., the gravitational field) or “contact interactions” (e.g.,
the normal force or the frictional force). Contact interactions are the result of the electromagnetic field and are, thus, truly field interactions (as all interactions are).
Impulse is the product of a force and a time interval over which that force acts, which is mathematically equivalent to the change in momentum (Impulse = $\vec{J} \equiv \vec{F} \Delta t$).
Sometimes, you might find it useful to think about the impulse applied to a system as being responsible for the change in momentum of the system. An impulse may be calculated for each force (e.g.,
impulse delivered by the gravitational force) or the total force (i.e., the “net” impulse applied to the system). | {"url":"https://msuperl.org/wikis/pcubed/doku.php?id=183_notes:momentum_principle","timestamp":"2024-11-05T19:22:09Z","content_type":"application/xhtml+xml","content_length":"41438","record_id":"<urn:uuid:dad35b19-b33e-4acf-a581-a7fa064055bf>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00276.warc.gz"} |
Vesselin Petkov, Introduction to Spacetime Physics (Minkowski Institute Press, Montreal), 250 pages (Lecture Notes Series, Volume 8).
ISBN: 978-1-927763-21-6 (ebook) - $7
ISBN: 978-1-927763-08-7 (softcover) - $18
ISBN: 978-1-998902-41-5 (hardtcover) - $26
Will appear in 2024
This textbook is built up from course notes on special relativity which was part of an advanced undergraduate course regularly given at the Physics Department of Concordia University in Montreal and
from notes used in a one-semester inter-university seminar on general relativity for advanced undergraduate and graduate students, which was given twice - in the fall semesters of 1994 and 1995 - and
attended mostly by graduate physics and mathematics students (and several colleagues) from the four Montreal Universities.
The major feature which makes this book fundamentally different from the existing textbooks on the subject is that by strictly following Minkowski's approach of geometrizing physics, gravitational
phenomena are shown to be nothing more than manifestations of the non-Euclidean geometry of spacetime, not manifestations of a physical (gravitational) interaction. In other words, the profound
physical meaning of Einstein's general relativity is that gravitation is not a physical interaction (as Minkowski, had he been alive in 1915, would have probably demonstrated, and as in 1921
Eddington even mentioned it explicitly - "gravitation as a separate agency becomes unnecessary") and therefore no gravitational energy and momentum are involved in gravitational phenomena.
Another distinct feature of the textbook is that it follows the opposite of the (in)famous "shut up and calculate" methodology (designed to suppress questions about open questions in quantum physics
in order to make its very learning possible) -- it follows an approach which clearly puts the emphasis on the physics behind the mathematical formalism. This approach enables students to overcome the
conceptual challenges of understanding the counter-intuitive properties of spacetime and to prepare them for more advanced courses on the subject. The main idea is to ensure that students have
perfect understanding of what they calculate. Another advantage of this approach is that it also enables students to deal adequately and confidently with misconceptions and controversies in spacetime
physics, e.g. the controversy over (rather the assault on) the concept of relativistic mass and the question of whether general relativity predicts the existence of singularities inside black holes
(a question recently revived by Roy Kerr [On 1 December 2023 Roy Kerr (who discovered the Kerr geometry of a rotating black hole) posted a very important paper "Do Black Holes have Singularities?"
(arXiv:2312.00841[gr-qc]) where he demonstrated that "There is no proof that black holes contain singularities when they are generated by real physical bodies."].
As this textbook does not require previous knowledge beyond calculus and as all new physical concepts and mathematical formalism are gently introduced, it can be also used for self-study by anyone
interested in the amazing physics of spacetime and its implications.
The textbook includes 7 appendices and a Glossary:
H. Minkowski, Space and Time (Minkowski's groundbreaking lecture)
Relativistic Mass is an Experimental Fact
Accelerating Spaceships Paradox and Physical Meaning of Length Contraction
Propagation of Light in Non-Inertial Reference Frames
Acceleration is Worldline Curvature (in flat and curved spacetime)
Common Misconceptions in Spacetime Physics
Open Questions in Spacetime Physics | {"url":"https://minkowskiinstitute.com/mip/books/spacetime.html","timestamp":"2024-11-05T11:05:39Z","content_type":"text/html","content_length":"24324","record_id":"<urn:uuid:6b08f883-838e-4b1d-853b-6cb3fb1f19d4>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00173.warc.gz"} |
(x When can a quadratic function have an inverse?
✔ x A When its range is restricted
I x B Whe...
Solved on Dec 07, 2023
When can a quadratic function have an inverse? A) When its range is restricted B) When its domain is restricted C) When all $x$-values are $>-1$ D) When all $y$-values are $>-1$
STEP 1
1. A quadratic function is a function of the form $f(x) = ax^2 + bx + c$ where $a$, $b$, and $c$ are constants and $a eq 0$.
2. An inverse function, denoted as $f^{-1}(x)$, exists if and only if the original function is one-to-one (bijective), meaning every $y$-value has exactly one corresponding $x$-value.
3. A function's domain is the set of all possible $x$-values, and its range is the set of all possible $y$-values that the function can output.
STEP 2
Understand the concept of a one-to-one function.
A function is one-to-one if it passes the horizontal line test, which means that no horizontal line intersects the graph of the function at more than one point.
STEP 3
Recognize the shape of a quadratic function's graph.
The graph of a quadratic function is a parabola. If $a > 0$, the parabola opens upwards, and if $a < 0$, it opens downwards.
STEP 4
Analyze the one-to-one nature of a quadratic function.
Because a parabola is symmetric with respect to its axis of symmetry (the line that passes through the vertex), it will not pass the horizontal line test for its entire domain. This means that a
quadratic function is not one-to-one over its entire domain.
STEP 5
Consider the effect of restricting the domain.
By restricting the domain of a quadratic function to one side of the vertex (either left or right), we can make the function one-to-one because it will then pass the horizontal line test.
STEP 6
Consider the effect of restricting the range.
Restricting the range of a quadratic function does not ensure that the function becomes one-to-one. Even if the range is restricted, the function can still fail the horizontal line test.
STEP 7
Evaluate the given options in the context of a quadratic function having an inverse.
A. When its range is restricted - This option is incorrect because restricting the range does not guarantee a one-to-one function. B. When its domain is restricted - This option is correct because
restricting the domain to one side of the vertex can make the function one-to-one. C. When all the $x$-values are greater than -1 - This option is incorrect because the specific value of $x$ does not
determine if the function is one-to-one. D. When all the $y$-values are greater than -1 - This option is incorrect because the specific value of $y$ does not determine if the function is one-to-one.
STEP 8
Select the correct answer.
The correct answer is B: When its domain is restricted.
A quadratic function can have an inverse when its domain is restricted to ensure that it is a one-to-one function. | {"url":"https://studdy.ai/learning-bank/problem/a-quadratic-function-can-have-an-PYkxwmDLLgkU7wfl","timestamp":"2024-11-05T15:50:16Z","content_type":"text/html","content_length":"140309","record_id":"<urn:uuid:a2391c8d-b142-4e15-b851-15935f410baa>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00569.warc.gz"} |
Effect of Loss Tangent on Characteristic Impedance
Created Thursday 12 October 2023
All simulation models contained on this page are available for download in a Simbeor project linked here.
The dielectric model type must be changed to see the variations explored.
This page is based on a question on the SI-List:
Date: Wed, 11 Oct 2023 16:14:34 -0700
Subject: [SI-LIST] reflection coefficient dependence on tangent loss
Hi Experts,
I wanted to get some insights about the reflection coefficient's (S11)
dependence on tangent loss.
My understanding is Dk/tangent loss doesn't affect the reflection
coefficient. Let me know what you think.
To answer it, we can use Simbeor to explore some basic transmission line structures. We know that the characteristic impedance can be described with the RLGC parameters as follows:
In this case, we reference Pozar's Microwave Engineering, Equation 2.7.
Since the loss tangent impacts dielectric loss (the G/conductance term), then we would perhaps expect that characteristic impedance will go down with a higher loss tangent. Using Simbeor, we can
configure a basic stripline structure.
We'll use a standard Dk of 4.2 and vary the loss tangent from 0 (lossless), to 0.02 (typical FR-4) to 0.05 (quite high). If we configure the tool to use a flat/non-causal dielectric model, then the
system behaves exactly as we expect. The figure below shows characteristic impedance and RLGC values. The Z[0] varies only slightly, from 49.527Ω with LT=0, to 49.497Ω with LT=0.05. This exactly
validates our expectation where Z[0] is reduced with higher loss tangent. Note that R, L and C are all totally unchanged for the three variations. The conductance, G, is increased with higher loss
tangent. Unfortunately, a flat, non-causal dielectric model should not be used in wideband simulations for signal integrity.
Green: LT = 0.05, Blue: LT = 0.02, Red: LT = 0.0
• Left: Characteristic impedance (left axis) and attenuation (right axis)
• Middle: R and L (left/right axes respectively)
• Right: G and C (left/right axes respectively)
We would want to use a causal, frequency dependent dielectric model for our simulations. In this case, we might select the Djordjevic-Sarkar model (also known as either Wideband Debye or Infinite
Pole Debye). This is discussed in great depth in Sections 6.3 and 6.4 of "Advanced Signal Integrity for High-Speed Digital Designs" by Stephen Hall and Howard Heck. Specifically, "the real part of
the permittivity (ε′) is a function of the imaginary part (ε′′), and vice versa" (from Pg. 270 of this book).
Recall that the loss tangent is related to the real and imaginary parts of the permittivity by:
Based on these definitions, we would expect to see some change in both C and G.
The plots below show the same characteristics, but with a Wideband Debye, causal, dielectric model. Here we must point out that while all of the markers are showing behavior at 16 GHz, the dielectric
properties were specified at 1 GHz. The plots following show the characteristic impedance and the capacitance zoomed into the 1 GHz point. At exactly that location, Z[0] does follow the same pattern
(higher loss tangent results in lower Z[0]). We see that capacitance at this point matches up exactly for all loss tangent variations. As we go up in frequency, there is a significant difference in
characteristic impedance depending on only a change in the loss tangent. The takeaway here is that we must be extremely aware of not only the material properties, but the methods used to model them
in simulation.
Green: LT = 0.05, Blue: LT = 0.02, Red: LT = 0.0
• Left: Characteristic impedance (left axis) and attenuation (right axis)
• Middle: R and L (left/right axes respectively)
• Right: G and C (left/right axes respectively)
Green: LT = 0.05, Blue: LT = 0.02, Red: LT = 0.0
• Left: Capacitance (C), (right axis)
• Right: Characteristic impedance (left axis)
Interestingly, nothing here yet answers the initial question about return loss or reflection coefficient.The plots below show that return loss is very high for the LT=0.05 case regardless of the
dielectric model used. Similarly, return loss matches well for LT=0 and LT=0.02 above ~15 GHz, though at low frequencies (under 5 GHz) there's significant difference.
Green: LT = 0.05, Blue: LT = 0.02, Red: LT = 0.0
• Left: Return loss of a 1" trace with a flat, non-causal dielectric model
• Right: Return loss of a 1" trace with a causal Wideband Debye dielectric model
This page makes extensive reference to the following:
1. "Advanced Signal Integrity for High-Speed Digital Designs" by Stephen Hall and Howard Heck
2. "Microwave Engineering" by David Pozar
Contact Stephen with any questions: Stephen@ShieldDigitalDesign.com © Shield Digital Design | {"url":"https://wiki.shielddigitaldesign.com/High_Speed_Design_Wiki/Signal_Integrity/Transmission_Lines/Effect_of_Loss_Tangent_on_Characteristic_Impedance.html","timestamp":"2024-11-09T23:40:06Z","content_type":"text/html","content_length":"15925","record_id":"<urn:uuid:b6436797-78c0-41dd-a9fb-d4d49cdb9574>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00383.warc.gz"} |
Directions for the following 5 Five items:Read the following information provided and answer the items that follow. Your answers to these items should be based on the information given only.I. The bar graph given above shows the sales of books in thousand number from six branches of a publishing company during two consecutive years 2000 and 2001 .Sales of Books in thousand numbers from Six Branches B1, B2, B3, B4, B5 and B6 of a publishing Company in 2000 and 2001 .Branches =2000=2001Question 16. Total sales of branches B1, B3 and B5 together for both the years in thousand numbers is?A. b 310B. c 435C. a 250D. d 560
Directions for the following 5 (Five) items:
Read the following information provided and answer the items that follow. Your answers to these items should be based on the information given only.
I.The bar graph given above shows the sales of books (in thousand number) from six branches of a publishing company during two consecutive years 2000 and 2001.
Sales of Books (in thousand numbers) from Six Branches - B1, B2, B3, B4, B5 and B6 of a publishing Company in 2000 and 2001.
Question 16. Total sales of branches B1, B3 and B5 together for both the years (in thousand numbers) is? | {"url":"https://byjus.com/question-answer/directions-for-the-following-5-five-items-read-the-following-information-provided-and-answer-the-4/","timestamp":"2024-11-13T23:12:03Z","content_type":"text/html","content_length":"144110","record_id":"<urn:uuid:de290514-fd33-49df-a281-35608b7160eb>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00570.warc.gz"} |
Sensitivity Analysis of the Information Gain in Infinite-Dimensional Bayesian Linear Inverse Problems
We study the sensitivity of infinite-dimensional Bayesian linear inverse problems governed by partial differential equations (PDEs) with respect to modeling uncertainties. In particular, we consider
derivative-based sensitivity analysis of the information gain, as measured by the Kullback-Leibler divergence from the posterior to the prior distribution. To facilitate this, we develop... Show more | {"url":"https://synthical.com/article/Sensitivity-Analysis-of-the-Information-Gain-in-Infinite-Dimensional-Bayesian-Linear-Inverse-Problems-1b30aad6-fea2-4467-aab5-781fa132325b?","timestamp":"2024-11-06T13:45:16Z","content_type":"text/html","content_length":"64708","record_id":"<urn:uuid:6400ac75-8b41-4468-9585-246cb0a908b3>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00308.warc.gz"} |
The RADON function implements the Radon transform, used to detect features within a two-dimensional image. This function can be used to return either the Radon transform, which transforms lines
through an image to points in the Radon domain, or the Radon backprojection, where each point in the Radon domain is transformed to a straight line in the image.
Radon Transform Theory
The Radon transform is used to detect features within an image. Given a function A(x, y), the Radon transform is defined as:
This equation describes the integral along a line s through the image, where ρ is the distance of the line from the origin and θ is the angle from the horizontal.
In medical imaging, each point R(θ, ρ) is called a ray-sum, while the resulting image is called a shadowgram. An image can be reconstructed from its ray-sums using the backprojection operator:
where the output, B(x, y), is an image of A(x, y) blurred by the Radon transform.
How IDL Implements the Radon Transform
To avoid the use of a two-dimensional interpolation and decrease the interpolation errors, the Radon transform equation is rotated by θ, and the interpolation is then done along the line s. The
transform is divided into two regions, one for nearly-horizontal lines (θ ≤ 45°; 135° ≤ θ ≤ 180°), and the other for steeper lines (45° < θ < 135°), where θ is assumed to lie on the interval [0°,
For nearest-neighbor interpolation (the default), the discrete transform formula for an image A(m, n) [m = 0, ..., M–1, n = 0, ..., N–1] is:
where brackets [ ] indicate rounding to the nearest integer, and the slope and offsets are given by:
For linear interpolation, the transform is:
where the slope and offsets are the same as above, and the lower-half brackets indicate flooring to the nearest lower integer. The weighting w is given by the difference between a'n + b' and its
floored value, or between a’n + b’ and its floored value.
How IDL Implements the Radon Backprojection
For the backprojection transform, the discrete formula for nearest-neighbor interpolation is:
with the nearest-neighbor for ρ given by:
For backprojection with linear interpolation:
and ρ is the same as in the nearest-neighbor.
Radon Transform:
Result = RADON( Array [, /DOUBLE] [, DRHO=scalar] [, DX=scalar] [, DY=scalar] [, /GRAY] [, /LINEAR] [, NRHO=scalar] [, NTHETA=scalar] [, RHO=variable] [, RMIN=scalar] [, THETA=variable] [, XMIN=
scalar] [, YMIN=scalar] )
Radon Backprojection:
Result = RADON( Array, /BACKPROJECT, RHO=variable, THETA=variable [, /DOUBLE] [, DX=scalar] [, DY=scalar] [, /LINEAR] [, NX=scalar] [, NY=scalar] [, XMIN=scalar] [, YMIN=scalar] )
Return Value
The result of this function is a two-dimensional floating-point array, or a complex array if the input image is complex. If Array is double-precision, or if the DOUBLE keyword is set, the result is
double-precision, otherwise, the result is single-precision.
The two-dimensional array of size M by N to be transformed.
If set, the backprojection is computed, otherwise, the forward transform is computed.
Note: The Radon backprojection does not return the original image. Instead, it returns an image blurred by the Radon transform. Because the Radon transform is not one-to-one, multiple (x, y) points
are mapped to a single (θ, ρ).
Set this keyword to force the computation to be done using double-precision arithmetic.
Set this keyword equal to a scalar specifying the spacing between ρ coordinates, expressed in the same units as Array. The default is one-half of the diagonal distance between pixels, 0.5[(DX^2 + DY^
2)]^1/2. Smaller values produce finer resolution, and are useful for zooming in on interesting features. Larger values may result in undersampling, and are not recommended. If BACKPROJECT is
specified, this keyword is ignored.
Set this keyword equal to a scalar specifying the spacing between the horizontal (x) coordinates. The default is 1.0.
Set this keyword equal to a scalar specifying the spacing between the vertical (y) coordinates. The default is 1.0.
Set or omit this keyword to perform a weighted Radon transform, with the weighting given by the pixel values. If GRAY is explicitly set to zero, the image is treated as a binary image with all
nonzero pixels considered as 1.
Set this keyword to use linear interpolation rather than the default nearest-neighbor sampling. Results are more accurate but slower when linear interpolation is used.
Set this keyword equal to a scalar specifying the number of ρ coordinates to use. The default is 2 CEIL([MAX(x^2 + y^2)]^1/2/ DRHO) + 1.
If BACKPROJECT is specified, this keyword is ignored.
Set this keyword equal to a scalar specifying the number of θ coordinates to use over the interval [0, π]. The default is CEIL(π[(M^2+ N^2)/2]^1/2). Larger values produce smoother results, and are
useful for filtering before backprojection. Smaller values result in broken lines in the transform, and are not recommended.
For the forward transform, if THETA is specified then this keyword is ignored.
If BACKPROJECT is specified, this keyword is ignored.
If BACKPROJECT is specified, set this keyword equal to a scalar specifying the number of horizontal coordinates in the output Result. The default is FLOOR(2 MAX(|RHO|)(DX^2 + DY^2)^-1/2 + 1). For the
forward transform this keyword is ignored.
If BACKPROJECT is specified, set this keyword equal to a scalar specifying the number of vertical coordinates in the output Result. The default is FLOOR(2 MAX(|RHO|)(DX^2 + DY^2)^-1/2 + 1). For the
forward transform, this keyword is ignored.
For the forward transform, set this keyword to a named variable that will contain the radial (ρ) coordinates, in units defined by the DX and DY keywords (pixels by default). If BACKPROJECT is
specified, this keyword must contain the ρ coordinates of the input Array. The ρ coordinates should be evenly spaced and in increasing order.
Set this keyword equal to a scalar specifying the minimum ρ coordinate to use for the forward transform. The default is –0.5(NRHO – 1) DRHO.
If BACKPROJECT is specified, this keyword is ignored.
For the forward transform, set this keyword to a named variable containing a vector of angular (θ) coordinates (in radians) to use for the transform. If NTHETA is specified, and THETA is set to an
undefined named variable, then on exit THETA will contain the θ coordinates. If BACKPROJECT is specified, this keyword must contain the θ coordinates of the input Array.
Set this keyword equal to a scalar specifying the x-coordinate of the lower-left corner of the input Array. The default is – (M–1)/2, where Array is an M by N array. If BACKPROJECT is specified, set
this keyword equal to a scalar specifying the x‑coordinate of the lower-left corner of the Result. In this case the default is -DX (NX-1)/2.
Set this keyword equal to a scalar specifying the y-coordinate of the lower-left corner of the input Array. The default is – (N–1)/2, where Array is an M by N array. If BACKPROJECT is specified, set
this keyword equal to a scalar specifying the y-coordinate of the lower-left corner of the Result. In this case, the default is -DY (NY-1)/2.
The code for this example is included in the file radon_doc.pro in the examples/doc/language subdirectory of the IDL distribution. Run this example procedure by entering radon_doc at the IDL command
prompt or view the file in an IDL Editor window by entering .EDIT radon_doc.pro.
This example displays the Radon transform and the Radon backprojection:
PRO radon_doc
DEVICE, DECOMPOSED=0
;Create an image with a ring plus random noise:
x = (LINDGEN(128,128) MOD 128) - 63.5
y = (LINDGEN(128,128)/128) - 63.5
radius = SQRT(x^2 + y^2)
array = (radius GT 40) AND (radius LT 50)
array = array + RANDOMU(seed,128,128)
;Create display window, set graphics properties:
WINDOW, XSIZE=440,YSIZE=700, TITLE='Radon Example'
!P.BACKGROUND = 255 ; white
!P.COLOR = 0 ; black
XYOUTS, .05, .94, 'Ring and Random Pixels', /NORMAL
;Display the image. 255b changes black values to white:
TVSCL, 255b - array, .05, .75, /NORMAL
;Calculate and display the Radon transform:
XYOUTS, .05, .70, 'Radon Transform', /NORMAL
result = RADON(array, RHO=rho, THETA=theta)
TVSCL, 255b - result, .08, .32, /NORMAL
PLOT, theta, rho, /NODATA, /NOERASE, $
POSITION=[0.08,0.32, 1, 0.68], $
XSTYLE=9,YSTYLE=9,XTITLE='Theta', YTITLE='R'
;For simplicity in this example, remove everything but
;the two stripes.
result[*,0:55] = 0
result[*,73:181] = 0
result[*,199:*] = 0
;Find the Radon backprojection and display the output:
XYOUTS, .05, .26, 'Radon Backprojection', /NORMAL
backproject = RADON(result, /BACKPROJECT, RHO=rho, THETA=theta)
TVSCL, 255b - backproject, .05, .07, /NORMAL
The following figure displays the program output. The top image is an image of a ring and random pixels, or noise. The center image is the Radon transform, and displays the line integrals through the
image. The bottom image is the Radon backprojection, after filtering all noise except for the two strong horizontal stripes in the middle image.
Note: A better (and more complicated) method than removing everything but the two stripes is to choose a threshold for the result at each value of theta, perhaps based on the average of the result
over the theta dimension.
1. Herman, Gabor T. Image Reconstruction from Projections. New York: Academic Press, 1980.
2. Hiriyannaiah, H. P. X-ray computed tomography for medical imaging. IEEE Signal Processing Magazine, March 1997: 42-58.
3. Jain, Anil K. Fundamentals of Digital Image Processing. Englewood Cliffs, NJ: Prentice-Hall, 1989.
4. Toft, Peter. The Radon Transform: Theory and Implementation. Denmark: Technical University; 1996. Ph.D. Thesis.
Version History
See Also | {"url":"https://www.nv5geospatialsoftware.com/docs/RADON.html","timestamp":"2024-11-13T20:47:21Z","content_type":"text/html","content_length":"97292","record_id":"<urn:uuid:2e155633-7a88-408a-a7ec-1d1cf639b272>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00485.warc.gz"} |
Sudan Hansraj - Vol. 30 Num. 2 (2010)
Calculus free optimisation 26-31
I argue for the inclusion of topics in high school mathematics curricula that are traditionally reserved for high achieving students preparing for mathematical contests. These include the
arithmetic mean - geometric mean inequality which has many practical applications in mathematical modelling. The problem of extremalising functions of more than one independent variable
usually falls outside the secondary school curriculum and is generally discussed in topics on single and multivariable calculus in undergraduate mathematics. In particular the method of
Lagrange multipliers is efficient in solving a wide class of problems where a function of many variables is to be maximised or minimised subject to a number of constraints on the variables. We
explore some non-calculus approaches to solving some classes of problems. The well known arithmetic mean - geometric mean inequality is discussed and is shown to be useful in solving types of
problems involving more than two variables but in a manner accessible to high school students. The method involves writing the constraint function as a sum of quantities with a finite sum and
whose product must be maximised or minimised in the target function. This is a powerful idea that should be be exploited in high school mathematics. | {"url":"https://flm-journal.org/index.php?do=details&lang=en&vol=30&num=2&pages=26-31&ArtID=910","timestamp":"2024-11-03T00:39:46Z","content_type":"text/html","content_length":"42897","record_id":"<urn:uuid:70cc3e07-3001-40a2-bc9e-a29676fcf613>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00152.warc.gz"} |
Journal of Korea Robotics Society
[ ARTICLE ]
The Journal of Korea Robotics Society - Vol. 19, No. 3, pp. 281-286
Abbreviation: J. Korea Robot. Soc.
ISSN: 1975-6291 (Print) 2287-3961 (Online)
Print publication date 31 Aug 2024
Received 19 Apr 2024 Revised 10 Jun 2024 Accepted 23 Jul 2024
DOI: https://doi.org/10.7746/jkros.2024.19.3.281
Puzzle Heuristics: Efficient Lifelong Multi-Agent Pathfinding Algorithm for Large-scale Challenging Environments
Wonjong Lee^1^, *
Joonyeol Sim^2^, *
Changjoo Nam^†
1Graduate Student, Dept. of Artificial Intelligence, Sogang University, Seoul, Korea (leewj1208@sogang.ac.kr)
2Graduate Student, Dept. of Electronic Engineering, Sogang University, Seoul, Korea (jysim@u.sogang.ac.kr)
퍼즐 휴리스틱스: 대규모 환경을 위한 효율적인 다중 에이전트 경로 탐색 알고리즘
이원종^1^, *
심준열^2^, *
Correspondence to : ^†Assistant Professor, Corresponding author: Dept. of Electronic Engineering, Sogang University, Seoul, Korea (cjnam@sogang.ac.kr)
Contributed by footnote: * Wonjong Lee and Joonyeol Sim contributed equally to this work.
Funding Information ▼
This paper describes the solution method of Team AIRLAB used to participate in the League of Robot Runners Competition which tackles the problem of Lifelong Multi-agent Pathfinding (MAPF). In
lifelong MAPF, multiple agents are tasked to navigate to their respective goal locations where new goals are consecutively revealed once they reach initial goals. The agents need to avoid collisions
and deadlock situations while they navigate to perform tasks. Our method consists of (i) Puzzle Heuristics, (ii) MAPF-LNS2, and (iii) RHCR. The Puzzle Heuristics is our own algorithm that generates a
compact heuristic table contributing to reduce memory consumption and computation time. MAPF-LNS2 and RHCR are state-of-the-art algorithms for MAPF. By combining these three algorithms, our method
can improve the efficiency of paths for all agents significantly.
Keywords: Multi-AMR Control, Path Planning, Search Algorithm
1. Introduction
Multi-agent Pathfinding (MAPF) is the problem that finds collision-free paths for multiple agents from their start locations to their goal locations while optimizing the sum of total path lengths or
the time taken for all tasks to complete^[1]. It has numerous applications in various domains, such as logistics, aircraft, and urban traffic systems. In many such applications, the agents are tasked
to visit multiple destinations consecutively.
Lifelong MAPF^[2] is an extended version of the MAPF problem. In contrast to the canonical form of MAPF, where each agent has only one task and remains at its goal location, lifelong MAPF requires
agents to generate paths to their respective new goal locations after reaching their current goals. The challenges lie in the fact that the next goals are not known to the agents beforehand and the
time that the agents finish their current tasks are not synchronized. Thus, it is paramount to have an efficient method that can find collision-free paths promptly for a dynamically varying task set
given a limited time budget.
The League of Robot Runners^[3] is a competition to tackle the lifelong MAPF in challenging environments in terms of the complexity of the environments as well as the scale of the agent team. Our
team AIRLAB (ranked 8th) developed a method employing one of the state-of-the-art algorithms MAPF-LNS2^[4] and Rolling Horizon Collision Resolution (RHCR) framework^[5] as the MAPF solver. To enhance
this approach, we develop Puzzle Heuristics that generates a compact heuristic table to reduce memory consumption and computation, and successfully integrated it with these existing methods.
2. Problem Definition
Our goal is to solve the lifelong MAPF instances presented in the League of Robot Runners competition. Therefore, the problem definition is dictated by the goal and rules of the competition.
A team of k agents A=a1,a2,⋯,ak perform errands (i.e., tasks) that appear in an online manner in a deterministic and a fully observable environment. The environment is represented by a grid map G = (
V, ε) where V is the set of vertices (representing grid cells) and ε is the set of edges representing the adjacency of the cells (i.e., G is a 4-connected graph). The agents operate on discrete time
steps, denoted as t = 0, 1, 2, ..., τ where a constant τ is the final time step of an instance. At each time step, each agent can perform one of the four actions: move forward, rotate left or right
90 degrees, or wait.
The position vit∈V is the position of agent a[i] on the grid map at time step t. Let Pit=vit,vit+1,⋯,vit+T be a path of agent a[i] from to t+T that the agent plans to move. Along the path, an agent
must avoid conflicts with other agents. Two types of conflicts are defined: (ⅰ) the vertex conflict, where more than one agent moves to the same position, and (ⅱ) the edge conflict, where two agents
swap their positions simultaneously. Specifically, a vertex conflict occurs at t if ∃i,j such that i ≠ j and vit=vjt. And edge conflict occurs at t if ∃i,j such that i ≠ j, vit=vjt-1 and vjt=vit-1.
An errand is located on a position in the map. It can be completed only if the agent assigned to it arrives at the position of the errand. An agent becomes to know the next errand only after it
finishes the current one.
Given the assumptions and definitions, the objective of the lifelong MAPF is to maximize the number of errands to be completed by A within t = 0, ⋯, τ.
3. Method
By integrating MAPF-LNS2, RHCR, and Puzzle Heuristics, our method can handle large-scale lifelong MAPF instances in challenging environments. [Fig. 1] represents an overview of our algorithm
[Fig. 1]
Illustration of the process of the MAPF-LNS2 and RHCR for Lifelong Multi-Agent Pathfinding (MAPF). Starting from the input MAPF instance, the solver plans w steps, which is the window size
representing the time horizon for conflict resolutions, and generates an action sequence. After each action is simulated, the system replans after h steps, where h is the replanning period indicating
the frequency of replanning. This windowed approach represents the RHCR method. The diagram shows the initial state, intermediate steps, and the continuous loop of planning and replanning to achieve
collision-free paths for agents in a grid map. The circles represent the positions of the robots, while the triangles indicate their target positions
We provide a detailed description of each component and how they contribute to the overall effectiveness and efficiency of our solution method.
3.1 Puzzle Heuristics
The shortest distance between a pair of cells is frequently queried in an MAPF instance. In a fast-paced situation where the quick online computation of paths is necessary, it is a common approach to
precalculate the distances of all pairs. However, as the size of the map grows, the memory required to store the (Manhattan) distances increases significantly. A naïve method has the space complexity
O(N^2) for a grid map with N cells. This is particularly problematic in the competition since the memory size allocated to our method is limited. If we reduce memory usage by storing only part of the
distances, some of the computation time needed to resolve inter-agent conflicts at runtime needs to be allocated to calculate the distances. Therefore, the quality of the solution should be
compromised to balance the trade-off between storage space and computational overhead.
The core idea of Puzzle Heuristics is to divide the grid map into areas such that each area contains contiguous cells within a depth d in a tree search from its central cell called the pivot point.
The algorithm begins by selecting the lowest indexed cell (the index increments from top to bottom and left to right like the pixel coordinates) in the grid map as the first pivot point. Using
Breadth-First Search (BFS), the area expands over non-obstacle cells from the pivot point, ensuring that the frontier of the BFS does not exceed a depth of d.
Cells within the area are labeled with the same unique puzzle index. The process selects the next unlabeled cell with the lowest index as the new pivot point to repeat the area expansion. This
procedure iterates until all cells in the grid map are labeled. The same-colored neighboring cells shown in the map of [Fig. 2] belong to the same area.
[Fig. 2]
An example result of the Puzzle Heuristics method for generating a compact heuristic table. The same-colored neighboring cells belong to the same area
Once all cells are formed into areas, the shortest distances between all pairs of areas (i.e., pivot points) are calculated and stored in the heuristic table. By storing only the distances between
pivot points instead of all cells, we could meet the memory constraint. When the distance between two cells is queried, the heuristic value in the table effectively approximates the actual distance.
The number of cells in an area can reach up to 2d^2+2d+1 so at most 2d^2+2d+1 cells can be represented by a single value. As a result, the size of the heuristic table using Puzzle Heuristics could be
reduced to (N/(2d^2+2d+1))^2, leading to significant memory savings compared to the naïve approach.
3.2 MAPF-LNS2
MAPF-LNS2 is one of the state-of-the-art algorithms for solving large MAPF instances by leveraging the powerful meta-heuristic technique Large Neighborhood Search^[6]. A key advantage of MAPF-LNS2 is
its ability to start from any of feasible or infeasible initial solutions. The algorithm iteratively selects a subset of agents using a neighborhood selection method. The method destroys the current
paths of agents and repairs them using an efficient single-agent path planner that minimizes the number of collisions. If the repaired solution reduces the number of collisions, the new paths replace
the old ones. This destroy-and-repair procedure continues until a stopping criterion is satisfied.
To achieve high efficiency, MAPF-LNS2 employs Safe Interval Path Planning with Soft Constraints (SIPPS), a variant of the Safe Interval Path Planning algorithm^[7] that can handle hard and soft
constraints. Integrating SIPPS has shown significant improvements of MAPF-LNS2. While MAPF-LNS2 can be used with many existing MAPF algorithms, we choose to use Prioritized Planning (PP)^[8] which is
simple to implement and fast.
3.3 RHCR
The RHCR is a framework for solving the lifelong MAPF by decomposing an instance into a sequence of Windowed MAPF instances. RHCR utilizes two user-specified parameters: the time horizon w and the
replanning period h. The time horizon w specifies the time window such that the MAPF solver must resolve conflicts within the next w steps (i.e., windowed MAPF solver). The replanning period h
determines the frequency of the solver to replan paths. To avoid collisions, w should be larger than or equal to h. The values of these parameters are critical for the performance of RHCR. Too small
values of w would lead to deadlocks whereas too large values could result in inefficient solutions and increased computation time.
3.4 Greedy Randomized Search
Although MAPF-LNS2 and RHCR can solve large instances quickly, they would lead to deadlocks (i.e., no solution found) as PP is not proven to be complete.
Consider a scenario where a[i] and a[j] face each other at t and attempt to move forward. Here a[i] has a higher priority. During the planning phase, a[i] plans its path Pit while ignoring the
presence of a[j] as a[j] has a lower priority. As a result, the planned path of a[i] goes through the current location of a[j] at t+1 (i.e., vit+1=vjt). Given this path Pit, a[j] begins to plan Pjt.
In this situation, all possible actions for a[j] at t+1 result in conflicts because Pit passes through vjt+1 in any case. Specific-ally, if a[j] chooses to move forward, then an edge conflict occurs
(i.e., vit+1=vjt=vjt+1 or vjt+1=vjt=vit+1), respectively). Thus, a[j] cannot find a conflict-free action at t+1. [Fig. 3] illustrates the three conflict situations incurring deadlocks.
[Fig. 3]
An illustration of deadlock situations where a[i] and a[j] face each other and attempt to move forward as shown in the grid map in the left. Blue (a[i]) has a higher priority so plans Pit which
passes through the current position of a[j] at t+1. If a[j] moves forward, an edge conflict occurs. If a[j] chooses to rotate or wait, a vertex conflict occurs
We develop a greedy randomized search that combines greediness and randomness to deal with deadlocks. We probabilistically multiply the number of edge collisions and vertex collisions by a factor of
two. By doing so, our method could have an opportunity to explore alternative paths rather than relying on routes determined by the heuristic values. This simple yet effective approach enables the
algorithm to explore a broader range of possibilities and escape from deadlocks. In our experiments, the version with the greedy randomized search recorded of 31 times higher number of errands than
the version without the greedy randomized search.
4. Experiments
We evaluate our method in five environments provided by the competition: Sortation, Warehouse, Game, City, and Random map, as shown in [Fig. 4]. Each instance has predetermined agent-task pairs. We
vary the number of agents from 50 to 3000 agents. The evaluation metric is the total number of errands completed in 5,000 seconds. The test system is with AMD Ryzen 5800X 3.8 GHz CPU and 32G RAM. The
source code is written in C++17.
[Fig. 4]
The test environments in the competition. White cells represent spaces that robots can occupy, while black cells indicate the obstacles that robots cannot occupy. From the top and left, (1) Random,
(2) City, and (3) Game, (4) Warehouse, and (5) Sortation maps
As shown in [Fig. 5], our method effectively solves instances in the random map environment up to 200 agents.
[Fig. 5]
The number of completed errands by the proposed method in five challenging environments: Sortation, Warehouse, Game, City, and Random, with the number of agents ranging from 50 to 3000
In other environments, the solution quality remains reasonable for instances with up to 1,200 agents but declines as the number of agents increases.
The ability of MAPF-LNS2 and RHCR to solve large instances allows our algorithm to handle scenarios with a significantly large number of agents. The Puzzle Heuristics plays a crucial role in reducing
the memory consumption and computation time for large-sized maps like Sortation, Warehouse, and Game. In the greedy randomized search allows us to escape deadlocks effectively even in dense
environments like Sortation, Warehouse, and Random.
The algorithm struggles to find individual paths of all agents within the time budget, particularly with a large number of agents owing to the characteristics of PP, which plans paths sequentially.
Since higher-rank agents are considered fixed obstacles to lower-rank agents, the feasible positions become scarce.
In the context of discussing the impact of the d value on the performance of our method, we analyzed how varying d affects both the number of completed errands and the associated file size in Small
Warehouse instance. [Fig. 6] illustrates these effects. As depicted, increasing the d value from 0 to 5 leads to a notable pattern: the number of completed errands rises initially, peaking at d = 2,
before declining as d continues to increase. This indicates the best d value for maximizing the number of errands completed. Concurrently, the file size decreases as d increases, suggesting that
higher d values lead to more efficient memory usage. This trade-off between errand completion and file size is critical for optimizing the performance of our method in various environments.
[Fig. 6]
The impact of varying the d value on the number of completed errands (blue line) and file size (red dashed line). As the d value increases from 0 to 5, the number of errands initially increases,
reaching a peak at d = 2, before declining. Conversely, the file size decreases with increasing d values
In [Fig. 7], we observe a trade-off in choosing the value of w, which is the planning horizon w in RHCR. A low value of w leads to efficient planning but could cause deadlocks because agents may not
be able to plan far enough ahead to avoid conflicts. If we choose a high value, the ability to avoid deadlocks is improved. Balancing between these two cases is critical for the performance of
[Fig. 7]
The impact of varying the window size w on the number of completed errands. As the w size increases from 5 to 25, the number of completed errands initially rises, reaching a peak at w = 20, before
showing a decline
5. Conclusion
In this paper, we presented a novel approach for solving the challenging lifelong MAPF problem in the context of the League of Robot Runners competition. Our method combines three key techniques:
Puzzle Heuristics, MAPF-LNS2, and RHCR. Additionally, we introduced a greedy randomized search to escape from deadlock. Experimental results demonstrated the effectiveness of our approach in various
environments, particularly in random maps with a moderate number of agents. Through participation in the competition, we found our future directions that include improving the scalability of the
method and implementing methods to avoid deadlock situations. | {"url":"https://jkros.org/_common/do.php?a=full&b=12&bidx=3777&aidx=41652","timestamp":"2024-11-10T20:40:40Z","content_type":"text/html","content_length":"75575","record_id":"<urn:uuid:e73d8d9b-6e23-49e3-87fa-1250c2ee8964>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00417.warc.gz"} |
How to Excel in Mathematics
There is no royal road to Mathematics. Mathematicians, like poets, cannot be made but they are born. Still I have firm conviction that the following guiding principles and cautions, if strictly
observed, shall convert Mathematics from a cold, unsociable stranger with knit brows and frowning countenance into a warm-hearted, cheerful and loving friend.
1(a) Never approach Mathematics just after taking heavy meals. Let the food be well digested, and then apply yourself to this subject. Otherwise you will find it a very dry and rather repulsive study
and most uninteresting.
(b) In days of hard Mathematical work you ought to take light simple food that you can digest very easily; and be temperate. Don’t take ghee in excess. High thinking and plain living should go side
by side.
2(a) Don’t attack Mathematical problems or hard pieces of book work when you are sleepy or when about to go to bed. You will in that state find them quite invincible and impregnable. Not only will
they offer passive resistance, but will then lay you flat down on your bed. Plainly speaking, you will in two or three minutes, after taking a difficult problem in hand, fall fast asleep. But you
may, with advantage, at such a time, revise that part of Mathematics which you are already thoroughly conversant with, or work easy sums and simple riders that require very little mental exertion.
(b) In, order to excel in Mathematics you should always give to sleep what is its due. We cannot have a clear brain if we do not have enough of sleep. It is said of a great Mathematician, Des Cartes,
that on account of his delicate health, he was permitted to lie in bed till late in the mornings; this was a custom which he always followed, and when he visited Pascal in 1647 he told him that the
only way to do good work in Mathematics and to preserve his health was never to allow anyone to make him get up in the morning before he felt inclined to do so.
3 (a) If, however, circumstances oblige you to study difficult portions of Mathematics or solve hard problems just after taking meals, or just before retiring to bed, you ought to keep standing as
you work, or be walking up and down while you think. Otherwise your efficiency of labour will be very small, and laziness will get the upper hand of you.
(b) Never neglect to take bodily exercise. This is a neglect which proves ruinous to most students. Irregular students waste the greater part of their time in idleness but overwork themselves just
before the examination, taking no exercise and setting at nought the laws of health. Thus they succeed very easily in breaking their health though not in passing the examination. Then, is imputed to
labour what is brought about in reality by laziness; the charge is laid at the door of hard work, whereas it was indolence that impaired their health. Remember it is not labour that kills a student,
bat it is laziness or neglect of exercise that does so. Workers are sadly wanted in India, but no lazy workers.
4. When you begin a new book, it is advisable, first, to go through the book-work of the whole, at the same time doing the easy sums which come out on the first or at most at the second trial. After
thus once passing through the book begin it anew, and omit no example. By adopting this system, you will, save a great deal of your time and labour and your work will, be most efficient.
5. As far as possible try to do everything with your own unaided efforts. Not only should you try to solve the examples by your own exertions, but try to do the book-work also without the aid of the
author. Try, as it were, to re-discover everything. This will do you immense good. Read the heading in the case of each Article or the enunciation in the case of each Proposition and then shut your
book, and try if you can give your own demonstration. Think over the subject for a time, if your exertions seem to be fruitless, read one or two sentences from the top in that Article or Proposition
and then closing the book try to complete the proof; if even then your attempts avail nothing, read one or two sentences from the bottom of the same Article or Proposition, and do your best to supply
the parts of the proof not seen by you. If, then also you fail, read a little more of the book, and try to fill up the gap yourself. Thus a part at least of each Article or Proposition must, by all
means, be drawn out from your own brain, if you want to acquire a sound knowledge of Mathematics. You may, at first read very little by this method, but whatever is not learnt in this way forms but a
very poor part of education. By and by your power will increase and this process will no longer be slow. Your progress will, after trying this method for a time, be both rapid and thorough, and you
will find yourself quick to perceive and slow to forget. It is to such readers that the Roman proverb applies: “Beware of the man of few books.”
“The great danger” says a Mathematician, “which all mathematical students have to guard against is that of learning off book-work without fully mastering the essential points of the methods.
Mathematics cannot be crammed; to be able to write out book-work faultlessly is not sufficient. The why and wherefore of each step must be fully grasped, and students must not rest content unless
they fully understand in every case what is the property to be proved, what known results are assumed, and what methods are to be applied. Otherwise their memory will be unfairly taxed, the work will
degenerate into mere drudgery, and all this will he, of little avail if the book-work so assiduously committed to memory should be set with some trifling alteration – a frequent artifice among
examiners for finding out whether candidates really know their work.”
The solution of easy problems and riders, which is also practically indispensable, also depends almost entirely on a thorough knowledge of fundamental principles and those who do not clearly realize
this are too often apt to rush on to results in their answers in the examination, and use the words “it is obvious” or “evident” to conceal their ignorance of the intermediate steps, which, however,
deceives no one but the candidates themselves. On the other hand, those who will trouble to realise fully the methods of the book-work and the framework of facts on which each Proposition is built
up, will possess sufficiency powerful machinery to solve any reasonable problems that may be set.
All that will then be required is readiness in applying their knowledge, and this can only be brought about by frequent practice in working examples.
6. Don’t disdain or pass over sums containing easy applications of the formulae, and never be satisfied with knowing merely the nay how to work out a rider; work it out actually, carry your theory
into practice. Never forget the precious maxim “The way to more light is the faithful use of what we have.” By so doing you will acquire practice which alone makes us perfect. You know the greater
part of your University Examination-papers will consist of such easy riders; and even those questions in which brainwork is most prominent, depend not a little for their full and leady solution on
practical applications of the formulae. If you are already practised in that work you will finish in a very short time the whole of the paper, except those portions which require thinking, and out of
the total amount of time allotted having got a great deal at your disposal for thinking only, you will most probably succeed in your efforts in this direction too, and thus do the whole of the paper.
As it is not enough for a man to know the theory of swimming but he ought to have practice in that art if he wants to swim across a river; so is practice necessary for you if you want to swim across
the troublous sea of University Examinations. Simple riders and easy sums are a great recreation to the student of Mathematics.
Most students when asked to work cut a sum, sometimes after making a few feeble efforts but frequently before making any, give up in despair ejaculating the words “It is very difficult, it will not
come out.” But the self-same students, after the problem has been explained to them, cannot help uttering “Oh, it was so easy!” I say, yes, it was so easy, but you could not get it cut because you
did not enter into it. You had no courage, no strong will, no patience, or no Mathematical virtue.
7. Frequently revise the portions which you have already read; otherwise your further progress will be very very slow, and you will find yourself no match for the examiners. “Every Mathematical book
that is worth anything” says Professor Chrystal, “must be read backwards and forwards. Go on but often return to strengthen your faith. When you come on a hard or dreary passage pass it over; and
come back to it after you have seen its importance or found the need for it further on.”
8. In order to attain dexterity in analysis and calculation and become expert in giving ready solutions to problems, it is desirable to acquire the habit of performing mathematical investigations
mentally. No other discipline is so effectual in strengthening the faculty of attention; it gives a facility of apprehension, an accuracy and steadiness to the conceptions; and what is a still more
valuable acquisition, it habituates the mind to arrangements in its reasonings and reflections. To give an illustration of how much it improves the intellectual powers I may cite the case of Euler,
who had always accustomed himself to that exercise; and having practised it with .assiduity he is an instance to what an astonishing degree it may be acquired.
“Two of Euler’s pupils had calculated a converging series as far as the seventeenth term, but found, on comparing the written results, that they differed one unit at the fiftieth figure; they
communicated this difference to their master, who went over the whole calculation by head, and his decision was found to be the true one. For the purpose of exercising his little grandson in the
extraction of roots, he has been known to form to himself the table of the first six powers of all numbers from 1 to 100, and to have preserved it actually in his memory.”
9. Mathematics requires of us a great deal of time and energy; we should be continually working at it. But though it requires our body to be always in motion ever working, and subject to the laws of
Dynamics; it demands our mind to be always at rest, in equilibrium and in a state subject, as it were, to the laws of Statics. A man wanting to excel in Mathematics, should banish care and anxiety
from his mind, think of nothing else bat his work, should have a serene and tranquil heart, should allow nothing to disturb his peace and calm of mind. His labour will bear little fruit unless he is
able to keep his mind in perfect solitude; which in most cases, will require his body also to be in loneliness.
One lesson, Nature, lot me learn of thee,
One lesson which in every wind is blown,
One lesson of two duties kept at one
Though the loud world proclaim then enmity—
Of toil unsever’d from tranquillity!
Of labour, that in lasting fruit outgrows
Far noisier schemes, accomplished in repose,
Too great for haste, too high for rivalry!
Yes, while on earth a thousand discords ring,
Man’s senseless uproar mingling with his toil,
Still do thy quiet ministers move on,
Their glorious tasks in silence perfecting;
Still working, blaming still oar vain turmoil;
Labourers that shall not fail, when man is gone.
(Matthew Arnold.)
10. A student of Mathematics should always have a humble heart and a docile spirit.
Carefully store in every piece of knowledge, gather every bit of Mathematical truth; what, if you can make no immediate use of them, and what, if no pleasing result seems likely to spring from them.
“….because right is right, to follow right
Were wisdom in the scorn of consequence”
What a noble spirit of research was betrayed by the great Mathematician when he spoke of himself as having been all his life but “a child gathering pebbles on the sea-shore” – a similitude expressing
not only his humility, but alluding likewise to “the spirit in which he had pursued his investigations, as having been that, not of selection and system -building, but of childlike alacrity in
seizing upon whatever contributions of knowledge Nature threw at his feet.”
These directions may be summed up m a single one: Love the subject, (Love conquers all, ) and try, by every means possible, to keep yourself in a state in which you may be able to concentrate your
mind and pay close and undivided attention to the subject. This is a faculty, which, if we consider the testimony of Newton sufficient evidence, is the great constituent of inventive power It is that
complete retirement of the mind within itself, during which the senses are looked up; that intense’ meditation on which no idea can intrude; that firm, straightforward progress of thought, deviating
into no irregular sally; that perfect yoga where the mind becomes one with the subject which can alone place Mathematical subjects in a light efficiently strong to illuminate them fully, and preserve
the perceptions of the mind’s eye in the right order.
In the end I shall lay before you the secret of success in the study of Mathematics as well as in that of any other undertaking. It is seeking not our own aggrandisement, but the glory of God; it is
like the Red Cross Knight to labour and struggle for the Faerie Queen Gloriana or the Glory of God. It is thus to make our whole life a continuous prayer by our acts. It is to carry into practice the
noble advice of Lord Sri Krishna –
“………………………………in thy thoughts
Do all thou dost for me! Renounce for Me!
Sacrifice heart and* mind arid will to Me
Live in the faith, of Me!”
Let me close with the following strictly true Shakespeare:
“Heaven doth with us as we with torches do,
Not light them for ourselves; for if our virtues
Did not go forth of us, it were all alike
As if we had them not;
Spirits are not finely touch’d,
But to fine issues; nor Nature never lends
The smallest scruple of her excellence
But like a thrifty goddess she determines
Herself the glory of a creator,
Both thanks and use.” | {"url":"https://english.sreyas.in/how-to-excel-in-mathematics/","timestamp":"2024-11-10T05:56:22Z","content_type":"text/html","content_length":"195361","record_id":"<urn:uuid:a4500d8f-41fb-4a3b-aec4-3b24450d7e8e>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00017.warc.gz"} |
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Reviews of Paul Halmos' books
We list some of Paul Halmos' books and give brief extracts from some reviews and some extracts from Prefaces of these books. We have not included
Naive set theory (1960)
, perhaps his most famous book, since we have devoted a separate page to reviews of this text.
For extracts from reviews of
Naive set theory
Click on a link below to go to that book
1. Finite dimensional vector spaces (1942), by Paul R Halmos.
1.1. Review by: E R Lorch.
Mathematical Reviews MR0033869 (11,504d).
A purpose of this study is to present n-dimensional transformation theory from the abstract point of view. Normally, due to the existence of a finite basis, elementary matrix theory has an aspect not
at all suggesting that the infinite dimensional theory of operators is a natural extension of it. The author exploits as completely as possible the methods and notions of the infinite in his
presentation of the finite case; such a program has long been needed.
1.2. Review by: Mark Kac.
Bull. Amer. Math. Soc. 49 (5) (1943), 349-350.
In this book the author presents the topics covered usually in an introductory course in algebra (matrices, linear equations, linear transformations, and so on) from the point of view of a modern
analyst interested in general vector spaces. The ever-growing interest in Hilbert and more general linear spaces makes the appearance of the book very timely, especially since it furnishes an
excellent introduction to the subject certainly within the grasp of a first-year graduate student or even a good senior or junior. The topics are treated in such a manner as to make future
generalizations look both natural and suggestive. This sometimes is done at the expense of the shortness of exposition. Some theorems, as the author himself confesses, could be proved in fewer lines.
He prefers, however, longer proofs that admit a generalization to shorter ones that do not.
2. Finite dimensional vector spaces, 2nd Edition (1958), by Paul R Halmos.
2.1. From the Preface.
My purpose in this book is to treat linear transformations on finite- dimensional vector spaces by the methods of more general theories. The idea is to emphasize the simple geometric notions common
to many parts of mathematics and its applications, and to do so in a language that gives away the trade secrets and tells the student what is in the back of the minds of people proving theorems about
integral equations and Hilbert spaces. The reader does not, however, have to share my prejudiced motivation. Except for an occasional reference to undergraduate mathematics the book is self-contained
and may be read by anyone who is trying to get a feeling for the linear problems usually discussed in courses on matrix theory or "higher" algebra. The algebraic, coordinate-free methods do not lose
power and elegance by specialization to a finite number of dimensions, and they are, in my belief, as elementary as the classical coordinatized treatment. I originally intended this book to contain a
theorem if and only if an infinite-dimensional generalization of it already exists. The tempting easiness of some essentially finite-dimensional notions and results was, however, irresistible, and in
the final result my initial intentions are just barely visible. They are most clearly seen in the emphasis, throughout, on generalizable methods instead of sharpest possible results. The reader may
sometimes see some obvious way of shortening the proofs I give. In such cases the chances are that the infinite-dimensional analogue of the shorter proof is either much longer or else non-existent.
2.2. Review by: J L B Cooper.
The Mathematical Gazette 44 (348) (1960), 142-143.
The purpose of this book is to give an account of the theory of linear operators and manifolds in finite-dimensional spaces suitable as an introduction to the theory of operators in Hilbert Space. It
adopts an approach based on operator theory, makes little use of matrices and determinants, though these are mentioned, and gives some indication of the use of transcendental methods, such as Zorn's
lemma. After an account of operators in general finite spaces, including a rather sketchy account of the Jordan canonical form, the scalar product spaces and the decomposition of hermitian and normal
operators in these are treated with great thoroughness. Finally convergence of operators, with an elementary ergodic theorem, are studied, and a brief account of Hilbert Space follows. The book is
written clearly and carefully, and has numerous examples well chosen to illustrate its point of view. It can be recommended strongly for the student of its subject.
2.3. Review by: Albert Wilansky.
Amer. Math. Monthly 66 (6) (1959), 528-529.
Foreign mathematicians are warned not to search in dictionaries for zeroish, zeroness, askable, .... In style, Professor Halmos follows G H Hardy in the role of a "missionary preaching to the
3. Measure theory (1950), by Paul R Halmos.
3.1. From the Preface.
My main purpose in this book is to present a unified treatment of that part of measure theory which in recent years has shown itself to be most useful for its applications in modern analysis. If I
have accomplished my purpose, then the book should be found usable both as a text for students and as a source of reference for the more advanced mathematician. I have tried to keep to a minimum the
amount of new and unusual terminology and notation. In the few places where my nomenclature differs from that in the existing literature of measure theory, I was motivated by an attempt to harmonize
with the usage of other parts of mathematics. There are, for instance, sound algebraic reasons (or using the terms "lattice" and "ring" for certain classes of sets - reasons which are more cogent
than the similarities that caused Hausdorff to use "ring" and "field." It might appear inconsistent that, in the text, many elementary notions are treated in great detail, while, in the exercises,
some quite refined and profound matters (topological spaces, transfinite numbers, Banach spaces, etc.) are assumed to be known. The material is arranged, however, so that when a beginning student
comes to an exercise which uses terms not defined in this book he may simply omit it without loss of continuity. The more advanced reader, on the other hand, might be pleased at the interplay between
measure theory and other parts of mathematics which it is the purpose of such exercises to exhibit.
3.2. Review by: J L B Cooper.
The Mathematical Gazette 35 (312) (1951), 142.
[The book] gives a comprehensive account of those aspects of the theory of measure and integration which are important in general measure spaces and in topological spaces and groups. ... The
exposition throughout is masterly. Proofs are clear, precise and elegant. Care has been taken to avoid the indigestion which accompanies the reading of abstract theories, by discussions in each
section aimed at giving the reader an intuitive grasp of the subject and an idea where he is going, and by well-chosen sets of examples. These last should not be overlooked by the reader: they
contain not only helpful illustrative matter on the main text, but also accounts of further developments of the subject.
3.3. Review by: Harry M Gehman.
Mathematics Magazine 26 (3) (1953), 173-174.
[The book] is intended for use as a textbook in graduate courses in Measure Theory. It should also prove useful for reference purposes with its collection of theorems, its well chosen problems, and
its penetrating comments on the subjects which it touches. ... Unlike some textbooks, the exercises in Halmos' book are not trivial. They contain well chosen special cases to illustrate the theory,
alternate methods of proofs of theorems, and additional definitions and theorems to extend the material of the text.
3.4. Review by: S Kakutani.
Mathematical Reviews MR0033869 (11,504d).
This book presents a unified theory of the general theory of measure and is intended to serve both as a text book for students and as a reference book for advanced mathematicians. This book is
written in a very clear style and will make an excellent text book for those graduate students who are already familiar with the theory of Lebesgue integration in a Euclidean space. But the treatment
of the subject is rather abstract so that the book is perhaps not to be recommended for beginners. On the other hand, this book contains many new results obtained in the last ten years, and will be a
very useful source of reference for research mathematicians.
3.5. Review by: J C Oxtoby.
Bull. Amer. Math. Soc. 59 (1) (1953), 89-91.
In this book Professor Halmos presents an account of the modern theory of measure and integration in the generality required for the study of measure in groups. Thus finiteness conditions are imposed
only where necessary, and algebraic and topological aspects are appropriately stressed. Although written primarily for the student, the many novel ideas in the book and its store of interesting
examples and counter examples have already made it an indispensable reference for the specialist. The clarity of expression and the sprightly style which are characteristic of the author make the
exposition a pleasure to follow. ... It seems likely that this book will come to be recognized as one of the few really good text books at its level. It can hardly fail to exert a stimulating
influence on the development of measure theory.
4. Introduction to Hilbert space and theory of spectral multiplicity (1951), by Paul R Halmos.
4.1. Review by: J L B Cooper.
The Mathematical Gazette 36 (317) (1952), 218-219.
The main purpose of this book is to make available in English the theory of the unitary invariants of normal operators in Hilbert Space, and the last of its three chapters is devoted to this subject.
The first two contain an excellently set out account of Hilbert Space and of the bounded operators, projectors and bounded hermitian and normal operators on it. The development of the theory of these
last operators is made to rest on measure theory: as a result, and by making full use of the latest work on the subject, the author gives a treatment in which the geometry of the space plays a large
part. ... The book can be recommended as an introduction to Hilbert Space theory, for a certain class of readers. Its merits are its accuracy, its concise and lively style and its concern to give the
reader an idea of the general direction of the arguments. However, it is clearly written for readers with a training in the abstract modes of mathematical reasoning and, particularly, in measure
theory. It would not be very suitable for readers interested in physical applications of the theory, partly because these applications involve only separable spaces, but mainly because no mention is
made of unbounded operators and consequently of differential operators ... These minor points should not obscure the fact that Professor Halmos has written a stimulating and useful book
4.2. Review by: B Sz-Nagy.
Mathematical Reviews MR0045309 (13,563a).
The main purpose of this book is to present the so-called multiplicity theory and the theory of unitary equivalence, for arbitrary spectral measures, in separable or not separable Hilbert space. This
is developed in chapter III; the preceding parts serve as an introduction.
4.3. Review by: E R Lorch.
Bull. Amer. Math. Soc. 58 (3) (1952), 412-415.
There is little doubt that the author of this book enjoyed himself thoroughly during its preparation. Reading the result afforded this reviewer considerable pleasure. In one hundred and nine
well-packed pages one finds an exposition which is always fresh, proofs which are sophisticated, and a choice of subject matter which is certainly timely. Some of the vineyard workers will say that P
R Halmos has become addicted to the delights of writing expository tracts. Judging from recent results one can only wish him continued indulgence in this attractive vice. The present work may
confidently be recommended. However, beginners in the field should be cautioned before they rush off to secure a copy. Unless one is equipped and in training, one should not attempt the expedition.
One must not be misled by the title. For this introduction to Hilbert space, one has to be an expert in measure theory. As a matter of fact it is best to have read the author's book on measure theory
or its equivalent. ... Most pages exhibit a zest for play as well as work which is refreshing. Indeed, at times one may have a vague apprehension that the author is preparing a prank or baiting a
trap; however it seldom turns out to be more than a friendly tweak given with a wink. Such an intimate style, in the present desert of works written with an unexceptionable scientific detachment, is
warmly welcome. It is certainly a facet to the general success enjoyed by Halmos, previous books.
5. Lectures on ergodic theory (1956), by Paul R Halmos.
5.1. Review by: Yael N Dowker.
Mathematical Reviews MR0097489 (20 #3958).
This book is the first work on ergodic theory in book form since E Hopf's 'Ergodentheorie' appeared on 1937. Its contents are based on a course of lectures given by the author at the University of
Chicago in 1956. The first of these facts makes the book very welcome; more so since the book is written in the pleasant, relaxed and clear style usually associated with the author. The material is
organised very well and painlessly presented. A number of remarks, ranging from the serious to the whimsical, add insight and enjoyment to the reading of the book.
5.2. Review by: Yael N Dowker.
Bull. Amer. Math. Soc. 65 (4) (1959), 253-254.
The author, in the apology (preface) to the book, asks the reader to regard these notes as "designed to rekindle" interest in the subject. From this point of view and considering the excellent and
effortless style of the book it is doubly regretful that the material discussed is so restricted in time and person. There is almost no indication of work done during the last decade, and the
reviewer cannot but be disappointed that the reader is left unaware of the recent sparks of interest found by workers in such branches of ergodic theory as, for instance, those related to probability
theory, number theory, abstract ergodic theory, dynamical systems in general and geodesic flows in particular.
6. Entropy in ergodic theory (1959), by Paul R Halmos.
6.1. From the Preface.
Shannon's theory of information appeared on the mathematical scene in 1948; in 1958 Kolmogorov applied the new subject to solve some relatively old problems of ergodic theory. Neither the general
theory nor its special application is as well known among mathematicians as they both deserve to be; the reason, probably, is faulty communication. Most extant expositions of information theory are
designed to make the subject palatable to non-mathematicians, with the result that they are full of words like "source" and "alphabet". Such words are presumed to be an aid to intuition; for the
serious student, however, who is anxious to get at the root of the matter, they are more likely to be confusing than helpful. As for the recent ergodic application of the theory, the communication
trouble there is that so far the work of Kolmogorov and his school exists in Doklady abstracts only, in Russian only. The purpose of these notes is to present a stop-gap exposition of some of the
general theory and some of its applications. While a few of the proofs may appear slightly different from the corresponding ones in the literature, no claim is made for the novelty of the results. As
a prerequisite, some familiarity with the ideas of the general, theory of measure is assumed; Halmos's Measure theory (1950) is an adequate reference.
7. Algebraic logic (1962), by Paul R Halmos.
7.1. From the Introduction.
It has often happened that a theory designed originally as a tool for the study of a physical problem came subsequently to have purely mathematical interest. When that happens, the theory is usually
generalized way beyond the point needed for applications, the generalizations make contact with other theories (frequently in completely unexpected directions) and the subject becomes established as
a new part of pure mathematics. The part of pure mathematics so created does not (and need not) pretend to solve the physical problem from which it arises; it must stand or fall on its own merits.
Physics is not the only external source of mathematical theories; other disciplines (such as economics and biology) can play a similar role. A recent (and possibly somewhat surprising) addition to
the collection of mathematical catalysts is formal logic; the branch of pure mathematics that it has precipitated will here be called algebraic logic. Algebraic logic starts from certain special
logical considerations, abstracts from them, places them into a general algebraic context, and, via the generalization, makes contact with other branches of mathematics (such as topology and
functional analysis). It cannot be overemphasized that algebraic logic is more algebra than logic. Algebraic logic does not claim to solve any of the vexing foundation problems that sometimes occupy
logicians. All that is claimed for it is that it is a part of pure mathematics in which the concepts that constitute the skeleton of modern symbolic logic can be discussed in algebraic language. The
discussion serves to illuminate and clarify those concepts and to indicate their connection with ordinary mathematics. Whether the subject as a whole will come to be considered sufficiently
interesting and sufficiently deep to occupy a place among pure mathematical theories remains to be seen.
7.2. Review by: Donald Monk.
Amer. Math. Monthly 71 (6) (1964), 708-709.
By algebraic logic the author means that branch of general algebra which deals with algebraic structures mirroring in some sense certain formal logics. ... As Halmos indicates, one of the main
problems in algebraic logic is to state and prove algebraically various important theorems of logic. ... Halmos' book is highly recommended as an introduction for those who wish to study logic from a
purely algebraic point of view.
7.3. Review by: Leon Henkin.
Science, New Series 138 (3543) (1962), 886-887.
Hopefully, the fragmentation of science caused by increasing specialization will be counterbalanced by the development of new modes of unification. In the contemporary development of mathematics one
such unifying influence is the massive intrusion of algebraic concepts and methods into all mathematical fields. Mathematical logic has been intertwined with algebra from its beginnings, through
Boole's discovery that simple laws of logic can be expressed symbolically as algebraic equations. But only in very recent years has the algebraic viewpoint in logic been systematized to the point
where an almost complete account of logic can be given in algebraic terms. Alfred Tarski pioneered this enterprise through successive exploration of Boolean, relation, and cylindric algebras. The
most detailed contributions are found in the papers of Paul Halmos on polyadic Boolean algebras. In this volume Halmos collects his papers (unchanged) and adds a brief preface, a bibliography, and an
7.4. Review by: The Editors.
Mathematical Reviews MR0131961 (24 #A1808).
The book consists of a collection of 10 papers of the author, together with a bibliography of 13 supplementary items, on algebraic logic.
8. Lectures on Boolean algebras (1963), by Paul R Halmos.
8.1. From the Preface.
In 1959 I lectured on Boolean algebras at the University of Chicago. A mimeographed version of the notes on which the lectures were based circulated for about two years; this volume contains those
notes, corrected and revised. Most of the corrections were suggested by Peter Crawley. To judge by his detailed and precise suggestions, he must have read every word, checked every reference, and
weighed every argument, and I am very grateful to him for his help.
8.2. Review by: A D Wallace.
Science, New Series 144 (3618) (1964), 531-532.
The literary style is a mature version of the author's earlier presentations and there is a trace of current Gallic hauteur as well as some Bourbakian pontification. ... But this is an entirely minor
and subjective opinion, and the book is most highly recommended as a generally elegant and perceptive introduction to the basic facts concerning Boolean algebra available to the very well-trained
senior and to the average second-year graduate student.
8.3. Review by: R Sikorski.
Mathematical Reviews MR0167440 (29 #4713).
This is a very good introduction to the theory of Boolean algebras. It contains a discussion of all fundamental notions like subalgebras, homomorphisms, ideals, filters, complete algebras, √-complete
algebras, etc. ... The book contains almost no references to the literature.
8.4. Review by: R S Pierce.
The Journal of Symbolic Logic 31 (2) (1966), 253-254.
The theory of Boolean algebras is one of the most attractive parts of mathematics. On the one hand, Boolean algebras arise naturally in such diverse fields as logic, measure theory, topology, and
ring theory, so that the study of these objects is motivated by important applications. At the same time, the theory which has been developed constitutes one of the most elegant parts of modern
algebra. Finally, the subject still poses many challenging questions, some of which have considerable importance. A graduate student who wishes to study Boolean algebras will find three excellent
books to smooth his way ... For an introduction, the book by Halmos is probably the best of these monographs. It offers a quick route to the most attractive parts of the theory ... Like all of the
books by Professor Halmos, Lectures on Boolean algebras is written with admirable style and clarity. This work is a welcome addition to the literature of mathematics.
9. A Hilbert space problem book (1967), by Paul R Halmos.
9.1. Review by: C R Putnam.
Mathematical Reviews MR0208368 (34 #8178).
The book is divided into three parts: problems, hints and solutions. According to the instructions, if one is unable to solve a problem, even with the hint, he should, at least temporarily, grant it
(if it is a statement) and proceed to another. He should be prepared, however, to use the as yet unproved statement in the hope of thereby getting some clues as to its solution. If he does solve a
problem he should look at the hint and the solution anyway for perhaps some other variations on the theme. The rules are simple and the advantages of following them to the conscientious and diligent
reader are surely obvious and incalculable. Nevertheless, there is the ever lurking temptation to combine each problem-hint-solution into a unified whole, to be read together and at one time.
Although this procedure admittedly thwarts, at least partially, the aim of the book, it is certainly not without its own rewards. Thus, a researcher thereby may quickly find a compact presentation of
a particular issue, often including some history, references, and enlightening side remarks.
10. A Hilbert space problem book (2nd edition) (1982), by Paul R Halmos.
10.1. Review by: Peter Fillmore and Nigel Higson.
Amer. Math. Monthly 91 (9) (1984), 592-594.
A recurring theme is the perturbation of an operator by a compact operator, beginning with the Fredholm alternative and a 1909 result of Weyl on the invariance of the limit points of the spectrum
under such perturbations. ... Because of the diversity of topics that it includes, single operator theory is ideally suited to a book of the nature of Halmos's (we can imagine a Homotopy Theory
Problem Book being less successful). The second edition contains 250 problems as against 199 in the first. It remains an excellent, compact survey of single operator theory, as well as a valuable
source for Hilbert space techniques. The form, content, and style of the second edition deserve the same high praise as the first.
10.2. Review by: Philip Maher.
The Mathematical Gazette 73 (465) (1989), 259-260.
It might seem somewhat bizarre to be reviewing here a book that was first published in 1967 and one, moreover, whose subject matter is at the upper end of accessibility for the Gazette readership.
Yet Halmos's 'A Hilbert space problem book' has never been reviewed in these pages before; the second edition (under review) differs quite a bit from the first; and, of course, anything by Halmos,
however technical, promises to be lucid and entertaining. In fact, this book - with its well-nigh unparalleled form - is an object lesson in the communication of mathematics. For it does not follow
that boring and so-often unilluminating format of definition, lemma, theorem, proof (a format which seems virtually de-rigeur for the exposition of advanced mathematics). Rather, this book consists
of problems; some 250 of them (and their solutions) which cover much of the theory of (single) linear operators on Hilbert space. Of course, I am aware that other books have appeared that approach
parts of mathematics in a problem-orientated (as distinct from, so to speak, a subject-orientated) way: one thinks of Burns's 'A pathway to number theory' for example. But Halmos's books is the first
to do this at an advanced (rather than at an elementary) level.
11. Bounded integral operators on $L^{2}$ spaces (1978), by Paul Richard Halmos and Viakalathur Shankar Sunder.
11.1. Review by: W A J Luxemburg.
Mathematical Reviews MR0517709 (80g:47036).
The book contains a very valuable amount of information on integral operators. The theory is richly illustrated by examples and counterexamples. Many open problems are discussed. A large number of
them have been solved recently by Schep and W Schachermayer, and the solutions will appear soon in print. Also, the history of the subject is well presented. The style of the book is lucid, lively
and entertaining, as we have come to expect from the senior author.
11.2. Review by: Adriaan Zaanen.
Bull. Amer. Math. Soc. (N.S.) 1 (6) (1979), 953-960.
The Halmos-Sunder book is written in a lively style, as was to be expected. It is a mine of information for any analyst interested in operators, in particular operators on L2 spaces. The theory is
illustrated by examples as well as by counterexamples; open problems are mentioned and sometimes analyzed. A list of bibliographical notes gives information about the history of the subject. The
preface ends with the remark that the book contains only a part of a large subject, with only one of several approaches, and with explicit mention of only a few of the many challenging problems that
are still open. I agree, but I hope that nevertheless it will be clear from my comments in this review that I believe the present contribution to operator theory by Halmos and Sunder is a valuable
12. I Want to Be a Mathematician: An Automathography (1985), by Paul R Halmos.
12.1. Review by: John Baylis.
The Mathematical Gazette 70 (453) (1986), 253-255.
P R Halmos has a well-deserved reputation as an expositor of mathematics. One of his favourite techniques is to excavate for the simplicity underlying layers of complexity, to extract it and to
display it in a strikingly illuminating way. He also of course has made important contributions to mathematics per se, but on his own self-assessment his research achievements rank only fourth,
behind writing, editing and teaching. These abilities as teacher of and doer of mathematics, combined with mild but significant eccentricity has made him a 'name' in mathematics, a name about whom
the mathematical public will be sufficiently curious to guarantee interest in his autobiography. ... this is a fascinating addition to recent mathematical culture by one of its makers. The main
message I absorbed from it was a set of conditions required for success in mathematics: talent, yes; single-mindedness, almost as obvious; sense of humour, essential when the going gets tough; and
love, yes that is the right word - you must love mathematics, and that means all the ingredients, passion, pain and loyalty.
12.2. Review by: Henry Helson.
Mathematical Reviews MR0789980 (86m:01059).
This autobiography is a frank, personal, witty commentary on mathematicians and mathematics by one of the most influential, and observant, mathematicians of our time. It is much more about the
profession of mathematics than about the personal life of its author. ... He makes two main points. The first is the importance of being literate. The ability to speak and write effectively,
preferably in more than one language, is essential to effectiveness in all professional activity. And second, a real professional must work in all aspects of the job: research of course, but also
teaching in several formats, exposition at several levels, refereeing and editing, departmental chores, and participation in meetings and conferences. His standard is high.
12.3. Review by: John A Dossey.
The Mathematics Teacher 79 (6) (1986), 481-482.
This autobiographical sketch details the professional aspects of the career of the distinguished American mathematician Paul Halmos. It gives the reader a delightfully candid view of the evolution of
his career of research, teaching, travel, editing, and service from his secondary school days to the present. ... The book is exciting, witty, and well worth the time invested in its study. It
communicates what it means to be a mathematician.
12.4. Review by: Gian-Carlo Rota.
Amer. Math. Monthly 94 (7) (1987), 700-702.
Every mathematician will rank other mathematicians in linear order according to their past accomplishments, while he rates himself on the promise of his future publications. Unlike most
mathematicians, Halmos has taken the unusual step of printing the results of his lifelong ratings. By and large, he is fair to everyone he includes in his lists (from first rate (Hilbert) to fifth
rate (almost everybody else)), except towards himself, to whom he is merciless (even in the choice of a title to the book: "I want to be a mathematician," as if there were any question in anybody's
mind as to his professional qualifications). ... The leading thread of his exposition, what makes his narration entertaining (rather than just interesting), is mathematical gossip, which is freely
allowed to unfold in accordance to its mysterious logic. The reader will be thankful for being spared the nauseating personal details that make most biographies into painful reading experiences ...
Whatever does not relate to the world of mathematics is ruthlessly and justly left out (we hardly even learn whether he has a wife and kids).
13. I Want to Be a Mathematician: An Automathography (2005), by Paul R Halmos. [American Mathematical Association Reprint]
13.1. Review by: Fernando Q Gouvea. http://www.maa.org/node/105650
Instead, the book is about his life as a mathematician among mathematicians. It shows us a little about how Halmos thinks about mathematics, about what interested and motivated him, and about how he
interacted with others. It includes a lot of what might, somewhat uncharitably, be described as "gossip": stories and anecdotes about mathematics, mathematics departments, and mathematicians. In my
experience, mathematicians love this sort of thing. Those of my colleagues who have read this book have enjoyed it. My students have liked it much less, partly because they aren't that interested in
the world of mathematics, partly because they feel "turned off" by what they describe as Halmos's "arrogance." I think what bothers them is Halmos's bluntness about what counts and what does not
count as significant in mathematics. That Halmos's harshness is mostly directed at his own work doesn't change my students' assessment. If all this famous guy can say, after trying for fifty years,
is "I want to be a mathematician," they argue, then we students have no chance at all.
14. I have a photographic memory (1987), by Paul R Halmos.
14.1. From the Preface.
Would you look at some of the snapshots I have taken in the last 40-odd years? If you put a penny into a piggy bank every day for 45 years, you'll end up with something over 16,000 pennies. I have
been a snapshot addict for more than 45 years, and I have averaged one snapshot a day. Over a third of the pictures so accumulated have to do with the mathematical world: they are pictures of
mathematicians, their spouses, their brothers and sisters and other relatives, their offices, their dogs, and their carillon towers. The pictures were taken at the universities where I worked, and
the places where I was a visitor (for a day or for a year), and, as you will see, many of them were taken over food and drink. That's rather natural, if you think about it. It is not easy, and often
just not possible, to snap mathematicians at work (in a professional conversation, thinking, lecturing, reading) - it is much easier to catch them at tea or at dinner or in a bar. In any event, the
result of my hobby was a collection of approximately 6000 "mathematical" pictures, and when it occurred to me to share them with the world I faced an extremely difficult problem of choice. ... The
people included are not necessarily the greatest mathematicians or the best known. If I think a picture is striking, or interesting, or informative, or nostalgic, then it is here, even if the
theorems its subject has proved are of less mathematical depth than those of a colleague whose office is two doors down the hall.
14.2. Review by: Arthur M Hobbs.
Mathematical Reviews MR0934204 (89f:01067).
This is a book of considerable merit in several different ways. Beginning with the most frivolous use, it is very nearly the perfect book for a mathematician's coffee table (it only lacks colour).
Less trivially, with 604 photographs of (mostly) mathematicians and two photographs of the old and new buildings at Oberwolfach, each picture accompanied by an informative caption, the book is
pleasant reading for any mathematician, for many mathematician's spouses, and for any other person interested in mathematics as it is lived. ... as a historical document, this book will be valuable
in at least two ways. It gives a fascinating cross-section of mathematical life in the mid-twentieth century, and it provides considerable insight into the personality and interests of Professor
Halmos himself.
15. Problems for Mathematicians Young and Old (1991), by Paul R Halmos.
15.1. Review by: Stan Wagon.
Amer. Math. Monthly 99 (9) (1992), 888-890.
What makes a problem interesting? Its statement should be simple, not requiring excessive explanations, and the solution should be readily understandable by the intended audience. [In] Halmos's book
... his problems ... meet these criteria admirably. ... Halmos has won several writing awards, and the reader won't be disappointed in the prose with which he wraps the problems ...The book lives up
to its title, which promises problems for both young and old. But is it the young or the old who are more likely to be impressed by the pretty and elegant elementary problems? I'm not sure. In any
event, for the experienced mathematician or beginning graduate student seeking meatier fare the book contains several chapters with problems for the more mature reader.
15.2. Review by: Lionel Garrison.
The Mathematics Teacher 85 (7) (1992), 592.
Buy this book. In fact, buy several. Give them to your students and colleagues, and save one for when your first copy wears out. ... [this book], impels the reader to get out a pencil and start doing
mathematics. Paul Halmos, one of the pre eminent mathematicians and teachers of our time, here shares his personal treasury of favourite problems. Perhaps a third of these 165 problems are at least
comprehensible to able high school students who have studied calculus and probability and who enjoy being challenged. The remainder generally assume the standard undergraduate pro gram in analysis,
topology, and abstract algebra.
15.3. Review by: G A Heuer.
Mathematical Reviews MR1143283 (92j:00009).
The 165 problems in this charming book are classified roughly by the mathematical subdiscipline into which they most naturally fit. ... While a good many of the problems are catchy enough to prick
the interest of most readers without further help, most are preceded by the kind of skilful discussion for which the author is so well known, to make the problem more appealing and to help make clear
why this is exactly the natural question to pose in this context. The title is apt: there is much here for the veteran professional mathematician; there is also a good bit for the budding
mathematician who is still a high school pupil.
16. Linear Algebra Problem Book (1995), by Paul R Halmos.
16.1. Review by: Nick Lord.
The Mathematical Gazette 81 (490) (1997), 168-170.
... it is the quality of Halmos' commentary that made [the book] such a treat to review and which means that I can recommend it without hesitation to all lecturers (who think they can teach linear
algebra) and all sufficiently mature students (who think they have learnt it). His quiet emphasis on details, his penetrating insights into what makes an approach to a proof plausible, or the mode of
construction of a counter-example clear, and his 'heart-on-sleeve' approach to problem-posing, in explaining why he cast the problems in the form he did, are matchless: you feel you are in the
company of a master expositor, striding together through the world of linear algebra while he points out the flora, fauna, topography and the way ahead. [It is] a book bristling with lovely touches
16.2. Review by: Robert Messer.
Amer. Math. Monthly 105 (6) (1998), 577-579.
For students going on in mathematics, linear algebra serves as a transition to upper-level mathematics courses. In addition to learning the subject matter of linear algebra itself, these students
must be fortified with a degree of mathematical maturity in working with axioms and definitions, basic proof techniques, and mathematical terminology and notation. These issues cannot be left to
chance; they must be addressed explicitly to prepare students for courses such as abstract algebra and real analysis. As a textbook for a linear algebra course, Paul Halmos's 'Linear Algebra Problem
Book' satisfies these criteria. ... The conversational style of writing in this book occasionally lapses into annoying chattiness. A definition can be guessable and an answer conjecturable. A
corollary can be unsurprising or minute but enchanting. Within three sentences the zero linear functional has two symbols and goes from most trivial to most important and ends up uninteresting.
16.3. Review by: Jaroslav Zemánek.
Mathematical Reviews MR1310775 (96e:15001).
Understanding simple things such as basic linear algebra does not seem to be an easy task. Indeed, the author offers original insights illuminating the essence of the associative and distributive
laws, and the underlying algebraic structures (groups, fields, vector spaces). The core of the book is, of course, the study of linear transformations on finite-dimensional spaces. The problems are
intended for the beginner, but some of them may challenge even an expert. ... Needless to say, more emphasis on the history of the subject would be attractive in a book of this type. In brief, the
reviewer regrets that the author chose not to go deeper into the subject; he shows a few trees, but makes little attempt to see the forest.
17. Logic as algebra (1998), by Paul Halmos and Steven Givant.
17.1. Review by: Stephen D Comer.
The Journal of Symbolic Logic 63 (4) (1998), 1604.
This slim book provides an introduction to logic with the goal, as suggested by the title, of demonstrating that logic can be profitably understood from an algebraic viewpoint. Instead of making the
point by providing the reader an extensive treatment of algebraic logic, the book takes a modest concrete approach by limiting its consideration to that part of logic most familiar to a general
audience, namely the propositional calculus and the monadic predicate calculus. The presentation reflects Halmos's pedagogical style. Topics are introduced first by logical considerations, then the
ideas are abstracted, and finally they are placed in an appropriate general algebraic context. The book reads like an essay. This is not to say the presentation is not rigorous. It is. Details are
presented when necessary to convey the ideas, but without overwhelming readers. The presentation is crisp and lucid yet informal. It is as if the principles of logic are being explained over a cup of
coffee. The book is directed towards a general (mathematically literate) audience with an interest in modern logic. Nevertheless, the prerequisite of "a working knowledge of the basic mathematical
notions that are studied in a first course in abstract algebra" should be taken seriously. It is not a textbook (in the usual sense) even though it is based on notes from a course in logic by Paul
17.3. Review by: Graham Hoare.
The Mathematical Gazette 84 (499) (2000), 172-173.
[The book] is intended 'to show that logic can (and perhaps should) be viewed from an algebraic perspective .... Moreover, the connection between the principal theorems of the subject and well-known
theorems in algebra become clearer.' Readers anticipating arguments based on truth tables or diagrams of switching circuits will be disappointed. In compensation they will be entertained by a rich
array of algebraic concepts such as prime and maximal ideals, filters, homomorphisms, equivalence classes, kernels, quotient algebras and duality, all in the service of logic. As the authors state,
'propositional logic and monadic predicate calculus - predicate logic with a single quantifier, are the principal topics treated'. ... the whole will serve as a neat, succinct, introduction to logic
particularly for readers very much at home with algebraic concepts.
17.4. Review by: Myra R Lipman.
The Mathematics Teacher 92 (4) (1999), 371.
The authors of this book have targeted a wide-ranging audience; however, the book cannot be all things to all people. ... Because of the scope and depth of material, this book would be most useful as
a classroom reference or supplement. If the book included many more examples and some exercises, it would be outstanding and certainly more helpful to students. Viewing logic from an algebraic
perspective is an intriguing concept, and the authors succeed in giving a concise overview of relatively complex material in an intelligible manner.
17.5. Review by: Natasha Dobrinen.
The Bulletin of Symbolic Logic 16 (2) (2010), 281-282.
This is an excellent and much-needed comprehensive undergraduate textbook on Boolean algebras. It contains a complete and thorough introduction to the fundamental theory of Boolean algebras. Aimed at
undergraduate mathematics students, the book is, in the first authors words, "a substantially revised version of Paul Halmos' Lectures on Boolean algebras." It certainly achieves its stated goal of
"steering a middle course between the elementary arithmetic aspects of the subject" and "the deeper mathematical aspects of the theory" of Boolean algebras.
17.6. Review by: Marcel Guillaume.
Mathematical Reviews MR1612588 (99m:03001).
... this booklet is a gem, whose reading the reviewer ... highly recommends, fundamentally because it gives a better designed, direct and concise overview of logic, going beyond the topics explicitly
treated, which are in fact reduced just to the minimum needed in order to explain and introduce the key notions, and to explain and prove fundamental theorems in the special cases considered. As to
the form, the style is vivid and clear, using simple words, and free of long and complex technicalities. The text is rich in brief comments explaining the ideas behind the reasoning and calculations,
and frequently refers to simple examples and to the basic notions of universal algebra. | {"url":"https://mathshistory.st-andrews.ac.uk/Extras/Halmos_books/","timestamp":"2024-11-02T20:02:54Z","content_type":"text/html","content_length":"73510","record_id":"<urn:uuid:9ab9af5d-b55a-4d98-932f-620a06af9948>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00422.warc.gz"} |
perplexus.info :: Just Math : Ay. ay. ay
First, I'm going to just assume this thing converges. Then we can do the trick of substituting the expression into itself to get y=sqrt(i*y).
Square each side and solve to get y=0 or y=i.
I would discard y=0 since we have no indication as to where a zero popped up in the original infinite radical.
But lets take another approach. Express each sqrt as a power of 1/2.
Then y = (i*(i*(i*(....)^1/2)^1/2)^1/2.
Again playing a bit loose with exponentiation (not well-defined for infinite products), lets distribute all the powers. Then that produces y=i^1/2*i^1/4*i^1/8*.....
(At this point I'll introduce another simplifying convention: the principal roots of i are the roots of i that are closest to the positive real axis from above.)
Everything has the same base, so lets collect all the exponents into a common power. Then y=i^(1/2+1/4+1/8+....)
Now the exponent is a well-known geometric series with a value of 1. Then y=i is our answer.
I suspect this approach, resulting in y=i, is what Ady is intending, based on the D1 rating. But what if we chose a different system to choose the roots? Lets take i to its trigonometric form: i =
cos (pi/2) + i*sin(pi/2).
If we were to choose some finite number of roots other than the principal root we will get something like y = cos (pi/2 + r/2^n) + i*sin(pi/2 + r/2^n), where n is the deepest root and r is some
integer between 0 and 2^n. An example, taking non-principal roots of i^1/2 and i^1/8 may give us y = cos (pi/2 + pi*5/8) + i*sin(pi/2 + pi*5/8) = -(sqrt[2+sqrt(2)])/2 - i*(sqrt[2-sqrt(2)])/2.
Things get worse as n approaches infinity as then we can get arbitrarily close to any angle we want, thus making the potential domain of y being the entire circumference of the unit circle centered
at the origin of the complex plane.
In summary, if we take "sensible" simplifying restrictions then y=i is the solution but there's plenty of complications that leave this problem as ill-defined.
I'm not saying this is a bad problem; actually it's really good for doing a deep dive into what chaos root extraction can do in the complex numbers if you aren't careful. Especially with the infinite
radicals/product mixed in, which have their own complications.
Posted by Brian Smith on 2023-04-23 11:05:12 | {"url":"http://perplexus.info/show.php?pid=13340&cid=68277","timestamp":"2024-11-07T07:55:44Z","content_type":"text/html","content_length":"14279","record_id":"<urn:uuid:74c86eb7-e094-44cb-90c0-a6c3cd0f4c24>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00318.warc.gz"} |
[Solved] If x sin (90° − θ) cot (90° − θ) = cos (90° − θ), then... | Filo
If x sin (90° − θ) cot (90° − θ) = cos (90° − θ), then x =
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We have: x sin(90°-θ) cot (90°-θ)=cos (90°-θ)
Here we have to find the value of x
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Question Text If x sin (90° − θ) cot (90° − θ) = cos (90° − θ), then x =
Topic Introduction to Trigonometry
Subject Mathematics
Class Class 10
Answer Type Text solution:1
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Mathematical modeling of a clinical waste incineration process
Arshad, Khairil Anuar and Ahmad, Tahir and Baharun, Sabariah (2006) Mathematical modeling of a clinical waste incineration process. Project Report. Faculty of Science. (Unpublished)
Official URL: https://www.researchgate.net/publication/279480976...
At present, there is no fundamentally based mathematical model of the incineration process of clinical waste. The development of an accurate mathematical model of the process is being hampered by the
lack of measured data for validation due to its complexity feature. Due to the wide applications of the graph theory, a study was carried out to model a clinical waste incineration process using
graph theory. Initially, a graphical model was constructed to explain the connections of the variables in the incinerator, which serves as a basis for a more dynamic graphical representation of the
system. This model was found to be an Autocatalytic Set (ACS), conforming to the key feature of the model proposed by Jain and Krishna. Their model was then adopted and modified to suit the
incineration process, which has resulted in a model explaining the two dynamics of the incineration process, that is, the concentration dynamics and the graph dynamics of the system. The former
described the rate of change of the variables or species and the latter represented the evolution of the species in the incineration process.
Item Type: Monograph (Project Report)
Uncontrolled Keywords: mathematical model; incineration process; clinical waste.
Subjects: Q Science > QA Mathematics
Divisions: Science
ID Code: 2674
Deposited By: Nor Azlin Nordin
Deposited On: 21 May 2007 07:47
Last Modified: 13 Jun 2017 03:46
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Guadalupe - math word problem (82596)
Guadalupe bought snacks for her team's practice. She bought a bag of chips for $1.93 and a 15-pack of juice bottles for a total cost before tax of $32.83. Write and solve an equation that can be used
to determine x, the cost of each bottle of juice.
Correct answer:
Did you find an error or inaccuracy? Feel free to
write us
. Thank you!
Tips for related online calculators
Do you have a linear equation or system of equations and are looking for its
? Or do you have
a quadratic equation
You need to know the following knowledge to solve this word math problem:
Units of physical quantities:
Grade of the word problem:
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Calculating hyperfocal distances using MTF50 ratios - the last word
This is a continuation of a report on new ways to look at depth of field. The series starts here:
I’m going to leave the object-field approach for a bit, and continue on with refinements to the more traditional image plane methods, although I freely admit that I’m taking those methods to places
their originators probably never intended.
A while back, I came up with a measure that I called the hyper-hyperfocal distance, which was a distance to set your lens to to allow for a degradation in sharpness at infinity equal to 10% of the
MTF50 that you’d get if you actually focused at infinity. In other words, the hyper-hyperfocal distance allow retention of 90% of the sharpness available at infinity, in return for greater sharpness
at distances closer than the hyper-hyperfocal distance.
Retaining 90% of the resolution is a pretty high bar, and the hyper-hyperfocal distances that I calculated were depressingly long.
Today, I’m going to generalize the concept, and drop the prefixing “hyper”, and talk about hyperfocal distances that are not based on the desired circle of confusion (CoC), but on the ratio of the
MTF50 reduction at infinity to that obtained when focused at infinity.
Without further ado, here is a set of curves for our top-notch simulated 55 mm lens on a Sony a7RII:
The way to use those curves in the field would be to decide on how much degradation at infinity you can tolerate, and find that ratio on the horizontal axis. Then run your finger upwards, and you’ll
dee the hyperfocal distance for each f-stop.
If you prefer tables, here is the data in that form:
Usage is similar: find the column corresponding to how much loss you can tolerate at infinity, and read the hyperfocal distances in meters for each aperture.
After a little prompting and some spot checking, I’ve been working under the assumption that, like CoC-based hyperfocal distances, these scale with the focal length squared. This provides an
opportunity for more checking.
Running the modeling program with a 110 mm lens:
We see the distances are approximately four times the 55 mm case, as expected. But, if we calculate the ratios of the two tables, there’s a minor surprise:
Instead of the expected distribution on both sides of 4, there is a bias towards a lower number. I don’t know why that is, but I’m not going to worry about it for a while, if ever, since it’s so
When I was running these curves, I came up with a way to look at DOF that may be even more useful. Rather than bury it down her, I’m going to write a new post just about it.
1. Wally says
The reason you do not get exactly 4 for the hyperfocal distance ratio for doubling the focal length has to do with short distance correction in the DOF calculations. In the standard formulas the
far distance for DOF reaches infinity when f^2 = Nc(Do – f), where f is focal length, N is the f-number, c circle of confusion, Do object (or focus) distance. That extra f being subtracted from
the object distance is what messes up the perfect square ratio.
□ Jim says
I hear what you’re saying, Wally, but I don’t think that’s all of it, because the ratio varies with f-stop and threshold.
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Analysis of Dose Distribution According to the Initial Electron Beam of the Linear Accelerator: A Monte Carlo Study
Modern radiation therapy techniques that use a linear accelerator (LINAC), such as intensity modulated radiation therapy (IMRT), enable the dose distribution to better conform to the target volume
with a small subfield sequence. To predict the patient dose, computerized treatment planning systems (TPS) have been used in the clinic. Most TPS use the pencil beam (PB) convolution algorithm, which
obtains patient dose distributions using dose kernels consisting of previously calculated dose maps. This algorithm accelerates the dose calculation. However, the uncertainty of PB-based dose
calculation has been reported to be high in heterogeneous tissue and in small-field beams due to the limited lateral scatter calculation [
]. For special cases requiring very accurate dose assessment, the TPS-based dose distribution must be verified with an independent dose evaluation method [
The Monte Carlo (MC) simulation is based on the premise that modeling is highly accurate. Thus, it has been regarded as the most accurate dose calculation method for radiation beams delivered into
patients because it calculates the dose by considering each transported particle. This approach enables highly accurate dose calculation using a process similar to that of the actual beam delivery [
]. In this regard, the MC method is an important verification tool in the clinic.
However, its high computational burden has hindered its adoption, and the MC method is not currently available in the clinic [
MC simulations for dose calculation begin by producing the accelerated electrons delivered to the target and the x-ray beam generated through the bremsstrahlung process. However, it is very difficult
to determine the characteristics of the electron beam of interest. For example, (1) manufacturers do not provide standardized information about the initial electron beam, (2) the electron beam in the
vacuum head cannot be directly measured, and (3) electron beams differ even when produced by the same model and also differ because of the aging effect [
]. To address these issues, indirect beam commissioning has been employed. This approach modifies various parameters of the initial electron beam to match the simulated MC beam quality with the
measured quality by comparing the percent depth dose (PDD) and the lateral beam profile [
In previous studies for LINAC beam commissioning, initial beam spot size and mean electron energy [
] were chosen empirically by matching the simulated beam to the measurement by changing the electron characteristics [
]. While this simple method is the most commonly used method in beam commissioning [
], the results and conclusions reached with this method depend significantly on the researcher. To accurately and efficiently determine the characteristics of the electron beam, it is important to
understand the relationship between the initial electron and the following Bremsstrahlung photon. This relationship is affected by the treatment head and is used as a decision-making standard during
beam commissioning.
Tzedakis et al.[
] previously investigated the initial electron effect on the 6 MV treatment beam. Tzedakis and colleagues based their work on the study by Bjork on the LINAC initial electron characteristics [
] and used the Electron Gamma Shower 4 (EGS4) code for the Philips SL LINAC model. However, for 4D simulation with IMRT and image-guided radiation therapy (IGRT), Geometry And Tracking 4 (Geant4) is
the preferred tool because Geant4 provides a flexible simulation setup including geometries and DICOM import for patient dose calculation, in addition to providing the same information as the TPS [
]. Moreover, precise modeling of the multileaf collimator (MLC), which determines the fluence map, is difficult without Geant4 because of the complex geometry of this instrument.
The purpose of this study was to simplify the beam commissioning procedure and develop an accurate and efficient process for determining the optimal parameters for MC commissioning. To this end, the
dose distributions, PDD values, and lateral profiles of the Varian Clinac 2300 IX machine were evaluated according to the characteristics of the initial electron beam using the Geant4 toolkit.
Materials and Methods
1. LINAC head modeling based on the Geant4 toolkit
The Geant4 toolkit (ver. 4.9.6.p02), developed with the C++ language, was employed to simulate a 6 MV photon beam produced by a Varian Clinac 2300 IX machine. The major components of the LINAC head,
including the target, primary collimator, Beryllium window, flattening filter, ion chamber, mirror, jaws, and MLC, were modeled as shown in
Figure 1
. To achieve detailed modeling of the treatment head, blueprint supported by the machine vender was used as the reference when modeling the geometry.
2. Evaluation of the dosimetric effect according to initial electron beam parameters
To investigate the influence of (a) mean energy, (b) radial intensity distribution, and (c) energy spread on curve attributes such as depth of the maximum dose (d[max]) on the PDD and profile dose
flatness, the PDDs and lateral dose profiles were calculated and analyzed with different initial electron beam conditions.
1) Mean energy of the initial electron pencil beam
To investigate the effect of mean energy on the PDD and the lateral beam profile, simulations were carried out while varying the mean initial electron beam energy from 5.6 MeV to 6.4 MeV in
increments of 0.2 MeV. To evaluate only the influence of the mean electron energy on the dose distribution, the pencil beam was used, and the standard deviation of the energy was set to zero.
2) Radial intensity distribution of the initial electron beam
In a LINAC, the initial electron beam has a radial intensity distribution and an angular distribution because of the bending magnet, steering coils, and focusing coils in the electron transport
system. The electron distribution was assumed to have a two-dimensional (2-D) Gaussian distribution, as suggested by the Keall et al. study. In this study, the distributions of the initial electrons
generating the bremsstrahlung photon in the target were defined [
]. Karzmark et al. recommended that the electron angle should range from 0.06 to 0.3 degrees for the initial electron beam transport, which is close to zero.
Therefore, we assumed that the beam was parallel to the beam axis and only varied the full width at half maximum (FWHM) of the radial distribution (1 mm increments from 0 mm to 4 mm). To assess the
effect of electron beam radial width, the mean electron energy and energy spread were fixed at 6 MeV and 0 MeV, respectively.
3) Energy spread of the initial electron pencil beam
The energy spread of the initial electron can be estimated with different waveguide types, and energy slits in the bending magnet orbit [
]. The electron spread from the accelerating waveguide has been reported to assume a Gaussian distribution according to Tanabe and Hamm’s measurement [
We modeled the energy distribution of the initial electron as a Gaussian distribution and varied the FWHM from 0 MeV to 1.019 MeV in 0.127 MeV increments. A pencil beam was used, and the mean
electron energy was set to 6 MeV.
4) Normalization and comparison of PDDs and profiles
To evaluate the relationships between the initial electron characteristics and the LINAC photon beam dose distributions, the PDDs and profiles of the various initial electron conditions were compared
with those of the reference initial electron beam. The reference conditions of the initial electron beam were set to a mean energy of 6 MeV, a radial intensity of 0 mm at the FWHM, and an energy
spread of 0 MeV.
The simulated PDDs and profiles were normalized to the dose values at d[max] and the distribution center, respectively. The PDD was assessed at 2 mm intervals at depths ranging from 0 to 50 mm to
observe the dose distribution in build-up region and dose distribution nearby d[max]. For depths greater than 50 mm, the depth dose was assessed at 15 mm intervals for depths greater than 50 mm to
reduce the statistical fluctuation. The depth before d[max] was not considered in the PDD comparison for treatment beam modeling because electron equilibrium was not reached, and statistically
significant fluctuation occurs in that region. Moreover, electron contamination affects the dose distribution before d[max].
In evaluating the profile, the simulated dose values were compared to those of the reference electron beam only in the in-field region in which the dose values showed more than 80% of the dose at the
center, due to relatively high dose difference from few distance that can be regarded as ignorable in the dose fall-off region and in the out-field region. Also, high statistical error caused in the
simulation because of the beam collimators such as jaws and the MLC. The relative dose difference was calculated to evaluate the PDD and lateral profile values. The relative dose difference was
determined as follows:
3. Simulation conditions for dose calculation
In the simulation, G$EmStandardPhysics_option3 was used because it is suitable for calculating transportations of photon and electron in medical applications [
]. To reduce the calculation time, variance reduction techniques were applied to components of the LINAC head, e.g., bremsstrahlung splitting in the target and Russian roulette in the downstream
collimators. In addition, the phase space files containing information on the particles reaching the MLC (position, direction, kinetic energy, and particle type) were used to reduce the simulation
time. The photon beam produced by more than 2.8×10
initial electrons was recorded in the phase space files and then used repeatedly four times to obtain the dose distribution with reduced statistical fluctuation.
In the simulation, the source-to-surface distance (SSD) was set to 100 cm. The beam was delivered into a box-shaped voxelized water phantom measuring 50×50×40 cm^3. Each voxel measured 5×5×2 mm^3.
The PDD was calculated with a field size of 10×10 cm
, which is generally considered to be the reference size for assessing the effects of initial electron parameters, while the profile was calculated at d
with a field size of 40×40 cm
. Since the profile with the larger beam field is more heavily impacted by changes to the initial electron conditions such as mean electron energy, energy spread, and beam radial width [
], the largest field size was selected for the profile assessment. Finally, the optimal parameters were determined by comparing the PDDs and profiles of golden beam data (GBD) achieved with beams of
4×4 cm
, 10×10 cm
, and 30×30 cm
1. Effects of the initial electron beam on the PDD and profile
1) Mean energy of the initial electron pencil beam
Figure 2
shows the PDDs and lateral dose profiles according to the initial electron energy. It also displays their differences compared to the reference initial electron beam. The electron beam with a mean
energy of 5.8 MeV showed a minimum PDD difference of about 0.24% from the reference electron beam, which had a mean energy of 6 MeV. Increasing the initial energy of the electron beam also increased
the PDD values after d
. Specifically, the dose values increased on average by 4.36%, while the mean energy increased from 5.6 to 6.4 MeV. This finding indicates that the overall PDD curve moved toward the beam direction
with the higher electron energy, even though the reference parameters of the initial electron beam were the same.
In the lateral dose profile, the relative dose differences from the reference condition were decreased on average by 4.81% as the electron beam energy increased from 5.6 MeV to 6.4 MeV. The maximum
dose difference in the horn with the reference energy of the electron beam, 6 MeV, was approximately 9.21% compared to that with the 6.4 MeV electron beam. Under the initial electron reference
conditions, the minimum difference in average dose between the two profiles was 0.56% and was found when the mean electron energy was 5.8 MeV (
Figure 2
Figure 2
indicates that the average initial electron energy is more sensitive to the lateral dose profile then the PDD, because dose difference change according to the mean electron energy increased more with
the profile.
2) Radial intensity distribution of the initial electron beam
The PDD and lateral dose profile were calculated according to various electron beam radial intensity distributions. Differences from those of the reference electron beam are shown in
Figure 3
. For the PDD, the electron beam with a radial width of 1 mm at the FWHM showed a maximum difference after d
of 0.49% from the reference condition. Moreover, the minimum dose difference from the reference beam (0.26%) was observed with the beam with a 3 mm radial width at the FWHM. In addition, no
significant difference was observed between the PDD curves with respect to beam radial width (
Figure 3
The lateral profile showed an increased horn as the electron beam radial width at the FWHM decreased (
Figure 3
Figure 3
also shows that the initial electron beam radial distribution is inversely proportional to the profile dose values. Compared with the reference beam, the average dose difference in the horn region
increased from 1.49% to 15.10% as the beam radial width increased from 1 mm to 4 mm at the FWHM. The maximum and minimum differences in the horn were estimated at 23.67% and 3.28% from the reference
beam when the FWHM of the beam width was 4 mm and 1 mm, respectively. In addition, the overall profile dose values decreased on average by 8.42% when the radial width increased from 0 mm to 4 mm at
the FWHM. The results are in agreement with those of previous studies reporting that the radial distribution of the electron beam was inversely proportional to beam flatness and affected the PDD less
than the profile [
3) Energy spread of the initial electron pencil beam
Figure 4
shows the changes in PDD and lateral dose profile as the initial electron energy spread was varied. Similar to the results described in section Results 1), the PDD values for regions deeper than d
increased as the energy spread broadened. The electron beam with an energy spread of 1.019 MeV at the FWHM showed the largest average difference from the reference beam, 2.77%. The minimum difference
was 0.17% and was found with an energy spread of 0.127 MeV at the FWHM.
For the lateral profile, the curves showed dose differences averagely about 3% compared with the reference beam (FWHM of 0 MeV), which are comparable to the results of the other electron parameters (
Figure 4
). Among the energy spread conditions, the beam with an energy spread of 0.51 MeV at the FWHM showed the maximum average relative difference of 3.18% from the reference condition. The minimum
difference of 1.48% was observed when the beam had an energy spread of 0.127 MeV at the FWHM. The beam with the largest energy spread in this study, an FWHM of 1.019 MeV, resulted in a difference of
2.51% on the average from the reference condition, which corresponded to the median of the energy spreads. The differences between the energy spreads could represent statistical fluctuation in the
simulation, considering the dose differences shown in
Figure 4
. These findings indicate that the energy spread cannot significantly affect the lateral dose distribution.
2. MC commissioning of the photon beam
Figure 5
presents the beam commissioning results carried out for field sizes of 4×4 cm
, 10×10 cm
, and 30×30 cm
. A total of 2.8×10
, 5.6×10
and 5.6×10
initial electrons, respectively, were generated by beams with field sizes of 4×4 cm
, 10×10 cm
, and 30×30 cm
. The calculation of the dose distributions for the three field sizes required 39, 270, and 860 hours, respectively, with a single CPU.
The initial electron characteristics were determined by matching the PDD and the lateral profile with those of the GBD. Since the mean energy affects both the PDD and the lateral profile
simultaneously, the mean electron energy was determined first, and all other parameters were determined later. The optimized initial electron parameters were 5.9 MeV, 2.5 mm, and 0.8268 MeV for the
electron energy, beam radial width, and energy spread, respectively.
To evaluate the availability of the beams commissioned in this study, the maximum local dose difference from the GBD was calculated for each field size (
Figure 5
). The build-up region was not included in the PDD comparison due to its high statistical fluctuation in the MC simulation. When comparing the lateral dose profiles, the penumbra and dose fall-off
region were not considered. The profile depth was set to 10 cm from the phantom surface, according to the recommendations in the AAPM Task Group No. 51 (TG-51) report [
]. Maximum local dose differences between the commissioning results and the GBD results were assessed to be less than 2% for both the PDDs and the lateral profiles.
To accurately simulate photon beams from a LINAC head and calculate the dose distribution, it is important to determine the appropriate initial electron parameters. The quality of the initial
electron beam depends on the geometric accuracy of the LINAC model. The initial electron characteristics can even differ because of minor differences in the geometry. The geometry information offered
by vendors is sufficient to satisfy the required geometric similarity between the MC LINAC model and the actual LINAC machine. However, little information is available regarding the initial electron
beam properties, which hinders decision-making for initial electron parameters.
Electron parameters are typically decided through empirical approaches, which are time-consuming. Since each initial electron parameter is optimized separately in the beam commissioning process, the
lack of parameter knowledge complicates this process. Thus, for accurate and efficient beam commissioning, the effects of various electron parameters on beam quality should be determined. In this
study, we assessed the influence of three initial electron characteristics (mean energy, energy spread, and radial intensity distribution) on PDD and lateral dose profile. We also outlined a process
for electron parameter determination. Based on our results, we commissioned the initial electron beam of the Geant4-modeled LINAC.
At depths deeper than d[max], PDD increased when the mean electron energy and the FWHM of the energy spread increased. The higher mean energy of the initial electron generates higher bremsstrahlung
photon energy in the target, enabling the photons to produce secondary electrons that penetrate water more efficiently and can deliver their energy to deeper locations along the beam direction.
Depths shallower then d[max], which corresponded to the dose difference from the reference condition, showed larger statistical uncertainty compared to deeper locations. In the case of the energy
spread, larger energy spread indicates a higher likelihood of producing electrons with both higher and lower energies. Electrons with lower energies produce relatively low energy bremsstrahlung
photons in the target that consists of tungsten. They showed less effect on the PDD compared to the photons generated by higher energy electrons because most of them cannot penetrate the target and
reach the phantom, even though larger energy spread results in electron production with much broader energy variation than smaller energy spread. With this respect, a larger energy spread results in
increased PDD values only in the region deeper than d[max]. However, the dose differences from the reference condition as the energy spread was varied were smaller than those observed when the mean
electron energy was varied, even though the mean energy was not varied in a wider range than that of the energy spread. Potential explanations for the findings include: (a) the energy spread follows
a Gaussian distribution; (b) the majority of electrons had energy near the mean value; and (c) the mean electron energy of the spread was not changed, although the standard deviation of the energy
distribution changed. The PDD was changed insignificantly according to the change of the intensity distribution of the electron beam. While changing the initial electron distribution on the target
surface results in changes of the treatment beam width, it does not affect the energy of the bremsstrahlung photons produced in the target. Since the depth dose distribution is mainly affected by the
bremsstrahlung photon energy, the radial intensity distribution of the initial electron beam could not have any significant effect on the PDD.
For the lateral dose profile, the lateral dose horn was inversely proportional to both the mean electron energy and the radial beam width. Compared with the reference initial electron beam, the dose
values in the horn region decreased. The average dose values of the overall profiles were also decreased due to the increased mean electron energy and radial width. The increased initial electron
beam energy delivered a decreased dose near the region at the edge of the field, referred to as the horn. One potential explanation for this finding is that the higher energy photons tend to be less
scattered in the flattening filter and to deliver higher intensity energy in the middle of the dose profile than in the horn region. In contrast, the energy spread of the initial electron could not
remarkably affect the shape of the dose profile. A broader energy spread indicates an increased initial electron energy range. As mentioned above, however, the energy spread follows a Gaussian
distribution, and many of the electrons delivered into the target have an energy near the mean energy, even though there is an increased possibility that electrons are produced with high or low
energies that are far from the center value.
Our results indicate that mean electron energy is the most important factor in determining the dose distribution; that is, the mean energy simultaneously affects the entire PDD shape and the profile.
Thus, we recommend matching the overall PDD shapes and profile to the desired measurement by adjusting the mean electron energy and correcting the other details using the beam radial width and energy
The optimal electron parameters determined through the MC beam commissioning process were as follows: mean energy, radial width, and energy spread; 5.9 MeV, 2.5 mm at the FWHM, and 0.8268 MeV at the
FWHM, respectively. The dose distribution of the commissioned beam was compared with that of the GBD to demonstrate the dose availability. As shown in
Figure 5
, the maximum local difference from the GBD was less than 2%, indicating that the beam yielded an available dose distribution on the MC simulation.
In this study, we investigated the relationship between the characteristics of the initial electron and the dose distribution of the LINAC beam. Our findings could be applicable for MC-based modeling
of LINAC beams in the same model and also in other models. Based on these results, future studies will focus on developing more efficient beam commissioning protocols in MC simulations. In addition,
the beam modeled in this study can be used for cutting-edge radiation therapy techniques using LINAC beams, such as IMRT and VMAT. | {"url":"https://jrpr.org/journal/view.php?number=1004","timestamp":"2024-11-10T01:47:50Z","content_type":"application/xhtml+xml","content_length":"109927","record_id":"<urn:uuid:af9988f0-5b4b-4e50-b933-63e52424da6d>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00498.warc.gz"} |
Largest number in an array that is at least twice of others – Study Algorithms
Question: Given an array, there is a largest element N. Check if that number is at least twice than all the other elements in the array. Return the index if it is, else return -1
Input: {3, 6, 1, 0}
Output: -1
6 is the largest integer, and for every other number in the array x,
6 is more than twice as big as x. The index of value 6 is 1, so we return 1.
Let us try to make one more example
Input: nums = [1, 2, 3, 4]
Output: -1
Explanation: 4 isn’t at least as big as twice the value of 3, so we return -1.
– Scan through the array to find the unique largest element ‘N’, keeping track of it’s index maxIndex.
– Scan through the array again. If we find some x != N with N < 2 * x, we should return -1.
– Otherwise, we should return maxIndex.
public int twiceIndex(int[] arr) {
int maxIndex = 0;
for (int i = 0; i < arr.length; ++i) {
if (arr[i] > arr[maxIndex])
maxIndex = i;
for (int i = 0; i < arr.length; ++i) {
if (maxIndex != i && arr[maxIndex] < 2 * arr[i])
return -1;
return maxIndex;
Code language: Java (java)
Time Complexity: O(N) where N is the length of array.
Space Complexity: O(1), the space used by our int variables.
A working solution can be found here:- https://ideone.com/uyEpv9
0 comments
a tech-savvy guy and a design buff... I was born with the love for exploring and want to do my best to give back to the community. I also love taking photos with my phone to capture moments in my
life. It's my pleasure to have you here.
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Container with maximum water
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This Is Why...
You need to study math.
( If I want to make Lemon Chicken for supper, but the recipe I always use, to great effect, calls for half a can of frozen lemonade concentrate... but I don't have any frozen lemonade... I could
convert the recipe from frozen lemonade concentrate to real, fresh lemonade, but calculate the water out, so as to concentrate it... therefore, if one full can is 355 ml, and if one whole can makes 4
times 355 ml- considering one usually has to add 4 cans of water... or would that actually equal 5 times 355 ml, because you've got 4 cans of water, plus one can of lemonade concentrate?.... and it
takes 125 ml of lemon juice and 125 ml of sugar plus 800 ml of water to make one litre of lemonade... then how much lemon juice and how much sugar, would be in half a can of frozen lemonade
concentrate...... so I should use the full amount of lemon juice and sugar that would be in half a can, plus half of 355 ml of water.... add to that the rest of the usual recipe ingredients... )
And you thought I had a cushy job.
2 comments:
1. I know the answer.
The answer is 'some'.
It can also be 'add to taste' in the tradition of good good cooks everywhere.
BTW 125g of sugar?
But IF I were to do it, looking at those figures I'd estimate you'd want around 900ml of lemonade equivalent, which means you'll need about 110ml of juice and 110g sugar. This is just routine
stuff really for working round here: how much complete HAT Doma Drive medium do I need for 18 T75 and 6 T25 flasks when a T75 requires 35ml and a T25 requires 10ml.
Answer - some.
Like I told Chris, cooking is JUST like working in the lab.
Laura - you should have been a scientist.
p.s. I always multiply the quantities by the number of covers required divided by the number a meal will feed FWIW.
2. or you could just run to the store and buy some lemonade concentrate | {"url":"http://www.lauralea.ca/2006/05/this-is-why_21.html","timestamp":"2024-11-13T14:24:15Z","content_type":"text/html","content_length":"53647","record_id":"<urn:uuid:a17478d0-6b62-487a-84e1-5f39d272339c>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00808.warc.gz"} |
Shane Lechler is Overrated...Or Is He?
I've been waiting for a good opportunity to write this article for a while. Recently, Carl Bialik at the WSJ asked me to take a look at some punting stats for a post on the 'Golden Age of Punting.'
For the last few years, Shane Lechler has been lauded for his super-human punting ability. For three of the last six years he's led the league in total punting yards, and for five of the last seven
years he's lead the league in yards per punt. He'll be going to his fourth pro-bowl in a few weeks.
But, there's a problem with all those gaudy stats. Lechler has the benefit of playing for the Oakland Raiders, who for the last few seasons have fielded a terrible offense. Lechler, more than most
other punters, gets to punt from deep in his own territory where the chance of a touchback doesn't shorten punts. We can call this the JaMarcus Effect.
In this post, I'll look at where Lechler's average field position is compared to the rest of the league. We'll see that this has a big effect on punt distances. But we'll also see that I'm wrong
about Lechler in the end. Despite this unfair advantage, he's still the NFL's best.
In the NFL this year, the average field position for punts has been a team's own 35.2 yard line. Oakland, however, sports an average of their own 31.2-yard line, the deepest of all 32 teams.
Lechler's net punt average is top in the league at 43.6 yds/punt. But for all teams, the correlation between average punt field position and average net punt distance is -0.45, meaning the deeper
your field position the longer the punt. It's no wonder Lechler tops the league most years. Below is a graph of expected net punt distance by field position to illustrate what I mean.
Accounting for the JaMarcus Effect, Lechler is still the best. Using Expected Points Added (EPA) and Win Probability Added (WPA) we can compare the expected value of a punter's kicks with the actual
values. In other words, if we know how many Expected Points each of Lechler's punts should gain, and he exceeds that, then we can say he's good. We can compare all punters this way, and see who
really had the best season. WPA can tell us which punter really made the biggest contribution to his team when it mattered most.
I'm going to look at average EPA per punt and total season WPA. I think a 'per punt' stat is most useful because obviously some punters are going to have more opportunities than others. But for WPA,
which measures impact on a game, I think a 'total' stat is more appropriate.
As it turns out, Shane Lechler is number one in both average EPA and total WPA. Below is a table of all 32 teams and their punting stats--Average EPA, total WPA, average field position for punts, and
average net punt distance.
I broke them out by team rather than punter because it's about 1,000 times easier with my data set. But this highlights an important point. We're really looking at total team punting performance, not
just the individual kicker. (Also, if someone is burning to know the stats for specific guy in the cases where a team changed punters mid-season, let me know. I'll generate the numbers.)
Congratulations Mr. Lechler, you just made the Advanced NFL Stats All-WPA team! (Psst. Your agent can reach me at 703-555-...)
│Punter│EPA Avg│WPA Sum│Avg YL (own) │Avg Net│
│OAK │0.25 │0.64 │31.2 │43.6 │
│BLT │0.22 │-0.33 │36.7 │39.9 │
│DAL │0.22 │0.35 │36.8 │41.1 │
│HST │0.20 │0.04 │36.4 │38.4 │
│NO │0.15 │-0.16 │37.2 │37.1 │
│SEA │0.15 │0.28 │33.0 │40.5 │
│KC │0.14 │0.47 │34.8 │40.9 │
│SL │0.13 │0.10 │36.1 │41.4 │
│ARZ │0.12 │0.64 │36.4 │39.7 │
│SF │0.09 │0.51 │32.0 │39.6 │
│ATL │0.06 │-0.26 │35.8 │36.6 │
│NYJ │0.04 │0.22 │36.2 │37.3 │
│BUF │0.03 │0.55 │33.3 │40.4 │
│NE │0.03 │-0.28 │39.7 │33.7 │
│SD │0.02 │-0.21 │34.7 │39.8 │
│MIN │0.00 │-0.40 │35.8 │39.2 │
│CLV │-0.01 │0.52 │37.5 │38.0 │
│MIA │-0.03 │0.01 │33.7 │39.2 │
│DEN │-0.05 │-0.59 │35.6 │37.4 │
│CAR │-0.06 │-0.21 │35.4 │36.0 │
│JAX │-0.06 │0.14 │34.4 │38.9 │
│NYG │-0.07 │-0.36 │35.8 │36.6 │
│DET │-0.08 │0.03 │33.7 │36.8 │
│IND │-0.09 │0.05 │35.2 │37.2 │
│PIT │-0.10 │-0.14 │38.0 │37.2 │
│CHI │-0.13 │0.10 │34.4 │38.1 │
│PHI │-0.15 │-0.19 │34.7 │36.0 │
│TEN │-0.15 │0.06 │33.1 │37.7 │
│TB │-0.17 │-0.23 │35.0 │36.1 │
│CIN │-0.22 │-0.09 │34.5 │37.1 │
│WAS │-0.23 │-0.50 │36.5 │36.1 │
│GB │-0.25 │-0.91 │35.5 │34.2 │
27 Responses to “Shane Lechler is Overrated...Or Is He?”
Another David says:
Accounting for the crappiness of a team is always so frustrating. During the Heisman race I was always rooting for Toby Gerhart, but a lot of my friends from Pennsylvania were pulling for
Dion Lewis, pointing to his 1799 yards. Of course he's got 1799 yards, he's Pitt's only decent running back! You put the ball in a guy's hands four times more often than the next three
running backs combined and he's bound to rack up yards.
Joe G says:
Interesting how WPA and EPA only have a correlation of .53. I thought it would be higher. Does it imply that quite a few kickers kick better and worse when the game is on the line?
Jake A says:
Did you isolate the effects of coaching decisions on punting WPA? In earlier articles, this site has made a case that choosing to punt in certain game situations is not the optimal choice
and may lower the team's win probability. Therefore, it seems certain punters would already be at a disadvantage because of a game decision that has nothing to do with their punting
Very nice article, as usual.
Martin says:
Colts is 9th worst in EPA(-0.09), but 14th in WPA (0.05). They is quite average in YL (number 17). How can that be?
Brian Burke says:
Joe-I'm surprised the correlation is that high. WPA can be erratic. One really bad or really good play in a very high-leverage situation can dominate a bunch of smaller plays in the final
Jake-Thanks. These numbers all assume conventional coaching decisions--punts outside the 35, field goals inside the 35. If a coach happens to punt inside the 35, it doesn't matter that
much because the EP values between FGs and punts are break-even in that region of the field. So no, punters are not penalized for sub-optimum coaching decisions. Punts are only compared
to other punts, and not to conversion attempts.
Martin-See my comment to Joe above. A -0.9 EP value translates to about a 1.5-yard/punt difference from the league average. One lesson here is that most teams' punt squads are
indistinguishable from one another. All teams are going to have good punt plays and bad punt plays. In the end it's a question of when the good ones and bad ones happen--in a critical
situation or when the game is already decided? If I were a GM, I wouldn't use a lot of salary or draft picks specifically for my punting unit.
Jonathan says:
"Accounting for the crappiness of a team is always so frustrating. During the Heisman race I was always rooting for Toby Gerhart, but a lot of my friends from Pennsylvania were pulling
for Dion Lewis, pointing to his 1799 yards. Of course he's got 1799 yards, he's Pitt's only decent running back! You put the ball in a guy's hands four times more often than the next
three running backs combined and he's bound to rack up yards."
That reminds me of how people want to put Chris Osgood in the Hall of Fame solely for winning three Stanley Cups and a high number of regular season games...on the Red Wings. Even a
mediocre goaltender (by NHL standards) could win a lot of games on the Red Wings.
Brian Burke says:
Jake-I take that back. There may be some contamination with fake punt conversion attempts for some teams. It depends on how I searched for the word 'PUNT' in each play description.
Joe G says:
What exactly does the WP for a punt mean? Say a team has a 4th and 3 from their own 40 with a WP of 62%. At that point the 62% assumes an average punt right? So an average punt gives the
opponebt the ball with a WP of 38%. But a better net punt might give the opponent only a 37% chance - so the punt gets credit for a 1% gain in WP. Is that how you are measuring this?
I guess what surprised me about the correlation between EPA and WPA was that punts don't seem to have the game changing potential that a last second pass from a QB could have. I mean
maybe a punter gets some great 25 yard bounce, or shanks a punt 15 yards, but those are rare. Punters seem pretty much a commodity to me.
Interesting stuff.
Brian Burke says:
I didn't do a good job explaining these stats.
WP for a punt means that the model sees that there is a 4th down outside field goal range and expects the coach to call for a punt. It calculates the expected net punt distance for the
given field position of the punt. Then it calculates the WP for current situation (time, score) and the opponent having a first down at the expected resulting field position. Call this WP
(avg punt).
The model also calculates the actual resulting WP after the punt. Call this WP(actual punt). The difference in WP between WP (actual punt) minus WP(avg punt) equals WPA.
Deeper punts result in high postive WPA. Shanks would be negative WPA.
WP and WPA are context-sensitive, meaning that WPA is magnified depending on how critical the situation is. Pinning a team on the 1 when the score is close would result in a very high
Joe G says:
Thanks - I'm with you on your method.
Slight tangent - but it relates to punting - in a tie game, under what conditions does possessing the ball have a WP < 50%?
It is prolly not linear with time remaining for a fixed yard line. IOW, if I have the ball on my own 5 in a tie game with 3 or more minutes remaining, is that WP < 50%? But if they were
just 45 seconds left, the WP > 50% - since for the most part I will win or go to OT.
I'm wondering if you have enough data to calculate when the WP flips for a given yard line.
Brian Burke says:
Joe-I can calculate that, roughly at least. In fact, it's already in the WP model, and just has to be pulled out. But you're right, it's variable depending on time remaining.
But in general, in 'normal football' when time is not a factor, the EP curve crosses zero at a team's own 15 yd line. So at a 1st and 10 on your own 15, you and your opponent are equally
likely to score the same amount of points.
Tarr says:
I'm a bit curious about the stats of Brett Kern. He was cut by the Denver Broncos after their 6-0 start, and signed by the 0-6 Titans. Of course, the teams' fortunes reversed right then,
with the Broncos finishing 2-8 while the Titans finished 8-2. I'm sure Kern had very little to do with this, but I'm curious if he was better than the other Bronco and Titan punters.
Joel says:
Nice piece - I think R^2 would be a better diagnostic than correlation since the relationship is non-linear and correlation only measures linear relationships. It would be nice to see
this done as a GLM analysis where you could first factor in the yard line, and then see the effect of adding punter as an additional factor. Then you're essentially answering the question
"Given the ball is on the X yard line, which punter is most likely to punt the deepest?"
Scott C. says:
EPA Average is the wrong stat here.
A kicker who kicks more often from farther away has an advantage in gaining EP value.
In other words, the EP gained needs to be normalized by the EP opportunity. If kicker A is forced to kick from the 50, where the BEST POSSIBLE RESULT is only 0.5 EP better than average
(just numbers i'm making up), it is not fair to compare to a kicker who kicks from his own 20 yard line, where the best possible result is 1.5EP.
In short EP is the wrong stat entirely. It is the right stat to measure the team's overall 'value from punting' but wrong for evaluating a player. The player who earned 0.5 EP by kicking
from the 50 might earn 1.0 EP by kicking from his 20 instead.
This study does NOT adjust for difference in opportunity properly.
Scott C. says:
Just to be clear, when I say "The player who earned 0.5 EP by kicking from the 50 might earn 1.0EP by kicking from his 20 instead", I mean 0.5 EP 'above average'.
It is not sufficient to just control for the average and aggregate. There is more opportunity to exceed the average at some field position locations than others.
The same thing holds for any WP analysis.
The above controls for value over average, but does not control for differences in opportunity.
Brian Burke says:
Scott-I see what you're suggesting. Some kicks have a greater opportunity for EP gain or loss then others. So do something like: this kick was +.5 SD given its field position. Another
kick was -1.0 SD given its field position, etc. Add them up or average them out to properly normalize.
Brian Burke says:
Scott-What would you suggest for field goals? I am planning to use similar methodology for kickers. So you see a similar issue?
Joe G says:
I think Scott is technically correct.
But Lechler's gap over Average YL is only 4 yard - or looking at the graph above - the 69 yards out vs. 65 yard out.
Since that gap doesn't come close to the bend in the curve, I would expect a significant difference in the results that were normalized for opportunity.
Joe G says:
I meant "would not expect" in above sentence.
Scott C. says:
Argh! I just lost a big post. I'll make a few small ones instead.
Part 1:
The sample size is too small to do this, but hypothetically you could build a "profile" for each player --
Expected value per kick as a function of distance.
Players would have different profiles, and you could translate players to other player's contexts:
How valuable would Lechler be if he kicked the same distance distribution as Scifres and vice-versa? What about against the league average kick distribution?
Scott C. says:
Part 2:
Well, there isn't enough data to do that, but we can bucket punts into a few types. Based on the chart above, there are three regions that make sense to me:
> 65 yards away (long). This is where the graph is flat and the far end zone doesn't come into play often.
> 50 and <=65 yards (medium). This is the 'knuckle' in the graph where the end zone sometimes comes into play.
<= 50 yards (short). This is where kick distance is almost linearly correlated with distance from the end zone.
Scott C. says:
Within these ranges you could do the same analysis you have already done for each player/team. You can also add the number of opportunities that each player had in each kick type, and the
average distance that player was at within each kick distance bucket.
Adding the cumulative standard deviation from the mean for a kicker as you suggest is useful too. It helps weigh each kick chance equally, but still has opportunity bias.
Alternatively, one could attempt to generate an expected kick value profile as a function of distance for each player if you assume or prove that they all ave the same sort of distance
curve: linearly correlated with distance from the goal line up until distance A, and constant kick length when farther than distance B.
Then, each punter could be 'rated' based on the four numbers that define that curve.
Scott C. says:
As I see it, this opportunity problem is similar to Place Kickers.
In each case, one "type" of player is better suited at certain types of kicks than others. If I had to make 10 punts from the 50 yard line (dumb idea given your other research!) I would
want a kicker like Scifres, not Lechler. Just because one kicker is asked to make a punt like that doesn't make them a worse punter, they are just being used differently.
Your work here is an excellent measurement of how much value the team gets from punting. But its hard to measure a player's quality because that isn't something you can do with one score.
I'm sure that we could dig further in to measure player's capability at different types of kicks. I'm sure Lechler is the king of the long kick, and among the top at the medium kicks. But
I'd wager that there are several kickers who are much better at the shorter kicks. The Raiders do get to use his strength more than most teams.
For place kickers you have the same sort of problem. One guy might be dependable from the 40 in, but not have the leg to hit a 55+. While another might have a well above average chance at
making a 60 yard kick but be expected to shank a 25 yarder more often than most.
If one team only ever kicked within the 40 all year long, the former is the 'better player'. But if that same team ended up trying a lot of end of game or end of half 55+ yard attempts
the latter player would be 'better'. Unfortunately, there are too few long kicks in a single year to measure a kicker's 'long distance' ability accurately, so such a study would have to
span several years.
Ben Morris says:
It seems to me this approach is still going to be biased. Take two punters who are both exactly 1 standard deviation above the mean. The one who attempts a higher percentage of punts from
positions on the field or at times in the game where/when +1 SD results are the most valuable will have significantly better numbers. And specifically, I suspect (though I haven't looked
at the data) that punts from fairly deep in your own territory -- which are more likely to be correct in the first place -- are probably also more important.
This could be figured out fairly easily by essentially repeating your graph above placing "line of scrimmage" on the X axis and "standard deviation of EPA" on the Y axis.
As for evaluating the skills of particular punters, there may be a better way, and this is off the top of my head, but I guess it could be fairly accurately approximated by taking their
averages from each yard line -- counted by standard deviations above the league mean for that yard line -- and then average those SDs, weighing by the league frequency of kicks from each
position AND the "importance" of each kick (i.e., the standard deviations of the EPA). Obviously there could be outlier issues, but those could be solved by smoothing out the data first.
Of course, you could go more accurate by breaking it down further for different game situations and such, but I think the benefits of that would be pretty marginal.
Buzz says:
Couple of quick questions on the EPA model as I was trying some things with the files you provided.
1) How are fumbles on punts handled. For example if above Lechler punted the ball to Cribbs and he fumbled it which led to an oakland TD the play would show as a huge +EP even though he
didn't have anything to do with it. Does Lechler get this credit or do you base the EP for the punt and return yards only? I guess this would be a similar question to QB's. If Jamarcus
Russell completes a pass to Heyward - Bey and Bey fumbles does Russell get penalized.
2) In one comment on an article I was reading last night said that TD's were worth 6.4pts because you have to subtract .6 points but most of the time you have said you subtract .7. Was
the .6 a mistake or have you modified this number? Sorry I forget which article it was commented on but it was a fairly recent one.
3)Do you subtract the .7 (or .6) from every play or just the scoring plays? For instance, i think you have said having the ball at the 27 yard line is worth .7points. Is this after
subtracting out .7 to where it would be worth 1.4 if there was no subsequent kickoff.
Thanks for your answers and all the great analysis and especially for posting all of the great play by play data.
Buzz says:
Sorry another question I forgot to put in this post. I know we are looking at the next score for all drives and assigning them back to the original plays but how do you account for drives
that don't have any points scored in them. For example, lets say two teams are scoreless at the half. Do you exclude all plays in the first half (1st qtr since you only used first qtr
plays) since there was no "next score" or do you use a zero as the "next score"? I can see some merit to both. Thanks again!
Brian Burke says:
1) Ideally, I'd only credit to the point where the punt lands, then adjust for the average expected return. But I just don't have the time to go into that level of depth, so this current
measure includes anything that happens on a punt play including turnovers.
2) The .6 or .7 is just a matter of rounding and how you calculate it. Once you adjust all the EP values accounting for kickoffs, now the kickoff EP value changes itself. So you have to
do it again. It's iterative. In the end it doesn't matter that much when measuring individual plays because the starting and ending point for each play are equally affected by whichever
value you chose.
3) The .6/.7 EP adjustment is only for scoring plays, but then that affects the value of every down at every yard line. So the answer is yes and no, both, but not.
4) If two teams are scoreless through the half, it would count as zero EP (in my current implementation). I could see where you might want to throw them out, but either way, that's
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Fluid Mechanics
McGraw-Hill Series in Mechanical Engineering CONSULTING EDITORS Jack P. Holman, Southern Methodist University John Lloyd, Michigan State University Anderson Computational Fluid Dynamics: The Basics
with Applications Anderson Modern Compressible Flow: With Historical Perspective Arora Introduction to Optimum Design Borman and Ragland Combustion Engineering Burton Introduction to Dynamic Systems
Analysis Culp Principles of Energy Conversion Dieter Engineering Design: A Materials & Processing Approach Doebelin Engineering Experimentation: Planning, Execution, Reporting Driels Linear Control
Systems Engineering Edwards and McKee Fundamentals of Mechanical Component Design Gebhart Heat Conduction and Mass Diffusion Gibson Principles of Composite Material Mechanics Hamrock Fundamentals of
Fluid Film Lubrication Heywood Internal Combustion Engine Fundamentals
Kimbrell Kinematics Analysis and Synthesis Kreider and Rabl Heating and Cooling of Buildings Martin Kinematics and Dynamics of Machines Mattingly Elements of Gas Turbine Propulsion Modest Radiative
Heat Transfer Norton Design of Machinery Oosthuizen and Carscallen Compressible Fluid Flow Oosthuizen and Naylor Introduction to Convective Heat Transfer Analysis Phelan Fundamentals of Mechanical
Design Reddy An Introduction to Finite Element Method Rosenberg and Karnopp Introduction to Physical Systems Dynamics Schlichting Boundary-Layer Theory Shames Mechanics of Fluids Shigley Kinematic
Analysis of Mechanisms Shigley and Mischke Mechanical Engineering Design Shigley and Uicker Theory of Machines and Mechanisms
Hinze Turbulence
Stiffler Design with Microprocessors for Mechanical Engineers
Histand and Alciatore Introduction to Mechatronics and Measurement Systems
Stoecker and Jones Refrigeration and Air Conditioning
Holman Experimental Methods for Engineers
Turns An Introduction to Combustion: Concepts and Applications
Howell and Buckius Fundamentals of Engineering Thermodynamics
Ullman The Mechanical Design Process
Jaluria Design and Optimization of Thermal Systems
Wark Advanced Thermodynamics for Engineers
Juvinall Engineering Considerations of Stress, Strain, and Strength
Wark and Richards Thermodynamics
Kays and Crawford Convective Heat and Mass Transfer
White Viscous Fluid Flow
Kelly Fundamentals of Mechanical Vibrations
Zeid CAD/CAM Theory and Practice
Fluid Mechanics Fourth Edition
Frank M. White University of Rhode Island
Burr Ridge, IL Dubuque, IA Madison, WI New York San Francisco St. Louis Bangkok Bogotá Caracas Lisbon London Madrid Mexico City Milan New Delhi Seoul Singapore Sydney Taipei Toronto
About the Author
Frank M. White is Professor of Mechanical and Ocean Engineering at the University of Rhode Island. He studied at Georgia Tech and M.I.T. In 1966 he helped found, at URI, the first department of ocean
engineering in the country. Known primarily as a teacher and writer, he has received eight teaching awards and has written four textbooks on fluid mechanics and heat transfer. During 1979–1990 he was
editor-in-chief of the ASME Journal of Fluids Engineering and then served from 1991 to 1997 as chairman of the ASME Board of Editors and of the Publications Committee. He is a Fellow of ASME and in
1991 received the ASME Fluids Engineering Award. He lives with his wife, Jeanne, in Narragansett, Rhode Island.
To Jeanne
General Approach
The fourth edition of this textbook sees some additions and deletions but no philosophical change. The basic outline of eleven chapters and five appendices remains the same. The triad of integral,
differential, and experimental approaches is retained and is approached in that order of presentation. The book is intended for an undergraduate course in fluid mechanics, and there is plenty of
material for a full year of instruction. The author covers the first six chapters and part of Chapter 7 in the introductory semester. The more specialized and applied topics from Chapters 7 to 11 are
then covered at our university in a second semester. The informal, student-oriented style is retained and, if it succeeds, has the flavor of an interactive lecture by the author.
Learning Tools
Approximately 30 percent of the problem exercises, and some fully worked examples, have been changed or are new. The total number of problem exercises has increased to more than 1500 in this fourth
edition. The focus of the new problems is on practical and realistic fluids engineering experiences. Problems are grouped according to topic, and some are labeled either with an asterisk (especially
challenging) or a computer-disk icon (where computer solution is recommended). A number of new photographs and figures have been added, especially to illustrate new design applications and new
instruments. Professor John Cimbala, of Pennsylvania State University, contributed many of the new problems. He had the great idea of setting comprehensive problems at the end of each chapter,
covering a broad range of concepts, often from several different chapters. These comprehensive problems grow and recur throughout the book as new concepts arise. Six more open-ended design projects
have been added, making 15 projects in all. The projects allow the student to set sizes and parameters and achieve good design with more than one approach. An entirely new addition is a set of 95
multiple-choice problems suitable for preparing for the Fundamentals of Engineering (FE) Examination. These FE problems come at the end of Chapters 1 to 10. Meant as a realistic practice for the
actual FE Exam, they are engineering problems with five suggested answers, all of them plausible, but only one of them correct. xi
New to this book, and to any fluid mechanics textbook, is a special appendix, Appendix E, Introduction to the Engineering Equation Solver (EES), which is keyed to many examples and problems
throughout the book. The author finds EES to be an extremely attractive tool for applied engineering problems. Not only does it solve arbitrarily complex systems of equations, written in any order or
form, but also it has builtin property evaluations (density, viscosity, enthalpy, entropy, etc.), linear and nonlinear regression, and easily formatted parameter studies and publication-quality
plotting. The author is indebted to Professors Sanford Klein and William Beckman, of the University of Wisconsin, for invaluable and continuous help in preparing this EES material. The book is now
available with or without an EES problems disk. The EES engine is available to adopters of the text with the problems disk. Another welcome addition, especially for students, is Answers to Selected
Problems. Over 600 answers are provided, or about 43 percent of all the regular problem assignments. Thus a compromise is struck between sometimes having a specific numerical goal and sometimes
directly applying yourself and hoping for the best result.
Content Changes
There are revisions in every chapter. Chapter 1—which is purely introductory and could be assigned as reading—has been toned down from earlier editions. For example, the discussion of the fluid
acceleration vector has been moved entirely to Chapter 4. Four brief new sections have been added: (1) the uncertainty of engineering data, (2) the use of EES, (3) the FE Examination, and (4)
recommended problemsolving techniques. Chapter 2 has an improved discussion of the stability of floating bodies, with a fully derived formula for computing the metacentric height. Coverage is
confined to static fluids and rigid-body motions. An improved section on pressure measurement discusses modern microsensors, such as the fused-quartz bourdon tube, micromachined silicon capacitive
and piezoelectric sensors, and tiny (2 mm long) silicon resonant-frequency devices. Chapter 3 tightens up the energy equation discussion and retains the plan that Bernoulli’s equation comes last,
after control-volume mass, linear momentum, angular momentum, and energy studies. Although some texts begin with an entire chapter on the Bernoulli equation, this author tries to stress that it is a
dangerously restricted relation which is often misused by both students and graduate engineers. In Chapter 4 a few inviscid and viscous flow examples have been added to the basic partial differential
equations of fluid mechanics. More extensive discussion continues in Chapter 8. Chapter 5 is more successful when one selects scaling variables before using the pi theorem. Nevertheless, students
still complain that the problems are too ambiguous and lead to too many different parameter groups. Several problem assignments now contain a few hints about selecting the repeating variables to
arrive at traditional pi groups. In Chapter 6, the “alternate forms of the Moody chart” have been resurrected as problem assignments. Meanwhile, the three basic pipe-flow problems—pressure drop, flow
rate, and pipe sizing—can easily be handled by the EES software, and examples are given. Some newer flowmeter descriptions have been added for further enrichment. Chapter 7 has added some new data on
drag and resistance of various bodies, notably biological systems which adapt to the flow of wind and water.
Chapter 8 picks up from the sample plane potential flows of Section 4.10 and plunges right into inviscid-flow analysis, especially aerodynamics. The discussion of numerical methods, or computational
fluid dynamics (CFD), both inviscid and viscous, steady and unsteady, has been greatly expanded. Chapter 9, with its myriad complex algebraic equations, illustrates the type of examples and problem
assignments which can be solved more easily using EES. A new section has been added about the suborbital X33 and VentureStar vehicles. In the discussion of open-channel flow, Chapter 10, we have
further attempted to make the material more attractive to civil engineers by adding real-world comprehensive problems and design projects from the author’s experience with hydropower projects. More
emphasis is placed on the use of friction factors rather than on the Manning roughness parameter. Chapter 11, on turbomachinery, has added new material on compressors and the delivery of gases. Some
additional fluid properties and formulas have been included in the appendices, which are otherwise much the same.
The all new Instructor’s Resource CD contains a PowerPoint presentation of key text figures as well as additional helpful teaching tools. The list of films and videos, formerly App. C, is now omitted
and relegated to the Instructor’s Resource CD. The Solutions Manual provides complete and detailed solutions, including problem statements and artwork, to the end-of-chapter problems. It may be
photocopied for posting or preparing transparencies for the classroom.
EES Software
The Engineering Equation Solver (EES) was developed by Sandy Klein and Bill Beckman, both of the University of Wisconsin—Madison. A combination of equation-solving capability and engineering property
data makes EES an extremely powerful tool for your students. EES (pronounced “ease”) enables students to solve problems, especially design problems, and to ask “what if” questions. EES can do
optimization, parametric analysis, linear and nonlinear regression, and provide publication-quality plotting capability. Simple to master, this software allows you to enter equations in any form and
in any order. It automatically rearranges the equations to solve them in the most efficient manner. EES is particularly useful for fluid mechanics problems since much of the property data needed for
solving problems in these areas are provided in the program. Air tables are built-in, as are psychometric functions and Joint Army Navy Air Force (JANAF) table data for many common gases. Transport
properties are also provided for all substances. EES allows the user to enter property data or functional relationships written in Pascal, C, C, or Fortran. The EES engine is available free to
qualified adopters via a password-protected website, to those who adopt the text with the problems disk. The program is updated every semester. The EES software problems disk provides examples of
typical problems in this text. Problems solved are denoted in the text with a disk symbol. Each fully documented solution is actually an EES program that is run using the EES engine. Each program
provides detailed comments and on-line help. These programs illustrate the use of EES and help the student master the important concepts without the calculational burden that has been previously
So many people have helped me, in addition to Professors John Cimbala, Sanford Klein, and William Beckman, that I cannot remember or list them all. I would like to express my appreciation to many
reviewers and correspondents who gave detailed suggestions and materials: Osama Ibrahim, University of Rhode Island; Richard Lessmann, University of Rhode Island; William Palm, University of Rhode
Island; Deborah Pence, University of Rhode Island; Stuart Tison, National Institute of Standards and Technology; Paul Lupke, Druck Inc.; Ray Worden, Russka, Inc.; Amy Flanagan, Russka, Inc.; Søren
Thalund, Greenland Tourism a/s; Eric Bjerregaard, Greenland Tourism a/s; Martin Girard, DH Instruments, Inc.; Michael Norton, Nielsen-Kellerman Co.; Lisa Colomb, Johnson-Yokogawa Corp.; K. Eisele,
Sulzer Innotec, Inc.; Z. Zhang, Sultzer Innotec, Inc.; Helen Reed, Arizona State University; F. Abdel Azim El-Sayed, Zagazig University; Georges Aigret, Chimay, Belgium; X. He, Drexel University;
Robert Loerke, Colorado State University; Tim Wei, Rutgers University; Tom Conlisk, Ohio State University; David Nelson, Michigan Technological University; Robert Granger, U.S. Naval Academy; Larry
Pochop, University of Wyoming; Robert Kirchhoff, University of Massachusetts; Steven Vogel, Duke University; Capt. Jason Durfee, U.S. Military Academy; Capt. Mark Wilson, U.S. Military Academy;
Sheldon Green, University of British Columbia; Robert Martinuzzi, University of Western Ontario; Joel Ferziger, Stanford University; Kishan Shah, Stanford University; Jack Hoyt, San Diego State
University; Charles Merkle, Pennsylvania State University; Ram Balachandar, University of Saskatchewan; Vincent Chu, McGill University; and David Bogard, University of Texas at Austin. The editorial
and production staff at WCB McGraw-Hill have been most helpful throughout this project. Special thanks go to Debra Riegert, Holly Stark, Margaret Rathke, Michael Warrell, Heather Burbridge, Sharon
Miller, Judy Feldman, and Jennifer Frazier. Finally, I continue to enjoy the support of my wife and family in these writing efforts.
Preface xi
2.6 2.7 2.8 2.9 2.10
Chapter 1 Introduction 3 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12 1.13 1.14
Preliminary Remarks 3 The Concept of a Fluid 4 The Fluid as a Continuum 6 Dimensions and Units 7 Properties of the Velocity Field 14 Thermodynamic Properties of a Fluid 16 Viscosity and Other
Secondary Properties 22 Basic Flow-Analysis Techniques 35 Flow Patterns: Streamlines, Streaklines, and Pathlines 37 The Engineering Equation Solver 41 Uncertainty of Experimental Data 42 The
Fundamentals of Engineering (FE) Examination Problem-Solving Techniques 44 History and Scope of Fluid Mechanics 44 Problems 46 Fundamentals of Engineering Exam Problems 53 Comprehensive Problems 54
References 55
Chapter 2 Pressure Distribution in a Fluid 59 2.1 2.2 2.3 2.4 2.5
Pressure and Pressure Gradient 59 Equilibrium of a Fluid Element 61 Hydrostatic Pressure Distributions 63 Application to Manometry 70 Hydrostatic Forces on Plane Surfaces 74
Hydrostatic Forces on Curved Surfaces 79 Hydrostatic Forces in Layered Fluids 82 Buoyancy and Stability 84 Pressure Distribution in Rigid-Body Motion 89 Pressure Measurement 97 Summary 100 Problems
102 Word Problems 125 Fundamentals of Engineering Exam Problems 125 Comprehensive Problems 126 Design Projects 127 References 127
Chapter 3 Integral Relations for a Control Volume 129 43
3.1 3.2 3.3 3.4 3.5 3.6 3.7
Basic Physical Laws of Fluid Mechanics 129 The Reynolds Transport Theorem 133 Conservation of Mass 141 The Linear Momentum Equation 146 The Angular-Momentum Theorem 158 The Energy Equation 163
Frictionless Flow: The Bernoulli Equation 174 Summary 183 Problems 184 Word Problems 210 Fundamentals of Engineering Exam Problems 210 Comprehensive Problems 211 Design Project 212 References 213
Chapter 4 Differential Relations for a Fluid Particle 215 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11
The Acceleration Field of a Fluid 215 The Differential Equation of Mass Conservation 217 The Differential Equation of Linear Momentum 223 The Differential Equation of Angular Momentum 230 The
Differential Equation of Energy 231 Boundary Conditions for the Basic Equations 234 The Stream Function 238 Vorticity and Irrotationality 245 Frictionless Irrotational Flows 247 Some Illustrative
Plane Potential Flows 252 Some Illustrative Incompressible Viscous Flows 258 Summary 263 Problems 264 Word Problems 273 Fundamentals of Engineering Exam Problems 273 Comprehensive Applied Problem 274
References 275
Chapter 5 Dimensional Analysis and Similarity 277 5.1 5.2 5.3 5.4 5.5
Introduction 277 The Principle of Dimensional Homogeneity 280 The Pi Theorem 286 Nondimensionalization of the Basic Equations 292 Modeling and Its Pitfalls 301 Summary 311 Problems 311 Word Problems
318 Fundamentals of Engineering Exam Problems 319 Comprehensive Problems 319 Design Projects 320 References 321
Chapter 6 Viscous Flow in Ducts 325 6.1 6.2 6.3 6.4
Reynolds-Number Regimes 325 Internal versus External Viscous Flows 330 Semiempirical Turbulent Shear Correlations 333 Flow in a Circular Pipe 338
6.5 6.6 6.7 6.8 6.9 6.10
Three Types of Pipe-Flow Problems 351 Flow in Noncircular Ducts 357 Minor Losses in Pipe Systems 367 Multiple-Pipe Systems 375 Experimental Duct Flows: Diffuser Performance 381 Fluid Meters 385
Summary 404 Problems 405 Word Problems 420 Fundamentals of Engineering Exam Problems 420 Comprehensive Problems 421 Design Projects 422 References 423
Chapter 7 Flow Past Immersed Bodies 427 7.1 7.2 7.3 7.4 7.5 7.6
Reynolds-Number and Geometry Effects 427 Momentum-Integral Estimates 431 The Boundary-Layer Equations 434 The Flat-Plate Boundary Layer 436 Boundary Layers with Pressure Gradient 445 Experimental
External Flows 451 Summary 476 Problems 476 Word Problems 489 Fundamentals of Engineering Exam Problems 489 Comprehensive Problems 490 Design Project 491 References 491
Chapter 8 Potential Flow and Computational Fluid Dynamics 495 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9
Introduction and Review 495 Elementary Plane-Flow Solutions 498 Superposition of Plane-Flow Solutions 500 Plane Flow Past Closed-Body Shapes 507 Other Plane Potential Flows 516 Images 521 Airfoil
Theory 523 Axisymmetric Potential Flow 534 Numerical Analysis 540 Summary 555
Contents Problems 555 Word Problems 566 Comprehensive Problems Design Projects 567 References 567
Problems 695 Word Problems 706 Fundamentals of Engineering Exam Problems Comprehensive Problems 707 Design Projects 707 References 708
Chapter 9 Compressible Flow 571 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 9.10
Introduction 571 The Speed of Sound 575 Adiabatic and Isentropic Steady Flow 578 Isentropic Flow with Area Changes 583 The Normal-Shock Wave 590 Operation of Converging and Diverging Nozzles 598
Compressible Duct Flow with Friction 603 Frictionless Duct Flow with Heat Transfer 613 Two-Dimensional Supersonic Flow 618 Prandtl-Meyer Expansion Waves 628 Summary 640 Problems 641 Word Problems 653
Fundamentals of Engineering Exam Problems 653 Comprehensive Problems 654 Design Projects 654 References 655
Chapter 10 Open-Channel Flow 659 10.1 10.2 10.3 10.4 10.5 10.6 10.7
Introduction 659 Uniform Flow; the Chézy Formula 664 Efficient Uniform-Flow Channels 669 Specific Energy; Critical Depth 671 The Hydraulic Jump 678 Gradually Varied Flow 682 Flow Measurement and
Control by Weirs Summary 695
Chapter 11 Turbomachinery 711 11.1 11.2 11.3 11.4 11.5 11.6
Introduction and Classification 711 The Centrifugal Pump 714 Pump Performance Curves and Similarity Rules 720 Mixed- and Axial-Flow Pumps: The Specific Speed 729 Matching Pumps to System
Characteristics 735 Turbines 742 Summary 755 Problems 755 Word Problems 765 Comprehensive Problems 766 Design Project 767 References 767
Appendix A
Physical Properties of Fluids 769
Appendix B
Compressible-Flow Tables 774
Appendix C
Conversion Factors 791
Appendix D
Equations of Motion in Cylindrical Coordinates 793
Appendix E
Introduction to EES 795
Answers to Selected Problems 806 687
Index 813
Hurricane Elena in the Gulf of Mexico. Unlike most small-scale fluids engineering applications, hurricanes are strongly affected by the Coriolis acceleration due to the rotation of the earth, which
causes them to swirl counterclockwise in the Northern Hemisphere. The physical properties and boundary conditions which govern such flows are discussed in the present chapter. (Courtesy of NASA/
Color-Pic Inc./E.R. Degginger/Color-Pic Inc.)
Chapter 1 Introduction
1.1 Preliminary Remarks
Fluid mechanics is the study of fluids either in motion (fluid dynamics) or at rest (fluid statics) and the subsequent effects of the fluid upon the boundaries, which may be either solid surfaces or
interfaces with other fluids. Both gases and liquids are classified as fluids, and the number of fluids engineering applications is enormous: breathing, blood flow, swimming, pumps, fans, turbines,
airplanes, ships, rivers, windmills, pipes, missiles, icebergs, engines, filters, jets, and sprinklers, to name a few. When you think about it, almost everything on this planet either is a fluid or
moves within or near a fluid. The essence of the subject of fluid flow is a judicious compromise between theory and experiment. Since fluid flow is a branch of mechanics, it satisfies a set of
welldocumented basic laws, and thus a great deal of theoretical treatment is available. However, the theory is often frustrating, because it applies mainly to idealized situations which may be
invalid in practical problems. The two chief obstacles to a workable theory are geometry and viscosity. The basic equations of fluid motion (Chap. 4) are too difficult to enable the analyst to attack
arbitrary geometric configurations. Thus most textbooks concentrate on flat plates, circular pipes, and other easy geometries. It is possible to apply numerical computer techniques to complex
geometries, and specialized textbooks are now available to explain the new computational fluid dynamics (CFD) approximations and methods [1, 2, 29].1 This book will present many theoretical results
while keeping their limitations in mind. The second obstacle to a workable theory is the action of viscosity, which can be neglected only in certain idealized flows (Chap. 8). First, viscosity
increases the difficulty of the basic equations, although the boundary-layer approximation found by Ludwig Prandtl in 1904 (Chap. 7) has greatly simplified viscous-flow analyses. Second, viscosity
has a destabilizing effect on all fluids, giving rise, at frustratingly small velocities, to a disorderly, random phenomenon called turbulence. The theory of turbulent flow is crude and heavily
backed up by experiment (Chap. 6), yet it can be quite serviceable as an engineering estimate. Textbooks now present digital-computer techniques for turbulent-flow analysis [32], but they are based
strictly upon empirical assumptions regarding the time mean of the turbulent stress field. 1
Numbered references appear at the end of each chapter.
Chapter 1 Introduction
Thus there is theory available for fluid-flow problems, but in all cases it should be backed up by experiment. Often the experimental data provide the main source of information about specific flows,
such as the drag and lift of immersed bodies (Chap. 7). Fortunately, fluid mechanics is a highly visual subject, with good instrumentation [4, 5, 35], and the use of dimensional analysis and modeling
concepts (Chap. 5) is widespread. Thus experimentation provides a natural and easy complement to the theory. You should keep in mind that theory and experiment should go hand in hand in all studies
of fluid mechanics.
1.2 The Concept of a Fluid
From the point of view of fluid mechanics, all matter consists of only two states, fluid and solid. The difference between the two is perfectly obvious to the layperson, and it is an interesting
exercise to ask a layperson to put this difference into words. The technical distinction lies with the reaction of the two to an applied shear or tangential stress. A solid can resist a shear stress
by a static deformation; a fluid cannot. Any shear stress applied to a fluid, no matter how small, will result in motion of that fluid. The fluid moves and deforms continuously as long as the shear
stress is applied. As a corollary, we can say that a fluid at rest must be in a state of zero shear stress, a state often called the hydrostatic stress condition in structural analysis. In this
condition, Mohr’s circle for stress reduces to a point, and there is no shear stress on any plane cut through the element under stress. Given the definition of a fluid above, every layperson also
knows that there are two classes of fluids, liquids and gases. Again the distinction is a technical one concerning the effect of cohesive forces. A liquid, being composed of relatively close-packed
molecules with strong cohesive forces, tends to retain its volume and will form a free surface in a gravitational field if unconfined from above. Free-surface flows are dominated by gravitational
effects and are studied in Chaps. 5 and 10. Since gas molecules are widely spaced with negligible cohesive forces, a gas is free to expand until it encounters confining walls. A gas has no definite
volume, and when left to itself without confinement, a gas forms an atmosphere which is essentially hydrostatic. The hydrostatic behavior of liquids and gases is taken up in Chap. 2. Gases cannot
form a free surface, and thus gas flows are rarely concerned with gravitational effects other than buoyancy. Figure 1.1 illustrates a solid block resting on a rigid plane and stressed by its own
weight. The solid sags into a static deflection, shown as a highly exaggerated dashed line, resisting shear without flow. A free-body diagram of element A on the side of the block shows that there is
shear in the block along a plane cut at an angle through A. Since the block sides are unsupported, element A has zero stress on the left and right sides and compression stress p on the top and
bottom. Mohr’s circle does not reduce to a point, and there is nonzero shear stress in the block. By contrast, the liquid and gas at rest in Fig. 1.1 require the supporting walls in order to
eliminate shear stress. The walls exert a compression stress of p and reduce Mohr’s circle to a point with zero shear everywhere, i.e., the hydrostatic condition. The liquid retains its volume and
forms a free surface in the container. If the walls are removed, shear develops in the liquid and a big splash results. If the container is tilted, shear again develops, waves form, and the free
surface seeks a horizontal configura-
1.2 The Concept of a Fluid Free surface
Static deflection
Fig. 1.1 A solid at rest can resist shear. (a) Static deflection of the solid; (b) equilibrium and Mohr’s circle for solid element A. A fluid cannot resist shear. (c) Containing walls are needed; (d
) equilibrium and Mohr’s circle for fluid element A.
A Solid
A Liquid
(c) p
σ1 θ
–σ = p
–σ = p
(1) 2θ
Hydrostatic condition
(d )
tion, pouring out over the lip if necessary. Meanwhile, the gas is unrestrained and expands out of the container, filling all available space. Element A in the gas is also hydrostatic and exerts a
compression stress p on the walls. In the above discussion, clear decisions could be made about solids, liquids, and gases. Most engineering fluid-mechanics problems deal with these clear cases,
i.e., the common liquids, such as water, oil, mercury, gasoline, and alcohol, and the common gases, such as air, helium, hydrogen, and steam, in their common temperature and pressure ranges. There
are many borderline cases, however, of which you should be aware. Some apparently “solid” substances such as asphalt and lead resist shear stress for short periods but actually deform slowly and
exhibit definite fluid behavior over long periods. Other substances, notably colloid and slurry mixtures, resist small shear stresses but “yield” at large stress and begin to flow as fluids do.
Specialized textbooks are devoted to this study of more general deformation and flow, a field called rheology [6]. Also, liquids and gases can coexist in two-phase mixtures, such as steam-water
mixtures or water with entrapped air bubbles. Specialized textbooks present the analysis
Chapter 1 Introduction
of such two-phase flows [7]. Finally, there are situations where the distinction between a liquid and a gas blurs. This is the case at temperatures and pressures above the socalled critical point of
a substance, where only a single phase exists, primarily resembling a gas. As pressure increases far above the critical point, the gaslike substance becomes so dense that there is some resemblance to
a liquid and the usual thermodynamic approximations like the perfect-gas law become inaccurate. The critical temperature and pressure of water are Tc 647 K and pc 219 atm,2 so that typical problems
involving water and steam are below the critical point. Air, being a mixture of gases, has no distinct critical point, but its principal component, nitrogen, has Tc 126 K and pc 34 atm. Thus typical
problems involving air are in the range of high temperature and low pressure where air is distinctly and definitely a gas. This text will be concerned solely with clearly identifiable liquids and
gases, and the borderline cases discussed above will be beyond our scope.
1.3 The Fluid as a Continuum
We have already used technical terms such as fluid pressure and density without a rigorous discussion of their definition. As far as we know, fluids are aggregations of molecules, widely spaced for a
gas, closely spaced for a liquid. The distance between molecules is very large compared with the molecular diameter. The molecules are not fixed in a lattice but move about freely relative to each
other. Thus fluid density, or mass per unit volume, has no precise meaning because the number of molecules occupying a given volume continually changes. This effect becomes unimportant if the unit
volume is large compared with, say, the cube of the molecular spacing, when the number of molecules within the volume will remain nearly constant in spite of the enormous interchange of particles
across the boundaries. If, however, the chosen unit volume is too large, there could be a noticeable variation in the bulk aggregation of the particles. This situation is illustrated in Fig. 1.2,
where the “density” as calculated from molecular mass m within a given volume is plotted versus the size of the unit volume. There is a limiting volume * below which molecular variations may be
important and
ρ Elemental volume
ρ = 1000 kg/m3
ρ = 1200
Fig. 1.2 The limit definition of continuum fluid density: (a) an elemental volume in a fluid region of variable continuum density; (b) calculated density versus size of the elemental volume.
Macroscopic uncertainty
ρ = 1100
Microscopic uncertainty
ρ = 1300 0
δ * ≈ 10-9 mm3
Region containing fluid (a) One atmosphere equals 2116 lbf/ft2 101,300 Pa.
1.4 Dimensions and Units
above which aggregate variations may be important. The density of a fluid is best defined as
The limiting volume * is about 109 mm3 for all liquids and for gases at atmospheric pressure. For example, 109 mm3 of air at standard conditions contains approximately 3 107 molecules, which is
sufficient to define a nearly constant density according to Eq. (1.1). Most engineering problems are concerned with physical dimensions much larger than this limiting volume, so that density is
essentially a point function and fluid properties can be thought of as varying continually in space, as sketched in Fig. 1.2a. Such a fluid is called a continuum, which simply means that its
variation in properties is so smooth that the differential calculus can be used to analyze the substance. We shall assume that continuum calculus is valid for all the analyses in this book. Again
there are borderline cases for gases at such low pressures that molecular spacing and mean free path3 are comparable to, or larger than, the physical size of the system. This requires that the
continuum approximation be dropped in favor of a molecular theory of rarefied-gas flow [8]. In principle, all fluid-mechanics problems can be attacked from the molecular viewpoint, but no such
attempt will be made here. Note that the use of continuum calculus does not preclude the possibility of discontinuous jumps in fluid properties across a free surface or fluid interface or across a
shock wave in a compressible fluid (Chap. 9). Our calculus in Chap. 4 must be flexible enough to handle discontinuous boundary conditions.
1.4 Dimensions and Units
A dimension is the measure by which a physical variable is expressed quantitatively. A unit is a particular way of attaching a number to the quantitative dimension. Thus length is a dimension
associated with such variables as distance, displacement, width, deflection, and height, while centimeters and inches are both numerical units for expressing length. Dimension is a powerful concept
about which a splendid tool called dimensional analysis has been developed (Chap. 5), while units are the nitty-gritty, the number which the customer wants as the final answer. Systems of units have
always varied widely from country to country, even after international agreements have been reached. Engineers need numbers and therefore unit systems, and the numbers must be accurate because the
safety of the public is at stake. You cannot design and build a piping system whose diameter is D and whose length is L. And U.S. engineers have persisted too long in clinging to British systems of
units. There is too much margin for error in most British systems, and many an engineering student has flunked a test because of a missing or improper conversion factor of 12 or 144 or 32.2 or 60 or
1.8. Practicing engineers can make the same errors. The writer is aware from personal experience of a serious preliminary error in the design of an aircraft due to a missing factor of 32.2 to convert
pounds of mass to slugs. In 1872 an international meeting in France proposed a treaty called the Metric Convention, which was signed in 1875 by 17 countries including the United States. It was an
improvement over British systems because its use of base 10 is the foundation of our number system, learned from childhood by all. Problems still remained because 3
The mean distance traveled by molecules between collisions.
Chapter 1 Introduction
even the metric countries differed in their use of kiloponds instead of dynes or newtons, kilograms instead of grams, or calories instead of joules. To standardize the metric system, a General
Conference of Weights and Measures attended in 1960 by 40 countries proposed the International System of Units (SI). We are now undergoing a painful period of transition to SI, an adjustment which
may take many more years to complete. The professional societies have led the way. Since July 1, 1974, SI units have been required by all papers published by the American Society of Mechanical
Engineers, which prepared a useful booklet explaining the SI [9]. The present text will use SI units together with British gravitational (BG) units.
Primary Dimensions
In fluid mechanics there are only four primary dimensions from which all other dimensions can be derived: mass, length, time, and temperature.4 These dimensions and their units in both systems are
given in Table 1.1. Note that the kelvin unit uses no degree symbol. The braces around a symbol like {M} mean “the dimension” of mass. All other variables in fluid mechanics can be expressed in terms
of {M}, {L}, {T}, and { }. For example, acceleration has the dimensions {LT 2}. The most crucial of these secondary dimensions is force, which is directly related to mass, length, and time by
Newton’s second law F ma
(1.2) 2
From this we see that, dimensionally, {F} {MLT }. A constant of proportionality is avoided by defining the force unit exactly in terms of the primary units. Thus we define the newton and the pound of
force 1 newton of force 1 N 1 kg m/s2
1 pound of force 1 lbf 1 slug ft/s2 4.4482 N
In this book the abbreviation lbf is used for pound-force and lb for pound-mass. If instead one adopts other force units such as the dyne or the poundal or kilopond or adopts other mass units such as
the gram or pound-mass, a constant of proportionality called gc must be included in Eq. (1.2). We shall not use gc in this book since it is not necessary in the SI and BG systems. A list of some
important secondary variables in fluid mechanics, with dimensions derived as combinations of the four primary dimensions, is given in Table 1.2. A more complete list of conversion factors is given in
App. C. Table 1.1 Primary Dimensions in SI and BG Systems
Primary dimension Mass {M} Length {L} Time {T} Temperature { }
SI unit
BG unit
Kilogram (kg) Meter (m) Second (s) Kelvin (K)
Slug Foot (ft) Second (s) Rankine (°R)
Conversion factor 1 1 1 1
slug 14.5939 kg ft 0.3048 m s1s K 1.8°R
4 If electromagnetic effects are important, a fifth primary dimension must be included, electric current {I}, whose SI unit is the ampere (A).
1.4 Dimensions and Units Table 1.2 Secondary Dimensions in Fluid Mechanics
Secondary dimension 2
Area {L } Volume {L3} Velocity {LT 1} Acceleration {LT 2} Pressure or stress {ML1T2} Angular velocity {T 1} Energy, heat, work {ML2T 2} Power {ML2T 3} Density {ML3} Viscosity {ML1T 1} Specific heat
{L2T 2 1}
SI unit
BG unit
Conversion factor
m m3 m/s m/s2
ft ft3 ft/s ft/s2
1 m 10.764 ft2 1 m3 35.315 ft3 1 ft/s 0.3048 m/s 1 ft/s2 0.3048 m/s2
Pa N/m2 s1
lbf/ft2 s1
1 lbf/ft2 47.88 Pa 1 s1 1 s1
JNm W J/s kg/m3 kg/(m s) m2/(s2 K)
ft lbf ft lbf/s slugs/ft3 slugs/(ft s) ft2/(s2 °R)
ft lbf 1.3558 J ft lbf/s 1.3558 W slug/ft3 515.4 kg/m3 slug/(ft s) 47.88 kg/(m s) m2/(s2 K) 5.980 ft2/(s2 °R)
EXAMPLE 1.1 A body weighs 1000 lbf when exposed to a standard earth gravity g 32.174 ft/s2. (a) What is its mass in kg? (b) What will the weight of this body be in N if it is exposed to the moon’s
standard acceleration gmoon 1.62 m/s2? (c) How fast will the body accelerate if a net force of 400 lbf is applied to it on the moon or on the earth?
Solution Part (a)
Equation (1.2) holds with F weight and a gearth: F W mg 1000 lbf (m slugs)(32.174 ft/s2) or
1000 m (31.08 slugs)(14.5939 kg/slug) 453.6 kg Ans. (a) 32.174 The change from 31.08 slugs to 453.6 kg illustrates the proper use of the conversion factor 14.5939 kg/slug.
Part (b)
The mass of the body remains 453.6 kg regardless of its location. Equation (1.2) applies with a new value of a and hence a new force F Wmoon mgmoon (453.6 kg)(1.62 m/s2) 735 N
Part (c)
Ans. (b)
This problem does not involve weight or gravity or position and is simply a direct application of Newton’s law with an unbalanced force: F 400 lbf ma (31.08 slugs)(a ft/s2) or 400 a 12.43 ft/s2 3.79
m/s2 31.08 This acceleration would be the same on the moon or earth or anywhere.
Ans. (c)
Chapter 1 Introduction
Many data in the literature are reported in inconvenient or arcane units suitable only to some industry or specialty or country. The engineer should convert these data to the SI or BG system before
using them. This requires the systematic application of conversion factors, as in the following example.
EXAMPLE 1.2 An early viscosity unit in the cgs system is the poise (abbreviated P), or g/(cm s), named after J. L. M. Poiseuille, a French physician who performed pioneering experiments in 1840 on
water flow in pipes. The viscosity of water (fresh or salt) at 293.16 K 20°C is approximately 0.01 P. Express this value in (a) SI and (b) BG units.
Solution Part (a)
1 kg [0.01 g/(cm s)] (100 cm/m) 0.001 kg/(m s) 100 0 g
Part (b)
1 slug [0.001 kg/(m s)] (0.3048 m/ft) 14.59 kg 2.09 105 slug/(ft s)
Ans. (a)
Ans. (b)
Note: Result (b) could have been found directly from (a) by dividing (a) by the viscosity conversion factor 47.88 listed in Table 1.2.
We repeat our advice: Faced with data in unusual units, convert them immediately to either SI or BG units because (1) it is more professional and (2) theoretical equations in fluid mechanics are
dimensionally consistent and require no further conversion factors when these two fundamental unit systems are used, as the following example shows.
EXAMPLE 1.3 A useful theoretical equation for computing the relation between pressure, velocity, and altitude in a steady flow of a nearly inviscid, nearly incompressible fluid with negligible heat
transfer and shaft work5 is the Bernoulli relation, named after Daniel Bernoulli, who published a hydrodynamics textbook in 1738: p0 p 12V2 gZ where p0 stagnation pressure p pressure in moving fluid
V velocity density Z altitude g gravitational acceleration 5
That’s an awful lot of assumptions, which need further study in Chap. 3.
1.4 Dimensions and Units
(a) Show that Eq. (1) satisfies the principle of dimensional homogeneity, which states that all additive terms in a physical equation must have the same dimensions. (b) Show that consistent units
result without additional conversion factors in SI units. (c) Repeat (b) for BG units.
Solution Part (a)
We can express Eq. (1) dimensionally, using braces by entering the dimensions of each term from Table 1.2: {ML1T 2} {ML1T 2} {ML3}{L2T 2} {ML3}{LT2}{L} {ML1T 2} for all terms
Part (b)
Ans. (a)
Enter the SI units for each quantity from Table 1.2: {N/m2} {N/m2} {kg/m3}{m2/s2} {kg/m3}{m/s2}{m} {N/m2} {kg/(m s2)} The right-hand side looks bad until we remember from Eq. (1.3) that 1 kg 1 N s2/
m. {N s2/m } {kg/(m s2)} {N/m2} {m s2}
Ans. (b)
Thus all terms in Bernoulli’s equation will have units of pascals, or newtons per square meter, when SI units are used. No conversion factors are needed, which is true of all theoretical equations in
fluid mechanics.
Part (c)
Introducing BG units for each term, we have {lbf/ft2} {lbf/ft2} {slugs/ft3}{ft2/s2} {slugs/ft3}{ft/s2}{ft} {lbf/ft2} {slugs/(ft s2)} But, from Eq. (1.3), 1 slug 1 lbf s2/ft, so that {lbf s2/ft}
{slugs/(ft s2)} {lbf/ft2} {ft s2}
Ans. (c)
All terms have the unit of pounds-force per square foot. No conversion factors are needed in the BG system either.
There is still a tendency in English-speaking countries to use pound-force per square inch as a pressure unit because the numbers are more manageable. For example, standard atmospheric pressure is
14.7 lbf/in2 2116 lbf/ft2 101,300 Pa. The pascal is a small unit because the newton is less than 14 lbf and a square meter is a very large area. It is felt nevertheless that the pascal will gradually
gain universal acceptance; e.g., repair manuals for U.S. automobiles now specify pressure measurements in pascals.
Consistent Units
Note that not only must all (fluid) mechanics equations be dimensionally homogeneous, one must also use consistent units; that is, each additive term must have the same units. There is no trouble
doing this with the SI and BG systems, as in Ex. 1.3, but woe unto
Chapter 1 Introduction
those who try to mix colloquial English units. For example, in Chap. 9, we often use the assumption of steady adiabatic compressible gas flow: h 12V2 constant where h is the fluid enthalpy and V2/2
is its kinetic energy. Colloquial thermodynamic tables might list h in units of British thermal units per pound (Btu/lb), whereas V is likely used in ft/s. It is completely erroneous to add Btu/lb to
ft2/s2. The proper unit for h in this case is ft lbf/slug, which is identical to ft2/s2. The conversion factor is 1 Btu/lb 25,040 ft2/s2 25,040 ft lbf/slug.
Homogeneous versus Dimensionally Inconsistent Equations
All theoretical equations in mechanics (and in other physical sciences) are dimensionally homogeneous; i.e., each additive term in the equation has the same dimensions. For example, Bernoulli’s
equation (1) in Example 1.3 is dimensionally homogeneous: Each term has the dimensions of pressure or stress of {F/L2}. Another example is the equation from physics for a body falling with negligible
air resistance: S S0 V0t 12gt2 where S0 is initial position, V0 is initial velocity, and g is the acceleration of gravity. Each term in this relation has dimensions of length {L}. The factor 21,
which arises from integration, is a pure (dimensionless) number, {1}. The exponent 2 is also dimensionless. However, the reader should be warned that many empirical formulas in the engineering
literature, arising primarily from correlations of data, are dimensionally inconsistent. Their units cannot be reconciled simply, and some terms may contain hidden variables. An example is the
formula which pipe valve manufacturers cite for liquid volume flow rate Q (m3/s) through a partially open valve: p Q CV SG
where p is the pressure drop across the valve and SG is the specific gravity of the liquid (the ratio of its density to that of water). The quantity CV is the valve flow coefficient, which
manufacturers tabulate in their valve brochures. Since SG is dimensionless {1}, we see that this formula is totally inconsistent, with one side being a flow rate {L3/T} and the other being the square
root of a pressure drop {M1/2/L1/2T}. It follows that CV must have dimensions, and rather odd ones at that: {L7/2/M1/2}. Nor is the resolution of this discrepancy clear, although one hint is that the
values of CV in the literature increase nearly as the square of the size of the valve. The presentation of experimental data in homogeneous form is the subject of dimensional analysis (Chap. 5).
There we shall learn that a homogeneous form for the valve flow relation is p Q Cd Aopening
where is the liquid density and A the area of the valve opening. The discharge coefficient Cd is dimensionless and changes only slightly with valve size. Please believe—until we establish the fact in
Chap. 5—that this latter is a much better formulation of the data.
1.4 Dimensions and Units
Meanwhile, we conclude that dimensionally inconsistent equations, though they abound in engineering practice, are misleading and vague and even dangerous, in the sense that they are often misused
outside their range of applicability.
Convenient Prefixes in Powers of 10
Engineering results often are too small or too large for the common units, with too many zeros one way or the other. For example, to write p 114,000,000 Pa is long and awkward. Using the prefix “M”
to mean 106, we convert this to a concise p 114 MPa (megapascals). Similarly, t 0.000000003 s is a proofreader’s nightmare compared to the equivalent t 3 ns (nanoseconds). Such prefixes are common
and convenient, in both the SI and BG systems. A complete list is given in Table 1.3.
Table 1.3 Convenient Prefixes for Engineering Units Multiplicative factor
tera giga mega kilo hecto deka deci centi milli micro nano pico femto atto
T G M k h da d c m n p f a
EXAMPLE 1.4 In 1890 Robert Manning, an Irish engineer, proposed the following empirical formula for the average velocity V in uniform flow due to gravity down an open channel (BG units): 1.49 V R2/
3S1/2 n
where R hydraulic radius of channel (Chaps. 6 and 10) S channel slope (tangent of angle that bottom makes with horizontal) n Manning’s roughness factor (Chap. 10) and n is a constant for a given
surface condition for the walls and bottom of the channel. (a) Is Manning’s formula dimensionally consistent? (b) Equation (1) is commonly taken to be valid in BG units with n taken as dimensionless.
Rewrite it in SI form.
Solution Part (a)
Introduce dimensions for each term. The slope S, being a tangent or ratio, is dimensionless, denoted by {unity} or {1}. Equation (1) in dimensional form is
T n{L L
This formula cannot be consistent unless {1.49/n} {L1/3/T}. If n is dimensionless (and it is never listed with units in textbooks), then the numerical value 1.49 must have units. This can be tragic
to an engineer working in a different unit system unless the discrepancy is properly documented. In fact, Manning’s formula, though popular, is inconsistent both dimensionally and physically and does
not properly account for channel-roughness effects except in a narrow range of parameters, for water only.
Part (b)
From part (a), the number 1.49 must have dimensions {L1/3/T} and thus in BG units equals 1.49 ft1/3/s. By using the SI conversion factor for length we have (1.49 ft1/3/s)(0.3048 m/ft)1/3 1.00 m1/3/s
Therefore Manning’s formula in SI becomes 1.0 V R2/3S1/2 n
Ans. (b) (2)
Chapter 1 Introduction with R in m and V in m/s. Actually, we misled you: This is the way Manning, a metric user, first proposed the formula. It was later converted to BG units. Such dimensionally
inconsistent formulas are dangerous and should either be reanalyzed or treated as having very limited application.
1.5 Properties of the Velocity Field
In a given flow situation, the determination, by experiment or theory, of the properties of the fluid as a function of position and time is considered to be the solution to the problem. In almost all
cases, the emphasis is on the space-time distribution of the fluid properties. One rarely keeps track of the actual fate of the specific fluid particles.6 This treatment of properties as
continuum-field functions distinguishes fluid mechanics from solid mechanics, where we are more likely to be interested in the trajectories of individual particles or systems.
Eulerian and Lagrangian Desciptions
There are two different points of view in analyzing problems in mechanics. The first view, appropriate to fluid mechanics, is concerned with the field of flow and is called the eulerian method of
description. In the eulerian method we compute the pressure field p(x, y, z, t) of the flow pattern, not the pressure changes p(t) which a particle experiences as it moves through the field. The
second method, which follows an individual particle moving through the flow, is called the lagrangian description. The lagrangian approach, which is more appropriate to solid mechanics, will not be
treated in this book. However, certain numerical analyses of sharply bounded fluid flows, such as the motion of isolated fluid droplets, are very conveniently computed in lagrangian coordinates [1].
Fluid-dynamic measurements are also suited to the eulerian system. For example, when a pressure probe is introduced into a laboratory flow, it is fixed at a specific position (x, y, z). Its output
thus contributes to the description of the eulerian pressure field p(x, y, z, t). To simulate a lagrangian measurement, the probe would have to move downstream at the fluid particle speeds; this is
sometimes done in oceanographic measurements, where flowmeters drift along with the prevailing currents. The two different descriptions can be contrasted in the analysis of traffic flow along a
freeway. A certain length of freeway may be selected for study and called the field of flow. Obviously, as time passes, various cars will enter and leave the field, and the identity of the specific
cars within the field will constantly be changing. The traffic engineer ignores specific cars and concentrates on their average velocity as a function of time and position within the field, plus the
flow rate or number of cars per hour passing a given section of the freeway. This engineer is using an eulerian description of the traffic flow. Other investigators, such as the police or social
scientists, may be interested in the path or speed or destination of specific cars in the field. By following a specific car as a function of time, they are using a lagrangian description of the
The Velocity Field
Foremost among the properties of a flow is the velocity field V(x, y, z, t). In fact, determining the velocity is often tantamount to solving a flow problem, since other prop6 One example where
fluid-particle paths are important is in water-quality analysis of the fate of contaminant discharges.
1.5 Properties of the Velocity Field 15
erties follow directly from the velocity field. Chapter 2 is devoted to the calculation of the pressure field once the velocity field is known. Books on heat transfer (for example, Ref. 10) are
essentially devoted to finding the temperature field from known velocity fields. In general, velocity is a vector function of position and time and thus has three components u, v, and w, each a
scalar field in itself: V(x, y, z, t) iu(x, y, z, t) jv(x, y, z, t) kw(x, y, z, t)
The use of u, v, and w instead of the more logical component notation Vx, Vy, and Vz is the result of an almost unbreakable custom in fluid mechanics. Several other quantities, called kinematic
properties, can be derived by mathematically manipulating the velocity field. We list some kinematic properties here and give more details about their use and derivation in later chapters:
V dt
Displacement vector:
dV a dt
(Sec. 4.1)
Volume rate of flow:
(Sec. 3.2)
Volume expansion rate:
Local angular velocity:
(V n) dA
1 d V dt
12 V
(Sec. 1.9)
(Sec. 4.2) (Sec. 4.8)
We will not illustrate any problems regarding these kinematic properties at present. The point of the list is to illustrate the type of vector operations used in fluid mechanics and to make clear the
dominance of the velocity field in determining other flow properties. Note: The fluid acceleration, item 2 above, is not as simple as it looks and actually involves four different terms due to the
use of the chain rule in calculus (see Sec. 4.1).
EXAMPLE 1.5 Fluid flows through a contracting section of a duct, as in Fig. E1.5. A velocity probe inserted at section (1) measures a steady value u1 1 m/s, while a similar probe at section (2)
records a steady u2 3 m/s. Estimate the fluid acceleration, if any, if x 10 cm. (1)
The flow is steady (not time-varying), but fluid particles clearly increase in velocity as they pass from (1) to (2). This is the concept of convective acceleration (Sec. 4.1). We may estimate the
acceleration as a velocity change u divided by a time change t x/uavg: (3.0 1.0 m/s)(1.0 3.0 m/s) u2 u1 velocity change 40 m/s2 ax 2(0.1 m) time change x/[12(u1 u2)]
A simple estimate thus indicates that this seemingly innocuous flow is accelerating at 4 times
Chapter 1 Introduction the acceleration of gravity. In the limit as x and t become very small, the above estimate reduces to a partial-derivative expression for convective x-acceleration: u u
ax,convective lim u t→0 t x In three-dimensional flow (Sec. 4.1) there are nine of these convective terms.
1.6 Thermodynamic Properties of a Fluid
While the velocity field V is the most important fluid property, it interacts closely with the thermodynamic properties of the fluid. We have already introduced into the discussion the three most
common such properties 1. Pressure p 2. Density 3. Temperature T These three are constant companions of the velocity vector in flow analyses. Four other thermodynamic properties become important when
work, heat, and energy balances are treated (Chaps. 3 and 4): 4. 5. 6. 7.
Internal energy e Enthalpy h û p/ Entropy s Specific heats cp and cv
In addition, friction and heat conduction effects are governed by the two so-called transport properties: 8. Coefficient of viscosity 9. Thermal conductivity k All nine of these quantities are true
thermodynamic properties which are determined by the thermodynamic condition or state of the fluid. For example, for a single-phase substance such as water or oxygen, two basic properties such as
pressure and temperature are sufficient to fix the value of all the others:
(p, T )
h h(p, T )
(p, T )
and so on for every quantity in the list. Note that the specific volume, so important in thermodynamic analyses, is omitted here in favor of its inverse, the density . Recall that thermodynamic
properties describe the state of a system, i.e., a collection of matter of fixed identity which interacts with its surroundings. In most cases here the system will be a small fluid element, and all
properties will be assumed to be continuum properties of the flow field: (x, y, z, t), etc. Recall also that thermodynamics is normally concerned with static systems, whereas fluids are usually in
variable motion with constantly changing properties. Do the properties retain their meaning in a fluid flow which is technically not in equilibrium? The answer is yes, from a statistical argument. In
gases at normal pressure (and even more so for liquids), an enormous number of molecular collisions occur over a very short distance of the order of 1 m, so that a fluid subjected to sudden changes
rapidly ad-
1.6 Thermodynamic Properties of a Fluid
justs itself toward equilibrium. We therefore assume that all the thermodynamic properties listed above exist as point functions in a flowing fluid and follow all the laws and state relations of
ordinary equilibrium thermodynamics. There are, of course, important nonequilibrium effects such as chemical and nuclear reactions in flowing fluids which are not treated in this text.
Pressure is the (compression) stress at a point in a static fluid (Fig. 1.1). Next to velocity, the pressure p is the most dynamic variable in fluid mechanics. Differences or gradients in pressure
often drive a fluid flow, especially in ducts. In low-speed flows, the actual magnitude of the pressure is often not important, unless it drops so low as to cause vapor bubbles to form in a liquid.
For convenience, we set many such problem assignments at the level of 1 atm 2116 lbf/ft2 101,300 Pa. High-speed (compressible) gas flows (Chap. 9), however, are indeed sensitive to the magnitude of
Temperature T is a measure of the internal energy level of a fluid. It may vary considerably during high-speed flow of a gas (Chap. 9). Although engineers often use Celsius or Fahrenheit scales for
convenience, many applications in this text require absolute (Kelvin or Rankine) temperature scales: °R °F 459.69 K °C 273.16 If temperature differences are strong, heat transfer may be important
[10], but our concern here is mainly with dynamic effects. We examine heat-transfer principles briefly in Secs. 4.5 and 9.8.
The density of a fluid, denoted by (lowercase Greek rho), is its mass per unit volume. Density is highly variable in gases and increases nearly proportionally to the pressure level. Density in
liquids is nearly constant; the density of water (about 1000 kg/m3) increases only 1 percent if the pressure is increased by a factor of 220. Thus most liquid flows are treated analytically as nearly
“incompressible.” In general, liquids are about three orders of magnitude more dense than gases at atmospheric pressure. The heaviest common liquid is mercury, and the lightest gas is hydrogen.
Compare their densities at 20°C and 1 atm: Mercury:
13,580 kg/m3
Hydrogen: 0.0838 kg/m3
They differ by a factor of 162,000! Thus the physical parameters in various liquid and gas flows might vary considerably. The differences are often resolved by the use of dimensional analysis (Chap.
5). Other fluid densities are listed in Tables A.3 and A.4 (in App. A).
Specific Weight
The specific weight of a fluid, denoted by (lowercase Greek gamma), is its weight per unit volume. Just as a mass has a weight W mg, density and specific weight are simply related by gravity:
Chapter 1 Introduction
The units of are weight per unit volume, in lbf/ft3 or N/m3. In standard earth gravity, g 32.174 ft/s2 9.807 m/s2. Thus, e.g., the specific weights of air and water at 20°C and 1 atm are
air (1.205 kg/m3)(9.807 m/s2) 11.8 N/m3 0.0752 lbf/ft3 water (998 kg/m3)(9.807 m/s2) 9790 N/m3 62.4 lbf/ft3 Specific weight is very useful in the hydrostatic-pressure applications of Chap. 2.
Specific weights of other fluids are given in Tables A.3 and A.4.
Specific Gravity
Specific gravity, denoted by SG, is the ratio of a fluid density to a standard reference fluid, water (for liquids), and air (for gases):
gas gas SGgas air 1.205 kg/m3
liquid liquid SGliquid water 998 kg/m3 For example, the specific gravity of mercury (Hg) is SGHg 13,580/998 13.6. Engineers find these dimensionless ratios easier to remember than the actual
numerical values of density of a variety of fluids.
Potential and Kinetic Energies
In thermostatics the only energy in a substance is that stored in a system by molecular activity and molecular bonding forces. This is commonly denoted as internal energy û. A commonly accepted
adjustment to this static situation for fluid flow is to add two more energy terms which arise from newtonian mechanics: the potential energy and kinetic energy. The potential energy equals the work
required to move the system of mass m from the origin to a position vector r ix jy kz against a gravity field g. Its value is mg r, or g r per unit mass. The kinetic energy equals the work required
to change the speed of the mass from zero to velocity V. Its value is 12mV2 or 12V2 per unit mass. Then by common convention the total stored energy e per unit mass in fluid mechanics is the sum of
three terms: e û 12V2 (g r)
Also, throughout this book we shall define z as upward, so that g gk and g r gz. Then Eq. (1.8) becomes e û 12V2 gz
The molecular internal energy û is a function of T and p for the single-phase pure substance, whereas the potential and kinetic energies are kinematic properties.
State Relations for Gases
Thermodynamic properties are found both theoretically and experimentally to be related to each other by state relations which differ for each substance. As mentioned,
1.6 Thermodynamic Properties of a Fluid
we shall confine ourselves here to single-phase pure substances, e.g., water in its liquid phase. The second most common fluid, air, is a mixture of gases, but since the mixture ratios remain nearly
constant between 160 and 2200 K, in this temperature range air can be considered to be a pure substance. All gases at high temperatures and low pressures (relative to their critical point) are in
good agreement with the perfect-gas law p RT
R cp cv gas constant
Since Eq. (1.10) is dimensionally consistent, R has the same dimensions as specific heat, {L2T2 1}, or velocity squared per temperature unit (kelvin or degree Rankine). Each gas has its own constant
R, equal to a universal constant divided by the molecular weight Rgas Mgas
where 49,700 ft2/(s2 °R) 8314 m2/(s2 K). Most applications in this book are for air, with M 28.97: Rair 1717 ft2/(s2 °R) 287 m2/(s2 K)
Standard atmospheric pressure is 2116 lbf/ft2, and standard temperature is 60°F 520°R. Thus standard air density is 2116 air 0.00237 slug/ft3 1.22 kg/m3 (1717)(520)
This is a nominal value suitable for problems. One proves in thermodynamics that Eq. (1.10) requires that the internal molecular energy û of a perfect gas vary only with temperature: û û(T).
Therefore the specific heat cv also varies only with temperature: û cv T
dû cv(T) dT
dû cv(T) dT
In like manner h and cp of a perfect gas also vary only with temperature: p h û û RT h(T) h cp T
dh cp(T) dT
dh cp(T) dT The ratio of specific heats of a perfect gas is an important dimensionless parameter in compressible-flow analysis (Chap. 9) cp k k(T) 1 cv
Chapter 1 Introduction
As a first approximation in airflow analysis we commonly take cp, cv, and k to be constant kair 1.4 R cv 4293 ft2/(s2 °R) 718 m2/(s2 K) k 1
kR cp 6010 ft2/(s2 °R) 1005 m2/(s2 K) k1 Actually, for all gases, cp and cv increase gradually with temperature, and k decreases gradually. Experimental values of the specific-heat ratio for eight
common gases are shown in Fig. 1.3. Many flow problems involve steam. Typical steam operating conditions are relatively close to the critical point, so that the perfect-gas approximation is
inaccurate. The properties of steam are therefore available in tabular form [13], but the error of using the perfect-gas law is sometimes not great, as the following example shows.
EXAMPLE 1.6 Estimate and cp of steam at 100 lbf/in2 and 400°F (a) by a perfect-gas approximation and (b) from the ASME steam tables [13].
Solution Part (a)
First convert to BG units: p 100 lbf/in2 14,400 lb/ft2, T 400°F 860°R. From Table A.4 the molecular weight of H2O is 2MH MO 2(1.008) 16.0 18.016. Then from Eq. (1.11) the gas constant of steam is
approximately 49,700 R 2759 ft2/(s2 °R) 18.016 whence, from the perfect-gas law, p 14,400 0.00607 slug/ft3 RT 2759(860)
Ans. (a)
From Fig. 1.3, k for steam at 860°R is approximately 1.30. Then from Eq. (1.17), kR 1.30(2759) cp 12,000 ft2/(s2 °R) k1 1.30 1
Part (b)
Ans. (a)
From Ref. 13, the specific volume v of steam at 100 lbf/in2 and 400°F is 4.935 ft3/lbm. Then the density is the inverse of this, converted to slugs: 1 1 0.00630 slug/ft3 (4.935 ft2/lbm)(32.174 lbm/
slug) v
Ans. (b)
This is about 4 percent higher than our ideal-gas estimate in part (a). Reference 13 lists the value of cp of steam at 100 lbf/in2 and 400°F as 0.535 Btu/(lbm °F). Convert this to BG units: cp [0.535
Btu/(lbm °R)](778.2 ft lbf/Btu)(32.174 lbm/slug) 13,400 ft lbf/(slug °R) 13,400 ft2/(s2 °R)
Ans. (b)
1.6 Thermodynamic Properties of a Fluid
This is about 11 percent higher than our ideal-gas estimate in part (a). The chief reason for the discrepancy is that this temperature and this pressure are quite close to the critical point and
saturation line of steam. At higher temperatures and lower pressures, say, 800°F and 50 lbf/in2, the perfect-gas law gives and cp of steam within an accuracy of 1 percent. Note that the use of
pound-mass and British thermal units in the traditional steam tables requires continual awkward conversions to BG units. Newer tables and disks are in SI units.
State Relations for Liquids
The writer knows of no “perfect-liquid law” comparable to that for gases. Liquids are nearly incompressible and have a single reasonably constant specific heat. Thus an idealized state relation for a
liquid is
cp cv const
dh cp dT
Most of the flow problems in this book can be attacked with these simple assumptions. Water is normally taken to have a density of 1.94 slugs/ft3 and a specific heat cp 25,200 ft2/(s2 °R). The steam
tables may be used if more accuracy is required.
1.7 Ar 1.6
Atmospheric pressure 1.5
1.4 cp k= c υ
Air and N2
Steam 1.2 CO2 1.1
Fig. 1.3 Specific-heat ratio of eight common gases as a function of temperature. (Data from Ref. 12.)
Temperature, ° R
Chapter 1 Introduction
The density of a liquid usually decreases slightly with temperature and increases moderately with pressure. If we neglect the temperature effect, an empirical pressuredensity relation for a liquid is
p (B 1) pa a
where B and n are dimensionless parameters which vary slightly with temperature and pa and a are standard atmospheric values. Water can be fitted approximately to the values B 3000 and n 7. Seawater
is a variable mixture of water and salt and thus requires three thermodynamic properties to define its state. These are normally taken as pressure, temperature, and the salinity Sˆ, defined as the
weight of the dissolved salt divided by the weight of the mixture. The average salinity of seawater is 0.035, usually written as 35 parts per 1000, or 35 ‰. The average density of seawater is 2.00
slugs/ft3. Strictly speaking, seawater has three specific heats, all approximately equal to the value for pure water of 25,200 ft2/(s2 °R) 4210 m2/(s2 K).
EXAMPLE 1.7 The pressure at the deepest part of the ocean is approximately 1100 atm. Estimate the density of seawater at this pressure.
Solution Equation (1.19) holds for either water or seawater. The ratio p/pa is given as 1100:
1100 (3001) a
3000 4100 3001 1.046 7
Assuming an average surface seawater density a 2.00 slugs/ft3, we compute
1.046(2.00) 2.09 slugs/ft3
Even at these immense pressures, the density increase is less than 5 percent, which justifies the treatment of a liquid flow as essentially incompressible.
1.7 Viscosity and Other Secondary Properties
The quantities such as pressure, temperature, and density discussed in the previous section are primary thermodynamic variables characteristic of any system. There are also certain secondary
variables which characterize specific fluid-mechanical behavior. The most important of these is viscosity, which relates the local stresses in a moving fluid to the strain rate of the fluid element.
When a fluid is sheared, it begins to move at a strain rate inversely proportional to a property called its coefficient of viscosity . Consider a fluid element sheared in one
1.7 Viscosity and Other Secondary Properties
plane by a single shear stress , as in Fig. 1.4a. The shear strain angle will continuously grow with time as long as the stress is maintained, the upper surface moving at speed u larger than the
lower. Such common fluids as water, oil, and air show a linear relation between applied shear and resulting strain rate
From the geometry of Fig. 1.4a we see that
u t tan y
In the limit of infinitesimal changes, this becomes a relation between shear strain rate and velocity gradient d du dt dy
From Eq. (1.20), then, the applied shear is also proportional to the velocity gradient for the common linear fluids. The constant of proportionality is the viscosity coefficient d du dt dy
Equation (1.23) is dimensionally consistent; therefore has dimensions of stress-time: {FT/L2} or {M/(LT)}. The BG unit is slugs per foot-second, and the SI unit is kilograms per meter-second. The
linear fluids which follow Eq. (1.23) are called newtonian fluids, after Sir Isaac Newton, who first postulated this resistance law in 1687. We do not really care about the strain angle (t) in fluid
mechanics, concentrating instead on the velocity distribution u(y), as in Fig. 1.4b. We shall use Eq. (1.23) in Chap. 4 to derive a differential equation for finding the velocity distribution u(y)
—and, more generally, V(x, y, z, t)—in a viscous fluid. Figure 1.4b illustrates a shear layer, or boundary layer, near a solid wall. The shear stress is proportional to the slope of the
y δu δt
u( y)
δθ δt
Velocity profile
u = δu du δθ
Fig. 1.4 Shear stress causes continuous shear deformation in a fluid: (a) a fluid element straining at a rate /t; (b) newtonian shear distribution in a shear layer near a wall.
τ = µ du dy
No slip at wall
τ (a)
Chapter 1 Introduction
Table 1.4 Viscosity and Kinematic Viscosity of Eight Fluids at 1 atm and 20°C
, kg/(m s)†
Ratio /(H2)
, kg/m3
Hydrogen Air Gasoline Water Ethyl alcohol Mercury SAE 30 oil Glycerin
8.8 E–6 1.8 E–5 2.9 E–4 1.0 E–3 1.2 E–3 1.5 E–3 0.29 1.5
00,0001.0 0,00002.1 00,0033 00,0114 0,00135 00,0170 033,000 170,000
00,000.084 00,001.20 0,0680 0,0998 0,0789 13,580 0,0891 01,264
m2/s† 1.05 1.51 4.22 1.01 1.52 1.16 3.25 1.18
E–4 E–5 E–7 E–6 E–6 E–7 E–4 E–3
Ratio /(Hg) 00,920 00,130 00,003.7 0000,8.7 000,13 0000,1.0 02,850 10,300
1 kg/(m s) 0.0209 slug/(ft s); 1 m2/s 10.76 ft2/s.
velocity profile and is greatest at the wall. Further, at the wall, the velocity u is zero relative to the wall: This is called the no-slip condition and is characteristic of all viscous-fluid flows.
The viscosity of newtonian fluids is a true thermodynamic property and varies with temperature and pressure. At a given state (p, T) there is a vast range of values among the common fluids. Table 1.4
lists the viscosity of eight fluids at standard pressure and temperature. There is a variation of six orders of magnitude from hydrogen up to glycerin. Thus there will be wide differences between
fluids subjected to the same applied stresses. Generally speaking, the viscosity of a fluid increases only weakly with pressure. For example, increasing p from 1 to 50 atm will increase of air only
10 percent. Temperature, however, has a strong effect, with increasing with T for gases and decreasing for liquids. Figure A.1 (in App. A) shows this temperature variation for various common fluids.
It is customary in most engineering work to neglect the pressure variation. The variation (p, T) for a typical fluid is nicely shown by Fig. 1.5, from Ref. 14, which normalizes the data with the
critical-point state ( c, pc, Tc). This behavior, called the principle of corresponding states, is characteristic of all fluids, but the actual numerical values are uncertain to 20 percent for any
given fluid. For example, values of (T) for air at 1 atm, from Table A.2, fall about 8 percent low compared to the “low-density limit” in Fig. 1.5. Note in Fig. 1.5 that changes with temperature
occur very rapidly near the critical point. In general, critical-point measurements are extremely difficult and uncertain.
The Reynolds Number
As we shall see in Chaps. 5 through 7, the primary parameter correlating the viscous behavior of all newtonian fluids is the dimensionless Reynolds number:
VL VL Re
where V and L are characteristic velocity and length scales of the flow. The second form of Re illustrates that the ratio of to has its own name, the kinematic viscosity:
It is called kinematic because the mass units cancel, leaving only the dimensions {L2/T}.
1.7 Viscosity and Other Secondary Properties
5 4 Dense gas 3 Two-phase region
µ µr = µ
Critical point 1 0.9 0.8 0.7 0.6
0.5 0.4
Fig. 1.5 Fluid viscosity nondimensionalized by critical-point properties. This generalized chart is characteristic of all fluids but is only accurate to 20 percent. (From Ref. 14.)
0.2 0.4
pr = 0.2 Low-density limit 0
Tr = T Tc
Generally, the first thing a fluids engineer should do is estimate the Reynolds number range of the flow under study. Very low Re indicates viscous creeping motion, where inertia effects are
negligible. Moderate Re implies a smoothly varying laminar flow. High Re probably spells turbulent flow, which is slowly varying in the time-mean but has superimposed strong random high-frequency
fluctuations. Explicit numerical values for low, moderate, and high Reynolds numbers cannot be stated here. They depend upon flow geometry and will be discussed in Chaps. 5 through 7. Table 1.4 also
lists values of for the same eight fluids. The pecking order changes considerably, and mercury, the heaviest, has the smallest viscosity relative to its own weight. All gases have high relative to
thin liquids such as gasoline, water, and alcohol. Oil and glycerin still have the highest , but the ratio is smaller. For a given value of V and L in a flow, these fluids exhibit a spread of four
orders of magnitude in the Reynolds number.
Flow between Plates
A classic problem is the flow induced between a fixed lower plate and an upper plate moving steadily at velocity V, as shown in Fig. 1.6. The clearance between plates is h, and the fluid is newtonian
and does not slip at either plate. If the plates are large,
Chapter 1 Introduction y Moving plate: u=V
u=V V
Fig. 1.6 Viscous flow induced by relative motion between two parallel plates.
Viscous fluid
Fixed plate
this steady shearing motion will set up a velocity distribution u(y), as shown, with v w 0. The fluid acceleration is zero everywhere. With zero acceleration and assuming no pressure variation in the
flow direction, you should show that a force balance on a small fluid element leads to the result that the shear stress is constant throughout the fluid. Then Eq. (1.23) becomes du const dy which we
can integrate to obtain u a by The velocity distribution is linear, as shown in Fig. 1.6, and the constants a and b can be evaluated from the no-slip condition at the upper and lower walls: u
V0 aa
b(0) b(h)
at y 0 at y h
Hence a 0 and b V/h. Then the velocity profile between the plates is given by y u V h
as indicated in Fig. 1.6. Turbulent flow (Chap. 6) does not have this shape. Although viscosity has a profound effect on fluid motion, the actual viscous stresses are quite small in magnitude even
for oils, as shown in the following example.
EXAMPLE 1.8 Suppose that the fluid being sheared in Fig. 1.6 is SAE 30 oil at 20°C. Compute the shear stress in the oil if V 3 m/s and h 2 cm.
Solution The shear stress is found from Eq. (1.23) by differentiating Eq. (1.26): du V h dy
1.7 Viscosity and Other Secondary Properties
From Table 1.4 for SAE 30 oil, 0.29 kg/(m s). Then, for the given values of V and h, Eq. (1) predicts [0.29 kg/(m s)](3 m/s) 43 kg/(m s2) 0.02 m 43 N/m2 43 Pa
Although oil is very viscous, this is a modest shear stress, about 2400 times less than atmospheric pressure. Viscous stresses in gases and thin liquids are even smaller.
Variation of Viscosity with Temperature
Temperature has a strong effect and pressure a moderate effect on viscosity. The viscosity of gases and most liquids increases slowly with pressure. Water is anomalous in showing a very slight
decrease below 30°C. Since the change in viscosity is only a few percent up to 100 atm, we shall neglect pressure effects in this book. Gas viscosity increases with temperature. Two common
approximations are the power law and the Sutherland law: T n power law T 0 (1.27) 0 (T/T0)3/2(T0 S) Sutherland law T S where 0 is a known viscosity at a known absolute temperature T0
(usually 273 K). The constants n and S are fit to the data, and both formulas are adequate over a wide range of temperatures. For air, n 0.7 and S 110 K 199°R. Other values are given in Ref. 3.
Liquid viscosity decreases with temperature and is roughly exponential, aebT; but a better fit is the empirical result that ln is quadratic in 1/T, where T is absolute temperature
T T ln a b 0 c 0 0 T T
For water, with T0 273.16 K, 0 0.001792 kg/(m s), suggested values are a 1.94, b 4.80, and c 6.74, with accuracy about 1 percent. The viscosity of water is tabulated in Table A.1. Curve-fit viscosity
formulas for 355 organic liquids are given by Yaws et al. [34]. For further viscosity data, see Refs. 28 and 36.
Thermal Conductivity
Just as viscosity relates applied stress to resulting strain rate, there is a property called thermal conductivity k which relates the vector rate of heat flow per unit area q to the vector gradient
of temperature T. This proportionality, observed experimentally for fluids and solids, is known as Fourier’s law of heat conduction q kT
which can also be written as three scalar equations T qx k x
T qy k y
T qz k z
Chapter 1 Introduction
The minus sign satisfies the convention that heat flux is positive in the direction of decreasing temperature. Fourier’s law is dimensionally consistent, and k has SI units of joules per
second-meter-kelvin. Thermal conductivity k is a thermodynamic property and varies with temperature and pressure in much the same way as viscosity. The ratio k/k0 can be correlated with T/T0 in the
same manner as Eqs. (1.27) and (1.28) for gases and liquids, respectively. Further data on viscosity and thermal-conductivity variations can be found in Ref. 11.
Nonnewtonian Fluids
Fluids which do not follow the linear law of Eq. (1.23) are called nonnewtonian and are treated in books on rheology [6]. Figure 1.7a compares four examples with a newtonian fluid. A dilatant, or
shear-thickening, fluid increases resistance with increasing applied stress. Alternately, a pseudoplastic, or shear-thinning, fluid decreases resistance with increasing stress. If the thinning effect
is very strong, as with the dashed-line curve, the fluid is termed plastic. The limiting case of a plastic substance is one which requires a finite yield stress before it begins to flow. The
linear-flow Bingham plastic idealization is shown, but the flow behavior after yield may also be nonlinear. An example of a yielding fluid is toothpaste, which will not flow out of the tube until a
finite stress is applied by squeezing. A further complication of nonnewtonian behavior is the transient effect shown in Fig. 1.7b. Some fluids require a gradually increasing shear stress to maintain
a constant strain rate and are called rheopectic. The opposite case of a fluid which thins out with time and requires decreasing stress is termed thixotropic. We neglect nonnewtonian effects in this
book; see Ref. 6 for further study.
Shear stress τ
Ideal Bingham plastic
Shear stress τ
Yield stress
Common fluids
Pseudoplastic Thixotropic
Constant strain rate
Fig. 1.7 Rheological behavior of various viscous materials: (a) stress versus strain rate; (b) effect of time on applied stress.
Shear strain rate dθ dt (a)
Time (b)
1.7 Viscosity and Other Secondary Properties
Surface Tension
A liquid, being unable to expand freely, will form an interface with a second liquid or gas. The physical chemistry of such interfacial surfaces is quite complex, and whole textbooks are devoted to
this specialty [15]. Molecules deep within the liquid repel each other because of their close packing. Molecules at the surface are less dense and attract each other. Since half of their neighbors
are missing, the mechanical effect is that the surface is in tension. We can account adequately for surface effects in fluid mechanics with the concept of surface tension. If a cut of length dL is
made in an interfacial surface, equal and opposite forces of magnitude dL are exposed normal to the cut and parallel to the surface, where is called the coefficient of surface tension. The dimensions
of are {F/L}, with SI units of newtons per meter and BG units of pounds-force per foot. An alternate concept is to open up the cut to an area dA; this requires work to be done of amount dA. Thus the
coefficient can also be regarded as the surface energy per unit area of the interface, in N m/m2 or ft lbf/ft2. The two most common interfaces are water-air and mercury-air. For a clean surface at
20°C 68°F, the measured surface tension is
0.0050 lbf/ft 0.073 N/m
0.033 lbf/ft 0.48 N/m
air-water air-mercury
These are design values and can change considerably if the surface contains contaminants like detergents or slicks. Generally decreases with liquid temperature and is zero at the critical point.
Values of for water are given in Fig. 1.8. If the interface is curved, a mechanical balance shows that there is a pressure difference across the interface, the pressure being higher on the concave
side, as illustrated in Fig. 1.9. In Fig. 1.9a, the pressure increase in the interior of a liquid cylinder is balanced by two surface-tension forces 2RL p 2L or
p R
We are not considering the weight of the liquid in this calculation. In Fig. 1.9b, the pressure increase in the interior of a spherical droplet balances a ring of surface-tension force
R2 p 2R or
2 p R
We can use this result to predict the pressure increase inside a soap bubble, which has two interfaces with air, an inner and outer surface of nearly the same radius R: 4 pbubble 2 pdroplet R
Figure 1.9c shows the general case of an arbitrarily curved interface whose principal radii of curvature are R1 and R2. A force balance normal to the surface will show that the pressure increase on
the concave side is p (R11 R21)
Chapter 1 Introduction
ϒ, N/m
Fig. 1.8 Surface tension of a clean air-water interface. Data from Table A.5.
0.050 0
πR2 ∆p
2RL ∆p
T, ° C
∆p dA
dL 1 2πR
dL 2
R2 R1 dL 2 L
dL 1
Fig. 1.9 Pressure change across a curved interface due to surface tension: (a) interior of a liquid cylinder; (b) interior of a spherical droplet; (c) general curved interface.
Equations (1.31) to (1.33) can all be derived from this general relation; e.g., in Eq. (1.31), R1 R and R2 . A second important surface effect is the contact angle which appears when a liquid
interface intersects with a solid surface, as in Fig. 1.10. The force balance would then involve both and . If the contact angle is less than 90°, the liquid is said to wet the solid; if 90°, the
liquid is termed nonwetting. For example, water wets soap but does not wet wax. Water is extremely wetting to a clean glass surface, with 0°. Like , the contact angle is sensitive to the actual
physicochemical conditions of the solid-liquid interface. For a clean mercury-air-glass interface, 130°. Example 1.9 illustrates how surface tension causes a fluid interface to rise or fall in a
capillary tube.
1.7 Viscosity and Other Secondary Properties
Gas Liquid
Fig. 1.10 Contact-angle effects at liquid-gas-solid interface. If 90°, the liquid “wets” the solid; if 90°, the liquid is nonwetting.
θ Solid
EXAMPLE 1.9 Derive an expression for the change in height h in a circular tube of a liquid with surface tension and contact angle , as in Fig. E1.9.
Solution θ
The vertical component of the ring surface-tension force at the interface in the tube must balance the weight of the column of fluid of height h 2R cos R2h Solving for h, we have the desired result h
2 cos h R
Fig. E1.9
Thus the capillary height increases inversely with tube radius R and is positive if 90° (wetting liquid) and negative (capillary depression) if 90°. Suppose that R 1 mm. Then the capillary rise for a
water-air-glass interface, 0°, 0.073 N/m, and 1000 kg/m3 is 2(0.073 N/m)(cos 0°) h 0.015 (N s2)/kg 0.015 m 1.5 cm (1000 kg/m3)(9.81 m/s2)(0.001 m) For a mercury-air-glass interface, with 130°, 0.48 N
/m, and 13,600 kg/m3, the capillary rise is 2(0.48)(cos 130°) h 0.46 cm 13,600(9.81)(0.001) When a small-diameter tube is used to make pressure measurements (Chap. 2), these capillary effects must be
corrected for.
Vapor Pressure
Vapor pressure is the pressure at which a liquid boils and is in equilibrium with its own vapor. For example, the vapor pressure of water at 68°F is 49 lbf/ft2, while that of mercury is only 0.0035
lbf/ft2. If the liquid pressure is greater than the vapor
Chapter 1 Introduction
pressure, the only exchange between liquid and vapor is evaporation at the interface. If, however, the liquid pressure falls below the vapor pressure, vapor bubbles begin to appear in the liquid. If
water is heated to 212°F, its vapor pressure rises to 2116 lbf/ft2, and thus water at normal atmospheric pressure will boil. When the liquid pressure is dropped below the vapor pressure due to a flow
phenomenon, we call the process cavitation. As we shall see in Chap. 2, if water is accelerated from rest to about 50 ft/s, its pressure drops by about 15 lbf/in2, or 1 atm. This can cause
cavitation. The dimensionless parameter describing flow-induced boiling is the cavitation number pa pv Ca 1V2 2
where pa ambient pressure pv vapor pressure V characteristic flow velocity Depending upon the geometry, a given flow has a critical value of Ca below which the flow will begin to cavitate. Values of
surface tension and vapor pressure of water are given in Table A.5. The vapor pressure of water is plotted in Fig. 1.11. Figure 1.12a shows cavitation bubbles being formed on the low-pressure
surfaces of a marine propeller. When these bubbles move into a higher-pressure region, they collapse implosively. Cavitation collapse can rapidly spall and erode metallic surfaces and eventually
destroy them, as shown in Fig. 1.12b.
No-Slip and No-TemperatureJump Conditions
When a fluid flow is bounded by a solid surface, molecular interactions cause the fluid in contact with the surface to seek momentum and energy equilibrium with that surface. All liquids essentially
are in equilibrium with the surface they contact. All gases are, too,
pv , kPa
Fig. 1.11 Vapor pressure of water. Data from Table A.5.
60 T, °C
1.7 Viscosity and Other Secondary Properties
Fig. 1.12 Two aspects of cavitation bubble formation in liquid flows: (a) Beauty: spiral bubble sheets form from the surface of a marine propeller. (Courtesy of the Garfield Thomas Water Tunnel,
Pennsylvania State University); (b) ugliness: collapsing bubbles erode a propeller surface. (Courtesy of Thomas T. Huang, David Taylor Research Center.)
Chapter 1 Introduction
except under the most rarefied conditions [8]. Excluding rarefied gases, then, all fluids at a point of contact with a solid take on the velocity and temperature of that surface Vfluid Vwall
Tfluid Twall
These are called the no-slip and no-temperature-jump conditions, respectively. They serve as boundary conditions for analysis of fluid flow past a solid surface (Chap. 6). Figure 1.13 illustrates the
no-slip condition for water flow past the top and bottom surfaces of a fixed thin plate. The flow past the upper surface is disorderly, or turbulent, while the lower surface flow is smooth, or
laminar.7 In both cases there is clearly no slip at the wall, where the water takes on the zero velocity of the fixed plate. The velocity profile is made visible by the discharge of a line of
hydrogen bubbles from the wire shown stretched across the flow. To decrease the mathematical difficulty, the no-slip condition is partially relaxed in the analysis of inviscid flow (Chap. 8). The
flow is allowed to “slip” past the surface but not to permeate through the surface Vnormal(fluid) Vnormal(solid)
while the tangential velocity Vt is allowed to be independent of the wall. The analysis is much simpler, but the flow patterns are highly idealized.
Fig. 1.13 The no-slip condition in water flow past a thin fixed plate. The upper flow is turbulent; the lower flow is laminar. The velocity profile is made visible by a line of hydrogen bubbles
discharged from the wire across the flow. [From Illustrated Experiments in Fluid Mechanics (The NCFMF Book of Film Notes), National Committee for Fluid Mechanics Films, Education Development Center,
Inc., copyright 1972.] 7
Laminar and turbulent flows are studied in Chaps. 6 and 7.
1.8 Basic Flow-Analysis Techniques
Speed of Sound
In gas flow, one must be aware of compressibility effects (significant density changes caused by the flow). We shall see in Sec. 4.2 and in Chap. 9 that compressibility becomes important when the
flow velocity reaches a significant fraction of the speed of sound of the fluid. The speed of sound a of a fluid is the rate of propagation of smalldisturbance pressure pulses (“sound waves”) through
the fluid. In Chap. 9 we shall show, from momentum and thermodynamic arguments, that the speed of sound is defined by p p a2 k s
c k p cv
This is true for either a liquid or a gas, but it is for gases that the problem of compressibility occurs. For an ideal gas, Eq. (1.10), we obtain the simple formula aideal gas (kRT)1/2
where R is the gas constant, Eq. (1.11), and T the absolute temperature. For example, for air at 20°C, a {(1.40)[287 m2/(s2 K)](293 K)}1/2 343 m/s (1126 ft/s 768 mi/h). If, in this case, the air
velocity reaches a significant fraction of a, say, 100 m/s, then we must account for compressibility effects (Chap. 9). Another way to state this is to account for compressibility when the Mach
number Ma V/a of the flow reaches about 0.3. The speed of sound of water is tabulated in Table A.5. The speed of sound of air (or any approximately perfect gas) is simply calculated from Eq. (1.39).
1.8 Basic Flow-Analysis Techniques
There are three basic ways to attack a fluid-flow problem. They are equally important for a student learning the subject, and this book tries to give adequate coverage to each method: 1.
Control-volume, or integral analysis (Chap. 3) 2. Infinitesimal system, or differential analysis (Chap. 4) 3. Experimental study, or dimensional analysis (Chap. 5) In all cases, the flow must satisfy
the three basic laws of mechanics8 plus a thermodynamic state relation and associated boundary conditions: 1. 2. 3. 4. 5.
Conservation of mass (continuity) Linear momentum (Newton’s second law) First law of thermodynamics (conservation of energy) A state relation like (p, T) Appropriate boundary conditions at solid
surfaces, interfaces, inlets, and exits
In integral and differential analyses, these five relations are modeled mathematically and solved by computational methods. In an experimental study, the fluid itself performs this task without the
use of any mathematics. In other words, these laws are believed to be fundamental to physics, and no fluid flow is known to violate them. 8 In fluids which are variable mixtures of components, such
as seawater, a fourth basic law is required, conservation of species. For an example of salt conservation analysis, see Chap. 4, Ref. 16.
Chapter 1 Introduction
A control volume is a finite region, chosen carefully by the analyst, with open boundaries through which mass, momentum, and energy are allowed to cross. The analyst makes a budget, or balance,
between the incoming and outgoing fluid and the resultant changes within the control volume. The result is a powerful tool but a crude one. Details of the flow are normally washed out or ignored in
control-volume analyses. Nevertheless, the control-volume technique of Chap. 3 never fails to yield useful and quantitative information to the engineering analyst. When the conservation laws are
written for an infinitesimal system of fluid in motion, they become the basic differential equations of fluid flow. To apply them to a specific problem, one must integrate these equations
mathematically subject to the boundary conditions of the particular problem. Exact analytic solutions are often possible only for very simple geometries and boundary conditions (Chap. 4). Otherwise,
one attempts numerical integration on a digital computer, i.e., a summing procedure for finite-sized systems which one hopes will approximate the exact integral calculus [1]. Even computer analysis
often fails to provide an accurate simulation, because of either inadequate storage or inability to model the finely detailed flow structure characteristic of irregular geometries or turbulent-flow
patterns. Thus differential analysis sometimes promises more than it delivers, although we can successfully study a number of classic and useful solutions. A properly planned experiment is very often
the best way to study a practical engineering flow problem. Guidelines for planning flow experiments are given in Chap. 5. For example, no theory presently available, whether differential or
integral, calculus or computer, is able to make an accurate computation of the aerodynamic drag and side force of an automobile moving down a highway with crosswinds. One must solve the problem by
experiment. The experiment may be full-scale: One can test a real automobile on a real highway in real crosswinds. For that matter, there are wind tunnels in existence large enough to hold a
full-scale car without significant blockage effects. Normally, however, in the design stage, one tests a small-model automobile in a small wind tunnel. Without proper interpretation, the model
results may be poor and mislead the designer (Chap. 5). For example, the model may lack important details such as surface finish or underbody protuberances. The “wind” produced by the tunnel
propellers may lack the turbulent gustiness of real winds. It is the job of the fluid-flow analyst, using such techniques as dimensional analysis, to plan an experiment which gives an accurate
estimate of full-scale or prototype results expected in the final product. It is possible to classify flows, but there is no general agreement on how to do it. Most classifications deal with the
assumptions made in the proposed flow analysis. They come in pairs, and we normally assume that a given flow is either Steady
As Fig. 1.14 indicates, we choose one assumption from each pair. We may have a steady viscous compressible gas flow or an unsteady inviscid ( 0) incompressible liquid flow. Although there is no such
thing as a truly inviscid fluid, the assumption 0 gives adequate results in many analyses (Chap. 8). Often the assumptions overlap: A flow may be viscous in the boundary layer near a solid surface
(Fig. 1.13) and effec-
1.9 Flow Patterns: Streamlines, Streaklines, and Pathlines 37 Fig. 1.14 Ready for a flow analysis? Then choose one assumption from each box.
Steady Unsteady
Inviscid Viscous
Incompressible Compressible
Gas Liquid
tively inviscid away from the surface. The viscous part of the flow may be laminar or transitional or turbulent or combine patches of all three types of viscous flow. A flow may involve both a gas
and a liquid and the free surface, or interface, between them (Chap. 10). A flow may be compressible in one region and have nearly constant density in another. Nevertheless, Eq. (1.40) and Fig. 1.14
give the basic binary assumptions of flow analysis, and Chaps. 6 to 10 try to separate them and isolate the basic effect of each assumption.
1.9 Flow Patterns: Streamlines, Streaklines, and Pathlines
Fluid mechanics is a highly visual subject. The patterns of flow can be visualized in a dozen different ways, and you can view these sketches or photographs and learn a great deal qualitatively and
often quantitatively about the flow. Four basic types of line patterns are used to visualize flows: 1. A streamline is a line everywhere tangent to the velocity vector at a given instant. 2. A
pathline is the actual path traversed by a given fluid particle. 3. A streakline is the locus of particles which have earlier passed through a prescribed point. 4. A timeline is a set of fluid
particles that form a line at a given instant. The streamline is convenient to calculate mathematically, while the other three are easier to generate experimentally. Note that a streamline and a
timeline are instantaneous lines, while the pathline and the streakline are generated by the passage of time. The velocity profile shown in Fig. 1.13 is really a timeline generated earlier by a
single discharge of bubbles from the wire. A pathline can be found by a time exposure of a single marked particle moving through the flow. Streamlines are difficult to generate experimentally in
unsteady flow unless one marks a great many particles and notes their direction of motion during a very short time interval [17, p. 35]. In steady flow the situation simplifies greatly: Streamlines,
pathlines, and streaklines are identical in steady flow. In fluid mechanics the most common mathematical result for visualization purposes is the streamline pattern. Figure 1.15a shows a typical set
of streamlines, and Fig. 1.15b shows a closed pattern called a streamtube. By definition the fluid within a streamtube is confined there because it cannot cross the streamlines; thus the streamtube
walls need not be solid but may be fluid surfaces. Figure 1.16 shows an arbitrary velocity vector. If the elemental arc length dr of a streamline is to be parallel to V, their respective components
must be in proportion: Streamline:
dx dy dz dr u v w V
Chapter 1 Introduction
V No flow across streamtube walls
Fig. 1.15 The most common method of flow-pattern presentation: (a) Streamlines are everywhere tangent to the local velocity vector; (b) a streamtube is formed by a closed collection of streamlines.
Individual streamline
If the velocities (u, v, w) are known functions of position and time, Eq. (1.41) can be integrated to find the streamline passing through the initial point (x0, y0, z0, t0). The method is
straightforward for steady flows (Example 1.10) but may be laborious for unsteady flow. The pathline, or displacement of a particle, is defined by integration of the velocity components, as mentioned
in Sec. 1.5:
x u dt
y v dt
z w dt
Given (u, v, w) as known functions of position and time, the integration is begun at a specified initial position (x0, y0, z0, t0). Again the integration may be laborious. Streaklines, easily
generated experimentally with smoke, dye, or bubble releases, are very difficult to compute analytically. See Ref. 18 for mathematical details.
V V w dr dz
dx u
Fig. 1.16 Geometric relations for defining a streamline.
1.9 Flow Patterns: Streamlines, Streaklines, and Pathlines 39
EXAMPLE 1.10 Given the steady two-dimensional velocity distribution u Kx
v Ky
where K is a positive constant, compute and plot the streamlines of the flow, including directions, and give some possible interpretations of the pattern.
Solution Since time does not appear explicitly in Eq. (1), the motion is steady, so that streamlines, pathlines, and streaklines will coincide. Since w 0 everywhere, the motion is two dimensional, in
the xy plane. The streamlines can be computed by substituting the expressions for u and v into Eq. (1.41): dx dy Kx Ky
dxx dyy
Integrating, we obtain ln x ln y ln C, or xy C
Ans. (2)
This is the general expression for the streamlines, which are hyperbolas. The complete pattern is plotted in Fig. E1.10 by assigning various values to the constant C. The arrowheads can be determined
only by returning to Eqs. (1) to ascertain the velocity component directions, assuming K is positive. For example, in the upper right quadrant (x 0, y 0), u is positive and v is negative; hence the
flow moves down and to the right, establishing the arrowheads as shown.
C = –3 +2
+1 +2 C=+3
Fig. E1.10 Streamlines for the velocity distribution given by Eq. (1), for K 0.
–2 –3
Chapter 1 Introduction Note that the streamline pattern is entirely independent of constant K. It could represent the impingement of two opposing streams, or the upper half could simulate the flow of
a single downward stream against a flat wall. Taken in isolation, the upper right quadrant is similar to the flow in a 90° corner. This is definitely a realistic flow pattern and is discussed again
in Chap. 8. Finally note the peculiarity that the two streamlines (C 0) have opposite directions and intersect. This is possible only at a point where u v w 0, which occurs at the origin in this
case. Such a point of zero velocity is called a stagnation point.
A streakline can be produced experimentally by the continuous release of marked particles (dye, smoke, or bubbles) from a given point. Figure 1.17 shows two examples. The flow in Fig. 1.17b is
unsteady and periodic due to the flapping of the plate against the oncoming stream. We see that the dash-dot streakline does not coincide with either the streamline or the pathline passing through
the same release point. This is characteristic of unsteady flow, but in Fig. 1.17a the smoke filaments form streaklines which are identical to the streamlines and pathlines. We noted earlier that
this coincidence of lines is always true of steady flow: Since the velocity never changes magnitude or direction at any point, every particle which comes along repeats the behavior of its earlier
neighbors. Methods of experimental flow visualization include the following: 1. 2. 3. 4.
Dye, smoke, or bubble discharges Surface powder or flakes on liquid flows Floating or neutral-density particles Optical techniques which detect density changes in gas flows: shadowgraph, schlieren,
and interferometer 5. Tufts of yarn attached to boundary surfaces 6. Evaporative coatings on boundary surfaces 7. Luminescent fluids or additives Uniform approach flow
Periodic flapping plate
Release point Streamline Pathline Streakline (a)
Fig. 1.17 Experimental visualization of steady and unsteady flow: (a) steady flow past an airfoil visualized by smoke filaments (C. A. A. SCIENTIFIC—Prime Movers Laboratory Systems); (b) unsteady
flow past an oscillating plate with a point bubble release (from an experiment in Ref. 17).
1.10 The Engineering Equation Solver
The mathematical implications of flow-pattern analysis are discussed in detail in Ref. 18. References 19 and 20 are beautiful albums of photographs. References 21 and 22 are monographs on flow
1.10 The Engineering Equation Solver
Most of the examples and exercises in this text are amenable to direct calculation without guessing or iteration or looping. Until recently, only such direct problem assignments, whether
“plug-and-chug” or more subtle, were appropriate for undergraduate engineering courses. However, the recent introduction of computer software solvers makes almost any set of algebraic relations
viable for analysis and solution. The solver recommended here is the Engineering Equation Solver (EES) developed by Klein and Beckman [33] and described in Appendix E. Any software solver should
handle a purely mathematical set of relations, such as the one posed in Ref. 33: X ln (X) Y3, X1/2 1/Y. Submit that pair to any commercial solver and you will no doubt receive the answer: X 1.467, Y
0.826. However, for engineers, in the author’s opinion, EES is superior to most solvers because (1) equations can be entered in any order; (2) scores of mathematical formulas are built-in, such as
the Bessel functions; and (3) thermophysical properties of many fluids are built-in, such as the steam tables [13]. Both metric and English units are allowed. Equations need not be written in the
traditional BASIC or FORTRAN style. For example, X Y 1 0 is perfectly satisfactory; there is no need to retype this as X Y 1. For example, reconsider Example 1.7 as an EES exercise. One would first
enter the reference properties p0 and 0 plus the curve-fit constants B and n: Pz 1.0 Rhoz 2.0 B 3000 n7 Then specify the given pressure ratio and the curve-fit relation, Eq. (1.19), for the equation
of state of water: P 1100*Pz P/Pz (B 1)*(Rho/Rhoz)^n B If you request an initial opinion from the CHECK/FORMAT menu, EES states that there are six equations in six unknowns and there are no obvious
difficulties. Then request SOLVE from the menu and EES quickly prints out Rho 2.091, the correct answer as seen already in Ex. 1.7. It also prints out values of the other five variables. Occasionally
EES reports “unable to converge” and states what went wrong (division by zero, square root of a negative number, etc.). One needs only to improve the guesses and ranges of the unknowns in Variable
Information to assist EES to the solution. In subsequent chapters we will illustrate some implicit (iterative) examples by using EES and will also assign some advanced problem exercises for which EES
is an ideal approach. The use of an engineering solver, notably EES, is recommended to all engineers in this era of the personal computer.
Chapter 1 Introduction
1.11 Uncertainty of Experimental Data
Earlier in this chapter we referred to the uncertainty of the principle of corresponding states in discussing Fig. 1.5. Uncertainty is a fact of life in engineering. We rarely know any engineering
properties or variables to an extreme degree of accuracy. Therefore, we need to know the uncertainty U of our data, usually defined as the band within which the experimenter is 95 percent confident
that the true value lies (Refs. 30, 31). In Fig. 1.5, we were given that the uncertainty of / c is U 20 percent. Fluid mechanics is heavily dependent upon experimentation, and the data uncertainty is
needed before we can use it for prediction or design purposes. Sometimes uncertainty completely changes our viewpoint. As an offbeat example, suppose that astronomers reported that the length of the
earth year was 365.25 days “give or take a couple of months.” First, that would make the five-figure accuracy ridiculous, and the year would better be stated as Y 365 60 days. Second, we could no
longer plan confidently or put together accurate calendars. Scheduling Christmas vacation would be chancy. Multiple variables make uncertainty estimates cumulative. Suppose a given result P depends
upon N variables, P P(x1, x2, x3, . . . , xN), each with its own uncertainty; for example, x1 has uncertainty x1. Then, by common agreement among experimenters, the total uncertainty of P is
calculated as a root-mean-square average of all effects:
P x1 x1
x2 x2
xN xN
2 1/2
This calculation is statistically much more probable than simply adding linearly the various uncertainties xi, thereby making the unlikely assumption that all variables simultaneously attain maximum
error. Note that it is the responsibility of the experimenter to establish and report accurate estimates of all the relevant uncertainties xi. If the quantity P is a simple power-law expression of
the other variables, for example, P Const x1n x2n x3n . . . , then each derivative in Eq. (1.43) is proportional to P and the relevant power-law exponent and is inversely proportional to that
variable. If P Const x1n x2n x3n . . . , then 1
P n P P n P P nP 1, 2, 3, x1 x1 x2 x2 x3 x3 Thus, from Eq. (1.43),
P P
n x n x n x 1
Evaluation of P is then a straightforward procedure, as in the following example. EXAMPLE 1.11 The so-called dimensionless Moody pipe-friction factor f, plotted in Fig. 6.13, is calculated in
experiments from the following formula involving pipe diameter D, pressure drop p, density , volume flow rate Q, and pipe length L:
2 D5p f 32 Q2L
1.12 The Fundamentals of Engineering (FE) Examination
Measurement uncertainties are given for a certain experiment: for D: 0.5 percent, p: 2.0 percent, : 1.0 percent, Q: 3.5 percent, and L: 0.4 percent. Estimate the overall uncertainty of the friction
factor f.
Solution The coefficient 2/32 is assumed to be a pure theoretical number, with no uncertainty. The other variables may be collected using Eqs. (1.43) and (1.44):
f U f
5D 1p 1 2Q 1L
2 1/2
[{5(0.5%)}2 (2.0%)2 (1.0%)2 {2(3.5%)}2 (0.4%)2]1/2 7.8%
By far the dominant effect in this particular calculation is the 3.5 percent error in Q, which is amplified by doubling, due to the power of 2 on flow rate. The diameter uncertainty, which is
quintupled, would have contributed more had D been larger than 0.5 percent.
1.12 The Fundamentals of Engineering (FE) Examination
The road toward a professional engineer’s license has a first stop, the Fundamentals of Engineering Examination, known as the FE exam. It was formerly known as the Engineer-in-Training (E-I-T)
Examination. This 8-h national test will probably soon be required of all engineering graduates, not just for licensure, but as a studentassessment tool. The 120-problem morning session covers many
general studies: Chemistry Electric circuits Materials science Statics
Computers Engineering economics Mathematics Thermodynamics
Dynamics Fluid Mechanics Mechanics of materials Ethics
For the 60-problem afternoon session you may choose chemical, civil, electrical, industrial, or mechanical engineering or take more general-engineering problems for remaining disciplines. As you can
see, fluid mechanics is central to the FE exam. Therefore, this text includes a number of end-of-chapter FE problems where appropriate. The format for the FE exam questions is multiple-choice,
usually with five selections, chosen carefully to tempt you with plausible answers if you used incorrect units or forgot to double or halve something or are missing a factor of , etc. In some cases,
the selections are unintentionally ambiguous, such as the following example from a previous exam: Transition from laminar to turbulent flow occurs at a Reynolds number of (A) 900 (B) 1200 (C) 1500
(D) 2100 (E) 3000 The “correct” answer was graded as (D), Re 2100. Clearly the examiner was thinking, but forgot to specify, Red for flow in a smooth circular pipe, since (see Chaps. 6 and 7)
transition is highly dependent upon geometry, surface roughness, and the length scale used in the definition of Re. The moral is not to get peevish about the exam but simply to go with the flow (pun
intended) and decide which answer best fits an
Chapter 1 Introduction
undergraduate-training situation. Every effort has been made to keep the FE exam questions in this text unambiguous.
1.13 Problem-Solving Techniques
Fluid flow analysis generates a plethora of problems, 1500 in this text alone! To solve these problems, one must deal with various equations, data, tables, assumptions, unit systems, and numbers. The
writer recommends these problem-solving steps: 1. Gather all the given system parameters and data in one place. 2. Find, from tables or charts, all needed fluid property data: , , cp, k, , etc. 3.
Use SI units (N, s, kg, m) if possible, and no conversion factors will be necessary. 4. Make sure what is asked. It is all too common for students to answer the wrong question, for example, reporting
mass flow instead of volume flow, pressure instead of pressure gradient, drag force instead of lift force. Engineers are expected to read carefully. 5. Make a detailed sketch of the system, with
everything clearly labeled. 6. Think carefully and then list your assumptions. Here knowledge is power; you should not guess the answer. You must be able to decide correctly if the flow can be
considered steady or unsteady, compressible or incompressible, onedimensional, or multidimensional, viscous or inviscid, and whether a control volume or partial differential equations are needed. 7.
Based on steps 1 to 6 above, write out the appropriate equations, data correlations, and fluid state relations for your problem. If the algebra is straightforward, solve for what is asked. If the
equations are complicated, e.g., nonlinear or too plentiful, use the Engineering Equation Solver (EES). 8. Report your solution clearly, with proper units listed and to the proper number of
significant figures (usually two or three) that the overall uncertainty of the data will allow.
1.14 History and Scope of Fluid Mechanics
Like most scientific disciplines, fluid mechanics has a history of erratically occurring early achievements, then an intermediate era of steady fundamental discoveries in the eighteenth and
nineteenth centuries, leading to the twentieth-century era of “modern practice,” as we self-centeredly term our limited but up-to-date knowledge. Ancient civilizations had enough knowledge to solve
certain flow problems. Sailing ships with oars and irrigation systems were both known in prehistoric times. The Greeks produced quantitative information. Archimedes and Hero of Alexandria both
postulated the parallelogram law for addition of vectors in the third century B.C. Archimedes (285–212 B.C.) formulated the laws of buoyancy and applied them to floating and submerged bodies,
actually deriving a form of the differential calculus as part of the analysis. The Romans built extensive aqueduct systems in the fourth century B.C. but left no records showing any quantitative
knowledge of design principles. From the birth of Christ to the Renaissance there was a steady improvement in the design of such flow systems as ships and canals and water conduits but no recorded
evidence of fundamental improvements in flow analysis. Then Leonardo da Vinci (1452–1519) derived the equation of conservation of mass in one-dimensional steady
1.14 History and Scope of Fluid Mechanics
flow. Leonardo was an excellent experimentalist, and his notes contain accurate descriptions of waves, jets, hydraulic jumps, eddy formation, and both low-drag (streamlined) and high-drag (parachute)
designs. A Frenchman, Edme Mariotte (1620–1684), built the first wind tunnel and tested models in it. Problems involving the momentum of fluids could finally be analyzed after Isaac Newton
(1642–1727) postulated his laws of motion and the law of viscosity of the linear fluids now called newtonian. The theory first yielded to the assumption of a “perfect” or frictionless fluid, and
eighteenth-century mathematicians (Daniel Bernoulli, Leonhard Euler, Jean d’Alembert, Joseph-Louis Lagrange, and Pierre-Simon Laplace) produced many beautiful solutions of frictionless-flow problems.
Euler developed both the differential equations of motion and their integrated form, now called the Bernoulli equation. D’Alembert used them to show his famous paradox: that a body immersed in a
frictionless fluid has zero drag. These beautiful results amounted to overkill, since perfect-fluid assumptions have very limited application in practice and most engineering flows are dominated by
the effects of viscosity. Engineers began to reject what they regarded as a totally unrealistic theory and developed the science of hydraulics, relying almost entirely on experiment. Such
experimentalists as Chézy, Pitot, Borda, Weber, Francis, Hagen, Poiseuille, Darcy, Manning, Bazin, and Weisbach produced data on a variety of flows such as open channels, ship resistance, pipe flows,
waves, and turbines. All too often the data were used in raw form without regard to the fundamental physics of flow. At the end of the nineteenth century, unification between experimental hydraulics
and theoretical hydrodynamics finally began. William Froude (1810–1879) and his son Robert (1846–1924) developed laws of model testing, Lord Rayleigh (1842–1919) proposed the technique of dimensional
analysis, and Osborne Reynolds (1842–1912) published the classic pipe experiment in 1883 which showed the importance of the dimensionless Reynolds number named after him. Meanwhile, viscous-flow
theory was available but unexploited, since Navier (1785–1836) and Stokes (1819–1903) had successfully added newtonian viscous terms to the equations of motion. The resulting Navier-Stokes equations
were too difficult to analyze for arbitrary flows. Then, in 1904, a German engineer, Ludwig Prandtl (1875–1953), published perhaps the most important paper ever written on fluid mechanics. Prandtl
pointed out that fluid flows with small viscosity, e.g., water flows and airflows, can be divided into a thin viscous layer, or boundary layer, near solid surfaces and interfaces, patched onto a
nearly inviscid outer layer, where the Euler and Bernoulli equations apply. Boundary-layer theory has proved to be the single most important tool in modern flow analysis. The twentiethcentury
foundations for the present state of the art in fluid mechanics were laid in a series of broad-based experiments and theories by Prandtl and his two chief friendly competitors, Theodore von Kármán
(1881–1963) and Sir Geoffrey I. Taylor (1886– 1975). Many of the results sketched here from a historical point of view will, of course, be discussed in this textbook. More historical details can be
found in Refs. 23 to 25. Since the earth is 75 percent covered with water and 100 percent covered with air, the scope of fluid mechanics is vast and touches nearly every human endeavor. The sciences
of meteorology, physical oceanography, and hydrology are concerned with naturally occurring fluid flows, as are medical studies of breathing and blood circulation. All transportation problems involve
fluid motion, with well-developed specialties in aerodynamics of aircraft and rockets and in naval hydrodynamics of ships and submarines. Almost all our electric energy is developed either from water
flow or from
Chapter 1 Introduction
steam flow through turbine generators. All combustion problems involve fluid motion, as do the more classic problems of irrigation, flood control, water supply, sewage disposal, projectile motion,
and oil and gas pipelines. The aim of this book is to present enough fundamental concepts and practical applications in fluid mechanics to prepare you to move smoothly into any of these specialized
fields of the science of flow—and then be prepared to move out again as new technologies develop.
Problems Most of the problems herein are fairly straightforward. More difficult or open-ended assignments are labeled with an asterisk as in Prob. 1.18. Problems labeled with an EES icon (for
example, Prob. 2.62), will benefit from the use of the Engineering Equation Solver (EES), while problems labeled with a computer disk may require the use of a computer. The standard end-of-chapter
problems 1.1 to 1.85 (categorized in the problem list below) are followed by fundamentals of engineering (FE) exam problems FE3.1 to FE3.10, and comprehensive problems C1.1 to C1.4. Problem
Distribution Section 1.1, 1.2, 1.3 1.4 1.5 1.6 1.7 1.7 1.7 1.7 1.8,9 1.10
Fluid-continuum concept Dimensions, units, dynamics Velocity field Thermodynamic properties Viscosity; no-slip condition Surface tension Vapor pressure; cavitation Speed of sound; Mach number Flow
patterns, streamlines, pathlines History of fluid mechanics
1.1–1.3 1.4–1.20 1.21–1.23 1.24–1.37 1.38–1.61 1.62–1.71 1.72–1.75 1.76–1.78 1.79–1.84 1.85
P1.1 A gas at 20°C may be considered rarefied, deviating from the continuum concept, when it contains less than 1012 molecules per cubic millimeter. If Avogadro’s number is 6.023 E23 molecules per
mole, what absolute pressure (in Pa) for air does this represent? P1.2 Table A.6 lists the density of the standard atmosphere as a function of altitude. Use these values to estimate, crudely— say,
within a factor of 2—the number of molecules of air in the entire atmosphere of the earth. P1.3 For the triangular element in Fig. P1.3, show that a tilted free liquid surface, in contact with an
atmosphere at pressure pa, must undergo shear stress and hence begin to flow. Hint: Account for the weight of the fluid and show that a no-shear condition will cause horizontal forces to be out of
pa Fluid density
P1.4 A beaker approximates a right circular cone of diameter 7 in and height 9 in. When filled with liquid, it weighs 70 oz. When empty, it weighs 14 oz. Estimate the density of this liquid in both
SI and BG units. P1.5 The mean free path of a gas, , is defined as the average distance traveled by molecules between collisions. A proposed formula for estimating of an ideal gas is
1.26 R T What are the dimensions of the constant 1.26? Use the formula to estimate the mean free path of air at 20°C and 7 kPa. Would you consider air rarefied at this condition? P1.6 In the {MLT }
system, what is the dimensional representation of (a) enthalpy, (b) mass rate of flow, (c) bending moment, (d) angular velocity, (e) modulus of elasticity; (f ) Poisson’s ratio? P1.7 A small village
draws 1.5 acre ft/day of water from its reservoir. Convert this average water usage to (a) gallons per minute and (b) liters per second. P1.8 Suppose we know little about the strength of materials
but are told that the bending stress in a beam is proportional to the beam half-thickness y and also depends upon the bending moment M and the beam area moment of inertia I. We also learn that, for
the particular case M 2900 in lbf, y 1.5 in, and I 0.4 in4, the predicted stress is 75 MPa. Using this information and dimensional reasoning only, find, to three significant figures, the only
possible dimensionally homogeneous formula y f(M, I ).
Problems 47 P1.9 The kinematic viscosity of a fluid is the ratio of viscosity where Q is the volume rate of flow and p is the pressure to density, /. What is the only possible dimensionrise produced
by the pump. Suppose that a certain pump less group combining with velocity V and length L? What develops a pressure rise of 35 lbf/in2 when its flow rate is is the name of this grouping? (More
information on this 40 L/s. If the input power is 16 hp, what is the efficiency? will be given in Chap. 5.) *P1.14 Figure P1.14 shows the flow of water over a dam. The volP1.10 The Stokes-Oseen
formula [18] for drag force F on a ume flow Q is known to depend only upon crest width B, sphere of diameter D in a fluid stream of low velocity V, acceleration of gravity g, and upstream water
height H density , and viscosity , is above the dam crest. It is further known that Q is proportional to B. What is the form of the only possible dimen9 F 3 DV V2D2 sionally homogeneous relation for
this flow rate? 16 Is this formula dimensionally homogeneous? P1.11 Engineers sometimes use the following formula for the volume rate of flow Q of a liquid flowing through a hole of diameter D in the
side of a tank: Q 0.68 D2g h
Water level
where g is the acceleration of gravity and h is the height of the liquid surface above the hole. What are the dimensions of the constant 0.68? P1.12 For low-speed (laminar) steady flow through a
circular pipe, as shown in Fig. P1.12, the velocity u varies with radius and takes the form p u B(r02 r2)
Dam B
where is the fluid viscosity and p is the pressure drop from entrance to exit. What are the dimensions of the constant B? Pipe wall r = r0
u (r) r=0
P1.15 As a practical application of Fig. P1.14, often termed a sharp-crested weir, civil engineers use the following formula for flow rate: Q 3.3BH3/2, with Q in ft3/s and B and H in feet. Is this
formula dimensionally homogeneous? If not, try to explain the difficulty and how it might be converted to a more homogeneous form. P1.16 Algebraic equations such as Bernoulli’s relation, Eq. (1) of
Ex. 1.3, are dimensionally consistent, but what about differential equations? Consider, for example, the boundary-layer x-momentum equation, first derived by Ludwig Prandtl in 1904: u u p u v gx x y
x y
P1.12 P1.13 The efficiency of a pump is defined as the (dimensionless) ratio of the power developed by the flow to the power required to drive the pump: Qp input power
where is the boundary-layer shear stress and gx is the component of gravity in the x direction. Is this equation dimensionally consistent? Can you draw a general conclusion? P1.17 The Hazen-Williams
hydraulics formula for volume rate of flow Q through a pipe of diameter D and length L is given by p Q 61.9 D 2.63 L
Chapter 1 Introduction
where p is the pressure drop required to drive the flow. What are the dimensions of the constant 61.9? Can this formula be used with confidence for various liquids and gases? For small particles at
low velocities, the first term in the Stokes-Oseen drag law, Prob. 1.10, is dominant; hence, F KV, where K is a constant. Suppose a particle of mass m is constrained to move horizontally from the
initial position x 0 with initial velocity V0. Show (a) that its velocity will decrease exponentially with time and (b) that it will stop after traveling a distance x mV0 /K. For larger particles at
higher velocities, the quadratic term in the Stokes-Oseen drag law, Prob. 1.10, is dominant; hence, F CV2, where C is a constant. Repeat Prob. 1.18 to show that (a) its velocity will decrease as 1/(1
CV0t/m) and (b) it will never quite stop in a finite time span. A baseball, with m 145 g, is thrown directly upward from the initial position z 0 and V0 45 m/s. The air drag on the ball is CV 2, as
in Prob. 1.19, where C 0.0013 N s2/m2. Set up a differential equation for the ball motion, and solve for the instantaneous velocity V(t) and position z(t). Find the maximum height zmax reached by the
ball, and compare your results with the classical case of zero air drag. A velocity field is given by V Kxti Kytj 0k, where K is a positive constant. Evaluate (a) and (b) V. According to the theory
of Chap. 8, as a uniform stream approaches a cylinder of radius R along the symmetry line AB in Fig. P1.22, the velocity has only one component: R2 for x R u U 1 x2
where U is the stream velocity far from the cylinder. Using the concepts from Ex. 1.5, find (a) the maximum flow deceleration along AB and (b) its location.
Q/Across-section and the dimensionless Reynolds number of the flow, Re VavgD/ . Comment on your results. Air at 1 atm and 20°C has an internal energy of approximately 2.1 E5 J/kg. If this air moves
at 150 m/s at an altitude z 8 m, what is its total energy, in J/kg, relative to the datum z 0? Are any energy contributions negligible? A tank contains 0.9 m3 of helium at 200 kPa and 20°C. Estimate
the total mass of this gas, in kg, (a) on earth and (b) on the moon. Also, (c) how much heat transfer, in MJ, is required to expand this gas at constant temperature to a new volume of 1.5 m3? When we
in the United States say a car’s tire is filled “to 32 lb,” we mean that its internal pressure is 32 lbf/in2 above the ambient atmosphere. If the tire is at sea level, has a volume of 3.0 ft3, and is
at 75°F, estimate the total weight of air, in lbf, inside the tire. For steam at 40 lbf/in2, some values of temperature and specific volume are as follows, from Ref. 13:
T, °F 3
v, ft /lbm
Is steam, for these conditions, nearly a perfect gas, or is it wildly nonideal? If reasonably perfect, find a least-squares† value for the gas constant R, in m2/(s2 K), estimate the percent error in
this approximation, and compare with Table A.4. P1.28 Wet atmospheric air at 100 percent relative humidity contains saturated water vapor and, by Dalton’s law of partial pressures, patm pdry air
pwater vapor
Suppose this wet atmosphere is at 40°C and 1 atm. Calculate the density of this 100 percent humid air, and compare it with the density of dry air at the same conditions. P1.29 A compressed-air tank
holds 5 ft3 of air at 120 lbf/in2 “gage,” that is, above atmospheric pressure. Estimate the y energy, in ft-lbf, required to compress this air from the atmosphere, assuming an ideal isothermal
process. P1.30 Repeat Prob. 1.29 if the tank is filled with compressed wau x ter instead of air. Why is the result thousands of times less A B R than the result of 215,000 ft lbf in Prob. 1.29?
*P1.31 The density of (fresh) water at 1 atm, over the temperature range 0 to 100°C, is given in Table A.1. Fit these valEES P1.22 ues to a least-squares† equation of the form a bT cT 2, with T in
°C, and estimate its accuracy. Use your forP1.23 Experiment with a faucet (kitchen or otherwise) to determine mula to compute the density of water at 45°C, and comtypical flow rates Q in m3/h,
perhaps timing the discharge of pare your result with the accepted experimental value of a known volume. Try to achieve an exit jet condition which 990.1 kg/m3. is (a) smooth and round and (b)
disorderly and fluctuating. † Measure the supply-pipe diameter (look under the sink). For The concept of “least-squares” error is very important and should be both cases, calculate the average flow
velocity, Vavg learned by everyone.
Problems 49 P1.32 A blimp is approximated by a prolate spheroid 90 m long and 30 m in diameter. Estimate the weight of 20°C gas within the blimp for (a) helium at 1.1 atm and (b) air at 1.0 atm. What
might the difference between these two values represent (see Chap. 2)? *P1.33 Experimental data for the density of mercury versus pressure at 20°C are as follows: p, atm
, kg/m
T, K
, kg/(m s) 2.27 E-5 2.85 E-5 3.37 E-5 3.83 E-5 4.25 E-5 4.64 E-5
Fit these value to either (a) a power law or (b) the Sutherland law, Eq. (1.30). P1.42 Experimental values for the viscosity of helium at 1 atm are as follows: T, K
, kg/(m s) 1.50 E-5 2.43 E-5 3.20 E-5 3.88 E-5 4.50 E-5 5.08 E-5
Fit this data to the empirical state relation for liquids, Fit these values to either (a) a power law or (b) the SutherEq. (1.22), to find the best values of B and n for mercury. land law, Eq.
(1.30). Then, assuming the data are nearly isentropic, use these *P1.43 Yaws et al. [34] suggest the following curve-fit formula values to estimate the speed of sound of mercury at 1 atm for
viscosity versus temperature of organic liquids: and compare with Table 9.1. 3 P1.34 If water occupies 1 m at 1 atm pressure, estimate the presB log10 A CT DT2 sure required to reduce its volume by 5
percent. T P1.35 In Table A.4, most common gases (air, nitrogen, oxygen, hydrogen) have a specific heat ratio k 1.40. Why do arwith T in absolute units. (a) Can this formula be criticized gon and
helium have such high values? Why does NH3 on dimensional grounds? (b) Disregarding (a), indicate anhave such a low value? What is the lowest k for any gas alytically how the curve-fit constants A,
B, C, D could be that you know of? found from N data points ( i, Ti) using the method of least P1.36 The isentropic bulk modulus B of a fluid is defined as the squares. Do not actually carry out a
calculation. isentropic change in pressure per fractional change in den- P1.44 The values for SAE 30 oil in Table 1.4 are strictly “repsity: resentative,” not exact, because lubricating oils vary
considerably according to the type of crude oil from which p B they are refined. The Society of Automotive Engineers [26] s allows certain kinematic viscosity ranges for all lubricatWhat are the
dimensions of B? Using theoretical p() reing oils: for SAE 30, 9.3 12.5 mm2/s at 100°C. SAE lations, estimate the bulk modulus of (a) N2O, assumed to 30 oil density can also vary 2 percent from the
tabulated be an ideal gas, and (b) water, at 20°C and 1 atm. value of 891 kg/m3. Consider the following data for an acP1.37 A near-ideal gas has a molecular weight of 44 and a speceptable grade of
SAE 30 oil: cific heat cv 610 J/(kg K). What are (a) its specific heat T, °C 0 20 40 60 80 100 ratio, k, and (b) its speed of sound at 100°C? P1.38 In Fig. 1.6, if the fluid is glycerin at 20°C and
the width be- , kg/(m s) 2.00 0.40 0.11 0.042 0.017 0.0095 tween plates is 6 mm, what shear stress (in Pa) is required How does this oil compare with the plot in Appendix Fig. to move the upper plate
at 5.5 m/s? What is the Reynolds A.1? How well does the data fit Andrade’s equation in number if L is taken to be the distance between plates? Prob. 1.40? P1.39 Knowing for air at 20°C from Table
1.4, estimate its viscosity at 500°C by (a) the power law and (b) the Suther- P1.45 A block of weight W slides down an inclined plane while lubricated by a thin film of oil, as in Fig. P1.45. The
film land law. Also make an estimate from (c) Fig. 1.5. Comcontact area is A and its thickness is h. Assuming a linear pare with the accepted value of 3.58 E-5 kg/m s. velocity distribution in the
film, derive an expression for *P1.40 For liquid viscosity as a function of temperature, a simthe “terminal” (zero-acceleration) velocity V of the block. plification of the log-quadratic law of Eq.
(1.31) is Andrade’s equation [11], A exp (B/T), where (A, B) are P1.46 Find the terminal velocity of the block in Fig. P1.45 if the block mass is 6 kg, A 35 cm2, 15°, and the film is curve-fit
constants and T is absolute temperature. Fit this 1-mm-thick SAE 30 oil at 20°C. relation to the data for water in Table A.1 and estimate the P1.47 A shaft 6.00 cm in diameter is being pushed axially
percent error of the approximation. through a bearing sleeve 6.02 cm in diameter and 40 cm P1.41 Some experimental values of the viscosity of argon gas at long. The clearance, assumed uniform, is
filled with oil 1 atm are as follows: EES
Chapter 1 Introduction sured torque is 0.293 N m, what is the fluid viscosity? Suppose that the uncertainties of the experiment are as follows: L (0.5 mm), M (0.003 N m), (1 percent), and ri or ro
(0.02 mm). What is the uncertainty in the measured viscosity? P1.52 The belt in Fig. P1.52 moves at a steady velocity V and skims the top of a tank of oil of viscosity , as shown. Assuming a linear
velocity profile in the oil, develop a simple formula for the required belt-drive power P as a function of (h, L, V, b, ). What belt-drive power P, in watts, is required if the belt moves at 2.5 m/s
over SAE 30W oil at 20°C, with L 2 m, b 60 cm, and h 3 cm?
Liquid film of thickness h W V θ
Block contact area A
whose properties are 0.003 m2/s and SG 0.88. Estimate the force required to pull the shaft at a steady veL locity of 0.4 m/s. V P1.48 A thin plate is separated from two fixed plates by very visMoving
belt, width b cous liquids 1 and 2, respectively, as in Fig. P1.48. The plate spacings h1 and h2 are unequal, as shown. The conOil, depth h tact area is A between the center plate and each fluid. (a)
Assuming a linear velocity distribution in each fluid, P1.52 derive the force F required to pull the plate at velocity V. (b) Is there a necessary relation between the two viscosities, 1 and 2?
*P1.53 A solid cone of angle 2, base r0, and density c is rotating with initial angular velocity 0 inside a conical seat, as shown in Fig. P1.53. The clearance h is filled with oil of viscosity .
Neglecting air drag, derive an analytical exh1 µ1 pression for the cone’s angular velocity (t) if there is no applied torque. F, V h2
Base radius r0
ω (t) Oil
P1.48 P1.49 The shaft in Prob. 1.47 is now fixed axially and rotated in2θ side the sleeve at 1500 r/min. Estimate (a) the torque (N m) and (b) the power (kW) required to rotate the shaft. h P1.50 An
amazing number of commercial and laboratory devices have been developed to measure the viscosity of fluids, as described in Ref. 27. The concentric rotating shaft of Prob. 1.49 is an example of a
rotational viscometer. Let the inner and outer cylinders have radii ri and ro, respectively, P1.53 with total sleeve length L. Let the rotational rate be (rad/s) and the applied torque be M. Derive a
theoretical relation for the viscosity of the clearance fluid, , in terms *P1.54 A disk of radius R rotates at an angular velocity inside of these parameters. a disk-shaped container filled with oil
of viscosity , as P1.51 Use the theory of Prob. 1.50 (or derive an ad hoc expresshown in Fig. P1.54. Assuming a linear velocity profile sion if you like) for a shaft 8 cm long, rotating at 1200 and
neglecting shear stress on the outer disk edges, derive r/min, with ri 2.00 cm and ro 2.05 cm. If the meaa formula for the viscous torque on the disk.
Problems 51
r4 p 0 8LQ
Ω Clearance h
Pipe end effects are neglected [27]. Suppose our capillary has r0 2 mm and L 25 cm. The following flow rate and pressure drop data are obtained for a certain fluid:
Q, m3/h
p, kPa
*P1.55 The device in Fig. P1.54 is called a rotating disk viscometer [27]. Suppose that R 5 cm and h 1 mm. If the torque required to rotate the disk at 900 r/min is 0.537 N m, what is the viscosity
of the fluid? If the uncertainty in each parameter (M, R, h, ) is 1 percent, what is the overall uncertainty in the viscosity? *P1.56 The device in Fig. P1.56 is called a cone-plate viscometer [27].
The angle of the cone is very small, so that sin , and the gap is filled with the test liquid. The torque M to rotate the cone at a rate is measured. Assuming a linear velocity profile in the fluid
film, derive an expression for fluid viscosity as a function of (M, R, , ).
*P1.61 Ω
R Fluid θ
P1.56 P1.63 *P1.57 For the cone-plate viscometer of Fig. P1.56, suppose that R 6 cm and 3°. If the torque required to rotate the cone at 600 r/min is 0.157 N m, what is the viscosity of the fluid? If
the uncertainty in each parameter (M, R, , ) is 1 percent, what is the overall uncertainty in the viscosity? *P1.58 The laminar-pipe-flow example of Prob. 1.12 can be used to design a capillary
viscometer [27]. If Q is the volume flow rate, L is the pipe length, and p is the pressure drop from entrance to exit, the theory of Chap. 6 yields a formula for viscosity:
What is the viscosity of the fluid? Note: Only the first three points give the proper viscosity. What is peculiar about the last two points, which were measured accurately? A solid cylinder of
diameter D, length L, and density s falls due to gravity inside a tube of diameter D0. The clearance, D0 D D, is filled with fluid of density and viscosity . Neglect the air above and below the
cylinder. Derive a formula for the terminal fall velocity of the cylinder. Apply your formula to the case of a steel cylinder, D 2 cm, D0 2.04 cm, L 15 cm, with a film of SAE 30 oil at 20°C. For
Prob. 1.52 suppose that P 0.1 hp when V 6 ft/s, L 4.5 ft, b 22 in, and h 7/8 in. Estimate the viscosity of the oil, in kg/(m s). If the uncertainty in each parameter (P, L, b, h, V) is 1 percent,
what is the overall uncertainty in the viscosity? An air-hockey puck has a mass of 50 g and is 9 cm in diameter. When placed on the air table, a 20°C air film, of 0.12-mm thickness, forms under the
puck. The puck is struck with an initial velocity of 10 m/s. Assuming a linear velocity distribution in the air film, how long will it take the puck to (a) slow down to 1 m/s and (b) stop completely?
Also, (c) how far along this extremely long table will the puck have traveled for condition (a)? The hydrogen bubbles which produced the velocity profiles in Fig. 1.13 are quite small, D 0.01 mm. If
the hydrogen-water interface is comparable to air-water and the water temperature is 30°C estimate the excess pressure within the bubble. Derive Eq. (1.37) by making a force balance on the fluid
interface in Fig. 1.9c. At 60°C the surface tension of mercury and water is 0.47 and 0.0662 N/m, respectively. What capillary height changes will occur in these two fluids when they are in contact
with air in a clean glass tube of diameter 0.4 mm? The system in Fig. P1.65 is used to calculate the pressure p1 in the tank by measuring the 15-cm height of liquid in the 1-mm-diameter tube. The
fluid is at 60°C (see Prob. 1.64). Calculate the true fluid height in the tube and the percent error due to capillarity if the fluid is (a) water and (b) mercury.
Chapter 1 Introduction mula for the maximum diameter dmax able to float in the liquid. Calculate dmax for a steel needle (SG 7.84) in water at 20°C. P1.70 Derive an expression for the capillary
height change h for a fluid of surface tension Y and contact angle between two vertical parallel plates a distance W apart, as in Fig. P1.70. What will h be for water at 20°C if W 0.5 mm?
15 cm p1
P1.65 P1.66 A thin wire ring, 3 cm in diameter, is lifted from a water surface at 20°C. Neglecting the wire weight, what is the force required to lift the ring? Is this a good way to meah sure
surface tension? Should the wire be made of any particular material? P1.67 Experiment with a capillary tube, perhaps borrowed from W the chemistry department, to verify, in clean water, the rise due
to surface tension predicted by Example 1.9. Add small P1.70 amounts of liquid soap to the water, and report to the class whether detergents significantly lower the surface tension. *P1.71 A soap
bubble of diameter D1 coalesces with another bubWhat practical difficulties do detergents present? ble of diameter D2 to form a single bubble D3 with the *P1.68 Make an analysis of the shape (x) of
the water-air intersame amount of air. Assuming an isothermal process, deface near a plane wall, as in Fig. P1.68, assuming that the rive an expression for finding D3 as a function of D1, D2, slope
is small, R1 d2/dx2. Also assume that the prespatm, and Y. sure difference across the interface is balanced by the specific weight and the interface height, p g. The P1.72 Early mountaineers boiled
water to estimate their altitude. If they reach the top and find that water boils at 84°C, apboundary conditions are a wetting contact angle at x 0 EES proximately how high is the mountain? and a
horizontal surface 0 as x → . What is the maxP1.73 A small submersible moves at velocity V, in fresh water imum height h at the wall? at 20°C, at a 2-m depth, where ambient pressure is 131 y kPa. Its
critical cavitation number is known to be Ca 0.25. At what velocity will cavitation bubbles begin to form y=h on the body? Will the body cavitate if V 30 m/s and the water is cold (5°C)? P1.74 A
propeller is tested in a water tunnel at 20°C as in Fig. 1.12a. The lowest pressure on the blade can be estimated by a form of Bernoulli’s equation (Ex. 1.3): pmin p0 12V 2
θ η (x) x x=0
P1.68 P1.69 A solid cylindrical needle of diameter d, length L, and density n may float in liquid of surface tension Y. Neglect buoyancy and assume a contact angle of 0°. Derive a for-
where p0 1.5 atm and V tunnel velocity. If we run the tunnel at V 18 m/s, can we be sure that there will be no cavitation? If not, can we change the water temperature and avoid cavitation? P1.75 Oil,
with a vapor pressure of 20 kPa, is delivered through a pipeline by equally spaced pumps, each of which increases the oil pressure by 1.3 MPa. Friction losses in the pipe are 150 Pa per meter of
pipe. What is the maximum possible pump spacing to avoid cavitation of the oil?
Fundamentals of Engineering Exam Problems
P1.76 An airplane flies at 555 mi/h. At what altitude in the stan- P1.82 A velocity field is given by u V cos , v V sin , and dard atmosphere will the airplane’s Mach number be exw 0, where V and are
constants. Derive a formula for actly 0.8? the streamlines of this flow. *P1.77 The density of 20°C gasoline varies with pressure ap- *P1.83 A two-dimensional unsteady velocity field is given by u
proximately as follows: x(1 2t), v y. Find the equation of the time-varying EES streamlines which all pass through the point (x0, y0) at p, atm 1 500 1000 1500 some time t. Sketch a few of these. ,
lbm/ft3 42.45 44.85 46.60 47.98 *P1.84 Repeat Prob. 1.83 to find and sketch the equation of the pathline which passes through (x0, y0) at time t 0. Use these data to estimate (a) the speed of sound
(m/s) P1.85 Do some reading and report to the class on the life and and (b) the bulk modulus (MPa) of gasoline at 1 atm. achievements, especially vis-à-vis fluid mechanics, of P1.78 Sir Isaac Newton
measured the speed of sound by timing (a) Evangelista Torricelli (1608–1647) the difference between seeing a cannon’s puff of smoke (b) Henri de Pitot (1695–1771) and hearing its boom. If the cannon
is on a mountain 5.2 mi away, estimate the air temperature in degrees Celsius if the (c) Antoine Chézy (1718–1798) time difference is (a) 24.2 s and (b) 25.1 s. (d) Gotthilf Heinrich Ludwig Hagen
(1797–1884) P1.79 Examine the photographs in Figs. 1.12a, 1.13, 5.2a, 7.14a, (e) Julius Weisbach (1806–1871) and 9.10b and classify them according to the boxes in Fig. (f) George Gabriel Stokes
(1819–1903) 1.14. (g) Moritz Weber (1871–1951) *P1.80 A two-dimensional steady velocity field is given by u 2 2 (h) Theodor von Kármán (1881–1963) x y , v 2xy. Derive the streamline pattern and
sketch a few streamlines in the upper half plane. Hint: The (i) Paul Richard Heinrich Blasius (1883–1970) differential equation is exact. (j) Ludwig Prandtl (1875–1953) P1.81 Repeat Ex. 1.10 by
letting the velocity components in(k) Osborne Reynolds (1842–1912) crease linearly with time: (l) John William Strutt, Lord Rayleigh (1842–1919) V Kxti Kytj 0k (m) Daniel Bernoulli (1700–1782) (n)
Leonhard Euler (1707–1783) Find and sketch, for a few representative times, the instantaneous streamlines. How do they differ from the steady flow lines in Ex. 1.10?
Fundamentals of Engineering Exam Problems FE1.1 The absolute viscosity of a fluid is primarily a function of (a) Density, (b) Temperature, (c) Pressure, (d) Velocity, (e) Surface tension FE1.2 If a
uniform solid body weighs 50 N in air and 30 N in water, its specific gravity is (a) 1.5, (b) 1.67, (c) 2.5, (d) 3.0, (e) 5.0 FE1.3 Helium has a molecular weight of 4.003. What is the weight of 2 m3
of helium at 1 atm and 20°C? (a) 3.3 N, (b) 6.5 N, (c) 11.8 N, (d) 23.5 N, (e) 94.2 N FE1.4 An oil has a kinematic viscosity of 1.25 E-4 m2/s and a specific gravity of 0.80. What is its dynamic
(absolute) viscosity in kg/(m s)? (a) 0.08, (b) 0.10, (c) 0.125, (d) 1.0, (e) 1.25 FE1.5 Consider a soap bubble of diameter 3 mm. If the surface tension coefficient is 0.072 N/m and external pressure
is 0 Pa gage, what is the bubble’s internal gage pressure?
(a) 24 Pa, (b) 48 Pa, (c) 96 Pa, (d) 192 Pa, (e) 192 Pa FE1.6 The only possible dimensionless group which combines velocity V, body size L, fluid density , and surface tension coefficient is (a) L/V,
(b) VL2/, (c) V2/L, (d) LV2/, (e) LV2/ FE1.7 Two parallel plates, one moving at 4 m/s and the other fixed, are separated by a 5-mm-thick layer of oil of specific gravity 0.80 and kinematic viscosity
1.25 E-4 m2/s. What is the average shear stress in the oil? (a) 80 Pa, (b) 100 Pa, (c) 125 Pa, (d) 160 Pa, (e) 200 Pa FE1.8 Carbon dioxide has a specific heat ratio of 1.30 and a gas constant of 189
J/(kg °C). If its temperature rises from 20 to 45°C, what is its internal energy rise? (a) 12.6 kJ/kg, (b) 15.8 kJ/kg, (c) 17.6 kJ/kg, (d) 20.5 kJ/kg, (e) 25.1 kJ/kg
Chapter 1 Introduction
FE1.9 A certain water flow at 20°C has a critical cavitation number, where bubbles form, Ca 0.25, where Ca 2(pa pvap)/V2. If pa 1 atm and the vapor pressure is 0.34 pounds per square inch absolute
(psia), for what water velocity will bubbles form? (a) 12 mi/h, (b) 28 mi/h, (c) 36 mi/h, (d) 55 mi/h, (e) 63 mi/h
FE1.10 A steady incompressible flow, moving through a contraction section of length L, has a one-dimensional average velocity distribution given by u U0(1 2x/L). What is its convective acceleration
at the end of the contraction, x L? (a) U02/L, (b) 2U02/L, (c) 3U02/L, (d) 4U02/L, (e) 6U02/L
Comprehensive Problems C1.1
Sometimes equations can be developed and practical problems can be solved by knowing nothing more than the dimensions of the key parameters in the problem. For example, consider the heat loss through
a window in a building. Window efficiency is rated in terms of “R value” which has units of (ft2 h °F)/Btu. A certain manufacturer advertises a double-pane window with an R value of 2.5. The same
company produces a triple-pane window with an R value of 3.4. In either case the window dimensions are 3 ft by 5 ft. On a given winter day, the temperature difference between the inside and outside
of the building is 45°F. (a) Develop an equation for the amount of heat lost in a given time period t, through a window of area A, with R value R, and temperature difference T. How much heat (in Btu)
is lost through the double-pane window in one 24-h period? (b) How much heat (in Btu) is lost through the triple-pane window in one 24-h period? (c) Suppose the building is heated with propane gas,
which costs $1.25 per gallon. The propane burner is 80 percent efficient. Propane has approximately 90,000 Btu of available energy per gallon. In that same 24-h period, how much money would a
homeowner save per window by installing triple-pane rather than doublepane windows? (d) Finally, suppose the homeowner buys 20 such triplepane windows for the house. A typical winter has the
equivalent of about 120 heating days at a temperature difference of 45°F. Each triple-pane window costs $85 more than the double-pane window. Ignoring interest and inflation, how many years will it
take the homeowner to make up the additional cost of the triple-pane windows from heating bill savings? C1.2 When a person ice skates, the surface of the ice actually melts beneath the blades, so
that he or she skates on a thin sheet of water between the blade and the ice. (a) Find an expression for total friction force on the bottom of the blade as a function of skater velocity V, blade
length L, water thickness (between the blade and the ice) h, water viscosity , and blade width W.
(b) Suppose an ice skater of total mass m is skating along at a constant speed of V0 when she suddenly stands stiff with her skates pointed directly forward, allowing herself to coast to a stop.
Neglecting friction due to air resistance, how far will she travel before she comes to a stop? (Remember, she is coasting on two skate blades.) Give your answer for the total distance traveled, x, as
a function of V0, m, L, h, , and W. (c) Find x for the case where V0 4.0 m/s, m 100 kg, L 30 cm, W 5.0 mm, and h 0.10 mm. Do you think our assumption of negligible air resistance is a good one? C1.3
Two thin flat plates, tilted at an angle !, are placed in a tank of liquid of known surface tension and contact angle , as shown in Fig. C1.3. At the free surface of the liquid in the tank, the two
plates are a distance L apart and have width b into the page. The liquid rises a distance h between the plates, as shown. (a) What is the total upward (z-directed) force, due to surface tension,
acting on the liquid column between the plates? (b) If the liquid density is , find an expression for surface tension in terms of the other variables.
References 55
Oil of viscosity and density drains steadily down the side of a tall, wide vertical plate, as shown in Fig. C1.4. In the region shown, fully developed conditions exist; that is, the velocity profile
shape and the film thickness are independent of distance z along the plate. The vertical velocity w becomes a function only of x, and the shear resistance from the atmosphere is negligible. (a)
Sketch the approximate shape of the velocity profile w(x), considering the boundary conditions at the wall and at the film surface. (b) Suppose film thickness , and the slope of the velocity profile
at the wall, (dw/dx)wall, are measured by a laser Doppler anemometer (to be discussed in Chap. 6). Find an expression for the viscosity of the oil as a function of , , (dw/dx)wall, and the
gravitational accleration g. Note
that, for the coordinate system given, both w and (dw/dx)wall are negative. Plate Oil film Air g z
References 1.
2. 3. 4. 5. 6. 7. 8. 9. 10. 11.
J. C. Tannehill, D. A. Anderson, and R. H. Pletcher, Computational Fluid Mechanics and Heat Transfer, 2d ed., Taylor and Francis, Bristol, PA, 1997. S. V. Patankar, Numerical Heat Transfer and Fluid
Flow, McGraw-Hill, New York, 1980. F. M. White, Viscous Fluid Flow, 2d ed., McGraw-Hill, New York, 1991. R. J. Goldstein (ed.), Fluid Mechanics Measurements, 2d ed., Taylor and Francis, Bristol, PA,
1997. R. A. Granger, Experiments in Fluid Mechanics, Oxford University Press, 1995. H. A. Barnes, J. F. Hutton, and K. Walters, An Introduction to Rheology, Elsevier, New York, 1989. A. E. Bergeles
and S. Ishigai, Two-Phase Flow Dynamics and Reactor Safety, McGraw-Hill, New York, 1981. G. N. Patterson, Introduction to the Kinetic Theory of Gas Flows, University of Toronto Press, Toronto, 1971.
ASME Orientation and Guide for Use of Metric Units, 9th ed., American Society of Mechanical Engineers, New York, 1982. J. P. Holman, Heat Transfer, 8th ed., McGraw-Hill, New York, 1997. R. C. Reid,
J. M. Prausnitz, and T. K. Sherwood, The Properties of Gases and Liquids, 4th ed., McGraw-Hill, New York, 1987. J. Hilsenrath et al., “Tables of Thermodynamic and Transport Properties,” U. S. Nat.
Bur. Stand. Circ. 564, 1955; reprinted by Pergamon, New York, 1960. R. A. Spencer et al., ASME Steam Tables with Mollier Chart, 6th ed., American Society of Mechanical Engineers, New York, 1993. O.
A. Hougen and K. M. Watson, Chemical Process Principles Charts, Wiley, New York, 1960.
15. 16. 17.
18. 19. 20. 21. 22. 23.
24. 25. 26. 27. 28. 29.
A. W. Adamson, Physical Chemistry of Surfaces, 5th ed., Interscience, New York, 1990. J. A. Knauss, Introduction to Physical Oceanography, Prentice-Hall, Englewood Cliffs, NJ, 1978. National
Committee for Fluid Mechanics Films, Illustrated Experiments in Fluid Mechanics, M.I.T. Press, Cambridge, MA, 1972. I. G. Currie, Fundamental Mechanics of Fluids, 2d ed., McGraw-Hill, New York, 1993.
M. van Dyke, An Album of Fluid Motion, Parabolic Press, Stanford, CA, 1982. Y. Nakayama (ed.), Visualized Flow, Pergamon Press, Oxford, 1988. W. J. Yang (ed.), Handbook of Flow Visualization,
Hemisphere, New York, 1989. W. Merzkirch, Flow Visualization, 2d ed., Academic, New York, 1987. H. Rouse and S. Ince, History of Hydraulics, Iowa Institute of Hydraulic Research, Univ. of Iowa, Iowa
City, 1957; reprinted by Dover, New York, 1963. H. Rouse, Hydraulics in the United States 1776–1976, Iowa Institute of Hydraulic Research, Univ. of Iowa, Iowa City, 1976. G. Garbrecht, Hydraulics and
Hydraulic Research: An Historical Review, Gower Pub., Aldershot, UK, 1987. 1986 SAE Handbook, 4 vols., Society of Automotive Engineers, Warrendale, PA. J. R. van Wazer, Viscosity and Flow
Measurement, Interscience, New York, 1963. SAE Fuels and Lubricants Standards Manual, Society of Automotive Engineers, Warrendale, PA, 1995. John D. Anderson, Computational Fluid Dynamics: The Basics
with Applications, McGraw-Hill, New York, 1995.
56 30.
Chapter 1 Introduction
H. W. Coleman and W. G. Steele, Experimentation and Uncertainty Analysis for Engineers, John Wiley, New York, 1989. 31. R. J. Moffatt, “Describing the Uncertainties in Experimental Results,”
Experimental Thermal and Fluid Science, vol., 1, 1988, pp. 3–17. 32. Paul A. Libby, An Introduction to Turbulence, Taylor and Francis, Bristol, PA, 1996. 33. Sanford Klein and William Beckman,
Engineering Equation Solver (EES), F-Chart Software, Middleton, WI, 1997.
C. L. Yaws, X. Lin, and L. Bu, “Calculate Viscosities for 355 Compounds. An Equation Can Be Used to Calculate Liquid Viscosity as a Function of Temperature,” Chemical Engineering, vol. 101, no. 4,
April 1994, pp. 119–128. 35. Frank E. Jones, Techniques and Topics in Flow Measurement, CRC Press, Boca Raton, FL, 1995. 36. Carl L. Yaws, Handbook of Viscosity, 3 vols., Gulf Publishing, Houston,
TX, 1994.
Roosevelt Dam in Arizona. Hydrostatic pressure, due to the weight of a standing fluid, can cause enormous forces and moments on large-scale structures such as a dam. Hydrostatic fluid analysis is the
subject of the present chapter. (Courtesy of Dr. E.R. Degginger/Color-Pic Inc.)
Chapter 2 Pressure Distribution in a Fluid
Motivation. Many fluid problems do not involve motion. They concern the pressure distribution in a static fluid and its effect on solid surfaces and on floating and submerged bodies. When the fluid
velocity is zero, denoted as the hydrostatic condition, the pressure variation is due only to the weight of the fluid. Assuming a known fluid in a given gravity field, the pressure may easily be
calculated by integration. Important applications in this chapter are (1) pressure distribution in the atmosphere and the oceans, (2) the design of manometer pressure instruments, (3) forces on
submerged flat and curved surfaces, (4) buoyancy on a submerged body, and (5) the behavior of floating bodies. The last two result in Archimedes’ principles. If the fluid is moving in rigid-body
motion, such as a tank of liquid which has been spinning for a long time, the pressure also can be easily calculated, because the fluid is free of shear stress. We apply this idea here to simple
rigid-body accelerations in Sec. 2.9. Pressure measurement instruments are discussed in Sec. 2.10. As a matter of fact, pressure also can be easily analyzed in arbitrary (nonrigid-body) motions V(x,
y, z, t), but we defer that subject to Chap. 4.
2.1 Pressure and Pressure Gradient
In Fig. 1.1 we saw that a fluid at rest cannot support shear stress and thus Mohr’s circle reduces to a point. In other words, the normal stress on any plane through a fluid element at rest is equal
to a unique value called the fluid pressure p, taken positive for compression by common convention. This is such an important concept that we shall review it with another approach. Figure 2.1 shows a
small wedge of fluid at rest of size x by z by s and depth b into the paper. There is no shear by definition, but we postulate that the pressures px, pz , and pn may be different on each face. The
weight of the element also may be important. Summation of forces must equal zero (no acceleration) in both the x and z directions.
Fx 0 pxb z pnb s sin Fz 0 pzb x pnb s cos 12 b x z
(2.1) 59
Chapter 2 Pressure Distribution in a Fluid z (up)
pn ∆s
θ Element weight: d W = ρ g ( 12 b ∆ x ∆ z)
∆z px ∆x
Fig. 2.1 Equilibrium of a small wedge of fluid at rest.
Width b into paper pz
but the geometry of the wedge is such that s sin z
s cos x
Substitution into Eq. (2.1) and rearrangement give px pn
pz pn 12 z
These relations illustrate two important principles of the hydrostatic, or shear-free, condition: (1) There is no pressure change in the horizontal direction, and (2) there is a vertical change in
pressure proportional to the density, gravity, and depth change. We shall exploit these results to the fullest, starting in Sec. 2.3. In the limit as the fluid wedge shrinks to a “point,’’ z → 0 and
Eqs. (2.3) become px pz pn p
Since is arbitrary, we conclude that the pressure p at a point in a static fluid is independent of orientation. What about the pressure at a point in a moving fluid? If there are strain rates in a
moving fluid, there will be viscous stresses, both shear and normal in general (Sec. 4.3). In that case (Chap. 4) the pressure is defined as the average of the three normal stresses ii on the element
p 13( xx yy zz)
The minus sign occurs because a compression stress is considered to be negative whereas p is positive. Equation (2.5) is subtle and rarely needed since the great majority of viscous flows have
negligible viscous normal stresses (Chap. 4).
Pressure Force on a Fluid Element
Pressure (or any other stress, for that matter) causes no net force on a fluid element unless it varies spatially.1 To see this, consider the pressure acting on the two x faces in Fig. 2.2. Let the
pressure vary arbitrarily p p(x, y, z, t)
An interesting application for a large element is in Fig. 3.7.
2.2 Equilibrium of a Fluid Element
p dy dz
(p+ dy
∂p d x) dy dz ∂x
Fig. 2.2 Net x force on an element due to pressure variation.
dx z
The net force in the x direction on the element in Fig. 2.2 is given by p p dFx p dy dz p dx dy dz dx dy dz x x
In like manner the net force dFy involves p/y, and the net force dFz concerns p/z. The total net-force vector on the element due to pressure is p p p dFpress i j k dx dy dz x y z
We recognize the term in parentheses as the negative vector gradient of p. Denoting f as the net force per unit element volume, we rewrite Eq. (2.8) as fpress ∇p
Thus it is not the pressure but the pressure gradient causing a net force which must be balanced by gravity or acceleration or some other effect in the fluid.
2.2 Equilibrium of a Fluid Element
The pressure gradient is a surface force which acts on the sides of the element. There may also be a body force, due to electromagnetic or gravitational potentials, acting on the entire mass of the
element. Here we consider only the gravity force, or weight of the element dFgrav g dx dy dz fgrav g
In general, there may also be a surface force due to the gradient, if any, of the viscous stresses. For completeness, we write this term here without derivation and consider it more thoroughly in
Chap. 4. For an incompressible fluid with constant viscosity, the net viscous force is 2V 2V 2V fVS ∇2V 2 2 x y z2
where VS stands for viscous stresses and is the coefficient of viscosity from Chap. 1. Note that the term g in Eq. (2.10) denotes the acceleration of gravity, a vector act-
Chapter 2 Pressure Distribution in a Fluid
ing toward the center of the earth. On earth the average magnitude of g is 32.174 ft/s2 9.807 m/s2. The total vector resultant of these three forces—pressure, gravity, and viscous stress—must either
keep the element in equilibrium or cause it to move with acceleration a. From Newton’s law, Eq. (1.2), we have
a f fpress fgrav fvisc ∇p g ∇2V
This is one form of the differential momentum equation for a fluid element, and it is studied further in Chap. 4. Vector addition is implied by Eq. (2.12): The acceleration reflects the local balance
of forces and is not necessarily parallel to the local-velocity vector, which reflects the direction of motion at that instant. This chapter is concerned with cases where the velocity and
acceleration are known, leaving one to solve for the pressure variation in the fluid. Later chapters will take up the more general problem where pressure, velocity, and acceleration are all unknown.
Rewrite Eq. (2.12) as ∇p (g a) ∇2V B(x, y, z, t)
where B is a short notation for the vector sum on the right-hand side. If V and a dV/dt are known functions of space and time and the density and viscosity are known, we can solve Eq. (2.13) for p(x,
y, z, t) by direct integration. By components, Eq. (2.13) is equivalent to three simultaneous first-order differential equations p Bx(x, y, z, t) x
p By(x, y, z, t) y
p Bz(x, y, z, t) z
Since the right-hand sides are known functions, they can be integrated systematically to obtain the distribution p(x, y, z, t) except for an unknown function of time, which remains because we have no
relation for p/t. This extra function is found from a condition of known time variation p0(t) at some point (x0, y0, z0). If the flow is steady (independent of time), the unknown function is a
constant and is found from knowledge of a single known pressure p0 at a point (x0, y0, z0). If this sounds complicated, it is not; we shall illustrate with many examples. Finding the pressure
distribution from a known velocity distribution is one of the easiest problems in fluid mechanics, which is why we put it in Chap. 2. Examining Eq. (2.13), we can single out at least four special
cases: 1. Flow at rest or at constant velocity: The acceleration and viscous terms vanish identically, and p depends only upon gravity and density. This is the hydrostatic condition. See Sec. 2.3. 2.
Rigid-body translation and rotation: The viscous term vanishes identically, and p depends only upon the term (g a). See Sec. 2.9. 3. Irrotational motion ( V 0): The viscous term vanishes identically,
and an exact integral called Bernoulli’s equation can be found for the pressure distribution. See Sec. 4.9. 4. Arbitrary viscous motion: Nothing helpful happens, no general rules apply, but still the
integration is quite straightforward. See Sec. 6.4. Let us consider cases 1 and 2 here.
2.3 Hydrostatic Pressure Distributions 63
p (Pascals) High pressure: p = 120,000 Pa abs = 30,000 Pa gage
120,000 30,000
Local atmosphere: p = 90,000 Pa abs = 0 Pa gage = 0 Pa vacuum
90,000 40,000
Vacuum pressure: p = 50,000 Pa abs = 40,000 Pa vacuum
50,000 50,000
Fig. 2.3 Illustration of absolute, gage, and vacuum pressure readings.
Gage Pressure and Vacuum Pressure: Relative Terms
Absolute zero reference: p = 0 Pa abs = 90,000 Pa vacuum
0 (Tension)
Before embarking on examples, we should note that engineers are apt to specify pressures as (1) the absolute or total magnitude or (2) the value relative to the local ambient atmosphere. The second
case occurs because many pressure instruments are of differential type and record, not an absolute magnitude, but the difference between the fluid pressure and the atmosphere. The measured pressure
may be either higher or lower than the local atmosphere, and each case is given a name: 1. p pa Gage pressure: 2. p pa Vacuum pressure:
p(gage) p pa p(vacuum) pa p
This is a convenient shorthand, and one later adds (or subtracts) atmospheric pressure to determine the absolute fluid pressure. A typical situation is shown in Fig. 2.3. The local atmosphere is at,
say, 90,000 Pa, which might reflect a storm condition in a sea-level location or normal conditions at an altitude of 1000 m. Thus, on this day, pa 90,000 Pa absolute 0 Pa gage 0 Pa vacuum. Suppose
gage 1 in a laboratory reads p1 120,000 Pa absolute. This value may be reported as a gage pressure, p1 120,000 90,000 30,000 Pa gage. (One must also record the atmospheric pressure in the laboratory,
since pa changes gradually.) Suppose gage 2 reads p2 50,000 Pa absolute. Locally, this is a vacuum pressure and might be reported as p2 90,000 50,000 40,000 Pa vacuum. Occasionally, in the Problems
section, we will specify gage or vacuum pressure to keep you alert to this common engineering practice.
2.3 Hydrostatic Pressure Distributions
If the fluid is at rest or at constant velocity, a 0 and ∇2V 0. Equation (2.13) for the pressure distribution reduces to ∇p g
This is a hydrostatic distribution and is correct for all fluids at rest, regardless of their viscosity, because the viscous term vanishes identically. Recall from vector analysis that the vector ∇p
expresses the magnitude and direction of the maximum spatial rate of increase of the scalar property p. As a result, ∇p
Chapter 2 Pressure Distribution in a Fluid
is perpendicular everywhere to surfaces of constant p. Thus Eq. (2.15) states that a fluid in hydrostatic equilibrium will align its constant-pressure surfaces everywhere normal to the local-gravity
vector. The maximum pressure increase will be in the direction of gravity, i.e., “down.’’ If the fluid is a liquid, its free surface, being at atmospheric pressure, will be normal to local gravity,
or “horizontal.’’ You probably knew all this before, but Eq. (2.15) is the proof of it. In our customary coordinate system z is “up.’’ Thus the local-gravity vector for smallscale problems is g gk
(2.16) where g is the magnitude of local gravity, for example, 9.807 m/s2. For these coordinates Eq. (2.15) has the components p 0 x
p 0 y
p g z
the first two of which tell us that p is independent of x and y. Hence p/z can be replaced by the total derivative dp/dz, and the hydrostatic condition reduces to dp dz p2 p1
Equation (2.18) is the solution to the hydrostatic problem. The integration requires an assumption about the density and gravity distribution. Gases and liquids are usually treated differently. We
state the following conclusions about a hydrostatic condition: Pressure in a continuously distributed uniform static fluid varies only with vertical distance and is independent of the shape of the
container. The pressure is the same at all points on a given horizontal plane in the fluid. The pressure increases with depth in the fluid. An illustration of this is shown in Fig. 2.4. The free
surface of the container is atmospheric and forms a horizontal plane. Points a, b, c, and d are at equal depth in a horizonAtmospheric pressure:
Free surface
Fig. 2.4 Hydrostatic-pressure distribution. Points a, b, c, and d are at equal depths in water and therefore have identical pressures. Points A, B, and C are also at equal depths in water and have
identical pressures higher than a, b, c, and d. Point D has a different pressure from A, B, and C because it is not connected to them by a water path.
Depth 1
Depth 2
2.3 Hydrostatic Pressure Distributions 65
tal plane and are interconnected by the same fluid, water; therefore all points have the same pressure. The same is true of points A, B, and C on the bottom, which all have the same higher pressure
than at a, b, c, and d. However, point D, although at the same depth as A, B, and C, has a different pressure because it lies beneath a different fluid, mercury.
Effect of Variable Gravity
For a spherical planet of uniform density, the acceleration of gravity varies inversely as the square of the radius from its center
r g g0 0 r
where r0 is the planet radius and g0 is the surface value of g. For earth, r0 3960 statute mi 6400 km. In typical engineering problems the deviation from r0 extends from the deepest ocean, about 11
km, to the atmospheric height of supersonic transport operation, about 20 km. This gives a maximum variation in g of (6400/6420)2, or 0.6 percent. We therefore neglect the variation of g in most
Hydrostatic Pressure in Liquids
Liquids are so nearly incompressible that we can neglect their density variation in hydrostatics. In Example 1.7 we saw that water density increases only 4.6 percent at the deepest part of the ocean.
Its effect on hydrostatics would be about half of this, or 2.3 percent. Thus we assume constant density in liquid hydrostatic calculations, for which Eq. (2.18) integrates to Liquids:
p2 p1 (z2 z1)
p p z1 z2 2 1
We use the first form in most problems. The quantity is called the specific weight of the fluid, with dimensions of weight per unit volume; some values are tabulated in Table 2.1. The quantity p/ is
a length called the pressure head of the fluid. For lakes and oceans, the coordinate system is usually chosen as in Fig. 2.5, with z 0 at the free surface, where p equals the surface atmospheric
pressure pa. When
Table 2.1 Specific Weight of Some Common Fluids
Specific weight at 68°F 20°C Fluid
Air (at 1 atm) Ethyl alcohol SAE 30 oil Water Seawater Glycerin Carbon tetrachloride Mercury
000.0752 049.2 055.5 062.4 064.0 078.7 099.1 846
000,011.8 007,733 008,720 009,790 010,050 012,360 015,570 133,100
Chapter 2 Pressure Distribution in a Fluid Z +b
p ≈ pa – b γ air Air
Free surface: Z = 0, p = pa 0
Water g
Fig. 2.5 Hydrostatic-pressure distribution in oceans and atmospheres.
p ≈ pa + hγ water
we introduce the reference value (p1, z1) (pa, 0), Eq. (2.20) becomes, for p at any (negative) depth z, Lakes and oceans:
p p a z
where is the average specific weight of the lake or ocean. As we shall see, Eq. (2.21) holds in the atmosphere also with an accuracy of 2 percent for heights z up to 1000 m.
EXAMPLE 2.1 Newfound Lake, a freshwater lake near Bristol, New Hampshire, has a maximum depth of 60 m, and the mean atmospheric pressure is 91 kPa. Estimate the absolute pressure in kPa at this
maximum depth.
Solution From Table 2.1, take 9790 N/m3. With pa 91 kPa and z 60 m, Eq. (2.21) predicts that the pressure at this depth will be 1 kN p 91 kN/m2 (9790 N/m3)(60 m) 1000 N 91 kPa 587 kN/m2 678 kPa
By omitting pa we could state the result as p 587 kPa (gage).
The Mercury Barometer
The simplest practical application of the hydrostatic formula (2.20) is the barometer (Fig. 2.6), which measures atmospheric pressure. A tube is filled with mercury and inverted while submerged in a
reservoir. This causes a near vacuum in the closed upper end because mercury has an extremely small vapor pressure at room temperatures (0.16 Pa at 20°C). Since atmospheric pressure forces a mercury
column to rise a distance h into the tube, the upper mercury surface is at zero pressure.
2.3 Hydrostatic Pressure Distributions 67
p1 ≈ 0 (Mercury has a very low vapor pressure.) z1 = h
p2 ≈ pa ( The mercury is in contact with the atmosphere.)
p h= γa M
z2 = 0
Fig. 2.6 A barometer measures local absolute atmospheric pressure: (a) the height of a mercury column is proportional to patm; (b) a modern portable barometer, with digital readout, uses the
resonating silicon element of Fig. 2.28c. (Courtesy of Paul Lupke, Druck Inc.)
From Fig. 2.6, Eq. (2.20) applies with p1 0 at z1 h and p2 pa at z2 0: pa 0 M(0 h) or
pa h M
At sea-level standard, with pa 101,350 Pa and M 133,100 N/m3 from Table 2.1, the barometric height is h 101,350/133,100 0.761 m or 761 mm. In the United States the weather service reports this as an
atmospheric “pressure’’ of 29.96 inHg (inches of mercury). Mercury is used because it is the heaviest common liquid. A water barometer would be 34 ft high.
Hydrostatic Pressure in Gases
Gases are compressible, with density nearly proportional to pressure. Thus density must be considered as a variable in Eq. (2.18) if the integration carries over large pressure changes. It is
sufficiently accurate to introduce the perfect-gas law p RT in Eq. (2.18) dp p g g dz RT
Chapter 2 Pressure Distribution in a Fluid
Separate the variables and integrate between points 1 and 2:
dp p g ln 2 p p1 R
dz T
The integral over z requires an assumption about the temperature variation T(z). One common approximation is the isothermal atmosphere, where T T0: g(z2 z1) p2 p1 exp RT0
The quantity in brackets is dimensionless. (Think that over; it must be dimensionless, right?) Equation (2.24) is a fair approximation for earth, but actually the earth’s mean atmospheric temperature
drops off nearly linearly with z up to an altitude of about 36,000 ft (11,000 m): T T0 Bz
Here T0 is sea-level temperature (absolute) and B is the lapse rate, both of which vary somewhat from day to day. By international agreement [1] the following standard values are assumed to apply
from 0 to 36,000 ft: T0 518.69°R 288.16 K 15°C B 0.003566°R/ft 0.00650 K/m
This lower portion of the atmosphere is called the troposphere. Introducing Eq. (2.25) into (2.23) and integrating, we obtain the more accurate relation
Bz p pa 1 T0
g where 5.26 (air) RB
in the troposphere, with z 0 at sea level. The exponent g/(RB) is dimensionless (again it must be) and has the standard value of 5.26 for air, with R 287 m2/(s2 K). The U.S. standard atmosphere [1]
is sketched in Fig. 2.7. The pressure is seen to be nearly zero at z 30 km. For tabulated properties see Table A.6. EXAMPLE 2.2 If sea-level pressure is 101,350 Pa, compute the standard pressure at
an altitude of 5000 m, using (a) the exact formula and (b) an isothermal assumption at a standard sea-level temperature of 15°C. Is the isothermal approximation adequate?
Solution Part (a)
Use absolute temperature in the exact formula, Eq. (2.27):
(0.00650 K/m)(5000 m) p pa 1 288.16 K
(101,350 Pa)(0.8872)5.26
101,350(0.52388) 54,000 Pa This is the standard-pressure result given at z 5000 m in Table A.6.
Ans. (a)
40 Altitude z, km
Altitude z, km
2.3 Hydrostatic Pressure Distributions 69
20.1 km
Eq. (2.24)
Eq. (2.27) 11.0 km
1.20 kPa
Eq. (2.26) Troposphere
Fig. 2.7 Temperature and pressure distribution in the U.S. standard atmosphere. (From Ref. 1.)
Part (b)
101.33 kPa
15°C 0
– 60
– 40
– 20 Temperature, °C
40 80 Pressure, kPa
If the atmosphere were isothermal at 288.16 K, Eq. (2.24) would apply: (9.807 m/s2)(5000 m) gz p pa exp (101,350 Pa) exp [287 m2/(s2 K)](288.16 K) RT
(101,350 Pa) exp( 0.5929) 60,100 Pa
Ans. (b)
This is 11 percent higher than the exact result. The isothermal formula is inaccurate in the troposphere.
Is the Linear Formula Adequate for Gases?
The linear approximation from Eq. (2.20) or (2.21), p z, is satisfactory for liquids, which are nearly incompressible. It may be used even over great depths in the ocean. For gases, which are highly
compressible, it is valid only over moderate changes in altitude. The error involved in using the linear approximation (2.21) can be evaluated by expanding the exact formula (2.27) into a series
Bz 1 T0
Bz n(n 1) Bz 1 n T0 2! T0
where n g/(RB). Introducing these first three terms of the series into Eq. (2.27) and rearranging, we obtain n 1 Bz p pa az 1 2 T0
Chapter 2 Pressure Distribution in a Fluid
Thus the error in using the linear formula (2.21) is small if the second term in parentheses in (2.29) is small compared with unity. This is true if 2T0 z 20,800 m (n 1)B
We thus expect errors of less than 5 percent if z or z is less than 1000 m.
2.4 Application to Manometry
From the hydrostatic formula (2.20), a change in elevation z2 z1 of a liquid is equivalent to a change in pressure (p2 p1)/. Thus a static column of one or more liquids or gases can be used to
measure pressure differences between two points. Such a device is called a manometer. If multiple fluids are used, we must change the density in the formula as we move from one fluid to another.
Figure 2.8 illustrates the use of the formula with a column of multiple fluids. The pressure change through each fluid is calculated separately. If we wish to know the total change p5 p1, we add the
successive changes p2 p1, p3 p2, p4 p3, and p5 p4. The intermediate values of p cancel, and we have, for the example of Fig. 2.8, p5 p1 0(z2 z1) w(z3 z2) G(z4 z3) M(z5 z4)
No additional simplification is possible on the right-hand side because of the different densities. Notice that we have placed the fluids in order from the lightest on top to the heaviest at bottom.
This is the only stable configuration. If we attempt to layer them in any other manner, the fluids will overturn and seek the stable arrangement.
A Memory Device: Up Versus Down
The basic hydrostatic relation, Eq. (2.20), is mathematically correct but vexing to engineers, because it combines two negative signs to have the pressure increase downward. When calculating
hydrostatic pressure changes, engineers work instinctively by simply having the pressure increase downward and decrease upward. Thus they use the following mnemonic, or memory, device, first
suggested to the writer by Professor John
z = z1 z2
z3 z4
Fig. 2.8 Evaluating pressure changes through a column of multiple fluids.
Known pressure p1 Oil, ρo Water, ρw Glycerin, ρG
Mercury, ρM
p2 – p1 = – ρog(z 2 – z1) p3 – p2 = – ρw g(z 3 – z 2) p4 – p3 = – ρG g(z 4 – z 3)
p5 – p4 = – ρM g(z 5 – z 4) Sum = p5 – p1
2.4 Application to Manometry
Open, pa
zA, pA
A z1, p1
Fig. 2.9 Simple open manometer for measuring pA relative to atmospheric pressure.
z 2 , p2 ≈ pa
ρ1 Jump across
p = p1 at z = z1 in fluid 2
Foss of Michigan State University: pdown pup z
Thus, without worrying too much about which point is “z1” and which is “z2”, the formula simply increases or decreases the pressure according to whether one is moving down or up. For example, Eq.
(2.31) could be rewritten in the following “multiple increase” mode: p5 p1 0z1 z2 wz2 z3 Gz3 z4 Mz4 z5 That is, keep adding on pressure increments as you move down through the layered fluid.
A different application is a manometer, which involves both “up” and “down” calculations. Figure 2.9 shows a simple open manometer for measuring pA in a closed chamber relative to atmospheric
pressure pa, in other words, measuring the gage pressure. The chamber fluid 1 is combined with a second fluid 2, perhaps for two reasons: (1) to protect the environment from a corrosive chamber fluid
or (2) because a heavier fluid 2 will keep z2 small and the open tube can be shorter. One can, of course, apply the basic hydrostatic formula (2.20). Or, more simply, one can begin at A, apply Eq.
(2.32) “down” to z1, jump across fluid 2 (see Fig. 2.9) to the same pressure p1, and then use Eq. (2.32) “up” to level z2: pA 1zA z1 2z1 z2 p2 patm
The physical reason that we can “jump across” at section 1 in that a continuous length of the same fluid connects these two equal elevations. The hydrostatic relation (2.20) requires this equality as
a form of Pascal’s law: Any two points at the same elevation in a continuous mass of the same static fluid will be at the same pressure. This idea of jumping across to equal pressures facilitates
multiple-fluid problems.
EXAMPLE 2.3 The classic use of a manometer is when two U-tube legs are of equal length, as in Fig. E2.3, and the measurement involves a pressure difference across two horizontal points. The typical
Chapter 2 Pressure Distribution in a Fluid Flow device (a)
(b) L
plication is to measure pressure change across a flow device, as shown. Derive a formula for the pressure difference pa pb in terms of the system parameters in Fig. E2.3.
Solution Using our “up-down” concept as in Eq. (2.32), start at (a), evaluate pressure changes around the U-tube, and end up at (b): pa 1gL 1gh 2gh 1gL pb pa pb ( 2 1)gh
The measurement only includes h, the manometer reading. Terms involving L drop out. Note the appearance of the difference in densities between manometer fluid and working fluid. It is a common
student error to fail to subtract out the working fluid density 1 —a serious error if both fluids are liquids and less disastrous numerically if fluid 1 is a gas. Academically, of course, such an
error is always considered serious by fluid mechanics instructors.
Although Ex. 2.3, because of its popularity in engineering experiments, is sometimes considered to be the “manometer formula,” it is best not to memorize it but rather to adapt Eq. (2.20) or (2.32)
to each new multiple-fluid hydrostatics problem. For example, Fig. 2.10 illustrates a multiple-fluid manometer problem for finding the ρ3
z 2, p2 zA, pA
Jump across
z 2, p2
Fig. 2.10 A complicated multiplefluid manometer to relate pA to pB. This system is not especially practical but makes a good homework or examination problem.
z1, p1
Jump across
z1, p1 z 3, p3
Jump across
z 3, p3
ρ2 ρ4
zB, pB
2.4 Application to Manometry
difference in pressure between two chambers A and B. We repeatedly apply Eq. (2.20), jumping across at equal pressures when we come to a continuous mass of the same fluid. Thus, in Fig. 2.10, we
compute four pressure differences while making three jumps: pA pB (pA p1) (p1 p2) (p2 p3) (p3 pB) 1(zA z1) 2(z1 z2) 3(z2 z3) 4(z3 zB)
The intermediate pressures p1,2,3 cancel. It looks complicated, but really it is merely sequential. One starts at A, goes down to 1, jumps across, goes up to 2, jumps across, goes down to 3, jumps
across, and finally goes up to B.
EXAMPLE 2.4 Pressure gage B is to measure the pressure at point A in a water flow. If the pressure at B is 87 kPa, estimate the pressure at A, in kPa. Assume all fluids are at 20°C. See Fig. E2.4.
SAE 30 oil
Gage B 6 cm
A Water flow
5 cm 11 cm 4 cm
Solution First list the specific weights from Table 2.1 or Table A.3:
water 9790 N/m3
mercury 133,100 N/m3
oil 8720 N/m3
Now proceed from A to B, calculating the pressure change in each fluid and adding: pA W(z)W M(z)M O(z)O pB or
pA (9790 N/m3)( 0.05 m) (133,100 N/m3)(0.07 m) (8720 N/m3)(0.06 m) pA 489.5 Pa 9317 Pa 523.2 Pa pB 87,000 Pa
where we replace N/m2 by its short name, Pa. The value zM 0.07 m is the net elevation change in the mercury (11 cm 4 cm). Solving for the pressure at point A, we obtain pA 96,351 Pa 96.4 kPa The
intermediate six-figure result of 96,351 Pa is utterly fatuous, since the measurements cannot be made that accurately.
Chapter 2 Pressure Distribution in a Fluid
In making these manometer calculations we have neglected the capillary-height changes due to surface tension, which were discussed in Example 1.9. These effects cancel if there is a fluid interface,
or meniscus, on both sides of the U-tube, as in Fig. 2.9. Otherwise, as in the right-hand U-tube of Fig. 2.10, a capillary correction can be made or the effect can be made negligible by using
large-bore ( 1 cm) tubes.
2.5 Hydrostatic Forces on Plane Surfaces
A common problem in the design of structures which interact with fluids is the computation of the hydrostatic force on a plane surface. If we neglect density changes in the fluid, Eq. (2.20) applies
and the pressure on any submerged surface varies linearly with depth. For a plane surface, the linear stress distribution is exactly analogous to combined bending and compression of a beam in
strength-of-materials theory. The hydrostatic problem thus reduces to simple formulas involving the centroid and moments of inertia of the plate cross-sectional area. Figure 2.11 shows a plane panel
of arbitrary shape completely submerged in a liquid. The panel plane makes an arbitrary angle with the horizontal free surface, so that the depth varies over the panel surface. If h is the depth to
any element area dA of the plate, from Eq. (2.20) the pressure there is p pa h. To derive formulas involving the plate shape, establish an xy coordinate system in the plane of the plate with the
origin at its centroid, plus a dummy coordinate down from the surface in the plane of the plate. Then the total hydrostatic force on one side of the plate is given by
F p dA (pa h) dA paA h dA
The remaining integral is evaluated by noticing from Fig. 2.11 that h sin and,
Free surface
p = pa
θ h (x, y) Resultant force: F = pCG A
ξ= y
Side view
Fig. 2.11 Hydrostatic force and center of pressure on an arbitrary plane surface of area A inclined at an angle below the free surface.
dA = dx dy
CP Plan view of arbitrary plane surface
h sin θ
2.5 Hydrostatic Forces on Plane Surfaces
by definition, the centroidal slant distance from the surface to the plate is
1 CG dA A
Therefore, since is constant along the plate, Eq. (2.35) becomes
F paA sin dA paA sin CGA
Finally, unravel this by noticing that CG sin hCG, the depth straight down from the surface to the plate centroid. Thus F pa A hCG A (pa hCG)A pCG A
The force on one side of any plane submerged surface in a uniform fluid equals the pressure at the plate centroid times the plate area, independent of the shape of the plate or the angle at which it
is slanted. Equation (2.38) can be visualized physically in Fig. 2.12 as the resultant of a linear stress distribution over the plate area. This simulates combined compression and bending of a beam
of the same cross section. It follows that the “bending’’ portion of the stress causes no force if its “neutral axis’’ passes through the plate centroid of area. Thus the remaining “compression’’
part must equal the centroid stress times the plate area. This is the result of Eq. (2.38). However, to balance the bending-moment portion of the stress, the resultant force F does not act through
the centroid but below it toward the high-pressure side. Its line of action passes through the center of pressure CP of the plate, as sketched in Fig. 2.11. To find the coordinates (xCP, yCP), we sum
moments of the elemental force p dA about the centroid and equate to the moment of the resultant F. To compute yCP, we equate
FyCP yp dA y(pa sin ) dA sin y dA
The term pay dA vanishes by definition of centroidal axes. Introducing CG y,
Pressure distribution
pav = pCG p (x, y)
Fig. 2.12 The hydrostatic-pressure force on a plane surface is equal, regardless of its shape, to the resultant of the three-dimensional linear pressure distribution on that surface F pCGA.
Centroid of the plane surface
Arbitrary plane surface of area A
Chapter 2 Pressure Distribution in a Fluid
we obtain
FyCP sin CG y dA y2 dA sin Ixx
where again y dA 0 and Ixx is the area moment of inertia of the plate area about its centroidal x axis, computed in the plane of the plate. Substituting for F gives the result Ix yCP sin x (2.41)
pCGA The negative sign in Eq. (2.41) shows that yCP is below the centroid at a deeper level and, unlike F, depends upon angle . If we move the plate deeper, yCP approaches the centroid because every
term in Eq. (2.41) remains constant except pCG, which increases. The determination of xCP is exactly similar:
sin xy dA sin I
FxCP xp dA x[pa (CG y) sin ] dA (2.42)
where Ixy is the product of inertia of the plate, again computed in the plane of the plate. Substituting for F gives Ixy xCP sin (2.43) pCGA For positive Ixy, xCP is negative because the dominant
pressure force acts in the third, or lower left, quadrant of the panel. If Ixy 0, usually implying symmetry, xCP 0 and the center of pressure lies directly below the centroid on the y axis.
L 2
A = bL
Ixx = L 2
b 2
A = π R2
12 R
Ix y = 0
Ixx =
π R4 4
Ix y = 0
b 2 (a)
(b) s
Ixx =
Fig. 2.13 Centroidal moments of inertia for various cross sections: (a) rectangle, (b) circle, (c) triangle, and (d) semicircle.
L 3 b 2
b 2 (c)
2 A = πR 2
A = bL 2
2L 3
bL3 36
Ixx = 0.10976R 4 y
b(b – 2s)L 2 Ix y = 72
Ix y = 0 x
R (d)
4R 3π
2.5 Hydrostatic Forces on Plane Surfaces
In most cases the ambient pressure pa is neglected because it acts on both sides of the plate; e.g., the other side of the plate is inside a ship or on the dry side of a gate or dam. In this case pCG
hCG, and the center of pressure becomes independent of specific weight
Gage-Pressure Formulas
Ixx sin yCP hCGA
F hCGA
Ixy sin xCP h C GA
Figure 2.13 gives the area and moments of inertia of several common cross sections for use with these formulas.
EXAMPLE 2.5 The gate in Fig. E2.5a is 5 ft wide, is hinged at point B, and rests against a smooth wall at point A. Compute (a) the force on the gate due to seawater pressure, (b) the horizontal force
P exerted by the wall at point A, and (c) the reactions at the hinge B. Wall
Seawater: 64 lbf/ft 3
15 ft
A pa Gate 6 ft B
θ 8 ft
Solution Part (a)
By geometry the gate is 10 ft long from A to B, and its centroid is halfway between, or at elevation 3 ft above point B. The depth hCG is thus 15 3 12 ft. The gate area is 5(10) 50 ft2. Neglect pa as
acting on both sides of the gate. From Eq. (2.38) the hydrostatic force on the gate is F pCGA hCGA (64 lbf/ft3)(12 ft)(50 ft2) 38,400 lbf
Part (b)
Ans. (a)
First we must find the center of pressure of F. A free-body diagram of the gate is shown in Fig. E2.5b. The gate is a rectangle, hence Ixy 0
bL3 (5 ft)(10 ft)3 Ixx 417 ft4 12 12
The distance l from the CG to the CP is given by Eq. (2.44) since pa is neglected. 6
(417 ft4)(10 ) Ixx sin 0.417 ft l yCP (12 ft)(50 ft2) hCGA
Chapter 2 Pressure Distribution in a Fluid A P
5 ft CG
B Bx
L = 10 ft
E2.5b The distance from point B to force F is thus 10 l 5 4.583 ft. Summing moments counterclockwise about B gives PL sin F(5 l) P(6 ft) (38,400 lbf)(4.583 ft) 0 P 29,300 lbf
Part (c)
Ans. (b)
With F and P known, the reactions Bx and Bz are found by summing forces on the gate
Fx 0 Bx F sin P Bx 38,400(0.6) 29,300 Bx 6300 lbf
Fz 0 Bz F cos Bz 38,400(0.8) Bz 30,700 lbf
Ans. (c)
This example should have reviewed your knowledge of statics.
EXAMPLE 2.6 A tank of oil has a right-triangular panel near the bottom, as in Fig. E2.6. Omitting pa, find the (a) hydrostatic force and (b) CP on the panel. pa Oil: ρ = 800 kg/m 3
11 m 4m 6m
2.6 Hydrostatic Forces on Curved Surfaces 79
Solution Part (a)
The triangle has properties given in Fig. 2.13c. The centroid is one-third up (4 m) and one-third over (2 m) from the lower left corner, as shown. The area is (6 m)(12 m) 36 m2
The moments of inertia are (6 m)(12 m)3 bL3 Ixx 288 m4 36 36 (6 m)[6 m 2(6 m)](12 m)2 b(b 2s)L2 Ixy 72 m4 72 72
The depth to the centroid is hCG 5 4 9 m; thus the hydrostatic force from Eq. (2.44) is F ghCGA (800 kg/m3)(9.807 m/s2)(9 m)(36 m2) 2.54 106 (kg m)/s2 2.54 106 N 2.54 MN
Part (b)
Ans. (a)
The CP position is given by Eqs. (2.44): (288 m4)(sin 30°) Ixx sin 0.444 m yCP (9 m)(36 m2) hCGA (72 m4)(sin 30°) Ixy sin xCP 0.111 m (9 m)(36 m2) hCGA
Ans. (b)
The resultant force F 2.54 MN acts through this point, which is down and to the right of the centroid, as shown in Fig. E2.6.
2.6 Hydrostatic Forces on Curved Surfaces
The resultant pressure force on a curved surface is most easily computed by separating it into horizontal and vertical components. Consider the arbitrary curved surface sketched in Fig. 2.14a. The
incremental pressure forces, being normal to the local area element, vary in direction along the surface and thus cannot be added numerically. We
Curved surface projection onto vertical plane
Fig. 2.14 Computation of hydrostatic force on a curved surface: (a) submerged curved surface; (b) free-body diagram of fluid above the curved surface.
F1 c
b W2
FH a
FV (a)
Chapter 2 Pressure Distribution in a Fluid
could sum the separate three components of these elemental pressure forces, but it turns out that we need not perform a laborious three-way integration. Figure 2.14b shows a free-body diagram of the
column of fluid contained in the vertical projection above the curved surface. The desired forces FH and FV are exerted by the surface on the fluid column. Other forces are shown due to fluid weight
and horizontal pressure on the vertical sides of this column. The column of fluid must be in static equilibrium. On the upper part of the column bcde, the horizontal components F1 exactly balance and
are not relevant to the discussion. On the lower, irregular portion of fluid abc adjoining the surface, summation of horizontal forces shows that the desired force FH due to the curved surface is
exactly equal to the force FH on the vertical left side of the fluid column. This left-side force can be computed by the planesurface formula, Eq. (2.38), based on a vertical projection of the area
of the curved surface. This is a general rule and simplifies the analysis: The horizontal component of force on a curved surface equals the force on the plane area formed by the projection of the
curved surface onto a vertical plane normal to the component. If there are two horizontal components, both can be computed by this scheme. Summation of vertical forces on the fluid free body then
shows that FV W1 W2 Wair
We can state this in words as our second general rule: The vertical component of pressure force on a curved surface equals in magnitude and direction the weight of the entire column of fluid, both
liquid and atmosphere, above the curved surface. Thus the calculation of FV involves little more than finding centers of mass of a column of fluid—perhaps a little integration if the lower portion
abc has a particularly vexing shape.
;; ;; ;;
EXAMPLE 2.7
A dam has a parabolic shape z/z0 (x/x0)2 as shown in Fig. E2.7a, with x0 10 ft and z0 24 ft. The fluid is water, 62.4 lbf/ft3, and atmospheric pressure may be omitted. Compute the pa = 0 lbf/ft2 gage
( (
x z = z0 x 0
2.6 Hydrostatic Forces on Curved Surfaces 81 forces FH and FV on the dam and the position CP where they act. The width of the dam is 50 ft.
Solution The vertical projection of this curved surface is a rectangle 24 ft high and 50 ft wide, with its centroid halfway down, or hCG 12 ft. The force FH is thus FH hCGAproj (62.4 lbf/ft3)(12 ft)
(24 ft)(50 ft) 899,000 lbf 899 103 lbf
The line of action of FH is below the centroid by an amount 1(50 ft)(24 ft)3(sin 90°) 12 Ixx sin yCP 4 ft (12 ft)(24 ft)(50 ft) hCGAproj
Thus FH is 12 4 16 ft, or two-thirds, down from the free surface or 8 ft from the bottom, as might have been evident by inspection of the triangular pressure distribution. The vertical component FV
equals the weight of the parabolic portion of fluid above the curved surface. The geometric properties of a parabola are shown in Fig. E2.7b. The weight of this amount of water is FV (23x0z0b) (62.4
lbf/ft3)(23)(10 ft)(24 ft)(50 ft) 499,000 lbf 499 103 lbf
z0 Area = 3z0 5
2 x0z 0 3
3x 0 8
x0 = 10 ft
This acts downward on the surface at a distance 3x0 /8 3.75 ft over from the origin of coordinates. Note that the vertical distance 3z0 /5 in Fig. E2.7b is irrelevant. The total resultant force
acting on the dam is F (FH2 FV2)1/2 [(499)2 (899)2]1/2 1028 103 lbf As seen in Fig. E2.7c, this force acts down and to the right at an angle of 29° tan1 489999 . The force F passes through the point
(x, z) (3.75 ft, 8 ft). If we move down along the 29° line until we strike the dam, we find an equivalent center of pressure on the dam at
xCP 5.43 ft
zCP 7.07 ft
This definition of CP is rather artificial, but this is an unavoidable complication of dealing with a curved surface.
Chapter 2 Pressure Distribution in a Fluid z Resultant = 1028 × 103 1bf acts along z = 10.083 – 0.5555x 3.75 ft 499
Parabola z = 0.24x2
7.07 ft
8 ft
2.7 Hydrostatic Forces in Layered Fluids
5.43 ft
The formulas for plane and curved surfaces in Secs. 2.5 and 2.6 are valid only for a fluid of uniform density. If the fluid is layered with different densities, as in Fig. 2.15, a single formula
cannot solve the problem because the slope of the linear pressure distribution changes between layers. However, the formulas apply separately to each layer, and thus the appropriate remedy is to
compute and sum the separate layer forces and moments. Consider the slanted plane surface immersed in a two-layer fluid in Fig. 2.15. The slope of the pressure distribution becomes steeper as we move
down into the denser z
F1= p
Plane surface
z=0 pa
p = pa – ρ1gz
ρ1 < ρ2 Fluid 1
z 1, p1 F 2= p
A CG 2 2
p1 = pa – ρ1gz1
ρ2 Fluid 2
z 2 , p2 p = p1 – ρ2 g(z – z 1)
Fig. 2.15 Hydrostatic forces on a surface immersed in a layered fluid must be summed in separate pieces.
p2 = p1 – ρ 2 g(z 2 – z 1)
2.7 Hydrostatic Forces in Layered Fluids
second layer. The total force on the plate does not equal the pressure at the centroid times the plate area, but the plate portion in each layer does satisfy the formula, so that we can sum forces to
find the total: F Fi pCGi Ai
Similarly, the centroid of the plate portion in each layer can be used to locate the center of pressure on that portion
ig sin i Ixx yCPi i pCGi Ai
ig sin i Ixy xCPi i pCGi Ai
These formulas locate the center of pressure of that particular Fi with respect to the centroid of that particular portion of plate in the layer, not with respect to the centroid of the entire plate.
The center of pressure of the total force F Fi can then be found by summing moments about some convenient point such as the surface. The following example will illustrate.
EXAMPLE 2.8 A tank 20 ft deep and 7 ft wide is layered with 8 ft of oil, 6 ft of water, and 4 ft of mercury. Compute (a) the total hydrostatic force and (b) the resultant center of pressure of the
fluid on the right-hand side of the tank.
Solution Part (a)
Divide the end panel into three parts as sketched in Fig. E2.8, and find the hydrostatic pressure at the centroid of each part, using the relation (2.38) in steps as in Fig. E2.8:
pa = 0
l: 5
7 ft
4 ft
11 ft
/ft 3 8 ft
6 ft
16 ft (2)
4 ft (3)
E2.8 PCG1 (55.0 lbf/ft3)(4 ft) 220 lbf/ft2 pCG2 (55.0)(8) 62.4(3) 627 lbf/ft2 pCG3 (55.0)(8) 62.4(6) 846(2) 2506 lbf/ft2
Chapter 2 Pressure Distribution in a Fluid These pressures are then multiplied by the respective panel areas to find the force on each portion: F1 pCG1A1 (220 lbf/ft2)(8 ft)(7 ft) 12,300 lbf F2 pCG2
A2 627(6)(7) 26,300 lbf F3 pCG3A3 2506(4)(7) 70,200 lbf F Fi 108,800 lbf
Part (b)
Ans. (a)
Equations (2.47) can be used to locate the CP of each force Fi, noting that 90° and sin 1 for all parts. The moments of inertia are Ixx1 (7 ft)(8 ft)3/12 298.7 ft4, Ixx2 7(6)3/12 126.0 ft4, and Ixx3
7(4)3/12 37.3 ft4. The centers of pressure are thus at
1gIxx (55.0 lbf/ft3)(298.7 ft4) yCP1 1 1.33 ft 12,300 lbf F1 846(37.3) yCP3 0.45 ft 70,200
62.4(126.0) yCP2 0.30 ft 26,300
This locates zCP1 4 1.33 5.33 ft, zCP2 11 0.30 11.30 ft, and zCP3 16 0.45 16.45 ft. Summing moments about the surface then gives
12,300(5.33) 26,300(11.30) 70,200(16.45) 108,800zCP
1,518,000 zCP 13.95 ft 108,800
Ans. (b)
The center of pressure of the total resultant force on the right side of the tank lies 13.95 ft below the surface.
2.8 Buoyancy and Stability
The same principles used to compute hydrostatic forces on surfaces can be applied to the net pressure force on a completely submerged or floating body. The results are the two laws of buoyancy
discovered by Archimedes in the third century B.C.: 1. A body immersed in a fluid experiences a vertical buoyant force equal to the weight of the fluid it displaces. 2. A floating body displaces its
own weight in the fluid in which it floats. These two laws are easily derived by referring to Fig. 2.16. In Fig. 2.16a, the body lies between an upper curved surface 1 and a lower curved surface 2.
From Eq. (2.45) for vertical force, the body experiences a net upward force FB FV (2) FV (1) (fluid weight above 2) (fluid weight above 1) weight of fluid equivalent to body volume
Alternatively, from Fig. 2.16b, we can sum the vertical forces on elemental vertical slices through the immersed body: FB
(p2 p1) dAH (z2 z1) dAH ()(body volume) (2.49)
2.8 Buoyancy and Stability
FV (1)
Surface 1
Horizontal elemental area d AH
z1 – z 2
Fig. 2.16 Two different approaches to the buoyant force on an arbitrary immersed body: (a) forces on upper and lower curved surfaces; (b) summation of elemental verticalpressure forces.
Surface 2 FV (2) (a)
p2 (b)
These are identical results and equivalent to law 1 above. Equation (2.49) assumes that the fluid has uniform specific weight. The line of action of the buoyant force passes through the center of
volume of the displaced body; i.e., its center of mass is computed as if it had uniform density. This point through which FB acts is called the center of buoyancy, commonly labeled B or CB on a
drawing. Of course, the point B may or may not correspond to the actual center of mass of the body’s own material, which may have variable density. Equation (2.49) can be generalized to a layered
fluid (LF) by summing the weights of each layer of density i displaced by the immersed body: (FB)LF ig(displaced volume)i
Each displaced layer would have its own center of volume, and one would have to sum moments of the incremental buoyant forces to find the center of buoyancy of the immersed body. Since liquids are
relatively heavy, we are conscious of their buoyant forces, but gases also exert buoyancy on any body immersed in them. For example, human beings have an average specific weight of about 60 lbf/ft3.
We may record the weight of a person as 180 lbf and thus estimate the person’s total volume as 3.0 ft3. However, in so doing we are neglecting the buoyant force of the air surrounding the person. At
standard conditions, the specific weight of air is 0.0763 lbf/ft3; hence the buoyant force is approximately 0.23 lbf. If measured in vacuo, the person would weigh about 0.23 lbf more. For balloons
and blimps the buoyant force of air, instead of being negligible, is the controlling factor in the design. Also, many flow phenomena, e.g., natural convection of heat and vertical mixing in the
ocean, are strongly dependent upon seemingly small buoyant forces. Floating bodies are a special case; only a portion of the body is submerged, with the remainder poking up out of the free surface.
This is illustrated in Fig. 2.17, where the shaded portion is the displaced volume. Equation (2.49) is modified to apply to this smaller volume FB ()(displaced volume) floating-body weight
Chapter 2 Pressure Distribution in a Fluid Neglect the displaced air up here.
W FB B
Fig. 2.17 Static equilibrium of a floating body.
(Displaced volume) × ( γ of fluid) = body weight
Not only does the buoyant force equal the body weight, but also they are collinear since there can be no net moments for static equilibrium. Equation (2.51) is the mathematical equivalent of
Archimedes’ law 2, previously stated.
EXAMPLE 2.9 A block of concrete weighs 100 lbf in air and “weighs’’ only 60 lbf when immersed in fresh water (62.4 lbf/ft3). What is the average specific weight of the block?
Solution 60 lbf
A free-body diagram of the submerged block (see Fig. E2.9) shows a balance between the apparent weight, the buoyant force, and the actual weight
Fz 0 60 FB 100 FB
FB 40 lbf (62.4 lbf/ft3)(block volume, ft3)
Solving gives the volume of the block as 40/62.4 0.641 ft3. Therefore the specific weight of the block is W = 100 lbf
100 lbf block 3 156 lbf/ft3 0.641 ft
Occasionally, a body will have exactly the right weight and volume for its ratio to equal the specific weight of the fluid. If so, the body will be neutrally buoyant and will remain at rest at any
point where it is immersed in the fluid. Small neutrally buoyant particles are sometimes used in flow visualization, and a neutrally buoyant body called a Swallow float [2] is used to track
oceanographic currents. A submarine can achieve positive, neutral, or negative buoyancy by pumping water in or out of its ballast tanks.
A floating body as in Fig. 2.17 may not approve of the position in which it is floating. If so, it will overturn at the first opportunity and is said to be statically unstable, like a pencil balanced
upon its point. The least disturbance will cause it to seek another equilibrium position which is stable. Engineers must design to avoid floating instabil-
2.8 Buoyancy and Stability Small ∆θ disturbance angle
Line of symmetry
Small disturbance angle
M G
G G
Fig. 2.18 Calculation of the metacenter M of the floating body shown in (a). Tilt the body a small angle . Either (b) B moves far out (point M above G denotes stability); or (c) B moves slightly
(point M below G denotes instability).
FB B'
Either (a)
Restoring moment (b)
M FB
Overturning moment (c)
ity. The only way to tell for sure whether a floating position is stable is to “disturb’’ the body a slight amount mathematically and see whether it develops a restoring moment which will return it
to its original position. If so, it is stable; if not, unstable. Such calculations for arbitrary floating bodies have been honed to a fine art by naval architects [3], but we can at least outline the
basic principle of the static-stability calculation. Figure 2.18 illustrates the computation for the usual case of a symmetric floating body. The steps are as follows: 1. The basic floating position
is calculated from Eq. (2.51). The body’s center of mass G and center of buoyancy B are computed. 2. The body is tilted a small angle , and a new waterline is established for the body to float at
this angle. The new position B of the center of buoyancy is calculated. A vertical line drawn upward from B intersects the line of symmetry at a point M, called the metacenter, which is independent
of for small angles. 3. If point M is above G, that is, if the metacentric height MG is positive, a restoring moment is present and the original position is stable. If M is below G (negative MG , the
body is unstable and will overturn if disturbed. Stability increases with increasing MG . Thus the metacentric height is a property of the cross section for the given weight, and its value gives an
indication of the stability of the body. For a body of varying cross section and draft, such as a ship, the computation of the metacenter can be very involved.
Stability Related to Waterline Area
Naval architects [3] have developed the general stability concepts from Fig. 2.18 into a simple computation involving the area moment of inertia of the waterline area about the axis of tilt. The
derivation assumes that the body has a smooth shape variation (no discontinuities) near the waterline and is derived from Fig. 2.19. The y-axis of the body is assumed to be a line of symmetry.
Tilting the body a small angle then submerges small wedge Obd and uncovers an equal wedge cOa, as shown.
Chapter 2 Pressure Distribution in a Fluid
y Original waterline area
Variable-width L(x) into paper dA = x tan dx
c a
Fig. 2.19 A floating body tilted through a small angle . The movement x of the center of buoyancy B is related to the waterline area moment of inertia.
e Tilted floating body
The new position B of the center of buoyancy is calculated as the centroid of the submerged portion aObde of the body:
x υabOde x dυ x dυ x dυ 0 x (L dA) x (L dA) cOdea
0 x L (x tan dx) xL (x tan dx) tan x2 dAwaterline IO tan Obd
where IO is the area moment of inertia of the waterline footprint of the body about its tilt axis O. The first integral vanishes because of the symmetry of the original submerged portion cOdea. The
remaining two “wedge” integrals combine into IO when we notice that L dx equals an element of waterline area. Thus we determine the desired distance from M to B: IO x MB MG GB υ submerged
I M G O G B υsub
The engineer would determine the distance from G to B from the basic shape and design of the floating body and then make the calculation of IO and the submerged volume υsub. If the metacentric height
MG is positive, the body is stable for small disturbances. Note that if G B is negative, that is, B is above G, the body is always stable.
EXAMPLE 2.10 A barge has a uniform rectangular cross section of width 2L and vertical draft of height H, as in Fig. E2.10. Determine (a) the metacentric height for a small tilt angle and (b) the
range of ratio L/H for which the barge is statically stable if G is exactly at the waterline as shown.
2.9 Pressure Distribution in Rigid-Body Motion
G O ●
Solution If the barge has length b into the paper, the waterline area, relative to tilt axis O, has a base b and a height 2L; therefore, IO b(2L)3/12. Meanwhile, υsub 2LbH. Equation (2.52) predicts
8bL3/12 I H L2 H MG O GB Ans. (a) 2LbH υsub 2 3H 2 The barge can thus be stable only if L2 3H2/2
2L 2.45H
Ans. (b)
The wider the barge relative to its draft, the more stable it is. Lowering G would help also.
Even an expert will have difficulty determining the floating stability of a buoyant body of irregular shape. Such bodies may have two or more stable positions. For example, a ship may float the way
we like it, so that we can sit upon the deck, or it may float upside down (capsized). An interesting mathematical approach to floating stability is given in Ref. 11. The author of this reference
points out that even simple shapes, e.g., a cube of uniform density, may have a great many stable floating orientations, not necessarily symmetric. Homogeneous circular cylinders can float with the
axis of symmetry tilted from the vertical. Floating instability occurs in nature. Living fish generally swim with their plane of symmetry vertical. After death, this position is unstable and they
float with their flat sides up. Giant icebergs may overturn after becoming unstable when their shapes change due to underwater melting. Iceberg overturning is a dramatic, rarely seen event. Figure
2.20 shows a typical North Atlantic iceberg formed by calving from a Greenland glacier which protruded into the ocean. The exposed surface is rough, indicating that it has undergone further calving.
Icebergs are frozen fresh, bubbly, glacial water of average density 900 kg/m3. Thus, when an iceberg is floating in seawater, whose average density is 1025 kg/m3, approximately 900/1025, or
seven-eighths, of its volume lies below the water.
2.9 Pressure Distribution in Rigid-Body Motion
In rigid-body motion, all particles are in combined translation and rotation, and there is no relative motion between particles. With no relative motion, there are no strains
Chapter 2 Pressure Distribution in a Fluid
Fig. 2.20 A North Atlantic iceberg formed by calving from a Greenland glacier. These, and their even larger Antarctic sisters, are the largest floating bodies in the world. Note the evidence of
further calving fractures on the front surface. (Courtesy of So/ ren Thalund, Greenland tourism a/s Iiulissat, Greenland.)
or strain rates, so that the viscous term ∇2V in Eq. (2.13) vanishes, leaving a balance between pressure, gravity, and particle acceleration ∇p (g a)
The pressure gradient acts in the direction g a, and lines of constant pressure (including the free surface, if any) are perpendicular to this direction. The general case of combined translation and
rotation of a rigid body is discussed in Chap. 3, Fig. 3.12. If the center of rotation is at point O and the translational velocity is V0 at this point, the velocity of an arbitrary point P on the
body is given by2 V V0 r0 where is the angular-velocity vector and r0 is the position of point P. Differentiating, we obtain the most general form of the acceleration of a rigid body: d dV a 0 ( r0)
r0 dt dt
Looking at the right-hand side, we see that the first term is the translational acceleration; the second term is the centripetal acceleration, whose direction is from point 2
For a more detailed derivation of rigid-body motion, see Ref. 4, Sec. 2.7.
2.9 Pressure Distribution in Rigid-Body Motion
P perpendicular toward the axis of rotation; and the third term is the linear acceleration due to changes in the angular velocity. It is rare for all three of these terms to apply to any one fluid
flow. In fact, fluids can rarely move in rigid-body motion unless restrained by confining walls for a long time. For example, suppose a tank of water is in a car which starts a constant acceleration.
The water in the tank would begin to slosh about, and that sloshing would damp out very slowly until finally the particles of water would be in approximately rigid-body acceleration. This would take
so long that the car would have reached hypersonic speeds. Nevertheless, we can at least discuss the pressure distribution in a tank of rigidly accelerating water. The following is an example where
the water in the tank will reach uniform acceleration rapidly.
EXAMPLE 2.11 A tank of water 1 m deep is in free fall under gravity with negligible drag. Compute the pressure at the bottom of the tank if pa 101 kPa.
Solution Being unsupported in this condition, the water particles tend to fall downward as a rigid hunk of fluid. In free fall with no drag, the downward acceleration is a g. Thus Eq. (2.53) for this
situation gives ∇p (g g) 0. The pressure in the water is thus constant everywhere and equal to the atmospheric pressure 101 kPa. In other words, the walls are doing no service in sustaining the
pressure distribution which would normally exist.
Uniform Linear Acceleration
In this general case of uniform rigid-body acceleration, Eq. (2.53) applies, a having the same magnitude and direction for all particles. With reference to Fig. 2.21, the parallelogram sum of g and a
gives the direction of the pressure gradient or greatest rate of increase of p. The surfaces of constant pressure must be perpendicular to this and are thus tilted at a downward angle such that ax
tan1 g az
a az
θ = tan –1
ax g + az
g ∆
Fig. 2.21 Tilting of constantpressure surfaces in a tank of liquid in rigid-body acceleration.
p µg – a
az ax
p2 p3
p = p1
Fluid at rest
Chapter 2 Pressure Distribution in a Fluid
One of these tilted lines is the free surface, which is found by the requirement that the fluid retain its volume unless it spills out. The rate of increase of pressure in the direction g a is
greater than in ordinary hydrostatics and is given by dp G ds
where G [a2x (g az)2]1/2
These results are independent of the size or shape of the container as long as the fluid is continuously connected throughout the container. EXAMPLE 2.12 A drag racer rests her coffee mug on a
horizontal tray while she accelerates at 7 m/s2. The mug is 10 cm deep and 6 cm in diameter and contains coffee 7 cm deep at rest. (a) Assuming rigidbody acceleration of the coffee, determine whether
it will spill out of the mug. (b) Calculate the gage pressure in the corner at point A if the density of coffee is 1010 kg/m3.
Solution Part (a)
The free surface tilts at the angle given by Eq. (2.55) regardless of the shape of the mug. With az 0 and standard gravity, a 7.0 tan1 x tan1 35.5° g 9.81 If the mug is symmetric about its central
axis, the volume of coffee is conserved if the tilted surface intersects the original rest surface exactly at the centerline, as shown in Fig. E2.12.
3 cm
∆z θ
7 cm ax = 7 m/s2 A
3 cm
Thus the deflection at the left side of the mug is z (3 cm)(tan ) 2.14 cm
Ans. (a)
This is less than the 3-cm clearance available, so the coffee will not spill unless it was sloshed during the start-up of acceleration.
Part (b)
When at rest, the gage pressure at point A is given by Eq. (2.20): pA g(zsurf zA) (1010 kg/m3)(9.81 m/s2)(0.07 m) 694 N/m2 694 Pa
2.9 Pressure Distribution in Rigid-Body Motion
During acceleration, Eq. (2.56) applies, with G [(7.0)2 (9.81)2]1/2 12.05 m/s2. The distance ∆s down the normal from the tilted surface to point A is s (7.0 2.14)(cos ) 7.44 cm Thus the pressure at
point A becomes pA G s 1010(12.05)(0.0744) 906 Pa
Ans. (b)
which is an increase of 31 percent over the pressure when at rest.
Rigid-Body Rotation
As a second special case, consider rotation of the fluid about the z axis without any translation, as sketched in Fig. 2.22. We assume that the container has been rotating long enough at constant for
the fluid to have attained rigid-body rotation. The fluid acceleration will then be the centripetal term in Eq. (2.54). In the coordinates of Fig. 2.22, the angular-velocity and position vectors are
given by k
r0 irr
Then the acceleration is given by ( r0) r2ir
as marked in the figure, and Eq. (2.53) for the force balance becomes
p p ∇p ir k (g a) (gk r2ir) r z
Equating like components, we find the pressure field by solving two first-order partial differential equations
p r2 r
p z
This is our first specific example of the generalized three-dimensional problem described by Eqs. (2.14) for more than one independent variable. The right-hand sides of z, k r, ir p = pa Ω a = –rΩ 2
–a Still-water level
Fig. 2.22 Development of paraboloid constant-pressure surfaces in a fluid in rigid-body rotation. The dashed line along the direction of maximum pressure increase is an exponential curve.
p = p1 Axis of rotation
p2 p3
Chapter 2 Pressure Distribution in a Fluid
(2.60) are known functions of r and z. One can proceed as follows: Integrate the first equation “partially,’’ i.e., holding z constant, with respect to r. The result is p 12 r 22 f(z)
(2.61) †
where the “constant’’ of integration is actually a function f(z). Now differentiate this with respect to z and compare with the second relation of (2.60):
p 0 f(z) z f(z) z C
where C is a constant. Thus Eq. (2.61) now becomes p const z 12 r 22
This is the pressure distribution in the fluid. The value of C is found by specifying the pressure at one point. If p p0 at (r, z) (0, 0), then C p0. The final desired distribution is p p0 z 12 r 22
The pressure is linear in z and parabolic in r. If we wish to plot a constant-pressure surface, say, p p1, Eq. (2.63) becomes p0 p1 r 22 z a br2 2g
Thus the surfaces are paraboloids of revolution, concave upward, with their minimum point on the axis of rotation. Some examples are sketched in Fig. 2.22. As in the previous example of linear
acceleration, the position of the free surface is found by conserving the volume of fluid. For a noncircular container with the axis of rotation off-center, as in Fig. 2.22, a lot of laborious
mensuration is required, and a single problem will take you all weekend. However, the calculation is easy for a cylinder rotating about its central axis, as in Fig. 2.23. Since the volume of a
paraboloid is
Still water level
Fig. 2.23 Determining the freesurface position for rotation of a cylinder of fluid about its central axis.
h 2
Volume =
π 2
h 2
2 2 h= Ω R 2g
† This is because f(z) vanishes when differentiated with respect to r. If you don’t see this, you should review your calculus.
2.9 Pressure Distribution in Rigid-Body Motion
one-half the base area times its height, the still-water level is exactly halfway between the high and low points of the free surface. The center of the fluid drops an amount h/2 2R2/(4g), and the
edges rise an equal amount.
EXAMPLE 2.13 The coffee cup in Example 2.12 is removed from the drag racer, placed on a turntable, and rotated about its central axis until a rigid-body mode occurs. Find (a) the angular velocity
which will cause the coffee to just reach the lip of the cup and (b) the gage pressure at point A for this condition.
Solution Part (a)
The cup contains 7 cm of coffee. The remaining distance of 3 cm up to the lip must equal the distance h/2 in Fig. 2.23. Thus 2(0.03 m)2 h 2R2 0.03 m 4(9.81 m/s2) 2 4g Solving, we obtain 2 1308
Part (b) z
36.2 rad/s 345 r/min
Ans. (a)
To compute the pressure, it is convenient to put the origin of coordinates r and z at the bottom of the free-surface depression, as shown in Fig. E2.13. The gage pressure here is p0 0, and point A is
at (r, z) (3 cm, 4 cm). Equation (2.63) can then be evaluated pA 0 (1010 kg/m3)(9.81 m/s2)(0.04 m) 12(1010 kg/m3)(0.03 m)2(1308 rad2/s2)
3 cm
396 N/m2 594 N/m2 990 Pa
Ans. (b)
This is about 43 percent greater than the still-water pressure pA 694 Pa. 0
7 cm Ω A 3 cm
3 cm
Here, as in the linear-acceleration case, it should be emphasized that the paraboloid pressure distribution (2.63) sets up in any fluid under rigid-body rotation, regardless of the shape or size of
the container. The container may even be closed and filled with fluid. It is only necessary that the fluid be continuously interconnected throughout the container. The following example will
illustrate a peculiar case in which one can visualize an imaginary free surface extending outside the walls of the container.
EXAMPLE 2.14 A U-tube with a radius of 10 in and containing mercury to a height of 30 in is rotated about its center at 180 r/min until a rigid-body mode is achieved. The diameter of the tubing is
negligible. Atmospheric pressure is 2116 lbf/ft2. Find the pressure at point A in the rotating condition. See Fig. E2.14.
Chapter 2 Pressure Distribution in a Fluid
10 in
B r
Convert the angular velocity to radians per second: 2 rad/r (180 r/min) 18.85 rad/s 60 s/min
30 in
From Table 2.1 we find for mercury that 846 lbf/ft3 and hence 846/32.2 26.3 slugs/ft3. At this high rotation rate, the free surface will slant upward at a fierce angle [about 84°; check this from Eq.
(2.64)], but the tubing is so thin that the free surface will remain at approximately the same 30-in height, point B. Placing our origin of coordinates at this height, we can calculate the constant C
in Eq. (2.62b) from the condition pB 2116 lbf/ft2 at (r, z) (10 in, 0): pB 2116 lbf/ft2 C 0 12(26.3 slugs/ft3)(1102 ft)2(18.85 rad/s)2
Imaginary free surface
C 2116 3245 1129 lbf/ft2
We then obtain pA by evaluating Eq. (2.63) at (r, z) (0, 30 in):
E2.14 pA 1129 (846 lbf/ft3)(3102 ft) 1129 2115 986 lbf/ft2
This is less than atmospheric pressure, and we can see why if we follow the free-surface paraboloid down from point B along the dashed line in the figure. It will cross the horizontal portion of the
U-tube (where p will be atmospheric) and fall below point A. From Fig. 2.23 the actual drop from point B will be (18.85)2(1102 )2 2R2 h 3.83 ft 46 in 2(32.2) 2g Thus pA is about 16 inHg below
atmospheric pressure, or about 1162(846) 1128 lbf/ft2 below pa 2116 lbf/ft2, which checks with the answer above. When the tube is at rest, pA 2116 846(3102 ) 4231 lbf/ft2 Hence rotation has reduced
the pressure at point A by 77 percent. Further rotation can reduce pA to near-zero pressure, and cavitation can occur.
An interesting by-product of this analysis for rigid-body rotation is that the lines everywhere parallel to the pressure gradient form a family of curved surfaces, as sketched in Fig. 2.22. They are
everywhere orthogonal to the constant-pressure surfaces, and hence their slope is the negative inverse of the slope computed from Eq. (2.64): dz 1 1 dr GL (dz/dr)pconst r2/g where GL stands for
gradient line or
dz g 2 dr r
2.10 Pressure Measurement
Separating the variables and integrating, we find the equation of the pressure-gradient surfaces 2z r C1 exp g
Notice that this result and Eq. (2.64) are independent of the density of the fluid. In the absence of friction and Coriolis effects, Eq. (2.66) defines the lines along which the apparent net
gravitational field would act on a particle. Depending upon its density, a small particle or bubble would tend to rise or fall in the fluid along these exponential lines, as demonstrated
experimentally in Ref. 5. Also, buoyant streamers would align themselves with these exponential lines, thus avoiding any stress other than pure tension. Figure 2.24 shows the configuration of such
streamers before and during rotation.
2.10 Pressure Measurement
Pressure is a derived property. It is the force per unit area as related to fluid molecular bombardment of a surface. Thus most pressure instruments only infer the pressure by calibration with a
primary device such as a deadweight piston tester. There are many such instruments, both for a static fluid and a moving stream. The instrumentation texts in Refs. 7 to 10, 12, and 13 list over 20
designs for pressure measurement instruments. These instruments may be grouped into four categories: 1. Gravity-based: barometer, manometer, deadweight piston. 2. Elastic deformation: bourdon tube
(metal and quartz), diaphragm, bellows, strain-gage, optical beam displacement. 3. Gas behavior: gas compression (McLeod gage), thermal conductance (Pirani gage), molecular impact (Knudsen gage),
ionization, thermal conductivity, air piston. 4. Electric output: resistance (Bridgman wire gage), diffused strain gage, capacitative, piezoelectric, magnetic inductance, magnetic reluctance, linear
variable differential transformer (LVDT), resonant frequency. The gas-behavior gages are mostly special-purpose instruments used for certain scientific experiments. The deadweight tester is the
instrument used most often for calibrations; for example, it is used by the U.S. National Institute for Standards and Technology (NIST). The barometer is described in Fig. 2.6. The manometer,
analyzed in Sec. 2.4, is a simple and inexpensive hydrostaticprinciple device with no moving parts except the liquid column itself. Manometer measurements must not disturb the flow. The best way to
do this is to take the measurement through a static hole in the wall of the flow, as illustrated for the two instruments in Fig. 2.25. The hole should be normal to the wall, and burrs should be
avoided. If the hole is small enough (typically 1-mm diameter), there will be no flow into the measuring tube once the pressure has adjusted to a steady value. Thus the flow is almost undisturbed. An
oscillating flow pressure, however, can cause a large error due to possible dynamic response of the tubing. Other devices of smaller dimensions are used for dynamic-pressure measurements. Note that
the manometers in Fig. 2.25 are arranged to measure the absolute pressures p1 and p2. If the pressure difference p1 p2 is de-
Chapter 2 Pressure Distribution in a Fluid
Fig. 2.24 Experimental demonstration with buoyant streamers of the fluid force field in rigid-body rotation: (top) fluid at rest (streamers hang vertically upward); (bottom) rigid-body rotation
(streamers are aligned with the direction of maximum pressure gradient). (From Ref. 5, courtesy of R. Ian Fletcher.)
2.10 Pressure Measurement
Fig. 2.25 Two types of accurate manometers for precise measurements: (a) tilted tube with eyepiece; (b) micrometer pointer with ammeter detector.
sired, a significant error is incurred by subtracting two independent measurements, and it would be far better to connect both ends of one instrument to the two static holes p1 and p2 so that one
manometer reads the difference directly. In category 2, elasticdeformation instruments, a popular, inexpensive, and reliable device is the bourdon tube, sketched in Fig. 2.26. When pressurized
internally, a curved tube with flattened cross section will deflect outward. The deflection can be measured by a linkage attached to a calibrated dial pointer, as shown. Or the deflection can be used
to drive electric-output sensors, such as a variable transformer. Similarly, a membrane or diaphragm will deflect under pressure and can either be sensed directly or used to drive another sensor.
A Section AA
Bourdon tube
A Pointer for dial gage
Flattened tube deflects outward under pressure
Fig. 2.26 Schematic of a bourdontube device for mechanical measurement of high pressures.
High pressure
Chapter 2 Pressure Distribution in a Fluid
Fig. 2.27 The fused-quartz, forcebalanced bourdon tube is the most accurate pressure sensor used in commercial applications today. (Courtesy of Ruska Instrument Corporation, Houston, TX.)
An interesting variation of Fig. 2.26 is the fused-quartz, forced-balanced bourdon tube, shown in Fig. 2.27, whose deflection is sensed optically and returned to a zero reference state by a magnetic
element whose output is proportional to the fluid pressure. The fused-quartz, forced-balanced bourdon tube is reported to be one of the most accurate pressure sensors ever devised, with uncertainty
of the order of 0.003 percent. The last category, electric-output sensors, is extremely important in engineering because the data can be stored on computers and freely manipulated, plotted, and
analyzed. Three examples are shown in Fig. 2.28, the first being the capacitive sensor in Fig. 2.28a. The differential pressure deflects the silicon diaphragm and changes the capacitancce of the
liquid in the cavity. Note that the cavity has spherical end caps to prevent overpressure damage. In the second type, Fig. 2.28b, strain gages and other sensors are diffused or etched onto a chip
which is stressed by the applied pressure. Finally, in Fig. 2.28c, a micromachined silicon sensor is arranged to deform under pressure such that its natural vibration frequency is proportional to the
pressure. An oscillator excites the element’s resonant frequency and converts it into appropriate pressure units. For further information on pressure sensors, see Refs. 7 to 10, 12, and 13.
This chapter has been devoted entirely to the computation of pressure distributions and the resulting forces and moments in a static fluid or a fluid with a known velocity field. All hydrostatic
(Secs. 2.3 to 2.8) and rigid-body (Sec. 2.9) problems are solved in this manner and are classic cases which every student should understand. In arbitrary viscous flows, both pressure and velocity are
unknowns and are solved together as a system of equations in the chapters which follow.
Cover flange
Seal diaphragm
High-pressure side
Low-pressure side
Filling liquid
Sensing diaphragm
Strain gages Diffused into integrated silicon chip
Wire bonding Stitch bonded connections from chip to body plug
Etched cavity Micromachined silicon sensor
(Druck, Inc., Fairfield, Connecticut) (b)
Temperature sensor On-chip diode for optimum temperature performance
Fig. 2.28 Pressure sensors with electric output: (a) a silicon diaphragm whose deflection changes the cavity capacitance (Courtesy of Johnson-Yokogawa Inc.); (b) a silicon strain gage which is
stressed by applied pressure; (c) a micromachined silicon element which resonates at a frequency proportional to applied pressure. [(b) and (c) are courtesy of Druck, Inc., Fairfield, CT.]
Chapter 2 Pressure Distribution in a Fluid P2.2 For the two-dimensional stress field shown in Fig. P2.1 suppose that
Problems Most of the problems herein are fairly straightforward. More difficult or open-ended assignments are indicated with an asterisk, as in Prob. 2.8. Problems labeled with an EES icon (for
example, Prob. 2.62), will benefit from the use of the Engineering Equation Solver (EES), while problems labeled with a disk icon may require the use of a computer. The standard endof-chapter
problems 2.1 to 2.158 (categorized in the problem list below) are followed by word problems W2.1 to W2.8, fundamentals of engineering exam problems FE2.1 to FE2.10, comprehensive problems C2.1 to
C2.4, and design projects D2.1 and D2.2.
xx 2000 lbf/ft2 yy 3000 lbf/ft2 n(AA) 2500 lbf/ft2
Problem Distribution Section
2.1, 2.2 2.3 2.3 2.4 2.5 2.6 2.7 2.8 2.8 2.9 2.9 2.10
Stresses; pressure gradient; gage pressure Hydrostatic pressure; barometers The atmosphere Manometers; multiple fluids Forces on plane surfaces Forces on curved surfaces Forces in layered fluids
Buoyancy; Archimedes’ principles Stability of floating bodies Uniform acceleration Rigid-body rotation Pressure measurements
2.1–2.6 2.7–2.23 2.24–2.29 2.30–2.47 2.48–2.81 2.82–2.100 2.101–2.102 2.103–2.126 2.127–2.136 2.137–2.151 2.152–2.158 None
P2.1 For the two-dimensional stress field shown in Fig. P2.1 it is found that
xx 3000 lbf/ft2 yy 2000 lbf/ft2 xy 500 lbf/ft2 Find the shear and normal stresses (in lbf/ft2) acting on plane AA cutting through the element at a 30° angle as shown.
σyy σyx
σxx 30°
σxy =
σyx σyy
Compute (a) the shear stress xy and (b) the shear stress on plane AA. Derive Eq. (2.18) by using the differential element in Fig. 2.2 with z “up,’’ no fluid motion, and pressure varying only in the z
direction. In a certain two-dimensional fluid flow pattern the lines of constant pressure, or isobars, are defined by the expression P0 Bz Cx2 constant, where B and C are constants and p0 is the
(constant) pressure at the origin, (x, z) (0, 0). Find an expression x f (z) for the family of lines which are everywhere parallel to the local pressure gradient V p. Atlanta, Georgia, has an average
altitude of 1100 ft. On a standard day (Table A.6), pressure gage A in a laboratory experiment reads 93 kPa and gage B reads 105 kPa. Express these readings in gage pressure or vacuum pressure (Pa),
whichever is appropriate. Any pressure reading can be expressed as a length or head, h p/ g. What is standard sea-level pressure expressed in (a) ft of ethylene glycol, (b) in Hg, (c) m of water, and
(d) mm of methanol? Assume all fluids are at 20°C. The deepest known point in the ocean is 11,034 m in the Mariana Trench in the Pacific. At this depth the specific weight of seawater is
approximately 10,520 N/m3. At the surface, 10,050 N/m3. Estimate the absolute pressure at this depth, in atm. Dry adiabatic lapse rate (DALR) is defined as the negative value of atmospheric
temperature gradient, dT/dz, when temperature and pressure vary in an isentropic fashion. Assuming air is an ideal gas, DALR dT/dz when T T0(p/p0)a, where exponent a (k 1)/k, k cp /cv is the ratio of
specific heats, and T0 and p0 are the temperature and pressure at sea level, respectively. (a) Assuming that hydrostatic conditions exist in the atmosphere, show that the dry adiabatic lapse rate is
constant and is given by DALR g(k 1)/(kR), where R is the ideal gas constant for air. (b) Calculate the numerical value of DALR for air in units of °C/km. For a liquid, integrate the hydrostatic
relation, Eq. (2.18), by assuming that the isentropic bulk modulus, B (p/ )s, is constant—see Eq. (9.18). Find an expression for p(z) and apply the Mariana Trench data as in Prob. 2.7, using
Bseawater from Table A.3. A closed tank contains 1.5 m of SAE 30 oil, 1 m of water, 20 cm of mercury, and an air space on top, all at 20°C. The absolute pressure at the bottom of the tank is 60 kPa.
What is the pressure in the air space?
Problems 103 P2.11 In Fig. P2.11, pressure gage A reads 1.5 kPa (gage). The fluids are at 20°C. Determine the elevations z, in meters, of the liquid levels in the open piezometer tubes B and C.
Gasoline 1.5 m
A B
C h 1m
1.5 m
Liquid, SG = 1.60
B Air
z= 0
P2.12 In Fig. P2.12 the tank contains water and immiscible oil at 20°C. What is h in cm if the density of the oil is 898 kg/m3?
4m Water
15 lbf/in2 abs
6 cm
A Air
12 cm
8 cm
2 ft
Oil 1 ft
P2.13 In Fig. P2.13 the 20°C water and gasoline surfaces are open to the atmosphere and at the same elevation. What is the height h of the third liquid in the right leg? P2.14 The closed tank in Fig.
P2.14 is at 20°C. If the pressure at point A is 95 kPa absolute, what is the absolute pressure at point B in kPa? What percent error do you make by neglecting the specific weight of the air? P2.15
The air-oil-water system in Fig. P2.15 is at 20°C. Knowing that gage A reads 15 lbf/in2 absolute and gage B reads 1.25 lbf/in2 less than gage C, compute (a) the specific weight of the oil in lbf/ft3
and (b) the actual reading of gage C in lbf/in2 absolute.
Oil 1 ft
2 ft C
P2.16 A closed inverted cone, 100 cm high with diameter 60 cm at the top, is filled with air at 20°C and 1 atm. Water at 20°C is introduced at the bottom (the vertex) to compress the air isothermally
until a gage at the top of the cone reads 30 kPa (gage). Estimate (a) the amount of water needed (cm3) and (b) the resulting absolute pressure at the bottom of the cone (kPa).
Chapter 2 Pressure Distribution in a Fluid
P2.17 The system in Fig. P2.17 is at 20°C. If the pressure at point A is 1900 lbf/ft2, determine the pressures at points B, C, and D in lbf/ft2. Mercury Air
3 ft
B A
C 2 ft
10 cm
10 cm
Air 4 ft 10 cm
P2.19 5 ft
2 ft
2000 lbf
3-in diameter
P2.18 The system in Fig. P2.18 is at 20°C. If atmospheric pressure is 101.33 kPa and the pressure at the bottom of the tank is 242 kPa, what is the specific gravity of fluid X?
1 in
15 in
1-in diameter
Oil SAE 30 oil
P2.20 Water
2m Air: 180 kPa abs
Fluid X
0.5 m
80 cm
P2.18 P2.21 P2.19 The U-tube in Fig. P2.19 has a 1-cm ID and contains mercury as shown. If 20 cm3 of water is poured into the righthand leg, what will the free-surface height in each leg be after the
sloshing has died down? P2.20 The hydraulic jack in Fig. P2.20 is filled with oil at 56 lbf/ft3. Neglecting the weight of the two pistons, what force F on the handle is required to support the
2000-lbf weight for this design? P2.21 At 20°C gage A reads 350 kPa absolute. What is the height h of the water in cm? What should gage B read in kPa absolute? See Fig. P2.21.
Mercury B
P2.22 The fuel gage for a gasoline tank in a car reads proportional to the bottom gage pressure as in Fig. P2.22. If the tank is 30 cm deep and accidentally contains 2 cm of water plus gasoline, how
many centimeters of air remain at the top when the gage erroneously reads “full’’? P2.23 In Fig. P2.23 both fluids are at 20°C. If surface tension effects are negligible, what is the density of the
oil, in kg/m3? P2.24 In Prob. 1.2 we made a crude integration of the density distribution (z) in Table A.6 and estimated the mass of the earth’s atmosphere to be m 6 E18 kg. Can this re-
Problems 105 Vent Air
Gasoline SG = 0.68
30 cm
mental observations. (b) Find an expression for the pressure at points 1 and 2 in Fig. P2.27b. Note that the glass is now inverted, so the original top rim of the glass is at the bottom of the
picture, and the original bottom of the glass is at the top of the picture. The weight of the card can be neglected.
2 cm Card
Top of glass
8 cm 6 cm
Bottom of glass
Original bottom of glass 10 cm 1●
2● sult be used to estimate sea-level pressure on the earth? Conversely, can the actual sea-level pressure of 101.35 kPa be used to make a more accurate estimate of the atmosP2.27b Card Original top
of glass pheric mass? P2.25 Venus has a mass of 4.90 E24 kg and a radius of 6050 km. Its atmosphere is 96 percent CO2, but let us assume it to (c) Estimate the theoretical maximum glass height such
be 100 percent. Its surface temperature averages 730 K, that this experiment could still work, i.e., such that the wadecreasing to 250 K at an altitude of 70 km. The average ter would not fall out of
the glass. surface pressure is 9.1 MPa. Estimate the atmospheric P2.28 Earth’s atmospheric conditions vary somewhat. On a cerpressure of Venus at an altitude of 5 km. tain day the sea-level
temperature is 45°F and the sea-level P2.26 Investigate the effect of doubling the lapse rate on atmospressure is 28.9 inHg. An airplane overhead registers an pheric pressure. Compare the standard
atmosphere (Table air temperature of 23°F and a pressure of 12 lbf/in2. EstiA.6) with a lapse rate twice as high, B2 0.0130 K/m. mate the plane’s altitude, in feet. Find the altitude at which the
pressure deviation is (a) 1 *P2.29 Under some conditions the atmosphere is adiabatic, p percent and (b) 5 percent. What do you conclude? (const)( k), where k is the specific heat ratio. Show that,
P2.27 Conduct an experiment to illustrate atmospheric pressure. for an adiabatic atmosphere, the pressure variation is Note: Do this over a sink or you may get wet! Find a drinkgiven by ing glass
with a very smooth, uniform rim at the top. Fill the glass nearly full with water. Place a smooth, light, flat (k 1)gz k/(k1) p p0 1 plate on top of the glass such that the entire rim of the kRT0
glass is covered. A glossy postcard works best. A small inCompare this formula for air at z 5000 m with the standex card or one flap of a greeting card will also work. See dard atmosphere in Table
A.6. Fig. P2.27a. (a) Hold the card against the rim of the glass and turn the P2.30 In Fig. P2.30 fluid 1 is oil (SG 0.87) and fluid 2 is glycerin at 20°C. If pa 98 kPa, determine the absolute
presglass upside down. Slowly release pressure on the card. sure at point A. Does the water fall out of the glass? Record your experi-
Chapter 2 Pressure Distribution in a Fluid
Air B
ρ1 SAE 30 oil 32 cm A
Liquid, SG = 1.45
3 cm
5 cm
10 cm
4 cm A Water
6 cm 8 cm 3 cm
P2.31 In Fig. P2.31 all fluids are at 20°C. Determine the pressure difference (Pa) between points A and B.
*P2.34 Sometimes manometer dimensions have a significant ef-
Kerosine Air
40 cm
9 cm
20 cm
fect. In Fig. P2.34 containers (a) and (b) are cylindrical and conditions are such that pa pb. Derive a formula for the pressure difference pa pb when the oil-water interface on the right rises a
distance h h, for (a) d D and (b) d 0.15D. What is the percent change in the value of p?
14 cm
8 cm Mercury
(b) (a)
P2.32 For the inverted manometer of Fig. P2.32, all fluids are at 20°C. If pB pA 97 kPa, what must the height H be in cm?
Meriam red oil, SG = 0.827
SAE 30 oil H
h 18 cm
d H
P2.34 35 cm
P2.32 P2.33 In Fig. P2.33 the pressure at point A is 25 lbf/in2. All fluids are at 20°C. What is the air pressure in the closed chamber B, in Pa?
P2.35 Water flows upward in a pipe slanted at 30°, as in Fig. P2.35. The mercury manometer reads h 12 cm. Both fluids are at 20°C. What is the pressure difference p1 p2 in the pipe? P2.36 In Fig.
P2.36 both the tank and the tube are open to the atmosphere. If L 2.13 m, what is the angle of tilt of the tube? P2.37 The inclined manometer in Fig. P2.37 contains Meriam red manometer oil, SG
0.827. Assume that the reservoir
Problems 107 with manometer fluid m. One side of the manometer is open to the air, while the other is connected to new tubing which extends to pressure measurement location 1, some height H higher in
elevation than the surface of the manometer liquid. For consistency, let a be the density of the air in the room, t be the density of the gas inside the tube, m be the density of the manometer
liquid, and h be the height difference between the two sides of the manometer. See Fig. P2.38. (a) Find an expression for the gage pressure at the measurement point. Note: When calculating gage
pressure, use the local atmospheric pressure at the elevation of the measurement point. You may assume that h H; i.e., assume the gas in the entire left side of the manometer is of density t. (b)
Write an expression for the error caused by assuming that the gas inside the tubing has the same density as that of the surrounding air. (c) How much error (in Pa) is caused by ignoring this density
difference for the following conditions: m 860 kg/m3, a 1.20 kg/m3, t 1.50 kg/m3, H 1.32 m, and h 0.58 cm? (d) Can you think of a simple way to avoid this error?
(2) 30 (1) h
50 cm
50 cm
Oil SG = 0.8
Water SG = 1.0
is very large. If the inclined arm is fitted with graduations 1 in apart, what should the angle be if each graduation corresponds to 1 lbf/ft2 gage pressure for pA?
t (tubing gas)
pa at location 1 a (air) H
1 in pA
U-tube manometer
h m
P2.38 Reservoir
P2.37 P2.38 An interesting article appeared in the AIAA Journal (vol. 30, no. 1, January 1992, pp. 279–280). The authors explain that the air inside fresh plastic tubing can be up to 25 percent more
dense than that of the surroundings, due to outgassing or other contaminants introduced at the time of manufacture. Most researchers, however, assume that the tubing is filled with room air at
standard air density, which can lead to significant errors when using this kind of tubing to measure pressures. To illustrate this, consider a U-tube manometer
P2.39 An 8-cm-diameter piston compresses manometer oil into an inclined 7-mm-diameter tube, as shown in Fig. P2.39. When a weight W is added to the top of the piston, the oil rises an additional
distance of 10 cm up the tube, as shown. How large is the weight, in N? P2.40 A pump slowly introduces mercury into the bottom of the closed tank in Fig. P2.40. At the instant shown, the air pressure
pB 80 kPa. The pump stops when the air pressure rises to 110 kPa. All fluids remain at 20°C. What will be the manometer reading h at that time, in cm, if it is connected to standard sea-level ambient
air patm?
Chapter 2 Pressure Distribution in a Fluid
10 cm
D = 8 cm Piston
d = 7 mm Meriam red oil, SG = 0.827
8 cm
Air: pB
9 cm
P2.44 Water flows downward in a pipe at 45°, as shown in Fig. P2.44. The pressure drop p1 p2 is partly due to gravity and partly due to friction. The mercury manometer reads a 6-in height difference.
What is the total pressure drop p1 p2 in lbf/in2? What is the pressure drop due to friction only between 1 and 2 in lbf/in2? Does the manometer reading correspond only to friction drop? Why?
Pump 10 cm
2 cm
P2.40 P2.41 The system in Fig. P2.41 is at 20°C. Compute the pressure at point A in lbf/ft2 absolute.
1 5 ft
Water Flow 2 Oil, SG = 0.85 5 in A
pa = 14.7
10 in
6 in
6 in Water
P2.44 P2.41
P2.42 Very small pressure differences pA pB can be measured accurately by the two-fluid differential manometer in Fig. P2.42. Density 2 is only slightly larger than that of the upper fluid 1. Derive
an expression for the proportionality between h and pA pB if the reservoirs are very large. *P2.43 A mercury manometer, similar to Fig. P2.35, records h 1.2, 4.9, and 11.0 mm when the water
velocities in the pipe are V 1.0, 2.0, and 3.0 m/s, respectively. Determine if these data can be correlated in the form p1 p2 Cf V2, where Cf is dimensionless.
P2.45 In Fig. P2.45, determine the gage pressure at point A in Pa. Is it higher or lower than atmospheric? P2.46 In Fig. P2.46 both ends of the manometer are open to the atmosphere. Estimate the
specific gravity of fluid X. P2.47 The cylindrical tank in Fig. P2.47 is being filled with water at 20°C by a pump developing an exit pressure of 175 EES kPa. At the instant shown, the air pressure
is 110 kPa and H 35 cm. The pump stops when it can no longer raise the water pressure. For isothermal air compression, estimate H at that time. P2.48 Conduct the following experiment to illustrate
air pressure. Find a thin wooden ruler (approximately 1 ft in
Problems 109 patm
50 cm
Air 20˚ C
Oil, SG = 0.85
75 cm
30 cm 45 cm
40 cm
Water Pump
P2.47 15 cm A
Ruler SAE 30 oil 10 cm
9 cm
Water 5 cm 7 cm
Fluid X
6 cm
4 cm
P2.50 P2.46
12 cm
length) or a thin wooden paint stirrer. Place it on the edge of a desk or table with a little less than half of it hang- P2.51 ing over the edge lengthwise. Get two full-size sheets of newspaper;
open them up and place them on top of the ruler, covering only the portion of the ruler resting on the *P2.52 desk as illustrated in Fig. P2.48. (a) Estimate the total force on top of the newspaper
due to air pressure in the room. (b) Careful! To avoid potential injury, make sure nobody is standing directly in front of the desk. Perform
a karate chop on the portion of the ruler sticking out over the edge of the desk. Record your results. (c) Explain your results. A water tank has a circular panel in its vertical wall. The panel has
a radius of 50 cm, and its center is 2 m below the surface. Neglecting atmospheric pressure, determine the water force on the panel and its line of action. A vat filled with oil (SG 0.85) is 7 m long
and 3 m deep and has a trapezoidal cross section 2 m wide at the bottom and 4 m wide at the top. Compute (a) the weight of oil in the vat, (b) the force on the vat bottom, and (c) the force on the
trapezoidal end panel. Gate AB in Fig. P2.51 is 1.2 m long and 0.8 m into the paper. Neglecting atmospheric pressure, compute the force F on the gate and its center-of-pressure position X. Suppose
that the tank in Fig. P2.51 is filled with liquid X, not oil. Gate AB is 0.8 m wide into the paper. Suppose that liquid X causes a force F on gate AB and that the moment of this force about point B
is 26,500 N m. What is the specific gravity of liquid X?
Chapter 2 Pressure Distribution in a Fluid pa
Oil, SG = 0.82
Water pa
4m h
A 1m
1.2 m
B 4 ft
F B
P2.53 Panel ABC in the slanted side of a water tank is an isosceles triangle with the vertex at A and the base BC 2 m, as in Fig. P2.53. Find the water force on the panel and its line of action.
200 kg h
m B
30 cm
A Water
P2.58 4m
B, C
*P2.59 Gate AB has length L, width b into the paper, is hinged at B, and has negligible weight. The liquid level h remains at the top of the gate for any angle . Find an analytic expression for the
force P, perpendicular to AB, required to keep the gate in equilibrium in Fig. P2.59.
P2.54 If, instead of water, the tank in Fig. P2.53 is filled with liqP uid X, the liquid force on panel ABC is found to be 115 kN. What is the density of liquid X? The line of action is found A to be
the same as in Prob. 2.53. Why? P2.55 Gate AB in Fig. P2.55 is 5 ft wide into the paper, hinged at A, and restrained by a stop at B. The water is at 20°C. Compute (a) the force on stop B and (b) the
reactions at h L A if the water depth h 9.5 ft. P2.56 In Fig. P2.55, gate AB is 5 ft wide into the paper, and stop B will break if the water force on it equals 9200 lbf. For Hinge what water depth h
is this condition reached? P2.57 In Fig. P2.55, gate AB is 5 ft wide into the paper. Suppose B P2.59 that the fluid is liquid X, not water. Hinge A breaks when its reaction is 7800 lbf, and the
liquid depth is h 13 ft. *P2.60 Find the net hydrostatic force per unit width on the recWhat is the specific gravity of liquid X? tangular gate AB in Fig. P2.60 and its line of action. P2.58 In Fig.
P2.58, the cover gate AB closes a circular opening 80 cm in diameter. The gate is held closed by a 200-kg *P2.61 Gate AB in Fig. P2.61 is a homogeneous mass of 180 kg, 1.2 m wide into the paper,
hinged at A, and resting on a mass as shown. Assume standard gravity at 20°C. At what smooth bottom at B. All fluids are at 20°C. For what wawater level h will the gate be dislodged? Neglect the
weight ter depth h will the force at point B be zero? of the gate.
Problems 111 P2.63 The tank in Fig. P2.63 has a 4-cm-diameter plug at the bottom on the right. All fluids are at 20°C. The plug will pop out if the hydrostatic force on it is 25 N. For this
condition, what will be the reading h on the mercury manometer on the left side?
1.8 m
1.2 m A
Water 2m
B 2m
P2.60 h 2 cm
Plug, D = 4 cm
P2.63 Glycerin h 2m A
*P2.64 Gate ABC in Fig. P2.64 has a fixed hinge line at B and is 2 m wide into the paper. The gate will open at A to release water if the water depth is high enough. Compute the depth h for which the
gate will begin to open.
1m C
P2.61 P2.62 Gate AB in Fig. P2.62 is 15 ft long and 8 ft wide into the paper and is hinged at B with a stop at A. The water is at EES 20°C. The gate is 1-in-thick steel, SG 7.85. Compute the water
level h for which the gate will start to fall.
Pulley A
20 cm
B h
1m Water at 20°C
10,000 lb
15 ft
*P2.65 Gate AB in Fig. P2.65 is semicircular, hinged at B, and held by a horizontal force P at A. What force P is required for equilibrium? P2.66 Dam ABC in Fig. P2.66 is 30 m wide into the paper and
made of concrete (SG 2.4). Find the hydrostatic force on surface AB and its moment about C. Assuming no seepage of water under the dam, could this force tip the dam over? How does your argument
change if there is seepage under the dam?
Chapter 2 Pressure Distribution in a Fluid
Oil, SG = 0.83
Water A
P Gate: Side view
;; ;;
50˚ B
P2.68 A
Water 20˚C
80 m
80 cm
Water, 20˚C
Hg, 20˚C
60 m
A 2m
P2.69 *P2.67 Generalize Prob. 2.66 as follows. Denote length AB as H, length BC as L, and angle ABC as . Let the dam material have specific gravity SG. The width of the dam is b. Assume no seepage of
water under the dam. Find an analytic relation between SG and the critical angle c for which the dam will just tip over to the right. Use your relation to compute c for the special case SG 2.4
(concrete). P2.68 Isosceles triangle gate AB in Fig. P2.68 is hinged at A and weighs 1500 N. What horizontal force P is required at point B for equilibrium? *P2.69 The water tank in Fig. P2.69 is
pressurized, as shown by the mercury-manometer reading. Determine the hydrostatic force per unit depth on gate AB. P2.70 Calculate the force and center of pressure on one side of the vertical
triangular panel ABC in Fig. P2.70. Neglect patm. *P2.71 In Fig. P2.71 gate AB is 3 m wide into the paper and is connected by a rod and pulley to a concrete sphere (SG
2 ft A
6 ft
B 4 ft
Problems 113 2.40). What diameter of the sphere is just sufficient to keep *P2.74 In “soft’’ liquids (low bulk modulus ), it may be necesthe gate closed? sary to account for liquid compressibility in
hydrostatic calculations. An approximate density relation would be
dp d a2 d
Concrete sphere, SG = 2.4
p p0 a2( 0)
where a is the speed of sound and (p0, 0) are the conditions at the liquid surface z 0. Use this approximation to show that the density variation with depth in a soft liq2 uid is 0egz/a where g is
the acceleration of gravity 8m A and z is positive upward. Then consider a vertical wall of width b, extending from the surface (z 0) down to depth z h. Find an analytic expression for the
hydrostatic Water 4m force F on this wall, and compare it with the incompressB ible result F 0gh2b/2. Would the center of pressure be below the incompressible position z 2h/3? P2.71 *P2.75 Gate AB in
Fig. P2.75 is hinged at A, has width b into the paper, and makes smooth contact at B. The gate has density s and uniform thickness t. For what gate density s, P2.72 The V-shaped container in Fig.
P2.72 is hinged at A and expressed as a function of (h, t, , ), will the gate just beheld together by cable BC at the top. If cable spacing is gin to lift off the bottom? Why is your answer indepen1
m into the paper, what is the cable tension? dent of gate length L and width b? Cable C
A 1m
P2.73 Gate AB is 5 ft wide into the paper and opens to let fresh water out when the ocean tide is dropping. The hinge at A is 2 ft above the freshwater level. At what ocean level h will the gate
first open? Neglect the gate weight.
A Tide range
10 ft h Seawater, SG = 1.025 Stop
P2.75 B
*P2.76 Consider the angled gate ABC in Fig. P2.76, hinged at C and of width b into the paper. Derive an analytic formula for the horizontal force P required at the top for equilibrium, as a function
of the angle . P2.77 The circular gate ABC in Fig. P2.77 has a 1-m radius and is hinged at B. Compute the force P just sufficient to keep the gate from opening when h 8 m. Neglect atmospheric
pressure. P2.78 Repeat Prob. 2.77 to derive an analytic expression for P as a function of h. Is there anything unusual about your solution? P2.79 Gate ABC in Fig. P2.79 is 1 m square and is hinged at
B. It will open automatically when the water level h becomes high enough. Determine the lowest height for which the
;; ;; ;;
Chapter 2 Pressure Distribution in a Fluid A
θ θ
Specific weight γ
1 atm 2m
SA E3
pa Water
cm Mercury
Panel, 30 cm high, 40 cm wide
P2.81 Gate AB in Fig. P2.81 is 7 ft into the paper and weighs 3000 lbf when submerged. It is hinged at B and rests against a smooth wall at A. Determine the water level h at the left which will just
cause the gate to open.
A 1m B 1m C
h 4 ft A
P2.77 Water 8 ft Water h
Water 6 ft
60 cm
40 cm
*P2.82 The dam in Fig. P2.82 is a quarter circle 50 m wide into P2.79 gate will open. Neglect atmospheric pressure. Is this result independent of the liquid density? P2.80 For the closed tank in Fig.
P2.80, all fluids are at 20°C, and the airspace is pressurized. It is found that the net outward hydrostatic force on the 30-by 40-cm panel at the bottom of the water layer is 8450 N. Estimate (a)
the pressure in the airspace and (b) the reading h on the mercury manometer.
the paper. Determine the horizontal and vertical components of the hydrostatic force against the dam and the point CP where the resultant strikes the dam. *P2.83 Gate AB in Fig. P2.83 is a quarter
circle 10 ft wide into the paper and hinged at B. Find the force F just sufficient to keep the gate from opening. The gate is uniform and weighs 3000 lbf. P2.84 Determine (a) the total hydrostatic
force on the curved surface AB in Fig. P2.84 and (b) its line of action. Neglect atmospheric pressure, and let the surface have unit width.
;;; ;;; ;;; 20 m
20 m
Problems 115
pa = 0
10 ft
2 ft
P2.87 The bottle of champagne (SG 0.96) in Fig. P2.87 is under pressure, as shown by the mercury-manometer reading. Compute the net force on the 2-in-radius hemispherical end cap at the bottom of the
r = 8 ft
B Water at 20° C z 1m
4 in
z = x3 2 in 6 in x
P2.84 P2.87
r = 2 in
P2.85 Compute the horizontal and vertical components of the hydrostatic force on the quarter-circle panel at the bottom of *P2.88 Gate ABC is a circular arc, sometimes called a Tainter gate, which
can be raised and lowered by pivoting about point the water tank in Fig. P2.85. O. See Fig. P2.88. For the position shown, determine (a) the hydrostatic force of the water on the gate and (b) its
line of action. Does the force pass through point O? 6m
5m Water
Water 2m 6m
2m 6m
P2.86 Compute the horizontal and vertical components of the hydrostatic force on the hemispherical bulge at the bottom of the tank in Fig. P2.86.
Chapter 2 Pressure Distribution in a Fluid
P2.89 The tank in Fig. P2.89 contains benzene and is pressurized to 200 kPa (gage) in the air gap. Determine the vertical hydrostatic force on circular-arc section AB and its line of action.
4m 60 cm p = 200 kPa
30 cm
Six bolts
Water Benzene at 20C
60 cm
P2.91 P2.89
A 2m
P2.90 A 1-ft-diameter hole in the bottom of the tank in Fig. P2.90 is closed by a conical 45° plug. Neglecting the weight of the plug, compute the force F required to keep the plug in the hole.
Water p = 3 lbf/in 2 gage
Bolt spacing 25 cm
P2.92 1 ft
Air :
Water 3 ft 1 ft
ρ, γ
45˚ cone
P2.91 The hemispherical dome in Fig. P2.91 weighs 30 kN and is filled with water and attached to the floor by six equally spaced bolts. What is the force in each bolt required to hold down the dome?
P2.92 A 4-m-diameter water tank consists of two half cylinders, each weighing 4.5 kN/m, bolted together as shown in Fig. P2.92. If the support of the end caps is neglected, determine the force
induced in each bolt. *P2.93 In Fig. P2.93, a one-quadrant spherical shell of radius R is submerged in liquid of specific gravity and depth h R. Find an analytic expression for the resultant
hydrostatic force, and its line of action, on the shell surface.
P2.94 The 4-ft-diameter log (SG 0.80) in Fig. P2.94 is 8 ft long into the paper and dams water as shown. Compute the net vertical and horizontal reactions at point C.
Problems 117 wall at A. Compute the reaction forces at points A and B.
Log 2ft
Water 2ft
Water C Seawater, 10,050 N/m3
*P2.95 The uniform body A in Fig. P2.95 has width b into the paper and is in static equilibrium when pivoted about hinge O. What is the specific gravity of this body if (a) h 0 and (b) h R?
2m 45°
B A
P2.97 h
P2.98 Gate ABC in Fig. P2.98 is a quarter circle 8 ft wide into the paper. Compute the horizontal and vertical hydrostatic forces on the gate and the line of action of the resultant force.
R R Water
r = 4 ft
P2.96 The tank in Fig. P2.96 is 3 m wide into the paper. Neglecting atmospheric pressure, compute the hydrostatic (a) horizontal force, (b) vertical force, and (c) resultant force on quarter-circle
panel BC.
45° 45°
P2.99 A 2-ft-diameter sphere weighing 400 lbf closes a 1-ft-diameter hole in the bottom of the tank in Fig. P2.99. Compute the force F required to dislodge the sphere from the hole.
4m B
3 ft 1 ft
P2.97 Gate AB in Fig. P2.97 is a three-eighths circle, 3 m wide into the paper, hinged at B, and resting against a smooth
1 ft
Chapter 2 Pressure Distribution in a Fluid
P2.100 Pressurized water fills the tank in Fig. P2.100. Compute the net hydrostatic force on the conical surface ABC.
2m A
P2.106 C
4m 7m B
150 kPa gage
P2.101 A fuel truck has a tank cross section which is approximately elliptical, with a 3-m horizontal major axis and a 2-m vertical minor axis. The top is vented to the atmosphere. If the tank is
filled half with water and half with gasoline, what is the hydrostatic force on the flat elliptical end panel? P2.102 In Fig. P2.80 suppose that the manometer reading is h 25 cm. What will be the net
hydrostatic force on the complete end wall, which is 160 cm high and 2 m wide? P2.103 The hydrogen bubbles in Fig. 1.13 are very small, less than a millimeter in diameter, and rise slowly. Their drag
in still fluid is approximated by the first term of Stokes’ expression in Prob. 1.10: F 3 VD, where V is the rise velocity. Neglecting bubble weight and setting bubble buoyancy equal to drag, (a)
derive a formula for the terminal (zero acceleration) rise velocity Vterm of the bubble and (b) determine Vterm in m/s for water at 20°C if D 30 m. P2.104 The can in Fig. P2.104 floats in the
position shown. What is its weight in N?
whether his new crown was pure gold (SG 19.3). Archimedes measured the weight of the crown in air to be 11.8 N and its weight in water to be 10.9 N. Was it pure gold? It is found that a 10-cm cube of
aluminum (SG 2.71) will remain neutral under water (neither rise nor fall) if it is tied by a string to a submerged 18-cm-diameter sphere of buoyant foam. What is the specific weight of the foam, in
N/m3? Repeat Prob. 2.62, assuming that the 10,000-lbf weight is aluminum (SG 2.71) and is hanging submerged in the water. A piece of yellow pine wood (SG 0.65) is 5 cm square and 2.2 m long. How many
newtons of lead (SG 11.4) should be attached to one end of the wood so that it will float vertically with 30 cm out of the water? A hydrometer floats at a level which is a measure of the specific
gravity of the liquid. The stem is of constant diameter D, and a weight in the bottom stabilizes the body to float vertically, as shown in Fig. P2.109. If the position h 0 is pure water (SG 1.0),
derive a formula for h as a function of total weight W, D, SG, and the specific weight 0 of water.
D SG = 1.0 h
Fluid, SG > 1 W
3 cm
8 cm
D = 9 cm
P2.105 It is said that Archimedes discovered the buoyancy laws when asked by King Hiero of Syracuse to determine
P2.110 An average table tennis ball has a diameter of 3.81 cm and a mass of 2.6 g. Estimate the (small) depth at which this ball will float in water at 20°C and sea level standard air if air buoyancy
is (a) neglected and (b) included. P2.111 A hot-air balloon must be designed to support basket, cords, and one person for a total weight of 1300 N. The balloon material has a mass of 60 g/m2. Ambient
air is at 25°C and 1 atm. The hot air inside the balloon is at 70°C and 1 atm. What diameter spherical balloon will just support the total weight? Neglect the size of the hot-air inlet vent. P2.112
The uniform 5-m-long round wooden rod in Fig. P2.112 is tied to the bottom by a string. Determine (a) the tension
Problems 119 in the string and (b) the specific gravity of the wood. Is it possible for the given information to determine the inclination angle ? Explain.
Hinge D = 4 cm B
= 30
D = 8 cm
2 kg of lead θ Water at 20°C
B String
8 ft
Wood SG = 0.6
P2.113 A spar buoy is a buoyant rod weighted to float and protrude vertically, as in Fig. P2.113. It can be used for measurements or markers. Suppose that the buoy is maple wood (SG 0.6), 2 in by 2
in by 12 ft, floating in seawater (SG 1.025). How many pounds of steel (SG 7.85) should be added to the bottom end so that h 18 in?
A Rock
P2.116 The homogeneous 12-cm cube in Fig. 2.116 is balanced by a 2-kg mass on the beam scale when the cube is immersed in 20°C ethanol. What is the specific gravity of the cube?
2 kg Wsteel
P2.114 The uniform rod in Fig. P2.114 is hinged at point B on the waterline and is in static equilibrium as shown when 2 kg of lead (SG 11.4) are attached to its end. What is the specific gravity of
the rod material? What is peculiar about the rest angle 30? P2.115 The 2-in by 2-in by 12-ft spar buoy from Fig. P2.113 has 5 lbm of steel attached and has gone aground on a rock, as in Fig. P2.115.
Compute the angle at which the buoy will lean, assuming that the rock exerts no moments on the spar.
12 cm
P2.117 The balloon in Fig. P2.117 is filled with helium and pressurized to 135 kPa and 20°C. The balloon material has a
Chapter 2 Pressure Distribution in a Fluid mass of 85 g/m2. Estimate (a) the tension in the mooring line and (b) the height in the standard atmosphere to which the balloon will rise if the mooring
line is cut.
P2.121 The uniform beam in Fig. P2.121, of size L by h by b and with specific weight b, floats exactly on its diagonal when a heavy uniform sphere is tied to the left corner, as shown. Show that this
can only happen (a) when b /3 and (b) when the sphere has size
Lhb D (SG 1) D = 10 m
Width b > Dpipes
θ θ
Vrel, 2 x
R0 T, P,ω
Q R1
Q 2
P3.113 Modify Example 3.14 so that the arm starts from rest and spins up to its final rotation speed. The moment of inertia of the arm about O is I0. Neglecting air drag, find d/dt and integrate to
determine the angular velocity (t), assuming 0 at t 0. P3.114 The three-arm lawn sprinkler of Fig. P3.114 receives 20°C water through the center at 2.7 m3/h. If collar friction is negligible, what is
the steady rotation rate in r/min for (a)
0° and (b) 40°? θ
P3.116 P3.117 A simple turbomachine is constructed from a disk with two internal ducts which exit tangentially through square holes, as in Fig. P3.117. Water at 20°C enters normal to the disk at the
center, as shown. The disk must drive, at 250 r/min, a small device whose retarding torque is 1.5 N m. What is the proper mass flow of water, in kg/s? 2 cm
d = 7 mm 2 cm
θ 32 cm
P3.115 Water at 20°C flows at 30 gal/min through the 0.75-in-diameter double pipe bend of Fig. P3.115. The pressures are p1 30 lbf/in2 and p2 24 lbf/in2. Compute the torque T at point B necessary to
keep the pipe from rotating. P3.116 The centrifugal pump of Fig. P3.116 has a flow rate Q and exits the impeller at an angle 2 relative to the blades, as shown. The fluid enters axially at section 1.
Assuming incompressible flow at shaft angular velocity , derive a formula for the power P required to drive the impeller.
P3.117 P3.118 Reverse the flow in Fig. P3.116, so that the system operates as a radial-inflow turbine. Assuming that the outflow into section 1 has no tangential velocity, derive an expression for
the power P extracted by the turbine.
Problems 201 P3.119 Revisit the turbine cascade system of Prob. 3.78, and derive a formula for the power P delivered, using the angular-momentum theorem of Eq. (3.55). P3.120 A centrifugal pump
impeller delivers 4000 gal/min of water at 20°C with a shaft rotation rate of 1750 r/min. Neglect losses. If r1 6 in, r2 14 in, b1 b2 1.75 in, Vt1 10 ft/s, and Vt2 110 ft/s, compute the absolute
velocities (a) V1 and (b) V2 and (c) the horsepower required. (d) Compare with the ideal horsepower required. P3.121 The pipe bend of Fig. P3.121 has D1 27 cm and D2 13 cm. When water at 20°C flows
through the pipe at 4000 gal/min, p1 194 kPa (gage). Compute the torque required at point B to hold the bend stationary.
Ω 4 ft
150 ft/s 75˚
50 cm
P3.123 C
P3.124 A rotating dishwasher arm delivers at 60°C to six nozzles, as in Fig. P3.124. The total flow rate is 3.0 gal/min. Each nozzle has a diameter of 136 in. If the nozzle flows are equal and
friction is neglected, estimate the steady rotation rate of the arm, in r/min.
V2 , p2 = pa 50 cm
2 B
5 in
V1, p1
*P3.122 Extend Prob. 3.46 to the problem of computing the center of pressure L of the normal face Fn, as in Fig. P3.122. (At the center of pressure, no moments are required to hold the plate at
rest.) Neglect friction. Express your result in terms of the sheet thickness h1 and the angle between the plate and the oncoming jet 1. V h2
ρ, V h1
L F n
5 in
6 in
P3.124 *P3.125 A liquid of density flows through a 90° bend as shown in Fig. P3.125 and issues vertically from a uniformly porous section of length L. Neglecting pipe and liquid weight, derive an
expression for the torque M at point 0 required to hold the pipe stationary. R
L Vw x
0 d 90° (Forward-curved)
β 2 = 90° (Radial blades) Head H
Fig. 11.5 Theoretical effect of blade exit angle on pump head versus discharge.
β 2 < 90° (Backward-curved)
Discharge Q
Chapter 11 Turbomachinery
The measured shutoff head of centrifugal pumps is only about 60 percent of the theoretical value H0 2r 22 /g. With the advent of the laser-doppler anemometer, researchers can now make detailed
three-dimensional flow measurements inside pumps and can even animate the data into a movie [40]. The positive-slope condition in Fig. 11.5 can be unstable and can cause pump surge, an oscillatory
condition where the pump “hunts’’ for the proper operating point. Surge may cause only rough operation in a liquid pump, but it can be a major problem in gascompressor operation. For this reason a
backward-curved or radial blade design is generally preferred. A survey of the problem of pump stability is given by Greitzer [41].
11.3 Pump Performance Curves and Similarity Rules
Since the theory of the previous section is rather qualitative, the only solid indicator of a pump’s performance lies in extensive testing. For the moment let us discuss the centrifugal pump in
particular. The general principles and the presentation of data are exactly the same for mixed-flow and axial-flow pumps and compressors. Performance charts are almost always plotted for constant
shaft-rotation speed n (in r/min usually). The basic independent variable is taken to be discharge Q (in gal/min usually for liquids and ft3/min for gases). The dependent variables, or “output,’’ are
taken to be head H (pressure rise p for gases), brake horsepower (bhp), and efficiency . Figure 11.6 shows typical performance curves for a centrifugal pump. The head is approximately constant at low
discharge and then drops to zero at Q Qmax. At this speed and impeller size, the pump cannot deliver any more fluid than Qmax. The positive-slope part of the head is shown dashed; as mentioned
earlier, this region can be unstable and can cause hunting for the operating point.
Positive slope may be unstable for certain system loss curves
Best efficiency point (BEP) or design point
Effect of cavitation or gas entrainment on liquid heads
Fig. 11.6 Typical centrifugal pump performance curves at constant impeller-rotation speed. The units are arbitrary.
Q* Flow rate Q
11.3 Pump Performance Curves and Similarity Rules
The efficiency is always zero at no flow and at Qmax, and it reaches a maximum, perhaps 80 to 90 percent, at about 0.6Qmax. This is the design flow rate Q* or best efficiency point (BEP), max. The
head and horsepower at BEP will be termed H* and P* (or bhp*), respectively. It is desirable that the efficiency curve be flat near max, so that a wide range of efficient operation is achieved.
However, some designs simply do not achieve flat efficiency curves. Note that is not independent of H and P but rather is calculated from the relation in Eq. (11.5), gQH/P. As shown in Fig. 11.6, the
horsepower required to drive the pump typically rises monotonically with the flow rate. Sometimes there is a large power rise beyond the BEP, especially for radial-tipped and forward-curved blades.
This is considered undesirable because a much larger motor is then needed to provide high flow rates. Backward-curved blades typically have their horsepower level off above BEP (“nonoverloading’’
type of curve).
Measured Performance Curves
Figure 11.7 shows actual performance data for a commercial centrifugal pump. Figure 11.7a is for a basic casing size with three different impeller diameters. The head curves H(Q) are shown, but the
horsepower and efficiency curves have to be inferred from the contour plots. Maximum discharges are not shown, being far outside the normal operating range near the BEP. Everything is plotted raw, of
course [feet, horsepower, gallons per minute (1 U.S. gal 231 in3)] since it is to be used directly by designers. Figure 11.7b is the same pump design with a 20 percent larger casing, a lower speed,
and three larger impeller diameters. Comparing the two pumps may be a little confusing: The larger pump produces exactly the same discharge but only half the horsepower and half the head. This will
be readily understood from the scaling or similarity laws we are about to formulate. A point often overlooked is that raw curves like Fig. 11.7 are strictly applicable to a fluid of a certain density
and viscosity, in this case water. If the pump were used to deliver, say, mercury, the brake horsepower would be about 13 times higher while Q, H, and would be about the same. But in that case H
should be interpreted as feet of mercury, not feet of water. If the pump were used for SAE 30 oil, all data would change (brake horsepower, Q, H, and ) due to the large change in viscosity (Reynolds
number). Again this should become clear with the similarity rules.
Net Positive-Suction Head
In the top of Fig. 11.7 is plotted the net positive-suction head (NPSH), which is the head required at the pump inlet to keep the liquid from cavitating or boiling. The pump inlet or suction side is
the low-pressure point where cavitation will first occur. The NPSH is defined as p pi V 2i NPSH g 2g g
where pi and Vi are the pressure and velocity at the pump inlet and p is the vapor pressure of the liquid. Given the left-hand side, NPSH, from the pump performance curve, we must ensure that the
right-hand side is equal or greater in the actual system to avoid cavitation.
Chapter 11 Turbomachinery
n = 1170 r/min
32-in dia.
NPSH, ft
87 %
36 34 -in dia.
87 % 88 %
Total head, ft
28-in dia.
U.S. gallons per minute × 1000 (a) n = 710 r/min
38-in dia.
88 %
% 88 %
Total head, ft
72% 80%
41 12 -in
35-in dia.
% 86 84
Fig. 11.7 Measured-performance curves for two models of a centrifugal water pump: (a) basic casing with three impeller sizes; (b) 20 percent larger casing with three larger impellers at slower speed.
(Courtesy of Ingersoll-Rand Corporation, Cameron Pump Division.)
NPSH, ft
bh p
U.S. gallons per minute × 1000 (b)
If the pump inlet is placed at a height Zi above a reservoir whose free surface is at pressure pa, we can use Bernoulli’s equation to rewrite NPSH as pa pv NPSH Zi hfi (11.20) g g where hf i is the
friction-head loss between the reservoir and the pump inlet. Knowing pa and hf i, we can set the pump at a height Zi which will keep the right-hand side greater than the “required’’ NPSH plotted in
Fig. 11.7.
11.3 Pump Performance Curves and Similarity Rules
If cavitation does occur, there will be pump noise and vibration, pitting damage to the impeller, and a sharp dropoff in pump head and discharge. In some liquids this deterioration starts before
actual boiling, as dissolved gases and light hydrocarbons are liberated.
Deviations from Ideal Pump Theory
The actual pump head data in Fig. 11.7 differ considerably from ideal theory, Eq. (11.18). Take, e.g., the 36.75-in-diameter pump at 1170 r/min in Fig. 11.7a. The theoretical shutoff head is
2r 22 [1170(2/60) rad/s]2[(36.75/2)/(12) ft]2 H0(ideal) 1093 ft g 32.2 ft/s2 From Fig. 11.7a, at Q 0, we read the actual shutoff head to be only 670 ft, or 61 percent of the theoretical value. This
is a sharp dropoff and is indicative of nonrecoverable losses of three types: 1. Impeller recirculation loss, significant only at low flow rates 2. Friction losses on the blade and passage surfaces,
which increase monotonically with the flow rate 3. “Shock’’ loss due to mismatch between the blade angles and the inlet flow direction, especially significant at high flow rates These are complicated
three-dimensional-flow effects and hence are difficult to predict. Although, as mentioned, numerical (CFD) techniques are becoming more important [42], modern performance prediction is still a blend
of experience, empirical correlations, idealized theory, and CFD modifications [45]. EXAMPLE 11.2 The 32-in pump of Fig. 11.7a is to pump 24,000 gal/min of water at 1170 r/min from a reservoir whose
surface is at 14.7 lbf/in2 absolute. If head loss from reservoir to pump inlet is 6 ft, where should the pump inlet be placed to avoid cavitation for water at (a) 60°F, p 0.26 lbf/in2 absolute, SG
1.0 and (b) 200°F, p 11.52 lbf/in2 absolute, SG 0.9635?
Solution Part (a)
For either case read from Fig. 11.7a at 24,000 gal/min that the required NPSH is 40 ft. For this case g 62.4 lbf/ft3. From Eq. (11.20) it is necessary that pa p NPSH Zi hf i g or
(14.7 0.26)(144) 40 ft Zi 6.0 62.4
Zi 27.3 40 12.7 ft
Ans. (a)
The pump must be placed at least 12.7 ft below the reservoir surface to avoid cavitation.
Part (b)
For this case g 62.4(0.9635) 60.1 lbf/ft3. Equation (11.20) applies again with the higher p (14.7 11.52)(144) 40 ft Zi 6.0 60.1
Chapter 11 Turbomachinery Zi 1.6 40 38.4 ft
Ans. (b)
The pump must now be placed at least 38.4 ft below the reservoir surface. These are unusually stringent conditions because a large, high-discharge pump requires a large NPSH.
Dimensionless Pump Performance
For a given pump design, the output variables H and brake horsepower should be dependent upon discharge Q, impeller diameter D, and shaft speed n, at least. Other possible parameters are the fluid
density , viscosity , and surface roughness . Thus the performance curves in Fig. 11.7 are equivalent to the following assumed functional relations:2 gH f1(Q, D, n, , , )
bhp f2(Q, D, n, , , )
This is a straightforward application of dimensional-analysis principles from Chap. 5. As a matter of fact, it was given as an exercise (Prob. 5.20). For each function in Eq. (11.21) there are seven
variables and three primary dimensions (M, L, and T); hence we expect 7 3 4 dimensionless pis, and that is what we get. You can verify as an exercise that appropriate dimensionless forms for Eqs.
(11.21) are gH Q nD2 , , 2 2 g1 nD nD3 D
bhp Q nD2 , , 3 5 g2 n D nD3 D
The quantities nD2/ and /D are recognized as the Reynolds number and roughness ratio, respectively. Three new pump parameters have arisen: Q Capacity coefficient CQ 3 nD gH Head coefficient CH n2D2
bhp Power coefficient CP n3D5 Note that only the power coefficient contains fluid density, the parameters CQ and CH being kinematic types. Figure 11.7 gives no warning of viscous or roughness
effects. The Reynolds numbers are from 0.8 to 1.5 107, or fully turbulent flow in all passages probably. The roughness is not given and varies greatly among commercial pumps. But at such high
Reynolds numbers we expect more or less the same percentage effect on all these pumps. Therefore it is common to assume that the Reynolds number and the roughness ratio have a constant effect, so
that Eqs. (11.23) reduce to, approximately, CH CH(CQ) 2
CP CP(CQ)
We adopt gH as a variable instead of H for dimensional reasons.
11.3 Pump Performance Curves and Similarity Rules
For geometrically similar pumps, we expect head and power coefficients to be (nearly) unique functions of the capacity coefficient. We have to watch out that the pumps are geometrically similar or
nearly so because (1) manufacturers put different-sized impellers in the same casing, thus violating geometric similarity, and (2) large pumps have smaller ratios of roughness and clearances to
impeller diameter than small pumps. In addition, the more viscous liquids will have significant Reynolds-number effects; e.g., a factor-of-3 or more viscosity increase causes a clearly visible effect
on CH and CP. The efficiency is already dimensionless and is uniquely related to the other three. It varies with CQ also CHCQ (CQ) CP
We can test Eqs. (11.24) and (11.25) from the data of Fig. 11.7. The impeller diameters of 32 and 38 in are approximately 20 percent different in size, and so their ratio of impeller to casing size
is the same. The parameters CQ, CH, and CP are computed with n in r/s, Q in ft3/s (gal/min 2.23 103), H and D in ft, g 32.2 ft/s2, and brake horsepower in horsepower times 550 ft lbf/(s hp). The
nondimensional data are then plotted in Fig. 11.8. A dimensionless suction-head coefficient is also defined g(NPSH) CHS CHS(CQ) n2D2
0.9 0.8 η D = 38 in D = 32 in
0.7 0.6
6 CH 5
Fig. 11.8 Nondimensional plot of the pump performance data from Fig. 11.7. These numbers are not representative of other pump designs.
0.7 C P 0.6
0.5 CHS
0.15 CQ
0.4 0.3 0.25
Chapter 11 Turbomachinery
The coefficients CP and CHS are seen to correlate almost perfectly into a single function of CQ, while and CH data deviate by a few percent. The last two parameters are more sensitive to slight
discrepancies in model similarity; since the larger pump has smaller roughness and clearance ratios and a 40 percent larger Reynolds number, it develops slightly more head and is more efficient. The
overall effect is a resounding victory for dimensional analysis. The best-efficiency point in Fig. 11.8 is approximately CQ* 0.115
max 0.88:
CP* 0.65 (11.27)
CH* 5.0
CHS* 0.37
These values can be used to estimate the BEP performance of any size pump in this geometrically similar family. In like manner, the shutoff head is CH(0) 6.0, and by extrapolation the shutoff power
is CP(0) 0.25 and the maximum discharge is CQ,max 0.23. Note, however, that Fig. 11.8 gives no reliable information about, say, the 28- or 35-in impellers in Fig. 11.7, which have a different
impeller-to-casing-size ratio and thus must be correlated separately. By comparing values of n2D2, nD3, and n3D5 for two pumps in Fig. 11.7 we can see readily why the large pump had the same
discharge but less power and head:
Fig. 11.7a Fig. 11.7b Ratio
D, ft
n, r/s
Discharge nD3, ft3/s
Head n2D2/g, ft
Power n3D5/550, hp
32/12 38/12 —
1170/60 710/60 —
370 376 1.02
0.84 0.44 0.52
3527 1861 0.53
Discharge goes as nD3, which is about the same for both pumps. Head goes as n2D2 and power as n3D5 for the same (water), and these are about half as much for the larger pump. The NPSH goes as n2D2
and is also half as much for the 38-in pump.
EXAMPLE 11.3 A pump from the family of Fig. 11.8 has D 21 in and n 1500 r/min. Estimate (a) discharge, (b) head, (c) pressure rise, and (d) brake horsepower of this pump for water at 60°F and best
Solution Part (a)
In BG units take D 21/12 1.75 ft and n 1500/60 25 r/s. At 60°F, of water is 1.94 slugs/ft3. The BEP parameters are known from Fig. 11.8 or Eqs. (11.27). The BEP discharge is thus Q* CQ*nD3 0.115(25)
(1.75)3 (15.4 ft3/s)(448.8) 6900 gal/min
Part (b)
Ans. (a)
Similarly, the BEP head is 5.0(25)2(1.75)2 CH*n2D2 H* 300-ft water g 32.2
Ans. (b)
11.3 Pump Performance Curves and Similarity Rules
Part (c)
Since we are not given elevation or velocity-head changes across the pump, we neglect them and estimate p gH 1.94(32.2)(300) 18,600 lbf/ft2 129 lbf/in2
Part (d)
Ans. (c)
Finally, the BEP power is P* C P*n3D5 0.65(1.94)(25)3(1.75)5 323,000 ft lbf/s 590 hp 550
Ans. (d)
EXAMPLE 11.4 We want to build a pump from the family of Fig. 11.8, which delivers 3000 gal/min water at 1200 r/min at best efficiency. Estimate (a) the impeller diameter, (b) the maximum discharge,
(c) the shutoff head, and (d) the NPSH at best efficiency.
Part (a)
3000 gal/min 6.68 ft3/s
1200 r/min 20 r/s. At BEP we have
Q* CQ*nD 6.68 ft3/s (0.115)(20)D3 3
Part (b)
6.68 D 0.115(20)
1.43 ft 17.1 in
Ans. (a)
The maximum Q is related to Q* by a ratio of capacity coefficients Q*CQ,max 3000(0.23) 6000 gal/min Qmax CQ* 0.115
Part (c)
From Fig. 11.8 we estimated the shutoff head coefficient to be 6.0. Thus 6.0(20)2(1.43)2 CH(0)n2D2 H(0) 152 ft g 32.2
Part (d)
Ans. (b)
Ans. (c)
Finally, from Eq. (11.27), the NPSH at BEP is approximately CHS*n2D2 0.37(20)2(1.43)2 NPSH* 9.4 ft g 32.2
Ans. (d)
Since this a small pump, it will be less efficient than the pumps in Fig. 11.8, probably about 85 percent maximum.
Similarity Rules
The success of Fig. 11.8 in correlating pump data leads to simple rules for comparing pump performance. If pump 1 and pump 2 are from the same geometric family and are operating at homologous points
(the same dimensionless position on a chart such as Fig. 11.8), their flow rates, heads, and powers will be related as follows:
Q2 n2 D2 Q1 n1 D1
H2 n2 H1 n1
D2 D1
Chapter 11 Turbomachinery
P2 2 n2 P1 1 n1
D D2
These are the similarity rules, which can be used to estimate the effect of changing the fluid, speed, or size on any dynamic turbomachine—pump or turbine—within a geometrically similar family. A
graphic display of these rules is given in Fig. 11.9, showing the effect of speed and diameter changes on pump performance. In Fig. 11.9a the size is held constant and the speed is varied 20 percent,
while Fig. 11.9b shows a 20 percent size change at constant speed. The curves are plotted to scale but with arbitrary units. The speed effect (Fig. 11.9a) is substantial, but the size effect (Fig.
11.9b) is even more dramatic, especially for power, which varies as D5. Generally we see that a given pump family can be adjusted in size and speed to fit a variety of system characteristics.
Strictly speaking, we would expect for perfect similarity that 1 2, but we have seen that larger pumps are more efficient, having a higher Reynolds number and lower roughness and clearance ratios.
Two empirical correlations are recommended for maximum efficiency. One, developed by Moody [43] for turbines but also used for pumps, is a size effect. The other, suggested by Anderson [44] from
thousands of pump tests, is a flow-rate effect: Size changes [43]: Flow-rate changes [44]:
1 2 D1 1 1 D2
Q 0.94 0.94 Q
(11.29a) 0.32
Anderson’s formula (11.29b) makes the practical observation that even an infinitely large pump will have losses. He thus proposes a maximum possible efficiency of 94 percent, rather than 100 percent.
Anderson recommends that the same formula be used
D = 10 = constant
bhp D = 12
n = 12 bhp
n = 10
H, bhp
H, bhp
n = 10 = constant
D = 10
Fig. 11.9 Effect of changes in size and speed on homologous pump performance: (a) 20 percent change in speed at constant size; (b) 20 percent change in size at constant speed.
n=8 0
0 Q
11.4 Mixed- and Axial-Flow Pumps: The Specific Speed
for turbines if the constant 0.94 is replaced by 0.95. The formulas in Eq. (11.29) assume the same value of surface roughness for both machines—one could micropolish a small pump and achieve the
efficiency of a larger machine.
Effect of Viscosity
Centrifugal pumps are often used to pump oils and other viscous liquids up to 1000 times the viscosity of water. But the Reynolds numbers become low turbulent or even laminar, with a strong effect on
performance. Figure 11.10 shows typical test curves of head and brake horsepower versus discharge. High viscosity causes a dramatic drop in head and discharge and increases in power requirements. The
efficiency also drops substantially according to the following typical results: /water
max, %
Beyond about 300water the deterioration in performance is so great that a positivedisplacement pump is recommended.
11.4 Mixed- and Axial-Flow Pumps: The Specific Speed
We have seen from the previous section that the modern centrifugal pump is a formidable device, able to deliver very high heads and reasonable flow rates with excellent efficiency. It can match many
system requirements. But basically the centrifugal pump is a high-head, low-flow machine, whereas there are many applications requiring low head and high discharge. To see that the centrifugal design
is not convenient for such systems, consider the following example.
EXAMPLE 11.5 We want to use a centrifugal pump from the family of Fig. 11.8 to deliver 100,000 gal/min of water at 60°F with a head of 25 ft. What should be (a) the pump size and speed and (b) brake
horsepower, assuming operation at best efficiency?
Solution Part (a)
Enter the known head and discharge into the BEP parameters from Eq. (11.27): 2
CH*n D 5.0n2D2 H* 25 ft 32.2 g Q* 100,000 gal/min 222.8 ft3/s CQ*nD3 0.115nD3 The two unknowns are n and D. Solve simultaneously for
D 12.4 ft
n 1.03 r/s 62 r/min
Ans. (a)
If you wish to avoid algebraic manipulation, simply program the above two simultaneous equations in EES, using English units: 25 5.0*n^2*D^2/32.2 222.8 0.115*n*D^3
Chapter 11 Turbomachinery
1.0 0.9
Ns 0
Axial flow
1.0 bhp
µ water = 10.0
Centrifugal pump
Mixed flow
H, bhp
η max µ
r/min (gal/min)1/2/(H, ft)3/4
Fig. 11.10 Effect of viscosity on centrifugal pump performance.
Specific speed Low
Fig. 11.11 (a) Optimum efficiency and (b) vane design of dynamicpump families as a function of specific speed.
Centrifugal 500
Mixed-flow 2000
5000 10,000–15,000
Specify in Variable Information that n and D are positive, and EES promptly returns the correct solution: D 12.36 ft and n 1.027 r/s.
Part (b)
The most efficient horsepower is then, from Eq. (11.27), 0.65(1.94)(1.03)3(12.4)5 bhp* CP*n3D5 720 hp 550
Ans. (b)
The solution to Example 11.5 is mathematically correct but results in a grotesque pump: an impeller more than 12 ft in diameter, rotating so slowly one can visualize oxen walking in a circle turning
the shaft. There are other dynamic-pump designs which do provide low head and high discharge. For example, there is a type of 38-in, 710 r/min pump, e.g., with the same input parameters as Fig.
11.7b, which will deliver the 25-ft head and 100,000 gal/min flow rate called for in Example 11.5. This is done by allowing the flow to pass through the impeller with an axial-flow component and less
centrifugal component. The passages can be opened up to the increased flow rate with very little size increase, but the drop in radial outlet velocity decreases the head produced. These are the
mixed-flow (part radial, part axial) and axialflow (propeller-type) families of dynamic pump. Some vane designs are sketched in Fig. 11.11, which introduces an interesting new “design’’ parameter,
the specific speed Ns or Ns.
11.4 Mixed- and Axial-Flow Pumps: The Specific Speed
The Specific Speed
Most pump applications involve a known head and discharge for the particular system, plus a speed range dictated by electric motor speeds or cavitation requirements. The designer then selects the
best size and shape (centrifugal, mixed, axial) for the pump. To help this selection, we need a dimensionless parameter involving speed, discharge, and head but not size. This is accomplished by
eliminating the diameter between CQ and CH, applying the result only to the BEP. This ratio is called the specific speed and has both a dimensionless form and a somewhat lazy, practical form:
Rigorous form:
C1/2 n(Q*)1/2 Q* Ns C3/4 (gH*)3/4 H*
Lazy but common:
(r/min)(gal/min)1/2 Ns [H (ft)]3/4
In other words, practicing engineers do not bother to change n to revolutions per second or Q* to cubic feet per second or to include gravity with head, although the latter would be necessary for,
say, a pump on the moon. The conversion factor is Ns 17,182Ns Note that Ns is applied only to BEP; thus a single number characterizes an entire family of pumps. For example, the family of Fig. 11.8
has Ns (0.115)1/2/(5.0)3/4 0.1014, Ns 1740, regardless of size or speed. It turns out that the specific speed is directly related to the most efficient pump design, as shown in Fig. 11.11. Low Ns
means low Q and high H, hence a centrifugal pump, and large Ns implies an axial pump. The centrifugal pump is best for Ns between 500 and 4000, the mixed-flow pump for Ns between 4000 and 10,000, and
the axialflow pump for Ns above 10,000. Note the changes in impeller shape as Ns increases.
Suction Specific Speed
If we use NPSH rather than H in Eq. (11.30), the result is called suction specific speed Rigorous:
nQ1/2 Nss (g NPSH)3/4
(r/min)(gal/min)1/2 Nss [NPSH (ft)]3/4
where NPSH denotes the available suction head of the system. Data from Wislicenus [4] show that a given pump is in danger of inlet cavitation if Nss 0.47
Nss 8100
In the absence of test data, this relation can be used, given n and Q, to estimate the minimum required NPSH.
Axial-Flow Pump Theory
A multistage axial-flow geometry is shown in Fig. 11.12a. The fluid essentially passes almost axially through alternate rows of fixed stator blades and moving rotor blades. The incompressible-flow
assumption is frequently used even for gases, because the pressure rise per stage is usually small.
Chapter 11 Turbomachinery Stator
r Rotor
ω, n (a)
α1 Stator w1
Vn 1
Vt 1
u (b) Rotor u = rω
Fig. 11.12 Analysis of an axialflow pump: (a) basic geometry; (b) stator blades and exit-velocity diagram; (c) rotor blades and exitvelocity diagram.
u (c)
The simplified vector-diagram analysis assumes that the flow is one-dimensional and leaves each blade row at a relative velocity exactly parallel to the exit blade angle. Figure 11.12b shows the
stator blades and their exit-velocity diagram. Since the stator is fixed, ideally the absolute velocity V1 is parallel to the trailing edge of the blade. After vectorially subtracting the rotor
tangential velocity u from V1, we obtain the velocity w1 relative to the rotor, which ideally should be parallel to the rotor leading edge. Figure 11.12c shows the rotor blades and their
exit-velocity diagram. Here the relative velocity w2 is parallel to the blade trailing edge, while the absolute velocity V2 should be designed to enter smoothly the next row of stator blades. The
theoretical power and head are given by Euler’s turbine relation (11.11). Since there is no radial flow, the inlet and exit rotor speeds are equal, u1 u2, and onedimensional continuity requires that
the axial-velocity component remain constant Q Vn1 Vn2 Vn const A
11.4 Mixed- and Axial-Flow Pumps: The Specific Speed
From the geometry of the velocity diagrams, the normal velocity (or volume flow) can be directly related to the blade rotational speed u: u rav Vn1(tan 1 tan 1) Vn2(tan 2 tan 2)
Thus the flow rate can be predicted from the rotational speed and the blade angles. Meanwhile, since Vt1 Vn1 cot 1 and Vt2 u Vn2 cot 2, Euler’s relation (11.11) for the pump head becomes gH uVn(cot 2
cot 1) u2 uVn(cot 1 cot 2)
the preferred form because it relates to the blade angles 1 and 2. The shutoff or noflow head is seen to be H0 u2/g, just as in Eq. (11.18) for a centrifugal pump. The blade-angle parameter cot 1 cot
2 can be designed to be negative, zero, or positive, corresponding to a rising, flat, or falling head curve, as in Fig. 11.5. Strictly speaking, Eq. (11.33) applies only to a single streamtube of
radius r, but it is a good approximation for very short blades if r denotes the average radius. For long blades it is customary to sum Eq. (11.33) in radial strips over the blade area. Such
complexity may not be warranted since theory, being idealized, neglects losses and usually predicts the head and power larger than those in actual pump performance.
Performance of an Axial-Flow Pump
At high specific speeds, the most efficient choice is an axial-flow, or propeller, pump, which develops high flow rate and low head. A typical dimensionless chart for a propeller pump is shown in
Fig. 11.13. Note, as expected, the higher CQ and lower CH compared with Fig. 11.8. The head curve drops sharply with discharge, so that a large system-head change will cause a mild flow change. The
power curve drops with head also, which means a possible overloading condition if the system discharge should sud1.0
η 0.8 3
CH, CP
2 0.4 CH 1
Fig. 11.13 Dimensionless performance curves for a typical axialflow pump, Ns 12,000. Constructed from data given by Stepanoff [8] for a 14-in pump at 690 r/min.
0.2 CP 0
0.4 CQ
Chapter 11 Turbomachinery
1.0 10,000 0.8
η max 10
Q = 5 gal/min 0.2
Fig. 11.14 Optimum efficiency of pumps versus capacity and specific speed. (Adapted from Refs. 4 and 31.)
denly decrease. Finally, the efficiency curve is rather narrow and triangular, as opposed to the broad, parabolic-shaped centrifugal pump efficiency (Fig. 11.8). By inspection of Fig. 11.13, CQ*
0.55, CH* 1.07, CP* 0.70, and max 0.84. From this we compute Ns (0.55)1/2/(1.07)3/4 0.705, Ns 12,000. The relatively low efficiency is due to small pump size: d 14 in, n 690 r/min, Q* 4400 gal/min. A
repetition of Example 11.5 using Fig. 11.13 would show that this propeller pump family can provide a 25-ft head and 100,000 gal/min discharge if D 46 in and n 430 r/min, with bhp 750; this is a much
more reasonable design solution, with improvements still possible at larger-Ns conditions.
Pump Performance versus Specific Speed
Specific speed is such an effective parameter that it is used as an indicator of both performance and efficiency. Figure 11.14 shows a correlation of the optimum efficiency of a pump as a function of
the specific speed and capacity. Because the dimensional parameter Q is a rough measure of both size and Reynolds number, increases with Q. When this type of correlation was first published by
Wislicenus [4] in 1947, it became known as the pump curve, a challenge to all manufacturers. We can check that the pumps of Figs. 11.7 and 11.13 fit the correlation very well. Figure 11.15 shows the
effect of specific speed on the shape of the pump performance curves, normalized with respect to the BEP point. The numerical values shown are representative but somewhat qualitative. The
high-specific-speed pumps (Ns 10,000) have head and power curves which drop sharply with discharge, implying overload or start-up problems at low flow. Their efficiency curve is very narrow.
11.5 Matching Pumps to System Characteristics 735
2 H H*
3 Ns = 10,000
Ns = 10,000
Ns = 10,000
bhp bhp*
Fig. 11.15 Effect of specific speed on pump performance curves.
1 Q Q*
1 Q Q*
1 Q Q*
A low-specific-speed pump (Ns 600) has a broad efficiency curve, a rising power curve, and a head curve which “droops’’ at shutoff, implying possible surge or hunting problems.
Computational Fluid Dynamics
The design of turbomachinery has traditionally been highly experimental, with simple theories, such as in Sec. 11.2, only able to predict trends. Dimensionless correlations, such as Fig. 11.15, are
useful but require extensive experimentation. Consider that flow in a pump is three-dimensional; unsteady (both periodic and turbulent); and involves flow separation, recirculation in the impeller,
unsteady blade wakes passing through the diffuser, and blade roots, tips, and clearances. It is no wonder that one-dimensional theory cannot give firm quantitative predictions. Modern computer
analysis can give realistic results and is becoming a useful tool for turbomachinery designers. A good example is Ref. 56, reporting combined experimental and computational results for a centrifugal
pump diffuser. A photograph of the device is shown in Fig. 11.16a. It is made of clear Perspex, so that laser measurements of particle tracking velocimetry (LPTV) and doppler anemometry (LDA) could
be taken throughout the system. The data were compared with a CFD simulation of the impeller and diffuser, using the grids shown in Fig. 11.16b. The computations used a turbulence formulation called
the k- model, popular in commercial CFD codes (see Sec. 8.9). Results were good but not excellent. The CFD model predicted velocity and pressure data adequately up until flow separation, after which
it was only qualitative. Clearly, CFD is developing a significant role in turbomachinery design.
11.5 Matching Pumps to System Characteristics
The ultimate test of a pump is its match with the operating-system characteristics. Physically, the system head must match the head produced by the pump, and this intersection should occur in the
region of best efficiency. The system head will probably contain a static-elevation change z2 z1 plus friction losses in pipes and fittings V2 Hsys (z2 z1) 2g
D K fL
Chapter 11 Turbomachinery
Fig. 11.16 Turbomachinery design now involves both experimentation and computational fluid dynamics (CFD): (a) a centrifugal impeller and diffuser (Courtesy of K. Eisele and Z. Zhang, Sulzer Innotec
Ltd.); (b) a three-dimensional CFD model grid for this system. (From Ref. 56 by permission of the American Society of Mechanical Engineers.)
Diffuser (b)
where K denotes minor losses and V is the flow velocity in the principal pipe. Since V is proportional to the pump discharge Q, Eq. (11.34) represents a system-head curve Hs(Q). Three examples are
shown in Fig. 11.17: a static head Hs a, static head plus laminar friction Hs a bQ, and static head plus turbulent friction Hs a cQ2. The intersection of the system curve with the pump performance
curve H(Q) defines
11.5 Matching Pumps to System Characteristics 737
Pump curves
Pump η (Q)
3 Turbulent friction
Η, η
Laminar friction Static head Pump H(Q)
System curves H(Q)
Fig. 11.17 Illustration of pump operating points for three types of system-head curves.
Operating points
the operating point. In Fig. 11.17 the laminar-friction operating point is at maximum efficiency while the turbulent and static curves are off design. This may be unavoidable if system variables
change, but the pump should be changed in size or speed if its operating point is consistently off design. Of course, a perfect match may not be possible because commercial pumps have only certain
discrete sizes and speeds. Let us illustrate these concepts with an example.
EXAMPLE 11.6 We want to use the 32-in pump of Fig. 11.7a at 1170 r/min to pump water at 60°F from one reservoir to another 120 ft higher through 1500 ft of 16-in-ID pipe with friction factor f 0.030.
(a) What will the operating point and efficiency be? (b) To what speed should the pump be changed to operate at the BEP?
Solution Part (a)
For reservoirs the initial and final velocities are zero; thus the system head is V2 fL V2 0.030(1500 ft) Hs z2 z1 120 ft 16 ft 2g D 2g 12 From continuity in the pipe, V Q/A Q/[ 14 ( 11 62 ft)2], and
so we substitute for V above to get Hs 120 0.269Q2
Q in ft3/s
Since Fig. 11.7a uses thousands of gallons per minute for the abscissa, we convert Q in Eq. (1) to this unit: Hs 120 1.335Q2
Q in 103 gal/min
We can plot Eq. (2) on Fig. 11.7a and see where it intersects the 32-in pump-head curve, as in Fig. E11.6. A graphical solution gives approximately H 430 ft
Q 15,000 gal/min
Chapter 11 Turbomachinery
490 ft
Hpump Operating point
430 ft
120 ft
15,000 gal/min
The efficiency is about 82 percent, slightly off design. An analytic solution is possible if we fit the pump-head curve to a parabola, which is very accurate Hpump 490 0.26Q2
Q in 103 gal/min
Equations (2) and (3) must match at the operating point: 490 0.26Q2 120 1.335Q2 or
Part (b)
490 120 Q2 232 0.26 1.335 Q 15.2 103 gal/min 15,200 gal/min
Ans. (a)
H 490 0.26(15.2)2 430 ft
Ans. (a)
To move the operating point to BEP, we change n, which changes both Q n and H n2. From Fig. 11.7a, at BEP, H* 386 ft; thus for any n, H* 386(n/1170)2. Also read Q* 20 103 gal/min; thus for any n, Q*
20(n/1170). Match H* to the system characteristics, Eq. (2), n H* 386 1170
n 120 1.335 20 1170
Ans. (b)
which gives n2 0. Thus it is impossible to operate at maximum efficiency with this particular system and pump.
Pumps Combined in Parallel
If a pump provides the right head but too little discharge, a possible remedy is to combine two similar pumps in parallel, i.e., sharing the same suction and inlet conditions. A parallel arrangement
is also used if delivery demand varies, so that one pump is used at low flow and the second pump is started up for higher discharges. Both pumps should have check valves to avoid backflow when one is
shut down. The two pumps in parallel need not be identical. Physically, their flow rates will sum for the same head, as illustrated in Fig. 11.18. If pump A has more head than pump B, pump B cannot
be added in until the operating head is below the shutoff head of pump B. Since the system curve rises with Q, the combined delivery QAB will be less than the separate operating discharges QA QB but
certainly greater than either one.
11.5 Matching Pumps to System Characteristics 739
H Pump A Pump B
Combined in parallel System curve
Fig. 11.18 Performance and operating points of two pumps operating singly and combined in parallel.
B A Operating points
For a very flat (static) curve two similar pumps in parallel will deliver nearly twice the flow. The combined brake horsepower is found by adding brake horsepower for each of pumps A and B at the
same head as the operating point. The combined efficiency equals g(QAB)(HAB)/(550 bhpAB). If pumps A and B are not identical, as in Fig. 11.18, pump B should not be run and cannot even be started up
if the operating point is above its shutoff head.
Pumps Combined in Series
If a pump provides the right discharge but too little head, consider adding a similar pump in series, with the output of pump B fed directly into the suction side of pump A. As sketched in Fig.
11.19, the physical principle for summing in series is that the two heads add at the same flow rate to give the combined-performance curve. The two H
System curve HB Combined in series
HA Pump A Pump B
Fig. 11.19 Performance of two pumps combined in series.
0 B
A A+B Operating points
Chapter 11 Turbomachinery
need not be identical at all, since they merely handle the same discharge; they may even have different speeds, although normally both are driven by the same shaft. The need for a series arrangement
implies that the system curve is steep, i.e., requires higher head than either pump A or B can provide. The combined operating-point head will be more than either A or B separately but not as great
as their sum. The combined power is the sum of brake horsepower for A and B at the operating point flow rate. The combined efficiency is
g(QAB)(HAB) 550 bhpAB similar to parallel pumps. Whether pumps are used in series or in parallel, the arrangement will be uneconomical unless both pumps are operating near their best efficiency.
Multistage Pumps
For very high heads in continuous operation, the solution is a multistage pump, with the exit of one impeller feeding directly into the eye of the next. Centrifugal, mixedflow, and axial-flow pumps
have all been grouped in as many as 50 stages, with heads up to 8000 ft of water and pressure rises up to 5000 lbf/in2 absolute. Figure 11.20 shows a section of a seven-stage centrifugal propane
compressor which develops 300 lbf/in2 rise at 40,000 ft3/min and 35,000 bhp.
Most of the discussion in this chapter concerns incompressible flow, that is, negligible change in fluid density. Even the pump of Fig. 11.7, which can produce 600 ft of head at 1170 r/min, will only
increase standard air pressure by 46 lbf/ft2, about a 2 percent change in density. The picture changes at higher speeds, p n2, and multiple stages, where very large changes in pressure and density
are achieved. Such devices are called compressors, as in Fig. 11.20. The concept of static head, H p/g, becomes inappropriate, since varies. Compressor performance is measured by (1) the pressure
ratio across the stage p2/p1 and (2) the change in stagnation enthalpy (h02 h01), where h0 h 12 V 2 (see Sec. 9.3). Combining m stages in series results in pfinal/pinitial (p2/p1)m. As density
increases, less area is needed: note the decrease in impeller size from right to left in Fig. 11.20. Compressors may be either of the centrifugal or axialflow type [21 to 23]. Compressor efficiency,
from inlet condition 1 to final outlet f, is defined by the change in gas enthalpy, assuming an adiabatic process: hf h01 Tf T01 comp h0f h01 T0f T01 Compressor efficiencies are similar to hydraulic
machines (max 70 to 80 percent), but the mass-flow range is more limited: on the low side by compressor surge, where blade stall and vibration occur, and on the high side by choking (Sec. 9.4), where
the Mach number reaches 1.0 somewhere in the system. Compressor mass flow is normally plotted using the same type of dimensionless function formulated in Eq. (9.47): m˙(RT0)1/2/(D2p0), which will
reach a maximum when choking occurs. For further details, see Refs. 21 to 23.
11.5 Matching Pumps to System Characteristics 741
Fig. 11.20 Cross section of a seven-stage centrifugal propane compressor which delivers 40,000 ft3/min at 35,000 bhp and a pressure rise of 300 lbf/in2. Note the second inlet at stage 5 and the
varying impeller designs. (Courtesy of DeLaval-Stork V.O.F., Centrifugal Compressor Division.)
EXAMPLE 11.7 Investigate extending Example 11.6 by using two 32-in pumps in parallel to deliver more flow. Is this efficient?
Solution Since the pumps are identical, each delivers 12 Q at the same 1170 r/min speed. The system curve is the same, and the balance-of-head relation becomes H 490 0.26( 12 Q)2 120 1.335Q2 or
490 120 Q2 1.335 0.065
Q 16,300 gal/min
This is only 7 percent more than a single pump. Each pump delivers 12 Q 8130 gal/min, for which the efficiency is only 60 percent. The total brake horsepower required is 3200, whereas a single pump
used only 2000 bhp. This is a poor design.
Chapter 11 Turbomachinery
EXAMPLE 11.8 Suppose the elevation change in Example 11.6 is raised from 120 to 500 ft, greater than a single 32-in pump can supply. Investigate using 32-in pumps in series at 1170 r/min.
Solution Since the pumps are identical, the total head is twice as much and the constant 120 in the system-head curve is replaced by 500. The balance of heads becomes H 2(490 0.26Q2) 500 1.335Q2 or
980 500 Q2 1.335 0.52
Q 16.1 103 gal/min
The operating head is 500 1.335(16.1)2 845 ft, or 97 percent more than that for a single pump in Example 11.5. Each pump is operating at 16.1 103 gal/min, which from Fig. 11.7a is 83 percent
efficient, a pretty good match to the system. To pump at this operating point requires 4100 bhp, or about 2050 bhp for each pump.
11.6 Turbines
A turbine extracts energy from a fluid which possesses high head, but it is fatuous to say a turbine is a pump run backward. Basically there are two types, reaction and impulse, the difference lying
in the manner of head conversion. In the reaction turbine, the fluid fills the blade passages, and the head change or pressure drop occurs within the impeller. Reaction designs are of the
radial-flow, mixed-flow, and axial-flow types and are essentially dynamic devices designed to admit the high-energy fluid and extract its momentum. An impulse turbine first converts the high head
through a nozzle into a highvelocity jet, which then strikes the blades at one position as they pass by. The impeller passages are not fluid-filled, and the jet flow past the blades is essentially at
constant pressure. Reaction turbines are smaller because fluid fills all the blades at one time.
Reaction Turbines
Reaction turbines are low-head, high-flow devices. The flow is opposite that in a pump, entering at the larger-diameter section and discharging through the eye after giving up most of its energy to
the impeller. Early designs were very inefficient because they lacked stationary guide vanes at the entrance to direct the flow smoothly into the impeller passages. The first efficient inward-flow
turbine was built in 1849 by James B. Francis, a U.S. engineer, and all radial- or mixed-flow designs are now called Francis turbines. At still lower heads, a turbine can be designed more compactly
with purely axial flow and is termed a propeller turbine. The propeller may be either fixed-blade or adjustable (Kaplan type), the latter being complicated mechanically but much more efficient at
low-power settings. Figure 11.21 shows sketches of runner designs for Francis radial, Francis mixed-flow, and propeller-type turbines.
Idealized Radial Turbine Theory
The Euler turbomachine formulas (11.11) also apply to energy-extracting machines if we reverse the flow direction and reshape the blades. Figure 11.22 shows a radial turbine runner. Again assume
one-dimensional frictionless flow through the blades. Adjustable inlet guide vanes are absolutely necessary for good efficiency. They bring the inlet flow to the blades at angle 2 and absolute
velocity V2 for minimum “shock’’ or
11.6 Turbines
Nsp = 20
10.0 CH 9.0 CH
0.3 (a) Nsp = 60
CQ CQ 0.2
0.8 η
Fig. 11.21 Reaction turbines: (a) Francis, radial type; (b) Francis mixed-flow; (c) propeller axialflow; (d) performance curves for a Francis turbine, n 600 r/min, D 2.25 ft, Nsp 29.
0.4 0.2
(c) Nsp = 140
CP (d)
directional-mismatch loss. After vectorially adding in the runner tip speed u2 r2, the outer blade angle should be set at angle 2 to accommodate the relative velocity w2, as shown in the figure. (See
Fig. 11.4 for the analogous radial-pump velocity diagrams.) Application of the angular-momentum control-volume theorem, Eq. (3.55), to Fig. 11.22 (see Example 3.14 for a similar case) yields an
idealized formula for the power P extracted by the runner: P T Q(r2Vt2 r1Vt1) Q(u2V2 cos 2 u1V1 cos 1) Adjustable guide vane u 2 = rω 2
Vt 2
α2 β2 Blade w2
Vn 2
V2 u1 r1
α1 Runner
Fig. 11.22 Inlet- and outlet-velocity diagrams for an idealized radialflow reaction turbine runner.
Chapter 11 Turbomachinery
where Vt2 and Vt1 are the absolute inlet and outlet circumferential velocity components of the flow. Note that Eq. (11.35) is identical to Eq. (11.11) for a radial pump, except that the blade shapes
are different. The absolute inlet normal velocity Vn2 V2 sin 2 is proportional to the flow rate Q. If the flow rate changes and the runner speed u2 is constant, the vanes must be adjusted to a new
angle 2 so that w2 still follows the blade surface. Thus adjustable inlet vanes are very important to avoid shock loss.
Power Specific Speed
Turbine parameters are similar to those of a pump, but the dependent variable is the output brake horsepower, which depends upon the inlet flow rate Q, available head H, impeller speed n, and
diameter D. The efficiency is the output brake horsepower divided by the available water horsepower gQH. The dimensionless forms are CQ, CH, and CP, defined just as for a pump, Eqs. (11.23). If we
neglect Reynolds-number and roughness effects, the functional relationships are written with CP as the independent variable: CH CH(CP)
CQ CQ(CP)
bhp (CP) gQH
Figure 11.21d shows typical performance curves for a small Francis radial turbine. The maximum efficiency point is called the normal power, and the values for this particular turbine are
max 0.89
CP* 2.70
CQ* 0.34
CH* 9.03
A parameter which compares the output power with the available head, independent of size, is found by eliminating the diameter between CH and CP. It is called the power specific speed: Rigorous form:
Lazy but common:
C*P1/2 n(bhp)1/2 Nsp 5/4 C*H 1/2(gH)5/4
(r/min)(bhp)1/2 Nsp [H (ft)]5/4
For water, 1.94 slugs/ft3 and Nsp 273.3Nsp. The various turbine designs divide up nicely according to the range of power specific speed, as follows: Turbine type
Nsp range
CH range
Impulse Francis Propeller: Water Gas, steam
1–10 10–110
15–50 5–25
100–250 25–300
1–4 10–80
Note that Nsp, like Ns for pumps, is defined only with respect to the BEP and has a single value for a given turbine family. In Fig. 11.21d, Nsp 273.3(2.70)1/2/(9.03)5/4 29, regardless of size.
11.6 Turbines
Like pumps, turbines of large size are generally more efficient, and Eqs. (11.29) can be used as an estimate when data are lacking. The design of a complete large-scale power-generating turbine
system is a major engineering project, involving inlet and outlet ducts, trash racks, guide vanes, wicket gates, spiral cases, generator with cooling coils, bearings and transmission gears, runner
blades, draft tubes, and automatic controls. Some typical large-scale reaction turbine designs are shown in Fig. 11.23. The reversible pump-and-turbine design of Fig. 11.23d requires special care for
adjustable guide vanes to be efficient for flow in either direction. The largest (1000-MW) hydropower designs are awesome when viewed on a human scale, as shown in Fig. 11.24. The economic advantages
of small-scale model testing are evident from this photograph of the Francis turbine units at Grand Coulee Dam.
Impulse Turbines
For high head and relatively low power, i.e., low Nsp, not only would a reaction turbine require too high a speed but also the high pressure in the runner would require a massive casing thickness.
The impulse turbine of Fig. 11.25 is ideal for this situation. Since Nsp is low, n will be low and the high pressure is confined to the small nozzle, which converts the head to an atmospheric
pressure jet of high velocity Vj. The jet strikes the buckets and imparts a momentum change similar to that in our controlvolume analysis for a moving vane in Example 3.10 or Prob. 3.51. The buckets
have an elliptical split-cup shape, as in Fig. 11.25b. They are named Pelton wheels, after Lester A. Pelton (1829–1908), who produced the first efficient design. From Example 3.10 the force and power
delivered to a Pelton wheel are theoretically F Q(Vj u)(1 cos ) P Fu Qu(Vj u)(1 cos )
where u 2nr is the bucket linear velocity and r is the pitch radius, or distance to the jet centerline. A bucket angle 180° gives maximum power but is physically impractical. In practice, 165°, or 1
cos 1.966 or only 2 percent less than maximum power. From Eq. (11.38) the theoretical power of an impulse turbine is parabolic in bucket speed u and is maximum when dP/du 0, or u* 2n*r 12 Vj
For a perfect nozzle, the entire available head would be converted to jet velocity Vj (2gH)1/2. Actually, since there are 2 to 8 percent nozzle losses, a velocity coefficient C is used Vj C (2gH)1/2
0.92 C 0.98
By combining Eqs. (11.36) and (11.40), the theoretical impulse turbine efficiency becomes
2(1 cos )(C ) where
u peripheral-velocity factor (2gH)1/2
Maximum efficiency occurs at 12 C 0.47.
Chapter 11 Turbomachinery
(a )
(b )
(c )
(d )
Fig. 11.23 Large-scale turbine designs depend upon available head and flow rate and operating conditions: (a) Francis (radial); (b) Kaplan (propeller); (c) bulb mounting with propeller runner; (d)
reversible pump turbine with radial runner. (Courtesy of Allis-Chalmers Fluid Products Company.)
11.6 Turbines
Fig. 11.24 Interior view of the 1.1million hp (820-MW) turbine units on the Grand Coulee Dam of the Columbia River, showing the spiral case, the outer fixed vanes (“stay ring’’), and the inner
adjustable vanes (“wicket gates’’). (Courtesy of Allis-Chalmers Fluid Products Company.)
Split bucket
Vj n,ω
β ≈ 165˚ (b)
Needle valve
Fig. 11.25 Impulse turbine: (a) side view of wheel and jet; (b) top view of bucket; (c) typical velocity diagram.
α Vj
u = 2π nr (a)
Figure 11.26 shows Eq. (11.41) plotted for an ideal turbine ( 180°, Cv 1.0) and for typical working conditions ( 160°, Cv 0.94). The latter case predicts max 85 percent at 0.47, but the actual data
for a 24-in Pelton wheel test are somewhat less efficient due to windage, mechanical friction, backsplashing, and nonuni-
Chapter 11 Turbomachinery
η 0.4
Fig. 11.26 Efficiency of an impulse turbine calculated from Eq. (11.41): solid curve ideal, 180°, Cv 1.0; dashed curve actual, 160°, Cv 0.94; open circles data, Pelton wheel, diameter 2 ft.
u (2gH)1/2
form bucket flow. For this test max 80 percent, and, generally speaking, an impulse turbine is not quite as efficient as the Francis or propeller turbines at their BEPs. Figure 11.27 shows the
optimum efficiency of the three turbine types, and the importance of the power specific speed Nsp as a selection tool for the designer. These efficiencies are optimum and are obtained in careful
design of large machines. The water power available to a turbine may vary due to either net-head or flow-rate changes, both of which are common in field installations such as hydroelectric plants.
The demand for turbine power also varies from light to heavy, and the operating response is a change in the flow rate by adjustment of a gate valve or needle valve (Fig. 11.25a). As shown in Fig.
11.28, all three turbine types achieve fairly uniform efficiency as a function of the level of power being extracted. Especially effective is the adjustable-blade (Kaplan-type) propeller turbine,
while the poorest is a fixed-blade propeller. The term rated power in Fig. 11.28 is the largest power delivery guaranteed by the manufacturer, as opposed to normal power, which is delivered at
maximum efficiency. For further details of design and operation of turbomachinery, the readable and interesting treatment in Ref. 33 is especially recommended. The feasibility of microhydropower is
discussed in [26]. See also Refs. 27 and 28.
1.0 Francis Propeller Impulse
η 0.9
Fig. 11.27 Optimum efficiency of turbine designs.
100 Nsp
11.6 Turbines
Kaplan (adjustable blade) 0.9
Impulse 0.8
Fixed-blade propeller
10° 0.7
20° 0.6
Fig. 11.28 Efficiency versus power level for various turbine designs at constant speed and head.
0.5 0
40 60 Rated power, percent
EXAMPLE 11.9 Investigate the possibility of using (a) a Pelton wheel similar to Fig. 11.26 or (b) the Francis turbine family of Fig. 11.21d to deliver 30,000 bhp from a net head of 1200 ft.
Solution Part (a)
From Fig. 11.27, the most efficient Pelton wheel occurs at about (r/min)(30,000 bhp)1/2 Nsp 4.5 (1200 ft)1.25 or
n 183 r/min 3.06 r/s
From Fig. 11.26 the best operating point is
D(3.06 r/s) 0.47 [2(32.2)(1200)]1/2 or
D 13.6 ft
Ans. (a)
This Pelton wheel is perhaps a little slow and a trifle large. You could reduce D and increase n by increasing Nsp to, say, 6 or 7 and accepting the slight reduction in efficiency. Or you could use a
double-hung, two-wheel configuration, each delivering 15,000 bhp, which changes D and n by the factor 21/2: 13.6 Double wheel: n (183)21/2 260 r/min D 9.6 ft Ans. (a) 21/2
Chapter 11 Turbomachinery
Part (b)
The Francis wheel of Fig. 11.21d must have (r/min)(30,000 bhp)1/2 Nsp 29 (1200 ft)1.25 or
n 1183 r/min 19.7 r/s
Then the optimum power coefficient is P 30,000(550) CP* 2.70 n3D5 (1.94)(19.7)3D5 or
D5 412
D 3.33 ft 40 in
Ans. (b)
This is a faster speed than normal practice, and the casing would have to withstand 1200 ft of water or about 520 lbf/in2 internal pressure, but the 40-in size is extremely attractive. Francis
turbines are now being operated at heads up to 1500 ft.
Wind Turbines
Wind energy has long been used as a source of mechanical power. The familiar fourbladed windmills of Holland, England, and the Greek islands have been used for centuries to pump water, grind grain,
and saw wood. Modern research concentrates on the ability of wind turbines to generate electric power. Koeppl [47] stresses the potential for propeller-type machines. Spera [49] gives a detailed
discussion of the technical and economic feasibility of large-scale electric power generation by wind. See also Refs. 46, 48, and 50 to 52. Some examples of wind turbine designs are shown in Fig.
11.29. The familiar American multiblade farm windmill (Fig. 11.29a) is of low efficiency, but thousands are in use as a rugged, reliable, and inexpensive way to pump water. A more efficient design is
the propeller mill in Fig. 11.29b, similar to the pioneering Smith-Putnam 1250-kW two-bladed system which operated on Grampa’s Knob, 12 mi west of Rutland, Vermont, from 1941 to 1945. The
Smith-Putnam design broke because of inadequate blade strength, but it withstood winds up to 115 mi/h and its efficiency was amply demonstrated [47]. The Dutch, American multiblade, and propeller
mills are examples of horizontalaxis wind turbines (HAWTs), which are efficient but somewhat awkward in that they require extensive bracing and gear systems when combined with an electric generator.
Thus a competing family of vertical-axis wind turbines (VAWTs) has been proposed which simplifies gearing and strength requirements. Figure 11.29c shows the “eggbeater’’ VAWT invented by G. J. M.
Darrieus in 1925, now used in several government-sponsored demonstration systems. To minimize centrifugal stresses, the twisted blades of the Darrieus turbine follow a troposkien curve formed by a
chain anchored at two points on a spinning vertical rod. An alternative VAWT, simpler to construct than the troposkien, is the straight-bladed Darrieus-type turbine in Fig. 11.29d. This design,
proposed by Reading University in England, has blades which pivot due to centrifugal action as wind speeds increase, thus limiting bending stresses.
11.6 Turbines
(a )
Fig. 11.29 Wind turbine designs: (a) the American multiblade farm HAWT; (b) propeller HAWT (Courtesy of Grumman Aerospace Corp.); (c) the Darrieus VAWT (Courtesy of National Research Council,
Canada); (d) modified straight-blade Darrieus VAWT (Courtesy of Reading University— Nat’l Wind Power Ltd.).
(c )
(d )
(b )
Chapter 11 Turbomachinery
Idealized Wind Turbine Theory
The ideal, frictionless efficiency of a propeller windmill was predicted by A. Betz in 1920, using the simulation shown in Fig. 11.30. The propeller is represented by an actuator disk which creates
across the propeller plane a pressure discontinuity of area A and local velocity V. The wind is represented by a streamtube of approach velocity V1 and a slower downstream wake velocity V2. The
pressure rises to pb just before the disk and drops to pa just after, returning to free-stream pressure in the far wake. To hold the propeller rigid when it is extracting energy from the wind, there
must be a leftward force F on its support, as shown. A control-volume–horizontal-momentum relation applied between sections 1 and 2 gives
Fx F m˙ (V2 V1) A similar relation for a control volume just before and after the disk gives
Fx F (pb pa)A m˙ (Va Vb) 0 Equating these two yields the propeller force F (pb pa)A m ˙ (V1 V2)
Assuming ideal flow, the pressures can be found by applying the incompressible Bernoulli relation up to the disk From 1 to b:
p 12 V 12 pb 12 V2
From a to 2:
pa 12 V 2 p 12 V 22
Subtracting these and noting that m ˙ AV through the propeller, we can substitute for pb pa in Eq. (11.42) to obtain pb pa 12 (V 12 V 22) V(V1 V2) V 12 (V1 V2)
Streamtube passing through propeller pa
V Wind
V1, p∞
Wake Swept area A
V2, p∞
F pb
Fig. 11.30 Idealized actuator-disk and streamtube analysis of flow through a windmill.
p∞ p pa
11.6 Turbines
Ideal Betz number
0.6 Ideal, propeller type
High-speed HAWT
0.4 American multiblade
Cp 0.3
Savonius rotor
Fig. 11.31 Estimated performance of various wind turbine designs as a function of blade-tip speed ratio. (From Ref. 53.)
Grumman windstream (Fig. 11.29 b)
Darrieus VAWT
Dutch, four-arm
3 4 5 Speed ratio ω r/V1
Continuity and momentum thus require that the velocity V through the disk equal the average of the wind and far-wake speeds. Finally, the power extracted by the disk can be written in terms of V1 and
V2 by combining Eqs. (11.42) and (11.43) P FV AV 2(V1 V2) 14 A(V 21 V 22)(V1 V2)
For a given wind speed V1, we can find the maximum possible power by differentiating P with respect to V2 and setting equal to zero. The result is P Pmax 28 7 AV 13
at V2 13 V1
which corresponds to V 2V1/3 through the disk. The maximum available power to the propeller is the mass flow through the propeller times the total kinetic energy of the wind Pavail 12 m ˙ V 12 12
AV31 Thus the maximum possible efficiency of an ideal frictionless wind turbine is usually stated in terms of the power coefficient P CP 1 AV 3 2 1 Equation (11.45) states that the total power
coefficient is Cp,max 12 67 0.593
This is called the Betz number and serves as an ideal with which to compare the actual performance of real windmills. Figure 11.31 shows the measured power coefficients of various wind turbine
designs. The independent variable is not V2/V1 (which is artificial and convenient only in
Chapter 11 Turbomachinery
Under 750
750 – 2250
2250 – 3750
3750 – 5000
Over 5000
Fig. 11.32 World availability of land-based wind energy: estimated annual electric output in kWh/kW of a wind turbine rated at 11.2 m/s (25 mi/h). (From Ref. 54.)
the ideal theory) but the ratio of blade-tip speed r to wind speed. Note that the tip can move much faster than the wind, a fact disturbing to the laity but familiar to engineers in the behavior of
iceboats and sailing vessels. The Darrieus has the many advantages of a vertical axis but has little torque at low speeds and also rotates more slowly at maximum power than a propeller, thus
requiring a higher gear ratio for the generator. The Savonius rotor (Fig. 6.29b) has been suggested as a VAWT design because it produces power at very low wind speeds, but it is inefficient and
susceptible to storm damage because it cannot be feathered in high winds. As shown in Fig. 11.32, there are many areas of the world where wind energy is an attractive alternative. Greenland,
Newfoundland, Argentina, Chile, New Zealand, Iceland, Ireland, and the United Kingdom have the highest prevailing winds, but Australia, e.g., with only moderate winds, has the potential to generate
half its electricity with wind turbines [53]. In addition, since the ocean is generally even windier than the land, there are many island areas of high potential for wind power. There have also been
proposals [47] for ocean-based floating windmill farms. The inexhaustible availability of the winds, coupled with improved low-cost turbine designs, indicates a bright future for this alternative.
Problems 755
Turbomachinery design is perhaps the most practical and most active application of the principles of fluid mechanics. There are billions of pumps and turbines in use in the world, and thousands of
companies are seeking improvements. This chapter discusses both positive-displacement devices and, more extensively, rotodynamic machines. With the centrifugal pump as an example, the basic concepts
of torque, power, head, flow rate, and efficiency are developed for a turbomachine. Nondimensionalization leads to the pump similarity rules and some typical dimensionless performance curves for
axial and centrifugal machines. The single most useful pump parameter is found to be the specific speed, which delineates the type of design needed. An interesting design application is the theory of
pumps combined in series and in parallel. Turbines extract energy from flowing fluids and are of two types: impulse turbines, which convert the momentum of a high-speed stream, and reaction turbines,
where the pressure drop occurs within the blade passages in an internal flow. By analogy with pumps, the power specific speed is important for turbines and is used to classify them into impulse,
Francis, and propeller-type designs. A special case of reaction turbine with unconfined flow is the wind turbine. Several types of windmills are discussed and their relative performances compared.
Problems Most of the problems herein are fairly straightforward. More difficult or open-ended assignments are labeled with an asterisk. Problems labeled with an EES icon will benefit from the use of
the Engineering Equations Solver (EES), while problems labeled with a computer icon may require the use of a computer. The standard end-of-chapter problems 11.1 to 11.103 (categorized in the problem
list below) are followed by word problems W11.1 to W11.10, comprehensive problems C11.1 to C11.3, and design project D11.1.
A PDP can deliver almost any fluid, but there is always a limiting very high viscosity for which performance will deteriorate. Can you explain the probable reason? Figure 11.2c shows a double-screw
pump. Sketch a single-screw pump and explain its operation. How did Archimedes’ screw pump operate? What type of pump is shown in Fig. P11.5? How does it operate?
Problem distribution Section
11.1 11.2 11.3 11.3 11.4 11.5 11.5 11.5 11.6 11.6
Introduction and classification Centrifugal pump theory Pump performance and similarity rules Net positive-suction head Specific speed: Mixed- and axial-flow pumps Matching pumps to system
characteristics Pumps in parallel or series Pump instability Reaction and impulse turbines Wind turbines
11.1–11.14 11.15–11.21 11.22–11.41 11.42–11.44 11.45–11.62 11.63–11.73 11.74–11.81 11.82–11.83 11.84–11.99 11.100–11.103
P11.1 P11.2
Describe the geometry and operation of a peristaltic positive-displacement pump which occurs in nature. What would be the technical classification of the following turbomachines: (a) a household fan,
(b) a windmill, (c) an aircraft propeller, (d) a fuel pump in a car, (e) an eductor, ( f ) a fluid-coupling transmission, and (g) a power plant steam turbine?
P11.5 P11.6
Figure P11.6 shows two points a half-period apart in the operation of a pump. What type of pump is this? How does it work? Sketch your best guess of flow rate versus time for a few cycles. A piston
PDP has a 5-in diameter and a 2-in stroke and operates at 750 r/min with 92 percent volumetric efficiency. (a) What is its delivery, in gal/min? (b) If the pump delivers SAE 10W oil at 20°C against a
head of 50 ft, what horsepower is required when the overall efficiency is 84 percent?
Chapter 11 Turbomachinery
Flow out
Flow out
P11.14 A
Flow in
Check valve
Flow in
P11.6 P11.8
P11.12 EES
A centrifugal pump delivers 550 gal/min of water at 20°C when the brake horsepower is 22 and the efficiency is 71 percent. (a) Estimate the head rise in ft and the pressure rise in lbf/in2. (b) Also
estimate the head rise and horsepower if instead the delivery is 550 gal/min of gasoline at 20°C. Figure P11.9 shows the measured performance of the Vickers model PVQ40 piston pump when delivering
SAE 10W oil at 180°F ( 910 kg/m3). Make some general observations about these data vis-à-vis Fig. 11.2 and your intuition about the behavior of piston pumps. Suppose that the piston pump of Fig.
P11.9 is used to deliver 15 gal/min of water at 20°C using 20 brake horsepower. Neglecting Reynolds-number effects, use the figure to estimate (a) the speed in r/min and (b) the pressure rise in lbf/
in2. A pump delivers 1500 L/min of water at 20°C against a pressure rise of 270 kPa. Kinetic- and potential-energy changes are negligible. If the driving motor supplies 9 kW, what is the overall
efficiency? In a test of the centrifugal pump shown in Fig. P11.12, the following data are taken: p1 100 mmHg (vacuum) and p2 500 mmHg (gage). The pipe diameters are D1 12 cm and D2 5 cm. The flow
rate is 180 gal/min of light oil (SG 0.91). Estimate (a) the head developed,
in meters; and (b) the input power required at 75 percent efficiency. A 20-hp pump delivers 400 gal/min of gasoline at 20°C with 75 percent efficiency. What head and pressure rise result across the
pump? A pump delivers gasoline at 20°C and 12 m3/h. At the inlet p1 100 kPa, z1 1 m, and V1 2 m/s. At the exit p2 500 kPa, z2 4 m, and V2 3 m/s. How much power is required if the motor efficiency is
75 percent? A lawn sprinkler can be used as a simple turbine. As shown in Fig. P11.15, flow enters normal to the paper in the center and splits evenly into Q/2 and Vrel leaving each nozzle. The arms
rotate at angular velocity and do work on a shaft. Draw the velocity diagram for this turbine. Neglecting friction, find an expression for the power delivered to the shaft. Find the rotation rate for
which the power is a maximum. Q, V rel 2 R
ω Q, V rel 2
P11.15 P11.16
For the “sprinkler turbine’’ of Fig. P11.15, let R 18 cm, with total flow rate of 14 m3/h of water at 20°C. If the nozzle exit diameter is 8 mm, estimate (a) the maximum power delivered in W and (b)
the appropriate rotation rate in r/min. A centrifugal pump has d1 7 in, d2 13 in, b1 4 in, b2 3 in, 1 25°, and 2 40° and rotates at 1160 r/min. If the fluid is gasoline at 20°C and the flow enters
the blades radially, estimate the theoretical (a) flow rate in gal/min, (b) horsepower, and (c) head in ft. A jet of velocity V strikes a vane which moves to the right at speed Vc, as in Fig. P11.18.
The vane has a turning angle .
ρ , V, A
65 cm
θ (1)
Problems 757
35 bar 70 bar 140 bar 210 bar
Fig. P11.9 Performance of the model PVQ40 piston pump delivering SAE 10W oil at 180°F. (Courtesy of Vickers Inc., PDN/PACE Division.)
Input power, kW
210 bar 140 bar 70 bar 35 bar
Pump displacement: 41 cm3/r
Derive an expression for the power delivered to the vane by the jet. For what vane speed is the power maximum? A centrifugal pump has r2 9 in, b2 2 in, and 2 35° and rotates at 1060 r/min. If it
generates a head of 180 ft, determine the theoretical (a) flow rate in gal/min and (b) horsepower. Assume near-radial entry flow. Suppose that Prob. 11.19 is reversed into a statement of the
theoretical power Pw 153 hp. Can you then compute the theoretical (a) flow rate and (b) head? Explain and resolve the difficulty which arises. The centrifugal pump of Fig. P11.21 develops a flow rate
of 4200 gal/min of gasoline at 20°C with near-radial absolute
1000 Speed, r/min
2 in 30° 1750 r / min
4 in
3 in
Delivery, L/min
Overall efficiency, percent
210 bar - 3000 lb/in2 140 bar - 2000 lb/in2 70 bar - 1000 lb/in2 35 bar - 500 lb/in2
210 bar 140 bar 70 bar 35 bar
Volumetric efficiency, percent
Chapter 11 Turbomachinery
FM Series 10
1160 RPM
Model 4013 15
Curve No. 806 Min. Imp. Dia. 10.0 Sizes 5 x 4 x13
45 L/s
6 NPSH, ft
12.95 in
60% 65%
74% 76%
25 79%
12.50 in
Head, ft
12.00 in
79% 78%
11.50 in 60
11.00 in
20 76% 74% 70%
10.50 in
10.00 in
40 5 bhp
10 50%
10 bhp
7.5 bhp
Curves based on clear water with specific gravity of 1.0
400 Flow, gal / min
Fig. P11.24 Performance data for a centrifugal pump. (Courtesy of Taco, Inc., Cranston, Rhode Island.)
inflow. Estimate the theoretical (a) horsepower, (b) head rise, and (c) appropriate blade angle at the inner radius. A 37-cm-diameter centrifugal pump, running at 2140 r/min with water at 20°C,
produces the following performance data: Q, m3/s
H, m
P, kW
(a) Determine the best efficiency point. (b) Plot CH versus CQ. (c) If we desire to use this same pump family to deliver 7000 gal/min of kerosine at 20°C at an input power of 400 kW, what pump speed
(in r/min) and impeller size (in cm) are needed? What head will be developed?
If the 38-in-diameter pump of Fig. 11.7b is used to deliver 20°C kerosine at 850 r/min and 22,000 gal/min, what (a) head and (b) brake horsepower will result? Figure P11.24 shows performance data for
the Taco, Inc., model 4013 pump. Compute the ratios of measured shutoff head to the ideal value U2/g for all seven impeller sizes. Determine the average and standard deviation of this ratio and
compare it to the average for the six impellers in Fig. 11.7. At what speed in r/min should the 35-in-diameter pump of Fig. 11.7b be run to produce a head of 400 ft at a discharge of 20,000 gal/min?
What brake horsepower will be required? Hint: Fit H(Q) to a formula. Determine if the seven Taco, Inc., pump sizes in Fig. P11.24 can be collapsed into a single dimensionless chart
Problems 759
1760 RPM
Model 4010
Curve No.756 Min. Imp. Dia. 7.70 Sizes 5 x 4 x10
CM & FM Series
70 L/s
NPSH, ft 50%
60% 65%
120 10.40 in
70% 74% 78%
10.00 in
Head, ft
30 80%
9.00 in
78% 76%
80 8.50 in
Head, m
83% 100 9.50 in
70% 8.00 in 65%
7.70 in
30 bhp 15 25 bhp 40
10 bhp
Curves based on clear water with specific gravity of 1.0
20 bhp 50%
625 Flow, gal / min
15 bhp 1000
Fig. P11.31 Performance data for a family of centrifugal pump impellers. (Courtesy of Taco, Inc., Cranston, Rhode Island.)
P11.27 EES
Q, ft3/s
of CH, CP, and versus CQ, as in Fig. 11.8. Comment on the results. The 12-in pump of Fig. P11.24 is to be scaled up in size to provide a head of 90 ft and a flow rate of 1000 gal/min at BEP.
Determine the correct (a) impeller diameter, (b) speed in r/min, and (c) horsepower required. Tests by the Byron Jackson Co. of a 14.62-in-diameter centrifugal water pump at 2134 r/min yield the
following data: 0
H, ft
What is the BEP? What is the specific speed? Estimate the maximum discharge possible.
P11.29 If the scaling laws are applied to the pump of Prob. 11.28 for the same impeller diameter, determine (a) the speed for which the shutoff head will be 280 ft, (b) the speed for which the BEP
flow rate will be 8.0 ft3/s, and (c) the speed for which the BEP conditions will require 80 hp. P11.30 A pump from the same family as Prob. 11.28 is built with D 18 in and a BEP power of 250 hp for
gasoline (not water). Using the scaling laws, estimate the resulting (a) speed in r/min, (b) flow rate at BEP, and (c) shutoff head. P11.31 Figure P11.31 shows performance data for the Taco, Inc.,
model 4010 pump. Compute the ratios of measured shutoff head to the ideal value U2/g for all seven impeller sizes. Determine the average and standard deviation of this ratio, and compare it to the
average of 0.58 0.02 for the seven impellers in Fig. P11.24. Comment on your results.
Chapter 11 Turbomachinery
Determine if the seven Taco, Inc., impeller sizes in Fig. P11.31 can be collapsed into a single dimensionless chart of CH, CP, and versus CQ, as in Fig. 11.8. Comment on the results. Clearly the
maximum efficiencies of the pumps in Figs. P11.24 and P11.31 decrease with impeller size. Compare max for these two pump families with both the Moody and the Anderson correlations, Eqs. (11.29). Use
the central impeller size as a comparison point. You are asked to consider a pump geometrically similar to the 9-in-diameter pump of Fig. P11.31 to deliver 1200 gal/min at 1500 r/min. Determine the
appropriate (a) impeller diameter, (b) BEP horsepower, (c) shutoff head, and (d) maximum efficiency. The fluid is kerosine, not water. An 18-in-diameter centrifugal pump, running at 880 r/min with
water at 20°C, generates the following performance data:
Q, gal/min
H, ft
P, hp
P11.38 EES
Determine (a) the BEP, (b) the maximum efficiency, and (c) the specific speed. (d) Plot the required input power versus the flow rate. Plot the dimensionless performance curves for the pump of Prob.
11.35 and compare with Fig. 11.8. Find the appropriate diameter in inches and speed in r/min for a geometrically similar pump to deliver 400 gal/min against a head of 200 ft. What brake horsepower
would be required? The efficiency of a centrifugal pump can be approximated by the curve fit aQ bQ3, where a and b are constants. For this approximation, (a) what is the ratio of Q* at BEP to Qmax?
If the maximum efficiency is 88 percent, what is the efficiency at (b) 13 Qmax and (c) 43 Q*? A 6.85-in pump, running at 3500 r/min, has the following measured performance for water at 20°C:
Q, gal/min
H, ft
, %
(a) Estimate the horsepower at BEP. If this pump is rescaled in water to provide 20 bhp at 3000 r/min, determine the appropriate (b) impeller diameter, (c) flow rate, and (d) efficiency for this new
condition. The Allis-Chalmers D30LR centrifugal compressor delivers 33,000 ft3/min of SO2 with a pressure change from
14.0 to 18.0 lbf/in2 absolute using an 800-hp motor at 3550 r/min. What is the overall efficiency? What will the flow rate and p be at 3000 r/min? Estimate the diameter of the impeller. P11.40 The
specific speed Ns, as defined by Eqs. (11.30), does not contain the impeller diameter. How then should we size the pump for a given Ns? Logan [7] suggests a parameter called the specific diameter Ds,
which is a dimensionless combination of Q, gH, and D. (a) If Ds is proportional to D, determine its form. (b) What is the relationship, if any, of Ds to CQ*, CH*, and CP*? (c) Estimate Ds for the two
pumps of Figs. 11.8 and 11.13. P11.41 It is desired to build a centrifugal pump geometrically similar to that of Prob. 11.28 to deliver 6500 gal/min of gasoline at 20°C at 1060 r/min. Estimate the
resulting (a) impeller diameter, (b) head, (c) brake horsepower, and (d) maximum efficiency. P11.42 An 8-in model pump delivering 180°F water at 800 gal/min and 2400 r/min begins to cavitate when the
inlet pressure and velocity are 12 lbf/in2 absolute and 20 ft/s, respectively. Find the required NPSH of a prototype which is 4 times larger and runs at 1000 r/min. P11.43 The 28-in-diameter pump in
Fig. 11.7a at 1170 r/min is used to pump water at 20°C through a piping system at 14,000 gal/min. (a) Determine the required brake horsepower. The average friction factor is 0.018. (b) If there is 65
ft of 12-indiameter pipe upstream of the pump, how far below the surface should the pump inlet be placed to avoid cavitation? P11.44 The pump of Prob. 11.28 is scaled up to an 18-in diameter,
operating in water at best efficiency at 1760 r/min. The measured NPSH is 16 ft, and the friction loss between the inlet and the pump is 22 ft. Will it be sufficient to avoid cavitation if the pump
inlet is placed 9 ft below the surface of a sea-level reservoir? P11.45 Determine the specific speeds of the seven Taco, Inc., pump impellers in Fig. P11.24. Are they appropriate for centrifugal
designs? Are they approximately equal within experimental uncertainty? If not, why not? P11.46 The answer to Prob. 11.40 is that the dimensionless “specific diameter” takes the form Ds D(gH*)1/4/Q*1/
2, evaluated at the BEP. Data collected by the author for 30 different pumps indicate, in Fig. P11.46, that Ds correlates well with specific speed Ns. Use this figure to estimate the appropriate
impeller diameter for a pump which delivers 20,000 gal/min of water and a head of 400 ft when running at 1200 r/min. Suggest a curve-fit formula to the data. Hint: Use a hyperbolic formula. P11.47 A
typical household basement sump pump provides a discharge of 5 gal/min against a head of 15 ft. Estimate (a) the maximum efficiency and (b) the minimum horsepower required to drive such a pump at
1750 r/min.
Problems 761
20 18 16 14 12 Ds 10 8 6 4 2 0
Data from 30 different pump designs
2 1.8 1.6 1.4 1.2 C* 1 P 0.8 0.6 0.4 0.2 0
Data from 30 different pump designs
Fig. P11.46 Specific diameter at BEP for 30 commercial pumps. P11.48
Compute the specific speeds for the pumps in Probs. 11.28, 11.35, and 11.38 plus the median sizes in Figs. P11.24 and P11.31. Then determine if their maximum efficiencies match the values predicted
in Fig. 11.14. Data collected by the author for flow coefficient at BEP for 30 different pumps are plotted versus specific speed in Fig. P11.49. Determine if the values of C* Q for the five pumps in
Prob. 11.48 also fit on this correlation. If so, suggest a curve-fitted formula for the data.
Data from 30 different pump designs
0.300 0.250 C* 0.200 Q
0.150 0.100 0.050 0.000
P11.56 0
Fig. P11.49 Flow coefficient at BEP for 30 commercial pumps. P11.50
Fig. P11.50 Power coefficient at BEP for 30 commercial pumps.
2000 Ns
Data collected by the author for power coefficient at BEP for 30 different pumps are plotted versus specific speed in Fig. P11.50. Determine if the values of C *P for the five pumps in Prob. 11.48
also fit on this correlation. If so, suggest a curve-fitted formula for the data. An axial-flow pump delivers 40 ft3/s of air which enters at 20°C and 1 atm. The flow passage has a 10-in outer radius
and an 8-in inner radius. Blade angles are 1 60° and 2 70°, and the rotor runs at 1800 r/min. For the first stage compute (a) the head rise and (b) the power required.
An axial-flow fan operates in sea-level air at 1200 r/min and has a blade-tip diameter of 1 m and a root diameter of 80 cm. The inlet angles are 1 55° and 1 30°, while at the outlet 2 60°. Estimate
the theoretical values of the (a) flow rate, (b) horsepower, and (c) outlet angle 2. If the axial-flow pump of Fig. 11.13 is used to deliver 70,000 gal/min of 20°C water at 1170 r/min, estimate (a)
the proper impeller diameter, (b) the shutoff head, (c) the shutoff horsepower, and (d) p at best efficiency. The Colorado River Aqueduct uses Worthington Corp. pumps which deliver 200 ft3/s at 450 r
/min against a head of 440 ft. What types of pump are these? Estimate the impeller diameter. We want to pump 70°C water at 20,000 gal/min and 1800 r/min. Estimate the type of pump, the horsepower
required, and the impeller diameter if the required pressure rise for one stage is (a) 170 kPa and (b) 1350 kPa. A pump is needed to deliver 40,000 gal/min of gasoline at 20°C against a head of 90
ft. Find the impeller size, speed, and brake horsepower needed to use the pump families of (a) Fig. 11.8 and (b) Fig. 11.13. Which is the better design? Performance data for a 21-in-diameter air
blower running at 3550 r/min are as follows:
p, inH2O
Q, ft3/min
Note the fictitious expression of pressure rise in terms of water rather than air. What is the specific speed? How does the performance compare with Fig. 11.8? What are *, C H*, and C *P? CQ
Chapter 11 Turbomachinery
P11.67 EES
The Worthington Corp. model A-12251 water pump, operating at maximum efficiency, produces 53 ft of head at 3500 r/min, 1.1 bhp at 3200 r/min, and 60 gal/min at 2940 r/min. What type of pump is this?
What is its efficiency, and how does this compare with Fig. 11.14? Estimate the impeller diameter. Suppose it is desired to deliver 700 ft3/min of propane gas (molecular weight 44.06) at 1 atm and
20°C with a single-stage pressure rise of 8.0 inH2O. Determine the appropriate size and speed for using the pump families of (a) Prob. 11.57 and (b) Fig. 11.13. Which is the better design? A 45-hp
pump is desired to generate a head of 200 ft when running at BEP with 20°C gasoline at 1200 r/min. Using the correlations in Figs. P11.49 and P11.50, determine the appropriate (a) specific speed, (b)
flow rate, and (c) impeller diameter. A mine ventilation fan, running at 295 r/min, delivers 500 m3/s of sea-level air with a pressure rise of 1100 Pa. Is this fan axial, centrifugal, or mixed?
Estimate its diameter in ft. If the flow rate is increased 50 percent for the same diameter, by what percentage will the pressure rise change? The actual mine ventilation fan discussed in Prob. 11.61
had a diameter of 20 ft [20, p. 339]. What would be the proper diameter for the pump family of Fig. 11.14 to provide 500 m3/s at 295 r/min and BEP? What would be the resulting pressure rise in Pa?
The 36.75-in pump in Fig. 11.7a at 1170 r/min is used to pump water at 60°F from a reservoir through 1000 ft of 12-in-ID galvanized-iron pipe to a point 200 ft above the reservoir surface. What flow
rate and brake horsepower will result? If there is 40 ft of pipe upstream of the pump, how far below the surface should the pump inlet be placed to avoid cavitation? In Prob. 11.63 the operating
point is off design at an efficiency of only 77 percent. Is it possible, with the similarity rules, to change the pump rotation speed to deliver the water near BEP? Explain your results. The 38-in
pump of Fig. 11.7a is used in series to lift 20°C water 3000 ft through 4000 ft of 18-in-ID cast-iron pipe. For most efficient operation, how many pumps in series are needed if the rotation speed is
(a) 710 r/min and (b) 1200 r/min? It is proposed to run the pump of Prob. 11.35 at 880 r/min to pump water at 20°C through the system in Fig. P11.66. The pipe is 20-cm-diameter commercial steel. What
flow rate in ft3/min will result? Is this an efficient application? The pump of Prob. 11.35, running at 880 r/min, is to pump water at 20°C through 75 m of horizontal galvanized-iron pipe. All other
system losses are neglected.
3m Pump 8m
20 m
12 m
Determine the flow rate and input power for (a) pipe diameter 20 cm and (b) the pipe diameter found to yield maximum pump efficiency. A 24-in pump is homologous to the 32-in pump in Fig. 11.7a. At
1400 r/min this pump delivers 12,000 gal/min of water from one reservoir through a long pipe to another 50 ft higher. What will the flow rate be if the pump speed is increased to 1750 r/min? Assume
no change in pipe friction factor or efficiency. The pump of Prob. 11.38, running at 3500 r/min, is used to deliver water at 20°C through 600 ft of cast-iron pipe to an elevation 100 ft higher.
Determine (a) the proper pipe diameter for BEP operation and (b) the flow rate which results if the pipe diameter is 3 in. The pump of Prob. 11.28, operating at 2134 r/min, is used with 20°C water in
the system of Fig. P11.70. (a) If it is operating at BEP, what is the proper elevation z2? (b) If z2 225 ft, what is the flow rate if d 8 in.? z2 z1 = 100 ft
1500 ft of cast-iron pipe
P11.70 P11.71
The pump of Prob. 11.38, running at 3500 r/min, delivers water at 20°C through 7200 ft of horizontal 5-in-diameter commercial-steel pipe. There are a sharp entrance, sharp exit, four 90° elbows, and
a gate valve. Estimate (a) the flow rate if the valve is wide open and (b) the valve closing percentage which causes the pump to operate at BEP. (c) If the latter condition holds continuously for 1
year, estimate the energy cost at 10 ¢/kWh. Performance data for a small commercial pump are as follows:
Q, gal/min
H, ft
Problems 763
This pump supplies 20°C water to a horizontal 58 -indiameter garden hose ( 0.01 in) which is 50 ft long. Estimate (a) the flow rate and (b) the hose diameter which would cause the pump to operate at
BEP. The piston pump of Fig. P11.9 is run at 1500 r/min to deliver SAE 10W oil through 100 m of vertical 2-cm-diameter wrought-iron pipe. If other system losses are neglected, estimate (a) the flow
rate, (b) the pressure rise, and (c) the power required. The 32-in pump in Fig. 11.7a is used at 1170 r/min in a system whose head curve is Hs (ft) 100 1.5Q2, with Q in thousands of gallons of water
per minute. Find the discharge and brake horsepower required for (a) one pump, (b) two pumps in parallel, and (c) two pumps in series. Which configuration is best? Two 35-in pumps from Fig. 11.7b are
installed in parallel for the system of Fig. P11.75. Neglect minor losses. For water at 20°C, estimate the flow rate and power required if (a) both pumps are running and (b) one pump is shut off and
It is proposed to use one 32- and one 28-in pump from Fig. 11.7a in parallel to deliver water at 60°F. The system-head curve is Hs 50 0.3Q2, with Q in thousands of gallons per minute. What will the
head and delivery be if both pumps run at 1170 r/min? If the 28-in pump is reduced below 1170 r/min, at what speed will it cease to deliver? Reconsider the system of Fig. P6.68. Use the Byron Jackson
pump of Prob. 11.28 running at 2134 r/min, no scaling, to drive the flow. Determine the resulting flow rate between the reservoirs. What is the pump efficiency? The S-shaped head-versus-flow curve in
Fig. P11.82 occurs in some axial-flow pumps. Explain how a fairly flat system-loss curve might cause instabilities in the operation of the pump. How might we avoid instability?
H z2 = 300 ft
P11.82 1 statute mile of cast-iron pipe, 24-in diameter
z 1 = 200 ft
Two pumps
P11.75 P11.76
P11.78 EES
Two 32-in pumps from Fig. 11.7a are combined in parallel to deliver water at 60°F through 1500 ft of horizontal pipe. If f 0.025, what pipe diameter will ensure a flow rate of 35,000 gal/min for n
1170 r/min? Two pumps of the type tested in Prob. 11.22 are to be used at 2140 r/min to pump water at 20°C vertically upward through 100 m of commercial-steel pipe. Should they be in series or in
parallel? What is the proper pipe diameter for most efficient operation? Suppose that the two pumps in Fig. P11.75 are modified to be in series, still at 710 r/min. What pipe diameter is required for
BEP operation? Two 32-in pumps from Fig. 11.7a are to be used in series at 1170 r/min to lift water through 500 ft of vertical cast-iron pipe. What should the pipe diameter be for most efficient
operation? Neglect minor losses.
The low-shutoff head-versus-flow curve in Fig. P11.83 occurs in some centrifugal pumps. Explain how a fairly flat system-loss curve might cause instabilities in the operation of the pump. What
additional vexation occurs when two of these pumps are in parallel? How might we avoid instability?
P11.83 P11.84
Turbines are to be installed where the net head is 400 ft and the flow rate 250,000 gal/min. Discuss the type, num-
Chapter 11 Turbomachinery
ber, and size of turbine which might be selected if the generator selected is (a) 48-pole, 60-cycle (n 150 r/min) and (b) 8-pole (n 900 r/min). Why are at least two turbines desirable from a planning
point of view? Turbines at the Conowingo Plant on the Susquehanna River each develop 54,000 bhp at 82 r/min under a head of 89 ft. What type of turbines are these? Estimate the flow rate and impeller
diameter. The Tupperware hydroelectric plant on the Blackstone River has four 36-in-diameter turbines, each providing 447 kW at 200 r/min and 205 ft3/s for a head of 30 ft. What type of turbines are
these? How does their performance compare with Fig. 11.21? An idealized radial turbine is shown in Fig. P11.87. The absolute flow enters at 30° and leaves radially inward. The flow rate is 3.5 m3/s
of water at 20°C. The blade thickness is constant at 10 cm. Compute the theoretical power developed at 100 percent efficiency.
Vr 2 35° V2 25° 0.8 m 30° 80 r / min
P11.90 P11.92
b = 10 cm
135 r / min
1.2 m
Vr 1
P11.93 30°
b = 20 cm
At a proposed turbine installation the available head is 800 ft, and the water flow rate is 40,000 gal/min. Discuss the size, speed, and number of turbines which might be suitable for this purpose
while using (a) a Pelton wheel and (b) a Francis wheel. Figure P11.93 shows a cutaway of a cross-flow or “Banki’’ turbine [55], which resembles a squirrel cage with slotted curved blades. The flow
enters at about 2 o’clock, passes through the center and then again through the blades, leaving at about 8 o’clock. Report to the class on the operation and advantages of this design, including
idealized velocity vector diagrams.
40 cm
70 cm
P11.87 P11.88
A certain turbine in Switzerland delivers 25,000 bhp at 500 r/min under a net head of 5330 ft. What type of turbine is this? Estimate the approximate discharge and size. A Pelton wheel of 12-ft pitch
diameter operates under a net head of 2000 ft. Estimate the speed, power output, and flow rate for best efficiency if the nozzle exit diameter is 4 in. An idealized radial turbine is shown in Fig.
P11.90. The absolute flow enters at 25° with the blade angles as shown. The flow rate is 8 m3/s of water at 20°C. The blade thickness is constant at 20 cm. Compute the theoretical power developed at
100 percent efficiency. The flow through an axial-flow turbine can be idealized by modifying the stator-rotor diagrams of Fig. 11.12 for energy absorption. Sketch a suitable blade and flow
arrangement and the associated velocity vector diagrams. For further details see chap. 8 of Ref. 25.
P11.93 P11.94
A simple cross-flow turbine, Fig. P11.93, was constructed and tested at the University of Rhode Island. The blades were made of PVC pipe cut lengthwise into three 120°arc pieces. When it was tested
in water at a head of 5.3 ft and a flow rate of 630 gal/min, the measured power
Word Problems
output was 0.6 hp. Estimate (a) the efficiency and (b) the power specific speed if n 200 rev/min. One can make a theoretical estimate of the proper diameter for a penstock in an impulse turbine
installation, as in Fig. P11.95. Let L and H be known, and let the turbine performance be idealized by Eqs. (11.38) and (11.39). Account for friction loss hf in the penstock, but neglect minor
losses. Show that (a) the maximum power is generated when hf H/3, (b) the optimum jet velocity is (4gH/3)1/2, and (c) the best nozzle diameter is Dj [D5/(2 fL)]1/4, where f is the pipe-friction
P11.98 Reservoir
(11.20), the empirical criterion given by Wislicenus [4] for cavitation is Impulse wheel
(r/min)(gal/min)1/2 11,000 Nss
[NPSH (ft)]3/4
Dj Penstock: L, D Vj
P11.95 P11.96
P11.97 EES
Apply the results of Prob. 11.95 to determining the optimum (a) penstock diameter and (b) nozzle diameter for the data of Prob. 11.92 with a commercial-steel penstock of length 1500 ft. Consider the
following nonoptimum version of Prob. 11.95: H 450 m, L 5 km, D 1.2 m, Dj 20 cm. The penstock is concrete, 1 mm. The impulse wheel diameter is 3.2 m. Estimate (a) the power generated by the wheel at
80 percent efficiency and (b) the best speed of the wheel in r/min. Neglect minor losses. Francis and Kaplan turbines are often provided with draft tubes, which lead the exit flow into the tailwater
region, as in Fig. P11.98. Explain at least two advantages in using a draft tube. Turbines can also cavitate when the pressure at point 1 in Fig. P11.98 drops too low. With NPSH defined by Eq.
Use this criterion to compute how high z1 z2, the impeller eye in Fig. P11.98, can be placed for a Francis turbine with a head of 300 ft, Nsp 40, and pa 14 lbf/in2 absolute before cavitation occurs
in 60°F water. One of the largest wind generators in operation today is the ERDA/NASA two-blade propeller HAWT in Sandusky, Ohio. The blades are 125 ft in diameter and reach maximum power in 19 mi/h
winds. For this condition estimate (a) the power generated in kW, (b) the rotor speed in r/min, and (c) the velocity V2 behind the rotor. A Darrieus VAWT in operation in Lumsden, Saskatchewan, that
is 32 ft high and 20 ft in diameter sweeps out an area of 432 ft2. Estimate (a) the maximum power and (b) the rotor speed if it is operating in 16 mi/h winds. An American 6-ft diameter multiblade
HAWT is used to pump water to a height of 10 ft through 3-in-diameter cast-iron pipe. If the winds are 12 mi/h, estimate the rate of water flow in gal/min. A very large Darrieus VAWT was constructed
by the U.S. Department of Energy near Sandia, New Mexico. It is 60 ft high and 30 ft in diameter, with a swept area of 1200 ft2. If the turbine is constrained to rotate at 90 r/min, use Fig. 11.31 to
plot the predicted power output in kW versus wind speed in the range V 5 to 40 mi/h.
Word Problems W11.1
We know that an enclosed rotating bladed impeller will impart energy to a fluid, usually in the form of a pressure rise, but how does it actually happen? Discuss, with sketches, the physical
mechanisms through which an impeller actually transfers energy to a fluid.
Dynamic pumps (as opposed to PDPs) have difficulty moving highly viscous fluids. Lobanoff and Ross [15] suggest the following rule of thumb: D (in) 0.015/water, where D is the diameter of the
discharge pipe. For example, SAE 30W oil ( 300water) should
Chapter 11 Turbomachinery
require at least a 4.5-in outlet. Can you explain some reasons for this limitation? The concept of NPSH dictates that liquid dynamic pumps should generally be immersed below the surface. Can you
explain this? What is the effect of increasing the liquid temperature? For nondimensional fan performance, Wallis [20] suggests that the head coefficient should be replaced by FTP/(n2D2), where FTP
is the fan total pressure change. Explain the usefulness of this modification. Performance data for centrifugal pumps, even if well scaled geometrically, show a decrease in efficiency with decreasing
impeller size. Discuss some physical reasons why this is so. Consider a dimensionless pump performance chart such as Fig. 11.8. What additional dimensionless parameters
might modify or even destroy the similarity indicated in such data? W11.7 One parameter not discussed in this text is the number of blades on an impeller. Do some reading on this subject, and report
to the class about its effect on pump performance. W11.8 Explain why some pump performance curves may lead to unstable operating conditions. W11.9 Why are Francis and Kaplan turbines generally
considered unsuitable for hydropower sites where the available head exceeds 1000 ft? W11.10 Do some reading on the performance of the free propeller that is used on small, low-speed aircraft. What
dimensionless parameters are typically reported for the data? How do the performance and efficiency compare with those for the axial-flow pump?
Comprehensive Problems C11.1 The net head of a little aquarium pump is given by the manufacturer as a function of volume flow rate as listed below: Q, m3/s 0 1.0 2.0 3.0 4.0 5.0
E-6 E-6 E-6 E-6 E-6
H, mH2O
1.10 1.00 0.80 0.60 0.35 0.0
What is the maximum achievable flow rate if you use this pump to pump water from the lower reservoir to the upper reservoir as shown in Fig. C11.1? Note: The tubing is
Q 0.80 m
Pump Q
smooth with an inner diameter of 5.0 mm and a total length of 29.8 m. The water is at room temperature and pressure. Minor losses in the system can be neglected. Reconsider Prob. 6.68 as an exercise
in pump selection. Select an impeller size and rotational speed from the Byron Jackson pump family of Prob. 11.28 which will deliver a flow rate of 3 ft3/s to the system of Fig. P6.68 at minimum
input power. Calculate the horsepower required. Reconsider Prob. 6.77 as an exercise in turbine selection. Select an impeller size and rotational speed from the Francis turbine family of Fig. 11.21d
which will deliver maximum power generated by the turbine. Calculate the turbine power output and remark on the practicality of your design. A pump provides a net head H which is dependent on the
volume flow rate Q, as follows: H a bQ2, where a 80 m and b 20 s2/m5. The pump delivers water at 20°C through a horizontal 30-cm-diameter cast-iron pipe which is 120 m long. The pressures at the
inlet and exit of the system are the same. Neglecting minor losses, calculate the expected volume flow rate in gal/min. In Prob. 11.23, estimate the efficiency of the pump in two ways: (a) Read it
directly from Fig. 11.7b (for the dynamically similar water pump); and (b) Calculate it from Eq. (11.5) for the actual kerosene flow. Compare your results and discuss any discrepancies.
References 767
Design Project D11.1 To save on electricity costs, a town water supply system uses gravity-driven flow from five large storage tanks during the day and then refills these tanks from 10 p.m. to 6 a.m.
at a cheaper night rate of 7 ¢/kWh. The total resupply needed each night varies from 5 E5 to 2 E6 gal, with no more than 5 E5 gallons to any one tank. Tank elevations vary from 40 to 100 ft. A single
constant-speed pump, drawing from a large groundwater aquifer and valved into five different cast-iron tank supply lines, does this job. Distances from the pump to the five tanks vary more or less
evenly from 1 to 3 mi. Each line averages one elbow every 100 ft and has four butterfly valves which can be controlled at any desirable angle. Select a suitable pump family from one of the six data
sets in this
chapter: Figs. 11.8, P11.24, and P11.31 plus Probs. 11.28, 11.35, and 11.38. Assume ideal similarity (no Reynolds-number or pump roughness effects). The goal is to determine pump and pipeline sizes
which achieve minimum total cost over a 5-year period. Some suggested cost data are 1. Pump and motor: $2500 plus $1500 per inch of pipe size 2. Valves: $100 plus $100 per inch of pipe size 3.
Pipelines: 50¢ per inch of diameter per foot of length Since the flow and elevation parameters vary considerably, a random daily variation within the specified ranges might give a realistic approach.
References 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.
D. G. Wilson, “Turbomachinery—From Paddle Wheels to Turbojets,” Mech. Eng., vol. 104, Oct. 1982, pp. 28–40. D. Japikse and N. C. Baines, Introduction to Turbomachinery, Oxford University Press, New
York, 1995. D. W. Childs, Turbomachinery Rotordynamics: Phenomena, Modeling, and Analysis, Wiley, New York, 1993. G. F. Wislicenus, Fluid Mechanics of Turbomachinery, 2d ed., McGraw-Hill, New York,
1965. S. L. Dixon, Fluid Mechanics and Thermodynamics of Turbomachinery, 2d ed., Butterworth, London, 1998. R. I. Lewis, Turbomachinery Performance Analysis, Wiley, New York, 1996. E. S. Logan, Jr.,
Turbomachinery: Basic Theory and Applications, 2d ed., Marcel Dekker, New York, 1993. A. J. Stepanoff, Centrifugal and Axial Flow Pumps, 2d ed., Wiley, New York, 1957. J. W. Dufour and W. E. Nelson,
Centrifugal Pump Sourcebook, McGraw-Hill, New York, 1992. Sam Yedidiah, Centrifugal Pump User’s Guidebook: Problems and Solutions, Chapman and Hall, New York, 1996. T. G. Hicks and T. W. Edwards,
Pump Application Engineering, McGraw-Hill, New York, 1971. Pump Selector for Industry, Worthington Pump, Mountainside, NJ, 1977. R. Walker, Pump Selection, 2d ed., Butterworth, London, 1979. R. H.
Warring, Pumps: Selection, Systems, and Applications, 2d ed., Gulf Pub., Houston, TX, 1984. V. L. Lobanoff and R. R. Ross, Centrifugal Pumps, Design and Application, Gulf Pub., Houston, TX, 1985. H.
L. Stewart, Pumps, 5th ed. Macmillan, New York, 1991.
18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29.
30. 31. 32.
A. B. McKenzie, Axial Flow Fans and Compressors: Aerodynamic Design and Performance, Ashgate Publishing, Brookfield, VT, 1997. W. C. Osborne, Fans, 2d ed., Pergamon, London, 1977. R. Jorgensen (ed.),
Fan Engineering, Buffalo Forge, Buffalo, NY, 1983. R. A. Wallis, Axial Flow Fans and Ducts, Wiley, New York, 1983. H. P. Bloch, A Practical Guide to Compressor Technology, McGraw-Hill, New York,
1996. H. Cohen et al., Gas Turbine Theory, Longman, London, 1996. N. A. Cumpsty, Compressor Aerodynamics, Longmans, London, 1989. G. C. Oates, Aerothermodynamics of Gas Turbine and Rocket Propulsion,
3d ed., AIAA, Washington, DC, 1998. W. W. Bathe, Fundamentals of Gas Turbines, 2d ed., Wiley, New York, 1995. J. Tong (ed.), Mini-Hydropower, Wiley, New York, 1997. C. C. Warnick, Hydropower
Engineering, Prentice-Hall, Englewood Cliffs, NJ, 1984. J. J. Fritz, Small and Mini Hydropower Systems, McGrawHill, New York, 1984. N. P. Cheremisinoff and P. N. Cheremisinoff, Pumps/Compressors/
Fans: Pocket Handbook, Technomic Publishing, Lancaster, PA, 1989. Hydraulic Institute, Hydraulic Institute Standards for Centrifugal, Rotating, and Reciprocal Pumps, New York, 1983. I. J. Karassik,
W. C. Krutzsch, W. H. Fraser, and J. P. Messina, Pump Handbook, 2d ed., McGraw-Hill, New York, 1985. J. S. Gulliver and R. E. A. Arndt, Hydropower Engineering Handbook, McGraw-Hill, New York, 1990.
Chapter 11 Turbomachinery
R. L. Daugherty, J. B. Franzini, and E. J. Finnemore, Fluid Mechanics and Engineering Applications, 8th ed., McGrawHill, New York, 1985. R. H. Sabersky, A. J. Acosta, and E. G. Hauptmann, Fluid Flow:
A First Course in Fluid Mechanics, 3d ed., Macmillan, New York, 1989. J. P. Poynton, Metering Pumps, Marcel-Dekker, New York, 1983. R. P. Lambeck, Hydraulic Pumps and Motors: Selection and
Application for Hydraulic Power Control Systems, Marcel Dekker, New York, 1983. T. L. Henshaw, Reciprocating Pumps, Van Nostrand Reinhold, New York, 1987. J. E. Miller, The Reciprocating Pump:
Theory, Design and Use, Wiley, NewYork, 1987. D. G. Wilson, The Design of High-Efficiency Turbomachinery and Gas Turbines, 2d ed., Prentice-Hall, Upper Saddle River, N.J., 1998. M. C. Roco, P.
Hamelin, T. Cader, and G. Davidson, “Animation of LDV Measurements in a Centrifugal Pump,” in Fluid Machinery Forum—1990, U.S. Rohatgi (ed.), FED, vol. 96, American Society of Mechanical Engineers,
New York, 1990. E. M. Greitzer, “The Stability of Pumping Systems: The 1980 Freeman Scholar Lecture,” J. Fluids Eng., vol. 103, June 1981, pp. 193–242. B. Lakshminarayana, Fluid Dynamics and Heat
Transfer in Turbomachinery, Wiley, New York, 1995. L. F. Moody, “The Propeller Type Turbine,” ASCE Trans., vol. 89, 1926, p. 628. H. H. Anderson, “Prediction of Head, Quantity, and Efficiency in
Pumps—The Area-Ratio Principle,” in Perfor-
35. 36.
37. 38. 39.
42. 43. 44.
47. 48. 49. 50. 51. 52. 53. 54.
mance Prediction of Centrifugal Pumps and Compressors, vol. I00127, ASME Symp., New York, 1980, pp. 201–211. B. Lakshminarayana and P. Runstadler, Jr. (eds.), “Measurement Methods in Rotating
Components of Turbomachinery,” ASME Symp. Proc., New Orleans, vol. I00130, 1980. M. Murakami, K. Kikuyama, and B. Asakura, “Velocity and Pressure Distributions in the Impeller Passages of Centrifugal
Pumps,” J. Fluids Eng., vol. 102, December 1980, pp. 420–426. G. W. Koeppl, Putnam’s Power from the Wind, 2d ed., Van Nostrand Reinhold, New York, 1982. R. L. Hills, Power from Wind: A History of
Windmill Technology, Cambridge Univ. Press, Cambridge, 1996. D. A. Spera, Wind Turbine Technology: Fundamental Concepts of Wind Turbine Engineering, ASME Press, New York, 1994. A. J. Wortman,
Introduction to Wind Turbine Engineering, Butterworth, Woburn, Mass., 1983. F. R. Eldridge, Wind Machines, 2d ed., Van Nostrand Reinhold, New York, 1980. D. M. Eggleston and F. S. Stoddard, Wind
Turbine Engineering Design, Van Nostrand, New York, 1987. M. L. Robinson, “The Darrieus Wind Turbine for Electrical Power Generation,” Aeronaut. J., June 1981, pp. 244–255. D. F. Warne and P. G.
Calnan, “Generation of Electricity from the Wind,” IEE Rev., vol. 124, no. 11R, November 1977, pp. 963–985. L. A. Haimerl, “The Crossflow Turbine,” Waterpower, January 1960, pp. 5–13; see also ASME
Symp. Small Hydropower Fluid Mach., vol. 1, 1980, and vol. 2, 1982. K. Eisele et al., “Flow Analysis in a Pump Diffuser: Part 1, Measurements; Part 2, CFD,” J. Fluids Eng,, vol. 119, December 1997,
pp. 968–984.
Appendix A Physical Properties of Fluids 0.5 0.4 0.3 0.2 0.1
Castor oil
SAE 10 oil
0.04 0.03
SAE 30 oil
Crude oil (SG 0.86)
Absolute viscosity µ, N ⋅ s / m2
0.01 6 4 3
Kerosine Aniline
Carb 1 × 10 – 3
on t
6 Ethyl alcohol
Water Gasoline (SG 0.68)
2 1 × 10 – 4 6 4 3
2 Carbon dioxide 1×
Fig. A.1 Absolute viscosity of common fluids at 1 atm.
10 – 5 5 – 20
Hydrogen 0
40 60 Temperature, °C
Appendix A 1 × 10 – 3 8 6 4 3 2
Glycerin Helium SAE 10 oil Hydrogen
Kinematic viscosity ν, m2 / s
1 × 10 – 4 8 6
SAE 30 oil
Air and oxygen
2 Carbon dioxide
1 × 10 – 5 8 6
Crude oil (SG 0.86)
4 3 2 1 × 10 – 6 8 6
Kerosine Benzene Ethyl alcohol Water
Gasoline (SG 0.68)
Carbon tetrachloride
Fig. A.2 Kinematic viscosity of common fluids at 1 atm.
1 × 10 – 7 –20
40 60 Temperature, °C
Physical Properties of Fluids Table A.1 Viscosity and Density of Water at 1 atm
T, °C
, kg/m3
, N s/m2 1.788 1.307 1.003 0.799 0.657 0.548 0.467 0.405 0.355 0.316 0.283
E-3 E-3 E-3 E-3 E-3 E-3 E-3 E-3 E-3 E-3 E-3
, m2/s 1.788 1.307 1.005 0.802 0.662 0.555 0.475 0.414 0.365 0.327 0.295
E-6 E-6 E-6 E-6 E-6 E-6 E-6 E-6 E-6 E-6 E-6
T, °F
, slug/ft3
1.940 1.940 1.937 1.932 1.925 1.917 1.908 1.897 1.886 1.873 1.859
, lb s/ft2 3.73 2.73 2.09 1.67 1.37 1.14 0.975 0.846 0.741 0.660 0.591
E-5 E-5 E-5 E-5 E-5 E-5 E-5 E-5 E-5 E-5 E-5
, ft2/s 1.925 E-5 1.407 E-5 1.082 E-5 0.864 E-5 0.713 E-5 0.597 E-5 0.511 E-5 0.446 E-5 0.393 E-5 0.352 E-5 0.318 E-5
Suggested curve fits for water in the range 0 T 100°C:
(kg/m3) 1000 0.0178T°C 4°C1.7 0.2% ln 1.704 5.306z 7.003z2 0 273 K z TK Table A.2 Viscosity and Density of Air at 1 atm
T, °C
, kg/m3
1.520 1.290 1.200 1.090 0.946 0.835 0.746 0.675 0.616 0.525 0.457
0 1.788 E-3 kg/(m s)
, N s/m2 1.51 1.71 1.80 1.95 2.17 2.38 2.57 2.75 2.93 3.25 3.55
E-5 E-5 E-5 E-5 E-5 E-5 E-5 E-5 E-5 E-5 E-5
, m2/s
T, °F
0.99 1.33 1.50 1.79 2.30 2.85 3.45 4.08 4.75 6.20 7.77
E-5 E-5 E-5 E-5 E-5 E-5 E-5 E-5 E-5 E-5 E-5
, slug/ft3 2.94 2.51 2.34 2.12 1.84 1.62 1.45 1.31 1.20 1.02 0.89
E-3 E-3 E-3 E-3 E-3 E-3 E-3 E-3 E-3 E-3 E-3
Suggested curve fits for air: p RT Power law: Sutherland law:
T 0 T0
T 0 T0
Rair 287 J/(kg K)
T0 S
TS 3/2
Sair 110.4 K
with T0 273 K, 0 1.71 E-5 kg/(m s), and T in kelvins.
, lb s/ft2 3.16 E-7 3.58 E-7 3.76 E-7 4.08 E-7 4.54 E-7 4.97 E-7 5.37 E-7 5.75 E-7 6.11 E-7 6.79 E-7 7.41 E-7
, ft2/s 1.07 1.43 1.61 1.93 2.47 3.07 3.71 4.39 5.12 6.67 8.37
E-4 E-4 E-4 E-4 E-4 E-4 E-4 E-4 E-4 E-4 E-4
Appendix A
Table A.3 Properties of Common Liquids at 1 atm and 20°C (68°F)
, kg/m3 , kg/(m s) , N/m*
Liquid Ammonia Benzene Carbon tetrachloride Ethanol Ethylene glycol Freon 12 Gasoline Glycerin Kerosine Mercury Methanol SAE 10W oil SAE 10W30 oil SAE 30W oil SAE 50W oil Water Seawater (30%) * †
13,608 13,881 31,590 13,789 31,117 31,327 13,680 31,260 13,804 13,550 13,791 13,870 13,876 13,891 13,902 13,998 31,025
2.20 E-4 6.51 E-4 9.67 E-4 1.20 E-3 2.14 E-2 2.62 E-4 2.92 E-4 1.49 1.92 E-3 1.56 E-3 5.98 E-4 1.04 E-1‡ 1.7 E-1‡ 2.9 E-1‡ 8.6 E-1‡ 1.00 E-3 1.07 E-3
2.13 E-2 2.88 E-2 2.70 E-2 2.28 E-2 4.84 E-2 — 2.16 E-2 6.33 E-2 2.83 E-2 4.84 E-1 2.25 E-2 3.63 E-2 — 3.53 E-2 — 7.28 E-2 7.28 E-2
p, N/m2 9.10 1.01 1.20 5.73 1.23
E5 E4 E4 E3 E1 — 5.51 E4 1.43 E-2 3.11 E3 1.13 E-3 1.34 E4 — — — — 2.34 E3 2.34 E3
Bulk modulus, N/m2
Viscosity parameter C†
— 1.43 E9 9.65 E8 9.03 E8 — — 9.58 E80 4.34 E90 1.63 E90 2.55 E10 8.33 E80 1.31 E90 — 1.38 E90 — 2.19 E90 2.33 E90
1.05 4.34 4.45 5.72 11.7 1.76 3.68 28.0 5.56 1.07 4.63 15.7 14.0 18.3 20.2 Table A.1 7.28
In contact with air. The viscosity-temperature variation of these liquids may be fitted to the empirical expression
293 K exp C 1 20°C TK
with accuracy of 6 percent in the range 0 T 100°C. ‡ Representative values. The SAE oil classifications allow a viscosity variation of up to 50 percent, especially at lower temperatures.
Table A.4 Properties of Common Gases at 1 atm and 20°C (68°F)
Molecular weight
R, m2/(s2 K)
g, N/m3
H2 He H2O Ar Dry air CO2 CO N2 O2 NO N2O Cl2 CH4
2.016 4.003 18.020 39.944 28.960 44.010 28.010 28.020 32.000 30.010 44.020 70.910 16.040
00.822 01.630 07.350 16.300 11.800 17.900 11.400 11.400 13.100 12.100 17.900 28.900 06.540
, N s/m2 9.05 1.97 1.02 2.24 1.80 1.48 1.82 1.76 2.00 1.90 1.45 1.03 1.34
E-6 E-5 E-5 E-5 E-5 E-5 E-5 E-5 E-5 E-5 E-5 E-5 E-5
Specific-heat ratio
Power-law exponent n†
1.41 1.66 1.33 1.67 1.40 1.30 1.40 1.40 1.40 1.40 1.31 1.34 1.32
0.68 0.67 1.15 0.72 0.67 0.79 0.71 0.67 0.69 0.78 0.89 1.00 0.87
The power-law curve fit, Eq. (1.27), /293K (T/293)n, fits these gases to within 4 percent in the range 250 T 1000 K. The temperature must be in kelvins.
Physical Properties of Fluids Table A.5 Surface T, °C Tension, Vapor 0 Pressure, and Sound 10 Speed of Water
, N/m
0.0756 0.0742 0.0728 0.0712 0.0696 0.0679 0.0662 0.0644 0.0626 0.0608 0.0589
120 140 160 180 200 220 240 260 280 300 320 340 360 374*
0.0550 0.0509 0.0466 0.0422 0.0377 0.0331 0.0284 0.0237 0.0190 0.0144 0.0099 0.0056 0.0019 0.0*19
*Critical point.
p, kPa 0.611 1.227 2.337 4.242 7.375 12.34 19.92 31.16 47.35 70.11 101.3 198.5 361.3 617.8 1,002 1,554 2,318 3,344 4,688 6,412 8,581 11,274 14,586 18,651 22,090*
a, m/s 1402 1447 1482 1509 1529 1542 1551 1553 1554 1550 1543 1518 1483 1440 1389 1334 1268 1192 1110 1022 920 800 630 370 0*
Table A.6 Propz, m erties of the 0500 Standard 0,0000 Atmosphere 0,0500 01,000 01,500 02,000 02,500 03,000 03,500 04,000 04,500 05,000 05,500 06,000 06,500 07,000 07,500 08,000 08,500 09,000 09,500
10,000 10,500 11,000 11,500 12,000 12,500 13,000 13,500 14,000 14,500 15,000 15,500 16,000 16,500 17,000 17,500 18,000 18,500 19,000 19,500 20,000 22,000 24,000 26,000 28,000 30,000 40,000 50,000
60,000 70,000
T, K
p, Pa
, kg/m3
a, m/s
291.41 288.16 284.91 281.66 278.41 275.16 271.91 268.66 265.41 262.16 258.91 255.66 252.41 249.16 245.91 242.66 239.41 236.16 232.91 229.66 226.41 223.16 219.91 216.66 216.66 216.66 216.66 216.66
216.66 216.66 216.66 216.66 216.66 216.66 216.66 216.66 216.66 216.66 216.66 216.66 216.66 216.66 218.60 220.60 222.50 224.50 226.50 250.40 270.70 255.70 219.70
107,508 101,350 095,480 089,889 084,565 079,500 074,684 070,107 065,759 061,633 057,718 054,008 050,493 047,166 044,018 041,043 038,233 035,581 033,080 030,723 028,504 026,416 024,455 022,612 020,897
019,312 017,847 016,494 015,243 014,087 013,018 012,031 011,118 010,275 009,496 008,775 008,110 007,495 006,926 006,401 005,915 005,467 004,048 002,972 002,189 001,616 001,197 000,287 000,080 000,022
1.2854 1.2255 1.1677 1.1120 1.0583 1.0067 0.9570 0.9092 0.8633 0.8191 0.7768 0.7361 0.6970 0.6596 0.6237 0.5893 0.5564 0.5250 0.4949 0.4661 0.4387 0.4125 0.3875 0.3637 0.3361 0.3106 0.2870 0.2652
0.2451 0.2265 0.2094 0.1935 0.1788 0.1652 0.1527 0.1411 0.1304 0.1205 0.1114 0.1029 0.0951 0.0879 0.0645 0.0469 0.0343 0.0251 0.0184 0.0040 0.0010 0.0003 0.0001
342.2 340.3 338.4 336.5 334.5 332.6 330.6 328.6 326.6 324.6 322.6 320.6 318.5 316.5 314.4 312.3 310.2 308.1 306.0 303.8 301.7 299.5 297.3 295.1 295.1 295.1 295.1 295.1 295.1 295.1 295.1 295.1 295.1
295.1 295.1 295.1 295.1 295.1 295.1 295.1 295.1 295.1 296.4 297.8 299.1 300.4 301.7 317.2 329.9 320.6 297.2
Appendix B Compressible-Flow Tables Table B.1 Isentropic Flow of a Perfect Gas, k 1.4
0.0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26 0.28 0.3 0.32 0.34 0.36 0.38 0.4 0.42 0.44 0.46 0.48 0.5 0.52 0.54 0.56 0.58 0.6 0.62 0.64 0.66 0.68 0.7 0.72
1.0 0.9997 0.9989 0.9975 0.9955 0.9930 0.9900 0.9864 0.9823 0.9776 0.9725 0.9668 0.9607 0.9541 0.9470 0.9395 0.9315 0.9231 0.9143 0.9052 0.8956 0.8857 0.8755 0.8650 0.8541 0.8430 0.8317 0.8201 0.8082
0.7962 0.7840 0.7716 0.7591 0.7465 0.7338 0.7209 0.7080
1.0 0.9998 0.9992 0.9982 0.9968 0.9950 0.9928 0.9903 0.9873 0.9840 0.9803 0.9762 0.9718 0.9670 0.9619 0.9564 0.9506 0.9445 0.9380 0.9313 0.9243 0.9170 0.9094 0.9016 0.8935 0.8852 0.8766 0.8679 0.8589
0.8498 0.8405 0.8310 0.8213 0.8115 0.8016 0.7916 0.7814
1.0 0.9999 0.9997 0.9993 0.9987 0.9980 0.9971 0.9961 0.9949 0.9936 0.9921 0.9904 0.9886 0.9867 0.9846 0.9823 0.9799 0.9774 0.9747 0.9719 0.9690 0.9659 0.9627 0.9594 0.9559 0.9524 0.9487 0.9449 0.9410
0.9370 0.9328 0.9286 0.9243 0.9199 0.9153 0.9107 0.9061
28.9421 14.4815 9.6659 7.2616 5.8218 4.8643 4.1824 3.6727 3.2779 2.9635 2.7076 2.4956 2.3173 2.1656 2.0351 1.9219 1.8229 1.7358 1.6587 1.5901 1.5289 1.4740 1.4246 1.3801 1.3398 1.3034 1.2703 1.2403
1.2130 1.1882 1.1656 1.1451 1.1265 1.1097 1.0944 1.0806
0.74 0.76 0.78 0.8 0.82 0.84 0.86 0.88 0.9 0.92 0.94 0.96 0.98 1.0 1.02 1.04 1.06 1.08 1.1 1.12 1.14 1.16 1.18 1.2 1.22 1.24 1.26 1.28 1.3 1.32 1.34 1.36 1.38 1.4 1.42 1.44 1.46
0.6951 0.6821 0.6690 0.6560 0.6430 0.6300 0.6170 0.6041 0.5913 0.5785 0.5658 0.5532 0.5407 0.5283 0.5160 0.5039 0.4919 0.4800 0.4684 0.4568 0.4455 0.4343 0.4232 0.4124 0.4017 0.3912 0.3809 0.3708
0.3609 0.3512 0.3417 0.3323 0.3232 0.3142 0.3055 0.2969 0.2886
0.7712 0.7609 0.7505 0.7400 0.7295 0.7189 0.7083 0.6977 0.6870 0.6764 0.6658 0.6551 0.6445 0.6339 0.6234 0.6129 0.6024 0.5920 0.5817 0.5714 0.5612 0.5511 0.5411 0.5311 0.5213 0.5115 0.5019 0.4923
0.4829 0.4736 0.4644 0.4553 0.4463 0.4374 0.4287 0.4201 0.4116
0.9013 0.8964 0.8915 0.8865 0.8815 0.8763 0.8711 0.8659 0.8606 0.8552 0.8498 0.8444 0.8389 0.8333 0.8278 0.8222 0.8165 0.8108 0.8052 0.7994 0.7937 0.7879 0.7822 0.7764 0.7706 0.7648 0.7590 0.7532
0.7474 0.7416 0.7358 0.7300 0.7242 0.7184 0.7126 0.7069 0.7011
1.0681 1.0570 1.0471 1.0382 1.0305 1.0237 1.0179 1.0129 1.0089 1.0056 1.0031 1.0014 1.0003 1.0000 1.0003 1.0013 1.0029 1.0051 1.0079 1.0113 1.0153 1.0198 1.0248 1.0304 1.0366 1.0432 1.0504 1.0581
1.0663 1.0750 1.0842 1.0940 1.1042 1.1149 1.1262 1.1379 1.1501
Compressible-Flow Tables 775 Table B.1 (Cont.) Isentropic Flow of a Perfect Gas, k 1.4
1.48 1.5 1.52 1.54 1.56 1.58 1.6 1.62 1.64 1.66 1.68 1.7 1.72 1.74 1.76 1.78 1.8 1.82 1.84 1.86 1.88 1.9 1.92 1.94 1.96 1.98 2.0 2.02 2.04 2.06 2.08 2.1 2.12 2.14 2.16 2.18 2.2 2.22 2.24 2.26 2.28
2.3 2.32 2.34 2.36 2.38 2.4 2.42 2.44 2.46 2.48 2.5 2.52 2.54
0.2804 0.2724 0.2646 0.2570 0.2496 0.2423 0.2353 0.2284 0.2217 0.2151 0.2088 0.2026 0.1966 0.1907 0.1850 0.1794 0.1740 0.1688 0.1637 0.1587 0.1539 0.1492 0.1447 0.1403 0.1360 0.1318 0.1278 0.1239
0.1201 0.1164 0.1128 0.1094 0.1060 0.1027 0.0996 0.0965 0.0935 0.0906 0.0878 0.0851 0.0825 0.0800 0.0775 0.0751 0.0728 0.0706 0.0684 0.0663 0.0643 0.0623 0.0604 0.0585 0.0567 0.0550
0.4032 0.3950 0.3869 0.3789 0.3710 0.3633 0.3557 0.3483 0.3409 0.3337 0.3266 0.3197 0.3129 0.3062 0.2996 0.2931 0.2868 0.2806 0.2745 0.2686 0.2627 0.2570 0.2514 0.2459 0.2405 0.2352 0.2300 0.2250
0.2200 0.2152 0.2104 0.2058 0.2013 0.1968 0.1925 0.1882 0.1841 0.1800 0.1760 0.1721 0.1683 0.1646 0.1609 0.1574 0.1539 0.1505 0.1472 0.1439 0.1408 0.1377 0.1346 0.1317 0.1288 0.1260
0.6954 0.6897 0.6840 0.6783 0.6726 0.6670 0.6614 0.6558 0.6502 0.6447 0.6392 0.6337 0.6283 0.6229 0.6175 0.6121 0.6068 0.6015 0.5963 0.5910 0.5859 0.5807 0.5756 0.5705 0.5655 0.5605 0.5556 0.5506
0.5458 0.5409 0.5361 0.5313 0.5266 0.5219 0.5173 0.5127 0.5081 0.5036 0.4991 0.4947 0.4903 0.4859 0.4816 0.4773 0.4731 0.4688 0.4647 0.4606 0.4565 0.4524 0.4484 0.4444 0.4405 0.4366
1.1629 1.1762 1.1899 1.2042 1.2190 1.2344 1.2502 1.2666 1.2836 1.3010 1.3190 1.3376 1.3567 1.3764 1.3967 1.4175 1.4390 1.4610 1.4836 1.5069 1.5308 1.5553 1.5804 1.6062 1.6326 1.6597 1.6875 1.7160
1.7451 1.7750 1.8056 1.8369 1.8690 1.9018 1.9354 1.9698 2.0050 2.0409 2.0777 2.1153 2.1538 2.1931 2.2333 2.2744 2.3164 2.3593 2.4031 2.4479 2.4936 2.5403 2.5880 2.6367 2.6865 2.7372
2.56 2.58 2.6 2.62 2.64 2.66 2.68 2.7 2.72 2.74 2.76 2.78 2.8 2.82 2.84 2.86 2.88 2.9 2.92 2.94 2.96 2.98 3.0 3.02 3.04 3.06 3.08 3.1 3.12 3.14 3.16 3.18 3.2 3.22 3.24 3.26 3.28 3.3 3.32 3.34 3.36
3.38 3.4 3.42 3.44 3.46 3.48 3.5 3.52 3.54 3.56 3.58 3.6 3.62
0.0533 0.0517 0.0501 0.0486 0.0471 0.0457 0.0443 0.0430 0.0417 0.0404 0.0392 0.0380 0.0368 0.0357 0.0347 0.0336 0.0326 0.0317 0.0307 0.0298 0.0289 0.0281 0.0272 0.0264 0.0256 0.0249 0.0242 0.0234
0.0228 0.0221 0.0215 0.0208 0.0202 0.0196 0.0191 0.0185 0.0180 0.0175 0.0170 0.0165 0.0160 0.0156 0.0151 0.0147 0.0143 0.0139 0.0135 0.0131 0.0127 0.0124 0.0120 0.0117 0.0114 0.0111
0.1232 0.1205 0.1179 0.1153 0.1128 0.1103 0.1079 0.1056 0.1033 0.1010 0.0989 0.0967 0.0946 0.0926 0.0906 0.0886 0.0867 0.0849 0.0831 0.0813 0.0796 0.0779 0.0762 0.0746 0.0730 0.0715 0.0700 0.0685
0.0671 0.0657 0.0643 0.0630 0.0617 0.0604 0.0591 0.0579 0.0567 0.0555 0.0544 0.0533 0.0522 0.0511 0.0501 0.0491 0.0481 0.0471 0.0462 0.0452 0.0443 0.0434 0.0426 0.0417 0.0409 0.0401
0.4328 0.4289 0.4252 0.4214 0.4177 0.4141 0.4104 0.4068 0.4033 0.3998 0.3963 0.3928 0.3894 0.3860 0.3827 0.3794 0.3761 0.3729 0.3696 0.3665 0.3633 0.3602 0.3571 0.3541 0.3511 0.3481 0.3452 0.3422
0.3393 0.3365 0.3337 0.3309 0.3281 0.3253 0.3226 0.3199 0.3173 0.3147 0.3121 0.3095 0.3069 0.3044 0.3019 0.2995 0.2970 0.2946 0.2922 0.2899 0.2875 0.2852 0.2829 0.2806 0.2784 0.2762
2.7891 2.8420 2.8960 2.9511 3.0073 3.0647 3.1233 3.1830 3.2440 3.3061 3.3695 3.4342 3.5001 3.5674 3.6359 3.7058 3.7771 3.8498 3.9238 3.9993 4.0763 4.1547 4.2346 4.3160 4.3990 4.4835 4.5696 4.6573
4.7467 4.8377 4.9304 5.0248 5.1210 5.2189 5.3186 5.4201 5.5234 5.6286 5.7358 5.8448 5.9558 6.0687 6.1837 6.3007 6.4198 6.5409 6.6642 6.7896 6.9172 7.0471 7.1791 7.3135 7.4501 7.5891
Appendix B
Table B.1 (Cont.) Isentropic Flow of a Perfect Gas, k 1.4
3.64 3.66 3.68 3.7 3.72 3.74 3.76 3.78 3.8 3.82 3.84 3.86 3.88 3.9 3.92 3.94 3.96 3.98 4.0 4.02 4.04 4.06 4.08 4.1 4.12 4.14 4.16 4.18 4.2 4.22 4.24 4.26 4.28 4.3 4.32
0.0108 0.0105 0.0102 0.0099 0.0096 0.0094 0.0091 0.0089 0.0086 0.0084 0.0082 0.0080 0.0077 0.0075 0.0073 0.0071 0.0069 0.0068 0.0066 0.0064 0.0062 0.0061 0.0059 0.0058 0.0056 0.0055 0.0053 0.0052
0.0051 0.0049 0.0048 0.0047 0.0046 0.0044 0.0043
0.0393 0.0385 0.0378 0.0370 0.0363 0.0356 0.0349 0.0342 0.0335 0.0329 0.0323 0.0316 0.0310 0.0304 0.0299 0.0293 0.0287 0.0282 0.0277 0.0271 0.0266 0.0261 0.0256 0.0252 0.0247 0.0242 0.0238 0.0234
0.0229 0.0225 0.0221 0.0217 0.0213 0.0209 0.0205
0.2740 0.2718 0.2697 0.2675 0.2654 0.2633 0.2613 0.2592 0.2572 0.2552 0.2532 0.2513 0.2493 0.2474 0.2455 0.2436 0.2418 0.2399 0.2381 0.2363 0.2345 0.2327 0.2310 0.2293 0.2275 0.2258 0.2242 0.2225
0.2208 0.2192 0.2176 0.2160 0.2144 0.2129 0.2113
7.7305 7.8742 8.0204 8.1691 8.3202 8.4739 8.6302 8.7891 8.9506 9.1148 9.2817 9.4513 9.6237 9.7990 9.9771 10.1581 10.3420 10.5289 10.7188 10.9117 11.1077 11.3068 11.5091 11.7147 11.9234 12.1354
12.3508 12.5695 12.7916 13.0172 13.2463 13.4789 13.7151 13.9549 14.1984
4.34 4.36 4.38 4.4 4.42 4.44 4.46 4.48 4.5 4.52 4.54 4.56 4.58 4.6 4.62 4.64 4.66 4.68 4.7 4.72 4.74 4.76 4.78 4.8 4.82 4.84 4.86 4.88 4.9 4.92 4.94 4.96 4.98 5.0
0.0042 0.0041 0.0040 0.0039 0.0038 0.0037 0.0036 0.0035 0.0035 0.0034 0.0033 0.0032 0.0031 0.0031 0.0030 0.0029 0.0028 0.0028 0.0027 0.0026 0.0026 0.0025 0.0025 0.0024 0.0023 0.0023 0.0022 0.0022
0.0021 0.0021 0.0020 0.0020 0.0019 0.0019
0.0202 0.0198 0.0194 0.0191 0.0187 0.0184 0.0181 0.0178 0.0174 0.0171 0.0168 0.0165 0.0163 0.0160 0.0157 0.0154 0.0152 0.0149 0.0146 0.0144 0.0141 0.0139 0.0137 0.0134 0.0132 0.0130 0.0128 0.0125
0.0123 0.0121 0.0119 0.0117 0.0115 0.0113
0.2098 0.2083 0.2067 0.2053 0.2038 0.2023 0.2009 0.1994 0.1980 0.1966 0.1952 0.1938 0.1925 0.1911 0.1898 0.1885 0.1872 0.1859 0.1846 0.1833 0.1820 0.1808 0.1795 0.1783 0.1771 0.1759 0.1747 0.1735
0.1724 0.1712 0.1700 0.1689 0.1678 0.1667
14.4456 14.6965 14.9513 15.2099 15.4724 15.7388 16.0092 16.2837 16.5622 16.8449 17.1317 17.4228 17.7181 18.0178 18.3218 18.6303 18.9433 19.2608 19.5828 19.9095 20.2409 20.5770 20.9179 21.2637 21.6144
21.9700 22.3306 22.6963 23.0671 23.4431 23.8243 24.2109 24.6027 25.0000
Table B.2 Normal-Shock Relations for a Perfect Gas, k 1.4
V1/V2 2/1
1.0 1.02 1.04 1.06 1.08 1.1 1.12 1.14 1.16 1.18 1.2 1.22 1.24
1.0000 0.9805 0.9620 0.9444 0.9277 0.9118 0.8966 0.8820 0.8682 0.8549 0.8422 0.8300 0.8183
1.0000 1.0471 1.0952 1.1442 1.1941 1.2450 1.2968 1.3495 1.4032 1.4578 1.5133 1.5698 1.6272
1.0000 1.0334 1.0671 1.1009 1.1349 1.1691 1.2034 1.2378 1.2723 1.3069 1.3416 1.3764 1.4112
1.0000 1.0132 1.0263 1.0393 1.0522 1.0649 1.0776 1.0903 1.1029 1.1154 1.1280 1.1405 1.1531
1.0000 1.0000 0.9999 0.9998 0.9994 0.9989 0.9982 0.9973 0.9961 0.9946 0.9928 0.9907 0.9884
1.0000 1.0000 1.0001 1.0002 1.0006 1.0011 1.0018 1.0027 1.0040 1.0055 1.0073 1.0094 1.0118
Compressible-Flow Tables 777 Table B.2 (Cont.) Normal-Shock Relations for a Perfect Gas, k 1.4
V1/V2 2/1
1.26 1.28 1.3 1.32 1.34 1.36 1.38 1.4 1.42 1.44 1.46 1.48 1.5 1.52 1.54 1.56 1.58 1.6 1.62 1.64 1.66 1.68 1.7 1.72 1.74 1.76 1.78 1.8 1.82 1.84 1.86 1.88 1.9 1.92 1.94 1.96 1.98 2.0 2.02 2.04 2.06
2.08 2.1 2.12 2.14 2.16 2.18 2.2 2.22 2.24 2.26 2.28 2.3 2.32
0.8071 0.7963 0.7860 0.7760 0.7664 0.7572 0.7483 0.7397 0.7314 0.7235 0.7157 0.7083 0.7011 0.6941 0.6874 0.6809 0.6746 0.6684 0.6625 0.6568 0.6512 0.6458 0.6405 0.6355 0.6305 0.6257 0.6210 0.6165
0.6121 0.6078 0.6036 0.5996 0.5956 0.5918 0.5880 0.5844 0.5808 0.5774 0.5740 0.5707 0.5675 0.5643 0.5613 0.5583 0.5554 0.5525 0.5498 0.5471 0.5444 0.5418 0.5393 0.5368 0.5344 0.5321
1.6855 1.7448 1.8050 1.8661 1.9282 1.9912 2.0551 2.1200 2.1858 2.2525 2.3202 2.3888 2.4583 2.5288 2.6002 2.6725 2.7458 2.8200 2.8951 2.9712 3.0482 3.1261 3.2050 3.2848 3.3655 3.4472 3.5298 3.6133
3.6978 3.7832 3.8695 3.9568 4.0450 4.1341 4.2242 4.3152 4.4071 4.5000 4.5938 4.6885 4.7842 4.8808 4.9783 5.0768 5.1762 5.2765 5.3778 5.4800 5.5831 5.6872 5.7922 5.8981 6.0050 6.1128
1.4460 1.4808 1.5157 1.5505 1.5854 1.6202 1.6549 1.6897 1.7243 1.7589 1.7934 1.8278 1.8621 1.8963 1.9303 1.9643 1.9981 2.0317 2.0653 2.0986 2.1318 2.1649 2.1977 2.2304 2.2629 2.2952 2.3273 2.3592
2.3909 2.4224 2.4537 2.4848 2.5157 2.5463 2.5767 2.6069 2.6369 2.6667 2.6962 2.7255 2.7545 2.7833 2.8119 2.8402 2.8683 2.8962 2.9238 2.9512 2.9784 3.0053 3.0319 3.0584 3.0845 3.1105
1.1657 1.1783 1.1909 1.2035 1.2162 1.2290 1.2418 1.2547 1.2676 1.2807 1.2938 1.3069 1.3202 1.3336 1.3470 1.3606 1.3742 1.3880 1.4018 1.4158 1.4299 1.4440 1.4583 1.4727 1.4873 1.5019 1.5167 1.5316
1.5466 1.5617 1.5770 1.5924 1.6079 1.6236 1.6394 1.6553 1.6713 1.6875 1.7038 1.7203 1.7369 1.7536 1.7705 1.7875 1.8046 1.8219 1.8393 1.8569 1.8746 1.8924 1.9104 1.9285 1.9468 1.9652
0.9857 0.9827 0.9794 0.9758 0.9718 0.9676 0.9630 0.9582 0.9531 0.9476 0.9420 0.9360 0.9298 0.9233 0.9166 0.9097 0.9026 0.8952 0.8877 0.8799 0.8720 0.8639 0.8557 0.8474 0.8389 0.8302 0.8215 0.8127
0.8038 0.7948 0.7857 0.7765 0.7674 0.7581 0.7488 0.7395 0.7302 0.7209 0.7115 0.7022 0.6928 0.6835 0.6742 0.6649 0.6557 0.6464 0.6373 0.6281 0.6191 0.6100 0.6011 0.5921 0.5833 0.5745
1.0145 1.0176 1.0211 1.0249 1.0290 1.0335 1.0384 1.0436 1.0492 1.0552 1.0616 1.0684 1.0755 1.0830 1.0910 1.0993 1.1080 1.1171 1.1266 1.1365 1.1468 1.1575 1.1686 1.1801 1.1921 1.2045 1.2173 1.2305
1.2441 1.2582 1.2728 1.2877 1.3032 1.3191 1.3354 1.3522 1.3695 1.3872 1.4054 1.4241 1.4433 1.4630 1.4832 1.5039 1.5252 1.5469 1.5692 1.5920 1.6154 1.6393 1.6638 1.6888 1.7144 1.7406
Appendix B
Table B.2 (Cont.) Normal-Shock Relations for a Perfect Gas, k 1.4
V1/V2 2/1
2.34 2.36 2.38 2.4 2.42 2.44 2.46 2.48 2.5 2.52 2.54 2.56 2.58 2.6 2.62 2.64 2.66 2.68 2.7 2.72 2.74 2.76 2.78 2.8 2.82 2.84 2.86 2.88 2.9 2.92 2.94 2.96 2.98 3.0 3.02 3.04 3.06 3.08 3.1 3.12 3.14
3.16 3.18 3.2 3.22 3.24 3.26 3.28 3.3 3.32 3.34 3.36 3.38 3.4
0.5297 0.5275 0.5253 0.5231 0.5210 0.5189 0.5169 0.5149 0.5130 0.5111 0.5092 0.5074 0.5056 0.5039 0.5022 0.5005 0.4988 0.4972 0.4956 0.4941 0.4926 0.4911 0.4896 0.4882 0.4868 0.4854 0.4840 0.4827
0.4814 0.4801 0.4788 0.4776 0.4764 0.4752 0.4740 0.4729 0.4717 0.4706 0.4695 0.4685 0.4674 0.4664 0.4654 0.4643 0.4634 0.4624 0.4614 0.4605 0.4596 0.4587 0.4578 0.4569 0.4560 0.4552
6.2215 6.3312 6.4418 6.5533 6.6658 6.7792 6.8935 7.0088 7.1250 7.2421 7.3602 7.4792 7.5991 7.7200 7.8418 7.9645 8.0882 8.2128 8.3383 8.4648 8.5922 8.7205 8.8498 8.9800 9.1111 9.2432 9.3762 9.5101
9.6450 9.7808 9.9175 10.0552 10.1938 10.3333 10.4738 10.6152 10.7575 10.9008 11.0450 11.1901 11.3362 11.4832 11.6311 11.7800 11.9298 12.0805 12.2322 12.3848 12.5383 12.6928 12.8482 13.0045 13.1618
3.1362 3.1617 3.1869 3.2119 3.2367 3.2612 3.2855 3.3095 3.3333 3.3569 3.3803 3.4034 3.4263 3.4490 3.4714 3.4937 3.5157 3.5374 3.5590 3.5803 3.6015 3.6224 3.6431 2.6636 3.6838 3.7039 3.7238 3.7434
3.7629 3.7821 3.8012 3.8200 3.8387 3.8571 3.8754 3.8935 3.9114 3.9291 3.9466 3.9639 3.9811 3.9981 4.0149 4.0315 4.0479 4.0642 4.0803 4.0963 4.1120 4.1276 4.1431 4.1583 4.1734 4.1884
1.9838 2.0025 2.0213 2.0403 2.0595 2.0788 2.0982 2.1178 2.1375 2.1574 2.1774 2.1976 2.2179 2.2383 2.2590 2.2797 2.3006 2.3217 2.3429 2.3642 2.3858 2.4074 2.4292 2.4512 2.4733 2.4955 2.5179 2.5405
2.5632 2.5861 2.6091 2.6322 2.6555 2.6790 2.7026 2.7264 2.7503 2.7744 2.7986 2.8230 2.8475 2.8722 2.8970 2.9220 2.9471 2.9724 2.9979 3.0234 3.0492 3.0751 3.1011 3.1273 3.1537 3.1802
0.5658 0.5572 0.5486 0.5401 0.5317 0.5234 0.5152 0.5071 0.4990 0.4911 0.4832 0.4754 0.4677 0.4601 0.4526 0.4452 0.4379 0.4307 0.4236 0.4166 0.4097 0.4028 0.3961 0.3895 0.3829 0.3765 0.3701 0.3639
0.3577 0.3517 0.3457 0.3398 0.3340 0.3283 0.3227 0.3172 0.3118 0.3065 0.3012 0.2960 0.2910 0.2860 0.2811 0.2762 0.2715 0.2668 0.2622 0.2577 0.2533 0.2489 0.2446 0.2404 0.2363 0.2322
1.7674 1.7948 1.8228 1.8514 1.8806 1.9105 1.9410 1.9721 2.0039 2.0364 2.0696 2.1035 2.1381 2.1733 2.2093 2.2461 2.2835 2.3218 2.3608 2.4005 2.4411 2.4825 2.5246 2.5676 2.6115 2.6561 2.7017 2.7481
2.7954 2.8436 2.8927 2.9427 2.9937 3.0456 3.0985 3.1523 3.2072 3.2630 3.3199 3.3778 3.4368 3.4969 3.5580 3.6202 3.6835 3.7480 3.8136 3.8803 3.9483 4.0174 4.0877 4.1593 4.2321 4.3062
Compressible-Flow Tables 779 Table B.2 (Cont.) Normal-Shock Relations for a Perfect Gas, k 1.4
V1/V2 2/1
3.42 3.44 3.46 3.48 3.5 3.52 3.54 3.56 3.58 3.6 3.62 3.64 3.66 3.68 3.7 3.72 3.74 3.76 3.78 3.8 3.82 3.84 3.86 3.88 3.9 3.92 3.94 3.96 3.98 4.0 4.02 4.04 4.06 4.08 4.1 4.12 4.14 4.16 4.18 4.2 4.22
4.24 4.26 4.28 4.3 4.32 4.34 4.36 4.38 4.4 4.42 4.44 4.46 4.48
0.4544 0.4535 0.4527 0.4519 0.4512 0.4504 0.4496 0.4489 0.4481 0.4474 0.4467 0.4460 0.4453 0.4446 0.4439 0.4433 0.4426 0.4420 0.4414 0.4407 0.4401 0.4395 0.4389 0.4383 0.4377 0.4372 0.4366 0.4360
0.4355 0.4350 0.4344 0.4339 0.4334 0.4329 0.4324 0.4319 0.4314 0.4309 0.4304 0.4299 0.4295 0.4290 0.4286 0.4281 0.4277 0.4272 0.4268 0.4264 0.4260 0.4255 0.4251 0.4247 0.4243 0.4239
13.4791 13.6392 13.8002 13.9621 14.1250 14.2888 14.4535 14.6192 14.7858 14.9533 15.1218 15.2912 15.4615 15.6328 15.8050 15.9781 16.1522 16.3272 16.5031 16.6800 16.8578 17.0365 17.2162 17.3968 17.5783
17.7608 17.9442 18.1285 18.3138 18.5000 18.6871 18.8752 19.0642 19.2541 19.4450 19.6368 19.8295 20.0232 20.2178 20.4133 20.6098 20.8072 21.0055 21.2048 21.4050 21.6061 21.8082 22.0112 22.2151 22.4200
22.6258 22.8325 23.0402 23.2488
4.2032 4.2178 4.2323 4.2467 4.2609 4.2749 4.2888 4.3026 4.3162 4.3296 4.3429 4.3561 4.3692 4.3821 4.3949 4.4075 4.4200 4.4324 4.4447 4.4568 4.4688 4.4807 4.4924 4.5041 4.4156 4.5270 4.5383 4.5494
4.5605 4.5714 4.5823 4.5930 4.6036 4.6141 4.6245 4.6348 4.6450 4.6550 4.6650 4.6749 4.6847 4.6944 4.7040 4.7135 4.7229 4.7322 4.7414 4.7505 4.7595 4.7685 4.7773 4.7861 4.7948 4.8034
3.2069 3.2337 3.2607 3.2878 3.3151 3.3425 3.3701 3.3978 3.4257 3.4537 3.4819 3.5103 3.5388 3.5674 3.5962 3.6252 3.6543 3.6836 3.7130 3.7426 3.7723 3.8022 3.8323 3.8625 3.8928 3.9233 3.9540 3.9848
4.0158 4.0469 4.0781 4.1096 4.1412 4.1729 4.2048 4.2368 4.2690 4.3014 4.3339 4.3666 4.3994 4.4324 4.4655 4.4988 4.5322 4.5658 4.5995 4.6334 4.6675 4.7017 4.7361 4.7706 4.8053 4.8401
0.2282 0.2243 0.2205 0.2167 0.2129 0.2093 0.2057 0.2022 0.1987 0.1953 0.1920 0.1887 0.1855 0.1823 0.1792 0.1761 0.1731 0.1702 0.1673 0.1645 0.1617 0.1589 0.1563 0.1536 0.1510 0.1485 0.1460 0.1435
0.1411 0.1388 0.1364 0.1342 0.1319 0.1297 0.1276 0.1254 0.1234 0.1213 0.1193 0.1173 0.1154 0.1135 0.1116 0.1098 0.1080 0.1062 0.1045 0.1028 0.1011 0.0995 0.0979 0.0963 0.0947 0.0932
4.3815 4.4581 4.5361 4.6154 4.6960 4.7780 4.8614 4.9461 5.0324 5.1200 5.2091 5.2997 5.3918 5.4854 5.5806 5.6773 5.7756 5.8755 5.9770 6.0801 6.1849 6.2915 6.3997 6.5096 6.6213 6.7348 6.8501 6.9672
7.0861 7.2069 7.3296 7.4542 7.5807 7.7092 7.8397 7.9722 8.1067 8.2433 8.3819 8.5227 8.6656 8.8107 8.9579 9.1074 9.2591 9.4131 9.5694 9.7280 9.8889 10.0522 10.2179 10.3861 10.5567 10.7298
Appendix B
Table B.2 (Cont.) Normal-Shock Relations for a Perfect Gas, k 1.4
Table B.3 Adiabatic Frictional Flow in a Constant-Area Duct for k 1.4
V1/V2 2/1
4.5 4.52 4.54 4.56 4.58 4.6 4.62 4.64 4.66 4.68 4.7 4.72 4.74 4.76 4.78 4.8 4.82 4.84 4.86 4.88 4.9 4.92 4.94 4.96 4.98 5.0
0.4236 0.4232 0.4228 0.4224 0.4220 0.4217 0.4213 0.4210 0.4206 0.4203 0.4199 0.4196 0.4192 0.4189 0.4186 0.4183 0.4179 0.4176 0.4173 0.4170 0.4167 0.4164 0.4161 0.4158 0.4155 0.4152
23.4583 23.6688 23.8802 24.0925 24.3058 24.5200 24.7351 24.9512 25.1682 25.3861 25.6050 25.8248 26.0455 26.2672 26.4898 26.7133 26.9378 27.1632 27.3895 27.6168 27.8450 28.0741 28.3042 28.5352 28.7671
4.8119 4.8203 4.8287 4.8369 4.8451 4.8532 4.8612 4.8692 4.8771 4.8849 4.8926 4.9002 4.9078 4.9153 4.9227 4.9301 4.9374 4.9446 4.9518 4.9589 4.9659 4.9728 4.9797 4.9865 4.9933 5.0000
4.8751 4.9102 4.9455 4.9810 5.0166 5.0523 5.0882 5.1243 5.1605 5.1969 5.2334 5.2701 5.3070 5.3440 5.3811 5.4184 5.4559 5.4935 5.5313 5.5692 5.6073 5.6455 5.6839 5.7224 5.7611 5.8000
0.0917 0.0902 0.0888 0.0874 0.0860 0.0846 0.0832 0.0819 0.0806 0.0793 0.0781 0.0769 0.0756 0.0745 0.0733 0.0721 0.0710 0.0699 0.0688 0.0677 0.0667 0.0657 0.0647 0.0637 0.0627 0.0617
10.9054 11.0835 11.2643 11.4476 11.6336 11.8222 12.0136 12.2076 12.4044 12.6040 12.8065 13.0117 13.2199 13.4310 13.6450 13.8620 14.0820 14.3050 14.5312 14.7604 14.9928 15.2284 15.4672 15.7902 15.9545
f L*/D
*/ V/V*
0.0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26 0.28 0.3 0.32 0.34 0.36 0.38 0.4 0.42
1778.4500 440.3520 193.0310 106.7180 66.9216 45.4080 32.5113 24.1978 18.5427 14.5333 11.5961 9.3865 7.6876 6.3572 5.2993 4.4467 3.7520 3.1801 2.7054 2.3085 1.9744
54.7701 27.3817 18.2508 13.6843 10.9435 9.1156 7.8093 6.8291 6.0662 5.4554 4.9554 4.5383 4.1851 3.8820 3.6191 3.3887 3.1853 3.0042 2.8420 2.6958 2.5634
1.2000 1.1999 1.1996 1.1991 1.1985 1.1976 1.1966 1.1953 1.1939 1.1923 1.1905 1.1885 1.1863 1.1840 1.1815 1.1788 1.1759 1.1729 1.1697 1.1663 1.1628 1.1591
0.0 0.0219 0.0438 0.0657 0.0876 0.1094 0.1313 0.1531 0.1748 0.1965 0.2182 0.2398 0.2614 0.2829 0.3043 0.3257 0.3470 0.3682 0.3893 0.4104 0.4313 0.4522
28.9421 14.4815 9.6659 7.2616 5.8218 4.8643 4.1824 3.6727 3.2779 2.9635 2.7076 2.4956 2.3173 2.1656 2.0351 1.9219 1.8229 1.7358 1.6587 1.5901 1.5289
Compressible-Flow Tables 781 Table B.3 (Cont.) Adiabatic Frictional Flow in a Constant-Area Duct for k 1.4
f L*/D
*/ V/V*
0.44 0.46 0.48 0.5 0.52 0.54 0.56 0.58 0.6 0.62 0.64 0.66 0.68 0.7 0.72 0.74 0.76 0.78 0.8 0.82 0.84 0.86 0.88 0.9 0.92 0.94 0.96 0.98 1.0 1.02 1.04 1.06 1.08 1.1 1.12 1.14 1.16 1.18 1.2 1.22 1.24
1.26 1.28 1.3 1.32 1.34 1.36 1.38 1.4 1.42 1.44 1.46 1.48 1.5
1.6915 1.4509 1.2453 1.0691 0.9174 0.7866 0.6736 0.5757 0.4908 0.4172 0.3533 0.2979 0.2498 0.2081 0.1721 0.1411 0.1145 0.0917 0.0723 0.0559 0.0423 0.0310 0.0218 0.0145 0.0089 0.0048 0.0021 0.0005
0.0000 0.0005 0.0018 0.0038 0.0066 0.0099 0.0138 0.0182 0.0230 0.0281 0.0336 0.0394 0.0455 0.0517 0.0582 0.0648 0.0716 0.0785 0.0855 0.0926 0.0997 0.1069 0.1142 0.1215 0.1288 0.1361
2.4428 2.3326 2.2313 2.1381 2.0519 1.9719 1.8975 1.8282 1.7634 1.7026 1.6456 1.5919 1.5413 1.4935 1.4482 1.4054 1.3647 1.3261 1.2893 1.2542 1.2208 1.1889 1.1583 1.1291 1.1011 1.0743 1.0485 1.0238
1.0000 0.9771 0.9551 0.9338 0.9133 0.8936 0.8745 0.8561 0.8383 0.8210 0.8044 0.7882 0.7726 0.7574 0.7427 0.7285 0.7147 0.7012 0.6882 0.6755 0.6632 0.6512 0.6396 0.6282 0.6172 0.6065
1.1553 1.1513 1.1471 1.1429 1.1384 1.1339 1.1292 1.1244 1.1194 1.1143 1.1091 1.1038 1.0984 1.0929 1.0873 1.0815 1.0757 1.0698 1.0638 1.0578 1.0516 1.0454 1.0391 1.0327 1.0263 1.0198 1.0132 1.0066
1.0000 0.9933 0.9866 0.9798 0.9730 0.9662 0.9593 0.9524 0.9455 0.9386 0.9317 0.9247 0.9178 0.9108 0.9038 0.8969 0.8899 0.8829 0.8760 0.8690 0.8621 0.8551 0.8482 0.8413 0.8344 0.8276
0.4729 0.4936 0.5141 0.5345 0.5548 0.5750 0.5951 0.6150 0.6348 0.6545 0.6740 0.6934 0.7127 0.7318 0.7508 0.7696 0.7883 0.8068 0.8251 0.8433 0.8614 0.8793 0.8970 0.9146 0.9320 0.9493 0.9663 0.9833
1.0000 1.0166 1.0330 1.0492 1.0653 1.0812 1.0970 1.1126 1.1280 1.1432 1.1583 1.1732 1.1879 1.2025 1.2169 1.2311 1.2452 1.2591 1.2729 1.2864 1.2999 1.3131 1.3262 1.3392 1.3520 1.3646
1.4740 1.4246 1.3801 1.3398 1.3034 1.2703 1.2403 1.2130 1.1882 1.1656 1.1451 1.1265 1.1097 1.0944 1.0806 1.0681 1.0570 1.0471 1.0382 1.0305 1.0237 1.0179 1.0129 1.0089 1.0056 1.0031 1.0014 1.0003
1.0000 1.0003 1.0013 1.0029 1.0051 1.0079 1.0113 1.0153 1.0198 1.0248 1.0304 1.0366 1.0432 1.0504 1.0581 1.0663 1.0750 1.0842 1.0940 1.1042 1.1149 1.1262 1.1379 1.1501 1.1629 1.1762
Appendix B
Table B.3 (Cont.) Adiabatic Frictional Flow in a Constant-Area Duct for k 1.4
f L*/D
*/ V/V*
1.52 1.54 1.56 1.58 1.6 1.62 1.64 1.66 1.68 1.7 1.72 1.74 1.76 1.78 1.8 1.82 1.84 1.86 1.88 1.9 1.92 1.94 1.96 1.98 2.0 2.02 2.04 2.06 2.08 2.1 2.12 2.14 2.16 2.18 2.2 2.22 2.24 2.26 2.28 2.3 2.32
2.34 2.36 2.38 2.4 2.42 2.44 2.46 2.48 2.5 2.52 2.54 2.56 2.58
0.1433 0.1506 0.1579 0.1651 0.1724 0.1795 0.1867 0.1938 0.2008 0.2078 0.2147 0.2216 0.2284 0.2352 0.2419 0.2485 0.2551 0.2616 0.2680 0.2743 0.2806 0.2868 0.2929 0.2990 0.3050 0.3109 0.3168 0.3225
0.3282 0.3339 0.3394 0.3449 0.3503 0.3556 0.3609 0.3661 0.3712 0.3763 0.3813 0.3862 0.3911 0.3959 0.4006 0.4053 0.4099 0.4144 0.4189 0.4233 0.4277 0.4320 0.4362 0.4404 0.4445 0.4486
0.5960 0.5858 0.5759 0.5662 0.5568 0.5476 0.5386 0.5299 0.5213 0.5130 0.5048 0.4969 0.4891 0.4815 0.4741 0.4668 0.4597 0.4528 0.4460 0.4394 0.4329 0.4265 0.4203 0.4142 0.4082 0.4024 0.3967 0.3911
0.3856 0.3802 0.3750 0.3698 0.3648 0.3598 0.3549 0.3502 0.3455 0.3409 0.3364 0.3320 0.3277 0.3234 0.3193 0.3152 0.3111 0.3072 0.3033 0.2995 0.2958 0.2921 0.2885 0.2850 0.2815 0.2781
0.8207 0.8139 0.8071 0.8004 0.7937 0.7869 0.7803 0.7736 0.7670 0.7605 0.7539 0.7474 0.7410 0.7345 0.7282 0.7218 0.7155 0.7093 0.7030 0.6969 0.6907 0.6847 0.6786 0.6726 0.6667 0.6608 0.6549 0.6491
0.6433 0.6376 0.6320 0.6263 0.6208 0.6152 0.6098 0.6043 0.5989 0.5936 0.5883 0.5831 0.5779 0.5728 0.5677 0.5626 0.5576 0.5527 0.5478 0.5429 0.5381 0.5333 0.5286 0.5239 0.5193 0.5147
1.3770 1.3894 1.4015 1.4135 1.4254 1.4371 1.4487 1.4601 1.4713 1.4825 1.4935 1.5043 1.5150 1.5256 1.5360 1.5463 1.5564 1.5664 1.5763 1.5861 1.5957 1.6052 1.6146 1.6239 1.6330 1.6420 1.6509 1.6597
1.6683 1.6769 1.6853 1.6936 1.7018 1.7099 1.7179 1.7258 1.7336 1.7412 1.7488 1.7563 1.7637 1.7709 1.7781 1.7852 1.7922 1.7991 1.8059 1.8126 1.8192 1.8257 1.8322 1.8386 1.8448 1.8510
1.1899 1.2042 1.2190 1.2344 1.2502 1.2666 1.2836 1.3010 1.3190 1.3376 1.3567 1.3764 1.3967 1.4175 1.4390 1.4610 1.4836 1.5069 1.5308 1.5553 1.5804 1.6062 1.6326 1.6597 1.6875 1.7160 1.7451 1.7750
1.8056 1.8369 1.8690 1.9018 1.9354 1.9698 2.0050 2.0409 2.0777 2.1153 2.1538 2.1931 2.2333 2.2744 2.3164 2.3593 2.4031 2.4479 2.4936 2.5403 2.5880 2.6367 2.6865 2.7372 2.7891 2.8420
Compressible-Flow Tables 783 Table B.3 (Cont.) Adiabatic Frictional Flow in a Constant-Area Duct for k 1.4
f L*/D
*/ V/V*
2.6 2.62 2.64 2.66 2.68 2.7 2.72 2.74 2.76 2.78 2.8 2.82 2.84 2.86 2.88 2.9 2.92 2.94 2.96 2.98 3.0 3.02 3.04 3.06 3.08 3.1 3.12 3.14 3.16 3.18 3.2 3.22 3.24 3.26 3.28 3.3 3.32 3.34 3.36 3.38 3.4
3.42 3.44 3.46 3.48 3.5 3.52 3.54 3.56 3.58 3.6 3.62 3.64 3.66
0.4526 0.4565 0.4604 0.4643 0.4681 0.4718 0.4755 0.4791 0.4827 0.4863 0.4898 0.4932 0.4966 0.5000 0.5033 0.5065 0.5097 0.5129 0.5160 0.5191 0.5222 0.5252 0.5281 0.5310 0.5339 0.5368 0.5396 0.5424
0.5451 0.5478 0.5504 0.5531 0.5557 0.5582 0.5607 0.5632 0.5657 0.5681 0.5705 0.5729 0.5752 0.5775 0.5798 0.5820 0.5842 0.5864 0.5886 0.5907 0.5928 0.5949 0.5970 0.5990 0.6010 0.6030
0.2747 0.2714 0.2682 0.2650 0.2619 0.2588 0.2558 0.2528 0.2498 0.2470 0.2441 0.2414 0.2386 0.2359 0.2333 0.2307 0.2281 0.2256 0.2231 0.2206 0.2182 0.2158 0.2135 0.2112 0.2090 0.2067 0.2045 0.2024
0.2002 0.1981 0.1961 0.1940 0.1920 0.1901 0.1881 0.1862 0.1843 0.1825 0.1806 0.1788 0.1770 0.1753 0.1736 0.1718 0.1702 0.1685 0.1669 0.1653 0.1637 0.1621 0.1616 0.1590 0.1575 0.1560
0.5102 0.5057 0.5013 0.4969 0.4925 0.4882 0.4839 0.4797 0.4755 0.4714 0.4673 0.4632 0.4592 0.4552 0.4513 0.4474 0.4436 0.4398 0.4360 0.4323 0.4286 0.4249 0.4213 0.4177 0.4142 0.4107 0.4072 0.4038
0.4004 0.3970 0.3937 0.3904 0.3872 0.3839 0.3807 0.3776 0.3745 0.3714 0.3683 0.3653 0.3623 0.3594 0.3564 0.3535 0.3507 0.3478 0.3450 0.3422 0.3395 0.3368 0.3341 0.3314 0.3288 0.3262
1.8571 1.8632 1.8691 1.8750 1.8808 1.8865 1.8922 1.8978 1.9033 1.9087 1.9140 1.9193 1.9246 1.9297 1.9348 1.9398 1.9448 1.9497 1.9545 1.9593 1.9640 1.9686 1.9732 1.9777 1.9822 1.9866 1.9910 1.9953
1.9995 2.0037 2.0079 2.0120 2.0160 2.0200 2.0239 2.0278 2.0317 2.0355 2.0392 2.0429 2.0466 2.0502 2.0537 2.0573 2.0607 2.0642 2.0676 2.0709 2.0743 2.0775 2.0808 2.0840 2.0871 2.0903
2.8960 2.9511 3.0073 3.0647 3.1233 3.1830 3.2440 3.3061 3.3695 3.4342 3.5001 3.5674 3.6359 3.7058 3.7771 3.8498 3.9238 3.9993 4.0763 4.1547 4.2346 4.3160 4.3989 4.4835 4.5696 4.6573 4.7467 4.8377
4.9304 5.0248 5.1210 5.2189 5.3186 5.4201 5.5234 5.6286 5.7358 5.8448 5.9558 6.0687 6.1837 6.3007 6.4198 6.5409 6.6642 6.7896 6.9172 7.0471 7.1791 7.3135 7.4501 7.5891 7.7305 7.8742
Appendix B
Table B.3 (Cont.) Adiabatic Frictional Flow in a Constant-Area Duct for k 1.4
Table B.4 Frictionless Duct Flow with Heat Transfer for k 1.4
f L*/D
*/ V/V*
3.68 3.7 3.72 3.74 3.76 3.78 3.8 3.82 3.84 3.86 3.88 3.9 3.92 3.94 3.96 3.98 4.0
0.6049 0.6068 0.6087 0.6106 0.6125 0.6143 0.6161 0.6179 0.6197 0.6214 0.6231 0.6248 0.6265 0.6282 0.6298 0.6315 0.6331
0.1546 0.1531 0.1517 0.1503 0.1489 0.1475 0.1462 0.1449 0.1436 0.1423 0.1410 0.1397 0.1385 0.1372 0.1360 0.1348 0.1336
0.3236 0.3210 0.3185 0.3160 0.3135 0.3111 0.3086 0.3062 0.3039 0.3015 0.2992 0.2969 0.2946 0.2923 0.2901 0.2879 0.2857
2.0933 2.0964 2.0994 2.1024 2.1053 2.1082 2.1111 2.1140 2.1168 2.1195 2.1223 2.1250 2.1277 2.1303 2.1329 2.1355 2.1381
8.0204 8.1691 8.3202 8.4739 8.6302 8.7891 8.9506 9.1148 9.2817 9.4513 9.6237 9.7990 9.9771 10.1581 10.3420 10.5289 10.7188
*/ V/V*
0.0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26 0.28 0.3 0.32 0.34 0.36 0.38 0.4 0.42 0.44 0.46 0.48 0.5 0.52 0.54 0.56 0.58 0.6
0.0 0.0019 0.0076 0.0171 0.0302 0.0468 0.0666 0.0895 0.1151 0.1432 0.1736 0.2057 0.2395 0.2745 0.3104 0.3469 0.3837 0.4206 0.4572 0.4935 0.5290 0.5638 0.5975 0.6301 0.6614 0.6914 0.7199 0.7470 0.7725
0.7965 0.8189
2.4000 2.3987 2.3946 2.3800 2.3787 2.3669 2.3526 2.3359 2.3170 2.2959 2.2727 2.2477 2.2209 2.1925 2.1626 2.1314 2.0991 2.0657 2.0314 1.9964 1.9608 1.9247 1.8882 1.8515 1.8147 1.7778 1.7409 1.7043
1.6678 1.6316 1.5957
0.0 0.0023 0.0092 0.0205 0.0362 0.0560 0.0797 0.1069 0.1374 0.1708 0.2066 0.2445 0.2841 0.3250 0.3667 0.4089 0.4512 0.4933 0.5348 0.5755 0.6151 0.6535 0.6903 0.7254 0.7587 0.7901 0.8196 0.8469 0.8723
0.8955 0.9167
0.0 0.0010 0.0038 0.0086 0.0152 0.0237 0.0339 0.0458 0.0593 0.0744 0.0909 0.1088 0.1279 0.1482 0.1696 0.1918 0.2149 0.2388 0.2633 0.2883 0.3137 0.3395 0.3656 0.3918 0.4181 0.4444 0.4708 0.4970 0.5230
0.5489 0.5745
1.2679 1.2675 1.2665 1.2647 1.2623 1.2591 1.2554 1.2510 1.2461 1.2406 1.2346 1.2281 1.2213 1.2140 1.2064 1.1985 1.1904 1.1822 1.1737 1.1652 1.1566 1.1480 1.1394 1.1308 1.1224 1.1141 1.1059 1.0979
1.0901 1.0826 1.0753
Compressible-Flow Tables 785 Table B.4 (Cont.) Frictionless Duct Flow with Heat Transfer for k 1.4
*/ V/V*
0.62 0.64 0.66 0.68 0.7 0.72 0.74 0.76 0.78 0.8 0.82 0.84 0.86 0.88 0.9 0.92 0.94 0.96 0.98 1.0 1.02 1.04 1.06 1.08 1.1 1.12 1.14 1.16 1.18 1.2 1.22 1.24 1.26 1.28 1.3 1.32 1.34 1.36 1.38 1.4 1.42
1.44 1.46 1.48 1.5 1.52 1.54 1.56 1.58 1.6 1.62 1.64 1.66 1.68
0.8398 0.8592 0.8771 0.8935 0.9085 0.9221 0.9344 0.9455 0.9553 0.9639 0.9715 0.9781 0.9836 0.9883 0.9921 0.9951 0.9973 0.9988 0.9997 1.0000 0.9997 0.9989 0.9977 0.9960 0.9939 0.9915 0.9887 0.9856
0.9823 0.9787 0.9749 0.9709 0.9668 0.9624 0.9580 0.9534 0.9487 0.9440 0.9391 0.9343 0.9293 0.9243 0.9193 0.9143 0.9093 0.9042 0.8992 0.8942 0.8892 0.8842 0.8792 0.8743 0.8694 0.8645
1.5603 1.5253 1.4908 1.4569 1.4235 1.3907 1.3585 1.3270 1.2961 1.2658 1.2362 1.2073 1.1791 1.1515 1.1246 1.0984 1.0728 1.0479 1.0236 1.0000 0.9770 0.9546 0.9327 0.9115 0.8909 0.8708 0.8512 0.8322
0.8137 0.7958 0.7783 0.7613 0.7447 0.7287 0.7130 0.6978 0.6830 0.6686 0.6546 0.6410 0.6278 0.6149 0.6024 0.5902 0.5783 0.5668 0.5555 0.5446 0.5339 0.5236 0.5135 0.5036 0.4940 0.4847
0.9358 0.9530 0.9682 0.9814 0.9929 1.0026 1.0106 1.0171 1.0220 1.0255 1.0276 1.0285 1.0283 1.0269 1.0245 1.0212 1.0170 1.0121 1.0064 1.0000 0.9930 0.9855 0.9776 0.9691 0.9603 0.9512 0.9417 0.9320
0.9220 0.9118 0.9015 0.8911 0.8805 0.8699 0.8592 0.8484 0.8377 0.8269 0.8161 0.8054 0.7947 0.7840 0.7735 0.7629 0.7525 0.7422 0.7319 0.7217 0.7117 0.7017 0.6919 0.6822 0.6726 0.6631
0.5998 0.6248 0.6494 0.6737 0.6975 0.7209 0.7439 0.7665 0.7885 0.8101 0.8313 0.8519 0.8721 0.8918 0.9110 0.9297 0.9480 0.9658 0.9831 1.0000 1.0164 1.0325 1.0480 1.0632 1.0780 1.0923 1.1063 1.1198
1.1330 1.1459 1.1584 1.1705 1.1823 1.1938 1.2050 1.2159 1.2264 1.2367 1.2467 1.2564 1.2659 1.2751 1.2840 1.2927 1.3012 1.3095 1.3175 1.3253 1.3329 1.3403 1.3475 1.3546 1.3614 1.3681
1.0682 1.0615 1.0550 1.0489 1.0431 1.0376 1.0325 1.0278 1.0234 1.0193 1.0157 1.0124 1.0095 1.0070 1.0049 1.0031 1.0017 1.0008 1.0002 1.0000 1.0002 1.0008 1.0017 1.0031 1.0049 1.0070 1.0095 1.0124
1.0157 1.0194 1.0235 1.0279 1.0328 1.0380 1.0437 1.0497 1.0561 1.0629 1.0701 1.0777 1.0856 1.0940 1.1028 1.1120 1.1215 1.1315 1.1419 1.1527 1.1640 1.1756 1.1877 1.2002 1.2131 1.2264
Appendix B
Table B.4 (Cont.) Frictionless Duct Flow with Heat Transfer for k 1.4
*/ V/V*
1.7 1.72 1.74 1.76 1.78 1.8 1.82 1.84 1.86 1.88 1.9 1.92 1.94 1.96 1.98 2.0 2.02 2.04 2.06 2.08 2.1 2.12 2.14 2.16 2.18 2.2 2.22 2.24 2.26 2.28 2.3 2.32 2.34 2.36 2.38 2.4 2.42 2.44 2.46 2.48 2.5
2.52 2.54 2.56 2.58 2.6 2.62 2.64 2.66 2.68 2.7 2.72 2.74 2.76
0.8597 0.8549 0.8502 0.8455 0.8409 0.8363 0.8317 0.8273 0.8228 0.8185 0.8141 0.8099 0.8057 0.8015 0.7974 0.7934 0.7894 0.7855 0.7816 0.7778 0.7741 0.7704 0.7667 0.7631 0.7596 0.7561 0.7527 0.7493
0.7460 0.7428 0.7395 0.7364 0.7333 0.7302 0.7272 0.7242 0.7213 0.7184 0.7156 0.7128 0.7101 0.7074 0.7047 0.7021 0.6995 0.6970 0.6945 0.6921 0.6896 0.6873 0.6849 0.6826 0.6804 0.6781
0.4756 0.4668 0.4581 0.4497 0.4415 0.4335 0.4257 0.4181 0.4107 0.4035 0.3964 0.3895 0.3828 0.3763 0.3699 0.3636 0.3575 0.3516 0.3458 0.3401 0.3345 0.3291 0.3238 0.3186 0.3136 0.3086 0.3038 0.2991
0.2945 0.2899 0.2855 0.2812 0.2769 0.2728 0.2688 0.2648 0.2609 0.2571 0.2534 0.2497 0.2462 0.2427 0.2392 0.2359 0.2326 0.2294 0.2262 0.2231 0.2201 0.2171 0.2142 0.2113 0.2085 0.2058
0.6538 0.6445 0.6355 0.6265 0.6176 0.6089 0.6004 0.5919 0.5836 0.5754 0.5673 0.5594 0.5516 0.5439 0.5364 0.5289 0.5216 0.5144 0.5074 0.5004 0.4936 0.4868 0.4802 0.4737 0.4673 0.4611 0.4549 0.4488
0.4428 0.4370 0.4312 0.4256 0.4200 0.4145 0.4091 0.4038 0.3986 0.3935 0.3885 0.3836 0.3787 0.3739 0.3692 0.3646 0.3601 0.3556 0.3512 0.3469 0.3427 0.3385 0.3344 0.3304 0.3264 0.3225
1.3746 1.3809 1.3870 1.3931 1.3989 1.4046 1.4102 1.4156 1.4209 1.4261 1.4311 1.4360 1.4408 1.4455 1.4501 1.4545 1.4589 1.4632 1.4673 1.4714 1.4753 1.4792 1.4830 1.4867 1.4903 1.4938 1.4973 1.5007
1.5040 1.5072 1.5104 1.5134 1.5165 1.5194 1.5223 1.5252 1.5279 1.5306 1.5333 1.5359 1.5385 1.5410 1.5434 1.5458 1.5482 1.5505 1.5527 1.5549 1.5571 1.5592 1.5613 1.5634 1.5654 1.5673
1.2402 1.2545 1.2692 1.2843 1.2999 1.3159 1.3324 1.3494 1.3669 1.3849 1.4033 1.4222 1.4417 1.4616 1.4821 1.5031 1.5246 1.5467 1.5693 1.5924 1.6162 1.6404 1.6653 1.6908 1.7168 1.7434 1.7707 1.7986
1.8271 1.8562 1.8860 1.9165 1.9476 1.9794 2.0119 2.0451 2.0789 2.1136 2.1489 2.1850 2.2218 2.2594 2.2978 2.3370 2.3770 2.4177 2.4593 2.5018 2.5451 2.5892 2.6343 2.6802 2.7270 2.7748
Compressible-Flow Tables 787 Table B.4 (Cont.) Frictionless Duct Flow with Heat Transfer for k 1.4
*/ V/V*
2.78 2.8 2.82 2.84 2.86 2.88 2.9 2.92 2.94 2.96 2.98 3.0 3.02 3.04 3.06 3.08 3.1 3.12 3.14 3.16 3.18 3.2 3.22 3.24 3.26 3.28 3.3 3.32 3.34 3.36 3.38 3.4 3.42 3.44 3.46 3.48 3.5 3.52 3.54 3.56 3.58
3.6 3.62 3.64 3.66 3.68 3.7 3.72 3.74 3.76 3.78 3.8 3.82 3.84
0.6761 0.6738 0.6717 0.6696 0.6675 0.6655 0.6635 0.6615 0.6596 0.6577 0.6558 0.6540 0.6522 0.6504 0.6486 0.6469 0.6452 0.6435 0.6418 0.6402 0.6386 0.6370 0.6354 0.6339 0.6324 0.6309 0.6294 0.6280
0.6265 0.6251 0.6237 0.6224 0.6210 0.6197 0.6184 0.6171 0.6158 0.6145 0.6133 0.6121 0.6109 0.6097 0.6085 0.6074 0.6062 0.6051 0.6040 0.6029 0.6018 0.6008 0.5997 0.5987 0.5977 0.5967
0.2030 0.2004 0.1978 0.1953 0.1927 0.1903 0.1879 0.1855 0.1832 0.1809 0.1787 0.1765 0.1743 0.1722 0.1701 0.1681 0.1660 0.1641 0.1621 0.1602 0.1583 0.1565 0.1547 0.1529 0.1511 0.1494 0.1477 0.1461
0.1444 0.1428 0.1412 0.1397 0.1381 0.1366 0.1351 0.1337 0.1322 0.1308 0.1294 0.1280 0.1267 0.1254 0.1241 0.1228 0.1215 0.1202 0.1190 0.1178 0.1166 0.1154 0.1143 0.1131 0.1120 0.1109
0.3186 0.3149 0.3111 0.3075 0.3039 0.3004 0.2969 0.2934 0.2901 0.2868 0.2835 0.2803 0.2771 0.2740 0.2709 0.2679 0.2650 0.2620 0.2592 0.2563 0.2535 0.2508 0.2481 0.2454 0.2428 0.2402 0.2377 0.2352
0.2327 0.2303 0.2279 0.2255 0.2232 0.2209 0.2186 0.2164 0.2142 0.2120 0.2099 0.2078 0.2057 0.2037 0.2017 0.1997 0.1977 0.1958 0.1939 0.1920 0.1902 0.1884 0.1866 0.1848 0.1830 0.1813
1.5693 1.5711 1.5730 1.5748 1.5766 1.5784 1.5801 1.5818 1.5834 1.5851 1.5867 1.5882 1.5898 1.5913 1.5928 1.5942 1.5957 1.5971 1.5985 1.5998 1.6012 1.6025 1.6038 1.6051 1.6063 1.6076 1.6088 1.6100
1.6111 1.6123 1.6134 1.6145 1.6156 1.6167 1.6178 1.6188 1.6198 1.6208 1.6218 1.6228 1.6238 1.6247 1.6257 1.6266 1.6275 1.6284 1.6293 1.6301 1.6310 1.6318 1.6327 1.6335 1.6343 1.6351
2.8235 2.8731 2.9237 2.9752 3.0278 3.0813 3.1359 3.1914 3.2481 3.3058 3.3646 3.4245 3.4854 3.5476 3.6108 3.6752 3.7408 3.8076 3.8756 3.9449 4.0154 4.0871 4.1602 4.2345 4.3101 4.3871 4.4655 4.5452
4.6263 4.7089 4.7929 4.8783 4.9652 5.0536 5.1435 5.2350 5.3280 5.4226 5.5188 5.6167 5.7162 5.8173 5.9201 6.0247 6.1310 6.2390 6.3488 6.4605 6.5739 6.6893 6.8065 6.9256 7.0466 7.1696
Appendix B
Table B.4 (Cont.) Frictionless Duct Flow with Heat Transfer for k 1.4
Table B.5 Prandtl-Meyer Supersonic Expansion Function for k 1.4
*/ V/V*
3.86 3.88 3.9 3.92 3.94 3.96 3.98 4.0
0.5957 0.5947 0.5937 0.5928 0.5918 0.5909 0.5900 0.5891
0.1098 0.1087 0.1077 0.1066 0.1056 0.1046 0.1036 0.1026
0.1796 0.1779 0.1763 0.1746 0.1730 0.1714 0.1699 0.1683
1.6359 1.6366 1.6374 1.6381 1.6389 1.6396 1.6403 1.6410
7.2945 7.4215 7.5505 7.6816 7.8147 7.9499 8.0873 8.2269
, deg
, deg
, deg
, deg
1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40 1.45 1.50 1.55 1.60 1.65 1.70 1.75 1.80 1.85 1.90 1.95 2.00 2.05 2.10 2.15 2.20 2.25 2.30 2.35 2.40 2.45 2.50 2.55 2.60 2.65 2.70 2.75 2.80 2.85 2.90 2.95
0.0 0.49 1.34 2.38 3.56 4.83 6.17 7.56 8.99 10.44 11.91 13.38 14.86 16.34 17.81 19.27 20.73 22.16 23.59 24.99 26.38 27.75 29.10 30.43 31.73 33.02 34.28 35.53 36.75 37.95 39.12 40.28 41.41 42.53 43.62
44.69 45.75 46.78 47.79 48.78 49.76
3.05 3.10 3.15 3.20 3.25 3.30 3.35 3.40 3.45 3.50 3.55 3.60 3.65 3.70 3.75 3.80 3.85 3.90 3.95 4.00 4.05 4.10 4.15 4.20 4.25 4.30 4.35 4.40 4.45 4.50 4.55 4.60 4.65 4.70 4.75 4.80 4.85 4.90 4.95 5.00
50.71 51.65 52.57 53.47 54.35 55.22 56.07 56.91 57.73 58.53 59.32 60.09 60.85 61.60 62.33 63.04 63.75 64.44 65.12 65.78 66.44 67.08 67.71 68.33 68.94 69.54 70.13 70.71 71.27 71.83 72.38 72.92 73.45
73.97 74.48 74.99 75.48 75.97 76.45 76.92
5.05 5.10 5.15 5.20 5.25 5.30 5.35 5.40 5.45 5.50 5.55 5.60 5.65 5.70 5.75 5.80 5.85 5.90 5.95 6.00 6.05 6.10 6.15 6.20 6.25 6.30 6.35 6.40 6.45 6.50 6.55 6.60 6.65 6.70 6.75 6.80 6.85 6.90 6.95 7.00
77.38 77.84 78.29 78.73 79.17 79.60 80.02 80.43 80.84 81.24 81.64 82.03 82.42 82.80 83.17 83.54 83.90 84.26 84.61 84.96 85.30 85.63 85.97 86.29 86.62 86.94 87.25 87.56 87.87 88.17 88.47 88.76 89.05
89.33 89.62 89.90 90.17 90.44 90.71 90.97
7.05 7.10 7.15 7.20 7.25 7.30 7.35 7.40 7.45 7.50 7.55 7.60 7.65 7.70 7.75 7.80 7.85 7.90 7.95 8.00 8.05 8.10 8.15 8.20 8.25 8.30 8.35 8.40 8.45 8.50 8.55 8.60 8.65 8.70 8.75 8.80 8.85 8.90 8.95 9.00
91.23 91.49 91.75 92.00 92.24 92.49 92.73 92.97 93.21 93.44 93.67 93.90 94.12 94.34 94.56 94.78 95.00 95.21 95.42 95.62 95.83 96.03 96.23 96.43 96.63 96.82 97.01 97.20 97.39 97.57 97.76 97.94 98.12
98.29 98.47 98.64 98.81 98.98 99.15 99.32
Compressible-Flow Tables 789
4.0 Ma 2
β= 20°
θ θ
Weak 30 shock
Mach line 10
Ma 2
θ = 0° 5 2.0
55 60 65 70 Strong 75 shock 80 90 85
Normal shock 0
Fig. B.1 Mach number downstream of an oblique shock for k 1.4.
3.0 Ma1
Appendix B 80° 70 β = 90 75 65 60 55 50
β Ma1, p 1
θ = 30°
8.0 Normal shock 7.0 35
Strong shock 6.0
p2 p1
5.0 20 4.0
Fig. B.2 Pressure ratio downstream of an oblique shock for k 1.4.
Weak shock 3.0
Appendix C Conversion Factors
During this period of transition there is a constant need for conversions between BG and SI units (see Table 1.2). Some additional conversions are given here. Conversion factors are given inside the
front cover. Length 1 1 1 1 1
ft 12 in 0.3048 m mi 5280 ft 1609.344 m nautical mile (nmi) 6076 ft 1852 m yd 3 ft 0.9144 m angstrom (Å) 1.0 E-10 m
Volume 1 1 1 1 1
ft 0.028317 m U.S. gal 231 in3 0.0037854 m3 L 0.001 m3 0.035315 ft3 U.S. fluid ounce 2.9574 E-5 m3 U.S. quart (qt) 9.4635 E-4 m3
ft 0.092903 m mi2 2.78784 E7 ft2 2.59 E6 m2 acre 43,560 ft2 4046.9 m2 hectare (ha) 10,000 m2
Mass 1 1 1 1
slug 32.174 lbm 14.594 kg lbm 0.4536 kg short ton 2000 lbm 907.185 kg tonne 1000 kg
Area 2
Velocity 1 ft/s 0.3048 m/s 1 mi/h 1.466666 ft/s 0.44704 m/s 1 kn 1 nmi/h 1.6878 ft/s 0.5144 m/s
Acceleration 1 ft/s 0.3048 m/s2 2
Mass flow 1 slug/s 14.594 kg/s 1 lbm/s 0.4536 kg/s
Volume flow 1 gal/min 0.002228 ft3/s 0.06309 L/s 1 106 gal/day 1.5472 ft3/s 0.04381 m3/s
Pressure 1 lbf/ft 47.88 Pa 1 lbf/in2 144 lbf/ft2 6895 Pa 1 atm 2116.2 lbf/ft2 14.696 lbf/in2 101,325 Pa 1 inHg (at 20°C) 3375 Pa 1 bar 1.0 E5 Pa 2
Force 1 1 1 1 1
lbf 4.448222 N 16 oz kgf 2.2046 lbf 9.80665 N U.S. (short) ton 2000 lbf dyne 1.0 E-5 N ounce (avoirdupois) (oz) 0.27801 N
Appendix C
1 ft lbf 1.35582 J 1 Btu 252 cal 1055.056 J 778.17 ft lbf 1 kilowatt hour (kWh) 3.6 E6 J
1 hp 550 ft lbf/s 745.7 W 1 ft lbf/s 1.3558 W
Specific weight 1 lbf/ft 157.09 N/m 3
Density 1 slug/ft 515.38 kg/m3 1 lbm/ft3 16.0185 kg/m3 1 g/cm3 1000 kg/m3
Kinematic viscosity
1 slug/(ft s) 47.88 kg/(m s) 1 poise (P) 1 g/(cm s) 0.1 kg/(m s)
1 ft /h 0.000025806 m2/s 1 stokes (St) 1 cm2/s 0.0001 m2/s 2
Temperature scale readings TF 95TC 32 TC 59(TF 32) TR TF 459.69 TK TC 273.16 where subscripts F, C, R, and K refer to readings on the Fahrenheit, Celsius, Kelvin, and Rankine scales, respectively
Specific heat or gas constant*
Thermal conductivity*
1 ft lbf/(slug °R) 0.16723 N m/(kg K) 1 Btu/(lb °R) 4186.8 J/(kg K)
1 Btu/(h ft °R) 1.7307 W/(m K)
*Although the absolute (Kelvin) and Celsius temperature scales have different starting points, the intervals are the same size: 1 kelvin 1 Celsius degree. The same holds true for the nonmetric
absolute (Rankine) and Fahrenheit scales: 1 Rankine degree 1 Fahrenheit degree. It is customary to express temperature differences in absolute-temperature units.
Appendix D Equations of Motion in Cylindrical Coordinates
The equations of motion of an incompressible newtonian fluid with constant , k, and cp are given here in cylindrical coordinates (r, , z), which are related to cartesian coordinates (x, y, z) as in
Fig. 4.2: x r cos
y r sin
z z
The velocity components are r, , and z. The equations are: Continuity: 1 1 (rr) ( ) (z) 0 r r r z
Convective time derivative: 1 V r z r z r
1 1 2 2 2 r 2 2 2 r r r r z
Laplacian operator:
The r-momentum equation: 1 1 p 2 r (V )r 2 gr 2 r 2r 2 t r r r r
The -momentum equation: 1 1 p 2 (V ) r g 2 2 2 r t r r r r
The z-momentum equation: 1 p z (V ) z gz 2 z t z
(D.7) 793
Appendix D
The energy equation: T 2 cp (V )T k2T [2(2rr
2zz) 2z 2rz 2r ] (D.8) t
rr r r
r r
zz z z
1 z z r z
rz r z z r
1 r r r r
Viscous stress components:
rr 2rr
zz 2zz
r r
z z
rz rz
Angular-velocity components: 1 r z r z r z z r 1 1 z (r ) r r r r
Appendix E Introduction to EES
EES (pronounced “ease”) is an acronym for Engineering Equation Solver. The basic function provided by EES is the numerical solution of nonlinear algebraic and differential equations. In addition, EES
provides built-in thermodynamic and transport property functions for many fluids, including water, dry and moist air, refrigerants, and combustion gases. Additional property data can be added by the
user. The combination of equation solving capability and engineering property data makes EES a very powerful tool. A license for EES is provided to departments of educational institutions which adopt
this text by WCB/McGraw-Hill. If you need more information, contact your local WCB/McGraw-Hill representative, call 1-800-338-3987, or visit our website at www.mhhe.com. A commercial version of EES
can be obtained from: F-Chart Software 4406 Fox Bluff Rd Middleton, WI 53562 Phone: (608)836-8531 Fax: (608)836-8536
Background Information
The EES program is probably installed on your departmental computer. In addition, the license agreement for EES allows students and faculty in a participating educational department to copy the
program for educational use on their personal computer systems. Ask your instructor for details. To start EES from the Windows File Manager or Explorer, double-click on the EES program icon or on any
file created by EES. You can also start EES from the Windows Run command in the Start menu. EES begins by displaying a dialog window which shows registration information, the version number, and
other information. Click the OK button to dismiss the dialog window. Detailed help is available at any point in EES. Pressing the F1 key will bring up a Help window relating to the foremost window.
(See Fig. E.1.) Clicking the Contents 795
Appendix E
Fig. E.1 EES Help index.
button will present the Help index shown below. Clicking on an underlined word (shown in green on color monitors) will provide help relating to that subject. EES commands are distributed among nine
pull-down menus as shown below. Many of the commands are accessible with the speed button palette that appears below the menu bar. A brief summary of their functions follows. (A tenth pull-down menu,
which is made visible with the Load Textbook command described below, provides access to problems from this text.)
The System menu appears above the File menu. The System menu is not part of EES but rather is a feature of the Windows operating system. It holds commands which allow window moving, resizing, and
switching to other applications.
Introduction to EES
The File menu provides commands for loading, merging, and saving work files; libraries; and printing. The Load Textbook command in this menu reads the problem disk developed for this text and creates
a new menu to the right of the Help menu for easy access to EES problems accompanying this text. The Edit menu provides the editing commands to cut, copy, and paste information. The Search menu
provides Find and Replace commands for use in the Equations window. The Options menu provides commands for setting the guess values and bounds of variables, the unit system, default information, and
program preferences. A command is also provided for displaying information on built-in and usersupplied functions. The Calculate menu contains the commands to check, format, and solve the equation
set. The Tables menu contains commands to set up and alter the contents of the parametric and lookup tables and to do linear regression on the data in these tables. The parametric table, which is
similar to a spreadsheet, allows the equation set to be solved repeatedly while varying the values of one or more variables. The lookup table holds user-supplied data which can be interpolated and
used in the solution of the equation set. The Plot menu provides commands to modify an existing plot or prepare a new plot of data in the parametric, lookup, or array tables. Curve-fitting capability
is also provided. The Windows menu provides a convenient method of bringing any of the EES windows to the front or to organize the windows. The Help menu provides commands for accessing the on-line
help documentation. A basic capability provided by EES is the solution of a set of nonlinear algebraic equations. To demonstrate this capability, start EES and enter this simple example problem in
the Equations window.
Text is entered in the same manner as for any word processor. Formatting rules are as follows: 1. Uppercase and lowercase letters are not distinguished. EES will (optionally) change the case of all
variables to match the manner in which they first appear. 2. Blank lines and spaces may be entered as desired since they are ignored.
Appendix E
3. Comments must be enclosed within braces { } or within quotation marks “ ”. Comments may span as many lines as needed. Comments within braces may be nested, in which case only the outermost set of
braces is recognized. Comments within quotes will also be displayed in the Formatted Equations window. 4. Variable names must start with a letter and consist of any keyboard characters except ( ) ‘*
/ ^ { } : " or ;. Array variables are identified with square braces around the array index or indices, e.g., X[5,3]. The maximum variable length is 30 characters. 5. Multiple equations may be entered
on one line if they are separated by a semicolon (;). The maximum line length is 255 characters. 6. The caret symbol (^) or ** is used to indicate raising to a power. 7. The order in which the
equations are entered does not matter. 8. The position of knowns and unknowns in the equation does not matter. If you wish, you may view the equations in mathematical notation by selecting the
Formatted Equations command from the Windows menu.
Select the Solve command from the Calculate menu. A Dialog window will appear indicating the progress of the solution. When the calculations are completed, the button will change from Abort to
Click the Continue button. The solution to this equation set will then be displayed.
Introduction to EES
A Pipe Friction Example Problem
Let us now solve Prob. 6.55 from the text, for a cast-iron pipe, to illustrate the capabilities of the EES program. This problem, without EES, would require iteration for Reynolds number, velocity,
and friction factor, a daunting task. State the problem: 6.55 Reservoirs 1 and 2 contain water at 20°C. The pipe is cast iron, with L 4500 m and D 4 cm. What will be the flow rate in m3/h if z 100 m?
This is a representative problem in pipe flow (see Fig. E.2), and, being water in a reasonably large (noncapillary) pipe, it will probably be turbulent (Re 4000). The steadyflow energy equation
(3.71) may be written between the surfaces of reservoirs 1 and 2: p1 V2 p2 V2 1 z1 2 z2 hf g 2g g 2g
2 L Vpipe hf f D 2g
Since p1 p2 patm and V1 V2 0, this relation simplifies to L V2 z f D 2g
where V Q/A is the velocity in the pipe. The friction factor f is a function of Reynolds number and pipe roughness ratio, if the flow is turbulent, from Eq. (6.64): 1 /D 2.51 1 2.0 log10 f /2 3.7 Re
f 1/2
if Re 4000
Finally, we need the definitions of Reynolds number and volume flow rate:
VD Re
Q VD2 4
where and are the fluid density and viscosity, respectively.
z 1 L, D, 2
Fig. E.2 Sketch of the flow system.
Appendix E
There are a total of 11 variables involved in this problem: (L, D, z, , g, , , V, Re, f, Q). Of these, seven can be specified at the start (L, D, z, , g, , ), while four (V, Re, f, Q) must be
calculated from relations (1) to (4) above. These four equations in four unknowns are well posed and solvable but only by laborious iteration, exactly what EES is designed to do. Start EES or select
the New command from the File menu if you have already been using the program. A blank Equations window will appear. Our recommendation is to always set the unit system immediately: Select Unit
System from the Options menu (Fig. E.3). We select SI and Mass units and trig Degrees, although we do not actually have trigonometric functions this time. We select kPa for pressure and Celsius for
temperature, which will be handy for using the EES built-in physical properties of water.
Fig. E.3 Unit Selection dialog window.
Now, onto the blank screen, enter the equations for this problem (Fig. E.4), of which five are known input values, two are property evaluations, and four are the relations (1) to (4) from above.
Fig. E.4 Equations window.
Introduction to EES
There are several things to notice in Fig. E.4. First, quantities in quotes, such as “m,” are for the user’s benefit and ignored by EES. Second, we changed Eps and D to meters right away, to keep the
SI units consistent. Third, we called on EES to input the viscosity and density of water at 20°C and 1 atm, a procedure well explained in the Help menu. For example, viscosity(water,T 20,P 101) meets
the EES requirement that temperature (T) and pressure (P) should be input in °C and kPa—EES will then evaluate in kg/(m s). Finally, note that EES recognizes pi to be 3.141593. In Fig. E.4 we used
only one built-in function, log10. There are many such functions, found by scrolling down the Function Information command in the Options menu. Having entered the equations, check the syntax by using
the Check/Format command in the Calculate menu. If you did well, EES will report that the 11 equations in 11 unknowns look OK. If not, EES will guess at what might be wrong. If OK, why not go for it?
Hit the Solve command in the Options menu. EES reports “logarithm of a negative number—try setting limits on the variables”. We might have known. Go to the Variable Information command in the Options
menu. A box, listing the 11 variables, will appear (Fig. E.5). All default EES “guesses” are unity; all default limits are to , which is too broad a range. Enter (as already shown in Fig. E.5)
guesses for f 0.02 and Re 10,000, while V 1 and Q 1 seem adequate, and other variables are fixed. Make sure that f, Re, V, and Q cannot be negative. The “display” columns normally say “A”, automatic,
satisfactory for most variables. We have changed “A” to “F” (fixed decimal) for Q and V to make sure they are displayed to four decimal places. The “units” column is normally blank—type in the
correct units and they will be displayed in the solution. Our guesses and limits are excellent, and the Solve command now iterates and reports success: “max residual 2E-10”, a negligible error. (The
default runs for 100
Fig. E.5 Variable Information window with units and guess values entered.
Appendix E
iterations, which can be modified by the Stop Criteria command in the Options menu.) Hit Continue, and the complete solution is displayed for all variables (Fig. E.6).
Fig. E.6 The Solutions window for Prob. 6.55.
This is the correct solution to Prob. 6.55: this cast-iron pipe, when subjected to a 100-m elevation difference, will deliver Q 3.17 m3/h of water. EES did all the iteration.
Parametric Studies with Tabular Input
Fig. E.7 New Parametric Table window showing selected variables (V is not shown).
One of the most useful features of EES is its ability to provide parametric studies. For example, suppose we wished to know how varying z changed the flow rate Q. First comment out the equation that
reads DELTAZ 100 by enclosing it within braces. (If you select the equation and press the right mouse button, EES will automatically enter the braces.) Select the New Parametric Table command in the
Options menu. A dialog will be displayed (Fig. E.7) listing all the variables in the problem. Highlight what you wish to vary: z. Also highlight variables to be calculated and tabulated: V, Q, Re,
and f.
Introduction to EES
Click the OK button and the new table will be displays (Fig. E.8). Enter 10 values of z that cover the range of interest—we have selected the linear range 10 z 500 m.
Fig. E.8 Parametric Table window.
Clearly the parametric table operates much like a spreadsheet. Select Solve Table from the Calculate menu, and the Solve Table dialog window will appear (Fig. E.9). These are satisfactory default
values; the author has changed nothing. Hit the OK button and the calculations will be made and the entire parametric table filled out, as in Fig. E.10.
Fig. E.9 Solve Table Dialog.
The flow rates can be seen in Fig. E.10, but as always, in the author’s experience, a plot is more illuminating. Select New Plot window from the Plot menu. The New Plot window dialog (Fig. E.11) will
appear. Choose z as the x-axis and Q as the y-axis. We added grid lines. Click the OK button, and the desired plot will appear in the Plot window (Fig. E.12). We see a nonlinear relationship, roughly
a square-root type, and learn that flow rate Q is not linearly proportional to head difference z. The plot appearance in Fig. E.12 can be modified in several ways. Double-click the mouse in the plot
rectangle to see some of these options.
Appendix E
Fig. E.10 Parametric Table window after calculations are completed.
Fig. E.11 New Plot Setup dialog window.
Introduction to EES
Fig. E.12 Plot window for flow rate versus elevation difference.
Loading a Textbook File
A problems disk developed for EES has been included with this textbook. Place the disk in the disk drive, and then select the Load Textbook command in the File menu. Use the Windows Open File command
to open the textbook problem index file which, for this book, is named WHITE.TXB. A new menu called Fluid Mechanics will appear to the right of the Help menu. This menu will provide access to all the
EES problem solutions developed for this book organized by chapter. As an example, select Chap. 6 from the Fluid Mechanics menu. A dialog window will appear listing the problems in Chap. 6. Select
Problem 6.55—Flow Between Reservoirs. This problem is a modification (smooth instead of cast-iron pipe) of the problem you just entered. It provides a diagram window in which you can enter z. Enter
other values, and then select the Solve command in the Calculate menu to see their effect on the flow rate. At this point, you should explore. Try whatever you wish. You can’t hurt anything. The
on-line help (invoked by pressing F1) will provide details for the EES commands. EES is a powerful tool that you will find very useful in your studies.
Answers to Selected Problems
Chapter 1 1.2 1.3 E44 molecules 1.4 1.63 slug/ft3, 839 kg/m3 1.6 (a) {L2/T 2}; (b) {M/T} 1.8 1.00 My/I 1.10 Yes, all terms are {ML/T 2} 1.12 {B} {L1} 1.14 Q Const B g1/2H3/2 1.16 All terms are {ML2T
2} 1.18 V V0emt/K 1.20 zmax 64.2 m at t 3.36 s 1.22 (a) 0.372U 2 /R; (b) x 1.291 R 1.24 e 221,000 J/kg 1.26 Wair 0.71 lbf 1.28 wet 1.10 kg/m3, dry 1.13 kg/m3 1.30 W1-2 21 ft lbf 1.32 (a) 76 kN; (b)
501 kN 1.34 1300 atm 1.36 (a) BN2O 1.33 E5 Pa; (b) Bwater 2.13 E9 Pa 1.38 1380 Pa, ReL 28 1.40 A 0.0016 kg/(m s), B 1903 K 1.42 /200K (T K/200 K)0.68 1.44 Data 50 percent higher; Andrade fit varies
50 percent 1.46 V 15 m/s 1.48 F (1/h1 2/h2)AV 1.50 M(ro ri)/(2ri3L) 1.52 P 73 W 1.54 M R4/h 1.56 3M sin /(2R3) 1.58 0.040 kg/(m s), last 2 points are turbulent flow 1.60 0.88 0.023 kg/(m s) 1.62
28,500 Pa 1.64 (a) 0.023 m; (b) 0.069 m 1.66 F 0.014 N 806
1.68 1.70 1.72 1.74 1.76 1.78 1.80 1.82 1.84
h (/g)1/2 cot h 2 cos /(gW) z 4800 m Cavitation occurs for both (a) and (b) z 7500 m (a) 25°C; (b) 4°C x2y y3/3 constant y x tan constant x x0{ln (y/y0) ln2 (y/y0)}
Chapter 2 xy 289 lb/ft2, AA 577 lb/ft2 2.2 2.4 x Const e2Cz/B 2.6 (a) 30.3 ft; (b) 30.0 in; (c) 10.35 m; (d) 13,100 mm 2.8 DALR 9.77°C/km 2.10 10,500 Pa 2.12 8.0 cm 2.14 74,450 Pa with air; 75,420 Pa
without air 2.16 (a) 21,526 cm3; (b) 137 kPa 2.18 1.56 2.20 14 lbf 2.22 0.94 cm 2.24 psealevel 117 kPa, mexact 5.3 E18 kg 2.26 (a) 2580 m; (b) 5410 m 2.28 4400 400 ft 2.30 101,100 Pa 2.32 22.6 cm
2.34 p h[ water(1 d 2/D2) oil(1 d 2/D2)] 2.36 25° 2.38 (a) p1,gage (m a)gh (t a)gH 2.40 21.3 cm 2.42 pA pB (2 1)gh 2.44 (a) 171 lb/ft2; (b) 392 lb/ft2; manometer reads friction loss
Answers to Selected Problems 2.46 2.48 2.50 2.52 2.54 2.56 2.58 2.60 2.62 2.64 2.66 2.68 2.70 2.72 2.74 2.76 2.78 2.80 2.82 2.84 2.86 2.88 2.90 2.92 2.94 2.96 2.98 2.100 2.102 2.104 2.106 2.108 2.110
2.112 2.114 2.116 2.118 2.120 2.122 2.124 2.126 2.128 2.130 2.132 2.134 2.136 2.138 2.140
1.45 F 39700 N (a) 524 kN; (b) 350 kN; (c) 100 kN 0.96 879 kg/m3 16.08 ft 0.40 m Fnet 23,940 N at 1.07 m above B 10.6 ft 1.35 m F 1.18 E9 N, MC 3.13 E9 N m counterclockwise, no tipping 18,040 N 4490
lbf at 1.44 ft to right and 1.67 ft up from point B 33,500 N F ba20{(a2/g)[exp(gh/a2) 1] h} P ( /24)h2b(3 csc2 ) P R3/4 (a) 58,800 Pa; (b) 0.44 m FH 97.9 MN, FV 153.8 MN FH 4895 N, FV 7343 N FH 0, FV
6800 lbf FH 176 kN, FV 31.9 kN, yes 467 lbf Fone bolt 11,300 N Cx 2996 lb, Cz 313 lbf (a) 940 kN; (b) 1074 kN; (c) 1427 kN FH 7987 lbf, FV 2280 lbf FH 0, FV 297 kN 124 kN 5.0 N 4310 N/m3 12.6 N h (a)
7.05 mm; (b) 7.00 mm (a) 39 N; (b) 0.64 0.636 19100 N/m3 (a) draft 7.24 in; (b) 25 lbf 34.3° a/b 0.834 6850 m h/H Z (Z2 1 )1/2, d/H, Z (2 )/2, pa/(gH) Yes, stable if S 0.789 Slightly unstable, MG
0.007 m Stable if R/h 3.31 (a) unstable; (b) stable MG L2/(3R) 4R/(3) 0 if L 2R 2.77 in deep; volume 10.8 fluid ounces ax (a) 1.96 m/s2 (deceleration); (b) 5.69 m/s2 (deceleration)
2.142 2.144 2.146 2.148 2.150 2.152 2.154 2.157 2.158
(a) 16.3 cm; (b) 15.7 N (a) ax 319 m/s2; (b) no effect, pA pB Leans to the right at 27° Leans to the left at 27° 5.5 cm; linear scale OK (a) 224 r/min; (b) 275 r/min (a) both are paraboloids; (b) pB
2550 Pa (gage) 77 r/min, minimum pressure halfway between B and C 10.57 r/min
Chapter 3 3.2 r position vector from point O 3.6 Q (2b/3)(2g)1/2[(h L)3/2 (h L)3/2] 3.8 Q K per unit depth 3.10 (/3)R2U0cpTw 3.12 (a) 44 m3/h; (b) 9.6 m/s 3.14 dh/dt (Q1 Q2 Q3)/(d2/4) 3.16 Qtop 3U0b/
8 3.18 (b) Q 16bhumax /9 3.20 (a) 7.97 mL/s; (b) 1.27 cm/s 3.22 (a) 0.06 kg/s; (b) 1060 m/s; (c) 3.4 3.24 h [3Kt2d2/(8 tan2 )]1/3 3.26 Q 2U0bh/3 3.28 t (1 1/3)h01/2/(2CA2g) 3.30 (a) dh/dt Q/(2hb cot
20°) 3.32 Vhole 6.1 m/s 3.34 V2 4660 ft/s 3.36 U3 6.33 m/s 3.38 V V0r/(2h) 3.40 500 N to the left 3.42 F (p1 pa)A1 1A1V 12[(D1/D2)2 1] 3.44 F U2Lb/3 3.46 (1 cos )/2 3.48 V0 2.27 m/s 3.50 102 kN 3.52
F WhV 12[1/(1 sin ) 1] to the left 3.54 163 N 3.56 2.45 N/m 3.58 40 N 3.60 2100 N 3.62 3100 N 3.64 980 N 3.66 8800 N 3.70 91 lbf 3.72 Drag 4260 N 3.74 Fx 0, Fy 17 N, Fz 126 N 3.76 77 m/s 3.80 F (/2)
gb(h 12 h 22) h1bV 12(h1/h2 1) 3.82 25 m/s 3.84 23 N 3.86 274 kPa
3.88 3.90 3.92 3.94 3.96 3.100 3.102 3.104 3.106 3.108 3.110 3.112 3.114 3.116 3.118 3.120 3.122 3.124 3.126 3.128 3.130 3.134 3.136 3.138 3.140 3.142 3.144 3.146 3.148 3.152 3.154 3.156 3.158 3.160
3.162 3.164 3.166 3.168 3.170 3.172 3.174 3.176 3.178 3.180 3.182
Answers to Selected Problems V [2 2Vj]1/2, Q/2k dV/dt g dV/dt gh/(L h) h 0 at t 70 s d 2Z/dt2 2gZ/L 0 (a) 507 m/s and 1393 m; (b) 14.5 km h2/h1 12 12[1 8V 12/(gh1)]1/2 (Ve /R) ln (1 m ˙ t/M0) final
75 rad/s (a) V V0 /(1 CV0t/M), C bh(1 cos ) (a) 0.113 ft lbf; (b) 250 r/min Tm ˙ R 02 (a) 414 r/min; (b) 317 r/min P Qr2[r2 Q cot 2/(2r2b2)] P QuVn(cot 1 cot 2) (a) 22 ft/s; (b) 110 ft/s; (c) 710 hp
L h1 (cot )/2 41 r/min 15.5 kW (work done on the fluid) 1.07 m3/s 34 kW 4500 hp 5.6 m3/h gd4(H L)/(128L Q) 2Q/(16L ) 1640 hp (a) 1150 gal/min; (b) 67 hp 26 kW h 3.6 ft hf 0.21 m (a) 85.9°; (b) 55.4°
h 0.133 m (a) 102 kPa; (b) 88 mi/h (a) 169.4 kPa; (b) 209 m3/h (a) 31 m3/s; (b) 54 kW Q 166 ft3/min, p 0.0204 lbf/in2 (a) 5.25 kg/s; (b) 2.9 cm (a) 60 mi/h; (b) 1 atm h 1.08 ft h 1.76 m D 0.132 ft
(a) 5.61 ft/s; (b) further constriction reduces V2 (a) 9.3 m/s; (b) 68 kN/m h2 2.03 ft (subcritical) or 0.74 ft (supercritical) V Vf tanh (Vf t/2L), Vf (2gh)1/2 kp/[(k 1)] V 2/2 gz constant
Chapter 4 4.2 (a) du/dt (2V 02/L)(1 2x/L) 4.4 At (2, 1), dT/dt 125 units 4.6 (a) 6V 02/L; (b) L ln 3/(2V0)
4.8 4.12 4.14 4.16 4.18 4.20 4.22 4.28 4.30 4.32 4.34 4.36 4.38 4.42 4.48 4.50 4.52 4.54 4.60 4.62 4.66 4.68 4.70 4.72 4.74 4.76 4.78 4.80 4.82 4.84 4.86 4.88 4.90
(a) 0.0196 V 2/L; (b) at t 1.05 L/U If 0, r r2 fcn (, ) fcn(r) only y2 3x2y fcn(x, z) 0L0/(L0 Vt) 0 const, {K} {L/T}, {a} {L1} xL 1.82 kg/m3 Exact solution for any a or b p const (K2/2)(x2 y2) f1
C1r; f2 C2/r p p(0) 4umaxz/R2 C g sin /(2) Cz yx xy Tmean (∫uT dy)/(∫u dy) Kxy const Inviscid flow around a 180° turn 4Q/(b) Q ULb Irrotational, z0 H 2R2/(2g) Vy2/(2h) const K sin /r m tan1[2xy/(x2
y2 a2)] cos /r 2, 2am (a) 8.8m; (b) 55 m Uy K ln r (a) 0.106 m from A; (b) 0.333 m above the wall (a) Vwall,max m/L ; (b) pmin at x L (a) w (g/2)(2x x2) Obsessive result: R2/r z (gb2/2) ln (r/a) (g/
4)(r2 a2) Q 0.0031 m3/(s m) z U ln (r/b)/[ln (a/b)] F 3.34 N
Chapter 5 5.2 1.21 m 5.4 V 1.55 m/s, F 1.3 N 5.6 F 450 N 5.10 (a) {ML2T 2}; (b) {MLT 2} 5.14 /x fcn (Ux/) 5.16 Stanton number h/(Vcp) 5.18 Q/[(p/L)b4] const 5.20 P/(3D5) fcn[Q/(D3), D2/] 5.22 D/V fcn
(N, H/L) 5.24 F/(V2L2) fcn(, VL/, L/D, V/a) 5.26 (a) indeterminate; (b) T 2.75 s 5.28 /L fcn[L/D, VD/, E/(V2)] 5.30 hL/k fcn(UL/, cp/k) 5.32 Q/(bg1/2H3/2) const 5.34 khydrogen 0.182 W/(m K)
Answers to Selected Problems (a) Qloss R/(A ) constant d/D fcn(UD/, U2D/Y) h/L fcn(gL2/Y, , ) (a) {} {L2} F 0.17 N; (doubling U quadruples F) (a) F/(UL ) constant U 5 ft/s, F 0.003 lbf/ft Power 7 hp
V 128 ft/s 87 mi/h V 2.8 m/s Prototype power 157 hp max 26.5 r/s; p 22,300 Pa aluminum 0.77 Hz (a) V 27 m/s; (b) z 27 m (a) F/( U) constant; (b) No, not plausible F 87 lbf V 25 ft/s Prototype moment
88 kN m Drag 107,000 lbf Weber no. 100 if Lm/Lp 0.0090 (a) 1.86 m/s; (b) 42,900; (c) 254,000 Speeds: 19.6, 30.2, and 40.8 ft/s; Drags: 14,600; 31,800; and 54,600 lbf 5.84 Vm 39 cm/s; Tm 3.1 s; Hm
0.20 m 5.88 At 340 W, D 0.109 m 5.90 pD/(V2L) 0.155(VD/)1/4 5.36 5.38 5.40 5.44 5.48 5.50 5.52 5.54 5.56 5.58 5.60 5.62 5.64 5.66 5.68 5.70 5.72 5.74 5.76 5.78 5.80 5.82
Chapter 6 6.2 (a) x 2.1 m; (b) x 0.14 m 6.4 (a) 39 m3/h; (b) 1.3 m3/h 6.6 (a) laminar; (b) laminar 6.8 (a) 3600 Pa/m; (b) 13,400 Pa/m 6.10 (a) from A to B; (b) hf 7.8 m 6.18 (a) 0.054 m3/s; (b) 8.5 m
/s; (c) 122 Pa; (d) 542 kPa 6.20 (a) 0.204 m; (b) 19,800 Pa/m; (c) 9980 Pa/m 6.22 (a) 39 kg/s; (b) 1430 6.24 Head loss 25 m 6.26 4 mm 6.28 Q 0.31 m3/h 6.30 F4N 6.32 (a) 127 MPa; (b) 127 kW 6.34
0.000823 kg/(m s) 6.36 p 65 Pa 6.38 (a) 19.3 m3/h; (b) flow is up 6.40 (a) flow is up; (b) 1.86 m3/h 6.44 hf 10.5 m, p 1.4 MPa 6.46 Input power 11.2 MW 6.48 r/R 1 e3/2 6.50 (a) 4000 Pa/m; (b) 50 Pa;
(c) 46 percent
6.52 6.54 6.56 6.58 6.64 6.66 6.68 6.70 6.72 6.74 6.76 6.78 6.80 6.82 6.84 6.86 6.90 6.92 6.94 6.96 6.98 6.102 6.104 6.106 6.108 6.110 6.112 6.114 6.116 6.118 6.120 6.122 6.124 6.126 6.128 6.130
6.132 6.134 6.136 6.138 6.140 6.142 6.144 6.146 6.148 6.150 6.152 6.154
p1 2.38 MPa D 0.118 m (a) 188 km; (b) 27 MW Power 870 kW Q 19.6 m3/h (laminar, Re 1450) (a) 56 kPa; (b) 85 m3/h; (c) u 3.3 m/s at r 1 cm Power 204 hp Q 2.21 ft3/s Optimum 90° (0.7 m rise) D 0.52 in Q
15 m3/h Q 25 m3/h (to the left) Q 0.905 m3/s D 0.394 m D 0.104 m (a) 3.0 m/s; (b) 0.325 m/m; (c) 2770 Pa/m Q 19.6 ft3/s (a) 1530 m3/h; (b) 6.5 Pa (vacuum) 260 Pa/m Cross section 0.106 m by 0.531 m
Approximately 128 squares (a) 5.55 hp; (b) 5.31 hp with 6° cone p 0.0305 lbf/in2 Q 0.0296 ft3/s Q 0.22 ft3/s 840 W Q 0.0151 ft3/s (a) Q1 0.0167 m3/s, Q2 0.0193 m3/s, p 774 kPa Q 0.027 m3/s p 131 lbf/
in2 Q1 0.0109 m3/s, Q2 0.0264 m3/s, Q3 0.0183 m3/s Increased /d and L/d are the causes Q1 2.09 ft3/s, Q2 1.61 ft3/s, Q3 0.49 ft3/s opening 35° QAB 3.47, QBC 2.90, QBD 0.58, QCD 5.28, QAC 2.38 ft3/s
(all) QAB 0.95, QBC 0.24, QBD 0.19, QCD 0.31, QAC 1.05 ft3/s (all) 2 6°, De 2.0 m, pe 224 kPa 2 10°, We 8.4 ft, pe 2180 lbf/ft2 (a) 25.5 m/s, (b) 0.109 m3/s, (c) 1.23 Pa 46.7 m/s p 273 kPa Q 18.6 gal
/min, dreducer 0.84 cm Q 54 m3/h (a) 0.00653 m3/s; (b) 100 kPa (a) 1.58 m; (b) 1.7 m p 27 kPa D 4.12 cm h 59 cm
Answers to Selected Problems
6.156 Q 0.924 ft3/s 6.158 (a) 49 m3/h; (b) 6200 Pa Chapter 7 7.2 Rec 1.5 E7 7.4 d 8 mm lies in the transition region 7.6 H 2.5 (versus 2.59 for Blasius) 7.8 Approximately 0.08 N 7.12 Does not satisfy
∂2u/∂y2 0 at y 0 7.14 C 0/ const 0 (wall suction) 7.16 (a) F 181 N; (b) 256 N 7.18 0.16°; Fdrag 0.024 N 7.20 x 0.91 m 7.22 ( xU)1/2 f(!) 7.24 h1 9.2 mm; h2 5.5 mm 7.26 Fa 2.83 F1, Fb 2.0 F1 7.28 (a)
Fdrag 2.66 N2(L)1/2U3/2a 7.30 (a) F 72 N; (b) 79 N 7.32 F 0.0245 1/7 L6/7 U013/7 7.34 F 725 N 7.36 7.2 m/s 14 kn 7.38 (a) 7.6 m/s; (b) 6.2 m/s 7.40 L 3.53 m, b 1.13 m 7.42 P4 blades 0.0321/7(C)6/7 20
/7 R27/7 7.44 Accurate to about 6 percent 7.46 9 mm, U 11.2 m/s 22 kn 7.48 Separation at x/L 0.158 (1 percent error) 7.50 Separation at x/R 1.80 rad 103.1° 7.52 CD (ReL )1/2 2.67 (by numerical
integration) 7.54 Moment 200,000 N m 7.56 (a) 10 N; (b) 80 N 7.58 (a) 3200 N/m; (b) 2300 N/m 7.60 Tow power 140 hp 7.62 Square side length 0.83 m 7.64 t1000–2000m 202 s 7.68 (a) 34 m/s; (b) no, only
67 percent of terminal velocity at impact 7.70 (a) 642 ft; (b) 425 ft 7.72 (a) L 6.3 m; (b) 120 m 7.78 p 100 Pa 7.80 72° 7.82 Vmin 138 ft/s; (b) Vmax 377 ft/s 7.84 V 9 m/s 7.86 Approximately 3.05 m
by 6.1 m 7.88 (a) 62 hp; (b) 86 hp 7.90 Voverturn 145 ft/s 99 mi/h 7.94 Torque (CD/4)2DR4, max 85 r/min 7.96 avg 0.21 U/D 7.98 (b) h 0.18 m 7.100 (b) Dmax 78 m 7.106 (a) 300 m; (b) 380 m
7.108 7.110 7.114 7.116 7.118 7.120 7.122 7.124
xball 13 m y 1.9 ft Vfinal 18.3 m/s 66 km/h (a) 87 mi/h; (b) 680 hp (a) 21 m/s; (b) 360 m (L/D)max 21; 4.8° (a) 6.7 m/s; (b) 13.5 m/s 26 kn crude theory 340 r/s
Chapter 8 8.2 " (R22 R12) 8.4 No, 1/r is not a proper two-dimensional potential 8.6 B(y2 x 2) 8.8 " 4B 8.12 "0 8.14 Irrotational outer, rotational inner; minimum p p 2R2 at r 0 8.18 From afar: a
single source 4m 8.20 Vortex near a wall (see Fig. 8.17b) 8.22 Same as Fig. 8.6 except upside down 8.24 Cp {2(x/a)/[1 (x/a)2]}2, Cp,min 1.0 at x a 8.26 Vresultant 9.4 m/s at 47° 8.28 Creates a source
in a square corner 8.34 Two stagnation points, at x a/3 8.36 U 12.9 m/s, 2L 53 cm, Vmax 22.5 m/s 8.42 K/(U a) 0.396, h/a 1.124 8.44 K 4.6 m2/s; (a) 218 kPa; (b) 214 kPa at upper shoulder, 6 kPa at
lower shoulder (cavitation) 8.46 F1-bolt 5000 N 8.50 h 3a/2, Umax 5U/4 8.52 Vboat 10.2 ft/s with wind at 44° 8.54 Fparallel 6700 lbf, Fnormal 2700 lbf, power 560 hp (very approximate) 8.56 CD 2.67
(too high, incorrect prear) 8.60 This is Fig. 8.15a, flow in a 60° corner 8.62 Stagnation flow near a “bump” 8.64 All favorable gradients: no separation 8.66 0.45m/(5m 1) if U Cx m 8.68 Flow past a
Rankine oval 8.70 Applied to wind-tunnel “blockage” 8.72 Adverse gradient for x a 8.74 VB,total (8Ki 4Kj)/(15a) 8.78 Need an infinite array of images 8.82 (a) 4.5 m/s; (b) 1.13; (c) 1.26 hp 8.84 (a)
0.21; (b) 1.9° 8.86 (a) 26 m; (b) 8.7; (c) 1600 N 8.88 Thrust1-engine 2900 lbf 8.90 (a) 4.0; (b) 4.8° 8.92 (a) 0.77 m; (b) V 4.5 m/s at (r, ) (1.81, 51°) and (1.11, 88°) 8.94 Yes, they are orthogonal
Answers to Selected Problems 8.98 Yes, a closed teardrop shape appears 8.100 V 14.1 m/s, pA 115 kPa 8.102 (a) 1250 ft; (b) 1570 ft (crudely) Chapter 9 9.2 (a) V2 450 m/s, s 515 J/(kg K); (b) V2 453 m
/s, s 512 J/(kg K) 9.4 About 50 m/s 9.6 Exit at about T2 54°C and V2 1445 m/s 9.8 410 K 9.10 Ma 0.78 9.12 (a) 2.13 E9 Pa and 1460 m/s; (b) 2.91 E9 Pa and 1670 m/s; (c) 2645 m/s 9.18 (a) 930 ft/s; (b)
878 ft/s 9.20 (a) air: 144 kPa and 995 m/s; (b) helium: 128 kPa and 2230 m/s 9.22 (a) 267 m/s; (b) 286 m/s 9.24 (b) at Ma 0.576 9.28 (a) 0.17 kg/s; (b) 0.90 9.30 (a) 262 m/s; (b) 0.563; (c) 0.905 kg/
m3 9.32 (a) 141 kPa; (b) 101 kPa; (c) 0.706 9.34 (a) 0.00424 slug/s; (b) 0.00427 slug/s 9.40 (a) 2.50; (b) 7.6 cm2; (c) 1.27 kg/s; (d) Ma2 1.50 9.42 (a) Ma 0.90, T 260 K, V 291 m/s 9.44 Ve 5680 ft/s,
pe 15.7 psia, Te 1587°R, thrust 4000 lbf 9.46 Rx 8 N (to the left) 9.48 (a) 313 m/s; (b) 0.124 m/s; (c) 0.00331 kg/s 9.50 (a) Dexit 5.8 cm 9.52 (a) 5.9 cm2; (b) 773 kPa 9.54 Ma2 0.648, V2 279 m/s, T2
461°K, p2 458 kPa, p02 607 kPa 9.56 At about A1 24.7 cm2 9.58 (a) 306 m/s; (b) 599 kPa; (c) 498 kPa 9.60 Upstream: Ma 1.92, V 585 m/s 9.62 C 19,100 ft/s, Vinside 15,900 ft/s 9.64 (a) 0.150 kg/s; (b,
c) 0.157 kg/s 9.66 h 1.09 m 9.68 patm 92.6 kPa; max flow 0.140 kg/s 9.70 (a) 388 kPa; (b) 19 kPa 9.72 Mass flow 0.5 kg/s, pe 185 kPa, Mae 0.407 9.74 (a) 1.096 MPa; (b) 2.24 kg/s 9.76 tshocks 23 s;
tchoking-stops 39 s 9.78 Case A: 0.0071 kg/s; B: 0.0068 kg/s 9.80 A* 2.4 E-6 ft2 or Dhole 0.021 in 9.82 Ve 110 m/s, Mae 0.67 (yes) 9.84 (a) 0.96 kg/s; (b) 0.27; (c) 435 kPa 9.86 V2 107 m/s, p2 371
kPa, T2 330 K, p02 394 kPa 9.88 L 2 m, yes, a shock at Ma2 2.14 9.90 (a) 0.764 kg/s; (b) 0.590 kg/s; (c) 0.314 kg/s 9.92 (a) 0.45; (b) 2.04 kg/s
9.98 9.100 9.102 9.104 9.106 9.108 9.112 9.116 9.118 9.120 9.122 9.126 9.128 9.130 9.132 9.134 9.136 9.138 9.140 9.142 9.146 9.148 9.150 9.152
(a) 430; (b) 0.12; (c) 0.00243 kg/h Lpipe 69 m Flow is choked at 0.69 kg/s ptank 99 kPa (a) 0.031 m; (b) 0.53 m; (c) 26 m Mass flow drops by about 32 percent (a) 105 m/s; (b) 215 kPa Vplane 2640 ft/s
V 204 m/s, Ma 0.6 P is 3 m ahead of the small circle, Ma 2.0, Tstag 518 K 23.13°, Ma2 2.75, p2 145 kPa (a) 25.9°; (b) 26.1° wedge 15.5° (a) 57.87°; (b) 21.82° (a) pA 18.0 psia; (b) pB 121 psia Ma3
1.02, p3 727 kPa, 42.8° (a) h 0.40 m; (b) Ma3 2.43 pr 21.7 kPa Ma2 2.75, p2 145 kPa (a) Ma2 2.641, p2 60.3 kPa; (b) Ma2 2.299, p2 24.1 kPa 9.47° (helium) CL 0.184 (approximately linear), CD 0.0193
(approximately parabolic) (a) 4.10°; (b) drag 2150 N/m Parabolic shape has 33 percent more drag
Chapter 10 10.2 (a) C 3.31 m/s; (b) V 0.030 m/s 10.4 These are piezometer tubes (no flow) 10.6 (a) Fr 3.8; (b) Vcurrent 7.7 m/s 10.8 ttravel 6.3 h 10.10 crit 2(/g)1/2 10.14 Flow must be fully rough
turbulent (high Re) for Chézy to be valid 10.16 20 percent less flow, independent of n 10.18 yn 0.993 m 10.20 Q 74 ft3/s 10.22 S0 0.00038 (or 0.38 m/km) 10.24 yn 0.56 m 10.26 (a) 17.8 m3/s; (b) 1.79
m 10.30 t 32 min 10.32 74,000 gal/min 10.34 If b 4 ft, y 9.31 ft, P 22.62 ft; if b 8 ft, y 4.07 ft, P 16.14 ft 10.36 y2 3.6 m 10.38 Dsemicircle 2.67 m (16 percent less perimeter) 10.42 P 41.3 ft (71
percent more than Prob. 10.39) 10.44 Hexagon side length b 2.12 ft 10.46 Best h0/b 0.53 0.03
Answers to Selected Problems
10.48 10.50 10.52 10.54 10.56 10.58 10.60 10.64 10.66 10.70 10.72 10.76 10.78 10.80 10.82 10.84 10.86 10.88 10.90 10.92 10.94 10.98 10.106 10.108 10.110 10.112 10.114 10.116 10.120 10.122 10.124
10.126 10.128
(a) 0.00634; (b) 0.00637 (a) 2.37; (b) 0.62 m; (c) 0.0026 W 2.06 m (a) 1.98 m; (b) 3.11 m/s; (c) 0.00405 (a) 1.02 m3/s; (b) 0.0205 Fr 0.628R1/6, R in meters (a) 0.052 m3/(m s); (b) 0.0765 m hmax 0.35
m (a) 1.47; (b) y2 1.19 m (a) 0.15 m; (b) 3.2; (c) 0.59 m3/(s m) (a) 0.046 m; (b) 4.33 m/s; (c) 6.43 H 0.011 m t 8.6 s (crude analysis) (a) 3.83 m; (b) 4.83 m3/(s m) (a) 0.88 m; (b) 17.6 m/s; (c)
2.89 m y2 0.82 ft; y3 5.11 ft; 47 percent (a) 6.07 m/s; (b) V 2.03 m/s (a) downstream; (b) 5.7 percent 0.0207 (or 1.19°) (a) 3370 ft3/s; (b) 7000 hp (a) 0.61 m; (b) 3.74 m/s; (c) 0.89 m (a) steep
S-3; (b) S-2; (c) S-1 No entry depth leads to critical flow (a, b) Both curves reach y yn 0.5 m at x 250 m (a) ycrest 0.782 m; (b) y(L) 0.909 m M-1 curve, with y 2 m at L 214 m Vexing! Flow chokes at
Q 17 m3/s Q 9.51 m3/s Y 0.64 m, 34° 5500 gal/min M-1 curve, y 10 ft at x 3040 ft At x 100 m, y 2.81 m At 300 m upstream, y 2.37 m
Chapter 11 11.6 This is a diaphragm pump 11.8 (a) H 112 ft and p 49 lb/in2; (b) H 112 ft (of gasoline); P 15 hp 11.10 (a) 1300 r/min; (b) 2080 lbf/in2 11.12 (a) 11.3 m; (b) 1520 W 11.14 1870 W
11.16 11.18 11.20
(a) 1450 W; (b) 1030 r/min Vvane (1/3)Vjet for max power (a) 2 roots: Q 7.5 and 38.3 ft3/s; (b) 2 roots; H 180 ft and 35 ft 11.22 (a) BEP 92 percent at Q 0.22 m3/s 11.26 Correlation is “fair,” not
geometrically similar 11.28 BEP at about 6 ft3/s; Ns 1430, Qmax 12 ft3/s 11.30 (a) 1700 r/min; (b) 8.9 ft3/s; (c) 330 ft 11.32 Correlation “fair,” not geometrically similar 11.34 (a) 11.5 in; (b) 28
hp; (c) 100 ft; (d) 78 percent 11.36 D 9.8 in, n 2100 r/min 11.38 (a) 18.5 hp; (b) 7.64 in; (c) 415 gal/min; (d) 81 percent 11.40 (a) Ds D(gH*)1/4/Q*1/2 11.42 NPSHproto 23 ft 11.44 No cavitation,
required depth is only 5 ft 11.46 Ds C/Ns, C 7800 7 percent 11.52 (a) 7.97 m3/s; (b) 14.6 kW; (c) 28.3° 11.54 Centrifugal pumps, D 7.2 ft 11.56 (a) D 5.67 ft, n 255 r/min, P 700 hp; (b) D 1.76 ft, n
1770 r/min, P 740 hp 11.58 Centrifugal pump, ! 67 percent, D 0.32 ft 11.60 (a) 623; (b) 762 gal/min; (c) 1.77 ft 11.62 D 18.7 ft, p 1160 Pa 11.64 No speed is able to get to BEP 11.66 Q 1240 ft3/min
11.68 Qnew 15,300 gal/min 11.70 (a) 212 ft; (b) 5.8 ft3/s 11.72 (a) 10 gal/min; (b) 1.3 in 11.74 (a) 14.9; (b) 15.9; (c) 20.7 kgal/min 11.76 Dpipe 1.70 ft 11.78 Dpipe 1.67 ft, P 2000 hp 11.80 Q32
22,900 gal/min; Q28 8400 gal/min, H 343 ft for both 11.84 Two turbines: (a) D 9.6 ft; (b) D 3.3 ft 11.86 Nsp 70, hence Francis turbines 11.88 Q 52 ft3/s, D 10.5 ft 11.90 P 800 kW 11.92 Pelton and
Francis wheels both OK 11.94 (a) 71 percent; (b) Nsp 19 11.96 (a) 1.68 ft; (b) 0.78 ft 11.100 (a) 190 kW; (b) 24 r/min; (c) 9.3 ft/s 11.102 Q 29 gal/min
A Acceleration of a particle, 15, 90, 216 centripetal, 90, 157 convective, 216 Coriolis, 157 local, 216 Ackeret airfoil theory, 635–636 Acoustics, 573 Actuator disk theory, 752–753 Added mass,
539–540 Adiabatic flow, 578–579 atmospheric lapse rate, 102, 105 with friction, 604–606, 776–780 Adverse pressure gradient, 430, 445–448, 501 Aerodynamic forces and moments, 452–453 NACA designs,
471, 528–530, 565 Air-cushion vehicle, 206 Airfoil description, 468, 529 Airfoil theory, 523–534 finite-span, 530–534 supersonic flow, 632–637 thick-cambered, 528–530 thin-plate, 524–528 Andrade’s
equation, 49 Anemometer cup, 387, 485 hot-wire and film, 387, 389 Angle of attack definition, 468 in inviscid flow, 499, 517
Angular momentum theorem, 130, 158–159, 230 Angular velocity of a fluid, 246–247 Annulus flow in, 362–364 laminar friction factors, 364 Answers to selected problems, 806–812 Archimedes’ laws of
buoyancy, 44, 84 Area, body reference, 453 Area change in a duct, 583–587 Aspect ratio of a diffuser, 384 of a wing, 472, 473, 530 Atmosphere isothermal, 68 U. S. standard, 69, 773 Automobile drag
forces, 461–463 Average velocity, 143, 26 in pipe flow, 144, 341, 344, 346 Avogadro’s number, 46 Axial-flow pumps, 730–734 Axisymmetric potential flow, 534–540
B Backwater curve, 693–695 Barometer, 66–67 Basic equations (see Differential equations of flow) Basic laws of fluid motion, 35, 129–131 Bend losses, 371 813
Index Bernoulli, Daniel, 10, 174 Bernoulli constant, 176, 248 Bernoulli obstruction meters, 397–404 Bernoulli’s equation, 10, 174–177, 230, 248–249 outside a boundary layer, 435 compared to the
energy equation, 176, 580 for irrotational flow, 230, 249, 496 for isentropic flow, 580–581 limitations and assumptions, 176, 177 in rotating coordinates, 717 for unsteady flow, 175, 248, 249, 496
Betz number, 753 BG units, 8 Bingham-plastic fluid, 28 Blasius flat-plate solution, 437–439, 478 Blasius pipe friction formula, 345 Blowdown analysis, 598, 643 Blower, 711 Blunt-body flows, 429 Body
forces, 61, 224 Bore in a channel, 702 Boundary conditions, 34, 234–236 for a boundary layer, 436 free surface or interface, 234–236, 497 at an inlet or outlet, 234, 496 for inviscid flow, 496, 497
kinematic, 235 no-slip, 24, 34, 234 Boundary-element method, 546–548 Boundary layer, 23, 45, 250, 431, 496, 578 displacement thickness, 433, 438, 443 equations of, 434–436 on a flat plate, 153–155,
266 momentum thickness, 431, 438 with pressure gradient, 445–450 with rough walls, 443–444 separation, 435, 447, 447–449 shape factor, 438, 443, 448 thickness, 428, 442 transition, 298, 432, 439
Bourdon tube gage, 99–100 Brinkman number, 268 Broad-crested weir, 689, 690 Buckingham pi theorem, 280, 286–288 Bulk modulus, 49, 102, 577 of various liquids, 772
Bump, channel flow over, 675–676 Butterfly valve, 370 Buoyancy, 84–86 Buoyant force, 85
C Cambered airfoil, 468, 471, 528–530 Capillary effect, 31, 299–300 Cauchy-Riemann equations, 249 Cavitation, 32–33, 552 of a pump, 720, 721 of a turbine, 765 Cavitation erosion, 33 Cavitation
number, 32, 294, 297 Center of buoyancy, 85, 88 Center of mass, 80 Center of pressure, 75–76 of an airfoil, 527, 636 Centrifugal pump, 161–162, 200, 714–718 dimensionless coefficients, 724–726
performance curves, 720–723, 758, 759 similarity rules, 727–729 Centripetal acceleration, 90, 157 Centroid, 75 of various cross-sections, 76 Channel flow (see Open channel flow) Chézy coefficient,
665 Chézy formulas, 664–666 Choked flow, 586, 598, 740 due to friction, 608–609 due to heat transfer, 617 in an open channel, 676 Chord line, 452, 468 Circular section open channel, 668–669 pipe
flow, 338–357 Circulation, 499 at airfoil trailing edge, 524 on a cylinder, 509, 511 Classification of flow, 36–37 Colebrook pipe-friction formula, 348 Complex-variable potential theory, 516–521
Index Composite channel flows, 687–688 Compressibility criterion, 35, 221, 571 Compressible flow, 220, 315, 571 with area change, 583–587, 774–776 with friction, 603–613, 780–784 with heat transfer,
613–618, 784–788 tables, 774–788 Compressor, 711, 740–741 Computational fluid dynamics, 3, 434, 465, 540–555, 735–736 commercial codes, 552–553 Concentric cylinder flows, 261–263 Cone flow,
supersonic, 638 Conformal mapping, 516, 562 Conical diffuser, 373, 386 Conjugate depths, 699 Conservation laws, 35, 130–131 for angular momentum, 130, 158–159 for energy, 131, 163–165, 231 for linear
momentum, 130, 146 for mass, 130, 141–146 for salt or species, 35, 315 Consistent units, 11–13 Contact angle, 30–31 Continuity, equation of, 218 cylindrical polar form, 219–220, 793 incompressible
flow, 220 spherical polar form, 265 turbulent flow, 334 Continuum, 6–7 Contraction losses, 372, 374 Control section of a channel, 687, 693 Control surface, 136 Control volume, 36, 133 arbitrary but
fixed, 135–136 deformable, 137–138, 140 differential-sized, 218 guidelines for selection, 183 moving, 133, 137, 152 one-dimensional, 134–135 Convective acceleration, 15–16 Converging-diverging
nozzle, 600–603 Converging nozzle, 598–600 Conversion factors, 8–9, 791–792 Coriolis acceleration, 156, 157, 250 Corner flow, inviscid, 242, 518–519 Correlations, turbulent, 334
Corresponding states, law of, 24 Couette flow between cylinders, 261–263 instability of, 262–263 between plates, 25–26, 258–259 nonnewtonian, 272 Couple, concentrated, 268 Creeping motion, 25, 298,
315, 317, 328, 483 past a sphere, 47, 457, 483 Critical channel flow, 671–674 Critical depth, 664, 672 Critical Reynolds number, 329–330 Critical slope of a channel, 674 Critical sonic-point
properties, 581, 605 Critical state, 6, 24 Crossflow turbine, 764 Crump weir, 705 Cup anemometer, 387, 485 Cup-mixing temperature, 268 Curl of a vector, 247 Current meter, 387, 389 Curved surface,
force on, 79–82 Curvilinear coordinate system, 267 Cylinder array of, 515 in inviscid flow, 245, 508–510 rotating, 512 in viscous flow, 261–263, 295–296, 298, 455 Cylindrical coordinates, 219
equations of motion, 793–794
D da Vinci, Leonardo, 44, 143 d’Alembert paradox, 45, 510 Darcy friction factor, 340, 342, 344, 604 Darcy’s law of porous flow, 420 Darcy-Weisbach equation, 340, 664 Decimal prefixes, 13 Deformable
control volume, 133, 137 Deformation of a fluid element, 245–247 Del operator, 216, 219
Index Density definition of, 6, 17 of various fluids, 771–772 Detached shock wave, 624 Diaphragm transducer, 101 Differential equations of flow, 36, 215, 682, 793 angular momentum, 230 continuity or
mass, 217–221 cylindrical coordinates, 793–794 energy, 231–233 incompressible, 220, 228, 236–237 linear momentum, 62, 223–227 Diffuser flows, 313, 381–385, 447 head loss, 373 performance maps, 385,
386 separation and stall, 383, 447–448 stability map, 382 subsonic versus supersonic, 584 Digital computer (see Computational fluid dynamics) Dilatant fluid, 28 Dimensional analysis, 7, 12, 277 of
the basic equations, 292–294 of the boundary conditions, 293–294 of pipe flow, 307–309 pitfalls of, 305–307 of turbomachines, 724–726, 744 Dimensional homogeneity, 11–12, 280 nonhomogeneity, 285–286
Dimensional matrix, 313 Dimensionless groups, list of, 297 Dimensions, 7–13, 278 list of, 8–9, 287 Discharge coefficient, 12, 179, 181, 398 flow nozzle, 401 orifice plate, 399–400 sluice gate, 677
venturi, 401, 402 weir, 690–692 Displacement thickness, 433 for a flat plate, 433, 438, 443 Dissipation function, 233 of a hydraulic jump, 680 Divergence of a vector, 220, 226 Dot product, 133
Doublet line, 270, 505–506 point, 536 Downwash on a wing, 531, 532 Draft tube, 765 Drag, 452–467 biological adaptation, 467–468 induced, 472, 533 Drag coefficient, 195, 297, 453, 468, 632 of
airfoils, 457, 468, 471–473 of a cylinder, 298, 455, 457 rotating, 511–512 on a flat plate, 438, 442–443 at high Mach numbers, 465–466 of road vehicles, 461–463 of a sphere, 298, 456, 457 spinning,
487, 488 of surface ships, 464–465 of three-dimensional bodies, 457, 460 of two-dimensional bodies, 457–458 Drag reduction, 456, 462, 464 Drowned channel flow, 678, 701 Duct flow, 325, 603
compressible, with friction, 603–613, 780 with heat transfer, 613–618, 784 Dynamic similarity, 304–305
E Eckert number, 297 Eddy viscosity, 406 Effective duct diameter, 360 Efficiency, 47, 715 of an open channel, 669–671 of a turbomachine, 715, 734 volumetric, 716, 757 of wind turbines, 753 Elbows,
losses in, 368, 369 Elliptical wing, 533 Energy, 18, 131, 163 Energy equation, 163–165, 231–233 steady flow, 167–168 Energy flux, 165, 231 Energy grade line, 177–178, 339 672
Index Engineering Equation Solver (EES), 41 Enthalpy, 19, 165, 574 Entrance length, 331 Entrance losses, 331, 371, 372 Entrance region, 330–331 Entropy, of an ideal gas, 574 Entropy change, 574
across a normal shock, 591 across a weak oblique shock, 627 Equations of motion (see Differential equations of flow) Equilibrium of forces, 61–62 Equivalent length, minor losses, 367 Erosion of
particles, 313, 698 Euler, Leonhard, 45, 174, 294 Euler number, 294, 297 Euler turbine equations, 161–162, 717 Eulerian description, 14, 216 Euler’s equation, 227, 237, 247 Exit pipe loss, 371, 372
Expansion losses, 372–373 Explicit numerical model, 549 External flow, 333, 427, 451–467
F Falling-body problem, 12 dimensional analysis of, 282–285, 289 Fanno line, 648 Favorable pressure gradient, 430, 445–447, 495, 502 Film of fluid draining down an inclined plane, 268 down a vertical
cylinder, 272 down a vertical plate, 271 Finite-difference method, 541–543 Finite-element method, 541 Finite-span wings, 472–473, 530–534 First law of thermodynamics, 131, 163 Fittings, losses in,
368 Flap, airfoil, 471, 472, 474 Flat-plate flow,153–155, 428, 431–433 Blasius solution, 437–439, 478
integral theory, laminar, 432 turbulent, 441–444 normal to the stream, 457–459, 519–521 with rough walls, 443–444 Flettner rotorship, 511–512, 559 Floating element shear measurement, 480 Flooding of
channels, 698 Flow between plates, 25–27, 258–260, 359– 360 Flow coefficient of a meter, 398 Flow meters (see Fluid meters) Flow net, 497 Flow nozzle, 399, 400–401 Flow straighteners, 479 Flow
visualization, 40, 426, 470, 515, 554 Fluctuation, turbulent, 326, 333–334 Fluid, definition of, 4 Fluid meters, 385–404 Coriolis type, 395, 396 electromagnetic, 389 flow nozzle, 399, 400–401 head
losses, 402 hot-wire, hot-film, 387, 389 laminar-flow element, 396–397 laser-doppler, 387, 389–390 obstruction type, 397–404 orifice plate, 398–400 pitot-static tube, 387, 388–389 rotameter, 395
Savonius rotor, 387, 486 turbine type, 392–393 ultrasonic, 394–395 venturi meter, 399, 401–402 volume flow type, 391 vortex type, 393–394 Fluid properties, 769–773 Force coefficient, 278, 310 Forces
hydrostatic, 74 –84 on a turning vane, 150–151 Fourier’s law of conduction, 27, 231 Francis turbine, 742 Free-body concept, 77, 133 Free overfall, 687, 688 Free-streamline theory, 520–521
Free-surface flows, 4–5, 236, 326, 497, 659
Index Free vortex, 253 Friction drag, 154, 453 Friction losses, 168 Friction factor, 42, 340, 342, 604 compressible flow, 606 Friction factor—Cont. laminar pipe-flow, 342 noncircular ducts, 364, 365
rocky channels, 665 turbulent pipe-flow, 345, 348 Friction velocity, 336 Frictionless flow, 227, 495, 613 Frontal area, 453, 462 Froude, William, 45, 294 Froude number, 294, 297, 465–466, 662, 679,
683 Froude scaling laws, 303–304 Fully developed flow, 258, 330–331 Fully rough flow in channels, 665 on a flat plate, 443–444 in pipes, 347, 348 Fundamentals of Engineering (FE) Exam, 43
G Gage pressure, 63, 77, 148 Gages, pressure, 97–101 Gas constant, 19, 573 of various gases, 772 Gas dynamics (see Compressible flow) Geometric similarity, 301–303 violation of, 302, 303, 307 Glide
angle, 489 Gradient operator, 61, 216, 219 Gradual contraction loss, 374 Gradual expansion loss, 373 Gradually varied flow, 662, 682–687 classification, 683–685 effect of width changes, 704 Grashof
number, 297 Gravity acceleration of, 9, 64 variation with radius, 65
Gravity force on an element, 61, 224 Grid, numerical, 542, 549, 553, 564
H Hagen, G. L. H., 329 Hagen-Poiseuille flow, 341 Half-body, plane, 256–258, 501–502 axisymmetric, 537–538 Halocline, 120 Hazen-Williams formula, 47, 285 Head loss, 168, 340, 661 of a hydraulic jump,
680 minor, 367–375 in pipe flow, 340 Heat addition, flow with, 613–618, 784–788 Heat conduction equation, 233 Heat flux through an element, 232 Heat transfer coefficient, 312 Hele-Shaw flow, 513–514
Herschel-type venturi, 401 High-lift devices, 474 History of fluid mechanics, 44–46, 280 Hodograph for an oblique shock, 623, 624 Homologous points, 302, 727 Honeycomb flow straightener, 408
Horseshoe vortex, 555 Hot-wire or hot-film anemometer, 387, 389 Hydraulic diameter, 358, 363, 661 Hydraulic efficiency, 716 Hydraulic grade line, 177–178, 339, 659 Hydraulic jump, 198–199, 664,
678–681, 702 classification, 679 sloping, 702 Hydraulic model, 307 Hydraulic radius, 13, 358, 661 Hydraulically smooth wall, 347 Hydrodynamic mass, 539–540 Hydrogen bubble technique, 34 Hydrometer,
118 Hydrostatic condition, 4, 59, 62, 63 in gases, 67–69 in liquids, 65–66
Index Hydrostatic forces on curved surfaces, 79–83 in layered fluids, 82–84 laboratory apparatus, 127 on plane surfaces, 74–79 Hydrostatic pressure distribution, 63–65 Hypersonic flow, 572, 639
I Icebergs, 89–90 Ideal gas (see Perfect-gas law) Images, 521–522 Implicit numerical model, 550 Impulse turbines, 745–749 Incompressible flow, 17, 142–143, 220, 236, 572 Induced drag, 533 Inertial
coordinate system, 156 Initial conditions, 234 Integral equations (see Control volume) Intensity of turbulence, 334, 405 Interface, 29 Internal energy, 18, 574 Internal flow, 330 Inviscid flow, 36,
496 Irrotational flow, 230, 247–249, 496 frictionless, 247, 579–582 Isentropic flow, 579–582, 529 with area change, 583–587, 774 compared to Bernoulli’s equation, 580–581 tables, 774–776 Isentropic
process, 574 Isothermal duct flow, 610–611 Isovelocity contours, 660
J Jet exit pressure condition, 149 Jet flow, laminar and turbulent, 326–327 Jet pump, 189, 712
Jet-turning vane, 150–151 Joukowski transformation, 562
K Kaplan turbine, 742, 746, 749 Kármán momentum-integral relation, 154, 431 Kármán vortex street, 295–296 Kelvin oval, 513–514 Kinematic properties, 15 Kinematic similarity, 303–304, 498 Kinematic
viscosity, 24 of various fluids, 24, 770–772 Kinetic energy, 18, 164 correction factor, 170–171 Kline-Fogleman airfoil, 473, 474 Kutta condition, 523–524 Kutta-Joukowski lift theorem, 510–511
L Lagrangian description, 14 Laminar flow, 34, 326, 327, 551 in a concentric annulus, 346–364 between parallel plates, 25–27, 258–260, 359–360 in a pipe, 341–344 between rotating cylinders, 261–263
Laplace’s equation, 239, 251, 496, 497, 516, 793 numerical simulation, 541–543 in polar coordinates, 498, 564, 793 Lapse rate, 68 Large-eddy simulation, 554 Laser-Doppler anemometer, 387, 389
Law-of-the-wall, 336 Lawn sprinkler analysis, 163 Layered fluids, 70, 82–84 Lift definition of, 452, 468 in flow past a cylinder, 510–511
Index Lift coefficient of airfoils, 470–474, 527, 529–530, 533 supersonic, 632–637 definition, 297, 468, 632 maximum, 473, 528 of a rotating cylinder, 511–512 of a rotating sphere, 488 Lift-drag
polar plot, 472 Lifting line theory, 532 Lifting vane, 150–151, 469 Linear momentum, 130, 146–158, 223–227 Liquids versus gases, 4–6 Local acceleration, 216 Local mass-flow function, 586–587
Logarithmic velocity profile, 336, 344 Loss minor, 367–375 in pumps, 723 Lubricating oil properties, 769, 770, 772 Lubrication theory, 271
M Mach angle, 619 Mach cone, 619 Mach number, 35, 221, 295, 297, 306, 572, 579, 592 effect on body drag, 467 Mach waves, 594, 618–621, 628 analogy to water waves, 663, 673, 681 Magnus effect, 510
Manifold flow, 422–423 Manning, Robert, 13, 665 Manning roughness factor, 13, 285, 665 for various channels, 667 Manometer, 70–73, 97, 99 two-fluid differential, 72, 108 Mass, units of, 8 Mass flow,
133, 142, 586, 599, 611 in choked flow, 586 MATLAB contouring, 506 Mean free path of a gas, 7, 46 Meniscus, 74 Metacenter, 87
Metacentric height, 87, 88 Meter (see Fluid meters) Minor losses in pipe flow, 367–375 Mixing-length theory, 406 Model-testing principles, 278, 301–307 pitfalls and discrepancies, 296–297 Mohr’s
circle, 4–5, 59 Molecular weight, 19, 573 of various gases, 772 Moment of inertia, 76, 88, 131 for various areas, 76 Momentum angular, 130, 158, 230 linear, 130, 146–148, 223–227 Momentum flux, 147,
224 correction factor, 155–156 Momentum integral theory, 155, 431–433, 448 Momentum thickness, 431 for a flat plate, 438, 442 Thwaites’ parameter, 448 Moody chart, 349, 443, 606, 665 Moody pump-size
formula, 728 Moving shock wave, 595–596 Multiple-pipe systems, 375–381
N NACA airfoils, 470–471, 528, 567 Nappe, 687, 689 Natural convection, 312, 315, 573 Navier-Stokes equations, 45, 228, 789 nonuniqueness of, 263 Net positive suction head, 721–722 Network, piping,
380–381 Neutral buoyancy, 86, 386 Newton, Sir Isaac, 23, 45, 577 Newtonian fluid, 23, 227–228 Newton’s second law, 8, 130, 146 for a fluid element, 62, 216, 224 in noninertial coordinates, 156–158
No-slip condition, 24, 34, 234, 259, 262, 340 No-temperature-jump condition, 34, 234 Noncircular duct flow, 357–366
Index Nondimensionalization (see Dimensional analysis) Noninertial coordinate system, 156–158 Nonnewtonian fluids, 28, 272 Nonwetting liquid, 30, 31 Normal channel depth, 662, 667 Normal shock wave,
590–595, 618 tables, 776–780 Normal stresses, 225 Nozzle flow, 598–601 analogy with a sluice gate, 664, 677 choked, 586, 600 converging-diverging, 600–603 design conditions, 600–601 Nozzle flow—Cont.
subsonic versus supersonic, 584 Numerical analysis, 3, 434, 540–555 instability, 549 of inviscid flow, 540–548 of open-channel flow, 685–687 of pumps, 735–736 of viscous flow, 548–555
O Oblique shock wave, 620, 621–628, 789–790 reflection of, 651 One-dimensional approximation, 134–135, 138, 147, 165, 583, 660–661 One-seventh power-law, 439, 442, 479 Open channel flow, 236, 659
analogy with gas dynamics, 663, 664, 673, 677, 681 classification of, 662–664 critical flow, 671–674 gradually varied flow, 682–687 most efficient section, 669–671 over weirs, 687–693 Orifice plate,
398–400 Orthogonality conditions, 249, 497–498, 516 Outer layer, turbulent, 335–336 Overlap layer, 335–336, 441 Overrelaxation, 543
P Parallel plates, 26, 258–259, 359–360 Pascal unit, 9, 11 Pascal’s law, 71 Pathline, 37–38 Pelton wheel turbine, 745, 747 Perfect-gas law, 19, 35, 67, 573, 585 Permeability of porous media, 315, 420
Physical properties of fluids, 769–773 Pi theorem, 286–288 Piezometer, 103 Pipe flow, 328, 338–357 bend loss, 371 compressible, 604–613, 780–788 flow rate determination, 352–355 head loss or pressure
drop, 307, 339, 351 laminar, 341–344 minor losses, 367–375 in a network, 380–381 noncircular, 357–366 in parallel, 376–378 with rough walls, 346–349 in series, 375–376 sizing problem, 355–357
turbulent, 344–348 Pipe standard sizes, 357 Pipelines, 168, 610 Pitching moment, 452, 527 Pitot-static tube, 387, 388, 642 Planform area, 299, 453, 468 Pode’s angle, 487 Poiseuille, J. L. M., 10, 341
Poiseuille flow, 260, 341–342 Poisson’s ratio, 577 Polar coordinates, 220, 243, 498, 499 Polar drag plot, 472 Positive-displacement pump, 711–714, 757 Potential energy, 18, 164 Potential flow,
252–257, 497 analog methods, 513–515 axisymmetric, 534–540 complex variable, 516–521 numerical analysis, 540–548 Potential lines, 248, 498 Potential vortex, 253, 498
Index Power coefficient, 724, 753, 761 Power-law correlation, for pipe flow, 309, 345 for velocity profile, 155, 439, 442, 479 for viscosity, 27, 772 Power product method, 286 Power specific speed,
744 Prandtl, Ludwig, 2, 45, 434, 629 flat-plate formulas, 479 lifting line theory, 532 Prandtl-Meyer angle, 630, 788 Prandtl-Meyer expansion waves, 628–631, 788 Prandtl number, 297, 579 Prefixes for
units, 13 Pressure, 17, 59–61 absolute versus gage, 63 at a point, 60 stagnation, 312, 580 vacuum, 63 vapor, 31–32, 772, 773 Pressure coefficient, 297, 454, 526, 546 Pressure condition at a jet exit,
149 Pressure distribution, 62, 89 hydrostatic, 63–65 in irrotational flow, 249 in a nozzle, 599, 601 in rigid-body translation, 91–93 in rotating rigid-body motion, 93–97 Pressure drop in pipes, 339,
345 Pressure drag, 453 Pressure force on a control volume, 147–148 on a curved surface, 79–82 on an element, 60–61 on a plane surface, 74–79 Pressure gradient, 61, 96, 259, 341 adverse and favorable,
430, 445–448, 501 Pressure head, 65, 102, 168 Pressure measurement, 97–101 Pressure recovery of a diffuser, 373, 382, 385, 386 Pressure transducers, 99–101 Primary dimensions, 8, 278 Principle of
corresponding sates, 24–25 Principle of dimensional homogeneity, 10–11, 280 Problem-solving techniques, 44 Product of inertia, 76
Propeller turbine, 742, 746, 749 Properties of fluids, 769–773 Propulsion, rocket, 158 Prototype, 36, 278 Pseudoplastic fluid, 28 Pump-system matching, 735–740 Pump-turbine system, 203, 746 Pumps,
47, 711 axial-flow, 729–733 centrifugal, 161–162, 714–718 dimensionless, 724 effect of blade angle, 719 effect of viscosity, 729, 730 multistage, 740 net positive-suction head, 721–722 in parallel,
738–739 performance curves, 204, 212, 714, 720– 722, 733–735, 758, 759 positive-displacement, 711–714 in series, 739–740 similarity rules, 727–728 size effects, 728
R Radius of curvature, 29, 235 Rankine half-body, plane, 256–258, 501–502 axisymmetric, 537–538 Rankine oval, plane, 507–508 axisymmetric, 563 Rankine-Hugoniot relations, 590 Rapidly varied channel
flow, 662, 687 Rarefaction shock, 593, 624 Rayleigh line, 649 Reaction turbines, 742 Rectangular duct flow, 365, 366 Relative roughness, 349 Relative velocity, 137–138, 152 Reversible adiabatic flow
(see Isentropic flow) Reynolds, Osborne, 45, 294, 330 Reynolds number, 24, 278, 294, 297, 325, 427 for an airfoil, 469 local, 429 Reynolds pipe-flow experiment, 330 Reynolds time-averaging concept,
Index Reynolds transport theorem, 133–141 Rheology, 5, 28 Rheopectic fluid, 28 Rigid-body fluid motion, plane, 89–97 Rocket motion, 158 Rolling moment, 452 Rotameter, 395 Rotating cylinder, 512
sphere, 488 Rotationality, generation of, 249–250 Rough-wall effects on channels, 665–667 on cylinder drag, 298 on a flat plate, 443–444 on pipe flow, 346–348 on pumps, 726 sand-grain tests, 347 on
sphere drag, 321, 456 Roughness of commercial pipes, 349 of open channels, 667
S Salinity, 22 Sandgrain roughness, 347 Savonius rotor, 387, 486 Saybolt viscosity, 286 Scaling laws, 278–279, 304, 306 Scaling parameters, 282 Schedule-40 pipe sizes, 357 Seawater properties, 22,
772 Second law of thermodynamics, 131, 233, 606, 624, 679, 680 Secondary dimensions, 8–9 Secondary flow, 365–366 Separated flow, 429, 455, 456, 502 Separation bubble, 469 Separation point on an
airfoil, 525 on a cylinder, 455 definition of, 447 in a diffuser, 447 in a laminar boundary layer, 449, 501–502 on a sphere, 456
Shaft work, 164 Shape factor, 438, 443, 448 Sharp-crested weir, 689–690 Shear stresses, 4–5, 23, 225 turbulent, 334 Shear work, 164 Shock-expansion theory, 632–637 Shock polar, 623, 624 Shock-tube
wind tunnel, 598 Shock wave, 250, 591, 521 detached, 624 linearized, 626 moving, 595–596 normal, 590–595, 776–780 oblique, 621–628, 789–790 rarefaction, 593, 624 strong versus weak, 624–626 Shut-off
head of a pump. 719–720 SI units, 7–8 Silicon resonance transducer, 67, 101 Similarity, 279, 301 dynamic, 304–305 geometric, 301–303 violations, 302, 303, 307 kinematic, 303–304, 498 for pumps,
727–728 Sink line, 253, 498, 563 point, 536 Siphon, 208, 406 Skin friction coefficient, 432, 438, 441, 442, 449 Slip conditions in inviscid flow, 237 Sluice gate, 664, 677–678 drowned, 678, 701
Smoke-flow visualization, 40, 470 Soap bubble, 29 Sonic boom, 620, 621 Sonic point, 581, 605 Source line, 253, 498, 517 point, 536 Spar buoy, 119 Specific diameter, 760–761 Specific energy, 671–672
Specific gravity, 12, 18 Specific heat, 19–20, 573
Index Specific-heat ratio, 19, 295, 297, 572 of common gases, 21, 772 Specific speed, 730–731, 734, 735 Specific weight, 17 of common fluids, 65, 772 Speed of sound, 35, 221, 575–577 in the
atmosphere, 773 of a perfect gas, 35, 577 of various materials, 577 of water, 577, 769 Sphere inviscid flow, 265, 538–539 viscous flow, 298, 321, 456, 457, 483, 488 Spherical droplet, 29 Spherical
polar coordinates, 265, 535–536 Stability of floating bodies, 86–89 Stability of a pump, 720, 763 Stability map of a diffuser, 382 Stagnation density, 580 Stagnation enthalpy, 167, 578, 614
Stagnation point, 38, 249, 252, 256, 519 plane flow near, 39–40, 265 Stagnation pressure, 312, 580 Stagnation properties, 578–580 Stagnation speed of sound, 579 Stagnation temperature, 579 Stall
angle of attack, 528 Stall speed, 473 Stalled airfoil, 470 Standard atmosphere, 69, 773 Stanton number, 312 Starting vortex, 469 State, equation of, 16, 18, 67, 131, 234, 573 for gases, 18–20,
573–574 for liquids, 21–22 van der Waals, 642 Static-pressure measurement, 97 Steady-flow energy equation, 167–168, 578, 661 Stokes flow past a sphere, 47, 457, 483, 487 Stokes’ stream function, 269,
535 Stopping vortex, 469 Strain rate, 23, 247 Stratified flow, 221, 250 Streakline, 37–38 Stream function, 238–244, 497 axisymmetric flow, 243–244, 535
compressible flow, 243 geometric interpretation, 240–241 irrotational flow, 239, 497 polar coordinates, 243 of Stokes, 269, 535 Streamline, 37–40, 240, 498 Streamline coordinates, 267 Streamlining of
bodies, 456 Streamtube, 38, 143, 174 Stress gradients, 225 Stress tensor, 225, 227, 794 symmetry condition, 231 Strouhal number, 295, 296, 297 Subcritical channel flow, 663, 672 Subsonic flow, 572
Substantial derivative, 216 Suction specific speed, 731 Sudden expansion or contraction, 192–193, 371–372 Supercritical channel flow, 663, 672–673 Superposition of potential flows, 254–257, 500–501
Supersonic airfoil theory, 632–637 Supersonic flow, 149, 572, 618 Surface forces, 61, 225 Surface tension, 29–31, 235, 693 of air-water, 30, 773 of various interfaces, 29, 772 Sutherland-law
viscosity formula, 27, 771 Swallow float, 86 System, 16, 130 System-matching of pumps, 735–740 Systems of units, 7
T Tainter gate, 115 Takeoff analysis for aircraft, 475 Taylor number, 262 Taylor vortices, 263 Tee-junction losses, 368 Temperature definition, 17 rise due to dissipation, 238 Terminal velocity, 309,
Index Thermal conductivity, 27, 231 Thermodynamic properties, 16, 131, 771–772 Thickness drag, supersonic, 635 Thin-airfoil theory, 524–527, 635 Thixotropic fluid, 28 Three-dimensional flow, 535–540,
637 compressible, 637–640 Three-reservoir pipe junction, 376, 379 Throat in a duct, 585 Thwaites’ integral method, 448–450 Time-averaging of turbulence, 333–334 Timeline, 37 Tornado flow model, 255,
501, 556 Torricelli’s formula, 179 Total head, 168, 177 Trailing vortex, 469, 531–532 Transition to turbulence, 326–330 in a boundary layer, 299, 432, 439 on a flat plate, 405, 439 in a jet exit
stream, 327 in pipe flow, 326, 328 in sphere flow, 310, 405 Transitional roughness, 347 Transonic flow, 572 Transport properties, 16 Trapezoidal channel, 668, 670–671 Triangular duct flow, 365, 366
Tri-diagonal matrix, 550 Trip wire, 405 Troposphere, 68 Tube bundle, 412 Turbines, 742–749 efficiency, 744, 748–749 impulse, 745–749 reaction, 742 windmills, 750–754 Turbomachine classification,
711–714 Turbulent flow, 3, 34, 376 on a flat plate, 441–444 fluctuations 326, 333–334 historical details, 328–330 intensity, 405 intermittency, 326 numerical models, 552–553 in a pipe, 328, 344–348
Turbulent puff, 328
Turbulent shear flow, 333–337 logarithmic overlap layer, 335–337 wall and outer layers, 335–336 Turbulent stresses, 334 Two-phase flow, 6, 219
U Ultrasonic flowmeter, 394 Uncertainty of data, 42–43 Uncoupling of velocity and temperature, 236–237 Uniform channel flow, 662, 664–667 Uniform stream, 252–253, 498, 517, 536 Unit normal vector,
132, 136 U. S. Standard Atmosphere, 69, 773 Units, 7–10 Universal gas constant, 19, 573 Unsteady Bernoulli equation, 175, 248, 249, 496 Unsteady flow, 36, 40, 549 Upwind differencing, 552
V V-notch weir, 692–693, 705 Vacuum pressure, 63 Valve flows, 12, 367–370 Van der Waals’ equation, 642 Vane flow, 150–151 Vapor pressure, 31–32 of various fluids, 772 of water, 32, 773 Varied flow,
662, 682–687 Vector differentiation, 215–217 Velocity-defect law, 336 Velocity diagrams, 717, 732, 743 Velocity field, 14–15, 215 Velocity gradient, 23, 246 Velocity head, 168, 367 Velocity
measurement, 385–390 Velocity of approach factor, 398 Velocity potential, 248, 496, 535
Index Velocity profile, 23, 34, 439, 449, 535 for the Blasius solution, 437, 439 Vena contracta, 391, 392, 397, 398, 678 VentureStar spacecraft, 639–640 Venturi flume, 704 Venturi meter, 180–181,
207, 399, 401–402 Virtual mass, 539–540 Viscometer, 50, 51, 203 Viscosity, 22–24 formula for liquids, 27 generalized chart, 25 Sutherland and power-law, 27, 772 of various fluids, 24, 769, 771–772
Viscous dissipation, 233 Viscous flow analysis, 258–263 Viscous force on an element, 226 Viscous stresses, 23, 225, 228 symmetry condition, 231 Viscous sublayer, 337–347 Viscous work, 164
Visualization of flow, 40 Volume expansion rate, 15 Volume flow, 132, 142 measurement of, 391–404 Von Kármán, Theodore, 45, 154, 406, 431, 567 Vortex, line, 253–254, 498, 503, 518 infinite row,
503–504 potential, 253, 498 starting and stopping, 469 trailing, 531–532 Vortex flowmeter, 393–394 Vortex shedding, 295–296 Vortex sheet, 504–505 for a thin airfoil, 524–527 Vorticity, 247
W Wake flow, 190, 195, 250, 426, 429, 455 Wall roughness, 339–340 Wall shear stress, 341, 342, 438, 442
Waterline area, 88 Wave drag of ships, 464–466 of supersonic bodies, 635 Wave motion, 494, 576, 663–664, 672 periodic, 696 Weber number, 294, 297, 693 Weirs, 47, 687–693 broad-crested, 689, 690 Crump
type, 705 drowned, 196, 705 inviscid flow model, 560 sharp-crested, 689–690 Wetted area, 453, 465 Wetted perimeter, 358, 661 Wind turbines, 750–754 performance of, 753 typical designs, 751 world
energy distribution, 754 Wing theory, two-dimensional, 523–534 finite span, 472, 530–534 Work, 164 due to viscous stresses, 232
X X-33 spacecraft, 639–640
Y Yawing moment, 452 Young’s modulus, 577
Z Zero-lift airfoil angle, 530 Zones of action and silence, 620, 672
StudyGuide for Fluid Mechanics Preface The following materials are provided as a study guide for the text Fluid Mechanics by Frank White. A brief summary of the key concepts and theory is presented
for each chapter along with the final form of basic equations (without detailed derivations) used in the various analyses being presented. In most cases, a detailed explanation for the physical
significance of each term in a fundamental governing equation is given (e.g., linear momentum, pg. III-8) to assist the student in identifying when a given term should be included in the analysis.
Example problems are provided for major sections. In each case, the starting general equation used in the solution is given followed by any necessary simplifications and the resulting complete
solution. Where appropriate, the control volume and coordinate system used in the analysis are shown with the problem schematic. In many cases, an explanation is given with the final numerical answer
to help the student understand the engineering significance of the answer (e.g., forces on curved surfaces, pg. II–18). In selected cases, computer based solutions to example problems are provided as
an example to the student in the use of computer based problem solving techniques (e.g., parallel pipe sections, pg. VI-23). For problems areas involving multiple steps in the solution, a summary of
the steps used in a typical problem solution sequence is provided and enclosed in a boxed border (e.g., rigid body motion, pg. II-22). Areas where the author’s experience has shown that mistakes in
the analysis can easily occur are noted as Key Points (e.g., laminar flat plate boundary layer, pg. VII-5) throughout the material. Finally, the author of this study guide appreciates the opportunity
to contribute to the instructional materials provided with one of the leading texts in the area of fluid mechanics and to collaborate with an educator with whom he has long has the highest respect
and had the privilege to further his education in fluid mechanics while a student at Georgia Tech. Jerry R. Dunn, P.E. Associate Professor, Department of Mechanical Engineering Texas Tech University
I. FLUID MECHANICS I.1 Basic Concepts & Definitions: Fluid Mechanics - Study of fluids at rest, in motion, and the effects of fluids on boundaries. Note: This definition outlines the key topics in
the study of fluids: (1) fluid statics (fluids at rest), (2) momentum and energy analyses (fluids in motion), and (3) viscous effects and all sections considering pressure forces (effects of fluids
on boundaries).
Fluid - A substance which moves and deforms continuously as a result of an applied shear stress. The definition also clearly shows that viscous effects are not considered in the study of fluid
statics. Two important properties in the study of fluid mechanics are: Pressure and Velocity These are defined as follows:
Pressure - The normal stress on any plane through a fluid element at rest. Key Point: The direction of pressure forces will always be perpendicular to the surface of interest. Velocity -
The rate of change of position at a point in a flow field. It is used not only to specify flow field characteristics but also to specify flow rate, momentum, and viscous effects for a fluid in
motion. I-1
I.4 Dimensions and Units This text will use both the International System of Units (S.I.) and British Gravitational System (B.G.). A key feature of both is that neither system uses gc. Rather, in
both systems the combination of units for mass * acceleration yields the unit of force, i.e. Newton’s second law yields S.I. 1 Newton (N) = 1 kg m/s2
B.G. 1 lbf = 1 slug ft/s2
This will be particularly useful in the following: Concept momentum
Units kg/s * m/s = kg m/s2 =N
& mV
slug/s * ft/s = slug ft/s2 = lbf manometry
kg/m3*m/s2*m = (kg m/s2)/ m2 =N/m2
slug/ft3*ft/s2*ft = (slug ft/s2)/ft2 = lbf/ft2 dynamic viscosity
N s /m2 = (kg m/s2) s /m2 = kg/m s lbf s /ft2 = (slug ft/s2) s /ft2 = slug/ft s
Key Point: In the B.G. system of units, the mass unit is the slug and not the lbm. and 1 slug = 32.174 lbm. Therefore, be careful not to use conventional values for fluid density in English units
without appropriate conversions, e.g., ρw = 62.4 lb/ft3 For this case the manometer equation would be written as
g h gc I-2
Example: Given: Pump power requirements are given by W& p = fluid density*volume flow rate*g*pump head = ρ Q g hp
For ρ = 1.928 slug/ft3, Q = 500 gal/min, and hp = 70 ft, Determine: The power required in kW. 3 W& p = 1.928 slug/ft3 * 500 gal/min*1 ft /s /448.8 gpm*32.2 ft/s2 * 70 ft
W& p = 4841 ft–lbf/s * 1.3558*10-3 kW/ft–lbf/s = 6.564 kW
Note: We used the following: 1 lbf = 1 slug ft/s2 to obtain the desired units Recommendation:
In working with problems with complex or mixed system units, at the start of the problem convert all parameters with units to the base units being used in the problem, e.g. for S.I. problems, convert
all parameters to kg, m, & s; for BG problems, convert all parameters to slug, ft, & s. Then convert the final answer to the desired final units.
Properties of the velocity Field Two important properties in the study of fluid mechanics are Pressure
and Velocity
The basic definition for velocity has been given previously, however, one of its most important uses in fluid mechanics is to specify both the volume and mass flow rate of a fluid. I-3
Volume flow rate:
& = V ⋅ n dA = Q ∫ ∫ Vn dA cs
where Vn is the normal component of velocity at a point on the area across which fluid flows. Key Point: Note that only the normal component of velocity contributes to flow rate across a boundary.
Mass flow rate:
& = ∫ ρV ⋅n d A = ∫ ρV d A m n cs
NOTE: While not obvious in the basic equation, Vn must also be measured relative to any flow area boundary motion, i.e., if the flow boundary is moving, Vn is measured relative to the moving
This will be particularly important for problems involving moving control volumes in Ch. III.
1.6 Thermodynamic Properties All of the usual thermodynamic properties are important in fluid mechanics P - Pressure
(kPa, psi)
T- Temperature
( C, F)
ρ ñ Density
(kg/m3, slug/ft3)
Alternatives for density γ - specific weight = weight per unit volume (N/m3, lbf/ft3) γ=ρg
γ = 9790 N/m3 = 62.4 lbf/ft3
γ = 11.8 N/m3 = 0.0752 lbf/ft3
S.G. - specific gravity = ρ / ρ (ref) where: ρ (ref) = ρ (water at 1 atm, 20˚C) for liquids = 998 kg/m3 = ρ (air at 1 atm, 20˚C) for gases = 1.205 kg/m3 Example: Determine the static pressure
difference indicated by an 18 cm column of fluid (liquid) with a specific gravity of 0.85. ∆P = ρ g h = S.G. γ h = 0.85* 9790 N/m3 0.18 m = 1498 N/m2 = 1.5 kPa I.7 Transport Properties Certain
transport properties are important as they relate to the diffusion of momentum due to shear stresses. Specifically: µ ≡ coefficient of viscosity (dynamic viscosity) {M / L t } ν ≡ kinematic viscosity
( µ / ρ ) I-5
{L /t}
This gives rise to the definition of a Newtonian fluid. Newtonian fluid: A fluid which has a linear relationship between shear stress and velocity gradient. dU dy The linearity coefficient in the
equation is the coefficient of viscosity µ .
τ =µ
Flows constrained by solid surfaces can typically be divided into two regimes: a. Flow near a bounding surface with 1. significant velocity gradients 2. significant shear stresses This flow region is
referred to as a "boundary layer." b. Flows far from bounding surface with 1. negligible velocity gradients 2. negligible shear stresses 3. significant inertia effects This flow region is referred to
as "free stream" or "inviscid flow region." An important parameter in identifying the characteristics of these flows is the Reynolds number = Re =
ρV L µ
This physically represents the ratio of inertia forces in the flow to viscous forces. For most flows of engineering significance, both the characteristics of the flow and the important effects due to
the flow, e.g., drag, pressure drop, aerodynamic loads, etc., are dependent on this parameter.
II. Fluid Statics
From a force analysis on a triangular fluid element at rest, the following three concepts are easily developed: For a continuous, hydrostatic, shear free fluid: 1. Pressure is constant along a
horizontal plane, 2. Pressure at a point is independent of orientation, 3. Pressure change in any direction is proportional to the fluid density, local g, and vertical change in depth. These concepts
are key to the solution of problems in fluid statics, e.g. 1. Two points at the same depth in a static fluid have the same pressure. 2. The orientation of a surface has no bearing on the pressure at
a point in a static fluid. 3. Vertical depth is a key dimension in determining pressure change in a static fluid. If we were to conduct a more general force analysis on a fluid in motion, we would
then obtain the following:
∇ P = ρ{ g − a} + µ ∇ V 2
Thus the pressure change in fluid in general depends on: effects of fluid statics (ρ g), Ch. II inertial effects (ρ a), Ch. III viscous effects ( µ ∇ 2 V ) Chs VI & VII Note: For problems involving
the effects v of v both (1) fluid statics and (2) inertial effects, it is the net g − a acceleration vector that controls both the magnitude and direction of the pressure gradient.
This equation can be simplified for a fluid at rest (ie., no inertial or viscous effects) to yield ∇p = ρ g
∂p ∂x
= 0;
∂p ∂y
∂p ∂z
dp dz
= −ρ g
P2 − P1 = − ∫ ρ g d Z 1
Free surface Pressure = Pa
For liquids and incompressible fluids, this integrates to
P1 – P2 = -ρg (Z2 – Z1)
=Z 2
P Z
Z2 – Z1 is positive for Z2 above Z1.
but P2 – P1 is negative for Z2 above Z1.
We can now define a new fluid parameter useful in static fluid analysis:
γ = ρg ≡ specific weight of the fluid With this, the previous equation becomes (for an incompressible, static fluid) P2 – P1 = - γ (Z2 – Z1) The most common application of this result is that of
Consider the U-tube, multifluid manometer shown on the right. If we first label all intermediate points between A & a, we can write for the overall pressure change
PA - Pa = (PA- P1) + (P1 - P2) + (P2 - Pa ) This equation was obtained by adding and subtracting each intermediate pressure. The total pressure difference now is expressed in terms of a series of
intermediate pressure differences. Substituting the previous result for static pressure difference, we obtain PA - PB = - ρ g(ZA- Z1) – ρ g (Z1 – Z2) – ρ g (Z2 - ZB )
Again note: Z positive up and ZA > Z1 , Z1 < Z2 , Z2 < Za
In general, follow the following steps when analyzing manometry problems: 1. On manometer schematic, label points on each end of manometer and each intermediate point where there is a fluid-fluid
interface: e.g., A – 1 – 2 - B 2. Express overall manometer pressure difference in terms of appropriate intermediate pressure differences. PA - PB = (PA- P1) + (P1 – P2) + (P2 - PB ) 3. Express each
intermediate pressure difference in terms of appropriate product of specific weight * elevation change (watch signs) PA - PB = - ρ g(zA- z1) – ρ g (z1 – z2) – ρ g (z2 - zB ) 4. Substitute for known
values and solve for remaining unknowns.
When developing a solution for manometer problems, take care to: 1. Include all pressure changes 2. Use correct ∆Z and γ with each fluid 3. Use correct signs with ∆ Z. If pressure difference is
expressed as PA – P1, the elevation change should be written as ZA – Z1 4. Watch units. Manometer Example: a
Given the indicated manometer, determine the gage pressure at A. Pa = 101.3 kPa. The fluid at A is Meriam red oil no. 3.
S.G. = .83 H 0
18 cm
ρgw = 9790 N/m
10 cm 1
ρg A = S.G.*ρgw = 0.83*9790 N/m
ρg A = 8126 N/m3 ρgair = 11.8 N/m3 With the indicated points labeled on the manometer, we can write PA - Pa = (PA- P1) + (P1 – P2) + (P2 - Pa ) Substituting the manometer expression for a static
fluid, we obtain PA - Pa = - ρgA(zA- z1) – ρgw(z1 – z2) – ρga(z2 - za ) Neglect the contribution due to the air column. Substituting values, we obtain PA - Pa = - 8126 N/m3 * 0.10 m – 9790 N/m3 *
-0.18 = 949.6 N/m2 Note why: (zA- z1) = 0.10 m and (z1 – z2) = -0.18 m, & did not use Pa Review the text examples for manometry.
Hydrostatic Forces on Plane Surfaces Consider a plane surface of arbitrary shape and orientation, submerged in a static fluid as shown: If P represents the local pressure at any point on the surface
and h the depth of fluid above any point on the surface, from basic physics we can easily show that
the net hydrostatic force on a plane surface is given by (see text for development): F = ∫ PdA = Pcg A A
The basic physics says that the hydrostatic force is a distributed load equal to the integral of the local pressure force over the area. This is equivalent to the following: The hydrostatic force on
one side of a plane surface submerged in a static fluid equals the product of the fluid pressure at the centroid of the surface times the surface area in contact with the fluid. Also: Since pressure
acts normal to a surface, the direction of the resultant force will always be normal to the surface. Note: In most cases since it is the net hydrostatic force that is desired and the contribution of
atmospheric pressure Pa will act on both sides of a surface, the result of atmospheric pressure Pa will cancel and the net force is obtained by
F = ρ gh cg A F = Pcg A
Pcg is now the gage pressure at the centroid of the area in contact with the fluid. Therefore, to obtain the net hydrostatic force F on a plane surface: 1. 2. 3.
Determine depth of centroid hcg for the area in contact with the fluid Determine the (gage) pressure at the centroid Pcg Calculate F = PcgA.
The following page shows the centroid, and other geometric properties of several common areas. It is noted that care must be taken when dealing with layered fluids. The required procedure is
essentially that the force on the plane area in each layer of fluid must be determined individually for each layer using the steps listed above. We must now determine the effective point of
application of F. This is commonly called the “center of pressure - cp” of the hydrostatic force. Define an x – y coordinate system whose origin is at the centroid, c.g, of the area. The location of
the resultant force is determined by integrating the moment of the distributed fluid load on the surface about each axis and equating this to the moment of the resultant force. Therefore, for the
moment about the x axis:
F y cp = ∫ y P dA A
Applying a procedure similar to that used previously to determine the resultant force, and using the definition (see text for detailed development), for Ixx defined as the
≡ moment of inertia, or 2nd moment of area we obtain
Ycp = −
ρ gsin θ I xx Pcg A
Therefore, the resultant force will always act at a distance ycp below the centroid of the surface ( except for the special case of sin θ = 0 ).
PROPERTIES OF PLANE SECTIONS Geometry
y x
Product of Inertia Ix y
Moment of Inertia Ix x
b ⋅L
y x
0, 0
y x
3 ,
bL3 3
b L
0 ,a =
π 8
− 9 π 8
s y
a =
L x
π R2
b ( b − 2s) L
b⋅ L
4 − R 9π 4
4 R − 8 9π 1
π R2 4
b 1
y x
a =
h( b + 2b 1 )
+ 4bb 1 + b 1
36 ( b + b1 )
3( b + b 1 )
( b + b 1)
Fluid Specific Weight N /m
N /m 3
Water Ethyl
1bf /ft 3
1bf/ft 3
h 2
Proceeding in a similar manner for the x location, and defining Ixy = product of inertia, we obtain X cp = −
ρ g sinθ I xy Pcg A
where Xcp can be either positive or negative since Ixy can be either positive or negative. Note: For areas with a vertical plane of symmetry (e.g., squares, circles, isosceles triangles, etc.)
through the centroid, i.e. the ( y - axis), the center of pressure is located directly below the centroid along the plane of symmetry, i.e., Xcp = 0. Key Points: The values Xcp and Ycp are both
measured with respect to the centroid of the area in contact with the fluid. Xcp and Ycp are both measured in the plane of the area; i.e., o Ycp is not necessarily a vertical dimension, unless θ = 90
Special Case: For most problems where (1) we have a single, homogeneous fluid ( i.e., not applicable to layers of multiple fluids) and (2) the surface pressure is atmospheric, the fluid specific
weight γ cancels in the equation for Ycp and Xcp and we have the following simplified expressions: F = ρ g h cg A
Ycp = −
I xx sinθ
X cp = −
h cg A
I xy sin θ h cg A
However, for problems where we have either (1) multiple fluid layers, or (2) a container with surface pressurization > Patm , these simplifications do not occur and the original, basic expressions
for F , Ycp , and Xcp must be used; i.e., take care to use the approximate expressions only for cases where they apply. The basic equations always work.
Summary: 1. The resultant force is determined from the product of the pressure at the centroid of the surface times the area in contact with the fluid 2. The centroid is used to determine the
magnitude of the force. This is not the location of the resultant force 3. The location of the resultant force will be at the center of pressure which will be at a location Ycp below the centroid and
Xcp as specified previously 4. Xcp = 0 for areas with a vertical plane of symmetry through the c.g. Example 2.5 Seawater 64 lbf/ft3
Given: Gate, 5 ft wide Hinged at B Holds seawater as shown Find: a. Net hydrostatic force on gate b. Horizontal force at wall - A c. Hinge reactions - B
9’ h c.g.
c. g.
θ 8’
a. By geometry: θ = tan-1 (6/8) = 36.87
Since plate is rectangular, hcg = 9 ft + 3ft = 12 ft
Neglect Patm A = 10 x 5 = 50 ft2
Pcg = γ hcg = 64 lbf/ft3 * 12 ft = 768 lbf/ft2 ∴
Fp = Pcg A = 768 lbf/ft2 * 50 ft2 = 38,400 lbf
Horizontal Reaction at A Must first find the location, c.p., for Fp
ycp = − ρ gsinθ
Ixx I sin θ = − xx Pcg A hcg A
Ixx = bh /12 Ixx = 5 * 103/12 = 417 ft4
For a rectangular wall:
• c.g. • c.p.
6 ft
8 ft
Note: The relevant area is a rectangle, not a triangle.
Note: Do not overlook the hinged reactions at B. y cp = −
417 ft 4 0.6 417 ft 4 0.6 = − 0.417 ft y = − = − 0.417 ft cp 12 ft 50 ft 2 12 ft 50 ft 2
below c.g.
xcp = 0 due to symmetry
∑M = 0
(5 − 0.417) ⋅38,400 − 6 P = 0 Bz
P = 29,330 lbf
• c.g. • c.p.
8 ft
6 ft
∑ Fx = 0, Bx + Fsinθ − P = 0 Bx + 38,400*0.6 - 29,330 = 0 Bx = 6290 lbf
∑ Fz = 0, Bz − Fcosθ = 0 Bz = 38,400 * 0.8 = 30,720 lbf
Note: Show the direction of all forces in final answers.
Summary: To find net hydrostatic force on a plane surface: 1. 2. 3. 4. 5.
Find area in contact with fluid. Locate centroid of that area. Find hydrostatic pressure Pcg at centroid, typically = γ hcg ( generally neglect Patm ). Find force F = Pcg A. Location will not be at
c.g., but at a distance ycp below centroid. ycp is in the plane of the area.
Review all text examples for forces on plane surfaces.
Forces on Curved Surfaces Since this class of surface is curved, the direction of the force is different at each location on the surface. Therefore, we will evaluate separate x and y components of
net hydrostatic force. Consider curved surface, a-b. Force balances in x & y directions yields Fh = FH Fv = Wair + W1 + W2 From this force balance, the basic rules for determining the horizontal and
vertical component of forces on a curved surface in a static fluid can be summarized as follows: Horizontal Component, Fh The horizontal component of force on a curved surface equals the force on the
plane area formed by the projection of the curved surface onto a vertical plane normal to the component.
hcg a’
The horizontal force will act through the c.p. (not the centroid) of the projected area.
ycp Fh b’
a Projected vertical plane Curved surface cp b
Therefore, to determine the horizontal component of force on a curved surface in a hydrostatic fluid:
1. Project the curved surface into the appropriate vertical plane. 2. Perform all further calculations on the vertical plane. 3. Determine the location of the centroid - c.g. of the vertical plane.
4. Determine the depth of the centroid - hcg of the vertical plane. 5. Determine the pressure - Pcg = g hcg at the centroid of the vertical plane. 6. Calculate Fh = Pcg A, where A is the area of the
projection of the curved surface into the vertical plane, ie., the area of the vertical plane. 7. The location of Fh is through the center of pressure of the vertical plane , not the centroid. Get
the picture?
All elements of the analysis are performed with the vertical plane. The original curved surface is important only as it is used to define the projected vertical plane.
Vertical Component - Fv
The vertical component of force on a curved surface equals the weight of the effective column of fluid necessary to cause the pressure on the surface. The use of the words effective column of fluid
is important in that there may not always actually be fluid directly above the surface. ( See graphic that follows.) This effective column of fluid is specified by identifying the column of fluid
that would be required to cause the pressure at each location on the surface.
Thus to identify the effective volume - Veff: 1. 2. 3. 4.
Identify the curved surface in contact with the fluid. Identify the pressure at each point on the curved surface. Identify the height of fluid required to develop the pressure. These collective
heights combine to form Veff.
a P
Fluid above the surface
No fluid actually above surface
These two examples show two typical cases where this concept is used to determine Veff. The vertical force acts vertically through the centroid (center of mass) of the effective column of fluid. The
vertical direction will be the direction of the vertical components of the pressure forces. Therefore, to determine the vertical component of force on a curved surface in a hydrostatic fluid: 1.
Identify the effective column of fluid necessary to cause the fluid pressure on the surface. 2. Determine the volume of the effective column of fluid. 3. Calculate the weight of the effective column
of fluid - Fv = ρgVeff. 4. The location of Fv is through the centroid of Veff.
Finding the Location of the Centroid A second problem associated with the topic of curved surfaces is that of finding the location of the centroid of Veff. Recall: Centroid = the location where the
first moment of a point area, volume, or mass equals the first moment of the distributed area, volume, or mass, e.g.
xcgV1 = ∫ x dV V1
This principle can also be used to determine the location of the centroid of complex geometries. For example: If Veff = V1 + V2
V1 a
xcgVeff = x1V1 + x2V2 b
or VT = V1 + Veff
Veff b
xTVT = x1V1 + xcgVeff
Note: In the figures shown above, each of the x values would be specified relative to a vertical axis through b since the cg of the quarter circle is most easily specified relative to this axis.
A •
Gate AB holds back 15 ft of water. Neglecting the weight of the gate, determine the magnitude (per unit width) and location of the hydrostatic forces on the gate and the resisting moment about B. a.
Horizontal component
Width - W
15 ft F
B •
γ = ρg = 62.4 lbf/ft3
Rule: Project the curved surface into the vertical plane. Locate the centroid of the projected area. Find the pressure at the centroid of the vertical projection. F = Pcg Ap
a h
A •
Note: All calculations are done with the projected area. The curved surface is not used at all in the analysis.
The curved surface projects onto plane a - b and results in a rectangle, (not a quarter circle) 15 ft x W. For this rectangle: hcg = 7.5,
Pcg = γhcg = 62.4 lbf/ft3 * 7.5 ft = 468 lbf/ft2
Fh = Pcg A = 468 lbf/ft2 * 15 ft*W= 7020 W lbf Location: Ixx = bh3/12 = W * 153 /12 = 281.25 W ft4 ycp = −
Ixx sin θ 281.25W ft 4 sin 90o =− = − 2.5 ft hcg A 7.5 ft 15W ft2
The location is 2.5 ft below the c.g. or 10 ft below the surface, 5 ft above the bottom.
B •
b. Vertical force:
Rule: Fv equals the weight of the effective column of fluid above the curved surface.
C •
F v
Q: What is the effective volume of fluid above the surface? What volume of fluid would result in the actual pressure distribution on the curved surface? Vol = A - B - C Vrec = Vqc + VABC,
VABC = Vrec - Vqc
VABC = Veff = 152 W - π 152/4*W = 48.29 W ft3 Fv = ρg Veff = 62.4 lbf/ft3 * 48.29 ft3 = 3013 lbf Note: Fv is directed upward even though the effective volume is above the surface. c. What is the
location? A
C •
Rule: Fv will act through the centroid of the “effective volume causing the force. b
We need the centroid of volume A-B-C.
F v •
How do we obtain this centroid?
Use the concept which is the basis of the centroid, the “first moment of an area.”
Since: Arec = Aqc + AABC
Mrec = Mqc + MABC
MABC = Mrec - Mqc
Note: We are taking moments about the left side of the figure, ie., point b. WHY? (The c.g. of the quarter circle is known to be 4 R/ 3 π w.r.t. b.) xcg A = xrec Arec - xqc Aqc xcg {152 - π*152/4} =
7.5*152 - {4*15/3/π}* π*152/4 xcg = 11.65 ft { distance to rt. of b to centroid } Q: Do we need a y location? Why? d. Calculate the moment about B needed for equilibrium.
clockwise positive.
MB + 5 Fh + (15− xv )Fv = 0 MB + 5 × 7020 W + (15 − 11.65 ) 3013 W = 0 MB + 5 × 7020 W + (15 − 11.65 ) 3013 W = 0 Pa ≠ ρ g y
MB + 35,100 W +10,093.6 W = 0
MB = − 45,194W ft −lbf
Why negative?
The hydrostatic forces will tend to roll the surface clockwise relative to B, thus a resisting moment that is counterclockwise is needed for static equilibrium. Always review your answer (all
aspects: magnitude, direction, units, etc.) to determine if it makes sense relative to physically what you understand about the problem. Begin to think like an engineer. II-18
Buoyancy An important extension of the procedure for vertical forces on curved surfaces is that of the concept of buoyancy. The basic principle was discovered by Archimedes.
It can be easily shown that (see text for detailed development) the buoyant force Fb is given by:
Fb = ρ g Vb
where Vb is the volume of the fluid displaced by the submerged body and ρ g is the specific weight of the fluid displaced.
Thus, the buoyant force equals the weight of the fluid displaced, which is equal to the product of the specific weight times the volume of fluid displaced. The location of the buoyant force is:
Through a vertical line of action, directed upward, which acts through the centroid of the volume of fluid displaced. Review all text examples and material on buoyancy.
Pressure distribution in rigid body motion All of the problems considered to this point were for static fluids. We will now consider an extension of our static fluid analysis to the case of rigid
body motion, where the entire fluid mass moves and accelerates uniformly (as a rigid body). The container of fluid shown below is accelerated uniformly up and to the right as shown.
From a previous analysis, the general equation governing fluid motion is ∇ P = ρ( g − a ) + µ ∇ V 2
For rigid body motion, there is no velocity gradient in the fluid, therefore
µ∇2 V = 0 The simplified equation can now be written as ∇ P = ρ( g − a ) = ρ G
G = g − a ≡ the net acceleration vector acting on the fluid.
This result is similar to the equation for the variation of pressure in a hydrostatic fluid. However, in the case of rigid body motion: * ∇ P = f {fluid density & the net acceleration vector- G = g −
a } * ∇ P acts in the vector direction of G = g − a *
Lines of constant pressure are perpendicular to G . The new orientation of the free surface will also be perpendicular to G .
The equations governing the analysis for this class of problems are most easily developed from an acceleration diagram.
Acceleration diagram:
Free surface
For the indicated geometry:
ax ax θ = tan−1 θ = tan−1 g + az g + az
dP ds
= ρG
ax s
where G = {a x + (g + a z ) 2
g G
P2 − P1 = ρ G(s 2 − s 1 )
Note: P2 − P1 ≠ ρ g(z 2 − z 1 )
Note: s is the depth to a given point perpendicular to the free surface or its extension. s is aligned with G .
and s2 – s1 is not a vertical dimension
In analyzing typical problems with rigid body motion: 1. Draw the acceleration diagram taking care to correctly indicate –a, g, and θ, the inclination angle of the free surface. 2. Using the
previously developed equations, solve for G and θ. 3. If required, use geometry to determine s2 – s1 (the perpendicular distance from the free surface to a given point) and then the pressure at that
point relative to the surface using P2 – P1 = ρ G (s2 – s1) . Key Point: Do not use ρg to calculate P2 – P1, use ρ G. ax
Example 2.12 Given: A coffee mug, 6 cm x 6 cm square, 10 cm deep, contains 7 cm of coffee. Mug is accelerated to the right with ax = 7 m/s2 . Assuming rigid body motion. ρc = 1010 kg/m3,
6 cm ∆z
10 cm
Determine: a. Will the coffee spill? 7 cm
b. Pg at “a & b”. c. Fnet on left wall. a. First draw schematic showing original orientation and final orientation of the free surface. ρc = 1010 kg/m3
ax = 7m/s2
g = 9.8907 m/s2
az = 0
Have a new free surface angle θ where -a
θ = tan−1 ax
g + az θ
7 = 35.5° θ = tan 9.807 −1
∆z = 3 tan 35.5 = 2.14 cm
hmax = 7 + 2.14 = 9.14 cm < 10 cm ∴ Will not spill. ax
b. Pressure at “ a & b.”
6 cm
Pa = ρ G ∆ sa 2
2 .5
2 .5
G = {a x + g } = { 7 + 9.807 } G = 12.05 m/s2
10 cm
∆ sa = {7 + z} cos θ ∆ sa = 9.14 cm cos 35.5 = 7.44 cm 3
7 cm
∆s a
Pa = 1010 kg/m *12.05m/s *0.0744 m Pa = 906 (kg m/s2)/m2 = 906 Pa
Note: Pa ≠ ρ g y G ≠ g Q: How would you find the pressure at b, Pb? c. What is the force on the left wall?
We have a plane surface, what is the rule?
6 cm ∆z
Find cg, Pcg, F = Pcg. A Vertical depth to cg is:
θ θ ∆ s cg
zcg = 9.14/2 = 4.57 cm
10 cm
cg •
∆scg = 4.57 cos 35.5 = 3.72 cm
7 cm
Pcg = ρ G ∆scg Pcg = 1010 kg/m3*12.05 m/s2* 0.0372 m Pcg = 452.7 N/m2
F = Pcg A = 452.7 N/m2*0.0914*0.06m2 F = 2.48 N ←
What is the direction? Horizontal, perpendicular to the wall; i.e., Pressure always acts normal to a surface. Q: How would you find the force on the right wall?
III. Control Volume Relations for Fluid Analysis From consideration of hydrostatics, we now move to problems involving fluid flow with the addition of effects due to fluid motion, e.g. inertia and
convective mass, momentum, and energy terms. We will present the analysis based on a control volume (not differential element) formulation, e.g. similar to that used in thermodynamics for the first
Basic Conservation Laws: Each of the following basic conservation laws is presented in its most fundamental, fixed mass form. We will subsequently develop an equivalent expression for each law that
includes the effects of the flow of mass, momentum, and energy (as appropriate) across a control volume boundary. These transformed equations will be the basis for the control volume analyses
developed in this chapter. Conservation of Mass: Defining m as the mass of a fixed mass system, the mass for a control volume V is given by
m sys =
∫ ρ dV
The basic equation for conservation of mass is then expressed as
dm =0 dt sys
The time rate of change of mass for the control volume is zero since at this point we are still working with a fixed mass system.
Linear Momentum: Defining P sys as the linear momentum of a fixed mass, the linear momentum of a fixed mass control volume is given by:
P sys = mV =
∫ V ρ dV
where V is the local fluid velocity and dV is a differential volume element in the control volume. The basic linear momentum equation is then written as
d (mV) dP ∑ F = dt = dt sys
Moment of Momentum: Defining H as the moment of momentum for a fixed mass, the moment of momentum for a fixed mass control volume is given by
H sys =
∫ r × V ρ dV
where r is the moment arm from an inertial coordinate system to the differential control volume of interest. The basic equation is then written as
dH sys
∑ M sys = ∑ r × F = dt Energy:
Defining E sys as the total energy of an element of fixed mass, the energy of a fixed mass control volume is given by
E sys =
∫ e ρ dV
where e is the total energy per unit mass ( includes kinetic, potential, and internal energy ) of the differential control volume element of interest. The basic equation is then written as dE Q& − W
& =
d t sys
(Note: written on a rate basis)
It is again noted that each of the conservation relations as previously written applies only to fixed, constant mass systems. However, since most fluid problems of importance are for open systems, we
must transform each of these relations to an equivalent expression for a control volume which includes the effect of mass entering and/or leaving the system. This is accomplished with the Reynolds
transport theorem. Reynolds Transport Theorem We define a general, extensive property ( an extensive property depends on the size or extent of the system) B sys where
B sys =
∫ β ρd V
Bsys could be total mass, total energy, total momentum, etc., of a system. and B sys per unit mass is defined as Thus,
β is the intensive equivalent of B sys
dB dm
Applying a general control volume formulation to the time rate of change of B sys , we obtain the following (see text for detailed development):
dB = d t sys
∂ β ρ dV ∂ t cv
ρi Vi d Ai
Rate of change of B in c.v.
System rate of change of B
ρ e Ve d Ae
Rate of B leaving c.v.
Rate of B entering c.v.
transient term
convective terms
where B is any conserved quantity, e.g. mass, linear momentum, moment of momentum, or energy. We will now apply this theorem to each of the basic conservation equations to develop their equivalent
open system, control volume forms. Conservation of mass For conservation of mass, we have that B = m
β =1
From the previous statement of conservation of mass and these definitions, Reynolds transport theorem becomes ∂ ρ dV ∂ t cv
ρe Ve d Ae
ρ i Vi d Ai
= 0
or ∂ ρ dV ∂ t cv
↓ Rate of change of mass in c.v.,
= 0 for steady-state
ρe Ve d Ae
ρ i Vi d Ai
↓ Rate of mass leaving c.v.,
↓ Rate of mass entering c.v.,
m& e
m& i
= 0
This can be simplified to dm &e − ∑m &i = 0 + ∑m d t cv
Note that the exit and inlet velocities Ve and Vi are the local components of fluid velocities at the exit and inlet boundaries relative to an observer standing on the boundary. Therefore, if the
boundary is moving, the velocity is measured relative to the boundary motion. The location and orientation of a coordinate system for the problem are not considered in determining these velocities.
Also, the result of Ve ⋅ dA e and Vi ⋅ dA i is the product of the normal velocity component times the flow area at the exit or inlet, e.g.
Ve,n dA e
Vi,n dA i
Special Case: For incompressible flow with a uniform velocity over the flow area, the previous integral expressions simplify to: & = m
∫ ρ V d A = ρ AV
Conservation of Mass Example Water at a velocity of 7 m/s exits a stationary nozzle with D = 4 cm and is o directed toward a turning vane with θ = 40 , Assume steady-state. Determine: a. Velocity and
flow rate entering the c.v. b. Velocity and flow rate leaving the c.v.
&1 a. Find V1 and m
Recall that the mass flow velocity is the normal component of velocity measured relative to the inlet or exit area. Thus, relative to the nozzle, V(nozzle) = 7 m/s and since there is no relative
motion of point 1 relative to the nozzle, we also have V1 = 7 m/s ans. From the previous equation: & = m
∫ ρ V d A = ρ AV
= 998 kg/m3*7 m/s*π*0.042/4
& 1 = 8.78 kg/s ans. m &2 b. Find V2 and m
Determine the flow rate first. Since the flow is steady state and no mass accumulates on the vane: & 1= m &2 , m & 2 = 8.78 kg/s ans. m
& 2 = 8.78 kg/s = ρ A V)2 m
Since ρ and A are constant, V2 = 7 m/s ans. Key Point: For steady flow of a constant area, incompressible stream, the flow velocity and total mass flow are the same at the inlet and exit, even though
the direction changes. or alternatively: Rubber Hose Concept: For steady flow of an incompressible fluid, the flow stream can be considered as a rubber hose and if it enters a c.v. at a velocity of
V, it exits at a velocity V, even if it is redirected.
Problem Extension: Let the turning vane (and c.v.) now move to the right at a steady velocity of 2 m/s (other values remain the same); perform the same calculations. Therefore: Given: Uc = 2 m/s
VJ = 7 m/s
For an observer standing at the c.v. inlet (point 1) V1 = VJ – Uc = 7 – 2 = 5 m/s & 1 = ρ1 V1 A1 = 998 kg/m3*5 m/s*π*0.042/4 = 6.271 kg/s m
Note: The inlet velocity used to specify the mass flow rate is again measured relative to the inlet boundary but now is given by VJ – Uc . Exit: & 1= m & 2 = 6.271 kg/s Again, since ρ and A are
constant, V2 = 5 m/s. m
Again, the exit flow is most easily specified by conservation of mass concepts. Note: The coordinate system could either have been placed on the moving cart or have been left off the cart with no
change in the results. Key Point: The location of the coordinate system does not affect the calculation of mass flow rate which is calculated relative to the flow boundary. It could have been placed
at Georgia Tech with no change in the results. Review material and work examples in the text on conservation of mass.
Linear Momentum For linear momentum, we have that
B = P = m V and β = V From the previous statement of linear momentum and these definitions, Reynolds transport theorem becomes
∑F =
d ( mV ) ∂ V ρ dV = ∂ t cv∫ d t sys
∂ V ρ dV ∂ t cv∫
= the ∑ of the external forces acting on the c.v.
= the rate of change of momentum in the c.v.
= body + point + distributed, e.g. (pressure) forces
= 0 for steady-state
∫V ρ
Ve ⋅ d Ae
∫ V d m&
↓ = the rate of momentum leaving the c.v.
∫V ρ V ⋅d A i
∫ V d m&
↓ = the rate of momentum entering the c.v.
and where V is the vector momentum velocity relative to an inertial reference frame. Key Point: Thus, the momentum velocity has magnitude and direction and is measured relative to the reference frame
(coordinate system) being used for the & i and m & e are scalars, as noted problem. The velocities in the mass flow terms m previously, and are measured relative to the inlet or exit boundary. Always
clearly define a coordinate system and use it to specify the value of all inlet and exit momentum velocities when working linear momentum problems. III-8
For the 'x' direction, the previous equation becomes
∂ V ρ dV ∂ t cv∫ x
x ,e
&e dm
x ,i
&i dm
Note that the above equation is also valid for control volumes moving at constant velocity with the coordinate system placed on the moving control volume. This is because an inertial coordinate
system is a nonaccelerating coordinate system which is still valid for a c.s. moving at constant velocity. Example: A water jet 4 cm in diameter with a velocity of 7 m/s is directed to a stationary
turning o vane with θ = 40 . Determine the force F necessary to hold the vane stationary.
Governing equation:
∂ V ρ dV ∂ t cv∫ x
x ,e
&e dm
x ,i
&i dm
Since the flow is steady and the c.v. is stationary, the time rate of change of momentum within the c.v. is zero. Also with uniform velocity at each inlet and exit and a constant flow rate, the
momentum equation becomes & e Ve − m & i Vi − Fb = m
Note that the braking force, Fb, is written as negative since it is assumed to be in the negative x direction relative to positive x for the coordinate system.
From the previous example for conservation of mass, we can again write & = m
∫ ρ V d A = ρ AV
= 998 kg/m3*7 m/s*π*0.042/4
& 1 = 8.78 kg/s and V1 = 7 m/s m
and for the exit: & 2 = 8.78 kg/s and V2 = 7 m/s inclined 40û above the horizontal. m
Substituting in the momentum equation, we obtain o
-Fb = 8.78 kg/s * 7 m/s *cos 40 - 8.78 kg/s * 7 m/s and -Fb = - 14.4 kg m/s
or Fb = 14.4 N ←
Note: Since our final answer is positive, our original assumption of the applied force being to the left was correct. Had we assumed that the applied force was to the right, our answer would be
negative, meaning that the direction of the applied force is opposite to what was assumed. Modify Problem: Now consider the same problem but with the cart moving to the right with a velocity Uc = 2 m
/s. Again solve for the value of braking force Fb necessary to maintain a constant cart velocity of 2 m/s. Note: The coordinate system for the problem has now been placed on the moving cart.
The transient term in the momentum equation is still zero. With the coordinate system on the cart, the momentum of the cart relative to the coordinate system is still zero. The fluid stream is still
moving relative to the coordinate system, however, the flow is steady with constant velocity and the time rate of change of momentum of the fluid stream is therefore also zero. Thus The momentum
equation has the same form as for the previous problem (However the value of individual terms will be different.) & e Ve − m & i Vi − Fb = m & 1 = ρ1 V1 A1 = 998 kg/m3*5 m/s*π*0.042/4 = 6.271 kg/s =
m &2 m
Now we must determine the momentum velocity at the inlet and exit. With the coordinate system on the moving control volume, the values of momentum velocity are V1 = VJ – Uc = 7 – 2 = 5 m/s
and V2 = 5 m/s inclined 40
The momentum equation ( x - direction ) now becomes o
-Fb = 6.271 kg/s * 5 m/s *cos 40 - 6.271 kg/s * 5 m/s and -Fb = - 7.34 kg m/s
or Fb = 7.34 N ←
Question: What would happen to the braking force Fb if the turning o o angle had been > 90 , e.g., 130 ? Can you explain based on your understanding of change in momentum for the fluid stream? Review
and work examples for linear momentum with fixed and nonaccelerating (moving at constant velocity) control volumes. Accelerating Control Volume The previous formulation applies only to an inertial
coordinate system, i.e., fixed or moving at constant velocity (non-accelerating). III-11
We will now consider problems with accelerating control volumes. For these problems we will again place the coordinate system on the accelerating control volume, thus making it a non-inertial
coordinate system. For coordinate systems placed on an accelerating control volume, we must account for the acceleration of the c.s. by correcting the momentum equation for this acceleration. This is
accomplished by including the term as shown below:
∑F − ∫ a
d mcv
∂ V ρ dV ∂ t cv∫
∫ V d m&
∫ V d m&
integral sum of the local c.v. (c.s.) acceleration * the c.v. mass The added term accounts for the acceleration of the control volume and allows the problem to be worked with the coordinate system
placed on the accelerating c.v. Note: Thus, all vector (momentum) velocities are then measured relative to an observer (coordinate system) on the accelerating control volume. For example, the
velocity of a rocket as seen by an observer (c.s.) standing on the rocket is zero and the time rate of change of momentum is zero in this reference frame even if the rocket is accelerating.
Accelerating Control Volume Example o
A turning vane with θ = 60 accelerates from rest due to a jet of water (VJ = 35 m/s, AJ = 0.003 m2 ). Assuming the mass of the cart mc, is 75 kg and neglecting drag and friction effects, find: a.
Cart acceleration at t = 0. b. Uc as a f(t)
Starting with the general equation shown above, we can make the following assumptions: 1. ∑ Fx = 0, no friction or body forces. 2. The jet has uniform velocity and constant properties. 3. The entire
cart accelerates uniformly over the entire control volume. 4. Neglect the relative momentum change of the jet stream that is within the control volume. With these assumptions, the governing equation
simplifies to & e Vx ,e − m & i Vx ,i − ac mc = m
We thus have terms that account for the acceleration of the control volume, for the exit momentum, and for the inlet momentum (both of which change with time.) Mass flow: As with the previous example
for a moving control volume, the mass flow terms are given by: &i = m &e = m & = ρ AJ (VJ – Uc) m
Note that since the cart accelerates, Uc is not a constant but rather changes with time. Momentum velocities: Ux,i= VJ - Uc
Ux,e = (VJ - Uc ) cos θ
Substituting, we now obtain 2
- ac mc = ρ AJ (VJ – Uc) cos θ - ρ AJ (VJ – Uc) Solving for the cart acceleration, we obtain
ρ AJ (1 − cos θ ) (VJ − Uc )
ac =
mc III-13
Substituting for the given values at t = 0, i.e., Uc = 0, we obtain 2
ac (t = 0) = 24.45 m/s = 2.49 g’s Note: The acceleration at any other time can be obtained once the cart velocity Uc at that time is known. To determine the equation for cart velocity as a function
of time, the equation for the acceleration must be written in terms of Uc (t) and integrated. dUc ρ AJ (1 − cosθ )(VJ − Uc ) = dt mc
Separating variables, we obtain Uc (t )
(VJ − U c )2
ρ AJ (1 − cosθ )
Completing the integration and rearranging the terms, we obtain a final expression of the form
Uc V bt = J VJ 1+ VJ b t
ρ AJ (1 − cosθ )
Substituting for known values, we obtain VJ b = 0.699 s Thus the final equation for Uc is give by
0.699 t Uc = VJ 1+0.699t
mc -1
The final results are now given as shown below: t (s) 0 2 5 10 15 ∞
Uc/VJ 0.0 0.583 0.757 0.875 0.912 1.0
Uc (m/s) 0.0 20.0 27.2 30.6 31.9 35
ac (m/s2) 24.45 4.49 1.22 0.39 0.192 0.0
Uc vs t 35 30 25 20 15 10 5 0 0
Note that the limiting case occurs when the cart velocity reaches the jet velocity. At this point, the jet can impart no more momentum to the cart, the acceleration is now zero, and the terminal
velocity has been reached. Review the text example on accelerating control volumes.
Moment of Momentum (angular momentum) For moment of momentum we have that
B = H = r × (m V ) and
β = r ×V
From the previous equation for moment of momentum and these definitions, Reynolds transport theorem becomes
↓ = the ∑ of all external moments acting on the c.v.
∂ r × V ρ dV ∂ t cv∫
∫ r × V d m&
↓ = the rate of change of moment of momentum in the c.v. = 0 for steady state
∫ r × V d m&
= the rate of moment of momentum leaving the c.v.
= the rate of moment of momentum entering the c.v.
For the special case of steady-state, steady-flow and uniform properties at any exit or inlet, the equation becomes
∑ M = ∑ m&
& i r ×Vi r ×Ve − ∑ m
For moment of momentum problems, we must be careful to correctly evaluate the moment of all applied forces and all inlet and exit momentum flows, with particular attention to the signs. Moment of
Momentum Example: A small lawn sprinkler operates as indicated. The inlet flow rate is 9.98 kg/min with an inlet pressure of 30 kPa. The two exit jets direct o flow at an angle of 40 above the
160 mm
For these conditions, determine the following: a. Jet velocity relative to the nozzle. b. Torque required to hold the arm stationary. c. Friction torque if the arm is rotating at 35 rpm. d. Maximum
rotational speed if we neglect friction.
D J = 5 mm
a. R = 160 mm, DJ = 5 mm, Therefore, for each of the two jets: 3
QJ = 0.5* 9.98 kg/min/998 kg/m = 0.005 m /min 2
AJ = π π 0.0025 = 1.963*10 m 3
VJ = 0.005 m /min / 1.963*10 m
/60 s/min
VJ = 4.24 m/s relative to the nozzle exit ans. b. Torque required to hold the arm stationary. First develop the governing equations and analysis for the general case of the arm rotating.
VJ cos θ
With the coordinate system at the center of rotation of the arm, a general velocity diagram for the case when the arm is rotating is shown in the adjacent schematic.
R + ω
Taking the moment about the center of rotation, the moment of the inlet flow is zero since the moment arm is zero for the inlet flow. The basic equation then becomes & e R (VJ cos α − R ω ) T0 = 2 m
Note that the net momentum velocity is the difference between the tangential component of the jet exit velocity and the rotational speed of the arm. Also note that the direction of positive moments
was taken as the same as for VJ and opposite to the direction of rotation. For a stationary arm R ω = 0. We thus obtain for the stationary torque III-17
To = 2 ρ QJ R VJ cos α To = 2 * 998
kg m3 1min m .005 m * 4.24 cos 4.160o 3 min 60 s m
To = 0.0864 N m clockwise.
A resisting torque of 0.0864 N m must be applied in the clockwise direction to keep the arm from rotating in the counterclockwise direction. c. At ω = 30 rpm, calculate the friction torque Tf ω = 30
To = 2 * 998
rev rad 1min rad 2π =π min rev 60 s
kg m3 1min m rad 0.005 0.16m 4.24 cos 40o − .16m * π 3 min 60 s s m
ans. Note; The resisting torque decreases as the speed increases. d. Find the maximum rotational speed. The maximum rotational speed occurs when the opposing torque is zero and all the moment of
momentum goes to the angular rotation. For this case, VJ cos θ – Rω = 0 V cosθ = ω= J R
rad = 193.8 rpm 4 m / s • cos 40 rad s = 20.3 = 193.8 rpm s 0.16 m
Review material and examples on moment of momentum.
Energy Equation (Extended Bernoulli Equation) For energy, we have that
B = E = ∫eρd V
β = e = u + V2 + g z
From the previous statement of conservation of energy and these definitions, Reynolds transport theorem becomes: dE ∂ Q& − W& = e ρ dV + = d t sys ∂ t cv∫
∫A ee ρe Ve ⋅ d Ae − A∫ ei ρi Vi ⋅ d Ai e
After extensive algebra and simplification (see text for detailed development), we obtain: P1 − P2 ρg
V2 − V1 2g
Pressure drop from 1 – 2, in the flow direction
Pressure drop due to acceleration of the fluid
Z2 − Z1
↓ Pressure drop due to elevation change
+ hf,1−2
↓ Pressure drop due to frictional head loss
↓ Pressure drop due to mechanical work on fluid
Note: this formulation must be written in the flow direction from 1 - 2 to be consistent with the sign of the mechanical work term and so that hf,1-2 is always a positive term. Also note the
following: ❑ ❑
The points 1 and 2 must be specific points along the flow path Each term has units of linear dimension, e.g., ft or meters, and z2 – z1 is positive for z2 above z1 The term hf,1-2 is always positive
when written in the flow direction and for internal, pipe flow includes pipe or duct friction losses and fitting or piping component (valves, elbows, etc.) losses,
The term hp is positive for pumps and fans ( i.e., pumps increase the pressure in the flow direction) and negative for turbines (turbines decrease the pressure in the flow direction)
For pumps:
hp =
ws g
where ws = the useful work per unit mass to the fluid
w s = g hp
& s = ρ Q g hp W& f = mw
W& f = the useful power delivered to the fluid
W& W& p = f
Water flows at 30 ft/s through a 1000 ft length of 2 in diameter pipe. The inlet pressure is 250 psig and the exit is 100 ft higher than the inlet.
Assuming that the frictional loss is 2 given by 18 V /2g, Determine the exit pressure.
is the pump efficiency
100 ft 1 250 psig
Given: V1 = V2 = 30 ft/s, L = 1000 ft, Z2 – Z1 = 100 ft, P1 = 250 psig Also, since there is no mechanical work in the process, the energy equation simplifies to
P1 − P2 = Z2 − Z1 ρg
+ hf 2
P1 − P2 30 ft / s = 100 ft + 18 ρg 64.4 ft / s 2
= 351.8 ft
P1 – P2 = 62.4 lbf/ft 351.8 ft = 21,949 psf = 152.4 psi P2 = 250 – 152.4 = 97.6 psig ans. Problem Extension A pump driven by an electric motor is now added to the system. The motor delivers 10.5 hp.
The flow rate and inlet pressure remain constant and the pump efficiency is 71.4 %, determine the new exit pressure. 2
Q = AV = π π (1/12) ft * 30 ft/s = 0.6545 ft /s Wf = ηp Wp= ρ Q g hp hp =
0.714 * 10.5 hp * 550ft − lbf / s / hp 62.4 lbm / ft 3 * 0.6545ft 3 / s
= 101ft
The pump adds a head increase equal to 101 ft to the system and the exit pressure should increase. Substituting in the energy equation, we obtain 2
P1 − P2 30 ft / s = 100 ft + 18 ρg 64.4 ft / s 2
− 101 ft = 250.8 ft
P1 – P2 = 62.4 lbf/ft 250.8 ft = 15,650 psf = 108.7 psi P2 = 250 – 108.7 = 141.3 psig ans. Review examples for the use of the energy equation
Ch. IV Differential Relations for a Fluid Particle This chapter presents the development and application of the basic differential equations of fluid motion. Simplifications in the general equations
and common boundary conditions are presented that allow exact solutions to be obtained. Two of the most common simplifications are 1). steady flow and 2). incompressible flow. The Acceleration Field
of a Fluid A general expression of the flow field velocity vector is given by:
V (r ,t) = iˆ u(x, y, z, t ) + ˆj v(x, y, z, t ) + kˆ w(x, y, z, t ) One of two reference frames can be used to specify the flow field characteristics: eulerian – the coordinates are fixed and we
observe the flow field characteristics as it passes by the fixed coordinates. lagrangian - the coordinates move through the flow field following individual particles in the flow. Since the primary
equation used in specifying the flow field velocity is based on Newton’s second law, the acceleration vector is an important solution parameter. In cartesian coordinates, this is expressed as
∂V ∂V ∂ V dV ∂ V ∂ V = + u +v +w + (V ⋅ ∇ )V = ∂t ∂ x ∂y ∂ z ∂t dt total
The acceleration vector is expressed in terms of three types of derivatives: Total acceleration = total derivative of velocity vector = local derivative + convective derivative of velocity vector
IV - 1
Likewise, the total derivative (also referred to as the substantial derivative ) of other variables can be expressed in a similar form, e.g.,
dP ∂P ∂P ∂P ∂ P ∂ P = + u +v +w + (V ⋅ ∇ )P = dt ∂ t ∂ x ∂y ∂ z ∂t Example 4.1 Given the eulerian velocity-vector field
V = 3 t iˆ + x z ˆj + t y 2 kˆ find the acceleration of the particle. For the given velocity vector, the individual components are u = 3t
v= xz
w = ty
Evaluating the individual components, we obtain
∂V 2 = 3i +y k ∂t ∂V = zj ∂x
∂V = 2tyk ∂y
∂V = xj ∂z
Substituting, we obtain
dV 2 2 = ( 3 i + y k) + (3 t) (z j) + (x z) (2 t y k) + (t y ) (x j) dt After collecting terms, we have
dV 2 2 = 3 i + (3 t z + t x y ) j + ( 2 x y z t + y ) k ans. dt
IV - 2
The Differential Equation of Conservation of Mass If we apply the basic concepts of conservation of mass to a differential control volume, we obtain a differential form for the continuity equation in
cartesian coordinates
∂ρ ∂ ∂ ∂ + (ρ u) + (ρ v) + (ρ w) = 0 ∂t ∂ x ∂y ∂z and in cylindrical coordinates
∂ρ 1 ∂ + (r ρ v r ) + 1 ∂ (ρ vθ ) + ∂ (ρ v z ) = 0 r ∂θ ∂t r ∂r ∂z Steady Compressible Flow For steady flow, the term
∂ = 0 and all properties are function of position only. ∂t
The previous equations simplify to Cartesian:
∂ ∂ ∂ (ρ u) + (ρ v ) + (ρ w) = 0 ∂x ∂y ∂z
1 ∂ (r ρ vr ) + 1 ∂ (ρ vθ ) + ∂ (ρ v z ) = 0 r∂r r ∂θ ∂z
Incompressible Flow For incompressible flow, density changes are negligible, ρ = const., and In the two coordinate systems, we have Cartesian:
∂u ∂ v ∂ w + + =0 ∂x ∂y ∂z
IV - 3
∂ρ = 0 ∂t
1 ∂ (r vr ) + 1 ∂ (vθ ) + ∂ (v z ) = 0 r∂r r ∂θ ∂z
Key Point: It is noted that the assumption of incompressible flow is not restricted to fluids which cannot be compressed, e.g. liquids. Incompressible flow is valid for (1) when the fluid is
essentially incompressible (liquids) and (2) for compressible fluids for which compressibility effects are not significant for the problem being considered. The second case is assumed to be met when
the Mach number is less than 0.3: Ma = V/c < 0.3
Gas flows can be considered incompressible
The Differential Equation of Linear Momentum If we apply Newton’s Second Law of Motion to a differential control volume we obtain the three components of the differential equation of linear momentum.
In cartesian coordinates, the equations are expressed in the form:
Inviscid Flow: Euler’s Equation If we assume the flow is frictionless, all of the shear stress terms drop out. The resulting equation is known as Euler’s equation and in vector form is given by:
ρg - ∇P = ρ
IV - 4
dV dt
dV is the total or substantial derivative of the velocity discussed dt previously and ∇ P is the usual vector gradient of pressure. This form of Euler’s
equation can be integrated along a streamline to obtain the frictionless Bernoulli’s equation ( Sec. 4.9). The Differential Equation of Energy The differential equation of energy is obtained by
applying the first law of thermodynamics to a differential control volume. The most complex element of the development is the differential form of the control volume work due to both normal and
tangential viscous forces. When this is done, the resulting equation has the form
du + P (∇⋅ V) = ∇⋅ (k ∇ T ) + Φ dt
where Φ is the viscous dissipation function. The term for the total derivative of internal energy includes both the transient and convective terms seen previously. Two common assumptions used to
simplify the general equation are: 1. du ≈ Cv dT
and 2. Cv, µ, k, ρ ≈ constants
With these assumptions, the energy equation reduces to
ρ Cv
dT = k ∇2 T + Φ dt
It is noted that the flow-work term was eliminated as a result of the assumption of constant density, ρ, for which the continuity equation becomes ∇ ⋅ V = 0 ,thus eliminating the term P (∇ ⋅V ) . We
now have the three basic differential equations necessary to obtain complete flow field solutions of fluid flow problems.
IV - 5
Boundary Conditions for the Basic Equations In vector form, the three basic governing equations are written as Continuity:
∂ρ + ∇ ⋅ (ρ V) = 0 ∂t
dV = ρ g − ∇ P + ∇ ⋅τi j dt
du + P (∇⋅ V) = ∇⋅ (k ∇ T ) + Φ dt
We have three equations and five unknowns: ρ, V, P, u, and T ; and thus need two additional equations. These would be the equations of state describing the variation of density and internal energy as
functions of P and T, i.e., ρ = ρ (P,T) and u = u (P,T) Two common assumptions providing this information are either: 1. Ideal gas:
ρ = P/RT and du = Cv dT
2. Incompressible fluid: ρ = constant and du = C dT Time and Spatial Boundary Conditions Time Boundary Conditions: If the flow is unsteady, the variation of each of the variables (ρ, V, P, u, and T )
must be specified initially, t = 0, as functions of spatial coordinates e.g. x,y,z. Spatial Boundary Conditions: The most common spatial boundary conditions are those specified at a fluid – surface
boundary. This typically takes the form of assuming equilibrium (e.g., no slip condition – no property jump) between the fluid and the surface at the boundary.
IV - 6
This takes the form: Vfluid = Vwall
Tfluid = Twall
Note that for porous surfaces with mass injection, the wall velocity will be equal to the injection velocity at the surface. A second common spatial boundary condition is to specify the values of V,
P, and T at any flow inlet or exit. Example 4.6 For steady incompressible laminar flow through a long tube, the velocity distribution is given by
r2 v z = U 1 − 2 R
vr = 0
vθ = 0
where U is the maximum or centerline velocity and R is the tube radius. If the wall temperature is constant at Tw and the temperature T = T(r) only, find T(r) for this flow. For the given conditions,
the energy equation reduces to
dT d vz k d d T ρ Cv v r = r + µ dr dr r dr dr
Substituting for vz and realizing the vr = 0, we obtain 2
k d d T d vz 4U 2 µ r 2 r = − µ =− 4 dr r dr dr R Multiply by r/k and integrate to obtain
IV - 7
dT µU r =− + C1 dr k R4 2 3
Integrate a second time to obtain
µ U 2 r4 4 k R4
+ C1 ln r + C2
Since the term, ln r, approaches infinity as r approaches 0, C1 = 0. Applying the wall boundary condition, T = Tw at r = R, we obtain for C2
C2 = Tw +
µ U2 4k
The final solution then becomes
µ U2
r4 T (r ) = Tw + 1 − 4 4k R The Stream Function The necessity to obtain solutions for multiple variables in multiple governing equations presents an obvious mathematical challenge. However,
the stream function, Ψ , allows the continuity equation to be eliminated and the momentum equation solved directly for the single variable, Ψ . The use of the stream function works for cases when the
continuity equation can be reduced to only two terms. For example, for 2-D, incompressible flow, continuity becomes
∂u ∂ v + =0 ∂x ∂y
IV - 8
Defining the velocity components to be
∂Ψ ∂y
v= −
∂Ψ ∂x
which when substituted into the continuity equation yields
∂ ∂ Ψ ∂ ∂ Ψ + − =0 ∂x∂y ∂y ∂x and continuity is automatically satisfied. Geometric interpretation of Ψ It is easily shown that lines of constant Ψ are flow streamlines. Since flow
does not cross a streamline, for any two points in the flow we can write 2
Q1→2 = ∫ (V ⋅n ) d A = ∫ d Ψ = Ψ2 − Ψ1 Thus the volume flow rate between two points in the flow is equal to the difference in the stream function between the two points.
Steady Plane Compressible Flow In like manner, for steady, 2-D, compressible flow, the continuity equation is
∂ ∂ (ρ u) + (ρ v ) = 0 ∂x ∂y For this problem, the stream function can be defined such that
ρu =
∂Ψ ∂y
IV - 9
∂Ψ ∂x
As before, lines of constant stream function are streamlines for the flow, but the change in stream function is now related to the local mass flow rate by 2
& 1− 2 = ∫ ρ (V ⋅ n ) d A = ∫ d Ψ = Ψ 2 − Ψ1 m
Vorticity and Irrotationality The concept of vorticity and irrotationality are very useful in analyzing many fluid problems. The analysis starts with the concept of angular velocity in a flow field.
Consider three points, A, B, & C, initially perpendicular at time t, that then move and deform to have the position and orientation at t + dt. The lines AB and BC have both changed length and
incurred angular rotation dα and dβ relative to their initial positions.
Fig. 4.10 Angular velocity and strain rate of two fluid lines deforming in the x-y plane We define the angular velocity ωz about the z axis as the average rate of counterclockwise turning of the two
lines expressed as
1 dα d β − 2 dt dt
ωz =
IV - 10
Applying the geometric properties of the deformation shown in Fig. 4.10 and taking the limit as ∆t → 0, we obtain
1 d v d u − 2 d x d y
ωz =
In like manner, the angular velocities about the remaining two axes are
1 d w d v − 2 d y d z
1 d u d w − 2 d z d x
ωx =
ωy =
From vector calculus, the angular velocity can be expressed as a vector with the form
ω = i ω x + j ω y + k ωz = 1/2 the curl of the velocity vector, e.g.
ω = (curl V) =
1 ∂ 2 ∂x u
∂ ∂ ∂y ∂z
The factor of 2 is eliminated by defining the vorticity, ξ , as follows: ξ = 2 ω = curl V Frictionless Irrotational Flows When a flow is both frictionless and irrotational, the momentum equation
reduces to Euler’s equation given previously by
ρg - ∇P = ρ
IV - 11
dV dt
As shown in the text, this can be integrated along the path, ds, of a streamline through the flow to obtain 2 ∂V dP 1 2 2 ds + + (V2 − V1 ) + g(z 2 − z1 ) = 0 ∫ ∫ 2 1 ∂t 1 ρ 2
For steady, incompressible flow this reduces to
1 2 V + gz = 2
constant along a streamline
IV - 12
V. Modeling, Similarity, and Dimensional Analysis To this point, we have concentrated on analytical methods of solution for fluids problems. However, analytical methods are not always satisfactory
due to: (1) limitations due to simplifications required in the analysis, (2) complexity and/or expense of a detailed analysis. The most common alternative is to: Use experimental test & verification
procedures. However, without planning and organization, experimental procedures can : (a) be time consuming, (b) lack direction, (c) be expensive. This is particularly true when the test program
necessitates testing at one set of conditions, geometry, and fluid with the objective to represent a different but similar set of conditions, geometry, and fluid. Dimensional analysis provides a
procedure that will typically reduce both the time and expense of experimental work necessary to experimentally represent a desired set of conditions and geometry. It also provides a means of
"normalizing" the final results for a range of test conditions. A normalized (non-dimensional) set of results for one test condition can be used to predict the performance at different but fluid
dynamically similar conditions ( including even a different fluid). The basic procedure for dimensional analysis can be summarized as follows:
1. Compile a list of relevant variables (dependent & independent) for the problem being considered, 2. Use an appropriate procedure to identify both the number and form of the resulting
non-dimensional parameters. This procedure is outlined as follows for the Buckingham Pi Theorem Definitions: n = the number of independent variables relevant to the problem j’ = the number of
independent dimensions found in the n variables j = the reduction possible in the number of variables necessary to be considered simultaneously k = the number of independent pi terms that can be
identified to describe the problem, k = n - j Summary of Steps: 1. List and count the n variables involved in the problem. 2. List the dimensions of each variable using {MLTΘ} or {FLTΘ}. Count the
number of basic dimensions ( j’) for the list of variables being considered. 3. Find j by initially assuming j = j’ and look for j repeating variables which do not form a pi product. If not
successful, reduce j by 1 and repeat the process. 4. Select j scaling, repeating variables which do not form a pi product. 5. Form a pi term by adding one additional variable and form a power
product. Algebraically find the values of the exponents which make the product dimensionless. Repeat the process with each of the remaining variables. 6. Write the combination of dimensionless pi
terms in functional form:
= f( Π1, Π2, …Πi)
Consider the following example for viscous pipe flow. The relevant variables for this problem are summarized as follows:
∆P = pressure drop µ = viscosity
ρ = density ε = roughness
Seven pipe flow variables:
V = velocity L = length
D = diameter
ρ, V, D, µ, ε, L } independent
Use of the Buckingham Pi Theorem proceeds as follows: 1. Number of independent variables: n = 7 2. List the dimensions of each variable ( use m L t Θ ): variables
-1 -2
V -3
dimensions mL t
mL t
-1 -1
The number of basic dimensions is j’ = 3. 3. Choose j = 3 with the repeating variables being ρ, V, and D. They do not form a dimensionless pi term. No combination of the 3 variables will eliminate
the mass dimension in density or the time dimension in velocity. 4. This step described in the above step. The repeating variables again are ρ, V, and D and j = 3. Therefore, k = n – j = 7 – 3 = 4
independent Π terms. 5. Form the Π terms:
ρa Vb Dc µ−1 = (mL-3)a ( Lt-1)b Lc ( mL-1t-1 )−1
In order for the Π term to have no net dimensions, the sum of the exponents for each dimension must be zero. Therefore, we have: mass: time: length:
a - 1 =0, a= 1 - b + 1 = 0, b = 1 -3a + b + c + 1 = 0, c = 3 – 1 – 1 = 1
We therefore have
ρ V D /µ = Re = Reynolds number
Repeating the process by adding the roughness ε
Π2 = ρa Vb Dc ε1 = (mL-3)a ( Lt-1)b Lc ( L )1 Solving: mass: time: Length:
Π2 = ε / D
a =0, a= 0 - b = 0, b = 0 -3a + b + c + 1 = 0, c = – 1 Roughness ratio
Repeat the process by adding the length L.
Π3 = ρa Vb Dc L1 = (mL-3)a ( Lt-1)b Lc ( L )1 Solving: mass: time: length:
Π3 = L / D
a =0, a= 0 - b = 0, b = 0 -3a + b + c + 1 = 0, c = – 1
length-to-diameter ratio
These three are the independent Π terms. Now obtain the dependent Π term by adding ∆P
Π4 = ρa Vb Dc ∆P1 = (mL-3)a ( Lt-1)b Lc ( mL-1t-2 )1 Solving: mass: time: length:
Π4 =
/ ρ V2
a + 1 = 0 , a = -1 - b - 2 = 0, b = -2 -3a + b + c - 1 = 0, c = 0
Pressure coefficient
Application of the Buckingham Pi Theorem to the previous list of variables yields the following non-dimensional combinations:
∆P ρ VD L ε = f , , ρ V2 µ D D or
Cp = f Re,L, ε
Thus, a non-dimensional pressure loss coefficient for viscous pipe flow would be expected to be a function of (1) the Reynolds number, (2) a nondimensional pipe length, and (3) a non-dimensional pipe
roughness. This will be shown to be exactly the case in Ch. VI, Viscous Internal Flow. A list of typical dimensionless groups important in fluid mechanics is given in the accompanying table. From
these results, we would now use a planned experiment with data analysis techniques to get the exact form of the relationship among these non dimensional parameters. The next major step is concerned
with the design and organization of the experimental test program Two key elements in the test program are: * design of the model * specification of the test conditions, particularly when the test
must be performed at conditions similar, but not the same as the conditions of interest. Similarity and non-dimensional scaling The basic requirement is in this process to achieve 'similarity'
between the 'experimental model and its test conditions' and the 'prototype and its test conditions' in the experiment.
Table 5.2 Dimensional Analysis and Similarity Parameter Reynolds number Mach number Froude number
Definition RE =
ρ UL µ
U A U2 Fr = gL
MA =
Weber number
We =
ρ U2 L γ
Cavitation number (Euler number)
Ca =
p - pv ρU 2
Prandtl number Eckert number Specific-heat ratio Strouhal number
Temperature ratio Pressure coefficient
Inertia Surface tension Pressure Inertia
C pµ k U2 Ec = c p To c γ = p cv ωL St = U
Dissipation Conduction Kinetic energy Enthalpy
Pr =
Enthalpy Internal energy Oscillation Mean speed Wall roughness Body length
Roughness ratio Grashof number
Qualitative ratio of effects Inertia Viscosity Flow speed Sound speed Inertia Gravity
Gr =
β ∆TgL3 ρ2 µ2
Tw To p − p∞ Cp = 1/ 2 ρ U2
Buoyancy Viscosity Wall temperature Stream temperature Static pressure Dynamic pressure
Lift coefficient
CL =
L 1/ 2 ρ U2 A
Lift force Dynamic force
Drag coefficient
CD =
D 1/ 2ρ U2 A
Lift force Dynamic force
Importance Always Compressible flow Free-surface flow Free-surface flow Cavitation Heat convection Dissipation Compressible flow Oscillating flow Turbulent,rough walls Natural convection Heat
transfer Aerodynamics, hydrodynamics Aerodynamics hydrodynamics Aerodynamics, hydrodynamics
In this context, “similarity” is defined as Similarity: All relevant dimensionless parameters have the same values for the model & the prototype. Similarity generally includes three basic
classifications in fluid mechanics: (1) Geometric similarity (2) Kinematic similarity (3) Dynamic similarity
Geometric similarity In fluid mechanics, geometric similarity is defined as follows:
Geometric Similarity
All linear dimensions of the model are related to the corresponding dimensions of the prototype by a constant scale factor SFG
Consider the following airfoil section (Fig. 5.4):
Fig. 5.4 Geometric Similarity in Model Testing For this case, geometric similarity requires the following:
rm L m Wm = = =⋅ ⋅ ⋅ rp L p Wp
SFG =
In addition, in geometric similarity, All angles are preserved. All flow directions are preserved. Orientation with respect to the surroundings must be same for the model and the prototype, ie.,
Angle of attack )m = angle of attack )p Kinematic Similarity In fluid mechanics, kinematic similarity is defined as follows: Kinematic Similarity
The velocities at 'corresponding' points on the model & prototype are in the same direction and differ by a constant scale factor SFk.
Therefore, the flows must have similar streamline pattterns Flow regimes must be the same. These conditions are demonstrated for two flow conditions, as shown in the following kinematically similar
flows (Fig. 5.6).
Fig. 5.6a Kinematically Similar Low Speed Flows
Fig. 5.6b Kinematically Similar Free Surface Flows The conditions of kinematic similarity are generally met automatically when geometric and dynamic similarity conditions are satisfied.
Dynamic Similarity In fluid mechanics, dynamic similarity is typically defined as follows: Dynamic Similarity
This is basically met if model and prototype forces differ by a constant scale factor at similar points.
This is illustrated in the following figure for flow through a sluice gate (Fig. 5.7).
Fig. 5.7 Dynamic Similarity for Flow through a Sluice Gate
This is generally met for the following conditions: 1. Compressible flows:
model & prototype Re, Ma, are equal
Rem = Rep, Mam= Map , γm = γp 2. Incompressible flows a. No free surface Rem = Rep b. Flow with a free surface Rem = Rep
Frm = Frp
Note: The parameters being considered, e.g., velocity, density, viscosity, diameter, length, etc., generally relate to the flow, geometry, and fluid characteristics of the problem and are considered
to be independent variables for the subject problem. The result of achieving similarity by the above means is that relevant non dimensional dependent variables, e.g., CD, Cp, Cf, or Nu, etc., are
then equal for both the model and prototype. This result would then indicate how the relevant dependent results, e.g. drag force, pressure forces, viscous forces, are to be scaled for the model to
the prototype. Equality of the relevant non-dimensional independent variables Re, Ma, x/L, etc., indicates how the various independent variables of importance should be scaled.
An example of this scaling is shown as follows: Example: The drag on a sonar transducer prototype is to be predicted based on the following wind tunnel model data and prototype data requirements.
Determine the model test velocity Vm necessary to achieve similarity and the expected prototype force Fp based on the model wind tunnel test results. Given:
1 ft
6 in
5 knots
5.58 lbf
ρ ν
slugs ft
2 ft -5 1.4 *10
1.56 * 10-4 ft
From dimensional analysis:
C D = f{ R e }
D 2 = f VD υ ρ V2
For the prototype, the actual operating velocity and Reynolds number are: Prototype: Vp = 5
na ⋅ mi ft 1hr ft na ⋅ mi ft 1hr ft = 8.44 Vp = 5 = 8.44 6080 6080 hr na ⋅ mi 3600 s s hr na ⋅ mi 3600 s s
Re = p
8. 44 ft * 1ft 5 s = = 6. 03 * 10 2 1. 4 *10 −5 ft s
υ p
Equality of Reynolds number then yields the required model test velocity of
Re = Re = m
υ m
= > Vm = 188 ft s
Based on actual test results for the model, i.e. measured Fm, equality of model and prototype drag coefficients yields
∴ C D = C D = > Fp = Fm p
Vm2 D 2m
D 2p
= 37. 4 lb f
Note: All fluid dynamic flows and resulting flow characteristics are not Re dependent. Example: The drag coefficient for bluff bodies with a fixed point of separation; e.g., radar antennae, generally
have a constant, fixed number for CD which is not a function of Re. CD = const. ≠ f (Re)
VI. VISCOUS INTERNAL FLOW To date, we have considered only problems where the viscous effects were either: a. known: i.e. - known FD or hf b. negligible: i.e. - inviscid flow This chapter presents
methodologies for predicting viscous effects and viscous flow losses for internal flows in pipes, ducts, and conduits. Typically, the first step in determining viscous effects is to determine the
flow regime at the specified condition. The two possibilities are: a. Laminar flow b. Turbulent flow The student should read Section 6.1 in the text, which presents an excellent discussion of the
characteristics of laminar and turbulent flow regions. For steady flow at a known flow rate, these regions exhibit the following: Laminar flow:
A local velocity constant with time, but which varies spatially due to viscous shear and geometry.
Turbulent flow:
A local velocity which has a constant mean value but also has a statistically random fluctuating component due to turbulence in the flow.
Typical plots of velocity time histories for laminar flow, turbulent flow, and the region of transition between the two are shown below.
Fig. 6.1 (a) Laminar, (b)transition, and (c) turbulent flow velocity time histories.
Principal parameter used to specify the type of flow regime is the Reynolds number - Re =
ρV D V D = µ ν
V - characteristic flow velocity D - characteristic flow dimension µ - dynamic viscosity
υ - kinematic viscosity =
µ ρ
We can now define the Recr ≡ critical or transition Reynolds number Recr ≡ Reynolds number below which the flow is laminar, above which the flow is turbulent While transition can occur over a range
of Re, we will use the following for internal pipe or duct flow:
Re cr ≅ 2300 =
ρVD VD = µ cr υ cr
Internal Viscous Flow A second classification concerns whether the flow has significant entrance region effects or is fully developed. The following figure indicates the characteristics of the
entrance region for internal flows. Note that the slope of the streamwise pressure distribution is greater in the entrance region than in the fully developed region. Typical criteria for the length
of the entrance region are given as follows: Laminar:
Le ≅ 0.06 Re D
Le ≅ 4.4Re1/ 6 D
where: Le = length of the entrance region
Note: Take care in neglecting entrance region effects. In the entrance region, frictional pressure drop/length > the pressure drop/length for the fully developed region. Therefore, if the effects of
the entrance region are neglected, the overall predicted pressure drop will be low. This can be significant in a system with short tube lengths, e.g., some heat exchangers.
Fully Developed Pipe Flow The analysis for steady, incompressible, fully developed, laminar flow in a circular horizontal pipe yields the following equations:
R2 dP r2 U (r ) = − 1 − 4 µ dx R 2 VI-3
U r2 = 1 − 2 , Umax = 2Vavg Umax R and
Q = A Vavg = π R2 Vavg Key Points: Thus for laminar, fully developed pipe flow (not turbulent): a. b. c. d.
The velocity profile is parabolic. The maximum local velocity is at the centerline (r = 0). The average velocity is one-half the centerline velocity. The local velocity at any radius varies only with
radius, not on the streamwise (x) location ( due to the flow being fully developed).
Note: All subsequent equations will use the symbol V (no subscript) to represent the average flow velocity in the flow cross section. Darcy Friction Factor: We can now define the Darcy friction
factor f as:
D ∆P f L f≡ V2 ρ 2
where ∆Pf = the pressure drop due to friction only. The general energy equation must still be used to determine total pressure drop.
Therefore, we obtain
L V2 ∆P f = ρ g h f = f ρ D 2 and the friction head loss hf is given as
L V2 hf = f D 2g Note: The definitions for f and hf are valid for either laminar or turbulent flow. However, you must evaluate f for the correct flow regime, laminar or turbulent.
Key Point: It is common in industry to define and use a “fanning” friction factor ff . The fanning friction factor differs from the Darcy friction factor by a factor of 4. Thus, care should be taken
when using unfamiliar equations or data since use of ff in equations developed for the Darcy friction factor will result in significant errors (a factor of 4). Your employer will not be happy if you
order a 10 hp motor for a 2.5 hp application. The equation suitable for use with ff I s
L V2 h f = 4 ff D 2g Laminar flow: Application of the results for the laminar flow velocity profile to the definition of the Darcy friction factor yields the following expression:
64 Re
laminar flow only (Re < 2300)
Thus with the value of the Reynolds number, the friction factor for laminar flow is easily determined. Turbulent flow: A similar analysis is not readily available for turbulent flow. However, the
Colebrook equation, shown below, provides an excellent representation for the variation of the Darcy friction factor in the turbulent flow regime. Note that the equation depends on both the pipe
Reynolds number and the roughness ratio, is transcendental, and cannot be expressed explicitly for f .
2.51 ε/D f = − 2 log + 1/ 2 3.7 Re f
turbulent flow only (Re > 2300)
ε = nominal roughness of pipe or duct being used.
(Note: Take care with units for ε;
(Table 6.1, text)
ε /D must be non-dimensional).
A good approximate equation for the turbulent region of the Moody chart is given by Haaland’s equation:
6.9 ε / D 1.11 f = −1.8log + 3.7 Re
Note again the roughness ratio ε/D must be non-dimensional in both equations. Graphically, the results for both laminar and turbulent flow pipe friction are represented by the Moody chart as shown
Typical roughness values are shown in the following table:
Table 6.1 Average roughness values of commercial pipe
Haaland’s equation is valid for turbulent flow (Re > 2300) and is easily set up on a computer, spreadsheet, etc. Key fluid system design considerations for laminar and turbulent flow a. Most internal
flow problems of engineering significance are turbulent, not laminar. Typically, a very low flow rate is required for internal pipe flow to be laminar. If you open your kitchen faucet and the outlet
flow stream is larger than a kitchen match, the flow is probably turbulent. Thus, check your work carefully if your analysis indicates laminar flow. b. The following can be easily shown: Laminar
W& f Turbulent flow:
∆Pf ~ µ , L,Q, D
~ µ, L,Q 2, D −4
∆Pf ~ {ρ , µ , L, Q , D 3/ 4
1/ 4
W& f ~ { ρ , µ , L, Q , D 3/4
1/ 4
Thus both pressure drop and pump power are very dependent on flow rate and pipe/conduit diameter. Small changes in diameter and/or flow rate can significantly change circuit pressure drop and power
requirements. VI-7
Example ( Laminar flow): Water, 20oC flows through a 0.6 cm tube, 30 m long, at a flow rate of 0.34 liters/min. If the pipe discharges to the atmosphere, determine the supply pressure if the tube is
inclined 10o above the horizontal in the flow direction.
L 10
L = 30 m
Water Properties:
D = 0.6 cm
Energy Equation
P1 − P2 V22 − V12 = + Z2 − Z1 + h f − hp ρg 2g
ρ = 998 kg/m
2 o
ρg = 9790 N/m 2
ν = 1.005 E-6 m /s which for steady-flow in a constant diameter pipe with P2 = 0 gage becomes, −3
P1 = Z2 −Z1 + hf = L sin 10o + hf ρg
Q 0.34 E m / min*1min/ 60 s V= = = 0.2 m / s A π (0.3 /100 )2 m2 Re = f=
0.2 * 0.006 = 1197 → laminar flow 1.005 E −6
64 64 = = 0.0535 Re 1197 2
0.2 LV 30m hf = f = 0.0535* = 0.545 m D 2g 0.006m 2 * 9.807m / s 2 P1 = 30 * sin10Þ+ 0.545 = 5.21 + 0.545 ρg gravity head 3
friction head
= 5.75m total head loss
P1 = 9790 N/m *5.75 m = 56.34 kN/m (kPa) ~ 8.2 psig ans.
Example: (turbulent flow) 3
Oil, ρ = 900 kg/m , ν = 1 E-5 m /s, 3 flows at 0.2 m /s through a 500 m length of 200 mm diameter, cast iron pipe. If the pipe slopes downward 10o in the flow direction, compute hf, total head loss,
pressure drop, and power required to overcome these losses. The energy equation can be written as follows where ht = total head loss.
L = 500 m
D = 200 mm
ht = Z2 − Z1 + h f
Q 0.2 m / s V= = = 6.4 m / s A π (.1)2 m 2 VD
P1 − P2 V22 − V12 = ht = +Z 2 −Z1 + h f − hp ρg 2g
which reduces to
Re =
Table 6.1, cast iron, ε = 0.26 mm
6.4 *.2 = 128, 000 → turbulent flow , 1 E −5
ε D
0.26 = 0.0013 200
Since flow is turbulent, use Haaland’s equation to determine friction factor (check your work using the Moody chart). −2
1.11 6.9 6.9 ε / D 1.11 0.0013 f = −1.8log + , f = −1.8log 128, 000 + 3.7 3.7 Re
6.4 LV 500m f = 0.02257 h f = f = 0.02257* = 116.6 m ans. D 2g 0.2m 2 * 9.807m / s 2 ht = Z2 – Z1 + hf = - 500 sin 0 + 116.6 = - 86.8 + 116.6 = 29.8 m ans.
Note that for this problem, there is a negative gravity head loss (i.e. a head increase) and a positive frictional head loss resulting in the net head loss of 29.8 m.
∆P = ρ g ht = 900 kg / m * 9.807 m / s * 29.8 m = 263 kPa ans. 3
W& = ρ Q g ht = Q ∆P = 0.2 m3 / s * 273,600 N / m 2 = 54.7 kw
Note that this is not necessarily the power required to drive a pump, as the pump efficiency will typically be less than 100%. These problems are easily set up for solution in a spreadsheet as shown
below. Make sure that the calculation for friction factor includes a test for laminar or turbulent flow with the result proceeding to the correct equation. Always verify any computer solution with
problems having a known solution. FRICTIONAL HEAD LOSS CALCULATION All Data are entered in S.I. Units e.g. (m, sec., kg), except as noted, ε Ex.
6.7 Input Data
Calculated Results
(m )
D= ε= ρ=
0.2 0.26 900
(m) (mm) (kg/m^3)
Re = e/D = f=
127324 0.0013 0.02258
hf =
sum Ki =
D1 =
hm =
D2 =
d KE =
ht =
(m) P1-P2 =
Solution Summary: To solve basic pipe flow frictional head loss problem, use the following procedure: 1. Use known flow rate to determine Reynolds number. 2. Identify whether flow is laminar or
turbulent. 3. Use appropriate expression to determine friction factor (w ε/D if necessary). 4. Use definition of hf to determine friction head loss. 5. Use general energy equation to determine total
pressure drop.
Unknown Flow Rate and Diameter Problems Problems involving unknown flow rate and diameter in general require iterative/ trial & error solutions due to the complex dependence of Re, friction factor,
and head loss on velocity and pipe size. Unknown Flow Rate: For the special case of known friction loss hf a closed form solution can be obtained for the problem of unknown Q. The solution proceeds
as follows: Given: Known values for D, L, hf, ρ, and µ
calculate V or Q.
Define solution parameter:
g D3 h f f Re = Lυ 2 2 D
Note that this solution does not contain velocity and the parameter ζ can be calculated from known values for D, L, hf, ρ, and µ. The Reynolds number and subsequently the velocity can be determined
from ζ and the following equations: Turbulent:
Re D = − (8ς )
1/ 2
ε / D 1.775 log + 3.7 ζ
Re D =
ς 32
and laminar to turbulent transition can be assumed to occur approximately at ζ = 73,600 (check Re at end of calculation to confirm). Note that this procedure is not valid (except perhaps for initial
estimates) for problems involving significant minor losses where the head loss due only to pipe friction is not known. For these problems a trial and error solution using a computer is best. Example
6.9 3
Oil, with ρ = 950 kg/m and ν = 2 E-5 m /s, flows through 100 m of a 30 cm diameter pipe. The pipe is known to have a head loss of 8 m and a roughness ratio ε/D = 0.0002. Determine the flow rate and
oil velocity possible for these conditions. Without any information to the contrary, we will neglect minor losses and KE head changes. With these assumptions, we can write:
g D3 hf 9.807 m / s 2 * 0.303 m3 *8.0 m ς= = = 5.3 E7 > 73,600; turbulent 2 2 2 Lυ 100 m* 2 E−5 m / s
Re D = − (8*5.3 E 7)
1/ 2
0.0002 1.775 log + = 72, 600 checks, turbulent 3.7 5.3 E 7 2
72,600*2 E −5 m / s m Re D = , V= = 4.84 ans. 0.3 m υ s VD
Q = A V = π .15 2 m 2 4.84 m/s = 0.342 m 3/s ans. This is the maximum flow rate and oil velocity that could be obtained through the given pipe and given conditions (hf = 8 m).
Note that this problem could have also been solved using a computer based trial and error procedure in which a value is assumed for the fluid flow rate until a flow rate is found which results in the
specified head loss. Note also that with this procedure, the problem being solved can include the effects of minor losses, KE, and PE changes with no additional difficulty. Unknown Pipe Diameter: A
similar difficulty arises for problems involving unknown pipe difficulty, except a closed form, analytical solution is not available. Again, a trial and error solution is appropriate for use to
obtain the solution and the problem can again include losses due to KE, PE, and piping components with no additional difficulty. Non-Circular Ducts: For flow in non-circular ducts or ducts for which
the flow does not fill the entire cross-section, we can define the hydraulic diameter Dh as
Dh =
4A P
where A = cross-sectional area of actual flow, P = wetted perimeter, i.e. the perimeter on which viscous shear acts
Cross sectional area - A
Perimeter - P
With this definition, all previous equations for the Reynolds number Re friction factor f and head loss hf are valid as previously defined and can be used on both circular and non-circular flow cross
sections. Minor Losses In addition to frictional losses for a length L of pipe, we must also consider losses due to various fittings (valves, unions, elbows, tees, etc.). These losses are expressed
V2 hm = Ki 2g VI-13
where hm = the equivalent head loss across the fitting or flow component. V = average flow velocity for the pipe size of the fitting Ki = the minor loss coefficient for given flow component; valve,
union, etc. See Sec. 6.7, Table 6.5, 6.6, Fig. 6.19, 6.20, 6.21, 6.22, etc. Table 6.5 shows minor loss K values for several common types of valves, fully open, and for elbows and tees. Table 6.5
Minor loss coefficient for common valves and piping components
Figure 6.18 shows minor loss K values for several types of common valves. Note that the K valves shown here are for the indicated fractional opening. Also, fully open values may not be consistent
with values indicated in Table 6.5 for fully open valves or for the valve of a particular manufacturer. In general, use specific manufacturer’s data when available.
Fig. 6.18 Average loss coefficients for partially open valves
Note that exit losses are K ≅ 1 for all submerged exits, e.g., fluid discharged into a tank at a level below the fluid surface. Also, for an open pipe discharge to the atmosphere, there is no loss
coefficient when the energy equation is written only to the end of the pipe. In general, do not take point 1 for an analysis to be in the plane of an inlet having an inlet loss. You do not know what
fraction of the inlet loss to consider.
Fig. 6.21 Entrance and exit loss coefficients (a) reentrant inlets; (b) rounded and beveled inlets
Note that the losses shown in Fig. 6.22 do not represent losses associated with pipe unions or reducers. These must be found in other sources in the literature. Also note that the loss coefficient is
always based on the velocity in the smaller diameter (d) of the pipe, irrespective of the direction of flow. Fig. 6.22 Sudden contraction and expansion losses.
Assume that this is also true for reducers and similar area change fittings.
These and other sources of data now provide the ability to determine frictional losses for both the pipe and other piping/duct flow components. The total frictional loss now becomes
V2 L V2 hf = f + ∑ Ki D 2g 2g or
L V2 h f = f + ∑ Ki D 2g These equations would be appropriate for a single pipe size (with average velocity V). For multiple pipe/duct sizes, this term must be repeated for each pipe
size. Key Point: The energy equation must still be used to determine the total head loss and pressure drop from all possible contributions.
Example 6.16
Water, ρ = 1.94 slugs/ft3 and ν = 1.1 E-5 ft2/s, is pumped between two reservoirs at 0.2 ft3/s through 400 ft of 2–in diameter pipe with ε/D = 0.001 having the indicated minor losses. Compute the
pump horsepower (motor size) required. Writing the energy equation between points 1 and 2 (the free surfaces of the two reservoirs), we obtain
P1 − P2 V22 − V12 = + Z2 − Z1 + h f − hp ρg 2g For this problem, the pressure (P1 = P2) and velocity (V1 = V2 = 0) head terms are zero and the equation reduces to
L V2 hp = Z2 − Z1 + h f = Z2 − Z1 + f + ∑ Ki D 2g For a flow rate Q = 0.2 ft3/s we obtain
Q 0.2 ft 3 / s V= = = 9.17 ft / s A π (1/ 12)2 ft 2 Re =
With ε/D = 0.001 and
9.17 ft / s ( 2 / 12) ft = 139,000 1.1E−5 ft 2 / s
the flow is turbulent and Haaland’s equation can be used to determine the friction factor:
6.9 .001 1.11 f = −1.8log + 139, 000 3.7
−2 = 0.0214
the minor losses for the problem are summarized in the following table: Note: The loss for a pipe bend is not the same as for an elbow fitting. If there were no tank at the pipe discharge and point 2
were at the pipe exit, there would be no exit loss coefficient. However, there would be an exit K.E. term.
Loss element Sharp entrance (Fig. 6.21) Open globe valve (Table 6.5) 12 " bend, R/D = 12/6 = 2 (Fig. 6.19) Threaded, 90Þ, reg. elbow, (Table 6.5) Gate valve, 1/2 closed (Fig. 6.18) Submerged exit
(Fig. 6.20) Ki =
Ki 0.5 6.9 0.15 0.95 2.7 1 12.2
Substituting in the energy equation we obtain
9.17 2 400 9.17 2 h p = (120 − 20) + 0.0214 + 12.2 2/12 64.4 64.4 h p = 100 + 67.1 + 15.9 = 183ft
Note the distribution of the total loss between static, pipe friction, and minor losses. The power required to be delivered to the fluid is give by 3
slug ft ft Pf = ρ Qg h p = 1.94 3 32.2 2 0.2 183 ft = 2286 ft lbf s ft s
Pf =
2286ftlbf = 4.2hp 550ftlbf / s/hp
If the pump has an efficiency of 70 %, the power requirements would be specified by
w& p = 4.2 hp = 6 hp 0.70
Solution Summary: To solve basic pipe flow pressure drop problem, use the following procedure: 1. Use known flow rate to determine Reynolds number. 2. Identify whether flow is laminar or turbulent.
3. Use appropriate expression to find friction factor (with ε/D if necessary). 4. Use definition of hf to determine friction head loss. 5. Tabulate and sum minor loss coefficients for piping
components. 6a. Use general energy equation to determine total pressure drop, or 6b.Determine pump head requirements as appropriate. 7. Determine pump power and motor size if required.
Multiple-Pipe Systems Basic concepts of pipe system analysis apply also to multiple pipe systems. However, the solution procedure is more involved and can be iterative. Consider the following: a.
Multiple pipes in series b. Multiple pipes in parallel Series Pipe System: The indicated pipe system has a steady flow rate Q through three pipes with diameters D1, D2, & D3. a
Two important rules apply to this problem. 1. The flow rate is the same through each pipe section. For incompressible flow, this is expressed as 2
Q1 = Q2 = Q3 = Q or D1 V1 = D2 V2 = D3 V3 2. The total frictional head loss is the sum of the head losses through the various sections.
h f ,a− b = h f ,1 + h f ,2 + h f ,3
h f ,a− b
2 2 V32 V1 L V2 L L = f + ∑ Ki + f + K + f + K D D1 2 g D ∑ i D2 2 g D ∑ i D3 2 g
Note: Be careful how you evaluate the transitions from one section to the next. In general, loss coefficients for transition sections are based on the velocity of the smaller section.
Example: Given a pipe system as shown in the previous figure. The total pressure drop is Pa – Pb = 150 kPa and the elevation change is Za – Zb = – 5 m. Given the following data, determine the flow
rate of water through the section. Pipe 1 2 3
L (m) 100 150 80
D (cm) 8 6 4
The energy equation is written as where hf is given by the sum of the total frictional losses for three pipe sections.
e (mm) 0.24 0.12 0.2
e/D 0.003 0.02 0.005
Pa − Pb Vb2 − Va2 = + Zb − Za + h f − h p ρg 2g
With no pump; hp is 0, Zb - Za = Pa − Pb 150,000 N / m 2 ht = = = 15.3m - 5 m and ht = 15.3 m for ∆P = ρg 9790 N / m2 150 kPa Since the flow rate Q and thus velocity is the only remaining variable,
the solution is easily obtained from a spreadsheet by assuming Q until ∆P = 150 kPa. Fluid
ρ(kg/m^3) = ν(m^2/s) =
1000 1.02E-06
inlet & exit dZ (m) = Da(m) = Db(m) =
-5 0.08 0.04
Assume Q (m^3/s)=
0.56 2.25 0.24 20.08 Actual 150
Vb(m/s)= dKE(m) = hf (net) = Pa - Pb (kPa)
100 0.08 0.24
150 0.06 0.12
80 0.04 0.20
V(m/s)= Re = f= hf =
0.56 44082.8 0.02872 0.58
1.00 58777.1 0.02591 3.30
2.25 88165.6 0.03139 16.18
Ki hm=
hf(calc) =
L(m) D(m) = ε(mm) =
Calculated 150.00
Q(m^3/hr) =
Thus it is seen that a flow rate of 10.17 m /hr produces the indicated head loss through each section and a net total ∆P = 150 kPa. A solution can also be obtained by writing all terms explicitly in
terms of a single velocity, however, the algebra is quite complex (unless the flow is laminar), and an iterative solution is still required. All equations used to obtained the solution are the same
as those presented in previous sections. Parallel Pipe Systems A flow rate QT enters the indicated parallel pipe system. The total flow splits and flows through 3 pipe sections, each with different
diameters and lengths.
a T
Two basic rules apply to parallel pipe systems; 1. The total flow entering the parallel section is equal to the sum of the flow rates through the individual sections, 2. The total pressure drop
across the parallel section is equal to the pressure drop across each individual parallel segment. Note that if a common junction is used for the start and end of the parallel section, the velocity
and elevation change is also the same for each section. Thus, the flow rate through each section must be such that the frictional loss is the same for each and the sum of the flow rates equals the
total flow. For the special case of no kinetic or potential energy change across the sections, we obtain: ht = ( hf + hm)1 = ( hf + hm)2 =( hf + hm)3 and QT = Q1 + Q2 + Q3
Again, the equation used for both the pipe friction and minor losses is the same as previously presented. The flow and pipe dimensions used for the previous example are now applied to the parallel
circuit shown above.
Example: A parallel pipe section consists of three parallel pipe segments with the lengths and diameters shown below. The total pressure drop is 150 kPa and the parallel section has an elevation drop
of 5 m. Neglecting minor losses and kinetic energy changes, determine the flow rate of water through each pipe section. The solution is iterative and is again presented in a spreadsheet. The net
friction head loss of 20.3 m now occurs across each of the three parallel sections. Fluid
ρ( kg/m^3) = ν(m^2/s) =
1000 1.02E-06
Db(m) =
-5 0.08 0.04
Q (m^3/hr)=
Da(m) =
Assume Q1 (m^3/s)= Q2(m^3/s)=
150 0.06 0.12
80 0.04 0.20
V(m/s)= Re = f= hf =
0.003 62.54 3.46 271083.5 0.02666 20.30
0.002 25.95 2.55 149977.8 0.02450 20.30
0.005 11.41 2.52 98919.7 0.03129 20.30
Ki = hm=
0 0.00
0 0.00
0 0.00
hf,net(m) = hf + ∆z =
inlet & exit dZ (m) =
100 0.08 0.24
L(m) D(m) = ε(mm) =
62.54 25.95
Pa - Pb (kPa)
ht(m) =
Q(m^3/hr) =
Q(m^3/hr) =
Total Flow, Qt(m^3/hr) =
The strong effect of diameter can be seen with the smallest diameter having the lowest flow rate, even though it also has the shortest length of pipe.
VII. Boundary Layer Flows The previous chapter considered only viscous internal flows. Viscous internal flows have the following major boundary layer characteristics: * An entrance region where the
boundary layer grows and dP/dx ≠ constant, * A fully developed region where: • The boundary layer fills the entire flow area. • The velocity profiles, pressure gradient, and τw are constant; i.e.,
they are not equal to f(x), • The flow is either laminar or turbulent over the entire length of the flow , i.e., transition from laminar to turbulent is not considered. However, viscous flow boundary
layer characteristics for external flows are significantly different as shown below for flow over a flat plate: U∞ y
free stream
x laminar
laminar to turbulent transition turbulent
edge of boundary layer δ(x)
Fig. 7.1 Schematic of boundary layer flow over a flat plate For these conditions, we note the following characteristics: • The boundary layer thickness δ grows continuously from the start of the
fluid-surface contact, e.g., the leading edge. It is a function of x, not a constant. • Velocity profiles and shear stress τ are f(x,y). • The flow will generally be laminar starting from x = 0. •
The flow will undergo laminar-to-turbulent transition if the streamwise dimension is greater than a distance xcr corresponding to the location of the transition Reynolds number Recr. • Outside of the
boundary layer region, free stream conditions exist where velocity gradients and therefore viscous effects are typically negligible. VII-1
As it was for internal flows, the most important fluid flow parameter is the local Reynolds number defined as Re x =
ρ U ∞ x U ∞x = µ υ
where ρ = fluid density µ = fluid dynamic viscosity ν = fluid kinematic viscosity U ∞ = characteristic flow velocity x = characteristic flow dimension It should be noted at this point that all
external flow applications will not use a distance from the leading edge x and the characteristic flow dimension. For example, for flow over a cylinder, the diameter will be used as the
characteristic dimension for the Reynolds number. Transition from laminar to turbulent flow typically occurs at the local transition Reynolds number which for flat plate flows can be in the range of
500, 000 ≤ Re cr ≤3,000, 00 With xcr = the value of x where transition from laminar to turbulent flow occurs, the typical value used for steady, incompressible flow over a flat plate is
Re cr =
ρ U∞ xcr = 500, 000 µ
Thus for flat plate flows for which:
x < xcr
the flow is laminar
x ≥ xcr
the flow is turbulent
The solution to boundary layer flows is obtained from the reduced “Navier – Stokes” equations, i.e., Navier-Stokes equations for which boundary layer assumptions and approximations have been applied.
Flat Plate Boundary Layer Theory Laminar Flow Analysis For steady, incompressible flow over a flat plate, the laminar boundary layer equations are: Conservation of mass:
∂u + ∂v = 0 ∂x ∂ y
'X' momentum:
'Y' momentum:
−∂p = 0
∂u ∂u 1 d p 1 ∂ ∂ u +v =− + µ ∂x ∂y ρ d x ρ ∂ y ∂ y ∂y
The solution to these equations was obtained in 1908 by Blasius, a student of Prandtl's. He showed that the solution to the velocity profile, shown in the table below, could be obtained as a function
of a single, non-dimensional variable defined as
η= y
U∞ υx
1/ 2
with the resulting ordinary differential equation:
f ′′ ′ + and
1 f f ′′ = 0 2 u f ′(η ) = U∞
Boundary conditions for the differential equation are expressed as follows: at y = 0, v = 0 → f (0) = 0 ; y component of velocity is zero at y = 0 at y = 0 , u = 0 → f ′(0 ) = 0 ; x component of
velocity is zero at y = 0 VII-3
The key result of this solution is written as follows:
∂2 f τw = 0.332 = ∂η2 y= 0 µ U∞ U∞ / υ x With this result and the definition of the boundary layer thickness, the following key results are obtained for the laminar flat plate boundary layer:
Local boundary layer thickness
δ (x) =
Local skin friction coefficient:
C fx =
0.664 Re x
CD =
1.328 Re x
CD =
FD / A 1 2 ρ U ∞ 2
(defined below) Total drag coefficient for length L ( integration of τw dA over the length of the plate, per unit 2 area, divided by 0.5 ρ U∞ ) where by definition
C fx =
τ w (x ) 1 2 ρ U ∞ 2
5x Re x
With these results, we can determine local boundary layer thickness, local wall shear stress, and total drag force for laminar flow over a flat plate. Example: Air flows over a sharp edged flat plate
with L = 1 m, a width of 3 m and U∞ = 2 m/s . For one side of the plate, find: δ(L), Cf (L), τw(L), CD, and FD. Air: First check Re:
ρ = 1.23 kg/m
Re L =
U∞ L
ν = 1.46 E-5 m /s
2 m / s* 2.15 m = 294,520 < 500,000 2 1.46E − 5m / s
Key Point: Therefore, the flow is laminar over the entire length of the plate and calculations made for any x position from 0 - 1 m must be made using laminar flow equations. VII-4
Boundary layer thickness at x = L:
δ (L ) =
5L 5* 2.15 m = = 0.0198 m = 1.98 cm 294, 520 Re L
Local skin friction coefficient at x = L:
C f ( L) =
0.664 0.664 = = 0.00122 Re L 294, 520
Surface shear stress at x = L:
τ w = 1 / 2 ρ U∞2 C f = 0.5 *1.23 kg / m3 * 2 2 m 2 / s 2 * 0.00122 τ w = 0.0030 N / m2 (Pa) Drag coefficient over total plate, 0 – L:
CD (L ) =
1.328 1.328 = = 0.00245 Re L 294, 520
Drag force over plate, 0 – L:
FD = 1/ 2 ρ U∞ CD A = 0.5 *1.23kg / m * 2 m / s * 0.00245 * 2 * 2.15 m FD = 0.0259 N 3
Two key points regarding this analysis: 1. Each of these calculations can be made for any other location on the plate by simply using the appropriate x location for any x ≤ L . 2. Be careful not to
confuse the calculation for Cf and CD. Cf is a local calculation at a particular x location (including x = L) and can only be used to calculate local shear stress, not drag force. CD is an integrated
average over a specified length (including any x ≤ L ) and can only be used to calculate the average shear stress and the integrated force over the length. VII-5
Turbulent Flow Equations While the previous analysis provides an excellent representation of laminar, flat plate boundary layer flow, a similar analytical solution is not available for turbulent flow
due to the complex nature of the turbulent flow structure. However, experimental results are available to provide equations for key flow field parameters. A summary of the results for boundary layer
thickness and local and average skin friction coefficient for a laminar flat plate and a comparison with experimental results for a smooth, turbulent flat plate are shown below. Laminar
δ (x) = C fx = CD =
5x Re x
δ (x ) =
0.664 Re x
1.328 Re L
C fx = where
τw 1 2 ρ U ∞ 2
C fx = CD =
0.074 .2 Re L
0.37 x .2 Re x
0.0592 .2 Re x for turbulent flow over entire plate, 0 – L, i.e. assumes turbulent flow in the laminar region.
local drag coefficient based on local wall shear stress (laminar or turbulent flow region).
and CD = total drag coefficient based on the integrated force over the length 0 to L
F/ A CD = 1 2 = ρ U ∞ 2
1 2 ρ U ∞ 2
−1 L
∫ τ w ( x ) w dx 0
A careful study of these results will show that in general, boundary layer thickness grows faster for turbulent flow and wall shear and total friction drag are greater for turbulent flow than for
laminar flow given the same Reynolds number. VII-6
It is noted that the expressions for turbulent flow are valid only for a flat plate with a smooth surface. Expressions including the effects of surface roughness are available in the text. Combined
Laminar and Turbulent Flow U y
laminar to turbulent transition
free stream
x laminar
edge of boundary layer
Flat plate with both laminar and turbulent flow sections For conditions (as shown above) where the length of the plate is sufficiently long that we have both laminar and turbulent sections: *
Local values for boundary layer thickness and wall shear stress for either the laminar or turbulent sections are obtained from the expressions for δ(x) and Cfx for laminar or turbulent flow as
appropriate for the given region.
The result for average drag coefficient CD and thus total frictional force over the laminar and turbulent portions of the plate is given by (assuming a transition Re of 500,000)
CD = *
0.074 .2 Re L
1742 Re L
Calculations assuming only turbulent flow can be made typically for two cases 1. 2.
when some physical situation (a trip wire) has caused the flow to be leading from the leading edge or if the total length L of the plate is much greater than the length xcr of the laminar section
such that the total flow can be considered turbulent from x = 0 to L. Note that this will overpredict the friction drag force since turbulent drag is greater than laminar.
With these results, a detailed analysis can be obtained for laminar and/or turbulent flow over flat plates and surfaces that can be approximated as a flat plate. VII-7
Example: Water flows over a sharp flat plate 2.55 m long, 1 m wide, with U∞ = 2 m/s. Estimate the error in FD if it is assumed that the entire plate is turbulent. 3
ν = 1.02 E- m /s
Water: ρ = 1000 kg/m
Re L =
Reynolds number:
U∞ L
2 m / s * 2.55 m = 5E 6 > 500, 000 1.02 E − 6 m 2 / s
with Re cr = 500,000 ⇒ x cr = 0.255 m ( or 10% laminar) a. Assume that the entire plate is turbulent
CD =
0.074 0.074 .2 = 0.00338 .2 = Re L 5 E 6 ( ) 2
kg m FD = 0.5 ρ U CD A = 0.5 *1000 3 * 2 2 2 * 0.00338 * 2.55 m 2 s m 2 ∞
FD = 17.26 N
This should be high since we have assumed that the entire plate is turbulent and the first 10% is actually laminar.
b. Consider the actual combined laminar and turbulent flow:
CD =
0.074 .2 Re L
1742 1742 = 0.00338 − = 0.00303 Re L 5 E6
Note that the CD has decreased when both the laminar and turbulent sections are considered. 2
kg m FD = 0.5 ρ U CD A = 0.5 *1000 3 * 2 2 2 * 0.00303* 2.55 m 2 s m 2 ∞
FD = 15.46 N {Lower than the fully turbulent value} Error =
17.26 − 15.46 * 100 = 11.6% high 15.46
Question: Since xcr = 0.255 m, what would your answers represent if you had calculated the Re, CD, and FD using x = xcr = 0.255 m? Answer:
You would have the value of the transition Reynolds number and the drag coefficient and drag force over the laminar portion of the plate (assuming you used laminar equations). If you used turbulent
equations, you would have red marks on your paper.
Von Karman Integral Momentum Analysis While the previous results provide an excellent basis for the analysis of flat plate flows, complex geometries and boundary conditions make analytical solutions
to most problems difficult. An alternative procedure provides the basis for an approximate solution which in many cases can provide excellent results. The key to practical results is to use a
reasonable approximation to the boundary layer profile, u(x,y). This is used to obtain the following: & = m
a. Boundary layer mass flow:
∫0 ρ u bdy
where b is the width of the area for which the flow rate is being obtained.
τw = µ
b. Wall shear stress:
d u d y y=0
You will also need the streamwise pressure gradient
dP dx
for many problems.
The Von Karman integral momentum theory provides the basis for such an approximate analysis. The following summarizes this theory. Displacement thickness: Consider the problem indicated in the
adjacent figure: A uniform flow field with velocity U∞ approaches a solid surface. As a result of viscous shear, a boundary layer velocity profile develops.
y=h +δ*
y Streamline
U∞ h
h u 0 x
Simulated effect
A viscous boundary layer is created when the flow comes in contact with the solid surface. Key point: Compared to the uniform velocity profile approaching the solid surface, the effect of the viscous
boundary layer is to displace streamlines of the flow outside the boundary layer away from the wall. *
With this concept, we define δ = displacement thickness *
δ = distance the solid surface would have to be displaced to maintain the same mass flow rate as for non-viscous flow. From the development in the text, we obtain *
δ = ∫ 1 − 0
u dy U∞ *
Therefore, with an expression for the local velocity profile we can obtain δ = f(δ) Example: Given:
u y y 2 − =2 δ δ U∞
determine an expression for δ = f(δ)
Note that for this assumed form for the velocity profile: 1. At y = 0, u = 0
correct for no slip condition
2. At y = δ, u = U∞ correct for edge of boundary layer 3. The form is quadratic To simplify the mathematics, let η = y/δ, Therefore:
at y = 0, η = 0 ;
u = 2η − η 2 U∞
at y = δ , η = 1; dy = δ dη
which yields
2η2 η3 δ = ∫ 1 − 2 η + η δ d η = δ η − + 2 3 0 0 *
δ* = 3δ
Therefore, for flows for which the assumed quadratic equation approximates the velocity profile, streamlines outside of the boundary layer are displaced approximately according to the equation 1
δ* = 3δ This closely approximates flow for a flat plate. Key Point: When assuming a form for a velocity profile to use in the Von Karman analysis, make sure that the resulting equation satisfies both
surface and free stream boundary conditions as well as has a form that approximates u(y). Momentum Thickness: The second concept used in the Von Karman momentum analysis is that of momentum thickness
- θ The concept is similar to that of displacement thickness in that θ is related to the loss of momentum due to viscous effects in the boundary layer. Consider the viscous flow regions shown in the
adjacent figure. Define a control volume as shown and integrate around the control volume to obtain the net change in momentum for the control volume.
If D = drag force on the plate due to viscous flow, we can write - D = ∑ ( momentum leaving c.v. ) - ∑ ( momentum entering c.v. ) Completing an analysis shown in the text, we obtain
u u 1 − dy U U 0 ∞ ∞
ρ U∞2 θ
CD =
Using a drag coefficient defined as
We can also show that
CD =
D/A 1 2 ρ U ∞ 2
2 θ (L) L
where: θ(L) is the momentum thickness evaluated over the length L. Thus, knowledge of the boundary layer velocity distribution u = f(y) allows the drag coefficient to be determined.
Momentum integral: The final step in the Von Karman theory applies the previous control volume analysis to a differential length of surface. Performing an analysis similar to the previous analysis
for drag D we obtain
τw d U∞ d * 2 = δ U∞ + U∞ θ ) ( dx ρ dx
This is the momentum integral for 2-D, incompressible flow and is valid for laminar or turbulent flow.
d U∞ δ* d P δ U∞ =− dx ρ dx *
Therefore, this analysis also accounts for the effect of freestream pressure gradient. For a flat plate with non-accelerating flow, we can show that
P = const., U∞ = const.,
d U∞ =0 dx
Therefore, for a flat plate, non-accelerating flow, the Von Karman momentum integral becomes
τw d 2 2 dθ = U∞ θ )= U∞ ( ρ dx dx From the previous analysis and the assumed velocity distribution of
u y y 2 2 =2 = 2η − η − δ δ U∞ The wall shear stress can be expressed as
τw = µ
d u 2 µ U∞ 2 2 y = 2U∞ − 2 = δ d y w δ δ y=0
Also, with the assumed velocity profile, the momentum thickness θ can be evaluated as
u u 1 − dy U U 0 ∞ ∞
θ=∫ or
θ = ∫ (2η − η 2 )(1 − 2η + η 2 )δ d η = 0
2δ 15
We can now write from the previous equation for τw
τ w = ρ U∞2
dθ 2 dδ = ρ U∞2 d x 15 dx
Equating this result to Eqn. A we obtain
τw = or
δ dδ =
15 µ dx ρ U∞
2 d δ 2 µ U∞ ρ U∞2 = 15 dx δ which after integration yields
1/ 2
30 µ x δ= ρ U ∞
5.48 Re x
Note that the this result is within 10% of the exact result from Blasius flat plate theory. Since for a flat plate, we only need to consider friction drag (not pressure drag), we can write
C fx =
τ w (x ) 2 µ U∞ 1 1 1 2 = 2 δ ρ U ρ U ∞ ∞ 2 2
Substitute for δ to obtain
C fx =
2 µ U∞ Re 0.73 = Re x 5.48 12 ρ U 2∞ VII-15
Exact theory has a numerical constant of 0.664 compared with 0.73 for the previous result. It is seen that the von Karman integral theory provides the means to determine approximate expressions for
δ, τw, and Cf using only an assumed velocity profile. Solution summary:
1. Assume an analytical expression for the velocity profile for the problem. 2. Use the assumed velocity profile to determine the solution for the displacement thickness for the problem. 3. Use the
assumed velocity profile to determine the solution for the momentum thickness for the problem. 4. Use the previous results and the von Karman integral momentum equation to determine the solution for
the drag/wall shear for the problem.
Bluff Body, Viscous Flow Characteristics ( Immersed Bodies) In general, a body immersed in a flow will experience both externally applied forces and moments as a result of the flow about its external
surfaces. The typical terminology and designation of these forces and moments are given in the diagram shown below. The orientation of the axis for the drag force is typically along the principal
body axis, although in certain applications, this axis is aligned with the principal axis of the free stream, approach velocity U.
Since in many cases the drag force is aligned with the principal axis of the body shape and not necessarily aligned with the approaching wind vector. Review all data carefully to determine which
coordinate system is being used: body axis coordinate system or a wind axis coordinate system. These externally applied forces and moments are generally a function of a. Body geometry b. Body
orientation c. Flow conditions
These forces and moments are also generally expressed in the form of a non-dimensional force/moment coefficient, e.g. the drag coefficient:
CD =
FD /A 1 2 2 ρ U∞
It is noted that it is common to see one of three reference areas used depending on the application: 1. Frontal (projected) area: Used for thick, stubby, non-aerodynamic shapes, e.g., buildings,
cars, etc. 2. Planform (top view, projected) area: Used for flat, thin shapes, e.g., wings, hydrofoils, etc. 3. Wetted area: The total area in contact with the fluid. Used for surface ships, barges,
etc. The previous, flat plate boundary layer results considered only the contribution of viscous surface friction to drag forces on a body. However, a second major (and usually dominant) factor is
pressure or form drag. Pressure drag is drag due to the integrated surface pressure distribution over the body. Therefore, in general , the total drag coefficient of a body can be expressed as
C D = CD,press + C D,friction or
CD =
FD, total /A 1 2
ρ U2∞
FD,press. /A 1 2
ρ U 2∞
FD,friction /A 1 2
ρ U2∞
Which factor, pressure or friction drag, dominates depends largely on the aerodynamics (streamlining) of the shape and to a lesser extent on the flow conditions.
Typically the most important factor in the magnitude and significance of pressure or form drag is the boundary layer separation and resulting low pressure wake region associated with flow around non
- aerodynamic shapes. Consider the two shapes shown below: large pressure drag boundary layer separation Low Pressure Wake
High Pressure
no separated flow region
Low pressure drag
The flow around the streamlined airfoil remains attached, producing no boundary layer separation and comparatively small pressure drag. However, the flow around the less aerodynamic circular cylinder
separates, resulting in an area of high surface pressure on the front side and low surface pressure on the back side and thus significant pressure drag. This effect is shown very graphically in the
following figures from the text.
Fig. 7.12 Drag of a 2-D, streamlined cylinder The previous figure shows the effect of streamlining and aerodynamics on the relative importance of friction and pressure drag. While for a thin flat
plate (t/c = 0), all the drag is due to friction with no pressure drag, for a circular cylinder (t/c = 1), only 3% of the drag is due to friction with 97% due to pressure. Likewise for most bluff,
non-aerodynamic bodies, pressure (also referred to as form drag) is the dominant contributor to the total drag. However, the magnitude of the pressure (and therefore the total) drag can also be
changed by reducing the size of the low pressure wake region. One way to do this is to change the flow conditions from laminar to turbulent. This is illustrated in the following figures from the text
for a circular cylinder.
Fig. 7.13 Circular cylinder with (a) laminar separation and (b) turbulent separation Note that for the cylinder on the left, the flow is laminar, boundary layer separation occurs at 82o and the CD is
1.2, whereas for the cylinder on the right, the flow is turbulent and separation is delayed and occurs at 120o. The drag coefficient CD is 0.3, a factor of 4 reduction due to a smaller wake region
and reduced pressure drag.
It should also be pointed out that the friction drag for the cylinder on the right is probably greater (turbulent flow conditions) than for the cylinder on the left (laminar flow conditions).
However, since pressure drag dominates, the net result is a significant reduction in the total drag. The pressure distribution for laminar and turbulent flow over a cylinder is shown in Fig. 7.13c to
the right. The front-to-rear pressure difference is greater for laminar flow, thus greater drag. Finally, the effect of streamlining on total drag is shown very graphically with the sequence of
modifications in Fig. 7.15. Two observations can be made: (1) As body shape changes from a bluff body with fixed points of separation to a more aerodynamic shape, the effect of pressure drag and the
drag coefficient will decrease. Fig. 7.15 The effect of streamlining on total drag (2) The addition of surface area from (a) to (b) and (b) to (c) increases the friction drag, however, since pressure
drag dominates, the net result is a reduction in the drag force and the CD .
The final two figures show results for the drag coefficient for two and three dimensional shapes with various geometries. Table 7.2 CD for Two-Dimensional Bodies at Re ≥ 10 First note that all values
in Table 7.2 are for 2-D geometries, that is, the bodies are very long (compared to the cross-section dimensions) in the dimension perpendicular to the page. Key Point: Non – aerodynamic shapes with
fixed points of separations (sharp corners) have a single value of CD, irrespective of the value of the Reynolds number, e.g. square cylinder, half-tube, etc. Aerodynamic shapes generally have a
reduction in CD for a change from laminar to turbulent flow as a result of the shift in the point of boundary layer separation, e.g. elliptical cylinder.
Table 7.3 Drag of three-dimensional bodies at Re ≥ 10
The geometries in Table 7.3 are all 3-D and thus are finite perpendicular to the page. Similar to the results from the previous table, bluff body geometries with fixed points of separation have a
single CD, whereas aerodynamic shapes such as slender bodies of revolution have individual values of CD for laminar and turbulent flow. VII-23
In summary, one must remember that broad generalizations such as saying that turbulent flow always increases drag, drag coefficients always depend on Reynolds number, or increasing surface area
increases drag are not always valid. One must consider carefully all effects (viscous and pressure drag) due to changing flow conditions and geometry. Example: A square 6-in piling is acted on by a
water flow of 5 ft/s that is 20 ft deep. Estimate the maximum bending stress exerted by the flow on the bottom of the piling. Water: ρ = 1.99 slugs/ft
ν = 1.1 E – 5 ft /s
Assume that the piling can be treated as 2-D and thus end effects are negligible. Thus for a width of 0.5 ft, we obtain:
Re =
5 ft / s.5 ft = 2.3 E 5 1.1E − 5 ft 2 / s
In this range, Table 7.2 applies for 2D bodies and we read CD = 2 .1. The 2 frontal area is A = 20*0.5 = 10 ft 2
slug ft FD = 0.5 ρ U CD A = 0.5 *1.99 3 *52 2 * 2.1*10 ft 2 = 522 lbf ft s 2 ∞
For uniform flow, the drag should be uniformly distributed over the total length with the net drag located at the mid-point of the piling. Thus, relative to the bottom of the piling, the bending
moment is given by Mo = F * 0.5 L = 522 lbf* 10 ft = 5220 ft-lbf VII-24
From strength of materials, we can write
M o c 5220 ft − lbf * 0.25 ft = 1 = 251,000 psf = 1740 psi 3 3 I 0.5 ft * 0.5 ft 12 3
where c = distance to the neutral axis, I = moment of inertia = b h /12 Question: Since pressure acts on the piling and increases with increasing depth, why wasn’t a pressure load considered? Answer:
Static pressure does act on the piling, but it acts uniformly around the piling at every depth and thus cancels. Dynamic pressure is considered in the drag coefficients of Tables 7.2 and 7.3 and does
not have to be accounted for separately.
Ch. VIII Potential Flow and Computational Fluid Dynamics Review of Velocity-Potential Concepts This chapter presents examples of problems and their solution for which the assumption of potential flow
is appropriate. For low speed flows where viscous effects are neglected, the flow is irrotational and
V = ∇φ
∂φ ∂x
∂φ ∂y
∂φ ∂z
The continuity equation , ∇ ⋅ V = 0 , now reduces to
∂2 φ ∂ 2φ ∂ 2φ ∇ V= + + =0 ∂ x2 ∂ y2 ∂ z2 2
The momentum equation reduces to Bernoulli’s equation:
∂φ P 1 2 + + V + g z = const ∂t ρ 2 Review of Stream Function Concepts For plane incompressible flow in x-y coordinates a stream function exists such that
∂Ψ ∂y
v= −
∂Ψ ∂x
The condition of irrotationality reduces to Lapace’s equation for Ψ and
∂2 Ψ ∂ 2 Ψ + =0 ∂ x2 ∂ y2 VIII - 1
Elementary Plane-Flow Solutions Three useful plane-flow solutions that are very useful in developing more complex solutions are: Uniform stream, iU, in the x direction:
Ψ = Uy
φ = Ux
Line source or sink:
Ψ = mθ
φ = m ln r
Line vortex:
Ψ = − K ln r
φ = Kθ
In these expressions, the source strength, ‘m’ and vortex strength, ‘ K ‘, have the 2 dimensions of velocity times length, or [L /t]. If the uniform stream is written in plane polar coordinates, we
Ψ = U r sin θ
Uniform stream, iU:
φ = U r c os θ
For a uniform stream moving at an angle, a , relative to the x-axis, we can write
u = Uc os α =
∂Ψ ∂φ = ∂y ∂x
v = Usin α = −
∂ Ψ ∂φ = ∂x ∂y
After integration, we obtain the following expressions for the stream function and velocity potential:
Ψ = U (y cos α − x sin α )
φ = U (x cos α + y sin α )
Circulation The concept of fluid circulation is very useful in the analysis of certain potential flows, in particular those useful in aerodynamics analyses. Consider Figure 8.3 shown below:
VIII - 2
We define the circulation, Γ , as the counterclockwise line integral of the arc length, ds times the velocity component tangent to the closed curve, C, e.g.
Γ = ∫ V cos α d s = ∫ V ⋅ds c
Γ = ∫ (u dx + v dy + wdz ) c
For most flows, this line integral around a closed path, starting and stoping at the same point, yields Γ = 0. However, for a vortex flow for which
the integral yields
Γ = 2πK
= Kθ
An equivalent calculation can by made by defining a circular path of radius r around the vortex center to yield 2π
Γ = ∫ vθ d s = ∫
K r d φ = 2π K r
Superposition of Potential Flows Due to the mathematical character of the equations governing potential flows, the principle of superposition can be used to determine the solution of the flow which
results from combining two individual potential flow solutions. Several classic examples of this are presented as follows:
VIII - 3
1. Source m at ( -a,0) added to an equal sink at (+a, 0).
2a y ψ = − m tan 2 x + y2 − a2 −1
x + a) + y 2 1 ( φ = mln 2 (x − a)2 + y 2 2
The streamlines and potential lines are two families of orthogonal circles (Fig. 4.13). 2. Sink m plus a vortex K, both at the origin.
ψ = mθ − K ln r
φ = m ln r + K θ
The streamlines are logarithmic spirals swirling into the origin (Fig. 4.14). They resemble a tornado or a bathtub vortex. 3. Uniform steam i U∞ plus a source m at the origin (Fig. 4.15), the Ranking
half body. If the origin contains a source, a plane half-body is formed with its nose to the left as shown below. If the origin contains a sink, m < 0, the half-body nose is to the right.. For both
cases, the stagnation point is at a position a = m / U∞ away from the origin.
VIII - 4
Example 8.1 3
An offshore power plant cooling water intake has a flow rate of 1500 ft /s in water 30 ft deep as in Fig. E8.1. If the tidal velocity approaching the intake is 0.7 ft/s, (a) how far downstream does
the intake effect extend and (b) how much width of tidal flow in entrained into the intake? The sink strength is related to the volume flow, Q and water depth by 3
1500 ft / s m= = = 7.96 ft 2 / s 2π b 2 π 30 ft Q
The lengths a and L are given by 2
m 7.96 ft / s a= = = 11.4 ft 0.7 ft / s U∞ L = 2 π a = 2 π 11.4 ft = 71 ft Flow Past a Vortex Consider a uniform stream, U∞ flowing in the x direction past a vortex of strength K with the center at
the origin. By superposition the combined stream function is
ψ = ψ stream + ψ vortex = U∞ r sin θ − K ln r The velocity components of this flow are given by
vr =
1 ∂ψ = U∞ cos θ r ∂θ
vθ = −
∂ψ K = − U∞ sin θ + ∂r r
Setting v r and vθ = 0, we find the stagnation point at θ = 90û, r = a = K/ U∞ or (x,y) = (0,a).
VIII - 5
An Infinite Row of Vortices Consider an infinite row of vortices of equal strength K and equal spacing a. A single vortex, i , has a stream function given by
Fig. 8.7 Superposition of vortices ∞
ψ i = − K ∑ ln ri i=1
This infinite sum can also be expressed as
1 1 2π y 2π x ψ = − K ln cosh − cosh a 2 a 2 The resulting left and right flow above and below the row of vortices is given by
∂ψ ∂y
=± y >a
VIII - 6
πK a
Plane flow past Closed-Body Shapes Various types of external flows over a closed-body can be constructed by superimposing a uniform stream with sources, sinks, and vortices. Key Point: The body shape
will be closed only if the net source of the outflow equals the net sink inflow. Two examples of this are presented below. The Rankine Oval A Rankine Oval is a cylindrical shape which is long
compared to its height. It is formed by a source-sink pair aligned parallel to a uniform stream. The individual flows used to produce the final result and the combined flow field are shown in Fig.
8.9. The combined stream function is given by
ψ = U∞ y − m tan−1
2a y x2 + y2 − a2
ψ = U∞ r sin θ + m(θ1 − θ 2 )
Fig. 8.9 The Rankine Oval
The oval shaped closed body is the streamline, ψ = 0 . Stagnation points occur at the front and rear of the oval, x = ± L, y = 0 . Points of maximum velocity and minimum pressure occur at the
shoulders, x = 0, y = ± h . Key geometric and flow parameters of the Rankine Oval can be expressed as follows:
VIII - 7
L 2m = 1 + a U∞ a
h h/a = cot a 2 m / (U∞ a)
1/ 2
2 m / (U∞ a) umax =1 + 2 2 U∞ 1+ h /a As the value of the parameter m / (U ∞ a) is increased from zero, the oval shape increases in size and transforms from a flat plate to a circular cylinder at the
limiting case of m / (U ∞ a) = ∞ . Specific values of these parameters are presented in Table 8.1 for four different values of the dimensionless vortex strength, K / (U ∞ a ).
m / (U∞ a) 0.0 0.01 0.1 1.0 10.0 10.0 ∞
Table 8.1 Rankine-Oval Parameters
umax /U∞
0.0 0.31 0.263 1.307 4.435 14.130 ∞
1.0 1.10 1.095 1.732 4.458 14.177 ∞
∞ 32.79 4.169 1.326 1.033 1.003 1.000
1.0 1.020 1.187 1.739 1.968 1.997 2.000
Flow Past a Circular Cylinder with Circulation It is seen from Table 8.1 that as source strength m becomes large, the Rankine Oval becomes a large circle, much greater in diameter than the
source-sink spacing 2a. Viewed, from the scale of the cylinder, this is equivalent to a uniform stream plus a doublet. To add circulation, without changing the shape of the cylinder, we place a
vortex at the doublet center. For these conditions the stream function is given by
VIII - 8
r a2 ψ = U∞ sin θ r − − K ln r a Typical resulting flows are shown in Fig. 8.10 for increasing values of nondimensional vortex strength K / U∞ a .
Fig. 8.10 Flow past a cylinder with circulation for values of K / U∞ a of (a) 0, (b) 1.0, (c) 2.0, and (d) 3.0 Again the streamline ψ = 0 is corresponds to the circle r = a. As the counterclockwise
circulation Γ = 2 π K increases, velocities below the cylinder increase and velocities above the cylinder decrease. In polar coordinates, the velocity components are given by
1 ∂ψ a2 vr = = U∞ cos θ 1 − 2 r r ∂θ
∂ψ a2 K vθ = − = − U∞ sin θ 1 + 2 + ∂r r r For small K, two stagnation points appear on the surface at angles
VIII - 9
θ s or for which
sin θ s =
K 2U∞ a
Thus for K = 0, θ s = 0 and 180o. For K / U∞ a = 1, θ s = 30 and 150o . Figure 8.10c is the limiting case for which with K / U∞ a = 2, θ s = 90o and the two stagnation points meet at the top of the
The Kutta-Joukowski Lift Theorem The development in the text shows that from inviscid flow theory, The lift per unit depth of any cylinder of any shape immersed in a uniform stream equals to ρ U ∞ Γ
where Γ is the total net circulation contained within the body shape. The direction of the lift is 90o from the stream direction, rotating opposite to the circulation. This is the well known
Kutta-Joukowski lift theorem.
VIII - 10
IX. Compressible Flow Compressible flow is the study of fluids flowing at speeds comparable to the local speed of sound. This occurs when fluid speeds are about 30% or more of the local acoustic
velocity. Then, the fluid density no longer remains constant throughout the flow field. This typically does not occur with fluids but can easily occur in flowing gases. Two important and distinctive
effects that occur in compressible flows are (1) choking where the flow is limited by the sonic condition that occurs when the flow velocity becomes equal to the local acoustic velocity and (2) shock
waves that introduce discontinuities in the fluid properties and are highly irreversible. Since the density of the fluid is no longer constant in compressible flows, there are now four dependent
variables to be determined throughout the flow field. These are pressure, temperature, density, and flow velocity. Two new variables, temperature and density, have been introduced and two additional
equations are required for a complete solution. These are the energy equation and the fluid equation of state. These must be solved simultaneously with the continuity and momentum equations to
determine all the flow field variables. Equations of State and Ideal Gas Properties: Two equations of state are used to analyze compressible flows: the ideal gas equation of state and the isentropic
flow equation of state. The first of these describe gases at low pressure (relative to the gas critical pressure) and high temperature (relative to the gas critical temperature). The second applies
to ideal gases experiencing isentropic (adiabatic and frictionless) flow. The ideal gas equation of state is
P RT
In this equation, R is the gas constant, and P and T are the absolute pressure and absolute temperature respectively. Air is the most commonly incurred compressible flow gas and its gas constant is
Rair = 1717 ft2/(s2-oR) = 287 m2/(s2K).
Two additional useful ideal gas properties are the constant volume and constant pressure specific heats defined as
Cv =
du dT
and C p =
dh dT
where u is the specific internal energy and h is the specific enthalpy. These two properties are treated as constants when analyzing elemental compressible flows. Commonly used values of the specific
heats of air are: cv = 4293 ft2/(s2-oR) = 718 m2/(s2-K) and cp = 6010 ft2/(s2-oR) = 1005 m2/(s2-K). Additional specific heat relationships are
R = C p − Cv
Cp Cv
The specific heat ratio k for air is 1.4. When undergoing an isentropic process (constant entropy process), ideal gases obey the isentropic process equation of state:
P k
= constant
Combining this equation of state with the ideal gas equation of state and applying the result to two different locations in a compressible flow field yields
P2 T2 = P1 T1
k / ( k −1)
Note: The above equations may be applied to any ideal gas as it undergoes an isentropic process. Acoustic Velocity and Mach Number The acoustic velocity (speed of sound) is the speed at which an
infinitesimally small pressure wave (sound wave) propagates through a fluid. In general, the acoustic velocity is given by IX-2
a2 =
The process experienced by the fluid as a sound wave passes through it is an isentropic process. The speed of sound in an ideal gas is then given by
a = k RT The Mach number is the ratio of the fluid velocity and speed of sound,
Ma =
V a
This number is the single most important parameter in understanding and analyzing compressible flows. Mach Number Example: An aircraft flies at a speed of 400 m/s. What is this aircraft’s Mach number
when flying at standard sea-level conditions (T = 289 K) and at standard 15,200 m (T = 217 K) atmosphere conditions?
k RT = (1.4)(287)( 289) = 341m / s and at 15,200 m, a = (1.4)(287)( 217) = 295 m / s . The aircraft Mach At standard sea-level conditions, a = numbers are then
V 400 = = 1.17 a 341 V 400 15,200 m : Ma = = = 1.36 a 295
sea − level : Ma =
Note: Although the aircraft speed did not change, the Mach number did change because of the change in the local speed of sound.
Ideal Gas Steady Isentropic Flow When the flow of an ideal gas is such that there is no heat transfer (i.e., adiabatic) or irreversible effects (e.g., friction, etc.), the flow is isentropic. The
steady-flow energy equation applied between two points in the flow field becomes
V12 V22 h1 + = h2 + = ho = constant 2 2 where h0 is called the stagnation enthalpy that remains constant throughout the flow field. Observe that the stagnation enthalpy is the enthalpy at any point
in an isentropic flow field where the fluid velocity is zero or very nearly so. The enthalpy of an ideal gas is given by h = Cp T over reasonable ranges of temperature. When this is substituted into
the adiabatic, steady-flow energy equation, we see that h o = C p To = constant and
To k −1 =1 + Ma 2 T 2 Thus, the stagnation temperature To remains constant throughout an isentropic or adiabatic flow field and the relationship of the local temperature to the field stagnation
temperature only depends upon the local Mach number. Incorporation of the acoustic velocity equation and the ideal gas equations of state into the energy equation yields the following useful results
for steady isentropic flow of ideal gases.
To k −1 =1 + Ma 2 T 2 1/2 ao To 1 / 2 k− 1 2 = = 1+ Ma a T 2 k / (k −1 ) Po To k / ( k −1) k −1 2 = = 1+ Ma P T 2 o
T = o T
1/ ( k −1)
1/ ( k −1) k −1 2 = 1 + Ma 2
The values of the ideal gas properties when the Mach number is 1 (i.e., sonic flow) are known as the critical or sonic properties and are given by
To k −1 = 1+ * T 2 ao To 1 / 2 k −1 1 / 2 = = 1+ a * T* 2 Po To k /( k −1) k −1 k /( k −1 ) = = 1+ P* T * 2 o *
T = o* T
1/ ( k −1)
k −1 1/ ( k −1) = 1 + 2
given by Isentropic Flow Example: Air flowing through an adiabatic, frictionless duct is supplied from a large supply tank in which P = 500 kPa and T = 400 K. What are the Mach number Ma. the
temperature T, density _, and fluid V at a location in this duct where the pressure is 430 kPa? The pressure and temperature in the supply tank are the stagnation pressure and temperature since the
velocity in this tank is practically zero. Then, the Mach number at this location is
2 Po ( k −1) / k Ma = − 1 k − 1 P 2 500 Ma = 0.4 430 Ma = 0.469 and the temperature is given by
− 1
To k −1 1+ Ma 2 2 400 T= 1 + 0.2 ( 0.469) 2 T = 383 K T=
The ideal gas equation of state is used to determine the density,
P 430,000 = = 3.91kg / m3 R T ( 287) (383)
Using the definition of the Mach number and the acoustic velocity,
V = Ma k RT = 0.469 (1.4)(287)( 383) = 184 m / s Solving Compressible Flow Problems Compressible flow problems come in a variety of forms, but the majority of them can be solved by 1. Use the
appropriate equations and reference states (i.e., stagnation and sonic states) to determine the Mach number at all the flow field locations involved in the problem. 2. Determine which conditions are
the same throughout the flow field (e.g., the stagnation properties are the same throughout an isentropic flow field). 3. Apply the appropriate equations and constant conditions to determine the
necessary remaining properties in the flow field. 4. Apply additional relations (i.e., equation of state, acoustic velocity, etc.) to complete the solution of the problem. Most compressible flow
equations are expressed in terms of the Mach number. You can solve these equations explicitly by rearranging the equation, by using tables, or by programming them with spreadsheet or EES software.
Isentropic Flow with Area Changes All flows must satisfy the continuity and momentum relations as well as the energy and state equations. Application of the continuity and momentum equations to a
differential flow (see textbook for derivation) yields:
dV 1 dA = V Ma 2 − 1 A This result reveals that when Ma < 1 (subsonic flow) velocity changes are the opposite of area changes. That is, increases in the fluid velocity require that the area decrease
in the direction of the flow. For supersonic flow (Ma > 1), the area must increase in the direction of the flow to cause an increase in the velocity. Changes in the fluid velocity dV can only be
finite in sonic flows (Ma = 1) when dA = 0. The effect of the geometry upon velocity, Mach number, and pressure is illustrated in Figure 1 below.
Figure 1
˙ = Combining the mass flow rate equation m preceding isentropic flow equations yields *
A V = constant with the
1 /( k −1)
2 k −1 2 = 1 + Ma 2 k + 1
V* 1 2 k −1 2 = 1 + Ma V Ma k + 1 2
( k + 1) / [ 2 (k −1 )]
A 1 1 + 0.5( k − 1) Ma2 = A* Ma 0.5(k +1)
where the sonic state (denoted with *) may or may not occur in the duct. If the sonic condition does occur in the duct, it will occur at the duct minimum or maximum area. If the sonic condition
occurs, the flow is said to be choked since ˙ = A V = * A* V * is the maximum mass flow rate the the mass flow rate m duct can accommodate without a modification of the duct geometry. Review Example
9.4 of the textbook. Normal Shock Waves Under the appropriate conditions, very thin, highly irreversible discontinuities can occur in otherwise isentropic compressible flows. These discontinuities
are known as shock waves which when they are perpendicular to the flow velocity vector are called normal shock waves. A normal shock wave in a one-dimensional flow channel is illustrated in Figure 2.
Figure 2
Application of the second law of thermodynamics to the thin, adiabatic normal shock wave reveals that normal shock waves can only cause a sharp rise in the gas pressure and must be supersonic
upstream and subsonic downstream of the normal shock. Rarefaction waves that result in a decrease in pressure and increase in Mach number are impossible according to the second law. Application of
the conservation of mass, momentum, and energy equations along with the ideal gas equation of state to a thin, adiabatic control volume surrounding a normal shock wave yields the following results.
k −1)Ma12 + 2 ( Ma = , Ma1 > 1 2 k Ma12 − ( k −1) 2 2
P2 1 + k Ma12 = P1 1 + k Ma 22 2 1
V1 ( k + 1) Ma12 = = V2 ( k − 1) Ma12 + 2
To1 = To 2 2 T2 2 2 k Ma1 − ( k − 1) = [2 + (k − 1) Ma1 ] T1 ( k +1) 2 Ma12
Po 2 = Po1
k / (k −1 )
o2 o1
( k + 1) Ma12 = 2 2 + ( k − 1) Ma1
A2* Ma 2 2 + (k −1) Ma12 = A1* Ma1 2 + (k − 1) Ma22
1/ ( k − 1)
k +1 2 2 k Ma1 − (k − 1)
( k + 1) / [ 2 ( k −1) ]
When using these equations to relate conditions upstream and downstream of a normal shock wave, keep the following points in mind:
1. Upstream Mach numbers are always supersonic while downstream Mach numbers are subsonic. 2. Stagnation pressures and densities decrease as one moves downstream across a normal shock wave while the
stagnation temperature remains constant. 3. Pressures increase greatly while temperature and density increase moderately across a shock wave in the downstream direction. 4. The effective throat area
increases across a normal shock wave in the downstream direction. 5. Shock waves are very irreversible causing the specific entropy downstream of the shock wave to be greater than the specific
entropy upstream of the shock wave. Moving normal shock waves such as those caused by explosions, spacecraft reentering the atmosphere, and others can be analyzed as stationary normal shock waves by
using a frame of reference that moves at the speed of the shock wave in the direction of the shock wave. Converging-Diverging Nozzle Example: Also see Example 9.6 of textbook Air is supplied to the
converging-diverging nozzle shown here from a large tank where P = 2 Mpa and T = 400 K. A normal shock wave in the diverging section of this nozzle forms at a point Po1 = Po2 = 2 MPa where the
upstream Mach number is 1.4. The ratio of the nozzle exit area to the throat area is 1.6. Determine (a) the Mach number downstream of the shock wave, (b) the Mach number at the nozzle exit, (c) the
pressure at the nozzle exit, and (d) the temperature at the nozzle exit. This flow is isentropic from the supply tank (1) to just upstream of the normal shock (2) and also from just downstream of the
shock (3) to the exit (4). Stagnation temperatures do not change in isentropic flows or across shock waves, To1 = To 2 = To3 = To 4 = 400 K . Stagnation pressures do not change in isentropic flows,
Po1 = Po2 = 2 MPa and Po 3 = Po 4 , but stagnation pressures change across shocks, Po 2 > Po3 . IX-10
Based upon the Mach number at 2 and the isentropic relations, 2 A2 A3 A2 1 (1 + 0.2 Ma2 ) = = = = 1.115 At At At* Ma2 1.728 3
The normal shock relations can be used to work across the shock itself. The answer to (a) is then:
(k − 1) Ma22 + 2 Ma3 = 2 2 k Ma2 − ( k − 1)
( 0.4)(1.4 )2 + 2 = 2 2 (1.4) (1.4 ) − 0.4
= 0.740
Continuing to work across the shock, k / ( k −1)
( k + 1) Ma22 Po 4 = Po3 = Po 2 2 2 + (k − 1) Ma2
1/ ( k −1)
k+ 1 2 k Ma 2 − (k − 1) 2
(2.4 )( 0.74) 2 2.4 Po 4 = Po3 = 2 2 2 2 + ( 0.4) ( 0.74) 2 (1.4) ( 0.74) − 0.4 A3* Ma3 2 + (k −1) Ma22 = A2* Ma 2 2 + (k − 1) Ma32
= 1.92 MPa
( k + 1) / [ 2 ( k −1) ]
= 1.044
Now, we know A4/At, and the flow is again isentropic between states 3 and 4. Writing an expression for the area ratio between the exit and the throat, we have
A4 A4 A4* A3* A2* A4 = 1.6 = * * * = * (1)(1.044)(1.115) At A4 A3 A2 At A4 Solving for
A4 we obtain A4*
A4 = 1.374 A4*
Using a previously developed equation for choked, isentropic flow, we can write
A4 1 1 + 0.5( k −1) Ma2 = 1.374 = A4* Ma 0.5( k + 1) or
(k + 1) / [ 2 ( k −1 )]
2 1 (1 + 0.2Ma4 ) 1.374 = Ma4 1.728
The solution of this equation gives answer (b) Ma4 = 0.483. Now that the Mach number at 4 is known, we can proceed to apply the isentropic relations to obtain answers (c) and (d).
P4 =
Po 4 1.92 MPa = k / ( k −1 ) 1 + 0.2(0.483) 2 [1 + 0.5 (k − 1) Ma42 ]
T4 =
= 1.637 MPa
To4 400 K = 382 K 2 = 1 + 0.5( k − 1) Ma4 1 + 0.2( 0.483)2
Note: Observe how the sonic area downstream from the shock is not the same as upstream of the shock. Also, observe the use of the area ratios to determine the Mach number at the nozzle exit. The
following steps can be used to solve most one-dimensional compressible flow problems. 1. Clearly identify the flow conditions:e.g., isentropic flow, constant stagnation temperature, constant
stagnation pressure, etc. 2. Use the flow condition relationships, tables, or software to determine the Mach number at major locations in the flow field. 3. Once the Mach number is known at the
principal flow locations, one can proceed to use the flow relations, tables, or software to determine other flow properties such as fluid velocity, pressure, and temperature. This may require the
reduction of property ratios to the product of several ratios, as was done with the area ratio in the above example to obtain the answer.
Operation of Converging-Diverging Nozzles A converging-diverging nozzle like that shown in Figure 3 can operate in several different modes depending upon the ratio of the discharge and supply
pressure Pd/Ps. These modes of operation are illustrated on the pressure ratio – axial position diagram of Figure 3.
Figure 3
Mode (a) The flow is subsonic throughout the nozzle, supply, and discharge chambers. Without friction, this flow is also isentropic and the isentropic flow equations may be used throughout the
nozzle. Mode (b) The flow is still subsonic and isentropic throughout the nozzle and chambers. The Mach number at the nozzle throat is now unity. At the throat, the flow is sonic, the throat is
choked, and the mass flow rate through the nozzle has reached its upper limit. Further reductions in the discharge tank pressure will not increase the mass flow rate any further. Mode (c) A shock
wave has now formed in the diverging section of the nozzle. The flow is subsonic before the throat, same as mode (b), the throat is choked, same as mode (b), and the flow is supersonic IX-13
and accelerating between the throat and just upstream of the shock. The flow is isentropic between the supply tank and just upstream of the shock. The flow downstream of the shock is subsonic and
decelerating. The flow is also isentropic downstream of the shock to the discharge tank. The flow is not isentropic across the shock. Isentropic flow methods can be applied upstream and downstream of
the shock while normal shock methods are used to relate conditions upstream to those downstream of the shock. Mode (d) The normal shock is now located at the nozzle exit. Isentropic flow now exists
throughout the nozzle. The flow at the nozzle exit is subsonic and adjusts to flow conditions in the discharge tank, not the nozzle. Isentropic flow methods can be applied throughout the nozzle. Mode
(e) A series of two-dimensional shocks are established in the discharge tank downstream of the nozzle. These shocks serve to decelerate the flow. The flow is isentropic throughout the nozzle, same as
mode (d). Mode (f) The pressure in the discharge tank equals the pressure predicted by the supersonic solution of the nozzle isentropic flow equations. The pressure ratio is known as the supersonic
design pressure ratio. Flow is isentropic everywhere in the nozzle, same as mode (d) and (e), and in the discharge tank. Mode (g) A series of two-dimensional shocks are established in the discharge
tank downstream of the nozzle. These shocks serve to decelerate the flow. The flow is isentropic throughout the nozzle, same as modes (d), (e), and (f).
Review Example 9.9 of the textbook.
Adiabatic, Constant Duct Area Compressible Flow with Friction When compressible fluids flow through insulated, constant-area ducts, they are subject to Moody-like pipe-friction which can be described
by an average DarcyWeisbach friction factor f . Application of the conservation of mass, momentum, and energy principles as well as the ideal gas equation of state yields the following set of working
f L* 1− Ma 2 k + 1 k + 1) Ma 2 ( = + ln D k Ma2 2k 2 + (k − 1) Ma 2 P 1 ( k + 1) = P* Ma 2 + ( k − 1) Ma 2
V* 1 2 + ( k − 1) Ma 2 = * = V Ma k +1 T a ( k +1) = = 2 T * a* 2 + ( k −1) Ma 2
P = Po*
o * o
1 2 + (k −1) Ma2 = Ma k +1
( k +1) / [ 2 ( k −1) ]
where the asterisk state is the sonic state at which the flow Mach number is one. This state is the same throughout the duct and may be used to relate conditions at one location in the duct to those
at another location. The length of the duct enters these calculations by
f ∆L f L* f L* = − D 1 D 2 D Thus, given the length ∆L of the duct and the Mach number at the duct entrance or exit, the Mach number at the other end (or location) of the duct can
be determined.
Compressible Flow with Friction Example: Air enters a 0.01-m-diameter duct ( f = 0.05) with Ma = 0.05. The pressure and temperature at the duct inlet are 1.5 MPa and 400 K. What are the (a) Mach
number, (b) pressure, and (c) temperature in the duct 50 m from the entrance? At the duct entrance, with f = 0.05, D = 0.01 m, and Ma = 0.05, we obtain
f L* 1 − Ma 2 k +1 ( k + 1) Ma 2 + ln = D 1 k Ma 2 2k 2 + (k − 1) Ma 2 1 f L* 1− 0.052 2.4 (2.4) 0.052 + ln = 280 = D 1 1.4 (0.05)2 2.8 2 + ( 0.4) 0.052 1
Then, at the duct exit we obtain
f L* f L* f ∆L ( 0.05) 50 = 280 − = 30 = − D 2 D 1 D 0.01 We can not write for the duct exit that
1− Ma2 k + 1 f L* ( k + 1) Ma 2 ln = 30 = 2 + D 2 2k 2 + (k − 1) Ma 2 2 k Ma or
1− Ma22 2.4 2.4 Ma22 30 = + ln 1.4 Ma22 2.8 2 + 0.4 Ma22 The solution of the second of these equations gives answer (a) Ma2 = 0.145. Writing the following expression for pressure ratios yields for
P2 P2* P1* P2 = P1 * * P2 P1 P1
1 P2 = (1.5) Ma2
( k +1) Ma1 2 + ( k −1) Ma12 2 + ( k −1) Ma 2 (1) 1 k +1 2 1/2
1 2.4 0.05 2 + ( 0.4) 0.052 P2 = (1.5) (1) 0.145 2 + (0.4 ) 0.1452 1 2.4
= 0.516
Application of the temperature ratios yields answer (c),
T1* T2* T2 2 + (k − 1) Ma12 2 + ( 0.4) 0.052 T2 = T1 = 400 = 400 = 399 T1 T1* T2* 2 + ( k − 1) Ma 22 2 + ( 0.4) 0.1452 This example demonstrates what happens when the flow at the inlet to the duct is
subsonic, the Mach number increases as the duct gets longer. When the inlet flow is supersonic, the Mach number decreases as the duct gets longer. A plot of the specific entropy of the fluid as a
function of the duct Mach number (length) is presented in Figure 4 for both subsonic and supersonic flow.
Figure 4
These results clearly illustrate that the Mach number in the duct approaches unity as the length of the duct is increased. Once the sonic condition exists at the duct exit, the flow becomes choked.
This figure also demonstrates that the flow can never proceed from subsonic to supersonic (or supersonic to subsonic) flow, as this would result in a violation of the second law of thermodynamics.
Other compressible flows in constant area ducts such as isothermal flow with friction and frictionless flow with heat addition may be analyzed in a similar manner using the equations appropriate to
each flow. Many of these flows also demonstrate choking behavior.
Oblique Shock Waves Bodies moving through a compressible fluid at speeds exceeding the speed of sound create a shock system shaped like a cone. The half-angle of this shock cone is given by
= sin−1
1 Ma
This angle is known as the Mach angle. The interior of the shock cone is called the zone of action. Inside the zone of action, it is possible to hear any sounds produced by the moving body. Outside
the Mach cone, in what is known as the zone of silence, sounds produced by the moving body cannot be heard. An oblique shock wave at angle with respect to the approaching compressible fluid whose
Mach number is supersonic is shown in Figure 5. Observe that the streamlines (parallel to the velocity vector) have been turned by the deflection angle by passing through the oblique shock wave.
Figure 5
This flow is readily analyzed by considering the normal velocity components Vn 1 = V1 sin and Vn2 = V2 sin ( − ) and the tangential components Vt 1 and Vt 2 . Application of the momentum principle in
the tangential direction (along which there are no pressure changes) verifies that tt 12 = VV
Vt 1 = Vt 2
By defining the normal Mach numbers as
Man1 =
Vn1 = Ma1 sin a1
and Man2 =
Vn 2 = Ma2 sin ( − a2
The simultaneous solution of the conservation of mass, momentum, and energy equations in the normal direction along with the ideal gas equation of state are the same as those of the normal shock wave
with Ma1 replaced with Man1 and Ma2 replaced with Man2. In this way, all the results developed in the normal shock wave section can be applied to two-dimensional oblique shock waves. Oblique Shock
Example: A two-dimensional shock wave is created at the leading edge of an aircraft flying at Ma = 1.6 through air at 70 kPa, 300 K. If this oblique shock forms a 55o angle with respect to the
approaching air, what is (a) the Mach number of the flow after the oblique shock (this is not the normal Mach number) and (b) the streamline deflection angle ? The velocity of the fluid upstream of
the oblique shock wave is
V1 = Ma1 a1 = Ma1 k R T = 1.6 (1.4)( 287)( 300) = 556 m / s whose components are
Vn 1 = V1 sin
= 5 5 6 s i n 5 5= 455 m / s
Vt 1 = Vt2 = V1 cos
= 556cos55= 319 m / s
The upstream normal Mach number is then
Man1 = Ma1 sin
= 1 . 6 s i n 5 5= 1.311
and the downstream normal Mach number is
( k −1)Man21 + 2 Man2 = 2 2 k Man 1 − ( k −1)
(0.4 )(1.311) 2 + 2 = 2 2 (1.4) (1.311) − 0.4
= 0.780
and the downstream temperature is
2 k Man12 − ( k − 1) 2 T2 = T1 [(k − 1) Man1 + 2] 2 2 ( k +1 ) Ma n1 2 (1.4) 1.3112 − 0.4 2 T2 = 300 [ (0.4)1.311 + 2] = 359 K 2 2 ( 2.4 ) 1.311 Now, the downstream normal
velocity is
Vn 2 = Man 2 a2 = Ma n2
k R T2 = 0.780 (1.4)(287)( 359) = 296 m / s
and the downstream fluid velocity is
V2 = Vn22 + Vt 22 = 296 2 + 319 2 = 435 m / s and the downstream Mach number is
Ma2 =
V2 435 = = 1.15 a2 (1.4 )( 287)(359)
According to the geometry of Figure 5,
− tan −1
Vn 2 296 = 55 − tan −1 = 12.1 Vt 2 319
Other downstream properties can be calculated in the same way as the downstream temperature by using the normal Mach numbers in the normal shock relations.
Prandl-Meyer Expansion Waves The preceding section demonstrated that when the streamlines of a supersonic flow are turned into the direction of the flow an oblique compression shock wave is formed.
Similarly, when the streamlines of a supersonic flow are turned away from the direction of flow as illustrated in Figure 6, an expansion wave system is established. Unlike shock waves (either normal
or oblique) which form a strong discontinuity to change the flow conditions, expansion waves are a system of infinitesimally weak waves distributed in such a manner as required to make the required
changes in the flow conditions.
Figure 6
The Mach waves that accomplish the turning of supersonic flows form an angle −1 with respect to the local flow velocity equal to the Mach angle = sin (1 /Ma) and are isentropic. Application of the
governing conservation equations and equation of state to an infinitesimal turning of the supersonic flow yields
k +1 − ( Ma) = (Ma) = k − 1
2 −1 ( k −1)( Ma − 1) tan k +1
− tan −1 ( Ma 2 −1)
where (Ma) is the Prandl-Meyer expansion function. The overall change in the flow angle as a supersonic flow undergoes a Prandl-Meyer expansion is then
( Ma1 ) − ( Ma2 )
where 1 refers to the upstream condition and 2 refers to the downstream condition. The flow through a Prandl-Meyer expansion fan is isentropic flow. The isentropic flow equations can then be used to
relate the fluid properties upstream and downstream of the expansion fan. Example: Air at 80 kPa, 300 K with a Mach number of 1.5 turns the sharp corner of an airfoil as shown here. Determine the
angles of the initial and final Mach waves, and the downstream pressure and temperature of this flow.
The initial angle between the flow velocity vector and the Prandtl-Meyer fan is the Mach angle. −1 1 = sin
1 1 = sin −1 = 41.8 0 Ma1 1.5
The upstream Prandtl-Meyer function is
( Ma1 ) = k +1 k −1
2 −1 ( k −1)( Ma1 −1) tan k +1
2 2.4 1 / 2 −1 ( 0.4)(1.5 −1) ( Ma1 ) = tan 0.4 2.4
− tan −1 ( Ma12 − 1)
− tan−1 (1.52 − 1)
( Ma1 ) = 11.90 0 The downstream Prandtl-Meyer function is then
( Ma2 ) = ( Ma1 ) − ∆
= 11.90 − 100 = 1.90 0
Solving the Prandtl-Meyer function gives the downstream Mach number Ma2 = 1.13 . The downstream Mach angle is then 2 = 62.2 0 . According to the geometry of the above figure, 2
− ∆ = 62.20 − 10 0 = 52.2 0
Since T0 and P0 remain constant, the isentropic flow relations yield
k −1 2 2 Ma 1 T01 T2 1+ 0.2(1.5) 2 T2 = T1 = T1 = 300 = 346 K k −1 2 T1 T02 1 + 0.2 ( 1.13 ) 1+ Ma2 2 1+
1+ k −1 Ma 2 1 P01 P2 2 P2 = P1 = P1 k −1 2 P1 P02 1+ Ma 2 2
k / ( k −1)
1 + 0.2(1.5)2 = 80 1+ 0.2(1.13)
= 132 MPa
Students are encouraged to examine the flow visualization photographs in Ch 9.
Ch. 10 Open-Channel Flow Previous internal flow analyses have considered only closed conduits where the fluid typically fills the entire conduit and may be either a liquid or a gas. This chapter
considers only partially filled channels of liquid flow referred to as open-channel flow. Open-Channel Flow: Flow of a liquid in a conduit with a free surface. Open-channel flow analysis basically
results in the balance of gravity and friction forces. One Dimensional Approximation While open-channel flow can, in general, be very complex ( three dimensional and transient), one common
approximation in basic analyses is the One-D Approximation: The flow at any local cross section can be treated as uniform and at most varies only in the principal flow direction.
This results in the following equations. Conservation of Mass (for ρ = constant) Q = V(x) A(x) = constant Energy Equation 2
V1 V + Z1 = 2 + Z 2 + h f 2g 2g
The equation in this form is written between two points ( 1 – 2 ) on the free surface of the flow. Note that along the free surface, the pressure is a constant, is equal to local atmospheric
pressure, and does not contribute to the analysis with the energy equation. The friction head loss hf is analogous to the head loss term in duct flow, Ch. VI, and can be represented by 2
x − x Vavg hf = f 2 1 D h 2g
where P = wetted perimeter Dh = hydraulic diameter =
4A P
Note: One of the most commonly used formulas uses the hydraulic radius:
1 A R h = Dh = 4 P Flow Classification by Depth Variation The most common classification method is by rate of change of free-surface depth. The classes are summarized as 1. Uniform flow (constant
depth and slope) 2. Varied flow a. Gradually varied (one-dimensional) b. Rapidly varied (multidimensional) Flow Classification by Froude Number: Surface Wave Speed A second classification method is
by the dimensionless Froude number, which is a dimensionless surface wave speed. For a rectangular or very wide channel we have
Fr = and
V V = 1/2 co (gy)
y is the water depth
co = the speed of a surface wave as the wave height approaches zero.
There are three flow regimes of incompressible flow. These have analogous flow regimes in compressible flow as shown below: X-2
Incompressible Flow Fr < 1 Fr = 1 Fr > 1
Compressible Flow
subcritical flow critical flow supercritical flow
Ma < 1 Ma = 1 Ma > 1
subsonic flow sonic flow supersonic flow
Hydraulic Jump Analogous to a normal shock in compressible flow, a hydraulic jump provides a mechanism by which an incompressible flow, once having accelerated to the supercritical regime, can return
to subcritical flow. This is illustrated by the following figure.
Fig. 10.5 Flow under a sluice gate accelerates from subcritical to critical to supercritical and then jumps back to subcritical flow. The critical depth
Q yc = 2 b g
is an important parameter in open-
channel flow and is used to determine the local flow regime (Sec. 10.4). Uniform Flow; the Chezy Formula Uniform flow
1. Occurs in long straight runs of constant slope 2. The velocity is constant with V = Vo 3. Slope is constant with So = tan θ
From the energy equation with V1 = V2 = Vo, we have
h f = Z1 − Z 2 = S o L Since the flow is fully developed, we can write from Ch. VI 1/2
L Vo hf = f Dh 2g
8g 1/2 1/2 Vo = Rh So f
8g For fully developed, uniform flow, the quantity is a constant f and can be denoted by C. The equations for velocity and flow rate thus become
Vo = C R h So 1/2
Q = CA R h So 1/2
The quantity C is called the Chezy coefficient, and varies from 60 ft /s for small 1/2 1/2 rough channels to 160 ft /s for large rough channels (30 to 90 m /s in SI). The Manning Roughness
Correlation The friction factor f in the Chezy equations can be obtained from the Moody chart of Ch. VI. However, since most flows can be considered fully rough, it is appropriate to use Eqn 6.64:
fully rough flow:
3.7Dh f ≈ 2.0 log ε
However, most engineers use a simple correlation by Robert Manning: S.I. Units
Vo (m/s ) ≈
B.G. Units
Vo (ft/s ) ≈
[R (m)] n h
[R (ft )] n h
S1/2 o
S1/2 o
where n is a roughness parameter given in Table 10.1 and is the same in both systems of units and α is a dimensional constant equal to 1.0 in S.I. units and 1.486 in B.G. units. The volume flow rate
is then given by
Uniform flow
Q = Vo A ≈
α n
1/2 A R 2/3 h So
Table 10.1 Experimental Values for Manning’s n Factor
Example 10.1 Given: Rectangular channel, finished concrete, slope = 0.5˚ water depth: y = 4 ft, width: b = 8 ft Find: 3 Volume flow rate (ft /s)
4 ft
8 ft
For the given conditions:
n = 0.012
A = b y = (8 ft) (4 ft) = 32 ft
A 32 ft Rh = = = 2 ft P 16 ft
So = tan 0.5˚ = 0.0873 P = b + 2 y = 8 + 2 (4) = 16 ft
D h = 4 R h = 8 ft
Using Manning’s formula in BG units, we obtain for the flow rate
1.486 2/3 32 ft 2 )(2 ft ) (0.00873)1.2 ≈ 590 ft 3 /s ans. ( 0.012
Alternative Problem The previous uniform problem can also be formulated where the volume flow rate Q is given and the fluid depth is unknown. For these conditions, the same basic equations are used
and the area A and hydraulic radius Rh are expressed in terms of the unknown water depth yn. The solution is then obtained using iterative or systematic trial and error techniques that are available
in several math analysis/ math solver packages such as EES (provided with the text) or Mathcad ®.
Uniform Flow in a Partly Full, Circular Pipe Fig. 10.6 shows a partly full, circular pipe with uniform flow. Since frictional resistance increases with wetted perimeter, but volume flow rate
increases with cross sectional flow area, the maximum velocity and flow rate occur before the pipe is completely full. For this condition, the geometric properties of the flow are given by the
equations below. Fig. 10.6 Uniform Flow in a Partly Full, Circular Channel
sin2θ 2 A=R θ− 2
P = 2 Rθ
Rh =
R sin2θ 1− 2θ 2
The previous Manning formulas are used to predict Vo and Q for uniform flow when the above expressions are substituted for A, P, and Rh.
α R
sin2θ Vo ≈ 1− n 2 2θ
2 /3 1/2
Q = Vo R
sin2θ θ− 2
These equations have respective maxima for Vo and Q given by
Vmax = 0.718 Q max = 2.129
α n
α n
R 2/3 S1/2 o
at θ = 128.73Þ and y = 0.813 D
R 8/3 S1/2 o
at θ = 151.21Þ and y = 0.938 D
Efficient Uniform Flow Channels A common problem in channel flow is that of finding the most efficient lowresistance sections for given conditions. This is typically obtained by maximizing Rh for a
given area and flow rate. This is the same as minimizing the wetted perimeter. Note: Minimizing the wetted perimeter for a given flow should minimize the frictional pressure drop per unit length for
a given flow. It is shown in the text that for constant value of area A and α = cot θ, the minimum value of wetted perimeter is obtained for
A = y 2 2 (1+ α 2 ) − α 1/ 2
P = 4 y (1+ α
2 1/ 2
− 2α y
Rh =
1 y 2
Note: For any trapezoid angle, the most efficient cross section occurs when the hydraulic radius is one-half the depth. For the special case of a rectangle (a = 0, q = 90˚), the most efficient cross
section occurs with
A = 2y
P = 4y
Rh =
1 y 2
b = 2y
Best Trapezoid Angle The general equations listed previously are valid for any value of α. For a given, fixed value of area A and depth y the best trapezoid angle is given by α = cot θ =
θ = 60
Example 10.3 What are the best dimensions for a rectangular brick channel designed to carry 5 3 m /s of water in uniform flow with So = 0.001? 2
Taking n = 0.015 from Table 10.1, A = 2 y , and Rh = 1/2 y ; Manning’s formula is written as 2/3 1.0 2 1 1/ 2 or 5 m / s = 2 y ) y (0.001) ( 2 0.015
1.0 2/3 1/2 Q ≈ A R h So n
This can be solved to obtain
8/ 3
= 1.882 m
y = 1.27 m
The corresponding area and width are
A = 2 y 2 = 3.21m2
A = 2.53 m y
Note: The text compares these results with those for two other geometries having the same area.
Specific Energy: Critical Depth 2
One useful parameter in channel flow is the specific energy E, where y is the local water depth.
V E = y+ 2g
Defining a flow per unit channel width as q = Q/b we write
q E = y+ 2 g y2
Fig. 10.8b is a plot of the water depth y vs. the specific energy E. The water depth for which E is a minimum is referred to as the critical depth yc.
Fig. 10.8 Specific Energy Illustration
q2 y = yc = g
Emin occurs at
Q2 = 2 b g
E min =
The value of Emin is given by
3 y 2 c
At this value of minimum energy and minimum depth we can write
Vc = (g y c ) = Co 1/2
and Fr = 1
Depending on the value of Emin and V, one of several flow conditions can exist. For a given flow, if
E < Emin
No solution is possible
E = Emin
Flow is critical, y = yc, V = Vc
E > Emin , V < Vc
Flow is subcritical, y > yc ,disturbances can propagate upstream as well as downstream
E > Emin , V > Vc
Flow is supercritical, y < yc , disturbances can only propagate downstream within a wave angle given by
C (g y ) µ = sin o = sin-1 V V
1/ 2
Nonrectangular Channels For flows where the local channel width varies with depth y, critical values can be expressed as
bo Q 2 Ac = g
Q g Ac Vc = = A c bo
where bo = channel width at the free surface. These equations must be solved iteratively to determine the critical area Ac and critical velocity Vc. For critical channel flow that is also moving with
constant depth (yc), the slope corresponds to a critical slope Sc given by
n gAc Sc = 2 α b o Rh, c
α = 1. for S I units and 2.208 for B. G. units
Example 10.5 Given: a 50˚, triangular channel has a 3 flow rate of Q = 16 m /s. Compute: (a) yc, (b) Vc, (c) Sc for n = 0.018 a. For the given geometry, we have P = 2 ( y csc 50˚)
A = 2[y (1/2 y cot 50˚)]
Rh = A/P = y/2 cos 50˚
bo = 2 ( y cot 50˚)
For critical flow, we can write
g Ac = bo Q 3
or g(yc cot 50Þ) = (2 yc cot 50Þ)Q 2
yc = 2.37 m
b. With yc, we compute Pc = 6.18 m
Ac = 4.70 m
bo,c = 3.97 m 3
The critical velocity is now
Q 16 m / s Vc = = = 3.41 m / s A 4.70 m
c. With n = 0.018, we compute the critical slope as
g n2 P 9.81(0.018)2 (6.18) Sc = 2 = 2 1 /3 = 0.0542 α b o R1/3 1.0 ( 3.97 ) ( 0.76 ) h
Frictionless Flow over a Bump Frictionless flow over a bump provides a second interesting analogy, that of compressible gas flow in a nozzle. The flow can either increase or decrease in depth
depending on whether the initial flow is subcritical or supercritical. The height of the bump can also change the results of the downstream flow. Fig. 10.9 Frictionless, 2-D flow over a bump Writing
the continuity and energy equations for two dimensional, frictionless flow between sections 1 and 2 in Fig. 10.10, we have
V1 y1 = V2 y2
V1 V + y1 = 2 + y2 + ∆ h 2g 2g
Eliminating V2, we obtain 2
V y V y − E2 y + 1 1 = 0 where E2 = 1 + y1 − ∆ h 2g 2g 3 2
The problem has the following solutions depending on the initial flow condition and the height of the jump:
Key Points: 1. The specific energy E2 is exactly ∆h less than the approach energy E1. 2. Point 2 will lie on the same leg of the curve as point 1. 3. For Fr < 1, subcritical approach
The water level will decrease at the bump. Flow at point 2 will be subcritical.
4. For Fr > 1, supercritical approach
The water level will increase at the bump. Flow at point 2 will be supercritical.
5. For bump height equal to ∆hmax = E1 - Ec
Flow at the crest will be exactly critical (Fr =1).
6. For ∆h > ∆hmax
No physically correct, frictionless solutions are possible. Instead, the channel will choke and typically result in a hydraulic jump.
Flow under a Sluice Gate A sluice gate is a bottom opening in a wall as shown below in Fig. 10.10a. For free discharge through the gap, the flow smoothly accelerates to critical flow near the gap and
the supercritical flow downstream.
Fig. 10.10 Flow under a sluice gate This is analogous to the compressible flow through a converging-diverging nozzle. For a free discharge, we can neglect friction. Since this flow has no bump (∆h =
0) and E1 = E2, we can write
V1 2 2 V12 y12 y − + y1 y2 + =0 2g 2g 3 2
This equation has the following possible solutions. Subcritical upstream flow and low to moderate tailwater (downstream water level)
One positive, real solution. Supercritical flow at y2 with the same specific energy E2 = E1. Flow rate varies as y2/y1. Maximum flow is obtained for y2/y1 = 2/3.
Subcritical upstream flow and high tailwater
The sluice gate is drowned or partially drowned (analogous to a choked condition in compressible flow). Energy dissipation will occur downstream in the form of a hydraulic jump and the flow
downstream will be subcritical.
The Hydraulic Jump The hydraulic jump is an irreversible, frictional dissipation of energy which provides a mechanism for supercritical flow to transition (jump) to subcritical flow analogous to a
normal shock in compressible flow.
The development of the theory is equivalent to that for a strong fixed wave (Sec 10.1) and is summarized for a hydraulic jump in the following section.
Theory for a Hydraulic Jump If we apply the continuity and momentum equations between points 1 and 2 across a hydraulic jump, we obtain 1 /2 2 y2 = −1 + (1 + 8 Fr12 ) y1
which can be solved for y2.
V2 =
We obtain V2 from continuity:
V1 y1 y2
The dissipation head loss is obtained from the energy equation as
V12 V22 h f = E1 − E2 = + y1 − + y2 2 g 2g or
(y − y1 ) = 2
4 y1 y2
Key points: 1. Since the dissipation loss must be positive, y2 must be > y1. 2. The initial Froude number Fr1 must be > 1 (supercritical flow). 3. The downstream flow must be subcritical and V2 < V1.
Example 10.7 3
Water flows in a wide channel at q = 10 m /(s m) and y1 = 1.25 m. If the flow undergoes a hydraulic jump, compute: (a) y2, (b) V2, (c) Fr2, (d) hf, (e) the percentage dissipation, (f) power
dissipated/unit width, and (g) temperature rise.
a. The upstream velocity is
3 q 10 m /(s ⋅ m) V1 = = = 8.0 m / s y1 1.25 m
Fr1 =
The upstream Froude number is
8.0 V1 1/ 2 = 1/ 2 = 2.285 (g y1 ) [9.81 (1.25)]
This is a weak jump and y2 is given by
2 y2 2 1/ 2 = −1 + (1 + 8 (2.285) ) = 5.54 y1 and
y2 = 1 / 2 y1 (5.54) = 3.46 m V2 =
b. The downstream velocity is
V1 y1 8.0 (1.25) = = 2.89 m / s y2 3.46
c. The downstream Froude number is
Fr2 =
2.89 V2 1/ 2 = 1/ 2 = 0.496 9.81 3.46 ( ) ] (g y2 ) [
and Fr2 is subcritical as expected. d. The dissipation loss is given by
(y − y1 ) = 2
4 y1 y2
3 3.46 − 1.25) ( = = 0.625 m
4 (3.46) (1.25)
e. The percentage dissipation is the ratio of hf/E1. 2
8.0 V E1 = 1 + y1 = 1.25 + = 4.51 m 2g 2 (9.81) X-17
The percentage loss is thus given by
% Loss =
hf E1
100 =
0.625 100 = 14% 4.51
f. The power dissipated per unit width is 3
Power = ρ Q g hf = 9800 M/m *10 m /(s m) * 0.625 m = 61.3 kw/m g. Using Cp = 4200 J/kg K, the temperature rise is given by & p ∆T Power dissipated = mC
or 61,300 W/m = 10,000 kg/s m * 4200 J/kg K* ∆T ∆T = 0.0015˚K negligible temperature rise
XI. Turbomachinery This chapter considers the theory and performance characteristics of the mechanical devices associated with the fluid circulation. General Classification: Turbomachine - A device
which adds or extracts energy from a fluid. Adds energy: Extracts energy:
Pump Turbine
In this context, a pump is a generic classification that includes any device that adds energy to a fluid, e.g. fans, blowers, compressors. We can classify pumps by operating concept: 1. Positive
displacement 2. Dynamic (momentum change) General Performance Characteristics Positive Displacement Pumps 1. Delivers pulsating or periodic flow (cavity opens, fluid enters, cavity closes, decreasing
volume forces fluid out exit opening. 2. Not sensitive to wide viscosity changes. 3. Delivers a moderate flow rate. 4. Produces a high pressure rise. 5. Small range of flow rate operation (fixed pump
speed). Dynamic Pumps 1. Typically higher flow rates than PD’s. 2. Comparatively steady discharge. 3. Moderate to low pressure rise. 4. Large range of flow rate operation. 5. Very sensitive to fluid
viscosity. XI - 1
Typical Performance Curves (at fixed impeller speed)
Fig. 11.2 Performance curves for dynamic and positive displacement pumps Centrifugal Pumps Most common turbomachine used in industry. Includes the general categories of (a) liquid pumps, (b) fans,
(c) blowers, etc. They are momentum change devices and thus fall within the dynamic classification. Typical schematic shown as
Fig. 11.3 Cutaway schematic of a typical centrifugal pump
XI - 2
Writing the energy equation across the device and solving for hp – hf ,we have
P − P1 V2 − V1 H = hp − hf = 2 + + Z 2 − Z1 ρg 2g 2
where H is the net useful head delivered to the fluid, the head that results in pressure, velocity, and static elevation change. Since for most pumps (not all), V1 = V2 and ∆Z is small, we can write
∆P ρg
Since friction losses have already been subtracted, this is the ideal head delivered to the fluid. Note that velocity head has been neglected and can be significant at large flow rates where pressure
head is small. Pw = ρ Q g H
The ideal power to the fluid is given by
The pump efficiency is given by
ρQgH ρQgH Pw = = BHP BHP ωT
where BHP = shaft power necessary to drive the pump ω = angular speed of shaft T = torque delivered to pump shaft Note that from the efficiency equation, pump efficiency is zero at zero flow rate Q
and at zero pump head,H.
XI - 3
Basic Pump Theory Development of basic pump theory begins with application of the integral conservation equation for moment-of-momentum previously presented in Ch. III. Applying this equation to a
centrifugal pump with one inlet, one exit, and uniform properties at each inlet and exit, we obtain & e r x Ve − m & i r x Vi T=m
where T is the shaft torque needed to drive the pump
Vi , Ve are the absolute velocities at the inlet and exit of the pump This is used to determine the change of angular momentum across the device.
Fig. 11.4 Inlet and exit velocity diagrams for an idealized impeller Since the velocity diagram is key to the analysis of the device, we will discuss the elements in detail.
XI - 4
1. At the inner radius r1 have two velocity components: a. the circumferential velocity due to the impeller rotation
u1 = r1 ω
blade tip speed at inner radius β1
b. relative flow velocity tangent to the blade
tangent to the blade angle
These combine to yield the absolute inlet velocity V1 at angle α1 The absolute velocity can be resolved into two absolute velocity components: 1. Normal ( radial ) component:
Vn1 = V1 sin α1 = w1 sin β1
Note that for ideal pump design,
Vn1 = V1 and α 1 = 90
2. Absolute tangential velocity:
Vt1 = V1 cos α 1 = u 1 - w1 cos β1
again, ideally Vt1 = 0
It is also important to note that Vn1 is use to determine the inlet flow rate, i.e.,
Q = A1 Vn1 = 2 π r1 b1 Vn1 where b1 is the inlet blade width
XI - 5
Likewise for the outer radius r2 we have the following: a. the circumferential velocity due to the impeller rotation
u 2 = r2 ω blade tip speed at outer radius b. relative flow velocity tangent to the blade
tangent to the blade angle
w2 β2
These again combine to yield the absolute outlet velocity V2 at angle α2 The exit absolute velocity can also be resolved into two absolute velocity components: 1. Normal ( radial ) component:
Vn2 = V2 sin α 2 = w2 sin β2 =
Q 2 π r2 b2
Note that Q is the same as for the inlet flow rate
2. Absolute tangential velocity:
Vt 2 = V2 cos α 2 = u 2 - w2 cos β2
Vt 2 = u 2 where
tan β 2
= u2 -
Q 2 π r2 b 2 tan β2
Q = A1 Vn1 = 2 π r1 b1 Vn1 = A 2 Vn2 = 2 π r2 b 2 Vn 2
Again, each of the above expressions follows easily from the velocity diagram, and the student should draw and use the diagram with each pump theory problem.
XI - 6
We can now apply moment - of – momentum equation.
T = ρ Q r2 * Vt 2 − r1 * Vt1
(again Vt1 is zero for the ideal design)
For a sign convention, we have assumed that Vt1 and Vt2 are positive in the direction of impeller rotation. The “ ideal” power supplied to the fluid is given by
Pw = ω T = ρ Q ω r2 Vt2 − ω r1 Vt1 or
Pw = ω T = ρ Q u 2 Vt 2 − u1 Vt1 = ρ Qg H Since these are ideal values, the shaft power required to drive a non-ideal pump is given by
BHP =
The head delivered to the fluid is
ρ Q{u 2 Vt2 − u1 Vt1 } ρQg
u { =
For the special case of purely radial inlet flow
H* =
XI - 7
u 2 Vt 2 g
Vt 2 − u 1 Vt1 g
From the exit velocity diagram, substituting for Vt2 we can show that
ωQ u H= 2 − g 2 π b 2 g tan β 2 2
has the form
C1 - C2 Q
shutoff head, the head produced at zero flow, Q = 0
u2 where: C1 = g Example:
A centrifugal water pump operates at the following conditions: speed = 1440 rpm, r1 = 4 in, r2 = 7 in, β1 = 30o, β2 = 20o, b1 = b2 = 1.75 in Assuming the inlet flow enters normal to the impeller
(zero absolute tangential velocity): find: (a) Q, (b) T, (c) Wp, (d) hp, (e) ∆P
ω = 1440
rev 2 π rad = 150.8 s min 60
Calculate blade tip velocities:
u1 = r1 ω =
4 rad ft ft150.8 = 50.3 12 s s
u 2 = r2 ω =
Since design is ideal, at inlet
7 rad ft ft150.8 = 88 12 s s
V1 = Vn1
α1 = 90 , Vt1 = 0 o
Vn1 = U1 tan 300 = 50.3 tan 30o = 29.04 ft/s
Q = 2 π r1 b1 Vn1
r1 •
XI - 8
30Þ u1
4 ft ft Q = 2 π ft1.75 ft 29.04 = 8.87 12 s s 3
s ft gal gal Q = 8.87 60 7.48 3 = 3981 ft min s min Repeat for the outlet:
ft 3 8.87 Q s Vn2 = = 2 π r2 b 2 2 π 7 ft 1.75 ft 12 12 ft Vn2 = 16.6 s Vn2 16.6 ft/s ft w2 = = = 48.54 s sin 20 o sin 20 o
20Þ u2
r2 •
Vt 2 = u 2 - w 2 cos β2 = 88 − 48.54 cos 20 o = 42.4
ft s
We are now able to determine the pump performance parameters. Since for the centrifugal pump, the moment arm r1 at the inlet is zero, the momentum equation becomes Ideal moment of momentum delivered
to the fluid:
T = ρ Q r2 * Vt2
slug ft 7 ft = 1.938 3 8.87 ft 42.4 = 425.1ft − lbf ft s s 12
Ideal power delivered to the fluid:
P = ω T = 150.8
rad ft − lbf 425.1ft − lbf = 64,103 = 116.5 hp s s XI - 9
Head produced by the pump (ideal):
P 64,103 ft − lbf/s = = 115.9 ft lbf ft 3 ρ gQ 62.4 3 8.87 ft s
Pressure increase produced by the pump: 3
ft ∆ P = ρ g H = 62.4 115.9 ft = 7226 psf = 50.2 psi s
Pump Performance Curves and Similarity Laws
Pump performance results are typically obtained from an experimental test of the given pump and are presented graphically for each performance parameter. • Basic independent variable - Q {usually gpm
or cfm } • Dependent variables typically H
– head pressure rise, in some cases ∆P
BHP – input power requirements (motor size) η
– pump efficiency
• These typically presented at fixed pump speed and impeller diameter Typical performance curves appear as
XI - 10
Fig. 11.6 Typical Centrifugal Pump Performance Curves at Fixed Pump Speed and diameter These curves are observed to have the following characteristics: 1. hp is approximately constant at low flow
rate. 2. hp = 0 at Qmax. 3. BHP is not equal to 0 at Q = 0. 4. BHP increases monotonically with the increase in Q. 5. ηp = 0 at Q = 0 and at Qmax. 6. Maximum pump efficiency occurs at approximately
Q* = 0.6 Qmax . This is the best efficiency point BEP. At any other operating point, efficiency is less, pump head can be higher or lower, and BHP can be higher or lower. 7. At the BEP, Q = Q*, hp =
hp*, BHP = BHP*. Measured Performance Data Actual pump performance data will typically be presented graphically as shown in Fig. 11.7. Each graph will usually have curves representing the pump head
vs flow rate for two or more impeller diameters for a given class/model of pumps having a similar design. The graphs will also show curves of constant efficiency and constant pump power (BHP) for the
impeller diameters shown. All curves will be for a fixed pump impeller speed.
XI - 11
Fig. 11.7 Measured performance curves for two models of a centrifugal water pump
XI - 12
How to Read Pump Performance Curves Care must be taken to correctly read the performance data from pump curves. This should be done as follows: (1) For a given flow rate Q (2) Read vertically to a
point on the pump head curve h for the impeller diameter D of interest. (3) All remaining parameters ( efficiency & BHP) are read at this point; i.e., graphically interpolate between adjacent curves
for BHP to obtain the pump power at this point. Note that the resulting values are valid only for the conditions of these curves: (1) pump model and design, (2) pump speed – N, (3) impeller size – D,
(4) fluid (typically water) Thus for the pump shown in Fig. 11.7a with an impeller diameter D = 32 in, we obtain the following performance at Q = 20,000 gpm: Q = 20,000 gpm, D = 32 in, N = 1170 rpm H
≅ 385 ft, BHP ≅ 2300 bhp, ηp ≅ 86.3 % Note that points that are not on an h vs. Q curve are not valid operating points. Thus for Fig. 11.7b, the conditions Q = 22,000 gpm, BHP = 1500 bhp, hp = 250 ft
do not correspond to a valid operating point because they do not fall on one of the given impeller diameter curves. However, for the same figure, the point Q = 20,000 gpm, BHP = 1250 bhp is a valid
point because it coincidentally also falls on the D = 38 in impeller curve at hp = 227 ft.
XI - 13
Net Positive Suction Head - NPH One additional parameter is typically shown on pump performance curves: NPSH = head required at the pump inlet to keep the fluid from cavitating. NPSH is defined as
follows: 2
P NPSH = i + ρg
Vi P − v 2g ρg
where Pi = pump inlet pressure Pv = vapor pressure of fluid Pump inlet
Considering the adjacent figure, write the energy equation between the fluid surface and the pump inlet to obtain the following:
P NPSH = i + ρg
Pi z=0
Vi P P P − v = a − Z i − h f,a−i − v 2g ρg ρ g ρg
For a pump installation with this configuration to operate as intended, the righthand-side of the above equation must be > the NPSH value for the operating flow rate for the pump. Example: A water
supply tank and pump are connected as shown. Pa = 13.6 psia and the water is at 20 o C with Pv = 0.34 psia. The system has a friction loss of 4.34 ft. Will the NPSH of the pump of Fig. 11.7a at
20,000 gpm work?
XI - 14
a 10 ft i
Applying the previous equation we obtain
NPSH =
Pa P − Z i − h f,a−i − v ρg ρg
13.6 − 0.34) lbf/in2 *144 in 2 /ft 2 ( NPSH = − (−10 ft) − 4.34 ft 3 62.4 lbf/ft
NPSH = 36.26 ft
The pump will work because the system NPSH as shown in Fig. 11.7a is 30 ft which provides a 6.3 ft safety margin. Conversely, the pump could be located as close as 3.7 ft below the water surface and
meet NPSH requirements.
Pump Similarity Laws Application of the dimensional analysis procedures of Ch. V will yield the following three dimensionless performance parameters: Dimensionless flow coefficient:
CQ =
Q ω D3
Dimensionless head coefficient:
CH =
gH ω 2 D2
Dimensionless power coefficient:
CP =
BHP ρω 3 D5
where ω is the pump speed in radians/time and other symbols are standard design and operating parameters with units that make the coefficients dimensionless. How are these used? These terms can be
used to estimate design and performance changes between two pumps of similar design.
XI - 15
Stated in another way: If pumps 1 and 2 are from the same geometric design family and are operating at similar operating conditions, the flow rates, pump head, and pump power for the two pumps will
be related according to the following expressions:
Q2 N2 = Q1 N1
D2 D1 2
H 2 N2 D2 = H1 N1 D1
Use to predict the new flow rate for a design change in pump speed N and impeller diameter D. 2
Used to predict the new pump head H for a design change in pump speed, N and impeller diameter D. 3
BHP2 ρ2 N 2 D2 = BHP1 ρ1 N1 D1
Used to predict the new pump power BHP for a design change in fluid, ρ, pump speed N and impeller diameter D.
Example It is desired to modify the operating conditions for the 38 in diameter impeller pump of Fig. 11.7b to a new pump speed of 900 rpm and a larger impeller diameter of 40 in.
• H(ft) BEP1•
Determine the new pump head and power for the new pump speed at the BEP.
XI - 16
BEP 2
For the D = 38 in impeller of Fig. 11.7b operating at 710 rpm, we read the best efficiency point (BEP) values as Q* = 20,000 gpm, H* = 225 ft, BHP * = 1250 hp
Applying the similarity laws for N2 = 900 rpm and D2 = D1 = 38 in, we obtain 3
3 Q2 N2 D2 900 40 = = 1.478 = 710 38 Q1 N1 D1
Q2 = 20,000*1.478 = 29,570 gpm 2
2 2 H 2 N2 D2 900 40 = = = 1.78 710 38 H1 N1 D1
H2 = 225*1.78 = 400.5 ft 3
3 5 BHP2 ρ2 N 2 D2 900 40 = = (1) = 2.632 710 38 BHP1 ρ1 N1 D1
BHP2 = 3290 hp
Thus, even small changes in the speed and size of a pump can result in significant changes in flow rate, head, and power. It is noted that every point on the original 38 in diameter performance curve
exhibits a similar translation to a new operating condition. The similarity laws are obviously useful to predict changes in the performance characteristics of an existing pump or to estimate the
performance of a modified pump design prior to the construction of a prototype.
XI - 17
Matching a Pump to System Characteristics The typical design/sizing requirement for a pump is to select a pump which has a pump head which matches the required system head at the design/operating
flow rate for the piping system.
Key Point
hp = hsys at Qdes.
It is noted that pump selection should occur such that the operating point of the selected pump should occur on the pump curve near or at the BEP. From the energy equation in Ch. VI, the system head
is typically expressed as 2
h sys
V2 P2 − P1 V2 − V1 L = + + Z 2 − Z 1 + f + ∑ K i D 2g ρg 2g
Thus the selection of a pump for a piping system design should result in a pump for which the pump head hp at the design flow rate Qdes is equal ( or very close) to the head
hp Hdes
requirements hsys of the piping system at the same flow rate, and this should occur at or near the point of maximum efficiency for the chosen pump.
• hsys
Other operating and performance requirements (such as NPSH) are obviously also a part of the selection criteria for a pump.
XI - 18
Pumping Systems: Parallel and Series Configurations For some piping system designs, it may be desirable to consider a multiple pump system to meet the design requirements. Two typical options include
parallel and series configurations of pumps. Specific performance criteria must be met when considering these options. Given a piping system which has a known design flow rate and head requirements,
Qdes, hdes. The following pump selection criteria apply. Pumps in Parallel: Assuming that the pumps are identical, each pump must provide the following: Q(pump) = 0.5 Qdes h(pump) = hdes
Pumps in Series: Assuming that the pumps are identical, each pump must provide the following: Q (pump) = Qdes h(pump) = 0.5 hdes For example, if the design point for a given piping system were Qdes =
600 gpm, and hsys = 270 ft, the following pump selection criteria would apply: 1. Single pump system
Q(pump) = 600 gpm, hp = 270 ft
2. Parallel pump system
Q(pump) = 300 gpm, hp = 270 ft for each of the two pumps
3. Series pump system
Q(pump) = 600 gpm, hp = 135 ft for each of the two pumps
XI - 19 | {"url":"https://silo.pub/fluid-mechanics-q-2249225.html","timestamp":"2024-11-02T15:22:56Z","content_type":"text/html","content_length":"837830","record_id":"<urn:uuid:bf2c3ccc-f3fa-4fe3-a99d-6a508a888283>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00425.warc.gz"} |
Quincunx Demonstrations
.: Quincunx Demonstrations :.
The uses of a Quincunx are only limited by the ingenuity of the instructor. Once you use a Quincunx and see how effective it is in communicating statistical concepts, you'll never teach without one.
The following is a sample of the types of demonstrations that are possible:
Can statistics really predict an outcome based on a sample? Demonstrate this by running a sample of 35 beads, calculate the mean and standard deviation and predict the six standard deviation limits.
Try taking bets with the trainees that the next 100 beads will fall between the six standard deviation limits calculated. Run the next hundred beads and collect your bets.
Draw specification limits on the face of the Quincunx at the 5th and 20th column. Set the funnel to the left or right and demonstrate how a certain percentage of the measurement will fall outside of
the specification. Now center the funnel and repeat the experiment. Watch the light bulbs go on with your students as they begin to understand the concept of centering a process' and how it could
apply to their own processes.
This demonstration is as effective with machine operators as it is useful in showing management how they are part of the problem. Point out current random sampling techniques that have been
determined by your company. i.e. check every 20th piece, every half hour, twice per shift. etc. Set up the Quincunx as in the process centering demonstration and shift the funnel near the upper
limit. Point out how many operators run to the side of the tolerance so they can reduce scrap (you can always rerun the part if it is oversize but it is scrap if its undersize). Drop 19 beads then
drop the 20th bead representing the inspected part and note whether it falls inside or outside the specification. Repeat the process five or six times. Chances are every bead checked will be inside
the specification with about 10% of all the other beads being outside the tolerance. If a bead does falls outside the tolerance during one of the checks remind the student that most operators would
run a second piece before adjusting the process. This demonstration drives home the reason why operators who are instructed to use random sampling techniques have trouble maintaining tolerance
Close the top gate on the Quincunx and without moving the funnel run and plot samples of five beads on an average-range control chart. Calculate and plot the control limits on the chart just as if
the quincunx were a machine. After creating the control chart move the funnel and plot another sample of five parts. Students will be amazed how quickly they can detect a shift in the funnel which is
analogous to a change in the process. To carry the point a little farther tape a piece of paper over the funnel portion of the Quincunx and then randomly move the funnel to different directions and
let the trainees guess whether the funnel has been moved or not based on their control chart plot. This technique is very helpful in building confidence in the value of control charting. Another
point can be made by sliding the funnel completely to the right and allow the beads to drop directly onto the pins. Plot another sample of 5 beads and demonstrate how the range portion of the chart
detects an out of control condition. This erratic behavior is analogous to machine conditions where bearings are worn, fixtures are loose, etc.
Run a sufficiently large sample to accurately calculate or predict the six standard deviation range. Mark the tolerance limits and pre-control lines on the face of the quincunx with the appropriate
tape or transparency marker. Drop a series of beads with the funnel centered and demonstrate the decision rules of the pre-control plan. Move the funnel to the left or right and show how pre-control
would re-center the process.
The WD-5, WD-6, and WD-7 models have a narrow distribution pin block for demonstrating the effect of process improvement. Since it has less variability, the control chart demonstration will have a
smaller range and consequently tighter control limits.
│ QT-5 "The Quincunx and Statistical Training" │
│ │
│The above described demonstrations and more are all described on our QT-5 Audio CD. This CD, which lasts over an hour, is one of our customer's favorites. Trainers keep telling us that it is a real│
│help in getting them to understand the power of the quincunx to illustrate the principles of variation and the need for quality improvement. │ | {"url":"https://www.qualitytng.com/quincunx-demonstrations/","timestamp":"2024-11-06T02:55:45Z","content_type":"application/xhtml+xml","content_length":"45318","record_id":"<urn:uuid:5adaf69d-d830-42ce-8975-4c38e03f00cc>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00002.warc.gz"} |
Applications of graded methods to cluster variables in arbitrary types
Booker-Price, Thomas and Grabowski, Jan (2017) Applications of graded methods to cluster variables in arbitrary types. PhD thesis, Lancaster University.
PDF (2017Booker-PricePhD)
2017Booker_PricePhD.pdf - Published Version
Available under License Creative Commons Attribution-NoDerivs. Download (1MB)
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This thesis is concerned with studying the properties of gradings on several examples of cluster algebras, primarily of infinite type. We start by considering two classes of finite type cluster
algebras: those of type Bn and Cn. We give the number of cluster variables of each occurring degree and verify that the grading is balanced. These results complete a classification in [16] for
coefficient-free finite type cluster algebras. We then consider gradings on cluster algebras generated by 3×3 skew-symmetric matrices. We show that the mutation-cyclic matrices give rise to gradings
in which all occurring degrees are positive and have only finitely many associated cluster variables (excepting one particular case). For the mutation-acyclic matrices, we prove that all occurring
degrees have infinitely many variables and give a direct proof that the gradings are balanced. We provide a condition for a graded cluster algebra generated by a quiver to have infinitely many
degrees, based on the presence of a subquiver in its mutation class. We use this to study the gradings on cluster algebras that are (quantum) coordinate rings of matrices and Grassmannians and show
that they contain cluster variables of all degrees in N. Next we consider the finite list (given in [9]) of mutation-finite quivers that do not correspond to triangulations of marked surfaces. We
show that A(X7) has a grading in which there are only two degrees, with infinitely many cluster variables in both. We also show that the gradings arising from Ee6, Ee7 and Ee8 have infinitely many
variables in certain degrees. Finally, we study gradings arising from triangulations of marked bordered 2- dimensional surfaces (see [10]). We adapt a definition from [24] to define the space of
valuation functions on such a surface and prove combinatorially that this space is isomorphic to the space of gradings on the associated cluster algebra. We illustrate this theory by applying it to a
family of examples, namely, the annulus with n + m marked points. We show that the standard grading is of mixed type, with finitely many variables in some degrees and infinitely many in the others.
We also give an alternative grading in which all degrees have infinitely many cluster variables.
?? cluster algebragraded cluster algebrainfinite typegrading ??
Deposited On:
21 Nov 2017 20:34
Last Modified:
22 Oct 2024 00:11 | {"url":"https://eprints.lancs.ac.uk/id/eprint/88866/","timestamp":"2024-11-11T06:50:53Z","content_type":"application/xhtml+xml","content_length":"30014","record_id":"<urn:uuid:2c39502d-1bad-4e15-8aff-6a7ab662ff83>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00607.warc.gz"} |
The perimeter of a triangle is 24 inches. The longest side of 4 inches is longer than the shortest side, and the shortest side is three-fourths the length of the middle side. How do you find the length of each side of the triangle? | Socratic
The perimeter of a triangle is 24 inches. The longest side of 4 inches is longer than the shortest side, and the shortest side is three-fourths the length of the middle side. How do you find the
length of each side of the triangle?
2 Answers
Well this problem is simply impossible.
If the longest side is 4 inches, there is no way that the perimeter of a triangle can be 24 inches.
You are saying that 4 + (something less than 4) + (something less than 4) = 24, which is impossible.
The sides are $6$ inches, $8$ inches and $10$ inches
I'd suggest that the question should read 'The perimeter of a triangle is 24 inches. The longest side is 4 inches longer than the shortest side, and the shortest side is three-fourths the length of
the middle side. How do you find the length of each side of the triangle?'
In this case the question can be answered. If $x$ is the length of the middle side, then the shortest side is $\frac{3}{4} x$ and the longest side is $\frac{3}{4} x + 4$
$x + \frac{3}{4} x + \frac{3}{4} x + 4 = 24$
$\frac{10}{4} x = 20$
$x = 8$
Then the shortest side is $6$ and the longest side is $10$
Impact of this question
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RE: Re: RE: Re: Dot Product in Cylindrical Coordinates
• To: mathgroup at smc.vnet.net
• Subject: [mg69362] RE: [mg69351] Re: [mg69301] RE: [mg69276] Re: Dot Product in Cylindrical Coordinates
• From: "David Park" <djmp at earthlink.net>
• Date: Sat, 9 Sep 2006 03:26:39 -0400 (EDT)
Your example does not work for me.
SetCoordinates[Cylindrical[r, theta, z]];
g = {gr[r, theta, z], gtheta[r, theta, z], gz[r, theta, z]};
f = {fr[r, theta, z], ftheta[r, theta, z], fz[r, theta, z]};
gr[r_, theta_, z_] := r
gtheta[r_, theta_, z_] := theta
gz[r_, theta_, z_] := z
fr[r_, theta_, z_] := r^2
ftheta[r_, theta_, z_] := theta
fz[r_, theta_, z_] := Cos[z]
If we use CrossProduct and DotProduct along with a standard identity using
Div and Curl, which should always be true, then it does not work.
Div[CrossProduct[g, f]] ==
DotProduct[Curl[g], f] - DotProduct[Curl[f], g] // Simplify
(r^2 + 3*r^4*z^2 - 5*r^3*z*Cos[z] +
Sqrt[r^2*((-r)*z + Cos[z])^2] + r^2*Cos[2*z])/
(r*Sqrt[r^2*((-r)*z + Cos[z])^2]) ==
(theta*(-z + Cos[z]))/r
However if we use the standard Dot and Cross functions the identity is
Div[Cross[g, f]] == Dot[Curl[g], f] - Dot[Curl[f], g] // Simplify
The point here is that the package functions DotProduct and CrossProduct are
rather special and have NOTHING to do with standard vector calculus. Users
should not stumble into using them where they do not apply.
In standard vector calculus, in a coordinate system, an orthonormal frame is
erected at each point in the space. The axes of the frame point along the
coordinate directions. The components of vectors are specified in terms of
this orthonormal frame. Since the frame is orthonormal, at any point we can
simple use the standard Dot and Cross product for combining two vectors at
that point. Curl and Div demand vector components in terms of an orthonormal
If businessmen can jump on computerized inventory control, lasers and
barcodes and DVDs and other technology all within a decade or so of their
invention, why or why can't engineering and physics schools dump the
misbegotten vector calculus of over a century ago?
David Park
djmp at earthlink.net
From: Pratik Desai [mailto:pratikd at wolfram.com]
To: mathgroup at smc.vnet.net
I think it is quite imperative to note here that
"There are often conflicting definitions of a particular coordinate
system in the literature. When you use a coordinate system with this
package, you should look at the definition given below to make sure it
is what you want." --Mathematica Documentation
So for cylindrical coordinate system one must define the system as:
g = {g?[r, theta, z], g?[r, theta, z], gz[r, theta, z]}
f = {f?[r, theta, z], f?[r, theta, z], fz[r, theta, z]}
g?[r_, theta_, z_] = r
g?[r_, theta_, z_] = theta
gz[r_, theta_, z_] = z
f?[r_, theta_, z_] = r^2
f?[r_, theta_, z_] = theta
fz[r_, theta_, z_] = Cos[z]
Then everything works fine:
Hope this helps | {"url":"http://forums.wolfram.com/mathgroup/archive/2006/Sep/msg00204.html","timestamp":"2024-11-14T22:13:09Z","content_type":"text/html","content_length":"32860","record_id":"<urn:uuid:1de575a7-814d-4fce-a542-54b7e3cd994d>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00620.warc.gz"} |
Moysey Brio
• Professor, Mathematics
• Professor, Applied Mathematics - GIDP
• Member of the Graduate Faculty
• Visiting Scholar
□ CNRS (French Center for Scientific Research), sabbatical at INSA, Rouen, France, Spring 2018
No activities entered.
2024-25 Courses
• Dissertation
APPL 920 (Fall 2024)
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• Dissertation
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• Formal Math Reasong+Wrtg
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• Independent Study
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• Research
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• Discrete Mathematics
MATH 243 (Fall 2023)
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• Math Analysis Engineers
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2022-23 Courses
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• Numerical Analysis Pde
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• Dissertation
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• Honors Thesis
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2020-21 Courses
• Adv Applied Mathematics
MATH 422 (Summer I 2021)
• Adv Applied Mathematics
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• Independent Study
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• Partial Diff Equations
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• Senior Capstone
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Scholarly Contributions
• Brio, M. (2023). Localized eigenvectors on metric graphs. Journal of Mathematics and Computers in Simulation, 215, 352-372.
• Brio, M. (2022). FDTD method with explicit non-iterative and second order treatment for Kerr nonlinearities. IEEE Journal on Multiscale and Multiphysics Computational Techniques,, 7, 195-199.
• Brio, M., Caputo, J., & Kravitz, H. (2022). Spectral solution of PDEs on networks. Applied Numerical Mathematics.
• Kano, P., Brio, M., & Bailey, J. (2021). "Optimal parameter selection in Week's method for numerical Laplace transform inversion based on machine learning". Journal of Algorithms and
Computational Technology, 15, 1-22.
• Brio, M., Caputo, J., Gwirtz, K., Liu, J., & Maimistov, A. (2019). Scattering of a short electromagnetic pulse from a Lorentz–Duffing film: Theoretical and numerical analysis. Wave Motion, 89,
• Brio, M., Nehls, J., Dineen, C., Poole, C., & Moloney, J. (2019). Second-order, stable, fully anisotropic FDTD. Int. J. Num. Modeling:Electronic Networks, Devices and Fields, 32(2), e2521.
• Brio, M. (2015). Exponential time-differencing with embedded Runge-Kutta adaptive step control. Journal of Computational Physics, 280, 579-601.
• Brio, M. (2014). Finite-Difference Time-Domain simulatio n of spacetime cloak. Optics Express, 22(10), 12087-12095.
• Liu, J., Brio, M., & Moloney, J. V. (2014). An overlapping Yee finite-difference time-domain method for material interfaces between anisotropic dielectrics and general dispersive or perfect
electric conductor media. International Journal of Numerical Modelling: Electronic Networks, Devices and Fields, 27(1), 22-33.
More info
Abstract: A novel stable anisotropic finite-difference time-domain (FDTD) algorithm based on the overlapping cells is developed for solving Maxwell's equations of electrodynamics in anisotropic
media with interfaces between different types of materials, such as the interface between anisotropic dielectrics and dispersive medium or perfect electric conductor (PEC). The previous proposed
conventional anisotropic FDTD methods suffer from the late-time instability due to the extrapolation of the field components near the material interface. The proposed anisotropic overlapping Yee
FDTD method is stable, as it relies on the overlapping cells to provide the collocated field values without any interpolation or extrapolation. Our method has been applied to simulate
electromagnetic invisibility cloaking devices with both anisotropic dielectrics and PEC included in the computational domain. Numerical results and eigenvalue analysis confirm that the
conventional anisotropic FDTD method is weakly unstable, whereas our method is stable. Copyright © 2013 John Wiley & Sons, Ltd. Copyright © 2013 John Wiley & Sons, Ltd.
• Liu, J., Brio, M., & Moloney, J. V. (2014). Transformation optics based local mesh refinement for solving Maxwell's equations. Journal of Computational Physics, 258, 359-370.
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Abstract: In this paper, a novel local mesh refinement algorithm based on transformation optics (TO) has been developed for solving the Maxwell's equations of electrodynamics. The new algorithm
applies transformation optics to enlarge a small region so that it can be resolved by larger grid cells. The transformed anisotropic Maxwell's equations can be stably solved by an anisotropic
FDTD method, while other subgridding or adaptive mesh refinement FDTD methods require time-space field interpolations and often suffer from the late-time instability problem. To avoid small time
steps introduced by the transformation optics approach, an additional application of the mapping of the material matrix to a dispersive material model is employed. Numerical examples on
scattering problems of dielectric and dispersive objects illustrate the performance and the efficiency of the transformation optics based FDTD method. © 2013 Elsevier Inc.
• Pasenow, B., Dineen, C., Hader, J., Brio, M., Moloney, J. V., Koch, S. W., Chen, S. H., Becker, A., & Jaroń-Becker, A. (2013). Anisotropic terahertz response from a strong-field ionized
electron-ion plasma. Physical Review E - Statistical, Nonlinear, and Soft Matter Physics, 87(3).
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Abstract: A microscopic theory is adapted to compute the time-resolved terahertz (THz) probe response of a dynamically evolving, strong-field ionized electron-ion plasma. The numerical solutions
show that the relaxation of the initially highly anisotropic carrier distributions leads to a polarization dependent short-time THz response. This THz polarization discrimination gradually
vanishes as the plasma approaches a thermodynamic equilibrium configuration via electron-electron and electron-ion scattering. The detailed carrier-relaxation dynamics causes a strongly
nonmonotonic time-development of the THz absorption. © 2013 American Physical Society.
• Kano, P. O., Brio, M., Dostert, P., & Cain, J. (2012). Dempster-Shafer evidential theory for the automated selection of parameters for Talbot's method contours and application to matrix
exponentiation. Computers and Mathematics with Applications, 63(11), 1519-1535.
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Abstract: In this paper, the Dempster-Shafer theory of evidential reasoning is applied to the problem of optimal contour parameters selection in Talbot's method for the numerical inversion of the
Laplace transform. The fundamental concept is the discrimination between rules for the parameters that define the shape of the contour based on the features of the function to invert. To
demonstrate the approach, it is applied to the computation of the matrix exponential via numerical inversion of the corresponding resolvent matrix. Training for the Dempster-Shafer approach is
performed on random matrices. The algorithms presented have been implemented in MATLAB. The approximated exponentials from the algorithm are compared with those from the rational approximation
for the matrix exponential returned by the MATLAB expm function. © 2012 Elsevier Ltd. All rights reserved.
• Liu, J., Brio, M., & Moloney, J. V. (2012). Subpixel smoothing finite-difference time-domain method for material interface between dielectric and dispersive media. Optics Letters, 37(22),
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Abstract: In this Letter, we have shown that the subpixel smoothing technique that eliminates the staircasing error in the finite-difference time-domain method can be extended to material
interface between dielectric and dispersive media by local coordinate rotation. First, we show our method is equivalent to the subpixel smoothing method for dielectric interface, then we extend
it to a more general case where dispersive/dielectric interface is present. Finally, we provide a numerical example on a scattering problem to demonstrate that we were able to significantly
improve the accuracy. © 2012 Optical Society of America.
• Kano, P. O., & Brio, M. (2010). Application of Post's formula to optical pulse propagation in dispersive media. Computers and Mathematics with Applications, 59(2), 629-650.
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Abstract: In this paper we consider the propagation of optical pulses in dielectric media with nontrivial dispersion relations. In particular, we implement Post's Laplace transform formula to
invert in time the Fourier-Laplace space coefficients which arise from the joint space solution of the optical dispersive wave equation. Due to the inefficiency of a direct application of this
formula, we have considered and present here two more efficient implementations. In the first, the Gaver-Post method, we utilize the well known Gaver functionals but store intermediate
calculations to improve efficiency. The second, the Bell-Post method, involves an analytic reformulation of Post's formula such that knowledge of the dispersion relation and its derivatives are
sufficient to invert the coefficients from Laplace space to time. Unlike other approaches to the dispersive wave equation which utilize a Debye-Lorentzian assumption for the dispersion relation,
our algorithm is applicable to general Maxwell-Hopkinson dielectrics. Moreover, we formulate the approach in terms of the Fourier-Laplace coefficients which are characteristic of a dispersive
medium but are independent of the initial pulse profile. They thus can be precomputed and utilized when necessary in a real-time system. Finally, we present an illustration of the method applied
to optical pulse propagation in a free space and in two materials with Cole-type dispersion relations, room temperature ionic liquid (RTIL) hexafluorophosphate and brain white matter. From an
analysis of these examples, we find that both methods perform better than a standard Post-formula implementation. The Bell-Post method is the more robust of the two, while the Gaver-Post is more
efficient at high precision and Post formula approximation orders. © 2009 Elsevier Ltd. All rights reserved.
• Liu, J., Brio, M., Zeng, Y., Zakharian, A. R., Hoyer, W., Koch, S. W., & Moloney, J. V. (2010). Generalization of the FDTD algorithm for simulations of hydrodynamic nonlinear Drude model. Journal
of Computational Physics, 229(17), 5921-5932.
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Abstract: In this paper we present a numerical method for solving a three-dimensional cold-plasma system that describes electron gas dynamics driven by an external electromagnetic wave
excitation. The nonlinear Drude dispersion model is derived from the cold-plasma fluid equations and is coupled to the Maxwell's field equations. The Finite-Difference Time-Domain (FDTD) method
is applied for solving the Maxwell's equations in conjunction with the time-split semi-implicit numerical method for the nonlinear dispersion and a physics based treatment of the discontinuity of
the electric field component normal to the dielectric-metal interface. The application of the proposed algorithm is illustrated by modeling light pulse propagation and second-harmonic generation
(SHG) in metallic metamaterials (MMs), showing good agreement between computed and published experimental results. © 2010 Elsevier Inc.
• Liu, J., Brio, M., & Moloney, J. V. (2009). A diagonal split-cell model for the overlapping Yee FDTD method. Acta Mathematica Scientia, 29(6), 1670-1676.
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Abstract: In this paper, we present a nonorthogonal overlapping Yee method for solving Maxwell's equations using the diagonal split-cell model. When material interface is presented, the diagonal
split-cell model does not require permittivity averaging so that better accuracy can be achieved. Our numerical results on optical force computation show that the standard FDTD method converges
linearly, while the proposed method achieves quadratic convergence and better accuracy. © 2009 Wuhan Institute of Physics and Mathematics.
• Liu, J., Brio, M., & Moloney, J. V. (2009). Overlapping Yee FDTD method on nonorthogonal grids. Journal of Scientific Computing, 39(1), 129-143.
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Abstract: We propose a new overlapping Yee (OY) method for solving time-domain Maxwell's equations on nonorthogonal grids. The proposed method is a direct extension of the Finite-Difference
Time-Domain (FDTD) method to irregular grids. The OY algorithm is stable and maintains second-order accuracy of the original FDTD method, and it overcomes the late-time instability of the
previous FDTD algorithms on nonorthogonal grids. Numerical examples are presented to illustrate the accuracy, stability, convergence and efficiency of the OY method. © 2008 Springer
Science+Business Media, LLC.
• Brio, M., Webb, G. M., & Zakharian, A. R. (2008). Bibliography. Mathematics in Science and Engineering, 213(C), 273-287.
• Brio, M., Webb, G. M., & Zakharian, A. R. (2008). Convergence theory for initial value problems. Mathematics in Science and Engineering, 213(C), 109-144.
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Abstract: The convergence theory for numerical methods approximating time-dependent problems parallels the theory of ordinary differential equations (ODEs) where two types of behavior are
studied, namely: (1) the finite time solution and (2) the long-time asymptotic behavior where the solution either passes through an initial transient state and sets into a steady state, or
evolves into a periodic or chaotic motion, or escapes to infinity. We describe the notions of consistency, stability, local and global error estimates, resolution and order of accuracy, followed
by Lax-Richtmyer equivalence theorem. The rest of the chapter is devoted to practical implications of the convergence theory in terms of the resolution and error estimates together with von
Neumann and CFL stability restrictions. © 2010 Elsevier Ltd.
• Brio, M., Webb, G. M., & Zakharian, A. R. (2008). Discretization methods. Mathematics in Science and Engineering, 213(C), 59-108.
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Abstract: Discretization of partial differential equations (PDEs) is based on the theory of function approximation, with several key choices to be made: an integral equation formulation, or
approximate solution operator; the type of discretization, defined by the function subspace in which the solution is approximated; the choice of grids, e.g. regular versus irregular grids to
conform to the geometry, or static versus solution adaptive grids. We explore some of the common approaches to the choice of form of the PDE and the space-time discretization, leaving discussion
of the grids for a later chapter. The goal is to introduce the reader to various forms of discretization and to illustrate the numerical performance of different methods. In particular, we will
address how to choose a method that is accurate, robust and efficient for the problem at hand. © 2010 Elsevier Ltd.
• Brio, M., Webb, G. M., & Zakharian, A. R. (2008). Numerical boundary conditions. Mathematics in Science and Engineering, 213(C), 145-174.
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Abstract: Boundary conditions arise in the process of numerical implementation of given physical boundary conditions on the original physical or truncated domain due to restrictions imposed by
computational resources. The latter situation is often encountered when an infinite domain is truncated or remapped onto a finite computational domain. Numerical implementation of the physical or
numerical interface boundary conditions should preserve the accuracy and stability of the inner numerical method. The inaccuracies and instabilities created by numerical implementation of the
interface and boundary conditions may be localized at the boundaries or interfaces, but more often they may propagate through out the whole computational domain. Examples include: (a) transparent
boundary conditions for hyperbolic and dispersive systems, (b) Berenger's perfectly matched layer (PML) boundary conditions applied to Maxwell's equations (c) stability analysis in the presence
of boundaries and interfaces, and (d) grid interfaces and material interfaces in semi-conductor device simulation models, and finite-difference time-domain (FDTD) discretization of Maxwell's
equations. © 2010 Elsevier Ltd.
• Brio, M., Webb, G. M., & Zakharian, A. R. (2008). Numerical grid generation. Mathematics in Science and Engineering, 213(C), 251-272.
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Abstract: In this chapter we overview the basic ideas behind numerical grid adaptation to geometrical features of the domain, its interfaces and boundaries, as well as adaptation to the solution
features, such as high gradients, that are often employed to improve the accuracy and efficiency of the computation. Through representative examples we discuss implementation and stability issues
for static and dynamic remapping grid generation methods. The last two sections describe Level Set and Front Tracking Methods for propagating moving interfaces and fronts. © 2010 Elsevier Ltd.
• Brio, M., Webb, G. M., & Zakharian, A. R. (2008). Overview of partial differential equations. Mathematics in Science and Engineering, 213(C), 1-57.
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Abstract: Chapter 1 begins with some examples of partial differential equations in science and engineering and their linearization and dispersion equations. The concepts of well-posedness,
regularity, and solution operator for systems of partial differential equations (PDE's) are discussed. Instabilities can arise from both numerical methods and from real physical instabilities.
Some physical instabilities are described, including: (a) the distinction between convective and absolute instabilities, (b) the Rayleigh-Taylor and Kelvin-Helmholtz instabilities in fluids, (c)
wave breaking and gradient catastrophe in gas dynamics and in conservation laws, (d) modulational or Benjamin Feir instabilities and nonlinear Schrödinger related equations, (e) three-wave
resonant interactions and explosive instabilities associated with negative energy waves. Basic wave concepts are described (e.g. wave-number surfaces, group velocity, wave action, wave
diffraction, and wave energy equations). A project from semiconductor transport modeling is described. © 2010 Elsevier Ltd.
• Brio, M., Webb, G. M., & Zakharian, A. R. (2008). Preface. Mathematics in Science and Engineering, 213(C), v-vii.
• Brio, M., Webb, G. M., & Zakharian, A. R. (2008). Problems with multiple temporal and spatial scales. Mathematics in Science and Engineering, 213(C), 175-249.
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Abstract: Problems with multiple disparate temporal and spatial scales are at the heart of most physical problems, as well as in mathematical modeling problems, in economics and social sciences.
Numerical treatment usually allows one to deal with two, or at most, three disparate scales due to limitations of computer resources. Problems with different scales arise when the interest is in
the description of the problem on the larger scales, for example the motion of weather fronts in the presence of fast gravity waves; the average trajectory of a fast rotating charged particle in
a magnetic field, in which one averages over the fast gyro-period of the particle; the behavior of society incorporating the dynamics of individuals; the role of singularities in solutions of
differential equations or the role of geometry; etc. The methods of dealing with such problems are based on either resolving (if ever possible) or not resolving the fine scales. In the latter
case, asymptotic analysis of the dominant terms is often used to separate scales or compute the cumulative effect of the small scale motion on the large scale dynamics. For problems where scales
are weakly coupled, numerical treatment may often produce physical results as long as the choice of the method has remnants of the smaller scale behavior such as: damping, diffusion, dispersion,
etc. For the second type of problem, under-resolved numerical computations usually never produce physical results as cumulative effects are not of the form of the truncation errors, in addition
to the fact that the details of the small scale motion are important and have to be accounted for on either a theoretical or experimental basis. In this Chapter we consider examples from stiff
ODEs, long-time integrators, Hamiltonian systems, multi-symplectic systems, hyperbolic conservation laws, Godunov methods, Riemann solvers with slope/flux limiters. © 2010 Elsevier Ltd.
• Kano, P., Barker, D., & Brio, M. (2008). Analysis of the analytic dispersion relation and density of states of a selected photonic crystal. Journal of Physics D: Applied Physics, 41(18).
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Abstract: In this paper, we consider an anisotropic three-dimensional photonic crystal for which an analytic solution to Maxwell's equations can be derived. We also discuss the reduction to a
two-dimensional variant which occurs for isotropic materials. For the two-dimensional case, we integrate an analytic dispersion relation to accurately calculate the density of states. A
comparison is made with a numerically determined density of states computed from the dispersion surface generated by the MIT photonic-bands package (MPB). Close agreement between the analytic and
numerical results is shown. Finally, we utilize the analytically calculated density of states to characterize the thermally pumped terahertz emission from the photonic crystal. © 2008 IOP
Publishing Ltd.
• Hongbo, L. i., Brio, M., Schülzgen, A., Peyghambarian, N., & Moloney, J. V. (2007). Multimode interference in circular step-index fibers studied with the mode expansion approach. Journal of the
Optical Society of America B: Optical Physics, 24(10), 2707-2720.
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Abstract: The mode expansion approach in vectorial form, using a complete set of guided modes of a circular step-index fiber (SIF), is developed and applied to analyze multimode interference in
multimode fibers (MMFs) for the first time, to the best of our knowledge. The complete set of guided modes of an SIF is defined based on its modal properties, and a suitable modal orthogonality
relation is identified to evaluate the coefficients in a mode expansion. An algorithm, adaptive to incident fields, is then developed to systematically and efficiently perform mode expansion in
highly MMFs. The mode expansion approach is successfully applied to investigate the mode-selection properties of coreless fiber segments incorporated in multicore fiber lasers and the
self-imaging in MMFs. © 2007 Optical Society of America.
• Sørensen, M., & Brio, M. (2007). The Maxwell-Lorentz model for optical pulses. European Physical Journal: Special Topics, 147(1), 253-264.
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Abstract: Dynamics of optical pulses, especially of ultra short femtosecond pulses, are of great technological and theoretical interest. The dynamics of optical pulses is usually studied using
the nonlinear Schrödinger (NLS) equation model. While such approach workssurprisingly well for description of pulse propagation, at least in the femtosecond regime,the full system posses a wealth
of new wave phenomena that we explore in this paper: envelope collapse regularization resulting in the orignal pulse splitting; development of infinite gradients in the carrier wave; existence of
the stable top hat traveling wave solutions formed by a pair of kink anti-kink shaped optical waves. © EDP Sciences/Societé Italiana di Fisica/Springer-Verlag 2007.
• Zakharian, A. R., Brio, M., Dineen, C., & Moloney, J. V. (2006). Second-order accurate FDTD space and time grid refinement method in three space dimensions. IEEE Photonics Technology Letters, 18
(11), 1237-1239.
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Abstract: We present an algorithm based on the finite-difference time-domain method for local refinement of a three-dimensional computational grid in space and time. The method has second-order
accuracy in space and time as verified in the numerical examples. A number of test cases with material traverse normal to the grid interfaces were used to assess the long integration time
stability of the algorithm. Resulting improvements in the computation time are discussed for a photonic crystal microcavity design that exhibits a sensitive dependence of the quality factor on
subwavelength geometrical features. © 2006 IEEE.
• Gundu, K. M., Brio, M., & Moloney, J. V. (2005). A mixed high-order vector finite element method for waveguides: Convergence and spurious mode studies. International Journal of Numerical
Modelling: Electronic Networks, Devices and Fields, 18(5), 351-364.
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Abstract: In this article we construct and analyse the performance of various combinations of high order edge elements for the transverse field component with high order node elements for the
longitudinal field component. The overall efficiency and mode resolution improve dramatically with high order elements. For example, quadratic/cubic elements provide accuracy of 10-4 in
eigenvalue 1000 times faster and resolve the discontinuities in the field very accurately. In the mixed FEM for waveguides we observed that edge elements may produce spurious modes. When the
order of the edge elements is equal or larger than the order of the node elements, spurious and degenerate modes are present. Any other combination of the orders did not produce spurious modes.
We study the spurious modes from the divergence point of view. We also study the convergence properties of high order elements and compare the results with the theory. Copyright © 2005 John Wiley
& Sons, Ltd.
• Sørensen, M. P., Webb, G. M., Brio, M., & Moloney, J. V. (2005). Kink shape solutions of the Maxwell-Lorentz system. Physical Review E - Statistical, Nonlinear, and Soft Matter Physics, 71(3).
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Abstract: In the limit of high amplitude oscillating electromagnetic fields, a sequence of kink antikink shaped optical waves has been found in the Maxwell's equations coupled to a single Lorentz
oscillator and with Kerr nonlinearity. The individual kinks and antikinks result from a traveling wave assumption and their stability has been assessed by numerical simulations. For typical
physical parameter values the kink width is of the order of tens of femtoseconds. © 2005 The American Physical Society.
• Zakharian, A. R., Moloney, J. V., Dineen, C., & Brio, M. (2005). Application of the Finite-Difference Time-Domain method with local grid refinement to nanostructure design. Progress in Biomedical
Optics and Imaging - Proceedings of SPIE, 5728, 154-163.
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Abstract: The Finite-Difference Time-Domain (FDTD) method is often a viable alternative to other computational methods used for the design of sub-wavelength components of photonic devices. We
describe an FDTD based grid refinement method, which reduces the computational cell size locally, using a collection of nested rectangular grid patches. On each patch, a standard FDTD update of
the electromagnetic fields is applied. At the coarse/fine grid interfaces the solution is interpolated, and consistent circulation of the fields is enforced on shared cell edges. Stability and
accuracy of the scheme depend critically on the update scheme, space and time interpolation, and a proper implementation of flux conditions at mesh boundaries. Compared to the conformal grid
refinement, the method enables better efficiency by using non-conformal grids around the region of interest and by refining both space and time dimensions, which leads to considerable savings in
computation time. We discuss the advantages and shortcomings of the method and present its application to the problem of computation of a quality factor of a 3-D photonic crystal microcavity.
• Gundu, K. M., Kano, P., Mafi, A., Moloney, J. V., & Brio, M. (2004). Numerical simulations of active microstructured fibers. Proceedings of the 4th International Conference on Numerical
Simulation of Optoelectronic Devices, NUSOD '04, 69-70.
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Abstract: The development of a state of the art numerical simulator based on beam propagation method (BPM) is discussed. A mixed edge-node full vectorial finite element method (VFEM) is proposed
for the numerical simulation of active microstructured fibers. The features of this simulator include mixed edge-node full VFE-BPM; accurate treatment of curved surfaces for finite differences
and a propagator based on numerical Laplace transform (NLT). The development of an algorithm for optimal parameter selection and automatic error control is also discussed. The main difficulty in
the development of a VFE-BPM is the instability due to the presence of complex modes.
• Kano, P., Kouznetsov, D., Moloney, J. V., & Brio, M. (2004). Slab delivery of incoherent pump light to double-clad fiber amplifiers: Numerical simulations. IEEE Journal of Quantum Electronics, 40
(9), 1301-1305.
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Abstract: Compact fiber amplifiers with tapered slab delivery of the pump are analyzed numerically. A procedure for the selection of the structural parameters of this device is presented together
with a simple asymptotic estimate for the efficiency. Simulations based on the scalar paraxial beam propagation equation demonstrate the validity of this estimate. © 2004 IEEE.
• Moloney, J. V., Zakharian, A., Dineen, C., & Brio, M. (2004). AMR FDTD solver for nanophotonic and plasmonic applications. Proceedings of SPIE - The International Society for Optical Engineering,
5451, 97-104.
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Abstract: Uniform, nonuniform and adaptive mesh refinement FDTD approaches to solving 3D Maxwell's equations are compared and contrasted. Specific applications of such schemes to optical memory,
nanophotonics and plasmonics problems will be illustrated.
• Webb, G., Sørensen, M. P., Brio, M., Zakharian, A. R., & Moloney, J. V. (2004). Variational principles, Lie point symmetries, and similarity solutions of the vector Maxwell equations in
non-linear optics. Physica D: Nonlinear Phenomena, 191(1-2), 49-80.
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Abstract: The vector Maxwell equations of non-linear optics coupled to a single Lorentz oscillator and with instantaneous Kerr non-linearity are investigated by using Lie symmetry group methods.
Lagrangian and Hamiltonian formulations of the equations are obtained. The aim of the analysis is to explore the properties of Maxwell's equations in non-linear optics, without resorting to the
commonly used non-linear Schrödinger (NLS) equation approximation in which a high frequency carrier wave is modulated on long length and time scales due to non-linear sideband wave interactions.
This is important in femto-second pulse propagation in which the NLS approximation is expected to break down. The canonical Hamiltonian description of the equations involves the solution of a
polynomial equation for the electric field E, in terms of the canonical variables, with possible multiple real roots for E. In order to circumvent this problem, non-canonical Poisson bracket
formulations of the equations are obtained in which the electric field is one of the non-canonical variables. Noether's theorem, and the Lie point symmetries admitted by the equations are used to
obtain four conservation laws, including the electromagnetic momentum and energy conservation laws, corresponding to the space and time translation invariance symmetries. The symmetries are used
to obtain classical similarity solutions of the equations. The traveling wave similarity solutions for the case of a cubic Kerr non-linearity, are shown to reduce to a single ordinary
differential equation for the variable y=E2, where E is the electric field intensity. The differential equation has solutions y=y(ξ), where ξ=z-st is the traveling wave variable and s is the
velocity of the wave. These solutions exhibit new phenomena not obtainable by the NLS approximation. The characteristics of the solutions depends on the values of the wave velocity s and the
energy integration constant ε. Both smooth periodic traveling waves and non-smooth solutions in which the electric field gradient diverges (i.e. solutions in which |Eξ|→∞ at specific values of E,
but where |E| is bounded) are obtained. The traveling wave solutions also include a kink-type solution, with possible important applications in femto-second technology. © 2003 Elsevier B.V. All
rights reserved.
• Sørensen, M. P., Brio, M., Webb, G. M., & Moloney, J. V. (2002). Solitary waves, steepening and initial collapse in the Maxwell-Lorentz system. Physica D: Nonlinear Phenomena, 170(3-4), 287-303.
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Abstract: We present a numerical study of Maxwell's equations in nonlinear dispersive optical media describing propagation of pulses in one Cartesian space dimension. Dispersion and nonlinearity
are accounted for by a linear Lorentz model and an instantaneous Kerr nonlinearity, respectively. The dispersion relation reveals various asymptotic regimes such as Schrödinger and KdV branches.
Existence of soliton-type solutions in the Schrödinger regime and light bullets containing few optical cycles together with dark solitons are illustrated numerically. Envelope collapse regimes of
the Schrödinger equation are compared to the full system and an arrest mechanism is clearly identified when the spectral width of the initial pulse broadens beyond the applicability of the
asymptotic behavior. We show that beyond a certain threshold the carrier wave steepens into an infinite gradient similarly to the canonical Majda-Rosales weakly dispersive system. The weak
dispersion in general cannot prevent the wave breaking with instantaneous or delayed nonlinearities. © 2002 Elsevier Science B.V. All rights reserved.
• Brio, M., Zakharian, A. R., & Webb, G. M. (2001). Two-Dimensional Riemann Solver for Euler Equations of Gas Dynamics. Journal of Computational Physics, 167(1), 177-195.
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Abstract: We construct a Riemann solver based on two-dimensional linear wave contributions to the numerical flux that generalizes the one-dimensional method due to Roe (1981, J. Comput. Phys. 43,
157). The solver is based on a multistate Riemann problem and is suitable for arbitrary triangular grids or any other finite volume tessellations of the plane. We present numerical examples
illustrating the performance of the method using both first- and second-order-accurate numerical solutions. The numerical flux contributions are due to one-dimensional waves and multidimensional
waves originating from the corners of the computational cell. Under appropriate CFL restrictions, the contributions of one-dimensional waves dominate the flux, which explains good performance of
dimensionally split solvers in practice. The multidimensional flux corrections increase the accuracy and stability, allowing a larger time step. The improvements are more pronounced on a coarse
mesh and for large CFL numbers. For the second-order method, the improvements can be comparable to the improvements resulting from a less diffusive limiter. © 2001 Academic Press.
• Webb, G. M., Brio, M., & Zank, G. P. (2001). Erratum: Steady MHD flows with an ignorable coordinate and the potential transonic flow equation (Journal of Plasma Physics (1994) 52 (141-188)).
Journal of Plasma Physics, 65(3), 255-256.
• Webb, G. M., Zakharian, A. R., Brio, M., & Zank, G. P. (2001). Parametric instabilities and wave coupling in Alfvén simple waves. Journal of Plasma Physics, 66(3), 167-212.
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Abstract: The parametric instabilities in the large-amplitude Alfv́en wave (pump waves) were studied. The study was based on the formulation of a wave coupling for magnetohydrodynamic (MHD) waves
in a non-uniform background flow. The small-amplitude MHD waves interact with the pump wave via wave coupling coefficients that were dependent on the gradient and time dependence of the
background flow. The dispersion equations for different short- and long-wavelength eigenmodes, based on initial value problem, were utilized in discussing the wave stability phenomenon.
• Hunter, J. K., & Brio, M. (2000). Weak shock reflection. Journal of Fluid Mechanics, 410, 235-261.
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Abstract: We present numerical solutions of a two-dimensional inviscid Burgers equation which provides an asymptotic description of the Mach reflection of weak shocks. In our numerical solutions,
the incident, reflected, and Mach shocks meet at a triple point, and there is a supersonic patch behind the triple point, as proposed by Guderley for steady weak-shock reflection. A theoretical
analysis indicates that there is an expansion fan at the triple point, in addition to the three shocks. The supersonic patch is extremely small, and this work is the first time it has been
• Webb, G. M., Zakharian, A. R., Brio, M., & Zank, G. P. (2000). Nonlinear and three-wave resonant interactions in magnetohydrodynamics. Journal of Plasma Physics, 63(5), 393-445.
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Abstract: Hamiltonian and variational formulations of equations describing weakly nonlinear magnetohydrodynamic (MHD) wave interactions in one Cartesian space dimension are discussed. For wave
propagation in uniform media, the wave interactions of interest consist of (a) three-wave resonant interactions in which high-frequency waves may evolve on long space and time scales if the wave
phases satisfy the resonance conditions; (b) Burgers self-wave steepening for the magnetoacoustic waves, and (c) mean wave field effects, in which a particular wave interacts with the mean wave
field of the other waves. The equations describe four types of resonant triads: slow-fast magnetoacoustic wave interaction, Alfven-entropy wave interaction, Alfven-magnetoacoustic wave
interaction, and magnetoacoustic-entropy wave interaction. The formalism is restricted to coherent wave interactions. The equations are used to investigate the Alfven-wave decay instability in
which a large-amplitude forward propagating Alfven wave decays owing to three-wave resonant interaction with a backward-propagating Alfven wave and a forward-propagating slow magnetoacoustic
wave. Exact solutions of the equations for Alfven-entropy wave interactions are also discussed.
• Zakharian, A. R., Brio, M., Hunter, J. K., & Webb, G. M. (2000). The von Neumann paradox in weak shock reflection. Journal of Fluid Mechanics, 422, 193-205.
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Abstract: We present a numerical solution of the Euler equations of gas dynamics for a weak-shock Mach reflection in a half-space. In our numerical solutions, the incident, reflected, and Mach
shocks meet at a triple point, and there is a supersonic patch behind the triple point, as proposed by Guderley. A theoretical analysis supports the existence of an expansion fan at the triple
point, in addition to the three shocks. This solution is in complete agreement with the numerical solution of the unsteady transonic small-disturbance equations obtained by Hunter & Brio (2000),
which provides an asymptotic description of a weak-shock Mach reflection. The supersonic patch is extremely small, and this work is the first time it has been resolved in a numerical solution of
the Euler equations. The numerical solution uses six levels of grid refinement around the triple point. A delicate combination of numerical techniques is required to minimize both the effects of
numerical diffusion and the generation of numerical oscillations at grid interfaces and shocks.
• Webb, G. M., Zakharian, A., Brio, M., & Zank, G. P. (1999). Wave interactions in magnetohydrodynamics, and cosmic-ray-modified shocks. Journal of Plasma Physics, 61(2), 295-346.
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Abstract: Multiple-scales perturbation methods are used to study wave interactions in magnetohydrodynamics (MHD), in one Cartesian space dimension, with application to cosmic-ray-modified shocks.
In particular, the problem of the propagation and interaction of short wavelength MHD waves, in a large-scale background flow, modified by cosmic rays is studied. The wave interaction equations
consist of seven coupled evolution equations for the backward and forward Alfven waves, the backward and forward fast and slow magnetoacoustic waves and the entropy wave. In the linear wave
regime, the waves are coupled by wave mixing due to gradients in the background flow, cosmic-ray squeezing instability effects, and damping due to the diffusing cosmic rays. In the most general
case, the evolution equations also contain nonlinear wave interaction terms due to Burgers self wave steepening for the magnetoacoustic modes, resonant three wave interactions, and mean wave
field interaction terms. The form of the wave interaction equations in the ideal MHD case is also discussed. Numerical simulations of the fully nonlinear cosmic ray MHD model equations are
compared with spectral code solutions of the linear wave interaction equations for the case of perpendicular, cosmic-ray-modified shocks. The solutions are used to illustrate how the different
wave modes can be generated by wave mixing, and the modification of the cosmic ray squeezing instability due to wave interactions. It is shown that the Alfven waves are coupled to the
magnetoacoustic and entropy waves due to linear wave mixing, only in background flows with non-zero field aligned electric current and/or vorticity (i.e. if B · Δ × B ≠ 0 and/or B · Δ × u ≠ 0,
where B and u are the magnetic field induction and fluid velocity respectively).
• Webb, G. M., Brio, M., & Zank, G. P. (1998). Lagrangian and Hamiltonian aspects of wave mixing in non-uniform media: Waves on strings and waves in gas dynamics. Journal of Plasma Physics, 60(2),
More info
Abstract: Hamiltonian and Lagrangian perturbation theory is used to describe linear wave propagation in inhomogeneous media. In particular, the problems of wave propagation on an inhomogeneous
string, and the propagation of sound waves and entropy waves in gas dynamics in one Cartesian space dimension are investigated. For the case of wave propagation on an inhomogeneous heavy string,
coupled evolution equations are obtained describing the interaction of the backward and forward waves via wave reflection off gradients in the string density. Similarly, in the case of gas
dynamics the backward and forward sound waves and the entropy wave interact with each other via gradients in the background flow. The wave coupling coefficients in the gas-dynamical case depend
on the gradients of the Riemann invariants R± and entropy S of the background flow. Coupled evolution equations describing the interaction of the different wave modes are obtained by exploiting
the Hamiltonian and Poisson-bracket structure of the governing equations. Both Lagrangian and Clebsch-variable formulations are used. The similarity of the equations to equations obtained by
Heinemann and Olbert describing the propagation of bidirectional Alfvén waves in the solar wind is pointed out.
• Webb, G. M., Ratkiewicz, R., Brio, M., & Zank, G. P. (1998). Multidimensional simple waves in gas dynamics. Journal of Plasma Physics, 59(3), 417-460.
More info
Abstract: A formalism for multidimensional simple waves in gas dynamics using ideas developed by Boillat is investigated. For simple-wave solutions, the physical variables depend on a single
function φ(r, t). The wave phase φ(r, t) is implicitly determined by an equation of the form(φ) = r·n(φ)-λ(φ)t, where n(φ) denotes the normal to the wave front, λ is the characteristic speed of
the wave mode of interest, r is the position vector, t is the time, and the function f(φ) determines whether the wave is a centred (f(φ) = 0) or a non-centred (f(φ) ≠ 0) wave. Examples are given
of time-dependent vortex waves, shear waves and sound waves in one or two space dimensions. The streamlines for the wave reduce to two coupled ordinary differential equations in which the wave
phase φ plays the role of a parameter along the streamlines. The streamline equations are expressed in Hamiltonian form. The roles of Clebsch variables, Lagrangian variables, Hamiltonian
formulations and characteristic surfaces are briefly discussed. © 1998 Cambridge University Press.
• Webb, G. M., Brio, M., Zank, G. P., & Story, T. (1997). Wave-wave interactions in two-fluid cosmic-ray hydrodynamics. Journal of Plasma Physics, 57(PART 3), 631-676.
More info
Abstract: Multiple-scales perturbation methods are used to derive equations describing wave-wave interactions in two-fluid cosmic-ray hydrodynamics in one Cartesian space dimension. Two problems
are considered: (a) the interaction between short-wavelength waves in a non-uniform large-scale background flow, and (b) wave interactions between long-wavelength waves propagating through a
uniform background medium. The short-wavelength wave equations describe the interaction between the backward and forward thermal gas sound waves and the contact discontinuity eigenmode via 'wave
mixing', in which the waves are reflected by gradients in the large scale background flow. The equations also contain quadratic wave interaction terms describing (i) the Burgers self-wave
interaction term, (ii) mean-wave-field interaction terms in which a specific wave interacts with the mean field of the other waves, and (iii) three-wave resonant interactions that describe how a
sound wave resonantly interacts with the contact discontinuity to generate a reverse sound wave. In the limit of no interaction between the cosmic rays and thermal gas, and for a uniform
background state, the equations reduce to two coupled, integro-differential Burgers equations derived by Majda and Rosales for resonantly interacting waves in adiabatic gasdynamics. The
short-wavelength equations also contain squeezing instability terms associated with the large-scale cosmic-ray pressure gradient, which were first investigated by Drury and Dorfi. A generalized
wave-action equation, or canonical wave-energy equation, variational principles, and WKB analyses of the linearized equations are used to investigate the modification of the cosmic-ray squeezing
instability by wave mixing. Coupled Burgers equations are also derived in the long-wavelength regime that describe resonant wave interactions for weak, diffusively smoothed cosmic-ray-modified
shocks. In the latter equations, the cosmic-ray-modified sound waves can resonantly interact with either the contact discontinuity or the cosmic-ray pressure-balance mode to generate a reverse
sound wave.
• Webb, G. M., Zakharian, A., Brio, M., & Zank, G. P. (1997). Hamiltonian aspects of three-wave resonant interactions in gas dynamics. Journal of Physics A: Mathematical and General, 30(12),
More info
Abstract: Equations describing three-wave resonant interactions in adiabatic gas dynamics in one Cartesian space dimension derived by Majda and Rosales are expressed in terms of Lagrangian and
Hamiltonian variational principles. The equations consist of two coupled integro-differential Burgers equations for the backward and forward sound waves that are coupled by integral terms that
describe the resonant reflection of a sound wave off an entropy wave disturbance to produce a reverse sound wave. Similarity solutions and conservation laws for the equations are derived using
symmetry group methods for the special case where the entropy disturbance consists of a periodic saw-tooth profile. The solutions are used to illustrate the interplay between the nonlinearity
represented by the Burgers self-wave interaction terms and wave dispersion represented by the three-wave resonant interaction terms. Hamiltonian equations in Fourier (p, t) space are also
obtained where p is the Fourier space variable corresponding to the fast phase variable θ of the waves. The latter equations are transformed to normal form in order to isolate the normal modes of
the system.
• Webb, G. M., Brio, M., & Zank, G. P. (1996). Lie-Bäcklund symmetries of dispersionless, magnetohydrodynamic model equations near the triple umbilic point. Journal of Physics A: Mathematical and
General, 29(16), 5209-5240.
More info
Abstract: Lie-Bäcklund symmetries and conservation laws are derived for weakly nonlinear magnetohydrodynamic (MHD) equations describing the interaction of the Alfvén and magnetoacoustic modes
propagating parallel to the ambient magnetic field, in the parameter regime near the triple umbilic point, where the gas sound speed ag matches the Alfvén speed VA. The dispersive form of the
equations can be expressed in Hamiltonian form and admit four Lie point symmetries and conservation laws associated with space-translation invariance (momentum conservation), time translation
invariance (energy conservation), rotational invariance about the magnetic field B (helicity conservation), plus a further symmetry that is associated with accelerating wave similarity solutions
of the equations. The main aim of the paper is a study of the symmetries and conservation laws of the dispersionless equations. The dispersionless equations are of hydrodynamic type and have
three families of characteristics analogous to the slow, intermediate and fast modes of MHD and the Riemann invariants for each of these modes are given in closed form. The dispersionless
equations are shown to be semi-Hamiltonian, and to possess two infinite families of symmetries and conservation laws. The analysis emphasizes the role of the Riemann invariants of the
dispersionless equations and a hodograph transformation for a restricted version of the equations. © 1996 IOP Publishing Ltd.
• Webb, G. M., Brio, M., Zank, G. P., & Story, T. (1995). Contact and pressure balance structures in two-fluid cosmic-ray hydrodynamics. Astrophysical Journal Letters, 442(2), 822-846.
More info
Abstract: The role of cosmic-ray-modified contact discontinuities and pressure balance structures in two-fluid cosmic-ray hydrodynamics in one Cartesian space dimension are investigated by means
of analytic and numerical solution examples, as well as by weakly nonlinear asymptotics. The fundamental wave modes of the two-fluid cosmic-ray hydrodynamic equations in the long-wavelength limit
consist of the backward and forward propagating cosmic-ray-modified sound waves, with sound speed dependent on both the cosmic-ray and thermal gas pressures; the contact discontinuity; and a
pressure balance mode in which the sum of the cosmic ray and thermal gas pressure perturbations is zero. The pressure balance mode, like the contact discontinuity is advected with the background
flow. The interaction of the pressure balance mode with the contact discontinuity is investigated by means of the method of multiple scales. The thermal gas and cosmic-ray pressure perturbations
satisfy a linear diffusion equation, and entropy perturbations arising from nonisentropic initial conditions for the thermal gas are frozen into the fluid. The contact discontinuity and pressure
balance eigenmodes both admit nonzero entropy perturbations in the thermal gas, whereas the cosmic-ray-modified sound waves are isentropic. The total entropy perturbation is shared between the
contact discontinuity and pressure balance eigenmodes, and examples are given in which there is a transfer of entropy between the two modes. In particular, N-wave type density disturbances are
obtained which arise as a result of the entropy transfer between the two modes. A weakly nonlinear geometric optics perturbation expansion is used to study the long timescale evolution of the
short-wavelength entropy wave and the thermal gas sound waves in a slowly varying, large-scale background flow. The weakly nonlinear geometric optics expansion is also used to generalize previous
studies of a squeezing instability for short-wavelength sound waves in the two fluid model, by including a weakly nonlinear wave steepening term that leads to shock formation, as well as the
effect of long time and space dependence of the background flow. Implications of cosmic-ray-modified pressure balance structures and contact discontinuities in models of the interaction of
traveling interplanetary shocks, and compression and rarefaction waves with the solar wind termination shock are briefly discussed.
• Brio, M., & Hunter, J. K. (1992). Asymptotic equations for conservation laws of mixed type. Wave Motion, 16(1), 57-64.
More info
Abstract: We derive canonical asymptotic equations for weakly nonlinear solutions of conservation laws of mixed type. When two real wave speeds coalesce and become complex, we obtain the
transonic small disturbance equation which changes type from hyperbolic to elliptic. When the coefficient of the nonlinear term in the transonic small disturbance equation vanishes, we derive
cubically nonlinear 2 × 2 asymptotic equations. These include canonical equations which are parabolic on a line in state space and strictly hyperbolic away from this line. © 1992.
• Brio, M., & Hunter, J. K. (1992). Mach reflection for the two-dimensional Burgers equation. Physica D: Nonlinear Phenomena, 60(1-4), 194-207.
More info
Abstract: We study shock reflection for the two 2D Burgers equation. This model equation is an asymptotic limit of the Euler equations, and retains many of the features of the full equations. A
von Neumann type analysis shows that the 2D Burgers equation has detachment, sonic, and Crocco points in complete analogy with gas dynamics. Numerical solutions support the detachment/sonic
criterion for transition from regular to Mach reflection. There is also strong numerical evidence that the reflected shock in the 2D Burgers Mach reflection forms a smooth wave near the Mach
stem, as proposed by Colella and Henderson in their study of the Euler equations. © 1992.
• Brio, M., & Wu, C. (1988). An upwind differencing scheme for the equations of ideal magnetohydrodynamics. Journal of Computational Physics, 75(2), 400-422.
More info
Abstract: Recently, upwind differencing schemes have become very popular for solving hyperbolic partial differential equations, especially when discontinuities exist in the solutions. Among many
upwind schemes successfully applied to the problems in gas dynamics, Roe's method stands out for its relative simplicity and clarity of the underlying physical model. In this paper, an upwind
differencing scheme of Roe-type for the MHD equations is constructed. In each computational cell, the problem is first linearized around some averaged state which preserves the flux differences.
Then the solution is advanced in time by computing the wave contributions to the flux at the cell interfaces. One crucial task of the linearization procedure is the construction of a Roe matrix.
For the special case γ = 2, a Roe matrix in the form of a mean value Jacobian is found, and for the general case, a simple averaging procedure is introduced. All other necessary ingredients of
the construction, which include eigenvalues, and a complete set of right eigenvectors of the Roe matrix and decomposition coefficients are presented. As a numerical example, we chose a coplanar
MHD Riemann problem. The problem is solved by the newly constructed second-order upwind scheme as well as by the Lax-Friedrichs, the Lax-Wendroff, and the flux-corrected transport schemes. The
results demonstrate several advantages of the upwind scheme. In this paper, we also show that the MHD equations are nonconvex. This is a contrast to the general belief that the fast and slow
waves are like sound waves in the Euler equations. As a consequence, the wave structure becomes more complicated; for example, compound waves consisting of a shock and attached to it a
rarefaction wave of the same family can exist in MHD. © 1988.
• Brio, M. (1987). Propagation of weakly nonlinear magnetoacoustic waves. Wave Motion, 9(5), 455-458.
More info
Abstract: Using the reductive perturbation method we show that the modified Burgers equation governs the propagation of a weakly nonlinear slow magnetoacoustic wave near the equilibrium state
with zero transverse magnetic field. © 1987. | {"url":"https://profiles.arizona.edu/person/brio","timestamp":"2024-11-09T12:22:32Z","content_type":"text/html","content_length":"135227","record_id":"<urn:uuid:191b7be3-c154-4753-bca3-b9ca26cb82a8>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00733.warc.gz"} |
EPSRC Reference: EP/L027399/1
Title: Workshop on Entanglement Entropy in Many Body Quantum Systems
Principal Investigator: Doyon, Professor B
Other Investigators:
Researcher Co-Investigators:
Project Partners:
Department: Mathematics
Organisation: Kings College London
Scheme: Standard Research
Starts: 01 May 2014 Ends: 31 October 2014 Value (£): 9,654
EPSRC Research Topic Classifications:
EPSRC Industrial Sector Classifications: No relevance to Underpinning Sectors
Related Grants:
│Panel Date │Panel Name │Outcome │
Panel History: ├───────────┼────────────────────────────────────────────────────┼─────────┤
│05 Mar 2014│EPSRC Mathematics Prioritisation Meeting March 2014 │Announced│
Summary on Grant Application Form
Quantum mechanics has very intriguing features. One of them is that we just cannot know all physical properties of any system for sure - there is always some fundamental uncertainty. Wherefore with
quantum mechanics, we only predict probabilities of measurements. Another intriguing feature is that what we thought were particles sometimes behave like waves, and what we thought were waves
sometimes behave like particles. Electrons will show interference patterns, as if there was a "probability wave".
But perhaps the most intriguing, and arguable the most "quantum" of all features, is quantum entanglement. This is the single property that displays most clearly the dichotomy between the description
as particles with probabilities, and as waves with interference. Entanglement was famously referred to as "spooky action at a distance" by Einstein, and led to what is still known as the
Einstein-Podolsky-Rosen "paradox". In simple terms, it says that, according to the rules of quantum mechanics, if two particles (or two quantum systems of any kind) are entangled, then a measurement
on one particle will instantaneously affect the actual physical state of the other particle. This is spooky because entanglement can in principle exist between particles that are as far as we want
from each other: for instance, pairs of particles spontaneously created at some point and traveling in opposite directions. Something happening here on one of these particles can affect
instantaneously the state of the other while it's on the other side of the galaxy!
Entanglement has led to many interpretative issues in quantum mechanics, especially with respect to the principle of locality that was so dear to Einstein (no information can travel faster than the
speed of light), and work as been done beyond the Copenhagen interpretation we implicitly referred to here. Perhaps most interestingly, however, as Feynman envisioned, quantum mechanics, and in
particular quantum entanglement, led to a revolution in information and computing science. Quantum entanglement, this very quantum correlation between particles, is nowadays perceived as a resource,
and gives rise to algorithm that are exponentially faster than their classical counterpart, like Shor's algorithm for prime factorization of large numbers; this may have very deep technological
In recent years, the quantum information viewpoint led to an unexpected direction: quantum entanglement, it turns out, is also at the basis of many phenomena of theoretical physics that occur when
many particles interact with each other. These many-body "emergent" phenomena are some of the most interesting and complex in theoretical physics, and have been known and studied for a long time; one
of the most well-known being the Kondo effect, by which magnetic impurities in metals drastically affect its conductivity at very small temperatures. In the quantum entanglement viewpoint, this is
simply because of the strong entanglement between metallic electrons and the magnetic impurities. Studying entanglement in many-body systems has led to surprising realizations, has challenged what we
thought we understood about many-body systems, and has led to new methods, new theoretical frameworks and even new classes of many-body behaviours. This is a very active research area, with
theoretical, numerical and potentially technological implications.
This workshop will bring together the leading researchers worldwide in the area of quantum entanglement in many-body systems, with an emphasis on, but not restricted to, the entanglement entropy, a
mathematical characterization of entanglement which has found deep underpinning in many-body systems. The workshop will provide the most recent research in the area, and will be a platform for
determining and disseminating the important problems and ideas to be developed in the near future.
Key Findings
This information can now be found on Gateway to Research (GtR) http://gtr.rcuk.ac.uk
Potential use in non-academic contexts
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Description This information can now be found on Gateway to Research (GtR) http://gtr.rcuk.ac.uk
Date Materialised
Sectors submitted by the Researcher
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Project URL:
Further Information:
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Hydrodynamic pressures
Penstocks, plugs of diversion tunnels, gates in bottom outlets, intake structures and spillways, valves and other hydromechanical components of the pressurised water system in hydro power plants have
rarely been designed for the hydrodynamic pressures, which may be caused during a strong earthquake. In general the hydrodynamic pressure according to Westergaard is assumed. This assumption is
acceptable for gates of surface spillways, but not for gates located in tunnels or valves in large diameter penstocks.
Although no case is known where a penstock has failed or been damaged during a strong earthquake, the question of earthquake safety still has to be addressed. As earthquakes affect all components of
a hydro power plant, hydrodynamic actions also have to be checked for all hydromechanical components. It may be argued that in the pressurised water system the water hammer is already investigated
and the emergency shut-down of a penstock may cause the maximum hydrodynamic pressures. As discussed in the subsequent sections, the hydrodynamic pressures due to valve regulation are larger than
those due to earthquakes in long penstocks or pressure tunnels with a natural period of vibration of the water of several seconds. However, in relatively short penstocks with fundamental
eigenfrequency of the oscillating water mass in the range of the dominant frequencies of the earthquake ground motion, the situation may be the opposite.
Damage and failure of buried water and gas mains has been caused by quite a number of earthquakes. The vulnerability of these lifelines is not a new phenomenon. However, the implications of these
observations may not yet have been fully realised elsewhere. Penstocks and gates of hydro power plants are not intrinsically safe against earthquakes. Earthquakes can cause damage to penstocks by
different mechanisms as discussed below:
• Hydrodynamic pressures due to ground shaking: The magnitude of the pressure depends on the response spectrum of the ground shaking and the natural period of the oscillating mass in the pressurized
water system.
• Relative displacements along active faults.
• Support movements due to earthquake-triggered slides.
• Rock falls on penstocks on the surface in steep terrain.
• Dynamic forces in penstock due to support movements and deformations of rigid anchor points.
• Dynamic soil pressures in buried penstocks; imposed (quasi-static) deformations due to soil liquefaction, differential settlements and creep movements of slopes, etc.
It is evident from the above list that the seismic actions, which can cause damage to a penstock, cover a wide spectrum. To ignore the earthquake action because it may not have been considered in the
past would not be an acceptable argument. Also, to simply declare the earthquake action on a penstock as an acceptable risk, does not comply with today’s state of knowledge practice, particularly in
view of the fact that rather sophisticated dynamic analyses are already carried out for the seismic design of dams.
In mountainous regions the main hazard is from rock falls and mass movements. For example, during the 1999 Chi-Chi earthquake in Taiwan (magnitude 7.5) and the 1990 Manjil earthquake in Iran
(magnitude 7.6) several thousand rock falls have occurred (Figs. 1 and 2). Therefore, rock falls and mass movements have to be expected and appropriate action shall be taken in the design of
penstocks (soil cover, concrete protection, pressure shaft instead of surface penstock etc.).
Fault crossings may require special solutions, as shown in the examples in Figure 3, where the Trans Alaska pipeline is crossing the Denali fault. During the magnitude 7.9 Denali earthquake of 3
November 2002, significant horizontal fault movements occurred at this location. Due to the sliding bearings, the fault movements could be absorbed without major damage and with a very short
interruption of the pipeline operation (Sorenson and Meyer, 2003).
Besides the penstocks, the gates of spillways, bottom outlets, intakes etc. can also be affected by high hydrodynamic forces. This is the case when the gate is located in a pressure tunnel or shaft.
For example, during the 1990 Manjil earthquake in Iran, one of the two radial gates of the intermediate level spillway of the Sefid Rud buttress dam was damaged due to hydrodynamic forces resulting
from a mass oscillation of the relatively short intake tunnel. One of the arms of the radial gates buckled and one of the trunnion plates cracked (Jalalzadeh, 2000). Due to the deformations, the
damaged gate could not be operated after the earthquake and was leaking.
So what should be done in order to improve the seismic safety of a penstock? First, it is necessary to identify the type of seismic hazard (see list above) and the weak elements in a penstock.Such
checks are not very costly and it is worthwhile to perform them in the context of an overall safety assessment involving other hazards from the natural and man-made environment.
In order to minimise the potential damage in the case of failure of a penstock, the proper functioning of valves during and after an earthquake must be guaranteed. Thus, the seismic safety of these
devices shall be carefully checked and assessed.
The rock fall and landslide hazard is best assessed together with an experienced geologist, who is familiar with the local conditions. The best protection of a penstock against rock falls is by
burying the penstock.
Anchor points in potentially unstable slopes need special solutions. Of course, the best solution would be to avoid such locations. However, for high head penstocks in the mountains, the slopes are
generally very steep with slope angles up to 45°, which exceeds the friction angle of the slope wash or the weathered rock, therefore, the slope stability of supports is a general problem. Earthquake
damage can be accepted when, due to the release of the water in the penstock, people are not affected and damage to the environment and property is limited and the economical losses can be covered by
insurance. Earthquake alarm systems, which allow the automatic shut-down of a penstock and the installation of additional valves in long penstocks may be more effective than designing a penstock for
the worst seismic actions (Griesser et al., 2004).
In the subsequent sections the earthquake-induced hydrodynamic pressures in penstocks are discussed, which for low head schemes can greatly exceed the hydrostatic pressures. These pressures may not
only jeopardise the safety of the penstock but also the gates and valves located in relatively short intake tunnels and penstocks, respectively. Moreover, the plugs in closed diversion tunnels may
experience very high hydrodynamic forces during an earthquake.
Hydrodynamic pressures in penstocks due to earthquake action
The hydrodynamic pressures in penstocks can be characterized by the one-dimensional wave equation, which can be expressed as follows:
where c is the wave propagation velocity of a pressure wave in an elastic pipe (depending on the flexibility of the pipe and the embedment, c varies between ca. 1000 to1300m/s), subscripts ’t’ und
’x’ denote partial derivation with respect to time and space, x is the coordinate along the pipe axis and u(x,t) is the time-dependent displacement of the elastic fluid in direction of the pipe axis.
In case of an earthquake ground acceleration üg(t) in direction of the pipe axis, the relative acceleration utt has to be replaced by the total acceleration, which is the sum of the ground
acceleration and the relative acceleration, i.e. the wave equation takes the following form:
The hydrodynamic pressure, p(x,t), can be obtained from the following relationship, assuming the fluid in the pipe being an elastic material:
where E = ρ c^2 represents the equivalent modulus of elasticity of the water and the elastic pipe and r is the mass density of water (1000 kg/m^3).
In a penstock with the length L, which is closed at one end and terminates at the surface of a reservoir or surge chamber, the hydrodynamic pressure due to a ground excitation can be determined with
the mode superposition method (Clough and Penzien, 1975). Accordingly, the modes of vibration X[n](x) (subscript n denotes the n-th mode) can be expressed as follows (note: x = 0 at the free surface
end of the pipe):
The modes of vibration satisfy the following orthogonality relationship:
Using u(x,t) = ∑ X[n](x) u[n](t) we obtain from equation 4
where u[n](t) represents the n-th modal coordinate.
By substituting equation 6 into equation 2, multiplication with Xn(x), introduction of a viscous damping term and integration along ther whole pipeline, the following equations of motion result:
where ξn: n-th damping ratio
ωn: n-th circular frequency of oscillating fluid in pipe in rad/s
d[n]: n-th modal partizipation factor,
üg(t): ground acceleration in direction of pipe axis.
The eigenfrequencies of the pressure system are as follows:
The fundamental frequency (n=1) can be taken as f[1] = c/(4L).
The maximum of u[n](t) can be obtained directly from the acceleration response spectrum, S[a](f[n], ξ[n]), i.e.
Using equation 3 the hydrodynamic pressure at the fixed end (x = L) of the n-th mode of vibration can be obtained in an analogous way.
By substituting equation. 11 and the expressions for f[n] and d[n] in equation 12, we obtain the following maximum n-th pressure at the fixed end of the pipe:
The total hydrodynamic pressure at the closed end of the pipe can be obtained by using the the following standard superposition method (SRSS-method) for the modal maxima:
If only the contribution of the first mode is taken into account then we obtain the following maximum hydrodynamic pressure at the fixed end:
In case of a pipe (or tunnel) filled with an incompressible fluid where the flexibility of the pipe is not accounted for, we obtain
where ao is the peak ground acceleration. This means the total mass in the pipe (or tunnel) behaves like a rigid body and the maximum inertia force is equal the total mass of water in the pipe of
length L multiplied with the peak ground acceleration.
The ratio between the maximum hydrodynamic pressures of the compressible and incompressible fluid, characterized by K, can be determined as follows:
If the design response spectra shown in Fig. 5 are used, then the maximum spectrum amplification factor for the absolute acceleration s = Sa,max/a0 varies between 2.5 (soil type A: rock) and 3.5
(soil type E) for a damping ratio of 5%. However, the damping ratio of the pressure oscillations in a penstock is usually less than 5%. For seismic design purposes a damping ratio of about 2% may be
assumed. The effect of the damping ratio on the elastic acceleration response spectra can be approximated by the following correction factor (Eurocode 8, 2004):
The response spectra for 5% damping have to be multiplied by the above factor for periods exceeding 0.15 s (soil type A). In the case of an undamped pressure oscillation, we obtain h = 1.414.
Therefore, for a rock foundation (s = 2.5) we obtain a maximum K-value of 2.9 (equation 17) for the undamped penstock. This means that the maximum hydrodynamic pressure in a compressible liquid in a
penstock can be more than twice as large as in an incompressible one.
In the case of soil type E (s = 3.5) we obtain a maximum value of K of 8×3.5×1.41/3.142 = 4.0.
If a pressure tunnel is located in the body of a massive concrete dam (e.g. orifice spillway in arch dam) then the hydrodynamic pressure must be calculated using the so-called floor response spectrum
at the location of the pressure tunnel rather than the acceleration response spectrum of the ground motion.
Example 1: 1000 m long penstock
Determine the maximum hydrodynamic pressure in a 1000 m long penstock with the following properties: rock site with a peak ground acceleration of 0.3 g (3m/s2); wave propagation velocity in the
elastic pipe of c=1100 m/s, damping ratio of 0%, mass density of water of 1000 kg/m3. The fundamental period of vibration according to equation 10 is T1 = 1/f1 = 4L/c = 3.64 s. The maximum
hydrodynamic pressure of the first mode of vibration is obtained from equation 15 using the response spectrum shown in Figure 5 for soil type A and a vibration period of 3.64 s (spectrum value: 0.264
for 5% damping). For an undamped pressure oscillation the 5% damped response spectrum has to be multiplied by h = 1.414 according to equation 18. Therefore, we obtain
p[1,max] = 8 x 1000 kg/m^3 x 1000 m x 1.414 x 0.264 x 3 m/s^2 / 3.14^2 = 0.91 MPa or 9.1 bar
This maximum hydrodynamic pressure is independent of the head or the average slope of the penstock. Therefore, for high head penstocks with a hydrostatic pressure of several MPa’s, a maximum dynamic
pressure of about 1 MPa is not a serious concern. However, for low head penstocks with static pressures in the order of the dynamic pressure – this applies mainly to pipes in relatively flat terrain
– the hydrodynamic pressure has to be checked carefully. For penstocks with a length of say 3000m the natural period of vibration is three times that of the 1000 m long one and increases to 10.9 s.
For this period the spectrum acceleration drops to roughly 1/9 of that of the 1000m long pipe as the tail of the response spectrum is proportional to one over the square of the period. This results
in a hydrodynamic pressure of ca. 0.1 MPa or 1 bar.
The maximum hydrodynamic pressure of the second mode of vibration (n=2) with a period of vibration of T2 = 4L/(3c) = 1.212 s can be determined from equation 13 (spectrum value for 1.212 s for soil
type A in Fig. 5: 0.825):
p[2,max] = 8 x 1000 kg/m^3 x 1000 m x 1.414 x 0.825 x 3 m/s^2 /(9 x 3.142) = 0.32 MPa (3.2 bar)
Similarly, for the third mode of vibration we obtain (period of vibration: 0.727 s, spectrum value: 1.375):
p[3,max] = 8 x 1000 kg/m^3 x 1000 m x 1.414 x 1.375 x 3 m/s^2 /(25 x 3.142) = 0.19 MPa (1.9 bar)
From these numerical values it can be seen that the maximum hydrodynamic pressures decrease with increasing mode number despite the fact that the response spectrum is still increasing for periods
down to 0.4 s. This is due to the fact that the hydrodynamic pressures are inversely proportional to (2n-1)2, where n is the mode number.
Finally the total hydrodynamic pressure has to be calculated using the SRSS superposition of the modal maxima according to equation 14. Thus for the first three modes we obtain:
p[tot] = (0.91^2 + 0.32^2 + 0.19^2)^1/2 = 0.98 MPa (9.8 bar)
By comparing the total hydrodynamic pressure of the first three modes with that of the first mode we can note that the total pressure is only 8% larger than that of the first mode contribution. In
view of the different assumptions made, this is a negligible difference. Therefore, for most applications it is sufficient to include only the contribution of the first mode.
Example 2: 130m long pressure tunnel
Determine the maximum hydrodynamic pressure in a 130m long pressure tunnel with the following properties: rock site with a peak ground acceleration of 0.3g (3m/s2); wave propagation velocity in the
elastic pipe of c=1300m/s, damping ratio of 5%, mass density of water of 1000kg/m3. The fundamental period of vibration according to equation 10 is T1 = 1/f1 = 4L/c = 0.4 s. The maximum hydrodynamic
pressure of the first mode of vibration is obtained from equation 15 using the response spectrum shown in Figure 5 for soil type A and a vibration period of 0.4 s (spectrum value: 2.5 for 5%
damping). Therefore, we obtain
p[max] = 8 x 1000 kg/m^3 x 130 m x 2.5 x 3 m/s^2 / 3.14^2 = 0.79 MPa (7.9 bar)
The maximum hydrodynamic pressure is of the same order as that of the 1000m long penstock in Example 1. If zero damping would be assumed then the maximum hydrodynamic pressure would reach 11.2 bar,
which is more than that in Example 1. For short penstocks with small head or gates and plugs in relatively short tunnels, this dynamic pressure is quite important. The example is selected in such a
way that for the given response spectrum the hydrodynamic pressure would become a maximum, i.e. if the pressure tunnel is longer or shorter than 130m then the hydrodynamic pressure will decrease.
The contributions of the higher modes of vibration play a smaller role in this example than in Example 1 as the spectrum values of the rock response spectrum do not increase further if the periods of
vibration decrease to below 0.4 s (in Example 2 the spectrum values of the 2nd and 3rd modes were larger than those of the 1st mode)..
In an incompressible pipe or tunnel the maximum hydrodynamic pressure acting on a gate or valve is given by equation 16. Accordingly in Example 1 we get a maximum dynamic pressure of
p[max] = 1000 kg/m^3 x 1000 m x 3 m/s^2 = 3 MPa (30 bar)
and in Example 2 we obtain
p[max] = 1000 kg/m^3 x 130 m x 3 m/s^2 = 0.39 MPa (3.9 bar).
We have to keep in mind that the slope of a penstock or pressure tunnel does not have any effect on the maximum hydrodynamic pressure caused by an earthquake.
Furthermore, during pressure oscillation no negative pressures are possible. If such pressures are calculated based on the linear analysis presented above then nonlinear processes would change the
nature of the oscillations. These phenomena are not discussed in this paper.
Conclusions drawn from numerical examples
Based on the above illustrative numerical examples the following conclusions may be drawn regarding the hydrodynamic pressures:
• The hydrodynamic pressures in short penstocks (typical situation: dam with power house at the base of the dam, see Figure 6) and short pressure tunnels can be quite high during strong earthquake
shaking. In this case resonance-like pressure oscillations can develop under earthquake shaking, i.e. the lowest eigenfrequency of the pressurised water system is in the range of the dominant
frequencies of ground shaking. In this case the assumption of an incompressible fluid leads to an underestimate of the hydrodynamic pressures.
• In penstocks with a length of several hundred meters, the lowest eigenfrequency of the pressurized water system is far below the dominant frequencies of ground shaking. In this case the maximum
hydrodynamic pressures in a compressible fluid are below those of an incompressible fluid.
• The contributions of higher modes of vibration of the pressurized water system are relatively small and may be neglected.
• The maximum hydrodynamic pressures depend on the shape of the acceleration response spectrum. In the case of long penstocks the fundamental period of vibration is of the order of several seconds,
which is beyond that of most buildings and bridges. Thus spectrum values result, which are rather low. Moreover, the design response spectra given in building codes may be rather inaccurate in that
range of periods. Therefore, the design response spectra used for such analyses have to be reviewed carefully.
Design of penstocks and gates in pressure tunnels
In the seismic design and safety check of penstocks and gates and valves in pressure tunnels, the following items have to be taken into consideration:
• The design earthquake (DE) or safety evaluation earthquake (SEE) occurs very seldom, i.e. every 475 years when the DE is used. However, for safety-relevant elements like the bottom outlet, which
must function after an earthquake, the SEE with a return period of say 10,000 years has to be taken into account.
• The high hydrodynamic pressures act during a relatively short period of time, i.e. for a sinusoidal oscillation with a period of 5 s the duration of high dynamic pressures is roughly 2 s in each
• In the case of extreme events with very low probability of occurrence average strength values may be taken into account together with load and safety factors of 1.0 if no specific regulations and
guidelines exist for such design situations.
• Local plastification in ductile steel elements is acceptable as long as the safety-relevant gates are still operable after the earthquake. This applies mainly to structural elements subjected to
high flexural stresses where, e.g., a small plastified zone in a flange does not cause any significant deformations, which would impair the serviceability of a gate.
Using the above concepts – if applicable – then penstocks or gates, which have been designed for a safety factor of 1.8 should also be able to cope with the increased stresses due to
earthquake-induced hydrodynamic pressures. In any case a check of the seismic stresses and deformations is necessary.
Penstocks and other elements of the pressurised water system in hydro power plants have seldom been designed for earthquake action.
Earthquakes can cause substantial hydrodynamic forces in penstocks and pressure tunnels. Depending on foundation conditions (soil type) and length of a penstock or pressure tunnel the maximum
hydrodynamic pressures are up to 1 MPa for a peak ground acceleration of 0.3g. These dynamic pressures are not a problem for high head penstocks in steep terrain as they amount to less than 20% of
the hydrostatic pressures in a penstock with a head of 500m. However, in pipes in rather flat terrain with low head, the maximum hydrodynamic pressures can be of the same magnitude as the hydrostatic
High dynamic pressures are also expected in short pressure tunnels located within the body of a dam, where at the location of the tunnel a higher acceleration response is expected than on the ground.
However, the main seismic hazard for long surface penstocks in mountainous regions results from mass movements and rock falls. In steep terrain large numbers of rock falls can be triggered by strong
earthquakes. This was, for example, the case during strong earthquakes in Taiwan (1999) and Iran (1990).
The analysis procedure discussed in this paper can be used to determine the maximum hydrodynamic pressures in penstocks at the location of closed valves, the dynamic pressures acting on valves, gates
and plugs installed in penstocks, pressure tunnels and diversion tunnels respectively.
Author Info:
Dr. Martin Wieland, Chairman, ICOLD Committee on Seismic Aspects of Dam Design, Electrowatt-Ekono Ltd. (Jaakko Pöyry Group), Hardturmstrasse 161, CH-8037 Zurich, Switzerland. Tel. 076 356 28 62; Fax
044 355 55 61; E-mail: martin.wieland@ewe.ch | {"url":"https://www.waterpowermagazine.com/analysis/hydrodynamic-pressures/","timestamp":"2024-11-14T23:49:54Z","content_type":"text/html","content_length":"115212","record_id":"<urn:uuid:03cde229-3ddf-48ed-9062-9718ba2fb7dc>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00539.warc.gz"} |
Further questions to the historiography of Arabic (but not only Arabic) mathematics from the perspective of Romance abbacus mathematics: Contribution to the "9<sup>ième</sup> Colloque Maghrébin sur l'Histoire des Mathématiques Arabes", Tipaza, 12-13-14 mai 2007
Since some years I have been engaged in a close reading of early Italian abbacus books and related material from the Ibero-Provençal orbit and in comparison of this material with Arabic mathematical
writings. At the 7th North African Meeting on the History of Arab Mathematics in Marrakesh in 2002 I presented the first outcome of this investigation: namely that early Italian abbacus algebra was
not influenced by the Latin algebraic writings of the 12th-13th centuries (neither by the translations of al-Khwa¯rizmı¯ nor by the works of Fibonacci); instead, it received indirect inspiration from
a so far unknown link to the Arabic world, viz to a level of Arabic algebra (probably integrated with mu a¯mala¯t mathematics) of which very little is known. At the 8th Meeting in Tunis in 2004
mathematics) of which very little is known.
At the 8th Meeting in Tunis in 2004 I presented a list of linguistic clues which, if applied to Arabic material, might enable us to say more about the links between the Romance abbacus tradition and
Arabic mu a¯mala¯t teaching.
Here I investigate a number of problem types and techniques which turn up in some but not necessarily in all of the following source types:
- Romance abbacus writings,
- Byzantine writings of abbacus type,
- Arabic mathematical writings of various kinds,
- Sanskrit mathematical writings,
in order to display the intricacies of the links between these - intricacies which force us to become aware of the shortcomings of our current knowledge, and hence formulate questions that go beyond
the answers I shall be able to present.
Originalsprog Engelsk
Antal sider 23
Status Udgivet - 2007 | {"url":"https://forskning.ruc.dk/da/publications/further-questions-to-the-historiography-of-arabic-but-not-only-ar","timestamp":"2024-11-03T18:17:55Z","content_type":"text/html","content_length":"46784","record_id":"<urn:uuid:45d29c83-a95a-424d-9434-d42afbbafcb7>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00176.warc.gz"} |
Extended hubbard model ground state energy not converging for negative V
Dear Sir / Ma'am,
I was calculating the ground state enrgy for extended hubbard model using the code provided in sample folder. I used the range for U1 to be -8 to 8, and the range for V1 (nearest neighbour spin-spin
interaction) to be in the range -6 to 6. in the above range, for some region of negative V1, the energy was not converging. It will be very helpful if someone can tell me what is going wrong here.
Some other points about the parameters I used:
1. I used a lattice site of 10, with half-filling
2. I used periodic boundary condition
3. The sweeps were following:
nsweeps = 60
maxdim mindim cutoff niter noise
50 10 1E-12 2 1E-7
100 20 1E-12 2 1E-8
200 20 1E-12 2 1E-10
400 20 1E-12 2 0
800 20 1E-12 2 1E-11
800 20 1E-12 2 0
Hi Sreetama,
Thanks for the question. I think before we can provide a good answer to you, we'll need to ask some questions of you.
1. what is the reason for choosing periodic boundary conditions? You may know that DMRG does not work nearly as well for periodic compared to open boundary conditions. It can make a very large
difference in the ability of DMRG to give accurate results. Would you be able to use open boundary conditions in your study or is there a special reason you need periodic?
2. when you say that the results were not converging, what did you see in the output? Was the energy still changing by a large amount in the last few sweeps? Or were you not reaching the truncation
error you desired? Or what was the criterion for convergence you were using? If you could show a printout of the last few sweeps that would be useful to see.
Best regards,
Hi Miles,
thanks for your reply. I will try to answer your questions below.
1. Yes I know that DMRG does not work well with periodic boundary conditions. I thought for small lattice size, it should not be a problem. But after the ground state energy did not converge for some
values of V, I removed the periodic boundary condition. But the non-convergence still persisted. The PBC is not necessary for my work, so I can work without it.
2. The energy was changing in the 6th or 7th decimal place. When I calculated the entanglement, it was showing an absurd change with respect to nearby regions.
3. The energy is not converging even in some regions where U and V both are positive.
Hi Sreetama,
Thanks for those details. It's hard to know for sure what's going wrong. Have you checked *very carefully* that any extra terms you added to your Hamiltonian are Hermitian? We don't check for that
automatically and sometimes adding non-Hermitian terms can happen by accident with fermions, which can lead to strange results in DMRG.
Other possibilities include:
1. you could just need to use a different initial state for DMRG. Please try experimenting with that.
2. if the main error you are getting is that some properties, such as entanglement, are coming out wrong, please check carefully that the measurement code itself that is measuring the entanglement is
correct and doesn't have a bug in it
Hi Streetama,
When I was computing E.H.M. ground states for V > U > 0, which nearly approaches a product state as V >>> U, the entanglement goes down.
This sounds to us as a good thing, but in fact with DMRG it is a bad thing because it takes an extremely large number of fast sweeps in order to converge. In essence, the small number of nonzero
entries in the SVD leads to poor communication of information across the lattice with each sweep.
For instance, I believe I needed to run hundreds of sweeps to get the energy to settle down, even for small system sizes, and I observed the entanglement entropy changed very slowly with each sweep.
In order to test this, I'd recommend changing your sweeps parameters so that you stay at a single maxdim for several hundred sweeps and wait until the energy settles down before increasing maxdim.
On the other hand, if your boundary conditions are incorrect (for instance, anti-periodic instead of periodic) you can also see wild changes in sweep to sweep energy; I experienced this error one
time due to a mistake in my definition of the Hamiltonian.
If you figure it out please post back here as I'm interested to know what you find, as I am continuing to use these methods.
Good luck!
Dear Jon,
So I was trying to get the ground state for U=0.4 and V=-6, and used about 600 sweeps with the final ones having maxdim 2400. The energy converged only upto 5th decimal place, but of course for
higher sweeps it will change in the 5th decimal place as well. So, I can say that the energy was not at all converged. As a result, when I am calculating two-site reduced density matrix, it is
providing a wrong state (because the probabilities are becoming negative).
I think the region you mentioned i.e. V>U>0, will have the same convergence problem.
Kindly let me know if you have any other insight about this. Were you able to get an well converged energy for V>U>0? Was the state a valid one?
With 32 sites and open boundaries, for V >> U, it required about 160 sweeps to get 6 decimal places on the energy.
I didn't do any explicit checks on the validity of the state because the energy was reasonable and settled to a definite value (as I said, 6 decimal place accuracy), and the bipartite entropy was
reasonable and real-valued. It is possible I missed something, however.
There may be some physical effects at play. With open boundaries, in the CDW phase (V >> U > 0) there are two equivalent ground states that crossover at a domain wall. My calculations took a long
time to move the domain wall to the middle of the lattice. You can fix this by pinning the boundaries to enforce one of the two degenerate ground states. There is also the possibility of phase
The essence of the difficulty of these effects for DMRG is that the ground state manifold hosts many states with similar energy but very different configuration -- and the different states are hard
for the DMRG local-update algorithm to access, at least in the CDW case, because of the low entanglement (low entanglement = few off-diagonal terms in density matrix). Using boundary pinning to
artificially select one of the ground states, or a well-chosen initial MPS helps.
I don't have much experience with negative V, although I did do a very limited number of calculations early in my project that came up with strange results, possibly for the reasons above. I didn't
go deeper into understanding the reasons at that time, because my project was focused on positive U,V, but also because I had no understanding of what was happening.
Another point I should mention:
If the reason for the slow convergence is low entanglement, you should find that increasing the bond dimension does not improve the convergence.
If this is the case, then try increasing the "noise" parameter to add random entries to the density matrix. | {"url":"https://www.itensor.org/support/2432/extended-hubbard-ground-state-energy-converging-negative","timestamp":"2024-11-10T22:28:03Z","content_type":"text/html","content_length":"36870","record_id":"<urn:uuid:81ef8a1b-282e-429e-8051-55b268d25e71>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00080.warc.gz"} |
The Matroid Union
Online Talk: Amena Assem
Tuesday, Jan 11, 3pm ET (8pm GMT, 9am Wed NZDT)
Amena Assem, University of Waterloo
Edge-Disjoint Linkage in Infinite Graphs
In 1980 Thomassen conjectured that, for odd $k$, an edge-connectivity of $k$ is enough for a graph to be weakly $k$-linked, meaning any $k$ pairs of terminals can be linked by $k$ edge-disjoint
paths. The best known result to date for finite graphs is from 1991, by Andreas Huck, and assumes an edge-connectivity of $k+1$ for odd $k$. In 2016, Ok, Richter, and Thomassen proved that, for odd
$k$, an edge-connectivity of $k+2$ implies weak $k$-linkage for $1$-ended locally finite graphs. An important auxiliary graph in edge-connectivity proofs is the lifting graph.
In this talk I will show how to reduce the connectivity condition in the result of Ok, Richter, and Thomassen to $k+1$, and then how to generalize to arbitrary infinite graphs, not necessarily
locally finite, and possibly with uncountably many ends. I will also prove an extension of a result of Ok, Richter, and Thomassen about characterizing lifting graphs, and show that if the $k$-lifting
graph of $G$ at $s$, $L(G,s,k)$, has a connected complement, then the graph $G$ has either a cycle-like or path-like structure around $s$ with $(k-1)/2$ edges between any two consecutive blobs.
Finally, will show how this structure might be used to prove that the conjecture for finite graphs implies the conjecture for infinite graphs.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | {"url":"http://matroidunion.org/?p=4537","timestamp":"2024-11-13T19:11:49Z","content_type":"text/html","content_length":"19957","record_id":"<urn:uuid:114e7053-b6dd-4dfa-8abf-ae7f0f0f8ff7>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00740.warc.gz"} |
This page will hopefully store my notes on various maths courses, and will hopefully grow over time.
Part II
Mathematics of Machine Learning [link], complete.
This is a new course, and I am not aware of any complete sets of notes available other than these at present. Highlights are the much nicer references to previous equations than in existing notes,
and also the elementary proof of the full form of Hoeffding’s Lemma.
Part IB
Geometry [link], near complete.
I found this course hard, and found a number of sections of the course covered at a rapid speed. My goal in these notes is to distill the important intuitions, but also fill in details in the
sections I was less clear about.
All sections of the course are covered in some detail, except the final section on hyperbolic geometry.
Analysis [link], very rough.
Currently covers some more background on the differentiation section of Analysis and Topology, as well as a technical detail that allows the general Cauchy’s Integral Formula for derivatives to be
Machine Learning
AI Safety Workshop (CUAI) [link]
A workshop (with exercises!) on the ‘off switch’ problem in AI. Explores ‘safe interruptibility’: one way of thinking about this problem in the context on current Reinforcement Learning systems.
Cambridge Societies
I spent some time at Cambridge in different societies:
Backpropagation and einsum
This post assumes familiarity with the einsum function. A great introduction can be found here which describes more than what we’ll need in this blog post.
The setup
I was recently preparing for machine learning interviews, and was told to have familiarity with Python 3.7 and numpy. It’s not surprising to need to code in Python for ML interviews, but using numpy?
It seemed a good idea to write some forwards and backwards passes for common functions in neural networks. It turns out that when you use einsum this is a lot less of a headache than you might have
Let’s suppose we have some linear module in a neural network with a weight W computing some example (intentionally complicated) operation
Y_{bij} = \sum_{k, l} X_{bik} W_{jkl}
This may seem scary, but of course we’ll just compute this as
Y = torch.einsum(
Now, how about writing the backwards method for this linear layer?
Previously, I would have gotten out pencil and paper to work out this mess. Now, we’ll use a result (proven later in this post) to make this process much faster. Recall that the backwards pass takes
some upstream gradient dL_dY as input and needs to calculate dL_dW to do backprop on this weight matrix and dL_dX the upstream gradient for further backprop (here, L is the loss we’re optimizing).
Indeed, we can immediately calculate
dL_dW = torch.einsum(
dL_dX = torch.einsum(
by simply permuting the three terms in the einsum string, and inserting the similarly shaped tensors (note that we only compute gradients from dL_dY, rather than Y).
Why is this true? Let’s start with the dL_dW expression. The key idea is to use the chain rule, so that
dL_dW = \sum dL_dY * dY_dW
where the * represents the product, and we sum this over all possible Y indices. I think the easiest way to see the result is then to imagine fixing one particular Y index and calculating the
gradient of that Y value with respect to the weight tensor W. Indeed, it’s now clear that we can compute dL_dY by looking at the relevant indices in the W tensor, and since we’re summing over
everything, we get the einsum expression at the end of the day. Similarly for dL_dX, here we can use dL_dX = \sum dL_dY * dY_dX and this explains the W that appears in the einsum expression for the
calculation here.
Additionally, even for functions such as convolutions, by representing these as matrix multiplications, we can calculate gradients in a far easier way. Consider the following implementation of a
minimal conv2D function from ARENA (this was implemented intentionally using numpy only, hence the ugly conversions!):
def conv1d_minimal(x: Float[Tensor, "b ic w"], weights: Float[Tensor, "oc ic kw"]) -> Float[Tensor, "b oc ow"]:
Like torch's conv1d using bias=False and all other keyword arguments left at their default values.
x: shape (batch, in_channels, width)
weights: shape (out_channels, in_channels, kernel_width)
Returns: shape (batch, out_channels, output_width)
x = x.numpy()
weights = weights.numpy()
output_width = x.shape[2] - weights.shape[2] + 1
xstrided = np.lib.stride_tricks.as_strided(
shape=(x.shape[0], x.shape[1], output_width, weights.shape[-1]), # batch, in_channels, output_width, kernel_width
strides=(x.strides[0], x.strides[1], x.strides[2], x.strides[2]),
return torch.tensor(
"batch in_channels output_width kernel_width, \
out_channels in_channels kernel_width \
-> batch out_channels output_width",
Then we can create a backwards pass by reconstructing the large xstrided function and then aggregating the contributions of X from this:
def conv2d_minimal_backwards(dL_dY: Float[Tensor, "b oc ow"])
dL_dW = torch.einsum(
"batch in_channels output_width kernel_width, \
batch out_channels output_width \
-> out_channels in_channels kernel_width",
dL_dXstrided = torch.einsum(
"out_channels in_channels kernel_width, \
batch in_channels output_width \
-> batch in_channels output_width kernel_width",
# calculate all the gradients with as_strided for the "middle" elements that contribute exactly kernel_width times to downstream convolutions
# calculate all the gradient manually for the "edge" elements which contribute less than kernel_width elements (there are few of these so this doesn't affect time complexity)
# tada! | {"url":"https://arthurconmy.github.io/notes/","timestamp":"2024-11-03T20:05:11Z","content_type":"text/html","content_length":"14254","record_id":"<urn:uuid:0ee9dfef-1dfa-48c3-b2c5-a09d4fcbdd6c>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00786.warc.gz"} |
Dijkstra Algorithm¶
You are given a directed or undirected weighted graph with $n$ vertices and $m$ edges. The weights of all edges are non-negative. You are also given a starting vertex $s$. This article discusses
finding the lengths of the shortest paths from a starting vertex $s$ to all other vertices, and output the shortest paths themselves.
This problem is also called single-source shortest paths problem.
Here is an algorithm described by the Dutch computer scientist Edsger W. Dijkstra in 1959.
Let's create an array $d[]$ where for each vertex $v$ we store the current length of the shortest path from $s$ to $v$ in $d[v]$. Initially $d[s] = 0$, and for all other vertices this length equals
infinity. In the implementation a sufficiently large number (which is guaranteed to be greater than any possible path length) is chosen as infinity.
$$d[v] = \infty,~ v \ne s$$
In addition, we maintain a Boolean array $u[]$ which stores for each vertex $v$ whether it's marked. Initially all vertices are unmarked:
$$u[v] = {\rm false}$$
The Dijkstra's algorithm runs for $n$ iterations. At each iteration a vertex $v$ is chosen as unmarked vertex which has the least value $d[v]$:
Evidently, in the first iteration the starting vertex $s$ will be selected.
The selected vertex $v$ is marked. Next, from vertex $v$ relaxations are performed: all edges of the form $(v,\text{to})$ are considered, and for each vertex $\text{to}$ the algorithm tries to
improve the value $d[\text{to}]$. If the length of the current edge equals $len$, the code for relaxation is:
$$d[\text{to}] = \min (d[\text{to}], d[v] + len)$$
After all such edges are considered, the current iteration ends. Finally, after $n$ iterations, all vertices will be marked, and the algorithm terminates. We claim that the found values $d[v]$ are
the lengths of shortest paths from $s$ to all vertices $v$.
Note that if some vertices are unreachable from the starting vertex $s$, the values $d[v]$ for them will remain infinite. Obviously, the last few iterations of the algorithm will choose those
vertices, but no useful work will be done for them. Therefore, the algorithm can be stopped as soon as the selected vertex has infinite distance to it.
Restoring Shortest Paths¶
Usually one needs to know not only the lengths of shortest paths but also the shortest paths themselves. Let's see how to maintain sufficient information to restore the shortest path from $s$ to any
vertex. We'll maintain an array of predecessors $p[]$ in which for each vertex $v \ne s$, $p[v]$ is the penultimate vertex in the shortest path from $s$ to $v$. Here we use the fact that if we take
the shortest path to some vertex $v$ and remove $v$ from this path, we'll get a path ending in at vertex $p[v]$, and this path will be the shortest for the vertex $p[v]$. This array of predecessors
can be used to restore the shortest path to any vertex: starting with $v$, repeatedly take the predecessor of the current vertex until we reach the starting vertex $s$ to get the required shortest
path with vertices listed in reverse order. So, the shortest path $P$ to the vertex $v$ is equal to:
$$P = (s, \ldots, p[p[p[v]]], p[p[v]], p[v], v)$$
Building this array of predecessors is very simple: for each successful relaxation, i.e. when for some selected vertex $v$, there is an improvement in the distance to some vertex $\text{to}$, we
update the predecessor vertex for $\text{to}$ with vertex $v$:
$$p[\text{to}] = v$$
The main assertion on which Dijkstra's algorithm correctness is based is the following:
After any vertex $v$ becomes marked, the current distance to it $d[v]$ is the shortest, and will no longer change.
The proof is done by induction. For the first iteration this statement is obvious: the only marked vertex is $s$, and the distance to is $d[s] = 0$ is indeed the length of the shortest path to $s$.
Now suppose this statement is true for all previous iterations, i.e. for all already marked vertices; let's prove that it is not violated after the current iteration completes. Let $v$ be the vertex
selected in the current iteration, i.e. $v$ is the vertex that the algorithm will mark. Now we have to prove that $d[v]$ is indeed equal to the length of the shortest path to it $l[v]$.
Consider the shortest path $P$ to the vertex $v$. This path can be split into two parts: $P_1$ which consists of only marked nodes (at least the starting vertex $s$ is part of $P_1$), and the rest of
the path $P_2$ (it may include a marked vertex, but it always starts with an unmarked vertex). Let's denote the first vertex of the path $P_2$ as $p$, and the last vertex of the path $P_1$ as $q$.
First we prove our statement for the vertex $p$, i.e. let's prove that $d[p] = l[p]$. This is almost obvious: on one of the previous iterations we chose the vertex $q$ and performed relaxation from
it. Since (by virtue of the choice of vertex $p$) the shortest path to $p$ is the shortest path to $q$ plus edge $(p,q)$, the relaxation from $q$ set the value of $d[p]$ to the length of the shortest
path $l[p]$.
Since the edges' weights are non-negative, the length of the shortest path $l[p]$ (which we just proved to be equal to $d[p]$) does not exceed the length $l[v]$ of the shortest path to the vertex $v$
. Given that $l[v] \le d[v]$ (because Dijkstra's algorithm could not have found a shorter way than the shortest possible one), we get the inequality:
$$d[p] = l[p] \le l[v] \le d[v]$$
On the other hand, since both vertices $p$ and $v$ are unmarked, and the current iteration chose vertex $v$, not $p$, we get another inequality:
$$d[p] \ge d[v]$$
From these two inequalities we conclude that $d[p] = d[v]$, and then from previously found equations we get:
$$d[v] = l[v]$$
Dijkstra's algorithm performs $n$ iterations. On each iteration it selects an unmarked vertex $v$ with the lowest value $d[v]$, marks it and checks all the edges $(v, \text{to})$ attempting to
improve the value $d[\text{to}]$.
The running time of the algorithm consists of:
• $n$ searches for a vertex with the smallest value $d[v]$ among $O(n)$ unmarked vertices
• $m$ relaxation attempts
For the simplest implementation of these operations on each iteration vertex search requires $O(n)$ operations, and each relaxation can be performed in $O(1)$. Hence, the resulting asymptotic
behavior of the algorithm is:
This complexity is optimal for dense graph, i.e. when $m \approx n^2$. However in sparse graphs, when $m$ is much smaller than the maximal number of edges $n^2$, the problem can be solved in $O(n \
log n + m)$ complexity. The algorithm and implementation can be found on the article Dijkstra on sparse graphs.
const int INF = 1000000000;
vector<vector<pair<int, int>>> adj;
void dijkstra(int s, vector<int> & d, vector<int> & p) {
int n = adj.size();
d.assign(n, INF);
p.assign(n, -1);
vector<bool> u(n, false);
d[s] = 0;
for (int i = 0; i < n; i++) {
int v = -1;
for (int j = 0; j < n; j++) {
if (!u[j] && (v == -1 || d[j] < d[v]))
v = j;
if (d[v] == INF)
u[v] = true;
for (auto edge : adj[v]) {
int to = edge.first;
int len = edge.second;
if (d[v] + len < d[to]) {
d[to] = d[v] + len;
p[to] = v;
Here the graph $\text{adj}$ is stored as adjacency list: for each vertex $v$ $\text{adj}[v]$ contains the list of edges going from this vertex, i.e. the list of pair<int,int> where the first element
in the pair is the vertex at the other end of the edge, and the second element is the edge weight.
The function takes the starting vertex $s$ and two vectors that will be used as return values.
First of all, the code initializes arrays: distances $d[]$, labels $u[]$ and predecessors $p[]$. Then it performs $n$ iterations. At each iteration the vertex $v$ is selected which has the smallest
distance $d[v]$ among all the unmarked vertices. If the distance to selected vertex $v$ is equal to infinity, the algorithm stops. Otherwise the vertex is marked, and all the edges going out from
this vertex are checked. If relaxation along the edge is possible (i.e. distance $d[\text{to}]$ can be improved), the distance $d[\text{to}]$ and predecessor $p[\text{to}]$ are updated.
After performing all the iterations array $d[]$ stores the lengths of the shortest paths to all vertices, and array $p[]$ stores the predecessors of all vertices (except starting vertex $s$). The
path to any vertex $t$ can be restored in the following way:
vector<int> restore_path(int s, int t, vector<int> const& p) {
vector<int> path;
for (int v = t; v != s; v = p[v])
reverse(path.begin(), path.end());
return path;
• Edsger Dijkstra. A note on two problems in connexion with graphs [1959]
• Thomas Cormen, Charles Leiserson, Ronald Rivest, Clifford Stein. Introduction to Algorithms [2005]
Practice Problems¶ | {"url":"https://gh.cp-algorithms.com/main/graph/dijkstra.html","timestamp":"2024-11-05T19:52:05Z","content_type":"text/html","content_length":"150275","record_id":"<urn:uuid:e6c7eb22-c79c-4190-bfa9-22d8e197d64e>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00129.warc.gz"} |
Whole Numbers
Whole numbers are 0, 1 , 2 , 3 , 4 , ... We have ten symbols representing numbers in
our system of arithmetic and all other numbers are represented by using these symbols.
What are these symbols? answer (in this web page click on " answer" to see
the answer)
One of the most significant developments in mathematics was the invention of zero. It was
developed by the Hindus and brought to Europe by the Arabs. Do you know the
difference between zero and nothing? answer
Another significant advancement in mathematics was the development of the concept of
place value. For example $4,764 = $ 4,000 + $ 700 + $60 +$4 each of the two 4s have
a different monetary value because of where it is located in the original number. Where
the number is located we call its place and the value of the number (to express this
another way, "the amount it represents") is determined by its location.
$7,986,756.87 Reading left to right what amount does each 7 represent? answer
You have 3,453 mangoes. How many mangoes are represented by the first
and last 3? answer
Each place in a number has a specific name. For example 349. The 3 is in the hundreds
place, the 4 is in the tens place and the 9 is in the ones place. Try and name the
first ten places in our number system located to the left of the decimal point.
Go back to table of contents click | {"url":"http://roconnell.net/wholenumbers/wholenumbers.htm","timestamp":"2024-11-07T02:41:06Z","content_type":"text/html","content_length":"2814","record_id":"<urn:uuid:6bba185b-81ba-42a1-8b3b-e7a9180dbbe8>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00105.warc.gz"} |
4 questions operations management 494809
Document Preview:
for the law firm. The desired service level is 95%. The corresponding Z =1.64. NOTE: Show all steps. Solve problems using EXCEL. Date: 7-19-2013 Time: 5pm to 9pm. is 2 days. The daily demand for
clean towels at the hotel is normally distributed with a mean of is a constant at 10000 pages to be copied. The mean lead time for the contractor is 4 days with a standard deviation of 1.25 days for
delivering copies. Find the re-order point ordering cost is $60 per order. The holding cost is $4 per unit per year. The lead time for this item is 3 days. Assume 250 working days per year. Find: b)
What is the optimal number of orders per year. c) What is the annual cost of ordering and holding inventory. d) What is the re-order point. 2. ABC Co. manufactures a subcomponent at the rate of 400
per day when required. Its annual demand for this item is 50000 units. Assume 250 working days per year. Holding cost is $10 per unit per year. Set-up cost is $150 per set-up. Find: 800 and a
standard deviation of 150. Management has set the service level 95%. For this service level, Z = 1.64 a) The Reorder point. 4. A printing company has a contract with a legal firm to copy their court
documents. Daily demand MGSC 303 Exam 2 SUM II 2013 1. ABC Co. purchases an item. The annual demand for this item is 75000 units. The 3. Hotel ABC has contracted out its towel laundering work. The
lead time for the contractor Find: b) Safety stock. d) What percentage of time during the year will this item be produced. a) Economic production quantity. b)How many production runs will be made per
year. c)What will be the maximum inventory level. e) What is the annual cost for ordering and holding inventory. a) Economic order quantity, EOQ. | {"url":"https://ibeehomeworksolutions.com/view/1649068/students/2021/08/29/4-questions-operations-management-494809/","timestamp":"2024-11-14T07:49:10Z","content_type":"text/html","content_length":"44680","record_id":"<urn:uuid:355b9d97-1c08-493a-833d-fc643bf43af5>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00561.warc.gz"} |
Re: Computing "first" in LR(1) parsing
max@gustavus.edu (Max Hailperin)
29 Nov 2001 23:09:46 -0500
From comp.compilers
| List of all articles for this month |
From: max@gustavus.edu (Max Hailperin)
Newsgroups: comp.compilers
Date: 29 Nov 2001 23:09:46 -0500
Organization: http://groups.google.com/
References: 01-11-123
Keywords: LR(1), parse
Posted-Date: 29 Nov 2001 23:09:46 EST
remove.haberg@matematik.su.se (Hans Aberg) wrote in message news:01-11-123...
> I want to compute the function "first" of LR(1) parsing: If x is a
> string of grammar symbols, first(x) is the set of terminals that begin
> the strings derivable from x with respect to the given grammar, plus
> the empty string, if derivable as well. In Aho et al, "Compilers", sec
> 4.4, p.189, 3, it says that if x is a nonterminal and there is a
> production x -> y1 ... yk, then to first(x) one should add first(yi)
> if the empty string is part of first(y1), ..., first(y{i-1}).
> But it does not say what happens if this procedure recurses, that is,
> if say when computing first(y1), one recurses back via some other
> relations to first(x). If that happens, is that not a LR(1) grammar,
> or how should this situation be handled?
What Aho, Sethi, and Ullman do not clearly explain is that this is an
iterative algorithm, convering on a least fixed point by working up
from below. So at each point in the iteration, you have some lower
bound on the eventual FIRST sets. Based on the current (iteration n)
lower bounds on the FIRST(yi), you calculate a new (iteration n+1)
lower bound on FIRST(x), in the manner you summarize above.
For an alternative presentation of this, you might see a pedagogical
paper I wrote on the topic, which is at
-Max Hailperin
Associate Professor of Computer Science
Gustavus Adolphus College
800 W. College Ave.
St. Peter, MN 56082
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Jim's Random Notes
Skip list revisited
In C# versus the data structure, I described the skip list data structure and the difficulty I saw with implementing it in .NET. I published a C# implementation a couple of months ago, and it did
indeed suffer from excessive memory usage. It performed quite well, but the memory requirement of 100 bytes per node blew my memory budget.
With a little thought and some indirection, I was able to reduce that to an average of about 20 bytes per node. I published that implementation in A More Memory-Efficient Skip List. The resulting
data structure is faster than my initial implementation and uses about one-fifth the memory. Overall, I’m happy with the way it works. I think the design is clever–perhaps even elegant–but the
implementation is, of necessity, somewhat involved and a little messy. That said, it works well and I’m glad to have it.
My friend and development team member (co-founder, business partner, whatever) Joe Langeway floated the idea one day of building something similar to a skip list, but that had only two pointers per
node: a Next pointer, and a FarForward pointer. The idea was to create a simple linked data structure that had much of the benefit of a skip list, but without all that complexity. We tossed ideas
around for a day or two, and then I sat down to code it up.
The node structure is very simple. It contains the data and two link references. I’ll illustrate here with a node that contains integers, but it’s easy enough to extend this to make it work with any
data type.
class Node
int Value;
Node Next;
Node FarForward;
The idea, then, is to have the data structure assign far forward pointers as items are inserted. How to assign those far forward pointers was (and perhaps still is) a matter of some disagreement.
Since I was implementing the thing, I chose what I thought was the simplest approach: when traversing the list to find the insertion spot, make the first unused far forward pointer encountered in the
traversal reference the new node.
Imagine you were inserting the numbers 1, 3, and 2, in that order. The first node is inserted trivially and its far forward pointer is null. When we try to insert the second node, the code has to
traverse the list (follow Next and FarForward pointers) to find the insertion spot. Along the way, the code chooses the first non-null forward pointer to reference the new node. When the value 3 is
inserted, the first node is modified so that its Next and FarForward pointers both reference the new node.
When 2 is inserted, there’s no unused forward pointer ahead of it, so the only thing pointing to the new node is the first node’s Next reference. The result is that node 1 has a Next value of 2, and
a FarForward value of 3. 2.Next is 3, and 2.FarForward is null.
This data structure has two nice properties. First, it’s already in sorted order. Traversing the list in sorted order is a simple matter of following the Next links. Finding a particular value,
though, isn’t necessarily a sequential search. The Find method can use the FarForward pointers to skip through the list much more quickly than examining every single node. In this particular case,
for example, it only has to examine two nodes to find the value 3. It never has to examine the second node.
Implementing the first version of this took very little time, and performance is surprisingly good in the general case. I found that search takes, on average, about 10 * (Log10(n) - 1) probes, where
n is the number of items in the list. So to find an item in a collection of one million takes approximately 50 probes (10 * (log10(1000000) – 1)). That’s not as fast as a skip list or a balanced
binary tree, both of which would be log2(n), or in this case about 20 probes. But it’s a huge improvement over sequential search (n/2, or 500,000 probes).
Unfortunately, the data structure has a few drawbacks. It exhibits pathological behavior when the input data is sorted or reverse-sorted, and it is pretty sensitive to ordered subranges. If you
insert items in sorted order, the FarForward link in every node points to the next node. And if you insert items in reverse order, all of the FarForward links are null. In either case, finding an
item becomes a sequential scan of the list, taking on average n/2 probes.
The other drawback is that the number of probes required to find an item in the general case is quite variable. The average is reasonable: about 2.5 times the worst case for binary search. But it
varies quite a bit. In my testing, the maximum number of probes required to find an item is approximately 3.5 times the average. So, whereas the average for a list of one million items is about 50,
the maximum is somewhere around 175.
I’ve explored several different ways to alleviate the problems associated with ordered data. Preventing pathological behavior when data is ordered is not very difficult, although the result is still
unacceptable, with a search time that’s about 10 times what it is for a list built from unordered data. Solving the ordered and reverse-ordered cases also complicates the code quite a bit. It’s still
manageable, but not nearly as elegant as the initial implementation.
You might rightly ask why I’m even bothering with a data structure that is almost certain to be slower than existing solutions that I already have access to. There are several reasons. First (and
perhaps foremost), it’s a really interesting puzzle. Just a little thought produced a simple data structure that has logarithmic complexity in the general case. It comes within spitting distance of
the skip list and is much easier to understand and implement. Also, there are certain benefits to having a linked data structure whose nodes are of a fixed size, even if that data structure is
somewhat slower than another data structure that has variable-sized nodes.
An interesting use of the insertion algorithm for this data structure would be to build a double-linked list from a stream of unordered data. The idea would be to treat the linked list node’s Prev
pointer as a far forward pointer during construction. When the last data item is received, you can make a single sequential pass through the list to set the Prev pointers to their correct values. You
could use a heap, but it would require more space. If your desired result is a double-linked list, this algorithm would use only O(n) memory. Building a heap would require O(n) extra memory.
I won’t post the code yet, since I’m still tinkering and the code is kind of ugly right now. But if you let me know that you’re interested in puzzling on this, you might give me some incentive to
clean things up sooner. I’d be interested to see if a combined effort could come up with solutions to some of the problems. | {"url":"https://blog.mischel.com/2011/12/14/skip-list-revisited/","timestamp":"2024-11-11T10:36:03Z","content_type":"text/html","content_length":"31318","record_id":"<urn:uuid:b4397e2b-310b-45e1-b33d-be952e1fda14>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00103.warc.gz"} |
Transitivity and Graphs for SQL
I have mentioned on a couple of prior occasions that basic graph operations ought to be integrated into the SQL query language.
The history of databases is by and large about moving from specialized applications toward a generic platform. The introduction of the DBMS itself is the archetypal example. It is all about
extracting the common features of applications and making these the features of a platform instead.
It is now time to apply this principle to graph traversal.
The rationale is that graph operations are somewhat tedious to write in a parallelize-able, latency-tolerant manner. Writing them as one would for memory-based data structures is easier but totally
unscalable as soon as there is any latency involved, i.e., disk reads or messages between cluster peers.
The ad-hoc nature and very large volume of RDF data makes this a timely question. Up until now, the answer to this question has been to materialize any implied facts in RDF stores. If a was part of b
, and b part of c, the implied fact that a is part of c would be inserted explicitly into the database as a pre-query step.
This is simple and often efficient, but tends to have the downside that one makes a specialized warehouse for each new type of query. The activity becomes less ad-hoc.
Also, this becomes next to impossible when the scale approaches web scale, or if some of the data is liable to be on-and-off included-into or excluded-from the set being analyzed. This is why with
Virtuoso we have tended to favor inference on demand ("backward chaining") and mapping of relational data into RDF without copying.
The SQL world has taken steps towards dealing with recursion with the WITH - UNION construct which allows definition of recursive views. The idea there is to define, for example, a tree walk as a
UNION of the data of the starting node plus the recursive walk of the starting node's immediate children.
The main problem with this is that I do not very well see how a SQL optimizer could effectively rearrange queries involving JOINs between such recursive views. This model of recursion seems to lose
SQL's non-procedural nature. One can no longer easily rearrange JOINs based on what data is given and what is to be retrieved. If the recursion is written from root to leaf, it is not obvious how to
do this from leaf to root. At any rate, queries written in this way are so complex to write, let alone optimize, that I decided to take another approach.
Take a question like "list the parts of products of category C which have materials that are classified as toxic." Suppose that the product categories are a tree, the product parts are a tree, and
the materials classification is a tree taxonomy where "toxic" has a multilevel substructure.
Depending on the count of products and materials, the query can be evaluated as either going from products to parts to materials and then climbing up the materials tree to see if the material is
toxic. Or one could do it in reverse, starting with the different toxic materials, looking up the parts containing these, going to the part tree to the product, and up the product hierarchy to see if
the product is in the right category. One should be able to evaluate the identical query either way depending on what indices exist, what the cardinalities of the relations are, and so forth —
regular cost based optimization.
Especially with RDF, there are many problems of this type. In regular SQL, it is a long-standing cultural practice to flatten hierarchies, but this is not the case with RDF.
In Virtuoso, we see SPARQL as reducing to SQL. Any RDF-oriented database-engine or query-optimization feature is accessed via SQL. Thus, if we address run-time-recursion in the Virtuoso query engine,
this becomes, ipso facto, an SQL feature. Besides, we remember that SQL is a much more mature and expressive language than the current SPARQL recommendation.
SQL and Transitivity
We will here look at some simple social network queries. A later article will show how to do more general graph operations. We extend the SQL derived table construct, i.e., SELECT in another SELECT's
FROM clause, with a TRANSITIVE clause.
Consider the data:
CREATE TABLE "knows"
("p1" INT,
"p2" INT,
PRIMARY KEY ("p1", "p2")
ALTER INDEX "knows"
ON "knows"
PARTITION ("p1" INT);
CREATE INDEX "knows2"
ON "knows" ("p2", "p1")
PARTITION ("p2" INT);
We represent a social network with the many-to-many relation "knows". The persons are identified by integers.
INSERT INTO "knows" VALUES (1, 2);
INSERT INTO "knows" VALUES (1, 3);
INSERT INTO "knows" VALUES (2, 4);
SELECT *
FROM (SELECT
T_IN (1)
T_OUT (2)
FROM "knows"
) "k"
WHERE "k"."p1" = 1;
We obtain the result:
The operation is reversible:
SELECT *
FROM (SELECT
T_IN (1)
T_OUT (2)
FROM "knows"
) "k"
WHERE "k"."p2" = 4;
Since now we give p2, we traverse from p2 towards p1. The result set states that 4 is known by 2 and 2 is known by 1.
To see what would happen if x knowing y also meant y knowing x, one could write:
SELECT *
FROM (SELECT
T_IN (1)
T_OUT (2)
FROM (SELECT
FROM "knows"
UNION ALL
FROM "knows"
) "k2"
) "k"
WHERE "k"."p2" = 4;
Now, since we know that 1 and 4 are related, we can ask how they are related.
SELECT *
FROM (SELECT
T_IN (1)
T_OUT (2)
T_STEP (1) AS "via",
T_STEP ('step_no') AS "step",
T_STEP ('path_id') AS "path"
FROM "knows"
) "k"
WHERE "p1" = 1
AND "p2" = 4;
p1 p2 via step path
The two first columns are the ends of the path. The next column is the person that is a step on the path. The next one is the number of the step, counting from 0, so that the end of the path that
corresponds to the end condition on the column designated as input, i.e., p1, has number 0. Since there can be multiple solutions, the last column is a sequence number allowing distinguishing
multiple alternative paths from each other.
For LinkedIn users, the friends ordered by distance and descending friend count query, which is at the basis of most LinkedIn search result views can be written as:
SELECT p2,
COUNT (*)
FROM "knows" "c"
WHERE "c"."p1" = "k"."p2"
FROM (SELECT
TRANSITIVE t_in (1) t_out (2) t_distinct "p1",
t_step ('step_no') AS "dist"
FROM "knows"
) "k"
WHERE "p1" = 1
ORDER BY "dist", 3 DESC;
p2 dist aggregate
The queries shown above work on Virtuoso v6. When running in cluster mode, several thousand graph traversal steps may be proceeding at the same time, meaning that all database access is parallelized
and that the algorithm is internally latency-tolerant. By default, all results are produced in a deterministic order, permitting predictable slicing of result sets.
Furthermore, for queries where both ends of a path are given, the optimizer may decide to attack the path from both ends simultaneously. So, supposing that every member of a social network has an
average of 30 contacts, and we need to find a path between two users that are no more than 6 steps apart, we begin at both ends, expanding each up to 3 levels, and we stop when we find the first
intersection. Thus, we reach 2 * 30^3 = 54,000 nodes, and not 30^6 = 729,000,000 nodes.
Writing a generic database driven graph traversal framework on the application side, say in Java over JDBC, would easily be over a thousand lines. This is much more work than can be justified just
for a one-off, ad-hoc query. Besides, the traversal order in such a case could not be optimized by the DBMS.
In a future blog post I will show how this feature can be used for common graph tasks like critical path, itinerary planning, traveling salesman, the 8 queens chess problem, etc. There are lots of
switches for controlling different parameters of the traversal. This is just the beginning. I will also give examples of the use of this in SPARQL.
Re:Transitivity and Graphs for SQL
"[...] meaning that all database access is parallelized and that the
algorithm is internally latency tolerant."
Do you have some deep magic voo-doo that works around the problem of having a straight chain? In the worst case, every single step of your search will cross nodes. Every one. That is in no way,
shape, or form "latency tolerant."
In the realm of parallel graph manipulation and searching, you need to be very, very precise in what you declare... Are your results for some general case, over some set of examples, or a technical
You certainly are not talking about worst-case performance. Some graphs contain no parallelism at all. Some pointer jumping or triangulation can add parallelism to otherwise sequential graphs but
at an explosive storage cost. There are some nifty techniques for pulling as much parallelism as possible out of a graph (splitting by the DFS tree, forgot the author, but it's referenced from
Reif's proof that DFS is sequential), but they are quite painful if your data is dynamic.
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Author: Orri Erling
Published: 09/08/2008 09:20 GMT
Modified: 08/20/2015 17:17 GMT
Tags: database , databases , rdf , jdbc , sql , semanticweb , sparql , howto , history , virtuoso
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The Physics behind the NASA Flyby Anomaly
1. Introduction
From 1990 to 2005 NASA did six flybys of Earth [1] in order to boost the energy of each spacecraft relative to the Sun, enabling them to go deeper into the solar system. Each of these fly-bys was
carefully monitored from Earth using Doppler radar in order to verify that the encounter went as planned. To their surprise, NASA found that these encounters appeared to violate the NASA models of
gravity and even more seriously violated conservation of energy in the center of mass of the Earth, where a spacecraft could leave the Earth with more energy than it came in with—100 sigmas more.
The NEAR mission passed the Earth in 1998 and gained 92.2 joules/kg of energy in the pass with an rms uncertainty of 0.9 joules/kg—about 100 sigmas of violation for energy conservation relative to
This conclusion of non-conservation of energy in the Earth coordinate system precipitated an in-depth evaluation at NASA of many candidate causes, all of which were discarded as negligible. This
paper explores a potential solution to this NASA mystery which eliminates the fundamental issues of non-conservation of energy that the NASA data appeared to show. It explores the possibility that
the vacuum field is being radiated by the Earth and Sun. Combining the NASA data with a previous experiment of annual variations in tunnel current on Earth, we found a simple physical model that
derives the radiation law for the vacuum field that matches the rather accurate NASA data available.
While variations in the vacuum field may sound unfamiliar, over the past 60 years many scientists have made the vacuum field vary temporally and spatially in our laboratories by factors of 10-100X.
This vast research has over 200 confirming experiments with well-developed theory and applications. Thus, the ability of the vacuum field to vary in both time and space is well documented in our
laboratories and accepted by our standard model of physics [2] [3].
The vacuum field has its strength based on a parameter h, called Planck’s constant. Despite its name suggesting that h is invariant, there is also considerable research supporting Planck’s constant h
varying across our solar system and across the cosmos. A number of papers [4] - [9], including US, German and Russian national labs, show up to 15 years of annual variations in strong and weak
radioactive decay rates—all peaking around January and minimizing around July. The researchers were not able to identify a cause.
The number of time varying radioactive decay reports caught DARPA’s interest, so Ref [10] was funded by DARPA to assemble the many experimental results related to time varying radioactive decay in
the hope that someone could integrate them into a single theory. DARPA sent that data to this author for his investigation.
Since radioactive decay is exquisitely sensitive to Planck’s constant in the denominator of the exponent, we hypothesized that these data could support a model where h varies linearly across the
Earth’s orbit, maximizing around January and minimizing around July.
To verify that variations in h were the cause, a separate tunnel diode experiment by the author had detected similar annual oscillations in tunnel current consistent with a gradient in h across the
Earth orbit of 231 ppm. The tunneling experiment also detected a radiated form of the vacuum field coming from the Sun with 13 sigma confidence as well as similar radiation aligned with the super red
giant Betelgeuse with 7 sigmas of confidence. In those published papers, we have chosen to call the radiated form of the vacuum field Casimir Radiation after the scientist who first proposed that the
vacuum field was real and how to verify it [11] [12] [13].
The success of this analysis below to explain the NASA Anomaly opens the possibility that there is a quantitative linkage between the vacuum field and the metric tensor of space.
2. The NASA Fly-By Anomaly
The six fly-bys of Earth were exquisitely measured and modeled by NASA, with accuracies as fine as 0.01 mm/sec for the velocity change in the Earth coordinate system. While the intent of the
monitoring was simply to determine how precisely the fly-bys were accomplished, the data showed, by up to 100 sigmas, that energy was not conserved in the fly-by encounters, and in most passes the
energy of the spacecraft increased.
In gravity theory, except for effects like atmospheric drag, an object enters and leaves the Earth space with the same energy. Thus, this anomaly led NASA to undertake an in-depth evaluation of all
the factors, such as atmospheric drag, relativity effects and tidal acceleration. Nothing explained the data, and getting more energy out than in was considered a serious challenge to physics [1].
Despite the lack of any identified cause, NASA did find an empirical formula for the energy gain or loss for these fly-by encounters as measured in the Earth coordinate system [1]. The spurious
velocity gain or loss in all six passes (ΔV[∞]) could be computed within the error bars by a simple formula.
$abla {V}_{\infty }=K{V}_{\infty }\left(\mathrm{cos}{\delta }_{in}-\mathrm{cos}{\delta }_{out}\right)$(1)
where V[∞] = the incident velocity of the spacecraft far from Earth
K = a dimensionless constant evaluated to be 3.099 × 10^−6.
δ[in] and δ[out] were the declination angles relative to the Earth-rotational plane of the incident and outgoing spacecraft trajectory, suggesting that Earth rotation was a significant factor.
This formula fits the data quite well as shown in Figure 1, but the challenge is to find a physical model that matches this effect. NASA was unable to find one in standard physics. This paper
presents a model that matches the NASA data, based on a coupling between the Earth’s rotation and the gravity metric. Such a coupling is well known for angular momentum [14], but this model
represents a potential new form of coupling between the Earth’s rotation and the polar angle θ, based on a radial component of the vacuum field radiated by a rotating warm Earth.
This paper derives the radiation equation for the vacuum field, accurate to about 1% based on the accuracy of the NASA data.
The NASA Fly-By Data
The NASA data in Figure 2 is quite complete, including all the parameters needed for precise modeling of the pass. When we look at the NASA data, we realize that:
1) Some effect makes the Earth’s gravity look stronger in the direction of the asymptotic spacecraft velocity, proportional to its velocity relative to the Earth.
Figure 1. (Color) The NEAR spacecraft showed 100 sigmas of unexplained energy increase when flying past the Earth. The model presented here matches this measurement for radar data backscattered from
the gold kapton, which covers the main body. Backscattered energy from dielectric surfaces is not predicted to show this anomaly as shown below. NASA did see a variation in the effect, ON or OFF in
some passes.
Figure 2. (Color) Here we plot the match between NASA’s best fit model and the measured data for six different fly-bys of the Earth. It shows a deviation from conservation of energy up to 100 sigmas,
which can be positive or negative. The deviation is linear in the cosine of the asymptotic velocities relative to the rotational plane of the Earth.
2) The increase in velocity is proportional to the magnitude of the velocity and not its sign since both incoming and outgoing portions of the trajectory have the effect of gravity increased.
3) Given that the sign of the velocity makes no difference, we conclude that the direction of Earth’s rotation is not a factor.[ ]
4) However, since the anomalous gravity effect is aligned with the Earth’s rotation axis, the magnitude of Earth’s rotation must be a significant causal factor.
3. Developing the Physical Model
First of all, adding additional or modified gravitational interactions to the GR gravity equations of motion fails. The author has applied numerous forms of additional terms with various power laws
with distance. They all fail for a fundamental reason. The effect NASA has found is proportional to the asymptotic velocity of the spacecraft at an infinite distance from the Earth. The velocity of
any spacecraft increases substantially as the spacecraft approaches Earth, and any Earth-induced gravity effects also increase. Thus, the added velocity when approaching Earth creates effects that
depend not just on the incoming velocity but also depend strongly on the distance of the closest approach—quite unable to match the NASA data which only depends on the spacecraft velocities far from
For this reason, we have concluded that modifications to the gravity equations that change gravity effects as the spacecraft gets closer to the Earth are unlikely to reproduce the effect reported by
We note that despite the many Casimir-type experiments [3], the vacuum field still offers mystery. Its ability to have no detectable power in one configuration and then to show the pressure of 19 psi
(2 terawatts/cm^2) in another configuration has only recently been given a simple physical model [15]. In that publication, the vacuum field is modeled as a complex field with only negative
frequencies, and this model is shown to match the theoretical intensity for spontaneous emission, making spontaneous emission into causal emission from the vacuum field.
In this paper, the vacuum field is central to the effects observed by NASA, which have no other identified cause in standard physics. The model [15], where the vacuum field is complex allows this
NASA Anomaly to be explained as shown below.
3.1. The Long-Range Doppler Measurements
Now we focus on the NASA measurements themselves, which measure the Doppler shift of radar signals both as the spacecraft approaches the Earth and then leaves the Earth. NASA has, of course, removed
the usual Doppler shift due to the Earth’s rotation. The asymptotic frequency shift far from the Earth is used to determine the NASA Anomaly.
Since these asymptotic measurements occur when the spacecraft is far from Earth, it is reasonable to assume that the spacecraft is not being significantly influenced by the Earth. Thus, we need to
postulate an effect on Earth at the Doppler radar site that can explain these results.
In this paper, we explore a modified metric tensor, which incorporates the effect of the Earth’s rotation. Specifically, we find that a new g[tr] term for the metric tensor, which couples time and
radial distance from the Earth, predicts measurements that exactly match the NASA data. Later in this paper, we explore the source for this new term.
We note that coupling time with another coordinate is not a new concept. The Kerr solution for the metric tensor outside a rotating planet links time and the equatorial angle θ. The exact form of the
extra term in the Kerr solution is shown in Equation (2), where θ is the angle of the path relative to the equatorial plane.
${g}_{t\phi }=\frac{4GJ}{{c}^{3}r}{\mathrm{sin}}^{2}\theta$(2)
where G = gravitational constant and J = angular momentum of the rotating planet, which for the Earth has $J=2/5{m}_{E}{R}_{E}^{2}{\omega }_{E}$. This modification to the metric tensor is many orders
of magnitude too small to explain the NASA anomaly and has the wrong dependence on the polar angle, but it shows other research that has coupled time to a spatial coordinate for a rotating planet.
3.2. What are the Physical Consequences of a g[tr] Modified Metric?
Let’s begin by asking what happens to light propagation with this hypothesized extra g[tr] term from a site on the Earth’s equator. In particular, we will compute the Stress-Energy-Momentum tensor
for a freely propagating E&M wave with the assumed modification to the metric tensor. We will find that the transmitted beam of known frequency has additional energy stored within it, due to the
modified metric tensor and resulting in the NASA-observed frequency shift. Even better, conservation of energy is restored in this model since the anomaly is a measurement effect, not a satellite
We begin by deriving the Stress-Energy-Momentum (SEM) tensor for a beam of light transmitted vertically into the equatorial plane from a site somewhere on the Earth. For convenience, we choose the
local lab coordinate system where x and y are parallel to the Earth sphere, and z is the radial vector to space. This model can apply to any latitude once we specify how the g[tr] term varies with
0 = ct = time
1 = x = local θ direction (East)
2 = y = local θ direction (North)
3 = z = r = radial distance from the center of the Earth (3)
To affect both polarizations of light equally, we hypothesize a coupling between the z (vertical) and time axes. To evaluate its impact, we simply denote it as ε, where ε is the order of 10^−6, and
our calculation will only keep the first order in ε, where g[μν] and g^μν are identical tensors.
${g}_{\mu u }={g}^{\mu u }=\left[\begin{array}{cccc}1& 0& 0& \epsilon \\ 0& -1& 0& 0\\ 0& 0& -1& 0\\ \epsilon & 0& 0& -1\end{array}\right]$(4)
Setting up the Initial Beam Leaving the Transmitter
We begin by launching a beam of light of spatial frequency k in the radial direction. Both polarizations have the same physics since the modification to the metric tensor is in the r direction and
thus orthogonal to both polarization axes. For mathematical convenience, we use a purely X polarized beam, which has only an A[1] component. This simple beam of light has only four components in the
E & M field tensor F[μν], and all are equal in magnitude. Note that we have set all the various physical constants to unity to make the structure of the equations as clear as possible, not unusual
for General Relativity derivations. [13]
$\begin{array}{l}{A}_{1}={A}_{x}\mathrm{sin}\left(k\left(r-ct\right)\right)\\ {F}_{10}={A}_{1|0}={A}_{x}\mathrm{cos}\left(k\left(r-ct\right)\right)=-{F}_{01}\\ {F}_{13}={A}_{1|3}=-{A}_{x}\mathrm{cos}
The E&M field tensor is shown in Equation (6).
${F}_{\mu u }=\left(\begin{array}{cccc}0& -1& 0& 0\\ 1& 0& 0& -1\\ 0& 0& 0& 0\\ 0& 1& 0& 0\end{array}\right){A}_{x}\mathrm{cos}\left(kr-ct\right)$(6)
To compute the Stress-Energy-Momentum (SEM) tensor for this E&M field we need to compute various forms of the field tensor and combine them properly.
$\begin{array}{l}{F}^{\mu }{}_{u }=\left[\begin{array}{cccc}0& \epsilon -1& 0& 0\\ -1& 0& 0& 1\\ 0& 0& 0& 0\\ 0& -\epsilon -1& 0& 0\end{array}\right]{A}_{x}\mathrm{cos}\left(k\left(r-ct\right)\right)
\\ {F}^{\mu u }=\left[\begin{array}{cccc}0& 1-\epsilon & 0& 0\\ -1+\epsilon & 0& 0& -1-\epsilon \\ 0& 0& 0& 0\\ 0& 1+\epsilon & 0& 0\end{array}\right]{A}_{x}\mathrm{cos}\left(k\left(r-ct\right)\
Now we assemble these terms into the Stress-Energy-Momentum (SEM) Tensor given by:
${T}^{\mu u }={F}^{\mu }{}_{\alpha }{F}^{\alpha u }+\frac{1}{4}{g}^{\mu u }{F}^{\alpha \beta }{F}_{\alpha \beta }$(8)
The result is a symmetric SEM tensor which multiplies the power density of the E&M wave, where we kept only the terms linear in ε. We see that the T^00 term (energy density) in Equation (10) has an
additional factor of 1 + 2ε which multiplies the usual power density.
$\begin{array}{c}{T}^{\mu u }=\left(\left[\begin{array}{cccc}{\left(\epsilon +1\right)}^{2}& 0& 0& -\left({\epsilon }^{2}-1\right)\\ 0& -2\epsilon & 0& 0\\ 0& 0& 0& 0\\ -\left({\epsilon }^{2}-1\
right)& 0& 0& {\left(\epsilon -1\right)}^{2}\end{array}\right]-4\epsilon \left[\begin{array}{cccc}1& 0& 0& -\epsilon \\ 0& -1& 0& 0\\ 0& 0& -1& 0\\ -\epsilon & 0& 0& -1\end{array}\right]\right)\\ \
text{\hspace{0.17em}}\text{\hspace{0.17em}}\cdot {A}_{x}^{2}{\mathrm{cos}}^{2}\left(k\left(r-ct\right)\right)\right)\end{array}$(9)
${T}^{\mu u }=\left[\begin{array}{cccc}1+2\epsilon & 0& 0& 1\\ 0& -2\epsilon & 0& 0\\ 0& 0& -4\epsilon & 0\\ 1& 0& 0& 1-2\epsilon \end{array}\right]{A}_{x}^{2}{\mathrm{cos}}^{2}\left(k\left(r-ct\
We note that the rotational acceleration is proportional to the cosine of the polar angle, which will make this effect also proportional to the cosine of the polar angle and match the NASA data.
3.3. The Candidate Model for the NASA Data
In our model, ε is positive from the additional vacuum field radiating in the radial direction, and the power in this transmitted beam is 2ε more than what is expected. In our model, the source of
the ε is the additional radiated vacuum field intensity from the Earth, which is a complex form of E&M radiation with zero mean power.
Since complex E&M radiation has zero mean real energy, if we are to see an effect from this complex radiation, something must happen to convert this complex vacuum radiation into a real signal that
can affect the backscatter frequency from the satellite.
Fortunately, the many Casimir experiments have given us some practical clues:
If we model the vacuum field as a complex E&M field [14], then its familiar properties can be deduced for Casimir experiments.
· The vacuum field has zero mean energy and is thus invisible for most optical interactions.
· If we set up two mirrors close together (nanometers) and parallel, then the real part of the vacuum field reflects as usual and thus has relatively few modes between the mirrors compared to
· The imaginary component does not interact with the mirror surfaces and simply transmits through the reflective surfaces without alteration.
· The reduced number of real modes between the plates is used to compute the vast pressure density due to the real component of radiation outside the pair of mirrors.
The key principle here is that only the real part of the vacuum field reflects from a mirror in the lab to create observable effects. The imaginary part simply transmits through the mirror. This
reflected real part of the vacuum field creates up to 19 psi of pressure—equivalent to 2 gigawatts/cm^2 of reflected light on the outside of the pair of mirrors.
This extreme vacuum pressure was a shock to most scientists when Casimir experiments first began, but is now accepted. The vacuum field has no net energy but can exert 19 psi radiation pressure on
two mirrors close together due to the extremely high power in the real part of the vacuum field.
Case 1: Reflecting off a Kapton Surface
Using this observed Casimir result, we hypothesize that the 2ε (1 + i) complex field component interacts with the transmitted beam as it exits the Earth’s environment to increase its energy by a
complex amount. When the beam reflects off a kapton metallic surface, the main beam and the real part of the Casimir term reflect together and in phase while the imaginary component does not reflect
as shown in Figure 3.
This reflected beam now has 2ε (~3 ppm) more fractional energy, which could mean either that new photons suddenly appear or that the energy of each photon is increased slightly. Given no obvious
mechanism to generate new photons of the same frequency at the reflection surface, we are left with the extra energy being added to the energy of each photon as it reflects from the metallic layer,
increasing the frequency of the light in the rest coordinate system of the satellite by the fraction 2ε. This increased frequency is in the center of mass of the satellite and thus increases the
return frequency at the NASA tracker by the change in frequency times the usual Doppler shift of 2v/c for the satellite. This is exactly what NASA reports.
What NASA observes is the asymptotic frequency shift for the incoming and outgoing spacecraft. According to modern gravity theory, they should be of equal magnitude and oppositely signed, indicating
equal incoming and outgoing speeds, but they are shifted by up to 3 ppm according to Equation (1).
Clearly, we don’t have field equations yet for such a process, but if we believe that conservation of photons and conservation of energy are relevant here, then we are guided to this conclusion.
Case 2: Scattering off a Dielectric Surface
When a dielectric surface is used, we have E&M backscatter that comes from the roughness of the surface and the change in refractive index, which we assume
Figure 3. (Color) Only the real part of the NASA radar beam reflects off a Kapton surface since the imaginary part cannot create imaginary reflective currents.
will affect the real and imaginary components equally. In this case, there would be no change in the energy or frequency of the return light.
If ε is proportional to the cosine of the direction relative to the equatorial plane due to the Earth radiation, then we have a model that exactly matches the NASA data with a metallic surface such
as kapton. The result is that the vehicle appears to be going faster than expected in proportion to the cosine of the latitude of the direction and in proportion to the speed of the spacecraft
divided by c—whether coming in or going out. This frequency shift due to the metric tensor around the Earth applies equally to both the incoming and exiting portions of the engagement, and since both
the incoming and outgoing velocities are identical in GR theory, the combined effect is proportional to the difference between their two cosine latitudes.
This is, of course, what NASA has found.
The key point of every Casimir experiment is that the vacuum field exerts no pressure unless you reflect it. The reason behind this observation is that the vacuum field is well modeled as a complex E
& M wave of the form e^ik^(^x^−ct). [14] As a complex form of electromagnetic radiation, it has zero momentum and zero energy density despite having a field intensity greater than the total power
output of the Earth in every square cm. However, when the vacuum field reflects off a mirror surface in a Casimir experiment, only the real part reflects, since the electrons do not have the ability
to oscillate in an imaginary direction. The reflection of the real part of the vacuum field creates the massive pressure of 19 psi observed in Casimir experiments even while the vacuum beam itself is
A single mirror shows no effect when reflecting the ubiquitous vacuum field because the reflected real part is joined by the transmitted imaginary part from the other side. However, Casimir
experiments use two mirrors close together so that the internal volume supports only a few wavelengths, making the internal pressure much less. Pressures as much as 19 psi are routinely reported due
to reflecting the real part of the Casimir radiation.
We suggest that the same effect is happening when NASA probes a spacecraft going by the Earth. The laser light has an added term in the Stress-Energy-Momentum tensor term T[00] equal to 2ε in
Equation (10), generated by the g[tr] term in the metric tensor. This term 2ε is complex since the Casimir radiation that generates this term is itself complex. The frequency of the light is thus
unaffected until the light reflects off a metalized surface (such as kapton) on the spacecraft. Upon reflection, only the real part of the Casimir light reflects just as in the Casimir lab
experiments, so the contribution of the radiation to the energy content of the light is now real and positive—increasing its frequency to match the NASA data. Unlike a usual Casimir experiment, this
reflected light is coherent with the reflected laser beam.
In summary, one way we know to make the vacuum field interact physically is to reflect it off a mirror, separating the real from the imaginary component. It works in the many Casimir experiments, and
we are proposing it here to explain the 100 sigmas of apparent violation of conservation of energy in the NASA anomaly. The effect of this process on gravity red shift experiments is discussed later
and shown to be negligible.
4. A Physical Model Derivation to Explain This Modified Metric Tensor
Since the new term in the metric tensor is so large in magnitude (~10^−6) compared to the usual gravity terms (10^−9), it appears not to have any dependence on the mass of the Earth or the
gravitational constant, and thus is not caused by any usual gravitational effect. So what would be its source?
We have some clues:
1) The source of this new term is associated with the rotation of the Earth and varies proportional to the cosine of the satellite direction with respect to the Earth’s plane of rotation.
2) An added metric tensor term that couples time and radius from the Earth matches the NASA data perfectly.
Uniting all these clues, we have chosen a new direction to explore.
Hypothesis: The effect we are looking for is a modification to a time term (g[tr]) in the metric tensor created by some form of radiation coming from a rotating body.
There are only two physical processes known to our physics that control time rate:
1) Gravity
2) Casimir radiation (the spectrum characterized by Planck’s constant h)
While Casimir radiation is typically specified as a particular spectrum computed from zero-point energy, it has been shown to vary from 10-100:1 in intensity in our labs both spatially and
temporally. [3] [4] Approximately 200 Casimir experiments have put two reflecting surfaces close together to reduce the Casimir radiation between the two. [4] The impact is huge, creating up to 19
psi radiation force squeezing the two plates together, and it all comes from a spatial non-uniformity in Casimir radiation. If variations in Casimir radiation exist in our labs, why not in the rest
of space?
Adding to the variations in h seen in laboratory Casimir experiments, we have two deep space studies showing with 3.4 and 4.2 sigma certainty that the fine structure constant alpha varies with an
apparent cosine distribution in the sky. That constant α = 2πk[e]e^2/hc includes h in its denominator, lending support that the same h variations we see in our labs may also be present in the larger
universe [15] [16] [17].
A completely different type of research on radioactive decay (strong and weak interactions) shows the fastest decay rates in January and minimum decay rates in July of each year [5] - [10].
The author verified annual variations in tunnel current with a purely electromagnetic test using Schottky tunnel diodes over two and a half years [11] [12].
All of these radiation and tunneling experiments share the common factor that they are exponentially affected by Planck’s constant h, and no other physical effect has been found as a probable cause.
4.1. Casimir Radiation Law Derived
We are hypothesizing that radial Casimir radiation from the Earth is coupling into the g[tr] term of the metric tensor. To explore this hypothesis in more detail, we would like to find a radiation
law for the vacuum field. Fortunately, previous papers by the author show how the Schottky tunnel diode experiment detected Casimir radiation coming from the Sun with 13 sigmas of confidence [11]
[12]. Using the measured data for the Sun radiation and the observed NASA data, we can search among the candidate radiation laws and see if any match the data for both effects.
To be precise:
· We hypothesize here that Casimir radiation (the vacuum field) is being continually emitted by stars and planets everywhere, and below we derive a candidate emission formula for that rate of Planck
radiation, which matches the observed NASA flyby anomaly as well as the detected Casimir radiation from the Sun.
· In searching for a radiation law, we hypothesize that the electromagnetic Planck spectrum can be generated by any warm source similar to the way that thermal light is radiated. However, to be
consistent with the NASA anomaly, the rate of Casimir radiation must include a rotation factor, which makes it quite distinct from simple thermal radiation.
4.2. How Is Casimir Radiation Produced?
There is currently no theoretical model available in standard physics for the emission of Casimir radiation, so we were guided by the NASA data and the measured magnitude of the Casimir radiation
from the Sun—published data from the author’s 2.5-year tunnel diode Planck experiment [11] [12].
We knew the NASA effect varied proportional to the cosine of the direction of the space vehicle velocity relative to the plane of rotation of the Earth. This cosine relationship required rotation of
the Earth to be a factor, quite different from standard blackbody emission of E&M light, and is a familiar angular dependence for radiated E&M amplitude from a spinning source. We discuss the physics
of this angular dependence more later.
Given both the NASA Earth data [1] and the previously detected Casimir radiation from the Sun [11] [12], we were able to test the candidate formulas to see which one best-matched the variations in
the vacuum field radiation for both Earth and Sun data simultaneously.
In searching for a physical formula for Planck radiation, we assumed it would be a simple power-law of local effects involving the temperature T and radial acceleration, and we allowed the
temperature to have an exponent in the range 0 - 4 and radial acceleration to have an exponent in the range 0 - 4. With these parameters, this created 5 * 5 = 25 possible power laws, which were
scored by how well they predicted the observed ratio between Casimir radiation measured for the Earth and the Sun. The functional form we chose to evaluate was a simple product of a constant times a
power of temperature times a power of radial acceleration.
$\text{Planck Radiation}\left(T,R{\omega }^{2}\right)=\text{Const}\cdot h{T}^{{n}_{1}}{\left(R{\omega }^{2}\right)}^{{n}_{2}}$(11)
For the truth value of Earth radiation, we used the metric value derived above (g[tr]= 1.547 ppm) that best matched the observed data in the NASA analysis discussed above.
4.3. Equation Evaluation
For the Sun, we used the experimentally measured value on Earth for the Casimir radiation from the Sun [11] [12] of 2.31 ppm divided by the estimated gain of the Schottky tunnel diode with respect to
h (40.85). Then we scaled that result by 1/r^2 from the Earth’s orbit at 1 AU from the Sun to the surface of the Sun’s equator (6.955 × 10^5 km). This gave us an estimate for the Casimir radiation at
the surface of the Sun on its equator. The tilt of the Sun relative to the Earth’s orbital plane is 7 deg, so depending on the time of year, there may be a cosine factor in the radiation varying from
cos(7 deg) = 0.993 to 1.0, which is a small enough correction to be ignored in this evaluation.
The accuracy of this estimate for solar Casimir radiation was set by our uncertainty in the fractional change in the tunnel diode signal per fractional change in the Planck radiation. We computed
this parameter Gain[sensor] = 40.85 with a ±25% uncertainty due to not knowing the exact shape of the barrier voltage profile inside the tunnel diode. [10] The uncertainty in the tunneling gain with
variations in h is the main uncertainty in setting the absolute radiation coefficient.
$\text{Sun Radiation}=h\frac{2.31\text{\hspace{0.17em}}\text{ppm}}{{\text{Gain}}_{\text{sensor}}}{\left(\frac{AU}{{R}_{\text{SUN}}}\right)}^{2}=2.616×{10}^{-3}\cdot h$(12)
The metric we used for evaluating various physical models for Casimir radiation was the ratio between the predicted emitted Planck intensity between the Sun and the Earth. The NASA radiance seems
quite well determined (<1% rms uncertainty), so our only significant uncertainty is with the Planck data for the sun as measured by our tunnel diode detector.
To find the meaningful candidates, we required the formula to have only locally impactful parameters. This meant that the local temperature T would be a valid parameter as would the rotational
acceleration Rω^2, but R by itself or ω by itself would not be a local parameter and would cause a candidate equation to be discarded. Thus, we chose only two local physical parameters, temperature T
and the local acceleration ω^2R to vary in our physical model of Casimir radiation, and we allowed their powers to vary from 0 - 4 for a total of 25 candidate radiation laws.
We then required that the formulas match the estimated ratio between the Sun and Earth Planck radiation ± 25%, our estimated maximum range of experimental error for that ratio based on uncertainty in
the barrier voltage profile across the tunnel diode. The scoring plot for all 25 candidate equations is shown in Figure 4, and then zoomed in to examine the more interesting candidates more closely
in Figure 5.
Expanding the vertical scale in Figure 4, we can see how the nearest candidate equations scored more clearly. Only two of our candidate equations (Equations (8) and (19)) scored well (within 3% of
the measured Sun-Earth radiation ratio), and all the rest were at least a factor of 4.3 away from the estimated Sun-Earth radiation ratio.
Figure 4. Of the 25 candidate formulas tested for matching the ratio between the measured Earth and Sun Planck radiation, only 2 fell within our ±25% error bounds.
Figure 5. Only formulas 8 and 19 (accidental pair) fell inside our ±25% error bars. They had an error of only 3% while the next closest formula had an error factor of 4.3, about 130 X worse.
An expanded scale plot (Figure 5) shows the two winning equations in more detail.
4.4. An Accidental Symmetry of Nature
The reader might notice that formula numbers 11 apart are almost identical. This is because of an accidental symmetry in the various parameters of the Earth and Sun as shown below. If you increase
the power of the temperature by 1 and the power of the rotation term by 2, the error changes by only 2.2%. When we have an accidental echo pair within our 25% target error, we need to use physical
logic to choose between them.
${\left(\frac{{T}_{\text{Earth}}}{{T}_{\text{SUN}}}\right)}^{1}{\left(\frac{{\omega }_{\text{Earth}}^{2}{R}_{\text{Earth}}\mathrm{cos}{\theta }_{\text{lat}}}{{\omega }_{\text{SUN}}^{2}{R}_{\text
Since the NASA-observed effect varies linearly with the cosine of the latitude for the distant spacecraft, we chose the formula that had acceleration to the unit power rather than the echo that
required acceleration cubed. This matches our experience with E&M radiation where the amplitude is emitted proportional to radial acceleration and varies linearly with the cosine of the latitude. The
rejected formula had acceleration cubed, which would be expected to produce a cosine-cubed angular distribution.
4.5. Result of the Equation Validation
Of the 25 candidate equations, only one and its accidental echo were within our ±25% constraints of the data. The radiation equation that varied linearly with the cosine of the latitude was selected
as our radiation model at 2.7% deviation from predictions: T^3(ω^2R[E]cos(θ[lat]))^1, where θ[lat] = the latitude of the transmitter (35.24 deg) and T = annual mean temperature for the ground site
(291 K).
The next best candidate was a factor of 4.3 from the green expectation line and far outside our estimated error bars. Our final choice for the Casimir radiation equation and the logic used to set the
coefficient is summarized in Table 1.
4.6. Is There a Thermal Cutoff to the Emission Frequency?
Since we have modeled Casimir radiation as a form of thermal emission coupled to acceleration, we considered adding an additional term, familiar in the standard blackbody radiation, i.e. the thermal
exponent—exp(−hν/k[B]T), but our frequency range is too small to show much effect. Since the vacuum field has zero real energy, it is expected that this term is not present. This zero real energy was
one of the reasons that it took so long to verify the existence of the vacuum field and why Casimir’s contribution was so important [2].
4.7. What about the Hotter Interior of the Earth?
If the full Earth volume or the full Sun volume were generating this radiation, then the effective temperatures would be much higher, since the core temperatures
Table 1. The logic to assemble the final equation for Casimir Radiation of the Vacuum Field from a warm rotating sphere.
R = radius of the star or planet, T = Surface temperature, ω = rotation rate (rad/sec), r = distance from the star or planet, θ = polar angle.
are vastly higher than the surface temperatures. Since the surface temperatures of the Sun and Earth provide good agreement with the NASA and Sun data, that would require that Casimir radiation
reaches an equilibrium with the local temperature and acceleration just like normal thermal radiation. This in turn would require a mechanism to equilibrate the Casimir radiation coming from a hotter
region into a cooler region.
We know from a large number of Casimir lab experiments that Casimir radiation is a form of E&M radiation that reflects off mirrors and very likely behaves like regular E&M energy in scattering off
atoms and other particles. Normal thermal light moving from a hotter to a cooler region of matter automatically adjusts itself into local equilibrium, and we expect the same behavior with Casimir
radiation. We note that Casimir radiation has zero energy and zero momentum, so it may be quite easy to equilibrate simply by multiple scattering.
4.8. Linkage between the Vacuum Field and the Metric Tensor
Based on the current model showing ag[tr] coupling between time and radius due to a warm sphere radiating the vacuum field, we point out that the physical process of generating a spatially varying
metric tensor has not been offered in physics. We simply say that every mass produces a metric tensor field around it with specific equations.
In the NASA data, we have found a coupling between time and radius which matches the experimental results when we have anisotropic vacuum radiation. If the vacuum radiation was isotropic, then we
would have only diagonal couplings into the metric tensor, offering the possible physics that the vacuum field is a potential cause of the metric tensor.
In discussing this conjecture, we point out that the main effect of gravity is variations in the rate of time, affecting the g[00] term of the metric tensor. Based on the current analysis of the NASA
Anomaly, we could also hypothesize that the metric tensor is the physical effect of the vacuum field radiating from a mass body. However, we would immediately point out that the vacuum field
radiation requires a warm body while gravity does not, allowing us to definitively conclude that the two effects on time (vacuum field and metric tensor) are independent.
5. Summary
A possible solution to the NASA conundrum is offered in this paper by hypothesizing that the effect is due to a change in the g[tr] term of the metric tensor, created by Casimir radiation (vacuum
field radiation) emitted by the Earth. When a complex g[tr] term was added to the metric tensor, it had no effects on clocks or orbital dynamics, but had exactly the effect needed to match the NASA
data while maintaining conservation of energy, which had appeared to be violated by up to 100 sigmas.
The effect of the added g[tr] term in the metric tensor is to change the beam energy by a small complex amount when the beam is launched and as it leaves the Earth. When the microwave beam travels to
the distant spacecraft and reflects off the surface, its frequency then increases as the real part of the energy stored in the beam is released into the light at the time the light reflects from the
satellite. However, if the beam primarily backscatters from a dielectric, no such effect is predicted. A process such as this can fully account for the observed NASA anomaly and restores conservation
of energy since there is no actual change in the vehicle velocity.
Once this physical principle had been suggested, we then explored how the extra term was created in the metric tensor. By combining the NASA solution with the previously detected radiation from the
Sun we were able to find a candidate radiation law for the Casimir radiation which matched the NASA and Sun data to <3%. All other candidate equations were far away from the observed NASA data except
one accidental echo, which was discarded to match the observed cosine distribution with spacecraft elevation.
Given that the NASA data appeared to be accurate to <1%, we have a potential radiation equation for the vacuum field, accurate to that same 1% uncertainty.
6. Discussion
While annually varying strong and weak radioactive decay plus similarly varying electromagnetic tunnel currents strongly pointed to a time-varying Planck constant h on the Earth, there was no basis
in our standard model to support Planck’s constant is a variable. In fact, there is no theory at all about how the vacuum field is generated since it is hypothesized to be a basic property of nature
that can never change.
Since numerous Casimir experiments in labs around the world have proven that the vacuum field does vary in space and time using mirrors, it is not unreasonable to assume that other factors in the
universe besides mirrors may also affect the vacuum field. Thus, a vast amount of data on annual variations in radioactive decay for strong and weak decays from three national labs and other groups
combined with similar time variations in tunnel diode current should be considered as strong support that the vacuum field radiation varies spatially and temporally all over the universe just as it
has been shown to do in our laboratories.
6.1. What about the NIST Experiment That Showed h Was Constant?
On the contrary side, one could then point to a recent experiment from NIST that showed Planck’s constant is extremely stable over several years [18]. Such a result is apparently inconsistent with
the data from radioactive decays and tunnel diodes [11] [12]. However, the answer is familiar in quantum mechanics. Since every linear physical process runs at a rate proportional to h, if h is
doubled, we would not know it except for the few processes that are nonlinear in h. Given that the NIST experiment was entirely linear in h, it could not measure a change in h, since its clocks would
also run faster or slower to cancel any variation.
Radioactive decay and tunneling are highly nonlinear in h, and all show a consistent time variation of h over a year. It is now reasonable for researchers to consider measuring and analyzing
spatially and temporally varying Casimir radiation, perhaps using tunneling sensors, which appear to be both accurate and convenient. Superconducting tunnel junctions may allow for exquisite
6.2. Impact on the Hubble Constant
The tunneling research presented above shows Casimir radiation being emitted by the Sun and the Earth, and by similarity all stars and planets. If Casimir radiation is continuously emitted by stars
and planets, then Planck’s constant h would be increasing with time. This could provide an alternative explanation for the Hubble constant, where the distant galaxies are redder simply because h is
smaller back in time, making local time move more slowly. In contrast to the expanding model of the universe, we could now consider whether our universe might simply be static, where gravity is
everywhere balanced on a large scale. Such a conclusion would end the search for dark energy since such a universe is essentially static while the usual red shift would still be observed.
We note that the Hubble constant would still be important because it would indicate the time constant of the universe as its vacuum field increases. Thus, a Hubble time constant of 16 billion years
may simply be the exponential coefficient of a universe constantly speeding up in its time rate, potentially eliminating the big bang model and replacing it with the quiet whisper model. In the quiet
whisper model of the universe, it starts with almost nothing, but increases its energy level for all processes as the Casimir radiation increases h. Such a conjectured universe may be trillions of
years old.
6.3. Liquid Earth and Mars Conundrum
This model of a time-varying Planck’s constant can also address another conundrum. How could our main sequence Sun have been so hot 3 - 4 billion years ago that it melted the ice on Earth and even on
Mars, creating vast oceans and rivers on both planets [18] [19]. Currently the average temperatures of the Earth and Mars are 15˚C and −63˚C respectively. Our model of a main-sequence star like our
Sun, says the Sun should be 30% dimmer 4 billion years ago than it is now, making the ancient Earth a block of ice. [18] Yet our Earth and even Mars had liquid oceans back then [19] [20]. Liquid
oceans on Mars would require the Sun to be much hotter than it is now. Even if Mars had an atmosphere in the past like present-day Earth (possible), it would have had about 0.70 * (93 × 10^6 miles/
142 × 10^6 miles)^2 = 30% of the current Earth’s sunlight. Even a 30% reduction of solar flux on Earth would make the Earth into an ice block, so Mars would have no chance for liquid oceans with 70%
less solar irradiance than Earth has now. And yet NASA data says ancient oceans existed on Mars in vast amounts.
In considering plausible explanations to the combined Earth/Mars liquid oceans in ancient times, it is important to realize that a star is one of the most sensitive responders to a change in Planck’s
constant with a gain of about ln(10^28) - ln(10^32) = 64 - 74, since it takes an estimated 10^28 - 10^32 proton-proton collisions to fuse into one deuterium nucleus, and this rate of fusion is
controlled by the intensity of the vacuum field at the frequency of the proton barrier. With this sensitivity, a 1.6% increase in the vacuum field would about triple the power output of our Sun and
warm Mars in the past to current Earth levels. Could a nearby star in our galaxy have fired up 4 billion years ago, flooding our solar system with 1.6% more Casimir radiation (vacuum field) and
heating up our Sun enough to melt Earth and Mars?
No one can say yet, but more research might consider that possibility since currently, we don’t have any answer for this well-documented mystery of a warm ancient Earth and even more, a warm and
liquid ancient Mars. [19] [20] [21]
Agreement with Gravity Red shift Experiments
Many readers would note that such a large frequency shift would dwarf the usual gravity red shift since this new term is several orders of magnitude larger. However, gravity red shift experiments are
not designed to look for complex changes in the metric tensor. For instance, transmitting a narrowband source to a narrow band absorber (i.e. Pound and Rebka [22] ) is unaffected by a complex
component in the photon energy, since the real frequency is unchanged. Also, the time drift of orbiting GPS clocks has no mechanism to be affected. Even a maser transmitted to Earth from space is not
a backscattered beam and has no obvious reason to show an effect, especially when the transmission occurred far out in space where the g[tr] term is much smaller.
Given that the NASA data used in the vacuum field analysis had an SNR up to 100, the effect is verified and quantified to about 1%, matching the theoretical model well, and allowing a radiation
equation that matches the separately measured vacuum field emissions of both the Earth and Sun.
How Can There Be Imaginary Field Components?
Physics has traditionally been double-minded about imaginary numbers. They are used throughout our many calculations but we also maintain they have no physical reality. This paper encourages us to
consider the option that imaginary numbers are real and physical in our universe. How can we explain the vast pressure in the vacuum field while it has no ability to excite or burn anything?
If we allow ourselves to consider the possibility that imaginary numbers are supported in our space-time coordinate system, then we must assume that those imaginary coordinates are tightly wrapped.
And yet they would strongly impact the properties of elementary particles. In particular, elementary particles might actually have complex masses. If so, then the worrisome inconsistencies in the
measurement of the gravitational constant G might occur simply because we have neglected the imaginary parts of atomic mass, which would interact in a repulsive manner and confuse the measurements. A
more careful gravity measurement design using several different materials in various pairings would potentially be able to detect the effect and verify whether we do indeed have imaginary mass and
energy and thus imaginary coordinates hidden within the fabric of our universe. | {"url":"https://www.scirp.org/journal/paperinformation?paperid=116178","timestamp":"2024-11-02T18:04:49Z","content_type":"application/xhtml+xml","content_length":"185217","record_id":"<urn:uuid:78fc1d67-ea0f-4bb2-bf9b-07643fe34a30>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00240.warc.gz"} |
seminars - Minimal dynamical systems, C*-algebras, and the classification program
Zoom 회의 ID: 356 501 3138 (암호: 471247)
This talk is based on joint work with Robin Deeley and Ian Putnam, where we study the existence of minimal dynamical systems, their orbit and minimal orbit-breaking equivalence relations, and their
applications to C*-algebras and K-theory. In particular, we show that given any finite CW-complex there exists a space with the same K-theory and cohomology that admits a minimal homeomorphism. The
proof relies on the existence of homeomorphisms on point-like spaces constructed by the authors in previous work, together with existence results for skew product systems due to Glasner and Weiss.
To any minimal dynamical system one can associate minimal equivalence relations by breaking orbits at small subsets. These were originally used by Putnam in his study of Cantor minimal systems, in
which case they are so-called AF relations. Using Renault's groupoid C*-algebra construction we can associate K-theory groups to minimal dynamical systems and orbit-breaking equivalence relations. We
show that given arbitrary countable abelian groups G_0 and G_1 we can find a minimal orbit-breaking relation such that the K-theory of the associated C*-algebra is exactly this pair.
These results have important applications to the Elliott classification program for C*-algebras. In particular, we make a step towards determining the range of the Elliott invariant of the
C*-algebras associated to minimal dynamical systems and their minimal orbit-breaking relations. | {"url":"http://www.math.snu.ac.kr/board/index.php?mid=seminars&sort_index=room&order_type=asc&l=en&page=85&document_srl=820250","timestamp":"2024-11-10T12:48:07Z","content_type":"text/html","content_length":"46944","record_id":"<urn:uuid:feba92b8-fd57-47ba-b8c8-545cd5a65eb7>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00779.warc.gz"} |
MeasurementsMeasurements — PennyLane
PennyLane can extract different types of measurement results from quantum devices: the expectation of an observable, its variance, samples of a single measurement, or computational basis state
For example, the following circuit returns the expectation value of the PauliZ observable on wire 1:
def my_quantum_function(x, y):
qml.RZ(x, wires=0)
qml.CNOT(wires=[0, 1])
qml.RY(y, wires=1)
return qml.expval(qml.PauliZ(1))
The available measurement functions are
expval(op) Expectation value of the supplied observable.
sample([op, wires]) Sample from the supplied observable, with the number of shots determined from the dev.shots attribute of the corresponding device, returning raw samples.
counts([op, wires, all_outcomes]) Sample from the supplied observable, with the number of shots determined from the dev.shots attribute of the corresponding device, returning the number of counts
for each sample.
var(op) Variance of the supplied observable.
probs([wires, op]) Probability of each computational basis state.
state() Quantum state in the computational basis.
density_matrix(wires) Quantum density matrix in the computational basis.
vn_entropy(wires[, log_base]) Von Neumann entropy of the system prior to measurement.
mutual_info(wires0, wires1 Mutual information between the subsystems prior to measurement:
[, log_base])
purity(wires) The purity of the system prior to measurement.
classical_shadow(wires[, seed]) The classical shadow measurement protocol.
shadow_expval(H[, k, seed]) Compute expectation values using classical shadows in a differentiable manner.
Combined measurements¶
Quantum functions can also return combined measurements of multiple observables. If the observables are not qubit-wise-commuting, then multiple device executions may occur behind the scenes.
Non-commuting oberservables can not be simultaneously measured in conjunction with non-observable type measurements such as sample(), counts(), probs(), state(), and density_matrix().
def my_quantum_function(x, y):
qml.RZ(x, wires=0)
qml.CNOT(wires=[0, 1])
qml.RY(y, wires=1)
return qml.expval(qml.PauliZ(0)), qml.var(qml.PauliX(0))
You can also use list comprehensions, and other common Python patterns:
def my_quantum_function(x, y):
qml.RZ(x, wires=0)
qml.CNOT(wires=[0, 1])
qml.RY(y, wires=1)
return [qml.expval(qml.PauliZ(i)) for i in range(2)]
As a full example of combined measurements, let us look at a Bell state \((|00\rangle + |11\rangle)/\sqrt{2}\), prepared by a Hadamard and CNOT gate.
import pennylane as qml
from pennylane import numpy as np
dev = qml.device("default.qubit", wires=2, shots=1000)
def circuit():
qml.CNOT(wires=[0, 1])
return qml.sample(qml.PauliZ(0)), qml.sample(qml.PauliZ(1))
The combined PauliZ-measurement of the first and second qubit returns a tuple of two arrays, each containing the measurement results of the respective qubit. sample() returns 1000 samples per
observable as defined on the device.
>>> results = circuit()
>>> results[0].shape
>>> results[1].shape
Since the two qubits are maximally entangled, the measurement results always coincide, and the lists are therefore equal:
>>> np.all(result[0] == result[1])
Tensor observables¶
PennyLane supports measuring the tensor product of observables, by using the @ notation. For example, to measure the expectation value of \(Z\otimes I \otimes X\):
def my_quantum_function(x, y):
qml.RZ(x, wires=0)
qml.CNOT(wires=[0, 1])
qml.RY(y, wires=1)
qml.CNOT(wires=[0, 2])
return qml.expval(qml.PauliZ(0) @ qml.PauliX(2))
Note that we don’t need to declare the identity observable on wire 1; this is implicitly assumed.
The tensor observable notation can be used inside all measurement functions that accept observables as arguments, including expval(), var(), and sample().
To avoid dealing with long arrays for the larger numbers of shots, one can use counts() rather than sample(). This performs the same measurement as sampling, but returns a dictionary containing the
possible measurement outcomes and the number of occurrences for each, rather than a list of all outcomes.
The previous example will be modified as follows:
dev = qml.device("default.qubit", wires=2, shots=1000)
def circuit():
qml.CNOT(wires=[0, 1])
return qml.counts(qml.PauliZ(0)), qml.counts(qml.PauliZ(1))
After executing the circuit, we can directly see how many times each measurement outcome occurred:
>>> circuit()
({-1: 496, 1: 504}, {-1: 496, 1: 504})
Similarly, if the observable is not provided, the count of the observed computational basis state is returned.
dev = qml.device("default.qubit", wires=2, shots=1000)
def circuit():
qml.CNOT(wires=[0, 1])
return qml.counts()
And the result is:
>>> circuit()
{'00': 495, '11': 505}
By default, only observed outcomes are included in the dictionary. The kwarg all_outcomes=True can be used to display all possible outcomes, including those that were observed 0 times in sampling.
For example, we could run the previous circuit with all_outcomes=True:
dev = qml.device("default.qubit", wires=2, shots=1000)
def circuit():
qml.CNOT(wires=[0, 1])
return qml.counts(all_outcomes=True)
>>> result = circuit()
>>> print(result)
{'00': 518, '01': 0, '10': 0, '11': 482}
Note: For complicated Hamiltonians, this can add considerable overhead time (due to the cost of calculating eigenvalues to determine possible outcomes), and as the number of qubits increases, the
length of the output dictionary showing possible computational basis states grows rapidly.
If counts are obtained along with a measurement function other than sample(), a tuple is returned to provide differentiability for the outputs of QNodes.
def circuit():
return qml.expval(qml.PauliZ(0)),qml.expval(qml.PauliZ(1)), qml.counts()
>>> circuit()
(-0.036, 0.036, {'01': 482, '10': 518})
You can also train QNodes on computational basis probabilities, by using the probs() measurement function. The function can accept either specified wires or an observable that rotates the
computational basis.
def my_quantum_function(x, y):
qml.RZ(x, wires=0)
qml.CNOT(wires=[0, 1])
qml.RY(y, wires=1)
qml.CNOT(wires=[0, 2])
return qml.probs(wires=[0, 1])
For example:
>>> dev = qml.device("default.qubit", wires=3)
>>> qnode = qml.QNode(my_quantum_function, dev)
>>> qnode(0.56, 0.1)
array([0.99750208, 0.00249792, 0. , 0. ])
The returned probability array uses lexicographical ordering, so corresponds to a \(99.75\%\) probability of measuring state \(|00\rangle\), and a \(0.25\%\) probability of measuring state \(|01\
Changing the number of shots¶
For hardware devices where the number of shots determines the accuracy of the expectation value and variance, as well as the number of samples returned, it can sometimes be convenient to execute the
same QNode with differing number of shots.
For simulators like default.qubit, finite shots will be simulated if we set shots to a positive integer.
The shot number can be changed on the device itself, or temporarily altered by the shots keyword argument when executing the QNode:
dev = qml.device("default.qubit", wires=1, shots=10)
def circuit(x, y):
qml.RX(x, wires=0)
qml.RY(y, wires=0)
return qml.expval(qml.PauliZ(0))
# execute the QNode using 10 shots
result = circuit(0.54, 0.1)
# execute the QNode again, now using 1 shot
result = circuit(0.54, 0.1, shots=1)
With an increasing number of shots, the average over measurement samples converges to the exact expectation of an observable. Consider the following circuit:
# fix seed to make results reproducible
dev = qml.device("default.qubit", wires=1)
def circuit():
return qml.expval(qml.PauliZ(0))
Running the simulator with shots=None returns the exact expectation.
>>> circuit(shots=None)
Now we set the device to return stochastic results, and increase the number of shots starting from 10.
>>> circuit(shots=10)
>>> circuit(shots=1000)
>>> circuit(shots=100000)
The result converges to the exact expectation. | {"url":"https://docs.pennylane.ai/en/stable/introduction/measurements.html","timestamp":"2024-11-11T16:22:54Z","content_type":"text/html","content_length":"94555","record_id":"<urn:uuid:a23be125-c6b0-4eae-aac5-488e18390d1b>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00673.warc.gz"} |
Provide refinement functions for a the raw data of a VASP calculation run in the current directory.
Usually one is not directly interested in the raw data that is produced but wants to produce either a figure for a publication or some post-processing of the data. This package contains multiple
modules that enable these kinds of workflows by extracting the relevant data from the HDF5 file and transforming them into an accessible format. The modules also provide plotting functionality to get
a quick insight about the data, which can then be refined either within python or a different tool to obtain publication-quality figures.
Generally, all modules provide a read function that extracts the data from the HDF5 file and puts it into a Python dictionary. Where it makes sense in addition a plot function is available that
converts the data into a figure for Jupyter notebooks. In addition, data conversion routines to_X may be available transforming the data into another format or file, which may be useful to generate
plots with tools other than Python. For the specifics, please refer to the documentation of the individual modules.
The raw data is read from the current directory. The Calculation class provides a more flexible interface with which you can determine the source directory or file for the VASP calculation manually.
That class exposes functions of the modules as methods of attributes, i.e., the two following examples are equivalent:
using calculation module
>>> from py4vasp import calculation
>>> calculation.dos.read()
using Calculation class
>>> from py4vasp import Calculation
>>> calc = Calculation.from_path(".")
>>> calc.dos.read()
In the latter example, you can change the path from which the data is extracted.
band The band structure contains the k point resolved eigenvalues.
bandgap This class describes the band extrema during the relaxation or MD simulation.
born_effective_charge The Born effective charge tensors couple electric field and atomic displacement.
CONTCAR CONTCAR contains structural restart-data after a relaxation or MD simulation.
density This class accesses various densities (charge, magnetization, ...) of VASP.
dielectric_function The dielectric function describes the material response to an electric field.
dielectric_tensor The dielectric tensor is the static limit of the dielectric function.
dos The density of states (DOS) describes the number of states per energy.
elastic_modulus The elastic modulus is the second derivative of the energy with respect to strain.
energy The energy data for one or several steps of a relaxation or MD simulation.
fatband BSE fatbands illustrate the excitonic properties of materials.
force The forces determine the path of the atoms in a trajectory.
force_constant Force constants are the 2nd derivatives of the energy with respect to displacement.
internal_strain The internal strain is the derivative of energy with respect to displacement and strain.
kpoint The k-point mesh used in the VASP calculation.
magnetism The local moments describe the charge and magnetization near an atom.
pair_correlation The pair-correlation function measures the distribution of atoms.
phonon_band The phonon band structure contains the q-resolved phonon eigenvalues.
phonon_dos The phonon density of states (DOS) describes the number of modes per energy.
piezoelectric_tensor The piezoelectric tensor is the derivative of the energy with respect to strain and field.
polarization The static polarization describes the electric dipole moment per unit volume.
potential The local potential describes the interactions between electrons and ions.
projector The projectors used for atom and orbital resolved quantities.
stress The stress describes the force acting on the shape of the unit cell.
structure The structure contains the unit cell and the position of all ions within.
system The SYSTEM tag in the INCAR file is a title you choose for a VASP calculation.
topology The topology of the crystal describes the ions of a crystal and their connectivity.
velocity The velocities describe the ionic motion during an MD simulation.
workfunction The workfunction describes the energy required to remove an electron to the vacuum. | {"url":"https://vasp.at/py4vasp/latest/calculation/","timestamp":"2024-11-14T10:42:19Z","content_type":"text/html","content_length":"77290","record_id":"<urn:uuid:27e9d113-36b2-451e-b849-87b475567a72>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00632.warc.gz"} |
How To Use Gann Angles Without Drawing A Single Line On The Chart
How to Use Gann Angles Without drawing a Single line on the Chart
Gann’s Square of Nine
Gann’s Law of Vibration
Gann’s Time Cycle Method
Gann’s Square of 360-degree
Above all the gann strategies, first, most gann learners rushed behind in learning the gann angles.
They spend thousands of dollars on costly software and invest another 2000 hours in learning how Gann is used to draw angles.
And in the end, they fail.
Yes, they cannot get the same result gann had with his angles.
And then they quit learning Gann’s angles concept and move ahead after wasting time and money.
Most Gann learners fail to learn Gann’s angle method because they think WD Gann used ANGLES only by drawing angles on the chart.
And that’s not just 100%, but 1000% wrong.
Gann had many ways to use and apply the Angles on the chart. In this article, we’ll learn one of those ways (method) to use gann angles, and that too without spending 1000s of dollars. So, let’s
First, let’s understand what Gann angles are in simple words.
A gann angle is a geometric tool WD Gann used to find support & resistance and identify the trend’s strength.
To use and draw angles on the charts, gann used the plastic overlay template, a protractor used in geometry lessons in schools.
Like a protractor has many angles, Gann’s plastic overlay template also has many angles. And out of all the angles, the 1×1 45-degree angle is important.
Gann said, “You can beat the market by trading against the 45° angle alone if you stick to the rule, wait to buy on the 45° angle or wait to sell it against the 45° angle.” – WD Gann.
Apart from the 1×1 45° angle, the other important angles are 2×1,3×1, 4×1, 8×1, 16×1 angles.
Now, the question is, can we use a plastic overlay template like Gann today?
Yes, we can use the protractor instead of the plastic overlay template.
The next important question emerges: Will we get the same results with angles as Gann does in his trades? The answer is no.
Because Gann used to make charts by hand, which were proper Square charts. In today’s time, we don’t have access to such charts, and no software can construct proper square charts.
We can make charts by hand, but that’s almost impossible because that exercise requires proper daily time. So, most of us can’t keep that chart updated.
But we can still use Gann angles to trade the markets—and too, with great accuracy. Yes, there is an option and handy way to use Gann angles.
It’s a mathematical approach. That requires a basic calculation that a 10-year-old kid can do.
It’s not the last dish, but what we are learning in getting there.
In cooking and trading, we rush through things because we’re trying to reach a foolproof zone. But along the way, we forget about the core meaning.
WD Gann was like a master chef. He perfected his knowledge of geometry and mathematics so well that he learned how to adjust things along the way. The same is true of his angle method. He directly
applied the degrees of angles as numbers.
We can use the same approach for using angles that work well and deliver excellent results in trading.
So, let’s understand the simple calculation that gann used.
First, we’ve to convert the 2×1,3×1, 4×1, 8×1, and 16×1 angles into degrees, as I have done below. You can save the image for your use.
Next, we’ve to add or subtract (as per the trend) the angle/degree number (s) from the major tops and bottoms. Then, finally, we’ll get the price levels that will act as support & resistance in
In simple words, we can use a 45-degree/angle or other degrees/angles as a number. The same gann did. But most gann students missed this view of gann and started working on tricky formulae.
We don’t have to use any complex formula to convert degrees into numbers. We just have to use degrees as numbers. And that works fine.
So, let’s discuss a few examples to understand how this Gann angle method works in real market charts.
Nifty Gann Angle Calculation: From the 19 October 2021 high (18604), if we subtract 262.5 points (2*1 Gann Angle), we’ll get 18341. That’s almost the exact price level that took resistance on 18
January 2022 and fell 14%.
Bank Nifty Gann Angle Calculation: From the 25 October 2021 high (41829), if we subtract 2625 points (2*1 Gann Angle), we’ll get 39204. Again, that’s almost the exact price level where Bank Nifty
took resistance on 03 & 10 February and fell 18%.
Maruti Gann Angle Calculation: In the 26 August 2021 low (6591), if we add 450 points (1*1 Gann Angle), we’ll get 7041. Then, between 29 November-to-20 December 2021, the stock Maruti took support
many times from the 7041 level and surged 26%.
MCX Crude Oil Gann Angle Calculation: In the 20 August 2021 low (4634), if we add 75 points (8*1 Gann Angle), we’ll get 4709. That’s almost the exact price level where MCX Crude Oil took support on
02 December 2021 and surged over 100%.
MCX Gold Gann Angle Calculation: In the 29 September 2021 low (47511), if we add 1500 points (4*1 Gann Angle 15-degree), we’ll get 47011. Then between November 2021-to-January 2022, MCX Gold took
support many times from near the 47011 level and surged 15%.
What’s the one thing you can do today?
Stop spending hard-earned money buying the software just for using the one Gann angle technique. That is a waste of money.
Instead, take the time to practice the Gann angle mathematical method we learned in this article. It will save you money and time.
Ready to look Deeper into WD Gann's Breakthrough work?
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For the analysis of multiple linear regression models statisticians apply Gauss-Markov theorem for the estimators of the regression to be best linear unbiased estimators (BLUE). The theorem includes
5 assumptions about heteroskedasticity, linearity, exogeneity, random sampling and non-collinearity.
One of the assumptions made about residuals/errors in OLS regression is that the errors have the same but unknown variance. This is known as constant variance or homoscedasticity. When this
assumption is violated, the problem is known as heteroscedasticity.Heteroskedasticity is one of 5 Gauss-Markov assumptions. It is tested by Breusch-Pagan and Goldfeld-Quandt tests.
Breusch-Pagan Test
Breusch Pagan Test was introduced by Trevor Breusch and Adrian Pagan in 1979. It is used to test for heteroskedasticity in a linear regression model and assumes that the error terms are normally
distributed. It tests whether the variance of the errors from a regression is dependent on the values of the independent variables. Null hypothesis states that error variances are constant.
model <- lm(real_gdp ~ imp + exp + poil + eurkzt,macroKZ)
#> Breusch-Pagan test
#> Heteroskedasticity presents.
#> Please use other tests additionally.
#> In case of opposite results study the case further.
#> --------------------
#> BP 15.152
#> p-value 0.004
#> --------------------
Goldfeld-Quandt Test
The Goldfeld Quandt Test is a test used in regression analysis to test for homoscedasticity. It compares variances of two subgroups; one set of high values and one set of low values. If the variances
differ, the test rejects the null hypothesis that the variances of the errors are not constant.
model <- lm(real_gdp ~ imp + exp+poil+eurkzt, macroKZ)
#> Goldfeld-Quandt test
#> Homoskedasticity presents.
#> Please use other tests additionally.
#> In case of opposite results study the case further.
#> -------------------
#> GQ 1.951
#> p-value 0.057
#> df1 24
#> df2 23
#> -------------------
Multiple regression assumes that the independent variables are not highly correlated with each other. This assumption is tested using Variance Inflation Factor (VIF) values and by Durbin-Watson and
Breusch-Godfrey tests for autocorrelation.
VIF Test
The VIF of the linear regression is defined as VIF = 1/T. With VIF > 5 there is an indication that multicollinearity may be present; with VIF > 10 there is certainly multicollinearity among the
model <- lm(real_gdp ~ imp + exp + poil + eurkzt,macroKZ)
#> Variance Inflation Factor
#> If statistics exceeds 5, please be aware of multicollinearity.
#> imp exp poil eurkzt
#> 2.062 14.815 16.244 2.328
#> This value 14.815 exceeds acceptable threshold.
#> This value 16.244 exceeds acceptable threshold.
Durbin-Watson Test
The Durbin Watson (DW) statistic is a test for autocorrelation in the residuals from a statistical model or regression analysis. The Durbin-Watson statistic will always have a value ranging between 0
and 4. A value of 2.0 indicates there is no autocorrelation detected in the sample. Values from 0 to less than 2 point to positive autocorrelation and values from 2 to 4 means negative
model <- lm(real_gdp ~ imp + exp + poil + eurkzt,macroKZ)
#> Durbin-Watson test
#> data: model
#> DW = 2.1296, p-value = 0.5525
#> alternative hypothesis: true autocorrelation is greater than 0
Breusch-Godfrey Test
Alternatively, there is Breusch-Godfrey Test for autocorrelation check.It tests for the presence of serial correlation that has not been included in a proposed model structure and which, if present,
would mean that incorrect conclusions would be drawn from other tests or that sub-optimal estimates of model parameters would be obtained.Null hypothesis states that there is no autocorrelation.
model <- lm(real_gdp ~ imp + exp + poil + eurkzt,macroKZ)
#> Breusch-Godfrey test for serial correlation of order up to 1
#> Residuals are not autocorrelated.
#> -------------------
#> LM test 0.324
#> p-value 0.569
#> -------------------
Normality refers to a specific statistical distribution called a normal distribution, or sometimes the Gaussian distribution or bell-shaped curve. The normal distribution is a symmetrical continuous
distribution defined by the mean and standard deviation of the data.
In AFR package 4 normality tests are compiled in one norm_test function referred to olsrr package.
Shapiro-Wilk statistic
The null-hypothesis of this test is that the population is normally distributed. Thus, if the p value is less than the chosen alpha level, then the null hypothesis is rejected and there is evidence
that the data tested are not normally distributed. On the other hand, if the p value is greater than the chosen alpha level, then the null hypothesis (that the data came from a normally distributed
population) can not be rejected.
Kolmogorov-Smirnov statistic
The Kolmogorov–Smirnov statistic quantifies a distance between the empirical distribution function of the sample and the cumulative distribution function of the reference distribution, or between the
empirical distribution functions of two samples. The null distribution of this statistic is calculated under the null hypothesis that the sample is drawn from the reference distribution (in the
one-sample case) or that the samples are drawn from the same distribution (in the two-sample case).Since the p-value is less than .05, we reject the null hypothesis.
Cramer-Von Mises test
Alternative to Kolmogorov-Smirnov test, Cramer-von Mises statistic is a measure of the mean squared difference between the empirical and hypothetical cumulative distribution functions. It is also
used as a part of other algorithms, such as minimum distance estimation.The Cramér–von Mises test can be seen to be distribution-free if empirical distribution is continuous and the sample has no
ties. Otherwise, statistic is not the true asymptotic distribution.
Anderson test
The Anderson-Darling test is used to test if a sample of data comes from a population with a specific distribution.The null hypothesis is that your data is not different from normal. Your alternate
or alternative hypothesis is that your data is different from normal. You will make your decision about whether to reject or not reject the null based on your p-value.
For additional information please address:
Wooldridge, Jeffrey M. 2012. Introductory Econometrics: A Modern Approach, Fifth Edition.
Hyndman, Rob J and George Athanasopoulos. 2018. Forecasting: Principles and Practice, 2nd Edition. | {"url":"http://rsync.jp.gentoo.org/pub/CRAN/web/packages/AFR/vignettes/Diagnostic-tests.html","timestamp":"2024-11-11T16:22:01Z","content_type":"text/html","content_length":"23334","record_id":"<urn:uuid:b8efb625-6d07-4fbd-ab34-5e54ebb41d8b>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00701.warc.gz"} |
Gr 4_NF_Fractions_BriefArgument_Construct | Bridging Practices Among Connecticut Mathematics Educators
Gr 4_NF_Fractions_BriefArgument_Construct
This task was intended for grade 4 students. The main objective was to have the students demonstrate understanding of equivalent fractions. The task required them to construct an argument which would
help explain whether they agreed or disagreed with the claim.
Microsoft Word version: 4_NF_Fractions_BriefArgument_Construct
PDF version: 4_NF_Fractions_BriefArgument_Construct | {"url":"https://bridges.education.uconn.edu/2015/06/19/gr-4_nf_fractions_briefargument_construct/","timestamp":"2024-11-11T16:27:40Z","content_type":"text/html","content_length":"52615","record_id":"<urn:uuid:20147e12-6ac7-4f77-8c69-a49035a6e983>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00095.warc.gz"} |
Compute and sketch vo and iR for the network given below.
Understand the Problem
The question is asking to compute the voltage (vo) and current (iR) for a given electrical network and to create a sketch representing these values. This involves analyzing the circuit and
potentially applying circuit theory concepts.
The specific computed values of $v_o$ and $i_R$ depend on the values of circuit elements provided.
Answer for screen readers
To fully answer your question, specific values for the circuit components (resistances, voltage sources) are needed. However, based on the steps given, you will derive values for voltage and current
from the circuit analysis.
Steps to Solve
1. Identify the Circuit Elements
First, identify all the components in the circuit given. Note down resistors (R), voltage sources (V), and any current sources (I). Make sure you understand how these components are connected (in
series or parallel).
2. Apply Ohm's Law
For calculating voltage ($v_o$) across a resistor, use Ohm's Law, which states:
$$ V = I \times R $$
where $V$ is the voltage, $I$ is the current, and $R$ is the resistance.
3. Calculate Total Current ($i_R$)
If the circuit has multiple resistors, use series and parallel rules to find the total resistance $R_{total}$:
• For resistors in series: $$ R_{total} = R_1 + R_2 + ... + R_n $$
• For resistors in parallel: $$ \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n} $$
4. Use Kirchhoff's Laws
Apply Kirchhoff’s Voltage Law (KVL) to analyze the loops in the circuit and solve for the unknowns. The sum of the voltages around any closed loop in a circuit must equal zero.
5. Sketch the Circuit with Calculated Values
Create a simple sketch of the circuit, labeling all components clearly. Include the computed values of $v_o$ and $i_R$ on the sketch.
To fully answer your question, specific values for the circuit components (resistances, voltage sources) are needed. However, based on the steps given, you will derive values for voltage and current
from the circuit analysis.
More Information
When calculating voltage and current in a circuit, it's helpful to visualize the components through a sketch, which makes it easy to understand the relationships. Circuit theory relies on several key
principles, including Ohm's Law and Kirchhoff's Laws, to solve for unknown values.
• Not properly identifying whether resistors are in series or parallel, which can lead to incorrect resistance calculations.
• Forgetting to account for all voltage sources in the circuit when applying KVL.
• Mislabeling circuit elements in the sketch, leading to confusion. | {"url":"https://quizgecko.com/q/compute-and-sketch-vo-and-ir-for-the-network-given-below-mksji","timestamp":"2024-11-11T03:37:34Z","content_type":"text/html","content_length":"170793","record_id":"<urn:uuid:102d468c-d81f-4169-a1a3-295a2eaf3375>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00560.warc.gz"} |
In Exercises \(1-8, Z\) is the standard normal variable. Find the indicated probabilities. $$ P(0 \leq Z \leq 1.5) $$
Short Answer
Expert verified
The probability that the standard normal variable Z falls between 0 and 1.5 is \(P(0 \leq Z \leq 1.5) = 0.4332\).
Step by step solution
Identify the given z-values
The problem gives us the range for Z, which is between 0 and 1.5. So, the given z-values are 0 and 1.5.
Find the area under the curve for z = 0
Since the standard normal distribution is symmetrical around the mean (which is 0 for Z), the area under the curve to the left of z = 0 is 0.5.
Use the z-table to find the area under the curve for z = 1.5
To find the area to the left of z = 1.5, we will look at a standard normal table (z-table). Locate the row that corresponds to the first two digits of the z-value (1.5). In this case, the row is 1.5.
Next, find the column that corresponds to the second decimal place; 0 in our case. Follow the row and column to find the intersecting value. For z = 1.5, the value we get from the table is 0.9332.
This means that the area under the curve to the left of z = 1.5 is 0.9332.
Calculate the area between z = 0 and z = 1.5
The area under the curve between two z-values is the difference between the areas to the left of each z-value. In this case, it is the area under the curve to the left of z = 1.5 minus the area under
the curve to the left of z = 0. So, the area between z = 0 and z = 1.5 is 0.9332 - 0.5 = 0.4332.
Write the probability
The probability that the standard normal variable, Z, falls between 0 and 1.5 is equal to the area under the curve between these z-values. In this case, \(P(0 \leq Z \leq 1.5) = 0.4332\).
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
The Z-table, often referred to as the standard normal table, is an essential tool in statistics that helps us determine the probability of a normal random variable not exceeding a particular value.
This table is built upon the standard normal distribution, a special normal distribution with a mean of 0 and a standard deviation of 1.
When you look at the Z-table, what you see is the area to the left of a Z-score. For example, if you want to find the probability that a normal variable is less than Z = 1.5, you would search the
table for the value 1.5. To improve the use of a Z-table in exercises and solutions, it's vital to ensure that students understand how to read and interpret this table accurately. They need to know
how to match their Z-score with the table properly to find the associated probability. It's also helpful to shed light on the significance of the right side of the curve, which represents probability
values greater than Z, often calculated by subtracting the table value from 1.
Area Under Normal Curve
The area under the normal curve is a representation of probability in the context of a normal distribution. Specifically, in a standard normal distribution, the total area under the curve equals 1,
signifying that the probability of a random variable falling within the full range of the distribution is 100%.
When we're concerned with a particular range, such as in the exercise where we examine the area from Z = 0 to Z = 1.5, the corresponding area represents the likelihood that the variable falls within
that range. The symmetrical nature of the curve around the mean allows for certain shortcuts. For instance, knowing that the area to the left of Z = 0 is 0.5 facilitates the computation of
probabilities for positive Z-scores without referencing the table. Improvements in describing this concept should focus on the graph and area visualization, helping students better understand how
probability relates to the area under the curve and the profound implications of symmetry on these calculations.
Probability Calculation
Probability calculation in the realm of normal distributions is integral for predicting the frequency of events within a dataset. The process involves finding the area under the curve for specific
Z-values using a combination of the Z-table and the understanding that the total area equals 1, corresponding to 100% probability.
In the given exercise, the computation between Z = 0 and Z = 1.5 involves subtracting the area to the left of Z = 0 from the area to the left of Z = 1.5, as derived from the Z-table. An essential
improvement in this aspect of education is to clarify why we subtract these areas and to explain how these areas correlate with probabilities, enhancing the student's ability to grasp the real-world
meaning behind these abstract concepts.
Normal Variable Z
The normal variable Z, often simply called the Z-score, is the standardized form of a random variable that follows a normal distribution. By standardizing, we subtract the mean of the distribution
and divide by the standard deviation, resulting in a Z-score with a mean of 0 and a standard deviation of 1.
This standardization process allows for comparison across different normal distributions and enables the utilization of the Z-table for probability calculations. Understanding the Z-score is a
foundational aspect of statistics as it is key to interpreting data, and gauging how unusual or typical a particular outcome is. Improvements in teaching about the normal variable Z should emphasize
the transformation process from regular scores to standardized ones and clearly explain why such standardization makes it easier to work with diverse data sets within a common framework. | {"url":"https://www.vaia.com/en-us/textbooks/math/finite-mathematics-and-applied-calculus-7-edition/chapter-8/problem-2-in-exercises-1-8-z-is-the-standard-normal-variable/","timestamp":"2024-11-14T04:00:20Z","content_type":"text/html","content_length":"248130","record_id":"<urn:uuid:c4a1fcc3-1d4b-4e97-bd5b-775da601e6cb>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00867.warc.gz"} |
1.3 Polar Coordinates - Calculus Volume 3 | OpenStax
• 1.3.1 Locate points in a plane by using polar coordinates.
• 1.3.2 Convert points between rectangular and polar coordinates.
• 1.3.3 Sketch polar curves from given equations.
• 1.3.4 Convert equations between rectangular and polar coordinates.
• 1.3.5 Identify symmetry in polar curves and equations.
The rectangular coordinate system (or Cartesian plane) provides a means of mapping points to ordered pairs and ordered pairs to points. This is called a one-to-one mapping from points in the plane to
ordered pairs. The polar coordinate system provides an alternative method of mapping points to ordered pairs. In this section we see that in some circumstances, polar coordinates can be more useful
than rectangular coordinates.
Defining Polar Coordinates
To find the coordinates of a point in the polar coordinate system, consider Figure 1.27. The point $PP$ has Cartesian coordinates $(x,y).(x,y).$ The line segment connecting the origin to the point
$PP$ measures the distance from the origin to $PP$ and has length $r.r.$ The angle between the positive $xx$-axis and the line segment has measure $θ.θ.$ This observation suggests a natural
correspondence between the coordinate pair $(x,y)(x,y)$ and the values $rr$ and $θ.θ.$ This correspondence is the basis of the polar coordinate system. Note that every point in the Cartesian plane
has two values (hence the term ordered pair) associated with it. In the polar coordinate system, each point also has two values associated with it: $rr$ and $θ.θ.$
Using right-triangle trigonometry, the following equations are true for the point $P:P:$
Each point $(x,y)(x,y)$ in the Cartesian coordinate system can therefore be represented as an ordered pair $(r,θ)(r,θ)$ in the polar coordinate system. The first coordinate is called the radial
coordinate and the second coordinate is called the angular coordinate. Every point in the plane can be represented in this form.
Note that the equation $tanθ=y/xtanθ=y/x$ has an infinite number of solutions for any ordered pair $(x,y).(x,y).$ However, if we restrict the solutions to values between $00$ and $2π2π$ then we can
assign a unique solution to the quadrant in which the original point $(x,y)(x,y)$ is located. Then the corresponding value of r is positive, so $r2=x2+y2.r2=x2+y2.$
Converting Points between Coordinate Systems
Given a point $PP$ in the plane with Cartesian coordinates $(x,y)(x,y)$ and polar coordinates $(r,θ),(r,θ),$ the following conversion formulas hold true:
These formulas can be used to convert from rectangular to polar or from polar to rectangular coordinates.
Converting between Rectangular and Polar Coordinates
Convert each of the following points into polar coordinates.
a. $(1,1)(1,1)$
b. $(−3,4)(−3,4)$
c. $(0,3)(0,3)$
d. $(53,−5)(53,−5)$
Convert each of the following points into rectangular coordinates.
e. $(3,π/3)(3,π/3)$
f. $(2,3π/2)(2,3π/2)$
g. $(6,−5π/6)(6,−5π/6)$
a. Use $x=1x=1$ and $y=1y=1$ in Equation 1.8:
Therefore this point can be represented as $(2,π4)(2,π4)$ in polar coordinates.
b. Use $x=−3x=−3$ and $y=4y=4$ in Equation 1.8:
The point $(-3, 4)(-3, 4)$ lies in Quadrant $IIII$. Subtract the value of the reference angle, $arctan43arctan43$, from $ππ$ to find the radian measure of $θθ$.
Therefore this point can be represented as $(5,2.21)(5,2.21)$ in polar coordinates.
c. Use $x=0x=0$ and $y=3y=3$ in Equation 1.8:
Direct application of the second equation leads to division by zero. Graphing the point $(0,3)(0,3)$ on the rectangular coordinate system reveals that the point is located on the positive y-axis.
The angle between the positive x-axis and the positive y-axis is $π2.π2.$ Therefore this point can be represented as $(3,π2)(3,π2)$ in polar coordinates.
d. Use $x=53x=53$ and $y=−5y=−5$ in Equation 1.8:
Therefore this point can be represented as $(10,−π6)(10,−π6)$ in polar coordinates.
e. Use $r=3r=3$ and $θ=π3θ=π3$ in Equation 1.7:
Therefore this point can be represented as $(32,332)(32,332)$ in rectangular coordinates.
f. Use $r=2r=2$ and $θ=3π2θ=3π2$ in Equation 1.7:
Therefore this point can be represented as $(0,−2)(0,−2)$ in rectangular coordinates.
g. Use $r=6r=6$ and $θ=−5π6θ=−5π6$ in Equation 1.7:
Therefore this point can be represented as $(−33,−3)(−33,−3)$ in rectangular coordinates.
Convert $(−8,−8)(−8,−8)$ into polar coordinates and $(4,2π3)(4,2π3)$ into rectangular coordinates.
The polar representation of a point is not unique. For example, the polar coordinates $(2,π3)(2,π3)$ and $(2,7π3)(2,7π3)$ both represent the point $(1,3)(1,3)$ in the rectangular system. Also, the
value of $rr$ can be negative. Therefore, the point with polar coordinates $(−2,4π3)(−2,4π3)$ also represents the point $(1,3)(1,3)$ in the rectangular system, as we can see by using Equation 1.8:
Every point in the plane has an infinite number of representations in polar coordinates. However, each point in the plane has only one representation in the rectangular coordinate system.
Note that the polar representation of a point in the plane also has a visual interpretation. In particular, $rr$ is the directed distance that the point lies from the origin, and $θθ$ measures the
angle that the line segment from the origin to the point makes with the positive $xx$-axis. Positive angles are measured in a counterclockwise direction and negative angles are measured in a
clockwise direction. The polar coordinate system appears in the following figure.
The line segment starting from the center of the graph going to the right (called the positive x-axis in the Cartesian system) is the polar axis. The center point is the pole, or origin, of the
coordinate system, and corresponds to $r=0.r=0.$ The innermost circle shown in Figure 1.28 contains all points a distance of 1 unit from the pole, and is represented by the equation $r=1.r=1.$ Then
$r=2r=2$ is the set of points 2 units from the pole, and so on. The line segments emanating from the pole correspond to fixed angles. To plot a point in the polar coordinate system, start with the
angle. If the angle is positive, then measure the angle from the polar axis in a counterclockwise direction. If it is negative, then measure it clockwise. If the value of $rr$ is positive, move that
distance along the terminal ray of the angle. If it is negative, move along the ray that is opposite the terminal ray of the given angle.
Plotting Points in the Polar Plane
Plot each of the following points on the polar plane.
a. $(2,π4)(2,π4)$
b. $(−3,2π3)(−3,2π3)$
c. $(4,5π4)(4,5π4)$
The three points are plotted in the following figure.
Plot $(4,5π3)(4,5π3)$ and $(−3,−7π2)(−3,−7π2)$ on the polar plane.
Polar Curves
Now that we know how to plot points in the polar coordinate system, we can discuss how to plot curves. In the rectangular coordinate system, we can graph a function $y=f(x)y=f(x)$ and create a curve
in the Cartesian plane. In a similar fashion, we can graph a curve that is generated by a function $r=f(θ).r=f(θ).$
The general idea behind graphing a function in polar coordinates is the same as graphing a function in rectangular coordinates. Start with a list of values for the independent variable $(θ(θ$ in this
case) and calculate the corresponding values of the dependent variable $r.r.$ This process generates a list of ordered pairs, which can be plotted in the polar coordinate system. Finally, connect the
points, and take advantage of any patterns that may appear. The function may be periodic, for example, which indicates that only a limited number of values for the independent variable are needed.
Plotting a Curve in Polar Coordinates
1. Create a table with two columns. The first column is for $θ,θ,$ and the second column is for $r.r.$
2. Create a list of values for $θ.θ.$
3. Calculate the corresponding $rr$ values for each $θ.θ.$
4. Plot each ordered pair $(r,θ)(r,θ)$ on the coordinate axes.
5. Connect the points and look for a pattern.
Watch this video for more information on sketching polar curves.
Graphing a Function in Polar Coordinates
Graph the curve defined by the function $r=4sinθ.r=4sinθ.$ Identify the curve and rewrite the equation in rectangular coordinates.
Because the function is a multiple of a sine function, it is periodic with period $2π,2π,$ so use values for $θθ$ between 0 and $2π.2π.$ The result of steps 1–3 appear in the following table. Figure
1.30 shows the graph based on this table.
$θθ$ $r=4sinθr=4sinθ$ $θθ$ $r=4sinθr=4sinθ$
0 0 $ππ$ 0
$π6π6$ $22$ $7π67π6$ $−2−2$
$π4π4$ $22≈2.822≈2.8$ $5π45π4$ $−22≈−2.8−22≈−2.8$
$π3π3$ $23≈3.423≈3.4$ $4π34π3$ $−23≈−3.4−23≈−3.4$
$π2π2$ $44$ $3π23π2$ $-4-4$
$2π32π3$ $23≈3.423≈3.4$ $5π35π3$ $−23≈−3.4−23≈−3.4$
$3π43π4$ $22≈2.822≈2.8$ $7π47π4$ $−22≈−2.8−22≈−2.8$
$5π65π6$ $22$ $11π611π6$ $−2−2$
$2π2π$ 0
This is the graph of a circle. The equation $r=4sinθr=4sinθ$ can be converted into rectangular coordinates by first multiplying both sides by $r.r.$ This gives the equation $r2=4rsinθ.r2=4rsinθ.$
Next use the facts that $r2=x2+y2r2=x2+y2$ and $y=rsinθ.y=rsinθ.$ This gives $x2+y2=4y.x2+y2=4y.$ To put this equation into standard form, subtract $4y4y$ from both sides of the equation and complete
the square:
$x 2 + y 2 − 4 y = 0 x 2 + ( y 2 − 4 y ) = 0 x 2 + ( y 2 − 4 y + 4 ) = 0 + 4 x 2 + ( y − 2 ) 2 = 4. x 2 + y 2 − 4 y = 0 x 2 + ( y 2 − 4 y ) = 0 x 2 + ( y 2 − 4 y + 4 ) = 0 + 4 x 2 + ( y − 2 ) 2 = 4.$
This is the equation of a circle with radius 2 and center $(0,2)(0,2)$ in the rectangular coordinate system.
Create a graph of the curve defined by the function $r=4+4cosθ.r=4+4cosθ.$
The graph in Example 1.12 was that of a circle. The equation of the circle can be transformed into rectangular coordinates using the coordinate transformation formulas in Equation 1.8. Example 1.14
gives some more examples of functions for transforming from polar to rectangular coordinates.
Transforming Polar Equations to Rectangular Coordinates
Rewrite each of the following equations in rectangular coordinates and identify the graph.
a. $θ=π3θ=π3$
b. $r=3r=3$
c. $r=6cosθ−8sinθr=6cosθ−8sinθ$
a. Take the tangent of both sides. This gives $tanθ=tan(π/3)=3.tanθ=tan(π/3)=3.$ Since $tanθ=y/xtanθ=y/x$ we can replace the left-hand side of this equation by $y/x.y/x.$ This gives $y/x=3,y/x=3,$
which can be rewritten as $y=x3.y=x3.$ This is the equation of a straight line passing through the origin with slope $3.3.$ In general, any polar equation of the form $θ=Kθ=K$ represents a
straight line through the pole with slope equal to $tanK.tanK.$
b. First, square both sides of the equation. This gives $r2=9.r2=9.$ Next replace $r2r2$ with $x2+y2.x2+y2.$ This gives the equation $x2+y2=9,x2+y2=9,$ which is the equation of a circle centered at
the origin with radius 3. In general, any polar equation of the form $r=kr=k$ where k is a positive constant represents a circle of radius k centered at the origin. (Note: when squaring both
sides of an equation it is possible to introduce new points unintentionally. This should always be taken into consideration. However, in this case we do not introduce new points. For example, $
(−3,π3)(−3,π3)$ is the same point as $(3,4π3).)(3,4π3).)$
c. Multiply both sides of the equation by $r.r.$ This leads to $r2=6rcosθ−8rsinθ.r2=6rcosθ−8rsinθ.$ Next use the formulas
This gives
To put this equation into standard form, first move the variables from the right-hand side of the equation to the left-hand side, then complete the square.
This is the equation of a circle with center at $(3,−4)(3,−4)$ and radius 5. Notice that the circle passes through the origin since the center is 5 units away.
Rewrite the equation $r=secθtanθr=secθtanθ$ in rectangular coordinates and identify its graph.
We have now seen several examples of drawing graphs of curves defined by polar equations. A summary of some common curves is given in the tables below. In each equation, a and b are arbitrary
A cardioid is a special case of a limaçon (pronounced “lee-mah-son”), in which $a=ba=b$ or $a=−b.a=−b.$ The rose is a very interesting curve. Notice that the graph of $r=3sin2θr=3sin2θ$ has four
petals. However, the graph of $r=3sin3θr=3sin3θ$ has three petals as shown.
If the coefficient of $θθ$ is even, the graph has twice as many petals as the coefficient. If the coefficient of $θθ$ is odd, then the number of petals equals the coefficient. You are encouraged to
explore why this happens. Even more interesting graphs emerge when the coefficient of $θθ$ is not an integer. For example, if it is rational, then the curve is closed; that is, it eventually ends
where it started (Figure 1.34(a)). However, if the coefficient is irrational, then the curve never closes (Figure 1.34(b)). Although it may appear that the curve is closed, a closer examination
reveals that the petals just above the positive r axis are slightly thicker. This is because the petal does not quite match up with the starting point.
Chapter Opener: Describing a Spiral
Recall the chambered nautilus introduced in the chapter opener. This creature displays a spiral when half the outer shell is cut away. It is possible to describe a spiral using rectangular
coordinates. Figure 1.35 shows a spiral in rectangular coordinates. How can we describe this curve mathematically?
As the point P travels around the spiral in a counterclockwise direction, its distance d from the origin increases. Assume that the distance d is a constant multiple k of the angle $θθ$ that the line
segment OP makes with the positive r-axis. Therefore $d(P,O)=kθ,d(P,O)=kθ,$ where $OO$ is the origin. Now use the distance formula and some trigonometry:
$d ( P , O ) = k θ ( x − 0 ) 2 + ( y − 0 ) 2 = k arctan ( y x ) x 2 + y 2 = k arctan ( y x ) arctan ( y x ) = x 2 + y 2 k y = x tan ( x 2 + y 2 k ) . d ( P , O ) = k θ ( x − 0 ) 2 + ( y − 0 ) 2 = k
arctan ( y x ) x 2 + y 2 = k arctan ( y x ) arctan ( y x ) = x 2 + y 2 k y = x tan ( x 2 + y 2 k ) .$
Although this equation describes the spiral, it is not possible to solve it directly for either x or y. However, if we use polar coordinates, the equation becomes much simpler. In particular, $d(P,O)
=r,d(P,O)=r,$ and $θθ$ is the second coordinate. Therefore the equation for the spiral becomes $r=kθ.r=kθ.$ Note that when $θ=0θ=0$ we also have $r=0,r=0,$ so the spiral emanates from the origin. We
can remove this restriction by adding a constant to the equation. Then the equation for the spiral becomes $r=a+kθr=a+kθ$ for arbitrary constants $aa$ and $k.k.$ This is referred to as an Archimedean
spiral, after the Greek mathematician Archimedes.
Another type of spiral is the logarithmic spiral, described by the function $r=a·bθ.r=a·bθ.$ A graph of the function $r=1.2(1.25θ)r=1.2(1.25θ)$ is given in Figure 1.36. This spiral describes the
shell shape of the chambered nautilus.
Suppose a curve is described in the polar coordinate system via the function $r=f(θ).r=f(θ).$ Since we have conversion formulas from polar to rectangular coordinates given by
it is possible to rewrite these formulas using the function
This step gives a parameterization of the curve in rectangular coordinates using $θθ$ as the parameter. For example, the spiral formula $r=a+bθr=a+bθ$ from Figure 1.31 becomes
Letting $θθ$ range from $−∞−∞$ to $∞∞$ generates the entire spiral.
Symmetry in Polar Coordinates
When studying symmetry of functions in rectangular coordinates (i.e., in the form $y=f(x)),y=f(x)),$ we talk about symmetry with respect to the y-axis and symmetry with respect to the origin. In
particular, if $f(−x)=f(x)f(−x)=f(x)$ for all $xx$ in the domain of $f,f,$ then $ff$ is an even function and its graph is symmetric with respect to the y-axis. If $f(−x)=−f(x)f(−x)=−f(x)$ for all
$xx$ in the domain of $f,f,$ then $ff$ is an odd function and its graph is symmetric with respect to the origin. By determining which types of symmetry a graph exhibits, we can learn more about the
shape and appearance of the graph. Symmetry can also reveal other properties of the function that generates the graph. Symmetry in polar curves works in a similar fashion.
Symmetry in Polar Curves and Equations
Consider a curve generated by the function $r=f(θ)r=f(θ)$ in polar coordinates.
i. The curve is symmetric about the polar axis if for every point $(r,θ)(r,θ)$ on the graph, the point $(r,−θ)(r,−θ)$ is also on the graph. Similarly, the equation $r=f(θ)r=f(θ)$ is unchanged by
replacing $θθ$ with $−θ.−θ.$
ii. The curve is symmetric about the pole if for every point $(r,θ)(r,θ)$ on the graph, the point $(r,π+θ)(r,π+θ)$ is also on the graph. Similarly, the equation $r=f(θ)r=f(θ)$ is unchanged when
replacing $rr$ with $−r,−r,$ or $θθ$ with $π+θ.π+θ.$
iii. The curve is symmetric about the vertical line $θ=π2θ=π2$ if for every point $(r,θ)(r,θ)$ on the graph, the point $(r,π−θ)(r,π−θ)$ is also on the graph. Similarly, the equation $r=f(θ)r=f(θ)$ is
unchanged when $θθ$ is replaced by $π−θ.π−θ.$
The following table shows examples of each type of symmetry.
Using Symmetry to Graph a Polar Equation
Find the symmetry of the rose defined by the equation $r=3sin(2θ)r=3sin(2θ)$ and create a graph.
Suppose the point $(r,θ)(r,θ)$ is on the graph of $r=3sin(2θ).r=3sin(2θ).$
i. To test for symmetry about the polar axis, first try replacing $θθ$ with $−θ.−θ.$ This gives $r=3sin(2(−θ))=−3sin(2θ).r=3sin(2(−θ))=−3sin(2θ).$ Since this changes the original equation, this test
is not satisfied. However, returning to the original equation and replacing $rr$ with $−r−r$ and $θθ$ with $π−θπ−θ$ yields
Multiplying both sides of this equation by $−1−1$ gives $r=3sin2θ,r=3sin2θ,$ which is the original equation. This demonstrates that the graph is symmetric with respect to the polar axis.
ii. To test for symmetry with respect to the pole, first replace $rr$ with $−r,−r,$ which yields $−r=3sin(2θ).−r=3sin(2θ).$ Multiplying both sides by −1 gives $r=−3sin(2θ),r=−3sin(2θ),$ which does
not agree with the original equation. Therefore the equation does not pass the test for this symmetry. However, returning to the original equation and replacing $θθ$ with $θ+πθ+π$ gives
Since this agrees with the original equation, the graph is symmetric about the pole.
iii. To test for symmetry with respect to the vertical line $θ=π2,θ=π2,$ first replace both $rr$ with $−r−r$ and $θθ$ with $−θ.−θ.$
Multiplying both sides of this equation by $−1−1$ gives $r=3sin2θ,r=3sin2θ,$ which is the original equation. Therefore the graph is symmetric about the vertical line $θ=π2.θ=π2.$
This graph has symmetry with respect to the polar axis, the origin, and the vertical line going through the pole. To graph the function, tabulate values of $θθ$ between 0 and $π/2π/2$ and then
reflect the resulting graph.
$θθ$ $rr$
$00$ $00$
$π6π6$ $332≈2.6332≈2.6$
$π4π4$ $33$
$π3π3$ $332≈2.6332≈2.6$
$π2π2$ $00$
This gives one petal of the rose, as shown in the following graph.
Reflecting this image into the other three quadrants gives the entire graph as shown.
Determine the symmetry of the graph determined by the equation $r=2cos(3θ)r=2cos(3θ)$ and create a graph.
Section 1.3 Exercises
In the following exercises, plot the point whose polar coordinates are given by first constructing the angle $θθ$ and then marking off the distance r along the ray.
$( 3 , π 6 ) ( 3 , π 6 )$
$( −2 , 5 π 3 ) ( −2 , 5 π 3 )$
$( 0 , 7 π 6 ) ( 0 , 7 π 6 )$
$( −4 , 3 π 4 ) ( −4 , 3 π 4 )$
$( 1 , π 4 ) ( 1 , π 4 )$
$( 2 , 5 π 6 ) ( 2 , 5 π 6 )$
$( 1 , π 2 ) ( 1 , π 2 )$
For the following exercises, consider the polar graph below. Give two sets of polar coordinates for each point.
For the following exercises, the rectangular coordinates of a point are given. Find two sets of polar coordinates for the point in $(0,2π].(0,2π].$ Round to three decimal places.
$( 3 , − 3 ) ( 3 , − 3 )$
For the following exercises, find rectangular coordinates for the given point in polar coordinates.
$( 2 , 5 π 4 ) ( 2 , 5 π 4 )$
$( −2 , π 6 ) ( −2 , π 6 )$
$( 5 , π 3 ) ( 5 , π 3 )$
$( 1 , 7 π 6 ) ( 1 , 7 π 6 )$
$( −3 , 3 π 4 ) ( −3 , 3 π 4 )$
$( 0 , π 2 ) ( 0 , π 2 )$
$( −4.5 , 6.5 ) ( −4.5 , 6.5 )$
For the following exercises, determine whether the graphs of the polar equation are symmetric with respect to the $xx$-axis, the $yy$-axis, or the origin.
$r = 3 sin ( 2 θ ) r = 3 sin ( 2 θ )$
$r 2 = 9 cos θ r 2 = 9 cos θ$
$r = cos ( θ 5 ) r = cos ( θ 5 )$
$r = 2 sec θ r = 2 sec θ$
$r = 1 + cos θ r = 1 + cos θ$
For the following exercises, describe the graph of each polar equation. Confirm each description by converting into a rectangular equation.
For the following exercises, convert the rectangular equation to polar form and sketch its graph.
$x 2 + y 2 = 16 x 2 + y 2 = 16$
$x 2 − y 2 = 16 x 2 − y 2 = 16$
For the following exercises, convert the rectangular equation to polar form and sketch its graph.
$3 x − y = 2 3 x − y = 2$
For the following exercises, convert the polar equation to rectangular form and sketch its graph.
$r = 4 sin θ r = 4 sin θ$
$r = 6 cos θ r = 6 cos θ$
$r = cot θ csc θ r = cot θ csc θ$
For the following exercises, sketch a graph of the polar equation and identify any symmetry.
$r = 1 + sin θ r = 1 + sin θ$
$r = 3 − 2 cos θ r = 3 − 2 cos θ$
$r = 2 − 2 sin θ r = 2 − 2 sin θ$
$r = 5 − 4 sin θ r = 5 − 4 sin θ$
$r = 3 cos ( 2 θ ) r = 3 cos ( 2 θ )$
$r = 3 sin ( 2 θ ) r = 3 sin ( 2 θ )$
$r = 2 cos ( 3 θ ) r = 2 cos ( 3 θ )$
$r = 3 cos ( θ 2 ) r = 3 cos ( θ 2 )$
$r 2 = 4 cos ( 2 θ ) r 2 = 4 cos ( 2 θ )$
$r 2 = 4 sin θ r 2 = 4 sin θ$
[T] The graph of $r=2cos(2θ)sec(θ).r=2cos(2θ)sec(θ).$ is called a strophoid. Use a graphing utility to sketch the graph, and, from the graph, determine the asymptote.
[T] Use a graphing utility and sketch the graph of $r=62sinθ−3cosθ.r=62sinθ−3cosθ.$
[T] Use a graphing utility to graph $r=11−cosθ.r=11−cosθ.$
[T] Use technology to graph $r=esin(θ)−2cos(4θ).r=esin(θ)−2cos(4θ).$
[T] Use technology to plot $r=sin(3θ7)r=sin(3θ7)$ (use the interval $0≤θ≤14π).0≤θ≤14π).$
Without using technology, sketch the polar curve $θ=2π3.θ=2π3.$
[T] Use a graphing utility to plot $r=θsinθr=θsinθ$ for $−π≤θ≤π.−π≤θ≤π.$
[T] Use technology to plot $r=e−0.1θr=e−0.1θ$ for $−10≤θ≤10.−10≤θ≤10.$
[T] There is a curve known as the “Black Hole.” Use technology to plot $r=e−0.01θr=e−0.01θ$ for $−100≤θ≤100.−100≤θ≤100.$
[T] Use the results of the preceding two problems to explore the graphs of $r=e−0.001θr=e−0.001θ$ and $r=e−0.0001θr=e−0.0001θ$ for $|θ|>100.|θ|>100.$ | {"url":"https://openstax.org/books/calculus-volume-3/pages/1-3-polar-coordinates","timestamp":"2024-11-02T03:13:38Z","content_type":"text/html","content_length":"571022","record_id":"<urn:uuid:888cc481-27f7-4b18-82ae-3d0517e19396>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00199.warc.gz"} |
Truncates a number to a specified number of digits.
Sample Usage
TRUNC(number, count)
• number - The number to be truncated.
• count - The number of decimal places which remain after truncation.
TRUNC(5.9) truncates the number and returns the integer part 5.
TRUNC(-5.9) truncates the number and returns the integer part 5.
• If count is greater than the number of significant digits in number, the number is returned without modification.
• count may be negative, in which case the specified number of digits to the left of the decimal place are changed to zero. All digits to the right of the decimal place are discarded.
• TRUNC performs no rounding, simply discarding unwanted digits.
• TRUNC and INT are similar in that both return integers. TRUNC removes the fractional part of the number. INT rounds numbers down to the nearest integer based on the value of the fractional part
of the number. INT and TRUNC are different only when using negative numbers. | {"url":"https://support.spreadsheet.com/hc/en-us/articles/360031018412-TRUNC","timestamp":"2024-11-12T00:28:55Z","content_type":"text/html","content_length":"44876","record_id":"<urn:uuid:1ca2fb57-ffb8-4295-91de-19237cfdff5a>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00506.warc.gz"} |
How Data is Represented on the Praxis Core Exam - dummies
For the Praxis Core exam, you need to become familiar with many ways to display or represent your data. Using lists, tables, graphs, charts, and plots to represent data is a surefire way to make sure
you aren’t tricked by the data. These methods of organizing data can also help you see patterns more readily. In the sections that follow, you become skilled at dissecting and interpreting different
types of data representations.
When you have gobs of data about a particular subject, you can sort, analyze, and display your data in a table. Tables only work if you have at least two sets of data to be organized into columns and
When working with tables, make sure to pay attention to the title of the table; it helps you understand what data to analyze. Next, notice the column and row titles.
In the following table, the data for the types of flowers and the number of each type of plant in Mary’s flower bed are listed. Make sure to read your question carefully and dissect the data
Which ratio compares the number of rose plants to the number of daffodil plants?
A. 3:2
B. 2:3
C. 4:3
D. 5:6
E. 3:4
The correct answer is Choice (B). The ratio of roses to daffodils is 8:12; when factored, the ratio is 2:3. The Praxis Core exam will expect answers in the simplest form.
Bar graphs and line graphs
A bar graph uses the length of vertical or horizontal bars to represent numbers and compare data. Bar graphs are good to use when your data is in categories. Bar graphs must contain a title, axis
labels for the horizontal and vertical axis, scales, and bars that represent numbers.
The following bar graph shows the number of canned goods collected by homerooms at Cardozo Middle School.
Mr. Smith’s homeroom collected more cans than how many other homerooms?
A. 3
B. 4
C. 5
D. 6
E. 7
The correct answer is Choice (A). Use the graph to compare the number of cans collected by each homeroom. According to the lengths of the vertical bars, Mr. Smith’s homeroom collected more cans than
Mr. Lewis, Mr. Davis, and Mrs. Reed’s classes.
Line graphs are graphs that show data that is connected in some way over a period of time. Suppose you’re preparing for a statistics test and each day you take a short online quiz to see how you’re
progressing. These are the results:
Day 1 30 percent
Day 2 20 percent
Day 3 50 percent
Day 4 60 percent
Day 5 80 percent
After you’ve created a table from your results, display them in a line graph. You can then decide, based on your progress on the practice quizzes, how likely you are to pass your statistics test.
What trends do you see in the following graph?
The graph indicates that as the days of practicing the online quizzes increase, your score increases; so, you will, more than likely, pass your statistics test.
Pie charts
Are you ready for a slice of pie? Pie charts are also known as circle graphs. These graphs focus on a whole set of data that is divided into parts. Each category represented in a pie chart is
represented by a part, called a sector, of the interior of the circle. The portion of the circle interior a category’s sector takes up is part of what represents the portion of the whole (population,
number of items sold, and so on) the category makes up. For example, if a pie chart represents categories of county government spending and 10 percent of the county government spending goes to road
maintenance, the category of road maintenance would be labeled in a sector that takes up 10 percent of the interior of the circle. The sector would also have “10%” presented in it, and the sector
would be labeled with “Road Maintenance.”
1. Move the decimal point two places to the right and add a percent symbol.
2. Multiply the percent by 360 to get the number of degrees for that slice of your pie chart.
3. Display all the slices together in the pie chart.
The following table shows the conversion of raw data to information that can be used to create a pie chart. The resulting pie chart is shown in the example question.
When reading a pie chart, the larger the value, the larger the piece of pie!
Stem-and-leaf plots
A stem-and-leaf plot blossoms into a useful graph when analyzed properly. You usually use this type of graph when you have large amounts of data to analyze. You can analyze data sets such as
classroom test results or scores of the basketball team using a stem-and-leaf plot.
Based on place value, each value in your data set is divided into a stem and leaf. What each stem and leaf plot represents is indicated by a Key. Draw a vertical line to separate the stem from the
leaf. The leaf is always the last digit in the number. The stem represents all other digits to the left of the leaf. To divide 105 into stem-and-leaf format, you draw a line to separate the stem from
the leaf, which indicates a stem of 10 and a leaf of 5.
Say you have the following numbers:
50, 65, 65, 60, 50, 50, 55, 70, 55
The first step is to arrange your data in least-to-greatest order, as follows:
50, 50, 50, 55, 55, 60, 65, 65, 70
Now arrange these numbers vertically in a table:
Math Test
Stem Leaf
Key: 5|0 means 50
This arrangement allows you to quickly identify your stems. Your stems in the data set are 5, 6, and 7. You have five data values in the list in the 50s: 50, 50, 50, 55, and 55. The leaves that go
along with the 5 stem are 0, 0, 0, 5, and 5. You have three data values in the 60s: 60, 65, and 65. The leaves that go with the 6 stem are 0, 5, and 5. Finally, you have one leaf with a data value of
0 to accompany the stem of 7.
When using a stem-and-leaf plot, you can quickly identify the least and greatest values in the data set (50 and 70), calculate the range (), and calculate the median or middle number (55).
The following data shows the number of people visiting a particular frozen yogurt shop per hour across a 12-hour day.
Hourly customers: 4, 17, 22, 31, 39, 40, 25, 43, 35, 40, 38, 13.
When this data is arranged in a stem-and-leaf plot, you get the following diagram. Use it to answer the questions that follow.
Hourly Customers
Stem Leaf
Key: 4|0 means 40
What was the largest number of people that entered the shop during an hour?
The correct answer is 43. Based on the diagram, the highest stem is 4 and the highest leaf in that stem is 3.
Box-and-whisker plots
Box-and-whisker plots, also known as box plots, show different parts of a data set using a line of numbers that are in order from least to greatest.
A box-and-whisker plot allows you to divide your data into four parts using quartiles. The median, or middle quartile, divides the data into a lower half and an upper half (for more on finding the
median, see the later section “Measuring arithmetic mean, median, or mode”). The median of the lower half is the lower quartile. The median of the upper half is the upper quartile. Your data set will
contain the following five parts:
• The least value: The smallest value in the data set.
• The lower quartile: The median of the lower half of the data set.
• The median: The median or middle quartile. This divides the data into a lower half and an upper half. The median is the number in the center. If two numbers are in the center, find the average of
the two middle values.
• The upper quartile: The median of the upper half of the data set.
• The greatest value: The largest value in the data set.
The following diagram shows how data is dissected using a box-and-whisker plot. To create a box-and-whisker plot, follow the diagram below.
The diagram begins with the data set. The data set is then put in least-to-greatest order. Underline the least value and the greatest value in the data set. Then find the median of the entire data
set. Remember, when calculating the median, if there are two values in the center of a data set, find their average. The median divides the data set into the lower set and higher set. You must then
find the median of the lower set. That is the lower quartile, or Q1. You also must find the median of the upper set. That median is the upper quartile, or Q3. The median of the entire set of data is
Q2. The word “quartile” can also be used to refer to a division of numbers marked off by Q1, Q2, Q3, or the highest number in the entire set.
After dissecting the data into the five values, graph the five values on a number line. Draw a box from Q1 to Q3. Draw a vertical line inside the box at Q2. The lines connecting the least and
greatest values to the box are called the whiskers.
Use the following graph to answer the following questions.
This box plot indicates the scores from yesterday’s math test. Approximately what percent of the students did not get above 65%?
A. 25%
B. 50%
C. 65%
D. 75%
E. 85%
The correct answer is Choice (A). The box-and-whisker plot indicates that the lowest score on the test was 60. The median of the lower quartile is 65, so about 25 percent of the students scored lower
than 65. It could be exactly 25 percent, depending on the number of students.
What is the median test score from yesterday’s math test?
A. 25%
B. 50%
C. 65%
D. 75%
E. 85%
The correct answer is Choice (D). The median of the box-and-whisker plot is indicated by a line drawn through the center of the box. The value graphed at this point is 75.
Venn diagrams
“Venn” you need to picture relationships between different groups of things, use a Venn diagram. A Venn diagram is an illustration where individual data sets are represented using basic geometry
shapes such as ovals, circles, or other shapes. Simply draw and label two or more overlapping circles to represent the sets you’re comparing. The sets overlap in an area called the intersection. When
an item is listed in both sets, it goes in the intersection. If an item doesn’t fit in either set, it falls outside the circles or other shapes.
Use the following graph to answer the following questions.
Football is the favorite sport of how many students in the Venn diagram?
A. 2
B. 5
C. 23
D. 25
E. 48
The correct answer is Choice (C). In the Venn diagram, 23 students picked football as their favorite sport. This is the only portion of the diagram reserved for football only.
Football and basketball are the favorite sports of how many students?
A. 2
B. 5
C. 23
D. 25
E. 48
The correct answer is Choice (B). Five students in the Venn diagram picked football and basketball. This is the portion of the diagram that football and basketball have in common.
How many students did not choose football or basketball?
A. 2
B. 5
C. 23
D. 25
E. 48
The correct answer is Choice (A). There are two students who did not fall inside the Venn diagram’s circles; therefore, they chose neither football nor basketball.
Scramble around scatter plots
If you want to determine the strength of your relationship, use a scatter plot. Scatter plots are graphical representations of two variables determining whether a positive, a negative, or no
correlation exists. A correlation is a relationship between two variables in which as one increases or decreases, the other one tends to increase or decrease. There is a correlation between time
studying and test scores, for example. As time studying increases, test scores tend to increase.
Data from two sets are plotted in scatter plots as ordered pairs (x, y). You can draw three conclusions from scatter plots:
• If the coordinates are close to forming a straight line that rises up from left to right, then a positive relationship or correlation exists.
• If the coordinates are close to forming a straight line (line of best fit) with one variable increasing as the other decreases, then a negative relationship or correlation exists.
• If the coordinates don’t come close to forming a line and are all over the place, then no relationship or correlation exists! Hence, the name scatter plot.
Here are the three types of correlations:
Make sure to give your plot a title and make sure to label your axes when you create a scatter plot.
Correlation alone does not prove causation. If a variable tends to increase as another variable increases, for example, it does not mean that the tendency is caused by the other variable. The number
of mud puddles tends to become greater when the occurrence of lightning rises. Does an increase in the occurrence of lightning cause the number of mud puddles to get higher? Do the additions of mud
puddles cause more lightning? The correlation between the two does not prove that one causes the other, and in fact neither causes the other. An increase in rain causes both.
Also, even when a change in one variable does cause a change in another, that alone does not prove which variable change is causing which. Studies have shown that changes in the closeness of the moon
cause shifting of the tides, but the correlation by itself is not the proof. If correlation worked that way, we could just as easily conclude that the shifting of the tides causes major changes in
the closeness of the moon.
Loiter around line plots
If you want to see your mode (the value that occurs the most) pop up quickly, use a line plot. A line plot, also known as a dot plot, allows you to identify the range, mode, and outliers in your data
set. Follow these simple steps:
1. Put your data in order from least to greatest.
2. Arrange your data on a number line.
3. Mark each value in the data set with an x or a dot.
Using the following line plot, determine the mode of the data set.
A. 50
B. 60
C. 70
D. 80
E. 90
The correct answer is Choice (E). The score of 90 appears in the data set the most.
About This Article
This article is from the book:
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SQL Window Functions Tutorial for Business Analysis - Cube Blog
Any person that has worked with data analytics has had a bad day when they sighed over a problem that was intuitively simple but practically hard to crack using pure SQL.
What is the revenue growth month over month and running total revenue? Can we trust the metric, or does the data have some accidental duplicates that affect it? What are the top N orders for every
month? What is the repeat purchase behavior? All these questions have to be translated from business language to programming language.
An intuitive solution in a countless number of cases like these is, “If only I could just loop over the results of my query I would be able to get the answer right away.” This opens a door to a world
of workarounds: writing complex joins, data ending up in another spreadsheet, using procedural extensions of SQL, or even moving data processing outside the database. Not all alternatives are viable,
others are just ugly.
At the same time, there’s a pure SQL implementation called “window functions” that is very readable, performant, and easy to debug.
In this SQL window functions tutorial, we will describe how these functions work in general, what is behind their syntax, and show how to answer these questions with pure SQL.
Learn by doing
For people unfamiliar with SQL window functions, the official database documentation can be hard to decipher, so we’ll go with real examples, starting from very basic queries and increasing the
degree of complexity.
Questions that were listed in the introduction above are real business questions that practically any company faces, and window functions are a huge help with answering them. Imagine we have the
following table of orders:
Data visualized with Statsbot
This dataset is very simple and small, however it is sufficient to illustrate the power of SQL window functions. Let’s start with the revenue growth.
Revenue growth
In business terms, revenue growth in month M1 is calculated as:
where m1 is the revenue in the given month and m0 is the revenue in the previous month. So, technically we would need to find what the revenue is for each month and then somehow relate every month to
the previous one to be able to do the calculation above. A very easy way to do this would be to calculate the monthly revenue in SQL:
date_trunc(‘month’,datetime) as month,
sum(amount) as revenue
FROM orders
GROUP BY 1
ORDER BY 1
then copy the output into the spreadsheet and use a formula to produce a growth metric:
When you copy the formula from cell C3 to cell C4 and so on, references are automatically shifted down, that’s what spreadsheets are good at. But what if you’d like to have this metric as a part of a
nice dashboard that is fed by data coming directly from the database? Such tools as Statsbot can help you with that:
Data visualized with Statsbot
In this case, a spreadsheet is definitely not what you want to end up with. Let’s try to calculate this in SQL. Without window functions, a query that gets you the final result would look like this:
monthly_revenue as (
date_trunc(‘month’,datetime)::date as month,
sum(amount) as revenue
FROM orders
GROUP BY 1
,prev_month_revenue as (
t2.revenue as prev_month_revenue
FROM monthly_revenue t1
LEFT JOIN monthly_revenue t2
ON datediff(month,t2.month,t1.month)=1
SELECT *,
round(100.0*(revenue-prev_month_revenue)/prev_month_revenue,1) as revenue_growth
FROM prev_month_revenue
ORDER BY 1
You’d have to calculate monthly revenue, then get the result of the previous month using self-join and use it in the final formula. The logic is broken down into 3 steps for clarity. You can’t break
it down further, but even so the second step can be quite confusing.
What you have to keep in mind about datediff is that that the minuend (i.e. what you subtract from) is the third parameter of the function and the subtrahend (i.e. what you subtract) is the second
parameter. I personally think that’s a bit counterintuitive for subtraction, and the self join concept itself is not basic. There’s actually a much better way to express the same logic:
monthly_revenue as (
date_trunc(‘month’,datetime)::date as month,
sum(amount) as revenue
FROM orders
GROUP BY 1
,prev_month_revenue as (
SELECT *,
lag(revenue) over (order by month) as prev_month_revenue
FROM monthly_revenue
SELECT *,
round(100.0*(revenue-prev_month_revenue)/prev_month_revenue,1) as revenue_growth
FROM prev_month_revenue
ORDER BY 1
Let’s break down the lag(… line of code:
• lag is a window function that gets you the previous row
• revenue is the expression of what exactly you would like to get from that row
• over (order by month) is your window specification that tells how exactly you would like to sort the rows to identify which row is the previous one (there’s plenty of options). In our case, we
told the database to get the previous month’s row for every given month.
This is a generic structure of all SQL window functions: function itself, expression and other parameters, and window specification:
function (expression, [parameters]) OVER (window specification)
It is very clean and powerful and has countless opportunities. For example, you can add partition by to your window specification to look at different groups of rows individually:
monthly_revenue as (
date_trunc(‘month’,datetime)::date as month,
sum(amount) as revenue
FROM orders
GROUP BY 1,2
,prev_month_revenue as (
SELECT *,
lag(revenue) over (partition by state order by month) as prev_month_revenue
FROM monthly_revenue
SELECT *,
round(100.0*(revenue-prev_month_revenue)/prev_month_revenue,1) as revenue_growth
FROM prev_month_revenue
ORDER BY 2,1
Calculating revenue by state in the first subquery and updating the window specification to take the new grouping into account, you can look at each state individually and see what the revenue growth
by state is.
Partitions are extremely useful when you need to calculate the same metric over different segments.
Running total
Another common request is to calculate running total over some period of time. This is the sum of the current element and all previous elements, for example, this is how the running total monthly
revenue would look in our dataset:
Data visualized with Statsbot
And the query to get this is below:
monthly_revenue as (
date_trunc(‘month’,datetime)::date as month,
sum(amount) as revenue
FROM orders
GROUP BY 1
SELECT *,
sum(revenue) over (order by month rows between unbounded preceding and current row) as running_total
FROM monthly_revenue
ORDER BY 1
The new thing here is the rows between unbounded preceding and current row part that is called “frame clause.” It’s a way to tell you which subset of other rows of your result set you’re interested
in, relative to the current row. The general definition of the frame clause is:
rows between frame_start and frame_end
where frame_start can be one of the following:
• unbounded preceding which is “starting from the first row or ending on the last row of the window”
• N preceding is N rows you’re interested in
• current row
and frame_end can be:
• unbounded following which is “starting from the first row or ending on the last row of the window”
• N following is N rows you’re interested in
• current row
It’s very flexible, except that you have to make sure the first part of between is higher than the second part, i.e. between 7 preceding and current row is totally fine, between 7 preceding and 3
preceding is fine too, but between 3 preceding and 7 preceding would throw an error. You can sum, count, and average values within the selected window. You can see a few examples in the query below:
,sum(amount) over () as amount_total
,sum(amount) over (order by order_id rows between unbounded preceding and current row) as running_sum
,sum(amount) over (partition by customer_id order by datetime rows between unbounded preceding and current row) as running_sum_by_customer
,avg(amount) over (order by datetime rows between 5 preceding and current row) as trailing_avg
FROM orders
ORDER BY 1
Every combination makes sense for particular use cases:
• amount_total is specified without a window and returns the entire total of $3400. Use it when you want to get the total in the same query as individual rows without grouping and joining back the
grouped result.
• running_sum is the running total. Use it when you want to see how some variable such as revenue or website visits is accumulated over a period of time.
• running_sum_by_customer is the same as above but broken down by segment (you can see how revenue from each individual customer grows, and on bigger data it can be cities or states).
• trailing_avg shows the average amount of the last 5 orders. Use trailing average when you want to learn the trend and disguise volatility.
The ordering is critical here since the database needs to know how to sum the values. The result can be completely different when different ordering is applied. The picture below shows the result of
the query:
Data visualized with Statsbot
The arrow next to running_sum tells us how the total amount is accumulated over time. The colored arrows next to running_sum_by_customer interconnect orders done by the same customer and the values
show the total order amount of the given customer at the point of every order. Finally, the grey brackets next to trailing_avg reference the moving window of the last 5 rows.
Dealing with duplicate data
If you paid attention to the dataset you would probably notice that both orders for customer D on 02–05 have the same order_id=5 which doesn’t look right. How did this happen? It turns out that the
original order was $250, then the customer decided to spend $50 more. The new record was inserted but the old record was not deleted. Such things happen in one or another table.
A long term solution in this case is to rectify the existing data flows and increase data awareness among developers, and what you can do right away is roll your sleeves up and clean the mess on your
SELECT *
FROM (
SELECT *,
row_number() over (partition by order_id order by datetime desc)
FROM orders
WHERE row_number=1
That would get you all order records except one that you’d like to filter out. Let’s review the function that allows us to do so: row_number() returns the incremental counter of the row within the
given window: 1,2,3, etc. It doesn’t take any parameters, that’s why it ends with empty brackets. The given window here is a set of rows that share a common order_id (partition by … is what separates
these sets from each other), sorted by datetime descending, so the intermediary result of the subquery in the middle looks like this:
Data visualized with Statsbot
Every partition here is represented by a single row, except order_id=5. Inside the partition, you sorted the rows by datetime descending, so the latest row gets 1 and the earlier row gets 2. Then you
filter only rows that have 1 to get rid of duplicates.
This is very useful for all sorts of duplicate problems. You might see that duplicates inflate the revenue number, so to calculate correct metrics we have to clean them out by combining the duplicate
filtering query with the revenue growth query like this:
orders_cleaned as (
SELECT *
FROM (
SELECT *,
row_number() over (partition by order_id order by datetime desc)
FROM orders
WHERE row_number=1
,monthly_revenue as (
date_trunc(‘month’,datetime)::date as month,
sum(amount) as revenue
FROM orders_cleaned
GROUP BY 1
,prev_month_revenue as (
SELECT *,
lag(revenue) over (order by month) as prev_month_revenue
FROM monthly_revenue
SELECT *,
round(100.0*(revenue-prev_month_revenue)/prev_month_revenue,1) as revenue_growth
FROM prev_month_revenue
ORDER BY 1
Now the data is clean and the revenue and growth metrics are correct. To avoid using orders_cleaned step in every query, you can create a view and use it as a table reference in other queries:
CREATE VIEW orders_cleaned AS
SELECT order_id, customer_id, state, “datetime”, amount
FROM (
SELECT *,
row_number() over (partition by order_id order by datetime desc)
FROM orders
WHERE row_number=1
Besides filtering duplicates, window partitions are very useful when you need to identify top N rows in every group.
Top N rows in every group
Finding top rows in every group is a typical task for a data analyst. Finding out who your best customers are and reaching out to them is a good way to know what people especially like about your
company and make it a standard. It is equally useful for employees, as leaderboards can be very good motivation for any team in the company. Let’s see how it works. Considering our dataset is tiny,
let’s get the top 2 orders for every month:
orders_ranked as (
date_trunc(‘month’,datetime)::date as month,
row_number() over (partition by date_trunc(‘month’,datetime) order by amount desc, datetime)
FROM orders_cleaned
SELECT *
FROM orders_ranked
WHERE row_number<=2
ORDER BY 1
The intermediary result of the orders_ranked statement would look like this (rows that appear in the final result are highlighted):
Data visualized with Statsbot
You can use any expression to separate partitions in window specification, not only column name (here we separated them by month, every partition is highlighted by its own color).
There are orders with the same month and amount, like order_id=1 and order_id=3, so we decided to resolve this by picking up the earliest order, adding datetime column to the sorting. If you’re
interested in pulling both rows in case of conflict you can use rank function instead of row_number.
Repeat purchase behavior
Repeat purchase behavior is the key for a successful business, and investors totally love companies that can retain customers and make them spend more and more. We can translate this to a more
precise data question: what is the repeat purchase rate and the typical difference between the first order and the second order amount? That would be expressed as:
customer_orders as (
SELECT *,
row_number() over (partition by customer_id order by datetime) as customer_order_n,
lag(amount) over (partition by customer_id order by datetime) as prev_order_amount
FROM orders_cleaned
round(100.0*sum(case when customer_order_n=2 then 1 end)/count(distinct customer_id),1) as repeat_purchases,
avg(case when customer_order_n=2 then 1.0*amount/prev_order_amount end) as revenue_expansion
FROM customer_orders
The result of the intermediary customer_orders statement would look like this:
Data visualized with Statsbot
Every customer’s partition is highlighted by its own color. The brackets next to customer_id outline our partitions, the arrows next to datetime show sorting direction, the arrows that point to
amount values show where prev_order_amount values come from, and customer_order_n values for repeat purchases are underlined. Since customers D, E, and F have only one order they are out of the
The final result looks like this:
Data visualized with Statsbot
This time we partitioned by customer_id to isolate sets of order rows that belong to the same customer, and identified the row order and the previous row’s amount value. In the final query, we have
used conditional aggregates to calculate desired metrics which tell us that half of customers buy again and they spend almost twice as much on the second order. Imagine it’s not a dummy dataset, what
a great business would that be!
Wrap up
As you have seen in our tutorial, SQL window functions are a powerful concept that allows advanced calculations. A typical window function consists of expression and window specification with
optional partitioning and frame clause, and the opportunities to slice, smooth, interconnect, and deduplicate your data are truly boundless. There are plenty of interesting use cases where these
functions give a great value. We’ll continue talking about them in the next few tutorials. Stay tuned! | {"url":"https://cube.dev/blog/sql-window-functions-tutorial","timestamp":"2024-11-03T20:24:33Z","content_type":"text/html","content_length":"138581","record_id":"<urn:uuid:aa3d4530-f862-49d9-a14a-cb74b1c0e21e>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00332.warc.gz"} |
Angles Between Unequal Edges of Kite are Congruent | Geometry Help
A kite is a quadrilateral with two pairs of sides that are equal. A kite has four internal angles, two of these are the opposite angles between the unequal edges, and two are the opposite angles
between the equal edges. It is fairly easy to show that the angles between the unequal edges of a kite are congruent. We will prove this by using congruent triangles.
On the other hand, the pair of opposite angles between the edges that are equal can be of any size, and have no special relationship to each other.
ABCD is a kite. Show that ∠ABC≅∠ADC
In general, a common tool for showing that two angles are congruent is to show that they are corresponding parts of congruent triangles. We will show that the two angles between the unequal sides of
the kite are congruent using this method. To do this, we'll create two triangles in which they are corresponding angles, by drawing the diagonal AC.
(1) ABCD is a kite //Given
(2) AB=AD //definition of a kite
(3) BC=CD //definition of a kite
(4) AC=AC //Common side, reflexive property of equality
(5) $\triangle DEG \cong \triangle FEG$
(6) ∠ABC≅∠ADC //Corresponding angles in congruent triangles (CPCTC)
The diagonal AC is known as the axis of symmetry of the kite. This means that if we "fold" the kite of this axis, or if we hold up a mirror on the line, the two halves will match. This is also known
as "Reflection Symmetry".
We know that the two sides will match in length, because that is the defintion of a kite - a quadrilateral with two pairs of adjacent sides of equal length. And we have now shown that because the
angles between the unequal edges of a kite are congruent, they will match as well. | {"url":"https://geometryhelp.net/kites-angles-between-unequal-edges-congruent/","timestamp":"2024-11-12T04:09:35Z","content_type":"text/html","content_length":"80899","record_id":"<urn:uuid:539a2078-cd96-44ea-af55-32ad440c4ad0>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00409.warc.gz"} |
Aerostructural Design Optimization of Wind Turbine Blades
Department of Mechanical & Aerospace Engineering, School of Engineering and Digital Sciences, Nazarbayev University, Astana 010000, Kazakhstan
Department of Mathematics, School of Humanities and Sciences, Nazarbayev University, Astana 010000, Kazakhstan
Department of Intelligent Systems and Cybersecurity, Astana IT University, Astana 010000, Kazakhstan
Author to whom correspondence should be addressed.
Submission received: 16 October 2023 / Revised: 7 December 2023 / Accepted: 11 December 2023 / Published: 21 December 2023
This study presents an aerostructural optimization process for wind turbine blades aimed at enhancing the turbine’s performance. The optimization framework integrates DAFoam as the computational
fluid dynamics (CFD) solver, TACS as the finite element method (FEM) solver, Mphys for fluid–structure coupling, and SNOPT as the optimizer within the OpenMDAO framework. The objective is to
simultaneously increase the torque generated by the wind turbine while decreasing the mass of the blade, thereby improving its efficiency. The design variables in this optimization process are the
blade shape and panel thickness. The aerodynamic objective function is torque, a key performance indicator for wind turbine efficiency. The structural objective function is the blade mass, as
reducing mass is essential to minimize material and manufacturing costs. The optimization process utilizes the integrated capabilities of DAFoam, TACS, Mphys, and SNOPT to iteratively evaluate and
modify the blade shape and panel thickness. The OpenMDAO framework facilitates seamless communication between the solvers and the optimizer, ensuring a well-coordinated, efficient optimization
process. The results of the optimization show a 6.78% increase in torque, which indicates a significant improvement in the wind turbine’s energy production capacity. Additionally, a 4.22% decrease in
blade mass demonstrates a successful reduction in material usage without compromising structural integrity. These findings highlight the potential of the proposed aerostructural optimization process
to enhance the performance and cost-effectiveness of wind turbine blades, contributing to the advancement of sustainable energy solutions. This work represents the first attempt to implement DAFoam
for wind turbine aerostructural design optimization.
1. Introduction
The development of renewable energy sources has become a priority for emerging nations because of the volatile energy market, the depletion of fossil fuels, and deteriorating environmental
conditions. The fundamental idea behind using renewable energy is that it comes from ongoing natural processes. As a result, emerging nations reject the use of fossil fuels and migrate to other
sources of energy like wind and solar. The majority of these renewable energy sources significantly reduce CO
emissions, as suggested by the Intergovernmental Panel on Climate Change (IPCC) [
Renewable energy is produced from natural resources that replenish themselves naturally and without human intervention. Wind energy is one of the renewable energy sources that is growing the
quickest. As a result, harnessing wind energy to generate electricity is more economical than using coal or gas-fired power plants. Despite all of its advantages, wind energy’s main disadvantage is
that power is intermittent. Consequently, another energy source that can be produced on demand must be combined with wind energy. Wind power should be considered an additional energy source rather
than the main one.
Only land-based wind power can economically meet humanity’s energy demands, claim Zhou et al. [
]. However, only a tiny fraction of this enormous potential for wind energy is now being used [
]. Three thousand years of human history have seen the utilization of wind energy [
]. The use of wind energy peaked during both World Wars, then decreased afterward and again rose as a result of the demise of the petroleum sector in the 1970s [
]. Moreover, modern wind turbines had been built in Denmark by the end of the seventeenth century [
]. The technological complexity of wind turbines makes them difficult to run with a variety of workloads [
Expanding the usage of carbon-free technologies that rely on high-potential, renewable energy sources like wind power would require lowering the levelized cost of energy. Modern, highly flexible wind
turbine rotors are defined by strongly coupled, multi-physics interactions, which must be captured using advanced computational approaches. As a consequence, computing methods may greatly improve the
time-consuming process that occurs in wind turbine blades.
The remainder of this paper starts with a literature review on the aerostructural optimization of wind turbine blades in
Section 2
, followed by the methodology on the aerodynamic optimization, structural optimization, and fluid–structural coupling processes of the aerostructural optimization in
Section 3
, and the main findings are presented and discussed in
Section 4
. Finally in
Section 5
, the conclusion is made.
2. Literature Review
Energy is crucial for social and economic progress, and because of rising greenhouse gas emissions and fuel costs, renewable energy sources are gaining popularity. Renewable energy does have certain
limitations, too, such intermittent generation and the need for sophisticated design, planning, and control optimization techniques.
Computational optimization techniques for renewable energy design, planning, and control difficulties were demonstrated by Baos et al. in 2011 [
]. The use of heuristic approaches, Pareto-based multi-objective optimization, and parallel processing are intriguing study areas in renewable and sustainable energy, according to one of the review’s
fascinating results.
A bibliography on wind speed and wind power projections was eventually provided by Lei et al. [
]. The incorporation and application of artificial intelligence techniques as well as the blending of diverse statistical models for both long- and short-term forecasts were seen as possible
prospects. The papers listed below [
] discuss forecasting techniques for wind energy.
Miller et al. [
] reviewed recent developments in numerical simulation techniques, technology, and applications to wind energy. They also looked at past numerical simulations.
The wind energy business heavily relies on computational fluid dynamics to research innovative turbine designs (CFD). New tools must be created in order to apply CFD early in the design phase, where
lower-fidelity methods like blade element momentum (BEM) are more common. Algorithmic optimization tools are extremely helpful since they reduce the dependency on trial-and-error design.
Over the past ten years, a number of tools for designing wind turbines have been created. The entire turbine system, including aerostructural rotor and tower models, the power generator, and
fixed-bottom foundations or floating platforms, is simulated using state-of-the-art software, such as Open-FAST [
], Cp-Max [
], HAWTopt2 [
], ATOM [
], SHARPy [
], Qblade [
], and MoWit [
Various traditional tools employ a mix of models of different accuracy levels to analyze different parts of a system. OpenFAST, a turbine analysis standard [
], integrates a modal and multibody dynamics framework that encompasses all major components. This system, utilizing BeamDyn and ElastoDyn modules, employs beam models to capture the turbine’s
dynamic response. However, these modules have limitations in terms of degrees of freedom (ElastoDyn) and accuracy of local stress values (ElastoDyn and BeamDyn). OpenFAST’s aerodynamic forecasts
through BEMT-based AeroDyn align well with traditional designs but face challenges with larger blades showing significant deflections and 3D effects.
Madsen et al. [
] and concurrent research employing a blade-resolved rotor model also overlooked the hub geometry. However, computational fluid dynamics (CFD) analyses effectively capture spanwise and tip flow
phenomena, significantly impacting in-plane and off-plane loads. Horcas et al. [
] demonstrated that a tool based on OpenFAST’s physical models, HAWC2, tends to overpredict loads on a 10 MW turbine at wind speeds below the rated threshold and underestimates the advantages of
curved wing tips, even with a steady power curve. Additionally, tools like OpenFAST lack accuracy in computing gradients, hindering their use in numerical optimization. They also rely on precomputed
wing section drag polars, limiting the optimization of airfoil shapes.
Efforts have been made to combine high- and low-fidelity tools to strike a balance between accuracy and computational expense. Ramos-García et al. [
] coupled finite-element structural models with a hybrid lifting-line–vortex method, showing agreement with Heinz et al. [
]. While this approach overcomes some limitations of BEMT at a lower cost than blade-resolved solvers, its application remains primarily limited to analyses due to challenges in computing
Other recent endeavors focused on a narrower but higher-fidelity approach to capture the intricate fluid–structure interaction of turbine blades [
]. Wainwright et al. [
] merged a commercial CFD software package and a finite element modal solver using GPU acceleration to study a rotor’s aerostructural behavior in the wake of an upstream turbine. Cheng et al. [
] validated a model coupling different solvers in OpenFOAM to study a floating wind turbine’s behavior. However, while high-fidelity codes offer accuracy, they suffer from computational cost and
implementation complexity, restricting their use for optimization.
Research on applying numerical analysis tools to wind turbine optimization problems has been undertaken by Gray et al. [
], Ning and Petch [
], and Ingersoll and Ning [
]. Bortolotti et al. [
] developed a design space exploration strategy efficiently sizing the rotor at a conceptual design stage, though it could not exploit tradeoffs between aerodynamic and structural design. Bottasso et
al. [
] proposed a bilevel design approach using a comprehensive turbine model, maximizing annual energy production over mass, but with limitations in accuracy and robustness.
Low-fidelity aerostructural optimization for wind turbine blades generally involves simplifications and approximations in representing the complexities of fluid–structure interactions and the
aerodynamics of the blades. These limitations impact the accuracy and robustness of the optimization process compared to high-fidelity approaches.
For high-fidelity aerodynamic shape optimization, the adjoint technique is an effective method for calculating the derivatives of an interest function with respect to a large number of design
In addition, proposed the theory and architecture of OpenMDAO in 2019 [
]. This open-source multidisciplinary design optimization (MDO) framework solves coupled systems utilizing Newton-type algorithms and takes use of problem structure via novel hierarchical
methodologies. The main goal of MDO is to build interconnected numerical models of challenging engineering systems. There are a number of MDO software frameworks; however, none of them completely
makes use of cutting-edge methods to solve connected models efficiently.
This study’s goal is to present an aerostructural optimization method for wind turbine blade design that aims to maximize torque while lowering mass because none of the earlier literature studies
have attempted to incorporate the associated performance optimization of wind turbines. While structural optimization aims to minimize bulk while maintaining structural integrity, aerodynamic
optimization concentrates on altering the shape of the blade. This approach is crucial for the widespread adoption of wind energy systems since it is both effective and affordable.
3. Materials and Methods
3.1. Baseline Wind Turbine Blades
The optimization is based on the NREL Phase VI geometry, which was created for applied CFD validation tests, with the two-bladed 10.058 m diameter NREL Phase VI Rotor based on the S809 airfoil [
Figure 1
illustrates the baseline geometry of the airfoil, which is replotted based on Hand et al. [
The design process involved thorough evaluations of tradeoffs, considering nonlinear changes in taper and twist patterns, as well as the integration of extra airfoils. The result is a blade featuring
a consistent taper and a twist distribution that varies in a nonlinear manner, using the S809 airfoil from the base to the tip. This initial design provides a good foundation for optimization and
offers the possibility of achieving even better performance.
3.2. Aerostructural Optimization Framework
In spite of having a higher computational cost and implementation effort when compared to low-fidelity solvers, the use of high-fidelity solvers in aerostructural design optimization of wind turbine
blades can provide improved accuracy and more complete model specification. To manage the high-dimensional and high-fidelity optimization formulations in a tractable numerical problem, gradient-based
optimization techniques are required, such as the use of adjoint-based derivative computations. Multiple aerostructural, hydrostructural, aerothermal, aeropropulsive, and aeroelastic MDO issues have
been incorporated in the OpenMDAO framework to handle interdisciplinary design optimization challenges in engineering systems [
]. The main functionality of the Python-based, open-source, high-performance computing platform known as OpenMDAO is gradient-based optimization using analytic derivatives. The implementation
presented in this paper is based on a two-way fluid–structure interaction (FSI) in the OpenMDAO framework that couples the discrete adjoint with OpenFOAM for high-fidelity multidisciplinary design
optimization (DAFoam) [
] and the Toolkit for the analysis of composite structures (TACS) [
] with Mphysics [
]. With competitive speed, scalability, and accuracy, DAFoam employs a discrete adjoint technique without Jacobians. It also offers a Python interface for linking interdisciplinary design
In this study, we optimize the aerostructural design of wind turbine blades to improve their aerodynamic performance and decrease their mass using the discrete CFD solver DAFoam in combination with
the OpenMDO framework. The aerodynamic and structural systems work together; the aerodynamics group receives a mesh as input and creates aerodynamic loads as an output, while the structural group
receives these loads as input and generates structural displacements as a result. This method has received a lot of previous research attention, with several studies examining its potential
applications in the optimization of aircraft aerodynamics and aerostructural design [
According to the extended design structure matrix (XDSM) schematic [
] in
Figure 2
, the MACH [
] is a group of closely connected submodules that enable geometry parametrization and deformation, coupled aerostructural analysis, and effective derivatives assessment in an optimization context.
The OpenMDAO framework, which is based on Mach-aero, is used to carry out the optimization procedure [
]. The creation of the structural mesh in MSC Patran and the aerodynamic mesh in pyHyp are both part of the preprocessing stage. In terms of aerodynamic optimization, during the preprocessing stage,
we create the CFD mesh for the baseline design surface geometry. To do this, we first create the surface mesh using ICEM CFD and then extrude it to create the volume mesh using pyHyp. Additionally,
we use the ICEM CFD to create the freeform deformation (FFD) control points for geometry parameterization. The first geometric design variables (FFD points displacement) will then be passed to the
geometry parameterization module pyGeo by the optimizer (SNOPT). An open-source FFD tool for parameterizing the geometry of the design surface is called PyGeo. A series of point clouds may be
embedded using pyGeo into the designated FFD box, and the point cloud can then be deformed by rearranging the FFD coordinates. Both organized and unstructured meshes may be used with pyGeo. The CFD
design surface and surface mesh will be entirely contained in the FFD box that is produced during the preprocessing stage. The updated design surface is subsequently sent to the IDWarp mesh
deformation module. Using an inverse-distance weighting method, IDWarp is an open-source mesh deformation tool that may be used to reshape high-quality volume mesh according to the new design
surface. As opposed to regenerating the volume mesh, we can eliminate numerical noise by using IDWarp, which is smooth while deforming the volume mesh. Updated volume mesh is passed to the CFD solver
via IDWarp (OpenFOAM). After solving the flow, OpenFOAM provides the adjoint solver component with the converged CFD state variables. The aerodynamic objective and constraint functions are likewise
computed using OpenFOAM (e.g., power, thrust). Ultimately, the adjoint solver returns the total derivatives of the constraint and goal functions to the optimizer after computing them. The function
values and derivatives are used by the optimizer to update the design variables for the subsequent optimization iteration. The optimization process keeps going until it converges.
OpenMDAO/MPhys is used in the aerostructural optimization framework. The aerodynamic-only and aerostructural optimization frameworks are comparable, with the following exceptions: The whole FEM mesh
and the CFD design surface are included in the FFD box. Therefore, in order to guarantee a constant fluid–structure interface, pyGeo deforms both the FEM mesh and the CFD surface mesh; the
aerostructural analysis module includes an integrated mesh deformation module; the aerostructural optimization conducts coupled aerostructural analysis and adjoint, which is needed to solve the CFD
and FEM systems utilizing block Gauss–Seidel techniques, as opposed to single-discipline analysis; aerodynamic and structural variables, such as power, thrust, aggregated von Mises stress, and
structural mass, are included in the objective and constraint functions that are returned to the optimizer.
TACS and OpenFOAM solvers are used in the aerostructural study. The baseline CFD and FEM meshes, CFD and FEM design variables, and the most recent design surface geometry (blade geometry) in the
optimization loop are the inputs (e.g., blade rotation speed). The converged aerostructural state variables, such as pressure, velocity, and structural displacement, as well as the objective and
constraint functions, are the outputs (e.g., power and thrust). The primary driving force behind the aerostructural analysis process is the computation of the updated CFD surface mesh using the
geometry of the design surface and the CFD surface mesh displacement derived from the displacement transfer component (in the first iteration, the displacement is zero). The updated CFD volume mesh
is then computed by the mesh deformation component using the updated CFD surface mesh. The updated CFD volume mesh is then sent to the CFD solver component, which runs flow field simulations,
extracts the CFD surface force, and sends it to the load transfer component. The FEM solver receives the force from the load transfer component, which interpolates the CFD surface force to the FEM
surface force. The structural displacement is then calculated by the FEM solver component, which also extracts the FEM surface displacement and feeds it to the displacement transfer component. After
interpolating the FEM surface displacement to the CFD surface displacement, the displacement transfer component delivers the data to the aerostructural analysis primary driver to initiate the
subsequent iteration. Until the residuals for the FEM and CFD solvers are smaller than the recommended tolerances, the aforementioned procedure is repeated.
The “Scenario” generic aerostructural template from MPhys makes the aforementioned interaction between CFD and FEM solvers easier. To expand the aerostructural situation, a Python interface known as
mphys dafoam is available that utilizes OpenFOAM and DAFoam as the CFD and adjoint solvers (the adjoint is elaborated on in the next section). An opensource inverse distance weighted mesh deformation
algorithm called IDWarp is used by the mesh deformation component. The adjoint-based gradient evaluation tool FUNtoFEM and a general aeroelastic analysis tool named Meld are the sources of the load
and displacement transfer components. As the FEM solver, TACS is employed. Using an Aitken relaxation technique, the nonlinear block Gauss–Seidel solver in OpenMDAO is used to do the CFD and FEM
Figure 3
depicts the entire aerostructural optimization profile.
3.3. RANS-Based Turbulent Simulation
For the conservation of mass and momentum, the Navier–Stokes equations are used to describe a three-dimensional steady incompressible flow.
$∂ u ∂ t + ∇ . u u = − 1 ρ ∇ p ¯ + μ ∇ . ∇ u .$
These equations are given as Equations (1) and (2), where
denotes the fluid’s velocity,
denotes its pressure,
denotes its density,
denotes its dynamic viscosity. The Spalart–Allmaras model [
] is employed as the turbulence model because it is reliable, effective, converges well, and is simple to use. In
Table 1
, the boundary conditions are displayed. The initial values for k, ε, νt, ω, ṽ are 0.8375 m
, 0.2 m
, 5 × 10
/s, 12.24 s
, and 5 × 10
/s, respectively. The inlet velocity of the fluid is 7 m/s, and the blade has a rotational velocity of 72 rpm. The initial values for the boundary conditions are calculated based on the turbulence
free-stream boundary conditions.
In our study, the computational domain is a spherical shape with the blade located at the center of the spherical far-field domain, as shown in
Figure 4
, where the radius of the blade is 5.029 m while the radius of the spherical domain is 20 times the blade’s radius, and the entire spherical domain is defined as inlet–outlet.
As for the turbulence model, as mentioned above, a one-equation Spalart–Allmaras model is used because we have limited computational power. It encompasses the resolution of just a single transport
equation, specifically the one for kinematic turbulent viscosity. As a result, the computational workload is reduced when compared to popular two-equation models such as k-epsilon. The S-A model has
undergone thorough validation for external flows and demonstrates strong alignment with experimental findings in aerospace applications.
3.4. Structural Model
The influence of temperature is not included in the structural simulation employed in this model. As a result, the simulation is be set to isothermal for this chapter. The state equation for the
momentum balance has the following structure:
$∂ 2 ρ u ∂ t 2 − ∇ · σ = 0$
are displacement vector and stress tensor, respectively, while
$μ s$
are the material properties. The stress tensor is specified using strain tensor
and there is transpose operation:
$σ = 2 μ s ε + λ t r ε I$
Equations (4) and (5) when combined provide the following results:
$∇ · σ = ∇ · μ s ∇ u + ∇ · μ s ( ∇ u T + λ I t r ∇ u )$
The tweak to the solution technique improves the convergence of a simulation by rearranging terms in Equation (6). The first component of Equation (3) is implicitly solved using OpenFoam, whereas the
second term is explicitly solved.
Therefore, by changing the order of phrases, the explicit and implicit components are more evenly distributed. Equation (3) in its modified form has the following form:
$∇ · σ = ∇ · 2 μ s + λ ∇ u + ∇ · μ s ( ∇ u T + λ I t r ∇ u − μ s + λ ∇ u )$
The traction force boundary condition has the following expression (where
is a surface normal to the boundary):
$T = σ · n = 2 μ s + λ ∇ u + ∇ · μ s ( ∇ u T + λ I t r ∇ u − μ s + λ ∇ u ) · n$
Only elastic deformation is described by Equations (3)–(8). Every substance in nature experiences plastic deformation after it has been subjected to a certain amount of stress. The following is the
format of the modified governing equation with plastic term:
$∇ · μ s ∇ d u + μ s ∇ d u T + λ I t r ∇ d u − 2 μ s d ε p + λ I t r d ε p = 0$
$d u$
$d ε p$
are incremental displacement vector and incremental plastic strain tensor, respectively. Equation (10) undergoes similar modification:
$T = σ · n → dT = d σ · n = 2 μ s + λ ∇ d u + μ s ∇ d u T + λ I t r ∇ d u − μ s + λ ∇ d u − 2 μ s d ε p + λ I t r d ε p$
3.5. Discrete Adjoint Derivative Computation
The torque in our example serves as the objective function $f$, and $x$ is the vector of design variables. The adjoint method is utilized to quickly determine the total derivatives $d f / d x$. In
the discrete technique, it is expected that the primary solver can provide a discretized form of the governing equations and that the design variable vectors $x ∈ ℝ n x$ and $ω ∈ ℝ n ω$ satisfy the
discrete residual equations $R ω , x$ = 0, where R $R ∈ ℝ n ω$ is the residual vector.
Therefore, the relevant functions are functions of the state variable as well as the design variable:
$f = R ω , x$
. Although there are several functions of relevance,
is regarded as a scalar in the following derivations to retain generality. Each new function demands the solution of an additional adjoint system, as becomes evident later. The total derivative
$d f / d x$
is derived using the chain rule:
$d f d x ⏟ 1 × n x = ∂ f ∂ x ⏟ 1 × n x + ∂ f ∂ ω ⏟ 1 × n ω d ω d x ⏟ n ω × n x$
when there are no implicit computations, making the estimation of the partial derivatives
$∂ f / ∂ x$
$∂ f / ∂ ω$
very simple. On the other hand, because it is implicitly specified by the residual equations
$R ω , x = 0$
, the total derivative matrix
$d ω / d x$
is expensive.
We may obtain
$d ω / d x$
using the chain rule for R. We may then exploit this information to our advantage because the governing equations should always hold. Therefore, the derivatives
$d R / d x$
must amount to zero:
$d R d x = ∂ R ∂ x + ∂ R ∂ ω d ω d x = 0 ⇒ d ω d x ⏟ n ω × n x = − ∂ R ∂ ω − 1 ⏟ n ω × n ω ∂ R ∂ x ⏟ n ω × n x$
In Equation (12), substituting
$d ω / d x$
from Equation (11) yields:
$d f d x ⏟ 1 × n x = ∂ f ∂ x ⏟ 1 × n x − ∂ f ∂ ω ⏟ 1 × n ω ∂ R ∂ ω − 1 ⏟ n ω × n ω ⏞ ψ T ∂ R ∂ x ⏟ n ω × n x$
We may solve the adjoint equation by transposing the state Jacobian matrix
$∂ R / ∂ ω$
and using,
$∂ f / ∂ ω T$
as the right-hand side.
$∂ R ∂ ω T ⏟ n ω × n ω ψ ⏟ n ω × 1 = ∂ f ∂ ω T ⏟ n ω × 1$
The adjoint vector is the
. When we enter the adjoint vector into Equation (13) after solving this equation, we obtain the total derivative, which is expressed as:
$d f d x = ∂ f ∂ x − ψ T ∂ R ∂ ω$
Since the design variable is not explicitly stated in Equation (14), we only need to solve the adjoint equations once for each function of interest. As a result, its computing cost is related to the
number of interesting functions rather than being independent of the number of design variables. This strategy, known as the adjoint technique, provides benefits for many design issues in
aeronautical engineering when just a small number of functions are of relevance but where several hundred design variables may be used.
The following four essential steps make up a discrete adjoint implementation, which involves computing the partial derivatives and resolving the adjoint equations:
Figuring out the partial derivatives $∂ f / ∂ ω T$ and $∂ R / ∂ ω T$
The adjoint vector $ψ$ in the linear Equation (13) solution.
The method of computing the partial derivatives $∂ f / ∂ x$ and $∂ R / ∂ x$.
Calculate the total derivative $d f / d x$ using Equation (15).
The four approaches outlined above do not need a specific residual function $R ω , x$; therefore, they may be used with any collection of discrete PDEs.
3.6. Aerostructural Coupling
The adjoint formulation indicated above assumes that the state variables of the finite element method (FEM) and computational fluid dynamics (CFD) are combined and solved concurrently, which results
in a bigger Jacobian matrix and increased memory use. To get around this, we solve the aerostructural adjoint in a linked way using a different technique called block Gauss–Seidel, as shown below.
$∂ R C F D ∂ w C F D T ∂ R F E M ∂ w C F D T ∂ R C F D ∂ w F E M T ∂ R F E M ∂ w F E M T Ψ C F D Ψ F E M = ∂ f ∂ w C F D T ∂ f ∂ w F E M T$
The abbreviations CFD and FEM, which stand for computational fluid dynamics and finite element method, respectively, refer to the residual and state variables for the solvers. We use DAFoam to solve
the CFD adjoint equation:
$∂ R C F D ∂ w C F D T Ψ C F D = ∂ f ∂ w C F D T$
According to Kenway et al.’s study [
], DAFoam uses a technique known as Jacobian-free adjoint, where automated differentiation is employed to compute partial derivatives and matrix–vector products. DAFoam employs the generalized
minimum residual (GMRES) iterative linear equation solver from the PETSc package to solve the adjoint equation [
]. DAFoam employs a layered technique using the additive Schwartz method for global preconditioning, and an incomplete lower and upper (ILU) factorization method with one level of fill-in is utilized
for local preconditioning.
The preconditioner matrix
$∂ R ∂ w R C T$
is produced to improve convergence by estimating the residuals and linearizing them [
]. Since constructing
$∂ R ∂ w R C T$
takes up around 30% of the adjoint runtime, this matrix is only created once and then utilized for the adjoint equation. This results in a considerable decrease in adjoint runtime. The adjoint
equation for the FEM component is solved using TACS:
$∂ R F E M ∂ w F E M T Ψ F E M = ∂ f ∂ w F E M T$
Without explicitly creating the matrix, automated differentiation is used to determine the nondiagonal portions of the matrix multiplied by a vector.
Using OpenMDAO and MPhys, the aerostructural system was created in a modular manner for adaptability. The components in
Figure 1
needed to multiply the state Jacobian matrix by a particular vector and implement ways to compute output based on input. The adjoint total derivative computation was unified with OpenMADO using the
MAUD algorithm [
]. In OpenMDAO, a linear block Gauss–Seidel solver with Aitken relaxation was employed to solve the coupled adjoint of CFD and FEM.
3.7. Baseline Geometry Configuration for Aerostructural Optimization
As in our earlier research [
], we used the NREL Phase VI, which has just two blades, as the baseline geometry in this investigation. Regarding the aerodynamic component, the surface mesh as shown in
Figure 5
c is constructed in ICEM CFD, and the spherical volume mesh as shown in
Figure 5
a is created by extruding the structured surface mesh using the opensource mesh generator pyHyp [
Figure 5
b depicts the hyperbolic expansion layer that surrounds the wind turbine blade. Additionally, MSC Patran generates a structural finite element mesh that matches the shape of the outer mold line and
incorporates an interior main shear web as seen in
Figure 5
d. The whole collection of structural elements is made up of thin CQUAD hexahedral shell elements. At the root, which is situated at a distance of about 0.66 from the rotating axis, two identical
blades are fastened. In
Figure 5
e, the FFD points for the aerodynamic optimization are presented, and the aerodynamic form optimization is put into practice. In contrast,
Figure 5
f shows the structural design factors as a total of 60 panels with random colors.
4. Mesh Convergence and Validation
Based on the multi-reference frame (MRF), a steady-state method used in computational fluid dynamics (CFD) to describe problems with rotating components, a mesh convergence study was conducted using
three levels of mesh: L0, L1, and L2, as shown in
Table 2
. The wind speed is 7 m/s and the rotor’s rotating speed is 72 pm. In order to maintain a value of y+ close to 1, the growth ratio is set to 1.2 and the initial cell height away from the turbine
surface is set to 0.00003 m. y+ is the nondimensional distance from the wall to the first mesh node, and the superscript cross denotes normalization with the axis that used to be the distance.
According to
Table 2
, the torque errors for the three L0-, L1-, and L2-based CFD mesh simulations are 5.35 percent, 10.19 percent, and 13.63 percent, respectively, when compared to the NREL experimental value. However,
owing to the high cost of computing, L2 mesh is utilized in our optimization.
As shown in
Figure 6
, the estimated Cp distribution from the CFD simulation using the L2 mesh at the spans 30%, 47, 63, 80, and 95% is compared with the experimental data from NREL. The experimental result and the CFD
findings agree.
Table 3
, the computer used is characterized and the computational time is listed.
5. Results and Discussion
The configuration for aerostructural optimization is summarized in
Table 4
. While the structural optimization seeks to minimize mass, the aerodynamic optimization is more concerned with torque.
Figure 5
e shows a body-fitted FFD box for the blade. Only the top six layers of FFD points are permitted to shift, creating 240 form variables for the blade as a consequence. This prevents mesh quality
issues. Several geometric and physical restrictions, such as a minimal blade volume restriction to prevent the blade volume from falling below three times its baseline volume and thickness
restrictions to prevent the blade from becoming too thin, are used to assure practical design. With a maximum nonorthogonality and skewness of 78 degrees and 5.8, respectively, two mesh quality
limits are placed on the volume mesh deformation to handle mesh deformation issues. The mesh quality values are calculated using the OpenFOAM checkMesh function, and the constraint derivatives are
automatically differentiated using DAFoam. In order to prevent structural failure, an aggregated von Mises stress limitation is applied together with a panel thickness constraint. The stress
constraint value, which is an approximate maximal stress value normalized by the material yield stress, is calculated using the Kreisselmeier–Steinhauser function.
Figure 7
shows the pressure and stress on the aerodynamic and structural side respectively after the optimization.
Figure 8
a shows that following optimization, the pressure coefficient varies more significantly closer to the tip while being more stable farther away from the root. The cross-section profile of the blade in
terms of the three separate sections along the blade is also shown in
Figure 8
b before and after the optimization. The leading edge and middle portion are improved to provide greater aerodynamic performance.
According to
Figure 9
the torque was increased by around 6.78 percent as the main goal of the aerodynamic portion of the aerostructural optimization. The torque production of the wind turbine blade increased by around
6.78 percent as a result of this optimization procedure. The optimization of the blade’s form is responsible for the increase in torque. By improving this aspect, the airflow around the blade can be
better managed, which improves the efficiency with which wind energy is transferred to the turbine blades.
The wind turbine’s total efficiency and power production significantly increased thanks to the 6.78 percent rise in torque output. Since the turbine can produce more electricity with the same
quantity of wind, this improved efficiency may result in lower energy prices. The optimization procedure may also result in lower maintenance costs due to decreased component wear and tear brought on
by better efficiency.
The term “optimization” in the context of blade design refers to the process of changing different blade design parameters in order to accomplish a certain objective. In this instance, the objective
is to reduce the blade’s bulk while preserving its structural integrity. In order to do this, Mphys is used to shift the aerodynamic load on the blade to its structural component. To distribute the
weight properly and lessen stress concentrations, the blades’ panel thicknesses must be changed.
The mass of the blade is continuously measured and recorded throughout the optimization process. The findings are shown in
Figure 10
, which indicates that with each optimization step, the mass of the blade reduces. After the sixth iteration, the rate of decline slows down, indicating that future tweaks would not result in
substantial mass reductions. The mass of the blade also decreased by 4.22 percent as a consequence of the optimization process, which has the potential to significantly increase the turbine’s
performance and efficiency.
6. Conclusions
This research offered an analysis of a wind turbine blade’s aerostructural optimization. For the aerodynamic portion of the study, DAFoam software (v3.0.3) was used, while TACs was used for the
structural portion. The OpenMDAO framework was used to implement the fluid–structure interaction between the CFD and FEM. Torque was the goal for the aerodynamic component, while mass reduction was
the goal for the structural component.
The findings of the optimization revealed a 6.78 percent increase in torque and a 4.22 percent decrease in mass. These findings show that the aerostructural optimization strategy can significantly
boost the efficiency of wind turbine blades. The weight of the turbine can be made lighter overall by lowering the mass of the blade, which lowers the cost of production and shipping. Additionally, a
higher torque can boost the turbine’s energy production, increasing its effectiveness.
The OpenMDAO framework’s integration of DAFoam, TACs, and Mphys made it possible to analyze the wind turbine blade in its entirety. The combination of these software tools made it easier to combine
the structural and aerodynamic properties of the blade, allowing for design optimization. The outcomes of the optimization show how crucial it is to take into account both the structural and
aerodynamic properties of wind turbine blades when designing them.
In conclusion, this work shows how aerostructural optimization can enhance the performance of wind turbine blades. The optimization’s results show that by reducing the bulk of the blade, there is the
possibility for considerable increases in energy output and cost savings. Future research might concentrate on improving the optimization method’s effectiveness and determining whether it can be used
with different kinds of wind turbines.
Author Contributions
Y.Z. oversaw the project and provides the fund as well as the equipment; S.B. developed the geometry, CFD analysis, and optimization process and wrote the manuscript. A.B. (Aigerim Baidullayeva)
assisted with the mesh generation and manuscript. A.B. (Akerke Baigarina), Y.S. and E.S. participated in the revision and review. Y.Z. and D.W. reviewed and edited the manuscript. All authors have
read and agreed to the published version of the manuscript.
This study is funded by Nazarbayev University through a FDCR grant no. 20122022FD4126.
Data Availability Statement
The authors would like to thank Nazarbayev University for the financial support for this work through a FDCR grant no. 20122022FD4126.
Conflicts of Interest
The authors declare no conflict of interest.
Figure 1.
Blade geometry [
Figure 5. (a) Spherical volume mesh; (b) hyperbolic expansion layer; (c) aerodynamic surface mesh; (d) structural mesh; (e) FFD points for aerodynamic optimization; (f) structural design variables
(in terms of panels with random colors).
Figure 6. Comparison of the pressure coefficients (Cp) from level 2 mesh and NREL experimental data.
Figure 7. (a) Pressure on the blade on the aerodynamic side (Pa); (b) stress on the structural side (Pa).
Figure 8. (a) Cp comparisons before and after the optimization in terms of 30%, 63%, and 95% spanwise section; (b) shape changes in the cross-section profiles accordingly.
Boundary Conditions Blade Inout
Epsilon epsilonWallFunction inletOutlet
Nut nutUSpaldingWallFunction fixedValue
nuTilda fixedValue inletOutlet
K kqRWallFunction inletOutlet
Omega omegaWallFunction inletOutlet
P zeroGradient fixedValue
U fixedValue inletOutlet
Table 2. Mesh convergence by comparing the torque result from simulation against experimental value.
Mesh# L0 L1 L2 NREL Exp.
Mesh type Fine mesh Medium mesh Course mesh -
Cells (million) 25.3 10.12 2.53 -
Torque (Nm) 743.2 705.1 678.4 785
Error (%) 5.35 10.19 13.63 -
HPC Workstation CPU Time (hours)
Model Intel^® Xeon(R) CPU E5-2699 v4 @ 2.20GHz CFD with L0 7.5
Processors 88 CFD with L 5.3
RAM 503.8GB CFD with L2 2.1
OS Ubuntu Optimization with L2 33
Aerostructural Optimization
Aerodynamic Optimization Structural Optimization
Objective function Torque Mass
Design variables Blade shape in terms of FFD points Panel thickness
Constraints Volume and thickness Panel thickness and von Mises stress
Quantity of the design variables 240 FFD points 60 panels
Optimization by percentage ↑ 6.78% ↓ 4.22%
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Batay, S.; Baidullayeva, A.; Zhao, Y.; Wei, D.; Baigarina, A.; Sarsenov, E.; Shabdan, Y. Aerostructural Design Optimization of Wind Turbine Blades. Processes 2024, 12, 22. https://doi.org/10.3390/
AMA Style
Batay S, Baidullayeva A, Zhao Y, Wei D, Baigarina A, Sarsenov E, Shabdan Y. Aerostructural Design Optimization of Wind Turbine Blades. Processes. 2024; 12(1):22. https://doi.org/10.3390/pr12010022
Chicago/Turabian Style
Batay, Sagidolla, Aigerim Baidullayeva, Yong Zhao, Dongming Wei, Akerke Baigarina, Erkhan Sarsenov, and Yerkin Shabdan. 2024. "Aerostructural Design Optimization of Wind Turbine Blades" Processes 12,
no. 1: 22. https://doi.org/10.3390/pr12010022
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How do you teach math in reading comprehension?
How do you teach math in reading comprehension?
Teach kids to visualize When using visualizing to help with reading comprehension, it’s a good idea to underline or circle visual words (adjectives). Kids can do this with math word problems as well.
Circle words that paint a picture and provide important information, and then use those words to draw a picture.
How do you teach reading comprehension to elementary students?
Best Strategies for Teaching Reading Comprehension to Elementary Students
1. Encourage openness.
2. Identify specific problem areas (and solutions)
3. Use visual aids to help them “see” structure and individual elements.
4. Have them summarize what they have read.
5. Compare and contrast to other assignments.
What are some math strategies?
Top 9 math strategies for engaging lessons
• Explicit instruction. You can’t always jump straight into the fun.
• Conceptual understanding.
• Using concepts in Math vocabulary.
• Cooperative learning strategies.
• Meaningful and frequent homework.
• Puzzle pieces math instruction.
• Verbalize math problems.
• Reflection time.
How does reading comprehension affect math?
After a course of reading comprehension practice, Nicely found that students in algebra II and pre-calculus improved their math scores by 14.8 percent and 5.4 percent, respectively, over the course
of a year. Those results show that it is never too late to improve high school students’ math skills, Nicely says.
What is mathematics comprehension?
Comprehension and Mathematics: Comprehension Strategies applied to Mathematics. In order for students to be successful in the maths classroom they must be able to find the meaning of a maths problem
and look for approaches to a possible solution. Students must analyse and make conjectures about information.
What are the seven reading comprehension strategies?
Monitoring comprehension. Students who are good at monitoring their comprehension know when they understand what they read and when they do not.
Metacognition. Metacognition can be defined as “thinking about thinking.” Good readers use metacognitive strategies to think about and have control over their reading.
Graphic and semantic organizers.
What are some examples of reading comprehension strategies?
Eliminate distractions. When you are distracted,your ability to comprehend what you are reading is negatively impacted.
Read a book below your reading level. Starting with books below your reading level will allow you to develop a baseline of your reading comprehension and build on that.
Re-read text to ensure understanding.
Read aloud.
What can teachers do to improve reading comprehension?
– Teach the Process of Writing, Text Structures for Writing, Paragraph or Sentence Construction Skills (Improves Reading Comprehension) – Teach Spelling & Sentence Construction Skills (Improves
Reading Fluency) – Teach Spelling Skills (Improves Word Reading Skills)
What are the best reading strategies?
Skimming. Skimming will help you grasp the general idea or gist of a text.
Scanning. Scanning allows you to locate precise information.
Detailed reading. Detailed reading allows you to critically consider aspects of the text.
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A comparison of Pearl indices and cumulative incidences for use in meta-analysis of contraception trials - European Society of Contraception (ESC)
A comparison of Pearl indices and cumulative incidences for use in
meta-analysis of contraception trials
P. O’Brien
Westside Contraceptive Services, London, UK
Objectives: One of the issues encountered when performing
meta-analysis of contraceptive trials is the variable way in which the outcomes
are reported, as cumulative rates (Kaplan Meier curves), single or multiple
decrement life-tables rates, or Pearl indices (events per 100 women-years). The
latter decrease with time, while the others increase. However, it is usually the
ratio of the rates that we use in meta-analysis. In this analysis we compare the
ratio of the Pearl rates to the ratio of the cumulative incidence and their
confidence intervals.
Design and methods: We used data from a large WHO randomised trial of
a frameless IUD and TCu380A, to compared the ratio of Pearl indices to the ratio
of cumulative incidences. We used the method described by Kleinbaum to calculate
the standard error of the ratio of cumulative incidences and the method of
Hasselblad for the standard error of the ratio of Pearl indices, to calculate
their confidence intervals.
Results: For accidental pregnancies the difference in the ratio of
Pearl ratios to ratio of incidence ratios was greatest at year 6 at 13% (Pearl
rate ratio 0.95, 95%CI 0.64 to 1.47; incidence rate ratios 0.85, 0.56 to 1.29).
In 4 of the 6 years of follow-up, the difference in ratios was less than 5%. The
difference was smaller for removals for bleeding and pain, never exceeding 3%
(Pearl ratio 1.10, 0.90 to 1.33; incidence ratio 1.13, 0.94 to 1.36 at 6 years).
For total use-related discontinuations, the difference never exceeded 4% (Pearl
ratio 1.26, 1.10 to 1.45; incidence ratio 1.22, 1.08 to 1.38 at 6 years).
Conclusions: The difference in rate ratios, whether we use the Pearl
indices or cumulative incidence rates, is usually small. The difference in rate
ratios from single and multiple decrement life-tables is likely to be smaller.
The ability to incorporate trials using different reporting methods in
meta-analyses enhances our capacity to systematically review the literature.
Further research is required to understand the determinants of the magnitude and
direction of the differences. | {"url":"https://contraception-esc.com/congress/abstracts/fc105.htm","timestamp":"2024-11-05T03:42:49Z","content_type":"text/html","content_length":"144668","record_id":"<urn:uuid:c5148e5c-d1bd-42da-aa32-1a69db0702b1>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00406.warc.gz"} |
Math.min has a red underline?
I’m trying to return the lowest number from a table
local LowestDistance = math.min(unpack(Distances))
However, I get a red underline, saying
‘This function must be called with self. Did you mean to use a colon instead of a dot?’ but a colon also shows a red underline. So what’s the correct function?
math | Documentation - Roblox Creator Hub shows math.min, so to me it should still work?
number math.min ( number x, number … )
Returns the minimum value among the numbers passed to the function.
1 Like
when i pasted this i did not get any red underline
Are you using the new Lua script analysis features in Beta Features? Try turning it off and see if you still get the red underline.
1 Like
Yeah that is it. Was barely gonna press reply but you beat me. It’s still in beta so there are still some bugs
1 Like | {"url":"https://devforum.roblox.com/t/mathmin-has-a-red-underline/569005","timestamp":"2024-11-10T23:03:51Z","content_type":"text/html","content_length":"27577","record_id":"<urn:uuid:8bfb4fac-7560-4508-8d65-668712f97bc6>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00171.warc.gz"} |
How Much Dirt Calculator – Fast & Accurate Results
This tool helps you calculate the amount of dirt you need for your project based on your specified area and depth.
How to Use the Calculator
To use the how much dirt calculator, follow these steps:
1. Enter the length of the area you want to fill with dirt in meters.
2. Enter the width of the area in meters.
3. Enter the depth in meters.
4. Enter the density of the soil in kilograms per cubic meter (kg/m³).
5. Click on the ‘Calculate’ button to see the volume and mass of the soil required.
How It Calculates the Results
The calculator uses the following steps to determine the amount of dirt needed:
1. It calculates the volume of soil required using the formula: Length × Width × Depth.
2. It calculates the mass of soil using the formula: Volume × Density.
The accuracy of the calculator depends on the correctness of the input values. Ensure that you use precise measurements for the most accurate results. The density of the soil can vary based on its
type, moisture content, and compaction, which may lead to variations in the calculated mass of soil required.
Use Cases for This Calculator
Garden Bed Preparation
When you’re ready to create a beautiful garden bed, knowing how much dirt you need is crucial. This calculator helps you determine the quantity of soil required based on the dimensions of your garden
bed, ensuring you have just the right amount without excess waste.
Creating a Raised Garden Bed
If you’re designing a raised garden bed, this calculator steps in to simplify your planning. By inputting the length, width, and depth of the bed, you’ll discover exactly how much soil you’ll need to
fill it, making the construction process a breeze.
Landscaping Projects
For landscaping projects like filling in low spots or creating new plant beds, this calculator is your best friend. Input your desired area specifications, and it will provide you with the volume of
dirt needed to enhance your landscape effectively.
Building Planters
If you’re building wooden or concrete planters, calculating the right amount of dirt is essential for healthy plants. By entering the dimensions of each planter into the calculator, you ensure your
plants have a proper environment to thrive.
Filling Pits or Holes
Whether it’s a large pit or minor holes around your property, knowing how much dirt to fill them is important. This calculator allows you to accurately assess the volume of dirt needed to restore the
area to its original condition.
Soil Replacement for Lawns
When it’s time to replace your lawn soil due to wear and tear, this calculator takes the guesswork out of the equation. By measuring the area of your lawn, you can find out precisely how much fresh
soil you’ll need for rejuvenation.
Adding Soil for New Trees
Planting new trees involves significant soil considerations to facilitate their growth. Use the calculator to determine how much dirt you need for the planting hole to create a nurturing environment
for your new additions.
DIY Home Projects
If you’re getting your hands dirty with DIY home improvement projects, you’ll want to know how much dirt you need for any outdoor undertakings. This tool provides accurate calculations to keep your
project on track and helps prevent last-minute runs to the supply store.
Creating a Pond or Water Feature
Enhancing your yard with a pond or water feature requires careful dirt calculations for proper shaping and leveling. Use the calculator to figure out the volume of dirt to remove or add for a
balanced and attractive design.
Constructing a Pathway
When building a pathway, the right amount of dirt is essential for a stable foundation. Enter the dimensions into the calculator, and you’ll receive an accurate figure to ensure your pathway is safe,
functional, and aesthetically pleasing. | {"url":"https://calculatorsforhome.com/how-much-dirt-calculator/","timestamp":"2024-11-06T17:19:00Z","content_type":"text/html","content_length":"144995","record_id":"<urn:uuid:effe92c9-1676-4b62-8f6a-3f02acc7bd35>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00149.warc.gz"} |
Agora Nomic - Office of Notary
The Cookie Jar
Recent statistics
1) The name of this public contract is The Cookie Jar.
2) Any party to this contract CAN amend it without party objection.
3) For each event defined by this contract as point-worthy:
a) Once per week, each contestant CAN by announcement guess
how many times that event will occur during the next week.
b) After the end of that next week, the contestants are ranked
based on the absolute value of the difference between their
guess and the actual number, smallest differences highest,
with ties broken in favor of earlier guesses.
c) As soon as possible after the end of that next week, the
contestmaster SHALL award points as follows along the specified
axis, in this order, where N is the maximum number of points that
can be awarded for this contest along that axis each week:
i) floor(4N/10) to the first-place contestant
ii) floor(3N/10) to the second-place contestant
iii) floor(2N/10) to the third-place contestant
iv) 1 to each contestant whose guess was exactly correct,
in order from high to low rank, then repeating this
step until no more points can be awarded
4) The distribution of a proposal is a point-worthy event (counting
at most 5 proposals per author per week) for the X axis.
5) The initiation of a judicial case by a player is a point-worthy event
(counting at most 5 cases per initiator per week) for the
Y axis.
page revision: 15, last edited: 12 Jul 2009 19:01 | {"url":"http://agora-notary.wikidot.com/the-cookie-jar","timestamp":"2024-11-11T17:06:08Z","content_type":"application/xhtml+xml","content_length":"30632","record_id":"<urn:uuid:f9132cde-81fd-4051-b942-4fd4e786f0d6>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00752.warc.gz"} |
Is curvature invariant?
In general relativity, curvature invariants are a set of scalars formed from the Riemann, Weyl and Ricci tensors – which represent curvature, hence the name, – and possibly operations on them such as
contraction, covariant differentiation and dualisation.
Is Ricci scalar invariant?
In Riemannian geometry, the scalar curvature (or the Ricci scalar) is the simplest curvature invariant of a Riemannian manifold.
What are tensors in general relativity?
In general relativity, the metric tensor (in this context often abbreviated to simply the metric) is the fundamental object of study. It may loosely be thought of as a generalization of the
gravitational potential of Newtonian gravitation.
How do you calculate Ricci tensor?
The general steps for calculating the Ricci tensor are as follows:
1. Specify a metric tensor (either in matrix form or the line element of the metric).
2. Calculate the Christoffel symbols from the metric.
3. Calculate the components of the Ricci tensor from the Christoffel symbols.
What kind of math did Einstein use for relativity?
At the time he was conceiving the General Theory of Relativity, he needed knowledge of more modern mathematicss: tensor calculus and Riemannian geometry, the latter developed by the mathematical
genius Bernhard Riemann, a professor in Göttingen. These were the essential tools for shaping Einstein’s thought.
Is spacetime a metric space?
The spacetime interval is a way of encoding the information contained in the metric g. Mathematically, the metric is a special kind of object called a two-tensor. Unfortunately, the word metric is
begin used in a different way here than in the context of metric spaces.
Who almost beat Einstein to general relativity?
mathematician David Hilbert
By late 1915 he knew that the influential German mathematician David Hilbert was close to finalizing a theory of general relativity himself. In a few feverish weeks in November 1915, Einstein
reverted to the equations of general relativity he’d had in hand for more than two years and applied the finishing touches.
How is curvature tensor defined?
The curvature tensor measures noncommutativity of the covariant derivative, and as such is the integrability obstruction for the existence of an isometry with Euclidean space (called, in this
context, flat space). The linear transformation.
How do you find the Ricci curvature tensor?
What is meant by curvature tensor?
The curvature tensor measures noncommutativity of the covariant derivative, and as such is the integrability obstruction for the existence of an isometry with Euclidean space (called, in this
context, flat space). The linear transformation. is also called the curvature transformation or endomorphism.
What mistake did Hilbert?
The equations of general relativity were first obtained by David Hilbert, the greatest mathematician of the time after Poincare. However, Hilbert made the mistake of sending the equations in a letter
to Einstein. At that time Einstein was lecturing and using the wrong equations. | {"url":"https://www.sheppard-arts.com/writing-prompts/is-curvature-invariant/","timestamp":"2024-11-14T11:53:31Z","content_type":"text/html","content_length":"73176","record_id":"<urn:uuid:f327d24d-7a1f-433f-b15a-e605b2199b6f>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00206.warc.gz"} |
A framework for safe, programmable, speculative parallelism
LTS Haskell 12.26: 1.5.0.3
Stackage Nightly 2018-09-28: 1.5.0.3
Latest on Hackage: 1.5.0.3
This version can be pinned in stack with:speculation-1.5.0.3@sha256:d8c7de7defa74ce883d4167b5e11595dc13fcfc5d7ccc56ce8265d3abbcbf2c2,4500
Module documentation for 1.5.0.3
A framework for safe, programmable, speculative parallelism, loosely based on:
• Prakash Prabhu, G. Ramalingam, and Kapil Vaswani, “Safe Programmable Speculative Parallelism”, In the proceedings of Programming Language Design and Implementation (PLDI) Vol 45, Issue 6 (June
2010) pp 50-61. http://research.microsoft.com/pubs/118795/pldi026-vaswani.pdf
This package provides speculative function application and speculative folds. Speculative STM transactions take the place of the transactional rollback machinery from the paper.
You can download it using cabal install speculation, if you have the Haskell Platform installed.
Speculative Function Application (Control.Concurrent.Speculation)
Various speculative function application combinators are provided. Two fairly canonical samples are described here.
spec :: Eq a => a -> (a -> b) -> a -> b
spec g f a evaluates f g while forcing a, if g == a then f g is returned. Otherwise f a is evaluated.
Furthermore, if the argument has already been evaluated, we avoid sparking the parallel computation at all.
If g is a good guess at the value of a, this is one way to induce parallelism in an otherwise sequential task.
However, if g isn't available more cheaply than a, then this saves no work, and if g is wrong, you risk evaluating the function twice. spec a f a = f $! a
The best-case timeline looks like: [–– f g ––] [—– a —–] [– spec g f a –]
The worst-case timeline looks like: [–– f g ––] [—– a —–] [–– f a ––] [—–– spec g f a ———–]
Compare these to the timeline of f $! a: [–– a —–] [–– f a ––]
specSTM provides a similar compressed timeline for speculated STM actions, but also rolls back side-effects.
Speculative Folds (Data.Foldable.Speculation)
A speculative version of the combinators from Data.Foldable is provided as Data.Foldable.Speculation.
Each combinator therein takes an extra argument that is used to speculate on the value of the list.
foldr :: (Foldable f, Eq b) => (Int -> b) -> (a -> b -> b) -> b -> f a -> b
Given a valid estimator g, foldr g f z xs yields the same answer as Foldable.foldr' f z xs.
g n should supply an estimate of the value returned from folding over the last n elements of the container.
As with spec, if the guess g n is accurate a reasonable percentage of the time and faster to compute than the ensuing fold, then this can provide increased opportunities for parallelism.
foldl :: (Foldable f, Eq b) => (Int -> b) -> (b -> a -> b) -> b -> f a -> b
foldl works similarly to Foldable.foldl', except that g n should provide an estimate for the first n elements.
Contact Information
Contributions and bug reports are welcome!
I can be reached through the user ekmett on github, as edwardk on irc.freenode.net #haskell channel, or by email at [email protected].
-Edward Kmett
• Support GHC 8
• Support transformers 0.5
• Build warning-free on GHC 7.10+
• Added an HLint configuration
• Removed a redundant constraint from the type of sequenceByA_.
• Removed a redundant Monad m constraint from instance MonadSpec (ContT r m).
• Removed the use of tag-bits. This enables the API to be Trustworthy.
• Removed old benchmark/test framework.
• Make numSparks support mandatory.
• Moved to Control.Concurrent from Data
• Added MonadSpec, so we can make instances for Codensity, etc. in other packages
• Removed the ContT r STM combinators
• Cleaned out old issues
• Made compatible with the removal of the Eq/Show superclasses of Num in GHC 7.3+
• Fixed name collision with the new Distribution.Simple.testHook in Setup.lhs
• Reorganized the module hierarchy into Data.Speculation
• Added support for numSparks
• Removed interim boxing in the unsafeIsEvaluated and unsafeGetTagBits check
• Added Data.List.Foldable
• Added Data.Traversable.Foldable
• Fixed an off-by-one error in the arguments to the speculative fold estimators
• changed tests and benchmarks to not build by default to work around corruption in the hackage db
• test suite now forces build | {"url":"https://www.stackage.org/lts-11.22/package/speculation-1.5.0.3","timestamp":"2024-11-06T05:54:15Z","content_type":"text/html","content_length":"20353","record_id":"<urn:uuid:5c1582a8-4fb3-40ea-a10c-1685e2aae10c>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00422.warc.gz"} |
Free Parlay Calculator For Sports Bettors
You can also see the probability of winning that the oddsmaker implies by setting the line at what it is. This is sometimes useful to help you understand what you are betting on. https://
shahr-oghli.com/connecticut-betting-on-horse-racing-for-dummies-sports-betting/ A parlay bet is a group of spread, moneyline, or total bets combined into one bet to increase the payout odds. In order
for the parlay to win, each separate bet has to win. Parlay odds offer bigger payouts than normal odds because they are riskier, since each individual bet has to win.
Three Types Of Odds Formats
Odds of 3-5 indicate that your profit will be three-fifths of a dollar. In other words, for every $5 you bet you can win $3 in informative post profit. To determine profit, multiply the amount you
bet by the fraction. W – Means that the horse has won a race on a track rated SLOW or HEAVY.
An Introduction To Betting
This wager is also sometimes referred to as a square bet or a quarter bet. – With a corner bet, you’re wagering on four numbers that form a kind of square on the table, like 22, 23, 25, and 26, for
instance. Use our roulette payout chart at the beginning of the article to help you with your chances. It’s referred to as “standard deviation”, which illustrates why some players leave the roulette
tables as winners.
Betting on 00 is known as a straight bet or a single number bet. So, always make the effort to find the best odds, in order to benefit from a slightly higher payout. And keep in mind that you will
have to achieve a high success rate to reap the benefits.
Therefore, with these new odds and probability figures, this is not a value bet. Therefore, any positive figure means the bet offers value as if the punter placed the same bet a 100 times, then they
would turn a profit. Let’s say you conduct some research into the Manchester United v Chelsea fixture and discover that all the stats point to a very good chance of a home win for Manchester United.
Recency Bias In Sports Betting
Another way in which bookmakers present this offer is as a ‘100% deposit bonus’. Armed with this new knowledge, you are ready to take on the world of sportsbooks. American sports bettors are
certainly some of the most capable bettors out there, as they deal with rather complex representations of probability and potential winnings.
American odds start with either a positive or negative sign (e.g. -200 or +200). A minus sign indicates a bookie’s favorite to win while a plus symbol indicates an underdog. A -250 favorite has a
better chance of winning an event than a -150 favorite, while a +200 underdog has a better chance of winning than a +500 underdog. Also known as moneyline odds, this format puts every bet relative to
$100. If an outcome has a negative number (e.g. -125), that’s how much money you have to bet to win $100 (e.g. a wager of $125). If the number is positive, that’s how much a bet of $100 will pay out.
Total Payout
Understanding betting odds and the probabilities they reflect is key to sports betting. In this guide we will take you through the basics of betting odds. If a sportsbook gets too many bets on one
side of a line, the book stands to lose big if that side wins. At most online sportsbooks, you can combine all types of NFL bets into a single parlay. For example, you can parlay a prop bet,
moneyline, and point spread bet into a single parlay. Some sportsbooks also let you do single-game parlays rather than multi-game parlays.
To do this we simply multiply the numerator by the number of times required to make it a round number. If your decimal odds are a round number, such as 7.0, then we simply subtract 1 from the decimal
odds and apply a denominator of 1. The outcome of this formula is 3, meaning we will express the odds as fractional odds, 3 /1. Multiplied then by 100, we get the implied probability percentage of | {"url":"https://www.madelac.com.ec/2021/07/27/free-parlay-calculator-for-sports-bettors/","timestamp":"2024-11-04T01:41:05Z","content_type":"text/html","content_length":"61112","record_id":"<urn:uuid:4c644d51-1efe-46c3-a970-57f5054191c7>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00558.warc.gz"} |
Grade 8 Math Quiz [PDF] Questions Answers | Grade 8 Math Quiz App Download & e-Book
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20 Mind-Blowing Fibonacci Sequence Facts You Won't Believe Are True | MathAware Nude AI Generators & NSFW
Key Takeaways
• Fibonacci Sequence Basics: The Fibonacci sequence is an infinite series where each number is the sum of the two preceding ones, starting from 0 and 1.
• Golden Ratio Connection: Closely linked to the Golden Ratio (~1.618), the Fibonacci sequence appears in various natural and human-made structures, adding aesthetic appeal.
• Historical Origins: First documented by Indian mathematician Pingala around 200 BC, it was later popularized in Europe by Leonardo Fibonacci through his book Liber Abaci.
• Natural Occurrences: This sequence manifests in numerous natural phenomena such as flower petals, pinecones, and sunflower seed arrangements which optimize growth patterns and space efficiency.
• Artistic Influence: Artists like Leonardo da Vinci and Salvador Dalí have used principles derived from this sequence to achieve balance and harmony in their works.
• Modern Technological Applications: From algorithms in computer science to financial market analysis tools, the Fibonacci sequence plays a crucial role due to its recursive properties.
Understanding the Fibonacci Sequence
The Fibonacci sequence is more than just a mathematical curiosity. Its profound impact spans various fields, from nature to computer science.
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What Is the Fibonacci Sequence?
The Fibonacci sequence follows a simple yet intriguing pattern: each number in the series is the sum of its two preceding numbers. Starting with 0 and 1, it progresses as follows: 0, 1, 1, 2, 3, 5,
and so on.
This pattern extends indefinitely since it’s an infinite sequence. The relationship between consecutive numbers remains constant throughout its expansion.
Interestingly enough not only does this sequence include positive integers but can also be extended to negative integers through a process known as negafibonacci numbers.
The Golden Ratio Connection
One of the most fascinating aspects of the Fibonacci sequence is its connection to the Golden Ratio, approximately valued at ( \frac{1 + \sqrt{5}}{2} ) or around 1.61803398875.
Historical Insights into the Fibonacci Sequence
The Fibonacci sequence has captivated mathematicians for centuries due to its unique properties and applications. It originated long before Leonardo Fibonacci introduced it to Europe.
Origins and Leonardo Fibonacci
Ancient Indian mathematician Pingala first described the sequence around 200 BC. He used it to enumerate patterns in Sanskrit poetry formed from syllables of two lengths. This early use showcased the
sequence’s versatility.
Leonardo Fibonacci, born around 1170 in Pisa, Italy, traveled extensively to study different numerical systems. In his 1202 book Liber Abaci (The Book of Calculation), he introduced the sequence to
Western European mathematics. The book popularized not only the sequence but also Indo-Arabic numerals by showing their practical applications.
Fibonacci’s life remains largely unknown beyond his mathematical contributions. His travels allowed him to gather knowledge from various cultures and incorporate these insights into his writings,
significantly impacting Western mathematics.
Fibonacci’s Impact on Mathematics
Leonardo Fibonacci’s introduction of Indo-Arabic numerals through Liber Abaci revolutionized calculations in Europe. Before this system, Roman numerals were prevalent but cumbersome for arithmetic
In addition to practical arithmetic improvements, Liber Abaci explored speculative mathematics with examples that demonstrated how numbers could solve real-world problems like trade calculations and
interest rates.
One famous example is a problem involving rabbit population growth that led directly to what is now known as the Fibonacci sequence: starting with a pair of rabbits and observing their reproduction
over time provided a natural context for understanding sequential growth patterns represented by 0,1,…n+1=Fn+Fn-1 .
This work laid foundational concepts that influenced later advancements in algebra and number theory while extending into modern computational algorithms where efficiency often hinges on recursive
sequences akin those found within classical texts such as Liber abacus itself illustrating timeless relevance inherent even today .
Through these contributions ,Fibonacci shaped future generations’ approach towards both theoretical explorations practical implementations alike .
Surprising Natural Occurrences of Fibonacci
The Fibonacci sequence appears astonishingly often in nature. From plants to animals, these numbers reveal a hidden mathematical harmony in the natural world.
Fibonacci in Flower Petals
Many flowers have petals that follow the Fibonacci sequence, such as 5, 8, 13, and 21. For instance:
• Lilies: Typically have 3 petals.
• Buttercups: Usually display 5 petals.
• Daisies: Often show up with either 34 or even more precisely arranged petal counts like 55 or even higher.
Two-thirds of the numbers within this famous sequence are odd. This fact explains why numerous flowers exhibit an odd number of petals.
Fibonacci and Pinecones
Pinecones also demonstrate fascinating occurrences of the Fibonacci sequence:
1. Spiral Patterns: The scales on pinecones form spiral patterns from their base to tip.
2. Number Count: When counted diagonally across each cone’s surface, these spirals often align with consecutive Fibonacci numbers such as:
• Small pinecone spirals typically count at around either 5 or sometimes jumping straight into double digits like hitting a perfect dozen (12).
• Larger cones may significantly showcase larger sets following through closer ranges around higher summations (upwards towards mid-twenties).
Fibonacci Sequence in Art and Architecture
The Fibonacci sequence’s influence extends beyond mathematics, significantly impacting art and architecture. Its connection to the golden ratio creates visually pleasing compositions.
Famous Artworks
Artists have long used the golden ratio, derived from the Fibonacci sequence, to achieve balance and harmony in their works. Leonardo da Vinci incorporated this principle into his masterpieces. For
• The Last Supper: Da Vinci utilized the golden ratio to structure this iconic painting, balancing elements between Jesus and his disciples.
• Vitruvian Man: This drawing exemplifies proportion by aligning human anatomy with geometric principles linked to the golden ratio.
Modern artists also embrace these concepts:
• Salvador Dalí: In “The Sacrament of the Last Supper,” Dalí employed a dodecahedron reflecting proportions related to the golden ratio.
These examples underscore how deeply rooted mathematical harmony is in artistic creation.
Architectural Marvels
Architects use Fibonacci sequences and the associated golden ratio for aesthetically pleasing designs. Historical structures demonstrate their application:
• Parthenon: Measurements of its facade align closely with phi (1.618), creating balanced proportions.
• Egyptian Pyramids: The Great Pyramid of Giza’s dimensions reflect a near-perfect alignment with phi.
Le Corbusier’s Modulor system stands out as a modern example:
• Modulor System by Le Corbusier: Developed post-WWII, it integrates human scale measurements based on phi for harmonious building designs.
Contemporary landmarks continue this tradition:
• Taj Mahal: The symmetry within its layout follows principles akin to those found in nature through Fibonacci numbers.
This intersection underscores that architectural elegance often stems from mathematical foundations ensuring structurally sound yet visually appealing edifices.
Practical Applications in Modern Technology
Fibonacci sequences appear in various technological applications, influencing fields from biology to design.
Nature’s Blueprint
The Fibonacci sequence manifests in the arrangement of leaves on a stem, the branching patterns of trees, and the structure of artichokes. These natural occurrences optimize light exposure and space
efficiency for plant growth.
Sunflower Seeds
Sunflower seeds arrange themselves in Fibonacci spirals to maximize packing density. This pattern ensures that each seed has optimal access to sunlight and nutrients, enhancing their survival rates.
Art and Design
Artists use Fibonacci sequences to create visually appealing compositions. In architecture, designers employ these ratios for balanced proportions. For example: The Parthenon uses these principles
for aesthetic harmony.
Biological systems often exhibit Fibonacci patterns. Sunflower seeds’ arrangements follow these series as do artichoke flowerings. These patterns contribute significantly to biological efficiency and
Experimental Western poetry sometimes uses structures based on the Fibonacci series similar to haiku forms but with a mathematical basis guiding line lengths or syllable counts.
Algorithms And Coding
Algorithms benefit significantly from incorporating Fibonacci numbers due to their recursive properties.
1. Sorting Algorithms: The quicksort algorithm can be optimized using pivot points derived from the Fibonacci sequence.
2. Search Techniques: Binary search algorithms leverage these numbers for efficient data partitioning.
3. Data Structures: Implementations like heaps or priority queues sometimes utilize this mathematical concept.
4.Math Libraries: Many programming languages include functions specifically designed around calculating terms within this sequence due its frequent use cases.
Financial Markets And Trading
Traders often rely on technical analysis tools rooted in the principles of the Fibonacci sequence.
1.Retracement Levels: Analysts identify potential reversal levels by applying retracement percentages (23%, 38%, 50%) derived directly from dividing sequential terms within this numeric progression
against market trends such as highs/lows over given intervals etc..
2.Extensions Projections : Beyond identifying potential pullback points during downtrends retracements projections help forecast where prices might extend upwards following bullish breakout movements
again utilizing similar percentage coefficients calculated relative previous price action movements periods being analyzed .
3Timing Cycles Some sophisticated quantitative models also factor timing cycles aligned based specifically around timeframes elapsed between significant high low turning events previously recorded
historical datasets further refining predictive accuracy trading strategies implemented accordingly .
The Fibonacci sequence continues to astonish with its widespread influence across different fields. Whether it’s the ancient roots traced back to Pingala or its introduction by Leonardo Fibonacci,
this mathematical marvel has left an indelible mark. Its presence in nature, art, and modern technology exemplifies how a simple numerical pattern can shape our world.
From optimizing growth in plants to creating visually stunning artworks and boosting technological efficiency, the Fibonacci sequence remains an essential tool. It merges mathematical precision with
practical applications seamlessly demonstrating that sometimes beauty indeed lies in numbers.
Frequently Asked Questions
What is the Fibonacci sequence?
The Fibonacci sequence is a series of numbers where each number is the sum of the two preceding ones, starting from 0 and 1. It often appears in nature, art, architecture, and technology.
Who discovered the Fibonacci sequence?
The origins trace back to ancient Indian mathematician Pingala. Leonardo Fibonacci introduced it to Western Europe through his book “Liber Abaci” in 1202.
How does the Fibonacci sequence appear in nature?
In nature, it optimizes growth patterns. Examples include sunflower seed arrangements and pine cone spirals which follow this mathematical pattern.
How is the Fibonacci sequence used in art?
Artists use it to create visually appealing compositions by employing proportions based on Fibonacci numbers for balance and harmony.
Why is the Fibonacci sequence important in architecture?
It ensures balanced proportions in structures like Greece’s Parthenon by aligning with aesthetic principles that are naturally pleasing to human eyes.
How does modern technology use the Fibonacci sequence?
It’s utilized for efficiency and predictive analysis across various fields such as biological studies, design algorithms, computer science sorting methods, and financial market trading strategies. | {"url":"https://www.mathaware.org/20-mind-blowing-fibonacci-sequence-facts-you-won-t-believe-are-true/","timestamp":"2024-11-08T05:23:26Z","content_type":"text/html","content_length":"107711","record_id":"<urn:uuid:735e5580-28ce-42c9-9285-c8288af5ccca>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00249.warc.gz"} |
Pascals to Pounds per square foot
Pounds per square foot to Pascals (Swap Units)
Note: Fractional results are rounded to the nearest 1/64. For a more accurate answer please select 'decimal' from the options above the result.
Note: You can increase or decrease the accuracy of this answer by selecting the number of significant figures required from the options above the result.
Note: For a pure decimal result please select 'decimal' from the options above the result.
1 Pascal = 1 N/m2 or 1 Kg / m.s2
Pascals to Pounds per square foot formula
1 psf is the pressure exterted by one pound-force of force being applied to an area of one square foot
Pascals to Pounds per square foot table
Pascals Pounds per square foot
0Pa 0.00psf
1Pa 0.02psf
2Pa 0.04psf
3Pa 0.06psf
4Pa 0.08psf
5Pa 0.10psf
6Pa 0.13psf
7Pa 0.15psf
8Pa 0.17psf
9Pa 0.19psf
10Pa 0.21psf
11Pa 0.23psf
12Pa 0.25psf
13Pa 0.27psf
14Pa 0.29psf
15Pa 0.31psf
16Pa 0.33psf
17Pa 0.36psf
18Pa 0.38psf
19Pa 0.40psf
Pascals Pounds per square foot
20Pa 0.42psf
21Pa 0.44psf
22Pa 0.46psf
23Pa 0.48psf
24Pa 0.50psf
25Pa 0.52psf
26Pa 0.54psf
27Pa 0.56psf
28Pa 0.58psf
29Pa 0.61psf
30Pa 0.63psf
31Pa 0.65psf
32Pa 0.67psf
33Pa 0.69psf
34Pa 0.71psf
35Pa 0.73psf
36Pa 0.75psf
37Pa 0.77psf
38Pa 0.79psf
39Pa 0.81psf
Pascals Pounds per square foot
40Pa 0.84psf
41Pa 0.86psf
42Pa 0.88psf
43Pa 0.90psf
44Pa 0.92psf
45Pa 0.94psf
46Pa 0.96psf
47Pa 0.98psf
48Pa 1.00psf
49Pa 1.02psf
50Pa 1.04psf
51Pa 1.07psf
52Pa 1.09psf
53Pa 1.11psf
54Pa 1.13psf
55Pa 1.15psf
56Pa 1.17psf
57Pa 1.19psf
58Pa 1.21psf
59Pa 1.23psf | {"url":"https://www.metric-conversions.org/pressure/pascals-to-pounds-per-square-foot.htm","timestamp":"2024-11-14T04:36:40Z","content_type":"text/html","content_length":"55254","record_id":"<urn:uuid:1b99cd8f-1456-47fe-9f63-74a4d5b1fc31>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00427.warc.gz"} |
Getting Started: X11 Procedure
The most common use of the X11 procedure is to produce a seasonally adjusted series. Eliminating the seasonal component from an economic series facilitates comparison among consecutive months or
quarters. A plot of the seasonally adjusted series is often more informative about trends or location in a business cycle than a plot of the unadjusted series.
The following example shows how to use PROC X11 to produce a seasonally adjusted series, , from an original series .
In the multiplicative model, the trend cycle component keeps the same scale as the original series , while , , and vary around 1.0. In all printed tables and in the output data set, these latter
components are expressed as percentages, and thus will vary around 100.0 (in the additive case, they vary around 0.0).
The naming convention used in PROC X11 for the tables follows the original U.S. Bureau of the Census X-11 Seasonal Adjustment program specification (Shiskin, Young, and Musgrave, 1967). Also, see the
section Printed Output. This convention is outlined in Figure 37.1.
The tables corresponding to parts A – C are intermediate calculations. The final estimates of the individual components are found in the D tables: table D10 contains the final seasonal factors, table
D12 contains the final trend cycle, and table D13 contains the final irregular series. If you are primarily interested in seasonally adjusting a series without consideration of intermediate
calculations or diagnostics, you only need to look at table D11, the final seasonally adjusted series.
For further details about the X-11-ARIMA tables, see Ladiray and Quenneville (2001). | {"url":"http://support.sas.com/documentation/cdl/en/etsug/65545/HTML/default/etsug_x11_gettingstarted.htm","timestamp":"2024-11-10T09:08:34Z","content_type":"application/xhtml+xml","content_length":"15808","record_id":"<urn:uuid:119b64d5-1ef0-415e-b246-269e501b1b2f>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00543.warc.gz"} |
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Herein, an averaging theory for the solutions to Cauchy initial value problems of arbitrary order,ε-dependent parabolic partial differential equations is developed. Indeed, by directly developing
bounds between the derivatives of the fundamental solution to such an equation and derivatives of the...
Kouritzin, Michael, Dawson, Donald
A general Hilbert-space-based stochastic averaging theory is brought forth herein for arbitrary-order parabolic equations with (possibly long range dependent) random coefficients. We use
regularity conditions onView the MathML sourcewhich are slightly stronger than those required to prove...
Heunis, A. J., Kouritzin, Michael
In this note we consider the almost sure convergence (as ϵ→0) of solution Xϵ(·), defined over the interval 0 ≤ τ ≤ 1, of the random ordinary differential equation View the MathML source Here {F
(x, t, ω), t ≥ 0} is a strong mixing process for each x and (x, t) → F(x, t, ω) is subject to regularity... | {"url":"https://era.library.ualberta.ca/search?direction=desc&facets%5Ball_contributors_sim%5D%5B%5D=Kouritzin%2C+Michael&facets%5Ball_subjects_sim%5D%5B%5D=Parabolic+partial+differential+equations&facets%5Ball_subjects_sim%5D%5B%5D=Mathematics&facets%5Bitem_type_with_status_sim%5D%5B%5D=article_published&sort=sort_year","timestamp":"2024-11-06T09:01:26Z","content_type":"text/html","content_length":"33725","record_id":"<urn:uuid:04ae4d63-3597-40ac-b190-ed5116f2278e>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00572.warc.gz"} |
Finding the Measure of an Angle given Its Supplementary Angle’s Measure
Question Video: Finding the Measure of an Angle given Its Supplementary Angle’s Measure Mathematics
Find the value of 𝑥.
Video Transcript
Find the value of 𝑥.
We’re given two angles on a straight line. We’re given the measurement of the first angle, which is 107 degrees, but we don’t know the measurement of angle 𝑥. You may have spotted that these two
angles are on a straight line. We call two angles like this supplementary angles. They always add together to total 180 degrees.
So, we know the sum of the given angle, 107 degrees, plus 𝑥 total 180 degrees. We can use the inverse to help us find the answer. 180 degrees take away 107 degrees equals the measurement of angle 𝑥.
180 take away 107 equals 73. So, the value of 𝑥 is 73. We realized our two angles are supplementary angles, whose sum is 180 degrees. So, to find the value of 𝑥, we subtracted 107 degrees from 180 to
give us our answer of 73. | {"url":"https://www.nagwa.com/en/videos/914198019584/","timestamp":"2024-11-04T01:53:33Z","content_type":"text/html","content_length":"240692","record_id":"<urn:uuid:7d4fbfc6-f2d8-4494-9f34-a0b2d5b21e97>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00184.warc.gz"} |
Assessments in AP Calculus
Sarah Stecher
At this point in the school year you’re probably getting ready to give your first AP Calculus test. What should the format of the test be? Where can you find good questions? How strict should you be
when you grade? Should scores be curved? These are all questions we have to consider when planning our course and writing our assessments. Here are some of our suggestions:
• Include calculator and non-calculator items
• Include multiple choice and free response items
• Write questions that reflect learning targets and success criteria
• Determine scoring rubric for FRQs before administering the assessment
On the last day of every unit in the 150 Days of AP Calculus curriculum, you’ll find a list of “Questions to Include”. These are not meant to be prescriptive or exhaustive, but to give you some ideas
for problem stems you can incorporate into a particular unit.
We like to include an error analysis question on each test. This is best done by labeling each line of a sample response and asking students in which step the sample student made his first mistake.
If assessing this in a multiple choice question, make sure one of the options is “there is no mistake”. For help on what common student error to highlight, check out the “Student Misconceptions”
heading on the lesson posts.
The goal is to match the style of questions that students will see on the AP Calculus Exam. This includes reasoning using multiple representations, and focusing on conceptual understanding, not just
procedures. One way to do this is by asking questions that address underlying structure and that can’t be solved using a known procedure--such as writing an equation for a function that has a
particular discontinuity at a certain x-value. These questions tend to ask students to be generative and there are usually many right answers. To make a question like that less open-ended, you can
have students solve for the value of a parameter in an equation so that a certain condition is met. (Ex: Find the value of k so that f(x) has a removable discontinuity at x=4). The best place to find
questions or models for questions is AP Classroom, or other released AP Calculus exam items.
We recommend that you prepare a rubric for the free response items before you begin grading your quizzes or tests. Know what information is necessary for a complete and correct response and award
points when a student presents that information. Many of the “Why did I get marked down?” questions are eliminated when you share the components that earn points.
Curving test grades in AP Calculus is common practice, and we do it on some, but not all, of our unit assessments. Generally, we don’t curve Units 1 and 2 (they’ve seen a lot of this content in
Precalculus) and then start curving tests in Unit 3 if necessary. While we acknowledge students’ learning progressions throughout the year and don’t use strict AP grading on day 1, we do push
students to be precise and rigorous in their communication. It takes time for students to get used to this caliber of assessment, so test scores tend to be lower than they’re used to from previous
math courses. Upholding this level of accountability in our grading is our primary reason for curving test scores. While there are many methods for curving scores, we prefer the method of linearizing
the data.
Going over tests
Assessments are most valuable for students when feedback is timely. For this reason, we return all quizzes and tests the day after students take them in class. Yes, that seems like a formidable night
of grading, but by batch-grading, I find I can get through them much more quickly. I record only the bare minimum on their papers and exclude lengthy comments. On a free response they will see how
many points they earned on each part and in class we will discuss the specifics of what was necessary for full points. We spend about 15-20 minutes going over assessments in class. First, students
work in groups to make test corrections on their own and I circulate and answer group questions only. Then I usually bring the class together to talk through the rubric I used for the free response
question or to highlight some frequently missed questions or common errors. | {"url":"https://blog.mathmedic.com/post/assessments-in-ap-calculus","timestamp":"2024-11-08T05:38:55Z","content_type":"text/html","content_length":"866729","record_id":"<urn:uuid:8605e08f-b194-49fb-8e95-aae21bbd91eb>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00681.warc.gz"} |
Where can I find assistance with my Electromagnetic Fields and Waves computational assignments? | Pay Someone To Do Electrical Engineering Assignment
Where can I find assistance with my Electromagnetic Fields and Waves computational assignments? Please help my colleagues by replying your queries to ask specific questions for help before answering
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first! Hi, I’m trying to get programming ideas to attach to a single piece model and I’m currently trying to do both of two models using the Electromagnetic Fields Equilibration and Waves
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stored, but I may be missing some more parameters, and be more selective with regards to what sets of values are encoded and how. My real world applications: Dynamic solar wind Kinematics of
spacecraft or space vessels Modeling solar systems that are heavy material Mass balance of solar wind in the solar cycle Syntherical control of solar system operations using Wind Electric Fields
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applications using wind or spacecraft in the solar wind and wouldWhere can I find assistance with my Electromagnetic Fields and Waves computational assignments? I tried to find any suggestions or
information related to Electromagnetic Field Units, from the Electromagnetic Domain Knowledge Group, but had no success. Could someone please keep me posted on any other, unrelated articles or
questions. A: The author has a solution: Find the electrical fields which can be interpreted modulated by radiation such as that from an electric charge. The electrical field is the most practical
source of radiation in terms of total thermal radiation, but it is expensive Continued some uses; see the CNC (charge center of mass) diagram here. So if you have a bunch of electric charge that have
total thermal radiation to carry and some fraction that is gamma; and they are modulated by radiation, and radiated by these thermal charges, then there is a total linear range of the total radiation
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a full Fourier plot and plot the result below as a graphical representation of that plot. For this calculation, the FouWhere can I find assistance with my Electromagnetic Fields and Waves
computational assignments? In general, I currently have a bunch of electromagnetic fields that I am using for my floating block (or floating gas) engine. Is there anything I need to add which will
help me understand how they develop into my floating block? Thanks! A: First off, the floating block only has the name “FEM” and the electronic name “electric” When an electronics schematic is made
up, do an electron backscatter and you will get the next address to what you want is the next energy state (that’s the state in which the charge transfer occurs). Next, what you should be looking for
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state). How to implement the electrons back to the Electromagnetic Fields for floating block? In my case I want a logic/output output every 60 seconds for the full I’m going to be in one particular | {"url":"https://electricalassignments.com/where-can-i-find-assistance-with-my-electromagnetic-fields-and-waves-computational-assignments","timestamp":"2024-11-09T07:46:17Z","content_type":"text/html","content_length":"145887","record_id":"<urn:uuid:dd30fbfc-8f34-46fb-a275-992841205177>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00371.warc.gz"} |
How to find all structures with the most equidistribution for a specific $f$ and $A$?
Suppose $f:A\to\mathbb{R}$ and $A$ is countably infinite subset of the real numbers. We can enumerate $A$ as $\left\{a_n\right\}_{n=1}^{\infty}$ and set $t$ as a natural number but the average $$\lim
\limits_{t\to\infty}\frac{f(a_1)+f(a_2)+\cdot\cdot\cdot+f(a_t)}{t}$$ can vary depending on the enumeration. In order, to reduce the number of values the average varies with, I did the following:
Split $A$ into a sequence of finite subsets $F_1,F_2,...$ such that $F_1\subset F_2\subset...$ and $\bigcup\limits_{n=1}^{\infty}F_n=A$. The sequence of subsets will be denoted as a "structure of
$A$" and a natural example (such as for $A=\left\{\frac{1}{m}:m\in\mathbb{N}\right\}$) would be $$\left\{F_n\right\}=\left\{\frac{1}{m}:m\in\mathbb{N},m\le n\right\}$$ Following this, I want a
structure that has an average and the most equidistribution as defined below.
Section 1: Definition Of Average From Structure
There exists an average or $\text{avg}\left(\left\{F_n\right\},f\right)$, if for every arbitrarily small positive $\epsilon$ there exists a sufficiently large integer $N$ such for all $n\ge N$. $$\
left|\frac{1}{|F_n|}\sum\limits_{x\in F_{n}}f(x)-\text{avg}\left(\left\{F_n\right\},f\right)\right|\le \epsilon$$
Depending on the structure we choose, we may get a different average. However, for an intuitive average, we want a structure with most equidistribution as shown below:
Section 2: Steps To Determining The Structure With The Most Equidistriubtion
1. Arrange the values in $F_n$ from least to greatest and take the difference between consecutive elements. Call this $\Delta F_n$. (Note $\Delta F_n$ could be a multi-set.)
Ex 1.1: If $A=\left\{\frac{1}{m}:m\in\mathbb{N}\right\}$ and $\left\{F_n\right\}=\left\{\frac{1}{m}:m\in\mathbb{N},m\le n\right\}$ then $F_4=\left\{1,1/2,1/3,1/4\right\}$, and $\Delta F_4=\left\{1-1/
2,1/2-1/3,1/3-1/4\right\}=\left\{1/2,1/6,1/12\right\}$. (While this is not a multi-set, the example below is)
Ex 1.2: If $A=\mathbb{Q}\cap[0,1]$ and $\left\{F_n\right\}=\left\{\frac{j}{k}:j,k\in\mathbb{N},0\le j\le k, k\le n\right\}$ then the elements of $F_4$, arranged from least to greatest is, $F_4=\left\
{0,1/4,1/3,1/2,2/3,3/4,1\right\}$ and $\Delta F_4=\left\{1/4-0,1/3-1/4,1/2-1/3,2/3-1/2,3/4-2/3,1-3/4\right\}=$ $\left\{1/4,1/12,1/6,1/6,1/12,1/4\right\}$. Here the elements repeat making this a
2. Divide $\Delta F_n$ by the sum of all its elements so we get a distribution where all the elements sum to 1. Call this $\Delta F_n/\sum\limits_{x\in\Delta F_n}x$ or "the information probability of
the structure"
Ex 2.1: From example 1.1 note $\sum\limits_{x\in\Delta F_3}x=1/2+1/6+1/12=3/4$ and $\Delta F_3/\sum\limits_{x\in\Delta F_3}x=4/3\cdot \Delta F_3=\left\{2/3,2/9,1/9\right\}$. Note the elements in this
set sum to $1$ and can hence act as a probability distribution (even though the elements are not acutal probabilties)
3. We now want to calculate the deviation of the information probability from being a discrete uniform distribution to find the structure with the most equidistribution. (The smaller the deviation
the greater the structure's equidistribution.). Since elements of the multiset $\Delta F_n/\sum\limits_{x\in\Delta F_n}x$ always sum to $1$ we can calculate the deviation using Entropy (from
Information Theory). This can be written as
$$E(F_n)=-\sum\limits_{j\in\Delta F_n/\sum\limits_{x\in \Delta F_n}x}j\log j$$
Ex 3.1: From example 2.1, $E(F_3)$ is the same as $-\sum\limits_{j\in\left\{2/3,2/9,1/9\right\}}j\log j=-\left(2/3\log \left(2/3\right)+2/9\log \left(2/9\right)+1/9\log\left( 1/9\right)\right)\approx
The structure whose information probability gives the highest entropy has the greatest equidistribution. However, with this comes problems.
1. In order for an average to exist, some structures add more than one element as $n$ increases by one. This means when comparing different structures with a defined average, it is hard to trust
which structure has the most equidistribution unless the structures have the same number of elements.
In order to fix this suppose $\mathbb{S}(A)$ represents all structures of $A$ with a defined average from function $f$ (see Section 1).
If $s,n,j\in\mathbb{N}$, I wish to find all $\left\{G_s\right\}\in \mathbb{S}(A)$ where (for each one):
$\forall\left(\left\{F_n\right\}\in\mathbb{S}(A)\right)\exists\left(s\in\mathbb{N}\right):\forall\left(j\ge s\right)\exists\left(n\in\mathbb{N}\right)\left(E(F_n)\le E(G_j)\Rightarrow|G_j|\le|F_n|\
Note structures $G_j$ and $F_n$ are equivelant when for $\left(\forall n\right)\left(\forall j\right) \left(\bigg[ \left|G_n\right|\le\left|F_j\right| \Rightarrow G_n\subseteq F_j \bigg] \lor \
bigg[\left|F_j\right|\le\left|G_n\right|\Rightarrow F_j\subseteq G_n \bigg]\right)$
We will call the statement above this one
Statement A
(not to be confused with set $A$).
Here is an example below:
Ex: I believe for $A=\left\{\frac{1}{m}:m\in\mathbb{N}\right\}$ and $$f(x)=\begin{cases} 2 & x\in\left\{\frac{1}{2m+1}:m\in\mathbb{N}\right\} \\ 1 & x\in\left\{\frac{1}{2m}:m\in\mathbb{N}\right\} \
end{cases}$$ the only $G_s$ that satisfies
Statement A
is $\left\{1/\left[2^s/m\right]:1\le m\le 2^s\right\}$ (where $\left[\cdot\right]$ is the nearest integer function) but I'm not sure how to prove this? (If I'm not correct, I need to adjust Statement
1) For $A=\left\{\frac{1}{m}:m\in\mathbb{N}\right\}$ and the $f$ shown above, is the assumption that the only $G_s$ that satisfies
Statement A
is $\left\{1/\left[2^s/m\right]:1\le m\le 2^s\right\}$, correct? If not, how do we correct
Statement A
so there is only one $G_s$?
2) For question 1), is there a simpler way to calculate the structure with the most equidistribution that gives the same result as my assumption for $G_s$ without using Entropy?
3) How do we determine $G_s$ for other types of $A$ and $f$?
• When you say: "In order for an average to exist, some structures add more than one element as t increases by one", I believe you mean "as n increases by one". where n is the index for the sets in
the structure.
• @Jip333 You’re right. I made the edits.
• Correct me if I am misunderstanding, but it appears that G_s is not a structure on A. e.g. G_1 = {1, 2}, and 2 is not an element of A.
• @Pxsort I thought G1={1,1/2} since G1={1/[2/1],1/[2/2]} where [] is the nearest integer function.
• Oh I see.
• You seem to be implicitly assuming a lot of properties about A. Really, A cannot be an arbitrary countable set, since its elements are totally ordered and can be added / subtracted / multiplied /
divided. Effectively, this requires that A be a countable subset of the real numbers.
• @Pxsort I made edits
• Why do you keep asking the same question? What is it that motivates you?
• @AlessandroIcari I thought if my question is clearer I would get a different response. Now my question is more mathematical. In most cases such an average would not exist but I do believe for the
rationals and nowhere dense countably infinite sets one can find a unique G_s that has the most equidistribution.
• I don't think the problem with your question is that it lacks mathematical precision, I think the problem with it is that there is no satisfactory solution. And I think I already showed you
enough counterexamples and cases in which your problem can't be solved. Sometimes maths is like this, we don't always get what we hope for.
• @AlexandroIsacri Can you apply your counterexample to the mathematics in my post. Show that a unique G_s cannot exist.
• @AlexandroIscari I take that back, it’s clear for cases such as n*(-1)^n for integer n that an average cannot exist. But for other functions/their domain (such as the example in my post) if one
or more averages exist (depending on the domains structure) does their exist a unique G_s?
• I think the answer is "no" unless you add a lot of hypotheses. But we've already had a similar discussion, here the problem is essentially the same, the differences are not relevant enough to
change the answer.
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Table of Contents
Arx Data Types
Scalar Data Types
The basic data types of Arx can be subdivided in scalar and vector data types.
The scalar types are given in the table below:
Data type Explanation
bit The data type for binary signals. Possible values are 0 and 1
boolean A data type with possible values true and false
integer The integer as found in most common programming languages.
real Floating-point number; becomes float in C
Remark: Arx code that uses the real data type, cannot be converted into VHDL!
Integer constants can be specified in decimal, hexadecimal or binary notation. A hexadecimal constant starts with 0h and may use the characters 'a' to 'f' in either upper or lower case. A binary
constant starts with 0b. Hexadecimal and binary constants always designate a positive integer, unless preceded by a minus sign. Real constants can be used to specify fixed-point constants. This may
lead to loss of precision as a consequence of quantization and overflow processing.
The code fragement below shows some examples of the use of constants.
# three registers initialized with the same value
bval1: bitvector(8) = 0b10101010
bval2: bitvector(8) = 0haa
bval3: bitvector(8) = 170
# more examples of constants
bval4: unsigned(8) = 0haa
bval5: unsigned(8,2) = 1.75 # no loss of precision
bval6: signed(8,2) = -1.5 # no loss of precision
bval7: signed(8,4) = 3.14 # will be converted to 3.125 = 50/16
Vector Data Types
The vector types all have in common that they consist of a sequence of bits. They differ in the interpretation of these bits. The vector types are listed below:
Data type Explanation
bitvector(<n>) A vector of <n> bits
unsigned(<n>) An <n>-bit vector, interpreted as an unsigned integer
signed(<n>) An <n>-bit vector, interpreted as a signed two's complement integer
unsigned(<n>, <m>) An <n>-bit vector, interpreted as an unsigned fixed-point number with <m> integer bits
signed(<n>, <m>) An <n>-bit vector, interpreted as a signed fixed-point number with <m> integer bits
The index that addresses individual positions in a vector type, is always in the range 0 to <n>–1. Square brackets are used in the syntax for indexing.
Two indices separated by a colon and enclosed between square brackets select a subrange or slice of a bitvector. As opposed to other HDLs the left index should always be smaller or equal to the right
one (when both indices are equal a single bit is selected). A subrange can be used both at the left as well as the right-hand side of an assignment. The example below illustrates the addressing of
bits in a vector.
component top
word_length : generic integer = 8
T_IO : generic type = bitvector(word_length)
data_in : in T_IO
data_out : out T_IO
storage: T_IO = 0
storage[0] = storage[word_length-1]
storage[1:word_length-1] = data_in[0:word_length-2]
data_out = storage
Quantization and Overflow in Fixed-Point Data Types
Consider the case that a signal carrying a fixed-point value is wired to another signal with fewer bits. In Arx, this amounts to a fixed-point signal being assigned to another fixed-point signal
(variable or register). In such an assignment the binary points of the fixed-point patterns are always aligned. If the signal at the left-hand side of the assignment has fewer bits at the
least-significant side of the bit pattern, quantization is necessary. If there are fewer bits available at the most-significant side, overflow occurs.
One can deal in many ways with overflow and quantization. The desired behavior should be indicacted as part of the fixed-point type declaration with two optional parameters, respectively <o-mode> and
unsigned(<n>, <m>, <o-mode>, <q-mode>)
signed(<n>, <m>, <o-mode>, <q-mode>)
The following quantization modes exist:
Quantization mode Explanation
trunc Truncate (default).
round Round to the closest value; break ties by going up.
round_zero As round, but break ties by going towards zero.
round_inf As round, but break ties by going away from zero.
The following overflow modes exist:
Overflow mode Explanation
wrap Wrap around (default).
sat Saturate to the extreme values.
sat_sym Saturate symmetrical, the minimal extreme value is the negative of the positive extreme value.
Enumeration Data Type
As in many other languages for hardware description or programming, Arx allows the definition of new data types with a finite number of values. The possible values are identifiers that are enumerated
at the time of the declaration of the so-called enumeration data type.
input_state = enum(start, processing, ready)
The value of an enumerated type is indicated by preceding the declared value by the type name and a dot. Example:
# a registered signal of type input_state with its reset value
current_state: input_state = input_state.start
# later on in the code
if current_state == input_state.start
current_state = input_state.processing
Arx supports 1-dimensional arrays. The array elements can be any of the basic types mentioned above as well as enumeration data types. Arrays declarations contain a <size> enclosed in square
brackets, which gives the number of elements that the array will have. An array index runs from 0 to <size> – 1. Individual array elements are selected by an index enclosed in square brackets.
Initial values of registers of an array type should be provided in a list delimited by curly braces and separated by commas in order of increasing index. When a single value is provided, this value
is used as the initial value of all vector elements.
When the base type of an array is a vector, an individual bits or a bit slice can be selected in a similar way as a single vector. In the syntax, there are two pairs of indices enclosed in square
brackets. The first one selects the array element, the second the bits within the vector. The code below, shows an example:
The following code fragment illustrates the use of arrays:
component top
T_IO : generic type = signed(10, 5, sat, round)
data_in : in T_IO
data_out : out T_IO
T_enum: enum(one, two, three)
T_ar1: array[3] of T_IO
T_ar2: array[3] of T_enum
v1 : T_ar1 = 0
v2 : T_ar2 = {T_enum.three, T_enum.two, T_enum.one}
v3 : array[5] of T_IO = {5, 4, 3, 2, 1}
v1[1] = data_in
for i in 0:1
v2[i] = v2[i+1]
# example of accessing individual bits in an array of vectors
v3[0][0:4] = v1[2][5:9]
v3[0][5:9] = v1[2][0:4]
# rest of description left out
Explicit Type Conversion
In a situations where a signal of some type is wired to one of another type, Arx is able to check whether the type conversion is possible and insert overflow and quantization logic if necessary (see
also above). There exists, however, situations where one would like to impose a type to an intermediate result in an expression. In such a case the Arx function convert can be used.
The example below illustrates the use of convert; without it, the addition would be performed at the resolution of the widest operand instead of the narrowest one.
component top
T_IO : generic type = signed(10, 5)
data_in : in T_IO
data_out : out T_IO
T_narrow: signed(6, 4, sat, round)
storage: T_narrow = 0
storage = storage + convert(T_narrow, data_in)
data_out = storage
A sequence of bits can be interpreted in many ways. There are situations in which the interpretation of the same pattern changes across the hardware. The hardware itself does not change but the type
change needs to be made explicit in the hardware description. For these situations, Arx uses the function reinterpret.
The example below illustrates the use of reinterpretation. The input to the hardware consists of a bit vector composed of two signed numbers. In the description, the two numbers are first extracted,
added together, and the numerical result is then interpreted again as a bit vector.
component top
word_length : generic integer = 8
T_in : generic type = bitvector(2*word_length)
T_out : generic type = bitvector(word_length+1)
data_in : in T_in
data_out : out T_out
T_num : signed(word_length)
T_num_p1: signed(word_length+1)
storage: T_out = 0
left, right: T_num
sum: T_num_p1
left = reinterpret(T_num, data_in[0:word_length-1])
right = reinterpret(T_num, data_in[word_length:2*word_length-1])
sum = left + right
storage = reinterpret(T_out, sum)
data_out = storage | {"url":"https://bibix.nl/dokuwiki/doku.php?id=arx:datatypes","timestamp":"2024-11-06T07:17:12Z","content_type":"application/xhtml+xml","content_length":"29423","record_id":"<urn:uuid:9ab0cf2c-c835-447b-89ed-3f681c37899a>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00645.warc.gz"} |
How do you find the equation of an osculating circle?
To find the equation of an osculating circle in two dimensions, we need find only the center and radius of the circle. Find the equation of the osculating circle of the curve defined by the function
y=x3−3x+1 at x=1. Figure 12.4.
What is meant by osculating circle?
The osculating circle of a curve at a given point is the circle that has the same tangent as at point as well as the same curvature. Just as the tangent line is the line best approximating a curve at
a point , the osculating circle is the best circle that approximates the curve at. (Gray 1997, p. 111).
What is the geometric location of the center of the osculating circle of the curve?
Its center lies on the inner normal line, and its curvature defines the curvature of the given curve at that point. This circle, which is the one among all tangent circles at the given point that
approaches the curve most tightly, was named circulus osculans (Latin for “kissing circle”) by Leibniz.
What is the radius of curvature of a circle?
In differential geometry, the radius of curvature, R, is the reciprocal of the curvature. For a curve, it equals the radius of the circular arc which best approximates the curve at that point. For
surfaces, the radius of curvature is the radius of a circle that best fits a normal section or combinations thereof.
What is a Osculation?
: the act of kissing also : kiss.
What is unit of curvature?
Let’s measure length in meters (m) and time in seconds (sec). Then the units for curvature and torsion are both m−1. Explanation #1 (quick-and-dirty, and at least makes sense for curvature): As you
probably know, the curvature of a circle of radius r is 1/r. = m−1.
How do you get curvature?
1. Step 1: Compute derivative. The first step to finding curvature is to take the derivative of our function,
2. Step 2: Normalize the derivative.
3. Step 3: Take the derivative of the unit tangent.
4. Step 4: Find the magnitude of this value.
5. Step 5: Divide this value by ∣ ∣ v ⃗′ ( t ) ∣ ∣ ||\vec{\textbf{v}}'(t)|| ∣∣v ′(t)∣∣
What is the parametric form of circle?
The equation of a circle in parametric form is given by x=acosθ , y=asinθ The locus of the point of intersection of the tangents to the circle, whose parametric angles differ by π2.
Is the osculating circle of a parabola disjoint?
Within any arc of a curve C within which the curvature is monotonic (that is, away from any vertex of the curve), the osculating circles are all disjoint and nested within each other. This result is
known as the Tait-Kneser theorem. The osculating circle of the parabola at its vertex has radius 0.5 and fourth order contact.
How to find the equation of an osculating circle?
To get to the center of the osculating circle, travel a distance ρ along this normal vector from the point ( 1, 1). Now you know the center and radius of the osculating circle, so you can write down
its equation. Thanks for contributing an answer to Mathematics Stack Exchange!
What is the radius of curvature of a parabola?
the radius of curvature is. At the vertex γ ( 0 ) = ( 0 0 ) {\\displaystyle \\gamma (0)={\\begin{pmatrix}0\\\\0\\end{pmatrix}}} the radius of curvature equals R(0)=0.5 (see figure). The parabola has
fourth order contact with its osculating circle there.
Which is the circle with Center at Q called the osculating circle?
The circle with center at Q and with radius R is called the osculating circle to the curve γ at the point P . If C is a regular space curve then the osculating circle is defined in a similar way,
using the principal normal vector N. | {"url":"https://diaridelsestudiants.com/how-do-you-find-the-equation-of-an-osculating-circle/","timestamp":"2024-11-15T01:21:24Z","content_type":"text/html","content_length":"47985","record_id":"<urn:uuid:5669c29a-858f-41df-aff8-108d7112cd35>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00760.warc.gz"} |
Yeni bir finansal araç : opsiyonlar
Fen Bilimleri Enstitüsü
Bir hisse senedi opsiyonu, alıcıya, prim ya da opsiyon fiyatı olarak bilinen bir ödeme yapması karşılığında, belli bir dönem içinde, o hisse senedini satıcıdan (opsiyonu düzenleyenden) satma İma ya
da ona satma hakkı veren devredilebilir bir sözleşmedir. Opsiyonlar "türetilmiş men kul kıymetler" olarak anılırlar, çünkü opsiyon ancak ilgili menkul kıymetin var olması sonucunda ortaya çıkabilir
ve vade sonunda değerini yitirir. Opsiyonlar yüzyıllardır kullanılmalarına rağmen son yıllara kadar karanlıkta kalmış f inansal araçlardır. Opsiyon pazarındaki gelişmenin en önemli aşaması, ilk defa
opsiyon kontratlarının standartlaştırıldığı ve mübadele edilebilir bir hale getirildiği 26 Nisan 1973 tarihinde Chicago Opsiyon Borsası 'nın açılmasıdır. Bu tarihten sonra opsiyon pazarının gelişmesi
hızlanmıştır. Yeni opsiyon kontratlarının yenilikçiliği ve hızlı gelişimi dikkate değer sonuçlar doğurmuş ve finansal pazarlarda önemli etkiler yapmıştır. Bu kontratlar finansal pazarların
karmaşıklığını arttırmakla birlikte yatırımcılar için yeni fırsatlar da yaratmışlardır. Günümüzde opsiyonlar oldukça yaygın bir kullanım alanına sahiptirler. Opsiyonlar; hisse senetleri, dövizler,
hisse senedi indeksleri, borçlanma aracı niteliğindeki men kul kıymetler üzerine yazılabilmektedirler. Bu tezde ise hisse senedi opsiyonlar ı incelenmiştir. Bölüm 2 'de kısaca opsiyon tanımı verilmiş
ve opsiyonlar m risk yönetimindeki ve finansmandaki önemine değinilmiştir. Kısaca opsiyonların tarihçesi ve ne amaçlarla kullanılabileceği yine bu bölümün bir konusunu oluşturmak tadır. Bölüm 3 'de
geniş anlamıyla opsiyon kavramı, opsiyon kontratlarının ana parametreleri ve opsiyonların kotasyonu incelenmiştir. Bölüm 4 'de önce opsiyon fiyatının temel özellikleri verilip, bileşenleri
gösterilmiştir. Daha sonra da opsiyon fiyatlarını etkileyen faktörler incelenmiş ve opsiyon fiyatlandırma modelleri anlatılmıştır. Bölüm 5' de en popüler opsiyon stratejileri incelenmiş ve
opsiyonların stratejik rolü belirlenmeye çalışılmıştır. Bölüm 6 ise opsiyonlar için gerekli olan ihtiyat rezervi ve komisyon ücretlerinin incelendiği bir bölümdür. Oldukça yeni bir finansal araç olan
opsiyonlar günümüzde Türkiye koşulların da kullanılabilir değillerdir. Gerekli koşulların oluşturulması durumunda ise opsiyonların kullanılmasının yararlı olacağı açıktır.
Historically, the first type of option contracts to be traded consisted of put options on tulip bulbs that Dutch growers would purchase to protect the price of their crops. This example, dating back
to the seventeeth century, has been followed an erratic development, including common stock written options trading Great Britain during the eighteenth century and in the United States since the
beginning of the nineteenth century. During those days, however, option contracts were not standardized. The mar kets were not regulated and were often manipulated, lead ing on several occasions to
suspensions of option trading. The major breakthrough in the development of option market happened in April 1973, when the Chicago Board of Options Exchange began treading standardized call option
contracts on 16 common stocks and thereby offered the first regulated structure capable of providing the volume, the liquidity, and the solvability necessary to create efficient option market. Since
then, the Chicago Board of Options Exchange has also developed transactions in put options, and several other exchanges in the United States, such as the American Stock Exchange, the Philadelphia
Stock Exchange, and the New York Stock Exchange, have begun trading standardized call and put contracts written on common stocks. In terms of volume, the success of these instruments has been
tremendous, as can be observed through the fact that the number of option contracts traded since the early 1980s exceed s-in terms of their underlying stock equivalent- the number of shares traded on
the New York Stock Exchange. However, the most remarkable feature of the recent development of standardized option contracts has to be its expansion in the variety of option contracts written on
different types of underlying instruments that have been introduced in different exchanges since the beginning of 1980s. Indeed, stock written options are now just one type of option contract among a
large population consisting of stock index options, stock index futures options, fixed -income securities written options, foreign currency options, commodity options, and, so on. - vx - The success
of standardized option contracts can be explained by the negotiability, the liquidity, and the smaller transactions costs their centralized, regulated, and standardized trading has been able to offer
market participants. But to an even greater extent this success has to be found in the very nature of option and thereby in the 'original1 payoff structure and various risk-monito ring strategies
this financial instrument can provide to its potential buyers or sellers. Finally, it should be noted that although options have indeed gained interest among the academic community since the
introduction of standardized option contracts -with the pioneering development of the Black and Scholes option pricing formula in 1973- the analysis of options and the development of pricing models
to assess their 'fair,' or theoritical, value extend far beyond the con-r cept of the standardized option contracts. Indeed, options can be found 'everywhere' j they can be standardized or not,
traded or not traded -the option embedded in a callable bond. Some securities or financial decisions that do not even look like options at first sight can easily be analy zed and priced in the
general setting of option pricing theory. The purpose of this study to provide with a general understanding essential features of an option contract, and analysis of option strategies, and analysis
of option pricing. Broadly speaking, an option is a contract that entitles its owner with the right to buy or sell a speci fic quantity of underlying instrument -in this study the stock- at a
specific price and during a specific time period. The first part of this general definition of an option contract is related to the fact that this instru ment conveys a 'right,' not an. obligation,
to its owner? in other words, the choice of buying or selling the under lying instrument is optional and cannot be imposed on the option owner. This is very different from a futures cont ract or a
forward contract since these financial instru ments 'oblige' both parties -the buyer and the seller- to meet their obligations at the contract's expiration date. However, the preceeding definition of
option is still to vague at this stage since it does not explicitly emphasize that there are two types of options, namely call and put options. The former type of option conveys to its owner the
right to buy, and the latter gives its owner the right to sell the underlying instrument. When the owner of a call or put option decides to use the right to buy or sell the underlying instrument, it
can be said that he or she is exercising his or her option. The owner has the choice of deciding whether or not to exercise that right, but note that the counterpart to the - vii call or the purt
contract is obliged to sell (in the case of an exercised put) the underlying instrument. Hence it can be seen that the structure of payoffs an option provides to its owner and to its seller -the
counterpart is generally is called the option writer- are not the same since the former bears a right while the lat ter has the resulting obligation to eventually -if the owner exercises the option-
buy or sell the underlying security. Since the option writers provide a valuable finan cial instrument to the option owners, they will have to be compansated by a cash amount equal to the market
price of the option, also called the option premium. This price should represent the 'fair' compansation to the option writers for bearing for counterpart's obligation of even tually having to buy or
to sell the underlying instruments. However, it should be distinguished between those options that enable the owner to exercise his or her right at any time during the contracts life time and those
that only enable him or her to exercise the option at the expi ration date of the contract. The first category is called American type. The second category contains options that can only be exercised
at expiration and that are generally referred to as Euro pean Options. This geographical distinction is somewhat artificial nowadays since most standardized option cont racts trading in the world are
of the American type. In particular, a call or a put option written on a stock is characterized by the following: The size of the contract is the number of shares of stock a single option contract
allows to buy or sell. Gene rally, a single contract enables its owner to buy or sell 100 shares of the underlying stock. The exercise price of an option represents the fixed price the owner of the
call (put) will have to pay (receive) for one share of stock when the option is exerci sed. The time to maturity of an option contract is also standardized so it is observed that each stock has
options with three different maturities. The payout -protection rule indicates whether the terms of the contract are adjusted for some type of events that can dilute the stock price. The underlying
stock must be identified through its name and must also meet certain requirements. - viii - Finally, it should be noted that standardized stock options can be exercised at any time since they are
almost always of the American type. The price of an option can consist of intrinsic value, time value, or a combination of both. Intrinsic value is the in-the-money portion of an option's price. Time
value is the portion of an option's price that is in excess of intrinsic value. If the stock price is above the strike price of a call option, the stock price minus the strike price repre sents the
intrinsic value of the call option. For put options, intrinsic value equals strike pri ce minus stock price because put options are in-the-money when the stock price is below the strike price. There
are six components of an option's theoretical value. These are as follows? 1. The price of the stock 2. The strike price of the option 3. The time remaining until the expiration date 4. The
prevailing interest rates 5. The expected volatility of the stock 6. The dividents The relationship of the stock's price to the opti on's strike price determines whether the option is called
in-the-money, at the money, or out-of-the-money. The value of an option is an increasing function of the time remaining until the expiration date. There are two reasons for this. First, the
leveraging advantage in creases with time. Second, the opportunity for the stock price to far exceed the exercise price increases over time, Another factor that affects an option's price is the
prevailing interest rates. The rising interest rates boost call prices and depress put prices. However the effect of changes in interest rates is small. Mathematically, volatility is a measure of
stock price fluctiations without regard to direction. Specifi cally, volatility is the annualized standart deviation of daily percentage changes in a stock's price. The rela tionship between
volatility and option prices is a direct one-as the volatility percentage increases, so do option prices. Of the other five components of an option's theore tical value,. the stock price, strike
price, time until expiration, prexailing interest rates, and dividents are readily observable. Only the expected volatility of the underlying stock is unknown. - ix - Among the others, the
Black-Scholes option pricing model is the industry standart, widely used with or with out modifications by many traders to guide their trading decisions. It was the first exact option pricing formula
to be derived. The formula is obtained by construction of a riskless hedge using the option and its underlying securities and then solving the resulting equation for either the option price or the
hedge ratio. The formula for the Black-Scholes option pricing method is given by -R-T C = S-Nfd^ - E.e x N(d2) ln(S/E) + [Rf-(l/2)a2]T where d-, = o/T" d2 = dx - a/T N(d,), N(d2) = cumulative normal
probability values of d, and d2, respectively S = Stock price E = exercise price Rf= the risk-free rate of interest a = volatility T = time remaining until expiration. Options give a unique advantage
to investors who want to achieve their financial goals. They increase the number of ways these investors can manage their financial assets by giving their power to create positions that precisely
reflect their expectations of the underlying security and at the some time, balance their risk-reward tolerance. From speculation to hedging, options can be used to construct scenarios that maximize
payoff for outcomes the investor considers most likely. There are ways to express the degree of one's opinion, each with its own particular risk-reward profile. The trading of options in the equities
markets has long been associated with a broad range of trading strate gies. Seme of these are simple and straight forward others are complex and subtle, some for hedging purposes and ot hers to
speculate. The basic option strategies are; buying calls, selling calls, buying puts and selling puts, the other trading strategies are created by combining these for strategies.
Tez (Yüksek Lisans) -- İstanbul Teknik Üniversitesi, Fen Bilimleri Enstitüsü, 1991
Anahtar kelimeler
İşletme, Hisse senetleri, Opsiyon, Business Administration, Stocks, Option | {"url":"https://polen.itu.edu.tr/items/2f60a499-391f-44ad-8ae4-389173228570","timestamp":"2024-11-04T01:42:45Z","content_type":"text/html","content_length":"184460","record_id":"<urn:uuid:2f2096c9-dc90-491b-a031-5501043ce17b>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00756.warc.gz"} |
Lattice Graph -- from Wolfram MathWorld
A lattice graph, also known as a mesh graph or grid graph, is a graph possessing an embedding in a Euclidean space tiling. Examples include grid graphs and triangular grid graphs.
Rook graphs are sometimes also known as lattice graphs (e.g., Brouwer). Another class of graph sometimes given this name are the "lattice graphs" of Ball and Coxeter (1987, p. 305) obtained by taking | {"url":"https://mathworld.wolfram.com/LatticeGraph.html","timestamp":"2024-11-05T23:32:44Z","content_type":"text/html","content_length":"52013","record_id":"<urn:uuid:16a60c8a-1932-4c51-bb0d-ab88e6983656>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00257.warc.gz"} |
Gennady Samorodnitsky
According to our database
, Gennady Samorodnitsky authored at least 36 papers between 1982 and 2024.
Collaborative distances:
Book In proceedings Article PhD thesis Dataset Other
On csauthors.net:
Spectral learning of multivariate extremes.
J. Mach. Learn. Res., 2024
Generalized Pareto GAN: Generating Extremes of Distributions.
Proceedings of the International Joint Conference on Neural Networks, 2024
Large deviations for subcomplex counts and Betti numbers in multiparameter simplicial complexes.
Random Struct. Algorithms, September, 2023
Extreme Bandits Using Robust Statistics.
IEEE Trans. Inf. Theory, March, 2023
Empirical Risk Minimization for Losses without Variance.
CoRR, 2023
Epsilon*: Privacy Metric for Machine Learning Models.
CoRR, 2023
A Cover Time Study of a non-Markovian Algorithm.
CoRR, 2023
Adaptive Privacy Composition for Accuracy-first Mechanisms.
Proceedings of the Advances in Neural Information Processing Systems 36: Annual Conference on Neural Information Processing Systems 2023, 2023
Kernel PCA for multivariate extremes.
CoRR, 2022
On Penalization in Stochastic Multi-armed Bandits.
CoRR, 2022
Catoni-style Confidence Sequences under Infinite Variance.
CoRR, 2022
Detection of Small Holes by the Scale-Invariant Robust Density-Aware Distance (RDAD) Filtration.
CoRR, 2022
Regret Analysis for RL using Renewal Bandit Feedback.
Proceedings of the IEEE Information Theory Workshop, 2022
Nearly Optimal Catoni's M-estimator for Infinite Variance.
Proceedings of the International Conference on Machine Learning, 2022
Minimax M-estimation under Adversarial Contamination.
Proceedings of the International Conference on Machine Learning, 2022
℘-MinHash Algorithm for Continuous Probability Measures: Theory and Application to Machine Learning.
Proceedings of the 31st ACM International Conference on Information & Knowledge Management, 2022
Consistent Sampling Through Extremal Process.
Proceedings of the WWW '21: The Web Conference 2021, 2021
Epidemic threshold and lifetime distribution for information diffusion on simultaneously growing networks.
Proceedings of the ASONAM '19: International Conference on Advances in Social Networks Analysis and Mining, 2019
Contact distribution in a thinned Boolean model with power-law radii.
J. Appl. Probab., 2017
Nonstandard regular variation of in-degree and out-degree in the preferential attachment model.
J. Appl. Probab., 2016
Poisson Process Driven Stochastic Differential Equations for bivariate heavy tailed distributions.
Proceedings of the 2016 American Control Conference, 2016
General inverse problems for regular variation.
J. Appl. Probab., 2014
Fractional Moments of Solutions to Stochastic Recurrence Equations.
J. Appl. Probab., 2013
Sign Stable Projections, Sign Cauchy Projections and Chi-Square Kernels.
CoRR, 2013
Sign Cauchy Projections and Chi-Square Kernel.
Proceedings of the Advances in Neural Information Processing Systems 26: 27th Annual Conference on Neural Information Processing Systems 2013. Proceedings of a meeting held December 5-8, 2013
Scaling Limits for Cumulative Input Processes.
Math. Oper. Res., 2007
Modeling teletraffic arrivals by a Poisson cluster process.
Queueing Syst. Theory Appl., 2006
Found. Trends Stoch. Syst., 2006
Limit Behavior of Fluid Queues and Networks.
Oper. Res., 2005
Variable heavy tails in Internet traffic.
Perform. Evaluation, 2004
Variable Heavy Tailed Durations in Internet Traffic, Part I: Understanding Heavy Tails.
Proceedings of the 10th International Workshop on Modeling, 2002
Optimal and Near-Optimal ( <i>s, S</i>) Inventory Policies for Levy Demand Processes.
RAIRO Oper. Res., 2001
Activity Periods of an Infinite Server Queue and Performance of Certain Heavy Tailed Fluid Queues.
Queueing Syst. Theory Appl., 1999
Heavy Tails and Long Range Dependence in On/Off Processes and Associated Fluid Models.
Math. Oper. Res., 1998
Performance Decay in a Single Server Exponential Queueing Model with Long Range Dependence.
Oper. Res., 1997
An efficient mixture method.
Oper. Res. Lett., 1982 | {"url":"https://www.csauthors.net/gennady-samorodnitsky/","timestamp":"2024-11-02T15:35:09Z","content_type":"text/html","content_length":"49779","record_id":"<urn:uuid:86815937-5e2b-40ef-a53a-506984145ac1>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00696.warc.gz"} |
How to Evaluate and Compare Experimental Designs
Experimental designs are evaluated and compared using the following criteria:
The most important test of any design is that the questions it generates are sensible and answerable. See Choice of Experimental Design Algorithm for an illustration of this approach.
A design is said to be balanced when all the attributes appear the same number of times. When comparing designs, an experimental design is more balanced if the frequencies are closer to this ideal.
When comparing designs it is useful to summarize all the balance information as a single metric. One measure is the Mean Version Balance, which is calculated as follows:
• The balance of an attribute is defined as the sum across levels of the absolute differences between the counts and the average count.
• This value is then scaled to give a value of 1 if all absolute differences are zero (i.e. the count of every level is the same as its mean, which implies perfect balance).
• The worst design consists of repeating the same levels for all alternatives and has a balance of zero.
• This is calculated for each version of the design and then averaged.
The Across Version Balance is the balance for the whole design across all versions, then averaged across all attributes.
All else being equal - and it isn't always - the more balanced a design, the better. For examples of this, see Choice of Experimental Design Algorithm.
Pairwise balance
Pairwise balance is balance, except that it relates to the balance of pairs of attribute levels.
The Mean Version Pairwise Balance is the same as the Mean Version Balance, except that is calculated for across all pairs of attributes. The Across Version Pairwise Balance is the pairwise version of
the Across Version Balance. These are not meant for designs that don't see balance (e.g., alternative specific and partial-profile designs).
Overlap refers to the extent to which an attribute level appears in the same question. For example, if there are there alternatives and each none of the alternatives have attribute levels in common,
then there is no overlap. Where there are more alternatives than attributes, there will always be some overlap.
The less overlap, the more information that is obtained from data using the design. However, designs with no overlap are cognitively harder and many researchers believe that a degree of overlap is
Standard errors
Designs can also be compared by calculating the expected standard errors of the parameter estimates. The lower better. A simple summary measure that can be used for comparing designs is the average
of each design's standard errors.
A better, but more complicated summary of standard errors is the d-error. A lower d-error indicates a better design. It is usual to compare d-errors for designs created using different algorithms,
rather than consider the number by itself.
The d-error is a similar idea to the effective sample size. A design with a smaller d-error requires fewer respondents to provide the same level of precision.
Although d-error is appealing in principle, the actual d-error that can be practically calculated assume a very simple model (either linear regression or multinomial logit), and such models are known
to be inappropriate for choice experiments. So, while there is value in comparing designs based on d-error it is by no means a definitive metric. This is also true of standard errors.
Using simulations to compare designs
A more complicated approach to comparing designs is to:
• Hypothesize the likely results of the experiment. A lazy but useful hypothesis is that all the utilities will be 0.
• Simulate data consistent with the hypothesized results.
• Fit models to the data (e.g., hierarchical Bayes), and evaluate:
□ The difference between key results (e.g., the estimated utilities and the hypothesized utilities).
□ Prediction accuracy.
For instructions on how to do this using Displayr, see:
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Engineering Mathematics by V.P. Mishra: A Must-Have for Ever
Engineering Mathematics by V.P. Mishra: A Must-Have for Every Engineer - Barnes & Noble
# Engineering Mathematics by V.P. Mishra: A Comprehensive Textbook for Engineering Students - ## Introduction - What is engineering mathematics and why is it important? - Who is V.P. Mishra and what
are his credentials? - What are the main features and benefits of his textbook? - ## Overview of the Book - How is the book organized and structured? - What are the topics and subtopics covered in
each chapter? - How are the concepts explained and illustrated with examples and exercises? - ## Chapter 1: Differential Calculus - What are the basic concepts and rules of differentiation? - How to
apply differentiation to engineering problems such as optimization, rate of change, and curve sketching? - What are the special functions and techniques of differentiation such as implicit,
logarithmic, exponential, and inverse trigonometric functions? - ## Chapter 2: Integral Calculus - What are the basic concepts and rules of integration? - How to apply integration to engineering
problems such as area, volume, work, and average value? - What are the special functions and techniques of integration such as integration by parts, substitution, partial fractions, and trigonometric
identities? - ## Chapter 3: Differential Equations - What are differential equations and how to classify them? - How to solve first-order differential equations using various methods such as
separation of variables, integrating factors, exact equations, and linear equations? - How to solve higher-order differential equations using various methods such as homogeneous equations,
undetermined coefficients, variation of parameters, and Cauchy-Euler equations? - ## Chapter 4: Series and Sequences - What are series and sequences and how to represent them using sigma notation? -
How to test the convergence and divergence of series and sequences using various criteria such as ratio test, root test, comparison test, and alternating series test? - How to find the sum of
infinite series using various formulas such as geometric series, arithmetic series, telescoping series, and power series? - ## Chapter 5: Complex Analysis - What are complex numbers and how to
perform arithmetic operations on them using rectangular and polar forms? - How to plot complex numbers on the complex plane and find their modulus and argument? - How to use De Moivre's theorem and
Euler's formula to find the roots and powers of complex numbers? - ## Chapter 6: Linear Algebra - What are matrices and vectors and how to perform arithmetic operations on them using addition,
subtraction, multiplication, scalar multiplication, and transpose? - How to find the determinant, inverse, rank, trace, eigenvalues, and eigenvectors of a matrix using various methods such as
cofactor expansion, row reduction, characteristic equation, and diagonalization? - How to solve systems of linear equations using various methods such as Gaussian elimination, matrix inversion,
Cramer's rule, and matrix factorization? - ## Chapter 7: Vector Calculus - What are vector functions and how to differentiate and integrate them using chain rule, product rule, quotient rule, and
fundamental theorem of calculus? - How to find the arc length, curvature, torsion, tangential and normal components of a vector function using various formulas such as arc length formula, curvature
formula, torsion formula, TNB frame, etc.? - How to use vector calculus to study scalar fields and vector fields using various concepts such as gradient, divergence, curl, directional derivative,
line integral, surface integral, volume integral, Green's theorem, Stokes' theorem, Divergence theorem, etc.? - ## Chapter 8: Fourier Analysis - What are Fourier series and how to find them for
periodic functions using trigonometric functions or complex exponentials? - How to use Fourier series to approximate non-periodic functions using half-range expansions or odd/even extensions? - How
to use Fourier transform to convert functions from time domain to frequency domain or vice versa using various properties such as linearity, scaling, shifting, convolution, modulation, etc.? - ##
Chapter 9: Laplace Transform - What is Laplace transform and how to find it for various functions using definition or tables of common transforms? - How to use Laplace transform to solve differential
equations or integral equations using various properties such as linearity, scaling, shifting, differentiation, integration, convolution, etc.? - How to use inverse Laplace transform to find the
original function from the transform using partial fraction decomposition or residue theorem? - ## Chapter 10: Probability and Statistics - What are probability and statistics and how to use them to
analyze data and make inferences? - How to use various concepts and tools of probability such as sample space, events, axioms of probability, conditional probability, Bayes' theorem, random
variables, probability distributions, expectation, variance, etc.? - How to use various concepts and tools of statistics such as descriptive statistics, inferential statistics, hypothesis testing,
confidence intervals, correlation, regression, etc.? - ## Chapter 11: Numerical Methods - What are numerical methods and why are they useful for solving engineering problems? - How to use various
numerical methods for finding roots of equations such as bisection method, Newton-Raphson method, secant method, etc.? - How to use various numerical methods for solving differential equations such
as Euler's method, Runge-Kutta method, predictor-corrector method, etc.? - ## Chapter 12: Optimization Techniques - What are optimization techniques and how to use them to find the optimal solution
for engineering problems? - How to use various optimization techniques for unconstrained problems such as gradient descent method, Newton's method, conjugate gradient method, etc.? - How to use
various optimization techniques for constrained problems such as Lagrange multipliers method, Kuhn-Tucker conditions, linear programming, etc.? - ## Conclusion - Summarize the main points and
benefits of the book - Provide some feedback and suggestions for improvement - Encourage the readers to buy the book and learn more about engineering mathematics - ## FAQs - Who is the target
audience of this book? - How can I access the solutions of the exercises in this book? - What are the prerequisites for reading this book? - How can I contact the author of this book? - Where can I
buy this book?
Engineering Mathematics By Vp Mishra
Download: https://www.google.com/url?q=https%3A%2F%2Fjinyurl.com%2F2ud3cO&sa=D&sntz=1&usg=AOvVaw1ic96f56sAtfj1eUKltoF4 | {"url":"https://www.weempowerleadership.com/group/mysite-200-group/discussion/365c5df5-b46d-4121-a305-8440439ab1df","timestamp":"2024-11-01T23:36:10Z","content_type":"text/html","content_length":"1050599","record_id":"<urn:uuid:d0055411-73a2-4819-b518-ed6260057ee2>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00849.warc.gz"} |
Cost per unit of volume
This online calculator calculates cost per unit of volume (cubic unit) given length, width and height of a strip
Let's suppose you look at the strip of material given its length, width and height, like 254x76.2x8. You know its cost, f.e. 19000.00.
To be able to compare this cost to cost of another strip of different size you need to calculate cost per unit of volume. For example, if you use millimeters for dimensions, you need the cost per
cubic millimeter for first and for the second strip. Then you can compare them.
Cost per unit of volume (cubic unit) can be obtained by multiplying the dimensions (to get the volume of a rectangular parallelepiped) and dividing the result by cost of strip.
$C_{cubic unit}=\frac{L \cdot W \cdot H}{C_{strip}}$
Of course, in the calculator below you can use any units, not only millimeters, as long as you enter all dimensions in the same units. Thus, if you use inches, you get cost per cubic inch, feet -
cost per cubic foot and so on.
Similar calculators
PLANETCALC, Cost per unit of volume | {"url":"https://embed.planetcalc.com/8614/","timestamp":"2024-11-09T06:38:18Z","content_type":"text/html","content_length":"33277","record_id":"<urn:uuid:fa001c3a-63b0-4092-9482-722f83f221e2>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00168.warc.gz"} |
What Math Do 3rd Graders Learn - 3rd Grade Math Worksheets
Math Graph Worksheets 3rd Grade – The saying goes that “A journey of a thousand miles starts with a single step.” This adage perfectly applies to the learning process of mathematics in 3rd grade.
This is an important phase in which students learn more complex math concepts. The reason 3rd grade Mathematics is Important The … Read more
3rd Grade Math Worksheets K5 Learning
3rd Grade Math Worksheets K5 Learning – The saying goes that “A journey that spans a thousand miles begins with a single step.” This saying is a perfect fit for the process of learning math in the
3rd grade. It’s a critical phase where students are beginning to understand more advanced math concepts, which are … Read more
Math Graphing Worksheets 3rd Grade
Math Graphing Worksheets 3rd Grade – The saying goes, “A journey that will be a thousand miles starts by taking just one step.” This saying is a perfect fit for the journey of learning mathematics in
3rd grade. It is a crucial time in which students begin to master more complex math concepts. Importance of … Read more
K12 Math Worksheets 3rd Grade
K12 Math Worksheets 3rd Grade – The old saying goes “A journey of one thousand miles starts with one step.” This is an adage that can be applied to the journey that students embark on while studying
math in the third grade. It is a crucial time for students to begin learning more complex math … Read more
Math Geometry Worksheets 3rd Grade
Math Geometry Worksheets 3rd Grade – According to the old saying, “A journey that will be a thousand miles starts by taking only one step.” This is an adage that perfectly describes the process of
learning math in the 3rd grade. In the third grade, students learn new math concepts that are more sophisticated. Third … Read more | {"url":"https://www.3stgrademathworksheets.com/tag/what-math-do-3rd-graders-learn/","timestamp":"2024-11-07T04:36:14Z","content_type":"text/html","content_length":"69383","record_id":"<urn:uuid:4abe4a2d-bfca-45b0-8830-eb52b7026a24>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00092.warc.gz"} |
The course studies theory of functions of one variable. In particular, we focus on the properties of continuity, derivability, and integration of functions. As soon as the student understands these
concepts, they are applied to the study of problems of interest in Economy, such as monotony and convexity, graphic representation, polynomial approximation, optimization and calculus of areas. The
program is divided in five big lessons: Lesson 1: elementary properties of functions. In particular, it is studied when a function is periodic, monotone, shows symmetries or has an inverse. Lesson 2:
continuity. In particular, it is studied when a function has limits and /or asymptotes, the calculus of intersection points of graphics and the existence of maxima and minima. Lesson 3:
differentiability, part one. We study the calculus of derivatives, stressing implicit differentiation. In the same way, we apply derivatives to study monotony and the calculus of maxima and minima.
Lesson 4: differentiability, part two. We use the concept of derivative to compute limits, to approximate locally a function by polynomials, to characterize concavity and convexity of a function and
for an introductory study of the income, cost and profit functions. Lesson 5: Integration. First of all, we introduce the concept of primitive of a function, and we study different methods of
computing them. Secondly, we introduce the concepts of area and integral, and its relationship with the concept of primitive function. In a third step, we study the calculus of areas. Finally, we
study improper integrals. | {"url":"https://aplicaciones.uc3m.es/cpa/generaFicha?est=201&anio=2023&plan=399&asig=14441&idioma=2","timestamp":"2024-11-06T02:50:37Z","content_type":"text/html","content_length":"13759","record_id":"<urn:uuid:ee049958-3d85-4084-ba7a-01bbf5e88669>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00312.warc.gz"} |
Convert Binary Number To One
Given a binary string S, the task is to print the numbers of steps required to convert it to one by the following operations:
If S is odd add 1 to it.
If S is even divide it by 2.
Input: S = “101110”
Output: 8
Number ‘101110’ is even, after dividing it by 2 we get an odd number ‘10111’ so we will add 1 to it. Then we’ll get ‘11000’ which is even and can be divide three times continuously in a row and get
’11’ which is odd, adding 1 to it will give us ‘100’ which is even and can be divided 2 times in a row. As, a result we get 1.
So we had to do any of the above two operations 8 times to reach 1 from the given binary number..
Let's take some examples and see how it works:
| /2
| +1
| /2
| /2
| /2
| +1
| /2
| /2
Few observations here:
• When we divide an even number we actually do a right shift. Which means each division will remove the right-most trailing zero.
So, given an even number how many times can we go on doing division by 2 ? As many trailing (right-most) zeroes we have. In the above example see how when we got 11000 it gets reduced to 11 by
doing division by 2 every time, ultimately making it an odd number. So when we have an even number we would take as many steps as there are trailing zero and reduce the number to a number without
the trailing zeroes. So 11000 reduces to 11. Basically each time we encounter an even number we remove the right-most 0 from the binary number.
• When we have ood number we add one to it. Let's see how it works:
111011 + 1 = 111100
11000011 + 1 = 11000100
10001 + 1 = 10010
11 + 1 = 100
Did you see a pattern here? Adding one to an odd number toggles all the trailing ones (by toggle I mean convert all the trailing ones to zeroes) and converts the first zero from right to one,
keeping all other left digits the same.
Interesting to see how 11 is converted to 100 by addition of one. It still follows the observation described above.
So when we encounter an odd number, we could combine these steps: combine addition of one and then required number of division by 2.
Java Code:
Please subscribe to access the complete working solution.
Python Code:
Please subscribe to access the complete working solution.
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Heilbronn Colloquium 2019: Zeev Rudnick
30 Jul 2019, by in Events
Thursday 5 September 2019 at 16:00
Organised in collaboration with the School of Mathematics, University of Bristol, UK
Quantum Chaos, Eigenvalue Statistics and the Fibonacci Sequence
Zeev Rudnick, Professor of Mathematics, Tel Aviv University, Israel
One of the outstanding insights in the field of “Quantum Chaos” is a conjectural description of local statistics of the energy levels of simple quantum systems according to crude properties of the
dynamics of classical limit, such as integrability, where one expects Poisson statistics, versus chaotic dynamics, where one expects Random Matrix Theory statistics. These insights were obtained by
physicists in the last quarter of the 20-th century (much of it in Bristol!). However, mathematicians are far behind in understanding the scope and validity of this theory. The first part of the
lecture will be dedicated to an introduction to these conjectures, which I believe deserve to be better known in the mathematics community.
In the second part, I will describe more recent work on statistics of the minimal gap between the first N eigenvalues for one such simple integrable system, a rectangular billiard having irrational
squared aspect ratio. When the aspect ratio is the “golden ratio”, the problem involves some curious and entertaining properties of the Fibonacci sequence.
About the Speaker: Zeev’s main interests are in number theory, mathematical physics and quantum chaos, and his world-leading research in these fields has attracted numerous awards and prizes. For
example, in 2001 he received the Erdos Prize of the Israel Mathematical Union and he was elected a Fellow of the American Mathematical Society in 2012. He was an invited speaker at the International
Congress of Mathematicians in 2014 and his current research is supported by an ERC Advanced Grant.
For more information please email the Heilbronn events team at heilbronn-coordinator@bristol.ac.uk
Information on past and future colloquia is available here
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What is Logical Representation in AI? | Kinnu
Logical representation provides machines with the capacity for representing their world and for logical reasoning
Statements in propositional logic can be either...
Predicate logic is able to...
Understand categories and variables
We can express the universal quantifier with which symbol?
You don’t always get what you want
Intelligent AI agents can perceive their environment and take actions autonomously to achieve their goals, but they require world knowledge to make good decisions.
A **‘knowledge base’** is a set of sentences expressed in a ‘knowledge representation language’ that tells the AI about its world. Knowledge-based agents use a process of drawing inferences
(reasoning) over their internal representation of knowledge to decide which actions to take.
Take shopping as an example. An online retailer typically tracks what you put in your basket, even if you later take it out, along with your address, the time of day, and your payment method. The
agent uses this highly personal data to constantly refine its performance, offering you new items to see what you will do.
You may have started out looking for a kettle but left with a nasal hair trimmer.
Knowledge representation languages
**Knowledge representation languages** capture assertions about the world in a form that allows AIs to reason in order to complete complex tasks and solve problems. They are defined by their syntax,
specifying the structure of the sentences, and their semantics, expressing the sentence truth in the world. These languages also allow for ‘entailment’ – the idea that one sentence logically follows
from another.
**For an AI agent to learn, it must be able to add to the language and derive new sentences from old ones.**
When an agent begins an action, it updates the knowledge base with what it perceives, asks what action it should perform, and is told what to do in response to reasoning about the current state of
its world.
‘Declarative methods’ involve telling the knowledge-based programs what they need to know to operate in their environment, while ‘procedural methods’ provide detailed coding to teach desired
Successful AI agents typically combine both declarative and procedural approaches within their design and include mechanisms for them to learn unsupervised.
Propositional logic
**‘Propositional logic’** is a branch of mathematical logic that uses a set of propositions, or statements, to represent objects in an imagined or real world and the relationships between them. Each
statement can be true or false. It combines rules for sentence structure, or syntax, with those for defining their logical truth, or semantics.
Put simply, propositional logic is a language for asserting things as **true or false**. It is also sometimes known as Boolean logic.
Such logic combines propositional symbols and logical connectives, such as ‘and’ and ‘or,’ and is used to analyze declarative statements, known as propositions, that make a statement about the world.
Individually or taken together, statements are assigned a truth value that cannot, under any circumstances, be ambiguous.
How to think about propositional logic
Here are some example statements of propositional logic:
*The sun sets in the west*
*The sun sets in the east*
*a2 + b2 = c2*
*a > b*
*b < a*
Each of these statements can either be true or false (or indeed unknown).
Propositional logic is **limited in what it can say**. It does not have the expressive power to concisely deal with time and space or represent universal patterns or relationships among objects.
As a result, it does not scale well; while it may be useful for specific, well-defined problem-solving, known as ‘bounded tasks’, it is less useful for open-ended and complex work, known as
‘unbounded tasks’.
Drawing inferences in propositional logic
Propositional logic allows for complex sentences to be constructed from simple ones, making use of a set of clearly defined operators known as ‘logical connectives.’
These logical connectives are essential for writing propositional statements. They are listed here with their logic symbols, and they include:
**not(¬)** – a way to ‘negate’ the statement
**and (∧)** – a ‘conjunction’ combines 2 sentences
**or (V)** – a ‘disjunction’ indicates the relationship between 2 distinct alternatives
**implies (=>)** – an ‘implication’ involves both a premise and a conclusion, similar to an if-then statement- ‘If I eat too much then I will be sick.’ This would be written as ‘Eating too much
implies me being sick.’
**If and only if (<=>)** – returns a true value only if both sides satisfy one another.
Inferences explained
The basic inference rule in propositional logic is known as **‘Modus Ponens.’** This is a kind of logic that works to infer a truth from two other propositions.
‘Modus Ponens’ can be expressed as follows, where P and Q both represent separate sentences:
*If P, then Q*
*Therefore Q*
To think about that more clearly, let’s imagine P is the sentence ‘It is a car’, and Q is the sentence ‘It has wheels’:
*If it is a car, then it has wheels*
*It is a car*
*Therefore, it has wheels*
This inference is logically sound in any world in which the premise of part one is true – so unless you’re in *Back to the Future*, you can rest assured that the inference made in this example is
Agents using propositional logic
Propositional logic allows computer code, or AI agents, to efficiently keep track of what they have learned about their world. They can store propositional statements such as ‘If it is a car, it has
wheels’. It can then apply logical inferences from these and similar known truths to enlarge its bank of knowledge – in this case, it now knows that any car it comes across will have wheels.
We begin by writing down the set of logical sentences, or **‘transition model,’** that define the AI agent’s world. These are written into the knowledge base. This is best imagined through examples
from game design.
We might have an AI agent that is an orc character in a game. The agent has the statement ‘If it is an orc, then it can move one square per turn’ stored. This means that any movement that the orc
makes will have this rule applied to it.
Effect axioms and frame problems
Once the agent has executed this logic, we can capture **‘effect axioms’** – they specify the outcome of the agent’s actions.
These effect axioms can also be expressed as propositional statements. For example, ‘my orc character is moving from square (5,6) to (6,6).’ This can be stored as a true or false value for the agent
to use in later inferences.
However, an agent working only on propositional logic will come across what’s known as ‘frame problems.’ **Frame problems** are where effect axioms don’t specify in full what has changed – we know
that the orc has moved its position, but what about the items it was carrying?
This is the limit of propositional logic – it can only create quite simple statements and inferences. So unless we specifically tell an agent that an orc’s items move along with them, it cannot infer
What is predicate logic?
First-order or predicate logic is another means of knowledge representation in artificial intelligence. While it is considered more powerful than propositional logic, it is an extension rather than a
**The key difference with predicate logic is that it can deal with properties, rather than just propositions**. In other words, in addition to propositions, predicate logic can identify specific
details about the subjects of those propositions.
It does this by understanding variables. In this context, variables mean things whose definitions can be changed by new declarations.
An example of predicate logic
Let’s consider the following set of inferences using propositional logic:
If Frodo puts on the ring, he’ll turn invisible (*If* P, *then* Q)
Frodo puts on the ring (P)
Therefore, he’ll turn invisible (*Therefore*, Q)
The conclusion that Frodo will turn invisible is inferable with propositional logic. However, these propositions make no distinction between the properties of their subjects. Propositional logic is
only able to make basic deductions – **it knows that P leads to Q, but nothing more**. That means that things that are obvious to a human reader will be impossible to infer under propositional logic.
For example, consider these two propositions:
If Frodo puts on the ring, he’ll turn invisible (*If* P, *then* Q)
Sauron can see invisible people (R)
Obviously, we can infer from this that ‘If Frodo puts on the ring, Sauron can see him’. However, it would be **impossible to compute this with propositional logic.**
Predicates and variables
The reason it would be impossible to infer that ‘Sauron can see Frodo when he puts on the ring’ from the two propositions in the example is that propositional logic does not understand that
‘invisible people’ is a category – one that can contain objects such as ‘Frodo’.
Predicate logic, on the other hand, is able to parse statements and store variables and the conditions governing them. It will place Frodo within the category ‘invisible’, and therefore infer that
Frodo is visible to Sauron. Parsing these statements with predicate logic would look like this – where ‘F’ represents Frodo, ‘S’, Sauron, and ‘IP(x)’ represents the categorical variable ‘invisible
If Frodo puts on the ring, he’ll turn invisible (*If* F *performs* P, IP(x) *will contain* F)
Sauron can see invisible people (S *can see whatever is in* IP(x))
If Frodo puts on the ring, Sauron can see him (*If* F *performs* P, S *can see* F)
By understanding this, **predicate logic can be used for far smarter inferences than propositional logic, and express more complex things in a more efficient way.**
Different items in predicate logic
The most basic form of predicate logic is atomic sentences. These are statements about a category, using the verb ‘to be’ as the modifier. The easiest way to express how this works is through putting
the objects contained in the categorical variable in brackets. For example:
*Bill and Bob are brothers => Brothers (Bill, Bob)*
*Ginger is a cat => Cat(Ginger)*
There are also more complex ways that predicate logic can work.
Like natural language, predicate logic assumes that the world contains facts, objects (people, numbers, theories, etc.), relations (unitary ones such as green or square), and n-any relations (such as
taller than or sits between), and functions (father of, daughter of, end of, etc.)
Predicate logic quantifiers
Another reason that predicate logic is valuable to AI is that it is able to use quantifiers. **This means it can understand the concepts of ‘all’, or ‘some’.** The quantifier ‘all’ is known as a
universal quantifier. When universal quantifiers are used, the agent will know that the statement will apply to every object in that category.
There are also existential quantifiers – words like ‘some’. Predicate logic is able to understand the concept that in a group of objects, only some of them may share a given quality.
Using predicate logic to create a knowledge base
Predicate logic provides AI researchers with an expressive logical language that allows them to represent knowledge in a particular domain. Predicate logic works by holding a ‘knowledge base’ – a
record of which variables contain what object.
For example, we can represent ‘Richard is a king, Jack is a person, and all kings are persons,’ using the following predicate logic:
Richard is a king = *TELL (KB, King(Richard))*
Jack is a person = *TELL (KB, Person(Jack))*
All kings are people = *TELL (KB, ∀x King(X) => Person(x))*
*KB* stands for ‘Knowledge Base’. *∀x* is the symbol we use to express ‘all’, the universal quantifier. *TELL* here is the instruction we would use to declare that the variable contains the object we
want it to.
Querying a knowledge base with predicate logic
Once something is stored in a knowledge base, we can query it using ASK. For example:
Q: ‘Is Richard a king?’ = *ASK (KB, King(Richard))*
‘True’ is returned. And yet, more often we wish to know what value(s) of x make a sentence true. For example:
Q: Are kings people? = *ASKVARS (KB, Person(x))*
This logic will tell us that Richard and Jack are both people.
These statements are difficult to read for those unfamiliar with the syntax, but once you’ve gotten used to it, predicate logic is **an extremely powerful tool for concisely representing complex
natural language descriptions.**
Using predicate logic for knowledge engineering
‘Knowledge engineering’ involves constructing knowledge bases for particular domains. While typically these knowledge bases are limited and their range of queries known in advance, general purpose
knowledge bases exist for other purposes, such as natural language processing.
This means that **most AI agents have large knowledge bases built-in**. If their job is natural language processing, this means the whole dictionary, plus many rules for syntax and grammar. This
means they don’t have to be taught, for example, that ‘people’ is a variable category, or that ‘to be’ is how variables are declared.
‘Knowledge engineering’ is the field of optimizing knowledge bases for specific programs and agents. If an AI agent is being built for analyzing designs for bridges, it needs to have a knowledge base
that covers bridge designs and common issues with them. If the AI is for quality checking new cars, it needs a similarly specialized knowledge base.
Acting with uncertainty
While knowledge engineering is highly important, in the real world, AI agents must handle uncertainty. After all, they may not have sufficient information regarding their present state or what things
will be like after a sequence of actions take place – whether successful or not.
Rational decisions might depend on the relative importance of various goals and the degree to which they can be achieved. Our primary tool for dealing with degrees of belief is **’probability
theory.’** It provides a helpful foundation for uncertain reasoning.
*Bayes’ rule* describes the probability of an event based on prior knowledge of the conditions. It is often used in probability theory to describe the likelihood of an event happening and underlies
most modern AI systems for probabilistic inference. Probability helps by expressing the agent’s inability to reach a decision regarding the truth of a sentence. On the other hand, ‘decision theory’
combines the agent’s beliefs and desires – the best action being the one that maximizes expected utility.
For AI to successfully interact with, and learn from its environment, it must be able to ‘see’ what’s going on
Start here for an overview of this exciting pathway
The rapidly advancing field of AI is challenging our understanding of what it means for a machine to display human-like intelligence
The potential of AI to transform what we mean by intelligence was recognized during the earliest days of computing
Machine learning has made AI more flexible and adaptive, able to modify its behavior in response to its learning and the environment
Without an accurate and helpful representation of its environment, the AI simply has a collection of abstract data | {"url":"https://kinnu.xyz/kinnuverse/technology/artificial-intelligence/logical-representation/","timestamp":"2024-11-03T18:33:00Z","content_type":"text/html","content_length":"121759","record_id":"<urn:uuid:9f3dc538-f88f-4a22-aa21-3c5222c4c965>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00186.warc.gz"} |
Zero Percent Loans ⇦ Calculate Loan ⇨ Buy The Best Car
How to calculate the payments and interest
Once the car salesperson has given you all the figures for the deal – CHECK THEM! You are free to use our Car Loan Calculator where you can enter any 3 of the four factors – Amount, Term, Repayment,
Rate – and it will calculate the missing one. For example, if you know the Amount, Term and Repayment it will show you the actual Flat rate you are being charged. Or, if you know the Term, Repayment
and Rate, it will show you how much you are really borrowing. Even if the salesperson tells you all four factors, enter any three of them into the Car Loan Calculator and make sure that the answer it
comes up with is the same as the salesperson is telling you – if it isn’t something is wrong!!!
Of course you can’t use our Car Loan Calculator while you are sitting in a car showroom, but you can calculate these figures yourself… any calculator will do… for all the following calculations the
Term is the number of months…
Calculate loan repayments
Do this when you know the Amount, Term and Rate of the loan:
Step Buttons to press on calculator Explanation in words
1 Rate x Term Multiply the Rate by the Term
2 ÷ 1200 Divide the answer by 1200
3 + 1 Add 1
4 x Amount Multiply by the amount
5 ÷ Term = Divide by the Term
Suppose you are borrowing £10000 over 3 years (36 months) at a Flat rate of 4.8%
Step press Calculator shows
1 4.8 x 36 = 172.8
2 ÷ 1200 = 0.144
3 + 1 1.144
4 x 10000 = 11440
5 ÷ 36 = 317.77777
So, when you calculate the repayments for this loan you will see that the repayments are £317.78 per month. Make sure that this agrees with what you are being told by the salesperson. If they are
telling you that the monthly repayments are different to this figure… something does not add up… at best - they have made a “mistake” with one of the figures… at worst – you are being fooled into
taking a finance deal with either a higher interest rate than they are telling you or it may be that you are actually going to be borrowing more than they are leading you to believe!
I am not saying for one minute that a car salesperson would openly lie about figures like this. It’s more likely that they would only tell you what you need to know, or be economical with the truth
by only answering your direct questions.
Another common tactic is where they say something like “we offer very competitive finance” without actually telling you the interest rate (Flat rate) of their loan.
So, when you are negotiating a deal you will get to the stage where you know the Amount you have to borrow. The salesperson might then say something like “that will cost you £xxx per month over 3
years or £xxx per month over 5 years, which is better for you?” In this case you need to check to see what interest rate they are trying to charge you…
Calculate loan rates
Do this when you know the Amount, Term and monthly Repayments of the loan:
Step Buttons to press on calculator Explanation in words
1 Term ÷ 1200 = Divide the Term by 1200
2 M+ Store the answer in your calculator’s memory (or write it down, for later!)
3 Repayment ÷ Amount = Divide the Repayment by the Amount
4 x Term Multiply the answer by the Term
5 - 1 = Subtract 1 from the answer
6 ÷ MR = Divide the answer by the result of step 1 above
Suppose you are borrowing £10000 over 3 years (36 months) and you are told the monthly Repayment is £330
Step Buttons to press on calculator Calculator shows
1 36 ÷ 1200 = 0.03
2 M+ Store or write down the answer for later
3 330 ÷ 10000 = 0.033
4 x 36 = 1.188
5 - 1 = 0.188
6 ÷ MR (or ÷ 0.03) = 6.2666666
So when you calculate the interest for this loan, it works out at 6.27% interest (Flat rate). Is that what the salesperson is telling you? If not, then again, something is wrong… you are not being
told the whole truth. It may be that they have “misquoted” the interest rate or that you are actually going to be borrowing MORE than you are being led to believe!
In any event, is it a good rate? How does it compare with current loan rates? Remember, you are not comparing this with the APR of a bank or personal loan – that will be about double. You need to
find out the sort of interest rates that are currently being offered for personal loans – this is constantly changing, so we can’t tell you. Just search the Internet for personal loans. You will only
find APR figures but it is reasonable to half the APR rate to find the current Flat rate.
If you want to be more precise - just search the internet using your favourite search engine. You should search for “personal loan calculator” and you will find plenty of reputable websites (banks,
finance companies etc) where you can enter an Amount and a Term and it will show you the APR and the Repayment. You can then enter the Amount, Term and Repayment into our Car Loan Calculator and it
will tell you what the Flat rate is for that loan.
You can then compare the Flat rate of the loan you are being offered by the car salesperson to what you know typical rates are. Obviously, if the rate is not good – don’t do the deal!
Getting a loan at a higher interest rate than necessary means that you are paying more than necessary and the dealer is making more money out of you than necessary.
Another situation you might find yourself in is where the salesperson isn’t exactly clear about how much the balance to change figure is, rather than tell you they might just say what the repayments
will be…
Calculate loan amount
Do this when you know the Rate, Term and monthly Repayments of the loan. This is also a very common situation: you get told how much they will allow you for your part exchange car and they proceed to
work out a deal, telling you how much it will cost you per month – without actually telling you how much you will end up borrowing. A surprising number of car buyers only focus on the monthly
repayment at this stage… very dangerous!
Step Buttons to press on calculator Explanation in words
1 Term ÷ 1200 Divide the Term by 1200
2 x Rate Multiply the answer by the Rate
3 + 1 = Add 1 to the answer
4 M+ Store the answer in your calculator’s memory (or write it down, for later!)
5 Repayment x Term Multiply the Repayment by the Term
6 ÷ MR Divide the answer by the result from step 4 above
Suppose the loan is over 3 years (36 months), the flat rate is 4.83% and you are told the monthly Repayment is £318
Step press Calculator shows
1 36 ÷ 1200 = 0.03
2 x 4.83 = 0.1449
3 + 1 = 1.1449
4 M+ Store or write down the answer for later
5 318 x 36 = 11448
6 ÷ MR (or ÷ 1.1449) = 9999.1265
So when you calculate the amount you are borrowing for this loan, it works out at £9999.13 Is that what the salesperson is telling you, or what you expected? If not, then again, something is wrong…
This figure is effectively the “balance to change” figure – the difference between what you are getting for your car and what you are being charged for your new car (assuming you haven’t put any cash
into the deal). Add this Amount to the Trade price for your car… the resulting figure is what you are paying for your new car and you will easily see what, if any, discount you are getting.
If you are borrowing more than you thought it obviously means that you are not getting as good a deal as you thought… you are not getting much discount on the car you are buying. You can do better!
Calculate the term of a loan
The final possibility is the least likely one – where you don’t know what Term the loan is over. It is sometimes worth checking though, as a loan can be arranged over any period of time, it doesn’t
have to be exact years, and an extra month or two on the Term means you are paying more than you thought…
Step Buttons to press on calculator Explanation in words
1 Amount x 0.000833333 = Enter as many 3’s on the end as you can
2 x Rate = Multiply the answer by the Rate
3 M+ Store the answer in your calculator’s memory (or write it down, for later!)
4 Repayment – MR = Subtract the result of step 3 above from the Repayment
5 MC then M+ Clear calculator’s memory then store the answer (or write it down, for later!)
6 Amount ÷ MR = Divide the Amount by the result from step 5 above
Suppose you are borrowing £15000 at 4.83% and you are told the monthly Repayment is £465.78
Step press Calculator shows
1 15000 x 0.0008333 = 12.4995
2 x 4.83 = 60.3725
3 M+
4 465.78 – MR (or - 60.3725) = 405.4074
5 MC then M+ Store or write down the answer for later
6 15000 ÷ MR (or ÷ 405.4074) = 36.9998
So when you calculate the term of this loan, it is 37 months. Is that what you thought? If not, then again, something is wrong…
It may be that another dealer is quoting a slightly higher repayment for exactly the same loan… so this finance, where the repayments are less, seems better. However, the other dealer may be quoting
over a Term of 36 months whereas this loan is over 37 months – that extra month has just cost you £465.78… you have just paid £465.78 more than you thought for this car!
Apart from a house, buying a car is probably the most expensive purchase you will ever make… don’t just accept whatever a car salesperson tells you about the finance – double-check the figures
because it’s so easy to be misled. Tell the salesperson to leave you alone for 10 minutes while you “work out your finances” (or whatever) and do the calculation yourself. You will only have to do
one of the above calculations: -
Remember there are 4 numbers – Amount, Rate, Term and Repayment. Just take the 3 numbers you are sure of (or in fact, ANY three of the four numbers!) and work out the missing one.
If your calculation produces an unexpected result tell the salesperson what you have found and get them to “reconsider” or do better. Do not accept a bad finance deal – if you do, you may as well
give up, lie down and ask them to fleece you… they are doing anyway!
So, when the time comes, check and double check the finance deal you are being offered but, how do you get to that point? How should you handle car salespeople and get them to do you the best deal?
We will see in the next section – but first, you need to examine your spending power…
The next section explains how you can check your buying power...
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In Statistics, Mean is the thing you normally know as "average". It's calculated by summing all numbers and then dividing the sum by the number of numbers.
// not the actual way to do it, just pointing out the idea
const mean = (arr) => arr.reduce((a,b) => a+b) / arr.length
const x = [1,2,3,4,5]
mean(x) // 3
Also see MedianMedian
In statistics, Median is the number sitting in the middle of a sorted set of numbers.
// not the actual way to do it, just pointing out the idea
const median = (sortedArray) => sortedArray[arr... and ModeMode
In statistics, Mode is the most common number in a data set.
// not the actual way to do it, just pointing out the idea
const mode = (arr) => arr.reduce((acc, el) => {
acc.counters[el] = a....
Status: #🌲 | {"url":"https://www.dobrica.sh/notes/Mean","timestamp":"2024-11-14T01:24:35Z","content_type":"text/html","content_length":"13143","record_id":"<urn:uuid:22941d5d-3494-4e0c-a2ed-5f41be524311>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00280.warc.gz"} |
Discover Where Your Profit Growth Comes From: A Power BI Tutorial
In this Power BI tutorial, I dive into what’s perhaps a very niche insight. But don’t let that close you off to using this technique — there are plenty of ways to implement this across a range of
scenarios. The key is to understand each different part and transfer them across similar elements in your own Power BI models and reports. You may watch the full video of this tutorial at the bottom
of this blog.
As with many things I showcase at Enterprise DNA and through this blog, learning aspects of techniques is the key. Everyone’s data will be different and the scenarios you are analyzing will be
different, but you can replicate a lot of what I’m doing over anything from an analytical perspective.
This Power BI tutorial will show you how you can showcase which grouping of customers contribute the most to the growth in your profits over any time period. Can you even imagine creating this
insight in Excel? It would be extremely difficult. In Power BI though, we can achieve this through just a few formulas, and use the data model to filter the results in many different ways.
We’re going to explore how we can discover products – or any dimension – is contributing the most to a company’s profit growth. We’re going to stick to products here just to keep things
simple, but you can use a range of different dimensions.
Determining Total Costs & Total Profits
As we walk through the steps in this Power BI tutorial, I am going to talk about the logic on how you can actually do this yourself. First of all, you obviously need to start with the profits. In
this example, we are going to show the profit growth from 2015 to 2016, then isolate the profits from those years as well. I’ve already placed a filter on 2015 and 2016 because the underlying data
will cover exactly those two years.
We already have our Total Sales, so the next step is to find our Profits. We are going to create a new measure and call it Total Costs. We are also going to use the SUMX function to look at the Sales
table, and within the Sales table find the Total Unit Cost times the Order Quantity.
Now we have our Total Cost and from this, we can create another measure to determine Total Profits. To do this, we can just deduct Total Costs from Total Sales.
So now we have all three: Total Costs, Total Profits, and Total Sales.
We drag our Total Profits into the canvas and we can now see that for 2015 and 2016, we have a profit of $44.77 million. So this is a pretty healthy business.
Isolating Product Growth Using The Filter Function
We need to see the product growth from 2015 to 2016, which is why we need to create some sort of filter here. We need to isolate the Total Profits from only each of those years and the way we do that
is by using a filter inside of the calculation.
To achieve this, we create a new measure and call it 2016 Profits. We’ll jump down another line, use the CALCULATE function, and put in our Total Profits measure. We jump down another line and use
the FILTER function. What we do with FILTER is to reference a table and then the filtering expression which is your additional filter, or what you want to isolate. In this case, we go to the Dates
table and then only return the total profits from 2016.
Now, we can isolate the 2016 profits, and drag it into the canvas. As you can see, we have 22.45 million which is about half of the Total Profits from those two years.
To get the 2015 profits, we just copy the previous measure, paste it into the new one, and then change the numbers from 2016 to 2015. Copy-paste is your new best friend in Power BI.
We started out with Sales then went to Costs. We were able to get Profits, and now we were able to break down these profits into years. So now we have our 2015 and 2016 profits using two different
Determining Profit Change
The next step is to determine what was the change from 2016 to 2015. All we have to do is create another measure and call it Profit Change. We are going to deduct the 2015 Profits from our 2016
Once we drag it into the canvas, we can see the Profit Change between 2015 and 2016 in the table.
The next step is to filter the Profit Change by the products and achieve a result for the change for every individual product. Let’s copy then paste this table and then add our Product Name
dimension to the mix. Now we can see the Profit Change for every single product.
Visually, we can represent this data using the Waterfall Chart, which is perfect for this type of analysis. Now we can see quite clearly what products from 2015 to 2016 have contributed to the Total
Profit Growth. We can also sort it out so the positives and negatives become a lot more obvious.
We can clearly see what products are contributing the most to our profit growth, which is Product 9. We can bring in other elements to our data model such as the difference between regions.
For example, I’m bringing in the Total Profits numbers from Island which breaks down the north and south of New Zealand and then create a visualisation out of this. If I wanted to isolate the North
Island and see what was the change there, I could click on the North and see what is affecting the profits in that region. Obviously, if you have more regions in your own data model, you can click on
any of those and the chart will update automatically for you.
Determining Percentage Of The Profit Change
Maybe we want to see not just the nominal numbers, but the percentage as well. So we just create another new measure and call it 2016 % Profit Growth, then divide Profit Change by 2016 Profits, with
0 as our alternative result.
We will then change the format to percentage and turn it into a visualisation.
Showcase Positive And Negative Changes In Different Visualisations
One last thing we can do in this Power BI tutorial is to only showcase the positive changes in one visualisation and the negative changes in another visualisation. To do this, we will create a new
measure and call it 2016 % Profit Growth Positive. Then, we’ll write some piece of logic here:
Then we copy this measure, create a new one and paste the old one. We just change the title to negative and the formula from greater than to less than.
We are now able to quickly isolate and compare how certain profits performed from one year to another.
We can also see the changes from a customer perspective, and see who from our customers are affecting these numbers.
***** Related Links *****
Calculating Percent Profit Margins Using DAX In Power BI
Learn Which Customer Groups Experienced The Greatest Growth – Power BI & DAX
Check Whether Revenue Growth Is Profitable – Analysis In Power BI w/DAX
In this Power BI tutorial, we were able to analyse — based on some sales information — which are the products that contributed the most to profit growth. We also discussed the products that
detracted from profit growth by looking up the negative numbers as well.
Perhaps, from a risk management perspective, you don’t want to see too much of your growth only coming from a few clients; alternatively, you do, and you want to see if it’s playing out how you want
it to. Well, this technique is going to be exactly what you need to showcase this.
In business, not only do you have to manage your entire makeup of sales or profits, but it’s equally important to understand where the growth, whether good or bad, is coming from. What you’re getting
is basically performance attribution on your results. It’s good insight.
Back in my portfolio management days, performance attribution was essential to show how you made the returns you did. This is exactly the same, but instead of bonds, stocks, and other assets, we are
analyzing customers, products, and sales people.
This project aims to implement a full data analysis pipeline using Power BI with a focus on DAX formulas to derive actionable insights from the order data.
A comprehensive guide to mastering the CALCULATETABLE function in DAX, focusing on practical implementation within Power BI for advanced data analysis.
An interactive web-based application to explore and understand various data model examples across multiple industries and business functions.
A comprehensive project aimed at enhancing oil well performance through advanced data analysis using Power BI’s DAX formulas.
Learn how to leverage key DAX table functions to manipulate and analyze data efficiently in Power BI.
Deep dive into the CALCULATETABLE function in DAX to elevate your data analysis skills.
One of the main reasons why businesses all over the world have fallen in love with Power BI is because...
A hands-on project focused on using the TREATAS function to manipulate and analyze data in DAX.
A hands-on guide to implementing data analysis projects using DAX, focused on the MAXX function and its combinations with other essential DAX functions.
Learn how to leverage the COUNTX function in DAX for in-depth data analysis. This guide provides step-by-step instructions and practical examples.
A comprehensive guide to understanding and implementing the FILTER function in DAX, complete with examples and combinations with other functions.
Learn how to implement and utilize DAX functions effectively, with a focus on the DATESINPERIOD function.
Comprehensive Data Analysis using Power BI and DAX
Exploring CALCULATETABLE Function in DAX for Data Analysis in Power BI
Data Model Discovery Library
Optimizing Oil Well Performance Using Power BI and DAX
Mastering DAX Table Functions for Data Analysis
Mastering DAX CALCULATETABLE for Advanced Data Analysis
Debugging DAX: Tips and Tools for Troubleshooting Your Formulas
Practical Application of TREATAS Function in DAX
MAXX in Power BI – A Detailed Guide
Leveraging the COUNTX Function In Power BI
Using the FILTER Function in DAX – A Detailed Guide With Examples
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Verifiable Random Functions from Non-interactive Witness-Indistinguishable Proofs
Verifiable random functions (VRFs) are pseudorandom functions where the owner of the seed, in addition to computing the function’s value y at any point x, can also generate a non-interactive proof π
that y is correct, without compromising pseudorandomness at other points. Being a natural primitive with a wide range of applications, considerable efforts have been directed toward the construction
of such VRFs. While these efforts have resulted in a variety of algebraic constructions (from bilinear maps or the RSA problem), the relation between VRFs and other general primitives is still not
well understood. We present new constructions of VRFs from general primitives, the main one being non-interactive witness-indistinguishable proofs (NIWIs). This includes: (1) a selectively secure VRF
assuming NIWIs and non-interactive commitments. As usual, the VRF can be made adaptively secure assuming subexponential hardness of the underlying primitives. (2) An adaptively secure VRF assuming
(polynomially hard) NIWIs, non-interactive commitments, and (single-key) constrained pseudorandom functions for a restricted class of constraints. The above primitives can be instantiated under
various standard assumptions, which yields corresponding VRF instantiations, under different assumptions than were known so far. One notable example is a non-uniform construction of VRFs from
subexponentially hard trapdoor permutations, or more generally, from verifiable pseudorandom generators (the construction can be made uniform under a standard derandomization assumption). This
partially answers an open question by Dwork and Naor (FOCS ’00). The construction and its analysis are quite simple. Both draw from ideas commonly used in the context of indistinguishability
• Foundations
• Non-interactive witness indistinguishable proofs
• Verifiable random functions
ASJC Scopus subject areas
• Software
• Computer Science Applications
• Applied Mathematics
Dive into the research topics of 'Verifiable Random Functions from Non-interactive Witness-Indistinguishable Proofs'. Together they form a unique fingerprint. | {"url":"https://nyuscholars.nyu.edu/en/publications/verifiable-random-functions-from-non-interactive-witness-indistin-2","timestamp":"2024-11-14T23:51:21Z","content_type":"text/html","content_length":"55566","record_id":"<urn:uuid:792e6a54-e6dd-4580-9461-c016982eff23>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00071.warc.gz"} |
Water Formulas
Excel Spreadsheet Tool - Download Now
Click here for an interactive Excel spreadsheet (XLS) for water use calculations. (Use your browser's "BACK" button to return to this page if you need to. You need not save changes to the spreadsheet
when leaving.) If you don't have Microsoft Excel, or simply want a bit more of a mathematical challenge, you can use the formulas below for these same calculations.
Calculating annual acre feet (AF) pumped (2 calculations)
The easiest way to calculate total acre feet used in a year is to multiply the gallons per minute rate you pumped the well, times the number of hours the well was pumped for the year, times the
numerical factor .0001841.
The Formula is: (RATE) X (HOURS) X (.0001841)
Example: (600 gpm) X (1500 Hours) X (.0001841) = 165.7 AF per year.
An alternative calculation for the same answer is: multiply gallons per minute (GPM) X 60 (yielding gallons per hour (GPH)), then multiply GPH X hours pumped for the year (yielding Gallons per Year
(GPY)), and finally dividing GPY by 325,851 (number of gallons in an AF)
The Formula is: (GPM) X (60) X (HOURS) / 325,851 (remember: do all multiplication first, then the division)
Example: (600 GPM) X (60) X (1500) / 325,851 = 165.7 AF per year.
Calculating hours to pump a known ammount
Figuring HOURS to pump a known amount of water at a known rate of diversion:
The question would be posed: "How many hours will it take me to pump 200 AF at 600 GPM?"
The Formula is: TOTAL GALLONS / GALLONS PER HOUR
Example: 65,170,200 / 36,000 = 1,810 hours
The trick to this calculation is converting AF into GALLONS, and GPM into GPH. Remember that 1 AF = 325,851 Gal, so 200 AF = 200 X 325,851, or 65,170,200 GAL. Also, 600 GPM = 600 X 60, or 36,000 GPH.
Helpful Conversion
Some conversions for water are:
1 Acre foot = 325,851 gal
1 cubic foot of water = 7.48 Gal = 62.4 Lbs
1 Gal of water weighs 8.33 Lbs
1 Acre foot = 12 Acre inches
1 Acre inch = 27,154 Gal
1 Acre inch per Hour = 449 GPM = 1 cubic foot per second (CFS) | {"url":"http://gmd4.org/WaterFormulas.html","timestamp":"2024-11-13T12:10:20Z","content_type":"text/html","content_length":"10128","record_id":"<urn:uuid:71bf8767-8ad3-4493-a8a0-73c44825fe3a>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00091.warc.gz"} |
In order to simplify any fraction, our Fraction Simplifier will calculate the greatest common divisor (GCD), also known as the greatest common factor (GCF), or the highest common factor (HCF) of
numerator and denominator that you entered. Then, fraction simplifier calculator will divide the numerator and denominator of the fraction by this number.
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Then, fraction simplifier calculator will divide the numerator and denominator of the fraction by this number. Here we will simplify 11/24 to its simplest form and convert it to a mixed number if
necessary. In the fraction 11/24, 11 is the numerator and 24 is the denominator. When you ask "What is 11/24 simplified?", we assume you want to know how to simplify the numerator and denominator to
their smallest values, while still keeping the same value of Using the steps above, here is the work involved in the solution for fraction 11/24 to simplest form .
Go. Rewrite f(x) in the form All skills learned lead eventually to the ability to solve equations and simplify the solutions. In previous capital reduction pursuant to Scheme of Arrangement
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24-bit opcode let us encode two million (221) different except for MOV whose iii = 111, in which case rr specifies the 24. (i) Calculate an estimate of the mean. You must show all your working. (iii)
Two students are chosen at random from those students that estimated the area 111-5):. (1) Notices of proposed contract actions shall follow the procedures in 5.704 for posting orders. (2) 20 Apr
2020 Here we present simplified arithmetic models of COVID-19 transmission, for monitoring and characterizing an epidemic,1,4,6,22,24,77 and for understanding 111.↵.
How do you simplify fractions? Example: For the fraction 33 / 99 the number 33 is the numerator and 99 is the denominator. First, find the greatest common factor of the two numbers. GCF = 33 GCF
Calculator. Second, divide both the numerator and denominator by the GCF. Numerator is found from 33/33 = 1
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Use this reduced improper fraction and divide 9 by 2: 9 ÷ 2 = 4 with remainder of 2; The whole number result is 4; The remainder is 1. With 1 as the numerator and 2 as the reduced denominator, the
fraction part of the mixed number is 1/2.
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We do this by first finding the greatest common factor of 24 and 111, which is 3. Therefore, 112/24 simplified to lowest terms is 14/3. MathStep (Works offline) Download our mobile app and learn to
work with fractions in your own time: Android and iPhone/ iPad. Equivalent fractions: 224 / 48 56 / 12 336 / 72 560 / 120 784 / 168. Steps to simplifying fractions Find the GCD (or HCF) of numerator
and denominator GCD of 24 and 111 is 3 Divide both the numerator and denominator by the GCD 24 ÷ 3 / 111 ÷ 3 Reduced fraction: 8 / 37 Therefore, 24/111 simplified to lowest terms is 8/37.
The denominator of the Simplify Rational Expressions Involving the Sum or Difference. PNAS June 17, 2014 111 (24) 8809-8814; first published June 3, 2014; oval, including a simplified view of the
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calculated in all the 100 patients 24 h The simplest form of 111/24 is 37/8. Steps to simplifying fractions. | {"url":"https://kopavguldtcgl.netlify.app/27159/27427","timestamp":"2024-11-07T23:22:43Z","content_type":"text/html","content_length":"10376","record_id":"<urn:uuid:56b9c164-e8a7-435b-a601-7d7d055182da>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00659.warc.gz"} |
From one perspective, mathematics is the art of creating knowledge by finding new and interesting relationships among existing mathematical objects. Sketchpad provides you with a rich set of such
mathematical objects and with many ways to connect them. It’s up to you to create those objects, set up the connections between them, and determine their attributes. Then you can investigate their
behavior, find new relationships, discover symmetry and patterns, and display and present your results.
Here’s a list of the objects that Sketchpad makes available:
Geometric Objects: Numeric and Algebraic Objects:
•Points •Measurements
•Segments, Rays, and Lines •Parameters Presentation Objects:
•Circles •Calculations •Captions
•Arcs •Tables •Angle Markers
•Polygons and Other Interiors •Coordinate Systems •Tick Marks
•Loci •Functions •Action Buttons
•Pictures and Drawings •Function Plots
•Iterations (some iterations are numeric) •Parametric Plots
Object Categories:
Certain categories of objects can be used for special purposes.
•Path Objects are objects on which you can construct and animate points.
•Straight Objects are path objects that you can use to construct parallels and perpendiculars. You can measure the slope of a straight object and the distance from a point to a straight object.
•Layered Objects occupy two-dimensional area and appear in a layered order when they overlap.
•Translucent Objects have adjustable opacity, allowing similar objects in layers below them to show through.
•Values are objects that you can use in situations when you need a numeric value.
•Plots are displayed by combining many small elements (called samples) that make up the full object. These objects include loci and function plots, among others.
•Text Objects include captions, values, functions, and action buttons. | {"url":"https://keyonline.keypress.com/gsp-reference-center/objects.htm","timestamp":"2024-11-14T13:34:20Z","content_type":"text/html","content_length":"18154","record_id":"<urn:uuid:7fed600b-6fee-4578-b674-227c15c8a796>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00260.warc.gz"} |
Distributive Property (Expanding) Quiz Online (6.EE.A.3) : 6th grade Math
#8 of 8: Spicy
Distributive property (expanding)
<p>Use the distributive property to write an equivalent, simplified expression.</p><p>`4.5 (5a - 9b)`</p>
#7 of 8: Spicy
Distributive property (expanding)
<p>Use the distributive property to write an equivalent, simplified expression.</p><p>`0.5 (18f - 34g + 12)`</p>
#6 of 8: Spicy
Distributive property (expanding)
<p>Use the distributive property to write an equivalent, simplified expression.</p><p>`1.5 (8j + 18)`</p>
#5 of 8: Medium
Distributive property (expanding)
<p>Use the distributive property to write an equivalent, simplified expression.</p><p>`9(8 - 5u + y)`</p>
#4 of 8: Medium
Distributive property (expanding)
<p>Use the distributive property to write an equivalent, simplified expression.</p><p>`5(7a + 8b - c)`</p>
#3 of 8: Medium
Distributive property (expanding)
<p>Use the distributive property to write an equivalent, simplified expression.</p><p>`2.5(4f + 22)`</p>
#2 of 8: Mild
Distributive property (expanding)
<p>Use the distributive property to write an equivalent, simplified expression.</p><p>`6(15k - 4j)`</p>
#1 of 8: Mild
Distributive property (expanding)
<p>Use the distributive property to write an equivalent, simplified expression.</p><p>`8(2x - 1)`</p>
Distributive property expanding is an important topic in math for the students to understand. These are the basic word problems that help the students to understand the concept of distributive
property exapnsion. This is majorly used to solve an equation using this property and mathematical operations involved. Teachers can share this amazing distributive property expanding quiz with their
students to...
Show all | {"url":"https://www.bytelearn.com/math-grade-6/quiz/distributive-property-expanding","timestamp":"2024-11-10T04:15:02Z","content_type":"text/html","content_length":"284605","record_id":"<urn:uuid:bc7fe889-0cfd-475a-bf3f-edc3c8ba581d>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00736.warc.gz"} |
Newest 'quadratic-forms' Questions
Questions tagged [quadratic-forms]
Algebraic and geometric theory of quadratic forms and symmetric bilinear forms, e.g., values attained by quadratic forms, isotropic subspaces, the Witt ring, invariants of quadratic forms, the
discriminant and Clifford algebra of a quadratic form, Pfister forms, automorphisms of quadratic forms.
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Let $O\subset\mathbb{R}^d$ be a bounded domain of the class $C^{1,1}$ (or $C^2$ for simplicity). Let the operator $A_D$ be formally given by the differential expression $A=-\mathrm{div}g(x)\nabla$ | {"url":"http://casinogallerian.com/quadratic-forms.html","timestamp":"2024-11-05T02:14:38Z","content_type":"text/html","content_length":"292308","record_id":"<urn:uuid:aa40ab04-ceaf-472d-8c73-2b0f96e76342>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00510.warc.gz"} |
A characterization of 3-i-critical graphs of connectivity two
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Ananchuen, N.
Ananchuen, W.
Caccetta, Louis
Ananchuen, N. and Ananchuen, W. and Caccetta, L. 2017. A characterization of 3-i-critical graphs of connectivity two. Quaestiones Mathematicae. 40 (7): pp. 937-965.
Source Title
Quaestiones Mathematicae
Department of Mathematics and Statistics
A subset S of V (G) is an independent dominating set of G if S is independent and each vertex of G is either in S or adjacent to some vertex of S. Let i(G) denote the minimum cardinality of an
independent dominating set of G. A graph G is k-i-critical if i(G) = k, but i(G + uv) < k for any pair of non-adjacent vertices u and v of G. The problem that arises is that of characterizing k-i
critical graphs. In this paper, we characterize connected 3-i-critical graphs with minimum vertex cutset of size 2. More specifically, we show that if G is a connected 3-i-critical graph with minimum
vertex cutset S of size 2 and the number of components of G - S is exactly two, then G is isomorphic to a graph in one of nine classes of connected 3-i-critical graphs. The results in this paper
together with results in [1] and [2] provide a complete characterization of connected 3-i-critical graphs with a minimum cutset of size at most 3.
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