Chapter 5, Problem 5.4CP

To determine

To rewrite:

Dimensionless function and To plot:

Using the pi theorem and plot the given data in dimensionless form.

Answer to Problem 5.4CP

The dimensionless function is $\frac{\Delta p}{\rho {\Omega }^2{D}^2}=fcn(\frac{Q}{\Omega D^3})$ and the plot is shown as follows.

Explanation of Solution

Given Information:

The Taco Inc. model 4013 centrifugal pump has an impeller of diameter D = 12.95 in. The measured flow rate Q and pressure rise $\Delta p$ are given by the manufacture as follows, when pumping 20°C water at $\Omega =1160r/\min$ :

Concept Used:

The number of pi groups are to be calculated:

$N=k-r$

Where k is the number of variables and r is the number of fundamental references.

On substituting 5 for k and 3 for r ,

$N=2$

According to tables, the density of water at 20°C is $\rho =1.94slug/f{t}^3$

Calculation:

Dimensional analysis is applied to find the pi groups.

First pi group:

${\pi }_1={\rho }^a{D}^b{\Omega }^{c}Q$

Where $\rho$ is the density, diameter is D, speed is $\Omega$ and the flow rate is Q.

On substituting $M^0{L}^0{T}^0$ for ${\pi }_1$, $[M{L}^{-3}]$ for $\rho$, $[L]$ for D, $[T^{-1}]$ for $\Omega$ and $[L^3{T}^{-1}]$ for Q ,

$M^0{L}^0{T}^0={[M{L}^{-3}]}^a{[L]}^b{[T^{-1}]}^c[L^3{T}^{-1}]$

$M^0{L}^0{T}^0=[M^a{L}^{-3a+b+3}{T}^{-c-1}]$

On equating M coefficients:

$a=0$

On equating T coefficients:

$\begin{array}{l}-c-1=0\\ c=-1\end{array}$

On equating L coefficients:

$\begin{array}{l}-3a+b+3=0\\ -3(0)+(b)+3=0\\ b=-3\end{array}$

Hence, a = 0, b = -3 and c = -1

Therefore, the first pi group is as follows:

${\pi }_1={\rho }^0{D}^{-3}{\Omega }^{-1}Q$

${\pi }_1=\frac{Q}{\Omega D^3}$

Second pi group:

${\pi }_2={\rho }^a{D}^b{\Omega }^c(\Delta p)$

Where $\rho$ is the density, diameter is D, speed is $\Omega$ and the pressure rise is $\Delta p$.

On substituting $M^0{L}^0{T}^0$ for ${\pi }_1$, $[M{L}^{-3}]$ for $\rho$, $[L]$ for D, $[T^{-1}]$ for $\Omega$ and $[M{L}^{-1}{T}^{-2}]$ for $\Delta p$,

$M^0{L}^0{T}^0={[M{L}^{-3}]}^a{[L]}^b{[T^{-1}]}^c[M{L}^{-1}{T}^{-2}]$

$M^0{L}^0{T}^0=[M^{a+1}{L}^{-3a+b-1}{T}^{-c-2}]$

On equating M coefficients:

$\begin{array}{l}a+1=0\\ a=-1\end{array}$

On equating T coefficients:

$\begin{array}{l}-c-2=0\\ c=-2\end{array}$

On equating L coefficients:

$\begin{array}{l}-3a+b-1=0\\ -3(-1)+b-1=0\\ b=-2\end{array}$ ]

Hence, a=- 1, b = -2 and c = -2

Therefore, the second pi group is as follows:

${\pi }_2={\rho }^{-1}{D}^{-2}{\Omega }^{-2}(\Delta p)$

${\pi }_2=\frac{\Delta p}{\rho {\Omega }^2{D}^2}$

Hence the choices are

${\pi }_2=fcn({\pi }_1)$

On substituting $\frac{\Delta p}{\rho {\Omega }^2{D}^2}$ for ${\pi }_2$ and $\frac{Q}{\Omega D^3}$ for ${\pi }_1$

$\frac{\Delta p}{\rho {\Omega }^2{D}^2}=fcn(\frac{Q}{\Omega D^3})$

Hence the dimensionless function is $\frac{\Delta p}{\rho {\Omega }^2{D}^2}=fcn(\frac{Q}{\Omega D^3})$

The units of angular velocity are converted from r/min to rev/s.

$\Omega =1160\frac{rev}{\min }\times (\frac{1\min }{60s})$

=19.33 rev/s

The units of diameter are converted into feet:

$D=12.96in.\times (\frac{1ft}{12in.})$

=1.079 ft

The flow rate in ft³ /s is calculated:

$Q=200gal/\min \times (\frac{0.00222801f{t}^3/s}{1gal/\min })$

$Q=0.4456f{t}^3/s$

The pressure in $lb/f{t}^2$ is calculated as:

$\Delta p=36\frac{lbf}{in{.}^2}\times (\frac{144in{.}^2}{1f{t}^2})$

$=5184lbf/f{t}^2$

The ${\pi }_1$ term is calculated:

${\pi }_1=\frac{Q}{\Omega D^3}$

On substituting $0.4456f{t}^3/s$ for Q, $19.33rev/s$ for $\Omega$ and $1.079ft$ for D ,

${\pi }_1=\frac{0.4456}{19.33\times D^3}$

${\pi }_1=\frac{0.02305225}{{1.079}^3}$

The ${\pi }_2$ term is calculated:

${\pi }_2=\frac{\Delta p}{\rho {\Omega }^2{D}^2}$

On substituting $5184lbf/f{t}^2$ for $\Delta p$, $1.94slug/f{t}^3$ for $\rho$, $19.33rev/s$ for $\Omega$ and $1.079ft$ for D ,

${\pi }_2=\frac{5184}{1.94\times {\Omega }^2{D}^2}$

${\pi }_2=\frac{2672.16495}{{19.33}^2\times {1.079}^2}$

=6.14

The values for pi groups at different values are calculated which are as follows:

Q(gal/min)	${\pi }_1=\frac{Q}{\Omega D^3}$	${\pi }_2=\frac{\Delta p}{\rho {\Omega }^2{D}^2}$
200	0.0183	6.14
300	0.0275	5.97
400	0.0367	5.8
500	0.0458	5.46
600	0.0550	4.95
700	0.0642	3.92

Hence, the plot between ${\pi }_1$ and ${\pi }_2$ is obtained.

Conclusion:

The dimensionless function is $\frac{\Delta p}{\rho {\Omega }^2{D}^2}=fcn(\frac{Q}{\Omega D^3})$ and the plot is shown as above.

To determine

To estimate:

The pressure rise $\Delta p$ is expected in lbf/in², according to the dimensionless correlation.

Answer to Problem 5.4CP

The pressure rise $\Delta p$ expected in lbf/in², according to the dimensionless correlation is 13 lbf/in².

Explanation of Solution

Given Information:

The Taco Inc. model 4013 centrifugal pump has an impeller of diameter D = 12.95 in. The measured flow rate Q and pressure rise $\Delta p$ are given by the manufacture as follows, when pumping 20°C water at $\Omega =1160r/\min$ :

A pump running at 900 r/min is used to pump 20°C gasoline at 400 gal/min.

Concept Used:

The units of angular velocity are converted from r/min to rev/s.

$\Omega =900\frac{rev}{\min }\times (\frac{1\min }{60s})$

=15 rev/s

The flow rate in ft³ /s is calculated:

$Q=400gal/\min \times (\frac{0.00222801f{t}^3/s}{1gal/\min })$

$Q=0.891f{t}^3/s$

According to the tables, the density of gasoline at 20°C is $1.32slug/f{t}^3$

Calculation:

The ${\pi }_1$ term is calculated:

${\pi }_1=\frac{Q}{\Omega D^3}$

On substituting $0.891f{t}^3/s$ for Q, $15rev/s$ for $\Omega$ and $1.079ft$ for D ,

${\pi }_1=\frac{0.891}{15\times {\left(1.079\right)}^3}$

${\pi }_1=\frac{0.0594}{{\left(1.079\right)}^3}$

=0.0473

According to the plot, for ${\pi }_1=0.0473$, the value of ${\pi }_2=5.4$.

The ${\pi }_2$ term is calculated:

${\pi }_2=\frac{\Delta p}{\rho {\Omega }^2{D}^2}$

On substituting 5.4 for ${\pi }_2$, $1.32slug/f{t}^3$ for $\rho$, $19.33rev/s$ for $\Omega$ and $1.079ft$ for D ,

$5.4=\frac{\Delta p}{1.32\times {15}^2\times {1.079}^2}$

$\Delta p=\left(1867.2\frac{lbf}{f{t}^2}\right)\times (\frac{1f{t}^2}{144i{n}^2})$

$\Delta p=13lbf/i{n}^2$

Conclusion:

The pressure rise $\Delta p$ expected in lbf/in², according to the dimensionless correlation is 13 lbf/in².