Chapter 14, Problem 28P

The performance data for a centrifugal water pump are shown in Table P14—28 for water at $20°C$
( $Lpm$ = liters per minute). (a) For each row of data, calculate the pump efficiency (percent). Show all units and units and unit conversions for full credit. (b) Estimate the volume flow rate ( $Lpm$ ) and net head ( $m$ ) at the $BEP$ of the pump.

To determine

(a)

The pump efficiency (percent) for each row.

Answer to Problem 28P

The pump efficiency for each row is tabulated below.

$\dot{V}\left(\text{Lpm}\right)$	$H\left(\text{m}\right)$	$\text{bhp}\left(\text{W}\right)$	$\eta \left(\%\right)$
$0$	$47.5$	$133$	$0$
$6.0$	$46.2$	$142$	$31.8$
$12.0$	$42.5$	$153$	$54.4$
$18.0$	$36.2$	$164$	$64.8$
$24.0$	$26.2$	$172$	$59.65$
$30.0$	$15$	$174$	$42.2$
$36.0$	$0$	$174$	$0$

Explanation of Solution

Given information:

The performance data for a centrifugal water pump is tabulated which is shown in the following figure.

  Figure-(I)

The figure shows the table containing the performance of a centrifugal pump with respect to its volume flow rate, head developed, and the power output.

Write the expression for the efficiency of the $i_{\text{th}}$ pump.

   ${\eta }_i=\frac{\left(\rho g{\dot{V}}_i{H}_i\right)}{{\text{bhp}}_i}$    ...... (I)

Here, the density of the water is $\rho$, the gravitational acceleration is $g$, the volume flow rate of the fluid in the $i_{\text{th}}$ row is ${\dot{V}}_i$, the power generation in the $i_{\text{th}}$ row is ${\text{bhp}}_i,$ and the head in the $i_{\text{th}}$ row is $H_i$.

Calculation:

Refer to Table A-3, “Properties of saturated water” to obtain the value of density $\left(\rho \right)$ as $998\text{kg}/{\text{m}}^{\text{3}}$ at $20°\text{C}$.

Substitute $998\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$, $0\text{Lpm}$ for ${\dot{V}}_1$, $47.5\text{m}$ for $H_1,$ and $133\text{W}$ for ${\text{bhp}}_1$ in Equation (I).

   $\begin{array}{c}{\eta }_1=\frac{\left(\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0\text{Lpm}\right)\left(47.5\text{m}\right)\right)}{133\text{W}}\\ =\frac{\left(\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0\text{Lpm}\left(\frac{1{\text{m}}^{\text{3}}/\text{s}}{60000\text{Lpm}}\right)\right)\left(47.5\text{m}\right)\right)}{133\text{W}}\\ =\frac{0{\text{kgm}}^{\text{2}}/{\text{s}}^{\text{3}}}{133\text{W}}\\ =0\%\end{array}$

Thus, the efficiency for the $1\text{st}$ row is $0\%$.

Substitute $998\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $6\text{Lpm}$ for ${\dot{V}}_2$, $46.2\text{m}$ for $H_2,$ and $142\text{W}$ for ${\text{bhp}}_2$ in Equation (I).

   $\begin{array}{c}{\eta }_2=\frac{\left(\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(6\text{Lpm}\right)\left(46.2\text{m}\right)\right)}{142\text{W}}\\ =\frac{\left(\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(6\text{Lpm}\left(\frac{1{\text{m}}^{\text{3}}/\text{s}}{60000\text{Lpm}}\right)\right)\left(46.2\text{m}\right)\right)}{142\text{W}}\\ =\frac{\left(\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left({10}^{-4}{\text{m}}^{\text{3}}/\text{s}\right)\left(46.2\text{m}\right)\right)}{142\text{W}}\\ =\frac{45.23{\text{kgm}}^{\text{2}}/{\text{s}}^{\text{3}}}{142\text{}\text{W}}\end{array}$

   $\begin{array}{c}{\eta }_2=\frac{45.23{\text{kgm}}^{\text{2}}/{\text{s}}^{\text{3}}\left(\frac{1{\text{kgm}}^{\text{2}}/{\text{s}}^{\text{3}}}{1\text{W}}\right)}{142\text{}\text{W}}\\ =\frac{45.23\text{W}}{142\text{W}}\\ =0.318\\ =31.8\%\end{array}$

Thus, the efficiency for the $2\text{nd}$ row is $31.8\%$.

Substitute $998\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $12\text{Lpm}$ for ${\dot{V}}_3$, $42.5\text{m}$ for $H_3,$ and $153\text{W}$ for ${\text{bhp}}_3$ in Equation (I).

   $\begin{array}{c}{\eta }_3=\frac{\left(\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(12\text{Lpm}\right)\left(42.5\text{m}\right)\right)}{153\text{W}}\\ =\frac{\left(\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(12\text{Lpm}\left(\frac{1{\text{m}}^{\text{3}}/\text{s}}{60000\text{Lpm}}\right)\right)\left(42.5\text{m}\right)\right)}{153\text{W}}\\ =\frac{\left(\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(2\times {10}^{-4}{\text{m}}^{\text{3}}/\text{s}\right)\left(42.5\text{m}\right)\right)}{153\text{W}}\\ =\frac{83.2{\text{kgm}}^{\text{2}}/{\text{s}}^{\text{3}}}{153\text{}\text{W}}\end{array}$

   $\begin{array}{c}{\eta }_3=\frac{83.2{\text{kgm}}^{\text{2}}/{\text{s}}^{\text{3}}\left(\frac{1{\text{kgm}}^{\text{2}}/{\text{s}}^{\text{3}}}{1\text{W}}\right)}{153\text{}\text{W}}\\ =\frac{83.2\text{W}}{142\text{W}}\\ =0.544\\ =54.4\%\end{array}$

Thus, the efficiency for the $3\text{rd}$ row is $54.4\%$.

Substitute $998\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $18\text{Lpm}$ for ${\dot{V}}_4$, $36.2\text{m}$ for $H_4,$ and $164\text{W}$ for ${\text{bhp}}_4$ in Equation (I).

   $\begin{array}{c}{\eta }_4=\frac{\left(\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(18\text{Lpm}\right)\left(36.2\text{m}\right)\right)}{164\text{W}}\\ =\frac{\left(\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(18\text{Lpm}\left(\frac{1{\text{m}}^{\text{3}}/\text{s}}{60000\text{Lpm}}\right)\right)\left(36.2\text{m}\right)\right)}{164\text{W}}\\ =\frac{\left(\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(3\times {10}^{-4}{\text{m}}^{\text{3}}/\text{s}\right)\left(36.2\text{m}\right)\right)}{164\text{W}}\\ =\frac{106.32{\text{kgm}}^{\text{2}}/{\text{s}}^{\text{3}}}{164\text{}\text{W}}\end{array}$

   $\begin{array}{c}{\eta }_4=\frac{106.32{\text{kgm}}^{\text{2}}/{\text{s}}^{\text{3}}\left(\frac{1{\text{kgm}}^{\text{2}}/{\text{s}}^{\text{3}}}{1\text{W}}\right)}{164\text{}\text{W}}\\ =\frac{106.32\text{W}}{164\text{W}}\\ =0.648\\ =64.8\%\end{array}$

Thus, the efficiency for the $4\text{th}$ row is $64.8\%$.

Substitute $998\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $24\text{Lpm}$ for ${\dot{V}}_5$, $26.2\text{m}$ for $H_5$ and $172\text{W}$ for ${\text{bhp}}_5$ in Equation (I).

   $\begin{array}{c}{\eta }_5=\frac{\left(\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(24\text{Lpm}\right)\left(26.2\text{m}\right)\right)}{172\text{W}}\\ =\frac{\left(\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(24\text{Lpm}\left(\frac{1{\text{m}}^{\text{3}}/\text{s}}{60000\text{Lpm}}\right)\right)\left(26.2\text{m}\right)\right)}{172\text{W}}\\ =\frac{\left(\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(4\times {10}^{-4}{\text{m}}^{\text{3}}/\text{s}\right)\left(26.2\text{m}\right)\right)}{172\text{W}}\\ =\frac{102.6{\text{kgm}}^{\text{2}}/{\text{s}}^{\text{3}}}{172\text{}\text{W}}\end{array}$

   $\begin{array}{c}{\eta }_5=\frac{102.6{\text{kgm}}^{\text{2}}/{\text{s}}^{\text{3}}\left(\frac{1{\text{kgm}}^{\text{2}}/{\text{s}}^{\text{3}}}{1\text{W}}\right)}{172\text{}\text{W}}\\ =\frac{102.6\text{W}}{172\text{W}}\\ =0.5965\\ =59.65\%\end{array}$

Thus, the efficiency for the $5\text{th}$ row is $59.65\%$.

Substitute $998\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $30\text{Lpm}$ for ${\dot{V}}_6$, $15\text{m}$ for $H_6,$ and $174\text{W}$ for ${\text{bhp}}_6$ in Equation (I).

   $\begin{array}{c}{\eta }_6=\frac{\left(\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(30\text{Lpm}\right)\left(15\text{m}\right)\right)}{174\text{W}}\\ =\frac{\left(\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(30\text{Lpm}\left(\frac{1{\text{m}}^{\text{3}}/\text{s}}{60000\text{Lpm}}\right)\right)\left(15\text{m}\right)\right)}{174\text{W}}\\ =\frac{\left(\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(5\times {10}^{-4}{\text{m}}^{\text{3}}/\text{s}\right)\left(15\text{m}\right)\right)}{174\text{W}}\\ =\frac{73.42{\text{kgm}}^{\text{2}}/{\text{s}}^{\text{3}}}{174\text{}\text{W}}\end{array}$

   $\begin{array}{c}{\eta }_6=\frac{73.42{\text{kgm}}^{\text{2}}/{\text{s}}^{\text{3}}\left(\frac{1{\text{kgm}}^{\text{2}}/{\text{s}}^{\text{3}}}{1\text{W}}\right)}{174\text{}\text{W}}\\ =\frac{73.42\text{W}}{174\text{W}}\\ =0.422\\ =42.2\%\end{array}$

Thus, the efficiency for the $6\text{th}$ row is $42.2\%$.

Substitute $998\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $36\text{Lpm}$ for ${\dot{V}}_7$, $0\text{m}$ for $H_7,$ and $174\text{W}$ for ${\text{bhp}}_7$ in Equation (I).

   $\begin{array}{c}{\eta }_7=\frac{\left(\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(36\text{Lpm}\right)\left(0\text{m}\right)\right)}{174\text{W}}\\ =\frac{\left(\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(36\text{Lpm}\left(\frac{1{\text{m}}^{\text{3}}/\text{s}}{60000\text{Lpm}}\right)\right)\left(0\text{m}\right)\right)}{174\text{W}}\\ =\frac{\left(\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(6\times {10}^{-4}{\text{m}}^{\text{3}}/\text{s}\right)\left(0\text{m}\right)\right)}{174\text{W}}\\ =0\%\end{array}$

Thus, the efficiency for the $7\text{th}$ row is $0\%$.

Conclusion:

Tabulate the calculated efficiencies for each row.

$\dot{V}\left(\text{Lpm}\right)$	$H\left(\text{m}\right)$	$\text{bhp}\left(\text{W}\right)$	$\eta \left(\%\right)$
$0$	$47.5$	$133$	$0$
$6.0$	$46.2$	$142$	$31.8$
$12.0$	$42.5$	$153$	$54.4$
$18.0$	$36.2$	$164$	$64.8$
$24.0$	$26.2$	$172$	$59.65$
$30.0$	$15$	$174$	$42.2$
$36.0$	$0$	$174$	$0$

To determine

(b)

The volume flow rate $\left(\text{Lpm}\right)$ and the net head $\left(\text{m}\right)$ at the BEP of the pump.

Explanation of Solution

The best efficiency point (BEP) of the pump is obtained at the $4\text{th}$ row which is $64.8\%$.

Therefore, the volume flow rate is $18.0\text{Lpm}$ and the net head is $36.2\text{m}$ at the BEP of the pump.