Chapter 14, Problem 31EP

The performance data for a centrifugal water pump arc shown in Table P14—31E for water at $77°F$ ( $gpm$ = gallons per minute). (a) For each row of data, calculate the pump efficiency (percent). Show all units and unit conversions for full credit. (b) Estimate the volume flow rate ( $gpm$ ) and net head $\left(ft\right)$ at the $\text{BEP}$ of the pump.

To determine

(a)

The centrifugal pump efficiency for each row.

Answer to Problem 31EP

The pimp efficiency for each row is tabulated below.

$\dot{V}\left(\text{gpm}\right)$	$H\left(\text{ft}\right)$	$\text{bhp}\left(\text{hp}\right)$	$\eta \left(\%\right)$
$0.0$	$19.0$	$0.06$	$0$
$4.0$	$18.5$	$0.064$	$29.2$
$8.0$	$17.0$	$0.069$	$49.8$
$12.0$	$14.5$	$0.074$	$59.4$
$16.0$	$10.5$	$0.079$	$53.7$
$20.0$	$6.0$	$0.08$	$37.9$
$24.0$	$0.0$	$0.078$	$0$

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 for the $i_{\text{th}}$ row is $\dot{V}$, the power generation in $i_{\text{th}}$ row is $\text{bhp,}$ and the head of the pump at the $i_{\text{th}}$ row is $H$.

Write the formula for the interpolation of two variables.

   $y_2=\frac{\left(x_2-x_1\right)\left(y_3-y_1\right)}{\left(x_3-x_1\right)}+y_1$    ...... (II)

Here, the density of the water is represented by the variable $y$ and the temperature is represented by the variable $x$.

Calculation:

Refer to Table A-3E, “Properties of saturated water” to obtain the values of $x_1$ as $70°\text{F}$, $x_2$ as $77°\text{F}$, $x_3$ as $80°\text{F}$, $y_1$ as $62.30\text{lbm}/{\text{ft}}^{\text{3}},$ and $y_3$ as $62.22\text{lbm}/{\text{ft}}^{\text{3}}$.

Prepare the table for pressure and temperature.

Temperature, $°\text{F}$	Density, $\text{lbm}/{\text{ft}}^{\text{3}}$
$70\left(x_1\right)$	$62.30\left(y_1\right)$
$77\left(x_2\right)$	$\rho \left(y_2\right)$
$80\left(x_3\right)$	$62.22\left(y_3\right)$

Substitute $70°\text{F}$ for $x_1$, $77°\text{F}$ for $x_2$, $80°\text{F}$ for $x_3$, $62.30\text{lbm}/{\text{ft}}^{\text{3}}$ for $y_1$, $62.22\text{lbm}/{\text{ft}}^{\text{3}}$ for $y_3,$ and $\rho$ for $y_2$ in Equation (II).

   $\begin{array}{c}{y}_2=\frac{\left(77°\text{F}-70°\text{F}\right)\left(62.22\text{lbm}/{\text{ft}}^{\text{3}}-62.30\text{lbm}/{\text{ft}}^{\text{3}}\right)}{\left(80°\text{F}-70°\text{F}\right)}+62.30\text{lbm}/{\text{ft}}^{\text{3}}\\ =-0.056\text{lbm}/{\text{ft}}^{\text{3}}+62.30\text{lbm}/{\text{ft}}^{\text{3}}\\ =62.24\text{lbm}/{\text{ft}}^{\text{3}}\left(\frac{1\text{slug}/{\text{ft}}^{\text{3}}}{32.17\text{lbm}/{\text{ft}}^{\text{3}}}\right)\\ =1.94\text{slug}/{\text{ft}}^{\text{3}}\end{array}$

Substitute $1.94\text{slug}/{\text{ft}}^{\text{3}}$ for $\rho$, $32.17\text{ft}/{\text{s}}^{\text{2}}$ for $g$, $0.0g\text{pm}$ for ${\dot{V}}_1$, $19\text{ft}$ for $H_1,$ and $0.06\text{hp}$ for ${\text{bhp}}_1$ in Equation (I).

   $\begin{array}{c}{\eta }_1=\frac{\left(\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.17\text{ft}/{\text{s}}^{\text{2}}\right)\left(0.0g\text{pm}\right)\left(19\text{ft}\right)\right)}{0.06\text{hp}}\\ =\frac{0}{0.06\text{hp}}\\ =0\\ =0\%\end{array}$

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

Substitute $1.94\text{slug}/{\text{ft}}^{\text{3}}$ for $\rho$, $32.17\text{ft}/{\text{s}}^{\text{2}}$ for $g$, $4.0g\text{pm}$ for ${\dot{V}}_2$, $18.5\text{ft}$ for $H_2,$ and $0.064\text{hp}$ for ${\text{bhp}}_2$ in Equation (I).

   $\begin{array}{c}{\eta }_2=\frac{\left(\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.17\text{ft}/{\text{s}}^{\text{2}}\right)\left(4.0g\text{pm}\right)\left(18.5\text{ft}\right)\right)}{0.064\text{hp}}\\ =\frac{\left(\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.17\text{ft}/{\text{s}}^{\text{2}}\right)\left(4.0g\text{pm}\left(\frac{1{\text{ft}}^{\text{3}}/\text{s}}{448.5\text{gpm}}\right)\right)\left(18.5\text{ft}\right)\right)}{0.064\text{hp}\left(\frac{550\text{lbf}\cdot \text{ft}/\text{s}}{1\text{hp}}\right)}\\ =\frac{\left(\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.17\text{ft}/{\text{s}}^{\text{2}}\right)\left(8.9\times {10}^{-3}{\text{ft}}^{\text{3}}/\text{s}\right)\left(18.5\text{ft}\right)\right)}{35.2\text{lbf}\cdot \text{ft}/\text{s}}\\ =\frac{10.27\text{slug}\cdot \text{ft}/{\text{s}}^{\text{2}}\left(\frac{1\text{lbf}}{1\text{slug}\cdot \text{ft}/{\text{s}}^{\text{2}}}\right)}{35.2\text{lbf}}\end{array}$

   $\begin{array}{c}{\eta }_2=\frac{10.27\text{lbf}}{35.2\text{lbf}}\\ =0.292\\ =29.2\%\end{array}$

Thus, the efficiency for the $\text{2nd}$ row is $29.2\%$.

Substitute $1.94\text{slug}/{\text{ft}}^{\text{3}}$ for $\rho$, $32.17\text{ft}/{\text{s}}^{\text{2}}$ for $g$, $8.0g\text{pm}$ for ${\dot{V}}_3$, $17\text{ft}$ for $H_3,$ and $0.069\text{hp}$ for ${\text{bhp}}_3$ in Equation (I).

   $\begin{array}{c}{\eta }_3=\frac{\left(\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.17\text{ft}/{\text{s}}^{\text{2}}\right)\left(8.0g\text{pm}\right)\left(17\text{ft}\right)\right)}{0.069\text{hp}}\\ =\frac{\left(\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.17\text{ft}/{\text{s}}^{\text{2}}\right)\left(8.0g\text{pm}\left(\frac{1{\text{ft}}^{\text{3}}/\text{s}}{448.5\text{gpm}}\right)\right)\left(17\text{ft}\right)\right)}{0.069\text{hp}\left(\frac{550\text{lbf}\cdot \text{ft}/\text{s}}{1\text{hp}}\right)}\\ =\frac{\left(\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.17\text{ft}/{\text{s}}^{\text{2}}\right)\left(0.0178{\text{ft}}^{\text{3}}/\text{s}\right)\left(17\text{ft}\right)\right)}{37.95\text{lbf}\cdot \text{ft}/\text{s}}\\ =\frac{18.88\text{slug}\cdot \text{ft}/{\text{s}}^{\text{2}}\left(\frac{1\text{lbf}}{1\text{slug}\cdot \text{ft}/{\text{s}}^{\text{2}}}\right)}{37.95\text{lbf}}\end{array}$

   $\begin{array}{c}{\eta }_3=\frac{18.88\text{lbf}}{37.95\text{lbf}}\\ =0.498\\ =49.8\%\end{array}$

Thus, the efficiency for the $\text{3rd}$ row is $49.8\%$.

Substitute $1.94\text{slug}/{\text{ft}}^{\text{3}}$ for $\rho$, $32.17\text{ft}/{\text{s}}^{\text{2}}$ for $g$, $12g\text{pm}$ for ${\dot{V}}_4$, $14.5\text{ft}$ for $H_4,$ and $0.074\text{hp}$ for ${\text{bhp}}_4$ in Equation (I).

   $\begin{array}{c}{\eta }_4=\frac{\left(\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.17\text{ft}/{\text{s}}^{\text{2}}\right)\left(12g\text{pm}\right)\left(14.5\text{ft}\right)\right)}{0.074\text{hp}}\\ =\frac{\left(\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.17\text{ft}/{\text{s}}^{\text{2}}\right)\left(12g\text{pm}\left(\frac{1{\text{ft}}^{\text{3}}/\text{s}}{448.5\text{gpm}}\right)\right)\left(14.5\text{ft}\right)\right)}{0.074\text{hp}\left(\frac{550\text{lbf}\cdot \text{ft}/\text{s}}{1\text{hp}}\right)}\\ =\frac{\left(\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.17\text{ft}/{\text{s}}^{\text{2}}\right)\left(0.0267{\text{ft}}^{\text{3}}/\text{s}\right)\left(14.5\text{ft}\right)\right)}{40.7\text{lbf}\cdot \text{ft}/\text{s}}\\ =\frac{24.16\text{slug}\cdot \text{ft}/{\text{s}}^{\text{2}}\left(\frac{1\text{lbf}}{1\text{slug}\cdot \text{ft}/{\text{s}}^{\text{2}}}\right)}{40.7\text{lbf}}\end{array}$

   $\begin{array}{c}{\eta }_4=\frac{24.16\text{lbf}}{40.7\text{lbf}}\\ =0.594\\ =59.4\%\end{array}$

Thus, the efficiency for the $\text{4th}$ row is $59.4\%$.

Substitute $1.94\text{slug}/{\text{ft}}^{\text{3}}$ for $\rho$, $32.17\text{ft}/{\text{s}}^{\text{2}}$ for $g$, $16g\text{pm}$ for ${\dot{V}}_5$, $10.5\text{ft}$ for $H_5,$ and $0.079\text{hp}$ for ${\text{bhp}}_5$ in Equation (I).

   $\begin{array}{c}{\eta }_5=\frac{\left(\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.17\text{ft}/{\text{s}}^{\text{2}}\right)\left(16g\text{pm}\right)\left(10.5\text{ft}\right)\right)}{0.079\text{hp}}\\ =\frac{\left(\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.17\text{ft}/{\text{s}}^{\text{2}}\right)\left(16g\text{pm}\left(\frac{1{\text{ft}}^{\text{3}}/\text{s}}{448.5\text{gpm}}\right)\right)\left(10.5\text{ft}\right)\right)}{0.079\text{hp}\left(\frac{550\text{lbf}\cdot \text{ft}/\text{s}}{1\text{hp}}\right)}\\ =\frac{\left(\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.17\text{ft}/{\text{s}}^{\text{2}}\right)\left(0.0356{\text{ft}}^{\text{3}}/\text{s}\right)\left(10.5\text{ft}\right)\right)}{43.45\text{lbf}\cdot \text{ft}/\text{s}}\\ =\frac{23.32\text{slug}\cdot \text{ft}/{\text{s}}^{\text{2}}\left(\frac{1\text{lbf}}{1\text{slug}\cdot \text{ft}/{\text{s}}^{\text{2}}}\right)}{43.45\text{lbf}}\end{array}$

   $\begin{array}{c}{\eta }_5=\frac{23.32\text{lbf}}{43.45\text{lbf}}\\ =0.537\\ =53.7\%\end{array}$

Thus, the efficiency for the $\text{5th}$ row is $53.7\%$.

Substitute $1.94\text{slug}/{\text{ft}}^{\text{3}}$ for $\rho$, $32.17\text{ft}/{\text{s}}^{\text{2}}$ for $g$, $20g\text{pm}$ for ${\dot{V}}_6$, $6\text{ft}$ for $H_6,$ and $0.08\text{hp}$ for ${\text{bhp}}_6$ in Equation (I).

   $\begin{array}{c}{\eta }_6=\frac{\left(\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.17\text{ft}/{\text{s}}^{\text{2}}\right)\left(20g\text{pm}\right)\left(6\text{ft}\right)\right)}{0.08\text{hp}}\\ =\frac{\left(\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.17\text{ft}/{\text{s}}^{\text{2}}\right)\left(20g\text{pm}\left(\frac{1{\text{ft}}^{\text{3}}/\text{s}}{448.5\text{gpm}}\right)\right)\left(6\text{ft}\right)\right)}{0.08\text{hp}\left(\frac{550\text{lbf}\cdot \text{ft}/\text{s}}{1\text{hp}}\right)}\\ =\frac{\left(\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.17\text{ft}/{\text{s}}^{\text{2}}\right)\left(0.0445{\text{ft}}^{\text{3}}/\text{s}\right)\left(\text{6}\text{ft}\right)\right)}{44\text{lbf}\cdot \text{ft}/\text{s}}\\ =\frac{16.66\text{slug}\cdot \text{ft}/{\text{s}}^{\text{2}}\left(\frac{1\text{lbf}}{1\text{slug}\cdot \text{ft}/{\text{s}}^{\text{2}}}\right)}{44\text{lbf}}\end{array}$

   $\begin{array}{c}{\eta }_6=\frac{16.66\text{lbf}}{44\text{lbf}}\\ =0.379\\ =37.9\%\end{array}$

Thus, the efficiency for the $\text{6th}$ row is $37.9\%$.

Substitute $1.94\text{slug}/{\text{ft}}^{\text{3}}$ for $\rho$, $32.17\text{ft}/{\text{s}}^{\text{2}}$ for $g$, $24g\text{pm}$ for ${\dot{V}}_7$, $0\text{ft}$ for $H_7,$ and $0.078\text{hp}$ for ${\text{bhp}}_7$ in Equation (I).

   $\begin{array}{c}{\eta }_7=\frac{\left(\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.17\text{ft}/{\text{s}}^{\text{2}}\right)\left(24g\text{pm}\right)\left(0\text{ft}\right)\right)}{0.078\text{hp}}\\ =\frac{0}{0.078\text{hp}}\\ =0\%\end{array}$

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

Conclusion:

Tabulate the calculated efficiencies for each row.

$\dot{V}\left(\text{gpm}\right)$	$H\left(\text{ft}\right)$	$\text{bhp}\left(\text{hp}\right)$	$\eta \left(\%\right)$
$0.0$	$19.0$	$0.06$	$0$
$4.0$	$18.5$	$0.064$	$29.2$
$8.0$	$17.0$	$0.069$	$49.8$
$12.0$	$14.5$	$0.074$	$59.4$
$16.0$	$10.5$	$0.079$	$53.7$
$20.0$	$6.0$	$0.08$	$37.9$
$24.0$	$0.0$	$0.078$	$0$

To determine

(b)

The volume flow rate $\left(\text{gpm}\right)$ and the net head $\left(\text{ft}\right)$ 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 $59.4\%$.

Therefore, the volume flow rate is $12.0\text{gpm}$ and the net head is $14.5\text{ft}$ at the BEP of the pump.