Chapter 14, Problem 52P

The performance data for a centrifugal water pump are shown in Table P14-47E for water at $77°\text{F}$ (gpm= gallons per minute). (a) For each row of data, calculate the pump efficiency (pe1tent). Show all units and unit conwrsions for full credit. (b) Estimate the volume flow rate (gpm) and net head (ft) at the BEP of the pump.

To determine

(a)

The pump efficiency for each data tabulated.

Answer to Problem 52P

The pump efficiency for each data is represented in Table-(2).

$\dot{V}\text{gpm}$	$H\text{ft}$	$bhp\text{hp}$	$\eta \%$
$0.0$	$19.0$	$0.06$	$0.0$
$4.0$	$18.5$	$0.064$	$29.2$
$8.0$	$17.0$	$0.069$	$49.8$
$12.0$	$14.5$	$0.074$	$59.5$
$16.0$	$10.5$	$0.079$	$58.3$
$20.0$	$6.0$	$0.08$	$37.9$
$24.0$	$0.0$	$0.078$	$0.0$

Explanation of Solution

Given information:

Table-(1) represents the volume flow rate, height and power of pump.

$\dot{V}\text{gpm}$	$H\text{ft}$	$bhp\text{hp}$
$0.0$	$19.0$	$0.06$
$4.0$	$18.5$	$0.064$
$8.0$	$17.0$	$0.069$
$12.0$	$14.5$	$0.074$
$16.0$	$10.5$	$0.079$
$20.0$	$6.0$	$0.08$
$24.0$	$0.0$	$0.078$

Write the expression for pump efficiency.

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

Here, density of water is $\rho$, acceleration due to gravity is $g$, volume flow rate is $\dot{V}$, head developed is $H$, brake power is $bhp$.

Calculation:

Refer to Table-A-3E "Properties of saturated water" to obtain density of water as $1.94\text{slug}/{\text{ft}}^{\text{3}}$.

Substitute $1.94\text{slug}/{\text{ft}}^{\text{3}}$for $\rho$, $32.2\text{ft}/{\text{s}}^{\text{2}}$for $g$, $0.0\text{gpm}$for $\dot{V}$, $19\text{ft}$for $H$and $0.06\text{hp}$for $bhp$in Equation (I).

  $\begin{array}{c}\eta =\frac{\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(0.0\text{gpm}\right)\left(19\text{ft}\right)}{0.06\text{hp}}\times 100\%\\ =\frac{\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(0.0\text{gpm}\times \frac{35.315{\text{ft}}^{\text{3}}/\text{s}}{15850\text{gpm}}\right)\left(19\text{ft}\right)}{0.06\text{hp}\times \frac{550\text{slug}\cdot {\text{ft}}^2/{\text{s}}^3}{1\text{hp}}}\times 100\%\\ =\frac{\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(0.0\times \frac{35.315}{15850}{\text{ft}}^{\text{3}}/\text{s}\right)\left(19\text{ft}\right)}{0.06\times 550\text{slug}\cdot {\text{ft}}^2/{\text{s}}^3}\times 100\%\\ =0\%\end{array}$

Substitute $1.94\text{slug}/{\text{ft}}^{\text{3}}$for $\rho$, $32.2\text{ft}/{\text{s}}^{\text{2}}$for $g$, $4.0\text{gpm}$for $\dot{V}$, $18.5\text{ft}$for $H$and $0.064\text{hp}$for $bhp$in Equation (I).

  $\begin{array}{c}\eta =\frac{\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(4.0\text{gpm}\right)\left(18.5\text{ft}\right)}{0.064\text{hp}}\times 100\%\\ =\frac{\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(4.0\text{gpm}\times \frac{35.315{\text{ft}}^{\text{3}}/\text{s}}{15850\text{gpm}}\right)\left(18.5\text{ft}\right)}{0.064\text{hp}\times \frac{550\text{slug}\cdot {\text{ft}}^2/{\text{s}}^3}{1\text{hp}}}\times 100\%\\ =\frac{\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(4.0\times \frac{35.315}{15850}{\text{ft}}^{\text{3}}/\text{s}\right)\left(18.5\text{ft}\right)}{0.064\times 550\text{slug}\cdot {\text{ft}}^2/{\text{s}}^3}\times 100\%\\ =29.2\%\end{array}$

Substitute $1.94\text{slug}/{\text{ft}}^{\text{3}}$for $\rho$, $32.2\text{ft}/{\text{s}}^{\text{2}}$for $g$, $8.0\text{gpm}$for $\dot{V}$, $17\text{ft}$for $H$and $0.069\text{hp}$for $bhp$in Equation (I).

  $\begin{array}{c}\eta =\frac{\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(8.0\text{gpm}\right)\left(17\text{ft}\right)}{0.069\text{hp}}\times 100\%\\ =\frac{\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(8.0\text{gpm}\times \frac{35.315{\text{ft}}^{\text{3}}/\text{s}}{15850\text{gpm}}\right)\left(17\text{ft}\right)}{0.069\text{hp}\times \frac{550\text{slug}\cdot {\text{ft}}^2/{\text{s}}^3}{1\text{hp}}}\times 100\%\\ =\frac{\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(8.0\times \frac{35.315}{15850}{\text{ft}}^{\text{3}}/\text{s}\right)\left(17\text{ft}\right)}{0.069\times 550\text{slug}\cdot {\text{ft}}^2/{\text{s}}^3}\times 100\%\\ =49.8\%\end{array}$

Substitute $1.94\text{slug}/{\text{ft}}^{\text{3}}$for $\rho$, $32.2\text{ft}/{\text{s}}^{\text{2}}$for $g$, $12.0\text{gpm}$for $\dot{V}$, $14.5\text{ft}$for $H$and $0.074\text{hp}$for $bhp$in Equation (I).

  $\begin{array}{c}\eta =\frac{\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(12.0\text{gpm}\right)\left(14.5\text{ft}\right)}{0.074\text{hp}}\times 100\%\\ =\frac{\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(12.0\text{gpm}\times \frac{35.315{\text{ft}}^{\text{3}}/\text{s}}{15850\text{gpm}}\right)\left(14.5\text{ft}\right)}{0.074\text{hp}\times \frac{550\text{slug}\cdot {\text{ft}}^2/{\text{s}}^3}{1\text{hp}}}\times 100\%\\ =\frac{\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(12.0\times \frac{35.315}{15850}{\text{ft}}^{\text{3}}/\text{s}\right)\left(14.5\text{ft}\right)}{0.074\times 550\text{slug}\cdot {\text{ft}}^2/{\text{s}}^3}\times 100\%\\ =59.5\%\end{array}$

Substitute $1.94\text{slug}/{\text{ft}}^{\text{3}}$for $\rho$, $32.2\text{ft}/{\text{s}}^{\text{2}}$for $g$, $16.0\text{gpm}$for $\dot{V}$, $10.5\text{ft}$for $H$and $0.079\text{hp}$for $bhp$in Equation (I).

  $\begin{array}{c}\eta =\frac{\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(16.0\text{gpm}\right)\left(10.5\text{ft}\right)}{0.079\text{hp}}\times 100\%\\ =\frac{\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(16.0\text{gpm}\times \frac{35.315{\text{ft}}^{\text{3}}/\text{s}}{15850\text{gpm}}\right)\left(10.5\text{ft}\right)}{0.079\text{hp}\times \frac{550\text{slug}\cdot {\text{ft}}^2/{\text{s}}^3}{1\text{hp}}}\times 100\%\\ =\frac{\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(16.0\times \frac{35.315}{15850}{\text{ft}}^{\text{3}}/\text{s}\right)\left(10.5\text{ft}\right)}{0.079\times 550\text{slug}\cdot {\text{ft}}^2/{\text{s}}^3}\times 100\%\\ =53.8\%\end{array}$

Substitute $1.94\text{slug}/{\text{ft}}^{\text{3}}$for $\rho$, $32.2\text{ft}/{\text{s}}^{\text{2}}$for $g$, $20.0\text{gpm}$for $\dot{V}$, $6\text{ft}$for $H$and $0.08\text{hp}$for $bhp$in Equation (I).

  $\begin{array}{c}\eta =\frac{\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(20.0\text{gpm}\right)\left(6\text{ft}\right)}{0.08\text{hp}}\times 100\%\\ =\frac{\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(20.0\text{gpm}\times \frac{35.315{\text{ft}}^{\text{3}}/\text{s}}{15850\text{gpm}}\right)\left(6\text{ft}\right)}{0.08\text{hp}\times \frac{550\text{slug}\cdot {\text{ft}}^2/{\text{s}}^3}{1\text{hp}}}\times 100\%\\ =\frac{\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(20.0\times \frac{35.315}{15850}{\text{ft}}^{\text{3}}/\text{s}\right)\left(6\text{ft}\right)}{0.08\times 550\text{slug}\cdot {\text{ft}}^2/{\text{s}}^3}\times 100\%\\ =37.9\%\end{array}$

Substitute $1.94\text{slug}/{\text{ft}}^{\text{3}}$for $\rho$, $32.2\text{ft}/{\text{s}}^{\text{2}}$for $g$, $24.0\text{gpm}$for $\dot{V}$, $0\text{ft}$for $H$and $0.78\text{hp}$for $bhp$in Equation (I).

  $\begin{array}{c}\eta =\frac{\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(24.0\text{gpm}\right)\left(0\text{ft}\right)}{0.078\text{hp}}\times 100\%\\ =\frac{\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(24.0\text{gpm}\times \frac{35.315{\text{ft}}^{\text{3}}/\text{s}}{15850\text{gpm}}\right)\left(0\text{ft}\right)}{0.078\text{hp}\times \frac{550\text{slug}\cdot {\text{ft}}^2/{\text{s}}^3}{1\text{hp}}}\times 100\%\\ =\frac{\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(24.0\times \frac{35.315}{15850}{\text{ft}}^{\text{3}}/\text{s}\right)\left(0\text{ft}\right)}{0.078\times 550\text{slug}\cdot {\text{ft}}^2/{\text{s}}^3}\times 100\%\\ =0\%\end{array}$

Table-(2) represents the efficiency of pump for different head, volume flow rate and power.

Table-(2)

$\dot{V}\text{gpm}$	$H\text{ft}$	$bhp\text{hp}$	$\eta \%$
$0.0$	$19.0$	$0.06$	$0.0$
$4.0$	$18.5$	$0.064$	$29.2$
$8.0$	$17.0$	$0.069$	$49.8$
$12.0$	$14.5$	$0.074$	$59.5$
$16.0$	$10.5$	$0.079$	$58.3$
$20.0$	$6.0$	$0.08$	$37.9$
$24.0$	$0.0$	$0.078$	$0.0$

To determine

(b)

The volume flow rate for BEP of the pump.

The net head for BEP of the pump.

Answer to Problem 52P

The volume flow rate for BEP of the pump is $12\text{gpm}$.

The net head for BEP of the pump is $14.5\text{ft}$.

Explanation of Solution

Given information:

Table-(2) represents the efficiency of pump for different head, volume flow rate and power.

Table-(2)

$\dot{V}\text{gpm}$	$H\text{ft}$	$bhp\text{hp}$	$\eta \%$
$0.0$	$19.0$	$0.06$	$0.0$
$4.0$	$18.5$	$0.064$	$29.2$
$8.0$	$17.0$	$0.069$	$49.8$
$12.0$	$14.5$	$0.074$	$59.5$
$16.0$	$10.5$	$0.079$	$58.3$
$20.0$	$6.0$	$0.08$	$37.9$
$24.0$	$0.0$	$0.078$	$0.0$

Here, from the Table-(2) the maximum efficiency of pump is $59.5\%$at the volume flow rate of $12\text{gpm}$and head of $14.5\text{ft}$.

Conclusion:

The volume flow rate for BEP of the pump is $12\text{gpm}$.

The net head for BEP of the pump is $14.5\text{ft}$.

To determine

(b)

The volume flow rate for BEP of the pump.

The net head for BEP of the pump.

Answer to Problem 52P

The volume flow rate for BEP of the pump is $12\text{gpm}$.

The net head for BEP of the pump is $14.5\text{ft}$.

Explanation of Solution

Given information:

Table-(2) represents the efficiency of pump for different head, volume flow rate and power.

Table-(2)

$\dot{V}\text{gpm}$	$H\text{ft}$	$bhp\text{hp}$	$\eta \%$
$0.0$	$19.0$	$0.06$	$0.0$
$4.0$	$18.5$	$0.064$	$29.2$
$8.0$	$17.0$	$0.069$	$49.8$
$12.0$	$14.5$	$0.074$	$59.5$
$16.0$	$10.5$	$0.079$	$58.3$
$20.0$	$6.0$	$0.08$	$37.9$
$24.0$	$0.0$	$0.078$	$0.0$

Here, from the Table-(2) the maximum efficiency of pump is $59.5\%$at the volume flow rate of $12\text{gpm}$and head of $14.5\text{ft}$.

Conclusion:

The volume flow rate for BEP of the pump is $12\text{gpm}$.

The net head for BEP of the pump is $14.5\text{ft}$.