Chapter 2.8, Problem 68P

Water is pumped from a lake to a storage tank 15 m above at a rate of 70 L/s while consuming 15.4 kW of electric power. Disregarding any frictional losses in the pipes and any changes in kinetic energy, determine (a) the overall efficiency of the pump–motor unit and (b) the pressure difference between the inlet and the exit of the pump.

FIGURE P2–69

To determine

The overall efficiency of the pump-motor unit.

Answer to Problem 68P

The overall efficiency of the pump-motor is $\underset{\_}{66.9\%}$.

Explanation of Solution

Write the mass flow rate of water.

$\dot{m}=\rho \dot{V}$ (I)

Here, the volumetric flow rate of water is $\dot{V}$ and density of water is $\rho$.

The potential energy at point 1 is considered as 0$\left(p{e}_1=0\right)$ because the lake surface is taken as the reference level $\left(z_1=0\right)$.

Write the equation of potential energy at point 2.

$p{e}_2=g{z}_2$ (II)

Here, acceleration due to gravity is g and lake surface at point 2 is $z_2$.

Calculate the rate at which mechanical energy of fluid supplied to the pump.

$\begin{array}{c}\Delta {\dot{E}}_{\text{mech,fluid}}=\dot{m}\left(e_{\text{mech,out}}-e_{\text{mech,in}}\right)\\ =\dot{m}\left(p{e}_2-p{e}_1\right)\end{array}$ (III)

Here, the mechanical energy of water inlet and outlet are $e_{\text{mech,in}}$ and $e_{\text{mech,out}}$ respectively.

Calculate the overall efficiency of the combined pump-motor.

${\eta }_{\text{pump-motor}}=\frac{\Delta {\dot{E}}_{\text{mech,fluid}}}{{\dot{W}}_{\text{elect,in}}}\times 100\%$ (IV)

Here, and electric power consumption is ${\dot{W}}_{\text{elect,in}}$.

Conclusion:

Substitute $1000{\text{kg/m}}^{\text{3}}$ for $\rho$ and $0.070{\text{m}}^{\text{3}}/\text{s}$ for $\dot{V}$ in Equation (I).

$\begin{array}{c}\dot{m}=1000{\text{kg/m}}^{\text{3}}\left(0.070{\text{m}}^{\text{3}}/\text{s}\right)\\ =70\text{kg/s}\end{array}$

Substitute $9.81{\text{m/s}}^{\text{2}}$ for g and 15 m for $z_2$ in Equation (II).

$\begin{array}{c}p{e}_2=9.81{\text{m/s}}^{\text{2}}\left(15\text{m}\right)\\ =9.81{\text{m/s}}^{\text{2}}\left(15\text{m}\right)\left(\frac{1\text{kJ/kg}}{1000{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\\ =0.1472\text{kJ/kg}\end{array}$

Substitute 70 kg/s for $\dot{m}$, 0 for $p{e}_1$, and 0.1472 kJ/kg for $p{e}_2$ in Equation (III).

$\begin{array}{c}\Delta {\dot{E}}_{\text{mech,fluid}}=\left(70\text{kg/s}\right)\left(0.1472\text{kJ/kg}-0\right)\\ =10.3\text{kW}\end{array}$

Substitute 10.3 kW for $\Delta {\dot{E}}_{\text{mech,fluid}}$ and 15.4 kW for ${\dot{W}}_{\text{elect,in}}$ in Equation (IV).

$\begin{array}{c}{\eta }_{\text{pump-motor}}=\frac{10.3\text{kW}}{15.4\text{kW}}\times 100\%\\ =0.669\times 100\%\\ =66.9\%\end{array}$

Thus, the overall efficiency of the pump-motor is $\underset{\_}{66.9\%}$.

To determine

The pressure difference between the inlet and the exit of the pump.

Answer to Problem 68P

The pressure difference between the inlet and the exit of the pump is $\underset{\_}{147\text{kPa}}$.

Explanation of Solution

Calculate the rate at which mechanical energy of fluid supplied by the pump.

$\begin{array}{l}\Delta {\dot{E}}_{\text{mech,fluid}}=\dot{m}\left(e_{\text{mech,out}}-e_{\text{mech,in}}\right)\\ \Delta {\dot{E}}_{\text{mech,fluid}}=\dot{m}\left(\frac{P_2-P_1}{\rho }\right)\\ \Delta {\dot{E}}_{\text{mech,fluid}}=\dot{V}\Delta P\\ \Delta P=\frac{\Delta {\dot{E}}_{\text{mech,fluid}}}{\dot{V}}\end{array}$ (V)

Here, change in the pressure of water is $\Delta P$, pressure at inlet and exit of the pump is $P_1$, and $P_2$.

Conclusion:

Substitute 10.3 kJ/s for $\Delta {\dot{E}}_{\text{mech,fluid}}$ and $0.070{\text{m}}^{\text{3}}/\text{s}$ for $\dot{V}$ in equation (V).

$\begin{array}{c}\Delta P=\frac{10.3\text{kJ/s}}{0.070{\text{m}}^{\text{3}}/\text{s}}\\ =\frac{10.3\text{kJ/s}}{0.070{\text{m}}^{\text{3}}/\text{s}}\left(\frac{1\text{kPa}\cdot {\text{m}}^{\text{3}}}{1\text{kJ}}\right)\\ =147\text{kPa}\end{array}$

Thus, the pressure difference between the inlet and the exit of the pump is $\underset{\_}{147\text{kPa}}$.