Chapter 2.8, Problem 73P

An oil pump is drawing 44 kW of electric power while pumping oil with ρ = 860 kg/m³ at a rate of 0.1 m³/s. The inlet and outlet diameters of the pipe are 8 cm and 12 cm, respectively. If the pressure rise of oil in the pump is measured to be 500 kPa and the motor efficiency is 90 percent, determine the mechanical efficiency of the pump.

FIGURE P2–73

To determine

The mechanical efficiency of the pump.

Answer to Problem 73P

The mechanical efficiency of the pump is $\underset{\_}{91.8\%}$.

Explanation of Solution

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

$\begin{array}{c}\Delta {\dot{E}}_{\text{mech,fluid}}=\dot{m}\left(e_{\text{mech,out}}-e_{\text{mech,in}}\right)\\ =\dot{m}\left({\left(Pv\right)}_2+\frac{V_2^2}{2}-{\left(Pv\right)}_1-\frac{V_1^2}{2}\right)\\ =\dot{V}\left(\left(P_2-P_1\right)+\rho \left(\frac{V_2^2-V_1^2}{2}\right)\right)\end{array}$ (I)

Here, the mechanical energy of water inlet and outlet are $e_{\text{mech,in}}$ and $e_{\text{mech,out}}$, mass flow rate of water is $\dot{m}$, Pressure of the pump at inlet and exit is $P_1$, $P_2$, density of the water is $\rho$, volume flow ate of water is $\dot{V}$, velocity at inlet and exit of the pump is $V_1$ and $V_2$ respectively.

Calculate the velocity of the pump at inlet.

$\begin{array}{c}{V}_1=\frac{\dot{V}}{A_1}\\ =\frac{\dot{V}}{\frac{\pi }{4}{D}_1^2}\end{array}$ (II)

Here, diameter of the pipe at inlet is $D_1$.

Calculate the velocity of the pump at exit.

$\begin{array}{c}{V}_2=\frac{\dot{V}}{A_2}\\ =\frac{\dot{V}}{\frac{\pi }{4}{D}_2^2}\end{array}$ (III)

Here, diameter of the pipe at exit is $D_2$.

Calculate the useful pumping power.

${\dot{W}}_{\text{pump,u}}=\Delta {\dot{E}}_{\text{mech,fluid}}$ (IV)

Calculate the shaft power.

${\dot{W}}_{\text{pump,shaft}}={\eta }_{\text{motor}}{\dot{W}}_{\text{electric}}$ (V)

Here, the efficiency of motor is ${\eta }_{\text{motor}}$ and electric power is ${\dot{W}}_{\text{electric}}$.

Calculate the mechanical efficiency of the pump.

${\eta }_{\text{pump}}=\frac{{\dot{W}}_{\text{pump,u}}}{{\dot{W}}_{\text{pump,shaft}}}\times 100\%$ (VI)

Conclusion:

Substitute $0.1{\text{m}}^{\text{3}}/\text{s}$ for $\dot{V}$, $500{\text{kN/m}}^{\text{2}}$ for $P_2-P_1$, $860{\text{kg/m}}^{\text{3}}$ for $\rho$, $8.84\text{m/s}$ for $V_2$, and 19.9 m/s for $V_1$ in Equation (I).

$\begin{array}{c}\Delta {\dot{E}}_{\text{mech,fluid}}=0.1{\text{m}}^{\text{3}}/\text{s}\left(\left(500{\text{kN/m}}^{\text{2}}\right)+860{\text{kg/m}}^{\text{3}}\left(\frac{{\left(\text{8}\text{.84}\text{m/s}\right)}^2-{\left(19.9\text{m/s}\right)}^2}{2}\right)\right)\\ =0.1{\text{m}}^{\text{3}}/\text{s}\left(\begin{array}{l}\left(500{\text{kN/m}}^{\text{2}}\right)+860{\text{kg/m}}^{\text{3}}\left(\begin{array}{l}\frac{{\left(\text{8}\text{.84}\text{m/s}\right)}^2-{\left(19.9\text{m/s}\right)}^2}{2}\\ \left(\frac{1\text{kN}}{1000\text{kg}\cdot {\text{m/s}}^{\text{2}}}\right)\end{array}\right)\\ \left(\frac{1\text{kW}}{1\text{kN}\cdot \text{m/s}}\right)\end{array}\right)\\ =36.3\text{kW}\end{array}$

Substitute $0.1{\text{m}}^{\text{3}}/\text{s}$ for $\dot{V}$ and 8 cm for $D_1$ in Equation (II).

$\begin{array}{c}{V}_1=\frac{0.1{\text{m}}^{\text{3}}/\text{s}}{\frac{\pi }{4}{\left(8\text{cm}\right)}^2}\\ =\frac{0.1{\text{m}}^{\text{3}}/\text{s}}{\frac{\pi }{4}{\left(8\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2}\\ =19.9\text{m/s}\end{array}$

Substitute $0.1{\text{m}}^{\text{3}}/\text{s}$ for $\dot{V}$ and 12 cm for $D_2$ in Equation (III).

$\begin{array}{c}{V}_2=\frac{0.1{\text{m}}^{\text{3}}/\text{s}}{\frac{\pi }{4}{\left(12\text{cm}\right)}^2}\\ =\frac{0.1{\text{m}}^{\text{3}}/\text{s}}{\frac{\pi }{4}{\left(12\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2}\\ =8.84\text{m/s}\end{array}$

Substitute 36.3 kW for $\Delta {\dot{E}}_{\text{mech,fluid}}$ in Equation (IV).

${\dot{W}}_{\text{pump,u}}=36.3\text{kW}$

Substitute 0.90 for ${\eta }_{\text{motor}}$ and 44 kW for ${\dot{W}}_{\text{electric}}$ in Equation (V).

$\begin{array}{c}{\dot{W}}_{\text{pump,shaft}}=\left(0.90\right)\left(44\text{kW}\right)\\ =39.6\text{kW}\end{array}$

Substitute 39.6 kW for ${\dot{W}}_{\text{pump,shaft}}$ and 36.3 kW for ${\dot{W}}_{\text{pump,u}}$ in Equation (VI).

$\begin{array}{c}{\eta }_{\text{pump}}=\frac{36.3\text{kW}}{39.6\text{kW}}\times 100\%\\ =0.918\times 100\%\\ =91.8\%\end{array}$

Thus, the mechanical efficiency of the pump is $\underset{\_}{91.8\%}$.