Chapter 3, Problem 72P

To determine

The mechanical efficiency of the pump.

Explanation of Solution

Given:

Volumetric flow rate of the oil $\left(\dot{V}\right)$ is $\text{0}\text{.1}{\text{m}}^3/\text{s}$.

Inlet diameter of the pipe $\left(D_1\right)$ is $8\text{cm}$.

Outlet diameter of the pipe $\left(D_2\right)$ is $12\text{cm}$.

Density of the oil $\left(\rho \right)$ is $860\text{kg}/{\text{m}}^{\text{3}}$.

Motor efficiency $\left({\eta }_{\text{motor}}\right)$ is $0.90$.

Power of the electric motor $\left({\dot{W}}_{\text{electric}}\right)$ is $44\text{kW}$.

Pressure rise of the oil in the pump $\left(P_2-P_1\right)$ is $\text{500}\text{kPa}$.

Acceleration due to gravity is $\left(g\right)$$9.81\text{m}/{\text{s}}^{\text{2}}$.

Calculation:

Calculate the inlet velocity of the pipe $\left(V_1\right)$.

  $V_1=\frac{\dot{V}}{A_1}$

  $\begin{array}{c}{V}_1=\frac{\dot{V}}{\frac{\pi }{4}{D}_1^2}\\ =\frac{0.1{\text{m}}^{\text{3}}/\text{s}}{\frac{\pi }{4}{\left(8\text{cm}\right)}^2}\\ =\frac{0.1{\text{m}}^{\text{3}}/\text{s}}{\frac{\pi }{4}{\left(0.08\text{m}\right)}^2}\\ =19.9\text{m}/\text{s}\end{array}$

Calculate the outlet velocity of the pipe $\left(V_2\right)$.

  $V_2=\frac{\dot{V}}{A_2}$

  $\begin{array}{c}{V}_2=\frac{\dot{V}}{\frac{\pi }{4}{D}_2^2}\\ =\frac{0.1{\text{m}}^{\text{3}}/\text{s}}{\frac{\pi }{4}{\left(12\text{cm}\right)}^2}\\ =\frac{0.1{\text{m}}^{\text{3}}/\text{s}}{\frac{\pi }{4}{\left(0.12\text{m}\right)}^2}\\ =8.84\text{m}/\text{s}\end{array}$

Calculate the rate at which the mechanical energy of the oil is supplied to the pump $\left(\Delta {\dot{E}}_{\text{mech,fluid}}\right)$.

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

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

Calculate the useful pumping power $\left({\dot{W}}_{\text{pump,u}}\right)$.

  $\begin{array}{c}{\dot{W}}_{\text{pump,u}}=\Delta {\dot{E}}_{\text{mech,fluid}}\\ =36.3\text{kW}\end{array}$

Calculate the shaft power $\left({\dot{W}}_{\text{pump,shaft}}\right)$.

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

Calculate the mechanical efficiency of the pump $\left({\eta }_{\text{pump}}\right)$.

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

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