Chapter 5, Problem 109P

An oil pump is drawing 18 kW of electric power while pumping oil with p = 860 kg/m³ at a rate of 0.1 m³/s. The inlet and outlet diameters of the pipe are 8cm and 12 cm. respectively. If the pressure rise of oil in the pump is measured to be 250 kPa and the motor efficiency is 95 percent, determine the mechanical efficiency of the pump. Take the kinetic energy correction factor to be 1.05.

To determine

The mechanical efficiency of the pump.

Answer to Problem 109P

The mechanical efficiency of the pump is $62.22\%$.

Explanation of Solution

Given information:

The input electric power is $18\text{kW}$, the density of oil is $860\text{kg}/{\text{m}}^3$, the volumetric flow rate of oil is $0.1{\text{m}}^3/\text{s}$, the inlet diameter is $8\text{cm}$, the outlet diameter is $12\text{cm}$, the pressure rise of oil is $250\text{kPa}$, the kinetic energy correction factor is $1.05,$ and the motor efficiency is $95\%$

Write the expression for the Bernoulli's equation.

  $\begin{array}{l}m\left(\frac{P_1}{\rho }+{\alpha }_1\frac{V_1^2}{2}+z_{1}g\right)+W_{pump}=m\left(\frac{P_2}{\rho }+{\alpha }_1\frac{V_2^2}{2}+z_{2}g\right)\\ W_{pump}=m\left(\frac{P_2}{\rho }+{\alpha }_1\frac{V_2^2}{2}+z_{2}g\right)-m\left(\frac{P_1}{\rho }+{\alpha }_1\frac{V_1^2}{2}+z_{1}g\right)\\ W_{pump}=m\left(\frac{P_2-P_1}{\rho }+{\alpha }_1\left(\frac{V_2^2-V_1^2}{2}\right)\right)\end{array}$  .......(I)

Here, the pressure of the air before enter in the tunnel is $P_1$, the pressure of air in the tunnel is $P_2$, the velocity of air before enter in the tunnel is $V_1$, the velocity of air in the tunnel is $V_2$, the density of oil is $\rho$

  $z_1$ and $z_1$ are the elevation points, the pump input is $W_{pump}$, the energy correction factor at point $1$ and point $2$ are ${\alpha }_1$ and ${\alpha }_2$ respectively, the mass flow rate is $m,$ and the gravity is $g$.

Write the expression for the mechanical efficiency of the pump.

  ${\eta }_{pump}=\frac{W_{pump}}{W_{in,pump}}$  .......(II)

Here, the work done by pump is $W_{pump}$, the inlet electric power is $W_{in,pump}$.

Write the expression for the mechanical efficiency of motor.

  ${\eta }_m=\frac{W_{mot,out}}{W_{mot,in}}$  .......(III)

Here, the work done by motor is $W_{mot,out}$, the inlet electric power to the motor is $W_{in,mot}$.

Write the expression for the velocity at inlet.

  $V_1=\frac{\dot{V}}{\frac{\pi }{4}{D}_1^2}$  .......(IV)

Here, the volumetric flow rate is $\dot{V}$ and the inlet diameter is $D_1$.

Write the expression for the velocity at outlet.

  $V_2=\frac{\dot{V}}{\frac{\pi }{4}{D}_2^2}$  .......(V)

Here, the outlet diameter is $D_2$.

Write the expression for the mass flow rate.

  $m=\rho \dot{V}$  .......(VI)

Calculation:

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

  $\begin{array}{c}{V}_1=\frac{0.1{\text{m}}^3/\text{s}}{\frac{\pi }{4}{\left(8\text{cm}\right)}^2}\\ =\frac{0.4{\text{m}}^3/\text{s}}{3.14\times 64{\text{cm}}^2{\left(\frac{1\text{m}}{100\text{cm}}\right)}^2}\\ =19.90\text{m}/\text{s}\end{array}$

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

  $\begin{array}{c}{V}_2=\frac{0.1{\text{m}}^3/\text{s}}{\frac{\pi }{4}{\left(12\text{cm}\right)}^2}\\ =\frac{0.4{\text{m}}^3/\text{s}}{3.14\times 144{\text{cm}}^2{\left(\frac{1\text{m}}{100\text{cm}}\right)}^2}\\ =8.84\text{m}/\text{s}\end{array}$

Substitute $0.1{\text{m}}^3/\text{s}$ for $\dot{V}$ and $860\text{kg}/{\text{m}}^3$ for $\rho$ in Equation (VI).

  $\begin{array}{l}m=\left(860\text{kg}/{\text{m}}^3\right)\left(0.1{\text{m}}^3/\text{s}\right)\\ m=86\text{kg}/\text{s}\end{array}$

Substitute $86\text{kg}/\text{s}$ for $m$, $250\text{kPa}$ for $P_2-P_1$, $860\text{kg}/{\text{m}}^3$ for $\rho$, $1.05$ for ${\alpha }_1$, $19.90\text{m}/\text{s}$ for $V_1$ and $8.84\text{m}/\text{s}$ in Equation (I).

  $\begin{array}{l}{W}_{pump}=86\text{kg}/\text{s}\left(\frac{250\text{kPa}}{860\text{kg}/{\text{m}}^3}+1.05\left(\frac{{\left(8.84\text{m}/\text{s}\right)}^2-\left(19.90\text{m}/\text{s}\right)2}{2}\right)\right)\\ W_{pump}=25\text{kPa}\cdot {\text{m}}^3/\text{s}\left(\frac{1000\text{N}/{\text{m}}^2}{1\text{kPa}}\right)-14351.83\text{kg}\cdot {\text{m}}^2/{\text{s}}^3\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^2}\right)\\ W_{pump}=25000\text{N}\cdot \text{m}/\text{s}-14351.83\text{N}\cdot \text{m}/\text{s}\\ W_{pump}=10648.17\text{N}\cdot \text{m}/\text{s}\end{array}$

  $\begin{array}{l}{W}_{pump}=10648.17\text{N}\cdot \text{m}/\text{s}\left(\frac{1\text{kW}}{1000\text{N}\cdot \text{m}/\text{s}}\right)\\ W_{pump}=10.64\text{kW}\end{array}$

Substitute $0.95$ for ${\eta }_m$ and $18\text{kW}$ for $W_{in,mot}$ in Equation (II).

  $\begin{array}{l}0.95=\frac{W_{mot,out}}{18\text{kW}}\\ W_{mot,out}=17.1\text{kW}\end{array}$

The output power from the motor is used to drive the motor.

  $W_{pump,in}=17.1\text{kW}$

Substitute, $10.64\text{kW}$ for $W_{pump}$, and $17.1\text{kW}$ for $W_{pump,in}$ in Equation (II).

  $\begin{array}{c}{\eta }_{pump}=\frac{10.64\text{kW}}{17.1\text{kW}}\\ =0.6222\\ =62.22\%\end{array}$

Conclusion:

The mechanical efficiency of the pump is $62.22\%$.