Chapter 5, Problem 27P

Water is pumped from a lake to a storage tank 18 m above at a rate of 70 L/s while consuming 20.4 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 P5−27

To determine

(a)

The overall efficiency of the pump motor unit.

Answer to Problem 27P

The overall efficiency of the pump motor system is $60.59\%$.

Explanation of Solution

Given information:

The height of tank water surface is from the lake surface is $18\text{m}$, the volume flow rate for the pump is $70\text{ft}/\text{s}$ and the electric power consumed by the motor of the system is $20.4\text{kW}$.

The following figure shows the free body diagram of the storage tank.

    

    Figure-(1)

Write the expression for the change in energy between the two points for unit mass.

    $\left|\Delta E_{mech,fluid}\right|=\left|\frac{P_1-P_2}{\rho }+\frac{V_1^2-V_2^2}{2}+g\left(z_1-z_2\right)\right|$   ..... (I)

The pressure at both points is same as both points are exposed to atmospheric pressure only which means the flow energy becomes zero and the velocity is $0$ at both points which mean kinetic energy is $0$ zero.

Substitute $P_1$ for $P_2$ and $0$ for $V_1$ and $0$ for $V_2$ in Equation (I).

    $\begin{array}{l}\left|\Delta E_{mech,fluid}\right|=\left|\frac{P_1-P_1}{\rho }+\frac{0-0}{2}+g\left(z_1-z_2\right)\right|\\ \left|\Delta E_{mech,fluid}\right|=\left|0+0+g\left(z_1-z_2\right)\right|\\ \left|\Delta E_{mech,fluid}\right|=\left|g\left(z_1-z_2\right)\right|\end{array}$   ..... (II)

Here, the pressure at point $1$ is $P_1$, the pressure at point $2$ is $P_2$, density of water is $\rho$, velocity at point $1$ is $V_1$, velocity at point $2$ is $V_2$, the height at point $1$ is $z_1$, the height at point $2$ is $z_2$ and the acceleration due to gravity is $g$.

Write the expression for the mass flow rate.

    $\dot{m}=\rho \dot{V}$   ..... (III)

Here, the volume flow rate is $\dot{V}$.

Write the expression for change in energy with respect to $\dot{m}$.

    $\Delta {\dot{E}}_{mech,fluid}=\dot{m}\times \Delta E_{mech,fluid}$   ..... (IV)

Here, the mass flow rate is $\dot{m}$.

Write the expression for the overall efficiency of the pump motor system.

    ${\eta }_{\text{pump-motor}}=\frac{\Delta {\dot{E}}_{mech,fluid}}{{\dot{W}}_{elect,in}}$   ..... (V)

Here, the electric power consumed by the system is ${\dot{W}}_{elect,in}$.

Calculation:

Substitute $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $0$ for $z_1$ and $18\text{m}$ for $z_2$ in Equation (II).

    $\begin{array}{l}\left|\Delta E_{mech,fluid}\right|=\left|\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0-18\text{m}\right)\right|\\ \left|\Delta E_{mech,fluid}\right|=\left|-176.58{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right|\\ \Delta E_{mech,fluid}=176.58{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\left(\frac{1\text{J}/\text{kg}}{1{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\\ \Delta E_{mech,fluid}=176.58\text{J}/\text{kg}\end{array}$

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $70\text{ft}/\text{s}$ for $\dot{V}$ in Equation (III).

    $\begin{array}{c}\dot{m}=\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(70\text{ft}/\text{s}\right)\\ =\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(70\text{ft}/\text{s}\left(\frac{1{\text{m}}^{\text{3}}/\text{s}}{1000\text{ft}/\text{s}}\right)\right)\\ =\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(0.070{\text{m}}^{\text{3}}/\text{s}\right)\\ =70\text{kg}/\text{sec}\end{array}$

Substitute $70\text{kg}/\text{s}$ for $\dot{m}$ and $176.58\text{J}/\text{kg}$ for $\Delta E_{mech,fluid}$ in Equation (IV).

    $\begin{array}{c}\Delta {\dot{E}}_{mech,fluid}=\left(70\text{kg}/\text{s}\right)\left(176.58\text{J}/\text{kg}\right)\\ =123606\text{J}/\text{s}\left(\frac{1\text{kW}}{1000\text{J}/\text{s}}\right)\\ =12.36\text{kW}\end{array}$

Substitute $12.36\text{kW}$ for $\Delta {\dot{E}}_{mech,fluid}$ and $20.4\text{kW}$ for ${\dot{W}}_{elect,in}$ in Equation (V).

    $\begin{array}{c}{\eta }_{\text{pump-motor}}=\frac{12.36\text{kW}}{20.4\text{kW}}\\ =0.6058\times 100\%\\ =60.59\%\end{array}$

Conclusion:

The overall efficiency of the pump motor system is $60.59\%$.

To determine

(b)

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

Answer to Problem 27P

The pressure difference between inlet and exit of the pump is $176.57\text{kPa}$.

Explanation of Solution

Given information:

Assume that the kinetic energy difference between the inlet and the exit is negligible. The height difference between the inlet and the exit pump is negligible.

Write the expression for the change in energy between inlet and exit.

    $\Delta E_{mech,fluid}=\frac{P_{inlet}-P_{exit}}{\rho }+\frac{V_{inlet}^2-V_{exit}^2}{2}+g\left(z_{inlet}-z_{exit}\right)$   ..... (VI)

Here, the flow energy is $\frac{P_{inlet}-P_{exit}}{\rho }$, the kinetic energy is $\frac{V_{inlet}^2-V_{exit}^2}{2}$ and the height difference is $z_{inlet}-z_{exit}$.

Substitute $0$ for $z_{inlet}-z_{exit}$ and $0$ for $\frac{V_{inlet}^2-V_{exit}^2}{2}$ in Equation (VI).

    $\begin{array}{l}\Delta E_{mech,fluid}=\frac{P_{inlet}-P_{exit}}{\rho }+0+g\left(0\right)\\ \Delta E_{mech,fluid}=\frac{P_{inlet}-P_{exit}}{\rho }\end{array}$   ..... (VII)

Here, the pressure difference is $P_{inlet}-P_{exit}$ and the density is $\rho$.

Substitute $\frac{P_{inlet}-P_{exit}}{\rho }$ for $\Delta E_{mech,fluid}$ in Equation (IV).

    $\Delta {\dot{E}}_{mech,fluid}=\dot{m}\times \frac{P_{inlet}-P_{exit}}{\rho }$   ..... (VIII)

Here, the mass flow rate is $\dot{m}$.

Calculation:

Substitute $12.36\text{kW}$ for $\Delta {\dot{E}}_{mech,fluid}$, $70\text{kg}/\text{s}$ for $\dot{m}$ and $1000\text{kg}/{\text{m}}^{\text{3}}$ in Equation (VIII).

    $\begin{array}{l}12.36\text{kW}=\left(70\text{kg}/\text{s}\right)\frac{\left(P_{inlet}-P_{exit}\right)}{\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)}\\ \left(P_{inlet}-P_{exit}\right)=\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\frac{\left(12.36\text{kW}\left(\frac{1000\text{W}}{1\text{kW}}\right)\right)}{\left(70\text{kg}/\text{s}\right)}\\ \left(P_{inlet}-P_{exit}\right)=176.571\times {10}^3\text{Ws}/{\text{m}}^{\text{3}}\left(\frac{1\text{Pa}}{1\text{Ws}/{\text{m}}^{\text{3}}}\right)\\ \left(P_{inlet}-P_{exit}\right)=176.571\times {10}^3\text{Pa}\end{array}$

    $\left(P_{inlet}-P_{exit}\right)=176.571\text{kPa}$

Conclusion:

The pressure difference between inlet and exit of the pump is $176.57\text{kPa}$.