Chapter 5, Problem 88P

Underground water is to be pumped by a 78 percent efficient 5-kW submerged pump to a pool whose free surface is 30 m above the underground water level. The diameter of the pipe is 7 cm on the intake side and 5 cm on the discharge side. Detennine (a) the maximum flow rate of water and (b) the pressure difference across the pump. Assume the elevation difference between the pump inlet and the outlet and the effect of the kinetic energy correction factors to be negligible.

To determine

(a)

The maximum discharge of water.

Answer to Problem 88P

The rate of flow of water is $0.01325{\text{m}}^{\text{3}}/\text{s}$.

Explanation of Solution

Given information:

The efficiency of the pump is $78\%$, rated power is $5\text{kW}$, head required to develop is $30\text{m}$, diameter of the suction pipe is $7\text{cm,}$ and diameter of the delivery pipe is $5\text{cm}$.

Write the expression for the maximum discharge.

  $\dot{Q}=\frac{\eta \times P}{\rho \times g\times h}$...........................................................................(I)

Here, discharge is $\dot{Q}$, efficiency is $\eta$, rated power is $P$, density of water is $\rho$, acceleration due to gravity is $g,$ and required head is $h$.

Calculation:

Substitute $78\%$ for $\eta$, $5\text{kW}$ for $P$, $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for acceleration due to gravity and $30\text{m}$ for head required.

  $\begin{array}{c}\dot{Q}=\frac{78\%\times 5\text{kW}}{\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\times \left(30\text{m}\right)}\\ =\frac{78\%\left(\frac{1}{100}\right)\times 5\text{kW}\times \left(\frac{1000\text{W}}{1\text{kW}}\right)}{\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\times \left(30\text{m}\right)}\\ =\frac{0.78\times 5000\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{3}}}{\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\times \left(30\text{m}\right)}\\ =0.01325{\text{m}}^{\text{3}}/\text{s}\end{array}$

Conclusion:

The maximum rate of flow of water is $0.01325{\text{m}}^{\text{3}}/\text{s}$.

To determine

(b)

The pressure difference across the pump.

Answer to Problem 88P

The pressure difference across the pump is

Explanation of Solution

Given information:

The efficiency of the pump is $78\%$, rated power is $5\text{kW}$, head required to develop is $30\text{m}$, diameter of the suction pipe is $7\text{cm,}$ and diameter of the delivery pipe is $5\text{cm}$.

Write the expression for the pressure difference across the pump.

  $p_2-p_1=\rho \left[\frac{P}{\rho \dot{Q}}-\left(\frac{V_2^2-V_1^2}{2}\right)\right]$  .......(I)

Here, pressure at suction side is $p_1$, pressure at delivery side is $p_2$. density is $\rho$, velocity in delivery pipe is $V_2$, velocity in suction pipe is $V_1,$ and discharge is $\dot{Q}$.

  $V=\frac{4\times \dot{Q}}{\pi \times d^2}$  .......(II)

Here, discharge is $\dot{Q}$, velocity is $V$, and diameter of the pipe is $d$.

Calculation:

Velocity in the suction pipe.

Substitute $0.01325{\text{m}}^{\text{3}}/\text{s}$ for $\dot{Q}$ and $7\text{cm}$ for $d$ in Equation (II).

  $\begin{array}{c}{V}_1=\frac{4\times \left(0.01325m^3/s\right)}{\pi \times {\left(7\text{cm}\right)}^2\times {\left(\frac{1\text{m}}{100\text{cm}}\right)}^2}\\ =3.443\text{m}/\text{s}\end{array}$

Velocity in the delivery pipe.

Substitute $0.01325m^3/s$ for $\dot{Q}$ and $5\text{cm}$ for $d$ in Equation (II)

  $\begin{array}{c}{V}_2=\frac{4\times \left(0.01325m^3/s\right)}{\pi \times {\left(5\text{cm}\right)}^2\times {\left(\frac{1\text{m}}{100\text{cm}}\right)}^2}\\ =6.748\text{m}/\text{s}\end{array}$

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $0.78\times 5\text{kW}$ for $P$, $0.01325{\text{m}}^{\text{3}}/\text{s}$ for $\dot{Q}$, $3.443\text{m}/\text{s}$ for $V_1$ and $6.748\text{m}/\text{s}$ for $V_2$.

  $\begin{array}{c}{p}_2-p_1=\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left[\begin{array}{l}\frac{0.78\times 5\text{kW}}{\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(0.01325{\text{m}}^{\text{3}}/\text{s}\right)}\\ -\left(\frac{\left({6.748}^2{m}^2/s^2\right)-\left(3.443m^2/s^2\right)}{2}\right)\end{array}\right]\\ =\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left[\begin{array}{l}\frac{0.78\times 5\text{kW}\left(\frac{1000\text{W}}{1\text{kw}}\right)}{\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(0.01325{\text{m}}^{\text{3}}/\text{s}\right)}\\ -\left(\frac{\left({6.748}^2{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right)-\left(3.443{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right)}{2}\right)\end{array}\right]\\ =\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left[\begin{array}{l}\frac{0.78\times 5000\text{W}}{\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(0.01325{\text{m}}^{\text{3}}/\text{s}\right)}\\ -\left(\frac{\left({6.748}^2{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right)-\left(3.443{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right)}{2}\right)\end{array}\right]\\ =\text{277}\text{.5}\text{kPa}\end{array}$

Conclusion:

The pressure difference across the pump is $277.5\text{kPa}$.