Chapter 5, Problem 95P

Reconsider Prob. 5-88. Determine the flow rate of water and the pressure difference across the pump if the irreversible head loss of the pipin2 system is 4 m.

To determine

The discharge of water and pressure difference across the pump.

Answer to Problem 95P

The discharge of water is $0.01169{\text{m}}^{\text{3}}/\text{s}$ and pressure difference across the pump is $320.511\text{kPa}$

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 $7\text{cm}$,head loss of the piping system is $4\text{m}$and diameter of the delivery pipe is $5\text{cm}$.

Write the energy balance equation.

  $\eta \times P=\rho \times \dot{Q}\times g\times \left(h+h_l\right)$   ....... (I)

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

Write the expression for the pressure difference across the pump.

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

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 $\dot{Q}$.

Write the expression for velocity in pipe.

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

Here, velocity in pipe is $V$, flow rate is $\dot{Q}$ and diameter of pipe is $d$.

Calculation:

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $0.78$ for $\eta$, $5\text{kW}$ for power, $30m$ for $h$ and $4\text{m}$for $h_l$.

  $\begin{array}{l}0.78\times 5\text{kW=}\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\times \dot{Q}\times \left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\times \left(30\text{m+4}\text{m}\right)\\ 0.78\times 5\text{kW}\times \left(\frac{1000\text{W}}{1\text{kW}}\right)=\left(333540\text{kg}/{\text{m×s}}^{\text{2}}\right)\times \dot{Q}\\ 3900\text{W=}\left(333540\text{kg}/{\text{m×s}}^{\text{2}}\right)\times \dot{Q}\\ 3900\text{W}\times \left(\frac{1\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{3}}}{1\text{W}}\right)\text{=}\left(333540\text{kg}/{\text{m×s}}^{\text{2}}\right)\times \dot{Q}\end{array}$

  $\begin{array}{c}\dot{Q}=\frac{3900\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{3}}}{333540\text{kg}/{\text{m×s}}^{\text{2}}}\\ =0.01169{\text{m}}^{\text{3}}/\text{s}\end{array}$

Velocity in suction pipe.

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

  $\begin{array}{c}{V}_1=\frac{4\times \left(0.01169{\text{m}}^{\text{3}}/\text{s}\right)}{\pi \times {\left(7\text{cm}\right)}^2}\\ =\frac{4\times \left(0.01169{\text{m}}^{\text{3}}/\text{s}\right)}{\pi \times {\left(7\text{cm}\right)}^2\times {\left(\frac{1\text{m}}{100\text{cm}}\right)}^2}\\ =3.038\text{m}/\text{s}\end{array}$

Velocity in delivery pipe.

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

  $\begin{array}{c}{V}_2=\frac{4\times \left(0.01169{\text{m}}^{\text{3}}/\text{s}\right)}{\pi \times {\left(5\text{cm}\right)}^2}\\ =\frac{4\times \left(0.01169{\text{m}}^{\text{3}}/\text{s}\right)}{\pi \times {\left(5\text{cm}\right)}^2\times {\left(\frac{1\text{m}}{100\text{cm}}\right)}^2}\\ =5.954\text{m}/\text{s}\end{array}$

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $0.78\times 5\text{kW}$ for $P$, $0.01169{\text{m}}^{\text{3}}/\text{s}$ for $\dot{Q}$, $3.038\text{m}/\text{s}$ for $V_1$ and $5.954\text{m}/\text{s}$ for $V_2$.in Equation (II).

  $p_2-p_1=\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left[\begin{array}{l}\frac{0.78\times 5\text{kW}}{\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(0.01169{\text{m}}^{\text{3}}/\text{s}\right)}\\ -\left(\frac{\left({5.954}^2{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right)-\left(3.038{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right)}{2}\right)\end{array}\right]$

  $=\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left[\begin{array}{l}\frac{0.78\times 5\text{kW}\left(\frac{1000\text{W}}{1\text{kw}}\right)}{\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(0.01325{\text{m}}^{\text{3}}/\text{s}\right)}\\ -\left(\frac{\left({5.954}^2{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right)-\left(3.038{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right)}{2}\right)\end{array}\right]$

  $=\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left[\begin{array}{l}\frac{0.78\times 5000\text{W}\times \left(\frac{1\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{3}}}{1\text{W}}\right)}{\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(0.01325{\text{m}}^{\text{3}}/\text{s}\right)}\\ -\left(\frac{\left({5.954}^2{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right)-\left(3.038{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right)}{2}\right)\end{array}\right]$

  $=\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left[\begin{array}{l}\frac{0.78\times 5000\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{3}}}{\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(0.01325{\text{m}}^{\text{3}}/\text{s}\right)}\\ -\left(\frac{\left({5.954}^2{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right)-\left(3.038{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right)}{2}\right)\end{array}\right]$

  $\begin{array}{c}{p}_2-p_1=320511\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\\ =320511\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\times \left(\frac{1\text{Pa}}{1\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}}\right)\\ =320511\text{Pa}\times \left(\frac{1\text{kPa}}{1000\text{Pa}}\right)\\ =320.511\text{kPa}\end{array}$

Conclusion:

The discharge is $0.01169{\text{m}}^{\text{3}}/\text{s}$ and pressure difference across the pump is $320.511\text{kPa}$.