Chapter 14, Problem 36EP

For the pump and piping system of Prob. 14-35E, plot the required pump head $H_{required}$ . (ft of water column) as a function of volume flow rate $\stackrel{\dot{}}{V}$ (gpm). On the same plot, compare the available pump head $H_{available}$ versus
$\stackrel{\dot{}}{V}$ . and mark the operating point. Discuss.

To determine

The plot between the available head and the capacity of the pump.

Answer to Problem 36EP

The plot between the available head and the capacity of the pump is

Explanation of Solution

Given information:

The shutoff head is $125\text{ft}$, the coefficient $a$is $2.50\text{ft}/{\left(\text{gpm}\right)}^{\text{2}}$, the elevation difference is $22\text{ft}$, the diameter of the pipe is $1.20\text{in}$, the loss coefficient at pipe entrance is $0.50$, the minor loss coefficient at valve 1 is $2.0$, the minor loss coefficient at valve 2 is $6.8$, the minor loss coefficient at each elbow is $0.34$, the minor loss coefficient at pipe exit is $1.05$, the total pipe length is $124\text{ft}$, the pipe roughness is $0.0011\text{in}$, the initial velocity of the water is zero, the final velocity of the water is zero and the pressure at exit is $1.01\times {10}^5\text{Pa}$and inlet is pressure is $1.01\times {10}^5\text{Pa}$.

Write the expression for the required head using the energy balance equation.

  $H_{required}=\frac{P_2-P_1}{\rho g}+\frac{V_2^2-V_1^2}{2g}+\left(z_2-z_1\right)+h_{turbine}+h_{Lfriction}$ (I)

Here, the initial pressure is $P_1$, the final pressure is $P_2$, the initial velocity is $V_1$, the final velocity is $V_2$, the density of the water is $\rho$, the acceleration due to gravity is$g$, the potential head of water at section 1 is $z_1$, the potential head of the water at section 2 is $z_2$, the turbine head is $h_{turbine}$, the frictional loss is $h_{Lfriction}$.

Write the expression for the roughness factor.

  $R=\frac{\epsilon }{D}$..... (II)

Here, the diameter of the pipe is $D$

Write the expression for the minor losses.

  ${\displaystyle \sum F_L=K_{L,entrance}+K_{L,valve1}+K_{L,valve2}+K_{L,elbow}+K_{L,exit}}$..... (III)

Here, the minor loss coefficient at pipe entrance is $K_{L,entrance}$, the minor loss coefficient at valve 1 is $K_{L,valve1}$, the minor loss coefficient at valve 2 is $K_{L,valve2}$, the minor loss coefficient at each elbow is $K_{L,elbow}$, the minor loss coefficient at pipe exit is $K_{L,exit}$.

Write the expression for the frictional loss head.

  $h_{Lfriction}=\left(f\frac{L}{D}+{\displaystyle \sum K_L}\right)\frac{V^2}{2g}$ (IV)

Here, the friction factor is $f$, the length of the pipe is $L$.

Write the expression for the available head.

  $H_{availabe}=H_o-a{\dot{V}}^2$ (V)

Here, the shutoff head is $H_o$, the coefficient is $a$the capacity is $\dot{V}$and the velocity of the water is $V$.

Write the expression for the capacity.

  $\dot{V}=\frac{\pi D^2}{4}V$ (VI)

Substitute $\frac{\pi D^2}{4}V$for $\dot{V}$in Equation (V).

  $\begin{array}{l}{H}_{availabe}=H_o-a{\left(\frac{\pi D^2}{4}V\right)}^2\\ H_{availabe}=H_o-a\left(\frac{{\pi }^2{D}^4}{16}{V}^2\right)\end{array}$ (VII)

Write the expression for the Reynolds number.

  $\mathrm{Re}=\frac{\rho VD}{\mu }$ (VIII)

Here, the density of the water is $\rho$and the dynamic viscosity is $\mu$.

Write the expression for the friction factor using the Colebrook equation.

  $\frac{1}{\sqrt{f}}=-2.0\log \left(\frac{R}{3.7}+\frac{2.51}{\mathrm{Re}\sqrt{f}}\right)$ (IX)

Calculation:

Refer to table A-3E Properties of saturated water to obtain the density of water as $1.940\text{slug}/{\text{ft}}^{\text{3}}$and viscosity of water as $1.002\times {10}^{-5}\text{lb}\cdot \text{s}/{\text{ft}}^2$at room temperature.

Substitute $0.0011\text{in}$for $\epsilon$and $1.20\text{in}$for $D$in Equation (II).

  $\begin{array}{c}R=\frac{0.0011\text{in}}{1.20\text{in}}\\ =9.16\times {10}^{-4}\end{array}$

Substitute $0.5$for $K_{L,entrance}$, $2.0$for $K_{L,valve1}$, $6.8$for $K_{L,valve2}$, $\left(0.34\times 3\right)$for $K_{L,elbow}$, $1.05$for $K_{L,exit}$in Equation (III).

  $\begin{array}{c}{\displaystyle \sum F_L}=0.5+2.0+6.8+\left(0.34\times 3\right)+1.05\\ =0.5+2.0+6.8+\left(1.02\right)+1.05\\ =11.37\end{array}$

Substitute $1.20\text{in}$for $D$, $2.50\text{ft}/{\left(\text{gpm}\right)}^{\text{2}}$for $a$and $125\text{ft}$for $H_o$in Equation (VII).

  $\begin{array}{c}{H}_{availabe}=125\text{ft}-2.50\text{ft}/{\left(\text{gpm}\right)}^{\text{2}}\left(\frac{{\pi }^2{\left(1.20\text{in}\right)}^4}{16}{V}^2\right)\\ =125\text{ft}-2.50\text{ft}/{\left(\text{gpm}\right)}^{\text{2}}\left(\frac{{\pi }^2{\left(1.20\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)\right)}^4}{16}{V}^2\right)\\ =125\text{ft}-\left(1.54\times {10}^{-4}{\text{ft}}^5/{\left(\text{gpm}\right)}^{\text{2}}\right)V^2\end{array}$

Substitute $0$for $h_{turbine}$

  $\left(f\frac{L}{D}+{\displaystyle \sum K_L}\right)\frac{V^2}{2g}$for $h_{L,friction}$, $0$for $V_1$, $0$for $V_2$, $1.01\times {10}^5\text{Pa}$for $P_1$and $1.01\times {10}^5\text{Pa}$for $P_2$in Equation (I).

  $\begin{array}{c}{H}_{required}=\frac{1.01\times {10}^5\text{Pa}-1.01\times {10}^5\text{Pa}}{\rho g}+0+\left(z_2-z_1\right)+0+\left(\left(f\frac{L}{D}+{\displaystyle \sum K_L}\right)\frac{V^2}{2g}\right)\\ =0+\left(z_2-z_1\right)+\left(\left(f\frac{L}{D}+{\displaystyle \sum K_L}\right)\frac{V^2}{2g}\right)\\ =\left(z_2-z_1\right)+\left(\left(f\frac{L}{D}+{\displaystyle \sum K_L}\right)\frac{V^2}{2g}\right)\end{array}$(X)

Since, the required head is equal to the available head that is $H_{available}=H_{required}$.

Substitute $H_{available}$for $H_{required}$in Equation (X).

  $H_{available}=\left(z_2-z_1\right)+\left(\left(f\frac{L}{D}+{\displaystyle \sum K_L}\right)\frac{V^2}{2g}\right)$ (XI)

Substitute $125\text{ft}-\left(1.54\times {10}^{-4}{\text{ft}}^5/{\left(\text{gpm}\right)}^{\text{2}}\right)V^2$for $H_{available}$, $22.0\text{ft}$for $\left(z_2-z_1\right)$, $124\text{ft}$for $L$, $1.20\text{in}$for $D$, $32.2\text{ft}/{\text{s}}^{\text{2}}$for $g$and $11.37$for ${\displaystyle \sum K_L}$in Equation (XI).

  $\left[\begin{array}{l}125\text{ft}-\\ \left(1.54\times {10}^{-4}{\text{ft}}^5/{\left(\text{gpm}\right)}^{\text{2}}\right)V^2\end{array}\right]=\left[\begin{array}{l}22.0\text{ft}+\\ \left(\left(f\frac{124\text{ft}}{1.20\text{in}}+\left(11.37\right)\right)\frac{V^2}{2\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)}\right)\end{array}\right]$

  $\left[\begin{array}{l}125\text{ft}-\\ \left(1.54\times {10}^{-4}{\text{ft}}^5/{\left(\text{gpm}\right)}^{\text{2}}\right)V^2\end{array}\right]=\left[\begin{array}{l}22.0\text{ft}+\\ \left(\left(f\frac{124\text{ft}}{1.20\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)}+\left(11.37\right)\right)\frac{V^2}{\left(64.4\text{ft}/{\text{s}}^{\text{2}}\right)}\right)\end{array}\right]$

  $V^2\left[\left(1.54\times {10}^{-4}{\text{ft}}^5/{\left(\text{gpm}\right)}^{\text{2}}\right)-\left(\left(f19.251/\text{ft}/{\text{s}}^{\text{2}}+\left(17.651/\text{ft}/{\text{s}}^{\text{2}}\right)\right)\right)\right]=-103\text{ft}$

  $V^2=\frac{-103\text{ft}}{\left[\left(1.54\times {10}^{-4}{\text{ft}}^5/{\left(\text{gpm}\right)}^{\text{2}}\right)-\left(\left(f19.251/\text{ft}/{\text{s}}^{\text{2}}+\left(17.651/\text{ft}/{\text{s}}^{\text{2}}\right)\right)\right)\right]}$

  $\begin{array}{l}{V}^2=\frac{-103\text{ft}}{\left[-1.611\times {10}^{-5}{\text{ft}}^4/{\left(\text{gpm}\right)}^{\text{2}}-f19.251/\text{ft}/{\text{s}}^{\text{2}}\right]}\\ f=\frac{5.53\text{ft}}{V^2}-0.0836\times {10}^{-5}{\text{ft}}^4/{\left(\text{gpm}\right)}^{\text{2}}\end{array}$

Substitute $1.940\text{slug}/{\text{ft}}^{\text{3}}$for $\rho$, $1.002\times {10}^{-5}\text{lb}\cdot \text{s}/{\text{ft}}^2$for $\mu$and $1.20\text{in}$for $D$in Equation (VII).

  $\begin{array}{c}\mathrm{Re}=\frac{\left(1.940\text{slug}/{\text{ft}}^{\text{3}}\right)\left(1.20\text{in}\right)V}{\left(1.002\times {10}^{-5}\text{lb}\cdot \text{s}/{\text{ft}}^2\right)}\\ =\frac{\left(1.940\text{slug}/{\text{ft}}^{\text{3}}\right)\left(1.20\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)\right)V}{\left(1.002\times {10}^{-5}\text{lb}\cdot \text{s}/{\text{ft}}^2\right)}\left(\frac{\text{1}\text{lb}\cdot \text{s}}{1\text{slug}}\right)\\ =\frac{0.194}{1.002\times {10}^{-5}}V\\ =19361.27V\end{array}$

Substitute $19361.27V$for $\mathrm{Re}$, $9.16\times {10}^{-4}$for $R$, $\frac{5.53\text{ft}}{V^2}-0.0836\times {10}^{-5}{\text{ft}}^4/{\left(\text{gpm}\right)}^{\text{2}}$for $f$in Equation (IX).

  $\begin{array}{l}\frac{1}{\sqrt{\left(\begin{array}{l}\frac{5.53\text{ft}}{V^2}-\\ 0.0836\times {10}^{-5}{\text{ft}}^4/{\left(\text{gpm}\right)}^{\text{2}}\end{array}\right)}}=-2.0\log \left(\begin{array}{l}\frac{\left(9.16\times {10}^{-4}\right)}{3.7}+\\ \frac{2.51}{\begin{array}{l}\left(19361.27V\right)\\ \sqrt{\left(\begin{array}{l}\frac{5.53\text{ft}}{V^2}-\\ 0.0836\times {10}^{-5}{\text{ft}}^4/{\left(\text{gpm}\right)}^{\text{2}}\end{array}\right)}\end{array}}\end{array}\right)\\ \frac{1}{\frac{2.35}{V}-9.143\times {10}^{-4}{\text{ft}}^2/\left(\text{gpm}\right)}=-2.0\log \left(\begin{array}{l}2.47\times {10}^{-4}+\\ \frac{2.51}{45498.98-17.70{\text{ft}}^2/\left(\text{gpm}\right)V}\end{array}\right)\\ \frac{V}{2.35-9.143\times {10}^{-4}{\text{ft}}^2/\left(\text{gpm}\right)V}=-2.0\log \left(\frac{-4.37\times {10}^{-3}{\text{ft}}^2/\left(\text{gpm}\right)V+13.74}{45498.98-17.70{\text{ft}}^2/\left(\text{gpm}\right)V}\right)\\ V=0.1798\text{ft}/\text{s}\end{array}$

Substitute $0.1798\text{ft}/\text{s}$for $V$and $1.20\text{in}$for $D$in Equation (VI).

  $\begin{array}{c}\dot{V}=\frac{\pi {\left(1.20\text{in}\right)}^2}{4}\left(0.1798\text{ft}/\text{s}\right)\\ =\frac{\pi {\left(1.20\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)\right)}^2}{4}\left(0.1798\text{ft}/\text{s}\right)\\ =\frac{0.05648}{4}{\text{ft}}^{\text{3}}/\text{s}\left(\frac{448.83\text{gpm}}{{\text{ft}}^{\text{3}}/\text{s}}\right)\\ =6.33\text{gpm}\end{array}$

Substitute $6.33\text{gpm}$for $\dot{V}$, $2.50\text{ft}/{\left(\text{gpm}\right)}^{\text{2}}$for $a$and $125\text{ft}$for $H_o$in Equation (V).

  $\begin{array}{c}{H}_{availabe}=125\text{ft}-\left(2.50\text{ft}/{\left(\text{gpm}\right)}^{\text{2}}\right){\left(6.33\text{gpm}\right)}^2\\ =125\text{ft}-\left(2.50\text{ft}/{\left(\text{gpm}\right)}^{\text{2}}\right)\left(40.068{\text{gpm}}^2\right)\\ =125\text{ft}-100.17\text{ft}\\ =24.82\text{ft}\end{array}$

The different values of the capacity and the available head is shown in the below Table.

S.No.	Capacity $\text{gpm}$	Available head $\text{ft}$
1	$6.33$	$24.82$
2	$6.5$	$19.37$
3	$6.75$	$11.09$
4	$7$	$2.55$

Draw the plot between the available head and the capacity of the pump for different values along with the required head.

  

  Figure-(1)

Conclusion:

The plot between the available head and the capacity of the pump is