Chapter 14, Problem 35EP

A water pump is used to pump water from one large reservoir to another large reservoir that is at a higher elevation. The free surfaces of both reservoirs are exposed to atmospheric pressure, as sketched in Fig. P14-35E. The dimensions and minor loss coefficients are provided in the figure. The pump's performance is approximated by the expression $H_{available}=H_0-a{\stackrel{\dot{}}{V}}^2$ where the shutoff head $H_0=125ft$ of water column, coefficient $a=2.50ft{\left(gpm\right)}^2$ , available pump head $H_{available}$ is in units of feet of water column, and capacity is in units of gallons per minute (gpm). Estimate the capacity delivered by the pump.

   $z_2-z_1=22.0ft$ (elevation difference)

   $D=1.20in$ (Pipe diameter)

   $K_L,entrance=0.50$ (Pipe entrance)

   $K_L,value1=2.0$ (value 1)

   $K_L,value2=6.8$ (value 2)

   $K_L,elbow=0.34$ (each elbow-the re are 3)

   $K_L,exit=1.05$ (pipe exit)

   $L=124ft$ (total pipe length)

   $\epsilon =0.0011in$ (pipe roughless)

To determine

The capacity delivered by the pump.

Answer to Problem 35EP

The capacity delivered by the pump is $6.33\text{gpm}$.

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).

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

Conclusion:

The capacity delivered by the pump is $6.33\text{gpm}$.