Chapter 3, Problem 3.181P

To determine

(a)

The flow rate.

Answer to Problem 3.181P

The flow rate is $1152.07\text{gal}/\text{min}$.

Explanation of Solution

Given information:

There is linear variation between the head and the flow rate.

The Figure-(1) shows the linear variation of the head with the flow rate.

Figure-(1)

Here, the point on the graph is $A$, the point on the graph is $B$, the coordinate on the head axis is $h_{p_1}$, the coordinate on the head axis is $h_{p_2}$, the coordinate on the flow rate axis is $Q_1$, the coordinate on the flow rate axis is $Q_2$.

Write the coordinates of the first point on the graph.

$A=\left(Q_1,h_{p_1}\right)$

Write the coordinates of the second point on the graph.

$B=\left(Q_2,h_{p_2}\right)$

Write the expression for the equation of straight line of the graph.

$\left(h_p-h_{p_1}\right)=\frac{\left(h_{p_2}-h_{p_1}\right)}{\left(Q_2-Q_1\right)}\left(Q-Q_1\right)$ ..... (I)

Here, the pump head is $h_p$ and the flow rate is $Q$.

Write the expression for the Bernoulli’s equation between the inlet and outlet of the pump.

$\begin{array}{l}\frac{p_1}{\rho g}+\frac{V_1^2}{2g}+z_1+h_p=\frac{p_2}{\rho g}+\frac{V_2^2}{2g}+z_2+h_f\\ \left(\frac{p_1-p_2}{\rho g}\right)+\left(\frac{V_1^2}{2g}-\frac{V_2^2}{2g}\right)+\left(z_1-z_2\right)=h_f-h_p\end{array}$ ..... (II)

Here, the pressure at inlet is $p_1$, the velocity at inlet is $V_1$, the height of the inlet from the datum is $z_1$, the pressure at outlet is $p_2$, the velocity at outlet is $V_2$, the height of the outlet from the datum is $z_2$, the density of water is $\rho$, the head loss due to friction is $h_f$, the head due to pump is $h_p$ and the acceleration due to gravity is $g$.

Write the expression for the head loss.

$h_f=27\frac{V^2}{2g}$

Here, the average velocity is $V$

Substitute in $27\frac{V^2}{2g}$ for $h_f$ Equation (II).

$\left(\frac{p_1-p_2}{\rho g}\right)+\left(\frac{V_1^2}{2g}-\frac{V_2^2}{2g}\right)+\left(z_1-z_2\right)=27\frac{V^2}{2g}-h_p$ ..... (III)

Write the expression for the flow rate.

$Q=AV$ ..... (IV)

Here, the area is $A$.

Write the expression for the area.

$A=\frac{\pi }{4}{d}^2$

Here, the diameter is $d$.

Substitute $\frac{\pi }{4}{d}^2$ for $A$ in Equation (IV).

$\begin{array}{l}Q=\left(\frac{\pi }{4}{d}^2\right)V\\ V=\frac{Q}{\frac{\pi }{4}{d}^2}\end{array}$

Substitute $\frac{Q}{\frac{\pi }{4}{d}^2}$ for $V$ in Equation (III).

$\begin{array}{l}\left(\frac{p_1-p_2}{\rho g}\right)+\left(\frac{V_1^2}{2g}-\frac{V_2^2}{2g}\right)+\left(z_1-z_2\right)=27\frac{{\left(\frac{Q}{\frac{\pi }{4}{d}^2}\right)}^2}{2g}-h_p\\ \left(\frac{p_1-p_2}{\rho g}\right)+\left(\frac{V_1^2}{2g}-\frac{V_2^2}{2g}\right)+\left(z_1-z_2\right)=27\left(\frac{8Q^2}{{\pi }^2{d}^{4}g}\right)-h_p\\ \left(\frac{p_1-p_2}{\rho g}\right)+\left(\frac{V_1^2}{2g}-\frac{V_2^2}{2g}\right)+\left(z_1-z_2\right)=\left(\frac{216Q^2}{g}\right)\left(\frac{1}{{\pi }^2{d}^4}\right)-h_p\end{array}$ ..... (V)

Calculation:

Substitute $\left(0{\text{ft}}^3/\text{s},300\text{ft}\right)$ for $A$ and $\left(4{\text{ft}}^3/\text{s},100\text{ft}\right)$ for $B$ in Equation (I).

$\begin{array}{l}\left[h_p-\left(300\text{ft}\right)\right]=\frac{\left\{\left(100\text{ft}\right)-\left(300\text{ft}\right)\right\}}{\left\{\left(4{\text{ft}}^3/\text{s}\right)-\left(0{\text{ft}}^3/\text{s}\right)\right\}}\left[Q-\left(0{\text{ft}}^3/\text{s}\right)\right]\\ h_p-\left(300\text{ft}\right)=-\left(50\text{s}/{\text{ft}}^2\right)\left(Q\right)\\ h_p=\left(300\text{ft}\right)-\left(50\text{s}/{\text{ft}}^2\right)\left(Q\right)\end{array}$

Since the pressure at the inlet and outlet are same so,

$p_1=p_2$

Since the diameter of the pipe is constant, so the velocity at section (1) and section (2) is same.

$V_1=V_2$

Substitute $p_2$ for $p_1$ and $V_2$ for $V_1$ in Equation (V).

$\begin{array}{l}\left(\frac{p_2-p_2}{\rho g}\right)+\left(\frac{V_2^2}{2g}-\frac{V_2^2}{2g}\right)+\left(z_1-z_2\right)=\left(\frac{216Q^2}{g}\right)\left(\frac{1}{{\pi }^2{d}^4}\right)-h_p\\ \left(z_1-z_2\right)=\left(\frac{216Q^2}{g}\right)\left(\frac{1}{{\pi }^2{d}^4}\right)-h_p\end{array}$ ..... (VI)

Substitute $\left(300\text{ft}\right)-\left(50\text{s}/{\text{ft}}^2\right)Q$ for $h_p$ in Equation (VI).

$\left(z_1-z_2\right)=\left(\frac{216Q^2}{g}\right)\left(\frac{1}{\pi d^4}\right)-\left[\left(300\text{ft}\right)-\left(50\text{s}/{\text{ft}}^2\right)Q\right]$ ..... (VII)

Substitute $150\text{ft}$ for $z_2$, $50\text{ft}$ for $z_1$, $32.2\text{ft}/{\text{s}}^2$ for $g$ and $6\text{in}$ for $d$ in Equation (VII).

$\begin{array}{l}\left[\left(50\text{ft}\right)-\left(150\text{ft}\right)\right]=\left(\frac{216Q^2}{\left(32.2\text{ft}/{\text{s}}^2\right)}\right)\left(\frac{1}{{\pi }^2{\left(6\text{in}\right)}^4}\right)-\left[\left(300\text{ft}\right)-\left(50\text{s}/{\text{ft}}^2\right)Q\right]\\ \left[\left(50\text{ft}\right)-\left(150\text{ft}\right)\right]=\left(\begin{array}{l}\left(\frac{216Q^2}{\left(32.2\text{ft}/{\text{s}}^2\right)}\right)\left(\frac{1}{{\pi }^2{\left(6\text{in}\right)}^4{\left(\frac{1\text{ft}}{12\text{in}}\right)}^4}\right)\\ -\left[\left(300\text{ft}\right)-\left(50\text{s}/{\text{ft}}^2\right)Q\right]\end{array}\right)\\ \left(10.87{\text{ft}}^3/{\text{s}}^2\right)Q^2+\left(50\text{s}/{\text{ft}}^2\right)Q-\left(200\text{ft}\right)=0\\ Q=2.567{\text{ft}}^3/\text{s}\end{array}$

Convert flow rate into $\text{gal}/\text{min}$.

$\begin{array}{c}Q=\left(2.567{\text{ft}}^3/\text{s}\right)\left(\frac{448.8\text{gal}/\text{min}}{1{\text{ft}}^3/\text{s}}\right)\\ =1152.07\text{gal}/\text{min}\end{array}$

Conclusion:

The flow rate is $1152.07\text{gal}/\text{min}$.

To determine

(b)

The horsepower required to drive the pump.

Answer to Problem 3.181P

The horsepower required to drive the pump is $66.72\text{hp}$.

Explanation of Solution

Given information:

The efficiency of the pump is $75\text{}\%$.

Write the expression for the pump head.

$h_p=300-50Q$

Write the expression for the power.

$P=\frac{\rho gQ{h}_p}{\eta }$ ..... (VIII)

Here, the power is $P$ and the pump efficiency is $\eta$.

Substitute $300-50Q$ for $h_p$ in Equation (VIII).

$P=\frac{\rho gQ\left(300-50Q\right)}{\eta }$ ..... (IX)

Calculation:

Substitute $1.94\text{slug}/{\text{ft}}^{\text{3}}$ for $\rho$, $32.2\text{ft}/{\text{s}}^2$ for $g$, $0.75$ for $\eta$ and $2.567{\text{ft}}^3/\text{s}$ for $Q$ in Equation (IX).

$\begin{array}{c}P=\frac{\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^2\right)\left(2.567{\text{ft}}^3/\text{s}\right)\left\{300\text{ft}-\left(50\text{s}/{\text{ft}}^2\right)\left(2.567{\text{ft}}^3/\text{s}\right)\right\}}{0.75}\\ =\left(36699.9\text{ft}\cdot \text{lbf}/\text{s}\right)\left(\frac{1\text{hp}}{550\text{ft}\cdot \text{lbf}/\text{s}}\right)\\ =66.72\text{hp}\end{array}$

Conclusion:

The horsepower required to drive the pump is $66.72\text{hp}$.