Chapter 3, Problem 3.159P

To determine

(a)

The absolute velocity $V_1$.

Answer to Problem 3.159P

The absolute velocity $V_1$ is $21.87\text{ft}/\text{s}$.

Explanation of Solution

Given information:

The flow rate of the pump is $4000\text{gal}/\text{m}$ . The temperature of the water is $20°\text{C}$ . The shaft rotation rate is $1750\text{r}/\text{min}$ . The radius of the impeller at section (1) is $6\text{in}$ and the radius of the pump at section (2) is $14\text{in}$ . The width of the blade is $1.75\text{in}$ . The tangential velocity at section (1) is $10\text{ft}/\text{s}$ and the tangential velocity at section (2) is $110\text{ft}/\text{s}$.

Write the expression for the absolute velocity at section (1).

$V_1=\sqrt{V_{t,1}^2+V_{n,1}^2}$ …… (I)

Here, the absolute velocity is $V_1$, the tangential velocity at section (1) is $V_{t,1}$ and the normal velocity is $V_{n,1}$.

Write the expression for the normal velocity at section (1).

$V_{n,1}=\frac{Q}{A_1}$ …… (II)

Here, the flow rate is $Q$ and the area at section (1) is $A_1$.

Write the expression for the area of the impeller at the section (1).

$A_1=2\pi r_1{b}_1$

Here, the width of the blade is $b_1$.

Substitute $2\pi r_1{b}_1$ for $A_1$ in Equation (II).

$V_{n,1}=\frac{Q}{2\pi r_1{b}_1}$ …… (III)

Calculation:

Substitute $4000\text{gal}/\text{m}$ for $Q$, $6\text{in}$ for $r_1$ and $1.75\text{in}$ for $b_1$ in Equation (III).

$\begin{array}{c}{V}_{n,1}=\frac{\left(4000\text{gal}/\text{m}\right)}{2\pi \left(6\text{in}\right)\left(1.75\text{in}\right)}\\ =\frac{\left(4000\text{gal}/\text{m}\right)\left(\frac{1{\text{ft}}^3}{7.48\text{gal}}\right)\left(\frac{1\text{m}}{60\text{s}}\right)}{\left(65.97{\text{in}}^2\right)\left(\frac{1{\text{ft}}^2}{144{\text{in}}^2}\right)}\\ =19.45\text{ft}/\text{s}\end{array}$

Substitute $19.45\text{ft}/\text{s}$ for $V_{n,1}$ and $10\text{ft}/\text{s}$ for $V_{t,1}$ in Equation (I).

$\begin{array}{c}{V}_1=\sqrt{{\left(10\text{ft}/\text{s}\right)}^2+{\left(19.45\text{ft}/\text{s}\right)}^2}\\ =\sqrt{478.3025{\text{ft}}^2/{\text{s}}^2}\\ =21.87\text{ft}/\text{s}\end{array}$

Conclusion:

The absolute velocity $V_1$ is $21.87\text{ft}/\text{s}$.

To determine

(b)

The absolute velocity $V_2$.

Answer to Problem 3.159P

The absolute velocity $V_2$ is $110.31\text{ft}/\text{s}$.

Explanation of Solution

Given information:

The flow rate of the pump is $4000\text{gal}/\text{m}$ . The temperature of the water is $20°\text{C}$ . The shaft rotation rate is $1750\text{r}/\text{min}$ . The radius of the impeller at section (1) is $6\text{in}$ and the radius of the pump at section (2) is $14\text{in}$ . The width of the blade is $1.75\text{in}$ . The tangential velocity at section (1) is $10\text{ft}/\text{s}$ and the tangential velocity at section (2) is $110\text{ft}/\text{s}$.

Write the expression for the absolute velocity at section (2).

$V_2=\sqrt{V_{t,2}^2+V_{n,2}^2}$ …… (IV)

Here, the absolute velocity is $V_2$, the tangential velocity at section (2) is $V_{t,2}$ and the normal velocity is $V_{n,2}$.

Write the expression for the normal velocity at section (2).

$V_{n,2}=\frac{Q}{A_2}$ …… (V)

Here, the flow rate is $Q$ and the area at section (2) is $A_2$.

Write the expression for the area of the impeller at the section (2).

$A_2=2\pi r_2{b}_2$

Here, the width of the blade is $b_2$.

Substitute $2\pi r_2{b}_2$ for $A_2$ in Equation (V).

$V_{n,2}=\frac{Q}{2\pi r_2{b}_2}$ …… (VI)

Calculation:

Substitute $4000\text{gal}/\text{m}$ for $Q$, $14\text{in}$ for $r_2$ and $1.75\text{in}$ for $b_2$ in Equation (VI).

$\begin{array}{c}{V}_{n,2}=\frac{\left(4000\text{gal}/\text{m}\right)}{2\pi \left(14\text{in}\right)\left(1.75\text{in}\right)}\\ =\frac{\left(4000\text{gal}/\text{m}\right)\left(\frac{1{\text{ft}}^3}{7.48\text{gal}}\right)\left(\frac{1\text{m}}{60\text{s}}\right)}{\left(153.94{\text{in}}^2\right)\left(\frac{1{\text{ft}}^2}{144{\text{in}}^2}\right)}\\ =8.3\text{ft}/\text{s}\end{array}$

Substitute $8.3\text{ft}/\text{s}$ for $V_{n,2}$ and $110\text{ft}/\text{s}$ for $V_{t,2}$ in Equation (VI).

$\begin{array}{c}{V}_2=\sqrt{{\left(110\text{ft}/\text{s}\right)}^2+{\left(8.3\text{ft}/\text{s}\right)}^2}\\ =\sqrt{12168.89{\text{ft}}^2/{\text{s}}^2}\\ =110.31\text{ft}/\text{s}\end{array}$

Conclusion:

The absolute velocity $V_2$ is $110.31\text{ft}/\text{s}$.

To determine

(c)

The horsepower required.

Answer to Problem 3.159P

The horsepower required is $894.25\text{hp}$.

Explanation of Solution

Given information:

The flow rate of the pump is $4000\text{gal}/\text{m}$ . The temperature of the water is $20°\text{C}$ . The shaft rotation rate is $1750\text{r}/\text{min}$ . The radius of the impeller at section (1) is $6\text{in}$ and the radius of the pump at section (2) is $14\text{in}$ . The width of the blade is $1.75\text{in}$ . The tangential velocity at section (1) is $10\text{ft}/\text{s}$ and the tangential velocity at section (2) is $110\text{ft}/\text{s}$.

Write the expression for the Bernoulli’s equation between the section (1) and section (2) of the given system.

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

Here, the pressure at section (1) is $p_1$, the height of the section (1) from the datum is $z_1$, the pressure at section (2) is $p_2$, the height of the section (2) from the datum is $z_2$, the density of water is $\rho$, the head loss due to friction is $h_f$ and the acceleration due to gravity is $g$.

Since the pressure at section (1) and section (2) is atmospheric, so the difference between them is zero.

$p_1-p_2=0$

Since the datum is at the same level so,

$z_1-z_2=0$

Substitute $0$ for $p_1-p_2$ and $0$ for $z_1-z_2$ in Equation (VII).

$\begin{array}{l}\left(\frac{0}{\rho g}\right)+\left(\frac{V_1^2}{2g}-\frac{V_2^2}{2g}\right)+\left(0\right)=-h_f\\ h_f=\left(\frac{V_2^2-V_1^2}{2g}\right)\end{array}$ …… (VIII)

Write the expression for the power loss due to friction head.

$P_{\text{loss}}=\rho Qg{h}_f$ …… (IX)

Here, the power loss is $P_{\text{loss}}$.

Write the expression for the ideal horsepower developed.

$P_{\text{ideal}}=\rho Q\left(r_2{V}_{t,2}-r_1{V}_{t,1}\right)\omega$ …… (X)

Here, the ideal power is $P_{\text{ideal}}$ the density is $\rho$ and angular revolutions is $\omega$.

Write the expression for the horsepower required.

$P_{\text{actual}}=P_{\text{ideal}}+P_{\text{loss}}$ …… (XI)

Here, the power required is $P_{\text{actual}}$.

Calculation:

Substitute $21.87\text{ft}/\text{s}$ for $V_1$, $110.31\text{ft}/\text{s}$ for $V_2$ and $32.2\text{ft}/{\text{s}}^2$ for $g$ in Equation (VIII).

$\begin{array}{c}{h}_f=\left(\frac{{\left(110.31\text{ft}/\text{s}\right)}^2-{\left(21.87\text{ft}/\text{s}\right)}^2}{2\left(32.2\text{ft}/{\text{s}}^2\right)}\right)\\ =\frac{11689.99{\text{ft}}^2/{\text{s}}^2}{64.4\text{ft}/{\text{s}}^2}\\ =181.52\text{ft}\end{array}$

Substitute $1.94\text{slug}/{\text{ft}}^{\text{3}}$ for $\rho$, $4000\text{gal}/\text{m}$ for $Q$, $32.2\text{ft}/{\text{s}}^2$ for $g$ and $181.52\text{ft}$ for $h_f$ in Equation (IX).

$\begin{array}{c}{P}_{\text{loss}}=\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(4000\text{gal}/\text{m}\right)\left(32.2\text{ft}/{\text{s}}^2\right)\left(181.52\text{ft}\right)\\ =\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(4000\text{gal}/\text{m}\right)\left(\frac{1{\text{ft}}^3}{7.48\text{gal}}\right)\left(\frac{1\text{m}}{60\text{s}}\right)\left(5844.94{\text{ft}}^2/{\text{s}}^2\right)\\ =\left(101062.31\text{lb}\cdot \text{ft}/\text{s}\right)\left(\frac{1\text{hp}}{550\text{lb}\cdot \text{ft}/\text{s}}\right)\\ =183.75\text{hp}\end{array}$

Substitute $1.94\text{slug}/{\text{ft}}^{\text{3}}$ for $\rho$, $4000\text{gal}/\text{m}$ for $Q$, $14\text{in}$ for $r_2$, $110\text{ft}/\text{s}$ for $V_{t,2}$, $6\text{in}$ for $r_1$, $10\text{ft}/\text{s}$ for $V_{t,1}$, $1750\text{r}/\text{min}$ for $\omega$ in Equation (X).

$\begin{array}{c}{P}_{\text{ideal}}=\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(4000\text{gal}/\text{m}\right)\left\{\left(14\text{in}\right)\left(110\text{ft}/\text{s}\right)-\left(6\text{in}\right)\left(10\text{ft}/\text{s}\right)\right\}\left(1750\text{r}/\text{min}\right)\\ =\left[\begin{array}{l}\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(4000\text{gal}/\text{m}\right)\left(\frac{1{\text{ft}}^3}{7.48\text{gal}}\right)\left(\frac{1\text{m}}{60\text{s}}\right)\times \\ \left\{\left(14\text{in}\right)\left(\frac{1\text{ft}}{12\text{in}}\right)\times \left(110\text{ft}/\text{s}\right)-\left(6\text{in}\right)\left(\frac{1\text{ft}}{12\text{in}}\right)\times \left(10\text{ft}/\text{s}\right)\right\}\times \\ \left(1750\text{r}/\text{min}\right)\left(\frac{2\pi \text{rad}}{1\text{rev}}\right)\left(\frac{1\text{min}}{60\text{s}}\right)\end{array}\right]\\ =\left(390801.31\text{lb}\cdot \text{ft}/\text{s}\right)\left(\frac{1\text{hp}}{550\text{lb}\cdot \text{ft}/\text{s}}\right)\\ =710.5\text{hp}\end{array}$

Substitute $183.75\text{hp}$ for $P_{\text{loss}}$ and $710.5\text{hp}$ for $P_{\text{ideal}}$ in Equation (XI).

$\begin{array}{c}P=\left(710.5\text{hp}\right)+\left(183.75\text{hp}\right)\\ =894.25\text{hp}\end{array}$

Conclusion:

The horsepower required is $894.25\text{hp}$.

To determine

(d)

The comparison of actual and ideal horsepower.

Answer to Problem 3.159P

The actual horsepower is greater than ideal horsepower.

Explanation of Solution

Given information:

The flow rate of the pump is $4000\text{gal}/\text{m}$ . The temperature of the water is $20°\text{C}$ . The shaft rotation rate is $1750\text{r}/\text{min}$ . The radius of the impeller at section (1) is $6\text{in}$ and the radius of the pump at section (2) is $14\text{in}$ . The width of the blade is $1.75\text{in}$ . The tangential velocity at section (1) is $10\text{ft}/\text{s}$ and the tangential velocity at section (2) is $110\text{ft}/\text{s}$.

Write the ideal horsepower required.

$P_{\text{ideal}}=710.5\text{hp}$

Write the actual horsepower required.

$P_{\text{actual}}=894.25\text{hp}$

Since due to frictional losses between the section (1) and section (2) there is an additional power required to overcome the friction.

$P_{\text{actual}}>P_{\text{ideal}}$

Conclusion:

The actual horsepower is greater than ideal horsepower.