Chapter 3, Problem 3.178P

To determine

The power delivered by the pump.

Answer to Problem 3.178P

The power delivered by the pump is $8.34\text{kW}$.

Explanation of Solution

Given information:

The discharge from the pump is $57{\text{m}}^3/\text{h}$ . The discharge temperature of the water is $20°\text{C}$ . The inward pressure is $120\text{kPa}$ and the outward pressure from the pump is $400\text{kPa}$ . The pipe diameter at the inlet end is $9\text{cm}$ and at the exit of pump is $3\text{cm}$.

Write the expression for the velocity at the inlet of the pump.

$V_1=\frac{Q}{A_1}$ .... (I)

Here, the velocity is $V_1$, the area at the inlet section of the pump is $A_1$ and the flow rate is $Q$.

Write the expression for the area.

$A_1=\frac{\pi }{4}{D}_1^2$

Here, the inlet diameter is $D_1$.

Substitute $\frac{\pi }{4}{D}_1^2$ for $A_1$ in Equation (I).

$V_1=\frac{4Q}{\pi D_1^2}$ .... (II)

Write the expression for velocity at section (2).

$V_2=\frac{Q}{A_2}$ .... (III)

Here, the velocity is $V_2$ and the area is $A_2$.

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

$A_2=\frac{\pi }{4}{D}_2^2$

Here, the diameter is $D_2$.

Substitute $\frac{\pi }{4}{D}_2^2$ for $A_2$ in Equation (III).

$\begin{array}{c}{V}_2=\frac{Q}{\frac{\pi }{4}{D}_2^2}\\ =\frac{4Q}{\pi D_2^2}\end{array}$ .... (IV)

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

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

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 oil is $\rho$, the head loss due to friction is $h_f$, the head loss due to pump is $h_p$ and the acceleration due to gravity is $g$.

Since both the section is at the same level from the datum plane, so the difference between the datum height at the two sections is zero.

$z_1-z_2=0$

Since the head loss due to friction is neglected.

$h_f=0$

Substitute $0$ for $z_1-z_2$ and $0$ for $h_f$ in Equation (V).

$\begin{array}{l}\left(\frac{p_1-p_2}{\rho g}\right)+\left(\frac{V_1^2}{2g}-\frac{V_2^2}{2g}\right)+0=0-h_p\\ h_p=\left(\frac{p_2-p_1}{\rho g}\right)+\left(\frac{V_2^2}{2g}-\frac{V_1^2}{2g}\right)\end{array}$ .... (VI)

Write the expression for the pump power.

$P=\rho Qg{h}_p$ .... (VII)

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

Calculation:

Substitute $57{\text{m}}^3/\text{h}$ for $Q$ and $9\text{cm}$ for $D_1$ in Equation (II).

$\begin{array}{c}{V}_1=\frac{4\left(57{\text{m}}^3/\text{h}\right)}{\pi {\left(9\text{cm}\right)}^2}\\ =\frac{4\left(57{\text{m}}^3/\text{h}\right)\left(\frac{1\text{h}}{3600\text{sec}}\right)}{\pi \times 81{\text{cm}}^2\left(\frac{1{\text{m}}^2}{10000{\text{cm}}^2}\right)}\\ \approx 2.48\text{m}/\text{s}\end{array}$

Substitute $57{\text{m}}^3/\text{h}$ for $Q$ and $3\text{cm}$ for $D_2$ in Equation (IV).

$\begin{array}{c}{V}_2=\frac{4\left(57{\text{m}}^3/\text{h}\right)}{\pi {\left(3\text{cm}\right)}^2}\\ =\frac{4\left(57{\text{m}}^3/\text{h}\right)\left(\frac{1\text{h}}{3600\text{sec}}\right)}{\pi \times 9{\text{cm}}^2\left(\frac{1{\text{m}}^2}{10000{\text{cm}}^2}\right)}\\ \approx 22.39\text{m}/\text{s}\end{array}$

Substitute $120\text{kPa}$ for $p_1$, $400\text{kPa}$ for $p_2$, $2.48\text{m}/\text{s}$ for $V_1$, $22.39\text{m}/\text{s}$ for $V_2$, $9.81\text{m}/{\text{s}}^2$ for $g$ and $998\text{kg}/{\text{m}}^3$ for $\rho$ in Equation (VI).

$\begin{array}{c}{h}_p=\left\{\frac{\left(400\text{kPa}\right)-\left(120\text{kPa}\right)}{\left(998\text{kg}/{\text{m}}^3\right)\left(9.81\text{m}/{\text{s}}^2\right)}\right\}+\left(\frac{{\left(22.39\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^2\right)}-\frac{{\left(2.48\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^2\right)}\right)\\ h_p=\left\{\frac{\left(280\text{kPa}\right)\left(\frac{1000\text{Pa}}{1\text{kPa}}\right)}{\left(998\text{kg}/{\text{m}}^3\right)\left(9.81\text{m}/{\text{s}}^2\right)}\right\}+25.24\text{m}\\ =\text{28}\text{.599}\text{m}+25.24\text{m}\\ \approx 53.83\text{m}\end{array}$

Substitute $57{\text{m}}^3/\text{h}$ for $Q$, $9.81\text{m}/{\text{s}}^2$ for $g$, $53.83\text{m}$ for $h_p$ and $998\text{kg}/{\text{m}}^3$ for $\rho$ in Equation (VII).

$\begin{array}{c}P=\left(998\text{kg}/{\text{m}}^3\right)\left(57{\text{m}}^3/\text{h}\right)\left(9.81\text{m}/{\text{s}}^2\right)\left(53.83\text{m}\right)\\ =\left(998\text{kg}/{\text{m}}^3\right)\left(57{\text{m}}^3/\text{h}\right)\left(\frac{1\text{h}}{3600\text{sec}}\right)\left(528.0723{\text{m}}^2/{\text{s}}^2\right)\\ =\left(8344.42\text{W}\right)\left(\frac{1\text{kW}}{1000\text{W}}\right)\\ \approx 8.34\text{kW}\end{array}$

Conclusion:

The power delivered by the pump is $8.34\text{kW}$.