Chapter 5, Problem 99P

A fireboat is to fight fires at coastal areas by drawing seawater with a density of 1030 kg/m³ through a 10-cm-diameter pipe at a rate of 0.04 m³/s and discharging it through a hose nozzle with an exit diameter of 5 cm. The total irreversible head loss of the system is 3 m. and the position of the nozzle is 3 m above sea level. For a pump efficiency of 70 percent, determine the required shaft power input to the pump and the water discharge velocity.

To determine

The required shaft input to the pump is $13.2\text{kW}$.

And the water discharge velocity is $20.371\text{m}/\text{s}$.

Answer to Problem 99P

The required shaft input to the pump.

And the water discharge velocity.

Explanation of Solution

Given information:

The density of the sea water is $1030\text{kg}/{\text{m}}^{\text{3}}$, diameter of the suction pipe is $10\text{cm}$, volume flow rate of water is $0.04{\text{m}}^{\text{3}}/\text{s}$, diameter of the delivery hose nozzle is $5\text{cm}$, irreversible head loss of the system is $3\text{m}$, position of the nozzle above the sea level is $3\text{m}$, and efficiency of the pump is $70\%$.

Write the expression for the volume flow rate through the suction pipe.

   $\dot{Q}=A_1{v}_1$    ...... (I)

Here, volume flow rate of water is $\dot{Q}$, area of the suction pipe is $A_1$, velocity of water in suction pipe is $v_1$.

Write the expression for the area of the suction pipe.

   $A_1=\frac{\pi }{4}{D}_1^2$    ...... (II)

Here, area of the suction pipe is $A_1$, and diameter of the suction pipe is $D_1$.

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

   $\dot{Q}=\left(\frac{\pi }{4}{D}_1^2\right)v_1$    ...... (III)

Here, volume flow rate of water is $\dot{Q}$, area of the suction pipe is $A_1$, velocity of water in suction pipe is $v_1$, and diameter of the suction pipe is $D_1$.

Writing the continuity equation between the inlet and exit points of the pipe.

   $\dot{Q}=A_2{v}_2$    ...... (IV)

Here, volume flow rate of water is $\dot{Q}$, area of the hose nozzle is $A_2$, velocity of water in hose nozzle is $v_2$.

Write the expression for the area of the hose nozzle.

   $A_2=\frac{\pi }{4}{D}_2^2$    ...... (V)

Here, area of the hose nozzle is

   $A_2$, and diameter of the hose nozzle is $D_2$.

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

   $\dot{Q}=\left(\frac{\pi }{4}{D}_2^2\right)v_2$    ...... (VI)

Here, volume flow rate of water is $\dot{Q}$, area of the hose nozzle is $A_2$, velocity of water in hose nozzle is $v_2$, and diameter of the hose nozzle is $D_2$.

Write the expression for the mechanical energy balance between the inlet and outlet of pipe.

   $\left(\frac{P_1}{\rho g}+\frac{v_1^2}{2g}+Z_1\right)+h_{pump}=\left(\frac{P_2}{\rho g}+\frac{v_2^2}{2g}+Z_2\right)+h_L$    ...... (VII)

Here, pressure at the suction point is $P_1$, pressure at the delivery point is $P_2$, velocity of water in suction pipe is $v_1$, velocity of water in hose nozzle is $v_2$, density of the seawater is $\rho$, acceleration due to gravity is $g$, height of reference point on suction side from sea level is $Z_1$, position of the nozzle from the sea level is $Z_2$, head delivered to the fluid by the pump is $h_{pump}$, and irreversible head loss of the system is $h_L$.

Write the expression for the pump power.

   $W_{pump}=\rho g\dot{Q}{h}_{pump}$    ...... (VIII)

Here, pump power is $W_{pump}$, density of the seawater is $\rho$, acceleration due to gravity is $g$, head delivered to the fluid by the pump is $h_{pump}$, and volume flow rate through the suction pipe is $\dot{Q}$.

Write the expression for the efficiency of the pump.

   $\eta =\frac{W_{pump}}{W_{shaft}}$    ...... (IX)

Here, efficiency of the pump is $\eta$, pump power is $W_{pump}$, and shaft work is $W_{shaft}$.

Calculation:

Substitute $0.04{\text{m}}^{\text{3}}/\text{s}$ for $\dot{Q}$, and

   $10\text{cm}$ for $D_1$ in Equation (III).

   $\begin{array}{l}\left(0.04{\text{m}}^{\text{3}}/\text{s}\right)=\frac{\pi }{4}{\left(10\text{cm}\right)}^2\\ \left(0.04{\text{m}}^{\text{3}}/\text{s}\right)=\left(\frac{\pi }{4}{\left(10\text{cm}\right)}^2\left(\frac{1{\text{m}}^{\text{2}}}{{10}^4{\text{cm}}^{\text{2}}}\right)\right)v_1\\ v_1=5.026\text{m}/\text{s}\end{array}$

Substitute $5\text{cm}$ for $D_2$, and $0.04{\text{m}}^{\text{3}}/\text{s}$ for $\dot{Q}$ in Equation (VI).

   $\begin{array}{l}\left(0.04{\text{m}}^{\text{3}}/\text{s}\right)=\frac{\pi }{4}{\left(5\text{cm}\right)}^2\\ \left(0.04{\text{m}}^{\text{3}}/\text{s}\right)=\left(\frac{\pi }{4}{\left(5\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\right)v_2\\ v_2=\left(\frac{0.04{\text{m}}^3/\text{s}}{1.963{\text{m}}^{\text{2}}}\right)\\ v_2=20.371\text{m}/\text{s}\end{array}$

Substitute $P_{atm}$ for $P_1$, $P_{atm}$ for $P_2$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $5.026\text{m}/\text{s}$ for $v_1$, $20.371\text{m}/\text{s}$ for $v_2$, $0\text{m}$ for $Z_1$, $3\text{m}$ for $Z_2$, and $3\text{m}$ for $h_L$ in Equation (VII).

   $\begin{array}{l}\left(\begin{array}{l}\left(\frac{P_{atm}}{\rho \left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}+\frac{{\left(5.026\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}+0\text{m}\right)+h_{pump}\\ =\left(\frac{P_{atm}}{\rho \left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}+\frac{{\left(20.371\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}+0\text{m}\right)+3\text{m}\end{array}\right)\\ h_{pump}=\left(\frac{\left({\left(20.371\text{m}/\text{s}\right)}^2-{\left(5.026\text{m}/\text{s}\right)}^2\right)}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\right)+\left(3\text{m}\right)\\ h_{pump}=22.86\text{m}\end{array}$

Substitute $22.86\text{m}$ for $h_{pump}$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $0.04{\text{m}}^{\text{3}}/\text{s}$ for $\dot{Q}$, and $1030\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ in Equation (VIII).

   $\begin{array}{c}{W}_{pump}=\left(1030\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.04{\text{m}}^{\text{3}}/\text{s}\right)\left(22.86\text{m}\right)\\ =\left(9241.08\text{kg}\cdot \text{m}/{\text{s}}^2\right)\left(\text{m}/\text{s}\right)\\ =\left(9241.08\text{kg}\cdot \text{m}/{\text{s}}^2\right)\left(\text{m}/\text{s}\right)\left(\frac{1\text{N}}{\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\left(\frac{1\text{J}}{1\text{N}\cdot \text{m}}\right)\left(\frac{1\text{W}}{1\text{J}/\text{s}}\right)\\ =9241.08\text{W}\end{array}$

Substitute $0.7$ for $\eta$, and $9241.08\text{W}$ for $W_{pump}$ in Equation (IX).

   $\begin{array}{l}0.7=\frac{\left(9241.08\text{W}\right)}{W_{shaft}}\\ 0.7=\frac{\left(9241.08\text{W}\right)}{W_{shaft}}\left(\frac{1\text{kW}}{{10}^3\text{W}}\right)\\ W_{shaft}=13.2\text{kW}\end{array}$

Conclusion:

The required shaft input to the pump is $13.2\text{kW}$.

And the water discharge velocity is $20.371\text{m}/\text{s}$.