A fire engine pump (shown in Fig. 2.21) delivers water to a nozzle at C located 30 m above the pump at the end of a 100 mm diameter delivery pipe. The pump draws water from a sump at A through a 150 mm diameter pipe. The pump is located at B at a height 2 m above the level of the sump.

Question

A fire engine pump (shown in Fig. 2.21) delivers water to a nozzle at C located 30 m above the pump at the end of a 100 mm diameter delivery pipe. The pump draws water from a sump at A through a 150 mm diameter pipe. The pump is located at B at a height 2 m above the level of the sump. The velocity at the nozzle is [latex]8.31 {m}/{s}[/latex] and jet diameter 75 mm. Determine the flow rate from the nozzle and the power required to drive the pump.

Accepted Answer

Applying equation of continuity [latex]Q=a imes V[/latex]

[latex] herefore Q = Velocity at C imes area of the jet[/latex]

[latex]Q = 8.31 imes \frac{\pi }{4} imes \left(75 imes 10^{-3} ight)^{2} = 0.0366 {m^{3}}/{s}[/latex]

Applying Bernoullis equation between points A and C

[latex]\frac{P_{A}}{\gamma } + \frac{V^{2}_{A}}{2g} + z_{A} = \frac{P_{C}}{\gamma } + \frac{V^{2}_{C}}{2g} + z_{C} - H_{P}[/latex]

At the free surface of the slimp, the pressure is atmospheric so that [latex]P_{A} = 0[/latex], the velocity [latex]V_{A}[/latex] is also zero as the slimp is considered very large and A can be taken as datum so that [latex]Z_{A} = 0 also P_{C} = atmospheric\quad pressure = 0 [/latex], the head developed by the pump is [latex]H_{P}[/latex], substituting proper values in the equation

[latex]\frac{P_{A}}{\gamma } + \frac{V^{2}_{A}}{2g} + z_{A} = \frac{P_{C}}{\gamma } + \frac{V^{2}_{C}}{2g} + z_{C} - H_{P}[/latex]

[latex]0 + 0 + 0 = 0 + \frac{8.31^{2}}{2 imes 9.8} + 32 - H_{P}[/latex]

[latex]H_{P} = 3.52 + 32 = 35.52 m[/latex]

Power required to drive the pump

[latex]P = \gamma QH_{P} = 9800 imes 0.0366 imes 35.52 = 12.75 KW[/latex]

[latex]P = 12.75 KW[/latex]

Question 2.10: A fire engine pump (shown in Fig. 2.21) delivers water to a ...

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