A water pump is driven by an electric motor during its operation. Water enters t

Question

A water pump is driven by an electric motor during its operation. Water enters the pump at 18 m/sec and 160 kPa and leaves it at 10 m/sec and 420 kPa. The elevation change between the entrance and exit locations of the pump is negligible. The volumetric flow rate of water through the pump is 0.36 m3/sec. The pump operates at an efficiency of 84% whereas the motor efficiency is 91%. Calculate the
a) shaft power required to run the pump
b) electric power required to run the motor (in hp - round the answer to the nearest higher whole number).

Explanation / Answer

change in head in pump = {density * 0.5*10^2 - 18^2 } + 420000-160000 = 148000 J/m^3 also pump output = change in head * volume flow rate = 0.36*148000 = 53.28 kW a) shaft power required to pump = Output / pump efficiency = 53.28 / 0.84 = 63.43 kW= 86.3 hp b) power required by motor = motor output / motor efficiency = 63.43 / 0.91 = 69.7 kW = 94.83 hp

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