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A hydraulic pump delivers 0.5 m³/s of water against a head of 5.5 m. If the...

A hydraulic pump delivers 0.5 m³/s of water against a head of 5.5 m. If the power needed to drive the pump is 33 KW, calculate the efficiency

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Answer #1

Solution - Given that water flow rate Q = 0.5 m³/s

Head H = 5.5 m

Power input to pump  P_{in} = 33 : kW

Now power output from the pump is

Part = pgHQ = ( 1000) (9.80) (5.5) (0.5) 26977.5 W-26.978 kW

Now efficiency of the pump is given by

26.978 × 100 Pai, × 100 =-33

eta = 81.75 %

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