Asked by jordan
The electric motor driving a pump is rated at 10 kW. The cooling water pump can lift 3.5 m3 of water to a height of 10 m each minute. What is the efficiency of the system
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Volume flow = 3.5 m^3/min = 3.5/60 = 0.05833 m^3/s.
Mass flow = ρ·Q = 1000·0.05833 = 58.33 kg/s.
Hydraulic power = ṁ g h = 58.33·9.81·10 = 5,722.5 W ≈ 5.72 kW.
Efficiency = output/input = 5.72 kW / 10 kW = 0.572 ≈ 57.2%.
(Assumes ρ = 1000 kg/m^3 and g = 9.81 m/s^2.)
Mass flow = ρ·Q = 1000·0.05833 = 58.33 kg/s.
Hydraulic power = ṁ g h = 58.33·9.81·10 = 5,722.5 W ≈ 5.72 kW.
Efficiency = output/input = 5.72 kW / 10 kW = 0.572 ≈ 57.2%.
(Assumes ρ = 1000 kg/m^3 and g = 9.81 m/s^2.)
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