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Question Number 136110 by physicstutes last updated on 18/Mar/21

An engine is pumping water from a well 25m deep. It discharges 0.4 m^3 of water each second with a velocity of 12 ms^(−1) . Find the power of the pump given that the density of water is 1000 kg m^(−3) . take g = 10 ms^(−2)

$$\mathrm{An}\:\mathrm{engine}\:\mathrm{is}\:\mathrm{pumping}\:\mathrm{water}\:\mathrm{from}\:\mathrm{a}\:\mathrm{well}\:\mathrm{25m}\:\mathrm{deep}.\:\mathrm{It}\:\mathrm{discharges} \\ $$$$\mathrm{0}.\mathrm{4}\:\mathrm{m}^{\mathrm{3}} \:\mathrm{of}\:\mathrm{water}\:\mathrm{each}\:\mathrm{second}\:\mathrm{with}\:\mathrm{a}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{12}\:\mathrm{ms}^{−\mathrm{1}} .\:\mathrm{Find}\:\mathrm{the}\: \\ $$$$\mathrm{power}\:\mathrm{of}\:\mathrm{the}\:\mathrm{pump}\:\mathrm{given}\:\mathrm{that}\:\mathrm{the}\:\mathrm{density}\:\mathrm{of}\:\mathrm{water}\:\mathrm{is}\:\mathrm{1000}\:\mathrm{kg}\:\mathrm{m}^{−\mathrm{3}} . \\ $$$$\mathrm{take}\:\boldsymbol{\mathrm{g}}\:=\:\mathrm{10}\:\mathrm{ms}^{−\mathrm{2}} \\ $$

Commented by mr W last updated on 18/Mar/21

P=Vρ(gh+(1/2)v^2 ) =0.4×1000×(10×25+(1/2)×12^2 ) =128800 Watt=128.8 kW

$${P}={V}\rho\left({gh}+\frac{\mathrm{1}}{\mathrm{2}}{v}^{\mathrm{2}} \right) \\ $$$$\:\:\:=\mathrm{0}.\mathrm{4}×\mathrm{1000}×\left(\mathrm{10}×\mathrm{25}+\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{12}^{\mathrm{2}} \right) \\ $$$$\:\:\:=\mathrm{128800}\:{Watt}=\mathrm{128}.\mathrm{8}\:{kW} \\ $$

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