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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-




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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