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-

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

$$\mathrm{An}\mathrm{engine}\mathrm{is}\mathrm{pumping}\mathrm{water}\mathrm{from}\mathrm{a}\mathrm{well}25\mathrm{m}\mathrm{deep}.\mathrm{It}\mathrm{discharges}$$$$0.4{\mathrm{m}}^3\mathrm{of}\mathrm{water}\mathrm{each}\mathrm{second}\mathrm{with}\mathrm{a}\mathrm{velocity}\mathrm{of}12{\mathrm{ms}}^{-1}.\mathrm{Find}\mathrm{the}$$$$\mathrm{power}\mathrm{of}\mathrm{the}\mathrm{pump}\mathrm{given}\mathrm{that}\mathrm{the}\mathrm{density}\mathrm{of}\mathrm{water}\mathrm{is}1000\mathrm{kg}{\mathrm{m}}^{-3}.$$$$\mathrm{take}\mathbf{g}=10{\mathrm{ms}}^{-2}$$

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

$$P=V\rho (gh+\frac{1}{2}{v}^2)$$$$=0.4\times 1000\times (10\times 25+\frac{1}{2}\times {12}^2)$$$$=128800Watt=128.8kW$$

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