What is the theoretical horsepower required to pump water at 6.30liters/second from a large reservoir to the surface of another reservoir 120m higher?
Explanations & Calculations
1W=0.00134 hpEnergy=mghpower(W)=mght=Vρ.ght=(Vt).ρgh=0.0063 m3s−1×1000 kgm−3×9.8 ms−2×120 m=7408.8 W=7408.8 W×0.00134 hp W−1=9.93 hp\qquad\qquad \begin{aligned} \small 1 W&=\small 0.00134\,hp\\ \\ \small Energy&=\small mgh\\ \small power(W)&=\small \frac{mgh}{t}=\frac{V\rho.gh}{t}=\Big(\frac{V}{t}\Big).\rho gh\\ &=\small 0.0063\,m^3s^{-1}\times1000\,kgm^{-3}\times9.8\,ms^{-2}\times120\,m\\ &=\small 7408.8\,W\\ &=\small 7408.8\,W\times0.00134\,hp\,W^{-1}\\ &=\small 9.93\,hp \end{aligned}1WEnergypower(W)=0.00134hp=mgh=tmgh=tVρ.gh=(tV).ρgh=0.0063m3s−1×1000kgm−3×9.8ms−2×120m=7408.8W=7408.8W×0.00134hpW−1=9.93hp
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