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(C) The power required to pump water is given by the rate of change of potential energy,$P = \frac{mgh}{t}$.Given:Depth,$h = 10 \ m$Volume flow rate,$

Vedclass · Accepted Answer

$9.8$

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(C) The power required to pump water is given by the rate of change of potential energy,$P = \frac{mgh}{t}$.Given:Depth,$h = 10 \ m$Volume flow rate,$\frac{V}{t} = 10^{-1} \ m^3/s$Density of water,$ho = 10^3 \ kg/m^3$Acceleration due to gravity,$g = 9.8 \ m/s^2$Mass of water pumped per unit time is $\frac{m}{t} = ho 	imes \frac{V}{t}$.$\frac{m}{t} = 10^3 \ kg/m^3 	imes 10^{-1} \ m^3/s = 100 \ kg/s$.Now,the power required is:$P = (\frac{m}{t}) 	imes g 	imes h$$P = 100 \ kg/s 	imes 9.8 \ m/s^2 	imes 10 \ m$$P = 9800 \ W$Converting to kilowatts:$P = \frac{9800}{1000} \ kW = 9.8 \ kW$.

Water is pumped from a depth of $10 \ m$ and delivered through a pipe of cross-section $10^{-2} \ m^2$. If it is needed to deliver a volume of $10^{-1} \ m^3$ per second,the power required will be ........ $kW$.

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