A pump on the ground floor can pump up water | vedclass

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Question

(A) The mass of water $m = 	ext{Volume} 	imes 	ext{Density} = 30 \ m^3 	imes 1000 \ kg/m^3 = 30,000 \ kg$.The output power required to lift the wa

Vedclass · Accepted Answer

$43.6$

Vedclass · Answer

(A) The mass of water $m = 	ext{Volume} 	imes 	ext{Density} = 30 \ m^3 	imes 1000 \ kg/m^3 = 30,000 \ kg$.The output power required to lift the water is $P_{	ext{out}} = \frac{mgh}{t}$.Given $h = 40 \ m$,$t = 15 \ minutes = 15 	imes 60 \ s = 900 \ s$,and $g = 9.8 \ m/s^2$.$P_{	ext{out}} = \frac{30,000 	imes 9.8 	imes 40}{900} = \frac{11,760,000}{900} = 13,066.67 \ W$.The efficiency $\eta = \frac{P_{	ext{out}}}{P_{	ext{in}}} 	imes 100\%$.$30 = \frac{13,066.67}{P_{	ext{in}}} 	imes 100$.$P_{	ext{in}} = \frac{13,066.67 	imes 100}{30} = 43,555.56 \ W \approx 43.6 \ kW$.

$A$ pump on the ground floor can pump up water to fill a tank of volume $30 \ m^3$ in $15 \ minutes$. If the tank is $40 \ m$ above the ground and the efficiency of the pump is $30\%$,then the electric power consumed by the pump is ....... $kW$.

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