An electric pump draws water from a well of depth $50 m$ at a rate of $2 m^{3}$ per second. The water is ejected from the pump with velocity, $v = 10 m s^{- 1}$ . If the efficiency of the pump $η = 80$ , the input power consumption of the pump is: $(g = 10 m s^{- 2})$

1100 k W

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880 k W

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1375 k W

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1080 k W

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Solution

The correct option is C $1375 k W$
Power supplied by pump $= \frac{W D}{t} = \frac{Δ (P . E)}{t} = \frac{m g h}{t}$
$= 2 \times 10^{3} \times 10 \times 50$
Power supplied by pump $= \frac{W . D}{t} = \frac{Δ K . E + Δ P . E}{t}$

$= \frac{\frac{1}{2} m v^{2} + m g h}{t}$

$= \frac{\partial m}{\partial t} (\frac{1}{2} \times 100 + 10 \times 50)$

$= P . Q \times 550$

$= 10^{3} \times 2 \times 550$

$= 1100 k W$
Power input $= \frac{1100}{0.8} = 1375 k W$

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An electric pump draws water from a well of depth 50m at a rate of 2m3 per second. The water is ejected from the pump with velocity, v=10ms−1. If the efficiency of the pump η=80, the input power consumption of the pump is: (g=10ms−2)

The correct option is C 1375 kWPower supplied by pump =WDt=Δ(P.E)t=mght =2×103×10×50Power supplied by pump =W.Dt=ΔK.E+ΔP.Et =12mv2+mght =∂m∂t(12×100+10×50) =P.Q×550 =103×2×550 =1100 kWPower input =11000.8=1375 kW

An electric pump draws water from a well of depth $50 m$ at a rate of $2 m^{3}$ per second. The water is ejected from the pump with velocity, $v = 10 m s^{- 1}$ . If the efficiency of the pump $η = 80$ , the input power consumption of the pump is: $(g = 10 m s^{- 2})$