A motor is used to lift water from a well of depth 10 m and to fill water tank of volume 30 m $^{3}$ 10 minutes. The tank is at height of 20 m above ground. If 20% of energy is wasted, the power of the motor is (in kW)

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Solution

$P o w e r = \frac{e n e r g y}{t i m e}$
$d P = \frac{d E}{d t}$
$= \frac{d m}{d t} \times g \times h$
As it will be useful power
Volume $= 30 m^{3}$
$\frac{m a s s}{v o l u m e} = d e n s i t y$
$m a s s = 1000 k g \times 30 m^{3}$
takes $10 m i n$ to fill
$∴ \frac{d m}{d t} = \frac{30000}{10 \times 60} k g / s$
$∴ \frac{d m}{d t} = 50 k g / s$
$∴ P = 50 \times 10 \times 30$
$= 15000 W = 15 k w$
$\frac{15}{t o t a l P} \times 100 = 80$ given by
total power $= \frac{150}{8} k w$ (question efficiency)
$= 18.75 k w$

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A motor is used to lift water from a well of depth 10 m and to fill water tank of volume 30 m3 10 minutes. The tank is at height of 20 m above ground. If 20% of energy is wasted, the power of the motor is (in kW)

Power=energytime dP=dEdt=dmdt×g×hAs it will be useful powerVolume =30m3massvolume=densitymass=1000kg×30m3takes 10min to fill∴dmdt=3000010×60kg/s∴dmdt=50kg/s∴P=50×10×30 =15000W=15kw 15totalP×100=80 given bytotal power =1508kw (question efficiency) =18.75kw

A motor is used to lift water from a well of depth 10 m and to fill water tank of volume 30 m $^{3}$ 10 minutes. The tank is at height of 20 m above ground. If 20% of energy is wasted, the power of the motor is (in kW)