Question

a pump can work 3.403 kJ per second in 70% efficiency. a motor is used to run the pump.efficiency of motor is 85%, what is the actual power of the pump?

Answer 1

Answer on Question #50732, Physics, Mechanics | Kinematics | Dynamics

A pump can work $3.403\ \mathrm{kJ}$ per second in $70\%$ efficiency. A motor is used to run the pump. Efficiency of motor is $85\%$, what is the actual power of the pump?

Solution:

$$\text{Power efficiency } (\%) = \frac{\text{power output}}{\text{actual power input}} \times 100$$

The power output is

$$\text{power output} = \frac{\text{work}}{\text{time}} = \frac{3.403 \times 10^{3}\ \mathrm{J}}{1\ \text{sec}} = 3403\ \mathrm{W}$$

Power input to pump is

$$\text{power input to pump} = \frac{\text{power output}}{\text{Power efficiency } (\%)} \times 100 = \frac{3403 \times 100}{70} = 4861.43\ \mathrm{W}$$ $$\text{actual power input} = \frac{\text{power input to pump}}{\text{Efficiency of motor} (\%)} \times 100 = \frac{4861.43}{85} \times 100 = 5719.33\ \mathrm{W}$$

Answer: 5719.33 W

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