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Work, Power and Energy

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Question : 15 of 30
Marks: +1, -0
A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m3 in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump.
Solution:  
As given that
Volume of water =30m3= 30 \text{m}^3
Time taken by pump to fill tank =15min=15×60=900s= 15 \text{min} = 15 \times 60 = 900 \text{s}
The height of tank = 40 m
The efficiency of pump = 30%
Consumption of power by pump = ?
Mass of water pumped = Volume × density =30×103= 30 \times 10^3 (Density of water =103kg/m3= 10^3 \text{kg/m}^3)
Output power, P0=Wt=mghtP_0 = \frac{W}{t} = \frac{m g h}{t} =30×103×9.8×40900= \frac{30 \times 10^3 \times 9.8 \times 40}{900} =13070=13070 watt
PiP_i is required power (input power), then efficiency η=P0Pi\eta = \frac{P_0}{P_i}
Pi=P0η=1307030100=43567wattP_i = \frac{P_0}{\eta} = \frac{13070}{\frac{30}{100}} = 43567 \text{watt} = 43.567 kW
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