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प्रश्न
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?
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उत्तर १
Time of operation, t = 15 min = 15 × 60 = 900 s
Height of the tank, h = 40 m
Efficiency of the pump, η = 30 %
Density of water, ρ = 103 kg/m3
Mass of water, m = ρV = 30 × 103 kg
Output power can be obtained as: P0 = `"Work done" / "Time" = "mgh" / "t"`
= `(30 × 10^3 × 9.8 × 40 )/ 900`
= 13.067 × 103 W
For input power Pi, efficiency η is given by the relation:
η = `P_0 /P_i` = 30%
Pi = `(13.067 × 10^3 × 100 )/ 30`
= 43.56 kW
उत्तर २
Here, volume of water = 30 m3 ; t = 15 min = 15 x 60 = 900s
h = 40 m ; n= 30%
As the density of water = p = 103 kg m-3
Mass of water pumped, m = volume x density = 30 x 103 kg
Actual power consumed or output power p0 = W/t = mgh/t
=>p0=(30 x 103 x 9.8 x 40)/900=13070 watt
If pi is input power (required), then as
η=p0/pi=> pi=p0/η = 13070/(30/100)=43567 W =43.56 KW
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