An electric pump on the ground floor of a building takes 10 minutes
to fill a tank of volume 30 mwith water. If the tank is 60 m above the
ground and the efficiency of the pump is 30% how much electric power is
consumed by the pump in filling the tank? (g = 10 ms 21
)
Answers
Answer:
Power of motor, P=
η
1
t
ρVgh
Where,
ρ=density, V=volume, h=heigth, t=time and η=efficiency of motor
P=
0.3
1
10×60
1000×30×10×60
=100 kW
Power is 100kW
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Answer:-
4.0
22
Answer:
4.05 kWh
Given that, A pump can fill a tank of 27 cm³ which is at a height of 40 m in 20 minutes. Also, The efficiency of the pump is 40%.
And, We have to find the electric power consumed by the pump in that process.
First, Let's find the work done which equal to change in potential energy ( because the pump has to work against gravity to pull water of 27m³ upto an height of 40 m )
⇒ Work Done = P E
⇒ Work Done = mgh
Because we don't know the mass of tank when it is completely filled with water, Let's apply the following formula here,
⇒ Density = Mass / Volume
Density of water = 1000 kg/m³
⇒ 1000 = Mass / 27 cm³
converting volume into m³,
⇒ 1000 = Mass / 2.7 × 10⁻⁵
⇒ Mass = 270 × 10⁻⁵ kg
Hence,
⇒ Work done = 270 × 10⁻⁵ × 10 × 40
g = 10 m/s², h = 40 m
Divide it by 1000 to convert it into kiloJoule (kJ)
⇒ Work Done = 27 × 4 × 10³ × 10⁻⁵
⇒ Work Done = 108 × 10⁻²
⇒ Work done = 1.08 kJ
Now, we know the time taken to pull 27000 kg of water upto an height of 40 m to be 20 minutes, Converting it into hour we get 2/3 hour
As we know,
⇒ Power = Work Done / time taken
⇒ Power = 1.08 / 2/3
⇒ Power = 1.62 kWh
Given the efficiency of the water pump is 40%,
⇒ Efficiency = Output / Input
(In terms of power)
⇒ 40/100 = 1.62 / Input
⇒ 0.4 × Input = 1.62
⇒ Input = 4.05 kWh
Hence, The electric power absorbed by the water pump is 4.05 kWh.
__________________________
100% correct answer
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