Question

A pump on the ground floor of a building can pump water to fill a tank of volume 30 m3 in 15 minutes. If the tank is 40 m above the ground and the efficiency of the pump is 30%, how munch electric power is consumed by the pump?

Answer 1

Hello mate ,

Your answer:-

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GIVEN-

Volume of the tank, V = 30 m^3

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 = 10^3 kg/m^3

Mass of water, m = ρV = 30 × 10^3 kg

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Output power can be obtained as:

P (output) = work done/ time

= mgh / t

= 30× 10^39 × 9.8 × 40 / 900

= 13.067× 10^3 W

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For input power P( input) ,efficiency, η is given by the relation:

η = P (output)/ P ( input)= 30%

=> P(input) = 13.067× 10^3× 100/ 30

=> P(input)= 0.436× 10^5 W= 43.6 kW

SO ELECTRIC POWER CONSUMED BY ELECTRIC MOTOR IS 43.6 kW.

I hope, this will help you.☺

Thank you______❤

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Answer 2

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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.

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100% correct answer

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