Question

A pump on
the ground floor of a building can pump up water to fill a tank of volume 30m'in 15 min. If the tank is 40m
above the ground and efficiency of the pump is 30%. How much power is consumed by the pump. (Density of water
10'kgm”. g=9.8ms

Answer 1

Answer:

Power consumed by pump is 43.6 kW

Explanation:

Volume of the tank $V=30 m^3$

Time taken to fill the tank $t = 15 min = 15\times 60 = 900 s$

Height of the tank $h = 40 m$

Efficiency of the pump, η = 30%

Density of water$\rho = 103 kg/m^3$

Mass of water is ,$m = (Density \times Volume) = \rho \times V= 30\times 1000 =30000 kg$

Output power can be obtained as:  

$Power = \frac{Work done }{Time}$=$\frac{mgh}{t}$ (here work done is change in potential energy)

$P=\frac{30000 \times 9.8\times 40 }{900} \\P=13.067 \times 10^3W =13.067 kW$

Power out put is 13.067 kW

For input power  efficiency η is given by the relation:

$\eta =\frac{P out}{Pin }\\\\Pin=\frac{13607}{0.3} \\\\Pin=43.6 kW$

Power consumed by pump is 43.6 kW

Answer 2

$\huge\bold{\mathtt{\purple{A{\pink{ N{\green{S{\blue{W{\red{E{\orange{R }}}}}}}}}}}}}$...

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

Hope it's helpful ✌️✌️

$\huge\color{red}\boxed{\colorbox{black}{please\:mark\:me\: brilliance}}$

$\huge\underline\purple{\underline {Follow\:Me}}$