Question

A overhead tank of capacity 1000 liter has to be filled in 1 /2 hour using water pump. Tank is kept at a height 10 m above ground and water level is 10 m below ground. The opening of inlet pipe inside tank is 1.11cm2 . Assuming the efficiency of motor to be 60%, the electric power used is (Neglect viscosity)

Answer 1

Answer:

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Answer 2

Answer:

The electric power used is 196.76W

Explanation:

Given the capacity of water tank, $V = 1000 l=1m^{3}$

Time taken to fill water tank, $t = 1 /2 hour =30min=30\times 60s$

Height of tank above the ground, $h_1 = 10 m$

Height of water level, $h_2= 10 m$

Cross sectional area of the inlet pipe inside tank, $a= 1.11cm^2=1.11\times10^{-4}m^2$

The rate of flow of water in filling the tank,

$q=\dfrac{V}{t} =\dfrac{1}{30\times 60}=\dfrac{1 }{1800}m^{3} /s$                                         ...(1)

Volume rate of water is also given by the product of velocity of flow of water and cross-sectional area of the pipe.

i.e., $q=Av=1.11\times v$

Comparing this with equation (1),

$1.11\times10^{-4}\times v=\dfrac{1 }{1800}$

$\implies v=\dfrac{1 }{1800\times1.11\times10^{-4}}=\dfrac{10^{2} }{18\times1.11}\approx 5m/s$

Water is pumped from the bottom of the ground to certain height above the ground.

Therefore, work done is sum of the potential energy and kinetic energy

$W=mgh+\dfrac{1}{2} mv^{2} =m\bigg(g(h_1+h_2)+\dfrac{1}{2} v^{2}\bigg)$                            ...(2)

Mass of the water =  density of the water x volume of the water

$=10^{3} \times 1=10^{3}g$

Substituting all the values in equation (2),

$W=10^{3} \bigg(10(10+10)+\dfrac{1}{2} (5)^{2}\bigg)$

    $=10^{3} \big(200+12.5\big)= 212.5\times 10^{3}J$

Power of the pump is given by the ratio of work done to time. i.e.,

$P_o=\dfrac{W}{t}= \dfrac{212.5\times 10^{3}}{1800}\approx 118.06W$

Given efficiency of motor,

$\eta= 60\%=\dfrac{60}{100}$

Efficiency is defined as the ratio of the output to input. Therefore,

$\eta=\dfrac{P_o}{P_i}= \dfrac{60}{100}$

$\implies \dfrac{118.06}{P_i}= \dfrac{6}{10}$

$\implies {P_i}= \dfrac{118.06\times 10}{6}=196.76W$

Therefore, the electric power used is 196.76W

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