A water reservoir tank of capacity $250\;m^3$ is situated at a height of $20 \; m$ from the water level. What will be the power of an electric motor to be used to fill the tank in $3$ hours? Efficiency of motor is $70\;\%$

Given

Volume $(v) = 250\; m^3$

Height $(h) = 20 \; m$

Time $(t) = 3 \; hrs = 10800 \; sec$

Efficiency of motor $(\eta) = 70 \%$

Power of the electric motor $(P_{in}) = ?$

We have, $\eta = $$\frac{P_{in}}{P_{out}}$$*100\%$ .......... (i)

 Firstly we have to find ,

$P_{out} = $$\frac{Work\;done}{time}$ ......... (ii)

But,

Work done to fill the tank $W = mgh = \rho * v * g * h = 1000 * 250 * 10 * 20 = 5 * 10^7\;J $

 [Since, density of water in SI system is $1000\;kg/m^3$]

Then from equation (ii), we get

$P_{out} = $$\frac{5 \;* \;10^7}{10800}$ $ = 4.63 * 10^3\; Watt$

Finally from equation (i), we get

$P_{in} = $$\frac{P_{out}}{\eta}$$ = $$\frac{4.63\; * \;10^3}{0.7}$$ = 6614\;Watt$

∴ Power of the electric motor $(P_{in}) = 6614\;Watt$

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