Question

A pump on the ground floor of a building can pump up water to fill a tank of volume $\text{30}{{\text{m}}^{\text{3}}}$ in 15 min. If the tank is 40m above the ground and the efficiency of the pump is 30%, how much electric power is consumed by the pump.
Density of water: $\text{1}{{\text{0}}^{\text{3}}}\text{kg}{{\text{m}}^{\text{-3}}}$

Answer 1

The efficiency of a motor is given as the ratio of output power to its input power. Power is basically defined as the amount of work done in a given interval of time. As the water pump takes the water to a particular height, the work done by it will be stored in the form of gravitational potential energy. Hence using the definition of power and efficiency of the pump, one can calculate the electric power (Input power)consumed by the pump.

Complete step-by-step solution:
To begin with, let us define the efficiency of the pump.
The efficiency of the pump mathematically is given by,
$\eta =\dfrac{\text{Output power}}{\text{Input power}}$ where output power is the rate of work done by the pump and the input power is the power drawn by the pump from an electric source which is equal to the electric power consumed.
The pump takes the water to a height of 40m from the ground. The mass of the water taken up to this height is,
$m=\rho V$where $\rho$ is the density of water and V is the volume of the water-filled in the tank
The density of water is given to be $\text{1}{{\text{0}}^{\text{3}}}\text{kg}{{\text{m}}^{\text{-3}}}$ and volume of water in the tank is $\text{30}{{\text{m}}^{\text{3}}}$. Hence the mass of water pumped by the pump into the tank is,
$\begin{aligned} & m=\rho V \\\ & m={{10}^{3}}\times 30=3\times {{10}^{4}}kg \\\ \end{aligned}$
The pump takes this mass of water to a height of 40 m from the ground. Hence the work done by the pump is,
$E=mgh$ where m is the mass of the water taken up, h is the height at which the water is released into the tank and g is the acceleration due to gravity.
Hence work done is,
$\begin{aligned} & E=mgh \\\ & E=3\times {{10}^{4}}\times 9.8\times 40 \\\ & E=1176\times {{10}^{4}}J \\\ \end{aligned}$
The power of the pump is basically the rate of work done. The pump delivers this amount of water in 15min. Hence the power of the pump is,
$P=\dfrac{E}{t}\text{where E is the work done by the pump and t is the time taken to do the work}\text{.}$
$P=\dfrac{1176\times {{10}^{4}}}{15\times 60}=13.066\times {{10}^{3}}\text{J per sec}$. This is the output power of the motor. Let us now use the definition of efficiency to calculate the power consumed by the pump.
The efficiency of the pump is given as 30%. Hence $\eta =0.3$.
Hence the power consumed that is the input power is given by,
$\begin{aligned} & \eta =\dfrac{\text{Output power}}{\text{Input power}} \\\ & \text{Input power}=\dfrac{\text{Output power}}{\eta } \\\ & \text{Input power}=\dfrac{13.06\times {{10}^{3}}}{0.3}=43.53\times {{10}^{3}}J \\\ \end{aligned}$ .

Note: The efficiency of the pump is not 100%. Hence some amount of energy is wasted in the form of heat generated in the coils of the pump. In the real-world, the efficiency of any electrical device cannot be equal to 100% as always some amount of energy is lost in the form of heat due to the heating effect of the electrical devices.