Question

A pump on the ground floor of a building can pump up water to fill a tank of volume in 49 min. If the tank is 40 m above the ground, and the efficiency of the pump is 50%, how much electric power (in kilowatt) is consumed by the pump?

Answer 1

$P_{\text{input}}=\frac{\rho V g h}{t\,\eta}=\frac{10^3\;V\times 9.8\times 40}{(49\times60)\times0.5}\approx0.267\,V\;\text{kW}$

Answer 2

Step 1. Define symbols and known data:

$$V\;(\text{tank volume}),\quad h=40\:\text{m},\quad t=49\:\text{min}=49\times60\:\text{s},\quad \eta=0.50,\quad \rho=10^3\:\mathrm{kg/m^3}.$$

Step 2. Mass of water:

$$m=\rho\,V=10^3\,V\;\mathrm{kg}.$$

Step 3. Useful work done (gravitational potential energy gain per second):

$$P_{\text{out}}=\frac{mgh}{t}=\frac{10^3\,V\times9.8\times40}{49\times60}\;\mathrm{W}.$$

Step 4. Account for efficiency to find electrical (input) power:

$$P_{\text{in}}=\frac{P_{\text{out}}}{\eta} =\frac{10^3\,V\times9.8\times40}{49\times60\times0.5}\;\mathrm{W} \approx0.267\,V\;\text{kW}.$$

Therefore, the electric power consumed is $\displaystyle0.267\,V$ kilowatts, where $V$ is the tank volume in cubic metres.