A pump on the ground floor of a building can pump up water to fill a tank of volume 30 $m^3$ in 15 min. If the tank is 40 $m$ above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump ?

Question

A body of mass $4$ kg is placed at a point $P$ having coordinates $ (3,4) $ m. Under the action of force $ \mathbf{F} = (2\hat{i} + 3\hat{j}) $ N, it moves to a new point $Q$ having coordinates $ (6,10) $ m in $4$ sec. The average power and instantaneous power at the end of $4$ sec are in the ratio:

Answer 1

Given Data

Parameter	Value
Tank volume	30 m³
Time	15 min = 900 s
Height	40 m
Efficiency ($\eta$)	30% = 0.30
Water density ($\rho$)	1000 kg/m³
$g$	9.8 m/s²

Step 1: Useful Power Output

Power = energy per unit time = rate of PE gain

$$\text{Mass flow rate} = \frac{\rho V}{t} = \frac{1000 \times 30}{900} = 33.33 \, \text{kg/s}$$

$$P_\text{useful} = \dot{m} g h = 33.33 \times 9.8 \times 40 = 13{,}064 \, \text{W}$$

Step 2: Electrical Power Input

Pump efficiency: $\eta = \frac{P_\text{useful}}{P_\text{input}}$

$$P_\text{input} = \frac{P_\text{useful}}{\eta} = \frac{13{,}064}{0.30} = 43{,}547 \, \text{W} \approx 43.5 \, \text{kW}$$

Electric Power Consumed

$P = \textbf{43.5 kW}$

Energy Breakdown

Type	Power	% of Input
Useful (PE gain)	13.1 kW	30%
Heat losses	30.4 kW	70%
Total Input	43.5 kW	100%

Verification

Volume flow: $Q = 30/900 = 0.0333$ m³/s
$P_\text{useful} = \rho g Q h = 1000 \times 9.8 \times 0.0333 \times 40 = 13{,}064$ W ✓
Reasonable for industrial pump

A pump on the ground floor of a building can pump up water to fill a tank of volume 30 \(m^3\) in 15 min. If the tank is 40 \(m\) above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump ?

Solution and Explanation