A pump raises 120 kg of water through a height of 6.0 m in 30 s. The input power to the pump is 400 W. What is its efficiency? (g = 9.81 m s−2)
- Useful work done = mgΔh = 120 × 9.81 × 6.0 = 7063 J
- Useful output power = W ÷ t = 7063 ÷ 30 = 235 W
- Efficiency = useful output power ÷ input power = 235 ÷ 400 = 0.59
Answer: Efficiency = 0.59, which is 59%