Efficiency in hydroelectric power: output power over gravitational potential energy input

Steps:

• Calculate input power from potential energy rate: P_in = ṁ g h, with ṁ = 1.5 × 10^6 kg/s, g = 9.8 m/s², h = 120 m.
• Compute g h = 9.8 × 120 = 1,176 m²/s².
• Then P_in = 1.5 × 10^6 × 1,176 = 1.764 × 10^9 W = 1,764 MW.
• Efficiency η = (P_out / P_in) × 100% = (100 / 1,764) × 100% ≈ 57%.

Why C is correct:

• 57% matches η = P_out / (ṁ g h) × 100%, the standard formula for hydroelectric efficiency.

Why the others are wrong:

• A: 5% ignores the decimal place in the ratio (actual ~5.7%, but rounded up).
• B: 43% from incorrect g ≈ 7 m/s² or miscomputed ṁ h product.
• D: 77% from inverting the ratio or using wrong height (e.g., 200 m).

Final answer: C