Answer to Question #121367 in Physics for Salihu Alpha

Question

Assuming 100% efficient energy conversion, how much water stored behind a 50 centimeters high hydroelectric dam would be required to charge the battery?

Answer 1

The power of the dam is

$$P = ρghQ,$$

where $Q$ - flow of water in m³/s.

Assume that the battery has $V=12 \text{ V}, q=50\text{ A}\cdot\text{h}.$ Its power will be

$$P=\frac{qV}{t}.$$

Equate:

$$ρghQ=\frac{qV}{3600\text{ s}},$$

the flow of water will be

$$Q=\frac{qV}{3600\rho gh}=\frac{50\cdot 12}{3600\cdot1000\cdot9.8\cdot0.5}=\frac{0.12\text{ m}^3}{3600\text{ s}},$$

the total volume for 1 hour is

$$v=Q\cdot3600=0.12 \text{ m}^3.$$