Answer to Question #121367 in Physics for Salihu Alpha

Question #121367
Assuming 100% efficient energy conversion, how much water stored behind a 50 centimeters high hydroelectric dam would be required to charge the battery?
1
Expert's answer
2020-06-10T18:24:59-0400

The power of the dam is


P=ρghQ,P = ρghQ,

where QQ - flow of water in m3/s.

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


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

Equate:


ρghQ=qV3600 s,ρghQ=\frac{qV}{3600\text{ s}},

the flow of water will be


Q=qV3600ρgh=5012360010009.80.5=0.12 m33600 s,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=Q3600=0.12 m3.v=Q\cdot3600=0.12 \text{ m}^3.

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