Finding the amount of water to charge a 50Ah car battery with a hydroelectric gen.

  • Thread starter Kevin2341
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Homework Statement


Assuming 100% efficient energy conversion, how much water
stored behind a hydroelectric plant with a head of 30m would
be required to charge the battery?

Homework Equations


?
MAYBE:
w=1000QgH j\s, where Q is the discharge rate @ m^3\s, 1000Q kg\s, H = net water head in m (30m), g is 9.8 m\s^2

The Attempt at a Solution


I don't even know where to begin with this question. My last question was a simple unit conversation question about how much energy was stored in a 12v 50Ah rated car battery, which I got 2,160,000 joules for my answer. I don't even know how to begin to tackle this question because the formula I gave was one I found on the internet that looked like it might apply.

What I'm thinking with that question however, is that perhaps I can use my figure from the total energy of the battery equaling:
2,160,000W = 1000*Q*9.8m\s^2*30m j (I borrowed the seconds unit from the j\s to turn my 2160000 joules into watts on the left hand side.)

2,160kW = 294Q kg\s m^2\s^2 j
7346.938 = Q kg m^2/s^3 j

The numbers look way off to me, and also the units have some serious funkiness going on with them which leads me to think this method isn't correct.

Any help?
 

Answers and Replies

  • #2
rude man
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1. How much energy is required to charge a dead 50Ahr battery (assumedly 12V)?
2. Potential energy of 1 kg of water 30m high is how many Joules? So how many of those Joules = 12V*50A*3600sec?
 

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