# Hydroelectric power plant

1. Oct 18, 2016

### BrainMan

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

What I attempted to do is

.80(1/2)mv2 = Ee

Where v is the velocity of the falling water, m is the mass of the falling water, and Ee is the electrical power generated by the electric plant.

So then I solved for m since you are given the Ee and the velocity and I got 3.86 x 10e5 kg/s.

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2. Oct 18, 2016

### kuruman

Your number is correct. To be strictly correct you should use dm/dt in your equation instead of m.

3. Oct 18, 2016

### haruspex

But then it would be dE/dt on the other side, assuming E stands for energy, not power.
The OP equation can be justified by prefixing it with "in the time in which a mass m of water passes through".

4. Oct 19, 2016

### kuruman

If E stands for energy, .80(1/2)(dm/dt)v2 = dE/dt is correct.
If Ee stands for power (as indicated by OP in attempt at solution), .80(1/2)(dm/dt)v2 = Ee is correct.