Is this the Maximum Power Generated from a Small Hydro Plant?

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Homework Statement

A small hydro plant is being built. A nearby river is 4m wide and 1m deep. The water flow is 1.5m/s over the brink of a 10m high waterfall. The turbine company advertises 25% efficiency in converting the potential energy of the water into electric energy. What's the power you can generate?

Homework Equations

Ug = mgh
K= 0.5mv²

The Attempt at a Solution

Since V = 1.5 m/s, I found the volume of water passing over the brink of the waterfall, every second:
Volume/ sec = LWH = 1.5(4)(1) = 6m³/ sec

Convert to mass:
1m³ = 1000 kg
6m³ = 6000 kg

Since I did the beginning calculations for "every second", I wrote the rest of the values as "per second" too

At the top of the waterfall, there is grav.potential and kinetic energy:
Ug = mgh = 6000(9.8)(10) = 588 000 J / s
AND
K = 0.5mv² = 0.5(6000)(1.5)² = 6750 J/s
SO
E_top = 6750 + 588 000 = 594 750 J/s


I'm not sure if it is right to assume that this system is conservative because there's a lot of friction in flowing water, but I did anyway...
At the bottom of the waterfall, there is kinetic energy. Assuming a conservative system:
E_bottom = E_top
E_bottom = 594 750 J/s

% efficiency is 25%, so:
25% of 594 750 = 148687.5 J/s = 148687.5 W = POWER

 

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I just need someone to check the answers because I don't know if they're right.