Generating Power from a Small Hydro Plant on a 10m High Waterfall

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A small hydro plant is being developed near a 10m high waterfall with a river flow of 1.5m/s, 4m wide, and 1m deep. The calculations show that the volume of water passing over the waterfall is 6m³/s, translating to a mass of 6000 kg. The total energy at the top of the waterfall is calculated to be 594,750 J/s, combining both potential and kinetic energy. Assuming a 25% efficiency in energy conversion, the power generated would be approximately 148,687.5 W. The discussion also touches on concerns about the assumption of a conservative system due to friction in flowing water.
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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
Etop = 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:
Ebottom = Etop
Ebottom = 594 750 J/s

% efficiency is 25%, so:
25% of 594 750 = 148687.5 J/s = 148687.5 W = POWER
 
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Looks right, though you might get some points off for having too many significant figures in your answer.
 
This is off topic, but how do I delete a post?
 
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