Power generated by water flowing over a dam

[mlb2358]

Homework Statement

The Grand Coulee Dam is 1270 m long and 170 m high. The electrical power output from generators at its base is approximately 2000 MW. How many cubic meters of water must flow
from the top of the dam per second to produce this amount ofpower if 92% of the work done on the water by gravity is converted to electrical energy? (Each cubic meter of water has a mass of 1000 kg.)

Homework Equations

The Attempt at a Solution

The main question I have regarding this problem is whether or not one can use P = FV to solve it. If so, is the force equivalent to the mass of the quantity of water that will produce 2000 MW times g? If so would that mean that the velocity is equivalent to the velocity of an object after it has undergone free-fall for 170m? I realize that the problem can be solved by using P = ΔW/Δt, but I am trying to solve it with P = FV so I will better understand the situation.
Thanks!

 

[CWatters]

Well you could assume the water was in free fall and work out the velocity but is it possible to work out the force F ? I suppose if you knew over what distance the turbine decelerates the water you might calculate the deceleration and hence calculate F from F=ma. However you don't so you would have to fall back on KE = 0.5mV².

Overall I think calculating power using P = ΔPE/T is the simplest approach

 

[Pkruse]

The simple answer to your question is that water flowing over a dam generates no power at all.

But to figure potential power generation, figure power flowing into the turbine from head height and mass flow rate going into the turbine. Then figure power of water flowing out by the flow rate and ambient pressure. Subtract the two and multiply by an assumes efficiency. If you know the actual power output, you can calculate the efficiency.