Calculating Max Power at Chute-Montmorency

[PVnRT81]

Homework Statement

At the Chute-Montmorency, water falls from a height of 83m. The average flow rate of the river is 35000 liters per second. Estimate the maximum power that a hydroelectric plant located at the bottom of the falls would provide.

Homework Equations

P = W/T

W = F x d

The Attempt at a Solution

W = (35000 x 9.8)(83)

To solve for time for water to fall:

d = (0.5)(a)(t^2)
83 = (0.5)(9.8)(t^2)
t = 4.12 seconds

Power = (35000 x 9.8 x 83) / (4.12 seconds)

Power = 6917002.427 Joules or 6917 KJ.

I am not entirely sure if this is right.

 

[rock.freak667]

They give you the volumetric flow rate as 35,000 L/s, what is the density of water and hence the mass flow rate of the water?

Now you are correct in that W=Fxd and F=mg to give you W=mgh (which is really saying that the energy is via gravitational potential energy).

Power = Work done/time

So P = W/t = (mgh)/t = (m/t)gh

and m/t is the flow rate you calculated above.

 

[PVnRT81]

They give you the volumetric flow rate as 35,000 L/s, what is the density of water and hence the mass flow rate of the water?

Now you are correct in that W=Fxd and F=mg to give you W=mgh (which is really saying that the energy is via gravitational potential energy).

Power = Work done/time

So P = W/t = (mgh)/t = (m/t)gh

and m/t is the flow rate you calculated above.

Yes, but wouldn't the time it takes for the water to get to the bottom make a difference in the power.

 

[rock.freak667]

Yes, but wouldn't the time it takes for the water to get to the bottom make a difference in the power.

Due to conservation of mass and energy, the flow rate at the top would be the flow rate at the bottom otherwise, the water would not be flowing through the hydroelectric plant.

And Power = Work done/time

the time in this case would be the duration in which the work is being done while the water is falling, no work is being done on the turbine within the hydroelectric plant so there will be no output of power from the plant.