How Much Water Must Flow Over the Grand Coulee Dam to Generate 2000 MW of Power?

[iamkristing]

[SOLVED] work and power

Homework Statement

the Grand Coulee Dam on the Columbia River is 1270 m long and 170 m high. The electrical power output from the generators at its base is approximately 2000 MW, How many cubic meters of water much flow from the top of the dam per second to produce this amount of power 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 1000kg)

Homework Equations

W=kf-ki=1/2mv^2
W=F*deltax*costheta
P=W/deltaT
F=m*a

The Attempt at a Solution

I attempted solving this problem using unknowns since I did not know another way to approach it. so I said

F=m(9.8)
W=m(9.8)(170 m) (costheta)
then I plugged Work into the Power equation:

P=[(m(9.8)(170 m) (costheta))/deltaT](.92)

and I multiplied the whole equation by .92 for the 92% in the problem

Now it seems that the only known in the problem is Power. I could solve for mass using the unknowns of time and theta, but I feel like that is the wrong approach and am not sure how to solve otherwise.

Thanks!

 

[lavalamp]

You seem to be making it a lot more difficult than it is.

You know that you get 2000 MW of electrical power, but that the generators are only 92% efficient. So firstly you need to divide that power figure by 0.92 and multiply by 1000000 to get the power of the water in W.

Next up, calculate the change in gravitational potential energy for 1 m^3 of water falling 170m (the height of the damn).

Then it's just division to get the number of m^3 / second of water required to generate that power.