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.

 

[Moetasim]

I would approach this problem by first understanding the definitions of work and power in the context of a river. Work is defined as the amount of energy transferred when a force is applied over a distance, while power is the rate at which work is done or energy is transferred. In this case, the work being done by the river is the gravitational force acting on the water as it flows down the dam, and the power is the rate at which this work is being converted into electrical energy.

To solve this problem, we can use the equation P = W/t, where P is power, W is work, and t is time. We know that the power output is 2000 MW, so we can plug this into the equation as P = 2000 MW. We also know that the work being done on the water is equal to the change in kinetic energy, which can be calculated using the equation W = 1/2mv^2, where m is the mass of the water and v is its velocity.

We can now set up an equation using the given information and solve for the unknown variables. First, we need to calculate the velocity of the water as it flows down the dam. We can use the equation v^2 = 2gh, where g is the acceleration due to gravity (9.8 m/s^2) and h is the height of the dam (170 m). This gives us a velocity of approximately 57.6 m/s.

Next, we can calculate the mass of the water using the given information that each cubic meter of water has a mass of 1000 kg. The volume of water flowing per second can be calculated by multiplying the length of the dam by its width (1270 m x 1 m) and its height (170 m), giving us a volume of 216,230 cubic meters. Multiplying this by the mass of water per cubic meter gives us a total mass of approximately 216,230,000 kg.

Now, we can plug in these values into our equation P = W/t and solve for time. We know that the work being done is equal to the change in kinetic energy, which can be calculated using the equation W = 1/2 mv^2. We also know that 92% of this work is being converted into electrical energy, so we can multiply our calculated work by 0.92 to get the actual work being converted into electrical energy.

Finally,