# Work, force, power turbine pump problem

1. Mar 11, 2009

### joseg707

1. The problem statement, all variables and given/known data
Some electric-power companies use water to store energy. Water is pumped by reversible turbine pumps from a low to high reservoir. To store the energy produced in 1.0 hour by a 120*106 electric-power plant, how many cubic meters of water will have to be pumped from the lower to the upper reservoir? Assume the upper reservoir is 520 m above the lower and we can neglect the small change in depths within each. Water has a mass of 1000 kg for every 1 m3

2. Relevant equations
P=W/t
W=Fd
F=ma

3. The attempt at a solution
I multiplied power by the total number of seconds in an hour to get work. I then divided the work by the distance of 520 m and I get force. This is where I'm stuck at. I have that force is equal to (120000000 kg*m2/s3)*(3600s)/(520m). I don't know how to solve for mass because it seems that the flow has a constant velocity and there is no acceleration. Can someone please point me in the right direction?

2. Mar 11, 2009

### Staff: Mentor

One has a 120 MW plant. One has to find how much water must be stored at 520 m, to provide the energy produced in one hour.

Energy = Power * time.

The mechanical energy in the turbine comes from the kinetic energy in the water flow, which is transformed from the gravitational potential energy of the water stored 520 m above the turbine. What is the GPE of the water at 520 m.

Once one finds the mass, the volume is easy to find given the density.

3. Mar 11, 2009

### joseg707

Are you saying that the kinetic energy is equal to the output energy of the turbine, 120MW, and the the potential energy is equal to the kinetic energy?
KE=120 MW

PE(2)=KE(2)
mgh=120W
m(9.8)(520)=120MW?

I don't feel like I understood that right.

4. Mar 11, 2009

### Staff: Mentor

Power = 120 MW

Energy = Power * time. How many J of energy is produced by 120 MW in one hour?

5. Mar 11, 2009

### joseg707

Oh! Haha yeah, I completely jumped right over that didn't I.
(120e6W/s)(3600s)=work
work=4.32e11J

6. Mar 11, 2009

### joseg707

So does this hold true?

m(9.8)(520)=4.32e11
m=84772370.49 kg*1m3/1000kg=84772 m3