# Energy and power of water reservoir

1. Nov 7, 2013

### oddjobmj

1. The problem statement, all variables and given/known data
Pumped Energy Storage.

A water reservoir has surface area A and depth D. The water flows down pipes and through turbines to generate electric power. The bottom of the reservoir is at height H above the turbines. The depth D of water in the reservoir decreases at a rate δ. (a) Calculate the total gravitational potential energy U. (b) Calculate the available power P = |dU/dt|, i.e., available for conversion to electric power.

DATA: A = 8 ×105 m2; D = 15 m; H = 108 m; δ = 0.5 m per hour; density = 1.0 ×103 kg/m3.

2. Relevant equations

U=mgh
Power=Energy/time

3. The attempt at a solution

To find the power I want to integrate mgh with respect to h from the initial 108 m to the total 123 m.

U=$\frac{3465gm}{2}$

The mass of the tank is A*D*Density=1.2*107 kg and g=9.8

U=2.03742*1011 J

As far as power goes can I not simply divide the total potential by the total time it takes to empty the reservoir? If I do that I get 1.89*106 J/s.

Alternatively if I integrate the function for U divided by t from t=0->108000 seconds (30 hours)

$\U/t$=$\int2.03742*10^{11}/t$=2.03742*1011Log[108000]=2.36135*1012 J/s

Those are wildly different answers and neither of them in combination with the first is correct. Both parts have to be correct for me to check my answers.

Where am I messing up? Thank you!

2. Nov 7, 2013

### haruspex

What dimensions would integrating mgh wrt h produce? Are they the right dimensions for power?
You correctly quoted Power=Energy/time, but don't seem to have used time. Shouldn't δ figure in the calculation?
It will make it a lot easier to follow what you are doing if you work entirely symbolically (δ, A etc.), only plugging in numbers as the final step. It will also help you find mistakes in your working.

3. Nov 7, 2013

### oddjobmj

It would produce Joules. Power would then by that divided by the time. I mis-typed when I typed power there. I meant potential energy.

We know that the depth of the pool is D=15 m. We also know we lose δ=.5 m per hour. So, it takes 30 hours for the reservoir to empty. 30 hours=108,000 seconds. Symbolically:

t=$\frac{D}{δ}$=$\frac{15}{.5}$=30 hours=108,000 seconds

My final string of equations was sort of jacked up, sorry. I meant to take the integral of U/t which you can probably see if you search for it but it is not very clear.

U=$\int$mgh, h=108->123

U=$\frac{3465gm}{2}$ Joules or $\frac{gh^2m}{2}$ if you do the indefinite integral.

Power=$\frac{U}{t}$=$\frac{3465gm}{2t}$=1.89*106 J/s

This pair of answers is incorrect.

4. Nov 7, 2013

### haruspex

No, integrating mgh wrt h would give mass*acceleration*distance*distance = energy*distance.
Please show the details of your working there. Use symbols, not numbers.
But there is a much easier way to find the total energy. You can just use the average height of the water above the turbine.
For the second part, note that the power will not be constant. I think they want P expressed as a function of t.

5. Nov 7, 2013

### oddjobmj

Oh my, yes it would. Thank you

I actually considered the shortcut using the average height but this was before I realized the tank was held high off the ground. Great point, thanks!

Yeah, the power bit is not quite clear. It does have to be a number, not a function. Perhaps the average power would work.

U=mghavg

Power=$\frac{U}{t}$=$\frac{mgh_{avg}}{t}$

That's as far as I can go symbolically.

For U I get 1.358*1010 J

For power I get 125800 J/s

6. Nov 7, 2013

### haruspex

Strange - they write dU/dt, suggesting a derivative, not ΔU/Δt, which would indicate an average.
I get the same digits, but a different power of 10. Did you factor in the density?

7. Nov 7, 2013

### oddjobmj

Ah, for some reason I remembered the density being 1 kg/m^3. Regardless, 1.358*10^13 J and the respective average power is also not correct. 1.257*10^8 J/s

Taking the absolute value of the derivative of U/t I get:

Power=$\frac{ghm}{t^2}$

The units in this case would be odd, no?

Also, I do know that the units have to be J/s because I've tried a number of incorrect answers and it will tell you if the units are correct or incorrect. It is clear that the power will change with time and I can't enter an answer that is proportional to a variable so it has to be the average or total available power.

8. Nov 7, 2013

### haruspex

It says 1.358*10^13 J is wrong? I cannot explain that. In practice, the water would have some residual KE after exiting the turbine, but since no aperture is specified you cannot take that into account.
No, that doesn't work. That would give you how quickly the average power would decrease if the tank took longer to empty.
To get the power at time t, consider what the head is (H+D(t)) at time t, how much water runs out in time dt, and what the loss of PE is in time dt.
The question seems to be broken in a couple of ways.

9. Nov 8, 2013

### oddjobmj

The energy was correct but it would not verify that until both were correct.

As you pointed out I had to take a sliver of the mass. (total mass/total time)*g*h

10. Nov 8, 2013

### haruspex

But.. that gives you P(t) = δρgA(h+D-tδ). How did you arrive at a single number that it would accept? If it's the average power, why isn't the answer 1.257*10^8 J/s?

11. Nov 8, 2013

### oddjobmj

Great question... I almost ran out of tries because of that.

It is the available power over the time period of a second which is the sliver of mass that passes each second times g and its height. This was a strange problem.

12. Nov 8, 2013

### haruspex

Yes, but that height changes... so what number did you plug in?

13. Nov 9, 2013

### oddjobmj

Confusing, right? I used 123 which was the maximum height. Why that and not at any time but t=1 I don't know.

14. Nov 9, 2013

### haruspex

OK, so you found the maximum power. If the original question doesn't specify that then I would have guessed it wanted either average or minimum. Poor question.