Determining Power Output of Pumped Energy Storage System

[kraigandrews]

Homework Statement

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 ×10^5 m2; D = 14 m; H = 102 m; δ = 0.55 m per hour; density = 1.0 ×10^3 kg/m3.

Homework Equations

U=mgh

P=U/T

The Attempt at a Solution

U=mgh
U=(density*area*depth)*g*h =1.1206E13 J

then 1.1206EJ/((1/3600)*.55)=7.3355E16 W

but that's not correct. what's wrong here

 

[gneill]

Only the water at the bottom of the reservoir is at height H above the turbine. The water at the surface starts out D above that.

Since the head is decreasing over time, the power generation won't be constant over time; I suppose they're just interested in the initial value? Or perhaps an average?

 

[kraigandrews]

so I'm confused, what would be the equation for U?
would it be density*area*(height-depth)*g?

 

[gneill]

so I'm confused, what would be the equation for U?
would it be density*area*(height-depth)*g?

I have to admit that I'm not certain at this point. Much depends upon what they mean by "available power". There will be more power provided when the reservoir is full as the pressure at the turbine will be higher. Unless, of course, there's some mechanism that's enforcing the .55 m/hr rate and the head height for the water running to the turbine is a constant H.

Would you happen to know what answer they're looking for?

 

[kraigandrews]

my guess would be the maximum available power

 

[gneill]

my guess would be the maximum available power

If that's the case, then it occurs when the reservoir is full. The rate of water provided will be:
$$A \delta \rho$$
in kg/hr. The height of the drop will be H plus the height to the current surface level of the reservoir.

 

[bacon]

Regarding part (a), your basic equation U=mgh is correct. But one must use integration for the reason gneill gave, the water varies from height H to H + D. Do you know calculus?

 

[kraigandrews]

yes.

 

[gneill]

Unless you want averages over the operating limits of the reservoir, I think you can make do without calculus for this one.

You can determine the mass delivery rate of the water and the height of the water head for a given instant. That should be sufficient to provide the "instantaneous" power output at that instant in time.