Pumped Energy Storage and Available Power

[derekbeau]

http://s1.lite.msu.edu/res/msu/stump/Energy101/pumpedstorage.gif

Consider the Ludington Pumped Storage Plant. Water may flow out of the reservoir, through the penstocks, and down into Lake Michigan, at a volume rate of 1710 m3/s. The water is directed through turbines to turn electric generators to make electric power. The height of the water in the reservoir is h = 100 meters above the surface of Lake Michigan. How much power ( = energy per unit time) would be available if the Plant is operating with these parameters?

Tip: In one second the height of the surface drops by Deltah where ρ×A×∆k=μ×(1 sec). (ρ is the density of water, 1.0×103 kg/m3.) How much has the gravitational potential energy changed?

Ok So what i have done so far is convert the volum to mass

1710 m^3/s * 1000 kg/m^3 = 1710000 kg/s

Then I thought id use the formula

P = Mgh
P = 1710000 * 9.81 * 100

and I got 1677510000 (watts?)

Well that wasnt the right answer, so i tried converting to kilowatt-hours

1677510000 W / 3600000 J = 465.975 kWh

Still not right.

I think my problem is because either the height of the water is changing, or i am using an incorrect height. I don't know. But any help would be great. (it is due tomorrow)

Thanks

 

[Clausius2]

http://s1.lite.msu.edu/res/msu/stump/Energy101/pumpedstorage.gif


P = Mgh
P = 1710000 * 9.81 * 100

and I got 1677510000 (watts?)

Well that wasnt the right answer, so i tried converting to kilowatt-hours

1677510000 W / 3600000 J = 465.975 kWh

Still not right.

I think my problem is because either the height of the water is changing, or i am using an incorrect height. I don't know. But any help would be great. (it is due tomorrow)

Thanks

The link does not work.

Anyway, the power obtained in the turbine is:

$$W=Q*\Delta P_o$$; where [Q]=[m^3/s] is the volumetric flow; [P_o]=[Pa] is the total pressure in both sides of the turbine.

In your case: $$P_{oentrance}-P_{oexit}=P_a+\rho g H$$-P_a[/tex]

So that: $$W=\rho g H Q$$ where the units are:

$$[W]=\frac{Kg}{m^3} * \frac{m}{s^2}* m *\frac{m^3}{s}=\frac{J}{s}=Watt$$

You said it doesn't work. Although the height of the reservoir is changing, in my opinion that change is cuasi-steady, so that the turbine power also changes with time W=W(H(t)). I need to view the drawing of your link to answer you better.

 

[krab]

A watt is a joule per second, which is a Newton metre per second, which is a kilogram metre per second squared metre per second, which is a kg m^2/s^3. Now look at your formula: you multiplied kg/s times m/s^2 times m. That's a watt alright.

The second try is totally wrong. A kwh is a power times a time, so is a unit of energy, not power.

 

[derekbeau]

here is the drawing