http://s1.lite.msu.edu/res/msu/stump/Energy101/pumpedstorage.gif(adsbygoogle = window.adsbygoogle || []).push({});

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 got1677510000 (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 dont know. But any help would be great. (it is due tomorrow)

Thanks

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Pumped Energy Storage and Available Power

**Physics Forums | Science Articles, Homework Help, Discussion**