Dismiss Notice
Join Physics Forums Today!
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

  1. Sep 27, 2004 #1
    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 dont know. But any help would be great. (it is due tomorrow)

    Thanks
     
    Last edited by a moderator: Apr 21, 2017
  2. jcsd
  3. Sep 28, 2004 #2

    Clausius2

    User Avatar
    Science Advisor
    Gold Member

    The link does not work.

    Anyway, the power obtained in the turbine is:

    [tex]W=Q*\Delta P_o[/tex]; 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: [tex]P_{oentrance}-P_{oexit}=P_a+\rho g H[/tex]-P_a[/tex]

    So that: [tex]W=\rho g H Q[/tex] where the units are:

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

    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.
     
    Last edited by a moderator: Apr 21, 2017
  4. Sep 28, 2004 #3

    krab

    User Avatar
    Science Advisor

    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.
     
  5. Sep 28, 2004 #4
    here is the drawing
     

    Attached Files:

    • phy.jpg
      phy.jpg
      File size:
      6 KB
      Views:
      115
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook