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Electrical Energy

  1. Feb 19, 2007 #1
    1. The problem statement, all variables and given/known data
    A hydroelectric dam holds back a lake of surface area 3.00×106 m^2 that has vertical sides below the water level. The water level in the lake is a height 130 m above the base of the dam. When the water passes through turbines at the base of the dam, its mechanical energy is converted into electrical energy with 93.0% efficiency.

    If gravitational potential energy is taken to be zero at the base of the dam, how much energy is stored in the top meter of the water in the lake? The density of water is = 1000 kg/m^3.

    What volume of water must pass through the dam to produce an amount of electrical energy totalling 1000 kilowatt-hours of electrical energy?

    What distance does the level of water in the lake fall when this much water passes through the dam?

    2. Relevant equations
    This is what I'm thinking but I'm not sure: deltaK + deltaU + delta(other energy) = 0


    3. The attempt at a solution
    I actually don't understand where to go with this problem. So far, I've used the surface area x height to get the volume and then divided by the density to get the mass. I don't know why I did that, it's probably not even relevant.

    Then I started with my book with problems that look a bit similar and then I tried to integrate it with the integral and the resulting eqn I got is:

    ((h_2)^2 - (h_1)^2)/2 * dAg

    d = density
    A = area
    g = gravity

    Plugging in everything, I got the wrong answer obviously.

    I know I'm completely lost on this problem, like I don't even understand the whole picture or what is electrical energy. Can someone pls shed some light for me?
     
  2. jcsd
  3. Feb 19, 2007 #2
    Youre given the surface area of the lake and asked to find the energy in the top meter of the lake. Multiply to find the volume and multiply volume with density to find the mass of the water in the first meter. This much mass held at a height of 130 meters contains mgh potential energy.

    Now, at 93% efficiency, 93% of the potential energy of the water is converted into electrical energy. Therefore, 0.93*mgh=1000KW*3600 (for kw-hour). This will give you the mass of the water required PER hour and from there you can find the volume required (you're given the density).

    You have the volume of water required and you have the surface area, so you can find the change in depth.
     
  4. Feb 19, 2007 #3
    The system keeps telling me it's wrong. I took the surface area x height = volume. Then I took volume x density = mass. I got 3.9x10^11. I used the mgh eqn but I have the wrong answer. Isn't there something else I'm missing...?

    Edit: Crap. >< I totally forgot to read the first meter. *hits head* I got it now I got it now. Many thanks!
     
    Last edited: Feb 19, 2007
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