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Pumped energy storage

  1. Jul 22, 2011 #1
    1. The problem statement, all variables and given/known data
    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.



    2. Relevant equations

    U=mgh

    P=U/T

    3. 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 thats not correct. whats wrong here
     
  2. jcsd
  3. Jul 22, 2011 #2

    gneill

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    Staff: Mentor

    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?
     
  4. Jul 22, 2011 #3
    so I'm confused, what would be the equation for U?
    would it be density*area*(height-depth)*g?
     
  5. Jul 22, 2011 #4

    gneill

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    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?
     
  6. Jul 22, 2011 #5
    my guess would be the maximum available power
     
  7. Jul 22, 2011 #6

    gneill

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    Staff: Mentor

    If that's the case, then it occurs when the reservoir is full. The rate of water provided will be:
    [tex] A \delta \rho [/tex]
    in kg/hr. The height of the drop will be H plus the height to the current surface level of the reservoir.
     
  8. Jul 22, 2011 #7
    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?
     
  9. Jul 22, 2011 #8
  10. Jul 22, 2011 #9

    gneill

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    Staff: Mentor

    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.
     
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