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Power Plant/Heat Problem

  1. May 30, 2007 #1
    1. The problem statement, all variables and given/known data
    An 800-MW electric power plant has an efficiency of 30%. It loses its waste heat in large cooling towers by evaporating a fine mist of water. If all waste heat went only into the evaporation of water, how many tons of water would be evaporated per day?


    2. Relevant equations

    W = |Qh|-|Qc|
    e = 1 - (|Qc|/|Qh|)
    Q = mL
    Lwater = 2.26x10^6 J/kg

    3. The attempt at a solution

    I can't seem to get any of the answers listed and I can't quite figure out what to do. Here's what I tried.

    W = 800MW
    e = .30
    Solving for |Qc| and |Qh| I got, |Qc| = 1866.69 MW, or 1866.69x10^6 J/s, |Qh| = 2666.67 MW
    Since the wasted heat going to the cold reservoir is what is evaporating the water I plugged |Qc|= 1866.69x10^6 J/s into the Q for the equation Q = mL, so |Qc| = mL. L is the latent heat of water. Solving for m I got m = 825.97 kg/s. Since it's per day, I multiply m by 86400seconds and get 71,363.72 tons of water per day.

    Would anybody like to take a crack at it and help me out? Thanks :)
     
  2. jcsd
  3. May 30, 2007 #2
    gsxrK3,

    Your solution seem correct to me.
    I got a waste heat of 161280 GJ/day.
    I like GigaJoules units because of my profession.
    For the same reason I remember that the heat of evaporation of water is about 2.25 GJ/ton.
    Therefore, your 71 ktons are ok.

    Note that the latent heat depends of the water and vapor temperature.
    But for the purpose here, it's fine.

    Quite a lot of water, isn't it?
     
    Last edited: May 30, 2007
  4. May 30, 2007 #3
    Yeah that is a ton of water, I mean a lot. :biggrin: Much thanks for the reply. May I ask what it is you do?
     
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