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Evaporation moisture

  1. Sep 7, 2011 #1
    1. The problem statement, all variables and given/known data
    - if I put 0.3 litres/squaremeter of hot water (50 C) on ice rink (-5 C), how long will take to the water to be 0 grade C ? How much evaporation moisture will be in the air (5 C) and how long will be the time for evaporation ?
    2. Relevant equations
    Heat load= 1000*0.3(liters/sm)*(4.2*(50-0)+333+2*(0-(-5))
    Heat load=165.9 kJ
    - evaporation heat of water =2270(kJ/kg)
    so the moisture will be: 165.9/2270=0.07 kg


    3. The attempt at a solution
     
    Last edited: Sep 7, 2011
  2. jcsd
  3. Sep 9, 2011 #2

    rude man

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    I would start by assuming a thin layer of water above the ice, so that it's of uniform temperature but still with finite heat capacity. Then you have to use the conductivity of water and ice, latent heat of the latter, and specific heats of both, to determine the rate at which heat leaves the water layer. Not easy, since the ice is melting to some extent. (Eventually, all the water should wind up as ice even though the ambient temperature is +5C.)

    Also, you should proably take into account the loss of heat of the water to radiatiion : eσ(T^4 - Ta^4) where Ta is +5 deg C and e for water has to be looked up (emissivity). T will vary from +50 to 0C.

    As for rate of evaporation, the formula I found is
    (mass loss rate)/(unit area) = (vapor pressure - ambient partial pressure)*sqrt( (molecular weight)/(2*pi*R*T) ). Has to be integrated since T is not constant. The rersult should give you both the amount of evaporate and the time to evaporate all the water.

    This is a very complicated problem which may explain why you're not getting any responses from anybody else, and why mine is so flimsy. :-)

    http://van.physics.illinois.edu/qa/listing.php?id=1440
     
    Last edited: Sep 9, 2011
  4. Sep 14, 2011 #3
    Thank You very much
     
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