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Heat exchange in active system

  1. Jul 3, 2012 #1
    I am trying to calculate the amount of water that would have to pass through a 200 litre tank (water inlet temp is 20degrees c) in order to cool water by 10degrees in a closed loop (in a coiled copper pipe) suspended in the tank the coil has a total length of 30 lineal m and a diametre of 20mm the water in the coil enters at 50 degrees and is traveling at approximately 3 litres per second. I am a geologist not a physician and this has me stumpt. Any help or advise would be much appreciated.
     
  2. jcsd
  3. Jul 4, 2012 #2

    CWatters

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    It’s been a while since I did these sorts of sums but..

    For the temperature in the tank to be constant the power entering must match that extracted so first work out the power in Watts (=Joules/second) entering the tank via the hot coil using:

    Heat capacity of water = 4180 J/Kg/K
    Temperature drop (10 K)
    Mass of water per second (3kg/S)

    = 4180 * 10 * 3
    = 125,400 J/S
    = 125 kW

    So that's how much has to be removed by the cold water flow.

    Quick check that the coil is big enough. Last time I did this I used this equation:

    Dt = (P * Wt)/(A * C)

    Where

    Dt = average delta T across wall of pipe in Centigrade
    C = thermal conductivity of copper = 401 W•m−1•K−1
    Wt = wall thickness in meters? Assume 2mm = 0.002m ?
    A = area of pipe wall in meters = Pi * 0,02 * 30 = 1.88sqm
    P = Power in watts

    Dt=(125*10^3 * 0.002)/(1.88 * 401)
    =0.33K

    So can assume no temperature drop across wall of copper pipe (unless it scales up!)

    Incoming hot is 50C, require 10C drop so it leaves at 40C. Average 45C

    Lets assume water temperature of the tank is maintained at 45C.

    Temperature rise of cooling water is 45-20C = 25C

    Flow rate required is

    = 125,400/(4180 * 25)
    = 1.2L/Sec

    Note that it will come out colder than required until the tank is upto temperature. that might take quite awhile.
     
  4. Jul 4, 2012 #3
    Thanks a lot for that.
    It all make sense to me and is far simpler than I thought. I think I will play with the variables to see if I can improve the effeciency and size etc of the system. I am using this to try and build a ground water cooled refrigerant system.

    Re the calc for the suitability of the coil length if i change the parameters at what point do you think that the temperature drop accross the pipe is of any concern to the whole system?

    Does the size of the tank have any bearing on any of this, or is it as I suspect that it will only vary the time it takes for the tank to reach the average water temp.

    Again thanks a bunch.
     
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