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Liquid Cooled Heat Exchanger Verification

  1. Jun 27, 2013 #1
    My friend and I are trying to design a simple liquid-cooled heat exchanger and I'm getting some odd results so I wanted to get another set of eyes on my math here. We're running four 1.5m sections of nylon tube in parallel through the heat source, which was a thermal input of 265W. The cooling loop will be supplied by a tank of cold water and pump. As the system runs, the water will gradually warm up, and once the the surface temperature of the nylon tube reaches 305K, we will consider that the point at which the water needs to be replaced/chilled, so I was trying to determine what the cold tank temperature would be at that point.

    My assumptions:
    *All thermal input is absorbed by the exchanger tube
    *Heat is transferred to the water and tube by conduction and convection
    *Neglect radiation and contact resistance
    *The thermal input is equally divided among the four tubes

    [itex]\dot{V}_{pump}=0.114 L/s[/itex]
    [itex]\dot{V}_{tube}=\frac{0.114}{4 tubes}=0.0285 L/s[/itex]
    [itex]OD_{tube}=0.953 cm[/itex]
    [itex]ID_{tube}=0.699 cm[/itex]
    [itex]k_{tube}=0.25 W/m \cdot K[/itex]
    [itex]\dot{Q}_{per tube}=265W/4 tubes=66.25W[/itex]

    Using the overall heat transfer equation:
    [itex]\dot{Q}=UA\Delta T[/itex]


    For my convection coefficient, I got the following (check attached document for more complete work):
    [itex]h=10.37 W/m^{2}\cdot K[/itex]
    [itex]A_{conv}=\pi (0.699cm)(1.5m)=0.033m^2[/itex]

    For the conduction term:
    [itex]k_{tube}=0.25 W/m\cdot K[/itex]
    [itex]d=r_o - r_i =\frac{0.953cm-0.699cm}{2}=0.127cm[/itex]
    [itex]A_{cond}=\pi (0.953cm)(1.5m)=0.045m^2[/itex]

    Plugging it in:
    [itex]\frac{1}{UA}=\frac{1}{(10.37)(0.033)} + \frac{0.127cm}{(0.25)(0.045)}=3.035[/itex]

    Substituting into the heat transfer equation:
    [itex]66.25W=(0.33W/K)\Delta T[/itex]
    [itex]\Delta T = 200.8K[/itex]
    [itex]200.8K=T_H - T_C=305K-T_C[/itex]

    Obviously a ridiculously low temperature. I thought that was odd, so I checked how much heat could be handled by conduction alone:
    [itex]\dot{Q}=k\frac{2\pi L\Delta T}{ln(r_o/r_i)}[/itex]
    [itex]66.25W=(0.25 W/m\cdot K)\frac{2\pi (1.5m)\Delta T}{ln(0.477/0.350)}=\frac{2.36\Delta T}{0.309}[/itex]
    [itex]\Delta T=\frac{(66.25)(0.309)}{2.36}=8.67K[/itex]

    So according to conduction alone, the system will be effective until the water is 296K, but with conduction AND convection, it's only good until the water is 104.2K, which would be ice. Any idea where I'm screwing up?

    Attached Files:

  2. jcsd
  3. Jun 27, 2013 #2


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    Gold Member

    The gives you the temperature ( 296K ) of the inner surface of the tube, not the water temp.
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