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Heat transferred from an insulated enclosure

  1. Jun 11, 2014 #1
    Thanks for checking out my post! I have a question for you heat transfer savvy folks.

    I have a pretty large enclosure (SA = 107,300 sq in.) made of 1/4" thick stainless steel, and has a 1-inch thick layer of insulation lining the inside of the wall. The purpose of this proposed enclosure is to protect water pipes from freezing during the winter. I found a 1.8 kW convection heater that I'd like to use, so I took a cut at a calculation to figure out if that'd be enough heat input to maintain 50 degrees inside the enclosure. I am assuming a worst case outside ambient air temperature of -10 degrees F, so that's a difference of 60 degrees. I also assumed that the steel plays no impact as a thermal barrier (to be conservative), and that the air provides a light film layer of insulation on either side of the wall. The resources I've looked at said that the heat transfer coefficient for the air film layer should be somewhere in the range of 5 - 37 W/m2K, so I plugged in both of those and found a lower and upper limit for the heat input required. The range I am getting is 3.7 kW to 6.6 kW, which shows that the 1.8 kW might be too small. Does this seem reasonable? I know it's a large surface area and a big delta T, but I don't have much experience here. I attached my calculation below, so any help would be appreciated.

    Thank you.


    Attached Files:

  2. jcsd
  3. Apr 17, 2015 #2
    Ignoring a bunch of factors, like internal and external convection, you can make a simple model on this one.

    Total resistance = thickness of steel/(conductivity of steel*SA) + thickness of insulation/(conductivity of insulation*SA)
    UA = 1/total resistance
    heat dissipation = UA*(T_high-T_low)

    What kind of steel is it and what kind of insulation is it? If you can specify what those are, I can probably find values of the conductivity for them (or maybe you already have in order to calculate those values?) Anyway, a good model with will include convection coefficients for the inside and outside surfaces.

    convective resistance = 1/(convection coefficient * SA)

    Then, just add them up like they're in series, invert it to get a value for UA, then calculate the heat transfer. Does that help any? This was a quick answer, not a thorough one. If anything needs clarification, let me know.
  4. Apr 17, 2015 #3
    This looks like exactly what he did, except for neglecting the thermal resistance of the steel.

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