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Thermal conductivity value of (.17-.20)W/m.K

  1. Feb 9, 2012 #1
    I was hoping someone could help me with a project I am doing. I have contacted a manufacturing company thatmakes thermal wrap material, and I need to see if this will work for my project. According to the manufacturer the material is 1/16" thick - this info was also given to me by the manufacturer:

    .17-.20 W/mK (1000 degrees C)

    I'm basically just trying to figure out how this formula and the data given to me, applies in the real world? I basically need to know to what extend thermal leakage will happen with the material. So if I have a pipe with 1000 degrees F flowing through it, and the pipe is wrapped with this thermal material, how much heat will actually be coming off the material into the surrounding environment. Basically what would the surface temperature of the thermal material be? Hope that makes sense. Thanks!
  2. jcsd
  3. Feb 9, 2012 #2
    This is a bit tricky due to convection. It works like this:

    The power going out of the pipe is (if the insulation is not very thick):
    [tex]P=0.2 \frac{A \Delta T}{d}[/tex]
    Where A is the surface area in sqare meters ΔT is the temperature difference in Celsius or Kelvin and d is the thickness of the wrapping (if it was tightly wrapped). The outside temperature depends a lot on how much power the air can carry away. When the pipe gets hot enough there wil be quite a bit of thermal radiation [tex]P=A \sigma T^4[/tex] where σ is the Stefan-Boltzmann constant, A is the surface area and T is the temperature in Kelvin. So you could calculate at what outside temperature the power going out of the pipe is radiated away at the same time. This only works for high temperatures and tight wrapping though, which is probably not what you want, but it would give an upper temperature limit.
    For "normal" temperatures most of the heat is carried away by convection -- the flow of the rising air around the warm pipe, which is very hard to calculate, so usually this quantity is measured. Additionally if you wrap your pipe not very tight the trapped layers of air will help to improve the insulation considerably.

    To make a long story short, you cannot really calculate this stuff all that well, and you have to rely on "best pratices". There is a bit about this in the wikipedia article: http://en.wikipedia.org/wiki/Pipe_insulation and in the end they have some links to free programs to calculate this stuff but to quote wikipedia: Equations for calculating this are based on empirical results and vary from standard to standard (both ASTM C 680 and EN ISO 12241 contain equations for estimating surface coefficients of heat transfer)

    I hope that this helps a bit.
  4. Feb 9, 2012 #3
    Thanks! I'll probably just order the material and experiment a little.
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