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Rate of heat transfer and heat conductors

  1. Jun 9, 2004 #1
    lets say you have the following situation: you have a cold pool of water, and hot air outside. how do you calculate the rate at which heat is exchanged (given Twater and Tair, the surface area of the pool, etc.)?
    Now, let's say you add a layer of a certain material x meters thick that has a known thermal conductivity between air and water. how is the rate of heat transfer modified?


  2. jcsd
  3. Jun 12, 2004 #2
    don't tell me nobody here knows how to do that...
  4. Jun 12, 2004 #3


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    Staff Emeritus
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    Gold Member

    It's not a straightforward calculation :
    If the water is contained in a vessel or cavity, what is the temperature of the walls of the vessel/cavity. Is this temperature going to change, ie : is the vessel infinitely thick compared to the volume of water ?)

    You need to set up a 2D heat flow equation inside the water, and you need to add heat loss due to evaporation (though this is probably small). Assume some temperature distribution T(r,z) for a cylindrical pool. Then :

    [tex]Q_r = -KA(r) {\frac {\partial{T}} {\partial{r}}}, and [/tex]
    [tex] A(r) = 2\pi rh[/tex]

    Similarly for the z-direction. Try a separable solution. Put in the boundary conditions and solve.
    Last edited: Jun 12, 2004
  5. Jun 12, 2004 #4


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    Staff: Mentor

    Its a little bit complicated for the pool because a little bit of wind makes a vast difference. Heat will leave the pool through all three typical ways: conduction (water touching the air), convection (air moving around above the water), and radiation (like a light bulb). The temperature difference is key though.

    Thermal conductivity in a single material is easier: you need temperature difference from one side to the other, thickness, and conductivity.

    You want some actual math? It gets pretty rough...
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