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Heat Transfer Convection Resistance Issue

  1. Jun 16, 2009 #1
    I am trying to calculate the length of a pipe needed for an air cooling system I have found the following formulas:
    R=1/(h*a)
    where h is the convection coefficient and a is the surface area.
    h=Nu*(k/L)
    where Nu is the nusult number and k is the thermal conductivity.
    a=π*d*L
    where d is the diameter and L is the length.

    Putting this together I get:
    R=1/[(Nu*(k/L))*(π*d*L)]=1/(Nu*k*π*d)
    This seems to imply that the length of the pipe has no impact on the amount of energy lost which I am pretty sure is wrong.

    Any help would be appreaciated.
     
  2. jcsd
  3. Jun 16, 2009 #2
    You've got it right. But first I think some clarification is needed.
    First, in your final equation, the "k" is the thermal conductivity of the fluid the pipe is in contact with. Second, the Nusselt number "Nu" has the length of the pipe incorporated in it (as can be seen in rearranging your second equation).
    The Nusselt number is generally used to determine an effect heat transfer coefficient "h", not the other way around.
    In order to solve for your "h", a second Nusselt equation is needed. The second equation for the Nusselt number depends on the Reynolds number, shape of object fluid is flowing over, and the state of the fluid (liquid or gas).
    The equation for the Reynolds number is
    Re = (density of fluid)(average fluid velocity)(diameter of the pipe)/(viscosity of the fluid)
    Even though this is the equation for the Reynolds number inside a pipe, the Nusselt number uses it.
    You'll also need to use the Prandtl number as well. Its equation is
    Pr = (viscosity of fluid)(specific heat capacity of fluid)/(thermal conductivity of fluid)
    The Nusselt number (for most reasonable gas flows around a pipe) is
    Nu = 0.193((Re)^0.618)((Pr)^(1/3))

    Doing these calculations takes a lot of time, but you'll get a heat transfer coefficient that's pretty close to the actual deal.

    By re-doing your last equation to account for "h" and "L", you'll get a good energy loss approximation.
     
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