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Heat conduction in hollow cilinder.

  1. Jun 19, 2011 #1
    1. The problem statement, all variables and given/known data
    Given is a hollow cilinder with inner radius R1 and outer radius R2. The heat conductivity of the material is k. The cilinder has length l and an inner temperature of T1 and outer temperature T2. Determine the temperature gradient in the cilinder and the heat flow that leaks away radially.

    2. Relevant equations
    Power = k*Area(Temperatureinside-Temperatureoutside)/(thickness)
    Area = 2*pi*R*l

    3. The attempt at a solution
    The reason I can't solve this problem is the changing area when going from inside to outside the cilinder. Obviously the heat flow going from the inside to the outside should be minus the heat flow from outside to inside, since the areas are different I think the only way this is possible if the temperature gradient falls off quicker going inside causing the heat flow to be the equal despite the smaller area.
    So much for intuition though since I've tried fruitlessly to write this down mathematically.
    I also think the trick is to use an intermediate temperature T and equate the powers going through an area between R2 and R1 since we've used that one before.
     
  2. jcsd
  3. Jun 19, 2011 #2

    Mapes

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    This equation assumes that the temperature profile is linear, which isn't the case here. A more general equation is [itex]\mathrm{Heat~flux}=kA(x)(dT/dx)[/itex], or since you're working with radius r, [itex]\mathrm{Heat~flux}=kA(r)(dT/dr)[/itex]. You should be able to find T(r) through integration.
     
  4. Jun 19, 2011 #3
    When I integrate the formula you've given me I get T = (heat flux)/(k*2pi*l)*ln(R)
    To find the heat flux should i simply fill in T1 and R1? I feel like I've missed something since the problem seems a bit simple this way.
     
  5. Jun 19, 2011 #4

    Mapes

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    Check your integration. A lone ln(R) can't be correct; you can't take the logarithm of a parameter with units.
     
  6. Jun 19, 2011 #5
    You're right, is this better? T = T1+ heatflux/(k*2*pi*l)*ln(r/R1)
    It seems to describe the right heat gradient but how do I get the heatflux from this?
     
  7. Jun 19, 2011 #6

    Mapes

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    You know the temperature at the outer radius too.
     
  8. Jun 19, 2011 #7
    Ah yes of course:blushing: My brain is all but melted right now because of all the studying I guess. Thanks for the help.
     
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