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Thermodynamics resistance proof

  1. Feb 9, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that thermal resistance is additive in series


    2. Relevant equations
    H=A(TH-Tc)/R
    where R=L/k


    3. The attempt at a solution

    For two slabs in thermal contact where TC is the outside cold temperature and TH is the outside hot temperature

    A(TH-Tc)/R=A(TH-T)/R1+A(T-Tc)/R2)

    The A's cancel out, and after a bit of math, I've gotten the equation down to

    R=(R1+R2)(TH-TC)/(R2(TH-T)+R1(T-Tc)

    How can I get rid of the T's with no subscript and just be left with R1+R2 on the right side?
     
  2. jcsd
  3. Feb 9, 2009 #2

    Mapes

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    Try considering an energy balance at the point between the two slabs.
     
  4. Feb 9, 2009 #3
    Are you suggesting that these slabs are in dynamic thermal equilibrium? If so, are you suggesting that I simply equate the two heat currents? I will not have "R" involved in my expression then, only R1 and R2...
    I truly want to understand this. Thank you.
     
  5. Feb 9, 2009 #4

    Mapes

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    Yes, the equation applies to steady state conditions only.
     
  6. Feb 9, 2009 #5

    LowlyPion

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    I'd approach it more simply.

    Heat Flow = ΔT/R

    So R*Heat Flow = ΔT

    The heat flow of R1 is then R1*H = ΔT = (Ti - T1)

    And through R2 is R2*H = (T1 - T2)

    Heat flow for the system then is

    R1*H + R2*H = (Ti - T1) + (T1 - T2) = Ti - T2

    If Rtotal*H for the system is Ti - T2, then Rtotal is (R1 + R2)
     
  7. Feb 10, 2009 #6
    Thank you very much! That is a lot more straightforward than how I was trying to prove it.
     
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