Thermodynamics resistance proof

AI Thread Summary
The discussion focuses on proving that thermal resistance is additive in series for two slabs in thermal contact. The initial equation presented involves heat flow and thermal resistance, leading to a complex expression that the user seeks to simplify. A suggestion is made to consider energy balance and equate heat currents at the interface of the slabs, emphasizing the need for steady-state conditions. The simplified approach shows that the total resistance is the sum of individual resistances, confirming the additivity of thermal resistance. This clarification highlights a more straightforward method to understand the concept.
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Homework Statement


Prove that thermal resistance is additive in series

Homework Equations


H=A(TH-Tc)/R
where R=L/k

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?
 
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Try considering an energy balance at the point between the two slabs.
 
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.
 
Yes, the equation applies to steady state conditions only.
 
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)
 
Thank you very much! That is a lot more straightforward than how I was trying to prove it.
 
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