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I am trying to work out how I would find the rate of heat transfer through two pipes made of the same material that are joined at their ends but which have different cross-sectional areas and lengths. One end of the combination is held at $T_1$ and the other at $T_3$ but I don't know the intermediate temperature $T_2$ where the two pipes are joined which is a problem since the thermal conductivity of the material that the pipes are made out of changes with temperature. I have formulated the problem thus:

$$ \dot Q = (A_1/l_1)\int_{T_1}^{T_2}K(T)\rm dt + (A_2/l_2)\int_{T_2}^{T_3}K(T)\rm dt $$

Where $A_1$ and $l_1$ are the dimensions of the first pipe and so on. I understand that you can use an analogy to electrical resistance and just add the thermal resistance for each pipe as if they were in series to eliminate the unknown intermediary temperature $T_2$, giving: $$\dot Q = T_3-T_1/(R_1 + R_2) $$ but the equation for resistivity doesn't seem to solve this problem of an unknown temperature since I believe:

$$ R_1 = l_1/KA_1 $$ and $$ R_2 = l_1/KA_2 $$ where K in each instance would still require an integral with an unknown temperature as one of its limits. ~ Am I over complicating things? Can I simply use the value of $(T_3-T_1)^{-1}\int_{T_1}^{T_3}K(T)\rm dT$ to work out the total integrated thermal conductivity for the total temperature change and use this to find both of the thermal resistivities? I suspect not, but I'm not sure how else to solve the problem.

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  • $\begingroup$ Is the assumption here a steady state heat transfer T3>T2>T1? The last equation you have is essentially finding an average k over the temperature range, which should give you a decent enough answer if the temperature range is small and k doesn't vary too much (depends on the material). $\endgroup$
    – Yandle
    Commented Oct 28, 2014 at 18:56
  • $\begingroup$ Hi Yandle, yes T3>T2>T1 is true for this question. I'm specifically looking at stainless steel over a large temperature range. If it helps the problem is related to materials used for cryogenic purposes. $\endgroup$
    – user46188
    Commented Oct 29, 2014 at 18:04
  • $\begingroup$ Are you calculating the total heat transfer along the length of the two pipes? In this case, your first equation is wrong. Alternatively, are you calculating the total heat transfer radially outward from the two pipes? In this case, you second equation is wrong. $\endgroup$
    – Jeffrey J Weimer
    Commented Dec 24, 2022 at 17:16

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From my heat transfer textbook by Cengel and Ghajar, they propose that you use the average thermal conductivity value as you specified in your last equation:

$k_{avg} = k(T_{avg}) = k_0(1+\beta(T_a+T_b)/2)$

where $\beta$ and $k_0$ are material properties, assuming that the material thermal conductivity follows a linear function w.r.t temperature. You simply plug in average k into the regular thermal conduction equation and solve. As far as whether the same average k value can be used for both sections of piping is up to you.

If you do not want to do the above then the only other way I can think of is to do this numerically since you have specified both boundary conditions.

Also, I do not believe your first equation is correct, with temperature boundary conditions the heat flux at location 1 and 3 should balance; I'm not sure why they are added. I think it should be:

$A_1/l_1k_{avg}(T_2-T_1)=A_2/l_2k_{avg}(T_3-T_2)$

This will give you T2 which will allow you to solve for the heat flux.

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