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Question is wrong. You can close it.

Actually I was learning about the conduction method of heat transfer. I found that there is thermal resistance for a system which can either be series or parallel. In a series case the net resistance is $$r_{tot=}\sum_{i=1}^{n} r_{i}$$

And this is totally analogous to the electric resistance.

The series case can only hold if the flow of current through different resistance is the same. And it turns out that it is true.

Because the heat flow is analogous to current and the current entered in must go out. Can the similar thing also hold for heat flow rate?

If yes I also have another doubt. Let me ask my doubt clearly with an example. Assume you have rods of length l,m,n and are made of material A,B,C. Now for simplicity we can assume the cross sectional area A is same for all the three rods and Temperature at the end of the C rod is 0. And take the temperature at the other end say A's end is $T_{A}$ Now their resistance will be $R_{tot}=R_{A}+R_{B}+R_{C}$. And the resistance is given as

$$R=\frac{L}{Ak}$$

k is the conductivity os the rod. So the first doubt is

Is $\dot{Q_{A}}=\dot{Q_{B}}=\dot{Q_{C}}$ If the system is in a series alignment?

And the second one is if the first one is right then the total rate is the same? I mean $\dot{Q}_{tot}=\dot{Q}_{A}$?

At first glance it seems obvious, but when I solved for a system analogous to my example, I got different solutions for the junction temperature (between A and B), when I used these results. The whole picture is hard to share but I'll share the important parts of it. Now I'd like to find the temperature at A and B rod's junction. So we can take

$$\dot{Q}_{A}=\dot{Q}_{B}$$

Now if we use the conduction equation:

$$\dot{Q}_{A}=\frac{k_{A}A(T_{A}-T)}{L_{A}}$$

Similarly we can do it for B

$$\dot{Q}_{B}=\frac{k_{B}A(T-T_{1})}{L_{B}}$$

Now we don't know what $T_{1}$, But it can found using another relationship

$$\dot{Q}_{B}=\dot{Q}_{C}$$

$$\dot{Q}_{C}=\frac{k_{C}AT_{1}}{L_{C}}$$

And solving we will get the equation of T as

$$T=\frac{k_{A}T_{A}}{L_{A}(\frac{k_{A}}{L_{A}}+\frac{k_{B}}{L_{B}}-\frac{k_{B}}{L_{B}(\frac{k_{C}L_{B}}{k_{B}L_{C}}+1)})}$$

All these were done through assumptions and I'm not whether it is right or not. Assume it is right. Then $\dot{Q_{A}}=\dot{Q_{B}}=\dot{Q_{C}}$. So the overall heat flow is equal. So the heat flow rate is same. That is heat flow through the system is same. So I found the resistance and applied that in the another form of heat flow of equation.

$$\dot{Q}_{overall}=\frac{T_{A}}{R_{Tot}}$$ Temperature at the end is 0.

So if we use the relationship

$$\dot{Q}_{overall}=\dot{Q}_{A}$$

So if we substitute accordingly we get,

$$T=T_{A}(1-\frac{L_{A}}{k_{A}(\frac{L_{A}}{k_{A}}+\frac{L_{B}}{k_{B}}+\frac{L_{C}}{k_{C}})})$$

Which is completely different from the first one. What was the right one? To check whether they are same I substituted some numerals and got clearly different answers. Can any pls check whether I'm right or not? Also pls check the algebra either. Whether I made a mistake or not. Pls explain me why when I used overall $\dot{Q}$ I got wrong answer.

(Sorry if made any mistake, I am new to this topic and I am so confused and tired of this. Also sorry if I made any mistake in Latex and about the ordering of my body. Also pls apologise me for any algebra errors. If there are any errors please mention that either. Thanks!!!)

Forget to mention this is a steady state flow.

Edit: Actually I got a question that is more or less similar to mine. And the correct answer is the first one not the second. Can you pls explain it why?? Ref: University physics, Richard stallman, ln 17.

Actually both the cases are right. So you can close the question as mine were wrong.

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    $\begingroup$ "At first glance it seems obvious, but when I solved for a system analogous to my example, I got different solutions for the junction temperature (between A and B)" Since you have not shown your work, it is impossible to know what your errors are. That said, your other assumptions about the analogy are correct for the case of one dimensional steady heat flow. $\endgroup$
    – Bob D
    Mar 2, 2022 at 12:51
  • $\begingroup$ Bro I have added some part of my calculations. Was that enough?? $\endgroup$
    – Sanjay
    Mar 2, 2022 at 14:55
  • $\begingroup$ The rate of heat flow in a series system is constant throughout the system for the case of steady state steady flow. $\endgroup$
    – John Darby
    Mar 2, 2022 at 15:04
  • $\begingroup$ @JohnDarby bro this is a steady state flow. Forget to mention. I'm sorry. $\endgroup$
    – Sanjay
    Mar 2, 2022 at 15:27
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    $\begingroup$ The second result is correct. There’s an algebra error somewhere in ”And solving we will get…” Since this is a calculation error rather than a conceptual question, I’m voting to close. $\endgroup$ Mar 2, 2022 at 16:10

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Just like for electrical circuits conservation of charge forces serial circuits to have the same current through every component, conservation of energy dictates that the heat flow through a 1-dimensional system in steady state must be equal for every component. If it wasn't, energy would continuously enter a region without exiting it, thus accumulating. It is important that this is only valid in the steady state, as temporary fluctuations can absolutely see some compenent absorb or emit more heat than it receives, raising or lowering its temperature.

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