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My textbook gives the thermal resistivity of a slab of a material having lateral thickness $L$, area of face $A$, and thermal conductivity $K$ by

$R$=$L/KA$

Using this formula, the formula for the rate of heat transfer ($Q/t$) b/w two ends of the same slab having temperature difference ∆$T$ is given as:

$Q/t$=($KA$∆$T$)/$L$=∆$T$/$R$

This formula resembles Ohm's law in the sense that $Q/t$ gives the "current" $i$, ∆$T$ gives the potential difference in terms of heat and $R$ gives the resistance. My textbook has then stated that

"Hence, results derived from Ohm's law are valid for thermal conduction also. For a slab at steady state, we have seen that the thermal current $i$ remains same at each cross-section. This is analogous to Kirchhoff's current law for electricity, which can now very conveniently be applied to thermal conductivity."

In an example problem following this, it is asked to find the temperature of a junction b/w 3 rods made of different materials, whose ends are maintained at different temperatures. The following diagram has been provided as a hint:

Diagram

My doubts are these; firstly, is the equation given for thermal resistivity correct? The formula that I find on other websites and my copy of Halliday and Resnick is the following:

$R=L/K$

Secondly, why do the currents in the figure all add up to zero?(As given in the solutions)? I am very confused about this analogous law.

Please do share your insights for it would be tremendously helpful for me. MUCH thanks in advance :) Regards.

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  • $\begingroup$ The formula for thermal resistivity is given by your textbook. So why do you ask if it is correct? Do you have some reason for doubting it? As with Conduction problem? you are asking questions about the problem before making an effort to think about it. $\endgroup$ – sammy gerbil Sep 14 '16 at 0:34
  • $\begingroup$ If I have made any aspect about my question unclear, it will do to let me know about this so that I may edit the question, without accusing me of being lazy, thanks. $\endgroup$ – user106570 Sep 14 '16 at 0:44
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Principle of the analogy

Thermal conductance, $\displaystyle{\frac{KA}{L}}$, is the amount of heat that passes through a slab of area ($A$) and thickness ($L$) in unit time, when the opposite faces of the surface differ in tmperature by $1K$. It's reciprocal is the thermal resistance $R_T$,

$$R_T=\frac{L}{KA}\longrightarrow (1)$$

Hence, the rate of heat transfer from one face to the other is given by:

$$\frac{Q}{t}=\frac{KA\Delta T}{L}\longrightarrow(2)$$

which follows from the definition of thermal conductance, when the opposite faces are having a difference in temperature by $\Delta T$. From equation $(1)$, we have

$$\frac{Q}{t}=\frac{\Delta T}{R_T}\longrightarrow(3)$$

Let, the heat current (amount of heat transferred in unit time) be denoted by $I_H$. Hence

$$I_H=\frac{Q}{t}=\frac{\Delta T}{R_T}\longrightarrow(4)$$

Now, comparing this equation with the Ohm's law, we have the electric current, which is the amount of charge flowing per second

$$I=\frac{V}{R}\longrightarrow(5)$$

Comparing both equations, we can make an analogy. We have in equation $(4)$, $\Delta T$, the difference in temperature in analogy with the electric potential $V$, and $R_T$, the thermal resistance in analogy with the electrical resistance $R$.

Just as like a voltage (or more precisely, potential difference) makes a current flow through a metallic conductor, the temperature difference between the two ends of a metallic rod makes the heat flow to happen from a point of higher temperature to that of lower temperature. This makes sense. Hence this analogy is useful.

Hence it can be extended to Kirchhoff's law in electricity also, as the only difference in the two cases are that in one heat is the energy transferred, while in the other, electrical energy is being transferred.

How to work out the problem using this analogy?

One of the ends of the three rods start from a common point, say a heat source maintained at a constant temperature $T$. The direction of current flow states that $T$ is the high temperature region.

The current starts from the junction. Hence at the junction, by the conservation of energy, the currents should be added to zero, just as in Kirchhoff's rule. How's that? Simple: For current flow to take place, there should be temperature difference between the two points. If you take the junction point alone, which is at a single temperature, there is no current flow. Hence the currents add up to zero at the junction. But it flows to the other ends, as obviously there is some temperature difference.

First, use the principle that at the junction $T$,

$$i_{Al}+i_{S}+i_{Cu}=0\longrightarrow(6)$$

If the temperature at the junction is $T$ and that at the end points of the rods are $T_{Al}$, $T_{S}$, and $T_{Cu}$, then the corresponding temperature differences become $(\Delta T)_{Al}=(T-T_{Al})$; $(\Delta T)_{S}=(T-T_{S})$; $(\Delta T)_{Cu}=(T-T_{Cu})$. Now using equation $(4)$, $(6)$ reads

$$\frac{(T-T_{Al})}{R_{Al}}+\frac{(T-T_{S})}{R_{S}}+\frac{(T-T_{Cu})}{R_{Cu}}=0$$

$$T\left[\frac{1}{R_{Al}}+\frac{1}{R_{S}}+\frac{1}{R_{Cu}}\right]=\frac{T_{Al}}{R_{Al}}+\frac{T_{S}}{R_{S}}+\frac{T_{Cu}}{R_{Cu}}\longrightarrow(7)$$.

Now, knowing the resistances, we can calculate $T$.

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  • $\begingroup$ Thank you for your detailed answer, although you needn't have explained just as much-I had understood the analogy :-) Secondly, I still don't quite understand exactly why the currents add up to zero. I changed the direction of $iAl$ and $iS$ in the question and then solved the question by conserving heat(/heat current) and was able to solve the question to get T=10°C but can you please elaborate as to why the currents taken in the direction given in the figure add up to zero? Thanks :-) $\endgroup$ – user106570 Sep 12 '16 at 3:59
  • $\begingroup$ I have mentioned that in the answer. For a current to flow between two points, the temperature at the two points should be different. Let's have an analogy. You have a pool. You have branches of the pool connecting to three beakers. The net water entering into the beakers will be the sum of the water leaving the pool. $\endgroup$ – UKH Sep 12 '16 at 8:59
  • $\begingroup$ Okay, it may be that I'm not able to directly correlate "heat current" with water. In this concept, exactly what are we conserving when we add the currents and equate to zero? $\endgroup$ – user106570 Sep 12 '16 at 10:00
  • $\begingroup$ we are conserving the total energy (heat) of the system. The summation of currents to zero can be viewed as a continuity equation $\endgroup$ – UKH Sep 12 '16 at 16:25
  • $\begingroup$ If it isn't too much to ask, can you please elaborate on this a little more? $\endgroup$ – user106570 Sep 13 '16 at 0:14

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