# Finding total thermal conductivity (Ksys) for n-number of connected materials

I was playing around with the idea of finding a "total" $K$ value for 2 solid, flush touching pieces of metal. The original problem involved the two touching inline bars with a two different maintained temperatures at the non-touching ends of each bar.

To find the rate of heat transfer, we first found the temperature at the point where the two bars touched and then plugged the temperature back into earlier equations to find the rate of heat transfer. I, however, wanted to find a total $K$ over total $L$ value for the system and was able to find an equation that worked for any number of bars connected (without having to find the temperature of the place where the bars are touching),$$\frac{1}{\displaystyle{\sum_{i=1}^{n}{\frac{L_n}{K_n}}}} \,,$$where $L$ is the length of a bar and $K$ is the thermal conductivity of each bar.

This simplifies to$$\frac{K_1 K_2}{L_1 K_2 + L_2 K_1}$$for a system of $n=2$ connected bars.

Questions:

1. I don't understand how my equation actually works. I started by dividing the length of the bar by the thermal conductivity and guess and checked my way to a final equation. It seems to work but I don't understand why or the theory behind it. Could anyone explain it?

2. Although I myself checked it for two-bar and three-bar problems, I might've messed up a step, so does it actually work?

3. Does this equation or a similar one appear elsewhere in physics or other science/math?

• 3. Resistance combination laws! Dec 30, 2019 at 8:59

When you write the thermal conductivity per length as $j_n=K_n/L_n$, which measures how well a slab conducts heat, then your formula looks suspiciously like a series circuit of 2 resistors:

$$R_{tot}=R_1+R_2=\frac{1}{G_{tot}}=\frac{1}{G_1}+\frac{1}{G_2} \Longleftrightarrow \frac{1}{j_{tot}}=\frac{1}{j_1}+\frac{1}{j_2},$$

where $R$ is the resistivity of the resistor and $G=1/R$ is the conductance of a resistor, where the latter measures how well a resistor conducts electricity. Both forms for $j$, the one I wrote and yours, are equivalent: all hail Wolfram Alpha, here and here.

It also makes sense physically: when you have two different temperatures at the end of the two bars, i.e. $$T_1 |\text{Bar}|\text{Bar}|T_2$$

then you can see it as a circuit where the temperature is the driving force, i.e. the "voltage", and the pieces of metal are the resistors to the heat flow. Then $j=K/L$ is the measure of how well a slab of metal conducts heat, analogous to the electrical conductance $G$. It's reciprocal, $L/K$ is then the thermal insulance, analogous to the electrical resistivity $R=1/G$.

When you have two slabs in series, this is then equivalent to two resistors in series. For two resistors in series you add their $R$'s, the same is then true for the insulances of the slabs by analogy. Adding the insulances is then the same as saying, with $j$ as defined earlier,

$$\frac{1}{j_{tot}}=\frac{1}{j_1}+\frac{1}{j_2},$$

which is exactly the same as your formula.

So your equation gives, when both $K$'s and $L$'s are the same

$$j_{tot}=K/2L,$$

so the thermal conductivity per length has halved, or in other words, if you take the reciprocal

$$\left(\frac{L}{K}\right)_{tot}=2\frac{L}{K},$$

which is just the sum of the two thermal inulsances of the two slabs.