Confusion about thermal resistance in parallel

This question is inspired by me insulating my front workshop door and realizing that I may have to leave 10% of the door uninsulated if I want the door to still work right.

If I have a wall where 90% of its surface is covered in a material that has thermal resistance $R_1$ and 10% of its surface has thermal resistance $R_2$ where $R_1 > R_2$, mathematically, these two resistances are in parallel and should be added as such:

$\left(\frac{1}{R_1} + \frac{1}{R_2}\right)^{-1} = \frac{R_1 R_2}{R_1 + R_2}$

So let's say for example that $R_1 = 10R_2$. The total thermal resistance of the wall is now:

$R_{tot} = \frac{10 R_2}{10+R_2}$

What is confusing is that this new resistance is now LOWER than the resistance of the wall made entirely of the lower resistance material, as can be seen by:

$\frac{10 R_2}{10+R_2} < R_2 \rightarrow 10R_2 < 10R_2 + R_2^2 \rightarrow 0<R_2$

which is a true statement in all cases.

This is confusing because I know that in real life, if I cover 90% of the door with a good insulator, the total heat transfer $Q$ is going to go down. However, these equations say that I am better off just leaving the door entirely uninsulated (which can be seen by $Q = \Delta T / R$), which can't be true.

• I believe that the error you are making is to ignore the area of the two materials. Usually, only the resistivity and not the resistance of the material is provided. So you should be dividing the resistances by the areas of the materials too. You can take them as $0.9A$ and $0.1A$. In an electrical conductor on the other hand, the resistance is indeed lowered, since now you have two parallel pathways for current to flow. But there, you don't have to worry about the area.
– Sidd
Jan 7, 2018 at 18:28
• One way to think about it: You always had two "resistors" in parallel. One of them was 90% of the door, and the other one was 10% of the door. You upped the value of the 90% one. Jan 7, 2018 at 18:30

1 Answer

Saying that $R_1 = 10R_2$ does not necessarily follow from your statement about the percentage of the wall which is covered.

The thermal conduction equation is equation $\dot Q = \dfrac{kA\Delta T}{L}$ where $\dot Q$ is the rate of heat flow, $\Delta T$ is the temperature difference, $k$ is the coefficient of thermal conductivity, $A$ is the cross-sectional area and $L$ id the thickness.

So the thermal resistance can be thought of as

$R_{\rm thermal} = \dfrac {\Delta T}{\dot Q} = \dfrac{L}{kA}$.

If we start with all the area $a+A$ being of a material with thermal conductivity $K$ the rate of flow of heat per unit temperature difference (the thermal conductance) is given by

$\dfrac {\dot Q_1}{\Delta T} = \dfrac{Ka}{L}+ \dfrac{KA}{L}$.

Now put a poorer conductor, thermal conductivity $k$, over the area $A$ and you get

$\dfrac {\dot Q_2}{\Delta T} = \dfrac{Ka}{L}+\dfrac{kA}{L}$ which shows that $\dot Q_2< \dot Q_1$ because $kA<KA$.

Note that both these equations are equivalent to $\dfrac {1}{R_{\rm thermal, total}} = \dfrac {1}{R_{\text{thermal, area a}}} + \dfrac {1}{R_{\text{thermal, area A}}}$

I think that this shows that it is easier to consider the thermal conductance $\dfrac {\dot Q}{\Delta T}$ rather than the thermal resistance $\dfrac {\Delta T}{\dot Q}$ in such a problem.