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.
Best 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.