Heat conduction through wall - alternatives?

Question

This is basically a physics question, though the motivation is DIY.

There is a wood burning stove on one side of a wall in a room. It is used for cooking. It has an external air supply.

On the other side of the wall in the adjacent room, there is an air-intake that moves air through a recirculating circuit of a mechanical ventilation unit. The purpose of that air intake and circuit is to mix warm air with other air supplies and redistribute heat around the house.

The air intake must be on the other side of the wall in order to avoid recirculating exhaust air from near the cook-stove.

The wall is 40cm thick and made of insulating porous brick

I have two options to increase the amount of heat available to that air intake, where the heat must come from the cook stove, which is next to the wall:

• I can cut a large hole in the wall behind the cook stove, about the size of the cook stove, and insert a large cast iron square, completely filling the hole, acting as a heat conductor better than the porous insulating brick currently in place (let us say 3 x 3 m of thickness 0.07m )
• I can place two thin copper sheets of the same size as the above cast iron square(3m x 3m of thickness 0.5mm) on either side of the wall connected by a thick copper rod of 10cm diameter through a 10cm hole in the wall.

Which of the two variants would deliver the most heat to the air intake on the other side of the wall?

Answer 1

Heat conduction through a wall, with convection at both sides, can be schematised as follow:

The heat flux $q$ (in $\mathrm{W/m^2}$) is given by:

$$q=\frac{T_1-T_2}{R}$$

where $R$ is the overall thermal resistance of the wall:

$$R=\frac{1}{h_1}+\frac{L}{k}+\frac{1}{h_2}$$

With:

• the $h_i$ the (convective) heat transfer coefficients of the respective convection zones,
• $k$ the thermal conductivity of the wall, with thickness $L$.

Multiply with the surface area $A$ to get the actual heat transfer (per unit of time), $Q=qA$.

Consider the wall itself to have a very high $R$ ($q \approx 0$), then compare the estimated heat transfers $Q$ for both solutions you have in mind, using relevant data.

It's sufficient to estimate $R$ for both scenarios: the lowest $\frac{A}{R}$-ratio wins the contest.

My best bet would be neither of your scenarios but rather a thinish (see the influence of $L$ on $R$) sheet of copper with, say, $3\mathrm{m} \times 3\mathrm{m}$ of surface area. And this solution will also be cheaper!

Further reading.