# Relationship between heat transfer and electricity How can it be proven that the heat transfer through a wall is the temperature difference divided by the total resistance? Can this be proven without using concepts related to electronics for example by using the fact that the rate of heat transfer by convection at the inner surface is the same as that by conduction in the wall, which is equal to the rate of heat transfer by convection through the outer surface?

• Thereisn't really a 'relationship' between heat transfer and electricity. There's an analogy between the application of Fourier (here) and Ohm's law of resistance/voltage. – Gert Jan 24 at 23:13
• This is the proof you're looking for: sfu.ca/~mbahrami/ENSC%20388/Notes/… – Gert Jan 25 at 11:28

## 1 Answer

If you have $$T_{in}$$ inside and $$T_{out}$$ outside fixed, the heat rate will be given by $$\dot{Q} \propto (T_{out}-T_{in})$$. A qualitative (but not correct) way of seeing it is to assume that the rate of heat transfer at one position $$\dot{Q}$$ is to a first approximation going to be proportional to the temperature [see later for a detailed explanation of why and when this is possible].

This means that you have a heat rate from inside (cold) given by roughly $$\dot{Q}_{in}=\alpha T_{in}$$ and the same thing from outside to inside $$\dot{Q}_{out}=\alpha T_{out}$$ where $$\alpha$$ is some constant given by your model (you can find it defined in detail later, I am just naming it $$\alpha$$ indicating "some constant").

You then have a net heat flow rate over the whole wall $$\dot{Q}=\dot{Q}_{in}-\dot{Q}_{out}=\alpha(T_{in}-T_{out})$$

In this approximation, the equation above looks like Ohm's law for a circuit in that the heat flow (=current) across a resistence (=wall, modelled by $$\alpha$$) is proportional to the drop in temperature (=voltage). If you call $$\alpha=1/R$$ then

$$\Delta \dot{Q}=\Delta T/R$$

which looks like Ohm's law. So the electrical analogy is just a similarity, coming from the fact that the two phaenomena share a similar equation if you consider the steady state ($$T(x)$$ does not change in time).

If you want to know precisely $$\alpha$$ you need to solve Fourier's heat transfer equation according to your model for the conductivity, the radiation etc. etc. but you will always find (in the steady state, where the heat flow is constant over time, and in simple geometries) something like $$\dot{Q}\approx \Delta T/R$$ because it stems directly from the steady state of the heat equation for temperature $$\nabla^2 T=0$$ which looks like Poisson's equation for the electric potential $$\nabla^2 \phi=0$$ which is at the basis of Ohm's law in a circuit.

Now we just need to derive why the heat transfer rate can be assumed, in a rough approximation, proportional to the temperature.

Assume a 1D model.

Consider Fourier's equation

$$q=-k\partial_x T$$

$$q$$ is the heat flux given by $$\dot{Q}/A$$ where $$A$$ is the area of your wall. SO

$$\dot{Q}=-Ak\partial_x T$$

You would have to solve this differential equation in general but we know from Fourier's equation that in the steady state (no changes in temperature over time) the temperature is linear (assuming the conductivity is constant) i.e. $$T(x)=T(0)+(T(L)-T(0))*x/L$$ where $$L$$ is the thickness of the wall and $$T(0)=T_{in}$$ and $$T(L)=T_{out}$$. This also comes from Fourier's equation +conservation of energy.

So we just differentiate using the expression of $$T(x)$$ we found

$$\dot{Q}=-Ak\partial_x (T_{in}+(T_{out}-T_{in})*x/L)$$ and get

$$\dot{Q} = Ak/L (T_{in}-T_{out})$$ so that is what we were looking for i.e. the heat transfer at the two sides of the wall is $$\Delta \dot{Q} = Ak/L \Delta T$$ which looks like Ohm's law as before and we also find the value we were looking for: $$\alpha=Ak/L$$.

This formula holds if you look for $$\dot{Q}$$ over the whole wall because we integrate $$T(x)$$ from 0 to $$L$$, otherwise you need to "break" the wall into little pieces of length $$l$$ and then the formula is $$\dot{Q} = Ak/l (T_{x}-T_{x+l})$$

(More in general, you can do the rough approximation $$\dot{Q}=-Ak\partial_x T\approx -Ak{\Delta T \over \Delta x}$$ and treat $$\Delta x$$ not as in infinitesimal quantity but as a thickness).

What we considered so far shows that the difference in heat flux over the wall is proportional to $$\Delta T$$ so to justify that the local rate of heat is also proportional to the local $$T$$ (in the small $$dx$$ limit) we can consider a tiny tiny slice of wall with thickness $$dx$$ where the temperature varies so little that it is constant, we can say that the heat rate through that slice is going to $$\dot{Q} = Ak/dx (T_{x}-T_{x+dx})\approx Ak/dx\bar{T}$$ where $$\bar{T}$$ is sort of an average temperature, justifying our initial assumption. If $$dx\to 0$$ of course you would need to take the gradient, but as a physicist you are fine with an average temperature so that you get the electrical analogy. This all fails if $$T(x)$$ is not linear and it decays very fast so that changing the gradient with a finite difference become a big error.

The electrical analysis also means that you can sum resitances in series and sum the inverse of resistances in parallel, but it stays a formal analogy for two kinds of phaneomena which are very different. They only share the fact that there is a "current" of some kind ($$I$$ for Ohm's law, $$\dot{Q}$$ for heat transfer) which is linear in the "drop" of somenthing else ($$\Delta T$$ or $$\Delta V$$). The proportionality coefficient is called $$R$$ (resistance).

Notice that this analysis holds for conduction in 1D and for conduction of heat. Similar reasoning can be done, if $$\dot{Q}\approx dT/dx$$ for other kinds of heat transfer and different geometries.

• If you have $T_{in}$ inside and $T_{out}$ outside, you can assume that the heat transfer q is going to be proportional to the temperature. Nope. The rate of heat flux is given by Fourier: $q=-k\partial_x T$. So it's proportional to the temperature gradient. – Gert Jan 24 at 23:08
• where α is some constant given by your model. No. It should really be $k$, the thermal conductivity of the material, not "some constant". – Gert Jan 25 at 1:02
• Because in the steady state $T(x)=T_A+(T_B-T_A)x$ if you integrate the heat equation between $A$ and $B$ you find $\partial_xT(x)=(T_B-T_A)$ so the heat rate between $A$ and $B$ is proportional to the difference in temperature. Which means that locally the rate is proportional to $T/dx$ where $dx$ is small. The answer was phrased poorly before, now it should be more clear – JalfredP Jan 25 at 1:41
• Your answer doesn't even remotely answer the OP's question. It's hotchpotch of ill digested ideas, half truths and nonsense. I will elaborate tomorrow. – Gert Jan 25 at 5:39
• sfu.ca/~mbahrami/ENSC%20388/Notes/… – Gert Jan 25 at 11:29