# How to describe heat transfer between two solid materials?

A general equation for dealing with heat transfer between one material and a region of insulating material. I've seen basic heat transfer equations for one material, but I'd love to see an explanation of how to do two.

-
Also look at "contact resistance"! – drN Mar 16 '13 at 2:01

The one-dimensional heat equation for a solid can be written as: $$\rho C_p\frac{\partial T}{\partial t}= -\frac{\partial}{\partial x} \left( k\frac{\partial T}{\partial x} \right) +\sigma$$ where $\sigma$ is the source term and $\dot q =-k\frac{\partial T}{\partial x}$ is the diffusive heat flux. At the boundary the temperature and the flux must be continuous (if we consider contact resistance negligible, otherwise a gap in temperature could be possible) that is: $$T_1=T_2$$ $$\dot q_1=\dot q_2$$

-
Are you certain that the temperature should be continuous across the interface between the two materials? In the steady state, I agree that the heat flux must be continuous (as long as there are no other sources of energy around). However, as written in my response here physics.stackexchange.com/questions/52900/… there is at least one example for which different thermal conductivities on either side of the interface leads to a discontinuity in temperature. – joshphysics Mar 15 '13 at 22:56
I fear your answer is wrong. Your example is pretty general, and since "natura non facit saltus" at least as long as we don't consider an atomic scale nor a contact resistance, I think, completely wrong. Notice also that you state that boundary condition is $T_0$ on both sides of the problem, which leads to a constant solution. Ergo your demonstration is at least misleading. – Lorenzo Mar 16 '13 at 7:45
I'm not sure I agree. See, for example, the addendum in my response. – joshphysics Mar 16 '13 at 17:44
I notice that in your last equation L is missing. But this is not the point. You have used fixed temperatures for both boundary condition, so the flux is not fixed. Can't it be that the flux is exactly $\frac{T_R-T_L}{L} \frac{k_b k_a}{k_a-k_b}$ ? – Lorenzo Mar 16 '13 at 18:54
What do you think about it? – Lorenzo Mar 17 '13 at 22:25

Basically you can use Fourier's Law $$q = -k\frac{dT}{dx}$$ with the appropriate boundary conditions between the two materials. The basic issue is that at the interface between the two materials, there is a jump discontinuity in the value of the thermal conductivity, and you have to take this into account in solving the equation.

I did a very detailed calculation in a related post Heat transfer between two surfaces that you might find useful in this regard.

Also, if you want to actually calculate something about something in the real world, you may find this list of thermal conductivities useful.

Consider two bars of length $L$ and of uniform (but unequal) thermal conductivities $k_a$ and $k_b$. Let the heat flow $q_0>0$ be constant along the bars, then Fourier's Law shows that the temperatures $T_a$ and $T_b$ of bars $a$ and $b$ satisfy $$T_a(x) = -\frac{q_0}{k_a} x + C_a, \qquad T_b(x) = - \frac{q_0}{k_b}x+C_b$$ for some constants $C_a$ and $C_b$. Now suppose that the left end of bar $a$ is located at $x=-L$ and the right end of bar $b$ is at $x=L$ so that they are joined at $x=0$. Suppose additionally that the left end of bar $a$ is at temperature $T_L$ and the right end of bar $b$ is at $T_R$, then we have the boundary conditions $$T_a(-L) = T_L, \qquad T_b(L) = T_R$$ which tells us that $$C_a = T_L -\frac{q_0}{k_a} L, \qquad C_b = T_R+\frac{q_0}{k_b}L$$ So that $$T_a(x) = T_L - \frac{q_0}{k_a}(x+L), \qquad T_b(x) = T_R -\frac{q_0}{k_b}(x-L)$$ In particular, at $x=0$ we find $$T_b(0)-T_a(0) = (T_R-T_L)+q_0\left(\frac{L}{k_a}-\frac{L}{k_b}\right)$$ In particular, there is, in general a jump discontinuity in the temperature at the interface between the two materials unless the temperatures at which the ends of the bars are being kept are related by $$T_R - T_L = q_0\left(\frac{L}{k_b}-\frac{L}{k_a}\right)$$
This is only true of $q$ is assumed constant throughout the domain, which it won't be in most of the cases. And if it is, it can be calculated from your last equation. – Nico Schlömer Apr 7 '13 at 20:14
@Nico Agreed. If $q$ varies with position, then my argument certainly breaks down. – joshphysics Apr 8 '13 at 1:33