The classical heat diffusion equation 1D is of the form:
$\begin{align} \rho C_p \frac{\partial T}{\partial t} = \frac{\partial (k \frac{\partial T}{\partial x})}{\partial x} \end{align} $
with $\rho$, the density, $C_p$, the specific heat capacity, $T$ the temperature, and $k$ the thermal conductivity.
I want to solve numerically a problem where I only have an expression for $\alpha$, the thermal diffusivity, that depends on T, so when it is not constant in space and in time.
$\alpha$ is formally simply equal to:
$\begin{align} \alpha = \frac{k}{\rho C_p} \end{align} $
I have seen papers where they solve:
$\begin{align} \frac{\partial T}{\partial t} = \frac{\partial (\alpha \frac{\partial T}{\partial x})}{\partial x} \end{align} $
My question is simple: what is the justification to put $\alpha$ in the derivative, when clearly $\rho$ and $C_p$ are not in it in the first expression? Is that fine to do so if $\rho$ and $C_p$ are also changing through space?
Thanks in advance,
If I understand correctly, in their model they assume $\kappa$ and $C_p$ depend on the temperature $T$ itself. So yes, in principle they will indirectly depend on space and time. Starting from your formula, it is clear that if $C_p$ depends on $T$, you will not end up with the equation of the paper. You would get an error term of the following form if my calculation is correct: $$ -\frac{k}{\rho}\frac{C_p’(T)}{C_p^2(T)} (\partial_xT)^2 =-\alpha\frac{C_p’(T)}{C_p(T)} (\partial_xT)^2. $$ I suspect they are making a simplification coming from the physics of the problem. One could try looking at the above error term and understand if it is negligible in the regime of the paper. It looks not so complicated, but I don’t see it immediately. You could try asking on PhysicsSE if you are interested in these considerations.