# Heating transient of a composite rod?

A straight rod is made of two parts, $[0,x_1]$ (green in the figure) with thermal diffusivity $\kappa_1$ and $[x_1,x_2]$ (blue) with thermal diffusivity $\kappa_2$. The rod is perfectly insulated. Zero $y$ and $z$ temperature gradients are assumed.

At $x=0$ temperature is maintained at constant $T_0$. At $x=x_2$ the rod is embedded into a perfect insulator ($\kappa=0$). At $t=0$ the rod has a uniform temperature $T(x,0)=T_i$.

Question: what is the temperature evolution of the rod?

1. The simple case where $\kappa_1=\kappa_2=\kappa$:

Let $u(x,t)=T(x,t)-T_0$.

Then Fourier's equation tells us:

$$u_t=\kappa u_{xx}$$

Boundary conditions:

$$u(0,t)=0$$ $$u_x(x_2,t)=0$$

Initial condition:

$$u(x,0)=u_i=T_i-T_0$$

Using the Ansatz $u(x,t)=X(x)\Gamma(t)$, separation constant $-k^2$ and the boundary conditions above, this solves easily to:

$$\Large{u(x,t)=\displaystyle \sum_{n=1}^{+\infty}B_n\sin\Bigg(\frac{n\pi x}{2x_2}\Bigg)e^{-\kappa \Big(\frac{n\pi }{2x_2}\Big)^2t}}$$

(for $n=1,3,5,7,...$)

The $B_n$ coefficients can easily be obtained from the initial condition with the Fourier sine series:

$$B_n=\frac{4u_i}{n\pi}$$

Back-substituting we get:

$$T(x,t)=T_0+\frac{4(T_i-T_0)}{\pi}\displaystyle \sum_{n=1}^{+\infty}\frac{1}{n} \sin\Bigg(\frac{n\pi x}{2x_2}\Bigg)e^{-\kappa \Big(\frac{n\pi }{2x_2}\Big)^2t}$$

(for $n=1,3,5,7,...$)

A plot for the first three terms at $t=0.1$:

2. The case where $\kappa_1\neq\kappa_2$:

We define two functions $u_1(x,t)$ for $[0,x_1]$ and $u_2(x,t)$ for $[x_1,x_2]$. We use the same Ansatz as under $1.$ We'll assume both functions have their own eigenvalues.

Boundary conditions:

$$u_1(0,t)=0\implies X_1(0)=0\tag{1}$$ $$\frac{\partial u_2(x_2)}{\partial x}=0\implies X_2'(x_2)=0\tag{2}$$

$$u_1(x_1,t)=u_2(x_1,t)\tag{3}$$

With Fourier, the heat flux is the same at $x=x_1$:

$$\alpha_1\frac{\partial u_1(x_1)}{\partial x}=\alpha_2\frac{\partial u_2(x_1)}{\partial x}\tag{4}$$

Where $\alpha_i$ are the thermal conductivities.

a. for $u_1(x,t)$:

$$X_1(x)=c_1\cos k_1x+c_2\sin k_1x$$ $$X_1(0)=0\implies c_1=0\implies X_1(x)=c_2\sin k_1x\tag{5}$$

b. for $u_2(x,t)$:

$$X_2(x)=c_3\cos k_2x+c_4\sin k_2x$$ $$X_2'(x_2)=0\tag{2}$$ $$\implies -c_3k_2\sin k_2x_2+c_4k_2\cos k_2x_2=0\tag{6}$$ Using the additional conditions $(3)$ and $(4)$:

$$c_2\sin k_1x_1=c_3\cos k_2x_1+c_4\sin k_2x_1\tag{7}$$ $$c_2\alpha_1k_1\cos k_1x_1=-c_3\alpha_2k_2\sin k_2x_1+c_4\alpha_2k_2\cos k_2x_1\tag{8}$$

Problem:

$(6)$, $(7)$ and $(8)$ form a system of three simultaneous equations but with five unknowns: $c_2$, $c_3$, $c_4$, $k_1$ and $k_2$.

I'm tempted to set $c_3=0$ as it would yield $k_2$ from $(6)$. I think this would yield also the remaining unknowns. But can I a priori assume $c_3=0$? Or is there another approach possible?

I'm also left wondering whether perhaps $k_1=k_2$. The eigenvalues do not depend on $\kappa$, so perhaps the eigenvalues $k$ are common to both functions. Due to $(4)$, $u_1$ and $u_2$ would then still be distinct.

• The time scale for both regions has to be the same. So, I think you are going to find that $$\frac{k_1}{k_2}=\sqrt{\frac{\kappa_2}{\kappa_1}}$$ I don't know whether this will help or not. This is as far as I've gotten into the problem so far. Commented Oct 13, 2016 at 12:41
• @ChesterMiller: interesting thought!
– Gert
Commented Oct 13, 2016 at 12:43
• Also. It looks like you are assuming that the thermal conductivities of the two regions are the same (even though the thermal diffusivities are different). Did you really want to do that? Commented Oct 13, 2016 at 12:45
• No, I'm not assuming that: see $(4)$.
– Gert
Commented Oct 13, 2016 at 12:47
• Oh. OK. I missed that. Commented Oct 13, 2016 at 12:48

I can see how to do this now. You have 3 BC equations, and 3 unknown c's plus the unknown time eigenvalue $\lambda^2$ (You know that $k_1=\lambda /\sqrt{\kappa_1}$ and $k_2=\lambda /\sqrt{\kappa_2}$). You can eliminate 2 of the c's. Your third equation will give you a relationship with the 3rd c out front and a zero on the other side of the equation. This will allow you to determine $\lambda$. The last c is then determined by making good on the initial condition, using an infinite series.

$$\frac{1}{T}\frac{dT}{dt}=\frac{\kappa}{X}\frac{d^2X}{dx^2}=-\lambda^2$$ So, $$\frac{d^2X}{dx^2}+\frac{\lambda^2}{\kappa}X=0$$ If the time scale for both regions is the same (as expected), then: $$k_1=\frac{\lambda}{\sqrt{\kappa_1}}$$and$$k_2=\frac{\lambda}{\sqrt{\kappa_2}}$$

• Ok. I've developed another approach, so now I need to compare the two. Will try and post tonight, if not tomorrow. Thanks for your help!
– Gert
Commented Oct 13, 2016 at 15:57
• A quick question: what's the rational basis for $\frac{k_1}{k_2}=\sqrt{\frac{\kappa_2}{\kappa_1}}$?
– Gert
Commented Oct 13, 2016 at 16:09
• This problem should also be analyzed in detail in Carslaw and Jaeger. Commented Oct 13, 2016 at 16:09
• "Carslaw and Jaeger" Huh?
– Gert
Commented Oct 13, 2016 at 16:26
• @Gert See ADDENDUM above. Conduction of Heat in Solids by Carslaw and Jaeger Commented Oct 13, 2016 at 16:58

This uses parameter symbols that I'm more accustomed to: k is thermal conductivity, $\alpha$ is thermal diffusivity, $\rho$ is density, and $c_p$ is heat capacity.

I followed the procedure which I outlined in my previous answer and obtained the following:

The characteristic equation for the n'th eigenvalue is: $$\tan\left[\frac{\lambda_n(x_2-x_1)}{\sqrt{\alpha_2}}\right]\tan\left[\frac{\lambda_nx_1}{\sqrt{\alpha_1}}\right]=\sqrt{\frac{k_1\rho_1c_{p1}}{k_2\rho_2c_{p2}}}$$ The n'th eigenfunction is: $$X_n=\cos\left[\frac{\lambda_n(x_2-x_1)}{\sqrt{\alpha_2}}\right]\sin\left[\frac{\lambda_nx}{\sqrt{\alpha_1}}\right]\tag{0 < x <x1}$$ $$X_n=\cos\left[\frac{\lambda_n(x_2-x)}{\sqrt{\alpha_2}}\right]\sin\left[\frac{\lambda_nx_1}{\sqrt{\alpha_1}}\right]\tag{x1<x<x2}$$ I leave it up to you to do the integation to get the n'th coefficient in the infinite series.