Heat equation: Convection

Homework Statement

A thin rod of length ∏ is heated at one end to temperature $T_0$. It is insulated along its length and cooled at the other end by convection in a fluid of temperature $T_f$ . Find the transient and steady-state temperature distribution in the rod, assuming unit thermal diff usivity (K = 1) and unit heat transfer coefficent ( $\gamma = 1$). Also assume arbitrary initial temperature distribution.

Homework Equations

The heat equation: $u_t = u_xx$

The Attempt at a Solution

For one thing, I'm not sure I'm setting this up entirely correctly. The equation I'm attempting to solve is

$$u_t = u_xx$$

Subject to

$$\begin{eqnarray*} u(0,t) &= & T_0 \\ u(x,0) & = & f(x) \\ u_x(\pi,t) & = & u(\pi,t) - T_f \end{eqnarray*}$$

That last equation comes from Fourier's law of heat conduction and ensures that the heat flux is properly matched at the boundary. If this is the correct equation with the correct set of initial/boundary conditions, them I'm still lost. Even if I try separation of variables, clearly the eigenvalue problem has to be the problem in x. However, since you don't have a pair of homogeneous boundary conditions. Even if you were to transform the first one into a homogeneous boundary condition, I have no idea what you'd do with the Robin one.

One last note: We did a similar example in class. I don't think that it was stated that the rod was insulated along its length (in particular, this example was a cooling fin). The equation we had in that case was:

$$\rho c u_t = u_xx - (p/A)(T-T_f)$$

where $\rho,c,p,A$ were density, specific heat, perimeter, and area, respectively. However, we arrived at this equation by considering a small slice of surface area ΔS, generated by looking at a small slice of length Δx, and equating the rate of increase of stored energy in the corresponding ΔV with heat flow by conduction - heat flow by convection through ΔS. But if it's insulated along its length, wouldn't conduction be the only mechanic at work?

Thank you so much!

Alright, I tried making a change of variables, and here's what I got:

I defined $u^0(x,t) = \frac{T_f-T_0}{1+\pi}x + T_f$. This satisfies both boundary conditions. Then define $v(x,t) = u(x,t) - \frac{T_f-T_0}{1+\pi}x - T_f$. Then v(x,t) satisfies the same old heat equation, with boundary conditions $v(0,t) = 0, v_x(\pi,t)-v(\pi,t) = 0$.

Then using separation of variables, you get

$$v(x,t) = \sum_{n=1}^{\infty}\ c_n e^{-\lambda^2t}\sin\lambda x$$

where $\lambda$ is given implicitly as the roots of the equation $\lambda = \tan\lambda\pi$ (which looks familiar, like something out of a PDE course I took last year). The coefficients are given in the usual Fourier series way, but you can't really come up with an analytic form for them, since it involves integrating $\sin\lambda x$. Then the final answer is

$$u(x,t) = \frac{T_f-T_0}{1+\pi}x + T_f + \sum_{n=1}^{\infty}\ (\int_0^{\pi}\ \sin\lambda_nx (f(x)-\frac{T_f-T_0}{1+\pi}x - T_f)\ dx) e^{-\lambda^2t}\sin\lambda x$$

But that solution feels weird

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