# How do I set up the differential equations for the finite length iron stick heat diffusion?

Consider an iron cylinder. The heat doesn't go through the cylinder surface, but at the circular cross-section in the endpoints. The temperature of the surroundings is $$15\ \mathrm{^\circ C}$$. The initial temperature of the iron stick is $$100\ \mathrm{^\circ C}$$. How to set up the differential equation? $$T(0,t)$$ and $$T(l,t)$$ doesn't fit anymore.

• Are you asking how the cylinder cools down, when its surface is insulated and heat is lost only through the cross sections at the ends of the cylinder? – Gert Mar 22 '20 at 15:48
• @Gert Yes, I cannot find an equation to describe the amount of heat into the cross-section. – yuanming luo Mar 22 '20 at 16:09
• OK, Will try and answer. – Gert Mar 22 '20 at 16:11
• The confusion is the endpoint of the iron cylinder loss heat and gets heat at the same point, I only know the rate of losing heat at a certain temperature but I cannot determine the rate of gaining heat. – yuanming luo Mar 22 '20 at 16:23
• Are you familiar with the transient heat conduction equation in 1D? – Chet Miller Mar 22 '20 at 17:40

So we have a cylinder of height $$H$$ and radius $$R$$:

Surrounding temperature is $$T_e=15\text{ C}$$ and initial, uniform temperature of the cylinder is $$T_0=100\text{ C}$$.

The cylinder is perfectly insulated, except for both circular ends, which lose heat through convection (only).

The go to equation for this kind of problem is Fourier's heat equation:

$$T_t=\alpha \nabla^2 T$$

Because of the geometry of the problem this equation begs for cylindrical coordinates:

$$\frac{\partial T}{\partial t}=\alpha \Big[\frac{1}{r}\frac{\partial}{\partial r}\Big(r\frac{\partial T}{\partial r}\Big)+\frac{1}{r^2}\frac{\partial^2 T}{\partial \phi^2}+\frac{\partial^2 T}{\partial x^2}\Big]$$

Due to symmetry considerations the $$\phi$$ term isn't needed and developing the partial derivatives, we get (using shorthand partials):

$$\frac{1}{\alpha}T_t=\frac{1}{r}T_r+T_{rr}+T_{xx}$$

$$T_r=T_{rr}=0$$

So:

$$\frac{1}{\alpha}T_t=T_{xx}$$

So we're looking for a function:

$$T(x,t)$$

which describes the spatial distribution of $$T$$, as well as its evolution in time $$t$$. Such a function would also allow to calculate the heat flux through the ends of the cylinder (as a function of time).

All real-world differential equations require boundary conditions and this case is no different.

Initial condition:

$$T(x,0)=100$$

Convection at ends:

$$T_t(r,+\frac{H}{2},t)=h(T-T_e)$$ and: $$T_t(r,-\frac{H}{2},t)=h(T-T_e)$$

where $$h$$ is the convection heat transfer coefficient.

It's advisable to make a small substitution:

$$u=T-T_e$$

So we have:

$$u(x,0)=85$$

$$u_t(+\frac{H}{2},t)=-hu$$

$$u_t(-\frac{H}{2},t)=-hu$$

Once $$u(x,t)$$ is found, we can convert it back to $$T(x,t)$$.

$$u=T-T_e$$

We assume (Ansatz):

$$u(x,t)=X(x)\Theta(t)$$ Inserting into the PDE we get:

$$X\Theta'=\alpha \Theta X''$$ Divide by $$u(x,t)=X(x)\Theta(t)$$:

$$\frac{\Theta''}{\alpha \Theta}=\frac{X''}{X}=-k^2$$

where $$k$$ is a Real number. So we have two ODEs:

$$\frac{\Theta''}{\alpha \Theta}=-k^2\tag{1}$$

$$\frac{X''}{X}=-k^2\tag{2}$$

$$(1)$$ solves to:

$$\Theta(t)=C \exp{(-k^2 \alpha t)}$$

And $$(2)$$ solves to:

$$X(x)=A \sin(kx)+B \cos(kx)\tag{3}$$

'All that is left to do' is determine the integration constants $$A$$, $$B$$ and $$C$$.

Unfortunately the BCs are of the Neumann type and thus non-homogeneous (non-zero). This generally creates an unsightly mess with no easy way from which to extricate $$A$$ and $$B$$.

Instead I'll try the much simpler case where the ends are kept at constant temperature $$u=0$$, so: $$u(-\frac{H}{2},t)=u(\frac{H}{2},t)=0$$ This also means: $$X(-\frac{H}{2})\Theta(t)=X(\frac{H}{2})\Theta(t)=0$$ Assume $$\Theta(t) \neq 0$$, thus: $$X(-\frac{H}{2})=X(\frac{H}{2})=0$$ Insert into $$(3)$$:

$$A \sin(k\frac{H}{2})+B \cos(k\frac{H}{2})\tag{4}=0$$

$$A \sin(-k\frac{H}{2})+B \cos(-k\frac{H}{2})=0$$ From the last equation: $$A \sin(k\frac{H}{2})-B \cos(k\frac{H}{2})=0\tag{5}$$ Now add $$(4)$$ to $$(5)$$:

$$2A\sin(k\frac{H}{2})=0$$

Assume $$A \neq 0$$, then:

$$\sin(k\frac{H}{2})=0$$

This happens for:

$$k\frac{H}{2}=n \pi$$ Or: $$k=\frac{2n\pi}{H}$$ For $$n=1,2,3,4,...$$

With $$\sin(k\frac{H}{2})=0$$, then $$B=0$$. So we get:

$$X_n(x)=A_n\sin\Big(\frac{2n\pi x}{H}\Big)$$

And:

$$u_n(x,t)=D_n \exp{\Big[-\Big(\frac{2n\pi}{H}\Big)^2 \alpha t\Big]}\sin\Big(\frac{2n\pi x}{H}\Big)$$

Applying the superposition principle:

$$u(x,t)=\sum_{n=1}^{+\infty}D_n \exp{\Big[-\Big(\frac{2n\pi}{H}\Big)^2 \alpha t\Big]}\sin\Big(\frac{2n\pi x}{H}\Big)$$

The coefficients $$D_n$$ are obtained with the initial condition and the Fourier series.

For $$t=0$$, $$u(x,0)=85$$ and:

$$85=\sum_{n=1}^{+\infty}D_n \sin\Big(\frac{2n\pi x}{H}\Big)$$

Thus:

$$D_n=\frac{2}{H}\int_{-H/2}^{+H/2}85 \sin\Big(\frac{2n\pi x}{H}\Big)\text{d}x$$

• Edit made: eliminated the confusing dual use o the symbol $T$. – Gert Mar 22 '20 at 18:12
• Now justify why the Ansatz yields a complete solution :p – user224659 Mar 22 '20 at 18:52
• I know that there is no easy justification for this in the general case.. Always left me kind of unsatisfied when first encountering this kind of problem in classical electrodynamics. – user224659 Mar 22 '20 at 18:55
• Trust me: the dreaded non-homogeneous BCs are a bigger problem! – Gert Mar 22 '20 at 19:48
• This is a 1D problem, and the temperature does not vary with r. It is only a function of x and t. – Chet Miller Mar 22 '20 at 21:45