1
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ 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? $\endgroup$ – Gert Mar 22 '20 at 15:48
  • $\begingroup$ @Gert Yes, I cannot find an equation to describe the amount of heat into the cross-section. $\endgroup$ – yuanming luo Mar 22 '20 at 16:09
  • $\begingroup$ OK, Will try and answer. $\endgroup$ – Gert Mar 22 '20 at 16:11
  • $\begingroup$ 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. $\endgroup$ – yuanming luo Mar 22 '20 at 16:23
  • $\begingroup$ Are you familiar with the transient heat conduction equation in 1D? $\endgroup$ – Chet Miller Mar 22 '20 at 17:40
2
$\begingroup$

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

Cylinder

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):

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

Furthermore, because of the insulation there's no radial temperature gradient, so:

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

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:

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

$\endgroup$
9
  • $\begingroup$ Edit made: eliminated the confusing dual use o the symbol $T$. $\endgroup$ – Gert Mar 22 '20 at 18:12
  • $\begingroup$ Now justify why the Ansatz yields a complete solution :p $\endgroup$ – user224659 Mar 22 '20 at 18:52
  • $\begingroup$ 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. $\endgroup$ – user224659 Mar 22 '20 at 18:55
  • $\begingroup$ Trust me: the dreaded non-homogeneous BCs are a bigger problem! $\endgroup$ – Gert Mar 22 '20 at 19:48
  • $\begingroup$ This is a 1D problem, and the temperature does not vary with r. It is only a function of x and t. $\endgroup$ – Chet Miller Mar 22 '20 at 21:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.