1
$\begingroup$

I have the following differential equation that is purported to represent the equilibrium temperature at a point $x\in [0,L]$ on an uninsulated rod of length $L$, whose end points are kept constant $T(0)=T_1$ and $T(L)=T_2$, in air at temperature $T_a$:

$$\frac{d^2T}{dx^2}+h'(T_a-T(x))=0, \qquad(\star)$$ where $h'$ is a constant (which when increases leads to more heat transfer between the rod and the air).

If I solve this I get a solution, with a u-shape when $T_a<\min(T_1,T_2)$, and an n-shape if $T_a>\max(T_1,T_2)$.

The problem is I have no idea where it comes from. It seems to be a mélange of the Heat Equation:

$$k\cdot \frac{\partial^2T}{\partial x^2}=\frac{\partial T}{\partial t},$$

and Newton Cooling: $$\frac{dT}{dt}=r\cdot (T_a-T(t)),$$

however I cannot seem to put the two sticks together. A totally naive approach gives:

$$\frac{d^2T}{dx^2}=\underbrace{\frac{r}{k}}_{=h'}(T_a-T(x)).$$

However this is wrong in that is gives the wrong concavity (and doesn't really make any sense --- for equilibrium surely $\displaystyle \frac{\partial T}{\partial t}=0$).

Perhaps this term is coming from some kind of boundary condition on the rod (+axial symmetry)? Perhaps there is a simple sign-change as the temperature change is going in some opposite direction.

Can anyone shed some light on equation ($\star$)? Is it just a toy model?

$\endgroup$
1
  • $\begingroup$ Someone had made pointless edit and made a mistake in that so I returned the question to its perfectly fine original form. $\endgroup$ Nov 12 '19 at 19:25
3
$\begingroup$

The equation comes from adding a source term to the diffusion equation:

$$k \frac{\partial^2 T}{\partial x^2} + r (T_a -T(x)) = \frac{\partial T}{\partial t} $$

and then assumes steady state:

$$\frac{\partial^2 T}{\partial x^2} + h^\prime (T_a - T(x)) = 0$$

where $h^\prime = r/k$. It is basically the addition of two models (diffusion, cooling) and assuming steady state. Your confusion may stem from trying to plug one model into another, rather than combining them.

$\endgroup$
8
  • 1
    $\begingroup$ Are we saying something like: the temperature at a point is changing with respect to time due to heat transfer in/along the rod PLUS Newton Cooling to the air? That makes sense. $\endgroup$ Nov 12 '19 at 17:49
  • $\begingroup$ This equation arose from the 3D model of heat exchange of the rod with the surrounding air $\lambda \nabla ^2 T=0$ with Neuman value on a surface $-\lambda \nabla T.\vec {n}=h(T_a-T)$. After averaging over the cross section of the rod in plane $(y,z)$ , we obtain a 1D model. $\endgroup$ Nov 12 '19 at 18:14
  • $\begingroup$ @Alex that sounds like an alternative answer. $\endgroup$ Nov 12 '19 at 19:26
  • 1
    $\begingroup$ @JPMcCarthy OK! I will expand my comment to an answer with an example of 3D and 1D model calculations. $\endgroup$ Nov 12 '19 at 20:08
  • $\begingroup$ @JPMcCarthy The approach Alex Trounev is referring to is correct also, but it comes from the exact opposite direction as this approach (but both will have the same end result). This approach is taking toy mathematical models and assembling them to be something physical that is "useful" (for some definition of useful). The other approach is to start at physical -- i.e. "energy must be conserved" -- and apply as many approximations as needed to come to a mathematical model that is tractable, while still being "useful." Both directions are useful and it just depends on your starting point. $\endgroup$
    – tpg2114
    Nov 12 '19 at 22:11
2
+50
$\begingroup$

This equation arose from the 3D model of heat exchange of the cylindrical rod with the surrounding air $$\lambda \nabla ^2 T=0$$ with Neuman value on a surface $$-\lambda \nabla T.\vec {n}=h(T_a-T), r=R$$ and with Dirichlet conditions at the ends $$T(r,0)=T_1, T(r,L)=T_2$$ What does the solution to this problem look like? Put $L=4, R=0.25,T_1=2,T2=1,T_a=1,h=0.25, \lambda =1$, then the 2D and 3D temperature distributions along the length of the rod look like Figure 1

Now we want to build a 1D model to describe the temperature distribution along the length of the rod. We use Laplace equation in cylindrical coordinates $$\lambda\frac {1}{r}\frac {\partial}{\partial r} (r\frac {\partial T}{\partial r})+\lambda \frac {\partial ^2T}{\partial z^2}=0 . (1) $$ We assume that the temperature distribution is almost uniform along the radial coordinate, therefore $$\int _0^R {T(r,z) 2\pi rdr}=\pi R^2 T(z)$$ We multiply equation (1) by $2\pi rdr$, integrate and use the boundary condition on the surface $$2 \pi \lambda R\frac {\partial T}{\partial r}|_{r=R}+\pi R^2\lambda \frac {\partial ^2T}{\partial z^2}=0 . (2) $$ Finally we have $$\frac {d^2T}{d z^2}+h'(T_a-T)=0$$ with $h'=\frac {2h}{\lambda R}$. Replacing here $z\rightarrow x$, we arrive at the equation under discussion. Now compare the two solutions on the axis (left) and on the surface (right). We see a good match. Figure 2

$\endgroup$
6
  • $\begingroup$ +1: Nice way of adding a source term into the equations via the boundary conditions. I wonder how this would look tho for a gas, where the radiaitve energy transport can be non-local. $\endgroup$ Nov 13 '19 at 12:58
  • $\begingroup$ @AtmosphericPrisonEscape Thank you. State exactly what you mean under "the radiaitve energy transport can be non-local". $\endgroup$ Nov 13 '19 at 13:14
  • $\begingroup$ Alex. Thank you. I will not unaccept the answer given above (it does my needs), but I will give you some bounty for this answer. $\endgroup$ Nov 13 '19 at 13:30
  • $\begingroup$ ... wait do ye not have bounties here? I cannot add a bounty (yet?). $\endgroup$ Nov 13 '19 at 13:30
  • 1
    $\begingroup$ We have, but only with sufficient points, I think:) $\endgroup$ Nov 13 '19 at 13:42

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.