# Heat distribution in a long cylindrical electrical resistive element

I want to know what the maximum temperature will be within a heating element. Quite a few assumptions can be made, such as constant thermal conductivity, constant electrical resistivity, and assume current flows in equal proportions across the cross section of the element.

How, for example, can I calculate the temperature at the centre of a heating element, where:

• power dissipated per metre is 100W
• surface temperature is 100°c
• thermal conductivity is 100W/mK

It would be interesting also to have a formula for determining temperature at any value within the radius.

This link gives formulas for heat flow in a "cylindrical shell", but since heat is generated throughout the heating element, rather than the centre, I don't think it can be used.

My calculus is beyond rusty but here's my progress so far:

Attempt 1: Take a simplified case: rather than deal with a 2d circle, solve a similar one dimensional problem: a wall of thickness 1cm, which throughout its thickness generates heat. Call the centre of the wall x=0.

Logically, if the wall evenly generates heat, the outer parts of its width must pass more heat than the inner parts. In fact, the power dissipated by a cross section of the wall is proportional to the distance from its centre.

And since the temperature across each unit width is proportional to the power it dissipates, the outer parts of the wall will have a steeper temperature gradient than the inner.

So if the wall generates 100W (per square metre) then 50W will dissipate to each side through a thickness of 5mm. Each 1mm of which will create 10W. The inner 5th will dissipate 10W, the 2nd 5th will dissipate 20W, and so on, with the outer 5th dissipating 50W.

Say each 1mm of thickness dissipates 1W/°C, the temperature across each 1mm of thickness would be, from inner to outer, 10°C, 20°C, 30°C, 40°C, and 50°C. So the inner temperature would be the sum of those greater than the outer temperature. (150°C hotter).

It seems that the drop in temperature at x relative to x=0 is proportional to the square of x. Not sure how I got to that.

So now I need to turn this into real maths and extend it to the case of the cylinder.

• I assume you know about Fourier's heat equation? May 22, 2016 at 12:09

Your question is: assuming a rod of radius $R=1$ cm with a fixed heat production per unit volume, totaling $Q=100$ W/m, surface temperature $T_0$ and thermal conductivity $\lambda$, what is the temperature $T(r)$ as a function of $r$?

First, observe that within a cylindrical shell of radius $r$, the total heat production is $q=Q(r/R)^2$. The temperature gradient at this $r$ value must satisfy $$\frac{dT}{dr} = -\frac{q}{2\pi\lambda r}=-\frac{Qr}{2\pi\lambda R^2}.$$ (The heat generated inside this shell over a length $L$ is $qL$. The area of the shell is $A=2\pi rL$. Use $dT=qLdr/(\lambda A)$.) Solve this differential equation; with the boundary condition $T(R)=T_0$, we find $$T(r) = T_0 + \frac{Q}{4\pi\lambda}\left(1-\frac{r^2}{R^2}\right).$$ With your values, the temperature at the center would be 100.08 °C, which sounds like not so much, but then, a conductivity of 100 W/m-K is rather high (brass rod).

• Thanks - that's really helpful. How did you get those formulas? Did you derive them in some way? If so, I'd love to know how?
– CL22
May 25, 2016 at 7:01
• The formula starting with "dT/dr" follows directly from the definition of thermal conductivity: $q = \lambda A\Delta T/\Delta r$, with $A=2\pi rL$. The one starting with T(r) is obtained by taking the primitive of the first one and then using the boundary condition to get the integration constant. May 28, 2016 at 9:24

If you have not just a uniform heat emitting cylinder but namely solid electrical conductor of certain non-zero diameter then consider that the current travels along its surface layer, not through inner layers. So happens because all electrons are charged '-' thus want to push away each other. Not too rough to say that in this system the heat is produced only at the outer layer, but nothing by the inner layers. Once you taken boundary condition of T0 at the surface then you get the same temperature all across the section because the only way the heat can get into the inner layers is conductivity that can only go from higher temperature to lower. For me this idea works if you have neither cooling nor heating inside the cylinder and if you deals with steady problem.

• If a conductor carries current, the amount of charge does not change. You may be confused with the skin effect, which is an issue with AC currents in good conductors and which has an entirely different origin than "repelling charges". May 23, 2016 at 6:44
• Oh yes-yes-yes, I was thinking about the skin effect, you are right. Agree, this is not the case. May 23, 2016 at 7:43