# Heat conduction through hollow cylinder with constant internal and external temperature

Consider a hollow cylinder of inner radius $$r_1$$ and outer radius $$r_2$$. The entire surface of the cylinder is kept at constant (but distinct) temperatures along the interior and exterior.

I found that the steady state temperature function is (in cylindrical coordinates) $$T = -kr_2 \log \left(\frac{r}{r_1}\right) + T_0.$$

I am then asked to find the heat flux through a Gaussian cylinder of radius $$r\in (r_1,r_2)$$. The answer given is that the flux per unit length $$\phi$$ is $$\phi = 2\pi r h_r,$$ where $$h_r$$ is the $$r$$ component of $$\nabla \mathbf{h}$$ (in cylindrical coordinates). My question is, how is this consistent with Gauss' Theorem (Divergence Theorem)? Because I found that $$\nabla \cdot \mathbf{h} = 0$$, so shouldn't the flux be $$\iint_S \mathbf{h}\cdot\mathrm{d}\mathbf{S} = \iiint_V \nabla\cdot \mathbf{h}\, \mathrm{d}V = \iiint_V 0\, \mathrm{d}V = 0?$$

Any help would be appreciated.

• You are that h is temperature T, right? You should be applying the divergence theorem to the gradient of h, not h. Commented Mar 10, 2023 at 12:00
• Wait, what do you mean? $\nabla\cdot \mathbf{h}$ is the divergence of $\mathbf{h}$. The divergence of the gradient of $\mathbf{h}$ is the Laplacian of $\mathbf{h}$; why do I want to use that? Commented Mar 10, 2023 at 12:04
• Because the steady state heat conduction equation is Laplacian of temperature equal to zero. Commented Mar 10, 2023 at 12:19
• But $\mathbf{h}$ is the heat, not the temperature... $T$ is the temperature. Commented Mar 10, 2023 at 14:09
• What are the units of h and phi? Commented Mar 10, 2023 at 15:27

Here is the correct development: $$\phi=2\pi r q_r$$where $$q_r$$ is the rate of heat flow per unit area in the radial direction: $$q_r=-k\frac{dT}{dr}$$where k is the thermal conductivity of the annular cylinder. These combine to give $$\frac{dT}{dr}=-\frac{\phi}{2\pi rk}$$integrating between $$r_1$$ and r gives:$$T-T_1=\frac{\phi}{2\pi k}\ln{(r_1/r)}$$The parameter $$\phi$$ can be obtained by applying the boundary condition at $$r_2$$: $$\phi=2\pi k\frac{(T_1-T_2)}{\ln{(r_2/r_1)}}$$
• Sorry, but I think you may have misunderstood my question. My problem isn't how we get from $\phi = 2\pi r h_r$ onwards; it is how we get to $\phi = 2\pi r h_r$ in the first place. Basically, I want to know how $\iiint_V \nabla\cdot \mathbf{h}\, \mathrm{d}V$ is evaluated (properly; not just handwaving that 'area is whatever') so that $\phi = 2\pi r h_r$. Commented Mar 11, 2023 at 14:44