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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.

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  • $\begingroup$ You are that h is temperature T, right? You should be applying the divergence theorem to the gradient of h, not h. $\endgroup$ Mar 10, 2023 at 12:00
  • $\begingroup$ 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? $\endgroup$
    – Trisztan
    Mar 10, 2023 at 12:04
  • $\begingroup$ Because the steady state heat conduction equation is Laplacian of temperature equal to zero. $\endgroup$ Mar 10, 2023 at 12:19
  • $\begingroup$ But $\mathbf{h}$ is the heat, not the temperature... $T$ is the temperature. $\endgroup$
    – Trisztan
    Mar 10, 2023 at 14:09
  • $\begingroup$ What are the units of h and phi? $\endgroup$ Mar 10, 2023 at 15:27

1 Answer 1

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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 $r_2$: $$\phi=2\pi k\frac{(T_1-T_2)}{\ln{(r_2/r_1)}}$$

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  • $\begingroup$ 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 $\phi = 2\pi r h_r$. $\endgroup$
    – Trisztan
    Mar 11, 2023 at 14:44

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