# How can I find the heat rate in a cylinder which has two temperatures?

#### Chemist116

The problem is as follows:

A coaxial cylinder has an interior radius $r_{1}$ and a temperature $T_{1}$ and a external radius $r_{2}$ and a temperature $T_{2}$ and has a height $h$. Assuming the thermal conductivity is $k$. Find the heat rate which is flowing radially.

The alternatives given are as follows:

$\begin{array}{ll} 1.&2\pi kh\frac{T_2-T_1}{\ln\frac{r_1}{r_2}}\\ 2.&2\pi kh^2\frac{T_1-T_2}{r_2-r_1}\\ 3.&2\pi kh\frac{T_1-T_2}{\ln\frac{r_2}{r_1}}\\ 4.&2\pi kh^2\frac{T_2-T_1}{r_2-r_1}\\ 5.&2\pi kh\frac{T_2-T_1}{\ln{r_2}{r_1}}\\ \end{array}$

I'm not exactly how to handle the integration for this problem. I'm assuming that the intended principle for this problem will be given by:

The Fourier's law is:

$q=-k A \frac{dT}{dx}$

Hence the area in the coaxial cylinder will be:

$A=(\pi r_2^2-\pi r_1^2)=\pi(r_2^2-r_1^2)$

$q=-k \pi(r_2^2-r_1^2) \frac{dT}{dx}$

I'm assuming that the integration is between $T_1$ and $T_2$ but I don't know how to assemble the Fourier Biot equation to adequately integrate it. Can someone help me here?. Since I dont know how this process is happening. Can someone include some sort of sketch or diagram to see how is the direction of the heat flowing?. As I don't understand how is the heat being integrated here.