# How should the integral be used when trying to find the heat rate in a cylinder which has two temperatures?

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.

In Conduction in the Cylindrical Geometry, your question (apart from them using $$L$$ for the cylinder (which they refer to specifically as a pipe) length instead of $$h$$) is explained and answered in quite of bit of detail (including several diagrams), with their result near the bottom of page $$4$$ being

$$Q = \ldots = 2\pi Lk\frac{T_1 - T_2}{\ln(r_2/r_1)}$$

This corresponds to your alternative #$$3$$. I'll outline what's written there.

As indicated, they use a balance of heat in & heat out approach using cylindrical shells, going the entire length $$h$$, of inner radius $$r$$ and outer radius $$r + \Delta r$$.

Have $$Q(r)$$ be the radial heat flow within the cylinder wall at a radial distance of $$r$$. Thus, the heat flow into the cylindrical shell is $$Q(r)$$ and the outward heat flow on the other side is $$Q(r + \Delta r)$$. At steady state, you have

\begin{aligned} Q(r + \Delta r) & = Q(r) \\ Q(r + \Delta r) - Q(r) & = 0 \\ \frac{Q(r + \Delta r) - Q(r)}{\Delta r} & = 0 \end{aligned}\tag{1}\label{eq1A}

Since $$Q(r)$$ is differentiable, taking the limit as $$\Delta r \to 0$$ gives the very simple differential equation of

$$\frac{dQ(r)}{dr} = 0 \tag{2}\label{eq2A}$$

$$Q(r) = C \tag{3}\label{eq3A}$$

for some constant $$C$$, i.e., it's independent of the radial location, i.e, $$r$$.

Next, note

$$Q(r) = q_rA \tag{4}\label{eq4A}$$

where $$q_r$$ is the heat flux in the radial direction and

$$A = 2\pi rh \tag{5}\label{eq5A}$$

is the area of the cylindrical surface normal to the $$r$$-direction along the cylinder of length $$h$$. Next, you have Fourier's law which states

$$q_r = -k\frac{dT}{dr} \tag{6}\label{eq6A}$$

Substituting \eqref{eq6A} into \eqref{eq4A}, plus using \eqref{eq3A} and \eqref{eq5A}, gives

\begin{aligned} \left(-k\frac{dT}{dr}\right)\left(2\pi rh\right) & = C \\ (-2k\pi h)r\frac{dT}{dr} & = C \\ r\frac{dT}{dr} & = \frac{C}{-2k\pi h} \\ r\left(\frac{dT}{dr}\right) & = C_1 \end{aligned}\tag{7}\label{eq7A}

where $$C_1$$ is a constant to be determined later using the boundary conditions. Note \eqref{eq7A} is a separable equation, so using separation of variables to solve it gives

$$T(r) = C_1\ln r + C_2 \tag{8}\label{eq8A}$$

where $$C_2$$ is another constant that can be solved for later, and is determined in the linked paper, but which I don't show here as it's not used in, and doesn't affect, the result.

Using the boundary conditions of $$T(r_1) = T_1$$ and $$T(r_2) = T_2$$ results in

$$C_1 = \frac{T_1 - T_2}{\ln(r_1/r_2)} \tag{9}\label{eq9A}$$

Using the above equations of \eqref{eq4A}, \eqref{eq6A}, \eqref{eq7A} and \eqref{eq8A},you now get

\begin{aligned} Q & = q_rA \\ & = \left(-k\frac{dT}{dr}\right)2\pi rh \\ & = \left(-kC_1\right)2\pi h \\ & = -2\pi hk\frac{T_1 - T_2}{\ln(r_1/r_2)} \\ & = 2\pi kh\left(\frac{T_1 - T_2}{\ln(r_2/r_1)}\right) \end{aligned}\tag{10}\label{eq10A}

This is basically the equation I quoted initially, except with their $$L$$ replaced with $$h$$.