1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

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

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

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

Integrating this leads to

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

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

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

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

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.