A long tube with a uniform heat source is insulated at its outer radius ro and cooled at its inner radius ri, and the one-dimensional, radial, steady-state heat transfer is calculated. Use buttons to view a cross section of the tube or plot the temperature as a function of the radius. The heat transfer rate, temperature at position r and the outer surface temperature To are shown on the cross section diagram. Use sliders to set the coolant temperature, tube thickness ro − ri, radial position, heat generation, convection coefficient and thermal conductivity.
For conduction through a cylinder with heat generation, the following assumptions are made:
The heat diffusion equation is solved to determine the radial temperature distribution \(T(r)\):
\[ \rho C_p \frac{\partial T}{\partial t} = \frac{1}{r}\frac{\partial}{\partial r}\left(k r \frac{\partial T}{\partial r}\right) + \frac{1}{r^2}\frac{\partial}{\partial \phi}\left(k \frac{\partial T}{\partial \phi}\right) + \frac{\partial}{\partial z}\left(k \frac{\partial T}{\partial z}\right) + q. \]
The above assumptions reduce this equation to:
\[ 0 = \frac{1}{r}\frac{d}{dr}\left(r \frac{dT}{dr}\right) + \frac{q}{k}. \]
Separating and integrating yields:
\[ (1) \quad r\frac{dT}{dr} = -\frac{q}{2k}r^2 + C_1, \]
\[ (2) \quad T(r) = -\frac{q}{4k}r^2 + C_1 \ln r + C_2. \]
Two boundary conditions are needed to determine constants \(C_1\) and \(C_2\); both conditions are evaluated at the outer radius \(r_o\):
\[ (4) \quad T(r_o) = T_o, \]
\[ (5) \quad \left.\frac{dT}{dr}\right|_{r_o} = 0. \]
Evaluating equation (1) at boundary condition (5), evaluating equation (2) at boundary condition (4), and solving for \(C_1\) and \(C_2\) yields:
\[ C_1 = \frac{q}{2k}r_o^2, \]
\[ C_2 = T_o + \frac{q}{4k}\left(r_o^2 - r^2\right) - \frac{q}{2k}r_o^2 \ln\left(\frac{r_o}{r}\right). \]
The general solution for the temperature distribution is then:
\[ T(r) = T_o + \frac{q}{4k}\left(r_o^2 - r^2\right) - \frac{q}{2k}r_o^2 \ln\left(\frac{r_o}{r}\right), \]
where \(T\) is temperature, \(r\) is radius, the subscript o refers to the outer surface, \(q\) is heat generation, and \(k\) is thermal conductivity.
From Fourier's law, the heat removal rate is:
\[ q' = -k \, 2\pi r \frac{dT}{dr}. \]
Substituting from \(T(r)\) and evaluating at the inner radius \(r_i\):
\[ q'(r_i) = -k \, 2\pi r_i \left(-\frac{q}{2k}r_i + \frac{q}{2k}\frac{r_o^2}{r_i}\right). \]
Because the tube is insulated at \(r_o\), the rate of heat generated in the tube must equal the rate of removal at \(r_i\). Simplifying yields:
\[ q' = -\pi q \left(r_o^2 - r_i^2\right). \]
The inner surface temperature \(T_i\) is calculated using the conservation of energy, since \(q'_{cond} = q'_{conv}\):
\[ \pi q \left(r_o^2 - r_i^2\right) = h \, 2\pi r_i \left(T_i - T_\infty\right), \]
\[ T_i = \frac{q}{2 h r_i}\left(r_o^2 - r_i^2\right) + T_\infty. \]
The outer surface temperature is found by evaluating \(T(r)\) at \(r_i\):
\[ T_o = T_i - \frac{q}{4k}\left(r_o^2 - r_i^2\right) + \frac{q}{2k}r_o^2 \ln\left(\frac{r_o}{r_i}\right), \]
where \(T_\infty\) is the coolant temperature, and the subscript i refers to the inner surface.
This simulation was created in the Department of Chemical and Biological Engineering at University of Colorado Boulder for LearnChemE.com by John L. Falconer using Claude AI. It is a JavaScript/HTML5 implementation of a Mathematica simulation by Rachael L. Baumann. It was prepared with financial support from the National Science Foundation (DUE 2336987 and 2336988) in collaboration with Washington State University. Address any questions or comments to LearnChemE@gmail.com.