Question 4 of 6: B2 — Conduction in a long hollow cylinder

Question 4: B2 — Conduction in a long hollow cylinder (25 marks: 7 + 12 + 6)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A long hollow cylinder (a pipe wall), length $L$, inner radius $r_i$ at $T_i$, outer radius $r_o$ at $T_o$, both ends insulated (no axial conduction) with rotational symmetry (no $\theta$-dependence); constant $k$, $\rho$, $c_p$; no heat generation.

Find. (a) the governing unsteady conduction equation; (b) the steady radial temperature profile $T(r)$; (c) the steady radial heat flow $q$.

Approach. This is a derivation: reduce the cylindrical energy equation to the radial direction for part (a); set the time derivative to zero and integrate twice with the two surface temperatures for part (b); then apply Fourier’s law over the cylindrical area for part (c).

(a) Unsteady-state equation [7 pts]

Starting from the energy equation in cylindrical coordinates (Appendix Table A.3) with no bulk motion ($\vec u=0$), no generation, and $T=T(r,t)$ only (insulated ends kill $\partial/\partial z$; symmetry kills $\partial/\partial\theta$), every $\theta$- and $z$-term drops and $$\boxed{\rho c_p\frac{\partial T}{\partial t}=\frac{1}{r}\frac{\partial}{\partial r}\!\left(k\,r\frac{\partial T}{\partial r}\right)\quad\Longleftrightarrow\quad \frac{\partial T}{\partial t}=\frac{\alpha}{r}\frac{\partial}{\partial r}\!\left(r\frac{\partial T}{\partial r}\right),}$$ with thermal diffusivity $\alpha=k/\rho c_p$. This is the transient radial-conduction (Fourier) equation for the cylinder.

(b) Steady-state temperature profile [12 pts]

At steady state $\partial T/\partial t=0$, so the right-hand side must vanish: $\dfrac{d}{dr}\!\left(r\dfrac{dT}{dr}\right)=0$. Integrating once, $r\dfrac{dT}{dr}=C_1$, i.e. $\dfrac{dT}{dr}=\dfrac{C_1}{r}$; integrating again, $$T(r)=C_1\ln r+C_2.$$ Applying the surface temperatures $T(r_i)=T_i$ and $T(r_o)=T_o$ gives $C_1=\dfrac{T_o-T_i}{\ln(r_o/r_i)}$ and $C_2=T_i-C_1\ln r_i$. Substituting, the profile is logarithmic: $$\boxed{\frac{T(r)-T_i}{T_o-T_i}=\frac{\ln(r/r_i)}{\ln(r_o/r_i)}.}$$

(c) Steady-state heat flow [6 pts]

Fourier’s law over the cylindrical conduction area $A(r)=2\pi rL$, using $dT/dr=C_1/r$: $$q=-k\,A\,\frac{dT}{dr}=-k(2\pi rL)\frac{C_1}{r}=-2\pi kL\,C_1.$$ The $r$ cancels — $q$ is the same through every radius, as steady conservation demands. With $C_1=(T_o-T_i)/\ln(r_o/r_i)$, $$\boxed{q=\frac{2\pi kL\,(T_i-T_o)}{\ln(r_o/r_i)}.}$$