Question 3 of 5: Steady Conduction Through a Hollow Sphere
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams / EGBC — May 2013 — 04-CHEM-B1 Transport Phenomena. Three-hour open-book examination; any non-communicating calculator permitted. Five questions, of unequal weight, and all five must be answered (Q1 15, Q2 30, Q3 25, Q4 15, Q5 15 marks). The conservation equations (continuity, Navier–Stokes, shear-stress/velocity-gradient, energy and species) are supplied as Tables 1–5 appended to the paper. All five are solved below: four are pure derivations (Q1, Q2, Q3, Q5) and one (Q4) closes with a numerical mass-transfer rate.
Reference texts: R. S. Brodkey & H. C. Hershey, Transport Phenomena — A Unified Approach (McGraw-Hill) — the paper’s own reference, source of the appended Tables 1–5; R. B. Bird, W. E. Stewart & E. N. Lightfoot, Transport Phenomena (2nd ed., Wiley) — the equations of change, differential momentum/energy balances and diffusion with reaction; J. R. Welty, C. E. Wicks, R. E. Wilson & G. L. Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (Wiley) — non-equimolar diffusion with a heterogeneous surface reaction; F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (Wiley) — steady conduction in a spherical wall.
Question 3: Steady Conduction Through a Hollow Sphere (25 marks)
Given. A hollow sphere, inner radius $R_1$ at $T_1$ and outer radius $R_2$ at $T_2$ (with $T_1>T_2$); steady state, no internal heat generation, constant thermal conductivity $k$, temperature a function of $r$ only.
Find. (i) the radial temperature profile $T(r)$; (ii) the steady heat loss $Q$ through the outer surface.
Radial conduction through a spherical wall: the inner surface ($R_1$) is held hot at $T_1$, the outer surface ($R_2$) cold at $T_2$. With no generation the same heat $Q$ crosses every concentric layer, so $Q$ is independent of $r$.
Approach. Reduce the spherical energy equation to the radial Laplace form $(r^2T')'=0$, integrate with the two surface temperatures to get $T(r)$, then apply Fourier’s law over the spherical area $4\pi r^2$ for the heat flow.
(i) Reduce the energy equation. For steady state, no generation and $T=T(r)$, the spherical energy equation collapses to $\dfrac{1}{r^2}\dfrac{d}{dr}\!\left(r^2\dfrac{dT}{dr}\right)=0$.
Apply the surface temperatures. From $T(R_1)=T_1$ and $T(R_2)=T_2$, $C_1=-\dfrac{(T_1-T_2)R_1R_2}{R_2-R_1}$ and $C_2=T_1-\dfrac{(T_1-T_2)R_2}{R_2-R_1}$.
Temperature distribution. Substituting and collecting terms, $$T(r)=T_1-(T_1-T_2)\frac{R_2(r-R_1)}{r(R_2-R_1)},$$ which correctly returns $T(R_1)=T_1$ and $T(R_2)=T_2$. $\boxed{\,T(r)=T_1-(T_1-T_2)\dfrac{R_2(r-R_1)}{r(R_2-R_1)}\,}$
(ii) Apply Fourier’s law over the spherical area. The radial heat flow is $Q=-kA\dfrac{dT}{dr}=-k(4\pi r^2)\dfrac{dT}{dr}$, and from step 2 $\dfrac{dT}{dr}=\dfrac{C_1}{r^2}$.
Heat loss. The $r^2$ factors cancel, giving $Q=-k(4\pi r^2)\dfrac{C_1}{r^2}=-4\pi k C_1=\boxed{\dfrac{4\pi k R_1 R_2 (T_1-T_2)}{R_2-R_1}}.$ Because $Q$ is independent of $r$, the same heat crosses every layer — this is precisely the heat loss at the outer surface. In the printed form, evaluating Fourier’s law directly at $r=R_2$ with $\dfrac{dT}{dr}\Big|_{R_2}=\dfrac{R_1R_2(T_1-T_2)}{R_2^2(R_1-R_2)}$ gives $\dot{Q}=-k\cdot4\pi R_2^2\cdot\dfrac{R_1R_2(T_1-T_2)}{R_2^2(R_1-R_2)}$; cancelling $R_2^2$ and using $-(R_1-R_2)=R_2-R_1$ recovers the same positive outward loss.
