Question 3 of 6: B1 — Conduction with temperature-dependent conductivity
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Exams, May 2015 — 04-CHEM-B1 Transport Phenomena; 3 hours, open book. Six problems in three sections (A Fluid Mechanics, B Heat Transfer, C Mass Transfer), each worth 25 marks; candidates attempt one problem from each section plus a fourth from any section (four of six marked). The worked solutions below cover all six problems. A summary of the conservation equations (continuity, Navier–Stokes, energy, species) is supplied as Appendix A in the paper and is used throughout.
Reference texts: R. B. Bird, W. E. Stewart & E. N. Lightfoot, Transport Phenomena (2nd ed., Wiley) — the equations of change and the differential momentum / energy / species balances used across every problem; J. R. Welty, C. E. Wicks, R. E. Wilson & G. L. Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (Wiley) — pipe-friction and boundary-layer mass-transfer correlations; F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (Wiley) — conduction with variable conductivity, exchanger analysis; D. Q. Kern, Process Heat Transfer (McGraw-Hill) — the Kern casing-side method and the LMTD–correction-factor chart for multiple casing passes; R. H. Perry & D. W. Green, Perry’s Chemical Engineers’ Handbook (9th ed.) — physical properties and the Moody chart.
Question 3: B1 — Conduction with temperature-dependent conductivity (25 marks)
Given. A hollow cylindrical wall, inner radius $R_i$ (surface temperature $T_i$) and outer radius $R_o$ (surface temperature $T_o$), length $L$; steady radial conduction with no generation; conductivity varies as $k(T)=aT^2+b$.
Find. (a) the temperature distribution $T(r)$ through the wall; (b) the radial heat-loss rate $Q$.
Fig. B1: Annular pipe wall. With no generation, the total radial heat rate $Q$ is the same across every cylindrical surface; the varying conductivity makes the temperature profile implicit rather than logarithmic.
Part (a) — temperature profile
Because there is no generation, energy conservation requires the same total heat rate $Q$ to cross every cylindrical surface of radius $r$. This single constant-flux statement, combined with Fourier’s law using the variable $k(T)$, integrates directly to an implicit profile.
Constant total heat rate. Fourier’s law on the surface at radius $r$ (area $2\pi rL$): $$Q=-k(T)\,(2\pi rL)\,\frac{dT}{dr}=\text{constant (independent of }r).$$
Separate variables. Rearranging and inserting $k=aT^2+b$, $$\big(aT^2+b\big)\,dT=-\frac{Q}{2\pi L}\,\frac{dr}{r}.$$
Integrate from the inner surface. Integrating from $(R_i,T_i)$ to a general $(r,T)$: $$\boxed{\;\frac{a}{3}\big(T^3-T_i^3\big)+b\big(T-T_i\big)=-\frac{Q}{2\pi L}\,\ln\!\frac{r}{R_i}\;}$$ This implicit relation is the temperature profile; for constant $k$ (i.e. $a=0$) it collapses to the familiar logarithmic profile $T-T_i=-\tfrac{Q}{2\pi bL}\ln(r/R_i)$.
Part (b) — heat loss
Apply the profile across the whole wall. Setting $r=R_o$, $T=T_o$ in the boxed relation removes the unknown $r$-dependence: $$\frac{a}{3}\big(T_o^3-T_i^3\big)+b\big(T_o-T_i\big)=-\frac{Q}{2\pi L}\,\ln\!\frac{R_o}{R_i}.$$
Solve for the heat rate. $$\boxed{\;Q=\frac{2\pi L}{\ln(R_o/R_i)}\left[\frac{a}{3}\big(T_i^3-T_o^3\big)+b\big(T_i-T_o\big)\right]\;}$$
Interpret as a mean-conductivity result. Factoring $T_i^3-T_o^3=(T_i-T_o)(T_i^2+T_iT_o+T_o^2)$ gives $$Q=\frac{2\pi L\,\bar k\,(T_i-T_o)}{\ln(R_o/R_i)},\qquad \bar k=\frac{a}{3}\big(T_i^2+T_iT_o+T_o^2\big)+b,$$ where $\bar k$ is exactly the temperature-average of $k(T)$ over $[T_o,T_i]$. So a variable-conductivity cylinder behaves like a constant-$k$ cylinder with $k\to\bar k$; the geometry factor $\ln(R_o/R_i)$ is untouched.
