Question 1 of 6: Conduction with Temperature-Dependent Conductivity
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Reference texts: Incropera, DeWitt, Bergman & Lavine, Fundamentals of Heat and Mass Transfer (Wiley) — variable-conductivity conduction, resistance networks, convection correlations; Coulson & Richardson, Chemical Engineering Vol. 1 (Butterworth-Heinemann) — double-pipe exchanger sizing, evaporator heat load, diffusion through stagnant films; Geankoplis, Transport Processes and Separation Process Principles (4th ed., Prentice Hall) — molecular-diffusion flux relations and two-film interphase transfer; Treybal, Mass-Transfer Operations (3rd ed., McGraw-Hill) — gas absorption. Property data from Perry's Chemical Engineers' Handbook (9th ed.).
Exam format: six problems worth 25 points each — three in Part A (Heat Transfer) and three in Part B (Mass Transfer, printed as Problems 1–3 of Part B and numbered Questions 4–6 here). At least two problems from each part must be attempted and only the first two from each section are marked. Open-book, three-hour paper. All six problems are solved here for completeness.
Paper note
This is the 16-CHEM-A3 Heat and Mass Transfer paper of May 2019. The printed paper is five pages (cover notes, Part A Problems 1–3, Part B Problems 1–3), followed by two sheets of graph paper and a periodic table.
Part A — Heat Transfer
Question 1: Conduction with Temperature-Dependent Conductivity (25 points)
Given. Two steady one-dimensional conduction geometries with a conductivity that is a known polynomial in temperature: a hollow sphere ($r_1\!\to\! r_2$, surfaces at $T_1,T_2$, $k=k_0(1+aT+bT^2)$) and a plane wall (thickness $L$, faces at $T_1,T_2$, $k=k_0(1+aT)$). The printed “$T_2$ at $x>0$” denotes the opposite face; it is placed at $x=L$, the wall thickness. No numerical data — the result is a closed-form expression in the constants.
Find. Closed-form expressions for the total heat flow $Q$ in each geometry.
Figure 1 — The two geometries. In each, the steady heat flow $Q$ is constant through every concentric layer; separating variables lets the temperature-dependent $k$ be handled by integrating $k(T)\,dT$ on one side and the pure-geometry factor on the other.
Approach. Both are 1-D steady conduction with no generation, so Fourier’s law gives a constant $Q$; separate the geometric factor ($dr/\text{area}$) from $k(T)\,dT$, integrate each side independently, and collect the temperature polynomial into a mean conductivity $k_m$.
(a) Set up Fourier’s law for the sphere. With spherical area $A=4\pi r^2$ and heat flowing outward,$$Q=-k(T)\,(4\pi r^2)\frac{dT}{dr}=\text{constant}\;\Rightarrow\;\frac{Q}{4\pi}\frac{dr}{r^2}=-k_0(1+aT+bT^2)\,dT.$$
Integrate both sides. The left side runs $r_1\!\to\! r_2$, the right side $T_1\!\to\! T_2$; flipping the right limits removes the minus sign:$$\frac{Q}{4\pi}\left(\frac{1}{r_1}-\frac{1}{r_2}\right)=k_0\!\int_{T_2}^{T_1}\!\!(1+aT+bT^2)\,dT=k_0\!\left[(T_1-T_2)+\tfrac{a}{2}(T_1^2-T_2^2)+\tfrac{b}{3}(T_1^3-T_2^3)\right].$$
Solve for $Q$ (sphere). Since $\left(\tfrac1{r_1}-\tfrac1{r_2}\right)=\tfrac{r_2-r_1}{r_1r_2}$,$$\boxed{\,Q=\frac{4\pi k_0\,r_1 r_2}{r_2-r_1}\left[(T_1-T_2)+\tfrac{a}{2}(T_1^2-T_2^2)+\tfrac{b}{3}(T_1^3-T_2^3)\right]\,}$$Factoring $(T_1-T_2)$ out of the bracket defines a mean conductivity $k_m=k_0\!\left[1+\tfrac{a}{2}(T_1+T_2)+\tfrac{b}{3}(T_1^2+T_1T_2+T_2^2)\right]$, so the result collapses to the constant-$k$ form $Q=\dfrac{4\pi k_m\,r_1 r_2\,(T_1-T_2)}{r_2-r_1}$.
(b) Plane wall. Here $A$ is constant and $Q=-k(T)A\,dT/dx=$ const, so $\dfrac{Q}{A}\,dx=-k_0(1+aT)\,dT$. Integrating $0\!\to\! L$ and $T_1\!\to\! T_2$,$$\frac{Q L}{A}=k_0\!\left[(T_1-T_2)+\tfrac{a}{2}(T_1^2-T_2^2)\right]\;\Rightarrow\;\boxed{\,Q=\frac{A k_0}{L}\left[(T_1-T_2)+\tfrac{a}{2}(T_1^2-T_2^2)\right]=\frac{A k_m}{L}(T_1-T_2)\,}$$with $k_m=k_0\!\left[1+\tfrac{a}{2}(T_1+T_2)\right]$ — the conductivity evaluated at the arithmetic-mean wall temperature.
