23-Chem-A2 Unit Operations and Separation Processes · December 2013
Question 4 of 6: Section B — Thermal Operations
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exam 04-Chem-A2 Mechanical and Thermal Operations, December 2013 — open-book, 3 hours, any non-communicating calculator. Two sections: Section A (Mechanical Operations, A1–A3) and Section B (Thermal Operations, B1–B3); every problem is 25 marks. The rubric asks candidates to attempt two problems per section, but all six are solved in full below.
Reference texts: McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed., McGraw-Hill) — pipe friction, loss coefficients, packed beds and centrifugal separation (Ch. 5–7); de Nevers, Fluid Mechanics for Chemical Engineers and Brodkey & Hershey, Transport Phenomena: A Unified Approach — the mechanical-energy balance and the appended friction-factor chart and fitting table; Geankoplis, Transport Processes and Separation Process Principles — tubular-centrifuge neutral-zone analysis; Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (7th ed., Wiley) and Çengel, Heat and Mass Transfer — conduction with generation, annular-fin efficiency (Fig. B1) and LMTD/correction-factor exchanger design (Fig. B2). Loss-coefficient, friction-factor, fin-efficiency and correction-factor data are read from the appended Table A1, Fig. A1, Table B1 and Figs. B1–B2.
Given. Radius $R=0.125$ m; uniform generation $\dot q=2000\ \mathrm{W/m^3}$; $k=0.1\ \mathrm{W/(m\cdot K)}$; surface temperature $T_s=40\ \mathrm{^\circ C}$.
Find. (a) the radial temperature profile $T(r)$; (b) the maximum (centre) temperature.
Figure B1 — Uniform internal generation produces a parabolic temperature profile, peaking at the centre and falling to the fixed surface temperature; the centre-to-surface rise scales as $\dot qR^2/k$.
Approach. Integrate the spherical steady-conduction equation with a uniform source, apply the finite-centre and surface-temperature conditions, then evaluate at $r=0$.
Integrate the conduction equation (part a). For spherically symmetric steady conduction with source $\dot q$, $$\frac{1}{r^2}\frac{d}{dr}\!\left(r^2\frac{dT}{dr}\right)=-\frac{\dot q}{k}.$$ Integrating once with a finite centre gradient ($C_1=0$) gives $dT/dr=-\dot q r/3k$; integrating again and applying $T(R)=T_s$: $$\boxed{T(r)=T_s+\frac{\dot q}{6k}\bigl(R^2-r^2\bigr)=40+3333\bigl(0.015625-r^2\bigr)\ \mathrm{^\circ C}}.$$
Maximum temperature (part b). The parabola peaks at the centre, $r=0$: $$T_{max}=T_s+\frac{\dot qR^2}{6k}=40+\frac{2000(0.125)^2}{6(0.1)}=40+52.1=\boxed{92.1\ \mathrm{^\circ C}}.$$ The centre runs 52 °C hotter than the surface — a large gradient set by the very low conductivity.
