23-Chem-A2 Unit Operations and Separation Processes · December 2014
Question 4 of 6: Furnace Wall with Added Insulation
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams — 04-CHEM-A2 Mechanical and Thermal Operations, December 2014. 3 hours, open book. Six problems (Section A Mechanical Operations: A1–A3; Section B Thermal Operations: B1–B3), each 25 marks; candidates attempt two from each section. All six are worked below for completeness.
Reference texts. McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.); Coulson & Richardson, Chemical Engineering Vol. 1 (fluid flow, heat transfer) and Vol. 2 (particle technology, filtration, crystallisation); Perry's Chemical Engineers' Handbook (9th ed.); Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (8th ed.).
Question B1: Furnace Wall with Added Insulation (25 marks)
Given. Two conduction layers in series discharge through a temperature-dependent convective film. The unknown outer temperature sets $h$, so the balance is implicit.
Quantity
Value
Refractory $t_1$ / $k_1$
0.150 m / 1.5 W/m·K
Insulation $t_2$ / $k_2$
0.025 m / 0.3 W/m·K
Inner face (held)
1400 K
Surroundings
290 K
$h(T)$ table
(370,4.2),(420,5.0),(470,6.1),(520,7.1)
Find. (a) insulation inner-surface and outer-surface temperatures; (b) fractional reduction in heat flux.
Figure B1 — Series conduction (refractory + insulation) discharging through a temperature-dependent convective film to the surroundings.
Approach. The conductive resistance is fixed, but the film coefficient depends on the unknown outer temperature; iterate the flux balance $q=(1400-T_o)/R_{\mathrm{cond}}=h(T_o)(T_o-290)$ to convergence, then march inward for the interface temperature. For (b) compare with the bare-refractory flux.
Conductive resistance and the $h(T)$ fit. $$R_{\mathrm{cond}}=\frac{t_1}{k_1}+\frac{t_2}{k_2}=\frac{0.150}{1.5}+\frac{0.025}{0.3}=0.1833\ \mathrm{m^2K/W}.$$ A linear fit of the table gives $h = 0.0196\,T - 3.122\ \mathrm{W/m^2K}$.
Iterate the flux balance. Equating conduction and convection, $\dfrac{1400-T_o}{0.1833}=(0.0196\,T_o-3.122)(T_o-290)$. Successive substitution converges to $$T_o = 677.7\ \mathrm{K},\qquad q = \frac{1400-677.7}{0.1833}=3.94\times10^{3}\ \mathrm{W/m^2}.$$ (Check: $h(677.7)=10.2$, $h(T_o-290)=10.2(387.7)=3.94\times10^3$ ✓.)
(a) Insulation surface temperatures. The outer surface is $T_o=677.7\ \mathrm{K}$. The inner surface of the insulation (refractory/insulation interface) follows from the refractory drop: $$T_i = 1400 - q\frac{t_1}{k_1}=1400-3940(0.100)=1006\ \mathrm{K}.$$ Insulation inner face $T_i = 1006\ \mathrm{K}$; outer face $T_o = 678\ \mathrm{K}$
(b) Reduction in heat loss. Bare refractory (inner 1400 K, outer 540 K) loses $$q_0=\frac{k_1(1400-540)}{t_1}=\frac{1.5(860)}{0.150}=8600\ \mathrm{W/m^2}.$$ The insulation cuts the flux from 8600 to 3940 $\mathrm{W/m^2}$, a reduction $$\frac{q_0-q}{q_0}=\frac{8600-3940}{8600}=54.2\%.$$ Heat loss falls by $\approx 54\%$
Check — extrapolated film coefficient The converged outer temperature (678 K) lies above the tabulated range (370–520 K), so $h$ is obtained by linear extrapolation of the four data points. The heat loss is only mildly sensitive to this ($q\propto h$ weakly through the implicit balance); if a measured $h$ near 680 K is available it should replace the extrapolated value.
