Section B — Thermal Operations: 23-Chem-A2 Unit Operations and Separation Processes, May 2013, Q4

Question B1: Composite Wall — Thermal Circuit, Heat Rate and Interface Temperatures (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Wall depth 1 m (into page), $H=3$ m.

Quantity	Value
Layer thicknesses $L_1/L_2/L_3$	0.05 / 0.10 / 0.05 m
Conductivities $k_A=k_D$ / $k_B$ / $k_C$	50 / 10 / 1 W m$^{-1}$K$^{-1}$
Convection $h_1$ / $h_2$	50 / 10 W m$^{-2}$K$^{-1}$
Fluid temperatures $T_{\infty,1}$ / $T_{\infty,2}$	200 / 25 °C
B, C sub-heights $H_B=H_C$	1.5 m each

Find. (a) the resistance network; (b) heat rate $q$ and interface temperatures $T_1$ (A–middle) and $T_2$ (middle–D); (c) replacement C-thickness for the same $q$.

Figure B1 — The wall and its resistance network: convection, A, the parallel pair B‖C, D and convection in series. B and C share the same two faces (isothermal surfaces), so they are in parallel.

Approach. Build the series–parallel resistance network (per-unit-depth areas), sum to a total resistance, then $q=\Delta T/R_{\text{tot}}$; interface temperatures follow by walking the network from either side.

(a) Individual resistances (area = height × 1 m depth; $R=L/kA$ or $1/hA$). Convection and end layers use $A=3\ \mathrm{m^2}$; B and C use $1.5\ \mathrm{m^2}$: $$R_{c1}=\tfrac{1}{50\cdot3}=6.67\times10^{-3},\;R_A=\tfrac{0.05}{50\cdot3}=3.33\times10^{-4},\;R_D=3.33\times10^{-4},\;R_{c2}=\tfrac{1}{10\cdot3}=3.33\times10^{-2}\ \mathrm{K/W}.$$
Parallel middle layer. $R_B=\tfrac{0.10}{10\cdot1.5}=6.67\times10^{-3}$, $R_C=\tfrac{0.10}{1\cdot1.5}=6.67\times10^{-2}$, so $$R_{BC}=\left(\frac{1}{R_B}+\frac{1}{R_C}\right)^{-1}=\frac{1}{150+15}=6.06\times10^{-3}\ \mathrm{K/W}.$$
(b) Total resistance and heat rate. $$R_{\text{tot}}=R_{c1}+R_A+R_{BC}+R_D+R_{c2}=4.673\times10^{-2}\ \mathrm{K/W},$$ $$q=\frac{T_{\infty,1}-T_{\infty,2}}{R_{\text{tot}}}=\frac{200-25}{0.04673}=\boxed{3745\ \mathrm{W}}\ \text{(per metre of depth)}.$$
Interface temperatures. Walk from the hot side: $T_1=T_{\infty,1}-q(R_{c1}+R_A)$ and $T_2=T_1-qR_{BC}$: $$T_1=200-3745(0.00700)=\boxed{173.8,{}^ rc\mathrm{C}},\qquad T_2=173.8-3745(0.006061)=\boxed{151.1,{}^ rc\mathrm{C}}.$$ Checking from the cold side, $25+3745(R_D+R_{c2})=151.1,{}^ rc$C — consistent.
(c) Equivalent single C-layer. Every resistance except the middle is unchanged, so for the same $q$ the middle resistance must equal $R_{BC}=6.06\times10^{-3}$. With C alone spanning the full area (3 m$^2$): $$L_C=R_{BC}\,k_C A=6.06\times10^{-3}(1)(3)=\boxed{18.2\ \mathrm{mm}}.$$

Quantity	Result
Total resistance $R_{\text{tot}}$	$0.0467\ \mathrm{K/W}$
(b) Heat rate $q$	3745 W per m depth (3.75 kW/m)
(b) Interface $T_1$ / $T_2$	173.8 / 151.1 °C
(c) Equivalent C thickness	18.2 mm

Concept & Theory

Topic. Steady 1-D conduction — series/parallel thermal-resistance networks with surface convection.

Key relations. $R_{\text{cond}}=L/(kA)$, $R_{\text{conv}}=1/(hA)$; series add, parallel combine as $1/R=\sum 1/R_i$; $q=\Delta T/R_{\text{tot}}$.

Why this works. With surfaces normal to the flow taken isothermal, each layer is a pure resistor and the electrical analogy holds exactly. B and C share the same left and right isothermal faces, so they carry the same driving $\Delta T$ — the definition of parallel — and the low-$k$ C layer dominates the pair.

Common pitfall. Treating B and C as series (they are stacked in the drawing) — heat flows horizontally, so they are parallel. Also, remember to use the correct sub-area (1.5 m$^2$) for B and C, not the full 3 m$^2$.

Source. Incropera & DeWitt, Fundamentals of Heat and Mass Transfer, §3.1–3.2 (composite walls, thermal circuits).

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Section A — Mechanical Operations

Section B — Thermal Operations

Question B1: Composite Wall — Thermal Circuit, Heat Rate and Interface Temperatures (25 marks)