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23-Chem-A3 Heat and Mass Transfer · December 2017

Question 2 of 6: Part A: Heat Loss Through Two Insulation Layers on a Pipe

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — December 2017 — 16-Chem-A3 Heat and Mass Transfer. Three-hour, open-book exam (one textbook of the candidate’s choice; any non-communicating calculator). Format: two parts — Part A — Heat Transfer (Q1–Q3) and Part B — Mass Transfer (Q4–Q6). All problems are worth 25 points; the rubric requires at least two problems attempted from each part and marks only the first two per part, so a complete paper is four questions (100 points). All six problems are solved below for completeness. Property data are quoted as stated in each problem; where a value applies to a film condition it is used at the stated film temperature.

Reference texts: F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (Wiley) — free/forced convection correlations, conduction shape factors, the heat–mass-transfer analogy; J. P. Holman, Heat Transfer (McGraw-Hill) — natural-convection constants $C,m$ and cylinder conduction; R. E. Treybal, Mass-Transfer Operations (McGraw-Hill) — wetted-wall $k_G$, gas absorption and the Chilton–Colburn analogy; R. B. Bird, W. E. Stewart & E. N. Lightfoot, Transport Phenomena (Wiley); supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.).

Question 2 — Part A: Heat Loss Through Two Insulation Layers on a Pipe (25 points)

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. Radial conduction through two cylindrical layers in series. The bare pipe outer radius is $r_1=0.15$ m; the inner insulation (50 mm) reaches $r_2=0.20$ m; the outer insulation (40 mm) reaches $r_3=0.24$ m. Both layers have $k=0.105$ W/m·K. The boundary temperatures are $T_1=350$ °C at $r_1$ and $T_3=50$ °C at $r_3$.

Surface / layerRadius / property
Inner surface (350 °C)$r_1 = 0.15$ m
Layer interface$r_2 = 0.20$ m
Outer surface (50 °C)$r_3 = 0.24$ m
Conductivity (both layers)0.105 W/m·K

Find. (a) $Q/L$ (W per metre); (b) the outer-surface flux $q''$ (W/m²); (c) the interface temperature $T_2$ at $r_2$.

pipeinner layerouter layerr₃ = 0.24 mouter surf. 50 °Cr₁ = 0.15 minner surf. 350 °Cinner insul. 50 mm, k = 0.105outer insul. 40 mm, k = 0.105interface at r₂ = 0.20 m
Figure 2 — End view of the lagged pipe: two concentric insulation layers between $r_1=0.15$ m (350 °C) and $r_3=0.24$ m (50 °C). Heat flows radially outward; the same $Q/L$ passes through both layers, which fixes the interface temperature.

Approach. Add the two cylindrical conduction resistances (per unit length) in series to get $Q/L$; divide by the outer circumference for the surface flux; then re-apply the conduction law across the inner layer alone, using the now-known $Q/L$, to solve for $T_2$.

  1. Series resistance and heat loss per metre. For radial conduction the resistance of a layer (per unit length) is $\ln(r_\text{out}/r_\text{in})/(2\pi k)$; adding the two layers, $$\frac{Q}{L} = \frac{2\pi\,(T_1-T_3)}{\dfrac{\ln(r_2/r_1)}{k_\text{in}}+\dfrac{\ln(r_3/r_2)}{k_\text{out}}} = \frac{2\pi\,(350-50)}{\dfrac{\ln(0.20/0.15)}{0.105}+\dfrac{\ln(0.24/0.20)}{0.105}}.$$ The bracket is $(0.2877+0.1823)/0.105 = 4.476\ \text{m}\cdot\text{K/W}$, so $$\frac{Q}{L} = \frac{1885}{4.476} = \boxed{421\ \text{W/m}}.$$ That is part (a).
  2. Flux at the outer surface. The heat leaving each metre spreads over the outer circumference $2\pi r_3 = 2\pi(0.24) = 1.508\ \text{m}^2$ per metre: $$q'' = \frac{Q/L}{2\pi r_3} = \frac{421}{1.508} = \boxed{279\ \text{W/m}^2}.$$ That is part (b).
  3. Interface temperature. The same $Q/L$ flows through the inner layer alone, so $Q/L = 2\pi k_\text{in}(T_1-T_2)/\ln(r_2/r_1)$. Solving for $T_2$: $$T_2 = T_1 - \frac{(Q/L)\,\ln(r_2/r_1)}{2\pi k_\text{in}} = 350 - \frac{(421)(0.2877)}{2\pi(0.105)} = 350 - 183.6 = \boxed{166\ \text{°C}}.$$ That is part (c).
QuantityResult
(a) Heat loss per metre $Q/L$≈ 421 W/m
(b) Outer-surface flux $q''$≈ 279 W/m²
(c) Interface temperature $T_2$≈ 166 °C
Check
The two layers are quoted with the same conductivity (0.105 W/m·K). The problem is nonetheless well posed — the interface temperature is fixed by the radius ratio of the layers, not by any conductivity contrast — so no assumption is needed; the equal-$k$ case simply behaves like a single layer from $r_1$ to $r_3$ for the total loss, while $T_2$ still follows from the inner-layer $\ln$-ratio.