Question 1 of 6: Ordering of Two Insulation Layers on a Steam Pipe
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) — conduction resistance networks, critical radius, convection correlations, radiation exchange between surfaces; Coulson & Richardson, Chemical Engineering Vol. 1 (Butterworth-Heinemann) — heat-exchanger sizing and diffusion through stagnant films; Treybal, Mass-Transfer Operations (3rd ed., McGraw-Hill) — differential (Rayleigh) distillation and penetration-theory absorption; Geankoplis, Transport Processes and Separation Process Principles (4th ed., Prentice Hall) — molecular-diffusion flux relations. Property data from Perry's Chemical Engineers' Handbook (9th ed.).
Exam format: six problems worth 25 points each — Part A (Heat Transfer, Q1–Q3) and Part B (Mass Transfer, Q4–Q6). The rubric asks for at least two problems from each part; every problem is solved here for completeness. Open-book, three-hour paper.
Part A — Heat Transfer
Question 1: Ordering of Two Insulation Layers on a Steam Pipe (25 points)
Given. Pipe outer radius $r_1=0.05$ m (10 cm OD), length $L=1$ m. Two 2.5 cm layers give interface radius $r_2=0.075$ m and outer radius $r_3=0.10$ m. One insulant has conductivity $k$, the other $3k$ (the "better" insulant is the low-$k$ one). The overall temperature difference $\Delta T$ across the lagging is the same in both arrangements.
Quantity
Value
Pipe outer radius $r_1$
0.05 m (10 cm OD)
Interface radius $r_2$
0.075 m (first 2.5 cm layer)
Outer radius $r_3$
0.10 m (second 2.5 cm layer)
Conductivity ratio
$k_{\text{better}} : k_{\text{poorer}} = 1 : 3$
Find. The ratio of heat losses (the "effectivity") for better-insulant-inside versus better-insulant-outside, and hence which arrangement loses less heat.
Figure 1 — Cross-section of the doubly-lagged pipe: bore and pipe wall at $r_1$, the interface between the two 2.5 cm layers at $r_2$, and the outer surface at $r_3$. Each layer contributes a log-resistance $\ln(r_{o}/r_{i})/k$; the inner annulus spans the smaller radii and therefore the larger logarithm.
Approach. Model the two layers as conduction resistances in series; because the same $\Delta T$, $L$ and $2\pi$ multiply every term, the heat-loss comparison collapses to a pure ratio of the dimensionless resistance sums, independent of the actual $k$ and $L$.
Write the series conduction resistance. For radial conduction through the two cylindrical layers,$$R=\frac{1}{2\pi L}\left[\frac{\ln(r_2/r_1)}{k_{\text{in}}}+\frac{\ln(r_3/r_2)}{k_{\text{out}}}\right],\qquad q=\frac{\Delta T}{R}.$$The two logarithms are $\ln(r_2/r_1)=\ln 1.5=0.4055$ and $\ln(r_3/r_2)=\ln(4/3)=0.2877$.
Better insulant on the inside. Take $k_{\text{in}}=k$ (better) and $k_{\text{out}}=3k$. Factoring out $1/(2\pi L k)$, the dimensionless resistance is$$\tilde R_{\text{in}}=\frac{0.4055}{1}+\frac{0.2877}{3}=0.4055+0.0959=\boxed{0.5014}.$$
Better insulant on the outside. Now $k_{\text{in}}=3k$, $k_{\text{out}}=k$:$$\tilde R_{\text{out}}=\frac{0.4055}{3}+\frac{0.2877}{1}=0.1352+0.2877=0.4228.$$
Form the effectivity ratio. Since $q\propto 1/R$ at fixed $\Delta T$, the ratio of heat losses is$$\frac{q_{\text{better in}}}{q_{\text{better out}}}=\frac{\tilde R_{\text{out}}}{\tilde R_{\text{in}}}=\frac{0.4228}{0.5014}=\boxed{0.843}.$$Placing the better insulant on the inside cuts the loss to 84.3% of the other arrangement — a 15.7% reduction. The better insulant belongs on the inside, because there it occupies the inner annulus where the radius ratio (and hence the log-resistance) is largest.
Quantity
Result
Dimensionless $R$, better inside
0.5014
Dimensionless $R$, better outside
0.4228
Effectivity $q_{\text{in}}/q_{\text{out}}$
0.843 (15.7% less loss with better insulant inside)
