Question 1 of 6: Critical Radius of Insulation on a Pipeline
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2017 — 16-Chem-A3 Heat and Mass Transfer. Three-hour, open-book exam; one textbook and any non-communicating calculator permitted. Format: two parts of three problems each (Part A — Heat Transfer, Part B — Mass Transfer), all problems worth 25 points; candidates attempt at least two per part and only the first two per part are marked. The six problems are renumbered 1–6 and all are solved below for completeness. Property data are taken from the problem statements and standard open-book tables (the paper supplies the air-property table A-9); values used are stated in each Given block.
Reference texts: Incropera, DeWitt, Bergman & Lavine, Fundamentals of Heat and Mass Transfer (Wiley) — conduction resistances, critical radius, fins, convection correlations; Coulson & Richardson, Chemical Engineering Vol. 1 (Butterworth-Heinemann) — the packed-bed, distillation and sublimation problems of this style; Treybal, Mass-Transfer Operations (3rd ed., McGraw-Hill) — differential distillation and gas adsorption; Geankoplis, Transport Processes and Separation Process Principles (4th ed., Prentice Hall) — packed-bed and convective mass-transfer coefficients; supporting data from Perry's Chemical Engineers' Handbook (9th ed.) and the paper's Table A-9 (properties of air).
Part A — Heat Transfer
Question 1: Critical Radius of Insulation on a Pipeline (25 points)
Given. Bare pipe outer radius $r_1 = 12.5$ mm ($= 0.0125$ m), length $L = 1$ m, outside film coefficient $h = 12\ \text{W/m}^2\text{K}$. First insulant $k = 0.25\ \text{W/m}\cdot\text{K}$; alternative insulant $k_2 = 0.04\ \text{W/m}\cdot\text{K}$; target heat loss for part (c) is $20.7\%$ of the bare pipe.
Quantity
Value
Bare outer radius $r_1$
0.0125 m (25 mm OD)
Outside coefficient $h$
12 W/m²·K
First insulant $k$
0.25 W/m·K
Alternative insulant $k_2$
0.04 W/m·K
Find. (a) whether the 0.25 W/m·K lagging actually reduces the loss; (b) the largest $k$ that would; (c) the thickness of the 0.04 W/m·K lagging needed to cut the loss to 20.7% of bare.
Figure 1 — Cross-section of the insulated pipeline: bore, pipe wall at $r_1$, lagging out to $r_2$, and convection to the surroundings. Adding lagging changes two competing resistances — conduction (grows with $r_2$) and convection (falls as the outer area grows).
Approach. Compare the bare radius with the critical radius of insulation $r_c = k/h$; below $r_c$ the falling convective resistance wins and lagging increases loss. For (c), set the ratio of total resistances equal to the target and solve for $r_2$.
Critical radius for the 0.25 W/m·K material (part a). For a cylinder the added heat loss peaks at
$$r_c = \frac{k}{h} = \frac{0.25}{12} = \boxed{0.0208\ \text{m} = 20.8\ \text{mm}}.$$
Since $r_c = 20.8\ \text{mm} > r_1 = 12.5\ \text{mm}$, the first layers of this lagging move the outer radius toward $r_c$, so the heat loss increases. The insulation is not effective — a poor choice on a 25 mm pipe.
Maximum conductivity for effectiveness (part b). Lagging reduces loss from the very first layer only if the critical radius does not exceed the bare radius, $r_c \le r_1$:
$$k_{\max} = h\,r_1 = 12 \times 0.0125 = \boxed{0.15\ \text{W/m}\cdot\text{K}}.$$
Any insulant with $k \le 0.15\ \text{W/m}\cdot\text{K}$ reduces the loss immediately.
Set up the resistance ratio (part c). The bare pipe has only the convective resistance, while the insulated pipe adds a conduction layer through the lagging:
$$R_{\text{bare}} = \frac{1}{h\,2\pi r_1 L} = \frac{1}{12\,(2\pi)(0.0125)(1)} = 1.061\ \text{K}\cdot\text{m/W},$$
$$R_{\text{ins}}(r_2) = \frac{\ln(r_2/r_1)}{2\pi k_2 L} + \frac{1}{h\,2\pi r_2 L}.$$
Because $q \propto 1/R$, requiring $q_{\text{ins}} = 0.207\,q_{\text{bare}}$ means $R_{\text{ins}} = R_{\text{bare}}/0.207 = 5.126\ \text{K}\cdot\text{m/W}$.
Solve for the outer radius. With $k_2 = 0.04\ \text{W/m}\cdot\text{K}$ the critical radius is $r_c = 0.04/12 = 3.3\ \text{mm} < r_1$, so $R_{\text{ins}}$ rises monotonically with $r_2$ and a unique root exists. Solving $R_{\text{ins}}(r_2) = 5.126$ numerically gives
$$r_2 = \boxed{41.9\ \text{mm}} \quad\Rightarrow\quad \text{thickness } = r_2 - r_1 = 41.9 - 12.5 = \boxed{29.4\ \text{mm}}.$$
That is part (c).
