25-Nav-A1 Fundamentals of Naval Architecture · May-98-Mar-A1 2016

Question 5 of 8: Insulating a Steam Tube to Cut Heat Loss by 95%

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

Notes on this paper

National Examinations, May 2016 — 98-Mar-A1 Applied Thermodynamics and Heat Transfer, 3 hours, open book (Part A: Thermodynamics, Part B: Heat Transfer; 5 of 8 questions required, all 8 answered below for full study coverage).

Reference texts: Cengel & Boles, Thermodynamics: An Engineering Approach; Sonntag, Borgnakke & Van Wylen, Fundamentals of Thermodynamics; Incropera & DeWitt, Fundamentals of Heat and Mass Transfer.

It is solved as the thermodynamics/heat-transfer exam it actually is.

Question 5: Insulating a Steam Tube to Cut Heat Loss by 95%

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. A bare copper tube loses heat to the surroundings by forced convection; insulation is added on the outside to cut that loss by 95%.

Given data
Quantity	Symbol	Value
Tube outside diameter	$D$	2.54 cm
Inside surface temperature	$T_i$	100°C
Room (air) temperature	$T_\infty$	27°C
Insulation conductivity	$k$	0.0875 W/m·K
Convection coefficient	$h$	55 W/m²·K

Find. Insulation thickness $t$ for a 95% reduction in heat loss per unit length, and the resulting outer-surface temperature $T_o$. (Copper's thermal conductivity is very high, so the tube-wall resistance is negligible — the insulation's inner surface is effectively at $T_i=100^\circ\text{C}$.)

Bare-tube heat loss per unit length. With the outer copper surface directly exposed to convection, $r_1=D/2=0.0127\text{ m}$: $$q'_{bare}=h\pi D(T_i-T_\infty)=55\pi(0.0254)(73)$$ $$\boxed{q'_{bare}\approx320.4\text{ W/m}}$$
Target heat loss with insulation. A 95% cut means only 5% remains: $q'_{target}=0.05(320.4)=16.02\text{ W/m}$.
Series conduction + convection resistance. With insulation to outer radius $r_2$: $$q'=\frac{T_i-T_\infty}{\dfrac{\ln(r_2/r_1)}{2\pi k}+\dfrac{1}{h\,2\pi r_2}}$$ Solving numerically for $r_2$ (the equation is implicit since $r_2$ appears in both the log term and the convection term) gives $r_2\approx0.1540\text{ m}$, so $$\boxed{t=r_2-r_1\approx0.1540-0.0127\approx0.1413\text{ m}\ (\approx14.1\text{ cm})}$$
Outer surface temperature. From the convection leg alone, $q'=h\,2\pi r_2(T_o-T_\infty)$: $$T_o=T_\infty+\frac{q'}{h\,2\pi r_2}=27+\frac{16.02}{55(2\pi)(0.1540)}$$ $$\boxed{T_o\approx27.3^\circ\text{C}}$$

Question 5 — final results
Quantity	Value
Bare-tube heat loss	$q'_{bare}\approx320.4\text{ W/m}$
Required insulation thickness	$t\approx14.1\text{ cm}$
Resulting outer surface temperature	$T_o\approx27.3^\circ\text{C}$

Topic: Cylindrical (radial) conduction through insulation in series with outer convection.

Key relations: radial conduction resistance $R'_{cond}=\ln(r_2/r_1)/(2\pi k)$; convection resistance $R'_{conv}=1/(h\,2\pi r_2)$; series thermal resistance $q'=\Delta T/\sum R'$.

Why this works: Adding insulation of low conductivity to a small-diameter pipe requires a surprisingly large radial buildup because the logarithmic conduction resistance grows slowly with radius while the convective resistance actually *shrinks* as the outer radius grows (larger surface area) — the classic "critical radius" trade-off. Here $k=0.0875\text{ W/m}\cdot\text{K}$ is a very modest insulator, so achieving a 95% cut demands roughly 14 cm of it around a 1.27 cm bare pipe.

Common pitfall: solving the implicit equation for $r_2$ by ignoring the convective term's radius dependence (dropping $1/r_2$ to a constant); forgetting that the tube-wall (copper) resistance is negligible only because copper's $k$ is orders of magnitude larger than the insulation's.

Source: Incropera & DeWitt, Fundamentals of Heat and Mass Transfer, Ch.3 (1-D radial conduction, composite cylindrical walls).

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