25-Nav-A1 Fundamentals of Naval Architecture · May-98-Mar-A1 2017

Question 5 of 8: Insulated Steam Tube

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

Notes on this paper

National Examinations, May 2017 — 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: Insulated Steam Tube (20 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.

Given data
Quantity	Symbol	Value
Tube inside diameter	$D_i$	150 mm ($r_i=75$ mm)
Wall thickness / tube $k$	$t,k_{tube}$	10 mm, 46.7 W/m°C ($r_o=85$ mm)
Steam / air temperatures	$T_i,T_\infty$	200°C, 25°C
Bare-tube heat loss	$q/L$	2000 W/m
Insulation thickness / $k$	$t_{ins},k_{ins}$	50 mm, 0.35 W/m°C
Convection coefficient ratio	$h_{metal}/h_{ins}$	1.40

Radial cross-section: bare tube (solid) vs. added insulation layer (dashed).

Find. Heat loss per unit length once the tube is insulated.

Approach. Back out the (unknown) bare-tube air-side coefficient $h_{metal}$ from the measured 2000 W/m loss using a series thermal-resistance network (steam side treated as isothermal at 200°C); derive $h_{ins}=h_{metal}/1.4$; rebuild the resistance network with the insulation layer added and the lower $h_{ins}$ on the new outer surface.

Tube-wall conduction resistance (per metre). $$R_{wall}=\dfrac{\ln(r_o/r_i)}{2\pi k_{tube}}=\dfrac{\ln(85/75)}{2\pi(46.7)}=4.27\times10^{-4}\ \text{K}\!\cdot\!\text{m/W}$$
Back out $h_{metal}$ from the bare-tube data. $R_{total}=\Delta T/(q/L)=175/2000=0.0875\ \text{K}\!\cdot\!\text{m/W}$, so $R_{conv}=R_{total}-R_{wall}=0.08707$, and $$h_{metal}=\dfrac{1}{R_{conv}\,(2\pi r_o)}=\dfrac{1}{0.08707(2\pi\times0.085)}=21.5\text{ W/m}^2{}^{\circ}\text{C}$$ $$h_{ins}=h_{metal}/1.4=15.4\text{ W/m}^2{}^{\circ}\text{C}$$
Insulated case — new resistances (per metre). Outer insulation radius $r_{ins}=85+50=135$ mm. $$R_{ins}=\dfrac{\ln(135/85)}{2\pi(0.35)}=0.2104\ \text{K}\!\cdot\!\text{m/W},\qquad R_{conv,2}=\dfrac{1}{h_{ins}(2\pi r_{ins})}=\dfrac{1}{15.4(2\pi\times0.135)}=0.0768\ \text{K}\!\cdot\!\text{m/W}$$ $$R_{new}=R_{wall}+R_{ins}+R_{conv,2}=0.000427+0.2104+0.0768=0.2876\ \text{K}\!\cdot\!\text{m/W}$$
Insulated heat loss. $$\boxed{\dfrac{q}{L}=\dfrac{\Delta T}{R_{new}}=\dfrac{175}{0.2876}=609\text{ W/m}}$$ a reduction of $(2000-609)/2000=69.6\%$ from the bare-tube loss.

Question 5 — final results
Quantity	Value
Air-side coefficient, bare metal	$h_{metal} = 21.5\text{ W/m}^2{}^{\circ}\text{C}$
Air-side coefficient, insulation	$h_{ins} = 15.4\text{ W/m}^2{}^{\circ}\text{C}$
Insulated heat loss	$q/L = 609\text{ W/m}$
Reduction from bare-tube loss	$69.6\%$

Topic: Radial conduction/convection resistance networks for insulated cylindrical pipes.

Key relations: cylindrical-wall conduction $R=\ln(r_o/r_i)/(2\pi k L)$; convective resistance $R=1/(hA)=1/(h\,2\pi rL)$; series resistances add.

Why this works: Because the tube-wall conduction resistance is three orders of magnitude smaller than the convective resistance here, the bare-tube measurement is almost entirely a probe of $h_{metal}$ — this is what lets a single measured heat-loss value back-calculate an unknown film coefficient, a common exam device when convection data isn't tabulated directly.

Common pitfall: Applying the outer convection coefficient at the wrong radius (the insulation's outer radius, not the bare tube's) once insulation is added, and forgetting that $h_{ins}$ (lower, per the given 40% ratio) replaces $h_{metal}$ entirely rather than being averaged with it.

Source: Incropera & DeWitt, Fundamentals of Heat and Mass Transfer, Ch. 3 (one-dimensional, steady-state conduction — cylindrical systems & the composite-wall/insulated-pipe resistance network).

← Question 4 Next: Question 6 →