Heat Loss from an Insulated Steam Pipe: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, May 2017, Q5

Question 5: Heat Loss from an Insulated Steam Pipe (equal value)

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.

Quantity	Value
Radii	$r_1=75$ mm (bore), $r_2=85$ mm (steel OD), $r_3=135$ mm (over 50 mm insulation)
Temperatures	Steam (inner surface) 200 °C, air 25 °C
Bare heat loss	2000 W/m (used to back out the bare-tube $h$)
Conductivities	$k_{tube}=46.7$, $k_{ins}=0.35$ W/m·°C
Convection	$h_{bare}=1.40\,h_{ins}$

Given. The geometry, temperatures and property data above, with the bare loss of 2000 W/m. Find. the heat loss per metre after adding the 50 mm insulation layer.

Figure 5 — Cross-section: steam bore ($r_1$), steel wall to $r_2$, 50 mm insulation to $r_3$, then convection to ambient air. The bare-pipe measurement fixes $h_{bare}$; the insulated pipe uses $h_{ins}=h_{bare}/1.4$.

Approach. Treat the loss as a series of radial-conduction and outer-convection resistances per metre; the bare-pipe measurement pins the bare convection coefficient, from which the insulated-surface coefficient follows, and the insulated loss is $\Delta T$ over the new resistance sum.

Steel conduction resistance (per metre). $R_{steel}=\dfrac{\ln(r_2/r_1)}{2\pi k_{tube}}=\dfrac{\ln(85/75)}{2\pi(46.7)}=4.27\times10^{-4}$ m·°C/W — negligibly small.
Back out the bare convection coefficient. Bare loss $q=\dfrac{\Delta T}{R_{steel}+R_{conv,bare}}=2000$ W/m with $\Delta T=175$ °C gives $R_{conv,bare}=\dfrac{175}{2000}-R_{steel}=0.08707$ m·°C/W, and since $R_{conv,bare}=\dfrac{1}{h_{bare}\,2\pi r_2}$, $$h_{bare}=\frac{1}{0.08707\,(2\pi)(0.085)}=21.5\ \text{W/m}^2\text{}\cdot\text{°C}$$
Insulated-surface convection coefficient. $h_{ins}=\dfrac{h_{bare}}{1.40}=\dfrac{21.5}{1.40}=15.36$ W/m²·°C.
Insulation conduction resistance. $R_{ins}=\dfrac{\ln(r_3/r_2)}{2\pi k_{ins}}=\dfrac{\ln(135/85)}{2\pi(0.35)}=0.2104$ m·°C/W (this now dominates).
Outer convection resistance (insulated). $R_{conv,ins}=\dfrac{1}{h_{ins}\,2\pi r_3}=\dfrac{1}{15.36\,(2\pi)(0.135)}=0.07675$ m·°C/W.
Insulated heat loss. With $R_{tot}=4.27\times10^{-4}+0.2104+0.07675=0.28755$ m·°C/W, $$q_{ins}=\frac{200-25}{0.28755}$$ $q_{ins}\approx609$ W/m — a ~70 % reduction from the bare 2000 W/m

Check — steam-side film.

No steam-side coefficient is given, so the inner surface is taken at the steam temperature (200 °C); the steam-side resistance is negligible for flowing/condensing steam and appears identically in the bare calibration and the insulated result, so it cancels. If a finite steam-side film were included, both $h_{bare}$ and $q_{ins}$ would shift slightly but the ~70 % reduction is unchanged.

Quantity	Result
Bare convection coefficient	$h_{bare}\approx21.5$ W/m²·°C
Insulated convection coefficient	$h_{ins}\approx15.4$ W/m²·°C
Total insulated resistance	0.2876 m·°C/W
Insulated heat loss	≈ 609 W/m

Concept & Theory

Topic. One-dimensional radial conduction through a composite cylinder (steel + insulation) with external convection, calibrated from a bare-pipe measurement.

Key relations. Cylindrical conduction $R=\dfrac{\ln(r_o/r_i)}{2\pi kL}$; convection $R=\dfrac{1}{h\,2\pi r L}$; series network $q=\dfrac{\Delta T}{\sum R}$.

Why this works. Because the pipe is long, heat flows radially and each layer is a resistance in series; the measured bare loss is the only data giving $h$, so it must be inverted first. With a thick, low-$k$ insulation the added conduction resistance ($R_{ins}$) far outweighs the reduced convection coefficient, so the loss drops sharply — the pipe is well above its critical radius $r_c=k_{ins}/h_{ins}\approx23$ mm, so insulation genuinely insulates here.

Common pitfall. Forgetting that adding insulation grows the outer radius, which by itself lowers the convective resistance — one must recompute $R_{conv}$ at $r_3$ with the reduced $h_{ins}$, not reuse the bare value; and neglecting to convert the "40 % greater" statement into $h_{ins}=h_{bare}/1.4$.

Source. Çengel & Ghajar, Heat and Mass Transfer §3 (steady conduction — thermal resistance networks, cylinders, critical radius of insulation).

Question 5 of 8: Heat Loss from an Insulated Steam Pipe

Question 5: Heat Loss from an Insulated Steam Pipe (equal value)