NivaarExam PrepOfficial exam papers ↗

17-Phys-B6 Applied Thermodynamics and Heat Transfer · December 2018

Question 5 of 8: Insulated Steam Pipe — Heat Loss Before and After Insulation

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

Notes on this paper

Paper format. 17-Phys-B6 Applied Thermodynamics and Heat Transfer, National Examination December 2018 — a three-hour open-book examination; candidates are expected to bring both a thermodynamics text and a heat-transfer text to make use of the property tables and graphs. A complete examination is five questions — either three from Part A (Thermodynamics, Q1–Q4) and two from Part B (Heat Transfer, Q5–Q8), or two from Part A and three from Part B — every question carrying equal value; all eight are solved below as a complete study set. Question 2 is solved as one connected narrative: the wet steam whose quality is measured by the throttling calorimeter in part (a) is the same steam entering the turbine in part (b), which is what makes part (c)'s "isentropic despite heat loss" observation checkable. Question 3 gives every cycle temperature directly from the printed diagram but no pressures, so it is solved purely from energy balances (constant specific heat, cold-air-standard) rather than isentropic pressure ratios — the intended reading, since no compressor/turbine pressure ratio is given anywhere on the page.

Reference texts. Y. A. Çengel and M. A. Boles, Thermodynamics: An Engineering Approach, 8th ed. (ideal-gas mixtures, air-standard Otto and Brayton cycles, throttling calorimeters, steam turbines, vapour-compression refrigeration); F. P. Incropera and D. P. DeWitt, Fundamentals of Heat and Mass Transfer, 7th ed. (composite cylindrical conduction with convection at both surfaces, heat generation in a solid cylinder, internal/external convection combined via an overall coefficient, effectiveness–NTU heat-exchanger analysis). Ammonia and steam saturation/superheat property values were computed (Bell et al., IAPWS-95 / REFPROP-quality equations of state) and cross-checked against the printed appendix tables on pages 5–6 of the source exam and standard steam tables, which they matched to 3–4 significant figures throughout.

Question 5: Insulated Steam Pipe — Heat Loss Before and After Insulation

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 steel tube ($r_1=75\text{ mm}$ I.D., $r_2=85\text{ mm}$ O.D. after adding the 10 mm wall thickness) carries steam at $200\,{}^{\circ}\text{C}$, losing 2000 W/m to surrounding air at $27\,{}^{\circ}\text{C}$. A 50 mm insulation layer ($k=0.35\text{ W/m}\cdot{}^{\circ}\text{C}$, outer radius $r_3=135\text{ mm}$) is then added over the pipe ($k=45\text{ W/m}\cdot{}^{\circ}\text{C}$); the metal–air convection coefficient is 40% larger than the insulation–air coefficient.

Given data
QuantitySymbolValue
Pipe I.D. / wall thickness$D_i,t$150 mm / 10 mm
Steam / air temperature$T_s,T_\infty$$200/27\,{}^{\circ}\text{C}$
Bare-pipe heat loss$Q/L$2000 W/m
Insulation thickness / conductivity$t_{ins},k_{ins}$50 mm / 0.35 W/m·°C
Pipe conductivity$k_{pipe}$45 W/m·°C
Coefficient ratio$h_{m}/h_{ins}$1.40

Find. The heat loss per metre after insulation is added.

steamr₁=75mmr₂=85mmr₃=135mmsteel pipe (k=45)insulation (k=0.35)air, 27°C (h_o)Bare pipe: Q/L = 2000 W/m. Insulated: Q/L = ? (find)
Radial resistance path: pipe wall conduction ($r_1\to r_2$), optional insulation conduction ($r_2\to r_3$), and outer convection to air — the bare-pipe case uses $h_m$ directly at $r_2$; the insulated case uses $h_{ins}$ at $r_3$.
Check: no inside (steam-side) heat transfer coefficient is given, so the inner wall is taken at the steam temperature directly (condensing wet steam gives a very large inside coefficient, making its convective resistance negligible next to the other three resistances) — the standard idealisation for this problem family.

Approach. Back out the metal–air coefficient $h_m$ from the BARE-pipe data (total resistance = pipe conduction + outer convection, from the known $Q/L$), get $h_{ins}=h_m/1.4$, then rebuild the resistance network with the insulation layer added (same pipe conduction + insulation conduction + outer convection at $r_3$ using $h_{ins}$) to find the new $Q/L$.

  1. Bare-pipe total resistance and pipe-wall conduction. $$R_{total,bare}=\frac{T_s-T_\infty}{Q/L}=\frac{200-27}{2000}=0.0865\text{ m}\cdot\text{K/W}$$ $$R_{pipe}=\frac{\ln(r_2/r_1)}{2\pi k_{pipe}}=\frac{\ln(85/75)}{2\pi\times45}$$ $$\boxed{R_{pipe}=0.000443\text{ m}\cdot\text{K/W}}$$
  2. Back out $h_m$, then $h_{ins}$. $$R_{conv,bare}=R_{total,bare}-R_{pipe}=0.0865-0.000443=0.08606\text{ m}\cdot\text{K/W}$$ $$h_m=\frac{1}{2\pi r_2R_{conv,bare}}=\frac{1}{2\pi\times0.085\times0.08606}$$ $$\boxed{h_m=21.76\text{ W/m}^2\cdot{}^{\circ}\text{C}}$$ $$h_{ins}=\frac{h_m}{1.4}=\frac{21.76}{1.4}$$ $$\boxed{h_{ins}=15.54\text{ W/m}^2\cdot{}^{\circ}\text{C}}$$
  3. Insulated-pipe resistance network. $$R_{ins}=\frac{\ln(r_3/r_2)}{2\pi k_{ins}}=\frac{\ln(135/85)}{2\pi\times0.35}=0.2104\text{ m}\cdot\text{K/W}$$ $$R_{conv,ins}=\frac{1}{2\pi r_3h_{ins}}=\frac{1}{2\pi\times0.135\times15.54}=0.0759\text{ m}\cdot\text{K/W}$$ $$R_{total,ins}=R_{pipe}+R_{ins}+R_{conv,ins}=0.000443+0.2104+0.0759$$ $$\boxed{R_{total,ins}=0.2867\text{ m}\cdot\text{K/W}}$$
  4. New heat loss. $$\frac{Q}{L}\bigg|_{insulated}=\frac{T_s-T_\infty}{R_{total,ins}}=\frac{173}{0.2867}$$ $$\boxed{Q/L=603.5\text{ W/m}}$$ The insulation cuts the heat loss to about 30% of the bare-pipe value, even though its own conductivity is more than 100× lower than the pipe's — because it dominates the total resistance at more than twice the pipe-wall term.
Question 5 — results
QuantityValue
Metal–air coefficient $h_m$ (from bare-pipe data)21.76 W/m²·°C
Insulation–air coefficient $h_{ins}$15.54 W/m²·°C
Total resistance, insulated0.2867 m·K/W
Heat loss after insulation, $Q/L$603.5 W/m (down from 2000 W/m)