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22-Mec-A1 Applied Thermodynamics and Heat Transfer · December 2014

Question 5 of 8: Insulating a Copper Steam Tube to Cut Heat Loss

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

Notes on this paper

Open-book, 3-hour paper. Part A (Thermodynamics, Q1–Q4) and Part B (Heat Transfer, Q5–Q8); the rubric grades any five (three from one part, two from the other), all of equal value. All eight questions are solved in full. Freon-12 property values are read from the saturated and superheated tables printed in the exam appendix (pages 4–5). Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed.); Çengel & Ghajar, Heat and Mass Transfer (6th ed.); Incropera et al., Fundamentals of Heat and Mass Transfer (8th ed.).

Question 5: Insulating a Copper Steam Tube to Cut Heat Loss (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.

Given. Copper OD $r_2=0.0127\ \text{m}$, inner $r_1=r_2-0.0054=0.0073\ \text{m}$; $T_i=100\ ^\circ\text{C}$, $T_\infty=27\ ^\circ\text{C}$; insulation $k=0.0875$; forced convection $h=55\ \text{W/m}^2\text{}\cdot\text{K}$; copper conduction negligible.

Find. Insulation thickness for a 95 % loss reduction (per unit length) and the outer surface temperature.

Tᵢ=100°Cr₁,r₂Rₕₙₓ=ln(r₃/r₂)/2πkTₒR₊ₒₙᵥ=1/2πr₃hT∞=27°Cseries: insulation conduction then outer convection
Figure 6 — Per-unit-length resistance network from the ~isothermal copper surface ($T_i$) through the insulation to the convective film at $T_\infty$.

Approach. Compute the bare-tube loss, target 5 % of it, then solve the insulation-plus-convection series resistance for the outer radius; the surface temperature follows from the convective leg.

  1. Bare-tube loss. Copper is nearly isothermal, so $q'_\text{bare}=h(2\pi r_2)(T_i-T_\infty)=55(2\pi\cdot0.0127)(73)=\boxed{320.4\ \text{W/m}}$.
  2. Target loss. A 95 % reduction means $q'=0.05(320.4)=16.0\ \text{W/m}$, so the required total resistance is $R'_\text{tot}=\dfrac{73}{16.0}=4.557\ \text{m}\cdot\text{K/W}$.
  3. Solve for outer radius. $\dfrac{\ln(r_3/r_2)}{2\pi k}+\dfrac{1}{2\pi r_3 h}=4.557$; numerically $r_3=0.154\ \text{m}$.
  4. Thickness. $t=r_3-r_2=0.154-0.0127=\boxed{0.141\ \text{m}\ (\approx14.1\ \text{cm})}$.
  5. Outer surface temperature. $T_o=T_\infty+\dfrac{q'}{2\pi r_3 h}=27+\dfrac{16.0}{2\pi(0.154)(55)}=\boxed{27.3\ ^\circ\text{C}}$.

The critical radius here is $r_c=k/h=0.0875/55=1.6\ \text{mm}$, far smaller than the 12.7 mm tube, so adding insulation always reduces the loss. Cutting the loss by 95 % still demands a thick blanket (~14 cm) because the low conductivity and modest film coefficient make each additional centimetre only marginally effective. With that blanket the outer face sits barely above room temperature — safe to touch.

QuantityResult
Bare-tube loss≈ 320 W/m
Target (5 %) loss≈ 16.0 W/m
Outer radius $r_3$≈ 0.154 m
Insulation thickness≈ 14.1 cm
Outer surface temperature≈ 27.3 °C