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17-Phys-B6 Applied Thermodynamics and Heat Transfer · Undated paper

Question 5 of 8: Minimum Composite-Window Thickness for a Self-Cleaning Oven

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 Examinations, May 2019 — 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 the exam supplies. 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.

Reference texts. Y. A. Çengel and M. A. Boles, Thermodynamics: An Engineering Approach, 8th ed. (polytropic closed-system processes, throttling, Rankine-cycle reheat/extraction turbines, air-standard Brayton-cycle energy balances, vapour-compression refrigeration); F. P. Incropera and D. P. DeWitt, Fundamentals of Heat and Mass Transfer, 7th ed. (composite plane-wall conduction with convection and radiation at both faces, combined entry-length internal convection, natural convection with radiation from a vertical plate, sh​ell-and-tube heat exchanger sizing via the LMTD correction-factor method). Ammonia, steam and R-134a property values were computed (Bell et al., IAPWS-95 / REFPROP-quality equations of state) and cross-checked against the printed saturated-ammonia appendix table on page 6 of the source exam, which it matched to 3–4 significant figures.

Question 5: Minimum Composite-Window Thickness for a Self-Cleaning Oven (10 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.

Check: the printed opening line names the oven-air temperature as $T_a=400\,{}^{\circ}\text{C}$, but the SAME sentence a few words later states the wall and air temperature that is actually used throughout the rest of the problem is $T_w=T_a=500\,{}^{\circ}\text{C}$.

Given. A two-layer plastic window separates the $500\,{}^{\circ}\text{C}$ oven interior from $25\,{}^{\circ}\text{C}$ room air; convection AND radiation act in PARALLEL on the hot (inside) face (same driving $\Delta T$, so their coefficients simply add), while only convection acts on the cool (outside) face.

Given data
QuantitySymbolValue
Oven air / wall temperature$T_a=T_w$$500\,{}^{\circ}\text{C}$
Room air temperature$T_o$$25\,{}^{\circ}\text{C}$
Max. outside surface temperature$T_{s,o}$$50\,{}^{\circ}\text{C}$
Inside convection coefficient$h_{ci}$25 W/m²·K
Inside radiation coefficient$h_r$25 W/m²·K
Outside convection coefficient$h_\infty$25 W/m²·K
Thermal conductivity, layer A$k_A$0.18 W/m·K
Thermal conductivity, layer B$k_B$0.10 W/m·K
Thickness relation$L_A$$3L_B$
OVEN AIRTₐ = Tᵤ = 500°CALₐ=3Lʙkₐ=0.18Bkʙ=0.10Tₛ,ₒ≤50°CROOM AIRTₒ = 25°CResistance path (per unit area): Tₐ → R_conv,in → R_A → R_B → R_conv,out → Tₒ
Composite plane wall: parallel convection+radiation on the hot face, series conduction through A then B, convection alone on the cool face.

Find. The minimum total window thickness $L_A+L_B$ that keeps $T_{s,o}\le50\,{}^{\circ}\text{C}$.

Approach. At the LIMITING design point $T_{s,o}=50\,{}^{\circ}\text{C}$ exactly, the outside convection alone fixes the heat flux; the same flux must cross the inside-convection resistance and both conduction resistances in series between the oven and that outside surface, which is one equation in the one unknown $L_B$ (via $L_A=3L_B$).

  1. Heat flux set by the outside-surface limit. $$q''=h_\infty(T_{s,o}-T_o)=25\times(50-25)$$ $$\boxed{q''=625\text{ W/m}^2}$$
  2. Total resistance needed between the oven and the outside surface. $$R''_{total}=\frac{T_w-T_{s,o}}{q''}=\frac{500-50}{625}$$ $$\boxed{R''_{total}=0.720\text{ m}^2\cdot\text{K/W}}$$
  3. Subtract the inside (parallel convection+radiation) resistance. $$R''_{conv,in}=\frac{1}{h_{ci}+h_r}=\frac{1}{50}=0.020\text{ m}^2\cdot\text{K/W}$$ $$R''_{cond,budget}=0.720-0.020=\boxed{0.700\text{ m}^2\cdot\text{K/W}}$$
  4. Solve the two-layer conduction budget for $L_B$ (with $L_A=3L_B$). $$\frac{L_A}{k_A}+\frac{L_B}{k_B}=\frac{3L_B}{0.18}+\frac{L_B}{0.10}=L_B\left(16.67+10.0\right)=0.700$$ $$\boxed{L_B=26.25\text{ mm}}$$ $$L_A=3L_B=\boxed{78.75\text{ mm}}$$
Question 5 — results
QuantityValue
Thickness of B, $L_B$26.25 mm
Thickness of A, $L_A$78.75 mm
Minimum total thickness $L_A+L_B$105.0 mm