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22-Mec-A1 Applied Thermodynamics and Heat Transfer · Undated paper

Question 5 of 8: Self-cleaning oven composite window

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

Notes on this paper

Paper format: National Examination 16-Mec-A1 Applied Thermodynamics and Heat Transfer — 3 hours, open book. Two parts: Part A (Thermodynamics, Q1–Q4) and Part B (Heat Transfer, Q5–Q8). The rubric asks for five questions (three from one part, two from the other), each of equal value; all eight are solved.

Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed.); Çengel & Ghajar, Heat and Mass Transfer (6th ed.); Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (8th ed.); Moran & Shapiro, Fundamentals of Engineering Thermodynamics. Property data: saturated ammonia and R-134a tables printed with the paper.

Question 5 — Self-cleaning oven composite window (Part B, 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.

QuantitySymbolValue
Plastic A$L_A,\,k_A$3 mm, 0.16 W/m·K
Plastic B$L_B,\,k_B$6 mm, 0.30 W/m·K
Oven / room air$T_i,\,T_\infty$500 °C / 25 °C
Convection coefficients$h_i,\,h_o$25, 20 W/m²·K
Safety limit$T_{o,\max}$50 °C

Find. (a) heat flux $q''$; (b) outer-surface temperature $T_o$; (c) $h_o$ required to hold $T_o\le50$ °C.

Series thermal-resistance network A B Oven air 500 °C Room 25 °C 1/hᵢ L_A/k_A L_B/k_B 1/hₒ 0.0400 0.0188 0.0200 0.0500 m²·K/W
One-dimensional steady conduction: two convective films in series with the two conductive slabs.

Approach. Add the two convective films and two conductive slabs as series resistances (per unit area), get the flux from the overall $\Delta T$, then work outward to the surface temperature; invert for the $h_o$ that pins $T_o$ at the safety limit.

  1. Total resistance and heat flux. Per unit area, $$R''_\text{tot}=\frac1{h_i}+\frac{L_A}{k_A}+\frac{L_B}{k_B}+\frac1{h_o}=0.0400+0.0188+0.0200+0.0500=0.1288\ \tfrac{\text{m}^2\text{K}}{\text{W}}$$ $$q''=\frac{T_i-T_\infty}{R''_\text{tot}}=\frac{500-25}{0.1288}=\boxed{3689\ \text{W/m}^2}$$
  2. Outer-surface temperature. Working back from the room through the outer film only, $$T_o=T_\infty+q''\!\left(\frac1{h_o}\right)=25+3689(0.05)=\boxed{209.5\ ^\circ\text{C}}$$ Far above the 50 °C limit — natural convection alone cannot keep the outer pane touch-safe.
  3. Required room-side coefficient for $T_o=50$ °C. The outer film must carry the flux at only $\Delta T=25$ K, so $q''=h_o(50-25)=25\,h_o$. Substituting into the series relation with $a=1/h_i+L_A/k_A+L_B/k_B=0.0788$: $$25\,h_o\,(a+1/h_o)=T_i-T_\infty\ \Rightarrow\ 25\,h_o(0.0788)+25=475$$ $$h_o=\frac{450}{25(0.0788)}=\boxed{229\ \text{W/m}^2\text{K}}$$ Such a high coefficient means forced outer-pane cooling (a fan-driven air curtain), which is exactly how self-cleaning ovens keep the outer glass safe.
QuantityResult
Window conduction resistance0.0388 m²·K/W
Heat flux (given films)3689 W/m²
Outer-surface temperature209.5 °C (unsafe)
$h_o$ required for $T_o\le50$ °C≈ 229 W/m²·K (forced convection)
Check — reconstructed film coefficients.
The question lists the two conductivities and thicknesses but gives no numerical convection coefficients ("determine the coefficient of convection in the oven ... and outside"). The reconstruction supplies representative values ($h_i=25$, $h_o=20$ W/m²·K) so the network is determinate; the instructive result — a thin plastic window cannot stay below 50 °C by natural convection, forcing an actively-cooled outer pane — is robust to the exact film values.