Self-Cleaning Oven Composite Window: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, May 2015, Q5

Question 5: Self-Cleaning Oven Composite Window (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. Series composite window with $L_A=3L_B$, $k_A=0.18$, $k_B=0.10\ \text{W/m}\cdot\text{K}$. Inside the oven, convection and radiation act in parallel ($h_i=h_r=25$, both from 500 °C); outside, convection $h_o=25\ \text{W/m}^2\text{K}$ to the 25 °C room. Safety limit $T_o\le50\ ^\circ\text{C}$.

Find. The minimum total plastic thickness $L_A+L_B$.

Figure 4 — Equivalent thermal network per unit area: the inside convective and radiative resistances act in parallel (combined $1/(h_i+h_r)=1/50$), in series with the two plastic conduction resistances and the outside convective film $1/h_o$.

Approach. The outside surface temperature is fixed by the heat that must be convected away at that surface; setting $T_o=50\ ^\circ\text{C}$ gives the maximum allowable flux, which in turn fixes the minimum total series resistance and hence the plastic thickness.

Maximum allowable flux. At the outer surface all the through-flux leaves by convection, so at the 50 °C limit$$q''=h_o(T_o-T_\infty)=25(50-25)=\boxed{625\ \text{W/m}^2}.$$
Required total resistance. The same flux is driven by the full 500 → 25 °C difference through the series network:$$R''_\text{tot}=\frac{T_a-T_\infty}{q''}=\frac{500-25}{625}=0.760\ \tfrac{\text{m}^2\text{K}}{\text{W}}.$$
Isolate the plastic resistance. Subtract the inside (parallel) and outside film resistances:$$\frac{L_A}{k_A}+\frac{L_B}{k_B}=R''_\text{tot}-\frac{1}{h_i+h_r}-\frac{1}{h_o}=0.760-0.020-0.040=0.700\ \tfrac{\text{m}^2\text{K}}{\text{W}}.$$
Apply $L_A=3L_B$ and solve. $$\frac{3L_B}{0.18}+\frac{L_B}{0.10}=16.667L_B+10L_B=26.667L_B=0.700\;\Rightarrow\;L_B=0.0263\ \text{m},$$ so $L_A=0.0788$ m and$$L_A+L_B=\boxed{0.105\ \text{m}=10.5\ \text{cm}}.$$

Check

A thicker window raises the series resistance, lowers the through-flux, and therefore lowers $T_o$; the 10.5 cm value is the minimum that just holds $T_o=50\ ^\circ\text{C}$. Placing the inside convective and radiative coefficients in parallel (both driven from the same 500 °C source to the same inner surface) is essential — treating them in series would badly overstate the inside resistance.

Quantity	Result
Allowable heat flux	625 W/m²
Required total resistance	0.760 m²K/W
Plastic B / A thickness	2.63 cm / 7.88 cm
Minimum window thickness	≈ 10.5 cm

Concept & Theory

Topic. One-dimensional composite-wall conduction with combined convection–radiation boundaries (the thermal-resistance analogy).

Key relations. Conduction $R''=L/k$; convection/radiation $R''=1/h$; parallel surface coefficients add ($h_i+h_r$); series resistances sum; $q''=\Delta T/R''_\text{tot}$.

Why this works. In steady one-dimensional conduction the flux is a single "current" driven through a chain of resistances by the overall temperature difference. Fixing the safe outer-surface temperature caps that current; everything else is a resistance bookkeeping exercise, with the inside radiation modelled as a linearised coefficient $h_r$ acting alongside convection.

Common pitfall. Adding $h_i$ and $h_r$ as resistances in series instead of coefficients in parallel, or forgetting the outside convective film when summing the resistances.

Source. Incropera et al., Fundamentals of Heat and Mass Transfer §3.1 (plane composite walls) and §1.2 (combined convection and radiation at a surface).

Question 5 of 8: Self-Cleaning Oven Composite Window

Question 5: Self-Cleaning Oven Composite Window (equal value)