Composite Plastic–Cork Wall with Radiant Heating: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, May 2014, Q5

Question 5: Composite Plastic–Cork Wall with Radiant Heating (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. Plastic layer $L_p=0.01\ \text{m}$, $k_p=2.3\ \text{W/m}\cdot\text{K}$; cork layer $L_c=0.03\ \text{m}$, $k_c=0.042\ \text{W/m}\cdot\text{K}$; convection coefficient $h=10\ \text{W/m}^2\text{}\cdot\text{K}$ at both outer surfaces; room $T_\infty=25\ ^\circ\text{C}$; required glue-interface temperature $T_i=30\ ^\circ\text{C}$; negligible glue resistance.

Find. The radiant heat flux $q''$ (per unit area) that must be supplied at the plastic surface.

Figure 4 — Equivalent thermal network. Applied flux splits at the plastic surface node $T_s$: one branch convects directly to the room ($R_{\text{conv},p}=1/h$); the other conducts through the plastic to the glue interface ($R_\text{plastic}=L_p/k_p$), then through the cork ($R_\text{cork}=L_c/k_c$) and out by convection ($R_{\text{conv},c}=1/h$) to the room.

Approach. Fix the interface at 30 °C, compute the heat leaving through the cork branch, use it to find the small plastic-surface overtemperature, then add the convective loss at the plastic surface to obtain the total applied flux.

Unit-area resistances. $$R_\text{plastic}=\frac{L_p}{k_p}=\frac{0.01}{2.3}=0.00435,\quad R_\text{cork}=\frac{L_c}{k_c}=\frac{0.03}{0.042}=0.714,\quad R_\text{conv}=\frac1h=0.10\ \tfrac{\text{m}^2\text{K}}{\text{W}}.$$
Heat through the cork branch. From the 30 °C interface out to the 25 °C room,$$q''_\text{cork}=\frac{T_i-T_\infty}{R_\text{cork}+R_\text{conv}}=\frac{30-25}{0.714+0.10}=\boxed{6.14\ \text{W/m}^2}.$$
Plastic-surface temperature. The same flux conducts through the thin, conductive plastic, so $T_s$ is barely above the interface:$$T_s=T_i+q''_\text{cork}R_\text{plastic}=30+6.14(0.00435)=30.03\ ^\circ\text{C}.$$
Total applied flux. At the plastic surface the applied heat both feeds the cork branch and makes up the direct convective loss to the room:$$q''=q''_\text{cork}+h(T_s-T_\infty)=6.14+10(30.03-25)=6.14+50.3=\boxed{56.4\ \text{W/m}^2}.$$

Check

Because the plastic is thin and highly conductive, almost the entire applied flux is lost by convection at the heated plastic surface (50.3 of 56.4 W/m²); only 6.14 W/m² actually crosses into the cork. If the intent were to insulate the plastic face against room convection, the required flux would drop to about 6.1 W/m². The result is quoted per unit bond area.

Quantity	Result
Heat through cork to room	6.14 W/m²
Plastic-surface temperature	30.0 °C
Convective loss at plastic face	50.3 W/m²
Required applied flux	≈ 56.4 W/m²

Concept & Theory

Topic. One-dimensional composite-wall conduction with the electrical (resistance) analogy and a fixed internal-node temperature.

Key relations. Conduction resistance $R''=L/k$; convection resistance $R''=1/h$; series resistances add; heat balance at a node: $\sum q''=0$.

Why this works. Steady conduction is mathematically identical to a resistor network ($q''=\Delta T/R''$, with temperature playing the role of voltage). Holding the glue interface at 30 °C makes it a fixed potential; the flux out through the cork side is set by the series cork-plus-convection resistance, while the plastic surface must also supply its own convective loss to the room — the applied flux is the sum of the two branch currents leaving that node.

Common pitfall. Forgetting the direct convective loss at the heated plastic face (which dominates here), or placing the plastic and cork resistances in series to the room as if all the applied heat flowed straight through both layers.

Source. Incropera et al., Fundamentals of Heat and Mass Transfer §3.1 (the thermal-resistance network for plane composite walls).

Question 5 of 8: Composite Plastic–Cork Wall with Radiant Heating

Question 5: Composite Plastic–Cork Wall with Radiant Heating (equal value)