Question 4 of 6: Heater Between Two Composite Walls

Question B1: Heater Between Two Composite Walls (25 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.

Given. One heater feeding two parallel conduction paths that share the same surface temperature $T_{\text{htr}}$ and the same ambient $T_\infty=25\ \mathrm{^\circ C}$; $A=(0.5)^2=0.25\ \mathrm{m^2}$.

Find. (a) left-side flux; (b) right-side flux; (c) interface temperatures $T_1\!\ldots\!T_6$; (d) qualitative discussion of the temperature drops.

[Figure not reproduced: Figure B1 — Although the layers are drawn stacked, the two composite walls are parallel resistances driven from the common heater surface $T_{\text{htr}}$ to a common ambient. $T_3$ is the heater itself; $T_1,T_2$ and $T_4,T_5,T_6$ march outward on the aluminum and copper sides, as labell. See the official exam paper.]

Approach. The heater surface is a single temperature feeding two independent series stacks that both dump to $25\ \mathrm{^\circ C}$; treat them as two resistances in parallel. Find $T_{\text{htr}}$ from the total flux and the parallel resistance, split the flux by each side's resistance, then march the temperatures outward layer by layer.

1. Total flux and side resistances. $q''_{\text{tot}}=Q/A=6500/0.25=26{,}000\ \mathrm{W/m^2}$. Summing $t/k$ (plus the film) on each side, $$R_{\text{Al}}=\tfrac{0.020}{237}+\tfrac{0.006}{1.4}+\tfrac{1}{50}=0.02437\ \mathrm{m^2K/W},\quad R_{\text{Cu}}=\tfrac{0.015}{97}+\tfrac{0.004}{30}+\tfrac{0.006}{0.35}+\tfrac{1}{150}=0.02410\ \mathrm{m^2K/W}.$$ The two sides turn out almost equally resistive: the Al side is dominated by its weak $h=50$ film, the Cu side by its 6 mm of Teflon.
2. Heater temperature. The two sides are parallel, so $q''_{\text{tot}}=(T_3-T_\infty)\big(\tfrac{1}{R_{\text{Al}}}+\tfrac{1}{R_{\text{Cu}}}\big)$: $$T_3=25+\frac{26{,}000}{41.03+41.50}=25+315.0=340.0\ \mathrm{^\circ C}.$$
3. (a),(b) Side fluxes. Each side carries $q''=(T_3-T_\infty)/R$: $$q''_{\text{Al}}=\frac{315.0}{0.02437}=1.293\times10^{4}\ \mathrm{W/m^2},\qquad q''_{\text{Cu}}=\frac{315.0}{0.02410}=1.307\times10^{4}\ \mathrm{W/m^2}.$$ They sum to $26{,}000\ \mathrm{W/m^2}$ ✓ (i.e. 3232 W and 3268 W through the 0.25 m$^2$ faces). Al side $\approx 12.9\ \mathrm{kW/m^2}$; Cu side $\approx 13.1\ \mathrm{kW/m^2}$
4. (c) Aluminum-side temperatures. Marching outward from $T_3=340.0\ \mathrm{^\circ C}$ with $\Delta T=q''_{\text{Al}}(t/k)$: $$T_2=340.0-12{,}927\tfrac{0.020}{237}=338.9\ \mathrm{^\circ C},\qquad T_1=338.9-12{,}927\tfrac{0.006}{1.4}=283.5\ \mathrm{^\circ C}.$$ (Convection check: $q''=h(T_1-T_\infty)=50(258.5)=12{,}925\ \mathrm{W/m^2}$ ✓.)
5. (c) Copper-side temperatures. From $T_3=340.0\ \mathrm{^\circ C}$ with $\Delta T=q''_{\text{Cu}}(t/k)$: $$T_4=340.0-13{,}073\tfrac{0.015}{97}=338.0,\quad T_5=338.0-13{,}073\tfrac{0.004}{30}=336.3,\quad T_6=336.3-13{,}073\tfrac{0.006}{0.35}=112.2\ \mathrm{^\circ C}.$$ (Convection check: $150(112.2-25)=13{,}080\approx13{,}073\ \mathrm{W/m^2}$ ✓.)
6. (d) Where the gradients land. The high-$k$ metals barely drop at all (aluminum $1.1\ \mathrm{^\circ C}$, copper $2.0\ \mathrm{^\circ C}$), while the low-$k$ glass drops $55.4\ \mathrm{^\circ C}$ and the Teflon $224.1\ \mathrm{^\circ C}$. But conductivity is not the only factor: each drop is $q''\,t/k$, so thickness and flux matter as well. The alumina ($k=30$, three times lower than copper) drops only $1.7\ \mathrm{^\circ C}$, less than the copper, because it is 4 mm thick against 15 mm. The convective films also act as resistances: the weak $h=50$ film on the aluminum side takes $258.5\ \mathrm{^\circ C}$, the largest drop anywhere in the system.