Question 4 of 6: B2 — Composite wall with a sandwiched heater
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Exams, May 2014 — 04-CHEM-B1 Transport Phenomena; 3 hours, open book. Six problems in three sections (A Fluid Mechanics, B Heat Transfer, C Mass Transfer); candidates attempt one from each section plus a fourth from any section (four of six marked, 25 marks each). The worked solutions below cover all six problems. A summary of the conservation equations (continuity, Navier–Stokes, energy, species) is provided as Appendix A in the paper and is used throughout.
Reference texts: R. S. Brodkey & H. C. Hershey, Transport Phenomena — A Unified Approach (McGraw-Hill) — the paper’s own reference and source of the appended conservation-equation tables; R. B. Bird, W. E. Stewart & E. N. Lightfoot, Transport Phenomena (2nd ed., Wiley) — equations of change and the differential momentum / energy / species balances; J. R. Welty, C. E. Wicks, R. E. Wilson & G. L. Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (Wiley) — differential balances and diffusion with reaction; F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (Wiley) — conduction with generation, composite walls, combined convection–radiation.
Question 4: B2 — Composite wall with a sandwiched heater
Given. One-dimensional composite wall, area $A=0.20\times0.20=0.04\ \text{m}^2$. Order from the insulated back face: copper → heater (500 W) → alumina → glass → open face at $T_4=55^\circ\text{C}$.
Fig. B2: Composite wall. Because the copper’s back face is perfectly insulated, no net heat can flow through the copper — it is isothermal at the heater temperature. All 500 W is driven rightward through the alumina and glass and dissipated from the open face by convection and radiation.
Approach. The perfectly-insulated copper back face is the key: copper carries no net flux, so it is isothermal and parts (a) and (b) collapse to the same temperature. All 500 W flows through alumina + glass to the open face; a combined convection–radiation balance there yields $h$.
Heat flux through the wall. Everything generated leaves the single open face: $q''=\dot Q/A=500/0.04=12{,}500\ \text{W}\,\text{m}^{-2}.$
(a) Heater temperature. Conduction from the heater through alumina then glass to the $55^\circ\text{C}$ face: $$T_{heater}=T_4+q''\left(\frac{L_2}{k_2}+\frac{L_3}{k_3}\right)=55+12{,}500\left(\frac{0.015}{30}+\frac{0.015}{1.5}\right).$$ With $L_2/k_2=5.0\times10^{-4}$ and $L_3/k_3=1.0\times10^{-2}$, the resistance is $0.0105\ \text{m}^2\text{K}\,\text{W}^{-1}$ and $\Delta T=131.25\ \text{K}$: $\boxed{T_{heater}=186.25^\circ\text{C}}.$
(b) Copper–insulation interface. The copper is bounded by perfect insulation (back) and the heater (front); with no generation inside it and zero flux entering from the insulated face, the flux through the copper is zero everywhere, so it is isothermal. Hence the copper–insulation interface sits at the heater temperature: $\boxed{T_{Cu,ins}=186.25^\circ\text{C}}.$ (No temperature drop occurs across the copper.)
(c) Split the open-face loss into radiation and convection. The surface at $T_s=55^\circ\text{C}=328.15$ K radiates to $T_{surr}=20^\circ\text{C}=293.15$ K: $$\dot Q_{rad}=\varepsilon\sigma A\left(T_s^4-T_{surr}^4\right)=0.8(5.67\times10^{-8})(0.04)\left(328.15^4-293.15^4\right)=7.64\ \text{W}.$$
Convective share and coefficient. $\dot Q_{conv}=\dot Q-\dot Q_{rad}=500-7.64=492.4\ \text{W}$, and with $T_\infty=25^\circ\text{C}$, $$h=\frac{\dot Q_{conv}}{A\,(T_s-T_\infty)}=\frac{492.4}{0.04(55-25)}=\boxed{410\ \text{W}\,\text{m}^{-2}\text{K}^{-1}}.$$ Radiation carries only about 1.5 % of the load, so the open face is convection-dominated.
