Topic: Heat Transfer
A furnace wall is constructed as shown in the figure. The interface temperature T1 will be (Given Inner temp = 1000 C, Outer temp = 120 C, L1 = 0.3m, k1 = 3 W/mK, L2 = 0.3m, k2 = 0.3 W/mK)
- A. 560 C
- B. 200 C
- C. 920 C
- D. 1120 C
Answer: C. 920 C
Explanation: Assuming steady-state heat conduction across the two-layer composite wall, the heat flux (Q/A) is constant through both sections. Thus, (1000 - T1) / (L1 / k1) = (T1 - 120) / (L2 / k2). Substituting the given values gives: (1000 - T1) / (0.3 / 3) = (T1 - 120) / (0.3 / 0.3). This simplifies to (1000 - T1) / 0.1 = (T1 - 120) / 1, which becomes 10000 - 10*T1 = T1 - 120. Rearranging terms gives 11*T1 = 10120, yielding T1 = 920 C.