Topic: Heat Transfer
Consider steady state heat conduction across the thickness in a plane composite wall as shown in the figure exposed to convection conditions on both sides. Given $h_i = 20 W/m^2K; h_o = 50 W/m^2K; T_{\infty,i} = 20^{\circ}C; T_{\infty,o} = -2^{\circ}C, k_1 = 20 W/mK, k_2 = 50 W/mK, L_1 = 0.30 m$ and $L_2 = 0.15 m$. Assuming negligible contact resistance between the wall surfaces, the interface temperature, T (in $^{\circ}C$) of the two walls will be
- A. 0.5
- B. 2.75
- C. 3.75
- D. 4.5
Answer: C. 3.75
Explanation: Using the thermal resistance network for steady-state heat transfer, the heat flux must be equal across both sections: $\frac{T_{\infty,i} - T}{1/h_i + L_1/k_1} = \frac{T - T_{\infty,o}}{L_2/k_2 + 1/h_o}$. Substituting the given values results in $\frac{20 - T}{1/20 + 0.3/20} = \frac{T - (-2)}{0.15/50 + 1/50}$. Simplifying the denominators yields $\frac{20 - T}{1.3/20} = \frac{T + 2}{1.15/50}$. Cross-multiplying leads to $460 - 26T = 65T + 130$. Solving this gives $330 = 88T$, which evaluates to $T = 3.75^{\circ}C$.