Junction Temperature of Two Rods – Rankers Physics

Heat Transfer - Conduction and Convection: Practice Problem & Solution

Two rods one made of material A and other made of material B of same length and same cross-sectional area are joined together. If thermal conductivity of material A is \(K_1\) while that of material B is \(K_2\) and the free end of rod made of material A is maintained at \(T_1\) while that of the rod of material B is maintained at \(T_2\), then the temperature of junction is (Where \(T_1 > T_2\))
\(\frac{T_1 + T_2}{2}\)
\(\frac{T_1 K_1 + T_2 K_2}{K_1 + K_2}\)
\(\frac{T_1 K_1 - T_2 K_2}{K_1 + K_2}\)
\(\frac{T_1 K_1 + T_2 K_2}{K_1 - K_2}\)

Solution Explained:

To solve this problem, we apply the core principles of Heat Transfer - Conduction and Convection. Understanding the underlying formula is key to arriving at the correct answer below:

Under steady state, the rate of heat flow is the same through both rods: \(\frac{K_1 A(T_1 - T_j)}{L} = \frac{K_2 A(T_j - T_2)}{L}\). Solving for \(T_j\) gives \(T_j = \frac{K_1 T_1 + K_2 T_2}{K_1 + K_2}\).

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