A plane composite wall (see figure) (cross-sectional area = 1 m²) is made up of two layers. One layer is made of material A (100 mm thick, thermal conductivity = 50 W/m-K) and another layer is made of material B (10 mm thick, thermal conductivity = 2 W/m-K). The thermal contact resistance at the interface is 0.003 m² K/W. The temperature of the open side of wall A is 300°C and that of open side of wall B is 50°C. What will be the rate of heat flow through wall?

Concept:

The heat transfer through a composite wall is governed by thermal resistance. The total thermal resistance is the sum of the resistances of individual layers and the thermal contact resistance at the interface.

The total thermal resistance is given by:

\( R_{\text{total}} = R_A + R_B + R_{\text{contact}} \)

The heat transfer rate is calculated using Fourier’s Law:

\( Q = \frac{T_1 - T_2}{R_{\text{total}}} \)

Given:

• Thickness of material A: \(L_A = 100 mm = 0.1 m\)
• Thermal conductivity of material A: \(k_A = 50 W/m·K\)
• Thickness of material B: L_B = 10 mm = 0.01 m
• Thermal conductivity of material B: \(k_B = 2 W/m·K\)
• Thermal contact resistance: \(R_{\text{contact}} = 0.003 ~m²K/W\)
• Temperature difference: \(T_1 = 300^\circ C ,~ T_2 = 50^\circ C\)
• Cross-sectional area: A = 1 m²

Calculation:

Step 1: Compute the thermal resistance of each material

Thermal resistance of material A:

\( R_A = \frac{L_A}{k_A A} = \frac{0.1}{50 \times 1} = 0.002~K/W\)

Thermal resistance of material B:

\( R_B = \frac{L_B}{k_B A} = \frac{0.01}{2 \times 1} = 0.005~K/W\)

Step 2: Compute total thermal resistance

\( R_{\text{total}} = R_A + R_B + R_{\text{contact}} \)

\( R_{\text{total}} = 0.002 + 0.005 + 0.003 = 0.01~K/W\)

Step 3: Compute heat transfer rate

\( Q = \frac{T_1 - T_2}{R_{\text{total}}} \)

\( Q = \frac{300 - 50}{0.01} = \frac{250}{0.01} = 25,000~W\)

\( Q = 25~kW\)