A wall of a furnace is made up of inside layer of silica brick 119 mm thick covered with a layer of magnesite brick 240 mm thick. The temperatures at the inside surface of silica brick wall and outside surface of magnesite brick wall are 780 °C and 102 °C respectively. The contact thermal resistance between the two walls at the interface is 0.003 °C/W per unit wall area. Thermal conductivities of silica and magnesite bricks are 1.7 W/m °C and 6 W/m °C. The rate of heat loss per unit area of the wall is

Concept:

To determine the rate of heat loss per unit area through a composite wall, we use the concept of thermal resistance in series.

Calculation:

Given:

• Thickness of silica brick, \( t_1 = 119 \, \text{mm} = 0.119 \, \text{m} \)
• Thickness of magnesite brick, \( t_2 = 240 \, \text{mm} = 0.24 \, \text{m} \)
• Temperature at the inside surface of silica brick, \( T_1 = 780 \, ^\circ \text{C} \)
• Temperature at the outside surface of magnesite brick, \( T_2 = 102 \, ^\circ \text{C} \)
• Thermal conductivity of silica brick, \( k_1 = 1.7 \, \text{W/m} \cdot ^\circ \text{C} \)
• Thermal conductivity of magnesite brick, \( k_2 = 6 \, \text{W/m} \cdot ^\circ \text{C} \)
• Contact thermal resistance, \( R_c = 0.003 \, ^\circ \text{C/W} \)

Calculation:

First, calculate the thermal resistance of the silica brick layer:

\( R_1 = \frac{t_1}{k_1} = \frac{0.119}{1.7} \, \text{m}^2 \cdot ^\circ \text{C/W} = 0.07 \, ^\circ \text{C/W} \)

Next, calculate the thermal resistance of the magnesite brick layer:

\( R_2 = \frac{t_2}{k_2} = \frac{0.24}{6} \, \text{m}^2 \cdot ^\circ \text{C/W} = 0.04 \, ^\circ \text{C/W} \)

Then, calculate the total thermal resistance:

\( R_{\text{total}} = R_1 + R_2 + R_c = 0.07 + 0.04 + 0.003 = 0.113 \, ^\circ \text{C/W} \)

Finally, calculate the rate of heat loss per unit area:

\( q = \frac{\Delta T}{R_{\text{total}}} = \frac{T_1 - T_2}{R_{\text{total}}} = \frac{780 - 102}{0.113} \, \text{W/m}^2 = \frac{678}{0.113} \, \text{W/m}^2 \approx 6000 \, \text{W/m}^2 \)

Therefore, the rate of heat loss per unit area of the wall is:

\( \text{6000 W/m}^2 \)