A fumace wall Consists of 200 mm layer of refractory bricks, 6 mm layer of steal plate and a 100 mm layer of insulation bricks. The maximum temperature of the wall is \( 1150^{\circ} \mathrm{C} \) on the furnace side and the minimum temperature is \( {4 O^{\circ}}^{\circ} \mathrm{C} \) on the outermost side of the wall. An accurate energy batance over the furnace shows that the hoat lass from the wall is \( 400 \mathrm{~W} / \mathrm{m}^{2} \). It is known that there is a thin layer of air between the layers of refractory bricks and the steel plate. Thermal conductivities for the three layers are \( 1,52,45 \) and \( 0,138 \mathrm{~W} / \mathrm{mC} \) respectively. Find (1) To how many millimeters of insulation brick is The air layer equivalent. (ii) What is the temperature of the outer surface of the cted plate.

To find the equivalent thickness of the air layer in terms of insulation bricks, we can use the formula for thermal resistance. Given the thermal conductivity of the layers, you can set up the equation to relate the resistance of the air layer to that of the insulation bricks. The resistance \( R \) is calculated as \( R = \frac{d}{k} \), where \( d \) is the thickness and \( k \) is the thermal conductivity. Hence, you find that \( R_{\text{air}} = \frac{d_{\text{air}}}{k_{\text{air}}} \) must equal \( R_{\text{insulation}} = \frac{0.1 \, \mathrm{m}}{0.138 \, \mathrm{W/mC}} \). Solving these equations gives you an equivalent thickness of the air layer in mm. For the temperature of the outer surface of the steel plate, you can use Fourier's Law of Heat Conduction. The heat transfer rate is given by \( q = \frac{\Delta T}{R} \), where \( R \) is the total thermal resistance of the wall. By calculating the temperature drop across each layer and using the known heat loss of \( 400 \, \mathrm{W/m^2} \), you can find the temperature at the steel plate's outer surface. It will be given by applying these principles iteratively from the inner to the outer layers.