A fumace wall Consists of 200 mm layer of refractory bricks, 6 mm layer of stcel 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 \( 40^{\circ} \mathrm{C} \) on the outermost side of the wall. An aceurate energ balance over the furnace shows that the hoat lass from the wall is \( 400 \mathrm{~W} / \mathrm{m}^{2} \). It is known that there is thin layer of air between the layers of refraciory bricks and the steel plate. Thernal conductivities for three layers are \( 1,52,45 \) and \( 0,138 \mathrm{~W} / \mathrm{m} \) 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 insulation brick for the air layer and the temperature of the outer surface of the steel plate, we can use the thermal resistance concept and the heat transfer equation. ### Equivalent Thickness Calculation: The thermal resistance (R) for each layer is given by the formula: \[ R = \frac{L}{k} \] where \( L \) is the thickness and \( k \) is the thermal conductivity. For air, with an approximate thickness \( x \): \[ R_{\text{air}} = \frac{x}{0.025} \] For insulation bricks (thickness = 100 mm): \[ R_{\text{insulation}} = \frac{0.1}{0.138} \approx 0.7246 \, m^2K/W \] Thus, the equivalent thickness \( x \) of insulation bricks is calculated by setting the resistances equal: \[ \frac{x}{0.025} = 0.7246 \] Solving for \( x \): \[ x \approx 0.7246 * 0.025 \approx 0.0181 \, m = 18.1 mm \] ### Outer Surface Temperature Calculation: Using the known heat loss \( q = 400 \, W/m^2 \): The temperature drop (ΔT) through the layers can be calculated using: \[ q = \frac{ΔT}{R_{\text{total}}} \] The total thermal resistance is: \[ R_{\text{total}} = R_{\text{refractory}} + R_{\text{air}} + R_{\text{steel}} + R_{\text{insulation}} \] Calculating each resistance, we get: - For the refractory layer: \( R_{\text{refractory}} = \frac{0.2}{1.52} \approx 0.1316 \, m^2K/W \) - For the steel plate: \( R_{\text{steel}} = \frac{0.006}{45} \approx 0.0001333 \, m^2K/W \) Thus, \[ R_{\text{total}} \approx 0.1316 + 0.0180 + 0.0001333 + 0.7246 \approx 0.8747 \, m^2K/W \] Now, calculate ΔT: \[ 400 = \frac{1150 - T_{\text{outer}}}{0.8747} \] Solving for \( T_{\text{outer}} \): \[ 1150 - T_{\text{outer}} = 400 * 0.8747 \] \[ 1150 - T_{\text{outer}} \approx 349.88 \] \[ T_{\text{outer}} \approx 1150 - 349.88 \approx 800.12^{\circ}C \] So, the temperature of the outer surface of the steel plate is approximately \( 800.12^{\circ}C \).