A fumace wall consists of 200 mm lager of refractory bricks. The moximum temperatur the wale is \( 1150^{\circ} \mathrm{C} \) on the furnace side and the minimum temperature is \( 40^{\circ} \mathrm{C} \) on the eutcrmat side of the wall. Ar accurate energy batance orer the furnace shows that the heat loss 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 steel plate. Thormal Conductivities for the three layers are; 1.52 , 45 and 0.138 WlmC respecti Find: (i) 10 how many millimeters of insulation brick is the air layer equivalent. (ii) What is the temperature of the outer Suface of The steel plate.

To solve this problem, we will use the concept of thermal resistance and heat transfer through the wall. We have three layers: refractory bricks, air, and steel plate. ### Known Conditions: 1. Thickness of refractory bricks, \( L_1 = 200 \, \text{mm} = 0.2 \, \text{m} \) 2. Maximum temperature on the furnace side, \( T_1 = 1150^{\circ} \text{C} \) 3. Minimum temperature on the outer side, \( T_4 = 40^{\circ} \text{C} \) 4. Heat loss from the wall, \( q = 400 \, \text{W/m}^2 \) 5. Thermal conductivities: - Refractory bricks, \( k_1 = 1.52 \, \text{W/mC} \) - Air, \( k_2 = 0.138 \, \text{W/mC} \) - Steel plate, \( k_3 = 45 \, \text{W/mC} \) ### Step 1: Calculate the thermal resistances The total thermal resistance \( R_{total} \) can be expressed as the sum of the resistances of each layer: \[ R_{total} = R_1 + R_2 + R_3 \] Where: - \( R_1 = \frac{L_1}{k_1} \) (Resistance of refractory bricks) - \( R_2 = \frac{L_2}{k_2} \) (Resistance of air) - \( R_3 = \frac{L_3}{k_3} \) (Resistance of steel plate) ### Step 2: Calculate the temperature drop across each layer Using the heat transfer equation: \[ q = \frac{T_1 - T_4}{R_{total}} \] We can rearrange this to find \( R_{total} \): \[ R_{total} = \frac{T_1 - T_4}{q} \] ### Step 3: Solve for the unknowns 1. **Calculate \( R_{total} \)**: \[ R_{total} = \frac{1150 - 40}{400} \] 2. **Express \( R_{total} \)** in terms of \( L_2 \) (the thickness of the air layer): \[ R_{total} = \frac{0.2}{1.52} + \frac{L_2}{0.138} + \frac{L_3}{45} \] Assuming \( L_3 \) (thickness of the steel plate) is negligible for this calculation, we can simplify our calculations. Now, let's calculate \( R_{total} \) and then find \( L_2 \). ### Performing the calculations Let's calculate \( R_{total} \) first. \[ R_{total} = \frac{1150 - 40}{400} \] Now, I will calculate this value. Calculate the value by following steps: - step0: Calculate: \(\frac{\left(1150-40\right)}{400}\) - step1: Remove the parentheses: \(\frac{1150-40}{400}\) - step2: Subtract the numbers: \(\frac{1110}{400}\) - step3: Reduce the fraction: \(\frac{111}{40}\) The total thermal resistance \( R_{total} \) is calculated as: \[ R_{total} = 2.775 \, \text{m}^2 \cdot \text{C/W} \] ### Step 4: Calculate the resistance of the refractory bricks Now, we can calculate the resistance of the refractory bricks: \[ R_1 = \frac{L_1}{k_1} = \frac{0.2}{1.52} \] Let's calculate \( R_1 \). Calculate the value by following steps: - step0: Calculate: \(\frac{0.2}{1.52}\) - step1: Convert the expressions: \(\frac{\frac{1}{5}}{\frac{38}{25}}\) - step2: Multiply by the reciprocal: \(\frac{1}{5}\times \frac{25}{38}\) - step3: Reduce the numbers: \(1\times \frac{5}{38}\) - step4: Multiply: \(\frac{5}{38}\) The resistance of the refractory bricks \( R_1 \) is calculated as: \[ R_1 \approx 0.1316 \, \text{m}^2 \cdot \text{C/W} \] ### Step 5: Set up the equation for total resistance Now we can express the total resistance in terms of the air layer thickness \( L_2 \): \[ R_{total} = R_1 + R_2 \] Substituting the known values: \[ 2.775 = 0.1316 + \frac{L_2}{0.138} \] ### Step 6: Solve for \( L_2 \) Rearranging the equation to solve for \( L_2 \): \[ \frac{L_2}{0.138} = 2.775 - 0.1316 \] Now, let's calculate the right side and then find \( L_2 \). Calculate the value by following steps: - step0: Calculate: \(2.775-0.1316\) - step1: Subtract the numbers: \(2.6434\) The calculation gives us: \[ 2.775 - 0.1316 \approx 2.6434 \, \text{m}^2 \cdot \text{C/W} \] Now, substituting this back into the equation for \( L_2 \): \[ \frac{L_2}{0.138} = 2.6434 \] ### Step 7: Solve for \( L_2 \) Rearranging gives: \[ L_2 = 2.6434 \times 0.138 \] Now, let's calculate \( L_2 \). Calculate the value by following steps: - step0: Calculate: \(2.6434\times 0.138\) - step1: Multiply the numbers: \(0.3647892\) The thickness of the air layer \( L_2 \) is calculated as: \[ L_2 \approx 0.3648 \, \text{m} = 364.8 \, \text{mm} \] ### Summary of Part (i) (i) The air layer is equivalent to approximately **364.8 mm** of insulation brick. ### Step 8: Calculate the temperature of the outer surface of the steel plate To find the temperature of the outer surface of the steel plate \( T_3 \), we can use the thermal resistance of the air layer \( R_2 \): \[ R_2 = \frac{L_2}{k_2} = \frac{0.3648}{0.138} \] Now, let's calculate \( R_2 \). Calculate the value by following steps: - step0: Calculate: \(\frac{0.3648}{0.138}\) - step1: Convert the expressions: \(\frac{\frac{228}{625}}{\frac{69}{500}}\) - step2: Multiply by the reciprocal: \(\frac{228}{625}\times \frac{500}{69}\) - step3: Reduce the numbers: \(\frac{76}{5}\times \frac{4}{23}\) - step4: Multiply the fractions: \(\frac{76\times 4}{5\times 23}\) - step5: Multiply: \(\frac{304}{115}\) The resistance of the air layer \( R_2 \) is calculated as: \[ R_2 \approx 2.6435 \, \text{m}^2 \cdot \text{C/W} \] ### Step 9: Calculate the temperature drop across the air layer Now, we can find the temperature drop across the air layer using the heat transfer equation: \[ \Delta T_2 = q \cdot R_2 \] Substituting the known values: \[ \Delta T_2 = 400 \cdot 2.6435 \] Let's calculate \( \Delta T_2 \). Calculate the value by following steps: - step0: Calculate: \(400\times 2.6435\) - step1: Multiply the numbers: \(1057.4\)