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 on the furnace side and the minimum temperature is on the outermost side of the wall. An aceurate energ balance over the furnace shows that the hoat lass from the wall is . 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 and 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.

Ask by Spencer King. in Zimbabwe

Jan 19,2025

Question

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 on the furnace side and the minimum temperature is on the outermost side of the wall. An aceurate energ balance over the furnace shows that the hoat lass from the wall is . 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 and 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.

Ask by Spencer King. in Zimbabwe

Jan 19,2025

UpStudy AI · Accepted Answer

Alright, I need to solve this thermodynamics problem involving a furnace wall with multiple layers and determine two specific values: the equivalent thickness of the air layer and the temperature of the outer surface of the steel plate. Let's break down the information provided and approach the problem step by step.

**Given:**
- **Layers and their properties:**
  - **Refractory bricks:** 200 mm thick, thermal conductivity $$ k_1 = 1.52 \, \mathrm{W/m \cdot K} $$
  - **Air layer:** Thickness $$ t_{\text{air}} $$ to be determined, thermal conductivity $$ k_2 = 0.138 \, \mathrm{W/m \cdot K} $$
  - **Steel plate:** 6 mm thick, thermal conductivity $$ k_3 = 45 \, \mathrm{W/m \cdot K} $$
  - **Insulation bricks:** 100 mm thick, thermal conductivity $$ k_4 = 0.138 \, \mathrm{W/m \cdot K} $$ (assuming the same as the air layer for consistency)
- **Temperatures:**
  - **Furnace side (hot side):** $$ T_{\text{hot}} = 1150^{\circ} \mathrm{C} $$
  - **Outermost side (cold side):** $$ T_{\text{cold}} = 40^{\circ} \mathrm{C} $$
- **Heat loss from the wall:** $$ q = 400 \, \mathrm{W/m}^2 $$

**Objective:**
1. Determine the equivalent thickness of the air layer in millimeters.
2. Find the temperature of the outer surface of the steel plate.

**Approach:**

1. **Understanding the Thermal Resistance:**
   The total thermal resistance $$ R_{\text{total}} $$ of the wall can be calculated by summing the thermal resistances of each layer. The formula for thermal resistance of a layer is:
   $$
   R = \frac{t}{k \cdot A}
   $$
   where:
   - $$ t $$ is the thickness of the layer (in meters),
   - $$ k $$ is the thermal conductivity (in $$ \mathrm{W/m \cdot K} $$),
   - $$ A $$ is the area (in $$ \mathrm{m}^2 $$).

Since the heat loss $$ q $$ is given per square meter, the area $$ A $$ cancels out in the calculations, simplifying the resistance to:
   $$
   R = \frac{t}{k}
   $$

2. **Calculating Thermal Resistances:**
   - **Refractory bricks:**
     $$
     R_1 = \frac{0.200 \, \mathrm{m}}{1.52 \, \mathrm{W/m \cdot K}} \approx 0.1316 \, \mathrm{K/W}
     $$
   - **Air layer:**
     $$
     R_2 = \frac{t_{\text{air}}}{0.138 \, \mathrm{W/m \cdot K}}
     $$
   - **Steel plate:**
     $$
     R_3 = \frac{0.006 \, \mathrm{m}}{45 \, \mathrm{W/m \cdot K}} \approx 0.000133 \, \mathrm{K/W}
     $$
   - **Insulation bricks:**
     $$
     R_4 = \frac{0.100 \, \mathrm{m}}{0.138 \, \mathrm{W/m \cdot K}} \approx 0.725 \, \mathrm{K/W}
     $$

3. **Total Thermal Resistance:**
   $$
   R_{\text{total}} = R_1 + R_2 + R_3 + R_4 \approx 0.1316 + R_2 + 0.000133 + 0.725 \approx 0.856733 + R_2
   $$

4. **Heat Loss Equation:**
   The heat loss $$ q $$ can be expressed as:
   $$
   q = \frac{T_{\text{hot}} - T_{\text{cold}}}{R_{\text{total}}}
   $$
   Plugging in the known values:
   $$
   400 = \frac{1150 - 40}{0.856733 + R_2}
   $$
   Simplifying:
   $$
   400 = \frac{1110}{0.856733 + R_2}
   $$
   Solving for $$ R_2 $$:
   $$
   0.856733 + R_2 = \frac{1110}{400} = 2.775
   $$
   $$
   R_2 = 2.775 - 0.856733 \approx 1.918267 \, \mathrm{K/W}
   $$

5. **Determining the Equivalent Thickness of the Air Layer:**
   Using the thermal resistance of the air layer:
   $$
   R_2 = \frac{t_{\text{air}}}{0.138} \approx 1.918267
   $$
   Solving for $$ t_{\text{air}} $$:
   $$
   t_{\text{air}} = 1.918267 \times 0.138 \approx 0.265 \, \mathrm{m} = 265 \, \mathrm{mm}
   $$
   So, the air layer is equivalent to approximately 265 mm.

6. **Finding the Temperature of the Outer Surface of the Steel Plate:**
   The temperature difference across the steel plate can be calculated using its thermal resistance:
   $$
   \Delta T_3 = q \times R_3 = 400 \times 0.000133 \approx 0.0532 \, \mathrm{K}
   $$
   Since the steel plate is on the cold side, the temperature of its outer surface $$ T_{\text{steel}} $$ is:
   $$
   T_{\text{steel}} = T_{\text{cold}} - \Delta T_3 = 40 - 0.0532 \approx 39.9468^{\circ} \mathrm{C}
   $$
   Rounding to two decimal places:
   $$
   T_{\text{steel}} \approx 39.95^{\circ} \mathrm{C}
   $$

**Conclusion:**
- The air layer is equivalent to approximately **265 mm**.
- The temperature of the outer surface of the steel plate is approximately **39.95°C**.

(1) The air layer is equivalent to approximately **265 mm**. (ii) The temperature of the outer surface of the steel plate is approximately **39.95°C**.

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