Calculate the conductive heat transport through the layered wall in kW of a furnace wall of 1 ft² consist of 0.5 in. thick stainless steel inner layer covered with 2 in. outside insulation layer of insulation board. The inside surface temperature of the steel is 800 K and the outside surface temperature of the insulation board is 350 K. The thermal conductivity of the stainless steel is 11 BTU/hr - ft.-°F and the thermal conductivity of the insulation board is 0.4 BTU/hr - ft.-°F.

To calculate the conductive heat transport through the layered wall, we can use the general formula for heat conduction through multiple layers:

\[
Q = \frac{A \cdot (T_1 - T_2)}{R_{\text{total}}}
\]

Where:
- \( Q \) is the rate of heat transfer (in BTU/hr),
- \( A \) is the area of the wall (1 ft²),
- \( T_1 \) is the temperature on the hot side (800 K),
- \( T_2 \) is the temperature on the cold side (350 K),
- \( R_{\text{total}} \) is the total thermal resistance.

### 1. Convert temperatures from Kelvin to Fahrenheit:
The conversion from Kelvin to Fahrenheit is done using the following formula:

\[
T_{\text{F}} = (T_{\text{K}} - 273.15) \times \frac{9}{5} + 32
\]

For the inside temperature (\(T_1 = 800 \, \text{K}\)):

\[
T_1 = (800 - 273.15) \times \frac{9}{5} + 32 = 980.33 \, \text{°F}
\]

For the outside temperature (\(T_2 = 350 \, \text{K}\)):

\[
T_2 = (350 - 273.15) \times \frac{9}{5} + 32 = 170.33 \, \text{°F}
\]

### 2. Calculate the thermal resistance for each layer:

The formula for thermal resistance is:

\[
R = \frac{L}{k \cdot A}
\]

Where:
- \( L \) is the thickness of the layer (in feet),
- \( k \) is the thermal conductivity of the material,
- \( A \) is the area (1 ft²).

#### a) Thermal resistance for the stainless steel layer:
- Thickness of stainless steel: 0.5 in = 0.04167 ft,
- Thermal conductivity of stainless steel: 11 BTU/hr-ft-°F.

\[
R_{\text{steel}} = \frac{0.04167}{11 \cdot 1} = 0.00379 \, \text{hr-ft²-°F/BTU}
\]

#### b) Thermal resistance for the insulation board layer:
- Thickness of insulation: 2 in = 0.1667 ft,
- Thermal conductivity of insulation: 0.4 BTU/hr-ft-°F.

\[
R_{\text{insulation}} = \frac{0.1667}{0.4 \cdot 1} = 0.41675 \, \text{hr-ft²-°F/BTU}
\]

### 3. Total thermal resistance:

\[
R_{\text{total}} = R_{\text{steel}} + R_{\text{insulation}} = 0.00379 + 0.41675 = 0.42054 \, \text{hr-ft²-°F/BTU}
\]

### 4. Calculate the heat transport:

\[
Q = \frac{1 \cdot (980.33 - 170.33)}{0.42054} = \frac{810}{0.42054} = 1926.55 \, \text{BTU/hr}
\]

### 5. Convert BTU/hr to kW:
1 BTU/hr = 0.000293071 kW.

\[
Q = 1926.55 \times 0.000293071 = 0.5648 \, \text{kW}
\]

### Final Answer:
The conductive heat transport through the layered wall is approximately **0.5648 kW**.

Would you like more details on any step or further questions?

Here are five related questions:
1. What factors influence the thermal resistance of materials?
2. How does the thickness of insulation impact heat transfer in a furnace?
3. What role does thermal conductivity play in heat transport calculations?
4. How would the result change if the surface temperatures were different?
5. Can thermal resistances in series and parallel behave differently in complex systems?

**Tip:** When dealing with layered walls, always account for each material’s unique thermal resistance to get an accurate estimate of heat transfer.

Learn how to calculate the conductive heat transport through a furnace wall with a 0.5 in thick stainless steel inner layer and a 2 in insulation board. This detailed example covers key thermodynamic concepts such as thermal resistance, heat conduction, and unit conversions. Perfect for university-level engineering students.

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