To determine the thickness of rock insulation needed to reduce the heat transfer through the wall by 85%, follow these steps:
1. Calculate the thermal resistance of the red brick layer:
- Thickness of red brick, [tex]\( d_{brick} \)[/tex] = 10 cm = 0.1 m
- Thermal conductivity of red brick, [tex]\( k_{brick} \)[/tex] = 0.9 W/m°C
[tex]\[ R_{brick} = \frac{d_{brick}}{k_{brick}} = \frac{0.1}{0.9} \approx 0.1111 \, \text{m}^2 \text{°C}/\text{W} \][/tex]
2. Calculate the thermal resistance of the plaster layer:
- Thickness of plaster, [tex]\( d_{plaster} \)[/tex] = 3 cm = 0.03 m
- Thermal conductivity of plaster, [tex]\( k_{plaster} \)[/tex] = 0.4 W/m°C
[tex]\[ R_{plaster} = \frac{d_{plaster}}{k_{plaster}} = \frac{0.03}{0.4} = 0.075 \, \text{m}^2 \text{°C}/\text{W} \][/tex]
3. Calculate the total thermal resistance without insulation:
[tex]\[ R_{total\_original} = R_{brick} + R_{plaster} = 0.1111 + 0.075 \approx 0.1861 \, \text{m}^2 \text{°C}/\text{W} \][/tex]
4. Determine the target total thermal resistance with insulation to achieve an 85% reduction in heat transfer:
- Reducing heat transfer by 85% means that only 15% of the original heat transfer remains. Therefore, the target reduction factor is 0.15.
[tex]\[ R_{total\_target} = \frac{R_{total\_original}}{0.15} = \frac{0.1861}{0.15} \approx 1.2407 \, \text{m}^2 \text{°C}/\text{W} \][/tex]
5. Determine the required thermal resistance of the rock insulation:
[tex]\[ R_{insulation\_required} = R_{total\_target} - R_{total\_original} = 1.2407 - 0.1861 \approx 1.0546 \, \text{m}^2 \text{°C}/\text{W} \][/tex]
6. Calculate the required thickness of the rock insulation:
- Thermal conductivity of rock insulation, [tex]\( k_{rock\_insulation} \)[/tex] = 0.07 W/m°C
[tex]\[ d_{insulation\_required} = R_{insulation\_required} \times k_{rock\_insulation} = 1.0546 \times 0.07 \approx 0.0738 \, \text{m} \][/tex]
Therefore, the thickness of rock insulation required to reduce the heat transfer through the wall by 85% is approximately 0.0738 meters, or 7.38 cm.
