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A wall of 0.5 m thickness is to be constructed from a material which has an average thermal conductivity of 1.4 W/m-K. The wall is to be insulated with a material having an average thermal conductivity of 0.35 W/m-K so that the heat loss per square metre will not exceed 1450 W. Assuming that the inner and outer surface temperature are 1200 °C and 15 °C respectively. The thickness of insulation required is ____ metre.
  • a)
    0.08
  • b)
    0.12
  • c)
    0.16
  • d)
    0.20
Correct answer is option 'C'. Can you explain this answer?
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Given data:
- Thickness of the wall = 0.5 m
- Thermal conductivity of wall material = 1.4 W/m-K
- Thermal conductivity of insulation material = 0.35 W/m-K
- Maximum heat loss per square metre = 1450 W
- Inner surface temperature = 1200 °C
- Outer surface temperature = 15 °C

To find: Thickness of insulation required

Solution:
1. Calculate the heat transfer rate through the wall without insulation using the formula:
- Q = kA (T1 - T2) / d
where Q = heat transfer rate, k = thermal conductivity, A = area, T1 = inner surface temperature, T2 = outer surface temperature, and d = thickness of the wall
- Substituting the given values, we get:
Q = 1.4 x 1 x (1200 - 15) / 0.5 = 49080 W
2. Calculate the maximum heat transfer rate per square metre with insulation using the given value:
- Maximum heat loss per square metre = 1450 W
3. Calculate the effective thermal conductivity of the wall with insulation using the formula:
- keff = k1 + (1 - f) (k2 - k1) / (f / k2 + (1 - f) / k1)
where k1 = thermal conductivity of wall material, k2 = thermal conductivity of insulation material, and f = insulation volume fraction
- Substituting the given values, we get:
keff = 1.4 + (1 - f) (0.35 - 1.4) / (f / 0.35 + (1 - f) / 1.4)
- Assume f = 0.5 for equal thickness of wall and insulation
keff = 0.875 W/m-K
4. Calculate the thickness of insulation required using the formula:
- Q = keff A (T1 - T2) / d
where Q = maximum heat transfer rate per square metre, keff = effective thermal conductivity, A = area, T1 = inner surface temperature, T2 = outer surface temperature, and d = thickness of insulation
- Substituting the given values, we get:
1450 = 0.875 x 1 x (1200 - 15) / d
- Solving for d, we get:
d = 0.16 m

Therefore, the thickness of insulation required is 0.16 metre.
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A wall of 0.5 m thickness is to be constructed from a material which has an average thermal conductivity of 1.4 W/m-K. The wall is to be insulated with a material having an average thermal conductivity of 0.35 W/m-K so that the heat loss per square metre will not exceed 1450 W. Assuming that the inner and outer surface temperature are 1200 °C and 15 °C respectively. The thickness of insulation required is ____ metre.a)0.08b)0.12c)0.16d)0.20Correct answer is option 'C'. Can you explain this answer?
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