The exhaust duct from a heater has an inside diameter of 114. mm with ceram. The average k =1.52 W/m· K. Outside this wall, an insulation or rock wool 102 The thermal conductivity of the rock wool is k= 0.046 + 1.56 × 104 T°C (W/m temperature of the ceramic is Tı= 588.7 K, and the outside surface temperature K. Calculate the heat loss for 1.5 m of duct and the interface temperature T2 be the insulation. [Hint: The correct value of km for the insulation is that evaluated of (T2 + T:)/2, Hence, for the first trial assume a mean temperature of, sav, 448

Elements Of Electromagnetics
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Author:Sadiku, Matthew N. O.
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6. The exhaust duct from a heater has an inside diameter of 114.3 mm with ceramic walls 6.4 mm thick.
The average k=1.52 W/m: K. Outside this wall, an insulation or rock wool 102 mm thick is installed.
The thermal conductivity of the rock wool is k= 0.046 + 1.56 × 10-4 T°C (W/m K). The inside surface
temperature of the ceramic is T1 = 588.7 K, and the outside surface temperature of the insulation is 311
K. Calculate the heat loss for 1.5 m of duct and the interface temperature T2 between the ceramic and
the insulation. [Hint: The correct value of km for the insulation is that evaluated at the mean temperature
of (T2 + T3)/2. Hence, for the first trial assume a mean temperature of, say, 448 K. Then calculate the
heat loss and T2 Using this new T2, calculate a new mean temperature and proceed as before.]
Transcribed Image Text:6. The exhaust duct from a heater has an inside diameter of 114.3 mm with ceramic walls 6.4 mm thick. The average k=1.52 W/m: K. Outside this wall, an insulation or rock wool 102 mm thick is installed. The thermal conductivity of the rock wool is k= 0.046 + 1.56 × 10-4 T°C (W/m K). The inside surface temperature of the ceramic is T1 = 588.7 K, and the outside surface temperature of the insulation is 311 K. Calculate the heat loss for 1.5 m of duct and the interface temperature T2 between the ceramic and the insulation. [Hint: The correct value of km for the insulation is that evaluated at the mean temperature of (T2 + T3)/2. Hence, for the first trial assume a mean temperature of, say, 448 K. Then calculate the heat loss and T2 Using this new T2, calculate a new mean temperature and proceed as before.]
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