Heat Loss with Trial-and-Error Solution. 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...

Extracted text: Heat Loss with Trial-and-Error Solution. 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 of rock wool 102 mm thick is installed. The thermal conductivity of the rock wool is k = 0.046 + 1.56 × 10- T°C(W/m·K). The inside surface temperature of the ceramic is T, = 588.7 K, and the outside surface temperature of the insulation is T, K. Calculate the heat loss for 1.5 m of duct and the interface tem perature T, between the ceramic and the insulation. [Hint : The correct value of k, for the insulation is that evaluated at the mean temperature of (T + T3)/2. Hence, for the first trial assume a mean temperature of, say, 448 K. Then calculate the heat loss and T,. Using this new T,, calculate a new mean temperature and proceed as before.] = 311