A long, horizontal, cylindrical steel reactor, 1 m in diameter, has a surface temperature of 300ºC. The emissivity of the steel is 0.6, and the heat transfer coefficient for natural convection is 5 W m−2 K−1 . Heat is lost by convection to the air at 15ºC, and also by radiation to the surroundings, which can be considered to be a black body at 15ºC. a) Calculate the total heat loss per metre length of the reactor, and the proportions lost by convection and radiation. b) The reactor is then insulated with a thin layer of insulation material to reduce the total heat loss to one-tenth of its original value. This causes the surface temperature of the steel to rise to 400ºC. The thermal conductivity of the insulation is 0.01 W m−1 K−1 , and its surface emissivity is 0.2. Show that the resulting surface temperature of the insulation is about 89ºC, and calculate the thickness of insulation required, stating any assumptions made. can you solve part b please?
A long, horizontal, cylindrical steel reactor, 1 m in diameter, has a surface temperature of 300ºC. The emissivity of the steel is 0.6, and the heat transfer coefficient for natural convection is 5 W m−2 K−1 . Heat is lost by convection to the air at 15ºC, and also by radiation to the surroundings, which can be considered to be a black body at 15ºC. a) Calculate the total heat loss per metre length of the reactor, and the proportions lost by convection and radiation.
b) The reactor is then insulated with a thin layer of insulation material to reduce the total heat loss to one-tenth of its original value. This causes the surface temperature of the steel to rise to 400ºC. The thermal conductivity of the insulation is 0.01 W m−1 K−1 , and its surface emissivity is 0.2. Show that the resulting surface temperature of the insulation is about 89ºC, and calculate the thickness of insulation required, stating any assumptions made.
can you solve part b please?
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how do you show the surface T of the insulation is 89
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