example 6-2 - Example 6-2 The left side of a large wall is...

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Example 6-2 The left side of a large wall is held at 300°F while the right side faces an evacuated chamber, as shown. Heat is transferred by conduction through the wall and by radiation from the right side of the wall, which has an emissivity of 0.89. The wall has a thermal conductivity of 0.041 Btu/h·ft·°F. The far side of the evacuated chamber is black at a temperature of 70°F. Calculate the heat transfer through a 1 ft 2 section of the wall by two methods: a) Compute the radiative heat transfer coefficient and add resistances in series. To find h rad , assume the right wall is at the average of T w and T surr . b) Calculate the exact temperature of the right side of the wall by equating conduction and radiation heat transfer rates. Use this temperature to compute the heat transferred. Compare to part a) Solution: a) Assume the right wall is at the average of the other two temperatures:
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() o 300 70 185 F 22 ws u r r s TT T + + == = 185 460 645R s T =+= 322 3 rad s s surr s surr surr hT T T T T T εσ =+++ 70 460 530R surr T =+ = ()() 8 24 3 33 Btu 0.89 0.1714 10 hf t R 645 645 530 645 530 530 R rad h ⎛⎞ ⎜⎟ ⋅⋅ ⎝⎠ ×+ + + 2 Btu 1.25 rad h = Calculate the resistances for radiation and conduction
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example 6-2 - Example 6-2 The left side of a large wall is...

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