Thermodynamics HW Solutions 907

# Thermodynamics HW Solutions 907 - to be 1 1 2 2 2 2 1 1 1 1...

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Chapter 12 Radiation Heat Transfer 12-9 A cylindrical enclosure is considered. The view factor from the side surface of this cylindrical enclosure to its base surface is to be determined. Assumptions The surfaces are diffuse emitters and reflectors. Analysis We designate the surfaces as follows: (2) (3) (1) L D Base surface by (1), top surface by (2), and side surface by (3). Then from Fig. 12-7 (or Table 12-1 for better accuracy) 38 . 0 1 1 21 12 2 2 2 1 1 1 = = = = = = F F r r L r r r r L 1 : rule summation 13 12 11 = + + F F F 62 . 0 1 38 . 0 0 13 13 = ⎯→ = + + F F () 0.31 = = π π = π π = = ⎯→ = ) 62 . 0 ( 2 1 2 2 : rule y reciprocit 13 1 1 2 1 13 1 2 1 13 3 1 31 31 3 13 1 F r r r F L r r F A A F F A F A Discussion This problem can be solved more accurately by using the view factor relation from Table 12-1
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Unformatted text preview: to be 1 1 2 2 2 2 1 1 1 1 = = = = = = r r L r R r r L r R 382 . 1 1 4 3 3 4 3 1 1 1 1 1 1 5 . 2 2 2 1 5 . 2 1 2 2 2 1 12 2 2 2 1 2 2 = ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − − = ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − = = + + = + + = R R S S F R R S 618 . 382 . 1 1 12 13 = − = − = F F ( ) 0.309 = = π π = π π = = ⎯→ ⎯ = ) 618 . ( 2 1 2 2 : rule y reciprocit 13 1 1 2 1 13 1 2 1 13 3 1 31 31 3 13 1 F r r r F L r r F A A F F A F A 12-4...
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