sm13_39
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PROBLEM 13.39 K NOWN: Long, inclined black su rfaces maintained at prescrib ed temperatures. FIND: (a) Net radiation exchange b etween the two surfaces per uni t length, (b) Net radiation t ransfer to surface A 2 with black, insulated surface positioned as shown below; determine temperature of this urface. s SCHEMATIC: ASSUMPTIONS: (1) Surfaces behave as blackbodies , (2) Surfaces are very long in direction normal o page. t ANALYSIS: (a) The net radiation exchang e between two black surfaces is q ij = A i F ij (J i - J j ) where J i = σ and Jj = σ . Therefore, 4 i T 4 i T ( ) 4 4 12 1 12 1 2 q A F T T σ = − Noting that A 1 = width×length ( and that from symmetry, F ) 12 = 0.5, find ( ) 8 2 4 4 4 4 12 12 q q 0.1 m 0.5 5.67 10 W/m K 1000 800 K 1680 W / m. − ′ = = × × × ⋅ − = < (b) With the insulated, black surface A 3 positioned as shown above, a three-surface enclosure i s formed. From an energy balance on the node representing A 2 , find 2 32 1 q q q ′ ′ ′ − = + 2 [ ] [ ] 2 3 32 b3 b2 1 12 b 1 b2 q A F E E A F E E − = − + − . To find E b3 , which at present is not known, perform an energy balance on the node representing A 3 . Note that A 3 is adiabatic and, hence q 3 = 0, q 13 = q 32 . [ ] [ ] 1 13 b 1 b3 3 32 b3 b2 A F E E A F E E − = − Since F 13 = F 23 = 0.5 and A 1 = A 3 , it follows that ( ) [ ] b3 b1 E 1 / 2 E E = + b2 and ( ) ( ) [ ] 2 3 32 b 1 b2 b2 12 q A / F E E / 2 E q ′ ′ − = + − + ( ) 8 2 4 4 4 4 2 q 0.1 m 0.5 5.67 10 W / m K 1000 800 /2 800 K − ′ − = × × × ⋅ + − 4 ⎡ ⎤ ⎢ ⎥ ⎣ ⎦ 1680 W/m 2517 W / m + = < Noting that E b3 = = (1/2) [E 4 3 T σ b1 +E b2 ], it follows that < ( ) ( ) 1 / 4 1 / 4 4 4 4 3 1 4 T T T /2 1000 800 / 2 K 916 K. 2 = + = + = ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or transla tion of this work beyond that permitted by S ections 107 or 108 of the 1 976 United States Copyright Act without the permission of the copyright owner is unlawful.
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Solution Manual - Fundamentals of Heat and Mass Transfer 6th Edition
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PROBLEM 13.39 K NOWN: Long, inclined black surfaces maintained at prescribed temperatures. FIND: (a) Net radiation exchange between the two surfaces per unit length, (b) Net radiation transfer to surface A2 with black, insulated surface positioned as shown below; determine temperature of this urface. s
SCHEMATIC:
ASSUMPTIONS: (1) Surfaces behave as blackbodies, (2) Surfaces are very long in direction normal o page. t
ANALYSIS: (a) The net radiation exchange between two black surfaces is qij = AiFij(Ji - Jj) where Ji = and Jj = . Therefore, 4iT 4iT ( )4 412 1 12 1 2q A F T T= Noting that A1 = widthlength ( and that from symmetry, F)A 12 = 0.5, find ( )8 2 4 4 4 41212 qq 0.1 m 0.5 5.67 10 W / m K 1000 800 K 1680 W / m. = = =A < (b) With the insulated, black surface A3 positioned as shown above, a three-surface enclosure is formed. From an energy balance on the node representing A2, find 2 32 1q q q = + 2 [ ] [ ]2 3 32 b3 b2 1 12 b1 b2q A F E E A F E E = + .
To find Eb3, which at present is not known, perform an energy balance on the node representing A3. Note that A3 is adiabatic and, hence q3 = 0, q13 = q32. [ ] [ ]1 13 b1 b3 3 32 b3 b2A F E E A F E E = Since F13 = F23 = 0.5 and A1 = A3, it follows that ( )[ ]b3 b1E 1/ 2 E E= + b2 and ( ) ( )[ ]2 3 32 b1 b2 b2 12q A / F E E / 2 E q = + +A ( )8 2 4 4 4 42q 0.1 m 0.5 5.67 10 W / m K 1000 800 / 2 800 K = + 4 1680 W / m 2517 W / m+ = < Noting that Eb3 = = (1/2) [E
43T b1 +Eb2], it follows that
< ( ) ( )1/ 4 1/ 44 4 43 1 4T T T / 2 1000 800 / 2 K 916 K.2= + = + =
Nancy ProyectText BoxExcerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
