Thermodynamics HW Solutions 578

Thermodynamics HW Solutions 578 - 000 , 282 . 521 1 6974 ....

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Chapter 7 External Forced Convection 7-49 A long aluminum wire is cooled by cross air flowing over it. The rate of heat transfer from the wire per meter length when it is first exposed to the air is to be determined. Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air is an ideal gas with constant properties. 4 The local atmospheric pressure is 1 atm. Properties The properties of air at 1 atm and the film temperature of (T s + T )/2 = (370+30)/2 = 200 ° C are (Table A-15) 6974 . 0 Pr /s m 10 455 . 3 C W/m. 03779 . 0 2 5 - = × = ° = υ k 370 ° C Aluminum wire D = 3 mm Analysis The Reynolds number is 0 . 521 /s m 10 455 . 3 m) m/s)(0.003 (6 Re 2 5 = × = = D V V = 6 m/s T = 30 ° C The Nusselt number corresponding this Reynolds number is determined to be () [] () [] 48 . 11
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Unformatted text preview: 000 , 282 . 521 1 6974 . / 4 . 1 ) 6974 . ( ) . 521 ( 62 . 3 . 000 , 282 Re 1 Pr / 4 . 1 Pr Re 62 . 3 . 5 / 4 8 / 5 4 / 1 3 / 2 3 / 1 5 . 5 / 4 8 / 5 4 / 1 3 / 2 3 / 1 5 . = + + + = + + + = = k hD Nu Then the heat transfer coefficient and the heat transfer rate from the wire per meter length become C . W/m 6 . 144 ) 48 . 11 ( m 003 . C W/m. 03779 . 2 = = = Nu D k h W 463.4 = C 30)-)(370 m 5 C)(0.00942 . W/m 6 . 144 ( ) ( m 0.009425 = m) m)(1 003 . ( 2 2 2 = = = = T T hA Q DL A s s conv s & 7-33...
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