Thermodynamics HW Solutions 569

Thermodynamics HW Solutions 569 - . 3 . 000 , 282 Re 1 Pr 4...

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Chapter 7 External Forced Convection 7-42E A person extends his uncovered arms into the windy air outside. The rate of heat loss from the arm 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 arm is treated as a 2-ft-long and 3-in.-diameter cylinder with insulated ends. 5 The local atmospheric pressure is 1 atm. Properties The properties of air at 1 atm and the film temperature of (T s + T )/2 = (86+54)/2 = 70 ° F are (Table A-15E) 7306 . 0 Pr /s ft 10 1643 . 0 F Btu/h.ft. 01457 . 0 2 3 - = × = ° = υ k Air V = 20 mph T = 54 ° F Analysis The Reynolds number is [ ] 4 2 3 10 463 . 4 /s ft 10 1643 . 0 ft (3/12) ft/s /3600) 5280 (20 Re × = × × = υ = D V Arm D = 3 in T s = 86 ° F The Nusselt number corresponding this Reynolds number is determined to be 6 . 129 000 , 282 10 463 . 4 1 7306 . 0 4 . 0 1 ) 7306 . 0 ( ) 10 463 . 4 ( 62
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Unformatted text preview: . 3 . 000 , 282 Re 1 Pr 4 . 1 Pr Re 62 . 3 . 5 / 4 8 / 5 4 4 / 1 3 / 2 3 / 1 5 . 4 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 arm becomes F . Btu/h.ft 557 . 7 ) 6 . 129 ( ft ) 12 / 3 ( F Btu/h.ft. 01457 . 2 = = = Nu D k h Btu/h 379.8 = F 54)-)(86 ft F)(1.571 . Btu/h.ft 557 . 7 ( ) ( ft 1.571 = ft) ft)(2 12 / 3 ( 2 2 2 = = = = T T hA Q DL A s s conv s & 7-24...
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