Thermodynamics HW Solutions 620

Thermodynamics HW Solutions 620 - . ( 871 ) 10 792 . 7 (...

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Chapter 7 External Forced Convection Review Problems 7-90 Wind is blowing parallel to the walls of a house. The rate of heat loss from the wall is to be determined. Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Re cr = 5 × 10 5 . 3 Radiation effects are negligible. 4 Air is an ideal gas with constant properties. 5 The pressure of air is 1 atm. Properties Assuming a film temperature of T f = 10 ° C for the outdoors, the properties of air are evaluated to be (Table A-15) 7336 . 0 Pr /s m 10 426 . 1 C W/m. 02439 . 0 2 5 - = × = ° = υ k Analysis Air flows along 8-m side. The Reynolds number in this case is [] 6 2 5 10 792 . 7 /s m 10 426 . 1 m) (8 m/s ) 3600 / 1000 50 ( Re × = × × = = L V L which is greater than the critical Reynolds number. Thus we have combined laminar and turbulent flow. Using the proper relation for Nusselt number, heat transfer coefficient is determined to be [ ] C . W/m 78 . 30 ) 096 , 10 ( m 8 C W/m. 02439 . 0 096 , 10 ) 7336
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Unformatted text preview: . ( 871 ) 10 792 . 7 ( 037 . Pr ) 871 Re 037 . ( 2 3 / 1 8 . 6 3 / 1 8 . = = = = = = = Nu L k h k L h Nu o L The thermal resistances are 2 m 24 = m) m)(8 3 ( = = wL A s Air V = 50 km/h T 2 = 4 C L = 8 m WALL T 1 = 22 C R i R insulation R o T 1 T 2 C/W 0014 . ) m C)(24 . W/m 78 . 30 ( 1 1 C/W 1408 . m 24 C/W . m 38 . 3 ) 38 . 3 ( C/W 0052 . ) m C)(24 . W/m 8 ( 1 1 2 2 2 2 2 2 = = = = = = = = = s o o s value insulation s i i A h R A R R A h R Then the total thermal resistance and the heat transfer rate through the wall are determined from W 122.1 = = = = + + = + + = C/W 1474 . C ) 4 22 ( C/W 1474 . 0014 . 1408 . 0052 . 2 1 total o insulation i total R T T Q R R R R & 7-75...
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This note was uploaded on 01/22/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.

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