Thermodynamics HW Solutions 623

Thermodynamics HW Solutions 623 - Using the proper relation...

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Chapter 7 External Forced Convection 7-93 Wind is blowing parallel to the walls of a house with windows. The rate of heat loss through the window 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 flow is turbulent because of the intense vibrations involved. 5 The minivan is modeled as a rectangular box. 6 Air is an ideal gas with constant properties. 6 The pressure of air is 1 atm. Properties Assuming a film temperature of 5 ° C, the properties of air at 1 atm and this temperature are evaluated to be (Table A-15) 7350 . 0 Pr /s m 10 382 . 1 C W/m. 02401 . 0 2 5 - = × = ° = υ k Analysis Air flows along 1.2 m side. The Reynolds number in this case is [] 6 2 5 10 447 . 1 /s m 10 382 . 1 m) (1.2 m/s ) 3600 / 1000 60 ( Re × = × × = = L V L Air V = 60 km/h T 2 = -2 ° C L = 1.2 m WINDOW T 1 = 22 ° C which is greater than the critical Reynolds number. Thus we have combined laminar and turbulent flow.
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Unformatted text preview: Using the proper relation for Nusselt number, heat transfer coefficient is determined to be [ ] C . W/m 93 . 40 ) 2046 ( m 2 . 1 C W/m. 02401 . 2046 ) 7350 . ( 871 ) 10 447 . 1 ( 037 . Pr ) 871 Re 037 . ( 2 3 / 1 8 . 6 3 / 1 8 . = = = = = = = Nu L k h k hL Nu L The thermal resistances are 2 m 5.4 = m) m)(1.5 2 . 1 ( 3 = s A C/W 0045 . ) m C)(5.4 . W/m 93 . 40 ( 1 1 C/W 0012 . ) m C)(5.4 W/m. (0.78 m 005 . C/W 0231 . ) m C)(5.4 . W/m 8 ( 1 1 2 2 , 2 2 2 , = = = = = = = = = s o o conv s cond s i i conv A h R kA L R A h R Then the total thermal resistance and the heat transfer rate through the 3 windows become W 833.3 = = = = + + = + + = C/W 0288 . C )] 2 ( 22 [ C/W 0288 . 0045 . 0012 . 0231 . 2 1 , , total o conv cond i conv total R T T Q R R R R & 7-78...
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