Thermodynamics HW Solutions 538

# Thermodynamics HW Solutions 538 - Assumptions 1 Steady...

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Chapter 6 Fundamentals of Convection 6-52 The windshield of a car is subjected to parallel winds. The drag force the wind exerts on the windshield is to be determined. Assumptions 1 Steady operating conditions exist. 2 The edge effects are negligible. Properties The properties of air at 0 ° C and 1 atm are (Table A-15) ρ = 1.292 kg/m 3 , C p =1.006 kJ/kg-K, Pr = 0.7362 Analysis The average heat transfer coefficient is C W/m 57 . 11 C ) 0 4 )( m 1.8 (0.6 W 50 ) ( ) ( 2 2 ° = ° × = = = T T A Q h T T hA Q s s s s The average friction coefficient is determined from the modified Reynolds analogy to be Air 0 ° C 80 km/h 0.6 m 1.8 m Windshield T s =4 ° C 0006534 . 0 C) J/kg m/s)(1006 6 . 3 / 80 )( kg/m 292 . 1 ( ) 7362 . 0 ( C) W/m (11.57 2 Pr 2 3 3 / 2 2 2/3 = ° ° = ρ = p f C h C V The drag force is determined from N 0.225 = × = ρ = 2 2 3 2 2 kg.m/s 1 N 1 2 ) m/s 6 . 3 / 80 )( kg/m (1.292 ) m 8 . 1 6 . 0 ( 0006534 . 0 2 V s f f A C F 6-53 An airplane cruising is considered. The average heat transfer coefficient is to be determined.
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Unformatted text preview: Assumptions 1 Steady operating conditions exist. 2 The edge effects are negligible. Properties The properties of air at -50 ° C and 1 atm are (Table A-15) C p =0.999 kJ/kg-K Pr = 0.7440 The density of air at -50 ° C and 26.5 kPa is 3 kg/m 4141 . 273)K 50 kJ/kg.K)(-287 . ( kPa 5 . 26 = + = = ρ RT P Analysis The average heat transfer coefficient can be determined from the modified Reynolds analogy to be C W/m 89.6 2 ⋅ = ° ⋅ = ρ = 3 / 2 3 2/3 ) 7440 . ( C) J/kg m/s)(999 6 . 3 / 800 )( kg/m 4141 . ( 2 0016 . Pr 2 p f C C h V Air -50 ° C 800 km/h 3 m 25 m Wing T s =4 ° C 6-54, 6-55 Design and Essay Problems KJ 6-27...
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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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