Thermodynamics HW Solutions 598

Thermodynamics HW Solutions 598 - Chapter 7 External Forced...

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Chapter 7 External Forced Convection kg/s 504 . 2 m) m)(1 .04 m/s)(10)(0 2 . 5 )( kg/m 204 . 1 ( ) ( 3 = = ρ = = L S N m m T T i i V Then the fluid exit temperature, the log mean temperature difference, and the rate of heat transfer become C 68 . 49 C) J/kg kg/s)(1007 (2.504 C) W/m 3 . 116 )( m 05 . 10 ( exp ) 20 100 ( 100 exp ) ( 2 2 ° = ° ° = = p s i s s e C m h A T T T T C 01 . 64 )] 68 . 49 100 /( ) 20 100 ln[( ) 68 . 49 100 ( ) 20 100 ( )] /( ) ln[( ) ( ) ( ln ° = = = Δ e s i s e s i s T T T T T T T T T W 74,836 = ° ° = Δ = ) C 01 . 64 )( m C)(10.05 W/m 3 . 116 ( 2 2 ln T hA Q s ( b ) For this staggered tube bank, the friction coefficient corresponding to Re D = 7713 and S T / D = 4/1.6 = 2.5 is, from Fig. 7-27b, f = 0.33. Also, χ = 1 for the square arrangements. Then the pressure drop across the tube bank becomes
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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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