Thermodynamics HW Solutions 595

# Thermodynamics HW Solutions 595 - Chapter 7 External Forced...

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Chapter 7 External Forced Convection kg/s 862 . 1 m) m)(1 05 m/s)(8)(0. 8 . 3 )( kg/m 225 . 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 42 . 28 C) J/kg kg/s)(1007 (1.862 C) W/m 5 . 87 )( m 222 . 4 ( exp ) 15 90 ( 90 exp ) ( 2 2 ° = ° ° = = p s i s s e C m h A T T T T C 07 . 68 )] 42 . 28 90 /( ) 15 90 ln[( ) 42 . 28 90 ( ) 15 90 ( )] /( ) ln[( ) ( ) ( ln ° = = = Δ e s i s e s i s T T T T T T T T T W 25,148 = ° ° = Δ = ) C 07 . 68 )( m C)(4.222 W/m 5 . 87 ( 2 2 ln T hA Q s For this square in-line tube bank, the friction coefficient corresponding to Re D = 9075 and S L / D = 5/2.1 = 2.38 is, from Fig. 7-27a, f = 0.22. Also,
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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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