Thermodynamics HW Solutions 632

# Thermodynamics HW Solutions 632 - W 4 27 m 28 51 C W/m 73...

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Chapter 7 External Forced Convection W 7361 ) m 81 . 52 )( C . W/m 71 . 11 ( 1 m) m)(2 C)(2.05 W/m. (0.035 4 m ) 2 05 . 2 ( C )] 196 ( 20 [ 1 4 m 81 . 52 ) m 1 . 4 ( 2 2 2 1 1 2 tan , tan , 2 2 2 = ° + ° ° = + = + = = = = π ππ s k s conv insulation k s s hA r kr r r T T R R T T Q D A The rate of evaporation of liquid nitrogen then becomes kg/s 0.0372 = = = ⎯→ = kJ/kg 198 kJ/s 361 . 7 if if h Q m h m Q ( c ) We use the dynamic viscosity value at the new estimated surface temperature of 0 ° C to be . Noting that D = D kg/m.s 10 729 . 1 5 × = μ 0 = 4.04 m in this case, the Nusselt number becomes [] 6 2 5 10 961 . 2 /s m 10 516 . 1 m) (4.04 m/s 1000/3600) (40 Re × = × × = υ = D V [] [] 1724 10 729 . 1 10 825 . 1 ) 7309 . 0 ( ) 10 961 . 2 ( 06 . 0 ) 10 961 . 2 ( 4 . 0 2 Pr Re 06 . 0 Re 4 . 0 2 4 / 1 5 5 4 . 0 3 / 2 6 5 . 0 6 4 / 1 4 . 0 3 / 2 5 . 0 = × × × + × + = μ μ + + = = s k hD Nu and C . W/m 73 . 10 ) 1724 ( m 04 . 4 C W/m. 02514 . 0 2 ° = ° = = Nu D k h The rate of heat transfer to the liquid nitrogen is
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Unformatted text preview: W 4 . 27 ) m 28 . 51 )( C . W/m 73 . 10 ( 1 m) m)(2 C)(2.02 W/m. (0.00005 4 m ) 2 02 . 2 ( C )] 196 ( 20 [ 1 4 m 28 . 51 ) m 04 . 4 ( 2 2 2 1 1 2 tan , tan , 2 2 2 = ° + ° − ° − − = + − − = + − = = = = ∞ ∞ s k s conv insulation k s s hA r kr r r T T R R T T Q D A & The rate of evaporation of liquid nitrogen then becomes kg/s 10 1.38 4-× = = = ⎯→ ⎯ = kJ/kg 198 kJ/s 0274 . if if h Q m h m Q & & & & 7-87...
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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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