Thermodynamics HW Solutions 764

# Thermodynamics HW Solutions 764 - 4 C W/m 530 3 55 20 m 15...

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Chapter 9 Natural Convection 1 - 2 5 K 003317 . 0 K ) 273 5 . 28 ( 1 1 7286 . 0 Pr /s m 10 594 . 1 C W/m. 02577 . 0 = + = = = × = ° = f T k β υ Air T =25 ° C 15 cm 180 W ε = 0.85 T s = 32 ° C 50 cm 50 cm Analysis Heat loss from the horizontal top surface: The characteristic length in this case is m 125 . 0 )] m 5 . 0 ( ) m 5 . 0 [( 2 ) m 5 . 0 ( 2 = + = = p A L s c . Then, 6 2 2 5 3 -1 2 2 3 10 275 . 1 ) 7286 . 0 ( ) /s m 10 594 . 1 ( ) m 125 . 0 )( K 25 32 )( K 003317 . 0 )( m/s 81 . 9 ( Pr ) ( × = × = = c s L T T g Ra 15 . 18 ) 10 275 . 1 ( 54 . 0 54 . 0 4 / 1 6 4 / 1 = × = = Ra Nu 2 2 2 m 25 . 0 ) m 5 . 0 ( C . W/m 741 . 3 ) 15 . 18 ( m 125 . 0 C W/m. 02577 . 0 = = ° = ° = = top c A Nu L k h and W 55 . 6 C ) 25 32 )( m 25 . 0 )( C . W/m 741 . 3 ( ) ( 2 2 = ° ° = = T T hA Q s top top Heat loss from vertical side surfaces: The characteristic length in this case is the height of the box L c = L = 0.15 m. Then, 6 2 2 5 3 -1 2 2 3 10 204 . 2 ) 7286 . 0 ( ) /s m 10 594 . 1 ( ) m 15 . 0 )( K 25 32 )( K 003317 . 0 )( m/s 81 . 9 ( Pr ) ( × = × = = L T T g Ra s 55 . 20 7286 . 0 492 . 0 1 ) 10 204 . 2 ( 387 . 0 825 . 0 Pr 492 . 0 1 Ra 387 . 0 825 . 0 2 27 / 8 16 / 9 6 / 1 7 2 27 / 8 16 / 9 6 / 1 = + × + = + + = Nu 2 2 m 3 . 0 ) m 5 . 0 )( m 15 .
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Unformatted text preview: ( 4 C . W/m 530 . 3 ) 55 . 20 ( m 15 . C W/m. 02577 . = = ° = ° = = side A Nu L k h and W 41 . 7 C ) 25 32 )( m 3 . )( C . W/m 530 . 3 ( ) ( 2 2 = ° − ° = − = ∞ T T hA Q s side side & The radiation heat loss is W 34 . 20 ] ) K 273 25 ( ) K 273 32 )[( .K W/m 10 67 . 5 )( m 3 . 25 . )( 85 . ( ) ( 4 4 4 2 8 2 4 4 = + − + × + = − = − surr s s rad T T A Q σε & Then the fraction of the heat loss from the outer surfaces of the box is determined to be 19.1% = = + + = 1906 . W 180 W ) 34 . 20 41 . 7 55 . 6 ( f 9-87...
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## This note was uploaded on 01/24/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.

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