Thermodynamics HW Solutions 320

Thermodynamics HW Solutions 320 - Chapter 4 Transient Heat...

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Chapter 4 Transient Heat Conduction ) 1881 . 0 ( 1 1 0 , 0 2 1 2 1 543 . 0 163 5 . 4 163 77 λ τλ θ = = ⎯→ = = e A e A T T T T i sph It is determined from Table 4-1 by trial and error that this equation is satisfied when Bi = 4.3, which corresponds to 7402 . 1 and 4900 . 2 1 1 = = A . Then the heat transfer coefficient can be determined from. C . W/m 22.5 2 ° = ° = = ⎯→ = ) m 08603 . 0 ( ) 3 . 4 )( C W/m. 45 . 0 ( o o r kBi h k hr Bi ( b ) The temperature at the surface of the rib is C 142.1 ° = ⎯→ = = = = ) , ( 132 . 0 163 5 . 4 163 ) , ( 49 . 2 ) 49 . 2 sin( ) 7402 . 1 ( / ) / sin( ) , ( ) , ( ) 1881 . 0 ( ) 49 . 2 ( 1 1 1 2 2 1 t r T t r T e r r r r e A T T T t r T t r o o o o o o i o sph o ( c ) The maximum possible heat transfer is kJ 2080 C ) 5 . 4 163 )( C kJ/kg. 1 . 4 )( kg 2 . 3 ( ) ( max = ° ° = = i p T T mC Q Then the actual amount of heat transfer becomes kJ 1512 = = = = = = kJ) 2080 )( 727 . 0 ( 727 . 0 727 . 0 ) 49 . 2 ( ) 49 . 2 cos( ) 49 . 2 ( ) 49 . 2 sin( ) 543 . 0 ( 3 1 ) cos( ) sin( 3 1 max 3 3 1 1 1 1 , max Q Q Q Q sph o ( d ) The cooking time for medium-done rib is determined to be hr 4 = = = × = α τ = = τ ⎯→ = ⎯→ = = θ τ τ λ min 0 . 240 s 403 , 14 /s) m 10 91 . 0 ( m) 08603 . 0 )( 177 . 0 ( 177 . 0 ) 7402 .
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This note was uploaded on 01/19/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.

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