Thermodynamics HW Solutions 387

Thermodynamics HW Solutions 387 - ° C becomes s 7.2 =...

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Chapter 4 Transient Heat Conduction 4-118 Internal combustion engine valves are quenched in a large oil bath. The time it takes for the valve temperature to drop to specified temperatures and the maximum heat transfer are to be determined. Assumptions 1 The thermal properties of the valves are constant. 2 The heat transfer coefficient is constant and uniform over the entire surface. 3 Depending on the size of the oil bath, the oil bath temperature will increase during quenching. However, an average canstant temperature as specified in the problem will be used. 4 The Biot number is Bi < 0.1 so that the lumped system analysis is applicable (this assumption will be verified). Properties The thermal conductivity, density, and specific heat of the balls are given to be k = 48 W/m. ° C, ρ = 7840 kg/m 3 , and C p = 440 J/kg. ° C. Oil T = 45 ° C Engine valve T i = 800 ° C Analysis ( a ) The characteristic length of the balls and the Biot number are 1 . 0 024 . 0 ) C W/m. 48 ( ) m 0018 . 0 )( C . W/m 650 ( m 0018 . 0 8 m) 008 . 0 ( 8 . 1 8 8 . 1 2 ) 4 / ( 8 . 1 2 2 < = ° ° = = = = = = = k hL Bi D DL L D A V L c s c π Therefore, we can use lumped system analysis. Then the time for a final valve temperature of 400
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Unformatted text preview: ° C becomes s 7.2 = ⎯→ ⎯ = − − ⎯→ ⎯ = − − = ° ° = = = − − ∞ ∞ t e e T T T t T D C h V C hA b bt i p p s )t s 10468 . ( 1-3 2-1 45 800 45 400 ) ( s 10468 . m) C)(0.008 J/kg. 440 )( kg/m 1.8(7840 C) . W/m 650 ( 8 8 . 1 8 ρ ( b ) The time for a final valve temperature of 200 ° C is s 15.1 = ⎯→ ⎯ = − − ⎯→ ⎯ = − − − − ∞ ∞ t e e T T T t T bt i )t s 10468 . (-1 45 800 45 200 ) ( ( c ) The time for a final valve temperature of 46 ° C is s 63.3 = ⎯→ ⎯ = − − ⎯→ ⎯ = − − − − ∞ ∞ t e e T T T t T bt i )t s 10468 . (-1 45 800 45 46 ) ( ( d ) The maximum amount of heat transfer from a single valve is determined from ) (per valve = J 564 , 23 C ) 45 800 )( C J/kg. 440 )( kg 0709 . ( ] [ kg 0709 . 4 m) 10 . ( m) 008 . ( 8 . 1 ) kg/m 7840 ( 4 8 . 1 2 3 2 kJ 23.56 = ° − ° = − = = π = π ρ = ρ = i f p T T mC Q L D V m 4-98...
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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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