Thermodynamics HW Solutions 763

Thermodynamics HW Solutions 763 - Chapter 9 Natural...

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Chapter 9 Natural Convection kg/s 10 066 . 3 kJ/kg 7 . 333 kJ/s 10 23 . 10 5 3 × = × = = = if if h Q m h m Q Therefore, the melting of the ice in the chest completely will take days 11.3 h 271.8 = = × = × = = Δ ⎯→ Δ = s 10 786 . 9 kg/s 10 066 . 3 kg 30 5 5 m m t t m m (b) The temperature drop across the styrofoam will be much greater in this case than that across thermal boundary layer on the surface. Thus we assume outer surface temperature of the styrofoam to be 19 ° . Radiation heat transfer will be neglected. The properties of air at 1 atm and the film temperature of ( T C s + T )/2 = (19+20)/2 = 19.5 ° C are (Table A-15) 1 - 2 5 K 00342 . 0 K ) 273 5 . 19 ( 1 1 7311 . 0 Pr /s m 10 512 . 1 C W/m. 0251 . 0 = + = = = × = ° = f T k β υ The characteristic length in this case is the width of the chest, L c = W = 0.4 m. Then, 538 , 367 /s m 10 512 . 1 ) m 4 . 0 ( m/s) 3600 / 1000 50 ( Re 2 5 = × × = = W V which is less than critical Reynolds number (
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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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