Heat Chap14-119

Heat Chap14-119 - Chapter 14 Mass Transfer 14-119 A person...

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Chapter 14 Mass Transfer 14-119 A person is standing outdoors in windy weather. The rates of heat loss from the head by radiation, forced convection, and evaporation are to be determined for the cases of the head being wet and dry. Assumptions 1 The low mass flux conditions exist so that the Chilton-Colburn analogy between heat and mass transfer is applicable since the mass fraction of vapor in the air is low (about 2 percent for saturated air at 300 K). 2 Both air and water vapor at specified conditions are ideal gases (the error involved in this assumption is less than 1 percent). 3 The head can be approximated as a sphere of 30 cm diameter maintained at a uniform temperature of 30 ° C. 4 The surrounding surfaces are at the same temperature as the ambient air. Properties The air-water vapor mixture is assumed to be dilute, and thus we can use dry air properties for the mixture. The properties of air at the free stream temperature of 25 ° C and 1 atm are, from Table A-15, /s m 10 56 . 1 s kg/m 10 85 . 1 73 . 0 Pr , C W/m 0255 . 0 2 5 5 - - × = × = = = ν μ k Also, s = = × ° - @ . 30 5 187 10 C kg / m s . The mass diffusivity of water vapor in air at the average temperature of (25 + 30)/2 = 27.5 ° C = 300.5 K is, from Eq. 14-15, ( 29 m²/s 10 55 . 2 atm 1 K 5 . 300 10 87 . 1 10 87 . 1 5 072 . 2 10 072 . 2 10 air - O H 2 - - - × = × = × = = P T D D AB The saturation pressure of water at 25 ° C is P sat@25 C kPa. ° = 3169 . Properties of water at 30 ° C are h P fg v = = 2431 4 246 kJ / kg and kPa . (Table A-9). The gas constants of dry air and water are R air = 0.287 kPa.m 3 /kg.K and R water = 0.4615 kPa.m 3 /kg.K (Table A-1). Also, the emissivity of the head is given to be 0.95. Analysis ( a ) When the head is dry, heat transfer from the head is by forced convection and radiation only. The radiation heat transfer is W 3 . 8 ] ) K 273 25 ( ) K 273 30 )[( K W/m 10 67 . 5 ]( m) 3 . 0 ( )[ 95 . 0 ( ) ( 4 4 4 2 8 2 4 surr 4 rad = + - + × = - = - π σ ε T T A Q s s The Reynolds number for flow over the head is 550 , 133 /s m 10 1.56 ) m 0.3 )( m/s 6 . 3 / 25 ( Re 2 5 = × = = - D V Then the Nusselt number and the heat transfer coefficient become [ ] ( 29 ( 29 [ ]( 29 269 10 87 . 1 10 85 . 1 73 . 0 550 , 133 06 . 0 550 , 133 4 . 0 2 Pr Re 06 . 0 Re 4 . 0 2 Nu 4 / 1 5 5 4 . 0 3 / 2 2 / 1 4 / 1 0.4 3 / 2 2 / 1 = × × + + = + + = - - s C W/m 9 . 22 (269) m 0.3 C W/m 0.0255 2 ° = ° = = Nu D k h Then the rate of convection heat transfer from the head becomes W 32.3 C ) 25 30 ]( ) m 3 . 0 ( )[ C . W/m 9 . 22 ( ) ( 2 2 conv = ° - ° = - = T T A h Q s s Therefore, 14-79 Air 25 ° C 1 atm 25 km/h D =30 cm Wet 30 ° C Evaporation
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Chapter 14 Mass Transfer W 40.6 = + = + = 3 . 8 32.3 rad conv dry total, Q Q Q ( b ) When the head is wet, there is additional heat transfer mechanism by evaporation. The Schmidt number is 612 . 0 /s m 10 2.55 /s m 10 56 . 1 Sc 2 5 2 5 = × × = = - - AB D ν The Sherwood number and the mass transfer coefficients are determined to be [ ] ( 29 ( 29 [ ]( 29 251 10 87 . 1 10 85 . 1 612 . 0 550 , 133 06 . 0 550 , 133 4 . 0 2 Sc Re 06 . 0 Re 4 . 0 2 Sh 4 / 1 5 5 4 . 0 3 / 2 2 / 1 4 / 1 0.4 3 / 2 2 / 1 = × × + + = + + = - - s μ m/s 0213 . 0
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This note was uploaded on 08/25/2009 for the course AET AET432 taught by Professor Rajadas during the Spring '06 term at ASU.

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Heat Chap14-119 - Chapter 14 Mass Transfer 14-119 A person...

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