Heat Chap14-107

# Heat Chap14-107 - Chapter 14 Mass Transfer 14-107 A...

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Chapter 14 Mass Transfer 14-107 A raindrop is falling freely in atmospheric air. The terminal velocity of the raindrop at which the drag force equals the weight of the drop and the average mass transfer coefficient are to be determined. Assumptions 1 The low mass flux model and thus the 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 The raindrop is spherical in shape. 3 The reduction in the diameter of the raindrop due to evaporation when the terminal velocity is reached is negligible. Properties Because of low mass flux conditions, we can use dry air properties for the mixture. The properties of air at 1 atm and the free-stream temperature of 25ºC (and the dynamic viscosity at the surface temperature of 9ºC) are (Table A-15) /s m 10 562 . 1 kg/m 184 . 1 2 5 3 - × = = υ ρ kg/m.s 10 759 . 1 kg/m.s 10 849 . 1 5 K 282 @ , 5 - - × = × = s μ At 1 atm and the film temperature of (25+9)/2 = 17ºC = 290 K, the kinematic viscosity of air is, from Table A-11, /s m 10 488 . 1 2 5 - × = ν , while the mass diffusivity of water vapor in air is, Eq. 14-15, D D T P AB = = × = × = × - - - H O-air 2 2 K atm m s 187 10 187 10 290 1 2 37 10 10 2 072 10 2 072 5 . . ( ) . / . . Analysis The weight of the raindrop before any evaporation occurs is N 10 38 . 1 ) m/s 8 . 9 ( 6 m) (0.003 ) kg/m 1000 ( 4 2 3 3 - × = = = = π Vg mg F D The drag force is determined from F C A u D D N = ρ 2 2 where drag coefficient C D is to be determined using Fig. 10-20 which requires the Reynolds number. Since we do not know the velocity we cannot determine the Reynolds number. Therefore, the solution requires a trial-error approach. We choose a velocity and perform calculations to obtain the drag force. After a couple trial we choose a velocity of 8 m/ s. Then the Reynolds number becomes 1536 /s m 10 562 . 1 m) m/s)(0.003 (8 Re 2 5 = × = = - D V The corresponding drag coefficient from Fig. 10-20 is 0.5. Then, 4 2 3 2 2 10 34 . 1 2 ) m/s 8 )( kg/m 184 . 1 ( 4 ) m 003 . 0 ( ) 5 . 0 ( 2 - × = = = u A C F N D D which is sufficiently close to the value calculated before. Therefore, the terminal velocity of the raindrop is V = 8 m/s. The Schmidt number is 628 . 0 /s m 10 2.37 /s m 10 488 . 1 Sc 2 5 2 5 AB = × × = = - - D Then the Sherwood number can be determined from the forced heat convection relation for a sphere by replacing Pr by the Sc number to be [ ] ( 29 ( 29 [ ]( 29 9 . 21 10 759 . 1 10 849 . 1 628 . 0 1536 06 . 0 1536 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 AB mass D D h Then the mass transfer coefficient becomes m/s 0.173 = × = = - m 0.003 ) /s m 10 37 . 2 )( 9 . 21 ( Sh 2 5 mass D D h AB 14-64 Air 25 ° C 1 atm Raindrop D = 3 mm

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Chapter 14 Mass Transfer 14-108 Wet steel plates are to be dried by blowing air parallel to their surfaces. The rate of evaporation from both sides of a plate is to be determined.
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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 Chap14-107 - Chapter 14 Mass Transfer 14-107 A...

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