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5 2 evapm 2.76 10 kg/s m . \u2212 \u2032\u2032 = × \u22c5\ufffd < Because evaporation is a convection mass transfer process, it also follows that evap Am n\u2032\u2032 \u2032\u2032=\ufffd or in terms of the rate equation, ( ) ( ) ( ) ( ) ( ) evap m A,s A, m A,sat s A,sat evap m A,sat m h h T T m h 305K 1 , \u3c1 \u3c1 \u3c1 \u3c6 \u3c1 \u3c1 \u3c6 \u221e \u221e \u221e \u221e \uf8ee \uf8f9\u2032\u2032 = \u2212 = \u2212\uf8f0 \uf8fb \u2032\u2032 = \u2212 \ufffd \ufffd and solving for the convection mass transfer coefficient, ( ) ( ) ( ) 5 2 evap m 3A,sat m 2.76 10 kg/s mh 305K 1 0.0336 kg/m 1 0.4\u3c1 \u3c6 \u2212 \u221e \u2032\u2032 × \u22c5 = = \u2212 \u2212 \ufffd 3 mh 1.37 10 m/s. \u2212 = × < COMMENTS: Conservation of species has been applied in exactly the same way as a conservation of energy. Note the sign convention. PROBLEM 6.48 KNOWN: CO2 concentration in air and at the surface of a green leaf. Convection mass transfer coefficient. FIND: Rate of photosynthesis per unit area of leaf. SCHEMATIC: ANALYSIS: Assuming that the CO2 (species A) is consumed as a reactant in photosynthesis at the same rate that it is transferred across the atmospheric boundary layer, the rate of photosynthesis per unit leaf surface area is given by the rate equation, ( )A m A, A,sn h .\u3c1 \u3c1\u221e\u2032\u2032 = \u2212 Substituting numerical values, find ( )2 -4 4 3An 10 m/s 6 10 5 10 kg/m\u2212 \u2212\u2032\u2032 = × \u2212 × 6 2 An 10 kg/s m . \u2212 \u2032\u2032 = \u22c5 < COMMENTS: (1) It is recognized that CO2 transport is from the air to the leaf, and (\u3c1A,s - \u3c1A,\u221e) in the rate equation has been replaced by (\u3c1A,\u221e - \u3c1A,s). (2) The atmospheric concentration of CO2 is known to be increasing by approximately 0.3% per year. This increase in \u3c1A,\u221e will have the effect of increasing the photosynthesis rate and hence plant biomass production. PROBLEM 6.49 KNOWN: Species concentration profile, CA(y), in a boundary layer at a particular location for flow over a surface. FIND: Expression for the mass transfer coefficient, hm, in terms of the profile constants, CA,\u221e and DAB. Expression for the molar convection flux, \u2032\u2032NA. SCHEMATIC: ASSUMPTIONS: (1) Parameters D, E, and F are constants at any location x, (2) DAB, the mass diffusion coefficient of A through B, is known. ANALYSIS: The convection mass transfer coefficient is defined in terms of the concentration gradient at the wall, ( ) )( ) A y=0 m AB A,s A, C / y h x D . C C \u2202 \u2202 \u221e = \u2212 \u2212 The gradient at the surface follows from the profile, CA(y), ( )2A y=0y=0 C Dy Ey F E. y y\u2202 \u2202\u2202 \u2202 \uf8f9 = + + = +\uf8fa\uf8fb Hence, ( ) ( ) ( )AB ABm A,s A, A, D E D Eh x . C C F C \u221e \u221e \u2212 = \u2212 = \u2212 \u2212 < The molar flux follows from the rate equation, ( ) ( ) ( )ABA m A,s A, A,s A,A,s A, D EN h C C C C . C C\u221e \u221e \u221e \u2212 \u2032\u2032 = \u2212 = \u22c5 \u2212 \u2212 A ABN D E.\u2032\u2032 = \u2212 < COMMENTS: It is important to recognize that the influence of species B is present in the property DAB. Otherwise, all the parameters relate to species A. PROBLEM 6.50 KNOWN: Steady, incompressible flow of binary mixture between infinite parallel plates with different species concentrations. FIND: Form of species continuity equation and concentration distribution. Species flux at upper surface. SCHEMATIC: ASSUMPTIONS: (1) Two-dimensional flow, (2) No chemical reactions, (3) Constant properties. ANALYSIS: For fully developed conditions, ¶CA/¶x = 0. Hence with v = 0, the species conservation equation reduces to 2 A 2 d C 0. dy = < Integrating twice, the general form of the species concentration distribution is ( )A 1 2C y C y C .= + Using appropriate boundary conditions and evaluating the constants, ( ) ( ) ( ) A A,2 2 A,2 A A,1 1 A,1 A,2 C 0 C C =C C L C C C C /L, = ® = ® = - the concentration distribution is ( ) ( ) ( )A A,2 A,1 A,2C y C y/L C C .= + - < From Fick\u2019s law, the species flux is ( ) AA AB y=L dC N L D dy ¢¢ = - ( ) ( )ABA A,2 A,1DN L C C .L¢¢ = - < COMMENTS: An analogy between heat and mass transfer exists if viscous dissipation is negligible. The energy equation is then d 2 T/dy 2 = 0. Hence, both heat and species transfer are influenced only by diffusion. Expressions for T(y) and ( )q L¢¢ are analogous to those for CA(y) and ( )AN L .¢¢ PROBLEM 6.51 KNOWN: Flow conditions between two parallel plates, across which vapor transfer occurs. FIND: (a) Variation of vapor molar concentration between the plates and mass rate of water production per unit area, (b) Heat required to sustain the process. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Fully developed, incompressible flow with constant properties, (3) Negligible body forces, (4) No chemical reactions, (5) All work interactions, including viscous dissipation, are negligible. ANALYSIS: (a) The flow will be fully developed in terms of the vapor concentration field, as well as the velocity and temperature fields. Hence ( ) ( )A A A C 0 or C x,y C y . x \u2202 \u2202 = = Also, with \u2202CA/\u2202t = 0, AN 0,=\ufffd v = 0 and constant DAB, the species conservation equation reduces to 2 A 2 d C 0. dy = Separating and integrating twice, ( ) ( )A 1 2C y C y C .= + Applying the boundary conditions, ( ) ( ) A A,0 2 A,0 A,0 A,L A A,L A,L 1 2 1 C 0 C C C C C C L C C C L C C L = \u2192 = \u2212 = \u2192 = + = \u2212 find the species concentration distribution, ( ) ( ) ( )A A,0 A,0 A,LC y C C C y/L .= \u2212 \u2212 < From Fick\u2019s law, Eq. 6.19, the species transfer rate is A,0 A,LA A A,s AB AB y=0 C C CN N D D . y L \u2202 \u2202 \u2212\uf8f9 \u2032\u2032 \u2032\u2032= = \u2212 =\uf8fa\uf8fb Continued \u2026.. PROBLEM 6.51 (Cont.) Multiplying by the molecular weight of water vapor, \u39c A, the mass rate of water production per unit area is A,0 A,L A A A A AB C C n N D . L \u2212 \u2032\u2032 \u2032\u2032= =\u39c \u39c < (b) Heat must be supplied to the bottom surface in an amount equal to the latent and sensible heat transfer from the surface, lat sen A,s fg y=0 q q q dTq n h k . dy \u2032\u2032 \u2032\u2032 \u2032\u2032= + \uf8ee \uf8f9 \u2032\u2032 \u2032\u2032= + \u2212\uf8ef \uf8fa\uf8f0 \uf8fb The temperature distribution may be obtained by solving the energy equation, which, for the prescribed conditions, reduces to 2 2 d T 0. dy = Separating and integrating twice, ( ) 1 2T y C y C .= + Applying the boundary conditions, ( ) ( ) ( )0 2 0L 1 1 0 T 0 T C T T L T C T T / L = \u2192 = = \u2192 = \u2212 find the temperature distribution, ( ) ( )0 0 LT y T T T y/L.= \u2212 \u2212 Hence, ( )0 L y=0 T TdTk k . dy L \u2212\uf8f9 \u2212 =\uf8fa\uf8fb Accordingly, ( )A,0 A,L 0 L A AB fg C C T T q D h k . L L \u2212 \u2212 \u2032\u2032 = +\u39c < COMMENTS: Despite the existence of the flow, species and energy transfer across the air are uninfluenced by advection and transfer is only by diffusion. If the flow were not fully developed, advection would have a significant influence on the species concentration and temperature fields and hence on the rate of species and energy transfer. The foregoing results would, of course, apply in the case of no air flow. The physical condition is an example of Poiseuille flow with heat and mass transfer. PROBLEM 6.52 KNOWN: The conservation equations, Eqs. E.24 and E.31. FIND: (a) Describe physical significance of terms in these equations, (b) Identify approximations and special conditions used to reduce these equations to the boundary layer equations, Eqs. 6.33 and 6.34, (c) Identify the conditions under which these two boundary layer equations have the same form and, hence, an analogy will exist. ANALYSIS: (a) The energy conservation equation, Eq. E.24, has the form i i T T p p u v k k u v q. x y x x y y x y 1a 1b 2a 2b 3 4 5 ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶r r m ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ é ù é ùé ù+ = + + + + F +ê ú ê úê úë û ë û ë û & The terms, as identified, have the following phnysical significance: 1. Change