PART 11- CONVECTIVE MASS TRANSFER 2.1 Introduction 2.2 Convective Mass Transfer coefficient 2.3 Significant parameters in convective mass transfer 2.4 The application of dimensional analysis to Mass Transfer 2.4.1 Transfer into a stream flowing under forced convection 2.4.2 Transfer into a phase whose motion is due to natural convection 2.5 Analogies among mass, heat, and momentum transfer 2.5.1 Reynolds analogy 2.5.2 Chilton – Colburn analogy 2.6 Convective mass transfer correlations 2.6.1 For flow around flat plat 2.6.2 For flow around single sphere 2.6.3 For flow around single cylinder 2.6.4 For flow through pipes 2.7 Mass transfer between phases 2.8 Simultaneous heat and mass transfer 2.8.1 Condensation of vapour on cold surface 2.8.2 Wet bulb thermometer 2.1 Introduction Our discussion of mass transfer in the previous chapter was limited to molecular diffusion, which is a process resulting from a concentration gradient. In system involving liquids or gases, however, it is very difficult to eliminate convection from the overall mass-transfer process. Mass transfer by convection involves the transport of material between a boundary surface (such as solid or liquid surface) and a moving fluid or between two relatively immiscible, moving fluids. There are two different cases of convective mass transfer: 1. Mass transfer takes place only in a single phase either to or from a phase boundary, as in sublimation of naphthalene (solid form) into the moving air. 2. Mass transfer takes place in the two contacting phases as in extraction and absorption. In the first few section we will see equation governing convective mass transfer in a single fluid phase.
2.2 Convective Mass Transfer Coefficient In the study of convective heat transfer, the heat flux is connected to heat transfer coefficient as Q A = q = ( − ) ——————– (2.1) h t t s m The analogous situation in mass transfer is handled by an equation of the form N A = k c (C As −C A ) ——————– (2.2) The molar flux N is measured relative to a set of axes fixed in space. The driving force is A the difference between the concentration at the phase boundary, CAS (a solid surface or a fluid interface) and the concentration at some arbitrarily defined point in the fluid medium, C A . The convective mass transfer coefficient kC is a function of geometry of the system and the velocity and properties of the fluid similar to the heat transfer coefficient, h. 2.3 Significant Parameters in Convective Mass Transfer Dimensionless parameters are often used to correlate convective transfer data. In momentum transfer Reynolds number and friction factor play a major role. In the correlation of convective heat transfer data, Prandtl and Nusselt numbers are important. Some of the same parameters, along with some newly defined dimensionless numbers, will be useful in the correlation of convective mass-transfer data. The molecular diffusivities of the three transport process (momentum, heat and mass) have been defined as: Momentum diffusivity ν = µ —————————– (2.3) ρ k Thermal diffusivity α = ————————— (2.4) ρC p and Mass diffusivity D AB ————————— (2.5) It can be shown that each of the diffusivities has the dimensions of L2 / t, hence, a ratio of any of the two of these must be dimensionless.
The ratio of the molecular diffusivity of momentum to the molecular diffusivity of heat (thermal diffusivity) is designated as the Prandtl Number Momentum diffusivity ν Cp µ ———————— (2.6) = Pr = = Thermal diffusivity α K The analogous number in mass transfer is Schmidt number given as Momentum diffusivity = Sc = ν = µ ————– (2.7) Mass diffusivity D AB ρD AB The ratio of the molecular diffusivity of heat to the molecular diffusivity of mass is designated the Lewis Number, and is given by Thermal diffusivity α k Le ————- (2.8) = = = Mass diffusivity D AB ρC p D AB Lewis number is encountered in processes involving simultaneous convective transfer of mass and energy. Let us consider the mass transfer of solute A from a solid to a fluid flowing past the surface of the solid. The concentration and velocity profile is depicted in Figure 2.1. Figure 2.1 The concentration and velocity profile of solid dissolving in fluid For such a case, the mass transfer between the solid surface and the fluid may be written as N A = k c (C As −C A ∞) ———————- (2.2 a)
Since the mass transfer at the surface is by molecular diffusion, the mass transfer may also described by d C A ————————- (2.9) N A = −D AB d y y =0 When the boundary concentration, CAs is constant, equation (2.9) may be written as d (C A −C A s ) N A = −D AB ———————- (2.10) d y y =0 Equation (2.2a) and (2.10) may be equated, since they define the same flux of component A leaving the surface and entering the fluid d ————— (2.11) k c (C A s −C A ∞)= −D AB (C A −C A s ) d y y =0 This relation may be rearranged into the following form: ( ) k c d C A −C A s d y ——————– (2.12) D AB = − (C A −C A ∞) y =0 Multiplying both sides of equation (2.12) by a characteristic length, L we obtain the following dimensionless expression: ( ) k c L d C A −C A s d y y =0 —————– (2.13) = − D AB (C A S −C A ∞) L The right hand side of equation (2.13) is the ratio of the concentration gradient at the surface to an overall or reference concentration gradient; accordingly, it may be considered as the ratio of molecular mass-transport resistance to the convective mass-transport resistance of the fluid. This ratio is generally known as the Sherwood number, Sh and analogous to the Nusselt number Nu, in heat transfer.
2.4 Application of Dimensionless Analysis to Mass Transfer One of the method of obtaining equations for predicting mass-transfer coefficients is the use of dimensionless analysis. Dimensional analysis predicts the various dimensionless parameters which are helpful in correlating experimental data. There are two important mass transfer processes, which we shall consider, the transfer of mass into a steam flowing under forced convection and the transfer of mass into a phase which is moving as the result of natural convection associated with density gradients. 2.4.1 Transfer into a stream flowing under forced convection Consider the transfer of mass from the walls of a circular conduit to a fluid flowing through the conduit. The mass transfer is due to the concentration driving force C As – C A . The result of the dimensional analysis of mass transfer by forced convection in a circular conduit indicates that a correlating relation could be of the form, ( ) ————————— (2.14) Sh =ψ Re, Sc k D Dνρ µ c where Re = and S = and Sherwood Number Sh = C µ ρ D AB D AB Which is analogous to the heat transfer correlation ( ) —————————- (2.14) Nu =ψ Re, Pr 2.4.2 Transfer into a phase whose motion is due to Natural Convection Natural convection currents develop if there exists any variation in density within the fluid phase. The density variation may be due to temperature differences or to relatively large concentration differences. The result of the dimensional analysis of mass transfer by natural convection indicates that a correlating relation could be of the form, ( ) —————————- (2.15) Sh =ψ Gr AB , Sc where the Grashof number in heat transfer by natural convection L 3 g ∆ ρ ρA Gr = AB 2 µ
2.5 Analysis among Mass, Heat and Momentum Transfer Analogies among mass, heat and momentum transfer have their origin either in the mathematical description of the effects or in the physical parameters used for quantitative description. To explore those analogies, it could be understood that the diffusion of mass and conduction of heat obey very similar equations. In particular, diffusion in one dimension is described by the Fick’s Law as d C A J A = −D AB —————————— (2.16) d z Similarly, heat conduction is described by Fourier’s law as d T q = −k ——————————— (2.17) d z where k is the thermal conductivity. The similar equation describing momentum transfer as given by Newton’s law is d ν τ =−µ —————————– (2.18) d z where τ is the momentum flux (or shear stress) and µ is the viscosity of fluid. At this point it has become conventional to draw an analogy among mass, heat and momentum transfer. Each process uses a simple law combined with a mass or energy or momentum balance. In this section, we shall consider several analogies among transfer phenomenon which has been proposed because of the similarity in their mechanisms. The analogies are useful in understanding the transfer phenomena and as a satisfactory means for predicting behavior of systems for which limited quantitative data are available. The similarity among the transfer phenomena and accordingly the existence of the analogies require that the following five conditions exist within the system 1. The physical properties are constant 2. There is no mass or energy produced within the system. This implies that there is no chemical reaction within the system
3. There is no emission or absorption of radiant energy. 4. There is no viscous dissipation of energy. 5. The velocity profile is not affected by the mass transfer. This implies there should be a low rate of mass transfer. 2.5.1 Reynolds Analogy The first recognition of the analogous behavior of mass, heat and momentum transfer was reported by Osborne Reynolds in 1874. Although his analogy is limited in application, it served as the base for seeking better analogies. Reynolds postulated that the mechanisms for transfer of momentum, energy and mass are identical. Accordingly, k c h f = = ——————————– (2.19) ν ∞ ρν ∞ C p 2 Here h is heat transfer coefficient f is friction factor ν is velocity of free stream ∞ The Reynolds analogy is interesting because it suggests a very simple relation between different transport phenomena. This relation is found to be accurate when Prandtl and Schmidt numbers are equal to one. This is applicable for mass transfer by means of turbulent eddies in gases. In this situation, we can estimate mass transfer coefficients from heat transfer coefficients or from friction factors. 2.5.2 Chilton – Colburn Analogy Because the Reynold’s analogy was practically useful, many authors tried to extend it to liquids. Chilton and Colburn, using experimental data, sought modifications to the Reynold’s analogy that would not have the restrictions that Prandtl and Schmidt numbers must be equal to one. They defined for the j factor for mass transfer as k c ( )2 3 ————————— (2.20) j D = Sc ν ∞ The analogous j factor for heat transfer is
j H = St Pr 2 3 —————————– (2.21) where St is Stanton number = Nu h = Re Pr ρϑ C ∞ p Based on data collected in both laminar and turbulent flow regimes, they found f —————————– (2.21) j D = j H = 2 This analogy is valid for gases and liquids within the range of 0.6 < Sc < 2500 and 0.6 < Pr < 100. The Chilton-Colburn analogy has been observed to hold for many different geometries for example, flow over flat plates, flow in pipes, and flow around cylinders. Example 2.1 A stream of air at 100 kPa pressure and 300 K is flowing on the top surface of a thin flat sheet of solid naphthalene of length 0.2 m with a velocity of 20 m/sec. The other data are: Mass diffusivity of naphthalene vapor in air = 6 x 10 –6 2 m /sec –5 2 Kinematic viscosity of air = 1.5 x10 m .sc –5 3 Concentration of naphthalene at the air-solid naphthalene interface = 1×10 kmol/m Calculate: (a) the overage mass transfer coefficient over the flat plate (b) the rate of loss of naphthalene from the surface per unit width Note: For heat transfer over a flat plate, convective heat transfer coefficient for laminar flow can be calculated by the equation. 1 2 1 3 Nu = 0.664 Re Pr L you may use analogy between mass and heat transfer.
Solution: Given: Correlation for heat transfer 1 2 1 3 Nu = 0.664 Re Pr L The analogous relation for mass transfer is Sh = 0.664 Re1 2 Sc 1 3 —————————————–(1) L where Sh = Sherwood number = kL/D AB Re L = Reynolds number = Lυρ/µ Sc = Schmidt number = µ / (ρ D AB ) k = overall mass transfer coefficient L = length of sheet D AB = diffusivity of A in B υ = velocity of air µ = viscosity of air ρ = density of air, and µ/ρ = kinematic viscosity of air. Substituting for the known quantities in equation (1) ( ) ( )( ) 1 2 −5 1 3 k 0.2 0.2 20 1.5×10 =0.664 6×10 −6 1.5×10 −5 6×10 −6 k = 0.014 m/sec Rate of loss of naphthalene = k (C Ai – C A∞) = 0.014 (1 x 10 –5 – 0) –7 2 = 1.4024 x 10 kmol/m sec Rate of loss per meter width = (1.4024 x 10 –7) (0.2) = 2.8048 x 10 –8 kmol/m.sec = 0.101 gmol/m.hr.
2.5 Convective Mass Transfer Correlations Extensive data have been obtained for the transfer of mass between a moving fluid and certain shapes, such as flat plates, spheres and cylinders. The techniques include sublimation of a solid, vaporization of a liquid into a moving stream of air and the dissolution of a solid into water. These data have been correlated in terms of dimensionless parameters and the equations obtained are used to estimate the mass transfer coefficients in other moving fluids and geometrically similar surfaces. 2.5.1 Flat Plate From the experimental measurements of rate of evaporation from a liquid surface or from the sublimation rate of a volatile solid surface into a controlled air-stream, several correlations are available. These correlation have been found to satisfy the equations obtained by theoretical analysis on boundary layers, 1 2 1 3 5 ————- (2.22) Sh = 0.664 ReL Sc (laminar) Re L <3 * 10 0.8 1 3 5 Sh = 0.036 ReL Sc (turbulent ) Re L >3 * 10 ———– (2.23) Using the definition of j factor for mass transfer on equation (2.22) and (2.23) we obtain −1 2 5 ————- (2.24) j D = 0.664 ReL (laminar) Re L <3 * 10 −0.2 5 J D = 0.037 ReL (turbulent ) Re L >3 * 10 ———– (2.25) These equations may be used if the Schmidt number in the range 0.6 < Sc < 2500. Example 2.2 If the local Nusselt number for the laminar boundary layer that is formed over a flat plate is 1 2 1/ 3 Nu = 0.332 Re Sc x x Obtain an expression for the average film-transfer coefficient k c, when the Reynolds number for the plate is a) Re L = 100 000 b) Re L = 1500 000 The transition from laminar to turbulent flow occurs at Re x = 3 x 10 5.
Derivation: L ∫k c dx By definition : o k c = L ∫dx o k x c x v ρ µ and Nu x = ; Re x = ; Sc = ; D AB µ ρD AB For Re L = 100 000 ; (which is less than the Reynolds number corresponding to Transition value of 3 x 10 5) 1 L x v ρ 2 1 D AB 0.332 ( ) ∫ Sc 3 d x o µ x k c = L 1 3 v ρ1 2 0.332 ( ) Sc L µ d x = L D AB ∫x 1 2 o 1 2 L 0.332 1 3 v ρ 1 2 [ ] = 1 Sc µ D AB x o L 2 (i.e.) k c L = 0.664 Re L1 2 Sc 1 3 [answer (a)] D AB For Re L = 1500 000 (> 3 x 10 5) Lt 1 2 1 3 d x L 4 5 1 3 d x ∫0.332 Re x Sc x + ∫0.0292 Re x Sc x o L t k c = D AB L where L t is the distance from the leading edge of the plane to the transition point where Re x = 3 x 10 5.
1 3 v ρ1 2 L t d x 1 3 v ρ4 5 L d x 0.332 Sc ∫ 1 2 + 0.0292 Sc ∫ 1 5 µ o x µ L t x k c = D AB L 4 5 k c L 1 2 1 3 0.0292 1 3 4 5 L V ρ = 0.664 Re + [ ] D AB t Sx 4 5 Sc x L t µ 1 2 1 3 1 3 4 5 4 5 =0.664 Re t Sc + 0.0365 Sc (Re L −Re t ) k c L 1 2 1 3 4 5 1 3 4 5 1 3 =0.664 Re Sc + 0.0365 Re Sc −0.0365 Re Sc D AB t L t 5 where Re t = 3 x 10 2.5.2 Single Sphere Correlations for mass transfer from single spheres are represented as addition of terms representing transfer by purely molecular diffusion and transfer by forced convection, in the form Sh = Sh + C Re m Sc n ———————- (2.26) o where C, m and n are constants, the value of n is normally taken as 1/3 For very low Reynold’s number, the Sherwood number should approach a value of 2. This value has been derived in earlier sections by theoretical consideration of molecular diffusion from a sphere into a large volume of stagnant fluid. Therefore the generalized equation becomes m 1 3 ————————– (2.27) Sh = 2 + C Re Sc For mass transfer into liquid streams, the equation given by Brain and Hales 1 2 2 3 ————————– (2.28) ( ) Sh = 4 + 1.21 Pe AB correlates the data that are obtained when the mass transfer Peclet number, Pe AB is less than 10,000. This Peclet number is equal to the product of Reynolds and Schmidt numbers (i.e.)
Pe AB = Re Sc ———————————- (2.29) For Peclet numbers greater than 10,000, the relation given by Levich is useful 1 3 Sh = 1.01 Pe ————————— (2.30) AB The relation given by Froessling Sh = 2 + 0.552 Re 1 2 Sc 1 3 ———————– (2.31) correlates the data for mass transfer into gases for at Reynold’s numbers ranging from 2 to 800 and Schmidt number ranging 0.6 to 2.7. For natural convection mass transfer the relation given by Schutz 1 4 ———————– (2.32) Sh 2 0.59 (Gr Sc) = + AB is useful over the range 2 x 10 8 < Gr AB Sc < 1.5 x10 10 Example 2.3 The mass flux from a 5 cm diameter naphthalene ball placed in stagnant air at 40°C and atmospheric pressure, is 1.47×10 –3 mol/m 2. sec. Assume the vapor pressure of naphthalene to be 0.15 atm at 40°C and negligible bulk concentration of naphthalene in air. If air starts blowing across the surface of naphthalene ball at 3 m/s by what factor will the mass transfer rate increase, all other conditions remaining the same? For spheres : Sh = 2.0 + 0.6 (Re) 0.5 (Sc)0.33 where Sh is the Sherwood number and Sc is the Schmids number. The viscosity and density of air are 1.8 x 10 –5 kg/m.s and 1.123 kg/m 3, respectively and the gas constant is 3 82.06 cm . atm/mol.K. Calculations: k L Sh = c where L is the characteristic dimension for sphere L = Diameter. D AB
µ Sc = ρD AB D v ρ R = c µ Mass flux, N A = K c ∆c ——————————(1) Sh = 2.0 + 0.6 (Re) 0.5 (Sc) 0.33 k c D = 2.0 + 0.6 D V ρ0.5 µ 0.33 ———————– (2) D AB ρD AB also N = K G ∆ p A k Therefore c = K G R T Given: −3 mol K c N =1.47×10 2 = ∆p A m .sec RT k c 0.15 −3 −4 mol −0 =(1.47×10 )10 2 RT 1 cm .sec 1.47×10 −7 ( ) k = x 82.06x 273 +40 c 0.15 = 0.0252 cm/sec k c = 2.517x 10 –4 m/sec ——————————(3) Estimation of D AB : From (2), 2.517 10 −4 )(5 10 −2 ) ( x x =2 (since v = 0) D AB
–6 2 Therefore D AB = 6.2925 x 10 m /sec. And k x −2 −2 0.5 −5 0.33 c (5 10 ) (5×10 )x 3×1.123 1.8×10 = + 2 0.6 −6 −5 −6 x x x 6.2925 10 1.8 10 1.123(6.2925 10 ) 7946 k c = 2 + [ 0.6 (96.74) (1.361)] k c = 0.0102 m/sec. —————————————————– (4) (4) N A 2 0.0102 ⇒ = =40.5 (3) N A 1 2.517×10 −4 Therefore, rate of mass transfer increases by 40.5 times the initial conditions. 2.5.3 Single Cylinder Several investigators have studied the rate of sublimation from a solid cylinder into air flowing normal to its axis. Bedingfield and Drew correlated the available data in the form k G P Sc 0.56 / −0.4 ( ) ———————— (2.33) = 0.281 Re G m / which is valid for 400 < Re < 25000 and 0.6 < Sc < 2.6 Where Re / is the Reynold’s number in terms of the diameter of the cylinder, G m is the molar mass velocity of gas and P is the pressure. 2.5.4 Flow Through Pipes Mass transfer from the inner wall of a tube to a moving fluid has been studied extensively. Gilliland and Sherwood, based on the study of rate of vaporization of nine different liquids into air given the correlation p B l m , 0.83 0.44 ————– (2.34) Sh = 0.023 Re Sc P
where p B, lm is the log mean composition of the carrier gas, evaluated between the surface and bulk stream composition. P is the total pressure. This expression has been found to be valid over the range 2000 < Re < 35000 0.6 < Sc < 2.5 Linton and Sherwood modified the above relation making it suitable for large ranges of Schmidt number. Their relation is given as 0.83 1 3 ——————— (2.35) Sh = 0.023 Re Sc and found to be valid for 2000 < Re < 70000 and 1000 < Sc < 2260 Example 2.4 A solid disc of benzoic acid 3 cm in diameter is spin at 20 rpm and 25°C. Calculate the rate of dissolution in a large volume of water. Diffusivity of benzoic acid in water is 1.0×10 –5 cm 2/sec, and solubility is 0.003 g/cc. The following mass transfer correlation is applicable: Sh = 0.62 Re ½ Sc 1/3 2 D ωρ Where Re = and ω is the angular speed in radians/time. µ Calculations: Dissolution rate = N A S —————————-(1) where N A = mass flux, and S = surface area for mass transfer N A = k c (C As – C A ∞) ——————————– (2) where C As is the concentration of benzoic and at in water at the surface of the dose. C A∞ is the concentration benzoic acid in wate for an from the surface of the disc. Given: Sh = 0.62 Re ½ Sc 1/3
1 1 (i.e.) k c D D 2 ωρ2 µ 3 —————— (3) = 0.62 D AB µ ρD AB 1 rotation = 2 π radian Therefore 20 rotation per minute = 20 x 2 π radian/min 20 = x 2π radian sec 60 3 For water ρ = 1 g/cm µ = 1 centipoise = 0.01 g/cm.sec. From (3), 1 1 ωρ2 µ 3 k c = 0.62 D AB µ ρD AB 1 1 =0.62(1.0×10 5 ) (40π 60)x12 0.01 −5 3 0.01 1(1.0×10 = 8.973 x 10 –4 cm/sec. From (2), N A = 8.973 x 10 –4 (0.003 – 0) = 2.692 x 10 –6 g/cm 2.sec From (1), N A S = N A (2πr 2) = (2.692 x 10 –6 ) (2π x 1.5 2) = 3.805 x 10 –5 g/sec = 0.137 g/hr. 2.7 Mass transfer between phases: Instead of a fluid in contact with a solid, suppose we now consider two immiscible fluids, designated 1 and 2, in contact with each other. If fluid 1 has dissolved in it a substance A that is also soluble in fluid 2, then as soon as the two fluids are brought together, substance A will begin to diffuse into fluid 2. As long as the two phases remain in contact, the transport of A will continue until a condition of equilibrium is reached.
The situation discussed here occurs in a variety of engineering processes such as gas absorption, stripping, and in liquid – liquid extraction. In all these separation processes, two immiscible fluids are brought into contact and one or more components are transferred from one fluid phase to the other. In the system of fluids 1 and 2 with A, the transported component, the concentration gradients in the region of the interface between the two fluids are illustrated in Figure 2.1. Figure 2.1: Concentration gradients near the interface between immiscible fluids 1 and 2 Concentration CA1 and CA2 are the bulk phase concentrations in fluids 1 and 2, respectively, CAi is the concentration of A at the interface, and NA is the molar flux of A. For steady state conditions, we can define the flux of A as N A = k c1 (C A1 −C Ai ) = k c2 (C A i −C A 2 ) = K c (C A 1 −C A 2 ) ___________ (1) where k c = individual mass transfer coefficient defined in terms of the concentration difference in a single phase. K c = overall mass transfer coefficient defined in terms of the overall difference in composition. Equation (1) is analogous to that in heat transfer, where the individual coefficients h are related to the overall coefficient U. From equation (1), 1 1 1 + = _____________ (2) k k K c1 c2 c In equation (1), the potential for mass transfer is exposed in terms of composition. However, this is not always the most convenient way to express it. For example, if fluid 1 is a gas and
fluid 2 a liquid, as in gas absorption, the potential in gas phase is often expressed in terms of partial pressures, while that in the liquid phase may be expressed in terms of concentrations. The expression for the molar flux is then written for the individual phases as: N A =Kp (P AG −P Ai ) = K c (C Ai −C AL ) __________ (3) where k p = individual mass transfer coefficient for the gas phase with the potential defined in terms of partial pressures. P AG , C AL = partial pressure and concentration of A in the bulk gas and liquid phases, respectively. P Ai , C Ai = partial pressure and concentration of A, respectively, at the interface. At the interface, it is usually assumed the two phases are in equilibrium. This means that P Ai and C Ai are related by an equilibrium relationship such as ‘Henry’s law : P A i =H C A i __________ (4) where H is Henry’s law constant. The flux N A can also be expressed in terms of overall mass transfer coefficients as, N A =K p (P AG −P AE )=K c (C AE −C AL ) __________ (5) where K p = overall mass transfer coefficient with the overall potential defined in terms of partial pressures. K c = overall mass transfer coefficient with the overall potential defined in terms of concentrations. P AE , C AE = equilibrium composition. P AE is related to the bulk liquid composition C AL AS P AE = H C AL __________ (6) similarly, P AG C AE = __________ (7) H The relationship between the individual and overall coefficients is readily obtained through the use of equations (3) to (7) as 1 1 H H __________ (8) = + = K k k K p p c c
In many system, mass transfer resistance is mainly in one phase. For example, gases such as nitrogen and oxygen do not dissolve much in liquids. Their Henry’s law constant H is very large, thus K c ≈ k c is a good approximation. In this case, the liquid phase controls the mass transfer press since mass transfer is slowest there. 2.8 Simultaneous Heat and Mass Transfer Diffusional mass transfer is generally accompanied by the transport of energy, even with in an isothermal system. Since each diffusing constituent carries its own individual enthalpy, the heat flux at a given plane is expressed as q = N H i __________ (1) ∑ i i where q is the heat flux due to diffusion of mass past the given plane, and H i is the partial molar enthalpy of constituent i in the mixture. When there is a temperature difference, energy transfer also occurs by one of the three heat transfer mechanisms (conduction, convection, radiation) ; for example, the equation for energy transport by convection and molecular diffusion becomes q =h ∆T + N H i __________ (2) ∑ i i If the heat transfer is by conduction, the first term on the right hand side of equation (2) becomes ∆T k L where L is the thickness of the phase through which conduction takes place. The most common examples of processes involving heat and mass transfer are condensation of mist on a cold surface and in wet bulb thermometer. There are a number of such processes involving simultaneous heat and mass transfer such as in formation of fog, and in cooling towers. 2.8.1 Condensation of vapor on cold surface: A process important in many engineering processes as well as in day – to – day events involve the condensation of a vapor upon a cold surface. Examples of this process include “sweating” on cold water pipes and the condensation of moist vapor on a cold surface.
Figure illustrates the process which involves a film of condensed liquid following down a cold surface and a film of gas through which the condensate is transferred by molecular diffusion. This process involves the simultaneous transfer of mass and energy. The heat flux passing through the liquid film is given by q = h l (T 2 −T 3 ) __________ (1) This flux is also equal to the total energy transported by convection and molecular diffusion in the gas film.(i.e.,) q = h c (T 1 −T 2 )+N A M A (H 1 −H 2 ) __________ (2) where M A is the molecular weight of the diffusing constituent A. H 1 and H 2 are enthalpies of the vapor at plane 1 and liquid at plane 2. From equation (1) and (2) q = ( − =) ( − )+ ( − ) ________ (3) h l T 2 T 3 h c T 1 T 2 N A M A H 1 H 2 The molar flux NA is calculated by diffusion through stagnant gas model as −C D AB d y A N A = 1−y A d Z substituting the appropriate limits, the integral form of equation is (C D AB ) (y A1 −y A2 ) N A = ( avg ) __________ (4) Z 2 −Z 1 y , B lm 2.8.2 The Wet – bulb Thermometer Another example of simultaneous heat and mass transfer is that taking place in wet-bulb thermometer. This convenient device for measuring relative humidity of air consists of two conventional thermometers, one of which is clad in a cloth nick wet with water. The unclad dry-bulb thermometer measures the air’s temperature. The clad wet-bulb thermometer measures the colder temperature caused by evaporation of the water. We want to use this measured temperature difference to calculate the relative humidity in air. This relative humidity is defined as the amount of water actually in the air divided by the
amount at saturation at the dry-bulb temperature. To find this humidity, we can write equation for the mass and energy fluxes as: N A =k c (C Ai −C A ) =k y (y Ai −y A ) __________ (1) and q = h (T i − T) __________ (2) Where C Ai and C A are the concentrations of water vapor at the wet bulb’s surface and in the bulk of air, y Ai and y A are the corresponding mole fractions ; T i is the wet-bulb temperature, and T is the dry-bulb temperature. It can be noted that y is the value at saturation at T . Ai i In the air-film surroundings the wet-bulb, the mass and energy fluxes are coupled as N A λ = −q __________ (3) Where λ is the latent heat of vaporization of water. Thus, ( ) ( T ) k y y Ai −y A λ = h T − i Rearranging, λk T i = T − y (y Ai − y A ) __________ (4) h From Chilton – colbum analogy, j H = j D or h 2 k c 2 ( ) 3 = ( ) 3 __________ (5) Pr Sc 2 ρν Cp να For gas Pr ≈ 1 and Sc ≈ 1. Therefore equation (5) becomes, k c 1 = (as k y C ≈ k y ρ = k c ) h C p Therefore equation (4) becomes
λ T i = T − (y Ai −y A ) C p where C is the bumid heat of air. By similar method, the other industrial processes of p importance involving simultaneous heat and mass transfer such as humidification and drying can be analysed. Example 2.5. Air at 1 atm is blown past the bulb of a mercury thermometer. The bulb is covered with a wick. The wick is immersed in an organic liquid (molecular weight = 58). The reading of the thermometer is 7.6 °C. At this temperature, the vapor pressure of the liquid is 5 kPa. Find the air temperature, given that the ratio fo heat transfer coefficient to the mass transfer coefficient (psychrometric ratio) is 2 kJ/kg. Assume that the air, which is blown, is free from the organic vapor. Solution: For simultaneous mass and heat transfer, heat flux q and mass flux N A are related as q = N A λ —————————————– (1) where λ is the latent heat of vaporization. Mass flux is given by ‘ N = k Y (Y −Y’) ———————————- (2) A ω where k Y = mass transfer coefficient ‘ Y = mass ratio of vapor in surrounding air at saturation; and ω Y’ = mass ratio of vapor in surrounding air. Convective heat flux is given by q = h (T −Tω) ———————————– (3) where h = heat transfer coefficient; T ω = wet bulb temperature of air; and T = dry bulb temperature of air. Substituting for N A and q from equation (2) and equation (3) in equation (1), ‘ ‘ ‘ ( ) ( ) h T −T = k T −Y λ ω y ω
( ‘ ‘ ) λ Y −Y ω ——————————————— (4) T −T = ω h k Y ’ Given: Y = 0; λ = 360 kJ/kg; h/k Y = 2 kJ/kg.K; and T ω = 7.6°C ‘ kg organic vapor at saturation Y = ω kg dry air 5 58 = = 0.1038 101.3 −5 29 Substituting these in equation (4) ( )( ) 360 0.1038 −0 T −7.6 = = 18.69 2 T = 18.69 + 7.6 = 26.29°C Temperature of air = 26.29°C FDE312- PARTII-CONVECTIVE MASS TRANSFER -24