Etiket Arşivleri: Mass Transfer

Absorption Operations in Packed Column

Absorption operations in packed column

A packed column is a cylindrical vessel under pressure, filled with a suitable packing material between the internal section 1 and 2.

The liquid flows down by gravity through the surface of the packing elements forming a thin film layer around it. The gas rises in counter-current mode with the liquid stream.

The mass transfer takes place through the interface film present between the gas and the liquid.

The column is now a continuous differential contacting devices. There is no distinguishable stage (like the plates in a tray column) present anymore.

Introduction to Mass Transfer

Introduction to Mass Transfer


Three fundamental transfer processes:

i)Momentum transfer

ii)Heat transfer

iii)Mass transfer

Mass transfer may occur in a gas mixture, a liquid solution or solid.

Mass transfer occurs whenever there is a gradient in the concentration of a species.

The basic mechanisms are the same whether the phase is a gas, liquid, or solid.

Definition of Concentration

  1. i) Number of molecules of each species present per unit volume (molecules/m3)

  2. ii) Molar concentration of species i = Number of moles of i per unit volume (kmol/m3)

iii) Mass concentration = Mass of i per unit volume (kg/m3)

Diffusion phenomena

Fick’s law: linear relation between the rate of diffusion of chemical species and the concentration gradient of that species.

Thermal diffusion: Diffusion due to a temperature gradient. Usually negligible unless the temperature gradient is very large.

Pressure diffusion: Diffusion due to a pressure gradient. Usually negligible unless the pressure gradient is very large.

Forced diffusion: Diffusion due to external force field acting on a molecule. Forced diffusion occurs when an electrical field is imposed on an electrolyte ( for example, in charging an automobile battery)

Knudsen diffusion: Diffusion phenomena occur in porous solids.

Whenever there is concentration difference in a medium,

  nature tends to equalize

  things by forcing a flow

  from the high to the low

  concentration region.

The molecular transport process of mass is characterized by the general equation:

    Rate of transfer process = driving force  resistance

Example of Mass Transfer Processes

Consider a tank that is divided into two equal parts by a partition.

Initially, the left half of the tank contains nitrogen N2 gas while the right half contains O2 at the same temperature and pressure.

When the partition is removed the N2  molecules will start diffusing into the air while the O2 molecules diffuse into the N2.

If we wait long enough, we will have a homogeneous mixture of N2 and O2 in the tank.

Liquid in open pail of water evaporates into air because of the difference in concentration of water vapor at the water surface and the surrounding air.

A drop of blue liquid dye is added to a cup of water. The dye molecules will diffuse slowly by molecular diffusion to all parts of the water.

Molecular Diffusion Equation

Fick’s Law

        is the molar flux of component A in the z     direction in kg mol A/s.m2.

      is the molecular diffusivity of the molecule A in B in m2/s

       is the concentration of A in kg mol/m3.

z  is the distance of diffusion in m

Convective Mass Transfer ( Dr. Filiz Alshanableh )

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/

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 ρ1 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 ρ1 2 L t d x 1 3 v ρ4 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 ρ0.5  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 ωρ2  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 ωρ2  µ  3 k c = 0.62 D AB  µ  ρD AB  1 1 =0.62(1.0×10 5 ) (40π 60)x12  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

Principles of Mass Transfer ( Dr. Filiz Alshanableh )


Mass Transfer à When a component in a mixture migrates in the same phase or from phase to phase because of a difference in concentration Examples of mass transfer • Evaporation of water in the open pail to the atmosphere • Coffee dissolves in water • Oxygen dissolves in the solution to the microorganism in the fermentation process • Reaction occurs when reactants diffuse from the surrounding medium to the catalyst surface Possible driving force for mass transfer • Concentration difference • Pressure difference • Electrical gradient, etc TYPES OF MASS TRANSFER Types of Mass Transfer 1. Molecular diffusion • Transfer of individual molecules through a fluid by random movement From high concentration to low concentration • E.g. a drop of blue liquid dye is added to a cup of water – the dye molecules will diffuse slowly by molecular diffusion to all parts of the water. • To increase this rate of mixing of the dye, the liquid can be mechanically agitated by a spoon and convective mass transfer will occur. 2. Convective mass transfer • Using mechanical force or action to increase the rate of molecular diffusion • E,g stirred the water to dissolve coffee during coffee making

PART I- MOLECULAR MASS TRANSPORT 1.1 Introduction to mass transfer 1.2 Properties of mixtures 1.2.1 Concentration of species 1.2.2 Mass Averaged velocity 1.3 Diffusion flux 1.3.1 Fick’s Law 1.3.2 Relation among molar fluxes 1.4 Diffusivity 1.4.1 Diffusivity in gases 1.4.2 Diffusivity in liquids 1.4.3 Diffusivity in solids 1.5 Steady state diffusion 1.5.1 Diffusion through a stagnant gas film 1.5.2 Pseudo – steady – state diffusion through a stagnant gas film. 1.5.3 Equimolar counter diffusion. 1.5.4 Diffusion into an infinite stagnant medium. 1.5.5 Diffusion in liquids 1.5.6 Mass diffusion with homogeneous chemical reaction. 1.5.7 Diffusion in solids 1.6 Transient Diffusion. 1.1 Introduction to Mass Transfer When a system contains two or more components whose concentrations vary from point to point, there is a natural tendency for mass to be transferred, minimizing the concentration differences within a system. The transport of one constituent from a region of higher concentration to that of a lower concentration is called mass transfer. The transfer of mass within a fluid mixture or across a phase boundary is a process that plays a major role in many industrial processes. Examples of such processes are: (i) Dispersion of gases from stacks (ii) Removal of pollutants from plant discharge streams by absorption (iii) Stripping of gases from waste water (iv) Neutron diffusion within nuclear reactors (v) Air conditioning

Many of day-by-day experiences also involve mass transfer, for example: (i) A lump of sugar added to a cup of coffee eventually dissolves and then eventually diffuses to make the concentration uniform. (ii) Water evaporates from ponds to increase the humidity of passing-air- stream (iii) Perfumes present a pleasant fragrance which is imparted throughout the surrounding atmosphere. The mechanism of mass transfer involves both molecular diffusion and convection. 1.2 Properties of Mixtures Mass transfer always involves mixtures. Consequently, we must account for the variation of physical properties which normally exist in a given system. The conventional engineering approach to problems of multicomponent system is to attempt to reduce them to representative binary (i.e., two components) systems. In order to understand the future discussions, let us first consider definitions and relations which are often used to explain the role of components within a mixture. 1.2.1 Concentration of Species: Concentration of species in multicomponent mixture can be expressed in many ways. For species A, mass concentration denoted by ρA is defined as the mass of A, mA per unit volume of the mixture. m ρ = A ———————————— (1) A V The total mass concentration density ρ is the sum of the total mass of the mixture in unit volume: ρ =∑ρi i where ρ is the concentration of species i in the mixture. i Molar concentration of, A, CA is defined as the number of moles of A present per unit volume of the mixture. By definition,

mass of A Number of moles = molecular weight of A m n = A —————————– (2) A M A Therefore from (1) & (2) n A ρ A C = = A V M A For ideal gas mixtures, P V A [ from Ideal gas law PV = nRT] n = A R T n P A A C = = A V R T where PA is the partial pressure of species A in the mixture. V is the volume of gas, T is the absolute temperature, and R is the universal gas constant. The total molar concentration or molar density of the mixture is given by C = ∑C i i 1.2.2 Velocities In a multicomponent system the various species will normally move at different velocities; and evaluation of velocity of mixture requires the averaging of the velocities of each species present. If νi is the velocity of species i with respect to stationary fixed coordinates, then mass-average velocity for a multicomponent mixture defined in terms of mass concentration is,

∑ρ ν ∑ρ ν i i i i i i ν = = ∑ρ i ρ i By similar way, molar-average velocity of the mixture ν * is ∑C i V i i ν * = C For most engineering problems, there will be title difference in ν * and ν and so the mass average velocity, ν, will be used in all further discussions. The velocity of a particular species relative to the mass-average or molar average velocity is termed as diffusion velocity (i.e.) Diffusion velocity = ν i – ν The mole fraction for liquid and solid mixture, x A ,and for gaseous mixtures, y A , are the molar concentration of species A divided by the molar density of the mixtures. C x = A (liquids and solids) A C C y A = A (gases). C The sum of the mole fractions, by definition must equal 1; (i.e.) ∑x i =1 i ∑y i = 1 i by similar way, mass fraction of A in mixture is; ρ A w = A ρ

Example 1.1 The molar composition of a gas mixture at 273 K and 1.5 x 10 5 Pa is: O 2 7% CO 10% CO 2 15% N 2 68% Determine a) the composition in weight percent b) average molecular weight of the gas mixture c) density of gas mixture d) partial pressure of O 2 . Calculations: Let the gas mixture constitutes 1 mole. Then O 2 = 0.07 mol CO = 0.10 mol CO 2 = 0.15 mol N 2 = 0.68 mol Molecular weight of the constituents is: O 2 = 2 x 16 = 32 g/mol CO = 12 + 16 = 28 g/mol CO 2 = 12 + 2 x 16 = 44 g/mol N 2 = 2 x 14 = 28 g/mol Weight of the constituents is: (1 mol of gas mixture) O 2 = 0.07 x 32 = 2.24 g CO = 0.10 x 28 = 2.80 g CO 2 = 0.15 x 44 = 6.60 g N 2 = 0.68 x 28 = 19.04 g Total weight of gas mixture = 2.24 + 2.80 + 6.60 + 19.04 = 30.68 g Composition in weight percent: 2.24 O = x100 =7.30% 2 30.68

2.80 CO = x100 =9.13% 30.68 6.60 CO = x100 =21.51% 2 30.68 19.04 N = x100 =62.06% 2 30.68 Weight of gas mixture Average molecular weight of the gas mixture M = Number of moles 30.68 M = =30.68 g mol 1 Assuming that the gas obeys ideal gas law, PV = nRT n P = V RT n =molar density =ρm or C V Therefore, density (or mass density) = ρ m M Where M is the molecular weight of the gas. PM (1.5×10 5 )x 30.68 3 Density =ρ M = = kg m m RT 8314x 273 3 = 2.03 kg/m Partial pressure of O 2 = [mole fraction of O 2 ] x Total pressure 7 5 (1.5 10 ) = x x 100 = 0.07 ( 1.5 x 10 5) = 0.105 x 10 5 Pa

1.3 Diffusion flux Just as momentum and energy (heat) transfers have two mechanisms for transport- molecular and convective, so does mass transfer. However, there are convective fluxes in mass transfer, even on a molecular level. The reason for this is that in mass transfer, whenever there is a driving force, there is always a net movement of the mass of a particular species which results in a bulk motion of molecules. Of course, there can also be convective mass transport due to macroscopic fluid motion. In this chapter the focus is on molecular mass transfer. The mass (or molar) flux of a given species is a vector quantity denoting the amount of the particular species, in either mass or molar units, that passes per given increment of time through a unit area normal to the vector. The flux of species defined with reference to fixed spatial coordinates, NA is N A =C Aν A ——————— (1) This could be written in terms of diffusion velocity of A, (i.e., ν A – ν) and average velocity of mixture, ν, as N C ν ν +C ν ———- (2) A = A ( A − ) A ∑C iν i By definition ν * i =ν = C Therefore, equation (2) becomes C A N = C (ν −ν) + ∑C ν A A A i i C i = C (ν −ν) + y ∑C ν A A A i i i For systems containing two components A and B, ( ) y (C C ) N C ν ν ν ν = − + + A A A A A A B B ( ) ( ) = C ν −ν + y N +N A A A A B N C + y N ——————————- (3) = (ν −ν) A A A A The first term on the right hand side of this equation is diffusional molar flux of A, and the second term is flux due to bulk motion.

1.3.1 Fick’s law: An empirical relation for the diffusional molar flux, first postulated by Fick and, accordingly, often referred to as Fick’s first law, defines the diffusion of component A in an isothermal, isobaric system. For diffusion in only the Z direction, the Fick’s rate equation is d C A J =−D A A B d Z where DAB is diffusivity or diffusion coefficient for component A diffusing through component B, and dCA / dZ is the concentration gradient in the Z-direction. A more general flux relation which is not restricted to isothermal, isobasic system could be written as d y A J A =−C DA B —————– (4) d Z using this expression, Equation (3) could be written as d y A N A =−C DA B + y A N ————— (5) d Z 1.3.2 Relation among molar fluxes: For a binary system containing A and B, from Equation (5), N A =J A +y A N or J A =N A −y A N ———————– (6) Similarly, J B = N B −y B N ——————– (7) Addition of Equation (6) & (7) gives, J N ———- (8) J + = +N −(y A +y B ) N A B A B By definition N = N + N and y + y = 1. A B A B Therefore equation (8) becomes,

J + J = 0 A B J A = -J B d y A d y B C D AB =−C D BA ————— (9) d z d Z From y + y = 1 A B dy A = – dy B Therefore Equation (9) becomes, D AB = D BA ———————————– (10) This leads to the conclusion that diffusivity of A in B is equal to diffusivity of B in A. 1.4 Diffusivity Fick’s law proportionality, DAB , is known as mass diffusivity (simply as diffusivity) or as the diffusion coefficient. DAB has the dimension of L 2 / t, identical to the fundamental dimensions of the other transport properties: Kinematic viscosity, νη = (µ / ρ) in momentum transfer, and thermal diffusivity, α (= k / ρ C ρ ) in heat transfer. 2 2 Diffusivity is normally reported in cm / s; the SI unit being m / s. Diffusivity depends on pressure, temperature, and composition of the system. In table, some values of DAB are given for a few gas, liquid, and solid systems. Diffusivities of gases at low density are almost composition independent, incease with the temperature and vary inversely with pressure. Liquid and solid diffusivities are strongly concentration dependent and increase with temperature. Table 1.1 : General range of values of diffusivity: Gases : 5 x 10 –6 -5 2 ————- 1 x 10 m / s –6 -9 2 Liquids : 10 ————- 10 m / s –14 -10 2 Solids : 5 x 10 ————- 1 x 10 m / s In the absence of experimental data, semi theoretical expressions have been developed which give approximation, sometimes as valid as experimental values, due to the difficulties encountered in experimental measurements.

1.4.1 Diffusivity in Gases: Pressure dependence of diffusivity is given by 1 D AB ∝ (for moderate ranges of pressures, up to 25 atm) P and temperature dependency is according to 3 2 D ∝T AB Diffusivity of a component in a mixture of components can be calculated using the diffusivities for the various binary pairs involved in the mixture. The relation given by Wilke is 1 D = 1−mixture ′ ′ ′ y 2 y 3 y n + +……….. + D D D − − −n 1 2 1 3 1 where D 1-mixture is the diffusivity for component 1 in the gas mixture D 1-n is the diffusivity for the binary pair, component 1 diffusing through component n y ′ is the mole fraction of component n in the gas mixture evaluated on a n component –1 – free basis, that is ′ y 2 y 2 = + + y 2 y 3 …….y n Example 1.2. Determine the diffusivity of CO (1), O (2) and N (3) in a gas mixture 2 2 2 having the composition: CO : 28.5 % 2 O : 15% 2 N 2 : 56.5% The gas mixture is at 273 K and 1.2 x 10 5 Pa. The binary diffusivity values are given as: (at 273 K) D 2 12 P = 1.874 m Pa/s 2 D 13 P = 1.945 m Pa/s D 2 23 P = 1.834 m Pa/s

Calculations: Diffusivity of CO 2 in mixture 1 D = 1m ′ ′ y 2 y 3 + D D 12 13 ′ y 2 0.15 where y 2 = = =0.21 y 2 +y 3 0.15+0.565 ′ y 3 0.565 y 3 = = =0.79 y 2 +y 3 0.15+0.565 1 Therefore D P = 1m 0.21 0.79 + 1.874 1.945 = 1.93 m 2.Pa/s Since P = 1.2 x 10 5 Pa, 1.93 −5 2 D = =1.61×10 m s 1m 5 1.2×10 Diffusivity of O 2 in the mixture, 1 D = 2m ′ ′ y 1 y 3 + D D 21 23 ′ y 1 0.285 Where y 1 = = =0.335 (mole fraction on-2 free bans). y 1 +y 3 0.285 +0.565 and ′ y 3 0.565 y 3 = = =0.665 y 1 +y 3 0.285 +0.565 and D 21 P = D 12 P = 1.874 m 2.Pa/s

1 2 Therefore D 2m P = = 1.847 m .Pa/s 0.335 0.665 + 1.874 1.834 1.847 −5 2 D m = =1.539×10 m sec 2 5 1.2×10 By similar calculations diffusivity of N 2 in the mixture can be calculated, and is found –5 2 to be, D 3m = 1.588 x 10 m /s. 1.4.2 Diffusivity in liquids: Diffusivity in liquid are exemplified by the values given in Table 1.1. Most of these -5 2 values are nearer to 10 cm / s, and about ten thousand times lower than those in dilute gases. This characteristic of liquid diffusion often limits the overall rate of processes accruing in liquids (such as reaction between two components in liquids). In chemistry, diffusivity limits the rate of acid-base reactions; in the chemical industry, diffusion is responsible for the rates of liquid-liquid extraction. Diffusion in liquids is important because it is slow. Certain molecules diffuse as molecules, while others which are designated as electrolytes ionize in solutions and diffuse as ions. For example, sodium chloride + – (NaCl), diffuses in water as ions Na and Cl . Though each ion has a different mobility, the electrical neutrality of the solution indicates the ions must diffuse at the same rate; accordingly it is possible to speak of a diffusion coefficient for molecular electrolytes such as NaCl. However, if several ions are present, the diffusion rates of the individual cations and anions must be considered, and molecular diffusion coefficients have no meaning. Diffusivity varies inversely with viscosity when the ratio of solute to solvent ratio exceeds five. In extremely high viscosity materials, diffusion becomes independent of viscosity. 1.4.3 Diffusivity in solids: Typical values for diffusivity in solids are shown in table. One outstanding characteristic of these values is their small size, usually thousands of time less than those in a liquid, which are in turn 10,000 times less than those in a gas. Diffusion plays a major role in catalysis and is important to the chemical/ food engineer. For metallurgists, diffusion of atoms within the solids is of more importance.

1.5 Steady State Diffusion In this section, steady-state molecular mass transfer through simple systems in which the concentration and molar flux are functions of a single space coordinate will be considered. In a binary system, containing A and B, this molar flux in the direction of z, as given by Eqn (5) is d y A N A =−C D AB + y A (N A + N B ) — (1) d z 1.5.1 Diffusion through a stagnant gas film The diffusivity or diffusion coefficient for a gas can be measured, experimentally using Arnold diffusion cell as shown in Fig 1.1. Fig 1.1 Arnold diffusion cell The narrow tube of uniform cross section which is partially filled with pure liquid A, is maintained at a constant temperature and pressure. Gas B which flows across the open end of the tub, has a negligible solubility in liquid A, and is also chemically inert to A. (i.e. no reaction between A & B). Component A vaporizes and diffuses into the gas phase; the rate of vaporization may be physically measured and may also be mathematically expressed in terms of the molar flux.

Consider the control volume S ∆ z, where S is the cross sectional area of the tube. Mass balance on A over this control volume for a steady-state operation yields [Moles of A leaving at z + ∆z] – [Moles of A entering at z] = 0. (i.e.) S N A − S N A =0. ————– (1) z +∆z z Dividing through by the volume, S∆Z, and evaluating in the limit as ∆Z approaches zero, we obtain the differential equation d N A = 0 ————————- (2) d z This relation stipulates a constant molar flux of A throughout the gas phase from Z1 to Z2 . A similar differential equation could also be written for component B as, d N B = 0, d Z and accordingly, the molar flux of B is also constant over the entire diffusion path from z and z . 1 2 Considering only at plane z1, and since the gas B is insoluble is liquid A, we realize that N , the net flux of B, is zero throughout the diffusion path; accordingly B is a B stagnant gas. From equation (1) (of section 1.5)

d y A N A =−C D AB + y A (N A +N B ) d z Since N B = 0, d y A N A =−C D AB + y A N A d z Rearranging, −C D AB d y A N A = ———— (3) 1−y A d z This equation may be integrated between the two boundary conditions: at z = z Y = Y 1 A A1 And at z = z Y = y 2 A A2 Assuming the diffusivity is to be independent of concentration, and realizing that NA is constant along the diffusion path, by integrating equation (3) we obtain Z2 y A 2 −d y A N d z =C D A ∫ AB ∫ Z1 y A 1 1−y A N = C D AB ln 1−y A 2  ————–(4) A   Z 2 −Z 1 1−y A 1  The log mean average concentration of component B is defined as y B 2 −y B 1 y B ,lm = ln y B 2   y B 1 Since y B =1−y A , (1 y ) (1 y ) y y − − − − y B ,lm = A 2 A 1 = A 1 A 2 ——- (5) ln y A 2  ln y A 2   y A 1  y A 1 Substituting from Eqn (5) in Eqn (4),

C D AB (y A 1 −y A 2) N A = ——————– (6) Z 2 −z 1 y B ,lm n p For an ideal gas C = = , and V R T P for mixture of ideal gases y A = A P Therefore, for an ideal gas mixture equation. (6) becomes D AB (p A 1 −p A 2) N = A RT (z2 −z1) p , B lm This is the equation of molar flux for steady state diffusion of one gas through a second stagnant gas. Many mass-transfer operations involve the diffusion of one gas component through another non-diffusing component; absorption and humidification are typical operations defined by these equation. Example 1.3 Oxygen is diffusing in a mixture of oxygen-nitrogen at 1 atm, 25°C. Concentration of oxygen at planes 2 mm apart are 10 and 20 volume % respectively. Nitrogen is non-diffusing. (a) Derive the appropriate expression to calculate the flux oxygen. Define units of each term clearly. (b) Calculate the flux of oxygen. Diffusivity of oxygen in nitrogen = 1.89 x 10 –5 2 m /s. Solution: Let us denote oxygen as A and nitrogen as B. Flux of A (i.e.) N A is made up of two components, namely that resulting from the bulk motion of A (i.e.), Nx A and that resulting from molecular diffusion J A : N =Nx +J ———————————- (1) A A A

From Fick’s law of diffusion, d C A —————————————– (2) J =−D A AB d z Substituting this equation (1) d C A ——————————— (3) N =Nx −D A A AB d z Since N = N A + N B and x A = C A / C equation (3) becomes C d C A A N A =(N A +N B ) −D AB C d z Rearranging the terms and integrating between the planes between 1 and 2, d z C A 2 dC A ∫ =−∫ ————– (4) cDAB C A 1 N A C −CA (N A +N B ) Since B is non diffusing N B = 0. Also, the total concentration C remains constant. Therefore, equation (4) becomes z C A 2 dC A =−∫ CD C A 1 N C −N C AB A A A 1 C −CA 2 = ln N C −C A A 1 Therefore, CD C −C AB A 2 N = ln —————————- (5) A z C −C A 1 Replacing concentration in terms of pressures using Ideal gas law, equation (5) becomes D P P −P AB t t A 2 N = ln ————————— (6) A RTz P −P t A 1 where

D AB = molecular diffusivity of A in B P T = total pressure of system R = universal gas constant T = temperature of system in absolute scale z = distance between two planes across the direction of diffusion P A1 = partial pressure of A at plane 1, and P A2 = partial pressure of A at plane 2 Given: –5 2 D AB = 1.89 x 10 m /s 5 2 P = 1 atm = 1.01325 x 10 N/m t T = 25°C = 273 + 25 = 298 K z = 2 mm = 0.002 m P A1 = 0.2 x 1 = 0.2 atm (From Ideal gas law and additive pressure rule) P A2 = 0.1 x 1 = 0.1 atm Substituting these in equation (6) (1.89×10 −5 )(1.01325×10 5 ) 1−0.1 N A = ln   ( )( )( ) 8314 298 0.002 1−0.2  = 4.55 x10 –5 kmol/m 2.s 1.5.2 Pseudo steady state diffusion through a stagnant film: In many mass transfer operations, one of the boundaries may move with time. If the length of the diffusion path changes a small amount over a long period of time, a pseudo steady state diffusion model may be used. When this condition exists, the equation of steady state diffusion through stagnant gas can be used to find the flux. If the difference in the level of liquid A over the time interval considered is only a small fraction of the total diffusion path, and t0 – t is relatively long period of time, at any given instant in that period, the molar flux in the gas phase may be evaluated by C D y y ( − ) N = AB A 1 A 2 —————— (1) A zy B , lm where z equals z – z , the length of the diffusion path at time t. 2 1 The molar flux NA is related to the amount of A leaving the liquid by

ρA ,L d z N A = ————————– (2) M d t A ρ , where A L is the molar density of A in the liquid phase M A under Psuedo steady state conditions, equations (1) & (2) can be equated to give ρ A ,L d z C D AB (y A 1 −y A 2) = ————— (3) M A d t z y B ,lm Equation. (3) may be integrated from t = 0 to t and from z = z to z = z as: t0 t t ρ y M Zt , , A L B lm A ∫dt = ∫z dz t=0 C D AB (y A 1 −y A 2) Z t 0 yielding ρA ,L y B ,lm M A z 2 −z 20  t =  t t  ——– (4) C D y y ( − ) 2 AB A 1 A 2   This shall be rearranged to evaluate diffusivity DAB as, 2 2 ρA ,L y B ,lm z t −z t 0  D = AB   M C y y t ( − ) 2 A A 1 A 2   Example 1.4 A vertical glass tube 3 mm in diameter is filled with liquid toluene to a depth of 20mm from the top opened. After 275 hrs at 39.4 °C and a total pressure of 760 mm Hg the level has dropped to 80 mm from the top. Calculate the value of diffusivity. Data: 2 vapor pressure of toluene at 39.4°C = 7.64 kN / m , density of liquid toluene = 850 kg/m3 Molecular weight of toluene = 92 (C H CH ) 6 6 3 2 2 D = ρA ,L y Blm Z t −Z t 0  AB ( )   M A C y A 1 −y A 2 t  2 

y B 2 −y B 1 where y B l m = , ln y B 2    y B 1  y B2 = 1 – y A2 y B1 = 1 – y A1 p A 1 7.64 2 y A 1 = = = 0.0754 (760 mm Hg = 101.3 kN/m ) P 101.3 y B1 = 1 – 0.0754 = 0.9246 y A2 = 0 y B = 1 – y A2 = 1 1−0.9246 Therefore y = =0.9618 , B lm  1  ln   0.9246 P 1.01325×10 5 3 C = = = 0.039 k mol /m ( ) R T 8314x 273 +39.4 850x 0.9618 0.08 2 −0.02 2    Therefore D = AB 92x 0.039x (0.0754 −0)x 275x 3600  2  –3 2 2 = 1.5262 x 10 (0.08 – 0.02 ) -6 2 = 9.1572 x 10 m /s 1.5.3 Equimolar counter diffusion: A physical situation which is encountered in the distillation of two constituents whose molar latent heats of vaporization are essentially equal, stipulates that the flux of one gaseous component is equal to but acting in the opposite direction from the other gaseous component; that is, NA = – NB. The molar flux NA , for a binary system at constant temperature and pressure is described by d y A N A =−C D AB + y A (N A +N B ) d z d C A or N A =−D AB + y A (N A +N B ) ——- (1) d z with the substitution of N B = – NA , Equation (1) becomes,

d C N =−D A —————– (2) A AB d z For steady state diffusion Equation. (2) may be integrated, using the boundary conditions: at z = z C = C 1 A A1 and z = z C = C 2 A A2 Giving, Z2 CA 2 N d z = −D d C A ∫ AB ∫ A Z1 CA 1 from which D N A = AB (C A 1 −C A 2) ——————- (3) z −z 2 1 n A p A For ideal gases, C A = = . V R T Therefore Equation. (3) becomes D N A = AB (P A 1 −P A 2) ———- (4) R T (z 2 −z 1) This is the equation of molar flux for steady-state equimolar counter diffusion. Concentration profile in these equimolar counter diffusion may be obtained from, d (N A ) = 0 (Since NA is constant over the diffusion path). d z And from equation. (2) d C A N =−D . A AB d z Therefore

d  d C  −D A = 0 . d z  AB d z  d 2 C A or =0. d z 2 This equation may be solved using the boundary conditions to give C A −C A 1 z −z 1 = ————– (5) C −C z −z A 2 1 2 A 1 Equation, (5) indicates a linear concentration profile for equimolar counter diffusion. Example 1.5. Methane diffuses at steady state through a tube containing helium. At point 1 the partial pressure of methane is p A1 = 55 kPa and at point 2, 0.03 m apart P A2 = 15 kPa. The total pressure is 101.32 kPa, and the temperature is 298 K. At this –5 2 pressure and temperature, the value of diffusivity is 6.75 x 10 m /s. i) calculate the flux of CH4 at steady state for equimolar counter diffusion. ii) Calculate the partial pressure at a point 0.02 m apart from point 1. Calculation: For steady state equimolar counter diffusion, molar flux is given by D N A = AB (p A 1 −p A 2 ) ————————— (1) R T z Therefore; 6.75 x 10 −5 kmol kmol N = (55 −15) = x −5 3.633 10 A 2 2 8.314x 298x 0.03 m .sec m sec And from (1), partial pressure at 0.02 m from point 1 is: 3.633×10 −5 = 6.75×10 −5 (55 −p A ) 8.314x 298x 0.02 p A = 28.33 kPa

Example 1.6. In a gas mixture of hydrogen and oxygen, steady state equimolar counter diffusion is occurring at a total pressure of 100 kPa and temperature of 20°C. If the partial pressures of oxygen at two planes 0.01 m apart, and perpendicular to the direction of diffusion are 15 kPa and 5 kPa, respectively and the mass diffusion flux of oxygen in the mixture is 1.6 x 10 –5 kmol/m 2.s, calculate the molecular diffusivity for the system. Solution: For equimolar counter current diffusion: D N A = AB (p A 1 −p A 2 ) ———————— (1) RTz where N A = molar flux of A (1.6 x 10 –5 kmol/m 2.s): D AB = molecular diffusivity of A in B R = Universal gas constant (8.314 kJ/kmol.k) T = Temperature in absolute scale (273 + 20 = 293 K) z = distance between two measurement planes 1 and 2 (0.01 m) P A1 = partial pressure of A at plane 1 (15 kPa); and P A2 = partial pressure of A at plane 2 (5 kPa) Substituting these in equation (1) D 1.6×10 −5 = AB (15−5) ( )( )( ) 8.314 293 0.01 –5 2 Therefore, D AB = 3.898 * 10 m /s Example 3.7. A tube 1 cm in inside diameter that is 20 cm long is filled with CO2 and H at a total pressure of 2 atm at 0°C. The diffusion coefficient of the CO – H 2 2 2 system under these conditions is 0.275 cm2/s. If the partial pressure of CO is 1.5 atm 2 at one end of the tube and 0.5 atm at the other end, find the rate of diffusion for: i) steady state equimolar counter diffusion (N A = – N B) ii) steady state counter diffusion where N B = -0.75 N A , and iii) steady state diffusion of CO through stagnant H (N = 0) 2 2 B

d y A i) N A =−C D AB +y A (N A +N B ) d z Given N B = – N A d y A d C A Therefore N =−C D =−D A AB AB d z d z (For ideal gas mixture CA = p A R T where pA is the partial pressure of A; such that p A + p B = P) Therefore N =−D d (p A RT ) A A B d z For isothermal system, T is constant −D AB d p A Therefore N = A RT d z Z P 2 D A 2 (i.e.) N A ∫d z =− AB ∫d p A RT Z P 1 A 1 D N A = AB (p A 1 −p A 2 ) ———————————- (1) RT z where Z = Z – Z 2 1 2 –4 2 Given: D AB = 0.275 cm /s = 0.275 x 10 m /s ; T = 0°C = 273 K 0.275x 10 −4 5 5 N A = (1.5×1.01325×10 −0.5×1.01325×10 ) 8314x 273 x 0.2 −6 k mol =6.138×10 2 m sec Rate of diffusion = N A S Where S is surface area -6 2 Therefore rate of diffusion = 6.138 x 10 ( π r ) –6 –2 2 = 6.138 x 10 π (0.5 x 10 ) = 4.821 x 10 –10 k mol/s

= 1.735 x 10 –3 mol/hr. ii) N A =−C D AB d y A +y A (N A +N B ) d z given: N B = – 0.75 N A Therefore N A =−C D AB d y A +y A (N A −0.75N A ) d z d y A =−C D A B +0.25 y A N A d z d y A N − 25 y N =−C D 0. A A A AB d z d y A N d z =−C D A AB 1−0.25 y A for constant N A and C Z 2 y A 2 d y A N d z =−CD A ∫ AB ∫ Z 1 y A 1 1−0.25 y A  d x 1  = ( + )  ln a b x ∫a +b x b  ( ) −1  ( )y A 2 N z = −C D [ − y ] A AB  ln 1 0.25 A 0.25 y A 1 4 CD 1−0.25 y  AB  A 2  N A =− ln ———————————- (2)  − 25  z 1 0. y A 1  Given: p 2×1.01325×10 5 3 C = = =0.0893 k mol m R T 8314x 273 p A 1 1.5 y A 1 = = =0.75 P 2 p A 2 0.5 y A 2 = = =0.25 P 2 Substituting these in equation (2), 4x 0.0893(0.275×10 −4 )  1−0.25x 0.25 N = ln A 0.2  1−0.25x 0.75

−6 kmol =7.028×10 2 m sec –6 –2 2 Rate of diffusion = N A S = 7.028 x10 π (0.5 x 10 ) = 5.52 x 10 –10 kmol/s = 1.987 x 10 –3 mol/hr. iii) N A =−CDAB d y A +y A (N A +N B ) d z Given: N B = 0 d y A Therefore N =−CD +y N A AB A A d z Z 2 y A 2 d y A N d z =−CD A ∫ AB ∫ Z 1 y A 1 1−y A CDAB 1−y A 2  = Z ln 1−y A 1  0.0893(0.275×10−4 )  1−0.25 = ln   0.2  1−0.75 kmol −5 =1.349 x 10 2 m .sec –5 –2 2 Rate of diffusion = 1.349 8 10 π (0.5 x 10 ) = 1.059 Kmol / s = 3.814 mol/hr 1.5.4 Diffusion into an infinite standard medium : Here we will discuss problems involving diffusion from a spherical particle into an infinite body of stagnant gas. The purpose in doing this is to demonstrate how to set up differential equations that describe the diffusion in these processes. The solutions, obtained are only of academic interest because a large body of gas in which there are no convection currents is unlikely to be found in practice. However, the solutions developed here for these problems actually represent a special case of the more common situation involving both molecular diffusion and convective mass transfer.

a) Evaporation of a spherical Droplet: As an example of such problems, we shall consider the evaporation of spherical droplet such as a raindrop or sublimation of naphthalene ball. The vapor formed at the surface of the droplet is assumed to diffuse by molecular motions into the large body of stagnant gas that surrounds the droplet. Consider a raindrop, as shown in figure. At any moment, when the radius of the drop is r 0, the flux of water vapor at any distance r from the center is given by N A =−C D AB d y A +y A (N A +N B ) d r Here N B = 0 (since air is assumed to be stagnant) Therefore, d y A N =−C D +y N A AB A A d r Rearranging, −C D AB d y A N = __________ (1) A 1−y A d r The flux N A is not constant, because of the spherical geometry; decreases as the distance from the center of sphere increases. But the molar flow rate at r and r + δr are the same. This could be written as, A N A =A N A __________ (2) r r +δr Where A = surface area of sphere at r or r + δr. Substituting for A = 4 π r 2 in equation (2), 2 2 4πr N − 4πr N =0 A A δ r + r r or r 2 N −r 2 N A A r +δr r lim =0 δr →0 δr d 2 as (r N A )=0 __________ (3) dr Integrating, r 2 N A =constant __________ (4) From equation (4), r 2 N A =r02 N A 0 Substituting for N A from equation (1),


Mass transfer is the net movement of a component in a mixture from one location to another where the component exists at a different concentration. Often, the transfer takes place between 2 phases across interface.

Distillation 1 ( Dr. Ali Coşkun DALGIÇ )

Distillation with Reflux and McCabe-Thiele Method


Distillation is the process of heating a liquid solution, or a liquid-vapour mixture, to derive off a vapour and then collecting and condensing this vapour. The products of a distillation process are most often limited to an overhead distillate and a bottoms, whose compositions differ from that of the feed.

Distillation 2 ( Dr. Ali Coşkun DALGIÇ )

FE 301 Mass Transfer

Distillation with Reflux and McCabe-Thiele Method

Part 2

Dr. Ali Coşkun DALGIÇ

Today we will discuss:

1) McCabe-Thiele graphical construction

2) Determination of N and XB

3) Minimum number of stages N

4) Minimum reflux

5) Example

6) Subcooled Reflux

7) Multiple Feeds

8) Side stream products

9) Open steam

10) Non-ideal distillation: Murphree efficiency

Fe 301 Mass Transfer 2011-2012 All Problem Sets ( Dr. Ali Coşkun DALGIÇ )

Alcohol vapor is being absorbed from a mixture of alcohol vapor and water vapor by means of a nonvolatile solvent in which alcohol is soluble but water is not. The temp. is 97 ºC, and Pt=760 mmHg. The alcohol vapor can be considered to be diffusing through a film of alcohol-water- vapor mixture 0.1 mm thick The molal concentration of the alcohol vapor at the outside of the film is 80 %, and that on the inside is, next to the solvent is, 10 %. D Et-OH- water =0.15 cm 2 /sec at 25 ºC and 1 atm. Calculate the rate of diffusion of alcohol vapor in pounds per hour if the area of the film is 100 ft 2 .