Principles of Mass Transfer ( Dr. Filiz Alshanableh )

PRINCIPLES OF MASS TRANSFER INTRODUCTION

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),

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