ch 3000 mass transfer
Popular in Department
This 36 page Class Notes was uploaded by Adam on Saturday June 4, 2016. The Class Notes belongs to ch 3000 at Clemson University taught by in Summer 2016. Since its upload, it has received 36 views.
Reviews for ch 3000 mass transfer
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 06/04/16
3. Molecular mass transport 3.1 Introduction to mass transfer 3.2 Properties of mixtures 3.2.1 Concentration of species 3.2.2 Mass Averaged velocity 3.3 Diffusion flux 3.3.1 Pick’s Law 3.3.2 Relation among molar fluxes 3.4 Diffusivity 3.4.1 Diffusivity in gases 3.4.2 Diffusivity in liquids 3.4.3 Diffusivity in solids 3.5 Steady state diffusion 3.5.1 Diffusion through a stagnant gas film 3.5.2 Pseudo – steady – state diffusion through a stagnant gas film. 3.5.3 Equimolar counter diffusion. 3.5.4 Diffusion into an infinite stagnant medium. 3.5.5 Diffusion in liquids 3.5.6 Mass diffusion with homogeneous chemical reaction. 3.5.7 Diffusion in solids 3.6 Transient Diffusion. 3.1 Introduction of 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 1 (v) Air conditioning Many of air daybyday 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 passingair stream (iii) Perfumes presents a pleasant fragrance which is imparted throughout the surrounding atmosphere. The mechanism of mass transfer involves both molecular diffusion and convection. 3.2 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. When a system contains three or more components, as many industrial fluid streams do, the problem becomes unwidely very quickly. The conventional engineering approach to problems of multicomponent system is to attempt to reduce them to representative binary (i.e., two component) 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. 3.2.1 Concentration of Species: Concentration of species in multicomponent mixture can be expressed in many ways. For species A, mass concentration denoted by isAdefined as the mass of A,m A per unit volume of the mixture. m A (1) A V The total mass concentration density is the sum of the total mass of the mixture in unit volume: i i where iis the concentration of species i in the mixture. 2 Molar concentration of, A,AC is defined as the number of moles of A present per unit volume of the mixture. By definition, mass of A es molecular weight of A (2) A M A Therefore from (1) & (2) C n A A A V M A For ideal gas mixtures, pAV A R T [ from Ideal gas law PV = nRT] n A pA A V RT 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 3.2.2 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 I is the velocity of species i with respect to stationary fixed coordinates, then massaverage velocity for a multicomponent mixture defined in terms of mass concentration is, 3 i i i i i i i i By similar way, molaraverage velocity of the mixture * is C i i * C For most engineering problems, there will be title difference in * and and so the mass average velocity, , will be used in all further discussions. The velocity of a particular species relative to the massaverage or molar average velocity is termed as diffusion velocity (i.e.) Diffusion velocity i The mole fraction for liquid and solid mixture,Ax ,and for gaseous mixtures,A y , are the molar concentration of species A divided by the molar density of the mixtures. C A x A (liquids and solids) C C y A A (gases). C The sum of the mole fractions, by definition must equal 1; x i 1 (i.e.) i i by similar way, mass fraction of A in mixture is; A 4 5 10. The molar composition of a gas mixture at 273 K and 1.5 * 10 Pa is: O 7% 2 CO 10% CO 215% 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 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 are: O 2 = 2 * 16 = 32 g/mol CO = 12 + 16 = 28 g/mol CO 2= 12 + 2 * 16 = 44 g/mol N 2 = 2 * 14 = 28 g/mol Weight of the constituents are: (1 mol of gas mixture) O 2 = 0.07 * 32 = 2.24 g CO = 0.10 * 28 = 2.80 g CO 2= 0.15 * 44 = 6.60 g N 2 = 0.68 * 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 2 *100 7.30% 30.68 5 2.80 CO 30.68*100 9.13% CO 6.60*100 21.51% 2 30.68 19.04 N 2 *100 62.06% 30.68 Weight of gas mixture Average molecular weight of the gas mixture Number of moles M 30.68 30.68 g mol 1 Assuming that the gas obeys ideal gas law, PV = nRT n P V RT n molardensity m V Therefore, density (or mass density) m M Where M is the molecular weight of the gas. PM 1.5 * 105 * 30.68 3 Density m M kg m RT 8314 * 273 = 2.03 kg/m Partial pressure of 2 = [mole fractio 2of O ] * total pressure 7 5 * 1.5 * 10 100 5 = 0.07 * 1.5 * 10 = 0.105 * 10 Pa 3.3 Diffusion flux Just as momentum and energy (heat) transfer have two mechanisms for transportmolecular 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 6 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,AN is A A A (1) This could be written interms of diffusion velocity of A, (i.A., ) and average velocity of mixture, , as N C ( ) C A A A A (2) By definition C i i * i C Therefore, equation (2) becomes C A A A (A ) C i i C i y A C i i i For systems containing two components A and B, A A (A ) y A(C A A C B B) A A ) y A (N A N B ) N C ( ) y N A A A A (3) 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. 3.3.1 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 7 component A in an isothermal, isobaric system. For diffusion in only the Z direction, the Fick’s rate equation is d C A A AB d Z where D ABis diffusivity or diffusion coefficient for component A diffusing through component B, and dC /AdZ is the concentration gradient in the Zdirection. A more general flux relation which is not restricted to isothermal, isobasic system could be written as d y A A D AB (4) d Z using this expression, Equation (3) could be written as d y A AB A y AN (5) d Z 3.3.2 Relation among molar fluxes: For a binary system containing A and B, from Equation (5), A A y AN or A A y A N (6) Similarly, B B y B N (7) Addition of Equation (6) & (7) gives, J J N N (y y )N A B A B A B (8) By definition N = NA + NB and A + B = 1. Therefore equation (8) becomes, A B J + J = 0 A B J = J 8 d y d y AB A CD BA B (9) d z d Z From y A+ y B= 1 dy = dy A B Therefore Equation (9) becomes, AB D BA (10) This leads to the conclusion that diffusivity of A in B is equal to diffusivity of B in A. 3.4 Diffusivity Fick’s law proportionality, D , AB known as mass diffusivity (simply as diffusivity) 2 or as the diffusion coefficient. D hAB the dimension of L / t, identical to the fundamental dimensions of the other transport properties: Kinematic viscosity, = ( / ) in momentum transfer, and thermal diffusivity, (= k / C ) in he transfer. Diffusivity is normally reported in cm / sec; the SI unit being m / sec. Diffusivity depends on pressure, temperature, and composition of the system. In table, some values of D areABiven 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. General range of values of diffusivity: –6 5 2 Gases : 5 X 10 1 X 10 m / sec. Liquids : 10 10 m / sec. Solids : 5 X 10 1 X 10 m / sec. In the absence of experimental data, semitheoretical expressions have been developed which give approximation, sometimes as valid as experimental values, due to the difficulties encountered in experimental measurements. 9 3.4.1 Diffusivity in Gases: Pressure dependence of diffusivity is given by 1 AB (for moderate ranges of pressures, upto 25 atm). p And temperature dependency is according to 3 D T 2 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 1mixture y2 y3 ........... n D D D 12 13 1n Where D is the diffusivity for component 1 in the gas mixture; D is the 1mixture 1n diffusivity for the binary pair, component 1 diffusing through component n; and y is the mole fraction of component n in the gas mixture evaluated on a n component –1 – free basis, that is y2 2 y2 y 3.......y n 9. Determine the diffusivity of Co 2(1), O 22) and N 2) in a gas mixture having the composition: Co 2: 28.5 %, O 2: 15%, N 256.5%, The gas mixture is at 273 k and 1.2 * 10 Pa. The binary diffusivity values are given as: (at 273 K) D P = 1.874 m Pa/sec 12 2 D 13= 1.945 m Pa/sec D 23= 1.834 m Pa/sec Calculations: Diffusivity of C 2 in mixture 10 1 D 1m y 2 y 3 D 12 D 13 y y2 0.15 0.21 where 2 y y 0.15 0.565 2 3 y3 0.565 y3 0.79 y 2 y 3 0.150.565 1 D 1m P Therefore 0.21 0.79 1.874 1.945 = 1.93 m .Pa/sec Since P = 1.2 * 10 Pa, 1.93 5 2 D1m 5 1.61*10 m sec 1.2*10 Diffusivity of 2 in the mixture, 1 D2m y1 y3 D D 21 23 y 1 0.285 Where y1 0.335 y1 y 3 0.285 0.565 (mole fraction on2 free bans). and y y3 3 0.565 0.665 y 1 y 3 0.285 0.565 and D 21 = D 12= 1.874 m .Pa/sec Therefore 1 D 2mP 0.335 0.665 1.874 1.834 2 = 1.847 m .Pa/sec 11 D m 1.847 1.539*10 5 m sec 2 1.2*10 5 By Similar calculations Diffusivity of N i 2the mixture can be calculated, and is found to be, D 3m.588 * 10 m /sec. 2 3.4.2 Diffusivity in liquids: Diffusivity in liquid are exemplified by the values given in table … Most of these values are nearer to 10 cm / sec, and about ten thousand times shower 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 acidbase reactions; in the chemical industry, diffusion is responsible for the rates of liquidliquid 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 ions 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. 3.4.3 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 inturn 10,000 times less than those in a gas. Diffusion plays a major role in catalysis and is important to the chemical engineer. For metallurgists, diffusion of atoms within the solids is of more importance. 3.5 Steady State Diffusion 12 In this section, steadystate 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 [section 3.3.1] d y A A CD AB d z y A (N A N )B (1) 3.5.1 Diffusion through a stagnant gas film The diffusivity or diffusion coefficient for a gas can be measured, experimentally using Arnold diffusion cell. This cell is illustrated schematically in figure. figure 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 interms 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 steadystate operation yields [Moles of A leaving at z + z] – [Moles of A entering at z] = 0. (i.e.) A z z S N A z 0. (1) Dividing through by the volume, SZ, and evaluating in the limit as Z approaches zero, we obtain the differential equation 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, 13 d N d Z and accordingly, the molar flux of B is also constant over the entire diffusion path from z 1and z 2. Considering only at plane z ,1and 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 B is a stagnant gas. From equation (1) (of section 3.5) d y A y (N N ) A AB d z A A B Since N B= 0, d y A A AB d z y AN A Rearranging, CD d y A AB A (3) 1y A d z This equation may be integrated between the two boundary conditions: at z = z Y = Y 1 A A1 And at2z = z A A2 Y = y Assuming the diffusivity is to be independent of concentration, and realizing that N Ais constant along the diffusion path, by integrating equation (3) we obtain Z2 yA2 d y A CD AB A Z y 1y A 1 A1 CD AB 1 y A2 A ln (4) Z 2 Z 1 1y A1 The log mean average concentration of component B is defined as 14 y y y B2 B1 y B2 ln y B1 Since y B 1 y A , (1y A2 ) (1y A1) y A1 y A2 y B,lm y A2 y A2 (5) ln y A1 ln y A1 Substituting from Eqn (5) in Eqn (4), A CD AB (y A1y A2 ) (6) Z 2 z 1 y B,lm For an ideal gas C n p , and V R T for mixture of ideal gasesy p A A P Therefore, for an ideal gas mixture equation. (6) becomes N D AB (p A1 p A2) A RT(z z ) p 2 1 B,lm This is the equation of molar flux for steady state diffusion of one gas through a second stagnant gas. Many masstransfer operations involve the diffusion of one gas component through another nondiffusing component; absorption and humidification are typical operations defined by these equation. The concentration profile (y Avs. z) for this type of diffusion is shown in figure: Figure 15 12. Oxygen is diffusing in a mixture of oxygennitrogen at 1 std atm, 25C. Concentration of oxygen at planes 2 mm apart are 10 and 20 volume % respectively. Nitrogen is nondiffusing. (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 * 10 2 m /sec. Solution: Let us denote oxygen as A and nitrogen as B. Flux of A (i.e.) N Ais made up of two components, namely that resulting from the bulk motion of A (i.e.), Nx Aand that resulting from molecular diffusion A : N A NxA J A (1) From Fick’s law of diffusion, d C A JA D AB (2) d z Substituting this equation (1) d C N A Nx A D AB A (3) d z Since N = N A N Bnd x A C AC equation (3) becomes C A dC A N A N A NB C DAB d z Rearranging the terms and integrating between the planes between 1 and 2, d z C dC A C A2 (4) cD AB A1 N A C A N A N B Since B is non diffusing B = 0. Also, the total concentration C remains constant. Therefore, equation (4) becomes 16 z C A2 dC A CD AB CA1 N CA N C A A 1 C C A2 N ln C C A A1 Therefore, CD AB C C A2 N A ln (5) z C C A1 Replacing concentration in terms of pressures using Ideal gas law, equation (5) becomes D AB P t P t P A2 N A RTz lnP P (6) t A1 where D ABmolecular 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 A1partial pressure of A at plane 1, and P A2partial pressure of A at plane 2 Given: D AB1.89 * 10 m /sec P = 1 atm = 1.01325 * 10 N/m 2 t T = 25C = 273 + 25 = 298 K z = 2 mm = 0.002 m P = 0.2 * 1 = 0.2 atm (From Ideal gas law and additive pressure rule) A1 P A20.1 * 1 = 0.1 atm Substituting these in equation (6) 5 5 N 1.89*10 1.01325*10 ln 1 0.1 A 8314 298 0.002 1 0.2 –5 2 = 4.55 * 10 kmol/m .sec 17 3.5.2 Psuedo 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. figure If the difference in the level of liquid A over the time interval considered is only a small fraction of the total diffusion path, an0 t – 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 AB (y A1 y A2 ) A zy (1) B,lm where z equals z 2– z 1 the length of the diffusion path at time t. The molar flux N A is related to the amount of A leaving the liquid by A,L d z A (2) M A d t A,L where 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 d z C D (y y ) A,L AB A1 A2 (3) M A d t z y B,lm Equation. (3) may be integrated from t = 0 to t and from z = z t0to z = t as: t A,Ly B,lm M A Z t zdz t0 C D AB (y A1 y A2) Zt0 yielding 18 A,L yB,lm M A z t z t0 (4) CD AB (y A1 y A2) 2 This shall be rearranged to evaluate diffusivity D as, AB 2 2 A,L y B,lm z t z t0 AB M C (y y )t 2 A A1 A2 1. A vertical glass tube 3 mm in diameter is filled with liquid toluene to a depth of 20mm from the top openend. 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: vapor pressure of toluene at 39.4C = 7.64 kN / m , 2 density of liquid toluene = 850 kg/m 3 Molecular weight of toluene = 92 (C 6 6H ) 3 A,L y Blm Z Z 2 D AB t t0 figure M CAy A1 y A2 t 2 y B2 y B1 y B,l m where yB2 ln y B1 y B2 1 – y A2 y B1 1 – y A1 p A1 7.64 2 y A1 (760 mm Hg = 101.3 kN/m ) P 101.3 = 0.0754 y B1 1 – 0.0754 = 0.9246 y A2 0 y B= 1 – y A2 y 1 0.9246 0.9618 Therefore B,lm 1 ln 0.9246 5 P 1.01325*10 C RT 8314* 273 39.4 3 = 0.039 k mol /m Therefore 19 850 * 0.9618 0.08 2 0.022 D AB 92 * 0.039 * 0.0754 0 * 275 * 3600 2 –3 2 2 = 1.5262 * 10 (0.08 – 0.02 ) = 9.1572 * 10 m /sec. 3.5.3 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, N = A N . B The molar flux N ,Afor a binary system at constant temperature and pressure is described by d y A A CD AB y A (N A N )B d z dC A or A D AB y A (N A N )B (1) d z with the substitution of NB = NA, Equation (1) becomes, dC A A D AB (2) d z For steady state diffusion Equation. (2) may be integrated, using the boundary conditions: at1 z = zA A1 C = C 2 AC =A2C Giving, Z C 2 A2 A z D AB dC A Z 1 C A1 from which D AB A (C A1 C A2) (3) z2z 1 20 For ideal gases, C n A pA . Therefore Equation. (3) becomes A V RT D AB A (P A1 P A2 ) (4) RT (z 2z ) 1 This is the equation of molar flux for steadystate equimolar counter diffusion. Concentration profile in these equimolar counter diffusion may be obtained from, d d z A ) 0 (Since NA is constant over the diffusion path). And from equation. (2) N D dC A A AB d z . Therefore dC A 0 . d z AB d z d 2 C or 0. d z 2 This equation may be solved using the boundary conditions to give C A C A1 zz 1 C C z z (5) A1 A2 1 2 Equation, (5) indicates a linear concentration profile for equimolar counter diffusion. 3. Methane diffuses at steady state through a tube containing helium. At point 1 the partial pressure of methane is p A155 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 * 10 m /sec. i) calculate the flux of CH at steady state for equimolar counter 4 diffusion. ii) Calculate the partial pressure at a point 0.02 m apart from point 1. 21 Calculation: For steady state equimolar counter diffusion, molar flux is given by D AB N A p A1 pA 2 (1) RT z Therefore; 5 N 6.75*10 55 15 kmol A 8.314 * 298 * 0.03 m .sec 3.633 * 105 kmol m sec And from (1), partial pressure at 0.02 m from point 1 is: 5 3.633 * 105 6.75 * 10 55 p 8.314 * 298 * 0.02 A p A= 28.33 kPa 11. 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 20C. 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 –5 2 diffusion flux of oxygen in the mixture is 1.6 * 10 kmol/m .sec, calculate the molecular diffusivity for the system. Solution: For equimolar counter current diffusion: D AB N A p A1p A2 (1) RTz where –5 2 N A= molar flux of A (1.6 * 10 kmol/m .sec): 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) 22 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 * 105 AB 155 8.314 293 0.01 –5 2 Therefore, D AB 3.898 * 10 m /sec 2. A tube 1 cm in inside diameter that is 20 cm long is filled with Co2 and H2 at a total pressure of 2 atm at 0C. The diffusion coefficient of the Co2 – H 2system 2 under these conditions is 0.275 cm /sec. If the partial pressure of Co2 is 1.5 atm 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 Aand iii) steady state diffusion of C2 through stagnant H2 (NB = 0) d y A i) A CD AB d z y AN A NB Given N B N A d y A d C A Therefore N A C D AB D AB d z d z p A (For ideal gas mixture C A where p Ais the partial pressure of A; such that RT p A+ p B= P) d p AT Therefore N A D AB d z For isothermal system, T is constant D d p Therefore N A AB A RT d z Z 2 D AB P A2 (i.e.) N A d z d pA Z RT P 1 A1 23 D AB N A p A1 pA2 (1) RT z where Z = Z 2Z 1 2 –4 2 Given: D AB0.275 cm /sec = 0.275 * 10 m /sec ; T = 0C = 273 k 4 N 0.275 *10 1.5 * 1.01325 * 105 0.5 * 1.01325 * 105 A 8314 * 273 * 0.2 6.138 * 10 6 k mol m 2 sec Rate of diffusion = N S A Where S is surface area Therefore rate of diffusion = 6.138 * 10 * r –6 –2 2 = 6.138 * 10 * (0.5 * 10 ) = 4.821 * 10 –1 k mol/sec –3 = 1.735 * 10 mol/hr. ii) C D d y A y N N A AB d z A A B given: N B 0.75 N A d y A Therefore N A C D AB d z y A N A 0.75N A d y C D A B A 0.25 y A N A d z N 0.25 y N C D d yA A A A AB d z d y A N A d z C D AB 1 0.25 y A for constant N And C Z 2 y A2 d y N d z CD A A AB 1 0.25 y A Z 1 y A1 d x 1 ln a b x a b x b 24 N z CD 1 ln 1 0.25y A2 A AB 0.25 A y A1 4CD AB 1 0.25 y A 2 N A ln (2) z 1 0.25 y A1 Given: p 2 * 1.01325 * 10 5 C 0.0893 K mol m 3 RT 8314 * 273 p y A1 1.5 0.75 A1 P 2 p y A2 0.5 0.25 A2 P 2 Substituting these in equation (2), 4 * 0.0893 * 0.275 * 10 4 1 0.25 * 0.25 N A ln 0.2 1 0.25 * 0.75 6 kmol 7.028 * 10 2 m sec Rate of diffusion = N S = 7.028 * 10 * * (0.5 * 10 ) –2 2 A –10 = 5.52 * 10 kmol/sec = 1.987 * 10 mol/hr. d y A iii) A CD AB y A N A N B d z Given: N B 0 d y A Therefore N A CD AB y A N A d z Z 2 yA2 d y N d z CD A A Z AB y 1 y A 1 A1 CD AB 1 y A2 ln Z 1 y A1 0.0893* 0.275*10 4 1 0.25 ln 0.2 1 0.75 5 kmol 1.349 * 10 2 m .sec 25 Rate of diffusion = 1.349 8 10 * * (0.5 * 10 ) 2 = 1.059 Kmol / sec = 3.814 mol/hr 3.5.4 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 Figure Evaporation of a raindrop of the drop is r , the flux of water vapor at any distance r from the center is given 0 by d y A N A C D AB y AN A N B d r Here N = 0 (since air is assumed to be stagnant) B Therefore, d y A N A C D AB y A N A d r Rearranging, C D AB d y A N A __________ (1) 1 y A d r The flux N Ais 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 + r are the same. This could be written as, A N A r A N A r r __________ (2) where A = surface area of sphere at r or r + r. 26 2 Substituting for A = 4 r in equation (2), 4 r 2 N A 4 r 2 N A 0 r r r or 2 2 r N A r r r N A r lim 0 r 0 r d 2 ss r N A 0 __________ (3) dr Integrating, 2 r N A constant __________ (4) 2 2 From equation (4), r N A r 0 N A 0 Substituting for N f Am equation (1), r 2C D d y AB A r 2 N A 1 y A d r 0 0 2 d r d y A r 0 N A 0 2 C D AB __________ (5) r 1 y A Boundary condition : At r = r 0 y A y AS And At r = y A y A Therefore equation (5) becomes, 2 1 y A r 0 N A 0 C D AB ln 1 y A y r r0 AS Simplifying, C D AB 1 y A N A ln __________ (6) 0 r 0 1 y A S Time required for complete evaporation of the droplet may be evaluated from making mass balance. Moles of water diffusing moles of water leaving the droplet unit time unit time 2 d 4 3 L 4 r 0 N A0 r 0 dt M A 2 L d r 0 4 r 0 __________ (7) M A d t Substituting for N from equation (6) in equation (7), A0 27 C D AB 1 y A L d r0 ln __________ (8) r0 1 y AS M A d t Initial condition : When t = 0 r 0= r 1 Integrating equation (8) with these initial condition, t L 1 1 0 d t r0 d r0 0 M A C D AB 1 y A r1 ln 1 y AS r 2 t L 1 1 M 2C D 1 y __________ (9) A AB ln A 1 y A S Equation (9) gives the total time t required for complete evaporation of spherical droplet of initial radius r 1 b) Combustion of a coal particle: The problem of combustion of spherical coal particle is similar to evaporation of a drop with the exception that chemical reaction (combustions) occurs at the surface of the particle. During combustion of coal, the reaction C + O 2 CO 2 cccurs. According to this reaction for every mole of oxygen that diffuses to the surface of coal (maximum of carbon), react with 1 mole of carbon, releases 1 mole of carbon dioxide, which must diffuse away from this surface. This is a case of equimolar counter diffusion of CO and 2. Norma 2y air (a mixture of N and 2 O 2s used for combustion, and in this case N does not 2es part in the reaction, and its flux is zero. i.e. N N 2 0 . The molar flux of O c 2ld be written as d y N CD O 2 y N N N O 2 O 2gas d r O 2 O 2 CO 2 N 2 __________ (1) Where D O 2gas is the diffusivity of 2 in the gas mixture. Since N N 2 0 , and from stoichiometry N O 2 N CO 2 , equation (1) becomes d yO N O C D O gas 2 __________ (2) 2 2 d r For steady state conditions, d r 2 N 0 __________ (3) d r O 2 Integrating, 28 r2 N O constant r02N O s __________ (4) 2 2 Where r is the radius of coal particle at any instant, and N O s is the flux of O 0 2 2 at the surface of the particle. Substituting for N O 2from equation (2) in equation (4), d yO 2 2 r 2CD O gas r N O s __________ (5) 2 d r 0 2 Boundary condition : r r0 yO y O s At 2 2 And At r yO 2 y O2 With these boundary condition, equation (5) becomes y 2 d r O 2 r0 NA 0 2 C D O 2gas d y O 2 r0 r yO s 2 which yields CD O2 gas N O 2 s yO 2 sy O 2 __________ (6) r 0 For fast reaction of O 2ith coal, the mole fraction of O 2 the surface of particle y O s 0 iz zero. (i.e.,) 2 . And also at some distance away from the surface of the particle y O 2 yO 2 0.21 (because air is a mixture of 21 mole % O an 279 mole % N 2) With these conditions, equation (6) becomes, 0.21 C D O 2 gas N O2 s ____________ (7) r0 Figure Combustion of a particle of coal 5. A sphere of naphthalene having a radius of 2mm is suspended in a large volume of shell air at 318 K and 1 atm. The surface pressure of the naphthalene can be assumed to be at 318 K is 0.555 mm Hg. The D ABof naphthalene in air –6 2 at 318 K is 6.92 * 10 /sec. Calculate the rate of evaporation of naphthalene from the surface. Calculation Steady state mass balance over a element of radius r and r + r leads to S N A r S N A r r 0 (1) 29 2 where S is the surface are (= 4 r ) dividing (1) by Sr, and taking the limit as r approaches zero, gives: d r N2 A 0 d r Integrating r N A= constant (or) 4 r N A= constant We can assume that there is a film of naphthalene – vapor / air film around naphthalene through which molecular diffusion occurs. Diffusion of naphthalene vapor across this film could be written as, N CD d y A y N N A AB d r A A B N B 0 (since air is assumed to be stagnant in the film) d y A N A CD AB y A N A d r y N CD d A A AB d r 1 y A dln 1 y A N A CD AB d r 2 W A Rate of evaporation = 4 r N A R = constant. 4 r 2CD d ln 1 y W AB A A d r W A d r 4 D AB C d ln 1 y A r 2 Boundary condition: 0.555 4 At r = R y A 7.303 * 10 760 –4 ln (1 – A) = 7.3 * 10 At r = y A= 0 ln (1 A) = 0 d r 0 Therefore W A 2 4 D AB C d ln 1 y A R r 7.3 *10 4 30 1 0 W A r 4 D AB C ln 1 y A 7.3 *104 R 1 4 W A 0 4 D AB C 0 7.3*10 R W A4 R D C * AB3 * 10 –4 5 C P 1.01325 * 10 Gas cons tant * T 8314 * 318 3 = 0.0383 kmol/m Initial rate of evaporation: Therefore W = A* 3.142 * 2 * 10 * 6.92 * 10 * 0.0383 * 7.3 * 10 –4 –12 = 4.863 * 10 –5 kmol/sec = 1.751 * 10 mol/hr. 3.5.5 Diffusion in Liquids: Equation derived for diffusion in gases equally applies to diffusion in liquids with some modifications. Mole fraction in liquid phases is normally written as ‘x’ (in gases as y). The concentration term ‘C’ is replaced by average molar density, . M av a) For steady – state diffusion of A through non diffusivity B: N A constant , N = B D AB N A xA1 x A2 z x BM M av where Z = Z – 2, th 1length of diffusion path; and X B2 X B1 X BM X B2 ln X B1 b) For steady – state equimolar counter diffusion : N A N = Bonst D AB D AB N A C A1 C A2 x A1 x A 2 Z Z M av 4. Calculate the rate of diffusion of butanol at 20C under unidirectional steady state conditions through
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'