PM3125: Lectures 1 to 5 Content: Mass transfer: concept and theory uploaded at http://www.rshanthini.com/PM3125.htm Prof. R. Shanthini 27 Feb & 05 Mar 2012

Reference books used for ppts C.J. Geankoplis Transport Processes and Separation Process Principles 4th edition, Prentice-Hall India J.D. Seader and E.J. Henley Separation Process Principles 2nd edition, John Wiley & Sons, Inc. 3. J.M. Coulson and J.F. Richardson Chemical Engineering, Volume 1 5th edition, Butterworth-Heinemann Prof. R. Shanthini 27 Feb & 05 Mar 2012

Modes of mass transfer Mass transfer could occur by the following three ways: Diffusion is caused by concentration gradient. Advection is caused by moving fluid. (It cannot therefore happen in solids.) Convection is the net transport caused by both diffusion and advection. (It occurs only in fluids.) Diffusion Advection Convection Prof. R. Shanthini 27 Feb & 05 Mar 2012

Stirring the water with a spoon creates forced convection. That helps the sugar molecules to transfer to the bulk water much faster. Diffusion (slower) Convection (faster) Prof. R. Shanthini 27 Feb & 05 Mar 2012

Diffusion Solvent B Solute A concentration of A concentration of A is high concentration of A is low Mass transfer by diffusion occurs when a component in a stationary solid or fluid goes from one point to another driven by a concentration gradient of the component. Prof. R. Shanthini 27 Feb & 05 Mar 2012

Air Blood Example of diffusion mass transfer At the surface of the lung: Air Blood Oxygen High oxygen concentration Low carbon dioxide concentration Low oxygen concentration High carbon dioxide concentration Carbon dioxide Prof. R. Shanthini 27 Feb & 05 Mar 2012

dCA JA = - DAB dz JA (1) CA CA + dCA dz Fick’s First Law of Diffusion for mass transfer in z-direction only CA A & B JA CA + dCA dz Prof. R. Shanthini 27 Feb & 05 Mar 2012

dCA JA = - DAB dz What is the unit of diffusivity? concentration gradient of A in z-direction (mass/moles per volume per distance) diffusion coefficient (or diffusivity) of A in B diffusion flux of A in relation to the bulk motion in z-direction (mass/moles per area per time) What is the unit of diffusivity? Prof. R. Shanthini 27 Feb & 05 Mar 2012

Unit and Scale of Diffusivity For dissolved matter in water: D ≈ 10-5 cm2/s For gases in air at 1 atm and at room temperature: D ≈ 0.1 to 0.01 cm2/s Diffusivity depends on the type of solute, type of solvent, temperature, pressure, solution phase (gas, liquid or solid) and other characteristics. Prof. R. Shanthini 27 Feb & 05 Mar 2012

Example 6.1.1 from Ref. 1 dCA JA = - DAB dz Molecular diffusion of Helium in Nitrogen: A mixture of He and N2 gas is contained in a pipe (0.2 m long) at 298 K and 1 atm total pressure which is constant throughout. The partial pressure of He is 0.60 atm at one end of the pipe, and it is 0.20 atm at the other end. Calculate the flux of He at steady state if DAB of He-N2 mixture is 0.687 x 10-4 m2/s. Solution: Use Fick’s law of diffusion given by equation (1) as JA = - DAB dz dCA Rearranging equation (1) and integrating gives the following: Prof. R. Shanthini 27 Feb & 05 Mar 2012

⌠ ⌡ ⌠ ⌡ JA = - DAB dz dCA (z2 – z1) (CA2 – CA1) JA = - DAB z1 z2 CA1 (2) At steady state, diffusion flux is constant. Diffusivity is taken as constant. Therefore, equation (2) gives (z2 – z1) (CA2 – CA1) JA = - DAB (3) DAB is given as 0.687 x 10-4 m2/s (z2 – z1) is given as 0.2 m (CA2 – CA1) = ? Prof. R. Shanthini 27 Feb & 05 Mar 2012

Combining the above we get CA = RT Even though CA is not given at points 1 and 2, partial pressures are given. We could relate partial pressure to concentration as follows: nA Number of moles of A CA = V Total volume Absolute temperature pA V = nA RT Gas constant Partial pressure of A pA Combining the above we get CA = RT Prof. R. Shanthini 27 Feb & 05 Mar 2012

(pA2 – pA1) (z2 – z1) JA = - DAB RT (pA2 – pA1) JA = - DAB RT(z2 – z1) Equation (3) can therefore be written as (pA2 – pA1) (z2 – z1) JA = - DAB RT which gives the flux as (pA2 – pA1) JA = - DAB RT(z2 – z1) (0.6 – 0.2) x 1.01325 x 105 Pa JA = - (0.687x10-4 m2/s) (8314 J/kmol.K) x (298 K) x (0.20–0) m JA = 5.63 x 10-6 kmol/m2.s Prof. R. Shanthini 27 Feb & 05 Mar 2012

Diffusion of gases A & B plus convection Diffusion is the net transport of substances in a stationary solid or fluid under a concentration gradient. Advection is the net transport of substances by the moving fluid, and so cannot happen in solids. It does not include transport of substances by simple diffusion. Convection is the net transport of substances caused by both advective transport and diffusive transport in fluids. JA is the diffusive flux described by Fick’s law, and we have already studied about it. Let us use NA to denote the total flux by convection (which is diffusion plus advection. Prof. R. Shanthini 27 Feb & 05 Mar 2012

dCA JA = - DAB (1) dz Molar diffusive flux of A in B: The velocity of the above diffusive flux of A in B can be given by JA (mol/m2.s) (4) vA,diffusion (m/s) = CA (mol/m3) The velocity of the net flux of A in B can be given by NA (mol/m2.s) (5) vA,convection (m/s) = CA (mol/m3) The velocity of the bulk motion can be given by (NA + NB) (mol/m2.s) (6) vbulk (m/s) = (CT) (mol/m3) Prof. R. Shanthini 27 Feb & 05 Mar 2012 Total concentration Ignore the derivation, if you wish

Multiplying the above by CA, we get vA,convection = vA,diffusion + vbulk Multiplying the above by CA, we get CA vA,convection = CA vA,diffusion + CA vbulk Using equations (4) to (6) in the above, we get (NA + NB) NA = JA + CA (7) CT Substituting JA from equation (1) in (7), we get dCA (NA + NB) NA = -DAB (8) + CA dz CT Prof. R. Shanthini 27 Feb & 05 Mar 2012 Ignore the derivation, if you wish

Let us introduce partial pressure pA into (8) as follows: nA pA (9a) CA = = V RT nT P (9b) CT = = V RT Total pressure Total number of moles Using (9a) and (9b), equation (8) can be written as DAB dpA pA (NA + NB) (10) NA = - + RT dz P Prof. R. Shanthini 27 Feb & 05 Mar 2012 Ignore the derivation, if you wish

Let us introduce molar fractions xA into (8) as follows: NA CA xA = = (11) (NA + NB) CT Using (11), equation (8) can be written as dxA NA = -CT DAB (12) + xA (NA + NB) dz Prof. R. Shanthini 27 Feb & 05 Mar 2012 Ignore the derivation, if you wish

Diffusion of gases A & B plus convection: Summary equations for (one dimensional) flow in z direction In terms of concentration of A: dCA CA NA = -DAB (NA + NB) (8) + dz CT Total concentration In terms of partial pressures (using pA = CART and P = CTRT): DAB dpA pA (NA + NB) (10) NA = - + dz P RT Total pressure In terms of molar fraction of A (using xA = CA /CT): dxA NA = -CT DAB xA (12) + (NA + NB) dz NA CA xA = = Molar fraction Prof. R. Shanthini 27 Feb & 05 Mar 2012 (NA + NB) CT

A diffusing through stagnant, non-diffusing B Evaporation of a pure liquid (A) is at the bottom of a narrow tube. Large amount of inert or non-diffusing air (B) is passed over the top. Vapour A diffuses through B in the tube. The boundary at the liquid surface (at point 1) is impermeable to B, since B is insoluble in liquid A. Hence, B cannot diffuse into or away from the surface. Therefore, NB = 0 Air (B) 2 z2 – z1 1 Liquid Benzene (A) Prof. R. Shanthini 27 Feb & 05 Mar 2012

⌠ ⌡ ⌠ ⌡ Substituting NB = 0 in equation (10), we get DAB dpA pA (NA + 0) NA = - + RT dz P Rearranging and integrating DAB dpA NA (1 - pA/P) = - RT dz ⌠ ⌡ z1 z2 ⌠ ⌡ pA1 pA2 DAB dpA NA dz = - RT (1 - pA/P) DAB P P - pA2 NA = ln (13) RT(z2 – z1) P – pA1 Prof. R. Shanthini 27 Feb & 05 Mar 2012

Introduce the log mean value of inert B as follows: (pB2 – pB1 ) (P – pA2 ) – (P – pA1 ) pB,LM = = ln(pB2 /pB1 ) ln[(P - pA2 )/ (P - pA1 )] (pA1 – pA2 ) = ln[(P - pA2 )/ (P - pA1 )] Equation (13) is therefore written as follows: DAB P NA = (14) (pA1 - pA2 ) RT(z2 – z1) pB,LM Equation (14) is the most used form. Prof. R. Shanthini 27 Feb & 05 Mar 2012

Using xA = CA /CT, pA = CART and P = CTRT, equation (13) can be converted to the following: NA = ln DAB CT 1 - xA2 (15) (z2 – z1) 1 – xA1 Introduce the log mean value of inert B as follows: (xB2 – xB1 ) (1 – xA2 ) – (1 – xA1 ) xB,LM = = ln(xB2 /xB1 ) ln[(1 - xA2 )/ (1 - xA1 )] (xA1 – xA2 ) = ln[(1 - xA2 )/ (1 - xA1 )] Therefore, equation (15) becomes the following: DAB CT NA = - (xA1 - xA2 ) (16) (z2 – z1) xB,LM Prof. R. Shanthini 27 Feb & 05 Mar 2012

Example 6.2.2 from Ref. 1 Diffusion of water through stagnant, non-diffusing air: Water in the bottom of a narrow metal tube is held at a constant temperature of 293 K. The total pressure of air (assumed to be dry) is 1 atm and the temperature is 293 K. Water evaporates and diffuses through the air in the tube, and the diffusion path is 0.1524 m long. Calculate the rate of evaporation at steady state. The diffusivity of water vapour at 1 atm and 293 K is 0.250 x 10-4 m2/s. Assume that the vapour pressure of water at 293 K is 0.0231 atm. Answer: 1.595 x 10-7 kmol/m2.s Prof. R. Shanthini 27 Feb & 05 Mar 2012

Data provided are the following: DAB = 0.250 x 10-4 m2/s; Solution: The set-up of Example 6.2.2 is shown in the figure. Assuming steady state, equation (14) applies. Air (B) DAB P NA = (pA1 - pA2 ) RT(z2 – z1) pB,LM 2 (14) where z2 – z1 (pA1 – pA2 ) pB,LM = ln[(P - pA2 )/ (P - pA1 )] 1 Data provided are the following: DAB = 0.250 x 10-4 m2/s; P = 1 atm; T = 293 K; z2 – z1 = 0.1524 m; pA1 = 0.0231 atm (saturated vapour pressure); pA2 = 0 atm (water vapour is carried away by air at point 2) Water (A) Prof. R. Shanthini 27 Feb & 05 Mar 2012

(8314 J/kmol.K) (293 K) (0.1524 m) (0.988 atm) Substituting the data provided in the equations given, we get the following: (0.0231 – 0 ) pB,LM = = 0.988 atm ln[(1 - 0 )/ (1 – 0.0231 )] (0.250x10-4 m2/s)(1x1.01325x105 Pa) (0.0231 - 0) atm NA = (8314 J/kmol.K) (293 K) (0.1524 m) (0.988 atm) = 1.595 x 10-7 kmol/m2.s Prof. R. Shanthini 27 Feb & 05 Mar 2012

Estimating Diffusivity Diffusivities for different systems could be estimated using the empirical equations provided in the following slides as well as those provided in other reference texts available in the library and other sources. Prof. R. Shanthini 27 Feb & 05 Mar 2012

Diffusivity of gases An example at 1 atm and 298 K: System Diffusivity (cm2/s) H2-NH3 0.783 H2-CH4 0.726 Ar-CH4 0.202 He-CH4 0.675 He-N2 0.687 Air-H2O 0.260 Air-C2H5OH 0.135 Air-benzene 0.0962 Prof. R. Shanthini 27 Feb & 05 Mar 2012

DAB is proportional to 1/P and T1.75 Binary Gas Diffusivity DAB - diffusivity in cm2/s P - absolute pressure in atm Mi - molecular weight T - temperature in K Vi - sum of the diffusion volume for component i DAB is proportional to 1/P and T1.75 Prof. R. Shanthini 27 Feb & 05 Mar 2012

Binary Gas Diffusivity Prof. R. Shanthini 27 Feb & 05 Mar 2012

Diffusivity in Liquids For very large spherical molecules (A) of 1000 molecular weight or greater diffusing in a liquid solvent (B) of small molecules: 9.96 x 10-12 T DAB = applicable for biological solutes such as proteins μ VA1/3 DAB - diffusivity in cm2/s T - temperature in K μ - viscosity of solution in kg/m s VA - solute molar volume at its normal boiling point in m3/kmol DAB is proportional to 1/μ and T Prof. R. Shanthini 27 Feb & 05 Mar 2012

Diffusivity in Liquids For smaller molecules (A) diffusing in a dilute liquid solution of solvent (B): 1.173 x 10-12 (Φ MB)1/2 T DAB = μB VA0.6 applicable for biological solutes DAB - diffusivity in cm2/s MB - molecular weight of solvent B T - temperature in K μ - viscosity of solvent B in kg/m s VA - solute molar volume at its normal boiling point in m3/kmol Φ - association parameter of the solvent, which 2.6 for water, 1.9 for methanol, 1.5 for ethanol, and so on DAB is proportional to 1/μB and T Prof. R. Shanthini 27 Feb & 05 Mar 2012

DAB is proportional to T Diffusivity of Electrolytes in Liquids For smaller molecules (A) diffusing in a dilute liquid solution of solvent (B): 8.928 x 10-10 T (1/n+ + 1/n-) DoAB = (1/λ+ + 1/ λ-) DoAB is diffusivity in cm2/s n+ is the valence of cation n- is the valence of anion λ+ and λ- are the limiting ionic conductances in very dilute solutions T is 298.2 when using the above at 25oC DAB is proportional to T Prof. R. Shanthini 27 Feb & 05 Mar 2012

Diffusion in solids Diffusion in solids are occurring at a very slow rate. In gas: DAB = 0.1 cm2/s Time taken 2.09 h In liquid: DAB = 10-5 cm2/s Time taken 2.39 year In solid: DAB = 10-9 cm2/s Time taken 239 centuries Prof. R. Shanthini 27 Feb & 05 Mar 2012

Diffusion in solids Diffusion in solids are occurring at a very slow rate. However, mass transfer in solids are very important. Examples: Leaching of metal ores Drying of timber, and foods Diffusion and catalytic reaction in solid catalysts Separation of fluids by membranes Treatment of metal at high temperature by gases. Prof. R. Shanthini 27 Feb & 05 Mar 2012

Diffusion in solids Diffusion in solids occur in two different ways: - Diffusion following Fick’s law (does not depend on the structure of the solid) - Diffusion in porous solids where the actual structure and void channels are important Prof. R. Shanthini 27 Feb & 05 Mar 2012

Diffusion in solids following Fick’s Law Start with equation (8): dCA CA NA = -DAB (NA + NB) (8) + dz CT Bulk term is set to zero in solids Therefore, the following equation will be used to describe the process: dCA NA = -DAB (17) dz Prof. R. Shanthini 27 Feb & 05 Mar 2012

Diffusion through a slab Applying equation (17) for steady-state diffusion through a solid slab, we get DAB (CA1 - CA2) NA = (18) z2 - z1 where NA and DAB are taken as constants. CA1 CA2 z2-z1 Similar to heat conduction. Prof. R. Shanthini 27 Feb & 05 Mar 2012

Relating the concentration and solubility The solubility of a solute gas in a solid is usually expressed by the notation S. Unit used in general is the following: m3 solute at STP m3 solid . atm partial pressure of solute Relationship between concentration and solubility: S pA CA = kmol solute /m3 solid 22.414 where pA is in atm STP of 0oC and 1 atm Prof. R. Shanthini 27 Feb & 05 Mar 2012

Relating the concentration and permeability The permeability of a solute gas (A) in a solid is usually expressed by the notation PM. in m3 solute at STP (0oC and 1 atm) diffusing per second per m2 cross-sectional area through a solid 1 m thick under a pressure difference of 1 atm. Unit used in general is the following: m3 solute at STP . 1 m thick solid s . m2 cross-sectional area . atm pressure difference Relationship between concentration and permeability: PM = DAB S where DAB is in m2/s and S is in m3/m3.atm Prof. R. Shanthini 27 Feb & 05 Mar 2012

Example 6.5.1 from Ref. 1 Diffusion of H2 through Neoprene membrane: The gas hydrogen at 17oC and 0.010 atm partial pressure is diffusing through a membrane on vulcanized neoprene rubber 0.5 mm thick. The pressure of H2 on the other side of neoprene is zero. Calculate the steady-state flux, assuming that the only resistance to diffusion is in the membrane. The solubility S of H2 gas in neoprene at 17oC is 0.051 m3 (at STP of 0oC and 1 atm)/m3 solid. atm and the diffusivity DAB is 1.03 x 10-10 m2/s at 17oC. Answer: 4.69 x 10-12 kmol H2/m2.s Prof. R. Shanthini 27 Feb & 05 Mar 2012

Example 6.5.2 from Ref. 1 Diffusion through a packging film using permeability: A polythene film 0.00015 m (0.15 mm) thick is being considered for use in packaging a pharmaceutical product at 30oC. If the partial pressure of O2 outside the package is 0.21 atm and inside it is 0.01 atm, calculate the diffusion flux of O2 at steady state. Assume that the resistances to diffusion outside the film and inside are negligible compared to the resistance of the film. Permeability of O2 in polythene at 303 K is 4.17 x 10-12 m3 solute (STP)/(s.m2.atm.m). Answer: 2.480 x 10-12 kmol O2/m2.s Would you prefer nylon to polythene? Permeability of O2 in nylon at 303 K is 0.029 x 10-12 m3 solute (STP)/(s.m2.atm.m). Support your answer. Prof. R. Shanthini 27 Feb & 05 Mar 2012

Diffusion through a cylinder wall Applying equation (17) for steady-state diffusion through a cylinder wall of inner radius r1 and outer radius r2 and length L in the radial direction outward, we get Mass transfer per time Mass transfer per area per time nA dCA (19) NA = = -DAB 2 π r L dr Area of mass transfer 2πL DAB(CA1 - CA2) CA2 nA = (20) CA1 r ln(r2 / r1) r2 r1 Similar to heat conduction. Prof. R. Shanthini 27 Feb & 05 Mar 2012

Diffusion through a spherical shell Applying equation (17) for steady-state diffusion through a spherical shell of inner radius r1 and outer radius r2 in the radial direction outward, we get Mass transfer per time Mass transfer per area per time nA dCA (21) NA = = -DAB 4 π r2 dr Area of mass transfer 4πr1r2 DAB(CA1 - CA2) nA = (22) CA2 CA1 (r2 - r1) r r2 r1 Similar to heat conduction. Prof. R. Shanthini 27 Feb & 05 Mar 2012

Microscopic (or Fick’s Law) approach: dCA JA = - DAB (1) dz good for diffusion dominated problems Macroscopic (or mass transfer coefficient) approach: NA = - k ΔCA (50) where k is known as the mass transfer coefficient good for convection dominated problems Prof. R. Shanthini 27 Feb & 05 Mar 2012

Mass Transfer Coefficient Approach NA = kc ΔCA = kc (CA1 – CA2 ) (51) kc is the liquid-phase mass-transfer coefficient based on a concentration driving force. CA1 A & B What is the unit of kc? NA CA2 Prof. R. Shanthini 27 Feb & 05 Mar 2012

Mass Transfer Coefficient Approach NA = kc ΔCA = kc (CA1 – CA2 ) (51) Using the following relationships between concentrations and partial pressures: CA1 = pA1 / RT; CA2 = pA2 / RT Equation (51) can be written as NA = kc (pA1 – pA2) / RT = kp (pA1 – pA2) (52) where kp = kc / RT (53) kp is a gas-phase mass-transfer coefficient based on a partial-pressure driving force. Prof. R. Shanthini 27 Feb & 05 Mar 2012 What is the unit of kp?

Models for mass transfer between phases Mass transfer between phases across the following interfaces are of great interest in separation processes: - gas/liquid interface - liquid/liquid interface Such interfaces are found in the following separation processes: - absorption - distillation - extraction - stripping Prof. R. Shanthini 27 Feb & 05 Mar 2012

Models for mass transfer at a fluid-fluid interface Theoretical models used to describe mass transfer between a fluid and such an interface: - Film Theory - Penetration Theory - Surface-Renewal Theory - Film Penetration Theory Prof. R. Shanthini 27 Feb & 05 Mar 2012

Film Theory Entire resistance to mass transfer in a given turbulent phase is in a thin, stagnant region of that phase at the interface, called a film. Liquid film For the system shown, gas is taken as pure component A, which diffuses into nonvolatile liquid B. pA Bulk liquid CAi In reality, there may be mass transfer resistances in both liquid and gas phases. So we need to add a gas film in which gas is stagnant. Gas CAb z=0 z=δL Prof. R. Shanthini 27 Feb & 05 Mar 2012 Mass transport

Two Film Theory There are two stagnant films (on either side of the fluid-fluid interface). Each film presents a resistance to mass transfer. Concentrations in the two fluid at the interface are assumed to be in phase equilibrium. Liquid film Gas film Gas phase Liquid phase pAb pAi CAi CAb Prof. R. Shanthini 27 Feb & 05 Mar 2012 Mass transport

Two Film Theory Interface Interface Liquid phase Liquid film Gas phase pAb CAi CAb pAi Gas film Mass transport Gas phase Liquid phase pAb pAi CAi CAb Mass transport Concentration gradients for the film theory More realistic concentration gradients Prof. R. Shanthini 27 Feb & 05 Mar 2012

Two Film Theory applied at steady-state Mass transfer in the gas phase: NA = kp (pAb – pAi) (52) Mass transfer in the liquid phase: Liquid film Gas film NA = kc (CAi – CAb ) (51) Gas phase Liquid phase Phase equilibrium is assumed at the gas-liquid interface. Applying Henry’s law, pAb pAi CAi CAb pAi = HA CAi (53) Prof. R. Shanthini 27 Feb & 05 Mar 2012 Mass transport, NA

Henry’s Law pAi = HA CAi at equilibrium, where HA is Henry’s constant for A Note that pAi is the gas phase pressure and CAi is the liquid phase concentration. Liquid film Gas film pAb pAi Unit of H: [Pressure]/[concentration] = [ bar / (kg.m3) ] CAi CAb Prof. R. Shanthini 27 Feb & 05 Mar 2012

Two Film Theory applied at steady-state We know the bulk concentration and partial pressure. We do not know the interface concentration and partial pressure. Therefore, we eliminate pAi and CAi from (51), (52) and (53) by combining them appropriately. Liquid film Gas film Gas phase Liquid phase pAb pAi CAi CAb Prof. R. Shanthini 27 Feb & 05 Mar 2012 Mass transport, NA

Two Film Theory applied at steady-state NA From (52): pAi = pAb - (54) kp NA From (51): CAi = CAb + (55) kc Substituting the above in (53) and rearranging: pAb - HA CAb NA = (56) HA / kc + 1 / kp The above expression is based on gas-phase and liquid-phase mass transfer coefficients. Let us now introduce overall gas-phase and overall liquid-phase mass transfer coefficients. Prof. R. Shanthini 27 Feb & 05 Mar 2012

Introducing overall gas-phase mass transfer coefficient: Let’s start from (56). Introduce the following imaginary gas-phase partial pressure: pA* ≡ HA CAb (57) where pA* is a partial pressure that would have been in equilibrium with the concentration of A in the bulk liquid. Introduce an overall gas-phase mass-transfer coefficient (KG) as 1 1 HA ≡ + (58) KG kp kc Combining (56), (57) and (58): NA = KG (pAb - pA* ) (59) Prof. R. Shanthini 27 Feb & 05 Mar 2012

Introducing overall liquid-phase mass transfer coefficient: Once again, let’s start from (56). Introduce the following imaginary liquid-phase concentration: pAb ≡ HA CA* (60) where CA* is a concentration that would have been in equilibrium with the partial pressure of A in the bulk gas. Introduce an overall liquid-phase mass-transfer coefficient (KL) as 1 1 1 ≡ + (61) KL HAkp kc Combining (56), (60) and (61): NA = KL (CA* - CAb) (62) Prof. R. Shanthini 27 Feb & 05 Mar 2012

CAb CAi CA* CA Gas-Liquid Equilibrium Partitioning Curve showing the locations of p*A and C*A pA pAb pAb = HACA* pAi pAi = HA CAi pA* pA* = HA CAb CAb CAi CA* CA Prof. R. Shanthini 27 Feb & 05 Mar 2012

Summary: NA = KL (CA* - CAb) (62) = KG (pAb - pA*) (59) where CA* = pAb / HA (60) pA* = HA CAb (57) 1 HA 1 HA = = + (58 and 61) KG KL kp kc Prof. R. Shanthini 27 Feb & 05 Mar 2012

Example 3.20 from Ref. 2 (modified) Sulfur dioxide (A) is absorbed into water in a packed column. At a certain location, the bulk conditions are 50oC, 2 atm, yAb = 0.085, and xAb = 0.001. Equilibrium data for SO2 between air and water at 50oC are the following: pA (atm) 0.0382 0.0606 0.1092 0.1700 CA (kmol/m3) 0.03126 0.04697 0.07823 0.10949 Experimental values of the mass transfer coefficients are kc = 0.18 m/h and kp = 0.040 kmol/h.m2.kPa. Compute the mass-transfer flux by assuming an average Henry’s law constant and a negligible bulk flow. Prof. R. Shanthini 27 Feb & 05 Mar 2012

Solution: Data provided: T = 273oC + 50oC = 323 K; PT = 2 atm; yAb = 0.085; xAb = 0.001; kc = 0.18 m/h; kp = 0.040 kmol/h.m2.kPa HA = 1.4652 atm.m3/kmol slope of the curve HA = 161.61 kPa.m3/kmol Prof. R. Shanthini 27 Feb & 05 Mar 2012

Equations to be used: NA = KL (CA* - CAb) (62) = KG (pAb - pA*) (59) where CA* = pAb / HA (60) pA* = HA CAb (57) 1 HA 1 HA = = + (58 and 61) KG KL kp kc Prof. R. Shanthini 27 Feb & 05 Mar 2012

Calculation of overall mass transfer coefficients: 1 HA 1 HA = = + (58 and 61) KG KL kp kc 1 1 = h.m2.kPa/kmol = 25 h.m2.kPa/kmol kp 0.040 HA 161.61 kPa.m3/kmol = = 897 h.m2.kPa/kmol kc 0.18 m/h KG = 1/(25 + 897) = 1/922 = 0.001085 kmol/h.m2.kPa KL = HA KG = 161.61/922 = 0.175 m/h Prof. R. Shanthini 27 Feb & 05 Mar 2012

(62) is used to calculate NA NA = KL (CA* - CAb) (62) is used to calculate NA CA* = pAb / HA = yAb PT / HA = 0.085 x 2 atm / 1.4652 atm.m3/kmol = 0.1160 kmol/m3 CAb = xAb CT = 0.001 CT CT = concentration of water (assumed) = 1000 kg/m3 = 1000/18 kmol/m3 = 55.56 kmol/m3 CAb = 0.001 x 55.56 kmol/m3 = 0.05556 kmol/m3 NA = (0.175 m/h) (0.1160 - 0.05556) kmol/m3 = 0.01058 kmol/m2.h Prof. R. Shanthini 27 Feb & 05 Mar 2012

(59) is used to calculate NA Alternatively, NA = KG (pAb - pA*) (59) is used to calculate NA pA* = CAb HA = xAb CT HA = 0.001 x 55.56 x 161.61 kPa = 8.978 kPa pAb = yAb PT = 0.085 x 2 x 1.013 x 100 kPa = 17.221 kPa NA = (1/922 h.m2.kPa/kmol) (17.221 - 8.978) kPa = 0.00894 kmol/m2.h Prof. R. Shanthini 27 Feb & 05 Mar 2012

Summary: Two Film Theory applied at steady-state = KG (pAb - pA*) = KL (CA* - CAb) NA = kp (pAb – pAi) = kc (CAi – CAb ) (52) (51) (59) (62) pAi = HA CAi (53) Liquid film Gas film pAb = HA CA* (60) pA* = HA CAb (57) Liquid phase pAb 1 HA 1 HA Gas phase pAi = = + CAi KG KL kp kc CAb (58 and 61) Prof. R. Shanthini 27 Feb & 05 Mar 2012 Mass transport, NA

Summary equations with mole fractions = Ky (yAb - yA*) = Kx (xA* - xAb) NA = ky (yAb – yAi) = kx (xAi – xAb ) (63) yAi = KA xAi (64) Liquid film Gas film yAb = KA xA* (65) yA* = KA xAb (66) Liquid phase yAb 1 KA 1 KA Gas phase yAi = = + (67) xAi Ky Kx ky kx xAb Prof. R. Shanthini 27 Feb & 05 Mar 2012 Mass transport, NA

Notations used: xAb : liquid-phase mole fraction of A in the bulk liquid yAb : gas-phase mole fraction of A in the bulk gas xAi : liquid-phase mole fraction of A at the interface yAi : gas-phase mole fraction of A at the interface xA* : liquid-phase mole fraction of A which would have been in equilibrium with yAb yA* : gas-phase mole fraction of A which would have been in equilibrium with xAb kx : liquid-phase mass-transfer coefficient ky : gas-phase mass-transfer coefficient Kx : overall liquid-phase mass-transfer coefficient Ky : overall gas-phase mass-transfer coefficient KA : vapour-liquid equilibrium ratio (or equilibrium distribution coefficient) Prof. R. Shanthini 27 Feb & 05 Mar 2012

xAb xAi xA* xA Gas-liquid equilibrium ratio (KA) curve yA yAb yAb = KAxA* yAi yAi = KA xAi yA* yA* = KA xAb xAb xAi xA* xA Prof. R. Shanthini 27 Feb & 05 Mar 2012

Gas & Liquid-side Resistances in Interfacial Mass Transfer 1 KG kp = + H kc fG = fraction of gas-side resistance = 1/KG 1/kp 1/kp = + H/kc kc = + H kp 1 KL H kp = + kc fL = fraction of liquid-side resistance = 1/KL 1/kc 1/Hkp 1/kc = + 1/kc + kc/H kp = Prof. R. Shanthini 27 Feb & 05 Mar 2012

Gas & Liquid-side Resistances in Interfacial Mass Transfer If fG > fL, use the overall gas-side mass transfer coefficient and the overall gas-side driving force. If fL > fG use the overall liquid-side mass transfer coefficient and the overall liquid-side driving force. Prof. R. Shanthini 27 Feb & 05 Mar 2012

1 H 1 H = = + KOG KOL KG KL 1 HA 1 HA = = + (58 and 61) KG KL kp kc = = + (58 and 61) KG KL kp kc The above is also written with the following notations: 1 H 1 H = = + KOG KOL KG KL Prof. R. Shanthini 27 Feb & 05 Mar 2012

Other Driving Forces Mass transfer is driven by concentration gradient as well as by pressure gradient as we have just seen. In pharmaceutical sciences, we also must consider mass transfer driven by electric potential gradient (as in the transport of ions) and temperature gradient. Transport Processes in Pharmaceutical Systems (Drugs and the Pharmaceutical Sciences, vol. 102), edited by G.L. Amidon, P.I. Lee, and E.M. Topp (Nov 1999) Encyclopedia of Pharmaceutical Technology (Hardcover) by James Swarbrick (Author) Prof. R. Shanthini 27 Feb & 05 Mar 2012

Example 1 Example 1 worked out An exhaust stream from a containing 3 mole% acetone and 90 mole% air is fed to a mass transfer column in which the acetone is stripped by a countercurrent, falling 293 K water stream. The tower is operated at a total pressure of 1.013x105 Pa. If a combination of Raoult-Dalton equilibrium relation may be used to determine the distribution of acetone between the air and aqueous phases, determine (a) The mole fraction of acetone within the aqueous phase which would be in equilibrium with the 3 mole% acetone gas mixture, and (b) The mole fraction of acetone in the gas phase which would be in equilibrium with 20 ppm acetone in the aqueous phase. Example 1 worked out Prof. R. Shanthini 27 Feb & 05 Mar 2012

Prof. R. Shanthini 27 Feb & 05 Mar 2012

Example 2 The Henry’s law constant for oxygen dissolved in water is 4.06x109 Pa/(mole of oxygen per total mole of solution) at 293 K. Determine the solution concentration of oxygen in water which is exposed to dry air at 1.013x105 Pa and 293 K. Prof. R. Shanthini 27 Feb & 05 Mar 2012

Example 2 worked out Henry’s law can be expressed in terms of the mole fraction units by pA = H’ xA where H’ is 4.06x109 Pa/(mol of oxygen/total mol of solution). Dry air contains 21 mole percent oxygen. By Dalton’s law pA = yA P = (0.21)(1.013x105 Pa) = 2.13 x 104 Pa Prof. R. Shanthini 27 Feb & 05 Mar 2012

Prof. R. Shanthini 27 Feb & 05 Mar 2012

Example 3 In an experiment study of the absorption of ammonia by water in a wetted-wall column, the overall mass-transfer coefficient, KG, was found to be 2.74 x 10-9 kgmol/m2.s.Pa. At one point in the column, the gas phase contained 8 mole% ammonia and the liquid-phase concentration was 0.064 kgmole ammonia/m3 of solution. The tower operated at 293 K and 1.013x105 Pa. At that temperautre, the Henry’s law constant is 1.358x103 Pa/(kgmol/m3). If 85% of the total resistance to mass transfer is encountered in the gas phase, determine the individual film mass-transfer coefficients and the interfacial compositions. Prof. R. Shanthini 27 Feb & 05 Mar 2012

Example 3 worked out Prof. R. Shanthini 27 Feb & 05 Mar 2012

Prof. R. Shanthini 27 Feb & 05 Mar 2012

Prof. R. Shanthini 27 Feb & 05 Mar 2012

Prof. R. Shanthini 27 Feb & 05 Mar 2012

Example 4 Example 4 worked out A wastewater stream is introduced to the top of a mass-transfer tower where it flows countercurrent to an air stream. At one point in the tower, the wastewater stream contains 10-3 mole A/m3 and the air is essentially free of any A. At the operation conditions within the tower, the film mass-transfer coefficients are KL = 5x10-4 kmole/m2.s.(kmole/m3) and KG = 0.01 kmole/m2.s.atm. The concentrations are in the henry’s law region where pA,i = H CA,i with H =1 0 atm/(kmole/m3). Determine the following: (a) The overall mass flux of A (b) The overall mass-transfer coefficients, KOL and KOG. Example 4 worked out Prof. R. Shanthini 27 Feb & 05 Mar 2012

Prof. R. Shanthini 27 Feb & 05 Mar 2012

Prof. R. Shanthini 27 Feb & 05 Mar 2012

Prof. R. Shanthini 27 Feb & 05 Mar 2012

Prof. R. Shanthini 27 Feb & 05 Mar 2012