1. Prof. R. Shanthini 21 Feb 2013 1 Course content of Mass transfer section LTA Diffusion Theory of interface mass transfer Mass transfer coefficients, overall coefficients and transfer units Application in absorption, extraction and adsorption 080205 Concept of continuous contacting equipment Simultaneous heat and mass transfer in gas- liquid contacting, and solids drying CSK 060105 CP302 Separation Process Principles Mass Transfer - Set 1

2. Prof. R. Shanthini 21 Feb 2013 2 CP302 Separation Process Principles Reference books used for ppts 1.C.J. Geankoplis Transport Processes and Separation Process Principles 4 th edition, Prentice-Hall India 2.J.D. Seader and E.J. Henley Separation Process Principles 2 nd edition, John Wiley & Sons, Inc. 3.J.M. Coulson and J.F. Richardson Chemical Engineering, Volume 1 5 th edition, Butterworth-Heinemann 4. Chapter 10 of R.P. Singh and D.R. Heldman Introduction to Food Engineering 3 rd edition, Academic Press

3. Prof. R. Shanthini 21 Feb 2013 3 Mass transfer could occur by the following three ways: 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. It cannot therefore 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. Modes of mass transfer

4. Prof. R. Shanthini 21 Feb 2013 4 forced convection 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) Modes of mass transfer

5. Prof. R. Shanthini 21 Feb 2013 5 Objectives of the slides that follow: Understanding diffusion in one dimension

6. Prof. R. Shanthini 21 Feb 2013 6 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. Solute A Solvent B concentration of A is high concentration of A is low Diffusion

7. Prof. R. Shanthini 21 Feb 2013 7 concentration of A is the same everywhere Diffusion Diffusion is the macroscopic result of random thermal motion on a micoscopic scale (Brownian motion). It occurs even when there is no concentration gradient (but there will be no net flux). Solute A Solvent B

8. Prof. R. Shanthini 21 Feb 2013 8 AB AB Liquids A and B are separated from each other. Separation removed. A goes from high concentration of A to low concentration of A. B goes from high concentration of B to low concentration of B. Equilibrium is reached Molecules of A and B are uniformly distributed everywhere in the vessel purely due to DIFFUSION. Equilibrium is reached

9. Prof. R. Shanthini 21 Feb 2013 9 Examples of Diffusion A diffusion problem that occurs in the field of microelectronics is the oxidation of silicon according to the reaction Si + O 2 SiO 2. When a slab of the material is exposed to gaseous oxygen, the oxygen undergoes a first-order reaction to produce a layer of the oxide. The task is to predict the thickness d of the very slowly- growing oxide layer as a function of time t. Silicon Silicon oxide layer Oxygen C A = C A0 at Z = 0 C A = C Aδ at Z = δ

10. Prof. R. Shanthini 21 Feb 2013 10 Examples of Diffusion If an insect flight muscle contains tracheal tubules which allow air to diffuse into all parts of the muscle, and the tracheal tubules make up 20% of the volume of the muscle, how large can the muscle be?

11. Prof. R. Shanthini 21 Feb 2013 11 Objectives of the slides that follow: Mathematical modelling of steady-state one dimensional diffusive mass transfer

12. Prof. R. Shanthini 21 Feb 2013 12 CACA C A + dC A dzdz Mixture of A & B For mass transfer occurring only in z-direction (1) Fick’s First Law of Diffusion JAJA What is J A ? J A = - D AB dzdz dCAdCA

13. Prof. R. Shanthini 21 Feb 2013 13 diffusion coefficient (or diffusivity) of A in B diffusion flux of A in relation to the bulk motion in z-direction Unit: mass (or moles) per area per time concentration gradient of A in z-direction Unit: mass (or moles) per volume per distance What is the unit of diffusivity? J A = - D AB dzdz dCAdCA Fick’s First Law of Diffusion

14. Prof. R. Shanthini 21 Feb 2013 14 For dissolved matter in water: D ≈ 10 -5 cm 2 /s For gases in air at 1 atm and at room temperature: D ≈ 0.1 to 0.01 cm 2 /s Diffusivity depends on the type of solute, type of solvent, temperature, pressure, solution phase (gas, liquid or solid) and other characteristics. Unit and Scale of Diffusivity

15. Prof. R. Shanthini 21 Feb 2013 15 Molecular diffusion of Helium in Nitrogen: A mixture of He and N 2 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 D AB of He-N 2 mixture is 0.687 x 10 -4 m 2 /s. Example 6.1.1 from Ref. 1 Solution: Use Fick’s law of diffusion given by equation (1) as J A = - D AB dzdz dCAdCA Rearranging the Fick’s law and integrating gives the following:

16. Prof. R. Shanthini 21 Feb 2013 16 J A = - D AB dz dC A D AB is given as 0.687 x 10 -4 m 2 /s (z 2 – z 1 ) is given as 0.2 m (C A2 – C A1 ) = ? ⌠ ⌡ z1z1 z2z2 ⌠ ⌡ C A1 C A2 At steady state, diffusion flux is constant. Diffusivity is taken as constant. (2) Therefore, equation (2) gives J A = - D AB (z 2 – z 1 )(C A2 – C A1 ) (3)

17. Prof. R. Shanthini 21 Feb 2013 17 Even though C A is not given at points 1 and 2, partial pressures are given. We could relate partial pressure to concentration as follows: C A = nAnA V Number of moles of A Total volume p A V = n A RT Partial pressure of A Absolute temperature Gas constant Combining the above we get C A = pApA RT

18. Prof. R. Shanthini 21 Feb 2013 18 Equation (3) can therefore be written as J A = - D AB (z 2 – z 1 ) (p A2 – p A1 ) RT which gives the flux as J A = - D AB (p A2 – p A1 ) RT(z 2 – z 1 ) J A = 5.63 x 10 -6 kmol/m 2.s J A = - (0.687x10 -4 m 2 /s) (0.6 – 0.2) x 1.01325 x 10 5 Pa (8314 J/kmol.K) x (298 K) x (0.20–0) m

19. Prof. R. Shanthini 21 Feb 2013 19 Summary: Modelling diffusion in z-direction = - D AB (p A2 – p A1 ) RT(z 2 – z 1 ) J A = - D AB (C A2 – C A1 ) (z 2 – z 1 ) z z1z1 z2z2 C A2 C A1 JAJA Longitudinal flow: Flow area perpendicular to the flow direction is a constant. C A1 and p A1 at z 1 and C A2 and p A2 at z 2 remain unchanged with time (steady state). D AB is constant

20. Prof. R. Shanthini 21 Feb 2013 20 Objectives of the slides that follow: Derivation of D AB = D BA under certain conditions D AB : diffusivity of A in B D BA : diffusivity of B in A

21. Prof. R. Shanthini 21 Feb 2013 21 J A = - D AB dCAdCA dzdz Consider steady-state diffusion in an ideal mixture of 2 ideal gases A & B at constant total pressure and temperature. J A and J B : molar diffusive flux of A and B, respectively (moles/area.time) C A and C B : concentration of A and B, respectively (moles/volume) D AB and D BA : diffusivities of A in B and of B in A, respectively z : distance in the direction of transfer Molar diffusive flux of A in B: Molar diffusive flux of B in A: (4) (5) dCBdCB dzdz J B = - D BA Equimolar counter-diffusion in gases

22. Prof. R. Shanthini 21 Feb 2013 22 For an ideal gas mixture at constant pressure, C A + C B = p A /RT + p B /RT = P/RT = constant Therefore, dC A + dC B = 0 J A + J B = 0 Since the total pressure remains constant, there is no net mass transfer. That is, (6) (7) Substituting (6) and (7) in (4) and (5), we get D AB = D BA = D (say) Therefore, (4) & (5) give J A = - D dCAdCA dzdz J B = - D dCBdCB dzdz (8a) (8b)

23. Prof. R. Shanthini 21 Feb 2013 23 This describes the mass transfer arising solely from the random motion of the molecules (i.e., only diffusion) stationary mediumfluid in streamline flow It is applicable to stationary medium or a fluid in streamline flow. This is known as equimolar counter diffusion. J A = - D dCAdCA dzdz J B = - D dCBdCB dzdz (8a) (8b)

24. Prof. R. Shanthini 21 Feb 2013 24 Objectives of the slides that follow: Mathematical modelling of steady-state one dimensional convective mass transfer

25. Prof. R. Shanthini 21 Feb 2013 25 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. Diffusion of gases A & B plus convection J A is the diffusive flux described by Fick’s law, and we have already studied about it. Let us use N A to denote the total flux by convection (which is diffusion plus advection.

26. Prof. R. Shanthini 21 Feb 2013 26 J A = - D AB dCAdCA dzdz Molar diffusive flux of A in B: (4) The velocity of the above diffusive flux of A in B can be given by v A,diffusion (m/s) = J A (mol/m 2.s) C A (mol/m 3 ) The velocity of the net flux of A in B can be given by v A,convection (m/s) = N A (mol/m 2.s) C A (mol/m 3 ) The velocity of the bulk motion can be given by v bulk (m/s) = (N A + N B ) (mol/m 2.s) (C T ) (mol/m 3 ) (9) (10) (11) Total concentration

27. Prof. R. Shanthini 21 Feb 2013 27 v A,diffusion +v A,convection =v bulk (N A + N B ) CTCT C A v A,diffusion +C A v A,convection =C A v bulk J A +N A =CACA (12) Substituting J A from equation (4) in (12), we get (N A + N B ) CTCT N A = + C A (13) dC A dz -D AB Multiplying the above by C A, we get Using equations (9) to (11) in the above, we get

28. Prof. R. Shanthini 21 Feb 2013 28 Using (14a) and (14b), equation (13) can be written as (N A + N B ) N A = - + (15) dp A dz D AB Let us introduce partial pressure p A into (13) as follows: C A = nAnA V (14a) = pApA RT C T = nTnT V (14b) = P RT Total number of moles Total pressure RT pApA P

29. Prof. R. Shanthini 21 Feb 2013 29 Using (16), equation (13) can be written as (N A + N B ) N A = + x A (17) dx A dz -C T D AB Let us introduce molar fractions x A into (13) as follows: x A = NANA (N A + N B ) = CACA CTCT (16)

30. Prof. R. Shanthini 21 Feb 2013 30 (N A + N B ) N A =(17) dx A dz -C T D AB Diffusion of gases A & B plus convection: Summary equations for (one dimensional) flow in z direction (N A + N B ) N A = - + (15) dp A dz D AB RT pApA P N A = + (13) dC A dz -D AB In terms of concentration of A: In terms of partial pressures (using p A = C A RT and P = C T RT): In terms of molar fraction of A (using x A = C A /C T ): xAxA (N A + N B ) CTCT CACA +

31. Prof. R. Shanthini 21 Feb 2013 31 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, N B = 0 A diffusing through stagnant, non-diffusing B Liquid Benzene (A) Air (B) 1 2 z 2 – z 1

32. Prof. R. Shanthini 21 Feb 2013 32 Substituting N B = 0 in equation (15), we get (N A + 0) N A = - + dp A dz D AB RT pApA P Rearranging and integrating N A (1 - p A /P) = - dp A dz D AB RT N A dz = - dp A (1 - p A /P) D AB RT ⌠ ⌡ z1z1 z2z2 ⌠ ⌡ p A1 p A2 N A = ln P - p A2 D AB P RT(z 2 – z 1 )P – p A1 (18)

33. Prof. R. Shanthini 21 Feb 2013 33 Introduce the log mean value of inert B as follows: N A = (p A1 - p A2 ) D AB P RT(z 2 – z 1 ) p B,LM (19) p B,LM = = (p B2 – p B1 ) ln(p B2 /p B1 ) (P – p A2 ) – (P – p A1 ) ln[(P - p A2 )/ (P - p A1 )] Equation (18) is therefore written as follows: Equation (19) is the most used form. (p A1 – p A2 ) ln[(P - p A2 )/ (P - p A1 )] =

34. Prof. R. Shanthini 21 Feb 2013 34 N A = - (x A1 - x A2 ) D AB C T (z 2 – z 1 ) x B,LM (21) N A = ln 1 - x A2 D AB C T (z 2 – z 1 ) 1 – x A1 (20) Using x A = C A /C T, p A = C A RT and P = C T RT, equation (18) can be converted to the following: Introduce the log mean value of inert B as follows: x B,LM = = (x B2 – x B1 ) ln(x B2 /x B1 ) (1 – x A2 ) – (1 – x A1 ) ln[(1 - x A2 )/ (1 - x A1 )] (x A1 – x A2 ) ln[(1 - x A2 )/ (1 - x A1 )] = Therefore, equation (20) becomes the following:

35. Prof. R. Shanthini 21 Feb 2013 35 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 m 2 /s. Assume that the vapour pressure of water at 293 K is 0.0231 atm. Answer: 1.595 x 10 -7 kmol/m 2.s Example 6.2.2 from Ref. 1