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CP302 Separation Process Principles

CP302 Separation Process Principles. Mass Transfer - Set 1. CP302 Separation Process Principles. Reference books used for ppts. C.J. Geankoplis Transport Processes and Separation Process Principles 4 th edition, Prentice-Hall India J.D. Seader and E.J. Henley

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CP302 Separation Process Principles

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  1. CP302 Separation Process Principles Mass Transfer - Set 1

  2. CP302 Separation Process Principles 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 • 4. Chapter 10 of R.P. Singh and D.R. Heldman • Introduction to Food Engineering • 3rd edition, Academic Press

  3. Modes of mass transfer Mass transfer could occur by the following three ways: Diffusion is the net transport of substancesin 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. Convectionis the net transport of substances caused by both advective transport and diffusive transport in fluids.

  4. Modes of mass transfer Stirring the water with a spoon creates forcedconvection. That helps the sugar molecules to transfer to the bulk water much faster. Diffusion (slower) Convection (faster)

  5. Objectives of the slides that follow: Understanding diffusion in one dimension

  6. Diffusion Solvent B Solute 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.

  7. Diffusion Solvent B Solute A concentration of A is the same everywhere 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).

  8. A B A B 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. Molecules of A and B are uniformly distributed everywhere in the vessel purely due to DIFFUSION. Equilibrium is reached

  9. Examples of Diffusion A diffusion problem that occurs in the field of microelectronics is the oxidation of silicon according to the reaction Si + O2    SiO2. 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. Oxygen CA = CA0 at Z = 0 Silicon oxide layer CA = CAδ at Z = δ Silicon

  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. Objectives of the slides that follow: Mathematical modelling of steady-state one dimensional diffusive mass transfer

  12. dCA JA= - DAB dz Fick’s First Law of Diffusion For mass transfer occurring only in z-direction (1) Mixture of A & B CA What is JA? JA CA + dCA dz

  13. dCA JA= - DAB dz Fick’s First Law of Diffusion concentration gradient of A in z-direction Unit: mass (or 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 Unit: mass (or moles) per area per time What is the unit of diffusivity?

  14. 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.

  15. dCA JA= - DAB dz Example 6.1.1 from Ref. 1 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 Rearranging the Fick’s law and integrating gives the following:

  16. ⌠ ⌠ ⌡ ⌡ CA2 z2 CA1 z1 JA= - DAB dz dCA (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) = ?

  17. 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

  18. 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

  19. Summary: Modelling diffusion in z-direction (CA2 – CA1) (pA2 – pA1) JA = - DAB = - DAB (z2 – z1) RT(z2 – z1) CA1 and pA1 at z1 and CA2 and pA2 at z2 remain unchanged with time (steady state). DAB is constant JA z1 z2 z CA1 CA2 Longitudinal flow: Flow area perpendicular to the flow direction is a constant.

  20. Objectives of the slides that follow: Derivation of DAB = DBA under certain conditions DAB: diffusivity of A in B DBA: diffusivity of B in A

  21. dCA JA = - DAB dz dCB JB = - DBA dz Equimolar counter-diffusion in gases Consider steady-state diffusion in an ideal mixture of 2 ideal gases A & B at constant total pressure and temperature. Molar diffusive flux of A in B: (4) Molar diffusive flux of B in A: (5) JA and JB : molar diffusive flux of A and B, respectively (moles/area.time) CA and CB : concentration of A and B, respectively (moles/volume) DAB and DBA : diffusivities of A in B and of B in A, respectively z: distance in the direction of transfer

  22. dCA JA = - D (8a) dz dCB JB = - D (8b) dz Since the total pressure remains constant, there is no net mass transfer. That is, (6) JA + JB = 0 For an ideal gas mixture at constant pressure, CA + CB = pA/RT + pB/RT = P/RT = constant Therefore, dCA + dCB = 0 (7) Substituting (6) and (7) in (4) and (5), we get DAB = DBA = D (say) Therefore, (4) & (5) give

  23. dCA JA = - D (8a) dz dCB JB = - D (8b) dz This is known as equimolar counter diffusion. This describes the mass transfer arising solely from the random motion of the molecules (i.e., only diffusion) It is applicable to stationary medium or a fluid in streamline flow.

  24. Objectives of the slides that follow: Mathematical modelling of steady-state one dimensional convective mass transfer

  25. Diffusion of gases A & B plus convection Diffusion is the net transport of substancesin 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. Convectionis 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.

  26. dCA JA = - DAB dz Molar diffusive flux of A in B: (4) The velocity of the above diffusive flux of A in B can be given by JA (mol/m2.s) (9) 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) (10) vA,convection (m/s) = CA (mol/m3) The velocity of the bulk motion can be given by (NA + NB)(mol/m2.s) (11) vbulk (m/s) = (CT)(mol/m3) Total concentration

  27. vA,convection = vA,diffusion + vbulk Multiplying the above by CA, we get CA vA,convection = CA vA,diffusion + CA vbulk Using equations (9) to (11) in the above, we get (NA + NB) NA = JA + CA (12) CT Substituting JA from equation (4) in (12), we get dCA (NA + NB) NA = -DAB (13) + CA dz CT

  28. Let us introduce partial pressure pA into (13) as follows: nA pA (14a) CA = = V RT nT P (14b) CT = = V RT Total pressure Total number of moles Using (14a) and (14b), equation (13) can be written as DAB dpA pA (NA + NB) (15) NA = - + RT dz P

  29. Let us introduce molar fractions xA into (13) as follows: NA CA (16) xA = = (NA + NB) CT Using (16), equation (13) can be written as dxA NA = -CT DAB (17) + xA (NA + NB) dz

  30. 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) (13) + dz CT In terms of partial pressures (using pA = CART and P = CTRT): DAB dpA pA (NA + NB) (15) NA = - + RT dz P In terms of molar fraction of A (using xA = CA /CT): dxA NA = -CT DAB xA (17) + (NA + NB) dz

  31. 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)

  32. ⌠ ⌠ ⌡ ⌡ z2 pA2 pA1 z1 Substituting NB = 0 in equation (15), we get DAB dpA pA (NA + 0) NA = - + RT dz P Rearranging and integrating DAB dpA NA (1 - pA/P) = - RT dz DAB dpA NA dz = - RT (1 - pA/P) DAB P P - pA2 NA = ln (18) RT(z2 – z1) P – pA1

  33. 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 (18) is therefore written as follows: DAB P NA = (19) (pA1 - pA2 ) RT(z2 – z1) pB,LM Equation (19) is the most used form.

  34. Using xA = CA /CT, pA = CART and P = CTRT, equation (18) can be converted to the following: NA = ln DAB CT 1 - xA2 (20) (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 (20) becomes the following: DAB CT NA = - (xA1 - xA2 ) (21) (z2 – z1) xB,LM

  35. 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

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