1. CHAPTER 1: DISTILLATION Definition& General description of the process Physicalconcept of distillation Vapor-liquidequilibriumrelationship Flash& Batch distillation Continuous Azeotropicdistillationdistillation Multicomponentdistillation

2. Definition & general description of the process Separating the various components of a liquid solution Depends upon the distribution of these components between a vapor phase & a liquid phase Distillation is done by vaporizing a definite fraction of a liquid mixture in a such way that the evolved vapor is in equilibrium with the residual liquid The equilibrium vapor is then separated from the equilibrium residual liquid by condensing 3. Laboratory / Testing 4. Physical Concept of distillationCarried out by either 2 principal methods First method: based on the production of a vapor by boiling the liquid mixture to be separated and condensing the vapors without allowing any liquid to return to the still - NO REFLUX (E.g. Flash, simple distillation) 5. Second method: based on the return part of the condensate to the still under such condition that this returning liquid is brought into intimate contact with the vapors on their way to the condenser conducted as continuous / batch process (E.g. continuous distillation) 6. Vapor liquid equilibrium DEFINITION: EVAPORATION: The phase transformation processes from liquid to gas or vapor phase VOLATILITY: The tendency of liquid to change form to gas VAPOR LIQUID EQUILIBRIUM OF AN ORDINARY BINARY LIQUID MIXTURE PREDICTION OF VAPOR LIQUID EQUILIBRIUM COMPOSITIONS FOR ORDINARY BINARY MIXTURES RELATIVE VOLATILITY OF A MIXTURE 7. VAPOR LIQUID EQUILIBRIUM OF AN ORDINARY BINARY LIQUID MIXTURE A binary liquid mixture consist of 2 different liquids. Can be classified into homogeneous mixtures or non-homogeneous (heterogeneous) mixtures.Low boiler liquid (A) liquid that vaporized easily) low boiling point or high vapor pressure)High boiler liquid (B) liquid which have higher boiling point or low vapor pressureHomogeneous mixtures mix at all proportions resulting in one continuous phaseHeterogeneous mixtures do not mix uniformly resulting in more than one distinct phases 8. VAPOR LIQUID EQUILIBRIUM OF AN ORDINARY BINARY LIQUID MIXTURE Equilibrium curve: shows the relationship between composition of residual liquid and vapor that are in dynamic phase equilibrium. The curve will be very useful in calculations to predict the number of stages required for a specified distillation process. Equilibrium mole fraction of A in vapor is larger than mole fraction of A in liquid phase. This is expected since that A has lower boiling point than B, A would vaporize more than B. 9. Prediction of vapor-liquid equilibrium compositions for ordinary binary mixtures Raoults Law for ideal solution & Daltons Law of partial pressure can be manipulated in order to calculate compostions of liquid and vapor, which are in equilibrium. Raoults Law the partial pressure of a component in the vapor phase is equal to the mole fraction of the component in the liquid multiplied by its pure vapor pressure at the temperature: pA = xA PAo pA = partial pressure of A in a vapor phase xA = mole fraction of A in liquid phase PAo = vapor pressure of A at the temperature 10. Prediction of vapor-liquid equilibrium compositions for ordinary binary mixtures For a mixture of the different gases inside a close container, Daltons law stated that the resultant total pressure of the container is the summation of partial pressures of each of all gases that make up the gas mixture: PT = pA + pB Dalton also state that the partial pressure of gas (pA) is: pA = y A PTpA = partial pressure of A in vapor phase yA = mole fraction of A in vapor phase PT = total pressure of the system 11. Example: Calculate the vapor and liquid compositions in equilibrium at 95oC (368.2K) for benzenetoluene using the vapor pressure. The vapor pressure of benzene = 155.7 kPa and toluene = 63.3 kPa. The boiling point of benzene = 80oC whereas toluene = 110oC. The pressure of the system is 101.32 kPa. 12. Relative volatility () of a mixture Separations of components by distillation process depends on the differences in volatilities of components that make up the solution to be distilled. Indicates the ease or difficulty of using distillation to separate the more volatile components from the less volatile components in a mixture. The greater difference in their volatility, the better is separation by heating (distillation). Conversely if their volatility differ only slightly, the separation by heating becomes difficult. 13. Relative volatility () of a mixture Useful in designing all types of distillation processes as well as other separation or absorption processes which involve the contacting of vapor and liquid phases in a series of equilibrium stages. However, cannot be used in separation or absorption processes that involve components reacting with each other (E.g. Absorption of CO2 in aqueous solutions of sodium hydroxide). 14. Relative volatility () of a mixture 15. Relative volatility () of a mixtureThe greater the distance between the equilibrium line & 45o line, the greater the difference the vapor composition and a liquid composition. Separation is more easily made. A numerical measure of how easy separation relative volatility, AB Relative volatility ratio of the concentration of A in the vapor to the concentration of A in liquid divided by the ratio of the concentration B in the vapor to the concentration of B in the liquid: 16. Relative volatility () of a mixture y A / xA ABy B / xBy A / xA (1 y A ) /(1 x A )AB relative volatility of A with respect to B in the binary system If the system obeys Raoults law for an ideal system: yAPA x A PTyBPB xB PTABPA PByASeparation is possible for > 1.0xA 1 ( 1)x A 17. Relative volatility () of a mixture Separation is possible for > 1.0 For non-ideal solution, the values of change with temperature. For ideal solution, the values of doesnt change with temperature. For solution that approaches ideal solution, its would fairly constant. 18. Exercise 1 A liquid mixture is formed by mixing n-heptane (A) & n-octane (B) in a closed container at constant pressure of 1 atm (101.3kPa). i. Calculate the equilibrium compositions of vapor & liquid ii. Plot a boiling point diagram for the system iii. Plot an equilibrium curve for the system iv. Calculate the ABv. What is the condition of the mixture? Use the following list if vapor pressure for pure n-heptane & noctane at various temperature. 19. Flash & batch distillation Flash (equilibrium) distillation Simple batch distillation 20. Concept of equilibrium Liquid molecules are continually vaporizing, while vapor molecules are continually condensing. If 2 chemical species are present, they will condense & vaporize at the same rates. If not in equilibrium, the liquid & vapor can be at different temperature and pressure & present in different mole fraction. At equilibrium, temperature & pressure & fractions stop to change. Although molecules continue to evaporate & condense, the rate at which each species condenses is equal to the rate yA, yB yA + yB = 1.0 at which it evaporates. AFigure : Vapor-liquid contacting systemPvap, TvapA Pliq, Tliq xA, xBBBxA + xB = 1.0 21. Flash (Equilibrium) Distillation Flash distillation a single stage process because it has only one vaporization stage (means one liquid phase is expected to one vapor phase) The vapor is allowed to come to equilibrium with the liquid The equilibrium vapor is then separated from the equilibrium residual liquid by condensing the vapor Flash distillation can be either by batch or continuous 22. Flash (Equilibrium) Distillation As illustrated in Figure 3, a liquid mixture feed, with initial mole fraction of A at XF, is pre-heated by a heater and its pressure is then reduced by an expansion valve. Because of the large drop in pressure, part of liquid vaporizes. The vapor is taken off overhead, while the liquid drains to the bottom of the drum The system is called flash distillation because the vaporization is extremely rapid after the feed enters the drum. Now, we interested to predict the composition (x and y) of these vapor and liquid that are in equilibrium with each other. 23. Flash (Equilibrium) Distillation 24. Flash (Equilibrium) Distillation 25. Exercise 2 A liquid mixture containing 70 mol% nheptane (A) and 30 mol % n-octane at 30oC is to be continuously flash at the standard atmospheric pressure vaporized 60 mol% of the feed. What will be the compositions of vapor and liquid and the temperature of the T (K) x separator for an equilibrium stage? y 371.6 374 377 380 383 386 389 392 395 398.21 0.825 0.647 0.504 0.387 0.288 0.204 0.132 0.068 01 0.92 0.784 0.669 0.558 0.449 0.342 0.236 0.132 0 26. Solution Basis = 100 moles of liquid feed (F) Given, xF = 0.7 V = 0.6(100) = 60 moles f = V/F = 60/100 = 0.6 We want to fine the equilibrium composition of liquid and liquid; y* & x* The operating line: y* = (0.6-1)x* + 0.7 0.6 0.6 = -0.667x* + 1.167 From the intersection of the operating line & the equilibrium curve as shown in the graph: equilibrium mol fraction of n-heptane in liquid, x* = 0.62 equilibrium mol fraction of n-heptane in vapor, y* = 0.76 the temperature of the separator at equilibrium 378oC 27. Determination of vapor-liquid equilibrium composition for a flash distillation of n-heptane/noctane mixture y, mol fraction of n-heptane in vapor1.210.80.60.40.20 00.20.40.60.81x, mol fraction of n-heptane in liquidFigure: Equilibrium curve and operating line1.2 28. Determination of equilibrium temperature for a flash distillation of n-heptane-n-octane mixture 400395395390390385385380380375375370Temperature (K)4003700x0.20.40.60.8Mol fraction of n-heptane in vapor (y) and liquid (x)1y 29. Single-stage equilibrium contact for VL system A single-stage process is defined as two different phases are brought into intimate contact with each other and then are separated. During the time contact, intimate mixing occurs and the various components diffuse and redistribute themselves between the 2 phases. If the time contact is long enough, they will reach and equilibrium condition. 30. Single-stage equilibrium contact for VL system The two entering phases, Lo and V2, of known amounts and compositions, enter the stage; mixing and equilibration occur; and the two exit streams, L1 and V1, leave in equilibrium with each other. If sensible heat effects are small and the latent heats of both compounds are the same, then when 1 mol of A condenses, 1 mol of B vaporize. Therefore, the total moles of vapor V2 = V1, also 2 = Lo. This case is 1called as L1 V V constant molal overflow. L1Lo 31. Single-stage equilibrium contact for VL system Total mass balance: Lo + V2 = L1 + V1 Component balance; LoXAo + V2yA2 = L1xA1 + V1yA1 Usually, need to find the xA1 and yA1 To solve, need to plot an equilibrium curve and make trial an error in the calculations. 32. Single-stage equilibrium contact for VL system Exersice A vapor at the dew point and 101.32 kPa containing a mole fraction of 0.40 benzene and 0.60 toluene and 100 kg mol total is brought into contact with 110 kg mol of a liquid at the boiling point containing a mole fraction of 0.30 benzene and 0.70 toluene. The two streams are contacted in a single stage, and the outlet streams leave in equilibrium with each other. Assume constant molal overflow. Calculate the amounts and compositions of the exit streams. 33. Simple batch distillation Simple batch distillation which is also known as differential distillation refer to a batch distillation in which only one vaporization stage (or one exposed liquid surface) is involved.Simple batch distillation is done by boiling a liquid mixture in a stream-jacketed-kettle (pot) and the vapor generated is withdrawn and condensed (distillate) as fast as it forms so that the vapor and the liquid do not have sufficient time to reach its equilibriumThe first portion of vapor condensed will be richest in the more volatile component A. As the vaporization proceeds, the vaporized product 34. Simple batch distillation 35. Simple batch distillation Raleigh equation for ideal and non-ideal mixtures Consider a typical differential distillation at an instant time, t1 as shown below: 36. Simple batch distillation 37. Simple batch distillation We know from definitions, yA xASinceABdA dA dB AxBA By A / xA y B / xBB A BdA dA dB dB dA dBAfter simplifying, ABdB dA dBis constant for an ideal mixture, AByBRearranging, dB BB dA A dBABdA AB A B A A B 38. Simple batch distillation Integrating within the limits of t1 and t2, B2B1dB BAB A1dA ASince AB is constant, B2 AB B1dB B AB lnA2A2A1dA AB2 B1ln B B2 AB B1lnA2 A1ln A A2 A1Eq .(5)Equation 5 known as simplified Raleigh equation for simple batch distillation which applicable for ideal solution. 39. Simple batch distillation Example 1 A mixture of 100 mol containing 50 mol% n-pentane and 50 mol% n-heptane is distilled under differential (batch) conditions at 101.3 kPa until 40 mol is distilled. What is the average composition of the total vapor distilled and the composition of the liquid left. The equilibrium data as follows, where x and y are mole fractions of n-pentane: xA yA 1.0001.0000.8670.9840.5940.9250.3980.8360.2540.7010.1450.5210.0590.27100 40. Solution Given L1 = 100 mol V (mol distilled) = 40 mol From material balance: L1L2 VL2L1 VL260 molSubstiting into Eq. (4) lnL1 L2x1x21 yxdx0.50.510 x2100 40ln 1yx100 600.5x21 yxdxdxThe unknown, x2 is the composition n of the liquid L2 at the end of batch distillation. 41. Solution0.511By plotting graphy x versus x, they x dx x2 value is referring to the value of area under the curve. From the graph, area under curve = 0.510 at x2 = 0.277. Composition of the liquid L2, x2 = 0.277. From material balance on more volatile component: x1L1x 2L2y avy avx1L1 x 2L2 Vy av0.5 100 0.277 60 400.835y av VAverage composition of total vapor distilled, yav = 0.835 42. Continuous / retrification distillation Retrification (fractionation) - or stage distillation with reflux can be considered as a process in which a series of flash vaporization stages are arranged in a series in such a manner that the vapor and liquid products from each stage flow counter-current to each other.Continuous distillation - the process is more suitable for mixtures of about the same volatility and the condensed vapor and residual liquid are more pure (since it is re-distilled)The fractionator consists of many trays which have holes to permit the vapor, V which rises up from the lower tray to bubble through and mixes with the liquid, L on the upper tray and equilibrated, and V 43. Continuous / retrification distillation During the mixing, the vapor will pick up more of component A from the liquid while the liquid will richer and richer in component B. As the vapor rises, it becomes richer and richer in component A but poorer with component B.Conversely, as the liquid falls, it becomes poorer with A but richer in B. Thus we obtain a bottom product and an overhead product of higher purity in comparison to those obtained by single-stage simple batch or flash distillation.NOTE: Fractionation refers to a process where a part or whole of distillate is being recycled to the fractionator. The recycled distillation (refulx) will supply the bulk of liquid need to mix with vapor. 44. Continuous / retrification distillation 45. Continuous / retrification distillation The feed stream is introduced on some intermediate tray where the liquid has approximately the same composition as the feed. The system is kept steady-state: quantities (feed input rate, output stream rates, heating and cooling rates, reflux ratio, and temperatures, pressures, and compositions at every point) related to the processdo not change as time passes during operation. With constant molal overflow assumption: Ln 1LnLn1....etc. Vn1VnVn1Conditions for constant molal overflow:....etc. Heat loses negligible (achieved more easily in industrial column) Negligible heat of mixing Equal or close heats of vaporization 46. Continuous / retrification distillationNumber of plates required in a distillation columnFour streams are involved in the transfer of heat and material across a plate, as shown in figure above: Plate n receives liquid Ln+1 from plate n+1 above, and vapor, Vn-1 from plate n-1 below.Plate n supplies liquid Ln to plate n-1, and vapor Vn to plate n+1Action of the plate is to bring about mixing so that the vapor Vn of composition yn reaches equilibrium with the liquid Ln of composition xn. 47. Continuous / retrification distillation Design and operation of a distillation column depends on the feed and desired products A continuous distillation is often a fractional distillation and can be a vacuum distillation or a steam distillation. Calculation for number of plates: Mc-Cabe & Thiele Lewis-Sorel Method 48. Continuous / retrification distillation 49. Mc-Cabe Thiele Method 50. The intersection of operating lines, q Feed enters as liquid at its boiling point that the two operating lines intersect at point having an x-coordinate of xF. The locus point of the intersection of the operating lines is considerable importance since it is dependent on the temperature and physical condition of feed. The condition of the feed (F) determines the relation between the vapor (Vm) in the stripping section and (Vn) in the enriching section, as well as between Lm and Ln. 51. The intersection of operating lines, needed to vaporize 1 mol of feed at entering conditions q heat qmolar latent heat of vaporizati on of feed qHv HvHF HLq also as the no. of moles of saturated liquid produced on the feed plate by each mole of feed added to tower. The relationship between qF Lm Ln flows above & below (1) entrance of feed: Vn Vm (1 q )F (2)Rewrite the equations ofn enriching3& stripping Vn y L x Dx D ( ) without the tray subscripts: Wxw (4) Vm y Lm xSubtractingVn )yfrom m Ln )x (DxD Wxw ) (5) (Vm (3) (L (4) 52. The intersection of operating lines, q Substituting:FxF will produce: y q q 1xDxD WxwxF q 1, Eq. (1) & (2) into (5)(q line equation )The equation locus of the intersection of the two operating lines Setting y = x in the equation, the intersection of the q-line equation with the 45o line is y = x = xF, where xF is the overall composition of the feed. Slope = q/(q-1). A convenient way to locate a stripping line operating line is 1st to plot the enriching operating line and then q-line. 53. The intersection of operating lines, q Dependingon the state of the feed, the feed lines will have different slopes: q = 0 (saturated vapour) q = 1 (saturated liquid) 0 < q < 1 (mix of liquid and vapour) q > 1 (subcooled liquid) q < 0 (superheated vapour) 54. Animation of the construction of enriching, stripping & q operating lines http://www.separationprocesses.com/Distillation/DT_ Animation/McCabeThiele.html 55. Example A mixture of benzene and toluene containing 40 mole% benzene is to be separated to give a product of 90 mole% benzene at the top, and a bottom product with not more than 10 mole% of benzene. The feed is heated so that it enters the column at its boiling point, and the vapor leaving the column is condensed but not cooled, and provides reflux and product. It is proposed to operate the unit with a reflux ratio of 3 kmol/kmol product. It is required to find the number of theoretical stages needed and the position of entry for the feed. 56. Example 57. Solution Feed, xF = 0.4 Product, xD = 0.9 Bottom, xw = 0.10 Taking basis; 100 kmol of feed. A total mass balance: hence; W = 100 D (Eq. 1)F=D+WA balance on MVC (benzene); F xF 40D xDW xw0.9D 0.1 W100(0.4)D(0.9) W (0.1)(Eq.2)From the calculations; D = 37.5 kmol, W = 62.5 kmol 58. Solution Using notation from reflux: R LnLn Ln D 112 .5Ln3DLn3(37.5)From material balance at the top stage; VnRD1LnDVn150kmol1Thus, the operating line equation: yn yn1Ln xn Vn 1D xD Vn 110.75 x n0.225yn1112 .5 xn 150D xD Vn 1 59. Solution Since the feed is all liquid at its boiling point, it will all run down as increased reflux to the plate below: LmFLm112.5 100Lm212 .5kmolThe material balance at the bottom:LmLnVm1WVm212 .5 62.5Vm150 kmolBottom operating line equation: ym ym1Lm xm Vm 1W xw Vm 111.417 x m0.0417ym1212 .5 xm 15062.5 (0.1) 150Vn 60. Example 11,200 kg/h of equal parts (in wt) of BenzeneToluene solution is to be distilled in a fractionating tower at atmospheric pressure. The liquid is fed as a liquid-vapor mixture in which the feed consist of 75% vapor. The distillate contains 94 wt% Benzene whereas the bottom products contains 98 wt% toluene. Determine; The flowrate of distillate and bottom product (kg/h) The minimum reflux ratio, Rm. The number of theoretical stages required if the reflux ratio used is 1.5 times the minimum reflux ratio The position of the feed tray The MW of Benzene = 78 The MW of Toluene = 92 61. Solution xF = 0.5 xD = 0.94Xw = 0.02 From the total & component material balance: D = 5739.1 kg/h, W = 5260.9 kg/h Convert mass fraction to mol fraction. (Basis of calculation = 100kg) Mol fraction: xF = 0.54, xD = 0.95, xw = 0.03 62. Solution Find q-line. Feed enters at 75% vapor. q0.75(qvapor ) 0.25(q liquid )q0.75(0) 0.25(1.0)q0.25q line equation yq yq Let xq q 1xf q 1xq0.25 xq 0.25 1 0.3, y q0.54 0.25 1yq0.333 (0.3) 0.72Plot (0.54, 0.54 ) and (0.3, 0.62 ) forq0.333 x qyq0.62line0.72 63. Solution From the graph, y intercept for q-line = x 0.95 0.36 0.36 0.36 R 1.64 DRm1Rm1mThe number of theoretical stages required if the reflux ratio used is 2 times R 2Rmin the minimum2(1.64 ) Rratio reflux 3.28 yn1yn1yn1At xR 1 xn xD R 1 R 1 3.28 1 xn (0.95 ) 3.28 1 3.28 1 0.766 x n 0.222 0.5,yn10.766 (0.5) 0.222yn10.605Plot (0.95, 0.95 ) and (0.5, 0.605 ) for enriching OL 64. Solution The number of theoretical stages required = 10.5 stages including boiler Feed plate location: 5 from top. 65. AZEOTROPIC DISTILLATION Azeotrope mixtures Minimum boiling point Maximum boiling point Azeotropic Distillation 66. Azeotrope mixtures Liquid and vapor are of exactly the same at a certain temperature It is a special class of liquid mixture that boils at a constant temperature at a certain composition Cannot be separated by a simple/conventional distillation 67. Azeotropic Distillation An introduction of a new component called entrainer is added to the original mixture to form an azeotrope with one or more of feed component The azeotrope is then removed as either the distillate or bottoms The purpose of the introduction of entrainer is to break an azeotrope from being formed by the original feed mixture Function of entrainer: To separate one component of a closely boiling point To separate one component of an azeotrope 68. Azeotropic Distillation Azeotropic distillation is a widely practiced process for the dehydration of a wide range of materials including acetic acid, chloroform, ethanol, and many higher alcohols. The technique involves separating close boiling components by adding a third component, called an entrainer, to form a minimum boiling. Normally ternary azeotrope which carries the water overhead and leaves dry product in the bottom. The overhead is condensed to two liquid phases; the organic, "entrainer rich" phase being refluxed while the aqueous phase is 69. Azeotropic Distillation A common example of distillation with an azeotrope is the distillation of ethanol and water. Using normal distillation techniques, ethanol can only be purified to approximately 89.4% Further conventional distillation is ineffective. Other separation methods may be used are azeotropic distillation or solvent extraction 70. Azeotropic Distillation The concentration in the vapor phase is the same as the concentration in the liquid phase (y=x) At this point, the mixture boils at constant temperature and doesnt change in composition This is called as minimum boiling point (positive deviation) 71. Azeotropic Distillation The characteristic of such mixture is boiling point curve goes through maximum phase diagram Example: Acetone-chloroform 72. Azeotropic Distillation The most common examples: Ethanol-water (89.4 mole%, 78.25 oC, 1 atm) Carbon Disulfide-acetone (61 mol% CS2, 39.25oC, 1 atm) Benzene-water (29.6 mol% water, 69.25 oC, 1 atm) 73. Azeotropic Distillation Let say binary mixture: A-B formed an azeotrope mixture Entrainer C is added to form a new azeotrope with the original components, often in the LVC, say A The new azeotrope (A-C) is separated from the other original component B This new azeotrope is then separated into entrainer C and original component A. Hence the seapration of A and B can be achieved 74. Azeotropic Distillation Example: Acetic acid-water using entrainer n-butyl acetate Boiling point of acetic acid is 118.1 oC, water is 100 oC & n-butyl acetate is 125 oC The addition of the entrainer results in the formation of a minimum boiling point azeotrope with water with a boiling point = 90.2 oC. The azeotropic mixture therefore be distilled over as a vapor product & acetic acid as a bottom product The distillate is condensed and collected in a decanter where it forms 2 insoluble layers 75. Azeotropic Distillation Example: Acetic acid-water using entrainer n-butyl acetate Top layer consist of nearly pure n-butyl acetate in water, whereas bottom layer of nearly pure water saturated with butyl acetate The liquid from top layer is returned to column as reflux and entrainer The liquid from bottom layer is sent to another column to recover the entrainer (by stream stripping)