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Introduction to Multicomponent Distillation

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Introduction to Multicomponent Distillation L/O/G/O 單元操作(三) Chapter 22 Introduction to Multicomponent Distillation 化學工程學系 李玉郎 Calculation of equilibrium stage : 1. mass balance : for each component, for each stage , and for the column. 2. enthalpy balance : for the column and for each stage. 3. vapor – liquid equilibria : more complex than for binary system, temperature dependent ( changes from stage to stage) Computer must be used with programs based squarely on the following principles. 2 Phase Equilibrium in Multicomponent Distillation distribution coefficient, or K factor y m x ie i component i ( y = K x ) Ki xie e at equilibrium If Raoult’s law and Dalton’s law hold P ' : vapor pressure of pure liquid Pi xi Pi (22.2) i P Pi : partial pressure of component i y i (22.3) i P : total pressure P yi Pi P xi Pi P Pi K (22.4) Pi i Ki xi xi xi P P K factor strongly temperature dependent 3 yi xi Ki Pi ij (22.5) y j x j K j Pj αij : relative volatility change only moderately with temperature Pi when Raoult’s law applies , ij (22.6) Pj Bubble – point and dew – point calculation Nc Nc Bubble point y i K i x i 1 . 0 (22.7) i1 i1 N N c c y Dew point x i 1 . 0 (22.8) i K i1 i1 i 4 Calculation procedure (1) Bubble point : ( for liquid mixture , given : x1 , x2 , x3 ,…, P ) find Pi' for each Pi choose T Ki yi Ki xi component P ( choose a lower T ) ∑ > 1 T too high No IF ∑ y = 1 ∑ < 1 T too low i ( choose a higher T ) yes T = Tb 5 (2) Dew point : ( for gas mixture, given : y1 , y2 ,…, P ) ( by published data ) choose T estimate K i ( by P i ) P (choose a higher T ) ∑ x > 1 T too low i No IF ∑ xi = 1 ∑ xi < 1 T too high (choose a lower T ) yes T = Td 6 when summation is close to 1.0 , the yi can be determined by : Ki xi yi (22.9) for liquid mixture Ki xi yi Ki similarity, x i for vapor mixture yi K i EXAMPLE 22.1. Find the bubble-point and the dew-point temperatures and the corresponding vapor and liquid compositions for a mixture of 33 mole percent n-hexane, 37 mole percent n-heptane, and 30 mole percent n- octane at 1.2 atm total pressure. 33% C6 37% C7 at P=1.2 atm 30% C8 7 (1) Bubble point choose T = 105oC find Pi' Hexane Pi' = 2.68 atm Heptane Pi' = 1.21 atm Octane Pi' = 0.554 atm calculate Ki P 2.68 K i 2.23 …Hexane i P 1.2 calculate yi = Ki xi y1 = 2.23 0.33 = 0.7359 … Hexane y2 = K2 0.37 = ? ∑ yi = 1 Bubble point. Choose T = 105oC, where the vapor pressure of heptane, the middle component, is 1.2 atm. Since ∑ yi is too large, try a lower temperature. Since the major contribution comes from the Hexane, pick a temperature where Ki is lower by a factor 1/1.24. Choose T = 96oC, where P ' atm. i ( P ' lower by a factor 1 ) 2.68 i 1.24 2.16 1.24 9 o o 96 C ∑ = 0.983, 105 C ∑ = 1.248 By interpolation, the bubble point is 97oC, close enough to 96oC so that the vapor compositions can be calculated using Eq. (22.9). – by normalization Dew point. The dew point is higher than the bubble point, so use 105oC as a first guess. o PC8= 0.554 at 105 C o Select PC8 = 1.17 0.554 ≒ 0.64 T = 110 C 10 Since the sum is too high, choose a higher temperature. Pick T = 110oC, where K3 is 17 percent higher. Find o Select PC8 = 1.17 0.554 ≒ 0.64 T = 110 C By extrapolation, the dew point is 110.5oC, and the composition of the liquid in equilibrium with vapor is obtained by dividing the values of y / K by 1.0162. i i 110oC ∑ = 1.016 o 105 C ∑ = 1.169 11 Flash distillation of multicomponent mixtures xF = f y + (1 – f ) x 1 f x For flash distillation of binary mixture , Eq (21.2) y x F f f can be written as : x 1 f y Fi x (22.10) Di f f Bi Eq (22.10) y 1 x (22.11) Di Fi Ki f 1 xBi x f x Bi Bi x solve xBi Fi xBi f Ki 11 Nc Nc x x 1 Fi then B i (22.12) i1 i1 f Ki 11 f (T) Eq(21.12) can be used to solve the temperature, and composition of liquid ( xi ) and vapor ( yi ) products. The method is the same as that for the dew – point calculation. 12 EXAMPLE 22.2. The mixture of Example 22.1 is subjected to a flash distillation at 1.2 atm pressure, and 60 percent of the feed is vaporized. (a) Find the temperature of the flash and the composition of the liquid and vapor products. (b) To what temperature must the feed liquid be heated for 60 percent vaporization on flashing ? Solution : o o xF1 ( Hexane ) = 0.33 , Tb = 97 C , Td = 110.5 C xF2 ( Heptane ) = 0.37 , f = 0.6 , P = 1.2 atm xF3 ( Octane ) = 0.30 flush temperature lie between Tb and Td , since f = 0.6 o assume T = Tb + 0.6 ( Td – Tb ) = 105 C , then use Eq.(22.12) with f = 0.6 xFi f Ki 11 13 0.33 0.37 0.30 x xFi Bi 0.62.2311 0.61.0111 0.60.462 11 f Ki 11 = 0.190 + 0.368 + 0.443 = 1.001 yDi o since K i , yDi = Ki xBi , flash T = 105 C xBi (b) Determine the temperature of feed before flashing ( liq. state , P > 1.2 atm ) use the enthalpy balance : enthalpy after flashing ( Hvapor + Hliq. )105oC = enthalpy before flashing ( Hliq )T ( T : to be determined ) 14 o o reference Temp. = 105 C , require : heat of vaporization ΔHv at 105 C o heat capacity of liq. C P ( 105 ~ 200 C ) Based on liquid at 105oC , the enthalpies of the product are Hvapor = 0.6 ( 0.424 6370 + 0.372 7510 + 0.204 8560 ) Hvapor = 4345 cal Hliquid = 0 For the feed , Hfeed = C P l ( T – 105 ) o C P = 0.33 62 + 0.37 70 + 0.30 78 = 69.8 cal/mol· C o 69.8 ( T0 – 105 ) = 4345 T0 = 167 C = preheat temp. For a more accurate answer , the liquid heat capacities could be reevaluated for the range 105 to 170oC 15 Fractionation of Multicomponent Mixture Calculations are made by too methods: 1. Assume desired separation, choose reflux ratio , calculate No. of plates above and below feed. ( used commonly for binary mixture ) 2. Assume : No. of plates above and below feed. choose reflux ( L ) and vapor rate from reboiler ( V ) calculate separation of component. 16 Key components choose 2 components whose concentration in the distillate and bottom products are a good index of the separation. C1 C LK L light key more volatile 2 C HK 3 H heavy key less volatile C4 C5 components lighter than the light key nearly completely recovered in the distillate components heavier than the heavy key nearly completely recovered in the bottom. small number are assigned to xB L ( light key in the bottom ) , or xD H ( heavy key in the distillate ) usually , L key and H key are adjacent in the rank order of volatility. such a choice is call a sharp separation. 17 Minimum number of plates ( total reflux , RD = ) The Fenske equation (21.42) n[(xDi xBi ) (xDj xBj )] Nmin 1 (22.13) n ij applied to any 2 components , i and j , in a multicomponent system. 3 ij Dij Fij Bij (22.14) αDij , αFij , αBij : relative volatility at temperature of distillate , feed plate , and bottom respectively. 18 Calculation of required reflux ratio and concentration profiles Lewis – Matheson method plate by plate calculation determine No. of ideal stage for a specified separation, at selected RD xD , xB V y1,y2,y3 …. L Dew point calculation, 1, T1 T1, x1, x2, x3, …,. x1, x2, x3 y1,y2,y3 …. Material balance: y2i V2 = L1 x1i + D xDi y ,y ,y …. 2, T2 1 2 3 x1, x2, x3 Dew point calculation, y1,y2,y3 …. T2, x1, x2, x3, …,. 19 specified : composition of distillate ( or vapor from top plate ) by trial Dew point calculation Temperature and xi on top plate Eq (22.8) yi ∑xi = 1.0 = ∑ Ki assume constant V , L ; if not , use enthalpy balance Material balance equation vapor composition from plate 2 , y2i y2i V2 = L1 x1i + D xDi (22.31) Dew point calculation temperature and xi on top plate 2. adjustthe product composition ( until composition is close to that of feed ) feed plate check compositions at the feed stage similar procedure (bubble point calculation) specified bottom composition Calculation can be repeated for different R at fixed No. of plate and feed plate. D 20 Azeotropic and extractive distillation complete separation may be impossible for : (1) azeotrope ( K = 1 ) (2) components those have nearly the same boiling point. Extractive distillation liquid – liquid extraction with an added vapor phase mixture of azeotrope + an additional component ( high – boiling liquid , or a solvent ) to alter the relative volatility of the original components solvent : miscible with both key components, but is chemically more similar to one of them, and the activity coefficient of the similar component is decreased.
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