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. – 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 of pure liquid Pi  xi Pi (22.2) i

P Pi : partial pressure of component i i (22.3) yi  P P : total pressure

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) i1 i1 N N c c y Dew point x  i  1 . 0 (22.8)  i  K i1 i1 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 : K x y  i i (22.9)  for liquid mixture i K x  i i

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 11 Nc Nc x x 1  Fi then  B i  (22.12) i1 i1 f Ki 11 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 11

13 0.33 0.37 0.30 x    xFi  Bi 0.62.2311 0.61.0111 0.60.462 11 f Ki 11 = 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 = ∑ K i 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.

adjust theadjustproduct 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) ( K = 1 ) (2) components those have nearly the same . 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.

21 furfural (solvent)  permit the separation of butadiene from mixture containing butane and butenes. lowers the activity of butadiene more than it does for the others.

22 Azeotropic distillation Adding a solvent that forms an azotrope with one of the key component. The azotrope is later separated into solvent and key component.

Entrainers (共沸劑，夾帶劑): the material added form a low boiling azeotrop and is taken overhead. Example : separation of ( Ethanol + water ) + ( benzene, or heptane , or cyclohexane )

azeotrope 95.6% entrainer

 the low boiling azotrope formed: H2O – Et-OH – benzene 23.3 22.8 53.9  feed : 95% Et-OH ( near middle of the dehydrating column )

23 Example: Et-OH/H2O Entrainer: Benzene

24 同學對將來就業應有的準備

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化學工程系 李玉郎