1

Multicomponent Flash Distillation

If there are more than two components, an analytical procedure is needed. The equation used are equilibrium, mass and energy balances, and stoichiometric relations. The mass and energy balances are very similar to those used in the binary case, but the equilibrium equations are usually written of K values. iii xKy where in general idrumdrumii xallpTKK ,,

These two equations are written once for each component.

2

Fortunately, for many systems the K values are approximately independent of composition. thus,

𝐾=𝐾 (𝑇 ,𝑝 )

For light hydrocarbons, the approximate K values can be determined from the monographs prepared by DePriester

3The DePriester Chart

4

5

McWilliams, M.L., “An Equation to Relate K-factors to Pressure and Temperature,” Chem. Eng., 80(25), 138, 1973.

The DePriester charts have been fit to the following equation (McWilliams, 1973):

Note that T is in oR and p is in psia. The constants are given in the following table. This equation is valid from –70o C (365.7o R) to 200o C (851.7o R) for pressure from 101.3 kPa (14.69 psia) to 6,000 kPa (870.1 psia)

p

a

p

apaa

T

a

T

aK pp

PTTT 3

2

213

221 lnln

See Table 2-3 p. 33 for constants

6

C is the number of components.

0.11

C

iiy 0.1

1

C

iix

The K values are used along with the stoichiometric equations which the mole fraction in liquid and vapor phases must equal to 1.0

7

For ideal system Raoult’s law holds

𝑝𝐴=𝑥𝐴 (𝑉𝑃 )𝐴

By Dalton’s law of partial pressure

𝑦 𝐴=𝑝𝐴/𝑝

Combining these equations

𝑦 𝐴= (𝑉𝑃 )𝐴 𝑥𝐴 /𝑝

iii xKy Comparing with

𝐾 𝐴=(𝑉𝑃 )𝐴 /𝑝

8

The mass balances,

and the energy balance,

iii VyLxFz

VLF

LVflashF LhVHQFh

Multicomponent Flash Distillation

9

Usually the feed, F , and the feed mole fractions z1 for C-1 of the components will be specified. If pdrum and Tdrum or one liquid or vapor composition are also specified, then a sequential procedure can be used.Suppose we have 10 components (C = 10). Then we must find 10 K’s, 10 x’s, 10 y’s one L and one V or 32 variables. To do this we must solve 32 equations [10 from 10 from 2 from , and 10 independence mass balances]

iii xKy idrumdrumii xallpTKK ,, 0.1

1

C

iiy

0.11

C

iix

10

How does one solve 32 simultaneous equations? However we will restrict ourselves to ideal solution where

drumdrumii pTKK ,

Since and are known, the 10 Ki can be determined easily (say, from the DePriester charts or using the McWilliams relation). Now there are only 22 equations to solve simultaneously.

drumT drump

11

To simplify the solution procedure, we first use equilibrium,

to remove ii xKy iy

iiii xVKLxFz

Substituting for L

i

ii VKVF

Fzx

Then upon rearranging we have

FV

K

zx

i

ii

11 Ci ,1 (2.38

)

i

ii VKL

Fzx

Solving for xi, we have

12

From iii xKy

and

0.1

1111

C

ii

iiC

ii

FV

K

zKy **

Then *

0.11111

C

ii

iC

ii

FV

K

zx

We obtain

FV

K

zKy

i

iii

11 (2.39

)

13

If C is greater than 3, a trial-and-error procedure or root finding technique must be used to find . Although the equations * and ** are both valid, they do not have the good convergence properties. That is, if the wrong is chosen, the that is chosen next may not be better.

FV

FVFV

Subtracting * from **

0

1111 11

C

ii

iC

ii

ii

FV

K

z

FV

K

zK

Subtracting the sum term by term, we have

011

1

1

C

ii

ii

FV

K

zK

F

Vf

14

011

1

1

C

ii

ii

FV

K

zK

F

Vf

This equation, which is known as the Rachford-Rice equation, has excellence convergence properties.

FV

Since the feed composition, are specified and can be calculated when and are given, the only variable is the fraction vaporize . This equation can be solved by many different convergence procedures. For instance, the Secant method can be used by selecting two values of and calculating the values of the summation (it will be zero only at the correct value of ).

iziK

drumTdrump

FV

FV

15

The Newton convergence procedure will

converge faster. Since is a function of

that should have a zero value, the equation

for the Newton convergence is

F

Vf FV

FVFVd

dfff k

kk 1

where is the value of the function for trial

k and is the value of the

derivative of the function for trial k. We desire

to have equal to zero, so we set

and solve for

kf

FVddfk

1kf

01 kf FV

16

FVddf

fFVFVFV

k

kkk 1

This equation gives us the best next guess for the fraction vaporized. To use it, however, we need equations for both the function and derivative. For fk, use the Rachford-Rice equation. Then the derivative is

C

i i

iik

FVK

zK

FVd

df

12

2

11

1(2-45)

(2-44)

Substituting the Rachford-Rice equation and (2-45) into (2-44)

C

i i

ii

C

i i

ii

kk

FVK

zK

FVK

zK

F

V

F

V

12

2

1

1

11

1

11

1

(2-46)

17

Once is calculated the value of the Rachford-

Rice equation can be determined. If it is close enough

to zero, the calculation is finished, otherwise, repeat

the Newton convergence for the next trial.

1kFV

Once has been found, and are calculated

from (2-38) and (2-39). L and V are determined

from the overall mass balances.

FV ix iy

The enthalpy and can now be calculated.For ideal solutions the enthalpies can be determined from the sum of the pure component enthalpies multiplied by the corresponding mole fractions:

Lh VH

1

,~

idrumdrumViV pTHyH

i(2-47a)

where and are enthalpies of the pure component. iVH

~iLh

~

(2-47b)

1

,~

idrumdrumiLiL pThxh

18

Example 2-2 Multicomponent Flash Distillation

A flash chamber operating at 50 oC and 200 kPa is separating 1,000 kg moles/hr of a feed that is 30 mole % propane, 10 mole % n-butane, 15 mole % n-pentane and 45 mole % n-hexane. Find the product compositions and flow rates.

F = 1,000 kg mole/hrz1 = .30z2 = .10z3 = .15z4 = .45

CT odrum 50

kPapdrum 200

?,, iyV

?,, ixL

19

Since and are given, a sequential solution can be used. We can use the Rachford-Rice equation to solve for and then find

drumT drumpFV

VLyx ii ,,,

Find from the DePriester charts (or use McWilliams equation)

iK

From the DePriester charts at 50 oC and 200 kPa we find

0.71 K

4.22 K

8.03 K

30.04 K

C3

n-C4

n-C5

n-C6

20

8785.03387.00306.01228.0125.1

Since f(0.1) is positive, a higher value for is required. Note that only term in the denominator of each term changes. Thus we can set up the equation so that only will change. Then equals

FV

FVFV

Calculate from the Rachford-Rice equation FVf

4

1 11

1

ii

ii

FV

K

zK

F

Vf

FV

FV

FV

FV

f13.01

45.013.0

18.01

15.018.0

14.21

1.014.2

10.71

3.010.71.0

Pick = 0.1 as first guess.

1.013.01

45.013.0

1.018.01

15.018.0

1.014.21

1.014.2

1.010.71

3.010.71.0

f

FV

21

The first derivative from (2-45)

2

2

22

22

1

12

1

1 11

1

11

1

FVK

zK

FVK

zK

FVd

df

24

42

42

3

32

3

11

1

11

1

FVK

zK

FVK

zK

22 4.11

196.0

0.61

8.10

FVFV

22 7.01

2205.0

2.01

006.0

FVFV

With = 0.1 this derivative is -4.631. From (2-46) the next guess for is

FVFV

29.0631.4

8785.01.0

2

F

V

22

Calculating the value of the Rachford-Rice equation, we have . This is still positive and V/F is still too low.

329.029.0 f

2nd Trial

891.12

FVd

df

46.0891.1

329.029.0

3

F

V

This is closer, but V/F is still too low. 066.046.0 f

3rd Trial

32.13

FVd

df51.0

32.1

066.046.0

3

F

V 00173.051.0 f

Thus 51.0F

V

23

Now we calculate from (2-38) and from ix iy iii xKy

0739.051.010.71

30.0

11 1

11

FV

K

zx

By similar calculations,

1400.0,0583.0 22 yx

1336.0,1670.0 33 yx

2099.0,6998.0 44 yx

Since F = 1,000 and = 0.51 FV

and kmol/h

51051.0 FV 4905101000 VFL

5172.00739.00.7111 xKy

24

Check and ix iy

999.04

1

i

ix 0007.14

1

i

iy

These are close enough.

25

Simultaneous Multicomponent Convergence

If the feed rate F, the feed composition consisting of (C-1)zi values, the flash drum pressure pdrum , and the feed temperature TF are specified, the hot liquid will vaporized when its pressure is dropped. This “flashing” cools the liquid to provide energy to vaporize some of the liquid. The result Tdrum is unknown; thus we must use a simultaneous solution procedure.

1

,i

c

F i F F Fi

h z h T p

For wide-boiling point feed, Tdrum cannot have much effect on V/F

(2-50)

26

Estimate Tdrum

Finished

Yes

No

Calculate ,i drum drumK T p

Estimate V/F

Solve Rachford-Rice eq.

( / ) 0f V F

Calculate xi, yi, L, V, H and h

Energy balance satisfied

Yes

NoEstimate Tdrum

Tbp < Tdrum < Tdp

27

Arranging the energy balance into the functional form

LVflashF LhVHQFh

( ) 0k drum V L F flashE T VH Lh Fh Q (2-51)

The Newtonian procedure estimates EK+1(Tdrum) from the derivative,

(2-52)𝐸𝑘+1−𝐸𝑘=𝑑𝐸𝑘

𝑑𝑇 𝑑𝑟𝑢𝑚( ∆𝑇 𝑑𝑟𝑢𝑚 )

- (2-53)

,

k v LPV PL

drum k drum drum

dE dH dhV L VC LC

dT dT dT (2-54)

28

Set Ek+1 =0

∆ 𝑇𝑑𝑟𝑢𝑚=−𝐸𝑘 (𝑇 𝑑𝑟𝑢𝑚 ,𝑘 )

𝑑𝐸𝑘

𝑑𝑇 𝑑𝑟𝑢𝑚 .𝑘

(2-55)

,, 1 ,

,

( )k drum kdrum k drum k

drum k

E TT T

dE

dT

Convergence when

= 0.2 or 0.01oC

(2-56a)

|∆𝑇 𝑑𝑟𝑢𝑚|<𝜀

29

Example We have a liquid feed that is 20 mole % methane, 45 mole % n-pentane and 35 mole % n-hexane. Feed rate is 1500 kmol/h and feed temperature is 45oC and pressure is 100 psia. The flash drum operates at 30 psia and is adiabatic. Find Tdrum, V/F, xi, yi, L, V. Give for Tdrum = 0.02 and for V/F = 0.005

Component (kcal/gmol) Tbp, normal, oC CPL, cal/gmoloCMethane 1.955 -161.48 11.0 n-pentane 6.160 36.08 39.66 n-hexane 6.896 68.75 45.58

Vapor heat capacities in cal/gmoloC; T in oC:7 2 9 3

, 8.20 0.01307 8.75*10 2.63*10PV MC T T T 5 2 9 3

, 27.45 0.08148 4.538*10 10.1*10PV PC T T T 5 2 9 3

, 32.85 0.09763 5.716*10 13.78*10PV HC T T T

30

Answer: Tdrum = 23.892, V/F = 0.2549

F = 1,500 kg mole/hrzM = .20zP = .45zH = .35TF = 45oCPF=100 psia

?odrumT C

30drump psia

?,, iyV

?,, ixL

31

IterationNo.

1 15.000 16 0.2361 27.903

2 27.903 13 0.2677 21.385

3 21.385 14 0.2496 25.149

4 25.149 7 0.2567 23.277

5 23.277 3 0.2551 24.128

6 24.128 2 0.2551 23.786

7 23.786 2 0.2550 23.930

8 23.930 2 0.2550 23.875

9 23.875 2 0.2549 23.900

10 23.900 2 0.2549 23.892

drumTFVFV

drumTInitial

Trials tofind

Calc