Design Of A Continuous Distillation Column For A Multi Component Mixture 4n1d2t

REPORT TO DEPARTMENT OF CHEMICAL ENGINEERING EGE UNIVERSITY FOR COURSE: CHE386 CONCEPTUAL DESIGN II DESIGN REPORT IV

DESIGN OF A CONTINUOUS DISTILLATION COLUMN FOR A MULTICOMPONENT MIXTURE

SUBMITTED TO Y.Doc.Dr. Serap CESUR

SUBMISSION DATE 25/05/2010

GROUP 3 05070008901 05070008103 05070008849 05060008091 05060008017

Ürün ARDA Berna KAYA Demet ACARGİL M. Serkan ACARSER Tayfun EVCİL

MAY 2010 Bornova-İZMİR

SUMMARY Separation of multicomponent mixtures is very hard and expensive method in industry. For high purity products and high percentage recovery, continuous distillation columns are used. In this study, a saturated vapor mixture which has 5 different components is tried to be separated and with the help of assumptions, simplifications the plate type distillation column design methods is tried to be explored. In this report you can find the information about shortcut design in ideal and real calculations, using the SRK model. The necessary assumptions were done for both real and ideal calculations, and with the help of Fenske, Underwood and Kirkbridge equations, the number of plates were found.

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TABLE OF CONTENTS

Summary ...................................................................................................................... i 1.0 Introduction .......................................................................................................... 1 2.0 Results..................................................................................................................... 3 2.1 Ideal System ..................................................................................................... 3 2.2 Real System (Non-Ideal) .................................................................................. 6 3.0 Discussion and Conclusion .................................................................................. 14 4.0 Nomenclature ...................................................................................................... 18 5.0 References ........................................................................................................... 19 6.0 Appendix ............................................................................................................. 20 6.1 Ideal System ................................................................................................... 20 6.2 Real System (Non-Ideal) ................................................................................ 27

1.0 INTRODUCTION Distillation is probably the most widely used separation process in the chemical and allied industries; its applications ranging from the rectification of alcohol, which has been practiced since antiquity, to the fractionation of crude oil. A good understanding of methods used for correlating vapour-liquid equilibrium data is essential to the understanding of distillation and other equilibrium-staged processes. Distillation column design The design of a distillation column can be divided into the following steps: 1. Specify the degree of separation required: set product specifications. 2. Select the operating conditions: batch or continuous; operating pressure. 3. Select the type of ing device: plates or packing. 4. Determine the stage and reflux requirements: the number of equilibrium stages. 5. Size the column: diameter, number of real stages. 6. Design the column internals: plates, distributors, packing s. 7. Mechanical design: vessel and internal fittings. The principal step will be to determine the stage and reflux requirements. This is a relatively simple procedure when the feed is a binary mixture, but a complex and difficult task when the feed contains more than two components (multicomponent systems).

Figure 1. Distillation column (a) Basic column (b) Multiple feeds and side streams

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Process Description The separation of liquid mixtures by distillation depends on differences in volatility between the components. The greater the relative volatilities, the easier the separation. The basic equipment required for continuous distillation is shown in Figure1. Vapor flows up the column and liquid counter-currently down the column. The vapor and liquid are brought into on plates, or packing. Part of the condensate from the condenser is returned to the top of the column to provide liquid flow above the feed point (reflux), and part of the liquid from the base of the column is vaporized in the reboiler and returned to provide the vapor flow.

In the section below the feed, the more volatile components are stripped from the liquid and this is known as the stripping section. Above the feed, the concentration of the more volatile components is increased and this is called the enrichment, or more commonly, the rectifying section. Figure1a shows a column producing two product streams, referred to as tops and bottoms, from a single feed. Columns are occasionally used with more than one feed, and with side streams withdrawn at points up the column, Figure1b. This does not alter the basic operation, but complicates the analysis of the process, to some extent. If the process requirement is to strip a volatile component from a relatively nonvolatile solvent, the rectifying section may be omitted, and the column would then be called a stripping column. In some operations, where the top product is required as a vapor, only sufficient liquid is condensed to provide the reflux flow to the column, and the condenser is referred to as a partial condenser. When the liquid is totally condensed, the liquid returned to the column will have the same composition as the top product. In a partial condenser the reflux will be in equilibrium with the vapor leaving the condenser. Virtually pure top and bottom products can be obtained in a single column from a binary feed, but where the feed contains more than two components; only a single “pure” product can be produced, either from the top or bottom of the column. Several columns will be needed to separate a multicomponent feed into its constituent parts. The problem of determining the stage and reflux requirements for multicomponent distillations is much more complex than for binary mixtures. With a multicomponent mixture, fixing one component composition does not uniquely determine the other component compositions and the stage temperature. Also when the feed contains more than two components it is not possible to specify the complete composition of the top and bottom products independently. The separation between the top and bottom products is specified by setting limits on two “key” components, between which it is desired to make the separation.

SHORT-CUT METHODS FOR STAGE AND REFLUX REQUIREMENTS Most of the short-cut methods were developed for the design of separation columns for hydrocarbon systems in the petroleum and petrochemical systems industries, and caution must be exercised when applying them to other systems. They usually depend on the assumption of constant relative volatility, and should not be used for severely non-ideal systems. In this project, during the shortcut calculations, Fenske, Underwood, Gilliand and Kirkbridge Equations were used. Trial and error procedures were all made in excel.

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2.0 RESULTS 2.1 Ideal System Table 1. Properties of Components Antoine Constants A B C

M w,i ρi [kg/kmol] [kg/m3]

Zi 1 [%]

λi [kJ/mol]

15

35.14

1592.864 226.184

25

38.58

1279.01

177.849

25

41.35

809.8

7.36366 1305.198 173.427

15

43.24

814.4

7.18246 1287.625

20

44.83

No

Name

1

Methanol

32.04

791.8

7.87863

1473.11

2

Ethanol

46.07

789

8.1122

88.15

812

7.27679

74.122 88.15

3 4 5

Neopentanol [Light Key] n-Butanol [Heavy Key] 1-Pentanol

230

161.33

Table 2. Calculated Values for the Feed Stream in Ideal System F [kmol/h]

100

T dew [oC]

113.66

y1

0.15

n1

15

K1

5.14

y2

0.25

n2

25

K2

3.50

y3

0.25

n3

25

K3

1.02

y4

0.15

n4

15

K 4 [ref.]

0.86

y5

0.2

n5

20

K5

0.42

Table 3. Calculated Values for the Top Product in Ideal System D [kmol/h]

63.6

T dew [oC]

96.84

P.R 1

0.98

n1

14.7

y1

0.231

K1

3.10

P.R 2

0.96

n2

24

y2

0.377

K2

2.00

P.R 3

0.95

n3

23.75

y3

0.373

K3

0.55

P.R 4

0.05

n4

0.75

y4

0.012

K 4 [ref.]

0.45

P.R 5

0.02

n5

0.4

y5

0.006

K5

0.21

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Table 4. Calculated Values for the Bottom Product in Ideal System W [kmol/h]

T bubble [oC]

36.4

122.39

y i =K i x i

P.R 1

0.02

n1

0.3

x1

0.008

K1

6.57

y1

0.054

P.R 2

0.04

n2

1

x2

0.027

K2

4.59

y2

0.126

P.R 3

0.05

n3

1.25

x3

0.034

K3

1.37

y3

0.047

P.R 4

0.95

n4

14.25

x4

0.391

K 4 [ref.]

1.18

y4

0.461

P.R 5

0.98

n5

19.6

x5

0.538

K5

0.58

y5

0.312

Table 5. Calculated Values of Vapor Pressures of Components in Ideal System Psat i [mmHg]

Feed

Top

Bottom

Psat Methanol [mmHg]

3909.69

2352.49

4991.92

Psat Ethanol [mmHg]

2661.98

1517.50

3487.52

Psat Neopentanol [mmHg]

774.99

417.46

1039.42

Psat n-Butanol [mmHg]

656.70

342.30

894.23

Psat 1-Pentanol [mmHg]

316.28

156.67

440.62

Table 6. Calculated Values of Relative Volatilities of Components in Ideal System Relative Volatility ,α ri No Name Feed Stream Top Product Bottom Product

Average

1

Methanol

5.95

6.87

5.58

6.11

2

Ethanol

4.05

4.43

3.90

4.12

1.18

1.22

1.16

1.19

1.00

1.00

1.00

1.00

0.48

0.46

0.49

0.48

3 4 5

Neopentanol [Light Key] n-Butanol [Heavy Key] 1-Pentanol

Table 7. Calculation of Minimum Number of Plates, θ, Minimum Reflux Ratio and Feed Location in Ideal System N min

34.33

θ

1.075

N R /N S

1.25

q

0

R D,min

3.577

λ ave, top [kJ/mol]

38.914

λ ave,bottom [kJ/mol]

43.836 -4-

Table 8. Calculation of Actual Reflux Ratio, Actual Number of Plates, Feed Location, Height of Column, Condenser Load (Q c ) and Reboiler Load (Q r ) in Ideal System R D /R D,min

1.5

1.8

2.4

R D,act

5.366

6.439

8.585

X

0.281

0.385

0.522

Y

0.376

0.313

0.238

N act

56

51

46

NS

25

23

20

NR

31

28

26

L [kmol/h]

341.25

409.50

546.00

G [kmol/h]

404.85

473.10

609.60

Q c [kW]

4376.1

5113.9

6589.3

L [kmol/h]

341.25

409.50

546.00

G [kmol/h]

304.85

373.10

509.60

Q r [kW] H c [m]

3712.1 28.5

4543.1 26.0

6205.2 23.5

Table 9. Required Calculations for the Fluid Velocity and Diameter of Column in Ideal System MW avg [kg/kmol]

73.34

Wdot [kg/s]

0.742

ρ vap [kg/m3]

2.261

ρ liq [kg/m3]

811.63

It [m]

0.5

u v [m/s]

0.856

D c [m]

0.698

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2.2 Real System (Non-Ideal) Table 10. Properties of Components No

Name

T c [K]

P c [bar]

ω

ψ

ε

1

Methanol

512.58

80.96

0.566

0.42748

0

2

Ethanol

516.25

63.84

0.637

Zc

σ

3

Neopentanol [Light Key]

550.00

38.80

0.604

0.333333

1

4

n-Butanol [Heavy Key]

562.93

44.13

0.595

Ω

R [cm3bar/molK]

5

1-Pentanol

586.15

38.80

0.594

0.08664

83.14

Table 11. Calculated Values of y i , n i and T r, i for the Feed Stream in Non-Ideal System F [kmol/h]

100

T dew [oC]

106.6

y1

0.15

n1

15

T r,1

0.741

y2

0.25

n2

25

T r,2

0.736

y3

0.25

n3

25

T r,3

0.690

y4

0.15

n4

15

T r,4

0.675

y5

0.2

n5

20

T r,5

0.648

Table 12. Calculated Values of α SRK,i , a i and b i for the Feed Stream in Non-Ideal System α SRK,1

1.400

a1

1.34*107

b1

45.61

α SRK,2

1.442

a2

1.78*107

b2

58.25

α SRK,3

1.515

a3

3.49*107

b3

102.11

α SRK,4

1.542

a4

3.27*107

b4

91.89

α SRK,5

1.598

a5

4.18*107

b5

108.82

Table 13. Calculated Values of β i , q i and Zv i for the Feed Stream in Non-Ideal System β1

0.00146

q1

9.321

Zv 1

0.9870

β2

0.00187

q2

9.672

Zv 2

0.9826

β3

0.00328

q3

10.829

Zv 3

0.9649

β4

0.00295

q4

11.281

Zv 4

0.9670

β5

0.00349

q5

12.166

Zv 5

0.9572

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Table 14. Calculated Values of I v i , Φv i and Zl i for the Feed Stream in Non-Ideal System I v1

0.00148

Φv 1

0.988

Zl 1

0.00157

I v2

0.00190

Φv 2

0.984

Zl 2

0.00215

I v3

0.00339

Φv 3

0.968

Zl 3

0.00395

I v4

0.00304

Φv 4

0.970

Zl 4

0.00345

I v5

0.00364

Φv 5

0.961

Zl 5

0.00370

Table 15. Calculated Values of I l i , Φl i and K i for the Feed Stream in Non-Ideal System Il 1

0.660

Φl 1

7.605

K1

7.70

Il 2

0.625

Φl 2

3.078

K2

3.13

Il 3

0.604

Φl 3

0.791

K3

0.82

l

l

I4

0.618

Φ4

0.693

K4

0.71

Il 5

0.665

Φl 5

0.545

K5

0.57

Table 16. Calculated Values of α r,i , x i and f v i for the Feed Stream in Non-Ideal System α r,1

10.78

x1

0.019

f v1

0.1501

α r,2

4.38

x2

0.080

f v2

0.2491

α r,3

1.14

x3

0.306

f v3

0.2451

α r,4

1.00

x4

0.210

f v4

0.1474

α r,5

0.79

x5

0.352

f v5

0.1947

Table 17. Calculated Values of f l i , Psat i , γ i and for the Feed Stream in Non-Ideal System f l1

0.1501

Psat 1 [bar]

4.237

γ1

1.818

f l2

0.2491

Psat 2 [bar]

2.822

γ2

1.105

f l3

0.2451

Psat 3 [bar]

0.804

γ3

0.997

f l4

0.1474

Psat 4 [bar]

0.672

γ4

1.044

f l5

0.1947

Psat 5 [bar]

0.317

γ5

1.741

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Table 18. Calculated Values of y i , n i and T r, i for the Top Product in Non-Ideal System D [kmol/h]

T dew [oC]

63.6

96.98

P.R 1

0.98

n1

14.7

y1

0.231

T r,1

0.722

P.R 2

0.96

n2

24

y2

0.377

T r,2

0.717

P.R 3

0.95

n3

23.75

y3

0.373

T r,3

0.673

P.R 4

0.05

n4

0.75

y4

0.012

T r,4

0.658

P.R 5

0.02

n5

0.4

y5

0.006

T r,5

0.631

Table 19. Calculated Values of α SRK,i , a i and b i for the Top Product in Non-Ideal System α SRK,1

1.434

a1

1.38*107

b1

45.61

α SRK,2

1.479

a2

1.82*107

b2

58.25

α SRK,3

1.551

a3

3.57*107

b3

102.11

α SRK,4

1.578

a4

3.35*107

b4

91.89

α SRK,5

1.633

a5

4.27*107

b5

108.82

Table 20. Calculated Values of β i , q i and Zv i for the Top Product in Non-Ideal System β1

0.00150

q1

9.798

Zv 1

0.9880

β2

0.00192

q2

10.181

Zv 2

0.9828

β3

0.00336

q3

11.373

Zv 3

0.9659

β4

0.00302

q4

11.840

Zv 4

0.9673

β5

0.00358

q5

12.758

Zv 5

0.9575

Table 21. Calculated Values of I v i , Φv i and Zl i for the Top Product in Non-Ideal System I v1

0.00152

Φv 1

0.987

Zl 1

0.00187

I v2

0.00195

Φv 2

0.982

Zl 2

0.00255

I v3

0.00347

Φv 3

0.965

Zl 3

0.00426

I v4

0.00312

Φv 4

0.967

Zl 4

0.00385

I v5

0.00373

Φv 5

0.958

Zl 5

0.00430

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Table 22. Calculated Values of I l i , Φl i and K i for the Top Product in Non-Ideal System I l1

0.590

Φl 1

3.108

K1

3.15

I l2

0.560

Φl 2

1.935

K2

1.97

I l3

0.582

Φl 3

0.551

K3

0.57

I l4

0.580

Φl 4

0.467

K4

0.48

I l5

0.606

Φl 5

0.226

K5

0.24

Table 23. Calculated Values of α r,i , x i and f v i for the Top Product in Non-Ideal System α r,1

6.52

x1

0.073

f v1

0.2310

α r,2

4.08

x2

0.192

f v2

0.3755

α r,3

1.18

x3

0.655

f v3

0.3651

v

α r,4

1.00

x4

0.024

f

4

0.0116

α r,5

0.49

x5

0.027

f v5

0.0061

Table 24. Calculated Values of f l i , Psat i , γ i and for the Top Product in Non-Ideal System f l1

0.2310

Psat 1 [bar]

3.149

γ1

1.000

f l2

0.3755

Psat 2 [bar]

2.032

γ2

0.964

f l3

0.3651

Psat 3 [bar]

0.559

γ3

0.997

f l4

0.0116

Psat 4 [bar]

0.459

γ4

1.031

f l5

0.0061

Psat 5 [bar]

0.210

γ5

1.089

Table 25. Calculated Values of x i , n i and T r, i for the Bottom Product in Non-Ideal System W [kmol/h]

T bubble [oC]

36.4

128.5

P.R 1

0.02

n1

0.3

x1

0.008

T r,1

0.784

P.R 2

0.04

n2

1

x2

0.027

T r,2

0.778

P.R 3

0.05

n3

1.25

x3

0.034

T r,3

0.730

P.R 4

0.95

n4

14.25

x4

0.391

T r,4

0.713

P.R 5

0.98

n5

19.6

x5

0.538

T r,5

0.685

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Table 26. Calculated Values of α SRK,i , a i and b i for the Bottom Product in Non-Ideal System α SRK,1

1.325

a1

1.27*107

b1

45.61

α SRK,2

1.361

a2

1.68*107

b2

58.25

α SRK,3

1.437

a3

3.31*107

b3

102.11

7

α SRK,4

1.465

a4

3.11*10

b4

91.89

α SRK,5

1.520

a5

3.98*107

b5

108.82

Table 27. Calculated Values of β i , q i and Zv i for the Bottom Product in Non-Ideal System β1

0.00138

q1

8.341

Zv 1

0.9892

β2

0.00177

q2

8.629

Zv 2

0.9853

β3

0.00310

q3

9.709

Zv 3

0.9704

v

β4

0.00279

q4

10.130

Z

4

0.9721

β5

0.00330

q5

10.946

Zv 5

0.9638

Table 28. Calculated Values of I v i , Φv i and Zl i for the Bottom Product in Non-Ideal System I v1

0.00140

Φv 1

0.990

Zl 1

0.00169

I v2

0.00179

Φv 2

0.987

Zl 2

0.00245

I v3

0.00319

Φv 3

0.973

Zl 3

0.00425

I v4

0.00286

Φv 4

0.975

Zl 4

0.00365

I v5

0.00342

Φv 5

0.967

Zl 5

0.00420

Table 29. Calculated Values of I l i , Φl i and K i for the Bottom Product in Non-Ideal System I l1

0.599

Φl 1

8.223

K1

8.31

I l2

0.543

Φl 2

4.982

K2

5.05

I l3

0.548

Φl 3

1.576

K3

1.62

I l4

0.567

Φl 4

1.365

K4

1.40

I l5

0.580

Φl 5

0.719

K5

0.74

- 10 -

Table 30. Calculated Values of α r,i , y i and f v i for the Bottom Product in Non-Ideal System α r,1

5.93

y1

0.068

f v1

0.0687

α r,2

3.60

y2

0.139

f v2

0.1386

α r,3

1.16

y3

0.056

f v3

0.0548

α r,4

1.00

y4

0.548

f v4

0.5415

α r,5

0.53

y5

0.400

f v5

0.3921

Table 31. Calculated Values of f l i , Psat i , γ i and for the Bottom Product in Non-Ideal System f l1

0.0687

Psat 1 [bar]

7.840

γ1

0.015

f l2

0.1386

Psat 2 [bar]

5.572

γ2

0.036

f l3

0.0548

Psat 3 [bar]

1.685

γ3

0.361

l

sat

4

0.5415

P

[bar]

1.464

γ4

0.481

f l5

0.3921

Psat 5 [bar]

0.732

γ5

1.801

f

4

Table 32. Calculated Values of Relative Volatilities of Components in Non-Ideal System Relative Volatility ,α ri No Name Feed Stream Top Product Bottom Product Average 1

Methanol

10.78

6.52

5.93

7.47

2

Ethanol

4.38

4.08

3.60

4.01

1.14

1.18

1.16

1.16

1.00

1.00

1.00

1.00

0.79

0.49

0.53

0.59

3 4 5

Neopentanol [Light Key] n-Butanol [Heavy Key] 1-Pentanol

Table 33. Calculation of Minimum Number of Plates, θ, Minimum Reflux Ratio and Feed Location in Non-Ideal System N min

39.56

θ

1.051

N R /N S

1.25

q

0

R D,min

3.516

λ ave, top [kJ/mol]

38.914

λ ave, bottom [kJ/mol]

43.836

- 11 -

Table 34. Calculation of Actual Reflux Ratio, Actual Number of Plates, Feed Location, Height of Column, Condenser Load (Q c ) and Reboiler Load (Q r ) in Ideal System R D /R D,min

1.5

1.8

2.4

R D,act

5.274

6.328

8.438

X

0.280

0.384

0.522

Y

0.376

0.313

0.238

N act

64

59

53

NS

28

26

24

NR

36

33

29

L [kmol/h]

335.40

402.48

536.64

G [kmol/h]

399.00

466.08

600.24

Q c [kW]

4312.9

5038.0

6488.2

L [kmol/h]

335.40

402.48

536.64

G [kmol/h]

299.00

366.08

500.24

Q r [kW] H c [m]

3640.9

4457.7

6091.3

32.5

30.0

27.0

Table 35. Required Calculations for the Fluid Velocity and Diameter of Column in Non-Ideal System MW avg [kg/kmol]

89.42

Wdot [kg/s]

0.904

ρ vap [kg/m3]

2.715

ρ liq [kg/m3]

811.63

It [m]

0.5

u v [m/s]

0.781

D c [m]

0.737

- 12 -

Table 36. Calculated Values by Using ChemCAD 6.0.2 for Ideal and Non-Ideal System Ideal System with Raoult’s Law

Non-Ideal System with SRK Method

R D /R D,min

1.5

1.8

2.4

1.5

1.8

2.4

R D,act

5.2821

6.3386

8.4514

62.0107

6.7886

9.0515

NS

29.2254

26.1033

23.2336

31.2342

27.9023

24.8464

N act

57.9572

51.6570

45.8660

5.6572

55.2872

49.1205

Q c [kJ/h]

-1.4678*107

-1.9449*107

-2.5049*107

-1.7721*107

-2.0733*107

-2.6756*107

Q r [kJ/h] R D,min

1.0981*107

1.5182*107

2.0782*107

1.3445*107

1.6457*107

2.2480*107

3.5214

3.5214

3.5214

3.7715

3.7715

3.7715

N min

35.0220

35.0220

35.0220

37.6120

37.6120

37.6120

- 13 -

3.0 DISCUSSION AND CONCLUSION The object of this report is to design a multi-component distillation column for ideal and non-ideal situations. Consider the case of the distillation of a multi-component mixture in a multi-stage distillation column. As before, the number of degrees of freedom is determined by the description rule (i.e., D.O.F. = number of variables set during construction or controlled during operation by independent means). Generally, all the variables associated with the feed, such as its composition, flow rate and enthalpy are set, as is the column pressure, which leaves four degrees of freedom and two basic types of problems (design and simulation), as was the case for a binary distillation. For a design problem, the goal is to determine the number of plates needed and the location of the feed plate, and the following is generally specified • • • • • •

Separation variable #1, the recovery of the light key component in the top (distillate) product. Separation variable #2, the recovery of the heavy key component in the bottom product. Condenser and reboiler loads Diameter and height of the column The fact that the optimal feed plate is used that minimizes the total number of plates. The reflux ratio.

Generally, in a multi-component distillation column, only two components will exist in significant quantities in both the bottom and top products. These are the two key components. The heavy non-key components will essentially all end up in the bottom product while the light non-key components will essentially all end up in the top product. This is why the separation variables described above are given in of the key components. In our process, the components were methanol, ethanol, neopentanol, n-butanol and 1pentanol. Our feed rate was 100 kmol/h; the light-key component was neopentanol and the heavy-key component was n-butanol. Our mole fractions in the feed were 15%, 25%, 25%,15% and 20% respectively (See Table 2). The separation targets for top and bottom products were 95%. In order to find temperature and equilibrium constants for the feed, a temperature was assumed. The Psat values were calculated at this temperature, by using Antoine equation (See Table 5). The equilibrium constants were calculated with this assumed temperature (See Table 2). By using Goal-seek, we equalized . The dew-point temperature was calculated as 113.66°C. According to the calculated temperature, the equilibrium constants were found as 5.14, 3.5, 1.02, 0.86, and 0.42. To calculate relative volatilities of the components, the ratio of K value of each component to the K value of the heavy-key product was taken. After this, the relative volatilities were found as, 5.95, 4.05, 1.18, 1 and 0.48 (See Table 6). - 14 -

In the ideal calculation, since we only had specific target separation for the light-key and heavy- key products were known (which was 95% for light-key at top and 95% for heavy at the bottom), the other separations were assumed 98% for methanol, 96% for ethanol, 95% for neopentanol, 5% for n-butanol and 2% 1-pentanol (See Table 3). The flow rate of top product was found as 63.6 kmol/h. As a result, the mole fractions for the top product were calculated as 0.231, 0.377, 0.373, 0.012, and 0.006 (See Table 3). In order to find temperature and equilibrium constants, a temperature was assumed. The Psat values were calculated at this temperature, by using Antoine equation (See Table 5).The equilibrium constants were calculated with this assumed temperature. By using Goal-seek, we equalized . The dew-point temperature was calculated as 96.84°C.According to the calculated temperature, the equilibrium constants were found as 3.10, 2, 0.55, 0.45, and 0.21 (See Table 3). To calculate relative volatilities of the components, the ratio of K value of each component to the K value of the heavy-key product was taken. After this, the relative volatilities were found as, 6.87, 4.43, 1.22, 1 and 0.46 (See Table 6). In the bottom product, the separations were assumed 2% for methanol, 4% for ethanol, 5% for neopentanol, 95% for n-butanol and 98% 1-pentanol (See Table 4). The flow rate of bottom product was found as 36.4 kmol/h. As a result, the mole fractions for the top product were calculated as 0.008, 0.027, 0.034, 0.391 and 0.538 (See Table 4). In order to find temperature and equilibrium constants, a temperature was assumed. The Psat values were calculated at this temperature, by using Antoine equation (See Table 5). The equilibrium constants were calculated with this assumed temperature. By using Goal-seek, we equalized . The dew-point temperature was calculated as 122.39°C. According to the calculated temperature, the equilibrium constants were found as 6.57, 4.59, 1.37, 1.18 and 0.58 (See Table 4). To calculate relative volatilities of the components, the ratio of K value of each component to the K value of the heavy-key product was taken. After this, the relative volatilities were found as, 5.58, 3.90, 1.16, 1 and 0.49 (See Table 6). The average relative volatilities of the components were determined by taking the geometric mean of the components’ volatilities at the feed, top product and bottom product. The values were 6.11, 4.12, 1.19, 1 and 0.48, respectively (See Table 6). Fenske equation was used to calculate the minimum number of trays. As a result, the minimum plate number was found as 34.33 (See Table 7). To find the minimum reflux ratio, the θ parameter was calculated at saturated vapor condition (q=0), and it must be between the relative volatilities of the light-key and heavy-key component, and was calculated using Goal-seek. As a result, the θ value was found as 1.075. Using Underwood equation, the minimum reflux ratio was determined as 3.57 (See Table 7). The ratio of

was assumed as 1.5, 1.8 and 2.4 respectively and the actual reflux

values were calculated and tabulated (See Table 8).

- 15 -

The Gilliland correlation was used to calculate actual plate number. For this, X and Y parameters were calculated and N actual was found as 56, 51 and 46, using the Y function and N min value (See Table 8). To determine the feed location, we used Kirkbridge equation. For N actual being 56, the entry location was found as the 26th tray. The N R and N S values were tabulated (See Table 8). To calculate reboiler and condenser duty, the average latent heat of vaporization was calculated. Our reboiler and condenser were selected as total reboiler and condenser. As a result, the reboiler duty was found as 3712.1kW for R min /R at 1.5. The condenser duty was found as 4376.1kW for the same ratio. Other values were tabulated (See Table 8). The height and diameter of the column were calculated lastly. The diameter was found as 0.698m and the height was 28.5 m. Other values were tabulated (See Table 8 and 9). In the case of the non-ideal system, the critical temperature, critical pressure, density, molecular weight and acentric factor were looked up from Ref.4 (See Table 10). The constants for the SRK Model were taken, a temperature was assumed and the reduced temperatures were calculated accordingly. The parameters α SRK , a i, b i , β i and q i were determined. Zv i was calculated using Goal-seek and Iv i and Φv i were determined by using the related equations in Ref.2.Same procedure was followed for ZL i , IL i and ΦL i . The equilibrium constants were determined using

. The liquid mole fractions were calculated to be used in

Modified Raoult’s Law. The P i sat values were determined by Antoine equation. The vaporliquid equilibrium was proven by the equality of fugacity coefficients of vapor and liquid phases. The activity coefficient was calculated from Modified Raoult’s Law. The assumed dew-point temperature was checked using the γ i and Psat values, and from Goal-seek, the actual dew-point temperatures was calculated as 106.6°C for feed, 96.98°C for top product. Same procedure was applied to the bottom product to find bubble-point temperature, which was found as 128.5°C. The average relative volatilities were calculated and found as 7.47, 4.01, 1.16, 1 and 0.59 respectively (See Table 32). The Fenske, Underwood, Gilliland and Kirkbridge equations were used like in the ideal system calculations. From these equations, N min was found as 39.56, θ was found as 1.051, R min was 3.516. For R min /R as 1.5, 1.8 and 2.4, our reflux ratios were found as 5.274, 6.328 and 8.438 respectively. For R min /R as 1.5, 1.8 and 2.4, N act were found as 64, 59, 53. The feed locations were found and tabulated (See Table 34). The condenser and reboiler duties were calculated like in the ideal system. Q c and Q r were calculated and tabulated (See Table 34). Finally, the diameter and height were calculated and were found as 0.737m and 32.5 m respectively. Other values were tabulated (See Table 34 and 35).

- 16 -

According to these results, it can be seen that the ideal system and non-ideal system are slightly different from each other. For both systems, actual plate numbers decrease with increasing R min /R. Also, height of the column decreases with increasing R min /R. Both Q c and Q r increase with increasing R min /R, as expected. In real systems, the relative volatilities of components are greater than the relative volatilities of ideal system. Also, the number of plates used are greater than that of ideal systems. The condenser and reboiler duties of real systems are less than the ideal system’s duties. The diameter and height of the column in real systems are greater than the diameter and height of the column in ideal systems. Finally, the most suitable reflux ratio for the real system can be selected within the calculation of total and operation costs. However, the optimum reflux ratio can be chosen without any calculations of costs and it can be seen in Table 34 as 1.8 for this project. In conclusion, the calculated values by using ChemCAD for the ideal and non ideal systems were tabulated in table 36.

- 17 -

4.0 NOMENCLATURE Symbol

Name

Unit

M w,i

Molecular Weight of i Component

[kg/kmol]

ρi

Density of i Component

[kg/m3]

ρ vap

Vapor Density of i Component

[kg/m3]

ρ liquid

Liquid Density of i Component

[kg/m3]

F

Mol Number of Feed Stream

[kmol/h]

D

Mol Number of Top Product

[kmol/h]

W

Mol Number of Bottom Product

[kmol/h]

ni

Mole Number of i Component

[kmol/h]

xi

Mole Fraction of i Component

P.R. i

Percentage Recovery of i Component

Ki

Distribution Coefficient of i Component

sat

P

Saturated Vapor Pressure

[bar]

Pt

Total Pressure

[bar]

αi

Relative Volatilities of i Component

N min

Minimum Number of Plate

N act

Actual Number of Plate

NS

Number of Plate of Stripping Section

NR

Number of Plate of Enriching Section

R D,min

Minimum Reflux Ratio

R D,act

Actual Reflux Ratio

λi

Latent Heat of Vaporization

MW uv

Molecular Weight of Mixing Fluid Velocity

[m/s]

Dc

Diameter of Distillation Column

[m]

Tr

Reduce Temperature

f

Fugacity

Φ

Fugacity Coefficient

T

Temperature

[oC]

Tc

Critical Temperature

[K]

Pc

Critical Pressure

[bar]

Qr

Reboiler Load

[kJ/s]

Qc

Condenser Load

[kJ/s]

- 18 -

5.0 REFERENCES 1.

Coulson, J.M., Richardson, J.F., Chemical Engineering Series - Chemical Engineering Design, Vol. 6, 4th Ed., Great Britain Pergamon Press, 1977.

2.

J.M. Smith, H.C. Van Ness, M.M. Abbott, Introduction to Chemical Engineering Thermodynamics, 2005, 7th Ed., Mc Graw Hill Company, Singapore.

3.

Felder, R.M., Rousseau, R.W., Elementary Principles of Chemical Process, 2nd Ed., John Willey and Sons Inc, USA, 1986.

4.

Yaws C.L., Yaws' Handbook of Thermodynamic and Physical Properties of Chemical Compounds, Lamar University, Beaumont, Texas, Norwich, New York, 2004

- 19 -

6.0 APPENDIX 6.1 Ideal System

G (See Table 4). L

F=100 kmol/h n1=15 kmol/h n2=25 kmol/h n3=25 kmol/h n4=15 kmol/h n5=20 kmol/h

G

L W

D

D=63.6 kmol/h n1=14.7 kmol/h n2=24 kmol/h n3=23.75 kmol/h n4=0.75 kmol/h n5=0.4 kmol/h

W=36.4 kmol/h n1=0.3 kmol/h n2=1 kmol/h n3=1.25 kmol/h n4=14.25 kmol/h n5=19.6 kmol/h

The mixture of methanol, ethanol, neopentanol, n-butanol and 1-pentanol was fed to the column as 100 kmol/h. The given data for the feed, top and bottom product was tabulated in Table 2, 3, 4.  Calculation of Mole Numbers of Components in Feed Stream F = 100 kmol h x1 = 0.15

n1 = 0.15 ×100 = 15 kmol h

x2 = 0.25

n2 = 0.25 ×100 = 25 kmol h

x3 = 0.25

n3 = 0.25 ×100 = 25 kmol h

x4 = 0.15

n4 = 0.15 ×100 = 15 kmol h

x5 = 0.20

n5 = 0.20 ×100 = 20 kmol h

 Assumption of Percentage Recoveries for Top Product P.R.1 = 0.98 P.R.2 = 0.96 P.R.3 = 0.95 P.R.4 = 0.05 P.R.5 = 0.02

n= nFi × P.Ri i 15 × 0.98 = 14.7 kmol / h n1 = 25 × 0.96 = 24kmol / h n2 = 25 × 0.95 = 23.75kmol / h n3 = 15 × 0.05 = 0.75kmol / h n4 = 20 × 0.02 = 0.4kmol / h n5 = - 20 -

= = / 63.6 0.231 x1 14.7

xdi = n i / D

= = / 63.6 0.377 x2 24

D = 14.7 + 24 + 23.75 + 0.75 + 0.4 D = 63.6kmol / h

= = / 63.6 0.373 x3 23.75 = = / 63.6 0.012 x4 0.75 = = / 63.6 0.006 x5 0.4

 Assumption of Percentage Recoveries for the Bottom Product

n= nFi × P.Ri i

P.R.1 = 1 − 0.98 = 0.02

n1 = 15 × 0.02 = 0.3kmol / h

1 − 0.96 = 0.04 P.R.2 =

n2 = 25 × 0.04 = 1kmol / h

P.R.3 = 1 − 0.95 = 0.05

n3 = 25 × 0.05 = 1.25kmol / h

P.R.4 = 1 − 0.05 = 0.95

n4 = 15 × 0.95 = 14.25kmol / h

P.R.5 = 1 − 0.02 = 0.98

n5 = 20 × 0.98 = 19.6kmol / h = = x1 0.3 / 36.4 0.008

xdi = n i /W W= 0.3 + 1 + 1.25 + 14.25 + 19.6 W = 36.4kmol / h

= x2 1/= 36.4 0.027 = = x3 1.25 / 36.4 0.034 = = x4 14.25 / 36.4 0.391 = = x5 19.6 / 36.4 0.538

 Calculation of Temperature, K-values & Relative Volatilities for the Feed Stream Dew point temperature for feed Dew point temperature for the feed was assumed as 100 oC. All calculations was done c yi = 1.00 value could not be ∑ =i 1 =i 1 K i c

by using this temperature in excel. However = ∑ xi

obtained. So the dew point temperature was calculated as 113.66 oC by using “Goal Seek” method in Excel and shown below. K i = yi / xi = ∑ xi i

yi

∑= K i

1.00

i

B T in oC ; P in mmHg ( A, B and C constants were shown inTable 2) T +C o Tdew = 113.66 C ( feed stream) log P sat= A −

Pt = 760mmHg 1473.11 sat = log Pmethanol 7.87863 − ; 113.66 + 230 For other components see in Table 5.

sat = Pmethanol 3909.69 mmHg

- 21 -

Pi sat Pt

= Ki

,

K= methanol

sat Pmethanol 3909.69 = = 5.14 760 Pt

sat Pethanol 2661.98 = = 3.5 K= ethanol 760 Pt

K= n-butanol

= αi

sat Pneopentanol 774.99 = = = 1.02 K neopentanol 760 Pt

,

sat Pn-butanol 656.70 = = 0.86 , 760 Pt

Ki K HK

,

= α methanol,n-butanol

K1-= pentanol

K methanol 5.14 = = 5.95 K n-butanol 0.86

= α ethanol ,n-butanol

K ethanol 3.5 = = 4.05 , K n-butanol 0.86

α

=

n-butanol , n-butanol

P1−satpentanol 316.28 = = 0.42 760 Pt

K n-butanol 0.86 = = 1 K n-butanol 0.86

,

K neopentanol 1.02 = = 1.18 K n-butanol 0.86

= α neopentanol,n-butanol

α 1- pentanol,n-butanol =

K1- pentanol 0.42 = = 0.48 K n-butanol 0.86

With excel calculation (goal seek) we found T dew =113.66 oC

 Calculation of Temperature, K-values & Relative Volatilities for the Top Product Dew point temperature for top Dew point temperature for the top was assumed as 90 oC. All calculations was done by c yi = 1.00 value could not be obtained. ∑ =i 1 =i 1 K i c

using this temperature in excel. However = ∑ xi

So the dew point temperature was calculated as 96.84 oC by using “Goal Seek” method in Excel and shown below. K i = yi / xi = ∑ xi i

yi

∑= K i

1.00

i

B T in oC ; P in mmHg ( A, B and C constants were shown inTable 2) T +C Tdew = 96.84 oC (top stream) log P sat= A −

Pt = 760mmHg 1473.11 sat sat = = log Pmethanol 7.87863 − ; Pmethanol 2352.49 mmHg 96.84 + 230 For other components see in Table 5.

- 22 -

Pi sat Pt

= Ki

K= methanol

,

sat Pmethanol 2352.49 = = 3.1 760 Pt

sat Pethanol 1517.50 = = 2 K= ethanol 760 Pt

K= n-butanol

= αi

sat Pneopentanol 417.46 = = = 0.55 K neopentanol 760 Pt

,

sat Pn-butanol 342.30 = = 0.45 , 760 Pt

Ki K HK

= α methanol,n-butanol

,

K1-= pentanol

K methanol 3.10 = = 6.87 K n-butanol 0.45

= α ethanol ,n-butanol

K ethanol 2 = = 4.43 , K n-butanol 0.45

α

=

n-butanol , n-butanol

P1−satpentanol 156.67 = = 0.21 760 Pt

K n-butanol 0.45 = = 1 K n-butanol 0.45

,

= α neopentanol,n-butanol

K neopentanol 0.55 = = 1.22 K n-butanol 0.45

α 1- pentanol,n-butanol =

K1- pentanol 0.21 = = 0.46 K n-butanol 0.45

With excel calculation (goal seek) we found T dew =96.84 0C

 Calculation of Temperature, K-values & Relative Volatilities for the Bottom Product Bubble point temperature for bottom Bubble point temperature for the top was assumed as 120 oC. All calculations was c

done by using this temperature in excel. However= ∑ yi

c

xK ∑=

=i 1 =i 1

i

i

1.00 value could not be

obtained. So the bubble point temperature was calculated as 122.39 oC by using “Goal Seek” method in Excel and shown below. K i = yi / xi

= ∑ yi i

xK ∑= i

i

i

1.00

B T in oC ; P in mmHg ( A, B and C constants were shown inTable 2) T +C Tbubble =122.39 oC (bottom stream) log P sat= A −

Pt = 760mmHg 1473.11 sat = log Pmethanol 7.87863 − ; 122.39 + 230 For other components see in Table 5.

sat = 4991.92 mmHg Pmethanol

- 23 -

= Ki

Pi sat Pt

,

K= methanol

sat Pmethanol 4991.92 = = 6.57 760 Pt

sat Pethanol 3487.52 = = 4.59 K= ethanol 760 Pt sat Pn-butanol 894.23 = = 1.18 , 760 Pt

K= n-butanol

= αi

,

Ki K HK

,

= α methanol,n-butanol

P1−satpentanol 440.62 = = 0.58 760 Pt

K1-= pentanol

K methanol 6.57 = = 5.58 K n-butanol 1.18

= α ethanol ,n-butanol

K ethanol 4.59 = = 3.90 , K n-butanol 1.18

α

=

n-butanol , n-butanol

sat Pneopentanol 1039.42 = = = 1.37 K neopentanol 760 Pt

K n-butanol 1.18 = = 1 K n-butanol 1.18

,

= α neopentanol,n-butanol = α 1- pentanol,n-butanol

K neopentanol 1.37 = = 1.16 K n-butanol 1.18 K1- pentanol 0.58 = = 0.49 K n-butanol 1.18

With excel calculation (goal seek) we found T bubble =122.39 0C

 Calculation of Average Relative Volatilities

α average,i = α ethanol= α n-butanol =

3 3

α top ,i × α bottom ,i × α feed ,i

,

4.05 × 4.43 × 3.9= 4.12 , 3

1 × 1 × 1= 1

,

α methanol =

3

α neopentanol= α 1- pentanol =

5.95 × 6.87 × 5.58 = 6.11 3

3

1.18 ×1.22 ×1.16= 1.19

0.48 × 0.46 × 0.49 = 0.48

 Calculation of Minimum Number of Plates by Using Fenske Equation x  x  0.373 0.391  log  LK , D × HK ,W  × log  x x 0.012 0.034  HK , D LK ,W    = = ; N min = 34.33 N min log α LK ,avg log1.19

 Calculation of q-Parameter We used saturated vapor for the feed stream; so q-parameter must be taken as zero (q=0).

- 24 -

 Calculation of θ-Parameter

α i ,ave × x f ,i =1 − q , α HK < θ < α LK i =1 α i , ave − θ c

∑

α met ,ave × x f ,met α eth ,ave × x f ,eth α neo ,ave × x f ,neo α n −but ,ave × x f ,n −but α1− pent ,ave × x f ,1− pent + + + + − 1 + q =0 α met ,ave − θ α eth ,ave − θ α neo ,ave − θ α n −but ,ave − θ α1− pent ,ave − θ 6.11× 0.15 4.12 × 0.25 1.19 × 0.25 1× 0.15 0.48 × 0.20 + + + + − 1 + 0 =0 6.11 − θ 4.12 − θ 1.19 − θ 1−θ 0.48 − θ

θ = 1.075

 Calculation of Minimum Reflux Ratio, R Dmin by Using Underwood Equation

α ir × xD ,i = RD min + 1 i =1 α ir − θ c

∑

6.11× 0.231 4.12 × 0.377 1.19 × 0.373 1× 0.012 0.48 × 0.006 + + + + = −1 0 6.11 − 1.075 4.12 − 1.075 1.19 − 1.075 1 − 1.075 0.48 − 1.075 = 3.577

= RD min RD min

= = 3.577 *1.5 5.366 RDactual R= D min * β1 = 3.577 *1.8 = 6.439 RDactual R= D min * β 2 RDactual R= = = 3.577 * 2.4 8.585 D min * β 3

 Calculation of Actual Plate Number by Using Gilliand Correlation = X

R - RDmin 5.366 − 3.577 = = 0.281 5.366 + 1 R +1

 1.805  1.805   = Y= 1 − exp 1.490 + 0.315 X − 0.1  = 1 − exp 1.490 + 0.315 × 0.281 − 0.376 X  (0.281)0.1    N − N min Y + N min = ⇒N N +1 1− Y 0.376 + 34.33 = = 56 plates N N= actual 1 − 0.376

= Y

For other reflux ratios, the calculated values were tabulated in Table 8. - 25 -

 Calculation of N R and N S for the Feed Location by Using Kirk Bridge Method  W x  NR     f . HK log = 0.206 log =   D   x f , LK  NS   NR 1.25 , = N R N act − N S = NS N= S

N act 56 = = 25 NR 1.25 1 + +1 NS

,

2  36.4  0.15  0.034 2    xb , LK    0.206 log  = =     0.09691      xd , HK    63.6  0.25  0.012  

56 − 25 = 31 N= R

 Calculation of Condenser and Reboiler Loads, Q c and Q r F= W +D ,

L= G +W ,

L 63.6

= RD 5.366 =

L 341.25kmol / h ,=

c

λi yi and λave,bottom ∑=

= λave,top

L RD = D

c

∑λ

=i 1 =i 1

G =+ L D

,

i

xi

( for results see in Table 7 )

G= 341.25 + 63.6 = 404.85kmol / h kmol kJ  1h 1000 mol  ,= Qc  404.85 × 38.914 × = 4376.1 kJ s ( kW ) × h mol  3600 s 1 kmol 

= Qc G * λave ,top

L−L L − 341.25 , q= 0= F 100 L= L= 341.25 , G =L − W =341.25 − 36.4 ,

q=

G =304.85kmol / h

kmol kJ  1h 1000 mol  = Qr G * λave ,bottom = , Qr  304.85 * 43.836 × 3712.1 kJ s ( kW ) × = h mol  3600 s 1 kmol   Calculation of Column Diameter and Height = MW

N

= MW * y , y ∑ i =1

i

Wi

Wi

xi K i

yWmethanol = xmethanol K methanol = 0.008 × 6.57 = 0.054

( for other components see Table 9 )

MW = (32.04*0.054) + (46.07 *0.126) + (88.15*0.047) + (74.122*0.461) + (88.15*0.312) = 73.34kg / kmol  *W = 73.34 kg × 36.4 kmol × 1 h = 0.742 kg s W = MV h 3600 s kmol P × MW = ρ × R × T

- 26 -

P × MW 1× 73.34 = = 2.261 kg m3 R ×T 0.082 × (122.39 + 273.15)

= ρvap

c

∑ρ x =

ρliq =

i =1

i i

791.8 × 0.008 + 789 × 0.027 + 812 × 0.034 + 809.8 × 0.391 + 814.4 × 0.538

= 811.63 kg m3 1/2

 ρ − ρv  uv = −0.171* l + 0.27 * lt − 0.047 *  l   ρv 

(

2 t

)

1/2

 811.63 − 2.261  uv = −0.171*0.52 + 0.27 *0.5 − 0.047 *   = 0.856 m s 2.261  4W 4 × 0.742 = =0.698 m , H c =lt × ( N + 1) =0.5 × (56 + 1) = 28.3 m Dc = πρv uv π × 2.261× 0.856

(

)

For other reflux ratios the results were in Table 9.

6.2 Real System (Non-Ideal)  For Feed Stream Dew point temperature for the feed was assumed as 100 oC. All calculations was done by c y 1.00 value could not be obtained. So using this temperature in excel. However Pt × ∑ i * = i =1 γ i Pi the dew point temperature was calculated as 106.6 oC (379.75 K) by using “Goal Seek” method in Excel and shown below.

Tr ,i =

T , T [ K ] , Tc [ K ] Tc ,i

379.75 379.75 379.75 = 0.741 = , Tr ,2 = 0.736 = , Tr ,3 = 0.690 512.58 516.25 550.00 379.75 379.75 , Tr ,5 = 0.648 = Tr ,4 = 0.675 = 562.93 586.15

= Tr ,1

1 + ( 0.480 + 1.574ωi − 0.176ωi2 )(1 − Tr1,i2 )  α SRK ,i =  

2

Tc ,i , ωi were taken in Reference 2 and tabulated in Table 10 1 ( 0.480 + 1.574 × 0.566 − 0.176 × 0.5662 )(1 − 0.7411 2 )  = α SRK ,1 =+   1.400 2

1 ( 0.480 + 1.574 × 0.637 − 0.176 × 0.637 2 )(1 − 0.7361 2 )  = 1.442 α SRK ,2 =+   2

2 12   1.515 α SRK ,3 =+ 1 ( 0.480 + 1.574 × 0.604 − 0.176 × 0.604 )(1 − 0.690 )  = 2

1 ( 0.480 + 1.574 × 0.595 − 0.176 × 0.5952 )(1 − 0.6751 2 )  = α SRK ,4 =+   1.542 2

1 ( 0.480 + 1.574 × 0.594 − 0.176 × 0.5942 )(1 − 0.6481 2 )  = α SRK ,5 =+   1.598 2

- 27 -

ai

α i R 2Tc ,i , Tc ,i [ K ] , Pc ,i [bar ] , R 83.14 cm3bar molK ψ= Pc ,i

ψ was taken in Reference 2 and tabulated in Table 10 1.400 × 83.142 × 512.58 0.42748 × 1.34 ×107 = a1 = 80.96 1.442 × 83.142 × 516.25 0.42748 × 1.78 ×107 = a2 = 63.84 1.515 × 83.142 × 550.00 = a3 0.42748 × = 3.49 ×107 38.80 1.542 × 83.142 × 562.93 a4 = 0.42748 × 3.27 ×107 = 44.13 1.598 × 83.142 × 586.93 a5 = 0.42748 × 4.18 ×107 = 38.80

RT 83.14 cm3bar molK bi = Tc ,i [ K ] , Pc ,i [bar ] , R = Ω c ,i , Pc ,i Ω was taken in Reference 2 and tabulated in Table 10 83.14 × 512.58 45.61 = 80.96 83.14 × 516.25 0.08664 × 58.25 = = 63.84 83.14 × 550.00 0.08664 × 102.11 = = 38.80 83.14 × 562.93 = 91.89 = 0.08664 × 44.13 83.14 × 586.15 = 0.08664 × = 108.82 38.80

0.08664 × b1 = b2 b3 b4 b5

bi P , T [ K ] , P [bar ] , R = 83.14 cm3bar molK RT 45.61×1.013 58.25 ×1.013 = = = 0.00187 β1 = 0.00146 , β2 83.14 × 379.75 83.14 × 379.75 102.11×1.013 91.89 ×1.013 = = = 0.00295 , β4 β3 = 0.00328 83.14 × 379.75 83.14 × 379.75 108.82 ×1.013 = 0.00349 β5 = 83.14 × 379.75

βi ≡

- 28 -

qi ≡

ai , bi RT

T [ K ] , R = 83.14 cm3bar molK

1.34 × 107 1.78 × 107 = = 9.321 , q2 ≡ 9.672 45.61× 83.14 × 379.75 58.25 × 83.14 × 379.75 3.49 × 107 3.27 × 107 = = q3 ≡ 10.829 , q4 ≡ 11.281 102.11× 83.14 × 379.75 91.89 × 83.14 × 379.75 4.18 × 107 q5 ≡ = 12.166 108.82 × 83.14 × 379.75 Z iv − βi 1 βi − qi βi Z iv =+ ( Ziv + εβi )( Ziv + σβi ) q1 ≡

0= 1 + 0.00146 − 9.321× 0.00146

Z iv − 0.00146 − Z iv v v ( Zi + 0 × 0.00146 )( Zi + 1× 0.00146 )

Z iv value for methanol was calculated as 0.987 by above equation from Goal Seek in Excel And for the other components, Z iv values were tabulated in Table 14, 21, 28. Z v + σβi 1 ln iv I iv = σ − ε Z i + εβi 1 0.9870 + 1× 0.00146 1 0.9826 + 1× 0.00187 = 0.00148 = = 0.00190 ln , I 2v ln 1− 0 0.9870 + 0 1− 0 0.9826 + 0 1 0.9649 + 1× 0.00328 1 0.9670 + 1× 0.00295 = ln 0.00339 , I 4v = ln 0.00304 I 3v = 1− 0 0.9649 + 0 1− 0 0.9670 + 0 1 0.9572 + 1× 0.00349 ln = 0.00364 I 5v = 1− 0 0.9572 + 0 ln Φ iv = Z iv − 1 − ln ( Z iv − βi ) − qi I iv ; Φ iv = exp  Z iv − 1 − ln ( Z iv − βi ) − qi I iv 

= I1v

0.00148 0.988 = Φ1v exp 0.9870 − 1 − ln ( 0.9870 − 0.00146 ) − 9.321× = 0.00190  0.984 = Φ v2 exp 0.9826 − 1 − ln ( 0.9826 − 0.00186 ) − 9.672 × = 0.968 = Φ 3v exp 0.9649 − 1 − ln ( 0.9649 − 0.00328 ) − 10.829 × 0.00339  = 0.00304  0.970 = Φ v4 exp 0.9670 − 1 − ln ( 0.9670 − 0.00295 ) − 11.281× = 0.00364  0.961 = Φ 5v exp 0.9572 − 1 − ln ( 0.9572 − 0.00349 ) − 12.166 × =  1 + βi − Z il  l l l Z i =βi + ( Z i + εβi )( Z i + σβi )    qi βi   1 + 0.00146 − Z il  l = 0 0.00146 + ( Z il + 0 × 0.00146 )( Z il + 1× 0.00146 )   − Zi  9.321× 0.00146  l Z i value for methanol was calculated as 0.00157 by above equation from Goal Seek in Excel And for the other components, Z il values were tabulated in Table 14, 21, 28.

- 29 -

I il =

Z l + σβi 1 ln il σ − ε Z i + εβi

1 0.00157 + 1× 0.00146 1 0.00215 + 1× 0.00187 = 0.660 = = 0.625 ln , I 2l ln 1− 0 0.00157 + 0 1− 0 0.00215 + 0 1 0.00395 + 1× 0.00328 1 0.00345 + 1× 0.00295 = ln 0.604 , I 4v = ln 0.618 I 3v = 1− 0 0.00395 + 0 1− 0 0.00345 + 0 1 0.00370 + 1× 0.00349 ln = 0.665 I 5v = 1− 0 0.00370 + 0 ln Φ li = Z il − 1 − ln ( Z il − βi ) − qi I il ; Φ li = exp  Z il − 1 − ln ( Z il − βi ) − qi I il 

= I1l

= Φ1l exp 0.00157 − 1 − ln ( 0.00157 − 0.00146 ) − 9.321 = × 0.660  7.605 = Φ l2 exp 0.00215 − 1 − ln ( 0.00215 − 0.00186 ) − 9.672 = × 0.625 3.078 0.791 = Φ 3l exp 0.00395 − 1 − ln ( 0.00395 − 0.00328 ) − 10.829 × 0.604  = = Φ l4 exp 0.00345 − 1 − ln ( 0.00345 − 0.00295 ) − 11.281 = × 0.618 0.693 = Φ 5l exp 0.00370 − 1 − ln ( 0.00370 − 0.00349 ) − 12.166 = × 0.665 0.545

= Ki

Φ li Φ iv

= K3

0.791 0.693 0.545 = 0.82 , = K 4 = 0.71 , = K 5 = 0.57 0.968 0.970 0.961

= αi

Ki K HK

, = K1

7.605 3.078 = 7.70 , = K 2 = 3.13 0.988 0.984

= α methanol,n-butanol

,

K methanol 7.70 = = 10.78 K n-butanol 0.71

= α ethanol ,n-butanol

K ethanol 3.13 = = 4.38 , K n-butanol 0.71

α

=

n-butanol , n-butanol

= xi

yi Ki

K n-butanol 0.71 = = 1 K n-butanol 0.71

= x1

,

,

= α neopentanol,n-butanol

α 1- pentanol,n-butanol =

K neopentanol 0.82 = = 1.14 0.71 K n-butanol K1- pentanol 0.57 = = 0.79 K n-butanol 0.71

0.15 0.25 = 0.019 , = x2 = 0.08 7.70 3.13

0.25 0.15 0.20 0.306 , x4 0.210 ,= 0.352 = = x3 = = = x5 0.82 0.71 0.57

fi v = yi Φ iv P ,

f1v = 0.15 × 0.988 ×1.013 = 0.1501

f 2v =0.25 × 0.984 ×1.013 =0.2491 ,

f3v =0.25 × 0.968 ×1.013 =0.2451

f 4v =0.15 × 0.970 ×1.013 =0.1474 ,

f5v =0.20 × 0.961×1.013 =0.1947

fi l = xi Φ li Pt

,

0.019 × 7.605 × 1.013 = 0.1501 f1l =

f 2l =0.08 × 3.078 ×1.013 =0.2491 ,

f3l =0.306 × 0.791×1.013 =0.2451

f 4l =0.21× 0.693 ×1.013 =0.1474 ,

f5l =0.352 × 0.545 ×1.013 =0.1947

- 30 -

In Vapor-Liquid Equilibrium, fugacity of liquid and vapor must be equal to each other. So, our results this condition. ( fi l = fi v )

B T in oC ; P in mmHg ( A, B and C constants were shown inTable 1) T +C Tdew =106.6 oC ( feed stream) log P sat= A −

Pt = 760mmHg sat = log Pmethanol 7.87863 −

1473.11 106.6 + 230

;

sat = Pmethanol 3178.72 mmHg ×

1 atm 1.013 bar × = 4.237 bar 760 mmHg 1 atm

For other components see in Table 17, 24, 31. * , γi = yi Φ iv Pt xi γ i P= i

yi Φ iv Pt xi Pi *

0.15 × 0.988 ×1.013 0.15 × 0.984 ×1.013 , γ2 = 1.818 = = 1.105 0.019 × 4.237 0.080 × 2.822 0.25 × 0.968 ×1.013 0.25 × 0.970 ×1.013 , γ4 = = 0.997 = = 1.044 γ3 0.306 × 0.804 0.210 × 0.672 0.20 × 0.961×1.013 = 1.741 γ5 = 0.352 × 0.317 yi With excel calculation (goal seek), Pt × ∑ = 1.00 we found T dew =106.6 oC sat i γ i Pi

γ1 =

 For Top Product Dew point temperature for the feed was assumed as 90 oC. All calculations was done by c yi 1.00 value could not be obtained. = using this temperature in excel. However Pt × ∑ sat i =1 γ i Pi So the dew point temperature was calculated as 96.98 oC (370.13 K) by using “Goal Seek” method in Excel and shown below.

Tr ,i =

T , T [ K ] , Tc [ K ] Tc ,i

370.13 370.13 370.13 = 0.722 = , Tr ,2 = 0.717 = , Tr ,3 = 0.673 512.58 516.25 550.00 370.13 370.13 , Tr ,5 = 0.631 = Tr ,4 = 0.658 = 562.93 586.15

= Tr ,1

- 31 -

1 + ( 0.480 + 1.574ωi − 0.176ωi2 )(1 − Tr1,i2 )  α SRK ,i =  

2

Tc ,i , ωi were taken in Reference 2 and tabulated in Table 10 1 ( 0.480 + 1.574 × 0.566 − 0.176 × 0.5662 )(1 − 0.7221 2 )  = α SRK ,1 =+   1.434 2

1 ( 0.480 + 1.574 × 0.637 − 0.176 × 0.637 2 )(1 − 0.7171 2 )  = 1.479 α SRK ,2 =+   2

2 12   1.551 α SRK ,3 =+ 1 ( 0.480 + 1.574 × 0.604 − 0.176 × 0.604 )(1 − 0.673 )  = 2

1 ( 0.480 + 1.574 × 0.595 − 0.176 × 0.5952 )(1 − 0.6581 2 )  = α SRK ,4 =+   1.578 2

1 ( 0.480 + 1.574 × 0.594 − 0.176 × 0.5942 )(1 − 0.6311 2 )  = α SRK ,5 =+   1.633 2

ai

α i R 2Tc ,i

, ψ= Tc ,i [ K ] , Pc ,i [bar ] , R 83.14 cm3bar molK Pc ,i

ψ was taken in Reference 2 and tabulated in Table 10 1.434 × 83.142 × 512.58 0.42748 × = 1.38 ×107 a1 = 80.96 1.479 × 83.142 × 516.25 0.42748 × = 1.82 ×107 a2 = 63.84 1.551× 83.142 × 550.00 = a3 0.42748 × = 3.57 ×107 38.80 1.578 × 83.142 × 562.93 a4 = 0.42748 × = 3.35 ×107 44.13 1.633 × 83.142 × 586.93 a5 = 0.42748 × = 4.27 ×107 38.80

RT 83.14 cm3bar molK bi = Tc ,i [ K ] , Pc ,i [bar ] , R = Ω c ,i , Pc ,i Ω was taken in Reference 2 and tabulated in Table 10 83.14 × 512.58 45.61 = 80.96 83.14 × 516.25 0.08664 × 58.25 = = 63.84 83.14 × 550.00 0.08664 × 102.11 = = 38.80 83.14 × 562.93 = 91.89 = 0.08664 × 44.13 83.14 × 586.15 = 0.08664 × = 108.82 38.80

0.08664 × b1 = b2 b3 b4 b5

- 32 -

bi P , T [ K ] , P [bar ] , R = 83.14 cm3bar molK RT 45.61×1.013 58.25 ×1.013 = β1 = 0.00150 = , β2 = 0.00192 83.14 × 370.13 83.14 × 370.13 102.11×1.013 91.89 ×1.013 = = , β4 = 0.00302 β3 = 0.00336 83.14 × 370.13 83.14 × 370.13 108.82 ×1.013 = 0.00358 β5 = 83.14 × 370.13

βi ≡

qi ≡

ai , bi RT

T [ K ] , R = 83.14 cm3bar molK

1.38 × 107 1.82 × 107 = = 9.798 , q2 ≡ 10.181 45.61× 83.14 × 370.13 58.25 × 83.14 × 370.13 3.57 × 107 3.35 × 107 = = q3 ≡ 11.373 , q4 ≡ 11.840 102.11× 83.14 × 370.13 91.89 × 83.14 × 370.13 4.27 × 107 q5 ≡ = 12.758 108.82 × 83.14 × 370.13 Z iv − βi 1 βi − qi βi Z iv =+ ( Ziv + εβi )( Ziv + σβi ) q1 ≡

0= 1 + 0.00150 − 9.798 × 0.00150

Z iv − 0.00150 − Z iv v v 0 0.00150 1 0.00150 Z Z + × + × ( i )( i )

Z iv value for methanol was calculated as 0.988 by above equation from Goal Seek in Excel And for the other components, Z iv values were tabulated in Table 13, 20, 27. Z v + σβi 1 ln iv I iv = σ − ε Z i + εβi 1 0.9880 + 1× 0.00150 1 0.9828 + 1× 0.00192 = 0.00152 = = 0.00195 ln , I 2v ln 1− 0 0.9880 + 0 1− 0 0.9828 + 0 1 0.9659 + 1× 0.00336 1 0.9673 + 1× 0.00302 = = 0.00347 = = 0.00312 ln , I 4v ln I 3v 1− 0 0.9659 + 0 1− 0 0.9673 + 0 1 0.9575 + 1 × 0.00358 ln = 0.00373 I 5v = 1− 0 0.9575 + 0 ln Φ iv = Z iv − 1 − ln ( Z iv − βi ) − qi I iv ; Φ iv = exp  Z iv − 1 − ln ( Z iv − βi ) − qi I iv  = I1v

0.00152  0.987 = Φ1v exp 0.9880 − 1 − ln ( 0.9880 − 0.00150 ) − 9.798 × = 0.00195 0.982 = Φ v2 exp 0.9828 − 1 − ln ( 0.9828 − 0.00192 ) − 10.181× = 0.965 = Φ 3v exp 0.9659 − 1 − ln ( 0.9659 − 0.00336 ) − 11.373 × 0.00347  = = Φ v4 exp 0.9673 − 1 − ln ( 0.9673 − 0.00302 ) − 11.840 × 0.00312 =  0.967 = Φ 5v exp 0.9575 − 1 − ln ( 0.9575 − 0.00358 ) − 12.758 × = 0.00373 0.958

- 33 -

 1 + βi − Z il  Z =βi + ( Z + εβi )( Z + σβi )    qi βi  l i

l i

l i

 1 + 0.00150 − Z il  l = 0 0.00150 + ( Z il + 0 × 0.00150 )( Z il + 1× 0.00150 )   − Zi  9.798 × 0.00150  l Z i value for methanol was calculated as 0.00187 by above equation from Goal Seek in Excel And for the other components, Z il values were tabulated in Table 14, 21, 28. I il =

Z l + σβi 1 ln il σ − ε Z i + εβi

1 0.00187 + 1× 0.00150 1 0.00255 + 1× 0.00192 = 0.590 = = 0.560 ln , I 2l ln 1− 0 0.00187 + 0 1− 0 0.00255 + 0 1 0.00426 + 1× 0.00336 1 0.00385 + 1× 0.00302 = = 0.582 = = 0.580 ln , I 4v ln I 3v 1− 0 0.00426 + 0 1− 0 0.00385 + 0 1 0.00430 + 1× 0.00358 ln = 0.606 I 5v = 1− 0 0.00430 + 0 ln Φ li = Z il − 1 − ln ( Z il − βi ) − qi I il ; Φ li = exp  Z il − 1 − ln ( Z il − βi ) − qi I il 

= I1l

= Φ1l exp 0.00187 − 1 − ln ( 0.00187 − 0.00150 ) − 9.798 = × 0.590  3.108 = Φ l2 exp 0.00255 − 1 − ln ( 0.00255 − 0.00192 ) − 10.181 = × 0.560  1.935 = Φ 3l exp 0.00426 − 1 − ln ( 0.00426 − 0.00336 ) − 11.373 × 0.582  = 0.551 = Φ l4 exp 0.00385 − 1 − ln ( 0.00385 − 0.00302 ) − 11.840 = × 0.580  0.467 = Φ 5l exp 0.00430 − 1 − ln ( 0.00430 − 0.00358 ) − 12.758 = × 0.606  0.226

= Ki

Φ li Φ iv

= K3

0.551 0.467 0.226 = 0.57 , = K4 = 0.48 , = K5 = 0.24 0.965 0.967 0.958

= αi

Ki K HK

, = K1

,

3.108 1.935 = 3.15 , = K 2 = 1.97 0.987 0.982

= α methanol,n-butanol

K methanol 3.15 = = 6.52 K n-butanol 0.48

= α ethanol ,n-butanol

K ethanol 1.97 = = 4.08 , K n-butanol 0.48

α

=

n-butanol , n-butanol

K n-butanol 0.48 = = 1 K n-butanol 0.48

,

= α neopentanol,n-butanol = α 1- pentanol,n-butanol

- 34 -

K neopentanol 0.57 = = 1.18 0.48 K n-butanol K1- pentanol 0.24 = = 0.49 K n-butanol 0.48

= xi

yi Ki

= x3

0.373 0.012 0.006 = 0.655 ,= x4 = 0.024 ,= x5 = 0.027 0.57 0.48 0.24

, = x1

fi v = yi Φ iv P ,

0.231 0.377 = 0.073 ,= x2 = 0.192 3.15 1.97

f1v = 0.231× 0.987 ×1.013 = 0.2310

f 2v = 0.377 × 0.982 ×1.013 = 0.3755 ,

f3v = 0.373 × 0.965 ×1.013 = 0.3651

f 4v = 0.012 × 0.967 ×1.013 = 0.0116 ,

f5v = 0.006 × 0.958 ×1.013 = 0.0061

fi l = xi Φ il Pt

,

f1l = 0.073 × 3.108 × 1.013 = 0.2310

f 2l = 0.192 ×1.935 ×1.013 = 0.3755 ,

f3l = 0.655 × 0.551×1.013 = 0.3651

f 4l = 0.024 × 0.467 ×1.013 = 0.0116 ,

f5l = 0.027 × 0.226 ×1.013 = 0.0061

In Vapor-Liquid Equilibrium, fugacity of liquid and vapor must be equal to each other. So, our results this condition. ( fi l = fi v )

B T in oC ; P in mmHg ( A, B and C constants were shown inTable 1) T +C Tdew = 96.98 oC (top stream) log P sat= A −

Pt = 760mmHg sat = log Pmethanol 7.87863 −

1473.11 96.98 + 230

;

sat = Pmethanol 2362.789 mmHg ×

1 atm 1.013 bar × = 3.149 bar 760 mmHg 1 atm

For other components see in Table 17, 24, 31. sat , γi = yi Φ iv Pt xi γ i P= i

yi Φ iv Pt xi Pi sat

0.231× 0.987 ×1.013 0.377 × 0.982 ×1.013 , γ2 = 1= = 0.964 0.073 × 3.149 0.192 × 2.032 0.373 × 0.965 ×1.013 0.012 × 0.967 ×1.013 , γ4 = = 0.997 = = 1.031 γ3 0.655 × 0.559 0.024 × 0.459 0.006 × 0.958 ×1.013 = 1.089 γ5 = 0.027 × 0.210 yi With excel calculation (goal seek), Pt × ∑ = 1.00 we found T dew =96.98 oC sat i γ i Pi = γ1

- 35 -

 For Bottom Product Bubble point temperature for the feed was assumed as 120 oC. All calculations was done c

by using this temperature in excel. However

∑x γ i =1

i

i

Pi sat

Pt

= 1.00 value could not be obtained.

So the bubble point temperature was calculated as 128.5 oC (401.65 K) by using “Goal Seek” method in Excel and shown below.

Tr ,i =

T , T [ K ] , Tc [ K ] Tc ,i

401.65 401.65 401.65 = 0.784 = , Tr ,2 = 0.778 = , Tr ,3 = 0.730 512.58 516.25 550.00 401.65 401.65 , Tr ,5 = 0.685 = Tr ,4 = 0.713 = 562.93 586.15

= Tr ,1

1 + ( 0.480 + 1.574ωi − 0.176ωi2 )(1 − Tr1,i2 )  α SRK ,i =  

2

Tc ,i , ωi were taken in Reference 2 and tabulated in Table 10 1 ( 0.480 + 1.574 × 0.566 − 0.176 × 0.5662 )(1 − 0.7841 2 )  = α SRK ,1 =+   1.325 2

1 ( 0.480 + 1.574 × 0.637 − 0.176 × 0.637 2 )(1 − 0.7781 2 )  = 1.361 α SRK ,2 =+   2

1 ( 0.480 + 1.574 × 0.604 − 0.176 × 0.6042 )(1 − 0.7301 2 )  = α SRK ,3 =+   1.437 2

1 ( 0.480 + 1.574 × 0.595 − 0.176 × 0.5952 )(1 − 0.7131 2 )  = α SRK ,4 =+   1.465 2

1 ( 0.480 + 1.574 × 0.594 − 0.176 × 0.5942 )(1 − 0.6851 2 )  = α SRK ,5 =+   1.520 2

ai

α i R 2Tc ,i , Tc ,i [ K ] , Pc ,i [bar ] , R 83.14 cm3bar molK ψ= Pc ,i

ψ was taken in Reference 2 and tabulated in Table 10 1.325 × 83.142 × 512.58 0.42748 × 1.27 ×107 = a1 = 80.96 1.361× 83.142 × 516.25 0.42748 × 1.68 ×107 = a2 = 63.84 1.437 × 83.142 × 550.00 = a3 0.42748 × = 3.31×107 38.80 1.465 × 83.142 × 562.93 a4 = 0.42748 × = 3.11×107 44.13 1.520 × 83.142 × 586.93 a5 = 0.42748 × 3.98 ×107 = 38.80

- 36 -

RT 83.14 cm3bar molK bi = Tc ,i [ K ] , Pc ,i [bar ] , R = Ω c ,i , Pc ,i Ω was taken in Reference 2 and tabulated in Table 10 83.14 × 512.58 45.61 = 80.96 83.14 × 516.25 0.08664 × 58.25 = = 63.84 83.14 × 550.00 0.08664 × 102.11 = = 38.80 83.14 × 562.93 = 91.89 = 0.08664 × 44.13 83.14 × 586.15 = 0.08664 × = 108.82 38.80

0.08664 × b1 = b2 b3 b4 b5

bi P , T [ K ] , P [bar ] , R = 83.14 cm3bar molK RT 45.61×1.013 58.25 ×1.013 = = = 0.00177 β1 = 0.00138 , β2 83.14 × 401.65 83.14 × 401.65 102.11×1.013 91.89 ×1.013 = = = 0.00279 , β4 β3 = 0.00310 83.14 × 401.65 83.14 × 401.65 108.82 ×1.013 = 0.00330 β5 = 83.14 × 401.65

βi ≡

qi ≡

ai , bi RT

T [ K ] , R = 83.14 cm3bar molK

1.27 ×107 1.68 ×107 = = 8.341 , q2 ≡ 8.629 45.61× 83.14 × 401.65 58.25 × 83.14 × 401.65 3.31× 107 3.11× 107 = = q3 ≡ 9.709 , q4 ≡ 10.130 102.11× 83.14 × 401.65 91.89 × 83.14 × 401.65 3.98 ×107 q5 ≡ = 10.946 108.82 × 83.14 × 401.65 Z iv − βi 1 βi − qi βi Z iv =+ ( Ziv + εβi )( Ziv + σβi ) q1 ≡

0= 1 + 0.00138 − 8.341× 0.00138

Z iv − 0.00138 − Z iv v v ( Zi + 0 × 0.00138)( Zi + 1× 0.00138)

Z iv value for methanol was calculated as 0.9892 by above equation from Goal Seek in Excel And for the other components, Z iv values were tabulated in Table 13 20, 27.

- 37 -

I iv =

Z v + σβi 1 ln iv σ − ε Z i + εβi

1 0.9892 + 1× 0.00138 1 0.9853 + 1× 0.00177 = = ln 0.00140 , I 2v = ln 0.00179 1− 0 0.9892 + 0 1− 0 0.9853 + 0 1 0.9704 + 1× 0.00310 1 0.9721 + 1× 0.00279 = = 0.00319 = = 0.00286 ln , I 4v ln I 3v 1− 0 0.9704 + 0 1− 0 0.9721 + 0 1 0.9638 + 1× 0.00330 ln = 0.00342 I 5v = 1− 0 0.9638 + 0 ln Φ iv = Z iv − 1 − ln ( Z iv − βi ) − qi I iv ; Φ iv = exp  Z iv − 1 − ln ( Z iv − βi ) − qi I iv  I1v

0.00140  0.990 = Φ1v exp 0.9892 − 1 − ln ( 0.9892 − 0.00138 ) − 8.341× = = Φ v2 exp 0.9853 − 1 − ln ( 0.9853 − 0.00177 ) − 8.629 × 0.00179 =  0.987 0.973 = Φ 3v exp 0.9704 − 1 − ln ( 0.9704 − 0.00310 ) − 9.709 × 0.00319  = = Φ v4 exp 0.9721 − 1 − ln ( 0.9721 − 0.00279 ) − 10.130 × 0.00286 =  0.975 = Φ 5v exp 0.9638 − 1 − ln ( 0.9638 − 0.00330 ) − 10.946 × 0.00342 =  0.967

Z il =βi + ( Z il + εβi

)( Z

l i

 1 + βi − Z il  + σβi )    qi βi 

 1 + 0.00138 − Z il  l = 0 0.00138 + ( Z + 0 × 0.00138 )( Z + 1× 0.00138 )   − Zi  8.341× 0.00138  Z il value for methanol was calculated as 0.00169 by above equation from Goal Seek in Excel l i

l i

And for the other components, Z il values were tabulated in Table 14, 21, 28. I il =

Z l + σβi 1 ln il σ − ε Z i + εβi

1 0.00169 + 1× 0.00138 1 0.00245 + 1× 0.00177 ln , I 2l ln = 0.599 = = 0.543 1− 0 0.00169 + 0 1− 0 0.00245 + 0 1 0.00425 + 1× 0.00310 1 0.00365 + 1× 0.00279 ln 0.548 , I 4v = ln 0.567 = I 3v = 1− 0 0.00425 + 0 1− 0 0.00365 + 0 1 0.00420 + 1× 0.00330 ln = 0.580 I 5v = 1− 0 0.00420 + 0 ln Φ li = Z il − 1 − ln ( Z il − βi ) − qi I il ; Φ li = exp  Z il − 1 − ln ( Z il − βi ) − qi I il 

= I1l

= Φ1l exp 0.00169 − 1 − ln ( 0.00169 − 0.00138 ) − 8.341 = × 0.599  8.223 = Φ l2 exp 0.00245 − 1 − ln ( 0.00245 − 0.00177 ) − 8.629 = × 0.543 4.982 = Φ 3l exp 0.00425 − 1 − ln ( 0.00425 − 0.00310 ) − 9.709 × 0.548 = 1.576 = Φ l4 exp 0.00365 − 1 − ln ( 0.00365 − 0.00279 ) − 10.130 = × 0.567  1.365 = Φ 5l exp 0.00420 − 1 − ln ( 0.00420 − 0.00330 ) − 10.946 = × 0.580  0.719

- 38 -

= Ki

Φ li Φ iv

, = K1

8.223 4.982 = 8.31 , = K 2 = 5.05 0.990 0.987

1.576 1.365 0.719 1.62 , = 1.40 , = 0.74 = = K3 = K4 = K5 0.973 0.975 0.967 = αi

Ki K HK

= α methanol,n-butanol

,

K methanol 8.31 = = 5.93 K n-butanol 1.40

= α ethanol ,n-butanol

K ethanol 5.05 = = 3.60 , K n-butanol 1.40

α

=

n-butanol , n-butanol

yi =K i xi

K n-butanol 1.40 = = 1 K n-butanol 1.40

,

= α neopentanol,n-butanol

α 1- pentanol,n-butanol =

y1 =8.31× 0.008 =0.068 ,

,

y3 =1.62 × 0.034 =0.056 , fi v = yi Φ iv P ,

K neopentanol 1.62 = = 1.16 K n-butanol 1.40 K1- pentanol 0.74 = = 0.53 K n-butanol 1.40

y2 =5.05 × 0.027 =0.139

y4 =1.40 × 0.391 =0.548 ,

y5 =0.74 × 0.538 =0.400

f1v = 0.068 × 0.99 ×1.013 = 0.0687

f 2v = 0.139 × 0.987 ×1.013 = 0.1386 ,

f3v = 0.056 × 0.973 × 1.013 = 0.0548

f 4v = 0.548 × 0.975 ×1.013 = 0.5415 ,

f5v = 0.400 × 0.967 ×1.013 = 0.3921

fi l = xi Φ li Pt

,

f1l = 0.008 × 8.223 × 1.013 = 0.0687

f 2l = 0.027 × 4.982 ×1.013 = 0.1386 ,

f3l = 0.034 ×1.576 ×1.013 = 0.0548

f 4l = 0.391×1.365 ×1.013 = 0.5415 ,

f5l = 0.538 × 0.719 ×1.013 = 0.3921

In Vapor-Liquid Equilibrium, fugacity of liquid and vapor must be equal to each other. So, our results this condition. ( fi l = fi v )

B T in oC ; P in mmHg ( A, B and C constants were shown inTable 1) T +C Tbubble = 128.5 oC (bottom stream) log P sat= A −

Pt = 760mmHg sat = log Pmethanol 7.87863 −

1473.11 128.5 + 230

;

sat = Pmethanol 5881.758 mmHg ×

For other components see in Table 17, 24, 31.

- 39 -

1 atm 1.013 bar × = 7.840 bar 760 mmHg 1 atm

sat , γi yi Φ iv Pt xi γ i P= = i

yi Φ iv Pt xi Pi sat

0.068 × 0.99 ×1.013 0.139 × 0.987 ×1.013 , γ2 = 0.015 = = 0.036 0.008 × 7.840 0.027 × 5.572 0.056 × 0.973 × 1.013 0.548 × 0.975 ×1.013 0.361 , γ 4 = 0.481 = γ3 = 0.034 ×1.685 0.391×1.464 0.400 × 0.967 ×1.013 = 1.801 γ5 = 0.538 × 0.732

γ1 =

c

∑x γ

With excel calculation (goal seek),

i =1

i

i

Pi sat

Pt

= 1.00 we found T bubble =128.5 oC

 Calculation of Average Relative Volatilities

α average,i= α ethanol = α n-butanol =

3 3

α top ,i × α bottom ,i × α feed ,i

,

4.38 × 4.08 × 3.60 = 4.01 , 3

1 × 1 × 1= 1

,

α methanol =

3

10.78 × 6.52 × 5.93= 7.47

α neopentanol = α 1- pentanol =

3 3

1.14 ×1.18 ×1.16 = 1.16

0.79 × 0.49 × 0.53 = 0.59

 Calculation of Minimum Number of Plates by Using Fenske Equation x  x  0.373 0.391  log  LK , D × HK ,W  × log  xHK , D xLK ,W  0.012 0.034    = = ; N min = 39.56 N min log α LK ,avg log1.16

 Calculation of q-Parameter We used saturated vapor for the feed stream; so q-parameter must be taken as zero (q=0).  Calculation of θ-Parameter

α i ,ave × x f ,i =1 − q , α HK < θ < α LK i =1 α i , ave − θ c

∑

α met ,ave × x f ,met α eth ,ave × x f ,eth α neo ,ave × x f ,neo α n −but ,ave × x f ,n −but α1− pent ,ave × x f ,1− pent + + + + − 1 + q =0 α met ,ave − θ α eth ,ave − θ α neo ,ave − θ α n −but ,ave − θ α1− pent ,ave − θ 7.47 × 0.15 4.01× 0.25 1.16 × 0.25 1× 0.15 0.59 × 0.20 + + + + − 1 + 0 =0 7.47 − θ 4.01 − θ 1.16 − θ 1−θ 0.59 − θ

θ = 1.051 - 40 -

 Calculation of Minimum Reflux, R Dmin by Using Underwood Equation

α ir × xD ,i = RD min + 1 i =1 α ir − θ c

∑

7.47 × 0.231 4.01× 0.377 1.16 × 0.373 1× 0.012 0.59 × 0.006 + + + + = −1 0 7.47 − 1.051 4.01 − 1.051 1.16 − 1.051 1 − 1.051 0.59 − 1.051 = 3.516

= RD min RD min

= = 5.274 3.516*1.5 RDactual R= D min * β1 = 3.516*1.8 = 6.328 RDactual R= D min * β 2 RDactual R= = = 3.516* 2.4 8.438 D min * β 3

 Calculation of Actual Plate Number by Using Gilliand Correlation R - RDmin 5.274 − 3.516 = = 0.280 5.274 + 1 R +1

= X

 1.805  1.805   = Y= 1 − exp 1.490 + 0.315 X − 0.1  = 1 − exp 1.490 + 0.315 × 0.280 − 0.376 X  (0.280)0.1    N − N min Y + N min = ⇒N N +1 1− Y 0.376 + 39.56 = = 64 plates N N= actual 1 − 0.376

= Y

For other reflux ratios, the calculated values were tabulated in Table 34.

 Calculation of N R and N S for the Feed Location by Using Kirk Bridge Method  W x  NR    f . HK log = =  0.206 log     D   x f , LK  NS   NR 1.25 , = N R N act − N S = NS N= S

N act 64 = = 28 NR 1.25 1 + +1 NS

,

2  36.4  0.15  0.034 2    xb , LK    0.206 log  = =     0.09691      xd , HK    63.6  0.25  0.012  

64 − 28 = 36 N= R

- 41 -

 Calculation of Condenser and Reboiler Loads, Q c and Q r F= W +D ,

L= G +W ,

L 63.6

= RD 5.274 =

L RD = D

L 335.4 kmol / h ,=

c

c

λ y and λ ∑= ∑λ

= λave,top

i i ave ,bottom =i 1 =i 1

G =+ L D

i

xi

( for results see in Table 33)

G= 335.4 + 63.6 = 399.00kmol / h

,

kmol kJ  1h 1000 mol  × 38.914 × = 4312.9 kJ s ( kW ) Qc  399.00 ,= × h mol  3600 s 1 kmol 

= Qc G * λave ,top

L−L L − 335.4 , q= 0= F 100 L= L= 335.4 , G =L − W =335.4 − 36.4 ,

q=

G =299.00 kmol / h

kmol kJ  1h 1000 mol  = = × Qr G * λave ,bottom , Qr  299.00 * 43.836 3640.9 kJ s ( kW ) × = h mol  3600 s 1 kmol   Calculation of Column Diameter and Height N

, y MW * y = ∑

MW =

i =1

i

Wi

Wi

xi K i

yWmethanol = xmethanol K methanol = 0.008 × 8.31 = 0.068

( for other components see Table 35)

MW = (32.04*0.068) + (46.07 *0.139) + (88.15*0.056) + (74.122*0.548) + (88.15*0.400) = 89.42 kg / kmol

 × W = 89.42 kg × 36.4 kmol × 1 h = 0.904 kg s W = MV kmol h 3600 s P × MW = ρ × R × T P × MW 1× 89.42 = = 2.715 kg m3 ρvap = R ×T 0.082 × (128.5 + 273.15) c

∑ρ x =

ρliq =

i =1

i i

791.8 × 0.008 + 789 × 0.027 + 812 × 0.034 + 809.8 × 0.391 + 814.4 × 0.538= 811.63 kg m3 1/2

 ρ − ρv  uv = −0.171* l + 0.27 * lt − 0.047 *  l   ρv 

(

2 t

)

1/2

 811.63 − 2.715  uv = −0.171*0.52 + 0.27 *0.5 − 0.047 *   = 0.781 m s 2.715  4W 4 × 0.904 = =0.737 m , H c =lt × ( N + 1) =0.5 × (64 + 1) =32.5 m Dc = πρvuv π × 2.715 × 0.781

(

)

For other reflux ratios the results were in Table 35.

- 42 -