Design of a Continuous Distillation Column for a Multi Component Mixture

REPORT TO DEP EPARTMENT ARTMENT 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 D ATE 

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.

i

TABLE OF  TABLE  OF CONTENTS  CONTENTS 

Summary

.......................................... ................................................................ ............................................ ............................................ ................................ .......... i

1.0 Introduction Introduction

.......................................... ................................................................ ............................................ .......................................... .................... 1

................................................................ ............................................ ............................................ ............................... ......... 2.0 Results ..........................................

2.1 Ideal System

............................................ .................................................................. ............................................ ................................... ............. 3

2.2 Real System (Non-Ideal) 3.0 Discussion and Conclusion 4.0 Nomenclature Nomenclature 5.0 References 6.0 Appendix

3

............................................ .................................................................. ...................................... ................ 6 ......................................... ............................................................... ......................................... ................... 14

............................................ .................................................................. ............................................ .................................... .............. 18

......................................... ............................................................... ............................................ ............................................ ...................... 19

........................................... ................................................................. ............................................ ............................................ ...................... 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 contacting 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 supports. 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 contai ns more than two components (multicomponent systems).

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

-1-

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 counter-currently down down the column. The vapor and liquid are   brought into contact 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.

-2-

2.0 RESULTS  2.1 Ideal System Table 1 . Properties of

 No

 Name

Components  Antoine Constants   A B

 M w,i  ρ i  w,i  3  [kg/kmol]   [kg/m  ] 

 Z i i  1 λ i   [%]  C   [kJ/mol] 

1

Methanol 

32.04

791.8

7.87863

1473.11

230

15

35.14

2

 Ethanol 

46.07

789

8.1122

1592.864

226.184

25

38.58

88.15

812

7.27679

1279.01

177.849

25

41.35

74.122

809.8

7.36366

1305.198

173.427

15

43.24

88.15

814.4

7.18246

1287.625

161.33

20

44.83

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

3 4

1-Pentanol 

5

Table 2 .

Calculated Values for the Feed Stream in Ideal System o

100

F [kmol/h] 

T dew  dew  [  C] 

113.66

 y 1

0.15

n1

15

K 1

5.14

 y 2

0.25

n2

25

K 2

3.50

 y 3

0.25

n3

25

K 3

1.02

 y 4 

0.15

n 4 

15

K 4  4 [ref.]   

0.86

 y 5

0.2

n5

20

K 5

0.42

Table 3 .

Calculated Values for the Top Product in Ideal System o

63.6

 D [kmol/h] 

T dew  dew  [  C] 

96.84

 P.R 1

0.98

n1

14.7

y1

0.231

K 1

3.10

 P.R 2

0.96

n2

24

y2

0.377

K 2

2.00

 P.R 3

0.95

n3

23.75

y3

0.373

K 3

0.55

 P.R 4 

0.05

n 4 

0.75

y 4 

0.012

K 4  4 [ref.]   

0.45

 P.R 5

0.02

n5

0.4

y5

0.006

K 5

0.21

-3-

Table 4 .

Calculated Values for the Bottom Product in Ideal System o

36.4

W [kmol/h] 

122.39

T bubble bubble [  C]

 P.R 1

0.02

n1

0.3

x 1

0.008

K 1

6.57

y1

 P.R 2

0.04

n2

1

x 2

0.027

K 2

4.59

y2

 P.R 3

0.05

n3

1.25

x 3

0.034

K 3

1.37

y3

 P.R 4 

0.95

n 4 

14.25

x 4 

0.391

 P.R 5

0.98

n5

19.6

x 5

0.538

Table 5 .

1.18

K 4  4  

0.126

0.047

y

4 

0.312

y5

Calculated Values of Vapor Pressures of Components in Ideal System

 sat 

 P  i  

[mmHg] Feed 

Top

 sat   Methanol  [mmHg] 

3909.69

2352.49

4991.92

 sat   Ethanol  [mmHg] 

2661.98

1517.50

3487.52

 sat   Neopentanol  [mmHg] 

774.99

417.46

1039.42

 sat  n-Butanol  [mmHg] 

656.70

342.30

894.23

 sat  1-Pentanol  [mmHg] 

316.28

156.67

440.62

 P 

 P   P 

 P   P 

Table 6 .

0.054

0.461

[ref.]

0.58

K 5

y i =K  x  i  i  i  i 

Bottom

Calculated Values of Relative Volatilities of Components in Ideal System  Relative Volatility ,α ri 

 No

 Name

1 2 3 4 5

Feed Stream

Top Product

Bottom Product

Average

 Methanol 

5.95

6.87

5.58

6.11

 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

 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

q

0

 λ ave,bottom [kJ/mol] 

43.836

θ   R D,min

-4-

1.075

 N  R /N  S 

1.25

3.577

 λ ave, top [kJ/mol] 

38.914

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

Table 8.

 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

 N  S 

25

23

20

 N  R

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/ [kmol/h]  h] 

341.25

409.50

546.00

[kmol/h] 

304.85

373.10

509.60

Q r  [kW] 

3712.1

4543.1

6205.2

 H c  [m] 

28.5

26.0

23.5

G 

Table 9 .

 Required Calculations for the Fluid Velocity and Diameter of Column in Ideal System

 MW avg  [kg/kmol]  dot 

W   [kg/s] 

73.34 0.742

3

2.261

 ρ liq [kg/m  ] 

3

811.63

 It [m] 

0.5

u v [m/s] 

0.856

 D c  [m] 

0.698

 ρ vap [kg/m  ] 

-5-

2.2 Real System (Non-Ideal) Table 10. Properties

 No

of Components

 

Name

c [K] 

 P c [bar]  T 

ω

ψ

ε

1

Methanol 

512.58

80.96

0.566

0.42748

0

2

 Ethanol 

516.25

63.84

0.637

 Z c 

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

5

1-Pentanol 

586.15

38.80

0.594

0.08664

Table 11.

100

T dew  dew  [  C] 

o

106.6

 y 1

0.15

n1

15

T r,1

0.741

 y 2

0.25

n2

25

T r,2

0.736

 y 3

0.25

n3

25

T r,3

0.690

 y 4 

0.15

n 4 

15

T r,4 

0.675

 y 5

0.2

n5

20

T r,5

0.648

Calculated Values of α SRK,i  , a i and b i for the Feed Stream in Non-Ideal System 7

α SRK,1

1.400

a1

1.34*10

α SRK,2

1.442

a2

1.78*10

α SRK,3

1.515

a3

3.49*10

α SRK,4 

1.542

a 4 

3.27*10

α SRK,5

1.598

a5

4.18*10

Table 13.

[cm3bar/molK] 

Calculated Values of y  , i n i and T r, r, i  for the Feed Stream in Non-Ideal System

F [kmol/h] 

Table 12.

R

7

7

7

7

b1

45.61

b2

58.25

b3

102.11

b 4 

91.89

b5

108.82

Calculated Values of  β   β i  , q i and Z v i for the Feed Stream in Non-Ideal System

 β 1

0.00146

q1

9.321

Z  1

v

0.9870

 β 2

0.00187

q2

9.672

Z  2

v

0.9826

 β 3

0.00328

q3

10.829

Z  3

v

0.9649

 β 4 

0.00295

q 4 

11.281

Z  4 

v

0.9670

 β 5

0.00349

q5

12.166

Z  5

v

0.9572

-6-

83.14

Table 14.

 I 

v

 I 

v

 I 

v

 I 

v

 I 

v

Calculated Values of I  v i , Φv i and Z l i  for the Feed Stream in Non-Ideal System v

1

0.00148

Φ

2

0.00190

Φ

3

0.00339

Φ

4 

0.00304

Φ

5

0.00364

Φ

Table 15.

v v v v

0.00395

Z 4 

l 

0.00345

l 

0.00370

Z 2

3

0.968

Z 3

4 

0.970

5

0.961

Z 5

Calculated Values of I  l i , Φl i and K i for the Feed Stream in Non-Ideal System

0.625

Φ2

l 

0.604

Φ3

 I 4 

l 

0.618

Φ 4 

l 

0.665

Φ5

Table 16.

l 

0.984

l 

 I 5

0.00215

2

Φ1

 I 3

l 

Z 1

0.660

 I 2

0.00157

0.988

l 

 I 1

l 

1

l 

7.605

K 1

7.70

l 

3.078

K 2

3.13

l 

0.791

K 3

0.82

l 

0.693

K 4 

0.71

l 

0.545

K 5

0.57

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

α r,1

10.78

x 1

0.019

f 

α r,2

4.38

x 2

0.080

f 

α r,3

1.14

x 3

0.306

f 

α r,4 

1.00

x 4 

0.210

f 

α r,5

0.79

x 5

0.352

f 

Table 17.

l 

0.1501

P  1 [bar] 

l 

0.2491

P  2 [bar] 

l 

0.2451

P  3 [bar] 

l 

0.1474

P  4  [bar] 

l 

0.1947

P  5 [bar] 

 f  2  f  3  f  4   f  5

v v v

0.1501

2

0.2491

3

0.2451

4 

0.1474

5

0.1947

 sat 

Calculated Values of f  i , P  i  , γ i and  for the Feed Stream in Non-Ideal System

l 

 f  1

v

1

 sat 

4.237

γ1

1.818

 sat 

2.822

γ2

1.105

 sat 

0.804

γ3

0.997

 sat 

0.672

γ 4 

1.044

 sat 

0.317

γ5

1.741

-7-

Table 18.

Calculated Values of y  , i n i and T r, r, i  for the Top Product in Non-Ideal System o

63.6

 D [kmol/h] 

96.98

T dew  dew  [  C] 

 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

n 4 

0.75

y 4 

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 7

α SRK,1

1.434

a1

1.38*10

α SRK,2

1.479

a2

1.82*10

α SRK,3

1.551

a3

3.57*10

α SRK,4 

1.578

a 4 

3.35*10

α SRK,5

1.633

a5

4.27*10

Table 20.

7

7

7

7

b1

45.61

b2

58.25

b3

102.11

b 4 

91.89

b5

108.82

v

Calculated Values of β i  , q i and Z  i for the Top Product in Non-Ideal System

 β 1

0.00150

q1

9.798

Z  1

v

0.9880

 β 2

0.00192

q2

10.181

Z  2

v

0.9828

 β 3

0.00336

q3

11.373

Z  3

v

0.9659

 β 4 

0.00302

q 4 

11.840

Z  4 

v

0.9673

 β 5

0.00358

q5

12.758

Z  5

v

0.9575

Table 21.

 I 

v

 I 

v

 I 

v

 I 

v

 I 

v

v

v

l 

Calculated Values of I  i , Φ i and Z i  for the Top Product in Non-Ideal System v

1

0.00152

Φ

2

0.00195

Φ

3

0.00347

Φ

4 

0.00312

Φ

5

0.00373

Φ

v v v v

l 

0.00187

l 

0.00255

l 

0.00426

Z 4 

l 

0.00385

l 

0.00430

1

0.987

Z 1

2

0.982

Z 2

3

0.965

Z 3

4 

0.967

5

0.958

Z 5

-8-

Table 22.

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

l 

0.590

Φ1

l 

0.560

Φ2

l 

0.582

Φ3

l 

0.580

Φ 4 

l 

0.606

Φ5

 I  1  I  2  I  3  I  4   I  5

Table 23.

l 

3.108

K 1

3.15

l 

1.935

K 2

1.97

l 

0.551

K 3

0.57

l 

0.467

K 4 

0.48

l 

0.226

K 5

0.24

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

α r,1

6.52

x 1

0.073

f 

α r,2

4.08

x 2

0.192

f 

α r,3

1.18

x 3

0.655

f 

α r,4 

1.00

x 4 

0.024

f 

α r,5

0.49

x 5

0.027

f 

Table 24.

0.2310

P  1 [bar] 

l 

0.3755

P  2 [bar] 

l 

0.3651

P  3 [bar] 

l 

0.0116

P  4  [bar] 

l 

0.0061

P  5 [bar] 

 f  2  f  3  f  4   f  5

Table 25.

v v v

0.2310

2

0.3755

3

0.3651

4 

0.0116

5

0.0061

Calculated Values of f  l i , P  sat i  , γ i and  for the Top Product in Non-Ideal System

l 

 f  1

v

1

 sat 

3.149

γ1

1.000

 sat 

2.032

γ2

0.964

 sat 

0.559

γ3

0.997

 sat 

0.459

γ 4 

1.031

 sat 

0.210

γ5

1.089

Calculated Values of x  ,  i , n i and T r,r, i  for the Bottom Product in Non-Ideal System o

36.4

W [kmol/h] 

128.5

T bubble bubble [  C] 

 P.R 1

0.02

n1

0.3

x 1

0.008

T r,1

0.784

 P.R 2

0.04

n2

1

x 2

0.027

T r,2

0.778

 P.R 3

0.05

n3

1.25

x 3

0.034

T r,3

0.730

 P.R 4 

0.95

n 4 

14.25

x 4 

0.391

T r,4 

0.713

 P.R 5

0.98

n5

19.6

x 5

0.538

T r,5

0.685

-9-

Table 26.

Calculated Values of α SRK,i  , a i and b i for the Bottom Product in Non-Ideal System 7

α SRK,1

1.325

a1

1.27*10

α SRK,2

1.361

a2

1.68*10

α SRK,3

1.437

a3

3.31*10

α SRK,4 

1.465

a 4 

3.11*10

α SRK,5

1.520

a5

3.98*10

Table 27.

7

7

7

7

b1

45.61

b2

58.25

b3

102.11

b 4 

91.89

b5

108.82

Calculated Values of β i  , q i and Z v i for the Bottom Product in Non-Ideal System

 β 1

0.00138

q1

8.341

Z  1

v

0.9892

 β 2

0.00177

q2

8.629

Z  2

v

0.9853

 β 3

0.00310

q3

9.709

Z  3

v

0.9704

 β 4 

0.00279

q 4 

10.130

Z  4 

v

0.9721

 β 5

0.00330

q5

10.946

Z  5

v

0.9638

Table 28.

 I 

v

 I 

v

 I 

v

 I 

v

 I 

v

Calculated Values of I  v i , Φv i and Z l i  for the Bottom Product in Non-Ideal System v

1

0.00140

Φ

2

0.00179

Φ

3

0.00319

Φ

4 

0.00286

Φ

5

0.00342

Φ

Table 29.

v v v v

0.599

Φ1

l 

0.543

Φ2

l 

0.548

Φ3

l 

0.567

Φ 4 

l 

0.580

Φ5

 I  2  I  3  I  4   I  5

0.00169

l 

0.00245

l 

0.00425

Z 4 

l 

0.00365

l 

0.00420

0.990

Z 1

2

0.987

Z 2

3

0.973

Z 3

4 

0.975

5

0.967

Z 5

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

l 

 I  1

l 

1

l 

8.223

K 1

8.31

l 

4.982

K 2

5.05

l 

1.576

K 3

1.62

l 

1.365

K 4 

1.40

l 

0.719

K 5

0.74

- 10 -

Table 30.

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

α r,1

5.93

y1

0.068

f 

α r,2

3.60

y2

0.139

f 

α r,3

1.16

y3

0.056

f 

α r,4 

1.00

y 4 

0.548

f 

α r,5

0.53

y5

0.400

f 

Table 31.

v v v v

1

0.0687

2

0.1386

3

0.0548

4 

0.5415

5

0.3921

Calculated Values of f  l i , P  sat i  , γ i and  for the Bottom Product in Non-Ideal System

l 

0.0687

P  1 [bar] 

l 

0.1386

P  2 [bar] 

l 

0.0548

P  3 [bar] 

l 

0.5415

P  4  [bar] 

l 

0.3921

P  5 [bar] 

 f  1  f  2  f  3  f  4   f  5

 sat 

7.840

γ1

0.015

 sat 

5.572

γ2

0.036

 sat 

1.685

γ3

0.361

 sat 

1.464

γ 4 

0.481

 sat 

0.732

γ5

1.801

Table 32.

Calculated Values of Relative Volatilities of Components in Non-Ideal System

 No

 Name

1 2

 Relative Volatility ,α ri 

3 4 5

Feed Stream

Top Product

Bottom Product

Average

 Methanol 

10.78

6.52

5.93

7.47

 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

 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

q

0

 λ ave, bottom  [kJ/mol] 

43.836

θ   R D,min

- 11 -

1.051

 N  R /N  S 

1.25

3.516

 λ ave, top  [kJ/mol] 

38.914

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

 N  S 

28

26

24

 N  R

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/ [kmol/h]  h] 

335.40

402.48

536.64

[kmol/h] 

299.00

366.08

500.24

Q r  [kW] 

3640.9

4457.7

6091.3

 H c  [m] 

32.5

30.0

27.0

G 

Table 35.

 Required Calculations for the Fluid Velocity and Diameter of Column in Non-Ideal System

 MW avg  [kg/kmol]  dot 

W   [kg/s] 

89.42 0.904

3

2.715

 ρ liq [kg/m  ] 

3

811.63

 It [m] 

0.5

u v [m/s] 

0.781

 D c  [m] 

0.737

 ρ vap [kg/m  ] 

- 12 -

Table 36. 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

 N  S 

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*10

Q r  [kJ/h] 

1.0981*10

7

7

-1.9449*10

7

7

-2.5049*10

7

7

7

-1.7721*10

-2.0733*10

7

7

7

-2.6756*10

1.5182*10

2.0782*10

1.3445*10

1.6457*10

2.2480*10

 R D,min

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 -

7

7

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 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 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 both the bottom and top products. These are the two key components. The heavy non-key components components will essentially all end up in the bottom product while the light non-key components will will essentially all end up in the top product. This is why the separation variables described above are given in terms of the key components. In our process, the components were methanol, ethanol, neopentanol, n-butanol and 1  pentanol. 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 constants for the feed, a sat temperature was assumed. The P values were calculated at this temperature, by using Antoine equation (See Table 5). The equilibrium constants were calculated with this assumed tem temperatu raturre (S (See Table 2) 2). By By us using Go Goal-s al-seeek, ek, we we eq equaliz alizeed

. Th The de dew-point int

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 P sat values were calculated at this temperature, by using Antoine equation (See Table 5).The equilibrium constants were calc calcu ulate lated d wit with h thi thiss as assumed umed tem tempera peratu ture re.. By By usi using ng Goaloal-se seek ek,, we we equ equal aliz ized ed

. Th 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 P sat 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 Nactual 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 26 th 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 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 I v i and Φv i were determined by using the related equations in Ref.2.Same procedure was followed for Z L 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 γi 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 min /R. Also, height of the column decreases with increasing R min /R. Both Q c and Q r  increase with increasing R min 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 of i Component

Psat

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

N R 

Number of Plate of Enriching Section

R D,min

Minimum Reflux Ratio

R D,act

Actual Reflux Ratio

λ i

Latent Heat of Vaporization

MW

Molecular Weight of Mixing

uv

Fluid Velocity

[m/s]

Dc

Diameter of Distillation Column

[m]

T r 

Reduce Temperature

f

Fugacity

Φ

Fugacity Coefficient

T

Temperature

[oC]

Tc

Critical Temperature

[K]

Pc

Critical Pressure

[bar]

Q r 

Reboiler Load

[kJ/s]

Qc

Condenser Load

[kJ/s]

- 18 -

5.0 REFERENCES 

1.

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

2.

J.M. Smith, H.C. Van Ness, M.M. Abbott, Abbott,  Introduction to Chemical Engineering  th Thermodynamics , 2005, 7 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 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

× P.Ri ni = n Fi Fi n1 =15 × 0.98 = 14 14.7 kmol / h n2 = 25 × 0.96 = 24 2 4 kmol / h n3 = 25 × 0.95 = 23 23.75 kmol / h n4 = 15 × 0.05 = 0.75 kmol / h n5 = 20 × 0.02 = 0.4 kmol / h

- 20 -

 x1 =14.7 / 63.6 = 0.231

xdi = ni / D

 x2 = 24 / 63.6 = 0.377

 D = 14.7 + 24 + 23.75 + 0.75 + 0.4

 x3 = 23.75 / 63.6 = 0.373

D= 63.6 kmo/l h

 x4 = 0.75 / 63.6 = 0.012  x5 = 0.4 / 63.6 = 0.006



 Assumption of Percentage Recoveries for the Bottom Product 

ni = n Fi Fi × P.Ri

 P.R.1 =1 − 0.98 = 0.02

n1 = 15 × 0.02 = 0. 0 .3kmol / h

 P.R.2 = 1 − 0.96 = 0.04

n2 = 25 × 0.04 = 1kmol / h

 P.R.3 =1 − 0.95 = 0.05

n3 = 25 × 0.05 =1.25 kmol / h

 P.R.4 = 1 − 0.05 = 0.95

n4 = 15 × 0.95 = 14 14.25 kmol / h

 P.R.5 = 1 − 0.02 = 0.98

n5 = 20 × 0.98 = 19 19.6 kmol / h  x1 = 0.3/ 36.4 36.4 = 0.00 0.008 8

xdi = ni / W  

 x2 = 1/ 36.4 = 0.027

W  = 0.3 + 1 + 1.25 + 14.25 +19.6

 x3 =1.25 1.25 / 36.4 36.4 = 0.03 0.034 4

W = 36.4kmol / h

 x4 = 14.2 14.25 5 / 36.4 36.4 = 0.3 0.391  x5 = 19.6 19.6 / 36.4 36.4 = 0.53 .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

∑

  by using this temperature in excel. However  xi = i =1

c

 yi

∑  K  i =1

= 1.00

value could not be

i

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

∑ x = ∑  K  = 1.00 i

i

i

i

 B log P = A−sat T in C; P oin mmHg ( A, B an a nd C c on onstants we w ere sh s hown in in Ta Table 2) T + C  Tdew = 113.66 o C t 

( feed stream)

= P 760 mmHg  

log

methanol

at 7.87863s− =P

1473.11 113. 113.66 66 + 230 230

;

=P3909.69sat

methanol  

  For other components see in Table 5.

- 21 -

 

mmHg

 

 K i =

sat

 Pi

 K ethanol =

 sat   P ethanol 

=

 Pt

 K n-butanol = αi =

K methanol =  

,

 Pt

2661.98 760

 sat   P n-butanol 

 P t 

 K i

,

 K

=

P methanol P t 

656.70

 K n-butanol

α  n-butanol ,n-butanol  =

 K n-butanol   K n-butanol

3.5 0.86

=

K 1-pentanol =  

K methanol

= 

K 

= 4.05 ,

0.86 0.86

=1

= 5.14

K neopentanol = 

,

α methanol,n-butanol  =

=

760

= 0.86 ,

760

 K ethanol 

=

 

3909.69

= 3.5

HK

α ethanol ,n-butanol =

sat  

 sat   P neopentanol 

P t 

 P 1 s−at   pentanol   P t 

=

774.99 760

316.28 760

= 1.02

= 0.42

5.14

= 5.95 0.86 n -butanol   α neopentanol,n-butanol   = α 1- pentanol,n-butanol  =

,

=

 K neopentanol  K n-butanol

 K 1- pentanol  pentanol  K n-butanol

=  

=  

1.02 0.86

0.42

= 1.18

= 0.48

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

using this temperature in excel. However

∑

 xi =

i =1

c

 yi

∑  K  i =1

= 1.00 value could not be obtained.

i

o

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

∑ x = ∑  K  = 1.00 i

i

i

i

 B log P = A−sat T in C; P oin mmHg ( A, B an a nd C c on onstants we w ere sh s hown in in Ta Table 2) T + C  Tdew = 96.84 o C t 

(top stream)

= P 760 mmHg  

log

7.87863sa−t methanol =P

1473.11 96.8 96.84 4 + 230 230

;

=P2352.49sat

methanol  

 For other components see in Table 5.

- 22 -

 

mmHg

 

 K i =

sat

 Pi

 K ethanol =

 sat   P ethanol  1517.50

=

 Pt

 K n-butanol =

αi =

K methanol =  

,

 Pt

760

 sat   P n-butanol 

 Pt 

 K i

P t 

342.30 760

 K ethanol   K n-butanol

α  n-butanol ,n-butanol  =

=

 K n-butanol   K n-butanol

=

  2352.49

2 0.45

=

760

K 1-pentanol =  

K methanol

= 

K 

= 4.43 ,

0.45 0.45

=1

= 3.1

K neopentanol = 

,

= 0.45 ,

HK

α ethanol ,n-butanol =

sat  

=2

α methanol,n-butanol  =

,

 K

=

P methanol

,

 sat   P neopentanol 

P t 

 P 1 sat  − pentanol 

=

=

P t 

417.46 760

156.67 760

= 0.55

= 0.21

3.10

= 6.87 0.45 n -butanol   α neopentanol,n-butanol   =

 K neopentanol  K n-butanol

 K 1-pentanol 

α 1- pentanol,n-butanol  =

K n-butanol

=  

=  

0.55 0.45

0.21 0.45

= 1.22

= 0.46

0

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



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

c

∑ y = ∑ x K  = 1.00 value could not be i

i =1

i

i

i =1

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

∑ y = ∑ x K  = 1.00 i

i

i

i

i

sat  B log P = A− T in C; P oin mmHg ( A, B an a nd C c on onstants w er ere sh s hown in in Ta Table 2) T + C 

Tbubble = 122.39 o C t 

= P 760

log

methanol

(bottom stream)

mmHg   at =P7.87863s−

1473.11 122. 122.39 39 + 230 230

;

=P4991.92sat

methanol  

 For other  components see in Table 5.

- 23 -

 

mmHg

 

 K i =

sat

 Pi

 K ethanol =

 sat   P ethanol 

 K n-butanol =

αi =

 K i  K

K methanol =  

,

 Pt

=

 Pt

3487.52 760

 sat   P n-butanol 

 P t  ,

=

P methanol P t 

894.23 760

 K n-butanol

α  n-butanol ,n-butanol  =

=

 K n-but anol   K n-butanol

760

4.59

=

K 1-pentanol =  

K methanol

= 

K 

= 3.90 ,

1.18

1.18 1.18

=1

= 6.57 K neopentanol = 

,

= 1.18 ,

α methanol,n-butanol  =

 K ethanol 

=

 

4991.92

= 4.59

HK

α ethanol ,n-butanol =

sat  

,

 sat   P neopentanol  1039.42

=

P t 

 P 1 sat  − pentanol   P t 

=

760

440.62

= 1.37

= 0.58

760

6.57

= 5.58 1.18 n -butanol   α neopentanol,n-butanol   = α 1- pentanol,n-butanol  =

 K neopentanol  K n-butanol

 K 1- pentanol  pentanol  K n-butanol

=  

=  

1.37 1.18

0.58 1.18

= 1.16

= 0.49

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



Calculation of Average Relative Volatilities

α average ,i = 3 α top ,i × α bottom ,i × α feed ,i

,

α methan ol   = 3 5.95 × 6.87 × 5.58 = 6.11

α ethanol = 3 4.05 × 4.43 × 3.9 = 4.12

,

3 α neopentanol = 1.18 ×1.22 ×1.16 = 1. 1.19  

α n-butanol = 3 1×1×1 = 1

,

α 1- pentanol  = 3 0.48 × 0.46 × 0.49 = 0.48



 N min =



Calculation of Minimum Number of Plates by Using Fenske Equation

 x  x log  LK, D × HK, W    x HK, D x LK, W   log α   LK, avg 

;

N min =

0.373 3 0.39 0.391 1  0.37 × 0.012 2 0.03 0.034 4   0.01 = 34.33

log 

log 1.19

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  -Parameter 

c

α i , ave × x f ,i

i =1

α i ,ave − θ 

∑

α met ,ave × fx,met α met ,ave − θ

6.11× 0.15 6.11 − θ

= 1− q ,

+

α eth ,ave × fx,eth α eth ,ave − θ

4.12 × 0.25

+

4.12 − θ

+

< θ < α 

α

HK

+

LK  

α neo ,ave × fx,neo α neo ,ave − θ

1.19× 0.25 1.19 − θ 

+

+

1× 0.15 1−θ

α n−but ,ave × fx,n−but α n−but ,ave − θ

+

0.48 × 0.2 0.20 0.48 − θ 

+

α 1− pent ,ave ×   fx,1− pent   α1− pent ,ave − θ

− 1+ q = 0

 

− 1+ 0 = 0

θ  = 1.075



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

c

α ir × xD ,i

i =1

α ir  − θ 

∑

 R D min =

6.11× 0.231

6.11 − 1.075 = 3.577

 R D min

4.12 × 0.377 4.12 − 1.075

1.19 × 0.373 1.19 −1.075

* β 1 = 3.577 *1 *1.5 = 5. 5.366

=R

mDin

* β 2 = 3.577*1. 7*1.8 = 6.43 6.439 9

=R

mDin

* β 3 = 3.577 * 2. 2.4 = 8. 8.585

 R

Dactual

 R

Dactual

+

1 × 0.012 1 −1.075

+

0.48 ×0.006 0.48 −1.075

−1= 0

Calculation of Actual Plate Number by Using Gilliand Correlation

  R - R Dmin  R+1

=

5.36 5.366 6 − 3.57 3.577 7 5.366 .366 + 1

 

Y = 1 − exp 1.490 + 0.315 X  −

Y=

+

mDin

Dactual



+

=R

 R

 X  =

=  R D min + 1

N − Nmin  N + 1

= Nactual N=

⇒ N  =

= 0.281

 1.805  = − + × − = 0.376 1 e x p 1 . 4 9 0 0 . 3 1 5 0 . 2 8 1  0.1   (0.281)  

1.805 

 X 0.1

Y + Nmin 

1 − Y  0.37 0.376 6 + 34.3 34.33 3 1 − 0.37 .376

= 56

plates

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  N  R      f . HK    xb , LK   log =  0 . 2 0 6 l o g =      D  xf , LK   xd , HK  NS    N  R

= 1.25

 N S 

 N S =

N R= N ac− t N S 

,

 N act  56 = = 25  N  R 1.25 + 1 +1  N S 



= 5.366 = R

λave ,top =

N R = 56 − 25 = 31

,

L= G+W  L

c

∑λ

i

G = L+ D

yi and λave ,bottom =

q=0=

,

y

i

i

kmol h

× 38.914

,

 

L − 341.25 100

G = L − W = 341.25 − 36.4

,

1h 1 000 mol   kJ  × × = 4376.1 kJ s ( kW )  1 kmol   mol  3600 s

 

Qr =   304.85

kmol h

,

G = 304.85 kmol / h

* 43.836

1h 1 000 mol   kJ  × × = 3712.1 kJ s ( kW )  1 kmol   mol  3600 s

Calculation of Column Diameter and Height   N 

∑ MW * y i

i =1 Wmethanol 

results see in Table 7 ) ∑ λ x   ( for re

 

Qr = G * λ ave ,bottom

MW =

c

Qc =  404.85

,

 L = L = 341.25



km/ol h

G = 341.25 + 63.6 = 404.85 kmol / h

Qc = G * λ ave ,top

 F 

 D

i =1

,

L−L

 L

=L341.25

,

63.6

R D =

,

i =1

q=

 

Calculation of Condenser and Reboiler Loads, Q c and Q r 

F=W+ D ,

 D

2  36.4   0.15   0.034 2       = 0.206 log  63.6   0.25   0.012   = 0.09691     

=

x

methanol

Wi

K

yWi = xi K i  

,

methanol  

= 0.008 × 6.57 = 0.054

(

for other components see Tabl9e)

MW = (32. (32.04*0.0 04*0.054 54)) + (46.0 (46.07* 7* 0.12 0.126) 6) + (88. (88.15*0.0 15*0.047 47)) + (74.1 (74.122*0.4 22*0.461 61)) + (88. (88.15*0.3 15*0.312 12)) = 73.3 73.34 4 kg / kmol  W = MV * W  = 73.34 

kg  kmol

× 36.4

kmol h

×

1h 3600 s

= 0.742 kg s

P× MW =  ρ × R× T  

- 26 -

 

P× MW  

 ρ vap =  ρ liq =

=

 R × T  c

∑ ρ  x

i i

1× 73.3 73.34 4 0.082 0.082 × (122.39 122.39 + 273.15 273.15))

= 2.261 kg m3

= 791.8 91.8× 0.008 .008 + 789 789 × 0.027 .027 + 812 × 0.034 .034 + 809. 809.8 8 × 0.3 0.391 + 814.4 14.4 ×0.53 0.538 8

i =1

= 811.63 kg m3  ρ − ρ v  uv = ( −0.171* l + 0.27 * l t  − 0.047 )*  l    ρ v 

1/2

2 t

811.63 3 − 2.261 2.261   811.6 uv = ( −0.17 .171*0.5 1*0.5 + 0.27* .27* 0.5 − 0.047 .047 )*   2.261  2

4W  

D c =

πρv uv

=

4 × 0.742 π  × 2.261× 0.856

= 0.698 m ,

1/2

= 0.856 m  s

Hc = t l× ( N+ 1) = 0.5 × (56 + 1) = 28.3 m

 For other reflux ratios the r esults were in Table 9.

6.2 Real 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 

using this temperature in excel. However  P t  ×

c

 yi

∑ γ   P  i =1

*

i

= 1.00 value could not be obtained. So

i

o

the dew point temperature was calculated as 106.6 C (379.75 K) by using “Goal Seek” method in Excel and shown below.

Tr ,i = Tr ,1 = Tr , 4 =

T 

, T [ K ] , Tc [ K ]  

T c ,i

379.75 512.58 379.75 562.93

= 0.741 , Tr ,2 = = 0.675 , T r ,5 =

379.75 516.25 379.75 586.15

= 0.736 , T r ,3 =

379.75 550.00

= 0.690

= 0.648

α SRK ,i = 1 + ( 0.480 + 1.574ωi − 0.176ω i2 ) (1 − T r1,i2 )

2

Tc ,i , ω i wer were take taken n in Refe Referrence ence 2 and and tabu tabula late ted d in Tab Table 10 2

α SRK ,1 = 1 + ( 0.480 + 1. 1.574 × 0. 0.566 − 0.176 × 0. 0.566 2 ) (1 − 0. 0.7411 2 ) = 1.400 2

α SRK ,2 = 1 + ( 0.480 + 1.574 × 0.637 − 0.176 × 0.637 2 ) (1 − 0.736 1 2 ) = 1.442





2

α SRK ,3 = 1 + ( 0.480 + 1. 1.574 × 0. 0.604 − 0.176 × 0. 0.604 2 ) (1 − 0. 0.690 1 2 ) = 1.515





2

α SRK ,4 = 1 + ( 0.480 + 1. 1.574 × 0. 0.595 − 0 .1 .176 × 0. 0.595 2 ) (1 − 0. 0.675 1 2 ) = 1.542 2

α SRK ,5 = 1 + ( 0.480 + 1. 1.574 × 0. 0.594 − 0.176 × 0. 0.594 2 ) (1 − 0. 0.648 1 2 ) = 1.598





- 27 -

ai = ψ 

2 α i  R T c ,i

 P c ,i

Tc ,i [ K ] , Pc ,i [bar ] , R = 83.14 cm 3bar molK  

,

ψ  was taken taken in Refer Referenc encee 2 and tabul tabulate ated d in Table Table 10 a1 = 0.42748 × a2 = 0.42748 × a3 = 0.42748 × a4 = 0.42748 × a5 = 0.42748 ×

bi = Ω

 RT c ,i  P c ,i

,

1.40 1.400 0 × 83.1 83.14 4 2 × 512. 512.58 58 80.96 1.44 1.442 2 × 83.1 83.14 4 2 × 516. 516.25 25 63.84 1.515 1.515 × 83.14 83.142 × 550.00 38.80 1.54 1.542 2 × 83.1 83.14 4 2 × 562. 562.93 93 44.13 1.59 1.598 8 × 83.1 83.14 4 2 × 586. 586.93 93 38.80

= 1.34 ×10 7 = 1.78 ×10 7 3.49 ×107 = 3.49

= 3.27 ×10 7 = 4.18 ×10 7

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

Ω was taken taken in Refe Referen rence ce 2 and tabul tabulate ated d in Table Table 10 83.14 83.14 × 512.58 512.58 = 45.61 b1 = 0.08664 × b2 = 0.08664 × b3 = 0.08664 × b4 = 0.08664 × b5 = 0.08664 ×  β i ≡  β1 =  β3 =  β 5 =

bi P 

,

80.96 83.14 83.14 × 516.25 516.25 63.84 83.14 83.14 × 550.00 550.00 38.80 83.14 83.14 × 562.93 562.93 44.13 83.14 83.14 × 586.15 586.15 38.80

= 58.25 = 102.11 = 91.89 = 108.82

3 T [ K ] , P [bar ] , R = 83.14 cm bar molK  

 RT  45.61×1.013

83.14 × 379.75 102.11×1.013 83.14 × 379.75 108.82 108.82 ×1.013 1.013 83.14 83.14 × 379.75 379.75

= 0.00146 , β 2 = = 0.00328 , β 4 =

58.25 ×1.013 83.14 × 379.75 91.89 ×1.013 83.14 × 379.75

= 0.00349

- 28 -

= 0.00187 = 0.00295

qi ≡ q1 ≡ q3 ≡

ai

,

bi RT 

T [ K ] , R = 83.14 cm 3bar molK  

1.34 ×10 7 45.61× 83.14 × 379.75

= 9.321

3.49 ×10 7 102.11 × 83.14 × 379.75

, q2 ≡

= 10.829 , q4 ≡

1.78 ×10 7 58.25 × 83.14 ×379.75 3.27 ×10 7 91.89 ×83.14 ×379.75

= 9.672 = 11.281

4.18 4.18×107

q5 ≡

= 12.166 108. 108.82 82 × 83.1 83.14 4 × 3 79.75  Z iv − β i v  Z i = 1 + β i − qi β i ( Z iv + εβi ) ( Z iv + σβi  ) 0 = 1 + 0.0 0.00146 0146 − 9.32 .321 × 0.00 0.0014 146 6

 Z iv − 0.00146

( Z

v i

+ 0 × 0.00146 ) ( Z  + 1 × 0.00146 ) v i

− Z iv

  Ziv valu valuee for for met metha hano noll was was calc calcul ulat ated ed as as 0.987 by abov abovee equa equati tion on from from Goal Goal Seek Seek in Exce Excel  l    And And for for the the oth other er comp compon onen ents ts, Z iv values were tabulated in Table 14, 14, 21, 21, 28.  I  = v i

 I1v =  I 3v =  I 5v =

1 σ −ε 1 1− 0 1 1− 0 1 1− 0

ln

ln ln ln

 Z iv + σβ i  Z iv + εβ i  

0.9870 + 1 × 0.00146 0.9870 + 0 0.9649 + 1 × 0.00328 0.9649 + 0 0.95 0.9572 72 + 1× 0.00 0.0034 349 9 0.9572 + 0

= 0.00148 ,

I 2v =

= 0.00339 ,

I 4v =

1 1−0 1 1−0

ln ln

0.9826 +1 × 0.00187 0.9826 + 0 0.9670 +1 × 0.00295 0.9670 + 0

= 0.00190 = 0.00304

= 0.00364

Φ iv = exp  Ziv − 1 − ln ( Ziv − β i ) − qi Iiv    Φ1v = exp  0.9 0.987 870 0 − 1 − ln ( 0.9 0.98 870 − 0.0 0.00146 0146 ) − 9.321 .321 ×0.00 0.0014 148 8  = 0.988 .988 ln Φ iv = Ziv − 1 − ln ( Ziv − βi ) − qi Iiv

;

Φ 2v = exp 0. 0.9826 − 1 − ln ( 0. 0.9826 − 0.00186 ) − 9.672 × 0.00190  = 0.984 Φ 3v = exp 0.964 0.00339  = 0.968 .9649 9 − 1 − ln (0.96 0.9649 49 − 0.0 0.00328) − 10.829 × 0.0 Φ 4v = exp 0.96 0.9670 70 −1 − ln ( 0.9 0.967 670 0 − 0.002 .00295 95 ) −11.28 1.281 1 × 0.0 0.00304 0304  = 0.9 0.970 Φ 5v = exp  0.9 0.957 572 2 − 1 − ln ( 0.9 0.95 572 − 0.0 0.00349 0349 ) −12.16 2.166 6 × 0.003 .00364 64  = 0.961 .961 Z=

l i

i

β+ ( Z+

εi β) (

l i

0 = 0.00146 + (

l i

 1 + β i − Z il   Z+   σi β)      qi  β i 

l  i

+Z0 × 0.00146 ) (

l i

 1 + 0.00 0.0014 146 6 − Z il   +Z 1 × 0.00146 )  − × 9.321 9.3 21 0.0014 0.0 0146 6  

l  i

Z

 

  Zil  valu alue for for meth ethanol anol was was calc calcu ulate lated d as as 0.00157 by above bove equ equatio ation n fro from Goal Goal Seek Seek in Exc Excel  component entss , Z il  values values were were tabu tabulate lated d in Table Table 14, 21, 28.   And And for the the other  other  compon

- 29 -

 I  = l  i

 I1l =  I 3v =  I 5v =

1 σ −ε 1 1− 0 1 1− 0 1 1− 0

ln

ln ln ln

 Z il  + σβ i  Z il  + εβi  

0.00157 + 1 × 0.00146 0.00157 + 0 0.00395 + 1× 0.00328 0.00395 + 0 0.00 0.0037 370 0 + 1 × 0.00 0.0034 349 9

= 0.660 ,

I 2l  =

= 0.604 ,

I 4v =

1 1−0 1 1−0

ln

0.00215 +1 × 0.00187

ln

= 0.625

0.00215 + 0 0.00345 +1 × 0.00295

= 0.618

0.00345 + 0

= 0.665

0.00370 + 0

Φ il = exp  Zil − 1 − ln ( Zil − β i ) − qi Iil    Φ1l  = exp 0. 0.00157 −1 − ln ( 0. 0.00157 − 0.00146 ) − 9.321 × 0.660  = 7.605 ln Φ li = Zil − 1 − ln ( Zil − βi ) − qi Iil

;

 

Φ l 2 = exp 0.002 .00215 15 −1 − ln (0.0 0.00215 215 − 0.001 .00186 86 ) − 9.672 .672 × 0.625 .625  = 3.078 .078 Φ l 3 = exp 0.0 0.604  = 0.791 0.00395 395 −1 − ln (0.00 0.0039 395 5 − 0.0 0.0 0328 ) − 10.829 × 0.6 Φ l 4 = exp 0.003 .00345 45 −1 − ln ( 0.0 0.00 0345 345 − 0.002 .00295 95 ) −11.2 1.281 ×0.6 0.618  = 0.693 .693 Φ l 5 = exp 0. 0.0037 00370 0 −1 − ln ( 0.0 0.003 037 70 − 0.00 .00349 349 ) −12.16 2.166 6 ×0.6 0.665  = 0.545 .545

i

3

Φ l i = v K Φi =K

αi =

,

0.791 0.968  K i

 K

1

 yi

3

=x

 K i 0.25 0.82

0.988

=K

4

= 7.70 , 0.693

HK

 K ethanol   K n-butanol

,

 K n-butanol   K n-butanol 1

=x

= 0.306 ,

fi v = yi Φ iv P ,

=

3.13 0.71

=

0.15 7.70 4

K methanol

= 

K 

0.71

=1

= 0.019 , 0.15 0.71

= K

3.078   = 3.13 0.984

,

=K

5

0.545 0.961

=  0.57

7.70

= 10.78 0.71 n -butanol  

= 4.38 ,

0.71

=x

2

= 0.71 ,

0.970

α methanol,n-butanol  =

,

α  n-butanol ,n-butanol  =

=x

7.605

= 0.82 ,

α ethanol ,n-butanol =

i

= K

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

2

=x

= 0.210 ,

0.25 3.13 5

 K neopentanol  K n-butanol

 K 1- pentanol  pentanol  K n-butanol

=  

=  

0.71

0.57 0.71

= 0.08

=x

0.20

0.57

= 0.352

f1v  = 0  .15 × 0.988 ×1.013 = 0. 0.1501

 f 2v = 0.25 × 0.9 0.984 × 1. 1.013 = 0.2491 ,

f  3v = 0.25 ×0. 0.968 ×1. 1.013 = 0.2451

 f 4v = 0.15 × 0. 0.970 × 1. 1.013 = 0.1474

f  5v = 0.20 ×0. 0.961 ×1.013 = 0.1947

fi l = xi Φ il Pt   ,

0.82

,

f1l  =  0 .019 × 7.605 ×1.013 = 0.1501

 f 2l = 0.08 × 3. 3.078 ×1.013 = 0.2491 ,

f  3l  = 0.306 ×0.7 0.791 ×1.013 = 0.2451

 f 4l = 0.21× 0.693 ×1.013 = 0.1474

f  5l  = 0.352 ×0. 0.545 ×1.013 = 0.1947

,

- 30 -

= 1.14

= 0.79

In Vapor-Liquid Equilibrium, fugacity of liquid and vapor must be equal to each other. So, our results support this condition. (  f i l = f  i v )  B log P = A−sat T in C; P oin mmHg ( A, B an a nd C constants w er ere sh s hown in in Ta Table1) T + C  Tdew = 106.6 o C t 

= P 760

log

methanol

( feed stream)

mmHg  

= 7P.87863sa−t

1473.11 106.6 + 230

3178.72sat =P

;

1 atm

× mmHg

 

methanol  

760 mmHg

×

1.013 bar   1 atm

= 4.237

17, 24, 24, 31. 31.   For ot her components see in Table 17,

yi Φ P = xi γ i Pi   , *

v i t

γ1 = γ3 = γ 5 =

γ i =

0.15 × 0.988 ×1.013 0.019 × 4.237 0.25 × 0.968 ×1.013 0.306 × 0.804 0.20 0.20 × 0.96 0.961 1 ×1.01 1.013 3 0.352 × 0.31 0.317 7

 yi Φiv P t   xi P i *

= 1.818 ,

γ 2 =

= 0.997 ,

γ 4 =

0.080 × 2.822 0.25 × 0.970 ×1.013

= 1.105

0.210 × 0.672

= 1.044

= 1.741

With excel calculation (goal seek),  P t  ×

 yi

∑ γ   P 

sat 

i



0.15 × 0.984 ×1.013

i

= 1.00 we found T dew =106.6 oC

i

For Top Product 

Dew point temperature for the feed was assumed as 90 oC. All calculations was done by using this temperature in excel. However  P t  ×

c

 yi

∑ γ   P  i =1

sat 

i

= 1.00 value could not be obtained.

i

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 = Tr ,1 = Tr , 4 =

T  T c ,i

, T [ K ] , Tc [ K ]  

370.13 512.58 370.13 562.93

= 0.722 , Tr ,2 = = 0.658 , T r ,5 =

370.13 516.25 370.13 586.15

= 0.717 , T r ,3 = = 0.631

- 31 -

370.13 550.00

= 0.673

bar

α SRK ,i = 1 + ( 0.480 + 1.574ωi − 0.176ω i2 ) (1 − T r1,i2 )

2

Tc ,i , ω i wer were take taken n in Refer eferen ence ce 2 and and tabu tabula late ted d in Tab Table 10 2

α SRK ,1 = 1 + ( 0.480 + 1. 1.574 × 0. 0.566 − 0.176 × 0. 0.566 2 ) (1 − 0. 0.722 1 2 ) = 1.434 2

α SRK ,2 = 1 + ( 0.480 + 1.574 × 0.637 − 0.176 × 0.637 2 ) (1 − 0.717 1 2 ) = 1.479





2

α SRK ,3 = 1 + ( 0.480 + 1. 1.574 × 0. 0.604 − 0.176 × 0. 0.604 2 ) (1 − 0. 0.6731 2 ) = 1.551





2

α SRK ,4 = 1 + ( 0.480 + 1. 1.574 × 0. 0.595 − 0 .1 .176 × 0. 0.595 2 ) (1 − 0. 0.658 1 2 ) = 1.578 2

α SRK ,5 = 1 + ( 0.480 + 1. 1.574 × 0. 0.594 − 0.176 × 0. 0.594 2 ) (1 − 0. 0.6311 2 ) = 1.633 ai = ψ 

α i  R 2T c ,i  P c ,i

,

Tc ,i [ K ] , Pc ,i [bar ] , R = 83.14 cm 3bar molK  

taken in Refer Referenc encee 2 and tabul tabulate ated d in Table Table 10 ψ  was taken a1 = 0.42748 × a2 = 0.42748 × a3 = 0.42748 × a4 = 0.42748 × a5 = 0.42748 ×

bi = Ω

 RT c ,i  P c ,i

,

1.43 1.434 4 × 83.1 83.14 4 2 × 512. 512.58 58 80.96 1.47 1.479 9 × 83.1 83.14 4 2 × 516. 516.25 25 63.84 2 1.551 1.551× 83.14 83.14 × 550.00

38.80 1.57 1.578 8 × 83.1 83.14 4 2 × 562. 562.93 93 44.13 1.63 1.633 3 × 83.1 83.14 4 2 × 586. 586.93 93 38.80

= 1.38 ×10 7 = 1.82 ×10 7 3.57 ×107 = 3.57

= 3.35 ×10 7 = 4.27 ×10 7

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

Ω was taken taken in Refe Referen rence ce 2 and tabul tabulate ated d in Table Table 10 83.14 83.14 × 512.58 512.58 = 45.61 b1 = 0.08664 × b2 = 0.08664 × b3 = 0.08664 × b4 = 0.08664 × b5 = 0.08664 ×

80.96 83.14 83.14 × 516.25 516.25 63.84 83.14 83.14 × 550.00 550.00 38.80 83.14 83.14 × 562.93 562.93 44.13 83.14 83.14 × 586.15 586.15 38.80

= 58.25 = 102.11 = 91.89 = 108.82

- 32 -

 β i ≡  β1 =  β3 =  β 5 =

qi ≡ q1 ≡ q3 ≡

bi P 

T [ K ] , P [bar ] , R = 83.14 cm3bar molK  

,

 RT  45.61×1.013

83.14 × 370.13 102.11×1.013 83.14 × 370.13 108.82 108.82 ×1.013 1.013 83.14 83.14 × 370.13 370.13

ai

,

bi RT 

= 0.00150 , β 2 = = 0.00336 , β 4 =

58.25 ×1.013

= 0.00192

83.14 × 370.13 91.89 ×1.013

= 0.00302

83.14 × 370.13

= 0.00358

T [ K ] , R = 83.14 cm 3bar molK  

1.38 ×10 7 45.61× 83.14 × 370.13

= 9.798

3.57 ×10 7 102.11 × 83.14 × 370.13

, q2 ≡

= 11.373 , q4 ≡

1.82 ×10 7 58.25 ×83.14 ×370.13 3.35 ×10 7 91.89 ×83.14 ×370.13

= 10.181 = 11.840

4.27 4.27 ×107

q5 ≡

= 12.758 108.82 108.82 × 83.14 83.14 × 370.13  Z iv − β i v  Zi = 1 + β i − qi β i ( Ziv + εβi ) ( Z iv + σβi  ) 0 = 1 + 0.00 .00150 150 − 9.798 .798 × 0.00 .00150 150

 Z iv − 0.00150

( Z

v i

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

− Z iv

v

  Zi valu valuee for for meth methan anol ol was was calc calcul ulat ated ed as as 0.988 by abov abovee equa equati tion on from from Goal Goal Seek Seek in Exce Excel  l  13, 20, 20, 27. 27.   And And for the other other com compon ponen ents ts, Z iv values were tabulated in Table 13,  I  = v i

 I1v =  I 3v =  I 5v =

1 σ −ε 1 1− 0 1 1− 0 1 1− 0

ln

ln ln ln

 Z iv + σβ i  Z iv + εβ i  

0.9880 + 1 × 0.00150 0.9880 + 0 0.9659 + 1 × 0.00336 0.9659 + 0 0.95 0.9575 75 + 1× 0.00 0.0035 358 8 0.9575 + 0

= 0.00152 ,

I 2v =

= 0.00347 ,

I 4v =

1 1−0 1 1−0

ln ln

0.9828 +1 × 0.00192 0.9828 + 0 0.9673 +1 × 0.00302 0.9673 + 0

= 0.00373

Φ iv = exp  Ziv − 1 − ln ( Ziv − β i ) − qi Iiv    Φ1v = exp 0. 0.9880 − 1 − ln ( 0. 0.9880 − 0.00150 ) − 9.798 × 0.00152  = 0.987 ln Φ iv = Ziv − 1 − ln ( Ziv − βi ) − qi Iiv

;

Φ 2v = exp 0.98 0.9828 28 − 1 − ln ( 0.9 0.98 828 − 0.00 .00192 192 ) −10.1 10.181 81 ×0.00 0.0019 195 5  = 0.982 .982 Φ 3v = exp 0.965 .9659 9 − 1 − ln (0.96 0.9659 59 − 0.0 0.00 336 ) − 11.373 × 0.0 0.00347  = 0.965 Φ 4v = exp 0.96 .9673 −1 − ln ( 0.9 0.967 673 3 − 0.00 0.0030 302 2 ) −11.8 11.840 40 × 0.00 0.0031 312 2  = 0.9 0.967 Φ 5v = exp 0.9 0.957 575 5 −1 − ln ( 0.9 0.957 575 5 − 0.003 .00358 58 ) −12.7 12.75 58 × 0.00 .00373 373  = 0.95 0.958 8

- 33 -

= 0.00195 = 0.00312

Z=

l i

i

β+ ( Z+

εi β) (

l i

0 = 0.00150 + (

l i

 1 + β i − Z il   Z+   σi β)      qi  β i 

l  i

+Z0 × 0.00150 ) (

l i

 1 + 0.00 0.0015 150 0 − Z il   +Z 1 × 0.00150 )  − × 9.798 9.7 98 0.0015 0.0 0150 0  

l  i

Z

 

  Zil  valu alue for for meth ethanol anol was was calc calcu ulate lated d as as 0.00187 by above bove equ equatio ation n fro from Goal Goal Seek Seek in Exc Excel    And And for the the other  other  compon component entss , Z il  values values were were tabu tabulate lated d in Table Table 14, 21, 28.  I  = l  i

 I1l =  I 3v =  I 5v =

1 σ −ε 1 1− 0 1 1− 0 1 1− 0

ln

ln ln ln

 Z il  + σβ i  Z il  + εβ i  

0.00187 + 1 × 0.00150 0.00187 + 0 0.00426 + 1 × 0.00336 0.00426 + 0 0.00 0.0043 430 0 + 1 × 0.00 0.0035 358 8

= 0.590 ,

I 2l  =

= 0.582 ,

I 4v =

1 1−0 1 1−0

ln

0.00255 +1 × 0.00192

ln

= 0.560

0.00255 + 0 0.00385 +1 × 0.00302

= 0.580

0.00385 + 0

= 0.606

0.00430 + 0

Φ il = exp  Zil − 1 − ln ( Zil − β i ) − qi Iil    Φ1l  = exp 0. 0.00187 −1 − ln ( 0. 0.00187 − 0.00150 ) − 9.798 × 0.590  = 3.108 ln Φ li = Zil − 1 − ln ( Zil − βi ) − qi Iil

;

 

Φ l 2 = exp  0.0 0.002 0255 55 −1 − ln ( 0.0 0.00 0255 255 − 0.001 .00192 92 ) −10.1 10.181 81 ×0.560 .560  =1.93 .935 Φ l 3 = exp 0.0 0.00426 426 −1 − ln (0.0 0.00426 0426 − 0.00336 ) − 11.373 × 0.5 0.582  = 0.551 Φ l 4 = exp 0.0 0.003 0385 85 −1 − ln ( 0. 0.0038 00385 5 − 0.0 0.0030 0302 ) −11.8 11.840 40 ×0.58 .580  = 0.467 .467 Φ l 5 = exp 0. 0.0043 00430 0 −1 − ln ( 0.0 0.00 0430 430 − 0.00 .00358 ) −12.7 12.758 58 × 0.606 .606  = 0.22 0.226 6

i

3

Φ l i = v K Φi =K

αi =

0.551 0.965  K i

 K

,

1

= K

3.108 0.987

= 0.57 ,

,

α ethanol ,n-butanol =

= 3.15 ,

4

=K

0.467 0.967

α methanol,n-butanol  = HK

 K ethanol   K n-butanol

α  n-butanol ,n-butanol  =

=

 K n-but anol   K n-butanol

1.97 0.48

=

2

= 

K 

= 4.08 ,

0.48

1.935

  = 1.97 0.982

= 0.48 ,

K methanol

0.48

= K

=1

,

5

=K

0.226 0.958

=  0.24

3.15

= 6.52 0.48 n -butanol   α neopentanol,n-butanol   = α 1- pentanol,n-butanol  =

- 34 -

 K neopentanol  K n-butanol

 K 1- pentanol  pentanol  K n-butanol

=  

=  

0.57 0.48

0.24 0.48

= 1.18

= 0.49

 yi

i

=x

3

=x

,

 K i 0.373 0.57

=x

1

0.231 3.15

0.012

= 0.655 ,

fi v = yi Φ iv P ,

= 0.073 ,

4

=x

0.48

2

=x

0.377 1.97

= 0.192 0.006

= 0.024 ,

5

=x

0.24

= 0.027

f1v  = 0  .231× 0.987 ×1.013 = 0.2310

 f 2v = 0.377 × 0.982 ×1. 1.013 = 0.3755

,

f  3v = 0.373 ×0. 0.965 ×1. 1.013 = 0.3651

 f 4v = 0.012 × 0.9 0.967 × 1. 1.013 = 0.0116

,

f  5v = 0.006 ×0.9 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

,

f  3l  = 0.655 ×0.5 0.551 ×1.013 = 0.3651

 f 4l = 0.024 × 0.4 0.467 ×1. 1.013 = 0.0116

,

f  5l  = 0.027 × 0.226 ×1. 1.013 = 0.0061

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

 B log P = A−sat T in C; P oin mmHg ( A, B an a nd C constants we w ere sh s hown in in Ta Table1) T + C  Tdew = 96.98 o C t 

= P 760

log

(top stream)

mmHg  

methanol

= 7P.87863sa−t

1473.11

= 2P362.78sa9t

;

96.98 + 230

 

methanol  

1 atm

× mmHg

760 mmHg

×

1.013 bar   1 atm

17, 24, 24, 31. 31.   For ot her components see in Table 17,

yi Φ P = xi γ i Pi   v i t

γ1 = γ3 = γ 5 =

sat

 

,

γ i =

0.231× 0.987 ×1.013 0.073 × 3.149 0.373 × 0.965 ×1.013 0.655 × 0.559 0.00 0.006 6 × 0.95 0.958 8 ×1.01 1.013 0.02 0.027 7 × 0.21 0.210 0

 yi Φiv P t   xi P i sat 

=1 ,

γ 2 =

= 0.997 ,

0.377 × 0.982 ×1.013

= 0.964 0.192 × 2.032 0.012 × 0.967 ×1.013 γ 4 = = 1.031 0.024 × 0.459

= 1.089

With excel calculation (goal seek),  P t  ×

 yi

∑ γ   P 

sat 

i

i

i

- 35 -

= 1.00 we found T dew =96.98 oC

= 3.149

ba

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

c

∑ x γ  P  i

  by using this temperature in excel. Howeveri =1

i

 sat  i

= 1.00

 P t 

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 = Tr ,1 = Tr , 4 =

T 

, T [ K ] , Tc [ K ]  

T c ,i

401.65 512.58 401.65 562.93

= 0.784 , Tr ,2 = = 0.713 , T r ,5 =

401.65 516.25 401.65 586.15

= 0.778 , T r ,3 =

401.65 550.00

= 0.730

= 0.685

α SRK ,i = 1 + ( 0.480 + 1.574ωi − 0.176ω i2 ) (1 − T r1,i 2 )

2

Tc ,i , ω i were were take taken n in Refe Referrence ence 2 and tab tabulat ulated ed in Tab Table 10 2

α SRK ,1 = 1 + ( 0.480 + 1. 1.574 × 0. 0.566 − 0.176 × 0. 0.566 2 ) (1 − 0. 0.784 1 2 )  = 1.325 2

α SRK ,2 = 1 + ( 0.480 + 1.574 × 0.637 − 0.176 × 0.637 2 ) (1 − 0.778 1 2 ) = 1.361





2

α SRK ,3 = 1 + ( 0.480 + 1. 1.574 × 0. 0.604 − 0 .1 .176 × 0. 0.604 2 ) (1 − 0. 0.730 1 2 ) = 1.437 2

α SRK ,4 = 1 + ( 0.480 + 1. 1.574 × 0. 0.595 − 0 .1 .176 × 0. 0.595 2 ) (1 − 0. 0.7131 2 )  = 1.465 2

α SRK ,5 = 1 + ( 0.480 + 1. 1.574 × 0. 0.594 − 0.176 × 0. 0.594 2 ) (1 − 0. 0.685 1 2 )  = 1.520



ai = ψ 



α i  R 2T c ,i  P c ,i

,

Tc ,i [ K ] , Pc ,i [bar ] , R = 83.14 cm 3bar molK  

taken in Refer Referenc encee 2 and tabul tabulate ated d in Table Table 10 ψ  was taken a1 = 0.42748 × a2 = 0.42748 × a3 = 0.42748 × a4 = 0.42748 × a5 = 0.42748 ×

1.32 1.325 5 × 83.1 83.14 4 2 × 512. 512.58 58 80.96 1.36 1.361 1× 83.1 83.14 4 2 × 516. 516.25 25 63.84 1.437 1.437 × 83.14 83.142 × 550.00 38.80 1.46 1.465 5 × 83.1 83.14 4 2 × 562. 562.93 93 44.13 1.52 1.520 0 × 83.1 83.14 4 2 × 586. 586.93 93 38.80

= 1.27 ×10 7 = 1.68 ×10 7 3.31×107 = 3.31

= 3.11 ×10 7 = 3.98 ×10 7

- 36 -

bi = Ω

 RT c ,i

,

 P c ,i

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

Ω was taken taken in Refe Referen rence ce 2 and tabul tabulate ated d in Table Table 10 83.14 83.14 × 512.58 512.58 = 45.61 b1 = 0.08664 × b2 = 0.08664 × b3 = 0.08664 × b4 = 0.08664 × b5 = 0.08664 ×  β i ≡  β1 =  β3 =  β 5 =

qi ≡ q1 ≡ q3 ≡ q5 ≡

bi P 

80.96 83.14 83.14 × 516.25 516.25 63.84 83.14 83.14 × 550.00 550.00 38.80 83.14 83.14 × 562.93 562.93 44.13 83.14 83.14 × 586.15 586.15 38.80

83.14 × 401.65 102.11×1.013 83.14 × 401.65 108.82 108.82 ×1.013 1.013 83.14 83.14 × 401.65 401.65

bi RT 

= 102.11 = 91.89 = 108.82

3 T [ K ] , P [bar ] , R = 83.14 cm bar molK  

,

 RT  45.61×1.013

ai

= 58.25

,

= 0.00138 , β 2 = = 0.00310 , β 4 =

58.25 ×1.013 83.14 × 401.65 91.89 ×1.013 83.14 × 401.65

= 0.00177 = 0.00279

= 0.00330

T [ K ] , R = 83.14 cm 3bar molK  

1.27 ×10 7 45.61× 83.14 × 401.65

= 8.341

3.31×10 7 102.11 × 83.14 × 401.65

, q2 ≡

= 9.709 , q4 ≡

1.68 ×10 7 58.25 ×83.14 × 401.65 3.11 ×10 7

91.89 ×83.14 × 401.65

= 8.629

= 10.130

3.98 3.98×107

= 10.946 108. 108.82 82 × 83.1 83.14 4 × 401.65  Z iv − β i v  Z i = 1 + β i − qi β i ( Ziv + εβi ) ( Z iv + σβi  ) 0 = 1 + 0.001 .0013 38 − 8.341 .341× 0.00 0.001 138

 Z iv − 0.00138

( Z

v i

+ 0 × 0.00138 ) ( Z  + 1 × 0.00138 ) v i

− Z iv

  Ziv valu valuee for for meth methan anol ol was was calc calcul ulat ated ed as 0.9892 by abov abovee equa equati tion on from from Goal Goal Seek Seek in Exce Excel  l    And And for for the the othe otherr compo compone nent ntss, Z iv values were tabulated in Table 13 20, 20, 27.

- 37 -

1

 I  = v i

σ −ε 1

 I1v =

1− 0 1

 I 3v =

1− 0 1

 I 5v =

1− 0

ln

ln ln ln

 Z iv + σβ i  Z iv + εβ i  

0.9892 + 1 × 0.00138 0.9892 + 0 0.9704 + 1 × 0.00310 0.9704 + 0 0.96 0.9638 38 + 1× 0.00 0.0033 330 0 0.9638 + 0

= 0.00140 ,

I 2v =

= 0.00319 ,

I 4v =

1 1−0 1 1−0

ln ln

0.9853 +1 × 0.00177

= 0.00179

0.9853 + 0 0.9721 +1 × 0.00279

= 0.00286

0.9721 + 0

= 0.00342

Φ iv = exp  Ziv − 1 − ln ( Ziv − β i ) − qi Iiv    Φ1v = exp  0.9 0.989 892 2 − 1 − ln ( 0.9 0.98 892 − 0.0 0.00138 0138 ) − 8.34 .341 × 0.00 0.0014 140 0  = 0.9 0.990 ln Φ iv = Ziv − 1 − ln ( Ziv − βi ) − qi Iiv

;

Φ 2v = exp 0. 0.9853 −1 − ln ( 0. 0.9853 − 0.00177 ) − 8.629 × 0.00179  = 0.987 Φ 3v = exp 0.970 .9704 4 − 1 − ln (0.97 0.9704 04 − 0.0 0.00310 ) − 9.709 × 0.0 0.00319  = 0.973 Φ 4v = exp 0.97 0.9721 21 −1 − ln (0.97 0.9721 21 − 0.0 0.00279 0279 ) −10.13 0.130 0 × 0.002 .00286 86  = 0.975 .975 Φ 5v = exp  0.9 0.963 638 8 −1 − ln (0.9 0.9638 638 − 0.003 .00330 30 ) −10.9 0.946 × 0.00 .00342 342  = 0.96 .967

Z=

l i

i

β+ ( Z+

εi β) (

l i

0 = 0.00138 + (

l i

 1 + β i − Z il   Z+   σi β)      qi  β i 

l  i

+ Z0 × 0.00138 ) (

l i

 1 + 0.00 0.0013 138 8 − Z il   +Z 1 × 0.00138 )  − × 8.341 8.3 41 0.0013 0.0 0138 8  

l  i

Z

 

  Zil  value alue for for metha ethan nol was was calc calcu ulate lated d as as 0.00169 by abov abovee equa equati tio on from from Goal oal See Seekk in Excel  xcel    And And for the the other  other  compon component entss , Z il  values values were were tabu tabulate lated d in Table Table 14, 21, 28.  I  = l  i

 I1l =  I 3v =  I 5v =

1 σ −ε 1 1− 0 1 1− 0 1 1− 0

ln

ln ln ln

 Z il  + σβ i  Z il  + εβ i  

0.00169 + 1× 0.00138 0.00169 + 0 0.00425 + 1 × 0.00310 0.00425 + 0 0.00 0.0042 420 0 + 1 × 0.00 0.0033 330 0 0.00420 + 0

= 0.599 ,

I 2l  =

= 0.548 ,

I 4v =

1 1−0 1 1−0

ln ln

0.00245 +1 × 0.00177

= 0.567

0.00365 + 0

= 0.580

Φ il = exp  Zil − 1 − ln ( Zil − β i ) − qi Iil    Φ1l  = exp 0.00 .00169 −1 − ln (0.00 0.0016 169 9 − 0.001 .00138 38 ) − 8.341 .341 ×0.5 0.599  = 8.223 .223 ln Φ li = Zil − 1 − ln ( Zil − βi ) − qi Iil

= 0.543

0.00245 + 0 0.00365 +1 × 0.00279

;

Φ l 2 = exp 0.002 .00245 45 −1 − ln (0.0 0.00245 245 − 0.001 .00177 77 ) − 8.629 .629 × 0.543 .543  = 4.98 .982 Φ l 3 = exp 0.0 0.00425 425 −1 − ln (0.00 0.0042 425 5 − 0.0 0.0 0310 ) − 9.709 × 0.548  = 1.576 Φ l 4 = exp 0.003 .00365 65 −1 − ln ( 0.0 0.00 0365 365 − 0.002 .00279 79 ) −10.1 10.130 30 ×0.56 .567  =1.365 .365 Φ l 5 = exp  0. 0.0042 00420 0 −1 − ln (0.0 0.0042 0420 − 0.00 .00330 330 ) −10.94 0.946 6 ×0.5 0.580  = 0.71 0.719 9

- 38 -

 

Φ l i K = v Φi

i

=K

3

αi =

1.576 0.973  K i

 K

= 1 K

,

4

=K

1.365 0.975

α methanol,n-butanol  =

,

HK

 K ethanol   K n-butanol

α  n-butanol ,n-butanol  =

3

= 8.31 ,

0.990

= 1.62 ,

α ethanol ,n-butanol =

yi = Ki xi

8.223

=

 K n-but anol   K n-butanol

5.05 1.40

=

= 2 K

= 1.40 ,

K methanol

= 

K 

= 3.60 ,

1.40 1.40

=1

,

fi v = yi Φiv P ,

4

=K

0.719

=  0.74

0.967

= 5.93 1.40 n -butanol   α neopentanol,n-butanol   = α 1- pentanol,n-butanol  =

1

= 1y.62 × 0.034 = 0.056 ,

5

8.31

y = 8.31 × 0.008 = 0.068

,

4.982   = 5.05 0.987

 K neopentanol  K n-butanol

 K 1- pentanol  pentanol  K n-butanol

=  

=  

1.62 1.40

0.74 1.40

= 1.16

= 0.53

y2 = 5.05 ×0.027 = 0. 0.139

,

0 .548 , = 1y.40 ×0.391 = 0.

5

= 0y.74 ×0.538 =0.400

f1v =   0 .068 × 0.99 ×1.013 = 0.0687

 f 2v = 0.139 × 0.9 0.987 × 1. 1.013 = 0.1386

,

f  3v = 0.056 ×0.9 0.973 ×1.013 = 0.0548

 f 4v = 0.548 × 0. 0.975 ×1.013 = 0.5415

,

f  5v = 0.400 ×0.9 0.967 ×1. 1.013 = 0.3921

fi l = xi Φ il Pt   ,

f1l  =  0 .008 × 8.223 ×1.013 = 0.0687

 f 2l = 0.027 × 4.9 4.982 ×1.013 = 0.1386

,

f  3l  = 0.034 ×1.576 ×1.013 = 0.0548

 f 4l = 0.391×1.365 ×1.013 = 0.5415

,

f  5l  = 0.538 ×0.7 0.719 ×1. 1.013 = 0.3921

In Vapor-Liquid Equilibrium, fugacity of liquid and vapor must be equal to each other. So, our results support this condition. (  f i l = f  i v )  B log P = A−sat T in C; P oin mmHg ( A, B an a nd C constants we w ere sh s hown in in Ta Table1) T + C  Tbubble = 128.5 o C t 

= P 760

log

methanol

(bottom stream)

mmHg  

= 7P.87863sa−t

1473.11 128.5 + 230

;

=P5881.75sa8t

methanol  

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

- 39 -

 

1 atm

× mmHg

760 mmHg

×

1.013 bar   1 atm

= 7.840

r 

ba

yi Φ P = xi γ i Pi   v i t

γ1 = γ3 = γ 5 =

sat

 

γ i =

,

0.068 × 0.99 ×1.013

 yi Φ iv P t   xi P i sat 

= 0.015 ,

0.008 × 7.840 0.056 × 0.973 ×1.013

0.034 ×1.685 0.40 0.400 0 × 0.96 0.967 7 ×1 .013 0.538 0.538 × 0.732 0.732

γ 2 =

= 0.361 ,

0.139 × 0.987 ×1.013

= 0.036

0.027 × 5.572 0.548 × 0.975 ×1.013

γ 4 =

= 0.481

0.391 ×1.464

= 1.801 c

∑ x γ  P  i

i =1

With excel calculation (goal seek),



 P t 

α methan ol   = 3 10.78 × 6.52 × 5.93 = 7.47

,

α ethanol = 3 4.38 × 4.08 × 3.60 = 4.01 , α n-butanol = 3 1×1×1 = 1

 N min =



= 1.00 we found T bubble =128.5 oC

Calculation of Average Relative Volatilities

α average ,i = 3 α top ,i × α bottom ,i × α feed ,i



 sat  i

i

,

3 α neopentanol = 1.14 ×1.18 ×1.16 = 1. 1.16  

α 1- pentanol  = 3 0.79 × 0.49 × 0.53 = 0.59

Calculation of Minimum Number of Plates by Using Fenske Equation

 x LK, D x HK, W   × log    x x H K , D L K , W    

N min =

;

log α   LK, avg 

0.373 3 0.39 0.391 1  0.37 × 0.012 2 0.03 0.034 4   0.01 = 39.56

log 

log 1.16

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  -Parameter 

c

α i , ave × x f ,i

i =1

α i ,ave − θ 

∑

α met ,ave × fx,met α met ,ave − θ

7.47 × 0.15 7.47 − θ

+

= 1− q ,

+

α eth ,ave × fx,eth α eth ,ave − θ

4.01× 0.25 4 .0 1 − θ

+

< θ < α 

α

HK

+

LK  

α neo ,ave × fx,neo α neo ,ave − θ

1.16× 0.25 1.16 − θ 

+

+

1× 0.15 1−θ

α n−but ,ave × fx,n−but α n−but ,ave − θ

+

0.59 × 0.20 0.59 − θ 

θ  = 1.051

- 40 -

+

α 1− pent ,ave ×   fx,1− pent   α1− pent ,ave − θ

−1+ 0 = 0

− 1+ q = 0

 



Calculation of Minimum Reflux, R Dmin by Using Underwood Equation

c

α ir × xD ,i

i =1

α ir  − θ 

∑

 R D min =

7.47 × 0.231

7.47 − 1.051 = 3.516

 R D min

4.01 × 0.377 4.01 −1.051

1.16 × 0.373 1 × 0.012 1.16 −1.051

* β 1 = 3.516 *1 *1.5 = 5. 5.274

=R

mDin

* β 2 = 3.516*1.8 6*1.8 = 6.32 6.328 8

=R

mDin

* β 3 = 3.516 * 2. 2.4 = 8. 8.438

 R

Dactual

 R

Dactual

+

1 −1.051

+

0.59 ×0.006 0.59 −1.051

−1= 0

Calculation of Actual Plate Number by Using Gilliand Correlation

  R - R Dmin

=

 R+1

5.27 5.274 4 − 3.51 3.516 6

 

N − Nmin  N + 1

= Nactual N=

= 0.280

5.27 .274 + 1

Y = 1 − exp 1.490 + 0.315 X  −

Y=

+

mDin

Dactual



+

=R

 R

 X  =

=  R D min + 1

⇒ N  =

 1.805  1 e x p 1 . 4 9 0 0 . 3 1 5 0 . 2 8 0 = − + × − = 0.376  0.1   (0.280)  

1.805 

 X 0.1

Y + Nmin 

1 − Y  0.37 0.376 6 + 39.5 39.56 6 1 − 0.37 .376

= 64

plates

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  N R      f . HK    xb , LK   log =  0 . 2 0 6 l o g =   D  x   x N   S    f , LK   d , HK  N  R  N S 

= 1.25

 N S =

,

N R= N ac− t N S 

 N act  64 = = 28  N  R 1.25 + 1 +1  N S 

,

2  36.4   0.15   0.034 2     0.206 log    0.25   0.012   = 0.09691   = 6 3 . 6         

 

N R = 64 − 28 = 36

- 41 -



Calculation of Condenser and Reboiler Loads, Q c and Q r 

F=W+ D ,

 D

= 5.274 = R

L = G+W  L

c

∑λ

λave,top =

=L335.4

,

63.6

R D =

,

y andλave ,bottom =

i

i

Qc = G * λ ave ,top

 

q =0=

,

y

kmol h

× 38.914

1h 1000 mol   kJ  × × = 4312.9 kJ s ( kW )  1 kmol   mol  3600 s

L− 335.4 100

 

Qr  =  299.00

,

kmol h

G = 299.00 kmol / h

,

* 43.836

1h 1000 mol   kJ  × × = 3640.9 kJ s ( kW )  1 kmol   mol  3600 s

 N 

∑ MW * y i

 

=

x

K

methanol

yWi = xi K i  

,

Wi

methanol  

(

= 0.008 × 8.31 = 0.068

for other components see Tabl3e5 )

MW = (32. (32.04*0.0 04*0.068 68)) + (46.0 (46.07* 7* 0.13 0.139) 9) + (88. (88.15*0.0 15*0.056 56)) + (74. (74.12 122* 2* 0.54 0.548) 8) + (88. (88.15*0.4 15*0.400 00)) = 89.4 89.42 2 kg / kmol

W = MV × W = 89.42 

 

Calculation of Column Diameter and Height 

i =1 Wmethanol 

x   ( for results see in Table33)

i

G = L − W = 335.4 − 36.4

,

Qr = G * λ ave ,bottom

MW =

i

Qc =  399.00

,

 L = L = 335.4



c

G = 335.4 + 63.6 = 399.00 kmol / h

,

 F 

km/ol h

i =1

G = L+D

L− L

 D

∑λ

i =1

q=

 L



kg kmol

× 36.4

kmol h

×

1h 3600 s

 

= 0.904 kg s

P× MW =  ρ × R× T    ρ vap =  ρ liq =

P× MW  

=

 R × T  c

∑ ρ 

i i

1× 89.4 89.42 2 0.082 0.082 × (128.5 128.5 + 273.15 273.15))

= 2.715 kg m3

x 0.008 = 791.8 91.8 × .008 + 789 789 × 0.02 0.027 7 + 812 × 0.03 .034 + 809 809.8 × 0.391 .391 + 814.4 14.4 ×0.53 0.538 8 = 811.6 11.63 3

i =1 1/2

 ρ − ρ v  uv = ( −0.171* l + 0.27 * l t  − 0.047 )*  l    ρ v  2 t

1/2

811.63 63 − 2.71 2.715 5  811. uv = ( −0.171 .171*0. *0.5 5 + 0.27*0. 0.27*0.5 5 − 0.047 .047 )*   2.715  2

D c =

4W  

πρv uv

=

4 × 0.904 π  × 2.715 × 0.781

= 0.737 m ,

Hc = t l× ( N+ 1) = 0.5 × (64 + 1) = 32.5 m

 For other reflux ratios the r esults were in Table 35.

- 42 -

= 0.781 m  s

3

kg