HENLEY 12.7

UNAM-FESC, Ingeniería Química, Transferencia de masa II, Destilación multicomponente, Henley 12.7 Para la operación de destilación que se indica seguidamente, calcúlese el número mínimo de etapas de equilibrio y la distribución de los componentes no clave utilizando la ecuación de fenske y los valores de Ki de la figura 7.5 de la página 302.

components

chemical formule

Zi

Kgmol/h

A

B

C

Pc

propane

C3H8

0.63613232

2500

5.353418

3371.084

414.488

617.4

i-butane

i-C4H10

0.10178117

400

5.611805

3870.419

409.949

529.01

n-butane LK

n-C4H10

0.15267176

600

5.741624

4126.385

409.5179

550.7

i-pentane HK

i-C5H12

0.02544529

100

5.49978

4221.154

387.287

483

n-pentane

n-C5H12

0.05089059

200

5.853654

4598.287

394.4148

489.5

n-hexane

n-C6H14

0.01017812

40

6.039243

5085.758

382.794

440

n-heptane

n-C7H16

0.01272265

50

5.98627

5278.902

359.5259

396.9

n-octane

n-C8H18

0.01017812

40

6.4141

5947.491

360.26

362.1

1

3930

TOTAL

destilledd components

chemical formule

yi

di=Kgmol/h

propane

C3H8

0.63613232

2500

i-butane

i-C4H10

0.10178117

400

n-butane LK

n-C4H10

0.15114504

594

i-pentane HK

i-C5H12

0.00381679

15

n-pentane

n-C5H12

0

0

n-hexane

n-C6H14

0

0

n-heptane

n-C7H16

0

0

n-octane

n-C8H18

0

0

0.89287532

3509

TOTAL

tails components

chemical formule

xi

bi=Kgmol/h

propane

C3H8

0

0

i-butane

i-C4H10

0

0

n-butane LK

n-C4H10

0.00152672

6

i-pentane HK

i-C5H12

0.0216285

85

n-pentane

n-C5H12

0.05089059

200

n-hexane

n-C6H14

0.01017812

40

n-heptane

n-C7H16

0.01272265

50

n-octane

n-C8H18

0.01017812

40

0.10712468

421

TOTAL

Para la t en las colas

 = ° −´+  °   =   Entonces













 = nC −´+   iC −´+   nC −´+   nC −´+   nC −´+   nC −´+   =  −´+    −´+    −´+    −´+    −´+    −´+  ∑nC nC  ∗  Resolviendo con Texas instriment voyage 200

´ = 254.9258027° = 123.8476682° = 124°

tails components

chemical formule

xi

bi=Kgmol/h

XiPc

xi*KI=1

ki

αi

propane

C3H8

0

0

0

0

0

0

i-butane

i-C4H10

0

0

0

0

0

0

n-butane LK

n-C4H10

0.01425178

6

7.84845606 0.03990499

2.8

2.24

i-pentane HK

i-C5H12

0.20190024

85

97.5178147 0.30285036

1.5

1.2

n-pentane

n-C5H12

0.47505938

200

232.541568 0.59382423

1.25

1

n-hexane

n-C6H14

0.09501188

40

41.8052257 0.06365796

0.67

0.536

n-heptane

n-C7H16

0.11876485

50

47.1377672 0.04038005

0.34

0.272

n-octane

n-C8H18

0.09501188

40

34.4038005 0.01900238

0.2

0.16

1

421

461.254632 1.05961995

ki

αi

TOTAL

Para el destilado

iC  

−´+  −´+  −´+  −´+      =           C Resolviendo con Texas instriment voyage 200

´ = 136.2113478° = 57.8951932° = 58° destilledd components

chemical formule

yi

di=Kgmol/h

propane

C3H8

0.71245369

2500

0.00115396 0.27402065

2.6

1

i-butane

i-C4H10

0.11399259

400

0.00021548 0.10856437

1.05

0.16

n-butane LK

n-C4H10

0.169279

594

0.00030739 0.62695925

0.27

0.2376

i-pentane HK

i-C5H12

0.00427472

15

8.8504E-06 0.00610675

0.7

0.006

n-pentane

n-C5H12

0

0

0

0

0

0

n-hexane

n-C6H14

0

0

0

0

0

0

n-heptane

n-C7H16

0

0

0

0

0

0

n-octane

n-C8H18

0

0

0

0

0

0

1

3509

TOTAL

Yi/Pc

yi/KI=1

0.00168568 1.01565102

Número mínimo de etapas

 , ]  log[  ,  ×  ,  = ,, Donde

  = −  ,  =   , ∗ , ,  = √2.5925 ∗ 2 = 2.2770

Por lo que te queda

47.2311956 = 2.778837 = 7.7757   =  log11.8777429∗  (2.277059) 0.357374