Balance

TRABAJO DE BALANCE DE MASA


1. El tren de separación de 4 unidades ha sido diseñado para separar 1000 Lb-mol/h de una alimentación de hidrocarburos que contiene 20 % de metano, 25% de etano, 40% de propano, 10% de butano y el resto pentano. Las combustiones son molares. Calcular los flujos de todas la corrientes suponiendo que la recirculación a la unidad 1 es el 50%.
F4
F4
(CH499.5%)
C2H6 0.5%


(CH4 99.5%)
C2H6 0.5%


IV
IV
III
III
II
II
I
I
F2
F2
F8
F8
F3
F3
F11
F11
F6
F6
C4H10 74%
C5 H12 26%



C4H10 74%
C5 H12 26%



C3H8 98%
(C4H10 2%)

C3H8 98%
(C4H10 2%)

C3H8
C4H10 27.3%
C5 H12

C3H8
C4H10 27.3%
C5 H12

C2H6
C3H8C4H10
C5 H12

C2H6
C3H8
C4H10
C5 H12

C2H6
C3H8
C4H10 0.2%

C2H6
C3H8
C4H10 0.2%

CH4 1%
(C2H6 89%)
C3H8 10%

CH4 1%
(C2H6 89%)
C3H8 10%

CH4 20%
C2H6 25%
C3H8 40%
C4H10 10%
C5 H12 5%

CH4 20%
C2H6 25%
C3H8 40%
C4H10 10%
C5 H12 5%

F5F5

F10
F10
F9
F9
F7
F7
F1
F1


CH4
C2H6
C3H8 3%

CH4
C2H6
C3H8 3%

GRADOS DE LIBERTAD |
| UNIDAD I | UNIDAD II | UNIDAD III | UNIDAD IV | DIVISOR | GLOBAL | PROCESO |
VI (+) | 15 | 8 | 10 | 7 | 9 | 17 | 33 |
BM(-) | 5 | 3 | 4 | 3 | 3 | 5 | 18 |
FC (-) | 1 | 0 | 0 | 0 | 0 | 1 | 1 |
CC (-) | 7 | 4 | 2 | 3 | 2 | 10 | 12 |
R (-) | 0 | 0 |0 | 0 | 1 | 0 | 1 |
REST. (-) | 0 | 0 | 0 | 0 | 2 | 0 | 2 |
TOTAL | 2 | 1 | 4 | 1 | 1 | 1 | -1 |

RESTRICCION:
(S-1)(C-1)
(2-1)(3-1) = 2

RELACION:
F6 = 0.5F5

B.C en el divisor = F5 = 1000 Lb-mol/h
F6 = 0.5F5
F6 = 0.5(1000)
F6 = 500 Lb-mol/h

BALANCE DE FLUJOS:

F5 = F6+ F7
F7= 1000 Lb-mol/h – 500 Lb-mol/h
F7= 500 Lb-mol/h

TORRE II
BALANCE DE C3 :

Xc3F2= Xc3F5(0.03)F2= (0.1)(1000)
F2 = 3333.33 Lb-mol/h

BALANCE DE FLUJOS:
F2= F4 +F5
F4= 3333.33 – 1000
F4= 2333.33 Lb- mol/h
BALANCE DE C2:
Xc2F2= Xc2F4+ Xc2F5
Xc2 (3333.33) = 0.5(2333.33) + 0.89(1000)
Xc2 F2= 0.2705 27.05%
Xc1 F2= 0.6995 69.95%

TORRE I

BALANCE DE FLUJOS:

F1 + F6= F2+ F3
F1 + 500= 3333.33+ F3

BALANCE DE C1:
Xc1F1+ Xc1F6= Xc1F20.2F1+ 0.01(500)= 0.89(1000)
F1= 11474.98 Lb-mol/h
F3= 8641.65 Lb-mol/h

BALANCE DE C2:

Xc2F1+ Xc2F6= Xc2F2+Xc2F3
0.25(11474.98)+0.89(500)=0.2705(3333.33)+XC2 (8641.65)
XC2 F3= 0.279 27.9%

BALANCE DE C3:

Xc3F1+ Xc3F6= Xc3F2+ Xc3F3
0.4 (11474.98) + 0.1 (500)= 0.03 (3333.33)+XC3 (8641.65)
XC3 F3= 0.525 52.5%

BALANCE DE C4:

Xc4F1= Xc4F3
0.1(11474.98) = XC4 (8641.65)
XC4 F3= 0.132 13.2%
XC5 F3= 0.068 6.8%

TORRE III

BALANCE DE FLUJOS:

F3= F8 +F9
8641.65 = F8 +F9

BALANCE DE C4:
XC4F3= XC4F8+XC4F9
0.132(8641.65) = 0.002F8+ 0.273F9
F8= 4496.2 Lb-mol/h
F9= 4145.4 Lb-mol/h

BALANCE DE C5:

XC5F3= XC5F9
0.068(8641.64)= XC5(4145.4)
XC5 F9= 0.1417 14.17%
XC3 F9= 0.585358.53%

BALANCE DE C3:

XC3F3= XC3F8+XC3F9
0.525(8641.65) = XC3(4496.2)+ 0.5853(4145.4)
XC3 F8= 0.4694 46.94%
XC2 F8= 0.5286 52.86%

TORRE IV

BALANCE DE FLUJOS:

F9= F10 +F11
4145.4= F10 +F11

BALANCE DE C4:

XC4F9= XC4F10+XC4F11
0.273(4145.4)= 0.02F10+0.74F11
F10=2688.7 Lb-mol/h
F11=1456.64 Lb.mol/h

FACTOR DE CONVERSIÓN:

F1= 11474.98100011474.98 = 1000 Lb-mol/h

F1= 1000 Lb-mol/h
F2= 290.4 Lb-mol/h
F3= 753.08 Lb-mol/h
F4= 203.3 Lb-mol/h
F5= 87.14 Lb-mol/h
F6= 43.57 Lb-mol/h
F7= 43.57 Lb-mol/h
F8= 391.8 Lb-mol/h
F9= 361.2 Lb-mol/h
F10=243.3 Lb-mol/h
F11= 126.9 Lb-mol/h

2. Halle todos los flujos. Se recircula el 40% de los productos de tope.
F1= 300 Lb-mol/h

A 8%
B 4%
C 2%
D 1%
E 85%
A 8%
B 4%
C 2%
D 1%...