Page 182 - Fluid Catalytic Cracking Handbook

154 Fluid Catalytic Cracking Handbook

Example 5-4
Calculation of Individual Components

= 0.155 x 16 MMSCFDx 2.02 0.08 x 3 MMSCFDx 2.02
2
~~" 379.5x24 379,5x24

_„ 0.358x16 MMSCFDx 16 0.472x3.0 MMSCFDx 16 - CM1Urt
CH 4 = --- = 7,585 Ib/hr
4
379.5x24 379.5x24
C = 0-171 xl6MMSCFDx30 0.149 x 3 MMSCFDx 30
2
~~~
379.5x24 379.5x24
„
/ _ 0.11x16 MMSCFDx 28 0.025 x 3 MMSCFD x 28
2
379.5x24 379.5x24
= 0.016x!6MMSCFDx44 | 0.179x 11,65]8BPDx 175.3
3 +
~~ 379.5x24 " 24
0.084x3 MMSCFDx 44
= 15,262 lb/hr
379.5 x 24

^_ 0.047x16 MMSCFDx 42 0.313x11,658 BPDx 181.8
• Cr = +
3
379.5 x 24 24
0.044x3 MMSCFDx 42 OACA/IIU/ ,
= 30,504 Ib/hr
379.5 x 24 '

^0.002x16MMSCFDx58 | 0.109x11,658BPDx204.6
4 +
"~ 379.5x24 24
0.02 x 30,230 x 204.6 MMSCFD x 42 0.032 x 3 MMSCFD x 58
24 379.5x24
=15,579 Ib/hr

= 0.007x16MMSCFDx58 [ 0.161x11658BPDx 197.2
4
' ~ 379.5x24 + 24
0.004x30,230x204.6 BPDx 197.2 0.009x3 MMSCFDx 58
24 379.5x24

= 16,95 8 Ib/hr