Page 130 - Applied Process Design For Chemical And Petrochemical Plants Volume II

Distillation 119

Operation Aio = 1.65/1.33 = 1.222
AN+l -Ai - (1.227)Nt1 - 1.227 = o.98
E,= -
Ah'+' - 1 (1.227)N'c' - 1 Theoretical trays at operating (LJVN + 1):
[ 1- 0.98 ] 1.222 - 0.98
1.227 - 0.98
(N + 1) log 1.227 = log (N + 1) log 1.222 - log [ 1-0.98 ]

ix = 11.22
N = 11.2 trays
7 Actual plates at operating (&/VN + 1):
Efficiency of oil at 90°F, and sp. gr. of 0.825 come Efficiency : (0.825) (62.3) = o.294
sponding to MI of 40. (1.35) (160) (0.81)
Viscosity = 0.81 centipoises
Reading curve 3, Eff. = 29%
For O'Connell's efficiency correlation, Figure 8-29. Actual trays = 11.1/0.29 = 38.6. Say 40 trays.
Lean oil rate = (1.222) (1.35) (1,756) = 2,900 mols/hr
0.825 (62.3 lb/fts) = o.034j
(11.5) (160) (0.81) GPM = (~900) (160) = 1,122
(8.33) (0.825) (60)
Reading curve (3), Eff. = 14% This is still a large quantity of oil to absorb the ethylene.
This value is low, but agrees generally with the spe- Under some circumstances it might be less expensive to
cific data in O'Connell's [49] report. separate the ethylene by low-temperature fractionation.
Although Drickamer's data are not so specific for
absorption, the graph of this correlation gives Eff. = Example 8-3s Determine Component Ahsorption in
20% for the 0.81 cp. Because no better information is Fixed-Tray Tower
available, use Eff. = 15%.
Actual trays, No = N/E, = 11.22/0.15 = 74.8 An existing 40-tray tower is to be examined to deter-
mine the absorption of a rich gas of the following analysis:
Use KO = 75 trays
8. Lean oil rate = L, = 4 (Ki) (Vx + l)o = (1.227) (11.25) 104"
(1,756) Component Mols/hr Mol wt I& 176 psia
Lo = 24,200 mols/hr . .. . -. -~ -
.
H2 500.0 2 59.0
24,200 (160) = 9,400 CH4 20.9 16 56.0
GPM oil =
(8.33) (0.825) (60) C2H4 131.5 26 8.1
CO 230.0 28 12.3
CsH4 3.5 40 0.07
This is unreasonable, and is due to the effect of c4H2 4.1 50 0.009
a. Operating pressure being too low, thus giving a ~- - 890.0
....
high K value
b. Ethylene being light component and difficult to 1. The key component is methyl acetylene, C3H4.
absorb Recovery will be based on 96.5% of this material.
c. Temperature too high.
9. Recalculation of Steps 3 thru 8. The tower average temperature will be assumed =
104°F.
Assume operating pressure is 700 psig, K = 1.35. The operating average pressure will be = 161 psig.
Note that this same K value could have been achieved
by lowering the operating temperature to -90°F. This 2. The & values are tabulated for the conditions of (1) ,
is also not practical from the oil standpoint or even and were determined from laboratory test data for
from the economics of operating the entire system and the special solvent oil being considered.
refrigeration system at this level, unless (1) the refng- & = 14.7 (Hi)/176, where Hi is Henry's constant
eration is available and (2) a suitable oil is available. expressed as atm/mol fraction for each component.

Min (L/V) = (1.35) (0.98) = 1.32 Note that conventional K charts are only applicable to
hydrocarbon oil systems, and do not apply for any
Operating (L/Vx + l)o = (1.25) (1.32) = 1.65 special solvents.