I N D U ST R I A L A N D E N G I N E E R I N G CH E MI ST R Y
692
(8) Heraog, J. Phys. Chem., 30, 457 (1926). (9) Hess, “Die Chemie der Zellulose und ihrer Begleiter,” p. 647,
Akademischen Verlagsgesellsohaft, Leipzig, Hyden, IND.ENQ.CHEM.,21, 405 (1929). Mack, J. Am. Chem. SOC.,54, 2141 (1932). Mark, Scientia, 51, 405 (1932).
(10) (11)’ (12)
1928.
(13)
Mark and Meyer, Ber., 61, 593
Vol. 23, No. 6
(1928); 2. phvsik. Chem., B2,
115 (1929). (14) (15)
Sponsler, Plant Physiol., 4, 32.9 (1929). Urquhart, J. Textile Inst., 20, 125 (1929).
RECEIVED November 28, 1932.
Equilibrium Conditions in Continuous Recycle System R. BARLOW SMITH, Sinclair Refining Company, East Chicago, Ind.
ORIGINAL
LIQUID CONDENSATE FROM COMPRESSOR
GAS
HYDROCARBONS-Mu METHANE 39.8064 ETHANE 19.8211 PROPANE 11.SOOO BUTANE IP.0110 PENTANE 8.3200 HEXANE 2.2805 T O T A L 100.0000
hYOROCARBONS-MOLS
,,
WETHANE
,0064
ETHANE
.02II
PROPANE BUTANE PENTANE HEXANE TOTAL
,0800 I420
.moo
,3005 l.0000
n =
G A S LEAV NG
RESIOUE GAS
the recycle part. The composition and quantity of liquid and gas that will result from compression can be calculated from the following well-known equi-
CCYPRESSOR -0TAL
GAS
ABSCRBER I,
HYDROCAR ON)
MOL5
HYDROCARBONS METHANE ETHANE PROPANE
MAS 39.6000 19.8000 17.8?00
RECYCLE SA5 hYDRCCARBJ\S
YOLS
\I
Where 2 n = N
XO
June, 1933
INDUSTRIAL AND ENGINEERING CHEMISTRY
on the other hand, there is not enough butane in the liquid condensate from the compressor to meet the specifications if much less than 90 per cent of the butane is absorbed. Therefore, 90 per cent butane absorption will be decided upon. Consequently, the remaining hydrocarbons will be absorbed in the following percentages:' methane 1.0, ethane 7.0, propane 20.00, pentane 100, hexane 100. The recycle gas arid also the other quantities that exist in this system can then be calculated by building up the system as it would actually occur. That is, calculate the amount of gas from the compressor that will be absorbed, combine the absorbed gas with the liquid condensate from the compressor, calculate the amount that will evolve as vapor from the accumulator, combine the evolved vapor with the gas from the compressor, and continue the calculations until there is no change in the composition and quantity of the vapor leaving the accumulator. These calculations can be carried out by using the equilibriuni flash formula previously mentioned. I n building up this system, the increase in the quantity of vapor leaving the accumulator will become less after each cycle, because each change in the composition :tnd quantity of vapor leaving the accumulator causes the distillate entering the accumulator to become heavier. Since the changes in the composition and quantity of vapor leaving the accumulator approach zero, they cannot build up into a vicious cycle, and the system is practicable. However, this system can be more readily understood if analyzed as a system already in operation. For instance, consider methane as shown in Figure 1: A moles of methane will leave the accumulator in the recycle gas. Therefore, 39.6 A moles will enter the absorber, 0.396 0.01A moles will be absorbed, and 0.396 0.0064 0.01.4 moles will enter the accumulator. If A moles of methane are leaving O.OLA - A , or the accumulator as vapor, then 0.4024 0.4024 - 0.998 moles nil1 leave the accumulator as a liquid. The same has been worked out in Figure 1for the other hydrocarbons in the system. Since the hydrocarbons in the vapor are in equilibrium with those in the liquid, and the properties and conditions under which they exist are known, it should be possible to calculate the quantity of each that exists in the liquid and vapor phase. By equating the partial pressure exerted by each hydrocarbon in the liquid phase to that of each in the vapor phase, the following formula has been developed:
+
+
+ +
+
SOMENCLATURE = mole fraction in liquid phase y = mole fraction in vapor phase p = partial pressure P A , P B ,etc. = vapor pressure of constituent A, B, etc. T = pressure on system L A , L B , etc. = constituent A, B , etc., in liquid phase, moles A , B , etc. = constituent in vapor phase, moles T O = total in vapor phase, moles TL = total in liquid phase, moles
z
The derivation is as follows: Px = p (by Raoult's law); ~y = p (by Dalton's lan-); P z = ~y (under equilibrium conditions).
-
L A P A T G- LBPBTG TLA TLB
1 Some of the most important papers on methods of calculating absorption are: Cox and Arnold, Refiner, 8, 63 (1929); Kremser, Oil &a J . , 29, 416 (May 22, 1930); Lewis, W. K., Trans. A m . Inst. Chen. Eng., 20, 1 (1927); Kaliam, F. L., Chem. & Met. Eng., 38, 78 (1931).
693
By substituting in the foregoing equation, the quantities required in the problem can be determined. Since 90 per cent butane recovery is required: Lo = 12.0720 X 0.90 = 10.8500 10.892 - 0.1OD = 10.8500 D = 0.42 PD at 38" C. = 3.33 atm. For hexane: PF at 38" C . = 0.355 atm. L F = 2.2805 L$F or 86.1 = 2.2805 X 0.355 10.8500 X 3.33 - 0.42 F F = 0.00942 For pentane: PE at 38" C . = 1 atm. LE = 8.3200 86.1 =
8.3200 X 1
E
E = 0.0965 For propane: PC at 38" C. = 12 atm. Lc = 3.6500 - 0.80C 86.1 =
c =
(3.6500 X 0.80C)12 0.457
Lc = 3.6500
-
(0.80 X 0.457) = 3.284
For ethane: PB at 38" C. = 50.6 atm. LB = 1.4061 - 0.93B
.
(1.4061 - 0.93B)50.6 86.1 = B B = 0.533 L B = 1.4061 - (0.93 X 0.533)
0.9101
For methane: P A at 38" C. = 333 atm. L A = 0.4024 - 0.99A (0.4024 - 0.99.4) 333 86.1 = A A = 0.322 = 0.4024
L*
- (0.99
X 0.322) = 0.0834
Characteristics of the liquid phase are as follows: P
HYDROCARBONS Methane Ethane Propane Butane Pentane Hexane
*
Total
Moles
Mole %
0.0834 0,9101 3.2840 10.8500 8.3200 2.2805
0.32 3.54 12.75 42.17 32.35 8.87
38' Atm. 1.065 1.790 1.530 1.405 0.324 0.031
AT
-
-
-
25.7280
100.00
6.145
C.
Therefore, it will be necessary to operate the accumulator under 6.145 atmospheres of pressure. Characteristics of the recycle gas are as follows: HYDROCARBONE Moles 0.3220 0.5330 0.4570 0.4200 0.0965 0.0094
Mole %
Methane Ethane Propane Butane Pentane Hexane Total
1.8379
100.00
17.50 29.04 24.90 22.80 5.25 0.51
-
Therefore the percentage of the recycle gas of total gas to 1.8379) = 1.825 per cent. the absorber will be 1.8379/(99 Since this is not excessive, the system is practical. These calculations have been carried out on the basis that the hydrocarbons follow Raoult's law exactly. However, the corrections for Raoult's law can be applied, although in a great deal of this type of work it has been found that Raoult's law is sufficiently accurate for engineering purposes.
+
RECEIVED November 26, 1932.
