October, 1929
I-VDCTSTRI-412 A;L‘D ENGIXEERIa\TG CHEIWSTRY
The licorice is used in largest amounts in chewing tobacco, although it also finds extensive application in cigars and in pipe and cigarette tobaccos. The remaining 10 per cent of licorice mass produced is used in about equal amounts by the pharmaceutical and confectionery industries. At present the chief role of licorice in pharmacy is to cover the acrid taste of various drugs. It is also used as a constituent of cough sirups, throat lozenges, and pastilles. Its use as a corrector of cathartics is still recognized in the form of compound licorice powder, a mixture of powdered senna leaves, sulfur, fennel seeds, sugar, and licorice, the active ingredients of which are the senna and sulfur. Powdered licorice root and powdered licorice extract are also used in pharmacy as medicaments, excipients, and dusting powders. Licorice confectionery has been somewhat under a cloud in the past, owing in large measure to the widespread substitution of oil of anise for the genuine licorice flavor. The use of licorice as a wholesome ingredient of high-grade confections is gaining increasing recognition. By-products
The by-products of the manufacture of licorice extract are especially interesting. A portion of the spent root from the primary extractors is subjected to a secondary extraction with a 5 per cent solution of caustic soda under LOO pounds steam pressure for several hours. This secondary extract is evaporated to 12” Bb., in which form it is known as Firefoam Liquid and is used as a foam stabilizer in the Foamite fire extinguisher. The solutions in this extinguisher have the following composition: the first, 11 parts aluminum sulfate and 89 parts water; the second, 8 parts sodium bicarbonate, 3 parts Firefoam Liquid (12’ BE.), and 89 parts water. On mixing these solutions a tough, durable foam, which resists heat and mechanical abuse, is produced. This method of attacking fires has proved to be of great value in extinguishing oil fires and is now in almost universal use in the protection of oil tanks. This use, however, requires only a small part of the available licorice resins. There are still about 5 million pounds of these resins which might be extracted annually a t very little cost. Recent work indicates that this secondary
917
extract may be valuable as a wetting agent and as a foaming agent in the flotation process for ore extraction and also as a wetting, spreading, and sticking agent in insecticides, The remaining root fiber is used in making boxboard, Jacquard cards for the weaving industries, Fiberlic wallboard, JIaftex, and Maflath insulating board. The fibers of the licorice root are tough and wiry and they impart this characteristic to these products. The boxboard is made from a mixture of paper and licorice root pulp and is used in making pasteboard cartons. The Fiberlic wallboard consists of several layers of boxboard cemented together with sodium silicate. RIaftex insulating board is a single-ply board made with one surface smooth and the other surface with a ripple finish. For interior decorating and paneling effects the smooth side is usually left exposed. The ripple side offers an unusually good bonding surface for plaster. Tests have shown this board t o have the following properties: (1) Thermal conductivity rating of 0.33 B. t . u.
( 2 ) An ability to hold four six-penny nails up to 250 pounds pull, and then have an average of only one nailhead out of four pull through the hlaftex. (3) A bonding strength with gypsum plaster of over 1000 pounds per square foot. (4) High water resistance, not becoming wet on complete immersion in water for 24 hours. Conclusion
The licorice root is thus a contributor to human activities in no small degree. The primary extract serves the tobacco, pharmaceutical, and confectionery industries; the secondary extract is a fire-extinguishing agent; and the residual fiber is a valuable raw material for cardboard boxes and provides a new structural material for the architect and builder. L i t e r a t u r e Cited Beal and Lacey, J . Am. Pharm. Assocn., 18, 145 (1929). Griffith and Thompson, “Demotic hfagic Papyrus of London and Leiden” (1904). Houseman, A m . J . Pharmacy, 84, 531 (1912); 88, 97 (1916); 43, 388, 455, 481 (1921); J . Assocn. Oficial Agr. Chem., 6, 191 (1922). Joachim, “Papyrus Ebers,” Berlin, 1890. Simmonds, Bull. Pharm., 8, 205 (1894). Tschirch, “Handbuch der Pharmacognosie,” Band 11. Lief. 20, p. 20 (1905).
The Compression of Refinery and Casinghead Gases’ Wallace J. M u r r a y ARTHCR
D. LITTLE,I N C . ,
S T H E compression of refinery and casinghead gases
partial liquefaction usually takes place in either the compressor or the after-cooler, or both. The purpose of this paper is to describe a means of calculating the composition and quantity of both the liquid and gaseous phases produced in this compression. Since the compressor and after-cooler ordinarily act as one unit, only the combined liquefaction taking place in both will be considered a t this time. Binary systems have been investigated by Calingaert and Hitchcock ( A ) , but the more general case with many components has not been presented so far as the writer knows. The equilibria between liquid and gaseous phases have been studied by Brown and Caine ( I ) , Lewis (6), and by the California Natural Gas Association ( 2 ) . The inethod of 1 Received April 16, 1929. Presented before the Division of Petroleum Chemistry a t the 77th Meeting of the American Chemical Society, Columbus, Ohio, April 29 t o M a y 3, 1929.
CAMBRIDGE, hfASS.
calculation presented in this paper is founded on the work of these men and is intended as a fairly simple method of checking up compressor action on casinghead and refinery gases. Development of Method of Calculation of Rich G a s Compression
(I) Take initial volume of gas in cubic feet. ( 2 ) Look up on chart pound-molecular volume at initial temperature. This may be most easily obtained from a chart showing pound-molecular volumes a t different temperatures and pressures, such as is shown herewith. (3) Divide the initial volume by the pound-molecular volume to get pound mols. For convenience, it is best to base all calculations on 100-pound mols and then a t the end multiply by the factor to cdnvert to true initial volume. All steps beyond this point are based on 100-pound mols.
918
INDUSTRIAL AKD ENGINEERING CHEMISTRY
Vol. 21, No. 10
where K is the deviation constant of Raoult's law. This K must be carried through all subsequent steps. In the PA . calculation, K will be substituted P PA for However, in most cases the P value 01 K is unknown but is fairly close to 1 ; the form given above with K omitted or made equal to 1 should be used.
-
-.
(9) From (4),(6), and (7), A~=A-A&andG=100-L (10) From (8) and (9), A PA AL - =-- AL 100-L P X L
(11) Solving for comes
&'ole-Steps (2), ( S ) , (4),and (6) are based on perfect gas laws. Calingaert and Hitchcock ( 4 ) have shown these laws to be substantially true with these gases a t atmospheric pressure. Podbielniak (7)gives data a t 1 and 10 atmospheres for saturated vapors. He recommends use of the modified Bertholet formula PV = RT (1 - K P ) , where K is a deviation constant. However, his data indicate t h a t this K is not constant, being 0.0226 a t 10 atmospheres and 0.068 a t 1 atmosphere for n-pentane. He also gives apparent molecular weights from densities. If these molecular weights are used instead of formula weights, some of these deviations will tend t o correct themselves.
(4) When 100-pound mols are taken, the volume per cent of each component is equal to the pound mols of the component (Avogadro's law). Designate the mols of component A by A. (5) After compression and partial liquefaction, equilibrium requires that the partial pressure of each component in the gas phase equal the vapor pressure of the same component in the liquid phase. (6) When there are G pound mols of gas and A G mols of component A in the gas, all a t pressure P , the partial pressure of component A will be
3P. A
(Dalton's law.)
(7) When there are L pound mols of liquid and A L pound mols of component A in the liquid and component A has PA vapor pressure in the pure state a t the temperature of operation, the vapor pressure of A in the liquid will be
PA. (Raoult's law. This law is followed quite closely L when all components present are the lower petroleum hydrocarbons of the same homologous series.) (8, 9) N0te-F'~ may most easily be obtained from Cox's chart ( 5 ) . For methane see Calingaert and Davis (3). See also Podbielniak (7). AL Brown and Caine ( I ) show t h a t the vapor pressure = K - PA. where
L K is a correction factor for Raoult's law. For the heavy naphthas they examined, K was about 0.9, but would probably be nearer 1 for lighter components. Calingaert and Hitchcock ( 4 ) give a 4-4 per cent deviation from Raoult's law a t atmospheric pressure or below for a 20 per cent butane and 80 per cent pentane mixture.
A - Gp = tA p G L Nole-In
A
case greater accuracy is desired, this may be written:
Nofe-The
2this be-
A
more exact form is A
This equation is the basis for the calculation of the condition of the material after compression. It is obvious that in treating a mixed gas me may substitute for A the quantities of other components as B, C, D, E , etc., thus obtaining Equations l l ~l l, c , etc. (12) It will be noted that this equation contains two unknowns, A L and L, while A , PA,and P are known. It cannot be solved directly, but the relationship
AL+BL
+ CL + DL + . . . . . . = L
provides the required extra relationship. (13) For calculation it is found easier to treat A L BL
z,
etc., as single unknowns and then Equation 12 becomes A - L+ T BL+ z CL+
. . . . . . . .=
1
L (14) Equations 11, 118, etc., for each component and Equation 13 may be solved simultaneously, but in general it is easier t o determine L roughly by inspection and substitute. After two approximations the value of L may be determined by interpolation. Steps in Numerical Calculation
Step 1. Calculate or obtain from tables, analyses or given data, the values of PA P B P A A , B , C, D , etc., P A ,PB, Pc,PD,P , 7 ,F , etc., 1 - 7 , 1 - PB -,P etc. For convenience these should be tabulated. PA Step 2. Substitute the values of A and 7 in Equation 11 AL and, putting L = 0, calculate the value of Note that in all calculations A L is always treated as a single unknown
r.
and that assumed values of L are never substituted in it.
INDUSTRIAL AND ENGINEERING CHEMISTRY
October, 1929
If
% is greater than 1, liquefaction will take place.
If less
CL ,etc., should be determined. At the dew 57 than 1, then BL point the sum of all these values is 1. If the sum is greater than 1, liquefaction will take place, if less than 1, no liquefaction will occur. Step 5. Assume values of L and solve. By taking two BL values of L giving A 2 L L . . . . . , respectively, slightly
+
greater and less than 1 and interpolating, a value of L can f BL , . . . . = 1. From this value of be found where A
+
L values of G, A L , BL, etc., can easily be calculated.
A sample of vent gas of the following composition: Per cent 10 53 26 8 3
+
CSHIZ residue C4Hio C~HE C2He CH4
B C
D
E
Step 1 = 90
A
10
B
S
PA =
53
PB =
PA/P =
0.089
57
PB/P =
0.63
+ 0.911 1- + 0.37 1
- PA 7=
PB = P PC = P
-
C = 26
Pc =
175
Pc/P =
1.95
1
-
D =
8
PD =
700
Po/P =
7.80
1
-= P
E
3
P E = 4000
44.50
1
- PE = P
=
Put L
=
PB/P
=
PD
-
0.95
6.80
-43.50
Step P 0 in Equation 11
This is greater than 1 and therefore liquefaction mill take place.
Step 3 Try L
=
60 in Equation 11
& = 8.9 4- 54.66 L BL--
L
63
L
= 0.157
+5322.2 = 0.623
CL -195
26 57.0
= 0.189
DL _ -= 0.011 L 780 - 40.8
-EL _ - 4450 - 261 = 0.001 L L
L
L
L
This shows that L = 60 is a little high. Try L this case solve as above A L +$+.... 7 B = 1.021
showing that L Interpolating:
=
50 is too low. 1.021 - 0.981 = 0.040 1.021 - 1.000 = 0.021
_ 21 40 Therefore, L
=
55.25
-- 0.525
cL
55.25 ' X 26 = 195 - 52.5
EL = 55'25 4450 - 240
= 10.0
=
0.04
__
CG = 26
EG =
- CL
= 16.0
3 - EL =
2.96 __
L = 55.0
G
= 45.0
(Check values differing from assumed by only 0.25 and thereby proving assumed values correct.)
+ higher
LIQUID Mols 9.3 35.1 10.0 0.6 0.04
GAS Mols 0.7 17.9 16.0 7.4
3.0
From our pound-mol volume chart the volume of residual gas may easily be obtained; for example, 45 pound mols of gas a t 75 pounds gage and 95' F. have a volume of 45 X 66.8 = 3008 cubic feet. The original volume of 100 mols a t 0 pounds gage and 95' F. was 40,540 cubic feet, or the volume after compression was 7.4 per cent of the original volume, assuming liquid volume as zero. A perfect gas would have gone to 15/90 = 16.7 per cent of the original volume. I n case steam is present in the original gas, it may be assumed that water and hydrocarbons are immiscible and that the water phase has the vapor pressure of pure water a t the given temperature. Subtracting the mater vapor pressure from the total pressure, one obtains the hydrocarbon partial pressure and may proceed as above. From the final vapor volume and the partial vapor pressure in it, the weight of water in the vapor phase may be calculated, and subtracting this from the total the figure for the amount liquefied may be obtained. If air is present an accurate calculation is almost impossible; however, if one assumes that the air follows Boyle's law and is insoluble.in gasoline no serious errors are introduced. In this case the mols of air in the liquid are assumed to be zero. The rest of the calculation is as above. From the mols liquefied and their latent heat and the heat of adiabatic compression of the unliquefied portion, it is possible to calculate the heat to be removed in the compressor jackets and after-coolers. The correct value of n to use in the adiabatic compression is rather uncertain, usually about 1.19 for many rich gases, but this heat is usually such a small percentage of the total that the error involved is slight. Literature Cited
A - - " +B L + - CL $ - $ LDL = 0 . 9E8 ]
L
= 44.76 (assumed value used in making calculations)
Pentane Butane Propane Ethane Methane
is to be compressed t o 90 pounds absolute (75 pounds gage) and 95" F. P
G
After compression there will be 55 mols of liquid and 45 mols of gas of the following composition:
Illustrative Example
A
919
=
50. In
(1) Brown and Caine, Trans. Am. Inst. Chem. Eng., 21, 91 (1928). (2) California Natural Gas Assocn., Oil Gas J., 27, 151 (1929). (3) Calingaert and Davis, IND. E K G . CHEM., 17, 1287 (1925). (4) Calingaert and Hitchcock, J . A m . Chem. Soc., 49, 760 (1927).
(5) Cox, IKD. ENG. CHEM., 16, 592 (1923). (6) Lewis, Trans. Am. Inst. Chem. Eng., 20, 1 (1927). (7) Podbielniak, Pelroleum World (Los Angeles) 13, No. 10, 114 (1928). ( 8 ) Podbielniak and Brown, IND. END. CHEM.,21, 773 (1929). (9) Podbielniak, Pefroleum W o r l d (Los A n g e l e s ) , 13, No. 11, 99 (1928).
Benzene exports from the United States during July amounted to 4,635,674 gallons, valued a t $1,105,793, an increase of 270 per cent in quantity and 325 per cent in value when compared with July, 1928.
