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T H E JO URXAL OF INDUSTRIAL A N D ENGINEERING CHEMISTRY
Vol. 14,No. 11
Condensation of Vapor from Mixtures of Vapors and Noncondensable Gas’ *
By Baxnett F. Dodge2 LEWISRECOVERY CORPORATION, CAMBRIDGE, MASS,
P
Equation 1 expresses it in a the The theory of the condensation of mixtures of uapors and nongeneral way so that it may be recovery of solvents condensable gas has been deueloped in such a way as to enable approxapplied directly to a wide used in industry, for imate calculations to be made of the amount of condensate obtainable the extraction of gasoline under giuen conditions. So f a r , the theory can only be applied to variety of conditions. The following example from natural gas, and for mixtures of uapors which roughly obey Raoult’s law when they exist illustrates the use of this the extraction of light Oil as liquids. For such mixtures the equations haue been deueloped for equation: from coke-oven gas and both “di$crential” and “simple” condensation, and it is shown that allied Processes, are all Prithe two quite diflercnt mathematical expressions giue nearly idcntiAssume that we have ai1 marib‘ o ~ ~ e r n ewith d the cal results for amount of condensate. The equafions for “simple” . containing 2.0 per cent by condensation of vapors from condensation are extended to include mixtures of three or more uavolume of benzene vapor and that we wish to know how mixtures of these vapors pors and noncondensable gas. Seueral illustrations of the pracmuch of the benzene can be and m ~ ~ o n d e n s a b l egas. tical annlications o f the eauations are giuen. condensed bv commessing. One of the main questions the mixture to a prekure ouf that arises, especially in con150 lbs. per sq. in. gage and nection with the design of apparatus for new plants, modifica- cooling the compressed mixture under constant pressure t o 20 C. From vapor pressure tables we find that p a t 20 C. = 75 mm. tion of present processes, and development of new ones, is: From the conditions of the problem, we see that How much of the given vapor or mixture of vapors, say gasoline, will be condensed under any given conditions of temperature, V = 2.0, a n d P = 150.+_14’7 X 760 = 8500 mm. 14. pressure, and concentration? It becomes important, then, to Substituting in Equation 1 we obtain 56.5 per cent condensed. have available methods of estimating the amount of recovery or of extraction under given conditions. It is the purpose of A similar calculation may be used for condensation of any single this article to develop equations which will enable this quantity vapor by any condition of compression or refrigeration. to be calculated for certain types of vapors. These equations Space does not permit the citing of the many practical apwill be developed in as general a form as possible consistent plications of this equation. It is hoped that this phase of the with simplicity, so that the desired result can be obtained di- subject can be taken in detail in a second paper. The curves rectly by merely substituting proper values of total pressure, shown in Fig. 1 are given, however, to show in a general way vapor pressures of pure liquid components and concentrations. how the equation may be applied to the study of a solvent recovery problem. These curves, which were constructed from SINGLE VAPORCOMPON~NT calculations made by the aid of the equation, enable one to preThe calculation of the amount of vapor which will condense dict immediately what percentage recovery would be obtained out of a mixture of a single vapor3 and noncondensable gas under by cooling different gasoline-air mixtures (gasoline was equivgiven conditions 01 concentration, temperature, and pressure, is alent to hexane in vapor pressure) to certain given temperatures a t a pressure of one atmosphere. For example, by reference to relatively simple. For example, the following equation4 may the figure it can be seen that a 15 per cent concentration by be used to calculate the per cent of the vapor that will condense: volume just begins to condense a t 20” C., is one-half condensed V P a t 4’ C., and over 90 per cent condensed a t -30” C. On the 100-v other hand, a 5 per cent concentration does not begin to condense Per cent condensation = D P x 100 (1) V until cooled to - 4” C. Such information, which may be closely 100-v checked in practice, is a valuable guide in the determination of where V = Volume per cent of vapor in the initial mixture the best process to use in the recovery of a given solvent under p = Vapor pressure of the liquid condensate a t the final given conditions. temperature of the system D = Final total pressure on the system Two VAPORCOMPONBNTS This same relation may be expressed in a number of ways but The usual problem that arises in practice involves the treat1 Received April 4, 1922. ment of gases containing, not a single vapor component, but a 2 Lecturer in Chemical Engineering, Harvard University. mixture of two or more vapors. Such a problem is usually handled 8 By the term “vapor” we mean a gas which is relatively near its boiling by assuming some sort of an average vapor with properties that point. represent some kind of a weighted average of the properties of the 4 The derivation of this equation is very simple. Assuming that the vapor and the noncondensable gas mixed with it obey the simple gas laws, single pure components. It is our object in the following discussion to develop the theory of the condensation of complex then --?!-= mols of vapor per mol of noncondensable gas in the ini100 - v mixtures and incidentally to show whether or not the use of an tial mixture. Assuming that the final gas mixture comes to equilibrium with average is justified. the condensed liquid at the final gas temperature in the system, which is As far as could be ascertained from a brief study of the literaalmost always the case, then 2 = mols of vapor per mol of nonconture, the condensation of mixtures of vapors in a noncondensable D - P gas has not been treated from a theoretical standpoint to develop densable gas in the final mixture (i. e . , since p is the vapor pressure of the condensate, it must also be the partial pressure of the vapor remaining in the equations of practical use. The theory of the condensation of gas, regardless of the total pressure on the system). Since the noncondensabinary mixtures not containing a noncondensable gas has been ble gas remains constant throughout, the mols of vapor per mol of noncondeveloped, but the resulting equations cannot be conveniently densable at the start minus the same ratio at the end divided by the ratio applied to this case, as will be shown later. a t the start gives the fraction condensed.
ROCESSES for
-.
O
a
- - _ _ .
Nov., 1922
T H E JOURNAL OF I N D U S T R I A L A N D ENGI,VEERING CHEMISTRY
It is obvioiis that the liquid which condenses from a mixture of vapors will not have the same composition as the initial vapor, and that the composition will vary throughout the process.
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gas and vapor were slowly cooled in a receiver and no condensate removed during the process, the condensation would be practically sim$Ze. If, on the other hand, the condensate were removed as fast as formed, the condensation would be practically differential. A differentid condensation is probably quite closely realized when the vapor from a mixture containing noncondensable gas is being continuously condensed in a surface condenser, particularly if the condensate is drawn off a t a number of points to minimize mixing. In the ordinary condensation of binary mixtures in connection with distillation, these two methods of condensation lead to quite different results, but it will be shown later that, as far as the amount of condensate obtained from the condensation of vapors from mixtures containing noncondensable gas is concerned, the results are practically the same. This is a fortunate fact as it greatly simplifies the mathematical treatment. We will now proceed to develop the theory for both types of condensation, in order to prove this point and obtain equations which can be applied to the solution of problems. SIMPLSCONDENSATION-In developing equations for simple condensation, the following nomenclature will be used : ’
x
Temperature “C FIG. PER CENT CONDENSATION AS A FUNCTION OR FINAI. TEMPERATURE FOR THREEDIFFERENT CONCENTRATIONS O F HEXANEA T .&TMOSPHERIC PRESSURE)
Consequently, it will be necessary, in deriving a formula for per cent condensation, to have a relation between liquid composition and vapor composition a t equilibrium for the components under consideration. Raoult’s law expresses this relation exactly for ideal solutions and approximately for a great many solutions, in which the components are somewhat related. For example, such binary mixtures as benzene and toluene, hexane and heptane, methanol and ethanol, and many others, obey Raoult’s law very closely. In the discussion that follows it will be assumed that we are interested only in mixtures that obey Raoult’s law a t least approximately.5 For mixtures not obeying Raoult’s law the vapor pressure relations are more complex and such mixtures will be considered beyond the scope of this paper. Before developing the theory, it is necessary to recognize that a condensation may take place in two ways.6 These two ways are limiting ones in that all condensations in practice are intermediate between the two, but approach one or other of the limiting types. Let us consider a condensation to be composed of a series of very small steps, each yielding a very small increment of condensate between given temperature limits. If we allow all these successive increments of condensate t o mix and remain in contact with the residual gas, then we have a simple condensation. The distinguishing characteristic ol simple condensation is, then, that a t any instant in the process the residual vapor is in equilibrium with the total amount of liquid originally associated with it. If, on the other hand, we keep the small increments of condensate separate during the process, then a t any instant the residual vapor will be in equilibrium only with the last increment of condensate, and the condensation may be said to be a differential one. Neither of these limiting types is exactly realized in practice but the common cases approach one or the other type. For instance, if a mixture of noncondensable 6 I t should be noted, in this connection, that some mixtures apparently deviate widely from Raoult’s law when the normal molecular weights are used, but approximately obey the law when a fictitious molecular weight is used for one of the components. The equations t o be developed apply to such mixtures. e W. K. Lewis, THIS JOURNAL, 1 (1909), 522: Alcan Hirsch, Ibid., 2 (1910),409.
= mol fraction of a component in the gas phase. y = mol fraction of a component in the liquid phase. The prime mark (’) and double prime (”) will represent initial and final states, respectively. D = total pressure, which is a constant in any given condensation. = vapor pressure of a pure liquid component a t the final temperature in the process. Subscripts a , b, c, etc., will refer to the components. W = number of mols of condensate. Thus yatr would represent a mol fraction of component “A” in the liquid in the final condition.
Taking one mol of initial gas as the basis for all calculations, we may establish the following condition equations for the case of two vapors mixed with noncondensable gas from Raoult’s law and that of conservation of mass :
equations 2 and 3 state that for each component the partial pressure of that component in the final vapor equals the vapor pressure of the same component over the final Condensate. In other words, we are equating the partial pressure givenby Dalton’s law to the vapor pressure given by Raoult’s law. Equations 4 and 5 state that the amount of a given component condensed from the gas equals the amount of the same component appearing in the liquid. Equation 6 states that the sum of the mol fractions of all the cbmponents in the liquid equals unity. There are five unknowns, the four final mol fractions and W, the amount condensed, and there are five equations; therefore, we can solve for W. Omitting intermediate steps, we obtain: xa
+
’
Xb‘
W ( l - g ) + E
For the sake of abbreviation, let P 5a = ai, = a2,1 a1 =
$
-
kl,
Pb
and 1
(7)
= 1
-
LYZ
=
kz.
The Equation 7 becomes:
By noting that Equation 8 is a quadratic of the form AW2 C = 0, where A = - ki kz B = %’a kz X b ‘ ki - LYI kz - LYZ kl
+
c
= X’a
012
+ Xb‘
LY1
-
LY1 (Yz
we can solve directly for W the amount of condensate.
+ BW
Vol. 14, No. 11
T H E JOUBNAL OF INDUSTRIAL A N D ENGINEERING CHEMIXTRY
1064
Thus W =
-BBdB2-4AC 2A
(9)
For any given set of conditions, A, B, and C may be determined and W obtained. W will equal the mols condensed per mol of original gas. If M a and Mb are the molecular weights of components A and B, respectively, then the lbs. condensed per mol of gas will equal (ya” W M a Yb” W M b ) . This can readily be determined after W has been calculated. Before applying this equation, we will next derive the analogous equation for differential condensation, after which we will compare the two on definite numerical problems. DIFFERENTIALCoNDENsATroN-We may again write the conditions for equilibrium, namely, the equality of partial pressure of a. component in the vapor and the vapor pressure of the same component over the liquid, omitting the primes t o indicate any instant.
+
It is more convenient for our purpose to express the mol fractions as a ratio’ of number of mols of each component to a total number of mols. Thus, if n = number of mols of a component in the gas, m = number of mols of a component in the liquid, N = total mols of gas, and M = total mols of liquid, then Equations LO and 11 become:
Dividing (12) by (13) we obtain
2
=
Pa
X
ma
(14)
If only a differential amount of liquid is to be in equilibrium with the gas at any point according to our definition of differential condensation, then we can replace ma and mb by dma and dmb, which are obviously equal to dna and dnb, respectively. There results:
To integrate this equation, we may assume that P,/Pb is a constant, c. This is only an approximation as c varies with the temperature throughout the condenser, but the variation in c is not great if the temperature range is narrow. It can also be shown that considerable variation in c affects the final result but slightly. The integral of :his equation is obviously C
log %a = log ‘iZb
+K
. (17)
where K is a constant involving initial and final conditions. 7 The equation given by Lewis8 for condensation of distilled liquids is, when expressed in the above nomenclature:
When dealing with a binary mixture of two condensable components with no noncondensable gas present, this equation can be solved by expressing y as a function of x and integrating, either by straight mathematical processes or by a graphical method. I n the case of the condensation of vapors from mixtures containing noncondensable gas, y is not only a function of 2, but also of W, if x is based on the total gas, condensable plus noncondensable. The differential equation obtained by expressing y as a function of x and W is not readily solved by ordinary methods. On the other hand, if x is based on the condensable gas only, then y is a definite function of x only, but in t h a t case the total pressure, which as will appear later, enters into the constant of integration, must also be the total pressure on the condensable components only, and hence will not be constant. Consequently, we were led to develop the equation in another form suggested by a similar equation for distillation in Young’s “Fractional Distillation.”
For the purpose of calculating W, the amount condensed, i t is simpler to integrate between limits. The upper limits are nu’ and nb’, respectively, and the lower limits, nu“ and nb”, will involve W. Consequently, we will obtain equations relating W with other variables which will be known for any specific problem and we can a t once solve for W. na”and q,”, the number of mols of the two condensable components in the gas at the end of the condensation, are not directly known but can be expressed in terms of the initial number of mols, and W, and certain known constants. The final equations are given as follows:
D and
(%a’
+ - W) - P a (1 - W) Wb’
D (1
nb’’ =
-
C)
(19)
These two equations are based on I mol of initial gas. Substituting these limits in the integrated form of the equation, we obtain as a final equation for differential condensation,
This equation contains a relation between the amount condensed during a differential condensation and the final temperature in the condenser, given the initial concentration and the total pressure. The temperature does not appear as such in the equation, but is involved in pa, the vapor pressure of pure A a t the final temperature. The relation is not a direct one, but the equation may be solved readily by trial and error, and it is just as well to leave it in the above form. The method of solution is fairly simple, usually involving only a few trials. APPLICATION OF THE EQUATIONS P\OR Two COMPONENTSWe have now two equations for calculating W, one for simple condensation and one for differential condensation. The former is much the simpler and easier to use, and it happens that the two give practically the same result for W over quite wide limits, so that the simpler one can generally be used for both cases.* The best way to show this is to take a number of cases and actually use both equations and compare the results. The following table shows the results of several calculations : PGR
CENT C O N D E N S E D
BY
WEIGRT~
Simple Condensation
CASE
Differential Condensation
MOLP E R
CENT OF A’ I N RESIDUAL VAPOR Simple Differential CondensaCondensation tion
THE
3 per cent benzene
and 1 per cent toluene compressed to 516Omm. (1001bs./ sa. in. abs.) and cdoled a t constant press to Z O O C.8 71.8 68.7 87.5 94.5 20 per cent benzene and 10 per cent toluene cooled to 20° C. a t 760 mm. 81.6 76.6 87.0 88.8 3 per cent benzene and 1 per cent toluene compressed to 35 Ibs./sq. in. abs. and cooled to 20’ C... 28.2 27.5 15 per cent methanol and 5 per cent ethanol cooled to 0“ C. a t 760 mm.. 88 0 86.0 1 Per cent figure based on oapov alone without the noncondenssble gas. 9 A refers t o the more volatile component. a Per cent by volume in mixtures with air or other noncondenssble gas.
....
.... . . . . . . . . .
....
....
.. .. .
....
....
It is quite evident that the two equations give practically the same result as far as weight of condensate is concerned, for a 8 It should be noted t h a t the two equations also apply to the case of binary vapor mixtures containing no noncondensable gas, but i t is by no means true t h a t the two different equations lead to even approximately the same amount condensed in many cases.
THE J0URATAL OF INDUSTRIAL A N D ENGINEERING CHEMISTRY
Nov., 1922
wide variety of conditions. As would be expected, the differential condensation leaves a residual vapor considerably richer in the more volatile component than does simple condensation. The mol per cent of A in the residual vapor is no" x 100
nu"
+ nb*
For differential condensation, nu'' is given by Equation 18 and nb" is given by Equation 19. For simple condensation we can obtain xav by combining Equations 2 and 4, giving:
and then n,"
=
xa" (1 - W)
These equations were used to obtain the mol per cent of A in the residual vapor, figures for which are given in the table. Let us now consider more in detail the application OF the equations to some more numerical problems. Having shown that the two different equations are identical for most practical purposes as far as the amount of condensate is concerned, we may use only the simpler one. Assume that we had a mixture of benzene and toluene vapor in air a t l atm., containing 3 per cent benzene and l per cent toluene by volume, and wished to know how much, if any, would condense a t 20' C. In the first place, it is obvious that x =:
V - where V 100'
=
volume per cent of the gas in ques-
From vapor pressure tables we find that fia = 75 mm. and 75 22 pb = 22 mm. Then CY, = - = 0.0986 and as = 760 760 -0.029; K1 = 0.9014 and kl = 0.971. From these values we may solve for A, B, and C, to be used in Equation 9. tion.
-
Thus we find that: A = -0.875; B = -0.0838; C = -0.001. Substituting in Equation 9 we find that both roots are negative. In other words, nothing will condense from this mixture under these conditions. Furthermore if C = 0 in Equation 9, W = 0 is one root and the other is negative. That is, if C = 0, the mixture is just a t the saturation point. Equating C t o 0, we see that: or
Xu' a2
f
xa' f i b
3. Xb' fia
7
(21)
f Xb'
Pa
