Analytically approximate solution to the VLE problem with the SRK equation of state

Hongqin Liu* Integrated High Performance Computing Branch, Shared Services Canada, Montreal, QC, Canada

Abstract

Since a transcendental equation is involved in vapor liquid equilibrium (VLE) calculations with a cubic equation of state (EoS), any exact solution has to be carried out numerically with an iterative approach [1,2]. This causes significant wastes of repetitive human efforts and computing resources. Based on a recent study [3] on the Maxwell construction [4] and the van der Waals EoS [5], here we propose a procedure for developing analytically approximate solutions to the VLE calculation with the Soave-Redlich-Kwong (SRK) EoS [6] for the entire coexistence curve. This procedure can be applied to any cubic EoS and thus opens a new area for the EoS study. For industrial applications, a simple databank can be built containing only the coefficients of a newly defined function and other thermodynamic properties will be obtained with analytical forms. For each system there is only a one-time effort, and therefore, the wastes caused by the repetitive efforts can be avoided. By the way, we also show that for exact solutions, the VLE problem with any cubic EoS can be reduced to solving a transcendental equation with one unknown, which can significantly simplify the methods currently employed [2,7].

*Emails: [email protected]; [email protected].

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Introduction contrast to the volumes of saturated liquid and vapor phases. By taking advantages of special features of the Vapor-liquid equilibrium (VLE) calculation plays a M-line, we will propose a procedure for developing central role in thermodynamics and process simulations analytically approximate solutions to the VLE problem [1,7]. With an equation of state (EoS), the chemical for a given cubic EoS. We demonstrate the procedure potential equilibrium condition, or the Maxwell with the Soave-Redlich-Kwong (SRK) EoS [6] by using the constructions [4], generates a transcendental function information at the critical point from Singley, Burns and with logarithm terms. As a result, exact solutions have Misovich (1997) [9]. to be carried out numerically by some iterative methods. The pressure equilibrium condition and the Exact solution to the VLE problem by a cubic EoS transcendental equation are usually solved together, Before targeting an analytical solution, here we first which is not a simple task and becomes a research topic show that the VLE problem with a cubic EoS can always by itself [2]. Worst of all, the efforts made by one be reduced to solving a transcendental equation with individual will have to be repeated by another even for one unknown. We use the SRK EoS as the example to the same system at the same conditions, unless prove the statement and it is applicable to any cubic numerical data are published, which is not practical. The EoS. The SRK EoS reads [6]: costs of computing resources are significant as well. As 푅푇 푎(푇 ) reported [7], the phase equilibrium computation is the 푃 = − (1) most expensive step in chemical process simulations. 푣 − 푏 푣(푣 + 푏) Some efforts have been made for developing analytical where, 푃 is pressure, 푇, temperature, 푣, molar volume, solutions, but mostly in the neighbourhood of the 푅, the gas constant, 푎(푇) is a function of temperature critical point [8,9]. The method aimed at the diameter (see the Appendix) and 푏 is a constant. For the SRK EoS, [10,11] of the coexistence curve seems pointing to a due to the adoption of a third parameter (species- right direction, but it is limited by the fact that both dependent), the acentric factor,휔, all the calculations saturated liquid and vapor volumes (densities) are have to be based on individual substance cases. From complex functions of temperature, which makes even the chemical potential equilibrium condition, we have: empirical correlations difficult. In practical applications, 푣 푣 휃푏 휃푏 − − + = engineers have to use empirical correlations for those 푣 − 푏 푣 − 푏 (푣 + 푏) (푣 + 푏) most useful properties. A well-known example is the 푣 − 푏 푣 + 푏 푣 + 푏 Antoine equation [13] for vapor pressure (the 푙푛 + 휃 푙푛 − 푙푛 (2) 푣 − 푏 푣 푣 equilibrium pressure of a pure system) correlations. where 휃 = 푎(푇 )⁄ (푅푇푏), 푣 and 푣 are molar volumes Others properties, including heat capacity, saturated for liquid and vapor phases, respectively. For the exact densities etc., have to be correlated separately [13]. solutions, we will generate a quadratic equation first. Theoretically, if an EoS works good enough for a given Throughout of this work, we only consider the VLE case system, the saturated volumes of the liquid and vapor where temperature is given. Applying the pressure phases will provide all the information needed for any equilibrium condition to liquid and vapor phases (see thermodynamic properties along entire the the Appendix, Eq.(S1)), we get: coexistence. 퐷 푣 − 푣 − (퐷 푏 +1− 휃)푏푣 − 휃푏 =0 (3) Moreover, for some applications where derivative properties, such as the vapor pressure vs temperature, where the liquid volume, 푣, has been replaced by a numerical solutions face difficulties. But with analytical dummy variable, 푣, and solutions, this task usually becomes minor, if not trivial. 1 휃푏 Consequently, analytical solution to the VLE problem is 퐷 = − (4) 푣 − 푏 푣(푣 + 푏) the ultimate goal for engineers and researchers. Eq.(3) is a cubic function. It is easy to prove that at When a cubic EoS is applied to the VLE calculation, temperature, 푇 ≥ 푇, there is only one root and as 푇 < subjected to the Maxwell construction, which is 푇 there are always three real (> 0) roots, and we denote equivalent to the combination of the pressure and them as 훼, 훽, 훾. According to the root-coefficient chemical potential equilibrium conditions, two real relations for a generic function, 푝푥 + 푞푥 + 푟푥 + 푠 = roots are for saturated volumes of liquid and vapor 0, we know that: phases, respectively, and the third “unphysical” root is 푞 1 the Maxwell crossover or the M-line [3]. It is found that 훼 + 훽 + 훾 =− = (5) 푝 퐷 the M-line is a smooth function of temperature, in

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푠 휃푏 1 훼훽훾 =− = (6) Cubic EoS 푝 퐷 푮 M-line 푟 퐷푏 +1− 휃 0.8 훼훽 + 훽훾 + 훼훾 = =− (7) 푪 푝 퐷 풆 푷 푻 푫 푩 Now we assume that 훼, 훽 represent two volumes: 푣, 푭 푷 0.6 푣, respectively, where 푣 is the liquid molar volume 푨 and 푣, the unphysical intermediate volume named as Liquid the volume of M phase. From Eq.(5)-(7), we can easily 0.4 푬 Vapor prove that 훾 = 푣, namely the vapor volume. Now we can construct a quadratic equation that has the same 풗푳 풗푴 풗푮 0.2 two roots (푣, 푣) as those from Eq.(3): 0 0.5 1 1.5 2 2.5 3 푣 + 푢푣 + 푤 =0 (8) 풗 where 1 Figure 1. The Maxwell construction and the M-line. At ( ) 푢 =− + 푣 9 equilibrium condition: area FEDF = area DBCD. 퐷 휃푏 The Maxwell can be written analytically as: 푤 = (10) 퐷 푣 1 푃 = 푃푑푣 (12) 푣 − 푣 Then we have the solution for liquid volume: −푢 − √푢 −4푤 where 푃 is the equilibrium pressure at a given 푣 = = 푓(푣 ) (11) 2 temperature. By re-arrangement, we have: where 푓(푣) is a given function of 푣. In Eq.(2), (푃 − 푃)푑푣 = (푃 − 푝 )푑푣 (13) replacing 푣 with Eq.(11), we have a transcendental equation with only one unknown, 푣, and this ends our which is the EQR. Eq.(13) means that area FEDF = area proof for previous statement. ∎ DBCD. Therefore, the EQR generates an intermediate With traditional method [3], the pressure and chemical volume, 푣. It should be mentioned that the Maxwell equilibrium conditions are solved at the time and high construction applies to any EoS, not necessarily a cubic quality initial guesses are demanded. The new method one. As a cubic EoS is adopted, the value of 푣 is the significantly simplifies the solution process. It is found “unphysical” third root since there is only one that the transcendental equation, Eq.(2) and (11), is intersection between the pressure curve 푃(푣) and the stable given a reasonable initial volume. It can be easily equilibrium pressure, 푃 . From Figure 1, we also see solved with the Excel Solver. All the exact solutions used that: in this work are obtained this way. 푃(푣) = 푃(푣) = 푃(푣) = 푝 (14)

The Maxwell construction and the M-line The state at 푣 represents an unstable state, since 휇(푣 ) ≠ 휇(푣 ) or 휇(푣 ), where 휇 is the chemical A full discussion on the Maxwell construction (known as potential. This means that a thermodynamic property at the equal-area rule, EAR) and the crossover is presented in [3] with the van der Waals EoS as an example. Figure this state is not a equilibrium property. Since we are only interested in the volume itself, which turns out to 1 illustrates the Maxwell construction and related be physically meaningful, the treatments below are quantities to help our presentations. physically sound. The trajectory of the intermediate volume, 푣, is called the Maxwell crossover or the M- line (Figure 1). The M-line can be easily calculated by the SRK EoS. In fact, the “unphysical” root of Eq.(8) is the intermediate volume as function of 푣: −푢 + √푢 −4푤 푣 = (15) 2 Therefore, as discussed above, the solution to the transcendental equation, Eq.(2), will give both 푣 and 푣 at a given temperature by Eq.(11) and (15),

3 respectively, as 푣 is known. Similarly, we can rewrite For the purpose of obtaining high accurate functions, Eq.(3) as the entire temperature range is divided into two regions: 0< 푇 ≤, 푇 and 푇 < 푇 ≤1. The 퐷푣 − 푣 − (퐷푏 +1− 휃)푏푣 − 휃푏 =0 (16) characteristic temperature, 푇 , is determined 퐷 퐷 where is defined the same way as by Eq.(4) with empirically, such that best results for both regions can 푣 푣 being replaced by . Therefore we have: be obtained. After some tests, the final definition is: −푢 ∓ √푢 −4푤 푣 = (17) 2 푇 | 2 푇 = (20) 5 푇 where the subscript, 퐿|퐺, correspond to ∓, respectively. 푇 Eq.(17) provides 푣 and 푣 as 푣 is known. From Eq.(11) where is the critical temperature of argon, and the and (17) we have the diameter of the coexistence empirical constant, 2/5, is only for the substances curve10: considered in this work (mainly hydrocarbons). One may choose other values with small deviations from Eq.(20), 휌 + 휌 1− 푣 퐷 퐷 푣 푣 (1− 푣 퐷 ) but other coefficients discussed below will have to be 푑 = = = (18) 2 2퐷 휃푏 2휃푏 changed accordingly. In the low temperature range, Or we can simply get 푣 as a function of 푑. This result 푇 ≤, 푇, from the pressure and chemical potential shows that the “unphysical root” does have a physical equilibrium conditions (see the Appendix), we can get meaning. Now we define an entropic function as analytically approximate solutions: following: 푣 1 푣 ≈ 휃 −1− 1−6휃 + 휃 (21) 푏 2 풮 = 푙푛(푣 −1) = 푙푛 −1 (19) 푏 푣 + 푏 Figure 2 plots the function, 풮, for three phases for 푣 ≈ 푒(푣 − 푏) (22) 푣 ethane, where M-phase is represented by the M-line. This figure shows a remarkable comparison between where 푒 is the natural constant. In the temperature the three phases: M-line behaves much smoother than range, 푇 < 푇 ≤1, we use a polynomial function: the other two. Therefore, we use the M-line as the base for developing our analytical solutions. 풮 = 푙푛(푣 −1) = 퐶푇 (23) 10 M phase where 푣 = 푣 ⁄푏. Eq.(23) has 6 coefficients, from 8 풮 Liquid phase which we have 푣 = 푏(1+ 푒 ). At low temperature 6 Vapor phase end point, 푇 = 푇, we impose the following 4 Ethane (SRK EoS) constraints generated from chemical potential condition: 퓢 2 푑 휇 푑 휇 = , 푛 = 0,1,2 (24) 0 푑푇 푑푇

-2 3 coefficients can be determined by Eq.(24), where 푣 and 푣 are calculated with Eq.(21) and (22), -4 respectively. Given SRK chemical potential and the 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 volume expressions discussed above, it is 푻풓 straightforward to use Eq.(24) (see the Appendix for

Figure 2. Plots of the function 풮 = 푙푛(푣 −1) for different details) for determining three coefficients. phases (ethane). All curves are calculated with exact solutions, For other 3 coefficients, we use the information at the Eq.(2), (11) and (15). critical point, which requires some elaboration. Singley, Burns and Misovich (1997) [9] have derived a series The analytical solution to the VLE problem with expansion at the critical point up to 10th order. The the SRK EoS reduced saturated densities of liquid and vapor phases are given by the following, respectively: 휌 =1+ 퐵(1− 푇) + 퐵(1− 푇) + 퐵(1− 푇) + 퐵(1− 푇) + 퐵(1− 푇) + 퐵(1− 푇) +⋯ (25)

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휌 =1− 퐵(1− 푇) + 퐵(1− 푇) − 퐵(1− 푇) + 퐵(1− 푇) − 퐵(1− 푇) + 퐵(1− 푇) −⋯ (26) We define the following functions: Ӽ= 휌 + 휌 =2+2퐵(1− 푇) +2퐵(1− 푇) +2퐵(1− 푇) +⋯ (27) Ỿ= 휌휌 =1+ (2퐵 − 퐵 )(1− 푇) + (2퐵 −2퐵퐵 + 퐵퐵)(1− 푇) + (2퐵 −2퐵퐵 +2퐵퐵 − 퐵퐵)(1− 푇) + (−2퐵퐵 +2퐵퐵 + 퐵퐵 +⋯ )(1− 푇) − 퐵퐵(1− 푇) + 퐵퐵(1− 푇) + 퐵퐵(1− 푇) … (28)

Where the coefficients, 퐵, 퐵 etc. are given functions Therefore, at the critical point we have three quantities, of the acentric factor [9]. Then we have: which are functions of 퐵 (i=1,2…): 푣 푣 (휌 + 휌)푣 푣Ӽ 푑 푣 푣 + 푣 = + = ≡ (29) , 푛 = 0,1,2 (40) 휌 휌 휌휌 Ỿ 푑푇 Eq.(17) leads to: The missing details about calculating the above derivatives are provided in the Appendix. Eq.(40) will be 1 1− 푣퐷 푣 + 푣 =−푢 = − 푣 = (30) used to determine 3 coefficients in Eq.(23) with linear 퐷 퐷 functions. For the SRK EoS, 풮 = 1.046384. In all the And calculations, the critical volume is calculated from 푣 = 휃푏 푍푅푇⁄푃. 푣푣 = 푤 = (31) 퐷푣 In total we have 6 equations and 3 are non-linear where functions, Eq.(24). The calculations have been carried in 1 휃푏 the Excel Solver. The derivatives required for the 퐷 = − (32) chemical potential constraints are derived from Eq.(17) 푣 − 푏 푣 (푣 + 푏) (Appendix). At the critical point, we have: In summary, in the low temperature range, 푇 ≤ 푇, 1 1 1 Eq.(21) and (22) will be used to calculate the volumes. 퐷 = + 휃 − (33) 푣 − 푏 푣 + 푏 푣 In the high temperature range, 푇 < 푇 ≤1, Eq.(23) 푑퐷 1 1 푑휃 and (17) will used to calculate the saturated volumes. = − (34) Finally in the entire temperature range, the equilibrium 푑푇 푣 + 푏 푣 푑푇 pressure is calculated by 푑 퐷 1 1 푑휃 푑푣 =2 − + ( ) ( ) 푅푇 푣 − 푏 푎 푇 푣 푣 + 푏 푑푇 푣 (푣 + 푏) 푑푇 푑푇 푃 = 푙푛 − 푙푛 (41) 푣 − 푣 푣 − 푏 푏(푣 − 푣 ) 푣 (푣 + 푏) 1 1 푑휃 − (35) 푣 + 푏 푣 푑푇 Which can be obtained from Eq.(1) and (12). From the analytical expressions for the volumes and pressure, we Where we have used the critical conditions: can derive all other thermodynamic properties along the entire coexistence curve analytically as they are 휕퐷 1 1 1 =− + 휃 − =0(36) 푣 푎푛푑 푣 휕푣 (푣 − 푏) 푣 (푣 + 푏) functions of . Here we provide a few of them. Enthalpy is given by: 휕 퐷 2 2 2 = + 휃 − + =0(37) 푎 푇푎 푣 + 푏 휕푣 (푣 − 푏) 푣 (푣 + 푏) 퐻 = 퐻∗ + 푅푇(푍 −1) − 1− 푙푛 (42) 푏 푎 푣 Finally: where 푎 = 푑푎⁄ 푑푇 (Appendix). Then we have the latent 푑푣 1 푑퐷 1 푑푣 푑푣 heat, ∆퐻 , from the enthalpy: =− − + (38) 푑푇 퐷 푑푇 푇 푑푇 푑푇 ∆퐻 푎 − 푎 푇 푣(푣 + 푏) = 푍 − 푍 − 푙푛 (43) 푑 푣 1 푑 퐷 2 푑퐷 푅푇 푅푇푏 푣(푣 + 푏) =− + − 푑푇 퐷 푑푇 퐷 푑푇 The isobaric heat capacity is given by: 1 푑푣 푑푣 + (39) 푇 푑푇 푑푇

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퐶 퐶 푎 (푇) 푣 + 푏 1 = −1+ 푙푛 + 푅 푅 푏푅 푣 0.9 Ethane 푣(푣 + 푏) −2푅푎(푇)푣(푣 − 푏) + 푅푎(푇)(푣 − 푏) (44) 0.8 푣(푣 + 푏) − 푎(푇)(푅푇)(2푣 + 푏)(푣 − 푏) Liquid: exact 0.7 Vapor: analytical where the heat capacity of the ideal gas, 퐶 , is from Ref Liquid: analytical 푻풓 0.6 [12]. 푎 (푇) is provided in Appendix. M-phase: analytical 0.5 Exp Liquid Another advantage of using the analytical solution is for Exp. Vapor derivative properties. For example, the well-know 0.4 M-phase: exact

Clapeyron equation correlates the latent heat with the 0.3 vapor pressure derivative [1,8]: 0 0.5 1 1.5 2 2.5 3 3.5 흆 풓 ∆퐻 푅푇 푣 − 푣 푑푃 = (45) 푃푣 푃푣 푅 푑푇 Figure 4. Calculation results with the SRK EoS for densities of A thermodynamic consistency testing is the comparison Ethane. Open triangles are experimental data [14]; open circles are exact solutions; solid lines are from analytical between the results from Eq.(43) and (45). By numerical solutions. solution, using Eq.(45) would be difficult since numerical differentiation is required. With analytical solution, 50.0 Ethane applying Eq.(45) is a minor effort. can be obtained 40.0 from Eq.(41) and the volume derivatives from Eq.(17) Exp Liquid (see Appendix for details). Exp Vapor 30.0 For demonstration purpose, in this work we totally 푪푷 Liquid: analytical 푹 considered 8 substances for vapor pressure predictions 20.0 Vapor: analytical and more details are presented for argon and ethane. The critical properties and other constants are listed in 10.0 Table 1. The coefficients of Eq.(23) obtained for those substances are listed in Table 2. Other data used for the 0.0 calculations are listed in Table S1. 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 푻풓 1 Figure 5. Isobaric heat capacity calculated by the analytical 0.9 Argon solutions, Eq.(44). The differences between the exact and analytical solutions are negligible compared to those between 0.8 Vapor: analytical the experimental data [15] and predictions. 0.7 vdW EoS Liquid 푻 vdW EoS Vapor 6 풓 Ethane 0.6 liquid: analytical liquid: exact 5 0.5 M-phase: analytical Exp Vapor 4 ∆푯 0.4 Exp Liquid 풗 Clapeyron equation 푹푻 M-phase: exact 푪 3 0.3 Exp 1991 2 0 0.5 1 1.5 2 2.5 3 3.5 EoS entropy 흆풓 1

Figure 3. Calculation results with the SRK EoS for densities of 0 Argon. Open triangles are experimental data [13]; open circles 0.5 0.6 0.7 0.8 0.9 1.0 푻 are exact solutions; solid lines are from analytical solutions; 풓 dashed lines are from vdW EoS, which obviously cannot serve Figure 6. Thermodynamic constancy testing for the latent quantitative prediction purpose. heat. The results from the Clapeyron equation, Eq.(45), are from the analytical solution . The differences between the exact and analytical solutions are again negligible. The experimental data are from ref [15].

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45 discrepancies between the analytical solutions and n-Butane 40 exact solutions are all negligible. Figure 5 depicts the Mathane isobaric heat capacities calculated by the analytical 35 Heptane solutions, Eq.(44), (17) and (23), vs experimental data. 30 Ar Figure 6 depicts the thermodynamic consistency testing Ethane 푷 (풃풂풓)25 with the SRK EoS, calculated by Eq.(43) and (45), cyclohexane respectively. The agreements between the two 20 n-Hexane methods are excellent. The application of the Clapeyron 15 Benzene equation requires 푑푃⁄푑푇, which is calculated with 10 Eq.(41) and the derivatives of saturated volumes are calculated from the analytical solutions, Eq.(17) and 5 (23). The details are provided in the Appendix. 0 0.5 0.6 0.7 0.8 0.9 1 Figure 7 presents the equilibrium pressure calculated by 푻풓 Eq.(41) with analytical expressions for the volumes. The agreements between the analytical solutions and exact Figure 7. The equilibrium pressure prediction results by the solutions are above 2-orders of magnitude higher than SRK EoS for all the systems considered in this work. The solid those when compared with the experimental data, curves are calculation results, from Eq.(41) with analytical therefore satisfy the needs of all practical applications. volume expressions, and points are experimental data. The data sources: Ar, [13]; Methane, [16]; Ethane, [14]; n-Butane, Finally, as an example, Table 3 lists a comparison [17]; cyclohexane, [19,20]; n-Hexane, [18]; n-Heptane, [21]; between the exact and analytical solutions for ethane. Benzene, [22]. The worst case is at 푇 = 0.6 and the low temperature Figure 3 and Figure 4 illustrate the densities calculated end at 푇 = 0.46. The table shows that for liquid volume by the analytical solution and the exact solutions for and the equilibrium pressure the agreements are Argon and Ethane, respectively. Compared with the excellent. For vapor volume the agreement is prediction errors against the experimental data, the expectable.

Table 1 Critical properties and constants used in the calculations

푇, 퐾 푃, 푏푎푟 푣, 푙/푚표푙 휔 푏 푎 Argon 150.8 48.7 0.0858 0.001 0.022306 1.3799 Methane 190.4 46 0.1147 0.011 0.029817 2.3289 Ethane 305.4 48.8 0.1734 0.099 0.045082 5.6480 n-Butane 425.2 38 0.3101 0.199 0.080605 14.0597 n-Hexane 507.5 30.1 0.4673 0.299 0.121457 25.2860 Cyclohexane 553.8 40.7 0.3771 0.212 0.098019 22.2682 n-Heptane 540.3 27.4 0.5465 0.349 0.142049 31.4842 Benzene 562.1 48.9 0.3186 0.212 0.082805 19.0938

푣 = 푍푅푇⁄푃 = 푅푇⁄3푃. The critical constants are from ref [12].

Table 2 Coefficients of Eq.(23) and AAD for pressure Substance AAD% 푻풓ퟎ 푪ퟎ 푪ퟏ 푪ퟐ 푪ퟑ 푪ퟒ 푪ퟓ Argon 0.40000 4.722378 -6.806245 4.570508 -1.460235 -0.123006 0.142984 0.0042 Methane 0.41910 4.662219 -6.239253 2.922949 0.763598 -1.575823 0.512695 0.0060 Ethane 0.46063 4.719780 -5.846706 1.998728 1.310195 -1.586006 0.450395 0.0041 n-Butane 0.49215 4.781632 -5.445759 1.037975 1.942411 -1.694007 0.424133 0.0043 n-Pentane 0.50205 4.536968 -3.398230 -4.296358 8.498408 -5.674155 1.379751 0.0013 cyclohexane 0.51886 4.935708 -6.329727 3.253134 -0.841432 0.017716 0.010986 0.0037 n-Hexane 0.50988 4.501827 -2.875797 -5.475596 9.607763 -6.181930 1.470117 0.0008

7 n-Heptane 0.51631 4.504431 -2.621153 -5.923348 9.760085 -6.085093 1.411463 0.0014 Benzene 0.52041 4.543005 -3.727193 -3.484854 7.684354 -5.266515 1.297588 0.0008 퐴퐴퐷%= ∑ , where the total number of data points, 푁 = 70, which equally divide a temperature range: 0.3 ≤ 푇 ≤1.

Table 3 A comparison between the results from the exact and analytic solutions for ethane 푇 푣 (exact) 푣 (exact) 푃 (exact) 푣 (analytic) 푣 (analytic) 푃 (analytic) 0.46 0.0523601 308.11 0.0378294 0.0523603 309.47 0.0378291 0.6 0.0571321 15.983 0.927126 0.0571319 15.704 0.926985

Conclusions and discussions The accuracies can be improved by selecting few points (exact solutions) in the middle range and re-fitting the This article presents a procedure for deriving explicit coefficients. After all, this is a one-time effort only and solutions to the VLE problem with a cubic EoS over the one can take other approaches. If the high order series entire temperature range. To author’s knowledge, no expansions are not available at the critical point, one such work has ever been reported before. This work will could only take some points from the exact solutions to change the way the VLE calculations are carried out with fit the coefficients. The functional forms of (23) are also a cubic EoS. The repetitive calculations are replaced a mater of choices, as mentioned. with one-time effort, namely finding the coefficients of Eq.(23) (or alike) by solving few non-linear equations. In Finally, for a thermodynamic-property databank addition, for the exact numerical solutions to the VLE establishment used in practical applications, the data problem using a cubic EoS, we show that the pressure needed are only the coefficients of the M-line, Eq.(23) and chemical potential equilibrium conditions can in this case. The rest will be some analytical equations always be reduced into a single transcendental function and the prediction results will be thermodynamically with one unknown. This makes the exact solution easier consistent. than solving two equations at the same time. Acknowledgement This work suggests some new areas in cubic EoS studies. The author is grateful to Dr. Misovich for providing his For demonstration purpose, we only discussed the SRK research information and publications. EoS. There are dozens of cubic EoS’s which are widely used in various applications1,23-25. Many of them are References more accurate than the original SRK EoS [6]. It is 1. Orbey, H. and Sandler, S. I. Modeling vapor-Liquid Equilibria: expected that the method developed here can be Cubic equations of state and their mixing rules. Cambridge applied to any cubic EoS. First of all, the series University Press, Cambridge, UK. 1998. expansions at the critical point need to be carried out 2. Sequra, H. & Wisniak, J. Calculation of pure saturation for the EoS interested. The work of Singley, Burns and properties using cubic equations of state. Comput. Chem. Eng. 21, 1339-1347 (1997). Misovich (1997) [9] provides such an example. 3. Liu, H. The Maxwell crossover and the van der Waals One of the objectives of using current procedure is that equation of state. To be published. the analytical equations possess prediction power, 4. Maxwell, J. C. On the dynamical evidence of the molecular which means that all properties can be predicted by constitution of bodies. Nature, 4, 357-359 (1875). using the properties at two “end” points only, namely 5. Van der Waals, J. D. On the continuity of the gaseous and the critical point and the characteristic low temperature liquid states, ed. By Rowlinson, J.S. Dover Pub. Inc. Mineola, N.Y. (1988). point, 푇. It is found that in a narrow “middle temperature (density) range the predicted vapor 6. Soave, G. Equilibrium constants from a modified Redlich- volumes are less accurate, sometimes with the average Kwong equation of state. Chem. Eng. Sci., 27, 1197-1203 (1972). absolute deviation (AAD) of 1+%, when compared with the exact solutions. The equilibrium pressure, Eq.(41), 7. Joseph, A., Sands, C. M., Hicks, P. D. & Chandler, H. W. and liquid density prediction are with much higher Convex hull method for the determination of vapour liquid equilibria (VLE) phase diagrams for binary and ternary accuracies (see Table 2). For industrial applications, the systems. Fluid Phase Equilibria, 431, 34-47 (2017). accuracies for predicted properties are good enough.

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8. Johnston, D. C. Advances in thermodynamics of the van de 17 Miyamoto, H. and Uematsu, M. Measurements of vapour Waals Fluid. Morgan and Claypool Publishers San Rafael, CA, pressures and saturated-liquid densities for n-butane at T = (2014). Summary: arXiv:1402.1205. (280 to 424) K. J. Chem. Thermodynamics 39, 827–832 (2007). 9. Singley, C. D., Burns, I. P. and Misovich, M. L. Evaluation of 18. Ewing, M. B. and Ochoa, J. C. S. Vapour pressures of n- series coefficients for classical equations of state. Fluid Phase hexane determined by comparative ebulliometry. J. Chem. Equil., 137, 99-109 (1997). Thermodynamics 38, 283–288 (2006). 10. Rowlinson, J. S., Are diameters rectilinear? Nature, 319, 19. Weclawski, J. and Bylicki, A. A new apparatus for total- 362 (1986). pressure measurements by the static methods: application to 11. Sushko, M. Y. and Olexandr M. Babiy, O.M. Asymmetry of the vapor pressure of cyclohexane, propan-2-ol and pyridine. the vapor–liquid coexistence curve: the asymptotic behavior Fluid Phase Equil., 12, 143-153 (1983). of the “diameter”. J. Mol. Fluids, 158, 68-74 (2011). 20. Dortmund data bank: Vapor Pressure of Cyclohexane. 12. Reid, R. C., Prausnitz, J. M. and Poling, B. E. The properties http://www.ddbst.com/en/EED/PCP/VAP_C50.php of gases and liquids.4th ed. McGraw-Hill, NY, 1987. 21. Weber, L. A. Vapor Pressure of Heptane from the Triple 13. Gilgen, R., Kleinrahm, R. and Wagner, W. Measurement Point to the Critical Point. J. Chem. Eng. Data, 45, 173-176 and correlation of the (pressure, density, temperature) (2000). relation of argon II. Saturated-liquid and saturated-vapor 22. Dortmund data bank: Vapor pressure of benzene. densities and vapour pressures along the entire coexistence http://www.ddbst.com/en/EED/PCP/VAP_C31.php. curve. J. Chem. Thermodynamics, 26, 399-413(1994). 23. Lopez-Echeverry, J. S., Simon Reif-Acherman, S. and 14. Funke,M., Kleinrahm, R. and Wagnera, W. Measurement Araujo-Lopez, E. Peng-Robinson equation of state: 40 years and correlation of the (p, q, T) relation of ethane II. Saturated- through cubics. Fluid phase Equil., 447, 39-71 (2017). liquid and saturated-vapour densities and vapour pressures 24. Wei, Y. S. & Sadus, R. J. Equation of state for the calculation along the entire coexistence curve. J. Chem. Thermodynamics, of fluid-phase equilibria. AIChE J. 46, 169-196 (2000). 34, 2017-2039 (2002). 25. Young, A. F., Pessoa, F. L. P. and Ahón, V. R. R., Comparison 15. Friend, D. G., Ingham, H., and Ely, J. F. Thermophysical of 20 Alpha Functions Applied in the Peng−Robinson Equaon properties of ethane. J. Phys. Chem. Ref. Data, 20, 275-285 of State for Vapor Pressure Estimation. I&EC Research, 55, (1991). 6506-6516 (2016). 16. Dortmund data bank: Vapor pressure of methane. http://www.ddbst.com/en/EED/PCP/VAP_C1051.php Appendix The SRK EoS and related relations The pressure equilibrium condition for three phases:

푅푇 푎(푇) 푅푇 푎(푇) 푅푇 푎(푇) 푃 = − = − = − (푆1) 푣 − 푏 푣(푣 + 푏) 푣 − 푏 푣(푣 + 푏) 푣 − 푏 푣(푣 + 푏) where 푎(푇) = 푎훼(푇) [6] and 훼(푇) = 1+ 푓(휔) 1− 푇 (푆2)

The function for the acentric factor is given by: 푓(휔) = 0.480 + 1.574휔 − 0.176휔 (푆3) and: 푅 푇 푅푇 푎 = 0.42747 , 푏 = 0.08664 (푆4) 푃 푃

푎(푇) 푎 휃 = , 휃 = = 4.933864 = 훺 (푆5) 푅푇푏 푅푇푏 From Eq.(S5), we get: 푑휃 푎(푇) 푎(푇) 푑휃 푎(푇) 2푎(푇) 2푎(푇) = − , = − − (푆6) 푑푇 푅푇푏 푅푇푏 푑푇 푅푇푏 푅푇푏 푅푇푏 ( ) 푑푎 푎푓 휔 푎 = =− 1+ 푓(휔) 1− 푇 (푆7) 푇푑푇 푇푇

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푎 푓(휔) 푓(휔) ( ) ( ) 푎 = 1+ 푓 휔 1− 푇 + 푆8 푇 2푇 푇 From Eq.(41), we have: 푑푃 푅 푣 − 푏 푎(푇) 푣 푣 + 푏 = 푙푛 − 푙푛 − 푙푛 푑푇 푣 − 푣 푣 − 푏 푏(푣 − 푣) 푣 푣 + 푏 푅푇 푣 푣 + 푏 푣 − 푏 푑푣 푑푣 + 휃 푙푛 − 푙푛 − 푙푛 − (푣 − 푣) 푣 푣 + 푏 푣 − 푏 푑푇 푑푇 푅푇 1 휃 휃 푑푣 1 휃 휃 푑푣 + − + − − + 푣 − 푣 푣 − 푏 푣 푣 + 푏 푑푇 푣 − 푏 푣 푣 + 푏 푑푇 (S9) Low temperature approximations For self completeness, few equations appeared in the main test may reappear here. From the SRK EoS: 푃 1 휃푏 1 휃푏 = − = 퐷 = − (푆10) 푅푇 푣 − 푏 푣(푣 + 푏) 푣 − 푏 푣(푣 + 푏) where 1 휃푏 퐷 = − (푆10푎) 푣 − 푏 푣(푣 + 푏) From Eq.(9) we have: 퐷푣 − 푣 − (퐷푏 +1− 휃)푏푣 − 휃푏 =0 (푆11)

Low temperature, 푇 ≤ 푇, 푣 ≫ 푏, 푣 ≫ 푏휃. In fact, since 휃 >1 the low temperature condition can be simply defined as 푣 ≫ 푏휃. From Eq.(10a) we have, 퐷 ≈1⁄푣 , which leads to (퐷푣 ~0, 퐷푏~0), then Eq.(S1) is simplified to: 푣 + (1− 휃)푏푣 + 휃푏 =0 (푆12) Therefore: 푏 푣 ≈ 휃 −1− 1−6휃 + 휃 (푆13) 2

This equation provides high accuracy for 푣, as 푇 ≤ 푇. However, equilibrium pressure cannot be calculated from it since the chemical potential equilibrium condition has not been involved. Therefore, for 푣, the chemical equilibrium condition, Eq.(2), has to be imposed. At low T, 푣 ≫ 푏, 푣 ≫ 푏휃, 푍 →1 (ideal gas), 푍 →0, finally we have:

푣 + 푏 푙푛푣 =1+ 휃푙푛 + 푙푛(푣 − 푏) (푆14) 푣

푣and 푣 calculated from the approximate equations agree with strict solution with high precisions. The differences between the equilibrium pressures from the strict and approximate solutions at low temperature are negligible (with ADD<0.01%). For example, for ethane, 푏 = 0.045083,, 푙/푚표푙, 푇 = 0.46063, at 푇 = 0.46, 푏휃 = 0.70098, strict 푣 = 308.11, 푙/푚표푙, therefore 푣 ≫ 푏휃 holds solidly. The equilibrium pressure (for all temperature range) can be calculated with Eq.(41). By the way, from Eq.(S13) and (S14) we get the derivatives:

푑푣 푏 3− 휃 푑휃 ≈ 1+ (푆15) 푑푇 2 √1−6휃 + 휃 푑푇

1 푑푣 푣 + 푏 푑휃 휃 휃 1 푑푣 = 푙푛 + − + (푆16) 푣 푑푇 푣 푑푇 푣 + 푏 푣 푣 − 푏 푑푇 푑 푣 푏 3− 휃 푑 휃 4푏 푑휃 = 1+ + (푆17) 푑푇 2 푑푇 푑푇 √1−6휃 + 휃 (1−6휃 + 휃)

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1 푑 푣 1 푑푣 푣 + 푏 푑 휃 2푏 푑휃 푑푣 = + 푙푛 − + 푣 푑푇 푣 푑푇 푣 푑푇 푣(푣 + 푏) 푑푇 푑푇 휃 휃 1 푑푣 휃 휃 1 푑 푣 − − + − + (푆18) 푣 (푣 + 푏) (푣 − 푏) 푑푇 푣 + 푏 푣 푣 − 푏 푑푇 The above equations can be used in the low temperature range as needed. For example for calculation of the pressure derivative, Eq.(S9), the first derivatives are required. Derivatives for coefficient evaluations for Eq.(23) As show in the main test, we define: 1 휃푏 퐷 = − (푆19) 푣 − 푏 푣(푣 + 푏) 1 푢 =− + 푣 (푆20) 퐷 휃푏 푤 = (푆21) 퐷푣 Then: −푢 ∓ √푢 −4푤 푣 = (푆22) | 2 From the above, we have the derivatives:

푑퐷 1 휃푏 1 1 푑푣 푏 푑휃 = − + + − (푆23) 푑푇 (푣 − 푏) 푣(푣 + 푏) 푣 푣 + 푏 푑푇 푣(푣 + 푏) 푑푇

푑푢 1 푑퐷 푑푣 = + (푆24) 푑푇 퐷 푑푇 푑푇 푑푤 휃푏 1 푑퐷 1 푑푣 푏 푑휃 =− + + (푆25) 푑푇 퐷푣 퐷 푑푇 푣 푑푇 퐷푣 푑푇

푑푣 1 푢 푑푢 1 푑푤 =− 1− − (푆26) 푑푇 2 √푢 −4푤 푑푇 √푢 −4푤 푑푇

푑푣 1 푢 푑푢 1 푑푤 =− 1+ + (푆27) 푑푇 2 √푢 −4푤 푑푇 √푢 −4푤 푑푇 푑 퐷 1 휃푏 휃푏 푑 푣 = − + + 푑푇 (푣 − 푏) 푣(푣 + 푏) 푣(푣 + 푏) 푑푇 2 2휃푏 2휃푏 2휃푏 푑푣 + − − − (푣 − 푏) 푣(푣 + 푏) 푣(푣 + 푏) 푣(푣 + 푏) 푑푇 푏 푏 푑푣 푑휃 푏 푑 휃 +2 + − (푆28) 푣(푣 + 푏) 푣(푣 + 푏) 푑푇 푑푇 푣(푣 + 푏) 푑푇 푑 푢 1 푑 퐷 2 푑퐷 푑 푣 = − + (푆29) 푑푇 퐷 푑푇 퐷 푑푇 푑푇 푑 푤 2푏 1 푑퐷 1 푑푣 푑휃 푏 휃 1 푑 퐷 1 푑 푣 1 푑 휃 = 푔 − + − + − (푆30) 푑푇 퐷푣 퐷 푑푇 푣 푑푇 푑푇 퐷푣 퐷 푑푇 푣 푑푇 휃 푑푇 2푏 휃 1 푑퐷 1 푑푣 푑퐷 1 푑푣 푔 = + + (푆31) 퐷푣 퐷 푑푇 퐷푣 푑푇 푑푇 푣 푑푇 From Eq.(S26) and (S27), we have: 푑푣 1 푑푤 =−푈 − 푈1− 푈2+ (푆32) 푑푇 ∆ 푑푇

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푑푣 1 푑푤 =−푈 + 푈1+ 푈2− (푆33) 푑푇 ∆ 푑푇 where: ∆= 푢 −4푤 (푆34푎) 1 푢 푑푢 푈 = 1− (푆34푏) 2 ∆ 푑푇 1 푑푢 푑푢 푢 푑푢 푑푤 푈1= − 푢 −2 (푆34푐) 2∆ 푑푇 푑푇 푢 −4푤 푑푇 푑푇 1 푑푢 푑푤 푑푤 푈2= 푢 −2 (푆34푑) 푑푇 푑푇 푑푇 (푢 −4푤)

The above equations will be used when applying the chemical potential constraints at the low temperature end, 푇 and at the critical point. The terms involving 푣 are given by Eq.(23) as functions of the unknown coefficients: 풮 푣 = 푏(1+ 푒 ) (푆35) Determining the coefficients of Eq.(23) 풮 = 푙푛(푣 −1) = 퐶 + 퐶푇 + 퐶푇 + 퐶푇 + 퐶푇 + 퐶푇 (푆36)

푑풮 = 퐶 +2퐶푇 +3퐶푇 +4퐶푇 +5퐶푇 (푆37) 푑푇 푑 풮 =2퐶 +6퐶푇 + 12퐶푇 + 20퐶푇 (푆38) 푑푇 At the critical point:

풮 = 퐶 + 퐶 + 퐶 + 퐶 + 퐶 + 퐶 (푆39)

푑풮 풮 = = 퐶 +2퐶 +3퐶 +4퐶 +5퐶 (푆40) 푑푇 푑 풮 풮 = =2퐶 +6퐶 + 12퐶 + 20퐶 (푆41) 푑푇

At the low temperature, 푇, we have: 풮 = 푙푛(푣 −1) = 퐶 + 퐶푇 + 퐶푇 + 퐶푇 + 퐶푇 + 퐶푇 (푆42)

푑풮 풮 = = 퐶 +2퐶푇 +3퐶푇 +4퐶푇 +5퐶푇 (푆43) 푑푇 푑풮 풮 = =2퐶 +6퐶 푇 + 12퐶 푇 + 20퐶 푇 (푆44) 푑푇 ( ) 푑푣 푣 − 푏 푑풮 푣 − 푏 = = 풮 (푆45) 푑푇 푇 푑푇 푇 푑 푣 (푣 − 푏) = (풮 + 풮 ) (푆46) 푑푇 푇

Chemical potential constraints at 푻풓ퟎ for determining the coefficients, 푪ퟑ, 푪ퟒ, 푪ퟓ. Zeroth derivative (condition 1):

푣 + 푏 푣 + 푏 푍 − 푙푛(푣 − 푏) − 휃푙푛 = 푍 − 푙푛(푣 − 푏) − 휃푙푛 (푆47) 푣 푣 Using the definition of the compressibility, we have Eq.(2). It’s first derivative (condition 2):

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휃 푣 휃푣 푑푣 푣 푏 푑휃 − − + 푙푛 − 푣 (푣 − 푏) (푣 + 푏) 푑푇 푣 + 푏 (푣 + 푏) 푑푇 휃 푣 휃푣 푑푣 푣 푏 푑휃 = − − + 푙푛 − 푣 (푣 − 푏) (푣 + 푏) 푑푇 푣 + 푏 (푣 + 푏) 푑푇 (S48) The second derivatives (condition 3): 휃 푣 휃푣 푑 푣 휃 푣 휃푣 푑 푣 − − − − − 푣 (푣 − 푏) (푣 + 푏) 푑푇 푣 (푣 − 푏) (푣 + 푏) 푑푇 휃 푣 + 푏 휃(푣 − 푏) 푑푣 1 푣 푑푣 푑휃 + − + + +2 − 푣 (푣 − 푏) (푣 + 푏) 푑푇 푣 (푣 + 푏) 푑푇 푑푇 휃 푣 + 푏 휃(푣 − 푏) 푑푣 1 푣 푑휃 푑푣 − − + + −2 − 푣 (푣 − 푏) (푣 + 푏) 푑푇 푣 (푣 + 푏) 푑푇 푑푇 푏 푏 푣 + 푏 푣 푑 휃 = − + 푙푛 − 푙푛 푣 + 푏 푣 + 푏 푣 + 푏 푣 푑푇 (S49)

The derivatives required in Eq.(S48) and (S49) are given by Eq.(S43)-Eq.(S46), which are functions of 퐶, 퐶, 퐶. Now we need to find 풮 , 풮 and 풮 for other three coefficients (linear functions) of Eq.(S36). Calculations of 퓢푪 and 퓢푪 For the critical volume, we calculate it from other two critical constants:

푍푅푇 푣 = (푆50) 푃 Then we have: 푣 푍 = = 3.84734 (푆51) 푏 0.08664 Therefore, the following constant is universal: 푣 풮 = 푙푛 −1 = 1.046384 (푆52) 푏 From definition, Eq.(29), we have

푑푣 푑푣 푑Ӽ 푑Ỿ + = 푣 −2 (푆53) 푑푇 푑푇 푑푇 푑푇 푑 푣 푑 푣 푑 Ӽ 푑 Ỿ 푑Ӽ 푑Ỿ 푑Ỿ + = 푣 −2 −2 +4 (푆54) 푑푇 푑푇 푑푇 푑푇 푑푇 푑푇 푑푇 where

푑푣 1 푑푣 푑푣 1 푑푣 = , = (푆55) 푑푇 푇 푑푇 푑푇 푇 푑푇 푑 푣 1 푑 푣 푑 푣 1 푑 푣 = , = (푆56) 푑푇 푇 푑푇 푑푇 푇 푑푇 Before moving on, we need to calculate Ӽ,Ỿ and their derivatives at the critical point. The work of Singley, Burns and Misovich (1997) [9] provides the key to this. For SRK EoS the authors used series expansion at the critical point for pressure and density, Eq.(25) and (26). At the critical point, from Eq.(27) and (28) we have: 푑Ӽ 푑Ӽ Ӽ = 2, = −2퐵 , =4퐵 (푆57) 푑푇 푑푇

13

푑Ỿ 푑Ỿ Ỿ = 1, =−(2퐵 − 퐵), =2(2퐵 −2퐵 퐵 + 퐵 퐵 ) (푆58) 푑푇 푑푇 Where the constants have been provided by Singley et al. [9], which are functions of 푓 = 푓(휔), Eq.(S3):

퐵 = 2.259921+ 푓, 퐵 = 0.98283(1+ 푓) (푆59)

퐵 =−1+ 푓(0.33227 + 1.17974푓) (푆60) 퐵 = −0.05345 − 0.84402푓 − 0.79057푓 (푆61) Finally, we can calculate 풮 and 풮 .

푑풮 푇 푑푣 푇 1 푑퐷 1 푑푣 푑푣 풮 = = =− + + (푆62) 푑푇 푣 − 푏 푑푇 푣 − 푏 퐷 푑푇 푇 푑푇 푑푇 푑 풮 푇 푑 푣 1 푑푣 풮 = = − (푆63) 푑푇 푣 − 푏 푑푇 푣 − 푏 푑푇

By using Eq.(S52) and the last two equations back to Eq.(S39)-(S41) to replace 퐶, 퐶 푎푛푑 퐶, we have three unknowns, 퐶, 퐶, 퐶. By solving the three equations, Eq.(S47)-(S49), we find all the coefficients for Eq.(23). In this work the Excel solver was used for the calculations. Table S1 lists calculated values for 퐵, 푖 = 1,…4, 풮 and 풮 for the substances discussed in this work. Table S1 Constants at the critical point 퐵 퐵 퐵 퐵 풮 풮 Argon 2.75077 1.45614 -1.09597 -0.64325 -1.82304 1.76321 Methane 2.76533 1.47158 -1.12446 -0.66868 -1.84238 1.77150 Ethane 2.88890 1.60604 -1.38103 -0.90652 -2.01072 1.83444 n-Butane 3.02040 1.75559 -1.68380 -1.20580 -2.19794 1.88498 n-Hexane 3.14356 1.90167 -1.99637 -1.53349 -2.38083 1.91457 Cyclohexane 3.03686 1.77477 -1.72392 -1.24684 -2.22196 1.88998 n-Heptane 3.20231 1.97341 -2.15571 -1.70722 -2.47065 1.92194 Benzene 3.03686 1.77477 -1.72392 -1.24684 -2.22196 1.88998

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