Page 273 - Characterization and Properties of Petroleum Fractions - M.R. Riazi

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6. THERMODYNAMIC RELATIONS FOR PROPERTY ESTIMATIONS 253
3
V
An alternative method for formulation of VLE of pure sub-
3
stances is to combine Eqs. (6.47) and (6.96), which gives the and V = 7055.8cm /mol. The volume translation parame-
ter c is calculated from Eq. (5.51) as c = 6.03 cm /mol, which
L
following relation in terms of fugacity: through use of Eq. (6.50) gives V = 20.84 and V = 7049.77
V
3
cm /mol. The specific volume is calculated as V(molar)/M,
V
L
(6.104) f = f L V
where for water M = 18. Thus, V = 1.158 and V = 391.653
L
3
V
3
where f V and f L are fugacity of a pure substance in vapor cm /g. Actual values of V and V are 1.093 and 374.7 cm /g,
V
L
and liquid phases at T and P sat . Obviously for solid–liquid respectively [1]. The errors for calculated V and V are +5.9
equilibrium, superscript V in the above relation is replaced and +4.6%, respectively. For a cubic EOS these errors are ac-
by S indicating fugacity of solid is the same as fugacity of ceptable, although without correction factor by volume trans-
L
liquid. Since at VLE pressure of both phases is the same, an lation the error for V is 36.7%. However, for calculation of
alternative form of Eq. (6.104) is volume translation a fourth parameter, namely Racket pa-
rameter is required. It is important to note that in calculation
L
V
(6.105) φ (T, P sat ) = φ (T, P sat ) of fugacity coefficients through a cubic EOS use of volume
An equation of state or generalized correlation may be used translation, c, for both vapor and liquid does not affect re-
L
V
to calculate both φ and φ if T and P sat are known. To cal- sults of vapor pressure calculation from Eq. (6.105). This has
culate vapor pressure (P sat ) from the above equation a trial- been shown in various sources [20].
and-error procedure is required. Value of P sat calculated from
Eq. (6.101) may be used as an initial guess. To terminate cal- Equation (6.105) is the basis of determination of EOS pa-
culations an error parameter can be defined as rameters from vapor pressure data. For example, coefficients
given in Table 5.8 for the LK EOS (Eqs. 5.109–5.111) or the
L

(6.106) ε = 1 − φ f ω relations for various cubic EOSs given in Table 5.1 were

φ found by matching predicted P and saturated liquid den-
V sat
−6
when ε is less than a small value (i.e., 10 ) calculations may sity with the experimental data for pure substances for each
be stopped. In each round of calculations a new guess for pres- equation.
V
L
sure may be calculated as follows: P new = P old (φ /φ ). The The same principle may be applied to any two-phase sys-
following example shows the procedure. tem in equilibrium, such as VSE or SLE, in order to derive
a relation between saturation pressure and temperature. For
example, by applying Eq. (6.96) for solid and vapor phases,
Example 6.7—Repeat Example 6.6 using Eq. (6.105) and the
SRK EOS to estimate vapor pressure of water at 151.84 C. a relation for vapor pressure curve for sublimation (i.e., see
◦
Also calculate V and V at this temperature. Fig. 5.2a) can be derived. The final resulting equation is sim-
L
V
sub
ilar to Eq. (6.101), where parameter B is equal to H /R
in which H sub is the heat of sublimation in J/mol as shown
Solution—For water T c = 647.3K, P c = 220.55 bar, ω = by Eq. (7.27). Then A and B can be determined by having
3
0.3449, and Z RA = 0.2338. Using the units of bar, cm /mol, two points on the sublimation curve. One of these points is
3
and kelvin for P, V, and T with R = 83.14 cm · bar/mol · K the triple point (Fig. 5.2a) as discussed in Section 7.3.4. The
and T = 423 K, SRK parameters are calculated using
same approach can be applied to SLE and derive a relation
relations given in Tables 5.1 and 6.1 as follows: a c = for melting (or freezing) point line (see Fig. 5.2a) of pure com-
3
2
5.6136 × 10 −6 bar (cm /mol) , α = 1.4163, a = 7.9504 × 10 −6
3 2 3 ponents. This is shown in the following example.
bar (cm /mol) , b = 21.1cm /mol, A = 0.030971, and B =
0.00291. Relation for calculation of φ for SRK is given in
A z Example 6.8—Effect of pressure on the melting point: Derive a
Table 6.1 as follows: ln φ = Z − 1 − ln (Z − B) + ln ( ),
B Z+B general relation for melting point of pure components versus
where Z for both saturated liquid and vapor is calculated M
3 2 pressure in terms of heat of melting (or fusion), H , and
from solution of cubic equation (SRK EOS): Z − Z + (A − volume change due to melting V , assuming both of these
M
2
B − B )Z − AB = 0. The first initial guess is to use the value
of P calculated in Example 6.6 from Eq. (6.101): P = 4.8 properties are constant with respect to temperature. Use this
bar, which results in ε = 1.28 × 10 −2 (from Eq. (6.106)) as equation to predict
shown in Table 6.9. The second guess for P is calculated a. melting point of n-octadecane (n-C 18 ) at 300 bar and
as P = 4.86 × (0.9848/0.97235) = 4.86, which gives a lower b. triple point temperature.
value for ε. Summary of results is shown in Table 6.9. The fi-
nal answer is P sat = 4.8637 bar, which differs by −2.7% from The following data are available from DIPPR data bank [13]:
◦
the actual value of 5 bar. Values of specific volumes of liquid Normal melting point, T M ◦ = 28.2 C; heat of melting at nor-
M
L
L
V
and vapor are calculated from Z and Z : Z = 0.003699 and mal melting point, H = 242.4597 kJ/kg; liquid density
L
3
S
V
Z = 0.971211. Molar volume is calculated from V = ZRT/P, at T Mo , ρ = 0.7755 g/cm ; solid density at T M , ρ = 0.8634
−5
3
L
where R = 83.14, T = 425 K, and P = 4.86 bar. V = 26.9 g/cm ; and triple point pressure, P tp = 3.39 × 10 kPa.
Solution—To derive a general relation for saturation pres-
TABLE 6.9—Estimation of vapor pressure of water at 151.8 C from sure versus temperature for melting/freezing point of pure
◦
SRK EOS (Example 6.7). compounds we start by applying Eq. (6.96) between solid and
P,bar Z L Z V φ L φ V ε liquid. Then Eq. (6.99) can be written as
4.8 0.00365 0.9716 0.9848 0.97235 1.2 × 10 −2
4.86 0.003696 0.9712 0.9727 0.972 7.4 × 10 −4 dP H M
4.8637 0.003699 0.971211 0.971981 0.971982 1.1 × 10 −6 dT M = T M V M
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