Page 233 - Characterization and Properties of Petroleum Fractions - M.R. Riazi

P2: IML/FFX
QC: IML/FFX
P1: IML/FFX
AT029-Manual
August 16, 2007
AT029-05
converted into a virial form. This is shown by the following
Considering the fact that a is a temperature-dependent
parameter one can see that the virial coefficients are all
example. AT029-Manual-v7.cls T1: IML 17:42 5. PVT RELATIONS AND EQUATIONS OF STATE 213
temperature-dependent parameters. With use of SRK EOS,
Example 5.4—Convert RK EOS into the virial form and ob- similar coefficients are obtained but parameter a also depends
tain coefficients B and C in terms of EOS parameters. on the acentric factor as given in Table 5.1. This gives bet-
ter estimation of the second and third virial coefficients (see
Solution—The RK EOS is given by Eq. (5.38). If both sides Problem 5.10)
of this equation are multiplied by V/RT we get
PV V a The following example shows application of truncated
(5.82) Z = = − virial equation for calculation of vapor molar volumes.
RT V − b RT(V + b)
Assume x = b/V and A = a/RT, then the above equation can Example 5.5—Propane has vapor pressure of 9.974 bar at
be written as 300 K. Saturated vapor molar volume is V = 2036.5cm /mol
3
V
1 1 1 [Ref. 8, p. 4.24]. Calculate (a) second virial coefficient from
(5.83) Z = − A ×
1 − x V 1 + x Eqs. (5.71)–(5.73), (b) third virial coefficient from Eq. (5.78),
V
(c) V from virial EOS truncated after second term using Eqs.
Since b < V, therefore, x < 1 and the terms in the RHS of V
the above equation can be expanded through Taylor series (5.65) and (5.66), (d) V from virial EOS truncated after third
V
[16, 17]: term using Eqs. (5.65) and (5.66), and (e) V from ideal gas
law.
∞ f (n) (x o )
n
(5.84) f (x) = (x − x o )
n! Solution—(a) and (b): For propane from Table 2.1 we
n=0
get T c = 96.7 C (369.83 K), P c 42.48 bar, and ω = 0.1523.
◦
n
where f (n) (x o ) is the nth order derivative d f (x)/dx evaluated T r = 0.811, P r = 0.23, and R = 83.14 cm · bar/mol · K. Second
n
3
at x = x o . The zeroth derivative of f is defined to be f itself virial coefficient, B, can be estimated from Eqs. (5.71) or
and both 0! and 1! are equal to 1. Applying this expansion rule (5.72) or (5.73) and the third virial coefficient from Eq. (5.78).
at x o = 0 we get: Results are given in Table 5.6. (c) Truncated virial equation
1 2 3 4 after second term from Eq. (5.65) is Z = 1 + B/V, which is
= 1 + x + x + x + x + ···
1 − x referred to as V expansion form, and from Eq. (5.66) is Z =
(5.85) 1 + BP/RT, which is the same as Eq. (5.75) and it is referred
1 2 3 4
= 1 − x + x − x + x − ··· to as P expansion form. For the V expansion (Eq. 5.65), V
1 + x
should be calculated through successive substitution method
It should be noted that the above relations are valid when or from mathematical solution of the equation, while in P
|x| < 1. Substituting the above two relations in Eq. (5.83) we expansion form (Eq. 5.66) Z can be directly calculated from
get T and P. Once Z is determined, V is calculated from Eq.
1 (5.15): V = ZRT/P. In part (d) virial equation is truncated
2
3
3
2
Z = (1 + x + x + x + ···) − A × (1 − x + x − x + ···) after the third term. The V expansion form reduces to Eq.
V
(5.86) (5.76). Summary of calculations for molar volume is given
in Table 5.6. The results from V expansion (Eq. 5.65) and P
If x is replaced by its definition b/V and A by a/RT we have expansion (5.66) do not agree with each other; however, the
3
2
2
b − a/RT b + ab/RT +b − ab /RT difference between these two forms of virial equation reduces
Z = 1 + + + + ··· as the number of terms increases. When the number of terms
V V 2 V 3
(5.87) becomes infinity (complete equation), then the two forms
of virial equation give identical results for V. Obviously for
A comparison with Eq. (5.65) we get the virial coefficients in truncated virial equation, the V expansion form, Eq. (5.65),
terms of RK EOS parameters as follows: gives more accurate result for V as the virial coefficients
a ab ab 2 are originally determined from this equation. As can be seen
(5.88) B = b − C = b + D = b − from Table 5.6, when B is calculated from Eq. (5.71) better
3
2
RT RT RT
TABLE 5.6—Prediction of molar volume of propane at 300 K and 9.974 bar from virial equation with different methods for second virial
coefficient (Example 5.5).
Virial equation with two terms Virial equation with three terms a
P expansion b V expansion c P expansion d V expansion e
Method of estimation of
3
3
3
3
3
second virial coefficient (B) B,cm /mol V,cm /mol %D V,cm /mol %D V,cm /mol %D V,cm /mol %D
Tsonopoulos (Eq. 5.71) −390.623 2110.1 3.6 2016.2 −1.0 2056.8 1.0 2031.6 −0.2
Normal fluids (Eq. 5.72) −397.254 2103.5 3.3 2005.3 −1.5 2048.1 0.6 2021.0 −0.7
McGlashan (Eq. 5.73) −360.705 2140.0 5.1 2077.8 2.0 2095.7 2.9 2063.6 1.3
3
The experimental value of vapor molar volume is: V = 2036.5cm /mol (Ref. [8], p. 4.24).
2
6
a In all calculations with three terms, the third virial coefficient C is calculated from Eq. (5.78) as C = 19406.21 cm /mol .
b Truncated two terms (P expansion) refers to pressure expansion virial equation (Eq. 5.66) truncated after second term (Eq. 5.75): Z = 1 + BP/RT.
c
Truncated two terms (V expansion) refers to volume expansion virial equation (Eq. 5.65) truncated after second term: Z = 1 + B/V.
2
2
2
d Truncated three terms (P expansion) refers to pressure expansion virial equation (Eq. 5.66) truncated after third term: Z = 1 + BP/RT + (C − B )P /(RT) .
e Truncated three terms (V expansion) refers to volume expansion virial equation (Eq. 5.65) truncated after third term (Eq. 5.76): Z = 1 + B/V + C/V .
2
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