Page 146 - Characterization and Properties of Petroleum Fractions - M.R. Riazi

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126 CHARACTERIZATION AND PROPERTIES OF PETROLEUM FRACTIONS
(3.77) and (3.78) as: x P = 3.2574 − 3.48148 SG + 0.011666 m
and x N =−1.9571 + 2.63853 SG − 0.03992m. Most of the data get x MA =−0.228 and x PA =−0.025. Since both numbers are
negative the actual estimated values are x MA = 0 and x PA = 0.
used in development of Eqs. (3.70)–(3.82) were in terms of From Eq. (3.85), x A = 0, which is consistent with the previous
volume fractions for x P , x N , and x A . Therefore, generally esti- result from Eqs. (3.77), (3.78), and (3.72).
mated values represent volume fractions; however, they can
be used as mole fractions without serious errors. 3.5.1.3 API Method
In all the above equations total aromatic content is cal- Since 1982 API has adopted the methods developed by Ri-
culated from Eq. (3.72). As discussed earlier for cases that azi and Daubert [36, 47] for prediction of the composition
aromatic content is high it should be split into two parts for of petroleum fractions. Equations (3.65)–(3.67) and similar
a more accurate representation of hydrocarbon types in a equations developed for light fractions in terms of R i and
petroleum mixture. Aromatics are divided into monoaromat- VGF by Riazi and Daubert in 1980 [47] were included in
ics (x MA ) and polyaromatics (x PA ) and the following relations the fourth edition of the API-TDB-82. However, after devel-
have been derived for fractions with molecular weights of less opment of Eqs. (3.70)–(3.74) in terms of viscosity and Eqs.
than 250 [36]:
(3.83)–(3.84) for prediction of the amount of different types
(3.83) x MA =−62.8245 + 59.90816R i − 0.0248335m of aromatics in 1986 [36], they were included in the fifth and
subsequent editions of the API-TDB [2]. The API methods for
(3.84) x PA = 11.88175 − 11.2213R i + 0.023745m
prediction of the composition of petroleum fractions require
(3.85) x A = x MA + x PA kinematic viscosity at 38 or 99 C and if not available, it should
◦
be estimated from Eq. (2.128) or (2.129) in Chapter 2.
The equations may be applied to fractions with total aromatic
content in the range of 0.05–0.96 and molecular weight range 3.5.1.4 n-d-M Method
of 80–250. Based on a data set for aromatic contents of 75 This method requires three physical properties of refractive
coal liquid sample, Eqs. (3.83)–(3.85) give AAD of about 0.055, index (n 20 ), density (d 20 ), and molecular weight (M). For this
0.065, and 0.063 for x MA , x PA , and x A , respectively. Maximum reason the method is called n-d-M method and is the oldest
AD is about 0.24 for x PA . Equations (3.83) and (3.84) have not method for prediction of the composition of petroleum frac-
been evaluated against petroleum fractions. For heavier frac- tions. The method is described in the book by Van Nes and
tions no detailed composition on aromatics of fractions were Van Westen in 1951 [30] and it is included in the ASTM man-
available; however, if such data become available expressions ual [4] under ASTM D 3238 test method. The method does
similar to Eqs. (3.83) and (3.84) may be developed for heavier not directly give the PNA composition, but it calculates the
fractions.
distribution of carbon in paraffins (%C P ), naphthenes (%C N ),
and aromatics (%C A ). However, since carbon is the dominant
Example 3.16—A gasoline sample produced from an Aus- element in a petroleum mixture it is assumed that the %C P ,
tralian crude oil has the boiling range of C 5 -65 C, specific %C N , and %C A distribution is proportional to %P, %N, and %A
◦
gravity of 0.646, and PNA composition of 91, 9, and 0 vol% distribution. In this assumption the ratio of carbon to hydro-
(Ref. [46], p. 302). Calculate the PNA composition from a suit- gen is considered constant in various hydrocarbon families.
able method and compare the results with the experimental Errors caused due to this assumption are within the range of
values.
uncertainty in experimental data reported on the PNA com-
position. Another input data required for this method is sulfur
Solution—For this fraction the only information available content of the fraction in wt% (%S) and should be known if
are boiling point and specific gravity. From Table 2.1 the boil- it exceeds 0.206 wt%. The method should not be applied to
ing point of n-C 5 is 36 C. Therefore, for the fraction the fractions with sulfur content of greater than 2%. This method
◦
characteristic boiling point is T b = (36 + 65)/2 = 50.5 C. The is applicable to fractions with boiling points above gasoline.
◦
other characteristic of the fraction is SG = 0.646. This is In addition this method should be applied to fractions with
a light fraction (low T b ) so we use Eq. (2.51) to calculate
ring percent, %C R (%C N + %C A ) up to 75% provided that %C A
molecular weight as 84.3. Since viscosity is not known, the (as determined from the n-d-M method) is not higher than
most suitable method to estimate composition is through 1.5 times %C N [7]. The n-d-M method also provides equations
Eqs. (3.77) and (3.78). They require parameter m, which
for calculation of total number of rings (R T ), number of
in turn requires refractive index, n. From Eq. (2.115), I = aromatic rings (R A ), and number of naphthenic rings (R N )
0.2216 and from Eq. (2.114), n = 1.3616. With use of M and in an average molecule in the fractions. The method is
n and Eq. (3.50), m =−9.562. From Eq. (3.77) and (3.78), expressed in two sets of equations, one for n 20 , d 20 (20 C)
◦
x P = 0.96 and x N = 0.083. From Eq. (3.72), x A =−4.3. Since and another set for n 70 and d 70 (70 C) as input data. In this
◦
x A is negative thus x A = 0 and x P , x N should be normalized section correlation in terms of n 20 and d 20 are presented. The
as x P = 0.96/(0.96 + 0.083) = 0.92 and x N = 1 − 0.92 = 0.08. other set of correlations for measurement of n and d at 70 C
◦
Therefore, the predicted PNA composition is 92, 8, 0% versus is given in the literature [7, 24].
the experimental values of 91, 9, and 0%.
The aromatic content for this fraction is zero and there %C A = av + 3660/M
is no need to estimate x MA and x PA from Eqs. (3.83) and
%C N = %C R − %C A
(3.84); however, to see the performance of these equations
for this sample we calculate x A from Eq. (3.85). From (3.86) %C P = 100 − %C R
Eq. (2.113), d = 0.6414 and from Eq. (2.14), R i = 1.0409. Us- R A = 0.44 + bvM
ing these values of R i and m in Eqs. (3.83) and (3.84), we R N = R T − R A
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