Equilibrium ratios: correlations and equations of state

Quick refresher

At low pressure the ratio is close to the ideal $p_{s,i}/p$ , where $p_{s,i}$ is the vapour pressure of the pure component, and it hardly depends on composition. As pressure rises, the compositions of the two phases move closer together, and the ratios of all components tend to one.
The convergence pressure $p_k$ is the pressure at which the ratios of a mixture at a given temperature would converge to one. It depends on composition and is not lower than the bubble point pressure (the pressure at which the first gas bubble appears). For a mixture at its critical temperature, it is the critical pressure.
Standing's correlation is a straight line, $\log_{10}(K_ip) = A_0 + A_1F_i$ , drawn in Hoffman coordinates ( $F_i$ is the Hoffman characterisation factor, explained below). Its intercept and slope depend on pressure. It is meant for separators: up to 6.9 MPa and 93 °C.
The Hoffman plot is also used to check analyses. The ratios $y_i/x_i$ of equilibrium gas and liquid samples must fall on a smooth line, and a point that drops out of the line shows an error in the analysis of that component.
The ratios of the C7+ fraction are the most uncertain. Correlations estimate them roughly, while an equation of state needs a characterisation of the fraction (a description of it by a few pseudo-components with calculated properties) and interaction coefficients.
In a separator calculation the methods differ in $R_{sb}$ and $B_{ob}$ by a fraction of a per cent, and by 1–2% at most. At the bubble point pressure and in reservoir conditions they differ by tens of per cent. There we need an equation-of-state calculation tuned to experiments.

Key relationships

F_i = b_i\left(\frac{1}{T_{b,i}} - \frac{1}{T}\right),\qquad b_i = \frac{\log_{10}(p_{c,i}/p_{\text{sc}})}{1/T_{b,i} - 1/T_{c,i}}

Hoffman characterisation factor

\log_{10}(K_ip) = A_0(p) + A_1(p)\,F_i

Standing's correlation: $p$ in psia, $T$ in °R (degrees Rankine, the absolute scale that goes with °F)

K_i = \left(\frac{p_{c,i}}{p_k}\right)^{A-1}\frac{p_{c,i}}{p}\,e^{5.373A(1 + \omega_i)(1 - T_{c,i}/T)},\quad A = 1 - \left(\frac{p - p_{\text{sc}}}{p_k - p_{\text{sc}}}\right)^{0.6}

Whitson–Torp formula: at $p = p_{\text{sc}}$ it is Wilson's formula, at $p = p_k$ it gives one

K_i = \frac{\varphi_i^{\,L}}{\varphi_i^{\,V}}

the ratio from an equation of state, found from the equality of fugacities

Common pitfalls

Applying the correlation for separators at reservoir pressure: its range is up to 6.9 MPa.
Putting gauge pressure (measured above atmospheric pressure) or degrees Celsius into the correlations. Standing's formulas are written in psia (absolute pressure) and °R (in the formula for C7+, in °F).
Assuming that fitting the convergence pressure to the bubble point pressure makes the ratios right at all pressures.
Describing the C7+ fraction by one pseudo-component in calculations near the bubble point pressure.
Accepting sample analyses without checking. Swapped peaks of two isomers (molecules with the same atoms arranged differently, such as isobutane and n-butane), or a component lost from a sample, show up at once on the Hoffman plot.

What the equilibrium ratios depend on

The equilibrium ratio $K_i = y_i/x_i$ shows how a component is divided between the gas and the liquid (definition). At low pressure the gas is almost ideal and the liquid is close to an ideal solution, so the ratio is close to the ideal one, $K_i \approx p_{s,i}(T)/p$ (here $p_{s,i}$ is the vapour pressure of the pure component). The ratio depends on pressure and temperature but hardly on composition. That is why correlations that are common to many oils are good enough for separators.

As pressure rises, the gas becomes denser and dissolves more and more heavy components, while the liquid dissolves light ones. The compositions of the two phases move closer together, and the ratios of all components tend to one. For light components they decrease. For heavy ones they pass through a minimum and then rise. If the mixture is at its critical temperature, the ratios reach one at the critical point, where gas and liquid cannot be told apart. At any other temperature, the two-phase region ends earlier, at the bubble point pressure. But if we continue the curves $K(p)$ further, they would converge to one at a pressure called the convergence pressure $p_k$ . It depends on the composition of the mixture: the heavier the liquid, the higher $p_k$ is, and the further the ratios are from one at the same pressure.

So the methods for calculating the ratios work in different pressure ranges. For separators, correlations are enough, where the ratio is a function of pressure, temperature and the properties of the component itself. At high pressures we must take composition into account. We can do this roughly through the convergence pressure, or strictly with an equation of state, which calculates the ratios from the equality of fugacities (fugacity). The critical point matters too. Near it the ratios are close to one and the equilibrium calculation converges badly. The critical point of a mixture is itself hard to measure, so it has to be predicted.

Methods of calculating equilibrium ratios

Method: Ideal mixture, Wilson's formula
What must be known: $p_c$ , $T_c$ , $\omega$ of the components
Where it is used: first guess for iterations; rough estimates at low pressure
Weak points: does not depend on composition; at high pressures it is wrong several-fold

Method: Low-pressure correlations (Standing's)
What must be known: pressure, temperature, the parameters $b$ and $T_b$ of the components
Where it is used: separators and stock tanks: up to 6.9 MPa and 93 °C
Weak points: describes an “average” oil; the ratios of nitrogen and of the C7+ fraction are rough

Method: Charts and formulas with a convergence pressure
What must be known: the convergence pressure of the mixture: from a pseudo-binary mixture, from a correlation or by fitting
Where it is used: hand calculations of past decades; initial ratios in programs
Weak points: composition is represented by a single number, $p_k$ ; near saturation the error is tens of per cent

Method: Equation of state
What must be known: $T_c$ , $p_c$ , $\omega$ of all the components, characterisation of C7+, interaction coefficients, experiments for tuning
Where it is used: any pressures, including the near-critical region; compositional models
Weak points: accuracy depends on the description of C7+ and on tuning; the calculation is iterative

Method	What must be known	Where it is used	Weak points
Ideal mixture, Wilson's formula	$p_c$ , $T_c$ , $\omega$ of the components	first guess for iterations; rough estimates at low pressure	does not depend on composition; at high pressures it is wrong several-fold
Low-pressure correlations (Standing's)	pressure, temperature, the parameters $b$ and $T_b$ of the components	separators and stock tanks: up to 6.9 MPa and 93 °C	describes an “average” oil; the ratios of nitrogen and of the C7+ fraction are rough
Charts and formulas with a convergence pressure	the convergence pressure of the mixture: from a pseudo-binary mixture, from a correlation or by fitting	hand calculations of past decades; initial ratios in programs	composition is represented by a single number, $p_k$ ; near saturation the error is tens of per cent
Equation of state	$T_c$ , $p_c$ , $\omega$ of all the components, characterisation of C7+, interaction coefficients, experiments for tuning	any pressures, including the near-critical region; compositional models	accuracy depends on the description of C7+ and on tuning; the calculation is iterative

Separators: the Hoffman plot and Standing's correlation

At low pressure it is convenient to compare the ratios of all components on one chart. For each component, draw a straight line of vapour pressure on a plot of $\log_{10} p_s$ against $1/T$ . The line goes through two points: the normal boiling point $T_b$ (the boiling temperature at standard pressure, where $p_s = p_{\text{sc}}$ ) and the critical point. Read the logarithm of the ratio $p_s/p_{\text{sc}}$ off this line at temperature $T$ . This number is the characterisation factor $F_i$ . For an ideal mixture $K_ip = p_{s,i}$ , so the points of all components on a chart of $\log_{10}(K_ip)$ against $F_i$ would fall on one straight line with a slope of 1. Measurements show that real ratios at low pressures also fall on a straight line, but its slope and intercept are different and depend on pressure. Such a chart is called the Hoffman plot.

Hoffman characterisation factor

F_i = b_i\left(\frac{1}{T_{b,i}} - \frac{1}{T}\right),\qquad b_i = \frac{\log_{10}\left(p_{c,i}/p_{\text{sc}}\right)}{1/T_{b,i} - 1/T_{c,i}}

where:

$F_i$ characterisation factor of the component at temperature $T$ , —
$b_i$ slope of the vapour pressure line; in Standing's correlation, for the light gases, a fitted value, °R
$T_{b,i}$ normal boiling point (at 0.101325 MPa), °R
$T_{c,i},\ p_{c,i}$ critical temperature and pressure of the component, °R, psia
$T$ absolute temperature of the system, °R
$p_{\text{sc}}$ standard pressure, 14.696 psia (0.101325 MPa), psia

Units: absolute temperatures in the same units (in Standing's correlation, °R); pressures in the same units

Holds when

A vapour pressure line through two points is an approximation: the real curve is slightly bent in these coordinates.
For components above their critical temperature (nitrogen and methane at separator temperature; CO₂ and ethane too, above 31–32 °C), $F_i$ is only the line continued beyond the critical point.

$F_i$ is an estimate of $\log_{10}(p_{s,i}/p_{\text{sc}})$ : at $T = T_b$ it is zero, at $T = T_c$ it is $\log_{10}(p_c/p_{\text{sc}})$ . The lighter the component, the greater its $F_i$ and the greater the ratio.

To get ratios without measuring them, the Hoffman line has been described by formulas. Standing's correlation was fitted to published measurements of ratios in oil–natural gas systems, at pressures up to 1000 psia (6.9 MPa) and temperatures up to 200 °F (93 °C). In it, the intercept and slope of the line are functions of pressure.

Standing's correlation for equilibrium ratios at low pressures

\log_{10}(K_ip) = A_0 + A_1F_i,\qquad A_0 = 1.2 + 4.5\cdot10^{-4}\,p + 15\cdot10^{-8}\,p^2,\qquad A_1 = 0.89 - 1.7\cdot10^{-4}\,p - 3.5\cdot10^{-8}\,p^2

where:

$K_i$ equilibrium ratio of the component, —
$p$ absolute pressure, psia
$A_0,\ A_1$ intercept and slope of the Hoffman line at this pressure, —
$F_i$ characterisation factor from $b_i$ and $T_{b,i}$ in the table below, $T$ in °R, —

Units: oilfield units only: $p$ in psia, temperatures in °R ( $p\,[\text{psia}] = 145.04\,p\,[\text{MPa}]$ , $T\,[{}^{\circ}\text{R}] = 1.8\,T\,[\text{K}]$ )

Holds when

The pressure is up to 6.9 MPa (1000 psia) and the temperature up to 93 °C (200 °F): the range of the data to which the correlation was fitted.
For nitrogen, CO₂, hydrogen sulfide, methane and ethane, $b$ and $T_b$ were fitted to measurements and differ from those calculated by Hoffman's formula.
The composition of the mixture does not enter the formula. The correlation describes an “average” oil, and for unusual mixtures the error grows.

As $p \to 0$ , the intercept tends to 1.2 and the slope to 0.89. For an ideal mixture they would be $\log_{10} 14.696 = 1.167$ and 1. As pressure rises, the slope decreases. This means the ratios of different components move closer together, as they should on the way to the convergence pressure.

Parameters of Standing's correlation

Component: N₂
$b$ , °R: 470
$T_b$ , °R: 109
$T_b$ , K: 60.6

Component: CO₂
$b$ , °R: 652
$T_b$ , °R: 194
$T_b$ , K: 107.8

Component: H₂S
$b$ , °R: 1136
$T_b$ , °R: 331
$T_b$ , K: 183.9

Component: C₁
$b$ , °R: 300
$T_b$ , °R: 94
$T_b$ , K: 52.2

Component: C₂
$b$ , °R: 1145
$T_b$ , °R: 303
$T_b$ , K: 168.3

Component: C₃
$b$ , °R: 1799
$T_b$ , °R: 416
$T_b$ , K: 231.1

Component: i-C₄
$b$ , °R: 2037
$T_b$ , °R: 471
$T_b$ , K: 261.7

Component: n-C₄
$b$ , °R: 2153
$T_b$ , °R: 491
$T_b$ , K: 272.8

Component: i-C₅
$b$ , °R: 2368
$T_b$ , °R: 542
$T_b$ , K: 301.1

Component: n-C₅
$b$ , °R: 2480
$T_b$ , °R: 557
$T_b$ , K: 309.4

Component: C₆ (hexanes)
$b$ , °R: 2738
$T_b$ , °R: 610
$T_b$ , K: 338.9

Component: n-C₆
$b$ , °R: 2780
$T_b$ , °R: 616
$T_b$ , K: 342.2

Component: n-C₇
$b$ , °R: 3068
$T_b$ , °R: 669
$T_b$ , K: 371.7

Component: n-C₈
$b$ , °R: 3335
$T_b$ , °R: 718
$T_b$ , K: 398.9

Component: n-C₉
$b$ , °R: 3590
$T_b$ , °R: 763
$T_b$ , K: 423.9

Component: n-C₁₀
$b$ , °R: 3828
$T_b$ , °R: 805
$T_b$ , K: 447.2

Component	$b$ , °R	$T_b$ , °R	$T_b$ , K
N₂	470	109	60.6
CO₂	652	194	107.8
H₂S	1136	331	183.9
C₁	300	94	52.2
C₂	1145	303	168.3
C₃	1799	416	231.1
i-C₄	2037	471	261.7
n-C₄	2153	491	272.8
i-C₅	2368	542	301.1
n-C₅	2480	557	309.4
C₆ (hexanes)	2738	610	338.9
n-C₆	2780	616	342.2
n-C₇	3068	669	371.7
n-C₈	3335	718	398.9
n-C₉	3590	763	423.9
n-C₁₀	3828	805	447.2

For nitrogen, CO₂, H₂S, methane and ethane the values are fitted: for example, for methane $T_b$ = 94 °R, whereas its normal boiling point is 201 °R. For the other components $T_b$ is the normal boiling point; for the hexanes as a whole (the C₆ row) it is a value between the boiling points of their isomers.

The C7+ fraction in Standing's correlation

n = 7.30 + 0.0075\,T_F + 0.0016\,p,\qquad b = 1013 + 324n - 4.256n^2,\qquad T_b = 301 + 59.85n - 0.971n^2

where:

$n$ effective carbon number of the C7+ fraction, —
$T_F$ temperature, °F
$p$ absolute pressure, psia
$b,\ T_b$ parameters of the fraction for the formula for $F$ , °R

Units: $T_F$ in °F ( $T_F = 1.8\,t + 32$ , $t$ in °C), $p$ in psia, $b$ and $T_b$ in °R

Holds when

The C7+ fraction is represented by an imaginary normal alkane (a straight-chain hydrocarbon) with $n$ carbon atoms. The molar mass and density of the fraction do not enter the formula.
The estimate is rough: in the lesson example the ratio of C7+ at atmospheric pressure comes out several times greater than from the equation of state (see the table below).

The number $n$ grows with pressure and temperature, because heavier and heavier parts of the fraction pass into the gas, so the “average” component that vaporises becomes heavier. Another simple estimate is Katz's rule: the ratio of the C7+ fraction is 15% of the ratio of heptane, $K_{7+} = 0.15\,K_{\mathrm{C_7}}$ .

Worked exampleStanding's ratios for the first separation stage

Given (assumed data)

Conditions: the first stage of the two-stage scheme from the lesson on separation (example): 1.2 MPa (absolute) and 40 °C
Components: propane, methane and the C7+ fraction of that lesson's assumed oil
For comparison: ratios from the Peng–Robinson equation from the same lesson

Solution

Units: $p = 1.2\cdot 145.04 = 174.05$ psia, $T = 1.8\cdot(40 + 273.15) = 563.67$ °R.
Intercept and slope: $A_0 = 1.2 + 4.5\cdot10^{-4}\cdot 174.05 + 15\cdot10^{-8}\cdot 174.05^2 = 1.2829$ , $A_1 = 0.89 - 1.7\cdot10^{-4}\cdot 174.05 - 3.5\cdot10^{-8}\cdot 174.05^2 = 0.8594$ .
Propane: $F = 1799\cdot(1/416 - 1/563.67) = 1.1329$ ; $\log_{10}(Kp) = 1.2829 + 0.8594\cdot 1.1329 = 2.2565$ ; $K = 10^{2.2565}/174.05 = 1.037$ .
Methane: $F = 300\cdot(1/94 - 1/563.67) = 2.6593$ ; $\log_{10}(Kp) = 1.2829 + 0.8594\cdot 2.6593 = 3.5683$ ; $K = 10^{3.5683}/174.05 = 21.26$ .
The C7+ fraction: $T_F = 104$ °F; $n = 7.30 + 0.0075\cdot 104 + 0.0016\cdot 174.05 = 8.3585$ ; $b = 1013 + 324\cdot 8.3585 - 4.256\cdot 8.3585^2 = 3423.8$ °R; $T_b = 301 + 59.85\cdot 8.3585 - 0.971\cdot 8.3585^2 = 733.4$ °R; $F = 3423.8\cdot(1/733.4 - 1/563.67) = -1.4057$ ; $\log_{10}(Kp) = 0.0748$ ; $K_{7+} = 10^{0.0748}/174.05 = 0.006825$ .
Katz's rule: for heptane $F = 3068\cdot(1/669 - 1/563.67) = -0.8570$ , $\log_{10}(Kp) = 0.5464$ , $K_{\mathrm{C_7}} = 0.02022$ ; $K_{7+} = 0.15\cdot 0.02022 = 0.003033$ .

Answer: Standing's correlation gives these ratios: propane 1.037 (the equation of state gives 1.123), methane 21.26 (19.55), and the C7+ fraction 0.006825 by the effective carbon number and 0.003033 by Katz's rule (0.002182).

What it means. The correlation's ratio for propane is 7.7% lower, and for methane 8.7% higher, than the equation of state gives. For a separator calculation such differences are small. The ratio of the C7+ fraction is uncertain. The two simple estimates differ by more than a factor of two, and both are greater than the “effective” ratio from the equation of state, which is the ratio of the C7+ fractions in the gas and in the liquid. At the first stage this hardly matters, since the heavy components make up only fractions of a per cent of its gas. In the stock tank the difference becomes noticeable (see the table below and the section on choosing a method). The data in this example are assumed for illustration.

Equilibrium ratios of the separation stages: Standing's correlation and the Peng–Robinson equation

Component: N₂
1.2 MPa, 40 °C: Standing: 107.5
1.2 MPa, 40 °C: equation: 54.91
0.101 MPa, 20 °C: Standing: 1190
0.101 MPa, 20 °C: equation: 645.9

Component: CO₂
1.2 MPa, 40 °C: Standing: 8.638
1.2 MPa, 40 °C: equation: 9.122
0.101 MPa, 20 °C: Standing: 84.21
0.101 MPa, 20 °C: equation: 81.79

Component: C₁
1.2 MPa, 40 °C: Standing: 21.26
1.2 MPa, 40 °C: equation: 19.55
0.101 MPa, 20 °C: Standing: 232.9
0.101 MPa, 20 °C: equation: 201.1

Component: C₂
1.2 MPa, 40 °C: Standing: 3.5
1.2 MPa, 40 °C: equation: 3.517
0.101 MPa, 20 °C: Standing: 29.33
0.101 MPa, 20 °C: equation: 27.87

Component: C₃
1.2 MPa, 40 °C: Standing: 1.037
1.2 MPa, 40 °C: equation: 1.123
0.101 MPa, 20 °C: Standing: 7.104
0.101 MPa, 20 °C: equation: 7.446

Component: i-C₄
1.2 MPa, 40 °C: Standing: 0.45
1.2 MPa, 40 °C: equation: 0.4985
0.101 MPa, 20 °C: Standing: 2.828
0.101 MPa, 20 °C: equation: 2.904

Component: n-C₄
1.2 MPa, 40 °C: Standing: 0.3373
1.2 MPa, 40 °C: equation: 0.3617
0.101 MPa, 20 °C: Standing: 2.04
0.101 MPa, 20 °C: equation: 2.009

Component: i-C₅
1.2 MPa, 40 °C: Standing: 0.1537
1.2 MPa, 40 °C: equation: 0.1571
0.101 MPa, 20 °C: Standing: 0.859
0.101 MPa, 20 °C: equation: 0.7649

Component: n-C₅
1.2 MPa, 40 °C: Standing: 0.1223
1.2 MPa, 40 °C: equation: 0.1221
0.101 MPa, 20 °C: Standing: 0.6601
0.101 MPa, 20 °C: equation: 0.5688

Component: n-C₆
1.2 MPa, 40 °C: Standing: 0.0481
1.2 MPa, 40 °C: equation: 0.0426
0.101 MPa, 20 °C: Standing: 0.2338
0.101 MPa, 20 °C: equation: 0.1679

Component: C₇₊ (fraction as a whole)
1.2 MPa, 40 °C: Standing: 0.006825
1.2 MPa, 40 °C: equation: 0.002182
0.101 MPa, 20 °C: Standing: 0.04135
0.101 MPa, 20 °C: equation: 0.006429

Component	1.2 MPa, 40 °C: Standing	1.2 MPa, 40 °C: equation	0.101 MPa, 20 °C: Standing	0.101 MPa, 20 °C: equation
N₂	107.5	54.91	1190	645.9
CO₂	8.638	9.122	84.21	81.79
C₁	21.26	19.55	232.9	201.1
C₂	3.5	3.517	29.33	27.87
C₃	1.037	1.123	7.104	7.446
i-C₄	0.45	0.4985	2.828	2.904
n-C₄	0.3373	0.3617	2.04	2.009
i-C₅	0.1537	0.1571	0.859	0.7649
n-C₅	0.1223	0.1221	0.6601	0.5688
n-C₆	0.0481	0.0426	0.2338	0.1679
C₇₊ (fraction as a whole)	0.006825	0.002182	0.04135	0.006429

The equation-of-state ratios for the stock tank are calculated for the liquid of the first stage. For the C7+ fraction, Standing's value comes from the effective carbon number, and the equation's value is the ratio of the total C7+ fractions in the gas and in the liquid. Katz's rule gives 0.003033 and 0.01334. For methane, Wilson's formula gives 32.1 and 304, which is one and a half times more than the equation of state. The data are assumed for illustration.

CalculatedRatios of the first stage (1.2 MPa, 40 °C) in Hoffman coordinates. The solid line is Standing's correlation. The dashed line is the ideal mixture (

\log_{10}(Kp) = \log_{10} p_{\text{sc}} + F

,

p

in psia). The dots are the ratios of the Peng–Robinson equation for the components from nitrogen to n-hexane and for the pseudo-component F1. The factor

F

is calculated from Standing's parameters, and for F1 by Hoffman's formula from its characterisation.

The equation-of-state points from methane to n-hexane lie near Standing's line. In $\log_{10}(Kp)$ they deviate from it by no more than 0.053, which means the ratios differ by no more than 13%. Nitrogen drops out: by the correlation its ratio is twice as great. But oils usually contain little nitrogen. The dashed line of the ideal mixture departs from the points at both ends: for light components it is above them, for heavy ones below. This is the same error as that of Wilson's formula, which gives 32.1 for methane instead of 19.55, and 0.0328 for n-hexane instead of 0.0426 (the advanced section of the lesson links Wilson's formula to the Hoffman plot).

High pressures: the convergence pressure

Above 5–7 MPa the ratios depend noticeably on composition, and a single line for all oils will not do. Before computers, engineers took the ratios from charts built for several convergence pressures (for example, 5000 psia, or 34.5 MPa). Each chart held a family of curves $K(p)$ for one component at different temperatures. The convergence pressure stood for the composition: the engineer estimated the $p_k$ of the mixture and took the charts with the nearest value. Today formulas use the same idea, with $p_k$ as the only parameter of the mixture.

How the convergence pressure used to be estimated: the pseudo-binary mixture method

1
Initial approximation
Assume a $p_k$ (for example, from a correlation) and run a flash calculation with the ratios for this $p_k$ .
2
Pseudo-binary mixture
Represent the liquid from the calculation by two components. One is the lightest component present in it in a noticeable amount, usually methane. The other is a “heavy pseudo-component”: everything else, with critical properties averaged by mass.
3
Critical locus
Use a chart of the critical loci of binary mixtures (the lines of the critical points of mixtures of two components). Find the pressure at which the critical locus of such a pair passes through the temperature of the system. This is the new $p_k$ .
4
Iterations
If the new $p_k$ differs from the assumed one, repeat the calculation with the new ratios until the values agree.

Hadden proposed the method in the 1950s. There are also correlations that give the convergence pressure from the properties of the C7+ fraction (molar mass, density) and from temperature. The simplest is Standing's linear estimate for oils.

Linear estimate of the convergence pressure of an oil from the molar mass of the C7+ fraction

p_k \approx 60\,M_{7+} - 4200

where:

$p_k$ convergence pressure, psia
$M_{7+}$ molar mass of the C7+ fraction, kg/kmol

Units: $p_k$ in psia; $p_k\,[\text{MPa}] = p_k\,[\text{psia}]/145.04$

Holds when

An empirical estimate for oils that gives only the order of magnitude; temperature does not enter it.

The heavier the C7+ fraction, the higher the convergence pressure. For the lesson's oil ( $M_{7+}$ = 214.1) it comes to 8646 psia, or 59.6 MPa, which is three times its bubble point pressure.

Equilibrium ratios with the convergence pressure (Whitson–Torp formula)

K_i = \left(\frac{p_{c,i}}{p_k}\right)^{A-1}\frac{p_{c,i}}{p}\exp\left[5.373\,A\,(1 + \omega_i)\left(1 - \frac{T_{c,i}}{T}\right)\right],\qquad A = 1 - \left(\frac{p - p_{\text{sc}}}{p_k - p_{\text{sc}}}\right)^{0.6}

where:

$p_k$ convergence pressure of the mixture, MPa
$A$ factor decreasing from 1 at $p = p_{\text{sc}}$ to 0 at $p = p_k$ , —
$p,\ T$ absolute pressure and temperature, MPa, K
$p_{c,i},\ T_{c,i},\ \omega_i$ critical pressure and temperature and acentric factor of the component, MPa, K, —
$p_{\text{sc}}$ standard pressure, 0.101325 MPa, MPa

Units: absolute pressures, in the same units (their differences are gauge pressures); temperatures in kelvins

Holds when

At $p = p_{\text{sc}}$ the formula coincides with Wilson's formula, and at $p = p_k$ it gives $K_i = 1$ . In between it is an interpolation, not a physical model.
Composition enters only through $p_k$ . The exponent 0.6 is empirical.
In the original, as with Wilson, the factor is 5.37. Here it is 5.373, as in the lesson on phase equilibrium.

The formula extends Wilson's ratios so that they converge to one at the convergence pressure. It is used as a first guess for an equation-of-state calculation near the bubble point pressure, where Wilson's ratios are poor, and for quick estimates.

CalculatedEquilibrium ratios of methane, propane and the pseudo-component F2 (C₁₀–C₁₉) of the lesson's oil at 80 °C. Solid lines: the Peng–Robinson equation up to the bubble point pressure of 19.4 MPa (vertical dashed line). Dashed lines of the same colour: the Whitson–Torp formula with a convergence pressure of 28.6 MPa, fitted to the bubble point pressure (the circle). Both axes are logarithmic.

Worked exampleConvergence pressure from the bubble point pressure

Given (assumed data)

Oil: the lesson's assumed oil at 80 °C; bubble point pressure from the Peng–Robinson equation $p_b$ = 19.40 MPa
Ratios: from the Whitson–Torp formula with an unknown $p_k$
Fitting condition: at $p = p_b$ , $\sum z_iK_i = 1$ holds (condition for the bubble point pressure)

Solution

The sum $\sum z_iK_i$ at $p = p_b$ depends on $p_k$ . Fitting $p_k$ by bisection (halving the interval again and again) gives $p_k$ = 28.63 MPa.
Check: $A = 1 - \left(\dfrac{19.40 - 0.101}{28.63 - 0.101}\right)^{0.6} = 1 - 0.6765^{0.6} = 0.2090$ .
Methane ( $p_c$ = 4.60 MPa, $T_c$ = 190.6 K, $\omega$ = 0.011): $(4.60/28.63)^{0.2090 - 1} = 4.247$ ; $4.60/19.40 = 0.2371$ ; $\exp[5.373\cdot 0.2090\cdot 1.011\cdot(1 - 190.6/353.15)] = 1.686$ ; $K = 4.247\cdot 0.2371\cdot 1.686 = 1.698$ . From the equation of state at the bubble point pressure: 1.974.
The same for the other components gives $\sum z_iK_i = 1.000$ : the bubble point pressure is reproduced.
At 10 MPa: the gas fraction from the Whitson–Torp ratios is $V$ = 0.421, from the equation of state 0.295; the ratio of the pseudo-component F2 is 0.0121 against 0.000788.
Linear estimate: $p_k = 60\cdot 214.1 - 4200 = 8646$ psia, or 59.6 MPa; with it the Whitson–Torp formula would give a bubble point pressure of 31.2 MPa instead of 19.40.

Answer: The convergence pressure at which the Whitson–Torp formula reproduces the bubble point pressure of the lesson's oil is 28.63 MPa; at 10 MPa these ratios give a gas fraction of 0.421 instead of 0.295.

What it means. One parameter is enough to hit the bubble point pressure, but not to describe the ratios at other pressures. The formula does not know that at 10 MPa the heavy components are still far from convergence. So it belongs as a first guess for iterations with an equation of state, not as a replacement for it. The linear estimate from $M_{7+}$ gives only the order of magnitude: with it the bubble point pressure is wrong by 61%. The data in this example are assumed for illustration.

Equilibrium ratios of the lesson's oil at 10 MPa and 80 °C

Component: C₁
Peng–Robinson equation: 3.249
Whitson–Torp: 3.927
Wilson: 5.606

Component: C₂
Peng–Robinson equation: 1.126
Whitson–Torp: 1.812
Wilson: 1.082

Component: C₃
Peng–Robinson equation: 0.5751
Whitson–Torp: 1.017
Wilson: 0.3169

Component: n-C₄
Peng–Robinson equation: 0.2961
Whitson–Torp: 0.5969
Wilson: 0.102

Component: n-C₆
Peng–Robinson equation: 0.08508
Whitson–Torp: 0.2364
Wilson: 0.01422

Component: F1 (C₇–C₉)
Peng–Robinson equation: 0.02675
Whitson–Torp: 0.1107
Wilson: 0.002833

Component: F2 (C₁₀–C₁₉)
Peng–Robinson equation: 0.0007878
Whitson–Torp: 0.01214
Wilson: 2.571 × 10⁻⁵

Component: F3 (C₂₀+)
Peng–Robinson equation: 1.197 × 10⁻⁶
Whitson–Torp: 0.0002695
Wilson: 7.818 × 10⁻⁹

Component: Gas fraction $V$
Peng–Robinson equation: 0.295
Whitson–Torp: 0.421
Wilson: 0.378

Component	Peng–Robinson equation	Whitson–Torp	Wilson
C₁	3.249	3.927	5.606
C₂	1.126	1.812	1.082
C₃	0.5751	1.017	0.3169
n-C₄	0.2961	0.5969	0.102
n-C₆	0.08508	0.2364	0.01422
F1 (C₇–C₉)	0.02675	0.1107	0.002833
F2 (C₁₀–C₁₉)	0.0007878	0.01214	2.571 × 10⁻⁵
F3 (C₂₀+)	1.197 × 10⁻⁶	0.0002695	7.818 × 10⁻⁹
Gas fraction $V$	0.295	0.421	0.378

The Whitson–Torp column uses a convergence pressure of 28.6 MPa, fitted to the bubble point pressure. For the heavy pseudo-components, its ratios are tens and hundreds of times greater than those of the equation of state, and Wilson's ratios are tens and hundreds of times smaller. The data are assumed for illustration.

The C7+ fraction in the calculation of ratios

The light components are the same in all oils, but every oil has its own C7+ fraction (the C7+ fraction and pseudo-components). A pseudo-component is an imaginary component that stands for a group of real ones. The ratios of the C7+ fraction are the weakest point of any method. Correlations get by with rough rules: Standing's effective carbon number or Katz's rule. Neither takes into account the molar mass and density of the C7+ fraction of the particular oil. An equation of state needs, for each pseudo-component, the critical temperature and pressure, the acentric factor and the interaction coefficients with the light components. Nobody measures these. They are calculated by correlations from the molar mass and density, and this is called the characterisation of the fraction.

In the lesson's assumed oil the C7+ fraction is divided into three pseudo-components. Their properties come from correlations. The normal boiling point of each comes from the Søreide correlation (from $M$ and $\gamma$ ). The critical temperature and pressure come from the Kesler–Lee correlations (from $T_b$ and $\gamma$ ). The acentric factor comes from the Lee–Kesler formula when $T_b/T_c < 0.8$ , and from the Kesler–Lee formula when $T_b/T_c \ge 0.8$ . The binary interaction coefficients $k_{ij}$ of methane with n-hexane and with the pseudo-components follow the formula $k_{ij} = 0.14\gamma_j - 0.0668$ , where $\gamma_j$ is the relative density of the second component of the pair. For nitrogen with all hydrocarbons, $k_{ij}$ is 0.12, and for CO₂ it is 0.15. These are assumptions of the example. For the heavy pseudo-components with $\omega > 0.49$ , we use the original dependence $m(\omega)$ of the PR equation, as in Separator calculations and optimum separator pressure, without the refinement for heavy substances from Cubic equations of state. With the refined dependence the bubble point pressure would come out about 0.6 MPa higher, and the solution gas–oil ratio and formation volume factor would change by no more than 0.3%.

Pseudo-components of the C7+ fraction of the lesson's oil and their properties for the equation of state

Pseudo-component: F1 (C₇–C₉)
$z$: 0.120
$M$ , kg/kmol: 110
$\gamma$: 0.750
$T_b$ , K: 395.7
$T_c$ , K: 576.1
$p_c$ , MPa: 2.866
$\omega$: 0.361
$k_{ij}$ with methane: 0.0382

Pseudo-component: F2 (C₁₀–C₁₉)
$z$: 0.155
$M$ , kg/kmol: 200
$\gamma$: 0.830
$T_b$ , K: 532.1
$T_c$ , K: 711.5
$p_c$ , MPa: 1.871
$\omega$: 0.631
$k_{ij}$ with methane: 0.0494

Pseudo-component: F3 (C₂₀+)
$z$: 0.075
$M$ , kg/kmol: 410
$\gamma$: 0.910
$T_b$ , K: 718.0
$T_c$ , K: 873.7
$p_c$ , MPa: 1.052
$\omega$: 1.073
$k_{ij}$ with methane: 0.0606

Pseudo-component: C₇₊ as one fraction
$z$: 0.350
$M$ , kg/kmol: 214.1
$\gamma$: 0.845
$T_b$ , K: 549.7
$T_c$ , K: 729.7
$p_c$ , MPa: 1.809
$\omega$: 0.663
$k_{ij}$ with methane: 0.0514

Pseudo-component	$z$	$M$ , kg/kmol	$\gamma$	$T_b$ , K	$T_c$ , K	$p_c$ , MPa	$\omega$	$k_{ij}$ with methane
F1 (C₇–C₉)	0.120	110	0.750	395.7	576.1	2.866	0.361	0.0382
F2 (C₁₀–C₁₉)	0.155	200	0.830	532.1	711.5	1.871	0.631	0.0494
F3 (C₂₀+)	0.075	410	0.910	718.0	873.7	1.052	1.073	0.0606
C₇₊ as one fraction	0.350	214.1	0.845	549.7	729.7	1.809	0.663	0.0514

The densities of the pseudo-components are relative densities (to water at 60 °F). The acentric factor of F3 is calculated by the Kesler–Lee formula (its $T_b/T_c$ = 0.82), and those of the others by the Lee–Kesler formula. The last row is the same fraction taken as one pseudo-component (see the example below). The data are assumed for illustration.

Worked exampleOne C7+ fraction instead of three

Given (assumed data)

Oil: the lesson's assumed oil; the C7+ fraction is described by one pseudo-component with its molar mass and density (the last row of the table)
Comparison: with the same oil in which C7+ is three pseudo-components
Calculations: bubble point pressure at 80 °C, gas fraction at 10 MPa, the two-stage separation from the lesson on separation

Solution

Characterisation of the single fraction: $T_b$ = 549.7 K, $T_c$ = 729.7 K, $p_c$ = 1.809 MPa, $\omega$ = 0.663, $k_{ij}$ with methane 0.0514.
Bubble point pressure: 20.88 MPa instead of 19.40, which is 1.48 MPa higher.
Gas fraction at 10 MPa: 0.311 instead of 0.295.
Separation: $R_{sb}$ = 144.8 and 144.8 m³/m³, practically the same. $B_{ob}$ = 1.3908 instead of 1.4056, which is 1.1% less, because the reservoir oil comes out denser (691.7 kg/m³ instead of 683.1).

Answer: Describing C7+ by one fraction does not change the solution gas–oil ratio in a separator calculation. The formation volume factor changes by 1.1%, through the density of the reservoir oil. But the single fraction overestimates the bubble point pressure by 1.48 MPa and the gas fraction at 10 MPa by 5%.

What it means. At separator pressure the heavy components stay almost entirely in the liquid however they are described, so the gas and the solution gas–oil ratio do not change. Near the bubble point pressure the solubility of methane matters. The light part of C7+ dissolves methane better than an “average” fraction does. When the parts are lumped together, this light part becomes “heavier”, and the bubble point pressure rises. Hence the rule: for reservoir-condition calculations, divide the C7+ fraction into several pseudo-components and tune their properties to experiments. The data in this example are assumed for illustration.

Ratios from an equation of state

An equation of state gives the ratios from the equilibrium condition: the fugacity of each component in the gas equals its fugacity in the liquid. From this, $K_i = \varphi_i^{\,L}/\varphi_i^{\,V}$ (fugacity). The fugacity coefficients depend on the compositions of the phases, and the compositions depend on the ratios, so the calculation is iterative. It starts from initial ratios (Wilson's or Whitson–Torp's) and uses successive substitution (repeating the cycle with updated ratios) until the fugacities are equal (equation-of-state equilibrium calculation). We shall not repeat the algorithm. Instead, let us see how such ratios differ from empirical ones.

The main difference is the dependence on composition. A correlation gives methane one ratio at a given pressure and temperature, for any oil. An equation of state gives a separate ratio for each mixture. It asks for little: the properties of the pure components and the binary interaction coefficients. Yet it describes a mixture of any composition, including the near-critical region (close to the critical point). That is why there is no other way for reservoir conditions.

Worked exampleThe same pressure and temperature, different ratios

Given (assumed data)

Conditions: 5 MPa, 90 °C
Mixture 1: the model oil of the lesson on phase equilibrium (Vapour–liquid equilibrium): methane 0.45, propane 0.15, n-hexane 0.15, n-decane 0.25
Mixture 2: the assumed oil of this lesson: 13 components, the C7+ fraction down to a pseudo-component with $M$ = 410

Solution

The Peng–Robinson equation gives gas fractions of 0.424 and 0.450 and methane ratios of 5.25 and 6.05, a difference of 15%.
Propane: 0.804 and 0.834 (4%); n-hexane: 0.0912 and 0.0894 (2%).
Standing's correlation (5 MPa = 725 psia and 90 °C, which is within its range) gives the same for both mixtures: methane 6.16, propane 0.834, n-hexane 0.0870; Wilson's formula gives 12.2, 0.758 and 0.0375.

Answer: At the same pressure and temperature the ratio of methane in the two oils differs by 15%, the ratios of propane and n-hexane by 4 and 2%.

What it means. The lesson's oil contains heavy pseudo-components in which methane dissolves less well than in a mixture with decane, so its ratio is higher. Standing's correlation turned out close to the lesson's oil, while for the model oil its methane ratio is wrong by 17%. Wilson's formula at 5 MPa is already wrong by a factor of two. At 5 MPa the differences are moderate. Closer to the bubble point pressure they grow, and there you cannot do without an equation of state. The data in this example are assumed for illustration.

Properties of pure components. Critical temperature and pressure, acentric factor and molar mass. They are reference values (Peng–Robinson equation).
Characterisation of C7+. Division into pseudo-components, with their properties taken from correlations (see the section above). For heavy oils and gas condensates, the analysis extends to C₂₀₊–C₃₀₊.
Binary interaction coefficients. The most important are the pairs of methane with the heavy pseudo-components, and the pairs with nitrogen, CO₂ and hydrogen sulfide.
Volume translation. It corrects the phase densities (volume translation) and does not affect the equilibrium ratios.
Experiments for tuning. The bubble point pressure; the constant-composition expansion (the sample expands in a closed cell with nothing removed); differential liberation (the pressure is lowered step by step and the gas is removed each time) or constant volume depletion; and separator tests (Laboratory PVT tests). For gas injection, swelling and miscibility experiments (Displacement of oil by gas).

What compositional data are needed

Whichever method you choose, it is no better than the data on composition. Here is what to get from the laboratory and what to check.

Detailed analysis. Report the components from nitrogen to the hexanes one by one. Report the heavy part by fractions with the same number of carbon atoms, at least up to C₁₀–C₂₀, and then a residue (C₂₀₊ or C₃₀₊). Analyses of gas condensates often end at C7+. That is not enough for an equation of state that has to be tuned to the dew point pressure (Retrograde condensation, wax, asphaltenes and high-pressure fluids).
Molar mass and density of the residue. Without them you can neither divide the fraction into pseudo-components nor characterise it.
Mole fractions from mass fractions. The laboratory analyses a liquid by mass, but the calculation is done in moles: $z_i \propto w_i/M_i$ . An error in the molar mass of a fraction goes into its own mole fraction and into the mole fractions of all the other components (see the example below).
Structure of the heavy fractions. A fraction with one carbon number contains paraffins (chains), naphthenes (rings) and aromatics (benzene-type rings). The properties of the pseudo-components and the interaction coefficients depend on their proportions (Watson factor). The heavier the fraction, the more isomers it contains, and the more approximate its “exact” properties are.
Samples of separator gas and liquid. Recombination (mixing the samples back together in the right proportions) needs their compositions and the gas–oil ratio (recombination). The consistency of the compositions is checked on the Hoffman plot (see the second example below).

Worked exampleA 5% error in the molar mass of the C7+ fraction

Given (assumed data)

Oil: the lesson's assumed oil; the mass composition is known exactly
Error: the molar mass of C7+ (and of each pseudo-component) is overestimated or underestimated by 5%, the densities are correct
Calculations: bubble point pressure at 80 °C and the two-stage separation from the lesson on separation

Solution

We recalculate the mole fractions from the mass fractions. With a molar mass 5% greater, the fraction of C7+ is 0.339 instead of 0.350; with a molar mass 5% smaller it is 0.362. The fractions of the light components change in the opposite direction.
We characterise the pseudo-components again from the new molar masses ( $M_{7+}$ = 224.8 and 203.4 instead of 214.1).
Bubble point pressure: 19.52 and 19.24 MPa instead of 19.40.
Separation: $R_{sb}$ = 145.0 and 144.6 m³/m³ instead of 144.8; $B_{ob}$ = 1.4072 and 1.4046 instead of 1.4056.

Answer: A 5% error in the molar mass of C7+ shifts the bubble point pressure by 0.8%, and the solution gas–oil ratio and formation volume factor by no more than 0.1%.

What it means. The solution gas–oil ratio and the formation volume factor are in essence mass quantities. The mass of gas per mass of oil is fixed by the analysis, and the number of moles it is divided into hardly matters. The bubble point pressure depends on the mole fraction of methane and on the properties of the pseudo-components, so it is more sensitive to the molar mass. Errors of a few per cent in the molar mass are common. For gas condensates, where the heavy fractions set the dew point pressure, the consequences are usually more serious. The data in this example are assumed for illustration.

Checking analyses with the Hoffman plot. If gas and liquid are taken from the same separator, the ratios $y_i/x_i$ are the equilibrium ratios at its pressure and temperature, and on the Hoffman plot they must fall on a smooth line. A point that drops out of the line points to an error in the analysis of that component. If the scatter is large for all the components, the samples are most likely not in equilibrium (droplets carried over, or the flow not yet steady), or the pressure and temperature of the separator were recorded wrongly.

Worked exampleChecking separator samples on the Hoffman plot

Given (assumed data)

Samples: separator gas and liquid at 1.2 MPa and 40 °C (174.05 psia, 563.67 °R); the analyses are in the table below
Comparison line: Standing's correlation at this pressure: $A_0$ = 1.2829, $A_1$ = 0.8594 (example above)

Solution

For each component, $K = y/x$ , $F$ comes from Standing's parameters, and the deviation is the difference between $\log_{10}(Kp)$ and the line $A_0 + A_1F$ (see the table).
Isobutane: $K = 0.0120/0.0123 = 0.976$ ; $\log_{10}(0.976\cdot 174.05) = 2.230$ ; the line at $F$ = 0.7110: $1.2829 + 0.8594\cdot 0.7110 = 1.8939$ . The point is higher by 0.34.
n-Butane: $K = 0.0061/0.0333 = 0.183$ ; $\log_{10}(0.183\cdot 174.05) = 1.503$ ; the line at $F$ = 0.5653: 1.7687. The point is lower by 0.27.
The other components deviate from the line by no more than 0.04. This is the usual disagreement between a correlation and a real oil.
If we swap the isobutane and n-butane values in the gas analysis, then $K_{\text{i-C}_4} = 0.0061/0.0123 = 0.496$ (deviation 0.04) and $K_{\text{n-C}_4} = 0.0120/0.0333 = 0.360$ (deviation 0.03). Both points are then on the line.

Answer: Isobutane and n-butane deviate from the line in opposite directions, each by about a factor of two in the ratio. In the gas analysis their peaks are swapped. After the correction all the points fall on the line.

What it means. A deviation of 0.3 in $\log_{10}(Kp)$ means an error of a factor of two in the ratio. With the usual scatter of a few hundredths, it is seen at once. Typical findings of such a check are swapped isomer peaks, light components lost from the liquid sample, and air contaminating the gas (the nitrogen point drops out). The analysis goes back to the laboratory for a repeat, and until then the doubtful components are not used for recombination or for tuning the equation of state. The data in this example are assumed for illustration.

Separator sample analyses (1.2 MPa, 40 °C) in Hoffman coordinates

Component: C₁
Liquid $x$: 0.0378
Gas $y$: 0.7399
$K = y/x$: 19.6
$F$: 2.6593
$\log_{10}(Kp)$: 3.533
Deviation from the line: −0.04

Component: C₂
Liquid $x$: 0.0364
Gas $y$: 0.1281
$K = y/x$: 3.52
$F$: 1.7475
$\log_{10}(Kp)$: 2.787
Deviation from the line: 0.00

Component: C₃
Liquid $x$: 0.0545
Gas $y$: 0.0611
$K = y/x$: 1.12
$F$: 1.1329
$\log_{10}(Kp)$: 2.290
Deviation from the line: 0.03

Component: i-C₄
Liquid $x$: 0.0123
Gas $y$: 0.0120
$K = y/x$: 0.976
$F$: 0.7110
$\log_{10}(Kp)$: 2.230
Deviation from the line: 0.34

Component: n-C₄
Liquid $x$: 0.0333
Gas $y$: 0.0061
$K = y/x$: 0.183
$F$: 0.5653
$\log_{10}(Kp)$: 1.503
Deviation from the line: −0.27

Component: i-C₅
Liquid $x$: 0.0181
Gas $y$: 0.0028
$K = y/x$: 0.155
$F$: 0.1680
$\log_{10}(Kp)$: 1.431
Deviation from the line: 0.00

Component: n-C₅
Liquid $x$: 0.0243
Gas $y$: 0.0030
$K = y/x$: 0.123
$F$: 0.0527
$\log_{10}(Kp)$: 1.331
Deviation from the line: 0.00

Component: n-C₆
Liquid $x$: 0.0366
Gas $y$: 0.0016
$K = y/x$: 0.0437
$F$: −0.4190
$\log_{10}(Kp)$: 0.881
Deviation from the line: −0.04

Component	Liquid $x$	Gas $y$	$K = y/x$	$F$	$\log_{10}(Kp)$	Deviation from the line
C₁	0.0378	0.7399	19.6	2.6593	3.533	−0.04
C₂	0.0364	0.1281	3.52	1.7475	2.787	0.00
C₃	0.0545	0.0611	1.12	1.1329	2.290	0.03
i-C₄	0.0123	0.0120	0.976	0.7110	2.230	0.34
n-C₄	0.0333	0.0061	0.183	0.5653	1.503	−0.27
i-C₅	0.0181	0.0028	0.155	0.1680	1.431	0.00
n-C₅	0.0243	0.0030	0.123	0.0527	1.331	0.00
n-C₆	0.0366	0.0016	0.0437	−0.4190	0.881	−0.04

Mole fractions; $p$ = 174.05 psia. The deviation is the difference between $\log_{10}(Kp)$ of the point and Standing's line at its $F$ ; positive means the point is above the line. The data are assumed for illustration: in the gas analysis isobutane and n-butane are deliberately swapped.

How the choice of ratios changes the result

Let us bring together the comparisons of the lesson. We begin with separation. The scheme is the same two-stage scheme as in the lesson on separation (1.2 MPa and 40 °C, then the stock tank at 0.101325 MPa and 20 °C), but the stage ratios come from different methods. In all variants the oil densities come from the equation of state, so the differences come only from the ratios.

Two-stage separation of the lesson's oil with the ratios of different methods

Ratios: Peng–Robinson equation
$V_1$: 0.5301
$V_2$: 0.1077
$R_{sb}$ , m³/m³: 144.8
$B_{ob}$ , m³/m³: 1.4056
Difference in $B_{ob}$ , %: —
C₇₊ in the stock-tank gas, mol %: 0.53

Ratios: Standing, C₇₊ by the effective carbon number
$V_1$: 0.5321
$V_2$: 0.1115
$R_{sb}$ , m³/m³: 147.2
$B_{ob}$ , m³/m³: 1.4202
Difference in $B_{ob}$ , %: 1.0
C₇₊ in the stock-tank gas, mol %: 3.44

Ratios: Standing, C₇₊ by Katz's rule
$V_1$: 0.5301
$V_2$: 0.1076
$R_{sb}$ , m³/m³: 145.1
$B_{ob}$ , m³/m³: 1.4089
Difference in $B_{ob}$ , %: 0.2
C₇₊ in the stock-tank gas, mol %: 1.11

Ratios: Wilson's formula
$V_1$: 0.5409
$V_2$: 0.0824
$R_{sb}$ , m³/m³: 143.9
$B_{ob}$ , m³/m³: 1.4019
Difference in $B_{ob}$ , %: −0.3
C₇₊ in the stock-tank gas, mol %: 0.69

Ratios	$V_1$	$V_2$	$R_{sb}$ , m³/m³	$B_{ob}$ , m³/m³	Difference in $B_{ob}$ , %	C₇₊ in the stock-tank gas, mol %
Peng–Robinson equation	0.5301	0.1077	144.8	1.4056	—	0.53
Standing, C₇₊ by the effective carbon number	0.5321	0.1115	147.2	1.4202	1.0	3.44
Standing, C₇₊ by Katz's rule	0.5301	0.1076	145.1	1.4089	0.2	1.11
Wilson's formula	0.5409	0.0824	143.9	1.4019	−0.3	0.69

$V_1$ , $V_2$ are the gas fractions at the first stage and in the stock tank (of the feed of each stage). The difference is measured from the variant with the equation of state. The data are assumed for illustration.

In a separator the methods differ by no more than 1.0% in $B_{ob}$ . At low pressure the light components go almost entirely into the gas whatever the ratios, and the intermediate ones are close to ideal. The variant in which the C7+ ratio comes from the effective carbon number differs most. In the stock tank its ratio is 6 times greater than from the equation, and the stock-tank gas carries away excess heavy components (3.4% C7+ instead of 0.5%). Wilson's formula is wrong by a factor of one and a half in the methane ratios, but this hardly affects the result, because the methane goes into the gas anyway. For separation calculations and for choosing stage pressures (optimum pressure), a correlation is enough if the C7+ ratio is reasonable.

The lesson's oil at 80 °C: bubble point pressure and gas fraction from the ratios of different methods

Ratios: Peng–Robinson equation
$p_b$ , MPa: 19.4
$V$ at 5 MPa: 0.437
$V$ at 10 MPa: 0.295
$V$ at 15 MPa: 0.146

Ratios: Whitson–Torp, $p_k$ = 28.6 MPa (fitted to $p_b$ )
$p_b$ , MPa: 19.4
$V$ at 5 MPa: 0.525
$V$ at 10 MPa: 0.421
$V$ at 15 MPa: 0.262

Ratios: Whitson–Torp, $p_k$ = 59.6 MPa (from $M_{7+}$ )
$p_b$ , MPa: 31.2
$V$ at 5 MPa: 0.523
$V$ at 10 MPa: 0.449
$V$ at 15 MPa: 0.373

Ratios: Wilson's formula
$p_b$ , MPa: 25.2
$V$ at 5 MPa: 0.480
$V$ at 10 MPa: 0.378
$V$ at 15 MPa: 0.274

Ratios	$p_b$ , MPa	$V$ at 5 MPa	$V$ at 10 MPa	$V$ at 15 MPa
Peng–Robinson equation	19.4	0.437	0.295	0.146
Whitson–Torp, $p_k$ = 28.6 MPa (fitted to $p_b$ )	19.4	0.525	0.421	0.262
Whitson–Torp, $p_k$ = 59.6 MPa (from $M_{7+}$ )	31.2	0.523	0.449	0.373
Wilson's formula	25.2	0.480	0.378	0.274

The gas fraction is the mole fraction in the flash of the reservoir oil. Standing's correlation is not included in the table: these pressures are outside its range. The data are assumed for illustration.

CalculatedMole fraction of gas of the lesson's oil at 80 °C against pressure. The solid line uses the ratios of the Peng–Robinson equation. The dashed lines are the Whitson–Torp formula, with a convergence pressure of 28.6 MPa, and Wilson's formula. Both empirical curves overestimate the gas fraction, and Wilson's formula also shifts the bubble point pressure to 25.2 MPa.

In reservoir conditions the picture is different. The gas fraction from empirical ratios is wrong by tens of per cent. The bubble point pressure from Wilson's formula, or from the linear estimate of the convergence pressure, is wrong by 30% and more. An error of this size passes into the gas–oil ratio, the volume of the gas cap (a free-gas zone on top of the oil) and the development forecast, so reservoir calculations are done only with an equation of state.

Terms and symbols

Convergence pressure $p_k$: The pressure at which the equilibrium ratios of all the components of a mixture, at a given temperature, would converge to one. It depends on composition. For a mixture at its critical temperature it equals the critical pressure.
Hoffman characterisation factor $F$: The quantity $b(1/T_b - 1/T)$ . It is the logarithm of the ratio of the vapour pressure of a component to the standard pressure, estimated from a line through the boiling point and the critical point.
Hoffman plot: A chart of $\log_{10}(Kp)$ against $F$ . At low pressures the points of the components fall on a straight line. It is used to correlate ratios and to check sample analyses.
Standing's correlation: Formulas for the intercept and slope of the Hoffman line as functions of pressure, with the parameters $b$ and $T_b$ of the components. Its range is up to 6.9 MPa and 93 °C.
Katz's rule: An estimate of the ratio of the C7+ fraction: 15% of the ratio of heptane.
Whitson–Torp formula: Equilibrium ratios that go from Wilson's formula at standard pressure to one at the convergence pressure.
Pseudo-binary mixture method: An estimate of the convergence pressure. The liquid is replaced by a pair, “lightest component and heavy pseudo-component”, and $p_k$ is found from the critical locus of this pair at the temperature of the system.
Characterisation of the C7+ fraction: The calculation of the critical properties, acentric factor and interaction coefficients of the pseudo-components from their molar mass and density.

Common mistakes and misconceptions

Wrong

“The equilibrium ratio is a tabulated property of a component, like molar mass.”

In fact

It depends on pressure and temperature, and at high pressure on the compositions of the phases too. In the lesson example at 5 MPa and 90 °C, the methane ratio in two oils differs by 15%.

Wrong

“The convergence pressure is the bubble point pressure of the mixture.”

In fact

The bubble point pressure is the boundary of the two-phase region. The convergence pressure is the point where the extended ratio curves would converge. For the lesson's oil they are 19.4 and 28.6 MPa.

Wrong

“A correlation for separators can be extended to reservoir pressure.”

In fact

Standing's correlation was fitted to data up to 6.9 MPa. Above that, the ratios depend on composition, and you need a method that takes composition into account.

Wrong

“If the bubble point pressure is reproduced, the ratios are right.”

In fact

The Whitson–Torp formula with a fitted convergence pressure reproduces the bubble point pressure of the lesson's oil, but at 10 MPa it gives a gas fraction of 0.42 instead of 0.29.

Wrong

“If one C7+ fraction is enough for a separator, it is enough for a reservoir.”

In fact

In a separator calculation one fraction does not change the solution gas–oil ratio. The formation volume factor changes by 1.1%, through the density of the reservoir oil. But the single fraction overestimates the bubble point pressure of the lesson's oil by 1.5 MPa.

Connections to other topics

Vapour–liquid equilibrium — the definition of the ratios, Wilson's formula, the flash calculation and iterations with an equation of state.
Cubic equations of state — the Peng–Robinson equation, fugacity and volume translation, on which the calculation of ratios rests.
Separator calculations and optimum separator pressure — where the ratios are applied: the calculation of separation and of the optimum pressure, for the same assumed oil.
Composition of reservoir fluids — the C7+ fraction, pseudo-components and the Watson factor.
Laboratory PVT tests — the experiments to which the equation of state is tuned.
Reservoir fluid sampling — sampling and recombination of samples, whose analyses are checked on the Hoffman plot.
How to read a PVT report — how to read the compositions and the results of the experiments in a PVT report.
Retrograde condensation, wax, asphaltenes and high-pressure fluids — gas condensates and near-critical fluids, where the requirements on data and on method are highest.

Check yourself

1
What happens to the equilibrium ratios of all the components of a mixture as the pressure approaches the convergence pressure?
1. A.They tend to zero
2. B.They become equal to the ideal ratios $p_{s,i}/p$
3. C.They cease to depend on pressure
4. D.They tend to one
Show the answer
Correct answer: D. They tend to one
As pressure rises, the compositions of the gas and the liquid move closer together, and the ratios of all components tend to one: for light components they decrease, for heavy ones they grow. The pressure at which the extended curves $K(p)$ would reach one is the convergence pressure. For a mixture at its critical temperature it is the critical pressure.
2
Find the equilibrium ratio of n-butane by Standing's correlation at 2 MPa (absolute) and 30 °C ( $b$ = 2153 °R, $T_b$ = 491 °R).
Show the answer
Units: $p = 2\cdot 145.04 = 290.08$ psia, $T = 1.8\cdot 303.15 = 545.67$ °R. $A_0 = 1.2 + 4.5\cdot10^{-4}\cdot 290.08 + 15\cdot10^{-8}\cdot 290.08^2 = 1.3432$ ; $A_1 = 0.89 - 1.7\cdot10^{-4}\cdot 290.08 - 3.5\cdot10^{-8}\cdot 290.08^2 = 0.8377$ . $F = 2153\cdot(1/491 - 1/545.67) = 0.4393$ . $\log_{10}(Kp) = 1.3432 + 0.8377\cdot 0.4393 = 1.7112$ ; $K = 10^{1.7112}/290.08 = 51.43/290.08 \approx 0.177$ .
3
Two-stage separation with Standing's ratios (C7+ by the effective carbon number) gave a $B_{ob}$ 1% greater than the calculation with an equation of state. What is the most probable cause?
1. A.An overestimated ratio of the C7+ fraction: in the stock tank excess heavy components leave with the gas
2. B.The correlation was applied above its pressure range
3. C.An error in the density of the reservoir oil
4. D.An underestimated methane ratio at the first stage
Show the answer
Correct answer: A. An overestimated ratio of the C7+ fraction: in the stock tank excess heavy components leave with the gas
The separator pressures are within the range of the correlation, and the densities in both variants are calculated in the same way. The ratios of methane and of the intermediate components from the correlation are close to the calculated ones, but the C7+ ratio by the effective carbon number in the stock tank is several times greater. The stock-tank gas carries away excess heavy components, so there is less stock-tank oil, and the solution gas–oil ratio and the formation volume factor grow.
4
A Hoffman plot was built for samples of separator gas and liquid. All the points fell near the line except n-pentane: its point is 0.30 higher. What does this mean and what should be done?
Show the answer
A deviation of 0.30 in $\log_{10}(Kp)$ means that the n-pentane ratio is $10^{0.3} \approx 2$ times greater than it should be. Its fraction in the gas is too high, or its fraction in the liquid too low, by a factor of two. The other components are consistent, so the problem is not how the separator was operated but the analysis of this component. The cause may be an error in processing the chromatogram (the record of the gas chromatograph, the instrument that separates the components of a sample), such as swapped or wrongly separated peaks, or an error in the sampling and storage of the sample. Ask for the analysis to be repeated, and until then do not use pentane for recombination or for tuning the equation of state.
5
What method should be used to calculate the dew point pressure of a gas condensate at 40 MPa?
1. A.The Whitson–Torp formula with a convergence pressure from the molar mass of C7+
2. B.Wilson's formula
3. C.An equation of state tuned to experiments on a sample
4. D.Standing's correlation
Show the answer
Correct answer: C. An equation of state tuned to experiments on a sample
At 40 MPa the ratios depend on the composition of both phases, and the heavy fractions set the dew point pressure. Standing's correlation is valid only up to 6.9 MPa. Wilson's formula takes into account neither the convergence pressure nor the composition. The formula with a convergence pressure gives at best a first guess. You need an equation-of-state calculation with a detailed characterisation of C7+, tuned to experiments on a sample.
6
Why does a 5% error in the molar mass of the C7+ fraction hardly change the solution gas–oil ratio calculated from composition, but shift the bubble point pressure more noticeably?
Show the answer
The analysis fixes the mass composition of the sample. The mass of gas dissolved in a unit mass of oil does not depend on the molar mass of C7+, so the solution gas–oil ratio and formation volume factor change by tenths of a per cent. The bubble point pressure depends on the mole fraction of methane and on the properties of the pseudo-components, which are calculated from the molar mass. In the lesson example it shifts by almost one per cent. For gas condensates, where the heavy fractions set the dew point, the consequences are usually stronger.

Equilibrium ratios: correlations and equations of state

Quick refresher

After this lesson you can

What the equilibrium ratios depend on

Separators: the Hoffman plot and Standing's correlation

High pressures: the convergence pressure

The C7+ fraction in the calculation of ratios

Ratios from an equation of state

What compositional data are needed

How the choice of ratios changes the result

Terms and symbols

Common mistakes and misconceptions

In depth: advanced detail

Connections to other topics

Check yourself