Separator calculations and optimum separator pressure

Quick refresher

Each stage is a flash calculation at its own pressure and temperature. The feed of a stage is the liquid of the previous stage. The last stage is the stock tank at atmospheric pressure.
The fraction of the reservoir oil that reaches the stock tank as liquid is the product of the liquid fractions of all the stages: $n_{\text{od}} = \prod_j (1 - V_j)$ . The gas of stage $j$ is the fraction $V_j$ of the liquid that reached that stage.
The gas–oil ratio of a stage is its gas at standard conditions per 1 m³ of stock-tank oil. The solution gas–oil ratio $R_{sb}$ is the sum over all stages, including the stock tank.
$B_{ob}$ is the volume of one kilomole of reservoir oil divided by the volume of the stock-tank oil that this kilomole yields. A mass balance checks it.
If the first-stage pressure is too low, the intermediate components (propane, butanes and pentanes) leave with its gas. If it is too high, they leave with the stock-tank gas, because a lot of methane reaches the stock tank. Between these extremes lies an optimum, with the smallest $R_{sb}$ and $B_{ob}$ and the greatest oil yield.
Each further stage adds less oil than the one before. In a good scheme, the pressure ratios of neighbouring stages are close to each other.

Key relationships

n_{\text{od}} = \prod_{j=1}^{N}(1 - V_j),\qquad n_{g,j} = V_j\prod_{k<j}(1 - V_k)

moles of stock-tank oil and of the gas of stage $j$ per 1 kmol of reservoir oil

R_j = \frac{n_{g,j}V_{m,\text{sc}}}{V_{\text{od}}},\qquad R_{sb} = \sum_j R_j

gas–oil ratio of a stage and solution gas–oil ratio; $V_{\text{od}} = n_{\text{od}}M_{\text{od}}/\rho_{\text{od}}$

B_{ob} = \frac{M_{\text{res}}/\rho_{ob}}{V_{\text{od}}}

formation volume factor at the bubble point pressure

B_{ob}\,\rho_{ob} = \rho_{\text{od}} + \sum_j R_j\,\gamma_{g,j}\,\rho_{\text{air}}

mass-balance check per 1 m³ of stock-tank oil

Common pitfalls

Forgetting the stock-tank gas: it was also dissolved in the reservoir oil and is part of $R_{sb}$ .
Adding up the mole fractions of gas of the stages: the fraction of the second stage refers to the liquid of the first, not to the reservoir oil.
Dividing the separator gas by the volume of oil that left the separator, and then comparing this number with the solution gas–oil ratio, which is per 1 m³ of stock-tank oil.
Putting the gauge pressures of the separators (measured above atmospheric pressure) into the equilibrium calculation instead of absolute ones.
Treating $R_{sb}$ and $B_{ob}$ as properties of the oil without stating the separation scheme and conditions.

After this lesson you can

describe multistage separation as a chain of flash calculations and keep the mole balance (the count of moles in and out) through the stages
calculate the gas fractions at the stages and the amount and composition of the stock-tank oil and of the gas of each stage
go from moles to volumes: find the gas–oil ratio of each stage, the solution gas–oil ratio $R_{sb}$ and the formation volume factor $B_{ob}$ , and check them with a mass balance
explain why the first-stage pressure has an optimum and find it from a calculated curve
estimate how the separation temperature and the number of stages affect the oil yield, and state the assumptions of the calculation

Separation stages

Oil arrives from the well at a pressure of several megapascals. In the field we do not reduce the pressure all at once but in stages. At each separation stage, the gas and the liquid are brought to equilibrium at a set pressure and temperature, and then sent down separate lines. The gas goes to the gas system. The liquid goes to the next stage, which is at a lower pressure. The last stage is the stock tank, at a pressure close to atmospheric. The liquid that remains in it is the stock-tank oil. An arrangement of “one separator + stock tank” is called two-stage, and “two separators + stock tank” is called three-stage.

SchematicThree-stage separation as a chain of flash calculations. Per 1 kmol of reservoir oil, each stage sends the fraction

V_j

of the liquid that came to it into the gas. The liquid that reaches the stock tank is the product of the liquid fractions of all the stages.

Why have stages if the pressure at the end is atmospheric anyway? The equilibrium ratio $K_i$ sets how a component is shared between the gas and the liquid (Vapour–liquid equilibrium). At a pressure of 1–2 MPa, the ratios of propane and the butanes are close to one or below it, while those of methane and ethane are large. So the gas separated at this pressure is mostly methane and ethane, and the intermediate components stay in the liquid. When this liquid reaches the stock tank, little dissolved methane is left in it, and the stock-tank gas carries away only a little propane and butanes. Now suppose the pressure is cut at once to atmospheric. All the methane is released at a pressure where the ratios of propane and the butanes are several times greater than one, so the intermediate components leave in the gas together with it. The lesson on separator tests explains this in words. Here we show how to calculate it.

Separation as a chain of flash calculations

Take 1 kmol of reservoir oil of composition $z_i$ . At the first stage we solve the Rachford–Rice equation (flash calculation) with the equilibrium ratios $K_{i,1}$ at pressure $p_1$ and temperature $T_1$ . $V_1$ kmol go into the gas, and $1 - V_1$ kmol of liquid of composition $x_{i,1}$ remain. This liquid is the feed of the second stage (the stream that enters it). There we solve the Rachford–Rice equation again, now with the composition of that liquid and the ratios $K_{i,2}$ , and so on up to the stock tank. The gas fraction of each stage refers to its own feed, so to get the amounts per kilomole of reservoir oil, we multiply.

Moles of gas and stock-tank oil in multistage separation

n_{g,j} = V_j\prod_{k=1}^{j-1}(1 - V_k),\qquad n_{\text{od}} = \prod_{k=1}^{N}(1 - V_k)

where:

$V_j$ mole fraction of gas at stage $j$ : the fraction of the feed of this stage that goes into the gas (the root of the Rachford–Rice equation), fraction
$n_{g,j}$ gas of stage $j$ per 1 kmol of reservoir oil, kmol/kmol
$n_{\text{od}}$ stock-tank oil per 1 kmol of reservoir oil, kmol/kmol
$N$ number of stages, including the stock tank, —

Units: kmol per 1 kmol of reservoir oil

Holds when

At each stage the gas and the liquid are in equilibrium and all the gas is removed.
The feed of the first stage is single-phase reservoir oil (for $B_{ob}$ , at the bubble point pressure); water is not included in the calculation.
The ratios $K_{i,j}$ of each stage are calculated for its pressure, temperature and feed composition (here from an equation of state, Equilibrium ratios: correlations and equations of state).

The check is the mole balance: $n_{\text{od}} + \sum_j n_{g,j} = 1$ . The balance for each component is $z_i = n_{\text{od}}x_{i,\text{od}} + \sum_j n_{g,j}y_{i,j}$ . It shows what share of a component has gone with the gas of each stage.

Order of calculation

1
Stage feed
For the first stage, take the composition of the reservoir oil $z_i$ . For each following stage, take the composition of the liquid of the previous stage.
2
Equilibrium ratios
Find them at the pressure and temperature of the stage: from an equation of state, and for separators also from correlations (Equilibrium ratios: correlations and equations of state).
3
Flash calculation
Solve the Rachford–Rice equation for $V_j$ , then find the compositions of the gas $y_{i,j}$ and of the liquid $x_{i,j}$ .
4
Transfer
Pass on the liquid of the stage, which is $1 - V_j$ of its feed. Enter the gas of the stage in the balance: $n_{g,j}$ per 1 kmol of reservoir oil.
5
Stock tank
The liquid of the last stage is the stock-tank oil. Record its amount $n_{\text{od}}$ and its composition $x_{i,\text{od}}$ .

Composition of the lesson's assumed reservoir oil

Component: N₂
Mole fraction $z_i$: 0.006
$M_i$ , kg/kmol: 28.01

Component: CO₂
Mole fraction $z_i$: 0.019
$M_i$ , kg/kmol: 44.01

Component: C₁
Mole fraction $z_i$: 0.410
$M_i$ , kg/kmol: 16.04

Component: C₂
Mole fraction $z_i$: 0.085
$M_i$ , kg/kmol: 30.07

Component: C₃
Mole fraction $z_i$: 0.058
$M_i$ , kg/kmol: 44.10

Component: i-C₄
Mole fraction $z_i$: 0.009
$M_i$ , kg/kmol: 58.12

Component: n-C₄
Mole fraction $z_i$: 0.022
$M_i$ , kg/kmol: 58.12

Component: i-C₅
Mole fraction $z_i$: 0.010
$M_i$ , kg/kmol: 72.15

Component: n-C₅
Mole fraction $z_i$: 0.013
$M_i$ , kg/kmol: 72.15

Component: n-C₆
Mole fraction $z_i$: 0.018
$M_i$ , kg/kmol: 86.18

Component: F1 (C₇–C₉)
Mole fraction $z_i$: 0.120
$M_i$ , kg/kmol: 110.00

Component: F2 (C₁₀–C₁₉)
Mole fraction $z_i$: 0.155
$M_i$ , kg/kmol: 200.00

Component: F3 (C₂₀+)
Mole fraction $z_i$: 0.075
$M_i$ , kg/kmol: 410.00

Component: C₇₊ as a whole
Mole fraction $z_i$: 0.350
$M_i$ , kg/kmol: 214.1

Component	Mole fraction $z_i$	$M_i$ , kg/kmol
N₂	0.006	28.01
CO₂	0.019	44.01
C₁	0.410	16.04
C₂	0.085	30.07
C₃	0.058	44.10
i-C₄	0.009	58.12
n-C₄	0.022	58.12
i-C₅	0.010	72.15
n-C₅	0.013	72.15
n-C₆	0.018	86.18
F1 (C₇–C₉)	0.120	110.00
F2 (C₁₀–C₁₉)	0.155	200.00
F3 (C₂₀+)	0.075	410.00
C₇₊ as a whole	0.350	214.1

The data are assumed for illustration. The C₇₊ fraction (heptane and all the heavier components taken together; relative density 0.845) is represented by three pseudo-components, which stand in for groups of real components. Their molar masses are 110, 200, 410 and their relative densities are 0.75, 0.83, 0.91. The equation-of-state properties and the binary interaction coefficients are given in the lesson on equilibrium ratios. The constants of the other components are reference values, as in the lesson on equations of state. For isopentane, $T_c$ = 460.4 K, $p_c$ = 3.38 MPa and $\omega$ = 0.229. The reservoir temperature is 80 °C, and the bubble point pressure by the Peng–Robinson equation is 19.40 MPa. This is not the same assumed oil as in the lessons on the black oil model and on PVT experiments.

Equilibrium ratios of the stages of the example (Peng–Robinson equation)

Component: N₂
Stage 1: 1.2 MPa, 40 °C: 54.91
Stock tank: 0.101 MPa, 20 °C: 645.9

Component: CO₂
Stage 1: 1.2 MPa, 40 °C: 9.122
Stock tank: 0.101 MPa, 20 °C: 81.79

Component: C₁
Stage 1: 1.2 MPa, 40 °C: 19.55
Stock tank: 0.101 MPa, 20 °C: 201.1

Component: C₂
Stage 1: 1.2 MPa, 40 °C: 3.517
Stock tank: 0.101 MPa, 20 °C: 27.87

Component: C₃
Stage 1: 1.2 MPa, 40 °C: 1.123
Stock tank: 0.101 MPa, 20 °C: 7.446

Component: i-C₄
Stage 1: 1.2 MPa, 40 °C: 0.4985
Stock tank: 0.101 MPa, 20 °C: 2.904

Component: n-C₄
Stage 1: 1.2 MPa, 40 °C: 0.3617
Stock tank: 0.101 MPa, 20 °C: 2.009

Component: i-C₅
Stage 1: 1.2 MPa, 40 °C: 0.1571
Stock tank: 0.101 MPa, 20 °C: 0.7649

Component: n-C₅
Stage 1: 1.2 MPa, 40 °C: 0.1221
Stock tank: 0.101 MPa, 20 °C: 0.5688

Component: n-C₆
Stage 1: 1.2 MPa, 40 °C: 0.0426
Stock tank: 0.101 MPa, 20 °C: 0.1679

Component: F1 (C₇–C₉)
Stage 1: 1.2 MPa, 40 °C: 0.006377
Stock tank: 0.101 MPa, 20 °C: 0.01885

Component: F2 (C₁₀–C₁₉)
Stage 1: 1.2 MPa, 40 °C: 1.291 × 10⁻⁵
Stock tank: 0.101 MPa, 20 °C: 1.359 × 10⁻⁵

Component: F3 (C₂₀+)
Stage 1: 1.2 MPa, 40 °C: 9.225 × 10⁻¹¹
Stock tank: 0.101 MPa, 20 °C: 1.287 × 10⁻¹¹

Component	Stage 1: 1.2 MPa, 40 °C	Stock tank: 0.101 MPa, 20 °C
N₂	54.91	645.9
CO₂	9.122	81.79
C₁	19.55	201.1
C₂	3.517	27.87
C₃	1.123	7.446
i-C₄	0.4985	2.904
n-C₄	0.3617	2.009
i-C₅	0.1571	0.7649
n-C₅	0.1221	0.5688
n-C₆	0.0426	0.1679
F1 (C₇–C₉)	0.006377	0.01885
F2 (C₁₀–C₁₉)	1.291 × 10⁻⁵	1.359 × 10⁻⁵
F3 (C₂₀+)	9.225 × 10⁻¹¹	1.287 × 10⁻¹¹

The pressures are absolute. The stock-tank ratios are calculated for the liquid of the first stage, because the composition of the feed also enters the calculation. The ratio of the heavy pseudo-component is negligible: it stays entirely in the oil.

Worked exampleTwo-stage separation: moles of gas and oil

Given (assumed data)

Feed: 1 kmol of the lesson's assumed reservoir oil at the bubble point pressure and 80 °C
First stage: 1.2 MPa (absolute), 40 °C
Second stage (stock tank): 0.101325 MPa, 20 °C
Equilibrium ratios: from the Peng–Robinson equation, the table above

Solution

First stage: the root of the Rachford–Rice equation is $V_1 = 0.53012$ . $n_{g,1} = 0.53012$ kmol goes into the gas, and $1 - V_1 = 0.46988$ kmol of liquid goes to the stock tank.
The gas of the first stage is 74.0% methane (by moles). It holds 6.1% propane, against 5.4% in the liquid. At 1.2 MPa the equilibrium ratio of propane, 1.12, is close to one, so propane has nearly the same mole fraction in both phases.
Stock tank: the feed is the liquid of the first stage, and the root is $V_2 = 0.10771$ , a fraction of the 0.46988 kmol. The stock-tank gas is $n_{g,2} = 0.46988\cdot 0.10771 = 0.05061$ kmol per 1 kmol of reservoir oil.
Stock-tank oil: $n_{\text{od}} = 0.46988\cdot(1 - 0.10771) = 0.41927$ kmol. Mole balance check: $0.53012 + 0.05061 + 0.41927 = 1.00000$ .
The stock-tank gas is much richer in heavy components: it contains 33.7% methane, 26.1% ethane, 23.9% propane and 9.0% butanes.

Answer: 1 kmol of reservoir oil gives 0.41927 kmol of stock-tank oil, 0.53012 kmol of first-stage gas and 0.05061 kmol of stock-tank gas.

What it means. Almost all the methane leaves with the first-stage gas (95.7%). The propane is divided: 56% goes into the first-stage gas, 21% into the stock-tank gas, and 23% stays in the oil. By moles there is 10 times less stock-tank gas than first-stage gas, but it carries away a noticeable share of the intermediate components. The optimum of the first-stage pressure rests on this (see the section below). The data in this example are assumed for illustration.

Compositions of the streams of two-stage separation, mole fractions

Component: N₂
Reservoir oil: 0.0060
Stage 1 gas: 0.0111
Stock-tank gas: 0.0019
Stock-tank oil: < 0.0001

Component: CO₂
Reservoir oil: 0.0190
Stage 1 gas: 0.0327
Stock-tank gas: 0.0302
Stock-tank oil: 0.0004

Component: C₁
Reservoir oil: 0.4100
Stage 1 gas: 0.7399
Stock-tank gas: 0.3375
Stock-tank oil: 0.0017

Component: C₂
Reservoir oil: 0.0850
Stage 1 gas: 0.1281
Stock-tank gas: 0.2606
Stock-tank oil: 0.0093

Component: C₃
Reservoir oil: 0.0580
Stage 1 gas: 0.0611
Stock-tank gas: 0.2393
Stock-tank oil: 0.0321

Component: i-C₄
Reservoir oil: 0.0090
Stage 1 gas: 0.0061
Stock-tank gas: 0.0295
Stock-tank oil: 0.0102

Component: n-C₄
Reservoir oil: 0.0220
Stage 1 gas: 0.0120
Stock-tank gas: 0.0603
Stock-tank oil: 0.0300

Component: i-C₅
Reservoir oil: 0.0100
Stage 1 gas: 0.0028
Stock-tank gas: 0.0142
Stock-tank oil: 0.0185

Component: n-C₅
Reservoir oil: 0.0130
Stage 1 gas: 0.0030
Stock-tank gas: 0.0145
Stock-tank oil: 0.0255

Component: n-C₆
Reservoir oil: 0.0180
Stage 1 gas: 0.0016
Stock-tank gas: 0.0067
Stock-tank oil: 0.0401

Component: F1 (C₇–C₉)
Reservoir oil: 0.1200
Stage 1 gas: 0.0016
Stock-tank gas: 0.0053
Stock-tank oil: 0.2835

Component: F2 (C₁₀–C₁₉)
Reservoir oil: 0.1550
Stage 1 gas: < 0.0001
Stock-tank gas: < 0.0001
Stock-tank oil: 0.3697

Component: F3 (C₂₀+)
Reservoir oil: 0.0750
Stage 1 gas: < 0.0001
Stock-tank gas: < 0.0001
Stock-tank oil: 0.1789

Component	Reservoir oil	Stage 1 gas	Stock-tank gas	Stock-tank oil
N₂	0.0060	0.0111	0.0019	< 0.0001
CO₂	0.0190	0.0327	0.0302	0.0004
C₁	0.4100	0.7399	0.3375	0.0017
C₂	0.0850	0.1281	0.2606	0.0093
C₃	0.0580	0.0611	0.2393	0.0321
i-C₄	0.0090	0.0061	0.0295	0.0102
n-C₄	0.0220	0.0120	0.0603	0.0300
i-C₅	0.0100	0.0028	0.0142	0.0185
n-C₅	0.0130	0.0030	0.0145	0.0255
n-C₆	0.0180	0.0016	0.0067	0.0401
F1 (C₇–C₉)	0.1200	0.0016	0.0053	0.2835
F2 (C₁₀–C₁₉)	0.1550	< 0.0001	< 0.0001	0.3697
F3 (C₂₀+)	0.0750	< 0.0001	< 0.0001	0.1789

Stream amounts per 1 kmol of reservoir oil: first-stage gas 0.5301, stock-tank gas 0.0506, stock-tank oil 0.4193 kmol. Calculation with the Peng–Robinson equation; the data are assumed for illustration.

Where the components of the reservoir oil go, % of their amount

Component: N₂
Stage 1 gas: 98.4
Stock-tank gas: 1.6
Stock-tank oil: 0.0

Component: CO₂
Stage 1 gas: 91.1
Stock-tank gas: 8.0
Stock-tank oil: 0.8

Component: C₁
Stage 1 gas: 95.7
Stock-tank gas: 4.2
Stock-tank oil: 0.2

Component: C₂
Stage 1 gas: 79.9
Stock-tank gas: 15.5
Stock-tank oil: 4.6

Component: C₃
Stage 1 gas: 55.9
Stock-tank gas: 20.9
Stock-tank oil: 23.2

Component: i-C₄
Stage 1 gas: 36.0
Stock-tank gas: 16.6
Stock-tank oil: 47.4

Component: n-C₄
Stage 1 gas: 29.0
Stock-tank gas: 13.9
Stock-tank oil: 57.2

Component: i-C₅
Stage 1 gas: 15.1
Stock-tank gas: 7.2
Stock-tank oil: 77.8

Component: n-C₅
Stage 1 gas: 12.1
Stock-tank gas: 5.6
Stock-tank oil: 82.2

Component: n-C₆
Stage 1 gas: 4.6
Stock-tank gas: 1.9
Stock-tank oil: 93.5

Component: C₇₊ as a whole
Stage 1 gas: 0.2
Stock-tank gas: 0.1
Stock-tank oil: 99.7

Component	Stage 1 gas	Stock-tank gas	Stock-tank oil
N₂	98.4	1.6	0.0
CO₂	91.1	8.0	0.8
C₁	95.7	4.2	0.2
C₂	79.9	15.5	4.6
C₃	55.9	20.9	23.2
i-C₄	36.0	16.6	47.4
n-C₄	29.0	13.9	57.2
i-C₅	15.1	7.2	77.8
n-C₅	12.1	5.6	82.2
n-C₆	4.6	1.9	93.5
C₇₊ as a whole	0.2	0.1	99.7

This is the two-stage scheme of the example. Methane, nitrogen and ethane go almost entirely with the gas. The hexanes and the C₇₊ fraction stay almost entirely in the oil. Propane, butanes and pentanes divide between all three streams, so the choice of stage pressures decides their fate.

From moles to cubic metres: gas–oil ratio and formation volume factor

Moles are convenient for balances, but the field measures volumes: cubic metres of gas at standard conditions and cubic metres of stock-tank oil. Gas at standard conditions is almost ideal. A kilomole of any gas occupies $V_{m,\text{sc}} = RT_{\text{sc}}/p_{\text{sc}}$ = 24.055 m³ at 0.101325 MPa and 20 °C (these are the standard conditions used in this library; at 60 °F and 14.696 psia, a pound-mole takes 379.5 standard cubic feet). We find the volume of a liquid from its molar mass and density.

Stage gas–oil ratio and solution gas–oil ratio from the separator calculation

R_j = \frac{n_{g,j}\,V_{m,\text{sc}}}{V_{\text{od}}},\qquad V_{\text{od}} = \frac{n_{\text{od}}\,M_{\text{od}}}{\rho_{\text{od}}},\qquad R_{sb} = \sum_{j=1}^{N} R_j

where:

$R_j$ gas–oil ratio of stage $j$ : its gas at standard conditions per 1 m³ of stock-tank oil, m³/m³
$V_{m,\text{sc}}$ volume of a kilomole of gas at standard conditions, 24.055 at 0.101325 MPa and 20 °C, m³/kmol
$V_{\text{od}}$ volume of the stock-tank oil obtained from 1 kmol of reservoir oil, at standard conditions, m³
$M_{\text{od}},\ \rho_{\text{od}}$ molar mass of the stock-tank oil $\sum x_{i,\text{od}}M_i$ and its density at standard conditions (in the lesson on the black oil model, $\rho_{o,\text{sc}}$ ), kg/kmol, kg/m³
$R_{sb}$ solution gas–oil ratio of the reservoir oil at the bubble point pressure for this separation scheme, m³/m³

Units: m³/m³: standard cubic metres of gas per cubic metre of stock-tank oil

Holds when

We treat gas at standard conditions as ideal. Its $Z$ (the compressibility factor, which is 1 for an ideal gas) is 0.99–1.00, and about 0.99 for the heavy stock-tank gas. The assumption overestimates the solution gas–oil ratio by less than one per cent: by 0.5% in the lesson example.
The density of the stock-tank oil is known. It is measured, or calculated from an equation of state with volume translation (a correction that improves the density of liquids), or calculated from the densities of the components, assuming that volumes add up.
The stock-tank gas is included in the sum: it too was dissolved in the reservoir oil.

The gas–oil ratio of a stage is counted per cubic metre of stock-tank oil, not per cubic metre of the oil that left that stage. Per 1 m³ of separator oil there is $B_{\text{sep}}$ times less separator gas, where $B_{\text{sep}}$ is the volume of separator oil needed for 1 m³ of stock-tank oil. The reason is that the separator oil will still shrink. We need this conversion when separator samples are recombined, that is, mixed back together in the right proportions (recombination).

In other unit systems (1)

Oilfield units

R\,[\text{scf/STB}] = 5.529\,R\,[\text{m}^3/\text{m}^3]

gas in m³ at 20 °C is converted to standard cubic feet at 60 °F. If both quantities refer to the same conditions, the factor is 5.6146.

5.6146 is the number of cubic feet in a barrel. In going from 20 °C to 60 °F, the gas contracts by about 1.5% (standard conditions).

Oil formation volume factor at the bubble point pressure from the separator calculation

B_{ob} = \frac{M_{\text{res}}/\rho_{ob}}{V_{\text{od}}}

where:

$M_{\text{res}}$ molar mass of the reservoir oil $\sum z_iM_i$ , kg/kmol
$\rho_{ob}$ density of the reservoir oil at the bubble point pressure and reservoir temperature, kg/m³
$V_{\text{od}}$ stock-tank oil from 1 kmol of reservoir oil, as above, m³

Units: m³/m³ (reservoir cubic metres per cubic metre of stock-tank oil)

Holds when

The feed is single-phase oil at the bubble point pressure; at a higher pressure the numerator is its molar volume at that pressure.
The density of the reservoir oil is measured or calculated from an equation of state with volume translation; such an estimate of liquid densities is off by a few per cent, and $B_{ob}$ by as much.

The numerator is the volume that one kilomole of oil occupies in the reservoir. The denominator is the volume of stock-tank oil obtained from it. A mass balance checks the result (where the formation volume factor comes from): $B_{ob}\rho_{ob} = \rho_{\text{od}} + \sum_j R_j\gamma_{g,j}\rho_{\text{air}}$ , where $\gamma_{g,j}$ is the specific gravity of the gas of stage $j$ (its density relative to air) and $\rho_{\text{air}}$ = 1.204 kg/m³ at 20 °C.

Worked exampleSolution gas–oil ratio and formation volume factor from two-stage separation

Given (assumed data)

Moles from the previous example: first-stage gas 0.53012, stock-tank gas 0.05061, stock-tank oil 0.41927 kmol per 1 kmol of reservoir oil
Stock-tank oil: $M_{\text{od}} = \sum x_iM_i = 189.18$ kg/kmol; density at 20 °C from the equation with volume translation 821.98 kg/m³
Gas of the stages: molar mass: first stage 21.951, stock tank 33.642 kg/kmol
Reservoir oil: $M_{\text{res}} = 92.657$ kg/kmol; density at the bubble point pressure and 80 °C from the equation 683.15 kg/m³

Solution

Volume of the stock-tank oil from 1 kmol of reservoir oil: $V_{\text{od}} = 0.41927\cdot 189.18/821.98 = 0.096496$ m³.
Gas at standard conditions: first stage $0.53012\cdot 24.055 = 12.752$ m³, stock tank $0.05061\cdot 24.055 = 1.2174$ m³.
Gas–oil ratios: $R_1 = 12.752/0.096496 = 132.15$ m³/m³, $R_2 = 1.2174/0.096496 = 12.62$ m³/m³; the solution gas–oil ratio is $R_{sb} = 144.77$ m³/m³, or $144.77\cdot 5.529 = 800$ scf/STB.
Molar volume of the reservoir oil $92.657/683.15 = 0.135632$ m³/kmol; formation volume factor $B_{ob} = 0.135632/0.096496 = 1.4056$ m³/m³.
Gas specific gravities: $\gamma_1 = 21.951/28.97 = 0.7577$ , $\gamma_2 = 33.642/28.97 = 1.1613$ .
Mass balance per 1 m³ of stock-tank oil: reservoir oil $1.4056\cdot 683.15 = 960.2$ kg; products $821.98 + 132.15\cdot 0.7577\cdot 1.204 + 12.62\cdot 1.1613\cdot 1.204 = 960.2$ kg.
Density of the stock-tank oil at 60 °F (assumed thermal expansion coefficient $8\cdot10^{-4}$ 1/°C): 824.90 kg/m³, that is 39.9 °API.

Answer: The two-stage scheme gives

R_{sb} = 144.77

m³/m³ and

B_{ob} = 1.4056

m³/m³: 1 m³ of reservoir oil yields 0.7114 m³ of stock-tank oil of density 821.98 kg/m³.

What it means. The stock-tank gas is 8.7% of the solution gas–oil ratio, but it is 1.5 times heavier than the first-stage gas, because it contains a lot of propane and butanes. It is either compressed and delivered with the rest of the gas, or collected by a vapour recovery unit (equipment that captures the vapours from the tank). If it is burned in a flare, both the gas and its valuable components are lost. The mass balance closes to within rounding. This is how both the calculation and the data of a separator test are checked. Densities from an equation of state with volume translation are off by a few per cent, so in a project $R_{sb}$ and $B_{ob}$ are checked against separator tests (separator test). The data in this example are assumed for illustration.

Optimum first-stage pressure

Let us change the first-stage pressure $p_1$ and keep its temperature and the stock tank as they were. At a very low $p_1$ , the first stage works almost like a single flash (one step straight down to stock-tank pressure), only at a higher temperature. Methane is released at a pressure where the ratios of propane and the butanes are large, so the intermediate components leave with the first-stage gas. At a very high $p_1$ , the first-stage gas is lean (it holds little of the heavier components), but a lot of dissolved methane stays in the liquid. It is released in the stock tank, where the ratios of propane and the butanes are 2–7 times greater than one, and the stock-tank gas carries them away. Between the two extremes lies the optimum separator pressure, where the amount of stock-tank oil is greatest and $R_{sb}$ and $B_{ob}$ are smallest.

CalculatedTwo-stage separation of the lesson's assumed oil. Top: the formation volume factor

B_{ob}

against first-stage pressure, at first-stage temperatures of 20, 40 and 60 °C (the dots are the optima). Bottom: the share of the propane and n-butane of the reservoir oil that stays in the stock-tank oil, at 40 °C. Stock tank: 0.101325 MPa and 20 °C. The pressure axis is logarithmic.

Worked exampleOptimum of the two-stage scheme at 40 °C

Given (assumed data)

Oil: the lesson's assumed reservoir oil at the bubble point pressure
Scheme: first stage at 40 °C and a pressure from 0.3 to 4 MPa, then the stock tank (0.101325 MPa, 20 °C)
For comparison: a single flash straight to stock-tank conditions

Solution

Single flash: $R_{sb} = 158.3$ m³/m³, $B_{ob} = 1.4555$ , stock-tank oil density 829.6 kg/m³.
First stage at 0.3 MPa: $R_{sb} = 153.9$ , $B_{ob} = 1.4400$ , 827.0 kg/m³; at 4 MPa: $R_{sb} = 147.9$ , $B_{ob} = 1.4169$ , 823.9 kg/m³.
We find the minimum of $B_{ob}$ by searching over pressure (a golden-section search on the logarithm of pressure, which narrows the interval step by step). The result is $p_1 = 1.43$ MPa, $B_{ob} = 1.4051$ , $R_{sb} = 144.7$ m³/m³ and density 821.9 kg/m³: the oil is also the lightest.
Yield of stock-tank oil from 1 m³ of reservoir oil: $1/1.4051 = 0.7117$ m³, against $1/1.4555 = 0.6870$ m³ for a single flash. This is 3.6% more by volume.
By mass the gain is smaller: $\dfrac{821.9/1.4051}{829.6/1.4555} - 1 = 2.6$ %, because the light propane and butanes stay in the oil.
The optimum is flat: at $p_1$ = 1.2 MPa (the example above), $B_{ob}$ exceeds the optimum by only 0.04%, and at 0.3 MPa by 2.5%.

Answer: The optimum first-stage pressure at 40 °C is about 1.43 MPa. Compared with a single flash, the scheme gives 3.6% more oil by volume and 2.6% by mass.

What it means. Take a field that produces 1 million t of oil a year by a single flash. The gain by mass is then about 26 thousand t a year from the same reservoir. The minimum is flat. In the range of pressures where $B_{ob}$ differs from the optimum by hundredths of a per cent, other things decide the pressure. These include the pressure of the gas pipeline and the capacity of the compressors for the gas of the low stages, the vapour pressure limit for the sales oil, and whether the gas goes to a plant that also recovers the propane and butanes. The data in this example are assumed for illustration.

Separation temperature and the number of stages

The colder the separator, the smaller the equilibrium ratios of the intermediate components, and the more of them stay in the liquid. The calculation (see the figure above) gives these results. At a first-stage temperature of 20 °C, the best $B_{ob}$ is 1.3980 at a pressure of 0.91 MPa. At 40 °C it is 1.4051 at 1.43 MPa, and at 60 °C it is 1.4128 at 2.13 MPa. As the temperature rises, the optimum shifts to higher pressures and the oil yield falls. But yield is not the only reason for choosing the temperature. Warm oil separates from water more easily and foams less. Oil that has kept more light components has a higher vapour pressure, so it has to be stabilised (its light components removed) for transport and storage.

Oil yield with different numbers of stages: separators at 40 °C, stock tank at 0.101325 MPa and 20 °C

Stages: 1
Separator pressures, MPa: — (straight to the stock tank)
$R_{sb}$ , m³/m³: 158.3
$B_{ob}$ , m³/m³: 1.4555
Oil from 1 m³ of reservoir oil, m³: 0.6870

Stages: 2
Separator pressures, MPa: 1.43
$R_{sb}$ , m³/m³: 144.7
$B_{ob}$ , m³/m³: 1.4051
Oil from 1 m³ of reservoir oil, m³: 0.7117

Stages: 3
Separator pressures, MPa: 2.62 and 0.527
$R_{sb}$ , m³/m³: 140.8
$B_{ob}$ , m³/m³: 1.3919
Oil from 1 m³ of reservoir oil, m³: 0.7184

Stages: 4
Separator pressures, MPa: 3.53; 0.925; 0.334
$R_{sb}$ , m³/m³: 139.2
$B_{ob}$ , m³/m³: 1.3865
Oil from 1 m³ of reservoir oil, m³: 0.7212

Stages	Separator pressures, MPa	$R_{sb}$ , m³/m³	$B_{ob}$ , m³/m³	Oil from 1 m³ of reservoir oil, m³
1	— (straight to the stock tank)	158.3	1.4555	0.6870
2	1.43	144.7	1.4051	0.7117
3	2.62 and 0.527	140.8	1.3919	0.7184
4	3.53; 0.925; 0.334	139.2	1.3865	0.7212

The separator pressures are chosen for the smallest $B_{ob}$ . The stock tank counts as a stage. The second stage adds 3.6% of oil, the third another 0.9%, the fourth 0.4%. The data are assumed for illustration.

Each further stage adds less oil than the one before, and it costs no less: it needs a separator, piping and, above all, a compressor to raise the pressure of the low-stage gas to the pipeline pressure. So in the field two or three stages are usually enough. The best stage pressures follow a simple pattern: the pressure ratios of neighbouring stages come out close to each other. This gives a technique for a first estimate: set the stage pressures with equal ratios.

Stage pressures with equal ratios (first estimate)

r = \left(\frac{p_1}{p_N}\right)^{1/(N-1)},\qquad p_{j+1} = \frac{p_j}{r}

where:

$p_1$ first-stage pressure (usually set by the wellhead pressure or by the gathering system), MPa (absolute)
$p_N$ pressure of the last stage, which is the stock tank, MPa (absolute)
$N$ number of stages, including the stock tank, —
$r$ pressure ratio of adjacent stages, —

Units: absolute pressures in the same units

Holds when

It is an empirical rule for the first choice, not a result of theory. The separator calculation and the economics refine the final pressures.
The temperatures of the stages are close. If the stages differ greatly in temperature, the rule is less reliable.

The example below shows that for the lesson's oil, stage pressures with equal ratios give almost the same yield as those found by search.

Worked exampleThree stages: equal ratios against search

Given (assumed data)

Scheme: two separators at 40 °C and the stock tank at 0.101325 MPa and 20 °C
Pressures by search: $p_1 = 2.62$ MPa and $p_2 = 0.527$ MPa (table above)
Task: for the same $p_1$ find $p_2$ by the equal-ratio rule and compare the oil yield

Solution

Ratios by search: $p_1/p_2 = 5.0$ and $p_2/p_{\text{tank}} = 5.2$ . They are indeed close.
The rule: $r = \sqrt{2.62/0.101325} = 5.085$ , $p_2 = 2.62/5.085 = 0.515$ MPa.
The separator calculation with this pressure gives $B_{ob} = 1.39187$ , against 1.39186 at the search pressures: a difference of 0.0007%.
For comparison, the second stage at half the pressure of the first, $p_2$ = 1.31 MPa: $B_{ob} = 1.39887$ , 0.50% more than at the search pressures.

Answer: The equal-ratio rule gave

p_2

= 0.515 MPa instead of 0.527 MPa and practically the same oil yield:

B_{ob}

is only 0.0007% above the smallest value. By contrast,

p_2

= 1.31 MPa costs 0.50%.

What it means. For a first estimate the rule is enough. Near the optimum, $B_{ob}$ hardly depends on $p_2$ , and refining by search adds nothing noticeable. The pressure of the gathering system and of the gas pipeline usually sets the first-stage pressure, and the rule picks the intermediate stages to match it. The final choice also depends on the cost of compressing the gas of the low stages. The data in this example are assumed for illustration.

Assumptions and sources of error

Equilibrium and completeness of separation. Droplets carried over with the gas make the calculated gas–oil ratio too high and the oil yield too low. Gas bubbles left in the liquid do the opposite. A short retention time and foam make a real stage worse than an equilibrium stage.
Equilibrium ratios. For separators, correlations and an equation of state usually give values of $R_{sb}$ and $B_{ob}$ that differ by one or two per cent. An error in the constants of the C₇₊ fraction has a stronger effect (Equilibrium ratios: correlations and equations of state).
Densities. Liquid densities from an equation of state, even with volume translation, are off by a few per cent. Where possible, replace them with measurements.
Stage temperature. Throttling (letting the pressure drop across a valve) and evaporation cool the stream. Use the actual stage temperature, not the design one.
Feed composition. If there is already free gas in the well, the separator feed is the oil together with that gas, and the gas–oil ratio is greater than the solution gas–oil ratio. An error in the composition of the sample (Reservoir fluid sampling) goes into the result in full.

Terms and symbols

Separation stage: The pressure and temperature at which gas is separated from liquid. It is a separator or the stock tank.
Stock tank (final stage): The last stage, at a pressure close to atmospheric. The liquid that remains in it is the stock-tank oil.
Stock-tank gas: The gas released in the stock tank from the oil of the last separator. It is heavy and rich in propane and butanes, and it is part of the solution gas–oil ratio.
Stage gas–oil ratio $R_j$: The gas of a stage at standard conditions per 1 m³ of stock-tank oil. The sum of these ratios is the solution gas–oil ratio $R_{sb}$ .
Optimum separator pressure: The pressure of the first stage (or a set of pressures) at which the yield of stock-tank oil is greatest and $R_{sb}$ and $B_{ob}$ are smallest.
Equal-ratio rule: A technique for a first estimate: the stage pressures are set so that the pressure ratios of neighbouring stages are equal.

Common mistakes and misconceptions

Wrong

“The solution gas–oil ratio is the first-stage gas per cubic metre of oil.”

In fact

The solution gas–oil ratio is the gas of all stages, including the stock tank, per 1 m³ of stock-tank oil. In the lesson example the stock-tank gas is about 9% of the solution gas–oil ratio.

Wrong

“The gas fractions of the stages add up: $V_1 + V_2$ .”

In fact

The fraction of the second stage refers to the liquid of the first. The total gas is $V_1 + V_2(1 - V_1)$ per 1 kmol of reservoir oil.

Wrong

“The higher the first-stage pressure, the better: gas is cheaper to transport.”

In fact

Beyond the optimum, more methane reaches the stock tank and carries the intermediate components away, so there is less oil. Choose the pressure within the flat minimum.

Wrong

“A gain in cubic metres is also a gain in tonnes.”

In fact

Stage separation keeps the light propane and butanes in the oil, so the gain by mass is smaller: in the lesson example 3.6% by volume and 2.6% by mass.

Wrong

“The optimum found for one oil is good for another.”

In fact

It depends on the composition of the oil and on temperature. For the lesson's oil the optimum is about 0.9 MPa at 20 °C and about 2.1 MPa at 60 °C.

In depth: advanced detail

Black oil table from composition

Below the bubble point pressure, the oil that stays in the reservoir has a different composition: part of the methane and ethane has gone into free gas. To find its solution gas–oil ratio and formation volume factor for the same treatment scheme, we calculate in two steps. First, flash the reservoir oil at pressure $p$ and reservoir temperature. Then pass the equilibrium liquid through the separation stages. This is how the tables $R_s(p)$ and $B_o(p)$ are built from an equation of state. The method avoids converting the data of the differential liberation test (a laboratory test that lowers the pressure step by step and removes the gas at each step), a conversion that the laboratory can do only approximately (conversion to separator conditions).

Solution gas–oil ratio and formation volume factor of the lesson's oil at 80 °C for the two-stage scheme of the example (1.2 MPa and 40 °C, stock tank)

Pressure, MPa: 19.40
Gas fraction in the reservoir $V$: 0.000
$R_s$ , m³/m³: 144.8
$B_o$ , m³/m³: 1.4056
Oil density, kg/m³: 683.15

Pressure, MPa: 15
Gas fraction in the reservoir $V$: 0.146
$R_s$ , m³/m³: 108.3
$B_o$ , m³/m³: 1.3195
Oil density, kg/m³: 704.07

Pressure, MPa: 10
Gas fraction in the reservoir $V$: 0.295
$R_s$ , m³/m³: 70.9
$B_o$ , m³/m³: 1.2295
Oil density, kg/m³: 728.98

Pressure, MPa: 5
Gas fraction in the reservoir $V$: 0.437
$R_s$ , m³/m³: 35.6
$B_o$ , m³/m³: 1.1399
Oil density, kg/m³: 757.27

Pressure, MPa	Gas fraction in the reservoir $V$	$R_s$ , m³/m³	$B_o$ , m³/m³	Oil density, kg/m³
19.40	0.000	144.8	1.4056	683.15
15	0.146	108.3	1.3195	704.07
10	0.295	70.9	1.2295	728.98
5	0.437	35.6	1.1399	757.27

The first row is the bubble point pressure. $V$ is the mole fraction of gas in the flash of the reservoir oil; $R_s$ and $B_o$ refer to the stock-tank oil obtained from the oil that remains in the reservoir at this pressure. Calculation with the Peng–Robinson equation with volume translation; the data are assumed for illustration.

The solution gas–oil ratio falls almost in proportion to pressure, and the formation volume factor falls with it, because the oil shrinks as it loses dissolved gas. The same technique is used for gas condensates and volatile oils (light oils that give off a lot of gas), where the reservoir gas itself yields noticeable liquid at the surface. For this oil such “condensate” is a fraction of a per cent and is not included in the table.

Connections to other topics

Vapour–liquid equilibrium — the flash calculation and the Rachford–Rice equation, which the separator calculation is built from.
Equilibrium ratios: correlations and equations of state — where the stage equilibrium ratios come from: correlations for separators and the equation of state.
Laboratory PVT tests — the separator test, the laboratory measurement of the same $R_{sb}$ and $B_{ob}$ .
Black oil model: solution gas–oil ratio and formation volume factors — solution gas–oil ratio and formation volume factor, their dependence on the separation scheme, and the mass balance.
Reservoir fluid sampling — separator samples, recombination and the separator gas–oil ratio.
Oil treating: separators, dehydration, desalting and stabilisation — separators as vessels: sizes, retention time, water separation, treatment of sales oil.

Check yourself

1
Two-stage separation: at the first stage 0.45 of the feed goes into the gas, in the stock tank 0.08 of the feed of that stage. How much stock-tank oil and gas does 1 kmol of reservoir oil give?
1. A.0.506 kmol of oil and 0.53 kmol of gas
2. B.0.47 kmol of oil and 0.53 kmol of gas
3. C.0.55 kmol of oil and 0.45 kmol of gas
4. D.0.506 kmol of oil and 0.494 kmol of gas
Show the answer
Correct answer: D. 0.506 kmol of oil and 0.494 kmol of gas
The liquid of the first stage is $1 - 0.45 = 0.55$ kmol. The stock-tank gas is $0.08\cdot 0.55 = 0.044$ kmol, and the oil is $0.55\cdot 0.92 = 0.506$ kmol. The total gas is $0.45 + 0.044 = 0.494$ kmol; together with the oil, exactly 1 kmol. The fractions 0.45 and 0.08 cannot be added, because the second refers to the liquid of the first stage.
2
1 kmol of reservoir oil gave 0.40 kmol of stock-tank oil, with a molar mass of 200 kg/kmol and a density of 850 kg/m³, and 0.60 kmol of gas from all the stages. Find the solution gas–oil ratio (gas at 20 °C).
Show the answer
Volume of oil: $0.40\cdot 200/850 = 0.0941$ m³. Gas: $0.60\cdot 24.055 = 14.43$ m³. Solution gas–oil ratio: $14.43/0.0941 \approx 153$ m³/m³.
3
Why is there less stock-tank oil when the first-stage pressure is too high, although less gas is released at the first stage?
Show the answer
A lot of dissolved methane stays in the liquid of the first stage. It is released in the stock tank, where the equilibrium ratios of propane and the butanes at atmospheric pressure are several times greater than one, so the stock-tank gas carries them away. In the lesson example, as the first-stage pressure rises from 1.4 to 6 MPa, the share of n-butane remaining in the oil falls from 57.5 to 41%.
4
How will the optimum first-stage pressure change if its temperature is raised from 40 to 60 °C?
1. A.It will increase
2. B.It will not change
3. C.The optimum will disappear
4. D.It will decrease
Show the answer
Correct answer: A. It will increase
At a higher temperature the equilibrium ratios of the intermediate components are greater, and a higher pressure is needed to keep them in the liquid. According to the calculation for the lesson's oil, the optimum shifts from about 1.4 to 2.1 MPa, and the smallest $B_{ob}$ rises, so there is less oil.
5
The separator calculation gave $B_{ob} = 1.40$ , $R_{sb} = 140$ m³/m³, a stock-tank oil density of 820 kg/m³ and a mean gas specific gravity of 0.80. The measured density of the reservoir oil is 700 kg/m³. Are the data consistent?
Show the answer
Mass balance per 1 m³ of stock-tank oil: reservoir oil $1.40\cdot 700 = 980$ kg; products $820 + 140\cdot 0.80\cdot 1.204 = 820 + 134.8 = 954.8$ kg. The discrepancy of about 2.6% is more than rounding allows. Look for the error in the densities, in the gas of one of the stages, or in $B_{ob}$ itself.

Updated: 17 October 2026