Reservoir fluid sampling

Quick refresher

A sample is representative if its composition equals the composition of the reservoir fluid. While the pressure at the bottom of the well is above the bubble point pressure (the dew point pressure, for a gas condensate), single-phase reservoir fluid flows to the well. Below that pressure, the well receives a mixture whose composition depends on the flow rate and on time.
Before sampling, the well is cleaned of filtrate and treatment fluids. Then the rate is reduced in steps. This goes on until the bottomhole pressure is above the bubble point pressure with a safety margin, and the gas–oil ratio, the pressures and the water cut (the share of water in the produced liquid) have stabilised.
An ordinary bottomhole sample in a rigid chamber becomes two-phase after cooling by a few degrees. The pressure of single-phase oil falls by $\beta_o/c_o$ (the thermal expansion coefficient divided by the compressibility), about 0.6 MPa per degree. A single-phase sampler keeps the pressure up with a nitrogen cushion.
OBM filtrate dissolves in the oil completely. The formation tester keeps pumping until its sensors show acceptable contamination, and the laboratory corrects for the remainder. In the stock-tank oil the share of filtrate is greater than in the reservoir volume, because the oil shrinks and the filtrate does not.
Separator gas and oil are recombined using the gas–oil ratio counted per volume of separator oil and corrected for the actual gas density. An error in the gas–oil ratio passes directly into the bubble point pressure.
A gas condensate is the hardest to sample. Condensate drops out of the gas near the bottom of the well, the fluids separate into layers in the wellbore, and the separator carries droplets over with the gas. The quality of the samples is checked at once at the well, and each sample comes with a record sheet.

Key relationships

q_{\max} = J\,(\bar p - p_{\text{sat}} - \delta)

the highest conditioning (pump-out) rate, with a margin $\delta$ above the saturation pressure

\left(\dfrac{\partial p}{\partial T}\right)_V = \dfrac{\beta_o}{c_o}

pressure change of single-phase oil in a rigid chamber per degree

x_{\text{sc}} = \dfrac{\eta}{\eta + (1 - \eta)/B_o}

fraction of filtrate in the stock-tank oil from its fraction $\eta$ in the reservoir volume

R_{\text{rec}} = R_{\text{sep}}/B_{\text{sep}}

separator gas per 1 m³ of separator oil (the recombination proportion)

Common pitfalls

Taking an oil sample at a bottomhole pressure below the bubble point pressure or just after the well has been produced with a large drawdown.
Recombining by a gas–oil ratio referred to the stock-tank oil as if it were referred to the separator oil.
Taking the gas–oil ratio from a flow meter without correcting for the actual gas density and the gas deviation factor ( $Z$ ).
Regarding a sample from a well drilled with OBM as clean without estimating the contamination.
Transporting a single-phase liquid sample in a cylinder without a gas cushion: on heating by a few degrees the pressure rises by megapascals.
Sending a sample without a record sheet: without the sampling conditions it is useless.

After this lesson you can

explain what makes a sample representative and why samples are taken at the start of development, when the bottomhole pressure is above the bubble point pressure
choose the well conditioning rate before sampling and estimate the time in which the production in the wellbore is replaced by reservoir fluid
compare bottomhole, formation tester and surface sampling and choose the one suited to an oil and to a gas condensate
explain why an ordinary bottomhole sample becomes two-phase on the way up and how a single-phase sampler prevents this
estimate the contamination of a sample by the filtrate of an oil-based mud (OBM) from pump-out data, and convert it into a fraction of stock-tank oil and a correction to the gas–oil ratio (the filtrate is the liquid that seeps from the mud into the rock)
calculate the proportion of separator gas and separator oil for recombination and check samples at the well site by the sampler pressure and the bubble point pressure

When and how samples are taken

The laboratory needs little material. For a full study of an oil, a litre or two of reservoir mixture in several cylinders is usually enough. A gas condensate needs more, because it holds little liquid. The real difficulty is different: the mixture must have the same composition as the fluid in the reservoir. A sample is called representative if it matches the reservoir fluid of the accumulation (the whole oil or gas deposit), or at least of the drainage area of the well (the part of the reservoir that feeds the well).

The main condition is pressure. While the pressure at the bottom of the well is above the bubble point pressure of an oil or the dew point pressure of a gas condensate (a gas that leaves liquid behind as its pressure falls), single-phase reservoir fluid flows to the well. Below that pressure, gas comes out of the oil, or condensate drops out of the gas, near the well. The phases move at different velocities, and a mixture of a different composition reaches the well (Phase behaviour: pure substances and binary mixtures). So samples are taken as early as possible: in exploration wells and the first production wells, before the reservoir pressure has fallen. They are also taken at the smallest drawdown, the pressure difference between the reservoir and the well (Types of reservoir fluids).

Sampling methods

Method: Bottomhole sample
Where it is taken: in the wellbore opposite the reservoir, with a sampler lowered on a cable or a thin wire (wireline or slickline)
When it suits: oil, if the pressure at the sampling point is above the bubble point pressure
Main risk: two-phase state near the bottom and on the way up, water in the wellbore

Method: Formation tester sample
Where it is taken: directly from the formation in open hole, a well section without casing (Pressure measurement with the wireline formation tester)
When it suits: exploration before the casing is run; oil, gas, water from selected beds
Main risk: contamination by drilling mud filtrate

Method: Separator samples
Where it is taken: separator gas and oil, then recombination
When it suits: gas condensates, oils with a high gas–oil ratio, watered-out wells
Main risk: an error in the gas–oil ratio, an unsteady flow

Method: Wellhead sample
Where it is taken: at the wellhead
When it suits: only if the flow at the wellhead is single-phase
Main risk: gas coming out of solution in the wellbore

Method	Where it is taken	When it suits	Main risk
Bottomhole sample	in the wellbore opposite the reservoir, with a sampler lowered on a cable or a thin wire (wireline or slickline)	oil, if the pressure at the sampling point is above the bubble point pressure	two-phase state near the bottom and on the way up, water in the wellbore
Formation tester sample	directly from the formation in open hole, a well section without casing (Pressure measurement with the wireline formation tester)	exploration before the casing is run; oil, gas, water from selected beds	contamination by drilling mud filtrate
Separator samples	separator gas and oil, then recombination	gas condensates, oils with a high gas–oil ratio, watered-out wells	an error in the gas–oil ratio, an unsteady flow
Wellhead sample	at the wellhead	only if the flow at the wellhead is single-phase	gas coming out of solution in the wellbore

A drill stem test (a temporary test of the formation through the drill pipe) also gives bottomhole and separator samples (Well testing: purposes and types of test). Formation water samples are needed to find the resistivity of the water, to check its compatibility with injected water, and to study scale (mineral deposits).

SchematicThree sampling methods (schematic): a bottomhole sampler on slickline opposite the perforated interval; a wireline formation tester pumps fluid contaminated with filtrate into the wellbore and then directs clean fluid into the sample chamber; samples of gas and oil from the separator lines.

Well conditioning before sampling

Well conditioning (preparing the well for sampling) begins with cleaning. The well flows back drilling mud filtrate, workover fluid (a fluid pumped in to control the well) and residues of acid from a treatment. The composition of the produced water shows when cleaning is finished: its chlorides, its pH, and the tracers added to the treatment fluids. Then the well is set to its sampling rate.

Order of preparation

1
Cleaning
Produce the well at a rate high enough to carry out the treatment fluids, and monitor the composition of the water and the impurities.
2
Reducing the rate in steps
Wait for stabilisation at each step. The bottomhole pressure must rise above the bubble point pressure with a margin for the inaccuracy of its estimate.
3
Stabilisation
The gas–oil ratio, the pressures at the wellhead and in the separator, the temperature and the water cut do not change, within the accuracy of the measurements, for several hours in a row.
4
Sampling
Take bottomhole samples at this rate or in the shut-in well. Take separator samples at this rate, gas and liquid at the same time, and record all the parameters.

Greatest conditioning rate before sampling

q_{\max} = J\,(\bar p - p_{\text{sat}} - \delta)

where:

$q_{\max}$ greatest rate at which the bottomhole pressure is not below $p_{\text{sat}} + \delta$ , m³/d (oil at standard conditions)
$J$ productivity index of the well for stock-tank oil (the rate the well gives per unit of drawdown), m³/(d·MPa)
$\bar p$ reservoir pressure in the drainage area, MPa
$p_{\text{sat}}$ saturation pressure ( $p_b$ in the notation of the lesson Black oil model: solution gas–oil ratio and formation volume factors; for a gas condensate, the dew point pressure), taken from analogues (similar fields), correlations or the first samples, MPa
$\delta$ margin for the inaccuracy of the saturation pressure, MPa

Units: metric oilfield units

Holds when

The inflow is single-phase and proportional to the drawdown: $q = J(\bar p - p_{wf})$ above the bubble point pressure (Well inflow: the well performance diagram).
$J$ was measured on this well after cleaning. While the skin factor is changing, $J$ changes with it. (The skin factor is a number that shows how much the zone around the well helps or hinders flow.)

The closer the reservoir pressure is to the bubble point pressure, the smaller the permissible rate and the longer the conditioning. In a saturated reservoir (for example, one under a gas cap, a free-gas zone on top of the oil), any drawdown pushes the oil near the bottom into the two-phase region. From such reservoirs, samples are taken at the surface at the lowest possible rate, or with a formation tester at a small drawdown.

Worked exampleChoosing the sampling rate from rate steps

Given (assumed data)

Oil: the same as in the lesson Black oil model: solution gas–oil ratio and formation volume factors: bubble point pressure about 18 MPa at 85 °C (an estimate, uncertainty ±1 MPa)
Reservoir pressure and productivity: $\bar p = 21.0$ MPa, $J = 10$ m³/(d·MPa)
Rate steps and gas–oil ratio: 60 m³/d — 152 m³/m³; 40 — 127; 25 — 111; 15 — 110
Tubing: inner diameter 62 mm, length to the perforated interval 2450 m

Solution

The bottomhole pressure at each step is $p_{wf} = \bar p - q/J$ : 15.0 MPa at 60 m³/d, 17.0 at 40, 18.5 at 25 and 19.5 MPa at 15. At the first two steps it is below the bubble point pressure (and lower still if we allow for two-phase inflow, in which the productivity falls). Gas was coming out near the bottom, so the gas–oil ratios of 152 and 127 m³/m³ are too high because of free gas.
The permissible rate with a margin of 1 MPa is $q_{\max} = 10\cdot(21.0 - 18 - 1.0) = 20$ m³/d. The 25 m³/d step does not meet this condition, even though its gas–oil ratio differs from the last one by only 0.9%, which is within the accuracy of the measurement.
The sampling rate is 15 m³/d, with $p_{wf} = 19.5$ MPa. The tubing volume is $\pi/4\cdot 0.062^2\cdot 2450 = 7.40$ m³. At the bottom the oil occupies $15\cdot 1.347 = 20.2$ m³/d, so one tubing volume is replaced in about 8.8 h.

Answer: Samples are taken at 15 m³/d. The bottomhole pressure is 19.5 MPa, 1.5 MPa above the estimate of the bubble point pressure. Before sampling, the well produces at this rate until several tubing volumes have been replaced and the gas–oil ratio and pressures have stabilised, which usually takes about a day.

What it means. The gas–oil ratio falls from 152 to 110 m³/m³ as the rate is reduced. This is a typical picture: at large drawdowns gas came out near the bottom, and free gas moves more easily than oil. The time to replace a tubing volume is estimated with a margin, because above the level where gas begins to come out in the tubing, the flow speeds up. If we had known the bubble point pressure more accurately, the 25 m³/d step would also have done. But a margin is cheaper than repeating the sampling. The data in this example are assumed for illustration.

Bottomhole samples

The bottomhole sampler, which takes the bottomhole samples, is a steel chamber of about half a litre with valves. It is lowered on a cable (wireline) or a thin wire (slickline) through a lubricator (a pressure-tight tube on top of the well). It is set opposite the reservoir, or a little above the perforated interval (the section where holes were made in the casing). While the tool goes down, fluid passes freely through the chamber. At the chosen depth the valves close, driven by a clockwork timer, by a command, or by a signal from built-in pressure and temperature gauges. The chamber then traps a portion of fluid at the pressure and temperature of the sampling point.

The sampling point is chosen so that water collected in the sump (the lowest part of the well, below the perforations) does not get in, and so that the pressure is above the bubble point pressure. Meanwhile the well is either at the sampling rate or shut in. Several samples are taken in a row. If their properties agree, this is the main argument for their quality (see the section on checking). At the well, the contents of the sampler are transferred into transport cylinders.

Bringing a sample to the surface is not a neutral operation. The chamber is rigid, but oil contracts as it cools. The volume that holds the oil does not change, so the pressure of the single-phase liquid falls very fast:

Pressure change of single-phase liquid in a rigid chamber

\left(\frac{\partial p}{\partial T}\right)_V = \frac{\beta_o}{c_o}

where:

$\beta_o$ thermal expansion coefficient of reservoir oil; usually $(0.7\text{–}1.2)\cdot 10^{-3}$ , 1/°C
$c_o$ oil compressibility above the bubble point pressure, 1/MPa
$(\partial p/\partial T)_V$ pressure change per degree at constant volume, MPa/°C

Units: MPa and °C

Holds when

The chamber is rigid. The thermal expansion of the steel, a few per cent of that of the oil, slightly softens the pressure drop.
The oil is single-phase. When gas comes out, the pressure falls more slowly, because the gas phase supports it.

For oils the ratio $\beta_o/c_o$ is of the order of 0.5–1 MPa per degree. Cooling by a few degrees uses up the whole margin of pressure above the bubble point pressure. Heating the trapped liquid by the same few degrees raises the pressure by megapascals.

Worked exampleWhen an ordinary bottomhole sample becomes two-phase

Given (assumed data)

Sample: taken at 19.5 MPa and 85 °C (the conditions of the previous example); bubble point pressure 18 MPa
Oil: $\beta_o = 1.0\cdot 10^{-3}$ 1/°C (assumed), $c_o = 1.6\cdot 10^{-3}$ 1/MPa
Bubble point pressure on cooling: falls by 0.06 MPa per degree (assumed)

Solution

The pressure of single-phase oil in a rigid chamber changes by $\beta_o/c_o = 1.0/1.6 = 0.625$ MPa per degree, the bubble point pressure by 0.06 MPa per degree.
The margin above the bubble point pressure at the moment of sampling is $19.5 - 18 = 1.5$ MPa. It shrinks by $0.625 - 0.06 = 0.565$ MPa per degree. It disappears after cooling by $1.5/0.565 = 2.7$ °C, that is, at 82.3 °C and 17.84 MPa.
After that, gas comes out. At the surface, at 20 °C, the bubble point pressure of this oil is about 14.1 MPa, and the sample is already two-phase: there is a gas cap above the oil in the chamber.
The opposite case: if the same single-phase liquid were heated by 10 °C in a rigid cylinder, the pressure would rise by 6.25 MPa.

Answer: An ordinary sample becomes two-phase after cooling by only 2.7 °C, while it is still in the wellbore.

What it means. That is why ordinary bottomhole samples arrive at the surface two-phase. This is not a defect but physics, and the laboratory restores such samples by heating and compressing them while stirring. Allowing for the thermal expansion of the steel chamber increases the permissible cooling only to 2.8 °C. The thermal expansion coefficient of the oil and the slope of the bubble point pressure are assumed values. The data in this example are assumed for illustration.

To restore the reservoir mixture, the laboratory heats such a sample to reservoir temperature and raises the pressure above the bubble point pressure, stirring until the gas has dissolved. This works for gas, but not always for asphaltenes (heavy, tar-like compounds in oil). Asphaltenes that dropped out as the pressure fell redissolve slowly and incompletely (Retrograde condensation, wax, asphaltenes and high-pressure fluids).

A single-phase sampler solves the problem differently. Behind the piston of its chamber there is nitrogen under high pressure. After sampling, the nitrogen squeezes the sample to a pressure above the reservoir pressure. On the way up, the piston follows the shrinking oil, and the pressure falls only moderately (see the advanced section). Such chambers are also fitted in wireline formation testers. Single-phase samples are needed where loss of gas (degassing) would spoil the sample for good: for studies of asphaltenes and of other properties that are sensitive to composition.

Bottomhole samples are poorly suited to gas condensates, because liquid almost always drops out in the wellbore and we do not know in what proportion the phases will get into the chamber. They are also poorly suited to strongly watered-out wells.

Formation tester samples and filtrate contamination

In an exploration well, before the casing is run, the most convenient way to sample is with a formation tester (Pressure measurement with the wireline formation tester). It takes fluid from exactly the chosen bed, and earlier than any other method. But the rock at the wall is soaked with drilling mud filtrate (Borehole environment: invaded zone and mudcake), and the first litres that reach the probe are almost pure filtrate. So the tool first pumps fluid into the wellbore. Meanwhile an analyser in the flowline follows the composition by optical density in different parts of the spectrum (the colour of the oil, the absorption bands of methane and water), fluorescence, density, viscosity and resistivity. When the contamination has become acceptable, the flow is directed into the sample chamber.

We pump at a drawdown that keeps the pressure at the probe above the bubble point pressure. Otherwise a mixture of phases will get into the chamber. Oil under a gas cap is sampled with special care: at the contact it is saturated with gas, and any noticeable drawdown takes it into the two-phase region. For tight rocks (rocks of very low permeability) a dual-packer module is used. For fast clean-up, focused probes with a guard ring are used: the ring takes the contaminated flow from the edges, and the centre takes cleaner fluid.

The filtrate of a water-based mud does not mix with oil, and it settles out of the sample. It mainly interferes with formation water samples, whose contamination is tracked by resistivity and by tracers added to the mud. The filtrate of an oil-based mud (OBM) is diesel fuel, mineral oil or a synthetic base, and it dissolves in the oil completely. It dilutes the sample. It lowers the gas–oil ratio and the bubble point pressure, changes the density and viscosity, and adds the hydrocarbons of the base to the composition.

We estimate contamination from the property in which the filtrate and the reservoir oil differ most. In the part of the spectrum where the oil is strongly coloured and the base is almost transparent, the optical density of the mixture is proportional to the fraction of oil. As pumping continues, the fraction of filtrate falls roughly as a power of the pumped volume. A widely used empirical power-law model, fitted to field measurements, has the exponent −5/12. We find the property of the clean oil by extrapolation (extending the trend to infinite volume):

Contamination from optical density and extrapolation to clean fluid

\eta = 1 - \frac{\mathrm{OD}}{\mathrm{OD}_\infty},\qquad \mathrm{OD}(V) = \mathrm{OD}_\infty - a\,V^{-5/12}

where:

$\eta$ fraction of filtrate in the inflowing fluid by volume at reservoir conditions, fraction
$\mathrm{OD}$ optical density of the flow in the chosen analyser channel, —
$\mathrm{OD}_\infty$ optical density of the clean reservoir fluid, the limit as $V \to \infty$ , —
$V$ pumped volume, L
$a$ a coefficient fitted to the measurements, —

Units: optical density is dimensionless; volume in any units

Holds when

The filtrate is transparent in the chosen channel; if not, $\eta = (\mathrm{OD}_\infty - \mathrm{OD})/(\mathrm{OD}_\infty - \mathrm{OD}_{\text{f}})$ , where $\mathrm{OD}_{\text{f}}$ is the optical density of the filtrate.
Mixing is linear by volume. The exponent 5/12 is empirical and depends on the geometry of the inflow and on how uneven the reservoir is. At the start of pumping the law does not hold.
For gas condensates and light oils the colour contrast is small, so the gas–oil ratio, density and methane bands are used instead.

On a plot of optical density against $V^{-5/12}$ , the points fall on a straight line, and its intercept on the axis gives $\mathrm{OD}_\infty$ . The fraction of filtrate falls as $V^{-5/12}$ , so to halve the contamination we have to pump $2^{12/5} \approx 5.3$ times more.

Contamination as a fraction of stock-tank oil

x_{\text{sc}} = \frac{\eta}{\eta + (1 - \eta)/B_o}

where:

$x_{\text{sc}}$ fraction of filtrate in the stock-tank liquid of the sample by volume at standard conditions, fraction
$\eta$ fraction of filtrate in the reservoir volume, fraction
$B_o$ formation volume factor of the reservoir oil at sampling conditions, m³/m³

Units: dimensionless

Holds when

The volume of the filtrate hardly changes on going to standard conditions, and there is no gas in it.

Reservoir oil shrinks by a factor of $B_o$ and the filtrate does not, so the share of filtrate is greater in the stock-tank oil of the sample: 10% in the reservoir at $B_o = 1.3$ is 13% in the stock tank. The laboratory usually reports contamination in mass per cent of stock-tank oil, found by chromatography (a method that separates a mixture into its components). The components of the base show up as a “hump” in the distribution by carbon number.

Worked exampleContamination of a sample by OBM filtrate: when to stop pumping and how to recalculate the gas–oil ratio

Given (assumed data)

Well: exploration, synthetic-base mud; a reservoir with the oil from the lesson Black oil model: solution gas–oil ratio and formation volume factors, 30 MPa, $B_o = 1.324$
Optical density during pumping: 10 L — 0.595; 20 L — 0.659; 40 L — 0.707; 80 L — 0.743; 120 L — 0.759
Pump: 8 cm³/s (28.8 L/h)
Densities: OBM base 780 kg/m³, stock-tank oil 850 kg/m³
Laboratory, on the sample taken at 120 L: gas–oil ratio 95.1 m³/m³

Solution

The variable for the line is $x = V^{-5/12}$ : for 40 L 0.2150, for 120 L 0.1360.
Slope: $(0.759 - 0.707)/(0.1360 - 0.2150) = -0.6582$ ; intercept on the axis as $V \to \infty$ : $\mathrm{OD}_\infty = 0.759 + 0.6582\cdot 0.1360 = 0.849$ .
Contamination at 120 L: $\eta = 1 - 0.759/0.849 = 10.6$ % of the reservoir volume of the flow.
To reach 5% we would have to pump $120\cdot(10.6/5)^{12/5} \approx 730$ L, which is another 21 h or so of pumping.
As a fraction of stock-tank oil: $x_{\text{sc}} = 0.106/(0.106 + 0.894/1.324) = 13.6$ %; by mass it is $13.6\cdot 780/(13.6\cdot 780 + 86.4\cdot 850) = 12.6$ %.
Gas–oil ratio of the clean oil: $95.1/(1 - 0.136) = 110$ m³/m³. The bubble point pressure of the contaminated sample from the solubility curve of this oil is about 15.3 MPa instead of 18 (a rough estimate: it assumes that the base dissolves gas in the same way as the stock-tank oil).

Answer: At 120 L the contamination is about 10.6% of the reservoir volume (13.6% of the stock-tank oil). Pumping down to 5% would take another 21 h or so. It is more sensible to take the sample now and recalculate the properties. The corrected gas–oil ratio is 110 m³/m³.

What it means. The $V^{-5/12}$ law shows the price of cleanliness. Each further halving of the contamination takes 5.3 times more pumping, and all that time the tool stays pressed against the formation under overbalance (the mud pressure in the well is higher than the formation pressure). Without allowing for the contamination, the solution gas–oil ratio would be too low by 14%, and the bubble point pressure by almost 3 MPa. When the contamination is known, the correction is simple. For a gas condensate this cannot be done, because the dew point is too sensitive to an admixture of the base. The exponent 5/12 is empirical. The data in this example are assumed for illustration.

CalculatedOptical density of the flow during the pump-out of the example: the dots are measurements, the line is the power-law clean-up curve, the dashed lines are clean oil and the 5% contamination level.

There is no general standard for permissible contamination. The operator and the laboratory set the limit according to what the sample is needed for. If the contamination is known and small, the properties of the oil are recalculated to clean oil. The bubble point pressure of many oils falls roughly in proportion to the contamination. For gas condensates, even a few per cent of the base noticeably changes the dew point and the condensate yield, so they are sampled as clean as possible.

Separator samples and recombination

Separator samples are the main method for gas condensates and for oils with a high gas–oil ratio. The well is brought to a steady flow (see the section on conditioning). Then gas is taken from the gas line of the separator and oil from the liquid line at the same time, and the separator pressure and temperature and the rates are recorded. Gas is collected into cylinders that have been evacuated or purged with the same gas. Liquid is collected by displacing water or brine out of a cylinder, or into a piston cylinder, so that it does not lose gas in the sampling line.

In the laboratory the gas and oil are mixed again, or recombined, in the proportion in which they left the separator. The field usually reports the gas–oil ratio per 1 m³ of stock-tank oil, from a measurement in the stock tank. But the laboratory has separator oil, which still contains dissolved gas. To convert, we use the formation volume factor of the separator oil, $B_{\text{sep}}$ . Part of the sample is degassed to stock-tank conditions, and we measure how much stock-tank oil and gas it gives. The field also calculates the gas rate from an assumed gas density, and this is corrected to the actual density measured on the sample.

Correction of the gas–oil ratio for gas density and gas deviation factor

R_{\text{sep}} = R_{\text{meas}}\,\sqrt{\frac{\gamma_{\text{set}}\,Z_{\text{set}}}{\gamma_g\,Z}}

where:

$R_{\text{meas}}$ separator gas–oil ratio from the field measurement: separator gas per 1 m³ of stock-tank oil, m³/m³
$R_{\text{sep}}$ the same gas–oil ratio after the correction, m³/m³
$\gamma_{\text{set}},\ Z_{\text{set}}$ gas specific gravity and gas deviation factor built into the flow meter calculation, —
$\gamma_g,\ Z$ actual gas specific gravity from the sample and its $Z$ at the pressure and temperature of the measurement (Ideal and real gas: the gas deviation factor (Z)), —

Units: gas at standard conditions (here 0.101325 MPa and 20 °C)

Holds when

The gas is measured by an orifice (differential-pressure) flow meter, whose reading is inversely proportional to $\sqrt{\gamma_g Z}$ . For other types of meter the correction is different.
Both volumes are referred to the same standard conditions.

If the meter calculation assumes a gas lighter than the actual one, the rate and the gas–oil ratio come out too high. If it ignores that $Z < 1$ , they come out too low.

Recombination proportion and total gas–oil ratio

R_{\text{rec}} = \frac{R_{\text{sep}}}{B_{\text{sep}}},\qquad R = R_{\text{sep}} + R_{\text{tank}}

where:

$R_{\text{rec}}$ separator gas (at standard conditions) per 1 m³ of separator oil at separator pressure and temperature (the recombination proportion), m³/m³
$B_{\text{sep}}$ volume of separator oil per 1 m³ of the stock-tank oil obtained from it, m³/m³
$R_{\text{tank}}$ gas released from the separator oil in the stock tank, per 1 m³ of stock-tank oil, m³/m³
$R$ total gas–oil ratio of the reservoir mixture: separator and stock-tank gas per 1 m³ of stock-tank oil, m³/m³

Units: gas at standard conditions

Holds when

The stock-tank gas is already dissolved in the separator oil and is not added separately in recombination.
The gas and oil samples were taken at the same time, in a steady flow, at a bottomhole pressure above the bubble point pressure.

For a sample taken above the bubble point pressure the total gas–oil ratio equals the solution gas–oil ratio of the reservoir oil $R_{sb}$ for the given separation scheme (Black oil model: solution gas–oil ratio and formation volume factors). The composition of the reservoir mixture from the compositions of the gas and the liquid is calculated as shown in the lesson on fluid types.

Worked exampleHow much separator gas to add to the separator oil

Given (assumed data)

Flow rates: 15.0 m³/d of stock-tank oil; separator gas by flow meter 1462 m³/d (20 °C)
Flow meter: designed for a gas with $\gamma = 0.70$ and $Z = 1$
Separator: 1.2 MPa, 35 °C; gas from the sample $\gamma_g = 0.78$
Laboratory degassing of the separator oil: 100.0 cm³ gave 93.0 cm³ of stock-tank oil and 1.48 L of gas (20 °C)
To be prepared: a sample of 500 cm³ of separator oil

Solution

Field gas–oil ratio: $1462/15.0 = 97.5$ m³/m³, that is, separator gas per 1 m³ of stock-tank oil.
Meter correction: $\sqrt{0.70/0.78} = 0.9473$ ; $Z$ of the gas at 1.2 MPa and 35 °C is 0.9627, $\sqrt{1/0.9627} = 1.0192$ ; the factor is 0.9655, $R_{\text{sep}} = 94.1$ m³/m³.
From degassing: $B_{\text{sep}} = 100.0/93.0 = 1.0753$ ; stock-tank gas $R_{\text{tank}} = 1.48\ \text{L}/93.0\ \text{cm}^3 = 15.9$ m³/m³.
Recombination proportion: $R_{\text{rec}} = 94.1/1.0753 = 87.5$ m³ of gas per 1 m³ of separator oil. For 500 cm³ of oil that is 43.8 L of gas at standard conditions, or 41.1 g at a density of $0.78\cdot 1.204 = 0.939$ kg/m³.
Check: the total gas–oil ratio $94.1 + 15.9 = 110.0$ m³/m³ is the solution gas–oil ratio of the reservoir oil for this separation scheme.
If by mistake the field 97.5 m³/m³ were used without the correction and without conversion to separator oil, the mixture would contain 9.7% more gas, and the bubble point pressure of the sample would come out at about 20.0 MPa instead of 18 (a rough estimate: the solubility curve of this oil extended above 110 m³/m³).

Answer: Add 43.8 L (41.1 g) of separator gas to 500 cm³ of separator oil. The recombination proportion is 87.5 m³/m³.

What it means. Two corrections together changed the gas–oil ratio by about 10% and the bubble point pressure by about 2 MPa. One is for the basis (stock-tank oil or separator oil), and the other is for the gas density in the flow meter. Errors of the same size come from liquid droplets in the gas that are not accounted for, and from an unsteady flow. That is why the record sheet of separator samples must include the rates, the separator pressure and temperature, and the parameters built into the flow meter. The data in this example are assumed for illustration.

Errors of separator sampling and how they shift the result

Error: The flow meter is designed for a gas lighter than the actual one
Gas–oil ratio: overestimated
Bubble point pressure of the sample: overestimated

Error: Liquid droplets go with the gas and are not counted in the liquid rate
Gas–oil ratio: overestimated
Bubble point pressure of the sample: overestimated

Error: Gas bubbles in the liquid line overstate the liquid rate
Gas–oil ratio: underestimated
Bubble point pressure of the sample: underestimated

Error: Separator oil taken with gas bubbles
Gas–oil ratio: —
Bubble point pressure of the sample: the bubble point pressure of the separator oil is above the separator pressure, which is a sign of a bad sample

Error	Gas–oil ratio	Bubble point pressure of the sample
The flow meter is designed for a gas lighter than the actual one	overestimated	overestimated
Liquid droplets go with the gas and are not counted in the liquid rate	overestimated	overestimated
Gas bubbles in the liquid line overstate the liquid rate	underestimated	underestimated
Separator oil taken with gas bubbles	—	the bubble point pressure of the separator oil is above the separator pressure, which is a sign of a bad sample

The sampling taps (valves) are placed at the side of the lines: a tap at the bottom of the gas line collects droplets, and a tap at the top of the liquid line collects gas bubbles.

Gas condensate: special difficulties

The reservoir pressure of a gas-condensate accumulation is often only slightly above the dew point. Then a noticeable drawdown makes condensate drop out near the bottom of the well (retrograde condensation). The condensate stays in the pores, and a leaner gas flows to the well. The gas–oil ratio rises, the condensate–gas ratio falls, and the recombined sample is leaner than the reservoir mixture, usually with a dew point that is too low.

The wellbore makes it hard to reduce the drawdown, because the gas must carry the liquid to the surface. At too low a rate, droplets run down and collect, and a flow of changing composition reaches the surface (Multiphase flow in tubing). So the sampling rate is a compromise. It must be high enough that the wellbore does not collect liquid, and low enough that the bottomhole pressure stays above the dew point, or as close to it as possible. After a shut-in, the liquid in the wellbore runs down. When the well starts up again, dry gas reaches the surface first, then an enriched mixture. We take a sample only when the gas–oil ratio has been stable again for several hours.

Cylinders with separator gas are heated above the separator temperature before the gas is transferred to the equipment. As the cylinders cool, heavy components condense in them, and without heating these would stay on the walls.
For accurate studies, gas with droplets is sampled isokinetically, that is, at the same velocity as the flow, so that the droplets enter the sample in the same proportion as in the flow.
A gas condensate is sampled with a formation tester at the smallest drawdown, and only when the contamination is low, because OBM filtrate is especially noticeable in it.
Check: the dew point of the recombined sample at reservoir temperature must not be above the reservoir pressure. If it equals the reservoir pressure, the accumulation may be saturated (with an oil rim, a thin oil zone around the gas), or the sample may be too rich. This is decided from the composition and the data of neighbouring wells.

Checking samples at the well and in the laboratory

We must find a defect in a sample before it leaves for the laboratory, because sampling can be repeated while the well is still at the sampling rate. The checks at the well are simple:

Sampler pressure at the surface (opening pressure): For an ordinary bottomhole sample it is below the bubble point pressure at surface temperature, because the sample is two-phase. For a single-phase sample it is above. Duplicates must open at close pressures. A noticeably lower pressure in one sample is a sign of a leak.
Bubble point pressure at surface temperature: We compress the sample in steps with a displacing fluid or a piston, stirring after each step. While there is gas in the sample, the pressure rises slowly. After the gas dissolves, it rises steeply. The break between the two is the bubble point pressure at this temperature (see the example below). For duplicates, it must agree within the accuracy of the check.
Separator samples: The bubble point pressure of the separator oil at separator temperature must be close to the separator pressure, because the oil there is saturated with gas. The pressure in the gas cylinder must agree with the separator pressure, corrected for temperature. Oxygen in the gas shows that air has leaked in.
In the laboratory: Composition, water and impurities, contamination by filtrate, and the bubble point pressure at reservoir temperature. For an undersaturated reservoir (one whose pressure is above the bubble point) this pressure must be below the reservoir pressure. If it is higher, the sample has captured free gas, or the gas–oil ratio is too high.

Worked exampleBubble point pressure of a sample at the well from the break of the compression curve

Given (assumed data)

Sample A: the ordinary bottomhole sample from the example above, chamber 600 cm³, 20 °C
Displacing fluid pumped in — pressure: 0 cm³ — 11.89 MPa; 10 — 12.38; 20 — 12.93; 30 — 13.54; 34 — 13.81; 36 — 13.96; 40 — 17.07; 41 — 18.56; 42 — 20.05
Sample B (duplicate): the same check gave 13.40 MPa
Expectation: bubble point pressure 18 MPa at 85 °C, falling by 0.06 MPa per degree

Solution

For the two-phase section, take the last two points before the break: the slope is $(13.96 - 13.81)/2 = 0.075$ MPa/cm³. For the single-phase section, take the two points after the break: $(20.05 - 17.07)/2 = 1.49$ MPa/cm³. This is 20 times steeper, because now only the liquid is being compressed.
The lines intersect where $13.96 + 0.075\,(V - 36) = 17.07 + 1.49\,(V - 40)$ . This gives $V = 38.0$ cm³ and a bubble point pressure of 14.11 MPa at 20 °C.
Expectation: $18 - 0.06\cdot 65 = 14.1$ MPa. This agrees. The sampler pressure before compression is 11.89 MPa, below the bubble point pressure. So there is a gas cap in the sample, as there should be.
Sample B gave 5.0% less. It has less dissolved gas, so probably some of the gas leaked out during lifting or transfer.

Answer: The bubble point pressure of sample A at 20 °C is 14.11 MPa, and the sample is fit for use. Sample B is not sent to the laboratory as the main sample.

What it means. The two-phase section is in fact slightly curved. A line through the two points nearest the break gives an estimate with an error of about 0.01 MPa. For volatile oils the break is smoothed, so the check is rougher. The result refers to surface temperature. So we compare samples with each other and with the expectation, and the laboratory determines the bubble point pressure at reservoir temperature. If the bubble point pressure had come out higher than expected, the sample could have captured free gas. The data in this example are assumed for illustration.

CalculatedChecking sample A from the example: the pressure during stepwise compression, and two lines, one for the two-phase section and one for the single-phase section. Their intersection is the bubble point pressure at 20 °C.

Record sheet, transport and custody of samples

Each sample comes with a sample record (a sampling report). It states the well, the reservoir and the interval, the method and place of sampling, the date and time, and the cylinder number. It gives the pressure and temperature at the sampling point, and the operating conditions of the well (rates, gas–oil ratio, pressures). It gives the separator conditions and the flow meter parameters, the sampler pressure at the surface, the results of the checks at the well, the hydrogen sulfide and carbon dioxide content measured on site, and an estimate of contamination. The number on the cylinder and in the record is the only link between the sample and its sampling conditions. A sample without a record is useless.

Chain of custody means that everyone who receives the sample (the carrier, the warehouse, the laboratory) records when and in what condition they received it, the pressure in the cylinder and whether the seals are intact. If the pressure has fallen on the way, this is seen at once, not a month later in the report.

Sample cylinders are pressure vessels with flammable contents. They are transported under the regulations for the carriage of dangerous goods, in cylinders with valid inspection certificates. Liquid samples must not be shipped “full to the brim”. Single-phase liquid heated by a few degrees raises the pressure by megapascals (see the example of the bottomhole sample). So a gas cushion is left in the cylinder, or piston cylinders with nitrogen are used. Hydrogen sulfide reacts with steel and is partly lost on the way. It is measured at the well, and samples are taken into cylinders made of resistant materials.

Terms and symbols

Representative sample: A sample whose composition equals that of the reservoir fluid of the accumulation or of the drainage area of the well.
Bottomhole sample: A sample taken in the wellbore opposite the reservoir with a sampler on wireline or slickline, at a pressure above the bubble point pressure.
Single-phase sampler: A sampler with a nitrogen cushion behind the piston. It keeps the pressure of the sample above the bubble point pressure during lifting and cooling.
Recombination: The mixing of separator gas and oil in the proportion in which they left the separator, to restore the reservoir mixture.
$B_{\text{sep}}$ , $R_{\text{rec}}$: The first is the volume of separator oil per 1 m³ of stock-tank oil. The second is the separator gas per 1 m³ of separator oil.
Contamination $\eta$ and $x_{\text{sc}}$: The fraction of drilling mud filtrate in the sample: $\eta$ in the reservoir volume and $x_{\text{sc}}$ in the stock-tank oil.
Sampler pressure at the surface: The pressure in the closed chamber at surface temperature. It is the first check of the integrity of the sample.

Common mistakes and misconceptions

Wrong

“The greater the rate during sampling, the sooner we get the sample.”

In fact

At a large drawdown the bottomhole pressure falls below the bubble point pressure, and the composition of the inflow changes. You need a small, steady rate and enough time for the fluid in the wellbore to be replaced.

Wrong

“The field gas–oil ratio can be used directly for recombination.”

In fact

It is counted per volume of stock-tank oil and calculated from an assumed gas density. Correct it to the actual density and convert it to separator oil. Together these corrections easily make 10% of gas and 2 MPa of bubble point pressure.

Wrong

“A formation tester sample from a well drilled with OBM is clean oil: the filtrate will separate.”

In fact

OBM filtrate dissolves in the oil. Contamination is monitored during pumping and determined in the laboratory, and the properties are recalculated to clean oil.

Wrong

“The bottomhole sample arrived two-phase, so it is a reject.”

In fact

An ordinary sample becomes two-phase on cooling, and the laboratory restores it. What marks a bad sample is a disagreement between the bubble point pressures of duplicates, or an abnormally low sampler pressure.

In depth: advanced detail

Why a nitrogen cushion keeps the sample single-phase

In a single-phase sampler a piston separates the oil from the nitrogen. Gas is compressible, so on cooling the piston follows the shrinking oil. The nitrogen pressure then falls only in proportion to the growth of its volume and the fall of the absolute temperature. The pressure drop in turn lets the oil expand a little, by $c_o$ for every megapascal:

Pressure of a sample with a nitrogen cushion after cooling

p_1 = p_0\,\frac{V_{\text{g}0}}{V_{\text{g}1}}\,\frac{T_1}{T_0},\qquad V_{\text{g}1} = V_{\text{g}0} + V_{\text{o}}\left[\beta_o\,(T_0 - T_1) - c_o\,(p_0 - p_1)\right]

where:

$p_0,\ p_1$ pressure of the sample after pressurisation by nitrogen at reservoir temperature and after cooling, MPa
$V_{\text{g}0},\ V_{\text{g}1}$ volume of nitrogen before and after cooling, cm³
$V_{\text{o}}$ volume of oil in the chamber, cm³
$T_0,\ T_1$ absolute temperatures of the reservoir and the surface, K
$\beta_o,\ c_o$ thermal expansion and compressibility of the oil, as above, 1/°C, 1/MPa

Units: MPa, cm³, K (the temperature difference in °C or K)

Holds when

Nitrogen is an ideal gas, the piston is frictionless, and the chamber is rigid.
The changes in the volume of the oil are small and linear in temperature and pressure. The equations are solved for $p_1$ by trial and error.

The more nitrogen there is relative to the oil and the higher the initial pressure, the less the pressure of the sample falls on cooling.

Worked examplePressure of a single-phase sample at the surface

Given (assumed data)

Chamber: 500 cm³ of oil and 200 cm³ of nitrogen; after sampling the nitrogen pressurised the sample to 40 MPa at 85 °C
Oil: the same: $\beta_o = 1.0\cdot 10^{-3}$ 1/°C, $c_o = 1.6\cdot 10^{-3}$ 1/MPa; bubble point pressure at 20 °C is 14.1 MPa
Surface: 20 °C

Solution

Cooling by 65 °C: the oil shrinks by $500\cdot 1.0\cdot 10^{-3}\cdot 65 = 32.5$ cm³. The pressure drop returns part of the volume.
Temperature ratio: $293.15/358.15 = 0.8185$ . By trial and error: $V_{\text{g}1} = 200 + 500\,[0.065 - 0.0016\,(40 - p_1)] = 223.9$ cm³ and $p_1 = 40\cdot(200/223.9)\cdot 0.8185 = 29.2$ MPa.
Margin above the bubble point pressure at 20 °C: $29.2 - 14.1 = 15.1$ MPa. Without the nitrogen, in a rigid chamber, single-phase oil cooled by 65 °C would lose $0.625\cdot 65 \approx 41$ MPa. That is, it would become two-phase within the first few degrees.

Answer: The sample arrives at the surface at 29.2 MPa and remains single-phase with a margin of about 15 MPa.

What it means. Nitrogen at 30–40 MPa is not ideal (its $Z$ is about 1.1–1.2), and an exact calculation gives a pressure about 10% lower. But the conclusion does not change: the margin stays above 10 MPa. The volumes and the nitrogen pressure are assumed. In real samplers they are set by the design and the initial charging. The data in this example are assumed for illustration.

Connections to other topics

Types of reservoir fluids — fluid types and recombination of gas and liquid compositions.
Black oil model: solution gas–oil ratio and formation volume factors — solution gas–oil ratio, formation volume factor and the separation scheme to which they refer.
Phase behaviour: pure substances and binary mixtures — bubble point pressure, dew point and the two-phase region.
Pressure measurement with the wireline formation tester — the wireline formation tester: pressure and mobility measurements before sampling.
Oil-based muds — what an oil-based mud and its filtrate consist of.
Laboratory PVT tests — what the laboratory does with samples: degassing, separator tests.
Retrograde condensation, wax, asphaltenes and high-pressure fluids — asphaltenes, waxes and retrograde gas condensates: what happens to them as pressure and temperature fall.
Well testing: purposes and types of test — the drill stem test, another source of samples.

Check yourself

1
The reservoir pressure is 24.0 MPa, the bubble point pressure is estimated at 22.5 ± 0.5 MPa, the productivity index is 6 m³/(d·MPa). What should the conditioning rate be before taking bottomhole samples?
Show the answer
With a margin of 0.5 MPa, $q_{\max} = 6\cdot(24.0 - 22.5 - 0.5) = 6$ m³/d, and the bottomhole pressure is then 23.0 MPa. A rate this low takes a long time to stabilise, but a higher rate risks taking the bottomhole pressure below the bubble point pressure.
2
The reservoir pressure of a gas-condensate accumulation is 1.5 MPa above the dew point. Which sampling method is preferable?
1. A.Separator samples at a steady rate, with the smallest drawdown at which the wellbore still carries the liquid out
2. B.A bottomhole sampler in a flowing well
3. C.A wellhead sample at the maximum rate
4. D.Any: a gas condensate is single-phase in the reservoir
Show the answer
Correct answer: A. Separator samples at a steady rate, with the smallest drawdown at which the wellbore still carries the liquid out
In the wellbore a gas condensate is almost always two-phase, so a bottomhole sample is unrepresentative, and a wellhead one even more so. The main method is separator samples at a steady rate with a small drawdown, followed by recombination using an accurately measured gas–oil ratio. If the well has not yet been cased, a formation tester sample at the smallest drawdown is also possible.
3
From pump-out data the contamination is 8% of the reservoir volume, $B_o = 1.5$ . What is the fraction of filtrate in the stock-tank oil, and how should the gas–oil ratio of 140 m³/m³ measured on this sample be corrected?
Show the answer
$x_{\text{sc}} = 0.08/(0.08 + 0.92/1.5) = 0.08/0.693 = 0.115$ , that is 11.5%. Only the reservoir oil gives gas, so $R = 140/(1 - 0.115) \approx 158$ m³/m³. The greater $B_o$ is, the more the contamination in the stock-tank oil exceeds that in the reservoir.
4
Three bottomhole samples taken at the same flow rate gave bubble point pressures at 20 °C of 15.2, 15.3 and 14.1 MPa. What should be done?
1. A.Regard the third sample as correct: it has less gas
2. B.Reject all three
3. C.Take the average, 14.9 MPa
4. D.Regard the two agreeing samples as good and the third as having lost gas, and exclude it
Show the answer
Correct answer: D. Regard the two agreeing samples as good and the third as having lost gas, and exclude it
Two samples agree within the accuracy of the check. The third is 1.1–1.2 MPa lower, which points to gas lost through a leaking valve or during transfer. An average would mix good samples with a bad one. If the third sample had been higher, we would suspect that it had captured free gas.
5
The separator gas–oil ratio after the flow meter correction is 150 m³ per 1 m³ of stock-tank oil, $B_{\text{sep}} = 1.12$ . How much separator gas should be added to 400 cm³ of separator oil?
Show the answer
$R_{\text{rec}} = 150/1.12 = 133.9$ m³/m³; for 0.4 L of oil, $0.4\cdot 133.9 = 53.6$ L of gas at standard conditions. If by mistake 150 m³/m³ were used per volume of separator oil, there would be 12% more gas, and the bubble point pressure of the sample would come out too high.
6
Why does an ordinary bottomhole oil sample arrive at the surface two-phase, although it was taken above the bubble point pressure?
1. A.Gas comes out because of vibration during lifting
2. B.The chamber is leaking
3. C.The bubble point pressure rises on cooling
4. D.On cooling in a rigid chamber the pressure of single-phase oil falls by about 0.6 MPa per degree and quickly drops below the bubble point pressure
Show the answer
Correct answer: D. On cooling in a rigid chamber the pressure of single-phase oil falls by about 0.6 MPa per degree and quickly drops below the bubble point pressure
Oil contracts on cooling while the chamber volume is constant, so the pressure falls by $\beta_o/c_o$ per degree, much faster than the bubble point pressure falls. A margin of 1–2 MPa disappears after cooling by a few degrees. This is not a defect: the laboratory restores the sample by heating and compressing it, and single-phase samplers prevent gas from coming out by using a nitrogen cushion.

Updated: 15 October 2026