How to read a PVT report

Quick refresher

First, the conditions and the base. Is the pressure absolute or gauge? Is the gas brought to 20 °C or to 60 °F (15.6 °C)? The difference is about 1.5%. Is the solution gas–oil ratio given per cubic metre or per tonne, and per stock-tank oil of the chosen separation scheme or per residual oil of differential liberation?
A sample is good if the duplicates agree, if it was taken at a bottomhole pressure above the bubble point pressure, and if the bubble point pressure at reservoir temperature is below the reservoir pressure. A bubble point pressure equal to the sampling pressure, or above the reservoir pressure, is a reason to doubt the sample or to look for a gas cap.
The tables of one report are linked. The residual oil and the gas of the stages add up, by mass, to the reservoir oil, and so do the products of the separator. The total formation volume factor of the constant-composition test agrees with the converted data of differential liberation. The Y-function of the two-phase readings lies on a straight line.
We build the property table for calculations from the separator test of the chosen scheme, the differential liberation converted to this scheme (below the bubble point pressure), the constant-composition test (above it), the viscosity of the oil and $B_g$ . Between neighbouring rows below the bubble point pressure, the gas that has come out must take more space than the oil loses: $B_g\,\Delta R_s > \Delta B_o$ .
In a gas condensate report the main things are the dew point against the reservoir pressure, and the constant volume test. In that test $p/Z_{2\text{ph}}$ falls linearly with the share of material taken out, and a row that falls off the line contains an error.

Key relationships

\rho_{ob}\,B_{oDb}\,V_{\text{r}} = \rho_{\text{r}}\,V_{\text{r}} + \rho_{\text{air}}\sum_j V_{g,j}\,\gamma_j

mass balance of differential liberation: reservoir oil = residual oil + the gas of all stages

B_g\,(R_{s,i} - R_{s,i+1}) > B_{o,i} - B_{o,i+1}

neighbouring rows of the table below the bubble point pressure: the apparent compressibility is positive

\frac{p}{Z_{2\text{ph}}} = \frac{p_d}{Z_d}\left(1 - \frac{n_p}{n_0}\right)

each row of the constant volume test lies on this line

p_{\text{abs}} = p_{\text{gauge}} + 0.101\ \text{MPa}

psia = psig + 14.7

Common pitfalls

Taking $R_{sD}$ and $B_{oD}$ of differential liberation instead of $R_s$ and $B_o$ of the chosen separation scheme.
Reading a pressure in psig as absolute. 0.1 MPa is hardly visible against the reservoir pressure, but it can reverse the conclusion about whether the oil is saturated.
Comparing solution gas–oil ratios at 20 °C and at 60 °F without a correction, or mixing up m³/t and m³/m³.
Not noticing that the numbers in a table are smoothed, not measured, or that the tests were run on a sample other than the one that passed the checks.
Passing to the reservoir simulation model a table in which, as the pressure falls, the oil together with the gas that came out takes less space.

After this lesson you can

find the needed sections in an oil report and in a gas condensate report, and for each number work out its conditions and base: absolute or gauge pressure, standard conditions of the gas, residual or stock-tank oil, cubic metre or tonne
check that the sample is representative (a true sample of the reservoir fluid): do the duplicates (samples taken in the same way, for comparison) agree, what is the bubble point pressure at laboratory temperature and at reservoir temperature, how far above the bubble point pressure was the pressure during sampling, and how does the bubble point pressure compare with the reservoir pressure
check that the report agrees with itself, using the mass balance of differential liberation and of the separator test (the mass put into the cell must equal the mass of the products), the Y-function (a combination of pressure and volume readings that should give a straight line) and the total formation volume factor found in two ways
assemble from the report the property table of the black oil model (the simple model of stock-tank oil plus dissolved gas) for the chosen separation scheme, and check that its values change steadily (it is monotonic) and that its apparent compressibility (the compressibility of oil together with the gas that comes out of it) is positive
read a gas condensate report: compare the dew point with the reservoir pressure, and check the rows of the constant volume test against the line $p/Z_{2\text{ph}}$ ( $Z$ is the gas deviation factor, a number that shows how far a real gas is from an ideal gas)

What is in the report and why

The report on a fluid study puts the tests described in the lesson Laboratory PVT tests together in one document with one aim: to give the engineers the properties of the reservoir fluid and to show how far they can be trusted. So besides the tables of results it contains information on where the sample came from and why the laboratory judged it good, and appendices with the raw measurements. The order of the sections differs between laboratories, but the set is almost always the same.

Sections of a reservoir oil report and what to check in them

Section: Summary
What it contains: reservoir pressure and temperature, bubble point pressure, solution gas–oil ratio, formation volume factor, density and viscosity of the oil
What to look at: under what conditions and on what base each number is given

Section: Samples
What it contains: bottles, sampling conditions, pressure in the sampler, bubble point pressure at laboratory temperature, choice of the sample
What to look at: do the duplicates agree; the margin above the bubble point pressure at sampling

Section: Composition
What it contains: molar composition of the reservoir oil up to C30+ (the group of components with 30 or more carbon atoms) and beyond, and of the gas and the oil of a single-stage flash; molar mass and density of the heavy residue
What to look at: the sum is 100%; agreement with the solution gas–oil ratio and the oil density

Section: Constant-composition expansion
What it contains: $V/V_b$ , bubble point pressure, compressibility, Y-function, density
What to look at: the kink of the curve; the straight line of the Y-function

Section: Differential liberation
What it contains: $R_{sD}$ , $B_{oD}$ , $B_g$ , $Z$ , density of the gas and of the oil at each stage
What to look at: the base is residual oil; the mass balance

Section: Separator tests
What it contains: $R_{sb}$ , $B_{ob}$ , density of the stock-tank oil, gas of the stages for several schemes
What to look at: the mass balance; does the scheme match the field one

Section: Viscosity
What it contains: viscosity of the oil at reservoir temperature and at different pressures
What to look at: minimum at the bubble point pressure

Section: Special tests
What it contains: waxes, asphaltenes, interfacial tension, mixing with injected gas
What to look at: conditions of the test; on which sample

Section	What it contains	What to look at
Summary	reservoir pressure and temperature, bubble point pressure, solution gas–oil ratio, formation volume factor, density and viscosity of the oil	under what conditions and on what base each number is given
Samples	bottles, sampling conditions, pressure in the sampler, bubble point pressure at laboratory temperature, choice of the sample	do the duplicates agree; the margin above the bubble point pressure at sampling
Composition	molar composition of the reservoir oil up to C30+ (the group of components with 30 or more carbon atoms) and beyond, and of the gas and the oil of a single-stage flash; molar mass and density of the heavy residue	the sum is 100%; agreement with the solution gas–oil ratio and the oil density
Constant-composition expansion	$V/V_b$ , bubble point pressure, compressibility, Y-function, density	the kink of the curve; the straight line of the Y-function
Differential liberation	$R_{sD}$ , $B_{oD}$ , $B_g$ , $Z$ , density of the gas and of the oil at each stage	the base is residual oil; the mass balance
Separator tests	$R_{sb}$ , $B_{ob}$ , density of the stock-tank oil, gas of the stages for several schemes	the mass balance; does the scheme match the field one
Viscosity	viscosity of the oil at reservoir temperature and at different pressures	minimum at the bubble point pressure
Special tests	waxes, asphaltenes, interfacial tension, mixing with injected gas	conditions of the test; on which sample

In Russian-language reports the constant-composition test is also called contact liberation, or the study of the “pressure–volume” relation. Gas viscosity and the two-phase gas deviation factor are usually not measured but calculated: the first by a correlation, the second from the amount of substance in the cell. The appendices hold the raw measurements. They show which numbers in the tables are measured and which are smoothed.

A gas condensate report is arranged differently, because the sample is usually taken at the surface (recombination: the separator gas and liquid are mixed again in the proportion in which they come out of the reservoir). It begins with the separator conditions, the corrected gas–oil ratio of the separator and the proportion in which the gas and the liquid were mixed. Then come the compositions of the separator gas and liquid and the composition of the reservoir mixture calculated from them, the constant-composition test with the dew point found by eye, and the gas deviation factor of the single-phase gas above it. The main table is constant volume depletion (tests on gas condensate). For each pressure step it gives the two-phase $Z$ , the share of material taken out, the volume of liquid that has dropped out, and the composition and condensate–gas ratio of the gas taken out.

Conditions and base of each number

The same property of an oil can be written with several numbers, and all of them are right, each for its own conditions. Before comparing a report with correlations, with field measurements or with another report, answer the questions in the table below for each number. The conversion factors and the standard conditions are explained in the lesson Systems of units: SI, oilfield and metric oilfield.

What to check before taking a number from a report

Quantity: Pressure
Possible variants: absolute (MPa, psia) or gauge (psig; Russian reports mark it with the abbreviation of the Russian word for “gauge”)
What the mix-up costs: 0.1 MPa: a small thing for the reservoir pressure, but not for the difference from the bubble point pressure

Quantity: Gas at standard conditions
Possible variants: 0.101325 MPa and 20 °C; 15 °C; 60 °F (15.6 °C)
What the mix-up costs: about 1.5% of the gas volume between 20 °C and 60 °F

Quantity: Base of the solution gas–oil ratio and $B_o$
Possible variants: stock-tank oil of a given separation scheme; residual oil of differential liberation; single-stage flash
What the mix-up costs: up to 10% or more in $R_s$ , several per cent in $B_o$

Quantity: Unit of the solution gas–oil ratio
Possible variants: m³/m³, m³/t, scf/STB (standard cubic feet of gas per barrel of stock-tank oil)
What the mix-up costs: a factor of 1.2 (m³/t against m³/m³ at 850 kg/m³), a factor of 5.5 (scf/STB)

Quantity: Oil density
Possible variants: at 20 °C, at 15.6 °C; at reservoir conditions
What the mix-up costs: about 0.35% between 20 and 15.6 °C, which is noticeable in °API

Quantity: Temperature of the test
Possible variants: reservoir, laboratory, separator
What the mix-up costs: the bubble point pressure changes by tenths of a megapascal for every few degrees

Quantity	Possible variants	What the mix-up costs
Pressure	absolute (MPa, psia) or gauge (psig; Russian reports mark it with the abbreviation of the Russian word for “gauge”)	0.1 MPa: a small thing for the reservoir pressure, but not for the difference from the bubble point pressure
Gas at standard conditions	0.101325 MPa and 20 °C; 15 °C; 60 °F (15.6 °C)	about 1.5% of the gas volume between 20 °C and 60 °F
Base of the solution gas–oil ratio and $B_o$	stock-tank oil of a given separation scheme; residual oil of differential liberation; single-stage flash	up to 10% or more in $R_s$ , several per cent in $B_o$
Unit of the solution gas–oil ratio	m³/m³, m³/t, scf/STB (standard cubic feet of gas per barrel of stock-tank oil)	a factor of 1.2 (m³/t against m³/m³ at 850 kg/m³), a factor of 5.5 (scf/STB)
Oil density	at 20 °C, at 15.6 °C; at reservoir conditions	about 0.35% between 20 and 15.6 °C, which is noticeable in °API
Temperature of the test	reservoir, laboratory, separator	the bubble point pressure changes by tenths of a megapascal for every few degrees

The orders of magnitude are for the assumed oil used in these lessons (850 kg/m³; the bubble point pressure changes by 0.06 MPa per degree).

Worked exampleTwo summaries of one oil

Given (assumed data)

Laboratory 1: bubble point pressure 18.0 MPa at 85 °C; solution gas–oil ratio 129.4 m³/t (gas at 20 °C, separation at 0.7 MPa and stock tank); stock-tank oil 850 kg/m³ at 20 °C; $B_{ob} = 1.350$ m³/m³
Laboratory 2: bubble point pressure 2611 psia at 185 °F; solution gas–oil ratio 610 scf/STB; $B_{ob} = 1.355$ rb/STB (reservoir barrels per stock-tank barrel; gas and oil at 60 °F, 14.696 psia); 34.2 °API; first separation stage 102 psia
Conversion: 1 psi = 0.006894757 MPa; gas from 20 °C to 60 °F (15.6 °C): factor 288.71/293.15; stock-tank oil expands by $8.0\cdot 10^{-4}$ 1/°C (assumed, as in the lesson Density and compressibility of reservoir oil); 1 m³/m³ = 5.615 scf/STB at the same conditions

Solution

Pressure: $2611\cdot 0.006894757 = 18.00$ MPa, absolute like that of laboratory 1. Temperature: $(185 - 32)/1.8 = 85.0$ °C. First stage: $102\cdot 0.006894757 = 0.703$ MPa, the same separation scheme.
Solution gas–oil ratio of laboratory 1 per cubic metre: $129.4\cdot 0.850 = 110.0$ m³/m³ at 20 °C.
We bring both the gas and the oil to 60 °F. The gas is $110.0\cdot 288.71/293.15 = 108.33$ m³. The same cubic metre of oil at 15.6 °C takes $1/(1 + 8.0\cdot 10^{-4}\cdot 4.4) = 0.9965$ m³. The solution gas–oil ratio is $108.33/0.9965 = 108.71$ m³/m³, or $108.71\cdot 5.615 = 610$ scf/STB.
Formation volume factor per barrel at 60 °F: $1.350/0.9965 = 1.355$ rb/STB.
Density: 850 kg/m³ at 20 °C is 34.2 °API (the oilfield density scale) after correction to 60 °F (example with °API).

Answer: Both summaries describe the same oil: 18.00 MPa at 85.0 °C, 610 scf/STB and 1.355 rb/STB. This is 110.0 m³/m³ and 1.350 m³/m³ at 20 °C for the same separation scheme.

What it means. The difference between 610 and 110 does not mean different oils. It is a different unit and a different base: stock-tank oil at 60 °F takes 0.35% less space than at 20 °C, so per unit of its volume there is that much more gas and reservoir volume. The short factor 5.53 takes only the gas into account and gives 608 scf/STB. For correlations that accuracy is enough (PVT correlations for oil and gas condensate), but when you compare two reports, you must also take the base of the oil into account. If laboratory 2 had written 2611 psig, the bubble point pressure would be 0.10 MPa higher, and such a difference has to be explained. If its separator had worked at a different pressure, we could not compare $B_{ob}$ , because it is different for different schemes (separator tests). The data in this example are assumed for illustration.

Is the sample good?

No check of the tables can repair a sample that does not represent the reservoir fluid, so read the samples section first. How samples are checked at the well and in the laboratory is explained in the lesson sampling. In the report you need to find the results of these checks and compare them with each other and with the conditions in the reservoir.

Duplicates. The bubble point pressures of the samples taken at the same flow rate, measured at one temperature, must agree within the accuracy of the check. A sample that differs clearly from the others has probably lost gas (if its pressure is lower) or picked up extra gas (if it is higher).
The path of the sample. Compare the bubble point pressure measured in the laboratory at surface temperature with the check at the well. If they agree, the sample arrived in the state in which it was taken. The pressure in the sampler (the opening pressure) of an ordinary bottomhole sample is below the bubble point pressure at this temperature. The sample is two-phase, and this is not a defect.
Sampling conditions. The pressure at the bottom of the well during sampling must be above the bubble point pressure of the sample at reservoir temperature. If they are almost equal, the oil at the bottom could already have lost gas, and the bubble point pressure of the sample reflects the sampling pressure, not the reservoir oil.
Reservoir pressure. For a single-phase reservoir the bubble point pressure is below the reservoir pressure. If it equals the reservoir pressure, the oil is saturated (it holds all the gas it can at this pressure): look for a gas cap (gas–oil contact from the bubble point pressure) or suspect the sample. If it is higher, a single-phase oil with such a bubble point pressure cannot exist in the reservoir, and the sample has picked up gas or was recombined with too high a gas–oil ratio.

Worked exampleWhich sample went into the tests, and is it good

Given (assumed data)

Well: samples A and B from the example of the lesson Reservoir fluid sampling: reservoir pressure 21.0 MPa, 85 °C; sampling at 15 m³/d and a bottomhole pressure of 19.5 MPa
Check at the well at 20 °C: bubble point pressure of A 14.11 MPa, of B 13.40 MPa
“Samples” section of the report, 20 °C: A: pressure in the sampler 11.89 MPa, bubble point pressure 14.10 MPa; B: 11.32 and 13.42 MPa; the tests were run on sample A
Constant-composition test, sample A: bubble point pressure 18.0 MPa at 85 °C
Expected slope: the bubble point pressure of this oil falls by 0.06 MPa per degree of cooling (assumed in the lesson Reservoir fluid sampling)

Solution

Path of the sample: sample A gave $14.10$ MPa in the laboratory against $14.11$ MPa at the well (−0.07%), and sample B gave $13.42$ against $13.40$ MPa (0.15%). Both samples arrived unchanged. The pressure in the sampler is below the bubble point pressure, so the samples are two-phase, as ordinary bottomhole samples should be.
Duplicates: B is 4.8% below A. B lost gas already at the well, and its pressure in the sampler is also lower. The main sample is A, as the laboratory chose.
From 20 to 85 °C: $14.10 + 0.06\cdot 65 = 18.00$ MPa, which agrees with the constant-composition test.
Pressure margins: at sampling, $19.5 - 18.0 = 1.5$ MPa above the bubble point pressure, so the inflow was single-phase. In the reservoir, $21.0 - 18.0 = 3.0$ MPa, so the oil is undersaturated.
If the tests had been run on sample B, the bubble point pressure would be about $13.42 + 3.90 = 17.32$ MPa. By the solubility curve of this oil, the solution gas–oil ratio would then be about 106.3 m³/m³, which is 3.4% less than the true value.

Answer: Sample A is good: it agrees with the check at the well, it was taken 1.5 MPa above the bubble point pressure, and the oil in the reservoir is undersaturated by 3.0 MPa. Sample B is not suitable for the tests.

What it means. The chain “well, laboratory at 20 °C, laboratory at 85 °C” closes, and this is a strong argument for the sample. A wrong choice of sample hides best of all. The tables obtained on sample B would be consistent inside and would pass all the checks of the next section, but they would describe an oil that does not exist in the reservoir. A margin of 1.5 MPa at sampling is small. The accuracy of the estimate of the bubble point pressure at the well was ±1 MPa (choosing the sampling rate), and the margin turned out to be enough. The slope of 0.06 MPa/°C is an assumption for this oil; for a real sample, the laboratory measures it. The data in this example are assumed for illustration.

Do the tables agree with each other?

The tests of a report are run on one sample but in different ways, so their results are linked by conservation laws (for example, the mass of the products equals the mass of the sample) and simple relations. If a link is broken, one of the numbers has an error: in the measurement, in a conversion, or in copying into the table. Most checks take a few minutes, and they are worth doing before the numbers go into calculations.

Consistency checks of an oil report

Check: Y-function (details)
What is compared: the two-phase readings of the constant-composition test with a straight line
What gives away an error: a point away from the line; a bend near the bubble point pressure

Check: Mass balance of differential liberation
What is compared: the mass of the reservoir oil in the cell with the mass of the residual oil and of the gas of all stages
What gives away an error: a discrepancy larger than the error of weighing and of volume measurement

Check: Mass balance of the separator test (example)
What is compared: $B_{ob}\,\rho_{ob}$ with the mass of the stock-tank oil and of the gas of the stages
What gives away an error: the same

Check: $B_t$ in two ways (example)
What is compared: $B_{ob}\,V/V_b$ of the constant-composition test with $B_o + B_g(R_{sb} - R_s)$ from the converted data
What gives away an error: a discrepancy of more than 1%

Check: Densities of the stock-tank oils
What is compared: the residual oil, the oil of single-stage flash and of stage separation
What gives away an error: the residual oil is lighter than the others

Check: Viscosity
What is compared: the viscosity curve with the bubble point pressure
What gives away an error: the minimum is not at the bubble point pressure

Check: Compositions of the gas and the liquid (Hoffman plot, a plot on which the components fall on a straight line when the analyses agree)
What is compared: equilibrium ratios of the separation stages
What gives away an error: components away from the line

Check	What is compared	What gives away an error
Y-function (details)	the two-phase readings of the constant-composition test with a straight line	a point away from the line; a bend near the bubble point pressure
Mass balance of differential liberation	the mass of the reservoir oil in the cell with the mass of the residual oil and of the gas of all stages	a discrepancy larger than the error of weighing and of volume measurement
Mass balance of the separator test (example)	$B_{ob}\,\rho_{ob}$ with the mass of the stock-tank oil and of the gas of the stages	the same
$B_t$ in two ways (example)	$B_{ob}\,V/V_b$ of the constant-composition test with $B_o + B_g(R_{sb} - R_s)$ from the converted data	a discrepancy of more than 1%
Densities of the stock-tank oils	the residual oil, the oil of single-stage flash and of stage separation	the residual oil is lighter than the others
Viscosity	the viscosity curve with the bubble point pressure	the minimum is not at the bubble point pressure
Compositions of the gas and the liquid (Hoffman plot, a plot on which the components fall on a straight line when the analyses agree)	equilibrium ratios of the separation stages	components away from the line

The order of the densities follows from the physics. Differential liberation at reservoir temperature sends the most intermediate components into the gas, and stage separation sends the fewest, so the residual oil is the heaviest.

Mass balance of differential liberation

\rho_{ob}\,B_{oDb}\,V_{\text{r}} = \rho_{\text{r}}\,V_{\text{r}} + \rho_{\text{air}}\sum_{j} V_{g,j}\,\gamma_j

where:

$\rho_{ob}$ density of the reservoir oil at the bubble point pressure, from the constant-composition test, g/cm³
$B_{oDb}\,V_{\text{r}}$ volume of the reservoir oil in the cell at the bubble point pressure, cm³
$\rho_{\text{r}},\ V_{\text{r}}$ density and volume of the residual oil at 20 °C, g/cm³, cm³
$V_{g,j},\ \gamma_j$ volume of the gas of stage $j$ at standard conditions and its specific gravity relative to air, L, —
$\rho_{\text{air}}$ density of air at the same standard conditions: 1.204 g/L at 0.101325 MPa and 20 °C, g/L

Units: grams on both sides: g/cm³·cm³ and g/L·L

Holds when

Only gas leaves the cell, and all the gas of each stage is collected and measured.
The density of the gas at standard conditions is $\rho_{\text{air}}\,\gamma_j$ , which means that we take the gas as ideal at 0.1 MPa. For the heavy gas of the last stages this slightly underestimates the mass.

The mass of the oil loaded into the cell must be found again in the products of the test. The same sum, cut off at step $k$ , gives the mass of the oil after that step. Divided by $B_{oD}(p_k)\,V_{\text{r}}$ , it gives the oil density. You can use this to check the oil densities at the steps that the laboratory gives in its table.

Worked exampleDoes the mass balance of differential liberation close

Given (assumed data)

Gas of the stages, L (standard conditions): 0.858; 0.876; 0.898; 0.929; 0.979; 1.510
Specific gravity of the gas of the stages: 0.715; 0.725; 0.745; 0.780; 0.860; 1.120
Residual oil: 50.00 cm³, 0.861 g/cm³ at 20 °C
Reservoir oil: $B_{oDb} = 1.392$ ; density at the bubble point pressure 0.708 g/cm³ (constant-composition test)
Stock-tank oil of the separator tests: 850 kg/m³ (two stages), 855 kg/m³ (single-stage flash)

Solution

Gas: $\sum V_{g,j}\gamma_j = 5.1753$ L, and its mass is $5.1753\cdot 1.204 = 6.231$ g. The last stage alone gives 2.036 g, which is 33% of the mass of the gas.
Residual oil: $50.00\cdot 0.861 = 43.050$ g. Products of the test: $43.050 + 6.231 = 49.281$ g.
Loaded: $1.392\cdot 50.00 = 69.60$ cm³ of oil at the bubble point pressure, with a mass of $69.60\cdot 0.708 = 49.277$ g. The discrepancy is 0.01%. The density of the reservoir oil from the balance is $49.281/69.60 = 0.7081$ g/cm³.
Densities of the stock-tank oils: 861 > 855 > 850 kg/m³. The residual oil is the heaviest, as it should be.
Sensitivity check: if the gas of the last stage were recorded 10% higher, the mass of the products would grow by $0.1\cdot 2.036 = 0.204$ g, and the discrepancy would be 0.42%.

Answer: The balance closes: the products weigh 49.281 g against 49.277 g of oil loaded (0.01%). The differential liberation table agrees with the constant-composition test.

What it means. A third of the mass of the gas comes from the last stage. It is the heaviest gas, which at atmospheric pressure comes out slowly and is easily lost. So the balance is especially sensitive to it. An error of 10% in its volume gives a discrepancy of 0.4%, tens of times more than for consistent data, and it is seen at once. A balance that closes proves that the measurements agree with each other, but not that the sample is representative: on sample B it would have closed in the same way. The data in this example are assumed for illustration.

A property table for calculations

The material balance and the reservoir simulation model need one property table of the oil and gas against pressure. It runs from a pressure above the initial reservoir pressure down to atmospheric pressure, on one base and for one separation scheme (Black oil model: solution gas–oil ratio and formation volume factors). A report usually has no such table. We assemble it from several tests.

How to assemble the table

1
Separation scheme
choose the separator test closest to the field treatment and take $R_{sb}$ and $B_{ob}$ from it. If the scheme has not been chosen yet, write which scheme the table refers to.
2
Below the bubble point pressure
$R_s$ and $B_o$ are the differential liberation converted to this scheme (conversion). At atmospheric pressure $R_s = 0$ , and $B_o$ is the volume of the stock-tank oil at reservoir temperature.
3
Above the bubble point pressure
$B_o = B_{ob}\,V/V_b$ from the constant-composition test; the solution gas–oil ratio is constant.
4
Gas
$B_g$ and $Z$ come from the gas of the stages of differential liberation; gas viscosity comes from a correlation (Gas formation volume factor, compressibility and viscosity).
5
Oil viscosity
from the viscometer at reservoir temperature; for the intermediate pressures, from a smooth curve.
6
Densities at standard conditions
of the stock-tank oil and of the separator gas of the same scheme. The program uses them to convert volumes to masses.
7
Check
$R_s$ and $B_o$ grow with pressure up to the bubble point pressure, and $B_g$ falls; the apparent compressibility is positive on every interval (below).

The condition of positive apparent compressibility between the rows of the table

c_o^{\text{app}} \approx \frac{\bar B_g\,(R_{s,i} - R_{s,i+1}) - (B_{o,i} - B_{o,i+1})}{\bar B_o\,(p_i - p_{i+1})} > 0

where:

$p_i > p_{i+1}$ pressures of two neighbouring rows below the bubble point pressure, MPa
$R_{s,i},\ B_{o,i}$ solution gas–oil ratio and oil formation volume factor in row $i$ , m³/m³
$\bar B_g$ gas formation volume factor at the mean pressure of the interval, m³/m³
$\bar B_o$ mean $B_o$ of the two rows, m³/m³
$c_o^{\text{app}}$ apparent compressibility of oil with gas coming out, over the interval, MPa⁻¹

Units: pressures in MPa, formation volume factors and solution gas–oil ratio in m³/m³

Holds when

This is the apparent compressibility $-\frac{1}{B_o}\frac{\partial B_o}{\partial p} + \frac{B_g}{B_o}\frac{\partial R_s}{\partial p}$ written for a finite interval: the derivatives are replaced by differences between two rows (apparent compressibility).
We take $R_s$ , $B_o$ and $B_g$ from one table, on one base.

When the pressure falls, the oil shrinks by $\Delta B_o$ , and the gas that has come out takes $\bar B_g\,\Delta R_s$ . The second must be larger. Otherwise the oil together with its gas would take less space as the pressure falls, which means it would have a negative compressibility. A reservoir simulation model cannot calculate such a table in a stable way, so we check the table before loading it.

Worked exampleA table of the assumed oil for two-stage separation

Given (assumed data)

Differential liberation, m³/m³: $R_{sDb} = 121.0$ , $B_{oDb} = 1.392$ ; 15 MPa — 103.8 and 1.345; 9 MPa — 68.4 and 1.248; 3 MPa — 30.2 and 1.144
Separator test, 0.7 MPa: $R_{sb} = 110.0$ m³/m³, $B_{ob} = 1.350$ m³/m³; factor $B_{ob}/B_{oDb} = 0.9698$
Constant-composition test: $V/V_b = 0.9810$ at 30 MPa
Gas of the stages: $B_g = 0.00653$ m³/m³ at 16.5 MPa, 0.02393 at 4.5 MPa ( $Z = 0.87$ )

Solution

New rows: at 15 MPa $R_s = 110.0 - 17.2\cdot 0.9698 = 93.3$ and $B_o = 1.345\cdot 0.9698 = 1.304$ ; at 9 MPa $R_s = 110.0 - 52.6\cdot 0.9698 = 59.0$ and $B_o = 1.210$ ; at 3 MPa $R_s = 110.0 - 90.8\cdot 0.9698 = 21.9$ and $B_o = 1.109$ m³/m³. The rows at 12 and 6 MPa are from the example of the lesson on PVT tests.
Above the bubble point pressure: $B_o = 1.350\cdot 0.9810 = 1.324$ m³/m³ at 30 MPa, and the solution gas–oil ratio is 110.0.
Interval 18–15 MPa: the gas is $0.00653\cdot(110.0 - 93.3) = 0.00653\cdot 16.7 = 0.10905$ m³/m³, against an oil shrinkage of $1.350 - 1.304 = 0.046$ m³/m³. That is 2.4 times more. $c_o^{\text{app}} = (0.10905 - 0.046)/(1.327\cdot 3) = 0.0158$ MPa⁻¹.
Interval 6–3 MPa: $0.02393\cdot(41.0 - 21.9) = 0.45706$ m³/m³, against $1.162 - 1.109 = 0.053$ . That is 8.6 times more. $c_o^{\text{app}} = 0.119$ MPa⁻¹.
The row at atmospheric pressure: the conversion formula would give $R_s = 110.0 - 121.0\cdot 0.9698 = -7.3$ m³/m³ and $B_o = 1.055\cdot 0.9698 = 1.023$ . In the table we put $R_s = 0$ and $B_o = 1.05$ , the volume of the stock-tank oil at 85 °C.

Answer: The table (below) is monotonic (its values change steadily), and on all intervals the gas that has come out takes more space than the oil loses. The margin is smallest at the bubble point pressure (2.4 times), and it grows towards low pressures.

What it means. At the bubble point pressure $B_g$ is small, because the gas is strongly compressed, and the condition is met with the smallest margin. Errors most often break it here: $R_s$ from one source and $B_o$ from another, a misprint in a row, or smoothing that spoiled the slope. The apparent compressibility just below the bubble point pressure is 0.0158 MPa⁻¹, which is 10 times more than the compressibility of undersaturated oil ( $1.6\cdot 10^{-3}$ MPa⁻¹). That is how it should be. The data in this example are assumed for illustration.

Property table of the lesson's assumed oil, 85 °C, separation at 0.7 MPa and stock tank

$p$ , MPa: 30
$R_s$ , m³/m³: 110.0
$B_o$ , m³/m³: 1.324
$B_g$ , m³/m³: —
$\mu_o$ , mPa·s: 1.20

$p$ , MPa: 25
$R_s$ , m³/m³: 110.0
$B_o$ , m³/m³: 1.335
$B_g$ , m³/m³: —
$\mu_o$ , mPa·s: 1.12

$p$ , MPa: 20
$R_s$ , m³/m³: 110.0
$B_o$ , m³/m³: 1.346
$B_g$ , m³/m³: —
$\mu_o$ , mPa·s: 1.03

$p$ , MPa: 18 ( $p_b$ )
$R_s$ , m³/m³: 110.0
$B_o$ , m³/m³: 1.350
$B_g$ , m³/m³: 0.00598
$\mu_o$ , mPa·s: 1.00

$p$ , MPa: 15
$R_s$ , m³/m³: 93.3
$B_o$ , m³/m³: 1.304
$B_g$ , m³/m³: 0.00718
$\mu_o$ , mPa·s: 1.12

$p$ , MPa: 12
$R_s$ , m³/m³: 76.3
$B_o$ , m³/m³: 1.258
$B_g$ , m³/m³: 0.00897
$\mu_o$ , mPa·s: 1.30

$p$ , MPa: 9
$R_s$ , m³/m³: 59.0
$B_o$ , m³/m³: 1.210
$B_g$ , m³/m³: 0.01197
$\mu_o$ , mPa·s: 1.55

$p$ , MPa: 6
$R_s$ , m³/m³: 41.0
$B_o$ , m³/m³: 1.162
$B_g$ , m³/m³: 0.01795
$\mu_o$ , mPa·s: 1.96

$p$ , MPa: 3
$R_s$ , m³/m³: 21.9
$B_o$ , m³/m³: 1.109
$B_g$ , m³/m³: 0.03590
$\mu_o$ , mPa·s: 2.79

$p$ , MPa: 0.1
$R_s$ , m³/m³: 0
$B_o$ , m³/m³: 1.050
$B_g$ , m³/m³: —
$\mu_o$ , mPa·s: 5.53

$p$ , MPa	$R_s$ , m³/m³	$B_o$ , m³/m³	$B_g$ , m³/m³	$\mu_o$ , mPa·s
30	110.0	1.324	—	1.20
25	110.0	1.335	—	1.12
20	110.0	1.346	—	1.03
18 ( $p_b$ )	110.0	1.350	0.00598	1.00
15	93.3	1.304	0.00718	1.12
12	76.3	1.258	0.00897	1.30
9	59.0	1.210	0.01197	1.55
6	41.0	1.162	0.01795	1.96
3	21.9	1.109	0.03590	2.79
0.1	0	1.050	—	5.53

The data are assumed for illustration. Gas at 0.101325 MPa and 20 °C; $B_g$ is calculated at $Z = 0.87$ , a simplification of the assumed oil (assumed oil). Gas viscosity from a correlation, from 0.0191 mPa·s at 18 MPa to 0.0132 mPa·s at 3 MPa. Densities at standard conditions: oil 850 kg/m³, separator gas 0.963 kg/m³ ( $\gamma_g = 0.80$ ).

Check of the table: gas that came out over the interval and oil shrinkage, m³/m³

Interval, MPa: 18–15
$\Delta R_s$: 16.7
$\bar B_g\,\Delta R_s$: 0.10905
$\Delta B_o$: 0.046
$c_o^{\text{app}}$ , MPa⁻¹: 0.0158

Interval, MPa: 15–12
$\Delta R_s$: 17.0
$\bar B_g\,\Delta R_s$: 0.13566
$\Delta B_o$: 0.046
$c_o^{\text{app}}$ , MPa⁻¹: 0.0233

Interval, MPa: 12–9
$\Delta R_s$: 17.3
$\bar B_g\,\Delta R_s$: 0.17750
$\Delta B_o$: 0.048
$c_o^{\text{app}}$ , MPa⁻¹: 0.0350

Interval, MPa: 9–6
$\Delta R_s$: 18.0
$\bar B_g\,\Delta R_s$: 0.25848
$\Delta B_o$: 0.048
$c_o^{\text{app}}$ , MPa⁻¹: 0.0592

Interval, MPa: 6–3
$\Delta R_s$: 19.1
$\bar B_g\,\Delta R_s$: 0.45706
$\Delta B_o$: 0.053
$c_o^{\text{app}}$ , MPa⁻¹: 0.119

Interval, MPa: 3–0.1
$\Delta R_s$: 21.9
$\bar B_g\,\Delta R_s$: 1.5210
$\Delta B_o$: 0.059
$c_o^{\text{app}}$ , MPa⁻¹: 0.467

Interval, MPa	$\Delta R_s$	$\bar B_g\,\Delta R_s$	$\Delta B_o$	$c_o^{\text{app}}$ , MPa⁻¹
18–15	16.7	0.10905	0.046	0.0158
15–12	17.0	0.13566	0.046	0.0233
12–9	17.3	0.17750	0.048	0.0350
9–6	18.0	0.25848	0.048	0.0592
6–3	19.1	0.45706	0.053	0.119
3–0.1	21.9	1.5210	0.059	0.467

$\bar B_g$ is at the mean pressure of the interval. All the values of $c_o^{\text{app}}$ are positive. On the lowest interval they are large, because gas at low pressure takes a lot of space.

CalculatedCheck of the table of the assumed oil on intervals of 3 MPa: the volume of gas that came out per megapascal of pressure decline (

\bar B_g\,\Delta R_s/\Delta p

), and the shrinkage of the oil (

\Delta B_o/\Delta p

). The vertical scale is logarithmic. The gas line is above the oil line everywhere, and the two are closest at the bubble point pressure.

The reservoir simulation model also asks for branches above the bubble point pressure, one for each solution gas–oil ratio. A branch shows how $B_o$ and viscosity change when the pressure of oil with that gas content rises above its own bubble point. Oil that has lost gas as the pressure fell may come under a higher pressure again, for example during water injection. It then stays undersaturated, with its own $R_s$ . If there are no tests for such branches, we build them with the same compressibility as was measured for the original oil. For $R_s = 110$ m³/m³ this gives $B_o$ from 1.350 at 18 MPa to 1.324 at 30 MPa, and the viscosity from 1.00 to 1.20 mPa·s. This is an assumption, because oil with a smaller solution gas–oil ratio usually has a smaller compressibility. State it next to the table.

The report on a gas condensate

A gas condensate report has fewer tests, but each number weighs more. Almost all the condensate that drops out in the reservoir stays there, and its amount depends on how far the reservoir pressure is above the dew point. So read the report in this order:

Recombination. Compare the corrected gas–oil ratio of the separator, the mixing proportion and the condensate–gas ratio of the reservoir mixture with the field measurements at the sampling rate (recombination). An error in the gas–oil ratio goes straight into the dew point.
Composition. The share of C7+ (the heavy components, with seven or more carbon atoms) in the reservoir mixture and the gas–oil ratio must point to one type of fluid (field signs).
Dew point against reservoir pressure. The dew point at reservoir temperature must not be above the reservoir pressure. If it equals the reservoir pressure, the reservoir may be saturated and have an oil rim (a thin layer of oil under the gas), or the sample is too rich, with too many heavy components (gas condensate). We decide from the composition and from the data of neighbouring wells. The difference between the reservoir pressure and the dew point is the drawdown (the pressure drop between the reservoir and the well) that a well can give without condensate dropping out at the bottom (Retrograde condensation, wax, asphaltenes and high-pressure fluids).
Constant-composition test. Compare the gas deviation factor of the single-phase gas above the dew point with a correlation based on the composition (Ideal and real gas: the gas deviation factor (Z)). The volumes of liquid below the dew point come from a closed cell. So they do not show how much condensate the reservoir loses during depletion. But they are just what we need to calculate the condensate zone around the wells, through which a mixture of the original composition flows (condensate banking: liquid condensate that builds up around a well and blocks the gas).
Constant volume depletion. The two-phase $Z$ and the share of material taken out are linked by the equation of state (the formula that links pressure, volume and temperature) of the whole mixture in the cell, so each row must lie on the line $p/Z_{2\text{ph}}$ (example below). From the compositions of the gas taken out we can balance each component and find the composition of the liquid that remains. A negative fraction of a component shows an error in the measurements.

Worked exampleChecking a constant volume depletion table

Given (assumed data)

Gas condensate: the same as in the lesson Laboratory PVT tests: dew point 36.0 MPa at 95 °C, $Z_d = 1.05$ ; reservoir pressure 38.0 MPa
Report: $p$ , MPa — $Z_{2\text{ph}}$ — taken out, % — liquid, %: 30 — 1.000 — 12.5 — 8.4; 24 — 0.955 — 24.6 — 11.6; 18 — 0.905 — 42.0 — 11.9; 12 — 0.855 — 59.1 — 10.9; 7 — 0.810 — 74.8 — 9.7

Solution

The line starts from $p_d/Z_d = 36.0/1.05 = 34.29$ MPa and comes to zero when all the material is taken out.
From the $Z_{2\text{ph}}$ of the report: $p/Z_{2\text{ph}}$ = 30.00; 25.13; 19.89; 14.04; 8.64 MPa.
From the share taken out: $34.29\,(1 - n_p/n_0)$ = 30.00; 25.85; 19.89; 14.02; 8.64 MPa. In the rows at 30, 18, 12 and 7 MPa the two numbers differ by no more than 0.02 MPa, which is rounding. In the row at 24 MPa it is 25.13 against 25.85, a discrepancy of −2.8%.
Which of the two numbers is wrong? The series of $Z_{2\text{ph}}$ is smooth: 1.000; 0.955; 0.905, in steps of 0.045–0.050. The increments of the share taken out are 12.5, then 12.1 and 17.4%, which are uneven. From $Z_{2\text{ph}}$ the share taken out at 24 MPa is $1 - 25.13/34.29 = 0.267$ , that is 26.7%, and the increments become even: 14.2 and 15.3%.
Margin above the dew point: $38.0 - 36.0 = 2.0$ MPa.

Answer: In the row at 24 MPa the share of material taken out is recorded with an error: it should be about 26.7%, not 24.6%. The other rows agree.

What it means. The check finds the row but does not say which number in it is wrong. The smoothness of the series and the raw measurements in the appendix give the hint. For calculations we take two things from the table. One is the two-phase $Z$ , for the material balance of the reservoir below the dew point (Material balance for gas reservoirs). The other is the share of liquid, which shows how much condensate stays in the reservoir in development by depletion. At 18 MPa, 42% of the material has left the cell, and the liquid that has dropped out takes 11.9% of the hydrocarbon pore volume. A margin of only 2.0 MPa above the dew point means that condensate will drop out at the bottom of the well even at a moderate drawdown. How this affects the well is explained in the lesson Retrograde condensation, wax, asphaltenes and high-pressure fluids. The data in this example are assumed for illustration.

CalculatedCheck of the constant volume depletion table of the example:

p/Z_{2\text{ph}}

against the share of material taken out. The line goes through the dew point and through zero at full withdrawal. The row at 24 MPa with the recorded 24.6% lies to the left of the line, and with the corrected 26.7% it lies on it.

The whole report: the order of reading

Let us put the checks in one order and go through the report on the lesson's assumed oil. Its summary looks like this:

Summary of the report on the assumed oil (sample A)

Quantity: Reservoir pressure and temperature
Value: 21.0 MPa; 85 °C
Source and base: measurement at sampling

Quantity: Bubble point pressure
Value: 18.0 MPa at 85 °C
Source and base: constant-composition test

Quantity: Solution gas–oil ratio
Value: 110.0 m³/m³ (129.4 m³/t)
Source and base: separation at 0.7 MPa and stock tank; gas at 20 °C

Quantity: The same from differential liberation
Value: 121.0 m³/m³
Source and base: per residual oil; not taken into calculations

Quantity: Formation volume factor at the bubble point pressure
Value: 1.350 m³/m³
Source and base: the same separation (1.392 is per residual oil)

Quantity: Formation volume factor at reservoir pressure
Value: 1.344 m³/m³
Source and base: $1.350\cdot V/V_b$

Quantity: Reservoir oil density
Value: 708 kg/m³ at the bubble point pressure; 711 kg/m³ at 21.0 MPa
Source and base: constant-composition test

Quantity: Reservoir oil viscosity
Value: 1.05 mPa·s at 21.0 MPa; 1.00 mPa·s at the bubble point pressure
Source and base: viscometer

Quantity: Compressibility above the bubble point pressure
Value: $1.6\cdot 10^{-3}$ MPa⁻¹
Source and base: 30–18 MPa

Quantity: Stock-tank oil
Value: 850 kg/m³ at 20 °C (34.2 °API)
Source and base: separation at 0.7 MPa and stock tank

Quantity	Value	Source and base
Reservoir pressure and temperature	21.0 MPa; 85 °C	measurement at sampling
Bubble point pressure	18.0 MPa at 85 °C	constant-composition test
Solution gas–oil ratio	110.0 m³/m³ (129.4 m³/t)	separation at 0.7 MPa and stock tank; gas at 20 °C
The same from differential liberation	121.0 m³/m³	per residual oil; not taken into calculations
Formation volume factor at the bubble point pressure	1.350 m³/m³	the same separation (1.392 is per residual oil)
Formation volume factor at reservoir pressure	1.344 m³/m³	$1.350\cdot V/V_b$
Reservoir oil density	708 kg/m³ at the bubble point pressure; 711 kg/m³ at 21.0 MPa	constant-composition test
Reservoir oil viscosity	1.05 mPa·s at 21.0 MPa; 1.00 mPa·s at the bubble point pressure	viscometer
Compressibility above the bubble point pressure	$1.6\cdot 10^{-3}$ MPa⁻¹	30–18 MPa
Stock-tank oil	850 kg/m³ at 20 °C (34.2 °API)	separation at 0.7 MPa and stock tank

The data are assumed for illustration; the tables of the tests are in the lesson Laboratory PVT tests.

The order of reading and what it gave for the assumed oil

1
1. Conditions and base
pressures are absolute, gas is at 20 °C, and the solution gas–oil ratio is given both per cubic metre and per tonne; the summary of the second laboratory in oilfield units agrees (example above).
2
2. Sample
the tests were run on sample A; it agrees with the check at the well, and it was taken 1.5 MPa above the bubble point pressure; sample B is rejected.
3
3. Bubble point pressure and reservoir pressure
the oil is undersaturated by 3.0 MPa; there are no signs of a gas cap.
4
4. Constant-composition test
a clear kink at 18.0 MPa; the Y-function revealed a wrong reading at 10 MPa, and it was replaced by a smoothed value (example).
5
5. Differential liberation
the mass balance closes to hundredths of a per cent; the residual oil is heavier than the stock-tank oil of the separator tests.
6
6. Separator tests
the mass balance closes; the 0.7 MPa scheme is chosen, and it is also the field one.
7
7. Link between the tests
$B_t$ at 12 MPa in two ways: 1.560 and 1.560 m³/m³.
8
8. Viscosity
minimum 1.00 mPa·s at the bubble point pressure.
9
9. Table for calculations
assembled on one base, monotonic, and the apparent compressibility is positive on all intervals.
10
10. What the report does not have
tests with waxes and asphaltenes, and mixing with gas. If they were not in the work order, this is recorded as a gap (Retrograde condensation, wax, asphaltenes and high-pressure fluids).

Terms and symbols

PVT report: A laboratory report on the study of reservoir fluid samples: the samples and their checks, the composition, the tests, the viscosity and special studies.
Base of a quantity: What the number refers to: stock-tank oil of a given separation scheme, residual oil, a tonne of oil, and for gas the standard conditions.
Smoothed values: Values of a table read from a smooth curve or from a straight line through the measurements, not the measurements themselves.
Mass balance of a test: A comparison of the mass of the fluid loaded into the cell with the mass of the products of the test. It checks that the measurements agree with each other.
Apparent compressibility between rows: The ratio $(\bar B_g\Delta R_s - \Delta B_o)/(\bar B_o\Delta p)$ . It must be positive on every interval of the table below the bubble point pressure.
The line $p/Z_{2\text{ph}}$: The link between the two-phase gas deviation factor and the share of material taken out in a constant volume test. The rows of a gas condensate report are checked against it.

Common mistakes and misconceptions

Wrong

“The summary has two solution gas–oil ratios. Let us take the larger one, to be safe.”

In fact

The 121.0 m³/m³ of differential liberation refers to the residual oil and is not suitable for calculations. Take the solution gas–oil ratio of the separator test of the scheme by which the oil is treated. For the lesson's assumed oil it is 110.0 m³/m³.

Wrong

“The pressure in the report is of course absolute.”

In fact

In oilfield units it is often given in psig. A difference of 14.7 psi (0.101 MPa) can decide whether the oil is saturated. Check the unit in every table.

Wrong

“If the mass balance closes, the report is right.”

In fact

The balance checks that the measurements on one sample agree with each other. A sample that is not representative gives equally consistent tables. Only the information on sampling and the comparison with reservoir conditions give it away.

Wrong

“A negative compressibility on one interval is a small thing; the model will cope.”

In fact

On such an interval the oil with the gas that has come out would be compressed as the pressure falls, which is physically impossible. The calculation becomes unstable or gives a wrong balance. Look for the source of the error, usually a mix of bases, and correct the table.

Wrong

“The share of liquid in the gas condensate test is the share of condensate that we will get.”

In fact

This is the liquid that stays in the “reservoir”, the cell. Condensate production in depletion is calculated from the compositions and the amount of gas taken out, and near the well the condensate saturation can be much higher (Retrograde condensation, wax, asphaltenes and high-pressure fluids).

In depth: advanced detail

Where the condition of positive compressibility comes from

Take a cubic metre of stock-tank oil together with all the gas dissolved in it, and lower the pressure by $\Delta p$ below the bubble point pressure without letting the gas out. The oil will lose $\Delta R_s$ standard cubic metres of gas and shrink by $\Delta B_o$ , and the gas at the current pressure will take $B_g\,\Delta R_s$ . The change of the total volume is $B_g\,\Delta R_s - \Delta B_o$ . Divided by the volume and by the fall of pressure, this is the apparent compressibility.

Any substance in equilibrium expands when the pressure falls; otherwise its state would be unstable. So the difference must be positive. The expansion of the gas that came out earlier does not enter this quantity. In the change of the total formation volume factor $B_t$ it gives a separate term, $(R_{sb} - R_s)\,\Delta B_g$ . So the condition for a table is stricter than the requirement that $B_t$ grows as the pressure falls: a table can give a growing $B_t$ and still break the condition on a single interval.

Conversion of differential data and the report

The formulas for converting differential data to separation conditions are approximate, and they fail near atmospheric pressure (why). A good report therefore says how it obtained the table for calculations: by the conversion formula, by separator tests on the oil of individual stages, or by a tuned equation of state. If the laboratory has already converted the table, do not convert it a second time. That is the same error as taking the differential data directly, only in the other direction.

The laboratory runs separator tests on the oil of the stages like this: at several pressures of differential liberation, it takes part of the equilibrium oil out of the cell through a separator of the chosen scheme and measures its $R_s$ and $B_o$ directly. This is more expensive, but it removes the assumption of a constant factor where that assumption is worst, at low pressures.

Connections to other topics

Laboratory PVT tests — the tests themselves, the Y-function and the conversion of differential data to separation conditions.
Reservoir fluid sampling — how samples are taken and checked at the well.
Black oil model: solution gas–oil ratio and formation volume factors — what $R_s$ , $B_o$ , $B_g$ and $B_t$ are and why they depend on the separation scheme.
Density and compressibility of reservoir oil — reservoir oil density from the mass balance, and apparent compressibility.
Systems of units: SI, oilfield and metric oilfield — absolute and gauge pressure, standard conditions and conversion factors.
Cubic equations of state — how an equation of state is tuned to the tests of the report.
Material balance for gas reservoirs — where the two-phase gas deviation factor of a gas condensate is used.
Retrograde condensation, wax, asphaltenes and high-pressure fluids — condensate dropout near wells, waxes and asphaltenes: the special sections of a report.

Check yourself

1
An oil is treated by the scheme “0.7 MPa and stock tank”. The report summary gives 121.0 m³/m³ from differential liberation, 125.0 m³/m³ from single-stage flash, 110.0 m³/m³ from separation at 0.7 MPa, and 129.4 m³/t. Which solution gas–oil ratio should be taken for the calculation of dissolved gas in place?
1. A.110.0 m³/m³
2. B.121.0 m³/m³
3. C.125.0 m³/m³
4. D.129.4 m³/t as it is
Show the answer
Correct answer: A. 110.0 m³/m³
The dissolved gas in place is $N R_{si}$ for the treatment scheme by which the gas will be obtained: 110.0 m³/m³. The value 121.0 refers to the residual oil, 125.0 to single-stage flash, and 129.4 m³/t is the same 110.0, only per tonne: $129.4\cdot 0.850 = 110.0$ m³/m³.
2
A foreign report gives a bubble point pressure of 2480 psig at reservoir temperature; the reservoir pressure from measurement is 17.15 MPa (absolute). Is the oil undersaturated? What would we get if we took 2480 psig as an absolute pressure?
Show the answer
The absolute bubble point pressure is $2480 + 14.7 = 2494.7$ psia, or $2494.7\cdot 0.006895 = 17.20$ MPa, which is above the reservoir pressure of 17.15 MPa. A single-phase oil with such a bubble point pressure cannot exist in the reservoir. Either the oil is saturated (a gas cap is possible) or the sample picked up gas, and this has to be sorted out. If we forget the 14.7 psi, we get $2480\cdot 0.006895 = 17.10$ MPa and the false conclusion that the oil is undersaturated by 0.05 MPa.
3
A differential liberation test of another oil: 40.00 cm³ of residual oil with a density of 0.872 g/cm³; 4.60 L of gas in total with a mean specific gravity of 0.86; at the bubble point pressure there was 54.40 cm³ of oil with a density of 0.7250 g/cm³ in the cell. Does the mass balance close?
Show the answer
Products: $40.00\cdot 0.872 + 4.60\cdot 0.86\cdot 1.204 = 34.880 + 4.763 = 39.643$ g; loaded: $54.40\cdot 0.7250 = 39.440$ g. The products are heavier by 0.203 g, that is by 0.5%. The density of the reservoir oil from the balance is $39.643/54.40 = 0.7287$ g/cm³ against the measured 0.7250. Such a discrepancy can be caused, for example, by an error of ten per cent or so in the volume of the gas of the last, heaviest stages. Check their volumes and densities, and the density of the residual oil.
4
Two neighbouring rows of a table below the bubble point pressure: 14 MPa — $R_s = 95$ m³/m³, $B_o = 1.300$ ; 11 MPa — $R_s = 80$ m³/m³, $B_o = 1.160$ . $B_g$ at 12.5 MPa is 0.0086 m³/m³. Can this table be loaded into the model?
1. A.Yes: both $R_s$ and $B_o$ decrease as the pressure decreases
2. B.No: on this interval the oil loses 0.140 m³/m³ of volume, and the gas that has come out takes only 0.129 m³/m³, so the compressibility is negative
3. C.Yes, if the oil viscosity rises as the pressure falls
4. D.It cannot be said without the bubble point pressure
Show the answer
Correct answer: B. No: on this interval the oil loses 0.140 m³/m³ of volume, and the gas that has come out takes only 0.129 m³/m³, so the compressibility is negative
$\bar B_g\,\Delta R_s = 0.0086\cdot 15 = 0.129$ m³/m³ is less than $\Delta B_o = 0.140$ : the oil with the gas that has come out would take less space as the pressure falls. Monotonic behaviour is not enough; this condition is needed too. Most likely $B_o$ and $R_s$ were taken from different sources, or there is a misprint in a row.
5
In a gas condensate report the dew point is 30.0 MPa and $Z_d = 0.98$ . At the 20 MPa step of constant volume depletion, $Z_{2\text{ph}} = 0.90$ and 30.0% of the material has been taken out. Do these numbers agree?
Show the answer
From $Z_{2\text{ph}}$ : $p/Z_{2\text{ph}} = 20/0.90 = 22.22$ MPa. From the share taken out: $(30.0/0.98)\cdot(1 - 0.300) = 30.61\cdot 0.700 = 21.43$ MPa. The discrepancy is almost 4%: the numbers do not agree. If $Z_{2\text{ph}}$ is right, the share taken out is $1 - 22.22/30.61 = 0.274$ , that is 27.4%. Which number is wrong is decided from the smoothness of the series and from the raw measurements.
6
The bubble point pressure of a sample at reservoir temperature is almost equal to the bottomhole pressure at which it was taken. What can this mean?
1. A.A gas cap is proven in the reservoir
2. B.The laboratory made a mistake
3. C.The oil at the bottom could already have lost gas, and the bubble point pressure of the sample reflects the sampling pressure, not the reservoir oil
4. D.The sample is ideal: it was taken exactly at the bubble point pressure
Show the answer
Correct answer: C. The oil at the bottom could already have lost gas, and the bubble point pressure of the sample reflects the sampling pressure, not the reservoir oil
If the pressure at the bottom fell to the bubble point pressure or below, part of the gas could stay in the reservoir, immobile, and the sample is in equilibrium with the sampling pressure. Then its bubble point pressure is too low. Check the sampling rate and the gas–oil ratio, and compare with other samples. If in doubt, take the samples again at a smaller drawdown.

Updated: 17 October 2026