Laboratory PVT tests

Quick refresher

The tests differ in what happens to the gas that comes out of the oil. In constant-composition expansion the gas stays in the cell. This is a closed system, like the path from the bottom of the well to the separator (the surface vessel that splits the gas from the oil). In differential liberation the gas of each stage is let out. This is how mobile gas moves away from its oil in the reservoir. The reservoir itself behaves somewhere in between.
Constant-composition expansion gives the bubble point pressure (the kink of the “pressure–volume” curve), the relative volume $V/V_b$ (the volume of the sample divided by its volume at the bubble point pressure) and the compressibility of the oil above the bubble point pressure. We check and smooth the two-phase readings with the Y-function.
Differential liberation gives, at each stage, the solution gas–oil ratio $R_{sD}$ and the oil formation volume factor $B_{oD}$ , the gas formation volume factor $B_g$ , the gas deviation factor $Z$ and the gas density. We give the oil values per 1 m³ of residual oil, not per 1 m³ of the stock-tank oil of the field. That is why $R_{sD}$ and $B_{oD}$ are larger than the field values.
The separator test gives the solution gas–oil ratio $R_{sb}$ and the formation volume factor $B_{ob}$ for one particular separation scheme. Separation in stages leaves more propane and butanes in the liquid than a single flash to atmospheric pressure does. There is a best pressure for the first stage.
For calculations we combine the data. Above the bubble point pressure, $B_o = B_{ob}\,V/V_b$ . Below it, we convert the differential data to separator conditions with the factor $B_{ob}/B_{oDb}$ . Close to atmospheric pressure this conversion gives wrong results.
For gas condensate the main test is constant volume depletion. It gives the condensate dropout curve (how much liquid drops out of the gas at each pressure), the amount and composition of the gas taken out, and the two-phase $Z$ for the material balance.

Key relationships

c_o = -\frac{1}{V}\left(\frac{\partial V}{\partial p}\right)_T,\qquad B_o = B_{ob}\,\frac{V}{V_b}

constant-composition test, single-phase oil ( $p \ge p_b$ )

Y = \frac{p_b - p}{p\,(V/V_b - 1)}

Y-function of the two-phase readings; almost linear in pressure for black oils

B_o = B_{oD}\,\frac{B_{ob}}{B_{oDb}},\qquad R_s = R_{sb} - (R_{sDb} - R_{sD})\,\frac{B_{ob}}{B_{oDb}}

conversion of differential data to separator conditions ( $p < p_b$ )

Z_{2\text{ph}} = \frac{p\,V_d}{(n_0 - n_p)\,R\,T}

two-phase gas deviation factor in a constant volume test

Common pitfalls

Taking $R_{sD}$ and $B_{oD}$ from the differential liberation table as $R_s$ and $B_o$ . They refer to residual oil and are too high.
Mixing up the relative volume $V/V_b$ of the constant-composition test (the oil together with the gas that came out) and the relative oil volume $B_{oD}$ of differential liberation (the liquid only).
Treating the results of the separator test as a property of the oil. They are valid for their own scheme and their own separator pressures.
Trusting the converted differential data at pressures below a few megapascals, and taking a negative solution gas–oil ratio seriously.
Looking for the dew point of a gas condensate (the pressure at which liquid first drops out of the gas) at the kink of the pressure–volume curve. There is almost no kink; you see the dew point through the window of the cell.

After this lesson you can

explain what a PVT cell measures, and which processes in the reservoir and in the field the three tests imitate: constant-composition expansion, differential liberation and the separator test
find the bubble point pressure, the relative volume and the compressibility of oil above the bubble point pressure from constant-composition expansion data, and check the two-phase readings (taken when the cell holds both gas and oil) with the Y-function (a number worked out from each reading; good readings give a straight line)
calculate, from differential liberation readings, the solution gas–oil ratio and the formation volume factor of the oil per 1 m³ of residual oil (the oil left at the end of the test), and the formation volume factor and the gas deviation factor (a number that shows how far a real gas is from an ideal gas) of the gas that came out
explain why the solution gas–oil ratio and the formation volume factor depend on the number of separation stages and on the separator pressure, and find them from the volumes of a separator test
convert differential liberation data to separator conditions and say where this conversion goes wrong
describe how viscosity is measured and how the constant volume depletion test works for gas condensate, and calculate the two-phase gas deviation factor

The PVT cell and the tests on a sample

The laboratory receives the sample in transport bottles (Reservoir fluid sampling). Before testing, it brings the sample back to its reservoir state: it heats it to reservoir temperature and compresses it above the bubble point pressure, mixing it until the gas that came out while the sample was brought up the well and transported has dissolved again. Then it checks that the sample is a good one: it compares the bubble point pressure of the sample at surface temperature with the check made at the well and with the duplicate samples (sample validation). It moves a portion of the good sample, tens of cubic centimetres, into the PVT cell.

The PVT cell is a thick-walled steel cylinder in a thermostat (a chamber that keeps the temperature constant), built for pressures of tens of megapascals. The laboratory changes the volume of the sample with a piston, or by pumping a displacement liquid in and out with a pump that has a precise volume scale (in the past, mercury was used for this). A gauge measures the pressure, and the contents are stirred so that the phases reach equilibrium (stop changing) faster. In a “blind” cell (one with no window) only the total volume is known. In a cell with a window of sapphire glass the volumes of the phases can be seen as well; such cells are needed for gas condensates and volatile oils (light oils that give off a lot of gas). The gas let out of the cell is expanded to standard conditions (a fixed reference pressure and temperature), and the laboratory measures its volume, density and composition.

Tests on an oil sample

Test: Constant-composition expansion
What is done to the sample: the pressure is lowered step by step at reservoir temperature; the gas stays in the cell
What it imitates: oil above the bubble point pressure; production in a closed system
Main results: $p_b$ , $V/V_b$ , $c_o$ , Y-function

Test: Differential liberation
What is done to the sample: at each pressure step all the gas that came out is let out
What it imitates: gas coming out of the oil in the reservoir below the bubble point pressure
Main results: $R_{sD}$ , $B_{oD}$ , $B_g$ , $Z$ , densities of the gas and of the oil

Test: Separator test
What is done to the sample: the oil at the bubble point pressure is pushed through the separation stages
What it imitates: treatment of the oil in the field
Main results: $R_{sb}$ , $B_{ob}$ , density of the stock-tank oil

Test: Viscosity
What is done to the sample: a viscometer at reservoir temperature and at different pressures
What it imitates: —
Main results: $\mu_o(p)$

Test: Composition
What is done to the sample: the sample is degassed, and the gas and the liquid are analysed with a chromatograph (an instrument that separates a mixture into its components)
What it imitates: —
Main results: molar composition up to C30+ (the last group holds all components with 30 or more carbon atoms), properties of the heavy fraction

Test	What is done to the sample	What it imitates	Main results
Constant-composition expansion	the pressure is lowered step by step at reservoir temperature; the gas stays in the cell	oil above the bubble point pressure; production in a closed system	$p_b$ , $V/V_b$ , $c_o$ , Y-function
Differential liberation	at each pressure step all the gas that came out is let out	gas coming out of the oil in the reservoir below the bubble point pressure	$R_{sD}$ , $B_{oD}$ , $B_g$ , $Z$ , densities of the gas and of the oil
Separator test	the oil at the bubble point pressure is pushed through the separation stages	treatment of the oil in the field	$R_{sb}$ , $B_{ob}$ , density of the stock-tank oil
Viscosity	a viscometer at reservoir temperature and at different pressures	—	$\mu_o(p)$
Composition	the sample is degassed, and the gas and the liquid are analysed with a chromatograph (an instrument that separates a mixture into its components)	—	molar composition up to C30+ (the last group holds all components with 30 or more carbon atoms), properties of the heavy fraction

In Russian-language practice the constant-composition test is also called contact liberation (the gas that came out stays in contact with the oil) or the study of the “pressure–volume” relation at reservoir temperature. The separator test is called stage separation of the sample, and its variant with one stage down to standard conditions is called single-stage (flash) liberation. The list of tests is fixed in the work order for the study.

SchematicThree tests on an oil sample (schematic): in the constant-composition test the gas stays in the cell; in differential liberation the gas of each stage is let out at constant pressure; in the separator test the oil at the bubble point pressure passes through the separation stages.

Besides the main tests, the laboratory finds the composition of the reservoir mixture. It works it out from the compositions of the gas and the liquid after a single flash of the sample (the sample is released to atmospheric pressure in one step). The laboratory separates the components up to C30+ and measures the molar mass and density of the heavy residue (Composition of reservoir fluids). These are the input data for the equation of state (a set of equations that predicts how the fluid behaves at different pressures and temperatures). If the work order asks for it, the laboratory also measures the wax appearance temperature (the temperature at which wax crystals begin to form) and the asphaltene onset pressure (the pressure at which asphaltenes, the heavy tar-like compounds, begin to drop out of the oil) (Retrograde condensation, wax, asphaltenes and high-pressure fluids). It can measure the interfacial tension (the tension at the boundary between two fluids), for example from the shape of a hanging drop (Oil viscosity and interfacial tension). How to read a complete report and check that its tables agree with each other is the subject of How to read a PVT report.

Constant-composition expansion: bubble point pressure and compressibility

Constant-composition expansion (CCE; also called constant mass expansion or the relative volume test) starts with a single-phase sample (all liquid, no free gas) at a pressure above the bubble point pressure. The laboratory heats the sample to reservoir temperature, keeping the pressure up; from the increase in volume it finds the thermal expansion coefficient of the oil (how much the oil swells when it is heated). Then it lowers the pressure in steps. At each step it increases the volume of the cell, stirs the contents until they reach equilibrium, and records the pressure and the volume. No gas is let out of the cell, so the composition of the system does not change.

While the oil is single-phase, it compresses very little, and a small increase in volume lowers the pressure a lot: on a graph of pressure against volume the points lie on a steep line. Below the bubble point pressure gas appears in the cell, the compressibility of the system rises sharply, and the line becomes flat. The kink is the bubble point pressure at reservoir temperature, $p_b$ , and the volume at the kink is $V_b$ . For a black oil (an ordinary oil with a moderate amount of dissolved gas) the kink is sharp. For a volatile oil the kink is smoothed out, and we refine the bubble point pressure with the Y-function (next section) or by watching through the window of the cell. We divide all volumes by $V_b$ , which gives the relative volume $V/V_b$ . It is less than one above the bubble point pressure and greater than one below it.

Oil compressibility above the bubble point pressure from two readings

c_o = -\frac{1}{V}\left(\frac{\partial V}{\partial p}\right)_T \approx \frac{V_2 - V_1}{\bar V\,(p_1 - p_2)}

where:

$c_o$ oil compressibility above the bubble point pressure, 1/MPa
$V_1,\ V_2$ volumes of the sample at pressures $p_1 > p_2$ (both above the bubble point pressure), cm³
$\bar V$ mean volume, $(V_1 + V_2)/2$ , cm³
$p_1,\ p_2$ pressures of the two readings, MPa

Units: MPa and 1/MPa; the volumes can be in any units, as long as both are in the same unit (relative volumes work too)

Holds when

The temperature is constant (reservoir temperature) and the oil is single-phase.
Between the two readings the compressibility changes little. Reports sometimes divide the change in volume not by the mean volume but by the volume at the higher pressure. The difference is about 1%, but you need to know it when you compare numbers.

Compressibility shows by what fraction the oil expands when the pressure falls by 1 MPa. For oils it is about $10^{-3}$ MPa⁻¹, and it usually grows as the pressure comes down towards the bubble point pressure. For simplicity, the lesson's assumed oil has a constant compressibility, $1.6\cdot 10^{-3}$ MPa⁻¹. We use compressibility to describe the elastic energy stored in an undersaturated reservoir (one whose pressure is above the bubble point pressure) (General material balance equation for oil reservoirs) and the density of the oil in the reservoir (Density and compressibility of reservoir oil).

In other unit systems (1)

Oilfield units

c_o\,[\text{psi}^{-1}] = 0.006895\;c_o\,[\text{MPa}^{-1}]

pressure in psi, compressibility in psi⁻¹.

1 psi = 0.006895 MPa: per 1 psi the volume changes 145 times less than per 1 MPa.

Oil formation volume factor above the bubble point pressure

B_o(p) = B_{ob}\,\frac{V(p)}{V_b},\qquad p \ge p_b

where:

$B_o(p)$ oil formation volume factor at pressure $p$ , m³/m³
$B_{ob}$ formation volume factor at the bubble point pressure from the separator test (section below), m³/m³
$V(p)/V_b$ relative volume from the constant-composition test, —

Units: m³ of reservoir oil per m³ of stock-tank oil

Holds when

Above the bubble point pressure the composition of the oil does not change, so the ratio of the volumes at $p$ and at $p_b$ is the same in the cell and in the reservoir.
$B_{ob}$ must be the value for the separation scheme to which the result should refer.

The constant-composition test gives the shape of the curve, and the separator test gives its level: the reservoir volume per 1 m³ of stock-tank oil at the bubble point pressure. Below the bubble point pressure the same product gives the total formation volume factor $B_t = B_{ob}\,V/V_b$ : in the cell there is the oil together with the gas that has come out.

Worked exampleCompressibility and formation volume factor of the oil above the bubble point pressure

Given (assumed data)

Sample: the oil from the lesson Black oil model: solution gas–oil ratio and formation volume factors; 55.16 cm³ loaded into the cell at 20 °C and 30 MPa
After heating to 85 °C at 30 MPa: 58.86 cm³
Readings at 85 °C: 25 MPa — 59.33 cm³; 20 MPa — 59.81 cm³; kink at 18.0 MPa and 60.00 cm³
Separator test: $B_{ob} = 1.350$ m³/m³ (see the section on separator tests)
Oil density at the bubble point pressure: 708 kg/m³

Solution

Thermal expansion on heating by 65 °C: $\beta = \dfrac{58.86 - 55.16}{57.01\cdot 65} = 1.00\cdot 10^{-3}$ 1/°C (the denominator uses the mean volume).
Relative volumes: $58.86/60.00 = 0.9810$ at 30 MPa, $59.33/60.00 = 0.9888$ at 25 MPa, $59.81/60.00 = 0.9968$ at 20 MPa.
Compressibility over the interval 30–18 MPa: $c_o = \dfrac{60.00 - 58.86}{59.43\cdot 12} = 1.60\cdot 10^{-3}$ MPa⁻¹, or $11.0\cdot 10^{-6}$ psi⁻¹. If the change in volume is divided by the volume at 30 MPa, the result is $1.61\cdot 10^{-3}$ MPa⁻¹.
Formation volume factor: $B_{oi} = 1.350\cdot 0.9810 = 1.324$ m³/m³ at 30 MPa and $1.350\cdot 0.9888 = 1.335$ m³/m³ at 25 MPa.
Oil density at 30 MPa: the mass of the sample does not change, so $\rho = 708/0.9810 = 722$ kg/m³.

Answer: The compressibility of the oil is about

1.60\cdot 10^{-3}

MPa⁻¹, the initial formation volume factor at 30 MPa is 1.324 m³/m³, and the density of the oil in the reservoir is about 722 kg/m³.

What it means. These are the same $B_{oi} = 1.324$ and $B_o = 1.335$ at 25 MPa that the examples of the lesson Black oil model: solution gas–oil ratio and formation volume factors use. The constant-composition test gave the shape of the curve above the bubble point pressure, and the separator test gave its level. The volumes are recorded to 0.01 cm³, so the compressibility from neighbouring readings, 5 MPa apart, has a scatter of about 1%; over the whole interval it is more precise. The thermal expansion coefficient is the one used for the samples of this oil in the lesson Reservoir fluid sampling. The data in this example are assumed for illustration.

Below the bubble point pressure: the Y-function

Below the bubble point pressure the cell holds oil and gas, and the relative volume grows quickly. For the lesson's assumed oil it is 1.16 at 12 MPa and 1.78 at 6 MPa. The two-phase readings are less accurate. Equilibrium is reached more slowly, and close to the bubble point pressure the increases in volume are small, so an error of a fraction of a cubic centimetre distorts the result noticeably. To check and smooth such readings, we use the Y-function, a number without units.

Y-function of two-phase readings

Y = \frac{p_b - p}{p\,\left(V/V_b - 1\right)}

where:

$Y$ Y-function, —
$p_b$ bubble point pressure at the temperature of the test, MPa
$p$ pressure of the reading, below the bubble point pressure, MPa
$V/V_b$ relative volume of the “oil + gas” system, —

Units: absolute pressures in any identical units; $Y$ is dimensionless

Holds when

Only for readings below the bubble point pressure.
For black oils $Y$ is almost linear in pressure. This is an empirical observation, not a law. For volatile oils, and close to the critical point, the line is curved.

The Y-function compares the increase in volume with the relative fall in pressure, so its values change smoothly, while the error of a single reading is magnified and shows up as a point away from the straight line. We draw the straight line $Y = a + b\,p$ through the reliable points and calculate the smoothed volumes from it: $V/V_b = 1 + (p_b - p)/(p\,Y)$ . A wrongly chosen bubble point pressure also shows up. If it is too high, the points close to $p_b$ bend steeply upward. If it is too low, they move down.

Worked exampleThe Y-function finds a wrong reading

Given (assumed data)

Constant-composition test, 85 °C: $p_b = 18.0$ MPa, $V_b = 60.00$ cm³
Readings below the bubble point pressure: 16 MPa — 62.06 cm³; 14 — 65.00; 12 — 69.34; 10 — 76.41; 8 — 86.95; 6 — 106.73 cm³
Separator test: $B_{ob} = 1.350$ m³/m³

Solution

Y for each reading. At 16 MPa: $V/V_b = 62.06/60.00 = 1.03433$ , $Y = \dfrac{18.0 - 16}{16\cdot 0.03433} = 3.641$ . In the same way at 14, 12, 10, 8 and 6 MPa: 3.429; 3.212; 2.925; 2.783; 2.568.
A straight line through the end points: $b = (3.641 - 2.568)/(16 - 6) = 0.1073$ MPa⁻¹, $a = 2.568 - 0.1073\cdot 6 = 1.9242$ .
The line at 14, 12, 10 and 8 MPa: 3.426; 3.212; 2.997; 2.783. Deviations of the readings from it: 0.1; 0.0; −2.4; 0.0%. Three points lie on the line; the reading at 10 MPa is 2.4% below it.
Smoothed volume at 10 MPa: $V/V_b = 1 + \dfrac{18.0 - 10}{10\cdot 2.997} = 1.2669$ , that is 76.01 cm³ instead of the recorded 76.41. The difference is 0.40 cm³, or 0.53% of the volume.
Total (two-phase) formation volume factor at 10 MPa: $B_t = 1.350\cdot 1.2669 = 1.710$ m³/m³ instead of 1.719 from the wrong reading.

Answer: The reading at 10 MPa is wrong: its Y-function is 2.4% below the line. The smoothed relative volume is 1.2669, and

B_t = 1.710

m³/m³.

What it means. An error of 0.53% in the volume gave a deviation of the Y-function 4.5 times larger. That is the point of the check. The cause may be a leak in the line of the displacement liquid, a mistake in reading, or a change in the temperature of the thermostat. The laboratory repeats the reading or replaces it with the smoothed value. The other points deviate from the line by no more than 0.1%; this is the reading noise of 0.01 cm³. The data in this example are assumed for illustration.

CalculatedConstant-composition test for the assumed oil at 85 °C. Top: the relative volume against pressure. Above the bubble point pressure it hardly changes; below it, it grows quickly as the pressure falls. Bottom: the Y-function of the two-phase readings and the straight line through the end points; the reading at 10 MPa lies below the line.

Differential liberation

Differential liberation (DL; also called differential vaporisation) starts with oil at the bubble point pressure and reservoir temperature. The laboratory lowers the pressure by one step and stirs the oil. Then it lets the gas that has come out leave the cell at constant pressure, by pushing in the piston or pumping in the displacement liquid. The distance the piston moves shows the volume the gas took in the cell. Outside the cell, the gas is expanded to standard conditions, and the laboratory measures its volume, density and composition. The oil that remains is no longer the same: it is a liquid that has lost some of its gas, and the next step starts with it. Usually the laboratory makes about ten steps, down to atmospheric pressure. The lesson's example has six, to keep it short.

After the last step the laboratory cools the oil, now at atmospheric pressure, to standard temperature. This is the residual oil. It is not the same as the stock-tank oil of the field (the oil left after the gas has come out at the surface). It comes from repeated degassing at reservoir temperature, not from separation at a low temperature, so there is less of it, it is heavier, and more gas came out per cubic metre of it. We give all the results of the test per 1 m³ of residual oil. That is why they carry the index D (differential): $R_{sD}$ , $B_{oD}$ .

Solution gas–oil ratio and oil volume in differential liberation

R_{sD}(p_k) = \frac{1}{V_{\text{r}}}\sum_{j > k} V_{g,j},\qquad B_{oD}(p_k) = \frac{V_o(p_k)}{V_{\text{r}}}

where:

$R_{sD}(p_k)$ gas that is still dissolved in the oil after step $k$ , at standard conditions, per 1 m³ of residual oil, m³/m³
$V_{g,j}$ gas let out at step $j$ , at standard conditions, cm³
$V_{\text{r}}$ volume of the residual oil at standard conditions, cm³
$V_o(p_k)$ volume of the oil in the cell after the gas is let out at step $k$ , at the pressure of the step and at reservoir temperature, cm³
$B_{oD}(p_k)$ relative volume of the oil per 1 m³ of residual oil, m³/m³

Units: gas at 0.101325 MPa and 20 °C; volume ratios in m³/m³

Holds when

The gas is let out completely and at the constant pressure of the step.
At the bubble point pressure $R_{sD} = R_{sDb}$ : all the gas that came out during the test. At the last step $R_{sD} = 0$ by definition.
The results refer to the residual oil, not to the stock-tank oil of separation.

The solution gas–oil ratio at a step is the gas that is still to come out at all the following steps. That is why we fill the table from the end: only after the last step do we know the volume of the residual oil and the total gas that came out.

Formation volume factor and gas deviation factor of the gas let out at a step

B_g = \frac{V_g(p, T)}{V_{g,\text{sc}}},\qquad Z = \frac{p\,V_g(p, T)\,T_{\text{sc}}}{p_{\text{sc}}\,V_{g,\text{sc}}\,T}

where:

$V_g(p, T)$ volume of the gas let out at the step, in the cell, at the pressure of the step and at reservoir temperature, cm³
$V_{g,\text{sc}}$ the same gas at standard conditions, cm³
$p,\ T$ pressure of the step and reservoir temperature, both absolute, MPa, K
$p_{\text{sc}},\ T_{\text{sc}}$ standard conditions: 0.101325 MPa and 293.15 K, MPa, K

Units: absolute pressures and temperatures; volumes in identical units

Holds when

At standard conditions we take the gas as ideal ( $Z_{\text{sc}} = 1$ ).
Only gas leaves the cell, with no drops of oil.

The formula for $Z$ is the equation of state of a real gas (Ideal and real gas: the gas deviation factor (Z)) written for one and the same amount of gas in two states. This is how the laboratory measures the gas deviation factor (a number that shows how far a real gas is from an ideal gas) and $B_g$ of the gas that came out in the reservoir, at each pressure (Gas formation volume factor, compressibility and viscosity).

Differential liberation of the lesson's assumed oil at 85 °C

$p$ , MPa: 18.0 ( $p_b$ )
$R_{sD}$ , m³/m³: 121.0
$B_{oD}$ , m³/m³: 1.392
$\gamma_g$ of the gas let out at the step: —

$p$ , MPa: 15.0
$R_{sD}$ , m³/m³: 103.8
$B_{oD}$ , m³/m³: 1.345
$\gamma_g$ of the gas let out at the step: 0.715

$p$ , MPa: 12.0
$R_{sD}$ , m³/m³: 86.3
$B_{oD}$ , m³/m³: 1.297
$\gamma_g$ of the gas let out at the step: 0.725

$p$ , MPa: 9.0
$R_{sD}$ , m³/m³: 68.4
$B_{oD}$ , m³/m³: 1.248
$\gamma_g$ of the gas let out at the step: 0.745

$p$ , MPa: 6.0
$R_{sD}$ , m³/m³: 49.8
$B_{oD}$ , m³/m³: 1.198
$\gamma_g$ of the gas let out at the step: 0.780

$p$ , MPa: 3.0
$R_{sD}$ , m³/m³: 30.2
$B_{oD}$ , m³/m³: 1.144
$\gamma_g$ of the gas let out at the step: 0.860

$p$ , MPa: 0.1
$R_{sD}$ , m³/m³: 0
$B_{oD}$ , m³/m³: 1.055
$\gamma_g$ of the gas let out at the step: 1.120

$p$ , MPa	$R_{sD}$ , m³/m³	$B_{oD}$ , m³/m³	$\gamma_g$ of the gas let out at the step
18.0 ( $p_b$ )	121.0	1.392	—
15.0	103.8	1.345	0.715
12.0	86.3	1.297	0.725
9.0	68.4	1.248	0.745
6.0	49.8	1.198	0.780
3.0	30.2	1.144	0.860
0.1	0	1.055	1.120

The data are assumed for illustration. $R_{sD}$ and $B_{oD}$ are per 1 m³ of residual oil at 20 °C, and $\gamma_g$ is the gas specific gravity (the density relative to air). At 0.1 MPa the oil is still hot (85 °C), so $B_{oD} > 1$ : thermal expansion remains. The gas let out at the last steps is heavier: at low pressure propane and butanes leave the oil. $B_g$ and $Z$ of the gas let out at a step are shown in the example below for 12 MPa.

Compare with the field values for the same oil: $R_{sDb} = 121.0$ against $R_{sb} = 110$ m³/m³, and $B_{oDb} = 1.392$ against $B_{ob} = 1.350$ . Repeated degassing at 85 °C sends more intermediate components (propane, butanes and pentanes) into the gas than two-stage separation at 30 and 20 °C, and less liquid is left. This is the usual result, which is why we do not take the differential data directly into calculations (see the section on conversion).

Worked exampleOne step of differential liberation

Given (assumed data)

Step 15 → 12 MPa, 85 °C: the gas that was let out took 7.86 cm³ in the cell and 0.876 L at standard conditions; 64.87 cm³ of oil remained in the cell
First step, 18 → 15 MPa: 0.858 L of gas let out
Result of the test: 6.050 L of gas let out in total; the residual oil at 20 °C is 50.00 cm³
Standard conditions: 0.101325 MPa and 20 °C (293.15 K); 85 °C = 358.15 K

Solution

Gas in the oil at the bubble point pressure: $R_{sDb} = 6050/50.00 = 121.0$ m³/m³.
After two steps the oil still holds $6050 - 858 - 876 = 4316$ cm³ of gas, so $R_{sD} = 4316/50.00 = 86.32$ m³/m³.
Relative volume of the oil: $B_{oD} = 64.87/50.00 = 1.297$ m³/m³.
Gas let out at the step: $B_g = 7.86/876 = 0.00897$ m³/m³; $Z = \dfrac{12\cdot 7.86\cdot 293.15}{0.101325\cdot 876\cdot 358.15} = 0.870$ .

Answer: At 12 MPa:

R_{sD} = 86.32

m³/m³,

B_{oD} = 1.297

m³/m³,

B_g = 0.00897

m³/m³,

Z = 0.870

What it means. This is the 12 MPa row of the table above. $B_g$ and $Z$ are the same as those used for this oil in the lesson Black oil model: solution gas–oil ratio and formation volume factors ( $B_g = 0.00897$ at $Z = 0.87$ ), but $R_{sD}$ and $B_{oD}$ are not: there $R_s \approx 76$ and $B_o = 1.258$ . The difference is not an error. It comes from a different base (residual oil) and a different path of degassing. How to convert to the field values is shown below. The data in this example are assumed for illustration.

The separator test: why the stages matter

The separator test shows how much stock-tank oil and gas the reservoir oil gives when it passes through a particular treatment scheme. The laboratory pushes the oil, at the bubble point pressure and reservoir temperature, out of the cell at constant pressure through one or several separation stages. A stage is a small vessel at a set pressure and temperature; the last stage works at atmospheric pressure, like the stock tank. At each stage the laboratory measures the volume and the density of the gas. At the end it measures the volume and the density of the stock-tank oil, brought to standard temperature.

Solution gas–oil ratio and formation volume factor from the separator test

R_{sb} = \frac{\sum_i V_{g,i}}{V_{\text{od}}},\qquad B_{ob} = \frac{V_b}{V_{\text{od}}}

where:

$V_{g,i}$ gas of the $i$ -th stage, including the stock tank, at standard conditions, cm³
$V_{\text{od}}$ stock-tank oil at standard conditions, cm³
$V_b$ oil pushed out of the cell, at the bubble point pressure and reservoir temperature, cm³
$R_{sb},\ B_{ob}$ solution gas–oil ratio and formation volume factor of the oil at the bubble point pressure for this separation scheme, m³/m³

Units: gas at 0.101325 MPa and 20 °C; m³/m³

Holds when

While it is pushed out, the oil in the cell stays single-phase: the pressure is kept at or above the bubble point pressure.
The pressures and temperatures of the stages are the same as in the field. Otherwise the result refers to another scheme.

These are the $R_{sb}$ and $B_{ob}$ of the black oil model (Black oil model: solution gas–oil ratio and formation volume factors) for the chosen treatment scheme. We add up the gas of all the stages, because the gas of the stock tank was also dissolved in the reservoir oil.

Why the stages matter. Propane, butanes and pentanes divide between the gas and the liquid according to pressure and temperature. When the pressure is lowered at once to atmospheric, a large share of them goes into the gas: there is more gas, less oil, and the oil is heavier. With two or three stages, the first one removes mostly methane and ethane at a moderate pressure, the intermediate components stay in the liquid, and less of them goes into the gas in the stock tank. But if the pressure of the first stage is too high, a lot of methane stays in the oil, comes out in the stock tank and carries the intermediate components away again. Between the two extremes there is an optimum pressure of the first stage: the solution gas–oil ratio and the formation volume factor are the smallest, the amount of oil is the largest, and the oil is the lightest (Separator calculations and optimum separator pressure).

Separator tests of the assumed oil: first stage at 30 °C, stock tank at 0.1 MPa and 20 °C; pressures are absolute

First stage, MPa: single flash to 0.1
$R_{sb}$ , m³/m³: 125.0
$B_{ob}$ , m³/m³: 1.380
Stock-tank oil, kg/m³: 855

First stage, MPa: 0.35
$R_{sb}$ , m³/m³: 112.0
$B_{ob}$ , m³/m³: 1.357
Stock-tank oil, kg/m³: 851

First stage, MPa: 0.7
$R_{sb}$ , m³/m³: 110.0
$B_{ob}$ , m³/m³: 1.350
Stock-tank oil, kg/m³: 850

First stage, MPa: 1.5
$R_{sb}$ , m³/m³: 110.5
$B_{ob}$ , m³/m³: 1.351
Stock-tank oil, kg/m³: 852

First stage, MPa: 3.0
$R_{sb}$ , m³/m³: 114.5
$B_{ob}$ , m³/m³: 1.360
Stock-tank oil, kg/m³: 855

First stage, MPa	$R_{sb}$ , m³/m³	$B_{ob}$ , m³/m³	Stock-tank oil, kg/m³
single flash to 0.1	125.0	1.380	855
0.35	112.0	1.357	851
0.7	110.0	1.350	850
1.5	110.5	1.351	852
3.0	114.5	1.360	855

The data are assumed for illustration. The single flash and the two-stage scheme with the first stage at 0.7 MPa are the ones compared in the lesson on the black oil model. The densities of the stock-tank oil agree with the mass balance at a reservoir oil density of 708 kg/m³ (example below).

Worked exampleHow much oil and gas two-stage separation gives

Given (assumed data)

Pushed out of the cell at 18.0 MPa and 85 °C: 60.00 cm³ of oil
First stage, 0.7 MPa and 30 °C: gas 4.133 L, specific gravity 0.745
Stock tank, 0.1 MPa and 20 °C: gas 0.756 L, specific gravity 1.10; stock-tank oil 44.44 cm³, 850 kg/m³
Densities: reservoir oil at the bubble point pressure 708 kg/m³; air at standard conditions 1.204 kg/m³

Solution

Gas per 1 m³ of stock-tank oil: first stage $4133/44.44 = 93.0$ , stock tank $756/44.44 = 17.0$ ; in total $R_{sb} = 110.0$ m³/m³.
Formation volume factor: $B_{ob} = 60.00/44.44 = 1.350$ m³/m³.
Mean gas specific gravity: $(4.133\cdot 0.745 + 0.756\cdot 1.10)/4.889 = 0.800$ .
Mass balance per 1 m³ of stock-tank oil: reservoir oil $1.350\cdot 708 = 955.8$ kg; products $850 + 110.0\cdot 0.800\cdot 1.204 = 956.0$ kg; the discrepancy is 0.02%.
A single flash of the same oil (table) gives $B_{ob} = 1.380$ . From 1 m³ of reservoir oil we get $1/1.350 = 0.7407$ m³ of stock-tank oil, against $1/1.380 = 0.7246$ m³ for the single flash. That is 2.2% more.

Answer: With two-stage separation

R_{sb} = 110.0

m³/m³ and

B_{ob} = 1.350

m³/m³. This scheme gives 2.2% more sales oil (oil ready for sale) than a single flash.

What it means. For a field that produces millions of tonnes a year, 2% is tens of thousands of tonnes of oil every year from the same reservoir. That is why engineers choose the pressures of the stages by calculation and check them with separator tests. The mass balance is a simple check of a test: if the two sides differ by a few tenths of a per cent or more, look for an error in the volumes or the densities. The data in this example are assumed for illustration.

Converting differential data to separator conditions

Calculations need $R_s$ and $B_o$ per 1 m³ of the field's stock-tank oil, at any reservoir pressure. The tests give them in parts. The separator test gives them only at the bubble point pressure. Differential liberation gives them at all pressures, but on a different base (per 1 m³ of residual oil, not of stock-tank oil). We combine them by assuming that, for the oil that remains in the reservoir at pressure $p$ , the ratio of residual oil to stock-tank oil is the same as for the oil at the bubble point pressure.

Conversion of differential data to separator conditions

B_o = B_{oD}\,\frac{B_{ob}}{B_{oDb}},\qquad R_s = R_{sb} - \left(R_{sDb} - R_{sD}\right)\frac{B_{ob}}{B_{oDb}},\qquad p < p_b

where:

$B_o,\ R_s$ formation volume factor and solution gas–oil ratio per 1 m³ of stock-tank oil of the chosen separation scheme, m³/m³
$B_{oD},\ R_{sD}$ the same quantities from differential liberation at pressure $p$ , per 1 m³ of residual oil, m³/m³
$B_{oDb},\ R_{sDb}$ the same at the bubble point pressure, m³/m³
$B_{ob},\ R_{sb}$ results of the separator test for the chosen scheme, m³/m³

Units: all quantities in m³/m³; in oilfield units (rb/STB, scf/STB) the formulas are the same

Holds when

We take the ratio $B_{ob}/B_{oDb}$ (the cubic metres of residual oil per cubic metre of stock-tank oil obtained from the same reservoir oil) to be the same at all pressures. This is an approximation, and close to atmospheric pressure it breaks down (advanced section).
We convert the gas that came out in the reservoir ( $R_{sDb} - R_{sD}$ per 1 m³ of residual oil) to stock-tank oil with the same factor.
Above the bubble point pressure $R_s = R_{sb}$ , and $B_o = B_{ob}\,V/V_b$ from the constant-composition test.

The factor $B_{ob}/B_{oDb}$ is less than one: the same volume of reservoir oil gives less residual oil than stock-tank oil in separation. The first formula converts the oil volume in the reservoir to the new base. The second subtracts from the field solution gas–oil ratio the gas that has already left the oil in the reservoir.

Worked exampleA table of solution gas–oil ratio and formation volume factor for calculations

Given (assumed data)

Differential liberation, 85 °C: $R_{sDb} = 121.0$ , $B_{oDb} = 1.392$ ; at 12 MPa $R_{sD} = 86.3$ , $B_{oD} = 1.297$ ; at 6 MPa $R_{sD} = 49.8$ , $B_{oD} = 1.198$ (all in m³/m³)
Separator test, first stage 0.7 MPa: $R_{sb} = 110.0$ m³/m³, $B_{ob} = 1.350$ m³/m³
Constant-composition test: $V/V_b = 0.9810$ at 30 MPa and 1.1557 at 12 MPa
Gas let out at the 12 MPa step: $B_g = 0.00897$ m³/m³

Solution

The factor: $B_{ob}/B_{oDb} = 1.350/1.392 = 0.9698$ .
12 MPa: $B_o = 1.297\cdot 0.9698 = 1.258$ m³/m³; $R_s = 110.0 - (121.0 - 86.3)\cdot 0.9698 = 76.3$ m³/m³.
6 MPa: $B_o = 1.198\cdot 0.9698 = 1.162$ m³/m³; $R_s = 110.0 - (121.0 - 49.8)\cdot 0.9698 = 41.0$ m³/m³.
30 MPa, above the bubble point pressure: $B_{oi} = 1.350\cdot 0.9810 = 1.324$ m³/m³.
Check at 12 MPa: from the converted data $B_t = 1.258 + 0.00897\cdot(110.0 - 76.3) = 1.560$ m³/m³; from the constant-composition test $B_{ob}\,V/V_b = 1.350\cdot 1.1557 = 1.560$ m³/m³.
The same formula at atmospheric pressure, where $R_{sD} = 0$ : $R_s = 110.0 - 121.0\cdot 0.9698 = -7.3$ m³/m³.

Answer: At 12 MPa

R_s = 76.3

m³/m³ and

B_o = 1.258

m³/m³; at 6 MPa, 41.0 m³/m³ and 1.162 m³/m³; at 30 MPa

B_{oi} = 1.324

m³/m³.

What it means. These are the values of the assumed oil in the lesson Black oil model: solution gas–oil ratio and formation volume factors (76 m³/m³ and 1.258; 41 m³/m³ and 1.162; 1.324), to within the rounding of the laboratory data. If we had taken the differential data directly, the solution gas–oil ratio at 12 MPa would be too high by 13%, the formation volume factor by 3.1%, and the oil in place calculated with $B_{oDb}\,V/V_b = 1.366$ instead of 1.324 would be too low by 3%. Two independent tests gave the same $B_t$ , which is a sign of consistent data. The negative solution gas–oil ratio at atmospheric pressure comes from the assumption of a constant factor (advanced section); there we take $R_s = 0$ . The data in this example are assumed for illustration.

CalculatedSolution gas–oil ratio (top) and oil formation volume factor (bottom) of the assumed oil: the differential liberation data per 1 m³ of residual oil (dashed lines) and the same data after conversion to two-stage separation, per 1 m³ of stock-tank oil (solid). Above the bubble point pressure of 18 MPa the solution gas–oil ratio is constant, and the formation volume factor follows the constant-composition test.

Oil viscosity

The laboratory measures the viscosity of reservoir oil at reservoir temperature and at several pressures. Above the bubble point pressure it uses a single-phase sample. Below it, it uses the oil from differential liberation: at each pressure it feeds the viscometer (the instrument that measures viscosity) with the equilibrium oil of the corresponding step, because this is the oil that stays in the reservoir. Gas viscosity is usually not measured but calculated from the composition or the density (Gas formation volume factor, compressibility and viscosity).

Three types of instruments are common. In a rolling-ball viscometer a ball rolls down an inclined tube filled with oil, and the rolling time is proportional to the viscosity (with a correction for the difference between the densities of the ball and the oil). In a capillary viscometer the oil is pumped at a constant rate through a thin tube, and the laboratory measures the pressure drop. In an electromagnetic viscometer a small piston moves in a chamber under a magnetic field, and the viscosity comes from the time of its stroke. All three need calibration with liquids of known viscosity.

Laminar flow in a capillary (Hagen–Poiseuille equation)

\mu = \frac{\pi\,d^4\,\Delta p}{128\,L\,q}

where:

$\mu$ dynamic viscosity of the liquid, Pa·s
$d$ inner diameter of the capillary, m
$\Delta p$ pressure drop across the capillary, Pa
$L$ length of the capillary, m
$q$ volumetric flow rate at the conditions of the test, m³/s

Units: SI

Holds when

The flow is laminar (smooth, in layers) and steady: the Reynolds number $\mathrm{Re} = 4\rho q/(\pi d\mu)$ , which shows whether a flow is smooth or turbulent, is much less than 2000.
The liquid is single-phase and Newtonian (its viscosity does not depend on the rate of shear). The effects of the inlet and the outlet of the capillary are small or are included in the calibration.
The pressure in the capillary is above the bubble point pressure of the liquid, otherwise gas comes out.

At a given flow rate the viscosity is proportional to the pressure drop. The diameter enters in the fourth power: an error of 0.5% in the diameter gives an error of about 2% in the viscosity. That is why we do not calculate the constant $\pi d^4/(128 L)$ from the data sheet of the capillary but find it by calibration.

Worked exampleOil viscosity from a capillary viscometer

Given (assumed data)

Capillary, from its data sheet: inner diameter 0.400 mm, length 2.000 m
Flow rate in both tests: 0.600 cm³/min
Calibration: a liquid of viscosity 0.700 mPa·s gives a pressure drop of 21.8 kPa
Sample: oil at 85 °C and 25 MPa gives a pressure drop of 34.8 kPa; oil density 716 kg/m³

Solution

At the same flow rate the viscosities are in the ratio of the pressure drops: $\mu = 0.700\cdot 34.8/21.8 = 1.117$ mPa·s.
From the diameter on the data sheet: $q = 0.600\ \text{cm}^3/\text{min} = 1.00\cdot 10^{-8}$ m³/s; $\mu = \dfrac{\pi\cdot (0.400\cdot 10^{-3})^4\cdot 34{,}800}{128\cdot 2.000\cdot 1.00\cdot 10^{-8}} = 1.093$ mPa·s, which is 2.2% lower.
The calibration “sees” the effective diameter: $d^4 = \dfrac{128\cdot 2.000\cdot 1.00\cdot 10^{-8}\cdot 0.700\cdot 10^{-3}}{\pi\cdot 21{,}800} = 2.617\cdot 10^{-14}$ m⁴, so $d = 0.402$ mm instead of 0.400 mm.
Flow regime: $\mathrm{Re} = \dfrac{4\cdot 716\cdot 1.00\cdot 10^{-8}}{\pi\cdot 4.02\cdot 10^{-4}\cdot 1.117\cdot 10^{-3}} = 20$ : the flow is laminar.

Answer: The viscosity of the oil at 85 °C and 25 MPa is 1.117 mPa·s.

What it means. Without calibration the error would be 2.2%, only because the diameter differs from the data sheet by 0.5%. The measured viscosity agrees with the curve of the assumed oil in the table below. The data in this example are assumed for illustration.

Viscosity of the assumed oil at 85 °C

$p$ , MPa: 30
Viscosity, mPa·s: 1.20
What happens: single-phase oil, compressed

$p$ , MPa: 25
Viscosity, mPa·s: 1.12
What happens: the oil expands as the pressure falls

$p$ , MPa: 18 ( $p_b$ )
Viscosity, mPa·s: 1.00
What happens: minimum: all the gas is dissolved

$p$ , MPa: 12
Viscosity, mPa·s: 1.30
What happens: gas leaves and the oil gets thicker

$p$ , MPa: 6
Viscosity, mPa·s: 1.96
What happens: less than half of the gas remains in the oil

$p$ , MPa: 0.1
Viscosity, mPa·s: 5.53
What happens: residual oil, still hot

$p$ , MPa	Viscosity, mPa·s	What happens
30	1.20	single-phase oil, compressed
25	1.12	the oil expands as the pressure falls
18 ( $p_b$ )	1.00	minimum: all the gas is dissolved
12	1.30	gas leaves and the oil gets thicker
6	1.96	less than half of the gas remains in the oil
0.1	5.53	residual oil, still hot

The data are assumed for illustration; the same curve is used in the lesson Performance prediction for solution gas drive reservoirs. The viscosity is lowest at the bubble point pressure: above it the oil is compressed, and below it the oil loses the dissolved gas that thins it. How to estimate the viscosity of oil without tests is the subject of the lesson Oil viscosity and interfacial tension.

Gas condensate: dew point and constant volume depletion

For a gas condensate the bubble point pressure is replaced by the dew point, the pressure at which the first drops of liquid come out of the gas (Phase behaviour: pure substances and binary mixtures). We can hardly find it from the “pressure–volume” curve, because the first drops of liquid hardly change the compressibility of the system. That is why the laboratory runs the constant-composition test in a cell with a window and finds the dew point by eye, from the appearance of a mist and a film of liquid on the glass. Above the dew point this test gives the gas deviation factor of the reservoir gas. Below it, the test gives the volume of the liquid that dropped out, which means little for the reservoir: in the reservoir the system does not stay closed.

Development by depletion (the field produces and the pressure falls, with no injection) is imitated by constant volume depletion (CVD; in Russian-language practice a test close in meaning is called differential condensation). The sample at the dew point takes the volume $V_d$ . The laboratory lowers the pressure by one step, expanding the cell, and then lets gas out at constant pressure until the volume is back to $V_d$ . The liquid that dropped out stays in the cell, like immobile condensate in the pores, and the gas that was let out plays the role of production: the laboratory measures its amount and composition. At each step it records the share of liquid in the volume $V_d$ . The test goes on down to a pressure of a few megapascals.

The condensate dropout curve is the share of liquid in the hydrocarbon pore volume, plotted against pressure (assumed curve). It shows how much condensate will stay in the reservoir.
The amount of material taken out, and its composition, at each step. From them we calculate the production of gas and condensate in development by depletion, and the fall of the condensate–gas ratio (the amount of condensate per unit volume of gas).
The two-phase gas deviation factor $Z_{2\text{ph}}$ of the whole mixture in the cell. It is used in the material balance of a gas-condensate reservoir below the dew point (Material balance for gas reservoirs).

Two-phase gas deviation factor in a constant volume test

Z_{2\text{ph}} = \frac{p\,V_d}{(n_0 - n_p)\,R\,T},\qquad n_0 = \frac{p_d\,V_d}{Z_d\,R\,T}

where:

$Z_{2\text{ph}}$ two-phase gas deviation factor of the mixture in the cell, —
$p$ pressure of the step, absolute, MPa
$V_d$ volume of the cell at the dew point, cm³
$n_0,\ n_p$ amount of substance in the cell at the dew point, and the accumulated amount of gas let out, mol
$R$ universal gas constant, 8.314, J/(mol·K)
$T$ temperature of the test, K
$p_d,\ Z_d$ dew point pressure and the gas deviation factor of the gas at it, MPa, —

Units: MPa·cm³ = J, so pressure in MPa and volume in cm³ give the amount of substance in moles

Holds when

After the gas is let out, the volume of the cell is again equal to $V_d$ , like the pore volume of a reservoir without water influx.
$n_0 - n_p$ includes both the gas and the liquid that remain in the cell.

This is the equation of state of a real gas written for the whole two-phase mixture. At constant volume it gives at once $p/Z_{2\text{ph}} = (p_d/Z_d)\,(1 - n_p/n_0)$ . In the coordinates “ $p/Z_{2\text{ph}}$ against the share of material taken out” the test lies on a straight line, like a gas reservoir on the $p/Z$ plot.

Constant volume depletion of the assumed gas condensate: dew point 36.0 MPa, 95 °C

$p$ , MPa: 36.0
$Z_{2\text{ph}}$: 1.050
Taken out, % of the material: 0
Liquid, % of the volume $V_d$: 0

$p$ , MPa: 30
$Z_{2\text{ph}}$: 1.000
Taken out, % of the material: 12.5
Liquid, % of the volume $V_d$: 8.4

$p$ , MPa: 24
$Z_{2\text{ph}}$: 0.955
Taken out, % of the material: 26.7
Liquid, % of the volume $V_d$: 11.6

$p$ , MPa: 18
$Z_{2\text{ph}}$: 0.905
Taken out, % of the material: 42.0
Liquid, % of the volume $V_d$: 11.9

$p$ , MPa: 12
$Z_{2\text{ph}}$: 0.855
Taken out, % of the material: 59.1
Liquid, % of the volume $V_d$: 10.9

$p$ , MPa: 7
$Z_{2\text{ph}}$: 0.810
Taken out, % of the material: 74.8
Liquid, % of the volume $V_d$: 9.7

$p$ , MPa	$Z_{2\text{ph}}$	Taken out, % of the material	Liquid, % of the volume $V_d$
36.0	1.050	0	0
30	1.000	12.5	8.4
24	0.955	26.7	11.6
18	0.905	42.0	11.9
12	0.855	59.1	10.9
7	0.810	74.8	9.7

The data are assumed for illustration. We take the share of liquid from the dropout curve of the lesson's assumed gas condensate (Types of reservoir fluids); at 36.0 MPa the table gives the single-phase $Z$ of the gas. We calculate the share of material taken out from the two-phase $Z$ with the relation under the equation above.

Worked exampleTwo-phase gas deviation factor at a depletion step

Given (assumed data)

Gas condensate: dew point 36.0 MPa at 95 °C (368.15 K); $Z_d = 1.05$ from the constant-composition test
Cell: $V_d = 250$ cm³
Step 18 MPa: 1.176 mol of gas has been let out since the start of the test; the liquid in the cell is 11.9% of the volume $V_d$

Solution

At the dew point: $n_0 = \dfrac{36.0\cdot 250}{1.05\cdot 8.314\cdot 368.15} = 2.800$ mol.
What remains is $2.800 - 1.176 = 1.624$ mol; the fraction taken out is $1.176/2.800 = 0.420$ , that is 42.0% of the material.
$Z_{2\text{ph}} = \dfrac{18.0\cdot 250}{1.624\cdot 8.314\cdot 368.15} = 0.905$ .
Check of the line: $18.0/0.905 = 19.89$ MPa and $(36.0/1.05)\cdot(1 - 0.420) = 19.89$ MPa.
The cell holds $0.119\cdot 250 = 29.8$ cm³ of liquid that has dropped out.

Answer: At the 18 MPa step

Z_{2\text{ph}} = 0.905

; 42.0% of the original material has been taken out, and 29.8 cm³ of condensate is left in the “reservoir”.

What it means. Below the dew point the single-phase $Z$ no longer describes the mixture in the reservoir: part of the material has turned into liquid, and the composition of the gas has changed. The two-phase $Z$ refers to all the material in the pore volume. On the $p/Z$ plot it also gives a straight line for a reservoir below the dew point, if the condensate in the reservoir cannot move and no water comes in, as in the test. The data in this example are assumed for illustration.

Terms and symbols

PVT cell: A high-pressure vessel kept at a constant temperature. The laboratory changes the volume of the sample in it and measures the pressure, the volumes of the phases and the amount of gas let out.
Constant-composition expansion (CCE): A test in which the pressure of the sample is lowered step by step at reservoir temperature, and no gas is let out. It gives the bubble point pressure, the relative volume and the compressibility of the oil.
Relative volume $V/V_b$: The volume of the sample (the oil together with the gas that came out) at pressure $p$ , divided by its volume at the bubble point pressure.
Y-function: The ratio $(p_b - p)/[p\,(V/V_b - 1)]$ . For black oils it is almost a straight line against pressure, and we use it to check and smooth the two-phase readings.
Differential liberation (DL): A test in which the pressure is lowered step by step at reservoir temperature, and all the gas that comes out is let out of the cell at each step.
Residual oil: The oil left after the last step of differential liberation, brought to standard temperature. It is the base of $R_{sD}$ and $B_{oD}$ .
Separator test: A test in which the oil at the bubble point pressure is pushed through separation stages at set pressures and temperatures. It gives $R_{sb}$ and $B_{ob}$ for this scheme.
Constant volume depletion (CVD): A test on gas condensate. The pressure is lowered in steps, and gas is let out so that the volume of the cell stays equal to the volume at the dew point.
Two-phase gas deviation factor $Z_{2\text{ph}}$: The value of $pV/(nRT)$ for the whole gas–liquid mixture in the volume. We use it in the material balance of a gas-condensate reservoir below the dew point.

Common mistakes and misconceptions

Wrong

“The solution gas–oil ratio of the oil is 121 m³/m³: that is what the differential liberation table says.”

In fact

121 m³/m³ is the gas per 1 m³ of residual oil after degassing at 85 °C. The field solution gas–oil ratio of the same oil with two-stage separation is 110 m³/m³. The table is converted to separator conditions.

Wrong

“The relative volume 1.1557 at 12 MPa, multiplied by $B_{ob}$ , is the formation volume factor of the oil at 12 MPa.”

In fact

Below the bubble point pressure the cell holds oil and gas, so $B_{ob}\,V/V_b = 1.560$ is the total (two-phase) formation volume factor $B_t$ . The formation volume factor of the oil alone at 12 MPa is 1.258.

Wrong

“The higher the pressure of the first separation stage, the more oil.”

In fact

If the pressure of the first stage is too high, a lot of methane stays in the oil, comes out in the stock tank and carries the intermediate components away. The yield of oil is greatest at the optimum pressure of the first stage, which for the lesson's assumed oil is about 0.7 MPa.

Wrong

“The dew point of a gas condensate is found at the kink of the pressure–volume curve, like the bubble point pressure of an oil.”

In fact

There is almost no kink: the first drops hardly change the compressibility of the system. The dew point is found by eye in a cell with a window.

Wrong

“A negative solution gas–oil ratio after conversion at low pressure is a laboratory error.”

In fact

It follows from the assumption of a constant factor $B_{ob}/B_{oDb}$ . At atmospheric pressure we take the solution gas–oil ratio as zero, and exact tables come from separator tests on the oil of individual steps or from an equation of state tuned to the test results.

In depth: advanced detail

Where the conversion formulas come from and why they fail near atmospheric pressure

Take an oil that occupies the volume $V_b$ at the bubble point pressure. The differential test turns it into $V_b/B_{oDb}$ of residual oil, and separation turns it into $V_b/B_{ob}$ of stock-tank oil. So $B_{ob}/B_{oDb}$ is the number of cubic metres of residual oil per cubic metre of stock-tank oil, obtained from the same reservoir oil. If at pressure $p$ there remain $B_{oD}$ cubic metres of oil in the reservoir per cubic metre of residual oil, then per cubic metre of stock-tank oil there are $B_{oD}\,B_{ob}/B_{oDb}$ . This is the first formula. The gas that came out in the reservoir down to pressure $p$ is $R_{sDb} - R_{sD}$ per cubic metre of residual oil, or $(R_{sDb} - R_{sD})\,B_{ob}/B_{oDb}$ per cubic metre of stock-tank oil. It is subtracted from the field $R_{sb}$ . This is the second formula.

The hidden assumption is that the factor $B_{ob}/B_{oDb}$ is the same at all pressures. But the less gas is left in the oil, the less its residual form and its stock-tank form differ. From an almost degassed oil, both the differential test and the separator get almost the same liquid, and the true factor approaches one. The formula keeps it constant, so at low pressures it underestimates both the solution gas–oil ratio (down to negative values) and the formation volume factor. In the lesson's example at atmospheric pressure $B_o = 1.055\cdot 0.9698 = 1.023$ , although a cubic metre of stock-tank oil at 85 °C and atmospheric pressure takes 1.05 m³, from thermal expansion alone.

That is why we trust the converted data less at pressures below a few megapascals, and at atmospheric pressure we take $R_s = 0$ . For exact tables, the laboratory runs separator tests on the oil of individual steps of differential liberation. Or it tunes an equation of state to all the tests (adjusts it until it matches them) and calculates the tables with it (Cubic equations of state). Other conversion formulas have been proposed, but they are also approximate. The report states which one was used (How to read a PVT report).

Connections to other topics

Reservoir fluid sampling — how the samples that the laboratory works with are taken and checked.
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.
PVT correlations for oil and gas condensate — how to estimate the same properties without samples, and how to check correlations against laboratory data.
How to read a PVT report — how to read a complete report and check that its tables agree with each other.
Types of reservoir fluids — how a gas condensate differs from an oil; the condensate dropout curve.
Cubic equations of state — how an equation of state is tuned to the results of the tests.
Material balance for gas reservoirs — where the two-phase gas deviation factor is used.
Separator calculations and optimum separator pressure — calculation of the separation stages and of the optimum pressure.

Check yourself

1
In which test does the gas that has come out stay in contact with the oil until the end of the test?
1. A.In constant-composition expansion
2. B.In the first stage of a separator test
3. C.In constant volume depletion
4. D.In differential liberation
Show the answer
Correct answer: A. In constant-composition expansion
In constant-composition expansion no gas is let out of the cell: the system is closed, like the path of the production from the bottom of the well to the separator. In differential liberation and in the constant volume test the gas is let out at each step, and in the separator test it is drawn off from each separation stage.
2
In the constant-composition test the relative volume of the oil at 28 MPa is 0.9832, and the bubble point pressure is 20 MPa. The separator test gave $B_{ob} = 1.42$ . Find $B_o$ at 28 MPa and the mean compressibility of the oil over the interval 20–28 MPa.
Show the answer
$B_o = 1.42\cdot 0.9832 = 1.396$ m³/m³. Compressibility: $c_o = \dfrac{1 - 0.9832}{0.9916\cdot 8} = 2.12\cdot 10^{-3}$ MPa⁻¹, where 0.9916 is the mean relative volume. The volumes can be relative ones: the units cancel.
3
Differential liberation: $R_{sDb} = 150$ , $B_{oDb} = 1.50$ ; at 10 MPa $R_{sD} = 95$ , $B_{oD} = 1.36$ (m³/m³). Separator test: $R_{sb} = 135$ m³/m³, $B_{ob} = 1.44$ . Find $R_s$ and $B_o$ at 10 MPa.
Show the answer
The factor is $1.44/1.50 = 0.96$ . $B_o = 1.36\cdot 0.96 = 1.306$ m³/m³; $R_s = 135 - (150 - 95)\cdot 0.96 = 135 - 52.8 = 82.2$ m³/m³. Without the conversion the solution gas–oil ratio would be too high by 16%, and the formation volume factor by 4%.
4
Why is $R_{sDb}$ of differential liberation larger than $R_{sb}$ of two-stage separation for the same oil?
1. A.In the separator the gas dissolves back into the oil
2. B.In repeated degassing at reservoir temperature more intermediate components go into the gas, and the base is a smaller volume of residual oil
3. C.The differential test is run at a higher pressure
4. D.The laboratory made a mistake in the volume of the gas
Show the answer
Correct answer: B. In repeated degassing at reservoir temperature more intermediate components go into the gas, and the base is a smaller volume of residual oil
High temperature and repeated degassing send more propane and butanes into the gas, so there is more gas and less liquid. Both effects increase the gas per cubic metre of residual oil. For the same reason $B_{oDb} > B_{ob}$ .
5
In a constant volume test on a gas condensate at 12 MPa the two-phase $Z = 0.855$ ; at the dew point of 36 MPa $Z_d = 1.05$ . What share of the original material has been taken out by this step?
Show the answer
At constant volume $p/Z_{2\text{ph}} = (p_d/Z_d)(1 - n_p/n_0)$ , so $n_p/n_0 = 1 - \dfrac{12}{36}\cdot\dfrac{1.05}{0.855} = 1 - 0.409 = 0.591$ , that is about 59%.
6
The Y-function of two-phase readings lies on a straight line, but the points nearest to the bubble point pressure go steeply up. What does this mean?
1. A.The oil is volatile
2. B.The readings at low pressure are wrong
3. C.The bubble point pressure was chosen too high
4. D.The bubble point pressure was chosen too low
Show the answer
Correct answer: C. The bubble point pressure was chosen too high
If $p_b$ is too high, the numerator $p_b - p$ at the nearest points does not become small, while the denominator is small, so Y grows sharply. If $p_b$ is too low, the points move down. We refine the bubble point pressure until the nearest points also lie on the line.

Updated: 16 October 2026