Density and compressibility of reservoir oil

Quick refresher

In Russian-language practice the density of stock-tank oil is given in kg/m³ at 20 °C. In international practice it is given as specific gravity $\gamma_o$ (density relative to water) at 60 °F (15.6 °C), or in degrees API. Correct it to the temperature you need before comparing values or using correlations.
Mass does not change on the way from the reservoir to the surface: $\rho_o = (\rho_{o,\text{sc}} + R_s\rho_{g,\text{sc}})/B_o$ . Reservoir oil is lighter than stock-tank oil, because the dissolved gas adds more to the volume than to the mass.
Above the bubble point pressure, oil is compressed and becomes denser as the pressure rises. Below it, as the pressure falls, the oil loses gas and also becomes denser. The density is lowest at the bubble point pressure, where $B_o$ is highest.
The pressure gradient in the oil zone is $\rho_o g$ . That is 9.81 kPa/m for every g/cm³ of density. For the oil of the lesson it is about 7.1 kPa/m.
The compressibility of undersaturated oil is $(0.7\text{–}3)\cdot 10^{-3}$ MPa⁻¹. It is higher for light, gas-rich oils, at high temperature and near the bubble point pressure. Below the bubble point pressure, the apparent compressibility, which includes the gas coming out of solution, is tens of times greater.
The thermal expansion of oil is about $10^{-3}$ 1/°C: heating by 10 °C increases the volume of oil by about 1%. A pressure drop of 6 MPa does the same.

Key relationships

\rho_o = \dfrac{\rho_{o,\text{sc}} + R_s\,\rho_{g,\text{sc}}}{B_o}

reservoir oil density from the mass balance; $\rho$ in kg/m³, $R_s$ and $B_o$ in m³/m³

G_o = \rho_o\,g

pressure gradient in the oil zone; 9.81 kPa/m per 1 g/cm³

c_o = -\dfrac{1}{B_o}\left(\dfrac{\partial B_o}{\partial p}\right)_T = \dfrac{1}{\rho_o}\left(\dfrac{\partial \rho_o}{\partial p}\right)_T

oil compressibility above the bubble point pressure

B_o = B_{ob}\left(\dfrac{p_b}{p}\right)^{A},\qquad A = c_o\,p

formation volume factor above the bubble point pressure from the Vasquez–Beggs correlation

c_o^{\text{app}} = -\dfrac{1}{B_o}\dfrac{\partial B_o}{\partial p} + \dfrac{B_g}{B_o}\dfrac{\partial R_s}{\partial p}

apparent compressibility below the bubble point pressure

Common pitfalls

Putting the density at 20 °C, instead of at 60 °F, into the °API formula. For the lesson oil this gives an error of about 0.6 °API.
Using the density of stock-tank oil, instead of reservoir oil, for the gradient and for the pressure of an oil column. This gives an error of 15–20%.
Confusing the instantaneous compressibility at pressure $p$ with the average over the interval from the bubble point pressure. When the compressibility depends on pressure, the two differ by tens of per cent.
Applying the compressibility of undersaturated oil below the bubble point pressure, where the release of gas is added to the compression of the oil.
Confusing the oil compressibility $c_o$ (MPa⁻¹) with the gas deviation factor $Z$ (dimensionless).

After this lesson you can

convert stock-tank oil density into specific gravity and API gravity, correcting to the temperature you need
calculate reservoir oil density from the mass balance with $B_o$ and $R_s$ , and explain why it is lowest at the bubble point pressure
convert oil density into a pressure gradient and use the gradient to estimate and check the formation volume factor
find oil compressibility above the bubble point pressure from laboratory data and from the Vasquez–Beggs correlation, tell instantaneous from average compressibility, and calculate $B_o$ above the bubble point pressure
estimate the apparent oil compressibility below the bubble point pressure and understand why it is tens of times greater
allow for the thermal expansion of oil when converting densities and volumes

Stock-tank oil density and API gravity

The first characteristic of produced oil is the density of the stock-tank oil $\rho_{o,\text{sc}}$ (oil at surface conditions, with its gas removed). We measure it at atmospheric pressure and convert it to a standard temperature. In Russian-language practice this is 20 °C. In international practice it is 60 °F (15.6 °C), and the density is given as the specific gravity $\gamma_o$ , which is the density of the oil divided by the density of water, both at 60 °F. It is also given in degrees API: ${}^{\circ}\mathrm{API} = 141.5/\gamma_o - 131.5$ . The API scale runs backwards (a lighter oil has a higher number). The oil classes on this scale and the reverse conversion are covered in the topic on the composition of fluids.

Density depends noticeably on temperature, because oil expands when heated while its mass stays the same. So a number without a temperature means little. You cannot use the density at 20 °C where the density at 60 °F is needed, for example in the PVT correlations (PVT correlations for oil and gas condensate) that are built on degrees API.

Correcting the density of stock-tank oil to another temperature

\rho(T_2) \approx \rho(T_1)\,\big[1 + \beta\,(T_1 - T_2)\big]

where:

$\rho(T_1),\ \rho(T_2)$ density of the stock-tank oil at temperatures $T_1$ and $T_2$ , kg/m³
$\beta$ thermal expansion coefficient of the stock-tank oil near these temperatures (the fraction by which the volume grows per degree); around 20 °C it is roughly $(0.6\text{–}1.0)\cdot 10^{-3}$ , higher for light oils, 1/°C
$T_1,\ T_2$ temperatures, °C

Units: kg/m³ and °C (or g/cm³ and °C)

Holds when

The pressure is atmospheric and the oil is degassed.
The temperature difference is a few degrees to a few tens of degrees. The formula is linear in $\beta$ , and $\beta$ itself changes a little with temperature.
In commercial measurement (oil sales), oil volumes and densities are converted with standard tables, not with a single coefficient.

Cooling makes oil denser, and heating makes it less dense. The difference between 20 °C and 15.6 °C is 4.4 °C, which is about 0.35% of the density. For hydrocarbons in place this is a small thing. For correlations it is not, as the example shows.

Worked exampleThe API gravity of the lesson oil and the cost of a forgotten correction

Given (assumed data)

Stock-tank oil of the lesson: 850 kg/m³ at 20 °C
Thermal expansion around 20 °C: $8.0\cdot 10^{-4}$ 1/°C (assumed, as in the topic PVT correlations for oil and gas condensate)
Density of water at 15.6 °C: 999.0 kg/m³

Solution

Correction to 15.6 °C. The oil cools by 4.4 °C and becomes denser: $\rho = 850\,[1 + 8.0\cdot 10^{-4}\cdot 4.4] = 852.99$ kg/m³.
$\gamma_o = 852.99/999.0 = 0.8538$ , and ${}^{\circ}\mathrm{API} = 141.5/0.8538 - 131.5 = 34.2$ .
Without the correction: $\gamma = 850/999.0 = 0.8509$ and $141.5/0.8509 - 131.5 = 34.8$ °API, which is 0.6 °API too high.
With this °API the bubble point pressure from the Standing correlation (formula) comes out too low, by 1.7% (0.29 MPa).

Answer: The lesson oil is 34.2 °API. If we forget the correction to 60 °F, we get 34.8 °API.

What it means. Six tenths of a degree API seem a small thing. But in the Standing correlation °API appears in the exponent, so this input error shifts the bubble point pressure by almost 2%, with no error in the correlation itself. The data in this example are assumed for illustration.

Oil density in the reservoir: the mass balance

Reservoir oil density can be measured in a PVT cell (a laboratory cell for fluid tests), but more often we calculate it from what is already in the property table. The reasoning is simple: no mass is lost on the way from the reservoir to the surface. $B_o$ cubic metres of reservoir oil are exactly 1 m³ of stock-tank oil plus the $R_s$ cubic metres of gas that it gives up in separation (Black oil model: solution gas–oil ratio and formation volume factors). So the mass of $B_o$ cubic metres of reservoir oil equals $\rho_{o,\text{sc}} + R_s\,\rho_{g,\text{sc}}$ .

Reservoir oil density from the mass balance

\rho_o = \frac{\rho_{o,\text{sc}} + R_s\,\rho_{g,\text{sc}}}{B_o}

where:

$\rho_o$ reservoir oil density, with the dissolved gas, at pressure $p$ and reservoir temperature, kg/m³
$\rho_{o,\text{sc}}$ density of the stock-tank oil at standard conditions, kg/m³
$\rho_{g,\text{sc}}$ density of the gas dissolved in the oil at standard conditions: $1.204\,\gamma_g$ at 0.101325 MPa and 20 °C, kg/m³
$R_s,\ B_o$ solution gas–oil ratio and oil formation volume factor at pressure $p$ , m³/m³

Units: SI: densities in kg/m³, $R_s$ and $B_o$ in m³/m³

Holds when

$R_s$ , $B_o$ , $\rho_{o,\text{sc}}$ and $\rho_{g,\text{sc}}$ refer to the same separation scheme and the same standard conditions.
All the gas freed in separation is counted in $R_s$ , and the oil is counted in $\rho_{o,\text{sc}}$ .
Below the bubble point pressure the heavy components of the gas stay in the oil, so the dissolved gas is heavier than the total separator gas. If we use $\gamma_g$ of the total gas, the density at low pressures comes out slightly too low.

The formula is the same balance that gave $B_o$ itself in the topic on the black oil model (mass balance). It works both ways. From $B_o$ it gives the density, and from a measured density or gradient it gives the formation volume factor.

In other unit systems (1)

Oilfield units

\rho_o\,[\text{lb/ft}^3] = \frac{62.4\,\gamma_o + 0.0136\,\gamma_g\,R_s\,[\text{scf/STB}]}{B_o}

$\gamma_o$ and $\gamma_g$ are the specific gravities of the oil (relative to water) and of the gas (relative to air) at 60 °F, $R_s$ is in scf/STB, $B_o$ in rb/STB, $\rho_o$ in pounds per cubic foot; 1 lb/ft³ = 16.02 kg/m³.

62.4 lb/ft³ is the density of water at 60 °F (more precisely 62.37); $0.0136 = 0.0764/5.615$ is the density of air at 60 °F and 14.696 psia in lb/ft³, divided by the number of cubic feet in a barrel.

Above the bubble point pressure the mass in the numerator is constant, while the volume $B_o$ decreases as the pressure rises. The oil is compressed and its density increases, by a fraction of a per cent for every megapascal.
Below the bubble point pressure, as the pressure falls, gas leaves the oil. The numerator decreases, but $B_o$ decreases faster, because the gas carries off a lot of volume and little mass. The density rises as degassing goes on, and at atmospheric pressure it approaches the density of stock-tank oil at reservoir temperature.

Worked exampleDensity of the lesson oil in the reservoir at three pressures

Given (assumed data)

Stock-tank oil: 850 kg/m³ at 20 °C
Separator gas: $\gamma_g = 0.80$ ; $\rho_{g,\text{sc}} = 0.80\cdot 1.204 = 0.963$ kg/m³ (0.101325 MPa, 20 °C)
Bubble point pressure 18 MPa: $R_{sb} = 110$ m³/m³, $B_{ob} = 1.350$
Initial pressure 30 MPa: $B_{oi} = 1.324$ , same solution gas–oil ratio
12 MPa (below the bubble point pressure): $R_s = 76.3$ m³/m³, $B_o = 1.258$

Solution

The mass that 1 m³ of stock-tank oil carries with it at the bubble point pressure is $850 + 110\cdot 0.963 = 955.9$ kg. Of this, gas is 105.9 kg, which is 11% of the mass.
Density at the bubble point pressure: $955.9/1.350 = 708$ kg/m³.
At 30 MPa the mass is the same and the volume is smaller: $955.9/1.324 = 722$ kg/m³.
At 12 MPa part of the gas has left, so the mass is $850 + 76.3\cdot 0.963 = 923.5$ kg and the density is $923.5/1.258 = 734$ kg/m³.

Answer: The density of the lesson oil in the reservoir is 722 kg/m³ at the initial pressure, 708 kg/m³ at the bubble point pressure and 734 kg/m³ at 12 MPa.

What it means. Reservoir oil is 17% lighter than stock-tank oil. Gas accounts for 11% of its mass but about 22% of its volume. Compare with the stock-tank oil at the same temperature, which takes up 1.05 m³. The dissolved 105.9 kg of gas added 0.30 m³. So in solution the gas behaves like a liquid of density about 353 kg/m³ (ignoring the slight compression of the oil by pressure). The density at 30 MPa is the same as the one found from the constant-composition experiment (the example in the topic on laboratory PVT tests). The data in this example are assumed for illustration.

CalculatedDensity of the lesson oil in the reservoir at 85 °C against pressure, calculated from the mass balance. The minimum, 708 kg/m³, is at the bubble point pressure of 18 MPa. Above it the oil is compressed. Below it, the oil loses gas.

Pressure gradient in the oil zone

In oil that is not moving, pressure increases with depth with a gradient $G_o = \rho_o g$ . This is 9.81 kPa/m for every gram per cubic centimetre of density (fluid gradients). For the lesson oil at the initial pressure it is $9.81\cdot 0.722 \approx 7.08$ kPa/m (0.313 psi/ft). If we take the density of the stock-tank oil, 850 kg/m³, by mistake, we get 8.34 kPa/m. That is 18% more, and over every hundred metres of oil column the pressure will be off by 0.13 MPa.

The inverse problem is just as useful. A formation tester gives a series of pressures at different depths. The slope of the line in the oil zone is the density of the oil in the reservoir (Pressure gradients, fluid contacts and capillary equilibrium). If we know the producing gas–oil ratio of the test and the density of the stock-tank oil, the mass balance gives us the formation volume factor from this slope. We can do this before the laboratory PVT data are available, or to check them.

Oil formation volume factor from the pressure gradient

G_o = \rho_o\,g,\qquad B_o = \frac{g\,\big(\rho_{o,\text{sc}} + R_s\,\rho_{g,\text{sc}}\big)}{G_o}

where:

$G_o$ measured pressure gradient in the oil zone, Pa/m
$g$ acceleration due to gravity, 9.81, m/s²
$\rho_{o,\text{sc}},\ \rho_{g,\text{sc}},\ R_s$ densities of the stock-tank oil and the gas at standard conditions, and the solution gas–oil ratio (the producing gas–oil ratio of a test above the bubble point pressure), kg/m³, kg/m³, m³/m³
$B_o$ oil formation volume factor at the reservoir pressure of the measurement zone, m³/m³

Units: SI; in practice $G_o$ is in kPa/m: $\rho_o\,[\text{g/cm}^3] = G_o\,[\text{kPa/m}]/9.81$

Holds when

The oil in the measurement zone is stationary and uniform in composition, and the measurements are of good quality.
The producing gas–oil ratio was measured at a bottomhole pressure above the bubble point pressure. Otherwise it contains free gas (Reservoir fluid sampling).
The relative error of $B_o$ equals the relative error of the gradient.

The more accurate the gradient, the more accurate the estimate. An error of 0.05 kPa/m at a gradient of about 7 kPa/m is 0.7% of the formation volume factor.

Worked exampleOil formation volume factor of a new area from the pressure gradient

Given (assumed data)

Formation tester in a new well: gradient in the oil-bearing part of the reservoir $7.25 \pm 0.05$ kPa/m
Test above the bubble point pressure: producing gas–oil ratio 95 m³/m³, separator gas $\gamma_g = 0.80$ ( $\rho_{g,\text{sc}} = 0.963$ kg/m³)
Stock-tank oil: 856 kg/m³ at 20 °C

Solution

Oil density in the reservoir: $\rho_o = 7.25/9.81 = 0.7390$ g/cm³ = 739.0 kg/m³.
Mass per 1 m³ of stock-tank oil: $856 + 95\cdot 0.963 = 947.5$ kg.
Formation volume factor: $B_o = 947.5/739.0 = 1.282$ m³/m³.
The gradient error of ±0.05 kPa/m is ±0.7% of the density, and $B_o$ lies between 1.273 and 1.291.

Answer:

B_o \approx 1.282 \pm 0.009

m³/m³ at reservoir pressure. This is an estimate made without a sample.

What it means. The oil of the new area carries less gas than the lesson oil (95 against 110 m³/m³), and its formation volume factor is smaller: 1.28 against 1.32. We use such an estimate for a first calculation of oil in place, and for checking a future PVT report. If the laboratory reports, say, 1.35, then either the sample or the producing gas–oil ratio is in doubt. The estimate is only as good as the gradient, which we find from many good-quality measurements (how many points are needed). The data in this example are assumed for illustration.

Isothermal compressibility above the bubble point pressure

While the pressure is above the bubble point pressure, the oil is single-phase. As the pressure falls, it simply expands a little. We measure this expansion by the isothermal compressibility, or oil compressibility, $c_o$ . In Russian-language literature it is also called the coefficient of oil compressibility. It has nothing in common with the gas deviation factor $Z$ (Ideal and real gas: the gas deviation factor (Z)), often called the compressibility factor, except a similar name.

Isothermal compressibility of oil

c_o = -\frac{1}{V}\left(\frac{\partial V}{\partial p}\right)_T = -\frac{1}{B_o}\left(\frac{\partial B_o}{\partial p}\right)_T = \frac{1}{\rho_o}\left(\frac{\partial \rho_o}{\partial p}\right)_T

where:

$c_o$ oil compressibility at pressure $p$ , MPa⁻¹
$V$ volume of a given portion of oil (composition unchanged), m³
$B_o,\ \rho_o$ formation volume factor and density of the oil, m³/m³, kg/m³
$p$ pressure, MPa

Units: MPa⁻¹ with pressure in MPa; in oilfield units, psi⁻¹

Holds when

The pressure is not below the bubble point pressure. The oil is single-phase and its composition does not change, so $V$ , $B_o$ and $1/\rho_o$ change in the same proportion.
The temperature is constant. The minus sign makes $c_o$ positive, because the volume decreases as the pressure rises.

Compressibility shows by what fraction the oil expands when the pressure falls by 1 MPa. Typical values are $(0.7\text{–}3)\cdot 10^{-3}$ MPa⁻¹ (reference values for reservoir fluids). Oil is usually several times more compressible than formation water (about $0.45\cdot 10^{-3}$ MPa⁻¹) and tens of times less compressible than gas.

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; $1.6\cdot 10^{-3}$ MPa⁻¹ $= 11\cdot 10^{-6}$ psi⁻¹.

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

Gas-rich and light oils are more compressible. Dissolved light molecules make the liquid “looser”. At high temperature the oil is also more yielding.
Compressibility decreases as pressure rises. A compressed liquid resists further compression more strongly, so compressibility is highest at the bubble point pressure.
It is measured in a constant-composition experiment. The sample is expanded in steps at reservoir temperature, and we find the compressibility from the volumes above the bubble point pressure (constant-composition expansion). For the lesson oil the experiment gave $1.60\cdot 10^{-3}$ MPa⁻¹ over the interval 18–30 MPa.

It is important to tell two values apart. The instantaneous (point) compressibility is the slope of the $\ln V(p)$ curve at a given pressure. We use it where the response to a small change in pressure matters: in the diffusivity equation and in well tests (Diffusivity equation). The average compressibility over an interval is the constant that gives the same change in volume between two pressures. We use it in the material balance of an undersaturated accumulation (special cases of the balance). If the compressibility does not depend on pressure, the two are the same. If it does, the average from the bubble point pressure to $p$ is greater than the instantaneous value at $p$ .

Formation volume factor of undersaturated oil in terms of compressibility

B_o(p) = B_{ob}\,\exp\!\left(-\int_{p_b}^{p} c_o\,\dd p\right) = B_{ob}\,\e^{-\bar c_o\,(p - p_b)},\qquad \bar c_o = \frac{1}{p - p_b}\int_{p_b}^{p} c_o\,\dd p

where:

$B_{ob},\ p_b$ formation volume factor at the bubble point pressure and the bubble point pressure itself, m³/m³, MPa
$c_o$ instantaneous compressibility, in general dependent on pressure, MPa⁻¹
$\bar c_o$ average compressibility over the interval from $p_b$ to $p$ , MPa⁻¹

Units: consistent units of pressure and compressibility

Holds when

$p \ge p_b$ , the oil is single-phase, and the temperature is constant.

This is the definition of compressibility, integrated. For a constant $c_o$ it gives the exponential that describes the lesson oil (oil formation volume factor): at 25 MPa, $B_o = 1.350\,\e^{-0.0016\cdot 7} = 1.335$ . For a compressibility that decreases with pressure, we take the integral with its dependence on pressure, as in the correlation of the next section.

Compressibility from the Vasquez–Beggs correlation

When there are no laboratory data, we estimate the compressibility of undersaturated oil from a correlation. The most widely used is the Vasquez–Beggs correlation for compressibility. It was built from 4036 laboratory measurements on oils from all over the world. This is the same database as for their correlations for the solution gas–oil ratio and the formation volume factor (the correlation equations). The signs of the main coefficients follow the physics. Compressibility is greater for oils with a higher solution gas–oil ratio, at high temperature and for light oils (higher °API), and it falls as the pressure rises.

Vasquez–Beggs correlation for the compressibility of undersaturated oil

c_o = \frac{-1433 + 5\,R_{sb} + 17.2\,T - 1180\,\gamma_{g100} + 12.61\,\mathrm{API}}{10^{5}\,p}

where:

$c_o$ instantaneous oil compressibility at pressure $p$ , psi⁻¹
$R_{sb}$ solution gas–oil ratio at the bubble point pressure, gas at 60 °F, scf/STB
$T$ reservoir temperature, °F
$\gamma_{g100}$ specific gravity of the gas, corrected to a first separator stage at 100 psig (gauge pressure), as in the same authors' correlation for the solution gas–oil ratio, —
$\mathrm{API}$ API gravity of the stock-tank oil, °API
$p$ absolute pressure, not below the bubble point pressure, psia

Units: oilfield units: scf/STB, °F, psia, psi⁻¹

Holds when

The oil is undersaturated, $p \ge p_b$ . The input data lie within the database of the correlation (roughly 15–60 °API).
The separator gas is corrected to 100 psig, and the pressure is absolute.

The numerator divided by $10^5$ is a dimensionless number $A$ . The correlation says that $c_o\,p = A$ does not depend on pressure. So we can use it directly in any pressure units: $c_o\,[\text{MPa}^{-1}] = A/p\,[\text{MPa}]$ . Integrating $c_o = A/p$ from the bubble point pressure gives a power-law formula for $B_o$ .

Formation volume factor above the bubble point pressure from the Vasquez–Beggs correlation

B_o = B_{ob}\left(\frac{p_b}{p}\right)^{A},\qquad \bar c_o = \frac{A\,\ln(p/p_b)}{p - p_b}

where:

$A$ dimensionless parameter of the correlation, $c_o\,p$ , —
$B_{ob},\ p_b,\ p$ formation volume factor at the bubble point pressure, bubble point pressure and current pressure (absolute, in the same units), m³/m³, MPa
$\bar c_o$ average compressibility between $p_b$ and $p$ , MPa⁻¹

Units: any pressure units (the ratio and the logarithm are dimensionless)

Holds when

The same as for the compressibility correlation.

The instantaneous compressibility decreases as $1/p$ , so the average from the bubble point pressure is always greater than the instantaneous one at the final pressure.

Worked exampleCompressibility of the lesson oil from the correlation and from the experiment

Given (assumed data)

Lesson oil in oilfield units: $R_{sb} = 608$ scf/STB, $T = 185$ °F, $\gamma_{g100} = 0.738$ , 34.2 °API (as in the topic PVT correlations for oil and gas condensate)
Bubble point pressure: 18 MPa (2611 psia), $B_{ob} = 1.350$ m³/m³
Constant-composition experiment: $c_o = 1.60\cdot 10^{-3}$ MPa⁻¹ over 18–30 MPa; $B_{oi} = 1.3243$ at 30 MPa
Water and pores of the lesson accumulation: $\dfrac{c_wS_w + c_f}{1 - S_w} = 1.2167\cdot 10^{-3}$ MPa⁻¹ (General material balance equation for oil reservoirs)

Solution

Terms of the numerator: $5\cdot 608 = 3040$ ; $17.2\cdot 185 = 3182$ ; $1180\cdot 0.738 = 870.84$ ; $12.61\cdot 34.2 = 431.26$ .
$A = (-1433 + 3040 + 3182 - 870.84 + 431.26)/10^5 = 0.04349$ .
Instantaneous compressibility is $c_o = A/p$ . At the bubble point pressure it is $0.04349/18 = 2.42\cdot 10^{-3}$ MPa⁻¹ (16.7·10⁻⁶ psi⁻¹ at 2611 psia). At 30 MPa it is $1.45\cdot 10^{-3}$ MPa⁻¹.
$B_o$ at 30 MPa: $1.350\cdot(18/30)^{0.04349} = 1.3203$ m³/m³, which is 0.3% below the experimental value.
Average compressibility over 18–30 MPa: $\bar c_o = \ln(1.350/1.3203)/12 = 1.85\cdot 10^{-3}$ MPa⁻¹, which is 16% above the experimental one. The expansion of the oil from 30 to 18 MPa is 2.25%, against 1.94% from the experiment.
Elastic expansion of the lesson accumulation from 30 to 18 MPa, per 1 m³ of oil in place. For the oil it is $1.350 - 1.3243 = 0.0257$ from the experiment and $1.350 - 1.3203 = 0.0297$ from the correlation. For water and pores it is $1.3243\cdot 1.2167\cdot 10^{-3}\cdot 12 = 0.0193$ m³/m³. The sums are 0.0450 and 0.0490. So the elastic withdrawal down to the bubble point pressure is 9% greater by the correlation.

Answer: The correlation gives the compressibility of the lesson oil as

2.42\cdot 10^{-3}

MPa⁻¹ at the bubble point pressure and

1.45\cdot 10^{-3}

at 30 MPa. The average is

1.85\cdot 10^{-3}

, which is 16% above the experimental value.

B_{oi}

is 1.3203 instead of 1.3243.

What it means. The correlation gave the formation volume factor almost exactly. The expansion of the oil itself is small, so an error in it hardly affects $B_o$ . But the elastic capacity is exactly a difference of volumes, and the error in it is already 16%. In the total elastic expansion of the accumulation, where oil accounts for more than half, the error is 9%. That is why, for undersaturated accumulations, we take the oil compressibility from the experiment and use the correlation only for first estimates. The correlation also shows something that an experiment with a constant compressibility does not: the compressibility of oil decreases with pressure. The data in this example are assumed for illustration.

CalculatedCompressibility of the lesson oil above the bubble point pressure. There are three curves: from the experiment (constant,

1.60\cdot 10^{-3}

MPa⁻¹), instantaneous from the Vasquez–Beggs correlation (

A/p

), and average from the correlation, from the bubble point pressure to

p

There are other compressibility correlations, built from regional databases. One example is Petrosky–Farshad, for Gulf of Mexico oils, in which compressibility decreases with pressure more slowly than $1/p$ . We choose and check them in the same way as bubble point pressure correlations: against laboratory data on our own samples (error and tuning).

Below the bubble point pressure: apparent compressibility

Below the bubble point pressure, a fall in pressure does two things to the oil at once. The oil loses gas and shrinks, so $B_o$ decreases. But the freed gas stays in the reservoir and takes up room: $B_g$ for every standard cubic metre. If we divide the total change in volume by the volume of the oil, we get the apparent oil compressibility. This is the quantity that replaces $c_o$ in the total compressibility $c_t$ when the reservoir works below the bubble point pressure.

Apparent oil compressibility below the bubble point pressure

c_o^{\text{app}} = -\frac{1}{B_o}\frac{\partial B_o}{\partial p} + \frac{B_g}{B_o}\frac{\partial R_s}{\partial p}

where:

$c_o^{\text{app}}$ apparent compressibility of the oil together with the gas coming out of it, MPa⁻¹
$B_o,\ B_g,\ R_s$ formation volume factors of oil and gas and the solution gas–oil ratio at pressure $p$ , m³/m³
$\partial B_o/\partial p,\ \partial R_s/\partial p$ slopes of the $B_o(p)$ and $R_s(p)$ curves below the bubble point pressure (both positive), MPa⁻¹, m³/(m³·MPa)

Units: consistent: pressure in MPa, compressibility in MPa⁻¹

Holds when

$p < p_b$ . The gas freed as the pressure falls stays within the volume of the reservoir that we are considering.
The expansion of gas that was already free before is not included here. The term $S_gc_g$ of the total compressibility accounts for it.
$B_o$ and $R_s$ come from a table for one degassing scheme. We take the derivatives from a smoothed curve or from differences between neighbouring points.

The first term is negative, because the oil shrinks as the pressure falls. The second is positive and usually several times larger, because the gas takes up far more room than the oil loses. So below the bubble point pressure the reservoir responds to a fall in pressure tens of times more strongly. In the material balance, the total formation volume factor accounts for the same effect (Black oil model: solution gas–oil ratio and formation volume factors).

Worked exampleApparent compressibility of the lesson oil at 12 MPa

Given (assumed data)

Property table of the lesson oil: 18 MPa: $R_s = 110$ , $B_o = 1.350$ ; 6 MPa: $R_s = 41.0$ , $B_o = 1.162$
At 12 MPa: $B_o = 1.258$ , $B_g = 0.00897$ m³/m³
Above the bubble point pressure: $c_o = 1.60\cdot 10^{-3}$ MPa⁻¹

Solution

Slopes by the central difference over the interval 6–18 MPa: $\Delta R_s/\Delta p = (110 - 41.0)/12 = 5.75$ m³/(m³·MPa), $\Delta B_o/\Delta p = (1.350 - 1.162)/12 = 0.01567$ MPa⁻¹.
Shrinkage of the oil: $-0.01567/1.258 = -0.01246$ MPa⁻¹.
Freed gas: $0.00897\cdot 5.75/1.258 = 0.04100$ MPa⁻¹, which is 3.3 times the shrinkage.
Apparent compressibility: $0.04100 - 0.01246 = 0.0285$ MPa⁻¹, which is 18 times the compressibility of undersaturated oil.

Answer: At 12 MPa the lesson oil, together with the gas coming out of it, behaves like a liquid with a compressibility of about 0.0285 MPa⁻¹. This is almost 18 times more than above the bubble point pressure.

What it means. A difference with a step of 6 MPa is only an estimate. From the smooth curve of the lesson oil at 12 MPa we get 0.0284 MPa⁻¹, which differs by 0.4%. We use this quantity in the interpretation of well tests below the bubble point pressure. It enters the total compressibility together with the compressibility of free gas, water and pores (total compressibility). The jump in compressibility on crossing the bubble point pressure is clearly visible in the chart below. The data in this example are assumed for illustration.

CalculatedCompressibility of the lesson oil on a logarithmic scale. Above the bubble point pressure it is

1.6\cdot 10^{-3}

MPa⁻¹. Below it, the curve shows the apparent compressibility, with the gas coming out of solution. For comparison, the dashed line is the compressibility of gas, close to

1/p

Thermal expansion of oil

Oil expands not only when the pressure falls but also when it is heated. In reservoir processes the temperature is usually constant, so thermal expansion does not act directly. But it matters wherever oil changes temperature. This happens when oil rises up the well, and when samples cool (bottomhole samples). It also matters when we convert density to a standard temperature, when we account for volumes at the surface, and in thermal recovery methods (methods that heat the reservoir).

Thermal expansion coefficient of oil

\beta_o = \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_p,\qquad \left(\frac{\partial p}{\partial T}\right)_V = \frac{\beta_o}{c_o}

where:

$\beta_o$ thermal expansion coefficient of oil at constant pressure, 1/°C
$V,\ T,\ p$ volume of a portion of oil, temperature and pressure, m³, °C, MPa
$c_o$ oil compressibility, MPa⁻¹

Units: 1/°C (the same as 1/K); the second formula is in MPa per °C

Holds when

The oil is single-phase and its composition does not change. For reservoir oil the coefficient is $(0.7\text{–}1.2)\cdot 10^{-3}$ 1/°C, and for stock-tank oil around 20 °C it is $(0.6\text{–}1.0)\cdot 10^{-3}$ 1/°C.
The second formula is for oil trapped in a rigid volume. Heating by one degree raises the pressure by $\beta_o/c_o$ .

For the lesson oil, $\beta_o = 1.0\cdot 10^{-3}$ 1/°C and $c_o = 1.6\cdot 10^{-3}$ MPa⁻¹. Heating by 1 °C expands the oil as much as a pressure decrease of 0.625 MPa. So a trapped liquid gains pressure sharply on heating. This is how pressure builds up in the sealed annular space (the gap around the tubing) of a well that is warming up. A bottomhole sample cooled by a few degrees, on the other hand, loses its margin of pressure above the bubble point pressure.

The lesson oil itself shows how much thermal expansion matters for volumes. Its stock-tank part is 1 m³ and 850 kg at 20 °C, but 1.05 m³ at the reservoir temperature of 85 °C and atmospheric pressure ( $B_{od}$ in the topic on the assumed oil). So the density falls to $850/1.05 = 809.5$ kg/m³. Five per cent of volume from heating by 65 °C is as much as the lesson oil would lose when compressed by a pressure of about 30 MPa.

Terms and symbols

Reservoir oil density $\rho_o$: The mass of a unit volume of oil, with its dissolved gas, at reservoir pressure and temperature. From the mass balance it is $(\rho_{o,\text{sc}} + R_s\rho_{g,\text{sc}})/B_o$ .
Oil specific gravity $\gamma_o$: The density of stock-tank oil divided by the density of water, both at 60 °F (15.6 °C). Degrees API are defined through it.
Isothermal oil compressibility $c_o$: The relative decrease in the volume of oil for each unit increase in pressure at constant temperature, in MPa⁻¹. It is defined for single-phase oil above the bubble point pressure.
Instantaneous and average compressibility: The instantaneous compressibility is the slope of the $\ln V(p)$ curve at a given pressure. The average compressibility is the constant that gives the same change in volume over a pressure interval. They are the same if the compressibility does not depend on pressure.
Apparent oil compressibility $c_o^{\text{app}}$: The change in the volume of the oil together with the gas coming out of it, for each unit fall in pressure, divided by the volume of the oil. Below the bubble point pressure it is tens of times greater than $c_o$ .
Thermal expansion coefficient $\beta_o$: The relative increase in the volume of oil when heated by 1 °C at constant pressure. It is about $10^{-3}$ 1/°C.

Common mistakes and misconceptions

Wrong

“The density of oil in the reservoir is the density of the stock-tank oil, 850 kg/m³.”

In fact

Reservoir oil with dissolved gas is lighter: 708–722 kg/m³ for the lesson oil. The pressure gradient in the oil zone is 7.08 kPa/m, not 8.34. Over a hundred metres of oil column the error would reach 0.13 MPa.

Wrong

“Below the bubble point pressure the oil becomes more compressible, so its density falls.”

In fact

As the pressure falls below the bubble point pressure, the oil loses gas and its density rises. The large apparent compressibility comes from the volume of the freed gas, not from an expansion of the oil itself.

Wrong

“The compressibility from the correlation at reservoir pressure can be substituted straight into $B_o = B_{ob}\e^{-c_o(p - p_b)}$ .”

In fact

The exponent contains the average compressibility over the interval from the bubble point pressure. The Vasquez–Beggs correlation gives the instantaneous value, $A/p$ . Integrate it to get $B_o = B_{ob}(p_b/p)^A$ .

Wrong

“Oil is almost incompressible, so the elastic capacity of an undersaturated accumulation can be ignored.”

In fact

In the lesson accumulation, the expansion of the oil, water and pores pushes about 3.3% of the oil in place out of the reservoir as the pressure falls from 30 to 18 MPa. This happens before any free gas appears, and with no injection (General material balance equation for oil reservoirs).

In depth: advanced detail

Does the oil density change with height in an accumulation

We draw the pressure line in the oil zone as a straight line. This means that we take the oil density to be the same over the whole height. But towards the base of the accumulation both pressure and temperature increase. The first compresses the oil, and the second expands it. It is easy to estimate how much this bends the line.

Worked exampleThe lesson oil in an accumulation 100 m high

Given (assumed data)

Density at the top: 722 kg/m³, gradient 7.08 kPa/m
Compressibility and thermal expansion of the reservoir oil: $1.6\cdot 10^{-3}$ MPa⁻¹ and $1.0\cdot 10^{-3}$ 1/°C
Geothermal gradient: 3 °C per 100 m (assumed)

Solution

Towards the base the pressure increases by $7.08\cdot 100 = 708$ kPa, and the oil is compressed. The density is greater by $722\cdot 1.6\cdot 10^{-3}\cdot 0.708 = 0.82$ kg/m³.
The temperature is 3 °C higher, and the oil expands. The density is smaller by $722\cdot 1.0\cdot 10^{-3}\cdot 3 = 2.17$ kg/m³.
Net result: at the base the density is lower than at the top by $2.17 - 0.82 = 1.35$ kg/m³, which is 0.19%.
If the density changes linearly with height, the pressure at the base departs from the straight line with the gradient of the top by $\tfrac12 g\,\Delta\rho\,L = 0.5\cdot 9.81\cdot 1.35\cdot 100 = 662$ Pa.

Answer: Over a height of 100 m the density of a uniform oil changes by 0.19%, and the pressure departs from the straight line by less than 1 kPa.

What it means. Compression and heating act in opposite directions and partly cancel each other (here heating is stronger). What is left is small. So in a uniform oil the pressure–depth line is straight, to within the accuracy of the measurements (Pressure gradients, fluid contacts and capillary equilibrium). A noticeable bend in the oil line, or a difference between the densities of samples from different depths, points to something else: a change in composition with height (compositional grading). The data in this example are assumed for illustration.

Why compressibility decreases with pressure

A liquid is compressed by using up the free volume between its molecules. The more it is already compressed, the less of this volume there is, and the more pressure we need to compress it by the same fraction again. This is why compressibility decreases with pressure. It is also the reason for the form $c_o = A/p$ in the Vasquez–Beggs correlation. Treat this form as a convenient approximation within the range of its data, not as a law: as the pressure tends to zero, it would give an infinite compressibility.

Dissolved gas increases the free volume, because the light molecules of methane and ethane “push apart” the heavy ones. That is why gas-rich oil is more compressible than stock-tank oil. It is also why compressibility rises with temperature, when thermal motion pushes the molecules apart too. Near the bubble point pressure the oil is closest to releasing gas, and its compressibility there is greatest. For volatile oils close to the critical point it is noticeably higher than the typical values.

Connections to other topics

Black oil model: solution gas–oil ratio and formation volume factors — the solution gas–oil ratio and the formation volume factor, through which the density and compressibility of oil are expressed.
Laboratory PVT tests — how compressibility and density are measured in the constant-composition experiment.
PVT correlations for oil and gas condensate — correlations of oil properties, their error and their tuning.
Pressure gradients, fluid contacts and capillary equilibrium — pressure gradients and fluid contacts from formation tester measurements.
General material balance equation for oil reservoirs — where the compressibility of oil determines the elastic capacity of an accumulation.
Flow regimes — total compressibility and its role in unsteady inflow.
Oil viscosity and interfacial tension — other properties of reservoir oil: viscosity and interfacial tension.

Check yourself

1
How does the density of reservoir oil change as the pressure falls from an initial value above the bubble point pressure to atmospheric?
1. A.It first falls down to the bubble point pressure, then grows
2. B.It grows steadily
3. C.It falls steadily
4. D.It does not change down to the bubble point pressure, then falls
Show the answer
Correct answer: A. It first falls down to the bubble point pressure, then grows
Above the bubble point pressure the oil expands as the pressure falls, so its density falls. Below it, gas leaves the oil, carrying off a lot of volume and little mass, so the density rises. The density is lowest at the bubble point pressure, where $B_o$ is highest.
2
Stock-tank oil is 870 kg/m³ at 20 °C, the solution gas–oil ratio is 60 m³/m³, the gas has a specific gravity of 0.75, and $B_o = 1.18$ . Find the density of the oil in the reservoir and the pressure gradient in the oil zone.
Show the answer
Gas: $\rho_{g,\text{sc}} = 0.75\cdot 1.204 = 0.903$ kg/m³, and its mass is $60\cdot 0.903 = 54.2$ kg. Then $\rho_o = (870 + 54.2)/1.18 = 783$ kg/m³. The gradient is $9.81\cdot 0.783 = 7.68$ kPa/m.
3
The Vasquez–Beggs correlation gives $A = 0.030$ for an oil; the bubble point pressure is 15 MPa, $B_{ob} = 1.25$ . Find $B_o$ and the instantaneous and average compressibility at 25 MPa. Why is the average greater?
Show the answer
$B_o = 1.25\cdot(15/25)^{0.030} = 1.25\,\e^{0.030\ln 0.6} = 1.231$ . Instantaneous: $c_o = 0.030/25 = 1.2\cdot 10^{-3}$ MPa⁻¹. Average: $0.030\ln(25/15)/10 = 1.53\cdot 10^{-3}$ MPa⁻¹. Compressibility decreases with pressure, so the average over the interval 15–25 MPa includes the larger values near the bubble point pressure.
4
Why is the apparent oil compressibility below the bubble point pressure tens of times greater than the compressibility of undersaturated oil?
1. A.Below the bubble point pressure the oil itself becomes more compressible
2. B.Below the bubble point pressure the viscosity of the oil increases
3. C.The volume of the gas coming out of solution, which is much greater than the shrinkage of the oil, is added to the change in the oil volume
4. D.Below the bubble point pressure the pores are compressed more strongly
Show the answer
Correct answer: C. The volume of the gas coming out of solution, which is much greater than the shrinkage of the oil, is added to the change in the oil volume
The oil itself shrinks as it loses gas when the pressure falls below the bubble point pressure. But the freed gas takes up $B_g\,\Delta R_s$ in the reservoir, which is several times more than the oil loses. The sum, divided by the volume of the oil, is the apparent compressibility.
5
Single-phase oil is trapped in a rigid chamber; $\beta_o = 0.9\cdot 10^{-3}$ 1/°C, $c_o = 1.5\cdot 10^{-3}$ MPa⁻¹. By how much will the pressure change on cooling by 4 °C? What will happen if the pressure was 1.5 MPa above the bubble point pressure?
Show the answer
$\beta_o/c_o = 0.6$ MPa per degree, so cooling by 4 °C lowers the pressure by 2.4 MPa. The margin of 1.5 MPa disappears after cooling by only about 2.5 °C (ignoring that the bubble point pressure also falls slightly on cooling), and then gas appears in the chamber.
6
The formation tester gives a gradient in the oil zone of 7.0 kPa/m. What is the density of the oil in the reservoir?
1. A.0.686 g/cm³
2. B.0.700 g/cm³
3. C.0.714 g/cm³
4. D.7.0 g/cm³
Show the answer
Correct answer: C. 0.714 g/cm³
$\rho_o = G_o/g = 7.0/9.81 = 0.714$ g/cm³ (714 kg/m³). The option 0.700 results from dividing by 10 instead of 9.81.

Updated: 28 October 2026