Black oil model: solution gas–oil ratio and formation volume factors

Quick refresher

In the black oil model, reservoir oil is stock-tank oil plus dissolved gas. Every quantity is given per 1 m³ (or one barrel) of stock-tank oil at standard conditions.
$R_s$ is the standard volume of gas dissolved in 1 m³ of stock-tank oil. Above the bubble point pressure it is constant ( $R_s = R_{sb}$ ). Below it, $R_s$ decreases together with the pressure.
$B_o$ is the reservoir volume of oil with its dissolved gas for each 1 m³ of stock-tank oil, usually 1.05–2. As the pressure falls above the bubble point pressure, $B_o$ grows slightly (the oil expands). Below the bubble point pressure it falls (gas leaves and the oil shrinks).
Below the bubble point pressure the produced gas is the sum of dissolved gas ( $R_s$ ) and free gas ( $R - R_s$ ). For every cubic metre of oil, the free gas takes up $(R - R_s)B_g$ in the reservoir.
$B_t = B_o + B_g(R_{sb} - R_s)$ is the volume of the oil together with all the gas that was originally dissolved in it, as if the gas had gone nowhere. It is a bookkeeping quantity of the material balance (the accounting of fluids in the reservoir).
$R_s$ and $B_o$ depend on the separation conditions (the pressures and temperatures of the surface separators). The same reservoir oil gives different values for different separation schemes.

Key relationships

B_o = \dfrac{V_o(p, T)}{V_{o,\text{sc}}},\qquad R_s = \dfrac{V_{g,\text{sc}}}{V_{o,\text{sc}}}

definitions; both in m³/m³

B_o = B_{ob}\,\e^{-c_o(p - p_b)}

undersaturated oil, $p \ge p_b$

B_t = B_o + B_g\,(R_{sb} - R_s)

total formation volume factor

q_{\text{res}} = q_o\big[B_o + (R - R_s)\,B_g\big] + q_w B_w

withdrawal of fluids from the reservoir, at reservoir conditions

Common pitfalls

Confusing the solution gas–oil ratio $R_s$ (dissolved gas) with the producing gas–oil ratio $R$ (all the produced gas per 1 m³ of oil). Below the bubble point pressure $R$ can be several times larger than $R_s$ .
Dividing by $B_o$ where you should multiply: reservoir volume = standard volume × $B_o$ .
Mixing a solution gas–oil ratio in m³/t with one in m³/m³, or gas converted to 20 °C with gas converted to 60 °F.
Treating $B_t$ as the real volume of fluids in the reservoir below the bubble point pressure. The gas moves away from its oil.
Taking $R_s$ and $B_o$ from a PVT report without checking what separation conditions they refer to.

After this lesson you can

explain what the solution gas–oil ratio $R_s$ and the formation volume factors $B_o$ , $B_g$ and $B_t$ describe, and what units they have
describe how $R_s$ and $B_o$ change as pressure falls, above and below the bubble point pressure (the pressure at which the first gas bubbles appear), and explain why
convert volumes and rates of oil and gas between reservoir and standard conditions
split the gas produced below the bubble point pressure into dissolved gas and free gas, and calculate how much fluid is withdrawn from the reservoir
calculate the total formation volume factor $B_t$ and know where it is needed
understand the limits of the black oil model, and the cases that need a compositional model (a model that tracks the composition of the fluid)

Reservoir oil as two components

Reservoir oil is a mixture of hundreds of hydrocarbons. While the pressure is high, the light components (methane, ethane, propane) are dissolved in the liquid. When the oil rises up the well and enters the separator (a vessel at the surface that splits gas from liquid), the pressure and temperature fall and gas comes out of the oil. What remains in the stock tank (the storage tank at the surface) is the stock-tank oil. The gas is taken off separately.

The black oil model does not track the composition of the mixture. It describes reservoir oil by two simplified components, which are defined by what we get at the surface: stock-tank oil and dissolved gas. All properties are given per 1 m³ of stock-tank oil at standard conditions (in oilfield units, per stock-tank barrel, STB). The name is historical. People called ordinary dark oils with a moderate gas content “black oils”, and the approach was created for them.

SchematicThe black oil model:

B_o

m³ of reservoir oil at pressure

p

and temperature

T

give, after separation, 1 m³ of stock-tank oil and

R_s

m³ of gas, both at standard conditions.

Standard conditions are the pressure and temperature to which we convert volumes measured at the surface. In Russian-language practice, gas is usually taken at 0.101325 MPa and 20 °C. In international practice it is taken at 0.101325 MPa (14.696 psia) and 15 °C or 60 °F (15.6 °C). For oil the difference is small. For gas it reaches 1.5–2%, so always state the conditions next to the number.

Bubble point pressure $p_b$: The pressure at which the first bubbles of gas come out of the oil as the pressure falls. It is measured at reservoir temperature.
Undersaturated oil: The reservoir pressure is above the bubble point pressure. The oil could dissolve more gas, and there is no free gas in the reservoir.
Saturated oil: The reservoir pressure is equal to or below the bubble point pressure. The slightest fall in pressure frees gas, and the reservoir may have a gas cap (a layer of free gas above the oil) or free gas.

Solution gas–oil ratio

How much gas can dissolve in oil depends on the pressure, the temperature and the compositions of the oil and the gas. Solubility rises with pressure and falls with temperature. In the black oil model a single quantity describes this.

Solution gas–oil ratio

R_s = \frac{V_{g,\text{sc}}}{V_{o,\text{sc}}}

where:

$R_s$ solution gas–oil ratio of the oil at pressure $p$ and reservoir temperature, m³/m³
$V_{g,\text{sc}}$ volume of the gas freed when the oil is degassed to standard conditions, measured at standard conditions, m³
$V_{o,\text{sc}}$ volume of the resulting stock-tank oil at standard conditions, m³

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

Holds when

At pressure $p$ and reservoir temperature the oil is single-phase and the gas in it is in equilibrium.
Both volumes come from the same method of degassing (the separation conditions). The result depends on this method.

The solution gas–oil ratio shows how much gas each cubic metre of stock-tank oil brings with it if the oil is lifted to the surface with no loss of gas on the way. At initial reservoir conditions we write it $R_{si}$ , and at the bubble point pressure $R_{sb}$ .

In other unit systems (1)

Oilfield units

R_s\,[\text{scf/STB}] = 5.615\;R_s\,[\text{m}^3/\text{m}^3]

scf is standard cubic feet of gas, STB is a barrel of stock-tank oil at standard conditions.

$5.615 = 0.158987/0.0283168$ , which is the volume of a barrel divided by the volume of a cubic foot (both in m³). The conversion holds when the standard conditions are the same. If the cubic metres of gas are converted to 20 °C and the cubic feet to 60 °F, the figure in cubic metres is larger by a further 1.5% or so.

Above the bubble point pressure all the gas present in the reservoir is already dissolved, and the oil could take up more. The solution gas–oil ratio is constant: $R_s = R_{si} = R_{sb}$ .
Below the bubble point pressure part of the gas goes into a free phase, and less of it stays in the oil. $R_s$ falls together with the pressure and approaches zero at atmospheric pressure.

Gas comes out in different ways, depending on whether it stays beside the oil. In flash (single-stage) liberation, the freed gas stays in contact with the oil until equilibrium is reached. In differential liberation, the gas is removed continuously. These processes give different $R_s$ values. This is why we convert the data of laboratory PVT studies (PVT stands for pressure, volume, temperature) to separation conditions (Laboratory PVT tests, How to read a PVT report). In the reservoir below the bubble point pressure, mobile gas moves away from its oil. On the way from the bottomhole to the separator, gas and oil travel together, which is closer to flash liberation.

Oil formation volume factor

A cubic metre of stock-tank oil with gas dissolved in it at reservoir conditions takes up more than a cubic metre. The dissolved gas “swells” the liquid, and the heat of the reservoir has also expanded the oil. Compression by pressure acts in the opposite direction, but more weakly. So $B_o$ is almost always greater than 1: 1.03–1.10 for heavy oils with a low gas content, 1.1–1.5 for typical oils, and 2 or more for light, gas-rich oils.

Oil formation volume factor

B_o = \frac{V_o(p, T)}{V_{o,\text{sc}}}

where:

$B_o$ oil formation volume factor, m³/m³
$V_o(p, T)$ volume of the oil together with its dissolved gas at pressure $p$ and reservoir temperature $T$ , m³
$V_{o,\text{sc}}$ volume of the stock-tank oil that comes from it, at standard conditions, m³

Units: m³/m³, which means reservoir cubic metres per cubic metre of stock-tank oil. In oilfield units it is rb/STB (reservoir barrels per stock-tank barrel), and the number is the same

Holds when

At $p$ and $T$ the oil is single-phase, so free gas is not part of $V_o$ .
The stock-tank oil is the oil left after separation under the stated conditions.

Multiplying by $B_o$ converts a standard volume of oil into a reservoir volume. Dividing converts a reservoir volume into a standard one. The inverse, $\theta = 1/B_o$ , is called the shrinkage factor. It is the fraction of the reservoir volume that will remain as oil in the stock tank. The shrinkage of the oil, in per cent, is $(B_o - 1)/B_o\cdot 100\,\%$ .

How $B_o$ changes as the reservoir pressure falls:

Above the bubble point pressure the composition of the oil does not change, and the oil simply expands as the pressure is relieved. $B_o$ grows slowly and reaches its maximum, $B_{ob}$ , at the bubble point pressure.
Below the bubble point pressure gas leaves the oil, and shrinkage outweighs expansion, so $B_o$ falls. At atmospheric pressure and reservoir temperature it is slightly above 1 (the thermal expansion remains). After cooling to standard temperature it equals 1 by definition.

Formation volume factor of undersaturated oil

B_o(p) = B_{ob}\,\e^{-c_o\,(p - p_b)},\qquad p \ge p_b

where:

$B_{ob}$ formation volume factor at the bubble point pressure, m³/m³
$c_o$ oil compressibility above the bubble point pressure, $c_o = -\dfrac{1}{B_o}\dfrac{\dd B_o}{\dd p}$ , MPa⁻¹
$p,\ p_b$ reservoir pressure and bubble point pressure (only their difference enters), MPa

Units: consistent units of pressure and compressibility: MPa and MPa⁻¹, or psi and psi⁻¹

Holds when

The pressure is not below the bubble point pressure and the oil is single-phase.
The compressibility is constant over the range considered (over a wide range, the average value is used).

The formula comes from integrating the definition of compressibility. The typical compressibility of undersaturated oil is $(0.7\text{–}3)\cdot 10^{-3}$ MPa⁻¹ ( $5\text{–}20\cdot 10^{-6}$ psi⁻¹). So when the pressure falls by 10 MPa, $B_o$ grows by only 0.7–3%. But in an undersaturated accumulation it is exactly this expansion of the oil, together with the expansion of the water and the shrinking of the pores, that pushes the oil to the wells (Density and compressibility of reservoir oil).

CalculatedSolution gas–oil ratio (top) and oil formation volume factor (bottom) against reservoir pressure for the assumed oil of the lesson's examples: bubble point pressure 18 MPa,

R_{sb} = 110

m³/m³,

B_{ob} = 1.35

m³/m³. Both curves bend sharply at the bubble point pressure.

Worked exampleFrom oil-filled pore volume to oil in place and dissolved gas

Given (assumed data)

Pore volume of the accumulation: 16.0 million m³
Initial water saturation: 0.25; no free gas
Initial pressure and bubble point pressure: 30 and 18 MPa
At initial pressure: $B_{oi} = 1.324$ m³/m³, $R_{si} = 110$ m³/m³ (gas at 20 °C)
Stock-tank oil density: 850 kg/m³

Solution

Oil in the reservoir: $V_{\text{res}} = 16.0\cdot(1 - 0.25) = 12.0$ million m³.
Stock-tank oil in place: $N = V_{\text{res}}/B_{oi} = 12.0/1.324 = 9.06$ million m³, or $9.06\cdot 0.85 = 7.70$ million t.
Dissolved gas: $G_s = N\,R_{si} = 9.06\cdot 10^6\cdot 110 = 0.997\cdot 10^9$ m³.
In oilfield units: $N = 9.06\cdot 10^6/0.158987 = 57.0$ million STB, $R_{si} = 110\cdot 5.615 = 618$ scf/STB (without a correction for the different standard temperatures).

Answer: The oil in place is 9.06 million m³ (7.70 million t) of stock-tank oil, with about 1.0 billion m³ of dissolved gas.

What it means. Of the 12.0 million m³ of reservoir oil, only 9.06 million m³ ends up in the stock tank. This is a shrinkage of 24%. An error in $B_{oi}$ passes straight into the oil in place. If we took $B_o = 1.25$ from another accumulation by mistake, we would overstate the oil in place by 5.9%. The data in this example are assumed for illustration.

Surface rates and withdrawal from the reservoir

We measure a well at the surface, in cubic metres (or barrels) per day of stock-tank oil, water and gas. The reservoir, however, “feels” the volume that has left it at reservoir conditions. We need this volume for three jobs. It tells us how much to inject to maintain the pressure. It closes the material balance. And it lets us compare a rate with the inflow equation, in which the flow rate is taken at reservoir conditions (see radial flow).

Withdrawal of fluids from the reservoir at reservoir conditions

q_{\text{res}} = q_o\,\big[B_o + (R - R_s)\,B_g\big] + q_w\,B_w

where:

$q_{\text{res}}$ total withdrawal of oil, gas and water from the reservoir at reservoir conditions, m³/d
$q_o,\ q_w$ rates of stock-tank oil and water at standard conditions, m³/d
$R$ producing gas–oil ratio (all the produced gas per cubic metre of oil), m³/m³
$R_s,\ B_o,\ B_g$ solution gas–oil ratio and the formation volume factors of oil and gas at the current reservoir pressure, m³/m³
$B_w$ water formation volume factor, usually 1.00–1.07, m³/m³

Units: m³/d; in oilfield units, bbl/d with $q_o$ in STB/d, $R$ in scf/STB and $B_g$ in rb/scf

Holds when

$R_s$ , $B_o$ and $B_g$ are taken at the average pressure in the drainage area of the well (the part of the reservoir that feeds the well).
All the gas in excess of the dissolved gas came from the reservoir as free gas. Gas injected into the well for gas lift (a way of lifting oil by injecting gas) is subtracted from the measurement.
Above the bubble point pressure, $R = R_s$ , and the free-gas term is zero.

This is the volume that must be replaced by injection to keep the pressure from falling. It also enters the material balance. Below the bubble point pressure the free gas quickly becomes the main part of the withdrawal.

Worked exampleHow much water to inject at a pressure above the bubble point pressure

Given (assumed data)

Reservoir pressure: 25 MPa (above the bubble point pressure of 18 MPa)
Production of the wells of the area: oil 120 m³/d, water 30 m³/d, producing gas–oil ratio 110 m³/m³
Formation volume factors: oil $B_o = 1.335$ , reservoir water $B_w = 1.02$ , injected water 1.01 m³/m³

Solution

Gas at the surface: $120\cdot 110 = 13{,}200$ m³/d. All of it is dissolved, because there is no free gas in the reservoir.
Oil in the reservoir: $120\cdot 1.335 = 160.2$ m³/d. Water: $30\cdot 1.02 = 30.6$ m³/d.
Withdrawal from the reservoir: $160.2 + 30.6 = 190.8$ m³/d.
To replace it fully, we must inject $190.8/1.01 = 188.9$ m³/d of water.
In oilfield units: oil 755 STB/d, gas 0.47 million scf/d.

Answer: The area withdraws 190.8 m³/d from the reservoir. To replace this fully, we must inject about 190 m³/d of water, which is more than one and a half times the oil produced at the surface.

What it means. If we inject water “a cubic metre for every cubic metre of liquid produced” (150 m³/d), the pressure will fall. This ignores that each cubic metre of stock-tank oil took up 1.335 m³ in the reservoir. The data in this example are assumed for illustration.

Below the bubble point pressure: dissolved and free gas

When the reservoir pressure falls below the bubble point pressure, free gas appears in the pores. While its saturation (the share of the pore space it fills) is below the critical value (end-point saturations), the gas is immobile. The producing gas–oil ratio may even fall a little, because part of the gas stays in the reservoir. Then the gas starts to flow towards the wells. It flows faster than the oil, because it is much less viscous, and the producing gas–oil ratio rises. The gas at the surface is the sum of the gas dissolved in the produced oil ( $R_s$ for every cubic metre) and the free gas that came from the reservoir ( $R - R_s$ ).

Gas formation volume factor

B_g = \frac{p_{\text{sc}}}{p}\cdot\frac{Z\,T}{T_{\text{sc}}}

where:

$B_g$ volume of the gas in the reservoir per unit of its volume at standard conditions, m³/m³
$p,\ T$ reservoir pressure and temperature, both absolute, MPa, K
$p_{\text{sc}},\ T_{\text{sc}}$ standard conditions: 0.101325 MPa and 293.15 K (20 °C), MPa, K
$Z$ gas deviation factor at $p$ and $T$ , —

Units: absolute pressures and temperatures in consistent units

Holds when

The gas is single-phase. The black oil model does not track its composition as the pressure falls.
$Z$ has been found for this gas at reservoir conditions (Ideal and real gas: the gas deviation factor (Z)).

Gas in the reservoir is squeezed tens to hundreds of times. At 10 MPa, 85 °C and $Z = 0.87$ , $B_g \approx 0.0108$ m³/m³. That is, a cubic metre of reservoir gas becomes about 93 m³ of gas at the surface. For more detail see Gas formation volume factor, compressibility and viscosity.

In other unit systems (1)

Oilfield units

B_g\,[\text{rb/scf}] = 0.00504\,\frac{Z\,T\,[{}^{\circ}\text{R}]}{p\,[\text{psia}]}

$T$ is in degrees Rankine, $T\,[{}^{\circ}\text{R}] = t\,[{}^{\circ}\text{F}] + 459.67$ ; $p$ is in psia; $B_g$ is in reservoir barrels per standard cubic foot.

$0.00504 = \dfrac{14.696}{519.67\cdot 5.6146}$ : standard conditions of 14.696 psia and 60 °F (519.67 °R), and the conversion of cubic feet into barrels.

Worked exampleWithdrawal from the reservoir when the producing gas–oil ratio is above the solution gas–oil ratio

Given (assumed data)

Reservoir pressure: 12 MPa (below the bubble point pressure of 18 MPa)
Well measurement: oil 100 m³/d, producing gas–oil ratio 300 m³/m³, no water
Properties at 12 MPa: $R_s = 76$ m³/m³, $B_o = 1.258$ m³/m³, $B_g = 0.00897$ m³/m³

Solution

Dissolved gas that came with the oil: $q_o R_s = 100\cdot 76 = 7600$ m³/d.
Free gas from the reservoir: $q_o (R - R_s) = 100\cdot(300 - 76) = 22{,}400$ m³/d at standard conditions. In the reservoir it took up $22{,}400\cdot 0.00897 = 200.9$ m³/d.
Oil with its dissolved gas in the reservoir: $q_o B_o = 100\cdot 1.258 = 125.8$ m³/d.
Withdrawal from the reservoir: $125.8 + 200.9 = 326.7$ m³/d. Free gas accounts for 61% of this.

Answer: The well withdraws about 327 m³/d of fluids from the reservoir, although at the surface this is 100 m³ of oil and 30,000 m³ of gas.

What it means. Free gas takes up more room in the reservoir than the oil. When the producing gas–oil ratio is four times the solution gas–oil ratio, free gas makes up 61% of the withdrawal. That is why, under solution gas drive (when the energy comes from gas coming out of the oil), the pressure falls quickly and the gas “carries off” reservoir energy. So operators try to keep the pressure in accumulations above the bubble point pressure (Reservoir drive mechanisms). The data in this example are assumed for illustration.

Total formation volume factor

Imagine that a cubic metre of stock-tank oil, together with all the gas dissolved in it at the bubble point pressure, is locked in a vessel at reservoir temperature. Above the bubble point pressure the vessel holds a single liquid of volume $B_o$ . Below it, the vessel holds oil of volume $B_o$ and the gas that has come out of it. That gas is $R_{sb} - R_s$ standard cubic metres, and at the current pressure it takes up $B_g(R_{sb} - R_s)$ .

Total (two-phase) formation volume factor

B_t = B_o + B_g\,(R_{sb} - R_s)

where:

$B_t$ volume of the oil and the gas freed from it per 1 m³ of stock-tank oil, m³/m³
$B_o,\ B_g,\ R_s$ formation volume factors of oil and gas and the solution gas–oil ratio at the current pressure, m³/m³
$R_{sb}$ solution gas–oil ratio at the bubble point pressure; for an undersaturated accumulation it equals the initial $R_{si}$ , m³/m³

Units: m³/m³; in oilfield units, rb/STB with $B_g$ in rb/scf and $R_s$ in scf/STB

Holds when

The freed gas stays with its oil (a closed system).
$B_o$ , $B_g$ and $R_s$ refer to the same method of degassing.

Above the bubble point pressure, $B_t = B_o$ . Below it, $B_t$ grows quickly, because the freed gas expands as the pressure falls. In the reservoir the gas moves away from its oil, towards the wells and upwards into the gas cap. So $B_t$ does not describe the volume at any particular place. It is a convenient bookkeeping quantity: the material balance uses it to write the expansion of the oil together with the gas freed from it.

CalculatedThe total formation volume factor

B_t

and the oil formation volume factor

B_o

for the assumed oil of the lesson. Above the bubble point pressure (18 MPa) they are equal. Below it,

B_t

grows quickly because of the expanding gas.

Worked exampleHow much the oil expands together with the gas freed from it

Given (assumed data)

Bubble point pressure: 18 MPa, $R_{sb} = 110$ m³/m³, $B_{ob} = 1.35$ m³/m³
At 12 MPa: $R_s = 76$ , $B_o = 1.258$ , $B_g = 0.00897$ m³/m³
At 6 MPa: $R_s = 41$ , $B_o = 1.162$ , $B_g = 0.01795$ m³/m³
Above the bubble point pressure: at 30 MPa $B_o = 1.324$ m³/m³

Solution

At 12 MPa: $B_t = 1.258 + 0.00897\cdot(110 - 76) = 1.563$ m³/m³.
At 6 MPa: $B_t = 1.162 + 0.01795\cdot(110 - 41) = 2.40$ m³/m³.
Expansion of the system from the bubble point pressure: to 12 MPa, $1.563/1.35 = 1.158$ , which is an expansion by 16%. To 6 MPa, the expansion is 78%.
For comparison, undersaturated oil expanded from 30 to 18 MPa by only $(1.35/1.324 - 1)\cdot 100 = 2.0$ %.

Answer:

B_t

is 1.563 m³/m³ at 12 MPa and 2.40 m³/m³ at 6 MPa. The oil with its freed gas expands by 16% at 12 MPa and by 78% at 6 MPa.

What it means. The expansion of the freed gas is the main source of energy under solution gas drive. Already at 12 MPa it is eight times greater, and at 6 MPa almost forty times greater, than the expansion of undersaturated oil above the bubble point pressure. But once the gas becomes mobile, it reaches the wells faster than the oil, and the energy is wasted. That is why oil recovery under this drive is low. The data in this example are assumed for illustration.

The solution gas–oil ratio and the formation volume factor depend on the path to the surface

Stock-tank oil and gas are not components that exist in the reservoir. They are the products of a particular process. If reservoir oil is degassed straight to atmospheric pressure, more of the intermediate components (propane and butanes) go into the gas, and less oil is left in the stock tank. With two- or three-stage separation, part of these components stays in the liquid. Then the solution gas–oil ratio and the formation volume factor come out smaller, and the amount of stock-tank oil comes out larger.

One reservoir oil, different separation schemes

Scheme: Single stage to atmospheric pressure
$R_s$ , m³/m³: 125
$B_o$ , m³/m³: 1.380
Stock-tank oil from 1 m³ of reservoir oil, m³: 0.725

Scheme: Two stages: 0.7 MPa and atmospheric pressure
$R_s$ , m³/m³: 110
$B_o$ , m³/m³: 1.350
Stock-tank oil from 1 m³ of reservoir oil, m³: 0.741

Scheme	$R_s$ , m³/m³	$B_o$ , m³/m³	Stock-tank oil from 1 m³ of reservoir oil, m³
Single stage to atmospheric pressure	125	1.380	0.725
Two stages: 0.7 MPa and atmospheric pressure	110	1.350	0.741

The data are assumed for illustration, but the pattern is typical: stage separation gives a few per cent more stock-tank oil. The pressures of the stages are chosen by calculation (Separator calculations and optimum separator pressure).

A reservoir oil has no single “correct” $R_s$ and $B_o$ . There are only values for a given separation scheme. In a PVT report, the data of laboratory degassing are converted to the actual field conditions (How to read a PVT report).
The black oil model describes oil by its outcome, which means by what we get at the surface. Sometimes several accumulations with different oils deliver their production into a common pipeline. The mixture then has no $R_s$ and $B_o$ of its own, and production is allocated between the reservoirs by composition or by separate measurements.

When the black oil model is not enough

The black oil model is still a working tool for most oil accumulations. The calculation of hydrocarbons in place, the material balance, the interpretation of well tests and most reservoir simulation models are all based on it. But it has limits.

Black oil model and compositional model

: Components
Black oil: two simplified ones: stock-tank oil and gas
Compositional model: ten or more: methane, ethane, … and heavy pseudo-components

: Properties
Black oil: $R_s$ , $B_o$ , $B_g$ and viscosities, given in tables against pressure
Compositional model: calculated from an equation of state, from the composition, pressure and temperature

: Composition of gas and oil
Black oil: does not change
Compositional model: changes as the pressure falls, with gas injection and when streams are mixed

: Direction of calculation
Black oil: from the products of a given separation backwards to the reservoir oil
Compositional model: from the composition of the reservoir mixture forwards to the products of any processing scheme

: Where it applies
Black oil: oils with a moderate gas content; waterflooding, solution gas drive
Compositional model: volatile oils (light oils with a lot of dissolved gas), gas condensate, gas injection and miscible displacement (the injected fluid mixes with the oil), design of fluid processing

: Cost
Black oil: fast calculations, little input data
Compositional model: needs a full PVT analysis, and calculations take several times longer

	Black oil	Compositional model
Components	two simplified ones: stock-tank oil and gas	ten or more: methane, ethane, … and heavy pseudo-components
Properties	$R_s$ , $B_o$ , $B_g$ and viscosities, given in tables against pressure	calculated from an equation of state, from the composition, pressure and temperature
Composition of gas and oil	does not change	changes as the pressure falls, with gas injection and when streams are mixed
Direction of calculation	from the products of a given separation backwards to the reservoir oil	from the composition of the reservoir mixture forwards to the products of any processing scheme
Where it applies	oils with a moderate gas content; waterflooding, solution gas drive	volatile oils (light oils with a lot of dissolved gas), gas condensate, gas injection and miscible displacement (the injected fluid mixes with the oil), design of fluid processing
Cost	fast calculations, little input data	needs a full PVT analysis, and calculations take several times longer

For gas condensate and volatile systems, people often use an extended black oil model, which also accounts for liquid hydrocarbons dissolved in the gas.

The topic Types of reservoir fluids shows how to tell these fluid types apart by their solution gas–oil ratio and by their behaviour as the pressure falls. Calculation with equations of state is in Cubic equations of state. Estimating $R_s$ , $B_o$ and the bubble point pressure without laboratory data is in PVT correlations for oil and gas condensate.

Terms and symbols

Stock-tank oil (STO): Oil from which gas has come out under the given separation, converted to standard conditions. $R_s$ and $B_o$ are given per unit volume of it.
Solution gas–oil ratio $R_s$: The standard volume of gas dissolved in 1 m³ of stock-tank oil at the given pressure and temperature. It is measured in m³/m³ or scf/STB.
Producing gas–oil ratio $R$: The ratio of all the produced gas to the produced oil (at standard conditions). Below the bubble point pressure it includes free gas.
Oil formation volume factor $B_o$: The reservoir volume of the oil with its dissolved gas per 1 m³ of stock-tank oil.
Shrinkage factor $\theta = 1/B_o$: The fraction of the reservoir volume of oil that remains oil in the stock tank.
Total formation volume factor $B_t$: The volume of the oil together with the gas freed from it, per 1 m³ of stock-tank oil. Above the bubble point pressure it equals $B_o$ .
Bubble point pressure $p_b$: The pressure at which the first bubbles of gas come out of the oil as the pressure falls.

Common mistakes and misconceptions

Wrong

“A producing gas–oil ratio of 300 m³/m³ means that 300 m³ of gas is dissolved in each cubic metre of oil.”

In fact

Below the bubble point pressure the producing gas–oil ratio includes free gas from the reservoir. The amount dissolved is whatever $R_s$ shows at the current pressure.

Wrong

“The formation volume factor is a constant for the oil of a field.”

In fact

$B_o$ depends on pressure (it is at its maximum at the bubble point pressure), temperature and separation conditions. For the estimate of oil in place, use $B_{oi}$ , the value at the initial pressure.

Wrong

“Reservoir cubic metres of oil can be compared directly with production in tonnes.”

In fact

First divide the reservoir volume by $B_o$ to get standard cubic metres. Only then multiply by the density of the stock-tank oil.

Wrong

“Since $B_t$ is so large below the bubble point pressure, the gas of this oil takes up a lot of room in the reservoir.”

In fact

$B_t$ describes a closed system. In the reservoir the gas moves to the wells and into the gas cap, so $B_t$ is a bookkeeping quantity of the material balance.

Wrong

“For gas it does not matter what temperature the standard conditions refer to.”

In fact

Going from 20 °C to 60 °F (15.6 °C) changes the volume of gas by about 1.5%. State the standard conditions next to every number.

In depth: advanced detail

The assumed oil of the examples and figures

All the examples and figures of the lesson are built for one assumed oil. Its bubble point pressure is 18 MPa at 85 °C, $R_{sb} = 110$ m³/m³, $B_{ob} = 1.35$ m³/m³, and its compressibility above the bubble point pressure is $1.6\cdot 10^{-3}$ MPa⁻¹. Below the bubble point pressure a simple approximation gives its properties:

Approximation of the properties of the assumed oil below the bubble point pressure

R_s = R_{sb}\left(\frac{p}{p_b}\right)^{0.9},\qquad B_o = B_{od} + (B_{ob} - B_{od})\,\frac{R_s}{R_{sb}}

where:

$B_{od}$ formation volume factor of the stock-tank oil at reservoir temperature and atmospheric pressure (thermal expansion only); we take 1.05, m³/m³
$p,\ p_b$ pressure and bubble point pressure, MPa
$R_s,\ R_{sb},\ B_o,\ B_{ob}$ as in the lesson, m³/m³

Units: MPa, m³/m³

Holds when

This is an approximation for one assumed oil, not a correlation. The exponent 0.9 and the linear link between $B_o$ and $R_s$ were chosen to reproduce the typical shape of laboratory curves.
A constant $Z = 0.87$ is taken for $B_g$ . This is a simplification that is acceptable for illustration in the range 5–18 MPa. The standard conditions are 0.101325 MPa and 20 °C.

The linear link between $B_o$ and the solution gas–oil ratio reflects the physics: the volume of the oil grows almost in proportion to the amount of dissolved gas. Take the real dependences from PVT studies of a sample of your own oil.

Worked exampleProperties of the assumed oil at 12 and 6 MPa

Given (assumed data)

Parameters: $p_b = 18$ MPa, $R_{sb} = 110$ m³/m³, $B_{ob} = 1.35$ , $B_{od} = 1.05$ m³/m³, $T = 85$ °C, $Z = 0.87$

Solution

At 12 MPa: $R_s = 110\cdot(12/18)^{0.9} = 76.4$ m³/m³, $B_o = 1.05 + 0.30\cdot 76.4/110 = 1.258$ m³/m³.
$B_g = \dfrac{0.101325}{12}\cdot\dfrac{0.87\cdot 358.15}{293.15} = 0.00897$ m³/m³.
At 6 MPa: $R_s = 40.9$ , $B_o = 1.162$ , $B_g = 0.01795$ m³/m³.
Above the bubble point pressure, at 30 MPa: $B_o = 1.35\,\e^{-0.0016\cdot 12} = 1.324$ m³/m³.

Answer: The examples of the lesson use these values, rounded: at 12 MPa

R_s \approx 76

B_o \approx 1.258

B_g \approx 0.00897

; at 6 MPa

R_s \approx 41

B_o \approx 1.162

B_g \approx 0.01795

What it means. The charts of $R_s$ , $B_o$ and $B_t$ in the lesson are calculated with the same formulas. The data are assumed for illustration.

Where

B_o

comes from: the mass balance

Mass does not change on the way from the reservoir to the surface. The mass of $B_o$ cubic metres of reservoir oil equals the mass of a cubic metre of stock-tank oil plus $R_s$ cubic metres of gas. From this we get the formation volume factor in terms of densities:

Oil formation volume factor from the mass balance

B_o = \frac{\rho_{o,\text{sc}} + R_s\,\rho_{g,\text{sc}}}{\rho_{o}(p, T)}

where:

$\rho_{o,\text{sc}}$ density of the stock-tank oil at standard conditions, kg/m³
$\rho_{g,\text{sc}}$ density of the freed gas at standard conditions: its specific gravity relative to air × 1.204 kg/m³ (the density of air at 20 °C), kg/m³
$\rho_o(p, T)$ density of the reservoir oil with its dissolved gas, kg/m³
$R_s$ solution gas–oil ratio, m³/m³

Units: SI (densities in kg/m³)

Holds when

Every component of the reservoir oil goes either into the stock-tank oil or into the gas that is counted; there are no losses.

The formula links the formation volume factor with the density of the reservoir oil (Density and compressibility of reservoir oil). It lets us check that PVT data are consistent. For example, take $\rho_{o,\text{sc}} = 850$ kg/m³, a gas of specific gravity 0.80 ( $\rho_{g,\text{sc}} = 0.963$ kg/m³), $R_s = 110$ m³/m³ and a reservoir oil density of 708 kg/m³. Then $B_o = (850 + 106)/708 \approx 1.35$ , the same as for the assumed oil of the lesson.

Connections to other topics

Types of reservoir fluids — how to tell black oil from volatile oil and gas condensate by its solution gas–oil ratio and its behaviour as the pressure falls.
Gas formation volume factor, compressibility and viscosity — the formation volume factor, compressibility and viscosity of gas.
Density and compressibility of reservoir oil — the density of reservoir oil and its compressibility above the bubble point pressure.
Laboratory PVT tests — how the solution gas–oil ratio, the formation volume factor and the bubble point pressure are measured in the laboratory.
General material balance equation for oil reservoirs — where $B_o$ , $B_t$ , $B_g$ and $R_s$ are put to work: the balance of withdrawal and fluid expansion.
PVT correlations for oil and gas condensate — estimating oil properties without laboratory data, and the limits of applicability of correlations.

Check yourself

1
Undersaturated oil: $B_o = 1.30$ m³/m³, $R_s = 90$ m³/m³. A well produces 200 m³/d of stock-tank oil. How much gas does it produce, and how much oil does it withdraw from the reservoir?
1. A.13,800 m³/d of gas; 260 m³/d from the reservoir
2. B.There is no gas, because the oil is undersaturated
3. C.18,000 m³/d of gas; 260 m³/d from the reservoir
4. D.18,000 m³/d of gas; 154 m³/d from the reservoir
Show the answer
Correct answer: C. 18,000 m³/d of gas; 260 m³/d from the reservoir
Gas: $200\cdot 90 = 18{,}000$ m³/d. From the reservoir: $200\cdot 1.30 = 260$ m³/d. Undersaturation only means that there is no free gas in the reservoir. The dissolved gas comes out at the surface.
2
How does $B_o$ change as the pressure falls from an initial value above the bubble point pressure to atmospheric?
1. A.It first grows slightly down to the bubble point pressure, then falls
2. B.It does not change down to the bubble point pressure, then grows
3. C.It falls steadily
4. D.It grows steadily
Show the answer
Correct answer: A. It first grows slightly down to the bubble point pressure, then falls
Above the bubble point pressure the oil is single-phase and simply expands as the pressure is relieved, so $B_o$ grows to its maximum $B_{ob}$ . Below it, gas comes out of the oil, the oil shrinks and $B_o$ falls.
3
What is the total formation volume factor $B_t$ above the bubble point pressure?
1. A. $B_o$
2. B. $B_o + B_g R_s$
3. C. $B_g$
4. D.1
Show the answer
Correct answer: A. $B_o$
Above the bubble point pressure, $R_s = R_{sb}$ , so the term $B_g(R_{sb} - R_s)$ is zero and $B_t = B_o$ . There is no free gas.
4
The reservoir pressure is 10 MPa, below the bubble point pressure. A well produces 50 m³/d of oil at a producing gas–oil ratio of 400 m³/m³; $R_s = 60$ m³/m³, $B_o = 1.20$ , $B_g = 0.0108$ m³/m³. Find the withdrawal from the reservoir and the share of free gas in it.
Show the answer
Free gas: $50\cdot(400 - 60) = 17{,}000$ m³/d, which in the reservoir is $17{,}000\cdot 0.0108 = 183.6$ m³/d. Oil: $50\cdot 1.20 = 60$ m³/d. The total is $243.6$ m³/d, and free gas is about 75% of it.
5
Why are the solution gas–oil ratio and the formation volume factor of one and the same reservoir oil greater for single-stage degassing to atmospheric pressure than for two-stage separation?
Show the answer
When the pressure is reduced abruptly in a single stage, more of the intermediate components (propane and butanes) go into the gas. There is more gas per cubic metre of stock-tank oil and less oil, so both $R_s$ and $B_o$ (the reservoir volume per cubic metre of stock-tank oil) are higher. With stage separation these components partly stay in the liquid.
6
A PVT report gives a solution gas–oil ratio of 95 m³/t and a stock-tank oil density of 840 kg/m³. Express it in m³/m³ and in scf/STB.
Show the answer
$95\cdot 0.840 = 79.8$ m³/m³, and $79.8\cdot 5.615 \approx 448$ scf/STB for identical standard conditions. If the cubic metres of gas are converted to 20 °C, but you need a conversion to 60 °F, reduce the result by a further 1.5% or so: about 441 scf/STB.

Updated: 19 October 2026