Types of reservoir simulation model

Quick refresher

Independent features tell models apart: the physics of the fluids, the description of the rock and the scale. Any combination is possible. For each feature we choose the simplest option that is enough for the question.
A black oil model keeps track of (conserves) three notional components, that is, the three surface products: oil, gas and water. Gas can dissolve in the oil, and the properties depend on pressure alone. This is enough for most problems: depletion (pressure decline), waterflooding and immiscible gas injection (gas that does not mix with the oil).
A compositional model keeps track of each component of the mixture. In each cell it calculates the phase equilibrium (how the mixture splits into oil and gas) from an equation of state. We need it when the composition of the phases changes appreciably: near-critical fluids (fluids close to the point where oil and gas become alike), gas cycling into a gas-condensate reservoir (injecting the produced gas back) and miscible displacement.
A thermal model adds temperature and an energy balance. In steam injection the main effect is that heating makes the viscosity of heavy oil fall by orders of magnitude (many times over). Chemical models add components in the water: polymer and surfactants (soap-like substances).
Dual porosity describes a fractured reservoir as two media laid over each other. The matrix (the rock itself, with its small pores) stores the oil, the fractures carry it, and oil passes between the two.
A full-field model answers for the field as a whole. A sector model gives more detail, but at the price of having to set conditions at its boundaries. A radial single-well model (a round grid around one well) shows coning and near-wellbore processes. The cost of a run grows with the number of cells, the unknowns per cell and the time steps. On the same grid, a compositional model costs several times as much as a black oil model, and often ten times as much.

Key relationships

N_o = \dfrac{V_p S_o}{B_o},\qquad G = V_p\left(\dfrac{S_g}{B_g} + \dfrac{R_s S_o}{B_o}\right)

oil and gas in a cell of a black oil model, m³ at standard conditions

m_i = V_p \sum_j S_j\,C_{ij}

mass of component $i$ in a cell of a compositional model; $C_{ij}$ is its concentration in phase $j$

\mu = A\,\e^{B/T}

the viscosity of oil falls quickly as the absolute temperature $T$ (in kelvin) rises

N_{\text{unk}} = N_{\text{cells}}\cdot n

the number of unknowns: the number of cells times the number of equations per cell ( $n = 3$ for black oil)

Common pitfalls

Choosing the most complex model “just in case”. It is slower, needs more data and leaves less time for the options.
Calculating the gas dissolved in the oil with the gas formation volume factor. We find it from the volume of oil at standard conditions, $R_s S_o/B_o$ .
Switching on dual porosity only because there are fractures in the core.
Carrying the conclusion of a sector model over to the field without checking the conditions at its boundaries.
Assuming that a compositional model is “more accurate” whatever the data. Without reliable analyses of fluid samples it is only slower.

After this lesson you can

name the independent features that tell models apart: the description of the fluids, the description of the rock and the scale
explain what is conserved in a cell of a black oil model and of a compositional model, and calculate the amounts of oil and gas in a cell
state when a black oil model is not enough and a compositional, thermal or chemical model is needed
explain the idea of dual porosity and when it is justified
choose the scale of a model — full-field, sector or single-well — to fit the task, and name the limits of each
estimate how the type of model affects the number of unknowns and the run time

How models differ from one another

Every reservoir simulator does the same thing. It divides the reservoir into cells, keeps a balance in each cell and calculates the flows by Darcy's law (What a reservoir simulation model is and why it is needed). But models of the same reservoir can be very different. Three questions tell them apart: what is conserved in a cell, how the rock is described and what part of the reservoir the grid covers. A fourth choice, about the calculation itself, comes on top of these three: an ordinary grid calculation, or a streamline calculation (a calculation along the paths that the fluid follows).

Features by which reservoir simulation models are distinguished

Feature: Fluids and processes
Options: black oil; compositional; thermal; chemical
What it affects: which displacement processes the model can reproduce

Feature: Rock
Options: a single porous medium; dual porosity or dual permeability
What it affects: how the model sees a fractured reservoir

Feature: Scale
Options: full-field; sector; pattern element (the repeating unit of a well pattern); cross-section; single-well
What it affects: which question the model answers and how much detail it has

Feature: Method of calculation
Options: ordinary grid-based; streamline
What it affects: how fast the calculation runs, and how much of the physics it includes

Feature	Options	What it affects
Fluids and processes	black oil; compositional; thermal; chemical	which displacement processes the model can reproduce
Rock	a single porous medium; dual porosity or dual permeability	how the model sees a fractured reservoir
Scale	full-field; sector; pattern element (the repeating unit of a well pattern); cross-section; single-well	which question the model answers and how much detail it has
Method of calculation	ordinary grid-based; streamline	how fast the calculation runs, and how much of the physics it includes

The lesson Grids in reservoir simulation models covers grids and the dimensions of models. Here we look at what is calculated on a grid and what part of the reservoir it covers.

The features are independent. There are compositional sector models, full-field black oil models with dual porosity, and thermal single-well models. So we choose the type of model separately for each feature, and the general rule is the same: the simplest option that is enough for the question.

SchematicWhat a cell stores in models of different types (schematics). A black oil model has three notional components, and the gas is partly dissolved in the oil. A compositional model has the components of the mixture in each phase. A thermal model adds temperature. With dual porosity, a matrix cell has a “twin”, a fracture cell.

The black oil model

The black oil model is the most widely used. It has three phases — oil, gas and water — and three notional components with the same names: stock-tank oil (oil with its gas removed), gas and water, in the form in which we measure them at the surface. The “gas” component can be in the gas phase or dissolved in the oil. The solution gas–oil ratio $R_s$ shows how much of it is dissolved. We give all the properties — $R_s$ , the formation volume factors, the viscosities — as tables against pressure (Black oil model: solution gas–oil ratio and formation volume factors).

Amounts of oil, water and gas in a cell of a black oil model

N_o = \frac{V_p S_o}{B_o}, \qquad N_w = \frac{V_p S_w}{B_w}, \qquad G = V_p\left(\frac{S_g}{B_g} + \frac{R_s\,S_o}{B_o}\right)

where:

$N_o,\ N_w,\ G$ amounts of the components “oil”, “water” and “gas” in the cell, m³ at standard conditions
$V_p$ pore volume of the cell: its volume multiplied by the porosity, m³
$S_o,\ S_w,\ S_g$ oil, water and free-gas saturations (the shares of the pore volume that each fluid fills); they add up to one, fraction
$B_o,\ B_w,\ B_g$ formation volume factors of oil, water and gas at the pressure in the cell (each is the volume of the fluid in the reservoir divided by its volume at the surface; surface volumes are at 0.101325 MPa and 20 °C), m³/m³
$R_s$ solution gas–oil ratio of the oil (the gas dissolved in a unit volume of oil) at the pressure in the cell, m³/m³

Units: metric; standard conditions are the ones that the property tables refer to

Holds when

The composition of the “oil” and the “gas” does not change. Their properties depend on pressure alone, and the reservoir temperature is constant.
Gas dissolves in oil, but oil does not dissolve in gas. The extended model adds the content of liquid hydrocarbons in the gas.
We neglect the gas dissolved in water. If necessary, we add the same kind of term for water.
We get the mass of a component by multiplying by its density at standard conditions.

The gas amount has two terms: free gas and gas dissolved in the oil. We find the dissolved gas from the volume of oil at standard conditions. That is why the second term has $B_o$ in its denominator, not $B_g$ . The simulator makes sure that the change in $N_o$ , $N_w$ and $G$ in a cell over a step equals the flows and withdrawals (the balance in a cell).

Worked exampleHow much oil and gas has left the cell

Given (assumed data)

Cell: 100 × 100 × 5 m, porosity 0.20; pore volume 10,000 m³ (we neglect its change as the pressure falls)
Oil: the assumed oil of the lesson Black oil model: solution gas–oil ratio and formation volume factors: bubble point pressure 18 MPa, $R_s = 110$ m³/m³ above it
Step 1: pressure 25 MPa: $S_w = 0.25$ , $S_o = 0.75$ , no free gas; $B_o = 1.3350$
Step 2: pressure 12 MPa: $S_w = 0.25$ , $S_o = 0.66$ , $S_g = 0.09$ ; $R_s = 76.37$ , $B_o = 1.2583$ , $B_g = 0.008975$ m³/m³

Solution

Step 1. Oil: $10{,}000\cdot 0.75/1.3350 = 5618$ m³. All the gas is dissolved: $5618\cdot 110 = 618$ thousand m³.
Step 2. Oil: $10{,}000\cdot 0.66/1.2583 = 5245$ m³.
Step 2. Free gas: $10{,}000\cdot 0.09/0.008975 = 100$ thousand m³; dissolved gas: $5245\cdot 76.37 = 401$ thousand m³; in all 501 thousand m³.
The cell has lost 373 m³ of oil (6.6% of the initial amount) and 117 thousand m³ of gas, that is, 314 m³ of gas per cubic metre of oil.

Answer: Between the two steps the cell gave up 373 m³ of oil and 117 thousand m³ of gas (at standard conditions). For each cubic metre of oil that left, 314 m³ of gas left too, against an initial solution gas–oil ratio of 110 m³/m³.

What it means. Almost three times as much gas left as was dissolved in the oil that left. Free gas moves more easily than oil, so it leaves the cell faster. This is how a black oil model reproduces the rise in the producing gas–oil ratio under solution gas drive (the drive from gas coming out of the oil) without knowing the composition of either the oil or the gas. Three tables are enough for it. The amount of water in the cell has stayed the same, because its saturation and formation volume factor have hardly changed. The data in this example are assumed for illustration.

A cell has three balance equations and three unknowns — the pressure and two saturations. When there is no free gas in the cell, the third unknown is the solution gas–oil ratio instead of the gas saturation. The simulator switches the variables by itself.

A black oil model can handle all the processes in which the oil and the gas do not change composition — and these are the majority. They include all the natural drive mechanisms (Reservoir drive mechanisms): expansion (elastic), solution gas, gas cap and water drive. They include waterflooding, together with imbibition (water soaking into the rock) and gravity segregation (water and oil separating by weight). They include the injection of gas that does not mix with the oil, also in alternation with water. Finally, they include gas reservoirs: for them the same model works with two phases, gas and water.

The model stops working when its main assumption — the constant composition of the “oil” and the “gas” — is broken. Three tables against pressure are then no longer enough. This happens if the gas released from the oil at the start of production differs appreciably from the gas released later. It happens if the injected gas is unlike the reservoir gas. It also happens if the oil and the gas exchange components so much that the boundary between the phases disappears.

The compositional model

The compositional model gives up the notional “oil” and “gas”. It keeps track of the components of the mixture itself — methane, ethane, propane and so on — while the heavy fractions are lumped into a few pseudo-components. Each component can be in both the oil phase and the gas phase. At each step, in each cell, the simulator works out how a component splits between the two phases at the given pressure, temperature and overall composition. It does this with an equation of state, a formula for the behaviour of a fluid mixture (Cubic equations of state, Vapour–liquid equilibrium). Then it finds the densities and viscosities of the phases from their composition.

Mass of a component in a cell of a compositional model

m_i = V_p \sum_{j} S_j\,C_{ij}

where:

$m_i$ mass of component $i$ in the cell, kg
$V_p$ pore volume of the cell, m³
$S_j$ saturation of phase $j$ — gas, oil or water, fraction
$C_{ij}$ mass concentration of component $i$ in phase $j$ : the density of the phase multiplied by the mass fraction of the component in it, kg/m³

Units: SI; instead of mass we often calculate the number of moles — then the concentrations are molar

Holds when

The phases in a cell are in equilibrium: the composition of each follows from the overall composition, the pressure and the temperature.
We sum over all the phases in which the component is present. We usually neglect hydrocarbon components in the water.

We write a balance for each component, so a cell has as many equations as there are components. A black oil model is a special case with two hydrocarbon components: the concentration of “gas” in the oil phase equals $\rho_{g,\text{sc}}R_s/B_o$ , and tables work in place of an equation of state.

Worked exampleOne component in two phases

Given (assumed data)

Cell: pore volume 10,000 m³; $S_g = 0.20$ , $S_o = 0.55$ , $S_w = 0.25$
Gas phase: density 180 kg/m³, methane mass fraction 0.70
Oil phase: density 650 kg/m³, methane mass fraction 0.08

Solution

Methane concentration in the gas phase: $180\cdot 0.70 = 126$ kg/m³; in the oil phase: $650\cdot 0.08 = 52$ kg/m³.
Methane in the gas phase: $10{,}000\cdot 0.20\cdot 126 = 252$ t; in the oil phase: $10{,}000\cdot 0.55\cdot 52 = 286$ t.
Total methane in the cell: 538 t; the share in the oil phase is 53%.

Answer: The cell contains 538 t of methane, and more than half of it — 53% — is in the oil phase.

What it means. A component and a phase are different things. Methane, a “gas” component, is mostly dissolved in the oil, although its share in the oil is small. When the pressure falls, or when the oil meets injected gas, these shares change, and so do the densities and viscosities of both phases. A compositional model tracks this for every component in every cell. The data in this example are assumed for illustration.

We need a compositional model where the composition of the phases changes so much that it changes the course of the process.

Fluids near the critical point (the point where oil and gas become alike) — volatile oils and rich gas-condensate mixtures (Types of reservoir fluids): a small change in pressure strongly changes the composition and properties of both phases.
Gas-condensate reservoirs with cycling of dry gas (gas cycling): the injected gas vaporises the condensate that has dropped out as liquid, and the composition of the produced mixture changes along the whole path from the injection well to the production well.
Miscible displacement — the injection of enriched gas or carbon dioxide: the gas and the oil exchange components until the boundary between them disappears. Miscibility usually does not arise at once but after repeated contacts, and we have to calculate this exchange (Displacement of oil by gas).
Reservoirs with variable composition — with depth or across the area — and the joint development of layers with different fluids.
Problems that need the composition of the production: the design of gas and condensate processing, and estimates of the yield of liquid hydrocarbons year by year.

Black oil and compositional models in a simulator

: What is conserved
Black oil: oil, gas, water
Compositional: each component or pseudo-component, and water

: Exchange between phases
Black oil: solution gas–oil ratio from a table against pressure
Compositional: phase equilibrium from an equation of state in every cell

: Unknowns per cell
Black oil: 3
Compositional: the number of components plus water — usually from 5 to 15

: Fluid data
Black oil: tables of $R_s$ , $B$ and viscosities
Compositional: the composition and an equation of state tuned to laboratory experiments

	Black oil	Compositional
What is conserved	oil, gas, water	each component or pseudo-component, and water
Exchange between phases	solution gas–oil ratio from a table against pressure	phase equilibrium from an equation of state in every cell
Unknowns per cell	3	the number of components plus water — usually from 5 to 15
Fluid data	tables of $R_s$ , $B$ and viscosities	the composition and an equation of state tuned to laboratory experiments

The number of pseudo-components is a compromise between the accuracy of the fluid description and the run time.

A compositional model costs twice. First, it costs data: we need representative samples, a full set of laboratory tests and an equation of state tuned to them (Laboratory PVT tests). Without these, the model will carefully calculate the wrong fluid. Second, it costs run time (see the section on cost below). In addition, miscibility depends on the composition in each cell, so numerical smearing of the front (numerical dispersion) distorts such calculations more than it distorts waterflood calculations, and we need a finer grid.

Thermal and chemical models

In an ordinary model the reservoir temperature is constant. A thermal model makes it an unknown, and adds an energy balance to the mass balances. Heat is carried by the fluids and by conduction (heat passing through the solid rock), is stored in the rock and in the fluids, and is lost to the rocks above and below the reservoir.

We need it first of all for heavy oil. Its viscosity (resistance to flow) is hundreds or thousands of mPa·s, and at reservoir temperature such oil hardly moves. But the viscosity of a liquid falls very fast on heating, and all thermal methods of production are based on this.

Dependence of oil viscosity on temperature (the Andrade equation)

\mu(T) = A\,\e^{B/T}

where:

$\mu$ dynamic viscosity of the oil, mPa·s
$T$ absolute temperature: $T\,[\text{K}] = t\,[{}^{\circ}\text{C}] + 273.15$ , K
$A$ a constant of the given oil, mPa·s
$B$ a constant of the given oil showing how steeply the viscosity falls with temperature, K

Units: viscosity in the units of the constant $A$ ; temperature in kelvin only

Holds when

It is an empirical relationship (fitted to measurements): we choose the constants to match the measured viscosity of this oil at two temperatures at least.
The oil is degassed (it has lost its gas), or its solution gas–oil ratio does not change. Pressure affects viscosity much less than temperature does.
Over a wide range of temperatures, the straight line on a plot of $\ln\mu$ against $1/T$ bends. So for heavy oils we use more accurate relationships, and we enter a table of laboratory values into the model.

The greater $B$ is, the more the oil “thins” on heating. For heavy oils $B$ is large, so heating by a hundred degrees cuts their viscosity tens or hundreds of times. For light oils it cuts it only a few times.

Worked exampleHow much heating reduces the viscosity of heavy oil

Given (assumed data)

Viscosity at 30 °C (reservoir temperature): 2000 mPa·s
Viscosity at 150 °C: 20 mPa·s
To find: the viscosity at 80 and 250 °C and the gain in rate

Solution

We find the constant $B$ from the two points: $\ln(2000/20)$ divided by $1/303.15 - 1/423.15$ gives 4923 K.
Then $A = 2000/\e^{4923/303.15}$ , that is, $1.77\cdot 10^{-4}$ mPa·s.
At 80 °C (353.15 K): $\mu = 201$ mPa·s. At 250 °C (523.15 K): $\mu = 2.16$ mPa·s.
By Darcy's law, at the same drawdown the rate is inversely proportional to the viscosity. Heating to 150 °C increases the rate by a factor of $2000/20 = 100$ . Heating to 250 °C increases it by a factor of $2000/2.16$ , that is, about 900.

Answer: Heating from 30 to 150 °C reduces the viscosity by a factor of 100, and heating to 250 °C by a factor of about 900.

What it means. The effect is huge, but we get it only where the heat has reached. The temperature in the reservoir changes from point to point and with time: the heated zone grows, the steam condenses, and part of the heat goes into the neighbouring rocks. So we cannot set the viscosity in advance with a single table against pressure. We have to calculate it from the temperature of each cell, and find the temperature from the energy balance. The data in this example are assumed for illustration.

CalculatedViscosity of the assumed heavy oil of the example against temperature (the viscosity axis is logarithmic): from 2000 mPa·s at 30 °C to 20 mPa·s at 150 °C and about 2 mPa·s at 250 °C.

Processes for which we build thermal models:

cyclic steam stimulation (huff-and-puff) in a production well: injection, a soak period to heat the near-wellbore zone (the rock right around the well), production of the heated oil — and injection again
steam drive: steam is injected continuously into injection wells, and the heating front moves towards the production wells
steam-assisted gravity drainage (SAGD): steam is fed into the upper of two parallel horizontal wells drilled one above the other, and the heated oil, together with the condensed steam, drains under gravity to the lower, production well
in-situ combustion (burning in the reservoir): air is injected into the reservoir, part of the oil burns, and the heat and the combustion gases drive out the rest. The model is supplemented by chemical reactions. This is the most complex and the rarest type of calculation

A thermal model calculates what an ordinary one does not: the evaporation of water and the condensation of steam, the dependence of viscosities and other properties on temperature, and heat losses through the top and the base of the reservoir. Thermal fronts are narrow, so such calculations need small cells. We most often run them on sectors and pattern elements rather than on the whole field.

Chemical flood models extend a black oil model with components dissolved in the water. In polymer flooding, the model carries the polymer with the water as an extra component. From the polymer concentration it recalculates the mobility of the water (how easily the water moves). The solution is more viscous, part of the polymer stays on the rock, and where the polymer has passed, the permeability to water is lower. Surfactants and alkalis lower the interfacial tension (the pull between oil and water at their boundary) and free part of the residual oil (the oil left behind after flooding). Foam reduces the mobility of gas. The same tools calculate how gels shut off water inflow in the near-wellbore zone. When the deformation of the rock itself matters — compaction of the reservoir, changes in permeability, the growth of fractures — we combine the flow calculation with a stress calculation in a geomechanical model.

Dual porosity: fractured reservoirs

A fractured reservoir has two systems of voids (empty spaces) with opposite properties. The matrix — the blocks of rock between the fractures — holds almost all of the pore volume, but its permeability is low. The fractures take up hundredths and tenths of a per cent of the rock volume, but carry fluid orders of magnitude better. The oil is stored in the matrix and travels to the wells along the fractures.

Worked exampleWho stores and who conducts

Given (assumed data)

Matrix: porosity 0.12; permeability 1 mD
Fractures: aperture (width of the opening) 0.1 mm, parallel, every metre (as in the example of the lesson Rock permeability)

Solution

Fracture porosity: $10^{-4}\ \text{m}/1\ \text{m} = 10^{-4}$ , that is, 0.01% (a 0.1 mm gap in every metre of rock). The share of the pore volume that belongs to the fractures: $10^{-4}/(0.12 + 10^{-4}) = 0.08$ %.
The permeability of the fracture itself is $w^2/12 = 844$ D, where $w$ is the aperture. Spread over the whole cross-section of the rock along the fractures, this is $844\cdot 10^{-4}$ D, that is, 84.4 mD.
Along the fractures, the matrix and the fractures work side by side (in parallel), so their permeabilities add: $1 + 84.4 = 85.4$ mD. The share of the flow that goes along the fractures is 98.8%.
If the matrix permeability were 50 mD, only 63% of the flow would go along the fractures.

Answer: The matrix holds 99.9% of the pore volume, and 98.8% of the flow goes along the fractures.

What it means. Storage and flow capacity sit in different media — this is what “dual porosity” means. A single cell with averaged properties would not show such behaviour. In it, the injected water would travel through the whole pore volume. In reality it passes quickly along the fractures and leaves the oil in the matrix. When the matrix is permeable, the difference is smoothed out, and a single medium with higher permeability along the fractures is enough. The data in this example are assumed for illustration.

We cannot describe each fracture with separate cells: there are millions of them. The dual porosity model replaces the real rock with two continuous media laid over each other. At every point of the reservoir there is a matrix cell and a fracture cell, each with its own pressure and its own saturations. The fracture cells are connected with each other as in an ordinary model, and the fluid travels to the wells along them. A matrix cell exchanges fluid only with “its own” fracture cell. If the matrix blocks are permeable enough for fluid to flow from block to block as well, we switch on the connections between the matrix cells too. This is the dual permeability model.

The main thing in such a model is the exchange between the matrix and the fractures. In waterflooding, the water fills the fractures quickly, but the oil leaves the matrix blocks slowly. Two things push it out. One is capillary imbibition (the rock soaks up water by capillary forces), which works if the rock is water-wet (Wettability and interfacial phenomena), that is, if its surface prefers water to oil. The other is the difference between the densities of water and oil. When gas is injected or a gas cap (free gas on top of the oil) expands, the oil drains from the blocks under gravity. The rate of these processes depends on the size and height of the blocks, the matrix permeability and the curves of capillary pressure, and it sets the recovery.

Dual porosity is justified when all three conditions are met at once:

the fractures form a connected network over distances about the size of a cell or more (Fractures and naturally fractured reservoirs)
the fractures carry fluid much better than the matrix, and the matrix stores the main share of the hydrocarbons in place
the process is multiphase: what matters is the rate at which oil leaves the matrix for the fractures

If the fractures are sparse or not connected, we describe the reservoir by a single medium with a higher permeability that depends on direction. We put individual large faults and fractures, known from wells and seismic data, into the grid explicitly. We must also remember the weak point of the approach: the exchange is calculated for idealised identical blocks, while the real sizes of the blocks are poorly known. So the exchange parameters are often just fitted numbers that we adjust in history matching — with all its non-uniqueness (there is no single answer).

Scale of the model: field, sector, well

The third choice is what part of the reservoir the grid covers. Run time limits the number of cells, so coverage and detail must be traded off against each other: the larger the area, the coarser the cells.

Full-field model: The whole reservoir with the aquifer and all the wells. Only this model can answer questions about the field as a whole: production levels, interference between wells (the way they affect each other), flows between areas, and the energy of the aquifer and the gas cap. It has no artificial internal boundaries. The price is coarse cells, tens to hundreds of metres wide.
Sector model: Part of the reservoir: a block between faults, or an area with several wells. With the same number of cells, the grid is much finer. The main difficulty is the boundaries, because in reality fluid flows across them. We either set these flows from a full-field model calculation, or we draw the boundaries where there is no flow: along sealing faults (faults that do not let fluid through) and along lines of symmetry between wells.
Pattern element: A repeating part of a regular pattern of wells, for example a quarter of a five-spot: a square with a quarter of an injection well in one corner and a quarter of a production well in the opposite one. The boundaries of the element are lines of symmetry, with no flow across them. It is convenient for studying how a process works and for expensive calculations, such as thermal and compositional ones.
Cross-sectional model: A vertical slice between an injection well and a production well, with the layering shown in detail. We use it to study the vertical sweep (how much of the thickness the injected fluid reaches) and to get averaged relationships for the coarse cells of a full-field model (Two-phase upscaling: pseudo-functions).
Single-well model: The vicinity of a single well on a radial r–z grid (distance from the well against depth) (an example of a grid). We use it to study water and gas coning (Water and gas coning), to choose the perforated interval and the allowable rate, to calculate condensate dropout near the wellbore and near-wellbore treatments, and to check the interpretation of well tests.

SchematicScale of the model (schematic): a full-field model covers the whole reservoir with coarse cells; a sector is part of it, with fine cells and conditions at the boundaries; a single-well model is the vicinity of one well on a radial grid.

Models of different scales do not compete. They work together. A cross-sectional model gives the coarse cells averaged relative permeabilities. A single-well model gives the dependence of water cut and producing gas–oil ratio on the rate, which a full-field model with its coarse cells cannot get by itself. We use a sector to check whether a coarse grid creates a false effect (the convergence check), and the full-field model, in turn, gives the sector the pressures and flows at its boundaries.

Worked exampleCoverage or detail

Given (assumed data)

Reservoir: 6 × 4 km, thickness 40 m; 70% of the cells are active (they take part in the calculation)
Full-field model: cells 50 × 50 × 2 m; one run takes 4 h
Sector: an area of 1.2 × 1.0 km over the whole thickness, cells 25 × 25 × 1 m, all active
Single-well model: 25 rings × 80 layers
Assumption: the run time is proportional to the number of active cells

Solution

Full-field model: $120\cdot 80\cdot 20\cdot 0.7 = 134{,}400$ active cells.
Sector: $48\cdot 40\cdot 40 = 76{,}800$ cells; the run takes about 2.3 h.
The whole reservoir with the cells of the sector: $240\cdot 160\cdot 40\cdot 0.7 = 1{,}075{,}200$ cells; the run takes 32 h.
Single-well model: $25\cdot 80 = 2000$ cells; the run takes about 3.6 min.
In one night (12 h) we have time for 3 runs of the full-field model, 5 runs of the sector, or about 200 runs of the single-well model.

Answer: The sector has cells with a quarter of the area and half the thickness, and a shorter run time. The whole reservoir with such cells would take 32 h to run.

What it means. History matching and the assessment of uncertainty need tens or hundreds of runs, so the time of one run limits us as much as the data do. The assumption of proportionality is also an optimistic one: small cells force us to use a smaller time step, so a detailed model runs longer than the number of cells implies. The data in this example are assumed for illustration.

The cost of a run and the choice of model

Four factors decide the run time: the number of cells, the number of unknowns in each cell, the number of time steps and the number of iterations per step. The type of model affects all four, but most of all the number of unknowns. In a fully implicit calculation (IMPES and fully implicit schemes), where all the unknowns of the new step are solved together, each cell adds to the system of equations a block of size $n \times n$ , where $n$ is the number of unknowns in the cell. The number of operations on such blocks grows as $n^2$ when we multiply a block by a vector, and as $n^3$ when we invert a block. This gives a rough estimate: the work per cell grows about as $n^2$ – $n^3$ . The actual ratio depends on the simulator, the solution method and the problem itself.

Worked exampleHow many unknowns, and how many times longer

Given (assumed data)

Grid: 300,000 active cells
Black oil: 3 unknowns per cell
Compositional: 8 hydrocarbon pseudo-components and water: 9 unknowns
Thermal: oil, water, gas and temperature: 4 unknowns
Dual porosity: black oil, twice as many cells

Solution

Number of unknowns. Black oil: $300{,}000\cdot 3 = 0.9$ million. Compositional: 2.7 million. Thermal: 1.2 million. Dual porosity: 1.8 million.
Compositional model: the work per cell is 9–27 times greater, from $(9/3)^2$ to $(9/3)^3$ .
Thermal model: 1.8–2.4 times greater, from $(4/3)^2$ to $(4/3)^3$ . This does not count the smaller cells and time steps that thermal fronts need.
Dual porosity: twice as many cells, so at least a factor of 2.
If we lump the components into 5 pseudo-components (6 unknowns), the factor falls to 4–8.

Answer: By this rough estimate, a compositional model with 9 unknowns per cell on the same grid costs 9–27 times as much as a black oil model at each iteration. The actual ratio depends on the simulator and the problem.

What it means. This estimate gives only the order of magnitude. It leaves out three things. Besides solving the equations, the simulator must calculate the phase equilibrium in each cell. Narrow fronts of composition and temperature force a smaller time step. And simulators cut the cost by solving only some of the unknowns implicitly. But the conclusion is robust: added physics multiplies the run time. So compositional and thermal calculations are usually run on sectors and with a small number of pseudo-components. The data in this example are assumed for illustration.

Which model is sufficient for the task

Task: Waterflooding an oil reservoir at pressures above the bubble point pressure
Sufficient model: black oil, often two-phase oil–water; full-field

Task: Solution gas drive, gas cap, immiscible gas injection
Sufficient model: three-phase black oil

Task: Gas-condensate reservoir under depletion
Sufficient model: extended black oil; compositional if the composition of the production matters

Task: Gas cycling into a gas-condensate reservoir, miscible displacement
Sufficient model: compositional, usually a sector or a pattern element with a fine grid

Task: Steam injection into a heavy-oil reservoir
Sufficient model: thermal, usually a pattern element or a sector

Task: Waterflooding a massive fractured reservoir with a tight matrix (a matrix of very low permeability)
Sufficient model: dual porosity

Task: Water or gas coning, choosing the perforated interval and the rate
Sufficient model: single-well r–z, black oil

Task: Quick comparison of many geological cases under waterflooding
Sufficient model: streamline simulation

Task	Sufficient model
Waterflooding an oil reservoir at pressures above the bubble point pressure	black oil, often two-phase oil–water; full-field
Solution gas drive, gas cap, immiscible gas injection	three-phase black oil
Gas-condensate reservoir under depletion	extended black oil; compositional if the composition of the production matters
Gas cycling into a gas-condensate reservoir, miscible displacement	compositional, usually a sector or a pattern element with a fine grid
Steam injection into a heavy-oil reservoir	thermal, usually a pattern element or a sector
Waterflooding a massive fractured reservoir with a tight matrix (a matrix of very low permeability)	dual porosity
Water or gas coning, choosing the perforated interval and the rate	single-well r–z, black oil
Quick comparison of many geological cases under waterflooding	streamline simulation

The table is a guide for a first choice. We check the final choice by running a small model both ways.

Streamlines. Streamline simulation (how it works) follows the fluid along its paths between the wells. It does not replace a full model. It is a fast tool for its own range of problems: waterflooding, where pressure differences and the uneven properties of the reservoir decide the flow. For such problems it runs many times faster than an ordinary calculation and hardly smears the front. It also gives a by-product: the share of the injection of each injection well that goes to each production well. This helps to manage the waterflood. Its speed lets us compare dozens of geological cases and select a few for a full calculation. Compressibility, gravity, capillary forces and complex phase behaviour are hard for this method.

Very large models. Since the late 1990s, calculations on multiprocessor systems have made it possible to run models of millions of cells, and then of far more. This has reduced the need for upscaling. It has let us keep in a full-field model the detail that only sectors could have before. It has also let us link the reservoir calculation with the calculation of wells and of the surface network (the pipes and facilities above ground). But the choice of the type of model remains. A million cells of a compositional model still cost several times more than a million cells of black oil, and the assessment of uncertainty still needs many runs.

How to choose

1
Name the process and the decision
What will happen in the reservoir — depletion, waterflooding, injection of gas or steam — and what decision the model must support.
2
Does the composition of the phases change
If tables against pressure can describe the oil and the gas throughout the development, use black oil. If gas of a different composition is injected, or the fluid is close to the critical point, use compositional.
3
Does the temperature change
If the process relies on heating, use a thermal model. For ordinary waterflooding, take the temperature as constant.
4
How the rock is built
A connected network of fractures, a tight matrix and a multiphase process: use dual porosity. Otherwise use a single medium.
5
What scale is needed
For a decision about the field, use a full-field model. For the mechanism of a process, a cone, or the choice of well design, use a sector, a pattern element or a single-well model.
6
Are there enough data and time
Check whether you have analyses of fluid samples for the equation of state, and block sizes for dual porosity. Estimate how many runs history matching and the assessment of uncertainty will need.
7
Check with a simple option
Compare a simple and a complex model on a sector. If the difference is smaller than the uncertainty of the input data, the added complexity is not justified.

Terms and symbols

Black oil model: A model with three notional components: oil, gas and water. Gas dissolves in the oil according to a table against pressure.
Compositional model: A model that keeps track of each component or pseudo-component of the mixture. It calculates the phase equilibrium from an equation of state.
Thermal model: A model that adds temperature and an energy balance. It is used for steam injection and in-situ combustion.
Chemical flood model: A black oil model with additional components in the water: polymer, surfactants, salt.
Dual porosity: A fractured reservoir described as two media laid over each other, the matrix and the fractures, with exchange between them. If the matrix blocks are also connected to each other, it is called dual permeability.
Full-field model: A model of the whole reservoir with the aquifer and all the wells.
Sector model: A model of part of the reservoir with a finer grid. It needs conditions at the boundaries.
Pattern element: A repeating part of a regular pattern of wells, bounded by lines of symmetry.

Common mistakes and misconceptions

Wrong

“A compositional model is more accurate, so it is better to build one.”

In fact

We need it only if the composition of the phases changes so much that it affects the process. In other cases it gives the same answer as black oil, but runs several times longer and needs an equation of state tuned to reliable laboratory data.

Wrong

“There are fractures in the core, so the model must have dual porosity.”

In fact

Dual porosity is justified when the fractures form a connected network, carry fluid much better than the matrix, and the multiphase exchange between the two matters. For sparse or unconnected fractures, we use the effective permeability of a single medium.

Wrong

“The sector showed a 15% increase in production — let us multiply by the number of areas.”

In fact

The result for a sector depends on the conditions at its boundaries and on how typical the area is. We check the carry-over to the field with a full-field model or with several sectors of different geology.

Wrong

“A black oil model is good only for oil.”

In fact

The name is historical. The same set of equations works for dry gas reservoirs (gas and water) and, with the extension for liquid hydrocarbons in the gas, for gas-condensate reservoirs under depletion.

Wrong

“For steam injection it is enough to give the oil a lower viscosity.”

In fact

Viscosity depends on temperature, and the temperature is different at each point of the reservoir and changes with time. We find it from the energy balance, which includes the transport of heat, the condensation of steam and the losses to the neighbouring rocks. This is what a thermal model does.

In depth: advanced detail

How we calculate the exchange between the matrix and the fractures

In a dual porosity model, we write the flow between a matrix cell and its fracture cell in the same way as Darcy's law. It is proportional to the matrix permeability, to the pressure difference between the two media, and to a shape factor that “hides” the geometry of the blocks.

Flow between the matrix and the fractures (single-phase)

q_{mf} = \sigma\,V_b\,\frac{k_m}{\mu}\,(p_m - p_f), \qquad \sigma = 4\left(\frac{1}{L_x^2} + \frac{1}{L_y^2} + \frac{1}{L_z^2}\right)

where:

$q_{mf}$ volumetric rate from the matrix into the fractures within a single cell, at reservoir conditions, m³/s
$\sigma$ shape factor of the matrix blocks, m⁻²
$V_b$ bulk volume of the cell (of the rock), m³
$k_m$ matrix permeability, m²
$\mu$ fluid viscosity, Pa·s
$p_m,\ p_f$ pressures in the matrix cell and in the fracture cell, Pa
$L_x,\ L_y,\ L_z$ dimensions of the matrix block — the distances between fractures in the three directions; if there are no fractures in one direction, we leave out that term, m

Units: SI

Holds when

All the matrix blocks in a cell are identical, and a single average value describes the pressure inside a block.
The expression given for $\sigma$ is one of the common ones. Other derivations give other numerical factors, and in history matching we often adjust it.
With several phases, we multiply the rate of each phase by its relative permeability and add capillary and gravity forces. These forces decide the exchange in displacement.

The smaller the blocks, the greater $\sigma$ and the faster the matrix responds to a change in pressure in the fractures. Setting this flow equal to the elastic storage of the block gives the characteristic time for the pressures to equalise, $\tau = \phi_m \mu c_t/(\sigma k_m)$ , where $\phi_m$ is the matrix porosity and $c_t$ is the total compressibility of the matrix and of the fluid that fills it.

Worked exampleHow long the pressure in a matrix block takes to catch up with the pressure in the fractures

Given (assumed data)

Matrix blocks: cubes of 2 × 2 × 2 m
Matrix: porosity 0.12; permeability 1 mD (variant: 0.01 mD)
Fluid and compressibility: viscosity 1 mPa·s; total compressibility $2\cdot 10^{-3}$ MPa⁻¹

Solution

Shape factor: $\sigma = 4\cdot 3/2^2 = 3$ m⁻².
Numerator: $\phi_m \mu c_t = 0.12\cdot 10^{-3}\cdot 2\cdot 10^{-9}$ s, that is, $2.4\cdot 10^{-13}$ s (viscosity in Pa·s, compressibility in Pa⁻¹).
At 1 mD ( $0.987\cdot 10^{-15}$ m²): $\tau = 81$ s, which is about 1.4 min.
At 0.01 mD: $\tau = 2.3$ h.

Answer: The pressure in a block equalises with the pressure in the fractures in 1.4 min at a matrix permeability of 1 mD, and in 2.3 h at 0.01 mD.

What it means. By the standards of field development this is instantaneous. So in single-phase flow (one fluid only), dual porosity shows only in the first minutes and hours after a change in operating conditions — on the curves of well tests. In a forecast over years, such a reservoir behaves as a single medium with the storage of the matrix and the flow capacity of the fractures. The slow and decisive process is not the equalising of pressure but the displacement of oil from the blocks by water or gas. It goes on for months and years. The data in this example are assumed for illustration.

Connections to other topics

What a reservoir simulation model is and why it is needed — what the simulator calculates in each cell and how we work with a model.
Grids in reservoir simulation models — dimensions and geometry of grids, numerical dispersion, streamline simulation.
History matching — how we match a model of any type to the development history.
Black oil model: solution gas–oil ratio and formation volume factors — solution gas–oil ratio and formation volume factors — the tables of a black oil model.
Types of reservoir fluids — which fluids need a compositional description.
Phase behaviour: pure substances and binary mixtures — phase diagrams and the critical point.
Rock permeability — permeability of a fracture and of fractured rock.
Capillary pressure and the transition zone — capillary forces that control the exchange between the matrix and the fractures.
Building a simulation model: input data and results — how to specify the chosen type of model in the input data.
IMPES and fully implicit schemes — methods of solving the equations and their effect on the cost of a run.

Check yourself

1
For which of these tasks is a black oil model not enough?
1. A.Depletion of an oil reservoir under solution gas drive
2. B.Waterflooding an oil reservoir at pressures above the bubble point pressure
3. C.Water influx from an aquifer into a gas reservoir
4. D.Cycling of dry gas into a gas-condensate reservoir
Show the answer
Correct answer: D. Cycling of dry gas into a gas-condensate reservoir
In gas cycling, the dry gas vaporises the condensate that has dropped out, and the composition of both phases changes along the flow path. Tables of properties against pressure are no longer enough, and we need a phase equilibrium calculation by components. Three notional components of unchanging composition can describe the other processes.
2
The pore volume of a cell is 5000 m³, $S_o = 0.6$ , $S_g = 0.1$ , $B_o = 1.25$ , $B_g = 0.008$ m³/m³, $R_s = 80$ m³/m³. How much oil and gas is in the cell at standard conditions?
Show the answer
Oil: $5000\cdot 0.6/1.25 = 2400$ m³. Free gas: $5000\cdot 0.1/0.008 = 62{,}500$ m³. Dissolved gas: $2400\cdot 80 = 192{,}000$ m³, which we calculate from the volume of oil at standard conditions. The total gas is 254,500 m³, and three quarters of it is dissolved in the oil.
3
A grid contains 200,000 active cells. How many unknowns are there in a black oil model and in a compositional model with 9 hydrocarbon pseudo-components and water?
1. A.600,000 and 1.8 million
2. B.600,000 and 2 million
3. C.200,000 and 2 million
4. D.200,000 and 1.8 million
Show the answer
Correct answer: B. 600,000 and 2 million
A black oil model has three unknowns per cell: $200{,}000\cdot 3 = 600{,}000$ . A compositional model has one for each component and for water: $200{,}000\cdot 10 = 2$ million. The run time grows faster than the number of unknowns.
4
Core and well tests show that the reservoir is fractured. What information do we need to decide whether to build a dual porosity model?
Show the answer
We need to know four things: whether the fractures form a connected network over distances about the size of a cell; how much better the fractures carry fluid than the matrix does; where most of the hydrocarbons in place are; and what process we have to calculate. Dual porosity is justified for a connected fracture network, a tight matrix that holds most of the volume in place, and a multiphase process — waterflooding or gas injection — where the rate at which oil leaves the blocks decides the recovery. Otherwise a single medium with an effective permeability is enough.
5
Where is it best to draw the boundaries of a sector model of an area with a regular pattern of wells?
1. A.As close as possible to the well being studied, so that the grid is finer
2. B.Anywhere: the boundaries do not affect the result
3. C.Through the production wells, to specify their rates more precisely
4. D.Along sealing faults and lines of symmetry between wells
Show the answer
Correct answer: D. Along sealing faults and lines of symmetry between wells
The model does not let fluid cross the boundary of a sector unless we set the flows explicitly. This is realistic only where there is no flow in reality either: on a sealing fault or on a line of symmetry between identically operated wells. In other cases we take the flows at the boundaries from a full-field model.
6
Why can steam injection not be calculated with a black oil model by simply giving the oil the viscosity at the steam temperature?
Show the answer
Because the temperature in the reservoir is not the same everywhere. The heated zone covers only part of the reservoir and grows with time, the steam condenses, and heat goes into the rocks above and below. The viscosity of the oil in each cell depends on its current temperature, and we can find the temperature only from the energy balance. A model with a single low viscosity would overestimate the rates many times over, and would show neither the heating time nor the steam consumption.

Updated: 19 October 2026