Reservoir drive mechanisms

Quick refresher

The reservoir volume of the fluids withdrawn always equals the sum of the expansion of the oil, gas, water and pores and the volume of the water that has entered the accumulation. This is the material balance. The drive shows which term of the balance is the main one.
Liquids and rock are only slightly compressible. Above the bubble point (the pressure at which gas starts to come out of the oil), the compressibility drive gives only a few per cent of the oil in place, while the pressure falls fast.
Gas is tens of times more compressible. In solution gas drive there is a lot of energy, but the mobile gas (gas that is free to flow) outruns the oil and goes to the wells. The gas–oil ratio rises sharply, the pressure falls, and the oil recovery is usually 5–30%.
A gas cap and an aquifer push on the oil from outside and use their energy more economically. The pressure falls slowly, and the oil recovery is usually 20–40% and 35–60% respectively. Water drive works only as long as the withdrawal does not exceed the water influx.
Gravity drainage is slow, but in thick, steeply dipping reservoirs with good vertical permeability it gives a high oil recovery. Compaction of weak rock adds energy at the price of subsidence of the ground surface.
Pure drives are rare. Usually several mechanisms work together, and their balance changes with time and with the withdrawal rate.

Key relationships

\frac{N_p}{N} = \frac{B_{oi}}{B_o}\,c_e\,(p_i - p)

oil recovery of the compressibility drive, pressure above the bubble point

c_e = \dfrac{S_o c_o + S_w c_w + c_f}{S_o}

effective compressibility of the oil-saturated part of the reservoir

\Delta V_g \approx V_g\,\dfrac{\Delta p}{p}

expansion of gas for a small pressure decline, if $Z$ is almost constant

\Delta W \approx (c_w + c_f)\,W\,\Delta p

elastic store of an aquifer with pore volume $W$

Common pitfalls

Counting on the compressibility drive as a serious source of production. Above the bubble point we usually recover only a few per cent of the oil in place.
Judging the drive from one sign. The gas–oil ratio rises in both gas cap drive and solution gas drive, and it is the pressure trend that tells them apart.
Relying on an aquifer without knowing its size and flow capacity. Before production starts they are almost never known exactly.
Keeping the withdrawal above the water influx. The pressure falls below the bubble point, and solution gas drive joins the water drive.
Confusing the drive mechanism of an accumulation with the flow regime (steady-state, pseudo-steady-state or transient).

After this lesson you can

list the sources of reservoir energy and compare their contributions per 1 MPa of pressure decline
describe the compressibility drive, solution gas drive, gas cap drive, water drive, gravity drainage and compaction drive
estimate the oil recovery of the compressibility drive and explain why it is small
from the trends of the reservoir pressure, gas–oil ratio and water cut, identify the dominant drive and name the typical oil recovery for it
explain why water drive is sensitive to the withdrawal rate and how this is taken into account in development

Where reservoir energy comes from

The fluids in a reservoir are squeezed: the reservoir pressure is tens of megapascals (Reservoir pressure and temperature). When a well withdraws fluid, the pressure around it falls, and everything in the reservoir expands a little: the oil with its dissolved gas, the connate water (the water that stays in the pores from the start), the gas and the rock itself. This expansion pushes fluid to the well. Another way is inflow from outside: water from the aquifer enters the accumulation and takes the place of the oil that has been withdrawn.

The material balance expresses this in numbers: the reservoir volume of the oil, gas and water withdrawn equals the sum of the expansions of all the components of the reservoir and the volume of the water that has entered. The drive mechanism of the accumulation is the answer to the question of which term of the balance is the main one.

SchematicSources of reservoir energy in one accumulation (a sketch, not to scale): 1 — expansion of the gas cap; 2 — release and expansion of dissolved gas; 3 — expansion of the oil and water, and compaction of the pores under the weight of the overlying rocks; 4 — water influx from the aquifer; 5 — gravity segregation: the gas rises and the oil drains down.

How much volume 1 m³ releases when the pressure falls by 1 MPa

What expands or contracts: Undersaturated oil (above the bubble point pressure)
Compressibility, MPa⁻¹: $(0.7\text{–}3)\cdot10^{-3}$
Volume, L: 0.7–3

What expands or contracts: Formation water
Compressibility, MPa⁻¹: about $0.45\cdot10^{-3}$
Volume, L: about 0.45

What expands or contracts: Pore space of a cemented rock
Compressibility, MPa⁻¹: $(0.3\text{–}1.5)\cdot10^{-3}$
Volume, L: 0.3–1.5

What expands or contracts: Gas at 10–20 MPa
Compressibility, MPa⁻¹: about $1/p$ , that is 0.05–0.1
Volume, L: 50–100

What expands or contracts	Compressibility, MPa⁻¹	Volume, L
Undersaturated oil (above the bubble point pressure)	$(0.7\text{–}3)\cdot10^{-3}$	0.7–3
Formation water	about $0.45\cdot10^{-3}$	about 0.45
Pore space of a cemented rock	$(0.3\text{–}1.5)\cdot10^{-3}$	0.3–1.5
Gas at 10–20 MPa	about $1/p$ , that is 0.05–0.1	50–100

For pores, this is the decrease of the pore volume per 1 m³ of pore space. The values are approximate; for loose sands and high-porosity chalk the pore compressibility is several times higher. More on compressibility in the lesson Flow regimes.

From this comes the main difference between the drives. For each megapascal, liquids and rock contract by hundredths and tenths of a per cent, and gas by several per cent. So an accumulation without free gas and without an aquifer loses pressure fast, while a gas cap or a large aquifer holds it for a long time.

Natural drives of oil accumulations

Drive mechanism: Compressibility drive
Name in Russian-language literature: elastic regime
Main source of energy: expansion of undersaturated oil and connate water, and decrease of the pore volume

Drive mechanism: Solution gas drive
Name in Russian-language literature: dissolved-gas regime
Main source of energy: release and expansion of the gas dissolved in the oil

Drive mechanism: Gas cap drive
Name in Russian-language literature: gas-pressure regime
Main source of energy: expansion of the gas cap

Drive mechanism: Water drive
Name in Russian-language literature: water-pressure regime
Main source of energy: water influx from the aquifer

Drive mechanism: Gravity drainage
Name in Russian-language literature: gravity regime
Main source of energy: oil draining downward under gravity

Drive mechanism: Compaction drive
Name in Russian-language literature: compaction regime
Main source of energy: irreversible decrease of the pore volume of weak rock

Drive mechanism: Combination drive
Name in Russian-language literature: mixed regime
Main source of energy: several sources at the same time

Drive mechanism	Name in Russian-language literature	Main source of energy
Compressibility drive	elastic regime	expansion of undersaturated oil and connate water, and decrease of the pore volume
Solution gas drive	dissolved-gas regime	release and expansion of the gas dissolved in the oil
Gas cap drive	gas-pressure regime	expansion of the gas cap
Water drive	water-pressure regime	water influx from the aquifer
Gravity drainage	gravity regime	oil draining downward under gravity
Compaction drive	compaction regime	irreversible decrease of the pore volume of weak rock
Combination drive	mixed regime	several sources at the same time

In English-language literature the compressibility drive and the solution gas drive are often combined under the name depletion drive: the energy comes only from the accumulation itself.

Worked exampleWhich source gives more energy

Given (assumed data)

Oil part of the accumulation: pore volume 16.0 million m³, connate water 0.25; the oil is the assumed oil of the lesson Black oil model: solution gas–oil ratio and formation volume factors: $p_b$ = 18 MPa, $R_{sb}$ = 110 m³/m³, $B_{ob}$ = 1.35 m³/m³
Compressibilities of oil above $p_b$ , of water and of pores: $1.6\cdot10^{-3}$ ; $0.45\cdot10^{-3}$ ; $0.8\cdot10^{-3}$ MPa⁻¹
Oil properties at 17 MPa: $R_s$ = 104.5 m³/m³; $B_o$ = 1.335 m³/m³; $B_g = 0.00634$ m³/m³
Cases: A: the oil is undersaturated. B: the pressure falls from 18 to 17 MPa. C: as B, but there is a gas cap with a volume of 0.3 of the oil part. D: the oil part is surrounded by an aquifer with a pore volume 10 times larger

Solution

A. The elastic store of the oil part per 1 MPa: $V_p\,(S_oc_o + S_wc_w + c_f) = 16.0\cdot10^6\cdot2.1125\cdot10^{-3} = 33.8$ thousand m³.
B. Below the bubble point, gas comes out of the oil. The reservoir contains $N = 12.0\cdot10^6/1.35 = 8.889$ million m³ of stock-tank oil (oil with its gas removed). The total formation volume factor at 17 MPa (the reservoir volume of the oil with all its gas, per unit volume of stock-tank oil) is $B_t = 1.335 + 0.00634\cdot(110 - 104.5) = 1.3699$ .
The increase in the volume of the oil with the gas released from it is $N\,(B_t - B_{ob}) = 176.6$ thousand m³. Water and pores add 14.6 thousand m³, so the total is 191.2 thousand m³, which is 5.7 times more than in Case A.
C. A gas cap of volume $0.3\cdot12.0 = 3.6$ million m³, when the pressure falls from 18 to 17 MPa, expands (at a constant gas deviation factor $Z$ ) by $3.6\cdot10^6\cdot(18/17 - 1) = 211.8$ thousand m³. This is on top of what the oil part gives, and it is 6.3 times more than in Case A.
D. An aquifer with a pore volume of 160 million m³ gives $(c_w + c_f)\,W\cdot1 = 1.25\cdot10^{-3}\cdot160\cdot10^6 = 200.0$ thousand m³. This is 5.9 times more than in Case A, if the pressure has fallen by 1 MPa in the whole aquifer.

Answer: The elastic store of the undersaturated oil part is about 33.8 thousand m³ per 1 MPa. The gas coming out of solution, the gas cap and the aquifer each give about six times more.

What it means. The dissolved gas holds almost as much energy as the gas cap, but it spends it badly. Once the released gas becomes mobile, it goes to the wells faster than the oil and carries its energy to the surface. The gas cap and the aquifer push on the oil from outside, through the contact, and work more economically. This is why the oil recovery differs. The aquifer also needs time: the pressure in it falls gradually, starting from the contact (Flow regimes). The data in this example are assumed for illustration, and the gas deviation factor is taken as constant.

Compressibility drive: above the bubble point pressure

Suppose the oil is undersaturated (its pressure is above the bubble point, so all the gas stays dissolved). Suppose also that the aquifer is small or cut off from the accumulation by faults. Then the only source of energy is the expansion of the oil and the connate water and the decrease of the pore volume as the pressure falls. This is the compressibility drive. No gas comes out in the reservoir, so the gas–oil ratio is constant and equals the initial solution gas–oil ratio $R_{si}$ (the gas dissolved in each cubic metre of oil at the start), and there is no water in the production.

Oil recovery of the compressibility drive

\frac{N_p}{N} = \frac{B_{oi}}{B_o}\,c_e\,(p_i - p),\qquad c_e = \frac{S_o c_o + S_w c_w + c_f}{S_o}

where:

$N_p$ cumulative oil production at standard conditions, m³
$N$ stock-tank oil initially in place (at standard conditions), m³
$B_{oi},\ B_o$ oil formation volume factors at the initial and the current pressure, m³/m³
$c_e$ effective compressibility per oil-saturated pore volume, 1/MPa
$p_i,\ p$ initial and current reservoir pressure, MPa
$S_o,\ S_w$ oil and water saturation, $S_o = 1 - S_w$ , fraction
$c_o,\ c_w,\ c_f$ compressibility of oil, water and pores, 1/MPa

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

Holds when

The pressure is above the bubble point, there is no free gas, and the aquifer is not working.
The compressibilities are constant over the pressure range considered.
The pressure is the same throughout the accumulation: a “tank” material balance.

The reservoir volume of the oil produced, $N_pB_o$ , equals the expansion of the oil-saturated part of the reservoir, $NB_{oi}\,c_e\,\Delta p$ . Since $c_e$ is usually $(2\text{–}4)\cdot10^{-3}$ MPa⁻¹, a pressure fall of 10 MPa gives only 2–4% of the oil in place.

Worked exampleHow much oil the compressibility drive gives down to the bubble point

Given (assumed data)

Pore volume; connate water: 16.0 million m³; 0.25
Initial pressure; bubble point pressure: 30 MPa; 18 MPa
Oil formation volume factors: $B_{oi}$ = 1.324 at 30 MPa; $B_{ob}$ = 1.35 at 18 MPa
Compressibilities of water and pores: $0.45\cdot10^{-3}$ and $0.8\cdot10^{-3}$ MPa⁻¹

Solution

The reservoir holds $16.0\cdot0.75 = 12.0$ million m³ of oil, so the oil in place is $N = 12.0/1.324 = 9.063$ million m³ (at standard conditions), as in the lesson Black oil model: solution gas–oil ratio and formation volume factors.
Expansion of the oil from 30 to 18 MPa: $N\,(B_{ob} - B_{oi}) = 9.063\cdot10^6\cdot0.026 = 235.6$ thousand m³.
Expansion of the connate water: $V_pS_wc_w\,\Delta p = 16.0\cdot10^6\cdot0.25\cdot0.45\cdot10^{-3}\cdot12 = 21.6$ thousand m³.
Decrease of the pore volume: $V_pc_f\,\Delta p = 16.0\cdot10^6\cdot0.8\cdot10^{-3}\cdot12 = 153.6$ thousand m³.
The total is 410.8 thousand m³ of reservoir volume. At the bubble point this is $410.8/1.35 = 304$ thousand m³ of oil at standard conditions, or 3.4% of the oil in place. Water and pores contribute 43%.

Answer: Down to the bubble point, the compressibility drive will give about 304 thousand m³ of oil, which is 3.4% of the oil in place.

What it means. The reservoir pressure fell by 12 MPa, which is 40%, and only about three per cent of the oil was recovered. This is typical of the compressibility drive: the pressure falls fast, and the oil recovery is small. Water and pores gave almost half of the energy, so the compressibility of the rock must be included in such calculations. After this, if the pressure is not maintained, the accumulation will change to solution gas drive. The data in this example are assumed for illustration.

The compressibility drive is the first stage of almost every accumulation with undersaturated oil. While it works in a closed block, after a short transient period the inflow to the wells becomes pseudo-steady-state: the pressure falls everywhere at one rate, proportional to the withdrawal (Flow regimes, Pseudo-steady-state flow and the shape factor).

Solution gas drive

When the pressure falls below the bubble point, gas comes out of the oil. At first these are separate immobile bubbles, but their volume grows fast, because gas is tens of times more compressible than oil. The expansion of the released gas pushes the oil to the wells. This is solution gas drive, also called dissolved gas drive.

This mechanism has a lot of energy (see the example above), but it spends it wastefully. When the gas saturation exceeds the critical gas saturation (usually a few per cent of the pore volume), the gas becomes mobile. Its viscosity is tens of times lower than that of oil, so it moves to the wells faster than the oil and carries its energy away. Free gas also lowers the relative permeability to oil (the share of the rock's permeability that is available to the oil; Relative permeability), and oil that has lost its gas becomes more viscous.

How the gas–oil ratio changes

1
Above the bubble point
No gas comes out in the reservoir. The gas–oil ratio is constant and equals the initial solution gas–oil ratio $R_{si}$ .
2
Just below the bubble point
The released gas is still immobile and stays in the reservoir, while the oil that reaches the well has less dissolved gas. The gas–oil ratio may fall a little.
3
The gas has become mobile
Free gas outruns the oil, and the gas–oil ratio rises fast. It may exceed $R_{si}$ several times.
4
Late stage
Little gas is left in the reservoir, and the pressure is low, so the same reservoir volume of gas matches a smaller volume at the surface. The gas–oil ratio passes a maximum and falls.

In this drive the pressure falls fast and continuously, the well rates decline, and artificial lift is needed early (Well inflow: the well performance diagram). There is almost no water in the production. The oil recovery is usually 5–30% of the oil in place. This is a guide, not a rule: it is higher for low-viscosity oil in reservoirs with good vertical permeability, where the gas can go up instead of to the wells.

If the vertical permeability is good, part of the gas rises to the top and forms a secondary gas cap, which then works like an ordinary one. To help this, the perforated intervals are placed lower in the section, but not too close to the oil–water contact, so as not to draw water up. The inflow performance curves of wells in this drive bend, and Vogel's equation (a standard formula for the curve of well rate against pressure) is used for them (Inflow performance relationship (IPR)). The forecast of pressure and gas–oil ratio from the material balance and the relative permeability curves is covered in the lesson Performance prediction for solution gas drive reservoirs.

Gas cap drive

Sometimes there is free gas above the oil from the very beginning: a gas cap. Then the oil at the gas–oil contact (GOC, the surface between the gas and the oil) is saturated with gas, and the pressure there equals the bubble point pressure (Types of reservoir fluids). As fluid is withdrawn, the gas cap expands, the contact moves down, and the gas pushes the oil down, towards the wells. This is gas cap drive. Near the wells the pressure falls below the bubble point, so solution gas drive works at the same time.

The size of a gas cap is described by the ratio $m$ of its reservoir volume to the reservoir volume of the oil. The larger $m$ is, the more slowly the pressure falls. In the example above, a cap of 0.3 of the oil part gave, for each megapascal, more than six times as much energy as the elastic store of the undersaturated oil.

At first the gas–oil ratio is close to the solution gas–oil ratio and grows slowly. When the gas front reaches the perforated interval, or a gas cone (gas drawn down to the well in the shape of a cone) breaks through to the well (Water and gas coning), the gas–oil ratio rises sharply. Then the well's rate must be cut back or its interval moved. This is why wells open the oil part far from the GOC, but not too close to the oil–water contact (OWC) either. There is almost no water in the production, the flowing period of the wells (when they flow on their own energy) is long, and the oil recovery is usually 20–40%.

This drive is sensitive to the withdrawal rate. At a slow rate gravity keeps the gas above the oil, and the contact moves down evenly. At a fast rate the gas breaks through along high-permeability layers and in cones, leaving oil behind.

Water drive

If the accumulation is connected with a large aquifer, then when the pressure falls, water enters the accumulation and replaces the oil that has been withdrawn (water influx). Most often the energy comes from the elastic expansion of the water and rock of the aquifer itself. In the Russian classification this is the elastic water drive. More rarely the aquifer is fed from outside, for example because it reaches the surface, and the pressure at the contact hardly falls. This is the rigid water drive. Together they are called water drive.

Edge water: The water comes to the accumulation along the reservoir from the flanks of the structure (edge water drive). This is typical of comparatively thin reservoirs.
Bottom water: The whole accumulation is underlain by water (bottom water drive). This is typical of thick reservoirs, and water cones are drawn up to the wells (Water and gas coning).

The compressibility of water and pores is small, so the aquifer must be huge. In the example above, an aquifer ten times larger than the oil part gave, per 1 MPa, about as much energy as a gas cap of 0.3 of the volume of the oil part. The flow capacity of the aquifer is just as important: the water must be able to reach the accumulation in time. A pressure disturbance spreads through the aquifer gradually, so the water influx lags behind the pressure fall at the contact, and it is calculated with models of transient flow (Aquifer models: water influx).

The signs of water drive: the pressure stays high or falls slowly; the gas–oil ratio is constant while the pressure is above the bubble point; water appears in the production early, and the water cut grows. Production is limited not by the fall of pressure but by the growth of the water cut: at an almost constant withdrawal of liquid, the share of oil in it decreases. Wells are placed higher on the structure, to delay the arrival of water. The oil recovery is usually 35–60%. In heterogeneous reservoirs it is lower: the water breaks through along high-permeability layers and leaves the oil in the tight ones.

Worked exampleThe limiting withdrawal under water drive

Given (assumed data)

Initial pressure at the OWC; bubble point pressure: 25 MPa; 18 MPa
Water influx (a simplified model): proportional to the pressure fall at the contact: $J_a\,(p_i - p)$ , with the aquifer productivity index $J_a$ = 350 m³/(d·MPa) at reservoir conditions
Oil formation volume factor $B_o$: 1.33 m³/m³; there is no water in the production yet

Solution

The pressure stops changing when the water influx equals the reservoir volume of the withdrawal: $q_oB_o = J_a\,(p_i - p)$ .
A withdrawal of 1000 m³/d of oil needs a pressure fall of $1000\cdot1.33/350 = 3.8$ MPa. The pressure holds at 21.2 MPa, above the bubble point.
The largest withdrawal at which the pressure does not fall below 18 MPa: $q_o = 350\cdot(25 - 18)/1.33 = 1842$ m³/d.
At a withdrawal of 2500 m³/d, the same formula says the pressure would have to fall to $25 - 2500\cdot1.33/350 = 15.5$ MPa. This means it will pass the bubble point earlier, and gas will start to come out in the reservoir.

Answer: The aquifer keeps the pressure above the bubble point only if the oil withdrawal is no more than 1842 m³/d.

What it means. This is the sensitivity of water drive to the withdrawal rate. If the withdrawal is too large, the pressure falls, free gas appears in the reservoir, the gas–oil ratio rises, and the accumulation changes to a combination drive. A lower withdrawal brings it back to water drive. A model with an influx proportional to the pressure drop is a simplification: a real aquifer responds with a delay, and its parameters are refined from the production history (History matching of aquifer parameters). The data in this example are assumed for illustration.

The size and flow capacity of an aquifer are almost never known before production starts, because there are few wells in it. They are refined from the production history, from how far the pressure in the accumulation lags behind a calculation without water influx. Pressure measurements with depth by a formation tester in new wells also help (Pressure measurement with the wireline formation tester). They show which thin layers are held up by the water and which are being depleted. Because of this uncertainty, the pressure is often maintained by water injection from the very beginning, without waiting for a natural water drive to show itself (Why water is injected: pressure maintenance and oil displacement).

Gravity drainage and compaction of the reservoir

Oil is heavier than gas, and water is heavier than oil. If the reservoir is thick or steeply dipping and the vertical permeability is good, the phases separate under gravity: the gas rises to the top, and the oil drains down, to wells in the low part. This is gravity drainage. It is slow: the draining velocity is proportional to $k\,k_{ro}\,\Delta\rho\,g/\mu_o$ , where $\Delta\rho$ is the difference between the densities of oil and gas, and it falls together with the relative permeability to oil $k_{ro}$ . But the oil keeps draining until its saturation becomes very small, so the oil recovery can be high. Usually this drive shows itself at a late stage, when the other sources of energy are exhausted, and it is especially effective in fractured, steeply dipping reservoirs and at a slow withdrawal rate.

When the reservoir pressure falls, the effective stress grows. This is the part of the weight of the overlying rocks that is carried by the rock skeleton. The rock compacts, and the decrease of the pore volume pushes out fluid. In strong cemented rocks this contribution is covered by the pore compressibility and is small. In weak ones (loose sands, high-porosity chalk) the pores can collapse irreversibly, and compaction drive gives a noticeable share of the reservoir energy. The price is subsidence of the reservoir and of the ground surface above it, deformation of the casing, and, at sea, sinking of the sea floor under the bases of platforms.

How to recognise the drive of an accumulation

The drive is assumed in the project even before production starts, from the geology and the fluid properties, and is refined from the first years of development. The sources of information are:

Fluid properties and contacts. How far the initial pressure is above the bubble point, whether there is a gas cap, and what the solution gas–oil ratio and the oil viscosity are (Black oil model: solution gas–oil ratio and formation volume factors).
Geology. Whether there is a permeable aquifer beyond the outline of the accumulation, how large it is, and whether it is cut off by faults; the thickness and dip of the reservoir, and the vertical permeability.
Pressure. Pressure measurements with depth in new wells, the fall of the average pressure per unit of withdrawal, and the response of observation wells outside the outline.
Production. The trends of the gas–oil ratio and the water cut in the wells, and the positions of the wells on the structure.
Material balance. Comparing the withdrawal with the expansion of the fluids and rocks shows whether water influx is needed to explain the pressure, and what the share of each mechanism is (General material balance equation for oil reservoirs, Straight-line material balance (Havlena–Odeh method)).

Signs of the main drives of oil accumulations (guide values)

Drive: Compressibility drive
Pressure: falls fast
Gas–oil ratio: constant, equal to $R_{si}$
Water: none
RF: a few per cent

Drive: Solution gas drive
Pressure: falls fast
Gas–oil ratio: a short fall, a sharp rise, then a decline
Water: almost none
RF: 5–30%

Drive: Gas cap drive
Pressure: falls slowly
Gas–oil ratio: grows slowly, sharply when the gas breaks through
Water: almost none
RF: 20–40%

Drive: Water drive
Pressure: stays high
Gas–oil ratio: constant
Water: appears early and grows
RF: 35–60%

Drive	Pressure	Gas–oil ratio	Water	RF
Compressibility drive	falls fast	constant, equal to $R_{si}$	none	a few per cent
Solution gas drive	falls fast	a short fall, a sharp rise, then a decline	almost none	5–30%
Gas cap drive	falls slowly	grows slowly, sharply when the gas breaks through	almost none	20–40%
Water drive	stays high	constant	appears early and grows	35–60%

RF is the recovery factor, the share of the oil in place that is recovered. The ranges are guide values for natural drives without injection. The actual value depends on the heterogeneity of the reservoir, the oil viscosity, the well spacing and the withdrawal rate. For the compressibility drive, the figure is the oil recovery down to the bubble point.

SchematicTypical trends of reservoir pressure, gas–oil ratio and water cut in solution gas drive, gas cap drive and water drive (a qualitative sketch, not to scale).

A combination drive is the rule, not the exception. An accumulation with a gas cap and an aquifer gets energy from two sides, and the balance depends on the withdrawal rate: the faster the withdrawal, the less the water has time to give. The drive also changes with time: the compressibility drive turns into solution gas drive, water drive at high withdrawal rates is joined by solution gas drive, and towards the end of development gravity matters more and more.

When the natural drive is not effective enough, it is supplemented. Water injection maintains the pressure and displaces the oil (Why water is injected: pressure maintenance and oil displacement), and gas injection maintains the gas cap (Displacement of oil by gas). The choice between these options is one of the main decisions of a field development plan, and it rests on an understanding of the natural drive of the accumulation.

Terms and symbols

Compressibility drive: Oil production by the expansion of undersaturated oil and connate water and the decrease of the pore volume above the bubble point.
Solution gas drive: The oil is pushed out by the gas that comes out of it below the bubble point.
Gas cap drive: The oil is pushed out by the expanding gas cap. $m$ is the ratio of the reservoir volume of the cap to the reservoir volume of the oil.
Water drive: The oil is replaced by the water of the aquifer. In the elastic water drive the energy comes from the elasticity of the aquifer, and in the rigid water drive the aquifer is fed from outside.
Gravity drainage: The oil drains under gravity to wells in the low part of the reservoir.
Compaction drive: Fluid is pushed out by an irreversible decrease of the pore volume of weak rock. It is accompanied by subsidence of the surface.
Secondary gas cap: A body of gas at the top of the reservoir that came out of the oil in solution gas drive.
Combination drive: Several sources of energy acting at the same time.

Common mistakes and misconceptions

Wrong

“The pressure in the accumulation is high, so the energy will last a long time.”

In fact

What matters is not the pressure itself but what expands when it falls. Undersaturated oil with no gas cap and no aquifer loses pressure fast: a fall of 40% in the lesson's example gave only about 3% of the oil in place.

Wrong

“The gas–oil ratio is rising, so the gas cap is working.”

In fact

The gas–oil ratio also rises in solution gas drive. We tell them apart by the pressure (with a gas cap it falls more slowly), by the positions of the wells relative to the contact, and by the material balance.

Wrong

“The aquifer will support the pressure at any withdrawal.”

In fact

Water influx is limited by the size and flow capacity of the aquifer. If the withdrawal is larger than the influx, the pressure falls, and solution gas drive joins the water drive.

Wrong

“The oil recovery of solution gas drive is small because the gas has little energy.”

In fact

There is a lot of energy: in the lesson's example, per megapascal almost as much as in a gas cap of 0.3 of the oil part. It is spent wastefully: the mobile gas outruns the oil and goes to the wells.

In depth: advanced detail

The material balance and the shares of the mechanisms

The general material balance equation of an oil accumulation brings all the energy sources together. In reservoir volumes it is written as follows:

General material balance equation (reservoir volumes)

\begin{aligned} N_p\big[B_t + (R_p - R_{si})B_g\big] + W_pB_w &= N\,(B_t - B_{ti}) + N m B_{ti}\left(\frac{B_g}{B_{gi}} - 1\right)\\ &+ N(1 + m)B_{ti}\,\frac{c_wS_w + c_f}{1 - S_w}\,\Delta p + W_e\end{aligned}

where:

$N_p,\ W_p$ cumulative oil and water production at standard conditions, m³
$R_p$ cumulative gas–oil ratio: all the gas produced per cubic metre of oil produced, m³/m³
$B_t,\ B_{ti}$ total (two-phase) formation volume factor at the current and the initial pressure, $B_t = B_o + B_g(R_{si} - R_s)$ , m³/m³
$B_g,\ B_{gi},\ B_w$ formation volume factors of gas at the current and initial pressure, and of water (surface volumes are at 0.101325 MPa and 20 °C), m³/m³
$R_{si}$ initial solution gas–oil ratio, m³/m³
$N$ oil initially in place, m³
$m$ ratio of the initial reservoir volume of the gas cap to the reservoir volume of the oil, —
$c_w,\ c_f,\ S_w$ compressibilities of water and pores, water saturation, 1/MPa, 1/MPa, —
$\Delta p$ fall of the average reservoir pressure from the initial pressure, MPa
$W_e$ cumulative water influx at reservoir conditions, m³

Units: consistent units; volumes in m³, gas in standard m³ (m³ of gas measured at standard conditions)

Holds when

The accumulation is treated as one “tank”: the pressure is the same throughout, and the phases are in equilibrium.
The fluid properties from PVT studies refer to the average pressure.
There is no injection; if there is, its volumes are added to the right side.

On the left is the reservoir volume of everything withdrawn. On the right, in order: the expansion of the oil with its dissolved and released gas, the expansion of the gas cap, the expansion of the connate water and the decrease of the pore volume in the whole accumulation, and the water influx. The full derivation and use are in the lessons General material balance equation for oil reservoirs and Straight-line material balance (Havlena–Odeh method).

If we divide each term on the right side by the reservoir volume of the oil and gas withdrawn, we get the drive indices: the shares given by the expansion of the oil, the gas cap, the elasticity of the water and pores, and the water influx (less the water produced). They add up to one. Indices calculated from the production history show how the drive of the accumulation changes with time, and they serve as a numerical check of the qualitative signs of the main section.

Connections to other topics

Black oil model: solution gas–oil ratio and formation volume factors — the solution gas–oil ratio and the formation volume factors through which the expansion of oil and gas is calculated.
Reservoir pressure and temperature — the initial reservoir pressure, the gradients and the contacts: the input data for judging the drive.
Flow regimes — flow regimes are a different idea; the compressibility drive in a closed block means pseudo-steady-state flow.
Relative permeability — why free gas outruns the oil.
Inflow performance relationship (IPR) — inflow to a well in solution gas drive.
General material balance equation for oil reservoirs — the numerical balance of all the energy sources.
Aquifer models: water influx — how water influx is calculated.
Why water is injected: pressure maintenance and oil displacement — pressure maintenance when natural energy is not enough.

Check yourself

1
In two years the pressure in an accumulation has fallen from 28 to 19 MPa (the bubble point pressure is 17 MPa). The gas–oil ratio is constant, there is no water in the production, and about 2.5% of the oil in place has been produced. Which drive is it?
1. A.Compressibility drive
2. B.Water drive
3. C.Gravity drainage
4. D.Gas cap drive
Show the answer
Correct answer: A. Compressibility drive
A fast fall of pressure with a small withdrawal, a constant gas–oil ratio above the bubble point, and no water are the signs of the compressibility drive. Soon the pressure will fall below the bubble point, and solution gas drive will begin if the pressure is not maintained by injection.
2
Why is the oil recovery in solution gas drive small, although the gas coming out of solution gives a lot of energy?
1. A.The gas raises the viscosity of the formation water
2. B.The gas dissolves back into the oil
3. C.The mobile gas outruns the oil and goes to the wells, and free gas lowers the permeability to oil
4. D.The gas coming out of solution raises the pressure in the reservoir
Show the answer
Correct answer: C. The mobile gas outruns the oil and goes to the wells, and free gas lowers the permeability to oil
The viscosity of gas is tens of times lower than that of oil. Once it is mobile, gas flows to the wells faster than the oil and carries energy to the surface. At the same time, free gas lowers the relative permeability to oil, and degassed oil becomes more viscous.
3
An undersaturated accumulation has $c_e = 3\cdot10^{-3}$ MPa⁻¹, $B_{oi}$ = 1.25 m³/m³, and $B_o$ = 1.27 m³/m³ at the bubble point. The initial pressure is 15 MPa above the bubble point. What share of the oil in place will the compressibility drive give?
Show the answer
$N_p/N = (1.25/1.27)\cdot3\cdot10^{-3}\cdot15 = 0.984\cdot0.045 \approx 0.044$ , about 4.4%. Even with a large margin of pressure above the bubble point, the compressibility drive gives only a few per cent.
4
The withdrawal from an accumulation with an active aquifer was doubled, and the gas–oil ratio began to rise. What is happening?
1. A.Water has broken through
2. B.Gas comes out of the water and forms a secondary gas cap
3. C.The withdrawal has exceeded the water influx, the pressure has fallen below the bubble point, and solution gas drive has started
4. D.Nothing special: the gas–oil ratio always rises
Show the answer
Correct answer: C. The withdrawal has exceeded the water influx, the pressure has fallen below the bubble point, and solution gas drive has started
Water drive is sensitive to the withdrawal rate. If the withdrawal is larger than the water influx, the pressure falls. Below the bubble point gas comes out, and the gas–oil ratio rises. A lower withdrawal brings the accumulation back to water drive.
5
Where should the oil part of the reservoir be opened in gas cap drive, and why?
Show the answer
In the lower part of the oil-saturated thickness, far from the gas–oil contact. This delays the breakthrough of gas from the expanding cap as long as possible. The interval should not be too close to the oil–water contact, so as not to draw up a water cone. The withdrawal rate is limited so that the contact moves down evenly and the gas does not break through along high-permeability layers.

Updated: 19 October 2026