Why water is injected: pressure maintenance and oil displacement

Quick refresher

Water in the reservoir does two jobs. It takes the place of the fluids that have been produced, so the pressure does not fall, the oil stays single-phase (no free gas) and the wells keep their productivity. And it displaces the oil towards the production wells.
The recovery factor is the product $E_D E_A E_V$ : the share of the oil displaced from the flushed pores (the pores the water has passed through), and the shares of the area and of the thickness of the formation that the water has passed through. $E_D$ is limited by the residual oil. The sweep (the two shares of area and thickness) is limited by heterogeneity (differences in the rock from place to place), the mobility ratio, gravity and the layout of the wells. All three grow with the volume of water pumped through.
When the mobility ratio is $M' \le 1$ , the displacement front is stable. When $M' > 1$ , water breaks through in tongues, the sweep at breakthrough is smaller, and a large part of the oil is produced only after breakthrough, at a rising water cut (the share of water in the produced liquid).
Injection is compared with withdrawal at reservoir conditions. The voidage replacement ratio (VRR) is their ratio. When VRR < 1, the pressure of a closed accumulation falls: every megapascal of decline corresponds to an unreplaced volume $c_tV_p$ .
Below the bubble point pressure (the pressure at which gas starts to come out of the oil), free gas sharply increases the underground withdrawal (the volume of the produced fluids at reservoir conditions) and carries energy away. So undersaturated accumulations (those whose oil still holds all its gas in solution) are usually flooded while the pressure is above the bubble point pressure. If injection starts late, the water first fills the volume of the free gas.
The flood pattern is chosen by the size and structure of the accumulation, its permeability, the oil viscosity, the dip of the formation (its slope) and the direction of the fractures. Peripheral flooding (injection wells outside the oil outline) and near-contour flooding (injection wells close to the oil–water contact) suit small accumulations with good conductivity (the rock passes fluid easily). Intra-contour flooding (injection wells inside the outline, in rows or in areal patterns) suits large accumulations and low-permeability ones.

Key relationships

E_R = E_D\,E_A\,E_V

recovery factor: displacement, areal sweep, vertical sweep

E_D = \dfrac{1 - S_{wi} - S_{or}}{1 - S_{wi}}

limiting displacement efficiency from the flushed pores

M' = \dfrac{k_{rw}'/\mu_w}{k_{ro}'/\mu_o}

mobility ratio at the end points of the relative permeability curves

\mathrm{VRR} = \dfrac{q_{wi}B_w}{q_oB_o + q_o(R - R_s)B_g + q_wB_w}

current voidage replacement ratio, reservoir volumes

\Delta\bar p = -\dfrac{(1 - \mathrm{VRR})\,Q\,\Delta t}{c_tV_p}

pressure decline of a closed accumulation above the bubble point pressure

Common pitfalls

Comparing injection with withdrawal in surface cubic metres. One cubic metre of oil takes $B_o$ cubic metres in the reservoir, and gas produced in addition to the dissolved gas takes even more.
Taking the displacement efficiency measured on a core (a rock sample from the well) as the recovery factor. The sweep of a heterogeneous formation is always less than one.
Putting off injection until the strength of the aquifer (the water-bearing rock around the reservoir) is clear, when the pressure is already approaching the bubble point pressure.
Keeping the replacement noticeably above 100% “just in case”. A pressure above the initial one risks fractures, crossflows (flows of fluid between layers) and water breakthrough.
Thinking that waterflooding is over once water breaks through. A large part of the oil is produced at a rising water cut.

After this lesson you can

explain the two jobs of water injection, pressure maintenance and oil displacement, and why together they give more oil than natural depletion
break the recovery factor into displacement efficiency (how much oil the water pushes out of the rock it passes through), areal sweep efficiency and vertical sweep efficiency (the shares of the area and of the thickness that the water reaches), and name what each depends on
calculate the mobility ratio (how much more easily water moves than oil) and explain how it affects the stability of the front (the boundary between the injected water and the oil), the water breakthrough time (when water first reaches the production wells) and injectivity (how much water a well accepts)
make a balance of injection and withdrawal at reservoir conditions, calculate the voidage replacement ratio (injection divided by withdrawal) and estimate how a shortfall will affect the reservoir pressure
justify when to start injection and choose, in outline, a pattern for the injection wells

The two jobs of injection: pressure and displacement

When fluids are produced from an accumulation with no inflow from outside, the pressure always falls. The place of the fluids that have been produced is taken by the expansion of the fluids that remain and by the shrinking of the pore volume (General material balance equation for oil reservoirs). Above the bubble point pressure, only the oil, the water and the rock expand, and the pressure falls fast. Below it, gas comes out of the oil. Once the gas becomes mobile, it flows to the wells and carries the reservoir energy away (Reservoir drive mechanisms). Water injection changes this balance: the place of the produced fluids is taken by the injected water, not by the expansion of the reservoir fluids.

Pressure maintenance: the water replaces the underground withdrawal. The pressure stays above the bubble point pressure, the oil stays single-phase and low in viscosity, the gas stays dissolved in it, and the wells keep their productivity and flow naturally for longer (Well inflow: the well performance diagram).
Displacement: water moves from the injection wells to the production wells and takes the place of the oil in the pores. Where the water has passed, only the residual oil $S_{or}$ is left (the oil that water cannot push out; Fluid saturation). The rest of the oil of the flushed zone goes to the production wells.

The water does both jobs at the same time, but not always to the same degree. In an accumulation with a strong aquifer, injection may be needed only in blocks that faults cut off from the aquifer. In a closed undersaturated accumulation without injection, the pressure reaches the bubble point pressure after only a few per cent of the oil in place has been produced.

In the past, injection started when the accumulation was already being depleted on its natural drive, and it was called a secondary recovery method. Now it is usually built into the field development plan from the very beginning, and the name is mostly historical. Methods that lower the residual oil saturation or improve the sweep beyond ordinary waterflooding are called enhanced oil recovery (EOR) methods. They include polymers (which thicken the water), surfactants (substances that lower the tension between oil and water), gas and heat.

Why water? There is a lot of it: sea water, water from shallow aquifers, and produced water (the water that comes up together with the oil). It is cheap and almost incompressible (it hardly changes in volume under pressure). In many reservoirs the rock is water-wet (its surface prefers water) or mixed-wet (some of its surface prefers water and some prefers oil), and water displaces oil effectively (Wettability and interfacial phenomena). The viscosity of water in the reservoir is usually 0.3–1 mPa·s, of the same order of magnitude as that of light oil. Gas is tens of times more mobile than oil and breaks through to the wells easily (Displacement of oil by gas). There are disadvantages too. Water is heavier than oil and tends to sink to the base of the formation. Sooner or later it also has to be produced, separated and injected again.

Recovery factor: displacement and sweep

To see how much oil waterflooding will give, it helps to break the answer into factors. Water does not displace the oil everywhere. It bypasses part of the area between the wells and part of the thickness of the formation, and where it has passed it leaves residual oil. So we write the recovery factor (RF) as a product.

Recovery factor in waterflooding

E_R = E_D\,E_A\,E_V

where:

$E_R$ recovery factor: the share of the initial oil in place that has been produced by the given time, fraction
$E_D$ displacement efficiency: the share of the initial oil displaced from the flushed pores; not more than $(1 - S_{wi} - S_{or})/(1 - S_{wi})$ , fraction
$E_A$ areal sweep efficiency: the share of the area of the formation that the water has passed through, fraction
$E_V$ vertical sweep efficiency (over the thickness): the share of the thickness of the formation, on the swept area, that the water has passed through, fraction

Units: dimensionless fractions

Holds when

The breakdown is approximate: the factors depend on each other and on the volume of water pumped through, and all three grow with the time of injection.
$E_R$ and $E_D$ refer to the oil at standard conditions. If the formation volume factor of the oil changes noticeably during the flood, the saturations have to be converted through $S_o/B_o$ .
The product $E_AE_V$ is called the volumetric sweep efficiency. In Russian-language practice the recovery factor is written as the product of the displacement efficiency and the sweep efficiency.

Displacement efficiency is a property of the pair “rock and fluids” and of the process. It is measured on core samples. The sweep efficiencies are properties of the formation and of the development system. They are set by heterogeneity, the mobility ratio, gravity, the layout of the wells and the volume of water pumped through. So two accumulations with the same core can differ in recovery factor by a factor of one and a half to two.

SchematicThe three factors of the recovery factor (a diagram, not to scale). Top left: the pores of the flushed zone. Water flows round the grains, and drops of residual oil are trapped in the larger pores (

E_D

). Top right: a quarter of a five-spot element (a production well surrounded by four injection wells at the corners of a square), seen from above, at the moment of water breakthrough. The area along the line between the wells is swept, and two corners are still oil-saturated (

E_A

). Bottom: a cross-section of a layered formation. Along the permeable layer the water has reached the production well, and along the tight layers (layers of low permeability) it has hardly moved away from the injection well (

E_V

What limits each factor and how to raise it

Factor: $E_D$
What limits it: oil trapped in the pores by capillary forces; wettability; the volume pumped through
How to raise it: a larger volume pumped through; enhanced oil recovery methods: surfactants, changing the wettability

Factor: $E_A$
What limits it: the layout of the wells, the mobility ratio, heterogeneity over the area, fractures
How to raise it: closer well spacing, converting wells to injection and adding injection wells, redistributing the injection

Factor: $E_V$
What limits it: layering and high-permeability layers, gravity, the mobility ratio
How to raise it: levelling the injection profile (how the injected water is shared between layers), separate injection into each layer, isolating flushed intervals, polymers

Factor	What limits it	How to raise it
$E_D$	oil trapped in the pores by capillary forces; wettability; the volume pumped through	a larger volume pumped through; enhanced oil recovery methods: surfactants, changing the wettability
$E_A$	the layout of the wells, the mobility ratio, heterogeneity over the area, fractures	closer well spacing, converting wells to injection and adding injection wells, redistributing the injection
$E_V$	layering and high-permeability layers, gravity, the mobility ratio	levelling the injection profile (how the injected water is shared between layers), separate injection into each layer, isolating flushed intervals, polymers

Rough guides for ordinary waterflooding: $E_D$ about 0.5–0.7; $E_AE_V$ from 0.4 to 0.9, depending on heterogeneity and the development system. The recovery factor is most often 30–50%, and up to 60% and more in the best formations with low-viscosity oil. The actual values come from core, the reservoir model and the production history.

Worked exampleWhat waterflooding gives for the same accumulation

Given (assumed data)

Accumulation: the closed oil part of the lessons Reservoir drive mechanisms and Performance prediction for solution gas drive reservoirs: oil in place $N$ = 9.06 million m³, initial pressure 30 MPa, bubble point pressure 18 MPa, $S_{wi}$ = 0.25
Without injection (forecast for solution gas drive): by the time the pressure reaches 5 MPa, 1.57 million m³ has been produced, or 17.3% of the oil in place
Residual oil saturation after displacement by water (core tests): $S_{or}$ = 0.25
Sweep with an areal pattern (an estimate from the reservoir model): $E_A$ = 0.85; $E_V$ = 0.70

Solution

Displacement efficiency: $E_D = (1 - 0.25 - 0.25)/(1 - 0.25) = 0.50/0.75 = 0.667$ . The pressure is maintained at about 24 MPa, and the formation volume factor of the oil changes by only 1%, so it is fine to calculate with saturations. In standard volumes $E_D$ would come out as 0.670, a little more.
Recovery factor: $E_R = 0.667\cdot0.85\cdot0.70 = 0.397$ .
Oil production: $0.397\cdot9.06 = 3.60$ million m³ against 1.57 million m³ by depletion. That is 2.3 times more, a gain of 2.03 million m³.
If the vertical sweep turns out to be 0.50 and not 0.70: $E_R = 0.667\cdot0.85\cdot0.50 = 0.283$ .

Answer: Waterflooding can give about 40% of the oil in place (3.60 million m³), against 17.3% by depletion down to 5 MPa.

What it means. The most uncertain factor is the sweep. $E_D$ is measured on core, while $E_A$ and $E_V$ are set by the heterogeneity of the formation, the layout of the wells and how much water can be pumped through before the limiting water cut (the highest water cut at which production can continue) is reached. Even with $E_V$ = 0.50, waterflooding gives noticeably more than depletion. And even a perfect sweep would leave a third of the oil in the formation: the residual oil, which only enhanced oil recovery methods can produce. We take the depletion forecast from the lesson on solution gas drive. The data in this example are assumed for illustration.

Mobility ratio

How stably water displaces oil is determined by the mobility ratio. It is the ratio of the mobility of the water behind the displacement front to the mobility of the oil ahead of it: $M' = (k_{rw}'/\mu_w)/(k_{ro}'/\mu_o)$ . (Mobility is how easily a fluid moves: the permeability to that fluid divided by its viscosity.) Behind the front, water flows at residual oil saturation, with the end-point relative permeability $k_{rw}'$ (the share of the rock's permeability that is open to water at the end of the displacement). Ahead of the front, oil flows at irreducible water saturation (the water that stays in the pores and does not move), with $k_{ro}'$ (relative permeability end points).

Why this matters. Suppose that a bump forms by chance on a flat front: in one place the water has moved a little further. If water is more mobile than oil ( $M' > 1$ ), the zone that the water occupies conducts better than the oil that it displaced. The resistance along the bump falls, the flow towards it grows, and the bump stretches into a “tongue” (viscous fingering). If $M' < 1$ , the bump meets more resistance, falls behind, and the front levels out. So at $M' \le 1$ the front stays sharp, and most of the mobile oil is produced before water breaks through (Buckley–Leverett theory and the Welge method). At $M' > 1$ water reaches the production wells sooner, the sweep at breakthrough is smaller, and a large part of the oil is produced at a high water cut.

Mobility ratio for different oil viscosities

Oil viscosity in the reservoir, mPa·s: 1
$M'$: 0.63
What to expect: the front is stable and sharp, and most of the mobile oil is produced before water breaks through

Oil viscosity in the reservoir, mPa·s: 5
$M'$: 3.1
What to expect: water breaks through earlier, and a noticeable part of the oil is produced after breakthrough

Oil viscosity in the reservoir, mPa·s: 20
$M'$: 12.5
What to expect: early breakthrough; a large part of the oil is produced at a water cut above 90%

Oil viscosity in the reservoir, mPa·s: 100
$M'$: 62.5
What to expect: ordinary waterflooding is ineffective: thickened water (polymers) or thermal methods (heating the reservoir) are needed

Oil viscosity in the reservoir, mPa·s	$M'$	What to expect
1	0.63	the front is stable and sharp, and most of the mobile oil is produced before water breaks through
5	3.1	water breaks through earlier, and a noticeable part of the oil is produced after breakthrough
20	12.5	early breakthrough; a large part of the oil is produced at a water cut above 90%
100	62.5	ordinary waterflooding is ineffective: thickened water (polymers) or thermal methods (heating the reservoir) are needed

The relative permeability end points are assumed to be the same: $k_{rw}'$ = 0.20, $k_{ro}'$ = 0.80. The water viscosity is 0.40 mPa·s (at about 85 °C). The data are assumed for illustration. We calculate how the fractional flow of water (the share of water in the flowing liquid) grows, and when breakthrough comes, with the fractional flow function (Fractional flow function and mobility ratio).

The oil of the accumulation in the example above has a low viscosity: about 1.1 mPa·s at 24 MPa. With the same end points, $M' \approx 0.69$ , so conditions for waterflooding are favourable. If the pressure is lowered to 10 MPa, the oil that has lost its gas becomes more viscous, to about 1.45 mPa·s, and $M'$ grows to 0.91. For viscous oils this effect is much stronger.

The mobility ratio also affects the injectivity of an injection well: the flow rate of water that it accepts at a given pressure difference. As the zone of water around the well grows, the resistance on the way to the production wells changes. At $M' > 1$ injection becomes easier with time, and at $M' < 1$ it becomes harder. So the injection pressure of the same well is different at different stages.

Voidage replacement

To keep the pressure from falling, injection must replace the underground withdrawal, the volume that the oil, gas and water occupied in the reservoir. The ratio of injection to withdrawal at reservoir conditions is called voidage replacement (the voidage replacement ratio, VRR). In Russian-language practice it is given as a percentage. The balance before water breakthrough, when only oil is produced, and the calculation of the number of injection wells are covered in the lesson Well inflow: the well performance diagram. Here we treat the general case, with produced water and free gas.

Current voidage replacement ratio

\mathrm{VRR} = \frac{q_{wi}B_w}{q_oB_o + q_o\,(R - R_s)\,B_g + q_wB_w}

where:

$\mathrm{VRR}$ current voidage replacement ratio, fraction (or %)
$q_{wi}$ injection rate of water at standard conditions, m³/d
$q_o,\ q_w$ production rates of oil and water at standard conditions, m³/d
$R$ instantaneous (current) producing gas–oil ratio: all the produced gas per cubic metre of oil, m³/m³
$R_s$ solution gas–oil ratio at the current average reservoir pressure, m³/m³
$B_o,\ B_g,\ B_w$ formation volume factors of oil, gas and water at the current average reservoir pressure (surface volumes are at 0.101325 MPa and 20 °C), m³/m³

Units: any identical units of rate; the gas–oil ratio and the solution gas–oil ratio are in m³ of gas per m³ of oil at standard conditions

Holds when

$q_o(R - R_s)$ is the free gas produced in addition to the gas dissolved in the oil. Above the bubble point pressure $R = R_s$ , and this term disappears.
Injected gas, if there is any, is added to the numerator as $q_{gi}B_g$ . Water influx from the aquifer is not part of VRR: it is a natural replacement, and we take it into account separately in the balance.
The cumulative replacement is the ratio of the sums of injection and withdrawal over the whole period, each converted to reservoir volumes on its own date.

At VRR = 1, injection replaces withdrawal, and the pressure of a closed accumulation is stable. At VRR < 1, the missing volume is supplied by the expansion of the fluids and the shrinking of the pore volume, so the pressure falls. At VRR > 1 it rises. The denominator is calculated at reservoir conditions. A cubic metre of produced oil frees $B_o$ cubic metres in the reservoir, and gas produced in addition to the dissolved gas frees $B_g$ cubic metres for every cubic metre at the surface. At a high gas–oil ratio there are hundreds of such cubic metres.

Pressure change under incomplete replacement (a closed accumulation)

\Delta\bar p = -\frac{(1 - \mathrm{VRR})\,Q\,\Delta t}{c_t V_p}

where:

$\Delta\bar p$ change in the average reservoir pressure over the time $\Delta t$ , MPa
$Q$ underground withdrawal per unit of time, the denominator of VRR, m³/d
$\Delta t$ time, d
$c_t$ total compressibility of rock and fluids, $S_oc_o + S_wc_w + c_f$ , at the current saturations (Flow regimes), 1/MPa
$V_p$ pore volume of the accumulation, m³

Units: MPa, m³/d, d, 1/MPa, m³

Holds when

The accumulation is closed (there is no aquifer), and the pressure is above the bubble point pressure, so there is no free gas.
The pressure equalises across the accumulation faster than it changes (the tank model: the whole accumulation is treated as one tank with a single pressure), and the compressibility is constant over the pressure interval.
With free gas the compressibility of the system is many times larger, and for the same deficit the pressure responds less, but the gas escapes into the wells.

The product $c_tV_p$ is the volume that the accumulation “gives up” for each megapascal of pressure decline. A shortfall in replacement of a few per cent seems small, but in a closed accumulation it builds up year after year. This is a special case of the material balance (General material balance equation for oil reservoirs). The derivation is in the advanced section.

Worked exampleBalance of injection and withdrawal over a month

Given (assumed data)

Accumulation: the same as in the example above: pore volume 16.0 million m³; the pressure is held at about 24 MPa, above the bubble point pressure of 18 MPa
Production, monthly average: oil 1200 m³/d, water 650 m³/d; gas–oil ratio 110 m³/m³, equal to the solution gas–oil ratio
Injection: 2100 m³/d
Formation volume factors at 24 MPa: $B_o$ = 1.337; $B_w$ = 1.02
Compressibilities of oil, water and pores; average water saturation: $1.6\cdot10^{-3}$ , $0.45\cdot10^{-3}$ and $0.8\cdot10^{-3}$ MPa⁻¹; $S_w$ = 0.35

Solution

Underground withdrawal: oil $1200\cdot1.337 = 1604.4$ m³/d, water $650\cdot1.02 = 663.0$ m³/d. There is no free gas ( $R = R_s$ ). The total is $Q = 2267.4$ m³/d.
Injection at reservoir conditions: $2100\cdot1.02 = 2142.0$ m³/d. $\mathrm{VRR} = 2142.0/2267.4 = 0.945$ , or 94.5%.
The shortfall is 125.4 m³/d. Over a year it is $125.4\cdot365 = 45.8$ thousand m³.
Total compressibility: $c_t = (0.65\cdot1.6 + 0.35\cdot0.45 + 0.8)\cdot10^{-3} = 2.00\cdot10^{-3}$ MPa⁻¹. For each 1 MPa the accumulation gives up $c_tV_p = 2.00\cdot10^{-3}\cdot16.0\cdot10^6 = 32.0$ thousand m³. Over a year the pressure falls by $45.8/32.0 = 1.43$ MPa.
For full replacement, injection must be raised to $2267.4/1.02 = 2223$ m³/d, that is by 123 m³/d.
For comparison, suppose that the pressure fell to 14 MPa and the gas–oil ratio grew to 262 m³/m³ (the forecast of the lesson Performance prediction for solution gas drive reservoirs). Each cubic metre of oil would then free $B_o + (R - R_s)B_g = 1.289 + (262 - 87.7)\cdot0.00769 = 2.63$ m³ in the reservoir instead of 1.337 m³, which is 2.0 times more.

Answer: The replacement is 94.5%. The shortfall is 125.4 m³/d, and the pressure of the closed accumulation will fall by about 1.43 MPa per year.

What it means. A replacement of 94.5% looks almost complete. But with this shortfall the pressure will fall from 24 MPa to the bubble point pressure in about 4.2 years, and after that gas will start to come out in the reservoir. So engineers track the replacement month by month and by area of the accumulation, and check it against pressure measurements. If the pressure falls while the calculated replacement is 100%, the formation volume factors or the injection records are wrong, or water is escaping outside the accumulation. If the pressure rises while the replacement is below 100%, an aquifer is working. The last step shows why the pressure is kept above the bubble point pressure: free gas almost doubles the underground withdrawal per cubic metre of oil. The data in this example are assumed for illustration.

After water breakthrough the same balance shows something different. To hold the pressure, we also have to inject enough to replace the water that is produced. If injection is limited by the capacity of the pumps and of water treatment, a rising water cut at full replacement inevitably reduces oil production.

Oil rate at constant injection and full replacement

q_o = \frac{Q_i}{B_o + B_w\,\dfrac{\text{WC}}{1 - \text{WC}}}

where:

$q_o$ oil production rate at standard conditions, m³/d
$Q_i$ injection at reservoir conditions, $q_{wi}B_w$ , m³/d
$\text{WC}$ water cut of the production: the share of water in the produced liquid at standard conditions, fraction
$B_o,\ B_w$ formation volume factors of oil and water, m³/m³

Units: m³/d; the formation volume factors are in m³/m³

Holds when

VRR = 1, and there is no aquifer.
The pressure is above the bubble point pressure, so there is no free gas.
Injection is limited by the capacity of the pumps and of water treatment.

For each cubic metre of oil there are $\text{WC}/(1 - \text{WC})$ cubic metres of water: one at a water cut of 50%, nine at 90%, nineteen at 95%. If injection is not increased, oil production at a late stage falls several-fold. So engineers design the capacity of injection and of water treatment from the water cut forecast.

CalculatedOil rate at constant injection and full replacement against the water cut of the production. Injection is 2223 m³/d, as needed for full replacement in the example above;

B_o

= 1.337,

B_w

= 1.02. The state of the example and a water cut of 90% are marked.

When to start injection

Three things decide when to start injection: how much natural energy the accumulation has, how far the initial pressure is above the bubble point pressure, and what is known about the aquifer (Reservoir drive mechanisms, Aquifer models: water influx).

Options

1
From the start of development
Typical for undersaturated accumulations without a strong aquifer, and for offshore fields, where an injection system is hard and costly to add later. In the oil part of the example, the elastic store down to the bubble point pressure (the most oil that the expansion of the oil, water and rock can push out) is about 3.3% of the oil in place. At a plateau rate (the steady top rate of production) of 1200 m³/d it would last about eight months.
2
After a period of observation
If the initial pressure is far above the bubble point pressure (for example in an accumulation with abnormally high pressure, Abnormal formation pressures: origin, prediction and recognition), or if water influx from the aquifer is expected, we first produce without injection and watch the pressure. The rate at which the pressure falls per unit of production shows the strength of the aquifer. Pressure measurements at different depths in new wells show which thin layers are being depleted and whether they are connected to each other (Pressure measurement with the wireline formation tester). We then build the injection system for the drive that the accumulation actually has.
3
After depletion
If the pressure is already below the bubble point pressure, there is free gas in the reservoir. The injected water first fills the volume that the gas occupies, so the production wells do not respond at once. Oil that has lost its gas is more viscous, the mobility ratio is worse, and the energy that the gas carried away is lost for good.

Worked exampleHow much water goes into filling the gas-filled volume

Given (assumed data)

State of the accumulation: the pressure has fallen to 14 MPa and 909 thousand m³ of oil has been produced; the pore volume, allowing for the shrinking of the pores, is 15.79 million m³, and the gas saturation is 0.080 (lesson Performance prediction for solution gas drive reservoirs)
Injection: exceeds the underground withdrawal by 1500 m³/d

Solution

Volume of free gas in the reservoir: $0.080\cdot15.79 = 1.26$ million m³.
Until this volume is filled with water, the gas is partly produced and partly compressed and dissolved in the oil. The pressure rises slowly, and the production wells hardly feel the injection. Time to fill: $1.26\cdot10^6/1500 = 840$ d, about 2.3 years.
If injection had started before the pressure fell below 18 MPa, there would be nothing to fill: the gas would have stayed dissolved in the oil.

Answer: About 1.26 million m³ of water will go into filling the gas-filled volume. With injection exceeding withdrawal by 1500 m³/d, this takes about 2.3 years.

What it means. Late injection does not only delay the response. During depletion the reservoir has lost gas, its main source of energy, and the oil has become more viscous. So accumulations with undersaturated oil are usually flooded before the pressure reaches the bubble point pressure. The estimate is rough, and it overestimates the volume: part of the gas will remain trapped in the pores, and the actual volume to fill will be somewhat smaller. The data in this example are assumed for illustration.

Patterns of injection wells

Where to place injection wells depends on the size and structure of the accumulation. The Russian-language classification of flood patterns has three main kinds. In peripheral flooding, the injection wells are outside the oil outline. In near-contour flooding, they are placed close to the oil–water contact. In intra-contour flooding, they are inside the oil outline. Intra-contour flooding can be line-drive (block), areal, selective, focal or barrier.

SchematicFlood patterns seen from above (a diagram, not to scale): peripheral, three-row, five-spot and nine-spot, with the injection well in the centre of the element. The line is the outer pool outline (the outer edge of the oil-bearing area); triangles are injection wells and dots are production wells; the dashed line marks an element of the areal pattern.

Flood patterns

Pattern: Peripheral
Where the injection wells are: in a ring outside the outer pool outline
When it is used: small and narrow accumulations with good permeability and good communication with the zone outside the outline

Pattern: Near-contour
Where the injection wells are: in the edge-water zone (where oil lies above water), near the outline
When it is used: communication with the zone outside the outline is poor

Pattern: Line-drive (block)
Where the injection wells are: in rows across the accumulation; between the rows there are one, three or five rows of production wells
When it is used: large elongated accumulations; the lower the permeability, the fewer rows in a block

Pattern: Areal
Where the injection wells are: in the elements of a grid: five-spot, seven-spot, nine-spot
When it is used: low-permeability and heterogeneous formations, viscous oil: when intensive injection is needed

Pattern: Selective, focal
Where the injection wells are: at chosen points, in addition to the main pattern
When it is used: heterogeneous formations: to reach places where water from the main pattern does not arrive

Pattern: Barrier
Where the injection wells are: in a row along the gas–oil contact
When it is used: accumulations with a gas cap: the water cuts the cap off from the oil

Pattern	Where the injection wells are	When it is used
Peripheral	in a ring outside the outer pool outline	small and narrow accumulations with good permeability and good communication with the zone outside the outline
Near-contour	in the edge-water zone (where oil lies above water), near the outline	communication with the zone outside the outline is poor
Line-drive (block)	in rows across the accumulation; between the rows there are one, three or five rows of production wells	large elongated accumulations; the lower the permeability, the fewer rows in a block
Areal	in the elements of a grid: five-spot, seven-spot, nine-spot	low-permeability and heterogeneous formations, viscous oil: when intensive injection is needed
Selective, focal	at chosen points, in addition to the main pattern	heterogeneous formations: to reach places where water from the main pattern does not arrive
Barrier	in a row along the gas–oil contact	accumulations with a gas cap: the water cuts the cap off from the oil

The names “direct” and “inverted” for seven-spot and nine-spot patterns are used in different ways in different places. Check which well stands in the centre of the element.

The ratio of injection wells to production wells shows how intensive a pattern is. In a five-spot it is one to one. In a seven-spot with the injection well in the centre of the element it is one to two, and in a nine-spot with the injection well in the centre it is one to three. In a line-drive pattern, each row of injection wells has one, three or five rows of production wells. The more injection wells per production well, the faster the pattern replaces the withdrawal, but the more it costs.

Size of the accumulation. The effect of peripheral injection fades towards the centre of a wide accumulation, and the central wells are left without support. There, intra-contour injection is needed.
Permeability and injectivity. In a low-permeability formation an injection well accepts little water and affects a small area, so a dense areal grid is needed. If the injectivity of the injection wells is lower than the productivity of the production wells, more injection wells are needed.
Dip of the formation and gravity. In steep formations, water is injected into the lower part: displacing the oil from the bottom upwards, it moves more stably.
Fractures and stresses. Engineers align the rows of injection wells with the direction of the maximum horizontal stress (the strongest sideways squeeze on the rock) and of the natural fractures, so that the fractures connect the injection wells with each other and not with the production wells (Rock mechanics for hydraulic fracturing).
Heterogeneity. Engineers refine the pattern as drilling and development go on: they add focal injection wells, convert production wells that have watered out to injection, and shut off flushed intervals.

What hinders waterflooding

High-permeability layers and fractures. Water takes the path of least resistance. It breaks through to the production wells along the best layer and bypasses the others. Then it circulates along the flushed path and hardly displaces any oil (Reservoir heterogeneity and permeability averaging).
Gravity. Water is heavier than oil, and when it advances slowly it sinks to the base of the formation and leaves the oil at the top.
Injectivity. Suspended solids, oil droplets and bacteria in the injected water, incompatibility of its salts with the formation water (sulfate scale, hard deposits that block the pores), and swelling of clays all reduce injectivity. The requirements for the water and its treatment are the subject of the lesson Water injection systems.
Cold water. Cooling of the rock near an injection well lowers the stresses in it, so fractures open at a lower pressure. Injectivity grows, but a fracture can carry the water all the way to a production well (Rock mechanics for hydraulic fracturing). If the wax appearance temperature of the oil (the temperature at which wax starts to crystallise out of it) is close to the reservoir temperature, cooling threatens to deposit wax in the pores. Such formations are flooded with heated water.
Isolated blocks. Faults and impermeable barriers separate parts of the accumulation from the injection wells. The pressure there keeps falling even though the overall replacement is 100%. This shows up in pressure measurements in new wells.
Produced water. Sooner or later water reaches the production wells, and it has to be lifted, separated, cleaned and injected back. Engineers design the capacity of these systems from the water cut forecast. An early breakthrough with limited capacity means a loss of oil production (see the figure in the section on voidage replacement).

Engineers monitor a waterflood by the replacement ratio in each area, maps of the reservoir pressure, injection profiles of the injection wells (which intervals accept water), tracer tests (a marker added to the injected water shows which production wells each injection well is connected to), and the growth of the water cut. A mismatch between these data and the forecast is the main reason to change the injection well by well.

Terms and symbols

Waterflooding: The injection of water into the formation to maintain the pressure and displace the oil.
Pressure maintenance system: The injection of water to maintain the reservoir pressure. The name covers both the injection and its whole system: the water intake, water treatment, pumping stations, water lines and injection wells.
$E_D$ , $E_A$ , $E_V$: Displacement efficiency, areal sweep efficiency and vertical sweep efficiency. Their product is the recovery factor.
$M'$: The ratio of the mobilities of water and oil at the end points of the relative permeability curves. At $M' > 1$ the displacement front is unstable.
VRR: Voidage replacement ratio: the ratio of injection to withdrawal at reservoir conditions.
Injectivity: The flow rate of water that an injection well accepts at a given pressure difference. The injectivity index is the flow rate per unit of pressure difference.
Fill-up (filling the gas-filled volume): The period after a late start of injection, when the water occupies the volume of the free gas and the pressure is restored.
Flood pattern: The arrangement of the injection wells relative to the production wells and to the outlines of the accumulation.

Common mistakes and misconceptions

Wrong

“We injected as many cubic metres of water as we produced of oil, so the pressure is maintained.”

In fact

We must compare reservoir volumes. Oil takes $B_o$ times more space in the reservoir, the produced water also has to be replaced, and gas produced in addition to the dissolved gas needs many cubic metres of water for each cubic metre of oil.

Wrong

“The displacement efficiency from the core is 0.67, so the recovery factor will be 67%.”

In fact

The core shows how much oil is displaced where the water has passed. The recovery factor equals the displacement efficiency times the areal and vertical sweeps. In a heterogeneous formation the sweep is noticeably less than one.

Wrong

“The more we inject, the better.”

In fact

A replacement well above 100% raises the pressure above the initial one. The risk of fractures, crossflow into neighbouring formations and water breakthrough along fractures grows. We keep injection near full replacement and share it out between areas.

Wrong

“Water has broken through, so the flood is over.”

In fact

After breakthrough the water keeps displacing oil, and the sweep keeps growing. With an unfavourable mobility ratio most of the oil is produced at a rising water cut. Production stops at the limiting water cut that is still economic.

Wrong

“There is an aquifer outside the outline, so injection is not needed.”

In fact

The strength of the aquifer is almost never known before production starts, part of the accumulation may be cut off by faults, and at a high withdrawal rate the water influx cannot keep up. Whether injection is needed is decided from pressure measurements and the material balance (Reservoir drive mechanisms).

In depth: advanced detail

Replacement and the material balance

The formula for the pressure change in the main section is a special case of the general material balance equation (General material balance equation for oil reservoirs). With injection of water and gas it is written like this:

Material balance with injection

F = N\,(E_o + mE_g + E_{fw}) + W_e + W_{inj}B_w + G_{inj}B_g

where:

$F$ underground withdrawal accumulated to date, $N_p[B_o + (R_p - R_s)B_g] + W_pB_w$ ; $R_p$ is the cumulative gas–oil ratio, m³
$N$ oil in place, m³
$E_o,\ E_g,\ E_{fw}$ expansion of the oil with its dissolved gas, of the gas cap (for $m = 1$ ) and of the irreducible water with the pores, per cubic metre of oil in place, m³/m³
$m$ ratio of the reservoir volume of the gas cap to the reservoir volume of the oil, —
$W_e$ cumulative water influx from the aquifer, m³
$W_{inj},\ G_{inj}$ cumulative water and gas injection at standard conditions, m³

Units: m³ (oil and gas at standard conditions, the rest at reservoir conditions)

Holds when

The assumptions of the tank material balance.
The injected volumes are converted to reservoir volumes at the current average pressure.

The ratio $(W_{inj}B_w + G_{inj}B_g)/F$ is the cumulative replacement. The closer it is to one, the less the pressure moves away from the initial one, and the less work is done by the other terms, the expansion of the fluids and the water influx.

Above the bubble point pressure, with no gas cap and no water influx, $E_o + E_{fw} \approx B_{oi}\,c_e\,\Delta p$ , where $c_e$ is the effective compressibility of the oil-saturated part (Reservoir drive mechanisms). The product $NB_{oi}$ is the oil-saturated pore volume $S_oV_p$ , so $NB_{oi}c_e = V_p(S_oc_o + S_wc_w + c_f) = c_tV_p$ . The balance becomes $F - W_{inj}B_w = c_tV_p\,\Delta p$ : the pressure decline is proportional to the withdrawal that is not replaced. For a short interval of time, with the current saturations in $c_t$ , this is the formula of the main section.

Areal sweep: what it depends on

Even in a uniform formation, water sweeps the area of an element unevenly. It travels fastest along the straight line between the injection well and the production well, and slowest towards the corners of the element. For a five-spot element of a uniform formation at $M' = 1$ , about 72% of the area is swept by the time of breakthrough. As $M'$ grows, the sweep at breakthrough gets smaller, and at $M' < 1$ it gets larger.

After breakthrough the sweep keeps growing. Each further pore volume of water pumped through (an amount of water equal to the volume of the pores) widens the swept area, but more and more slowly and at an ever higher water cut. The sweep also depends on the pattern. In a line-drive pattern it is higher the farther apart the rows are compared with the distance between wells in a row, because the flow is then closer to linear.

These findings come from models of uniform formations, both physical (laboratory) and numerical (computer) ones. In a real formation the areal sweep depends more strongly on the position of high-permeability zones, fractures and faults. So engineers refine it with the reservoir model and with the production history (History matching).

Connections to other topics

Reservoir drive mechanisms — natural drives and why their energy is often not enough.
General material balance equation for oil reservoirs — the balance of withdrawal, expansion and injection in reservoir volumes.
Performance prediction for solution gas drive reservoirs — the depletion forecast for the same accumulation, the point of comparison for waterflooding.
Relative permeability — the end points of the relative permeability curves and the mobility ratio.
Fluid saturation — residual oil saturation and displacement efficiency.
Well inflow: the well performance diagram — the injection balance before water breakthrough and the number of injection wells.
Reservoir heterogeneity and permeability averaging — why water goes along the best layers, and how to describe heterogeneity.
Fractional flow function and mobility ratio — the fractional flow of water and the growth of the water cut.
Water injection systems — water treatment and the equipment of the pressure maintenance system.

Check yourself

1
An accumulation is undersaturated: the initial pressure is 26 MPa, the bubble point pressure is 21 MPa, and the aquifer is weak. Why is water injection planned from the start of development?
1. A.Without injection the reservoir pressure cannot be measured
2. B.To displace the residual oil at once
3. C.The elastic store is enough for only a small share of the oil, and below the bubble point pressure gas will come out: it will carry the energy away and reduce the oil recovery
4. D.The injected water lowers the viscosity of the oil
Show the answer
Correct answer: C. The elastic store is enough for only a small share of the oil, and below the bubble point pressure gas will come out: it will carry the energy away and reduce the oil recovery
Above the bubble point pressure, only the oil, the water and the rock expand. This is only 1–2% of the oil in place for a pressure decline of 5 MPa. After that, solution gas drive will begin, with a fast fall in pressure and a growing gas–oil ratio. Injection from the start keeps the pressure above the bubble point pressure. The oil stays single-phase, and the water also displaces it.
2
Over a month, on average 800 m³/d of oil ( $B_o$ = 1.25) and 400 m³/d of water ( $B_w$ = 1.02) were produced, with the pressure above the bubble point pressure, and 1300 m³/d of water was injected. What is the replacement ratio, and what will happen to the pressure if there is no aquifer?
Show the answer
Withdrawal at reservoir conditions: $800\cdot1.25 + 400\cdot1.02 = 1000 + 408 = 1408$ m³/d. Injection: $1300\cdot1.02 = 1326$ m³/d. VRR = 1326/1408 ≈ 0.94, that is 94%. The shortfall is 82 m³/d, and the pressure will fall slowly. For full replacement, about 1380 m³/d must be injected.
3
The oil viscosity in the reservoir is 12 mPa·s, the water viscosity is 0.5 mPa·s, and the relative permeability end points are $k_{rw}' = 0.25$ and $k_{ro}' = 0.75$ . What is the mobility ratio, and what should we expect in waterflooding?
1. A.8: the front is unstable, early water breakthrough
2. B.1: piston-like displacement (the water pushes the oil ahead of it like a piston)
3. C.It cannot be found without the permeability of the formation
4. D.0.125: the front is stable
Show the answer
Correct answer: A. 8: the front is unstable, early water breakthrough
$M' = (0.25/0.5)/(0.75/12) = 0.5/0.0625 = 8$ . Water is eight times more mobile than oil. The front stretches into tongues, water breaks through early, and a large part of the oil is produced at a high water cut. The permeability of the formation does not enter the ratio: it is the same for both phases and cancels out.
4
From core, the displacement efficiency is 0.62. From the reservoir model, the areal sweep is 0.80 and the vertical sweep is 0.55. Estimate the recovery factor. Which factor should be refined first, and how?
Show the answer
$E_R = 0.62\cdot0.80\cdot0.55 \approx 0.27$ . The smallest and most uncertain factor is the vertical sweep: it is set by layering and high-permeability layers. It is refined with a detailed description of the section (core, well logs), injection and production profiles, pressure measurements in the separate layers, and by matching the model to the production history.
5
In an areal pattern, an injection well stands in the centre of a nine-spot element. How many production wells are there per injection well?
1. A.One
2. B.Three
3. C.Eight
4. D.Two
Show the answer
Correct answer: B. Three
There are eight production wells around the injection well, but they are shared with neighbouring elements: each corner well belongs to four elements, and each well at the midpoint of a side belongs to two. Per element there are $4\cdot\tfrac14 + 4\cdot\tfrac12 = 3$ production wells for one injection well.
6
The injection pumps are designed for full replacement at a water cut of 50%. By how many times will oil production fall if the water cut grows to 80% and injection cannot be increased? $B_o$ = 1.3, $B_w$ = 1.0; the pressure is above the bubble point pressure.
Show the answer
At full replacement, $q_o = Q_i/(B_o + B_w\,\text{WC}/(1 - \text{WC}))$ . At 50% the denominator is $1.3 + 1 = 2.3$ , and at 80% it is $1.3 + 4 = 5.3$ . The oil rate will fall by a factor of $5.3/2.3 \approx 2.3$ , to 43% of the former rate. So engineers design the capacity of injection and water treatment from the water cut forecast.

Updated: 19 October 2026