Displacement of oil by gas

Quick refresher

We write the fractional flow of gas in the same way as that of water: $f_g = (\lambda_g/\lambda_t)(1 - N_g\,k_{ro}/k_{ro}')$ . Here $\lambda$ is mobility (how easily a fluid flows), $k_{ro}$ is the relative permeability of the oil (the share of the rock's permeability that is available to the oil), and $N_g$ is the gravity number (how strong gravity is compared with the viscous force). For gas that moves down the dip (down the slope of the formation), $N_g > 0$ : gravity holds the gas back.
In the reservoir, gas is tens of times less viscous than oil, so the mobility ratio $M'$ (how much more easily gas flows than oil) is in the tens. In horizontal displacement the gas saturation at the front (the share of the pore space filled with gas) is low. Breakthrough comes after only a small fraction of a pore volume (the total volume of the pores in the zone being displaced) has been injected. After that, the producing gas–oil ratio rises by an order of magnitude (about ten times).
Gravity and a low rate make the front stable. Gas injected at the crest (the highest part of the structure), or gas from an expanding gas cap, keeps a stable front as long as the rate is below the critical rate. For gas, the critical rate is almost equal to the gravity rate, the rate at which the gravity number equals one (Segregated flow and gravity displacement).
What decides is the flow cross-section, the area that the flow passes through. A gas cap that moves down over the whole area of the accumulation (the oil deposit) is usually stable at ordinary withdrawal rates (the rates at which fluids are taken out of the reservoir). Gas that moves along a thin formation runs ahead as a tongue along the top of the formation.
To hold the pressure, we usually have to inject more gas than is produced. For each cubic metre of oil we need $B_o/B_g$ standard cubic metres of gas (gas measured at surface conditions). On top of that we must inject all the free gas that is produced (gas that is not dissolved in the oil). Here $B_o$ and $B_g$ are the formation volume factors: the volume that one cubic metre of oil or gas measured at the surface takes up in the reservoir. Reinjecting only the produced gas covers just a part of the withdrawal.
The residual oil in gas displacement (the oil left behind in the swept rock) is often lower than after waterflooding, yet gas usually recovers less oil. The cause is the unfavourable mobility ratio, not the residual saturation. Above the minimum miscibility pressure (the lowest pressure at which gas and oil mix completely), the boundary between the phases disappears, and the immiscible calculation no longer applies.

Key relationships

f_g = \dfrac{\lambda_g}{\lambda_t}\left(1 - N_g\,\dfrac{k_{ro}}{k_{ro}'}\right)

fractional flow of gas; $N_g > 0$ for gas moving down the dip

M' = \dfrac{k_{rg}'/\mu_g}{k_{ro}'/\mu_o}

mobility ratio of gas to oil at the end points of the curves (the highest relative permeabilities)

G_{iD,BT} = \dfrac{1}{f_g'(S_{gf})},\qquad N_{pD} = S_{ge} + (1 - f_{ge})\,G_{iD}

the Welge method for gas; there is no initial free gas

R = R_s + \dfrac{B_o}{B_g}\cdot\dfrac{f_g}{1 - f_g}

producing gas–oil ratio from the fractional flow of gas in the reservoir flow; $R_s$ is the solution gas–oil ratio (the gas dissolved in the oil)

q_{\text{cr}} = q_{\text{gr}}\,\dfrac{M'}{M' - 1} \approx q_{\text{gr}}

the critical rate from the lesson Segregated flow and gravity displacement; for gas $M'$ is large

q_{g,\text{inj}} = q_o\,\dfrac{B_o}{B_g} + q_o\,(R - R_s)

gas injection to hold the pressure when no water is produced, in sm³/d (standard m³ per day)

Common pitfalls

Counting the pore volumes of injected gas in standard cubic metres. The displacement calculation uses reservoir volumes: $G_{iD} = G_{inj}B_g/V_p$ .
Mixing up the producing gas–oil ratio with the fractional flow of gas in the reservoir flow. At $f_g \approx 0.6$ , the producing gas–oil ratio is already four times the solution gas–oil ratio.
Injecting gas into the low part of the formation, or along a thin formation, and counting on gravity to stabilise the front.
Taking the limiting displacement efficiency of gas as the forecast. This limit (the largest share of the oil that the gas could push out) follows from the residual oil saturation. With an unfavourable mobility ratio, the real value is very far below it.
Using the immiscible calculation above the minimum miscibility pressure, when gas and oil become one phase.

After this lesson you can

write the fractional flow of gas with gravity included, and explain why gas breaks through early in horizontal displacement
use the Welge method (finding the saturation at the front from a tangent to the fractional flow curve) to calculate the gas saturation at the front (the share of the pore space filled with gas), the amount of gas injected by breakthrough (the moment the gas reaches the producing wells), the oil production and the producing gas–oil ratio after breakthrough
apply the critical rate (the highest rate at which gravity still keeps the front stable) to crestal gas injection (injection at the top of the structure) and to gas cap expansion, and decide whether the front is stable
calculate the volume of gas needed to maintain pressure, and the share of it that the reinjection of produced gas covers
explain how the residual oil saturation (the share of the pore space that still holds oil after displacement) in gas displacement differs from that after waterflooding, and when gas becomes miscible with oil (mixes with it completely)

How gas differs from water as a displacing agent

We inject gas into oil accumulations for three purposes. The first is pressure maintenance: keeping the reservoir pressure from falling. While the pressure stays above the bubble point pressure (the pressure at which gas starts to come out of the oil), no gas comes out of the oil in the reservoir. The oil does not become more viscous. The accumulation also does not slide into the inefficient solution gas drive, in which the gas that comes out of the oil does the pushing (Performance prediction for solution gas drive reservoirs). The second is displacement: the gas pushes the oil out of the part of the reservoir that it reaches. The third is use of the produced gas, which cannot be sold or is not worth selling. We return it to the reservoir instead of burning it or cleaning it (gas injection).

Water and gas in the reservoir

Property: Viscosity in the reservoir
Water: 0.3–1 mPa·s
Gas: 0.01–0.03 mPa·s
What it means for displacement: the mobility ratio of gas to oil is in the tens: the front is unstable

Property: Density in the reservoir
Water: 1.0–1.2 g/cm³
Gas: 0.1–0.3 g/cm³
What it means for displacement: gas rises: injection at the crest helps, injection along the formation hurts

Property: Wettability
Water: wets a water-wet rock
Gas: does not wet the rock
What it means for displacement: gas goes through the large pores; oil spreads in films

Property: Compressibility (how much the volume changes with pressure)
Water: small
Gas: large
What it means for displacement: the volume of the gas in the reservoir is set by $B_g$ ; the calculation is made at constant pressure

Property: Exchange of components with the oil
Water: practically none
Gas: possible
What it means for displacement: at high pressure the gas can become miscible

Property	Water	Gas	What it means for displacement
Viscosity in the reservoir	0.3–1 mPa·s	0.01–0.03 mPa·s	the mobility ratio of gas to oil is in the tens: the front is unstable
Density in the reservoir	1.0–1.2 g/cm³	0.1–0.3 g/cm³	gas rises: injection at the crest helps, injection along the formation hurts
Wettability	wets a water-wet rock	does not wet the rock	gas goes through the large pores; oil spreads in films
Compressibility (how much the volume changes with pressure)	small	large	the volume of the gas in the reservoir is set by $B_g$ ; the calculation is made at constant pressure
Exchange of components with the oil	practically none	possible	at high pressure the gas can become miscible

Approximate ranges. For gas, density and viscosity rise strongly with pressure (Gas formation volume factor, compressibility and viscosity).

The calculations in this lesson are for immiscible displacement. The gas and the oil stay separate phases with a boundary between them, and they do not exchange components. This is a good approximation when the gas is injected into saturated oil at its bubble point pressure and the pressure is kept constant. Saturated oil already holds all the gas it can dissolve. In that case the gas in the reservoir behaves as an incompressible phase (its volume does not change with pressure) with a constant $B_g$ . The calculation ignores the exchange of components. Usually this exchange lowers the residual oil a little. Either the gas evaporates intermediate components (hydrocarbons of medium weight) out of the oil, or, the other way round, it gives its own components to the oil, and the oil swells. So in this respect the calculation errs on the safe side. If the pressure is above the minimum miscibility pressure, the approximation does not apply (see the section on miscibility).

The lesson's accumulation

The examples use the accumulation of the lessons General material balance equation for oil reservoirs, Performance prediction for solution gas drive reservoirs and Why water is injected: pressure maintenance and oil displacement: pore volume 16.0 million m³, connate water 0.25 (the water that has stayed in the pores since the rock was formed), bubble point pressure 18 MPa, production plateau 1200 m³/d (the steady oil rate that the field holds for a period). The gas is injected at the bubble point pressure. The gas–oil relative permeability curves (they show how easily each phase flows when both are present) are those of the lesson on solution gas drive: $k_{ro} = 0.80\,((0.55 - S_g)/0.55)^2$ , $k_{rg} = 0.30\,((S_g - 0.03)/0.52)^{1.5}$ . The properties at 18 MPa are $\mu_o = 1.00$ and $\mu_g = 0.01912$ mPa·s, $B_o = 1.35$ , $B_g = 0.005983$ m³/m³ and $R_s = 110$ m³/m³. The data are assumed for illustration.

Fractional flow of gas

We derive the equation for the fractional flow of gas in the same way as the equation for the fractional flow of water in the lesson Fractional flow function and mobility ratio. We write Darcy's law (the law of flow through rock) for the gas and for the oil, subtract one equation from the other, and write the gas velocity in terms of the total velocity. Gas takes the place of water, and the density difference of water and oil is replaced by $\rho_g - \rho_o$ , which is negative. If we neglect the capillary term (the part that comes from capillary pressure), we get:

Fractional flow of gas

f_g = \frac{q_g}{q_g + q_o} = \frac{\lambda_g}{\lambda_t}\left(1 - N_g\,\frac{k_{ro}}{k_{ro}'}\right),\qquad N_g = \frac{k\,k_{ro}'\,(\rho_g - \rho_o)\,g\sin\alpha}{\mu_o\,u_t}

where:

$f_g$ the share of gas in the total reservoir flow rate of gas and oil, —
$\lambda_g,\ \lambda_o,\ \lambda_t$ the mobilities of the gas, $k\,k_{rg}/\mu_g$ , of the oil, $k\,k_{ro}/\mu_o$ , and the total mobility, m²/(Pa·s)
$k_{rg},\ k_{ro}$ relative permeabilities of the gas and of the oil at the current gas saturation (the water does not move), —
$N_g$ the gravity number of the lesson Fractional flow function and mobility ratio, with the density difference of gas and oil, —
$\alpha$ angle of the flow axis to the horizontal: positive if the gas moves up the dip, rad
$u_t$ total Darcy velocity of gas and oil (flow rate per unit area), $q_t/A$ , m/s

Units: SI (the metric oilfield forms are the same as for the gravity number in the lesson Fractional flow function and mobility ratio)

Holds when

The flow is one-dimensional, and the saturation is the same over the cross-section (this is called diffuse, or spread-out, flow). The connate water does not move.
The phases are immiscible, the pressure is constant, and the capillary term is neglected.

The sign of $N_g$ shows whether gravity helps. Gas that moves down the dip ( $\alpha < 0$ ) has $N_g > 0$ . Its fractional flow is smaller than in a horizontal formation, and the front is more stable. Gas that moves up the dip has $N_g < 0$ and runs ahead of the oil even more. In a horizontal formation, $f_g = 1/(1 + (k_{ro}/k_{rg})(\mu_g/\mu_o))$ .

At the surface, one cubic metre of reservoir oil gives $1/B_o$ cubic metres of stock-tank oil (oil that has lost its dissolved gas) and $R_s/B_o$ cubic metres of dissolved gas. One cubic metre of reservoir gas gives $1/B_g$ cubic metres of gas at the surface. For each cubic metre of reservoir oil in the flow there are $f_g/(1 - f_g)$ cubic metres of reservoir gas. From this we get the producing gas–oil ratio:

Producing gas–oil ratio from the fractional flow of gas

R = R_s + \frac{B_o}{B_g}\cdot\frac{f_g}{1 - f_g}

where:

$R$ producing gas–oil ratio: the gas per cubic metre of oil at standard conditions (producing gas–oil ratio), m³/m³
$R_s$ solution gas–oil ratio of the oil at reservoir pressure, m³/m³
$B_o,\ B_g$ formation volume factors of the oil and of the gas, m³/m³
$f_g$ fractional flow of gas in the reservoir flow at the bottom of the well, —

Units: m³ of gas per m³ of oil at standard conditions (0.101325 MPa and 20 °C)

Holds when

The oil is saturated with gas at reservoir pressure. The free gas and the dissolved gas are converted to the same standard conditions.

In a horizontal formation, $f_g/(1 - f_g) = (k_{rg}/\mu_g)/(k_{ro}/\mu_o)$ , and the formula is the same as the equation for the instantaneous gas–oil ratio (the ratio at one moment) in the lesson Performance prediction for solution gas drive reservoirs. At 18 MPa the factor $B_o/B_g$ is about 226. When the fractional flow of gas in the reservoir flow is 0.5, each cubic metre of oil at the surface comes with 226 m³ of free gas on top of the dissolved gas.

The gas displacement front and breakthrough

The Buckley–Leverett theory (Buckley–Leverett theory and the Welge method) carries over to gas without change. Each gas saturation moves at a speed proportional to the slope of the curve $f_g(S_g)$ . A sudden jump in saturation (a shock) forms at the front. The tangent drawn from the point of the initial state gives the saturation at the front. When gas is injected at the bubble point pressure, there is no free gas in the reservoir at the start. So we draw the tangent from the origin. We measure the injection in pore volumes, counting the gas at reservoir conditions: $G_{iD} = G_{inj}B_g/V_p$ . Here $G_{inj}$ is the cumulative gas injection (the total injected so far), measured at standard conditions, as in the lesson General material balance equation for oil reservoirs.

Gas breakthrough and oil production after it (Welge method)

f_g'(S_{gf}) = \frac{f_{gf}}{S_{gf}},\qquad G_{iD,BT} = N_{pD,BT} = \frac{1}{f_g'(S_{gf})},\qquad N_{pD} = S_{ge} + (1 - f_{ge})\,G_{iD},\quad G_{iD} = \frac{1}{f_g'(S_{ge})}

where:

$S_{gf},\ f_{gf}$ the gas saturation at the front and the fractional flow of gas at that saturation, —
$G_{iD}$ the gas injected, in pore volumes, at reservoir conditions, —
$N_{pD}$ the oil produced, in pore volumes, at reservoir conditions, —
$S_{ge},\ f_{ge}$ the gas saturation at the outlet after breakthrough and the fractional flow of gas at that saturation, —

Units: fractions and pore volumes

Holds when

The assumptions of the Buckley–Leverett theory. There is no initial free gas (the tangent starts at the origin).
The pressure is constant, and the injection rate at reservoir conditions is constant.

The displacement efficiency, the share of the initial oil that has been displaced, is $E_D = N_{pD}/(1 - S_{wi})$ , where $S_{wi}$ is the connate water saturation. Before breakthrough, the formation gives up oil in a volume equal to the volume of gas injected (at reservoir conditions). After breakthrough, the producing gas–oil ratio rises, and oil production grows more and more slowly.

CalculatedFractional flow of gas for the oil and gas of the lesson's accumulation at 18 MPa: a horizontal formation (

N_g = 0

) and gas moving down the dip, at

N_g = 0.4

and

0.8

. The dashed tangent from the origin gives the front for the horizontal formation. The dots mark the saturations at the front.

Worked exampleGas breakthrough in horizontal displacement

Given (assumed data)

Gas–oil relative permeability curves: $k_{ro} = 0.80\,((0.55 - S_g)/0.55)^2$ , $k_{rg} = 0.30\,((S_g - 0.03)/0.52)^{1.5}$ ; connate water 0.25
Properties at 18 MPa: $\mu_o = 1.00$ , $\mu_g = 0.01912$ mPa·s; $B_o = 1.35$ , $B_g = 0.005983$ m³/m³, $R_s = 110$ m³/m³
Accumulation: pore volume 16.0 million m³; gas is injected at a reservoir rate equal to the withdrawal at the plateau of 1200 m³/d of oil
For comparison: waterflooding of the same rock: $S_{or} = 0.25$ , $k_{rw}' = 0.20$ , water 0.40 mPa·s

Solution

Mobility ratio: $M' = (0.30/0.01912)/(0.80/1.00) = 19.6$ . At the end points the gas is almost twenty times more mobile than the oil.
We look for the tangent from the origin. At the tangent point, the slope of the curve equals the slope of the chord, $f_g/S_g$ (the chord is the straight line from the origin to the point on the curve). At $S_g = 0.08$ : $f_g = 0.445$ , and the slope of the chord is $0.445/0.08 = 5.56$ . For Corey curves (curves built from power laws, see Buckley–Leverett theory and the Welge method), the slope of the curve is $f_g(1 - f_g)(1.5/0.05 + 2/0.47) = 8.46$ . This is more than the slope of the chord, so the tangent point lies further to the right.
At $S_g = 0.13$ : $f_g = 0.739$ , the slope of the chord is 5.68, and the slope of the curve is $f_g(1 - f_g)(1.5/0.10 + 2/0.42) = 3.81$ . This is less than the slope of the chord, so the tangent point lies further to the left. The model gives the point in between: $S_{gf} = 0.103$ , $f_{gf} = 0.612$ , slope 5.92.
Breakthrough: $G_{iD,BT} = 1/5.92 = 0.169$ of a pore volume. The same volume of oil (at reservoir conditions) has been produced, so $E_{D,BT} = 0.169/0.75 = 0.225$ .
Time: the withdrawal at the plateau is $1200\cdot1.35 = 1620$ m³/d at reservoir conditions. One pore volume is injected in $16.0\cdot10^6/1620 = 9877$ d, so breakthrough comes after $0.169\cdot9877 = 1669$ d (4.6 years). By then, $0.169\cdot16.0/0.005983 = 452$ million m³ of gas at standard conditions has been injected.
Producing gas–oil ratio at breakthrough: $R = 110 + (1.35/0.005983)\cdot0.612/(1 - 0.612) = 110 + 225.6\cdot1.577 = 466$ m³/m³.
Waterflooding of the same rock at the same pressure and rate (Welge method, model values): $M'$ = 0.625, breakthrough after 0.419 of a pore volume, $E_{D,BT}$ = 0.559, after 11.3 years.

Answer: The gas will break through to the wells after 0.169 of a pore volume has been injected, in about 4.6 years. By then 23% of the oil has been displaced. The producing gas–oil ratio will jump from 110 to 466 m³/m³. Water in the same rock would break through after 0.419 of a pore volume, having displaced 56%.

What it means. The gas saturation at the front is only about 0.1. The gas takes only the large pores and races through them, with no time to push the oil out of the rest. So by breakthrough the gas has displaced less than half as much oil as water, although the residual oil saturation to gas is lower in this rock (see the section on residual oil). The calculation is one-dimensional. It treats the whole pore volume of the accumulation as one linear element that the gas sweeps through. In a real accumulation the gas also rises to the top and goes through the best layers, so breakthrough will come earlier. The estimate is optimistic. The data in this example are assumed for illustration.

The lesson's accumulation after gas breakthrough

$S_{ge}$: 0.15
$G_{iD}$: 0.363
$E_D$ by gas: 0.295
Producing gas–oil ratio, m³/m³: 1040
$E_D$ by water at the same injection: 0.484

$S_{ge}$: 0.20
$G_{iD}$: 0.768
$E_D$ by gas: 0.369
Producing gas–oil ratio, m³/m³: 2150
$E_D$ by water at the same injection: 0.586

$S_{ge}$: 0.25
$G_{iD}$: 1.50
$E_D$ by gas: 0.438
Producing gas–oil ratio, m³/m³: 4200
$E_D$ by water at the same injection: 0.609

$S_{ge}$: 0.30
$G_{iD}$: 2.77
$E_D$ by gas: 0.501
Producing gas–oil ratio, m³/m³: 8120
$E_D$ by water at the same injection: 0.625

$S_{ge}$: 0.35
$G_{iD}$: 5.01
$E_D$ by gas: 0.559
Producing gas–oil ratio, m³/m³: 16,270
$E_D$ by water at the same injection: 0.637

$S_{ge}$: 0.40
$G_{iD}$: 9.22
$E_D$ by gas: 0.611
Producing gas–oil ratio, m³/m³: 35,830
$E_D$ by water at the same injection: 0.646

$S_{ge}$	$G_{iD}$	$E_D$ by gas	Producing gas–oil ratio, m³/m³	$E_D$ by water at the same injection
0.15	0.363	0.295	1040	0.484
0.20	0.768	0.369	2150	0.586
0.25	1.50	0.438	4200	0.609
0.30	2.77	0.501	8120	0.625
0.35	5.01	0.559	16,270	0.637
0.40	9.22	0.611	35,830	0.646

Horizontal immiscible gas displacement at 18 MPa (Welge method with the curves of this lesson). The producing gas–oil ratio is rounded to 10 m³/m³. The last column is waterflooding of the same rock at the same injection in pore volumes. The data are assumed for illustration.

After breakthrough, the producing gas–oil ratio rises by an order of magnitude (about ten times), while oil production grows slowly. To raise the displacement efficiency from 0.44 to 0.50, we must pump more than one extra pore volume of gas, and the producing gas–oil ratio almost doubles. In practice, injection is limited not by the residual oil but by the capacity of the compressors and of the gas treatment plants. Almost all the gas produced after breakthrough has to be compressed and injected again.

Gravity stabilisation: crestal injection and gas cap expansion

Only gravity can outweigh an unfavourable mobility ratio. Suppose the gas is injected into a high part of the accumulation (at the crest or into the gas cap) and moves down the dip. Then it stays above the oil. If the rate is low enough, the boundary between the gas and the oil holds, and it moves down as a whole. The lesson on segregated flow derives the condition for stability and the critical rate. In segregated flow, gravity separates the phases across the thickness of the formation. For gas the mobility ratio is large, so the factor $M'/(M' - 1)$ is close to one. The critical rate is then practically equal to the gravity rate:

Critical rate for displacement of oil by gas from above

q_{\text{cr}} = \frac{0.0853\,k\,k_{ro}'\,A\cdot0.00981\,(\rho_o - \rho_g)\,|\sin\alpha|}{\mu_o}\cdot\frac{M'}{M' - 1}

where:

$q_{\text{cr}}$ critical rate: the largest reservoir flow rate of gas (and of the oil it displaces) through the cross-section $A$ at which the boundary stays stable, m³/d
$A$ flow cross-section: thickness × width of the front when the gas moves along the formation; the area of the gas–oil contact (the boundary between the gas and the oil), seen from above, when the cap moves down vertically, m²
$k$ permeability along the direction of gas movement (the vertical permeability when the cap moves down), mD
$\rho_o - \rho_g$ difference between the densities of oil and gas at reservoir conditions, g/cm³
$\alpha$ dip angle of the gas path; when the gas–oil contact moves vertically, $|\sin\alpha| = 1$ , °

Units: metric oilfield units: $k$ in mD, $A$ in m², densities in g/cm³, $\mu_o$ in mPa·s, rate in m³/d at reservoir conditions

Holds when

The gas moves down the dip or from top to bottom. Vertical equilibrium holds (the fluids separate over the thickness faster than the front moves along the formation), and the boundary is sharp. These conditions are easier to meet for gas than for water.
The pressure is constant, the phases are immiscible, and the formation is uniform.

The constants are the same as in the lesson Segregated flow and gravity displacement. The critical rate is proportional to the flow cross-section. So the same accumulation can keep a stable gas cap, but it cannot keep a stable front of injection along the formation.

Worked exampleIs the gas front stable in the lesson's accumulation?

Given (assumed data)

Accumulation: 8 km², an elongated structure 4 × 2 km, $k = 50$ mD, thickness 10 m; withdrawal at the plateau 1620 m³/d at reservoir conditions
Crestal injection (assumed): injection wells on a crest 4 km long; the gas moves down both flanks (the sloping sides of the structure), which dip at 4°, to producing wells 1 km away; $k_v/k_h = 0.1$ (the ratio of vertical to horizontal permeability)
Gas cap over the whole area (assumed): the contact moves down vertically; $k_v = 5$ mD
Densities at 18 MPa: oil 0.708 g/cm³; gas 0.151 g/cm³ (from the real gas law, with the gas deviation factor $Z = 0.87$ , at 85 °C)

Solution

Density difference: $0.708 - 0.151 = 0.557$ g/cm³. The factor is $M'/(M' - 1) = 19.6/18.6 = 1.054$ .
Crestal injection: the cross-section is $A = 2\cdot10\cdot4000 = 80{,}000$ m². The pressure gradient from gravity is $0.00981\cdot0.557\cdot\sin4° = 0.00981\cdot0.557\cdot0.0698 = 3.814\cdot10^{-4}$ MPa/m. The gravity rate is $q_{\text{gr}} = 0.0853\cdot50\cdot0.80\cdot80{,}000\cdot3.814\cdot10^{-4}/1.00 = 104.1$ m³/d, and the critical rate is $q_{\text{cr}} = 104.1\cdot1.054 = 109.7$ m³/d. That is 15 times less than the withdrawal.
Do the phases have time to separate on the flanks? First, $N_g = 104.1/1620 = 0.0643$ . Then we use the ratio of times from the lesson Segregated flow and gravity displacement: the time the front needs to travel along the formation from the injection wells to the producing wells, divided by the time the phases need to separate across the thickness. Here it is $(1000/10)\cdot0.1\cdot0.0643/0.0698 = 9.2$ , about ten. For the gas–oil pair this estimate is closer to reality than for water: the gas rises to the top, and breakthrough along the top comes after $0.55/(19.6\cdot(1 - 0.0643)) = 0.030$ of a pore volume. The diffuse calculation would have given 0.18 (model value).
Gas cap over the whole area: $A = 8.0\cdot10^6$ m² and $|\sin\alpha| = 1$ . The pressure gradient from gravity is $0.00981\cdot0.557 = 5.4642\cdot10^{-3}$ MPa/m. The gravity rate is $q_{\text{gr}} = 0.0853\cdot5\cdot0.80\cdot8.0\cdot10^6\cdot5.4642\cdot10^{-3}/1.00 = 14{,}915$ m³/d, and the critical rate is $q_{\text{cr}} = 15{,}720$ m³/d. That is 9.7 times more than the withdrawal.
For comparison, take the thick, steep formation of the lesson Segregated flow and gravity displacement (400 mD, cross-section 25 × 600 m, dip 12°) with the same oil. The critical rate $q_{\text{cr}}$ is ≈ 490 m³/d for one such element (model value).

Answer: Along the thin formation of the flanks, the critical rate is about 109.7 m³/d. That is 15 times less than the withdrawal, so the gas will run ahead as a tongue along the top. A cap that moves down over the whole area is stable up to 15,720 m³/d.

What it means. The fluid properties are the same in both cases. The cross-section of the flow and the path of the gas decide everything. Crestal injection into a thin, gently dipping formation does not stabilise the front. The gas will break through to the wells along the top after only a few per cent of a pore volume has been injected. The one-dimensional diffuse calculation ignores the rising of the gas, and its 18% of a pore volume is far too optimistic. A gas cap over the whole area is a different matter. The vertical velocities are small, and the front as a whole is stable with a tenfold margin. What limits it is not the stability of the front but gas cones at individual wells (Water and gas coning): the gas is drawn down towards the well in a funnel shape. So the wells are perforated (opened to the formation) well away from the gas–oil contact. Thick, permeable formations with steep flanks are the best candidates for crestal gas injection. The data in this example are assumed for illustration.

CalculatedDisplacement efficiency of the lesson's accumulation against the injection (in pore volumes, at reservoir conditions): waterflooding; horizontal gas displacement with diffuse flow; gas that has run ahead as a tongue along the top in crestal injection at a rate above the critical rate (the flanks of the example, calculated with thickness-averaged relative permeabilities, which is optimistic after breakthrough); gas with a stable front below the critical rate (piston-like displacement, in which the gas pushes the oil ahead of it down to the residual oil saturation to gas; the wedge of oil that the tilted boundary leaves behind is ignored).

The rate is the main lever. Permeability, density difference and viscosities are set by nature. The engineer chooses the injection and withdrawal rates, the places of the injection wells and the perforated intervals.
Heterogeneity across the thickness. The layers of the formation differ from one another. If the permeability grows towards the base (the formation gets coarser downwards), the best paths for the gas lie low down and work against its rising. This is favourable for gas injection, but water in such a formation runs ahead along the base. A high-permeability layer near the top does the opposite and speeds up gas breakthrough (Displacement in heterogeneous reservoirs).
Gas cap expansion. While the withdrawal is less than the critical rate for the whole area of the gas–oil contact, the cap moves down steadily and pushes the oil ahead of it. At a higher withdrawal, the gas breaks through along high-permeability channels and as cones (gas cap drive).

Gas injection to maintain pressure

The pressure does not fall as long as we inject the same reservoir volume as we take out (voidage replacement). When gas maintains the pressure, the withdrawal consists of oil, produced free gas and water. The injection is gas, and its formation volume factor at reservoir pressure is $B_g$ :

Gas injection that fully replaces the withdrawal

q_{g,\text{inj}} = q_o\,\frac{B_o}{B_g} + q_o\,(R - R_s) + q_w\,\frac{B_w}{B_g}

where:

$q_{g,\text{inj}}$ rate of the injected gas at standard conditions, m³/d
$q_o,\ q_w$ rates of oil and water at standard conditions, m³/d
$R,\ R_s$ producing gas–oil ratio and solution gas–oil ratio of the oil at reservoir pressure, m³/m³
$B_o,\ B_g,\ B_w$ formation volume factors of oil, gas and water, m³/m³

Units: m³/d at standard conditions

Holds when

The pressure is constant and not above the bubble point pressure, so the injected gas stays free (does not dissolve in the oil).
The injected gas has the same composition as the reservoir gas (the same $B_g$ ).

The first term is the gas that will take the place of the produced oil. The second is the produced free gas, which must be put back. The third is the gas that will take the place of the produced water. The produced gas, $q_oR$ , is usually not enough. The shortfall is $q_o(B_o/B_g - R_s) + q_wB_w/B_g$ (for the lesson's oil at 18 MPa, $B_o/B_g \approx 226$ against $R_s = 110$ ). This make-up gas has to be taken from other fields or from the gas cap, or else we must accept a fall in pressure.

Worked exampleHow much gas is needed to hold the pressure

Given (assumed data)

The lesson's accumulation at 18 MPa: $B_o = 1.35$ , $B_g = 0.005983$ m³/m³, $R_s = 110$ m³/m³; oil production 1200 m³/d, no water
Two points in time: before gas breakthrough ( $R = R_s$ ) and after it, when $R = 1000$ m³/m³

Solution

The ratio is $B_o/B_g = 1.35/0.005983 = 225.6$ m³ of gas per m³ of oil. That many standard cubic metres of gas will take the place of one cubic metre of produced oil.
Before breakthrough: the injection is $1200\cdot225.6 = 270.7$ thousand m³/d. The gas produced is $1200\cdot110 = 132.0$ thousand m³/d. Injecting it back replaces $132.0/270.7 = 0.488$ of the withdrawal. The make-up gas is $270.7 - 132.0 = 138.7$ thousand m³/d.
After breakthrough, at $R = 1000$ m³/m³: the free gas produced is $1200\cdot(1000 - 110) = 1068.0$ thousand m³/d. The injection is $270.7 + 1068.0 = 1338.7$ thousand m³/d, the gas produced is $1200.0$ thousand m³/d, and the make-up gas is still 138.7 thousand m³/d.
The volume that the compressors must handle has grown by a factor of $1338.7/270.7 = 4.95$ .

Answer: To hold the bubble point pressure at an oil rate of 1200 m³/d, we must inject about 270.7 thousand m³/d of gas. Injecting back the produced gas gives only 48.8%. The other 138.7 thousand m³/d is make-up gas. After breakthrough the injection rises to 1338.7 thousand m³/d, and the make-up gas does not change.

What it means. At 18 MPa the gas is compressed about 170 times. Even so, the gas needed to fill the room left by one cubic metre of oil is more than the gas that the oil itself brings along in solution. So the pressure cannot be held without gas from outside. After breakthrough, almost all the produced gas goes back into the reservoir, and the project turns into pumping the same gas round and round. The cost of compression grows fivefold for the same oil production. If the pressure is not held, the accumulation goes into solution gas drive, which gives only about 17% of the oil in place (the oil that was in the reservoir at the start) by the time the pressure has fallen to 5 MPa (Performance prediction for solution gas drive reservoirs). We choose between water, gas and a combination of the two by looking at the gas reserves available, the compressor capacity and a comparison of the options over time. The data in this example are assumed for illustration.

Residual oil in gas displacement

The residual oil saturation in gas displacement, $S_{org}$ (residual oil saturation to gas), and the residual oil saturation after waterflooding, $S_{or}$ , are different quantities, even for the same rock (gas and oil). In a hydrophilic (water-wet) rock with connate water, the water coats the grains, the gas takes the largest pores, and the oil is left between them. Many oils spread over the water in thin films when gas is present. The films stay connected, so the oil keeps draining to the wells even at a low saturation. In waterflooding the opposite happens: the oil breaks up into trapped drops. So $S_{org}$ is often lower than $S_{or}$ , especially in slow gravity drainage (oil flowing down under gravity).

One rock, two agents: waterflooding and gas displacement at 18 MPa

Quantity: Residual oil saturation
Water: 0.25
Gas: 0.20

Quantity: Limiting $E_D$ from the end points
Water: 0.667
Gas: 0.733

Quantity: $M'$ at 18 MPa
Water: 0.63
Gas: 19.6

Quantity: At breakthrough: pore volumes; $E_D$
Water: 0.42; 0.56
Gas: 0.17; 0.23

Quantity: $E_D$ after 1 pore volume
Water: 0.60
Gas: 0.40

Quantity: $E_D$ after 3 pore volumes
Water: 0.63
Gas: 0.51

Quantity	Water	Gas
Residual oil saturation	0.25	0.20
Limiting $E_D$ from the end points	0.667	0.733
$M'$ at 18 MPa	0.63	19.6
At breakthrough: pore volumes; $E_D$	0.42; 0.56	0.17; 0.23
$E_D$ after 1 pore volume	0.60	0.40
$E_D$ after 3 pore volumes	0.63	0.51

Horizontal displacement, Welge method. The water end points are from the lesson Why water is injected: pressure maintenance and oil displacement, and the gas curves from the lesson Performance prediction for solution gas drive reservoirs. The displacement efficiencies are fractions of the initial oil. The data are assumed for illustration.

The limiting displacement efficiency of gas is higher than that of water. But with an unfavourable mobility ratio the real value is very far from this limit: after three pore volumes, gas has displaced noticeably less oil than water. The oil stays behind not because capillary forces hold it, but because the gas goes past it. Only gravity-stable displacement and slow gravity drainage come close to the limit (gravity drainage).

When gas mixes with oil

At high pressure the gas and the oil exchange components so strongly that the boundary between them disappears. The interfacial tension (the tension at the boundary between two fluids) falls to zero. Capillary forces no longer hold the oil, and the residual oil saturation in the flushed zone (the zone the gas has swept) becomes very small. This kind of displacement is called miscible (the fluids mix completely). The lowest pressure at which the injected gas, at a given temperature, becomes miscible with a given oil is the minimum miscibility pressure (minimum miscibility pressure, MMP).

Miscibility can occur at the first contact, when the gas and the oil mix in any proportion. It can also develop after many contacts. A light (lean) gas evaporates intermediate components out of the oil, while a rich gas gives its own intermediate components to the oil. Step by step, the compositions at the front become the same.
The more intermediate hydrocarbons (ethane to butane), carbon dioxide and hydrogen sulfide the gas contains, the lower the MMP. Dry gas (almost pure methane) needs the highest pressures.
We measure the MMP in a slim tube, a long, thin tube packed with sand. We can also calculate it with an equation of state (a formula that links the pressure, temperature and volume of a fluid mixture), after we adjust the formula to fit laboratory experiments on fluid samples (Cubic equations of state). We simulate this kind of displacement with compositional models, which follow the composition of the fluids (compositional model).
Miscibility does not remove the unfavourable mobility ratio or the rising of the gas. The solvent (the injected gas) stays light and of low viscosity, so the sweep (the share of the reservoir that the gas reaches) is still decided by gravity, heterogeneity and rate.

Gas-based methods of enhanced oil recovery (EOR), such as miscible displacement and the alternate injection of water and gas, are not covered in this lesson. In gas-condensate accumulations (gas that partly turns into liquid condensate when the pressure falls), the reinjection of dry gas maintains the pressure, so that the condensate does not drop out in the reservoir. This is miscible displacement of the reservoir gas by dry gas (Types of reservoir fluids).

Gas injection in the fields of Kazakhstan

Gas reinjection is used in the largest fields of the Pre-Caspian Basin, where the gas contains a lot of hydrogen sulfide, a toxic and corrosive gas (Composition of reservoir fluids). There it does two jobs. It maintains the reservoir pressure, and it gives a way to use raw gas that contains hydrogen sulfide (raw gas is gas straight from the wells, before treatment). Otherwise this gas would have to be cleaned at the surface or burned.

Tengiz: the operator completed the second-generation plant and raw gas injection project in 2008. In 2022–2024, Tengiz injected 3.5–3.6 billion m³ of raw gas a year back into the reservoir at very high pressure. That is about 22–23% of the associated gas produced (the gas that comes up with the oil); in 2024 it was 3.52 of 15.06 billion m³. In 2025, after the third plant started up, the figure was 11.05 of 22.57 billion m³, almost half.
Karachaganak: one of the largest oil, gas and condensate fields in the world (Types of reservoir fluids). The operator returns most of the raw gas it produces (about 4% hydrogen sulfide) to the reservoir at a pressure of up to 550 bar (55 MPa): about 59% of all produced gas in 2024 and about 66% in 2025. The operator links the increase in reinjection to maintaining the reservoir pressure, to the rising gas–oil ratio and to extending the plateau of liquid hydrocarbon production.

Terms and symbols

Fractional flow of gas $f_g$: The share of gas in the total reservoir flow rate of gas and oil through a cross-section of the formation.
Mobility ratio of gas to oil $M'$: $(k_{rg}'/\mu_g)/(k_{ro}'/\mu_o)$ . For gas it is in the tens.
$G_{iD}$: The gas injected, in pore volumes, at reservoir conditions: $G_{inj}B_g/V_p$ .
Crestal injection: Injection of gas into the high part of the accumulation or into the gas cap, so that the gas displaces the oil from top to bottom and gravity stabilises the front.
Make-up gas: The gas needed in addition to the produced gas to replace the withdrawal fully when gas maintains the pressure.
$S_{org}$: The residual oil saturation in gas displacement. It is often lower than the residual saturation after waterflooding.
Minimum miscibility pressure (MMP): The lowest pressure at which the injected gas becomes miscible with the given oil.

Common mistakes and misconceptions

Wrong

“Gas is lighter than oil, so when it is injected at the crest it displaces the oil from above at any rate.”

In fact

Only below the critical rate, which for gas is almost equal to the gravity rate. In the lesson's example, along the flanks it is about 110 m³/d against a withdrawal of 1620 m³/d, so the gas will run ahead as a tongue along the top.

Wrong

“Injecting back all the produced gas maintains the pressure.”

In fact

At the bubble point pressure, a cubic metre of produced oil frees room in the reservoir for $B_o/B_g$ ≈ 226 standard cubic metres of gas, but it brings only 110. The produced gas replaces about half of the withdrawal. The rest is make-up gas, or a fall in pressure.

Wrong

“The residual oil to gas is lower than to water, so gas will recover more oil.”

In fact

The limit on displacement is higher, but at a mobility ratio of about 20 the gas goes past the oil. In the lesson's example, after three pore volumes the gas has displaced about 0.51 of the oil, against 0.63 for water.

Wrong

“Gas breakthrough can be estimated from the average speed of the gas in the pores: one pore volume, and the gas is at the well.”

In fact

A front with a low gas saturation moves $f_g'(S_{gf})$ times faster than the average speed of the gas in the pores. In the lesson's example it is almost six times faster: breakthrough comes after 0.17 of a pore volume.

Wrong

“Injected gas always stays a separate phase.”

In fact

Above the minimum miscibility pressure, gas and oil form one phase, and the immiscible calculations of this lesson do not apply. We simulate this kind of displacement with a compositional model.

In depth: advanced detail

Injection into an accumulation that already has free gas

If the pressure has already fallen below the bubble point pressure, there is free gas in the pores, with saturation $S_{gi}$ . While $S_{gi}$ is below the critical gas saturation $S_{gc}$ (the saturation at which gas starts to move), the free gas does not move. Then we draw the tangent from the point $(S_{gi}, 0)$ , just as in waterflooding with immobile initial water (initial water). If $S_{gi}$ is above the critical saturation, the gas moves from the very start. Then we draw the tangent from the point $(S_{gi}, f_g(S_{gi}))$ . Before the front arrives, the wells already give a higher producing gas–oil ratio, and the gain in production from injection is smaller. In addition, when the pressure recovers, part of the injected gas dissolves in the oil. So the immiscible calculation at constant pressure applies to such injection only roughly.

Gas displacement under vertical equilibrium

For gas, vertical equilibrium is easier to reach than for water. The density difference is larger, and the gas offers hardly any resistance to the counter-current flow (oil moving down while gas moves up), because it is many times more mobile than the oil. So when gas is injected along a formation with good vertical permeability, we use relative permeabilities averaged over the thickness. These are straight lines between the end points (calculation of segregated displacement). The range of mobile saturation here is $1 - S_{wi} - S_{org}$ , and the formulas for breakthrough and full sweep are the same. Above the critical rate, breakthrough comes after $(1 - S_{wi} - S_{org})/(M'(1 - N_g))$ of a pore volume has been injected. At $M' \approx 20$ and a small $N_g$ this is about 3% of a pore volume. The mobile oil is fully swept out only after $M'(1 - S_{wi} - S_{org})/(1 + M'N_g)$ pore volumes. Below the critical rate the boundary is stable, and almost all the mobile oil is displaced before gas breakthrough, except for the wedge of oil that the tilted boundary leaves behind (tilt of the boundary).

Connections to other topics

Segregated flow and gravity displacement — the condition for stability, the critical rate and gravity drainage.
Fractional flow function and mobility ratio — the equation for the fractional flow of a phase, with gravity included.
Buckley–Leverett theory and the Welge method — the displacement front, the Welge tangent and production after breakthrough.
Performance prediction for solution gas drive reservoirs — the gas–oil relative permeability curves, the producing gas–oil ratio, and the drive that gas injection helps to avoid.
Why water is injected: pressure maintenance and oil displacement — voidage replacement and the comparison with water injection.
Reservoir drive mechanisms — gas cap drive and gravity drainage.
Water and gas coning — gas cones at wells when the gas cap expands.
Types of reservoir fluids — gas-condensate accumulations and the reinjection of dry gas.

Check yourself

1
Gas will be injected into an accumulation with steep flanks and good vertical permeability. Which option gives the most stable front?
1. A.The rate does not matter: the gas will rise anyway
2. B.Even injection over the whole area at the highest rate
3. C.Injection at the crest at a rate below the critical rate
4. D.Injection into the low part of the accumulation at a high rate
Show the answer
Correct answer: C. Injection at the crest at a rate below the critical rate
Gas injected at the crest moves down the dip, and gravity holds it above the oil ( $N_g > 0$ ). The front is stable as long as the rate is below the critical rate, which for gas is almost equal to the gravity rate. Gas from the low part rises and runs ahead of the oil along the top at any rate.
2
At a certain gas saturation $k_{rg} = 0.05$ and $k_{ro} = 0.30$ . The gas viscosity is 0.025 mPa·s and the oil viscosity is 1.5 mPa·s; $B_o/B_g = 200$ , $R_s = 90$ m³/m³. Find the fractional flow of gas in the reservoir flow of a horizontal formation and the producing gas–oil ratio.
Show the answer
The mobilities per unit of permeability are $0.05/0.025 = 2.0$ for the gas and $0.30/1.5 = 0.20$ for the oil, in 1/(mPa·s). So $f_g = 2.0/2.2 \approx 0.909$ . The relative permeability of the gas is six times smaller than that of the oil, but the gas is 60 times less viscous. The producing gas–oil ratio is $R = 90 + 200\cdot0.909/0.091 \approx 90 + 2000 = 2090$ m³/m³, about 23 times the solution gas–oil ratio.
3
Oil production is 800 m³/d, $B_o = 1.25$ , $B_g = 0.0070$ m³/m³, and the solution gas–oil ratio is 80 m³/m³. There is no free gas or water in the production. How much gas must be injected to replace the withdrawal fully, and what share of it does the reinjection of the produced gas cover?
Show the answer
The injection is $800\cdot1.25/0.0070 \approx 142.9$ thousand m³/d. The gas produced is $800\cdot80 = 64.0$ thousand m³/d, about 45% of what is needed. The make-up gas is about 78.9 thousand m³/d.
4
A gas cap is expanding over the whole area of the accumulation. The withdrawal is 2000 m³/d at reservoir conditions, and the critical rate of vertical displacement over the whole area is 20,000 m³/d. Where is gas most likely to appear in the production, and why?
1. A.At individual wells, as a cone, if the intervals are open close to the gas–oil contact
2. B.Over the whole accumulation at once: the withdrawal is above the critical rate
3. C.Nowhere: the front is stable, and no gas will come
Show the answer
Correct answer: A. At individual wells, as a cone, if the intervals are open close to the gas–oil contact
The front as a whole moves down at a tenth of the limiting rate for stability. But near a well the flow converges, the velocities and pressure drops are large, and the gas is drawn towards the interval as a cone. We solve this problem with the critical rate of the well, the highest rate at which no gas comes in (Water and gas coning).
5
Why is the residual oil in the flushed zone small above the minimum miscibility pressure, while the sweep of the accumulation can still stay poor?
Show the answer
Miscibility removes the boundary between the phases and the interfacial tension, so capillary forces no longer hold the drops of oil. The flushed zone is washed almost completely clean. But the solvent is still of low viscosity and light. It runs ahead of the oil along the best layers and rises to the top, and it does not reach the unswept parts of the accumulation. The sweep is decided by the same things as in immiscible displacement: the mobility ratio, gravity and heterogeneity.
6
Gas injection has started in an accumulation where the pressure is already below the bubble point pressure and the gas saturation is above the critical one. From which point is the Welge tangent drawn?
1. A.From the origin
2. B.From the point $(S_{gc}, 0)$
3. C.No tangent is drawn
4. D.From the point $(S_{gi}, f_g(S_{gi}))$
Show the answer
Correct answer: D. From the point $(S_{gi}, f_g(S_{gi}))$
The tangent is drawn from the point of the initial state. The initial gas is mobile, so this point lies on the curve: $(S_{gi}, f_g(S_{gi}))$ . The tangent is drawn from the origin only when there is no free gas, and from $(S_{gi}, 0)$ when there is free gas but it does not move.

Updated: 28 October 2026