Segregated flow and gravity displacement

Quick refresher

In diffuse flow the saturation is the same over the thickness. In segregated flow the phases are separated: water at the base, gas at the top, oil between them, and the interface is sharp. Which picture is closer to a given formation depends on the ratio of two times: the time the front needs to travel along the formation, divided by the time the phases need to separate across the thickness.
The interface is stable if $M'N_g > M' - 1$ . Here $M'$ is the mobility ratio, and $N_g$ is the gravity number of the lesson Fractional flow function and mobility ratio (how strong gravity is compared with the viscous forces). When gravity holds the displacing phase back (water going up the dip, gas going down the dip), this holds at any rate if $M' \le 1$ , and for $M' > 1$ only below the critical rate $q_{\text{cr}} = q_{\text{gr}}M'/(M' - 1)$ . Here $q_{\text{gr}}$ is the gravity rate, the rate at which the gravity number equals one.
Below the critical rate the interface is tilted to the bedding (the plane of the layers). At a very low rate it is horizontal. As the rate grows, for $M' > 1$ it lies closer and closer to the bedding and stretches along the formation, and at $q_{\text{cr}}$ it becomes parallel to it. Above $q_{\text{cr}}$ water runs ahead as a tongue along the base, and gas along the top.
Under vertical equilibrium the thickness-averaged relative permeabilities are straight lines between the end points. With them the fractional flow curve bends upwards (its slope grows, and the front is piston-like) when the interface is stable, and bends downwards (the slope falls, and the tongue breaks through early) when it is not. Two routes give the same condition.
Just above the critical rate the tongue is still short, and breakthrough is even later than in diffuse flow. In the example of this lesson ( $M' = 1.76$ ) segregated displacement breaks through earlier at a rate twice the critical rate or more. For a larger $M'$ it does so already a little above $q_{\text{cr}}$ , and for $M'$ close to one it may never get ahead of the diffuse case. But after breakthrough it washes out the mobile oil with a much smaller volume of water. Below the critical rate it is close to piston-like, but slow.
Gravity drainage cannot give more than $q_{\text{gr}}$ , the rate at which $N_g = 1$ . For gas $M'$ is in the tens, so the critical rate of a gas front is only a few per cent above this limit.

Key relationships

N_g = \dfrac{q_{\text{gr}}}{q_t},\qquad q_{\text{gr}} = \dfrac{k\,k_{ro}'\,A\,\Delta\rho\,g\sin\alpha}{\mu_o}

the gravity number of the lesson Fractional flow function and mobility ratio through the gravity rate

M'N_g > M' - 1

stability condition of a sharp interface

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

critical rate for $M' > 1$

\tan\beta = \tan\alpha\left(1 - \dfrac{q_t}{q_{\text{cr}}}\right)

steady tilt of the interface to the bedding, $M' > 1$

\bar k_{rw} = k_{rw}'\,h_D,\qquad \bar k_{ro} = k_{ro}'\,(1 - h_D)

thickness-averaged relative permeabilities; $h_D$ is the fraction of the thickness occupied by water

\dfrac{t_x}{t_v} = \dfrac{L}{h}\cdot\dfrac{k_v\,k_{ro}'\,|\Delta\rho|\,g}{\mu_o\,u_t}

vertical equilibrium is likely if the ratio of times is of the order of ten or more

Common pitfalls

Applying the one-dimensional fractional flow curve to a thick formation with good vertical permeability. Water runs along the base, and breakthrough comes earlier than calculated.
Assuming that the front is compact below the critical rate. Near $q_{\text{cr}}$ the interface is stretched along the formation over hundreds of metres or more.
Mixing up the critical rate of stable displacement with the critical water-free rate of a well (the highest rate at which the well still produces no water). The first refers to a front between rows of wells, and the second to a cone (water drawn up under the well in the shape of a cone) near a single well (Water and gas coning).
Putting the velocity (the flow rate per unit area) into the critical rate formula instead of the total reservoir flow rate through the whole cross-section of the formation, or using the relative permeability at the current saturation instead of the end points.
Injecting gas into the lower part of the formation or water into the upper part and counting on a stable front. In these directions gravity drives the displacing phase forward.

After this lesson you can

tell diffuse flow from segregated flow, and use the ratio of times to estimate whether vertical equilibrium holds
derive the stability condition of the sharp interface between the displacing phase and the oil, and calculate the critical rate for water and gas moving up the dip or down the dip
calculate the tilt of the interface below the critical rate and the length of the zone where the displacing phase and the oil lie one above the other
build thickness-averaged relative permeabilities (average values over the whole thickness) and use them to find breakthrough and production in segregated displacement
estimate the limiting rate of gravity drainage and explain why for gas it almost coincides with the critical rate

Diffuse and segregated flow

The Buckley–Leverett theory (Buckley–Leverett theory and the Welge method) treats diffuse (spread-out) flow: the saturation is the same over the thickness of the formation, as in a core sample. In a thick formation with good vertical permeability this is not always so. Water is heavier than oil and sinks to the base, oil floats above it, and gas rises to the top. If the phases have time to separate across the formation faster than the front covers a noticeable part of the way between wells, the displacement goes in two layers with an interface between them. This is segregated flow.

The calculation model of segregated flow rests on three assumptions:

Vertical equilibrium. Across the formation the pressure in each phase changes hydrostatically, following the weight of its column. Vertical crossflow has time to equalise the pressure, so the flow along the formation decides the whole picture.
A sharp interface. The capillary transition zone is thin compared with the thickness of the formation, and the displacing phase is separated from the oil by a surface, not by a smeared band.
End-point saturations on both sides. Behind the interface the displacing phase is already at its highest saturation (water at residual oil $S_{or}$ , gas at the residual oil saturation to gas), and ahead of it there is oil with irreducible water. Each zone passes its own phase with the end-point relative permeability: $k_{rw}'$ (or $k_{rg}'$ ) behind the interface and $k_{ro}'$ ahead of it (Relative permeability).

SchematicWater displaces oil up the dip of a formation inclined at an angle

\alpha

(a diagram, with the thickness exaggerated). Top: stable displacement. The interface is tilted to the bedding at an angle

\beta

and moves without changing its shape. Bottom: a rate above the critical one. Water runs ahead as a tongue along the base. Gas injected into the upper part of the formation behaves as a mirror image: a stable interface or a tongue along the top.

Gravity-stable displacement: the interface is tilted but moves as a whole. Almost all the mobile oil leaves ahead of it, and the water (gas) reaches the wells later than the main mass of the oil.
Tongue of the displacing phase: a thin layer of water along the base or of gas along the top (a gravity tongue, or gravity override) that runs ahead of the main mass. Breakthrough comes early, and the oil above a water tongue (below a gas tongue) has to be produced together with water or gas (gravity tongue).
Diffuse (spread-out) flow: the phases have no time to separate over the thickness. The displacement is described by the one-dimensional fractional flow curve with the gravity number along the formation (dip of the formation).

When vertical equilibrium holds

Vertical equilibrium requires that the phases have time to separate over the thickness while the front covers the way between wells. Let us compare two times. The first is how long the density difference takes to move the phases across a formation of thickness $h$ . The speed of such separation is not more than $u_v = k_v\,k_{ro}'\,|\Delta\rho|\,g/\mu_o$ . The second is how long the front takes to cover the distance $L$ along the formation at the velocity $u_t = q_t/A$ .

Ratio of the time to travel along the formation to the time to separate across the thickness

\frac{t_x}{t_v} = \frac{L}{h}\cdot\frac{k_v\,k_{ro}'\,|\Delta\rho|\,g}{\mu_o\,u_t} = \frac{L}{h}\cdot\frac{k_v}{k}\cdot\frac{N_g}{\sin\alpha}

where:

$t_v$ time the phases need to separate across the formation, $\phi\,\Delta S\,h/u_v$ ; $u_v$ is the speed of separation, an upper estimate, s
$t_x$ time in which the front covers the distance $L$ along the formation, $\phi\,\Delta S\,L/u_t$ , s
$L,\ h$ distance along the formation between the rows of injection wells and production wells, and thickness of the formation, m
$k_v,\ k$ vertical permeability and permeability along the formation, m²
$\Delta S$ range of mobile saturation, $1 - S_{wi} - S_{or}$ , —
$\Delta\rho$ density difference between the displacing phase and the oil, kg/m³
$u_t$ total velocity along the formation, $q_t/A$ , m/s
$N_g,\ \alpha$ gravity number and dip angle of the formation, as in the lesson Fractional flow function and mobility ratio, —, rad

Units: SI (any consistent system); the ratio is dimensionless

Holds when

This is an order-of-magnitude estimate, not an exact criterion.
The speed of separation is taken with the end-point relative permeability of oil, which is an upper estimate. Where the phases move towards each other, their combined mobility is smaller: for the pair water–oil with the curves of the examples of this lesson, by about an order of magnitude or more. This is why a ratio of the order of ten is required for vertical equilibrium. Gas is much more mobile than oil, and for the pair gas–oil the estimate is closer to reality.
The formation is uniform vertically. Shale layers cut $k_v$ sharply and make the separation slower.

If the ratio is of the order of ten or more, the phases have time to separate, and the displacement is calculated as segregated. If it is of the order of one or less, the flow is closer to diffuse, and the one-dimensional calculation of the lesson Buckley–Leverett theory and the Welge method is suitable. Between these limits the flow is intermediate, and we calculate it with a two-dimensional model of a vertical cross-section (Types of reservoir simulation model). The ratio grows with vertical permeability, density difference and the distance between wells, and falls with the injection rate, the thickness of the formation and the oil viscosity.

In other unit systems (1)

Metric oilfield units

u_v\,[\text{m/d}] = \frac{0.0853\,k_v\,k_{ro}'\cdot0.00981\,|\Delta\rho|}{\mu_o},\qquad u_t\,[\text{m/d}] = \frac{q_t}{A}

$k_v$ in mD, $\Delta\rho$ in g/cm³, $\mu_o$ in mPa·s, $q_t$ in m³/d at reservoir conditions, $A$ in m²; times in days.

$0.0853$ is the constant of Darcy's law in metric units (Rock permeability), and $0.00981$ converts the density difference in g/cm³ to a pressure gradient in MPa/m, as in the lesson Fractional flow function and mobility ratio.

Worked exampleDo the phases have time to separate

Given (assumed data)

Formation (lesson Fractional flow function and mobility ratio): 400 mD, thickness 25 m, width 600 m, dip 12°
Added as an assumption: porosity 0.20; 1000 m along the formation from the row of injection wells to the row of production wells
Vertical permeability: clean massive sandstone $k_v/k = 0.5$ ; layered $k_v/k = 0.05$
Fluids and relative permeability: oil 3.0 mPa·s, densities 1.04 and 0.80 g/cm³; $S_{wi} = 0.22$ , $S_{or} = 0.25$ , $k_{ro}' = 0.85$
Injection rate at reservoir conditions: 600 and 150 m³/d

Solution

Range of mobile saturation: $\Delta S = 1 - 0.22 - 0.25 = 0.53$ . Cross-section: $A = 25\cdot600 = 15{,}000$ m². Velocity: $u_t$ = 0.04 m/d at 600 m³/d and 0.01 m/d at 150 m³/d.
Clean sandstone, $k_v = 200$ mD: $u_v = 0.0853\cdot200\cdot0.85\cdot0.00981\cdot0.24/3.0 = 0.01138$ m/d, and $t_v = 0.20\cdot0.53\cdot25/0.01138 = 233$ d.
Transport over 1000 m: $t_x = 0.20\cdot0.53\cdot1000/0.04 = 2650$ d at 600 m³/d, and $10{,}600$ d at 150 m³/d.
Ratios: $2650/233 = 11.4$ and $10{,}600/233 = 45.5$ . A check through the gravity number of the lesson Fractional flow function and mobility ratio ( $N_g = 0.118$ at 600 m³/d): $(1000/25)\cdot0.5\cdot0.118/0.2079 = 11.4$ .
Layered sandstone, $k_v = 20$ mD: $u_v$ is ten times smaller, 0.001138 m/d; $t_v = 2329$ d; the ratios are 1.14 and 4.55.

Answer: In clean sandstone

t_x/t_v

≈ 11.4 at 600 m³/d and 45.5 at 150 m³/d. In layered sandstone it is 1.14 and 4.55.

What it means. At 150 m³/d in clean sandstone the ratio is more than four times above the threshold: the phases separate with a margin, and the displacement is segregated. At 600 m³/d the ratio is about ten, and allowing for the fact that the countercurrent flow of water and oil is slower than the estimate, it is borderline. The real displacement will lie between diffuse and segregated, and the example below calculates both limits. In layered sandstone at 600 m³/d the phases do not have time to separate, and the one-dimensional calculation of the lesson Buckley–Leverett theory and the Welge method with the gravity number of the lesson Fractional flow function and mobility ratio describes the displacement better. A rate four times lower gives a ratio four times larger. The data in this example are assumed for illustration.

Stability of the interface and the critical rate

Let a formation of thickness $h$ be inclined at an angle $\alpha$ . The axis $x$ runs along the formation in the direction of displacement, and $z$ is the height of the interface above the base, measured along the normal to the bedding (at right angles to the layers). The subscript $d$ denotes the displacing phase: water below the interface or gas above it. The interface moves without changing its shape only if both phases move along the formation with the same velocity $u_t$ (the reason is in the advanced section). Then Darcy's law gives the pressure gradient in each phase, and each flows with its own end-point relative permeability:

Pressure gradients in the oil and in the displacing phase at the interface

-\frac{\partial p_o}{\partial x} = \frac{\mu_o\,u_t}{k\,k_{ro}'} + \rho_o\,g\sin\alpha,\qquad -\frac{\partial p_d}{\partial x} = \frac{\mu_d\,u_t}{k\,k_{rd}'} + \rho_d\,g\sin\alpha

where:

$p_o,\ p_d$ pressure in the oil and in the displacing phase, Pa
$k_{ro}',\ k_{rd}'$ end-point relative permeabilities of the oil (at irreducible water) and of the displacing phase (at residual oil), —
$\rho_o,\ \rho_d$ densities of the oil and of the displacing phase at reservoir conditions, kg/m³
$\alpha$ angle of the axis $x$ to the horizontal: positive if the displacing phase moves up the dip, rad

Units: SI

Holds when

Both phases at the interface move along the formation with the same velocity $u_t$ , so the shape of the interface does not change.
Incompressible phases, a uniform formation of constant thickness.

Across the formation the pressure in each phase is hydrostatic: on rising by $\dd z$ along the normal to the bedding it falls by $\rho\,g\cos\alpha\,\dd z$ . The pressure difference between the oil and the displacing phase at a sharp interface is the same everywhere. So if we move along the interface by $\dd x$ , the change of this difference due to the flow along the formation must be exactly compensated by its change due to the rise or fall of the interface by $\dd z$ :

Pressure difference between the phases along a sharp interface

\frac{\partial (p_o - p_d)}{\partial x} = -(\rho_d - \rho_o)\,g\cos\alpha\,\frac{\dd z}{\dd x}

where:

$z$ height of the interface above the base of the formation along the normal to the bedding, m
$\dd z/\dd x$ slope of the interface relative to the bedding, —

Units: SI

Holds when

Vertical equilibrium: the pressure across the formation is hydrostatic in each phase.
The capillary pressure at the sharp interface is constant.

Substitute the gradients from Darcy's law and divide both sides by $\mu_o u_t/(k\,k_{ro}')$ . The ratio of the phase mobilities at the end points is the mobility ratio $M'$ , and the ratio of gravity along the formation to the viscous resistance of the oil is the gravity number $N_g$ of the lesson Fractional flow function and mobility ratio. We get the link between the tilt of the interface and these two numbers:

Slope of a sharp interface in segregated displacement (Dietz analysis)

1 - \frac{1}{M'} = N_g\left(1 + \frac{1}{\tan\alpha}\cdot\frac{\dd z}{\dd x}\right),\qquad M' = \frac{k_{rd}'/\mu_d}{k_{ro}'/\mu_o},\qquad N_g = \frac{k\,k_{ro}'\,(\rho_d - \rho_o)\,g\sin\alpha}{\mu_o\,u_t}

where:

$M'$ mobility ratio of the displacing phase and the oil at the end points, —
$N_g$ gravity number of the lesson Fractional flow function and mobility ratio, with the density difference between the displacing phase and the oil. It is positive when gravity holds the displacing phase back: water moves up the dip or gas moves down the dip, —
$z,\ x,\ \alpha$ height of the interface above the base, coordinate along the formation and dip angle, as above, m, m, rad

Units: dimensionless

Holds when

Vertical equilibrium and a sharp interface, with end-point saturations on both sides of it.
The interface moves without changing shape (a steady tilt).
A uniform formation of constant thickness, incompressible phases, a constant flow rate.

The left side is the viscous “pull” of the displacing phase forward: it is larger the more mobile this phase is compared with the oil. The right side is gravity along the formation and the contribution of the tilt of the interface. At $M' = 1$ the left side is zero and $\dd z/\dd x = -\tan\alpha$ : if gravity holds the displacing phase back, the interface is horizontal at any rate, as at rest.

The interface can be stable only if the layer of the displacing phase gets thinner towards the oil: for water below the interface $\dd z/\dd x < 0$ , and for gas above it $\dd z/\dd x > 0$ . Putting this into the relation above and taking account of the sign of $N_g$ , we get the same condition in both cases:

Condition of gravity stability of the front

M'N_g > M' - 1

where:

$M',\ N_g$ mobility ratio and gravity number with its sign, as in the relation above, —

Units: dimensionless

Holds when

The same as for the slope of the interface.

If $M' \le 1$ and gravity holds the displacing phase back ( $N_g > 0$ ), the condition holds at any rate: the viscous forces themselves do not let the displacing phase overtake the oil. If $M' > 1$ , enough gravity is needed: $N_g > 1 - 1/M'$ . For $N_g < 0$ (water from top to bottom, gas from bottom to top), stability is possible only for $M' < 1$ and a high enough rate, and for $M' \ge 1$ never.

When the interface of the displacing phase is stable

Displacement: Water from bottom to top, up the dip (injection into the lower part, edge water)
Sign of $N_g$: $> 0$
The interface is stable: always for $M' \le 1$ ; for $M' > 1$ below $q_{\text{cr}}$

Displacement: Gas from top to bottom, down the dip (injection into the crest, the top of the structure; gas cap expansion)
Sign of $N_g$: $> 0$
The interface is stable: below $q_{\text{cr}}$ ; for gas $M'$ is large, and $q_{\text{cr}}$ is close to $q_{\text{gr}}$

Displacement: Water from top to bottom, down the dip
Sign of $N_g$: $< 0$
The interface is stable: only for $M' < 1$ and a high enough rate; otherwise the water runs down as a tongue along the base

Displacement: Gas from bottom to top, up the dip
Sign of $N_g$: $< 0$
The interface is stable: practically never: gas floats up and overtakes the oil along the top

Displacement: A horizontal formation
Sign of $N_g$: $0$
The interface is stable: only for $M' < 1$ ; for $M' > 1$ there is a tongue at any rate

Displacement	Sign of $N_g$	The interface is stable
Water from bottom to top, up the dip (injection into the lower part, edge water)	$> 0$	always for $M' \le 1$ ; for $M' > 1$ below $q_{\text{cr}}$
Gas from top to bottom, down the dip (injection into the crest, the top of the structure; gas cap expansion)	$> 0$	below $q_{\text{cr}}$ ; for gas $M'$ is large, and $q_{\text{cr}}$ is close to $q_{\text{gr}}$
Water from top to bottom, down the dip	$< 0$	only for $M' < 1$ and a high enough rate; otherwise the water runs down as a tongue along the base
Gas from bottom to top, up the dip	$< 0$	practically never: gas floats up and overtakes the oil along the top
A horizontal formation	$0$	only for $M' < 1$ ; for $M' > 1$ there is a tongue at any rate

The rate enters only through $N_g$ . Lowering the rate increases $|N_g|$ . It helps where gravity holds the displacing phase back, and it harms where gravity drives it on.

For $M' > 1$ the interface loses stability when $N_g$ falls to $1 - 1/M'$ . The gravity number is inversely proportional to the flow rate: $N_g = q_{\text{gr}}/q_t$ , where $q_{\text{gr}}$ is the gravity rate, the rate at which the gravity number equals one. Hence the limiting rate:

Critical rate of gravity-stable displacement

q_{\text{cr}} = q_{\text{gr}}\,\frac{M'}{M' - 1} = \frac{k\,k_{rd}'\,A\,|\Delta\rho|\,g\,|\sin\alpha|}{\mu_d\,(M' - 1)},\qquad q_{\text{gr}} = \frac{k\,k_{ro}'\,A\,|\Delta\rho|\,g\,|\sin\alpha|}{\mu_o}

where:

$q_{\text{cr}}$ critical rate: the largest total reservoir flow rate through the cross-section of the formation at which the interface is still stable, m³/s
$q_{\text{gr}}$ gravity rate: the rate at which the gravity number equals one, m³/s
$A$ area of the cross-section of the formation across the flow: thickness × width of the front, m²
$k_{rd}',\ \mu_d$ end-point relative permeability and viscosity of the displacing phase, —, Pa·s
$\Delta\rho$ density difference between the displacing phase and the oil, kg/m³

Units: SI (any consistent system)

Holds when

$M' > 1$ and gravity holds the displacing phase back. For $M' \le 1$ such displacement is stable at any rate.
Vertical equilibrium, a sharp interface, a uniform formation.
The rate is at reservoir conditions, through the whole cross-section $A$ : for an element of a line-drive pattern it is the injection of one row over its width.

The two forms are equal because $M'k_{ro}'/\mu_o = k_{rd}'/\mu_d$ . The critical rate grows with permeability, cross-section, density difference and dip angle, and falls as $M'$ grows. For a large $M'$ , as for gas, the factor $M'/(M' - 1)$ is close to one, and $q_{\text{cr}} \approx q_{\text{gr}}$ .

In other unit systems (2)

Metric oilfield units

q_{\text{cr}}\,[\text{m}^3/\text{d}] = \frac{0.0853\,k\,k_{ro}'\,A\cdot0.00981\,|\Delta\rho|\,|\sin\alpha|}{\mu_o}\cdot\frac{M'}{M' - 1}

$k$ in mD, $A$ in m², $\Delta\rho$ in g/cm³, $\mu_o$ in mPa·s; the rate is at reservoir conditions.

$0.0853$ is the constant of Darcy's law (Rock permeability); $0.00981\,|\Delta\rho|\,|\sin\alpha|$ is the gradient due to the density difference along the formation, in MPa/m, as in the lesson Fractional flow function and mobility ratio.

Oilfield units

q_{\text{cr}}\,[\text{bbl/d}] = 4.886\cdot10^{-4}\,\frac{k\,k_{rd}'\,A\,\Delta\gamma\,|\sin\alpha|}{\mu_d\,(M' - 1)}

$k$ in mD, $A$ in ft², $\Delta\gamma$ the difference between the relative densities of the phases (fresh water = 1), $\mu_d$ in cP, the rate in reservoir barrels per day.

$4.886\cdot10^{-4} = 1.127\cdot10^{-3}\cdot0.4335$ , as in the oilfield form of the gravity number in the lesson Fractional flow function and mobility ratio.

Worked exampleCritical rate for water injection up the dip

Given (assumed data)

Formation and fluids: as in the previous example: 400 mD, cross-section 25 × 600 m, 12°; densities 1.04 and 0.80 g/cm³; oil 3.0 mPa·s, water 0.5 mPa·s
End-point relative permeabilities: $k_{ro}' = 0.85$ , $k_{rw}' = 0.25$
Compare with the rates of the lesson Fractional flow function and mobility ratio: 600 and 150 m³/d at reservoir conditions

Solution

Gradient due to the density difference along the formation: $0.00981\cdot0.24\cdot\sin12° = 0.00981\cdot0.24\cdot0.2079 = 4.895\cdot10^{-4}$ MPa/m.
Gravity rate: $q_{\text{gr}} = 0.0853\cdot400\cdot0.85\cdot15{,}000\cdot4.895\cdot10^{-4}/3.0 = 71.0$ m³/d.
Mobility ratio: $M' = (0.25/0.5)/(0.85/3.0) = 1.765$ . The factor is $M'/(M' - 1) = 1.765/0.765 = 2.307$ .
Critical rate: $q_{\text{cr}} = 71.0\cdot2.307 = 163.8$ m³/d at reservoir conditions, or $163.8/1.02 = 161$ m³/d of injected water at standard conditions.
Through the gravity number: stability requires $N_g > 1 - 1/1.765 = 0.433$ . At 150 m³/d, $N_g = 71.0/150 = 0.473$ , which is above the limit. At 600 m³/d, $N_g = 0.118$ , which is 3.7 times below it.

Answer: The critical rate is about 163.8 m³/d at reservoir conditions. An injection of 150 m³/d is below it, and 600 m³/d is 3.7 times above it: water will run ahead as a tongue along the base.

What it means. The lesson Fractional flow function and mobility ratio calculated the same formation as diffuse and found that gravity changes the fractional flow of water by only a few per cent. If the phases do have time to separate over the thickness, the dip decides more. Below 163.8 m³/d the water at the base holds as a single front, and above it runs ahead as a tongue. The factor $M'/(M' - 1)$ here is more than two, because the water is only 1.765 times more mobile than the oil. With a more viscous oil, $M'$ grows, the factor approaches one, and $q_{\text{gr}}$ itself falls in inverse proportion to the viscosity, so the critical rate falls fast. The data in this example are assumed for illustration.

Tilt of the interface below the critical rate

Let $\beta$ be the angle between the interface and the bedding, measured so that for $\beta > 0$ the layer of the displacing phase thins out towards the oil: for water $\tan\beta = -\dd z/\dd x$ . From the relation between the slope of the interface and $M'$ and $N_g$ , replacing $N_g$ by $q_{\text{gr}}/q_t$ , we get the steady tilt:

Steady tilt of the interface to the bedding

\tan\beta = \tan\alpha\left(1 - \frac{M' - 1}{M'N_g}\right) = \tan\alpha\left(1 - \frac{q_t}{q_{\text{cr}}}\right)

where:

$\beta$ angle between the interface and the bedding, rad
$\alpha$ dip angle of the formation (in absolute value: one formula serves for water going up the dip and for gas going down the dip), rad
$q_t,\ q_{\text{cr}}$ reservoir flow rate through the cross-section and the critical rate, m³/s

Units: dimensionless

Holds when

The interface has had time to reach the steady tilt.
The second form is for $M' > 1$ ; for $M' \le 1$ use the first.
The assumptions of the stability analysis: vertical equilibrium, a sharp interface, a uniform formation.

At a very low rate $\beta = \alpha$ : the interface is horizontal, like a fluid contact at rest. For $M' > 1$ , as the rate grows, $\beta$ decreases: the interface lies down towards the bedding, and at $q_t = q_{\text{cr}}$ it becomes parallel to it. For $M' < 1$ , $\beta > \alpha$ : the interface tilts more steeply than the horizontal, and the displacement is even more stable. For $M' = 1$ it is horizontal at any rate. Water and oil lie one above the other over a length of $h/\tan\beta$ along the formation.

CalculatedThe angle between the interface and the bedding against the flow rate divided by the gravity rate

q_{\text{gr}}

, for a formation dip of 12° and different values of

M'

. The curves for

M' > 1

end at the critical rate. The dots are the rates of 75 and 150 m³/d of the example below (

M' = 1.76

q_{\text{gr}} \approx 71

m³/d).

Worked exampleTilt of the interface and the length of the interface zone

Given (assumed data)

Formation and fluids: as in the examples above: thickness 25 m, dip 12°, $q_{\text{cr}}$ = 163.8 m³/d
Injection rates: 75 and 150 m³/d at reservoir conditions; for comparison, rest (no flow)
Distance between the rows: 1000 m along the formation

Solution

At rest: $\tan\beta = \tan12° = 0.2126$ . The interface is horizontal and crosses the thickness over a length of $25/0.2126 = 118$ m.
75 m³/d: $q_t/q_{\text{cr}} = 75/163.8 = 0.458$ ; $\tan\beta = 0.2126\cdot(1 - 0.458) = 0.1152$ and $\beta = 6.6°$ . Length of the zone: $25/0.1152 = 217$ m.
150 m³/d: $q_t/q_{\text{cr}} = 0.916$ ; $\tan\beta = 0.2126\cdot(1 - 0.916) = 0.0179$ and $\beta = 1.0°$ . Length of the zone: $25/0.0179 = 1397$ m, more than the distance between the rows.
Suppose that at 75 m³/d the interface has had time to turn to the steady tilt. By the time its lower edge at the base reaches the production row, the tilted interface leaves behind a wedge of oil: $217/(2\cdot1000) = 0.1085$ of the mobile volume, about 11%.

Answer: At 75 m³/d the interface is tilted to the bedding by about 6.6° and takes 217 m along the formation. At 150 m³/d it is tilted by about 1.0° and takes about 1.4 km.

What it means. A fluid contact at rest is horizontal. As production goes on, it turns to the steady tilt, and the water at the base is pushed forward. At half the critical rate the interface zone takes a fifth of the way between the rows, and water reaches the production wells when about 11% of the mobile oil is still above the interface. At 150 m³/d the displacement is formally stable, but the steady interface is longer than the distance between the rows. It turns to this tilt more slowly the closer the rate is to the critical one, and it keeps stretching all the time while the front moves to the wells. So engineers choose the rate with a margin, usually two times or more below $q_{\text{cr}}$ , the more so as $q_{\text{cr}}$ itself is not known precisely. The data in this example are assumed for illustration.

Calculating displacement under vertical equilibrium

With a sharp interface, in each cross-section of the formation a part of the thickness, $h_D$ , is occupied by water at the saturation $1 - S_{or}$ , and the rest by oil with irreducible water $S_{wi}$ . The average water saturation over the thickness grows linearly with $h_D$ , and the thickness-averaged relative permeabilities change linearly in the same way: water flows only in its own layer, with $k_{rw}'$ , and oil flows only in its own layer, with $k_{ro}'$ .

Thickness-averaged relative permeabilities under vertical equilibrium

h_D = \frac{\bar S_w - S_{wi}}{1 - S_{wi} - S_{or}},\qquad \bar k_{rw} = k_{rw}'\,h_D,\qquad \bar k_{ro} = k_{ro}'\,(1 - h_D)

where:

$\bar S_w$ average water saturation over the thickness in the cross-section, —
$h_D$ fraction of the thickness occupied by water (by gas, for gas), —
$\bar k_{rw},\ \bar k_{ro}$ thickness-averaged relative permeabilities of water and oil, —
$k_{rw}',\ k_{ro}'$ end points of the core relative permeability curves, —

Units: fractions

Holds when

Vertical equilibrium, a sharp interface, a formation uniform over the thickness.
Behind the interface the oil is already at its residual saturation.

These are straight lines between the end points: the result does not depend on the shape of the core relative permeability curves, and only $S_{wi}$ , $S_{or}$ , $k_{rw}'$ and $k_{ro}'$ are needed. Such averaged curves are the simplest example of pseudo relative permeabilities (effective curves that stand in for the detailed flow across the thickness). General methods of building them for heterogeneous formations are the subject of Two-phase upscaling: pseudo-functions.

Putting the straight lines into the fractional flow equation of the lesson Fractional flow function and mobility ratio with the gravity correction, $f_w = (\lambda_w/\lambda_t)(1 - N_g\,k_{ro}/k_{ro}')$ , we get the fractional flow of water through the cross-section as a function of $h_D$ :

Fractional flow of water in segregated displacement

f_w = \frac{M'h_D\,\bigl(1 - N_g\,(1 - h_D)\bigr)}{1 + (M' - 1)\,h_D}

where:

$f_w$ share of water in the reservoir flow rate through the cross-section, —
$h_D,\ M',\ N_g$ fraction of the thickness under water, mobility ratio at the end points, gravity number, —

Units: dimensionless

Holds when

The averaged relative permeabilities above. The tilt of the interface itself does not enter the formula (it is taken into account in the advanced section).

The curve bends downwards (its slope falls as $h_D$ grows) if $M'N_g < M' - 1$ , and bends upwards (the slope grows) if $M'N_g > M' - 1$ . This is the same condition as for the stability of the interface. For a curve that bends downwards, the Welge tangent from the point $(S_{wi}, 0)$ touches it at the very beginning: the thinnest part of the tongue reaches the wells first. For a curve that bends upwards, the tangent goes up to $h_D = 1$ : the front is piston-like.

Breakthrough and full sweep in unstable segregated displacement

W_{iD,BT} = \frac{\Delta S}{M'(1 - N_g)},\qquad W_{iD}\big|_{h_D = 1} = \frac{M'\,\Delta S}{1 + M'N_g},\qquad \Delta S = 1 - S_{wi} - S_{or}

where:

$W_{iD,BT}$ water injection up to breakthrough in pore volumes; the same amount of oil has been produced by breakthrough, —
$W_{iD}\big|_{h_D = 1}$ injection after which water occupies the whole thickness at the outlet and all the mobile oil has been produced, —
$\Delta S$ range of mobile saturation, —

Units: pore volumes (reservoir conditions), as in the lesson Buckley–Leverett theory and the Welge method

Holds when

$M'N_g < M' - 1$ (an unstable interface, a tongue). With a stable interface, breakthrough and full sweep coincide: $W_{iD} = \Delta S$ .
Injection at a constant rate, a linear element, the assumptions of the averaged relative permeabilities.

The first formula is the reciprocal of the slope of the curve at $h_D = 0$ , and the second is the reciprocal at $h_D = 1$ (the reciprocal is one divided by the slope). Between them, we find production with the Welge formulas (production after breakthrough) for an outlet saturation given through $h_D$ .

Worked exampleSegregated and diffuse displacement in one formation

Given (assumed data)

Element: the formation of the examples above: 1000 m × 600 m × 25 m, porosity 0.20, $\Delta S = 0.53$ , $S_{wi} = 0.22$
Injection rates: 600 m³/d (above the critical rate) and 75 m³/d (below it)
From the examples above: $M' = 1.765$ ; $q_{\text{gr}} = 71.0$ m³/d; $1 - 1/M' = 0.433$

Solution

Pore volume: $V_p = 1000\cdot600\cdot25\cdot0.20 = 3.0$ million m³. At 600 m³/d one pore volume is injected in 5000 d, and at 75 m³/d in 40,000 d.
600 m³/d: $N_g = 71.0/600 = 0.118$ and $M'N_g = 0.208$ , which is less than $M' - 1 = 0.765$ , so the interface is unstable. Breakthrough: $W_{iD,BT} = 0.53/(1.765\cdot0.882) = 0.53/1.5567 = 0.340$ pore volume, after $0.340\cdot5000 = 1700$ d (4.7 years). Displacement efficiency at breakthrough: $0.340/0.78 = 0.436$ .
All the mobile oil is produced at $W_{iD} = 1.765\cdot0.53/(1 + 0.208) = 0.93545/1.208 = 0.774$ pore volume, after 10.6 years. The displacement efficiency is $0.53/0.78 = 0.679$ .
A diffuse calculation at the same $N_g$ (the fractional flow curve of the lesson Fractional flow function and mobility ratio and the Welge method, model values): breakthrough at 0.399 pore volume ( $E_D$ = 0.512, after 5.5 years). At an injection of 0.774 pore volume $E_D$ = 0.552, and at 2 pore volumes 0.597.
75 m³/d: $N_g = 71.0/75 = 0.947 > 0.433$ , so the interface is stable. The calculation with the averaged relative permeabilities gives a piston-like front: water would come after $\Delta S = 0.53$ pore volume has been injected, after $0.53\cdot40{,}000 = 21{,}200$ d (about 58 years). Allowing for the wedge of oil that the tilted interface leaves behind (the previous example), about $0.679\cdot(1 - 0.1085) = 0.61$ of the initial oil has been produced by breakthrough. Before breakthrough injection equals production, $0.53\cdot(1 - 0.1085) = 0.472$ pore volume, so the water arrives earlier, after $0.472\cdot40{,}000 = 18{,}880$ d (about 52 years).

Answer: At 600 m³/d segregated displacement gives breakthrough after 4.7 years (

E_D

= 0.436) and washes out all the mobile oil in 10.6 years. Diffuse displacement gives breakthrough after 5.5 years (

E_D

= 0.512), but by the same time

E_D

is only 0.552. At 75 m³/d breakthrough comes at

E_D

≈ 0.61, but only after about 52 years.

What it means. The example illustrates three conclusions. At a rate 3.7 times the critical rate, the tongue brings water earlier than a diffuse front (at a rate only slightly above the critical one it is the opposite: later, because the tongue is still short). But then the averaged relative permeabilities are linear: the oil in the upper part of the formation flows with its end-point relative permeability, and the mobile oil is washed out fast. The diffuse front gives noticeably less in the same time, because behind the front the oil flows with a low relative permeability. Below the critical rate the displacement is almost piston-like, but 75 m³/d per element means decades of development. Delayed production is worth less (Net present value (NPV)), and engineers choose the rate by comparing options. At 600 m³/d vertical equilibrium is borderline (the first example), so the real picture lies between the two calculations. The sharp-interface model is optimistic after breakthrough: it assumes that behind the interface the oil is at once at its residual saturation, while a high-permeability layer at the base speeds up the tongue (Displacement in heterogeneous reservoirs). The data in this example are assumed for illustration.

CalculatedDisplacement efficiency against injection for the element of the example: segregated and diffuse displacement at 600 m³/d and segregated displacement at 75 m³/d (a piston-like front, ignoring the wedge of oil that the tilted interface leaves behind). Before breakthrough all the lines coincide: production equals injection.

Gravity drainage and its limiting rate

In gravity drainage oil drains under its own weight to wells in the lower part of the formation, and its place is taken by gas: gas of a gas cap, gas that came out of the oil, or gas injected into the crest (gravity drive). The pressure in the gas is distributed almost hydrostatically, and the driving force for the oil is the density difference between oil and gas. If the gas does not have to push the oil with a pressure difference, the oil rate cannot exceed what gravity alone gives:

Limiting rate of gravity drainage

q_{\text{gd}} = \frac{k\,k_{ro}\,A\,(\rho_o - \rho_g)\,g\sin\alpha}{\mu_o}

where:

$q_{\text{gd}}$ largest reservoir flow rate of oil that drainage under gravity along the formation can give, m³/s
$k_{ro}$ relative permeability of oil where it drains: in the oil part below the gas–oil contact it is the end-point value $k_{ro}'$ , and in a zone already occupied by gas it is many times smaller, —
$\rho_o - \rho_g$ difference between the densities of oil and gas at reservoir conditions, kg/m³
$\alpha$ dip angle of the formation. For drainage in the vertical direction $\sin\alpha = 1$ , and $k$ and $A$ are the vertical permeability and the area in plan, rad

Units: SI (any consistent system)

Holds when

The pressure in the gas is hydrostatic: the gas only takes the place of the draining oil.
Vertical equilibrium, a uniform formation.

With the end-point relative permeability this is the same gravity rate $q_{\text{gr}}$ , only with the density difference between oil and gas. If production is higher, the gas has to push the oil with a pressure difference, and such displacement is stable only below $q_{\text{cr}} = q_{\text{gr}}M'/(M' - 1)$ . For gas $M'$ is in the tens, and the critical rate exceeds the drainage limit by only a few per cent: gas cannot advance stably faster than the oil has time to drain away from it.

In other unit systems (1)

Metric oilfield units

q_{\text{gd}}\,[\text{m}^3/\text{d}] = \frac{0.0853\,k\,k_{ro}\,A\cdot0.00981\,(\rho_o - \rho_g)\sin\alpha}{\mu_o}

$k$ in mD, $A$ in m², densities in g/cm³, $\mu_o$ in mPa·s; the rate is at reservoir conditions.

The constants are as in the formula for the critical rate.

Worked exampleHow much oil gravity drainage can give

Given (assumed data)

Formation: the same: 400 mD, cross-section 25 × 600 m, dip 12°; there is a gas cap above the oil, and the production wells are in the lower part
Fluids: oil 0.80 g/cm³, 3.0 mPa·s, $B_o = 1.25$ ; gas at reservoir conditions 0.20 g/cm³ (assumed)
Relative permeability of oil: $k_{ro}' = 0.85$ below the gas–oil contact; 0.10 for drainage through a zone already occupied by gas (assumed)

Solution

Gradient due to the density difference along the formation: $0.00981\cdot(0.80 - 0.20)\cdot0.2079 = 1.2237\cdot10^{-3}$ MPa/m, which is 2.5 times more than for the pair water–oil.
Limiting rate: $q_{\text{gd}} = 0.0853\cdot400\cdot0.85\cdot15{,}000\cdot1.2237\cdot10^{-3}/3.0 = 177.4$ m³/d at reservoir conditions, or $177.4/1.25 = 142$ m³/d of oil at standard conditions.
A liquid production of 600 m³/d at reservoir conditions is above the limit by a factor of $600/177.4 = 3.4$ .
When the oil drains through a zone already occupied by gas, with $k_{ro} = 0.10$ , the limit is $177.4\cdot0.10/0.85 = 20.9$ m³/d, which is 8.5 times less.

Answer: Gravity drainage of this element can give no more than 177.4 m³/d at reservoir conditions (142 m³/d of oil). For drainage through a gas-saturated zone, it is about 20.9 m³/d.

What it means. The density difference between oil and gas is two and a half times larger than between water and oil, so drainage goes faster than the rise of the water at the base, but it is still slow. At a production rate of 600 m³/d the gas can no longer just take the place of the draining oil. It will break through along the top to the production wells, and the energy of the cap will leave with the gas. So in a gravity drive engineers place the production wells in the lower part and open them to the formation (perforate them) far from the gas–oil contact, keep the production rate below the limit, and maintain the pressure by injecting gas into the crest (Displacement of oil by gas). The last oil from the gas-saturated zone drains in thin films over years, but the residual oil saturation can become low. The data in this example are assumed for illustration.

Terms and symbols

Segregated flow: Displacement in which gravity has separated the phases over the thickness of the formation: water at the base, gas at the top, and a sharp interface with the oil.
Vertical equilibrium (VE): The pressure in each phase across the formation is hydrostatic. The phases have time to redistribute over the thickness faster than the front covers the way between wells.
Critical rate $q_{\text{cr}}$: The largest reservoir flow rate through the cross-section of the formation at which the interface between the displacing phase and the oil is stable. For $M' > 1$ it equals $q_{\text{gr}}M'/(M' - 1)$ .
Gravity rate $q_{\text{gr}}$: The rate at which the gravity number equals one. With the relative permeability of the oil where it drains, it is the limit of gravity drainage.
Angle $\beta$: The angle between the interface of the displacing phase and the bedding. At rest it equals the dip angle, and at the critical rate it is zero.
Thickness-averaged relative permeabilities: Straight lines between the end points that replace the core curves under vertical equilibrium with a sharp interface.
Gravity tongue: A layer of water at the base or of gas at the top that runs ahead of the main mass of the displacing phase at a rate above the critical one.

Common mistakes and misconceptions

Wrong

“Injection goes up the dip, so the front is stable at any rate.”

In fact

Only for $M' \le 1$ . For $M' > 1$ gravity holds the front only below $q_{\text{cr}}$ . In the example of this lesson this is about 164 m³/d, and 600 m³/d is 3.7 times more.

Wrong

“Below the critical rate the water reaches the wells together with the front.”

In fact

The interface is tilted, and the water at the base is ahead of the water at the top by $h/\tan\beta$ . In the example of this lesson it is about 220 m at 75 m³/d, and near the critical rate it is more than the distance between the rows of wells.

Wrong

“Segregated displacement is worse than diffuse: the water goes as a tongue.”

In fact

In the example of this lesson, at a rate much above $q_{\text{cr}}$ breakthrough is indeed earlier (and just above $q_{\text{cr}}$ it is even later than for diffuse flow), but then the averaged relative permeabilities are linear, and the mobile oil is washed out with a smaller volume of water. Below $q_{\text{cr}}$ the displacement is close to piston-like, better than diffuse, although slower.

Wrong

“A gas cap can be expanded at any rate: gas is lighter than oil and will stay on top anyway.”

In fact

Gas stays on top only while the oil has time to drain away from it. At a production rate above the gravity rate, the gas pushes the oil with a pressure difference, its mobility is tens of times higher, and it breaks through to the wells along the top.

Wrong

“Vertical equilibrium is a property of the formation: this formation has it.”

In fact

The ratio of the times depends on the rate too. In the example of this lesson, clean sandstone at 150 m³/d separates the phases with a margin, and at 600 m³/d only just.

In depth: advanced detail

Why both phases at the interface move at the same velocity

Let the interface move along the formation at a speed $V$ without changing its shape: the thickness of the layer of the displacing phase is $z(x, t) = z(x - Vt)$ . Under vertical equilibrium the flow in each layer is almost parallel to the bedding, and we write the balance of the displacing phase in a layer of thickness $z$ as $\phi\,\Delta S\,\partial z/\partial t = -\partial(u_d z)/\partial x$ , where $u_d$ is its velocity (the flow rate per unit area). For a shape that travels without change, $\partial z/\partial t = -V\,\partial z/\partial x$ , and after integrating over $x$ we get $u_d z = \phi\,\Delta S\,V z + C$ .

At the tip of the layer, where $z = 0$ , the flux of the displacing phase is also zero, so $C = 0$ and $u_d = \phi\,\Delta S\,V$ in the whole layer. The same argument for the oil (its layer thins out on the other side of the interface) gives $u_o = \phi\,\Delta S\,V$ . The velocities of the phases are equal, and equal to the total velocity $u_t$ . This is the condition used in the derivation. If instead the displacing phase at the base (or the top) moves faster than the oil above it, the tip of the layer runs ahead, and the interface keeps stretching: a tongue forms.

Curvature of the fractional flow curve and the stability of the interface

Write the fractional flow with the averaged relative permeabilities as $f(h) = M'h\,(a + bh)/(1 + ch)$ , with $a = 1 - N_g$ , $b = N_g$ , $c = M' - 1$ . Its slope is $f'(h) = M'(a + 2bh + bch^2)/(1 + ch)^2$ , and the sign of the second derivative is the sign of $b - ac = M'N_g - (M' - 1)$ . So the curve bends downwards (its slope falls) exactly when the interface is unstable, and bends upwards (its slope grows) when it is stable. The values of the slope at the ends give the formulas for breakthrough and full sweep: $f'(0) = M'(1 - N_g)$ and $f'(1) = (1 + M'N_g)/M'$ .

The fractional flow equation of the lesson Fractional flow function and mobility ratio has one more term, with the gradient of the capillary pressure. With a sharp interface its role is played by the difference of the hydrostatic pressures of the phases due to the tilt of the interface itself: $\partial(p_o - p_w)/\partial x = -\Delta\rho\,g\cos\alpha\,\dd z/\dd x$ . If we put the steady tilt into this term, the fractional flow becomes simply $f_w = h_D$ : all fractions of the thickness move at the same speed, and the interface is carried as a whole, the same result as in the stability analysis. A calculation with the averaged relative permeabilities without this term replaces the tilted interface by a jump. So for stable displacement it gives a piston-like front and does not allow for the wedge of oil that the tilted interface leaves behind (of length $h/\tan\beta$ ), and its estimate of the production up to breakthrough is too high.

A horizontal formation: production in unstable displacement

For $N_g = 0$ and $M' > 1$ the curve $f = M'h/(1 + (M' - 1)h)$ bends downwards (its slope falls) over the whole range of mobile saturation, and the Welge formulas can be solved explicitly. If injection $w$ and production $N$ are measured in mobile volumes $V_p\,\Delta S$ (the volume of oil that can be moved), breakthrough comes at $w = 1/M'$ , and all the mobile oil has been produced at $w = M'$ . Between them:

Production in segregated displacement in a horizontal formation

N = \frac{2\sqrt{M'w} - w - 1}{M' - 1},\qquad \frac{1}{M'} \le w \le M'

where:

$N$ produced oil in fractions of the mobile volume $V_p\,\Delta S$ , —
$w$ injected water in fractions of the mobile volume (reservoir conditions), —
$M'$ mobility ratio at the end points, $M' > 1$ , —

Units: fractions of the mobile volume

Holds when

A horizontal formation, vertical equilibrium, a sharp interface.
$M' > 1$ . For $M' \le 1$ the interface is stable, and a calculation with the averaged relative permeabilities gives a piston-like front.

Derivation: the saturation at the outlet is found from $f'(h_e) = 1/w$ , which gives $1 + (M' - 1)h_e = \sqrt{M'w}$ . Putting this into the Welge formula $N = h_e + (1 - f_e)\,w$ gives the expression above. For example, for $M' = 2$ half of the mobile oil has been produced by breakthrough, and all of it after two mobile volumes of water have been injected.

Vertical equilibrium with a capillary transition zone

A sharp interface is the limit in which the capillary transition zone, whose height is of the order of $\Delta p_c/(\Delta\rho\,g)$ , is small compared with the thickness of the formation (transition zone: the zone where capillary forces make the water saturation change gradually from water to oil). In thin formations and in low-permeability rocks with a high capillary pressure this is not so. Vertical equilibrium can still hold (capillary forces even help to level out the saturation vertically), but the distribution of saturation over the thickness is set by the balance of capillary forces and the weight of the phases. The thickness-averaged relative permeabilities are then no longer straight lines: we obtain them by integrating the core curves over this distribution. Such pseudo functions, and also dynamic pseudo functions for formations without vertical equilibrium, are the subject of Two-phase upscaling: pseudo-functions.

Connections to other topics

Fractional flow function and mobility ratio — the gravity number and the fractional flow of water in an inclined formation in diffuse flow.
Buckley–Leverett theory and the Welge method — the displacement front, the Welge tangent and production after breakthrough: the same tools, with averaged relative permeabilities.
Displacement of oil by gas — the critical rate for gas injection into the crest and gas cap expansion.
Water and gas coning — water and gas drawn up towards a single well: the radial analogue of a tongue.
Reservoir drive mechanisms — gas cap drive and gravity drive of accumulations.
Displacement in heterogeneous reservoirs — displacement in layered formations, crossflow and vertical equilibrium between layers.
Two-phase upscaling: pseudo-functions — pseudo relative permeabilities for models with large cells.
Relative permeability — end points of relative permeability curves.

Check yourself

1
Water displaces oil up the dip, and the mobility ratio at the end points is $M' = 0.8$ . At what rate is the interface stable under vertical equilibrium?
1. A.Only if the formation is horizontal
2. B.At none
3. C.At any rate
4. D.Only below the critical rate
Show the answer
Correct answer: C. At any rate
For $M' < 1$ the right side of the condition $M'N_g > M' - 1$ is negative, and for water moving up the dip $N_g > 0$ , so the condition holds at any rate. Water is not more mobile than oil and does not tend to overtake it, and gravity also holds it at the base. There is no critical rate.
2
A formation has a permeability of 250 mD, a thickness of 15 m, a front width of 800 m, a dip of 8°, and a density difference between water and oil of 0.25 g/cm³; $k_{ro}' = 0.75$ , oil 2.0 mPa·s; $k_{rw}' = 0.30$ , water 0.45 mPa·s. Find the critical rate of water injection into the lower part of the formation.
Show the answer
Cross-section: $A = 15\cdot800 = 12{,}000$ m². Gradient: $0.00981\cdot0.25\cdot\sin8° = 0.00981\cdot0.25\cdot0.1392 = 3.414\cdot10^{-4}$ MPa/m. $q_{\text{gr}} = 0.0853\cdot250\cdot0.75\cdot12{,}000\cdot3.414\cdot10^{-4}/2.0 \approx 32.8$ m³/d. $M' = (0.30/0.45)/(0.75/2.0) = 0.667/0.375 \approx 1.78$ , and the factor is $1.78/0.78 \approx 2.28$ . $q_{\text{cr}} \approx 32.8\cdot2.28 \approx 75$ m³/d at reservoir conditions. The formation is thinner, less steeply dipping and less permeable than in the examples of the lesson, and the limit is half as high, although the oil here is less viscous.
3
Why does the critical rate of a gas front almost coincide with the limiting rate of gravity drainage of oil?
1. A.Because the mobility ratio of gas and oil is large, and the factor $M'/(M' - 1)$ is close to one
2. B.Because gas wets the rock
3. C.Because the viscosity of gas equals the viscosity of oil
4. D.Because gas is incompressible
Show the answer
Correct answer: A. Because the mobility ratio of gas and oil is large, and the factor $M'/(M' - 1)$ is close to one
$q_{\text{cr}} = q_{\text{gr}}M'/(M' - 1)$ , and $q_{\text{gr}}$ with the end-point relative permeability of the oil is exactly the limit of the draining of oil under gravity. For gas $M'$ is in the tens (the gas viscosity is tens of times lower), so the factor is only a few per cent above one: gas can advance stably only at the speed at which oil drains away from it.
4
A horizontal uniform formation, vertical equilibrium, $M' = 3$ . What share of the mobile oil will be produced by water breakthrough, and how many mobile volumes of water must be injected to produce all of it?
Show the answer
In a horizontal formation $N_g = 0$ , and for $M' > 1$ the interface is unstable: water runs ahead as a tongue along the base. By the formulas of the advanced section, breakthrough comes after $1/M' = 1/3$ of a mobile volume has been injected, and the same amount of oil, a third of the mobile oil, has been produced. All the mobile oil is produced after $M' = 3$ mobile volumes have been injected.
5
What raises the likelihood of vertical equilibrium, other things being equal?
1. A.A higher injection rate
2. B.A higher vertical permeability
3. C.A smaller distance between the rows of wells
4. D.A more viscous oil
Show the answer
Correct answer: B. A higher vertical permeability
The ratio of the time to travel along the formation to the time to separate across the thickness is proportional to $k_v$ , the density difference and the distance between wells, and inversely proportional to the rate, the oil viscosity and the thickness of the formation. Shale layers that lower $k_v$ are the main reason why a thick formation behaves as a diffuse one.
6
A formation 20 m thick dips at 10°, and injection goes at 0.6 of the critical rate. What is the steady angle between the interface and the bedding, and over what length along the formation do water and oil lie one above the other?
Show the answer
$\tan\beta = \tan10°\cdot(1 - 0.6) = 0.1763\cdot0.4 = 0.0705$ , so $\beta \approx 4.0°$ , against 10° for a horizontal interface: the water at the base is pushed forward. The length of the zone is $20/0.0705 \approx 284$ m. At rest it would be $20/0.1763 \approx 113$ m: as the rate grows, the zone stretches.

Updated: 15 October 2026