Displacement in heterogeneous reservoirs

Quick refresher

In a layered formation each layer is flooded at its own speed. With the same end points of the relative permeability curves, the speed is proportional to $k/\phi$ (permeability divided by porosity). The first water arrives along the fastest layer, the share of water in the production grows in steps as the next layers break through, and the vertical sweep $E_V$ (the share of the mobile oil of the section that the water has passed through) grows with the injection.
If the layers are linked only through the wells, the methods of Stiles (the fronts move as at $M' = 1$ ) and of Dykstra–Parsons (with the mobility ratio) apply. For $M' > 1$ a flushed layer conducts more and more easily and takes more and more of the injection, so the sweep is worse. For $M' < 1$ it is the other way round.
At $M' = 1$ all the layered methods agree, and the Lorenz curve of the section (the cumulative flow capacity $kh$ , how much fluid the layers can pass, plotted against the cumulative storage capacity $\phi h$ , how much fluid they can hold, layer by layer) plays the role of its fractional flow curve: the Stiles formula for the sweep is the Welge formula.
If the layers communicate ( $R_L \gg 1$ , a large effective length-to-thickness ratio), the pressure in each cross-section levels out over the thickness: vertical equilibrium. The section is replaced by a single “formation” with pseudo functions and calculated by the Buckley–Leverett method. Crossflow improves the sweep for $M' < 1$ and worsens it for $M' > 1$ .
Gravity pulls water towards the base. If the best layers are at the bottom, gravity acts together with permeability and the water breaks through along the base. If they are at the top, gravity levels the front. The ratio $N_{gv}$ helps to compare gravity with the viscous forces.
In the field, heterogeneity shows itself by early first water and a long growth of the water cut. From the history of injection, production and water cut, we build the fractional flow curve of the formation. It lies far to the left of the core curve (the curve of a small rock sample) and shows how much water will be needed to reach the limiting water cut.

Key relationships

E_V = \dfrac{\sum_i \phi_i h_i\,\xi_i}{\sum_i \phi_i h_i}

vertical sweep for piston-like displacement in the layers; $\xi_i$ is the fraction of the length covered by the front

E_{V,j} = C_j + (1 - F_j)\,\dfrac{\bar v}{v_j},\qquad f_{we} = \dfrac{M'F_j}{M'F_j + 1 - F_j}

Stiles method: the moment when layer $j$ breaks through

\xi_i = \dfrac{M' - \sqrt{M'^2 + a_i\,(1 - M'^2)}}{M' - 1},\qquad a_i = \dfrac{v_i}{v_j}

Dykstra–Parsons method: the front of layer $i$ when layer $j$ has broken through

\hat S_w = S_{wi} + (1 - S_{wi} - S_{or})\,C_n,\quad \hat k_{rw} = k_{rw}'F_n,\quad \hat k_{ro} = k_{ro}'\,(1 - F_n)

pseudo functions under vertical equilibrium after $n$ layers have been flushed

S_{we} = S_{wi} + N_{pD} - (1 - f_{we})\,W_{iD}

the fractional flow curve of the formation from the production history

Common pitfalls

Treating the vertical sweep as a constant number. It grows with the volume injected and depends on the oil viscosity.
Applying the no-crossflow method to a formation with good vertical permeability, or, the other way round, applying vertical equilibrium to layers separated by continuous shales. For $M' \ne 1$ the results differ noticeably.
Explaining an early breakthrough in a simulation model by a high-permeability layer without first checking the numerical dispersion (smearing caused by the calculation) of a coarse grid.
Counting on the isolation of a high-permeability layer in a well to make the water go through the tight layers when the layers communicate. The water will bypass the isolated interval inside the formation.
Comparing the fractional flow of water in the reservoir flow with the water cut at the surface without converting through $B_o$ and $B_w$ .

After this lesson you can

explain why in a layered formation water reaches the wells earlier and the water cut grows gradually, and define the vertical sweep efficiency
calculate the flooding of layers without crossflow (flow of fluid from one layer to another) by the methods of Stiles and of Dykstra–Parsons: the positions of the fronts, the vertical sweep, the share of water in the production and the injection in pore volumes
estimate how the vertical sweep depends on the volume injected, the mobility ratio and the heterogeneity (measured by the Dykstra–Parsons coefficient $V_{DP}$ )
decide from the effective length-to-thickness ratio $R_L$ and from the ratio of gravity to viscous forces which limit is closer, layers without crossflow or vertical equilibrium, and build the pseudo functions of the section for vertical equilibrium
explain why crossflow improves the sweep for $M' < 1$ and worsens it for $M' > 1$ , and how the order of the layers in the section together with gravity changes the flood
compare the methods with each other and with the field water cut, and build the fractional flow curve of the formation from the production history

A layered formation and the vertical sweep

The Buckley–Leverett theory (Buckley–Leverett theory and the Welge method) describes a uniform linear element: water advances as one front, and the saturation is the same over the thickness. A real formation consists of layers whose permeability differs by a factor of tens (Reservoir heterogeneity and permeability averaging). The injected water is shared among them in proportion to their flow capacity ( $kh$ , how much each layer can pass), and it moves at its own speed in each. Along the best layer it reaches the production wells while the tight layers have been flushed only near the injection wells. After that the wells produce water from the flushed layers together with oil from the others.

The share of the section that the water has covered is called the vertical sweep efficiency $E_V$ (sweep efficiency). In the lesson Why water is injected: pressure maintenance and oil displacement it is a factor of the recovery factor $E_R = E_DE_AE_V$ . The layered methods of this lesson treat the displacement inside a layer as piston-like (piston-like displacement: the water pushes the oil ahead of it like a piston): behind the front only residual oil is left, and ahead of it the saturation is the initial one. Then $E_V$ is the share of the mobile oil of the section that has been displaced by the given moment, and the oil production in pore volumes of the section is $N_{pD} = (1 - S_{wi} - S_{or})\,E_V$ .

Vertical sweep for piston-like displacement in the layers

E_V = \frac{\sum_i \phi_i h_i\,\xi_i}{\sum_i \phi_i h_i},\qquad \xi_i = \frac{x_{f,i}}{L}

where:

$E_V$ vertical sweep: the share of the mobile oil of the section between the rows of wells that the water has passed through, —
$\phi_i,\ h_i$ porosity and thickness of layer $i$ , —, m
$x_{f,i}$ distance covered by the front in layer $i$ ; for a flushed layer $x_{f,i} = L$ , m
$L$ distance between the rows of injection and production wells, m

Units: dimensionless

Holds when

The displacement in each layer is piston-like; the layers are uniform along their length.
The end-point saturations $S_{wi}$ and $S_{or}$ are the same in all layers. If they are not, the weight of a layer is $\phi_ih_i(1 - S_{wi,i} - S_{or,i})$ .
A linear element between the rows of wells: the areal sweep is not taken into account ( $E_A = 1$ ).

The weight of a layer is its pore volume. A thin high-permeability layer adds almost nothing to the sweep, but it gives almost all the early water. In the lesson Why water is injected: pressure maintenance and oil displacement $E_V$ is described as the share of the thickness that the water has passed through. For layers of equal porosity this is the same thing.

Five things decide how the flood of a layered formation will go:

The permeability contrast: the set of $k_i$ , $\phi_i$ , $h_i$ and its measures $V_{DP}$ (the Dykstra–Parsons coefficient) and $L_c$ (the Lorenz coefficient) (measures of heterogeneity).
The mobility ratio $M'$ (mobility ratio): whether flushing a layer makes it easier or harder for water to move along it.
Communication between the layers: whether water and oil flow between them (crossflow) or the layers are linked only through the wells.
Gravity and the order of the layers in the section: where the best layers lie, near the top or near the base.
Capillary forces: the thickness of the capillary transition zone (the zone where the saturation changes gradually from water to oil) compared with the thickness of the formation.

For the calculation we need, for each layer, the thickness, the permeability, the porosity, the irreducible water saturation, the residual oil saturation and the end-point relative permeabilities. The examples of this lesson use the eight layers of the section from the lesson Reservoir heterogeneity and permeability averaging (16 m thick, $V_{DP} \approx 0.75$ ). All the layers have the relative permeability curves of the lesson Relative permeability ( $S_{wi} = 0.22$ , $S_{or} = 0.25$ , $k_{ro}' = 0.85$ , $k_{rw}' = 0.25$ ). The rows of injection and production wells are 500 m apart, and the three oils are those of the lesson Buckley–Leverett theory and the Welge method: 1.2, 3.0 and 30 mPa·s with water of 0.5 mPa·s, that is $M'$ = 0.706, 1.765 and 17.65.

Layers of the section in order of decreasing flooding speed

Layer: 4
$k$ , mD: 900
$k/\phi$ , mD: 3600
Share of $\phi h$: 0.085
Share of $kh$: 0.403

Layer: 1
$k$ , mD: 310
$k/\phi$ , mD: 1348
Share of $\phi h$: 0.117
Share of $kh$: 0.208

Layer: 8
$k$ , mD: 200
$k/\phi$ , mD: 909
Share of $\phi h$: 0.075
Share of $kh$: 0.090

Layer: 3
$k$ , mD: 120
$k/\phi$ , mD: 571
Share of $\phi h$: 0.214
Share of $kh$: 0.161

Layer: 6
$k$ , mD: 75
$k/\phi$ , mD: 395
Share of $\phi h$: 0.129
Share of $kh$: 0.067

Layer: 2
$k$ , mD: 45
$k/\phi$ , mD: 250
Share of $\phi h$: 0.122
Share of $kh$: 0.040

Layer: 5
$k$ , mD: 18
$k/\phi$ , mD: 120
Share of $\phi h$: 0.127
Share of $kh$: 0.020

Layer: 7
$k$ , mD: 8
$k/\phi$ , mD: 62
Share of $\phi h$: 0.132
Share of $kh$: 0.011

Layer	$k$ , mD	$k/\phi$ , mD	Share of $\phi h$	Share of $kh$
4	900	3600	0.085	0.403
1	310	1348	0.117	0.208
8	200	909	0.075	0.090
3	120	571	0.214	0.161
6	75	395	0.129	0.067
2	45	250	0.122	0.040
5	18	120	0.127	0.020
7	8	62	0.132	0.011

The layer number counts from the top down, as in the lesson Reservoir heterogeneity and permeability averaging. The shares of storage capacity $\phi h$ and of flow capacity $kh$ are the same as in the Lorenz curve of the section. The data are assumed for illustration.

Whether the layers communicate is checked by measurements

Before development starts, the pressure over the section lies on one line of the hydrostatic gradient (the rise of pressure with depth because of the weight of the fluid). After some years of production and injection, measurements with a wireline formation tester (a tool lowered on a cable that measures the pressure in the rock at chosen depths) in a new well (Pressure measurement with the wireline formation tester) show how the layers have worked. If the pressure still lies on one line, only shifted, the layers communicate. If the profile is “stepped” (some layers are depleted and others are supported by injection), there are barriers between them. How to read such profiles is covered in the lesson Pressure measurements in producing reservoirs: barriers and accuracy. These measurements, and the production and injection profiles, are the main data for choosing a calculation method.

The front in a single layer

Let us start with layers that do not communicate with each other. They are separated by continuous shale barriers and are linked only through the injection and production wells. Then the same pressure difference $\Delta p$ between the rows acts on all the layers, and each layer is a separate linear “pipe” of length $L$ .

Let the front in layer $i$ have covered the distance $x_{f,i}$ . Behind it water flows with the mobility $\lambda_w = k_{rw}'/\mu_w$ , and ahead of it oil flows with $\lambda_o = k_{ro}'/\mu_o$ . The resistances of the two sections add up, as for sections of a formation connected in series (averaging over layers), and the whole flow rate of the layer goes into advancing the front, that is, into filling the mobile pore volume:

Front speed in a layer without crossflow

\phi_i\,\Delta S\,\frac{\dd x_{f,i}}{\dd t} = \frac{k_i\,\Delta p}{x_{f,i}/\lambda_w + (L - x_{f,i})/\lambda_o},\qquad \Delta S = 1 - S_{wi} - S_{or}

where:

$x_{f,i}$ position of the front in layer $i$ , m
$k_i,\ \phi_i$ permeability and porosity of the layer, m², —
$\Delta p$ pressure difference between the rows of injection and production wells, common to all the layers, Pa
$\lambda_w,\ \lambda_o$ end-point mobilities of water and oil, $k_{rw}'/\mu_w$ and $k_{ro}'/\mu_o$ , 1/(Pa·s)
$\Delta S$ range of mobile saturation: the amount by which the water saturation grows when the front passes, —

Units: SI (any consistent system); in metric oilfield units the right side is multiplied by 0.0853, the constant of Darcy's law

Holds when

Piston-like displacement in the layer; the rock and the fluids are incompressible.
There is no crossflow between the layers; the pressure difference $\Delta p(t)$ is common to all the layers.
Capillary pressure and gravity are not taken into account.

The front speed is proportional to $k_i/\phi_i$ and changes as the front advances. For $M' = \lambda_w/\lambda_o > 1$ the flushed part conducts more easily than the oil part, the resistance of the layer falls, and the front speeds up. For $M' < 1$ it slows down. For $M' = 1$ the speed is constant. So the layers break through in order of decreasing $k/\phi$ . If the end points of the layers differ, the order is also set by their end-point relative permeabilities and the range of mobile saturation.

If we write this equation for two layers and divide one by the other, the pressure difference $\Delta p(t)$ cancels. The ratio of the front positions does not depend on whether the pressure difference or the total injection is held constant. Integration (the advanced section) gives the position of the front in layer $i$ at the moment when the water has broken through along the faster layer $j$ :

Position of the front in a layer when a faster layer breaks through (Dykstra–Parsons)

\xi_i = \frac{M' - \sqrt{M'^2 + a_i\,(1 - M'^2)}}{M' - 1},\qquad a_i = \frac{v_i}{v_j} = \frac{k_i/\phi_i}{k_j/\phi_j} \le 1

where:

$\xi_i$ fraction of the length $L$ covered by the front in layer $i$ at the moment when layer $j$ breaks through, —
$v_i,\ v_j$ layer speeds $k/\phi$ , mD
$M'$ mobility ratio at the end points, $(k_{rw}'/\mu_w)/(k_{ro}'/\mu_o)$ , —

Units: dimensionless

Holds when

The assumptions of the front speed equation; the end points are the same in all layers.
At $M' = 1$ the formula turns into $\xi_i = a_i$ .

At $M' = 1$ the front lags behind exactly in proportion to the speed of the layer. For $M' > 1$ it lags more, because the fast layer sped up as it was flushed. For $M' < 1$ it lags less. The Dykstra–Parsons method is built on this equation.

CalculatedThe section of the lesson at the moment when water has broken through along layer 4 (the layers do not communicate). The shaded part is the flushed part of each layer for oil of 3.0 mPa·s (

M'

= 1.765). The marks are the front positions for oil of 1.2 mPa·s (

M'

= 0.706) and 30 mPa·s (

M'

= 17.65). The layer thicknesses are to scale.

The Stiles method

The simplest of the layered methods, the Stiles method (1949), assumes that the front in each layer moves at a constant speed proportional to $k/\phi$ , as at $M' = 1$ . But it calculates the fractional flow of water (the share of water in the flow) with the real mobilities. The flushed layers give water with the mobility $\lambda_w$ , and the others give oil with $\lambda_o$ , each in proportion to its flow capacity $kh$ . The layers are numbered in the order of breakthrough, by decreasing $v = k/\phi$ .

Stiles method: the sweep and the fractional flow of water when layer j breaks through

E_{V,j} = C_j + (1 - F_j)\,\frac{\bar v}{v_j},\qquad f_{we,j} = \frac{M'F_j}{M'F_j + 1 - F_j},\qquad \bar v = \frac{\sum_i k_ih_i}{\sum_i \phi_ih_i}

where:

$C_j$ share of the pore volume (storage capacity $\phi h$ ) of the section in the $j$ layers flushed by this moment; here the layers are numbered in the order of breakthrough, —
$F_j$ share of the flow capacity $kh$ of the section in the same layers, —
$v_j$ speed of layer $j$ , $k_j/\phi_j$ , mD
$\bar v$ average speed of the section: flow capacity divided by storage capacity, mD
$f_{we,j}$ fractional flow of water at the outlet, at reservoir conditions, just after layer $j$ breaks through, —

Units: dimensionless

Holds when

Piston-like displacement in the layers, no crossflow, the same end points.
The fronts move at constant speeds. This is exact at $M' = 1$ ; for other $M'$ it is an approximation, and the method is not internally consistent: $M'$ is taken into account in the share of water but not in the movement of the fronts.
Gravity and capillary pressure are not taken into account.

The first term of the sweep is the flushed layers taken whole, and the second is the unflushed ones. The front of layer $i$ has covered the fraction $v_i/v_j$ of the length, and the sum $\sum \phi_ih_iv_i/v_j$ equals $\sum k_ih_i/v_j$ . The fractional flow of water is the share of the flow capacity in the flushed layers, “weighted” by the mobility ratio.

Worked exampleSweep by the Stiles method: eight layers, oil of 3.0 mPa·s

Given (assumed data)

Section: eight layers (the table above): $\sum\phi h$ = 2.95 m, $\sum kh$ = 2234 mD·m
Mobility ratio: $M'$ = 1.765 (oil 3.0 mPa·s, water 0.5 mPa·s, $k_{ro}'$ = 0.85, $k_{rw}'$ = 0.25)
Formation volume factors: $B_o$ = 1.25, $B_w$ = 1.02

Solution

Average speed of the section: $\bar v = 2234/2.95 = 757.3$ mD.
Layer 4 breaks through first ( $v = 900/0.25 = 3600$ mD): $C_1 = 0.25/2.95 = 0.0847$ and $F_1 = 900/2234 = 0.4029$ .
Sweep: $E_V = 0.0847 + (1 - 0.4029)\cdot757.3/3600 = 0.0847 + 0.1256 = 0.210$ . The oil produced is $0.53\cdot0.210 = 0.111$ pore volume, and the same amount of water has been injected.
Fractional flow of water: $\dfrac{1.765\cdot0.4029}{1.765\cdot0.4029 + 1 - 0.4029} = 0.544$ , and the water cut at the surface is 59.4%.
Layer 1 breaks through second ( $v = 310/0.23 = 1347.8$ mD): $C_2 = 0.595/2.95 = 0.2017$ and $F_2 = 1365/2234 = 0.6110$ . $E_V = 0.2017 + (1 - 0.6110)\cdot757.3/1347.8 = 0.2017 + 0.2186 = 0.420$ . The fractional flow of water is 0.735, and the water cut at the surface is 77.3%.
The other breakthroughs are in the table below.

Answer: By the first water, 21% of the mobile oil of the section is swept, and just after breakthrough water makes up 59.4% of the production at the surface. By the breakthrough of the second layer it is 42% and 77.3%.

What it means. In a homogeneous element with the same relative permeability curves the water would arrive after 0.396 pore volume had been injected (water breakthrough). Here it arrives after 0.111, which is 3.6 times earlier. Layer 4 is 6% of the thickness and 8% of the pore volume of the section, but it carries 40% of its flow capacity, and it determines the first water. The Stiles method does not say how much water has been pumped through by the next breakthroughs: the flushed layers already pass water, and the method does not calculate its volume. The data in this example are assumed for illustration.

The Stiles method: eight breakthroughs, oil of 3.0 mPa·s

Breakthrough of layer: 4
$C$ ; $F$: 0.085; 0.403
$E_V$: 0.210
$f_{we}$: 0.544
Water cut: 59.4%

Breakthrough of layer: 1
$C$ ; $F$: 0.202; 0.611
$E_V$: 0.420
$f_{we}$: 0.735
Water cut: 77.3%

Breakthrough of layer: 8
$C$ ; $F$: 0.276; 0.701
$E_V$: 0.526
$f_{we}$: 0.805
Water cut: 83.5%

Breakthrough of layer: 3
$C$ ; $F$: 0.490; 0.862
$E_V$: 0.673
$f_{we}$: 0.917
Water cut: 93.1%

Breakthrough of layer: 6
$C$ ; $F$: 0.619; 0.929
$E_V$: 0.755
$f_{we}$: 0.958
Water cut: 96.5%

Breakthrough of layer: 2
$C$ ; $F$: 0.741; 0.969
$E_V$: 0.834
$f_{we}$: 0.982
Water cut: 98.5%

Breakthrough of layer: 5
$C$ ; $F$: 0.868; 0.989
$E_V$: 0.936
$f_{we}$: 0.994
Water cut: 99.5%

Breakthrough of layer: 7
$C$ ; $F$: 1.000; 1.000
$E_V$: 1.000
$f_{we}$: 1.000
Water cut: 100.0%

Breakthrough of layer	$C$ ; $F$	$E_V$	$f_{we}$	Water cut
4	0.085; 0.403	0.210	0.544	59.4%
1	0.202; 0.611	0.420	0.735	77.3%
8	0.276; 0.701	0.526	0.805	83.5%
3	0.490; 0.862	0.673	0.917	93.1%
6	0.619; 0.929	0.755	0.958	96.5%
2	0.741; 0.969	0.834	0.982	98.5%
5	0.868; 0.989	0.936	0.994	99.5%
7	1.000; 1.000	1.000	1.000	100.0%

$C$ and $F$ are the shares of storage capacity and of flow capacity of the flushed layers. $f_{we}$ is the fractional flow of water in the reservoir flow just after breakthrough. The water cut is at the surface, converted from $f_{we}$ for $B_o = 1.25$ , $B_w = 1.02$ . The data are assumed for illustration.

The Dykstra–Parsons method: sweep, injection and the mobility ratio

The Dykstra–Parsons method (1950) does not assume $M' = 1$ . The positions of the fronts at each breakthrough come from the equation of the previous section, and the fractional flow of water takes into account that the resistance of the unflushed layers changes as they are flushed. The authors of the method proposed the same coefficient $V_{DP}$ as a measure of heterogeneity, to generalise the calculations for log-normal sections (sections whose permeabilities follow a log-normal distribution, a skewed distribution that is typical of permeability; see the advanced section).

Fractional flow of water in the production of a layered section without crossflow

f_{we} = \frac{\sum_{\text{flushed}} M'k_ih_i}{\sum_{\text{flushed}} M'k_ih_i + \sum_{\text{unflushed}} k_ih_i\big/\left[1 - \xi_i\,(1 - 1/M')\right]}

where:

$\sum_{\text{flushed}},\ \sum_{\text{unflushed}}$ sums over the flushed layers and over the layers where the front has not yet reached the production wells, —
$\xi_i$ fraction of the length covered by the front in an unflushed layer, —
$k_i,\ h_i$ permeability and thickness of the layer, mD, m

Units: dimensionless; the fractional flow of water at reservoir conditions

Holds when

The assumptions of the front speed equation: piston-like displacement, no crossflow, a common pressure difference.
The common factor $\lambda_o\Delta p/L$ has been cancelled.

The expression in square brackets is the relative resistance of an unflushed layer: from 1 at the start to $1/M'$ when it breaks through. For $M' > 1$ , between breakthroughs the unflushed layers give up their oil faster and faster, and the fractional flow of water in the production can even fall. This is a feature of the piston-like model. In real layers the transition zone behind the front smooths it out.

We find the injection at each moment by summation: an unflushed layer has taken as much water as the oil it displaced, and a flushed layer, in addition, passes water with its own mobility $\lambda_w$ all the time after breakthrough. The derivation is in the advanced section. Below are the results for the section of the lesson.

Worked exampleThe same section by the Dykstra–Parsons method, for three oils

Given (assumed data)

Section: the eight layers of the lesson, with the end points and layer speeds as in the Stiles example
Oils: 1.2, 3.0 and 30 mPa·s: $M'$ = 0.706, 1.765 and 17.65

Solution

The moment of the first breakthrough (layer 4), oil of 3.0 mPa·s. For layer 1, $a = 1347.8/3600 = 0.3744$ . With $1.765^2 = 3.115$ we get $\xi = \dfrac{1.765 - \sqrt{3.115 + 0.3744\cdot(1 - 3.115)}}{1.765 - 1} = \dfrac{1.765 - \sqrt{2.3231}}{0.765} = 0.315$ , which is less than the 0.3744 of the Stiles method.
The same for the other layers (the figure above). The sweep is $E_V = \sum \phi_ih_i\xi_i/2.95 = 0.187$ , against 0.210 by the Stiles method. The injection and the production are $0.53\cdot0.187 = 0.099$ pore volume. The fractional flow of water just after breakthrough is 0.522, against 0.544 by the Stiles method.
First water for the other oils: after 0.121 pore volume for 1.2 mPa·s and 0.082 for 30 mPa·s.
After one pore volume has been injected (a step-by-step calculation): $E_V$ = 0.779, 0.684 and 0.344 for oils of 1.2, 3.0 and 30 mPa·s, and the fractional flow of water is 0.92, 0.89 and 0.95.
Production after one pore volume for oil of 3.0 mPa·s: $N_{pD} = 0.53\cdot0.684 = 0.363$ , against 0.438 in a homogeneous element with the core curves.

Answer: After one pore volume has been injected, water has swept 78% of the mobile oil of the section for low-viscosity oil, 68% for oil of 3.0 mPa·s and only 34% for viscous oil.

What it means. For $M' > 1$ a flushed layer conducts more and more easily and takes a larger and larger share of the injection. The fast layers “intercept” the water, and the sweep falls, the more so the more viscous the oil. The Stiles method does not see this: its fronts are the same for any oil. For $M' > 1$ it overstates the sweep at breakthrough, and for $M' < 1$ it understates it. A comparison with the homogeneous element shows the price of heterogeneity: for the same injection, 17% less oil is produced. Piston-like displacement in the layers is a simplification on the optimistic side. In reality, mobile oil also remains behind the front in each layer, and the water picks it up gradually (Buckley–Leverett theory and the Welge method). The data in this example are assumed for illustration.

CalculatedVertical sweep against the injection for the section of the lesson without crossflow (the Dykstra–Parsons method), for three oils. The dashed line is a homogeneous formation with piston-like displacement: no water leaves the formation until all the mobile oil has been displaced (0.53 pore volume).

Crossflow between layers and vertical equilibrium

If the layers are not separated by continuous barriers, water and oil flow between them: crossflow. How well the crossflow levels out the pressure over the thickness is shown by the effective length-to-thickness ratio $R_L = (L/H)\sqrt{k_v/k_h}$ , where $k_v$ and $k_h$ are the vertical and horizontal permeability, from the lesson on anisotropy (permeability that differs with direction). For $R_L \gg 1$ the pressure in each cross-section is practically the same over the whole thickness. This is vertical equilibrium (VE). For $R_L \ll 1$ crossflow can be neglected, and the methods of the previous sections work. In between, a two-dimensional model of the section is needed.

Worked exampleWhich limit is closer for the section of the lesson

Given (assumed data)

Section: thickness 16 m, distance between the rows 500 m; average permeabilities of the layers (lesson Reservoir heterogeneity and permeability averaging): arithmetic 139.6 mD, harmonic 25.8 mD
Case 1: the layers are uniform and isotropic (the same permeability in all directions), with no shales between them: $k_v$ of the unit (a group of neighbouring layers) is the harmonic mean (the average that suits flow across layers)
Case 2: there are thin shale streaks in the layers: $k_v/k_h$ of the unit = 0.01
Case 3: there are continuous shale barriers between the layers: $k_v/k_h$ of the unit = 0.0001

Solution

Case 1: $k_v/k_h = 25.8/139.6 = 0.1848$ ; $R_L = (500/16)\sqrt{0.1848} = 31.25\cdot0.4299 = 13.4$ .
Case 2: $R_L = 31.25\cdot\sqrt{0.01} = 3.1$ .
Case 3: $R_L = 31.25\cdot\sqrt{0.0001} = 0.31$ .

Answer:

R_L

≈ 13.4, 3.1 and 0.31: the same set of layers can behave as vertical equilibrium, as an intermediate case, or as layers without crossflow.

What it means. The set of permeabilities does not decide which method to use. The barriers decide, and they may not be visible on the permeability diagram. We check communication by pressure measurements in the layers, by comparing the injection and production profiles, and by correlating the shale layers between wells (matching the same layer from one well to the next; Well correlation). For $R_L$ of about 1–5 neither limit is accurate, and we calculate the section with a two-dimensional model with several options for the conductivity of the barriers. The data in this example are assumed for illustration.

Under vertical equilibrium, if gravity and capillary forces are small, the section behaves as a single formation with pseudo functions: the saturation and the relative permeabilities averaged over the thickness (pseudo relative permeability). The pressure gradient in each cross-section is the same in all layers, so water fills the layers with the highest speed $k/\phi$ first. The order of flushing is the same as without crossflow. When the first $n$ layers have been flushed:

Pseudo functions of the section under vertical equilibrium

\hat S_w = S_{wi} + (1 - S_{wi} - S_{or})\,C_n,\qquad \hat k_{rw} = k_{rw}'F_n,\qquad \hat k_{ro} = k_{ro}'\,(1 - F_n)

where:

$\hat S_w$ water saturation averaged over the pore volume of the section in the given cross-section, —
$\hat k_{rw},\ \hat k_{ro}$ pseudo relative permeabilities of the section for water and oil, —
$C_n,\ F_n$ shares of storage capacity $\phi h$ and of flow capacity $kh$ in the $n$ flushed layers, —

Units: dimensionless

Holds when

Vertical equilibrium ( $R_L \gg 1$ ); gravity and capillary pressure are negligible.
Piston-like displacement in the layers; the end points are the same (in the general case the layers enter the sums with their own $S_{wi}$ , $S_{or}$ , $k_{rw}'$ , $k_{ro}'$ ).
The order of flushing is by decreasing $k/\phi$ ; under strong segregation it is from the bottom up (the next section).

The pseudo functions describe a homogeneous formation of the same thickness that behaves as the layered section. The fractional flow of water that they give, $\hat f_w = M'F_n/(M'F_n + 1 - F_n)$ , is the same expression as in the Stiles method, but now it is the curve $\hat f_w(\hat S_w)$ for a one-dimensional Buckley–Leverett problem. Its points form a broken line. Instead of the Welge tangent, we stretch a “string” over the points, from $(S_{wi}, 0)$ to $(1 - S_{or}, 1)$ . It forms an upper envelope that bends downwards (its slope falls). Each straight segment of it is a separate front, which reaches the wells after $1/\text{slope}$ pore volumes have been injected.

Pseudo functions of the section of the lesson, oil of 3.0 mPa·s

Layer flushed: 4
$\hat S_w$: 0.265
$\hat k_{rw}$: 0.101
$\hat k_{ro}$: 0.508
$\hat f_w$: 0.544

Layer flushed: 1
$\hat S_w$: 0.327
$\hat k_{rw}$: 0.153
$\hat k_{ro}$: 0.331
$\hat f_w$: 0.735

Layer flushed: 8
$\hat S_w$: 0.366
$\hat k_{rw}$: 0.175
$\hat k_{ro}$: 0.255
$\hat f_w$: 0.805

Layer flushed: 3
$\hat S_w$: 0.480
$\hat k_{rw}$: 0.215
$\hat k_{ro}$: 0.118
$\hat f_w$: 0.917

Layer flushed: 6
$\hat S_w$: 0.548
$\hat k_{rw}$: 0.232
$\hat k_{ro}$: 0.060
$\hat f_w$: 0.958

Layer flushed: 2
$\hat S_w$: 0.613
$\hat k_{rw}$: 0.242
$\hat k_{ro}$: 0.026
$\hat f_w$: 0.982

Layer flushed: 5
$\hat S_w$: 0.680
$\hat k_{rw}$: 0.247
$\hat k_{ro}$: 0.009
$\hat f_w$: 0.994

Layer flushed: 7
$\hat S_w$: 0.750
$\hat k_{rw}$: 0.250
$\hat k_{ro}$: 0.000
$\hat f_w$: 1.000

Layer flushed	$\hat S_w$	$\hat k_{rw}$	$\hat k_{ro}$	$\hat f_w$
4	0.265	0.101	0.508	0.544
1	0.327	0.153	0.331	0.735
8	0.366	0.175	0.255	0.805
3	0.480	0.215	0.118	0.917
6	0.548	0.232	0.060	0.958
2	0.613	0.242	0.026	0.982
5	0.680	0.247	0.009	0.994
7	0.750	0.250	0.000	1.000

The layers are flushed in order of decreasing $k/\phi$ . Before injection starts, $\hat S_w$ = 0.22 and $\hat k_{ro}$ = 0.85. The data are assumed for illustration.

Worked exampleThe same section under vertical equilibrium

Given (assumed data)

Section and oil: the eight layers of the lesson, oil of 3.0 mPa·s ( $M'$ = 1.765); the pseudo functions are in the table above
Comparison: the no-crossflow calculation, the Dykstra–Parsons example

Solution

After layer 4 is flushed: $\hat S_w = 0.22 + 0.53\cdot0.0847 = 0.2649$ , $\hat k_{rw} = 0.25\cdot0.4029 = 0.1007$ , $\hat k_{ro} = 0.85\cdot(1 - 0.4029) = 0.5075$ , $\hat f_w = 0.5435$ .
The slopes of the segments of the broken line fall from the first to the last (12.10, 3.09, 1.77, …). The line already bends downwards, and each layer gives its own front.
The first front reaches the wells after $1/12.10 = 0.0826$ pore volume has been injected, against 0.0993 without crossflow. The fractional flow of water just after it is 0.5435.
After one pore volume has been injected: $E_V$ = 0.64 against 0.68 without crossflow. For oil of 1.2 mPa·s it is 0.80 against 0.78, and for viscous oil 0.23 against 0.34.

Answer: Crossflow brought the first water earlier and reduced the sweep for

M'

= 1.765 and 17.65, and it increased the sweep for

M'

= 0.706.

What it means. The reason is where the crossflow takes the water. At the front of a slow layer, for $M' > 1$ , part of the water that comes to it goes into a neighbouring layer that is already flushed, where the resistance is smaller, and the front lags even more. For $M' < 1$ the water, on the contrary, flows from the flushed layer into the slow one and pushes its front on (the advanced section). At $M' = 1$ the pressure is distributed in the same way in all layers, there is nothing to flow across, and all the methods give one result. The data in this example are assumed for illustration.

Gravity, $k_v/k_h$ and the order of the layers

Water is heavier than oil. If the layers communicate, water that gets into the section tends to move down, to the base, and oil tends to move up. How fast this separation is compared with the advance of water between the rows of wells is shown by the ratio of two times: the time the injected water needs to cover the distance $L$ , divided by the time the oil needs to rise (and the water to sink) through the thickness $H$ .

Ratio of gravity to viscous forces in the section

N_{gv} = \frac{k_v\,k_{ro}'\,\Delta\rho\,g\,L}{\mu_o\,u_t\,H}

where:

$N_{gv}$ ratio of the time to cover the distance $L$ to the time the phases need to separate vertically over the thickness $H$ , —
$k_v$ vertical permeability of the unit (the group of layers), m²
$\Delta\rho$ density difference between water and oil at reservoir conditions, kg/m³
$g$ acceleration of gravity, m/s²
$u_t$ total velocity along the formation, m/s
$L,\ H$ distance between the rows of wells and thickness of the unit, m
$k_{ro}',\ \mu_o$ end-point relative permeability and viscosity of the oil, —, Pa·s

Units: SI

Holds when

This is an order-of-magnitude estimate, not an exact criterion. Vertical movement is possible only if the layers communicate ( $R_L$ is not small).
$k_v k_{ro}'\Delta\rho g/\mu_o$ is the velocity at which oil would rise under gravity alone.

For $N_{gv} \gg 1$ gravity has time to separate the phases: water collects at the base and fills the section from the bottom up. For $N_{gv} \ll 1$ the distribution of water among the layers is set by the viscous forces, in the order of $k/\phi$ . In the lesson Segregated flow and gravity displacement the same ratio is written as $t_x/t_v$ (the time to travel along the formation divided by the time to separate across the thickness) and serves as the sign that the phases are segregated, in vertical equilibrium. The gravity number $N_g$ of the lesson on the fractional flow of water describes gravity along an inclined formation, and $N_{gv}$ describes it across the formation.

In other unit systems (1)

Metric oilfield units

N_{gv} = \frac{0.0853\,k_v\,k_{ro}'\,(0.00981\,\Delta\rho)\,L}{\mu_o\,u_t\,H}

$k_v$ in mD, $\Delta\rho$ in g/cm³, $\mu_o$ in mPa·s, $u_t$ in m/d, $L$ and $H$ in m.

0.0853 is the constant of Darcy's law in metric oilfield units (Rock permeability); $0.00981\,\Delta\rho$ is the pressure gradient in MPa/m produced by the density difference (Capillary pressure and the transition zone).

Worked exampleGravity and capillary forces in the section of the lesson

Given (assumed data)

Vertical permeability: 25.8 mD (layers without shales, case 1 of the previous example); with thin shale streaks, 1.40 mD ( $k_v/k_h$ = 0.01)
Fluids: $\Delta\rho$ = 1.04 − 0.80 = 0.24 g/cm³; oil 3.0 or 1.2 mPa·s; $k_{ro}'$ = 0.85
Velocity: 0.0708 m/d, as in the element of the lesson Buckley–Leverett theory and the Welge method
Imbibition capillary pressure: from 0.04 MPa at $S_{wi}$ down to zero (the assumed curve of the lesson Two-phase flow equations)

Solution

Speed at which oil rises under gravity: $0.0853\cdot25.8\cdot0.85\cdot0.00981\cdot0.24/3.0 = 0.00147$ m/d.
$N_{gv} = 0.00147\cdot500/(0.0708\cdot16) = 0.65$ . For oil of 1.2 mPa·s the rising speed is 2.5 times higher, and $N_{gv}$ = 1.6.
With thin shale streaks $k_v$ is smaller by a factor of $25.8/1.40 \approx 18.4$ , and $N_{gv}$ = 0.035 for oil of 3.0 mPa·s.
Height of the capillary transition zone during imbibition: $0.04/(0.00981\cdot0.24) = 17$ m, as much as the whole thickness of the section.

Answer: With good vertical permeability

N_{gv}

≈ 0.65–1.6: gravity is comparable with the viscous forces. With shale streaks it is 0.035, and gravity can be neglected.

What it means. For the section of the lesson neither limit is accurate. With good vertical permeability the water sinks noticeably towards the base but does not have time to separate completely, and the capillary transition zone is comparable with the thickness of the formation. So the flow is neither diffuse (the same saturation over the thickness) nor segregated with a sharp interface. We calculate such a section with a two-dimensional model. Capillary equilibrium can also be included in the pseudo functions if $R_L$ is large. The slower the injection, the larger $N_{gv}$ and the stronger the effect of gravity. The data in this example are assumed for illustration.

In the limit of strong segregation, water fills the section from the bottom up, and the order of flushing is set not by the speed of the layers but by their position. The pseudo functions are the same, only the layers enter the sums from the bottom up. Then it matters where the best layers lie:

Permeability grows downwards. This is how channel sandstones (sandstones laid down by rivers) are often built, with the grain getting finer from the bottom up. Gravity and permeability drive the water to the base together, the water breaks through along the lower layers, and unproduced oil is left near the top.
Permeability grows upwards. This is typical of the sandstones of bars and beach ridges (banks of sand built by waves and currents), in which the grain gets coarser upwards. Gravity pulls the water down, permeability pulls it up, and the front levels out.

Strong segregation: the same eight layers in different orders, oil of 1.2 mPa·s

Order of flushing: no gravity: order by $k/\phi$
First water, $W_{iD}$: 0.139
Fronts: 8
$E_V$ after 0.53 pore volume: 0.65

Order of flushing: from the bottom up, permeability grows downwards
First water, $W_{iD}$: 0.139
Fronts: 8
$E_V$ after 0.53 pore volume: 0.65

Order of flushing: from the bottom up, the order of the layers of the lesson
First water, $W_{iD}$: 0.530
Fronts: 1
$E_V$ after 0.53 pore volume: 1.00

Order of flushing: from the bottom up, permeability grows upwards
First water, $W_{iD}$: 0.530
Fronts: 1
$E_V$ after 0.53 pore volume: 1.00

Order of flushing	First water, $W_{iD}$	Fronts	$E_V$ after 0.53 pore volume
no gravity: order by $k/\phi$	0.139	8	0.65
from the bottom up, permeability grows downwards	0.139	8	0.65
from the bottom up, the order of the layers of the lesson	0.530	1	1.00
from the bottom up, permeability grows upwards	0.530	1	1.00

Calculated with the pseudo functions under vertical equilibrium. Under strong segregation the layers are flushed from the bottom up. “Permeability grows downwards” and “upwards” are the same eight layers, rearranged in order of increasing permeability towards the base or towards the top. A homogeneous formation with piston-like displacement would give the first water after 0.53 pore volume with $E_V$ = 1. The data are assumed for illustration.

When the best layers are at the bottom, gravity drives the water to the same place as permeability does. The result is the same as without gravity: early water and a stepped growth of the water cut. When the best layers are at the top, water goes through the whole section as one front. The same calculation gives one front also for oil of 3.0 mPa·s ( $M' > 1$ ): the tight layers lie at the base, and water cannot run ahead along them as a tongue. Only with viscous oil ( $M'$ = 17.65) does it arrive earlier along the lower part of the section. In a homogeneous horizontal formation with $M' > 1$ the water–oil interface is unstable, and water runs ahead as a tongue along the base. When the interface is stable, how it tilts and at what injection rate are covered in the lesson Segregated flow and gravity displacement.

Comparison of the methods, and the field water cut

Methods for calculating the flooding of a layered formation

Method: Homogeneous element (Buckley–Leverett theory and the Welge method)
Crossflow: one layer
Mobility ratio: through the core $f_w$ curve
What it gives: breakthrough and production in diffuse flow

Method: Stiles
Crossflow: none
Mobility ratio: only in the fractional flow of water
What it gives: the sweep and the water cut at the breakthrough of each layer; not the injection

Method: Dykstra–Parsons
Crossflow: none
Mobility ratio: in the movement of the fronts and in the fractional flow of water
What it gives: the sweep, the water cut and the injection; generalisations by $V_{DP}$

Method: Pseudo functions, vertical equilibrium
Crossflow: full
Mobility ratio: in the pseudo functions
What it gives: the fractional flow curve of the section for a one-dimensional calculation; segregation when the order is from the bottom up

Method: Two-dimensional model of the section
Crossflow: any
Mobility ratio: fully
What it gives: intermediate $R_L$ and $N_{gv}$ , the capillary transition zone, variation along the strike

Method	Crossflow	Mobility ratio	What it gives
Homogeneous element (Buckley–Leverett theory and the Welge method)	one layer	through the core $f_w$ curve	breakthrough and production in diffuse flow
Stiles	none	only in the fractional flow of water	the sweep and the water cut at the breakthrough of each layer; not the injection
Dykstra–Parsons	none	in the movement of the fronts and in the fractional flow of water	the sweep, the water cut and the injection; generalisations by $V_{DP}$
Pseudo functions, vertical equilibrium	full	in the pseudo functions	the fractional flow curve of the section for a one-dimensional calculation; segregation when the order is from the bottom up
Two-dimensional model of the section	any	fully	intermediate $R_L$ and $N_{gv}$ , the capillary transition zone, variation along the strike

In all the layered methods the displacement inside a layer is piston-like. As the number of layers grows, and for a smooth distribution of permeability, the calculations approach a continuous model.

The section of the lesson with oil of 3.0 mPa·s: what the methods give

Calculation: Homogeneous element, core curves (Buckley–Leverett theory and the Welge method)
First water, $W_{iD}$: 0.396
$f_{we}$ just after it: 0.869
$N_{pD}$ after 0.5 pore volume: 0.408
$N_{pD}$ after 1 pore volume: 0.438

Calculation: Stiles
First water, $W_{iD}$: 0.111
$f_{we}$ just after it: 0.544
$N_{pD}$ after 0.5 pore volume: —
$N_{pD}$ after 1 pore volume: —

Calculation: Dykstra–Parsons
First water, $W_{iD}$: 0.099
$f_{we}$ just after it: 0.522
$N_{pD}$ after 0.5 pore volume: 0.264
$N_{pD}$ after 1 pore volume: 0.363

Calculation: Vertical equilibrium
First water, $W_{iD}$: 0.083
$f_{we}$ just after it: 0.544
$N_{pD}$ after 0.5 pore volume: 0.239
$N_{pD}$ after 1 pore volume: 0.341

Calculation	First water, $W_{iD}$	$f_{we}$ just after it	$N_{pD}$ after 0.5 pore volume	$N_{pD}$ after 1 pore volume
Homogeneous element, core curves (Buckley–Leverett theory and the Welge method)	0.396	0.869	0.408	0.438
Stiles	0.111	0.544	—	—
Dykstra–Parsons	0.099	0.522	0.264	0.363
Vertical equilibrium	0.083	0.544	0.239	0.341

$W_{iD}$ and $N_{pD}$ are in pore volumes (reservoir conditions); $f_{we}$ is the fractional flow of water in the reservoir flow. We calculate the homogeneous element from the relative permeability curves, with a transition zone behind the front. The layered methods assume piston-like displacement in the layers. The data are assumed for illustration.

All the layered calculations agree on the main point: the first water arrives 3.5–5 times earlier than in a homogeneous element, and after one pore volume has been injected, about a fifth less oil has been produced. They differ in the details, and the more so the further $M'$ is from one. The Stiles method is simple and is good for an estimate when $M'$ is about 1. The Dykstra–Parsons method takes the oil viscosity into account. With good vertical communication, the pseudo functions of vertical equilibrium are the better choice. In intermediate cases, a model of the section is the better choice.

CalculatedFractional flow of water at the outlet against oil production (oil of 3.0 mPa·s): a homogeneous element with the core curves; the section of the lesson without crossflow and under vertical equilibrium; and a log-normal section with the same

V_{DP} \approx 0.75

made of many thin layers (without crossflow).

In the field the water cut rarely behaves “by Buckley–Leverett”, with a jump. Water appears early and grows gradually. A well is open to many layers, they break through one after another, and over the area water comes to a single well along streamlines of different length. The earlier the first water and the longer the “tail” of the growth of the water cut, the stronger the heterogeneity or the less favourable $M'$ . We can see the steps in the water cut when separate layers break through in the production profiles (how much each layer produces) of the production wells, and we compare them with the injection profiles of the injection wells.

The areal sweep adds its own uncertainty (areal sweep), and it is often larger than the vertical one. A practical route is to calculate the vertical sweep with a layered method and to adjust the areal sweep so that the calculation reproduces the production and water cut history (History matching: adjusting the model until it reproduces the measured history).

The fractional flow curve of the formation from the production history

The Welge formulas can be applied in reverse, to the history of a whole formation or area. The average water saturation follows from the balance, $\bar S_w = S_{wi} + N_{pD}$ . The fractional flow of water at the outlet, $f_{we}$ , is given by measurements of the water cut. And the saturation “at the outlet” $S_{we}$ comes from the Welge formula after breakthrough (production after breakthrough):

A point of the fractional flow curve of the formation from the production history

S_{we} = S_{wi} + N_{pD} - (1 - f_{we})\,W_{iD},\qquad N_{pD} = \frac{N_pB_o}{V_p},\qquad W_{iD} = \frac{W_{inj}B_w}{V_p}

where:

$N_p,\ W_{inj}$ cumulative oil production and water injection at standard conditions, m³
$V_p$ pore volume of the flooded formation or area, m³
$f_{we}$ fractional flow of water in the reservoir flow at the outlet, found from the water cut of the production through $B_o$ and $B_w$ , —
$S_{we}$ equivalent saturation at the outlet (a calculated value, not a measured one): the abscissa (horizontal coordinate) of a point of the fractional flow curve of the formation, —

Units: dimensionless; the volumes are in the same units

Holds when

The pressure is maintained and there is no free gas: injection at reservoir conditions replaces production.
Injection and production refer to the same area, and no water flows out of it.
$V_p$ is known with an error, which shifts all the points in the same way.

The points $(S_{we}, f_{we})$ form the fractional flow curve of the formation, a pseudo curve of the whole formation that includes both the heterogeneity and the sweep. Its slope at each point equals $1/W_{iD}$ . This follows from the water balance and is not an assumption. So the tangent at the last point, extended to $f_w = 1$ , gives the average saturation, and the continuation of the curve to the limiting water cut gives a forecast of production and injection.

Worked exampleThree years of waterflooding of an area

Given (assumed data)

Area: pore volume 1.5 million m³ (reservoir conditions), $S_{wi}$ = 0.22, $B_o$ = 1.25, $B_w$ = 1.02; the core relative permeability curves are those of the lesson Relative permeability, oil 3.0 mPa·s
Cumulative water injection: 294,100, 588,200, 882,400 m³ at the end of the 1st, 2nd and 3rd year (standard conditions)
Cumulative oil production: 153,000, 234,000, 289,400 m³ (standard conditions)
Water cut of the production: 62.3, 76.6, 83.3%

Solution

Third year: $W_{iD} = 882{,}400\cdot1.02/(1.5\cdot10^6) = 0.6000$ and $N_{pD} = 289{,}400\cdot1.25/(1.5\cdot10^6) = 0.2412$ .
Fractional flow of water at reservoir conditions: $f_{we} = \dfrac{0.833\cdot1.02}{0.833\cdot1.02 + 0.167\cdot1.25} = 0.8028$ .
$\bar S_w = 0.22 + 0.2412 = 0.4612$ and $S_{we} = 0.4612 - (1 - 0.8028)\cdot0.6000 = 0.3429$ .
The same for the first and second years: $S_{we}$ = 0.2623 and 0.3060 at $f_{we}$ = 0.5742 and 0.7276.
Check: the slope between the second and the third year is $(0.8028 - 0.7276)/(0.3429 - 0.3060) = 2.04$ , and $1/W_{iD}$ in the middle of the interval is $1/0.500 = 2.00$ . This is almost the same: the slope of a chord is close to the slope of the curve in the middle of the segment.
On the core fractional flow curve at the same saturations, water would make up only 0.001, 0.012 and 0.041 of the flow.

Answer: The fractional flow curve of the formation passes through the points (0.2623; 0.5742), (0.3060; 0.7276) and (0.3429; 0.8028), far to the left of the core curve.

What it means. In a homogeneous element with the same curves, water would not have come before 0.40 pore volume had been injected, but here already at 0.2 it makes up more than half of the reservoir flow. This is how layering shows itself. The continuation of the curve is a forecast. In the example the history was built from a model of a log-normal section without crossflow with $V_{DP} \approx 0.75$ , like the section of the lesson. This model, fitted to three points, predicts a water cut of 98% after 3.52 pore volumes have been injected. At the same injection rate as before, that takes about 18 years, and by then 0.439 pore volume of oil will have been produced. A homogeneous element with the core curves would reach the same water cut after 1.44 pore volumes (Buckley–Leverett theory and the Welge method). At the same injection rate, that would take about 7 years, with 0.451 pore volume of oil produced. Here heterogeneity costs not so much oil as time and water: 2.4 times more water has to be pumped through. In a real field, fitting a model to the history does not give a single answer, and we check the forecast against new measurements. The data in this example are assumed for illustration.

CalculatedThe fractional flow curve of the formation from three years of the history of the area (dots) and its continuation by the fitted model of a log-normal section (solid line), against the core fractional flow curve for the same oil (dashed). The curve of the formation lies far to the left: at the same “saturation at the outlet” the water comes along the best layers.

Terms and symbols

Vertical sweep $E_V$: The share of the mobile oil of the section between the injection and production wells that the water has passed through. It grows with the injection.
Layers without crossflow: Layers separated by continuous barriers and linked only through the wells. One pressure difference acts on all of them.
Crossflow: Movement of water and oil between neighbouring layers across the bedding.
Vertical equilibrium: A state in which the pressure (allowing for hydrostatics) has time to level out over the thickness in each cross-section. It holds when $R_L \gg 1$ .
Pseudo functions: The saturation and relative permeabilities averaged over the thickness of the section: the properties of a homogeneous formation that behaves as a layered one.
$N_{gv}$: The ratio of gravity to viscous forces across the formation: the time to cover the distance between wells divided by the time the phases need to separate vertically.
Fractional flow curve of the formation: The fractional flow of water at the outlet as a function of the equivalent saturation $S_{we}$ , built from the history of injection, production and water cut.

Common mistakes and misconceptions

Wrong

“The one-dimensional calculation gave breakthrough after 0.4 pore volume, so it will be the same in a layered formation.”

In fact

Breakthrough is determined by the fastest layer. In the section of the lesson the first water arrives after 0.08–0.11 pore volume, 3.5–5 times earlier, and then the water cut grows gradually.

Wrong

“Crossflow between layers always helps: the water levels out over the section.”

In fact

For $M' < 1$ yes. For $M' > 1$ crossflow takes water into the flushed layers, and the sweep is worse than without crossflow: in the example of this lesson 0.64 against 0.68 after one pore volume has been injected.

Wrong

“We isolate the high-permeability layer in the wells, and the water will go through the tight ones.”

In fact

This works only if the layers do not communicate. With crossflow, water bypasses the isolation inside the formation. Treatment deep in the formation or another injection system is needed.

Wrong

“An early breakthrough in the model proves that there is a high-permeability layer in the formation.”

In fact

A coarse model grid also gives an early breakthrough: numerical dispersion (Numerical dispersion). Before introducing such a layer, check that the result stays the same when the grid is refined, and check the production profiles.

Wrong

“The core fractional flow curve is good for forecasting the water cut of an accumulation.”

In fact

A core describes displacement in the flushed part. The water cut of an accumulation is also determined by the sweep: the fractional flow curve of the formation from the history lies far to the left of the core curve.

In depth: advanced detail

Derivation of the Dykstra–Parsons equation and of the injection

Let $\xi = x_f/L$ , and take $\lambda_o$ out of the denominator of the front speed equation: $x_f/\lambda_w + (L - x_f)/\lambda_o = (L/\lambda_o)\left[1 - \xi(1 - 1/M')\right]$ . Then $\left[1 - \xi(1 - 1/M')\right]\dd\xi = (k/\phi)\,\dd T$ , where $T = \lambda_o\int\Delta p\,\dd t/(\Delta S\,L^2)$ is a “reduced time” common to all layers. Integrating from zero, we get $\xi - (1 - 1/M')\,\xi^2/2 = (k/\phi)\,T$ .

Layer $j$ breaks through at $\xi_j = 1$ , that is, at $(k_j/\phi_j)\,T = (1 + 1/M')/2$ . Putting this $T$ into the equation for layer $i$ and multiplying by $2M'$ , we get the quadratic equation $(M' - 1)\,\xi_i^2 - 2M'\xi_i + a_i(M' + 1) = 0$ with $a_i = (k_i/\phi_i)/(k_j/\phi_j)$ . Its root between zero and one is the formula of the main section.

After breakthrough a layer passes water with the mobility $\lambda_w$ over its whole length, and its flow rate $k\lambda_w\Delta p/L$ no longer goes into displacement. In the same variables the volume of water that has passed through the layer, in fractions of its mobile pore volume, is $1 + (M' + 1)(a - 1)/2$ , where $a = (k/\phi)\,T\cdot 2/(1 + 1/M') \ge 1$ . Summing over the layers with the weights $\phi_ih_i$ gives the injection $W_{iD}$ , and summing the positions of the fronts gives the sweep $E_V$ . This is how the figure and the examples of the lesson were built.

Where the crossflow goes

Take two layers under vertical equilibrium: a fast one (1) and a slow one (2), with flow capacities $k_1h_1$ and $k_2h_2$ . Along the length of the formation there are three stretches: behind both fronts (both layers are flushed), between the fronts (only the fast one is flushed) and ahead of both. In each stretch the pressure gradient is the same in both layers and is set by the total mobility of the column. Behind the slow front, its layer carries $q/(k_1h_1 + k_2h_2)$ per unit of flow capacity, and ahead of it $q/(M'k_1h_1 + k_2h_2)$ , where $q$ is the total flow rate.

For $M' > 1$ the second is smaller than the first: the slow layer ahead of its front carries less than behind it, and the difference flows across into the fast layer at the front. Part of the water that comes up to the front goes into the flushed layer, and the front lags more than without crossflow. For $M' < 1$ it is the other way round: at the slow front, water comes from the fast layer and pushes the front on. At $M' = 1$ the flows are equal, there is nothing to flow across, and vertical equilibrium gives the same as the no-crossflow calculation. The same conclusion follows from the shape of the pseudo curve $\hat f_w = M'F/(M'F + 1 - F)$ . For $M' > 1$ it bends downwards (its slope falls) more strongly than the Lorenz curve, and the fronts diverge more. For $M' < 1$ it does so less, and for a small enough $M'$ the segments of the envelope merge and the layers are flushed by a common front. In the section of the lesson, at $M'$ = 0.1 there are two fronts left, and at 0.05 one.

A log-normal section: the sweep as a function of

V_{DP}

and

M'

If there are many layers and their permeability follows a log-normal distribution (log-normal distribution), the sums of the Dykstra–Parsons method become integrals over the distribution of $\ln k$ , and the result depends only on $\sigma_{\ln k}$ (that is, on $V_{DP} = 1 - e^{-\sigma}$ ) and on $M'$ . The general charts of the method are built on this: the sweep at a given water cut or injection as a function of $V_{DP}$ and $M'$ .

Vertical sweep after one pore volume has been injected: a log-normal section without crossflow

$V_{DP}$: 0.50
$M'$ = 0.706: 0.93
$M'$ = 1.765: 0.86
$M'$ = 17.65: 0.51

$V_{DP}$: 0.75
$M'$ = 0.706: 0.68
$M'$ = 1.765: 0.57
$M'$ = 17.65: 0.25

$V_{DP}$: 0.90
$M'$ = 0.706: 0.37
$M'$ = 1.765: 0.28
$M'$ = 17.65: 0.10

$V_{DP}$	$M'$ = 0.706	$M'$ = 1.765	$M'$ = 17.65
0.50	0.93	0.86	0.51
0.75	0.68	0.57	0.25
0.90	0.37	0.28	0.10

Layers of equal thickness and porosity, the end points of the lesson, piston-like displacement in the layers. The integrals over the distribution were calculated numerically. The data are assumed for illustration.

The sweep falls both as heterogeneity grows and as $M'$ grows, and in relative terms the effect of $M'$ is stronger the more heterogeneous the section. The section of the lesson, with eight layers ( $V_{DP} \approx 0.75$ ), gave 0.68 after one pore volume for oil of 3.0 mPa·s, while a log-normal section with the same $V_{DP}$ gave about 0.57. A coarse division into eight layers cuts off the “tail” of the most permeable and the tightest parts, just as the Lorenz curve through eight points does in the lesson Reservoir heterogeneity and permeability averaging. So the number of layers in the calculation is not a formality.

Connections to other topics

Buckley–Leverett theory and the Welge method — one-dimensional displacement, the Welge formulas and the breakthrough of a homogeneous element.
Reservoir heterogeneity and permeability averaging — measures of heterogeneity, the Lorenz curve, $R_L$ and the eight layers of the section.
Why water is injected: pressure maintenance and oil displacement — displacement and sweep efficiencies, the areal sweep.
Fractional flow function and mobility ratio — the mobility ratio and gravity along the formation.
Segregated flow and gravity displacement — segregated displacement, the stability of the water–oil interface and the critical rate.
Pressure measurement with the wireline formation tester — pressure measurements in the layers, which show whether they communicate.
Two-phase upscaling: pseudo-functions — pseudo functions for the cells of a simulation model.
Numerical dispersion — why a coarse grid also gives an early breakthrough.
History matching — refining the sweep and the heterogeneity from the production history.
Reservoir architecture and performance — how the architecture of sand bodies determines the communication between layers.

Check yourself

1
Two layers of equal thickness and porosity, 400 and 100 mD, do not communicate; $M' = 1$ . Where is the front in the tight layer when the water has broken through along the permeable one, and what is the vertical sweep?
1. A.At the outlet; sweep 1
2. B.At half the length; sweep 0.75
3. C.At a sixteenth of the length; sweep 0.53
4. D.At a quarter of the length; sweep 0.625
Show the answer
Correct answer: D. At a quarter of the length; sweep 0.625
At $M' = 1$ the front speed is constant and proportional to $k/\phi$ : the tight layer has covered $100/400 = 0.25$ of the length. The sweep is the average over the pore volume: $(1 + 0.25)/2 = 0.625$ .
2
The same layers, but $M' = 3$ . Where is the front in the tight layer when the permeable one breaks through, and why does it lag more?
Show the answer
$\xi = \dfrac{3 - \sqrt{9 + 0.25\cdot(1 - 9)}}{3 - 1} = \dfrac{3 - \sqrt{7}}{2} = \dfrac{3 - 2.646}{2} \approx 0.18$ , and the sweep is $(1 + 0.18)/2 \approx 0.59$ . For $M' > 1$ the flushed part of the permeable layer conducts better than its oil part, its resistance falls, and it takes a larger and larger share of the total pressure difference. Its front speeds up, and the front of the tight layer lags more than at $M' = 1$ .
3
The layers communicate well ( $R_L \approx 15$ ), the oil has a low viscosity ( $M' = 0.5$ ), and gravity can be neglected. How will crossflow change the sweep compared with the no-crossflow calculation?
1. A.It depends only on the order of the layers in the section
2. B.It will not change it
3. C.It will increase it
4. D.It will decrease it
Show the answer
Correct answer: C. It will increase it
For $M' < 1$ the flushed layers conduct worse than the unflushed ones. At the front of a slow layer, water flows into it from the flushed fast layer and pushes its front on, so the fronts of the layers come closer together and the sweep grows. For $M' > 1$ it would be the other way round, and for $M' = 1$ crossflow changes nothing.
4
A unit 30 m thick, with 600 m between the rows of wells, a vertical permeability of 50 mD, $k_{ro}'$ = 0.8, a density difference between water and oil of 0.2 g/cm³, oil of 2 mPa·s and a total velocity of 0.05 m/d. Does gravity matter?
Show the answer
The speed at which oil rises is $0.0853\cdot50\cdot0.8\cdot0.00981\cdot0.2/2 \approx 0.00335$ m/d. $N_{gv} = 0.00335\cdot600/(0.05\cdot30) \approx 1.3$ : gravity is comparable with the viscous forces, and water sinks noticeably towards the base. If the best layers are at the bottom, this speeds up breakthrough along the base, and if they are at the top, it levels the front. As the injection rate is lowered, the effect of gravity grows.
5
Over four years, 980 thousand m³ of water ( $B_w$ = 1.0) was injected into an area with a pore volume of 2.0 million m³, 400 thousand m³ of oil ( $B_o$ = 1.2) was produced, the water cut of the production is 70%, and $S_{wi}$ = 0.25. Find the point of the fractional flow curve of the formation.
Show the answer
$W_{iD} = 980{,}000/2{,}000{,}000 = 0.49$ and $N_{pD} = 400{,}000\cdot1.2/2{,}000{,}000 = 0.24$ . The fractional flow of water at reservoir conditions is $f_{we} = 0.7/(0.7 + 0.3\cdot1.2) = 0.7/1.06 \approx 0.660$ . $\bar S_w = 0.25 + 0.24 = 0.49$ and $S_{we} = 0.49 - (1 - 0.660)\cdot0.49 \approx 0.49 - 0.167 = 0.323$ . The point is (0.323; 0.66). At such a saturation at the outlet the core curve would normally give much less water, which is a sign of heterogeneity.
6
Which feature in the water cut history most likely points to strong heterogeneity over the section?
1. A.The water cut does not change for years
2. B.A long water-free period, then a fast growth of the water cut
3. C.Early first water and a long gradual growth of the water cut
4. D.The water cut falls as the injection grows
Show the answer
Correct answer: C. Early first water and a long gradual growth of the water cut
High-permeability layers bring water to the wells early, and the others break through in turn, so the water cut starts to grow almost at once and keeps growing for a long time. A long water-free period with a jump in the water cut is a sign of a homogeneous formation and a favourable mobility ratio.

Updated: 15 October 2026

Quick refresher

After this lesson you can

A layered formation and the vertical sweep

The front in a single layer

The Stiles method

The Dykstra–Parsons method: sweep, injection and the mobility ratio

Crossflow between layers and vertical equilibrium

Gravity, kv/khk_v/k_hkv​/kh​ and the order of the layers

Comparison of the methods, and the field water cut

The fractional flow curve of the formation from the production history

Terms and symbols

Common mistakes and misconceptions

In depth: advanced detail

Connections to other topics

Check yourself

Gravity, $k_v/k_h$ and the order of the layers