Buckley–Leverett theory and the Welge method

Quick refresher

In linear displacement at a constant rate, each water saturation moves at a constant speed $(q_t/\phi A)\,f_w'(S_w)$ . The fastest are the saturations near the inflection point of the $f_w$ curve (where the curve stops bending upwards, with its slope growing, and starts bending downwards, with its slope falling).
A solution without a jump would give three saturations at once at one point of the formation. In reality a front forms: a jump from $S_{wi}$ to $S_{wf}$ . Its speed is set by the water balance at the jump, and the saturation $S_{wf}$ is found with a tangent to the $f_w$ curve from the point $(S_{wi}, 0)$ .
Behind the front the saturation grows smoothly towards the injection side. We read its average value where the tangent crosses the line $f_w = 1$ : $\bar S_w = S_{wi} + 1/f_w'(S_{wf})$ .
Before breakthrough, oil production equals water injection in reservoir volumes. Breakthrough comes after $1/f_w'(S_{wf})$ pore volumes have been injected, and the water cut jumps from zero to the value that matches $f_{wf}$ .
After breakthrough the saturation at the outlet $S_{we}$ grows, and we find production with the same Welge formula at an injection of $W_{iD} = 1/f_w'(S_{we})$ . The last per cent of the mobile oil needs tens of pore volumes of water.
Viscous oil gives a low saturation at the front, early breakthrough and long production at a high water cut. When the mobility ratio is much less than one, the displacement is close to piston-like (all the mobile oil leaves before any water arrives).

Key relationships

v_{S_w} = \dfrac{q_t}{\phi A}\,f_w'(S_w)

speed of advance of the saturation $S_w$

f_w'(S_{wf}) = \dfrac{f_{wf}}{S_{wf} - S_{wi}}

tangency condition: the saturation at the front

\bar S_w = S_{wf} + \dfrac{1 - f_{wf}}{f_w'(S_{wf})} = S_{wi} + \dfrac{1}{f_w'(S_{wf})}

average saturation behind the front before breakthrough (Welge)

W_{iD,BT} = N_{pD,BT} = \dfrac{1}{f_w'(S_{wf})},\qquad t_{BT} = \dfrac{W_{iD,BT}V_p}{q_i}

breakthrough: injection and production in pore volumes, and the time

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

production after breakthrough

Common pitfalls

Mixing up the total velocity $u_t$ (the flow rate per unit area of the cross-section) with the speed of the front. The front moves faster than $u_t/\phi$ by a factor of $f_w'(S_{wf})$ .
Taking the saturation at the front for the average saturation of the flushed zone (the part of the formation that the water has swept). Behind the front there is more water.
Drawing the tangent from the origin and not from the point of the initial saturation $(S_{wi}, 0)$ .
Taking the result of a one-dimensional calculation for the recovery factor of the accumulation. It gives the displacement efficiency, while we account for the areal and vertical sweep separately.
Converting pore volumes to time with the surface water rate without $B_w$ , and converting oil production to standard cubic metres without $B_o$ .

After this lesson you can

derive the equation for the speed at which a saturation advances from the water balance, and explain why different saturations move at different speeds
explain where the saturation jump comes from, and find the saturation at the front with the Welge tangent
calculate the average saturation behind the front, the water breakthrough time and the oil production up to breakthrough, in pore volumes (one pore volume of injected water equals the volume of the pores) and in cubic metres
calculate the oil production, the water cut (the share of water in the produced liquid) and the volume injected after breakthrough
estimate how the oil viscosity changes the picture of displacement, list the assumptions of the theory, and explain what changes in a real formation where they do not hold

The problem and its assumptions

Consider a linear element of a formation between a row of injection wells and a row of production wells (or a core sample) with length $L$ , cross-section area $A$ and porosity $\phi$ . At the start it holds oil and irreducible water $S_{wi}$ (the water that stays in the pores and does not move). From the moment $t = 0$ we inject water at the inlet at a constant rate $q_i$ at reservoir conditions and withdraw liquid at the outlet. The question is how the water will be spread along the length, when it will reach the outlet, and how much oil will have been produced at each moment.

The flow is one-dimensional and linear. The saturation is the same across the section: diffuse (spread-out) flow (Fractional flow function and mobility ratio).
The rock and the fluids are incompressible: the total flow rate $q_t = q_o + q_w$ is the same in every cross-section and equals the injection rate $q_i$ .
The formation is uniform. The initial water saturation is equal to the irreducible one, $S_{wi}$ , everywhere (the water does not move), and there is no free gas.
The fractional flow of water depends only on saturation. We neglect the capillary term, and, if needed, we include the dip and the rate in the curve $f_w(S_w)$ .
The phases do not mix. The relative permeabilities are imbibition curves (the curves for water soaking into the rock) if the rock is water-wet.

It is convenient to measure injection not in cubic metres but in pore volumes: $W_{iD} = W_{inj}B_w/V_p$ . Here $W_{inj}$ is the cumulative water injection at standard conditions (as in the lessons General material balance equation for oil reservoirs and Why water is injected: pressure maintenance and oil displacement), $W_{inj}B_w$ is the same volume at reservoir conditions, and $V_p = \phi A L$ is the pore volume of the element. We express oil production in pore volumes too, as $N_{pD}$ (at reservoir conditions). Then the solution does not depend on the size of the element or on the injection rate, and the same curve works for a core and for a reservoir (pore volumes injected).

The speed at which a saturation advances

Take a thin slice of the formation with thickness $\dd x$ . More water enters it than leaves it if the fractional flow of water falls along the flow. The difference builds up in the pores, and the water saturation of the slice grows. The water flow rate is $q_w = f_w q_t$ , and $q_t$ does not change along the length, so we write the water balance of the slice like this:

Water balance in a slice of a linear formation

\phi A\,\frac{\partial S_w}{\partial t} = -\,q_t\,\frac{\partial f_w}{\partial x}

where:

$\phi,\ A$ porosity and cross-section area, —, m²
$S_w$ water saturation in the section $x$ at the moment $t$ , —
$q_t$ total flow rate at reservoir conditions, equal to the injection rate, m³/s
$f_w$ fractional flow of water, —

Units: SI (any consistent system)

Holds when

Incompressible fluids and rock, a constant total flow rate.
One-dimensional flow, with the same saturation across the section.

This is the saturation equation of the lesson Two-phase flow equations for one-dimensional linear flow without a capillary term. It is the Buckley–Leverett equation in differential form.

Follow the section where the saturation equals a given value $S_w$ . The fractional flow depends only on saturation, so $\partial f_w/\partial x = f_w'(S_w)\,\partial S_w/\partial x$ , where $f_w' = \dd f_w/\dd S_w$ is the slope of the fractional flow curve. Put this into the balance, and require that the saturation does not change as we move together with the section. We get the speed at which the section moves:

Saturation advance equation (Buckley–Leverett)

v_{S_w} = \left(\frac{\dd x}{\dd t}\right)_{S_w} = \frac{q_t}{\phi A}\,f_w'(S_w)

where:

$v_{S_w}$ speed of the section with water saturation $S_w$ , m/s
$f_w'(S_w)$ slope of the fractional flow curve at this saturation, —
$q_t,\ \phi,\ A$ as in the water balance, SI

Units: SI (any consistent system)

Holds when

The assumptions of the theory (the section above).
The curve $f_w(S_w)$ does not change in time.

Each saturation moves at a constant speed proportional to the slope of the $f_w$ curve. Here $q_t/\phi A = u_t/\phi$ is the average speed of the liquid in the pores, and $f_w'$ shows how many times faster or slower a given saturation moves. The fastest are the saturations near the inflection point of the curve, where the slope is largest. This is the Buckley–Leverett equation in the form of the speed of advance (the frontal advance equation).

In other unit systems (1)

Metric oilfield units

v_{S_w}\,[\text{m/d}] = \frac{q_t\,[\text{m}^3/\text{d}]}{\phi\,A\,[\text{m}^2]}\,f_w'(S_w)

$q_t$ in m³/d at reservoir conditions ( $q_t = q_{wi}B_w$ for an injection rate $q_{wi}$ in standard m³/d), $A$ in m², $v$ in m/d.

There is no constant: the units are consistent.

Integrating the speed over time, we get the position of each saturation after $q_t t = W_{inj}B_w$ reservoir cubic metres have been injected: $x_{S_w} = q_t t\,f_w'(S_w)/(\phi A)$ . In fractions of the length of the element it is very simple: $x_{S_w}/L = W_{iD}\,f_w'(S_w)$ . For the Corey model the slope of the curve can be written explicitly:

Slope of the fractional flow curve for the Corey model

f_w' = \frac{f_w\,(1 - f_w)}{1 - S_{wi} - S_{or}}\left(\frac{n_w}{S_n} + \frac{n_o}{1 - S_n}\right)

where:

$S_n$ normalised water saturation $(S_w - S_{wi})/(1 - S_{wi} - S_{or})$ , —
$n_w,\ n_o$ Corey exponents for water and oil (Corey model), —

Units: dimensionless

Holds when

The Corey model, a horizontal formation without a capillary term (the same slope is used in the lesson IMPES and fully implicit schemes).
At the ends of the range of mobile saturation, where only one phase flows, the slope is zero.

The saturation jump and the Welge tangent

Apply the equation to an S-shaped $f_w$ curve. The saturation $1 - S_{or}$ at the inlet is almost immobile ( $f_w' = 0$ ). The intermediate saturations near the inflection point race ahead. The saturations just above $S_{wi}$ move slowly again. It turns out that higher saturations overtake lower ones, and at one point of the formation there are three saturation values at once (the dashed curve in the figure below). Physically this is impossible.

In reality the saturations do not overtake each other. They “pile up” into a jump, the displacement front (shock front): over a very short distance the water saturation falls from $S_{wf}$ to $S_{wi}$ . Ahead of the front only oil flows, and behind it oil and water flow together. Capillary forces smear the jump a little, but at the scale of the reservoir the width of the smearing is from tens of centimetres to metres, much less than the distance between wells (capillary term).

Front speed from the water balance at the jump

v_f = \frac{q_t}{\phi A}\cdot\frac{f_{wf} - f_w(S_{wi})}{S_{wf} - S_{wi}} = \frac{q_t}{\phi A}\cdot\frac{f_{wf}}{S_{wf} - S_{wi}}

where:

$v_f$ speed of the displacement front, m/s
$S_{wf},\ f_{wf}$ water saturation just behind the front and the fractional flow of water at it, —
$f_w(S_{wi})$ fractional flow of water ahead of the front; it is zero if the initial water does not move, —

Units: SI

Holds when

The water that reaches the front during the time $\dd t$ all goes to raising the saturation from $S_{wi}$ to $S_{wf}$ in the slice that the front passes in that time.

The speed of the front is proportional to the slope of the chord (the straight line) that joins the points of the $f_w$ curve ahead of the front and behind it.

It remains to find $S_{wf}$ . The saturations just behind the front cannot overtake it, since otherwise they would go ahead of the jump. They cannot lag behind either, since otherwise a gap would open between them and the jump. So the speed of the saturation $S_{wf}$ from the advance equation equals the speed of the front: the slope of the curve at the point $S_{wf}$ equals the slope of the chord from $(S_{wi}, 0)$ . This is the tangency condition.

Saturation at the front: the tangency condition

f_w'(S_{wf}) = \frac{f_{wf}}{S_{wf} - S_{wi}}

where:

$f_w'(S_{wf})$ slope of the fractional flow curve at the front point, —
$S_{wf},\ f_{wf},\ S_{wi}$ saturation at the front, the fractional flow at it, the initial water saturation, —

Units: dimensionless

Holds when

The initial water does not move. If the initial saturation $S_{w0}$ is above the irreducible one and the water is mobile, we draw the tangent from the point $(S_{w0}, f_w(S_{w0}))$ .
The $f_w$ curve is S-shaped. If over the whole range of mobile saturation it lies below the straight line joining the points $(S_{wi}, 0)$ and $(1 - S_{or}, 1)$ , the jump goes up to $1 - S_{or}$ and the displacement is piston-like.

Graphically: from the point $(S_{wi}, 0)$ we draw a tangent to the $f_w$ curve, and the tangent point is the front (Welge method). Of all the chords drawn from $(S_{wi}, 0)$ , the tangent is the steepest. So the front is the fastest of the possible “steps”.

CalculatedThe Welge tangent for the relative permeability curves of the lesson Relative permeability, oil of 3.0 mPa·s and water of 0.5 mPa·s. The tangent from the point

(S_{wi}, 0)

touches the curve at the front (

S_{wf}

≈ 0.564,

f_{wf}

≈ 0.869) and crosses the line

f_w = 1

at the average saturation behind the front,

\bar S_w

≈ 0.616.

Worked exampleSaturation at the front from the tangent

Given (assumed data)

Relative permeability curves (lesson Relative permeability): $S_{wi} = 0.22$ , $S_{or} = 0.25$ , $k_{ro}' = 0.85$ , $k_{rw}' = 0.25$ , $n_o = 2.5$ , $n_w = 3$
Viscosities: oil 3.0 mPa·s, water 0.5 mPa·s; the formation is horizontal

Solution

Tangency condition: the slope of the curve $f_w'$ equals the slope of the chord $f_w/(S_w - 0.22)$ . Let us test two saturations.
$S_w = 0.55$ : $f_w = 0.830$ , the chord is $0.830/0.33 = 2.52$ , and the slope of the curve from the formula for the Corey model is $3.05$ . This is more than the chord, so the tangent point is further to the right.
$S_w = 0.58$ : $f_w = 0.905$ , the chord is $0.905/0.36 = 2.51$ , and the slope is $1.99$ . This is less than the chord, so the tangent point is further to the left.
Between them, at $S_{wf} \approx$ 0.564, we have $f_{wf} = 0.869$ and the slope $f_w' = 2.525$ . The chord is $0.869/(0.564 - 0.22) = 2.526$ , which is the same within rounding.
Average saturation behind the front: $\bar S_w = 0.22 + 1/2.525 = 0.22 + 0.396 = 0.616$ . Check: $0.564 + (1 - 0.869)/2.525 = 0.616$ .
The front moves 2.525 times faster than the average speed of the liquid in the pores, $u_t/\phi$ . Without a jump, the saturations near the inflection point (about 0.47, slope 5.22) would move about twice as fast again. It is they that overtake the lower saturations and give rise to the jump.

Answer: At the front the water saturation jumps from 0.22 to 0.564, the fractional flow behind the front is 0.869, and the average saturation of the flushed zone is 0.616.

What it means. The chord from the point $(S_{wi}, 0)$ at 0.55 and at 0.58 is almost the same: near the tangent point its slope reaches a maximum, so we find the tangent point by several trials or graphically. The saturation at the front is not the same as the average in the flushed zone. Behind the front water keeps arriving, and the average saturation is higher by 0.052. The data in this example are assumed for illustration.

CalculatedDistribution of water saturation along the length of the element after 0.15 and 0.30 pore volumes of injection and at breakthrough (0.396), for the same data. The dashed line is the solution without a jump for 0.15 pore volume: there are three saturations in one section. The jump cuts off the “extra” part so that the areas to the left and to the right of it are equal (the water balance).

Average saturation behind the front

How much water is left in the formation before breakthrough matters more than the saturation at the front itself, because this water is the displaced oil. Until the front reaches the outlet, no water leaves the element, and all the injected water, $W_{inj}B_w$ reservoir cubic metres, is in the flushed zone of length $x_f$ : $W_{inj}B_w = \phi A x_f(\bar S_w - S_{wi})$ . The position of the front comes from the advance equation: $x_f = W_{inj}B_w f_w'(S_{wf})/(\phi A)$ . Substituting one into the other, we get:

Average saturation behind the front before breakthrough (Welge)

\bar S_w = S_{wi} + \frac{1}{f_w'(S_{wf})} = S_{wf} + \frac{1 - f_{wf}}{f_w'(S_{wf})}

where:

$\bar S_w$ average water saturation of the flushed zone, from the inlet to the front, —
$S_{wf},\ f_{wf},\ f_w'(S_{wf})$ saturation, fractional flow and slope of the curve at the front, —

Units: dimensionless

Holds when

Before breakthrough; incompressible phases; the initial water does not move.

The second form follows from the first one and the tangency condition. Graphically: extend the tangent to the line $f_w = 1$ , and the crossing point lies at $S_w = \bar S_w$ . The average saturation behind the front does not change until breakthrough: the flushed zone grows, but its profile only stretches.

Water breakthrough

Before breakthrough the element gives only oil, and its reservoir volume equals the volume of the injected water: the incompressible phases simply swap places. Breakthrough comes when the front reaches the outlet, $x_f = L$ . At this moment $W_{iD,BT} = 1/f_w'(S_{wf})$ pore volumes have been injected. The same number is the average rise in the water saturation of the element and so the oil production up to breakthrough in pore volumes (breakthrough time).

Injection, oil production and time to water breakthrough

W_{iD,BT} = N_{pD,BT} = \bar S_w - S_{wi} = \frac{1}{f_w'(S_{wf})},\qquad t_{BT} = \frac{W_{iD,BT}\,V_p}{q_i}

where:

$W_{iD,BT}$ water injection up to breakthrough, in pore volumes of the element, —
$N_{pD,BT}$ oil production up to breakthrough, in pore volumes, at reservoir conditions, —
$V_p$ pore volume of the element, $\phi A L$ , m³
$q_i$ injection rate at reservoir conditions, m³/d
$t_{BT}$ time from the start of injection to water breakthrough, d

Units: volumes in m³ at reservoir conditions, time in days

Holds when

The assumptions of the theory. Injection starts into an element with no free gas and with immobile water.
Oil is converted to standard cubic metres by dividing by $B_o$ , and the water injection is converted by dividing by $B_w$ .

The displacement efficiency at breakthrough, $E_{D,BT} = N_{pD,BT}/(1 - S_{wi})$ , is the share of the initial oil that has been displaced by the time the water arrives. It is less than the limiting value $(1 - S_{wi} - S_{or})/(1 - S_{wi})$ from the lesson Why water is injected: pressure maintenance and oil displacement, and equals it only for piston-like displacement.

Worked exampleWater breakthrough in an element of a line-drive pattern

Given (assumed data)

Element: distance between the rows 500 m, width 300 m, net thickness 12 m, porosity 0.20
Injection: 250 m³/d of water at standard conditions, $B_w = 1.02$
Oil: 3.0 mPa·s, $B_o = 1.25$ ; the relative permeability curves of the lesson Relative permeability
Front (the example above): $S_{wf}$ = 0.564, $f_{wf}$ = 0.869, $f_w'(S_{wf})$ = 2.525, $\bar S_w$ = 0.616

Solution

Pore volume: $V_p = 500\cdot300\cdot12\cdot0.20 = 360{,}000$ m³. Injection at reservoir conditions: $q_i = 250\cdot1.02 = 255$ m³/d. One pore volume is injected in $360{,}000/255 \approx 1412$ d.
Injection up to breakthrough: $W_{iD,BT} = 1/2.525 = 0.396$ pore volume. Time: $t_{BT} = 0.396\cdot1412 = 559$ d, or 1.53 years.
Oil production up to breakthrough: at reservoir conditions $0.396\cdot360{,}000 = 142.6$ thousand m³, at standard conditions $142.6/1.25 = 114$ thousand m³. Oil rate before breakthrough: $255/1.25 = 204$ m³/d.
Displacement efficiency at breakthrough: $E_{D,BT} = 0.396/(1 - 0.22) = 0.508$ , against the limiting value $0.53/0.78 = 0.679$ .
Front speed: $500/559 = 0.894$ m/d. Average speed of the liquid in the pores: $255/(300\cdot12\cdot0.20) = 0.354$ m/d. The ratio 2.53 is $f_w'(S_{wf})$ .
At the moment of breakthrough the fractional flow of water at the outlet jumps to 0.869. The water cut at the surface is $\dfrac{0.869/1.02}{0.869/1.02 + 0.131/1.25} = 0.890$ , and the oil rate falls to $0.131\cdot255/1.25 = 26.7$ m³/d.

Answer: Water will reach the production row after about 559 d (1.53 years). By that time 114 thousand m³ of oil will have been produced, with a displacement efficiency of 0.508.

What it means. The one-dimensional theory predicts a jump in the water cut from zero to 89%: behind the front oil and water flow together in the ratio set by the $f_w$ curve. In a real well the water cut rises more smoothly, because water arrives along different layers and streamlines (the paths that the water follows) at different times (Displacement in heterogeneous reservoirs). By breakthrough 51% of the initial oil of the element has been displaced, and up to the limit of 68% there is still a noticeable share left, which is produced together with water. The data in this example are assumed for illustration.

Production after breakthrough

After breakthrough, the saturation at the outlet $S_{we}$ gradually grows from $S_{wf}$ towards $1 - S_{or}$ : higher and higher saturations arrive at the outlet one after another from behind the front. The saturation $S_{we}$ arrives when $x_{S_{we}} = L$ , that is, after an injection of $W_{iD} = 1/f_w'(S_{we})$ pore volumes. The water balance of the element (injected minus produced) gives the average saturation. It is the same Welge formula, but for the point at the outlet:

Average saturation and oil production after breakthrough

\bar S_w = S_{we} + (1 - f_{we})\,W_{iD},\qquad N_{pD} = \bar S_w - S_{wi},\qquad W_{iD} = \frac{1}{f_w'(S_{we})}

where:

$S_{we},\ f_{we}$ water saturation at the outlet and the fractional flow of water at it, which is the share of water in the production at reservoir conditions, —
$W_{iD}$ cumulative water injection in pore volumes, —
$\bar S_w$ average water saturation of the whole element, —
$N_{pD}$ cumulative oil production in pore volumes (reservoir conditions), —

Units: dimensionless

Holds when

After breakthrough; the assumptions of the theory.
Graphically: the tangent to the $f_w$ curve at the point $S_{we}$ crosses the line $f_w = 1$ at $\bar S_w$ .

Oil production grows more and more slowly after breakthrough: each further portion of injection carries out more and more water. At constant injection the oil rate is proportional to $1 - f_{we}$ .

Calculation procedure after breakthrough

1
Choose the saturation at the outlet
several values of $S_{we}$ between $S_{wf}$ and $1 - S_{or}$ .
2
Fractional flow and slope
$f_{we}$ from the curve, $f_w'(S_{we})$ from the formula for the Corey model or graphically.
3
Injection
$W_{iD} = 1/f_w'(S_{we})$ ; time $t = W_{iD}V_p/q_i$ .
4
Production
$\bar S_w = S_{we} + (1 - f_{we})W_{iD}$ , $N_{pD} = \bar S_w - S_{wi}$ ; in cubic metres, $N_{pD}V_p/B_o$ .
5
Water cut
from $f_{we}$ through the formation volume factors; the oil rate is $(1 - f_{we})\,q_i/B_o$ .

Worked exampleThe element after breakthrough

Given (assumed data)

Element and fluids: as in the breakthrough example: $V_p$ = 360 thousand m³, $q_i$ = 255 m³/d, one pore volume in 1412 d
Saturations at the outlet: 0.60 and 0.65

Solution

$S_{we} = 0.60$ : $S_n = 0.38/0.53 = 0.7170$ and $f_{we} = 0.93851$ . Slope: $f_{we}(1 - f_{we})/0.53 = 0.1089$ , the bracket is $3/0.7170 + 2.5/(1 - 0.7170) = 13.018$ , and $f_w' = 1.418$ .
Injection: $W_{iD} = 1/1.418 = 0.7052$ pore volume, reached after $0.7052\cdot1412 = 996$ d (2.7 years). Average saturation: $0.60 + (1 - 0.93851)\cdot0.7052 = 0.643$ . Production: $N_{pD} = 0.643 - 0.22 = 0.423$ , and the displacement efficiency is $0.423/0.78 = 0.542$ .
$S_{we} = 0.65$ : $S_n = 0.8113$ , $f_{we} = 0.98386$ , $f_w' = 0.02996\cdot16.946 = 0.5077$ . Then $W_{iD} = 1.970$ (after 7.6 years), $\bar S_w = 0.65 + (1 - 0.98386)\cdot1.970 = 0.682$ , $N_{pD} = 0.462$ and $E_D = 0.592$ .
Water cut at the surface: 94.9% at 0.60 and 98.7% at 0.65. If the economic limit is 98% water, the fractional flow of water at the outlet must reach 0.976. This is a saturation of about 0.636, an injection of 1.44 pore volume (5.6 years) and a displacement efficiency of 0.58.

Answer: In the first 2.7 years after the start of injection the element gives 0.423 pore volume of oil (

E_D

= 0.542). Up to a water cut of 98%, it gives about 0.45 pore volume (

E_D

≈ 0.58) in 5.6 years.

What it means. The gain in production after breakthrough is expensive: between outlet saturations of 0.60 and 0.65, production grew by 0.039 pore volume while injection grew by 1.26. The one-dimensional theory reaches the limiting displacement efficiency of 0.679 only asymptotically (it comes closer and closer but never gets there): the last few per cent of the mobile oil need tens of pore volumes of water (the table below). So in practice operators stop a waterflood at the economic limit of the water cut, not when the residual oil saturation is reached. The data in this example are assumed for illustration.

The element after breakthrough: injection, production and water cut

$S_{we}$: 0.57
$W_{iD}$: 0.431
$N_{pD}$ ; $E_D$: 0.400; 0.513
Water cut: 90.3%
Time, years: 1.7

$S_{we}$: 0.60
$W_{iD}$: 0.705
$N_{pD}$ ; $E_D$: 0.423; 0.542
Water cut: 94.9%
Time, years: 2.7

$S_{we}$: 0.62
$W_{iD}$: 1.03
$N_{pD}$ ; $E_D$: 0.439; 0.563
Water cut: 96.9%
Time, years: 4.0

$S_{we}$: 0.64
$W_{iD}$: 1.57
$N_{pD}$ ; $E_D$: 0.454; 0.582
Water cut: 98.2%
Time, years: 6.1

$S_{we}$: 0.65
$W_{iD}$: 1.97
$N_{pD}$ ; $E_D$: 0.462; 0.592
Water cut: 98.7%
Time, years: 7.6

$S_{we}$: 0.67
$W_{iD}$: 3.27
$N_{pD}$ ; $E_D$: 0.477; 0.612
Water cut: 99.3%
Time, years: 12.6

$S_{we}$: 0.70
$W_{iD}$: 8.56
$N_{pD}$ ; $E_D$: 0.498; 0.638
Water cut: 99.8%
Time, years: 33.1

$S_{we}$	$W_{iD}$	$N_{pD}$ ; $E_D$	Water cut	Time, years
0.57	0.431	0.400; 0.513	90.3%	1.7
0.60	0.705	0.423; 0.542	94.9%	2.7
0.62	1.03	0.439; 0.563	96.9%	4.0
0.64	1.57	0.454; 0.582	98.2%	6.1
0.65	1.97	0.462; 0.592	98.7%	7.6
0.67	3.27	0.477; 0.612	99.3%	12.6
0.70	8.56	0.498; 0.638	99.8%	33.1

Calculated with the Welge formulas for the data of the examples. Production is in pore volumes of the element (reservoir conditions), the water cut is at the surface for $B_o = 1.25$ , $B_w = 1.02$ , and the time is for an injection of 255 m³/d at reservoir conditions. Breakthrough comes at 0.396 pore volume. The data are assumed for illustration.

CalculatedOil production (in pore volumes) and the water cut of the production against the water injected, for the element of the examples. Before breakthrough, production equals injection (the straight line

N_{pD} = W_{iD}

). After breakthrough the water cut jumps to almost 90%, and production grows more and more slowly, approaching the mobile oil volume of 0.53.

Oil viscosity and the picture of displacement

The shape of the $f_w$ curve sets the whole picture of displacement (viscosity and the shape of the curve). The more viscous the oil, the further to the left lie the inflection point and the tangent point. The saturation at the front is lower, the flushed zone at breakthrough is “poorer” in water, breakthrough comes earlier, and after it more oil has to be produced with water. With a very favourable mobility ratio, the tangent reaches almost to $1 - S_{or}$ : the saturation at the front is close to the limiting one, and the displacement is close to piston-like.

One rock, three oils: the front, breakthrough and the limit of 98% water

$\mu_o$ , mPa·s: 1.2
$M'$: 0.71
$S_{wf}$ ; $f_{wf}$: 0.617; 0.905
At breakthrough: $W_{iD}$ ; $E_D$: 0.439; 0.563
At 98% water: $W_{iD}$ ; $E_D$: 1.17; 0.606

$\mu_o$ , mPa·s: 3.0
$M'$: 1.76
$S_{wf}$ ; $f_{wf}$: 0.564; 0.869
At breakthrough: $W_{iD}$ ; $E_D$: 0.396; 0.508
At 98% water: $W_{iD}$ ; $E_D$: 1.44; 0.579

$\mu_o$ , mPa·s: 30
$M'$: 17.6
$S_{wf}$ ; $f_{wf}$: 0.427; 0.782
At breakthrough: $W_{iD}$ ; $E_D$: 0.264; 0.339
At 98% water: $W_{iD}$ ; $E_D$: 1.97; 0.473

$\mu_o$ , mPa·s	$M'$	$S_{wf}$ ; $f_{wf}$	At breakthrough: $W_{iD}$ ; $E_D$	At 98% water: $W_{iD}$ ; $E_D$
1.2	0.71	0.617; 0.905	0.439; 0.563	1.17; 0.606
3.0	1.76	0.564; 0.869	0.396; 0.508	1.44; 0.579
30	17.6	0.427; 0.782	0.264; 0.339	1.97; 0.473

The relative permeability curves of the lesson Relative permeability, water 0.5 mPa·s, a horizontal formation. The water cut of 98% is at the surface, for $B_o = 1.25$ and $B_w = 1.02$ for all three oils. The limiting displacement efficiency for this rock is 0.679. The data are assumed for illustration.

With oil of 1.2 mPa·s, more than nine tenths of the oil that is produced before the water cut reaches 98% comes out before breakthrough. With 3.0 mPa·s it is about 88%, and with 30 mPa·s only about 70%. The rest has to be produced together with large volumes of water. And even after almost two pore volumes have been pumped through, the displacement efficiency of the viscous oil stays below 0.5. This is why viscous oils are developed at a high water cut, and why operators thicken the injected water or lower the oil viscosity by heating (mobility ratio).

The accumulation of the waterflooding lesson: what displacement gives over time

The lesson Why water is injected: pressure maintenance and oil displacement estimated the recovery factor of the accumulation as the product $E_D E_A E_V$ , where it took the displacement efficiency at its limit, from the end-point saturations: 0.667. The Buckley–Leverett theory shows how fast this limit is reached in the flushed part of the formation.

Worked exampleDisplacement efficiency of the accumulation over time

Given (assumed data)

Relative permeability curves of the accumulation: the end points of the lesson Why water is injected: pressure maintenance and oil displacement: $S_{wi} = 0.25$ , $S_{or} = 0.25$ , $k_{ro}' = 0.80$ , $k_{rw}' = 0.20$ ; the Corey exponents are taken as for the curves of the lesson Relative permeability: $n_o = 2.5$ , $n_w = 3$
Fluids at 24 MPa: oil 1.10 mPa·s, water 0.40 mPa·s ( $M'$ ≈ 0.69)
Injection and pore volume of the accumulation: 2100 m³/d at standard conditions ( $B_w = 1.02$ ), $V_p$ = 16.0 million m³
Sweep (the waterflooding lesson): $E_A = 0.85$ , $E_V = 0.70$

Solution

Tangent from the point $(0.25;\ 0)$ : the front is at $S_{wf} \approx$ 0.626, with $f_{wf} = 0.906$ and slope 2.407. Average saturation behind the front: $0.25 + 1/2.407 = 0.665$ . Displacement efficiency at breakthrough: $(0.665 - 0.25)/0.75 \approx 0.55$ .
After one pore volume has been injected, the saturation at the outlet is 0.667 (the slope of the curve there equals 1) and $f_{we} = 0.9724$ . Then $\bar S_w = 0.667 + (1 - 0.9724)\cdot1 = 0.695$ and $E_D = (0.695 - 0.25)/0.75 \approx 0.59$ .
After two pore volumes: $S_{we}$ = 0.690, $f_{we} = 0.9896$ , $\bar S_w = 0.690 + (1 - 0.9896)\cdot2 = 0.711$ and $E_D \approx 0.61$ .
Time: one pore volume of the accumulation is $16.0\cdot10^6/(2100\cdot1.02) = 7470$ d, about 20.5 years.
Water does not pass through the whole pore volume, only through the swept part, $E_AE_V = 0.595$ . So in 20.5 years the swept part receives $1/0.595 = 1.68$ of its own pore volume, and $E_D$ there is ≈ 0.61. The recovery factor is $0.61\cdot0.85\cdot0.70 \approx 0.36$ , against 0.40 with the limiting $E_D$ .

Answer: The displacement efficiency of the accumulation grows from 0.55 at breakthrough to 0.59 after one pore volume and 0.61 after two pore volumes of injection. The limit of 0.667 is reached only with a very large volume pumped through.

What it means. The oil of this accumulation has a low viscosity, and the displacement is stable ( $M' < 1$ ). And yet in the first two decades of injection the flushed part of the formation gives about 0.61 of the initial oil, not 0.667. And 0.61 is an upper estimate: the calculation treats the swept part as flushed equally and from the very start, while areas that the water reached later are flushed less. An estimate of the recovery factor (RF) with the limiting $E_D$ is an upper bound. We build a realistic forecast over time. In it the one-dimensional calculation sets $E_D$ , and models of a heterogeneous formation give the sweep, which also grows with injection (Displacement in heterogeneous reservoirs). The data in this example are assumed for illustration; the Corey exponents of the accumulation are an assumption.

The assumptions and the real formation

The Buckley–Leverett theory is exact for its own model: a uniform linear element with diffuse flow. It describes core tests and thin uniform formations well. In other cases it gives the displacement efficiency, and everything connected with sweep needs other methods.

What breaks the assumptions and where it leads

Violation: Layering, high-permeability layers
What happens: water goes along the best layer, breakthrough comes earlier, and the water cut grows smoothly
Where it is covered: Displacement in heterogeneous reservoirs

Violation: A thick formation with good vertical permeability
What happens: water sinks to the base, and the saturation is not the same over the thickness
Where it is covered: Segregated flow and gravity displacement

Violation: An areal pattern of wells
What happens: streamlines of different length, and the areal sweep is less than one
Where it is covered: areal sweep

Violation: Free gas in the formation before injection
What happens: water first fills the volume of the free gas, and the response of production to injection is delayed
Where it is covered: when to start injection

Violation: Gas injection instead of water
What happens: a very unfavourable mobility ratio, and gravity matters more
Where it is covered: Displacement of oil by gas

Violation	What happens	Where it is covered
Layering, high-permeability layers	water goes along the best layer, breakthrough comes earlier, and the water cut grows smoothly	Displacement in heterogeneous reservoirs
A thick formation with good vertical permeability	water sinks to the base, and the saturation is not the same over the thickness	Segregated flow and gravity displacement
An areal pattern of wells	streamlines of different length, and the areal sweep is less than one	areal sweep
Free gas in the formation before injection	water first fills the volume of the free gas, and the response of production to injection is delayed	when to start injection
Gas injection instead of water	a very unfavourable mobility ratio, and gravity matters more	Displacement of oil by gas

Terms and symbols

Displacement front (saturation jump): The cross-section where the water saturation drops abruptly from $S_{wf}$ to the initial value. Only oil flows ahead of it.
Welge tangent: The tangent to the $f_w$ curve from the point of the initial saturation. The tangent point gives the saturation at the front, and the crossing with the line $f_w = 1$ gives the average saturation behind the front.
$W_{iD}$ , $N_{pD}$: Cumulative water injection and oil production in pore volumes (reservoir conditions).
Water breakthrough: The moment when the front reaches the outlet of the element. Before it, oil production equals water injection in reservoir volumes.
$S_{we}$: The saturation at the outlet of the element after breakthrough. The fractional flow of water at it is the share of water in the production.
Diffuse flow: Flow in which the saturation is the same over the thickness of the formation. It is the condition for using the one-dimensional theory.

Common mistakes and misconceptions

Wrong

“The water front moves at the Darcy velocity $q/A$ (the flow rate per unit area of rock) or at the pore velocity $q/\phi A$ (the average speed of the liquid in the pores).”

In fact

The front moves at the speed $(q_t/\phi A)\,f_w'(S_{wf})$ , which in the example of this lesson is 2.5 times faster than the average speed of the liquid in the pores. If we calculate with $q/\phi A$ , breakthrough comes out 2.5 times later.

Wrong

“By breakthrough the formation has been flushed down to the residual oil saturation.”

In fact

Behind the front the saturation only gradually grows towards $1 - S_{or}$ . In the example of this lesson the average is 0.616 instead of 0.75. The displacement efficiency at breakthrough is 0.51, against the limiting 0.68.

Wrong

“You draw the tangent from the origin.”

In fact

You draw it from the point of the initial state of the formation: $(S_{wi}, 0)$ if the initial water does not move, or $(S_{w0}, f_w(S_{w0}))$ if the water is mobile from the start.

Wrong

“The one-dimensional calculation gave $E_D = 0.58$ at a water cut of 98%, so the recovery factor of the accumulation will be 0.58.”

In fact

This is the displacement efficiency of the flushed part. The recovery factor equals it multiplied by the areal and vertical sweep, which for real formations are noticeably less than one (Why water is injected: pressure maintenance and oil displacement).

In depth: advanced detail

Derivation of the saturation advance equation

Saturation is a function of two variables, $S_w(x, t)$ . Its total differential (the small change of $S_w$ when both $x$ and $t$ change) is $\dd S_w = (\partial S_w/\partial x)\,\dd x + (\partial S_w/\partial t)\,\dd t$ . Along a line of constant saturation, $\dd S_w = 0$ , so $\partial S_w/\partial t = -(\partial S_w/\partial x)\,(\dd x/\dd t)_{S_w}$ .

In the water balance $\phi A\,\partial S_w/\partial t = -q_t\,\partial f_w/\partial x$ write $\partial f_w/\partial x = f_w'\,\partial S_w/\partial x$ . Substituting the expression for $\partial S_w/\partial t$ , we get $\phi A\,(\dd x/\dd t)_{S_w}\,\partial S_w/\partial x = q_t f_w'\,\partial S_w/\partial x$ , and after cancelling, the advance equation. The lines of constant saturation on the plane “distance and time” are straight lines with the slope (the speed) $q_t f_w'/(\phi A)$ . They are called characteristics, and the method itself is the method of characteristics.

Why a tangent: equal areas

The multi-valued profile that the characteristics give without a jump contains the right amount of water: the water balance holds for any profile obtained from the advance equation. The jump must be placed so that this amount does not change. If the “loop” is cut off by a vertical line, the area cut off to the right of it (water that would have gone ahead) must equal the area added on the left (water that stays behind the jump). It can be shown that this equal-areas condition is the same as the tangency condition.

Why not another jump? A jump higher than $S_{wf}$ , for example straight up to $1 - S_{or}$ , would move ahead at the speed of its chord, while the saturation behind it, whose slope on the curve is smaller, would lag behind. Intermediate saturations would enter the gap between them, and the jump itself would become smaller. A jump lower than $S_{wf}$ , on the contrary, would be caught up by the saturations behind it, whose slope on the curve is larger than the slope of the chord, and it would grow. Only a jump behind which the saturation moves at exactly the speed of the jump is stable, and the tangent gives exactly that jump.

An inclined formation and mobile initial water

The construction does not change if the $f_w$ curve includes gravity (dip of the formation). We apply the advance equation, the tangent and the Welge formulas to the curve with the gravity correction. For the examples of this lesson, injection up the dip at a gravity number of about 0.47 raises the saturation at the front from 0.564 to 0.576 and the displacement efficiency at breakthrough from 0.508 to 0.524. Injection down the dip lowers them to 0.553 and 0.490.

If the initial water saturation $S_{w0}$ is above the irreducible one and the water is mobile (for example, the element lies in the transition zone, above the oil–water contact, where the water saturation changes gradually), we draw the tangent from the point $(S_{w0}, f_w(S_{w0}))$ . The wells give some water from the very start, the jump is lower, and production up to breakthrough is smaller. The formula for the average saturation is then $\bar S_w = S_{wf} + (1 - f_{wf})/f_w'(S_{wf})$ (the second form). The first form, through $S_{wi}$ , is correct only for immobile initial water.

An unsteady-state core test

The Welge formulas also work in reverse. In an unsteady-state test (a test in which the saturations keep changing; how relative permeability is measured), we pump water through a sample at a constant rate and record the oil production and the pressure drop over time. From the slope of the curve of production against injection after breakthrough, we find the fractional flow of water at the outlet. From the Welge formula we find the saturation at the outlet, and from the pressure drop we find the mobilities of the phases. This is how relative permeabilities are obtained without waiting for a steady state. The method requires the test to match the assumptions of the theory. In particular, the capillary smearing must be small compared with the length of the sample (capillary length). Preparing such data for a model is the subject of Petrophysical input for simulation: relative permeability and capillary pressure.

Connections to other topics

Fractional flow function and mobility ratio — the fractional flow curve on which the whole solution is built.
Relative permeability — relative permeability curves and end points.
Why water is injected: pressure maintenance and oil displacement — displacement and sweep efficiencies, the mobility ratio.
Two-phase flow equations — the saturation equation, of which the Buckley–Leverett equation is a special case.
IMPES and fully implicit schemes — the numerical solution of the same equation, and the smearing of the front in a model.
Displacement in heterogeneous reservoirs — displacement in a layered formation.
Segregated flow and gravity displacement — displacement with the phases separated over the thickness.
Numerical dispersion — why a model smears the saturation jump.

Check yourself

1
How do we find the water saturation at the displacement front from the curve $f_w(S_w)$ when the initial water does not move?
1. A.With a tangent to the curve from the point $(S_{wi}, 0)$
2. B.From the inflection point of the curve
3. C.With a tangent from the origin
4. D.From the point where $f_w = 0.5$
Show the answer
Correct answer: A. With a tangent to the curve from the point $(S_{wi}, 0)$
The saturations just behind the front must move at the speed of the front: the slope of the curve at this point equals the slope of the chord from $(S_{wi}, 0)$ . The inflection point only marks the fastest saturations of the solution without a jump.
2
A tangent from the point $(0.18;\ 0)$ touches the $f_w$ curve at $S_w = 0.58$ , where $f_w = 0.80$ . Find the average saturation behind the front, the injection up to breakthrough and the displacement efficiency at breakthrough.
Show the answer
The slope of the tangent is $0.80/(0.58 - 0.18) = 2.0$ . $\bar S_w = 0.18 + 1/2.0 = 0.68$ . The injection up to breakthrough is $1/2.0 = 0.50$ pore volume, and the same amount of oil has been produced. $E_{D,BT} = 0.50/(1 - 0.18) \approx 0.61$ . Check with the second form: $0.58 + (1 - 0.80)/2.0 = 0.68$ , which agrees.
3
The pore volume of an element is 2.0 million m³, the injection is 800 m³/d of water at standard conditions, $B_w = 1.03$ , and by calculation breakthrough comes after 0.35 pore volume. When will the water arrive?
Show the answer
Injection at reservoir conditions: $800\cdot1.03 = 824$ m³/d. $t_{BT} = 0.35\cdot2.0\cdot10^6/824 \approx 850$ d, about 2.3 years. If we forget $B_w$ , the error is about 3%. If we assume that the front moves at the average speed of the liquid in the pores, so that breakthrough comes after one pore volume, the error is almost a factor of three.
4
What happens when we move to a more viscous oil in the same rock?
1. A.The saturation at the front grows, and breakthrough comes later
2. B.The saturation at the front falls, breakthrough comes earlier, and more oil is produced after breakthrough
3. C.Only the breakthrough time changes, but not the saturation at the front
Show the answer
Correct answer: B. The saturation at the front falls, breakthrough comes earlier, and more oil is produced after breakthrough
The $f_w$ curve shifts to the left, and so does the tangent point. In the table of the lesson, with oil of 30 mPa·s the saturation at the front is 0.43 against 0.56 at 3 mPa·s, the injection up to breakthrough is 0.26 against 0.40 pore volume, and to reach 98% water we must pump 1.97 pore volumes against 1.44, and displace less oil.
5
Why is the average saturation behind the front higher than the saturation at the front itself, and why does it not change before breakthrough?
Show the answer
Behind the front the saturation grows towards the injection side: more water has passed there, and the saturations up to $1 - S_{or}$ move more slowly than the front. Each saturation moves at a constant speed, so the profile behind the front only stretches in proportion to the injection and keeps its shape, and its average value $S_{wi} + 1/f_w'(S_{wf})$ is constant.
6
A one-dimensional calculation gave a displacement efficiency of 0.58 at a water cut of 98%. Can we take the recovery factor of the accumulation to be 0.58?
1. A.No: the recovery factor equals this coefficient multiplied by the areal and vertical sweep
2. B.Yes, that is the recovery factor
3. C.No: the recovery factor is always larger than the displacement efficiency
Show the answer
Correct answer: A. No: the recovery factor equals this coefficient multiplied by the areal and vertical sweep
The Buckley–Leverett theory describes the flushed part of the formation. In a real accumulation the areal and vertical sweeps are less than one, and the recovery factor is the product of the displacement and sweep efficiencies.

Updated: 14 October 2026