Fractional flow function and mobility ratio

Quick refresher

$f_w$ is the share of water in the flow, not in the pores. At the same saturation, water can carry 20% or 85% of the flow rate. What decides it is the ratio of the mobilities, $k_{rw}/\mu_w$ and $k_{ro}/\mu_o$ .
In a horizontal formation without capillary forces, $f_w = \lambda_w/\lambda_t$ and it does not depend on the injection rate. Gravity and capillary pressure add corrections that are inversely proportional to the total velocity $u_t$ (the flow rate of both phases per unit area of the cross-section).
The more viscous the oil, the further to the left the curve lies. Water starts to dominate the flow at a lower saturation, and most of the oil is produced at a high water cut. Half of the flow is water where $k_{rw}/k_{ro} = \mu_w/\mu_o$ .
In an oil-wet rock (one whose surface prefers oil), water flows through the large pores and its relative permeability is high. The curve shifts to the left, and displacement is worse than in a water-wet rock.
Water is heavier than oil. When it moves up the dip (upwards along the slope of the formation), gravity holds it back, and the curve shifts to the right. When it moves down the dip, the curve shifts to the left. The gravity number $N_g$ measures the effect, and it grows when the rate is reduced.
In a water-wet rock the capillary term increases $f_w$ in the front zone and smears the front over a length of the order of $\lambda_o\Delta p_c/u_t$ . In a reservoir this is from tens of centimetres to metres. In a core test it is comparable to the length of the sample.

Key relationships

f_w = \dfrac{q_w}{q_w + q_o} = \dfrac{1 + \dfrac{k k_{ro}}{\mu_o u_t}\left(\dfrac{\partial p_c}{\partial x} - \Delta\rho\,g\sin\alpha\right)}{1 + \dfrac{k_{ro}}{k_{rw}}\cdot\dfrac{\mu_w}{\mu_o}}

general form for one-dimensional flow; $\alpha > 0$ if the water moves up the dip

f_w = \dfrac{\lambda_w}{\lambda_t}\left(1 - N_g\,\dfrac{k_{ro}}{k_{ro}'}\right),\qquad N_g = \dfrac{k\,k_{ro}'\,\Delta\rho\,g\sin\alpha}{\mu_o u_t}

without the capillary term: the viscous part and the correction for gravity

f_w = 0.5\ \Longleftrightarrow\ \dfrac{k_{rw}}{k_{ro}} = \dfrac{\mu_w}{\mu_o}

horizontal formation: equal flows of water and oil

\ell_c = \dfrac{k\,k_{ro}\,\Delta p_c}{\mu_o u_t}

capillary length: the scale of the smearing of the front

Common pitfalls

Mixing up the fractional flow of water in the reservoir with the water cut of the production at the surface. They are linked by the formation volume factors of oil and water.
Thinking that the injection rate changes the $f_w$ curve of a horizontal formation. Without gravity and capillary forces, it cancels out.
Carrying the conclusion “gravity hardly matters” from one-dimensional flow along the dip over to a thick formation. Across a thick formation water and oil separate, and the one-dimensional model stops working.
Comparing the curves of different rocks only by water saturation, without looking at their ranges of mobile saturation $1 - S_{wi} - S_{or}$ (the part of the pore space in which both fluids can move).
Judging how much oil is left near a well from the water cut alone. The same water cut with viscous oil means much more oil left.

After this lesson you can

derive the fractional flow equation from Darcy's law for two phases (the flow rate of each phase is proportional to its pressure gradient) with gravity and capillary pressure, and explain what each term means
plot the curve $f_w(S_w)$ from relative permeability curves and viscosities, and link its shape to the mobility ratio (how much more easily water moves than oil)
estimate how the oil viscosity, the wettability of the rock (which fluid its surface prefers), the dip of the formation (its slope) and the injection rate shift the curve, and what this means for displacement
calculate the gravity number (how strong gravity is compared with the viscous forces that push the fluids along) and the capillary length (the distance over which capillary forces smear the front), and decide whether gravity and capillary forces can be neglected
use the curve as a tool: convert the fractional flow into water cut, find the saturation at a well from its water cut, and compare development options

What the fractional flow of water shows

Where water displaces oil, both phases move in the pores. Each flows under the push of its own pressure gradient (the change of pressure per metre), with its own mobility $\lambda = k\,k_r/\mu$ : the permeability of the rock times the relative permeability of the phase, divided by its viscosity (Relative permeability). So the flow rate is shared between the phases not in proportion to their saturations, but in proportion to their mobilities.

The fractional flow of water $f_w = q_w/(q_w + q_o)$ is the result of this sharing at a given point of the reservoir. We take the flow rates at reservoir conditions. At the exit from the reservoir, at a production well, $f_w$ sets the composition of the well's production. Inside the reservoir it sets how fast water is carried forward and how the displacement front moves.

Three quantities that are often confused

Quantity: Water saturation $S_w$
What it shows: the share of the pore volume occupied by water
What it depends on: the history of saturation: how much water has come in and how much oil has gone out

Quantity: Fractional flow of water $f_w$
What it shows: the share of water in the reservoir flow rate through a cross-section
What it depends on: $S_w$ , the relative permeability curves and the viscosities; in an inclined formation also the rate and the dip

Quantity: Water cut of the production
What it shows: the share of water in the produced liquid at standard conditions
What it depends on: $f_w$ at the bottomhole and the formation volume factors $B_o$ , $B_w$

Quantity	What it shows	What it depends on
Water saturation $S_w$	the share of the pore volume occupied by water	the history of saturation: how much water has come in and how much oil has gone out
Fractional flow of water $f_w$	the share of water in the reservoir flow rate through a cross-section	$S_w$ , the relative permeability curves and the viscosities; in an inclined formation also the rate and the dip
Water cut of the production	the share of water in the produced liquid at standard conditions	$f_w$ at the bottomhole and the formation volume factors $B_o$ , $B_w$

Water cut is the share of water in the production. Its link with $f_w$ is derived in the lesson on relative permeability.

The fractional flow equation with gravity and capillary pressure

Consider one-dimensional flow along an axis $x$ from the injection side to the production side. The axis is inclined to the horizontal at an angle $\alpha$ . The fluids and the rock are incompressible (their volume does not change with pressure), so the total velocity $u_t = u_o + u_w = q_t/A$ is the same in every cross-section. We take the saturation to be the same across the section. This is called diffuse (spread-out) flow. For each phase we write Darcy's law with its pressure and its weight (two-phase Darcy's law), express the pressure gradients, subtract one equation from the other, and use the fact that the difference of the phase pressures is the capillary pressure $p_c = p_o - p_w$ . The derivation is in the advanced section. The result is:

Fractional flow of water with gravity and capillary pressure

f_w = \frac{1 + \dfrac{k\,k_{ro}}{\mu_o\,u_t}\left(\dfrac{\partial p_c}{\partial x} - \Delta\rho\,g\sin\alpha\right)}{1 + \dfrac{k_{ro}}{k_{rw}}\cdot\dfrac{\mu_w}{\mu_o}}

where:

$f_w$ share of water in the total flow at reservoir conditions, —
$k$ absolute permeability of the rock, m²
$k_{ro},\ k_{rw}$ relative permeabilities of oil and water at the current water saturation, —
$\mu_o,\ \mu_w$ viscosities of oil and water at reservoir conditions, Pa·s
$u_t$ total velocity $q_t/A$ : the flow rate of both phases per unit area of the cross-section, m/s
$p_c$ capillary pressure $p_o - p_w$ , Pa
$x$ coordinate along the flow, from the injection side to the production side, m
$\Delta\rho$ difference between the densities of water and oil, $\rho_w - \rho_o$ , at reservoir conditions, kg/m³
$g$ acceleration of gravity, 9.81, m/s²
$\alpha$ angle of the axis $x$ to the horizontal: positive if the water moves up the dip, negative if it moves down the dip, rad

Units: SI (any consistent system of units)

Holds when

One-dimensional linear flow, with the same saturation across the section (diffuse flow).
The rock and the fluids are incompressible, so $u_t$ does not change along the flow. There is no free gas.
The phases do not mix. The relative permeabilities and the capillary pressure depend only on the saturation and on the direction of its change (for the waterflooding of a water-wet formation, the imbibition curves, which describe water soaking into the rock).
Darcy's law holds for each phase: the flow is laminar (smooth, not turbulent).

The denominator is the viscous part: how the flow would be shared in a horizontal formation without capillary forces. The numerator contains two corrections. The capillary one depends on how the saturation changes along the flow. The gravity one depends on the dip and on the density difference. Both corrections are divided by $u_t$ , so they are stronger the slower the injection.

In other unit systems (1)

Metric oilfield units

f_w = \frac{1 + \dfrac{0.0853\,k\,k_{ro}\,A}{q_t\,\mu_o}\left(\dfrac{\partial p_c}{\partial x} - 0.00981\,\Delta\rho\sin\alpha\right)}{1 + \dfrac{k_{ro}}{k_{rw}}\cdot\dfrac{\mu_w}{\mu_o}}

$k$ in mD, $A$ in m², $q_t$ in m³/d at reservoir conditions, $\mu$ in mPa·s, $\partial p_c/\partial x$ in MPa/m, $\Delta\rho$ in g/cm³.

$0.0853$ is the constant of Darcy's law in metric oilfield units (Rock permeability). $0.00981 = 9.81\cdot1000/10^6$ converts g/cm³ to kg/m³ and Pa to MPa, as in the lesson Capillary pressure and the transition zone.

It is convenient to write the same formula with the mobilities $\lambda_o = k\,k_{ro}/\mu_o$ , $\lambda_w = k\,k_{rw}/\mu_w$ and the total mobility $\lambda_t = \lambda_o + \lambda_w$ . Then we see that the viscous part is the ratio $\lambda_w/\lambda_t$ , and the corrections are multiplied by the factor $\lambda_o\lambda_w/\lambda_t$ , which is zero where only one phase is mobile:

Fractional flow of water through the phase mobilities

f_w = \frac{\lambda_w}{\lambda_t} + \frac{\lambda_o\lambda_w}{\lambda_t}\cdot\frac{1}{u_t}\left(\frac{\partial p_c}{\partial x} - \Delta\rho\,g\sin\alpha\right)

where:

$\lambda_o,\ \lambda_w$ mobilities of oil and water, $k\,k_{ro}/\mu_o$ and $k\,k_{rw}/\mu_w$ , m²/(Pa·s)
$\lambda_t$ total mobility $\lambda_o + \lambda_w$ , m²/(Pa·s)
$u_t,\ p_c,\ \Delta\rho,\ g,\ \alpha$ as in the general form above, SI

Units: SI

Holds when

The same as for the general form.

This is the same expression as the water flux through the total velocity in the lesson Two-phase flow equations, divided by $u_t$ . There, the direction of gravity is set by the gradient of depth; here, by the angle of the flow axis. In a reservoir simulator, the fractional flow without the corrections, $\lambda_w/\lambda_t$ , carries the saturation, and the corrections are calculated as separate terms.

Viscosity of oil and water: the shape of the curve

In a horizontal formation without capillary forces, only the viscous part $f_w = \lambda_w/\lambda_t$ remains: the part set by the denominator of the general formula. The curve depends only on the ratio of the relative permeabilities (a property of the rock) and the ratio of the viscosities (a property of the fluids). Viscous oil flows worse, so at the same saturation water carries the larger part of the flow, and the curve shifts to the left and up.

A convenient reference point is the saturation at which water and oil flow in equal amounts. From the formula, $f_w = 0.5$ when $k_{rw}/k_{ro} = \mu_w/\mu_o$ : the ratio of the relative permeabilities must make up for the ratio of the viscosities. If the oil is 60 times more viscous than the water, it is enough for the relative permeability of water to be 60 times smaller than that of oil, and this happens soon after the water becomes mobile.

CalculatedFractional flow of water for the relative permeability curves of the lesson Relative permeability (a water-wet rock), with a water viscosity of 0.5 mPa·s and three oil viscosities. The dots are the saturations at which water and oil flow in equal amounts.

Worked exampleOne rock, three oils

Given (assumed data)

Relative permeability curves (lesson Relative permeability): $S_{wi} = 0.22$ , $S_{or} = 0.25$ , $k_{ro}' = 0.85$ , $k_{rw}' = 0.25$ , $n_o = 2.5$ , $n_w = 3$
Water viscosity $\mu_w$: 0.5 mPa·s
Oil viscosity $\mu_o$: 1.2, 3.0 and 30 mPa·s
Water saturation for comparison: $S_w = 0.45$

Solution

Mobility ratio at the end points, $M' = (0.25/0.5)/(0.85/\mu_o)$ : 0.71, 1.76 and 17.6.
Relative permeabilities at $S_w = 0.45$ : $S_n = (0.45 - 0.22)/0.53 = 0.4340$ (the normalised saturation, scaled to run from 0 to 1 over the range of mobile saturation); $k_{rw} = 0.25\cdot0.4340^3 = 0.02044$ ; $k_{ro} = 0.85\cdot(1 - 0.4340)^{2.5} = 0.2049$ . The ratio is $k_{ro}/k_{rw} = 0.2049/0.02044 = 10.02$ .
Fractional flow: $f_w = 1/(1 + 10.02\cdot\mu_w/\mu_o)$ . For 1.2 mPa·s it is $1/(1 + 10.02\cdot0.4167) = 0.193$ . For 3.0 it is $1/(1 + 10.02\cdot0.1667) = 0.374$ . For 30 it is $1/(1 + 10.02\cdot0.0167) = 0.857$ .
We find the saturation at which $f_w = 0.5$ , that is $k_{rw}/k_{ro} = \mu_w/\mu_o$ , by trial on the curves: 0.518, 0.474 and 0.373. A check for 3.0 mPa·s: at $S_w$ = 0.474, $k_{rw} = 0.0275$ and $k_{ro} = 0.1663$ , and their ratio is 0.165. Within rounding, this is $0.5/3.0 \approx 0.167$ .
A fractional flow of 0.98, a level close to the limit of economic production, is reached at $S_w$ of 0.670, 0.643 and 0.550.

Answer: At

S_w = 0.45

the fractional flow of water is 19%, 37% and 86% for oil with a viscosity of 1.2, 3.0 and 30 mPa·s. Half of the flow becomes water at

S_w

of 0.518, 0.474 and 0.373.

What it means. The rock is the same, but the picture is different. With low-viscosity oil ( $M' < 1$ ), water dominates the flow only after more than half of the mobile oil has been displaced: the point $f_w = 0.5$ lies at 56% of the range of mobile saturation. With oil of 30 mPa·s it lies already at 29%. After that the oil comes together with a large amount of water, and the fractional flow of 0.98 is reached long before the saturation $1 - S_{or} = 0.75$ . Hence the rule: the larger $M'$ , the larger the part of the oil that is produced after water breakthrough, and the more water has to be pumped through. The data in this example are assumed for illustration.

Late rise: with very low-viscosity oil or thickened water, when $M'$ is much less than one: the fractional flow stays small almost to the end of the range of mobile saturation and then rises sharply. The displacement is close to piston-like (the water pushes the oil ahead of it like a piston): almost all the mobile oil is produced before the water arrives (piston-like displacement).
S-shaped curve: the typical case, and all three curves of the figure belong to it. At first the fractional flow grows faster and faster. After the inflection point (where the curve stops bending upwards, with its slope growing, and starts bending downwards, with its slope falling) it approaches one more and more slowly. A jump of saturation forms at the front, and behind it water and oil flow together.
Early rise: the more viscous the oil, the closer the inflection is to $S_{wi}$ (about 0.36 for the 30 mPa·s curve). The fractional flow rises steeply almost as soon as the water becomes mobile. Water reaches the wells early, and most of the oil is produced at a high water cut.

Wettability of the rock

Wettability decides which pores the water occupies. In a water-wet rock, water coats the grains with a film and occupies the small pores, and oil occupies the large ones. Injected water moves for a long time along narrow paths beside the grains, and its relative permeability stays low. In an oil-wet rock it is the other way round: water is the non-wetting phase (the fluid that the rock surface does not prefer), it goes through the large pores, its relative permeability grows fast, and the oil stays in the small pores and in films on the grains. The lesson on relative permeability collects the signs of wettability in the curves into rules.

Relative permeability curves of two rocks for comparison

Parameter: $S_{wi}$ ; $S_{or}$
Water-wet (lesson on relative permeability): 0.22; 0.25
Oil-wet (assumed): 0.12; 0.20

Parameter: $k_{ro}'$ ; $k_{rw}'$
Water-wet (lesson on relative permeability): 0.85; 0.25
Oil-wet (assumed): 0.80; 0.60

Parameter: $n_o$ ; $n_w$
Water-wet (lesson on relative permeability): 2.5; 3
Oil-wet (assumed): 3.5; 1.8

Parameter: Crossover of the relative permeability curves
Water-wet (lesson on relative permeability): $S_w \approx 0.56$
Oil-wet (assumed): $S_w \approx 0.40$

Parameter	Water-wet (lesson on relative permeability)	Oil-wet (assumed)
$S_{wi}$ ; $S_{or}$	0.22; 0.25	0.12; 0.20
$k_{ro}'$ ; $k_{rw}'$	0.85; 0.25	0.80; 0.60
$n_o$ ; $n_w$	2.5; 3	3.5; 1.8
Crossover of the relative permeability curves	$S_w \approx 0.56$	$S_w \approx 0.40$

The parameters of the oil-wet rock are assumed in line with Craig's rules: a low irreducible water saturation, a crossover of the curves to the left of 0.5, and a high end-point relative permeability of water.

CalculatedFractional flow of water in a water-wet and an oil-wet rock with the same fluids (oil 3.0 mPa·s, water 0.5 mPa·s). The dots are the values at

S_w = 0.40

from the example below.

Worked exampleThe same water saturation in two rocks

Given (assumed data)

Water-wet rock: the curves of the lesson Relative permeability
Oil-wet rock (assumed): $S_{wi} = 0.12$ , $S_{or} = 0.20$ , $k_{ro}' = 0.80$ , $k_{rw}' = 0.60$ , $n_o = 3.5$ , $n_w = 1.8$
Fluids: oil 3.0 mPa·s, water 0.5 mPa·s
Water saturation: $S_w = 0.40$

Solution

Water-wet: $S_n = (0.40 - 0.22)/0.53 = 0.3396$ ; $k_{rw} = 0.25\cdot0.3396^3 = 0.00979$ , $k_{ro} = 0.85\cdot(1 - 0.3396)^{2.5} = 0.301$ . Mobilities per unit of permeability: $\lambda_w = 0.00979/0.5 = 0.01958$ and $\lambda_o = 0.301/3.0 = 0.10033$ in 1/(mPa·s). Then $f_w = 0.01958/(0.01958 + 0.10033) = 0.163$ .
Oil-wet: $S_n = (0.40 - 0.12)/0.68 = 0.4118$ ; $k_{rw} = 0.60\cdot0.4118^{1.8} = 0.1215$ and $k_{ro} = 0.80\cdot(1 - 0.4118)^{3.5} = 0.1249$ . Then $\lambda_w = 0.1215/0.5 = 0.2430$ and $\lambda_o = 0.1249/3.0 = 0.04163$ , and $f_w = 0.2430/(0.2430 + 0.04163) = 0.854$ .
Half of the flow is water at $S_w$ = 0.474 in the water-wet rock and 0.289 in the oil-wet rock.

Answer: At

S_w = 0.40

, water makes up 16% of the flow in the water-wet rock and 85% in the oil-wet rock.

What it means. In the oil-wet rock, water becomes the main phase of the flow almost at once: it goes through the large pores and bypasses the oil. So when such a formation is flooded, water breaks through earlier, and the oil is produced for longer and with a large amount of water. A calculation of the front from these curves (Buckley–Leverett theory and the Welge method) gives a displacement efficiency at breakthrough of 0.36, against 0.51 in the water-wet rock. This does not necessarily mean that the oil-wet formation will give less oil in the end. In the assumed example its range of mobile saturation is even wider (0.68 against 0.53), but it gives up this oil slowly, with a large volume of water pumped through. Most formations have mixed or intermediate wettability, and their curves lie between these two. The data in this example are assumed for illustration.

Dip of the formation and gravity

Water is heavier than oil. When it moves up the dip, for example when it is injected into the lower part of the accumulation or outside the oil outline, its own weight holds it back, and the oil, which is lighter, gets an extra “lift”. The water lags behind, the fractional flow at the same saturation is smaller, and the displacement is better. When the water moves down the dip, gravity does the opposite and drives the water on. In the formula this corresponds to the term $-\Delta\rho\,g\sin\alpha$ . If we neglect the capillary term, it is convenient to collect this term into one dimensionless number.

Gravity number

N_g = \frac{k\,k_{ro}'\,\Delta\rho\,g\sin\alpha}{\mu_o\,u_t}

where:

$N_g$ gravity number: the ratio of two pressure gradients. One is due to the difference between the weights of the phases along the formation, $\Delta\rho\,g\sin\alpha$ . The other is the gradient needed to pump oil alone through the rock at the velocity $u_t$ , —
$k_{ro}'$ end-point relative permeability of oil at irreducible water saturation, —
$k,\ \Delta\rho,\ g,\ \alpha,\ \mu_o,\ u_t$ as in the fractional flow equation, SI

Units: SI (a dimensionless number)

Holds when

Diffuse one-dimensional flow along the dip: the saturation is the same over the thickness.
Sign: $N_g > 0$ when the water moves up the dip.

When $N_g$ is of the order of hundredths, gravity hardly affects the displacement. When $N_g$ is of the order of tenths or more, it must be taken into account. The number grows with permeability, density difference and angle, and is inversely proportional to the velocity and to the oil viscosity.

In other unit systems (2)

Metric oilfield units

N_g = \frac{0.0853\,k\,k_{ro}'\,A\cdot 0.00981\,\Delta\rho\sin\alpha}{q_t\,\mu_o}

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

$0.00981\,\Delta\rho\sin\alpha$ is the pressure gradient along the formation due to the density difference, in MPa/m. $0.0853$ is the constant of Darcy's law (Rock permeability).

Oilfield units

N_g = 4.886\cdot10^{-4}\,\frac{k\,k_{ro}'\,A\,\Delta\gamma\sin\alpha}{q_t\,\mu_o}

$k$ in mD, $A$ in ft², $\Delta\gamma$ the difference between the relative densities (specific gravities) of water and oil (fresh water = 1), $q_t$ in reservoir barrels per day, $\mu_o$ in cP.

$4.886\cdot10^{-4} = 1.127\cdot10^{-3}\cdot0.4335$ : the constant of Darcy's law in oilfield units times the pressure gradient of a column of fresh water in psi/ft.

Fractional flow of water in an inclined formation without the capillary term

f_w = \frac{\lambda_w}{\lambda_t}\left(1 - N_g\,\frac{k_{ro}}{k_{ro}'}\right)

where:

$\lambda_w/\lambda_t$ viscous part: the fractional flow of water in a horizontal formation, —
$k_{ro}/k_{ro}'$ relative permeability of oil divided by its end-point value: 1 at $S_{wi}$ , 0 at $1 - S_{or}$ , —

Units: dimensionless

Holds when

We neglect the capillary term.
$N_g$ is taken from the formula above, with the same sign of the dip.

The correction is proportional to $k_{ro}/k_{ro}'$ . In relative terms it is largest where the oil is still highly mobile: at low and medium water saturations, that is, in the front zone. At $1 - S_{or}$ the oil is immobile, and $f_w = 1$ at any dip.

Worked exampleGravity number of a dipping formation

Given (assumed data)

Permeability $k$: 400 mD
Flow cross-section: thickness 25 m, width 600 m
Dip angle of the formation: 12°, injection into the lower part: the water moves up the dip
Densities in the reservoir: water 1.04 g/cm³, oil 0.80 g/cm³ (as in the lesson Capillary pressure and the transition zone)
Oil: 3.0 mPa·s; the relative permeability curves of the lesson Relative permeability, $k_{ro}' = 0.85$
Total flow rate at reservoir conditions: 600 m³/d; for comparison, 150 m³/d

Solution

Cross-section area: $A = 25\cdot600 = 15{,}000$ m². Density difference: $\Delta\rho = 1.04 - 0.80 = 0.24$ g/cm³.
Gravity gradient along the formation: $0.00981\cdot0.24\cdot\sin 12° = 0.00981\cdot0.24\cdot0.2079 = 4.89\cdot10^{-4}$ MPa/m.
Gravity number: $N_g = \dfrac{0.0853\cdot400\cdot0.85\cdot15{,}000\cdot4.89\cdot10^{-4}}{600\cdot3.0} = 0.118$ . At a flow rate of 150 m³/d it is four times larger: 0.472.
At $S_w = 0.40$ the viscous part is $\lambda_w/\lambda_t = 0.163$ (from the example on wettability), and $k_{ro}/k_{ro}' = 0.301/0.85 = 0.354$ .
Up the dip at 600 m³/d: $f_w = 0.163\cdot(1 - 0.118\cdot0.354) = 0.156$ . At 150 m³/d: $0.163\cdot(1 - 0.472\cdot0.354) = 0.136$ .
If the same water were injected into the upper part and moved down the dip (150 m³/d): $0.163\cdot(1 + 0.472\cdot0.354) = 0.190$ .

Answer:

N_g \approx

0.118 at 600 m³/d and 0.472 at 150 m³/d. At

S_w = 0.40

, gravity changes the fractional flow from 0.163 to 0.136 (up the dip) or to 0.190 (down the dip).

What it means. Even in a permeable, steep formation, at the ordinary rate gravity along the formation changes the fractional flow by only a few per cent. At a rate four times lower, it changes it by 17%. A front calculation shows that with injection up the dip at 150 m³/d, the displacement efficiency at breakthrough grows from 0.508 to 0.524. In thin formations with a permeability of tens of millidarcies and a gentle dip, $N_g$ does not exceed hundredths, and we usually neglect the dip in a one-dimensional calculation. The data in this example are assumed for illustration.

CalculatedFractional flow of water for oil of 3.0 mPa·s in a horizontal formation (solid line) and in a formation with a dip of 12° at 150 m³/d from the example (

N_g

about 0.47). The water moves up the dip or down the dip (dashed lines).

If $N_g\,k_{ro}/k_{ro}' > 1$ , the formula gives $f_w < 0$ . At a very low rate up the dip, water sinks downward against the general flow, and oil rises faster than the flow. This is countercurrent flow, a sign that gravity is stronger than the viscous forces, so that gravity separates the phases during the displacement.

Injection rate

The injection rate enters the equation only through $u_t$ , in the denominators of the capillary and gravity corrections. In a horizontal formation without capillary forces the velocity cancels out, and the curve $f_w(S_w)$ does not depend on the rate: fast and slow injection go through the same states, only over a different time. The rate matters when other forces are at work.

Up the dip. A lower rate strengthens the stabilising action of gravity: the curve shifts to the right, and the displacement is closer to piston-like.
Down the dip. A lower rate, on the contrary, lets the water “roll” forward: the curve shifts to the left.
Capillary forces. At a lower rate the front is smeared over a greater length (see the section below). In a reservoir this is usually not important; in a core test it is.
Heterogeneous formation. At a lower rate the water has time to cross between the layers under capillary forces and gravity. This is already beyond the one-dimensional curve (Displacement in heterogeneous reservoirs).

The capillary term: a smeared front

The capillary correction in the numerator of the general formula is proportional to the gradient of the capillary pressure along the flow. It is convenient to split it into two factors:

Gradient of the capillary pressure along the flow

\frac{\partial p_c}{\partial x} = \frac{\dd p_c}{\dd S_w}\cdot\frac{\partial S_w}{\partial x}

where:

$\dd p_c/\dd S_w$ slope of the capillary pressure curve (the imbibition curve in waterflooding), Pa
$\partial S_w/\partial x$ change in water saturation along the flow, 1/m

Units: SI

Holds when

The capillary pressure is a function of water saturation only.

In a water-wet rock, the less water there is, the higher the capillary pressure: $\dd p_c/\dd S_w < 0$ . At the displacement front the water saturation falls along the flow: $\partial S_w/\partial x < 0$ . The product is positive, and the capillary term increases the fractional flow of water in the front zone. The rock “soaks up” water ahead of the front (capillary imbibition), and the front is smeared.

We usually neglect this term, for a simple reason: the saturation gradient along the flow is not known in advance, since it is exactly what the displacement calculation is looking for. The capillary length shows how justified this is. It is the distance over which the capillary correction is comparable to one.

Capillary length: the scale of the smearing of the front

\ell_c = \frac{k\,k_{ro}\,\Delta p_c}{\mu_o\,u_t}

where:

$\ell_c$ length over which the capillary term in the fractional flow equation is of the order of one, m
$\Delta p_c$ capillary pressure drop across the front: between the water saturation ahead of the front and behind it, Pa
$k,\ k_{ro},\ \mu_o,\ u_t$ as in the fractional flow equation; $k_{ro}$ is a typical relative permeability of oil in the front zone, SI

Units: SI

Holds when

This is a rough, order-of-magnitude estimate: if the front zone is longer than $\ell_c$ , the capillary term is small compared with the viscous part.
With the end-point relative permeability $k_{ro}'$ instead of the average over the front, the estimate comes out too high.

Capillary forces smear the displacement front over a length of the order of $\ell_c$ . This smeared region of the front is called the stabilized zone. If $\ell_c$ is much smaller than the distance between wells, the front can be treated as a jump.

In other unit systems (1)

Metric oilfield units

\ell_c\,[\text{m}] = \frac{0.0853\,k\,k_{ro}\,\Delta p_c}{\mu_o\,u_t}

$k$ in mD, $\Delta p_c$ in MPa, $\mu_o$ in mPa·s, $u_t$ in m/d.

$0.0853$ is the constant of Darcy's law (Rock permeability): $q = 0.0853\,kA\,\Delta p/(\mu L)$ , hence $L = 0.0853\,k\,\Delta p/(\mu u)$ .

Worked exampleReservoir and core: how smeared is the front

Given (assumed data)

Rock: 50 mD, the relative permeability curves of the lesson Relative permeability, $k_{ro}' = 0.85$ ; oil 3.0 mPa·s
Imbibition capillary pressure: the assumed curve of the lesson Two-phase flow equations: $p_c = 0.04\,(1 - S_n)^2$ MPa
Saturations on both sides of the front: $S_{wi} = 0.22$ and the saturation at the front 0.564 (Buckley–Leverett theory and the Welge method)
Velocity in the reservoir: injection of the element 255 m³/d through a cross-section of 300 × 12 m: 0.0708 m/d
Core test: a sample 7 cm long, velocity 3.0 m/d

Solution

Capillary pressure drop across the front: $p_c(S_{wi}) = 0.040$ MPa, $p_c(0.564) = 0.0049$ MPa, $\Delta p_c = 0.0351$ MPa.
Numerator with the end-point relative permeability (an upper estimate): $0.0853\cdot50\cdot0.85\cdot0.0351 = 0.1272$ .
In the reservoir: $\ell_c = 0.1272/(3.0\cdot0.0708) = 0.60$ m, which is 0.12% of the distance of 500 m between the rows of wells.
In the core: $\ell_c = 0.1272/(3.0\cdot3.0) = 1.4$ cm, which is 20% of the length of the sample.
If the core were flooded at the reservoir velocity of 0.0708 m/d, $\ell_c$ would be 0.60 m, which is 8.6 times longer than the sample.

Answer: In the reservoir the front is smeared over 0.60 m out of 500 m. In the core test at 3 m/d it is smeared over 1.4 cm out of 7 cm.

What it means. At the scale of the reservoir, capillary forces smear the front over tens of centimetres (over metres in a more permeable rock or at a slower injection), and in a displacement calculation we can safely treat it as a jump. In the core the same forces are comparable to the length of the sample. At the reservoir velocity the test would show mainly capillary imbibition, and at the outlet end the water would be held back by capillary forces. This is the capillary end effect (how relative permeability is measured). So engineers run unsteady-state tests (tests in which the saturations are still changing) at higher rates or on long composite samples (several core plugs joined end to end), and they check that the result does not depend on the rate. The estimate with the end-point relative permeability overstates $\ell_c$ , because $k_{ro}$ is smaller in the front zone. The data in this example are assumed for illustration.

The fractional flow curve as a tool

Once the engineer has plotted the curve $f_w(S_w)$ from the relative permeability curves and the viscosities, it answers several practical questions at once.

Water cut of a well: the share of water at the bottomhole equals $f_w$ at the saturation near the well. At the surface we convert it through the formation volume factors (the water cut formula). In reverse, from the measured water cut we find $f_w$ , and from the curve the saturation near the well (example below).
Oil rate at a given liquid rate: at reservoir conditions, oil makes up the share $1 - f_w$ of the flow: $q_o = (1 - f_w)\,q_t/B_o$ in standard cubic metres.
Velocity of a saturation: the slope of the curve, $f_w' = \dd f_w/\dd S_w$ , is proportional to the speed at which a given saturation advances through the formation. A tangent to the curve from the point $(S_{wi}, 0)$ gives the saturation at the displacement front, the moment of breakthrough and the average saturation behind the front (Buckley–Leverett theory and the Welge method).
Comparing options: anything that shifts the curve to the right (less viscous oil or more viscous water, injection up the dip, a water-wet rock) improves the displacement, and a shift to the left makes it worse. This is how polymer flooding, hot water and the choice of the injection side are compared on one graph.

Worked exampleA water cut of 90%: how much oil is left near the well

Given (assumed data)

Production well: water cut of the production 90% (standard conditions)
Fluids: oil 3.0 mPa·s, $B_o = 1.25$ ; water 0.5 mPa·s, $B_w = 1.02$ (as in the lesson Relative permeability)
Relative permeability curves: of the lesson Relative permeability; the formation is horizontal
For comparison: the same water cut with oil of 30 mPa·s

Solution

Fractional flow of water at the bottomhole (reservoir conditions): $f_w = \dfrac{0.90\cdot1.02}{0.90\cdot1.02 + 0.10\cdot1.25} = \dfrac{0.918}{1.043} = 0.880$ .
On the curve for oil of 3.0 mPa·s, $f_w = 0.880$ at $S_w \approx$ 0.569. The oil saturation near the well is $1 - 0.569 = 0.431$ , of which the mobile oil is $0.431 - 0.25 = 0.181$ , that is 34% of the range of mobile saturation, 0.53.
For oil of 30 mPa·s, the same fractional flow is at $S_w \approx$ 0.460. The mobile oil still near the well is $0.75 - 0.460 = 0.290$ , that is 55% of the range.
Oil rate at a liquid rate of 255 m³/d at reservoir conditions: $(1 - 0.880)\cdot255/1.25 = 24.5$ m³/d.

Answer: A water cut of 90% corresponds to a fractional flow of water at the bottomhole of 0.880 and a water saturation near the well of about 0.569 (about 0.460 with viscous oil).

What it means. The water cut alone does not tell how well the formation near the well has been flushed. With oil of 3.0 mPa·s, 90% water means that about two thirds of the mobile oil has been displaced. With oil of 30 mPa·s it means less than half, and most of the oil is still to come, although it will have to be produced with a very large amount of water. The $f_w$ curve converts one into the other. The data in this example are assumed for illustration.

Terms and symbols

Fractional flow of water $f_w$: The share of water in the total reservoir flow rate of oil and water through a cross-section of the formation.
Viscous part $\lambda_w/\lambda_t$: The fractional flow of water in a horizontal formation without capillary forces. It depends only on the saturation, the relative permeabilities and the viscosities.
Mobility ratio $M'$: It equals $(k_{rw}'/\mu_w)/(k_{ro}'/\mu_o)$ : the ratio of the mobility of the water behind the front to the mobility of the oil ahead of it. It is a quick measure of the shape of the curve.
Gravity number $N_g$: The ratio of gravity along the formation to the viscous resistance of the oil at a given velocity. It is positive when the water moves up the dip.
Capillary length $\ell_c$: The scale over which capillary forces smear the displacement front. It is inversely proportional to the velocity.
Diffuse flow: Flow with the same saturation over the thickness of the formation. It is the condition for using the one-dimensional $f_w$ curve.
Countercurrent flow: Movement of the phases in opposite directions (water down, oil up) under gravity or capillary forces.

Common mistakes and misconceptions

Wrong

“The fractional flow of water equals the water saturation: if the pores hold 50% water, then about half of the flow is water.”

In fact

The flow is shared according to the mobilities. In the example of this lesson, at $S_w = 0.45$ water carries from 19 to 86% of the flow depending on the oil viscosity, and in the oil-wet rock at $S_w = 0.40$ it carries 85%.

Wrong

“If we inject faster, the water will break through sooner and the $f_w$ curve will get worse.”

In fact

In a horizontal formation without capillary forces the curve does not depend on the rate: faster injection takes the flow through the same states in less time. The rate acts through gravity (in an inclined formation) and capillary forces, and in a heterogeneous formation through crossflow between layers.

Wrong

“The capillary pressure is small compared with the pressure drop between the wells, so it affects displacement nowhere.”

In fact

In the reservoir it smears the front over only tens of centimetres to metres. But in a core test it is comparable to the length of the sample and distorts the result, and in the transition zone it sets the initial saturation and the water cut of wells.

Wrong

“Gravity turned out to be small in the one-dimensional calculation, so it can be ignored in this formation.”

In fact

A one-dimensional calculation includes only gravity along the formation. In a thick formation water sinks to the base across the bedding, and this affects the displacement more strongly. Such a case needs a different calculation (Segregated flow and gravity displacement).

In depth: advanced detail

Derivation of the fractional flow equation

Let us write Darcy's law for the phases, for one-dimensional flow along an axis $x$ inclined to the horizontal at an angle $\alpha$ (two-phase Darcy's law), in terms of the pressure gradients:

Pressure gradients of the phases from Darcy's law

-\frac{\partial p_o}{\partial x} = \frac{\mu_o\,u_o}{k\,k_{ro}} + \rho_o g\sin\alpha,\qquad -\frac{\partial p_w}{\partial x} = \frac{\mu_w\,u_w}{k\,k_{rw}} + \rho_w g\sin\alpha

where:

$u_o,\ u_w$ flow velocities of oil and water, m/s
$p_o,\ p_w$ pressures in the oil and in the water, Pa
$\rho_o,\ \rho_w$ densities of oil and water, kg/m³

Units: SI

Holds when

The same as for the fractional flow equation.

Subtract the first equality from the second. On the left we get $\partial(p_o - p_w)/\partial x = \partial p_c/\partial x$ . On the right we get the difference of the viscous terms and $\Delta\rho\,g\sin\alpha$ :

Difference of the Darcy equations for oil and water

\frac{\partial p_c}{\partial x} = \frac{\mu_w\,u_w}{k\,k_{rw}} - \frac{\mu_o\,u_o}{k\,k_{ro}} + \Delta\rho\,g\sin\alpha

where:

$p_c$ capillary pressure $p_o - p_w$ , Pa
$\Delta\rho$ $\rho_w - \rho_o$ , kg/m³

Units: SI

Holds when

The same as for the fractional flow equation.

Substitute $u_o = u_t - u_w$ and collect the terms with $u_w$ on the left:

Water velocity through the total velocity

u_w\left(\frac{\mu_w}{k\,k_{rw}} + \frac{\mu_o}{k\,k_{ro}}\right) = \frac{\mu_o\,u_t}{k\,k_{ro}} + \frac{\partial p_c}{\partial x} - \Delta\rho\,g\sin\alpha

where:

$u_w,\ u_t$ velocity of water and total velocity, m/s

Units: SI

Holds when

The same as for the fractional flow equation.

Multiply both sides by $k\,k_{ro}/(\mu_o u_t)$ and note that $f_w = u_w/u_t$ . On the left we are left with $f_w$ multiplied by the denominator of the general formula, and on the right with its numerator. Dividing by the denominator gives the general form. If instead we write the sum in the bracket on the left through the mobilities, $1/\lambda_w + 1/\lambda_o = \lambda_t/(\lambda_o\lambda_w)$ , and divide by it, we get the form through the mobilities.

The formula for gas displacing oil is derived in the same way: gas in place of water, the difference of the oil and gas pressures in place of $p_o - p_w$ , and the density difference $\rho_g - \rho_o$ is negative. When gas is injected into the upper part of the accumulation and displaces the oil down the dip, gravity stabilises it. When gas moves up the dip, gravity drives it forward (Displacement of oil by gas).

The inflection point and the slope of the curve

For Corey curves in a horizontal formation, the slope of the fractional flow can be written explicitly (it is also used in the lesson IMPES and fully implicit schemes):

Slope of the fractional flow curve for the Corey model

f_w' = \frac{\dd f_w}{\dd S_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:

$f_w'$ slope of the fractional flow curve, —
$S_n$ normalised water saturation $(S_w - S_{wi})/(1 - S_{wi} - S_{or})$ , —
$n_w,\ n_o$ Corey exponents for water and oil, —

Units: dimensionless

Holds when

The Corey model, a horizontal formation, no capillary term.
At the ends of the range of mobile saturation, where only one phase flows, the slope is zero.

As the oil viscosity grows, the inflection point shifts to lower saturations. The largest slope hardly changes for oils from 1 to 3 mPa·s and grows for more viscous ones. For the curves of this lesson, with oil of 1.2 mPa·s the inflection is at about $S_w = 0.52$ with a slope of 5.2. With 3.0 mPa·s it is at about 0.47 (5.2), and with 30 mPa·s at about 0.36 (6.7). The saturations near the inflection move through the formation faster than any others, and this is why a jump forms at the front (saturation jump). The largest slope also sets the limit on the time step of the explicit scheme (a calculation that finds the new saturation straight from the old one) in a simulation model (IMPES and fully implicit schemes).

Connections to other topics

Relative permeability — relative permeability curves, end points and the simplest form of the fractional flow of water.
Why water is injected: pressure maintenance and oil displacement — the mobility ratio, displacement efficiency and sweep efficiencies.
Buckley–Leverett theory and the Welge method — how the displacement front, water breakthrough and the production after it are found from the $f_w$ curve.
Two-phase flow equations — the water flux through the total velocity in the equations of two-phase flow.
Capillary pressure and the transition zone — capillary pressure of imbibition and drainage, the transition zone.
Wettability and interfacial phenomena — why water-wet and oil-wet rocks give different curves.
Segregated flow and gravity displacement — when gravity separates the phases across the formation.
Displacement of oil by gas — the fractional flow of gas and gravity stabilisation in gas injection.

Check yourself

1
A horizontal uniform formation, with capillary forces neglected. The injection rate has been doubled. What will happen to the curve $f_w(S_w)$ ?
1. A.It will shift to the right: the water will displace the oil better
2. B.It will shift to the left: the water will break through sooner
3. C.It will not change: under these conditions the curve does not depend on the rate
4. D.It will get steeper in the middle
Show the answer
Correct answer: C. It will not change: under these conditions the curve does not depend on the rate
Without gravity and capillary pressure, $f_w = \lambda_w/\lambda_t$ depends only on saturation, relative permeabilities and viscosities: the velocity cancels out. Faster injection passes through the same states in less time.
2
At a certain water saturation $k_{ro} = 0.12$ and $k_{rw} = 0.05$ , the oil viscosity is 4.0 mPa·s and the water viscosity is 0.6 mPa·s. Find the fractional flow of water in a horizontal formation.
Show the answer
Mobilities per unit of permeability: $\lambda_o = 0.12/4.0 = 0.030$ and $\lambda_w = 0.05/0.6 \approx 0.0833$ in 1/(mPa·s). $f_w = 0.0833/(0.0833 + 0.030) \approx 0.74$ . The relative permeability of water is almost two and a half times smaller than that of oil, but water is 6.7 times less viscous, and almost three quarters of the flow is water.
3
Water can be injected into the lower part of an inclined formation (it will move up the dip) or into the upper part (it will move down the dip). Where is the fractional flow of water smaller at the same saturation, and why?
1. A.The same: gravity acts on both phases
2. B.With injection into the lower part: the weight of the water holds back its upward movement
3. C.With injection into the upper part: the water sinks and does not take part in the flow
Show the answer
Correct answer: B. With injection into the lower part: the weight of the water holds back its upward movement
Water is heavier than oil. When it moves up the dip, gravity holds it back, and the oil “floats up”: the term $-\Delta\rho\,g\sin\alpha$ with $\alpha > 0$ reduces $f_w$ . Gravity acts on both phases, but what matters is the density difference.
4
The gravity number of a formation at a flow rate of 800 m³/d is 0.2. What will it be at 200 m³/d, and by how many times will the fractional flow of water be smaller than in a horizontal formation, at a saturation where $k_{ro}/k_{ro}' = 0.5$ (the water moves up the dip)?
Show the answer
$N_g$ is inversely proportional to the flow rate: $0.2\cdot800/200 = 0.8$ . The factor is $1 - N_g\,k_{ro}/k_{ro}' = 1 - 0.8\cdot0.5 = 0.6$ : the fractional flow will be 60% of that in a horizontal formation, that is, about 1.7 times smaller. At 800 m³/d the factor would have been 0.9.
5
In which rock, with the same fluids and the same water saturation, is the fractional flow of water most likely to be higher?
1. A.In the water-wet rock
2. B.In the oil-wet rock
3. C.Wettability does not affect the fractional flow
Show the answer
Correct answer: B. In the oil-wet rock
In an oil-wet rock, water is the non-wetting phase and occupies the large pores, so its relative permeability is high, and that of the oil left in the small pores and films is low. The $f_w$ curve is shifted to the left: in the example of this lesson, at $S_w = 0.40$ it is 85% water, against 16% in the water-wet rock.
6
Why is the capillary term in the fractional flow equation usually neglected in calculating the waterflooding of a formation, but not in processing a test on a core 5 cm long that was flooded at the reservoir velocity?
Show the answer
Capillary forces smear the front over a length of the order of $\ell_c = k\,k_{ro}\Delta p_c/(\mu_o u_t)$ . At reservoir velocities this is from tens of centimetres to metres: small compared with the hundreds of metres between wells, but more than the length of the sample. In a core, capillary imbibition and the end effect would decide the result, so engineers run tests at higher rates and check that the result does not depend on the rate.

Updated: 14 October 2026