Fluid saturation · MunaiHub

Quick refresher

$S_x = V_x/V_p$ is a fraction of the pore volume, not of the whole rock. $S_w + S_o + S_g = 1$ . The volume of hydrocarbons in a unit volume of rock is $\phi\,(1 - S_w)$ .
Irreducible (connate) water saturation $S_{wirr}$ is the water that oil or gas could not push out when the trap (the structure that holds oil and gas) filled. Residual oil saturation $S_{or}$ is the oil that stays trapped after water or gas has pushed out the rest.
Above the free water level (where the oil and water pressures are equal), water saturation falls with height along the capillary pressure curve. At the bottom is a transition zone with mobile water. Above it is the zone of irreducible water, from which the inflow to a well is free of water.
On core, saturation is measured by extraction (the Dean–Stark method, which drives the fluids out of the sample with hot solvent vapour) or by retort distillation (heating the sample in a furnace). But drilling the core and bringing it up distorts the saturation. Water saturation is given most reliably by core taken with oil-based mud (drilling fluid with an oil base).
From logs, water saturation is calculated from the electrical resistivity of the rock, and is calibrated against core and capillary pressure data.
Average saturation with the pore volume $\phi h$ as the weight, not the thickness.

Key relationships

S_x = \dfrac{V_x}{V_p},\qquad S_w + S_o + S_g = 1

definition; $x$ is water, oil or gas

p_c = (\rho_w - \rho_o)\,g\,h

capillary pressure at height $h$ above the free water level

\bar S_w = \dfrac{\sum_i S_{w,i}\,\phi_i\,h_i}{\sum_i \phi_i\,h_i}

average saturation, weighted by pore volume

E_D = \dfrac{1 - S_{wi} - S_{or}}{1 - S_{wi}}

displacement efficiency: the upper limit of recovery from the swept volume

Common pitfalls

Referring saturation to the rock volume instead of the pore volume.
Confusing the initial water saturation with the irreducible one. In the transition zone the initial one is higher, and the water there is mobile.
Using “residual water saturation” as the English term. In English-language literature this water is called irreducible (connate) water; “residual” is used for oil and gas.
Taking the oil saturation of a core brought to the surface as the reservoir oil saturation. In the core the oil has shrunk, and part of it has been pushed out by gas coming out of solution.
Averaging saturation over thickness without taking porosity into account.

After this lesson you can

define water, oil and gas saturation, and relate them to the pore volume and the volume of hydrocarbons
tell initial, irreducible and residual saturations apart, and explain what causes each of them physically
explain why water saturation changes with height above the free water level, and what the transition zone is
calculate saturation from core extraction data, and know why core gives a distorted picture of reservoir saturations
average saturation correctly over the well section, and calculate the moveable oil and the displacement efficiency

What saturation is

Porosity tells us how many voids a rock has. Saturation tells us what fills them. In an oil or gas reservoir the pores are almost never filled with hydrocarbons alone. Water left over from the time the sediment was deposited is always there too. In a gas cap (the layer of gas on top of an oil reservoir), or below the bubble point pressure (the pressure at which gas starts to come out of the oil), there is also free gas.

Saturation

S_w = \frac{V_w}{V_p},\qquad S_o = \frac{V_o}{V_p},\qquad S_g = \frac{V_g}{V_p},\qquad S_w + S_o + S_g = 1

where:

$S_w,\ S_o,\ S_g$ water, oil and gas saturation, fraction (or % of pore volume)
$V_w,\ V_o,\ V_g$ volumes of water, oil and gas in the pores at reservoir conditions, m³, cm³
$V_p$ pore volume (as a rule the open, that is connected, pores), m³, cm³

Units: any units of volume, the same for all; saturations have no units

Holds when

The phase volumes and the pore volume refer to the same conditions. For reservoir saturations, these are reservoir pressure and temperature.
Saturation is measured against the same pore volume as the porosity you use (open or total). Otherwise the bound water will be counted wrongly.

Saturation is a fraction of the pore volume, not of the whole rock volume. The volume of oil in 1 m³ of rock is $\phi\,S_o$ : at $\phi = 0.20$ and $S_o = 0.75$ it is 0.15 m³. The sum $S_o + S_g = 1 - S_w$ is called the hydrocarbon saturation.

SchematicHow the fluids are distributed in a water-wet rock (one that water sticks to more strongly than oil does). On the left is the oil-bearing part of the accumulation before production: water coats the grains in thin films and fills the narrow corners of the pores, and oil fills the pore centres,

S_w = S_{wirr}

. On the right is the same rock after waterflooding (injecting water to push the oil out): water has filled the connected pores, and the oil is left as separate droplets,

S_o = S_{or}

Initial, irreducible and residual saturation

Oil and gas form outside the trap and move into it, pushing out the water that used to fill the pores. They cannot push out all of it. Most reservoir rocks are wetted better by water than by oil. In such a rock, water stays as thin films on the grains, in the corners of the pores and in the smallest pores. No pressure difference we can realistically apply will squeeze it out of there. Later, during production, water or gas pushes the oil out in turn — and again not completely.

Initial water saturation $S_{wi}$: The water saturation at a given point of the reservoir before production starts. In the main part of the accumulation it equals the irreducible saturation. In the transition zone it is greater.
Irreducible (connate) water saturation $S_{wirr}$: The lowest water saturation that can be reached by pushing the water out with oil or gas. A related term is connate water, the “buried” water that has stayed in the pores since the rock formed. This water cannot move, so this part of the reservoir produces no water.
Residual oil saturation $S_{or}$: The oil left in the swept part of the reservoir after water ( $S_{orw}$ ) or gas ( $S_{org}$ ) has pushed out the rest. Capillary forces (forces in the narrow pores) hold it as separate droplets, and in ordinary waterflooding it no longer moves.
Critical gas saturation $S_{gc}$: The gas saturation below which the gas that comes out of the oil stays as bubbles that cannot move. Until this saturation is reached, almost no gas flows into the well.
Residual gas saturation $S_{gr}$: The gas trapped by water as the water advances into a gas accumulation (also called trapped gas). This gas stays behind in the reservoir after it has been flooded with water.

Approximate values of end-point (limiting) saturations

Quantity: $S_{wirr}$
Typical range: 0.10–0.35 in good sandstones; 0.4–0.6 and above in fine-grained and shaly rocks
What it depends on: pore size and specific surface area (the grain surface per unit of rock volume; fine grains and clays hold more water), wettability

Quantity: $S_{orw}$ after waterflooding
Typical range: 0.15–0.40
What it depends on: wettability, structure of the pore space, initial oil saturation

Quantity: $S_{gc}$
Typical range: 0.02–0.10
What it depends on: pore structure, rate of pressure decline

Quantity: $S_{gr}$
Typical range: 0.15–0.35
What it depends on: pore structure, initial gas saturation

Quantity	Typical range	What it depends on
$S_{wirr}$	0.10–0.35 in good sandstones; 0.4–0.6 and above in fine-grained and shaly rocks	pore size and specific surface area (the grain surface per unit of rock volume; fine grains and clays hold more water), wettability
$S_{orw}$ after waterflooding	0.15–0.40	wettability, structure of the pore space, initial oil saturation
$S_{gc}$	0.02–0.10	pore structure, rate of pressure decline
$S_{gr}$	0.15–0.35	pore structure, initial gas saturation

These are rough values for guidance. For calculations, end-point saturations are measured in the laboratory on core from the reservoir in question.

The end-point saturations set limits on production from two sides. Irreducible water reduces the initial oil in place. Residual oil reduces the part of the oil in place that can be displaced. They are also the end points of the relative permeability curves (curves that show how easily each fluid flows when two fluids share the pores): at $S_{wirr}$ water does not flow, and at $S_{or}$ oil stops flowing. Wettability decides which fluid is held on the surface of the grains.

Saturation with height in the accumulation and the transition zone

In an accumulation that has come to equilibrium (nothing is moving any more), the difference between the oil and water pressures at every height is balanced by capillary forces in the pores. At the free water level (FWL) the oil and water pressures are equal. Above it, the pressure in the oil is higher than the pressure in the water, and the higher the point, the bigger the difference. This is because a column of oil is lighter than a column of water of the same height. The pressure difference is the capillary pressure $p_c$ at that height.

Capillary pressure at a height above the free water level

p_c(h) = (\rho_w - \rho_o)\,g\,h

where:

$p_c$ capillary pressure: the difference between the oil and water pressures at height $h$ , Pa
$\rho_w,\ \rho_o$ densities of formation water and oil at reservoir conditions (for a gas accumulation, the gas density instead of $\rho_o$ ), kg/m³
$g$ acceleration due to gravity, 9.81, m/s²
$h$ height above the free water level, m

Units: SI

Holds when

The accumulation is in hydrostatic and capillary equilibrium. This is the case before production starts, or after a long shut-in (wells closed for a long time).
Water and oil each form a continuous (unbroken) phase over the whole height of the accumulation.

Each height has its own capillary pressure, and each capillary pressure has its own water saturation. The pore-size distribution sets this saturation (it is shown by the capillary pressure curve measured on core). This is how a laboratory curve becomes a profile of saturation with height.

In other unit systems (2)

Metric oilfield units

p_c\,[\text{MPa}] = 0.00981\,(\rho_w - \rho_o)\,[\text{g/cm}^3]\;h\,[\text{m}]

densities in g/cm³, height in m, pressure in MPa.

$0.00981 = 9.81\cdot 1000/10^6$ : conversion of g/cm³ to kg/m³ and of Pa to MPa.

Oilfield units

p_c\,[\text{psi}] = 0.433\,(\gamma_w - \gamma_o)\;h\,[\text{ft}]

$\gamma_w,\ \gamma_o$ are the specific gravities of water and oil (relative to fresh water), $h$ is in feet, $p_c$ in psi.

$0.433$ psi/ft is the pressure gradient of a column of fresh water.

CalculatedCalculated profile of water saturation above the free water level (FWL,

h = 0

), for an assumed rock with an irreducible water saturation of 0.20. The model and its parameters are in the section “The saturation profile model used in the plot”. The horizontal dashed line is the oil–water contact; the vertical one is the irreducible water saturation.

From bottom to top there are several zones:

Below the free water level. The water-bearing part of the formation, $S_w = 1$ .
From the free water level to the oil–water contact (OWC). The pores are still completely filled with water, because oil cannot enter even the largest pores until the capillary pressure exceeds a threshold value (the entry pressure). So the OWC that we see on logs and in formation testing lies above the free water level.
The transition zone. The water saturation decreases quickly with height. Both fluids can move, and the well produces oil together with water.
The main part of the accumulation. The water saturation is close to irreducible, so the water cannot move and the inflow has no water in it.

The topic Capillary pressure and the transition zone covers capillary pressure curves in detail: how they are converted from core to reservoir conditions, and how the saturation profile is built. The topic Pressure gradients, fluid contacts and capillary equilibrium compares the profile with pressure measurements through the section.

Saturation from core

The direct way is to take the fluids out of a core sample and measure how much there is of each. Then the pore volume is measured on the same cleaned sample (see measuring porosity).

How it is measured

1
Dean–Stark extraction
Heat the sample in the vapour of a boiling solvent, usually toluene. The water evaporates together with the solvent, condenses in a cooler and collects in a graduated trap (a tube with a volume scale), where you read its volume directly. The oil dissolves in the toluene. To find the mass of oil, take the mass of the sample before extraction, subtract its mass after extraction and drying, and then subtract the mass of the water.
2
Retort analysis
Heat the sample in a furnace, in steps, up to about 650 °C. The water and oil evaporate and are collected in a graduated receiver. Read the water volume before the water locked inside minerals (clays, gypsum), the “water of crystallisation”, starts to come off. Correct the oil volume for the losses caused by cracking (oil breaking down when it is hot), using a calibration curve. The method is fast, but it destroys the sample.
3
Gas
In a fresh sample, the volume of the pores not filled with liquid can be measured by forcing mercury into them. More often the gas saturation is found by difference: $S_g = 1 - S_w - S_o$ .

Saturation from extraction data

S_w = \frac{V_w}{V_p},\qquad S_o = \frac{m_1 - m_2 - \rho_w V_w}{\rho_o\,V_p}

where:

$V_w$ volume of water collected in the trap, cm³
$m_1$ mass of the sample before extraction, g
$m_2$ mass of the sample after extraction and drying, g
$\rho_w,\ \rho_o$ densities of the collected water and of the stock-tank oil (oil with its gas removed), g/cm³
$V_p$ pore volume of the sample measured after extraction, cm³

Units: g, cm³, g/cm³

Holds when

The water and oil are fully extracted and the solvent has been removed by drying.
The salts of the formation water stay in the sample when the water evaporates, and they are not accounted for. For salty waters a correction is applied to the water volume.
The oil density is known (from a sample of stock-tank oil).

The water is measured directly; the oil is found from a mass balance. The saturations you get are for the core at the surface, not for the reservoir. To compare them with reservoir saturations, you must understand what happened to the core while it was being cut and brought up.

Worked exampleSaturation of a sample by the Dean–Stark method

Given (assumed data)

Mass of the sample before extraction $m_1$: 125.40 g
Mass after extraction and drying $m_2$: 116.80 g
Water in the trap $V_w$: 3.10 cm³, density 1.00 g/cm³
Stock-tank oil density $\rho_o$: 0.85 g/cm³
Pore volume of the sample $V_p$: 13.00 cm³ (measured with a helium porosimeter after extraction)
Sampling conditions: drilled with oil-based mud. The sample was cut from the central part of the core, which the filtrate hardly reached. The accumulation is undersaturated (reservoir pressure above the bubble point, so no free gas). Oil formation volume factor 1.30 m³/m³

Solution

Water: $S_w = 3.10/13.00 = 0.238$ .
Mass of oil: $m_1 - m_2 - \rho_w V_w = 125.40 - 116.80 - 3.10 = 5.50$ g; volume $V_o = 5.50/0.85 = 6.47$ cm³; $S_o = 6.47/13.00 = 0.498$ .
The remaining pores, $1 - S_w - S_o = 0.264$ , are occupied by gas and air at the surface.
There was no free gas in the reservoir, so the oil saturation there was about $1 - S_w = 0.762$ . If the oil had only shrunk on its way up, it would have left $0.762/1.30 = 0.586$ of the pore volume in the sample.

Answer: At the surface,

S_w = 0.238

and

S_o = 0.498

. 26% of the pores have been emptied of liquid.

What it means. Oil-based mud adds no water to the core. The irreducible water in the reservoir cannot move, so it stays in the sample. So $S_w \approx 0.238$ is a good estimate of the irreducible water saturation in the reservoir. The oil saturation of the core is distorted. The oil shrank as gas came out of solution, and the expanding gas pushed part of the oil out of the sample. We measured 0.498. Shrinkage alone would have left 0.586, and the reservoir had 0.762. The 26% of the pores that emptied is not reservoir gas. It is a result of bringing the core to the surface. The data in this example are assumed for illustration.

Why core distorts saturation

Filtrate invasion. When the well is drilled with water-based mud, the filtrate (the water that seeps out of the mud) pushes the mobile oil out of the core. The water saturation reads too high, and the oil saturation falls almost to the residual value.
Bringing the core to the surface. The pressure falls. The dissolved gas comes out of solution and expands, and it pushes out oil and water. The oil itself also shrinks.
Storage and handling. Unprotected core loses water, and the light fractions of the oil (its lighter components), by evaporation.
That is why core for reservoir water saturation is taken with oil-based mud. In special cases it is taken with a pressure-retaining core barrel, which keeps the core at reservoir pressure on the way up. Such core is frozen, and it is thawed only during analysis, when all the fluids that come out are collected.

Saturation from logs

The mineral framework of most rocks practically does not conduct electric current, and neither do hydrocarbons. Salty formation water does. So the resistivity of the formation (how strongly it resists current) rises as the water saturation falls. For clean reservoirs, with no shale, the relationship is described by the empirical Archie equation (a formula found from experiments):

Water saturation from Archie's equation (clean reservoir)

S_w = \left(\frac{a\,R_w}{\phi^m\,R_t}\right)^{1/n}

where:

$R_t$ resistivity of the uninvaded part of the formation, beyond the invaded zone (the zone around the well that mud filtrate has entered), Ω·m
$R_w$ resistivity of the formation water at reservoir temperature, Ω·m
$\phi$ porosity, fraction
$a,\ m,\ n$ empirical coefficients (found from experiments). For sandstones they are often $a \approx 1$ and $m \approx n \approx 2$ , but they should be measured on core, —

Units: resistivities in the same units (Ω·m); the other quantities have no units

Holds when

Only the formation water conducts current: there are no clays or conductive minerals. (Other models are used for shaly rocks: Water saturation in shaly sands.)
$R_t$ has been corrected for the effect of the borehole and the invaded zone, and $R_w$ has been converted to reservoir temperature.

At $n = 2$ , the resistivity is inversely proportional to the square of the water saturation: half the water means four times the resistivity. Porosity is in the formula, so an error in porosity becomes an error in $S_w$ . The topic Archie's equation and saturation of clean reservoirs covers how to choose the coefficients and estimate the uncertainty.

Saturation is also estimated from other logs besides resistivity. Nuclear magnetic resonance (NMR) logging separates bound water (held in the rock) from moveable water (Nuclear magnetic resonance (NMR) logging). In cased wells (wells lined with steel pipe), pulsed neutron methods let us follow how saturation changes during production. All log estimates are calibrated against core and capillary pressure curves. Three independent sources should give a consistent profile of saturation with height.

Average saturation and the volume of hydrocarbons

Saturation is a fraction of the pore volume, so we average it with the pore volume $\phi h$ as the weight: a layer with higher porosity contains more fluid. A simple average over thickness gives too much weight to the tight layers (layers with low porosity and permeability), which usually hold more water too.

Average water saturation over a section

\bar S_w = \frac{\sum_i S_{w,i}\,\phi_i\,h_i}{\sum_i \phi_i\,h_i}

where:

$S_{w,i}$ water saturation of the $i$ -th layer, fraction
$\phi_i$ porosity of the $i$ -th layer, fraction
$h_i$ net thickness of the $i$ -th layer, m

Units: all thicknesses in the same units; porosity and saturation as fractions

Holds when

Only reservoir layers enter the sum.
Porosity and saturation have been obtained consistently, from the same definition of porosity.

With this average, the volume of hydrocarbons per unit area, $\sum_i \phi_i h_i\,(1 - S_{w,i})$ , is exactly equal to $\bar\phi\,h\,(1 - \bar S_w)$ . This is the product that goes into the volumetric calculation of hydrocarbons in place.

Worked exampleAverage weighted by pore volume and average over thickness

Given (assumed data)

Layer 1: thickness 4.0 m, porosity 0.25, water saturation 0.20
Layer 2: thickness 6.0 m, porosity 0.12, water saturation 0.55

Solution

Pore volumes per 1 m² of area: $\phi_1 h_1 = 0.25\cdot 4.0 = 1.00$ m³, $\phi_2 h_2 = 0.12\cdot 6.0 = 0.72$ m³.
Average weighted by pore volume: $\bar S_w = \dfrac{0.20\cdot 1.00 + 0.55\cdot 0.72}{1.00 + 0.72} = 0.347$ .
Average over thickness (incorrect): $\dfrac{0.20\cdot 4.0 + 0.55\cdot 6.0}{10.0} = 0.41$ .
Hydrocarbon volume per 1 m² of area: $\sum \phi_i h_i\,(1 - S_{w,i}) = 1.124$ m³. With the thickness average we would get $(1.00 + 0.72)(1 - 0.41) = 1.015$ m³, which is 9.7% less.

Answer: The average water saturation is 0.347, not 0.41. The hydrocarbon volume is 1.124 m³ per 1 m² of area, at reservoir conditions.

What it means. The tight layer takes up 60% of the thickness but only 42% of the pore volume. The thickness average spreads its high water saturation over the whole reservoir, so it underestimates the hydrocarbons in place by almost a tenth. The data in this example are assumed for illustration.

Moveable oil and displacement efficiency

The end-point saturations show what part of the oil can be displaced where the water has arrived. At the start, the oil takes up $1 - S_{wi}$ of the pore volume. After complete flushing (washing by water) it takes up $S_{or}$ . The difference is the moveable oil.

Displacement efficiency

E_D = \frac{S_{oi} - S_{or}}{S_{oi}} = \frac{1 - S_{wi} - S_{or}}{1 - S_{wi}}

where:

$E_D$ displacement efficiency: the share of the initial oil displaced from a completely swept volume, fraction
$S_{oi} = 1 - S_{wi}$ initial oil saturation (no free gas), fraction
$S_{wi}$ initial water saturation; in the main part of the accumulation it equals the irreducible one, fraction
$S_{or}$ residual oil saturation for the given displacing agent (water or gas), fraction

Units: fractions

Holds when

The volume considered is completely swept: the injected water has reached every part of it (the sweep efficiency equals one).
The pressure is maintained, so the oil formation volume factor is the same at the start and at the end. If it is not, do the calculation in standard volumes (volumes at surface conditions).
$S_{or}$ was measured in the laboratory for the same displacing agent, at conditions close to reservoir conditions.

This is the upper limit of recovery from the swept part of the reservoir. The actual recovery factor is the displacement efficiency multiplied by the areal and vertical sweep efficiencies (the shares of the reservoir area and thickness that the water reaches). So it is noticeably lower.

Worked exampleMoveable oil of a reservoir block

Given (assumed data)

Pore volume of the block: 2.0 million m³
Initial (irreducible) water saturation $S_{wi}$: 0.25
Residual oil saturation after waterflooding $S_{or}$: 0.30 (displacement experiments on core)
Oil formation volume factor: 1.25 m³/m³, pressure maintained by injection

Solution

Initial oil saturation: $S_{oi} = 1 - 0.25 = 0.75$ . The oil in the reservoir is $2.0\cdot 0.75 = 1.50$ million m³. At standard conditions this is $1.50/1.25 = 1.20$ million m³.
Moveable oil: $1 - S_{wi} - S_{or} = 0.45$ of the pore volume. That is $2.0\cdot 0.45/1.25 = 0.72$ million m³ at standard conditions.
Displacement efficiency: $E_D = 0.45/0.75 = 0.60$ .
Suppose $S_{or} = 0.20$ were taken by mistake. Then $E_D = (1 - 0.25 - 0.20)/0.75 = 0.73$ , and the recoverable oil would be overestimated by 22%.

Answer: No more than 60% of the initial oil can be displaced from the swept part of the block: 0.72 million m³ out of 1.20 million m³.

What it means. Even a perfectly swept volume keeps 40% of its oil. This is the target of enhanced oil recovery (EOR) methods, which lower the interfacial tension or change the wettability. In a real reservoir the water bypasses part of the volume, so the recovery factor is lower still. An error of 0.1 in $S_{or}$ changes the recoverable oil by about a fifth. So the residual oil saturation is measured on core from the reservoir in question, not taken by analogy. The data in this example are assumed for illustration.

The end-point saturations are also the end points of the relative permeability curves and of the fractional flow function (the share of water in the flowing fluid). They are used to calculate the displacement front (shock front) in Buckley–Leverett theory.

Terms and symbols

Saturation $S$ ( $K_{\text{v}}$ , $K_{\text{n}}$ , $K_{\text{g}}$ in Russian-language notation): The fraction of the pore volume taken up by a given fluid. The sum over all the phases equals one.
Irreducible water saturation $S_{wirr}$: The lowest water saturation that can be reached by pushing water out with oil or gas. Such water cannot move.
Residual oil saturation $S_{or}$: The oil trapped in the swept volume after water or gas has pushed out the rest.
Free water level: The level at which the oil and water pressures are equal and the capillary pressure is zero. The OWC lies above it.
Transition zone: The interval above the OWC in which the water saturation decreases with height and both fluids are mobile.
Displacement efficiency $E_D$: The share of the initial oil displaced from a completely swept volume. It does not depend on how much of the reservoir is swept.

Common mistakes and misconceptions

Wrong

“Water saturation is 30%, so water occupies 30% of the rock volume.”

In fact

It is 30% of the pore volume. At a porosity of 20%, water takes up 6% of the rock volume.

Wrong

“The well produces water, so the formation is water-bearing.”

In fact

In the transition zone both oil and water can move, so an oil-bearing formation can still produce water along with the oil. You need the saturation profile with height, and the position of the interval relative to the OWC.

Wrong

“The oil saturation of a core is the oil saturation of the reservoir.”

In fact

As the core comes up, the oil shrinks and is pushed out by gas coming out of solution. With water-based mud, the filtrate pushes it out too. Core taken with oil-based mud gives the water saturation reliably, but not the oil saturation.

Wrong

“Residual oil can be produced by injecting more water.”

In fact

Capillary forces hold residual oil as separate droplets, and pumping more water hardly moves it. You need methods that change the interfacial tension, the wettability or the viscosities of the fluids.

Wrong

“I calculate the average water saturation as the average over thickness.”

In fact

Saturation is a fraction of the pore volume, so it is averaged with the weight $\phi h$ . A thickness average gives too much weight to the tight layers, which hold more water.

In depth: advanced detail

The saturation profile model used in the plot

The capillary pressure curve for drainage (oil pushing water out) is often approximated by the Brooks–Corey power law. If we put the capillary pressure at height $h$ into it, we get the water saturation profile:

Water saturation profile from the Brooks–Corey model

S_w(h) = S_{wirr} + (1 - S_{wirr})\left(\frac{p_d}{(\rho_w - \rho_o)\,g\,h}\right)^{\lambda},\qquad h > h_d = \frac{p_d}{(\rho_w - \rho_o)\,g}

where:

$p_d$ threshold (entry) pressure: the capillary pressure at which oil begins to enter the largest pores, Pa
$\lambda$ pore-size distribution index: the larger it is, the more uniform the pores and the sharper the transition, —
$h_d$ height of the OWC above the free water level; below it $S_w = 1$ , m
$S_{wirr},\ \rho_w,\ \rho_o,\ g,\ h$ as above in the lesson, —, kg/m³, m/s², m

Units: SI

Holds when

A drainage curve: the accumulation formed as oil displaced water and has not been flushed since.
The rock is the same over the whole height. For a layered reservoir, build the profile for each rock type separately.
The parameters $p_d$ , $\lambda$ and $S_{wirr}$ are fitted to the capillary pressure curves of core, after conversion to reservoir conditions.

The transition zone gets taller as the threshold pressure rises (small pores) and shorter as the density difference between the fluids grows. The index $\lambda$ sets how steeply the saturation approaches the irreducible value.

Worked exampleSaturation profile for an assumed rock (the data of the plot)

Given (assumed data)

Rock: $S_{wirr} = 0.20$ , $p_d = 5$ kPa at reservoir conditions, $\lambda = 1.5$
Fluids: $\rho_w = 1050$ kg/m³, $\rho_o = 780$ kg/m³

Solution

Capillary pressure rises with each metre of height by $(\rho_w - \rho_o)\,g = 270\cdot 9.81 = 2649$ Pa/m.
Height of the OWC above the free water level: $h_d = 5000/2649 = 1.9$ m.
At a height of 5 m: $p_c = 13{,}244$ Pa, $S_w = 0.20 + 0.80\,(5000/13{,}244)^{1.5} = 0.39$ .
At a height of 10 m: $S_w = 0.27$ ; at 20 m 0.22; at 40 m 0.21.
The height at which $S_w$ falls to 0.25: $(5000/p_c)^{1.5} = (0.25 - 0.20)/0.80 = 0.0625$ , so $p_c = 31{,}748$ Pa and $h = 31{,}748/2649 = 12$ m.

Answer: The OWC lies 1.9 m above the free water level. The transition zone with mobile water extends to about 12 m, and above that the water practically cannot move.

What it means. An interval in the lowest metres above the OWC will produce a lot of water. The higher the perforations (the holes shot through the casing), the lower the water cut (the share of water in the produced liquid). Water-free inflow can be expected only from about 12 m above the free water level. The data in this example are assumed for illustration.

Bulk volume water and the indicator of irreducible water

The product $\phi\,S_w$ is the volume of water per unit volume of rock (bulk volume water, BVW). In the zone of irreducible water it is nearly constant for one rock type. The smaller the pores, the lower the porosity, and the larger the share of the pores that holds bound water. On a plot of $S_w$ against $\phi$ , such points lie on the hyperbola $\phi\,S_w = \text{const}$ .

This is a handy check. If the log gives almost the same $\phi\,S_w$ throughout an interval, the interval most likely lies above the transition zone and will give water-free production. If $\phi\,S_w$ scatters and grows downwards through the section, this points to a transition zone. Archie's equation with $m = n = 2$ has a downside too. In it, $\phi\,S_w = \sqrt{a R_w/R_t}$ does not depend on porosity. So the whole error in porosity passes into the hydrocarbon volume $\phi - \phi S_w$ , and is amplified.

Connections to other topics

Porosity — the storage capacity of the rock; every saturation refers to its pore volume.
Capillary pressure and the transition zone — capillary pressure curves, and how the saturation profile above the OWC is built.
Archie's equation and saturation of clean reservoirs — calculating water saturation from resistivity and choosing the coefficients of Archie's equation.
Relative permeability — how the permeability to each phase depends on saturation; the end points are $S_{wirr}$ and $S_{or}$ .
Wettability and interfacial phenomena — why water or oil is held on the grain surfaces and how this changes the end-point saturations.
Volumetric estimation of hydrocarbons in place — how saturation enters the calculation of hydrocarbons in place.

Check yourself

1
In a sample with a porosity of 25%, water occupies 5% of the rock volume and the remaining pores are filled with oil. What is the oil saturation?
1. A.5%
2. B.20%
3. C.80%
4. D.75%
Show the answer
Correct answer: C. 80%
Saturation refers to the pore volume: $S_w = 0.05/0.25 = 0.20$ , so $S_o = 1 - 0.20 = 0.80$ , that is 80%.
2
One well penetrated the formation 30 m above the OWC and produces oil with no water. Another well, 3 m above the OWC, produces oil with water. How can this be explained?
1. A.The second well has poor-quality cementing
2. B.The second well penetrated the transition zone, where both fluids are mobile
3. C.The first well has a higher irreducible water saturation
4. D.The oil in the first well is lighter
Show the answer
Correct answer: B. The second well penetrated the transition zone, where both fluids are mobile
Above the OWC the water saturation decreases with height. Not far above the contact it is greater than the irreducible value, so the water is mobile and the well produces a mixture. High above the contact the water is bound and cannot move. Poor cement can also let water in, but that is a separate hypothesis, and it is checked by other surveys.
3
Why is core taken with oil-based mud preferred for finding the reservoir water saturation, while it is unsuitable for the residual oil saturation?
Show the answer
The filtrate of such mud pushes the mobile oil out of the core. But the bound water cannot move, so it stays, and the core keeps the irreducible water saturation. The oil in the core, however, is mixed with filtrate, has shrunk, and was partly pushed out by gas on the way up. So this core tells us nothing about the reservoir or residual oil saturation. The residual oil saturation is measured in laboratory displacement experiments on core.
4
A clean sandstone: porosity 0.18, $R_t = 25$ Ω·m, $R_w = 0.04$ Ω·m, $a = 1$ , $m = n = 2$ . Find the water saturation and the volume of oil in 1 m³ of rock.
Show the answer
$S_w = \sqrt{0.04/(0.18^2\cdot 25)} = \sqrt{0.04/0.81} = 0.222$ . The oil saturation is $1 - 0.222 = 0.778$ , and the volume of oil in 1 m³ of rock is $0.18\cdot 0.778 = 0.14$ m³ at reservoir conditions.
5
The initial water saturation is 0.20, the residual oil saturation after waterflooding is 0.35. What is the displacement efficiency?
1. A.0.45
2. B.0.56
3. C.0.35
4. D.0.80
Show the answer
Correct answer: B. 0.56
$E_D = (1 - 0.20 - 0.35)/(1 - 0.20) = 0.45/0.80 = 0.56$ . The number 0.45 is the moveable oil as a fraction of the pore volume, not as a fraction of the initial oil.
6
How will the height of the transition zone change if gas rather than oil lies above the water in the same rock?
Show the answer
Capillary pressure builds up with the density difference. For the water–gas pair this difference is roughly three times greater than for the water–oil pair, so the same capillary pressure is reached at a height three times smaller. The gas–water interfacial tension is higher than the oil–water one, which partly offsets the effect. Overall, though, the transition zone of gas accumulations is noticeably thinner.

Updated: 22 September 2026