Two-phase flow and reservoir description from formation-tester data

Quick refresher

If both phases are mobile, each flows by its own Darcy's law, and they share one pressure (if we ignore capillary pressure). So the gradient equals $g$ times the density weighted by the mobilities, plus the total vertical Darcy velocity divided by the total mobility.
In countercurrent flow there is no total flow, and the permeability drops out of the answer. Only the densities of the phases and the ratio of their mobilities set the gradient. Gas is tens of times less viscous than oil. So the gradient has moved half of the way from the oil gradient to the gas gradient as soon as the gas saturation exceeds the critical value (the gas saturation at which gas starts to move) by a few per cent, and it becomes almost a gas gradient long before the top of the zone.
So in a gassing zone the tester sees the gas line almost down to the base of the zone. The crossing of the gas line and the oil line marks the base of the two-phase zone, not its top, and there is still a lot of mobile oil above it.
In co-current flow a viscous part is added to the weighted density. In a good reservoir rock this is a few hundredths to tenths of a kPa/m. In a laminated unit with low vertical permeability it is a few kPa/m, and it is largest in the middle of the two-phase zone, where the total mobility is lowest.
In a water-wet rock (one in which water coats the grains) the tool shows the pressure of the oil as long as the oil is continuous. If the two-phase zone is not thicker than the capillary height (the height to which surface forces hold the water up), the capillary term cancels the weighting, and the profile looks like the oil line down to the current free water level (FWL, the level where the oil and water pressures are equal). An intermediate gradient is seen only when the zone is much thicker.
Residual oil in the swept zone (the part of the accumulation that water has passed through) has no effect on the pressure: the tool sees water, while the log sees oil. The rise of the current FWL, multiplied by the area, the porosity and the range of mobile saturation (the share of the pore space in which water can replace oil), is the volume of water that has entered the accumulation and stayed in it.

Key relationships

\dfrac{\dd p}{\dd z} = g\bar\rho + \dfrac{u_t}{\lambda_t},\qquad \bar\rho = \dfrac{\lambda_1\rho_1 + \lambda_2\rho_2}{\lambda_1 + \lambda_2}

gradient with two phases moving: $z$ is downward, $u_t > 0$ is upward flow, $\lambda_j = k_zk_{rj}/\mu_j$

\bar\rho = \dfrac{\rho_o + M\rho_g}{1 + M},\qquad M = \dfrac{k_{rg}/\mu_g}{k_{ro}/\mu_o}

countercurrent flow of gas and oil: the density from the gradient

G_{\text{oil}} = G + f_w\,\dfrac{\dd p_c}{\dd z},\qquad f_w = \dfrac{\lambda_w}{\lambda_t}

gradient of the oil pressure with the capillary term; $p_c = p_o - p_w$

W_e - W_pB_w \approx A\,\phi\,(1 - S_{wi} - S_{or})\,\Delta z

water that has entered the accumulation and stayed in it, from the rise of the FWL with bottom water (water that lies under the whole oil zone)

Common pitfalls

Taking an intermediate gradient for the density of a “third fluid” or for a separate pressure system.
Taking the crossing of the gas line and the oil line in a gassing zone for the top of the oil: there can be a lot of mobile oil above it.
Assuming a hydrostatic gradient for the water line in a producing accumulation without checking whether the water is flowing.
Taking the oil that the log shows below the current FWL for mobile oil or for oil that has not been swept.
Treating vertical pressure equilibrium during depletion as the same thing as vertical saturation equilibrium during displacement.

After this lesson you can

derive the pressure gradient in a zone where two phases move from Darcy's law for each phase, and explain why it lies between the gradients of the phases and is weighted by mobilities, not by saturations
calculate the density from the gradient (the density that the measured gradient corresponds to) in countercurrent flow, where the phases move in opposite directions (a gassing zone, separation by gravity), and the gradient in co-current flow, where both phases move the same way, including the viscous part; estimate at what saturation the profile is already almost a gas profile or a water profile
allow for capillary pressure (the pressure difference between the phases that surface forces in the small pores cause): which phase's pressure the tool shows, and when a two-phase zone on the profile cannot be told from the transition zone (the interval where the water saturation changes gradually) of an accumulation at rest (one that is not producing)
find the current contacts in a producing accumulation, tell them from the contacts seen on logs and from the zone of residual oil, and estimate how far a contact shifts when we assume a hydrostatic gradient where water is in fact flowing
know what vertical interference tests (tests that watch how a pressure change at one depth reaches another depth) tell us and how they add to pressure profiles
convert the rise of a contact into water influx, and use it to check the material balance and to tune the aquifer model

Where two phases move in a producing accumulation

In an accumulation that is not producing, the phases are at rest. Each continuous phase gives its own straight line, and the transition zone at a contact (the zone where the saturation changes gradually from oil to water) is in capillary equilibrium (which pressure the tester sees). Production breaks this equilibrium in several places, and a new well may cross an interval where two phases are mobile at the same time.

Two-phase intervals in a producing accumulation

Where: Gassing zone below a secondary gas cap
What moves: gas up, oil down
How: countercurrent flow
What to expect on the profile: from the oil gradient at the base to the gas gradient; it quickly becomes almost a gas gradient

Where: Swept zone below an OWC that has risen
What moves: water; the oil is residual and does not move
How: one phase
What to expect on the profile: a water gradient, with a viscous part if the water flows

Where: Displacement zone at the water front
What moves: water and oil
How: co-current or countercurrent flow
What to expect on the profile: intermediate; steeper than the water gradient if $k_z$ is low

Where: Gas cap into which water has pushed oil
What moves: gas and oil
How: co-current or countercurrent flow
What to expect on the profile: steeper than the gas gradient: oil moves into the cap

Where: Gas accumulation with rising water
What moves: water and gas
How: co-current flow
What to expect on the profile: from the water gradient to the gas gradient

Where	What moves	How	What to expect on the profile
Gassing zone below a secondary gas cap	gas up, oil down	countercurrent flow	from the oil gradient at the base to the gas gradient; it quickly becomes almost a gas gradient
Swept zone below an OWC that has risen	water; the oil is residual and does not move	one phase	a water gradient, with a viscous part if the water flows
Displacement zone at the water front	water and oil	co-current or countercurrent flow	intermediate; steeper than the water gradient if $k_z$ is low
Gas cap into which water has pushed oil	gas and oil	co-current or countercurrent flow	steeper than the gas gradient: oil moves into the cap
Gas accumulation with rising water	water and gas	co-current flow	from the water gradient to the gas gradient

For a water-wet rock and a well drilled with water-based mud. OWC is the oil–water contact, and FWL is the free water level. The single-phase profiles of a producing accumulation (vertical flow, barriers, tight units) are covered in the lesson Pressure measurements in producing reservoirs: barriers and accuracy.

A two-phase zone is not the same thing as the transition zone at a contact in an accumulation at rest. In a transition zone both phases can be continuous, but they do not flow. The capillary pressure balances the density difference, and each phase follows its own hydrostatics. In a two-phase zone the phases flow, and the structure of the zone is set by the withdrawal, by gravity and by the mobilities of the phases. There are two limiting pictures of the flow. One is countercurrent flow, where the phases move towards each other and there is no total flow. The other is co-current flow, where both move in the same direction, as when rising water displaces oil.

Pressure gradient when two phases move

Write Darcy's law for each phase in the vertical direction (two-phase Darcy's law). We measure the depth $z$ downward and take the vertical Darcy velocity $u$ as positive upward, as in the lesson on vertical flow. A phase flows upward if the pressure grows with depth faster than its hydrostatic pressure: $u_j = \lambda_j(\dd p/\dd z - \rho_j g)$ , where $\lambda_j = k_zk_{rj}/\mu_j$ is its vertical mobility (how easily it flows). While we ignore capillary pressure, the two phases share one pressure. Adding the velocities gives the total velocity $u_t$ , and from it the gradient.

Pressure gradient with vertical flow of two phases

\frac{\dd p}{\dd z} = g\bar\rho + \frac{u_t}{\lambda_t},\qquad \bar\rho = \frac{\lambda_1\rho_1 + \lambda_2\rho_2}{\lambda_1 + \lambda_2},\qquad \lambda_j = \frac{k_z\,k_{rj}}{\mu_j}

where:

$\dd p/\dd z$ pressure gradient with depth, as given by the pretests, Pa/m
$z$ true vertical depth, measured downward, m
$\bar\rho$ density weighted by the mobilities of the phases, which is the density shown by the gradient, kg/m³
$\rho_1,\ \rho_2$ densities of the phases at reservoir conditions, kg/m³
$\lambda_j,\ \lambda_t$ vertical mobility of phase $j$ , and total mobility $\lambda_1 + \lambda_2$ , m²/(Pa·s)
$k_z,\ k_{rj},\ \mu_j$ vertical permeability, relative permeability (the share of the permeability that the phase can use) and viscosity of phase $j$ , m², —, Pa·s
$u_t$ total vertical Darcy velocity $u_1 + u_2$ , positive for upward flow, m/s
$g$ acceleration of gravity, 9.81, m/s²

Units: SI

Holds when

one-dimensional vertical flow in a uniform unit; the phases are incompressible over the measured interval
the pressures of the phases are equal, so capillary pressure is ignored (the correction is in the section on capillary pressure)
the formula is local: the mobilities are taken at the saturation of the point in question. If the saturation changes with depth, the gradient changes with it, and the profile bends

The first term is the hydrostatic pressure of the mixture. But the mixture is weighted not by the saturations but by the mobilities: $\bar\rho = \rho_1 + f_2(\rho_2 - \rho_1)$ , where $f_2 = \lambda_2/\lambda_t$ is the share of phase 2 in the total mobility. The second term is the viscous part, as in single-phase flow (gradient with vertical flow), but with the total mobility. With only one mobile phase the formula turns into the single-phase one.

In other unit systems (1)

Metric oilfield units

G\,[\text{kPa/m}] = 9.81\,\bar\rho\,[\text{g/cm}^3] + \frac{1000\,u_t\,[\text{m/d}]}{0.0853\,\lambda_t\,[\text{mD/(mPa·s)}]}

$\bar\rho$ is in g/cm³, $u_t$ is in m/d (cubic metres per day through a square metre), $\lambda_t = k_z(k_{r1}/\mu_1 + k_{r2}/\mu_2)$ is in mD/(mPa·s), $G$ is in kPa/m.

9.81 kPa/m is the pressure of a one-metre column of liquid with a density of 1 g/cm³. 0.0853 is the constant of Darcy's law in metric oilfield units (Darcy's law). 1000 converts MPa to kPa.

This gives two limiting cases. In countercurrent flow there is no total flow ( $u_t = 0$ ), and the gradient is pure hydrostatics of the weighted density. The vertical permeability drops out of the answer, because it enters both mobilities in the same way. In co-current flow the viscous part $u_t/\lambda_t$ is added to it. It becomes noticeable only when the vertical permeability is low or the flow is large, as in the single-phase case.

Countercurrent flow: the gassing zone

When the pressure in the accumulation falls below the bubble point pressure, gas comes out of the oil. While its saturation is below the critical value (the gas saturation at which gas starts to move), the gas does not move. After that the gas starts to rise towards the top and forms a secondary gas cap, while the oil drains down towards it. Between the cap, where the oil is already residual (left behind and immobile), and the oil zone, where the gas is still immobile, lies the gassing zone, the interval of countercurrent flow. Its thickness is set by the rate of pressure decline and by the vertical permeability: the faster the phases separate, the thinner it is (gravity drainage, Performance prediction for solution gas drive reservoirs).

Density from the gradient in countercurrent flow of gas and oil

\bar\rho = \frac{\rho_o + M\rho_g}{1 + M},\qquad M = \frac{k_{rg}/\mu_g}{k_{ro}/\mu_o}

where:

$\bar\rho$ density from the pressure gradient in the countercurrent zone, g/cm³
$\rho_o,\ \rho_g$ densities of oil and gas at reservoir conditions, g/cm³
$M$ ratio of the mobilities of gas and oil at the current gas saturation, —
$k_{rg},\ k_{ro}$ relative permeabilities of gas and oil at the current gas saturation, —
$\mu_g,\ \mu_o$ viscosities of gas and oil at reservoir conditions, mPa·s

Units: densities in any identical units; the gradient is $g\bar\rho$

Holds when

there is no total vertical flow: the gas rises and the oil drains down at the same volumetric rate
the gas–oil capillary pressure is negligible (the interfacial tension between the two fluids is low); the water is immobile at irreducible saturation (the water that stays in the pores and cannot be displaced)

It is convenient to read the formula as a fraction of a path. $\bar\rho$ moves from $\rho_o$ towards $\rho_g$ by the fraction $M/(1 + M)$ , which is the share of the gas in the total mobility. Half of the way is covered at $M = 1$ , that is when $k_{rg}/k_{ro} = \mu_g/\mu_o$ . Gas is tens of times less viscous than oil. So it is enough for $k_{rg}$ to be tens of times smaller than $k_{ro}$ , and this happens already a few per cent above the critical gas saturation (with the relative permeabilities and fluids of the example below, at $S_g \approx 0.08$ ). The oil gradient stays only where the gas is barely mobile, and the gradient becomes almost a gas gradient long before the top of the zone.

Worked exampleDensity from the gradient in a gassing zone

Given (assumed data)

Rock and relative permeabilities (Performance prediction for solution gas drive reservoirs): $k_{ro} = 0.80\,[(0.55 - S_g)/0.55]^2$ , $k_{rg} = 0.30\,[(S_g - 0.03)/0.52]^{1.5}$ ; critical gas saturation 0.03, irreducible water 0.25
Fluids at 16 MPa: oil 1.08 mPa·s and 0.716 g/cm³; gas 0.018 mPa·s and 0.134 g/cm³
Flow: countercurrent flow with no total vertical flow

Solution

At $S_g = 0.06$ : $k_{ro} = 0.80\cdot(0.49/0.55)^2 = 0.635$ and $k_{rg} = 0.30\cdot(0.03/0.52)^{1.5} = 0.00416$ .
Mobility ratio: $M = \dfrac{0.00416/0.018}{0.635/1.08} = 0.393$ . Density from the gradient: $\bar\rho = \dfrac{0.716 + 0.393\cdot 0.134}{1 + 0.393} = 0.552$ g/cm³. Gradient: $9.81\cdot 0.552 = 5.42$ kPa/m.
In the same way at $S_g$ = 0.04, 0.10 and 0.20: $M$ = 0.070, 1.66 and 10.4, and $\bar\rho$ = 0.678, 0.353 and 0.185 g/cm³.
Fraction of the way from the oil gradient to the gas gradient, $(0.716 - \bar\rho)/(0.716 - 0.134)$ : 0.07, 0.28, 0.62 and 0.91 at $S_g$ = 0.04, 0.06, 0.10 and 0.20.
For comparison, if the mixture were weighted by the saturations (the oil saturation at $S_g = 0.10$ is 0.65), we would get $(0.65\cdot 0.716 + 0.10\cdot 0.134)/0.75 = 0.638$ g/cm³.

Answer: At

S_g = 0.10

, only 0.07 above the critical value, the gradient has covered 62% of the way from the oil gradient to the gas gradient. At 0.20 it has covered 91%.

What it means. Here the gas is 60 times less viscous than the oil. So as soon as it becomes mobile, it “takes over” the gradient. By saturations, the mixture at $S_g = 0.10$ would be almost oil (0.638 g/cm³), but by mobilities it is already closer to gas (0.353). Two conclusions follow. The oil gradient stays only at the very base of the zone, where the gas is barely mobile. Above that, a gassing zone is hard to tell from a gas cap by its gradient: at $S_g = 0.20$ the gradient is steeper than the gas gradient by only 0.5 kPa/m. With a more viscous oil the change is even sharper. With a higher critical gas saturation it shifts towards higher gas saturation. The data in this example are assumed for illustration.

CalculatedDensity from the gradient in countercurrent flow, against saturation. Top: gas and oil (relative permeabilities and fluids of the example, 16 MPa). Bottom: water and oil (rock and fluids of the field with bottom water, 19.6 MPa). The dashed horizontal lines are the densities of the phases. The dotted line at the top is the density of the mixture weighted by saturations, for comparison.

What will the tester see in a well that crosses a secondary gas cap and a gassing zone? Below is a survey of an observation well in such a column. The distribution of gas in the zone is assumed: the gas saturation decreases linearly from 0.55 at the top (2190 m) to the critical value of 0.03 at the base (2205 m).

Pretests of an observation well in a column with a secondary gas cap

Depth, m: 2181.0
Pressure, MPa: 15.9568
Where: secondary gas cap

Depth, m: 2184.0
Pressure, MPa: 15.9598
Where: secondary gas cap

Depth, m: 2187.0
Pressure, MPa: 15.9645
Where: secondary gas cap

Depth, m: 2191.0
Pressure, MPa: 15.9692
Where: gas on the log, weaker with depth

Depth, m: 2194.0
Pressure, MPa: 15.9740
Where: gas on the log, weaker with depth

Depth, m: 2197.0
Pressure, MPa: 15.9772
Where: gas on the log, weaker with depth

Depth, m: 2200.0
Pressure, MPa: 15.9827
Where: gas on the log, weaker with depth

Depth, m: 2203.0
Pressure, MPa: 15.9894
Where: gas on the log, weaker with depth

Depth, m: 2207.0
Pressure, MPa: 16.0145
Where: oil

Depth, m: 2210.0
Pressure, MPa: 16.0345
Where: oil

Depth, m: 2213.0
Pressure, MPa: 16.0564
Where: oil

Depth, m: 2216.0
Pressure, MPa: 16.0772
Where: oil

Depth, m	Pressure, MPa	Where
2181.0	15.9568	secondary gas cap
2184.0	15.9598	secondary gas cap
2187.0	15.9645	secondary gas cap
2191.0	15.9692	gas on the log, weaker with depth
2194.0	15.9740	gas on the log, weaker with depth
2197.0	15.9772	gas on the log, weaker with depth
2200.0	15.9827	gas on the log, weaker with depth
2203.0	15.9894	gas on the log, weaker with depth
2207.0	16.0145	oil
2210.0	16.0345	oil
2213.0	16.0564	oil
2216.0	16.0772	oil

All the readings are valid (they passed the quality checks), the pressures are absolute and the depths are true vertical depths. The scatter is about 0.5 kPa. “Gas on the log” means that the neutron and density curves cross over, which happens when gas fills the pores (gas effect). The data are assumed for illustration.

Worked exampleWhere the tester “sees” the gas–oil contact

Given (assumed data)

Survey: the table above
Fluids: as in the previous example
From the logs: gas is seen from the top of the formation down to 2205 m, getting weaker with depth

Solution

The points at 2181–2197 m fall on one straight line: $G = 1.310$ kPa/m, a density of $1.310/9.81 = 0.134$ g/cm³, which is gas. The scatter is 0.5 kPa. The oil line through the four points below 2205 m is $G = 7.000$ kPa/m (0.714 g/cm³), with a scatter of 0.6 kPa.
At 2200 m the gas line gives 15.98132 MPa and the oil line gives 15.96515 MPa. The crossing is at $z^* = 2200 + (15.98132 - 15.96515)\cdot 1000/(7.000 - 1.310) = 2202.8$ m.
The points at 2200 and 2203 m lie to the right of the gas line by 1.4 and 4.1 kPa. This is the only trace of the two-phase zone. The gradient between them is $(15.9894 - 15.9827)\cdot 1000/3 = 2.23$ kPa/m, and the density is 0.228 g/cm³. On the curve of the previous example this corresponds to $S_g \approx 0.16$ .
The base of the gassing zone is at 2205 m and its top at 2190 m. The crossing lies 2.2 m above the base and 12.8 m below the top.

Answer: The gas–oil contact from the pressure is at 2202.8 m, at the base of the two-phase zone, not at its top.

What it means. By pressure, the gas “cap” extends almost down to 2203 m. But between 2190 and 2203 m the oil saturation in the model of the example is from 0.20 to 0.65, and the oil is mobile. This is a zone where gas and oil are separating, not a cap. So we compare the contact from pressure with the logs, and we choose the perforation interval with a margin below the zone: a well right below the crossing of the lines would produce a lot of gas. The local gradient at the base of the zone is sensitive to saturation, and it confirms the picture (about 0.16 against 0.15 in the model at 2201.5 m). But with a scatter of 0.5 kPa over three metres, its error is about 0.24 kPa/m. The distribution of gas in the zone is assumed here. In reality it is set by the release of gas and the rate of phase separation, and it is calculated with the reservoir model. The data in this example are assumed for illustration.

CalculatedSurvey of an observation well in a column with a secondary gas cap. Left: the gas saturation of the model. Right: the pressure excess over the oil line. The dots are pretests, the solid line is the calculated profile, and the dashed line is the gas line through the points at 2181–2197 m. The band marks the gassing zone.

Co-current flow: displacement from below and the viscous part

When edge water or injected water rises under the oil zone, both phases move upward. Above the front, oil is pushed upward. In the swept zone, water flows with residual oil, at the end-point relative permeability $k_{rw}'$ (the largest value of the water curve). Between them lies the two-phase zone. In each of the three zones the gradient follows the formula of the previous section, but the total mobility differs: in the oil zone it is $k_zk_{ro}'/\mu_o$ , in the swept zone it is $k_zk_{rw}'/\mu_w$ , and in the two-phase zone the phases hinder each other, and $\lambda_t$ passes through a minimum (Two-phase flow equations). The same flow needs the largest pressure difference there.

Worked exampleDisplacement from below in clean sandstone and in a laminated unit

Given (assumed data)

Rock (Why water is injected: pressure maintenance and oil displacement): $S_{wi} = S_{or} = 0.25$ ; $k_{ro}' = 0.80$ , $k_{rw}' = 0.20$ ; curve exponents 2.5 and 3
Fluids at 19.6 MPa: oil 1.03 mPa·s and 0.710 g/cm³; formation water 0.40 mPa·s and 1.060 g/cm³
Flow: edge water enters from below through an OWC area of 6.0 km², 2000 m³/d at reservoir conditions: $u_t = 2000/(6.0\cdot10^6) = 3.33\cdot10^{-4}$ m/d
Vertical permeability: clean sandstone 150 mD, laminated unit 3 mD

Solution

Mobilities per 1 mD at $S_w$ = 0.25 (oil zone), 0.50 (two-phase zone) and 0.75 (swept zone): oil 0.7767, 0.1373 and 0; water 0, 0.0625 and 0.5000 1/(mPa·s).
Densities weighted by mobilities: 0.710, $0.710 + \dfrac{0.0625}{0.1998}\cdot 0.350 = 0.8195$ and 1.060 g/cm³. Their hydrostatic parts $9.81\bar\rho$ are 6.97, 8.04 and 10.40 kPa/m.
Viscous part $1000\,u_t/(0.0853\,\lambda_t)$ at $k_z = 3$ mD ( $\lambda_t$ = 2.330, 0.599 and 1.500 mD/(mPa·s)): 1.68, 6.52 and 2.60 kPa/m. The gradients are 8.65, 14.56 and 13.00 kPa/m.
At $k_z = 150$ mD the viscous part is 50 times smaller: 0.034, 0.130 and 0.052 kPa/m.
The largest gradient of the laminated unit is at $S_w \approx 0.58$ : 16.1 kPa/m. The lowest total mobility is at $S_w \approx 0.54$ , and it is 2.7 times lower than in the swept zone.

Answer: In clean sandstone the viscous part is no larger than 0.14 kPa/m, and the profile is hydrostatic. In the laminated unit the gradient in the two-phase zone reaches 16.1 kPa/m, which is steeper than in the swept zone below it (13.00).

What it means. The “density” 16.1/9.81 = 1.64 g/cm³ is heavier than any formation water. Such a gradient is a sign of flow, not of hydrostatics. In a unit with low vertical permeability the displacement zone shows on the profile as the interval with the steepest gradient. In a good reservoir rock it does not: the viscous part there is at the level of the error of the gradient (about 0.05 kPa/m). The flow is taken as the average over the OWC area. Near producing wells and under injection wells it is larger. The data in this example are assumed for illustration.

CalculatedPressure gradient when rising water displaces oil, against water saturation (the data of the example), for a vertical permeability of 3 mD and of 150 mD. The dashed horizontal lines are the hydrostatic gradients of the oil and of the water.

Capillary pressure: which phase's pressure the tool sees

The pressures of the phases differ by the capillary pressure $p_c = p_o - p_w$ , which depends on saturation. The tester measures the pressure of the liquid at the probe. In a water-wet rock with water-based mud, this is the pressure of the continuous oil, in the oil zone and in the transition zone (which pressure the tester sees). The same holds in a two-phase zone as long as the oil in it is continuous. In the swept zone the oil is residual and broken up into separate droplets, and the tool shows the pressure of the water.

Gradient of the oil pressure in a two-phase zone with the capillary term

G_{\text{oil}} = \frac{\dd p_o}{\dd z} = G + f_w\,\frac{\dd p_c}{\dd z},\qquad f_w = \frac{\lambda_w}{\lambda_t}

where:

$G_{\text{oil}}$ gradient of the oil pressure, which is what the pretests show in a water-wet rock, Pa/m
$G$ gradient without capillary pressure, the formula of the section on the two-phase gradient, Pa/m
$f_w$ share of water in the total mobility, —
$\dd p_c/\dd z$ change of the capillary pressure with depth: $(\dd p_c/\dd S_w)(\dd S_w/\dd z)$ . In a zone where there is more water towards the bottom, it is negative, Pa/m

Units: SI (in practice kPa/m)

Holds when

a water-wet rock, with continuous oil in the zone; the capillary pressure curve is the imbibition curve (the curve for water entering the rock) if the zone is being flooded (imbibition profile)
the other assumptions of the two-phase gradient; the derivation is in the advanced section

The capillary term moves the oil gradient towards the oil line: $\dd p_c/\dd z < 0$ . In a transition zone at rest, $\dd p_c/\dd z = -(\rho_w - \rho_o)g$ and $G = g\bar\rho = g[\rho_o + f_w(\rho_w - \rho_o)]$ , and the sum gives exactly $\rho_og$ . A transition zone in equilibrium shows the oil line, whatever the mobilities are. The water pressure meanwhile follows its own hydrostatics.

This gives the yardstick for measuring a two-phase zone: the capillary height $h_c = \Delta p_c/[(\rho_w - \rho_o)g]$ , where $\Delta p_c$ is the drop of the imbibition capillary pressure across the zone. If the two-phase zone is not thicker than $h_c$ , the saturation in it is distributed almost as in capillary equilibrium. This is the stabilized zone of the front (a zone whose shape no longer changes as the front moves). Then the tool sees the continuation of the oil line down to the current FWL. An intermediate gradient appears only when the zone is much thicker than $h_c$ : with low vertical permeability, with heterogeneity or with fast displacement.

Worked exampleA thin and a thick two-phase zone

Given (assumed data)

Zone: the rock and fluids of the previous example; no total flow; in the middle of the zone $S_w = 0.50$ , and the mobilities per 1 mD are 0.1373 (oil) and 0.0625 (water) 1/(mPa·s)
Imbibition capillary pressure: from 8 kPa at the upper boundary of the zone to zero at the lower one (assumed)
Thickness of the two-phase zone: (a) equal to the capillary height; (b) 12 m

Solution

$f_w = 0.0625/0.1998 = 0.313$ . The gradient without capillary pressure is $9.81\cdot 0.8195 = 8.04$ kPa/m.
Capillary height: $h_c = 8/(9.81\cdot 0.350) = 2.33$ m. (a) $\dd p_c/\dd z = -8/2.33 = -3.43$ kPa/m, and $G_{\text{oil}} = 8.04 + 0.313\cdot(-3.43) = 6.97$ kPa/m. The oil gradient is $9.81\cdot 0.710 = 6.97$ .
(b) $\dd p_c/\dd z = -8/12 = -0.667$ kPa/m, and $G_{\text{oil}} = 8.04 + 0.313\cdot(-0.667) = 7.83$ kPa/m.

Answer: A zone as thick as the capillary height (2.33 m) gives the oil gradient, 6.97 kPa/m. A zone of 12 m gives 7.83 kPa/m, which is 0.86 kPa/m steeper than the oil gradient.

What it means. If the water and oil in the zone are distributed almost as in capillary equilibrium, then the tool, which shows the pressure of the oil, sees on the profile what an accumulation at rest would show in a transition zone: the oil line down to the current FWL. We can tell such a zone from an ordinary transition zone only from the logs (imbibition leaves more water at the same height) and from the history of the contact. In a thick zone the capillary term only slightly softens the weighting by mobilities, and the gradient is clearly steeper than the oil gradient. For the gas–oil pair the interfacial tension is low and the capillary height is decimetres. So a gassing zone metres thick shows the gradient by mobilities almost without a correction. The data in this example are assumed for illustration.

Current contacts in a producing accumulation

The main practical case follows from the previous section. Suppose the phases separate quickly (vertical equilibrium of saturation, see when the phases have time to separate). Then the two-phase zone is not thicker than the capillary height, the capillary term cancels the weighting, and the profile looks as it does in an accumulation at rest: the oil line above and the water line below. Their crossing is the current FWL, the level where the pressures of the phases are equal now. With bottom water it rises as production goes on. When a gas cap expands, the gas–oil contact moves down.

Below the current FWL lies the swept zone. The oil in it is residual, and the pressure there is the water pressure. The resistivity log and the pulsed-neutron log see the residual oil saturation (the oil left behind in the pores), about a quarter of the pore volume. Both levels show on the logs. At the top is the transition to the swept zone, where the water saturation rises to $1 - S_{or}$ . At the bottom is the former contact, below which there is no oil at all. If we put the contact from the logs where the oil disappears, it stays at the former level, while the contact from the pressure has risen. Between them lies the swept interval, a kind of young paleo-contact (an old contact that the fluids have since left). This is not a contradiction. It is a measure of the rise.

We find the lines from the points of each zone, with their own gradients. In a producing accumulation we can set the gradient of the water line to the hydrostatic one, using the density from a sample (Forced-gradient technique and multi-well analysis), only after making sure that the water is not flowing noticeably. Upward flow makes the true line steeper. A hydrostatic line that we set in this way, drawn through the water points, then ends up to the right of the true line at the contact.

Shift of the contact if the water line is set to the hydrostatic gradient

\delta z^* = -\frac{G_{\text{visc}}\,L}{G_w - G_o}

where:

$\delta z^*$ shift of the contact that we find; plus means downward, as in the lesson Forced-gradient technique and multi-well analysis. With upward flow ( $G_{\text{visc}} > 0$ ) it is negative: the contact is found above the true one. With downward flow it is found below, m
$G_{\text{visc}}$ viscous component of the water gradient over the interval of its points, Pa/m
$L$ vertical distance from the mean depth of the water points to the contact, m
$G_w - G_o$ difference between the hydrostatic gradient that we set for the water and the gradient of the oil, Pa/m

Units: any consistent units (kPa/m and m)

Holds when

the line with the set gradient passes through the middle point of the water readings, and the true oil and water lines are straight. Then the formula is exact. With the measured, steeper water gradient in the denominator, it would underestimate the shift
the oil line is found from its own points, without a shift

The error grows with the viscous part and with the length of the extrapolation. In a good reservoir rock it is about ten centimetres, and in a laminated unit it is metres (see the example below).

Pretests of a new well in the fifth year of production (a field with bottom water)

Depth, m: 1949.5
Pressure, MPa: 19.3571
Where: oil zone

Depth, m: 1954.0
Pressure, MPa: 19.3869
Where: oil zone

Depth, m: 1958.5
Pressure, MPa: 19.4196
Where: oil zone

Depth, m: 1963.0
Pressure, MPa: 19.4502
Where: oil zone

Depth, m: 1967.5
Pressure, MPa: 19.4828
Where: oil zone

Depth, m: 1972.0
Pressure, MPa: 19.5125
Where: oil zone

Depth, m: 1978.5
Pressure, MPa: 19.5631
Where: residual oil on the log

Depth, m: 1981.0
Pressure, MPa: 19.5904
Where: residual oil on the log

Depth, m: 1984.0
Pressure, MPa: 19.6210
Where: water leg

Depth, m: 1987.5
Pressure, MPa: 19.6563
Where: water leg

Depth, m: 1991.0
Pressure, MPa: 19.6941
Where: water leg

Depth, m	Pressure, MPa	Where
1949.5	19.3571	oil zone
1954.0	19.3869	oil zone
1958.5	19.4196	oil zone
1963.0	19.4502	oil zone
1967.5	19.4828	oil zone
1972.0	19.5125	oil zone
1978.5	19.5631	residual oil on the log
1981.0	19.5904	residual oil on the log
1984.0	19.6210	water leg
1987.5	19.6563	water leg
1991.0	19.6941	water leg

The well was drilled with water-based mud. The water leg is the water-filled part of the section below the oil. The readings are valid, the pressures are absolute and the depths are true vertical depths. The FWL at discovery (the first well) is 1982.0 m. Densities from the samples: oil 0.710 and water 1.060 g/cm³ at reservoir conditions. The data are assumed for illustration.

Worked exampleCurrent FWL in a new well and its rise

Given (assumed data)

Survey: the table above
Initial FWL: 1982.0 ± 0.3 m; the depths of the two wells are matched to ±0.3 m
Flow: average water inflow 2000 m³/d through 6.0 km²; vertical permeability of the sandstone 150 mD; $k_{rw}' = 0.20$

Solution

Oil line through six points: $\bar z = 1960.75$ m, $\sum(z - \bar z)^2 = 354.4$ m², $G_o = 6.954$ kPa/m (0.709 g/cm³), scatter $s = 0.85$ kPa. Water line through five points, two in the swept zone and three in the water leg: $\bar z = 1984.4$ m, $\sum(z - \bar z)^2 = 99.7$ m², $G_w = 10.407$ kPa/m (1.061 g/cm³), $s = 0.81$ kPa.
At 1977 m: $p_o = 19.54786$ and $p_w = 19.54797$ MPa. The crossing is at $z^* = 1977 + (19.54786 - 19.54797)\cdot 1000/3.453 = 1976.97$ m.
Errors of the lines at this depth (contact uncertainty): $0.85\sqrt{1/6 + (z^* - 1960.75)^2/354.4} = 0.81$ kPa for the oil line and 0.70 kPa for the water line. Then $\sigma_{z^*} = \sqrt{0.81^2 + 0.70^2}/3.453 = 0.31$ m.
Rise: $\Delta z = 1982.0 - 1976.97 = 5.03$ m. With the errors of the initial FWL and of the depth matching, the error is $\sqrt{0.31^2 + 0.3^2 + 0.3^2} = 0.53$ m.
Is the water flowing? $u_t = 3.33\cdot10^{-4}$ m/d. The viscous part in the swept zone is $1000\cdot3.33\cdot10^{-4}\cdot 0.40/(0.0853\cdot150\cdot0.20) = 0.052$ kPa/m, and in the water leg it is 0.010 kPa/m, which is less than the error of the slope, 0.081 kPa/m. A water line set to the hydrostatic gradient would shift the contact by no more than $0.052\cdot7.4/(10.399 - 6.954) = 0.11$ m ( $10.399 = 9.81\cdot 1.060$ is the water gradient set from the sample). In a laminated unit with $k_z$ = 3 mD the viscous part is 2.60 kPa/m, and the shift is already 5.6 m.

Answer: The current FWL is at 1976.97 ± 0.31 m. In five years it has risen by 5.03 ± 0.53 m.

What it means. The points at 1978.5 and 1981.0 m lie on the water line, although the log sees residual oil there. In the swept zone the oil is disconnected, and the tool shows the water pressure. The accuracy of the rise is set not by the pressure gauge but by the initial contact and by the depth matching of the two wells. One well gives the rise at one place. Near producing wells the contact is higher (water is drawn up in cones towards the well; Water and gas coning), at the edges of the accumulation it is lower, and in an aquifer with flow across the area the contact is even tilted (Perched and tilted contacts, trapped water). We take the average over the accumulation from several new wells and from repeat logging of cased wells. The data in this example are assumed for illustration.

CalculatedA new well of a field with bottom water in the fifth year. Left: the water saturation (assumed): the oil zone, a thin transition zone of imbibition, the swept zone with residual oil, and the water leg. Right: the pressure excess over the water line. The points of the oil zone fall on a straight line that reaches zero at the current FWL. The points of the swept zone lie at zero together with those of the water leg.

Vertical interference tests in brief

A pressure profile shows a parting between units only when production forces fluid to flow through it. A vertical interference test checks the connection directly: we change the pressure in one interval and watch how another interval responds. The time and the size of the response give the transmissibility of the rock or of the parting between them.

Methods of vertical interference testing

Method: Multi-probe tester: a sink probe (it draws fluid) and an observation probe
Scale: 0.5–1 m
What it gives: $k_h$ and $k_v$ of the rock between the probes
Limitations: open hole; small volume

Method: Dual-packer module (two rubber seals that cut off a short interval of the hole) and a probe on the other side of a thin layer
Scale: metres
What it gives: transmissibility of the thin layer near the well
Limitations: small flow rates; storage capacity between the packers

Method: Cased well: production from one interval, a gauge in another, a packer between them
Scale: metres to tens of metres
What it gives: transmissibility of the parting around the well
Limitations: flow behind the casing; long records

Method: A pulse from a neighbouring well and a tester survey
Scale: hundreds of metres
What it gives: the parting over the area between wells
Limitations: needs a model of two units with crossflow

Method	Scale	What it gives	Limitations
Multi-probe tester: a sink probe (it draws fluid) and an observation probe	0.5–1 m	$k_h$ and $k_v$ of the rock between the probes	open hole; small volume
Dual-packer module (two rubber seals that cut off a short interval of the hole) and a probe on the other side of a thin layer	metres	transmissibility of the thin layer near the well	small flow rates; storage capacity between the packers
Cased well: production from one interval, a gauge in another, a packer between them	metres to tens of metres	transmissibility of the parting around the well	flow behind the casing; long records
A pulse from a neighbouring well and a tester survey	hundreds of metres	the parting over the area between wells	needs a model of two units with crossflow

The first method is covered in the lesson on the observation probe, and the last in the lesson on the pulse survey. Interference testing between wells (interference test) is a variety of the same principle.

In a two-phase zone the response is slower and harder to interpret. Free gas raises the total compressibility several times, and with it the time that a disturbance takes to cross a unit. The speed of the response is set by the total mobility of the phases, not by the single-phase permeability. So the “vertical permeability” found from it is converted through the relative permeabilities and compared with the model, not directly with the core.

Profiles and contacts in tuning the reservoir model

A reservoir model that has been matched to the history must reproduce three things in each new well on the date of the survey: the pressures of the units, the position of the current contacts, and the structure of the two-phase zones. The first checks the links between units and blocks (profiles and the model). The second and the third check where the water and the gas have gone.

Which phase's pressure. The model stores the pressure of one phase and the capillary pressure. We compare it with what the tool shows: in the oil-saturated part and in the two-phase zone of a water-wet rock, the oil pressure; in the swept zone, the water pressure (what to compare with what).
Contacts are saturation data. The rise of the FWL tests the water influx and the vertical sweep (how much of the thickness the water has covered). If the contact in the model at the new well is higher than the measured one, the water in the model rises too readily: the vertical permeability or the connection with the aquifer is too high.
Two-phase zones. The position of the base of the gas line tests the critical gas saturation and the gas relative permeability. Steep gradients in the displacement zone test the vertical permeability of the laminated units.

Water that has entered the accumulation, from the rise of the contact

W_e - W_pB_w \approx A\,\phi\,(1 - S_{wi} - S_{or})\,\Delta z

where:

$W_e$ cumulative water influx from the aquifer at reservoir conditions (water influx), m³
$W_pB_w$ produced water at reservoir conditions, m³
$A$ area of the oil–water contact under which the bottom water rises, m²
$\phi$ porosity of the rocks through which the contact has risen, —
$S_{wi},\ S_{or}$ irreducible water saturation and residual oil saturation of the swept zone, —
$\Delta z$ rise of the current FWL relative to the initial one, found from pressure, m

Units: SI (m, m², m³)

Holds when

the water rises evenly over the whole area $A$ : bottom water, vertical saturation equilibrium. With cones near wells or inflow from the flanks, the rise in one well is not representative
the swept zone stays at residual oil saturation; the change in the shape of the transition zone and the compressibility of the pores and the water are ignored
the initial and the current FWL are both found from pressure, with common depth matching of the wells

Each metre of rise means $A\phi(1 - S_{wi} - S_{or})$ cubic metres of water that has stayed in the accumulation. Adding the produced water gives the influx. In this way the position of the contact links the influx that the balance of production and pressure requires (General material balance equation for oil reservoirs) with the place where this water has gone.

Worked exampleRise of the contact, water influx and the aquifer coefficient

Given (assumed data)

Rise of the FWL: from the previous example; we assume that the water rises evenly over the whole OWC area (bottom water, vertical equilibrium)
Accumulation: OWC area 6.0 km²; porosity 0.20; $S_{wi} = S_{or} = 0.25$
Five years of production: $N_p$ = 2.40 million m³, $W_p$ = 0.30 million m³ ( $B_w$ = 1.02); the average pressure has fallen from 21.0 to 19.6 MPa (above the bubble point pressure); $\int(p_i - p)\,\dd t$ = 2300 MPa·d
Oil (General material balance equation for oil reservoirs): $B_{oi}$ = 1.3435 at 21.0 MPa, $B_o$ = 1.3465 at 19.6 MPa; $(c_wS_w + c_f)/(1 - S_w) = 1.2167\cdot10^{-3}$ MPa⁻¹; volumetric oil in place 20.0 million m³

Solution

Water that has entered the accumulation and stayed in it: $6.0\cdot10^6\cdot 0.20\cdot 0.50\cdot 5.03 = 3.02$ million m³, with an error of $0.60\cdot 0.53 = 0.32$ million m³.
Balance with the volumetric oil in place (shorthand form): $F = 2.40\cdot 1.3465 + 0.30\cdot 1.02 = 3.538$ million m³. $E_o = 1.3465 - 1.3435 = 0.0030$ , $E_{fw} = 1.3435\cdot 1.2167\cdot10^{-3}\cdot 1.4 = 0.00229$ , $E_t = 0.00529$ . $NE_t = 20.0\cdot 0.00529 = 0.106$ million m³.
Influx from the balance: $W_e = 3.538 - 0.106 = 3.432$ million m³. The water that stayed in the accumulation is $3.432 - 0.306 = 3.126$ million m³, which means a rise of 5.21 m. From the pressure the value is 3.02 ± 0.32. The difference is 0.3 of the error, so the estimates agree.
Steady-state aquifer model (Aquifer models: water influx): $W_e = 3.02 + 0.306 = 3.326$ million m³, and $J_a = 3.326\cdot10^6/2300 = 1446$ m³/(d·MPa), with an error of ±139. The current influx is $1446\cdot 1.4 = 2024$ m³/d.
Oil in place from the rise of the contact: $N = (F - W_e)/E_t = (3.538 - 3.326)/0.00529 = 40$ million m³, and the influx error of ±0.32 million m³ gives ±60 million m³.

Answer: The rise of the contact confirms the influx that the balance requires, and it gives the aquifer coefficient 1446 ± 139 m³/(d·MPa). The oil in place cannot be found from it.

What it means. Two independent estimates agree: the balance of production and pressure with the volumetric oil in place, and an almost direct “measurement” of the water from the position of the contact. The agreement tests the area and the porosity under the contact, the range of mobile saturation and the even rise of the water. The expansion of the oil and of the pores gave only 3% of the withdrawal, and the rest is water. So the oil in place cannot be found from the balance here: an influx error of 0.32 million m³ is equal to 60 million m³ of oil in place. If the contact had risen noticeably less than the balance requires, the water would be entering not evenly from below, but from the flanks or through high-permeability units, leaving oil behind. Then we change the connection with the aquifer and the vertical permeability in the model, and not the oil in place. The $J_a$ that we found is the starting point for a pressure forecast, and we check it with the next survey. If there are several blocks, we keep the balance for each of them (Lumped multi-tank material balance model). The data in this example are assumed for illustration.

Terms and symbols

Two-phase pressure profile: A profile recorded by the tester where two phases are mobile. Its gradient lies between the gradients of the phases.
Mobility-weighted density $\bar\rho$: $(\lambda_1\rho_1 + \lambda_2\rho_2)/\lambda_t$ , the density that the gradient shows in countercurrent flow. It is close to the density of the more mobile phase.
Countercurrent and co-current flow: Flow of the phases towards each other with no total flow (separation by gravity), and flow in the same direction (displacement). In co-current flow a viscous part $u_t/\lambda_t$ is added to the hydrostatic gradient.
Gassing zone: The interval between a secondary gas cap and the oil zone, where gas rises and oil drains down. On a pressure profile almost all of it looks like gas.
Capillary height $h_c$: $\Delta p_c/[(\rho_w - \rho_o)g]$ . A two-phase zone that is not thicker than this shows the oil line on the profile.
Current FWL: The level where the oil pressure and the water pressure are equal now. We find it from the crossing of the lines in a new well.

Common mistakes and misconceptions

Wrong

“A gradient of 3.5 kPa/m below a gas cap means that there is a very light oil with a density of 0.36 g/cm³.”

In fact

Oil like this does not exist at ordinary reservoir conditions. This is a two-phase countercurrent zone, where the gradient is weighted by the mobilities of gas and oil. We take the oil density from a sample, and we read an intermediate gradient as a sign of mobile gas.

Wrong

“The crossing of the gas line and the oil line is the top of the oil: above it is the gas cap.”

In fact

In a gassing zone the gas line extends almost to the base of the zone, and there is still a lot of mobile oil above the crossing. We compare the contact from the pressure with the logs, and we choose the perforation interval from both.

Wrong

“The log sees oil down to 1982 m, so the OWC has not risen, and the points with a water gradient above it are a measurement error.”

In fact

Below the current FWL the residual oil is disconnected and immobile. The log sees it, but the pressure shows water. The difference by the height of the swept interval is the rise of the contact.

Wrong

“We draw the water line from the density of the sample, because that is more accurate than using the points.”

In fact

Only if the water is not flowing. In a producing accumulation upward flow makes the line steeper, and a hydrostatic line that we set moves the contact that we find up by $G_{\text{visc}}L/(G_w - G_o)$ . In laminated units this is metres.

Wrong

“During depletion the profile was hydrostatic, so during waterflooding water and oil will quickly separate vertically.”

In fact

These are different conditions. For a hydrostatic profile it is enough that the viscous part is smaller than the error of the gradient. For the phases to separate, the counterflow driven by the density difference must not be weaker than the displacement. And the combined mobility of the phases in a two-phase zone is more than ten times smaller than the mobility of a single phase (see the advanced section; when the phases have time to separate).

In depth: advanced detail

Derivation of the two-phase gradient and the capillary term

For oil and water: $u_o = \lambda_o(\dd p_o/\dd z - \rho_og)$ , $u_w = \lambda_w(\dd p_w/\dd z - \rho_wg)$ , and $\dd p_w/\dd z = \dd p_o/\dd z - \dd p_c/\dd z$ . Adding them gives $u_t = \lambda_t\,\dd p_o/\dd z - g(\lambda_o\rho_o + \lambda_w\rho_w) - \lambda_w\,\dd p_c/\dd z$ . From this, $\dd p_o/\dd z = g\bar\rho + u_t/\lambda_t + f_w\,\dd p_c/\dd z$ , which are the formulas of the main sections. Without capillary pressure this is the two-phase gradient, and with it we get the capillary correction.

Putting the gradient back in, we get the water velocity: $u_w = f_wu_t - (\lambda_o\lambda_w/\lambda_t)\big[(\rho_w - \rho_o)g + \dd p_c/\dd z\big]$ . The first term is the fractional flow of water (the share of water in the total flow). The second is the counterflow driven by gravity and capillary forces (general equation for the fractional flow of water). In countercurrent flow ( $u_t = 0$ ) without capillary pressure, the water drains down at the velocity $(\lambda_o\lambda_w/\lambda_t)(\rho_w - \rho_o)g$ , and the oil rises at the same velocity. This flow creates the intermediate gradient.

Vertical equilibrium of pressure and of saturation

Both conditions follow from the equations of this lesson, but they test different things. Vertical pressure equilibrium is a single-phase idea: the viscous part of the gradient, $u/\lambda$ , is smaller than its error (vertical flow). In the water leg of the field in the example the water flows upward. But at $k_z$ = 150 mD the viscous part is 0.010 kPa/m, which is 8 times smaller than the error of the slope (0.081 kPa/m), and the profile is hydrostatic.

Vertical saturation equilibrium needs more. The density difference must separate the phases faster than the displacement mixes them. In the derivation above, the second term of the water velocity is responsible for this: the counterflow $(\lambda_o\lambda_w/\lambda_t)(\rho_w - \rho_o)g$ . The phases have time to separate if it is comparable with the total velocity $u_t$ or larger. The combined mobility $\lambda_o\lambda_w/\lambda_t$ is more than ten times smaller than the mobility of a single phase: in the middle of the two-phase zone of the example it is 0.043 per 1 mD, against 0.78 for the oil alone. With the flow of the example, the counterflow in a sandstone of 150 mD is 5.7 times larger than $u_t$ . The phases are separated, and the zone is thin, as in the example with the current FWL. In a unit of 3 mD it is only 0.11 of $u_t$ . The zone is wide, with a steep gradient, as in the example of co-current flow. In horizontal waterflooding the displacement speed is many times larger than the vertical crossflow, and separation needs an even higher vertical permeability (comparison of times).

Two conclusions follow for reading profiles. A hydrostatic profile during depletion tests only the first, softer condition, and it says almost nothing about future separation by gravity. The reverse route is more reliable. If, in an interval that is being flooded, new wells see clear oil and water lines with a thin zone between them, then the counterflow is not weaker than the displacement. With known $u_t$ and relative permeabilities, this condition gives a lower estimate of $k_z$ . Intermediate or too steep gradients, on the other hand, are a sign that separation lags behind and the displacement is closer to diffuse flow (spread out, with no sharp front) (Buckley–Leverett theory and the Welge method).

Connections to other topics

Pressure measurements in producing reservoirs: barriers and accuracy — single-phase profiles of a producing accumulation: vertical flow, barriers, the pulse survey.
Pressure gradients, fluid contacts and capillary equilibrium — contacts of an accumulation at rest, and which phase's pressure the tester sees.
Relative permeability — relative permeability and the two-phase Darcy's law from which the gradient is derived.
Segregated flow and gravity displacement — when the phases have time to separate, and gravity drainage.
Capillary pressure and the transition zone — the imbibition profile and the residual oil under a risen contact.
Performance prediction for solution gas drive reservoirs — release of gas below the bubble point pressure, and the secondary gas cap.
Aquifer models: water influx — the steady-state aquifer model and its coefficient.
General material balance equation for oil reservoirs — the balance with water influx, against which the rise of the contact is checked.
History matching — tuning a model to pressures and contacts.
Lumped multi-tank material balance model — a balance for several connected blocks.
Perched and tilted contacts, trapped water — tilted contacts in aquifers with flow, and perched water (a pocket of water cut off from the main aquifer).
Comparing formation-tester, core and well-test data — permeability from the tester, core and well tests.

Check yourself

1
In a countercurrent zone of gas and oil, $k_{ro} = 0.60$ and $k_{rg} = 0.010$ , the viscosities of oil and gas are 1.5 and 0.020 mPa·s, and the densities are 0.72 and 0.15 g/cm³. Find the pressure gradient.
Show the answer
$M = (0.010/0.020)/(0.60/1.5) = 0.50/0.40 = 1.25$ . $\bar\rho = (0.72 + 1.25\cdot0.15)/2.25 \approx 0.403$ g/cm³. The gradient is $9.81\cdot0.403 \approx 3.95$ kPa/m. The relative permeability of gas is 60 times smaller than that of oil, but gas is 75 times less viscous. So the gradient is already closer to the gas gradient (1.47 kPa/m) than to the oil gradient (7.06 kPa/m).
2
A straight line with the gas gradient was drawn through the points of a secondary gas cap and of the upper part of a gassing zone. Its crossing with the oil line is 2 m above the base of the zone picked on the logs. What is this crossing?
1. A.The top of the oil: the gas cap is above it
2. B.An error: the lines must be drawn through all the points at once
3. C.Roughly the base of the two-phase zone: there is still mobile oil above it
4. D.The free water level
Show the answer
Correct answer: C. Roughly the base of the two-phase zone: there is still mobile oil above it
In countercurrent flow the gradient moves half of the way from the oil gradient to the gas gradient as soon as the gas saturation exceeds the critical value by a few per cent, and it becomes almost a gas gradient long before the top of the zone. So the gas line extends almost to the base of the zone. Above the crossing the gas saturation rises gradually, and the oil is mobile. We cannot find the top of the oil from pressure: the logs and the model give it.
3
A laminated unit, with water rising: the viscous part of the water gradient is 0.8 kPa/m, the mean depth of the water points is 10 m below the contact, and the difference between the water and oil gradients is 3.4 kPa/m. By how much and in which direction will the contact shift if the water line is set to the hydrostatic gradient from the sample density?
Show the answer
$\delta z^* = -0.8\cdot10/3.4 \approx -2.4$ m. The contact that we find will be 2.4 m above the true one. At the contact the line with the set gradient is to the right of the true, steeper water line, and the oil line crosses it higher up.
4
In a water-wet rock, a water–oil two-phase zone has a thickness of about the capillary height. What gradient will the pretests show in it?
1. A.Weighted by mobilities, between the oil and the water gradients
2. B.Steeper than the water gradient
3. C.The oil gradient, down to the current FWL
4. D.The water gradient
Show the answer
Correct answer: C. The oil gradient, down to the current FWL
The tool shows the pressure of continuous oil. In a zone close to capillary equilibrium, the capillary term $f_w\,\dd p_c/\dd z$ cancels the weighting by mobilities, and the gradient of the oil pressure equals the oil gradient. An intermediate gradient is seen only in zones much thicker than the capillary height.
5
Bottom water, OWC area 4.0 km², porosity 0.22, $S_{wi} = 0.20$ , $S_{or} = 0.30$ . From a new well the FWL has risen by 3.0 m. 0.20 million m³ of water has been produced (at reservoir conditions). What water influx does this mean?
Show the answer
The water that has stayed in the accumulation is $4.0\cdot10^6\cdot0.22\cdot(1 - 0.20 - 0.30)\cdot3.0 = 1.32$ million m³. Together with the produced water, the influx is $W_e \approx 1.32 + 0.20 = 1.52$ million m³. The estimate is right if the water rises evenly over the whole area. An error of 0.5 m in the rise gives ±0.22 million m³.
6
Why can the gradient in the middle of the displacement zone of a laminated unit be steeper than in the swept zone below it, although the water there is heavier than the mixture?
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In co-current flow a viscous part $u_t/\lambda_t$ is added to the hydrostatic gradient. In the middle of the two-phase zone both phases hinder each other, and the total mobility is 2–3 times lower than in the swept zone, where only water flows. With low vertical permeability this part is a few kPa/m, and it outweighs the difference in hydrostatics.

Updated: 19 October 2026