Water and gas coning

Quick refresher

A stationary cone (one that stays in place) holds as long as the weight of the raised column of water, $\Delta\rho\,g\,h_c$ , balances the pressure drop in the oil beneath the well. With a density difference of 300 kg/m³ this is only about 0.003 MPa for each metre from the interval to the contact.
The critical water-free rate of a vertical well is usually small, a few cubic metres per day. So most wells work above it. We do not do the calculation to restrict the well. We do it to choose the interval, to estimate the breakthrough time and to compare the options.
Methods of calculating the critical rate differ several times over. The simple estimate from the bottomhole drawdown is too low, and solutions for the pressure field give more. What matters is the order of magnitude and the trends, not the third digit.
Low vertical permeability holds back the rise of the cone and lengthens the water-free period, although it increases the pseudo-skin and lowers the productivity.
The breakthrough time is inversely proportional to the vertical permeability, and it gets shorter quickly as the rate grows. After breakthrough the water cut rises roughly with the logarithm of time, towards a limit set by the mobilities (how easily each fluid flows) and thicknesses of the water and oil zones.
The same formulas describe a gas cone, with the density difference between oil and gas. After breakthrough, gas has a low viscosity and quickly takes over the inflow. In an oil rim between gas and water, the stability condition puts the interval closer to the gas, because a gas cone is “stiffer” than a water cone.

Key relationships

p_e - p_w \le \Delta\rho\,g\,h_c

critical drawdown of a stationary cone (pressures at the level of the bottom perforations)

q_c = \dfrac{2\pi k_h h\,\Delta\rho\,g\,h_c}{\mu B\left(\ln\dfrac{r_e}{r_w} + S_{pp}\right)}

estimate from hydrostatic equilibrium, consistent units

q_c = \dfrac{\pi k_h\,\Delta\rho\,g\,(h^2 - h_w^2)}{\mu B\ln(r_e/r_w)}

cone at vertical equilibrium

Z = \dfrac{2\pi k_h\,\Delta\rho\,g\,h\,h_c}{q\,\mu B},\qquad t_{D,BT} = \dfrac{Z}{3 - 0.7Z}

dimensionless cone height and breakthrough time

f_{w,\lim} = \dfrac{M' h_{aq}}{M' h_{aq} + h}

limiting water cut

Common pitfalls

Seeing low vertical permeability as harmful only because the simple formula gives a lower critical rate for it.
Taking the critical rate of one method as an exact number. The methods differ several times over.
Putting in densities at standard conditions instead of reservoir conditions. This is especially dangerous for gas.
Converting breakthrough-time correlations to other units without checking the constants. Published versions differ by about a factor of two.
Blaming a cone for watering-out without first ruling out breakthrough along a high-permeability streak (a thin layer of very permeable rock), flow behind the casing (along the outside of the casing, through poor cement) and water from injection wells.

After this lesson you can

explain how a cone works, as a balance between the pressure drop in the oil and the weight of the raised column of water (or of the lowered column of gas)
calculate the critical water-free rate from the hydrostatic condition, including the pseudo-skin of partial penetration, and by two other methods; and explain why the results differ several times over
show how the open interval, vertical permeability, density difference and oil viscosity affect the critical rate
use dimensionless correlations to estimate the water breakthrough time at a rate above the critical rate, and the rise in water cut after breakthrough, and understand how accurate they are
calculate the critical rate and the position of the interval in an oil rim between a gas cap and water
list ways of delaying breakthrough and limiting its consequences

How a cone forms

Consider an oil reservoir with water underneath. While the well is not producing, the oil–water contact (OWC) is horizontal. Each phase is in hydrostatic equilibrium (its pressure grows with depth only because of its own weight), and at the level of the contact the oil and water pressures match (Capillary pressure and the transition zone). To keep the interval away from the water, the well is perforated in the upper part of the oil-saturated section. So the well opens the reservoir only partly (Partial penetration and well deviation).

After the well starts, a funnel of lowered pressure (the drawdown cone) forms around it. At the level of the contact beneath the well, the pressure in the oil becomes lower than far away. The water under the oil is at rest, and its pressure is still hydrostatic. So the water rises to where the pressure has fallen, until the weight of the raised column of water balances the pressure decrease. The interface bulges into a dome, a cone. The closer to the well, the deeper the funnel of lowered pressure and the higher the water rises. So the tip of the cone lies directly beneath the interval.

The equilibrium condition is simple. Suppose the water has risen by a height $z$ above the original contact. It has replaced oil, so the column is heavier by $\Delta\rho\,g\,z$ . Here $\Delta\rho = \rho_w - \rho_o$ is the density difference between water and oil at reservoir conditions. A cone of height $z$ is stable if the pressure drop in the oil at that place is exactly equal to $\Delta\rho\,g\,z$ . As long as such equilibrium is possible all the way up to the interval, the cone stays in place. Now suppose the pressure drop at the bottom perforations is larger than the weight that water raised right up to the interval can give. Then there is no equilibrium, and the water breaks through into the well.

SchematicWater and gas cones at a vertical well (schematic, not to scale). The open interval

h_w

h_c

away from the original OWC; away from the well the contacts are horizontal. Below the interval the water rises as a cone, and above it the gas goes down.

With a gas cap the picture is a mirror image. The gas above the oil goes down to the interval if the interval is close to the gas–oil contact (GOC), and the density difference is $\rho_o - \rho_g$ . If the oil-saturated section is squeezed between gas and water (an oil rim), two cones are pulled towards the interval at once.

A cone is not the only way for water to reach a well. Edge water (water from the aquifer around the oil zone) comes through the reservoir from beyond the edges of the oil zone. Injected water comes from injection wells along high-permeability streaks (thief zones). Water from a neighbouring reservoir comes behind the casing when the cement is poor. Cone water has a typical sign: it “feels” the rate. If you reduce production, the cone sinks and the water cut falls. For water that has come along a streak, this link is almost absent.

Critical water-free rate: an estimate from hydrostatic equilibrium

The simplest estimate comes from writing the pressure at the level of the original OWC. Beneath the well there is a column of water of height $h_c$ , from the tip of the cone to the contact. At the critical rate the tip stands at the bottom perforations. We take the pressure at the tip to be equal to the bottomhole pressure $p_w$ . Far from the well, at the same level, there is oil, and the pressure at the height of the bottom perforations equals the pressure at the outer boundary $p_e$ . Setting the two pressures equal, $p_w + \rho_w g h_c = p_e + \rho_o g h_c$ , gives the critical drawdown:

Critical drawdown of a stationary cone

p_e - p_w = \Delta\rho\,g\,h_c,\qquad \Delta\rho = \rho_w - \rho_o

where:

$p_e$ pressure in the oil at the outer boundary at the level of the bottom perforations, Pa
$p_w$ bottomhole pressure at the level of the bottom perforations, Pa
$\rho_w,\ \rho_o$ densities of water and oil at reservoir conditions, kg/m³
$g$ acceleration due to gravity, 9.81, m/s²
$h_c$ distance from the bottom perforations of the interval to the original OWC: the critical height of the cone, m

Units: SI (any consistent system of units)

Holds when

The cone is stationary, and the water beneath it is in hydrostatic equilibrium. The interface is sharp: we ignore the capillary transition zone (the zone where oil and water change over gradually).
We take the pressure at the tip of the cone to be equal to the bottomhole pressure. In fact the tip is below the interval, and the pressure there is higher than the bottomhole pressure. So the estimate understates the permissible drawdown, which errs on the safe side.
For a gas cone $\Delta\rho = \rho_o - \rho_g$ , and $h_c$ is the distance from the GOC to the top perforations.

The critical drawdown is the weight of a column of height $h_c$ , “lightened” by the density difference. It depends neither on permeability nor on viscosity. These set what flow rate we can get at that drawdown.

In other unit systems (2)

Metric oilfield units

\Delta p_c\,[\text{MPa}] = 9.81\cdot10^{-6}\,\Delta\rho\,[\text{kg/m}^3]\;h_c\,[\text{m}]

$\Delta\rho$ in kg/m³, $h_c$ in m.

$9.81\cdot10^{-6}$ combines the acceleration due to gravity and the conversion of pascals to megapascals.

Oilfield units

\Delta p_c\,[\text{psi}] = 0.433\,\Delta\rho\,[\text{g/cm}^3]\;h_c\,[\text{ft}]

$\Delta\rho$ in g/cm³, $h_c$ in ft.

$0.433$ psi/ft is the hydrostatic gradient of a fluid of density 1 g/cm³.

It remains to link the drawdown to the rate. Bottom water maintains the pressure at the boundary, so we treat the inflow as steady-state. We use the Dupuit equation with the pseudo-skin of partial penetration $S_{pp}$ (Steady-state radial flow, Partial penetration and well deviation). Putting in the critical drawdown, we get the critical rate:

Critical water-free rate from hydrostatic equilibrium

q_c = \frac{2\pi k_h h\,\Delta\rho\,g\,h_c}{\mu B\left(\ln\dfrac{r_e}{r_w} + S_{pp}\right)}

where:

$q_c$ critical water-free oil rate at standard conditions, m³/s
$k_h$ horizontal permeability of the reservoir to oil, m²
$h$ oil-saturated thickness above the OWC, m
$h_c$ distance from the bottom perforations to the OWC; for an interval open from the top $h_c = h - h_w$ , m
$h_w$ open interval, m
$\Delta\rho,\ g$ density difference between water and oil at reservoir conditions, and the acceleration due to gravity, kg/m³, m/s²
$\mu,\ B$ oil viscosity and formation volume factor, Pa·s, m³/m³
$r_e,\ r_w$ radius of the outer boundary and radius of the well, m
$S_{pp}$ pseudo-skin of partial penetration (from the penetration ratio $h_w/h$ and the anisotropy), —

Units: SI; $\ln$ is the natural logarithm

Holds when

The inflow of single-phase oil to a partially penetrating well is steady-state, in a homogeneous reservoir. The pressure at the outer boundary is maintained.
The water is stationary, and the cone does not distort the pressure field in the oil. The tip “feels” the whole drawdown at the bottomhole, including the resistance of the flow converging on the interval. This is why the result is too low.
Damage skin does not enter the formula, because the pressure drop in the thin damaged zone at the wellbore does not reach the formation below the interval.

The critical rate is the productivity index of a partially penetrating well multiplied by the critical drawdown. It grows with the flow capacity $k_h h$ , the density difference and the distance to the contact, and it falls with oil viscosity. A longer interval reduces $h_c$ but also lowers the pseudo-skin. So there is an optimum interval, about half of the oil-saturated thickness (figure below).

In other unit systems (2)

Metric oilfield units

q_c\,[\text{m}^3/\text{d}] = \frac{5.26\cdot10^{-6}\,k_h\,h\,\Delta\rho\,h_c}{\mu\,B\left(\ln(r_e/r_w) + S_{pp}\right)}

$k_h$ in mD, $h$ and $h_c$ in m, $\Delta\rho$ in kg/m³, $\mu$ in mPa·s, $B$ in m³/m³.

$5.26\cdot10^{-6} = 0.536\cdot9.81\cdot10^{-6}$ . It combines the constant of the Dupuit equation with the conversion of $\Delta\rho\,g\,h_c$ from pascals to megapascals.

Oilfield units

q_c\,[\text{STB/d}] = \frac{3.07\cdot10^{-3}\,k_h\,h\,\Delta\rho\,h_c}{\mu\,B\left(\ln(r_e/r_w) + S_{pp}\right)}

$k_h$ in mD, $h$ and $h_c$ in ft, $\Delta\rho$ in g/cm³, $\mu$ in cP, $B$ in rb/STB.

$3.07\cdot10^{-3} = 7.08\cdot10^{-3}\cdot0.433$ .

Worked exampleCritical water-free rate of a well that has opened the top of the reservoir

Given (assumed data)

Well: from the lesson on partial penetration: oil-saturated thickness $h$ = 32 m above bottom water, the upper $h_w$ = 10 m perforated; $k_h$ = 45 mD, $k_v/k_h$ = 0.15; $r_w$ = 0.108 m, $r_e$ = 350 m
Fluids (reservoir conditions): $\rho_w$ = 1070 kg/m³, $\rho_o$ = 770 kg/m³; $\mu$ = 1.4 mPa·s, $B$ = 1.25 m³/m³

Solution

Density difference: $\Delta\rho = 1070 - 770 = 300$ kg/m³. Distance to the contact: $h_c = 32 - 10 = 22$ m. Critical drawdown: $\Delta p_c = 300\cdot9.81\cdot22 = 64{,}746$ Pa $= 0.0647$ MPa.
The pseudo-skin of partial penetration (calculated in the lesson on partial penetration) is $S_{pp} = 11.04$ . Also $\ln(350/0.108) = 8.084$ .
Productivity index: $J = \dfrac{0.536\cdot45\cdot32}{1.4\cdot1.25\cdot(8.084 + 11.04)} = \dfrac{771.8}{1.75\cdot19.124} = 23.1$ m³/(d·MPa).
Critical rate: $q_c = J\,\Delta p_c = 23.1\cdot0.0647 = 1.49$ m³/d.
For comparison: at an ordinary drawdown of 2.5 MPa the well would deliver $23.1\cdot2.5 = 58$ m³/d, which is 39 times the critical rate.

Answer: The critical water-free rate is about 1.49 m³/d. A stationary cone withstands a drawdown of only 0.0647 MPa.

What it means. A water-free regime needs a drawdown of less than 0.1 MPa and a rate of about one and a half cubic metres per day. It does not pay to run a well like that. So the well will be put on production at a rate tens of times higher, and sooner or later water will break through. The estimate is also too low, because in it the tip of the cone “feels” the whole drawdown, including the resistance of the flow converging on the interval. The following sections show what other methods give, when the water will come, and what happens afterwards. The data in this example are assumed for illustration.

Other methods and the scatter of the estimates

The hydrostatic estimate rests on two strong assumptions. The tip of the cone “feels” the whole drawdown at the bottomhole, and the cone itself does not distort the pressure distribution in the oil. Other methods relax these assumptions in different ways. So the results for one and the same well differ several times over.

Methods of calculating the critical water-free rate of a vertical well

Method: Hydrostatic equilibrium with a pseudo-skin
How it works: drawdown $\Delta\rho g h_c$ in the inflow formula
Effect of $k_v$: lower $k_v$ — lower $q_c$
Result: the lowest, on the safe side

Method: Vertical equilibrium
How it works: oil flows horizontally above the cone
Effect of $k_v$: not taken into account
Result: higher

Method: Pressure-field solutions
How it works: the flow field to a partially penetrating well
Effect of $k_v$: lower $k_v$ — higher $q_c$
Result: higher still

Method: Empirical formulas
How it works: model experiments and calculations
Effect of $k_v$: isotropic reservoir
Result: of the same order

Method: Two-phase simulation
How it works: a radial model with relative permeability and capillary pressure
Effect of $k_v$: taken into account
Result: the basis of design decisions

Method	How it works	Effect of $k_v$	Result
Hydrostatic equilibrium with a pseudo-skin	drawdown $\Delta\rho g h_c$ in the inflow formula	lower $k_v$ — lower $q_c$	the lowest, on the safe side
Vertical equilibrium	oil flows horizontally above the cone	not taken into account	higher
Pressure-field solutions	the flow field to a partially penetrating well	lower $k_v$ — higher $q_c$	higher still
Empirical formulas	model experiments and calculations	isotropic reservoir	of the same order
Two-phase simulation	a radial model with relative permeability and capillary pressure	taken into account	the basis of design decisions

The vertical-equilibrium method is the Meyer–Garder formula. The pressure-field solutions are the methods of Muskat and Wyckoff, Chierici, Chaperon, Høyland and others. The empirical formula is that of Schols. The formulas of the first three rows are given below. The other methods give charts or cumbersome expressions. The order in the last column holds for ordinary short intervals at the top. When more than half of the thickness is open, the curves in the figure of the next section cross.

Cone at vertical equilibrium. Suppose the pressure in the oil is constant along the vertical. Then the oil flows to the well horizontally, through a layer that gets thinner above the cone, and the pressure drop at each point sets the height of the cone there. Integrating Darcy's law from the outer boundary to the wellbore gives the Meyer–Garder formula (the derivation is in the advanced section):

Critical rate of a cone at vertical equilibrium (Meyer–Garder)

q_c = \frac{\pi k_h\,\Delta\rho\,g\,(h^2 - h_w^2)}{\mu B\ln\dfrac{r_e}{r_w}}

where:

$q_c$ critical water-free rate at standard conditions, m³/s
$h,\ h_w$ oil-saturated thickness and the interval open from the top, m
$k_h,\ \Delta\rho,\ g,\ \mu,\ B,\ r_e,\ r_w$ as in the hydrostatic estimate, m², kg/m³, m/s², Pa·s, m³/m³, m, m

Units: SI; $\ln$ is the natural logarithm

Holds when

Vertical equilibrium: the pressure in the oil is constant along the vertical, and the oil flows horizontally. Close to the well this assumption is rough.
The reservoir is isotropic, so $k_v$ does not enter the formula. The resistance of the flow converging on the interval is not taken into account.
The interval is open from the top. At the critical rate the thickness of oil above the cone at the wellbore equals the interval $h_w$ .

The rate given by the formula grows as the interval gets shorter, and it is greatest for the shortest interval. The model does not see that a short interval needs a larger drawdown for the same rate. For a gas cone, $h_w$ is replaced by the thickness of oil from the bottom of the oil column to the top perforations, and $\Delta\rho = \rho_o - \rho_g$ .

In other unit systems (2)

Metric oilfield units

q_c\,[\text{m}^3/\text{d}] = \frac{2.63\cdot10^{-6}\,k_h\,\Delta\rho\,(h^2 - h_w^2)}{\mu\,B\,\ln(r_e/r_w)}

$k_h$ in mD, $\Delta\rho$ in kg/m³, $h$ and $h_w$ in m, $\mu$ in mPa·s.

$2.63\cdot10^{-6} = \pi\cdot0.0853\cdot9.81\cdot10^{-6}$ .

Oilfield units

q_c\,[\text{STB/d}] = \frac{2.46\cdot10^{-5}\,k_h\,\Delta\rho\,(h^2 - h_w^2)}{\mu\,B\,\ln(r_e/r_w)}

$\Delta\rho$ in pounds per cubic foot (lb/ft³), $h$ and $h_w$ in ft, $k_h$ in mD, $\mu$ in cP.

$2.46\cdot10^{-5} = \pi\cdot1.127\cdot10^{-3}/144$ (144 converts lb/ft² to psi).

Pressure-field solutions. These methods first calculate the pressure distribution for inflow to a partially penetrating well in an anisotropic reservoir. Then they look for the highest rate at which the tip of the cone still finds equilibrium below the interval. The results are put into charts or into simple formulas. One of the formulas takes anisotropy into account through a dimensionless radius:

Critical rate in an anisotropic reservoir (Chaperon's approximation)

q_c = \frac{k_h\,\Delta\rho\,g\,(h - h_w)^2}{\mu B}\,F,\qquad F = 0.7311 + \frac{1.943}{r_{eD}},\qquad r_{eD} = \frac{r_e}{h}\sqrt{\frac{k_v}{k_h}}

where:

$F$ dimensionless critical rate (an approximation fitted to calculations), —
$r_{eD}$ dimensionless radius of the outer boundary with allowance for anisotropy, —
$k_v$ vertical permeability, m²
$h - h_w$ distance from the bottom perforations to the OWC ( $h_c$ ), m
$k_h,\ \Delta\rho,\ g,\ \mu,\ B,\ r_e$ as above, m², kg/m³, m/s², Pa·s, m³/m³, m

Units: SI (any consistent system of units)

Holds when

A homogeneous anisotropic reservoir. Steady-state inflow, with a constant pressure at the outer boundary. Stationary water. An interval open from the top.
The formula approximates theoretical calculations of a stable cone for a well that is open at the top of the reservoir.
$F$ grows as $r_{eD}$ decreases, so a lower vertical permeability raises the critical rate.

The “gravity” quantity $k_h\Delta\rho g h_c^2/(\mu B)$ sets the scale of the rate. The geometry and anisotropy enter only through $F$ , which is usually about 1.

In other unit systems (2)

Metric oilfield units

q_c\,[\text{m}^3/\text{d}] = 8.37\cdot10^{-7}\,\frac{k_h\,\Delta\rho\,(h - h_w)^2}{\mu\,B}\,F

$k_h$ in mD, $\Delta\rho$ in kg/m³, $h$ and $h_w$ in m, $\mu$ in mPa·s.

$8.37\cdot10^{-7} = 0.0853\cdot9.81\cdot10^{-6}$ .

Oilfield units

q_c\,[\text{STB/d}] = 7.83\cdot10^{-6}\,\frac{k_h\,\Delta\rho\,(h - h_w)^2}{\mu\,B}\,F

$\Delta\rho$ in lb/ft³, $h$ and $h_w$ in ft, $k_h$ in mD, $\mu$ in cP.

$7.83\cdot10^{-6} = 1.127\cdot10^{-3}/144$ .

Worked exampleThree estimates for one well

Given (assumed data)

Well and fluids: from the previous example: $h$ = 32 m, $h_w$ = 10 m, $k_h$ = 45 mD, $k_v/k_h$ = 0.15, $\Delta\rho$ = 300 kg/m³, $\mu$ = 1.4 mPa·s, $B$ = 1.25, $r_e$ = 350 m, $r_w$ = 0.108 m
Hydrostatic estimate: $q_c$ = 1.49 m³/d

Solution

Vertical equilibrium: $h^2 - h_w^2 = 1024 - 100 = 924$ m². Then $q_c = \dfrac{2.63\cdot10^{-6}\cdot45\cdot300\cdot924}{1.75\cdot8.084} = 2.32$ m³/d.
Approximation for an anisotropic reservoir: $r_{eD} = (350/32)\sqrt{0.15} = 4.236$ and $F = 0.7311 + 1.943/4.236 = 1.190$ .
$q_c = 8.37\cdot10^{-7}\cdot\dfrac{45\cdot300\cdot22^2}{1.75}\cdot1.190 = 3.72$ m³/d.
Spread of the estimates: $3.72/1.49 = 2.5$ .

Answer: The three methods give 1.49, 2.32 and 3.72 m³/d. The estimates differ by a factor of 2.5.

What it means. The assumptions explain the spread. The hydrostatic estimate gives the tip of the cone the whole drawdown. Vertical equilibrium does not see the resistance of the flow converging on the interval. The pressure-field approximation takes anisotropy into account, and it gives the most. But all three estimates are a few cubic metres per day, while the well can deliver tens. The practical conclusion does not depend on the method: the well will work above the critical rate, so it is more important to estimate the breakthrough time and the rise in water cut. The data in this example are assumed for illustration.

Open interval and vertical permeability

CalculatedCalculated critical rate of the well in the examples against the penetration ratio

h_w/h

, by three methods (

k_v/k_h

= 0.15). The dots mark the interval of the example (10 m of 32).

The methods differ not only in the numbers but also in the interval they point to. The hydrostatic estimate ties the critical rate to the productivity of the well, so it has an optimum. A long interval brings the water closer, and a short one creates a large pseudo-skin. The highest critical rate comes with an interval of about half the thickness. The other two models describe only the stability of the cone, and they give the highest critical rate for the shortest interval. But such a well needs an enormous drawdown for any rate. In practice we choose the interval by weighing productivity, the expected breakthrough time and the economics, not by the maximum of a single formula.

CalculatedCalculated critical rate of the same well (the upper 10 m open) against the ratio

k_v/k_h

. The hydrostatic estimate and the pressure-field approximation show opposite trends. Vertical equilibrium does not take anisotropy into account.

Shale layers that stay continuous over a wide area between the interval and the OWC help especially strongly. Suppose a layer is impermeable and extends farther than the zone of noticeable pressure drop. Then the cone cannot rise through it, and the water arrives only by a roundabout path. So before choosing the interval, we study the section of the well (its rock layers) and the correlation of layers with neighbouring wells (matching the same layers from well to well). The reverse is also true: fractures and zones of high vertical permeability below the interval speed up breakthrough.

Breakthrough time at a rate above the critical rate

If the rate is above the critical rate, the cone grows until it reaches the interval. Hydrostatics cannot tell us how long that takes. For that we need the dynamics of the water's rise, which can be reduced to two dimensionless groups. The first compares the rate with the “gravity” scale: the larger the rate is compared with this scale, the more strongly the viscous forces pull the water upwards. The second is the dimensionless time: the fraction of the reservoir thickness that water, driven by the density difference, would travel in time $t$ at a vertical permeability $k_v$ .

Dimensionless cone height and dimensionless time

Z = \frac{2\pi k_h\,\Delta\rho\,g\,h\,h_c}{q\,\mu B},\qquad t_D = \frac{k_v\,\Delta\rho\,g\left(1 + M'^{\,\alpha}\right)}{2\,\phi\,\mu\,h}\,t

where:

$Z$ dimensionless cone height: the ratio of the “gravity” scale of the rate to the actual rate, —
$q$ oil rate at standard conditions, m³/s
$t_D,\ t$ dimensionless time and the time since the well was started, —, s
$k_v$ vertical permeability, m²
$M'$ end-point mobility ratio $(k_{rw}'/\mu_w)/(k_{ro}'/\mu_o)$ : water over oil, each as end-point relative permeability divided by viscosity (Relative permeability), —
$\alpha$ 0.5 for $M' \le 1$ and 0.6 for $1 < M' \le 10$ (in the Bournazel–Jeanson correlation discussed below it is 0.7 for $0.14 \le M' \le 7.3$ ), —
$\phi$ porosity, —
$k_h,\ \Delta\rho,\ g,\ h,\ h_c,\ \mu,\ B$ as in the critical rate formulas, m², kg/m³, m/s², m, m, Pa·s, m³/m³

Units: SI (any consistent system of units)

Holds when

The groups and the exponent $\alpha$ come from a correlation built on model experiments and calculations, for a homogeneous reservoir with bottom water and an interval open from the top.
In oilfield units ( $\Delta\rho$ in g/cm³, $h$ in ft, $k$ in mD, $q$ in STB/d, $t$ in days) the factors are 0.00307 (for $Z$ ) and 0.00137 (for $t_D$ ). Some reference books write the dimensionless time with the same factor as $Z$ (0.00307 instead of 0.00137). Then the breakthrough time comes out 2.2 times too small. The original definition gives 0.00137, which matches the form in consistent units above. If you use other units, work out the factor again from that form. A check of the units will not catch this error.

When $Z$ is greater than a few units, the rate is small compared with the gravity scale, and the cone stops below the interval. At small $Z$ the viscous forces dominate: the larger the rate, the earlier breakthrough comes.

In other unit systems (1)

Metric oilfield units

Z = 5.26\cdot10^{-6}\,\frac{k_h\,\Delta\rho\,h\,h_c}{q\,\mu\,B},\qquad t_D = 4.18\cdot10^{-7}\,\frac{k_v\,\Delta\rho\left(1 + M'^{\,\alpha}\right)}{\phi\,\mu\,h}\,t

$k$ in mD, $\Delta\rho$ in kg/m³, $h$ and $h_c$ in m, $q$ in m³/d, $\mu$ in mPa·s, $t$ in days.

$5.26\cdot10^{-6} = 2\pi\cdot0.0853\cdot9.81\cdot10^{-6}$ ; $4.18\cdot10^{-7} = 0.0853\cdot9.81\cdot10^{-6}/2$ .

Empirical correlations give the breakthrough time. They link the dimensionless breakthrough time $t_{D,BT}$ to $Z$ . The simplest of them is:

Dimensionless breakthrough time (Bournazel–Jeanson correlation)

t_{D,BT} = \frac{Z}{3 - 0.7Z},\qquad Z < \frac{3}{0.7} \approx 4.29

where:

$t_{D,BT}$ dimensionless breakthrough time; we get the time with units from the definition of $t_D$ , —
$Z$ dimensionless cone height, —

Units: dimensionless

Holds when

This is an empirical correlation of model experiments. An earlier correlation with the same groups (Sobocinski–Cornelius) is $t_{D,BT} = (4Z + 1.75Z^2 - 0.75Z^3)/(7 - 2Z)$ for $Z < 3.5$ .
For $Z$ above 4.29 (3.5 in the second correlation) the water does not reach the interval. This is one more, “dynamic”, estimate of the critical rate.
The reservoir is homogeneous. Layers, fractures and vertical heterogeneity change the result several times over, so the accuracy is at best about a factor of two.

At small $Z$ the dimensionless time is proportional to $Z$ , that is, inversely proportional to the rate: a rate twice as high means breakthrough twice as early. As $Z$ approaches 4.29 the breakthrough time grows without bound.

Worked exampleWhen the water will come

Given (assumed data)

Well and fluids: from the examples above; $\phi$ = 0.19; $k_v = 0.15\cdot45 = 6.75$ mD
End-point relative permeability: $k_{rw}'$ = 0.30 at a water viscosity of 0.5 mPa·s; $k_{ro}'$ = 0.80
Rates: 40 and 100 m³/d of oil

Solution

Scale: $Z\,q = 5.26\cdot10^{-6}\cdot45\cdot300\cdot32\cdot22/(1.4\cdot1.25) = 28.57$ m³/d. So $Z = 28.57/40 = 0.714$ at 40 m³/d, and $Z = 28.57/100 = 0.286$ at 100 m³/d.
Mobilities: $M' = (0.30/0.5)/(0.80/1.4) = 1.050 > 1$ , so $\alpha = 0.6$ and $1 + M'^{0.6} = 2.030$ . The Bournazel–Jeanson correlation uses the exponent 0.7. With it the factor would be 2.035. At $M' \approx 1$ the difference is less than 0.3%, so we use 0.6 for both correlations.
Days per unit of dimensionless time: $\dfrac{\phi\mu h}{4.18\cdot10^{-7}\,k_v\,\Delta\rho\,(1 + M'^{\alpha})} = \dfrac{0.19\cdot1.4\cdot32}{4.18\cdot10^{-7}\cdot6.75\cdot300\cdot2.030} = 4954$ d.
40 m³/d: $t_{D,BT} = 0.714/(3 - 0.7\cdot0.714) = 0.2856$ , $t_{BT} = 0.2856\cdot4954 = 1415$ d, about 3.9 years.
100 m³/d: $t_{D,BT} = 0.286/(3 - 0.7\cdot0.286) = 0.1022$ , $t_{BT} = 506$ d, about 1.4 years.
The Sobocinski–Cornelius correlation for the same $Z$ gives $t_{D,BT}$ = 0.6237 and 0.1975, that is, 3090 and 978 d, which is roughly twice as late.
The limits of the correlations, $Z \approx 4.29$ and $Z = 3.5$ , correspond to rates of $28.57/4.29 = 6.66$ and $28.57/3.5 = 8.16$ m³/d.

Answer: By the first correlation, the water arrives in about 3.9 years at 40 m³/d and in about 1.4 years at 100 m³/d. The second correlation gives times about twice as long.

What it means. A rate 2.5 times higher has brought breakthrough almost three times closer. Two correlations with the same groups differ by a factor of two. The “dynamic” limits, 6.66 and 8.16 m³/d, are higher still than the three estimates of the critical rate. This is normal accuracy for a homogeneous reservoir model. For a project, we refine the breakthrough time with a radial two-phase model and with data from analogue wells (similar wells that have already produced). The data in this example are assumed for illustration.

After breakthrough: the rise in water cut

After breakthrough, water enters the bottom perforations, the cone widens, and the share of water in the production rises. This rise has a limit. When the cone is fully developed, the well draws liquid from both the oil zone and the water zone, roughly as from two layers working side by side (in parallel). The rates are proportional to mobility and thickness (Rock permeability, fractional flow). This gives the limiting water cut at reservoir conditions:

Limiting water cut with a developed cone

f_{w,\lim} = \frac{M' h_{aq}}{M' h_{aq} + h}

where:

$f_{w,\lim}$ limiting water fraction in the flow, —
$M'$ end-point mobility ratio of water to oil, —
$h_{aq}$ thickness of the water-saturated part of the reservoir below the oil within the drainage area, m
$h$ current oil-saturated thickness, m

Units: dimensionless; thicknesses in the same units

Holds when

Both zones deliver liquid to the well side by side, in proportion to $k\,k_r h/\mu$ . The pressure and the permeability are the same in both zones.
As fluid is produced, the OWC rises. So $h_{aq}$ grows, $h$ decreases, and the limit rises with time.
The water fraction is at reservoir conditions. At standard conditions the water cut is somewhat higher, because when oil is brought to the surface its volume shrinks more than the volume of water (the oil formation volume factor is greater).

A thick aquifer and a large mobility ratio $M'$ raise the limit. This happens above all with viscous oil, and then the limit is close to 1.

Numerical calculations of cones show how the water cut approaches the limit. A simple correlation (Kuo and DesBrisay) sums up their results: the ratio of the water cut to the limiting water cut grows linearly with the logarithm of time.

Growth of water cut after breakthrough (Kuo–DesBrisay correlation)

\frac{f_w}{f_{w,\lim}} = 0.29 + 0.94\,\lg\frac{t}{t_{BT}},\qquad 0.5 \le \frac{t}{t_{BT}} \le 5.7

where:

$f_w$ water fraction in the flow, —
$t$ time since the well was started, days
$t_{BT}$ breakthrough time from the correlation, days
$\lg$ decimal logarithm, —

Units: dimensionless

Holds when

At $t/t_{BT} < 0.5$ the ratio is zero, and at $t/t_{BT} > 5.7$ it is 1. The reservoir is homogeneous with bottom water, and the liquid rate is constant.
Formally the water appears already at half the calculated breakthrough time. This comes from how the correlation is scaled to the estimate of $t_{BT}$ , and it is one more reminder of how accurate that estimate is.
We allow separately for the rise of the limit $f_{w,\lim}$ as the OWC rises, by recalculating $h$ and $h_{aq}$ from the cumulative production (the total produced so far).

At the breakthrough time the ratio equals 0.29. At two and a half times the breakthrough time it is about 0.66, and at 5.7 times the breakthrough time it is 1.

Worked exampleWater cut after 1100 and 2200 days

Given (assumed data)

Well: from the examples above; 40 m of water-saturated reservoir below the oil; $M'$ = 1.050
Breakthrough time: 506 d at 100 m³/d and 1415 d at 40 m³/d (from the previous example)

Solution

Limiting water cut: $f_{w,\lim} = 1.050\cdot40/(1.050\cdot40 + 32) = 0.568$ .
100 m³/d, 1100 d: $t/t_{BT} = 2.174$ , $0.29 + 0.94\lg2.174 = 0.607$ , $f_w = 0.607\cdot0.568 = 0.345$ .
100 m³/d, 2200 d: $t/t_{BT} = 4.348$ , ratio 0.890, $f_w = 0.506$ . The oil rate is $100\cdot(1 - 0.506) = 49.4$ m³/d.
40 m³/d, 2200 d: $t/t_{BT} = 1.555$ , ratio 0.470, $f_w = 0.267$ . The oil rate is $40\cdot(1 - 0.267) = 29.3$ m³/d.

Answer: After 2200 days at a liquid rate of 100 m³/d the water cut is about 51%, and at 40 m³/d it is about 27%.

What it means. A lower rate has delayed breakthrough and keeps the water cut lower, but the well also delivers less oil: after 2200 days it gives 29.3 m³/d against 49.4 m³/d. So we choose the rate by the economics of the whole life of the well: the cumulative oil, the costs of lifting and treating the water, and the time value of money (Net present value (NPV)). We do not choose it just to delay breakthrough. The limit here is constant. In reality it grows as the OWC rises, so the late water cut will be higher than calculated. We split the production by the water fraction at reservoir conditions. At the surface the water cut will be somewhat higher and the oil rate lower. The data in this example are assumed for illustration.

CalculatedCalculated water cut (the water fraction at reservoir conditions) of the well in the examples, at a constant liquid production of 40 and 100 m³/d. The breakthrough time comes from the Bournazel–Jeanson correlation, and the rise comes from the Kuo–DesBrisay correlation, at a constant limiting water cut (dashed line). The dots mark the calculated moments of breakthrough.

Gas coning and oil rims

The same formulas describe a gas cone. We put $\rho_o - \rho_g$ in place of $\rho_w - \rho_o$ , and the distance from the GOC to the top perforations in place of the distance to the OWC. We take the gas density at reservoir conditions. At a pressure of 20–35 MPa it is of the order of 150–250 kg/m³, not about one kilogram per cubic metre as at the surface (Gas formation volume factor, compressibility and viscosity). Usually $\rho_o - \rho_g$ is greater than $\rho_w - \rho_o$ . So for each metre of distance to the contact, a gas cone withstands a larger drawdown.

After breakthrough the picture is the reverse. The viscosity of gas in the reservoir is tens of times lower than that of oil, or even more. So the gas quickly takes over the inflow, and the gas–oil ratio jumps (gas–oil ratio). The energy of the gas cap is wasted, and the capacity of the system that gathers and treats the gas often limits production. So gas breakthrough usually costs more than water breakthrough.

In an oil rim the interval is squeezed between two contacts. Treat both cones with the vertical-equilibrium approximation. The derivation (advanced section) then gives the same formula with a combined density difference. It also gives the position of the interval at which both cones reach its edges at the same rate:

Critical rate and position of the interval in an oil rim

\begin{gathered} q_c = \frac{\pi k_h\,\Delta\rho_{\text{eff}}\,g\,(h^2 - h_w^2)}{\mu B\ln(r_e/r_w)},\qquad \Delta\rho_{\text{eff}} = \frac{(\rho_w - \rho_o)(\rho_o - \rho_g)}{\rho_w - \rho_g}\\ h_g = (h - h_w)\,\frac{\rho_w - \rho_o}{\rho_w - \rho_g}\end{gathered}

where:

$\Delta\rho_{\text{eff}}$ effective density difference of the two cones, kg/m³
$\rho_w,\ \rho_o,\ \rho_g$ densities of water, oil and gas at reservoir conditions, kg/m³
$h_g$ distance from the GOC to the top perforations; $h - h_w - h_g$ remains to the OWC, m
$h,\ h_w$ thickness of the rim and the open interval, m
$k_h,\ g,\ \mu,\ B,\ r_e,\ r_w$ as in the Meyer–Garder formula, m², m/s², Pa·s, m³/m³, m, m

Units: SI; $\ln$ is the natural logarithm

Holds when

Vertical equilibrium and an isotropic reservoir, as in the formula with one cone. Both phases, below and above the oil, are stationary.
We find the position of the interval from the stability condition. We do not take into account the movement of the contacts during development (the cap expands, the water rises).

The effective density difference is smaller than either of the two. So a rim is more vulnerable than a reservoir with only water or only gas. We shift the interval towards the contact with the larger density difference, which is usually the gas.

In other unit systems (1)

Metric oilfield units

q_c\,[\text{m}^3/\text{d}] = \frac{2.63\cdot10^{-6}\,k_h\,\Delta\rho_{\text{eff}}\,(h^2 - h_w^2)}{\mu\,B\,\ln(r_e/r_w)}

$k_h$ in mD, $\Delta\rho_{\text{eff}}$ in kg/m³, $h$ and $h_w$ in m, $\mu$ in mPa·s.

as in the Meyer–Garder formula.

Worked exampleWhere to perforate a rim

Given (assumed data)

Rim: oil thickness 24 m between the gas cap and the water; interval 6 m; $k_h$ = 150 mD
Densities (reservoir conditions): water 1050, oil 740, gas 190 kg/m³
Oil and well: $\mu$ = 0.9 mPa·s, $B$ = 1.3; $r_e$ = 300 m, $r_w$ = 0.108 m

Solution

Density differences: water–oil 310, oil–gas 550, water–gas 860 kg/m³. Then $\Delta\rho_{\text{eff}} = 310\cdot550/860 = 198.3$ kg/m³.
Position of the interval: $h_g = (24 - 6)\cdot310/860 = 6.49$ m from the GOC to the top perforations. The distance to the OWC is $18 - 6.49 = 11.51$ m.
Critical rate: $\ln(300/0.108) = 7.929$ , so $q_c = \dfrac{2.63\cdot10^{-6}\cdot150\cdot198.3\cdot(576 - 36)}{0.9\cdot1.3\cdot7.929} = 4.55$ m³/d.
For comparison: the same interval at the top of a reservoir without a gas cap would give $7.1$ m³/d (density difference 310). At the bottom of a reservoir without water it would give $12.6$ m³/d (density difference 550).

Answer: The interval is placed 6.49 m below the GOC and 11.51 m above the OWC. The critical rate is about 4.55 m³/d.

What it means. By the stability condition the interval is closer to the gas, because a gas cone is “stiffer”. But this is a static estimate. If the cap expands and the GOC moves down, and gas breakthrough does more harm, the interval is placed lower. This is done in reservoirs with a gas cap drive (the expanding gas cap pushes the oil to the wells) (Reservoir drive mechanisms). We choose the final position and rate with a two-phase model that includes the movement of both contacts. In long rims, engineers often prefer horizontal wells drilled at a set distance from the contacts (Horizontal wells). The data in this example are assumed for illustration.

How to delay breakthrough and what to do after it

Rate. Running the well below the critical rate almost never pays. Choose the rate by the economics of the whole life of the well, including the costs of lifting and treating the water. If the water cut is rising because of a cone, a lower rate reduces it. This is both a way to control the water cut and a sign of a cone.
Open interval. Keep it away from the contact, but do not make it so short that the pseudo-skin eats up the productivity (Partial penetration and well deviation). In a rim, allow for both contacts and for which of them moves faster.
Natural and artificial barriers. A continuous shale layer below the interval is the best protection. If there is none, engineers pump a gel-forming fluid (a fluid that turns into a gel in the rock) or cement into the rock below the interval, to make an impermeable screen. The screen works only if it is wide enough. Otherwise the water goes round it.
Horizontal well. For the same production the drawdown is several times smaller. Instead of a cone, a water crest (a ridge of water along the well) forms, and it rises more slowly (Horizontal wells).
Separate water production. A second interval below the OWC, with its own lift, produces water and holds the contact below the well. The method suits a thick aquifer and cheap water disposal.
Shut-off after breakthrough. Engineers shut off the watered-out lower part of the interval with cement, a bridge plug or a packer (devices that seal the well at a chosen depth). If needed, they perforate again higher up. After gas breakthrough, they shut off the top perforations.

We check design decisions (interval, rate, barrier, horizontal wellbore) on a two-phase, single-well, radial model with fine layers at the OWC and at the interval (Types of reservoir simulation model, Grids in reservoir simulation models). This model accounts for what simple formulas do not see: relative permeabilities, the capillary transition zone, layers in the rock, and a rate that changes in time. We tune the model to the actual water cut.

Terms and symbols

Coning: Bottom water, or gas from the gas cap, is pulled towards the open interval of a producing well by the drawdown cone around the well.
Critical water-free (gas-free) rate $q_c$: The highest rate at which a stationary cone still does not reach the open interval.
Critical drawdown $\Delta\rho g h_c$: The weight of a column of height $h_c$ , lightened by the density difference. It is the largest drawdown that a stationary cone withstands.
Dimensionless cone height $Z$: The ratio of the “gravity” scale of the rate, $2\pi k_h\Delta\rho g h h_c/(\mu B)$ , to the actual rate.
Breakthrough time $t_{BT}$: The time from the start of the well to the first appearance, in the production, of water (or gas) that has come by coning.
Limiting water cut: The water fraction in the flow when the cone is fully developed. Then the well draws liquid from the oil and water zones in proportion to their mobility and thickness.
Oil rim: The oil-saturated part of a reservoir between the gas cap and the water. It is vulnerable to two cones at once.

Common mistakes and misconceptions

Wrong

“A rate below the critical rate means that water will never come.”

In fact

The critical rate refers to a stationary contact. If the OWC rises throughout the reservoir, the water will reach the interval even without a cone. Also, estimates of the critical rate differ several times over.

Wrong

“The simple formula gave a lower critical rate at low $k_v$ , so low vertical permeability is harmful.”

In fact

This is a result of the assumption that the tip of the cone “feels” the whole drawdown at the bottomhole. A low $k_v$ holds back the rise of the cone and lengthens the water-free period. Pressure-field methods give a higher critical rate for it.

Wrong

“The density difference between oil and gas is greater than between water and oil, so a gas cone need not be feared.”

In fact

Before breakthrough, a gas cone is indeed “stiffer”. But after breakthrough, gas that is tens of times less viscous quickly takes over the inflow. The gas–oil ratio jumps, and the energy of the cap is lost.

Wrong

“The well has watered out, so a cone is to blame.”

In fact

Water can come along a high-permeability streak, behind the casing through poor cement, or from an injection well. We recognise a cone by the link between water cut and rate, by water entering through the bottom perforations, and by the composition of the water.

Wrong

“The correlation gave a breakthrough time of four years, so we put four years of water-free production in the plan.”

In fact

Breakthrough-time correlations are accurate at best to within a factor of two (in the lesson's example, two correlations differed by a factor of two). The heterogeneity of the reservoir can bring breakthrough even closer. Refine the estimate by simulation and with data from analogues.

In depth: advanced detail

Derivation of the vertical equilibrium formulas

Let the oil–water interface at a distance $r$ from the well have risen by $z(r)$ above the original OWC. The water is stationary, so the pressure drop in the oil relative to the outer boundary (at the same level) is balanced by the weight of the raised water: $p_e - p(r) = \Delta\rho\,g\,z(r)$ . From this, $\dd p/\dd r = -\Delta\rho\,g\,\dd z/\dd r$ . At vertical equilibrium the pressure in the oil is constant along the vertical, and the oil flows to the well horizontally through a layer of thickness $h - z$ : $q B = 2\pi r\,(h - z)\,\dfrac{k_h}{\mu}\,\dfrac{\dd p}{\dd r}$ .

We put in the pressure gradient and separate the variables: $\dfrac{qB\mu}{2\pi k_h\Delta\rho g}\cdot\dfrac{\dd r}{r} = -(h - z)\,\dd z$ . Now we integrate from the wellbore to the outer boundary, where $z = 0$ . At the critical rate, exactly the interval is left above the cone at the wellbore ( $z = h - h_w$ ). So: $\dfrac{qB\mu}{2\pi k_h\Delta\rho g}\ln\dfrac{r_e}{r_w} = \displaystyle\int_0^{h - h_w}(h - z)\,\dd z = \frac{h^2 - h_w^2}{2}$ . This is the Meyer–Garder formula.

In a rim both interfaces respond to the pressure drop. The water rises by $z_w = \Delta p/(\Delta\rho_{wo}g)$ , the gas goes down by $z_g = \Delta p/(\Delta\rho_{og}g)$ , and the oil thickness is $h - z_w - z_g = h - \dfrac{\Delta p}{g}\left(\dfrac{1}{\Delta\rho_{wo}} + \dfrac{1}{\Delta\rho_{og}}\right)$ . This is the same problem with a density difference “added in series”, $1/\Delta\rho_{\text{eff}} = 1/\Delta\rho_{wo} + 1/\Delta\rho_{og}$ . The two cones at the wellbore are in the ratio $z_g/z_w = \Delta\rho_{wo}/\Delta\rho_{og}$ . We divide the unperforated thickness $h - h_w$ in this proportion, and this gives $h_g$ .

What Z and the dimensionless time mean

The quantity $2\pi k_h\Delta\rho g h h_c/(\mu B)$ is the rate at which the “unit” of radial pressure drop, $qB\mu/(2\pi k_h h)$ , equals the weight of the column $\Delta\rho g h_c$ . Suppose the actual rate is much greater ( $Z \ll 1$ ). Then the water rises almost as if the flow carried it to the well. The correlation gives $t_{D,BT} \approx Z/3$ , and the breakthrough time $t_{BT} \approx \dfrac{4\pi}{3}\cdot\dfrac{\phi\,h^2 h_c}{qB\,(1 + M'^{\alpha})}\cdot\dfrac{k_h}{k_v}$ does not depend on the density difference. The oil viscosity enters it only through the mobility ratio $M'$ . The time is set by how fast the well draws off the pore volume above the contact, “stretched” by the anisotropy $k_h/k_v$ . For the lesson's example at 100 m³/d, this estimate gives about 470 d against 506 d by the full formula.

In these terms the hydrostatic estimate is the condition $Z = \ln(r_e/r_w) + S_{pp}$ . For the well of the lesson this is $Z \approx 19$ , but the breakthrough-time correlations already stop the cone at $Z \approx 3.5$ –4.3. The discrepancy of 4.5–5.5 times measures how cautious the assumption is that the tip of the cone “feels” the whole drawdown at the bottomhole.

Connections to other topics

Partial penetration and well deviation — the pseudo-skin of partial penetration, which enters the hydrostatic estimate, and the choice of the interval.
Steady-state radial flow — the Dupuit equation on which the estimate of the critical rate is built.
Horizontal wells — the water crest at a horizontal wellbore and why a horizontal well delays breakthrough.
Relative permeability — end-point relative permeabilities and the mobility ratio $M'$ .
Capillary pressure and the transition zone — contacts, the free water level and the transition zone.
Reservoir drive mechanisms — bottom water and the gas cap as sources of energy and of breakthroughs.
Fractional flow function and mobility ratio — the water fraction in the flow and the water cut.
Types of reservoir simulation model — single-well radial models for calculating cones.
Segregated flow and gravity displacement — gravity segregation of the phases in the reservoir.
Net present value (NPV) — choosing the rate by the economics of the whole life, not by the critical rate.

Check yourself

1
Which change raises the critical water-free rate of a vertical well, other things being equal?
1. A.A more viscous oil
2. B.A lower horizontal permeability
3. C.A greater density difference between water and oil
4. D.An open interval lowered closer to the OWC
Show the answer
Correct answer: C. A greater density difference between water and oil
In all the formulas the critical rate is proportional to $k_h\,\Delta\rho$ and inversely proportional to viscosity. The heavier the water is relative to the oil, the heavier the raised column is, and the cone withstands a larger drawdown. Lowering the interval reduces $h_c$ , and with it the critical rate.
2
Densities in the reservoir: water 1060, oil 780 kg/m³; from the bottom perforations to the OWC 15 m. What drawdown will a stationary cone withstand, and is that a lot?
Show the answer
$\Delta p_c = (1060 - 780)\cdot9.81\cdot15 \approx 41{,}200$ Pa $\approx 0.041$ MPa. Ordinary drawdowns are from one to several megapascals, which is tens of times greater. So unless production is restricted, the cone will reach the interval.
3
The hydrostatic estimate with a pseudo-skin gave a lower critical rate at $k_v/k_h$ = 0.01 than at $k_v/k_h$ = 1. How should we understand this?
1. A.It follows from the assumption: the whole drawdown, including the resistance of the flow converging at the interval, is given to the tip of the cone; in reality a low $k_v$ holds the cone back
2. B.The whole thickness should be perforated
3. C.Anisotropy does not affect coning
4. D.Low vertical permeability hastens watering-out
Show the answer
Correct answer: A. It follows from the assumption: the whole drawdown, including the resistance of the flow converging at the interval, is given to the tip of the cone; in reality a low $k_v$ holds the cone back
The pseudo-skin grows at a low $k_v$ , and a formula that gives the tip of the cone the whole drawdown at the bottomhole reduces $q_c$ . But this resistance is concentrated at the interval. A low $k_v$ passes the pressure drop downwards less easily and slows the rise of the water. Pressure-field methods and simulation show a higher critical rate and a later breakthrough.
4
A well works at a dimensionless cone height $Z = 0.5$ . By what factor does the breakthrough time change if the rate is doubled? Use the correlation $t_{D,BT} = Z/(3 - 0.7Z)$ .
Show the answer
When the rate is doubled, $Z = 0.25$ . Before, $t_{D,BT} = 0.5/2.65 \approx 0.189$ . Now it is $0.25/2.825 \approx 0.0885$ . So the breakthrough time gets shorter by a factor of about 2.13, slightly more than two. At the initial, larger $Z$ , the weight of the water held the cone back more strongly, and this stretched the breakthrough time more.
5
In a rim the density difference between water and oil is 250 kg/m³, between oil and gas 600 kg/m³. Where, by the stability condition, should the interval be placed?
1. A.Right up against the GOC
2. B.In the middle of the unperforated thickness
3. C.Closer to the GOC: from it to the top perforations is about 0.29 of the unperforated thickness
4. D.Closer to the OWC: about 0.29 of the unperforated thickness from it
Show the answer
Correct answer: C. Closer to the GOC: from it to the top perforations is about 0.29 of the unperforated thickness
$h_g/(h - h_w) = (\rho_w - \rho_o)/(\rho_w - \rho_g) = 250/850 \approx 0.29$ . A gas cone is “stiffer”, so the interval can come closer to the GOC. If the cap is going to expand, shift the interval lower.
6
After water breakthrough the rate of the well was halved, and the water cut fell from 40 to 22%. What does this say, and what can be done?
Show the answer
A strong dependence of the water cut on the rate is a sign of a cone: at a smaller drawdown the cone sinks. Next, compare the economics of the lower-rate regime with the other options: shutting off the lower part of the interval and perforating again higher up, a screen below the interval, or a horizontal sidetrack (a new horizontal branch drilled from the existing well). If the water cut hardly depended on the rate, we would look for breakthrough along a streak or flow behind the casing.

Updated: 10 October 2026