Horizontal wells · MunaiHub

Quick refresher

Inflow to a horizontal well meets two resistances one after the other. The first is flow in the horizontal plane to a wellbore of length $L$ , as to a slot. The second is the flow converging on the wellbore in the vertical plane. The second resistance is proportional to $I h/L$ , where $I = \sqrt{k_h/k_v}$ is the anisotropy index (it shows how much easier fluid flows sideways than up and down).
The gain grows with the length of the wellbore and falls as anisotropy grows. When $k_v \ll k_h$ , it is hard for fluid to reach the wellbore from above and below. In a thick layered reservoir, a horizontal well can lose to a vertical well with a hydraulic fracture.
The damage skin of a horizontal wellbore (the extra resistance of a damaged zone around it) enters the resistance with the factor $I h/L$ . So the same damaged zone costs less than in a vertical well. But damage in a horizontal wellbore is often greater, and clean-up is worse.
At a small drawdown (high permeability, a long wellbore), friction losses in the wellbore are comparable with the drawdown. The inflow shifts towards the heel, and water arrives there earlier. Inflow control devices even out the inflow, at the cost of an additional pressure drop.
A water crest under a horizontal wellbore rises more slowly than a cone (a cone-shaped rise of water under a vertical well) at the same production rate. The critical water-free rate is several times higher, but it is usually still below the economically justified rate.
The standard productivity models differ mainly in the vertical term. The difference is smaller than the uncertainty of $k_v/k_h$ .

Key relationships

J = \dfrac{2\pi k_h h}{\mu B\,R}

productivity index of a horizontal well, consistent units

R = \ln\dfrac{a + \sqrt{a^2 - (L/2)^2}}{L/2} + \dfrac{I h}{L}\left[\ln\dfrac{I h}{r_w(I + 1)} + S\right]

resistance: the horizontal plane plus the vertical one with the wellbore skin

a = \dfrac{L}{2}\sqrt{0.5 + \sqrt{0.25 + \left(\dfrac{2r_{eh}}{L}\right)^4}},\qquad I = \sqrt{\dfrac{k_h}{k_v}}

semi-axis of the drainage ellipse and the anisotropy index

\text{PIF} = \dfrac{J_H}{J_V},\qquad S_{eq} = R - \ln\dfrac{r_{eh}}{r_w}

productivity improvement factor and equivalent skin

q_c \approx \dfrac{k_h\,\Delta\rho\,g\,h^2 L}{\mu B\,y_e}

critical rate of the crest at vertical equilibrium, wellbore at the top of the formation

Common pitfalls

Expecting a gain proportional to the length of the wellbore. With a small $k_v$ and a large thickness, the gain is much smaller.
Taking $k_v$ from core samples only. A thin shale layer between the samples can cut the wellbore off from most of the thickness.
Comparing horizontal and vertical wells with different drainage areas or drawdowns, and attributing the whole difference to the length of the wellbore.
Neglecting friction in the wellbore of a high-permeability reservoir. A calculation with an infinitely conductive wellbore will overestimate the rate and will not show early water breakthrough at the heel.
Installing inflow control devices without packers in the annulus of an open hole. The flow will bypass them along the annulus.

After this lesson you can

explain where a horizontal well gives a gain, what the gain comes from, and where there is none
calculate the productivity index of a horizontal well in an anisotropic reservoir (one where permeability differs by direction) with the standard model, know its assumptions, and compare it with other models
estimate the productivity improvement factor, the equivalent skin (a number that shows how much the zone around the well helps or hinders flow), the effective wellbore radius and the cost of damage to the wellbore
compare options for completing a horizontal wellbore, and understand when inflow control devices are needed
estimate the frictional pressure loss along the wellbore and its effect on the distribution of inflow (the heel–toe effect)
estimate the critical water-free rate of a water crest (a ridge of water that rises under the wellbore) and compare it with that of a vertical well

Where a horizontal well gives a gain

A vertical well is in contact with the formation over a length equal to the formation thickness, from a few metres to tens of metres. A horizontal wellbore opens the formation over hundreds of metres. The longer the contact, the smaller the resistance to inflow. At the same drawdown the rate is greater, and at the same rate the drawdown is smaller. A smaller drawdown has other benefits. It draws in water and gas more slowly (Water and gas coning). It less often destroys weak rock (Sand production: types, prediction and cost of the problem). And it lets less gas come out of the oil near the wellbore. One horizontal well drains a strip of the formation and replaces several vertical wells. Only long-term development shows for certain how many.

Where horizontal wells are used

Conditions: Thin formation (a few metres)
Source of the gain: contact with the formation tens and hundreds of times greater than the thickness
Limitation: accurate steering of the wellbore through the formation is needed

Conditions: Low-permeability formation
Source of the gain: length of the wellbore and multistage hydraulic fracturing
Limitation: without fracturing the gain is small

Conditions: Bottom water, gas cap, oil rims
Source of the gain: lower drawdown, a crest instead of a cone
Limitation: the distance to the contacts must be maintained

Conditions: Viscous oil, high permeability
Source of the gain: a high rate at a small drawdown
Limitation: friction in the wellbore, sand production

Conditions: Vertical natural fractures
Source of the gain: a wellbore across the fractures intersects many of them
Limitation: their direction must be known

Conditions: Fault blocks, lenses
Source of the gain: one wellbore opens several blocks
Limitation: different pressures in the blocks

Conditions: Limited number of sites (offshore, artificial islands)
Source of the gain: fewer wells for the same production
Limitation: risk is concentrated in one well

Conditions	Source of the gain	Limitation
Thin formation (a few metres)	contact with the formation tens and hundreds of times greater than the thickness	accurate steering of the wellbore through the formation is needed
Low-permeability formation	length of the wellbore and multistage hydraulic fracturing	without fracturing the gain is small
Bottom water, gas cap, oil rims	lower drawdown, a crest instead of a cone	the distance to the contacts must be maintained
Viscous oil, high permeability	a high rate at a small drawdown	friction in the wellbore, sand production
Vertical natural fractures	a wellbore across the fractures intersects many of them	their direction must be known
Fault blocks, lenses	one wellbore opens several blocks	different pressures in the blocks
Limited number of sites (offshore, artificial islands)	fewer wells for the same production	risk is concentrated in one well

The summary is qualitative; in each case the gain is estimated by calculation and by the experience of neighbouring wells.

A horizontal wellbore does not pay off in three cases. The first is a thick formation with low vertical permeability or with continuous shale layers. There the wellbore drains only its own layer, and a vertical well with a hydraulic fracture will open the whole thickness more cheaply. The second is a formation whose properties are so poorly known that the wellbore may be put in the wrong place. The third is where the lift or the gathering system limits the rate, not the formation.

Profile and steering. The crew drills the well in the ordinary way to a point above the formation, and then builds the angle up to horizontal (Directional wells: applications and profile, Deflection tools and downhole motors). The production casing is often run at a high angle to the top of the formation, and then the horizontal section is drilled. If the structure of the formation is poorly known, the crew first drills and logs a pilot hole (a trial hole through the formation). Then they abandon it and kick off (start a new direction) into the horizontal section. We tell several types of profile apart by the radius of curvature of the build arc (the curved part where the angle builds up):

Profiles by build radius (approximate)

Profile: Long radius
Build-up rate, °/30 m: 2–6
Radius, m: 290–860
Features: conventional tools; long horizontal section

Profile: Medium radius
Build-up rate, °/30 m: 6–20
Radius, m: 85–290
Features: the main option for horizontal wells

Profile: Short radius
Build-up rate, °/30 m: 40–170
Radius, m: 10–40
Features: special assemblies; short section, simple completion

Profile: Ultra-short radius
Build-up rate, °/30 m: —
Radius, m: fractions of a metre
Features: radial channels jetted out from a cased well

Profile	Build-up rate, °/30 m	Radius, m	Features
Long radius	2–6	290–860	conventional tools; long horizontal section
Medium radius	6–20	85–290	the main option for horizontal wells
Short radius	40–170	10–40	special assemblies; short section, simple completion
Ultra-short radius	—	fractions of a metre	radial channels jetted out from a cased well

Radius $R = 1718.9/i$ for a build-up rate $i$ in degrees per 30 m (Directional wells: applications and profile). The boundaries of the classes vary from one practice to another.

We correct the position of the wellbore in the formation while drilling. This is geosteering. Logging while drilling (LWD) tools measure resistivity, gamma ray and density while the bit drills. They, together with the cuttings (the rock chips), show where the bit is relative to the top of the formation, a shale layer or a contact (Open-hole logging: why, how and when). In this way we keep the wellbore a few metres from the gas–oil or oil–water contact, directly under the seal (the rock that caps the reservoir), or in a high-permeability sandstone, where there is less residual water and the resistivity is higher. Vertically we can hold the wellbore to within a metre. The horizontal section can be from hundreds of metres to kilometres long. Drilling is not usually what limits its length. More often it is limited by inflow, friction in the wellbore and the possibility of servicing it. Wells whose horizontal departure is more than twice the true vertical depth are called extended-reach wells.

SchematicA horizontal well (schematic, not to scale). Top: plan view (seen from above). It shows a wellbore of length

L

from the heel to the toe, the drainage ellipse with semi-axis

a

, and a circle of the same area with radius

r_{eh}

. Bottom: cross-section across the wellbore. Near the wellbore the flow is radial. Farther away it is plane-parallel between the top and the base.

Inflow to a horizontal well

Far from the wellbore the fluid flows in the horizontal plane, as to a slot of length $L$ . The lines of equal pressure are ellipses with their foci at the ends of the wellbore. Near the wellbore, at distances of the order of the thickness of the formation, the flow also converges on the wellbore vertically. In the cross-section across the wellbore it is radial, and farther away it is plane-parallel between the top and the base. These two resistances are connected in series, so they add up. We take anisotropy into account by “stretching” the formation vertically by a factor of $I = \sqrt{k_h/k_v}$ . In the stretched formation the permeability is the same in all directions, the thickness is $I h$ , and the round wellbore becomes an ellipse (Partial penetration and well deviation).

Productivity index of a horizontal well (the standard model corrected for anisotropy)

\begin{gathered} J = \frac{q}{p_e - p_w} = \frac{2\pi k_h h}{\mu B\,R}\\ R = \ln\frac{a + \sqrt{a^2 - (L/2)^2}}{L/2} + \frac{I h}{L}\left[\ln\frac{I h}{r_w(I + 1)} + S\right]\\ a = \frac{L}{2}\sqrt{0.5 + \sqrt{0.25 + \left(\frac{2r_{eh}}{L}\right)^4}},\qquad I = \sqrt{\frac{k_h}{k_v}}\end{gathered}

where:

$J$ productivity index in terms of the rate at standard conditions, m³/(s·Pa)
$R$ dimensionless resistance to inflow (the analogue of $\ln(r_e/r_w) + S$ of a vertical well), —
$L$ length of the horizontal section that opens the formation, m
$a$ semi-major axis of the drainage ellipse with its foci at the ends of the wellbore and the area of a circle of radius $r_{eh}$ , m
$r_{eh}$ radius of the circular drainage area of the horizontal well, m
$I$ anisotropy index, —
$k_h,\ k_v$ horizontal and vertical permeability, m²
$h,\ r_w$ thickness of the formation and radius of the wellbore, m
$S$ damage skin around the horizontal wellbore, —
$\mu,\ B,\ p_e,\ p_w$ viscosity and formation volume factor of the oil, pressure at the boundary and in the wellbore, Pa·s, m³/m³, Pa, Pa

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

Holds when

The inflow is steady-state (it does not change with time), at constant pressure at the boundary of a circular area, which is replaced by an ellipse of the same area.
The formation is homogeneous, with $k_h$ the same in all horizontal directions. The wellbore is in the middle of the thickness, parallel to the top.
The pressure along the wellbore is the same (there is no friction), and the flow is single-phase.
$I h < L$ and $L/2 < 0.9\,r_{eh}$ . If these conditions are not met, we need numerical models.

The first term is the resistance to flow in the horizontal plane. When $L$ is comparable with the size of the area, it is about one, against eight for a vertical well. The second is the resistance of the flow converging on the wellbore in the vertical direction. The factor $I h/L$ is small for a thin isotropic formation (one with the same permeability in all directions), and it grows with thickness and anisotropy. The wellbore skin enters with the same factor.

In other unit systems (2)

Metric oilfield units

J\,[\text{m}^3/(\text{d}\cdot\text{MPa})] = \frac{0.536\,k_h\,h}{\mu\,B\,R}

$k_h$ in mD, $h$ in m, $\mu$ in mPa·s, $B$ in m³/m³.

$0.536 = 2\pi\cdot 0.0853$ , as in the Dupuit equation.

Oilfield units

J\,[\text{STB/(d·psi)}] = \frac{7.08\cdot 10^{-3}\,k_h\,h}{\mu\,B\,R}

$k_h$ in mD, $h$ in ft, $\mu$ in cP, $B$ in rb/STB.

$7.08\cdot 10^{-3} = 2\pi\cdot 1.127\cdot 10^{-3}$ .

Worked exampleA horizontal well against a vertical well

Given (assumed data)

Formation and oil: from the lesson on steady-state flow: $k_h$ = 60 mD, $h$ = 18 m, $\mu$ = 1.8 mPa·s, $B$ = 1.25; $r_w$ = 0.108 m; drainage radius $r_e = r_{eh}$ = 400 m
Horizontal wellbore: $L$ = 600 m in the middle of the thickness; $k_v/k_h$ = 0.2 (assumed); no damage

Solution

Anisotropy index: $I = \sqrt{1/0.2} = 2.236$ , and $I h = 40.25$ m. This is less than $L$ , so the model applies.
Semi-axis of the ellipse: $a = 300\sqrt{0.5 + \sqrt{0.25 + (800/600)^4}} = 459.6$ m. The horizontal term is $\ln\dfrac{459.6 + \sqrt{459.6^2 - 300^2}}{300} = 0.9905$ .
Vertical term: $\dfrac{40.25}{600}\ln\dfrac{40.25}{0.108\cdot (2.236 + 1)} = 0.0671\cdot 4.7464 = 0.3185$ . The resistance is $R = 0.9905 + 0.3185 = 1.3090$ .
Productivity: $J_H = \dfrac{0.536\cdot 60\cdot 18}{1.8\cdot 1.25\cdot 1.3090} = 196.5$ m³/(d·MPa). For the vertical well, $J_V = \dfrac{578.88}{2.25\cdot 8.217} = 31.3$ .
Productivity improvement factor $\text{PIF} = 196.5/31.3 = 6.28$ .
At a drawdown of 4.5 MPa, the horizontal well gives $196.5\cdot 4.5 = 884$ m³/d, against $31.3\cdot 4.5 = 141$ m³/d for the vertical well. The horizontal well gives the rate of 141 m³/d at a drawdown of only $141/196.5 = 0.72$ MPa.

Answer: A horizontal wellbore 600 m long is 6.28 times more productive than a vertical well: 884 m³/d against 141 m³/d at the same drawdown.

What it means. The horizontal term accounts for three quarters of the resistance, and the convergence in the vertical direction accounts for a quarter. In the vertical well the whole resistance, about 8.2, is radial flow. The horizontal well gives the same rate at a drawdown of about 0.7 MPa instead of 4.5. For cones, sand and gas coming out of the oil, this is sometimes more important than the increase in rate. We compared the wells for the same drainage area. In practice a horizontal well gets a larger area. The data in this example are assumed for illustration.

CalculatedCalculated productivity improvement factor of a horizontal well in the formation of the example (

h

= 18 m,

r_{eh}

= 400 m, no damage) against the ratio

k_v/k_h

for three wellbore lengths. The dot is the example.

The figure shows two patterns. Doubling the length of the wellbore from 300 to 600 m increases the productivity not twofold but by a factor of 1.7–1.9. This is because the horizontal term falls more slowly than $L$ grows. And lowering $k_v/k_h$ from 1 to 0.01 halves the gain or more. That is why vertical permeability is the main thing to know about the formation before drilling a horizontal well.

Other models. The widely used formulas for the productivity of a horizontal well are built in the same way: the sum of a horizontal and a vertical resistance. They differ in detail:

Models of horizontal-well productivity and the resistance

R

for the wellbore of the example

Model: Standard model corrected for anisotropy (used in this lesson)
Vertical term: $\frac{I h}{L}\ln\frac{I h}{r_w(I + 1)}$
$R$ at $k_v = k_h$: 1.123
$R$ at $k_v/k_h$ = 0.2: 1.309

Model: Original formulation of anisotropy
Vertical term: $\frac{I h}{L}\ln\frac{I h}{2r_w}$
$R$ at $k_v = k_h$: 1.123
$R$ at $k_v/k_h$ = 0.2: 1.341

Model: Line sink between top and base (method of images)
Vertical term: $\frac{I h}{L}\ln\frac{I h}{\pi r_w(I + 1)}$
$R$ at $k_v = k_h$: 1.089
$R$ at $k_v/k_h$ = 0.2: 1.232

Model: Yu. P. Borisov (isotropic formation; horizontal term $\ln(4r_e/L)$ )
Vertical term: $\frac{h}{L}\ln\frac{h}{2\pi r_w}$
$R$ at $k_v = k_h$: 1.079
$R$ at $k_v/k_h$ = 0.2: —

Model	Vertical term	$R$ at $k_v = k_h$	$R$ at $k_v/k_h$ = 0.2
Standard model corrected for anisotropy (used in this lesson)	$\frac{I h}{L}\ln\frac{I h}{r_w(I + 1)}$	1.123	1.309
Original formulation of anisotropy	$\frac{I h}{L}\ln\frac{I h}{2r_w}$	1.123	1.341
Line sink between top and base (method of images)	$\frac{I h}{L}\ln\frac{I h}{\pi r_w(I + 1)}$	1.089	1.232
Yu. P. Borisov (isotropic formation; horizontal term $\ln(4r_e/L)$ )	$\frac{h}{L}\ln\frac{h}{2\pi r_w}$	1.079	—

All the models describe steady-state inflow in a homogeneous formation. The spread of $R$ between the models is about 4% in an isotropic formation and up to 9% at $k_v/k_h$ = 0.2. A change of $k_v/k_h$ from 0.05 to 0.5, by contrast, changes $R$ by 40%.

The formula with $\ln(4r_e/L)$ is the classical one in Russian-language practice. Its horizontal term is the limit of the standard one for a large drainage area ( $r_e \gg L$ ). Its vertical term coincides with the vertical term of the method of images for an isotropic formation. For a rectangular area and the pseudo-steady-state regime (flow in which the pressure falls at the same rate everywhere), we use rectangular-block models, in which the position of the wellbore in the block is specified explicitly. The advanced section treats the special case where the wellbore passes through the block from side to side.

Position of the wellbore across the thickness. The method of images also gives the cost of moving the wellbore away from the middle of the formation. The resistance increases by $-\dfrac{I h}{L}\ln\sin\dfrac{\pi z_w}{h}$ , where $z_w$ is the height of the wellbore above the base. For the wellbore of the example, at 85% of the thickness from the base (2.7 m below the top), the addition is only about 0.053 on a resistance of 1.31. This is a loss of productivity of about 4%. So we choose the position of the wellbore across the thickness by its distance from the contacts, not for the sake of productivity.

Direction of the wellbore. Permeability may differ in the horizontal plane as well, because of vertical fractures or oriented sand bodies. Then the permeability across the wellbore matters, because inflow to a long wellbore goes mainly in that direction. So we drill the wellbore across the direction of greatest permeability and across the strike of the fractures (the direction in which they run along the surface). It then intersects more fractures, and the flow to it follows the better direction. The stresses at the time the natural fractures formed set their direction. The ones that conduct best are usually those parallel to the present-day greatest horizontal stress (Rock mechanics for hydraulic fracturing).

Skin of a horizontal wellbore and the effective radius

A horizontal wellbore stays in contact with the drilling fluid longer than a vertical one, and it works at a small drawdown. So the mudcake (the layer of mud solids on the wall) and the contamination are removed less well. The damage is distributed unevenly, because at the heel the wellbore has been open longest. We reduce it with special drill-in fluids (fluids for drilling through the reservoir), whose mudcake is removed at a small drawdown. Treatments of a long wellbore require acid diversion (steering the acid into all the intervals, not only the easiest one) (Formation damage and damage skin, Acid treatments: principle, economics and candidate selection).

In the standard model the damage enters as the skin $S$ of radial inflow to the wellbore, with the factor $I h/L$ . The skin comes from Hawkins' formula for the zone around the wellbore (Skin factor), which is only approximate in an anisotropic formation. The flow through the damaged zone is spread along the whole length of the wellbore, so each metre carries only a little of it. That is why a skin of 8 in a horizontal well “costs” much less than in a vertical one.

It is convenient to compare a horizontal well with a vertical one through the equivalent skin and the effective wellbore radius. These describe the vertical well that gives the same rate in the same area.

Equivalent skin and effective wellbore radius of a horizontal well

S_{eq} = R - \ln\frac{r_{eh}}{r_w},\qquad r_w' = r_w\,e^{-S_{eq}} = r_{eh}\,e^{-R}

where:

$S_{eq}$ skin of a vertical well of radius $r_w$ that is equivalent to the horizontal one, —
$r_w'$ radius of a vertical well with no skin that is equivalent to the horizontal one, m
$R,\ r_{eh},\ r_w$ resistance to inflow, radius of the area and radius of the wellbore, —, m, m

Units: dimensionless; radii in the same units

Holds when

The same drainage area and the same boundary conditions as in the productivity model.

For an undamaged horizontal well $S_{eq}$ is usually from −3 to −8, and the effective radius is from a few metres in a thick anisotropic formation to a hundred metres or more. A long wellbore behaves like a well of enormous radius.

Worked exampleThe cost of damage to a horizontal wellbore

Given (assumed data)

Well: from the previous example: $R$ = 1.3090, $I h/L$ = 0.0671, $J_H$ = 196.5 m³/(d·MPa); $\ln(r_e/r_w)$ = 8.217
Damage: skin of the zone around the wellbore $S$ = 8 (estimated from a test)

Solution

Addition to the resistance: $0.0671\cdot 8 = 0.537$ . The new resistance is $R = 1.3090 + 0.537 = 1.8460$ .
Productivity: $J_H = 578.88/(2.25\cdot 1.8460) = 139.4$ . That is 71% of the undamaged value.
A vertical well with the same skin keeps $8.217/(8.217 + 8) = 51$ % of its productivity.
Equivalent skin of the undamaged horizontal well: $S_{eq} = 1.3090 - 8.217 = -6.91$ , and the effective radius is $r_w' = 400\,e^{-1.3090} = 108$ m. With the damage, the equivalent skin is $-6.37$ and the effective radius is 63 m.

Answer: A skin of 8 takes away about 29% of the productivity of the horizontal well and about 49% of that of the vertical well.

What it means. Damage to a horizontal wellbore costs less, but it is not free: it takes almost a third of the productivity. And in long wellbores opened with an unsuitable fluid, the skin can be greater than 8. The effective radius is convenient for comparison with vertical wells: an undamaged wellbore 600 m long is equivalent to a vertical well of radius about 108 m. The data in this example are assumed for illustration.

Completing a horizontal wellbore

The section to be completed in a horizontal wellbore is ten to a hundred times longer than in a vertical well. All the tasks of completion become harder with the length: holding the walls and the sand, dividing the wellbore into intervals, shutting off water or gas that has broken through, carrying out treatments, and giving instruments access. The lesson Well completion: opening the reservoir and the tubing string describes the main bottomhole designs. Here we look at what changes in them for a horizontal wellbore.

Completion options for a horizontal wellbore

Option: Open hole
Advantages: cheap; full diameter; no perforation resistance
Disadvantages: in weak rock the wellbore will collapse; an interval cannot be shut off

Option: Uncemented screen liner (slotted, wire-wrapped, mesh)
Advantages: holds the walls; retains sand (Sand control methods); instruments can pass through it
Disadvantages: the annulus connects the whole wellbore: an interval cannot be shut off

Option: Screens with packers in the annulus (swellable or mechanical)
Advantages: the wellbore is divided into sections; a watered-out section can be closed
Disadvantages: more expensive; the packer positions are chosen from logs

Option: Inflow control devices on the screen
Advantages: even out the inflow, limit breakthroughs
Disadvantages: an additional pressure drop; useless without packers

Option: Cemented liner with perforations
Advantages: reliable isolation; selective perforating; multistage hydraulic fracturing
Disadvantages: expensive; good cement is required; perforation resistance

Option: Intelligent completion
Advantages: pressure and temperature sensors, controllable interval valves
Disadvantages: expensive; the number of controllable zones is limited

Option	Advantages	Disadvantages
Open hole	cheap; full diameter; no perforation resistance	in weak rock the wellbore will collapse; an interval cannot be shut off
Uncemented screen liner (slotted, wire-wrapped, mesh)	holds the walls; retains sand (Sand control methods); instruments can pass through it	the annulus connects the whole wellbore: an interval cannot be shut off
Screens with packers in the annulus (swellable or mechanical)	the wellbore is divided into sections; a watered-out section can be closed	more expensive; the packer positions are chosen from logs
Inflow control devices on the screen	even out the inflow, limit breakthroughs	an additional pressure drop; useless without packers
Cemented liner with perforations	reliable isolation; selective perforating; multistage hydraulic fracturing	expensive; good cement is required; perforation resistance
Intelligent completion	pressure and temperature sensors, controllable interval valves	expensive; the number of controllable zones is limited

In an open hole, and in a liner without packers, fluid can flow along the wellbore through the annulus. That is why we have to divide the wellbore into sections with packers in the annulus. Swellable packers have an elastomer (a rubber-like material) that swells in oil or water, and we set them directly in the open hole without cement. If the tubing is run inside the horizontal section, the point of greatest drawdown shifts from the heel to the middle of the wellbore. This is a simple way to even out the inflow.

In low-permeability formations, a horizontal wellbore is almost always supplemented with multistage hydraulic fracturing. We divide the wellbore into sections with a cemented liner or with packers, and we create a fracture in each section (Hydraulic fracturing: purpose and well selection, Fracture design: size and conductivity). The lesson Multilateral wells covers several wellbores from one well.

Smart wells (intelligent completions) are fitted with downhole pressure and temperature gauges and with valves that open and throttle individual intervals from the surface. A multiphase flowmeter at the wellhead measures the total rate. Smart wells appeared because repairing a horizontal well is slow, expensive and risky. The repairs include isolating an interval and re-perforating. This is true particularly for subsea wells at offshore fields. Coiled tubing (a long continuous pipe that is unreeled into the well) and downhole tractors give instruments and tools access into a long wellbore.

Wellbore friction and inflow control devices

The standard model assumes that the pressure is the same along the wellbore. In reality the flow in the wellbore builds up from the toe to the heel, and some pressure is spent on friction. So the pressure at the heel is lower than at the toe. This means that the drawdown on the formation is greater at the heel, the inflow there is stronger, and water or gas arrives there first. This is the heel–toe effect. It is important when friction losses are comparable with the drawdown: with high permeability, a long wellbore, a small diameter or a high rate.

Inflow and friction along a horizontal wellbore

\frac{\dd q}{\dd x} = \frac{J}{L}\,\bigl(p_e - p(x)\bigr),\qquad -\frac{\dd p}{\dd x} = \lambda\,\frac{\rho\,v^2}{2d},\quad v = \frac{qB}{\pi d^2/4}

where:

$x$ distance along the wellbore from the toe to the heel, m
$q(x)$ rate accumulated in the wellbore up to the point $x$ (standard conditions), m³/s
$J/L$ specific productivity per metre of wellbore (from the standard model), m³/(s·Pa·m)
$p(x),\ p_e$ pressure in the wellbore and at the boundary, Pa
$\lambda$ Darcy friction factor (Haaland formula, Multiphase flow in tubing), —
$\rho,\ v,\ d$ density of the oil in the wellbore, mean velocity, internal diameter of the liner, kg/m³, m/s, m

Units: SI

Holds when

The oil is single-phase, and the specific productivity is the same along the whole length. This underestimates the inflow to the ends of the wellbore.
Friction is the same as in a pipe with no inflow through the wall (inflow through the wall changes it slightly). We neglect the acceleration of the flow.
The system is solved numerically. We specify the drawdown at the heel and choose the drawdown at the toe so that the pressures agree.

Inflow decreases where the pressure in the wellbore is higher, which is towards the toe. The greater $J/L$ is compared with the “conductivity” of the wellbore (how easily fluid flows along it), the more uneven the inflow.

Worked exampleFriction losses in a long wellbore in a high-permeability formation

Given (assumed data)

Formation: $k_h$ = 800 mD, $k_v/k_h$ = 0.25, $h$ = 15 m, $r_{eh}$ = 600 m; oil 780 kg/m³, $\mu$ = 0.9 mPa·s, $B$ = 1.2
Wellbore: $L$ = 1000 m, $r_w$ = 0.108 m; liner with an internal diameter of 124 mm, roughness 0.05 mm; drawdown at the heel 0.25 MPa

Solution

Without friction: $R = 1.0257$ , and $J = 0.536\cdot 800\cdot 15/(0.9\cdot 1.2\cdot 1.0257) = 5806$ m³/(d·MPa). The rate is $5806\cdot 0.25 = 1452$ m³/d.
Joint calculation of inflow and friction: the rate is 1271 m³/d, the drawdown at the toe is 0.208 MPa, and the inflow per metre at the heel is 1.20 times that at the toe.
Check of friction at the heel: the cross-sectional area is 0.012076 m², and the velocity is $v = 1.2\cdot 1271/(86{,}400\cdot 0.012076) = 1.46$ m/s. The Reynolds number (it shows how turbulent the flow is) is $Re = 780\cdot 1.46\cdot 0.124/0.0009 = 156{,}901$ , the friction factor is $\lambda = 0.0186$ , and the gradient is $\lambda\rho v^2/(2d) = 125$ Pa/m.
Estimate of the total losses: if the rate grows linearly along the wellbore, the gradient grows as the square of the rate, and on average it is a third of the gradient at the heel: $125\cdot 1000/3 = 0.042$ MPa. This agrees with the difference of the drawdowns at the heel and the toe ( $0.25 - 0.208$ ).

Answer: Friction takes about 0.042 MPa, which is 17% of the drawdown at the heel. The rate is 12% lower than without friction, and the inflow at the heel is 20% greater than at the toe.

What it means. At a drawdown of a fraction of a megapascal, friction in the wellbore is already a noticeable part of the balance. A calculation with an infinitely conductive wellbore would overestimate the rate, and it would not show that water arrives at the heel first. A larger diameter reduces the losses roughly in inverse proportion to the fifth power of the diameter. In a low-permeability formation the drawdown is measured in megapascals, and friction there hardly matters. The data in this example are assumed for illustration.

Inflow control devices (ICDs) are calibrated restrictions between the formation and the inner pipe of the screen. Fluid from the annulus of each section enters the pipe through a nozzle, a channel or a tube. The pressure drop across them grows with the inflow of the section, roughly as the square of the rate. A section where the inflow is high, for example at the heel, in a high-permeability layer or in a fracture, meets a greater resistance, and the inflow along the length evens out.

Inflow of a section through an inflow control device

p_e - p(x) = \frac{q'}{J/L} + c\,q'^2,\qquad \Delta p_{\text{ICD}} = c\,q'^2 \approx \frac{\rho\,v_n^2}{2C_d^2}

where:

$q'$ inflow per metre of wellbore, m³/(s·m)
$c$ constant of the devices, reduced to a metre of wellbore, Pa·s²/m⁴
$v_n,\ C_d$ velocity in the nozzle and discharge coefficient (for nozzle-type devices), m/s, —
$p_e,\ p(x),\ J/L$ as above, Pa, Pa, m³/(s·Pa·m)

Units: SI

Holds when

Packers divide the annulus into sections, and all the inflow of a section passes through its device.
The flow is single-phase. The pressure drop of nozzle-type devices is almost independent of viscosity, while that of channel-type devices depends on it appreciably.
We choose the constant so that the pressure drop across the devices at the design rate is comparable with the drawdown on the formation, or greater.

As long as $c\,q'^2$ is small compared with the drawdown on the formation, the device hardly interferes. When the two are comparable, the inflow of each section depends less on the local drawdown and permeability. The price is a greater overall pressure drop at the same rate.

Worked exampleWhat inflow control devices give

Given (assumed data)

Well: from the previous example; the rate is held at 1500 m³/d
Devices: pressure drop of 0.4 MPa at an inflow of 24 m³/d per 12 m section, that is, $c$ = 0.1 MPa per (m³/(d·m))²
Heterogeneity (second case): a layer with five times the specific productivity 300–400 m from the toe; the total productivity is the same

Solution

Without devices, the drawdown at the heel is 0.30 MPa and at the toe it is 0.24 MPa. The inflow at the heel is 1.23 times that at the toe.
With devices, the drawdown at the heel is 0.53 MPa. (The devices lose $0.1\cdot 1.5^2 = 0.225$ MPa at the mean inflow of 1.5 m³/(d·m).) The inflow at the heel exceeds that at the toe only 1.08 times.
The layer without devices gives 34% of the inflow with 10% of the length. With devices, it gives 14%.

Answer: The devices evened out the inflow along the length (heel-to-toe ratio 1.08 instead of 1.23) and reduced the share of the high-permeability layer from 34 to 14%. But they required a drawdown at the heel of 0.53 MPa instead of 0.30 MPa.

What it means. Inflow control devices do not increase the rate. At the same drawdown the rate is even lower. Their benefit is even drainage of the wellbore and a delay in breakthrough. Water that arrives through a layer or at the heel will not take all the inflow. Autonomous devices (AICDs) restrict the inflow of low-viscosity water and gas more strongly than the inflow of oil. After breakthrough they close the watered-out section by themselves. The data in this example are assumed for illustration.

CalculatedCalculated inflow per metre of wellbore for the examples at a rate of 1500 m³/d: without friction (uniform), with friction in a 124 mm liner, and with inflow control devices. The toe is on the left, the heel on the right.

Water and gas cresting

Under a vertical well, water rises as a cone. Under a horizontal wellbore, it rises as a crest, a ridge elongated along the wellbore. Gas from above descends in the same way. The mechanism is the same as for a cone (Water and gas coning): water rises until the weight of the raised column balances the pressure drop in the oil. But the drawdown is spread over hundreds of metres of wellbore, so at the same rate it is several times smaller. So the crest grows more slowly, and the critical water-free rate is higher. We choose the distance from the wellbore to the contact by the same logic as the interval of a vertical well. In oil rims, geosteering holds the wellbore at a set height between the contacts.

Critical water-free rate of a crest at vertical equilibrium

q_c = \frac{k_h\,\Delta\rho\,g\,\bigl[h^2 - (h - z_w)^2\bigr]\,L}{\mu B\,y_e}

where:

$q_c$ critical water-free rate at standard conditions, m³/s
$z_w$ height of the wellbore above the oil–water contact (OWC); for a wellbore at the top, $z_w = h$ , m
$y_e$ distance from the wellbore to the line where the OWC stays in place (half the distance between neighbouring wellbores), m
$k_h,\ \Delta\rho,\ g,\ h,\ L,\ \mu,\ B$ horizontal permeability, density difference between water and oil, acceleration due to gravity, oil thickness, wellbore length, viscosity and formation volume factor, m², kg/m³, m/s², m, m, Pa·s, m³/m³

Units: SI (any consistent system of units)

Holds when

There is vertical equilibrium: oil flows to the wellbore horizontally from both sides through the layer above the crest. This is a plane problem, and the ends of the wellbore are not taken into account.
The water is stationary, and the contact stays in place at the distance $y_e$ . The formation is isotropic.
Right at the wellbore the assumption is crude, but it hardly affects the critical rate. Exact solutions of this plane free-boundary problem (a flow problem with a moving surface) for $y_e \gg h$ practically coincide with the formula. An approximation built from the pressure field without allowing for the crest (see below) gives several times more.

The critical rate is proportional to the length of the wellbore and inversely proportional to the distance to the line of constant contact. The derivation is the same as for a cone at vertical equilibrium. Only the flow is plane and not radial.

In other unit systems (1)

Metric oilfield units

q_c\,[\text{m}^3/\text{d}] = 8.37\cdot 10^{-7}\,\frac{k_h\,\Delta\rho\,\bigl[h^2 - (h - z_w)^2\bigr]\,L}{\mu\,B\,y_e}

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

$8.37\cdot 10^{-7} = 0.0853\cdot 9.81\cdot 10^{-6}$ .

Worked exampleA crest against a cone in the same formation

Given (assumed data)

Formation and fluids: from the lesson on coning: 32 m of oil above bottom water, $k_h$ = 45 mD, $k_v/k_h$ = 0.15, $\Delta\rho$ = 300 kg/m³, $\mu$ = 1.4 mPa·s, $B$ = 1.25
Horizontal wellbore: $L$ = 500 m at the top ( $z_w = h$ = 32 m); the contact is held at a distance $y_e$ = 350 m on both sides
Vertical well in the same formation: critical rate 1.49–3.72 m³/d by three methods

Solution

Vertical equilibrium: $q_c = 8.37\cdot 10^{-7}\cdot\dfrac{45\cdot 300\cdot 32^2\cdot 500}{1.75\cdot 350} = 9.45$ m³/d.
An approximation of the pressure-field solutions (Chaperon) for a wellbore at the top: we multiply the vertical-equilibrium result by $F_H = 3.9625 + 0.0616\,\alpha - 0.00054\,\alpha^2$ , where $\alpha = (y_e/h)\sqrt{k_v/k_h} = 4.236$ . The approximation is valid for $1 \le \alpha < 70$ and for a distance between neighbouring wellbores of less than four of their lengths. This gives $F_H = 4.214$ and $q_c = 4.214\cdot 9.45 = 39.8$ m³/d.
The cone formula at vertical equilibrium, with the effective radius of the horizontal well: $\ln(r_{eh}/r_w') = R = 1.924$ (the standard model with $r_{eh} = y_e$ = 350 m), and $q_c = \dfrac{2.63\cdot 10^{-6}\cdot 45\cdot 300\cdot 32^2}{1.75\cdot 1.924} = 10.8$ m³/d.
Comparison with the vertical well: by vertical equilibrium the ratio is $9.45/2.32 = 4.1$ times, and by the pressure-field solutions it is $39.8/3.72 = 10.7$ times.
Productivity of the wellbore: $J_H = 229$ m³/(d·MPa). At a drawdown of only 1 MPa it would give 229 m³/d.

Answer: The critical rate of the crest is from 9.45 to 39.8 m³/d by different methods (the upper value comes from the approximation, which overestimates here). This is 4–11 times that of a vertical well, but still several times below the rate the wellbore could deliver.

What it means. A horizontal wellbore moves the water away, but it does not prevent the water from arriving. Operating below the critical rate is unprofitable here too. Its real gain is a smaller drawdown at the same rate, slower growth of the crest and of the water cut, and more oil produced before water breakthrough. The methods again differ several-fold. For this idealised plane problem, vertical equilibrium is closer to the exact solution, and the pressure-field approximation overestimates it. Friction in the wellbore and heterogeneity bring the water first to the heel or to a high-permeability section. The data in this example are assumed for illustration.

We estimate the time to breakthrough of the crest, and the growth of the water cut of a horizontal well, with dimensionless correlations similar to those for cones, or by simulation. Water cut develops differently with a horizontal wellbore. Water first arrives at one point, at the heel or in a fractured or high-permeability zone, and spreads from there along the wellbore. That is why sections with packers, inflow control devices and monitoring of the inflow along the wellbore are so important for managing the crest.

Limitations and economics

Drilling. We need a rig with enough hoisting capacity and torque, a steerable drilling system and experienced directional-drilling specialists. The first wells in a field cost noticeably more than the later ones.
Clean-up. A small drawdown is poor at removing mudcake and contamination. We need special drill-in fluids to open the formation.
Servicing. Surveys of the inflow along the length, treatments and interval isolation need coiled tubing or tractors, and they cost more than in a vertical well.
Uncertainty. The actual rate of individual horizontal wells often differs noticeably from the forecast, by tens of per cent or more, in both directions. The forecast for the field as a whole may still come true. A high initial rate does not always turn into an increase in recoverable reserves, because we also have to deal with water and gas breakthroughs along the length of the wellbore.
Concentrated risk. With fewer wells, the shutdown of one well has a greater effect on production.

Terms and symbols

Heel and toe: The start of the horizontal section of the wellbore in the formation (the heel) and its end (the toe).
Anisotropy index $I = \sqrt{k_h/k_v}$: The factor by which the formation is “stretched” vertically to make it isotropic.
Productivity improvement factor (PIF): The ratio of the productivity index of a horizontal well to that of a vertical well in the same formation (productivity improvement factor).
Equivalent skin $S_{eq}$ and effective radius $r_w'$: The skin and radius of a vertical well that has the same productivity as the horizontal one.
Heel–toe effect: Uneven inflow along the wellbore because of frictional pressure losses. The drawdown and the inflow are greater at the heel.
Inflow control device (ICD): A calibrated restriction between the formation and the screen pipe that evens out the inflow along the wellbore.
Crest (cresting): The rise of water (or the descent of gas) along a horizontal wellbore. It is the analogue of the cone of a vertical well.
Geosteering: Control of the trajectory of the wellbore in the formation, using logging-while-drilling data and cuttings.

Common mistakes and misconceptions

Wrong

“A wellbore twice as long gives twice the rate.”

In fact

The horizontal resistance falls more slowly than the length grows. In the example in this lesson, lengthening the wellbore from 300 to 600 m increased the productivity 1.77 times. With strong friction in the wellbore the gain is smaller still.

Wrong

“A horizontal well is always better than a vertical one.”

In fact

In a thick formation with low vertical permeability or with continuous shale layers, the wellbore drains only part of the thickness, and a vertical well with a hydraulic fracture can be more profitable.

Wrong

“A skin of 10 in a horizontal well is a catastrophe, as in a vertical one.”

In fact

The wellbore skin enters the resistance with the factor $I h/L$ , usually 0.05–0.2. A skin of 10 adds 0.5–2 to a resistance of about one. The loss is noticeable but not catastrophic. Moreover, the damage along the length is usually uneven.

Wrong

“Inflow control devices increase the rate.”

In fact

At the same drawdown they reduce it. Their job is to even out the inflow along the length, to delay the breakthrough of water and gas, and to restrict the inflow from a watered-out section.

Wrong

“The longer the wellbore, the better: we drill as far as the rig allows.”

In fact

In high-permeability formations, friction in the wellbore eats up the gain of the far part of the wellbore, and water arrives at the heel earlier. The cost, the risks and the difficulty of servicing grow with length. We find the best length by calculating the inflow with friction, and by looking at the economics.

In depth: advanced detail

Where the terms of the model come from

The horizontal term. Suppose we neglect the thickness of the formation. Then the wellbore is a slot of length $L$ in the plane. The pressure field around a slot is known exactly. The lines of equal pressure are confocal ellipses (ellipses that share the same foci), with their foci at the ends of the slot. The pressure difference between the slot and an ellipse with the semi-major axis $a$ is $\dfrac{qB\mu}{2\pi k_h h}\,\operatorname{arcosh}\dfrac{2a}{L}$ , where $\operatorname{arcosh} x = \ln\bigl(x + \sqrt{x^2 - 1}\bigr)$ . We replace a circular area of radius $r_{eh}$ by an ellipse of the same area: $\pi a b = \pi r_{eh}^2$ with $b = \sqrt{a^2 - L^2/4}$ . This gives the formula for $a$ . For $r_{eh} \gg L$ the ellipse is almost a circle, $a \approx r_{eh}$ , and the term tends to $\ln(4r_{eh}/L)$ .

The vertical term. The impermeable top and base act as mirrors, and the wellbore between them is reflected in them. The result is an infinite series of image sinks (mirror images of the wellbore that also draw fluid in) above and below the formation. For a wellbore in the middle of the thickness, the spacing is $h$ . The sum of their potentials gives plane-parallel flow far from the wellbore and radial flow near it. The difference between the two is the exact “convergence resistance” $\dfrac{h}{L}\ln\dfrac{h}{2\pi r_w}$ (in units of $qB\mu/(2\pi k_h h)$ ) for a wellbore in the middle of an isotropic formation. In an anisotropic formation, after stretching, the thickness is $I h$ , the permeability is $\sqrt{k_h k_v}$ , and the wellbore is an ellipse with semi-axes $r_w$ and $I r_w$ . This ellipse is equivalent to a circle of radius $r_w(I + 1)/2$ . This gives $\dfrac{I h}{L}\ln\dfrac{I h}{\pi r_w(I + 1)}$ . The standard model uses a cruder approximation with $\ln\dfrac{I h}{r_w(I + 1)}$ , which is greater by $\dfrac{I h}{L}\ln\pi$ , that is, slightly on the safe side. The same series of images for a wellbore at height $z_w$ gives the addition $-\dfrac{I h}{L}\ln\sin\dfrac{\pi z_w}{h}$ .

Pseudo-steady-state inflow: a wellbore along a rectangular block

Let the wellbore pass right through a closed block of width $y_e$ (for example, between two neighbouring parallel wellbores or faults). The flow is then plane, with the same picture in any cross-section across the wellbore. In the cross-section the wellbore is a “vertical well” in a rectangle of width $y_e$ and height $I h$ (after stretching). For it, the pseudo-steady-state formula with the shape factor of a rectangle is suitable (approximate shape factor of a rectangle):

A wellbore passing right through a rectangular block (pseudo-steady state)

J = \frac{q}{\bar p - p_w} = \frac{2\pi k_h h}{\mu B\,R_b},\qquad R_b = \frac{\pi y_e}{6L} + \frac{I h}{L}\left[\ln\frac{I h}{\pi r_w(I + 1)\sin(\pi z_w/h)} + S\right]

where:

$\bar p$ average pressure in the block, Pa
$y_e$ full width of the block across the wellbore (in the crest formula $y_e$ is half the distance between wellbores); the wellbore is in the middle of the width, m
$z_w$ height of the wellbore above the base, m
$R_b$ resistance: linear inflow to the wellbore plus convergence in the vertical direction, —
$k_h,\ h,\ I,\ L,\ r_w,\ S,\ \mu,\ B$ as in the standard model, m², m, —, m, m, —, Pa·s, m³/m³

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

Holds when

The block is closed, and the regime is pseudo-steady state. The wellbore passes right through the block, so the flow is plane.
The wellbore is at least $I h$ from the lateral boundaries of the block. Then the approximation of the shape factor is accurate to hundredths.
If the wellbore is shorter than the block, we add the pseudo-skin of partial penetration of the block. Programs calculate such rectangular-block models.

The first term is the resistance of the linear inflow to the wellbore from both sides. The second is the same convergence as in the standard model, written in the form of the method of images. The width of the block acts linearly, not under a logarithm. In a narrow block the productivity is high, but the hydrocarbons in place are also small.

Worked exampleA wellbore in a block 400 m wide

Given (assumed data)

Formation and wellbore: from the first example: $h$ = 18 m, $k_v/k_h$ = 0.2 ( $I h$ = 40.25 m), $L$ = 600 m in the middle of the thickness, $r_w$ = 0.108 m
Block: width $y_e$ = 400 m, the wellbore passes right through it

Solution

Linear term: $\pi\cdot 400/(6\cdot 600) = 0.3491$ .
Convergence: $\dfrac{40.25}{600}\ln\dfrac{40.25}{\pi\cdot 0.108\cdot 3.236} = 0.0671\cdot 3.6016 = 0.2417$ . The resistance is $R_b = 0.5908$ . (The shape factor of a rectangle gives the same result: $C_A = 0.0266$ .)
Productivity from the average pressure: $J = 578.88/(2.25\cdot 0.5908) = 435$ m³/(d·MPa).
Wellbore at 85% of the thickness from the base: the addition is $-0.0671\ln\sin(0.85\pi) = 0.0530$ , and $R_b = 0.6438$ .

Answer: The productivity of the wellbore in the block is about 435 m³/(d·MPa). Moving the wellbore towards the top reduces it by 8%.

What it means. About 60% of the resistance here is linear inflow through a strip 200 m wide on each side, and about 40% is convergence to the wellbore. That is why the position of the wellbore across the thickness costs more here than in the first example, where convergence is only a quarter of the resistance. The productivity is higher than in the first example because the block is narrow and limited in area. We cannot compare the two examples, because the shape of the area and the regime are different. The data in this example are assumed for illustration.

Flow regimes in a test of a horizontal well

After a horizontal well is started up or shut in, the shape of the flow changes several times. Each flow regime gives its own formation parameter (Flow regimes, Diagnosis with the pressure derivative):

Flow regimes to a horizontal wellbore

Regime: Early radial (in the cross-section across the wellbore)
When: before the pressure disturbance (the pressure change spreading through the formation) has reached the top and the base
Signature on the derivative: plateau
What it determines: $\sqrt{k_h k_v}\,L$ and the wellbore skin

Regime: Early linear
When: the disturbance has filled the thickness but is small compared with the length of the wellbore
Signature on the derivative: slope ½
What it determines: $L\sqrt{k_h}$ (the length of the working wellbore)

Regime: Late pseudoradial
When: the disturbance has travelled far compared with the length
Signature on the derivative: second plateau
What it determines: $k_h h$ and the total skin

Regime: Boundary regimes
When: limited width or a closed area
Signature on the derivative: slope ½ or 1
What it determines: the width of the channel, the volume

Regime	When	Signature on the derivative	What it determines
Early radial (in the cross-section across the wellbore)	before the pressure disturbance (the pressure change spreading through the formation) has reached the top and the base	plateau	$\sqrt{k_h k_v}\,L$ and the wellbore skin
Early linear	the disturbance has filled the thickness but is small compared with the length of the wellbore	slope ½	$L\sqrt{k_h}$ (the length of the working wellbore)
Late pseudoradial	the disturbance has travelled far compared with the length	second plateau	$k_h h$ and the total skin
Boundary regimes	limited width or a closed area	slope ½ or 1	the width of the channel, the volume

Wellbore effects (the flow in and around the wellbore itself) often mask the early radial regime. In a low-permeability formation, the pseudoradial regime may not be reached during the whole test. We interpret the test with a horizontal-well model. Parallel boundaries are treated as in the lesson Several boundaries: parallel faults and wedge-shaped blocks.

Connections to other topics

Steady-state radial flow — the Dupuit equation and the formation of the first example.
Partial penetration and well deviation — convergence of flow, anisotropy and stretching of the formation vertically.
Water and gas coning — water and gas cones, the critical rate and the breakthrough time of a vertical well.
Skin factor — skin, Hawkins' formula and the effective wellbore radius.
Directional wells: applications and profile — well profile, build-up rate and radius of curvature.
Deflection tools and downhole motors — how the angle is built and the wellbore is steered through the formation.
Well completion: opening the reservoir and the tubing string — open hole, slotted liner and cemented casing.
Multiphase flow in tubing — friction factor and pressure losses in pipes.
Pseudo-steady-state flow and the shape factor — shape factor of a rectangular area.
Multilateral wells — several wellbores from one well.
Hydraulic fracturing: purpose and well selection — hydraulic fracturing as an alternative to a horizontal wellbore and a complement to it.

Check yourself

1
How will the productivity improvement factor of a horizontal well change if the vertical permeability turns out to be 10 times smaller than expected?
1. A.It will not change: the horizontal permeability does the work
2. B.It will increase: there is less crossflow
3. C.Only the skin will change
4. D.It will fall: the vertical resistance grows with $I h/L$
Show the answer
Correct answer: D. It will fall: the vertical resistance grows with $I h/L$
The anisotropy index $I = \sqrt{k_h/k_v}$ will grow by a factor of $\sqrt{10} \approx 3.2$ , and the vertical term $\frac{I h}{L}\ln\frac{I h}{r_w(I+1)}$ will grow by a factor of 3–4. In the example in this lesson, the factor is 6.3 at $k_v/k_h$ = 0.2 and about 4 at 0.02.
2
A horizontal wellbore 800 m long in an isotropic formation 10 m thick, $r_w$ = 0.1 m. What is the vertical term of the standard model, and what will it be at $k_v/k_h$ = 0.01?
Show the answer
At $k_v = k_h$ : $I = 1$ , and $\frac{10}{800}\ln\frac{10}{0.1\cdot 2} = 0.0125\cdot\ln 50 \approx 0.049$ . This is negligible beside the horizontal term of about one. At $k_v/k_h$ = 0.01: $I = 10$ , and $\frac{100}{800}\ln\frac{100}{0.1\cdot 11} = 0.125\cdot\ln 90.9 \approx 0.56$ . This is 11 times greater and already a noticeable part of the resistance.
3
The damage skin of a horizontal wellbore is 10, $I h/L$ = 0.1, and the resistance without damage is $R$ = 1.2. By what factor will the productivity fall?
1. A.10 times
2. B.About 1.8 times
3. C.11 times
4. D.It will not change
Show the answer
Correct answer: B. About 1.8 times
The skin adds $0.1\cdot 10 = 1$ , and $R$ grows from 1.2 to 2.2. The productivity falls by a factor of $2.2/1.2 \approx 1.8$ . In a vertical well with a logarithmic term of about 8, the same skin would reduce the productivity by a factor of $(8 + 10)/8 \approx 2.25$ .
4
Screens with inflow control devices were installed in an open hole, but no packers were set in the annulus. Will the devices help to even out the inflow?
1. A.Hardly: the fluid will flow along the wellbore through the annulus to the sections with less resistance
2. B.Yes, if the rate is increased
3. C.Yes, but only for gas
4. D.Yes, completely
Show the answer
Correct answer: A. Hardly: the fluid will flow along the wellbore through the annulus to the sections with less resistance
A device evens out the inflow only if all the inflow of its section passes through it. Without packers the annulus connects the sections. The flow is redistributed along the annulus, and it bypasses the resistances of the devices.
5
Why in a high-permeability formation with a long horizontal wellbore does water often arrive at the heel first, and what is done about it?
Show the answer
Because of friction in the wellbore, the pressure at the heel is lower, the drawdown and the inflow there are greater, and the crest grows faster. What helps is a larger liner diameter, tubing run deep into the wellbore, and inflow control devices with packers in the annulus. After breakthrough, we can close the watered-out section.
6
A vertical well has a critical water-free rate of 3 m³/d. What is the critical rate of a horizontal wellbore 500 m long in the same formation most likely to be?
1. A.The same: it depends only on the density difference
2. B.Infinite: a horizontal well has no crest
3. C.Several times higher, but probably still below the economic rate
4. D.Lower: the wellbore is closer to the water
Show the answer
Correct answer: C. Several times higher, but probably still below the economic rate
The drawdown is spread along the length of the wellbore, so at the same rate it is several times smaller. In the example in this lesson, the critical rate of the crest is 4–11 times that of a vertical well (the upper bound is from the approximation, which overestimates). But this too is only ten to forty cubic metres per day, against the hundreds that the wellbore can deliver.

Updated: 10 October 2026