Partial penetration and well deviation

Quick refresher

When only an interval $h_w$ out of a thickness $h$ is open, the flow at the wellbore passes through an area $h/h_w$ times smaller. The streamlines converge vertically, and the resistance of this convergence is the pseudo-skin $S_{pp} > 0$ .
$S_{pp}$ grows as the penetration ratio $h_w/h$ falls. It also grows with the thickness of the reservoir and with the anisotropy $k_h/k_v$ . An interval at the top or bottom gives a larger pseudo-skin than the same interval in the middle of the reservoir.
Damage in the open interval enters the measured skin multiplied by $h/h_w$ : $S = S_d\,h/h_w + S_{pp}$ . A treatment removes only $S_d$ .
On a pressure record, partial penetration gives three regimes. First comes early radial flow to the interval (it gives $k_h h_w$ ). Then comes spherical or hemispherical flow (a slope of −½ on the pressure derivative curve; it gives the spherical permeability). Last comes late radial flow to the whole thickness (it gives $k_h h$ and the full skin).
A deviated wellbore of length $h/\cos\theta$ gives $S_\theta < 0$ . Anisotropy greatly reduces this gain. The damage skin of a deviated well enters the measured skin multiplied by $\cos\theta$ in an isotropic reservoir, and by $\cos\theta'$ (closer to 1) in an anisotropic one.
Pseudo-skins are estimates from models of a homogeneous reservoir. Their accuracy is limited above all by how well we know the ratio $k_v/k_h$ .

Key relationships

S_{pp} = \left(\dfrac{h}{h_w} - 1\right)\left[\ln h_D - G\!\left(\dfrac{h_w}{h}\right)\right],\quad h_D = \dfrac{h_s}{r_w}\sqrt{\dfrac{k_h}{k_v}}

pseudo-skin of partial penetration (analytical approximation)

S = S_d\,\dfrac{h}{h_w} + S_{pp}

measured skin of a partially penetrating well with damage

k_s = \left(k_h^2\,k_v\right)^{1/3}

spherical permeability

S_\theta = -\left(\dfrac{\theta'}{41}\right)^{2.06} - \left(\dfrac{\theta'}{56}\right)^{1.865}\lg\dfrac{h_D}{100}

pseudo-skin of a deviated wellbore, $\theta'$ in degrees

S = S_d\cos\theta' + S_\theta

measured skin of a deviated well with damage; in an isotropic reservoir $\theta' = \theta$

Common pitfalls

Blaming damage for a large skin in a partially penetrating well, and ordering an acid treatment.
Dividing the flow capacity found in the early radial section by the whole thickness of the reservoir.
Taking a section of spherical flow for radial flow, and calculating $kh$ and the skin from it.
Calculating the pseudo-skin with the isotropic formula when $k_v$ is much smaller than $k_h$ . It comes out too low by several units.
Taking a skin close to zero in a deviated well as proof that there is no damage.

After this lesson you can

explain why opening only part of the thickness creates extra resistance, and why the vertical permeability decides how large it is
calculate the pseudo-skin of partial penetration from the penetration ratio, the thickness, the anisotropy (the difference between vertical and horizontal permeability) and the position of the interval, and estimate the loss of productivity
break the measured skin of a partially penetrating well down into the pseudo-skin and the damage skin, and estimate what a treatment will give
recognise early radial flow, spherical or hemispherical flow, and late radial flow on a pressure record; estimate how long each lasts; and find the spherical and vertical permeability
calculate the negative pseudo-skin of a deviated wellbore, allowing for anisotropy, and explain why deviation hides damage
list the interpretation errors that appear if the geometry of penetration is ignored

What happens at a partially penetrating well

The radial inflow formulas are derived for an ideal well. In an ideal well, a vertical wellbore is open to the reservoir over its whole thickness, and the fluid approaches it horizontally (Steady-state radial flow). In practice the well often opens only part of the reservoir:

deliberately — the upper part of the reservoir is perforated to keep the perforations away from the bottom water (the water under the oil), or the lower part to keep them away from the gas cap, the gas layer above the oil (Water and gas coning)
because of drilling and testing conditions — the well stops at the top of a thick reservoir, or a drill stem test (a test of the formation made through the drill pipe) is run over a short interval
accidentally — some of the perforation tunnels have not cleaned up and do not work, so the inflow goes through a smaller interval than the one that was perforated

At the wall of the well the flow is radial, but it passes only through the open interval $h_w$ . The inflow area is $h/h_w$ times smaller than for a well that opens the whole thickness $h$ , and the velocities are larger by the same factor. Fluid from the parts of the reservoir above and below the interval has to move vertically as well. So the resistance depends on the vertical permeability $k_v$ . In layered reservoirs $k_v$ is usually several times lower than the horizontal permeability $k_h$ , and sometimes as much as ten times lower (permeability anisotropy). At a distance from the wellbore of the order of $h\sqrt{k_h/k_v}$ the streamlines straighten out. Farther on, the flow is again radial over the whole thickness.

SchematicA partially penetrating well (schematic, not to scale): the upper part of the reservoir is perforated. The streamlines converge on the interval. The dashed outlines show where the pressure disturbance has reached after the well starts, in three flow regimes, one after another.

In Russian-language practice, a well's imperfection is divided into two kinds. The first is imperfection in the degree of penetration: only part of the thickness is open, which is the subject of this lesson. The second is imperfection in the character of penetration: the inflow goes through perforation tunnels and not through an open wall (Effect of completion on inflow, Shaped-charge perforating). The extra resistances are expressed by coefficients of additional flow resistance. In meaning these are the same pseudo-skins (pseudo-skin).

Pseudo-skin of partial penetration

The zone where the flow converges is small compared with the drainage area. The flow in it becomes steady-state long before the pressure disturbance reaches the boundaries of the area. So, like the influence of a damaged zone, its influence is reduced to a single number: the pseudo-skin $S_{pp}$ . This is the additional pressure drop in “units” of radial flow, $qB\mu/(2\pi k_h h)$ (Skin factor). It is calculated numerically for a homogeneous anisotropic reservoir, and the results are put into an approximate formula:

Pseudo-skin of partial penetration

\begin{gathered} S_{pp} = \left(\frac{h}{h_w} - 1\right)\left[\ln h_D - G\!\left(\frac{h_w}{h}\right)\right],\qquad h_D = \frac{h_s}{r_w}\sqrt{\frac{k_h}{k_v}}\\ G(x) = 2.948 - 7.363\,x + 11.45\,x^2 - 4.675\,x^3\end{gathered}

where:

$S_{pp}$ pseudo-skin of partial penetration, referred to the whole thickness $h$ , —
$h,\ h_w$ net thickness of the reservoir and the working open interval, m
$h_w/h$ penetration ratio, —
$h_D$ dimensionless thickness of the symmetry element (the part of the reservoir that repeats by symmetry), with allowance for anisotropy, —
$h_s$ thickness of the symmetry element: $h$ for an interval at the top or bottom, $h/2$ for an interval in the middle of the reservoir, m
$r_w$ well radius, m
$k_h,\ k_v$ horizontal and vertical permeability; $k_h$ is the one written as $k$ in the radial inflow formulas, mD

Units: dimensionless; lengths and permeabilities in identical units; $\ln$ is the natural logarithm

Holds when

A homogeneous reservoir with constant $k_h$ and $k_v$ between an impermeable top and bottom; a vertical well, one continuous interval.
The formula approximates numerical solutions. For $h_D$ from 100 to tens of thousands and $h_w/h \ge 0.1$ , it differs from the numerical solution with uniform pressure along the interval by no more than 0.6 (a few per cent), usually on the low side. For smaller $h_D$ (a thin reservoir) it underestimates the pseudo-skin: at $h_D = 20$ and a small penetration ratio, by a third. Then a numerical calculation is needed.
For an interval in an in-between position, the value lies between the values for an interval at the top and an interval in the middle. For greater accuracy, use a numerical model.

The factor $(h/h_w - 1)$ shows that the pseudo-skin grows sharply when the penetration ratio is small, and disappears at full penetration. Anisotropy enters through $\ln h_D$ : cutting $k_v$ ten times adds $1.15\,(h/h_w - 1)$ . An interval in the middle of the reservoir gives a pseudo-skin smaller by $(h/h_w - 1)\ln 2$ than the same interval at the top, because the flow approaches it from above and from below.

Worked examplePseudo-skin and productivity of a well that has opened the top of the reservoir

Given (assumed data)

Reservoir: oil-saturated thickness $h$ = 32 m above bottom water; $k_v/k_h$ = 0.15 from core
Penetration: the upper $h_w$ = 10 m are perforated
Well: $r_w$ = 0.108 m; drainage radius $r_e$ = 350 m

Solution

Penetration ratio: $h_w/h = 10/32 = 0.3125$ . Then $G(0.3125) = 1.6226$ .
The interval is at the top, so $h_s = h$ : $h_D = (32/0.108)\cdot\sqrt{1/0.15} = 765.0$ , $\ln h_D = 6.640$ .
$S_{pp} = (32/10 - 1)\cdot(6.640 - 1.6226)$ $= 2.2\cdot5.017 = 11.04$ .
For comparison: in an isotropic reservoir ( $h_D = 296.3$ ), $S_{pp} = 8.95$ . The same interval in the middle of the reservoir ( $h_s$ = 16 m) gives $9.51$ .
Productivity in the pseudo-steady-state regime relative to full penetration: $X = \ln(350/0.108) - 0.75 = 7.334$ , $J/J_{\text{full}} = X/(X + S_{pp}) = 0.399$ .

Answer: The pseudo-skin of partial penetration is about 11.04. The well delivers about 40% of the flow rate of the same well when it is open over the whole thickness.

What it means. This is the price of keeping the interval away from the water. Two thirds of the thickness are not perforated, and the productivity has fallen by more than half. Of the pseudo-skin 11.04, about 2.1 comes from anisotropy. The figure below shows how much a longer interval would restore. We decide together with a coning calculation (water or gas drawn towards the well in a cone; Water and gas coning). The data in this example are assumed for illustration.

CalculatedCalculated productivity of the well in the example (thickness 32 m,

r_w

= 0.108 m,

r_e

= 350 m, interval at the top, no damage), relative to full penetration, against the penetration ratio, for three ratios

k_v/k_h

. The dot is the example.

Damage to the near-wellbore zone in partial penetration

The damaged zone surrounds only the open interval, and all the flow passes through it. The velocity in it is $h/h_w$ times higher than it would be at full penetration, and the pressure drop in it is higher by the same factor. Suppose we describe the damage of the open interval by a skin $S_d$ from the Hawkins formula (Skin factor). Then it enters the skin counted over the whole thickness, multiplied by $h/h_w$ :

Measured skin of a partially penetrating well

S = S_d\,\frac{h}{h_w} + S_{pp},\qquad S_d = (S - S_{pp})\,\frac{h_w}{h}

where:

$S$ skin from the well test (late radial flow), counted over the whole thickness, —
$S_d$ damage skin of the open interval, from the Hawkins formula, —
$S_{pp}$ pseudo-skin of partial penetration, —
$h,\ h_w$ thickness of the reservoir and the working (open and flowing) interval, m

Units: dimensionless

Holds when

The damaged zone is thin: its radius is small compared with the interval and with the zone where the flow converges.
The flow through the damaged zone is radial towards the interval. In a gas well an inertial skin is added to $S$ , and partial penetration also magnifies it (Inflow at high rates: non-Darcy flow).

The second form tells us how to interpret a test: subtract the pseudo-skin from the measured skin, and multiply the result by the penetration ratio. An acid treatment can reduce only $S_d$ . The pseudo-skin changes only if the well is opened differently.

Worked exampleWhat an acid treatment will give a partially penetrating well

Given (assumed data)

Well: from the previous example: $h$ = 32 m, the upper 10 m opened, $S_{pp}$ = 11.04, $X = \ln(r_e/r_w) - 0.75 = 7.334$
Pressure build-up curve: late radial section: flow capacity $k_h h$ = 1440 mD·m, skin $S$ = 14.6

Solution

Damage accounts for this part of the measured skin: $S - S_{pp} = 14.6 - 11.04 = 3.56$ .
Damage skin of the open interval: $S_d = 3.56\cdot10/32 = 1.11$ .
If the treatment removes the damage completely, the productivity grows by a factor of $(X + 14.6)/(X + S_{pp})$ $= 1.19$ .
If we took the whole skin of 14.6 to be damage, we would expect a growth by a factor of $(X + 14.6)/X = 3.0$ .

Answer: The damage skin is about 1.11. A treatment can raise the productivity by no more than 19%, not by a factor of 3.0.

What it means. A skin of 14.6 looks like severe damage, but three quarters of it is the geometry of penetration. A treatment is justified only if a gain of 19% pays for its cost. Extending the interval would give noticeably more, if the distance to the water allows it. The data in this example are assumed for illustration.

Flow regimes on the pressure record

After a partially penetrating well starts, the pressure disturbance spreads out from the interval, and the shape of the flow changes three times. On the diagnostic plot (Diagnosis with the pressure derivative) we recognise each regime by how the pressure derivative $\Delta p'$ behaves. The pressure derivative is the slope of the pressure curve against the logarithm of time.

Early radial flow to the interval. While the disturbance has not gone beyond the interval vertically, the well behaves as if the reservoir had thickness $h_w$ . The derivative levels off on a plateau (a flat section). The plateau gives $k_h h_w$ , not $k_h h$ .
Spherical or hemispherical flow. The disturbance has gone beyond the interval vertically, but it has not yet reached the top and the bottom. For an interval in the middle of the reservoir the flow is spherical. For an interval at the top or bottom it is hemispherical, because the boundary acts as a “mirror”. The pressure drop tends to a constant value, and the derivative falls with a slope of −½.
Late radial flow. The disturbance has reached the top and the bottom, and from then on the flow is radial over the whole thickness. The derivative levels off on a second, lower plateau. This plateau gives $k_h h$ , and the position of the pressure curve gives the full skin $S = S_d\,h/h_w + S_{pp}$ .

CalculatedCalculated diagnostic plot of the well in the example below (the upper 4 m of a reservoir 36 m thick are open; no wellbore storage, which is the effect of the well's own volume, and no damage). It shows the pressure drop and its derivative for

k_v

= 3.5 mD, and the derivative for

k_v

= 35 mD. The dashed horizontal lines are the plateaus of early and late radial flow.

The plateau levels are linked to permeability in the same way as the ordinary plateau of radial flow ( $kh = 0.933\,qB\mu/m_e$ in metric units). The only difference is that the early plateau has $h_w$ in place of $h$ . The ratio of the plateau levels is $h/h_w$ . If both plateaus are visible, we use this ratio to check the working interval.

Pressure derivative in spherical and hemispherical flow

\Delta p' = \frac{c\,qB\mu\sqrt{\phi\mu c_t}}{8\pi^{3/2}\,k_s^{3/2}\sqrt{t}},\qquad k_s = \left(k_h^2\,k_v\right)^{1/3}

where:

$\Delta p'$ logarithmic derivative of the pressure drop, $t\,\dd\Delta p/\dd t$ , Pa
$c$ 1 for spherical flow (an interval in the middle of the reservoir), 2 for hemispherical flow (an interval at the top or bottom), —
$k_s$ spherical permeability; in the Hawkins formula (skin of a zone of altered permeability) the same letter means the permeability of the altered zone, m²
$q,\ B,\ \mu$ flow rate at standard conditions, formation volume factor, viscosity, m³/s, m³/m³, Pa·s
$\phi,\ c_t$ porosity and total compressibility, —, Pa⁻¹
$t$ time since the start of production (since shut-in, for a pressure build-up), s

Units: SI (any consistent system of units)

Holds when

A homogeneous anisotropic reservoir; the interval is short compared with the thickness.
The time is later than the end of early radial flow and earlier than the start of late radial flow. Wellbore storage has ended.

On a log–log plot (both axes logarithmic) the derivative is a straight line with a slope of −½, while the pressure drop itself tends to a constant value. Spherical flow does not “accumulate” resistance as the radius grows. One point on this section gives the spherical permeability $k_s$ . Then $k_h$ from the late plateau gives $k_v = k_s^3/k_h^2$ .

In other unit systems (2)

Metric oilfield units

\Delta p'\,[\text{MPa}] = 4.417\,c\,\frac{q\,B\,\mu\sqrt{\phi\,\mu\,c_t}}{k_s^{3/2}\sqrt{t}}

$q$ in m³/d, $\mu$ in mPa·s, $c_t$ in MPa⁻¹, $k_s$ in mD, $t$ in h.

$4.417$ is the unit conversion divided by $8\pi^{3/2}$ .

Oilfield units

\Delta p'\,[\text{psi}] = 1227\,c\,\frac{q\,B\,\mu\sqrt{\phi\,\mu\,c_t}}{k_s^{3/2}\sqrt{t}}

$q$ in STB/d, $\mu$ in cP, $c_t$ in psi⁻¹, $k_s$ in mD, $t$ in h.

The pressure drop itself is $\Delta p = \Delta p_\infty - 2\Delta p'$ . So the slope of the line of $\Delta p$ against $1/\sqrt t$ is twice this constant, that is, $2453\,c$ .

Duration of the regimes (estimate)

t_{\text{early}} \approx a_1\,\frac{\phi\mu c_t\,h_w^2}{k_v},\qquad t_{\text{late}} \approx a_2\,\frac{\phi\mu c_t\,h^2}{k_v}

where:

$t_{\text{early}}$ end of early radial flow, h
$t_{\text{late}}$ start of late radial flow, h
$a_1,\ a_2$ interval at the top or bottom: 2.2 and 100; interval in the middle of the reservoir: 0.55 and 25, —
$\phi,\ \mu,\ c_t,\ k_v$ porosity, viscosity, total compressibility, vertical permeability, —, mPa·s, MPa⁻¹, mD

Units: metric oilfield units: $t$ in h, $h$ and $h_w$ in m

Holds when

These are estimates from a model of uniform inflow along the interval, for a penetration ratio of up to 0.3. We take the boundaries of the regimes where the derivative departs from the plateau by 5%. If we allow 10% instead, the factor $a_1$ is four times larger and $a_2$ is about 20% smaller.
Wellbore storage is not taken into account: it often hides early radial flow completely.

Both durations are inversely proportional to the vertical permeability. The length of the interval sets the scale of the early regime, and the thickness of the reservoir sets the scale of the late one. So in thick reservoirs with low $k_v$ , late radial flow may not be reached during the whole test. The factors 2.2 and 100 correspond to $k_vt/(\phi\mu c_t h_w^2) \approx 0.008$ and $k_vt/(\phi\mu c_t h^2) \approx 0.37$ in consistent units.

Worked exampleVertical permeability from a section of hemispherical flow

Given (assumed data)

Exploration well: oil reservoir 36 m thick, the upper 4 m perforated; $r_w$ = 0.108 m
Test conditions: $q$ = 90 m³/d, $B$ = 1.2, $\mu$ = 1.3 mPa·s; $\phi$ = 0.17, $c_t = 1.5\cdot10^{-3}$ MPa⁻¹
Derivative from the pressure record: 0.269 MPa at 1.5 h and 0.152 MPa at 5 h; plateau 0.104 MPa after 20 h. Wellbore storage ended in the first minutes

Solution

Slope between 1.5 and 5 h: $\lg(0.152/0.269)/\lg(5/1.5)$ $= -0.47$ , which is close to −½. The interval is at the top, so the flow is hemispherical ( $c = 2$ ).
Late plateau: $k_h h = 0.933\cdot90\cdot1.2\cdot1.3/0.104$ $= 1260$ mD·m, $k_h = 35.0$ mD.
Spherical permeability from the 5 h point: $\sqrt{\phi\mu c_t} = 0.01821$ , and $qB\mu\sqrt{\phi\mu c_t} = 140.4\cdot0.01821 = 2.557$ . Then $k_s^{3/2} = 4.417\cdot2\cdot2.557/(0.152\cdot\sqrt5)$ $= 66.5$ , so $k_s = 16.4$ mD.
Vertical permeability: $k_v = k_s^3/k_h^2 = 16.4^3/35.0^2 = 3.60$ mD, $k_v/k_h = 0.10$ .
Check: late radial flow should begin at about $100\cdot0.17\cdot1.3\cdot1.5\cdot10^{-3}\cdot36^2/3.60$ $= 11.9$ h. This agrees with the record. Early radial flow ended after 12 s, and wellbore storage hides it.
Pseudo-skin of the interval for the anisotropy we found: $h_D = (36/0.108)\sqrt{35.0/3.60} = 1039$ , $\ln h_D = 6.946$ , $G(4/36) = 2.265$ . Then $S_{pp} = 8\cdot(6.946 - 2.265) = 37.45$ .

Answer:

k_h \approx 35.0

mD,

k_s \approx 16.4

mD,

k_v \approx 3.60

mD. The pseudo-skin of penetration is about 37.

What it means. A short interval in a thick reservoir gives an enormous pseudo-skin. A skin of about 40 from such a test would be almost entirely geometry, not damage. The section with a slope of −½ spans about one logarithmic cycle (a tenfold range of time), so the $k_v$ estimate has limited accuracy. From the 1.5 h point, where the slope has not yet reached −½, we would get $k_v \approx 3.8$ mD. So take the estimate from the middle of the section, and check it against the time when the late plateau starts. The data in this example are assumed for illustration.

How ignoring the geometry of penetration distorts interpretation

Interpretation errors and how to avoid them

Situation: Only the early plateau is visible
What results: you get $k_h h_w$ ; dividing by $h$ makes $k_h$ too low by a factor of $h/h_w$
What to do: divide by the working interval, extend the test

Situation: A straight line is drawn through the −½ section
What results: a random $kh$ and skin, because the slope there is not constant
What to do: first diagnose the flow regime from the derivative

Situation: The late skin is taken for damage
What results: damage overestimated, a useless treatment
What to do: subtract $S_{pp}$ , multiply by $h_w/h$

Situation: The perforated interval is used instead of the working one
What results: $S_{pp}$ underestimated, damage overestimated
What to do: check the working interval by production logging

Situation: A deviated well is treated as vertical
What results: damage underestimated, because $S_\theta < 0$ hides it
What to do: subtract $S_\theta$ , divide by $\cos\theta'$

Situation: A gas well, a single rate
What results: the term $Dq$ (the inertial part of the skin) enters the skin, magnified by partial penetration
What to do: test at several rates (Inflow at high rates: non-Darcy flow)

Situation	What results	What to do
Only the early plateau is visible	you get $k_h h_w$ ; dividing by $h$ makes $k_h$ too low by a factor of $h/h_w$	divide by the working interval, extend the test
A straight line is drawn through the −½ section	a random $kh$ and skin, because the slope there is not constant	first diagnose the flow regime from the derivative
The late skin is taken for damage	damage overestimated, a useless treatment	subtract $S_{pp}$ , multiply by $h_w/h$
The perforated interval is used instead of the working one	$S_{pp}$ underestimated, damage overestimated	check the working interval by production logging
A deviated well is treated as vertical	damage underestimated, because $S_\theta < 0$ hides it	subtract $S_\theta$ , divide by $\cos\theta'$
A gas well, a single rate	the term $Dq$ (the inertial part of the skin) enters the skin, magnified by partial penetration	test at several rates (Inflow at high rates: non-Darcy flow)

Deviated wellbore

A directionally drilled well crosses the reservoir at an angle $\theta$ to the normal (the perpendicular) to the bedding. For a horizontal reservoir this angle is the inclination (Directional wells: applications and profile). The length of the wellbore in the reservoir is $h/\cos\theta$ , which is greater than the thickness. So the inflow area is larger, the velocities at the wall are lower, and the inflow is easier: the slant pseudo-skin is negative. We calculate it from a correlation (a formula fitted to results) built on numerical solutions:

Pseudo-skin of a deviated wellbore

\begin{gathered} S_\theta = -\left(\frac{\theta'}{41}\right)^{2.06} - \left(\frac{\theta'}{56}\right)^{1.865}\lg\frac{h_D}{100}\\ \theta' = \arctan\left(\sqrt{\frac{k_v}{k_h}}\,\tan\theta\right),\qquad h_D = \frac{h}{r_w}\sqrt{\frac{k_h}{k_v}}\end{gathered}

where:

$S_\theta$ slant pseudo-skin, counted over the thickness $h$ , —
$\theta$ angle between the wellbore and the normal to the reservoir, degrees
$\theta'$ effective angle in a reservoir “stretched” vertically until it is isotropic, degrees
$h_D$ dimensionless thickness with allowance for anisotropy, —
$h,\ r_w,\ k_h,\ k_v$ reservoir thickness, well radius, horizontal and vertical permeability, m, m, mD, mD

Units: dimensionless; $\lg$ is the decimal logarithm, angles in degrees

Holds when

The wellbore opens and perforates the whole thickness of a homogeneous reservoir. The limits are $\theta' \le 75^\circ$ and $h_D > 40$ .
In an isotropic reservoir $\theta' = \theta$ . Allowing for anisotropy through the effective angle is a common approximation.
The value refers to late, pseudo-radial flow. Other models describe horizontal wells (Horizontal wells).

The size of the pseudo-skin grows with the angle and with the ratio of thickness to well radius. Anisotropy “straightens” the wellbore: at $k_v/k_h = 0.2$ an angle of 50° acts like 28°, and the gain from deviation falls almost three times.

CalculatedCalculated slant pseudo-skin for a reservoir 18 m thick and a well of radius 0.108 m, against the angle, for three ratios

k_v/k_h

. The dots mark the angle of the example (50°).

The damaged zone surrounds the whole wellbore in the reservoir. The flow through it, per unit length, is $1/\cos\theta$ times smaller than in a vertical well. So in an isotropic reservoir damage enters the skin counted over the thickness $h$ , multiplied by $\cos\theta$ . A deviated well “dilutes” damage. In an anisotropic reservoir the flow to the wall of a deviated wellbore partly crosses the bedding, where the permeability is lower. The same damaged zone then costs more: the factor is $\cos\theta'$ , with the same effective angle as in the pseudo-skin formula:

Measured skin of a deviated well

S = S_d\cos\theta' + S_\theta,\qquad S_d = \frac{S - S_\theta}{\cos\theta'}

where:

$S$ skin from the well test, counted over the thickness $h$ , —
$S_d$ damage skin from the Hawkins formula for the zone around the wellbore, —
$S_\theta$ slant pseudo-skin, —
$\theta'$ effective angle from the slant pseudo-skin formula; in an isotropic reservoir $\theta' = \theta$ , degrees

Units: dimensionless

Holds when

The damaged zone is thin and runs along the whole wellbore in the reservoir. The wellbore is perforated along its whole length in the reservoir.
$\cos\theta'$ equals $\cos\theta$ (the ratio of the thickness to the length of the wellbore in the reservoir) multiplied by the ratio of $k_h$ to the effective permeability across the wellbore. In an anisotropic reservoir that permeability is smaller than $k_h$ , so $\cos\theta' > \cos\theta$ .

A negative $S_\theta$ and a factor $\cos\theta'$ smaller than 1 together make the measured skin of a deviated well smaller than its damage skin. A skin close to zero in such a well does not mean that there is no damage.

Worked exampleIs a deviated well with a skin close to zero damaged?

Given (assumed data)

Well: crosses a horizontal reservoir 18 m thick at an inclination of 50°, perforated along its whole length in the reservoir; $r_w$ = 0.108 m
Test: skin from the pressure build-up curve $S$ = +0.3
Anisotropy: not measured; we consider $k_v/k_h$ = 1 and 0.2

Solution

Isotropic reservoir: $h_D = 18/0.108 = 166.7$ . Then $(50/41)^{2.06} = 1.505$ , $(50/56)^{1.865} = 0.8095$ and $\lg(166.7/100) = 0.2218$ , so $S_\theta = -1.505 - 0.8095\cdot0.2218 = -1.685$ .
Damage skin: $S_d = (0.3 - (-1.685))/\cos50^\circ$ $= 1.985/0.6428 = 3.09$ .
At $k_v/k_h = 0.2$ : $\theta' = \arctan(\sqrt{0.2}\,\tan50^\circ) = 28.1^\circ$ , $h_D = 373$ , $S_\theta = -0.615$ .
Damage skin: $\cos\theta' = 0.8825$ , $S_d = (0.3 - (-0.615))/0.8825 = 1.04$ .

Answer: The damage skin is from 1.04 to 3.09, depending on the anisotropy. The well is damaged, although the measured skin is close to zero.

What it means. Deviation has “hidden” the damage: in a vertical well the same $S_d$ would give a skin of 1.04–3.09. To decide whether a treatment is needed, we must know the anisotropy, because the result changes almost three times with it. The data in this example are assumed for illustration.

On a pressure record a deviated well also passes through several regimes. First the flow is radial around the deviated wellbore, along its whole length in the reservoir. In an isotropic reservoir the early derivative plateau is about $\cos\theta$ times the late one. When the disturbance reaches the top and the bottom, pseudo-radial flow through the whole thickness begins. It gives $k_h h$ and the skin, which includes $S_\theta$ . Suppose a deviated well also opens only part of the thickness. Then the pseudo-skins of partial penetration and of deviation affect each other. Simply adding them gives only a rough estimate, so for the calculation we use tables of numerical solutions or a numerical model.

How to break down the measured skin

From the test skin to the damage skin

1
Remove the effect of the flow rate
In gas wells and high-rate wells the skin depends on the flow rate. Tests at several rates give the skin $S$ without the inertial component (Inflow at high rates: non-Darcy flow).
2
Refine the geometry
Get the working interval (from production logging), its position in the reservoir, the angle of the wellbore in the reservoir, the thickness and the radius of the well.
3
Estimate the anisotropy
Get $k_v/k_h$ from core, from spherical flow on the pressure record, or from analogues (similar reservoirs). Take a plausible range.
4
Calculate the pseudo-skins
Calculate $S_{pp}$ of partial penetration, $S_\theta$ of deviation and the pseudo-skin of perforation (Effect of completion on inflow), at the extreme values of anisotropy.
5
Find the damage skin
For a partially penetrating vertical well, $S_d = (S - S_{pp})\,h_w/h$ . For a deviated well, $S_d = (S - S_\theta)/\cos\theta'$ .
6
Compare the options
Work out how much each option would give: a treatment (removing $S_d$ ), additional perforation of the interval (a smaller $S_{pp}$ ) or hydraulic fracturing. Then see which of them is acceptable given the risk of water and gas breakthrough.

Terms and symbols

Partial penetration: A well that is open to the reservoir over only part of its thickness. It causes the flow to converge and gives a positive pseudo-skin.
Penetration ratio $h_w/h$: The ratio of the working interval to the thickness of the reservoir.
Pseudo-skin of partial penetration $S_{pp}$: The additional resistance to the convergence of the flow on the interval. It grows when the penetration ratio is small and the anisotropy is strong.
Symmetry element $h_s$: The part of the thickness to which the problem is reduced by symmetry. It is the whole thickness for an interval at the top or bottom, and half of it for an interval in the middle.
Spherical and hemispherical flow: Flow to a short interval after the disturbance has gone beyond the interval vertically, but before it has reached the top and bottom. The pressure derivative falls with a slope of −½.
Spherical permeability $k_s$: The permeability $(k_h^2k_v)^{1/3}$ , which a section of spherical flow determines.
Slant pseudo-skin $S_\theta$: The negative component of the skin of a deviated wellbore that crosses the reservoir at an angle to the normal.

Common mistakes and misconceptions

Wrong

“A skin of 14.6 from the build-up curve is severe damage; an acid treatment is needed.”

In fact

In the well of the lesson's example, which is open over a third of the thickness, 11 units out of the 14.6 are the pseudo-skin of partial penetration. A treatment can give only about 19% of growth.

Wrong

“Since the vertical permeability is low, partial penetration does not matter — the fluid goes horizontally anyway.”

In fact

The opposite is true. The lower $k_v$ is, the harder it is for fluid from the unopened part of the reservoir to reach the interval, and the larger the pseudo-skin.

Wrong

“The flow capacity from the first derivative plateau is the $kh$ of the whole reservoir.”

In fact

With partial penetration, the first plateau belongs to radial flow to the interval and gives $k_h h_w$ . The late plateau gives $k_h h$ , and it is lower by a factor of $h/h_w$ .

Wrong

“A deviated well with a skin close to zero is not damaged.”

In fact

The negative slant pseudo-skin and the factor $\cos\theta'$ hide the damage. In the lesson's example, a skin of +0.3 means a damage skin of 1.0 to 3.1.

In depth: advanced detail

The exact solution for uniform inflow and the accuracy of the formula

Suppose the inflow along the interval is uniform. Then we can solve the problem exactly by splitting the source into vertical harmonics $\cos(n\pi z/h)$ (simple wave-like components). The zero harmonic is ordinary radial flow to the whole thickness. Each following harmonic adds a term to the pressure that dies away with distance like the Macdonald function $K_0(n\pi r/h')$ , where $h' = h\sqrt{k_h/k_v}$ is the thickness “stretched” until the reservoir is isotropic. So the zone where the flow converges extends from the wellbore by about $h'$ . The pressure averaged over the interval gives the pseudo-skin:

Pseudo-skin for uniform inflow along the interval

S_{pp} = \frac{2}{\pi^2 x^2}\sum_{n=1}^{\infty}\frac{\left[\sin(n\pi z_2) - \sin(n\pi z_1)\right]^2}{n^2}\,K_0\!\left(\frac{n\pi r_w}{h}\sqrt{\frac{k_v}{k_h}}\right)

where:

$z_1,\ z_2$ boundaries of the interval as fractions of the thickness, counted from the top, —
$x = z_2 - z_1$ penetration ratio, —
$K_0$ modified Bessel function of the second kind of order zero (the Macdonald function), —

Units: dimensionless

Holds when

A homogeneous anisotropic reservoir. A line source with uniform inflow along the interval. The pressure is averaged over the interval.

At full penetration ( $z_1 = 0$ , $z_2 = 1$ ) all the sines are zero, and the pseudo-skin vanishes. A real well is closer to the case of equal pressure along the interval, where the inflow to the ends of the interval is greater than to its middle. The pseudo-skin of that case generally lies between the exact value for uniform inflow and the approximate formula of the main section.

Worked exampleThe approximation against the exact solution

Given (assumed data)

The well of the first example: $h$ = 32 m, the upper 10 m opened, $r_w$ = 0.108 m, $k_v/k_h$ = 0.15
The well of the third example: $h$ = 36 m, the upper 4 m opened, $k_v/k_h$ = 0.1

Solution

First well: the approximation gives $S_{pp} = 11.04$ , and the series for uniform inflow gives $11.34$ . The difference is 0.30.
Third well: $37.56$ and $38.62$ . The difference is 1.05.

Answer: The discrepancy is from 0.30 to 1.05, that is, about 3% of the pseudo-skin. The solution with equal pressure along the interval lies between them.

What it means. For a productivity calculation this accuracy is more than enough. An uncertainty of a factor of two in $k_v/k_h$ changes the pseudo-skin by $0.35\,(h/h_w - 1)$ , which for the third well is almost three units. The data in this example are assumed for illustration.

Where the plateaus and the −½ slope come from

Take the same uniform source in a reservoir that has no boundaries in plan. The logarithmic derivative of the dimensionless pressure then takes a simple form: the product of the radial-flow factor and a vertical factor $F$ :

Pressure derivative of a partially penetrating well

p_D' = \tfrac12\,e^{-1/(4t_D)}\,F(t_D),\qquad F(t_D) = 1 + \sum_{n=1}^{\infty}\frac{a_n^2}{2}\,e^{-n^2\pi^2 t_D/h_D^2}

where:

$p_D'$ dimensionless derivative, $p_D = 2\pi k_h h\,\Delta p/(qB\mu)$ (over the whole thickness, without skin), —
$t_D$ dimensionless time $k_h t/(\phi\mu c_t r_w^2)$ , —
$h_D$ $(h/r_w)\sqrt{k_h/k_v}$ , —
$a_n$ $2[\sin(n\pi z_2) - \sin(n\pi z_1)]/(n\pi x)$ — the amplitudes of the harmonics of the source, —

Units: dimensionless

Holds when

A line source with uniform inflow along the interval; a reservoir unbounded in plan; no wellbore storage.

At the start all the harmonics are “alive”, and $F = h/h_w$ , so the early plateau is $\tfrac12 h/h_w$ . At the end they have died away, $F \to 1$ , and the late plateau is $\tfrac12$ . In between, the sum over many harmonics behaves like $h_D/\sqrt{\pi t_D}$ for an interval at the top (half as large for an interval in the middle). This is the −½ slope. Putting this limit into quantities with units gives the spherical derivative formula of the main section.

Connections to other topics

Skin factor — skin and its components; the Hawkins formula for a damaged zone.
Diagnosis with the pressure derivative — the diagnostic plot, plateaus and the −½ slope.
Flow regimes — flow geometries: radial, spherical, linear.
Water and gas coning — why part of the thickness is opened and how to choose the interval.
Inflow at high rates: non-Darcy flow — inertial skin and its growth when only part of the thickness is opened.
Effect of completion on inflow — pseudo-skin of perforation and of a gravel pack.
Directional wells: applications and profile — inclination and the profile of a directionally drilled well.
Horizontal wells — inflow to horizontal wellbores, the limiting case of deviation.
Acid treatments: principle, economics and candidate selection — how to choose wells for treatment by their damage skin.

Check yourself

1
A reservoir 40 m thick, isotropic, $r_w$ = 0.1 m. A 10 m interval is perforated either at the top or in the middle of the reservoir. How do the pseudo-skins differ?
1. A.They are the same: only the penetration ratio matters
2. B.At the top it is larger by $(h/h_w - 1)\ln 2 \approx 2.1$
3. C.At the top it is twice as large
4. D.In the middle it is larger by 2.1
Show the answer
Correct answer: B. At the top it is larger by $(h/h_w - 1)\ln 2 \approx 2.1$
For an interval in the middle, the symmetry element is half the thickness, so $h_D$ is half as large. Then $\ln h_D$ is smaller by $\ln 2$ , and the pseudo-skin is smaller by $3\ln 2 \approx 2.1$ (12.7 at the top against 10.6 in the middle). The flow approaches an interval in the middle from above and from below.
2
The skin from the build-up curve is 12, the calculated pseudo-skin of partial penetration is 9, a quarter of the thickness is open, and $\ln(r_e/r_w) - 3/4 = 7$ . What is the damage skin, and what would its complete removal give?
Show the answer
$S_d = (12 - 9)\cdot0.25 = 0.75$ . The productivity grows by a factor of $(7 + 12)/(7 + 9) \approx 1.19$ , that is, by 19%. It does not grow by a factor of 2.7, which is what we would expect if we took the whole skin to be damage: $(7 + 12)/7 \approx 2.7$ .
3
On the diagnostic plot of a partially penetrating well the early derivative plateau is 0.6 MPa and the late one is 0.15 MPa. What part of the thickness is working?
1. A.A half
2. B.It cannot be found
3. C.A sixteenth
4. D.A quarter
Show the answer
Correct answer: D. A quarter
The plateau levels are inversely proportional to the thicknesses through which the radial flow passes: $h/h_w = 0.6/0.15 = 4$ . This checks the working interval, as long as wellbore storage has not distorted the early plateau.
4
Why does a lower vertical permeability increase the pseudo-skin of partial penetration, but reduce the gain from a deviated wellbore?
Show the answer
If we stretch the reservoir vertically by a factor of $\sqrt{k_h/k_v}$ , it becomes isotropic. With partial penetration, the effective thickness grows, and so does the path that the fluid takes vertically to the interval. So the pseudo-skin increases. A deviated wellbore in the stretched reservoir becomes steeper: the effective angle $\theta'$ is smaller than $\theta$ , and the wellbore works almost like a vertical one. So the negative pseudo-skin gets smaller in size.
5
A well deviated at 60° to the normal to an isotropic reservoir has $h/r_w$ = 200. The test gave a skin of −0.5. What can we say about the near-wellbore zone?
1. A.It is most likely damaged: with $S_\theta \approx -2.5$ the damage skin is about 4
2. B.The test is wrong: the slant skin is never lower than −1
3. C.There is no damage
4. D.It is improved: the skin is negative
Show the answer
Correct answer: A. It is most likely damaged: with $S_\theta \approx -2.5$ the damage skin is about 4
$S_\theta = -(60/41)^{2.06} - (60/56)^{1.865}\lg 2$ $\approx -2.5$ , and $S_d = (-0.5 + 2.5)/\cos60^\circ \approx 4$ . The geometry of the wellbore creates the negative skin here, and the damage is substantial.
6
A reservoir is 25 m thick, and the upper 5 m are perforated. $k_v$ = 2 mD, $\phi$ = 0.2, $\mu$ = 1.0 mPa·s, $c_t = 1.4\cdot10^{-3}$ MPa⁻¹. The test lasted 6 h. Can we find the $kh$ of the reservoir and the full skin from the end of the test?
Show the answer
Late radial flow begins after about $100\cdot0.2\cdot1.0\cdot1.4\cdot10^{-3}\cdot25^2/2 \approx 8.8$ h, which is later than the end of the test. The end of the record falls in hemispherical flow, or in the transition to the late regime. A straight line on a semilog plot has no meaning there. We must extend the test, or interpret the record with a model that includes partial penetration, using the known $h_w$ and an estimated $k_v$ .

Updated: 19 October 2026