Effect of completion on inflow

Quick refresher

The reference for comparison is an undamaged open hole of the same radius ( $S = 0$ ). Completion skin can be positive or negative. Dense, deep tunnels work like a wellbore of a larger radius.
Perforation skin has three parts. The first is $s_H$ , from the flow converging on the “star” of tunnels in the horizontal plane (usually negative). The second is $s_V$ , from the flow converging in the vertical direction between the levels of the shots. The third is $s_{wb}$ , from the flow going around the cased wellbore.
A higher shot density mainly reduces the vertical part. Longer tunnels reduce both the plane part and the vertical part. A phasing of 0° increases the flow around the wellbore. The tunnel diameter matters little. If the vertical permeability is low, the perforations must be dense.
The crushed zone around the tunnels can easily add +1 or more. Underbalanced perforating and cleaning the tunnels cut its contribution several times.
If the tunnels end inside the damaged zone, the perforation skin is multiplied by $k/k_s$ , and the result is usually worse than for an open hole. If the tunnels pass right through the zone, the damage is almost bypassed.
In a gravel pack inside the casing, all the flow goes through narrow tunnels filled with gravel. This adds a skin $S_g$ and an inertial skin $D_gq$ . The inertial skin grows with the rate and is especially large in gas wells. In an open hole the gravel pack hardly hinders the inflow.

Key relationships

s_p = s_H + s_V + s_{wb}

perforation skin by the component method

s_c = \dfrac{h}{l_p}\left(\dfrac{k}{k_c} - \dfrac{k}{k_s}\right)\ln\dfrac{r_c}{r_p}

skin of the crushed zone around the tunnels

s_{dp} = \left(\dfrac{k}{k_s} - 1\right)\ln\dfrac{r_s}{r_w} + \dfrac{k}{k_s}\,s_p

tunnels end inside the damaged zone

S_g = \dfrac{8\,k\,h\,L}{k_g\,n\,d^2}

skin of the tunnels filled with gravel (any consistent units)

E_c = \dfrac{X}{X + S_c},\quad X = \ln\dfrac{r_e}{r_w} - \dfrac34

completion efficiency

Common pitfalls

Comparing the tunnel length from charge tests on a surface target with the depth of the damage. In the reservoir the tunnels are shorter.
Expecting a big benefit from dense perforation at 0° phasing. The flow still goes around the casing.
Treating the skin of a gravel pack as constant. Its inertial part grows with the rate, and its viscous part grows as the gravel gets clogged.
Blaming the whole measured skin on damage and ordering an acid treatment, without first subtracting the skins from the opening of the interval and from the completion.

After this lesson you can

calculate the perforation skin by the component method, and explain how the shot density, phasing, tunnel length, tunnel diameter and the anisotropy of the reservoir (the difference between its horizontal and vertical permeability) affect it
estimate the skin of the crushed zone around the tunnels and the gain from underbalanced perforating
calculate the perforation skin in a damaged reservoir for tunnels that end inside the damaged zone and for tunnels that pass right through it, and compare it with an open hole
calculate the pressure loss in tunnels filled with gravel, with its viscous and inertial parts, and compare a gravel pack in cased hole with one in open hole
find the completion efficiency and split a measured skin into the parts caused by how the interval is opened, by the completion and by damage

What we compare with, and where the losses come from

The radial-flow formulas (Steady-state radial flow, Pseudo-steady-state flow and the shape factor) describe an ideal well. In an ideal well the fluid enters the wellbore through its whole wall. A real bottomhole completion (Well completion: opening the reservoir and the tubing string) adds resistance of its own. Like damage and an incomplete opening of the interval, this resistance is expressed as a skin (a number that shows how much the zone around the well helps or hinders the flow; see components of skin). The reference for comparison is an undamaged open hole of the same radius. Its completion skin is zero.

Perforation tunnels. The fluid enters only through separate tunnels. In the horizontal plane it converges on a “star” of tunnels. In the vertical direction it converges on the levels where the shots were fired. Part of the flow also goes around the impermeable casing.
Crushed zone. The jet from the shaped charge crushes and packs the rock around the tunnel. A ring of lower permeability stays around every tunnel (crushed zone).
Damaged zone. Tunnels that end inside the zone damaged by the drilling mud work in poor rock. Tunnels that pass right through it connect the wellbore with the undamaged reservoir (Formation damage and damage skin).
Gravel pack. Sometimes gravel is pumped into the space around a screen inside the cased well. This is a gravel pack: a layer of sorted gravel that holds back the sand. Then the fluid must pass through the tunnels in the casing and the cement, and these tunnels are filled with gravel. The flow area is small and the velocity is high. Inertial losses (losses that grow with the square of the velocity) are added to the viscous ones.

SchematicPaths of the fluid in a cased and perforated well (a diagram, not to scale). In the horizontal plane the flow converges on four tunnels (90° phasing) and goes around the casing. In the vertical section it converges on the shot levels, which are a distance

h = 1/n_s

apart. Around each tunnel there is a crushed zone. In a gravel pack inside the casing, the flow goes along a tunnel filled with gravel.

A convenient way to show how much the completion lowers the productivity of a well is the ratio of its productivity to that of the reference open hole. This ratio is the completion efficiency (also called the productivity ratio):

Completion efficiency

E_c = \frac{J}{J_{\text{OH}}} = \frac{X}{X + S_c},\qquad X = \ln\frac{r_e}{r_w} - \frac34

where:

$E_c$ completion efficiency (also called the productivity ratio, PR), —
$J,\ J_{\text{OH}}$ productivity of the well, and of an undamaged open hole of the same radius in the same drainage area (the part of the reservoir that feeds the well), m³/(d·MPa)
$S_c$ completion skin: the sum of the parts in this lesson (with $Dq$ if there are inertial losses), —
$X$ the logarithmic term of an ideal well in pseudo-steady-state flow, —

Units: dimensionless; natural logarithm

Holds when

pseudo-steady-state flow (the pressure falls at the same steady rate everywhere in the drainage area) in a circular area. For another shape, replace $X$ with $\tfrac12\ln\big(4A/(\gamma C_A r_w^2)\big)$ , $\gamma = 1.781$ (Pseudo-steady-state flow and the shape factor)
unlike the flow efficiency (FE), it includes only the completion skins, not damage and not the incomplete opening of the interval

When $X \approx 7$ , a completion skin of +1 takes away about 12% of the productivity, and +3 takes about 30%. In Russian-language practice the same thing is expressed by the “well perfection coefficient for the character of opening”.

How this was calculated in the past

In the middle of the 20th century in the USSR, engineers studied the extra resistance of perforation on electrolytic models. An electric current in an electrolyte (a salt solution that conducts current) follows the same equations as flow through rock. The results were put into charts of the coefficient of additional flow resistance for the character of opening, $C_2$ . The charts show it against the number of holes per metre, their depth and their diameter. Such charts can still be found in Russian-language calculations. In essence, $C_2$ is the perforation skin: in the inflow formula it is added to $\ln(r_e/r_w)$ . The component method in this lesson was developed later, from numerical solutions of the three-dimensional problem.

Perforation skin: the component method

The exact three-dimensional flow to a perforated well is solved numerically (by computer). The component method replaces it with three simpler problems. It reduces the results of the numerical calculations to formulas with constants from a table for each phasing. The perforation is described by four parameters. The first is the tunnel length in the rock $l_p$ , measured beyond the wellbore wall, without the casing and the cement. The second is the tunnel radius $r_p$ . The third is the shot density $n_s$ , the number of holes per metre. The fourth is the phasing $\theta$ , the angle between the directions of neighbouring charges.

Perforation skin by the component method

s_p = s_H + s_V + s_{wb}

where:

$s_p$ perforation skin relative to an open hole of the same radius, —
$s_H$ plane part: the flow converging on the tunnels in the horizontal plane, —
$s_V$ vertical part: the flow converging on the levels of the shots, —
$s_{wb}$ wellbore part: the flow going around the impermeable casing, —

Units: dimensionless

Holds when

a uniform reservoir, perforated over the whole thickness; the tunnels are clean, with no crushed zone and no damage (these are added separately)
the flow follows Darcy's law (the pressure drop is proportional to the rate); at high velocities in the tunnels an inertial skin is added (Inflow at high rates: non-Darcy flow)

The parts are chosen so that their sum approximates the numerical solutions of the three-dimensional problem. By themselves they are a convenient way to split the problem, not exact physical quantities. A check of the plane part against the exact solution is in the advanced section.

Plane part

s_H = \ln\frac{r_w}{r_w'(\theta)},\qquad r_w'(\theta) = \begin{cases} l_p/4, & \theta = 0^\circ\\ \alpha_\theta\,(r_w + l_p), & \theta \ne 0^\circ \end{cases}

where:

$r_w$ wellbore radius (the radius of the bit), m
$l_p$ tunnel length in the rock beyond the wellbore wall, m
$r_w'(\theta)$ equivalent radius of the “star” of tunnels in the plane problem, m
$\alpha_\theta$ phasing constant (table below), —

Units: any units of length

Holds when

a plane problem: tunnels of infinite conductivity, seen in a horizontal cross-section

With 180° phasing, two tunnels act like a fracture of length $2(r_w + l_p)$ , and $r_w'$ is a quarter of that length. This is the same result as $x_f/2$ for a hydraulic fracture of half-length $x_f$ . With dense phasings, $\alpha_\theta \to 1$ : the tunnels work like a wellbore of radius $r_w + l_p$ . With 0° phasing all the tunnels point in one direction, and $s_H$ is usually positive.

Vertical part

\begin{gathered} s_V = 10^{a}\,h_D^{\,b-1}\,r_D^{\,b},\qquad a = a_1\lg r_D + a_2,\qquad b = b_1 r_D + b_2\\ h_D = \frac{h}{l_p}\sqrt{\frac{k_H}{k_V}},\qquad r_D = \frac{r_p}{2h}\left(1 + \sqrt{\frac{k_V}{k_H}}\right),\qquad h = \frac{1}{n_s}\end{gathered}

where:

$h$ distance between shots in the vertical direction, m
$n_s$ shot density, shots/m
$r_p$ radius of the perforation tunnel, m
$k_H,\ k_V$ horizontal and vertical permeability, mD
$h_D,\ r_D$ dimensionless distance between shots and dimensionless tunnel radius, —
$a_1,\ a_2,\ b_1,\ b_2$ phasing constants (table below), —

Units: dimensionless; $\lg$ is the base-10 logarithm; lengths in the same units

Holds when

an approximation of numerical solutions for $h_D \le 10$ and $r_D \ge 0.01$ ; outside these limits a numerical calculation is needed

$s_V$ grows with the distance between shots $h$ . It grows roughly in proportion at 60–90° phasing, and faster at 0° and 180°. It also grows more strongly the smaller $k_V$ is: when the perforations are sparse, the fluid from the rock between the shots has to travel in the vertical direction. The tunnel diameter has little effect.

Wellbore part

s_{wb} = c_1\,e^{c_2 r_{wD}},\qquad r_{wD} = \frac{r_w}{l_p + r_w}

where:

$c_1,\ c_2$ phasing constants (table below), —
$r_{wD}$ the share of the wellbore radius in the radius of the “star”, —

Units: dimensionless

Holds when

the same model as for $s_H$ ; the casing and the cement are impermeable

The flow to the tunnels goes around the cased wellbore. This is noticeable with short tunnels and with 0° phasing, when the whole flow goes to one side of the casing. At 60–90° phasing this part is very small.

Constants of the plane part and the wellbore part

Phasing: 0°
$\alpha_\theta$: 0.250
$c_1$: 0.160
$c_2$: 2.675

Phasing: 180°
$\alpha_\theta$: 0.500
$c_1$: 0.026
$c_2$: 4.532

Phasing: 120°
$\alpha_\theta$: 0.648
$c_1$: $6.6\cdot10^{-3}$
$c_2$: 5.320

Phasing: 90°
$\alpha_\theta$: 0.726
$c_1$: $1.9\cdot10^{-3}$
$c_2$: 6.155

Phasing: 60°
$\alpha_\theta$: 0.813
$c_1$: $3.0\cdot10^{-4}$
$c_2$: 7.509

Phasing: 45°
$\alpha_\theta$: 0.860
$c_1$: $4.6\cdot10^{-5}$
$c_2$: 8.791

Phasing	$\alpha_\theta$	$c_1$	$c_2$
0°	0.250	0.160	2.675
180°	0.500	0.026	4.532
120°	0.648	$6.6\cdot10^{-3}$	5.320
90°	0.726	$1.9\cdot10^{-3}$	6.155
60°	0.813	$3.0\cdot10^{-4}$	7.509
45°	0.860	$4.6\cdot10^{-5}$	8.791

For 0°, $s_H$ uses $r_w' = l_p/4$ , not $\alpha_\theta$ . The constants are published approximations of numerical solutions.

Constants of the vertical part

Phasing: 0°
$a_1$: −2.091
$a_2$: 0.0453
$b_1$: 5.1313
$b_2$: 1.8672

Phasing: 180°
$a_1$: −2.025
$a_2$: 0.0943
$b_1$: 3.0373
$b_2$: 1.8115

Phasing: 120°
$a_1$: −2.018
$a_2$: 0.0634
$b_1$: 1.6136
$b_2$: 1.7770

Phasing: 90°
$a_1$: −1.905
$a_2$: 0.1038
$b_1$: 1.5674
$b_2$: 1.6935

Phasing: 60°
$a_1$: −1.898
$a_2$: 0.1023
$b_1$: 1.3654
$b_2$: 1.6490

Phasing: 45°
$a_1$: −1.788
$a_2$: 0.2398
$b_1$: 1.1915
$b_2$: 1.6392

Phasing	$a_1$	$a_2$	$b_1$	$b_2$
0°	−2.091	0.0453	5.1313	1.8672
180°	−2.025	0.0943	3.0373	1.8115
120°	−2.018	0.0634	1.6136	1.7770
90°	−1.905	0.1038	1.5674	1.6935
60°	−1.898	0.1023	1.3654	1.6490
45°	−1.788	0.2398	1.1915	1.6392

Worked examplePerforation skin at 0° and 90° phasing

Given (assumed data)

Wellbore: 216 mm bit, $r_w$ = 0.108 m
Tunnels: length in the rock $l_p$ = 0.30 m, diameter 10 mm ( $r_p$ = 0.005 m), 12 shots/m
Reservoir: $k_H/k_V$ = 4; drainage area of radius $r_e$ = 300 m

Solution

The distance between shots is $h = 1/12 = 0.0833$ m. Then $h_D = \dfrac{0.0833}{0.30}\sqrt4 = 0.556$ and $r_D = \dfrac{0.005}{2\cdot0.0833}\,(1 + \sqrt{0.25}) = 0.045$ .
Phasing 0°: $s_H = \ln\dfrac{0.108}{0.30/4} = 0.365$ . Then $a = -2.091\lg 0.045 + 0.0453 = 2.861$ , $b = 2.098$ and $s_V = 0.569$ . Next, $r_{wD} = 0.108/0.408 = 0.265$ and $s_{wb} = 0.16\,e^{2.675\cdot0.265} = 0.325$ . The total is $s_p = 1.26$ .
Phasing 90°: $s_H = \ln\dfrac{0.108}{0.726\cdot0.408} = -1.009$ . Then $a = 2.669$ , $b = 1.764$ and $s_V = 1.255$ . Next, $s_{wb} = 0.010$ . The total is $s_p = 0.26$ .
Completion efficiency, with $X = \ln(300/0.108) - 0.75 = 7.179$ : $E_c = 7.179/(7.179 + s_p)$ = 0.85 for 0° and 0.97 for 90°.

Answer: The same 12 shots per metre give a skin of 1.26 at 0° phasing and 0.26 at 90°. That is 85% and 97% of the productivity of an open hole.

What it means. At 90° the plane part is strongly negative: four tunnels of 0.3 m work almost like a wellbore of radius 0.3 m. The vertical part takes most of this gain back, because the shots that point the same way are 0.33 m apart. The result is close to zero. At 0° all the tunnels point one way, the flow goes around the casing, and the skin is five times bigger. The data in this example are assumed for illustration.

Perforation skin of the lesson's tunnels (0.30 m, 10 mm,

k_H/k_V

= 4) at different shot densities and phasings

Shots/m: 6
0°: 2.55
180°: 1.81
90°: 1.68
60°: 1.99

Shots/m: 12
0°: 1.26
180°: 0.35
90°: 0.26
60°: 0.37

Shots/m: 20
0°: 0.87
180°: −0.17
90°: −0.33
60°: −0.31

Shots/m: 30
0°: 0.75
180°: −0.38
90°: −0.63
60°: −0.65

Shots/m	0°	180°	90°	60°
6	2.55	1.81	1.68	1.99
12	1.26	0.35	0.26	0.37
20	0.87	−0.17	−0.33	−0.31
30	0.75	−0.38	−0.63	−0.65

Calculated with the formulas of this section. The crushed zone and damage are not included. In all rows $h_D$ and $r_D$ are within the limits where the formula applies.

The crushed zone around the tunnels

The jet of a shaped charge does not take the rock out of the tunnel. It pushes the rock aside, crushing and packing the grains. A ring about a centimetre thick stays around the tunnel. Its permeability $k_c$ is several times lower than the reservoir permeability. Inside the tunnel there are fragments of rock and of the charge. Underbalanced perforating (the well pressure is lower than the reservoir pressure) and the inflow that follows wash out part of the fragments and of the crushed rock (Perforating operations: gun conveyance and procedures). With overbalanced perforating (the well pressure is higher than the reservoir pressure) the opposite happens: well fluid is forced into the tunnels.

Skin of the crushed zone

s_c = \frac{h}{l_p}\left(\frac{k}{k_c} - \frac{k}{k_s}\right)\ln\frac{r_c}{r_p}

where:

$s_c$ skin of the crushed zone, —
$h,\ l_p$ distance between shots and tunnel length, m
$k_c$ permeability of the crushed zone, mD
$k_s$ permeability of the rock around the crushed zone: $k$ if the tunnel is in the undamaged reservoir, and the permeability of the damaged zone if the tunnel is in it, mD
$r_c,\ r_p$ radius of the crushed zone ( $r_p$ plus its thickness) and radius of the tunnel, m

Units: dimensionless; natural logarithm

Holds when

the inflow into the tunnel is uniform along its length and radial through the thin ring. In reality it is concentrated nearer the end of the tunnel, so the formula gives a lower limit for the skin
the thickness and permeability of the zone are taken from charge tests under reservoir conditions, or estimated; in this lesson's examples they are assumed

This is Hawkins' formula (the skin of a ring of changed permeability around a hole) for one tunnel, multiplied by $h/l_p$ . Each tunnel carries the share of the flow that belongs to an interval of height $h$ , and the area of its wall is proportional to $l_p$ . So sparse, short tunnels suffer most from the crushed zone.

Worked exampleOverbalanced and underbalanced perforating

Given (assumed data)

Perforation: from the previous example, 90° phasing: $s_p$ = 0.26
Crushed zone: thickness 12 mm ( $r_c$ = 0.017 m)
Permeability of the zone: $k_c/k$ = 0.2 after overbalanced perforating, and 0.5 after underbalanced perforating with clean-up

Solution

The multiplier is $\dfrac{h}{l_p}\ln\dfrac{r_c}{r_p} = \dfrac{0.0833}{0.30}\ln\dfrac{0.017}{0.005} = 0.340$ .
Overbalanced: $s_c = 0.340\cdot(5 - 1) = 1.36$ . Underbalanced: $s_c = 0.340\cdot(2 - 1) = 0.34$ .
The completion skin $s_p + s_c$ is 1.62 and 0.60. The completion efficiency is 0.82 and 0.92.

Answer: The crushed zone adds 1.36 with overbalanced perforating and 0.34 with underbalanced perforating. The completion efficiency is 82% against 92%.

What it means. The crushed zone gave a skin 5 times bigger than the whole geometry of the perforation. So the way the well is perforated matters as much as the choice of charges. With the same pattern of tunnels, underbalanced perforating gives about 13% more productivity. The data in this example are assumed for illustration.

Perforation and the damaged zone

The drilling mud leaves a damaged zone around the wellbore. Its thickness is $l_s$ and its permeability is $k_s$ (Formation damage and damage skin). In that lesson its permeability and radius are written $k_d$ and $r_d$ . Here they are $k_s$ and $r_s = r_w + l_s$ , as in the lesson on skin factor. In an open hole all the flow passes through this zone, and the skin equals the Hawkins skin (the skin of a damaged ring around an open hole; see the Hawkins formula). In perforated casing everything depends on where the tunnels end.

Tunnels end inside the damaged zone

s_{dp} = \left(\frac{k}{k_s} - 1\right)\ln\frac{r_s}{r_w} + \frac{k}{k_s}\,s_p,\qquad r_s = r_w + l_s

where:

$s_{dp}$ skin of a perforated completion in a damaged reservoir (without the crushed zone), —
$k,\ k_s$ permeability of the reservoir and of the damaged zone, mD
$r_s,\ l_s$ radius and thickness of the damaged zone (the thickness is measured from the wellbore wall), m
$s_p$ perforation skin of the same tunnels in an undamaged reservoir, —

Units: dimensionless; natural logarithm

Holds when

$l_p < l_s$ : the whole zone where the flow converges on the tunnels lies in damaged rock
this is an upper limit: when the ends of the tunnels are close to the edge of the zone, part of the flow reaches them from undamaged rock, and the skin is smaller

The first term is the Hawkins skin, the same as for an open hole. The second term is the perforation skin multiplied by $k/k_s$ , because the tunnels work in rock whose permeability is $k/k_s$ times lower. By this formula a perforated well is worse than an open hole with the same damage if $s_p > 0$ . In reality the difference is smaller, and for tunnels that almost reach the edge of the zone it can disappear.

Tunnels pass right through the damaged zone

\begin{gathered} s_{dp} \approx s_H + s_V(\tilde l_p) + s_{wb}\!\left(\frac{\tilde r_w}{r_w + l_p}\right)\\ \tilde l_p = l_p - \left(1 - \frac{k_s}{k}\right)l_s,\qquad \tilde r_w = r_w + \left(1 - \frac{k_s}{k}\right)l_s\end{gathered}

where:

$\tilde l_p$ equivalent tunnel length: the part that lies in the damaged zone counts with a weight of $k_s/k$ , m
$\tilde r_w$ equivalent wall radius: the casing is as if “extended” into the damaged zone (do not confuse it with the equivalent radius $r_w'$ ), m
$s_H,\ s_V,\ s_{wb}$ parts of the method: $s_H$ is the same as without damage (at 0° phasing, with $\tilde l_p$ instead of $l_p$ ), $s_V$ is calculated with $\tilde l_p$ , and $s_{wb}$ with $r_{wD} = \tilde r_w/(r_w + l_p)$ , —

Units: dimensionless

Holds when

$l_p > l_s$ . The tunnel ends are in undamaged rock, so the plane part does not change at phasings of 45–180°. At 0° its equivalent radius $l_p/4$ depends on the tunnel length beyond the wall, and there $\tilde l_p$ is used
an approximation. When $k_s \to 0$ , it is the component method for a casing of radius $r_w + l_s$ , with the skin counted from the true radius $r_w$ (an impermeable ring is the same as a thicker casing). When $k_s = k$ , it gives the usual $s_p$ . A check against a numerical solution of the plane problem shows a difference of no more than 0.15
$h_D$ calculated with the length $\tilde l_p$ must stay at 10 or less

The damaged ring is almost impermeable, but the flow can bypass it through the tunnels, so it works like a thicker casing. This is why the tunnels almost bypass damage that is strong but shallow, and why an acid treatment would give such a well little.

CalculatedCalculated perforation skin (90° phasing, 12 shots/m, 10 mm tunnels,

k_H/k_V

= 4) against the tunnel length in the rock. One curve is for no damage, and one for a reservoir with a damaged zone 0.2 m thick and

k/k_s

= 5. The horizontal line is the skin of an open hole with the same damage. The break at the edge of the zone is where both formulas are only rough: the true skin falls sharply, but smoothly, between their values.

Worked exampleShort and long tunnels in a damaged reservoir

Given (assumed data)

Perforation: as in the first example: 90° phasing, 12 shots/m
Damage: a zone with $l_s$ = 0.20 m beyond the wellbore wall, $k/k_s$ = 5
Tunnels: 0.15 m (strong rock, shorter) or 0.30 m

Solution

Open hole: $S = (5 - 1)\ln\dfrac{0.308}{0.108} = 4.19$ .
The 0.15 m tunnels end inside the zone. The perforation skin without damage is $s_H = -0.551$ , $s_V = 2.131$ ( $h_D$ = 1.111) and $s_{wb} = 0.025$ , so $s_p = 1.61$ . With damage: $s_{dp} = 4.19 + 5\cdot1.61 = 12.2$ .
The 0.30 m tunnels pass through the zone: $\tilde l_p = 0.30 - 0.8\cdot0.20 = 0.14$ m and $\tilde r_w = 0.108 + 0.8\cdot0.20 = 0.268$ m.
$s_H$ is the same as without damage, −1.009. With $h_D = \dfrac{0.0833}{0.14}\sqrt4 = 1.190$ , $s_V$ is 2.246. With $r_{wD} = 0.268/0.408 = 0.657$ , $s_{wb}$ is 0.108. The total is $s_{dp} = 1.35$ .
Productivity as a fraction of an undamaged open hole, $X/(X + S)$ : open hole 0.63, 0.15 m tunnels 0.37, 0.30 m tunnels 0.84.

Answer: With the same damage, the open hole has a skin of 4.19, the short tunnels about 12.2 and the long tunnels 1.35.

What it means. Tunnels that do not reach undamaged rock give a skin three times bigger than the open hole (although the formula overestimates here). Tunnels that pass right through the zone almost remove the damage: of the open-hole skin of 4.19, about 1.1 is left on top of the skin of the perforation itself. So the length of the tunnels in the rock, and not only the shot density, decides the fate of a well with a damaged near-wellbore zone. The crushed zone is not included here. If the tunnels are in damaged rock, its $k_c$ is measured relative to $k_s$ . The data in this example are assumed for illustration.

Open hole or perforated casing: which is better for inflow

The bottomhole design is chosen first of all for the stability of the rock, the need to isolate intervals and the need to fight sand (Well completion: opening the reservoir and the tubing string). But the options also differ in inflow, and the calculations of this lesson let us compare them in numbers.

Completion skin in this lesson's examples

Conditions: No damage
Open hole: 0
Perforated casing: +1.3 (0°) … +0.3 (90°); dense and deep: below zero

Conditions: Crushed zone
Open hole: —
Perforated casing: +0.3…+1.4 depending on how the well is perforated

Conditions: Shallow damage (0.2 m, $k/k_s$ = 5)
Open hole: +4.2
Perforated casing: +1.3 with 0.30 m tunnels

Conditions: Tunnels do not reach undamaged rock
Open hole: +4.2
Perforated casing: +12 (upper limit)

Conditions: Gravel pack, 500 mD
Open hole: ≈ 0
Perforated casing: +0.6 (clean gravel) … +5.7 (clogged) at 400 m³/d

Conditions	Open hole	Perforated casing
No damage	0	+1.3 (0°) … +0.3 (90°); dense and deep: below zero
Crushed zone	—	+0.3…+1.4 depending on how the well is perforated
Shallow damage (0.2 m, $k/k_s$ = 5)	+4.2	+1.3 with 0.30 m tunnels
Tunnels do not reach undamaged rock	+4.2	+12 (upper limit)
Gravel pack, 500 mD	≈ 0	+0.6 (clean gravel) … +5.7 (clogged) at 400 m³/d

The values come from the lesson's examples and depend on the input data. The skin of an open hole with gravel is the resistance of a thin layer of highly permeable gravel around the screen. It is a few hundredths.

In short: with shallow damage and tunnels that are long enough, perforated casing is as good as an open hole, and often better. But the damage may be deeper than the tunnels. This can happen after a long time of drilling with overbalance in a highly permeable reservoir, when the mud pressure was above the reservoir pressure. Then an open hole, or hydraulic fracturing that goes around the damage (Hydraulic fracturing: purpose and well selection), will give much more. For a gravel pack, the open hole wins on inflow, because it has no narrow tunnels filled with gravel.

Gravel pack: linear flow in tunnels filled with gravel

In weak sandstones a screen is placed opposite the reservoir, and the space between the screen and the wall is filled with sorted gravel. This is a gravel pack: the gravel holds the sand back, and the screen holds the gravel back. If the pack is made inside cemented perforated casing, the gravel fills both the ring-shaped space between the screen and the casing and the perforation tunnels. Then the fluid from the reservoir reaches the screen through these tunnels. Each tunnel has a length $L$ , from the reservoir wall through the cement and the casing. The cross-section of all the tunnels together is small and the velocity is high. So the losses in them have two parts: a viscous part (from the friction of the fluid in the pores), and an inertial part that grows as the square of the rate. In an open hole the gravel lies as a layer around the screen. The flow passes through it radially, across the whole area of the wall, and the losses are very small.

Skin of the tunnels filled with gravel

S_g = \frac{8\,k\,h\,L}{k_g\,n\,d^2}

where:

$S_g$ skin from the viscous losses in the tunnels with gravel, —
$k,\ h$ permeability and thickness of the reservoir, mD, m
$L$ length of a tunnel with gravel: from the screen through the casing and cement to the reservoir wall, m
$k_g$ permeability of the gravel in the tunnels, mD
$n$ number of open tunnels in the interval, —
$d$ tunnel diameter, m

Units: any consistent units: $k$ and $k_g$ in the same units, and $h$ , $L$ , $d$ in the same units too

Holds when

linear flow along tunnels of equal length and diameter; the whole inflow is divided equally among the $n$ tunnels
the gravel is uniform; clogging by fine particles, mixing with the reservoir sand and leftovers of the completion fluid are allowed for by a reduced $k_g$

It follows from Darcy's law: the pressure drop along the tunnels, $qB\mu L/(k_g\,n\,\pi d^2/4)$ , divided by $qB\mu/(2\pi kh)$ . The skin is proportional to the flow capacity $kh$ of the reservoir. So a pack inside the casing costs most in highly permeable reservoirs, which are exactly where sand control is usually needed. The number of tunnels and the square of their diameter are in the denominator. This is why wells with a gravel pack are perforated densely, with large holes.

In other unit systems (1)

Oilfield units

S_g = \frac{96\,k h\,L}{k_g\,n\,d^2}

$kh$ in mD·ft, $L$ and $d$ in inches, $k_g$ in mD.

$96 = 8\cdot12$ : it converts the thickness from feet to inches.

Inertial losses in the tunnels with gravel

D_g = \frac{32\,k h\,\beta\,\rho\,B\,L}{\pi\,\mu\,n^2 d^4}

where:

$D_g$ non-Darcy coefficient of the tunnels with gravel; the inertial skin is $D_g q$ , s/m³ (for a rate at standard conditions)
$\beta$ inertial resistance coefficient of the gravel, m⁻¹
$\rho$ fluid density at reservoir conditions, kg/m³
$B,\ \mu$ formation volume factor and viscosity, m³/m³, Pa·s
$k,\ h,\ L,\ n,\ d$ as in the formula for $S_g$ , m², m, m, —, m

Units: SI

Holds when

the Forchheimer equation in each tunnel (Darcy's law plus a term that grows with the square of the velocity; see Inflow at high rates: non-Darcy flow). For gas, $\rho B$ is replaced by the gas density at standard conditions, and the viscosity is taken at bottomhole conditions
the $\beta$ of the gravel grows as it gets clogged. We estimate it from tests or from correlations with the permeability of the gravel

The velocity in the tunnels is inversely proportional to $nd^2$ , and the inertial losses are proportional to the square of the velocity. This gives $n^2d^4$ . With half as many open tunnels, the viscous skin is twice as big and the inertial skin is four times as big. This is how a growth of $D_g$ from one well test to the next shows that the tunnels are getting plugged.

In other unit systems (3)

Metric oilfield units (liquid)

D_g\,[1/(\text{m}^3/\text{d})] = 1.164\cdot10^{-16}\,\frac{k h\,\beta\,\rho\,B\,L}{\mu\,n^2 d^4}

$k$ in mD, $h$ , $L$ , $d$ in m, $\beta$ in m⁻¹, $\rho$ in kg/m³, $\mu$ in mPa·s.

$1.164\cdot10^{-16} = 64\cdot1.818\cdot10^{-18}$ , where $1.818\cdot10^{-18}$ is the constant of the coefficient $D$ in the lesson Inflow at high rates: non-Darcy flow.

Metric oilfield units (gas)

D_g\,[1/(\text{thousand m}^3/\text{d})] = 1.164\cdot10^{-13}\,\frac{k h\,\beta\,\rho_{sc}\,L}{\mu\,n^2 d^4}

$\rho_{sc}$ is the gas density at standard conditions, in kg/m³; $\mu$ is the gas viscosity at the bottomhole, in mPa·s; the rest is as for liquid.

Oilfield units (liquid)

D_g\,[1/(\text{STB/d})] = 1.808\cdot10^{-11}\,\frac{k h\,\beta\,\rho\,B\,L}{\mu\,n^2 d^4}

$kh$ in mD·ft, $\beta$ in ft⁻¹, $\rho$ in lb/ft³, $L$ and $d$ in inches, $\mu$ in cP, $B$ in rb/STB.

The oil density must be included: without it the dimensions do not match.

Worked exampleLosses in a gravel pack inside the casing: clean and clogged gravel

Given (assumed data)

Reservoir: $k$ = 500 mD, $h$ = 10 m
Perforation: 12 shots/m over the whole thickness, $n$ = 120, $d$ = 20 mm; length of a tunnel with gravel $L$ = 0.06 m
Gravel: clean: $k_g$ = 150 D, $\beta = 8\cdot10^4$ m⁻¹; clogged: $k_g$ ten times smaller and $\beta$ ten times bigger
Oil: $\rho$ = 850 kg/m³, $B$ = 1.15, $\mu$ = 2 mPa·s; rate 400 m³/d

Solution

Viscous skin: $S_g = \dfrac{8\cdot500\cdot10\cdot0.06}{150{,}000\cdot120\cdot0.02^2} = \dfrac{2400}{7200} = 0.333$ .
Inertial: $D_g = 1.164\cdot10^{-16}\,\dfrac{500\cdot10\cdot8\cdot10^4\cdot850\cdot1.15\cdot0.06}{2\cdot120^2\cdot0.02^4} = 5.9\cdot10^{-4}$ 1/(m³/d). At 400 m³/d, $D_gq = 0.24$ .
Pressure drop across the pack: $\Delta p = 1.866\,\dfrac{400\cdot1.15\cdot2}{500\cdot10}\,(0.333 + 0.24) = 0.20$ MPa.
Clogged gravel: $S_g = 3.33$ , $D_gq = 2.37$ and $\Delta p = 1.96$ MPa.
Completion efficiency (the pack only), with $X = 7.179$ : 0.93 for clean gravel and 0.56 for clogged gravel.

Answer: A clean pack loses about 0.20 MPa of pressure and 7% of the productivity. A clogged pack loses 1.96 MPa and 44%.

What it means. In the clean pack the inertial part is already comparable to the viscous part. In the clogged pack both have grown ten times. Some operators limited the pressure drop across the pack to about 2 MPa (300 psi), because they believed that larger drops destroy the pack more often. Others work successfully with larger drops. Here the clogged pack at 400 m³/d is right at this limit. The perforation skin in the rock behind the cement is added to these figures. The data in this example are assumed for illustration.

CalculatedCalculated pressure drop across the tunnels with gravel against the rate, for the pack in the example: clean and clogged gravel. The dashed horizontal line is the limit of about 2 MPa that some operators used. The curves bend upwards because of the inertial part, which grows as the square of the rate.

Skin at the screen and at the boundary with the reservoir. We include clogging of the screen, gravel mixed with reservoir sand at the boundary, and leftover fluid in the pores of the reservoir behind the tunnels as an extra skin at the screen or at the reservoir wall. If the gravel was pumped behind the cement into voids in the reservoir, and there is no damage, this skin can be negative.
How to find the pressure drop across the pack. A nodal analysis (Nodal analysis of the production system, a way to work out the rate from the pressure losses in the reservoir and in the well) with the node at the reservoir wall and no pack gives the highest possible rate. The pressure drop across the pack at the actual rate is the difference between the inflow and outflow curves at that rate. We compare it with the result of the formulas in this section.
What to choose. Use the coarsest gravel that still holds the sand back, dense perforation with large holes, and clean fluids when pumping the gravel. In highly permeable reservoirs, use a gravel pack in an open hole, or a short, wide fracture packed with gravel behind the casing. The design of the pack is the subject of Sand control methods.

How to read a composite skin

A well test gives one skin: the sum of everything that makes the well different from an ideal one (components of skin). To decide what to do, we split it into parts. Some parts act only in the opened interval, of thickness $h_w$ . In the skin for the whole thickness $h$ they carry a multiplier $h/h_w$ (Partial penetration and well deviation):

Measured skin of a perforated well

S = S_{pp} + \frac{h}{h_w}\left(s_p + s_c + S_{d,\text{eff}}\right) + S_g + Dq

where:

$S$ skin from the well test, for the whole thickness, —
$S_{pp}$ pseudo-skin of partial penetration (the well opens only part of the thickness), —
$s_p,\ s_c$ skin of the perforation and of the crushed zone in the opened interval, —
$S_{d,\text{eff}}$ the remaining contribution of damage, after allowing for how the tunnels bypass it, —
$S_g,\ Dq$ skin of the gravel pack and the inertial skin, —

Units: dimensionless

Holds when

simple addition is an approximation: the parts interact (damage and perforation, as in the section on the damaged zone)
$Dq$ is separated out by well tests at several rates (Inflow at high rates: non-Darcy flow); the other terms do not depend on the rate

An acid treatment reduces only $S_{d,\text{eff}}$ (and partly the clogging of a pack). Re-perforating reduces $s_p$ , $s_c$ and $S_{d,\text{eff}}$ . Perforating more of the interval reduces $S_{pp}$ . Hydraulic fracturing bypasses all of them at once.

Signs and likely causes

Sign: The skin is large and does not depend on the rate; the tunnels are shorter than the damaged zone
Likely cause: damage that the tunnels did not pass through; a crushed zone

Sign: The skin and $D$ are large, and $D$ grows from one test to the next
Likely cause: plugged tunnels or clogged gravel: $D$ grows faster than the viscous skin

Sign: The skin is large when only part of the thickness is perforated
Likely cause: pseudo-skin of partial penetration, and parts multiplied by $h/h_w$

Sign: The skin of the gravel pack grows with time
Likely cause: fine particles of the reservoir carried into the gravel, deposits on the screen

Sign: The skin after overbalanced perforating is larger than the one calculated from the geometry
Likely cause: a crushed zone and debris in the tunnels

Sign	Likely cause
The skin is large and does not depend on the rate; the tunnels are shorter than the damaged zone	damage that the tunnels did not pass through; a crushed zone
The skin and $D$ are large, and $D$ grows from one test to the next	plugged tunnels or clogged gravel: $D$ grows faster than the viscous skin
The skin is large when only part of the thickness is perforated	pseudo-skin of partial penetration, and parts multiplied by $h/h_w$
The skin of the gravel pack grows with time	fine particles of the reservoir carried into the gravel, deposits on the screen
The skin after overbalanced perforating is larger than the one calculated from the geometry	a crushed zone and debris in the tunnels

Worked exampleSplitting a skin and choosing between acid and re-perforating

Given (assumed data)

Reservoir: $h$ = 30 m, $k_V/k_H$ = 0.25, $r_w$ = 0.108 m, $r_e$ = 300 m
Opening: the top 15 m perforated overbalanced: 90°, 12 shots/m, tunnels 0.30 m long and 10 mm in diameter
Well test: skin +9.0, the same at two rates
Crushed zone: 12 mm, $k_c/k$ = 0.2 (after overbalanced perforating, an estimate)

Solution

Pseudo-skin of partial penetration (from the lesson on the incomplete opening of the interval, with the opened interval at the top): $S_{pp} = 4.78$ .
Skin in the opened interval: $(9.0 - 4.78)\cdot\dfrac{15}{30} = 2.11$ .
Of this, the perforation is $s_p$ = 0.26 and the crushed zone is $s_c$ = 1.36. What is left for damage is $S_{d,\text{eff}} = 2.11 - 0.26 - 1.36 = 0.49$ .
An acid treatment (it removes $S_{d,\text{eff}}$ ): the skin becomes $9.0 - 2\cdot0.49 = 8.02$ , and the productivity grows by a factor of $\dfrac{7.179 + 9.0}{7.179 + 8.02} = 1.06$ .
Re-perforating the same interval underbalanced at 20 shots/m: $s_p$ = −0.33 and $s_c = \dfrac{0.05}{0.30}\ln\dfrac{0.017}{0.005}\cdot(2 - 1) = 0.20$ . With the same $S_{d,\text{eff}}$ , the skin is $4.78 + 2\cdot(-0.33 + 0.20 + 0.49) = 5.50$ , and the productivity grows by a factor of 1.28.

Answer: Of the skin of +9.0, about 4.78 comes from partial penetration, 3.2 from the perforation with its crushed zone, and only 1.0 from damage. Acid will give +6%, and re-perforating +28%.

What it means. A skin of +9 looked like severe damage, but the acid has almost nothing to dissolve here. The main room for improvement is in the crushed zone and in the sparse perforation of the opened half of the thickness. Perforating the lower part of the reservoir would give even more, if the nearby water allows it. The data in this example are assumed for illustration.

Terms and symbols

Perforation skin $s_p$: The extra resistance from the flow converging on the perforation tunnels, compared with an open hole. It can be negative.
Shot density $n_s$: The number of holes per metre of interval. The distance between shots is $h = 1/n_s$ .
Phasing $\theta$: The angle between the directions of neighbouring charges: 0°, 60°, 90°, 120° or 180°.
Crushed zone: A ring of crushed and packed rock of lower permeability around a perforation tunnel.
Gravel pack: A layer of sorted gravel between the screen and the reservoir that holds the sand back. Inside casing, the gravel also fills the perforation tunnels.
Completion efficiency $E_c$: The ratio of the productivity of the well to that of an undamaged open hole of the same radius.

Common mistakes and misconceptions

Wrong

“Perforated casing is always worse than an open hole.”

In fact

Dense, deep tunnels give a skin near zero or below. With shallow damage that the tunnels pass right through, the casing beats the open hole: in the lesson's example +1.3 against +4.2.

Wrong

“The more shots per metre the better, and the direction they point does not matter.”

In fact

Shot density mainly reduces the vertical part. With 0° phasing the flow still goes around the casing. In the example, 12 shots/m at 0° phasing give +1.3, and at 90° phasing about +0.3.

Wrong

“The skin of a gravel pack is a constant.”

In fact

Its inertial part grows with the rate, and its viscous part grows as the gravel gets clogged. If the number of open tunnels is halved, the viscous part doubles and the inertial part becomes four times bigger.

Wrong

“The tunnels passed through the damaged zone, so we can forget about the damage.”

In fact

The damage is almost bypassed, but not completely: in the lesson's example about +1.1 is left on top of the skin of the perforation itself. The crushed zone also stays, and so does the damage if the tunnels in the reservoir are shorter than calculated.

Wrong

“A skin of +9 means severe damage, so we need an acid treatment.”

In fact

First subtract the pseudo-skin of partial penetration and the skins of the perforation and the crushed zone, multiplied by $h/h_w$ . In the lesson's example only about 1 of the 9 was due to damage.

In depth: advanced detail

The plane problem exactly: two tunnels and a casing

For phasings of 0° and 180° the plane problem can be solved exactly by a conformal mapping (a change of coordinates that keeps the flow equations the same). The casing of radius $r_w$ is impermeable. The tunnels, out to the radius $R = r_w + l_p$ , have infinite conductivity. The Joukowski mapping $w = z + r_w^2/z$ turns the circle of the casing into a segment on the line of symmetry, where the flow does not cross the axis anyway. It turns the tunnels into segments of the same line. The equivalent radius is expressed through the logarithmic capacity of these segments:

Exact equivalent radius of tunnels in a cased well

r_w'(180^\circ) = \frac{R^2 - r_w^2}{2R},\qquad r_w'(0^\circ) = \frac{l_p^2}{4R},\qquad R = r_w + l_p

where:

$r_w'$ equivalent radius in the plane problem, including the flow around the casing, m
$R$ radius of the tunnel ends, m
$r_w,\ l_p$ wellbore radius and tunnel length, m

Units: any units of length

Holds when

a plane problem; the casing and the cement are impermeable, and the tunnels have infinite conductivity

When $r_w \to 0$ , both formulas give the values of the component method: $R/2$ and $l_p/4$ . For a finite $r_w$ the equivalent radius is smaller, and this is what the part $s_{wb}$ describes.

A check for the lesson's tunnels ( $r_w$ = 0.108 m, $l_p$ = 0.30 m, $R$ = 0.408 m). For 180° phasing, exactly, $r_w' = (0.1665 - 0.0117)/0.816 = 0.190$ m and $\ln(r_w/r_w') = -0.56$ . By the component method, $s_H + s_{wb} = -0.636 + 0.086 = -0.55$ . For 0° phasing, exactly, $r_w' = 0.09/1.632 = 0.0551$ m and $\ln(r_w/r_w') = 0.67$ . By the method, $0.365 + 0.325 = 0.69$ . The differences are only hundredths, so the approximation fits both the plane problem and the flow around the casing.

Derivation of the formulas for tunnels with gravel

The rate at reservoir conditions, $qB$ , is divided among $n$ tunnels of area $A = \pi d^2/4$ each, so the flow velocity in a tunnel is $v = qB/(nA)$ . By the Forchheimer equation, the pressure drop over a length $L$ is $\Delta p = \dfrac{\mu L}{k_g}v + \beta\rho L v^2$ .

We divide by the “unit” of radial flow, $qB\mu/(2\pi kh)$ (Skin factor). The viscous term gives $S_g = 2\pi khL/(k_g nA) = 8khL/(k_g n d^2)$ . The inertial term gives a skin that is proportional to the rate: $D_g q$ with $D_g = 2\pi kh\,\beta\rho B L/(\mu n^2A^2) = 32\,kh\beta\rho BL/(\pi\mu n^2d^4)$ . For gas, the product $\rho B$ is the mass flow per unit of standard rate, and it equals the gas density at standard conditions.

The constants in oilfield units come from unit conversion. With $kh$ in mD·ft and $L$ and $d$ in inches, the factor 8 becomes $8\cdot12 = 96$ . For the inertial part, with $\beta$ in ft⁻¹, $\rho$ in lb/ft³, $\mu$ in cP and the rate in STB/d, the constant is $1.808\cdot10^{-11}$ . For gas, with the rate in thousand ft³/d (Mscf/d) and $\rho_{sc} = 0.0764\,\gamma_g$ lb/ft³, it is $2.46\cdot10^{-10}\,\gamma_g$ , where $\gamma_g$ is the gas specific gravity (the density relative to air).

Connections to other topics

Skin factor — skin, Hawkins' formula and the components of a measured skin.
Partial penetration and well deviation — pseudo-skin of partial penetration and the multiplier $h/h_w$ for the parts that act in the opened interval.
Inflow at high rates: non-Darcy flow — the Forchheimer equation, the coefficient $D$ and well tests at several rates.
Formation damage and damage skin — damage of the near-wellbore zone and the choice of drilling fluid for a well that will be perforated.
Well completion: opening the reservoir and the tubing string — open hole, slotted liner and perforated casing: the choice of bottomhole design.
Inflow performance relationship (IPR) — how completion skin enters the inflow performance relationship through the completion efficiency.
Shaped-charge perforating — shaped charges, and the length and diameter of their tunnels.
Sand control methods — design of gravel packs and screens.

Check yourself

1
A perforating gun with 12 shots/m can be assembled with 0° or 90° phasing. The tunnels are long and the reservoir is uniform. Which option gives the smaller skin, and why?
1. A.The same: skin depends only on shot density
2. B.0°: all tunnels are in one plane, so the vertical part is smaller
3. C.0°: the casing is damaged less
4. D.90°: the tunnels surround the wellbore, the plane part is strongly negative and there is almost no flow around the casing
Show the answer
Correct answer: D. 90°: the tunnels surround the wellbore, the plane part is strongly negative and there is almost no flow around the casing
At 90° the four directions of tunnels work almost like a wellbore of radius $r_w + l_p$ , and $s_H$ is negative. The part for the flow around the casing is very small. At 0° the flow goes to one side of the casing, and $s_H$ and $s_{wb}$ are positive. In the lesson's example the skin is +1.3 against +0.3 at the same shot density.
2
The damaged zone is 0.25 m thick, and the charges make tunnels 0.15 m long in the reservoir. Which is better for inflow: an open hole or this perforation? What can be changed?
Show the answer
The tunnels end inside the damaged zone, so the perforation skin is multiplied by $k/k_s$ . Perforated casing is worse than an open hole with the same damage. Longer tunnels would help (deep-penetrating charges, underbalanced perforating), as would a drilling fluid that causes only shallow damage, or a hydraulic fracture that goes around the zone.
3
In a gravel pack inside the casing, half of the tunnels have become plugged. How will the viscous skin of the pack $S_g$ and the coefficient $D_g$ change?
1. A.Both double
2. B. $S_g$ doubles and $D_g$ becomes four times bigger
3. C. $S_g$ becomes four times bigger and $D_g$ doubles
4. D.They do not change: the gravel is the same
Show the answer
Correct answer: B. $S_g$ doubles and $D_g$ becomes four times bigger
$S_g \propto 1/n$ and $D_g \propto 1/n^2$ . The velocity in the remaining tunnels is twice as high, and inertial losses are proportional to the square of the velocity. So a growth of $D$ from one well test to the next is a sensitive sign that the tunnels are getting plugged.
4
After overbalanced perforating, the permeability of the crushed zone is 0.2 of the reservoir permeability. After underbalanced perforating it is 0.5. By what factor does the skin of the crushed zone fall with the same geometry of tunnels?
1. A.4 times
2. B.2.5 times
3. C.1.6 times
4. D.10 times
Show the answer
Correct answer: A. 4 times
The skin is proportional to $k/k_c - 1$ : $(5 - 1)/(2 - 1) = 4$ . The geometric multiplier $(h/l_p)\ln(r_c/r_p)$ is the same.
5
The skin from a well test is +7. Half of the thickness is perforated ( $h/h_w$ = 2), the pseudo-skin of partial penetration is 3.0, and the calculated skin of the perforation with its crushed zone is 1.2. How much is due to damage, and is an acid treatment worth doing?
Show the answer
In the opened interval the skin is $(7 - 3)/2 = 2.0$ . After subtracting 1.2, 0.8 is left for damage, and in the total skin it “weighs” $2\cdot0.8 = 1.6$ . An acid treatment will cut the skin from 7 to 5.4. With $X \approx 7.2$ this is about +13% of productivity. Re-perforating (which would reduce the 1.2·2 = 2.4) or perforating more of the interval will give more.

Updated: 19 October 2026