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ProductionPerforating Intermediate 19 min read

Shaped-charge perforating

Shaped-charge perforating is the opening of a reservoir in a cased well (a well lined with steel pipe, the casing) with directional charges, which act in one direction. The explosion squeezes the thin metal liner of the charge into a narrow jet. The jet flies at a speed of several kilometres per second and pierces the casing, the cement sheath (the ring of cement behind the casing) and the rock. The result is tunnels (channels), usually tens of centimetres long, through which the reservoir is connected to the well.

Why it matters

Perforating is the only path for fluid from the reservoir into a cemented casing. The length and diameter of the tunnels, their number and layout, and how clean they stay decide several things. They decide the skin factor of the well (a number that shows how much the zone right around the well helps or hinders flow). They decide whether a gravel pack (a filter of selected gravel around the well) or a hydraulic fracture (a crack opened in the rock by pumping fluid) is possible. And they decide how much sand the well produces. The tunnel length in the reservoir cannot be taken from a catalogue. The data-sheet figures are measured in concrete at the surface, while in the reservoir the rock has its own strength and is squeezed by stresses.

Quick refresher

  • A shaped charge is a case, an explosive and a metal liner in a conical cavity. The detonation squeezes the liner towards the axis. The smaller part of the metal goes into a fast jet, and the larger part forms a slow slug (a thick lump of metal).
  • The jet pierces the target by the pressure of impact. At its head this pressure is tens of gigapascals, tens to hundreds of times the strength of steel and rock. The metal and the rock do not melt: they flow aside and get compacted.
  • By the density law (the denser the target, the less the jet pierces), a continuous jet of length LL pierces a depth of Lρj/ρtL\sqrt{\rho_j/\rho_t}. A millimetre of casing steel “costs” almost two millimetres of tunnel in sandstone, and cement and water cost less. The law ignores the strength of the rock and the stresses in it, yet these are what make tunnels in the reservoir shorter than the data-sheet values.
  • Deep-penetrating charges make a narrow, long tunnel that passes through the damaged zone (the part of the rock near the well whose permeability the drilling has reduced). Big-hole charges make a wide, short hole for gravel and proppant (grains that prop a fracture open). The hole must be at least about six times the diameter of the largest grain.
  • A carrier gun (the charges sit in a sealed steel tube) withstands the well pressure and takes the debris out with it. An expendable gun holds larger charges but leaves debris. A small through-tubing gun shoots best into the near wall.
  • A crushed zone and debris remain around the tunnel. In underbalanced perforating (well pressure lower than reservoir pressure), the instant inflow from the reservoir carries them out. In overbalanced perforating (well pressure higher), the well fluid is pushed into the tunnels.
Key relationships
u=v1+ρt/ρj,p=12 ρt u2u = \dfrac{v}{1 + \sqrt{\rho_t/\rho_j}},\qquad p = \tfrac12\,\rho_t\,u^2

speed at which the tunnel deepens and the pressure at the front of the jet (ideal hydrodynamic model)

P=Lρj/ρtP = L\sqrt{\rho_j/\rho_t}

density law: the depth that a continuous jet of length LL pierces

t ρ/ρrt\,\sqrt{\rho/\rho_r}

tunnel length in the rock that a layer of thickness tt and density ρ\rho “eats up”

lp=Lres−tcas−tceml_p = L_{\text{res}} - t_{\text{cas}} - t_{\text{cem}}

tunnel in the rock beyond the wall of the wellbore, from the forecast of the total tunnel length in the reservoir

dhole≳6 dgrd_{\text{hole}} \gtrsim 6\,d_{\text{gr}}

the hole through which gravel and proppant pass

Common pitfalls
  • Taking the data-sheet tunnel length in a concrete target as the tunnel length in the reservoir.
  • Choosing deep-penetrating charges for a gravel pack or for a frac-pack (fracturing with gravel). The hole is small, and the grains will get stuck.
  • Shooting a small gun in a wide casing without pressing it against the wall and aiming the charges at the near wall.
  • Perforating overbalanced in a fluid with solids and hoping that the tunnels will clean up during production.

After this lesson you can

  • explain how an explosion turns the liner of a charge into a jet and a slug, and why the jet pierces steel and rock by pressure and not by heat
  • estimate the pressure at the front of the jet and, by the density law, what part of the jet is used up by the casing, the cement and the water in the clearance (the gap between the gun and the casing)
  • compare deep-penetrating and big-hole charges, and choose a charge for an ordinary completion, a gravel pack and hydraulic fracturing
  • compare carrier guns, expendable guns and through-tubing guns, and explain the role of the clearance, the phasing (the angle between neighbouring shots) and the shot density (the number of shots per metre)
  • explain how the standard tests of charges differ from conditions in the reservoir, and go from the data sheet to the tunnel length in the rock
  • explain where the crushed zone comes from, and why an underbalance (well pressure lower than reservoir pressure) at the moment of perforating cleans the tunnels

Why we perforate, and how

In a cemented casing (bottomhole designs), the reservoir is separated from the well by steel and cement. The connection is made by perforating: tunnels (channels) through the pipe and the cement sheath into the rock. All the inflow goes through them, and acid, fracturing fluid and gravel are pumped through them. Perforating lets us choose the intervals: we open the oil-bearing part of the reservoir and leave alone the water-bearing part, or a weak part that tends to produce sand. Later we can perforate new intervals.

There are several methods. Bullet perforating, in which powder charges fire steel bullets, was the first, and in strong rocks it makes shallow tunnels. Abrasive-jet perforating cuts holes with a jet of liquid carrying sand through nozzles, and drilling perforators drill holes without any impact load on the casing and the cement. But the great majority of wells are opened with shaped charges (shaped charges): this is fast, comparatively cheap and makes deep tunnels. In Russian-language practice, perforating, together with other work with explosives in a well, is called “perforating and blasting operations”.

How a charge is built and how the jet forms

A shaped perforating charge is a charge of explosive in a case, with a thin metal liner lining a conical cavity on its front side. The charges of a gun are set off almost at the same time by a detonating cord (a cord filled with explosive that carries the detonation) that runs along the whole gun.

Shaped charge in a gun, the jet on its way to the casing, and the tunnel after the shotCharge in the guncharge caseexplosive and linercordgun bodyA few microseconds laterhead: 6–8 km/stail: 1–2 km/sslugcasing and cementTunnel after the shottunnel in the rock lpwellcasingcementcrushed zonedebrisburrnot to scale

Two diagrams. At the top left is a cross-section of a charge in the wall of the gun body: the charge case, the explosive, the conical liner, the detonating cord behind, and a thinned part of the gun wall opposite the charge. To the right of the charge is a thin long jet with an arrow. A label at its head says about 6–8 km/s, and one at its tail says about 1–2 km/s. Behind it is a short thick slug. Further on is the wall of the casing. At the bottom is a cross-section of the well wall: casing, cement and rock. A tunnel goes through them, narrowing towards its end. Around the tunnel in the rock is a ring of crushed zone, and at the end of the tunnel is debris. At the entrance to the casing there is a burr pointing inward.

SchematicTop: a charge in the body of a gun and the jet on its way to the casing (diagram, not to scale). The head of the jet is faster than its tail, so the jet stretches, and the slow slug lags behind. Bottom: the tunnel after the shot: a hole in the casing with a burr (a rough raised edge) pointing inward, a tunnel through the cement into the rock, a crushed zone around it, and debris at the end.

The detonation runs through the explosive at about 8–9 km/s and squeezes the liner towards the axis. The metal does not melt. Under a pressure of tens of gigapascals it flows like a liquid. The inner layer of the liner is squeezed forward as a thin shaped-charge jet, while the outer, more massive layer moves slowly and forms the slug. A smaller part of the metal goes into the jet, usually not more than a third. The head of the jet flies at a speed of about 6–8 km/s, and the tail is several times slower. Because of this difference the jet stretches on its way and then breaks into particles. From the detonation to the stop of the jet in the rock, only fractions of a millisecond pass.

Liners are either solid, usually of copper, or pressed from metal powders. A solid liner leaves a massive slug that can plug the tunnel, while the slug of a powder liner crumbles into small particles. The jet needs a short distance to form. This distance is given by the air gap inside the case and by the wall of the gun body. But water between the gun and the casing slows the jet down: the larger the clearance (the gap between the gun and the casing), the shorter the tunnel, especially with deep-penetrating charges.

How the jet pierces the casing, the cement and the rock

The pressure at the front of the jet is much larger than the strength of steel and rock. So, as a first approximation, strength is ignored, and the jet and the target are treated as ideal incompressible fluids (fluids that cannot be compressed and have no strength). Imagine that we ride along with the bottom of the growing tunnel. The metal of the jet then runs onto the bottom at a speed v−uv - u, and the material of the target runs onto it at a speed uu. The stagnation pressures on the two sides must be equal (the stagnation pressure is the pressure that a flow reaches when it is brought to rest).

Speed of the tunnel deepening and pressure at the front of the jet
12 ρj (v−u)2=12 ρt u2  ⇒  u=v1+ρt/ρj,p=12 ρt u2\tfrac12\,\rho_j\,(v - u)^2 = \tfrac12\,\rho_t\,u^2 \;\Rightarrow\; u = \frac{v}{1 + \sqrt{\rho_t/\rho_j}},\qquad p = \tfrac12\,\rho_t\,u^2
where:
  • vvspeed of an element of the jet, m/s
  • uuspeed at which the tunnel deepens, m/s
  • ρj\rho_jdensity of the metal of the jet, kg/m³
  • ρt\rho_tdensity of the target: steel, cement, rock, kg/m³
  • pppressure at the front of the jet, Pa

Units: SI

Holds when
  • The jet and the target are incompressible fluids with no strength (the pressure is much larger than the strength).
  • The penetration is steady: the speed of the element of the jet is constant.
  • The jet of a powder liner consists of particles, and its effective density is lower than the density of the metal.

The pressure grows as the square of the jet speed. The head presses with tens of gigapascals, and the tail, which is several times slower, with 10 to 100 times less (one to two orders of magnitude). So the strength of the target hardly matters for the head of the jet, and it becomes important for the tail.

Worked exampleImpact pressure against the strength of steel and rock
Given (assumed data)
Jet
copper, 8900 kg/m³; head 6.0 km/s, tail 1.0 km/s
Casing
steel 7850 kg/m³, yield strength 552 MPa (the stress at which steel starts to deform for good; grade N-80)
Rock
sandstone 2350 kg/m³, uniaxial compressive strength 50 MPa (the stress that crushes a rock sample squeezed in one direction)
Solution
  1. Steel, head of the jet: u=60001+7850/8900=3094.1u = \dfrac{6000}{1 + \sqrt{7850/8900}} = 3094.1 m/s, p=0.5⋅7850⋅3094.12=37.58p = 0.5\cdot7850\cdot3094.1^2 = 37.58 GPa (about 5.45 million psi), which is 68 times the yield strength.
  2. Sandstone, head of the jet: u=60001+2350/8900=3963.4u = \dfrac{6000}{1 + \sqrt{2350/8900}} = 3963.4 m/s, p=18.46p = 18.46 GPa, which is 369 times the strength.
  3. Tail of the jet, 1.0 km/s: on steel p=1.04p = 1.04 GPa (1.9 times the yield strength), on sandstone p=513p = 513 MPa (10 times the strength).
Answer: At the head of the jet the pressure is 37.58 GPa on steel and 18.46 GPa on sandstone, which is 68 and 369 times their strength. At the tail the margin falls to 1.9 and 10 times.

What it means. The head of the jet pierces the casing and the rock without “noticing” their strength. The tail works at its limit. The resistance of rock to penetration is several times its uniaxial compressive strength, and it grows when the rock is squeezed by stresses in the reservoir. For the slow parts of the jet in strong, loaded rock, this is almost enough to stop them. So strength and stress shorten the tunnel, although they do not affect the head of the jet. The model is an ideal one: it does not take into account the compressibility of the metal and the rock, or the breakup of the jet into particles. The data in this example are assumed for illustration.

Pressure at the front of the jet against its speed0246810100100010,000100,000Speed of an element of the jet, km/sPressure, MPasteelsandstoneyield strength of steelstrength of sandstone

Graph with a logarithmic vertical axis: pressure from 10 to 100,000 megapascals. The horizontal axis is the jet speed from 0 to 8 kilometres per second. Two curves grow as the square of the speed: for steel, from about 260 MPa at 0.5 km/s to 67,000 MPa at 8 km/s, and for sandstone, about half as high. Horizontal dashed lines: 552 MPa is the yield strength of steel, and 50 MPa is the strength of sandstone. The points of the example at 6 km/s are marked.

CalculatedPressure at the front of the jet against its speed, for casing steel and for sandstone (ideal hydrodynamic model, copper jet of 8900 kg/m³), and, for comparison, the yield strength of N-80 steel and the strength of the sandstone of the example. The head of the jet presses tens of times (steel) and hundreds of times (sandstone) harder than the strength of the target. At the tail the margin falls to a few times (steel) and about ten times (sandstone).

How deep will the jet pierce? Each element of it is “used up” at the bottom of the tunnel during the time while the metal runs onto it at a speed v−uv - u. During this time the bottom moves forward at a speed uu. The ratio u/(v−u)=ρj/ρtu/(v - u) = \sqrt{\rho_j/\rho_t} follows from the previous formula, and for a continuous jet of length LL it gives the density law (the derivation is in the advanced section):

Density law: depth of the tunnel from the length of the jet
P=L ρjρtP = L\,\sqrt{\frac{\rho_j}{\rho_t}}
where:
  • PPdepth of the tunnel in the target, mm
  • LLlength of the continuous jet, mm
  • ρj, ρt\rho_j,\ \rho_tdensities of the jet and of the target, kg/m³

Units: any consistent units of length and density

Holds when
  • The jet is continuous and has no strength, and all its elements penetrate in the steady regime.
  • It is an upper estimate: the stretching and breakup of the jet, the resistance of the target and water on the way reduce the depth.

The depth does not depend on the speed, only on the length of the jet and the ratio of the densities. A layer of thickness tt and density ρ\rho uses up tρ/ρjt\sqrt{\rho/\rho_j} of the length of the jet. That is, it “costs” tρ/ρrt\sqrt{\rho/\rho_r} of tunnel in a rock of density ρr\rho_r, whatever the density of the jet is.

Worked exampleHow much tunnel the casing, the cement and the water “eat up”
Given (assumed data)
Casing
177.8 × 10.36 mm, inside diameter 157.08 mm; cement sheath 25 mm
Gun
carrier gun, 86 mm, in the centre of the casing; water in the clearance
Densities, kg/m³
steel 7850, cement 1900, water 1000, sandstone 2350, copper jet 8900
Solution
  1. Clearance with water: (157.08−86)/2=35.5(157.08 - 86)/2 = 35.5 mm.
  2. Use of the length of the jet: the casing wall 10.367850/8900=9.710.36\sqrt{7850/8900} = 9.7 mm, the cement 251900/8900=11.625\sqrt{1900/8900} = 11.6 mm, the water 35.51000/8900=11.935.5\sqrt{1000/8900} = 11.9 mm.
  3. Converted into tunnel in sandstone: the casing 10.367850/2350=18.910.36\sqrt{7850/2350} = 18.9 mm, the cement 251900/2350=22.525\sqrt{1900/2350} = 22.5 mm, the water 35.51000/2350=23.235.5\sqrt{1000/2350} = 23.2 mm, in all 64.6 mm.
Answer: By the ideal theory, the casing, the cement sheath and the water in the clearance cost the tunnel about 64.6 mm. A millimetre of steel costs 1.83 mm of tunnel in sandstone, and a millimetre of water costs 0.65 mm.

What it means. Steel is the most “expensive” layer. Each extra millimetre of the wall, and even more a second casing string, shortens the tunnel by almost twice its thickness. By the theory, water is cheaper than rock, but in practice the clearance does more harm: the jet stretches and breaks in it, and the particles are slowed down. So for deep tunnels we take the gun of the largest diameter that passes safely in the casing, and a small gun is pressed against the wall. There is one more consequence. The concrete of the standard target is close to sandstone in density, so by the density law the tunnels in them would be almost the same. The difference between the data sheet and the reservoir comes from strength and stresses, not from density. The data in this example are assumed for illustration.

Deep-penetrating and big-hole charges

The same mass of explosive can give a narrow, long tunnel or a wide, short hole. The shape and the material of the liner decide which. So charges are divided into two classes.

Two classes of perforating charges
Feature
Liner
Deep-penetrating
usually pressed from powders
Big-hole
usually solid metal
Feature
Hole in the casing
Deep-penetrating
about a centimetre
Big-hole
up to 2–2.5 cm
Feature
Tunnel in the rock
Deep-penetrating
long
Big-hole
short
Feature
Purpose
Deep-penetrating
inflow through the damaged zone
Big-hole
gravel, proppant, weak sands

In the industry these are called deep penetrating (DP) and big hole (BH) charges. The actual sizes of the hole and the tunnel are in the data sheet of the charge.

For an ordinary well with natural inflow, the length of the tunnel matters: it must pass through the zone damaged by the drilling fluid and the cement (tunnels and the damaged zone). For a gravel pack and for hydraulic fracturing, the hole matters more: liquid with a high concentration of gravel or proppant goes through it at high speed. If the hole is smaller than about six diameters of the largest grain, the grains form an arch and block it, and pumping stops. For 20/40 gravel (a sieve size: the largest grains are 0.84 mm) this is about 5 mm, and for 12/20 proppant (up to 1.68 mm) about 10 mm. Big-hole charges give holes with a margin. A gravel pack inside the casing also needs a high shot density, because the whole flow goes through the tunnels filled with gravel (losses in tunnels filled with gravel).

For hydraulic fracturing, the tunnels must point in the direction of the future fracture, perpendicular to the smallest horizontal stress (orientation of the fracture). So we use a phasing of 60° or less, so that at least one tunnel is close to the needed direction, or oriented perforating. The interval for one fracture is made short, a few metres, so that the fracture starts at the chosen place. When pumping must be shared between several intervals, the opposite is done: few holes are made, and the pressure drop across them evens out the rate (limited entry).

Special low-power charges pierce only the tubing, without damaging the casing behind it. They are used to connect the tubing with the annulus if there is no sliding sleeve (a circulation valve) or if it does not open (circulation valves).

Perforating guns: carrier, expendable and through-tubing

The charges are assembled in a perforating gun, a device that holds them at the needed depth and in the needed direction and protects them until the shot.

Carrier gun (hollow carrier)
The charges sit inside a sealed steel tube at atmospheric pressure. The carrier takes the pressure and the temperature of the well, protects the charges from the liquid, and after the shot takes most of the debris out of the well. The wall of the carrier is thinned opposite the charges, with scallops (recesses), so that the jet spends less energy and the burrs stay in the recess, and the carrier can be pulled out freely. After the shot the carrier swells a little, and this is allowed for when we check that the gun will pass through the casing or tubing. Carriers can be reusable (the carrier is reloaded) or single-use.
Expendable gun (exposed)
Each charge is in its own sealed capsule, and the charges are fixed on a strip, a wire or a rod. With no carrier, a larger charge fits in the same outside size, and the tunnels are longer. But the charges and the detonating cord are exposed to the pressure, the temperature and the liquid, the remains of the capsules and the strip stay in the well, and the blast hits the casing directly.
Through-tubing gun
A gun of small diameter, from about 35 to 90 mm depending on the tubing bore. It passes through the tubing string and shoots into the casing below its end. It lets us open the reservoir when the tubing string is already run and the wellhead is installed, including underbalanced. The price is small charges, short tunnels and a large clearance to the far wall.
Casing gun
A carrier gun of the largest diameter that passes in the casing, usually from 86 to 127 mm (3⅜–5 in), and larger in wide casings. It is run before the tubing string is installed, on a cable or on pipe (Perforating operations: gun conveyance and procedures). It has large charges, any phasing, and a shot density of up to several tens of holes per metre.
Guns of three diameters in a 177.8 mm casing: position and directions of the shotsThrough-tubing, 43 mm0°, to the near wallto the far wall 114 mmCarrier 86 mm90°, centredclearance 35.5 mmCarrier 114 mm60°, centredclearance 21.5 mm

Three circles of the same size: casing with cement outside. In the first, a small circle of the gun is pressed against the wall at the bottom of the picture. One jet arrow points down into the near wall, and a dashed arrow up through a wide layer of water shows the clearance to the far wall, 114 mm. In the second, a medium-sized circle is in the centre, with four arrows in a cross through a clearance of 35.5 mm. In the third, a large circle is in the centre, with six arrows through a narrow clearance of about 21.5 mm.

SchematicGuns in a 177.8 mm casing (157 mm inside), cross-section, diagram. A small 43 mm gun is pressed against the wall and shoots with 0° phasing into the near wall. An 86 mm carrier gun in the centre has 90° phasing and a clearance of 35.5 mm all round. A 114 mm carrier gun has 60° phasing and a clearance of about 21.5 mm.

A gun in the casing is seldom exactly in the centre. A small gun in a wide casing ends up at one of the walls (the low side in a deviated well). The clearance to the near wall is almost zero, and to the far wall there is more than 10 cm of water. So such guns usually shoot with 0° phasing and are turned, by a magnetic or eccentric device that presses the gun against the wall, so that the charges face the wall. A large gun leaves a small clearance all round, and any phasing suits it.

Explosives slowly decompose when heated, so the allowed temperature depends on how long the charges will stay in the well: hours when run on a cable, a day or more when run on pipe. For hot wells, heat-resistant explosives are used, and for each charge the manufacturer states the allowed combination of temperature and time.

Shot density and phasing

The shot density nsn_s is the number of holes per metre, from one or two up to several tens. English-language data sheets give shots per foot (spf; 1 shot per foot ≈ 3.28 shots per metre). The phasing is the angle between the directions of neighbouring charges: 0° means all in one direction, 180° in two opposite directions, and 120°, 90°, 60° or 45° in a spiral.

Layout of the tunnels at a shot density of 12 holes per metre
Phasing
0°
Directions
1
Between shots, m
0.083
Between tunnels of the same direction, m
0.083
Phasing
180°
Directions
2
Between shots, m
0.083
Between tunnels of the same direction, m
0.167
Phasing
120°
Directions
3
Between shots, m
0.083
Between tunnels of the same direction, m
0.250
Phasing
90°
Directions
4
Between shots, m
0.083
Between tunnels of the same direction, m
0.333
Phasing
60°
Directions
6
Between shots, m
0.083
Between tunnels of the same direction, m
0.500

The vertical distance between shots is h=1/nsh = 1/n_s. Tunnels of the same direction repeat every 360∘/θ360^\circ/\theta shots, where θ\theta is the phasing angle.

The shot density and the phasing decide the perforation skin (calculation by the method of components), the flow speed in each tunnel (and so the sand production and the inertial losses, which are the extra pressure losses at high flow speed), and the strength of the casing. Dense perforating with a phasing of 45–90° comes closest to an open hole. A phasing of 0° makes the flow go around the casing and gives a larger skin, but for a gun pressed to the wall it gives the longest tunnels. Low vertical permeability and thin layers call for a higher shot density: a thin layer between two shots otherwise hardly works.

Each hole weakens the casing, and the burrs inside get in the way of running packers (seals for the well) and tools. At ordinary shot densities and hole diameters in a well-cemented casing, the strength falls only moderately. But very dense perforating with large holes, thin-walled pipes and poor cement can lead to cracks in the casing. The allowed combinations are set by the gun manufacturer and by calculation.

The data sheet of the charge and the tunnel in the reservoir

Manufacturers compare charges by the standard method API RP 19B (this role was earlier played by the document API RP 43). It includes:

  • firing a section of a gun into a casing pipe cast in concrete in a barrel, at surface conditions: the diameter of the holes in the pipe, the total length of the tunnels and the height of the burrs are measured, and this is the “data-sheet” tunnel length (penetration)
  • a single shot into a sandstone sample squeezed by pressure as in the reservoir, followed by pumping fluid through the tunnel to estimate its productivity
  • firing after the charges have been held at a high temperature
  • measuring the debris that the gun leaves in the well, and the swelling of the carrier

A concrete target is convenient for comparing charges with each other, but it does not predict the tunnel in the reservoir. The tunnel length in the reservoir becomes smaller when:

  • the casing is stronger and thicker, because more of the jet goes into the steel
  • the rock is stronger: the uniaxial compressive strength of different reservoirs differs by tens of times (rock strength)
  • the effective stress is larger (the difference between the overburden pressure, which is the weight of the rock above, and the pore pressure, which is the fluid pressure in the pores; it grows as the reservoir pressure falls during production)
  • the clearance between the gun and the casing is larger
  • the charges have lost part of their power from a long stay at high temperature

So the tunnel length for the inflow calculation is taken not from the catalogue but from a forecast for the particular reservoir. The forecast comes from tests on rock samples under stress and from manufacturers' programs that convert the data sheet for the rock strength, the stresses, the casing and the clearance. The length in the data sheet and in the forecast is usually counted from the inside wall of the casing, but the skin calculation needs the length of the tunnel in the rock beyond the wall of the wellbore, lpl_p. So the casing wall and the cement sheath are subtracted from it.

Worked exampleFrom the charge data sheet to the tunnel for the inflow calculation
Given (assumed data)
Gun
12 holes per metre, 90° phasing; by the data sheet in a concrete target, the tunnel is 640 mm and the hole in the pipe is 11 mm
Forecast for the reservoir
335 mm from the inside wall of the 177.8 × 10.36 mm casing; the diameter of the tunnel in the rock is about 10 mm
Cement sheath and reservoir
sheath 25 mm; kH/kVk_H/k_V = 4 (the ratio of horizontal to vertical permeability), drainage area 300 m
Solution
  1. Tunnel in the rock: lp=335−10.36−25=300l_p = 335 - 10.36 - 25 = 300 mm. The total length in the reservoir is 52% of the data-sheet length.
  2. Layout: between shots 1/12=0.08331/12 = 0.0833 m, four directions, and tunnels of the same direction every 0.333 m.
  3. Similar tunnels (0.30 m long, 5 mm radius, 12 holes per metre, 90°) are worked out in the lesson Effect of completion on inflow; in a 216 mm bore their skin is 0.26. Here the bore is wider (177.8 mm casing with a 25 mm sheath, rw=88.9+25=113.9r_w = 88.9 + 25 = 113.9 mm), and the skin by the method of components is 0.30.
  4. If we took the data-sheet length less the casing and the cement, lp=640−35.36=605l_p = 640 - 35.36 = 605 mm, the skin would be −0.78.
Answer: The inflow calculation uses a tunnel of about 300 mm in the rock, not 640 mm. The perforation skin is 0.30, not −0.78.

What it means. The error is more than one unit of skin: with the data-sheet length, the productivity would come out about 17% higher than with the forecast. And near a damaged zone it becomes decisive. Tunnels that pass through the zone completely by the data sheet may end inside it in the reservoir, and then the skin grows many times over (tunnels in the damaged zone). The forecast figure is also assumed: it is checked by well tests after perforating. The data in this example are assumed for illustration.

The crushed zone and cleaning the tunnels

The jet pushes the rock aside; it does not remove it. Around the tunnel a ring of crushed and compacted grains is left, about a centimetre thick, with a permeability several times lower than that of the reservoir. This is the crushed zone. The tunnel is left with debris of rock, particles of the liner and of the charge case, and, with a solid liner, the slug. How much this lowers the productivity is calculated in the lesson skin of the crushed zone: the crushed zone can easily give a larger skin than all the geometry of the perforations.

If at the moment of the shot the pressure in the well is higher than the reservoir pressure (overbalanced perforating), the well fluid, together with the solid particles of the drilling fluid, at once rushes into the fresh tunnels and plugs them together with the debris. Later, during production, the flow in each tunnel is small, and the plugged tunnels never clean up. Only some of them work.

In underbalanced perforating, the pressure in the well is lower than the reservoir pressure. In the first fractions of a second after the shot, the reservoir gives the tunnel a short, strong pulse of inflow. The pressure difference across the loose, crushed rock and the force of the flow tear off the debris and part of the crushed zone and carry them into the well. It is this instant pulse that does the work, not the production afterwards. The lower the permeability of the reservoir, the larger the pressure difference needed for cleaning. A similar pulse can be created without a permanent underbalance, from empty spaces in the gun that fill with liquid at once after the shot and for a moment lower the pressure at the tunnels (dynamic underbalance).

The underbalance cannot be as large as we like. In weak rocks it destroys the walls of the tunnels and causes sand production (Sand production: types, prediction and cost of the problem), and too large a pressure difference is also dangerous for the casing. How the underbalance is chosen, and how to perforate underbalanced when the gun is run on a cable or on pipe, is the subject of Perforating operations: gun conveyance and procedures.

Terms and symbols

Shaped charge
A case, a charge of explosive and a metal liner in a conical cavity. The explosion turns the liner into a jet.
Shaped-charge jet and slug
The fast, thin jet from the inner layer of the liner, and the slow, massive part from the outer layer.
Density law
P=Lρj/ρtP = L\sqrt{\rho_j/\rho_t}: the depth that a continuous jet pierces in a target with no strength.
Perforating gun
A carrier, expendable or small through-tubing gun. It holds the charges at the depth and in the direction needed.
Tunnel length (penetration) and hole diameter
The main characteristics of a charge. By the data sheet they are for a concrete target, and in the reservoir the tunnel is shorter.
Underbalanced perforating
Perforating with the well pressure below the reservoir pressure. The pulse of inflow cleans the tunnels.

Common mistakes and misconceptions

Wrong

“The jet burns through the casing and the rock.”

In fact

The target is destroyed by the pressure of impact, tens of gigapascals. The metal and the rock do not melt but flow aside and get compacted. So a crushed zone is left around the tunnel, not a melted crust.

Wrong

“The tunnel length in the reservoir is the figure from the catalogue.”

In fact

The catalogue gives the length in a concrete target at the surface. In the reservoir the tunnel is usually shorter: the strength of the rock and the stresses in it, the thickness and strength of the casing, and the clearance all get in the way. For the calculation we take a forecast for the reservoir and subtract the casing wall and the cement.

Wrong

“The deeper the tunnel, the better, for any well.”

In fact

For a gravel pack and for hydraulic fracturing, the diameter and the number of holes matter more: grains get stuck in a hole smaller than about six of their diameters.

Wrong

“The underbalance in perforating should be as large as possible.”

In fact

A sufficient underbalance cleans the tunnels, but an excessive one destroys them in weak rocks and causes sand production, and a large pressure difference is dangerous for the casing. It is chosen from the permeability and the strength of the rock.

Wrong

“Phasing does not matter, as long as there are many holes.”

In fact

With 0° the flow goes around the casing and the skin is larger. With 60–90° the tunnels surround the wellbore. A small gun lying on the wall needs 0° with the charges aimed at the near wall, and a large gun needs 60–90°.

In depth: advanced detail

Where the density law comes from and where it stops working

From the equality of the stagnation pressures 12ρj(v−u)2=12ρtu2\tfrac12\rho_j(v - u)^2 = \tfrac12\rho_t u^2 it follows that u/(v−u)=ρj/ρtu/(v - u) = \sqrt{\rho_j/\rho_t}. An element of the jet of length  dL\dd L is used up at the bottom of the tunnel in a time  dt= dL/(v−u)\dd t = \dd L/(v - u), and in this time the bottom moves by  dP=u  dt= dL ρj/ρt\dd P = u\,\dd t = \dd L\,\sqrt{\rho_j/\rho_t}. Integrating over the whole length of the jet, we get P=Lρj/ρtP = L\sqrt{\rho_j/\rho_t}. The jet speed has cancelled out.

A real jet differs from the ideal one in three ways. First, its speed changes along its length (a speed gradient). It stretches and breaks into particles, and particles penetrate worse than a continuous jet. This is why there is an optimum distance to the target, and why water in the clearance does harm. Second, the target has strength. The resistance of the target to penetration adds to its stagnation pressure, and the slow parts of the jet stop without deepening the tunnel. The stronger the rock, and the more it is squeezed, the earlier this happens. Third, metal and rock are compressible, and at pressures of tens of gigapascals this is noticeable. So the density law explains well why steel “costs” more than rock and why a dense liner pierces deeper. But the tunnel length in the reservoir is predicted from tests, not from this law.

Connections to other topics

Check yourself

  1. 1

    A well is completed with a frac-pack (hydraulic fracturing with a gravel pack), with 12/20 proppant (the largest grains are 1.68 mm). Which charge should be chosen?

    1. A.Any: the proppant will pass through any hole
    2. B.Big-hole, with a hole of about 20 mm and a high shot density
    3. C.Deep-penetrating, with an 8 mm hole: the tunnel is longer
    4. D.Deep-penetrating, with 0° phasing
    Show the answer

    Correct answer: B. Big-hole, with a hole of about 20 mm and a high shot density

    The hole must be at least about six diameters of the largest grain: 6⋅1.68≈106\cdot1.68 \approx 10 mm. An 8 mm hole risks being blocked by an arch of grains. The tunnel length hardly matters here, because the fracture and the gravel bypass the damaged zone anyway, and a high shot density is needed so that the flow through the tunnels filled with gravel does not lose extra pressure.

  2. 2

    A 177.8 × 10.36 mm casing is perforated where another casing string with an 11.05 mm wall runs behind it. How much tunnel in sandstone (2350 kg/m³) will the second wall “eat up” by the density law?

    Show the answer

    11.057850/2350=11.05⋅1.83≈2011.05\sqrt{7850/2350} = 11.05\cdot1.83 \approx 20 mm, plus the cement between the strings. In practice it is more: by this point the jet is already stretched and has partly broken up, and such jets pierce worse than the ideal theory says. So through two casing strings we perforate with the most powerful charges, and where possible we avoid it.

  3. 3

    A small 43 mm gun is run through the tubing into a 177.8 mm casing. How is it best positioned, and with what phasing should it shoot?

    1. A.Press it against the wall and shoot with 0° phasing into the near wall
    2. B.In the centre of the casing, with 60° phasing
    3. C.The position does not matter
    4. D.Press it against the wall and shoot into the far wall, where there is more space
    Show the answer

    Correct answer: A. Press it against the wall and shoot with 0° phasing into the near wall

    A small charge has a short jet, and the water in the clearance quickly eats it up: there is more than 10 cm of water to the far wall. The gun is pressed against the wall (with a magnetic or eccentric device) and shoots into the near wall, where the clearance is almost zero.

  4. 4

    Why does underbalanced perforating give cleaner tunnels than overbalanced perforating followed by production?

    1. A.The tunnels are longer underbalanced
    2. B.At the moment of the shot, the pulse of inflow from the reservoir carries out debris and part of the crushed zone. Overbalanced, well fluid with solid particles is pushed into the tunnels
    3. C.The jet is hotter underbalanced
    4. D.There is no difference, because the tunnels clean up in the same way during production
    Show the answer

    Correct answer: B. At the moment of the shot, the pulse of inflow from the reservoir carries out debris and part of the crushed zone. Overbalanced, well fluid with solid particles is pushed into the tunnels

    What cleans is the instant, strong inflow in the first fractions of a second. During later production the flow in each tunnel is small and does not clean the plugged tunnels, and overbalanced they are plugged by particles from the well fluid right after the shot.

  5. 5

    The data sheet of a charge says 900 mm in a concrete target. Can we assume that in the reservoir the tunnel will be 0.85 m beyond the wall of the wellbore?

    Show the answer

    No. The data sheet was obtained in concrete at the surface, with no stresses. In the reservoir the rock may be stronger than concrete and squeezed by effective stress, the casing may be thicker and stronger, and the clearance larger. The tail of the jet hardly works in strong, loaded rock. The length is taken from a forecast for the reservoir, with the casing wall and the cement subtracted, and it is then checked by well tests.

  6. 6

    A gun has 16 holes per metre with 60° phasing. Find the distance between shots and between tunnels of the same direction. What changes with 0° phasing?

    Show the answer

    Between shots it is 1/16=0.06251/16 = 0.0625 m. At 60° there are six directions, and tunnels of the same direction repeat every 6⋅0.0625=0.3756\cdot0.0625 = 0.375 m. At 0° all the tunnels point in one direction, every 0.0625 m. The flow to the tunnels along the vertical is good, but it goes around the casing, and the skin is larger.

Updated: 19 October 2026