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ProductionFormation damage and stimulation Intermediate 29 min read

Sources of formation damage

Sources of formation damage are the operations and processes that, during the life of a well, reduce the permeability of the near-wellbore zone (the rock right around the well). They are drilling and opening the reservoir (drilling into the reservoir rock), cementing, perforating (shooting holes through the casing into the rock), production, injection, well killing (stopping the well with a heavy fluid) and workovers (repair jobs), and treatments. Each source has its own set of mechanisms, its own depth of damage, and its own moment when the damage shows up. So each has its own way to prevent the damage or to remove it.

Why it matters

To remove damage, we need to know what exactly has blocked the pores, and where. Acid, solvent, surfactants (surface-active agents) and hydraulic fracturing (cracking the rock with fluid pumped at very high pressure) each work on something different. To avoid creating the damage again, we need to know which operation causes it. This lesson goes through the sources of damage operation by operation. It teaches you to turn laboratory tests into a skin (a number for the extra resistance to flow near the well), to estimate the rate at which fine particles start to move, and to judge how likely a mixture of waters is to form scale (mineral deposits). It also shows how to trace the source of damage from the history of the well.

Quick refresher

  • Almost any operation can damage the reservoir, and each leaves its own “signature”. Drilling leaves a cake (a layer of mud solids on the wall of the hole) and a zone invaded by filtrate (the liquid that leaks out of the mud into the rock), from centimetres to tens of centimetres deep. Cementing leaves an alkaline filtrate (the opposite of acidic). Perforating leaves a compacted zone and debris in the tunnels (the channels that the perforation charges make in the rock). Production leaves fine particles, salts, paraffins and asphaltenes that build up with time. Injection leaves suspended solids and incompatible waters. A workover leaves the solids of the kill fluid in the tunnels.
  • The return permeability measured on core is an average over the length of the sample. If the loss is concentrated in the first centimetre, the local permeability there is many times lower. The skin is then calculated by rings: a thin, badly damaged zone plus the filtrate zone.
  • Fine particles are torn loose when the velocity of the phase that wets them, usually water, exceeds a critical value. In a perforated well the flow enters through the walls of the tunnels, whose area can be several times smaller than the wall of an open hole. So the critical rate is lower. Oil with no water hardly moves the particles.
  • The saturation index is SI=lg⁡(IAP/Ksp)\mathrm{SI} = \lg(\text{IAP}/K_{sp}), where IAP is the ion activity product (the product of the “effective” concentrations of the ions) and KspK_{sp} is the solubility product (the equilibrium constant of dissolving the salt). When SI > 0 the solution is supersaturated (it holds more salt than it can keep dissolved), and the salt can come out. When barium-rich formation water mixes with sulfate-rich injected water, the index is highest at equal shares. The mass of precipitate is usually highest at a small share of injected water, that is, just after the injected water breaks through.
  • Clean fluids matter a great deal. A few cubic metres of “dirty” kill fluid bring kilograms of solids into the tunnels, a volume comparable with the volume of the tunnels themselves.
  • We find the source of damage from a set of signs: when productivity fell and after which operation, what the deposits, waters and core showed, and whether the decline grows with the rate.
Key relationships
LR=∑iLiRi\dfrac{L}{R} = \sum_i \dfrac{L_i}{R_i}

return permeability of a sample made of sections that work in series

Sd=∑i(1Ri−1)ln⁡riri−1S_d = \sum_i \left(\dfrac{1}{R_i} - 1\right)\ln\dfrac{r_i}{r_{i-1}}

damage skin by rings with return coefficients Ri=ki/kR_i = k_i/k, r0=rwr_0 = r_w

qcr=ucr Ainq_{\text{cr}} = u_{\text{cr}}\,A_{\text{in}}

critical water rate per metre: the inflow area is 2πrw2\pi r_w in an open hole, or nπdln\pi d l for perforation tunnels

SI=lg⁡γBa[Ba2+] γSO4[SO42−]Ksp\mathrm{SI} = \lg\dfrac{\gamma_{\text{Ba}}[\mathrm{Ba^{2+}}]\,\gamma_{\text{SO}_4}[\mathrm{SO_4^{2-}}]}{K_{sp}}

saturation index of barium sulfate

Common pitfalls
  • Applying the return permeability measured on a short sample to the whole zone invaded by filtrate.
  • Judging the danger of fines migration from the oil rate. The velocity of the water matters, and after water breakthrough the same liquid rate can become “supercritical”.
  • Assuming that SI > 0 means that salt comes out at once, or, the other way round, that if there are no deposits in the pipes there are none in the reservoir.
  • Killing a well with unfiltered fluid from unwashed tanks, and pumping treatment fluids through old tubing without cleaning it first.
  • Taking for damage something that is not damage: depletion, gas coming out of solution near the wellbore, partial penetration.

After this lesson you can

  • list, for drilling, cementing, perforating, production, injection, well killing and workovers, the main damage mechanisms, the typical depth of damage and its signs
  • convert the return permeability measured on core (rock samples taken from the well) into a damage skin, taking into account where in the sample the loss is concentrated and how deep the filtrate has gone into the reservoir
  • estimate the critical rate at which fine particles start to move (fines migration), for an open hole and for perforated casing, and explain why the movement often starts after water breakthrough (the moment water first reaches the well)
  • calculate the saturation index of barium sulfate (a number that shows how likely a salt is to come out of water) and the mass of the precipitate (the solid that settles out) that can form when formation water and injected water mix
  • estimate how much solid material a kill fluid brings into the perforation tunnels, and choose measures to prevent it
  • build a diagnosis from the history of the well, laboratory tests, and analyses of deposits and waters, and choose preventive measures for each operation

Damage at every stage of the life of a well

The mechanisms by which rock loses permeability are covered in the lesson on formation damage. They are plugging of channels by particles, swelling and migration of clays, deposits of salts and organic matter, and water and emulsion blocks. That lesson also shows how damage is expressed as a skin. This lesson answers a different question: where and when damage happens. The answer decides how to treat it and how not to create it again.

Sources of damage by operation
Operation
Drilling and opening the reservoir
Main mechanisms
particles of drilling mud and cuttings (chips of drilled rock); filtrate: swelling and migration of clays, water and emulsion blocks, deposits
Depth
cake: millimetres and centimetres; filtrate: centimetres and tens of centimetres
How it shows
a large skin from the first test
Operation
Cementing
Main mechanisms
alkaline filtrate with calcium; cement in fractures
Depth
centimetres; deeper in fractures
How it shows
usually bypassed by the perforation tunnels
Operation
Perforating
Main mechanisms
compacted zone, debris in the tunnels, wellbore fluid in the tunnels under overbalance (well pressure above reservoir pressure)
Depth
about a centimetre around a tunnel
How it shows
completion skin, tunnels that do not work
Operation
Production
Main mechanisms
fine particles, salts, paraffins and asphaltenes, sand, bacteria
Depth
near the wellbore; salts and paraffin deeper
How it shows
a gradual fall in productivity, often after water breakthrough
Operation
Injection
Main mechanisms
suspended solids, oil, corrosion products, bacteria; incompatible waters
Depth
the walls of the wellbore and the tunnels, then deeper
How it shows
a fall in injectivity
Operation
Well killing and workovers
Main mechanisms
solids and polymers of the kill fluid, debris from the pipes, incompatible brines (salt-water fluids)
Depth
the tunnels and the first centimetres
How it shows
productivity after the workover is lower than before it
Operation
Treatments
Main mechanisms
secondary precipitates (new deposits formed by the reaction products), emulsions, sludge (a thick, sticky mass), destruction of the rock's own cement (the mineral that binds the grains)
Depth
tens of centimetres
How it shows
the treatment did not help or did harm

The depths are orders of magnitude, for guidance only. In practice they depend on the rock, the fluids and the way the work is done.

Two practical conclusions follow from the table. First, damage from different operations adds up. A well can carry the cake from drilling, the compacted zone around the tunnels, and salts that came out after water breakthrough. So we break the measured skin down not only into its components (Skin factor) but also into sources. Second, the depth of the damage decides whether the perforation will bypass it (Formation damage and damage skin) and how much fluid we need to remove it (Acid systems and treatment design).

Drilling and opening the reservoir: particles, filtrate and sensitive minerals

While the reservoir is being opened, the pressure of the drilling fluid is higher than the reservoir pressure. Large particles of mud and cuttings stop at the wall and build an external cake. Smaller particles enter the pores and form an internal cake, from millimetres to a few centimetres deep. This plugging of pores by particles is called bridging. Particles that are added to the mud on purpose, to seal the pores quickly, are called bridging agents. Which particles go where depends on their size compared with the pore throats (the narrow channels between the pores) (the rule for choosing bridging agents). In fractures and in very permeable rock, the cake has no time to form, and the whole drilling fluid goes into the reservoir.

Filtrate flows through the cake into the reservoir. Its volume depends on the properties of the cake, the overbalance (the amount by which the well pressure exceeds the reservoir pressure), and the time the hole stays open. Most of it goes in during circulation (while the drilling fluid is pumped round the well), when the flow erodes the cake (Borehole environment: invaded zone and mudcake). The depth of invasion depends on porosity (the share of the rock that is pores). The same volume goes further in tight rock (Borehole environment: invaded zone and mudcake). The filtrate does harm in several ways:

  • Swelling of clays. Smectite (a clay that swells in water) and mixed-layer minerals (clays built of layers of different kinds) take in water between their layers and narrow the channels. They do this most with fresh water, and when sodium is the main exchangeable cation (the ion held loosely on the clay surface) (hydration and swelling of clays).
  • Detachment and transport of clay particles (clay dispersion). When the salinity of the water in the pores (how salty it is) falls below a critical value, electrical repulsion tears particles of kaolinite and illite (two common clay minerals) away from the grains, and the flow carries them to the narrow channels. This “salt shock” is dangerous even where there is no smectite. An alkaline filtrate (high pH, the opposite of acid) makes the detachment stronger.
  • Water block. The filtrate raises the water saturation (the share of the pore space filled with water) near the wellbore, and the phase permeability to oil and gas (how easily they flow when water is also in the pores) falls. In tight and gas reservoirs, capillary forces (the forces that hold liquid in tiny pores) hold the water for months.
  • Emulsions and a change of wettability. Surfactants in oil-based muds (Oil-based muds), and the mixing of filtrate with oil, give thick emulsions (mixtures of tiny droplets of one liquid in another). Surfactants that stick to the grains make the rock hydrophobic (water-repelling).
  • Deposits. The filtrate may be incompatible with the formation water. Carbonates and sulfates from the filtrate, together with calcium, barium and strontium from the formation water, form insoluble salts.
  • Polymers. The large molecules of thickeners stick to the pore walls and stay in the channels as films and clumps.
Rock sensitivity: what upsets it and how we test it
What changes
The flow velocity rises
What happens
particles break away and block the channels
What is sensitive
kaolinite, illite, small fragments
How it is tested on core
in steps of flow rate, to find the critical velocity
What changes
The salinity of the water falls
What happens
swelling, detachment of particles
What is sensitive
smectite, mixed-layer clays, kaolinite
How it is tested on core
in steps of salinity, to find the critical salt concentration
What changes
The pH rises (cement, alkali)
What happens
detachment of particles, dissolving of silica (the material of quartz grains), deposits
What is sensitive
clays, quartz dust
How it is tested on core
by pumping an alkaline solution
What changes
Acid
What happens
minerals dissolve out of the rock, iron goes into solution, the rock's own cement is weakened
What is sensitive
chlorite, zeolites, carbonate cement
How it is tested on core
by pumping acid and analysing the spent acid

The tests are run at reservoir temperature, and the permeability is compared before and after each step. We find out beforehand which minerals the rock contains and how the clays sit in it (clay minerals of reservoirs).

We prevent this damage by the choice of drilling fluid and of the way the reservoir is opened. Keep the overbalance small, within the limits that the well control rules allow. Keep the time the reservoir is open short. Use a drilling fluid for opening the reservoir with bridging agents that dissolve in acid or water. Use a filtrate that is compatible with the clays and the formation water: one with potassium or ammonium, and with enough salinity (inhibitive muds, which keep clays from swelling). Where the stability of the hole and well control allow it, drill underbalanced (with the well pressure below the reservoir pressure). The fluid is chosen together with the design of the well bottom (Formation damage and damage skin).

Laboratory tests: from core to skin

Laboratory tests answer two questions: will this fluid damage the reservoir, and by how much? The main one is the test of return permeability. First, reservoir fluid is pumped through a core sample. Then the sample is exposed to the fluid under test. After that, reservoir fluid is pumped through it again, in the direction of inflow.

How the test is run (in outline)
  1. 1
    Sample

    A core of the reservoir, prepared to reservoir conditions: saturation, confining stress (the squeeze of the surrounding rock), temperature.

  2. 2
    Initial permeability

    Measured with reservoir oil or water, in the direction of inflow to the well.

  3. 3
    Exposure

    The fluid under test is applied to the end face of the sample under overbalance, as in the well. It is either static or flowing along the face. The time is comparable with the time in the well.

  4. 4
    Flowing back

    Reservoir fluid is pumped again in the direction of inflow, in steps of flow rate. We also record the pressure at which the cake begins to lift off.

  5. 5
    Result and analysis

    The ratio of the final permeability to the initial one is the return permeability. Pressure sensors along the sample show in which section the loss is. The sample is cut and studied under a microscope.

The return permeability is an average over the length of the sample, but the damage usually sits at the end face that the fluid acted on. The sections of the sample work in series, like layers when flow crosses the bedding (the layering of the rock) (Rock permeability). So the overall coefficient is their harmonic mean (an average in which the lowest values count most), weighted by length:

Return permeability of a sample made of several sections
LR=∑iLiRi,Ri=kiafterkibefore\frac{L}{R} = \sum_i \frac{L_i}{R_i},\qquad R_i = \frac{k_i^{\text{after}}}{k_i^{\text{before}}}
where:
  • RRreturn permeability of the whole sample, fraction
  • L, LiL,\ L_ilength of the sample and of the ii-th section between the pressure sensors, mm
  • RiR_ireturn permeability of the ii-th section, fraction
  • kibefore, kiafterk_i^{\text{before}},\ k_i^{\text{after}}permeability of the section before and after exposure to the fluid, mD

Units: any consistent units: lengths in the same units

Holds when
  • The flow is one-dimensional and the sections work in series.
  • The initial permeability is the same along the whole sample. Otherwise the sections are weighted not by the length LiL_i but by Li/kibeforeL_i/k_i^{\text{before}}.
  • The permeability before and after is measured with the same fluid at the same saturation.

A short, badly damaged section gives far more than its fair share of the resistance, but in the overall coefficient it is “diluted” by the length of the sample. So a single overall number cannot tell us where the damage is or how severe it is. To convert it into a skin, we need the sections.

Worked exampleDamage skin from a return permeability test
Given (assumed data)
Core sample
length 50 mm; a pressure sensor 10 mm from the end face that the drilling fluid acted on
Return permeability by section
first 10 mm: 0.05; remaining 40 mm: 0.88
Well (open hole)
rw=0.108r_w = 0.108 m, re=300r_e = 300 m, porosity 0.20
Filtrate that entered the reservoir while it was open
35 L per metre of thickness; it fills 75% of the pores (Sxo=0.75S_{xo} = 0.75)
Internal cake in the reservoir
to the same depth as in the sample: 10 mm beyond the wall
Solution
  1. The sample as a whole: R=5010/0.05+40/0.88=50200+45.45=0.204R = \dfrac{50}{10/0.05 + 40/0.88} = \dfrac{50}{200 + 45.45} = 0.204. In a report this reads “return of 20%”.
  2. The radius of the filtrate zone from a volume balance (Borehole environment: invaded zone and mudcake): rinv=0.1082+0.035π⋅0.20⋅0.75=0.2931r_{inv} = \sqrt{0.108^2 + \dfrac{0.035}{\pi\cdot 0.20\cdot 0.75}} = 0.2931 m. The filtrate went 18.5 cm beyond the wall.
  3. Skin of the cake, the ring from 0.108 to 0.118 m: S1=(10.05−1)ln⁡0.1180.108=19⋅0.0886=1.683S_1 = \left(\dfrac{1}{0.05} - 1\right)\ln\dfrac{0.118}{0.108} = 19\cdot 0.0886 = 1.683.
  4. Skin of the filtrate zone, the ring out to 0.2931 m: S2=(10.88−1)ln⁡0.29310.118=0.136⋅0.910=0.124S_2 = \left(\dfrac{1}{0.88} - 1\right)\ln\dfrac{0.2931}{0.118} = 0.136\cdot 0.910 = 0.124.
  5. The damage skin is Sd=1.683+0.124=1.81S_d = 1.683 + 0.124 = 1.81. The share of the rate that is kept (Formation damage and damage skin): X=ln⁡3000.108−0.75=7.179X = \ln\dfrac{300}{0.108} - 0.75 = 7.179, XX+Sd=0.80\dfrac{X}{X + S_d} = 0.80.
  6. For comparison, here are two “naive” readings of the overall coefficient. Applying 0.204 to the whole invaded zone gives (10.204−1)ln⁡0.29310.108=3.9\left(\dfrac{1}{0.204} - 1\right)\ln\dfrac{0.2931}{0.108} = 3.9. Applying it to the first 10 mm only gives 0.35.
Answer: The damage skin is about 1.81. A well completed in open hole gives about 80% of the possible rate. The naive readings of the same test give 3.9 and 0.35.

What it means. “Return of 20%” sounds like a verdict, but almost all of the loss sits in the first centimetre, where the permeability fell 20 times. Beyond it, the filtrate reduced the permeability by only 12%. The skin turned out moderate, and the error of the “naive” readings is more than a factor of two in both directions. If the well is cased and perforated, tunnels 0.3 m long would pass through both rings (Formation damage and damage skin). We took the depth of the internal cake in the reservoir to be the same as in the sample, because it is set by the sizes of the particles and the channels, not by the volume of filtrate. We check the depth of the filtrate with resistivity logs (records of how strongly the rock resists an electric current) that read to different depths into the rock. The data in this example are assumed for illustration.

Besides return permeability, we run compatibility tests. We mix the fluids with formation water and oil in different proportions at reservoir temperature. We look for a precipitate, an emulsion or a thick sludge. Tests with oil need a fresh sample whose properties have not changed in storage. We also test the rock for sensitivity to velocity, salinity, pH and acid (the table in the previous section). We find the mineral composition of the rock, and the place of the clays in the pores, by X-ray analysis, in thin sections (thin slices of rock examined under a microscope) and under a scanning electron microscope.

Cementing and perforating

Before cementing, flushes and spacers (fluids pumped ahead of the cement) wash off the mudcake. Without this the cement would not bond well with the rock (Primary cementing). Without the cake, the filtrate of the cement slurry goes into the reservoir more freely. It is strongly alkaline (pH above 12) and rich in calcium. It disperses clays and dissolves part of the silica, and when it meets formation water, calcium carbonate and hydroxide come out of it. Additives limit the fluid loss of the cement slurry (how much liquid it loses into the rock) (Oilwell cements and additives), first of all for the sake of the cementing itself. So there is little filtrate, and the perforation tunnels usually pass through the zone it damages.

Two situations are more dangerous. In a fractured reservoir, the cement slurry goes into the fractures and blocks the main paths of inflow. Such reservoirs are often completed with an open hole or with an uncemented liner (a shorter casing string hung inside the one above) (Fractures and naturally fractured reservoirs). Opposite caverns (enlarged sections of the hole) the cement ring is thick, the tunnels lose length in the cement, and some of them do not reach the rock (Shaped-charge perforating).

Perforating creates a compacted zone around each tunnel, with fragments of rock and remains of the charge. How much it adds to the skin is calculated in the lesson Effect of completion on inflow. How it is cleaned by drawdown (the pressure difference that pushes fluid from the rock into the well) is covered in the lessons Shaped-charge perforating and Perforating operations: gun conveyance and procedures. For the topic of sources of damage, something else matters. When we perforate with overbalance, the fluid in the wellbore, with its solids, is the first thing to enter the fresh tunnels. So before such perforating we fill the well with clean, filtered fluid. It is even better to perforate underbalanced.

Production: fine particles, salts and organic deposits

During production the damage usually builds up gradually. Productivity falls over months, often after some event: water breakthrough, a drop in pressure, a rate change. The main sources are fine particles, salts, paraffins and asphaltenes, and sand.

Fine particles (fines) are fragments of clay minerals and feldspars (common rock-forming minerals), and quartz dust. Their size is from fractions of a micrometre to tens of micrometres. They lie on the surface of the grains and in the pores. The flow pulls on them with a force that grows with velocity. Electrical and molecular forces hold them in place, and these depend on the salinity and pH of the water. While the velocity is low, the particles stay put. Above the critical velocity they are torn loose and get stuck in narrow channels, most of all near the wellbore, where the velocity is highest (fines migration).

The particles are usually wetted by water and sit in a water film. So it is the flow of water, not of oil, that moves them. While the well gives oil with no water, the particles hardly move. After water breakthrough the same liquid rate can be above the critical value. We measure the critical velocity on core, by raising the flow rate in steps until the permeability starts to fall. We then apply it to the well, using the area where the flow enters the wellbore:

Critical water rate per metre of interval
qcr=ucr Ain,Ain=2πrw  or  n π dt ltq_{\text{cr}} = u_{\text{cr}}\,A_{\text{in}},\qquad A_{\text{in}} = 2\pi r_w\ \ \text{or}\ \ n\,\pi\,d_{\text{t}}\,l_{\text{t}}
where:
  • qcrq_{\text{cr}}critical water rate per metre of interval (at reservoir conditions; for water it is almost the same as the rate at the surface), m³/s per m
  • ucru_{\text{cr}}critical filtration velocity from the core test: the flow rate at which the permeability began to fall, divided by the cross-sectional area of the sample, m/s
  • AinA_{\text{in}}area through which the flow enters the well, per metre of interval, m²/m
  • rwr_wradius of the open hole, m
  • nnnumber of perforation holes per metre, 1/m
  • dt, ltd_{\text{t}},\ l_{\text{t}}diameter and length of a tunnel in the rock, m

Units: SI; for a rate in m³/d, multiply by 86,400

Holds when
  • The inflow is spread evenly over the area. In reality it is stronger towards the ends of the tunnels and towards the few tunnels that are open, and the local velocity is higher.
  • The criterion is the velocity of water in the test, where brine flows alone. When oil and water flow together, the water takes up only part of the pores and moves faster in them. So the allowable water rate is smaller, and the estimate is optimistic.
  • The ends of the tunnels and tunnels that do not work are not counted.

In an open hole the flow enters through the whole wall. In a cased well it enters only through the walls of the tunnels. With sparse, short perforation their area is several times smaller, and the velocity at the walls of the tunnels is several times higher. Dense, deep perforation, and even more a hydraulic fracture, lower this velocity. The critical liquid rate at a water cut WC\text{WC} equals qcrh/WCq_{\text{cr}}h/\text{WC}.

Worked exampleCritical rate for fine particles per metre of perforated interval
Given (assumed data)
Critical velocity test
a core 38 mm in diameter saturated with formation water; the permeability began to fall at a flow rate of 12 cm³/min
Well
rw=0.108r_w = 0.108 m, interval 12 m; perforation of 12 holes per metre, tunnels 10 mm in diameter and 0.30 m long in the rock
Operating plan
90 m³/d of liquid at a water cut of 50%
Solution
  1. Cross-sectional area of the core: π⋅3.82/4=11.3411\pi\cdot 3.8^2/4 = 11.3411 cm². Critical filtration velocity: ucr=12/6011.3411=0.01763u_{\text{cr}} = \dfrac{12/60}{11.3411} = 0.01763 cm/s, that is, 0.1763 mm/s.
  2. Inflow area per metre: open hole, 2π⋅0.108=0.67862\pi\cdot 0.108 = 0.6786 m²; walls of the tunnels, 12⋅π⋅0.010⋅0.30=0.113112\cdot\pi\cdot 0.010\cdot 0.30 = 0.1131 m², which is 6.0 times smaller.
  3. Critical water rate per metre: q=ucrA⋅86,400q = u_{\text{cr}}A\cdot 86{,}400. It is 10.34 m³/d for the open hole and 1.723 m³/d for the perforation.
  4. For the 12 m interval: 124 and 20.68 m³/d of water.
  5. The plan has 90⋅0.5=4590\cdot 0.5 = 45 m³/d of water. This is 2.2 times the critical rate for this perforation, and 36% of the critical rate for the open hole.
  6. The critical liquid rate for the same perforation is 41 m³/d at a water cut of 50% and 103 m³/d at 20%.
  7. Perforation of 40 holes per metre with tunnels 0.45 m long: the area is 40⋅π⋅0.010⋅0.45=0.565540\cdot\pi\cdot 0.010\cdot 0.45 = 0.5655 m²/m, and the critical water rate for the interval is 103 m³/d.
Answer: With 12 holes per metre, fine particles start to move at a water rate of only 20.68 m³/d. The plan of 45 m³/d is 2.2 times higher. With dense, deep perforation the limit is 103 m³/d.

What it means. While the water cut was 20%, the same liquid rate was safe. After the water cut grew it became “supercritical”, and productivity will start to fall without any new operation. Several things help: a smooth start-up, a limit on how much water is produced, and a larger inflow area, which means dense, deep perforation or hydraulic fracturing. The estimate is optimistic, because it ignores the higher water velocity when oil and water flow together and the uneven inflow to the tunnels. The data in this example are assumed for illustration.

Critical liquid rate for fines migration against water cut102040608010010205010020050010002000Water cut, %Critical rate, m³/dplanopen hole, 40 holes/m12 holes/m

Graph with a logarithmic vertical axis: the critical liquid rate, from 10 to 2000 m³/d, against the water cut, from 10 to 100%. There are three falling curves. The top one is the open hole, from 1240 m³/d at 10% to 124 at 100%. A little lower is the dense perforation, from 1030 to 103. The bottom one is the sparse perforation, from 207 to 21. The point of the plan, 90 m³/d at 50%, lies above the bottom curve (41 m³/d) and below the two upper ones.

CalculatedThe critical liquid rate for fine particles for the well of the example, against the water cut. The lines are: open hole (grey line); perforation of 40 holes per metre with tunnels 0.45 m long (dashed, almost the same as the open hole); and 12 holes per metre with tunnels 0.30 m long (solid, teal). The dot is the plan: 90 m³/d at a water cut of 50%. The calculation uses the critical velocity of water, and the inflow to the tunnels is even.

If the calculation shows a danger, we bring the well up to its working rate smoothly, in steps. The particles are then carried out a little at a time, and they have no chance to collect into plugs at the tunnels. We also limit how much water the well produces, and we increase the inflow area. Clay stabilisers (chemicals that fix the particles in place) hold the particles, but their effect is usually temporary. Particles that do break loose are better carried into the well than held back in the pores. Screens and gravel packs are chosen to hold back sand and let the fines pass (Sand control methods).

Salts come out when the water in the pores or in the wellbore becomes supersaturated (holds more dissolved salt than it can keep). There are three causes. The first is a fall in pressure: carbon dioxide leaves the water, the pH rises, and calcium carbonate comes out. The second is the mixing of incompatible waters: formation water rich in barium and strontium, and injected water rich in sulfate. The third is cooling and evaporation: in gas wells, halite (rock salt) comes out of salty water. Hydrogen sulfide reacts with iron and gives iron sulfides. Salt blocks the pores near the wellbore and the perforation tunnels, and higher up it blocks the tubing, pumps and safety valves. We judge how likely a water is to leave salt deposits (scale) with the saturation index (SI):

Saturation index (for barium sulfate as an example)
SI=lg⁡γBa [Ba2+]  γSO4 [SO42−]Ksp\mathrm{SI} = \lg\frac{\gamma_{\text{Ba}}\,[\mathrm{Ba^{2+}}]\;\gamma_{\text{SO}_4}\,[\mathrm{SO_4^{2-}}]}{K_{sp}}
where:
  • SI\mathrm{SI}saturation index, —
  • [Ba2+], [SO42−][\mathrm{Ba^{2+}}],\ [\mathrm{SO_4^{2-}}]molar concentrations of barium and sulfate ions in the water, mol/L
  • γBa, γSO4\gamma_{\text{Ba}},\ \gamma_{\text{SO}_4}activity coefficients of the ions: how many times the “effective” concentration is smaller than the actual one, —
  • KspK_{sp}solubility product: the equilibrium constant of dissolving the salt at the given temperature and pressure, (mol/L)²

Units: concentrations in mol/L; the logarithm is decimal (base 10)

Holds when
  • SI > 0 means the water is supersaturated and salt can come out. SI < 0 means it is undersaturated. SI = 0 means equilibrium.
  • In brines the activity coefficients are much smaller than one. They, and KspK_{sp} at reservoir temperature and pressure, are calculated with brine models in special software.
  • The index is thermodynamic: it tells whether the salt can come out, but not how fast.

The numerator is the ion activity product (IAP), that is, the product of the “effective” concentrations of the ions. If it is 10 times larger than the solubility product, SI = 1. If it is 1000 times larger, SI = 3. For calcium carbonate the index is calculated in the same way, but it also includes the pH and the content of dissolved carbon dioxide. So it grows when the pressure falls.

A balance of the ions tells how much salt can come out if the mixture reaches equilibrium. Barium and sulfate leave the solution in equal amounts, xx mol/L each, until the product of the remaining concentrations equals Ksp′=Ksp/(γBaγSO4)K'_{sp} = K_{sp}/(\gamma_{\text{Ba}}\gamma_{\text{SO}_4}):

How much barium sulfate can come out of a mixture
x=(a+b)−(a−b)2+4Ksp′2x = \frac{(a + b) - \sqrt{(a - b)^2 + 4K'_{sp}}}{2}
where:
  • xxamount of barium sulfate that comes out of a litre of the mixture, mol/L
  • a, ba,\ bconcentrations of Ba2+\mathrm{Ba^{2+}} and SO42−\mathrm{SO_4^{2-}} in the mixture before precipitation, mol/L
  • Ksp′K'_{sp}solubility product in terms of concentrations: KspK_{sp} divided by the product of the activity coefficients, (mol/L)²

Units: mol/L; the mass of the precipitate is xx multiplied by the molar mass of barium sulfate (233.4 g/mol)

Holds when
  • The mixture has reached equilibrium. Other ions and precipitates are not counted.
  • The activity coefficients do not change as the salt comes out.

This is the smaller root of the equation (a−x)(b−x)=Ksp′(a - x)(b - x) = K'_{sp}. When there is much more of one ion than of the other, almost all of the second one comes out: x≈min⁡(a,b)x \approx \min(a, b).

Worked exampleScaling tendency when formation water and injected water mix
Given (assumed data)
Formation water
barium Ba2+\mathrm{Ba^{2+}} 220 mg/L, practically no sulfate
Injected water
sulfate SO42−\mathrm{SO_4^{2-}} 2800 mg/L, no barium
Constants
Ksp=1.08⋅10−10K_{sp} = 1.08\cdot 10^{-10} (mol/L)² at 25 °C; molar masses Ba\mathrm{Ba} 137.33, SO4\mathrm{SO_4} 96.06, BaSO4\mathrm{BaSO_4} 233.38 g/mol; density of barite (the mineral form of barium sulfate) 4.48 g/cm³
Product of the activity coefficients
0.03 for all mixtures (assumed for a brine of this salinity)
Production well
50 m³/d of water, of which 10% is injected water; interval 12 m, porosity 0.20
Solution
  1. The mixture with 10% injected water: [Ba2+]=0.9⋅220/137.33=1.442[\mathrm{Ba^{2+}}] = 0.9\cdot 220/137.33 = 1.442 mmol/L, [SO42−]=0.1⋅2800/96.06=2.915[\mathrm{SO_4^{2-}}] = 0.1\cdot 2800/96.06 = 2.915 mmol/L.
  2. The ion activity product: 0.03⋅1.442⋅10−3⋅2.915⋅10−3=1.26⋅10−70.03\cdot 1.442\cdot 10^{-3}\cdot 2.915\cdot 10^{-3} = 1.26\cdot 10^{-7}. The index is SI=lg⁡1.26⋅10−71.08⋅10−10=3.07\mathrm{SI} = \lg\dfrac{1.26\cdot 10^{-7}}{1.08\cdot 10^{-10}} = 3.07.
  3. There is twice as much sulfate as barium, so almost all the barium comes out: x=1.439x = 1.439 mmol/L, or 1.439⋅233.38=3361.439\cdot 233.38 = 336 g of barite per cubic metre of mixture.
  4. A 50:50 mixture has SI = 3.51. This is higher, but there is less barium in it, and only 187 g/m³ come out. The most precipitate, about 348 g/m³, forms at a share of injected water of about 6%. This is a little more than the share where barium and sulfate are equal in moles (5.2%). A small excess of sulfate also precipitates the barium that would stay in solution at equality.
  5. Per day: 50⋅336/1000=16.850\cdot 336/1000 = 16.8 kg of barite, which is 16.8/4.48=3.7516.8/4.48 = 3.75 L of solid, and 1.37 m³ per year.
  6. For comparison, the pore volume of a ring of radius 0.5 m around the wellbore over the 12 m interval is π(0.52−0.1082)⋅0.20⋅12=1.80\pi(0.5^2 - 0.108^2)\cdot 0.20\cdot 12 = 1.80 m³.
Answer: The saturation index is 3.07. The mixture is strongly supersaturated, and up to 336 g of barite can come out of each cubic metre. At the present inflow, that is up to 1.37 m³ per year.

What it means. The saturation index tells how strong the “driving force” is. The mass of precipitate tells how much salt can come out at all. They reach their maximum at different shares. By the amount of precipitate, the most dangerous mixture appears just after the injected water breaks through. The possible precipitate in a year is comparable with the pore volume of the near-wellbore zone, so even a small part of it that settles in the reservoir near the wellbore is dangerous. The solubility of barite grows with salinity and pressure. In fresh water, as the temperature rises, it reaches a maximum at about 100 °C. So the constant at 25 °C overstates the index for a reservoir at 60–90 °C, and the estimate is on the safe side. The data in this example are assumed for illustration.

Mass of barium sulfate that can come out of a mixture of formation water and injected water0204060801000100200300400Share of injected water in the mixture, %Barite, g per m³ of mixtureSI 3.07SI 3.51maximum 348 g/m³ at 6%

Graph. The horizontal axis is the share of injected water, from 0 to 100%. The vertical axis is the mass of barite, from 0 to 400 g per cubic metre of mixture. The curve rises steeply from zero, reaches a maximum of about 348 g/m³ at 6% injected water, and then falls almost linearly to zero at 100%. Two points are marked: 10%, about 336 g/m³, saturation index 3.07; and 50%, about 187 g/m³, saturation index 3.51.

CalculatedThe mass of barium sulfate that can come out of a cubic metre of a mixture of the formation water and injected water of the example, against the share of injected water. The labels at the points are the saturation index of the mixture. The calculation is at 25 °C with a constant product of the activity coefficients.

The place where the water mixes decides where the salt comes out. In the reservoir, the injected water pushes the formation water ahead of it as a front, and the two mix only in a narrow band that moves away from the injection well. When the water breaks through to the production well, the flows from different layers, with and without injected water, mix right at the wellbore and in the tubing. That is where the deposits grow. So we squeeze scale inhibitors into the reservoir in advance (we pump them into the rock, where they are held and released slowly). At concentrations of a few to tens of parts per million, they slow down the start and growth of crystals, and they are washed out slowly with the water. Calcium carbonate dissolves in hydrochloric acid, but barium sulfate hardly dissolves in acid at all. We remove it with chelating solutions (solutions that bind the metal ions), or mechanically, or we bypass it by re-perforating. Deposits of barium and strontium sulfates can be weakly radioactive, and they are handled under the radiation safety rules. The chemistry of produced water and its treatment are covered in the lesson Produced water: composition, treatment and disposal.

Paraffins are dissolved in the oil at reservoir temperature. They come out as crystals when the temperature falls below the wax appearance temperature (WAT). This usually happens in the tubing and flowlines (the pipes that carry production from the wellhead to the plant). It can also happen in the reservoir near the wellbore, if that zone is cooled by injecting cold water or flushing fluids. An example is the Uzen field on the Mangyshlak peninsula. Its oil is rich in paraffin (according to studies of the 2020s, about 17–19% paraffin by mass). Decades of cold-water injection have lowered the reservoir temperature, and now the wax appearance temperature is comparable with the present reservoir temperature. So paraffin can come out already in the reservoir and the near-wellbore zone. Paraffin is removed with hot oil and solvents, and it is prevented by keeping the temperature up and by inhibitors.

Asphaltenes are heavy, polar components of the oil. They stay in the oil as a colloidal suspension (very fine particles spread through the liquid) while resins keep them stable. When the pressure falls, but is still above the bubble point pressure (the pressure at which gas starts to come out of the oil), the oil expands, its power to dissolve falls, and the suspension loses its stability. This is strongest near the bubble point pressure. Below it, gas leaves the oil, and the oil again becomes a better solvent. Asphaltenes that have come out dissolve poorly, and only in aromatic solvents (such as toluene). So it is easier to prevent them: keep the pressure above the level at which they start to come out, and use inhibitors. Acid, especially acid with ferric iron (iron in the +3 state), forms a sludge with such oil, and the sludge is almost impossible to remove (Acid systems and treatment design).

Sand is carried out by the flow when the stresses at the wall of the wellbore and of the tunnels exceed the strength of the rock. When and why this happens is covered in the lesson Sand production: types, prediction and cost of the problem. As the reservoir is depleted, the effective stress (the stress that the rock grains carry) rises throughout the reservoir. In rocks that are sensitive to stress, the permeability falls and fractures close (Rock compressibility and the effect of stress on core properties). Gas coming out of solution near the wellbore below the bubble point pressure, and condensate dropping out, also lower productivity. But these are not damage to the rock. They are a component of skin that depends on the operating regime (Skin factor).

Bacteria enter the well with injected fluids. Their slime and waste products plug the pores. Sulfate-reducing bacteria produce hydrogen sulfide. It causes corrosion, and it reacts with iron to form a precipitate of iron sulfide. The defence is biocides (chemicals that kill bacteria) in all injected fluids and clean equipment.

Injection: water quality and compatibility

An injection well takes tens and hundreds of thousands of cubic metres of water a year. That is far more than any fluid pumped into a production well. So even a small content of suspended solids (particles of sand and clay, corrosion products, drops of oil, bacteria) slowly builds a cake on the walls of the tunnels and plugs the pores near the wellbore. Then the injectivity (how much water the well takes per unit of pressure) falls. Particles larger than about a third of the pore throats build an external cake. Smaller particles go deeper (Formation damage and damage skin).

Produced water that is returned to the reservoir adds oil and fine suspended solids, and the mixing of waters from different sources adds salts. Fresh water in a clayey reservoir lowers the salinity in the pores below the critical value, and clay particles detach and migrate (see the advanced section). Cold water cools the near-wellbore zone. In a high-paraffin oil this threatens paraffin deposits. But it also lowers the stresses in the rock, and during injection fractures often open. These fractures raise the injectivity and make the well less sensitive to suspended solids. The requirements for water quality and its treatment are covered in the lesson Water injection systems.

Well killing and workovers

Before a workover we usually kill the well. We fill it with kill fluid, whose column balances the reservoir pressure with a margin. Part of the fluid goes into the reservoir, and with it goes everything that was in it: particles from unwashed tanks and lines, rust and mill scale (flakes of oxide from the steel) from old pipes, thread grease, remains of drilling mud and cement slurry, and thickeners. Work above the reservoir brings an especially large amount of dirt: drilling out cement and packers (seals that close the gap around the tubing), scraping the casing, and milling (cutting away metal). If the brine is incompatible with the formation water or the clays, deposits and swelling add to the plugging, and in gas reservoirs a water block adds too.

The sign of such damage is productivity after the workover that is lower than before it, at the same reservoir pressure. It can be noticed already during the workover. If the daily loss of fluid into the reservoir drops sharply after some operation, the tunnels have most likely become blocked.

Worked exampleHow many solids a kill fluid brings into the tunnels
Given (assumed data)
Loss into the reservoir during the workover
8 m³ of kill fluid
Suspended particles in the fluid
500 mg/L (unwashed tanks, rust, thread grease); after filtering, 10 mg/L
Particles
density 2650 kg/m³; in the tunnels they pack to a porosity of 0.4
Perforation
as in the example on fine particles: 12 holes per metre over 12 m, tunnels 10 mm × 0.30 m in the rock
Solution
  1. Mass of particles: 8⋅500/1000=4.008\cdot 500/1000 = 4.00 kg.
  2. The volume they will take when packed: 4.00/2650/(1−0.4)=0.002524.00/2650/(1 - 0.4) = 0.00252 m³, that is, 2.52 L.
  3. The volume of the tunnels in the rock: 144⋅π⋅0.0052⋅0.30=0.00339144\cdot\pi\cdot 0.005^2\cdot 0.30 = 0.00339 m³, that is, 3.39 L.
  4. The particles can fill 74% of the volume of the tunnels.
  5. With filtered fluid: 0.08 kg of particles, which is 1.5% of the volume of the tunnels.
Answer: A “dirty” kill fluid brings about 4.00 kg of particles. That is enough to fill about 74% of the volume of the perforation tunnels. Filtered fluid brings 50 times less.

What it means. This is an upper estimate: part of the particles will stay in the wellbore or come out when the well is brought on stream. But the particles do not go equally into all the tunnels. They go into the few that take the most fluid, and they block these completely. So clean fluids and clean tanks are one of the cheapest measures against damage. The data in this example are assumed for illustration.

  • Clean fluids. Filtered brines with no solids. Clean tanks, lines and pipes. We check the quality by the content of suspended solids and by turbidity (cloudiness). On the threads of pipes that go down to the reservoir, we use a minimum of grease.
  • Pills instead of weighting material. If the losses are large, we place pills (small volumes of special fluid) with bridging agents of the right size, soluble in acid or water (calcium carbonate, salt), or viscous pills with a breaker that destroys the thickener.
  • Compatible brines. Solutions of potassium or ammonium chloride for clayey reservoirs. Calcium and zinc brines are checked for compatibility with the sulfates, carbonates and hydrogen sulfide of the reservoir fluids.
  • Work without killing. Where possible, we work under pressure: through coiled tubing (a continuous flexible pipe), on wireline (a cable), or with a snubbing unit (equipment that runs pipe into a well under pressure). Or we shut off the reservoir with a plug in the tail pipe of the tubing (its lowest section).
  • Clean pipes. Before treatments are pumped through old tubing, the pipes are washed with acid, with the returns taken to the surface. Otherwise the fluid will drive rust and deposits from the walls into the reservoir.

Diagnosis and prevention

The signs of damage are productivity below what we expect, a large skin after the pseudo-skins are subtracted, and intervals that do not work. They are listed in the lesson Formation damage and damage skin. To choose a cure, we need the next step: to understand which operation and which mechanism created the damage.

From observation to source of damage
Observation
A large skin from the first test, open hole or short tunnels
Likely source
the drilling fluid used to open the reservoir
How to check
tests of the fluid on core, depth of invasion from logs
Observation
Productivity fell right after a workover or a treatment
Likely source
the kill fluid, the treatment
How to check
daily fluid losses, analysis of the particles carried out and of the spent fluids
Observation
Productivity falls after water breakthrough
Likely source
fine particles, salts
How to check
whether the decline depends on the rate; barium and sulfate in the produced water
Observation
There is less barium in the produced water than the share of formation water implies
Likely source
barite is coming out in the reservoir or in the wellbore
How to check
an ion balance, analysis of deposits
Observation
The injectivity of an injection well falls
Likely source
suspended solids, oil, bacteria in the water
How to check
the Hall plot (a graph that follows how an injection well takes water; see the advanced section), analysis of the water and deposits, a falloff test (a pressure test after injection stops)
Observation
Deposits in the tubing and on the pump
Likely source
salts, paraffins, asphaltenes
How to check
X-ray and chemical analysis of the deposits

The diagnosis almost always has several sources, and we test it by experiment: laboratory tests with the fluids that were pumped into the well, and the result of the first treatment. A failed treatment is information too. If acid did not help, the damage was perhaps not the kind that it was designed for (Acid treatments: principle, economics and candidate selection).

A programme for preventing damage
  1. 1
    Properties of the reservoir

    The mineralogy and the position of the clays, the permeability and the sizes of the pore throats, the compositions of the formation water and oil, the bubble point pressure, the wax appearance temperature, and the onset pressure of asphaltene precipitation.

  2. 2
    Fluids

    For each operation (the drilling fluid for opening the reservoir, the cement slurry and spacers, the perforating and kill fluids, the treatment fluids, the injection water), run tests of compatibility and of return permeability.

  3. 3
    Technology

    A small overbalance, a short time with the hole open, underbalanced perforating, a smooth start-up, and work without killing where it is possible.

  4. 4
    Control at the well

    Filtering and checking of fluids, clean tanks and lines, and daily records of losses into the reservoir.

  5. 5
    Monitoring during production

    The skin from repeated well tests, analyses of produced water and of deposits, and injectivity. Inhibitors go in before salts start to come out.

Terms and symbols

Return permeability RR
The ratio of the permeability of a sample after exposure to a fluid to its original permeability. The overall value for a sample is the harmonic mean over its sections.
Critical velocity
The filtration velocity of the wetting phase above which fine particles are torn off the grains. It is measured on core in steps of flow rate.
Critical salt concentration
The salinity of water below which clay particles detach from the grains. It exists for monovalent cations (ions with a charge of one, such as sodium).
Saturation index SI
The decimal logarithm of the ratio of the ion activity product to the solubility product. When SI > 0, salt can come out.
Scale inhibitor
A chemical that, in small concentrations, slows down the start and growth of salt crystals. It is squeezed into the reservoir or dosed into the well.
Wax appearance temperature
The temperature below which the first crystals of paraffin come out of the oil.
Kill fluid
A fluid whose column balances the reservoir pressure during a workover. It damages the reservoir if it carries solids or is incompatible with it.

Common mistakes and misconceptions

Wrong

“A return permeability of 20% means that the permeability near the wellbore fell five times over the whole depth of invasion.”

In fact

This is the average over the sample. If the loss is concentrated in the first centimetre, the permeability fell 20 times there and hardly changed beyond it. We calculate the skin by rings. In the example of the lesson it is half of the “naive” one.

Wrong

“While the well produced no water, fine particles were no trouble, so they will not be trouble later.”

In fact

Particles that are wetted by water are moved by the flow of water. After water breakthrough the same liquid rate can be above the critical value, especially in a perforated well with a small area of tunnels.

Wrong

“The saturation index is above zero, so the salt has already come out.”

In fact

The index says that precipitation is possible, but not where or how fast. The mass of the precipitate is set by the amount of ions, and the speed by the kinetics of crystallisation (how fast the crystals form and grow). Inhibitors act on exactly this speed.

Wrong

“We will remove barite with hydrochloric acid, just as we do calcium carbonate.”

In fact

Barium sulfate hardly dissolves in acid at all. We prevent it with inhibitors. We remove it with chelating solutions or mechanically, or we bypass it by re-perforating.

Wrong

“Well killing is a routine operation and does not harm the reservoir.”

In fact

The kill fluid goes into the tunnels together with all its dirt. A few kilograms of suspended solids can fill most of the volume of the tunnels. Clean fluids and work without killing pay for themselves.

In depth: advanced detail

Why particles break away: forces, salinity and pH

Two groups of forces act on a particle that is stuck to a grain. One is the force of the flow. It pulls the particle along the surface and tends to tip it over. The others are the forces that hold the particle: van der Waals attraction, and the electrical interaction of the double layers (clouds of charged ions around each surface). This is the idea of the DLVO theory, a theory of the forces between charged particles in a liquid. Clay and quartz are negatively charged in water, and their double layers repel each other. While the water is salty, the layers are thin, the repulsion is weak, and the particle holds on. In fresh water the layers are thicker and the repulsion is stronger. When the salinity falls below the critical salt concentration, particles break away even in a weak flow.

Several consequences follow. The critical concentration exists for monovalent cations (positive ions with one charge), such as sodium and potassium. Divalent calcium and magnesium (two charges) hold the particles even at a much lower salinity. A higher pH strengthens the negative charge and the detachment, and a higher temperature usually does too. A sudden change of salinity is more dangerous than a gradual one. Many particles break away at once, meet in narrow channels and form plugs. With a gradual change they are carried out one at a time. For the same reason a sudden start-up of a well is more dangerous than a stepwise one.

The critical velocity and the critical concentration are linked: the closer the salinity is to the critical value, the weaker the flow that is needed to tear a particle loose. So on core we measure them with the waters that will be in the reservoir: formation water, injected water and filtrate.

Injection well: the Hall plot

It is convenient to track damage in an injection well from data that are recorded every day anyway: the injection rate and pressure. If injection is close to steady state, the pressure drop is proportional to the rate (Steady-state radial flow), and the accumulated sum of the pressure drops grows in proportion to the accumulated volume injected:

Hall plot
∑j(pw,j−pˉ)Δtj=μB(ln⁡rerw+S)2πkh W\sum_j \left(p_{w,j} - \bar p\right)\Delta t_j = \frac{\mu B\left(\ln\dfrac{r_e}{r_w} + S\right)}{2\pi k h}\,W
where:
  • pw,jp_{w,j}average bottomhole injection pressure over period jj, Pa
  • pˉ\bar preservoir pressure in the injection area, Pa
  • Δtj\Delta t_jduration of period jj, s
  • WWaccumulated volume injected by the end of the last period, m³
  • μ, B\mu,\ Bviscosity and formation volume factor of the injected water, Pa·s, m³/m³
  • k, h, re, rw, Sk,\ h,\ r_e,\ r_w,\ Spermeability, thickness, radii of the area and of the well, skin, m², m, m, m, —

Units: SI. In metric oilfield units, 2π2\pi is replaced by 0.536, and the slope comes out in MPa·d/m³ with kk in mD, hh in m and μ\mu in mPa·s

Holds when
  • Injection is close to steady state: the pressure at the distance rer_e equals the reservoir pressure pˉ\bar p and hardly changes. The mobilities of the water and of the oil it displaces (how easily each fluid moves) are close.
  • The wellhead pressure is often used instead of the bottomhole pressure. At a constant rate, hydrostatic pressure and friction give a constant correction.
  • The slope also changes for other reasons: a rise of the reservoir pressure, or water saturation of the zone around the well. The most accurate measurement of the skin is a falloff test (a test that records how the pressure falls after injection stops).

A straight line means that the condition of the well is not changing. A rising slope means that the skin is increasing, so the near-wellbore zone is clogging. A falling slope means that the skin is decreasing, after a treatment or when the injection pressure has exceeded the fracture pressure and a fracture has opened.

Worked exampleSkin of an injection well from the slopes of the Hall plot
Given (assumed data)
Reservoir and water
kh=600kh = 600 mD·m, μ=0.5\mu = 0.5 mPa·s, B=1.0B = 1.0
Radii
re=250r_e = 250 m, rw=0.1r_w = 0.1 m
Slopes of the plot
first year: 0.0165 MPa·d/m³; second year: 0.0261 MPa·d/m³
Solution
  1. From the slope of the plot mHm_H: ln⁡rerw+S=mH⋅0.536⋅khμB\ln\dfrac{r_e}{r_w} + S = \dfrac{m_H\cdot 0.536\cdot kh}{\mu B}; ln⁡2500=7.824\ln 2500 = 7.824.
  2. First year: 0.0165⋅0.536⋅600/0.5=10.60.0165\cdot 0.536\cdot 600/0.5 = 10.6, skin S=10.6−7.824=2.8S = 10.6 - 7.824 = 2.8.
  3. Second year: 0.0261⋅0.536⋅600/0.5=16.80.0261\cdot 0.536\cdot 600/0.5 = 16.8, skin S=9.0S = 9.0. It has grown by 6.2.
Answer: The skin has grown from about 2.8 to 9.0. The near-wellbore zone is clogging, and at the same pressure the well takes 1.6 times less water.

What it means. We look for the cause in the water (suspended solids, oil, corrosion products, bacteria) and confirm it by analysing the deposits and by a falloff test. The estimate of skin from the plot is approximate. If the reservoir pressure in the injection area has been rising, part of the increase in slope did not come from the near-wellbore zone. The data in this example are assumed for illustration.

Connections to other topics

Check yourself

  1. 1

    The skin of a well from well tests: +1.0 in January; +1.2 in May, two months after the water cut rose from 5 to 9%; +6.5 in October, a month after a pump replacement for which the well was killed. The reservoir pressure hardly changed during the year. Which source of damage is the most likely?

    1. A.A compacted zone from perforating
    2. B.Asphaltene precipitation during depletion
    3. C.Fine particles after the rise in water cut
    4. D.Kill fluid that went into the tunnels
    Show the answer

    Correct answer: D. Kill fluid that went into the tunnels

    After the water cut rose, the skin hardly changed. After the workover with well killing, it grew by 5.3. The pressure is the same, so depletion and asphaltenes are unlikely. A compacted zone from perforating would have been visible from the first test. Kill fluid brings particles from tanks and pipes into the tunnels, and, if it is incompatible, deposits. We check this from the daily fluid losses during the workover and from the analysis of the particles carried out. We cure it with a treatment chosen for the composition of these particles.

  2. 2

    A sample is 60 mm long. After exposure to the drilling fluid, the first 6 mm returned 10% of their permeability and the remaining 54 mm returned 95%. Find the overall return permeability. Can it be applied to the whole zone invaded by filtrate?

    Show the answer

    R=60/(6/0.10+54/0.95)=60/(60+56.8)≈0.51R = 60/(6/0.10 + 54/0.95) = 60/(60 + 56.8) \approx 0.51. It cannot be applied. Almost all of the loss is concentrated in the first 6 mm, where the permeability fell 10 times. We calculate the skin by rings: a thin, badly damaged ring at the wall, and a ring that is only slightly damaged out to the radius of invasion.

  3. 3

    The perforation was made half as dense: 6 holes per metre instead of 12, with the same length and diameter of tunnels. How will the critical rate for fine particles change?

    1. A.It will double
    2. B.It will fall four times
    3. C.It will halve
    4. D.It will not change
    Show the answer

    Correct answer: C. It will halve

    The critical rate is proportional to the inflow area, and the area of the tunnel walls is proportional to the number of tunnels. Half as many tunnels means half the area, twice the velocity at the walls at the same rate, and half the critical rate.

  4. 4

    The formation water contains 300 mg/L of barium, and the injected water contains 2000 mg/L of sulfate. At what share of injected water in the mixture can the most barite come out, and why is the saturation index not the highest at that share?

    Show the answer

    If the solubility of barite is negligible, the precipitate is limited by the ion that is in shorter supply. It is the largest when barium and sulfate are equal in moles. In the original waters there are 300/137.3=2.18300/137.3 = 2.18 and 2000/96.1=20.82000/96.1 = 20.8 mmol/L. The share of injected water is 2.18/(2.18+20.8)≈0.0952.18/(2.18 + 20.8) \approx 0.095, about 9.5%. If we allow for the remaining solubility, the maximum shifts a little to a larger share. The saturation index depends on the product of the concentrations, (1−f)f(1 - f)f, which is largest at f=0.5f = 0.5. But in a 50:50 mixture the barium is already halved, and so there is less precipitate.

  5. 5

    40 m³ of cold flushing fluid was pumped into a well with high-paraffin oil. What should we worry about first?

    1. A.Paraffin coming out in the near-wellbore zone
    2. B.Nothing: paraffin comes out only in the tubing
    3. C.Barite coming out
    4. D.Swelling of smectite
    Show the answer

    Correct answer: A. Paraffin coming out in the near-wellbore zone

    If the temperature near the wellbore falls below the wax appearance temperature, paraffin will come out right in the pores. So for such oils, flushing fluids are heated or solvents are added, and after pumping we let the zone warm up.

  6. 6

    An injection well takes 1200 m³/d of water with 2 mg/L of suspended solids. How many solids will it receive in a year, and why does it matter at such a small concentration?

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    1200⋅365⋅0.002≈8801200\cdot 365\cdot 0.002 \approx 880 kg per year. This is about 0.33 m³ of solid at a density of 2650 kg/m³, and about 0.55 m³ in a loose cake with a porosity of 0.4. The pore volume of a ring 0.3 m thick around the wellbore over a 10 m interval, at a porosity of 0.20, is about 1 m³. The particles settle at the wall and in the tunnels. So even such water will noticeably lower the injectivity within a year or two, unless fractures appear.

Updated: 19 October 2026