Acid treatment of carbonates and special wells

Quick refresher

Hydrochloric acid dissolves calcite so fast that the dissolving front does not stay even. The acid goes into the most conductive pores, widens them and cuts wormholes. The pattern depends on the ratio of the speed of the reaction to the speed of the acid supply, which the Damköhler number expresses. The pattern can be face dissolution, conical channels, one dominant wormhole, ramified (branching) channels, or almost uniform dissolution.
We measure the efficiency of an acid by the number of pore volumes to breakthrough, $PV_{bt}$ , in a core test (how many pore volumes of acid must pass through a rock sample before a wormhole breaks out of its far end). The curve of $PV_{bt}$ against the pore velocity has a minimum. At lower velocities the acid is spent at the inlet, and at higher velocities on branching. At the minimum, a fraction of a pore volume is enough for a wormhole.
In a well the velocity at the front of the wormholes falls in inverse proportion to the radius, $v_i = q/(2\pi r h\phi)$ . So the acid is pumped at the highest rate that is allowed below the fracture pressure. At the wall the velocity is above the optimum anyway, and a metre or two from the wellbore it is already below it.
Wormholes use tens or hundreds of times less acid than dissolving the rock solidly. A fraction of a cubic metre of 15% acid per metre takes them 1–2 m out, and the skin falls to −2…−3. The volume per metre is $V/h = \int 2\pi r\phi PV_{bt}\,dr$ .
Acid goes where it is easiest, and it makes its own path easier. Without diversion, a permeable layer, a fracture or the heel of a horizontal well (the end nearest the vertical section) takes almost all of it. Foam, self-diverting fluids, ball sealers (balls that block the holes that take fluid), packers (seals that close off part of the well) and limited-entry perforating (a few small holes that force the acid to spread out) bring the acid back into the tight intervals.
A campaign is evaluated by the extra production against the forecast without treatment (the declining baseline), by the share of successful jobs, and by the cost per cubic metre of gain. Failures are analysed to improve the selection of wells and the technology.

Key relationships

v_i = \dfrac{q}{2\pi r h\phi}

pore velocity at the wormhole front at a distance $r$ from the axis

PV_{bt} = PV_{bt,opt}\,\dfrac{(v_i/v_{opt})^{1/3}}{\left[1 - e^{-4v_i^2/v_{opt}^2}\right]^2}

the efficiency curve of the acid (the Buijse and Glasbergen model)

\dfrac{V}{h} = \int_{r_w}^{r_{wh}} 2\pi r\phi\,PV_{bt}\,dr \approx \pi\phi\,PV_{bt}\,(r_{wh}^2 - r_w^2)

acid per metre so that the wormholes reach $r_{wh}$

S \approx -\ln\dfrac{r_{wh}}{r_w}

skin when the wormholes have passed through the damaged zone

\dfrac{q_j}{q} = \dfrac{k_jh_j/(\ln(r_e/r_w) + S_j)}{\sum_i k_ih_i/(\ln(r_e/r_w) + S_i)}

the share of the rate that goes to layer $j$ at a common bottomhole pressure

Common pitfalls

Calculating the volume of acid for a carbonate as if the rock were dissolved solidly. The answer comes out tens or hundreds of times too high.
Pumping slowly “so that the acid reacts better”. At a low velocity it dissolves the wall and does not cut wormholes. Slow pumping is needed only for acid baths, when the damage is right at the wall.
Transferring the optimum velocity from core to the well directly. At the wormhole front the velocity falls with radius, and on larger samples the optimum is lower than on standard ones.
Treating a long interval without diversion. The acid will go into a fracture, a permeable layer, or the first part of a horizontal well.
Estimating the gain by comparing the rate after a treatment with the rate before it. If we ignore the decline of the baseline and the faster decline after a treatment, the annual gain is overstated: in the example of the lesson, by almost one and a half times.

After this lesson you can

explain why hydrochloric acid forms wormholes in carbonate, how the dissolving regime (the pattern in which the acid eats the rock) depends on the injection velocity, and what the Damköhler number (the ratio of the speed of the reaction to the speed of the acid supply) and the curve of “pore volumes to breakthrough against velocity” show
estimate the volume of acid per metre that wormholes need to reach a given radius at a chosen rate, and the expected skin (the extra resistance to flow near the well); choose the rate with the fracture pressure in mind
calculate how the rate is divided between layers and how a diverter redistributes it; choose a method of diversion
plan the treatment of a long horizontal well, an injection well, a gravel-packed well and a fractured reservoir, and understand when acid fracturing is needed instead of a matrix treatment
evaluate a campaign of treatments: the extra production against a declining baseline (what the wells would have produced without the treatments), the share of successful jobs, the cost per cubic metre of gain, and the conclusions for the next wells

How carbonate differs: the reaction and the choice of acid

The lesson Acid treatments: principle, economics and candidate selection explains what acid does in sandstone and in carbonate, and why the value of a treatment is set by the skin that it removes. The reactions of hydrochloric acid with calcite and dolomite and the dissolving power are in the lesson on acid systems: one cubic metre of 15% hydrochloric acid dissolves about 0.08 m³ of calcite. This lesson covers how treating a carbonate differs from treating a sandstone, and how to calculate it.

The main difference is speed. The reaction of hydrochloric acid with calcite on the surface is so fast that the overall speed of dissolving is limited not by the reaction itself but by the delivery of acid to the wall of the pores. Hydrogen ions cannot come up to the surface as fast as it can take them in. Dolomite reacts more slowly. At a low temperature the speed is limited by the reaction on the surface itself, and it grows fast as the temperature rises. The products of the reaction (soluble calcium and magnesium chlorides, and carbon dioxide) give no precipitates. So a carbonate has none of the secondary precipitates of hydrofluoric acid, and its stages are not needed. Mud acid is not used in carbonates at all, because with calcium it gives insoluble calcium fluoride.

A fast reaction is both an advantage and a difficulty. The acid dissolves the rock and can go deep, but only if it is delivered in time to where it is needed. At high temperature it is spent right at the wellbore, and in a long interval, in the first conductive layer. This gives two tasks, which this lesson deals with: to control the speed of supply (the sections on wormholes and on the rate), and to control the distribution along the interval (the sections on diversion and on special wells). For both of them the fluid itself is changed as well.

Acid fluids for carbonates

Fluid: Hydrochloric acid 15%
Why: the main fluid for limestones and mixed carbonates
Limits: at high temperature it is spent at the wall; a corrosion inhibitor (a chemical that protects steel from the acid) is needed

Fluid: Hydrochloric acid 20–28%
Why: more rock dissolved per cubic metre; dolomites
Limits: stronger corrosion; the spent fluid is heavier and more viscous

Fluid: Emulsified acid
Why: drops of acid in an oil outer phase reach the wall more slowly, so wormholes go deeper at low rates
Limits: viscous: friction in the pipes is higher; the emulsion is chosen by temperature

Fluid: Gelled and self-diverting acid
Why: gelled acid reaches the wall more slowly and less of it leaks off into the rock; the viscosity of a self-diverting acid grows as it is neutralised and steers the next batch away: viscoelastic surfactants (chemicals whose solution thickens as the acid is spent), polymers that crosslink in the reservoir
Limits: a breaker (a chemical that destroys the gel) and a flow-back are needed; a polymer can leave a residue

Fluid: Organic acids (acetic, formic)
Why: react slowly and hardly corrode steel: for hot wells
Limits: dissolve less rock; the reaction of acetic acid does not go to completion

Fluid: Chelating agents (chemicals that bind metal ions)
Why: dissolve calcite slowly and almost without corrosion: for very high temperatures
Limits: expensive; the dissolving power is low

Fluid	Why	Limits
Hydrochloric acid 15%	the main fluid for limestones and mixed carbonates	at high temperature it is spent at the wall; a corrosion inhibitor (a chemical that protects steel from the acid) is needed
Hydrochloric acid 20–28%	more rock dissolved per cubic metre; dolomites	stronger corrosion; the spent fluid is heavier and more viscous
Emulsified acid	drops of acid in an oil outer phase reach the wall more slowly, so wormholes go deeper at low rates	viscous: friction in the pipes is higher; the emulsion is chosen by temperature
Gelled and self-diverting acid	gelled acid reaches the wall more slowly and less of it leaks off into the rock; the viscosity of a self-diverting acid grows as it is neutralised and steers the next batch away: viscoelastic surfactants (chemicals whose solution thickens as the acid is spent), polymers that crosslink in the reservoir	a breaker (a chemical that destroys the gel) and a flow-back are needed; a polymer can leave a residue
Organic acids (acetic, formic)	react slowly and hardly corrode steel: for hot wells	dissolve less rock; the reaction of acetic acid does not go to completion
Chelating agents (chemicals that bind metal ions)	dissolve calcite slowly and almost without corrosion: for very high temperatures	expensive; the dissolving power is low

The strength and the additives are chosen from the temperature, the mineralogy and core tests. The additives (a corrosion inhibitor, an iron control agent that keeps iron dissolved, non-emulsifying surfactants) are the same as for sandstones (lesson on acid systems).

Wormholes and dissolving regimes

Why does the acid not dissolve a carbonate with an even front? The pore channels differ in size, and more acid enters a large channel. The acid widens it, the channel takes even more acid, and so on. This is positive feedback, as with viscous fingering in a displacement (a thin liquid pushing into a thick one in narrow fingers). The acid that has entered a channel is spent on its walls and at its end, and hardly touches the neighbouring pores. The result is a network of thin channels, whose permeability is incomparably higher than that of the rock. These are wormholes. How thin and how long they are depends on which is faster: the reaction or the supply of acid.

SchematicDissolving regimes of a carbonate when acid is injected into a sample at different velocities (a diagram, not to scale). The acid enters from the top, and the velocity grows from left to right. The least acid is needed to break through in the regime of the dominant wormhole.

Face dissolution: At a very low velocity the acid has time to react right at the inlet and dissolves the rock with a solid front. An enormous volume of acid is spent to advance by a centimetre.
Conical channels: At a slightly higher velocity the acid enters the large pores but is spent on their walls near the inlet. The channels come out wide at the base and short.
Dominant wormhole: Just enough acid enters the channel for it to reach the end of the channel and extend it. The channel is thin and long, with few branches, and the least acid is needed to break through.
Ramified (branching) channels: The velocity is high. The acid has no time to react in the main channel and seeps into neighbouring pores, and the channels branch. The acid is spent on the branches.
Uniform dissolution: At a very high velocity the acid passes through the sample almost unreacted and dissolves the rock a little everywhere, as in a sandstone treatment.

The ratio of reaction to supply is expressed by the dimensionless Damköhler number. For a wormhole it is written as the ratio of the amount of acid that its walls can use up in a unit of time to the amount that the flow supplies:

Damköhler number for a wormhole

Da = \frac{\pi\,d\,l\,\kappa}{q_{wh}}

where:

$Da$ Damköhler number, —
$d,\ l$ diameter and length of the wormhole; $\pi d l$ is the area of its walls, m
$\kappa$ overall rate constant of dissolving: it covers both the delivery of acid to the wall and the reaction on it; the slower of the two decides it, m/s
$q_{wh}$ flow rate of acid through the wormhole, m³/s

Units: SI (dimensionless)

Holds when

The acid is spent on the walls at a rate proportional to its concentration.
This is an estimate. The diameter and the length of a wormhole are measured only approximately in a test.

A large $Da$ means that the acid is spent at the inlet (face dissolution, conical channels). A small $Da$ means that it passes almost unreacted (branching, uniform dissolution). In laboratory tests the lowest acid consumption to break through came at about the same $Da \approx 0.29$ for strong and weak acids and for chelating agents. The practical conclusion is this: the faster a fluid reacts (the higher $\kappa$ , for example at a higher temperature), the larger the rate that each wormhole needs. The slower it reacts (emulsified or organic acid), the smaller the rate.

In practice the efficiency of an acid is measured in a core test. Acid is pumped through a sample 15–30 cm long at a constant rate, and we note the volume at which a wormhole came out of the other end. This is the number of pore volumes to breakthrough, $PV_{bt}$ : the volume of acid divided by the pore volume of the sample (the volume of its pores). The tests are repeated at different pore (interstitial) velocities $v_i$ , that is, the flow rate divided by the cross-sectional area and the porosity, and we draw the efficiency curve. It always has a minimum: the optimum velocity $v_{opt}$ and the smallest volume $PV_{bt,opt}$ . For field calculations the curve is described by a formula. A widely used semi-empirical model is that of Buijse and Glasbergen:

Wormhole growth velocity and the efficiency curve of the acid

v_{wh} = \frac{v_{opt}^{1/3}}{PV_{bt,opt}}\,v_i^{2/3}\left[1 - e^{-4v_i^2/v_{opt}^2}\right]^2,\qquad PV_{bt} = \frac{v_i}{v_{wh}} = PV_{bt,opt}\,\frac{(v_i/v_{opt})^{1/3}}{\left[1 - e^{-4v_i^2/v_{opt}^2}\right]^2}

where:

$v_{wh}$ velocity at which the wormhole front advances, m/s
$v_i$ pore velocity at the wormhole front: the flow rate divided by the cross-sectional area and the porosity, m/s
$v_{opt},\ PV_{bt,opt}$ coordinates of the minimum of the laboratory curve: the optimum velocity and the pore volumes to breakthrough at it, m/s, —
$PV_{bt}$ pore volumes to breakthrough at the velocity $v_i$ , —

Units: any consistent units of velocity

Holds when

The constants $v_{opt}$ and $PV_{bt,opt}$ are taken from core tests of the same rock with the same acid at reservoir temperature.
The formula smooths the minimum: at $v_i = v_{opt}$ it gives $PV_{bt} \approx 1.04\,PV_{bt,opt}$ . At high velocities $PV_{bt}$ grows as $v_i^{1/3}$ , and at low velocities very steeply.
On larger samples the optimum velocity and $PV_{bt}$ are smaller than on standard ones. So laboratory values are transferred to the reservoir with a correction, or are calibrated against the results of treatments.

To the right of the minimum the acid is spent on branching, but only moderately. To the left it dissolves the face, and $PV_{bt}$ grows by orders of magnitude. So it is much cheaper to err towards a higher velocity than towards a lower one.

CalculatedEfficiency curves of the acid in the model of this lesson: 15% hydrochloric acid (solid line, optimum 1.5 cm/min) and emulsified acid in the same rock (dashed line, optimum 0.3 cm/min). Both axes are logarithmic. The points on the solid curve are the velocities at the wormhole front in the example below: at the wall of the well and at a distance of 1.5 m. Emulsified acid wins at low velocities, to the left of about 1 cm/min. The laboratory values are assumed.

Rate and volume: from core to well

In a core test the pore velocity is the same along the whole length of the sample. In a well the acid spreads out radially. The cross-section of the flow grows with distance from the axis, and the velocity at the wormhole front falls in inverse proportion to the radius:

Pore velocity at the wormhole front in radial injection

v_i = \frac{q}{2\pi r\,h\,\phi},\qquad r_{opt} = \frac{q}{2\pi h\phi\,v_{opt}}

where:

$v_i$ pore velocity at the radius $r$ , m/s
$q$ rate of acid into the interval, m³/s
$r$ distance of the front from the well axis, m
$h,\ \phi$ thickness of the treated interval and porosity, m, —
$r_{opt}$ radius out to which the velocity at the front is not lower than the optimum, m

Units: SI

Holds when

The acid enters the interval evenly over the thickness and around the circumference.
The wormhole front is taken as a cylinder. The wormholes themselves take a negligible share of the pore volume.

At the wall of the well, at ordinary rates, the velocity is tens of times higher than the optimum, and already a metre or two out it is lower. The larger the rate, the further from the wellbore the wormholes grow in a good regime. So carbonates are treated at the highest rate allowed below the fracture pressure. This method is known as the maximum pressure-differential and rate method (MAPDIR). The limit on the rate grows as the skin falls (formula for the limit on the rate), and the rate is raised in steps after it.

How much acid is needed for the wormhole front to reach the radius $r_{wh}$ ? At each step of the front the acid fills the pores of a new ring, multiplied by $PV_{bt}$ at the velocity at that radius. Adding up the rings, we get:

Volume of acid per metre for wormholes out to the radius r_wh

\frac{V}{h} = \int_{r_w}^{r_{wh}} 2\pi r\,\phi\,PV_{bt}\big(v_i(r)\big)\,dr \;\approx\; \pi\,\phi\,PV_{bt}\,\left(r_{wh}^2 - r_w^2\right)

where:

$V/h$ volume of acid per metre of interval, m³/m
$r_{wh},\ r_w$ radius of the wormhole front and the well radius, m
$PV_{bt}(v_i)$ pore volumes to breakthrough at the velocity at the front (the efficiency curve), —
$\phi$ porosity, —

Units: SI (m, m³/m)

Holds when

The wormhole front advances at the velocity $v_{wh}(v_i)$ , as in a linear test at the same pore velocity. The rate is constant.
The second form is a “volumetric model” with one value of $PV_{bt}$ . It is good if the velocity along the whole path is close to the optimum. Otherwise the integral is calculated by rings.
The rock is uniform. Natural fractures and vugs take acid faster and change the picture.

Compare this with the formula for a solid front (volume of a stage per metre). There the brackets hold $\phi + (1 - \phi)\,x/X$ , which for pure limestone with 15% acid is about 11 volumes of rock. Here $\phi\,PV_{bt}$ is about a tenth. Wormholes dissolve only a small share of the rock along their path, and that is why they go so far.

Worked exampleVolume of acid and rate for wormholes out to 1.5 m

Given (assumed data)

Reservoir: limestone; interval 24 m, porosity 0.10, $k$ = 40 mD, $r_w$ = 0.108 m, $r_e$ = 300 m; damage skin +6, the damage reaches 0.4 m from the axis
Pressures: reservoir 30.0 MPa, fracture opening 48.0 MPa, margin 2.0 MPa; viscosity of the fluid in the reservoir ahead of the acid about 1.0 mPa·s
Acid: 15% HCl; core tests at reservoir temperature: $v_{opt}$ = 1.5 cm/min $= 2.5\cdot10^{-4}$ m/s, $PV_{bt,opt}$ = 0.7
Aim: wormholes out to 1.5 m from the axis

Solution

The limit on the rate at a skin of +6 (formula for the limit on the rate): $q_{\max} = \dfrac{0.536\cdot 40\cdot 24\cdot(46.0 - 30.0)}{1.0\cdot(7.929 + 6)} = 591$ m³/d, or 0.41 m³/min. We start at 0.40 m³/min.
The velocity at the wall: $v_i = \dfrac{0.40/60}{2\pi\cdot 0.108\cdot 24\cdot 0.10} = 4.093\cdot10^{-3}$ m/s, or 24.6 cm/min, which is 16.4 times the optimum.
The optimum velocity holds out to $r_{opt} = \dfrac{0.40/60}{2\pi\cdot 24\cdot 0.10\cdot 2.5\cdot10^{-4}} = 1.77$ m. For it to hold out to 3 m we would need 0.68 m³/min, which is more than the limit on the rate.
We divide the path into three rings and take $PV_{bt}$ at the velocity in the middle of each ring. The ring 0.108–0.5 m: the middle is 0.304 m, $v_i/v_{opt}$ = 5.817, $PV_{bt}$ = 1.259, and the volume is $\pi\cdot0.10\cdot(0.5^2 - 0.108^2)\cdot 1.259 = 0.094$ m³/m.
The ring 0.5–1.0 m: $v_i/v_{opt}$ = 2.358, $PV_{bt}$ = 0.932, and the volume is $\pi\cdot0.10\cdot(1.0^2 - 0.5^2)\cdot 0.932 = 0.220$ m³/m. The ring 1.0–1.5 m: $v_i/v_{opt}$ = 1.415, $PV_{bt}$ = 0.786, and the volume is $\pi\cdot0.10\cdot(1.5^2 - 1.0^2)\cdot 0.786 = 0.309$ m³/m.
The total is 0.623 m³/m (a calculation with finer rings gives 0.620). For the interval, $0.623\cdot 24 = 15.0$ m³, which is about 38 min of pumping at 0.40 m³/min.
For comparison, a solid front (one cubic metre of 15% acid dissolves 0.0815 m³ of calcite): $r = \sqrt{0.108^2 + 0.623/\big(\pi\,(0.10 + 0.90/0.0815)\big)} = 0.172$ m, only 6.4 cm beyond the wall.
The wormholes have passed through the damaged ring, and the skin is close to $-\ln(1.5/0.108) = -2.63$ . At the same drawdown (the pressure difference that pushes fluid into the well), productivity will grow by a factor of $\dfrac{7.179 + 6}{7.179 - 2.63} = 2.90$ , where $7.179 = \ln(300/0.108) - 0.75$ .

Answer: About 0.623 m³ of 15% hydrochloric acid per metre (15.0 m³ for the interval) takes the wormholes 1.5 m out. The skin will fall from +6 to about −2.63, and productivity will grow 2.90 times.

What it means. Half of the acid is spent in the last ring, because the volume grows as the square of the radius. The rate decides how far we can go. At 0.2 m³/min, out to 1.5 m we need almost as much acid (0.59 m³/m), but out to 3 m we need 6.8 m³/m already. After 0.9 m the velocity at the front falls below the optimum, and the further out, the closer the regime gets to face dissolution. At 0.4 m³/min it is 2.4 m³/m, and at 1.2 m³/min it is 2.9 (this rate is above the limit for this well even after the treatment, and it is here only for comparison). The estimate is approximate. A curve from samples of standard size usually overstates $PV_{bt}$ for the reservoir, and the wormhole zone does not conduct infinitely well, so the skin of −2.63 is a limit that the treatment approaches. The volumes are refined from well tests after the first treatments. The data in this example are assumed for illustration.

CalculatedThe radius of the wormhole front against the volume of acid per metre at three rates into the 24 m interval (the model of this lesson, data of the example; 1.2 m³/min is above the limit on the rate of this well and is shown for comparison), and the radius of a solid dissolving front at the same volume. At a low rate the wormholes quickly reach the first metre and then slow down sharply. At a high rate more acid is spent on branching at the wall.

Acid bath: when wormholes are not needed

Wormholes are needed when the damage lies in the reservoir. Damage right at the wall (mudcake of the drilling fluid, cement, and debris in the perforations) is only bypassed by acid that has run off along the wormholes. What works against it is exactly the regime that the rest of the lesson treats as a mistake: face dissolution. What to dissolve depends on the make-up of the damage. Hydrochloric acid dissolves cement and the carbonate filler of the cake. But acid does not wet the cake of an oil-based drilling fluid, so the cake is first removed with a mutual solvent (a liquid that mixes with both oil and water). The acid is placed opposite the interval with coiled tubing and left for a few hours, or it is pumped so slowly that the velocity at the wall is far below the optimum. The standing time is limited by the corrosion inhibitor: the stronger and hotter the acid, the fewer hours the inhibitor protects the pipes. Afterwards the spent acid, with dissolved iron and suspended solids, is washed out or taken out, and not left in the well.

Acid diversion in carbonates

In sandstone the acid is spent in the first tens of centimetres, and a permeable layer takes it only in proportion to its flow capacity. In carbonate the feedback is stronger. The acid that has gone into a layer cuts wormholes there, the skin of the layer becomes negative, and the layer takes even more acid. It is worst with natural fractures and vuggy layers (layers full of small cavities), which take acid almost without resistance. So diversion in carbonates is not an extra but a part of the design. At a common bottomhole pressure the rate is divided between the layers in proportion to how much each layer will take (its injectivity):

Share of the rate of a layer at injection with a common bottomhole pressure

\frac{q_j}{q} = \frac{k_jh_j\big/\left(\ln(r_e/r_w) + S_j\right)}{\sum_i k_ih_i\big/\left(\ln(r_e/r_w) + S_i\right)}

where:

$q_j,\ q$ rate into layer $j$ and the total rate, m³/min
$k_j,\ h_j$ permeability and thickness of the layer, mD, m
$S_j$ skin of the layer: damage, wormholes, a slug of diverting agent, —
$r_e,\ r_w$ radius of the drainage area and well radius, m

Units: fractions; the units of $k$ and $h$ are the same for all layers

Holds when

Steady injection, as in the formula for the limit on the rate. The reservoir pressure and the viscosity of the fluid are the same in all layers, and there is no crossflow between the layers at the wellbore.
A slug of diverting agent (a batch of it pushed into the layer) adds $(\mu_{\text{v}}/\mu - 1)\ln(r_{\text{v}}/r_w)$ to the skin of the layer, where $\mu_{\text{v}}$ and $r_{\text{v}}$ are its apparent viscosity and radius (skin of a viscous diverter).

The share of a layer is proportional to its flow capacity divided by the sum of the logarithm and the skin. Lowering the skin of a layer from +4 to −2.5 doubles its weight. This is the feedback that the diverter breaks.

Methods of diverting acid in carbonates

Method: Packers, a straddle packer on coiled tubing
How it works: the interval is treated in parts
Where it fits: cased hole; expensive and slow

Method: Ball sealers
How it works: balls sit on the perforations that take fluid and close them
Where it fits: cased and perforated hole; the rate must be high enough to press the balls on

Method: Diverting particles
How it works: a cake of particles on the wall of the layer that takes fluid: salt, benzoic acid, degradable fibres
Where it fits: short and medium intervals; the particles must dissolve afterwards

Method: Self-diverting fluids
How it works: the viscosity grows where the acid is neutralised, that is, in the working wormholes
Where it fits: long intervals, open hole

Method: Foam
How it works: in permeable, water-bearing rock the foam is stronger, and in oil-bearing rock it breaks down
Where it fits: high heterogeneity, watered-out layers

Method	How it works	Where it fits
Packers, a straddle packer on coiled tubing	the interval is treated in parts	cased hole; expensive and slow
Ball sealers	balls sit on the perforations that take fluid and close them	cased and perforated hole; the rate must be high enough to press the balls on
Diverting particles	a cake of particles on the wall of the layer that takes fluid: salt, benzoic acid, degradable fibres	short and medium intervals; the particles must dissolve afterwards
Self-diverting fluids	the viscosity grows where the acid is neutralised, that is, in the working wormholes	long intervals, open hole
Foam	in permeable, water-bearing rock the foam is stronger, and in oil-bearing rock it breaks down	high heterogeneity, watered-out layers

Besides these, the highest allowable rate is used (all layers get acid faster), and limited-entry perforating in horizontal wells (the next section). A diverter raises the injection pressure, and this rise does not mean new damage.

Worked exampleWhere the acid goes: two layers without diversion and with foam

Given (assumed data)

Interval of 24 m: layer A: 6 m, 100 mD, skin +4; layer B: 18 m, 20 mD, skin +8; porosity 0.10, $\ln(r_e/r_w)$ = 7.929
First batch of acid: the wormholes in A pass through the damage, and its skin falls to −2.5; B is hardly treated
Foam: 2.4 m³ (0.1 m³ per metre of interval); apparent viscosity relative to the acid is 40 in A and 10 in B (assumed: foam is stronger in more permeable rock)

Solution

At the start the shares are proportional to $k_jh_j/(\ln(r_e/r_w) + S_j)$ : A is $600/(7.929 + 4) = 50.30$ and B is $360/(7.929 + 8) = 22.60$ . The shares are 69.0% and 31.0%. Per metre of thickness, A gets 6.7 times more acid.
After the first batch: A is $600/(7.929 - 2.5) = 110.52$ . The shares are 83.0% and 17.0%, and per metre A already gets 14.6 times more.
The foam is divided in the same way: 1.99 m³ into A (0.332 m³ per metre) and 0.41 m³ into B (0.0228 m³/m). In A it will fill the pores out to $\sqrt{0.108^2 + 0.332/(\pi\cdot 0.10)} = 1.034$ m, and in B out to 0.290 m.
The skin of the foam slug: in A, $39\ln(1.034/0.108) = 88.1$ , and in B, $9\ln(0.290/0.108) = 8.89$ .
The next batch of acid: A is $600/(7.929 - 2.5 + 88.1) = 6.42$ , and B is $360/(7.929 + 8 + 8.89) = 14.51$ . The shares are 30.7% and 69.3%, which is 5.12% and 3.85% of the rate per metre of thickness.

Answer: Without diversion, layer A, a quarter of the thickness of the interval, takes 83.0% of the acid. Behind a batch of foam it takes 30.7%, and per metre of thickness the layers get almost equal amounts: 5.12% and 3.85% of the rate.

What it means. What decided it is that foam is stronger in the permeable layer. A viscous diverter of the same viscosity in both layers (30) would also go mostly into A, but it would raise the skin of B noticeably too. The shares would be 51% and 49%, and per metre A would get 3.1 times more. Foam breaks down with time, especially in contact with oil, so batches of acid and diverter are alternated. The viscosities of the foam that we took are an assumption of the example. In practice they are chosen from core tests. The data in this example are assumed for illustration.

CalculatedThe share of the acid rate per metre of thickness of layers A and B of the example: at the start of the treatment, after the wormholes have passed through the damage in A, and behind a batch of foam. The calculation uses the formula for the division of the rate with the data of the example.

Why, then, does a treatment without diversion not always fail? To pass a shallow damaged zone the wormholes need only a little acid. At the rate of the earlier wormhole example, the front reaches the edge of the damage (0.4 m) after about 0.06 m³ per metre (see the advanced section). So with a moderate difference between layers, the tight layer also gets its share in time and is cleaned. Diversion decides the outcome when the difference is large (fractures, vugs, a difference of flow capacity of many times), when the interval is long and there is little acid per metre, and when the tight layer gets so little acid that the velocity at its front falls into the face dissolution regime.

Long horizontal wells

A horizontal well in carbonate that is hundreds of metres or kilometres long sets three tasks at once. There is little acid per metre: even 0.25 m³ per metre over a kilometre of wellbore is 250 m³, a whole row of tanks. The well is usually open hole (bare rock) or has an uncemented liner, a pipe that is not cemented in place (completion options). The space between the liner and the rock, the annulus, joins the whole interval, and a packer in the casing cuts nothing off. Finally, acid pumped down the tubing meets the heel of the well (the end nearest the surface) first, and on the way to the toe (the far end) it loses pressure to friction (friction along the wellbore). Without special measures almost all the acid goes into the first part of the well and into fracture zones. The ways of delivery are:

Pumping down the tubing with diverting fluids. It is cheap and fast, but the distribution is hard to predict. Self-diverting acids and foam help.
Coiled tubing. The acid is delivered opposite the section that needs it, while the tubing is pulled from the toe towards the heel, sometimes through a rotating jetting nozzle that also strips off the cake. The rate through coiled tubing is low, and the velocity at the wormhole front may turn out to be below the optimum. So emulsified acid, whose optimum is lower, is often pumped through it.
A liner with limited entry. A few holes are made in the uncemented liner, sized so that the pressure loss in them is larger than the difference in injectivity between the sections (limited-entry perforating). The acid is pumped down the tubing at a high rate, and the holes themselves distribute it along the well. The holes are placed more densely towards the toe, to make up for the friction along the liner.
Sections with annular packers and sleeves. The well is divided in advance by swellable or mechanical packers, and acid is pumped into each section in turn, with the sleeves (valves in the liner wall) opened by dropped balls. This is already a staged treatment, close to multistage hydraulic fracturing (clusters and limited entry).

A well that was opened with a drilling fluid that contains carbonate bridging particles is covered with a cake that the acid dissolves by itself. But this also works against an even spread of the acid: where the cake dissolved first, the rest of the acid goes there. So the cake is often removed first with a small volume of acid through coiled tubing, and the reservoir is treated afterwards.

Worked exampleA liner with limited entry against an open liner

Given (assumed data)

Well: horizontal, 800 m, eight sections of 100 m each; injectivity of a section 0.06 m³/(min·MPa); section 3 crosses a fracture zone, 0.50 m³/(min·MPa)
Injection: 2.4 m³/min of 15% HCl (1073 kg/m³) down the tubing; acid 0.25 m³ per metre, 200 m³ in total; allowable excess of pressure over the reservoir pressure 10 MPa
Limited entry: 4 holes of 6 mm diameter in each section, discharge coefficient 0.6; friction along the liner is ignored

Solution

Open liner: the pressure above the reservoir pressure is $\Delta p = 2.4/(7\cdot0.06 + 0.50) = 2.61$ MPa. Section 3 takes $0.50\cdot 2.61 = 1.305$ m³/min (54%), and each of the others takes 0.157 m³/min (6.5%).
The loss in the holes of a section (formula for the loss in the holes): $\Delta p_{\text{perf}} = \dfrac{\rho q^2}{2C_d^2 n^2 A^2}$ , $A = \pi\cdot0.006^2/4$ . With the rate in m³/min and the pressure in MPa this is $a q^2$ with $a = 32.36$ .
The rate of a section at a pressure $\Delta p_L$ in the liner above the reservoir pressure is $q = I(\Delta p_L - a q^2)$ , where $I$ is the injectivity of the section. This gives $q = \dfrac{-1 + \sqrt{1 + 4aI^2\Delta p_L}}{2aI}$ . By trial we find $\Delta p_L$ = 7.20 MPa. An ordinary section takes $\dfrac{-1 + \sqrt{1 + 4\cdot 32.36\cdot 0.06^2\cdot 7.20}}{2\cdot 32.36\cdot 0.06} = 0.280$ m³/min, and section 3 takes 0.442 m³/min. Check: $7\cdot 0.280 + 0.442 = 2.402$ .
The shares: section 3 gets 18.4%, and each of the others 11.7%. The losses in the holes are $32.36\cdot 0.280^2 = 2.54$ MPa for an ordinary section and 6.32 MPa for section 3.
Acid per metre (200 m³): with the open liner, section 3 gets 1.09 m³/m and each of the others 0.13. With limited entry it is 0.37 and 0.23 m³/m.

Answer: Limited entry lowers the share of the fracture zone from 54 to 18.4%, and each ordinary section gets 11.7% instead of 6.5%. The pressure in the liner above the reservoir pressure is 7.20 MPa instead of 2.61, which is within the allowable 10 MPa.

What it means. The holes work because the loss in them grows as the square of the rate: a section that tries to take more “chokes” itself. The price is pressure: 7.20 MPa above the reservoir pressure instead of 2.61, and the margin to the fracture opening pressure is checked for each section. As the treatment goes on, the injectivity of the sections grows, and limited entry keeps the distribution as long as the loss in the holes is larger than the difference between the sections. Friction along the liner gives more rate to the sections near the heel, and it is taken into account when the holes are placed. The data in this example are assumed for illustration.

CalculatedThe share of the acid rate by section of the horizontal well of the example: an open liner and a liner with limited entry (four holes of 6 mm per section). Section 3 crosses a fracture zone.

Injection wells, gravel packs and fractured reservoirs

Injection wells lose injectivity because of suspended solids (fine particles carried in the water), oil, corrosion products and bacteria in the injected water, and because of salts when waters mix (damage during injection). The Unified Rules for the Rational and Integrated Use of Subsoil (Order of the Minister of Energy of 15 June 2018 No. 239) require the injectivity of an injection well to be maintained over the whole specified thickness. That means restoring the filtration properties of the near-wellbore zone, and improving them where needed (paragraph 248). Among the additional measures for unfavourable reservoir properties, the rules name treating the near-wellbore zone with acids and solutions of surfactants (paragraph 251). The water used to prepare the injected solutions of acids and other reagents must not destroy the reagents or form precipitates with them, and it must be compatible with the formation water (paragraph 253).

Usually there is no clean-up. The spent acid is pushed into the reservoir with the injected water. In carbonate the reaction products are soluble, but the iron stays in solution and comes out when the acid is neutralised far from the wellbore, so an iron control agent is essential. In sandstone the spent mud acid is moved away with an overflush (secondary precipitates).
Pipes first, then the reservoir. The water line and the tubing of an injection well carry rust and salts. They are washed first with acid, with the returns taken to the surface. Otherwise the treatment will pump new damage into the reservoir.
Fractures from cold water. Many injection wells work with open fractures. The acid will go into them, and this is no longer a matrix treatment. Injectivity profiles before and after show which intervals took the acid.
Evaluation is from the injectivity index and the slope of the Hall plot (a graph that shows how hard it is to push water into the well). After a successful treatment the slope falls (monitoring of injection).

Wells with a gravel pack (how a pack is built) are rare in carbonates, because packs are placed in weak sandstones. But the order of reasoning is the same: first find where the pressure is being lost, and only then choose the fluid. In a cased well most of the loss in the pack happens in the perforation tunnels filled with gravel. Their cross-section is small and the velocity is high. Fines (tiny rock particles) or polymer residues of the carrier fluid in the gravel of the tunnels increase these losses several times (losses in the pack). A well test gives only the total skin. Three things suggest where exactly the plugging is: the calculated skin of a clean pack, the history of the well (did the rate fall right after the pack was placed, or later, as fines were carried out?), and the samples brought up during flushing. Polymer residues are broken down with oxidisers or enzymes, carbonate deposits with hydrochloric acid, and clay fines of the sandstone with mud acid. Damage to the reservoir behind the pack is treated with acid by the general rules (Acid systems and treatment design). The fluid is placed with coiled tubing opposite the interval, so that it washes the place of the plugging and does not go away through the first permeable section. There is almost nothing to divert the acid with: packers and ball sealers do not work through a pack, and the gravel will stop large particles. What remains is foam, self-diverting fluids, and particles smaller than the pores of the gravel.

Fractured carbonates. The acid goes by the same paths along which the oil comes to the well, into a few open fractures (types of fractured reservoirs). The matrix between them can be almost impermeable. A calculation of the volume from the growth of wormholes does not apply here. The acid works on the walls of the fractures, as in acid fracturing (see below), only without splitting the rock. So we first check what will remain in the fracture after the acid has etched (eaten into) its walls. The solubility of the rock in hydrochloric acid is measured on cuttings or core. If there is a lot of insoluble residue (clays, quartz), it will settle in the etched channel and may plug it. Then additives that hold the particles in suspension are put into the acid, and the well is cleaned up right after the treatment. The gain near the wellbore is explained by the cubic law: the conductivity of a fracture grows as the cube of its aperture, so an aperture twice as large gives eight times the conductivity, but only where the acid has arrived unspent. If the reservoir took drilling fluid in during drilling, the fractures near the wellbore are blocked with its solids and with cement. The acid dissolves the carbonate filler of the drilling fluid, but not barite (the heavy mineral powder that makes the fluid heavier), which is removed only by flushing and by inflow. And remember water: a fracture that is connected to the water-bearing part of the reservoir will pass water after the treatment as readily as oil.

Acid fracturing or matrix treatment

A matrix treatment of a carbonate gives a skin of down to −2…−3. Wormholes do not go deeper, and their length sets the limit (limits on lowering the skin). In a tight reservoir this is not enough, and a fracture is needed. In a carbonate we can get a fracture without proppant (the sand or ceramic grains that hold a fracture open): acid fracturing. A viscous fluid opens a fracture, and then acid is pumped in. It flows along the walls and dissolves them unevenly: deeper where the rock is cleaner and more permeable, less where it is tighter. When the pressure is released, the ridges rest against each other, and open channels remain between them. The ideal etched width shows how much rock the acid can remove from the walls:

Ideal etched width of an acid fracture

w_i = \frac{X\,V}{(1 - \phi)\cdot 2x_f h_f}

where:

$w_i$ ideal width: the volume of dissolved rock spread over the plane of the fracture, m
$X$ volume dissolving power of the acid (lesson on acid systems), m³/m³
$V$ volume of acid that has reacted in the fracture, m³
$\phi$ porosity of the rock: the dissolved mineral took the share $(1 - \phi)$ of the volume, —
$x_f,\ h_f$ half-length and height of the etched fracture; $2x_fh_f$ is the area of the plane of both wings, m

Units: SI

Holds when

All the acid has reacted on the walls within the fracture, with no leak-off (loss of acid into the reservoir rock) and no use near the wellbore. In reality the width is smaller.
The etching is even. But the conductivity is created by unevenness, so $w_i$ is a measure of the volume of work that the acid has done, not the width of the channel after closure.

Even a large acid treatment removes only millimetres of rock from the walls. The conductivity after closure depends on how unevenly the walls are etched and on whether the ridges will withstand the closure pressure (the stress that squeezes the fracture shut). It is estimated from tests on samples and from experimental correlations that take the strength of the rock into account, for example those of Nierode and Kruk.

Worked exampleHow much rock acid removes in a fracture

Given (assumed data)

Acid: 40 m³ of 15% HCl; dissolving power for calcite 0.0815 m³/m³
Fracture: half-length 50 m, height 20 m; limestone, porosity 0.08

Solution

The volume of dissolved calcite: $0.0815\cdot 40 = 3.26$ m³. The volume of rock with its pores: $3.26/(1 - 0.08) = 3.543$ m³.
The area of the plane of the fracture (both wings): $2\cdot 50\cdot 20 = 2000$ m². The ideal width is $w_i = 3.543/2000 = 1.77$ mm.

Answer: If all the acid reacted on the walls, it would remove a layer with a total thickness of about 1.77 mm, which is less than a millimetre from each wall.

What it means. In practice a large part of the acid leaks off into the reservoir, all the faster because wormholes grow from the fracture walls, and it is used up near the wellbore. So the etched length is shorter than the created length, especially in hot reservoirs. To make the acid go further, gelled, emulsified and self-diverting acids are used, and the acid is alternated with a viscous fluid that reduces leak-off (leak-off of the fracturing fluid). The data in this example are assumed for illustration.

Matrix treatment, acid fracturing or propped fracturing

Condition: Permeability
Matrix: medium and high
Acid fracturing: low
Propped fracturing: low

Condition: Rock
Matrix: any carbonate
Acid fracturing: strong, heterogeneous
Propped fracturing: any, including soft chalk

Condition: Water or gas nearby
Matrix: acceptable
Acid fracturing: dangerous: the fracture will reach it
Propped fracturing: dangerous: the fracture will reach it

Condition: Expected skin
Matrix: down to −2…−3
Acid fracturing: −3…−5
Propped fracturing: −3…−6

Condition: Complexity and cost
Matrix: low
Acid fracturing: medium
Propped fracturing: high

Condition	Matrix	Acid fracturing	Propped fracturing
Permeability	medium and high	low	low
Rock	any carbonate	strong, heterogeneous	any, including soft chalk
Water or gas nearby	acceptable	dangerous: the fracture will reach it	dangerous: the fracture will reach it
Expected skin	down to −2…−3	−3…−5	−3…−6
Complexity and cost	low	medium	high

The guide values are qualitative. In soft rocks the ridges of the etched walls are crushed, and a fracture without proppant almost closes. In strong heterogeneous limestones acid fracturing works without proppant and without the risk of proppant being carried out of the fracture. The choice between fractures is made from a calculation of inflow and from economics (choosing wells for hydraulic fracturing).

Campaigns of treatments and evaluating the result

A single treatment teaches little. Whether it worked or not, we cannot tell the cause from chance. So treatments are run in campaigns: series in one field, in which the cycle “selection, diagnosis, design, execution, evaluation” (treatment cycle) is repeated, and each next job uses the experience of the previous ones. The Unified Rules for the Rational and Integrated Use of Subsoil require the method of treating the near-wellbore zone to be chosen from studies of its condition and of the composition of the rocks and fluids. The choice must also rest on a systematic summary of the results of the methods by well and by reservoir unit (paragraph 315). They also require a set of measurements before and after each measure: treatment of the near-wellbore zone, hydraulic fracturing and others (paragraph 336). The rules do not count such jobs as major or routine well repair (paragraph 320).

To evaluate each job we collect the rate, the water cut (the share of water in the produced fluid) and the bottomhole pressure before and after, the skin from well tests, the record of the injection (rates, pressures, apparent skin; evaluation during pumping), and the costs. The main question is what to compare the rate after the treatment with. Without a treatment the well would not stand still either: its rate would decline. After a treatment it declines faster, because the well takes the same oil faster (faster production). So the gain is counted against the declining baseline, the forecast of production without a treatment:

Extra production over a period against the declining baseline

\Delta N_T = \frac{q_2\left(1 - e^{-a_2T}\right)}{a_2} - \frac{q_1\left(1 - e^{-a_1T}\right)}{a_1},\qquad a_2 = a_1\,\frac{X + S_1}{X + S_2}

where:

$\Delta N_T$ extra production over the period $T$ , m³
$q_1,\ q_2$ rate before the treatment and right after it, m³/yr
$a_1,\ a_2$ exponential decline rate without a treatment and after it, 1/yr
$X,\ S_1,\ S_2$ the logarithmic term $\ln(r_e/r_w) - 3/4$ , as in the lesson Acid treatments: principle, economics and candidate selection (not the dissolving power $X$ of the section on acid fracturing), and the skins before and after, —
$T$ period of evaluation, yr

Units: consistent units: the rate and the decline rate in the same units of time

Holds when

A closed drainage area with no pressure support and a constant bottomhole pressure (the drawdown falls together with the reservoir pressure). The decline is exponential, and its rate is proportional to productivity.
With pressure support by injection the gain lasts longer, and the formula understates it. With water breakthrough after the treatment, it overstates it.

The first term is the production of the treated well, and the second is the production of the same well without a treatment. If we compare with a constant rate before the treatment and ignore the faster decline after it, the annual gain comes out overstated.

Campaign: input data by well

Well: 1
$kh$ , mD·m: 400
Skin before → after: 9 → 1
Cost, US$ thousand: 95

Well: 2
$kh$ , mD·m: 150
Skin before → after: 12 → 11
Cost, US$ thousand: 90

Well: 3
$kh$ , mD·m: 600
Skin before → after: 3 → 2.9
Cost, US$ thousand: 100

Well: 4
$kh$ , mD·m: 300
Skin before → after: 8 → 0
Cost, US$ thousand: 95

Well: 5
$kh$ , mD·m: 350
Skin before → after: 10 → −1.5
Cost, US$ thousand: 120

Well: 6
$kh$ , mD·m: 200
Skin before → after: 7 → −2
Cost, US$ thousand: 115

Well: 7
$kh$ , mD·m: 500
Skin before → after: 6 → −2.5
Cost, US$ thousand: 125

Well: 8
$kh$ , mD·m: 120
Skin before → after: 11 → −1
Cost, US$ thousand: 110

Well	$kh$ , mD·m	Skin before → after	Cost, US$ thousand
1	400	9 → 1	95
2	150	12 → 11	90
3	600	3 → 2.9	100
4	300	8 → 0	95
5	350	10 → −1.5	120
6	200	7 → −2	115
7	500	6 → −2.5	125
8	120	11 → −1	110

Wells 1–4 were treated with the first technology: acid down the tubing without diversion. Wells 5–8 were treated after the first jobs were analysed: selection by breaking the skin down into components, foam, and the limit on the rate. The skins are from well tests before the treatment and after clean-up. The data are assumed.

Worked exampleEvaluating a campaign of eight treatments

Given (assumed data)

Wells: a carbonate reservoir; the skins and costs are in the table above
Common data: drawdown at the time of treatment 5.0 MPa, $\mu$ = 1.2 mPa·s, $B$ = 1.3; $X = \ln(300/0.108) - 0.75 = 7.179$
Decline: without a treatment 0.25 1/yr; after it, faster by the same factor by which productivity grew
Price: net income of US$251.6 per cubic metre of oil, the assumed price from the example on the gain

Solution

Well 1: $q_1 = \dfrac{0.536\cdot 400\cdot 5.0}{1.2\cdot 1.3\cdot(7.179 + 9)} = 42.5$ m³/d. After the treatment (skin 1.0), $q_2 = 84.0$ m³/d. The ratio is $F = \dfrac{7.179 + 9}{7.179 + 1} = 1.978$ , and the decline rate after the treatment is $0.25\cdot 1.978 = 0.4945$ 1/yr.
Production in a year without a treatment: $42.5\cdot 365\cdot\dfrac{1 - e^{-0.25}}{0.25} = 13{,}725$ m³. With the treatment: $84.0\cdot 365\cdot\dfrac{1 - e^{-0.4945}}{0.4945} = 24{,}188$ m³. The gain is 10,463 m³.
Without allowing for the decline the gain would be $(84.0 - 42.5)\cdot 365 = 15{,}148$ m³, which is 45% more.
In the first days the gain brings in $(84.0 - 42.5)\cdot 251.6/1000 = 10.44$ US$ thousand per day, and the US$95 thousand of costs are returned in about 9 days.
The same for all the wells (the table below): treatments 2 and 3 did not pay back in a year. In the first half of the campaign 2 of 4 jobs were successful, and in the second half 4 of 4. The gain in a year is 20,350 and 57,553 m³ at costs of US$380 thousand and US$470 thousand, that is, US$18.7 and US$8.2 per cubic metre of gain.
The whole campaign: 6 successful jobs out of 8, a gain of 77,903 m³, and costs of US$10.9 per cubic metre (US$1.73 per barrel). Well 7 gives 35% of the whole gain.

Answer: The campaign paid back many times over: 77,903 m³ of extra oil in a year at US$10.9 of cost per cubic metre. The second half gave 2.8 times more oil than the first, at a cost per cubic metre 2.3 times lower.

What it means. The two failures of the first half are different. In well 2 the acid went into a watered-out permeable layer, and the skin hardly changed. In well 3 there was almost no damage: most of the skin came from the pseudo-skins of the opening (extra skin from the way the well was completed, not from damage), which acid does not remove. Both causes are removed by selection and diversion, which is what was done in the second half. The averages of a campaign hide the spread: well 7 alone gave about a third of the gain. So we look both at the share of successful jobs and at every failure. The decline model is right for a closed volume without pressure support. With waterflooding (water injected to push the oil towards the wells) the gain lasts longer. The data in this example are assumed for illustration.

Campaign: results for a year

Well: 1
Rate before → after, m³/d: 42.5 → 84.0
Gain in a year, m³: 10,463
Payback: 9 d

Well: 2
Rate before → after, m³/d: 13.4 → 14.2
Gain in a year, m³: 228
Payback: more than a year

Well: 3
Rate before → after, m³/d: 101.3 → 102.3
Gain in a year, m³: 283
Payback: more than a year

Well: 4
Rate before → after, m³/d: 34.0 → 71.8
Gain in a year, m³: 9376
Payback: 10 d

Well: 5
Rate before → after, m³/d: 35.0 → 105.9
Gain in a year, m³: 15,816
Payback: 7 d

Well: 6
Rate before → after, m³/d: 24.2 → 66.3
Gain in a year, m³: 9708
Payback: 11 d

Well: 7
Rate before → after, m³/d: 65.2 → 183.6
Gain in a year, m³: 27,047
Payback: 4 d

Well: 8
Rate before → after, m³/d: 11.3 → 33.4
Gain in a year, m³: 4982
Payback: 20 d

Well	Rate before → after, m³/d	Gain in a year, m³	Payback
1	42.5 → 84.0	10,463	9 d
2	13.4 → 14.2	228	more than a year
3	101.3 → 102.3	283	more than a year
4	34.0 → 71.8	9376	10 d
5	35.0 → 105.9	15,816	7 d
6	24.2 → 66.3	9708	11 d
7	65.2 → 183.6	27,047	4 d
8	11.3 → 33.4	4982	20 d

The gain is measured against the declining baseline, by the formula above. The payback is the time in which the income from the gain, at US$251.6/m³, returns the costs. The rounded values from the table give the totals of the example.

CalculatedThe skin of the wells of the campaign before and after treatment: the light circle is before, and the dark one is after. Treatments that did not pay back in a year are marked in orange. The second technology (wells 5–8) lowers the skin well below zero.

Acids, pressure and hydrogen sulfide

The Rules for ensuring industrial safety at hazardous production facilities of the oil and gas industries (Order of the Minister for Investment and Development of 30 December 2014 No. 355) require the following. Stimulation (a treatment that raises the rate of a well) is carried out under a work organisation plan (paragraph 733). The plan lists the reagents, their maximum allowable concentrations and their hazard class (paragraph 734). The near-wellbore zone is not treated in wells with leaking wellhead equipment or casing, flow behind the casing, or pressure between the casings (paragraph 736). The lines are pressure-tested (paragraph 737). A danger zone with a radius of at least 50 m is set up (paragraph 739). Corrosion inhibitors are used in acid fracturing (paragraph 760). Personnel are provided with protective equipment, and the air of the work area is monitored (paragraph 761). The description in this lesson is for teaching. It is not a work procedure. Treatments are run by trained crews under approved plans.

Terms and symbols

Wormhole: A thin, branching channel that acid cuts in a carbonate by dissolving the rock along the most conductive pores.
Pore volumes to breakthrough $PV_{bt}$: The volume of acid until a wormhole comes out of the sample, divided by the pore volume of the sample. A measure of the efficiency of an acid.
Optimum velocity $v_{opt}$: The pore velocity at which $PV_{bt}$ is the smallest: one dominant wormhole grows.
Pore velocity $v_i$: The flow rate divided by the cross-sectional area of the flow and the porosity. In a well it falls in inverse proportion to the distance from the axis.
Damköhler number $Da$: The ratio of the speed at which acid is used up on the walls to the speed at which it is supplied. The optimum of dissolving is at about 0.29.
Emulsified acid: Acid dispersed in an oil phase. It reacts more slowly, and its optimum velocity is lower.
Ball sealers: Balls that are pumped with the acid. They sit on the perforations that take fluid and divert the flow into others.
Limited entry: A small number of holes in which the pressure loss evens out the rate between sections.
Acid bath: Acid placed opposite an interval with no injection, or with a very low rate, to dissolve cake and cement on the wall.
Acid fracturing: A fracture in a carbonate whose conductivity is created by uneven etching of the walls by acid, without proppant.
Declining baseline: The forecast of the production of a well without a treatment. The extra production is counted as the difference from it.

Common mistakes and misconceptions

Wrong

“The slower we pump the acid, the better it reacts and the deeper the treatment goes.”

In fact

In a carbonate at a low velocity the acid has time to react at the wall and dissolves it with a solid front, which is the most wasteful regime. Only acid baths are placed slowly, when the damage is right at the wall.

Wrong

“We calculate the volume of acid for limestone from the dissolving power, as for sandstone.”

In fact

Wormholes dissolve only a small share of the rock along their path. In the example of the lesson, a solid front with the same acid would reach 0.17 m instead of 1.5 m. The volume is calculated from the efficiency curve and the velocity at the front.

Wrong

“The skin after the treatment is negative, so there was enough acid for the whole interval.”

In fact

An average skin can be negative when one permeable layer is treated with an excess and the tight ones are hardly touched. The distribution is checked with inflow or injectivity profiles, and diversion is designed in advance.

Wrong

“An injection well can be treated by simply pumping acid down the water line.”

In fact

The water line and the tubing carry rust and salts. Without a preliminary wash with the returns taken to the surface, the acid will pump them into the reservoir. Also, the acid will go into the fractures if the well works with them.

Wrong

“Acid fracturing is a stronger acid treatment, and it fits everywhere that a matrix treatment is not enough.”

In fact

The conductivity of an acid fracture holds only in strong, heterogeneous rock. In soft chalk the ridges are crushed under the closure pressure, and a propped fracture is needed.

In depth: advanced detail

How wormholes pass through the damaged zone

The formula for the volume per metre is an integral along the path of the front. It can also be written in time. The front moves at the velocity $dr_{wh}/dt = v_{wh}\big(v_i(r_{wh})\big)$ , and in the time $dt$ the amount of acid $q\,dt$ is pumped. Since $q = 2\pi r h\phi v_i$ , we get $dV = 2\pi rh\phi\,(v_i/v_{wh})\,dr$ , that is, $dV = 2\pi rh\phi\,PV_{bt}\,dr$ , the same formula. While the front is inside the damaged ring, the wormhole zone works as a wellbore of a larger radius, and the rest of the ring still resists. If we take the conductivity of the wormhole zone as infinite, the skin is made of two parts:

Skin of a well with wormholes inside the damaged ring

S = \begin{cases} \dfrac{k}{k_d}\ln\dfrac{r_d}{r_{wh}} - \ln\dfrac{r_d}{r_w}, & r_{wh} < r_d\\[2mm] -\ln\dfrac{r_{wh}}{r_w}, & r_{wh} \ge r_d \end{cases}

where:

$k,\ k_d$ permeability of the reservoir and of the damaged ring, mD
$r_d$ radius of the damaged ring, m
$r_{wh},\ r_w$ radius of the wormhole front and the well radius, m

Units: dimensionless; the logarithm is natural

Holds when

The wormhole zone conducts infinitely well, and the pressure in it equals the bottomhole pressure. So the skin comes out as a limit, and in reality it is somewhat higher.
The damage is uniform in the ring from $r_w$ to $r_d$ . At $r_{wh} = r_w$ the formula turns into the Hawkins formula (Skin factor).

While the front has not reached $r_d$ , the skin falls fast: the resistance of the ring falls together with its remaining thickness. Beyond that the skin decreases only logarithmically.

Worked exampleHow much acid is needed to pass through the damage

Given (assumed data)

Well and acid: the example on the rate: interval 24 m, porosity 0.10, rate 0.40 m³/min, $v_{opt}$ = 1.5 cm/min, $PV_{bt,opt}$ = 0.7
Damage: skin +6 in the ring out to 0.4 m: $k/k_d = 1 + 6/\ln(0.4/0.108) = 5.582$

Solution

The front at 0.2 m: 0.014 m³/m of acid, skin $5.582\ln(0.4/0.2) - \ln(0.4/0.108) = 2.56$ .
The front at 0.3 m: 0.035 m³/m, skin 0.30.
The front at 0.4 m: 0.061 m³/m, skin $-\ln(0.4/0.108) = -1.31$ . The damage has been passed. For the interval this is $0.061\cdot 24 = 1.46$ m³, or about 4 min of pumping.

Answer: Wormholes pass through damage 0.3 m deep beyond the wall of the well with about 0.061 m³ of acid per metre, a tenth of the volume of the whole treatment.

What it means. Two conclusions follow. First, even a layer that gets a small share of the acid is cleaned in time, when the difference of flow capacities is moderate. Second, most of the acid goes not on removing the damage but on creating a negative skin beyond it, and this is the part that is cut when there is little acid. The data in this example are assumed for illustration.

Why the optimum shifts: temperature, retarded acids and sample size

The Damköhler number explains where the optimum moves when conditions change. If the best $Da$ is the same for a wormhole of given size, the optimum rate through it is proportional to the dissolving constant: $q_{wh,opt} = \pi d l\kappa/0.29$ . Heating speeds up the reaction of dolomite and the delivery of acid to the walls in limestone. So $\kappa$ grows, and at the same rate in the well the regime moves to the left, towards conical channels and face dissolution. Emulsified acid, on the contrary, reaches the wall more slowly. Its $\kappa$ is smaller, and the optimum moves to low velocities. This is why it is chosen for hot reservoirs, long intervals and pumping through coiled tubing.

The second shift is a scale effect: it comes from the size of the sample. In a core test the wormhole grows in a narrow cylinder and soon comes out of its end. In a reservoir the fronts of many wormholes compete, and as they grow, the leading channels take the flow from the lagging ones. Tests on samples of different length and diameter show that the optimum velocity and $PV_{bt}$ both fall as the sample grows. There are models that transfer the laboratory curve to a well while taking this effect into account. But the most reliable way is to calibrate against the results of the first treatments in the field: from the skin after clean-up we find which radius of wormholes was reached with the volume that was pumped.

Connections to other topics

Acid treatments: principle, economics and candidate selection — what acid does in sandstone and carbonate, the economics of a treatment, and the choice of candidate wells.
Acid systems and treatment design — reactions and dissolving power, additives, the limit on the rate, and the apparent skin.
Carbonate sediments, environments and carbonate porosity — carbonate reservoirs, their voids, and the subsalt buildups of the Pre-Caspian Basin.
Horizontal wells — completion of horizontal wells and friction along the wellbore.
Perforating operations: gun conveyance and procedures — pressure losses in the holes, and limited entry.
Water injection systems — the quality of the injected water, injectivity and the Hall plot.
Sources of formation damage — damage during drilling, production and injection.
Fractures and naturally fractured reservoirs — natural fractures and fractured reservoirs.
Fracturing fluids and tip-screenout fracturing — fracturing fluids, leak-off and viscoelastic surfactants.
Hydraulic fracturing: purpose and well selection — when a fracture is needed instead of acid.

Check yourself

1
Core tests of limestone with hydrochloric acid gave $PV_{bt}$ = 9 at 0.3 cm/min and 0.8 at 2 cm/min, and 2 cm/min is close to the optimum. What will happen if the velocity is raised to 20 cm/min?
1. A. $PV_{bt}$ will keep falling
2. B.Wormholes will stop forming
3. C. $PV_{bt}$ will grow by an order of magnitude, as to the left of the optimum
4. D. $PV_{bt}$ will grow moderately, about twice: the acid will go into branching
Show the answer
Correct answer: D. $PV_{bt}$ will grow moderately, about twice: the acid will go into branching
To the right of the optimum $PV_{bt}$ grows slowly. In the model of the lesson it grows roughly as $v_i^{1/3}$ : $0.8\cdot 10^{1/3} \approx 1.7$ . To the left the growth is steep: at a velocity 7 times lower than the optimum, $PV_{bt}$ is already 11 times larger. So it is cheaper to err towards a higher velocity.
2
Acid is pumped at 0.6 m³/min into a 30 m interval with a porosity of 0.12. The optimum velocity from core is 1.0 cm/min. Out to what radius is the velocity at the wormhole front not lower than the optimum? What rate is needed to keep it out to 4 m?
Show the answer
$v_{opt} = 1.0$ cm/min $= 1.667\cdot10^{-4}$ m/s. The denominator is $2\pi\cdot 30\cdot 0.12\cdot 1.667\cdot10^{-4} = 3.770\cdot10^{-3}$ m²/s, and $r_{opt} = (0.6/60)/(3.770\cdot10^{-3}) \approx 2.65$ m. For 4 m: $q = 2\pi\cdot 4\cdot 30\cdot 0.12\cdot 1.667\cdot10^{-4}\cdot 60 \approx 0.90$ m³/min, which is one and a half times more. If the fracture pressure does not allow it, beyond 2.65 m the wormholes will grow more slowly than at the optimum, and more acid per metre will be needed. The way out is a retarded acid (one made to react more slowly) with a lower optimum velocity.
3
An open horizontal well of 1200 m in a carbonate, with an uncemented liner, crosses a fracture zone. Which method of distributing the acid is the most sensible here?
1. A.A packer in the casing above the interval
2. B.Lower the rate so that the acid distributes itself better
3. C.A liner with limited entry, self-diverting fluids, or delivery with coiled tubing
4. D.Ball sealers
Show the answer
Correct answer: C. A liner with limited entry, self-diverting fluids, or delivery with coiled tubing
Balls have nothing to sit on, because there are no perforations, and a packer in the casing does not shut off the annulus of the liner. The methods that work are those that act inside the well. Limited entry distributes the rate through the pressure loss in the holes, self-diverting fluids thicken in the sections that take fluid, and coiled tubing delivers the acid opposite the place that needs it. Lowering the rate will only make things worse: the acid will go into the face dissolution regime, and the fractures will still take most of it.
4
The interval consists of layer A (4 m, 300 mD, skin +2) and layer B (16 m, 15 mD, skin +6); $\ln(r_e/r_w) = 8$ . What share of the acid will each layer get at the start of pumping, and how many times more per metre of thickness does A get?
Show the answer
The weights: A is $1200/(8 + 2) = 120$ , and B is $240/(8 + 6) \approx 17.1$ . The shares: A is $120/137.1 \approx 87.5$ %, and B is 12.5%. Per metre: A gets 21.9%/m and B gets 0.78%/m, 28 times less. Layer B, which holds 80% of the thickness, will hardly get any acid, and when the wormholes lower the skin of A, its share will grow even more. Diversion is needed.
5
After a treatment the rate of a well grew from 30 to 66 m³/d (productivity grew 2.2 times). Without a treatment the rate would decline at 0.20 1/yr. After the treatment the decline rate grows in proportion to productivity. What is the gain in a year, and by how much would a comparison with the rate before the treatment overstate it?
Show the answer
The decline rate after the treatment is $0.20\cdot 2.2 = 0.44$ 1/yr. With the treatment: $66\cdot 365\cdot(1 - e^{-0.44})/0.44 \approx 19{,}490$ m³. Without it: $30\cdot 365\cdot(1 - e^{-0.20})/0.20 \approx 9{,}920$ m³. The gain is about 9,570 m³. The comparison “ $(66 - 30)\cdot 365 = 13{,}140$ m³” overstates it by about 37%, because it takes into account neither the decline of the baseline nor the faster decline of the treated well.
6
In which rock is acid fracturing most likely not to give lasting conductivity?
1. A.Strong, heterogeneous limestone
2. B.Fractured limestone
3. C.Soft chalk with high porosity
4. D.Dolomite with vugs
Show the answer
Correct answer: C. Soft chalk with high porosity
The conductivity of an acid fracture is held by the ridges of unevenly etched walls. In soft chalk they are crushed under the closure pressure, and the fracture almost closes, so hydraulic fracturing with proppant is needed here. In strong, heterogeneous carbonates the ridges carry the load.

Updated: 28 October 2026