Acid systems and treatment design

Quick refresher

Hydrochloric acid dissolves carbonates, and the products of its reactions are soluble. Hydrofluoric acid dissolves silica and aluminosilicates (the minerals of clays and feldspars): clays and feldspars quickly, quartz slowly. But its products with calcium, sodium and potassium come out as precipitates. This is why a sandstone treatment has three stages: hydrochloric acid in front, then mud acid, then an overflush.
The dissolving power is $X = C\beta\rho_{\text{a}}/\rho_{\text{ma}}$ . One cubic metre of 15% hydrochloric acid dissolves about 0.08 m³ of calcite (the main mineral of limestone). One cubic metre of mud acid with 3% HF dissolves only about 0.01 m³ of kaolinite (a clay mineral).
The volume of a stage per metre is the pore volume of the ring plus the acid needed to dissolve the mineral in it. The hydrochloric acid must remove carbonates further out than the hydrofluoric acid will reach. The overflush must push the spent mud acid 1–1.5 m away from the wellbore.
Secondary precipitates (a silica gel that forms from fluorosilicates, the salts of silicon and fluorine) cannot be avoided. They do less harm the further they are from the wellbore: the same amount of precipitate 1.5 m from the axis gives a skin several times smaller than at the wall. This is why we use an overflush and a quick clean-up.
A corrosion inhibitor is always needed. An iron control agent (a chemical that keeps iron in solution), non-emulsifying surfactants, a mutual solvent and a clay stabiliser are used after compatibility tests. Each additive is justified separately.
The bottomhole pressure is kept below the fracture pressure with a margin. The limit on the rate grows as the zone is cleaned, and the progress of the treatment shows in the fall of the apparent skin (the skin calculated from the pumping rate and pressure).

Key relationships

\beta = \dfrac{\nu_{\text{ma}}M_{\text{ma}}}{\nu_{\text{a}}M_{\text{a}}},\quad X = \dfrac{C\beta\rho_{\text{a}}}{\rho_{\text{ma}}}

mass and volume dissolving power

\dfrac{V}{h} = \pi(r^2 - r_w^2)\left[\phi + (1 - \phi)\dfrac{x}{X}\right]

volume of a stage per metre, so that the dissolving front reaches the radius $r$

q_{\max} = \dfrac{2\pi kh\,(p_f - \Delta p_{\text{mar}} - \bar p)}{\mu\,(\ln(r_e/r_w) + S)}

limit on the rate of matrix injection

p_{\text{wh}} = p_w - \rho g H + \Delta p_{\text{fr}}

wellhead pressure during injection

Common pitfalls

Pumping mud acid without a preflush, or diluting it with seawater and sodium and potassium brines. Calcium fluoride and fluorosilicates will come out.
Using mud acid in a rock where more than about 20% of the minerals dissolve in hydrochloric acid. Hydrofluoric acid is not needed there, and it is dangerous.
Leaving spent mud acid near the wellbore: giving no overflush, or waiting a long time before the clean-up.
Calculating the volume of acid from the pore volume. Dissolving the minerals takes several times more.
Limiting the injection by the wellhead pressure without converting it to the bottom of the hole. The hydrostatic pressure changes when the fluids in the pipes change.

After this lesson you can

write the reactions of hydrochloric and hydrofluoric acids with carbonates, quartz and clays, and explain where calcium fluoride, fluorosilicates and silica gel (a jelly-like form of silica) come from
calculate the dissolving power of an acid and the volume of rock that one cubic metre of the fluid dissolves
choose the strength of the mud acid and the additives from the mineralogy, the temperature and the composition of the reservoir fluids
calculate the volumes of the preflush (the acid pumped first), the main acid and the overflush (the fluid that pushes the spent acid away from the wellbore) per metre of interval, and show how the stages prevent secondary and tertiary precipitates (new deposits that form from the reaction products)
find the limits of the pumping rate and pressure below the fracture pressure, and convert the pressure to the wellhead
evaluate the result of the treatment from the skin (the extra resistance to flow near the well), both during pumping and from a well test after clean-up

What hydrochloric and hydrofluoric acids dissolve

The lesson Acid treatments: principle, economics and candidate selection explains how acid works in sandstone and in carbonate, and why the value of a treatment is set by the damage skin that it removes. The lesson Sources of formation damage explains which kinds of damage appear, and where. This lesson covers the chemistry and the design: which reactions take place, how much rock one cubic metre of fluid dissolves, how much fluid to pump and how fast.

Hydrochloric acid (HCl) quickly dissolves carbonates: calcite, dolomite and siderite (an iron carbonate). It forms chlorides that dissolve well, and carbon dioxide, which at reservoir pressure mostly dissolves in the liquid:

Reactions of hydrochloric acid with carbonates

\begin{aligned} 2\,\mathrm{HCl} + \mathrm{CaCO_3} &\rightarrow \mathrm{CaCl_2} + \mathrm{CO_2} + \mathrm{H_2O}\\ 4\,\mathrm{HCl} + \mathrm{CaMg(CO_3)_2} &\rightarrow \mathrm{CaCl_2} + \mathrm{MgCl_2} + 2\,\mathrm{CO_2} + 2\,\mathrm{H_2O}\end{aligned}

where:

$\mathrm{CaCO_3}$ calcite, the main mineral of limestones and of the carbonate cement of sandstones, —
$\mathrm{CaMg(CO_3)_2}$ dolomite, —
$\mathrm{CaCl_2},\ \mathrm{MgCl_2}$ calcium and magnesium chlorides, products of the reaction that dissolve well in water, —

Units: the coefficients are numbers of moles

Holds when

The acid is fully spent. The reaction with calcite is very fast, and with dolomite it is slower, especially at low temperature.
Siderite, $\mathrm{FeCO_3}$ , reacts in the same way as calcite, but it puts iron into solution.

Two moles of HCl are used per mole of calcite, and four per mole of dolomite. Spent hydrochloric acid is a solution of chlorides. It gives no precipitates as long as it contains no iron.

The usual strength of hydrochloric acid is 15% by mass. For carbonates, 28% is also used, and for the preflush of sandstones, 5–15%. The main danger is iron. The acid dissolves rust in the pipes, siderite and iron-bearing chlorite (a clay mineral). When the acid is spent and the pH rises (the solution becomes less acidic), the iron comes out as hydroxides.

Hydrofluoric acid (HF) is the only common acid that dissolves silica and aluminosilicates (minerals made of aluminium, silicon and oxygen, such as clays and feldspars). It reacts slowly with quartz and fast with clays and feldspars, because small particles have an enormous surface. In the end the silicon goes into solution as the fluorosilicate ion, and the aluminium as fluoride complexes:

Reactions of hydrofluoric acid

\begin{aligned} \mathrm{SiO_2} + 4\,\mathrm{HF} &\rightarrow \mathrm{SiF_4} + 2\,\mathrm{H_2O},\qquad \mathrm{SiF_4} + 2\,\mathrm{HF} \rightarrow \mathrm{H_2SiF_6}\\ \mathrm{Al_4Si_4O_{10}(OH)_8} + 32\,\mathrm{HF} &\rightarrow 4\,\mathrm{AlF_2^{+}} + 4\,\mathrm{SiF_6^{2-}} + 18\,\mathrm{H_2O} + 4\,\mathrm{H^{+}}\\ \mathrm{CaCO_3} + 2\,\mathrm{HF} &\rightarrow \mathrm{CaF_2}\downarrow + \mathrm{CO_2} + \mathrm{H_2O}\end{aligned}

where:

$\mathrm{SiO_2}$ quartz, —
$\mathrm{H_2SiF_6},\ \mathrm{SiF_6^{2-}}$ fluorosilicic acid and its anion, the fluorosilicate ion, —
$\mathrm{Al_4Si_4O_{10}(OH)_8}$ kaolinite, a clay mineral, —
$\mathrm{AlF_2^{+}}$ one of the fluoride complexes of aluminium, —
$\mathrm{CaF_2}$ calcium fluoride, an insoluble precipitate, —

Units: the coefficients are numbers of moles

Holds when

The acid is in excess: silicon goes into the fluorosilicate ion and aluminium into complexes with two fluorides. If fluorine is in short supply the products are different, and less HF is used per mole of mineral.
The reactions with clays and feldspars are fast, and those with quartz are slow. The speed grows with temperature.
With calcium minerals HF gives insoluble calcium fluoride. This is why they are removed with hydrochloric acid before the hydrofluoric acid.

A lot of hydrofluoric acid is used per mineral: 6 moles per mole of silica, and 32 per formula unit of kaolinite. So it is quickly used up near the wellbore. Its products, fluorosilicates and aluminium fluorides, stay in solution as long as it is acidic, and can enter new reactions.

Mud acid is a mixture of hydrochloric and hydrofluoric acids. The hydrochloric acid keeps the solution acidic: at low pH the products of the HF reactions stay dissolved. The strength is written as a pair of numbers. “12–3” means 12% HCl and 3% HF by mass (full-strength mud acid), and “6–1.5” is half strength. Mixtures with 0.5–1.5% HF and an HCl-to-HF ratio of 6 to 9 are used when the rock has a lot of clays and feldspars. Hydrofluoric acid is often made right in the solution from ammonium bifluoride, a solid salt (in Russian-language practice, ammonium bifluoride-fluoride, known by its Russian abbreviation BFA). The bifluoride reacts with hydrochloric acid and gives two molecules of HF and ammonium chloride. About 1.43 kg of bifluoride and 0.91 kg of HCl are used per kilogram of HF. So 3% HF “eats” about 2.7% HCl, and for “12–3” we start from an acid of about 15% strength.

How mineralogy changes the choice of mud acid

Rock: Solubility in HCl above ~20%
What we fear: hydrofluoric acid is not needed; a lot of calcium fluoride
Main acid: hydrochloric acid only
Preflush: —

Rock: Clean quartz sandstone, few clays, high permeability
What we fear: almost nothing
Main acid: 12–3
Preflush: HCl 15%

Rock: A lot of feldspars
What we fear: fluorosilicates and fluoroaluminates of sodium and potassium
Main acid: less HF, more HCl: about 13.5–1.5
Preflush: HCl 15%

Rock: A lot of clays
What we fear: secondary precipitates, fines migration (movement of very small rock particles)
Main acid: weak: about 6.5–1
Preflush: weak HCl with an iron control agent

Rock: Iron-bearing chlorite
What we fear: iron hydroxides
Main acid: very weak: about 3–0.5
Preflush: weak HCl or acetic acid with an iron control agent

Rock: Low permeability
What we fear: precipitates and fines in narrow channels
Main acid: weaker than for high permeability, for example 6–1.5
Preflush: HCl 7.5% or an organic acid

Rock	What we fear	Main acid	Preflush
Solubility in HCl above ~20%	hydrofluoric acid is not needed; a lot of calcium fluoride	hydrochloric acid only	—
Clean quartz sandstone, few clays, high permeability	almost nothing	12–3	HCl 15%
A lot of feldspars	fluorosilicates and fluoroaluminates of sodium and potassium	less HF, more HCl: about 13.5–1.5	HCl 15%
A lot of clays	secondary precipitates, fines migration (movement of very small rock particles)	weak: about 6.5–1	weak HCl with an iron control agent
Iron-bearing chlorite	iron hydroxides	very weak: about 3–0.5	weak HCl or acetic acid with an iron control agent
Low permeability	precipitates and fines in narrow channels	weaker than for high permeability, for example 6–1.5	HCl 7.5% or an organic acid

Guidelines of this kind were published in industry recommendations in the 1980s and have been refined since then. At high temperature the strength is reduced even more, and part of the hydrochloric acid is replaced by an organic acid. The final choice is checked on core.

Dissolving power: how much rock one cubic metre of acid dissolves

The equation of the reaction tells us how much rock an acid will dissolve. The mass dissolving power $\beta$ is the mass of mineral per unit mass of pure acid. The volume dissolving power $X$ is the volume of mineral per unit volume of a solution of a given strength (dissolving power):

Dissolving power of an acid

\beta = \frac{\nu_{\text{ma}} M_{\text{ma}}}{\nu_{\text{a}} M_{\text{a}}},\qquad X = \frac{C\,\beta\,\rho_{\text{a}}}{\rho_{\text{ma}}}

where:

$\beta$ mass of mineral that a unit mass of pure acid dissolves, kg/kg
$\nu_{\text{ma}},\ \nu_{\text{a}}$ coefficients of the mineral and of the acid in the reaction equation, —
$M_{\text{ma}},\ M_{\text{a}}$ molar masses of the mineral and of the acid, g/mol
$X$ volume of mineral that a unit volume of acid solution dissolves (do not confuse it with the logarithmic term $X = \ln(r_e/r_w) - 3/4$ in the lesson Acid treatments: principle, economics and candidate selection), m³/m³
$C$ mass fraction of acid in the solution, fraction
$\rho_{\text{a}},\ \rho_{\text{ma}}$ density of the acid solution and density of the mineral, g/cm³

Units: ratios of masses and volumes; the units of densities and molar masses are the same in the numerator and the denominator

Holds when

The acid is fully spent, and only on this mineral.
For HF the coefficient $\nu_{\text{a}}$ depends on which fluorides form. Here silicon goes into the fluorosilicate ion. In tables where $\mathrm{SiF_4}$ is taken as the product, the values are 1.3–1.5 times larger.
The density of the solution is taken at the temperature at which it is prepared.

$\beta$ depends only on the reaction. $X$ also depends on the strength of the solution and on the densities. The reciprocal $1/X$ is how many cubic metres of solution are needed to dissolve one cubic metre of mineral.

Worked exampleHow much rock hydrochloric acid and mud acid dissolve

Given (assumed data)

Fluids: 15% HCl with a density of 1.073 g/cm³ (20 °C); mud acid 12–3 with a density of about 1.07 g/cm³
Densities of minerals: calcite 2.71, dolomite 2.87, quartz 2.65, kaolinite 2.60 g/cm³
Molar masses: HCl 36.46, HF 20.006, calcite 100.09, dolomite 184.40, quartz 60.08, kaolinite $\mathrm{Al_4Si_4O_{10}(OH)_8}$ 516.31 g/mol

Solution

Calcite: $\beta = \dfrac{100.09}{2\cdot 36.46} = 1.373$ ; $X = \dfrac{0.15\cdot 1.373\cdot 1.073}{2.71} = 0.0815$ m³/m³.
By mass: one cubic metre of the solution weighs 1073 kg, of which 161 kg is HCl. It dissolves $161\cdot 1.373 = 221$ kg of calcite and releases 53 m³ of carbon dioxide (at 20 °C and atmospheric pressure).
Dolomite: $\beta = \dfrac{184.40}{4\cdot 36.46} = 1.264$ ; $X = \dfrac{0.15\cdot 1.264\cdot 1.073}{2.87} = 0.0709$ m³/m³.
Quartz, 3% HF: $\beta = \dfrac{60.08}{6\cdot 20.006} = 0.5005$ ; $X = \dfrac{0.03\cdot 0.5005\cdot 1.07}{2.65} = 0.00606$ m³/m³.
Kaolinite, 3% HF: $\beta = \dfrac{516.31}{32\cdot 20.006} = 0.8065$ ; $X = \dfrac{0.03\cdot 0.8065\cdot 1.07}{2.60} = 0.00996$ m³/m³.
Comparison: $0.0815/0.00996 = 8.2$ . By volume, hydrochloric acid dissolves 8.2 times more calcite than mud acid dissolves kaolinite.

Answer: One cubic metre of 15% hydrochloric acid dissolves 221 kg (0.0815 m³) of calcite. One cubic metre of 12–3 mud acid dissolves 0.00996 m³ of kaolinite, which is 8.2 times less by volume.

What it means. These are the main proportions of design: about 12 m³ of hydrochloric acid per cubic metre of carbonate, and about 100 m³ of mud acid per cubic metre of clay. Mud acid cannot “dissolve the reservoir”. It is used up on the clays and particles right at the wall, and its volume is set by how much fast-reacting mineral lies in its path. At reservoir pressure the carbon dioxide dissolves in the liquid, and during clean-up it comes out and helps to carry the fluid out. The density of the mud acid is an estimate. 12% HCl has a density of 1.057 g/cm³, and 3% HF adds about 0.01. If the HF is made from bifluoride, ammonium chloride adds another 0.01 or so, and $X$ changes by less than 1%. The other data are from reference tables.

Stages of a sandstone treatment and volumes per metre

A sandstone treatment is run in stages, and each stage has its own task:

Preflush: Hydrochloric acid of 5–15%; with sensitive minerals, weaker, or an organic acid. It dissolves carbonates where the hydrofluoric acid will reach later, and it pushes the formation water, with its sodium, potassium and calcium, further away from the wellbore. Sometimes a hydrocarbon solvent is pumped before it, to push away oil that tends to form sludge (a thick, sticky mass) with acid and to remove organic films.
Main acid: Mud acid of a chosen strength. It dissolves clays, particles of drilling fluid and fines in the damaged zone.
Overflush: A solution of ammonium chloride, weak hydrochloric acid or a hydrocarbon liquid, and in gas wells, with nitrogen. It pushes the spent mud acid 1–1.5 m away from the wellbore, where precipitates from it have little effect on inflow. It is also called the final flush. In Russian-language practice this stage, together with the displacement out of the tubing, is often called simply the displacement, and the fluid is called the displacement fluid.
Displacement: The fluid that pushes the last stage out of the tubing and into the reservoir. It is usually the same as for the overflush.

We calculate the volume of each stage from the radius to which it must act. If the reaction is fast, the acid dissolves the mineral completely behind a narrow front. To reach the radius $r$ we then need enough acid to fill the pores of the ring and to dissolve all the mineral in it:

Volume of acid per metre of interval for a dissolving front at radius r

\frac{V}{h} = \pi\,(r^2 - r_w^2)\left[\phi + (1 - \phi)\,\frac{x}{X}\right]

where:

$V/h$ volume of fluid per metre of interval, m³/m
$r$ radius of the dissolving front from the well axis, m
$r_w$ well radius, m
$\phi$ porosity, fraction
$x$ volume fraction of the mineral that dissolves, in the solid part of the rock, fraction
$X$ volume dissolving power of the fluid for this mineral, m³/m³

Units: SI (m, m³/m)

Holds when

The acid enters the reservoir evenly over the thickness and around the circumference.
The mineral dissolves completely behind a sharp front (a fast reaction). The slow reaction of HF with quartz, mixing and the growth of porosity are not counted. The volume is underestimated, so a margin is added to the calculation.
For mud acid, $x$ is the fraction of fast-reacting minerals that the acid can reach: clays on the pore walls, fines, and particles of drilling fluid.

The first term in the brackets is the pore volume of the ring, and the second is the acid spent on the mineral. When $X$ is small, the second is many times larger than the first. For hydrochloric acid this is the usual rule for calculating the preflush. For mud acid it is the limit of a fast reaction (see the advanced section).

The overflush reacts with nothing. Its volume is the pore volume out to the radius $r_c$ to which the spent acid must be pushed, $V/h = \pi\phi(r_c^2 - r_w^2)$ .

Worked exampleVolumes of the stages of a sandstone treatment per metre of interval

Given (assumed data)

Well: sandstone, $r_w = 0.108$ m, interval 10 m, porosity 0.18; 73 mm tubing (62 mm inside) down to 2500 m
Damage: clay particles from the drilling fluid and fines, to a depth of about 0.20 m beyond the wall
Mineralogy (X-ray analysis): calcite, 5% of the solid part; minerals that react fast with HF and that the acid can reach, 5% (for the HF consumption they are treated as kaolinite)
Fluids: preflush: 15% HCl; main acid: mud acid 12–3; overflush and displacement: 3% ammonium chloride solution
Targets: hydrofluoric acid to 0.22 m beyond the wall (with a margin over the damage); hydrochloric acid another 0.15 m further; the spent mud acid to 1.5 m from the axis

Solution

Radii of the fronts: $r_{\mathrm{HF}} = 0.108 + 0.22 = 0.328$ m, $r_{\mathrm{HCl}} = 0.328 + 0.15 = 0.478$ m.
Main acid: $\pi(0.328^2 - 0.108^2) = 0.3013$ m², $\dfrac{V}{h} = 0.3013\left(0.18 + \dfrac{0.82\cdot 0.05}{0.009957}\right) = 0.3013\cdot 4.298 = 1.295$ m³/m.
Preflush: $\pi(0.478^2 - 0.108^2) = 0.6812$ m², $\dfrac{V}{h} = 0.6812\left(0.18 + \dfrac{0.82\cdot 0.05}{0.0815}\right) = 0.6812\cdot 0.683 = 0.465$ m³/m.
Overflush: $\pi\cdot 0.18\cdot(1.5^2 - 0.108^2) = 1.266$ m³/m.
For the 10 m interval: 4.65 m³ of hydrochloric acid, 12.95 m³ of mud acid, 12.66 m³ of overflush fluid. The displacement is the volume of the tubing, $\pi\cdot 0.062^2/4\cdot 2500 = 7.55$ m³. The total is 37.8 m³.
A check: the front of the hydrochloric acid (0.478 m) is further out than the front of the hydrofluoric acid (0.328 m), so the HF will not meet carbonate. The main acid fluid itself will fill the pores out to $\sqrt{0.108^2 + 1.295/(\pi\cdot 0.18)} = 1.52$ m. Without an overflush, the spent mud acid with its fluorosilicates would stay right at the wellbore.

Answer: Per metre of interval: 0.465 m³ of hydrochloric acid, 1.295 m³ of mud acid and 1.266 m³ of overflush fluid. For the whole treatment, with the displacement, about 37.8 m³ of fluids.

What it means. The main acid is almost entirely spent on the fast-reacting minerals. The pore volume of the ring out to 0.328 m is only 4% of its volume. So the volume depends strongly on the share of these minerals and on the depth of the damage. If the HF had to reach not 0.22 m but 0.30 m beyond the wall, the volume would grow 1.6 times. A calculation with a sharp front underestimates the volume, because it does not count the reaction with quartz and the uneven intake of acid by the layers. So in practice a margin is added, and the volume is checked with core tests and with the experience of treatments in the field. The data in this example are assumed for illustration.

Secondary and tertiary precipitates, and how the stages prevent them

Hydrofluoric acid inevitably creates products that can come out of solution. We divide the reactions into three groups, by when and where these products form (secondary precipitates):

Primary reactions and precipitates on contact: HF itself dissolves minerals. A precipitate comes out here if it meets calcium (calcium fluoride), or sodium and potassium from formation water or seawater, from brines and from salt diverters (salt particles that steer the acid). Then fluorosilicates and fluoroaluminates come out.
Secondary reactions: Spent acid that is rich in fluorosilicates reacts slowly with clays and feldspars further from the wellbore. Fluorine moves from silicon to aluminium, and the silicon comes out as hydrated silica, a gel. These reactions are slower than the primary ones, and they speed up with temperature. In hot reservoirs they start already during pumping.
Tertiary reactions: Aluminium fluoride complexes go on reacting with clays. They pull aluminium out of the clays and leave more silica. When the acid is exhausted and the pH rises, aluminium fluorides and hydroxides can come out, and iron hydroxides from iron.

Secondary reactions cannot be avoided completely, but they can be made harmless. This is what the stages are for:

The preflush removes carbonates and pushes away formation water with its sodium, potassium and calcium. Calcium fluoride and fluorosilicates do not form where the hydrofluoric acid goes.
The hydrochloric acid in the mud acid, and a high HCl-to-HF ratio, keep the pH low. The reaction products stay in solution longer.
The overflush pushes the spent acid to where a precipitate has little effect on inflow.
A quick clean-up carries the spent acid out before the secondary reactions go far.
Ammonium salts only. For flushes and displacement we use ammonium chloride solution, not sodium or potassium chlorides. Mud acid is not diluted with seawater, and rock salt is not used as a diverter together with hydrofluoric acid.

The Hawkins formula shows why pushing the acid away pays off (Skin factor). Spent mud acid takes the same pore volume wherever it is pushed: a ring from $r_1$ to $r_2$ , where $r_2^2 - r_1^2 = V/(\pi\phi h)$ . If the precipitate lowers the permeability of the ring by a factor of $k/k_{\text{pr}}$ , the skin of the ring is:

Skin of a ring with a precipitate from spent acid

S_{\text{pr}} = \left(\frac{k}{k_{\text{pr}}} - 1\right)\ln\frac{r_2}{r_1} = \frac12\left(\frac{k}{k_{\text{pr}}} - 1\right)\ln\left(1 + \frac{V}{\pi\phi h\,r_1^2}\right)

where:

$S_{\text{pr}}$ skin of the ring with the precipitate, —
$k,\ k_{\text{pr}}$ permeability of the reservoir and of the ring with the precipitate, mD
$r_1,\ r_2$ inner and outer radii of the ring occupied by spent acid, m
$V$ volume of the main acid, m³
$h,\ \phi$ thickness of the interval and porosity, m, fraction

Units: any consistent units; the logarithm is natural

Holds when

The acid is displaced like a piston, without mixing. The precipitate is spread evenly over the ring and lowers the permeability by the same amount wherever the ring ends up.
The flow is radial, and the damage near the wellbore has already been removed.

A ring of the same volume pushed further out is thinner, and its weight in the skin is smaller. When $r_1$ is much larger than the thickness of the ring, the skin falls roughly as $1/r_1^2$ .

Worked exampleWhy we need an overflush: the same precipitate at the wellbore and at 1.5 m

Given (assumed data)

From the previous example: main acid 1.295 m³/m, porosity 0.18, $r_w = 0.108$ m; overflush 1.266 m³/m
Precipitate: lowers the permeability of the ring of spent acid by half, $k/k_{\text{pr}} = 2$ (assumed)

Solution

Without an overflush the ring starts at the wall and reaches 1.52 m: $S_{\text{pr}} = (2 - 1)\ln\dfrac{1.52}{0.108} = 2.64$ .
With an overflush the ring starts at 1.5 m: $r_2 = \sqrt{1.5^2 + 1.295/(\pi\cdot 0.18)} = 2.13$ m, $S_{\text{pr}} = \ln\dfrac{2.13}{1.5} = 0.35$ .
The same precipitate gives a skin 7.5 times smaller.

Answer: Without an overflush the precipitate would add about 2.64 to the skin, and with an overflush about 0.35.

What it means. A skin of +2.6 would bring back a noticeable part of what the treatment removed. A skin of +0.35 is almost unnoticeable. This gives the rule: push the spent mud acid 1–1.5 m away, or carry it out quickly. The calculation is idealised. In reality the spent acid mixes with the flush, and part of the precipitate stays closer to the wellbore, so the overflush is taken with a margin. The data in this example are assumed for illustration.

CalculatedWhere the fluids of the stages end up at the end of the treatment in the example, without an overflush and with one. The radius scale is logarithmic. The calculation uses the volumes of the stages. The displacement is piston-like, without mixing.

Additives

Acid is never pumped without additives. It corrodes metal, puts iron into solution, and forms emulsions and sludge with oil. Additives solve these problems, but they can do harm themselves: they can react with each other, stabilise an emulsion, or stick to the rock. So each one is justified separately, and the whole mixture is tested for compatibility with the oil, the formation water and the rock (Sources of formation damage).

The main additives to acid mixtures

Additive: Corrosion inhibitor
Why: protects pipes and equipment: acid dissolves steel and releases hydrogen
What to remember: its effect weakens with temperature and time; chromium and duplex steels need special formulations; spent acid is also corrosive

Additive: Iron control agent
Why: keeps iron in solution: chelating agents (chemicals that bind the iron; citric acid, EDTA) or reducing agents that turn ferric (trivalent) iron into ferrous (divalent) iron
What to remember: ferric hydroxide comes out already at a pH of about 2–3, ferrous hydroxide only at about 7–8

Additive: Non-emulsifying surfactants and demulsifiers (chemicals that break emulsions)
Why: stop the acid and the spent fluid from forming an emulsion with oil
What to remember: chosen by tests with the oil of the reservoir; too much surfactant stabilises an emulsion itself

Additive: Anti-sludge additives
Why: prevent asphaltenes from coming out as sludge when acid meets oil
What to remember: especially needed if the acid contains ferric iron

Additive: Mutual solvent
Why: dissolves in both water and oil: removes oil films, helps make the rock water-wet and carry out the spent acid
What to remember: an example is ethylene glycol monobutyl ether; it can weaken the corrosion inhibitor

Additive: Clay stabiliser
Why: fixes clay particles in place and reduces their migration
What to remember: its effect is often temporary

Additive	Why	What to remember
Corrosion inhibitor	protects pipes and equipment: acid dissolves steel and releases hydrogen	its effect weakens with temperature and time; chromium and duplex steels need special formulations; spent acid is also corrosive
Iron control agent	keeps iron in solution: chelating agents (chemicals that bind the iron; citric acid, EDTA) or reducing agents that turn ferric (trivalent) iron into ferrous (divalent) iron	ferric hydroxide comes out already at a pH of about 2–3, ferrous hydroxide only at about 7–8
Non-emulsifying surfactants and demulsifiers (chemicals that break emulsions)	stop the acid and the spent fluid from forming an emulsion with oil	chosen by tests with the oil of the reservoir; too much surfactant stabilises an emulsion itself
Anti-sludge additives	prevent asphaltenes from coming out as sludge when acid meets oil	especially needed if the acid contains ferric iron
Mutual solvent	dissolves in both water and oil: removes oil films, helps make the rock water-wet and carry out the spent acid	an example is ethylene glycol monobutyl ether; it can weaken the corrosion inhibitor
Clay stabiliser	fixes clay particles in place and reduces their migration	its effect is often temporary

The concentrations and grades of additives are chosen from the reservoir temperature, the contact time and tests. Nitrogen is added to the flushes of gas wells and depleted wells to help carry out the spent acid.

Iron is the main source of trouble. Ferric iron comes from rust and mill scale (flakes of oxide from the steel) in pipes and tanks. Ferrous iron comes mainly from the iron-bearing minerals of the reservoir: siderite, chlorite and pyrite. Ferric iron also provokes sludge with asphaltene oils. So before pumping, the tubing is washed with acid and the returns are taken to the surface. The acid is delivered through clean lines or through coiled tubing (a continuous flexible pipe run into the well), bypassing the old tubing.

Placement and diversion

Acid goes where the resistance is lower: into the most permeable and least damaged layers, and when it is pumped down the tubing, first into the upper perforations. After it dissolves the damage there, it opens an even wider path for itself, and the damaged layers stay untreated. So on an interval longer than a few metres we need a method of diversion (steering the acid into all the layers):

Mechanical. Packers and bridge plugs (devices that seal off part of the well), a straddle packer (two packers that isolate one interval between them) on coiled tubing, treating the interval section by section, and ball sealers (balls that sit on the perforations that take fluid and close them).
Chemical. Viscous gels and foams that temporarily raise the resistance of the layers that take fluid. Diverting particles that form a cake on the wall, which dissolves later: benzoic acid or salt in carbonates, and oil-soluble resins in sandstones.
Coiled tubing. It delivers acid exactly opposite the interval, without driving dirt from the tubing or fluid from the wellbore into the reservoir.

Usually the batches of acid and diverter alternate, roughly one pair for every 5–8 m of interval. The diverter raises the injection pressure, and this rise should not be taken for new damage. Diversion, treatments of horizontal wells and of carbonates are covered in more detail in the lesson Acid treatment of carbonates and special wells.

Injection rate and pressure below the fracture pressure

A matrix treatment is run so that the bottomhole pressure stays below the pressure at which a fracture opens and grows in the reservoir (Fracture gradient and leak-off testing). A step-rate injection test finds this pressure: the rate is raised in steps while the pressure is recorded (Rock mechanics for hydraulic fracturing). A margin is kept for the uncertainty of this pressure and for the fluctuations when the pumps work. The limit on the rate follows from the formula for steady-state inflow (Steady-state radial flow), written for injection:

Limit on the rate of matrix injection

q_{\max} = \frac{2\pi k h\,(p_f - \Delta p_{\text{mar}} - \bar p)}{\mu\left(\ln\dfrac{r_e}{r_w} + S\right)}

where:

$q_{\max}$ limit on the rate at reservoir conditions; for water-based acid solutions it is almost equal to the rate at the surface, m³/s
$k,\ h$ permeability of the reservoir and thickness of the treated interval, m², m
$p_f$ pressure at the depth of the reservoir at which a fracture opens and grows, Pa
$\Delta p_{\text{mar}}$ margin for the uncertainty of $p_f$ and for pressure fluctuations, Pa
$\bar p$ reservoir pressure, Pa
$\mu$ viscosity of the injected fluid in the reservoir (close to the viscosity of the reservoir fluid), Pa·s
$r_e,\ r_w,\ S$ radius of the drainage area, well radius and the current skin, m, m, —

Units: SI; $\ln$ is the natural logarithm

Holds when

Steady radial flow, with the pressure at the distance $r_e$ equal to the reservoir pressure. Injection lasts hours, and the pressure disturbance has no time to reach $r_e$ . So the pressure difference is overstated, and the rate comes out with a safety margin.
The viscosities of the acid and of the reservoir fluid are close. If the reservoir fluid is much more viscous, the rate calculated from the viscosity of the acid is too high.
The reservoir is uniform. In a layered interval the rate is divided unevenly between the layers.

As the damage dissolves, the skin falls and the limit on the rate grows. So pumping starts at a low rate, and the rate is raised in steps, while we watch that the bottomhole pressure stays below the limit.

In other unit systems (2)

Metric oilfield units

q_{\max}\,[\text{m}^3/\text{d}] = \frac{0.536\,k\,h\,(p_f - \Delta p_{\text{mar}} - \bar p)}{\mu\left(\ln(r_e/r_w) + S\right)}

$k$ in mD, $h$ in m, pressures in MPa, $\mu$ in mPa·s; for a rate in m³/min the constant is $3.72\cdot 10^{-4}$ .

$0.536 = 2\pi\cdot 0.0853$ , as in the Dupuit equation; $3.72\cdot 10^{-4} = 0.536/1440$ .

Oilfield units

q_{\max}\,[\text{bbl/min}] = \frac{4.917\cdot 10^{-6}\,k\,h\,(p_f - \Delta p_{\text{mar}} - \bar p)}{\mu\left(\ln(r_e/r_w) + S\right)}

$k$ in mD, $h$ in ft, pressures in psi, $\mu$ in cP; the rate is in barrels per minute.

$4.917\cdot 10^{-6} = 7.08\cdot 10^{-3}/1440$ .

The pump “sees” not the bottomhole pressure but the wellhead pressure. The two are linked by the hydrostatic pressure (the pressure of the column of fluid in the pipes) and by the friction in the pipes:

Wellhead pressure during injection

p_{\text{wh}} = p_w - \rho g H + \Delta p_{\text{fr}}

where:

$p_{\text{wh}}$ wellhead pressure in the pipes through which we pump, Pa
$p_w$ bottomhole pressure opposite the interval, Pa
$\rho$ density of the fluid in the pipes, kg/m³
$g$ acceleration of gravity, m/s²
$H$ vertical depth of the interval, m
$\Delta p_{\text{fr}}$ friction loss in the pipes at the given rate, from the Fanning equation (Circulating pressure losses), Pa

Units: SI; in metric units $\rho g H = 0.00981\,\rho H$ (MPa, with $\rho$ in g/cm³ and $H$ in m)

Holds when

The pipes are filled with one fluid. When fluids of different densities replace each other in the pipes, the hydrostatic pressure changes during pumping.
Friction grows almost as the square of the rate. Losses in the wellhead equipment and in the perforations are small, or are counted separately.

The higher the rate, the larger the friction, and the higher the allowable wellhead pressure for the same limit at the bottom. So the limit is set not as one number but as a line of “rate against wellhead pressure”.

Worked exampleLimits on the injection rate and pressure

Given (assumed data)

Reservoir: interval at 2500 m vertical depth, $k = 50$ mD, $h = 10$ m, $r_w = 0.108$ m, $r_e = 300$ m; reservoir pressure 23.0 MPa
Fracture opening pressure: 41.0 MPa from a step-rate test in a neighbouring well; margin 2.0 MPa
Fluids: viscosity of the acid and of the oil in the reservoir about 0.9 mPa·s; in the pipes 15% HCl, 1.073 g/cm³, viscosity 1.0 mPa·s
Well: skin before the treatment +8; tubing with an inside diameter of 62 mm, roughness 0.03 mm; the volume of the whole programme from the previous example 37.8 m³

Solution

The limit on the bottomhole pressure is $41.0 - 2.0 = 39.0$ MPa; the allowable overbalance (the excess of pressure over the reservoir pressure) is $39.0 - 23.0 = 16.0$ MPa.
$\ln(300/0.108) = 7.929$ . At $S = 8$ : $q_{\max} = \dfrac{0.536\cdot 50\cdot 10\cdot 16.0}{0.9\cdot(7.929 + 8)} = 299$ m³/d, or 0.2076 m³/min (1.31 bbl/min).
The hydrostatic pressure of the acid: $0.00981\cdot 1.073\cdot 2500 = 26.315$ MPa.
Friction at 0.2076 m³/min: the velocity is 1.146 m/s, the Reynolds number (a number that shows how turbulent the flow is) is about 76 thousand, the Fanning friction factor from the Colebrook equation is 0.00526, and the loss is 0.598 MPa.
The limit on the wellhead pressure: $39.0 - 26.315 + 0.598 = 13.28$ MPa.
When the skin falls to +2: $q_{\max} = 0.333$ m³/min, the friction is 1.444 MPa, and the limit at the wellhead is $39.0 - 26.315 + 1.444 = 14.13$ MPa.
At an average rate of 0.25 m³/min, the whole programme will take about $37.8/0.25 = 151$ min, which is 2.5 h.

Answer: At the start, no more than 0.208 m³/min at a wellhead pressure of up to 13.28 MPa. As the zone is cleaned, up to 0.333 m³/min and 14.13 MPa.

What it means. The limit at the wellhead hardly changes, while the allowable rate grows 1.6 times as the zone is cleaned. That is why the rate is raised in steps while we watch the pressure. While the pipes hold a lighter fluid (oil, ammonium chloride solution), the hydrostatic pressure is smaller, and the same limit at the bottom corresponds to a higher pressure at the wellhead. The limit is recalculated at every change of fluid. If the oil in the reservoir were much more viscous than the acid, the rate calculated from the viscosity of the acid would be too high. The data in this example are assumed for illustration.

CalculatedThe bottomhole pressure during injection into the well of the example, against the rate, for several values of skin. The limit on pressure is 39 MPa (the fracture opening pressure of 41 MPa minus the margin). The points are the limits on the rate at a skin of +8 and +2.

Clean-up and evaluating the result from the skin

After the overflush, the well is cleaned up as quickly as possible. The spent acid goes on reacting, and the longer it stands in the reservoir, the more secondary precipitates form. Gas wells and depleted wells are cleaned up with nitrogen, and oil wells with a pump or by swabbing (lifting liquid out with a tool lowered on a cable). The spent fluid is acidic and corrosive, and it can carry hydrogen sulfide. It is collected in separate tanks and neutralised. Samples of the spent acid (its acidity and its content of iron, calcium and silicon) show what has dissolved and how much.

The progress of the treatment can be seen already during pumping. If the rate and pressure are recorded and converted to the bottom of the hole, we can solve the formula for the limit on the rate for the skin. The result is the apparent skin:

Apparent skin during injection

S_{\text{app}} = \frac{2\pi k h\,(p_w - \bar p)}{q\,\mu} - \ln\frac{r_e}{r_w}

where:

$S_{\text{app}}$ apparent skin at the moment of measurement, —
$p_w$ bottomhole pressure: wellhead pressure plus hydrostatic pressure minus friction, Pa
$q$ current injection rate, m³/s
$k,\ h,\ \mu,\ \bar p,\ r_e,\ r_w$ as in the formula for the limit on the rate, m², m, Pa·s, Pa, m, m

Units: SI; in metric oilfield units $2\pi$ is replaced by 0.536 (the rate in m³/d)

Holds when

The flow is steady. Transient effects at the start of pumping and after a change of rate distort the value.
The apparent skin includes the pseudo-skins, the viscous slug of acid (a batch of thickened acid), and the action of the diverter.

What matters is not the value but the trend. While the apparent skin falls, the acid is working. When it has stopped falling, the stage can be ended. Batches of diverter raise it temporarily. Methods that take the transient regime into account give a more accurate estimate.

Worked exampleEvaluating the treatment: the apparent skin and a test after clean-up

Given (assumed data)

Measurement in the middle of the mud acid injection: rate 0.30 m³/min, wellhead pressure 12.6 MPa; the pipes hold acid, and we calculate the hydrostatic pressure and the friction as in the previous example, for 15% HCl (the density of the mud acid is almost the same)
Well: from the previous example: $k = 50$ mD, $h = 10$ m, reservoir pressure 23.0 MPa, $\mu = 0.9$ mPa·s
Well tests: before the treatment, skin +8.0, of which the perforation pseudo-skin is +1.0; after clean-up, skin +1.4

Solution

Friction at 0.30 m³/min is 1.187 MPa. The bottomhole pressure is $12.6 + 26.315 - 1.187 = 37.73$ MPa, which is below the limit of 39.0 MPa.
The apparent skin: $\dfrac{0.536\cdot 50\cdot 10\cdot(37.73 - 23.0)}{0.30\cdot 1440\cdot 0.9} - 7.929 = 2.2$ , down from +8 at the start of pumping.
The damage skin before the treatment is $8.0 - 1.0 = 7.0$ , and after it $1.4 - 1.0 = 0.4$ . So 94% has been removed.
The growth in productivity at the same drawdown (the pressure difference that pushes fluid into the well; Formation damage and damage skin): $\ln(300/0.108) - 0.75 = 7.179$ ; $\dfrac{7.179 + 8.0}{7.179 + 1.4} = 1.77$ .

Answer: By the middle of pumping the apparent skin had fallen to 2.2. According to the test after clean-up, about 94% of the damage skin was removed, and productivity at the same drawdown grew about 1.77 times.

What it means. The apparent skin during pumping shows a trend. The final result comes from a test after clean-up and from the rate. The result is compared with the forecast and with the costs (Acid treatments: principle, economics and candidate selection). If the skin has not fallen, we look into the reasons: the wrong fluid, acid that went into the wrong layer, precipitates, an emulsion. Then we improve the technology for the next wells. The data in this example are assumed for illustration.

Terms and symbols

Mud acid: A mixture of hydrochloric and hydrofluoric acids for sandstones. The strength is written as a pair of numbers, for example “12–3”.
Dissolving power $\beta$ , $X$: The mass of mineral per unit mass of pure acid, and the volume of mineral per unit volume of solution.
Preflush: Hydrochloric acid before the mud acid. It removes carbonates and pushes away formation water.
Overflush: A solution of ammonium chloride, a weak acid or a hydrocarbon liquid after the mud acid. It pushes the spent acid away from the wellbore.
Secondary and tertiary precipitates: Silica gel, and aluminium fluorides and hydroxides, that come out in further reactions of the spent mud acid with minerals.
Acid diversion: Methods that make the acid enter all the layers of the interval.
Apparent skin: The skin calculated from the current injection rate and pressure. Its fall is used to follow the progress of the treatment.

Common mistakes and misconceptions

Wrong

“Hydrofluoric acid dissolves quartz, so the more of it the better.”

In fact

Strong HF is used up faster at the wall, destroys more of the rock's own cement (the mineral that binds the grains), and gives more secondary precipitates. In rocks rich in clays and feldspars, 0.5–1.5% HF is used.

Wrong

“A hydrochloric acid preflush is unnecessary: the mud acid already contains hydrochloric acid.”

In fact

A carbonate that the mud acid reaches reacts with both the hydrochloric and the hydrofluoric acid, and calcium fluoride comes out. Also, without a preflush, the HF meets formation water with sodium and potassium. The preflush removes both in advance.

Wrong

“Any brine will do for the displacement, for example a potassium chloride solution.”

In fact

Sodium and potassium form insoluble salts with fluorosilicates. For flushes and displacement after mud acid we use ammonium chloride.

Wrong

“After the acid it is better to leave the well for a day so that the acid can finish its work.”

In fact

Spent mud acid does not “finish its work”. It forms secondary precipitates. The well is cleaned up as quickly as possible.

Wrong

“We keep the wellhead pressure below the fracture pressure, so we will not fracture the reservoir.”

In fact

The limit applies to the bottom of the hole. The pressure at the bottom is larger than at the wellhead by the hydrostatic pressure, and smaller by the friction. At 2500 m the column of acid adds about 26 MPa, so 13 MPa at the wellhead is already almost 39 MPa at the bottom.

In depth: advanced detail

Fast and slow minerals: why the front of hydrofluoric acid is smeared

The calculation of the volume with a sharp front assumes that the HF dissolves the fast-reacting minerals completely and at once. In reality there are two speeds. Clays and feldspars react fast, and quartz reacts slowly, but there is a great deal of quartz in the rock. So on its way the acid is also spent on quartz. The front where the fast minerals dissolve is smeared out, and partly spent acid runs ahead of it. The ratio of the speed of the reaction to the speed of the acid supply is described by the Damköhler number. When the reaction is fast compared with the supply (high temperature, low rate), the front is sharp and short. When it is slow, the front is smeared and long, but the acid reaches further, in a weakened state.

Acid capacity number and the volume of a stage

N_{\text{ac}} = \frac{\phi\,X}{(1 - \phi)\,x},\qquad \frac{V}{h} = \pi\,(r^2 - r_w^2)\,\phi\left(1 + \frac{1}{N_{\text{ac}}}\right)

where:

$N_{\text{ac}}$ acid capacity number: how much mineral the acid that fills the pores will dissolve, relative to the mineral in the same volume of rock, —
$\phi,\ X,\ x,\ r,\ r_w$ as in the formula for the volume of a stage, —, m³/m³, —, m, m

Units: dimensionless

Holds when

The same as for the calculation of the volume with a sharp front.

The volume of a stage is the pore volume of the ring multiplied by $1 + 1/N_{\text{ac}}$ . For the mud acid of the example of this lesson, $N_{\text{ac}} \approx 0.044$ : for each pore volume we need about 23 more to dissolve the fast minerals. For hydrochloric acid and calcite, $N_{\text{ac}} \approx 0.36$ .

At high temperature all reactions are faster, the HF is used up closer to the wellbore, and secondary reactions start already during pumping. So for hot reservoirs the strength of the mud acid is reduced, part of the hydrochloric acid is replaced by an organic acid, or we use mixtures that release HF gradually, for example mixtures based on fluoboric acid, which is hydrolysed (broken down by water) in the reservoir and forms HF.

What a viscous diverter “costs”

A viscous diverter works as a zone of reduced mobility (a zone where the fluid moves less easily). Suppose the pores out to the radius $r_{\text{v}}$ are filled with a fluid of viscosity $\mu_{\text{v}}$ , and beyond it with a fluid of viscosity $\mu$ . The extra pressure drop is then the same as from a zone with the permeability $k\mu/\mu_{\text{v}}$ . By analogy with the Hawkins formula:

Skin of a zone of viscous diverter

S_{\text{v}} = \left(\frac{\mu_{\text{v}}}{\mu} - 1\right)\ln\frac{r_{\text{v}}}{r_w}

where:

$S_{\text{v}}$ extra skin of the zone filled with the diverter, —
$\mu_{\text{v}},\ \mu$ viscosity of the diverter and viscosity of the fluid to which the skin is referred, mPa·s
$r_{\text{v}},\ r_w$ radius of the diverter zone and well radius, m

Units: any consistent units; the logarithm is natural

Holds when

The viscosity is constant. In reality gels and foams thin under shear: near the wellbore, where the velocity is high, their viscosity is lower, and deeper in it is higher. The diversion is based on this.
The permeability of the zone does not change.

A gel that is 50 times more viscous than the acid and has gone 0.3 m beyond the wall of a well of radius 0.108 m gives $S_{\text{v}} = 49\ln(0.408/0.108) \approx 65$ . So batches of diverter sharply raise the apparent skin. This skin goes away when the gel breaks down or is pushed out.

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 wells.
Sources of formation damage — which kinds of damage operations create, and how to recognise them.
Formation damage and damage skin — the damage skin and the model of a zone of reduced permeability.
Acid treatment of carbonates and special wells — wormholes, treatments of carbonates and of horizontal wells, and treatment campaigns.
Fracture gradient and leak-off testing — the fracture pressure and how it is estimated.
Rock mechanics for hydraulic fracturing — the fracture opening pressure from a step-rate injection test.
Steady-state radial flow — the formula for steady-state inflow on which the limit on the rate is based.
Skin factor — the Hawkins formula, the components of skin, and pseudo-skins.
Circulating pressure losses — friction in pipes.
Hydraulic fracturing: purpose and well selection — when a fracture is needed instead of acid.

Check yourself

1
Why is hydrochloric acid pumped before the mud acid?
1. A.To dissolve quartz
2. B.To remove carbonates and to push away the formation water with its sodium, potassium and calcium
3. C.To reduce the viscosity of the oil
4. D.To open fractures
Show the answer
Correct answer: B. To remove carbonates and to push away the formation water with its sodium, potassium and calcium
Hydrofluoric acid with calcium gives insoluble calcium fluoride, and its products with sodium and potassium are fluorosilicates. The preflush dissolves carbonates where the HF will reach, and it moves the formation water away from the wellbore. Hydrochloric acid does not dissolve quartz, and fractures must not be opened in a matrix treatment.
2
How many kilograms of dolomite will 1 m³ of 10% hydrochloric acid with a density of 1.047 g/cm³ dissolve, and what volume is this at a dolomite density of 2.87 g/cm³?
Show the answer
One cubic metre of the solution holds $0.10\cdot 1047 = 104.7$ kg of HCl. For dolomite $\beta = 184.40/(4\cdot 36.46) = 1.264$ , so $104.7\cdot 1.264 \approx 132$ kg of dolomite will dissolve, or $132/2870 \approx 0.046$ m³. For comparison, 15% acid dissolves about 0.071 m³ of dolomite.
3
Which fluid should we choose for the overflush after mud acid?
1. A.Seawater
2. B.Calcium chloride brine
3. C.3% potassium chloride solution
4. D.Ammonium chloride solution
Show the answer
Correct answer: D. Ammonium chloride solution
The potassium and sodium of seawater form insoluble salts with fluorosilicates, and calcium forms calcium fluoride. Ammonium salts dissolve well, so for flushes and displacement after mud acid we use ammonium chloride (or weak hydrochloric acid, or a hydrocarbon liquid).
4
It was decided to push the spent mud acid not 1.5 m but only 1.0 m from the well axis (porosity 0.18, $r_w = 0.108$ m, main acid 1.30 m³/m). How much overflush fluid per metre is needed, and what do we lose?
Show the answer
The volume is $\pi\cdot 0.18\cdot(1.0^2 - 0.108^2) \approx 0.56$ m³/m, which is 2.3 times less than 1.27 m³/m for 1.5 m. But the ring of spent acid will lie between 1.0 m and $\sqrt{1 + 1.30/(\pi\cdot 0.18)} \approx 1.82$ m, and with permeability reduced by half its skin will be $\ln 1.82 \approx 0.60$ instead of 0.35. The saving on the flush turns into a larger leftover skin.
5
The pipes are already full of acid, the rate is constant, and the wellhead pressure has started to fall smoothly. What does this most likely mean?
1. A.The diverter has reached the perforations
2. B.The acid is dissolving the damage and the skin is falling
3. C.The reservoir has fractured
4. D.The pump is faulty
Show the answer
Correct answer: B. The acid is dissolving the damage and the skin is falling
At a constant rate and with the same fluid in the pipes, the hydrostatic pressure and the friction do not change. So a fall in the wellhead pressure means a fall in the bottomhole pressure, that is, in the resistance of the near-wellbore zone. A diverter, on the contrary, would raise the pressure. A sharp drop in pressure right at the limit would be a warning, because it is a sign of a fracture that has opened.
6
Why does one cubic metre of 12–3 mud acid dissolve eight times less clay by volume than one cubic metre of 15% hydrochloric acid dissolves calcite, and what follows from this for the volumes of a treatment?
Show the answer
Mud acid has only 3% HF, and a lot of it is needed to dissolve clays: 32 moles per formula unit of kaolinite, against two moles of HCl per mole of calcite. So $X$ is about 0.01 for kaolinite and about 0.08 for calcite. The volume of mud acid is spent almost entirely on the fast-reacting minerals. It grows in proportion to their share and to the square of the radius of the treatment, so a deep zone cannot be treated with mud acid. It needs a fracture.

Updated: 19 October 2026