Acid treatments: principle, economics and candidate selection

Quick refresher

A matrix treatment is pumped at a pressure below the fracture pressure. The fluid spreads out radially from the wellbore, through the pores, and works in the near-wellbore zone, where most of the drawdown (the pressure difference that drives the flow into the well) is spent.
In sandstone, acid dissolves what has plugged the pores but hardly touches the quartz framework. At best, the damage skin returns to zero. In carbonate, hydrochloric acid dissolves the rock itself and cuts wormholes (thin branching channels) beyond the damaged zone. The skin can become negative, usually down to −2…−3.
At the same drawdown the rate is inversely proportional to the sum $X + S$ , where $X = \ln(r_e/r_w) - 3/4 \approx 7$ . How many times the rate grows depends only on the skins. The absolute gain also depends on the transmissibility $kh/\mu$ (permeability times thickness, divided by viscosity) and on the drawdown. So the skin alone is not a criterion for choosing a well.
A treatment is justified if the income from the extra production is larger than its cost. The income must take into account the decline of the gain, the probability of success, the production lost during the job, and the time value of money. A simple guide is the payback time.
Acid does not remove the pseudo-skins (extra resistance to flow that is not caused by damage) of partial penetration (the well opens only part of the reservoir thickness) and perforation. It does not remove inertial losses (extra pressure losses at high velocity), limits set by the lift system, or depletion of the reservoir (its pressure and oil running down as it is produced).
Acid can itself damage the reservoir. Preventing damage is almost always cheaper than curing it.

Key relationships

\dfrac{q_2}{q_1} = \dfrac{X + S_1}{X + S_2},\quad X = \ln\dfrac{r_e}{r_w} - \dfrac34

how many times the rate grows at the same drawdown if the skin falls from $S_1$ to $S_2$

\Delta q = \dfrac{0.536\,kh\,\Delta p}{\mu B}\left(\dfrac{1}{X + S_2} - \dfrac{1}{X + S_1}\right)

absolute gain, m³/d ( $k$ in mD, $h$ in m, $\Delta p$ in MPa, $\mu$ in mPa·s)

t_{\text{pb}} = -\dfrac{1}{a}\ln\left(1 - \dfrac{aC}{\Delta q_0\,v}\right)

payback time when the gain declines exponentially

S \ge -\ln\dfrac{r_s}{r_w}

the limit of the skin after treating a zone of radius $r_s$

Common pitfalls

Ranking wells by skin. A large skin in a well with low flow capacity gives a smaller gain than a moderate skin in a good reservoir.
Planning a treatment from the total skin without subtracting the pseudo-skins of partial penetration and perforation.
Expecting a negative skin from acid in sandstone, or a several-fold gain from treating an undamaged well.
Forgetting that a gain in productivity becomes a gain in production only if the lift system and the surface equipment can pass it.
Raising the injection pressure above the fracture pressure. The fluid will go into the fracture and miss the damaged zone.

After this lesson you can

explain what acid does in sandstone and in carbonate, and why in sandstone it only removes damage, while in carbonate it can give a negative skin (a skin below zero, which means the zone around the well helps the flow)
calculate how many times the rate grows, and by how much in absolute terms, when the skin is lower, and explain why the gain depends not only on the skin but also on the flow capacity of the reservoir and on the drawdown
estimate the payback time of a treatment, taking into account the cost of the job, the production lost during the job, the decline of the gain and the probability of success
select candidate wells: screen out those where the rate is limited by something other than damage, and rank the rest by the gain and by the cost per unit of gain
choose the kind of chemical treatment from the place of the damage, and describe the cycle “selection, diagnosis, design, execution, evaluation”

What an acid treatment does

Formation damage adds resistance exactly where the flow already converges on the wellbore and spends most of the drawdown. There are three ways to deal with it: prevent it, remove it, or bypass it (lesson on formation damage). An acid treatment is the main way to remove it. The fluid passes through the damaged zone and dissolves what narrows the pore channels. Hydraulic fracturing (cracking the rock by pumping fluid at very high pressure) bypasses the damage: the fracture takes the flow far beyond it.

All methods of well stimulation (ways to raise the rate of a well) act on one of the quantities in the formula for radial inflow, $q = 2\pi kh\,\Delta p/[\mu B(X + S)]$ , where $X = \ln(r_e/r_w) - 3/4$ (lesson on the skin factor), or on the outflow (the flow from the bottom of the well up to the surface). It helps to see at once which quantity each method changes. This rules out options that cannot work.

What can be improved and by which methods

Quantity: Damage skin $S_d$
How to improve it: removed by acid and other chemical treatments; bypassed by hydraulic fracturing or re-perforation

Quantity: Effective wellbore radius (the radius of an undamaged well that would take the same flow)
How to improve it: increased by a hydraulic fracture, a horizontal well or a multilateral well

Quantity: Flow capacity $kh$
How to improve it: perforating intervals that were not perforated; a fracture that links thin layers

Quantity: Oil viscosity $\mu$
How to improve it: heating the reservoir, solvents, for viscous oil

Quantity: Outflow, not inflow
How to improve it: tubing (the pipe that carries the fluid up the well) of another diameter, artificial lift (a pump, or gas injection, that helps to raise the fluid), cleaning deposits out of the tubing

Quantity	How to improve it
Damage skin $S_d$	removed by acid and other chemical treatments; bypassed by hydraulic fracturing or re-perforation
Effective wellbore radius (the radius of an undamaged well that would take the same flow)	increased by a hydraulic fracture, a horizontal well or a multilateral well
Flow capacity $kh$	perforating intervals that were not perforated; a fracture that links thin layers
Oil viscosity $\mu$	heating the reservoir, solvents, for viscous oil
Outflow, not inflow	tubing (the pipe that carries the fluid up the well) of another diameter, artificial lift (a pump, or gas injection, that helps to raise the fluid), cleaning deposits out of the tubing

This summary is qualitative. Methods that change the flow capacity and the effective radius are more expensive and more complex than chemical treatments, but they also work where there is no damage.

The main condition of a matrix treatment is that the bottomhole pressure during pumping stays below the fracture pressure (Fracture gradient and leak-off testing). Then the fluid seeps into the pores radially and passes in turn through the whole ring of damaged rock. If the pressure goes above the fracture pressure, the rock splits, and the fluid goes into the fracture. It hardly touches the damaged ring, and the fracture may reach a water-bearing layer. We estimate the allowable pumping rate from the same formula for radial flow, written for injection, with a bottomhole pressure a little below the fracture pressure. As the acid dissolves the damage, the resistance of the near-wellbore zone falls, and at the same rate the injection pressure falls. We use this to follow the progress of the treatment.

Sandstone and carbonate: removing damage or improving the rock

In sandstone, the framework of the rock is quartz. Hydrochloric acid does not dissolve quartz, and hydrofluoric acid dissolves it slowly. The pore channels are narrowed by clay minerals, fine particles, particles of drilling fluid and carbonate cement (the material that glues the grains together). So we use a sequence of fluids. First comes hydrochloric acid, which removes carbonates and prepares the rock. Then comes mud acid, a mixture of hydrochloric and hydrofluoric acids that dissolves clays, feldspars (common minerals of sandstone) and fine particles. The reactions are fast, and the acid is spent in the first tens of centimetres from the wall. The best we can achieve is to bring the permeability of the near-wellbore zone back to its original value, that is, to bring the damage skin to zero.

In carbonate, hydrochloric acid dissolves the rock itself, calcite fast and dolomite more slowly, and gives off carbon dioxide. The acid front does not stay even. The fluid goes into the most conductive pores, widens them, and cuts thin, branching channels of high conductivity. These are wormholes. They pass through the damaged zone into the undamaged rock. So the treatment not only removes the damage but also improves the inflow compared with an ideal well: the skin becomes negative. In fractured carbonates, acid also widens the natural fractures.

SchematicTop view of the wellbore and the near-wellbore zone. On the left is sandstone: the acid has dissolved the damage in the ring around the wellbore (the dashed line is the edge of the former damaged zone), and the skin tends to zero. On the right is carbonate: wormholes reach beyond the damaged zone, and the skin becomes negative. The diagram is not to scale.

Acid treatment in sandstone and in carbonate

: Main fluid
Sandstone: hydrochloric acid, then mud acid (hydrochloric with hydrofluoric)
Carbonate: hydrochloric acid, usually 15%, sometimes stronger

: What dissolves
Sandstone: clays, feldspars, fine particles, carbonate cement, particles of drilling fluid; quartz hardly dissolves
Carbonate: the rock itself (calcite, dolomite) and carbonate deposits

: How the acid moves
Sandstone: in an almost even front, spent in the first tens of centimetres
Carbonate: in wormholes that reach a metre and further

: What can be achieved
Sandstone: restoring the original permeability: skin to zero
Carbonate: removing the damage and getting a negative skin, usually down to −2…−3

: Main risks
Sandstone: secondary precipitates (new deposits formed by the reaction products), weakening of the cementation and sand production
Carbonate: acid goes only into the most permeable intervals; at high temperature it is spent right at the wall

	Sandstone	Carbonate
Main fluid	hydrochloric acid, then mud acid (hydrochloric with hydrofluoric)	hydrochloric acid, usually 15%, sometimes stronger
What dissolves	clays, feldspars, fine particles, carbonate cement, particles of drilling fluid; quartz hardly dissolves	the rock itself (calcite, dolomite) and carbonate deposits
How the acid moves	in an almost even front, spent in the first tens of centimetres	in wormholes that reach a metre and further
What can be achieved	restoring the original permeability: skin to zero	removing the damage and getting a negative skin, usually down to −2…−3
Main risks	secondary precipitates (new deposits formed by the reaction products), weakening of the cementation and sand production	acid goes only into the most permeable intervals; at high temperature it is spent right at the wall

Fluids, additives and volumes are covered in the lesson Acid systems and treatment design, and the features of carbonates in the lesson Acid treatment of carbonates and special wells.

The Hawkins formula explains why the skin in carbonate usually does not fall below −2…−3. Even an infinitely permeable treated zone of radius $r_s$ gives a skin no lower than $-\ln(r_s/r_w)$ , as if the wellbore had been enlarged to $r_s$ . Wormholes 1–2 m long around a wellbore of radius 0.1 m give a limit of about $-\ln 11 \approx -2.4$ … $-\ln 21 \approx -3.0$ . A more strongly negative skin comes only from a fracture that reaches tens of metres.

The value of a removed skin

At the same drawdown the rate of a well is inversely proportional to the sum of the logarithmic term of an ideal well, $X$ , and the skin. This gives two measures of the effect of a treatment: how many times the rate grows, and by how much.

How many times the rate grows when the skin is reduced

\frac{q_2}{q_1} = \frac{X + S_1}{X + S_2},\qquad X = \ln\frac{r_e}{r_w} - \frac34

where:

$q_1,\ q_2$ rate before and after the treatment at the same drawdown, m³/d
$S_1,\ S_2$ skin before and after the treatment; the treatment reduces only the damage skin, and the pseudo-skins stay, —
$X$ logarithmic term of an ideal well in pseudo-steady-state flow (the pressure falls at the same pace everywhere in the drainage area), —
$r_e,\ r_w$ radius of the drainage area (the part of the reservoir that feeds the well) and the well radius, m

Units: dimensionless; the logarithm is natural

Holds when

The average reservoir pressure and the drawdown are the same before and after the treatment: the lift system holds the same bottomhole pressure at the higher rate.
Pseudo-steady-state, single-phase flow in a circular drainage area with the well at the centre.
The treatment changes only the skin. The permeability and thickness of the reservoir are the same.

The ratio depends only on the skins and on $X$ , and for most wells $X$ is about 7–8. Removing a skin of +7 almost doubles the rate. Removing a skin of +2 gives 25–30%. The same formula with $S_2 = 0$ is the reciprocal of the flow efficiency (the productivity of the well divided by the productivity of an undamaged well).

Absolute gain in rate after the skin is reduced

\Delta q = \frac{2\pi kh\,\Delta p}{\mu B}\left(\frac{1}{X + S_2} - \frac{1}{X + S_1}\right)

where:

$\Delta q$ gain in rate at standard conditions, m³/s
$k,\ h$ permeability and net thickness of the reservoir, m², m
$\Delta p$ drawdown $\bar p_R - p_{wf}$ , the same before and after the treatment, Pa
$\mu,\ B$ viscosity of the oil in the reservoir and its formation volume factor (how much the oil shrinks between the reservoir and the surface), Pa·s, m³/m³
$X,\ S_1,\ S_2$ as in the formula for the ratio, —

Units: SI (any consistent system of units)

Holds when

The same as for the formula for the ratio.

For the same skins, the gain is proportional to the transmissibility $kh/\mu$ and to the drawdown. A large skin in a poor reservoir can give a smaller gain than a moderate skin in a good one. The ratio and the gain are different things, and we should choose a well by the gain.

In other unit systems (2)

Metric oilfield units

\Delta q\,[\text{m}^3/\text{d}] = \frac{0.536\,kh\,\Delta p}{\mu B}\left(\frac{1}{X + S_2} - \frac{1}{X + S_1}\right)

$k$ in mD, $h$ in m, $\Delta p$ in MPa, $\mu$ in mPa·s, $B$ in m³/m³; the rate at standard conditions.

$0.536 = 2\pi\cdot 0.0853$ , where $0.0853$ is the constant of Darcy's law in these units (see Rock permeability).

Oilfield units

\Delta q\,[\text{STB/d}] = \frac{7.08\cdot 10^{-3}\,kh\,\Delta p}{\mu B}\left(\frac{1}{X + S_2} - \frac{1}{X + S_1}\right)

$k$ in mD, $h$ in ft, $\Delta p$ in psi, $\mu$ in cP, $B$ in rb/STB (reservoir barrels per stock-tank barrel).

$7.08\cdot 10^{-3} = 2\pi\cdot 1.127\cdot 10^{-3}$ .

Worked exampleHow much removing the damage skin will give

Given (assumed data)

Reservoir: sandstone: $k = 40$ mD, $h = 10$ m
Oil in the reservoir: $\mu = 1.4$ mPa·s, $B = 1.2$ m³/m³
Well: $r_w = 0.108$ m, $r_e = 300$ m, drawdown 4.0 MPa
Skin from well testing: +9
Pseudo-skin of partial penetration and perforation (estimate): +2

Solution

Damage skin: $S_d = 9 - 2 = 7$ . The best that we can expect from acid in sandstone is to remove it completely. Then $S_2 = +2$ .
The logarithmic term: $X = \ln\dfrac{300}{0.108} - 0.75 = 7.179$ .
Rate before the treatment: $q_1 = \dfrac{0.536\cdot 40\cdot 10\cdot 4.0}{1.4\cdot 1.2\cdot(7.179 + 9)} = 31.5$ m³/d.
After the treatment ( $S_2 = 2$ ): $q_2 = 55.6$ m³/d; the ratio is $\dfrac{7.179 + 9}{7.179 + 2} = 1.76$ .
The gain is $\Delta q = 24.0$ m³/d, or 151 bbl/d (1 m³ = 6.29 barrels). The assumed price in the example project that runs through the economics lessons (Project cash flow) is US$50 per barrel. After a 10% payment on revenue (a royalty-type payment) and US$5 of variable costs per barrel, US$40 per barrel is left. For simplicity, we ignore income tax. The gain brings in about US$6.1 thousand per day.
If we wrongly treat the whole skin as removable (+9 → 0), we get $q = 71.1$ m³/d and a gain of 39.5 m³/d, which is 64% more than the real one.

Answer: Removing the damage skin will raise the rate from about 31.5 to 55.6 m³/d, which is 1.76 times, or 24.0 m³/d more.

What it means. The estimate is optimistic. It assumes that the fluid is matched to the type of damage, that the acid reaches the whole damaged zone and creates no new deposits, and that the pump holds the same bottomhole pressure at the higher rate. The mistake with the pseudo-skin would overstate the expected gain by more than one and a half times. So before we count the benefit, we break the skin into components (lesson on the skin factor). The data in this example are assumed for illustration.

Choosing candidate wells

Selecting candidates answers two questions. The first is what limits the productivity of this particular well: damage, the properties of the reservoir, the geometry of the opening, or something else. The second is whether a rise in productivity will turn into a rise in production. A well whose rate is limited by the pump, the tubing or the processing plant will give almost the same after the treatment as before it (Nodal analysis of the production system).

There is something to produce. The drainage area still holds enough producible oil or gas, and the reservoir pressure is high enough, for the gain to last and to pay for the job. The oil or gas saturation near the well is high, and the water cut is moderate.
The system can pass the gain. The lift system, the flowline (the surface pipe that carries the production to the plant) and the processing plant must be able to take the extra liquid, so they need spare capacity.
The skin comes from damage. We subtract the pseudo-skins from the skin of the well test. If damage accounts for only a small part of the skin, it is better to add perforations, re-perforate, or change the design of the well bottom.
The permeability is not too low. In a tight reservoir, even complete removal of the damage gives little, and the candidate is more likely to be hydraulic fracturing.
The type of damage is known. If the history of the well, the deposits and core tests show what has plugged the pores, the fluid can be chosen to target it, and the probability of success is much higher.
The risk of doing harm is small. A nearby oil–water contact (the boundary between oil and water), flow behind the casing, a weak sandstone that tends to produce sand, or a gravel pack (a sand filter built in the well) all lower the chances of success and call for special care.
The job can be done. The casing and tubing are in good order, the interval is accessible, the well can be shut in without harm to its neighbours, and the equipment and materials are available.

Candidates are found in two ways. A new well is treated right after completion, if the first test shows a large skin from the drilling fluid. Among producing wells, candidates are looked for during reservoir surveillance (routine monitoring of how the field performs). We compare the cumulative production or the productivity of each well with the flow capacity $kh$ in its area. A well that gives noticeably less than neighbours with a similar $kh$ is suspicious. But before we list it as a candidate, we check other explanations. It was started later. Its drainage area is smaller (near the edge of the oil accumulation or between neighbours). It has no pump, or the pump works badly. Its reservoir is depleted or watered out.

When acid will not help

What is observed: Only part of the thickness is open, few holes per metre
The real cause: a pseudo-skin of penetration and perforation
What to do: add perforations, re-perforate

What is observed: The skin of a gas well grows with the rate
The real cause: inertial losses near the wellbore
What to do: more holes, a fracture

What is observed: A gas well loads up with liquid (liquid collects in the well and holds the gas back); an oil well has a very high water cut
The real cause: the outflow is not enough
What to do: tubing of another diameter, artificial lift

What is observed: The rate falls together with the reservoir pressure
The real cause: depletion
What to do: pressure maintenance (injecting water to hold up the reservoir pressure), a pump

What is observed: The rate and the wellhead pressure fall, and there are deposits in the tubing
The real cause: narrowing of the tubing
What to do: washing the tubing without pumping into the reservoir

What is observed	The real cause	What to do
Only part of the thickness is open, few holes per metre	a pseudo-skin of penetration and perforation	add perforations, re-perforate
The skin of a gas well grows with the rate	inertial losses near the wellbore	more holes, a fracture
A gas well loads up with liquid (liquid collects in the well and holds the gas back); an oil well has a very high water cut	the outflow is not enough	tubing of another diameter, artificial lift
The rate falls together with the reservoir pressure	depletion	pressure maintenance (injecting water to hold up the reservoir pressure), a pump
The rate and the wellhead pressure fall, and there are deposits in the tubing	narrowing of the tubing	washing the tubing without pumping into the reservoir

In all these cases the skin from a test may be large while the damage skin is small or zero.

Worked exampleWhich well to treat first

Given (assumed data)

Common to the wells: oil with $\mu = 1.4$ mPa·s, $B = 1.2$ ; $r_w = 0.108$ m, $r_e = 300$ m; the lift systems and the gathering system can pass the gain
Well 1: $k = 8$ mD, $h = 12$ m, drawdown 6.0 MPa; skin +14, of which pseudo-skin +1
Well 2: $k = 60$ mD, $h = 15$ m, drawdown 3.0 MPa; skin +5, no pseudo-skin
Well 3: $k = 30$ mD, $h = 20$ m, drawdown 3.5 MPa; skin +12, of which +9 is partial penetration
Cost of the treatment: US$80 thousand plus US$4 thousand per 1 m³ of fluid; 1.5 m³ of fluid per metre of thickness

Solution

By skin, the order would be: 1 (+14), 3 (+12), 2 (+5).
The damage skin that acid can remove is 13, 5 and 3. After the treatment the pseudo-skins +1, 0 and +9 remain.
The rate before and after the treatment at the same drawdown ( $X = 7.179$ ): well 1, 8.7 → 22.5 m³/d (2.6 times); well 2, 70.7 → 119.9; well 3, 34.9 → 41.4.
The gain: 13.8; 49.2 and 6.5 m³/d. By gain, the order is 2, 1, 3.
The cost of the treatments: US$152, US$170 and US$200 thousand. Per 1 m³/d of gain: US$11.0, US$3.5 and US$30.9 thousand.
The simple payback time is the cost divided by the income from the gain per day. At a net income of US$252 per cubic metre of oil, it is 44, 14 and 123 days.

Answer: Well 2 is treated first. It has the smallest skin, but the largest gain and the cheapest unit of gain. Well 3, with a “large” skin, is the worst candidate.

What it means. Well 1 is badly damaged, but its reservoir gives little. So even a rise in rate of 2.6 times is only 13.8 m³/d. In well 3, three quarters of the skin is partial penetration. Acid will not remove it, and perforating the part of the reservoir that was left unperforated is likely to give more. The measure “cost per unit of gain” is also convenient because the alternative to a treatment is drilling a new well or a sidetrack (a new branch drilled from an existing well), and it is easy to compare them in the same units. The simple payback time overstates the benefit here, because it takes into account neither the decline of the gain nor the risk. We deal with this in the next section. The data in this example are assumed for illustration.

The economics of a treatment

A treatment is justified only if the income from the extra production, brought back to today's value (Discounting and present value), is larger than its cost. Three things decide the income: the initial gain, how long it lasts, and the net income from each extra tonne or cubic metre. This is the price less payments on revenue, variable costs and taxes. We estimate the initial gain from the skin and the Hawkins formula if the depth and degree of damage are known. But the most reliable way is from the results of similar treatments in the same field.

Decline of the gain. After a treatment a well usually declines faster than before it, because it takes the same oil out of its drainage area faster. A large part of the extra production is oil produced earlier, not oil that would otherwise never be produced (more detail in the advanced section).
Probability of success. Not every treatment succeeds. The fluid may not suit the damage, or the acid may go into the wrong interval. The expected gain is the gain of a successful treatment multiplied by the probability of success, which is clearly less than one.
Losses during the job. The well is shut in for preparation, treatment and clean-up (flowing the well to remove the spent acid). Sometimes the spent acid cannot be accepted at once at the processing plant, so it is trucked away or the intake of production is limited.
Costs. Some depend on the kind of job: delivery and hire of pumping units, tanks and coiled tubing (a continuous flexible pipe run into the well), the crew, and bringing the well back to flow after the treatment (for example, with nitrogen). Others depend on the volume: acid and additives.

The simplest criterion is the payback period: the time in which the income from the gain returns the costs. Companies usually rank treatments by it, and they require a treatment to pay back within a few months. For a full comparison with other investments we calculate the NPV, the net present value (Net present value (NPV)). If the gain declines exponentially, the payback period follows from a simple integral:

Payback time when the gain declines exponentially

t_{\text{pb}} = -\frac{1}{a}\ln\left(1 - \frac{aC}{\Delta q_0\,v}\right)

where:

$t_{\text{pb}}$ payback time, d
$C$ total costs: the cost of the job plus the value of the production not received during the job, US$
$\Delta q_0$ initial gain in the oil rate; with risk, the expected gain $P_{\text{succ}}\,\Delta q_0$ , where $P_{\text{succ}}$ is the probability of success, m³/d
$a$ rate of decline of the gain: $\Delta q(t) = \Delta q_0\,\e^{-at}$ ; the gain halves in the time $\ln 2/a$ , 1/d
$v$ net income per cubic metre of extra oil, US$/m³

Units: any consistent units: here days, m³/d and dollars

Holds when

The gain declines exponentially at a constant rate $a$ , and the net income per cubic metre is constant.
Discounting is not applied. For payback times of up to a year it changes the answer little.
A solution exists only if $aC < \Delta q_0 v$ . Otherwise all the future income from the gain, $\Delta q_0 v/a$ , is smaller than the costs, and the treatment will never pay back.

The cumulative income from the gain grows as $\Delta q_0 v\,(1 - \e^{-at})/a$ and tends to $\Delta q_0 v/a$ . The payback time is the moment when it equals the costs. Without decline ( $a \to 0$ ) the formula turns into the simple ratio $C/(\Delta q_0 v)$ . The closer $aC/(\Delta q_0 v)$ is to one, the more the decline pushes the payback back.

Worked exampleWill the treatment of a doubtful candidate pay back

Given (assumed data)

Well: No. 3 from the previous example: gain 6.5 m³/d, rate before the treatment 34.9 m³/d, cost of the job US$200 thousand
Shut-in during the job and clean-up: 3 d
Decline of the gain: halves in 240 d (8 months); in the unfavourable case, in 120 d
Probability of success: 0.75
Net income: US$252 per cubic metre of oil, as in the previous example

Solution

The production not received in 3 days is worth $3\cdot 34.9\cdot 252$ US dollars, which is US$26.4 thousand. The total costs are $C = 200 + 26.4 = 226$ US$ thousand.
Without decline and without risk: $C/(\Delta q_0 v) = 139$ d.
The rate of decline: $a = \ln 2/240 = 0.00289$ 1/d; $\dfrac{aC}{\Delta q_0 v} = 0.40$ , which gives $t_{\text{pb}} = 178$ d.
To allow for risk, we put the expected gain $0.75\,\Delta q_0$ into the formula. The ratio is 0.54, and $t_{\text{pb}} = 265$ d, which is about 8.7 months.
If the gain declines twice as fast ( $a = 0.00578$ 1/d), the ratio is 1.07, which is more than one. All the expected income, US$212 thousand, is smaller than the costs, and the treatment will not pay back.
For comparison, for well 2 under the same assumptions the expected payback time is about 25 d.

Answer: Allowing for risk, the treatment of well 3 pays back in 8.7 months if the decline is slow, and does not pay back at all if it is fast. Well 2 returns its costs in less than a month.

What it means. The decline of the gain and the risk turned a simple payback time of 139 d into 265 d, and with a fast decline into a loss. For a good candidate the margin is so large that these corrections change almost nothing. For a doubtful one they decide the matter. Before spending money on well 3, we should evaluate a cheap alternative: perforating the rest of the reservoir. The data in this example are assumed for illustration.

CalculatedThe cumulative cash flow of the treatment of well 3 of the example: the costs at the start, and the income from the gain, which declines with time. A: a successful treatment, the gain halves in 8 months. B: the same, allowing for the probability of success of 0.75. C: allowing for risk, the gain halves in 4 months. The points on the zero line are the payback times. Curve C does not rise to zero.

The kind of treatment and the work cycle

We choose the fluid by two things: what exactly limits the flow, and where it is: in the pipes, on the wall of the wellbore, in the near-wellbore zone, or deeper. The place decides whether the fluid is pumped into the reservoir at all.

Where the problem is and which treatment is needed

Where the problem is: Deposits in the tubing: salts, paraffin
Treatment: washing the tubing with acid or a solvent, with circulation
Fluid into the reservoir?: no: the fluid is returned to the surface

Where the problem is: Cake and deposits on the wall of an open hole, in a screen (a filter that holds back sand), in perforation tunnels
Treatment: an acid bath: the fluid is left standing opposite the interval
Fluid into the reservoir?: hardly any

Where the problem is: The near-wellbore zone, up to a metre from the wall
Treatment: matrix acid treatment
Fluid into the reservoir?: yes

Where the problem is: Emulsion, water block, ARPD in the near-wellbore zone
Treatment: solvents, mutual solvents (liquids that dissolve in both oil and water), surfactants
Fluid into the reservoir?: yes

Where the problem is: Future salt deposits
Treatment: squeezing an inhibitor a few metres into the reservoir
Fluid into the reservoir?: yes; the inhibitor is slowly carried out with the production

Where the problem is	Treatment	Fluid into the reservoir?
Deposits in the tubing: salts, paraffin	washing the tubing with acid or a solvent, with circulation	no: the fluid is returned to the surface
Cake and deposits on the wall of an open hole, in a screen (a filter that holds back sand), in perforation tunnels	an acid bath: the fluid is left standing opposite the interval	hardly any
The near-wellbore zone, up to a metre from the wall	matrix acid treatment	yes
Emulsion, water block, ARPD in the near-wellbore zone	solvents, mutual solvents (liquids that dissolve in both oil and water), surfactants	yes
Future salt deposits	squeezing an inhibitor a few metres into the reservoir	yes; the inhibitor is slowly carried out with the production

ARPD means asphaltene, resin and paraffin deposits (the sticky solids that the heavy parts of the oil leave behind). The damage mechanisms that these treatments target are covered in the lesson Formation damage and damage skin.

The treatment cycle

1
Selection
A well with a removable damage skin, enough oil or gas left, enough pressure and spare capacity in the system, as described above.
2
Diagnosis
The type and place of the damage: the history of operations on the well, analysis of deposits and of particles carried out, and core tests with the fluids we plan to use.
3
Fluids and volumes
The main acid, the preflush and the overflush (fluids pumped before and after the acid), additives (a corrosion inhibitor, iron control agents, surfactants, a mutual solvent), and volumes set by the thickness of the interval and the depth of the damage (Acid systems and treatment design).
4
Pumping regime
A rate and pressure below the fracture pressure, and diversion methods (ways to steer the acid), so that the acid enters all the intervals and not only the most permeable one.
5
Execution
Preparing the site, pressure-testing the lines, pumping with a record of pressure and rate, and pushing the fluid into the reservoir.
6
Clean-up
Quick removal of the spent acid before secondary precipitates come out of it. If needed, we start the flow with nitrogen or by swabbing (lowering a tool on a cable to lift liquid out of the well).
7
Evaluation and conclusions
Measuring the rate and testing the well after the treatment, and comparing with the forecast. We update the selection rules and the fluids for the next wells.

A treatment can also do harm. Secondary precipitates (new deposits that form from the reaction products) come out of the spent acid: iron hydroxides, fluorides and fluorosilicates. With oil rich in asphaltenes, acid forms a thick sludge and emulsions. In a weak sandstone it dissolves the cement between the grains and provokes sand production. Without an inhibitor it eats the pipes. So we check the fluid for compatibility with the reservoir fluids and the rock, and we remove the spent acid quickly. And most important: preventing damage (lesson on formation damage) is almost always cheaper than curing it afterwards.

Terms and symbols

Matrix treatment: Pumping a fluid into the reservoir at a pressure below the fracture pressure. The fluid seeps into the pores radially and acts in the near-wellbore zone.
Hydrochloric acid treatment: A matrix treatment with hydrochloric acid. It is the main kind of treatment for carbonates and the first stage of the treatment of sandstones.
Mud acid: A mixture of hydrochloric and hydrofluoric acids for dissolving clays, feldspars and fine particles in sandstones.
Wormhole: A thin, branching channel of high conductivity that hydrochloric acid cuts in carbonate rock.
Acid diversion: Methods that make the fluid enter all the intervals and not only the most permeable one: mechanical plugs, viscous and foam systems, and diverting particles.
Payback period: The time in which the income from the extra production returns the cost of the treatment.
Candidate well: A well whose removable damage skin, remaining oil or gas, pressure and spare capacity of the system promise a gain that pays back.

Common mistakes and misconceptions

Wrong

“The larger the skin, the better the candidate for an acid treatment.”

In fact

The gain is also proportional to the flow capacity $kh$ of the reservoir and to the drawdown. In the example of the lesson, a skin of +14 in a reservoir with $kh$ of about 100 mD·m gave a gain three and a half times smaller than a skin of +5 in a reservoir with $kh$ of 900 mD·m.

Wrong

“Acid in sandstone will give a negative skin, as in carbonate.”

In fact

In sandstone, acid dissolves what has plugged the pores but hardly dissolves the framework. At best, the damage skin returns to zero. A negative skin comes from wormholes in carbonates and from fractures.

Wrong

“Productivity doubled, so production will double.”

In fact

Only if the lift system and the surface equipment can pass the extra liquid. Also, after a treatment a well usually declines faster, and part of the gain is only faster production.

Wrong

“Let us raise the injection pressure so that the acid reaches the damage faster.”

In fact

Above the fracture pressure the rock splits, and the fluid goes into the fracture and bypasses the damaged ring. A matrix treatment is run below the fracture pressure.

Wrong

“If the treatment did not help, there was no damage.”

In fact

Perhaps the fluid did not suit the type of damage, the acid went into one permeable interval, or it created deposits itself. We analyse a failure to improve the diagnosis and the technology.

In depth: advanced detail

Faster production or extra reserves

Imagine a well that drains a closed volume with no pressure support (no water injection or influx to hold up the pressure) and works at a constant bottomhole pressure. Its rate declines exponentially (by the same share in each period of time), $q = q_1\e^{-D_1 t}$ , and the decline rate $D_1$ is proportional to productivity. The faster the well takes out liquid, the faster the pressure in the drainage area falls. A treatment that raises productivity $F$ times raises both the initial rate and the decline rate by the same factor. Up to the economic limit, which is the rate $q_{\text{el}}$ below which production loses money, the well produces $(q_{\text{i}} - q_{\text{el}})/D$ . This is the oil that can be produced before the well becomes uneconomic (the reserves of the well). The difference between the treated and the untreated well is:

Extra production up to the economic limit after a treatment

\Delta N = \frac{q_{\text{el}}}{D_1}\left(1 - \frac{1}{F}\right)

where:

$\Delta N$ extra cumulative production up to the economic limit, m³
$q_{\text{el}}$ rate at the economic limit, m³/yr
$D_1$ decline rate before the treatment: $q = q_1\e^{-D_1 t}$ , 1/yr
$F$ factor by which the treatment raises productivity, —

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

Holds when

A closed drainage area, with no water influx and no injection. Single-phase inflow. A constant bottomhole pressure.
The decline is exponential, and its rate is proportional to productivity. The economic limit in terms of rate is the same in both cases.

Both wells would take out the same total movable volume, $q_1/D_1$ . The treated well takes it out faster and reaches the economic limit sooner, but at a lower reservoir pressure, and this is where the extra production comes from. It is larger when the rate at the economic limit is higher (wells that are expensive to run, offshore fields) and when the factor is larger. The main value of a treatment is in time: the oil is received earlier, and early money is worth more than late money.

Worked exampleHow many reserves a treatment adds

Given (assumed data)

Rate before the treatment $q_1$: 20 m³/d
Decline rate $D_1$: 0.15 1/yr
Productivity improvement factor $F$: 1.7
Rate at the economic limit: 2 m³/d

Solution

Without treatment: $N_1 = \dfrac{(20 - 2)\cdot 365}{0.15} = 43{,}800$ m³ in $\ln(20/2)/0.15 = 15.4$ years.
After the treatment: the initial rate is $1.7\cdot 20 = 34$ m³/d, the decline rate is $1.7\cdot 0.15 = 0.255$ 1/yr; $N_2 = 45{,}804$ m³ in 11.1 years.
The extra production: $\Delta N = \dfrac{2\cdot 365}{0.15}\left(1 - \dfrac{1}{1.7}\right) = 2004$ m³, which is 4.6% of the production of the untreated well.
In the first year alone the treated well gives 4175 m³ more, which is 2.1 times more than the whole extra production.

Answer: The treatment adds about 2004 m³ to the production up to the economic limit (4.6%), but the gain of the first year is 4175 m³. Most of the benefit is faster production.

What it means. This explains the faster decline after a treatment that is seen in practice, and it gives the main conclusion for the economics: extra oil must be valued with time taken into account, and it is better to treat a well early, while there is something to speed up. If the pressure in the reservoir is maintained by injection, the picture is different: the decline rate grows less, and the share of new reserves is higher. The data in this example are assumed for illustration.

Connections to other topics

Formation damage and damage skin — damage mechanisms, the damage skin, and the choice between preventing, removing and bypassing.
Skin factor — the skin factor, its components, and the limit of a negative skin for a treated zone.
Acid systems and treatment design — hydrochloric acid and mud acid, additives, volumes and pumping rates.
Acid treatment of carbonates and special wells — wormholes, acid diversion, and treatments of horizontal and injection wells.
Hydraulic fracturing: purpose and well selection — when a fracture is needed instead of acid.
Nodal analysis of the production system — why a gain in productivity does not always become a gain in production.
Cumulative cash flow and payback period — payback time as a criterion, and its limits.
Net present value (NPV) — how to compare a treatment with other investments, taking the time value of money into account.

Check yourself

1
Well A has a skin of +12 and $kh = 50$ mD·m. Well B has a skin of +4 and $kh = 1500$ mD·m. The drawdowns, the fluids and $X = 7$ are the same, and the whole skin is damage. Which treatment will give the larger gain in rate?
1. A.Treating well A: its skin is three times larger
2. B.Treating well B
3. C.The same: the gain depends only on the skin
4. D.We cannot tell without the bubble point pressure (the pressure at which gas starts to come out of the oil)
Show the answer
Correct answer: B. Treating well B
The gain is proportional to $kh\,[1/X - 1/(X + S)]$ . For A: $50\cdot(1/7 - 1/19) \approx 4.5$ . For B: $1500\cdot(1/7 - 1/11) \approx 78$ in the same units, which is about 17 times more. The ratio is higher for A ( $19/7 \approx 2.7$ against $11/7 \approx 1.6$ ), but the gain in cubic metres decides, and it is B's.
2
What is it reasonable to expect from an acid treatment of an undamaged well in sandstone?
1. A.A skin of −3 or lower, as after hydraulic fracturing
2. B.About a doubling of the rate
3. C.A small gain, at best 10–20%
4. D.A rise in the permeability of the whole reservoir
Show the answer
Correct answer: C. A small gain, at best 10–20%
In sandstone, acid dissolves the plugging material, not the quartz framework. If there is nothing plugging the pores, only a little can be gained. Even a fivefold rise in permeability in a ring out to 0.4 m from the axis would give a skin of about −1 and a gain of about 17%. The value of the treatment is in the damage skin that it removes.
3
The gain after a treatment is 8 m³/d of oil, the net income is US$250/m³, and the costs are US$150 thousand. Find the payback time without decline of the gain, and with a decline at the rate $a = 2$ 1/yr. Will the treatment pay back if $a = 5$ 1/yr?
Show the answer
The income from the gain is $8\cdot 250 = 2000$ US$/d, and without decline the treatment pays back in $150{,}000/2000 = 75$ d. At $a = 2$ 1/yr, that is $0.00548$ 1/d, the ratio is $aC/(\Delta q_0 v) \approx 0.411$ , and the payback formula gives $t_{\text{pb}} \approx 97$ d. At $a = 5$ 1/yr the ratio is about 1.03, which is more than one. All the future income, $2000\cdot 365/5 = 146{,}000$ US$, is smaller than the costs, and the treatment will not pay back.
4
The skin of a well from a well test is +10, of which +7 is a pseudo-skin of partial penetration. What should we expect from an acid treatment?
1. A.The skin will become negative
2. B.The skin will fall to about +7, and no lower
3. C.The skin will fall to zero
4. D.The skin will not change under any conditions
Show the answer
Correct answer: B. The skin will fall to about +7, and no lower
Acid removes only the damage skin, here about +3. The pseudo-skin of partial penetration stays. It can be reduced only by adding perforations or by a different design of the opening. If +3 gives little, it is better to start with perforating.
5
Why is a matrix acid treatment run at a pressure below the fracture pressure of the reservoir?
Show the answer
Above the fracture pressure the rock splits, and the fluid goes into the fracture. In sandstone it then hardly passes through the damaged ring around the wellbore, and the fracture may reach a water-bearing layer. Below the fracture pressure the fluid seeps radially and passes through the whole damaged zone. Acid fracturing of carbonates is a separate technology with a different aim.
6
Why does the rate of a well often decline faster after a successful treatment than before it?
Show the answer
A well with higher productivity takes oil out of its drainage area faster, and the pressure there falls faster. In the simple model of a closed volume, the decline rate grows by the same factor as the productivity. A large part of the gain is faster production, not new reserves. So the economics of a treatment are evaluated with time taken into account.

Updated: 1 October 2026