Water injection systems

Quick refresher

A pressure maintenance system consists of a source, treatment, pump stations, water lines, distribution blocks and injection wells. We design it for the whole life of the field: the share of produced water in the injection grows with time, and the share of source water falls.
Each source brings its own troubles: produced water brings oil, suspended solids, hydrogen sulfide and bacteria; underground water brings gases and iron; sea water brings oxygen, living organisms and sulfates; fresh surface water brings seasonal turbidity (cloudiness) and the swelling of clays. Incompatible waters mixed at the surface form a precipitate right in the pipes, filters and wells.
The reservoir sets the water quality: particles a third of the pore throat size or larger build a cake (a layer of solids) on the wall of the well, and those from a tenth to a third plug the pores from inside. The lower the permeability, the smaller the allowed particles.
We remove oxygen first mechanically, by vacuum or by stripping with gas (blowing gas through the water), which lowers its partial pressure. Then we bind what is left with a scavenger. Without deaeration, we need more than a hundred times as much scavenger.
The bottomhole pressure of an injection well is the wellhead pressure plus the weight of the water column minus the friction in the tubing. The injection limit is set by the fracture propagation pressure at the bottom of the hole. We can convert it to a wellhead pressure only for a given rate, because at a high rate the friction amounts to megapascals.
The pump station keeps the pressure needed by the “tightest” well (the one that takes water with the most difficulty), the others are throttled, and part of the power is lost in chokes. A Hall plot (a plot of cumulative pressure × time against cumulative injected volume) built from daily data shows plugging (the slope grows), cleaning and the opening of fractures (the slope falls): the slope equals $1/I$ .

Key relationships

I = \dfrac{q}{p_w - \bar p}

injectivity index of an injection well

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

bottomhole pressure in injection: wellhead pressure plus the water column minus the friction in the tubing

q_{\max} = I\,(p_{\text{frac}} - \Delta p_{\text{margin}} - \bar p)

the largest rate at which the bottomhole pressure stays below the fracture propagation pressure with a margin

d_{\text{th}} \approx 2\sqrt{8k/\phi}

pore throat diameter by the model of a bundle of straight capillaries (a lower estimate)

\dfrac{\Delta\Sigma(p_w - \bar p)\Delta t}{\Delta W} = \dfrac{1}{I}

slope of the Hall plot in steady injection

Common pitfalls

Mixing waters of different sources without a compatibility analysis: the precipitate will form in your own system.
Converting the fracture pressure limit to the wellhead with the hydrostatic pressure alone, or carrying a wellhead pressure limit found at a high rate over to a low rate.
Counting on an oxygen scavenger to replace the deaerator.
Judging plugging by the rise of the injection pressure without taking the rate into account. The right way is to look at the injectivity and at the slope of the Hall plot.
Keeping the pressure of the whole station for one “tight” well, instead of stimulating that well (treating the zone around it) or giving it a separate pump.

After this lesson you can

compare the sources of water for injection (produced, underground, sea and surface water) and name the treatment that each one needs
judge the compatibility of waters that are mixed in one system, and estimate the mass of salt that can drop out in it in a year
link the allowed particle size in the water to the pore throats (the narrow openings between pores) of the reservoir, and list the other quality requirements
calculate what partial pressure of oxygen a deaerator (a unit that removes oxygen) must leave and how much scavenger (a chemical that binds oxygen) is needed for the remainder
calculate the wellhead pressure of an injection well, counting hydrostatic pressure and friction, find the largest rate without opening a fracture, and estimate the power losses in distributing water among wells
read the injection history from a Hall plot: plugging, treatment and the opening of a fracture

What an injection system consists of

The field development project decides how much water to inject, and where. It sets the voidage replacement (how much of the volume of oil and water taken out is put back by injection), the pattern of injection wells, and the pressure in the reservoir (Why water is injected: pressure maintenance and oil displacement). The injection system has to deliver this: it takes the water, treats it, raises its pressure and distributes it to the wells, day after day, for decades. The volumes are large. One injection well takes from tens to thousands of cubic metres a day, and a field takes tens of thousands (loads on the facilities).

SchematicA pressure maintenance system on land (schematic): sources of water, treatment, the water injection pump station (WIPS), high-pressure water lines, the water distribution block (WDB) and the injection wells. The circles mark the points where the water quality is checked. In the boxes on the left, PWRU is the preliminary water removal unit and CPF is the central processing facility. WIPS, WDB, PWRU and CPF are short labels used only in the figure.

Links of the pressure maintenance system

1
Source and water intake
Water supply wells with submersible pumps, a sea water intake with screens, and produced water from the preliminary water removal units and the central processing facility (Produced water: composition, treatment and disposal).
2
Treatment
Removal of oil and suspended solids, filtration, removal of oxygen and hydrogen sulfide, and dosing of chemicals. Each source has its own scheme.
3
Pump stations
Water injection pump stations, often in packaged form (assembled from factory-made blocks), with multistage centrifugal pumps. Booster pumps give the main pumps enough pressure at their inlet.
4
Water lines
Low-pressure lines go from the treatment to the pump stations, and high-pressure lines from the stations to the wells. Coatings protect them from corrosion, or they are made of fibreglass.
5
Distribution
Water distribution blocks have a flowmeter and a flow regulator for each well. Injection is metered well by well here.
6
Injection wells
A wellhead with valves and pressure gauges, and tubing (the pipe that carries the water down the well) with a packer (a seal that closes the space around the tubing). The packer protects the casing (the steel pipe that lines the well) from the injection pressure. When water is injected into several formations, equipment for separate injection is used (Multiple-zone completions: commingled and segregated production).

The pump stations are based on sectional multistage centrifugal pumps (in Russian practice they have the type name CNS). Such a pump has many impellers (spinning wheels with blades) one after another on one shaft, and each stage adds to the pressure. The designation gives the delivery in cubic metres per hour and the head in metres (the height to which the pump could lift the water). For example, the pump CNS 180-1900 delivers 180 m³/h at a head of 1900 m. The head is converted to pressure through the density of the liquid pumped, $\Delta p = \rho g H$ : 1900 m of fresh water is about 18.6 MPa, and 1900 m of brine with a density of 1040 kg/m³ is about 19.4 MPa. A pump develops the same head, and the pressure depends on what it pumps. For individual wells with low injectivity (wells that take little water for a given pressure), we install separate pumps: plunger pumps (piston pumps, good for high pressure at a low rate), or horizontal multistage pumps at the wellhead.

Injection is one of the largest consumers of energy in a field. The hydraulic power of the pumps equals the rate times the pressure rise (hydraulic power), and it grows together with the injected volumes. So we keep the pressure in the system no higher than needed, and we adjust the power of the pumps with variable speed drives (devices that change the pump speed), not by throttling (closing a valve to hold the flow back). By the affinity laws the delivery of a pump is proportional to the rotation speed, and its head is proportional to the square of it.

Where the water comes from

Sources of water for injection

Source: Produced water
Advantages: available in any field with a water cut (water in the produced liquid); injection into its own reservoir
Difficulties: oil, suspended solids, hydrogen sulfide, bacteria; the volume is set by production
Treatment: removal of oil and suspended solids, a closed system, biocides (chemicals that kill bacteria)

Source: Underground water of aquifers (water-bearing layers)
Advantages: constant composition, few suspended solids, usually no oxygen
Difficulties: iron, carbon dioxide or hydrogen sulfide; water supply wells are needed
Treatment: removal of gases and iron, filtration

Source: Sea water
Advantages: unlimited volume at offshore fields
Difficulties: saturated with oxygen, full of living organisms, rich in sulfate
Treatment: screens and filters, chlorination, deaeration (removal of dissolved oxygen), and if needed removal of sulfates

Source: Fresh surface water (rivers, lakes)
Advantages: available near the shore
Difficulties: turbidity and composition change with the seasons; fresh water harms shaly (clay-rich) reservoirs
Treatment: settling with coagulants (chemicals that make fine particles clump together), filtration, deaeration

Source: Crossflow from an aquifer in the same well
Advantages: no surface facilities
Difficulties: the rate is hard to measure and control; a strong aquifer is needed
Treatment: only a check of the compatibility of the formations

Source	Advantages	Difficulties	Treatment
Produced water	available in any field with a water cut (water in the produced liquid); injection into its own reservoir	oil, suspended solids, hydrogen sulfide, bacteria; the volume is set by production	removal of oil and suspended solids, a closed system, biocides (chemicals that kill bacteria)
Underground water of aquifers (water-bearing layers)	constant composition, few suspended solids, usually no oxygen	iron, carbon dioxide or hydrogen sulfide; water supply wells are needed	removal of gases and iron, filtration
Sea water	unlimited volume at offshore fields	saturated with oxygen, full of living organisms, rich in sulfate	screens and filters, chlorination, deaeration (removal of dissolved oxygen), and if needed removal of sulfates
Fresh surface water (rivers, lakes)	available near the shore	turbidity and composition change with the seasons; fresh water harms shaly (clay-rich) reservoirs	settling with coagulants (chemicals that make fine particles clump together), filtration, deaeration
Crossflow from an aquifer in the same well	no surface facilities	the rate is hard to measure and control; a strong aquifer is needed	only a check of the compatibility of the formations

The last option (a dump flood) is possible when an aquifer with a higher pressure lies above or below the productive formation: the water flows across in the same well by itself or with the help of a submersible pump.

How much water is needed from the source depends on the year of development. With full voidage replacement it is the reservoir volume of the produced oil, $q_oB_o/B_w$ . In the example of the lesson on produced water it is about 7.8 thousand m³/d on the plateau and about 0.8 thousand in year 20, when the injection consists almost entirely of produced water (the water balance of a field). So we size the water intake and the treatment of source water for the first years. Later we take them out of service in parts or reassign them, while we build up the treatment of produced water.

The law of Kazakhstan also affects the choice. Under the Water Code, taking underground water or water from surface water bodies is, as a general rule, special water use, which needs a permit. Drinking water must not be used in industry if water of another quality will do and the process does not require drinking water. Under the Environmental Code, injecting produced formation water, sea water, desalinated water, or industrial water with a salinity of 2000 mg/L or more for pressure maintenance is not counted as a discharge (as of the texts of the codes of October 2026).

Water compatibility and scale

The injected water must be compatible with three things: with the reservoir rock (it must not cause swelling or movement of clays), with the formation water (it must not form precipitates where they meet), and with the other waters with which it is mixed in the same system. The first two checks are covered in the lesson on formation damage in injection. The same lesson has the saturation index and the mass of precipitate for the example of barium sulfate. Here we take the third: mixing at the surface, where the precipitate (a solid that drops out of the water) forms not in the reservoir but in the system's own pipes, filters and wells.

While there is little produced water, we mix it with the water of the source, and the share of each changes from year to year. If the produced water contains barium or strontium and the source water contains sulfate, the mixture becomes supersaturated with barium or strontium sulfate: it holds more of the dissolved salt than the water can keep, and the salt starts to settle out. A similar trouble is calcium carbonate: on heating and loss of carbon dioxide, a mixture rich in calcium and bicarbonate deposits carbonate on the walls of heat exchangers and pipes.

Worked exampleCan produced water be mixed with water from an aquifer?

Given (assumed data)

Produced water: barium 15 mg/L, practically no sulfate
Aquifer water (the source): sulfate 900 mg/L, no barium
Shares in the injection: from the water balance of the field in the lesson on produced water: in year 8, 14,040 m³/d are injected, of which 6740 is source water; in year 20, 12,947 m³/d, of which 799 is source water
Constants: as in the lesson on formation damage: the solubility product of barite $K_{sp} = 1.08\cdot 10^{-10}$ (mol/L)² at 25 °C, the product of the activity coefficients 0.03; molar masses: barium 137.33, sulfate 96.06, barite 233.38 g/mol

Solution

Year 8. The share of source water is $6740/14{,}040 = 0.480$ . In the mixture there is barium $(1 - 0.480)\cdot 15/137.33 = 0.0568$ mmol/L and sulfate $0.480\cdot 900/96.06 = 4.497$ mmol/L.
The saturation index is $\mathrm{SI} = \log_{10}\dfrac{0.03\cdot 0.0568\cdot 10^{-3}\cdot 4.497\cdot 10^{-3}}{1.08\cdot 10^{-10}}$ $= 1.85$ . The mixture is strongly supersaturated.
By the formula for the mass of precipitate, $x = 0.0560$ mmol/L drops out, which is 98.6% of the barium of the mixture; there is tens of times more sulfate. This is $0.0560\cdot 233.38 = 13.07$ g of barite per cubic metre, or $13.07\cdot 14{,}040/1000 = 183.5$ kg a day, or 67.0 t a year.
Year 20. The share of source water is $799/12{,}947 = 0.0617$ ; barium 0.1025 and sulfate 0.578 mmol/L; $\mathrm{SI} = 1.22$ , which is lower than in year 8.
But $x = 0.0950$ mmol/L drops out (92.7% of the barium), 22.18 g/m³, which is $22.18\cdot 12{,}947/1000 = 287.2$ kg a day, or 104.8 t a year. For comparison, all the barium of the produced water expressed as barite is $12{,}148\cdot 15\cdot 233.38/137.33/1000$ $= 309.7$ kg a day.

Answer: No, they cannot be mixed. In year 8 about 183.5 kg of barite a day (67.0 t a year) can drop out in the system, and in year 20 about 287.2 kg a day (104.8 t a year), although the saturation index fell from 1.85 to 1.22.

What it means. While there is more sulfate in the mixture than barium, almost all the barium that the produced water brings drops out, and the mass of precipitate grows together with the produced water, although the saturation index falls. The index tells about the “driving force”, not about the amount. Tens of tonnes of barite a year means plugged filters, pumps, water lines and perforation tunnels of the injection wells (the channels shot through the casing into the reservoir), and barite hardly dissolves in acid. There are three ways out. We can inject the waters separately, into different groups of wells (but after breakthrough they will meet at the producing wells; Sources of formation damage). We can choose a source without sulfate. Or we can remove the sulfate with membranes (nanofiltration: membranes with very fine pores). An inhibitor delays the precipitation but does not change the equilibrium. With such supersaturation and tonnes of possible precipitate, it only moves the precipitation further along the path of the water, to the injection well and into the reservoir near it. The composition of the produced water was taken as constant, although after breakthrough of the injected water it has less barium, so for the late years the calculation overestimates the precipitate. The constant is taken for 25 °C and the product of the activity coefficients is assumed. With a large excess of sulfate (year 8) the barium sets the mass of precipitate, and refining the constants hardly changes it. With a smaller excess (year 20) the mass is more sensitive to them, so we refine it with a calculation at the working temperature and salinity. The data in this example are assumed for illustration.

CalculatedBarite that can drop out in the injection system when produced water (barium 15 mg/L) is mixed with aquifer water (sulfate 900 mg/L), by year. The dashed line is all the barium of the produced water expressed as barite. The shares of the waters follow the water balance of the field at full voidage replacement.

Water analysis: the full ion composition of each water, including barium, strontium, iron, bicarbonate, sulfate and pH, in several samples. The composition of produced water changes with the breakthrough of the injected water.
Calculation of mixtures: the saturation index and the mass of precipitate for all mixing ratios, at the temperature and pressure of each part of the system, from the water intake to the bottom of the hole.
Laboratory check: mixing the waters in different proportions at working temperature and watching for cloudiness and precipitate, and flowing the water through core of the reservoir.
Decision: separate systems, another source, removal of sulfates or softening (removing calcium and magnesium), and inhibitors where the supersaturation is small.

Water quality requirements

An injection well takes hundreds of thousands of cubic metres of water in a year, and everything that is suspended in the water stays near the wellbore. Even a few milligrams per litre means hundreds of kilograms of particles a year (plugging in injection). Not only particles of sand and clay are dangerous, but also oil droplets, which glue particles together and reduce the permeability to water, corrosion products, iron sulfide and the slime of bacteria. Which particles are dangerous for a particular reservoir is set by the size of its pore throats.

An empirical rule (a rule found from experience; selection of bridging agents) says that particles a third of the throat diameter or larger are held at the wall of the well and build an external cake. Particles from a tenth to a third penetrate the reservoir and plug the pores from inside, and still smaller ones pass through. We measure the throat size on core by mercury injection (forcing mercury into a sample under rising pressure). For an estimate, a simple model is enough: a bundle of capillaries (the bundle of capillaries model):

Pore throat diameter by the model of a bundle of straight capillaries

d_{\text{th}} \approx 2\sqrt{\frac{8k}{\phi}}

where:

$d_{\text{th}}$ characteristic diameter of the pore throat, m
$k$ permeability of the reservoir (1 mD = $0.987\cdot 10^{-15}$ m²), m²
$\phi$ porosity, fraction

Units: SI

Holds when

The rock is represented by a bundle of identical straight tubes. With tortuosity $\tau$ (the winding of the flow paths) the radius is $\tau$ times larger, so the formula gives a lower estimate, and the allowed particle sizes derived from it have a margin.
This is an order-of-magnitude estimate. Real throats have a wide distribution of sizes, and the design is based on the large ones, which carry the main flow.

The throats grow as the root of the permeability: a reservoir ten times more permeable has throats about three times wider. So a tight reservoir needs water cleaned of finer particles.

Worked exampleWhich particles are dangerous for two reservoirs

Given (assumed data)

Reservoir A: permeability 50 mD, porosity 0.18
Reservoir B: permeability 500 mD, porosity 0.22

Solution

Reservoir A: $k = 50\cdot 0.987\cdot 10^{-15} = 4.935\cdot 10^{-14}$ m², $d_{\text{th}} = 2\sqrt{8\cdot 4.935\cdot 10^{-14}/0.18}$ $= 2.96$ µm.
Particles of a third of the throat, $2.96/3 = 0.99$ µm, build a cake on the wall. Particles from a tenth, $0.30$ µm, up to 0.99 µm, penetrate and plug the pores.
Reservoir B: $k = 4.935\cdot 10^{-13}$ m², $d_{\text{th}} = 8.47$ µm. The cake is built by particles of 2.82 µm and larger, and particles from 0.85 to 2.82 µm penetrate inside.

Answer: The throats of reservoir A are about 2.96 µm, and its pores are plugged by particles from 0.30 to 0.99 µm. The throats of reservoir B are about 8.47 µm, and the particles dangerous for its pores are three times larger, from 0.85 to 2.82 µm.

What it means. Water that is fit for reservoir B can plug reservoir A: a filter that holds back particles from 5 µm will let into it exactly the particles that plug the pores from inside. This is the logic of the industry standards: the lower the permeability, the lower the allowed content of suspended solids and the finer the particles. The estimate is rough and made with a margin, because with tortuosity the throats are wider. We refine it with the throat size distribution from mercury injection tests and by flowing water through core. The data in this example are assumed for illustration.

What is limited in injection water and why

Indicator: Suspended particles and their size
Why it is dangerous: a cake on the wall, plugging of pores
How it is met: settling, filtration, coagulants

Indicator: Oil
Why it is dangerous: glues particles together and reduces the permeability to water
How it is met: treatment of produced water

Indicator: Dissolved oxygen
Why it is dangerous: corrosion and its products, which are new particles
How it is met: a closed system, deaeration, a scavenger

Indicator: Hydrogen sulfide and iron
Why it is dangerous: corrosion, a precipitate of iron sulfide
How it is met: degassing, removal of iron, separate systems

Indicator: Bacteria (sulfate-reducing bacteria, SRB, and others)
Why it is dangerous: slime, hydrogen sulfide, infection of the reservoir
How it is met: chlorination, biocides, cleanliness

Indicator: pH, ion composition
Why it is dangerous: scale, swelling of clays
How it is met: compatible sources, inhibitors

Indicator	Why it is dangerous	How it is met
Suspended particles and their size	a cake on the wall, plugging of pores	settling, filtration, coagulants
Oil	glues particles together and reduces the permeability to water	treatment of produced water
Dissolved oxygen	corrosion and its products, which are new particles	a closed system, deaeration, a scavenger
Hydrogen sulfide and iron	corrosion, a precipitate of iron sulfide	degassing, removal of iron, separate systems
Bacteria (sulfate-reducing bacteria, SRB, and others)	slime, hydrogen sulfide, infection of the reservoir	chlorination, biocides, cleanliness
pH, ion composition	scale, swelling of clays	compatible sources, inhibitors

The exact standards for a field are set by its project documents on the basis of tests of the waters and of core.

The industry standard for flooding water

In the USSR the requirements for flooding water were brought together in the industry standard OST 39-225-88 “Water for flooding of oil reservoirs. Quality requirements”. It was approved by the Ministry of the Oil Industry in 1988 and came into force on 1 July 1990, and it is still referred to in design. The allowed content of suspended solids and oil in it depends on the permeability and the fracturing of the reservoir, from a few milligrams per litre for tight porous formations to tens for permeable fractured ones. For porous reservoirs the same standard says that 90% of the particles must be no larger than 5 µm at a permeability above 0.1 µm² (about 100 mD), and no larger than 1 µm at a lower permeability. Dissolved oxygen must be no more than 0.5 mg/L, and the pH must be from 4.5 to 8.5. For formations without hydrogen sulfide, the water must contain no sulfate-reducing bacteria. If the injectivity has fallen by 20% since the start of injection, the standard calls for restoring the permeability of the near-wellbore zone and, if needed, improving the water treatment.

Sea water is saturated with oxygen, so it is treated more strictly. In offshore projects we bring dissolved oxygen down to tens of parts per billion (ppb) or less, for example no more than 20 ppb at the pump outlet, because steel in salt water with oxygen quickly gets covered with pits. We check the quality at several points: after the filters, after the deaerator, at the pump outlet and at the wellheads of the most distant wells, where corrosion products and bacteria have got into the water on its way along the water lines.

Treatment of water for injection

Treatment of sea water (an example of a full scheme)

1
Water intake
Screens hold back debris and living organisms. We choose the place and depth of the intake so that the water is as clean and as constant in composition as possible. Lift pumps deliver it to treatment.
2
Chlorination
Sodium hypochlorite, often made by electrolysis of the sea water itself, kills algae and bacteria and keeps them from fouling filters and pipes.
3
Coarse and fine filtration
Granular filters (sand, anthracite, which is a hard coal, and garnet, a heavy mineral sand) with coagulants, then cartridge filters. We wash the filters with a reverse flow.
4
Deaeration
Oxygen is removed in a column, under vacuum or by stripping (a stream of gas, nitrogen or associated gas, carries the oxygen out of the water), and an oxygen scavenger takes care of what is left.
5
Removal of sulfates
If the formation water is rich in barium or strontium, nanofiltration membranes hold back sulfate and let sodium chloride pass.
6
Chemicals and pumps
Biocides in shock doses from time to time, scale and corrosion inhibitors, and booster and injection pumps.

We treat produced water by the scheme of the lesson on produced water: settling tanks, hydrocyclones, flotation, filters. We keep it in a closed system without air. Underground water usually needs only to be freed of gases and iron and filtered. If we remove the iron by oxidation, we must remove the oxygen again afterwards. We check each scheme for mixing with all the other waters of the system.

Deaeration (removing dissolved oxygen) rests on Henry's law: the equilibrium concentration of a dissolved gas is proportional to its partial pressure over the water. Water saturated with air holds about 7–10 mg/L of oxygen (less in sea water than in fresh water, and the less, the warmer the water), and the partial pressure of oxygen in air is about 21 kPa. To leave only tens of parts per billion in the water, the partial pressure of oxygen over the water must be hundreds of times lower. We do this by lowering the total pressure (vacuum) or by replacing the gas over the water with an oxygen-free one (stripping).

Equilibrium oxygen content of water (Henry's law)

C = C_{\text{sat}}\,\frac{p_{\mathrm{O_2}}}{p_{\mathrm{O_2}}^{\text{air}}}

where:

$C$ equilibrium concentration of dissolved oxygen, mg/L
$C_{\text{sat}}$ concentration in water saturated with air at atmospheric pressure, at the same temperature and salinity, mg/L
$p_{\mathrm{O_2}}$ partial pressure of oxygen in the gas over the water (the share of oxygen times the absolute pressure), kPa
$p_{\mathrm{O_2}}^{\text{air}}$ partial pressure of oxygen in air: $0.2095\cdot 101.325 = 21.23$ kPa, kPa

Units: concentrations in identical units, pressures in identical units

Holds when

Equilibrium: the water is in contact with the gas long enough. In a column it is approached gradually, stage by stage, so in practice more oxygen is left.
The temperature and salinity are the same as for $C_{\text{sat}}$ . With heating, the solubility of oxygen falls.

A deaerator cannot leave less in the water than corresponds to the partial pressure of oxygen in the gas over it. This sets the requirements for the vacuum and for the purity of the stripping gas. The remainder is bound chemically.

Worked exampleA deaerator and an oxygen scavenger for sea water

Given (assumed data)

Water: sea water, 20,000 m³/d at 15 °C; saturated with air, about 8 mg/L of oxygen
Requirement: not more than 20 ppb (0.020 mg/L) of oxygen in the injection water
Deaerator: a vacuum column at 4 kPa (abs.) or a gas stripping column at 150 kPa (abs.); 50 ppb is left after the column
Scavenger: sodium bisulfite: $2\,\mathrm{NaHSO_3} + \mathrm{O_2} \to \mathrm{Na_2SO_4} + \mathrm{H_2SO_4}$ ; molar masses $\mathrm{NaHSO_3}$ 104.06, $\mathrm{O_2}$ 32.00, $\mathrm{SO_4^{2-}}$ 96.06 g/mol

Solution

The partial pressure of oxygen in equilibrium with 0.020 mg/L is $21.23\cdot 0.020/8 = 0.0531$ kPa, which is 400 times less than in air.
In a vacuum column at 4 kPa, the share of oxygen in the gas over the water must be not more than $0.0531/4 = 1.33$ %. With stripping at a pressure of 150 kPa, the stripping gas must contain not more than $0.0531/150$ , that is 354 ppm of oxygen (by moles).
Scavenger: for 1 mg of oxygen, $2\cdot 104.06/32.00 = 6.50$ mg of bisulfite. After the column: $0.05\cdot 6.50 = 0.325$ mg/L, that is 6.5 kg a day (with no margin for incomplete reaction).
Without a deaerator: $8\cdot 6.50 = 52.0$ mg/L, which is 1040 kg a day, or 380 t a year.
Each milligram of oxygen turns into $2\cdot 96.06/32.00 = 6.00$ mg of sulfate: after the column this is 0.30 mg/L, without it 48.0 mg/L.

Answer: The deaerator must keep the partial pressure of oxygen at not more than 0.0531 kPa: in a vacuum column at 4 kPa, not more than 1.33% of oxygen in the gas. Then about 6.5 kg of scavenger a day are needed, and without a deaerator about 1040 kg, which is 160 times more.

What it means. Mechanical deaeration removes almost all the oxygen, and chemistry cleans up the remainder. Full chemical removal would take hundreds of tonnes of chemical a year, and the chemical itself would add almost 50 mg/L of sulfate to the water, one more reason for barite if the formation water contains barium. We inject the scavenger after the column and far from the chlorination point, because chlorine and bisulfite react with each other and are wasted. The calculation is an equilibrium one: a column works worse than equilibrium, so we take the vacuum and the stripping gas with a margin. The data in this example are assumed for illustration.

Bacteria need special attention. Sulfate-reducing bacteria brought in with the water, especially sea water, which is rich in sulfate, multiply in the reservoir near the injection wells and produce hydrogen sulfide. Years later it appears in the production of producing wells that earlier gave “sweet” oil (oil without hydrogen sulfide). So we treat the injected water with biocides, and we monitor the bacteria and the hydrogen sulfide in the production from the first years of waterflooding.

Pumps, water lines and distribution to the wells

From the pump station the water goes along high-pressure water lines to the water distribution blocks, and from there to the wells. The pressure at the outlet of the station is one for all, but the wells need different pressures: each has its own injectivity and its own design rate. The station keeps the pressure needed by the “tightest” well, and for the others it is lowered by regulators and chokes (valves or narrow openings that take off the extra pressure). The energy that the pump spent on this excess pressure turns into heat.

Power lost in throttling

N_{\text{thr}} = \sum_i \frac{q_i\,(p_{\text{hdr}} - p_{\text{wh},i})}{86.4}

where:

$N_{\text{thr}}$ hydraulic power that the pump spends on pressure that the regulators then take off, kW
$q_i$ injection rate into well $i$ , m³/d
$p_{\text{hdr}}$ pressure in the header of the water distribution block, MPa
$p_{\text{wh},i}$ wellhead pressure that well $i$ needs for its rate, MPa

Units: metric oilfield units: m³/d, MPa, kW

Holds when

$86.4$ is the same constant as in the formula for the hydraulic power of a pump: the seconds in a day divided by a thousand.
The losses in the water lines from the station to the block are the same for all wells and are not included here.
The electric power is $1/\eta$ times the hydraulic power, where $\eta$ is the efficiency of the pump unit.

The losses are greater the larger the rate of the “easy” wells and the more their pressure differs from that of the “tight” one. We reduce them by splitting the wells into pressure levels, by installing separate pumps for the “tight” wells, or by stimulating those wells (treating the near-wellbore zone) so that the common pressure is enough for them.

Wells of one water distribution block

Well: IW-1
Rate, m³/d: 600
$I$ , m³/(d·MPa): 60
$p_w$ , MPa: 32.00
$p_{\text{wh}}$ , MPa: 9.68

Well: IW-2
Rate, m³/d: 450
$I$ , m³/(d·MPa): 150
$p_w$ , MPa: 25.00
$p_{\text{wh}}$ , MPa: 1.76

Well: IW-3
Rate, m³/d: 250
$I$ , m³/(d·MPa): 22
$p_w$ , MPa: 33.36
$p_{\text{wh}}$ , MPa: 9.29

Well: IW-4
Rate, m³/d: 700
$I$ , m³/(d·MPa): 140
$p_w$ , MPa: 27.00
$p_{\text{wh}}$ , MPa: 5.44

Well	Rate, m³/d	$I$ , m³/(d·MPa)	$p_w$ , MPa	$p_{\text{wh}}$ , MPa
IW-1	600	60	32.00	9.68
IW-2	450	150	25.00	1.76
IW-3	250	22	33.36	9.29
IW-4	700	140	27.00	5.44

$I$ is the injectivity, $p_w$ the bottomhole pressure and $p_{\text{wh}}$ the wellhead pressure. The reservoir pressure is 22 MPa. The depth, tubing and water are as for the well of the next section: the water column gives 24.49 MPa, and the friction in the tubing is calculated from the rate. The data are assumed for illustration.

Worked exampleHow much power is lost in the chokes of a water distribution block

Given (assumed data)

Wells: four; the rates and the wellhead pressures needed are in the table above
Pump station: loss in the water line to the block 0.6 MPa, pressure at the pump inlet 0.3 MPa, efficiency of the pump unit 0.75

Solution

An example of the calculation of the wellhead pressure for IW-1: the bottomhole pressure is $22 + 600/60 = 32.00$ MPa. The wellhead pressure is $32.00 - 24.49 + 2.17 = 9.68$ MPa, where 2.17 MPa is the friction in the tubing at 600 m³/d (the next section).
The “tightest” well is IW-1: it needs 9.68 MPa. The pressure at the outlet of the station is $9.68 + 0.6 = 10.28$ MPa, and the pumps raise it by $10.28 - 0.3 = 9.98$ MPa. The hydraulic power is $2000\cdot 9.98/86.4 = 231.0$ kW.
Losses in the regulators: IW-2, $450\cdot(9.68 - 1.76)$ ; IW-3, $250\cdot(9.68 - 9.29)$ ; IW-4, $700\cdot(9.68 - 5.44)$ . The sum is 6629.5 MPa·m³/d, and $6629.5/86.4 = 76.7$ kW, which is 33% of the pump power.
If IW-2 and IW-4 are moved to a separate header with the pressure of IW-4, 5.44 MPa, the saving is $1150\cdot(9.68 - 5.44)/86.4 = 56.4$ kW of hydraulic power, or $56.4/0.75 = 75.2$ kW of electric power, that is 659 MWh a year.

Answer: About 76.7 kW is lost in the chokes, which is 33% of the hydraulic power of the pumps. Two pressure levels save about 75.2 kW of electric power, or 659 MWh a year.

What it means. A third of the power goes into first raising the pressure and then taking it off. Note IW-3 too: with its low injectivity it needs almost the same wellhead pressure as IW-1, and its bottomhole pressure is 33.36 MPa, close to the fracture propagation pressure. Stimulating the near-wellbore zone of IW-3 and giving IW-2 and IW-4 a separate “easy” header is often cheaper than keeping the pressure of the whole system for the worst well. The calculation is for one operating mode. The rates of the wells change, so we check the decision over the whole range. The data in this example are assumed for illustration.

Injectivity and injection pressure

How much water a well takes at a given pressure is shown by the injectivity index: the rate divided by the amount by which the bottomhole pressure exceeds the reservoir pressure. It is the mirror image of the productivity index of a producing well (productivity index). In steady state (when the pressure no longer changes with time) it is set by the transmissibility of the reservoir (permeability times thickness over viscosity), the skin (a number that shows how much the zone around the well helps or hinders flow) and the dimensions of the area. In time it changes. As the water zone around the well grows, it rises or falls depending on the ratio of the mobilities of water and oil (mobility ratio), and when the near-wellbore zone is plugged, it falls.

Injectivity index

I = \frac{q}{p_w - \bar p}

where:

$I$ injectivity index, m³/(d·MPa)
$q$ injection rate at standard conditions, m³/d
$p_w$ bottomhole pressure during injection (for producing wells the bottomhole pressure is written $p_{wf}$ ), MPa
$\bar p$ reservoir pressure in the injection area, MPa

Units: metric oilfield units

Holds when

The injection is close to steady state. Right after a change of the rate, the pressure has not settled yet.
The reservoir pressure in the injection area is known, from a falloff test (pressure falloff after shut-in) or from neighbouring wells.
The fracture has not opened. After it opens, the rate grows with almost no rise in pressure, and the injectivity stops being a constant.

For a homogeneous reservoir in steady state, the injectivity is given by the same formula as the productivity, but with the viscosity and the formation volume factor of water (how much its volume changes between the reservoir and the surface; the form is below). By comparing $I$ over time, we can see plugging and the results of well treatments.

In other unit systems (1)

Homogeneous reservoir, steady state (metric oilfield units)

I = \frac{0.536\,k\,h}{\mu_w B_w\left(\ln\dfrac{r_e}{r_w} + S\right)}

$k$ in mD, $h$ in m, $\mu_w$ in mPa·s, $B_w$ in m³/m³, $r_e$ and $r_w$ in m, $I$ in m³/(d·MPa). Here $\bar p$ is the pressure at the boundary $r_e$ , as in the Dupuit equation and in the derivation of the Hall plot. If $\bar p$ is the average pressure of the area, the brackets contain $\ln(r_e/r_w) - 1/2 + S$ (average pressure).

$0.536 = 2\pi\cdot 0.0853$ , as in the Dupuit equation (Steady-state radial flow).

The pressure is measured at the wellhead, but the reservoir “sees” the bottomhole pressure. In injection the two are linked by the weight of the water column, which helps the pump, and by the friction in the tubing, which hinders it (friction factor):

Bottomhole pressure of an injection well

p_w = p_{\text{wh}} + \rho g H - \Delta p_{\text{fr}},\qquad \Delta p_{\text{fr}} = \lambda\,\frac{H}{d}\,\frac{\rho v^2}{2}

where:

$p_{\text{wh}}$ wellhead pressure, Pa
$\rho$ density of the injected water, kg/m³
$H$ vertical depth of the middle of the injection interval (for friction, the length of the tubing along the hole; in a vertical well they coincide), m
$\lambda$ Darcy friction factor (from the Haaland formula), —
$d,\ v$ inner diameter of the tubing and the mean velocity of the water in it, $v = 4q/(\pi d^2)$ , m, m/s

Units: SI; to get MPa, divide by $10^6$

Holds when

A single-phase flow of water. The viscosity and density are at the mean temperature in the tubing.
Friction grows roughly as the square of the rate, so at a low rate it can be neglected, and at a high rate it cannot.
Injection down the annulus (the space between the tubing and the casing) or into a deviated well needs its own lengths and diameters.

At a constant wellhead pressure the bottomhole pressure rises when the rate falls: the friction disappears and the water column stays. That is why a pressure limit found for the bottom of the hole is converted to the wellhead for a specific rate.

The upper limit of the bottomhole pressure is set by the rock: above the fracture opening (propagation) pressure, the water goes into the reservoir through a fracture. We find this pressure with a step-rate test (step-rate test): we inject water at a series of rising rates, and a bend in the line of pressure against rate shows that a fracture has opened. We keep the bottomhole pressure below it with a margin. The limit is not constant. Cold water cools the rock near the well and lowers the fracture pressure (stresses at the wall of the well), while the rise of the reservoir pressure during injection raises it. Sometimes in tight reservoirs the injection is deliberately above the fracture opening pressure. Then the spacing of the wells is matched with the direction of the fractures, so that the water does not go along them straight to the producing wells (well patterns).

Worked exampleWhat rate and what wellhead pressure are allowed

Given (assumed data)

Well: vertical, the middle of the interval at 2400 m; tubing with an inner diameter of 62 mm, roughness 0.05 mm
Water: density 1040 kg/m³, viscosity in the tubing 0.9 mPa·s
Reservoir: reservoir pressure 22 MPa, injectivity 60 m³/(d·MPa), injection plan 600 m³/d
Fracture propagation pressure: 36 MPa at the bottom of the hole from a step-rate test; the margin adopted is 1 MPa

Solution

The water column: $1040\cdot 9.81\cdot 2400 = 24{,}485{,}760$ Pa, that is 24.49 MPa. The bottomhole pressure at 600 m³/d is $22 + 600/60 = 32.00$ MPa.
Friction: $v = 600/86{,}400/(\pi\cdot 0.062^2/4) = 2.30$ m/s, the Reynolds number (it shows how turbulent the flow is) is about 165 thousand, $\lambda = 0.0204$ , and $\Delta p_{\text{fr}} = 0.0204\cdot\dfrac{2400}{0.062}\cdot\dfrac{1040\cdot 2.30^2}{2}$ $= 2.17$ MPa.
The wellhead pressure is $32.00 - 24.49 + 2.17 = 9.68$ MPa.
The limit: the bottomhole pressure is not higher than $36 - 1 = 35$ MPa, so $q_{\max} = 60\cdot(35 - 22) = 780$ m³/d. At this rate $v = 2.99$ m/s, $\lambda = 0.0200$ , the friction is 3.60 MPa, and the wellhead pressure is $35 - 24.49 + 3.60 = 14.11$ MPa.
If the limit is converted to the wellhead with hydrostatic pressure alone, $35 - 24.49 = 10.51$ MPa, the well will be allowed only about 635 m³/d, which is 19% less than it can really take.
The opposite mistake: 14.11 MPa is held at the wellhead, but the injectivity has halved because of plugging. The rate will fall to about 459 m³/d and the friction to 1.30 MPa, and without a fracture the bottomhole pressure would be $14.11 + 24.49 - 1.30 = 37.30$ MPa, which is above the fracture propagation pressure: the fracture will open.

Answer: The plan of 600 m³/d needs 9.68 MPa at the wellhead. Without opening a fracture, the well will take up to 780 m³/d at 14.11 MPa at the wellhead. A wellhead pressure limit is valid only for its own rate.

What it means. The friction in the tubing at 600–780 m³/d is two to four megapascals, comparable to the margin to the fracture pressure. So converting between the wellhead and the bottom of the hole without friction is wrong in both directions: with hydrostatic pressure alone the limit is too low at a high rate, and a wellhead limit found at a high rate is dangerous at a low rate. It is more reliable to limit the bottomhole pressure, calculated from the wellhead pressure and the rate, or to set the wellhead pressure limit as a curve against the rate (the figure below). The cooling of the rock lowers the fracture opening pressure with time, so we review the limit with new step-rate tests. The data in this example are assumed for illustration.

CalculatedWellhead pressure of the well of the example against the rate: the pressure needed for injection (solid line) and the limit at which the bottomhole pressure equals 35 MPa (dashed line). The horizontal line is the same limit converted with hydrostatic pressure alone. The lines meet at the largest allowed rate. Up to 141 m³/d the water column is heavier than needed, and the well takes water with no pressure at the wellhead.

Monitoring injection: the Hall plot and other signs

Rate and pressure of each well: daily, from the flowmeters of the water distribution blocks and the pressure gauges at the wellheads. From them come the injectivity and the Hall plot.
Water quality: suspended solids, oil, oxygen, iron and bacteria, after each treatment stage and at the wellheads of distant wells.
Corrosion: coupons and probes in the water lines, wall thickness, iron in the water.
Well tests: a pressure falloff after shut-in (falloff test) gives the reservoir pressure, the permeability and the skin. Injection profiles along the hole (production logging) show which layers take water. Step-rate tests give the fracture propagation pressure.
The reservoir: voidage replacement by area (Why water is injected: pressure maintenance and oil displacement), pressure maps, and tracer tests that show which producing wells an injection well is connected with.

The cheapest and most sensitive tool is the Hall plot. Its horizontal axis is the cumulative injection $W$ (the total injected so far), and its vertical axis is the cumulative sum of the excesses of the bottomhole pressure over the reservoir pressure multiplied by time, $\Sigma(p_w - \bar p)\Delta t$ . The link of the slope with permeability and skin is derived in the lesson on formation damage. For monitoring it is enough that in steady injection, in each period, $(p_w - \bar p)\Delta t = q\Delta t/I$ , that is, the sum grows in proportion to the injection with a slope $1/I$ .

Slope of the Hall plot and injectivity

m_H = \frac{\Delta\!\left[\sum (p_w - \bar p)\,\Delta t\right]}{\Delta W} = \frac{1}{I}

where:

$m_H$ slope of the Hall plot on a segment, MPa·d/m³
$\Delta W$ injection over the segment, m³
$\Delta t$ length of the averaging period (a day, a month), d
$I$ average injectivity index over the segment, m³/(d·MPa)

Units: metric oilfield units

Holds when

The injection in each period is close to steady state, and the reservoir pressure in the injection area hardly changes. Its rise also increases the slope.
We calculate the bottomhole pressure from the wellhead pressure, counting the hydrostatic pressure and the friction at the actual rate.
The slope over a segment is an average. If the injectivity changed within the segment, the slope is the average value of $1/I$ , weighted by the injection.

A straight line means that the state of the well does not change. A growing slope means that the injectivity falls (plugging, a rise of the reservoir pressure). A falling slope means that the injectivity grows, after a treatment or when a fracture opens. In the second case the pressure hardly rises when the rate grows and stands at the fracture opening pressure.

Injection history of an injection well (monthly averages)

Month: 1
Rate, m³/d: 600
Bottomhole pressure, MPa: 32.0
W, m³: 18,000
Σ, MPa·d: 300

Month: 2
Rate, m³/d: 600
Bottomhole pressure, MPa: 32.0
W, m³: 36,000
Σ, MPa·d: 600

Month: 3
Rate, m³/d: 600
Bottomhole pressure, MPa: 32.0
W, m³: 54,000
Σ, MPa·d: 900

Month: 4
Rate, m³/d: 600
Bottomhole pressure, MPa: 32.0
W, m³: 72,000
Σ, MPa·d: 1200

Month: 5
Rate, m³/d: 570
Bottomhole pressure, MPa: 32.0
W, m³: 89,100
Σ, MPa·d: 1500

Month: 6
Rate, m³/d: 535
Bottomhole pressure, MPa: 32.0
W, m³: 105,150
Σ, MPa·d: 1800

Month: 7
Rate, m³/d: 500
Bottomhole pressure, MPa: 32.0
W, m³: 120,150
Σ, MPa·d: 2100

Month: 8
Rate, m³/d: 470
Bottomhole pressure, MPa: 32.0
W, m³: 134,250
Σ, MPa·d: 2400

Month: 9
Rate, m³/d: 640
Bottomhole pressure, MPa: 31.0
W, m³: 153,450
Σ, MPa·d: 2670

Month: 10
Rate, m³/d: 640
Bottomhole pressure, MPa: 31.0
W, m³: 172,650
Σ, MPa·d: 2940

Month: 11
Rate, m³/d: 640
Bottomhole pressure, MPa: 31.0
W, m³: 191,850
Σ, MPa·d: 3210

Month: 12
Rate, m³/d: 640
Bottomhole pressure, MPa: 31.0
W, m³: 211,050
Σ, MPa·d: 3480

Month: 13
Rate, m³/d: 1050
Bottomhole pressure, MPa: 36.1
W, m³: 242,550
Σ, MPa·d: 3903

Month: 14
Rate, m³/d: 1150
Bottomhole pressure, MPa: 36.2
W, m³: 277,050
Σ, MPa·d: 4329

Month: 15
Rate, m³/d: 1250
Bottomhole pressure, MPa: 36.2
W, m³: 314,550
Σ, MPa·d: 4755

Month	Rate, m³/d	Bottomhole pressure, MPa	W, m³	Σ, MPa·d
1	600	32.0	18,000	300
2	600	32.0	36,000	600
3	600	32.0	54,000	900
4	600	32.0	72,000	1200
5	570	32.0	89,100	1500
6	535	32.0	105,150	1800
7	500	32.0	120,150	2100
8	470	32.0	134,250	2400
9	640	31.0	153,450	2670
10	640	31.0	172,650	2940
11	640	31.0	191,850	3210
12	640	31.0	211,050	3480
13	1050	36.1	242,550	3903
14	1150	36.2	277,050	4329
15	1250	36.2	314,550	4755

A month is taken as 30 days, and the reservoir pressure is 22 MPa. Σ is the cumulative sum of $(p_w - \bar p)\Delta t$ . The bottomhole pressure is calculated from the wellhead, as in the previous example. The data are assumed for illustration.

Worked exampleWhat happened to the well: reading the Hall plot

Given (assumed data)

History: the table above: months 1–4, normal injection; 5–8, the rate falls at the same pressure; at the start of month 9, an acid treatment; from month 13 the rate was raised to 1050–1250 m³/d
Fracture propagation pressure: 36.0 MPa at the bottom of the hole from a step-rate test

Solution

Months 1–4: $\Delta\Sigma = 1200$ MPa·d, $\Delta W = 72{,}000$ m³. The slope is $1200/72{,}000 = 0.01667$ , and the injectivity is $72{,}000/1200 = 60.0$ m³/(d·MPa).
Months 5–8: $\Delta\Sigma = 1200$ , $\Delta W = 62{,}250$ . The slope is 0.01928, 16% larger, and the injectivity is 51.9: the near-wellbore zone is getting plugged.
Months 9–12: $\Delta\Sigma = 1080$ , $\Delta W = 76{,}800$ . The slope is 0.01406 and the injectivity is 71.1, higher than the initial one: the treatment removed not only the new plugging but also part of the old.
Months 13–15: $\Delta\Sigma = 1275$ , $\Delta W = 103{,}500$ . The slope is 0.01232 and the apparent injectivity is 81.2. The rate rose from 1050 to 1250 m³/d, while the bottomhole pressure stays at 36.1–36.2 MPa, at the fracture propagation pressure. Without a fracture, a rate of 1050 m³/d at an injectivity of 71.1 would need $22 + 1050/71.1 = 36.8$ MPa.

Answer: Plugging in months 5–8 (the injectivity from 60.0 to 51.9), a successful treatment (71.1), and from month 13 injection through a fracture: an apparent injectivity of 81.2 at a pressure that has run up against the fracture opening pressure.

What it means. The fall of the slope in the last segment looks like a success, but it is a fracture. At an injectivity of 71.1, the well without a fracture would take not more than $71.1\cdot(35 - 22) = 924$ m³/d with a margin of 1 MPa. The geology decides what to do next. If the fracture will go towards a producing well, we lower the rate. If the project provides for injection above the fracture opening pressure, we monitor the water cut of the nearest producing wells. We look for the cause of the plugging in months 5–8 in the water (suspended solids, oil, corrosion products) and confirm it with a falloff test. The data in this example are assumed for illustration.

CalculatedThe Hall plot of the injection history in the table: the points are the ends of the months, and the dashed line continues the slope of months 1–4. The slope grows with plugging (months 5–8, the points go above the dashed line), falls after the treatment (9–12) and falls still more with injection through a fracture (13–15).

Terms and symbols

Pressure maintenance system: Injection of water to keep up the reservoir pressure, and the whole system of facilities that does it.
Water injection pump station: A pump station (often packaged, that is, assembled from ready-made blocks) that delivers water under pressure to a group of injection wells.
Sectional multistage centrifugal pump: A pump with many impellers on one shaft. In Russian practice its type name, such as CNS 180-1900, gives the delivery in m³/h (180) and the head in metres (1900).
Water distribution block: A block with a flowmeter and a flow regulator for each well. It distributes the water among the wells.
Water compatibility: The ability of a water to mix with another water and to contact the rock without precipitates, swelling or movement of clays.
Deaeration: Removal of oxygen from water by lowering its partial pressure, by vacuum or by gas stripping.
Oxygen scavenger: A chemical that binds the oxygen left after deaeration, for example sodium bisulfite.
Injectivity index $I$: The injection rate per unit of excess of the bottomhole pressure over the reservoir pressure, m³/(d·MPa).
Fracture propagation pressure: The bottomhole pressure above which the reservoir takes water through a growing fracture. It is found with a step-rate test.
Hall plot: The cumulative sum of $(p_w - \bar p)\Delta t$ against the cumulative injection. The slope equals $1/I$ .

Common mistakes and misconceptions

Wrong

“Both waters are fit for injection separately, so they can be mixed.”

In fact

Water with barium and water with sulfate are each clean on their own, but together they give barite in the pipes and wells, tens of tonnes a year in the example of the lesson. We check compatibility for all mixing ratios.

Wrong

“The wellhead pressure limit is the fracture pressure minus the weight of the water column.”

In fact

This is true only without friction. At a high rate the friction in the tubing is megapascals, and such a limit underestimates the allowed rate. A limit found at a high rate is dangerous at a low rate. What we limit is the bottomhole pressure.

Wrong

“An oxygen scavenger is more reliable than a deaerator, so let us just add more of it.”

In fact

Full chemical treatment of sea water would take hundreds of tonnes of chemical a year, and the chemical also adds sulfate. The scavenger cleans up the remainder after mechanical deaeration.

Wrong

“The injection pressure has grown, so the well is getting plugged.”

In fact

The pressure may have been raised for the sake of the rate, or the reservoir pressure may have risen. Plugging is shown by a fall of the injectivity, that is, a growth of the slope of the Hall plot.

Wrong

“The slope of the Hall plot has fallen, so the treatment worked.”

In fact

The slope also falls when a fracture opens. If the bottomhole pressure stays at the fracture opening pressure when the rate grows, the water is going through a fracture.

Connections to other topics

Produced water: composition, treatment and disposal — treatment of produced water and the water balance of a field.
Why water is injected: pressure maintenance and oil displacement — why and how much water is injected: voidage replacement, well patterns.
Sources of formation damage — plugging in injection, the saturation index, the mass of precipitate and the derivation of the Hall plot.
Rock mechanics for hydraulic fracturing — step-rate tests and the fracture opening pressure, the effect of cold water.
Fracture gradient and leak-off testing — the fracturing pressure of rocks and how it is estimated.
Steady-state radial flow — the productivity index, the same formula as for injectivity.
Multiphase flow in tubing — friction in pipes and the friction factor.
Field development concepts: surface facilities — the place of the pressure maintenance system and the water injection pump stations in the field facilities, the injection volumes over the life of the field, and the field's power plant running on associated gas.

Check yourself

1
Sea water is going to be injected into a reservoir with barium in the formation water. Which of the following reduces the risk of sulfate scale first of all?
1. A.Removal of sulfates from the sea water by membranes
2. B.Heating the water
3. C.Chlorination
4. D.Deaeration of the water
Show the answer
Correct answer: A. Removal of sulfates from the sea water by membranes
Barite forms from the barium of the formation water and the sulfate of the injected water. Deaeration and chlorination fight corrosion and bacteria, and heating has little effect on the solubility of barite. Removing the sulfate means removing one of the two ions. Other ways are another source of water, or inhibitors at moderate supersaturation.
2
An injection well 2000 m deep takes 500 m³/d at a wellhead pressure of 6.0 MPa. The water density is 1050 kg/m³, the friction in the tubing is 1.5 MPa and the reservoir pressure is 20 MPa. Find the bottomhole pressure and the injectivity index.
Show the answer
The water column is $1050\cdot 9.81\cdot 2000 = 20.6$ MPa. The bottomhole pressure is $6.0 + 20.6 - 1.5 = 25.1$ MPa, the excess over the reservoir pressure is 5.1 MPa, and the injectivity is $500/5.1 \approx 98$ m³/(d·MPa). If friction is forgotten, the injectivity would come out as $500/6.6 \approx 76$ , almost a quarter less.
3
The slope of the Hall plot over the last three months has grown by 30%, and the reservoir pressure in the injection area, according to a test, has not changed. What does this mean and what should be done?
1. A.The injectivity has fallen by about a quarter: the near-wellbore zone is getting plugged, so check the water and treat the well
2. B.A fracture has opened: lower the pressure
3. C.Nothing: the slope of the Hall plot is not linked with injectivity
4. D.The injectivity has grown by 30%: the injection can be increased
Show the answer
Correct answer: A. The injectivity has fallen by about a quarter: the near-wellbore zone is getting plugged, so check the water and treat the well
The slope equals $1/I$ . A growth of 30% means that the injectivity has fallen to $1/1.3 \approx 0.77$ of the former value, almost a quarter. With an unchanged reservoir pressure the cause is plugging of the near-wellbore zone. The water quality is checked (suspended solids, oil, iron, bacteria), the diagnosis is confirmed with a falloff test, and a treatment is planned.
4
Why is it not enough to remove oxygen from water by simply pumping the air over it down to 10 kPa (abs.), if the water must end up with 20 ppb of oxygen and water saturated with air holds 9 mg/L?
Show the answer
By Henry's law the partial pressure of oxygen that is needed is $21.2\cdot 0.020/9 \approx 0.047$ kPa. If there is air over the water at 10 kPa, the oxygen in it is $0.21\cdot 10 = 2.1$ kPa, 45 times more. Over the water there must be almost pure water vapour or an oxygen-free gas: either a deeper vacuum (the water then boils, and the vapour pushes out the air) or stripping with nitrogen, and a scavenger for the remainder.
5
A pump station serves four wells. One of them, with low injectivity, needs 5 MPa more than the others. Which solutions will reduce the energy losses?
Show the answer
The pumps keep the pressure needed by the “tightest” well, the others are throttled, and the power $q\,\Delta p/86.4$ kW goes into the chokes. What helps: stimulating the “tight” well (acid, hydraulic fracturing) so that the common pressure is enough for it; a separate pump at its wellhead; splitting the wells into pressure levels; and variable speed control of the pumps instead of throttling.
6
What does the Water Code of Kazakhstan say about the injection into the subsoil of formation water produced with oil?
1. A.It needs only a notification
2. B.It is allowed only at sea
3. C.It is prohibited
4. D.The condition of having environmental permits and a permit for special water use, set for the discharge and injection of such waters, does not apply to injection for pressure maintenance
Show the answer
Correct answer: D. The condition of having environmental permits and a permit for special water use, set for the discharge and injection of such waters, does not apply to injection for pressure maintenance
Under paragraph 4 of Article 123 of the Water Code, the discharge of underground water taken incidentally in subsoil use, and its injection into the subsoil, are allowed only with environmental permits and permits for special water use, except for the injection for pressure maintenance of formation water produced with hydrocarbons. Injection into disposal horizons (layers used only to get rid of water) does not fall under the exception. The other requirements for the field remain in force: the project and the environmental permit for the facility.

Updated: 19 October 2026