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ProductionFacilities and processing of production Intermediate 28 min read

Produced water: composition, treatment and disposal

Produced water is the water that wells bring up together with oil and gas. It is the formation water of the accumulation and, after breakthrough (when the injected water reaches the wells), also the water that was injected into the reservoir. At the surface we separate it from the oil, clean out the oil droplets, suspended solids (solid particles that float in the water) and dissolved gases, and return it to the reservoir or dispose of it. In a field with a high water cut (a large share of water in the produced liquid) it is the biggest stream of the field: several times more water is produced than oil.

Why it matters

Water treatment decides how much oil leaves with the water as waste, and whether the reservoir gets plugged when we inject the water back. It decides how fast corrosion destroys pipes and scale (a hard mineral deposit) coats equipment. And it decides how much the oil of the late years costs, when the field mainly lifts, separates and injects water. This lesson explains what produced water contains and how oil is separated from it, from Stokes' law to hydrocyclones, flotation and filters. It also explains how oil in water is measured, how the system is protected from scale and corrosion, where the water goes under the law of Kazakhstan, and how to draw up the water balance of a field.

Quick refresher

  • Produced water is formation water and injected water, with dissolved salts and gases, oil droplets, suspended solids, bacteria and the remains of chemicals. We do not remove the salts, but we take them into account. We remove the oil droplets, suspended solids and dissolved gases.
  • An oil droplet rises at the Stokes velocity (the rising speed that Stokes' law gives), which is proportional to the square of the diameter: a droplet half the size rises four times more slowly. So the breaking up of droplets in chokes and pumps, and the cooling of the water, which raises its viscosity, make the treatment worse.
  • An ideal settling tank catches every droplet whose rising velocity is greater than the surface loading q/Aq/A (the water rate per unit of area), whatever the depth of the tank. A plate pack (a stack of inclined plates) adds area through the projections of its plates, and in a volume of a few cubic metres it replaces a tank of hundreds of cubic metres.
  • The treatment stages go from large droplets to small ones: settling tanks take droplets from hundreds down to a few tens of micrometres, hydrocyclones and flotation down to about ten, and filters down to a few micrometres. The oil content of water can be compared only when one measurement method is used.
  • We choose materials by the partial pressure of hydrogen sulfide, the share of the gas multiplied by the absolute pressure: from 0.3 kPa the steel must resist sulfide stress cracking (cracking of steel in the presence of hydrogen sulfide). For people, what is dangerous is the concentration of the gas, not its partial pressure.
  • With full voidage replacement (injecting as much water as the volume taken out of the reservoir), the water that we need from the source equals the space that the produced oil occupied in the reservoir, qoBo/Bwq_oB_o/B_w. By the end of the field's life the injection consists almost entirely of produced water. If edge water (water from the aquifer around the field) supports the pressure in part, then in time there is more produced water than needs to be injected, and the excess goes into disposal wells (wells whose only job is to take water).
Key relationships
vt=g d2(ρw−ρo)18 μwv_t = \dfrac{g\,d^2(\rho_w - \rho_o)}{18\,\mu_w}

rising velocity of an oil droplet in water (Stokes' law)

dmin⁡=18 μw qg (ρw−ρo) Ad_{\min} = \sqrt{\dfrac{18\,\mu_w\,q}{g\,(\rho_w - \rho_o)\,A}}

the smallest droplet that an ideal settling tank of area AA catches completely at a water rate qq

pi=yi pp_i = y_i\,p

partial pressure: the mole fraction of the component in the gas times the absolute pressure

qsrc=VRR Boqo+BwqwBw−qwq_{\text{src}} = \mathrm{VRR}\,\dfrac{B_oq_o + B_wq_w}{B_w} - q_w

water from the source; a negative value is an excess of produced water that must be disposed of

Common pitfalls
  • Assuming that a deep settling tank cleans better than a shallow one: the droplet that is caught is set by the area, while the depth gives room for sediment and a reserve of volume.
  • Putting a centrifugal pump or a throttling valve before the settling tank: it breaks the droplets up and cancels the work of the tank.
  • Comparing the oil content of water measured by different methods.
  • Letting air into the treatment system: oxygen in water with hydrogen sulfide and iron causes both corrosion and a solid sediment.
  • Mixing the produced waters of different horizons (producing layers), or produced water with water from another source, without a compatibility check.

After this lesson you can

  • list what produced water contains, and for each component say how it causes trouble and what is done about it
  • calculate the rising velocity of an oil droplet by Stokes' law and the smallest droplet that a settling tank or a plate pack catches completely, and explain why the area, not the depth, matters for a settling tank
  • use the droplet size distribution to estimate how much oil stays in the water after a settling tank, and explain how the water temperature and the breaking up of droplets affect this
  • describe the treatment stages (settling tanks, hydrocyclones, flotation, filters) and the methods for measuring the oil content of water
  • use the partial pressure of hydrogen sulfide to judge where materials resistant to it are needed, and tell this criterion apart from the danger of the gas to people
  • draw up the water balance of a field: how much produced water can be returned to the reservoir, how much must be taken from a source, and when disposal wells will be needed

Where produced water comes from and how much there is

There is water in every reservoir. In the oil-bearing part it fills some of the pores and does not move; this is irreducible water. Below the oil–water contact (the level under which there is only water) and outside the oil-bearing contour (the edge of the oil zone), the water fills the pores completely. While a well gives water-free oil, no water appears in the production. It appears when bottom water (water from below the oil) or edge water (water from around the oil zone) is drawn up to the well, or when the front of the injected water arrives (Why water is injected: pressure maintenance and oil displacement, Water and gas coning). Gas wells also bring up condensation water: vapour that was in the reservoir gas and condensed on cooling. It is almost fresh.

The share of water in the produced liquid is called the water cut WC\text{WC}, and the ratio of water to oil is the water–oil ratio: WOR=WC/(1−WC)\text{WOR} = \text{WC}/(1 - \text{WC}). At a water cut of 50% there is one cubic metre of water for every cubic metre of oil, at 90% there are nine, and at 95% nineteen. So in a mature field the main work is to lift, separate and inject water.

In a field under waterflooding, with time there is several times more water than oil (loads on the facilities over the life of the field). In the example of the lesson on field facilities, in year 20 of the project the field produces 12,148 m³/d of water and 652 m³/d of oil: the water cut is 94.9%, and there are 18.6 m³ of water for every cubic metre of oil. The water is separated in stages. Most of it is removed first, at the preliminary water removal unit at the booster pump station (the station that pumps the liquid onward), and the rest during oil treatment at the central processing facility. From there the water (in Russian-language practice often called “water from under the oil”) goes to treatment, and then to the water injection pump stations of the pressure maintenance system.

Other waste streams of the field join the produced water: the wash water of the oil desalting unit, the water drained from the bottom of sales tanks, rain and snowmelt runoff from the sites, and at offshore facilities the water used to wash equipment and the water displaced from oil storage. We treat them together with the produced water, but first we check that they are compatible with it (that mixing does not form deposits).

What produced water contains

Formation water is a solution of salts, from fractions of a gram to more than three hundred grams per litre. The lesson on formation water covers how salinity (the amount of dissolved salt) is expressed, and how grams per litre differ from ppm. Most of the salt is sodium and chloride ions (charged particles of dissolved salt). For the field, more important are the ions that form poorly soluble compounds with each other: calcium, barium, strontium, iron, bicarbonate and sulfate. Besides salts, the water carries everything it met on the way from the reservoir to the separator.

What produced water contains and what is done about it
Component
Dissolved salts: sodium, chloride, calcium, magnesium
What trouble it causes
they speed up corrosion; halite (rock salt) drops out when the water evaporates
What is done
not removed: we allow for them when choosing materials and for injection
Component
Ions that form poorly soluble salts: calcium, barium, strontium, iron, bicarbonate, sulfate
What trouble it causes
scale (hard deposits) forms when the pressure falls, on heating and when waters are mixed
What is done
scale inhibitors; incompatible waters are not mixed
Component
Oil droplets (dispersed oil)
What trouble it causes
they plug the reservoir during injection and pollute the environment when discharged
What is done
settling tanks, hydrocyclones, flotation, filters
Component
Dissolved hydrocarbons and organic matter: aromatic hydrocarbons (such as benzene), phenols, organic acids
What trouble it causes
toxic when discharged; cannot be separated by settling
What is done
usually do not interfere with injection; for discharge, biological treatment (bacteria that eat the organic matter) and sorption (uptake by a solid material)
Component
Suspended solids: sand, clay, corrosion products, iron sulfide, salt crystals
What trouble it causes
they plug the reservoir and filters and stabilise emulsions
What is done
settling, filtration
Component
Dissolved gases: carbon dioxide and hydrogen sulfide; oxygen only when air is sucked in
What trouble it causes
corrosion, danger to people
What is done
degassing; a closed system under a gas blanket (a cushion of inert gas above the water)
Component
Bacteria, above all sulfate-reducing bacteria (they turn sulfate into hydrogen sulfide)
What trouble it causes
hydrogen sulfide, corrosion under deposits, slime
What is done
biocides (chemicals that kill bacteria), cleaning of equipment
Component
Remains of chemicals: demulsifiers, inhibitors, methanol
What trouble it causes
they hinder the separation of oil and are toxic when discharged
What is done
choice of chemicals and control of the doses
Component
Natural radionuclides (radium)
What trouble it causes
they pass into sulfate scale and sludge
What is done
monitoring and handling under the radiation safety rules

The composition of a particular water is found by analysing samples. The set of indicators that are monitored depends on where the water will go: into the reservoir, into a disposal well or to discharge.

There are two kinds of oil in water. Dispersed oil is droplets from fractions of a micrometre to several hundred micrometres in size. They can be separated, because oil is lighter than water. Dissolved hydrocarbons and organic matter are individual molecules, above all aromatic hydrocarbons and organic acids. Settling cannot separate them. For injection into the reservoir, dispersed oil matters most. For discharge into a water body, both kinds matter.

The droplet size decides almost everything. Large droplets form when the phases separate in a separator. Small ones form when the flow passes a choke, a control valve or a centrifugal pump at high velocity: in the shear (the rubbing of fast layers of liquid against each other), droplets break up, and one droplet of 100 µm turns into hundreds of droplets of 10–20 µm. Small droplets are also stabilised by fine solids, asphaltenes (heavy components of oil) and surfactants (surface-active substances), which coat them and keep them from merging. This gives the first rule of water treatment: separate the oil before the flow passes through devices that break up droplets, and choose pumps and valves with low shear.

How an oil droplet separates from the water

Two forces act on an oil droplet in still water: the buoyancy force (the difference between the weights of water and oil in the volume of the droplet) and the resistance of the water, which grows with velocity. Very soon they balance, and the droplet rises at a constant velocity. For small droplets, around which the water flows slowly and smoothly, this velocity is given by Stokes' law, the same law that describes the settling of drilled cuttings in tanks and centrifuges (drilling fluid cleaning).

Rising velocity of an oil droplet in water (Stokes' law)
vt=g d2 (ρw−ρo)18 μwv_t = \frac{g\,d^2\,(\rho_w - \rho_o)}{18\,\mu_w}
where:
  • vtv_tsteady rising velocity of the droplet relative to the water, m/s
  • ggacceleration due to gravity, 9.81 m/s²; in a hydrocyclone and a centrifuge the centrifugal acceleration takes its place, m/s²
  • dddiameter of the droplet, m
  • ρw, ρo\rho_w,\ \rho_odensities of water and oil at the water temperature, kg/m³
  • μw\mu_wdynamic viscosity of the water, Pa·s

Units: SI

Holds when
  • The droplet is a sphere and moves like a solid ball. For oil, which is more viscous than water, the circulation inside the droplet speeds up the rise by only a few per cent.
  • The flow around the droplet is laminar (smooth): the Reynolds number of the droplet, ρwvtd/μw\rho_w v_t d/\mu_w, is less than about 1. For oil droplets in water this holds up to sizes of the order of 150 µm.
  • The droplets do not hinder each other and do not merge, and the water is at rest. In a vessel, flow and mixing hinder the rise, while the merging of droplets helps it.

The velocity grows as the square of the diameter: a droplet of 100 µm rises 100 times faster than one of 10 µm. The velocity is proportional to the density difference and inversely proportional to the viscosity. The viscosity of water depends strongly on temperature, so heating the water, or at least protecting it from cooling, speeds up the separation of oil.

Now suppose that the water does not stand still but flows through a settling tank of area AA (seen from above) at a rate qq, and that we skim the collected layer of oil off the top. The droplet must rise to this layer before the water carries it to the outlet. The deeper the tank, the longer the droplet must rise, but also the longer the water stays in it: the retention time AH/qAH/q grows with the depth HH in exactly the same way as the path of the droplet. The depth cancels, and a droplet that started at the bottom has time to rise if its rising velocity is not less than the surface loading (or overflow rate) vc=q/Av_c = q/A, the flow rate divided by the area. The loading is often expressed in metres per hour: this is cubic metres of water per hour per square metre of area.

The smallest droplet that an ideal settling tank catches completely
dmin⁡=18 μw qg (ρw−ρo) Ad_{\min} = \sqrt{\frac{18\,\mu_w\,q}{g\,(\rho_w - \rho_o)\,A}}
where:
  • dmin⁡d_{\min}diameter of the droplet whose rising velocity equals the surface loading q/Aq/A; droplets of this size and larger are all caught, m
  • qqwater rate through the tank, m³/s
  • AAplan area of the tank (for a plate pack, the total horizontal projection of the plates), m²
  • μw, ρw, ρo, g\mu_w,\ \rho_w,\ \rho_o,\ gas in Stokes' law, Pa·s, kg/m³, kg/m³, m/s²

Units: SI

Holds when
  • An ideal settling tank: the water moves evenly over the whole area, the droplets enter evenly over the height, and they neither break up nor merge.
  • With a horizontal flow, droplets smaller than dmin⁡d_{\min} are caught in part: those that entered close enough to the surface. Their share is (d/dmin⁡)2(d/d_{\min})^2. In a vertical tank, where the water moves down against the droplets, all the small droplets leave with the water (the derivation is in the section for advanced study).
  • In a real vessel part of the flow moves faster than the average (“short-circuiting”), and thermal currents and eddies interfere, so it catches less than the ideal one. The calculation gives the best case: the real smallest droplet that is caught is larger than the calculated one.

The depth does not enter the formula: the droplet that is caught is set by the area. The area can be increased without enlarging the vessel by dividing the flow into thin layers with inclined plates. Each plate works as the bottom of a separate shallow tank, and a droplet needs to rise only through the height of the gap between the plates.

Paths of oil droplets in an ideal settling tank and in a plate packoil layerwaterwater outletH1231droplet of size dmin: from the bottom it reaches the oil2smaller droplet, entered low: leaves with the water3the same droplet, entered high: it is caughtArea A = length × width of the tank in plan viewplate packa dropletrises onlyup to theplateabove it

A diagram. On the left is a rectangular settling tank. Water enters on the left and leaves at the bottom right, and there is a layer of oil at the top. Three straight lines show the paths of droplets. The steepest starts at the bottom and reaches the oil layer at the outlet. A gentler line from the bottom goes to the outlet without reaching the oil. An equally gentle line that started higher reaches the oil layer in time. On the right is a pack of inclined parallel plates. Short arrows show that droplets rise to the nearest plate above them and go up along its lower side.

SchematicDroplet paths in an ideal settling tank (schematic): a droplet of size dmin⁡d_{\min} that entered at the bottom has time to rise to the oil layer right at the outlet. A smaller droplet is caught only if it entered high enough. On the right is a plate pack: a droplet needs to rise only as far as the plate above it.

This is how plate separators (thin-layer settlers, also called corrugated plate interceptors) are built: a pack of parallel plates inclined at 45–60°, with a gap of a few centimetres. The horizontal projection (the area covered when seen from above) of a pack of nn plates of length LL and width bb at an inclination θ\theta is nLbcos⁡θnLb\cos\theta. The plates are made corrugated: the oil collects on the ridges, the droplets merge there into large ones and rise along the plates, and the suspended solids slide down.

Worked exampleWhich droplets a settling tank and plate packs will catch
Given (assumed data)
Produced water
4800 m³/d (200 m³/h) at 40 °C; density 1040 kg/m³, viscosity 0.65 mPa·s
Oil
density 860 kg/m³ at 40 °C
Settling tank
round, inner diameter 12 m, water depth 8 m; the water is fed to the centre over the whole depth and spreads towards the walls
Plate packs (an option)
four identical packs in parallel; each has 48 plates 1.75 m long and 1.0 m wide, inclined at 45°, with a gap between plates of 30 mm
Solution
  1. The area of the tank is A=π⋅122/4=113.10A = \pi\cdot 12^2/4 = 113.10 m². The surface loading is q/A=200/113.10=1.77q/A = 200/113.10 = 1.77 m/h, that is 0.491 mm/s.
  2. The smallest droplet caught: the expression under the root is 18⋅0.00065⋅0.00049129.81⋅180\dfrac{18\cdot 0.00065\cdot 0.0004912}{9.81\cdot 180} =3.255⋅10−9= 3.255\cdot 10^{-9} m², so dmin⁡=57d_{\min} = 57 µm.
  3. The volume of water in the tank is 113.10⋅8=904.8113.10\cdot 8 = 904.8 m³, and the water passes through it in 904.8/200=4.52904.8/200 = 4.52 h. In exactly this time a droplet of 57 µm rises 8 m, from the bottom to the surface.
  4. If, because of an uneven flow, only half of the area works well, the loading is twice as large, and the smallest droplet is 2\sqrt 2 times larger: about 81 µm.
  5. A plate pack: the projection of the plates is 48⋅1.75⋅1.0⋅cos⁡45∘48\cdot 1.75\cdot 1.0\cdot\cos 45^\circ =59.40= 59.40 m². Each pack takes 1200 m³/d, and the loading is 1200/(86,400⋅59.40)1200/(86{,}400\cdot 59.40) =0.0002338= 0.0002338 m/s. The expression under the root is 1.549 × 10⁻⁹ m², and dmin⁡=39d_{\min} = 39 µm.
  6. Check of the flow regime between the plates: the velocity is 1200/(86,400⋅48⋅0.030⋅1.0)1200/(86{,}400\cdot 48\cdot 0.030\cdot 1.0) =0.0096= 0.0096 m/s. The Reynolds number, based on the hydraulic diameter of the slot (two gaps, 0.06 m), is 1040⋅0.0096⋅0.06/0.000651040\cdot 0.0096\cdot 0.06/0.00065 =922= 922, so the flow is laminar. A droplet of 39 µm needs to rise only 4.2 cm vertically to the plate above it, which takes about 3.0 min. The volume of the plate pack is 1.75⋅1.0⋅48⋅0.030=2.51.75\cdot 1.0\cdot 48\cdot 0.030 = 2.5 m³.
Answer: A tank 12 m in diameter catches all droplets larger than 57 µm (with an uneven flow, larger than 81 µm). Four plate packs of about 2.5 m³ each catch those larger than 39 µm.

What it means. The tank holds 904.8 m³ of water for more than four hours, while the plate packs catch even smaller droplets although the water passes through them in minutes. The gain comes not from the rising velocity but from the area of the plate projections and the short path of the droplet. On the other hand, the tank smooths out the swings of the flow rate and of the composition of the water and collects sediment, while plates get clogged with suspended solids and wax and have to be cleaned. The calculation is for an ideal vessel: short-circuiting of the flow, eddies and gas bubbles make a real settling tank worse, and the merging of droplets on the plates makes it a little better. The data in this example are assumed for illustration.

How much oil stays in the water: the role of droplet sizes

The smallest droplet that is caught does not yet say how much oil will leave with the water. This depends on what share of the oil is in small droplets. We measure the droplet size distribution on samples, under a microscope or with laser particle analysers, and express it as shares of the oil volume in size classes. An ideal settling tank with a horizontal flow catches the droplets of each class in the share (d/dmin⁡)2(d/d_{\min})^2, but not more than one. Adding up the contributions of the classes gives the share of the oil that it will catch.

Oil at the tank inlet by droplet size, and the share that the tank catches
Droplets, µm
up to 10 (5)
Share of the oil, %
5
Caught in summer, %
1
Caught in winter, %
0
Droplets, µm
10–20 (15)
Share of the oil, %
10
Caught in summer, %
7
Caught in winter, %
4
Droplets, µm
20–40 (30)
Share of the oil, %
20
Caught in summer, %
28
Caught in winter, %
14
Droplets, µm
40–60 (50)
Share of the oil, %
20
Caught in summer, %
77
Caught in winter, %
39
Droplets, µm
60–90 (75)
Share of the oil, %
25
Caught in summer, %
100
Caught in winter, %
88
Droplets, µm
more than 90 (120)
Share of the oil, %
20
Caught in summer, %
100
Caught in winter, %
100
Droplets, µm
All the oil
Share of the oil, %
100
Caught in summer, %
66.75
Caught in winter, %
53.00

In brackets is the droplet size that represents the class. The share of a class that is caught is (d/dmin⁡)2(d/d_{\min})^2, rounded to a per cent and not more than 100%, with dmin⁡d_{\min} = 57 µm in summer (40 °C) and 80 µm in winter (15 °C). The row “All the oil” is the sum of the class shares multiplied by the shares caught. The distribution is assumed for illustration.

Worked exampleHow much oil leaves the settling tank in summer and in winter
Given (assumed data)
Tank and flow rate
as in the previous example: 12 m in diameter, 4800 m³/d
Oil at the inlet
700 mg/L; the droplet size distribution is in the table above
Summer, 40 °C
density difference between water and oil 180 kg/m³, water viscosity 0.65 mPa·s
Winter, 15 °C
density difference 170 kg/m³, water viscosity 1.20 mPa·s (the water has cooled in the pipes and the tank)
Solution
  1. In summer the smallest droplet caught is 57 µm (the previous example).
  2. In winter dmin⁡d_{\min} grows as the root of the ratio of the viscosity to the density difference: (1.20/0.65)⋅(180/170)=1.398\sqrt{(1.20/0.65)\cdot(180/170)} = 1.398, and dmin⁡=57⋅1.398=80d_{\min} = 57\cdot 1.398 = 80 µm.
  3. For each class the share of droplets caught is (d/dmin⁡)2(d/d_{\min})^2. For example, for droplets of 50 µm it is (50/57)2=77(50/57)^2 = 77% in summer and (50/80)2=39(50/80)^2 = 39% in winter. The sum of the contributions of the classes (the table) gives 66.75% of the oil caught in summer and 53.00% in winter.
  4. What stays in the water: in summer 700⋅(1−0.6675)=233700\cdot(1 - 0.6675) = 233 mg/L, in winter 700⋅(1−0.5300)=329700\cdot(1 - 0.5300) = 329 mg/L, which is 329/233=1.41329/233 = 1.41 times more, or 41% more.
  5. In summer the tank catches droplets larger than 60 µm completely, and only small ones leave. Of the oil that is left, 43% is in droplets smaller than 20 µm and 57% is in droplets of 20–60 µm.
Answer: In summer about 233 mg/L of oil leaves the tank, and in winter about 329 mg/L. Cooling the water by 25 °C raises the oil content at the outlet by about 41%.

What it means. A settling tank removes large droplets and smooths out the swings, but small droplets pass through it almost entirely. They need the next stages: hydrocyclones, flotation, filters. Winter in a sharply continental climate (very cold winters and hot summers) makes separation noticeably worse, so we insulate pipes and tanks and heat the water when necessary. The breaking up of droplets has the same effect as cooling: if a pump before the tank shifts the whole distribution to small droplets, there will be noticeably more oil at the outlet for the same concentration at the inlet. The calculation is for an ideal settling tank; a real one will catch less. The data in this example are assumed for illustration.

Share of droplets caught by an ideal settling tank against droplet size0255075100125150020406080100Droplet size, µmCaught, %1231 — plates2 — tank in summer3 — tank in winter

A chart: the droplet size from 0 to 150 micrometres on the horizontal axis, and the share caught from 0 to 100 per cent on the vertical axis. Three curves rise like a parabola and reach 100 per cent: the plate pack at 39 micrometres, the tank in summer at 57, and the tank in winter at 80. For a droplet of 30 micrometres, the pack catches about 58 per cent, the tank in summer about 28, and in winter about 14.

CalculatedThe share of droplets that an ideal settling tank catches, against their size: the tank of the example in summer (dmin⁡d_{\min} = 57 µm) and in winter (80 µm), and a plate pack in summer. Calculated with the formula of the lesson.

Treatment stages: from the settling tank to the filter

The water separated in the separators and at the preliminary water removal unit (Oil treating: separators, dehydration, desalting and stabilisation) carries from hundreds to several thousand milligrams of oil per litre. One vessel cannot clean it. Each stage takes the droplets that the previous one did not hold, from large to small. The oil that is caught (recovered oil) goes back to oil treatment, and the sediment and sludge are collected and disposed of separately.

Treatment stages of produced water: settling tank, hydrocyclones, flotation, filters; the recovered oil returns to oil treatmentwater fromseparatorssettlingtankhydro-cyclonesflotationpumphundreds totens of µmdroplets fromabout 10 µmfiltersbuffertankpump station:into the reservoirdisposalwellsexcess, if injection is smallera few µmrecovered oil — to oil treatmenttanks are under a gas blanket; air is not allowed into the system

A diagram in two rows. In the upper row, from left to right: water from the separators enters a settling tank, and then a pump feeds it to the hydrocyclones and to a flotation unit. From there an arrow leads down to the filters, and the lower row runs from right to left: filters, a buffer tank, and the water injection pump station for injection into the reservoir. A dashed line from the buffer tank leads to the disposal wells, for the excess water if the injection is smaller. Labels at the stages: above the settling tank, from hundreds to tens of micrometres; above the hydrocyclones and the flotation unit, droplets from about 10 micrometres; below the filters, a few micrometres. Dashed arrows from the settling tank, the hydrocyclones and the flotation unit lead to the line of recovered oil, which returns to oil treatment. A note at the bottom says: tanks are under a gas blanket, and air is not allowed into the system.

SchematicTreatment stages of produced water (schematic): settling tank, hydrocyclones, flotation, filters. The recovered oil returns to oil treatment, and the cleaned water goes to the water injection pump station or to disposal wells.
Treatment stages
  1. 1
    Settling tanks and skim tanks

    Tanks and horizontal vessels in which the water moves slowly down or along, while the oil collects at the top and goes into a trap. They catch droplets from hundreds down to a few tens of micrometres, and they also smooth out the swings of the flow rate and the composition and collect sediment.

  2. 2
    Plate separators

    The same principle, but with the area multiplied by packs of plates. They are compact and catch droplets from a few tens of micrometres. The plates get clogged with suspended solids and wax and have to be cleaned.

  3. 3
    Hydrocyclones

    A deoiling hydrocyclone is a narrow conical vessel with no moving parts. The water enters along the side (tangentially) and starts to spin. The light oil collects near the axis and leaves through the top outlet together with a few per cent of water, and the cleaned water leaves through the bottom outlet. It needs a pressure drop. It works only while water is the continuous phase (water surrounds the oil droplets).

  4. 4
    Flotation

    In gas flotation, small gas bubbles stick to droplets and particles and carry them up into a foam, which we skim off at the top. We introduce the gas with a stirrer or an ejector (a jet device that sucks in gas), or we dissolve it under pressure and release it when the pressure drops (dissolved gas flotation). We use associated gas or nitrogen, not air.

  5. 5
    Filters

    Granular filters (sand, anthracite, which is a hard coal, and crushed nutshell) catch suspended solids and droplets down to a few micrometres. We wash them regularly with a reverse flow (backwashing). Cartridge filters give the water a final cleaning before injection. We treat the wash water and the sludge again.

  6. 6
    Degassing and buffer tanks

    The dissolved gas is released before the pumps, and the tanks are kept under a gas blanket: air is not allowed into the system.

A hydrocyclone replaces gravity with centrifugal force (the outward push in spinning water). In a vessel about 35 mm in diameter the water spins at a velocity of the order of 10 m/s, and at a radius of 1 cm the centrifugal acceleration v2/rv^2/r = 10,000 m/s². This is about 1019 times the acceleration due to gravity. By Stokes' law a droplet moves towards the axis that many times faster. A droplet of 10 µm, which in a settling tank rises 5.4 cm in an hour, shifts by a centimetre in 0.65 s in a hydrocyclone. So a few tens of such vessels in one casing do the work of a tank, and the rolling of an offshore platform does not disturb them. The estimate is rough: the acceleration and the path of the droplet change along the length of the vessel.

Treatment stages: which droplets they catch (approximate)
Stage
Settling tank, skim tank
Oil droplets
from hundreds down to a few tens of µm
Features
cheap, they smooth out the swings; large volume
Stage
Plate separator
Oil droplets
from a few tens of µm
Features
compact; plates get clogged with suspended solids and wax
Stage
Hydrocyclone
Oil droplets
of the order of 10 µm and larger
Features
needs a pressure drop; the continuous phase must be water
Stage
Flotation
Oil droplets
of the order of 10 µm and larger, and also suspended solids
Features
the gas is associated gas or nitrogen; flocculant chemicals help
Stage
Granular and cartridge filters
Oil droplets
a few µm, and suspended solids
Features
backwashing; the wash water is treated

Orders of magnitude for guidance. The limit depends on the design, the operating mode, the properties of the oil and water and the chemicals. It is confirmed by tests on the water of a particular field.

Chemicals help too. We dose demulsifiers, which break the emulsion of water in oil, already into the production of the wells. For the reverse emulsion (oil droplets in water) we use other chemicals: reverse demulsifiers and flocculants. They neutralise the electric charges of droplets and particles and gather them into flocs (loose clumps), which rise more easily or are caught by a filter. We choose the chemicals by tests on fresh samples, because an unsuitable chemical can even make the emulsion more stable.

How oil in water is measured

The oil in water content (OiW) is the main indicator of how well the treatment works. We measure it at the inlet and the outlet of each stage and before injection or discharge, and we express it in milligrams per litre. For almost fresh water this is the same as ppm, and for brines the numbers differ by the density of the solution. The difficulty is that “oil” in water is a mixture of substances, and different methods see different things in it. In Russian-language analysis methods and standards this indicator is usually called the content of petroleum products.

Methods for measuring oil in water
Method
Gravimetric
What is measured
the mass of everything that was extracted by a solvent and did not evaporate when the solvent was removed
Features
light fractions are lost; part of the dissolved organic matter is also picked up
Method
Infrared spectrometry of the extract
What is measured
the absorption of light by carbon–hydrogen bonds
Features
calibrated with the oil of the field or with a standard mixture
Method
Fluorescence (fluorimetric)
What is measured
the glow of aromatic hydrocarbons in ultraviolet light
Features
very sensitive, suitable for inline analysers; calibrated with the local oil
Method
Gas chromatography
What is measured
hydrocarbons in a given range of boiling temperatures
Features
selective and accurate, but slow; often used as the reference method

The result depends on the solvent, on which substances are counted as oil, and on the calibration. So a standard for oil content is always tied to a measurement method.

  • Place of sampling. We take the sample after a section where the flow is mixed, not from the upper or stagnant (still) part of the pipe, where droplets collect.
  • Containers and time. Oil sticks to plastic and to walls, droplets rise in the bottle, and bacteria break the oil down. So we take the whole sample into glass, extract it in the same container, and analyse it quickly or preserve it.
  • Inline analysers (usually fluorescence ones) show changes in real time, but their readings are checked regularly against the laboratory method accepted for this water.
  • Average and instantaneous values. Discharge limits are usually set on the average over a period, weighted by flow rate. A short peak during filter washing or start-up can be several times higher than the average.

Scale and corrosion in the treatment system

Scale starts when the water becomes supersaturated with a salt (it holds more of the salt than it can keep dissolved; salts, saturation index and the mass of the precipitate). In a treatment system there are three causes. The first is a fall of pressure: carbon dioxide leaves the water, the pH (the acidity of the water) rises, and calcium carbonate drops out, in separators, at chokes and in heat exchangers, where heating also lowers its solubility. The second is the mixing of incompatible waters, either produced waters of different horizons (producing layers) or produced water with water from another source: barium and strontium sulfates drop out right in the pipes and vessels (water compatibility). The third is hydrogen sulfide and iron: iron sulfide drops out as a fine black suspension, which clogs filters and stabilises emulsions.

Against salts, we dose scale inhibitors, in parts per million but into all of the water, so a field uses tonnes of them a year. We inject the inhibitor before the place where supersaturation starts. We choose it for the salt, the temperature and the other chemicals, and we monitor its residual concentration and the ion balance: if there is less barium at the outlet of the system than at the inlet, barite (barium sulfate) is dropping out somewhere. Deposits of barium and strontium sulfates can be weakly radioactive, and they are handled under the radiation safety rules.

Corrosion in a produced water system is caused by dissolved gases and bacteria. Carbon dioxide forms carbonic acid with water. The corrosion rate grows with its partial pressure and with temperature, until at high temperatures a protective film of iron carbonate forms on the steel. Hydrogen sulfide gives a film of iron sulfide, pitting corrosion (corrosion that eats small deep holes) and, for high-strength steels, sulfide stress cracking. There is no oxygen in formation water. It gets into the water through open tanks, pump seals and breather valves (valves through which a tank “breathes” air in and out), and it is the most dangerous: even tenths of a milligram per litre noticeably speed up corrosion and cause pits, and together with hydrogen sulfide, oxygen produces elemental sulfur and speeds up corrosion still more. Sulfate-reducing bacteria (SRB) live under deposits, produce hydrogen sulfide and cause pitting corrosion.

  • A closed system: tanks under a gas blanket, sealed pumps, and no contact of the water with air.
  • Materials: fibreglass pipes and pipes with an internal coating for water, and resistant steels where needed. Outside, buried pipelines are protected by a coating and by cathodic protection (a small electric current that keeps the steel from corroding).
  • Chemicals: corrosion inhibitors that form a film on the metal, and biocides, either continuously in small doses or from time to time in shock doses.
  • Monitoring: corrosion coupons (small metal samples left in the flow and weighed later) and corrosion rate probes, wall thickness measurements, and water analyses for iron and bacteria.

We decide where materials resistant to hydrogen sulfide are needed from its partial pressure in the gas that the steel and the water are in contact with. The international standard ISO 15156 (NACE MR0175) requires carbon and low-alloy steels to resist sulfide stress cracking when the partial pressure of hydrogen sulfide is 0.3 kPa (0.05 psi) or more. Partial pressure is not concentration: with the same share of hydrogen sulfide in the gas, it grows with pressure.

Partial pressure of a gas component
pi=yi pp_i = y_i\,p
where:
  • pip_ipartial pressure of the component: the pressure it would create if it occupied the whole volume alone, kPa
  • yiy_imole fraction of the component in the gas (100 ppm = 10−410^{-4}, 1% = 0.01), —
  • ppabsolute pressure of the gas, kPa

Units: any units of pressure, the same for pip_i and pp

Holds when
  • The gas is treated as a mixture of ideal gases (Dalton's law). At high pressures the standards also allow more accurate methods.
  • The pressure is absolute: the gauge pressure from a pressure gauge gives too low a value, especially at low pressures (Systems of units: SI, oilfield and metric oilfield).
  • For the water, we take the gas that is in equilibrium with it: in the separator or in the vapour space of the tank.

The same gas may need hydrogen-sulfide-resistant materials at high pressure and not at low pressure. The danger to people is set not by the partial pressure but by the concentration of the gas in the air that they breathe.

Worked exampleWhere materials resistant to hydrogen sulfide are needed
Given (assumed data)
First-stage separator
absolute pressure 1.8 MPa; the gas contains 200 ppm of hydrogen sulfide and 2.5% of carbon dioxide (by moles)
Settling tank
under a gas blanket at 105 kPa (abs.); in the vapour space there is 0.2% of hydrogen sulfide and 6% of carbon dioxide, gas that came out of the water when the pressure fell
Limit by ISO 15156
0.3 kPa of hydrogen sulfide
Solution
  1. Separator: pH2S=200⋅10−6⋅1800=0.36p_{\mathrm{H_2S}} = 200\cdot 10^{-6}\cdot 1800 = 0.36 kPa, which is more than 0.3 kPa. The steel of the separator, of its pipes and of the water line up to the first pressure drop must resist sulfide stress cracking. The partial pressure of carbon dioxide is 0.025⋅1800=450.025\cdot 1800 = 45 kPa.
  2. Tank: pH2S=0.002⋅105=0.21p_{\mathrm{H_2S}} = 0.002\cdot 105 = 0.21 kPa, which is less than 0.3 kPa. By this criterion the requirements for the steel are milder, although the share of hydrogen sulfide is 10 times larger than in the separator. The partial pressure of carbon dioxide is 0.06⋅105=6.30.06\cdot 105 = 6.3 kPa.
  3. The concentration of hydrogen sulfide in the vapour space of the tank is 0.2%, that is 2000 ppm. Such gas is deadly dangerous and must not be released into the atmosphere.
Answer: Materials resistant to hydrogen sulfide are needed in the separator (0.36 kPa). For the tank (0.21 kPa) this criterion does not require them. But the gas of the tank, with 2000 ppm of hydrogen sulfide, is deadly dangerous to people.

What it means. High pressure “amplifies” a small share of hydrogen sulfide, and low pressure “weakens” a large one. So we choose materials by the partial pressure at each point of the scheme, and we protect people by the concentration of the gas. The limit of the standard refers to sulfide stress cracking. General corrosion and pitting from hydrogen sulfide, carbon dioxide and bacteria do not disappear in the tank, so we protect it with a coating, an inhibitor and a gas blanket without oxygen. The data in this example are assumed for illustration.

Where to put the water: injection, disposal wells, discharge

  • Injection into the productive reservoir for pressure maintenance is the main route on land: the water is disposed of and at the same time displaces oil. The requirements for its quality and compatibility are set by the reservoir (Water injection systems).
  • Injection into disposal horizons (disposal wells) means injection into non-productive formations that are reliably isolated from fresh underground water. A formation that takes water, a reliable well casing and pressure monitoring are needed. The water brings no benefit to production, and energy is spent on injection.
  • Discharge into water bodies and into the sea is allowed only after treatment to the set standards, and under permits.
  • Storage facilities: storage ponds, filtration fields (fields where the water seeps into the ground) and evaporation ponds (this is what the Water Code calls them). They need permits and protection of the groundwater, because evaporation leaves salts and petroleum products on the surface.
  • Reuse after desalination (removing the salt), for industrial needs or irrigation. It is rare, because desalination of brines is expensive.
Water Code (2025), Article 123, paragraph 4
Underground water taken out incidentally in subsoil use may be discharged, or injected into the subsoil, only with environmental permits and permits for special water use, except for the injection of formation water produced with hydrocarbons for pressure maintenance.
Environmental Code (2021), Article 213, paragraph 3
The injection for pressure maintenance of formation water produced with hydrocarbons, of sea water and desalinated water, and of industrial water with a salinity of 2000 mg/L or more is not considered a discharge of pollutants, and no limits on permissible discharge are set for it.
Subsoil Code, Article 148
When subsoil users under different contracts develop fields jointly, one of them may inject its treated formation water into the reservoir of the other's field to maintain pressure. A subsoil user may entrust the treatment of produced and other water, including sea water, for injection into the reservoir to the owners of the necessary infrastructure.
Subsoil Code, Article 159
In offshore exploration and production, the discharge of waste into the sea and its burial on the seabed are prohibited. Industrial and other wastewater is discharged into the sea only with the permission and under the control of the authorised body in the field of environmental protection, and after treatment to the set standards.
Environmental Code, Article 273, subparagraphs 8 and 9
In the state nature reserve zone in the northern part of the Caspian Sea, the discharge of waste into the sea is prohibited, and so is the discharge of wastewater, except for a limited list of treated waters (for example, sea water cleaned of oil). They are discharged only under permits of the authorised bodies.

A practical conclusion follows from these rules: returning produced water to the reservoir to maintain pressure is the simplest route as far as permits go, while injection into disposal horizons and discharge need separate permits and standards. This summary is short and follows the texts of the codes as of October 2026. The project of a particular field and the permits for its facilities set the exact requirements (State regulation of field development).

The water balance of a field

The pressure maintenance system injects water that replaces the underground withdrawal (the volume of fluid taken out of the reservoir; voidage replacement). While there is little produced water, almost all the water for injection is taken from a source: an aquifer (an underground layer that holds water), the sea or a river. As the water cut grows there is more produced water, and it displaces the water of the source. The water balance shows where the boundary will lie and whether there will be more produced water than needs to be injected.

Water from the source at a given voidage replacement
qsrc=VRR Bo qo+Bw qwBw−qwq_{\text{src}} = \mathrm{VRR}\,\frac{B_o\,q_o + B_w\,q_w}{B_w} - q_w
where:
  • qsrcq_{\text{src}}water that must be taken from the source; a negative value is an excess of produced water that must be disposed of, m³/d
  • VRR\mathrm{VRR}voidage replacement ratio: the share of the underground withdrawal that is replaced by injection, fraction
  • qo, qwq_o,\ q_woil and water rates at standard conditions, m³/d
  • Bo, BwB_o,\ B_wformation volume factors of oil and water at reservoir pressure (the reservoir volume that gives one cubic metre at standard conditions), m³/m³

Units: rates at standard conditions, m³/d

Holds when
  • The pressure is above the bubble point pressure (the pressure at which gas starts to come out of the oil), and there is no free gas in the reservoir. Otherwise the gas is also counted in the voidage (Why water is injected: pressure maintenance and oil displacement).
  • All the produced water is returned to the reservoir. Losses in treatment (water that left with the oil, the reject of hydrocyclones, the washing of filters) increase the water from the source a little.
  • The produced water is compatible with the reservoir into which it is injected.

With full voidage replacement (VRR=1\mathrm{VRR} = 1) the water needed from the source is qoBo/Bwq_oB_o/B_w, exactly the space that the produced oil occupied in the reservoir. The reservoir has already “given up” all the water that the field returns. If part of the volume is replaced by edge water (VRR<1\mathrm{VRR} < 1), there is more produced water than needs to be injected when the water–oil ratio exceeds VRR (Bo/Bw)/(1−VRR)\mathrm{VRR}\,(B_o/B_w)/(1 - \mathrm{VRR}).

Worked exampleThe water balance of a field over its life
Given (assumed data)
Field
from the lesson on development concepts: oil plateau rate 6400 m³/d, liquid capacity 12,800 m³/d; production continues until year 21 of the project
Fluids
Bo=1.25B_o = 1.25, Bw=1.02B_w = 1.02; the pressure is kept above the bubble point pressure
Base option
injection replaces the withdrawal in full, VRR=1\mathrm{VRR} = 1; the source is an aquifer
Another option
edge water replaces 30% of the withdrawal, and injection the other 70% (VRR=0.7\mathrm{VRR} = 0.7), with the same production profile
Solution
  1. Year 8: oil 5500, water 7300 m³/d. The injection is (1.25⋅5500+1.02⋅7300)/1.02(1.25\cdot 5500 + 1.02\cdot 7300)/1.02 =14,040= 14{,}040 m³/d. From the source 14,040−7300=674014{,}040 - 7300 = 6740 m³/d is needed, and the produced water is 52.0% of the injection.
  2. Year 20: oil 652, water 12,148 m³/d. The injection is (1.25⋅652+1.02⋅12,148)/1.02(1.25\cdot 652 + 1.02\cdot 12{,}148)/1.02 =12,947= 12{,}947 m³/d. From the source 12,947−12,148=79912{,}947 - 12{,}148 = 799 m³/d is needed, and the produced water gives 93.8% of the injection.
  3. Check: what is needed from the source is qoBo/Bwq_oB_o/B_w. In year 20 this is 652⋅1.25/1.02=799652\cdot 1.25/1.02 = 799 m³/d.
  4. The option with edge water: there is more produced water than needs to be injected when the water–oil ratio exceeds 0.7⋅(1.25/1.02)/0.3=2.860.7\cdot(1.25/1.02)/0.3 = 2.86, that is at a water cut of 2.86/(1+2.86)=74.12.86/(1 + 2.86) = 74.1%. The yearly average water cut crosses this limit in year 11.
  5. In year 20 in this option 0.7⋅12,947=90630.7\cdot 12{,}947 = 9063 m³/d is injected, while the produced water is 12,148. The excess 12,148−9063=308512{,}148 - 9063 = 3085 m³/d goes into disposal wells.
Answer: With full voidage replacement, water from the source is needed mostly on the plateau (7843 m³/d in year 5), and in year 20 it falls to 799 m³/d. If edge water supports the pressure in part, an excess of produced water appears from year 11, and by year 20 it is 3085 m³/d.

What it means. The water intake and the source water treatment are needed at full capacity only in the first years, and later stand idle. So we design them for a phased shutdown or reassignment, and we design the produced water treatment unit for its volume to grow to 12,148 m³/d and more. Source water and produced water are mixed in one system, and we check their compatibility before design (Water injection systems). In the second option, disposal wells and permits for them are needed. The number of wells depends on their injectivity (how much water they take). The production profile is taken to be the same in both options, and the water losses in treatment are not counted: they slightly increase the water from the source and reduce the excess. The data in this example are assumed for illustration.

Water balance of the assumed field, m³/d
Year
5
Produced water
0
Injection at VRR 1
7843
From the source
7843
Excess at VRR 0.7
—
Year
8
Produced water
7300
Injection at VRR 1
14,040
From the source
6740
Excess at VRR 0.7
—
Year
12
Produced water
10,395
Injection at VRR 1
13,342
From the source
2947
Excess at VRR 0.7
1055
Year
16
Produced water
11,578
Injection at VRR 1
13,076
From the source
1498
Excess at VRR 0.7
2425
Year
20
Produced water
12,148
Injection at VRR 1
12,947
From the source
799
Excess at VRR 0.7
3085

Yearly averages. Production continues until year 21 of the project (the economic limit in the lesson on field facilities). “Excess” is the produced water over the injection at VRR=0.7\mathrm{VRR} = 0.7, which is sent to disposal wells. The data are assumed for illustration.

Water balance of the field by year468101214161820220481216Project yearThousand m³/dinjection at VRR 1injection at VRR 0.7produced waterwater from the source

A chart in thousand cubic metres per day, from 0 to 16, by project year from 4 to 21. The injection at full voidage replacement grows from 3.9 to 14.0 by year 8 and then slowly falls to 12.9. Produced water appears in year 6 and grows to 12.1 by year 20. Water from the source is highest on the plateau, about 7.8, and falls to 0.8 by year 20. The dashed injection line at a replacement of 0.7 runs 30 per cent below the line of full replacement and crosses the produced water line between year 10 and year 11.

CalculatedWater balance of the field of the example by year: produced water, injection at full voidage replacement, and water from the source. The dashed line is the injection if edge water replaces 30% of the withdrawal. Where the dashed line is below the produced water line (from year 11), part of the water has to be disposed of in disposal wells. VRR in the labels is the voidage replacement ratio.

Terms and symbols

Produced water
Water produced together with oil and gas: formation water and injected water. The laws of Kazakhstan classify it as wastewater.
Water–oil ratio (WOR)
The ratio of produced water to produced oil, WC/(1−WC)\text{WC}/(1 - \text{WC}).
Surface loading q/Aq/A
The water rate through a settling tank divided by its plan area. An ideal settling tank catches completely the droplets that rise faster.
dmin⁡d_{\min}
The smallest droplet that a settling tank catches completely. With a horizontal flow, smaller droplets are caught in the share (d/dmin⁡)2(d/d_{\min})^2.
Plate separator
A vessel with a pack of inclined plates. Its area is the sum of the horizontal projections of the plates.
Hydrocyclone
A vessel in which a spinning flow drives the light oil to the axis by centrifugal force.
Flotation
Separation of droplets and particles by gas bubbles that carry them up into a foam.
Recovered oil
Oil caught in water treatment and returned to oil treatment.
Oil in water (OiW)
The oil content of water, mg/L. The number has meaning only together with the measurement method.
Partial pressure pip_i
The share of a component in the gas multiplied by the absolute pressure. It is used to choose materials for a hydrogen sulfide environment.
Sulfate-reducing bacteria (SRB)
Bacteria that produce hydrogen sulfide, cause corrosion under deposits and infect the reservoir.

Common mistakes and misconceptions

Wrong

“Let us make the settling tank deeper, and the water will be cleaner.”

In fact

For an ideal settling tank the depth does not affect the droplet that is caught: what decides is the area AA and the loading q/Aq/A. The depth gives room for sediment and a reserve of volume. The area is increased with plates or with a second vessel.

Wrong

“The oil in water is 20 mg/L, so the water is fit both for discharge and for injection.”

In fact

The number depends on the method. For discharge, dissolved hydrocarbons and chemicals also matter, and for injection, suspended solids, particle size, oxygen and compatibility. The requirements for the water are set by where it is going.

Wrong

“There is no oxygen in formation water, so the system is protected from oxygen corrosion.”

In fact

Oxygen gets into the water through open tanks, pumps and breather valves. That is why the tanks are kept under a gas blanket and the system is kept closed.

Wrong

“The share of hydrogen sulfide in the tank is ten times that in the separator, so resistant steels are needed there too.”

In fact

What matters for the choice of steel is not the percentage but the partial pressure. At a pressure close to atmospheric, even 0.2% gives less than 0.3 kPa. For people, however, what is dangerous is the concentration: 2000 ppm is deadly.

Wrong

“Produced water can be injected into any reservoir: it comes from a reservoir anyway.”

In fact

It is compatible with the reservoir from which it was produced, and even then not always after mixing with injected water. In another horizon its ions can form salts, and the suspended solids and oil will plug the pores. In addition, injection other than for pressure maintenance needs permits.

In depth: advanced detail

Why the area, not the depth, matters for a settling tank

Suppose that the water passes evenly through a rectangular settling tank of length LL, width bb and depth HH, at a velocity u=q/(bH)u = q/(bH), and that a droplet rises at a velocity vtv_t at the same time. The path of the droplet is a straight line with a slope vt/uv_t/u. A droplet that entered at the bottom needs the time H/vtH/v_t to rise through the whole depth HH, and the water needs the time L/uL/u to pass through the tank. The droplet is caught if H/vt≤L/uH/v_t \le L/u, that is vt≥uH/L=q/(bL)=q/Av_t \ge uH/L = q/(bL) = q/A. The depth cancels: the condition depends only on the plan area (the area seen from above).

A smaller droplet, with vt<q/Av_t < q/A, rises during the retention time only through the height vtL/uv_tL/u. It is caught only if it entered no deeper than vtL/uv_tL/u below the surface. With an even distribution of droplets over the height, the share of such droplets is vtL/(uH)=vt/(q/A)v_tL/(uH) = v_t/(q/A). The Stokes velocity is proportional to d2d^2, so the share is (d/dmin⁡)2(d/d_{\min})^2. In a vertical tank, where the water moves down and the droplets rise against it, the condition vt≥q/Av_t \ge q/A follows at once, but there is no partial catching: a droplet smaller than dmin⁡d_{\min} is carried down, wherever it entered. The share (d/dmin⁡)2(d/d_{\min})^2 refers to a horizontal flow, as in the tank of the examples of the lesson, where the water spreads from the centre to the walls.

For a pack of plates, the velocity between the plates is u=q/(nsb)u = q/(n s b), where ss is the gap, and a droplet needs to rise vertically only through s/cos⁡θs/\cos\theta along a path of length LL. If the plates are long compared with the gap, the condition for catching reduces to vt≥q/(nLbcos⁡θ)v_t \ge q/(nLb\cos\theta), the loading on the total projection of the plates. The exact formula also takes into account the component of the rise along the plates, vtsin⁡θv_t\sin\theta. If the water goes down along the plates, against the rising oil (counter-current flow), this component keeps the droplet in the pack longer. For the plates of the example of the lesson, the rising velocity needed is then only 2% smaller. If the water goes up, together with the oil, it is 2% larger.

How a bubble speeds up the rise of a droplet

Flotation works not because the gas “dissolves” the oil, but because a bubble that has stuck to a droplet sharply lowers the density of the clump that rises and increases its size. The Stokes velocity grows with the square of the diameter and with the density difference, and both factors grow at once.

Worked exampleA 10 µm droplet with and without a bubble
Given (assumed data)
Oil droplet
10 µm, density 860 kg/m³
Gas bubble
60 µm, gas density about 1 kg/m³ (negligibly small)
Water
1040 kg/m³, viscosity 0.65 mPa·s
Solution
  1. The droplet alone: vt=9.81⋅(10−5)2⋅18018⋅0.00065v_t = \dfrac{9.81\cdot(10^{-5})^2\cdot 180}{18\cdot 0.00065} =1.509⋅10−5= 1.509\cdot 10^{-5} m/s, which is 5.4 cm per hour.
  2. The clump “droplet + bubble”: the equivalent diameter (the diameter of a sphere with the same volume) is (603+103)1/3=60.09(60^3 + 10^3)^{1/3} = 60.09 µm, and the density is 860⋅103603+103=3.96\dfrac{860\cdot 10^3}{60^3 + 10^3} = 3.96 kg/m³ (the gas is neglected).
  3. The Stokes velocity is proportional to d2(ρw−ρ)d^2(\rho_w - \rho). Because of the size it is larger by a factor of (60.09/10)2=36.11(60.09/10)^2 = 36.11, because of the density by (1040−3.96)/180(1040 - 3.96)/180 =5.756= 5.756, and in all by 36.11⋅5.756=20836.11\cdot 5.756 = 208 times: vt=0.003137v_t = 0.003137 m/s.
  4. The Reynolds number of the clump is 1040⋅0.003137⋅60.09⋅10−6/0.000651040\cdot 0.003137\cdot 60.09\cdot 10^{-6}/0.00065 =0.30= 0.30, so Stokes' law still applies.
Answer: With a bubble the droplet rises 208 times faster: in a minute it rises 18.8 cm instead of fractions of a millimetre.

What it means. In essence it is the bubble that rises, and the droplet only needs to stick to it. So the size of the bubbles, their number, and how often they collide with droplets and hold them decide how well a flotation unit works. This is the reason for stirrers, water recirculation and flocculant chemicals. We treated the bubble as a solid ball. For real bubbles in water with surfactants this is close to the truth. The data in this example are assumed for illustration.

Connections to other topics

Check yourself

  1. 1

    A settling tank catches droplets from 80 µm. What will improve the treatment most at the same water rate?

    1. A.Put a second identical tank in parallel
    2. B.Raise the water level in the tank from 8 to 10 m
    3. C.Put a centrifugal pump before the tank to speed up the flow
    4. D.Cool the water so that the oil becomes denser
    Show the answer

    Correct answer: A. Put a second identical tank in parallel

    For an ideal settling tank the droplet that is caught is set by the loading q/Aq/A. A second tank halves it, and dmin⁡d_{\min} falls by a factor of 2\sqrt 2, to about 57 µm. The water level does not affect dmin⁡d_{\min}. A pump would break up the droplets. Cooling would raise the viscosity of the water and slow the rise more than it changes the density difference.

  2. 2

    Water passes through a settling tank of area 50 m² at a rate of 2160 m³/d. The density difference between water and oil is 200 kg/m³, and the water viscosity is 0.8 mPa·s. Find the smallest droplet that it catches completely.

    Show the answer

    The loading is q/A=2160/(86,400⋅50)q/A = 2160/(86{,}400\cdot 50) =5⋅10−4= 5\cdot 10^{-4} m/s. The expression under the root is 18⋅0.0008⋅5⋅10−4/(9.81⋅200)18\cdot 0.0008\cdot 5\cdot 10^{-4}/(9.81\cdot 200) =3.67⋅10−9= 3.67\cdot 10^{-9} m², and dmin⁡≈6.1⋅10−5d_{\min} \approx 6.1\cdot 10^{-5} m, that is about 61 µm. Droplets of 30 µm will be caught by this tank, if the flow in it is horizontal, only in the share (30/61)2≈0.24(30/61)^2 \approx 0.24.

  3. 3

    Why does a hydrocyclone a few centimetres in diameter catch smaller droplets than a tank of hundreds of cubic metres?

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    In a spinning flow the centrifugal acceleration v2/rv^2/r is hundreds and thousands of times the acceleration due to gravity, and by Stokes' law the droplet moves towards the axis that many times faster than it rises in a settling tank. And the distance it must travel is only a few millimetres. So in the seconds that the water spends in the vessel, droplets of the order of ten micrometres have time to reach the axis.

  4. 4

    The gas in a separator at an absolute pressure of 0.6 MPa contains 400 ppm of hydrogen sulfide, and the gas in another separator at 3.0 MPa contains 120 ppm. Where must the steel resist sulfide stress cracking by the ISO 15156 criterion?

    1. A.In neither
    2. B.In both
    3. C.Only in the first
    4. D.Only in the second
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    Correct answer: D. Only in the second

    The partial pressures are 400⋅10−6⋅600=0.24400\cdot 10^{-6}\cdot 600 = 0.24 kPa and 120⋅10−6⋅3000=0.36120\cdot 10^{-6}\cdot 3000 = 0.36 kPa. The limit is 0.3 kPa, so resistant steel is needed only in the second separator, although the share of hydrogen sulfide in it is more than three times smaller.

  5. 5

    A field produces 1500 m³/d of oil and 9000 m³/d of water; Bo=1.2B_o = 1.2, Bw=1.0B_w = 1.0. Edge water replaces 40% of the withdrawal, and the rest is injection. Is the produced water enough for injection?

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    The underground withdrawal is 1.2⋅1500+1.0⋅9000=10,8001.2\cdot 1500 + 1.0\cdot 9000 = 10{,}800 m³/d, and the injection is 0.6⋅10,800=64800.6\cdot 10{,}800 = 6480 m³/d. There is 9000 m³/d of produced water, 2520 m³/d more than needed, and the excess is sent to disposal wells. This can also be seen from the rule of the lesson: the water–oil ratio 9000/1500=69000/1500 = 6 is larger than 0.6⋅1.2/0.4=1.80.6\cdot 1.2/0.4 = 1.8.

  6. 6

    Two laboratories obtained 18 and 31 mg/L of oil for one water sample. What is the most likely explanation, and what should be done?

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    Most likely they work by different methods: different solvents, a different set of substances counted as oil, a different calibration. Sampling and storage of the sample also affect the result. Only numbers from one method can be compared. So a field should fix one method, the one in which the limit is set, and check the inline analysers and the other laboratories against it.

Updated: 19 October 2026