Oil treating: separators, dehydration, desalting and stabilisation

Quick refresher

The treatment is set by the requirements for sales oil: water, chloride salts, hydrogen sulfide and mercaptans, vapour pressure, organic chlorides. Each requirement has its own operation: dehydration, desalting, stabilisation, stripping or neutralisation, and the choice of chemicals.
A separator splits the phases by gravity, so it needs a density difference, a calm flow and time. For gas, the limit is the carry-over of droplets (liquid droplets that the gas carries away), that is, the allowed gas velocity in the Souders–Brown equation. For liquid, the limits are the retention time and the distance that a droplet must travel to the interface (the boundary between the oil and water layers).
By Stokes' law, the settling velocity of a droplet is proportional to the square of its diameter and to the density difference, and inversely proportional to the viscosity of the continuous phase (the liquid around the droplet). Water droplets settle in viscous oil tens of times more slowly than oil droplets rise in water.
A film of asphaltenes and resins (the heaviest components of oil) and fine particles on the surface of the droplets keeps an emulsion stable. Heat, demulsifiers and an electric field break it. We remove free water before heating, so that we do not heat ballast.
Salts are dissolved in the residual water: the salt content equals the water fraction multiplied by its salinity (the salt content of the water). If there is already little water, but it is very salty, we dilute it with fresh wash water and separate it again.
Stabilisation lowers the vapour pressure by removing the dissolved light hydrocarbons. The price is a few per cent of the oil volume, which goes into the gas and can be recovered as liquid hydrocarbons in gas processing.

Key relationships

v_{\max} = K\sqrt{\frac{\rho_{\text{L}} - \rho_{\text{g}}}{\rho_{\text{g}}}}

allowed gas velocity in a separator (Souders–Brown)

v = \frac{g\,d^2\,\Delta\rho}{18\,\mu_c}

settling velocity of a droplet (Stokes' law, $\mathrm{Re} < 1$ )

t = \frac{V}{q}

retention time of the liquid in the vessel

S = \varphi\,C_{\text{w}}

salt content of the oil: the fraction of residual water times its salinity

m_{\text{net}} = m_{\text{gross}}\left(1 - \frac{W_{\text{w}} + W_{\text{cs}} + W_{\text{imp}}}{100}\right)

net mass of the oil: the gross mass less the ballast

Common pitfalls

Calculating a separator for gas only: in oil separators the liquid usually sets the size.
Putting the gas rate at standard conditions into the Souders–Brown equation, instead of the rate at the pressure and temperature of the separator.
Assuming that dehydrated oil is automatically desalted: with salty produced water, even a few tenths of a per cent of water give hundreds of milligrams of salts per cubic decimetre (one litre).
Heating the whole production together with the free water: the heat capacity of water (the heat needed to warm it) is twice that of oil, and the heat goes into ballast.
Assuming that oil is stable because it does not boil in the tank at 20 °C: at 37.8 °C its vapour pressure can be higher than atmospheric.

After this lesson you can

list the requirements for sales oil and link each of them to a treatment operation
describe how two-phase and three-phase, horizontal and vertical separators are built, and what their internal parts do
check a separator for gas (the Souders–Brown equation) and for liquid (retention time and the settling of droplets by Stokes' law)
explain why water-in-oil emulsions form and how heat, demulsifiers (chemicals that break emulsions) and an electric field break them, and estimate the heat needed
calculate the content of chloride salts and the amount of wash water needed for desalting
explain oil stabilisation through vapour pressure, and calculate the net mass of oil from the metering data

Requirements for sales oil and the route of its treatment

A well does not deliver oil alone but a mixture: oil with dissolved and free gas; produced water, often a brine (very salty water) with tens or hundreds of grams of salts per litre; sand and clay particles; and sometimes hydrogen sulfide and mercaptans (sulfur compounds with a bad smell). The buyer and the pipeline company accept only oil that meets the requirements for water, salts, sulfur compounds and vapour pressure (how readily the oil gives off vapour). Everything that a treatment unit does serves one aim: to meet these requirements with the smallest losses of oil and at the lowest cost.

Requirements for oil in the Technical Regulation of the Eurasian Economic Union TR EAEU 045/2017 “On the Safety of Oil Prepared for Transportation and (or) Use”

Indicator: Mass fraction of hydrogen sulfide, ppm (parts per million)
Not more than: 100
On delivery to a main pipeline or for processing: 20

Indicator: Methyl and ethyl mercaptans together, ppm
Not more than: 100
On delivery to a main pipeline or for processing: 40

Indicator: Mass fraction of water, %
Not more than: 1
On delivery to a main pipeline or for processing: 0.5

Indicator: Chloride salts, mg/dm³
Not more than: 900
On delivery to a main pipeline or for processing: 100

Indicator: Vapour pressure, kPa
Not more than: 66.7 (500 mm Hg)
On delivery to a main pipeline or for processing: 66.7

Indicator: Organic chlorides in the fraction that boils off below 204 °C, ppm
Not more than: 6
On delivery to a main pipeline or for processing: 6

Indicator	Not more than	On delivery to a main pipeline or for processing
Mass fraction of hydrogen sulfide, ppm (parts per million)	100	20
Methyl and ethyl mercaptans together, ppm	100	40
Mass fraction of water, %	1	0.5
Chloride salts, mg/dm³	900	100
Vapour pressure, kPa	66.7 (500 mm Hg)	66.7
Organic chlorides in the fraction that boils off below 204 °C, ppm	6	6

The regulation was adopted by Decision No. 89 of the Council of the Eurasian Economic Commission of 20 December 2017 (the requirements are in the version of 29 August 2023). It has been in force since 1 July 2019. It is binding in Kazakhstan, which is a member state of the Eurasian Economic Union. The vapour pressure is not measured if the pour point of the oil (the lowest temperature at which it still flows) is 10 °C or higher. The regulation does not apply to oil supplied for export beyond the customs territory of the Union. Contracts and the requirements of the pipeline systems set the quality of that oil. The interstate specification standard GOST 31378-2009 divides oil into classes, types, groups and kinds, by sulfur content, density, degree of treatment, hydrogen sulfide and light mercaptans. The regulation's list of standards names this standard and the Kazakhstan standard ST RK 1347-2024 “Oil. General specifications”. One cubic decimetre (dm³) is one litre.

Water and chloride salts: dehydration and desalting. The salts are dissolved in the water, so removing the water removes them too. When little water is left but it is very salty, we dilute it with fresh water and separate it again.
Vapour pressure: stabilisation. We remove the dissolved light hydrocarbons from the oil, because otherwise they would boil off in tanks, tank cars and tankers.
Hydrogen sulfide and mercaptans: stripping (blowing gas or steam through the oil to carry them away) in a column, and chemical scavengers (neutralisers) for what is left.
Organic chlorides (compounds of chlorine with carbon) are not formed in the reservoir; chemicals bring them in. The regulation expressly forbids the use of chemicals that contain organochlorine compounds in oil production and transport (paragraph 11). At a refinery they break down when heated and form hydrogen chloride, which causes corrosion.

SchematicThe route of oil through a treatment unit (schematic): after each operation gas or water leaves the oil, and heat, chemicals and wash water are added to it.

The route of oil through a treatment unit

1
Gas separation
We lower the pressure in one to three stages and take away the gas that comes out. The lesson Separator calculations and optimum separator pressure explains how much gas each stage gives and why there are several stages. The first stage is often at the booster pump station (the path of production in the field).
2
Removal of free water
Water that is not bound in an emulsion settles in minutes. We separate it as early as possible, so that we do not heat or pump ballast. It goes to treatment and injection (Produced water: composition, treatment and disposal).
3
Dehydration
We heat the emulsion, add a demulsifier (a chemical that breaks the emulsion) and leave it to settle. The last fractions of a per cent of water are separated in an electric field.
4
Desalting
We mix the oil with fresh wash water and separate the water again. Most of the salts leave with it.
5
Stabilisation
Hot separation or a stabiliser column drives off the light hydrocarbons. Their gas goes to gas processing (Gas processing and compression).
6
Storage, metering and hand-over
The oil collects in tanks. We measure its quantity and quality and hand it over to the main oil pipeline or for processing.

Separators: design and types

A separator is a pressure vessel in which the flow becomes calm and the phases (gas, oil and water) move apart under gravity. At 1 MPa the density of gas is about 9 kg/m³, of oil 800–900 kg/m³ and of produced water 1000–1200 kg/m³. The greater the density difference and the longer the mixture stays calm, the more complete the separation. A two-phase separator separates gas from liquid. A three-phase separator also separates the water from the oil.

SchematicA horizontal three-phase separator (schematic, not to scale). The inlet device slows the jet. In the gas space, liquid droplets settle. In the liquid, water droplets settle through the oil layer and oil droplets rise out of the water. The oil flows over a weir (an overflow wall), and the water is drawn off before the weir.

Inlet device: an impingement plate (a plate that the jet hits), a vane or a cyclonic inlet (which spins the flow). It slows the jet, at once separates the main mass of liquid, and spreads the flow over the cross-section.
Gravity settling zone: in the gas, liquid droplets settle; in the liquid, water droplets settle and gas bubbles and oil droplets rise.
Coalescing plate packs: packs of inclined or corrugated plates. A droplet needs to travel only a few centimetres to a plate, where it merges with other droplets.
Mist extractor: a mesh, vane or cyclone device. It catches the small droplets that had no time to settle, from a few to tens of micrometres in size.
Overflow weir and level controllers: they keep the oil level and the oil–water interface level, while the pressure controller on the gas outlet keeps the pressure in the vessel.
Vortex breakers (fittings that stop a whirlpool from forming) on the liquid outlets, so that gas is not drawn into the liquid lines; a safety valve, manholes, and a flushing system that washes sand off the bottom.

Horizontal or vertical

Property: Path of a droplet to the interface
Horizontal: short: across the flow, about half the diameter
Vertical: long: the droplet falls against the gas flow

Property: Interface area between the phases
Horizontal: large: gas leaves foam, and water leaves oil, more easily
Vertical: small

Property: Liquid slugs
Horizontal: takes them worse: there is little room between the working and the emergency level
Vertical: takes them better: there is spare height

Property: Sand and sediment
Horizontal: builds up along the long bottom
Vertical: collects in the conical bottom

Property: Area on the site
Horizontal: large
Vertical: small

Property	Horizontal	Vertical
Path of a droplet to the interface	short: across the flow, about half the diameter	long: the droplet falls against the gas flow
Interface area between the phases	large: gas leaves foam, and water leaves oil, more easily	small
Liquid slugs	takes them worse: there is little room between the working and the emergency level	takes them better: there is spare height
Sand and sediment	builds up along the long bottom	collects in the conical bottom
Area on the site	large	small

Horizontal vessels are usually chosen for large flows of oil and gas, for foaming oil and for three-phase separation. Vertical vessels are chosen for gas with a small amount of liquid (gas scrubbers before compressors, fuel gas separators), for production with sand, and where there is little room.

Treatment also uses other vessels besides the separators of the separation stages. Gas scrubbers are vertical vessels that protect compressors and burners from droplets. Slug catchers stand at the inlet of long multiphase pipelines and take the liquid slugs (large plugs of liquid) (flow patterns). In test separators we separate the production of individual wells one at a time, so that we can measure their rates (group metering stations). Hydrocyclones (devices that spin the water to throw out the oil) clean the water (Produced water: composition, treatment and disposal).

Gas capacity

A liquid droplet in gas falls under gravity, and the gas slows it down with a drag force (the resistance of the gas). Very soon the velocity stops growing. The droplet reaches its settling velocity, at which its weight, less the buoyancy force (the upward push of the gas), equals the drag. In a vertical separator the gas moves upward, and the droplet settles only if the gas is slower than its settling velocity. In a horizontal separator the gas moves along the vessel, and the droplet must drop through the height of the gas space while the gas passes along its length. So the allowed gas velocity is higher here. Both pictures lead to one rule: the gas velocity must not exceed a limit that is proportional to the square root of the density difference divided by the gas density.

Allowed gas velocity (Souders–Brown equation)

v_{\max} = K\sqrt{\frac{\rho_{\text{L}} - \rho_{\text{g}}}{\rho_{\text{g}}}}

where:

$v_{\max}$ highest allowed gas velocity in the free cross-section of the vessel, at its pressure and temperature, m/s
$K$ capacity coefficient: a value found from experience and tests. It depends on the mist extractor, the orientation and size of the vessel, the pressure and the tendency of the liquid to foam, m/s
$\rho_{\text{L}},\ \rho_{\text{g}}$ density of the liquid and of the gas at separator conditions, kg/m³

Units: any consistent units: $K$ and $v_{\max}$ in the same units of velocity, the densities in the same units

Holds when

The form follows from the balance of the weight and the drag of a droplet (the derivation is in the advanced part of the lesson). Everything that depends on the droplet size and on the way the gas flows around it is collected in $K$ .
Rough guides for an estimate: for a vertical vessel with a mesh mist extractor $K$ is about 0.05–0.1 m/s, and for a horizontal one about 0.12–0.15 m/s (0.40–0.50 feet per second). The manufacturers give exact values for their internal parts.
At high pressure, with foaming oil and with a heavy liquid load on the mist extractor, we reduce $K$ .
Without a mist extractor the allowed velocity is clearly lower: everything that has not settled leaves with the gas.

The denser the gas, the more slowly it must flow. When the pressure grows 4 times, the gas density grows about four times, and $v_{\max}$ falls by half. But the same amount of gas, measured at standard conditions, takes four times less volume. So in standard cubic metres the vessel passes more gas at high pressure.

In other unit systems (1)

Oilfield units

v_{\max}\,[\text{ft/s}] = K\,[\text{ft/s}]\sqrt{\frac{\rho_{\text{L}} - \rho_{\text{g}}}{\rho_{\text{g}}}}

densities in any identical units (lb/ft³ or kg/m³); $K$ in feet per second, 1 ft/s = 0.3048 m/s.

There is no constant: the root of the density ratio is dimensionless.

Gas rate at separator conditions

Q = Q_{\text{sc}}\,\frac{p_{\text{sc}}}{p}\cdot\frac{T}{T_{\text{sc}}}\cdot Z

where:

$Q$ actual volume rate of gas at the pressure and temperature of the separator, m³/s
$Q_{\text{sc}}$ gas rate at standard conditions, m³/s
$p_{\text{sc}},\ T_{\text{sc}}$ standard conditions: 0.101325 MPa and 293.15 K (20 °C), MPa, K
$p,\ T$ absolute pressure (measured from zero, not from atmospheric pressure) and temperature in the separator, MPa, K
$Z$ gas deviation factor at $p$ and $T$ (taken as 1 at standard conditions), —

Units: absolute pressures, and temperatures in kelvins

Holds when

The gas at standard conditions is treated as ideal. $Z$ at separator conditions comes from a correlation or an equation of state (Ideal and real gas: the gas deviation factor (Z)).

Mistakes with this conversion are the most frequent ones in separator calculations. For example, if the standard rate is put into the Souders–Brown equation, at 1 MPa it is about 10 times larger than the actual rate.

Worked exampleChecking a first-stage separator for gas

Given (assumed data)

Gas: 250,000 m³/d at 0.101325 MPa and 20 °C, specific gravity relative to air $\gamma = 0.75$
Separator: horizontal, $D = 2.4$ m, the liquid fills half of the cross-section; 1.0 MPa (absolute), 35 °C
Liquid: oil with a density of 840 kg/m³ at separator conditions
Coefficient: $K = 0.12$ m/s, a mesh mist extractor (assumed)

Solution

The molar mass of the gas is $M = 0.75\cdot 28.97 = 21.73$ kg/kmol. The gas deviation factor (how far the gas is from an ideal gas) at 1.0 MPa and 35 °C is $Z = 0.9712$ (from a correlation for natural gas, as in the lesson on real gas).
Gas density: $\rho_{\text{g}} = \dfrac{pM}{ZRT} = \dfrac{1.0\cdot 10^6\cdot 21.73\cdot 10^{-3}}{0.9712\cdot 8.314\cdot 308.15} = 8.73$ kg/m³.
Actual rate: $Q = 250{,}000\cdot\dfrac{0.101325}{1.0}\cdot\dfrac{308.15}{293.15}\cdot 0.9712/86{,}400 = 0.299$ m³/s, which is 9.7 times smaller than the standard rate.
Allowed velocity: $v_{\max} = 0.12\sqrt{(840 - 8.73)/8.73} = 1.17$ m/s.
The cross-section area needed for the gas is $A = Q/v_{\max} = 0.299/1.17 = 0.256$ m². In the vessel the upper half of the cross-section is left for the gas: $\pi D^2/8 = 2.26$ m².
The actual gas velocity is $0.299/2.26 = 0.132$ m/s, which is 11% of the allowed one.

Answer: For gas the vessel is loaded only to 11%: the gas needs 0.256 m² of cross-section, and it has 2.26 m².

What it means. This often happens in oil separators: the size is set not by the gas but by the liquid, that is, by the time that gas bubbles and water droplets need to leave it (the next section). In gas separators and gas scrubbers before compressors it is the other way round: there is little liquid, and the gas sets the diameter. The margin for gas is useful with foam, slugs and a growing gas–oil ratio. But the coefficient $K$ is only an estimate: for foaming oil we reduce it, sometimes by half. The data in this example are assumed for illustration.

Retention time and the separation of oil and water

The liquid must stay in the separator long enough for the gas bubbles to leave it. In a three-phase vessel it must also stay long enough for the water droplets to settle through the oil layer and for the oil droplets to rise out of the water. The measure of this is the retention time: the volume that a phase occupies in the vessel divided by its flow rate.

Retention time of the liquid

t = \frac{V}{q}

where:

$t$ average retention time of the phase in the vessel, min
$V$ volume occupied by the phase in the working part of the vessel, m³
$q$ flow rate of the phase at the conditions of the vessel, m³/min

Units: any consistent units

Holds when

The flow goes through the whole volume evenly. Dead zones (places where the liquid hardly moves) and short paths of the flow (where part of the liquid runs straight to the outlet) reduce the real time, which is why the inlet and distribution devices are so important.
Guides from industry practice: in two-phase separators from 1 to 3–4 min; in three-phase separators 3–10 min for light oil and 10–30 min or more for heavy and cold oil. For a particular oil, a laboratory settling test of a sample gives a better value.

For a given retention time, the volume of the vessel is proportional to the liquid rate. So the rise of the water cut over the life of the field needs ever larger settling volumes (loads on the facilities).

Share of the cross-section of a horizontal vessel that the liquid occupies

\frac{A_{\text{L}}}{A} = \frac{\theta - \sin\theta}{2\pi},\qquad \theta = 2\arccos\!\left(1 - \frac{2h}{D}\right)

where:

$A_{\text{L}}$ cross-section area occupied by the liquid up to the level $h$ (a circular segment), m²
$A$ area of the whole cross-section, $\pi D^2/4$ , m²
$h$ level of the liquid (or of the interface) above the bottom, m
$D$ inner diameter of the vessel, m
$\theta$ central angle of the segment, rad

Units: geometry; angles in radians

Holds when

The cylindrical part of the vessel. The volume of the heads (the rounded ends of the vessel) is counted separately or included in the margin.

At $h = D/2$ the share is 1/2. Near the bottom the area grows slowly, and near the middle it grows fast. This formula is used to find the oil–water interface level from the phase volumes, and the other way round.

For a water droplet to separate, it must fall through the oil layer of thickness $h_{\text{o}}$ to the interface while the oil passes through the vessel. Small droplets in a viscous liquid move slowly, and the resistance of the liquid is described by Stokes' law, the same law that explains the slow settling of silt in water.

Settling or rising velocity of a droplet (Stokes' law)

v = \frac{g\,d^2\,\Delta\rho}{18\,\mu_c}

where:

$v$ settling (or rising) velocity of the droplet relative to the liquid, m/s
$g$ acceleration due to gravity, 9.81, m/s²
$d$ droplet diameter, m
$\Delta\rho$ absolute value of the difference between the densities of the droplet and of the continuous phase, kg/m³
$\mu_c$ dynamic viscosity of the continuous phase, the one in which the droplet moves, Pa·s

Units: SI

Holds when

Slow (creeping) flow around the droplet: the Reynolds number of the droplet, $\mathrm{Re} = \rho_c v d/\mu_c$ (a number that shows how turbulent the flow around it is), is less than about 1. At larger $\mathrm{Re}$ the drag grows faster, and the formula overestimates the velocity.
The droplet is a rigid sphere, and the droplets are far from each other (at a high concentration they hinder each other and settle more slowly).
The flow in the vessel is calm. Turbulence and convection from uneven heating hinder settling.

The velocity grows as the square of the diameter: a droplet twice as large settles four times faster. So the main task of dehydration is to make the droplets larger. The viscosity is in the denominator, and water droplets in oil, whose viscosity is 10–20 times that of water, settle that many times more slowly than the same oil droplets rise in water.

In other unit systems (1)

Oilfield units

v\,[\text{ft/s}] = 1.79\cdot 10^{-6}\,\frac{\Delta\gamma\;d^2\,[\text{$\mu$m}^2]}{\mu_c\,[\text{cP}]}

$d$ in micrometres, $\Delta\gamma$ is the difference of specific gravities (relative to water), $\mu_c$ in centipoise, $v$ in feet per second.

$1.79\cdot10^{-6} = 9.81\cdot 10^{-12}\cdot 1000/(18\cdot 10^{-3})/0.3048$ : micrometres to metres, specific gravity to kg/m³, cP to Pa·s, metres to feet.

CalculatedThe time that a water droplet needs to pass through the oil layer (0.59 m thick) in the separator of the example below, against its diameter. The curves are for 35 °C and for oil heated to 60 °C (Stokes' law). Droplets that need less time than the retention time of the oil (10 min) reach the water.

Worked exampleA three-phase separator: length, interface level and the smallest droplet that settles

Given (assumed data)

Vessel: the separator of the previous example, $D = 2.4$ m, liquid up to half of the cross-section
Rates at vessel conditions: oil 1500 m³/d, water 1000 m³/d
Retention time: 10 min each for the oil and for the water (from a laboratory settling test, assumed)
Properties at 35 °C: oil 840 kg/m³, 10 mPa·s; water 1080 kg/m³, 0.75 mPa·s
Properties at 60 °C: oil 825 kg/m³, 4.5 mPa·s; water 1070 kg/m³

Solution

Phase volumes: $V_{\text{o}} = 1500/1440\cdot 10 = 10.42$ m³, $V_{\text{w}} = 1000/1440\cdot 10 = 6.94$ m³ (a day has 1440 minutes).
The liquid area is half of the cross-section: $\pi\cdot 2.4^2/8 = 2.262$ m². The working length is $L = (10.42 + 6.94)/2.262 = 7.67$ m, that is $L/D = 3.2$ . With room for the inlet device and for the compartment behind the weir, the length of the vessel is about 9 m.
The water takes the share $6.94/(10.42 + 6.94) = 0.400$ of the liquid, that is $0.200$ of the cross-section. By the segment formula this corresponds to $h_{\text{w}}/D = 0.254$ . The interface level is $h_{\text{w}} = 0.61$ m, and the thickness of the oil layer is $h_{\text{o}} = 1.2 - 0.61 = 0.59$ m.
The smallest water droplet that has time to pass through the whole oil layer in 10 min (600 s) has the diameter $d = \sqrt{\dfrac{18\mu_{\text{o}} h_{\text{o}}}{g\,\Delta\rho\,t}} = \sqrt{\dfrac{18\cdot 0.010\cdot 0.59}{9.81\cdot 240\cdot 600}} = 274$ µm.
Check of the flow regime: the settling velocity of this droplet is $0.59/600 = 0.983$ mm/s, and $\mathrm{Re} = 840\cdot 0.983\cdot 10^{-3}\cdot 274\cdot 10^{-6}/0.010 = 0.023$ . This is much less than 1, so Stokes' law applies.
After heating to 60 °C: $d = \sqrt{\dfrac{18\cdot 0.0045\cdot 0.59}{9.81\cdot 245\cdot 600}} = 182$ µm.
The smallest oil droplet that rises through the water layer $h_{\text{w}} = 0.61$ m in 10 min at 35 °C has the diameter $d =\sqrt{\dfrac{18\cdot 0.00075\cdot 0.61}{9.81\cdot 240\cdot 600}} = 76$ µm.

Answer: The working length is 7.67 m at a diameter of 2.4 m. At 35 °C water droplets larger than 274 µm are sure to settle, and after heating to 60 °C those larger than 182 µm. Oil droplets larger than 76 µm rise out of the water.

What it means. Gravity settling removes free water and large droplets in minutes, but not an emulsion. In an emulsion the water droplets are a few or tens of micrometres in size, and at 35 °C a droplet of 30 µm would take about 14 h to pass through the oil layer instead of 10 min. Heating lowers the viscosity of the oil by more than half and reduces the limiting diameter (the smallest droplet that settles) by a third. But it does not solve the problem: we must make the droplets larger, with chemicals, coalescing plate packs and an electric field. Oil leaves water more easily than water leaves oil: the viscosity of water is 0.75 mPa·s against 10 mPa·s for the oil. The calculation is optimistic. A droplet that starts at the surface of the oil is the worst case, but an uneven flow and short flow paths reduce the real retention time. The data in this example are assumed for illustration.

Emulsions and dehydration

A water-in-oil emulsion consists of water droplets, from fractions of a micrometre to tens of micrometres in size, suspended in oil. Oil and water break each other into droplets where the flow speeds up sharply or is mixed: in chokes, valves and pumps (especially electric submersible pumps, Electric submersible pumps (ESP)) and where flows join. A small droplet alone would merge with its neighbours sooner or later. But oil contains natural emulsifiers (substances that help an emulsion to form): asphaltenes, resins, naphthenic acids and very fine particles of clay and wax. They gather on the surface of the droplets and form a strong film that keeps the droplets from merging. There is also an emulsion of the other type, with oil droplets in water. We deal with it when we clean the water (Produced water: composition, treatment and disposal).

Time and gravity. Free water and large droplets settle in settling tanks. Small droplets do not (previous section).
Heating. It lowers the viscosity of the oil, so droplets settle faster. It weakens the film on the droplets. It increases the density difference. But it costs fuel and drives light hydrocarbons out of the oil.
Demulsifiers are surface-active chemicals (chemicals that act on the surface of the droplets). They push the natural emulsifiers off the surface and let the droplets merge. The doses are small, usually tens of grams per tonne of oil. We choose the chemical by tests on samples, and too much of it can even make the emulsion more stable.
Electric field. In an electrostatic coalescer (an electric dehydrator), between electrodes at a voltage of tens of kilovolts, the water droplets become polarised (each gets a positive and a negative end), attract each other and merge. The unit works when there is already little water, a few per cent: at a high water cut, chains of droplets short-circuit the electrodes.
Coalescing packings: packs of plates or fibres on which droplets collect and grow larger.

It is best to add the demulsifier as early as possible, at the wellhead or before the booster pump station. While the production flows through the pipelines, the chemical has time to mix, and the emulsion is partly broken even before it reaches the unit (in-pipe demulsification). Heating, settling and the electric field can be combined in different ways. In the fields of the CIS (the former Soviet states) these are usually fired heaters or heat exchangers, settling tanks and electrostatic coalescers. In other countries a heater-treater is often installed: one vessel in which the production is heated, left to settle and passed through a coalescing section. Recirculating hot water and heating with the heat of the outgoing stable oil reduce fuel consumption.

Worked exampleWhy remove the water before heating

Given (assumed data)

Flow: oil and water from the separator of the previous examples: 1500 m³/d of oil (840 kg/m³) and 1000 m³/d of water (1080 kg/m³)
Heating: from 35 to 60 °C; heat capacity of oil 2.0 kJ/(kg·K), of produced water 3.9 kJ/(kg·K)
Water after the three-phase separator: the water cut of the flow at the oil outlet is 4% (assumed)
Fired heater: efficiency 0.80; lower heating value of the fuel gas (the heat released when it burns) 36 MJ/m³ (assumed)

Solution

Mass rate of the oil: $1500\cdot 840/86{,}400 = 14.58$ kg/s. Heating it takes $Q = \dot m\,c_p\,\Delta T = 14.58\cdot 2.0\cdot 25 = 729$ kW.
The water that comes with the oil after the separator: $1500\cdot 0.04/0.96 = 62.5$ m³/d, or $62.5\cdot 1080/86{,}400 = 0.781$ kg/s. Heating it takes $0.781\cdot 3.9\cdot 25 = 76.1$ kW. The total is $729 + 76.1 = 805$ kW.
Fuel: $805/(0.80\cdot 36{,}000)\cdot 86{,}400 = 2415$ m³/d of gas.
If all the water is heated: $1000\cdot 1080/86{,}400 = 12.5$ kg/s, and $12.5\cdot 3.9\cdot 25 = 1219$ kW. The total is $729 + 1219 = 1948$ kW and $5844$ m³/d of fuel.

Answer: After the free water is removed, the heater uses about 805 kW (2415 m³/d of gas). Without that, it uses 1948 kW, which is 2.4 times more.

What it means. Water is ballast with twice the heat capacity of oil, and the higher the water cut, the stronger this effect. So we remove the free water before heating, if possible already at the booster station, and inject it back into the reservoir nearby. The calculation ignores heat losses to the surroundings and heat recovery. In practice we recover part of the heat by warming the cold feed with the hot treated oil. The data in this example are assumed for illustration.

Desalting

The salts in oil are mostly chlorides of sodium, calcium and magnesium, dissolved in the droplets of produced water. At a refinery, when heated, magnesium and calcium chlorides break down and form hydrogen chloride, which together with water corrodes the equipment. Salts also deposit in heat exchangers and poison catalysts (the materials that speed up the refining reactions). So a main pipeline or a refinery accepts oil with not more than 100 mg of chloride salts per cubic decimetre (the table at the start of the lesson).

Content of chloride salts in oil

S = \varphi\,C_{\text{w}}

where:

$S$ content of chloride salts in the oil, mg/dm³
$\varphi$ volume fraction of residual water in the oil, fraction
$C_{\text{w}}$ salinity of this water: the concentration of chloride salts (expressed as NaCl), mg/dm³

Units: mg/dm³; with a small fraction of water, the volume of the mixture is almost equal to the volume of the oil

Holds when

The salts are only dissolved in the water. Salt crystals and salts on solid particles are not counted.

With a brine of 100–200 g/L, to stay within 100 mg/dm³ the water that is left must be not more than 0.05–0.1%, and this is hard to achieve by dehydration alone. It is easier to make the water less salty: mix the oil with fresh wash water and separate the water again.

Wash water for desalting

w = \frac{1}{E}\left(\frac{\varphi_1 C_{\text{w}}}{C_{\text{allow}}} - \varphi_1\right),\qquad C_{\text{allow}} = \frac{S_{\text{req}}}{\varphi_2}

where:

$w$ rate of fresh wash water as a fraction of the oil volume, fraction
$\varphi_1,\ \varphi_2$ fraction of residual water before washing and after the water is separated again, fraction
$C_{\text{w}}$ salinity of the produced water, mg/dm³
$C_{\text{allow}}$ highest allowed salinity of the water that stays in the oil after desalting, mg/dm³
$S_{\text{req}}$ required salt content of the oil, mg/dm³
$E$ mixing efficiency: the share of the wash water that really meets the droplets of produced water, from 0 to 1

Units: fractions and mg/dm³

Holds when

The wash water is fresh, and the droplets that have mixed merge and average their salinity.
The residual water after washing has the salinity of the mixture. In practice, small droplets of produced water that have not mixed are harder to remove from the oil than the mixed ones. So a model with a single fraction $E$ is a simplification that underestimates the wash water needed.
$E$ is chosen from the performance of a particular unit: the mixing valve, the pressure drop across it, the number of washing stages.

The wash water dilutes the produced water. For the salinity to fall by a factor of $C_{\text{w}}/C_{\text{allow}}$ , the water in the oil after mixing must occupy a volume that many times larger than the produced water did. Incomplete mixing needs more water. In two-stage units we use the water of the second stage to wash in the first stage, so less fresh water is needed.

Worked exampleHow much fresh water is needed to desalt the oil

Given (assumed data)

After dehydration: 0.4% of water by volume of the oil; salinity of the produced water 130 g/L (130,000 mg/dm³, expressed as NaCl)
After desalting: 0.2% of water remains in the oil (electrostatic coalescer)
Requirement: not more than 100 mg/dm³ of chloride salts
Mixing efficiency: $E = 0.5$ (assumed)

Solution

Without washing: $S = 0.004\cdot 130{,}000 = 520$ mg/dm³, which is 5.2 times the limit.
The allowed salinity of the residual water is $C_{\text{allow}} = 100/0.002 = 50{,}000$ mg/dm³.
After mixing, the water must occupy $0.004\cdot 130{,}000/50{,}000 = 0.0104$ of the oil volume. So the fresh water that really mixes must be $0.0104 - 0.004 = 0.0064$ .
Allowing for the mixing efficiency: $w = 0.0064/0.5 = 0.0128$ , or 12.8 L per 1 m³ of oil. For 1500 m³/d this is about 19 m³/d of fresh water.

Answer: About 1.28% of fresh water is needed by volume of the oil, which is 12.8 L per cubic metre.

What it means. This is a lower estimate. In practice, treatment units use wash water with a margin, usually a few per cent of the oil volume, and repeat the washing in a second stage. The formula shows the strongest levers: reduce the residual water after the vessel ( $\varphi_2$ in the denominator of $C_{\text{allow}}$ ) and improve the mixing. The wash water must be compatible with the produced water. Otherwise calcium and barium salts will precipitate (form solid crystals) in the vessels (Produced water: composition, treatment and disposal). The data in this example are assumed for illustration.

Stabilisation of oil

Oil from the last separation stage is in equilibrium with gas at the pressure and temperature of that stage: dissolved methane, ethane, propane and butanes stay in it. If the oil warms up, in a tank in the sun, in a tank car or in a tanker, they start to come out. Losses from evaporation grow, an explosive mixture forms above the oil, and pumps lose pressure (head) through cavitation (the formation of vapour bubbles in the pump). The measure of this tendency is the vapour pressure of the oil, and the regulation limits it to 66.7 kPa (500 mm Hg). Stabilisation lowers it.

It matters how this pressure is measured. The true vapour pressure (TVP) is the pressure at which the oil starts to boil at a given temperature, when there is an infinitely small amount of vapour. The standard methods work differently. At 37.8 °C (100 °F) they put the sample into a chamber with several times more vapour space than liquid. In the Reid method, the chamber contains air and its volume is four times the volume of the sample. In the expansion method, a sample without air is expanded to a set ratio of vapour to liquid, often 4 : 1. A significant part of the lightest components goes into the large vapour volume, the pressure falls, and the instrument shows less than the true value. So the same number, 66.7 kPa, means different requirements for the true pressure and for the measured one.

Hot separation. We heat the oil before the last stage (in treatment units, with the same fired heater that is used for dehydration) and lower the pressure almost to atmospheric. The light hydrocarbons go into the gas. This is simple, but part of the butanes and pentanes that could have stayed in the oil leave with them.
Stabiliser column. A distillation column (a tall tower in which a mixture is separated by boiling) with a reboiler (a heater) at the bottom. The oil flows down over trays against rising vapours, and the light components are separated from the heavy ones much more sharply than in a single flash (one-step evaporation at a fixed pressure). For the same vapour pressure, we get more stable oil. The top product is gas and a wide fraction of light hydrocarbons (natural gas liquids, NGL), which go to gas processing (Gas processing and compression).
Stripping of hydrogen sulfide. In the same column or in a separate column we blow a sulfur-free gas through the oil. The hydrogen sulfide passes into this gas, because its concentration in the oil is much higher than the concentration that would be in balance with clean gas. We bind the hydrogen sulfide and mercaptans that are left with chemical scavengers.

Worked exampleIs the oil from the stage separation example stable?

Given (assumed data)

Oil: the stock-tank oil (the oil after degassing) of the two-stage scheme of the lesson on separator calculations: the first stage at 1.2 MPa and 40 °C, then a tank at 0.101325 MPa and 20 °C
Calculation: the Peng–Robinson equation of state (a formula for the properties of fluid mixtures) with the same components and coefficients as in that lesson; equilibrium stages (stages in which gas and liquid reach balance)
Vapour pressure: the true one is the pressure at which boiling starts at 37.8 °C; “by the expansion method” is the pressure of a sample without air at 37.8 °C, expanded so that the volume of the chamber is 5 times the initial volume of the liquid (vapour to liquid 4 : 1)
Stabilisation option: the last stage (at the same atmospheric pressure) is run at a raised temperature

Solution

At 20 °C the oil in the tank boils at 0.101 MPa, which is how we got it. At 37.8 °C its true vapour pressure is 136.4 kPa. That is higher than atmospheric, and in a warm tank it will “breathe” gas.
By the expansion method it is 67.1 kPa against the limit of 66.7 kPa. The oil is at the edge of the requirement, with no margin. The difference from the true pressure is a factor of 2.0: almost all the dissolved methane, more than half of the ethane and almost a third of the propane go into the vapour space of the chamber.
The last stage is run at 40 °C. The true vapour pressure of the oil at 37.8 °C falls to 97.7 kPa, and by the expansion method to 49.2 kPa. The margin to the limit is 26%.
The price: from 1 kmol of reservoir oil we get 0.4065 kmol of stable oil instead of 0.4193, which is 1.25% less by volume. More than half of the molecules that left are propane and butanes. The rest are ethane, pentanes and a little of heavier ones. All of them go into the gas of the last stage.

Answer: Without heating, the oil is right at the limit: 67.1 kPa by the expansion method against the limit of 66.7 kPa, that is, formally a little above it. A hot stage at 40 °C gives 49.2 kPa, at the price of 1.25% of the oil volume.

What it means. Stabilisation trades oil volume for the oil's fitness for storage and transport. The lost components are not wasted: if we compress and cool the gas of the last stage, a wide fraction of light hydrocarbons condenses from it. A stabiliser column separates them more sharply than a hot stage: for the same vapour pressure, more pentanes stay in the oil. The calculation is idealised: the stages are equilibrium stages, and the equation of state and the different standard measurement methods give differences of a few kPa. So for a real scheme we check the result by measuring the vapour pressure of the stable oil. The data in this example are assumed for illustration.

Stabilisation by a hot last stage: the oil of the example at 37.8 °C

Stage temperature, °C: 20
True vapour pressure, kPa: 136.4
By the expansion method, kPa: 67.1
Oil yield by volume, %: 100.00

Stage temperature, °C: 30
True vapour pressure, kPa: 115.4
By the expansion method, kPa: 57.6
Oil yield by volume, %: 99.41

Stage temperature, °C: 40
True vapour pressure, kPa: 97.7
By the expansion method, kPa: 49.2
Oil yield by volume, %: 98.75

Stage temperature, °C: 50
True vapour pressure, kPa: 82.8
By the expansion method, kPa: 41.8
Oil yield by volume, %: 98.03

Stage temperature, °C: 60
True vapour pressure, kPa: 70.3
By the expansion method, kPa: 35.4
Oil yield by volume, %: 97.22

Stage temperature, °C	True vapour pressure, kPa	By the expansion method, kPa	Oil yield by volume, %
20	136.4	67.1	100.00
30	115.4	57.6	99.41
40	97.7	49.2	98.75
50	82.8	41.8	98.03
60	70.3	35.4	97.22

Calculated with the Peng–Robinson equation for the assumed oil of the lesson on separator calculations. The yield is the volume of stable oil at 20 °C relative to the option with a tank at 20 °C. The data are assumed for illustration.

CalculatedVapour pressure of the stable oil at 37.8 °C against the temperature of the last stage: true, and by the expansion method (4 : 1), calculated with the Peng–Robinson equation. The dashed line is the limit of 66.7 kPa.

More detail:How the pressure “by the expansion method” is calculated

We take 1 kmol of stable oil at 37.8 °C, find its initial molar liquid volume $v_0$ , and look for the pressure $p$ at which, after a one-step evaporation calculation (flash calculation), the vapour and the remaining liquid together occupy $5v_0$ : $V\,v_{\text{g}}(p) + (1 - V)\,v_{\text{L}}(p) = 5v_0$ , where $V$ is the mole fraction of vapour. As the pressure falls there is more vapour, so we find the root (the solution) by bisection, which halves the interval again and again. The air in the chamber is not counted: this is how the expansion method is built. In the Reid method there is air in the chamber, and the result depends on it too.

For the oil of the example, almost all the dissolved methane, more than half of the ethane and almost a third of the propane go into the vapour phase of the chamber. So the pressure by the expansion method (67.1 kPa) is about half the true one (136.4 kPa). The difference is larger when methane and ethane create a bigger share of the vapour pressure. A hot stage lowers the vapour pressure almost without changing this share, so in the table above the ratio of the two pressures stays near two. But if methane and ethane are driven off selectively, as in a stabiliser column, and the vapour pressure is created mainly by propane and butanes, the two pressures come noticeably closer.

Storage, metering and hand-over

The treated oil collects in tanks, usually vertical steel tanks. A tank farm (a group of storage tanks) smooths out the swings in production and pumping out, and gives time for analyses before hand-over. When a tank is filled and emptied, and with daily changes of temperature, its gas space “breathes” and pushes vapours out into the atmosphere. The higher the vapour pressure of the oil, the greater the losses. Stabilisation reduces them. So do pontoons and floating roofs that lie directly on the oil, and vapour recovery units that collect gas from the tanks and send it to a compressor.

The oil metering system measures the quantity and quality of the oil that is handed over for sale (a custody-transfer system; similar systems in the USA are called LACT units). It consists of filters; flowmeters, either Coriolis mass meters (which measure the mass of the flow directly) or turbine meters (which measure its volume) together with an inline densitometer (a density meter in the pipe); inline water meters; an automatic sampler; and a flow computer. From time to time we check turbine meters against a prover. A ball or a piston pushes a known volume of oil between two sensors, and the ratio of this volume to the meter reading gives the meter's correction factor (the meter factor). In the laboratory we analyse the samples for the mass fraction of water, the concentration of chloride salts and the solid impurities. These are what we deduct from the mass of the oil.

In Kazakhstan, the oil produced in a subsoil area is counted separately for each well, in tonnes. The measuring instruments must have a valid verification certificate (proof that they have been checked against a reference) and be entered in the register of the state system for ensuring the uniformity of measurements. Direct methods are allowed (scales and mass meters) and indirect ones (the volume-mass and the hydrostatic method). The net mass equals the gross mass less the ballast, that is, water, chloride salts and solid impurities (the Rules for measuring and weighing oil produced by a subsoil user in a subsoil area, approved by Order No. 163 of the Minister of Energy of 5 May 2018). The data of the oil and gas metering devices are sent to the accounting information system of the Ministry of Energy (reporting, accounting and inspections).

Net mass of oil

m_{\text{net}} = m_{\text{gross}}\left(1 - \frac{W_{\text{w}} + W_{\text{cs}} + W_{\text{imp}}}{100}\right),\qquad W_{\text{cs}} = \frac{0.1\,\varphi_{\text{cs}}}{\rho_{\text{o}}}

where:

$m_{\text{net}},\ m_{\text{gross}}$ net mass and gross mass of the oil, t
$W_{\text{w}},\ W_{\text{imp}}$ mass fractions of water and of solid impurities, %
$W_{\text{cs}}$ mass fraction of chloride salts, %
$\varphi_{\text{cs}}$ concentration of chloride salts, mg/dm³
$\rho_{\text{o}}$ density of the oil at the conditions to which the salt concentration refers, kg/m³

Units: t, %, mg/dm³, kg/m³

Holds when

The gross mass is measured by a mass meter, or worked out from the volume and the density at the same conditions.
The factor 0.1 converts units: mg/dm³ divided by kg/m³ (that is, by g/dm³) gives thousandths of a fraction, which is tenths of a per cent.

The buyer pays for the net mass. Every tenth of a per cent of water in the oil handed over is one tonne in every thousand tonnes that will not be paid for.

Worked exampleNet mass for one day

Given (assumed data)

Mass meter: gross mass for the day 1236.8 t
Laboratory: water 0.35% (by mass), chloride salts 92 mg/dm³, solid impurities 0.012% (by mass), oil density 852.4 kg/m³

Solution

Mass fraction of salts: $W_{\text{cs}} = 0.1\cdot 92/852.4 = 0.0108$ %.
Ballast: $0.35 + 0.0108 + 0.012 = 0.3728$ %.
Net mass: $1236.8\cdot(1 - 0.3728/100) = 1232.19$ t; the ballast is 4.61 t.

Answer: 1232.19 t of oil are paid for. 4.61 t of the gross mass are water, salts and impurities.

What it means. Almost all of the ballast is water. So the quality of dehydration shows directly in money: lowering the water from 0.35 to 0.2% at the same gross mass would increase the net mass by almost 1.9 t per day. Chloride salts are negligible by mass, but the regulation limits them to protect refinery equipment. The data in this example are assumed for illustration.

Terms and symbols

Oil treatment unit: A set of vessels in which the oil is brought to the requirements for sales oil by separation, dehydration, desalting and stabilisation.
Three-phase separator: A vessel in which gas is separated from the production, and oil and water are separated from each other.
Souders–Brown equation: The allowed gas velocity in a separator: the coefficient $K$ multiplied by the square root of the density difference divided by the gas density.
Retention time: The volume of a phase in the vessel divided by its flow rate. It decides whether the gas has time to come out and the droplets to settle.
Stokes' law: The settling velocity of a small droplet is proportional to the square of its diameter and to the density difference, and inversely proportional to the viscosity of the continuous phase.
Water-in-oil emulsion: Small droplets of water in oil, protected by a film of natural emulsifiers.
Demulsifier: A surface-active chemical that pushes natural emulsifiers off the surface of the droplets.
Electrostatic coalescer: A vessel in which water droplets merge in a high-voltage electric field.
Desalting: Washing the oil with fresh water and separating the water again, to lower the salt content.
Oil stabilisation: Removal of dissolved light hydrocarbons to lower the vapour pressure.
Oil metering system: The system that measures the quantity and quality of the oil handed over for sale: the custody-transfer metering unit.
Net mass: The gross mass of the oil less water, chloride salts and solid impurities.

Common mistakes and misconceptions

Wrong

“A separator that is enough for gas will cope with the liquid too.”

In fact

In oil separators the retention time of the liquid and the separation of water usually set the sizes. For gas, such vessels are often loaded only to 10–30%.

Wrong

“The more the emulsion is heated, the better.”

In fact

Heating speeds up the separation, but it costs fuel, and extra degrees drive off light hydrocarbons and reduce the oil yield. We choose the temperature together with the chemical and the settling time, and we remove the free water before heating.

Wrong

“The oil is dehydrated to 0.5%, so the salts are within the limit too.”

In fact

The salts are dissolved in the residual water: 0.5% of brine with 150 g/L gives 750 mg/dm³, which is 7.5 times the limit for a main pipeline. We need to wash the oil with fresh water.

Wrong

“The oil does not boil in the tank, so its vapour pressure is within the limit.”

In fact

Oil in a tank boils exactly at atmospheric pressure and its own temperature. At 37.8 °C its vapour pressure is higher. We check the limit with a standard method and, if needed, stabilise the oil.

In depth: advanced detail

Where the Souders–Brown equation comes from

A droplet of diameter $d$ settles at a constant velocity when its weight, less the buoyancy force, equals the drag force of the gas (the equation on the left of the arrow below; $C_D$ is the drag coefficient). From this, the settling velocity is:

Settling velocity of a droplet for any flow around it

\frac{\pi d^3}{6}\,(\rho_{\text{L}} - \rho_{\text{g}})\,g = C_D\,\frac{\pi d^2}{4}\cdot\frac{\rho_{\text{g}} v_t^2}{2}\quad\Rightarrow\quad v_t = \sqrt{\frac{4\,g\,d}{3\,C_D}}\cdot\sqrt{\frac{\rho_{\text{L}} - \rho_{\text{g}}}{\rho_{\text{g}}}}

where:

$v_t$ steady settling velocity of the droplet in the gas, m/s
$C_D$ drag coefficient of a sphere; it depends on the Reynolds number, —
$d$ droplet diameter, m
$g$ acceleration due to gravity, m/s²
$\rho_{\text{L}},\ \rho_{\text{g}}$ densities of the droplet and of the gas, kg/m³

Units: SI

Holds when

A single spherical droplet; the gas is at rest on average. With creeping flow $C_D = 24/\mathrm{Re}$ , and the formula turns into Stokes' law.

The first square root is the coefficient $K$ of the Souders–Brown equation. It grows with the diameter of the droplet that has to be settled and falls as the drag grows. Droplets of 100–150 µm in high-pressure gas are no longer surrounded by a creeping flow, and $C_D$ depends on the Reynolds number. So instead of calculating for every droplet, we use values of $K$ found by experience for the type of vessel and mist extractor.

In a horizontal vessel the droplet must drop through the height of the gas space $h_{\text{g}}$ while the gas passes the working length $L$ : $h_{\text{g}}/v_t \le L/v_{\text{g}}$ . So the allowed gas velocity is $v_{\text{g}} \le v_t\,L/h_{\text{g}}$ , which is $L/h_{\text{g}}$ times the settling velocity. That is why $K$ is larger in horizontal vessels, and a long vessel with the same cross-section area passes more gas, until the velocity becomes so high that the gas starts to tear liquid off the surface.

Connections to other topics

Separator calculations and optimum separator pressure — how much gas and oil the separation stages give and how their pressures are chosen.
Field development concepts: surface facilities — where separators, settling tanks and storage tanks stand in the facilities of a field.
Gas processing and compression — what is done with the separation gas and with the light hydrocarbons from stabilisation.
Produced water: composition, treatment and disposal — cleaning of the separated water before injection or disposal.
Oil viscosity and interfacial tension — how the viscosity of oil depends on temperature, the main parameter of droplet settling.
Retrograde condensation, wax, asphaltenes and high-pressure fluids — wax and asphaltenes, the causes of deposits and stable emulsions.
State regulation of field development — accounting of production, metering devices and the reporting of the subsoil user.

Check yourself

1
The oil was heated, and its viscosity fell from 10 to 4.5 mPa·s, while the density difference between water and oil hardly changed. How many times faster do the same water droplets settle?
1. A.About 2.2 times
2. B.About 4.9 times
3. C.About 1.5 times
4. D.The velocity will not change: it depends on the droplet size
Show the answer
Correct answer: A. About 2.2 times
By Stokes' law the settling velocity is inversely proportional to the viscosity of the continuous phase: $10/4.5 \approx 2.2$ . The square dependence applies to the droplet diameter: to speed up settling 2.2 times without heating, the droplets would have to be made $\sqrt{2.2} \approx 1.5$ times larger.
2
The pressure in a separator was raised from 1 to 4 MPa. The gas density grew about four times, and the oil density hardly changed. How will the allowed actual gas velocity and the gas capacity in standard cubic metres change?
1. A.The velocity will fall by half, and the capacity in standard cubic metres will grow about two times
2. B.The velocity will not change, and the capacity will grow four times
3. C.The velocity and the capacity will both fall by half
4. D.The velocity will fall four times, and the capacity will not change
Show the answer
Correct answer: A. The velocity will fall by half, and the capacity in standard cubic metres will grow about two times
By the Souders–Brown equation, $v_{\max} \propto 1/\sqrt{\rho_{\text{g}}}$ (for $\rho_{\text{g}} \ll \rho_{\text{L}}$ ): four times the density gives half the velocity. But a standard cubic metre of gas at 4 MPa takes about four times less volume, so through the same cross-section about $4/2 = 2$ times more gas will pass in standard cubic metres (ignoring the change of $Z$ and the lowering of $K$ at high pressure).
3
After the electrostatic coalescer, 0.3% of water with a salinity of 150 g/L is left in the oil. What is the content of chloride salts, and is the oil fit for delivery to a main pipeline?
Show the answer
$S = 0.003\cdot 150{,}000 = 450$ mg/dm³, which is 4.5 times the limit of 100 mg/dm³. Washing with fresh water is needed. For example, to meet the limit with the same fraction of residual water, the salinity of this water must fall to $100/0.003 \approx 33{,}000$ mg/dm³, that is, by a factor of 4.5.
4
Why do oil droplets leave the water in a three-phase separator more easily than water droplets leave the oil, although the density difference is the same?
Show the answer
Stokes' law contains the viscosity of the continuous phase. An oil droplet rises in water with a viscosity of about 0.7–1 mPa·s, while a water droplet settles in oil with a viscosity of a few or tens of mPa·s. At the same diameter an oil droplet in water moves 10–20 times faster, so in the same time much smaller droplets leave the water (76 and 274 µm in the example of the lesson).
5
Oil has left a tank in which at 20 °C and atmospheric pressure it was in equilibrium with gas. Why can its vapour pressure at 37.8 °C exceed atmospheric pressure, and what do we do about it at the treatment unit?
Show the answer
In the tank the oil boils at 0.101 MPa and 20 °C: this is its boiling-start pressure at 20 °C. As the temperature rises, the vapour pressure of the dissolved light components grows, and at 37.8 °C it is higher than atmospheric (136 kPa in the example of the lesson). We stabilise the oil: we run the last stage at a higher temperature, or we install a stabiliser column, which drives propane, butanes and the remains of methane and ethane into the gas.
6
The gross mass of the oil handed over is 500 t, with water 0.5% (by mass), chloride salts 100 mg/dm³ at a density of 850 kg/m³, and solid impurities 0.02% (by mass). What is the net mass?
1. A.495.0 t
2. B.497.3 t
3. C.500.0 t
4. D.499.9 t
Show the answer
Correct answer: B. 497.3 t
$W_{\text{cs}} = 0.1\cdot 100/850 = 0.0118$ %; ballast $0.5 + 0.0118 + 0.02 = 0.5318$ %; net mass $500\cdot(1 - 0.005318) = 497.34$ t. Almost all of the ballast is water.

Updated: 18 October 2026