Oil-based muds · MunaiHub

Quick refresher

An OBM is most often an invert emulsion. The continuous phase is oil (diesel, mineral or synthetic), and in it are droplets of calcium chloride brine a few micrometres across. Emulsifiers hold the droplets, and wetting agents make the solid particles oil-wet (hydrophobic: they repel water).
The oil–water ratio (OWR) is found from the retort (an instrument that boils the liquids out of a sample): $V_o/(V_o + V_w)$ , usually from 70/30 to 90/10. To raise it, add base oil. To lower it, add brine. The volumes turn out large, and the density is then restored with weighting material.
The water-phase salinity is found from the chlorides of the whole mud and the share of water from the retort: there is 1.565 times more calcium chloride than chloride ions. Through it we set the water activity of the mud, equal to the activity of the clays or a little lower.
Electrical stability is the breakdown voltage of a sample between electrodes, an indicator of how stable the emulsion is. If it falls together with water in the filtrate (the liquid that leaves the mud) at high pressure and high temperature and with a “grainy” look of the mud, the emulsion is broken or the solids are wetted by water.
Excess lime, $3.7\,P_{om}$ kg/m³, is a reserve of alkali for the emulsifiers and a protection against carbon dioxide and hydrogen sulfide.
Gas dissolves in the base oil. An inflow gives a smaller gain in volume, and the gas does not expand on the way up. It comes out of solution only where the pressure falls below the saturation pressure: in the upper part of the hole, fast and in a large volume.

Key relationships

\text{OWR} = \dfrac{V_o}{V_o + V_w}

oil–water ratio by volumes from the retort

w = \dfrac{1.565\,[\mathrm{Cl^-}]}{1.565\,[\mathrm{Cl^-}] + 10^6 f_w}

mass fraction of CaCl₂ in the water phase; chlorides in mg/L of mud

f_b = \dfrac{f_w}{\rho_b\,(1 - w)}

share of brine from the share of water from the retort

\Delta V_o = V\left(\dfrac{f_w R}{1 - R} - f_o\right)

base oil that raises the ratio to $R$

p_{\text{sat}} = \dfrac{G}{f_o\,k}

saturation pressure of the mud with gas (linear solubility model)

Common pitfalls

Confusing the water from the retort with brine: the salt stays in the retort cup and goes into the “solids”. So the shares of brine and solids are corrected for the salinity.
Judging the emulsion by one value of electrical stability: it depends on the share of water, on the solids and on the temperature. The trend and the combination with other tests matter.
Raising the salinity “with a margin”: too low an activity dries the clays out, and calcium chloride beyond saturation crystallises.
Expecting a gas kick in an OBM to give the same gain in volume and the same expansion of gas on the way up as in a water-based mud.
Adding base oil or water without recalculating the density, the salinity and the concentration of emulsifiers: in an OBM everything is linked.

After this lesson you can

describe the structure of an invert emulsion and the roles of its components: base fluid, brine, emulsifiers and wetting agents, organophilic clay, lime, fluid-loss reducers and weighting material
calculate the oil–water ratio, the water-phase salinity, and the shares of brine and solids from the retort and the chlorides of the mud
calculate the amounts of base fluid, brine and weighting material needed to change the oil–water ratio while keeping the density
explain how the water activity of the mud (a measure of how strongly the water is held; salt lowers it) is balanced with the activity of the clays, and calculate the calcium chloride that must be added
explain the meaning of electrical stability, of fluid loss at high pressure and high temperature, and of excess lime
compare the advantages and limitations of oil-based muds, gas solubility and the detection of kicks included, and describe general ways of handling cuttings

How an invert emulsion is built

An emulsion is a mixture of two liquids that do not mix, one of which is broken into droplets in the other. In the early “emulsion” water-based muds, droplets of oil were spread in water. In an oil-based mud it is the other way round: water (more exactly, brine) is spread in oil. This is why the emulsion is called invert, that is, turned around. The continuous phase decides what the walls of the well, the cuttings, the metal and the logging tools “see”. In an invert emulsion this is oil.

The brine droplets tend to merge by themselves: this reduces the surface between the liquids and its energy. They are kept apart by an emulsifier. This is a substance whose molecule is polar at one end and reaches for the water, and at the other end is a long hydrocarbon chain that dissolves in oil. Such molecules line up in a film on the surface of the droplets, lower the interfacial tension (the pull of the surface between the two liquids) and mechanically keep the droplets apart. The smaller and more even the droplets, the more stable the emulsion. So fresh mud is stirred hard, and its droplets are a few micrometres across.

SchematicStructure of an invert emulsion (a schematic, not to scale). The continuous phase is oil. In it are droplets of calcium chloride brine, each in a shell of emulsifier molecules (polar “heads” in the water, hydrocarbon “tails” in the oil), particles of barite (the usual weighting material) coated with a wetting agent, and plates of organophilic clay. At the bottom left is a particle wetted by water: it sticks to the droplets, and they clump together.

In the emulsion the brine droplets behave like small solid particles: the more of them there are, the higher the viscosity and the gel strength of the mud, and the easier it is for the droplets to collide and merge. So the share of water is one of the main “levers” on the properties. Adding base oil thins the mud and makes the emulsion more stable. Adding brine thickens it. The droplets also work for the mud: together with the organophilic clay they keep the weighting material from settling, and when they lodge in the pores of the mudcake (the layer of solids on the wall of the hole) they reduce the filtration.

The second concern is the wettability of the solids (Wettability and interfacial phenomena). Barite and drilled rock are hydrophilic (water-loving) by nature. When a hydrophilic particle gets into oil, it attracts brine droplets, they stick together around it in clumps, the clumps settle, and the emulsion loses water. To prevent this, wetting agents are added to the mud: surface-active substances that coat the particles and make them hydrophobic. Several signs show it at once: the electrical stability falls, the barite begins to settle, the mud turns dull and “grainy”, and sticky cuttings block the screens of the shale shakers (the vibrating screens; see the table in the section on tests).

Classes of oil-based muds

Class: All-oil muds
Water in the mud: up to a few per cent, accidental
Features: the water is a contaminant; they were used for drilling into formations and for coring

Class: Invert emulsions
Water in the mud: from 5–10 to 40–50% of the liquid phase
Features: the main class; the oil–water ratio is chosen for the density and the temperature

Class: Synthetic-based muds (SBM)
Water in the mud: as in invert emulsions
Features: the base is olefins, paraffins or esters; lower toxicity, higher price

Class	Water in the mud	Features
All-oil muds	up to a few per cent, accidental	the water is a contaminant; they were used for drilling into formations and for coring
Invert emulsions	from 5–10 to 40–50% of the liquid phase	the main class; the oil–water ratio is chosen for the density and the temperature
Synthetic-based muds (SBM)	as in invert emulsions	the base is olefins, paraffins or esters; lower toxicity, higher price

The boundaries of the classes differ a little between sources and suppliers. SBM stands for synthetic-based mud.

Composition: base fluid, brine, emulsifiers and additives

Components of an invert-emulsion mud

Component: Base fluid
Examples: diesel fuel; mineral oils; olefins, paraffins, esters
What for: the continuous phase; toxicity, viscosity, fire safety and price depend on it

Component: Brine
Examples: calcium chloride, less often sodium chloride
What for: the internal phase; the salt sets the water activity

Component: Emulsifiers
Examples: fatty-acid soaps, polyamides and other surfactants
What for: hold the brine droplets

Component: Wetting agents
Examples: surface-active substances
What for: make barite and cuttings hydrophobic

Component: Organophilic clay, rheology modifiers
Examples: bentonite or hectorite treated with amines; fatty acids, polymers
What for: viscosity and gel strength in the oil

Component: Lime
Examples: calcium hydroxide
What for: activates soap emulsifiers; a reserve of alkali against CO₂ and H₂S

Component: Fluid-loss reducers
Examples: asphalt and gilsonite (a natural solid asphalt), lignite treated with amines, polymers
What for: a thin cake of low permeability

Component: Weighting material
Examples: barite, less often hematite or calcium carbonate
What for: density

Component	Examples	What for
Base fluid	diesel fuel; mineral oils; olefins, paraffins, esters	the continuous phase; toxicity, viscosity, fire safety and price depend on it
Brine	calcium chloride, less often sodium chloride	the internal phase; the salt sets the water activity
Emulsifiers	fatty-acid soaps, polyamides and other surfactants	hold the brine droplets
Wetting agents	surface-active substances	make barite and cuttings hydrophobic
Organophilic clay, rheology modifiers	bentonite or hectorite treated with amines; fatty acids, polymers	viscosity and gel strength in the oil
Lime	calcium hydroxide	activates soap emulsifiers; a reserve of alkali against CO₂ and H₂S
Fluid-loss reducers	asphalt and gilsonite (a natural solid asphalt), lignite treated with amines, polymers	a thin cake of low permeability
Weighting material	barite, less often hematite or calcium carbonate	density

Suppliers use their own trade names; the table gives classes of substances. A surfactant is a surface-active substance.

Base fluid. Diesel fuel is cheap and easy to get, but it holds a lot of aromatic hydrocarbons. They are toxic, they smell, they irritate the skin and they dissolve rubber. Mineral oils with little aromatic content are less toxic. Synthetic bases (linear and internal olefins, paraffins, esters) are even less toxic and break down better in nature, but they cost more. At high temperature and in a strongly alkaline medium, esters hydrolyse (are broken down by water), so little lime is kept in such muds. The base is chosen by four properties. The first is viscosity: the lower it is, the lower the plastic viscosity of the mud. The second is the flash point, the temperature at which its vapour can catch fire (fire safety). The third is the pour point, the temperature at which it stops flowing. It matters for work in the cold and in cold marine drilling risers, the large pipes between the seabed and the rig. The fourth is the aniline point.

Brine is almost always made with calcium chloride. With it we can get both a moderate and a very low water activity (see the section on activity), while sodium chloride cannot give an activity below 0.75, even at saturation. The concentration is set by the water-phase salinity, usually 20–35% calcium chloride by mass. Organophilic clay is bentonite in which the exchangeable cations are replaced by large organic cations (amines). Such a clay does not swell in water, but it swells in oil and builds a structure. Lime is needed by soap emulsifiers: with it, fatty acids form calcium soaps right in the mud. Its undissolved excess is a buffer against acid gases.

The retort, the oil–water ratio and the water-phase salinity

A sample of OBM is distilled in the same retort as a water-based mud (solids), and we get the volumes of oil and water. From them we calculate the oil–water ratio (OWR): the share of base oil in the liquid phase. It is written as a pair of numbers, for example 80/20.

Oil–water ratio

\text{OWR} = \frac{V_o}{V_o + V_w} = \frac{f_o}{f_o + f_w}

where:

$V_o,\ V_w$ volumes of base oil and water collected in the retort, cm³
$f_o,\ f_w$ shares of base oil and water in the volume of the mud, fraction

Units: dimensionless; in per cent, multiply by 100

Holds when

By established practice, the ratio is calculated from the water from the retort, not from the brine. Some programmes set the ratio “base oil — brine”, and then the water is recalculated into brine (the formulas are below).
No formation oil or condensate gets into the retort. Otherwise it is counted as base oil.

A high ratio (85/15–95/5) means a stable emulsion, low viscosity and little sensitivity to temperature, but it costs more and holds barite worse without extra thickeners. A low ratio (60/40–70/30) is cheaper, and the viscosity and the gel strength are higher. For heavy muds the ratio is raised: barite takes up volume, and there is less room for brine droplets.

Chlorides in an OBM are titrated (measured by adding a reagent drop by drop) not in the filtrate, because it is oil, but in the whole sample of mud. The sample is diluted with a solvent, water is added, and it is titrated with silver nitrate. The result is milligrams of chloride ions per litre of mud. If the brine holds only calcium chloride, its mass is $110.98/(2\cdot 35.45) = 1.565$ times the mass of the chloride ions, and the mass of water comes from the retort.

Water-phase salinity from the chlorides of the mud

w = \frac{1.565\,[\mathrm{Cl^-}]}{1.565\,[\mathrm{Cl^-}] + 10^6\,f_w}

where:

$w$ mass fraction of calcium chloride in the water phase (water-phase salinity), fraction
$[\mathrm{Cl^-}]$ chlorides of the whole mud, mg/L of mud
$f_w$ share of water from the retort, fraction
$10^6 f_w$ mass of water in a litre of mud at a water density of 1 g/cm³, mg

Units: mixed: chlorides in mg/L, mass of water in mg per litre of mud

Holds when

All the chlorine is in calcium chloride. If rock salt has got into the brine, we need the calcium of the mud too, to separate CaCl₂ from NaCl.
All the salt is dissolved. With oversaturation, undissolved crystals also give chlorides, and the salinity comes out too high.

In practice the salinity is also given in parts per million by mass (ppm): 20% is 200,000 ppm of CaCl₂. From it we find the density and the activity of the brine.

Share of brine from the water in the retort

f_b = \frac{f_w}{\rho_b\,(1 - w)}, \qquad f_s = 1 - f_o - f_b

where:

$f_b$ share of brine in the volume of the mud, fraction
$\rho_b$ density of the brine; for CaCl₂ about 1.08 at 10%, 1.18 at 20%, 1.28 at 30%, g/cm³
$f_s$ corrected share of solids, fraction

Units: densities in g/cm³, shares as fractions

Holds when

In a litre of brine there are $1000\rho_b(1 - w)$ grams of water. This is the same correction as for salty water-based muds (the retort in a salty mud), written through the mass fraction of salt.

Brine is heavier and takes up more volume than water. Without the correction, the dissolved salt goes into the “solids” of the retort, and the mud looks dirtier than it is. With a concentrated brine in an OBM the error reaches several per cent of the volume.

Worked exampleAnalysis of an OBM: ratio, salinity and solids

Given (assumed data)

Density (pressurised balance): 1.45 g/cm³
Retort, sample of 20 cm³: base oil 12.7 cm³, water 3.2 cm³
Chlorides of the mud: 2.55 cm³ of 0.282 N silver nitrate per 1 cm³ of mud (10 mg of Cl⁻ per cm³)
Density of the base oil: 0.80 g/cm³

Solution

Shares: $f_o = 12.7/20 = 0.635$ , $f_w = 3.2/20 = 0.160$ . The ratio is $0.635/(0.635 + 0.160) = 79.9$ %, that is about 80/20. Without the correction for salt, the solids are 20.5%.
Chlorides: $10{,}000\cdot 2.55 = 25{,}500$ mg/L. Calcium chloride: $1.565\cdot 25{,}500 \approx 39{,}900$ mg/L, that is 39.9 kg/m³. Water: 160 kg/m³. Water-phase salinity: $39.9/(39.9 + 160) = 20.0$ %.
A brine of 20% calcium chloride has a density of about 1.177 g/cm³. Its share is $f_b = 0.160/(1.177\cdot 0.800) = 0.1699$ . The solids are $f_s = 1 - 0.635 - 0.1699 = 0.1951$ .
Average density of the solids: $\dfrac{1.45 - 0.635\cdot 0.80 - 0.1699\cdot 1.177}{0.1951} = 3.803$ g/cm³.
Low-gravity solids: $19.51\cdot(4.2 - 3.803)/1.6 = 4.84$ % (126 kg/m³). Barite: $19.51 - 4.84 = 14.67$ % (616 kg/m³).

Answer: The oil–water ratio is about 80/20, the water-phase salinity is 20.0% calcium chloride, the brine takes up 17% of the volume, and the solids are 19.51%: 4.84% of clay and cuttings and 14.67% of barite.

What it means. This is a typical weighted mud: low-gravity solids are under five per cent, and the rest is barite. The correction for salt is small here, about one per cent of the volume. With a saltier brine and a larger share of water it grows. A salinity of 20% gives a water activity of about 0.84 (see the section on activity), and we compare it with the activity of the clays being drilled. The data in this example are assumed for illustration.

How to change the oil–water ratio

The ratio changes by itself: water comes with the cuttings, from the washing of equipment and from the formations, and base oil leaves with the cuttings and with the filtration. The programme sets limits of the ratio for the interval, and when the mud goes outside them, we add base oil or brine. The calculation is simple: base oil and water do not dissolve in each other, and their volumes add up.

How much base oil or water to add to get the ratio R

\Delta V_o = V\left(\frac{f_w R}{1 - R} - f_o\right), \qquad \Delta V_w = V\left(\frac{f_o\,(1 - R)}{R} - f_w\right)

where:

$\Delta V_o,\ \Delta V_w$ volume of base oil to add (to raise the ratio) or of water (to lower it), m³
$V$ volume of the mud, m³
$f_o,\ f_w$ shares of base oil and water from the retort, fraction
$R$ the required ratio: the share of base oil in the liquid phase, fraction

Units: volumes in m³, shares as fractions

Holds when

The volumes add up, and the base oil and the water in the mud are not lost during the treatment.
Water is added as brine of the same salinity. Otherwise the water activity changes. For $\Delta V_w$ cubic metres of water we need $\Delta V_w\,w/(1 - w)$ tonnes of salt.
The formulas follow from the condition $(V f_o + \Delta V_o)/(V f_o + \Delta V_o + V f_w) = R$ and a similar one for water.

The larger the difference between the required ratio and the present one, the larger the volume to add. Base oil and water are lighter than the mud, so after the addition the density falls, and it is restored with weighting material (the weighting-up formula).

Worked exampleBefore a hot interval: from 80/20 to 85/15

Given (assumed data)

Mud: the example above: 1.45 g/cm³, $f_o = 0.635$ , $f_w = 0.160$ ; 150 m³ in circulation
Required ratio: 85/15, the same density
Base oil and barite: 0.80 and 4.2 g/cm³

Solution

Base oil per 1 m³ of mud: $\Delta V_o = \dfrac{0.160\cdot 0.85}{0.15} - 0.635 = 0.2717$ m³. For the whole volume, 41 m³.
Density after adding the base oil: $\dfrac{1.45 + 0.2717\cdot 0.80}{1 + 0.2717} = 1.3111$ g/cm³.
Barite to bring back 1.45 g/cm³: $(1 + 0.2717)\cdot\dfrac{1.45 - 1.3111}{4.2 - 1.45} = 0.0642$ m³, or 0.270 t for each cubic metre of the original mud. For the whole volume, 40.5 t.
Volume of the mud: $1 + 0.2717 + 0.0642 = 1.336$ m³ per 1 m³, that is 200 m³ instead of 150.
Check: $\dfrac{0.635 + 0.2717}{0.635 + 0.2717 + 0.160} = 85.0$ %.

Answer: We need about 41 m³ of base oil and 40.5 t of barite. There will be 200 m³ of mud, a third more.

What it means. Five points of the ratio cost tens of cubic metres of base oil and tens of tonnes of barite, and the tanks must hold a third more on top. So the ratio is changed gradually: fresh mud with the needed ratio is added to the circulation, and the surplus is sent to the mud store. Together with the base oil, the emulsifiers and the lime are diluted, and their concentration is restored from tests. Lowering the ratio is cheaper: we add brine. But then the viscosity rises. The data in this example are assumed for illustration.

Water activity and the stability of clays

Why do brine droplets hidden in oil affect the clays at all? Oil, together with the film of emulsifier, hardly lets dissolved ions pass, but it lets water molecules pass. So it works as a semipermeable membrane, and a much better one than a clay in a water-based mud. Water passes through such a membrane to where its activity (water activity, $a_w$ ) is lower: from the less salty liquid into the more salty one (osmosis).

The pore water of clay rocks has its own activity. The denser the clay and the saltier its pore water, the lower the activity, usually from 0.7 to 0.95. If the water activity of the mud is higher, water goes into the clay, which swells and loses strength. If it is lower, the mud draws water out of the clay. This gives the principle of balanced activity: the water activity of the mud is made equal to the activity of the clays being drilled, or a little lower. The activity of the clay is measured on fresh cuttings: by the relative humidity of the air above the sample in a closed vessel (with an electric hygrometer). The mud is measured in the same way.

Osmotic pressure difference between the mud and the clay

\Delta\Pi = \frac{R\,T}{\bar V_w}\,\ln\frac{a_{\text{m}}}{a_{\text{cl}}}

where:

$\Delta\Pi$ osmotic difference; positive means that water tends to go from the mud into the clay, MPa
$a_{\text{m}},\ a_{\text{cl}}$ water activity of the mud and of the pore water of the clay, —
$R T/\bar V_w$ at 25 °C about 137 MPa ( $R$ is the gas constant, $T$ the absolute temperature, $\bar V_w$ the molar volume of water), MPa

Units: MPa; natural logarithm

Holds when

An ideal membrane. A real one also lets part of the ions pass, and the acting difference is smaller. But in an OBM the membrane is more efficient than in water-based muds.
This is the difference of the osmotic pressures from the lesson on water-based muds: $\Pi = -(RT/\bar V_w)\ln a_w$ for each liquid.

Even a difference of activity of a few hundredths gives megapascals, which is comparable to the overbalance on the formation (the excess of the mud pressure over the pore pressure). So the activity of the mud is kept under control like the density.

CalculatedThe calculated water activity of calcium chloride and sodium chloride brines at 25 °C, against the concentration. Sodium chloride does not lower the activity below 0.75 (saturation at 26.4%), and calcium chloride goes down to 0.39 at 40%. The points are the mud of the example (20%, 0.84) and the concentration at which the activity equals that of the clay, 0.80 (22.3%).

Worked exampleBalancing the mud with the clay

Given (assumed data)

Mud: the example of the analysis: 20.0% CaCl₂ in the water phase, 160 kg/m³ of water, 150 m³ in circulation
Activity of fresh cuttings of the clay: 0.80

Solution

The water activity of the mud at 20% calcium chloride (the graph, the advanced section): 0.836, higher than that of the clay.
The ideal osmotic difference: $137\cdot\ln(0.836/0.80) \approx 6$ MPa towards the clay: water tends to go from the mud into the clay.
An activity of 0.80 gives a brine of 22.3% calcium chloride (2.58 mol per kilogram of water).
Salt per kilogram of water: it was $20.0/80.0 = 0.250$ kg, and we need $22.3/77.7 = 0.2870$ kg. The addition is $0.0370\cdot 160 = 5.9$ kg per 1 m³ of mud, and for the whole volume 0.89 t of anhydrous calcium chloride.

Answer: The salinity is raised from 20 to 22.3%: about 0.89 t of calcium chloride per 150 m³ of mud.

What it means. The addition is small by mass, but it changes the direction of osmosis: the mud stops giving water to the clay. The salt is added gradually, checking the activity and the electrical stability: a concentrated brine noticeably changes both the density of the water phase and the behaviour of the emulsifiers. Commercial calcium chloride contains 77 to 98% of the main substance, and the amount is recalculated for the real purity. If the clay is much “drier” than the mud, the salinity is raised within the limits set by the programme, because too much dehydration can also weaken the rock. The data in this example are assumed for illustration.

Electrical stability and other tests of an OBM

The density, the rheology and the retort of an OBM are measured with the same instruments as for a water-based mud (Field testing of drilling fluids). The rheology is measured at an agreed temperature, usually 50 or 65 °C, because the viscosity of the base oil depends strongly on temperature. The rest of the set is specific to an OBM.

Tests that are specific to an OBM

Test: Electrical stability
How it is done: an alternating voltage is raised between electrodes in a sample at 50 °C until breakdown
What it shows: the stability of the emulsion and the wettability of the solids, from the trend

Test: HPHT fluid loss
How it is done: usually 150 °C and a pressure difference of 3.45 MPa; the volume is doubled, as for a water-based mud
What it shows: the filtrate must be only oil; water or emulsion in it means that the emulsion is broken

Test: Alkalinity of the mud $P_{om}$
How it is done: a sample in a solvent, titration with 0.1 N sulfuric acid to pH 8.3
What it shows: the excess lime

Test: Chlorides and calcium of the mud
How it is done: chlorides in the same sample with silver nitrate, calcium in a new sample with EDTA (a reagent that binds calcium)
What it shows: the salinity of the water phase and the composition of the salts

Test: Water activity
How it is done: an electric hygrometer above a sample of mud and above fresh cuttings
What it shows: how well the mud is balanced with the clays

Test	How it is done	What it shows
Electrical stability	an alternating voltage is raised between electrodes in a sample at 50 °C until breakdown	the stability of the emulsion and the wettability of the solids, from the trend
HPHT fluid loss	usually 150 °C and a pressure difference of 3.45 MPa; the volume is doubled, as for a water-based mud	the filtrate must be only oil; water or emulsion in it means that the emulsion is broken
Alkalinity of the mud $P_{om}$	a sample in a solvent, titration with 0.1 N sulfuric acid to pH 8.3	the excess lime
Chlorides and calcium of the mud	chlorides in the same sample with silver nitrate, calcium in a new sample with EDTA (a reagent that binds calcium)	the salinity of the water phase and the composition of the salts
Water activity	an electric hygrometer above a sample of mud and above fresh cuttings	how well the mud is balanced with the clays

The order of the tests is standardised (the international standard ISO 10414-2; in English-language practice API RP 13B-2). HPHT stands for high pressure, high temperature.

Electrical stability (ES). Oil does not conduct current, and brine does. A probe with two flat electrodes, separated by a gap of 1.55 mm, is lowered into a sample at 50 °C, and an alternating (sine-wave) voltage that rises smoothly is applied. While the brine droplets are separated by oil, almost no current flows. When the voltage lines the droplets up in chains between the electrodes, the emulsion “breaks down”, and the current rises sharply. The breakdown voltage in volts is the electrical stability.

The absolute value says little. The electrical stability is lower in a mud with more water. It depends on the content and the wettability of the solids, on the temperature and on the base oil. The trend and the combinations are what count: a fall in electrical stability at an unchanged composition, water in the HPHT filtrate, a “grainy” look of the mud, and barite settling.

Excess lime in an OBM

c_{\text{lime}} = 3.7\,P_{om}

where:

$c_{\text{lime}}$ excess of undissolved lime, kg/m³
$P_{om}$ alkalinity of the mud: cm³ of 0.1 N sulfuric acid per 1 cm³ of mud up to pH 8.3, cm³/cm³

Units: kg/m³; in oilfield units $1.295\,P_{om}$ lb/bbl

Holds when

All the alkali of the sample is lime (its equivalent mass is 37.05 g): $0.1$ eq/L $\times\ 37.05$ g/eq $= 3.7$ g/L.
The acid is five times stronger than in the test of a water-based mud (there it is 0.02 N and the coefficient is 0.742). This is why the coefficient is five times larger.

Lime is used up in forming soaps and in neutralising carbon dioxide and hydrogen sulfide ( $\mathrm{Ca(OH)_2 + CO_2 \to CaCO_3 + H_2O}$ ). A fall in the excess lime for no visible reason is a reason to check the gas readings and the presence of acid gases.

Worked exampleThe lime went on carbon dioxide

Given (assumed data)

Alkalinity of the mud $P_{om}$: 2.4 cm³/cm³ in the morning; 1.2 cm³/cm³ after drilling into a bed with carbon dioxide
Mud in circulation: 150 m³

Solution

Excess lime in the morning: $3.7\cdot 2.4 = 8.88$ kg/m³ (3.1 lb/bbl). After: $3.7\cdot 1.2 = 4.44$ kg/m³.
Used up: $8.88 - 4.44 = 4.44$ kg/m³, and for the whole volume 666 kg of lime.
By the reaction with lime (74.1 g of lime for 44.0 g of carbon dioxide), this is about 395 kg of absorbed carbon dioxide.

Answer: To restore the reserve we need about 666 kg of lime per 150 m³ of mud.

What it means. The lime has absorbed an acid gas: this is what it is there for. But the reserve is limited, and without it carbon dioxide and hydrogen sulfide start to destroy the emulsifiers, and hydrogen sulfide may come out at the surface. Where the section holds hydrogen sulfide (many subsalt fields of the Pre-Caspian Basin are sour), the excess lime is watched especially closely, and the actions on finding hydrogen sulfide are set by the rules of the company. The data in this example are assumed for illustration.

What changes in the readings of an OBM tell us

What changed: Electrical stability falls, there is water in the HPHT filtrate
Probable cause: there is not enough emulsifier or lime, and the droplets merge
What is done: add emulsifier and lime, as chosen on a pilot sample

What changed: The mud is “grainy”, cuttings stick, the screens of the shale shakers clog
Probable cause: the solids are wetted by water
What is done: add wetting agent

What changed: The share of water rises, the salinity of the water phase falls
Probable cause: fresh or formation water is coming in
What is done: check the volumes in the tanks (it may be an inflow); then brine and base oil

What changed: Viscosity and gel strength rise
Probable cause: a lot of fine cuttings or water; when cooled, the base oil thickens
What is done: cleaning, dilution with base oil, check the temperature of the measurement

What changed: The excess lime falls
Probable cause: carbon dioxide or hydrogen sulfide; heating
What is done: gas readings, lime

What changed	Probable cause	What is done
Electrical stability falls, there is water in the HPHT filtrate	there is not enough emulsifier or lime, and the droplets merge	add emulsifier and lime, as chosen on a pilot sample
The mud is “grainy”, cuttings stick, the screens of the shale shakers clog	the solids are wetted by water	add wetting agent
The share of water rises, the salinity of the water phase falls	fresh or formation water is coming in	check the volumes in the tanks (it may be an inflow); then brine and base oil
Viscosity and gel strength rise	a lot of fine cuttings or water; when cooled, the base oil thickens	cleaning, dilution with base oil, check the temperature of the measurement
The excess lime falls	carbon dioxide or hydrogen sulfide; heating	gas readings, lime

The combinations are only a guide; the treatment is chosen on pilot samples. The signs of an inflow are checked against the well control procedures (Well control: primary barrier and kick indicators).

Advantages and limitations

Strengths of oil-based muds

Property: Stability of clays
Why: there is no free water, and the water activity is controlled
Where it matters: active clays, long times with the hole open

Property: Salt does not dissolve
Why: the continuous phase is oil
Where it matters: salt-bearing sequences, for example in the Pre-Caspian Basin

Property: Low friction
Why: an oil film on the steel and on the walls
Where it matters: long deviated and horizontal holes

Property: A thin cake, low filtration
Why: only oil filters through; droplets and fluid-loss reducers plug the pores
Where it matters: fewer cases of differential sticking (the pipe held against the wall by the pressure difference)

Property: Heat resistance
Why: there are no polymers that break down in water when heated
Where it matters: deep hot wells

Property: Protection of metal
Why: the steel is wetted by oil
Where it matters: corrosion, acid gases (with lime)

Property: Repeated use
Why: the mud is cleaned and put back to work
Where it matters: series of wells

Property	Why	Where it matters
Stability of clays	there is no free water, and the water activity is controlled	active clays, long times with the hole open
Salt does not dissolve	the continuous phase is oil	salt-bearing sequences, for example in the Pre-Caspian Basin
Low friction	an oil film on the steel and on the walls	long deviated and horizontal holes
A thin cake, low filtration	only oil filters through; droplets and fluid-loss reducers plug the pores	fewer cases of differential sticking (the pipe held against the wall by the pressure difference)
Heat resistance	there are no polymers that break down in water when heated	deep hot wells
Protection of metal	the steel is wetted by oil	corrosion, acid gases (with lime)
Repeated use	the mud is cleaned and put back to work	series of wells

Limitations

Limitation: Cost
What the problem is: the base oil and the chemicals cost more; mud lost in lost-circulation zones is expensive

Limitation: Gas dissolves in the base oil
What the problem is: a kick is harder to notice (see the section below)

Limitation: Cuttings and waste
What the problem is: the cuttings contain base oil; the waste must be collected, hauled and treated

Limitation: Logging
What the problem is: the mud does not conduct current, so logging methods that need a conducting mud (the spontaneous potential, SP, log and the laterolog) do not work, and induction and special tools are needed

Limitation: Rubber and people
What the problem is: the base oil damages elastomers; the vapours and contact with skin are harmful, and there is a fire risk

Limitation: Density “floats”
What the problem is: the base oil is compressed and expands more than water: the density in the hole depends on pressure and temperature

Limitation: Formation and cement
What the problem is: surfactants may make the rock hydrophobic and create an emulsion block (pores plugged by droplets of emulsion); an oil film hinders the bonding of cement

Limitation	What the problem is
Cost	the base oil and the chemicals cost more; mud lost in lost-circulation zones is expensive
Gas dissolves in the base oil	a kick is harder to notice (see the section below)
Cuttings and waste	the cuttings contain base oil; the waste must be collected, hauled and treated
Logging	the mud does not conduct current, so logging methods that need a conducting mud (the spontaneous potential, SP, log and the laterolog) do not work, and induction and special tools are needed
Rubber and people	the base oil damages elastomers; the vapours and contact with skin are harmful, and there is a fire risk
Density “floats”	the base oil is compressed and expands more than water: the density in the hole depends on pressure and temperature
Formation and cement	surfactants may make the rock hydrophobic and create an emulsion block (pores plugged by droplets of emulsion); an oil film hinders the bonding of cement

In more detail: logging, Overview of logging tools; formation damage, Formation damage and damage skin.

Before cementing, the oil film is washed off the walls and the casing with spacers (buffer fluids) that contain surface-active substances. Otherwise the cement bonds badly to hydrophobic surfaces. For productive formations an OBM cuts both ways. The clays do not swell, and the filtrate does not create a water block (water trapped in the pores that stops oil from flowing). But emulsifiers and wetting agents may change the wettability of the rock. On the other hand, a core cut with an OBM keeps the formation water almost untouched, and this is used to find the water saturation (Fluid saturation).

Gas in an oil-based mud

Methane and other hydrocarbon gases dissolve well in oil and synthetic bases, the better the higher the pressure. At high pressure and high temperature, gas and base oil may mix in any proportion. Carbon dioxide and hydrogen sulfide dissolve even more. In water, hydrocarbon gas dissolves tens of times less. This is the main difference of a kick in an OBM (Well control: primary barrier and kick indicators).

A smaller gain in volume. Dissolved gas takes up much less room in the base oil than the same gas in free form. So the gain in volume in the tanks, the main direct sign of a kick, turns out smaller.
No expansion on the way up. In a water-based mud the gas bubbles expand as they rise, and the volume in the tanks grows faster and faster. Dissolved gas does not expand while the pressure is above the saturation pressure.
A sudden breakout near the surface. When the mud with gas rises to a depth where the pressure falls below the saturation pressure, the gas comes out of solution. It does so fast and at once in a large volume, and it pushes the mud out of the annulus.
A hidden slug of gas. Gas that dissolved in the mud at the bottom during a stop or a trip can travel to the surface with the mud at the next circulation, without any warning.

Saturation pressure of the mud with gas (linear solubility)

p_{\text{sat}} = \frac{G}{f_o\,k}

where:

$p_{\text{sat}}$ pressure below which gas begins to come out of the mud, MPa (absolute)
$G$ amount of gas in the mud: the volume at standard conditions (gas measured at surface pressure and temperature) per 1 m³ of mud, m³/m³
$f_o$ share of base oil in the mud, fraction
$k$ solubility of gas in the base oil: the volume at standard conditions per 1 m³ of base oil for each megapascal, m³/(m³·MPa)

Units: pressure in MPa (absolute), gas volumes at standard conditions

Holds when

Solubility is proportional to pressure (Henry's law) and does not depend on temperature. In reality the dependence is not linear, and the saturation pressure for a given base oil and gas is taken from laboratory data.
Gas dissolves only in the base oil; the brine and the solids do not take it up.
The gas and the mud are in equilibrium. In practice, dissolving takes time and mixing.

Solubility is proportional to the share of base oil, so muds with a high oil–water ratio “hide” gas better. The depth at which gas begins to come out is the depth where the pressure in the annulus equals $p_{\text{sat}}$ .

CalculatedThe calculated volume of free gas per 1 m³ of mud by depth, for the same gas content (10 m³ at standard conditions per 1 m³ of mud, mud of 1.45 g/cm³). In a water-based mud the gas is seen at any depth and expands smoothly towards the surface. In an oil-based mud (base oil share 0.635, solubility 3 m³/(m³·MPa)) there is no free gas below 362 m, and above that it comes out like an avalanche. Both scales are logarithmic, and the depth starts from 10 m.

Worked exampleWhere the gas comes out of the mud

Given (assumed data)

Well: vertical, 3000 m; mud 1.45 g/cm³, base oil share 0.635
Gas in a portion of mud at the bottom: 10 m³ at standard conditions per 1 m³ of mud
Solubility of gas in the base oil (assumed): 3 m³ per 1 m³ of base oil for each MPa

Solution

Absolute pressure at the bottom: $0.101 + 0.00981\cdot 1.45\cdot 3000 = 42.77$ MPa.
How much gas can dissolve in 1 m³ of mud at the bottom: $0.635\cdot 3\cdot 42.77 = 81.5$ m³, which is much more than 10: all the gas is dissolved.
Saturation pressure: $p_{\text{sat}} = 10/(0.635\cdot 3) = 5.25$ MPa. Depth: $(5.25 - 0.101)/(0.00981\cdot 1.45) = 362$ m.
The same gas in a water-based mud (the gas does not dissolve): at the bottom $10\cdot 0.101/42.77 = 0.024$ m³ per 1 m³ of mud; at a depth of 362 m, 0.19 m³. In the OBM at this depth, zero.
At a depth of 100 m (pressure 1.52 MPa): in the water-based mud 0.66 m³; in the OBM $(10 - 0.635\cdot 3\cdot 1.52)\cdot 0.101/1.52 = 0.47$ m³ per 1 m³.

Answer: In an oil-based mud, free gas appears only in the upper 362 m of the hole. Below that it is fully dissolved.

What it means. For most of the way, the gas in an OBM does not show itself: it does not expand, and it hardly changes the volume in the tanks or the density at the outlet. But the mud covers the last hundreds of metres in minutes, and in this time the gas comes out and pushes the mud out of the annulus. So with an OBM, exact records of volumes are especially important. We also watch the early indirect signs: a rise in the rate of penetration (how fast the bit drills), the gas readings, and a change in the standpipe pressure (the pump pressure at the surface). And we shut the well in at any suspicion. The gas is circulated out through the choke manifold (a set of valves that controls the flow from a closed well) and the mud–gas separator (a vessel that takes the gas out of the mud), without waiting for it to reach the wellhead. The numbers of the example are an illustration: the solubility is assumed and linear, and the temperature is not taken into account. The data in this example are assumed for illustration.

Cuttings and waste

The cuttings that are separated on the shale shakers (Solids control in drilling fluids) carry away base oil, usually about ten per cent of their mass or more. Its share is measured with the same retort on a weighed sample of cuttings. It is expressed as the oil on cuttings, as a share of the mass of the wet or the dry cuttings, whichever the rules for the site require. Dryers and centrifuges bring the content of base oil down to a few per cent and return it to the mud.

Treatment. Thermal desorption: the cuttings are heated without air, the base oil evaporates and condenses and is used again, and the dry residue is disposed of.
Injection. The cuttings are ground, mixed with water and pumped through a special well into an isolated formation.
Hauling and burial. The cuttings are taken to landfills that have a permit to accept such waste.
Discharge. In some sea areas, cuttings with synthetic bases may be discharged after cleaning down to a set residual content. Cuttings with diesel fuel and mineral oils are, as a rule, not discharged.

The mud itself is not thrown away: at the end of the work it is collected, cleaned of solids and put back to work on the next well. Which way of handling waste is allowed is decided by the environmental law of the country and by the permits of the particular project. The requirements for offshore work are usually stricter than for work on land. When choosing a mud, the cost of handling the cuttings is counted together with the cost of the mud itself. Sometimes it decides the choice in favour of a highly inhibitive water-based mud (Water-based muds and clay chemistry).

Terms and symbols

Invert emulsion: An emulsion of brine in oil, the basis of most oil-based muds.
Emulsifier: A surface-active substance whose molecules form a film on the brine droplets and keep them from merging.
Wetting agent: A surface-active substance that makes the solid particles hydrophobic.
Organophilic clay: A clay treated with amines that swells and builds a structure in oil.
Oil–water ratio (OWR): The share of base oil in the liquid phase of the mud, from the retort, for example 80/20.
Water-phase salinity: The mass fraction of salt (usually CaCl₂) in the brine of the mud, from the chlorides of the mud and the retort.
Water activity, $a_w$: The relative humidity of the air above the mud at equilibrium. It sets the direction of osmosis.
Balanced activity: The principle by which the water activity of the mud is equal to the activity of the clays or a little lower.
Electrical stability (ES): The breakdown voltage of a sample of mud between electrodes; an indicator of the stability of the emulsion.
Excess lime: Undissolved lime in an OBM, $3.7\,P_{om}$ kg/m³.
Saturation pressure: The pressure below which the gas dissolved in the base oil begins to come out.

Common mistakes and misconceptions

Wrong

“There is no water in an oil-based mud.”

In fact

In an invert emulsion, water is tens of per cent of the liquid phase. It is hidden in droplets and does not touch the walls, but its share and its salinity decide the viscosity, the stability of the emulsion and osmosis.

Wrong

“The saltier the water phase, the better for the clays.”

In fact

The activity must be balanced: a mud that is too “dry” dehydrates the clay, and one that is too “wet” wets it. The activity of the mud is chosen from the activity of the clays being drilled.

Wrong

“An electrical stability of 400 V means a bad emulsion.”

In fact

The absolute value depends on the share of water, the solids, the base oil and the temperature. We judge by the trend and by the combination with the HPHT fluid loss and the look of the mud.

Wrong

“The gain in volume in the tanks is small, so it is not a kick.”

In fact

In an OBM the gas dissolves, and the gain in volume is smaller than the volume of the inflow. A small gain, a rise in gas readings or a change in the standpipe pressure is a reason to check the well by the well control procedures.

Wrong

“The oil–water ratio is easy to correct with a couple of cubic metres of base oil.”

In fact

To move the ratio by a few points we need tens of cubic metres of base oil and then weighting material. So the ratio is kept gradually, with fresh mud of the right composition.

In depth: advanced detail

Where the activity of a brine comes from

The water activity of a salt solution is expressed through the osmotic coefficient $\varphi$ . It is a correction to an ideal solution, and it is measured and tabulated for each salt. In a dilute solution each particle (ion) lowers the activity by the same amount. In a concentrated solution the ions bind water in hydration shells, and the lowering grows faster than the concentration. The osmotic coefficient becomes greater than one.

Water activity in a salt solution

\ln a_w = -\nu\,m\,M_w\,\varphi(m), \qquad m = \frac{1000\,w}{M_s\,(1 - w)}

where:

$\nu$ number of ions in the formula of the salt: 3 for CaCl₂, 2 for NaCl, —
$m$ molality: moles of salt per 1 kg of water, mol/kg
$M_w$ molar mass of water, 0.018015, kg/mol
$\varphi(m)$ osmotic coefficient (tabulated data for 25 °C), —
$w,\ M_s$ mass fraction of salt and its molar mass (110.98 g/mol for CaCl₂, 58.44 for NaCl), —, g/mol

Units: natural logarithm

Holds when

A solution of one salt. For a mixture of CaCl₂ and NaCl the activity is calculated by special methods or measured.
25 °C. The activity of concentrated brines depends only weakly on temperature, but the electric hygrometer is calibrated, and samples are compared at one temperature.

At 20% CaCl₂ the molality is 2.25 mol/kg, $\varphi \approx 1.47$ , $\ln a_w = -3\cdot 2.25\cdot 0.018015\cdot 1.47 \approx -0.179$ and $a_w \approx 0.84$ . For a saturated NaCl brine (6.14 mol/kg) we get 0.75, the known relative humidity of the air above a saturated solution of table salt.

Water activity of brines at 25 °C

Salt, % by mass: 5
CaCl₂: 0.98
NaCl: 0.97

Salt, % by mass: 10
CaCl₂: 0.95
NaCl: 0.94

Salt, % by mass: 15
CaCl₂: 0.90
NaCl: 0.89

Salt, % by mass: 20
CaCl₂: 0.84
NaCl: 0.84

Salt, % by mass: 25
CaCl₂: 0.75
NaCl: 0.77

Salt, % by mass: 26.4
CaCl₂: 0.72
NaCl: 0.75 (saturation)

Salt, % by mass: 30
CaCl₂: 0.64
NaCl: —

Salt, % by mass: 35
CaCl₂: 0.52
NaCl: —

Salt, % by mass: 40
CaCl₂: 0.39
NaCl: —

Salt, % by mass	CaCl₂	NaCl
5	0.98	0.97
10	0.95	0.94
15	0.90	0.89
20	0.84	0.84
25	0.75	0.77
26.4	0.72	0.75 (saturation)
30	0.64	—
35	0.52	—
40	0.39	—

Calculated from the osmotic coefficients; the accuracy is about 0.01. For calculations at the rig, suppliers' tables are used, or the activity is measured with an electric hygrometer.

Connections to other topics

Drilling fluids: functions and properties — the functions of the mud, the density, weighting up and the types of mud.
Water-based muds and clay chemistry — the chemistry of clays, osmosis and inhibitive water-based muds, the alternative to an OBM.
Field testing of drilling fluids — the retort, the pressurised balance, the viscometer, the filter press and the correction of the retort for salt.
Well control: primary barrier and kick indicators — the signs of a kick and the behaviour of gas in the hole.
Well kill methods — how an inflow is circulated out of a closed well.
Drilling fluid rheology and flow regimes — the rheology of the mud and its dependence on temperature and pressure.
Circulating pressure losses — pressure losses and the ECD; why the properties at bottomhole pressure and temperature matter for an OBM.
Wettability and interfacial phenomena — wettability and the contact angle, the basis of how wetting agents work.
Formation damage and damage skin — emulsion block and the change of wettability in the formation.
Overview of logging tools — which logging methods work in a non-conducting mud.
Solids control in drilling fluids — shale shakers, centrifuges and cuttings dryers.

Check yourself

1
What is the continuous phase of an invert emulsion?
1. A.Organophilic clay
2. B.Calcium chloride brine
3. C.Barite
4. D.The hydrocarbon base fluid
Show the answer
Correct answer: D. The hydrocarbon base fluid
In an invert emulsion the brine is broken into droplets, and the continuous phase is oil. So the walls, the cuttings and the metal are in contact with oil and not with water. The clay and the barite are solids suspended in the oil.
2
A 20 cm³ retort gave 14.4 cm³ of base oil and 2.8 cm³ of water. For 1 cm³ of mud, 2.2 cm³ of 0.282 N silver nitrate was used. Find the oil–water ratio and the water-phase salinity.
Show the answer
$f_o = 0.72$ , $f_w = 0.14$ ; the ratio is $0.72/0.86 \approx 83.7/16.3$ . Chlorides: $10{,}000\cdot 2.2 = 22{,}000$ mg/L. Calcium chloride: $1.565\cdot 22{,}000 \approx 34{,}400$ mg/L, that is 34.4 kg/m³. Water: 140 kg/m³. Salinity: $34.4/(34.4 + 140) \approx 19.7$ %.
3
Over a day the electrical stability of the mud fell from 800 to 350 V, water appeared in the HPHT filtrate, and the look of the mud did not change. What is most likely, and what is done?
1. A.The mud was weighted up: electrical stability always falls as the density rises
2. B.The emulsion is getting weaker: there is not enough emulsifier or lime; this is checked on pilot samples, and they are added
3. C.Nothing: electrical stability has nothing to do with the emulsion
4. D.Gas has come in: the mud must be weighted up
Show the answer
Correct answer: B. The emulsion is getting weaker: there is not enough emulsifier or lime; this is checked on pilot samples, and they are added
A fall in electrical stability together with water in the filtrate is a sign that the brine droplets have begun to merge. Usually there is not enough emulsifier or lime (or a lot of water has got into the mud). The look of the mud has not changed, so wetting of the solids by water is less likely. The treatment is chosen on pilot samples.
4
In 100 m³ of mud $f_o = 0.68$ and $f_w = 0.12$ (ratio 85/15). How much water (as brine) must be added to get a ratio of 80/20?
Show the answer
$\Delta V_w = V\,(f_o(1 - R)/R - f_w) = 100\cdot(0.68\cdot 0.20/0.80 - 0.12) = 100\cdot(0.170 - 0.120) = 5$ m³ of water. We need more brine of the same salinity, by the volume of the salt. The density of the mud will change after this, and it is checked.
5
Why can a gas kick in an oil-based mud stay unnoticed for a long time?
1. A.The volume sensors do not work in such a mud
2. B.Gas dissolves in the base oil: the gain in volume is small, the gas does not expand while the pressure is above the saturation pressure, and it comes out only in the upper part of the hole
3. C.Gas cannot enter a well that is filled with such a mud
4. D.An oil-based mud is lighter than water, so the gas rises faster
Show the answer
Correct answer: B. Gas dissolves in the base oil: the gain in volume is small, the gas does not expand while the pressure is above the saturation pressure, and it comes out only in the upper part of the hole
Methane dissolves well in the base oil. Dissolved gas takes up little room and does not expand on the way up, so the usual signs, the gain in volume and its acceleration, are weak. The gas comes out of the mud where the pressure falls below the saturation pressure, usually in the last hundreds of metres, fast and in a large volume.
6
The activity of fresh cuttings of a clay is 0.90, and the water phase of the mud is 30% calcium chloride. What happens, and what can be done?
Show the answer
At 30% CaCl₂ the water activity is about 0.64, much lower than that of the clay. The ideal osmotic difference $137\cdot\ln(0.90/0.64) \approx 47$ MPa pulls water from the clay into the mud. The clay is dehydrated, and too strong dehydration can weaken it. The salinity is lowered to the balanced one, which for 0.90 is about 15% CaCl₂. This is done by adding fresh water (remembering that water also lowers the oil–water ratio) or fresh mud with a less salty brine.

Updated: 8 October 2026