Field testing of drilling fluids

Quick refresher

Density is measured with a lever balance. Air or gas bubbles make the result too low. A balance with a sealed cup under pressure gives the true density, and the share of bubbles in the sample is $1 - \rho_{\text{atm}}/\rho_{\text{press}}$ .
Funnel viscosity is only an indicator of changes. The rheology (how the mud flows) is described by the readings of a rotational viscometer at about 50 °C: PV $= \theta_{600} - \theta_{300}$ , YP $= \theta_{300} -$ PV, and the gel strength after 10 s and 10 min. A rise in YP and gel strength with an unchanged PV is a sign of flocculation. A rise in PV at the same density is a sign of fine cuttings.
The standard filter press works at a pressure difference of 0.69 MPa, at room temperature, for 30 min, with an area of 45.8 cm². The volume of filtrate grows as $V_{sp} + C\sqrt t$ . Twice the volume at 7.5 min overstates the volume at 30 min by exactly the spurt loss $V_{sp}$ . The high pressure, high temperature (HPHT) cell has half the area, so its volume is doubled.
The retort (an instrument that boils the liquids out of a sample) gives the shares of water, oil and solids. Together with the density, they split the solids into low-gravity solids (clay and cuttings, about 2.6 g/cm³) and barite (4.2 g/cm³). The methylene blue test counts all the active clay, drilled clay included, as “bentonite equivalent”.
Titrating the filtrate (adding acid to it drop by drop) up to two end points ( $P_f$ and $M_f$ ) tells hydroxides from carbonates and bicarbonates. $P_m$ and $P_f$ give the excess lime. Chlorides show salt and formation water. The hardness shows calcium from cement, gypsum and anhydrite.
The cause of the changes is usually shown by a combination of readings and by the time of the event, not by one number. A rise in YP and gel strength together with a rise in pH and calcium after drilling out the cement in the shoe track (the cement left inside the lowest joints of the casing) means that there is cement in the mud.

Key relationships

x_{\text{g}} = 1 - \dfrac{\rho_{\text{atm}}}{\rho_{\text{press}}}

share of air or gas in the sample at atmospheric pressure

V(t) = V_{sp} + C\sqrt{t}, \quad V_{sp} = 2V_{7.5} - V_{30}

static filtration; volumes at 7.5 and 30 min

\bar\rho_s = \dfrac{\rho_m - f_w\rho_w - f_o\rho_o}{f_s}

average density of the solids from the density of the mud and the retort

c_{\text{bent}} = 14.25\,\mathrm{MBC}

bentonite equivalent, kg/m³, from the methylene blue test

c_{\text{lime}} = 0.742\,(P_m - f_w P_f)

excess lime, kg/m³

[\mathrm{Cl^-}] = 1000\,V_{\mathrm{AgNO_3}}/V_{\text{mf}}

chlorides, mg/L, with 0.0282 N silver nitrate

Common pitfalls

Writing down the density of a foamed or gas-cut mud from an ordinary balance and “correcting” it by weighting up.
Comparing readings that were measured at different temperatures or with instruments of different standards: funnels of different sizes, gel strength after 10 s and after 1 min.
Taking the bentonite equivalent for the amount of bentonite that was added: it includes drilled clay too.
Taking chlorides in mg/L as equal to ppm: in a saturated brine the difference is about 16%.
Treating a mud that has thickened with water, without looking at the chemistry of the filtrate.

After this lesson you can

measure the density of the mud and explain why mud with air or gas in it needs a balance that works under pressure
record the funnel viscosity, the plastic viscosity (PV), the yield point (YP) and the gel strength from the funnel and the rotational viscometer, and tell a build-up of solids from flocculation (clay particles clumping together) by how these numbers change
find the spurt loss (the fast first loss of liquid) from a filter-press test, estimate the permeability of the cake, and explain how the standard test differs from a test at high pressure and high temperature
calculate the content of barite, bentonite and drilled rock from the density, the retort and the methylene blue test
find the ions in the filtrate, the excess lime and the source of contamination from the pH, the alkalinity, the chlorides and the hardness
read a mud report as a diagnostic picture, not as a set of separate numbers

What is measured, where and how often

The drilling programme sets the properties of the mud. It gives not one number but limits for each interval (Drilling fluids: functions and properties). The job of keeping the mud within these limits belongs to the mud engineer. The engineer takes samples, runs the tests, writes the results in the daily mud report and, from the results, orders a treatment: to add weighting material or water, to add a chemical, or to change how the cleaning equipment works.

A sample is taken in two places. The mud coming out of the well is taken from the flowline (the channel that carries the mud from the well to the shale shakers, the vibrating screens that remove cuttings). It shows what the well has done to the mud: whether it has picked up gas, formation water or cuttings, and how hot it has become. The mud in the suction tank of the mud pumps shows what goes into the well after cleaning and treatment. Comparing “in” and “out” is one of the most sensitive indicators: a difference in density or in chlorides points to what is happening in the hole.

Density and funnel viscosity are checked often during drilling. Programmes usually set an interval of about 15–30 min, and the rig crew does the checks. The engineer runs the full set of tests several times a day and after every notable event: drilling out the cement in the shoe track, drilling into salt, or adding a large volume of water. The order of the tests is standardised (the international standard ISO 10414-1; in English-language practice API RP 13B-1). So numbers from different rigs can be compared, if they were obtained by the same method and at the same temperature.

Main tests of a water-based mud

Test: Density
Instrument: mud balance (a lever balance with a cup); balance with a sealed cup under pressure
What it gives: density of the mud
Units: g/cm³, kg/m³, ppg

Test: Funnel viscosity
Instrument: funnel
What it gives: run-out time
Units: s

Test: Rheology
Instrument: rotational viscometer (a sleeve that rotates in the mud)
What it gives: readings at six speeds; PV, YP, gel strength
Units: mPa·s; lb/100 ft², dPa

Test: Filtration
Instrument: filter press (forces mud through filter paper); HPHT cell
What it gives: volume of filtrate in 30 min, thickness of the cake
Units: cm³; mm

Test: Liquids and solids
Instrument: retort (boils the liquids out of a sample)
What it gives: shares of water, oil and solids
Units: % of volume

Test: Sand
Instrument: 74 µm sieve and a measuring tube
What it gives: particles larger than 74 µm
Units: % of volume

Test: Active clay
Instrument: titration with methylene blue (a dye added drop by drop)
What it gives: methylene blue capacity, bentonite equivalent
Units: cm³/cm³; kg/m³

Test: Chemistry of the filtrate
Instrument: pH meter; titration
What it gives: pH, alkalinity, chlorides, hardness
Units: —; cm³/cm³; mg/L

Test	Instrument	What it gives	Units
Density	mud balance (a lever balance with a cup); balance with a sealed cup under pressure	density of the mud	g/cm³, kg/m³, ppg
Funnel viscosity	funnel	run-out time	s
Rheology	rotational viscometer (a sleeve that rotates in the mud)	readings at six speeds; PV, YP, gel strength	mPa·s; lb/100 ft², dPa
Filtration	filter press (forces mud through filter paper); HPHT cell	volume of filtrate in 30 min, thickness of the cake	cm³; mm
Liquids and solids	retort (boils the liquids out of a sample)	shares of water, oil and solids	% of volume
Sand	74 µm sieve and a measuring tube	particles larger than 74 µm	% of volume
Active clay	titration with methylene blue (a dye added drop by drop)	methylene blue capacity, bentonite equivalent	cm³/cm³; kg/m³
Chemistry of the filtrate	pH meter; titration	pH, alkalinity, chlorides, hardness	—; cm³/cm³; mg/L

HPHT stands for high pressure, high temperature: a test at high pressure and high temperature. Volumes of filtrate are written in cm³ (mL) per 30 min.

Density: the mud balance and the pressurised balance

The mud balance is a beam with a cup of known volume at one end and a movable weight (a rider) on a scale. The cup is filled to the brim with mud and closed with a lid that has a small hole. The extra mud runs out through the hole, and exactly the set volume stays in the cup. The beam is put on its support, and the rider is moved until the bubble of the level sits in the middle. The density is read on the scale at the rider. The beam usually carries two scales: g/cm³ and ppg (pounds per US gallon). Some beams also have lb/ft³ and the pressure gradient. One division is 0.01 g/cm³ (0.1 ppg).

The balance is checked with fresh water. At room temperature it must show 1.00 g/cm³ (8.33 ppg). If it does not, it is calibrated with the counterweight. An error of 0.01 g/cm³ at a depth of 3000 m is about 0.3 MPa of pressure on the bottom of the hole (pressure of the mud column).

The mud balance weighs everything that is in the cup, bubbles included. The mud takes in air when it is stirred in the tanks, especially if it has surface-active additives, lubricants or detergents. Gas may also be in the mud at the outlet of the well. Bubbles take up volume and weigh almost nothing, so the balance shows less than the mud itself weighs. In the well, under a pressure of tens of megapascals, the bubbles are squeezed hundreds of times smaller. Almost only the “heavy” part of the mud presses on the bottom (gas-cut mud). The pressurised mud balance solves the problem. A piston pushes more mud into a sealed cup, the pressure in it rises, the air is squeezed to a tiny volume, and the balance shows the density without bubbles.

Share of air or gas in the sample

x_{\text{g}} = 1 - \frac{\rho_{\text{atm}}}{\rho_{\text{press}}}

where:

$x_{\text{g}}$ volume fraction of bubbles in the sample at atmospheric pressure, —
$\rho_{\text{atm}}$ density from an ordinary mud balance, g/cm³
$\rho_{\text{press}}$ density of the same sample from the pressurised balance, g/cm³

Units: any identical units of density

Holds when

The mass of the bubbles is negligible compared with the mass of the liquid and the solids.
In the pressurised balance the gas is squeezed to a negligible volume, and the compression of the mud itself can be neglected.
Both samples are taken at the same place and at the same time.

In a sample of volume $V$ the bubbles take up $x_{\text{g}}V$ , and mud of the true density $\rho_{\text{press}}$ takes up the rest. So $\rho_{\text{atm}} = (1 - x_{\text{g}})\,\rho_{\text{press}}$ . A few per cent of air make the mud “lighter” on the balance by hundredths of g/cm³. This is the same amount that separates a normal mud from a mud that needs weighting up.

Worked exampleFoamed mud: which balance is right

Given (assumed data)

Density by the programme: 1.28 g/cm³
Sample from the suction tank, ordinary mud balance: 1.24 g/cm³
The same sample, pressurised balance: 1.28 g/cm³
True vertical depth: 2600 m

Solution

Share of air in the sample: $x_{\text{g}} = 1 - 1.24/1.28 = 0.0313$ , that is 3.1% of the volume.
The true density of the mud is 1.28 g/cm³, as the programme requires, so there is no need to weight up. In oilfield units the ordinary balance showed $1.24\cdot 8.345 = 10.35$ ppg, and the pressurised balance showed $1.28\cdot 8.345 = 10.68$ ppg.
What would happen if the mud were “brought” to 1.28 g/cm³ by the ordinary balance, with the same air content: the true density would be $1.28/(1 - 0.0313) = 1.321$ g/cm³, which is 0.0413 g/cm³ more than needed.
The extra pressure on the bottom from such weighting up: $0.00981\cdot 0.0413\cdot 2600 = 1.05$ MPa.

Answer: The sample contains 3.1% of air, and the true density is 1.28 g/cm³. Weighting up by the ordinary balance would give a mud of 1.321 g/cm³ and an extra 1.05 MPa on the bottom.

What it means. An extra megapascal is not a reserve but a risk. At such a depth it may use up the whole margin up to the lost-circulation pressure of the rocks (the pressure at which they begin to take mud; Fracture gradient and leak-off testing). The right actions are to measure the density with the pressurised balance, to find why the mud foams, and to break the foam with a defoamer or by changing how the mud is stirred. Weighting up is wrong. Suppose the ordinary and the pressurised balances disagree on a sample from the outlet, and there is no foam in the tanks. Then there is gas in the mud. Check the gas readings and the signs of a kick (formation fluid entering the well). The data in this example are assumed for illustration.

Funnel viscosity and rotational viscometer readings

Funnel viscosity is the time that the mud takes to run out of a funnel of standard size. The Marsh funnel is filled through a screen with 1500 cm³ of mud, and we time how many seconds it takes for one quart (946 cm³) to run out. Fresh water at 21 °C runs out in 26 s. In Russian-language practice a funnel of another standard is common. 700 cm³ is poured into it, and we time the run-out of 500 cm³. Water takes about 15 s. Seconds from different funnels cannot be compared, so the report states which funnel was used.

The shear rate (how fast the layers of mud slide past each other) in the stream that runs out keeps changing, and the density of the mud also affects the run-out time. So funnel viscosity cannot be converted into plastic viscosity or yield point. Its advantage is speed: the crew can measure it every half hour and notice a jump at once. A rise in funnel viscosity at the same density is a reason to make a full analysis.

The readings of a rotational viscometer describe the rheology of the mud. How the instrument works, and how flow models are built from its readings, is covered in the lesson Drilling fluid rheology and flow regimes. At the rig, the order of the measurements matters. The sample is heated in a thermostat cup to the agreed temperature, which for water-based muds is about 50 °C. Then we take the steady readings at 600, 300, 200, 100, 6 and 3 rpm. After that the mud is stirred at 600 rpm and left at rest for 10 s and for 10 min. Each time we note the highest reading at 3 rpm. This is the gel strength (the strength of the gel that forms when the mud stands still).

The report gives the plastic viscosity (PV), $\theta_{600} - \theta_{300}$ , in mPa·s. It gives the yield point (YP), $\theta_{300} -$ PV, in lb/100 ft² or in dPa (1 lb/100 ft² ≈ 4.79 dPa), and the gel strength after 10 s and 10 min. More and more often it also gives the YP at low shear rates, $2\theta_3 - \theta_6$ , which comes closest to the real yield stress (the stress that the mud must exceed to start flowing). Russian-language reports often give the gel strength after 1 and 10 min of rest, measured by another method and with another instrument. Such values cannot be compared with readings after 10 s and 10 min.

What changes in the rheological readings tell us

What changed: PV rises at the same density
Probable cause: fine drilled solids are building up
What to check: the retort and the methylene blue test; how the cleaning equipment is working

What changed: PV rises together with the density
Probable cause: weighting up: more barite means more friction between particles
What to check: whether the PV fits the programme for this density

What changed: YP and gel strength rise, PV about the same
Probable cause: flocculation: calcium, salt, carbonates, high temperature
What to check: pH, alkalinity, calcium, chlorides

What changed: The 10-min gel strength is much higher than the 10-s one, and the gap grows
Probable cause: progressive gel (a gel that keeps getting stronger the longer the mud rests): flocculation or too much active clay
What to check: the methylene blue test, the chemistry of the filtrate

What changed: YP and gel strength fall
Probable cause: dilution with water, too much thinner (a chemical that lowers the viscosity), breakdown of the polymer
What to check: the log of treatments, the temperature

What changed	Probable cause	What to check
PV rises at the same density	fine drilled solids are building up	the retort and the methylene blue test; how the cleaning equipment is working
PV rises together with the density	weighting up: more barite means more friction between particles	whether the PV fits the programme for this density
YP and gel strength rise, PV about the same	flocculation: calcium, salt, carbonates, high temperature	pH, alkalinity, calcium, chlorides
The 10-min gel strength is much higher than the 10-s one, and the gap grows	progressive gel (a gel that keeps getting stronger the longer the mud rests): flocculation or too much active clay	the methylene blue test, the chemistry of the filtrate
YP and gel strength fall	dilution with water, too much thinner (a chemical that lowers the viscosity), breakdown of the polymer	the log of treatments, the temperature

The limits of PV, YP and gel strength are set by the drilling programme for each density and interval. The values are compared with these limits and with the previous measurements.

Worked exampleThe mud before and after drilling out the cement in the shoe track

Given (assumed data)

Sample A (before): readings at 600 / 300 / 200 / 100 / 6 / 3 rpm: 46 / 28 / 21 / 14 / 4 / 3
Sample A: gel strength after 10 s / 10 min: 4 / 7 lb/100 ft²
Sample B (after): readings at the same speeds: 61 / 42 / 34 / 25 / 10 / 9
Sample B: gel strength after 10 s / 10 min: 11 / 29 lb/100 ft²
Mud: fresh bentonite–polymer mud, 1.28 g/cm³; measured at 50 °C

Solution

Sample A: PV $= 46 - 28 = 18$ mPa·s; YP $= 28 - 18 = 10$ lb/100 ft², or $10\cdot 0.479 = 4.79$ Pa (48 dPa); YP at low shear rates $2\cdot 3 - 4 = 2$ lb/100 ft².
Sample B: PV $= 61 - 42 = 19$ mPa·s; YP $= 42 - 19 = 23$ lb/100 ft², or 11.0 Pa (110 dPa); YP at low shear rates $2\cdot 9 - 10 = 8$ lb/100 ft².
The ratio of YP to PV is 0.56 for sample A and 1.21 for sample B. In 10 min of rest the gel becomes 1.75 times stronger in sample A and 2.64 times stronger in sample B.
The plastic viscosity has hardly changed (+1 mPa·s), but the YP has grown 2.3 times, and the YP at low shear rates has grown 4 times.

Answer: Sample A: PV 18 mPa·s, YP 10 lb/100 ft² (4.79 Pa). Sample B: PV 19 mPa·s, YP 23 lb/100 ft² (11.0 Pa), gel strength 11 / 29 lb/100 ft². Judging by the PV, there are no more solids, but the particles attract each other more strongly: the clay has flocculated.

What it means. The picture “same PV, YP and gels sharply up” is typical of flocculation, and the time of the event suggests the cause. Drilled-out cement gives calcium ions to the mud and raises the pH. The chemical tests of the filtrate confirm this (the example below). The cure is a chemical treatment that ties up the calcium, together with a thinner (Water-based muds and clay chemistry). It is not dilution with water: dilution lowers the density and does not remove the cause. A progressive gel is dangerous. After a stop, a higher pressure is needed to restart circulation, and the pressure swings are stronger when pipe is run in or pulled out. The data in this example are assumed for illustration.

Filtration: the standard test and the test at high pressure and high temperature

The filter press is a cell with the mud in it. At the bottom of the cell, filter paper lies on a screen. The cell is closed, and compressed gas (air, nitrogen or carbon dioxide) is applied with a pressure difference of 0.69 MPa (100 psi). For half an hour the filtrate is collected in a measuring cylinder. The filtering area of the standard cell is 45.8 cm² (7.1 in²), and the test is made at room temperature. The result is the fluid loss in cm³ (mL) per 30 min. Then the cell is taken apart, the remains of the mud are rinsed off the cake, and its thickness is measured: in millimetres, or in English-language reports in 1/32 in (0.8 mm). The cake is also described: thin and dense, loose, soft, “rubbery”.

How the volume of the filtrate grows is explained by the theory of static filtration (the derivation is in the lesson on the invaded zone). The cake gets thicker in proportion to the filtrate that has passed, its resistance grows with it, and the volume grows as the square root of time. Until the cake has formed, part of the liquid, and fine particles with it, pass through the filter almost freely. This first portion is called the spurt loss.

Volume of filtrate in a static test

V(t) = V_{sp} + C\sqrt{t}, \qquad C = A\sqrt{\frac{2\,k_{mc}\,\Delta p}{\mu_f\,c}}

where:

$V(t)$ volume of filtrate by the time $t$ , cm³
$V_{sp}$ spurt loss: the volume that passes before the cake forms, cm³
$t$ time from the start of the test, min
$C$ slope of the line on a plot of volume against the square root of time, cm³/min $^{1/2}$ ; in the expression through the properties of the cake, m³/s $^{1/2}$
$A$ filtering area, m²
$k_{mc}$ permeability of the cake, m²
$\Delta p$ pressure difference across the cake, Pa
$\mu_f$ viscosity of the filtrate, Pa·s
$c$ volume of cake formed per unit volume of filtrate, —

Units: $V$ in cm³, $t$ in min; the expression for $C$ is written in consistent SI units (m³, m², Pa, Pa·s, s)

Holds when

It is a static test: the mud above the cake does not move and does not erode it, and the cake only grows.
The cake is incompressible, and its composition and permeability are constant. In real cakes the permeability falls as the pressure difference grows.
The viscosity of the filtrate and the temperature are constant during the test.

On a plot of volume against the square root of time, the readings lie on a straight line. Its slope characterises the cake, and its intercept on the volume axis gives the spurt loss. The volume is proportional to the area of the filter, so results from cells of different areas are recalculated.

Since $\sqrt{30/7.5} = 2$ , the fluid loss at 30 min is sometimes estimated as twice the volume at 7.5 min. Doubling also multiplies the spurt loss by two, so the estimate is too high by exactly the spurt loss: $2V_{7.5} - V_{30} = V_{sp}$ . The same fact gives a simple way to find the spurt loss from two readings.

From one test we can also estimate the permeability of the cake itself. The more filtrate has passed, the thicker the cake is. Its permeability links the thickness, the volume and the time:

Permeability of the cake from the test results

k_{mc} = \frac{\mu_f\,h_{mc}\,V}{2\,A\,\Delta p\,t}

where:

$k_{mc}$ permeability of the cake, m²
$\mu_f$ viscosity of the filtrate at the temperature of the test, Pa·s
$h_{mc}$ thickness of the cake at the end of the test, m
$V$ volume of filtrate that has passed through the cake: $V_{30} - V_{sp}$ , m³
$A$ filtering area, m²
$\Delta p$ pressure difference, Pa
$t$ duration of the test, s

Units: SI (any consistent system)

Holds when

The same as for static filtration: the cake is incompressible and uniform in thickness.
The resistance of the paper and the screen is small compared with the resistance of the cake.
The thickness is measured after the liquid mud above the cake has been removed. An error in the thickness goes straight into an error in $k_{mc}$ .

The 2 in the denominator is there because the cake grew together with the filtrate. On average, each portion of the filtrate passed through a cake half as thick as the final one (the derivation is in the advanced section). The result is the average permeability of the cake at the pressure difference of the test. At a larger pressure difference it is usually smaller.

The standard test is made at room temperature and at a small pressure difference. For deep wells it is repeated in a cell for high pressure and high temperature (HPHT). The mud is heated, most often to 150 °C (300 °F), and a pressure difference of 3.45 MPa (500 psi) is applied. The filtrate receiver is kept under a back pressure of about 0.7 MPa, so that the hot filtrate does not boil. The filter area of this cell is half the standard one (about 22.6 cm²). So the volume collected in 30 min is doubled and recorded as the HPHT fluid loss.

The HPHT filtrate is almost always larger than the standard one. When heated, the viscosity of the filtrate falls (for water, by more than five times between 20 and 150 °C). The pressure difference is five times greater. Part of the polymers, and the clay itself, work worse at high temperature. So for hot intervals and for drilling into productive formations the programme sets limits for both the standard fluid loss and the HPHT fluid loss.

CalculatedThe standard test of the example below, plotted as the volume of filtrate against the square root of time. The readings at 7.5 and 30 min lie on a straight line. Its intercept on the volume axis is the spurt loss, 0.6 cm³. Twice the reading at 7.5 min (8.2 cm³) overstates the volume at 30 min by exactly this amount.

Worked exampleThe standard test and the HPHT test for sample A

Given (assumed data)

Standard test (0.69 MPa, 45.8 cm²): volume at 7.5 and at 30 min: 4.1 and 7.6 cm³
Thickness of the cake: 1.6 mm (2/32 in)
Viscosity of the filtrate at the temperature of the test: 1.0 mPa·s
HPHT test (150 °C, 3.45 MPa, a cell of half the area): collected in 30 min: 8.6 cm³

Solution

Spurt loss: $V_{sp} = 2\cdot 4.1 - 7.6 = 0.6$ cm³. Slope: $C = \dfrac{7.6 - 4.1}{\sqrt{30} - \sqrt{7.5}} = 1.28$ cm³/min $^{1/2}$ .
The estimate by doubling: $2\cdot 4.1 = 8.2$ cm³, which is 0.6 cm³ (8%) more than the measured volume at 30 min.
The volume that passed through the cake is $V = 7.6 - 0.6 = 7.0$ cm³. In SI units: $\mu_f = 10^{-3}$ Pa·s, $h_{mc} = 1.6\cdot10^{-3}$ m, $V = 7.0\cdot10^{-6}$ m³, $A = 4.58\cdot10^{-3}$ m², $\Delta p = 6.9\cdot10^{5}$ Pa, $t = 1800$ s.
Permeability of the cake: $k_{mc} = \dfrac{\mu_f h_{mc} V}{2A\,\Delta p\,t} = 9.845\cdot 10^{-19}$ m², that is 0.0010 mD, or about 1.0 µD.
HPHT fluid loss: $2\cdot 8.6 = 17.2$ cm³ in 30 min, which is 2.3 times more than the standard value.

Answer: The fluid loss is 7.6 cm³ in 30 min, with a spurt loss of 0.6 cm³. A cake 1.6 mm thick has a permeability of about 1.0 µD. The HPHT fluid loss is 17.2 cm³ in 30 min.

What it means. The cake is a hundred thousand times less permeable than a sandstone of 100 mD. So once it has formed, it, and not the formation, decides the filtration. The spurt loss is small: the mud has enough fine particles to plug the pores of the filter quickly. For drilling into a productive formation, the plugging ability is also tested on ceramic discs with pores close to those of the formation. The HPHT fluid loss is compared with the limit of the programme for the hot interval. Suppose it grows from one measurement to the next while the standard fluid loss does not change. Then the cause is most likely the temperature: the polymers are breaking down, or the clay is flocculating. The data in this example are assumed for illustration.

Solids: the retort, sand and methylene blue

The retort is a small still (an apparatus that boils a liquid and condenses its vapour). An exactly measured sample of mud (10, 20 or 50 cm³) is heated to about 500 °C. Water and oil evaporate, the vapours condense and run into a measuring cylinder. From the volumes of the condensate we find the shares of water $f_w$ and of oil $f_o$ . The share of solids comes from the difference: $f_s = 1 - f_w - f_o$ . The “solids” of the retort also include the salts that were dissolved in the water. After the water evaporates, they stay in the retort cup. In a fresh mud this is a fraction of a per cent. In a salty mud it is a noticeable error, and it is corrected (see the advanced section).

The retort alone is not enough. A solids content of 12% may belong to a good weighted mud, or to a mud that is overloaded with drilled clay. The density helps to tell them apart. The solids are split into two groups. Low-gravity solids (LGS, about 2.6 g/cm³) are bentonite, drilled clay and sand, and carbonates. High-gravity solids (HGS) are the weighting material, usually barite (4.2 g/cm³). From the density of the mud and the shares from the retort we find the average density of the solids, and from it the shares of the two groups.

Average density of the solids and the share of low-gravity solids

\bar\rho_s = \frac{\rho_m - f_w\,\rho_w - f_o\,\rho_o}{f_s}, \qquad f_{LGS} = f_s\,\frac{\rho_{HGS} - \bar\rho_s}{\rho_{HGS} - \rho_{LGS}}

where:

$\rho_m$ density of the mud (from the pressurised balance), g/cm³
$f_w,\ f_o,\ f_s$ shares of water, oil and solids in the volume of the mud, from the retort, fraction
$\rho_w,\ \rho_o$ densities of water and oil: 1.0 and about 0.8, g/cm³
$\bar\rho_s$ average density of the solids, g/cm³
$\rho_{LGS},\ \rho_{HGS}$ densities of the low-gravity solids and of the weighting material: 2.6 and 4.2, g/cm³
$f_{LGS}$ share of low-gravity solids in the volume of the mud; barite is $f_s - f_{LGS}$ , fraction

Units: densities in g/cm³, shares as fractions (or all shares in per cent)

Holds when

The volumes of the components add up.
The solids consist of two groups with known densities. If the weighting material is hematite or calcium carbonate, its density is used.
The mud is fresh. Otherwise the share of water is replaced by the share of brine (the correction is in the advanced section).
There is no air in the sample.

These are balances of volume and mass. The closer the average density of the solids is to 2.6 g/cm³, the more clay and cuttings the mud contains and the less barite. Low-gravity solids thicken the mud more than barite does: their particles are finer, and the clay is also active. So their share is kept as low as possible, usually within a few per cent of the volume.

Sand in a drilling mud is a term of size, not of mineralogy. It means everything that does not pass through a sieve with a mesh of 74 µm (200 mesh). The sample is diluted with water and washed through the sieve. The particles that stay on it are washed into a measuring tube, and after they settle we read their volume in per cent. Coarse particles wear the pumps, the valves and the bit. A rise in their content says that the desander (a hydrocyclone that removes sand) works badly, or that it is time to clean the tanks (Solids control in drilling fluids).

The methylene blue test (MBT) estimates the content of active clay. The dye methylene blue is a large organic cation (a positively charged particle). Clays take it up and give away their own exchangeable cations in return, in proportion to their cation exchange capacity (how it arises is explained in the lesson Water-based muds and clay chemistry). First the sample of mud is boiled with hydrogen peroxide and a little sulfuric acid. This oxidises the polymers and lignites (thinners made from brown coal), which also take up the dye. Then the dye is added in portions of 0.5 cm³. It is 0.01 N: 1 cm³ of it holds 0.01 meq (milliequivalent, a unit that counts electric charge). After each portion a drop of the suspension is put on filter paper. While the clay is not yet saturated, all the dye sits on the particles. The end point is a pale blue halo of free dye around the dark spot of the solids.

Methylene blue capacity and the bentonite equivalent

\mathrm{MBC} = \frac{V_{MB}}{V_m}, \qquad c_{\text{bent}} = 14.25\,\mathrm{MBC}

where:

$\mathrm{MBC}$ methylene blue capacity of the mud, cm³ of dye per 1 cm³ of mud
$V_{MB}$ volume of 0.01 N dye up to the end point, cm³
$V_m$ volume of the sample of mud, cm³
$c_{\text{bent}}$ bentonite equivalent: how much bentonite would give the same uptake, kg/m³

Units: volumes in cm³, $c_{\text{bent}}$ in kg/m³

Holds when

All the active clay of the sample is counted as bentonite with a cation exchange capacity of about 70 meq/100 g. The coefficient comes from this.
The organic substances that take up the dye have been oxidised with the peroxide.
The dye is 0.01 N: 1 cm³ carries 0.01 meq of the dye cation.

1 cm³ of dye per 1 cm³ of mud is 0.01 meq per cm³, or 10 meq per litre of mud. If bentonite takes up 70 meq per 100 g, then $10/0.70 \approx 14.3$ g of bentonite per litre will bind this much dye. That is about 14.3 kg/m³. In the standard the coefficient is rounded to 14.25.

In other unit systems (1)

Oilfield units

c_{\text{bent}}\,[\text{lb/bbl}] = 5\,\mathrm{MBC}

the bentonite equivalent in pounds per barrel, $\mathrm{MBC}$ in cm³/cm³.

$5 \approx 14.25/2.853$ (1 lb/bbl = 2.853 kg/m³).

Worked exampleHow much barite, bentonite and cuttings are in the mud

Given (assumed data)

Density (pressurised balance): 1.28 g/cm³
Retort, sample of 10 cm³: water 8.80 cm³, no oil
Chlorides of the filtrate: 2400 mg/L: the mud is fresh, no correction for salt is needed
Methylene blue test: 9.0 cm³ of dye used for 2.0 cm³ of mud

Solution

Shares from the retort: water $f_w = 8.80/10 = 0.880$ , solids $f_s = 1 - 0.880 = 0.120$ , that is 12.0%.
Average density of the solids: $\bar\rho_s = \dfrac{1.28 - 0.880\cdot 1.00}{0.120} = 3.33$ g/cm³.
Low-gravity solids: $f_{LGS} = 12.0\cdot\dfrac{4.20 - 3.33}{4.20 - 2.60} = 6.5$ % of the volume, and by mass $0.065\cdot 2600 = 169$ kg/m³. Barite: $12.0 - 6.5 = 5.5$ %, or $0.055\cdot 4200 = 231$ kg/m³.
Methylene blue test: $\mathrm{MBC} = 9.0/2.0 = 4.5$ ; bentonite equivalent $14.25\cdot 4.5 = 64.1$ kg/m³ (22.5 lb/bbl).
Drilled rock: $169 - 64.1 = 105$ kg/m³. The ratio of drilled rock to bentonite is 1.64.
Check by mass: $880 + 169 + 231 = 1280$ kg in 1 m³. The density of 1.28 g/cm³ agrees.

Answer: The mud holds 5.5% of barite (231 kg/m³) and 6.5% of low-gravity solids (169 kg/m³). Of these solids, about 64.1 kg/m³ is bentonite equivalent and about 105 kg/m³ is drilled rock.

What it means. Barite takes up only 5.5% of the volume, but it gives almost a fifth of the mass of the mud. There is already more than one and a half times as much drilled rock as bentonite. If part of the “bentonite” is really active drilled clay, there is even more (the refined calculation is in the advanced section). The low-gravity solids, 6.5%, are rather a lot for a mud of 1.28 g/cm³. The PV will rise, the cake will get thicker and the rate of penetration (how fast the bit drills) will fall. The solution is to improve the cleaning and to dilute with fresh mud, not with water, which would lower the density. The data in this example are assumed for illustration.

Chemistry of the filtrate: pH, alkalinity, chlorides and hardness

pH is the measure of acidity: the negative decimal logarithm of the activity (the effective concentration) of hydrogen ions. It is measured with a pH meter with a glass electrode (accuracy 0.1; the instrument is calibrated with buffer solutions and corrected for temperature) or with indicator paper (accuracy about 0.5). Water-based muds are usually kept alkaline, with a pH of about 8.5–10.5. Lime muds (muds with extra lime added) are kept above 11. At a high pH the clays carry a negative charge and flocculate less, many chemicals work better, and corrosion is slower. Hydrogen sulfide also stays in the mud as ions and does not escape as a gas while the pH is high (Water-based muds and clay chemistry). But pH is a logarithm. At pH 12 there are a hundred times more hydroxide ions than at pH 10. And the pH alone does not say which ions make the mud alkaline.

This is shown by the alkalinity: the amount of acid needed to neutralise the alkali in the filtrate. The filtrate is titrated with 0.02 N sulfuric acid up to two end points. Up to pH 8.3 (the indicator phenolphthalein, whose crimson colour disappears) we use $P_f$ cm³ of acid per 1 cm³ of filtrate. During this stage the hydroxide ions are neutralised, and the carbonate ions turn into bicarbonate. The full consumption up to pH 4.3 (the indicator methyl orange) is $M_f$ cm³. By this point the bicarbonate has turned into carbonic acid too. By comparing $P_f$ and $M_f$ we find which ions give the alkalinity.

SchematicWhat the acid neutralises when the filtrate is titrated. Up to the phenolphthalein point (pH 8.3) it neutralises the hydroxide ions and the first step of the carbonates. Up to the methyl orange point (pH 4.3) it neutralises the bicarbonate that has formed and the bicarbonate that was there at the start. So

P_f =

OH⁻ + CO₃²⁻ and

M_f =

OH⁻ + 2·CO₃²⁻ + HCO₃⁻ (in cm³ of acid). A schematic.

Alkalinity ions from

P_f

and

M_f

, mg/L

Relation: $P_f = 0$
OH⁻: 0
CO₃²⁻: 0
HCO₃⁻: $1220\,M_f$

Relation: $2P_f < M_f$
OH⁻: 0
CO₃²⁻: $1200\,P_f$
HCO₃⁻: $1220\,(M_f - 2P_f)$

Relation: $2P_f = M_f$
OH⁻: 0
CO₃²⁻: $1200\,P_f$
HCO₃⁻: 0

Relation: $2P_f > M_f$
OH⁻: $340\,(2P_f - M_f)$
CO₃²⁻: $1200\,(M_f - P_f)$
HCO₃⁻: 0

Relation: $P_f = M_f$
OH⁻: $340\,M_f$
CO₃²⁻: 0
HCO₃⁻: 0

Relation	OH⁻	CO₃²⁻	HCO₃⁻
$P_f = 0$	0	0	$1220\,M_f$
$2P_f < M_f$	0	$1200\,P_f$	$1220\,(M_f - 2P_f)$
$2P_f = M_f$	0	$1200\,P_f$	0
$2P_f > M_f$	$340\,(2P_f - M_f)$	$1200\,(M_f - P_f)$	0
$P_f = M_f$	$340\,M_f$	0	0

$P_f$ and $M_f$ are cm³ of 0.02 N acid per 1 cm³ of filtrate, and $M_f$ is the full consumption up to pH 4.3. The coefficients convert 0.02 eq/L into mg/L. We multiply by the equivalent mass of the ion (17 g for OH⁻, 61 g for HCO₃⁻) or by the molar mass of carbonate (60 g), because up to pH 8.3 a carbonate takes one proton out of two. Hydroxide and bicarbonate cannot exist together: they react with each other.

Hydroxide ions are the “useful” alkalinity, which is made by caustic soda and lime. Carbonates, and especially bicarbonates, appear when carbon dioxide from the formation gets into the mud, when too much soda ash has been added, or when organic chemicals break down at high temperature. They flocculate the clay and make the mud thicken, especially after heating. Thinners remove this thickening only for a short time. The bicarbonate is turned into carbonate by raising the pH, and the carbonate is precipitated with lime or gypsum (chemical treatment).

The mud itself, and not the filtrate, is also titrated with the same acid. This gives the alkalinity of the mud $P_m$ . The difference between $P_m$ and the alkalinity of the water in the sample ( $f_w P_f$ ) is the alkalinity of the solid particles that have not dissolved yet, first of all of lime $\mathrm{Ca(OH)_2}$ . In lime muds a reserve of it is kept on purpose. In a fresh mud it appears after cement is drilled out.

Excess of undissolved lime

c_{\text{lime}} = 0.742\,(P_m - f_w P_f)

where:

$c_{\text{lime}}$ excess of undissolved lime, kg/m³
$P_m$ alkalinity of the mud: cm³ of 0.02 N acid per 1 cm³ of mud up to pH 8.3, cm³/cm³
$P_f$ alkalinity of the filtrate up to pH 8.3, cm³/cm³
$f_w$ share of water in the mud, from the retort, fraction

Units: $c_{\text{lime}}$ in kg/m³; $P_m$ , $P_f$ in cm³ per cm³

Holds when

All the difference between the alkalinity of the mud and of the filtrate is given by lime (carbonates and other alkaline particles make it too high).
The filtrate has the same composition as the water inside the mud.

1 cm³ of acid per 1 cm³ of mud is 0.02 eq/L, or $0.02\cdot 37.05 \approx 0.74$ g of $\mathrm{Ca(OH)_2}$ per litre (37.05 g is the equivalent mass of lime).

In other unit systems (1)

Oilfield units

c_{\text{lime}}\,[\text{lb/bbl}] = 0.26\,(P_m - f_w P_f)

excess lime in pounds per barrel.

$0.26 \approx 0.742/2.853$ .

Chlorides are found by titrating the filtrate with silver nitrate. First the filtrate is neutralised with acid until the colour of phenolphthalein disappears, because in an alkaline solution the silver would precipitate as an oxide. Then the indicator potassium chromate is added, and silver nitrate is added drop by drop. The chloride ions precipitate as white silver chloride. When they run out, red-brown silver chromate begins to precipitate, and the yellow colour turns orange-red.

Chlorides of the filtrate

[\mathrm{Cl^-}] = 1000\,\frac{V_{\mathrm{AgNO_3}}}{V_{\text{mf}}}

where:

$[\mathrm{Cl^-}]$ mass concentration of chloride ions, mg/L
$V_{\mathrm{AgNO_3}}$ volume of 0.0282 N silver nitrate up to the end point, cm³
$V_{\text{mf}}$ volume of filtrate in the sample, cm³

Units: mg/L; volumes in cm³

Holds when

Silver nitrate is 0.0282 N: 1 cm³ precipitates 1 mg of chloride. For salty muds 0.282 N is used, and the coefficient becomes 10,000.
There are no other ions in the filtrate that silver precipitates: bromides, iodides, sulfides.

If all the chlorine is in sodium chloride, NaCl is 1.65 times more ( $58.44/35.45$ ). The result is in milligrams per litre. To get parts per million by mass (ppm), we divide it by the density of the filtrate in g/cm³. In a fresh mud the difference is tiny. In a saturated NaCl brine (about 190,000 mg/L of chlorides at a density of about 1.2 g/cm³) it is about 16%.

Hardness is the content of calcium and magnesium ions. The filtrate is titrated with a solution of EDTA, a reagent that binds calcium and magnesium (it is the disodium salt of ethylenediaminetetraacetic acid; in Russian practice it is called Trilon B). A buffer keeps the pH at about 10, and a metal indicator (a dye that changes colour when it binds metal ions) is added. While free calcium and magnesium ions are in the solution, the colour is wine-red. After they are bound, it turns blue. With a 0.01 M EDTA solution the hardness, counted as calcium, is $400\,V_{\text{EDTA}}/V_{\text{mf}}$ mg/L. Calcium alone, apart from magnesium, is found at pH 12–13, when the magnesium precipitates as hydroxide. A dark filtrate with lignites and lignosulfonates (thinners made from wood pulp) makes it hard to see the colour change, so the sample is diluted or decolourised.

Worked exampleChemistry of the filtrate after cement is drilled out

Given (assumed data)

pH (pH meter): 12.4; before drilling out it was 9.8
Alkalinity of the filtrate $P_f$ / $M_f$: 1.6 / 1.6 cm³/cm³
Alkalinity of the mud $P_m$: 5.2 cm³/cm³
Share of water from the retort $f_w$: 0.88
Hardness: 0.01 M EDTA per 1.0 cm³ of filtrate: 0.9 cm³; before drilling out it was 80 mg/L

Solution

We compare: $P_f = M_f = 1.6$ , so all the alkalinity of the filtrate is hydroxide. There are no carbonates or bicarbonates. It cannot be otherwise. Next to hundreds of milligrams of calcium per litre, carbonate ions do not stay in solution: they drop out as calcium carbonate.
Hydroxide ions: $340\cdot 1.6 = 544$ mg/L.
A check with the pH: 544 mg/L of OH⁻ is $544/17{,}007 = 0.0320$ mol/L, and $\text{pH} \approx 14 + \log_{10} 0.0320 = 12.5$ . This is close to the measured 12.4 (a calculation without a correction for the activity of the ions overstates the pH a little).
Excess lime: $0.742\cdot(5.2 - 0.88\cdot 1.6) = 0.742\cdot 3.792 = 2.81$ kg/m³ (0.99 lb/bbl).
Calcium: $400\cdot 0.9/1.0 = 360$ mg/L, which is 4.5 times more than before drilling out.

Answer: The filtrate holds 544 mg/L of hydroxide ions (no carbonates) and 360 mg/L of calcium. The mud holds about 2.81 kg/m³ of undissolved lime. The mud is contaminated with cement.

What it means. A high pH, a rise in $P_m$ with a moderate $P_f$ , and a rise in calcium, all together, are the “signature” of cement. Cement gives lime to the mud, and the undissolved reserve of lime goes on feeding the filtrate with calcium. This is what flocculated the clay in the example with the viscometer. The treatment must tie up the calcium and lower the pH at the same time. Sodium bicarbonate does this. Soda ash, which is used to treat contamination with gypsum and anhydrite, would raise the pH even more here (calculation of the treatment). The data in this example are assumed for illustration.

Worked exampleChlorides have risen: salt or formation water?

Given (assumed data)

Titration of 1.0 cm³ of filtrate with 0.0282 N silver nitrate: 2.4 cm³ in the morning; 6.9 cm³ after 40 m of drilling
Mud in circulation: 180 m³, share of water 0.88
Chlorides of the formation water (from neighbouring wells): 120,000 mg/L
Diameter of the hole: 215.9 mm

Solution

Chlorides: $1000\cdot 2.4/1.0 = 2400$ mg/L and $1000\cdot 6.9/1.0 = 6900$ mg/L. As NaCl they are 3960 and 11,385 mg/L.
Water in the mud: $0.88\cdot 180 = 158.4$ m³. This much formation water would raise the chlorides to 6900 mg/L: $V_{\text{fw}} = 158.4\cdot\dfrac{6900 - 2400}{120{,}000 - 6900} = 6.3$ m³.
The same rise would come from this much rock salt: $1.65\cdot 4500 = 7425$ mg/L of NaCl, that is $7.425\cdot 158.4 = 1176$ kg, or $1176/2165 = 0.543$ m³ of salt with a density of 2.165 g/cm³.
That is $0.543/0.0366 = 15$ m of a hole of 215.9 mm diameter (cross-section area 0.0366 m²) drilled through salt.
In parts per million by mass, with a filtrate density of 1.005 g/cm³: $6900/1.005 = 6866$ ppm. In a fresh mud the difference from mg/L is negligible.

Answer: The chlorides have risen from 2400 to 6900 mg/L. Such a rise would come from about 6.3 m³ of formation water or from about 0.543 m³ of dissolved salt, which is about 15 m of hole through a salt bed.

What it means. More than six cubic metres of inflow would not go unnoticed: the volume in the tanks would grow, and the density at the outlet would fall. Suppose there is no gain in volume and salt is visible in the cuttings. Then the chlorides came from a salt bed that has dissolved. We then decide whether to change the mud to a salt mud before the hole is washed out (Water-based muds and clay chemistry). Suppose instead there is both a gain in volume and a fall in density at the outlet. These are signs of a kick, and we must act by the well control procedures (Well control: primary barrier and kick indicators). The calculation is simplified. It does not take into account the water added in the same time and the treatments, and dilution reduces the rise in chlorides. So the calculation underestimates the volume of the source. The data in this example are assumed for illustration.

How to read a mud report

Single tests answer narrow questions. The cause of changes is usually shown by a combination of readings and by the time at which they changed. Below are the typical “signatures” of events in a water-based mud.

Typical combinations of changes

Event: Build-up of fine cuttings
What changes: PV rises at the same density, YP rises moderately; the cake is thicker
How to confirm: the retort and the methylene blue test: more low-gravity solids

Event: Drilling out cement
What changes: YP and gel strength rise sharply, the gel is progressive
How to confirm: pH above 11; $P_m$ and calcium rise

Event: Anhydrite or gypsum
What changes: YP, gel strength and fluid loss rise
How to confirm: calcium rises, but the pH does not rise (it may fall); anhydrite in the cuttings

Event: Salt or salty formation water
What changes: YP, gel strength and fluid loss rise
How to confirm: chlorides rise; salt in the cuttings, or a gain in volume in the tanks

Event: Carbon dioxide, too much soda ash, breakdown of chemicals
What changes: YP and gel strength rise, especially after heating
How to confirm: $M_f$ is much greater than $P_f$ , the pH falls

Event: High bottomhole temperature
What changes: after a trip (pulling the string out and running it back), the mud from the well is thicker than the mud in the tanks
How to confirm: the HPHT fluid loss rises; the properties are worse after the sample is heated

Event: Air or gas
What changes: the rheology hardly changes
How to confirm: the pressurised balance shows more than the ordinary one

Event	What changes	How to confirm
Build-up of fine cuttings	PV rises at the same density, YP rises moderately; the cake is thicker	the retort and the methylene blue test: more low-gravity solids
Drilling out cement	YP and gel strength rise sharply, the gel is progressive	pH above 11; $P_m$ and calcium rise
Anhydrite or gypsum	YP, gel strength and fluid loss rise	calcium rises, but the pH does not rise (it may fall); anhydrite in the cuttings
Salt or salty formation water	YP, gel strength and fluid loss rise	chlorides rise; salt in the cuttings, or a gain in volume in the tanks
Carbon dioxide, too much soda ash, breakdown of chemicals	YP and gel strength rise, especially after heating	$M_f$ is much greater than $P_f$ , the pH falls
High bottomhole temperature	after a trip (pulling the string out and running it back), the mud from the well is thicker than the mud in the tanks	the HPHT fluid loss rises; the properties are worse after the sample is heated
Air or gas	the rheology hardly changes	the pressurised balance shows more than the ordinary one

The combinations are only a guide: one event can hide another. The treatment is chosen on pilot samples, which are small portions of the mud into which the chemicals are added and the tests are repeated.

Besides the main set of tests, the programme may require other measurements:

the resistivity of the mud, the filtrate and the cake, to interpret logs (Borehole environment: invaded zone and mudcake, Determining formation water resistivity)
potassium ions in potassium muds, where the chlorides cannot tell potassium chloride from sodium chloride that has come in
sulfides in regions with hydrogen sulfide: the hydrogen sulfide is driven off from the acidified filtrate and measured with an indicator tube
the particle size of the solids and the plugging ability on ceramic discs, when drilling into productive formations (Formation damage and damage skin)
the temperature of the mud at the outlet: its trend with depth helps to recognise the entry into a zone of abnormally high pressure (Abnormal formation pressures: origin, prediction and recognition)

Terms and symbols

Pressurised mud balance: A lever balance with a sealed cup in which the mud is pressed by a piston. The bubbles are squeezed, and the balance shows the true density.
Funnel viscosity: The time in seconds that the mud takes to run out of a funnel of standard size. It is an indicator of changes, not a physical viscosity.
Spurt loss, $V_{sp}$: The volume of filtrate that passes before the cake forms. It is the intercept of the line $V(\sqrt t)$ on the volume axis.
HPHT fluid loss: The volume of filtrate in 30 min at high pressure and high temperature (usually 3.45 MPa and 150 °C), doubled because the cell has half the area.
Retort: An instrument for distilling a sample of mud. It gives the shares of water, oil and solids by volume.
Low-gravity and high-gravity solids (LGS, HGS): Clay, cuttings and sand (about 2.6 g/cm³) and the weighting material (barite, 4.2 g/cm³). Their shares are found from the density and the retort.
Methylene blue capacity (MBC), bentonite equivalent: The cm³ of dye per 1 cm³ of mud, and the content of bentonite that matches it, $14.25\,\mathrm{MBC}$ kg/m³.
Alkalinity $P_f$ , $M_f$ , $P_m$: The consumption of 0.02 N acid per 1 cm³ of filtrate up to pH 8.3 and 4.3, and per 1 cm³ of mud up to pH 8.3.
Excess lime: Undissolved lime in the mud, $0.742\,(P_m - f_w P_f)$ kg/m³.
Hardness: The content of calcium and magnesium in the filtrate, in mg/L as calcium. It is found by titration with EDTA.

Common mistakes and misconceptions

Wrong

“The density from the mud balance is always the true one.”

In fact

Bubbles of air or gas make it too low. A foamed or gas-cut mud needs a pressurised balance. Weighting up by the ordinary balance makes the mud heavier than it should be.

Wrong

“Twice the volume of filtrate at 7.5 min is the fluid loss at 30 min.”

In fact

Doubling also multiplies the spurt loss by two, so the estimate is too high by exactly $V_{sp}$ . The standard fluid loss is the volume collected in the full 30 min.

Wrong

“The methylene blue test shows how much bentonite was added to the mud.”

In fact

It shows all the active clay, counted as bentonite, drilled clay included. To separate bentonite from cuttings we need the cation exchange capacity of the drilled rock.

Wrong

“Chlorides in mg/L are the same as ppm.”

In fact

Milligrams per litre are divided by the density of the filtrate to give parts per million by mass. In a fresh mud the difference is tiny. In a saturated brine it is about 16%.

Wrong

“The pH is high, so the alkalinity is fine.”

In fact

The same pH can come from hydroxides or from carbonates. $P_f$ and $M_f$ tell them apart: when $M_f$ is much greater than $P_f$ , the mud is contaminated with carbonates and bicarbonates.

In depth: advanced detail

Permeability and compressibility of the cake

The formula for the permeability of the cake comes from two relations of static filtration (the lesson on the invaded zone). The first is the balance of the solids: the more filtrate has passed, the thicker the cake, $h_{mc} = cV/A$ . The second is the law of filtration through a growing cake: $V^2 = 2A^2 k_{mc}\,\Delta p\,t/(\mu_f c)$ . We put $c = A h_{mc}/V$ from the first into the second and get $V = 2A\,k_{mc}\,\Delta p\,t/(\mu_f h_{mc})$ . This gives the formula for $k_{mc}$ . The 2 appears because the rate of filtration is inversely proportional to the current thickness of the cake, and the thickness grew in proportion to the filtrate that had passed. On average, each portion of the filtrate passed through a cake half as thick as the final one.

Real cakes are compressible. Under a large pressure difference the particles pack more tightly, and the permeability falls. So the volume of filtrate grows with the pressure difference more slowly than $\sqrt{\Delta p}$ , which is what the model of an incompressible cake predicts. This is described by a power law $V \propto \Delta p^{\,x}$ . For an incompressible cake $x = 0.5$ , and for good bentonite cakes it is much smaller. The exponent is found from two tests at one temperature and at different pressure differences: $x = \ln(V_2/V_1)/\ln(\Delta p_2/\Delta p_1)$ . For example, 7.6 and 10.0 cm³ at 0.69 and 3.45 MPa give $x = \ln(10.0/7.6)/\ln 5 \approx 0.17$ . The mud “compacts” its own cake in response to the rise in pressure.

During circulation the flow of mud erodes the cake. Its thickness stops growing, and the filtration goes on at an almost constant rate. The static test does not reproduce this. So it compares muds with each other and does not predict the volume of filtrate in the formation. Dynamic filtration is studied in laboratory setups with a flow along the filter.

The retort in a salty mud

The retort collects pure water, and the salt stays in the cup and is counted as “solids”. In fact the liquid phase of the mud is a brine, and its volume is larger than the volume of the water. In a litre of brine with a density $\rho_b$ (g/cm³) and a salt concentration $c_s$ (g/L) there are $1000\rho_b - c_s$ grams of water, that is $(1000\rho_b - c_s)/1000$ litres. This gives the correction.

Share of brine from the volume of water in the retort

f_b = \frac{f_w}{\rho_b - c_s/1000}

where:

$f_b$ share of brine in the volume of the mud, fraction
$f_w$ share of water from the retort, fraction
$\rho_b$ density of the brine (the water phase), g/cm³
$c_s$ concentration of the dissolved salt in the brine; for NaCl from the chlorides it is $1.65\,[\mathrm{Cl^-}]/1000$ , g/L

Units: $\rho_b$ in g/cm³, $c_s$ in g/L

Holds when

The density of the water in the retort is 1.0 g/cm³.
There is one salt, and it is known (usually NaCl). For a mixture of salts we need reference data on the density of the brine.

The corrected share of solids is $1 - f_b - f_o$ . In the formula for the average density of the solids we put $f_b\rho_b$ in place of $f_w\rho_w$ . In a fresh mud the denominator is almost 1. In a saturated NaCl brine (26.4% by mass, 1.197 g/cm³, 316 g/L) it is 0.881, and there is 13.5% more brine than water.

Worked exampleSalt-saturated mud: what the retort hides

Given (assumed data)

Density of the mud: 1.40 g/cm³
Retort: water 78.0%, no oil
Water phase: saturated NaCl brine: 1.197 g/cm³, 316 g/L of salt

Solution

Without the correction: $f_s = 0.220$ , $\bar\rho_s = (1.40 - 0.780)/0.220 = 2.818$ g/cm³, $f_{LGS} = 22.0\cdot(4.20 - 2.818)/1.60 = 19.0$ %.
Share of brine: $f_b = 0.780/(1.197 - 0.316) = 0.780/0.881 = 0.8854$ ; the solids are $1 - 0.8854 = 0.1146$ , that is 11.46%.
With the correction: $\bar\rho_s = (1.40 - 0.8854\cdot 1.197)/0.1146 = 2.968$ g/cm³, $f_{LGS} = 11.46\cdot(4.20 - 2.968)/1.60 = 8.8$ %, barite 2.6%.

Answer: Without the correction the solids are 22%, of which 19.0% is “clay and cuttings”. In fact the solids are 11.46%, the low-gravity solids are 8.8% and the barite is 2.6%.

What it means. Without the correction, the salt that stays in the retort cup looks like drilled rock, and the mud seems twice as dirty as it is. The engineer would start to dilute it and lose the expensive brine. So in salty muds the retort is always recalculated from the chlorides. The data in this example are assumed for illustration.

When drilled clay takes up the dye too

The bentonite equivalent assigns all the uptake of the dye to bentonite. But drilled clay takes up the dye too. Per unit of mass it takes up less than bentonite, but there may be a lot of it. Suppose we measure the cation exchange capacity of washed cuttings, with the same methylene blue test on a sample of dried rock. Then the bentonite and the drilled rock can be separated. The dye taken up by a litre of mud is the sum of the contributions of the two clays, and their total mass is the low-gravity solids.

Bentonite and drilled rock from the methylene blue test and the low-gravity solids

B = \frac{1000\,\mathrm{MBC} - E_d\,c_{LGS}}{E_b - E_d}, \qquad D = c_{LGS} - B

where:

$B,\ D$ content of bentonite and of drilled rock, kg/m³
$c_{LGS}$ low-gravity solids from the density and the retort, kg/m³
$E_b,\ E_d$ cation exchange capacity of bentonite (about 70) and of the drilled rock (from the cuttings), meq/100 g
$\mathrm{MBC}$ methylene blue capacity of the mud, cm³/cm³

Units: kg/m³ = g/L; capacities in meq/100 g

Holds when

The dye is 0.01 N: $\mathrm{MBC}$ cm³/cm³ means $10\,\mathrm{MBC}$ meq per litre of mud, and 1 kg/m³ of clay with a capacity $E$ takes up $E/100$ meq per litre.
Sand and carbonates among the low-gravity solids take up almost no dye. If their share is large, it is included in the $E_d$ of the cuttings.
The sample of cuttings is representative of the fine drilled rock in the mud.

At $E_d = 0$ the formula becomes the ordinary bentonite equivalent $1000\,\mathrm{MBC}/70 \approx 14.3\,\mathrm{MBC}$ . The more active the drilled clay, the more the simple calculation overstates the bentonite and understates the cuttings.

Worked exampleA refined split for sample A

Given (assumed data)

Methylene blue capacity of the mud: 4.5 cm³/cm³ (the example above)
Low-gravity solids: 169 kg/m³
Cation exchange capacity: bentonite / washed cuttings: 70 / 12 meq/100 g

Solution

Bentonite: $B = \dfrac{1000\cdot 4.5 - 12\cdot 169}{70 - 12} = \dfrac{2472}{58} = 42.6$ kg/m³.
Drilled rock: $D = 169 - 42.6 = 126.4$ kg/m³; the ratio to bentonite is 2.97.

Answer: Bentonite 42.6 kg/m³ instead of 64 by the bentonite equivalent, drilled rock 126.4 kg/m³; the ratio is 2.97 instead of 1.6.

What it means. About a third of the dye that was taken up really belongs to the drilled clay, and the simple calculation counted it as bentonite. The mud carries noticeably more cuttings than it seemed, and the decision to clean and dilute becomes even clearer. The data in this example are assumed for illustration.

Connections to other topics

Drilling fluids: functions and properties — the functions and properties of the mud, the units of density and weighting up.
Drilling fluid rheology and flow regimes — flow models from the viscometer readings and their conversion into SI units.
Water-based muds and clay chemistry — the chemistry of clays and the treatment of the mud on the basis of the analyses.
Well control: primary barrier and kick indicators — density as the primary barrier; gas-cut mud and the signs of a kick.
Borehole environment: invaded zone and mudcake — the cake, the filtrate and their resistivities in logging; the theory of static filtration.
Solids control in drilling fluids — how cuttings and sand are removed when the analysis shows too much of them.
Oil-based muds — tests of oil-based muds.
Formation damage and damage skin — why filtration and particle size matter especially when drilling into a formation.

Check yourself

1
The mud in the tank is foaming. The ordinary mud balance showed 1.18 g/cm³, and the pressurised balance on the same sample showed 1.22 g/cm³. Which statement is true?
1. A.The mud must be weighted up so that the ordinary balance shows 1.22 g/cm³
2. B.A mud with a density of 1.18 g/cm³ will press on the bottom
3. C.The true density is 1.22 g/cm³, and the sample holds about 3% of air; there is no need to weight up
4. D.The pressurised balance is wrong, because the pressure compresses the mud itself
Show the answer
Correct answer: C. The true density is 1.22 g/cm³, and the sample holds about 3% of air; there is no need to weight up
The share of air is $1 - 1.18/1.22 \approx 0.033$ , that is about 3.3%. Under the pressure in the balance (and even more in the well) the bubbles are squeezed, while the mud itself is almost incompressible. So the true density is 1.22 g/cm³. Weighting up “to 1.22 by the ordinary balance” would give a mud of about 1.26 g/cm³.
2
In a standard filter-press test, 3.0 cm³ of filtrate was collected in 7.5 min and 5.2 cm³ in 30 min. Find the spurt loss and the error of the estimate by doubling.
Show the answer
Since $\sqrt{30} = 2\sqrt{7.5}$ , $V_{sp} = 2\cdot 3.0 - 5.2 = 0.8$ cm³. Doubling gives 6.0 cm³, which is 0.8 cm³ (about 15%) more than the measured value. The slope of the line is $C = 2.2/(\sqrt{30} - \sqrt{7.5}) \approx 0.80$ cm³/min $^{1/2}$ .
3
Over one shift the PV of the mud did not change, the YP rose from 9 to 21 lb/100 ft², the gel strength became 10 / 31 lb/100 ft², the pH rose from 9.6 to 12.1, and the calcium rose from 60 to 420 mg/L. What most likely happened?
1. A.Fresh formation water entered the mud
2. B.The mud is contaminated with cement
3. C.Fine cuttings built up
4. D.There is air in the sample
Show the answer
Correct answer: B. The mud is contaminated with cement
The PV is the same, so there are no more solids. The YP and the progressive gel rose, so the clay flocculated. The pH and the calcium rising together is the “signature” of cement. Fine cuttings would first of all raise the PV. Fresh water would dilute the mud and lower the YP. Air would change only the readings of the ordinary balance.
4
A fresh mud with no oil: density 1.45 g/cm³, 80% water by the retort. Find the shares of low-gravity solids and barite (2.6 and 4.2 g/cm³). Is the mud good?
Show the answer
$f_s = 0.20$ ; $\bar\rho_s = (1.45 - 0.80)/0.20 = 3.25$ g/cm³; $f_{LGS} = 20\cdot(4.2 - 3.25)/1.6 \approx 11.9$ %, and barite is about 8.1% (about 340 kg/m³). Almost 12% of clay and cuttings is too much: the mud will be thick, the cake thick and the drilling slow. We need cleaning (a centrifuge) and dilution with fresh mud.
5
Alkalinity of the filtrate: $P_f = 0.1$ , $M_f = 3.4$ cm³/cm³. Which ions dominate, and what does it mean?
1. A.Calcium ions: the mud is contaminated with cement
2. B.Hydroxide ions: the mud is fine
3. C.Bicarbonate ions: probably carbon dioxide or too much soda ash; the bicarbonate is tied up with lime
4. D.Chloride ions: salt has got into the mud
Show the answer
Correct answer: C. Bicarbonate ions: probably carbon dioxide or too much soda ash; the bicarbonate is tied up with lime
$2P_f = 0.2 < M_f$ , so there is no hydroxide. Carbonates are $1200\cdot 0.1 = 120$ mg/L, and bicarbonates are $1220\cdot(3.4 - 0.2) \approx 3900$ mg/L. Such a mud tends to thicken, especially when heated. Lime turns the bicarbonate into carbonate and precipitates it as calcium carbonate, which raises the pH. Chlorides and calcium are not measured by the alkalinity.
6
In a salt-saturated mud, 9.6 cm³ of 0.282 N silver nitrate was used for 0.5 cm³ of filtrate. Find the chlorides in mg/L and in ppm for a filtrate density of 1.20 g/cm³. Is the brine saturated?
Show the answer
For a 0.282 N solution the coefficient is 10,000: $[\mathrm{Cl^-}] = 10{,}000\cdot 9.6/0.5 = 192{,}000$ mg/L, or $192{,}000/1.20 = 160{,}000$ ppm. NaCl is about $1.65\cdot 192{,}000 \approx 317{,}000$ mg/L. This is about the limit of solubility at room temperature (about 316 g/L), so the brine is saturated.

Updated: 19 October 2026