Solids control in drilling fluids

Quick refresher

Drilled rock is a useless solid: unlike bentonite and barite, it only thickens the mud. The pumps, the bit and the flow in the pipes grind it at every cycle, and the finer the particles, the larger their surface and the harder they are to separate. So the cuttings are removed in the first cycle, while they are still coarse.
Particles are separated by screening (shale shakers) and by faster settling (hydrocyclones and centrifuges), and the rest of the fines is diluted. By Stokes' law the settling velocity is proportional to $d^2(\rho_s - \rho_f)$ : fine particles hardly settle in the tanks, and a centrifugal acceleration of hundreds and thousands of $g$ makes them separable.
The cut point $d_{50}$ is the size of the particles that the device sends to waste half of the time. For barite it is smaller than for drilled rock, in water by a factor of about $\sqrt2$ . So hydrocyclones throw barite out, and a weighted mud is cleaned with shale shakers, mud cleaners and centrifuges.
The system is arranged by decreasing size of the particles to be removed: shale shakers, degasser (it takes gas out of the mud), desander (it removes sand), desilter (it removes silt), centrifuge. For a weighted mud, a mud cleaner (hydrocyclones over a fine screen) stands in place of the plain hydrocyclones, and the centrifuge returns the barite.
Whatever the equipment has not removed must be diluted: $V_{\text{dil}} = (1 - E)V_{ds}/f_{ds}$ . Raising the cleaning efficiency from 0.4 to 0.8 cuts the dilution threefold, and the new mud and the waste in the lesson's example by 2.6–2.7 times. The extra equipment usually pays for itself.
We check the work of the system by the actual dilution: how much mud was made, judging by the use of water, and how much drilled rock is left in it, judging by the retort and the methylene blue test.

Key relationships

V_{ds} = \frac{\pi}{4}D^2 L\,k_w(1-\phi)

volume of drilled rock over the interval

v_s = \dfrac{a\,d^2(\rho_s - \rho_f)}{18\,\mu}

settling velocity by Stokes; $a = g$ in a tank, $a = \omega^2 r$ in a centrifuge

d_{50,b} = d_{50}\sqrt{\dfrac{\rho_{LGS} - \rho_f}{\rho_{HGS} - \rho_f}}

cut point for barite particles

V_{\text{dil}} = \dfrac{(1 - E)\,V_{ds}}{f_{ds}}

mud needed to dilute the rock that was not removed

Common pitfalls

Putting hydrocyclones on a weighted mud: together with sand and silt they throw out barite.
Fitting screens finer than the shakers can pass: the mud runs over the screen together with the cuttings.
Diluting a weighted mud with water: the density falls, and barite and chemicals are used to restore it.
Letting the mud bypass the equipment, or feeding the next stage from the same tank section into which it discharges the cleaned mud.
Judging the cleaning by the look of the cuttings on the screens, not by the analysis of the mud and the solids balance.

After this lesson you can

explain where drilled rock in the mud comes from, why it gets ground finer, and why fine particles do more harm than coarse ones
describe how shale shakers and screens, hydrocyclones, mud cleaners and centrifuges work, read a partition curve (a graph of what share of each particle size a device removes) and explain why the cut point is smaller for barite
calculate the volume of drilled rock and make a solids balance: how much the equipment removes, how much mud goes to dilution and how much waste is left
judge the efficiency of the cleaning system from the actual use of water and from the mud analysis, and compare the cost of cleaning and of dilution
choose a layout of the cleaning system for an unweighted mud, a weighted mud and an oil-based mud

Drilled rock in the mud

The bit turns rock into cuttings, and all of it gets into the mud. The cuttings have a density of about 2.6 g/cm³. In the mud analysis they are joined with bentonite as low-gravity solids. They come to the surface as fragments from fractions of a millimetre to a centimetre. On the way they are crushed by the bit, by the turbulent flow in the pipes and by the pumps, and clays also soften and break up in water. Every new cycle makes the particles smaller, and removing them becomes harder.

There are two useful kinds of solids in a mud: bentonite (or another clay that builds structure) and the weighting material, barite. Drilled rock is never useful. It raises the plastic viscosity, the ECD and the pump pressure. It makes the cake thicker and more permeable, and with it the risk of differential sticking (the pipe held against the wall by the pressure difference) grows. It wears the pumps and the valves, and it lowers the rate of penetration. It is impossible to remove it completely. So the drilling programme sets an allowed content, usually a few per cent of the volume, and a way to keep to it.

Solid particles of drilling mud by size

Class: Sand
Size: larger than 74 µm
What it usually is: coarse cuttings, sand, fragments
What separates it: shale shakers, desanders

Class: Silt
Size: 2–74 µm
What it usually is: ground cuttings, most of the barite
What separates it: desilters, centrifuges

Class: Colloidal particles
Size: smaller than 2 µm
What it usually is: bentonite, the finest clay
What separates it: hardly separated at all: dilution

Class	Size	What it usually is	What separates it
Sand	larger than 74 µm	coarse cuttings, sand, fragments	shale shakers, desanders
Silt	2–74 µm	ground cuttings, most of the barite	desilters, centrifuges
Colloidal particles	smaller than 2 µm	bentonite, the finest clay	hardly separated at all: dilution

The boundaries follow the classification of the American Petroleum Institute (API). 74 µm is the opening of the No. 200 sieve on which the sand content is measured (tests of the mud).

Volume of drilled rock

V_{ds} = \frac{\pi}{4}\,D^2\,L\,k_w\,(1 - \phi), \qquad q_{ds} = \frac{\pi}{4}\,D^2\,v_{\text{ROP}}\,k_w\,(1 - \phi)

where:

$V_{ds}$ volume of drilled rock (its mineral matrix) over the interval, m³
$q_{ds}$ the same per unit of drilling time, m³/h
$D$ bit diameter, m
$L$ length of the interval drilled, m
$v_{\text{ROP}}$ rate of penetration (ROP), m/h
$k_w$ washout factor: the ratio of the actual volume of the hole, enlarged by washouts, to its nominal volume; not less than 1, —
$\phi$ porosity of the rock, fraction

Units: SI (metres and hours)

Holds when

The pore fluid of the rock is not part of the solids balance.
All the destroyed rock goes into the mud. Caving of the walls would add more to it, and lost circulation would carry part of it into the formation.

The inflow of rock is proportional to the area of the bottom and to the rate of penetration: fast drilling with a large bit gives the cleaning system the most work.

In other unit systems (1)

Oilfield units

q_{ds}\,[\text{bbl/h}] = \frac{(1-\phi)\,d^2\,[\text{in}^2]\;\text{ROP}\,[\text{ft/h}]}{1029}

the bit diameter is in inches, the rate of penetration in feet per hour, and the result in barrels per hour (without washouts).

$1029 = 144\cdot 5.615\cdot 4/\pi$ : square inches to square feet and cubic feet to barrels.

Worked exampleHow much rock one run brings

Given (assumed data)

Interval: 2600–3000 m, bit 215.9 mm: the run from the lesson on bit hydraulics
Rate of penetration: 25 m/h
Washouts and porosity: $k_w = 1.10$ ; $\phi = 0.15$
Mud in circulation: about 250 m³ (the tanks and the hole)

Solution

Area of the bottom: $\frac{\pi}{4}\cdot 0.2159^2 = 0.03661$ m². Volume of the hole in the interval with washouts: $0.03661\cdot 400\cdot 1.10 = 16.11$ m³.
Drilled rock: $16.11\cdot(1 - 0.15) = 13.69$ m³, or $13.69\cdot 2.6 = 35.6$ t.
Per hour of drilling: $0.03661\cdot 25\cdot 1.10\cdot 0.85 = 0.856$ m³ (about 2.2 t). The whole interval is 16 h of drilling.
If all this rock stayed in 250 m³ of mud, its share would grow by $13.69/(250 + 13.69) = 5.2$ % of the volume.

Answer: In one run, about 13.69 m³ (35.6 t) of rock gets into the mud: 0.856 m³ for each hour of drilling.

What it means. More than five per cent of the volume in sixteen hours is more than is usually allowed for the whole content of drilled rock in the mud. So almost all the rock must be taken out of the mud, and best of all in the first cycle, before it is crushed. The data in this example are assumed for illustration.

How particles are separated: screening, settling, dilution

There are three ways. Screening holds back everything larger than the openings of a screen. This is the work of the shale shakers. Settling separates particles by the speed at which they settle in a liquid, that is, by size and density. In a tank, gravity is too weak for this, so hydrocyclones and centrifuges speed up the settling many times with centrifugal force. Finally, there is dilution. What could not be separated is held within the allowed limits by adding fresh mud and discarding part of the old.

Settling velocity of a particle (Stokes' law) and the separation factor

v_s = \frac{a\,d^2\,(\rho_s - \rho_f)}{18\,\mu}, \qquad a = g\ \text{or}\ a = \omega^2 r, \qquad G = \frac{\omega^2 r}{g}

where:

$v_s$ settling velocity of the particle relative to the liquid, m/s
$a$ acceleration: gravity $g$ in a tank, or the centrifugal $\omega^2 r$ in a hydrocyclone and a centrifuge, m/s²
$d$ particle diameter, m
$\rho_s,\ \rho_f$ densities of the particle and of the liquid around it, kg/m³
$\mu$ viscosity of the liquid around the particle, Pa·s
$\omega,\ r$ angular velocity of rotation and radius, s⁻¹, m
$G$ separation factor: how many times the centrifugal acceleration is greater than $g$ , —

Units: SI

Holds when

A spherical particle with laminar flow around it: the Reynolds number of the particle, $\rho_f v_s d/\mu$ , is below about 1.
The liquid is Newtonian (its viscosity is constant). For a drilling mud we take the effective viscosity at the shear rate around the particle, and at rest the gel may hold fine particles altogether.
The particles do not hinder each other. In a thick suspension, settling is slower.

The settling velocity grows as the square of the size: a particle of 5 µm settles 16 times more slowly than one of 20 µm. The acceleration and the density difference enter linearly. Hydrocyclones and centrifuges increase the acceleration many times.

Worked exampleSettling in a tank and in a centrifuge

Given (assumed data)

Particles: drilled rock 2.6 g/cm³, sizes 20 and 5 µm
Liquid: 1.00 g/cm³; effective viscosity 5 mPa·s (assumed)
Centrifuge: bowl of diameter 360 mm, 2500 rpm; liquid layer at the wall 50 mm

Solution

In a tank: $v_s = \dfrac{9.81\cdot(20\cdot10^{-6})^2\cdot 1600}{18\cdot 0.005}\cdot 3600 = 0.251$ m/h, so the particle takes 4.0 h to pass one metre of the layer. A particle of 5 µm settles 16 times more slowly: 64 h per metre.
Centrifuge: $\omega = 2\pi\cdot 2500/60 = 261.8$ s⁻¹. At the wall $a = \omega^2 r = 261.8^2\cdot 0.18 = 12{,}337$ m/s², that is $G = 12{,}337/9.81 = 1258$ .
The velocities in the centrifuge are 1258 times greater: 0.088 and 0.0055 m/s. The particles cross the 50 mm layer in 0.6 and 9 s.
Check: the Reynolds number of a 20 µm particle in the centrifuge is $1000\cdot 0.088\cdot 20\cdot10^{-6}/0.005 = 0.35$ , below 1, so Stokes' law applies.

Answer: In a tank, a particle of 20 µm sinks one metre in 4.0 h and a particle of 5 µm in 64 h. In the centrifuge they cross the layer of liquid in 0.6 and 9 s.

What it means. While a silt particle sinks in a tank, the mud has long gone to the pumps, and the gel holds the particle in suspension altogether. Settling tanks do not remove silt. In a centrifuge the same particle is separated in seconds, while the mud passes through the bowl. The acceleration is taken at the wall of the bowl, and nearer the axis it is smaller. So in reality the particles take about 15–20% longer to cross the layer. The viscosity here is assumed: in a thickened mud, settling is slower, and the cut point of the centrifuge moves to larger particles. The data in this example are assumed for illustration.

Neither a screen, nor a hydrocyclone, nor a centrifuge cuts the particles at exactly one size. Each device is described by a partition curve: what share of the particles of a given size goes to waste. The size at which half of the particles go to waste is called the cut point ( $d_{50}$ ). Part of the fines goes to waste at any size, together with the liquid that goes with the separated particles.

CalculatedCalculated partition curves for drilled rock: a shale-shaker screen with an opening of about 75 µm, a desander, a desilter and a centrifuge. The dashed curve is the same desilter for barite particles. They send half of the particles to waste at sizes of about 65, 44, 17 and 5 µm, and half of the barite at about 12 µm. At the bottom the curves of the hydrocyclones and the centrifuge do not go down to zero: a few per cent of liquid goes to waste with the fine particles. Because of this the cut point is a little smaller than without this carry-over (45, 18 and 5 µm). The parameters are assumed.

Cut point for particles of another density

d_{50,b} = d_{50}\sqrt{\frac{\rho_{LGS} - \rho_f}{\rho_{HGS} - \rho_f}}

where:

$d_{50}$ cut point of the device for drilled rock, µm
$d_{50,b}$ cut point of the same device for barite, µm
$\rho_{LGS},\ \rho_{HGS},\ \rho_f$ densities of drilled rock (2.6), barite (4.2) and the liquid; the symbols are the same as in the analysis of solids, g/cm³

Units: any identical units

Holds when

The particles are separated by their settling velocity in the Stokes regime, and the working conditions of the device are the same.
In a weighted mud a large particle is surrounded by a mixture of liquid and fine solids, not by water, so the estimate is approximate.

A barite particle settles like a rock particle that is larger by a factor of $\sqrt{(4.2 - 1)/(2.6 - 1)} = \sqrt2$ (in water). A desilter with a cut point of 18 µm for rock sends to waste half of the barite particles of about 13 µm, and most of the larger ones. A large share of the barite is in these sizes.

Shale shakers and screens

The mud from the well comes along the flowline (the channel from the well) to the receiving box of the shale shaker and spreads over inclined screens. The frame with the screens vibrates. The liquid and the fine particles pass through the openings, and the cuttings jump and move to the edge, from which they fall into waste. Shale shakers are the first and cheapest stage of cleaning. They take out the coarse cuttings before they are crushed, and they protect the rest of the equipment from them. The whole flow from the well passes through them.

On old machines the frame moved in a circle or an ellipse. On modern machines it usually moves in a straight line at an angle to the screen. This way the cuttings can be driven up an inclined screen, and the liquid stays on the screen longer, which allows finer screens. The intensity of the vibration is estimated by the acceleration in units of $g$ : for a full stroke $s$ and an angular frequency $\omega$ it is $G = s\omega^2/(2g)$ . For example, at a stroke of 4 mm and 1800 vibrations a minute, $G \approx 7.2$ . Modern machines usually give several $g$ . Shaking harder makes the cuttings leave faster and raises the capacity, but the screens tear sooner.

Earlier, screens were named by the number of wires per inch (mesh), for example 80 × 80. But for the same density of wires the opening depends on the thickness of the wire, and multi-layer screens have different openings altogether. So since the mid-2000s a screen is tested and given an API number. This is the number of the standard sieve whose opening matches the size of the particles that the screen holds back almost completely. The label also gives the conductance (how easily liquid passes through the screen) and the non-blanked area (the part of the panel that is not covered, so that the flow can go through it). In Russian-language practice, woven screens are named by the nominal size of the mesh in millimetres.

API screen number and the opening of the matching standard sieve

Number: API 40
Opening, µm: 425
in mm: 0.425

Number: API 60
Opening, µm: 250
in mm: 0.250

Number: API 80
Opening, µm: 180
in mm: 0.180

Number: API 120
Opening, µm: 125
in mm: 0.125

Number: API 140
Opening, µm: 106
in mm: 0.106

Number: API 200
Opening, µm: 75
in mm: 0.075

Number: API 325
Opening, µm: 45
in mm: 0.045

Number	Opening, µm	in mm
API 40	425	0.425
API 60	250	0.250
API 80	180	0.180
API 120	125	0.125
API 140	106	0.106
API 200	75	0.075
API 325	45	0.045

The larger the number, the finer the screen and the less mud it passes per unit of area. API 200 matches the limit of sand (74–75 µm).

Fineness against capacity. We choose the finest screen of those that pass the whole flow. Usually the aim is that the mud drains through the screen while it covers about three quarters of the screen's length. If the mud runs over the edge, mud is lost together with the cuttings. Then we put in more shakers, not coarser screens.
Blinding. Particles almost as large as the opening get stuck in it, and the screen “blinds” (becomes blocked). Sticky clay (gumbo) coats the screens, so coarse scalping shakers (a first stage that takes out the bulk of the cuttings) are put in front of the main ones.
Weighted mud. By the requirements for barite, it holds no more than 3% by mass of particles larger than 75 µm. But there is a noticeable amount of particles of 45–75 µm, so screens finer than API 200 begin to hold back barite. The finest screen on a weighted mud is a compromise between cleaning and loss of barite.
Oil-based mud. The cuttings carry base oil with them (cuttings and waste), and diluting an expensive mud does not pay. Fine screens do not let the fines into the mud, and cuttings dryers (machines that squeeze the oil out of the cuttings) return part of the base oil.

Hydrocyclones and mud cleaners

A hydrocyclone is a cone with no moving parts. A centrifugal pump feeds the mud tangentially (along the side) into its upper cylindrical part, and the mud starts to spin. Large and heavy particles are thrown to the wall and slide down in a spiral to the narrow outlet, the underflow. The liquid with fine particles turns towards the axis, rises and leaves through the pipe in the cover, the overflow. Hydrocyclones with cones of 150–300 mm are called desanders, and their cut point for drilled rock is about 40 µm and more. Those with cones of about 100 mm are called desilters, and their cut point is about 20 µm. The smaller the cone, the smaller the cut point and the less mud it passes. So desilters are assembled in batteries of many cones.

SchematicA hydrocyclone and a decanting centrifuge. In the hydrocyclone the spinning flow throws large and heavy particles to the wall and down, to the underflow, and the liquid with fines leaves upwards along the axis. In the centrifuge the mud is fed inside the screw conveyor. The particles settle on the wall of the rotating bowl, and the screw, turning with a small difference in speed, moves the sediment to the narrow end, while the liquid with fines drains over weirs at the wide end. A schematic, not to scale.

A hydrocyclone is designed for a certain head (a pressure given as the height of a column of mud). Many designs need about 23 m (75 feet) of mud column. The pressure gauge at the inlet shows pressure, not head: $p = \rho g H$ . So for 23 m of head, a mud of 1.10 g/cm³ needs 0.25 MPa, and a mud of 1.30 g/cm³ needs 0.29 MPa. With too small a head, the cut point moves to larger particles. With too large a head, the cone wears faster. The work of a cone is watched at the underflow. Normally it flies out in a widening spray (spray discharge). A dense jet, a rope (rope discharge), means that more solids arrive than the outlet opening can pass. The separation gets worse, some particles slip through into the overflow, and the opening is about to block.

A hydrocyclone also loses liquid: a few per cent of the feed goes out with the underflow together with fine particles. For an unweighted mud this is acceptable, and a desander and a desilter are the usual stages of cleaning. In a weighted mud a hydrocyclone would throw out barite: its particles are the same size as silt, but heavier than rock (the cut point for barite is smaller, see above).

For a weighted mud, the underflow of the hydrocyclones is sent to a fine vibrating screen (usually API 140–200; a finer one would hold back barite too). This is a mud cleaner. Barite and fine particles pass through the screen and return to the mud, and the sand and coarse silt that are held back go to waste. In this way the drilled particles larger than barite are removed. At a mud density above about 1.8 g/cm³ the underflow is almost all barite, the screen is overloaded, and a centrifuge is better.

Centrifuges

A decanting centrifuge is a horizontal bowl, cylindrical at one end and conical at the other, that turns at a speed of the order of thousands of revolutions per minute. Inside is a screw conveyor that turns in the same direction with a small difference in speed. The mud is fed inside the screw. A layer of liquid forms at the wall of the bowl, and in it the particles settle onto the wall with an acceleration of hundreds and thousands of $g$ . The screw moves the sediment to the narrow end and lifts it out above the liquid layer. The liquid, with the fines that had no time to settle, drains over weirs at the wide end. The cut point is a few micrometres. It is shifted by the rotation speed, the feed rate and the depth of the liquid layer.

Weighted water-based mud: barite recovery. Part of the flow goes through the centrifuge. The sediment, which is barite and the larger drilled particles, is returned to the mud. The liquid with the colloidal drilled clay is discarded. This lowers the viscosity, but chemicals and fine barite leave with the liquid. When it is set up correctly, the centrifuge returns about 90% of the barite or more to the mud. The liquid that was discarded is replaced with fresh liquid, which is also dilution.
Unweighted mud: removal of silt. The centrifuge treats the underflow of the desilters or a part of the flow. The sediment is thrown away, and the liquid with the bentonite is returned.
Two centrifuges. The first, at a lower speed, returns the barite. The second, a fast one, separates the fine drilled rock from the liquid of the first, and the liquid goes back into the mud and is not discarded.
Oil-based mud. The base oil is expensive, so the centrifuges work to return the liquid. Cuttings dryers are added to them: centrifuges that squeeze the base oil out of the cuttings from the shale shakers (see the advanced section).

How the cleaning system is built

The equipment is placed in a chain by decreasing size of the particles to be removed. Large particles overload and block fine equipment, so they are removed first. The tanks of the circulating system are divided into sections. Each device takes mud from one section and discharges the cleaned mud into the next one downstream.

SchematicThe order of cleaning for an unweighted and a weighted mud. In an unweighted mud, after the shale shakers and the degasser come the desander, the desilter and a centrifuge that removes silt. In a weighted mud, a mud cleaner stands in place of the hydrocyclones, and the centrifuge returns the barite and discards the liquid with the colloids. A schematic, without pumps and transfers between the sections.

No bypasses. The whole flow from the well passes the shale shakers, and no stage works in a bypass.
The degasser comes before the hydrocyclones. Centrifugal pumps handle gas-cut mud badly, and hydrocyclones separate worse with it.
Feed and discharge go to different sections. The pump of a hydrocyclone takes mud from the section into which the previous stage discharges, and the cleaned mud is discharged into the next one. The capacity of each stage is taken with a margin above the flow of the mud pumps. Then part of the cleaned mud flows back and is cleaned again, and uncleaned mud does not slip through.
Dilution and additives come after the cleaning. Fresh mud, barite and chemicals are added in the sections after the equipment. Otherwise they are partly thrown away. In an unweighted mud, water is sometimes added before the hydrocyclones: in a less viscous mud they separate particles better.
Waste. The cuttings from the shale shakers, the underflows and the sediment of the centrifuges, and the discarded mud are drilling waste. How it is collected, treated and disposed of is decided by the environmental requirements and the permits of the project.

Cleaning or dilution: the solids balance

Everything that is drilled must go somewhere. Part of the rock is taken out by the equipment, and the rest stays in suspension. To keep its share from growing above the limit, the mud has to be diluted: new mud is made, it fills the new hole, and the surplus old mud is discarded. This gives a simple balance.

Mud needed to dilute the rock that was not removed

V_{\text{dil}} = \frac{(1 - E)\,V_{ds}}{f_{ds}}, \qquad V_{\text{new}} = V_{\text{dil}} + V_{\text{carr}} - (1 - E)\,V_{ds}

where:

$V_{\text{dil}}$ volume of mud in which the remaining rock is at the allowed share: it goes into the new hole or is discarded, m³
$E$ cleaning efficiency: the share of drilled rock that the equipment removes, —
$V_{ds}$ volume of drilled rock, m³
$f_{ds}$ allowed share of drilled rock in the mud by volume, —
$V_{\text{new}}$ volume of new mud that must be prepared, m³
$V_{\text{carr}}$ mud carried away with the waste of the equipment, m³

Units: any identical units of volume

Holds when

The content of drilled rock is held constant, at the limit $f_{ds}$ .
The fine particles in the mud carried away with the waste of the equipment are counted as removed.
No mud goes into the formation. Lost circulation carries away part of the rock itself, and it is counted separately.
The expression for $V_{\text{new}}$ is true while $V_{\text{dil}}$ is larger than the volume of the new hole. Otherwise the new mud is needed first of all to fill it.

Each cubic metre of rock that is not removed needs $1/f_{ds}$ cubic metres of mud: at a limit of 4%, 25 m³. So raising the cleaning efficiency pays back fast, and a stricter limit (half as large) doubles the dilution.

Worked exampleThree ways of cleaning on one interval

Given (assumed data)

Interval: the example “How much rock one run brings”: $V_{ds} = 13.69$ m³, volume of the new hole 16.11 m³
Limit of drilled rock: $f_{ds} = 4$ % by volume
Waste of the equipment: half solids, half mud carried away
Options: A: shale shakers with coarse screens, $E = 0.40$ ; B: fine screens and a mud cleaner, $E = 0.65$ , rental 3000 c.u. (conventional units); C: the same and a centrifuge, $E = 0.80$ , rental 8000 c.u.
Prices (conventional units, c.u.): new mud of 1.30 g/cm³ 300 c.u./m³, hauling and disposal of waste 70 c.u./m³

Solution

Option B: the dilution is $V_{\text{dil}} = 0.35\cdot 13.69/0.04 = 119.8$ m³. Of this, 16.11 m³ fill the new hole, and $119.8 - 16.11 = 103.7$ m³ must be discarded.
Waste of the equipment: the removed rock is $0.65\cdot 13.69 = 8.90$ m³, and the same volume of mud is carried away with it.
New mud: $119.8 + 8.90 - 0.35\cdot 13.69 = 123.9$ m³. Waste: $103.7 + 8.90 + 8.90 = 121.5$ m³.
Costs: $123.9\cdot 300 + 121.5\cdot 70 + 3000 \approx 48{,}700$ c.u., or 48.7 thousand c.u. Options A and C are calculated in the same way: 74.8 and 36.2 thousand c.u. (the table below).

Answer: The higher the cleaning efficiency, the less new mud and waste. At

E = 0.40

there are 203 m³ of new mud and 74.8 thousand c.u. At

E = 0.80

there are 77 m³ and 36.2 thousand c.u., even with the rental of the centrifuge.

What it means. The difference is set by the factor $1/f_{ds} = 25$ . Each extra cubic metre of rock removed saves 25 m³ of dilution. In the end it saves about 23 m³ of new mud and the same amount of waste: a little mud leaves with the removed rock, and the rock that stays would itself add volume to the mud. If the system removed nothing, then over 400 m we would need 329 m³ of new mud, more than all there is in circulation. The calculation is simplified: in the waste of the equipment of a weighted mud, barite is lost too, and with fine cleaning the share of mud leaving with the waste grows. Both effects reduce the gain, but do not change the conclusion. The data and the prices of the example are assumed.

Three ways of cleaning on the interval 2600–3000 m

Option: A
$E$: 0.40
New mud, m³: 202.6
Waste, m³: 200.2
Cost, thousand c.u.: 74.8

Option: B
$E$: 0.65
New mud, m³: 123.9
Waste, m³: 121.5
Cost, thousand c.u.: 48.7

Option: C
$E$: 0.80
New mud, m³: 76.7
Waste, m³: 74.3
Cost, thousand c.u.: 36.2

Option	$E$	New mud, m³	Waste, m³	Cost, thousand c.u.
A	0.40	202.6	200.2	74.8
B	0.65	123.9	121.5	48.7
C	0.80	76.7	74.3	36.2

The cost is the new mud, hauling and disposal of waste, and the rental of the extra equipment (0, 3000 and 8000 c.u.). The prices are assumed.

CalculatedNew mud over the interval 2600–3000 m against the cleaning efficiency, for a limit of drilled rock of 4% (solid) and 6% (dashed). The points are options A, B and C. The waste is almost as much as the new mud. At an efficiency of about 0.95 the curves meet: almost no dilution is needed any more, and the new mud goes to fill the hole and to replace what is carried away with the waste.

How to check the work of the system

The daily analysis of the mud shows whether the cleaning copes. We look at whether the content of low-gravity solids from the retort is growing, and the drilled rock after subtracting the bentonite equivalent (the calculation from the retort and the methylene blue test). We look at whether the plastic viscosity grows at an unchanged density, and how much sand there is. Single devices are checked by their waste: whether the cuttings on the screens are dry, whether the underflow of the hydrocyclones comes out in a spray, and what the density of the underflow and of the centrifuge sediment is. But the best overall measure is the efficiency of the whole system. It can be found from how much mud had to be made.

Cleaning efficiency from the actual dilution

E = 1 - \frac{f_{ds}\,V_{\text{dil}}}{V_{ds}}, \qquad V_{\text{dil}} \approx \frac{V_w}{f_w}

where:

$V_w$ volume of water (the base of the mud) added over the interval to make the mud, m³
$f_w$ share of water in the mud from the retort, —
$f_{ds}$ average share of drilled rock in the mud over the interval, —
$V_{ds}$ volume of drilled rock over the interval, m³

Units: any identical units of volume

Holds when

The water is used only to make the mud. Evaporation and leaks are not taken into account.
The share of water in the new mud is the same as in the circulating mud.
No mud goes into the formation. In case of lost circulation, that volume is counted separately.
The mud that was made, together with what is carried away in the waste, is for simplicity counted as dilution mud. This slightly understates $E$ .

If the equipment removed nothing, all the rock would stay in the mud that was made, and $E = 0$ . The less mud was needed for the same volume of rock at the same content, the higher the efficiency.

Worked exampleWhat efficiency the system showed

Given (assumed data)

Interval: 2600–3000 m, $V_{ds} = 13.69$ m³
Water added over the interval: 98 m³
Average analysis of the mud: 1.30 g/cm³; retort: water 87.5%; methylene blue test: 8.0 cm³ of dye for 2.0 cm³ of mud

Solution

The solids from the retort are 12.5%, and their average density is $(1.30 - 0.875)/0.125 = 3.40$ g/cm³. Low-gravity solids: $12.5\cdot(4.2 - 3.40)/1.6 = 6.25$ %, or 162.5 kg/m³.
Bentonite equivalent: $14.25\cdot 8.0/2.0 = 57.0$ kg/m³. Drilled rock: $162.5 - 57.0 = 105.5$ kg/m³, and by volume $105.5/2600 = 0.0406$ .
Mud that was made: $98/0.875 = 112.0$ m³. Drilled rock in it: $0.0406\cdot 112.0 = 4.55$ m³.
Efficiency: $E = 1 - 4.55/13.69 = 0.67$ .

Answer: The system removed about 67% of the drilled rock.

What it means. This lies between options B and C of the previous example. The equipment works, but fines build up, and the limit is held at the cost of almost a hundred cubic metres of water and the matching amount of barite and chemicals. If the mud holds a lot of active drilled clay, it also takes up the dye and is counted as bentonite. Then there is in fact more drilled rock, and the efficiency is lower (the refinement of the methylene blue test). The data in this example are assumed for illustration.

The plastic viscosity and the share of low-gravity solids grow at the same density. The cleaning cannot keep up. Check the screens, the head and the discharge of the hydrocyclones, and the work of the centrifuge.
The sand content grows. Screens are torn, or the mud goes round the shale shakers. Sand quickly wears the pumps.
The bentonite equivalent and the yield point (YP) grow without any bentonite being added. Active drilled clay is building up in the mud. It is removed with a centrifuge, and the mud is inhibited if needed (inhibitive muds).
The cuttings are too wet, or the mud runs over the screens. The screens are overloaded, and the loss of mud with the waste grows.

Terms and symbols

Shale shaker: The first stage of cleaning: vibrating screens that hold back coarse cuttings. The whole flow from the well passes through them.
API screen number: The name of a screen by the size of the particles it holds back almost completely, through the number of the standard sieve (API 200 is 75 µm).
Hydrocyclone: A cone in which a spinning flow separates large and heavy particles. Desanders (about 40 µm) and desilters (about 20 µm) are hydrocyclones.
Mud cleaner: Hydrocyclones over a fine screen: barite passes through the screen and returns, and drilled particles larger than barite go to waste.
Decanting centrifuge: A rotating bowl with a screw. It separates particles of a few micrometres and returns barite or liquid.
Cut point, $d_{50}$: The size of particles of which the device sends half to waste. For barite it is smaller than for rock.
Cleaning efficiency, $E$: The share of drilled rock that the system removes from the mud. The rest has to be diluted.
Dilution: Making new mud and discarding part of the old, so that the share of drilled rock does not exceed the limit.
Overflow and underflow: The outlets of a hydrocyclone: at the top the liquid with fine particles, at the bottom the large and heavy particles with a small share of liquid.

Common mistakes and misconceptions

Wrong

“The finer the screens on the shale shakers, the better.”

In fact

A fine screen passes less mud. If the flow cannot get through it, mud goes to waste together with the cuttings. On a weighted mud a screen that is too fine holds back barite too. We choose the finest screen of those that pass the whole flow.

Wrong

“Hydrocyclones suit any mud.”

In fact

Barite particles are the same size as silt, and heavier than rock: a hydrocyclone on a weighted mud will throw out barite. A weighted mud is cleaned with a mud cleaner and a centrifuge.

Wrong

“A settling tank will replace a desilter.”

In fact

A particle of 20 µm settles in a tank at about a quarter of a metre per hour, and the gel of the mud holds it altogether. Silt can be separated only by centrifugal acceleration.

Wrong

“It is easier to dilute than to clean.”

In fact

At a limit of 4%, each cubic metre of rock that is not removed needs 25 m³ of new mud, which then becomes waste. Dilution is needed only for what the equipment cannot remove.

Wrong

“A centrifuge in the barite-recovery mode cleans the mud of drilled rock.”

In fact

It returns to the mud, together with the barite, the drilled particles of the same size, and removes only the colloidal fines. The shale shakers and the mud cleaner must remove the coarse rock.

In depth: advanced detail

The partition curve in more detail

The partition curves in the lesson's figure are built from a common empirical form. First a “corrected” curve is described, without the fines that the liquid carries away, and then this carry-over is added.

An empirical partition curve with carry-over of fines

E(d) = R_f + (1 - R_f)\,\frac{e^{\alpha x} - 1}{e^{\alpha x} + e^{\alpha} - 2}, \qquad x = \frac{d}{d_{50c}}

where:

$E(d)$ share of particles of size $d$ that go to waste, —
$d_{50c}$ corrected cut point, without the carry-over, µm
$\alpha$ sharpness of the separation: the larger it is, the steeper the curve, —
$R_f$ share of liquid that goes to waste with the fine particles, —

Units: dimensionless

Holds when

The form was fitted to tests and describes hydrocyclones and centrifuges. The parameters are found by testing a particular device on a particular mud.
In the lesson's figure: the screen has $d_{50c} = 65$ µm, $\alpha = 8$ , $R_f = 0$ ; the desander 45 µm, 3 and 0.05; the desilter 18 µm, 3 and 0.05; the centrifuge 5 µm, 3 and 0.02.

At $x = 1$ the fraction is 1/2. At large $x$ it tends to 1, and at small $x$ to 0. Because of the carry-over $R_f$ , the actual cut point is a little smaller than the corrected one, and the finest solids, bentonite and fine barite, go to waste in part at any size.

The cut point depends on the conditions. In a more viscous mud it moves to larger particles roughly as $\sqrt\mu$ (by Stokes' law), and when the solids are overloaded it also moves to larger particles. In a centrifuge, at a lower feed rate the particles stay longer in the bowl, and the cut point gets smaller. In a hydrocyclone a lower feed rate means a lower head and a weaker spin of the flow, and the cut point moves, on the contrary, to larger particles. So the cut points in the data sheets are a guide, not a constant of the device.

Oil-based mud: base oil on cuttings

In an oil-based mud each cubic metre of liquid is expensive, and cuttings with base oil are a special kind of waste. So the cleaning system is built to return the liquid. The centrifuges work to return it, and the cuttings after the shale shakers are squeezed in dryers, which are centrifuges for wet cuttings.

Worked exampleHow much base oil a cuttings dryer saves

Given (assumed data)

Cuttings from the shale shakers if the same interval is drilled with an oil-based mud: 60% of the drilled rock of the example “How much rock one run brings”: $0.60\cdot 35.6$ t of dry rock
Base oil on cuttings (per mass of wet cuttings): 12% after the shale shakers, 4% after the dryer (assumed)
Base oil: density 0.80 g/cm³: the mud of the lesson on oil-based muds

Solution

Dry rock in the cuttings of the shale shakers: $0.60\cdot 35.6 = 21.36$ t.
If the base oil is a share $w$ of the mass of the wet cuttings, its mass is $w/(1 - w)$ per tonne of dry rock. After the shale shakers: $0.12/0.88\cdot 21.36 = 2.91$ t, or 3.64 m³.
After the dryer: $0.04/0.96\cdot 21.36 = 0.89$ t, or 1.11 m³.

Answer: The dryer returns about 2.5 m³ of base oil to the mud over the interval and reduces the mass of the waste by 2.0 t.

What it means. The base oil that is returned pays back part of the cost of the dryer. But the main point is usually another. After the dryer, the content of base oil on the cuttings may meet the requirements for disposal, and the mass of waste is smaller. The water of the brine in the cuttings is not taken into account here, and the share of base oil on the cuttings is measured with a retort on a weighed sample of cuttings (cuttings and waste). The data in this example are assumed for illustration.

Connections to other topics

Drilling fluids: functions and properties — the functions of the mud, useful and harmful solids.
Field testing of drilling fluids — the retort, the methylene blue test and the sand content: they show whether the cleaning copes.
Water-based muds and clay chemistry — inhibition: the less the cuttings soften, the easier they are to remove.
Oil-based muds — base oil on cuttings, dryers and handling of waste.
Circulating pressure losses — how cuttings and fine solids raise the ECD.
Bit hydraulics optimisation — the same run: flow rate, nozzles and carrying out cuttings.
Formation damage and damage skin — fine particles of the mud as one cause of damage to the formation.
Rig circulating and rotary systems — pumps and tanks of the circulating system of the rig.

Check yourself

1
A mud is weighted with barite to 1.60 g/cm³. What equipment removes drilled sand and silt larger than the barite particles while keeping the barite?
1. A.Dilution with water
2. B.A desilter
3. C.A mud cleaner
4. D.A centrifuge in the barite-recovery mode
Show the answer
Correct answer: C. A mud cleaner
In a mud cleaner the underflow of the hydrocyclones goes onto a fine screen: barite passes through it and returns, and the large drilled particles are held back. A desilter would throw out barite too. A centrifuge in the barite-recovery mode would return the coarse rock as well. Dilution with water lowers the density.
2
500 m of hole were drilled with a 311.2 mm bit. The washout factor is 1.05 and the porosity of the rock is 0.20. How much mud is needed for dilution at a cleaning efficiency of 0.6 and a drilled-rock limit of 5%? And at an efficiency of 0.8?
Show the answer
Volume of drilled rock: $\frac{\pi}{4}\cdot 0.3112^2\cdot 500\cdot 1.05\cdot 0.80 \approx 31.9$ m³. At $E = 0.6$ : $V_{\text{dil}} = 0.4\cdot 31.9/0.05 \approx 255$ m³. At $E = 0.8$ it is half as much, about 128 m³. Raising the efficiency by 0.2 saved more than 120 m³ of mud.
3
The cut point of a desilter for drilled rock (2.6 g/cm³) is 24 µm. At what size of barite particles (4.2 g/cm³) does it send half of them to waste, if the liquid is water?
1. A.24 µm
2. B.12 µm
3. C.17 µm
4. D.34 µm
Show the answer
Correct answer: C. 17 µm
$24\cdot\sqrt{(2.6 - 1)/(4.2 - 1)} = 24\cdot\sqrt{0.5} \approx 17$ µm. Barite is heavier and settles like rock that is $\sqrt2$ times larger. This is why a hydrocyclone loses barite.
4
Over an interval, 60 m³ of water were used to make the mud. The share of water in the mud from the retort is 0.90, the drilled rock in it is on average 3.5% of the volume, and the drilling produced 9.0 m³ of drilled rock. What is the cleaning efficiency?
Show the answer
Mud made: $60/0.90 \approx 66.7$ m³. Drilled rock in it: $0.035\cdot 66.7 \approx 2.33$ m³. Efficiency: $E = 1 - 2.33/9.0 \approx 0.74$ . The system removed about three quarters of the rock.
5
The underflow of a desilter does not come out as a spray but as a dense rope. What does this mean?
1. A.The head at the inlet is too high
2. B.There is too much gas in the mud
3. C.The desilter works perfectly
4. D.The cone is overloaded with solids: some particles go into the overflow, and the underflow will soon block
Show the answer
Correct answer: D. The cone is overloaded with solids: some particles go into the overflow, and the underflow will soon block
A rope is a sign that there is more solid than the lower opening can pass. The separation gets worse, and the waste outlet blocks. It helps to reduce the load, by cleaning the mud better at the previous stage or by adding cones, or to correct the lower opening.
6
Why do fine drilled particles thicken a mud more than the same volume of coarse ones, and what follows for cleaning?
Show the answer
The surface area per unit volume of particles is inversely proportional to their size: crushing ten times finer makes the surface ten times larger. Water is held on the surface, and the particles interact through it, so the viscosity and the strength of the structure grow. Hence the rule: remove the cuttings in the first cycle, while they are coarse and are held on the screens. Do not wait until they are ground into silt and colloids, which can be removed only by a centrifuge or by dilution.

Updated: 10 October 2026