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Drilling E2University of SalfordDrilling E2 Section 1 Drilling Fluid

DRILLING FLUID

BY

G.C.ENYISchool of Computing, Science and EngineeringUniversity of Salford, ManchesterDrilling E2University of SalfordDrilling E2 Section 1 Drilling Fluid INTRODUCTIONDrilling fluid is a fluid used to aid the drilling of boreholes. It is used while drilling oil and gas wells.The three main types of drilling fluids are water-based muds, oil-based muds and gaseous drilling fluid in which a wide range of gases can be used.Liquid drilling fluid is called drilling mud.

COMPOSITION OF A DRILLING FLUIDA drilling fluid consists of:

The Base Fluid water fresh or saline oil diesel or crude mineral oil or other synthetic fluids The Dispersed Solids Colloidal particles which are suspended particles of various sizes.The Dissolved Solids Usually salts and their effects on colloids are importantDrilling E2University of SalfordDrilling E2University of SalfordDrilling E2 Section 1 Drilling Fluid TYPES OF DRILLING FLUIDDrilling fluid may be classified generally as Air, Gas or Mud. Depending on the liquid base, drilling mud are two types:

Water base mud Oil base mud

(A) Water Base Muds ( Water base muds are most commonly used)

Fresh water mudsChemical treated mudsCalcium treated mudsSalt water treated mudsOil-emulsion muds

(B) Oil Base MudsOil base mudsInvert emulsion mudsDrilling E2University of SalfordDrilling E2 Section 1 Drilling Fluid TYPES OF DRILLING FLUIDGasesLiquids

Gas-Liquid MixturesWater-BaseMudsOil-BaseMudsFoam(Mostly Gas)Aerated Water(Mostly Water)AirNatural GasDrilling E2University of SalfordDrilling E2 Section 1 Drilling Fluid TYPES OF DRILLING FLUID(C) Gas Drilling FluidsAir or natural gasAerated muds

(D) Foam

The criteria for selection of a particular drilling fluid with specific design properties are the minimum cost in conjunction with the following factors:Type of formation to be penetratedFormation pore pressureFormation evaluation methods to be usedFormation temperature and permeabilityEcological considerationsCorrosionDrilling E2University of SalfordDrilling E2 Section 1 Drilling Fluid TYPES OF DRILLING FLUIDWATER BASED MUD (WBM)The water base mud system begins with water, then clay and other chemical additives are mixed with the water to create a homogenous blend depending on the viscosity. The most common of the clay used is BENTONITE.

OIL BASED MUD (OBM)This is mud where the base fluid is a petroleum product such as diesel fuel. OBMs are used for many reasons such as increased lubrication and greater cleaning abilities with less viscosity. They withstand greater heat without breaking down.

SYNTHETIC BASED FLUID (SBM)This is mud where the base fluid is a synthetic oil. Often used on offshore rigs. It has the properties of OBM but the toxicity of the fluid fumes are much less than OBM. OIL BASED MUDSOil muds find application under the following conditions:Drilling under deep condition and hot formation (t > 375oF) Drilling salt formation or active shale formation containing H2S or CO2.Drilling formations easily damaged by water based mud.4) Corrosion control.

Drilling E2University of SalfordOIL BASED MUDS5) Directional drilling or slim holes where high torque is a problem.6) Preventing stuck pipe.7) Drilling weak formations of subnormal pore pressures.Drilling E2University of SalfordOIL BASED MUDSADVANTAGES

Good rheological properties at temperature as high as 500oFMore inhibitive than the water base mudsEffective against all types of corrosionSuperior lubricating characteristicsPermits mud densities as low as 7.5 Ib/gal.Drilling E2University of SalfordOIL BASED MUDSDISADVANTAGES

Higher initial costRequires more stringent pollution control proceduresSlower drilling rateReduced effectiveness of some logging toolsRemedial treatment for lost circulation is more difficultDetection of gas kicks is more difficultDrilling E2University of SalfordFUNCTIONS OF DRILLING FLUIDSThe main functions of a drilling fluid can be summarised as follows: Removal of cuttings from the boreholeCooling and lubrication of the bit and drill stringControl of subsurface pressureWall buildingMinimise formation damageSeal permeable formationsTransmit hydraulic energy to tools and bitEnsure adequate formation evaluation Facilitate cementing and completionSuspend and release cuttings

.Drilling E2University of SalfordDRILLING FLUIDS RELATED PROBLEMSMajor hole problems which are directly or indirectly related to drilling mud are:

Lost circulationCorrosionShale problemsSome of these problems are:Pipe stickingIncreased mud volume and continuous treatmentIneffective hole cleaningDifficult loggingDrilling E2University of SalfordDRILLING FLUIDS RELATED PROBLEMS4) High borehole temperature Excessive temperature increase may lead to the following problemsReduction in penetration rateLost circulationLogging and downhole testing difficultiesExpensive workover operations

Drilling E2University of SalfordDRILLING FLUIDS ADDITIVESMany substances are added to drilling fluids to perform specialized functions. The most common functions are:Alkalinity and pH controlBactericidesFiltrate and Calcium ReducersCorrosion InhibitorsFoaming Agents and deformerEmulsifiersFlocculantsSurfactants and LubricantsWeighting and Pipe freeing agentsDrilling E2University of SalfordPROPERTIES OF DRILLING FLUIDSAll drilling fluids have essentially the same properties, only the magnitude varies. These properties include:DensityViscosityGel strengthFilter cakeWater lossElectrical resistance

Drilling E2University of SalfordDIAGNOSTIC TESTSThe test equipment needed to perform the diagnostic tests include:

A mud balance for determining drilling fluid density.A Marsh funnel for checking drilling fluid consistency.A rotational viscometer for determining gel strength and apparent viscosity at various shear ratesA filter press for determining mud filtration rate and mud cake characteristics.A high-pressure, high temperature filter press for determining mud filtration rate and mud cake characteristics at elevated temperature and pressure.A pH meter for determining hydrogen ion concentration.A sand screen for determining sand contentA mud still for determining solids, oil and water contentsA titration apparatus for chemical analysisDrilling E2University of SalfordDIAGNOSTIC TESTSDrilling E2University of SalfordThe density of fresh water is 8.33 Ibm/gal.

The Marsh FunnelThe test consists essentially of filling the funnel with a mud sample and then measuring the time required for 1 quart of the sample to flow from the initially full funnel into the mud cup.

DIAGNOSTIC TESTSThe Rotational Viscometer

This provides a more meaningful measurement of the rheological characteristics (deformation & flow of matter: non-Newtonian fluid) of the mud than the marsh funnel.Drilling E2University of Salford

DIAGNOSTIC TESTSThe mud is sheared at a constant rate between an inner bob and an outer rotating sleeve. The dimensions of the bob and the rotor are chosen so that the dial reading is equal to the apparent Newtonian viscosity (cp) at a rotor speed of 300 rpm. At other rotor speeds, the apparent viscosity a is given by:

Where N = dial reading in degrees and N is the rotor speed in rpm

Drilling E2University of Salford

DIAGNOSTIC TESTSDrilling E2University of SalfordThe viscometer also can be used to determine rheological parameters that describe non-Newtonian fluid behaviour.Two parameters are required to characterise fluids that follow the Bingham plastic model. These parameters are called the plastic viscosity (p) and the yield point (y) of the fluid.

p = 600 300

600 = dial reading with viscometer operating at 600 300 = dial reading with viscometer operating at 300

DIAGNOSTIC TESTSDrilling E2University of SalfordThe yield point is computed using:

y = 300 - p (Ibf/100 ft2)

Gel strength (Ibf/100 ft2) is obtained by noting the maximum dial deflection when the rotational viscometer is turned at a low rotor speed (usually 3 rpm) after the mud has remained static for some period of time. WORKED EXAMPLEDrilling E2University of Salford A mud sample in a rotational viscometer equipped with a standard torsion spring gives a dial reading of 46 when operated at 600 rpm and a dial reading of 28 when operated at 300 rpm. Compute the apparent viscosity of the mud at each rotor speed. Also compute the plastic viscosity and yield point.SOLUTIONDrilling E2University of Salford

For the 300 rpm dial reading:

a = 300(28)/300 = 28 cp For the 600 rpm dial reading:

a = 300(46)/600 = 23 cp

Note: The apparent viscosity does not remain constant but decreases as the rotor speed is increased. This type of non- Newtonian behaviour is shown by all drilling muds.

SOLUTIONDrilling E2University of Salford The plastic viscosity of the mud can be computedas:

p = 600 300 = 46 28 = 18 cp

The yield point can be computed as: y = 300 - p = 28 18 = 10 Ibf/100 ft2

PILOT TESTSDrilling E2University of SalfordPilot testing involves evaluation of mixtures of given concentrations and densities. It is assumed in mixing calculations that the resulting mixture is ideal (ie the total volume is equal to the sum of component volumes): Vt = V1 + V2 + . . . + VnAlso Vi = mi/i

PILOT TESTSDrilling E2University of SalfordThe mixture density can be computed from a knowledge of the total mass and total volume added to the mixture. Thus, the mixture density is given by:

The units of measure most commonly used when treating the active drilling fluid system are pounds for weight and barrels for volume.The units of measure most commonly used for pilot tests are grams for weight and cubic centimeters for volume. Converting from Ibm/bbl to g/cm3 gives:

1.0 Ibm/1 bbl x 454g/Ibm x 1bbl/42 gal x 1 gal/3785 mL = 1g/350 mL

Thus adding 1g of material to 350 mL of fluid is equivalent to adding 1 Ibm of material to 1 bbl of fluid.WORKED EXAMPLEDrilling E2University of SalfordCompute the volume and density of a mud composed of 25 Ibm of bentonite clay, 60 Ibm of API barite and 1 bbl of fresh water.

SOLUTIONThe densities of clay and barite are 910 Ibm/bbl and 1470 Ibm/bbl respectively. Density of water is 8.33 lbm/gal = 350 lbm/bbl) The total volume is given by: Vt = V1 + V2 + V3 = 1.0 + 25/910 + 60/1470 = 1.0683 bblThe mixture density is: = (350 + 25 + 60)/1.0683 = 407 Ibm/bbl = 9.7 Ibm/galDENSITY CONTROL ADDITIVESDrilling E2University of SalfordFor ideal mixing the volume of mud V1 and weight material VB must sum to the desired new volume V2.

V2 = V1 + VB = V1 + mB/B

Likewise, the total mass of mud and weight material must sum to the desired density-volume product:

2V2 = 1V1 + mB

DENSITY CONTROL ADDITIVESDrilling E2University of SalfordSolving these simultaneous equations for V1 and mB yields:

and

When the final volume of mud is not limited, the final can be calculated from the initial volume as:

DENSITY CONTROL ADDITIVESDrilling E2University of SalfordThe reduction in solids concentration by dilution is much less expensive before the conversion is made because the mud discarded during dilution does not contain any API barite.When mud dilution comes before (precedes) weighting operations, the proper volume of old mud V1, dilution water Vw and mass of weight material mB, that should be combined to obtain a desired volume of new mud V2 can be calculated by:

V2 = V1 + Vw + mB/B

Similarly, a mass balance on mixing procedure requires:

2V2 = 1V1 + wVw + mB

DENSITY CONTROL ADDITIVESDrilling E2University of SalfordA third equation can be obtained by performing a volume balance on the low-specific-gravity solids. This can be expressed in terms of the present volume fraction fc1 and the desired new volume fraction fc2: fc2 V2 = fc1 V1

Solving these three equations simultaneously yields:

V1 = V2 (fc2 / fc1)

and

WORKED EXAMPLEDrilling E2University of SalfordIt is desired to increase the density of 200bbl of 11 Ibm/gal mud to 11.5 Ibm/gal using API barite. The final volume is not limited. Compute the weight of API barite required.

SOLUTIONThe density of API barite is 35.0 Ibm/gal.

= 204.255 bbl The weight of API barite required is: mB = (V2 V1)B = (204.255 200)(42)(35) = 6,255 Ibm

EXAMPLEDrilling E2University of Salford After cementing casing in the well, it is desirable to increase the density of the 9.5 Ibm/gal mud to 14 Ibm/gal before resuming drilling operations. It also is desired to reduce the volume fraction of low-specific-gravity solids from 0.05 to 0.03 by dilution with water. The present mud volume is 1000 bbl but a final mud volume of 800 bbl is considered adequate. Compute the amount of original mud that should be discarded and the amount of water and API barite that should be added.

MUD RHEOLOGYDrilling E2University of SalfordNEWTONIAN FLUIDS

Viscosity is defined as the resistance to flow and in the field, it is routinely measured with marsh funnel. Meaningful results are obtained with rotational viscometers.

Gel Strength is a measure of the gelling properties or thixotropic property of the mud under static conditions.MUD RHEOLOGYDrilling E2University of SalfordConsider a fluid contained between two large parallel plates of area A, which are separated by a small distance L. The upper plate which is initially at rest is set in motion in the x-direction at a constant velocity v.

After sufficient time has passed for steady motion to be achieved, a constant force F is required to keep the upper plate moving at constant velocity.

The magnitude of the force F was found experimentally to be:

MUD RHEOLOGYDrilling E2University of SalfordThe term F/A is called the shear stress exerted on the fluid. Shear stress is defined as:

The area of the plate A is the area in contact with the fluid. The velocity gradient v/L is an expression of the shear rate.

Hence the Newtonian model states that the shear stress is directly proportional to the shear rate. The constant of proportionality is known as the viscosity of the fluid.

MUD RHEOLOGYDrilling E2University of SalfordThe linear relationship between the shear stress and shear rate is valid only for laminar flow. This is true at relatively low rate of shear.At high rates of shear, the flow pattern changes from laminar to turbulent in which the fluid particles are in chaotic and unpredictable movement.

MUD RHEOLOGYDrilling E2University of SalfordNON-NEWTONIAN FLUIDS

Most drilling fluids are too complex to be characterised by a single value for viscosity. The apparent viscosity measured depends upon the shear rate at which the measurement is made and the prior shear rate history of the fluid.

Fluids which do not exhibit a direct proportionality between the shear stress and shear rate are called Non-Newtonian fluids.MUD RHEOLOGYDrilling E2University of SalfordNON-NEWTONIAN FLUIDSNon-Newtonian fluids that are shear rate dependent are called Pseudo-plastics if the apparent viscosity decreases with increasing shear rate.They are called Dilatants if the apparent viscosity increases with increasing shear rate. Generally drilling fluids are pseudo-plastics.

MUD RHEOLOGYDrilling E2University of SalfordNON-NEWTONIAN FLUIDS

Two models are used to approximate the behaviour of non-Newtonian fluids:

Bingham modelPower Law model

A Bingham plastic will not flow until the applied shear stress exceeds a certain minimum value (y) known as the yield point. After the yield point has been exceeded, changes in the shear stress are proportional to changes in the shear rate and the constant of proportionality is called the plastic viscosity (p).

MUD RHEOLOGYDrilling E2University of SalfordThe Bingham Plastic Model is defined by:

= p + y ( > y )

The Power Law Model is defined by: = K||n-1

The power law model requires two parameters for fluid characterisation. The parameter K is called the consistency index of the fluid. The unit of K depends on the value of n. It has units of g/cm.s2-nThe parameter n is called the power law exponent or the flow-behaviour index.

WORKED EXAMPLE 1Drilling E2University of SalfordAn upper plate of 20-cm2 area is spaced 1cm above a stationary plate. Compute the viscosity (cp) of a fluid between the plates if a force of 100 dyne is required to move the upper plate at a constant velocity of 10 cm/s. (1 dyne.s/cm^2 = 100 cp)

SOLUTION

The shear stress is given by:

= 100 dyne/20 cm2 = 5 dyne/cm2.

The shear rate is given by: = 10 cm/s 1 cm = 10 s-1 = / = 5/10 = 0.5 dyne.s-1/cm2 = 50 cpWORKED EXAMPLE 2Drilling E2University of Salford An upper plate of 20-cm2 area is spaced 1 cm above a stationary plate. Compute the yield point and plastic viscosity of a fluid between the plates if a force of 200 dynes is required to cause any movement of the upper plate and a force of 400 dynes is required to move the upper plate at a constant velocity of 10 cm/s.SOLUTION 2Drilling E2University of SalfordThe yield point y is given as: = p + y with = 0 = y = 200 dyne/20 cm2 = 10 dyne/cm2.

In Oilfield units: y = 10/4.79 = 2.09 Ibf/100 ft2.

The plastic viscosity p is given as:

= p + y with = 10cm/s 1cm = 10 s-1.

p = [(400/20) 10] 10 = 1.0 dyne-s/cm2. = 100 cp

WORKED EXAMPLE 3Drilling E2University of SalfordAn upper plate of 20 cm2 is placed 1cm above a stationary plate. Compute the consistency index and flow-behaviour index if a force of 50 dyne is required to move the upper plate at a constant velocity of 4 cm/s and a force of 100 dyne is required to move the upper plate at a constant velocity of 10 cm/s.