2 Drilling Fluids

DRILLING FLUIDS

Agip KCO

Well Area Operations ENGINEERING CALCULATIONS-MAP-1 Drilling Muds Drilling Supervisors Training Course

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INDEX BASIC CHEMISTRY……….................…3 RHEOLOGY AND HYDRAULICS……..63 ENGINEERING CALCULATIONS……142

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INTRODUCTION •

Mass – The unit of measurement normally used for the mass are grams (g) and pounds.

•

Volume – Gallons (gal), barrels (bbl), cubic feet (ft3), litres (l) cubic metre (m3).

•

Density – Pounds per gallon (lb/gal), pounds per cubic foot (lb/ft3), kilograms per cubic metre (kg/m3) and grams per cubic centimetre (g/cm3). Specific Weight is a particular definition of density often used for solids and liquids. It is a ratio between the weight of a substance and the weight of the distilled water at a temperature of 4 centigrades.

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CLASSIFICATION OF MATTER There are three different states in which matter can exist in the universe: – Solid – Liquid – Gaseous •

SOLIDS – Characterized by higher density than liquids. – Solids keep their own shape and volume, independent of the container.

•

LIQUIDS – Lower density than solids but higher than gases. Liquids take the shape of the container they are in.

•

GASES – Lack of definite shape and volume, easy contraction and expansion.

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CLASSIFICATION OF MATTER All the substances can be divided in one of the following two categories: – Homogeneous substances (pure substances). Like table salt where each grain is identical in chemical composition. – Heterogeneous substances (mixture of substances) as the riverbed gravel, mixture of rocks having different chemical composition. It is possible to separate the pure substances in two categories: – Elements. Cannot be decomposed into simpler substances by ordinary chemical methods. – Compounds. Can be reduced into two or more simpler substances.

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CLASSIFICATION OF MATTER COMMON ELEMENTS Element

Symbol

Atomic Weight

Common Valence

Alluminium

Al

26,98

3+

Arsenic

As

74,92

5+

Barium

Ba

137,34

2+

Boron

B

10,81

3+

Bromine

Br

79,90

1-

Cadmium

Cd

112,40

2+

Calcium

Ca

40,08

2+

Carbon

C

12,01

4+

Caesium

Cs

132,91

1+

Chlorine

Cl

35,45

1-

Chromium

Cr

52,00

6+

Copper

Cu

63,55

2+

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CLASSIFICATION OF MATTER Element

Symbol

Atomic Weight

Common Valence

Fluorine

F

19,00

1-

Hydrogen

H

1,01

1+

Iodine

I

126,90

1-

Iron

Fe

55,85

3+

Lead

Pb

207,19

2+

Lithium

Li

6,94

1+

Magnesium

Mg

24,31

2+

Manganese

Mn

54,94

2+

Mercury

Hg

200,59

2+

Nickel

Ni

58,71

2+

Nitrogen

N

14,00

5+

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CLASSIFICATION OF MATTER Element

Symbol

Atomic Weight

Common Valence

Oxygen

O

16,00

2-

Phosphorus

P

30,97

5+

Potassium

K

39,10

1+

Silicon

Si

28,09

4+

Silver

Ag

107,87

1+

Sodium

Na

22,99

1+

Sulphur

S

32,06

2-

Tin

Sn

118,69

2+

Titanium

Ti

47,90

4+

Zirconium

Zr

91,22

4+

Zinc

Zn

65,37

2+

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ATOMIC STRUCTURE Atoms are made up of three subatomic particles:

Diffuse cloud of electrons (-) orbiting the nucleus in structured shells

– Protons – Neutrons – Electrons

Nucleus is compact and dense, containing protons (+) and neutrons (neutral) Atom structure

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ATOMIC STRUCTURE •The atoms have two distinct zones: a small dense nucleus, which contains the protons and neutrons, surrounded by a diffuse cloud of electrons •The nucleus is spherical in shape, 10-4 angstrom (Å) in diameter. •The nucleus contains only protons and neutrons. •The proton has positive charge. •The neutron has no charge. •Surrounding the nucleus is the electron cloud, in which electrons orbit the nucleus in specific orbits. The electron cloud is also approximately spherical, 1 Å (10-10m) in diameter and contains only electrons. •The electron has a negative charge equal in strenght to the positive charge of the proton. •Every atom has a specific number of electrons that surround the nucleus and if the atom is neutral (has no charge) that number is equal to the number of protons in the atom’s nucleus

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ATOMIC STRUCTURE Certain atoms can acquire or loose electrons = name ion Ion with positive charge = cation When electron is lost. Ion with negative charge = anion When electron is gained. The mass, that belongs to a proton or a neutron is approximatively 1837 times greater than the mass of an electron Proton and neutrons mass ≃ atom mass (electron mass is irrilevant) Mass and charge of subatomic particles

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Particle

Charge

Mass (g)

Proton

Positive (1+)

1,6724 X 10–24

Neutron

None (neutral)

1,6757 X 10–24

Elettron

Negative (1-)

0,000911 X 10–24


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ATOMIC STRUCTURE The nucleus of an atom has a very high density: around 1,770 tons/in.3 (98.000 kg/cm3). The electrons cloud has a diameter 10.000 times larger than that of the nucleus. The average density of the electron cloud is between 2 and 20 g/cm3. The hydrogen that is the lightest and simplest element has only one proton in each nucleus. Atoms of other elements contain 1 to 93 protons in their nucleus. 92 elements occur in nature which in various combinations form the physical world. The number of protons in the nucleus is used to define each element and is called atomic number Number of protons = (p+) Atomic number = (z) Atomic mass a = p+ + n. Isotopics atomic number (z), atomic mass number (a) (azX). The hydrogen has three isotopes. (11H), (21H) (31H).

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ATOMIC STRUCTURE Most recurrent compounds in the drilling activity Name

Formula

Silver nitrate

AgNO3

Aluminium oxide

Al2O3

Alumina

Barium sulphate

BaSO4

Barite

Barium carbonate

BaCO3

Mineral whiterite

Barium hydroxide

Ba(OH)2

Calcium hydroxide

Ca(OH)2

Hydrated Lime

Calcium sulphate (anhydrous)

CaSO4

Anhydrite

Calcium sulphate (hydrous)

CaSO4 • 2H2O

Gypsum

Calcium carbonate

CaCO3

Calcite

Calcium chloride

CaCl2

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Common name


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ATOMIC STRUCTURE

Agip KCO

Name

Formula

Common name

Calcium oxide

CaO

Quick lime, Hot lime

Hydrochloric acid

HCl

Muriatic acid

Hydrogen oxide

H2O

Water

Sulphuric acid

H2SO4

Hydrogen sulphide

H2S

Magnesium oxide

MgO

Mag ox

Magnesium hydroxide

Mg(OH)2

—

Nitric acid

HNO3

Aqua fortis

Potassium chloride

KCl

Muriate of potash

Sodium hydroxide

NaOH

Caustic soda


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ATOMIC STRUCTURE

Name

Formula

Common name

Sodium bicarbonate

NaHCO3

Baking soda

Sodium chloride

NaCl

Salt

Sodium carbonate

Na2CO3

Soda ash

Sodium sulphate

Na2SO4•10H2O

Salt cake, Glauber’s salt

Sodium acid pyrophosphate

Na2H2P2O7

SAPP

Sodium tetraphosphate

Na6P4O13

Phos

Silicon dioxide

SiO2

Quartz, silica

Zinc carbonate

ZnCO3

Zinc sulphide

ZnS

Zinc oxide

ZnO

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VALENCE The valence of an element or of a ion represents the number of elettrons that it can acquire loose or share to become a stable, neutral charged compound.

Symbols of elements and compounds and their valences

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Element

Symbol

Valence

Hydrogen

H

1+

Oxygen

O

2-

Potassium

K

1+

Sodium

Na

1+

Calcium

Ca

2+

Magnesium

Mg

2+

Aluminum

Al

3+


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VALENCE Element

Symbol

Zinc

Zn

Iron

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Fe

Valence 2+ 3+, 2+

Silver

Ag

Carbon

C

4+

Phosphorus

P

5+

Sulphur

S

2+,4+,6+

Chlorine

Cl

1+,3+,5+,7+


1+

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VALENCE

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Ion or group

Symbol

Valence

Hydroxide

OH

1-

Oxide

O

2-

Carbonate

CO3

2-

Bicarbonate

HCO3

1-

Sulphate

SO4

2-

Sulphite

SO3

2-

Sulphide

S

2-


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VALENCE

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Ion or group

Symbol

Valence

Nitrate

NO3

1-

Nitrite

NO2

1-

Phosphate

PO4

3-

Ammonium

NH4

3-

Acetate

C2H3O2

1-

Formate

CHO2

1-

Thiocyanate

SCN

1-


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ELECTRONIC STRUCTURE •

Electrons orbit around the nucleus of an atom in an ordered disposition called electron shell.

•

Each shell can contain only a maximum and defined number of electrons, and in general each following shell cannot contain more than 8 electrons.

•

The first orbital or shell cannot contain more than two electrons. Completely filled shells form stable structure; they tend not to accept or give up electrons.

•

N=1

•

N=2

•

N=3

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IONIC BONDING Electron shells and ionic bonding

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COVALENT BONDING •

Hydrogen bond:

Water: 2 Hydrogen atoms

Oxygen atoms

Covalent bond Shared electrons

Water and hydrogen gas covalent bond

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HYDROGEN BONDING

Polar molecule and hydrogen bonding in water.

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IONIC BOND Clay hydration by means of water through hydrogen bond

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COMPOUNDS

A compound is a substance composed of elements in definite proportions. The following rules are applied to all compounds: – The composition of a compound is always the same; it is uncheangeable and exact. – The elements loose their own identity (and their peculiar characteristics) when they combine to form a compound. – A compound is homogeneous.

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FORMULA

•

The formula of a compound represents a molecule of the same compound.

•

The weight is used to measure the quantity of chemical substances involved in chemical reactions.

•

A sodium atom weights 22,99 a.m.u.

•

A chlorine atom weights 35,45 a.m.u.

•

22,99 g of sodium will combine with a definite number of atoms in 35,45 g of chlorine to originate salt (sodium chloride)

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FORMULA

• The atomic weight corresponds to 6,023 x 1023 atoms. (Avogadro’s number) • This value corresponds to the weight of a ‘gram-atom’ molecular or ‘mole’ • A mole is a quantitative unit of measurement which contains a definite number of atoms. • A mole is a quantitative unit of measure that contains the exact number of atoms, molecules or formula units which have a mass in grams, equal to the atomic, molecular or formula weight. The normal usage of the mole is the formula weight expressed in grams. For sodium chloride, the formula weight is 58.44 (sodium 22.99 a.m.u., chlorine 35.45 a.m.u.), so one mole of Sodium Chloride would be 58.44 g.

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FORMULA The number of atoms of an element in the formula of a compound is equal to the number of moles of that element needed to form one mole of the compound. •Water: -Hydrogen (atomic weight 1,01) -Oxygen (atomic weight 16,00) •Ratio: -2,02 g of hydrogen for 16,00 g of oxygen -The formula is H2O.

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STOICHIOMETRY – STOICHIOMETRIC REACTIONS •

Atoms react with the other atoms according to their valence. This is the reason why the compounds show a fixed ratio between the elements.

•

Atoms react according to these ratios based on predetermined weights of each atom involved.

•

Stoichiometry deals with the quantities and the ratios among reacting substances

•

Stoichiometric calculations permit to fix exactly the weight and ratio of the chemical elements which will react to be fixed exactly and will determine the desired result.

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EQUIVALENT WEIGHT •

Often, chemical test and reactions are carried out with unknown materials.

•

Since we do not know the correct composition, it is often convenient to express the results in terms of “equivalents” of a standard compound instead of moles.

•

The equivalent weight is the molecular weight of an element, molecule or ion divided by its valence.

– H2SO4 + 2OH– Æ 2H2O + SO42– – HCl + OH– Æ H2 O + Cl –

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BALANCE OF THE CHEMICAL EQUATION – To balance the chemical equation is the first step to determine the stoichiometric reaction. Considering this non balanced equation involving the reaction between iron (Fe3+) and oxygen (O2–) which generates iron oxide: - Fe3+ + O22– Æ Fe23+O32– We can observe that the equation is not balanced in the number of atoms and in the valence charges: there are 2 oxygen atoms on the left side and 3 on the right; there are 4 neg. charges (2 x 2-) on the left and 6 neg. charges (3 x 2-) on the right. The correct equation becomes: 1) Fe3+ + 3O22– Æ 2Fe23+O32– 2) 4Fe3+ + 3O22– Æ 2Fe23+O32– 3) 4Fe + 3O2 Æ 2Fe2O3 Stoichiometrically, 4 moles of Iron combine with 3 moles of Oxigen to yield two moles of Iron Oxide.

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BALANCE OF THE CHEMICAL EQUATION Atomic weight Fe = 55,85 4 moles Fe = 4 x 55,85 = 223,4 g Atomic weight O ≈ 16 3 moles O2 = 3 x 2 x 16 = 96 g How many grams of Oxigen would be required to react with 140 gr of Iron to produce iron oxide? •

Since only 140 g of iron are used (and not 223,4)

•

The ratio given by 140 and divided 223,4 must be multiplied by 96 g of oxigen to determine the quantity of oxygen needed to react with 140 g of iron.

•

Oxygen needed = (140 g Fe / 223,4 g Fe) x 96 g O2 = 60,2 g O2

•

Hence, 60.2 g of oxygen are necessary to react with 140 g of iron to produce iron oxide.

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SOLUBILITY •

SOLUTIONS – Sugar + Water

solution of sugar in water.

– Sugar = Solute (Substance dissolved). – Water = Solvent (Substance that does the dissolving). – A saturated solution is a solution that contains as much dissolved materials as it can hold at a given temperature.

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SOLUBILITY BONDS EFFECTS

Sodium chloride ionization in water

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SOLUBILITY Solubility quantification •The quantity of solute which dissolves in a quantity of solvent to have a saturates solution. •Unit grams of solute per 100 grams of water. Compound

Name

Solubility (g per 100 g of water)

NaOH

Causic soda

119

CaCl2

Calcium chloride

47,5

NaCl

Sodium chloride (kitchen salt)

36

KCl

Potassium chloride

34,7

Na2CO3

Sodium carbonate

21,5

NaHCO3

Sodium bicarbonate

9,6

CaSO4

Anhydride

0,290

Ca(OH)2

Lime

0,185

MgCO3

Magnesium carbonate

0,129

CaCO3

Limestone

0,0014

Mg(OH)2

Milk of magnesia

0,0009

BaSO4

Barite

0,0002

ZnO

Zinc oxide

0,00016

Solubility of common chemical compounds

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SOLUBILITY Factors affecting solubility

Temperature pH (acid or basic) Ionic ambient (salinity) Pressure

Temperature For the majority of solids and liquids, the solubility increases as the temperature increases. Generally, gas solubility decreases as the temperature increases.

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SOLUBILITY pH pH (potential (of) hydrogen) is a measure of the relative acid or basic condition of a solution. The solubility of many chemicals is a function of pH. Some salts of hydroxide or carbonate are more soluble in acid condition. Others are soluble only in neutral pH range and others like organic acids and ligno-sulfonate are more soluble when pH is over 9.5.

Calcium solubility decreasing as the pH increases

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SOLUBILITY

Carbonate-bicarbonate equilibrium

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SOLUBILITY Ionic Ambient (salinity) • Chlorides concentrations or salinity. A salinity increase generally causes an increase of the solubility of other salts and additives and will affect both the chemical reactions and precipitations. This trend decreases as the salinity approaches saturation Pressure • A pressure increase, increases the solubility of a gas in a liquid, but it has no effect on the solubility of the liquids and solids.

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pH AND ALKALINITY pH – The pH is used to define the acidity or the basicity of the solutions. – pH is defined as the negative logarithm of the concentration of the hydrogen ions. – Low values of the pH correspond to an increase of the acidity and high vlues of the pH correspond to a high basicity.

Water is in equilibrium with the ions according to the following equation: 2H2O ⇌ H3O+ (aq) + OH– (aq) The expression of equilibrium for the process of autoionization is: Kw = [H+] [OH–]

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pH AND ALKALINITY • AT 25°C, Kw = l,0 x 10–14 • Kw, the product [H+] e [OH–] • It is constant provided that the temperature is constant. • In a neutral solution, the concentration of hydrogen [H+] is equal to the concentration of hydroxide[OH–]; hence, each will have a concentration of 1,0 x 10–7, and the solution will have a pH of 7,0. • If the [H+] increases, the [OH–] decreases and the solution becomes more acid. Similarly, if the [OH–] increases, then the [H+] must decrease and the solution becomes more basic. • The terms pH and pOH are defined as: - pH = - log [H+] - pOH = - log [OH–]

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pH AND ALKALINITY

• Ratio between pH and pOH: pKw = -log Kw = -log [H+] -log [OH–] Through the above definitions of pH and pOH, we find that at 25° C: pKw = pH + pOH because Kw = l,0 x 10–14 pKw = - log Kw = 14 pH + pOH = 14

pH range, acids and bases

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pH AND ALKALINITY ALKALINITY - Alkalinity titrations define the OH–, HCO3– and CO32– concentrations with the measurement of the quantity of acid required to reduce the pH. - Alkalinity is the aggregating force of a base measured by the quantity of acid which can react to form a salt. - Phenolphtalein alkalinity (P) is reported as the number of millimetres of 0,02 N H2SO4 (water-based mud) required to titrate a millimetre of filtrate (Pf) or mud (Pm), reducing the pH at 8,3. - The alkalinity of the methylorange filtrate (Mf) measures the acid required to reduce the pH to 4,3.

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pH AND ALKALINITY Ratio between the pH and the alkalinity for the distilled water.

• Alkalinity measurements (Pf, Mf and other values) are used to calculate the concentrations of hydroxide, bicarbonate and carbonate.

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ACIDS, BASES AND SALTS ACIDS - Substances with a sour taste - Their pH is in the range from 0 to 7 - Effervescence when in contact with bicarbonates - React with bases and alkalis to form salts - All the acids contain hydrogen . - Acids are defined “strong” or “weak” according to the concentrations of hydrogen ion (H+)

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ACIDS, BASES ANS SALTS BASES - Bases taste sour - Bases have pH in the range from 7 to 14 - React with the acids to form the salts - A base is defined strong or weak according to the number of molecules that dissociate into hydroxide ion (OH–) in the solution. SALTS - Salts are simply the combination of an anion (negative ion), of an acid with a cation (positive ion) of a base. The combination of a weak acid and a strong base form an alkaline salt. The combination of strong acid and a weak base form an acidic salt. The combination of strong acid and a strong base in a neutral salt.

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ACIDS, BASES AND SALTS Acids, bases and common salts

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Chemical noun

Usual name

Formula

Type

Chloridric acid

Muriatic acid

HCl

Acid (strong)

Sulphuric acid

—

H2SO4

Acid (strong)

Nitric acid

Aqua fortis

HNO3

Acid (strong)

Phosphoric acid

Ortho-phosphoric

H3PO4

Acid (weak)

Carbonic acid

Soda (effervescent)

H2CO3

Acid (weak)

Citric acid

—

H3C6H5O7

Acid (weak)

Sodium hydroxide

Caustic soda

NaOH

Base (strong)

Potassium hydroxide

Caustic potash

KOH

Base (strong)

Magnesium hydroxide

Magnesium hydrate

Mg(OH)2

Base

Sodium carbonate

Soda ash

Na2CO3

Base (weak)

Calcium hydroxide

Slaked Lime

Ca(OH)2

Base (strong)

Calcium oxyde

Lime

CaO

Base (strong)

Spdium chloride

Salt

NaCl

Salt

Potassium chloride

Muriate of Potash

KCl

Salt

Calcium chloride

—

CaCl2

Salt

Calcium sulphate

Anhydrite (gypsum)

CaSO4 ( • 2H2O )

Salt


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ACIDS, BASES AND SALTS BUFFER SOLUTIONS - Certain solutions called buffer solutions, resist large pH changes when a base or an acid is added to a solution. - Many drilling liquids and chemical compounds to treat the muds are buffer solutions. Buffering can be highly beneficial to maintain stable fluid properties ELECTROLYTE - An electrolyte is an element or compound that, when dissolved or melted in water or other solvent, dissociates into ions and is able to conduct electric current - All the solutions of ionic compounds are electrolytes - Acids, bases, and salts are common electrolytes. Salt: NaCl = Na+ + Cl– Lime: Ca(OH)2 = Ca2+ + 2OH– Electrolyts that are excellent conductors, contain high concentration of ions in solution and are called strong electrolytes. In general, salts are strong electrolytes as well as some strong acids and hydroxides. Pure water is not a good electrolyte

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OSMOSIS •

The process takes place when two solutions of different salinity are reported by a semi-permeable membrane. There is a movement of solvent (water) from the solution with the lower salinity to the solution of high concentration of the solute.

•

Transfer of water through a membrane from a low solute concentration to one of higher concentration occurs until the two solutions have a similar solute concentration (salinity)

•

The ‘activity’ of a solution is a measure of the vapour pressure or ‘relative humidity’ and it is connected to the concentration of the solute (salinity)

•

In drilling reactive shale it is important to provide similar activity for mud and formation, to minimize the transfer of water from mud to shales. This is also more important when drilling with OBM or synthetic base fluids

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TITRATION The chemical tests made in the mud are called titrations. •

The fundamental equation which refers to this quantitative analysis is: V2 x N2 = V1 x N1 where N1 is a solution of known concentration and N2 is the unknown concentration V1 the volume required to reach the end-Point and V2 the sample of known volume

•

Solving this equation N2 will be obtained as follows: N2 = (V1/ V2) x N1

INDICATORS The chemical compounds used to determine the end-Point in the titrations are called indicators. The indicators are compounds that change colour when change the pH or the chemical concentration. This change in colour happens at a proper pH (particular and exact for every indicator).

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TITRATION Indicator

Original colour

Colour change

Titration

Chemical compound for titration

Phenolphthalein

Pink/red: pH >8,3

Absence of colour: pH <8,3

Pm Pf Pom

Sulphuric acid

Methyl orange/ Green bromic cresol

Green: pH >4,3

Yellow: pH <4,3

Mf

Sulphuric acid

Methyl orange

Yellow/orange: pH >4,3

Rose/red: pH <4,3

Mf

Sulphuric acid

Green bromic cresol

Blue: pH >3,8

Yellow: pH <3,8

—

Sulphuric acid

Thymolphthalein

Absence of colour: pH <9,5

Blue: pH >9,5

—

Caustic solution

Methyl red

Yellow: pH >5,4

Rose/Red: pH <5,4

PHPA

Erio Crome T

Red wine: Ca2+ and Mg2+ presence

Blue/deep red: Ca2+ and Mg2+ absence

Total hardness

Standard Versenate (0.01m EDTA)

CalVer II or Calcon

Red wine: Ca2+ presence

Blue/deep red: Ca2+ absence

Calcium

Standard Versenate (0.01m EDTA)

Potassium chromate solution

Yellow

Orange/red: AgNO3 excess

Chlorides

Silver nitrate solution

Sulphuric acid

Indicators

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SOLUTIONS CONCENTRATIONS MOLALITY (m). - A molal solution is a solution which contains a mole of solute per kilogram of solvent. e.g. 1 m solution of NaOH would be 40 g of NaOH per 1000g of water. MOLARITY (M) - A solution which contains a mole of solute per litre of solution is called molar solution. - e.g. 0,1 M solution of HCl would be 1/10 mole=3,646 of HCl per liter of solution - Molarity = Normality ÷ net positive valence If the net positive valence is 1, N and M will have the same numerical value

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SOLUTIONS CONCENTRATIONS NORMALITY (N) A normal (N) solution 1,0 is defined as a solution with a concentration that contains 1 gweight equivalent of a substance per litre of solution. E.g. 1.0 N solution of HCl has 365g of solute per liter of solution. Molarity = Normality ÷ net positive valence. MILLIGRAMS PER LITRE (mg/l) If the net positive valence is 1, N and M will have the same numerical value. Milligrams per litre is a ratio weight-volume. 100 mg/l solution contains 100 mg of solute per liter of solution.

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SOLUTIONS CONCENTRATIONS PARTS PER MILION (ppm). - Parts per milion : “ppm”, is a ratio weight/weight. It is the weight fraction times 1 million or the weight percent times 10.000. e.g. saturated salt water is 26% salt by weight, therefore it would contain 26x10.000=260.000 ppm salt. - Mg/l can be converted to ppm (if the fluid density is known) by dividing the mg/l value by the specific gravity of the solution.

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SOLUTIONS CONCENTRATIONS EQUIVALENT PARTS PER MILLION (EPM). - The EPM is the unit chemical weight of solute per million unit weights of solution. The EPM is equal to the ppm divided by the E.W. Solutions Concentrations.

Concentration*

Solute weight

Solvent weight

Solution Volume

Solution Weight

1 m (molal)

1 g mole wt

1.000 g

—

1 g mole

1 M (Molar)

1 g mole wt

—

1 liter

—

1 N (Normal)

1 g equiv. wt

—

1 ”

—

100.000 mg/l

100.000 mg

—

1 “

—

100.000 ppm

100.000 mg

900.000 mg

—

1.000 g

* There’s no standard volume of solvent for each of these concentrations of solutions. A specific weight of solute is added to a specific weight of solvent or a volume of solvent is added to obtain the final volume of the solution

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SOLUTIONS CONCENTRATIONS Example: Given a solution of sodium chloride of specific weight 1.148 g/cm3 (SG) with a volume of 1.000 cm3 which contains 230 g of dissolved salt, calculate the following ompositions and concentrations: a) % composition in weight b) % composition in volume c) Molality d) Molarity e) Normality f) mg/l of sodium chloride g) ppm of sodium chloride h) EPM i) Weight ratio between NaCl to H2O, (pounds/pounds) Total solution weight = 1,000 cm3 x 1.148 g/cm3 = 1.148 g Water weight = total weight – salt weight = 1.148 - 230 = 918 g

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SOLUTIONS CONCENTRATIONS a) % Composition in weight: % weight NaCl = (230 ÷ 1.148) x 100 = 20.0% % weights H2O = (918 ÷ 1.148) x 100 = 80.0% b) % Composition in volume: Distilled water specific weight (20°C) = 0,998 g/cm3 Distilled water volume = (918 ÷ 0,998) = 920 cm3 % Volume of water = (920 ÷ 1.000) x 100 = 92% Volume of sodium chloride = 1.000 - 920 = 80 cm3 % Volume of sodium chloride = (80 ÷ 1.000) x 100 = 8%

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SOLUTIONS CONCENTRATIONS c) Molality — moles of solute per kilogram of solvent: Molecular weight of the sodium chloride: 58,44 Grams-mole NaCl = (230 ÷ 58.44) = 3,94 Since there are only 918 g of water, not 1 kg (1.000 g), then the number of grams-mole will have to be calculated for the value of 1 kg. Molality = 3,94 x (1.000 ÷ 918) = 4,287 grams-mole NaCl per kg d) Molarity — moles of solute per 1 litre of solvent: From value c) above, there are 3,94 mole grams of NaCl in the original solution 1.000-cm3 (1-l). Molarity = 3,94 grams-mole-per-litre

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SOLUTIONS CONCENTRATIONS e) Normality — equivalent weight in grams per litre of solution: Since sodium chloride has a net positive valence of 1, the normality has the same value as molarity. Normality = 3,94 grams-equivalent weight per litre f) mg/l of sodium chloride: Since 230 g (230.000 mg) of sodium chloride are contained in 1.000-cm3 (1-l) of solution: mg/l of sodium chloride = 230.000 mg ÷ 1 l = 230.000 mg/l

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SOLUTIONS CONCENTRATIONS g) ppm sodium chloride — ratio weight x 1.000.000: ppm of sodium chloride = (230 ÷ 1.148) x 1.000.000 = 200.350 ppm h) EPM — ppm divided per equivalent weight: EPM sodium chloride = (200.350 ÷ 58,44) = 3.428 EPM

i) Weight ratio between NaCl and H2O : Grams NaCl ÷ grams H2O = (230 ÷ 918) = 0,2505

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MIXTURES – SOLUTIONS – EMULSIONS - DISPERSIONS MIXTURES A mixture is a combination of two or more substances with no consistent composition.

SOLUTIONS A solution is a homogeneous mixture of two or more substances. A solution has a consistent composition. A solution can consist of: - solids - liquids - gas

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MIXTURES – SOLUTIONS – EMULSIONS - DISPERSIONS DISPERSIONS - A dispersion is a two component system in which one component consists of solid particles of rocks distributed in a second component. EMULSIONS - An emulsion is a stable misture of immiscible liquids tenuti assieme da un emulsifier.

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RHEOLOGY AND HYDRAULICS

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INTRODUCTION The physical properties of the drilling fluids, such as density and rheological properties, are monitored in order to optimize the drilling process. These physical properties directly characterize different aspects which are necessary for the good result of the drilling operation.

¾ To guarantee the control on pressure in order to avoid inflows of formation fluids. ¾ To guarantee henergy to the bit to maximise the rate of penetration (ROP).

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INTRODUCTION ¾

To guarantee the stability of the wellbore by means of an adequate distribution of the stresses, either hydrostatic or mechanical on the well walls (pressured or mechanically stressed zones).

¾

To maintain drilling cuttings and weighting materials in suspension during static periods (immobility periods without circulation); for instance, the electrical logs, trips for bit changing etc.

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INTRODUCTION ¾ To allow the separation of the drilling cuttings and gas on the surface.

¾ To remove drilling cuttings from the well .

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RHEOLOGY Rheology is the science of deformation of materials (if they are solid), or the science of the flow (if they are liquid) under applied stress. In the case of fluid, applied force induces flow.

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RHEOLOGY VISCOSITY Viscosity is a measure of the resistence of a fluid to flow. The following terms are used to describe the drilling fluids viscosity and their rheological properties: 1)

Funnel Viscosity (Marsh)(sec/qt o sec/l).

2)

Apparent Viscosity (cP or mPa.sec).

3)

Effective Viscosity (cP or mPa.sec).

4)

Plastic Viscosity (cP or mPa.sec).

5)

Yield point (lb/100 ft2 or Pa).

6)

Gel Strengths (lb/100 ft2 or Pa).

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RHEOLOGY Marsh funnel viscosity The funnel viscosity (or Marsh viscosity), is measured through the Marsh funnel viscosimeter.

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RHEOLOGY

V=0

V max

Speed profile- Shear rate or Shear stress

Laminar Annular flow

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RHEOLOGY Shear Stress (τ) τ = Force / Area It is usually expressed in Dyne/cm2 or in lb/100 ft2. Note: 1 Dyne/cm2 = 4.79 lb/100ft2 Shear Rate (γ) γ = Velocity / Distance = (V2-V1) / d = (cm/sec) / cm = 1 / sec = sec-1 It is expressed in sec-1 (reciprocal seconds).

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RHEOLOGY Where: γ = Shear rate in reciprocal seconds V2 = Velocity of the B layer (cm/sec) V1 = Velocity of the A layer (cm/sec) d = Distance between A and B (cm)

Fig. 2 Shear rate and shear stress Note: 1.703 Sec-1 = 1 x RPM. (Where the RPM refer to the Fann VG meter viscometer.)

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RHEOLOGY Viscosity (µ) The concept of viscosity was introduced by Newton: It is a measure of the resistance of a fluid to flow. The shear stress between adjacent fluid layers is proportional to the negative value of the velocity gradient between the two layers.

Poise – The visosity of 1 poise is equivalent to the force in dynes required to move 2 adjacent layers of fluid with a surface area of 1 cm2 and distance between them 1 cm in 1 second. Viscosity (µ) = shear stress (τ) / shear rate (y) = τ / y

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RHEOLOGY The result of the previous definitions is: µ = (Dyne.sec) / cm2 (poise difinition) Because one poise is a larger value, in practical a centipoise, is used for mud measurements (100 centipoise = 1 poise)

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RHEOLOGY Plastic viscosity Drilling muds are usually composed of a continuous fluid phase in which solids are dispersed. The plastic viscosity indicates the resistance to the flow produced by mechanical friction. Mechanical friction is caused by: • Solids concentration • Solids shape and size • Viscosity of the fluid phase • Presence of polymers carboxymethylcellulose (CMC)) • The ratio oil to water in the inverted emulsions • The type of emulsifier

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RHEOLOGY The value of the plastic viscosity (PV) is obtained subtracting the reading at 300 RPM from the reading at 600 RPM. The plastic viscosity in centipoise (cP) or in milliPascal seconds (mPa s), therefore, is calculated according to the formula: PV (cP) = θ 600 – θ 300 There are three methods to remove the drilled solids: 1. Mechanical control of the solids;(continuous). 2. Settling. 3. Dilution or substitution. (expensive).

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RHEOLOGY Yield Point (YP) The yield point, is the second component of the drilling fluids flow resistance. It is a measure of the electrochemical and attractive forces in the fluid. These forces are determined by positive or negative charges, localized or near the surface of the particles. The yield point is a measure of the whole effect of these forces. It depends on:

1) The superficial properties of the clayey solids dispersed in the fluid; 2) The volumetric concentration of the solids; 3) The electrical ambient of the solids (the concentration and types of ions present in the fluid phase).

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RHEOLOGY A high viscosity deriving from a high yeld point or from attractive forces can be cause by: •

The introduction of soluble contaminants (ions) such as salts, cement, anhydrite or gypsum, which cause the formation of flocculated clay and reactive solids;

•

The crushing of particles from formation clay by the action of the bit or by the drill pipes produces new residual forces on the surface of the crushed particles. Due to these forces, the particles attract mutually and gather in shapes or disorganized aggregations.

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RHEOLOGY •

The introduction of inert solids (barite) into the mud with the consequent increasing of the yield point. This increase is the consequence of the fact that the solid particles are closer one to another. This decreases the interparticle distance causing an increase in the values of the attractive forces between them.

•

The drilling of hydratable clay formation. Such situation introduces new active solids in the mud. Again this brings the particles closer together and increases the total number of charges.

•

Insufficient deflocculant treatment.

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RHEOLOGY The yield point values can be controlled through adequate chemical treatments. Reducing the attractive forces (by chemical means), will cause a decrease of the yield point. The yield point can be lowered by the following methods: The charges present on the edges of the particles can be neutralized by the absorption of negative ions (great quantities) on the surface of the clay particles. These residual charges are neutralized by chemical substances such as: tannines, lignines, complex phosphates, lignosulfonates, etc.. The attractive forces existing before the treatment, are cancelled by the chemical substance introduced and the negative charges on the clays are predominant. As a consequence, the solids repel each other

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RHEOLOGY •

In case of calcium or magnesium contamination, the ions which cause the development of the attractive forces are removed making them precipitate in the form of insoluble compounds. In this way, the attractive forces and the mud YP decrease.

•

Dilution with water can lower/decrease the yield point but, unless the solids concentration is really high, this method is ineffective and can be expensive. Water can alter mud properties in an undesirable manner. This is the case of weighted muds where the water addition can increase the filtrate and decrease the mud weight

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RHEOLOGY The yield point (YP) is calculated by the reading at the rotational viscometer (lb/100 ft2) as follows:

YP = θ 300 – (θ 600 – θ 300) O

YP = θ 300 – PV

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RHEOLOGY Apparent viscosity (µa) The apparent viscosity of a fluid, measured by the rotational viscometer Fann VG meter, is the viscosity of a mud at 600 RPM (1022 sec-1). It is the consequence of the combined action of the plastic viscosity and the yield point. The increasing of one or the other will give an increase in the apparent viscosity (and probably of the funnel viscosity). The latter is sometimes called single point viscosity. The equation for the apparent viscosity is:

µa = (300)(θ 600) / ( 600 ) = (θ600) / 2

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RHEOLOGY Effective viscosity (µe) The effective viscosity read on the VG meter, is the mud viscosity at the determining RPM, i.e. at the speed of the viscometer, It is calculated through the following equation . µe = (300)( θ at a determined RPM) / RPM NOTE: One unit on the reading /dial = 1.067 lb/100 ft2 = 5.11 dyne/cm2 and one RPM = 1.703 sec-1

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RHEOLOGY Gel force Thixotropy is the property exhibited by some fluids to form a gelled structure over time when not subject to shear and then to liquefy when agitated. Most water-base drilling fluids exhibit thixotropy, because of the presence of electrically charged particles or special polymers which form a rigid matrix.

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RHEOLOGY •

The readings of the gel strength through the rotational viscometer (velocity = 3 RPM) at intervals of 10 seconds, 10 minutes and critical situations at intervals of 30 minutes give a measure of the thixotropy degree of the fluid. The gel strength is a function of the quantity and type of solids in suspension, of the time, of the temperature and of the chemical treatment. In other words, all that promotes or prevents the bonding of the particles will increase or decrease the tendence of a fluid to gelatinize.

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RHEOLOGY Excessive gel strength can cause problems :

1. Excessive pressure at the start of circulation after a trip or a break; 2. Reduction of the efficiency of the equipment for solids removal; 3. Swabbing or surging phenomena when the drillstring is pulled out or runs in with escessive velocity,

the consequent reduction or increase of the hydrostatic

weight can cause blow out or loss of circulation.

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RHEOLOGY

¾

Excessive pressures waves as the bit is lowered;

¾

Impossibility for the logging tools to run to the bottom;

¾

Trapped air or gas in the mud.

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RHEOLOGY

Types of gels

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RHEOLOGY EFFECTS OF TEMPERATURE AND PRESSURE ON VISCOSITY

TYPES OF FLUIDS

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RHEOLOGY In Newtonian fuids, if the temperature remains constant, the viscosity does not vary as the shear rate changes. The base fluids of the majority of drilling muds (fresh water, sea water, diesel, mineral and synthetical oils) are Newtonian. In such fluids the shear stress is directly proportional to the shear rate. Graphically the Newtonian fluids are represented by a line passing through the origin. The viscosity is the angular gradient of the line. The initial yield stress (the stress necessary to cause flow to start) of a Newtonian fluid is always equal to zero

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RHEOLOGY

Example of viscosity on a newtonian fluid

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RHEOLOGY τ = µ xγ This general definition is independent from the unit of measurement chosen. µ = (1.0678 x θ) / (1.703 xω)

The viscosity obtained is in English units of measurement ( ft, lb, etc.), but viscosity is reported in centipoise on the API Daily Mud Report. µ (cP) = [478.9 x (1.0678 / 1.703)] x (θ / ω) If the values of the numbers are approximated the formula is: µ (cP) = 300 x (θ / ω)

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RHEOLOGY Newtonian Laminar velocity profile

Maximum velocity profile

Profile inclination = Minimum shear rate

dv

Average velocitya

dr

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RHEOLOGY The flow velocity profile increases with the distance of the wall- The more distant the wall of the tube the higher the flow velocity

The shear rate (sec–1) at the wall of a cylindirc pipe can be calculated with the following equation: γ = 8V / D Where: V = average velocity of the fluid in the pipe (ft/sec) D = diameter of the pipe (ft)

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RHEOLOGY This equation becomes different in case of concentric pipes. γ = 12V / (DH – DP) Where: V = average velocity of the fluid in the pipe (ft/sec) DH = hole diameter (ft) DP = external diameter of the pipe (ft) The DH – DP difference is also called hydraulic diameter. (Annular space).

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RHEOLOGY

Laminar Annular Newtonian velocity profile

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RHEOLOGY

Average velocity

Non Newtonian fluids (Muds)

Laminar flow in non-Newtonian velocity profile

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RHEOLOGY

Shear rate effect on the effective viscosity for a non-Newtonian fluid

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RHEOLOGY

“Shear thinning” effect (pseudoplastic behaviour) in non-Newtonian fluids

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RHEOLOGY Non-Newtonian fluids are classified in two categories: ¾ Fluids with properties independent of time. ¾ Fluids with properties dependent on time.

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RHEOLOGY Non-Newtonian fluids independent of time:

¾ Bingham plastic fluids. ¾ Pseudoplastic fluids. ¾ Dilatant fluids.

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RHEOLOGY Non-Newtonian fluids dependent on time:

•

Thyxotropic fluids

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RHEOLOGY

Rheogram showing the behaviour of the Bingham plastic, pseudoplastic and dilatant fluids

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RHEOLOGY Bingham’s plastic fluids In Bingham plastic fluids, the shear stress also varies linearly with shear rate but, unlike Newtonian fluids, a minimum force must be applied to impart motion to them. This force is known as the yield point or yield value.

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RHEOLOGY The equation for the Bingham plastic fluids is: τ = PV (γ/300) + YP Where: Plastic Viscosity (PV) = θ 600 - θ 300 (cps) Yield Point (YP) = θ 300 – PV

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RHEOLOGY

Curve of shear stress versus shear rate

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RHEOLOGY The common terms associated with the model of the Bingham’s plastic fluids are: Plastic viscosity (plastic viscosity – PV), apparent viscosity (µa), yield point and gel strength.

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RHEOLOGY

Viscosity –Shear Rate Curve

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RHEOLOGY

Parameters of the Bingham plastic model

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RHEOLOGY

Place

Shear Rate, sec-1

drill pipe

00 - 500

heavy drill pipe

00 - 3000

bit nozzles

0,000 - 100,000

annulus

0 - 500

mud tanks

-5 Shear rate in a flow system

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RHEOLOGY Pseudoplastic fluids (Power Law Fluids))

τ = Kγ n Pseudoplastic fluids, like Newtonian, will flow under any applied stress, however small. But, as distinct from Newtonian fluids, the shear stress is not proportional to the shear rate, but to its “nth” power; hence the name “power-law-fluids”. “K” is the consistency index in Pa·sn or in lb·sn/100ft2, and “n” is the dimensionless flow behaviour index which is unity or smaller than unity.

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RHEOLOGY Depending on the n value, there are three different types of flow and rheological behaviour profiles: n < 1:The fluid is non-Newtonian pseudoplastic. n = 1: The fluid is Newtonian. n > 1: the fluid is dilatant, shear thickening (drilling fluids do not belong to this category) (shear thickening – increasing of the stress viscosity, expanding behaviour. This behaviour, is opposite to the shear thinning, and is characterized by an increase in viscosity, as the shear rate increases)

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RHEOLOGY For the pseudoplastic fluids: τ = K (γ)n; (0 < n < 1) where: τ = shear stress γ = shear rate K = fluid consistency index (constant characteristic of a fluid) n = flow behaviour index

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RHEOLOGY Dilatant fluids τ = K (γ)n; (n > 1)

“n” effects on fluids behaviour

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RHEOLOGY Thyxotropic fluids

Rheogram of the hysteresis cycle

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RHEOLOGY Law of modified power (YPL- yield power law model)

Comparison on rheological models

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RHEOLOGY

Comparison on rheological models in logarithmic coordinates

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RHEOLOGY The Herschel-Bulkley equation is expressed as follows: τ = τо+ K(γ)n Where: τ = Shear stress τо = yield stress or stress to begin the motion K = Consistency index γ= Shear rate n = Power law index

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FLOW RUNNING Stage 1 — No flow

Stage 2 — Plug flow

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FLOW RUNNING Stage 3 —Transition from plug flow to laminar flow

Stage 4 — Laminar flow

Stage 5 — Transition from laminar flow to turbulent flow

Stage 6 — Turbulent flow

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FLOW RUNNING

or

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laminar flow

Flow stages


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HYDRAULIC RHEOLOGICAL MODELS Bingham Plastic Due to its semplicity, the Bingham Plastic is the most used model to describe the rheological properties of a drilling fluid. It assumes that the shear stress would be a linear function of the shear rate and it is expressed as follows:

YP = yield point, lbf /100 ft2, PV = plastic viscosity, cp (centipoise).

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HYDRAULIC RHEOLOGICAL MODELS As this model is developed by the determination of the data of shear rates from 500 to 1000 sec-1, it allows a better identification of a fluid with higher shear rates. The PV and YP are calculated from data obtained by the conventional rotating viscometer, at 300 and at 600 rpm with the following formula: PV = θ600 – θ300 Where:

PV = plastic viscosity, cp θ600 = 600 rpm (instrument reading) θ300 = 300 rpm (instrument reading)

YP = θ300 – PV Multiplying the rpm by 1.702 the shear rate in sec-1 is obtained. Once having determined the PV and YP values, the model can be used to determine the shear stress at whatever shear rate

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HYDRAULIC RHEOLOGICAL MODELS Power Law The Power Law model describes a fluid where the shear stress meets the shear rate in a straight line. As it is less approximate than the Bingham Plastic model, this module is often used to keep the characteristics of suspension under control and to calculate the pressure losses of the mud in the annular space. The Power Law formula is: τ = Kγⁿ where:

K = flow consistency index (lb-secn/100 ft2),

n = flow behaviour index (dimensionless).

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HYDRAULIC RHEOLOGICAL MODELS The rheological parameters n and K can be calculated through two points of whatever shear-rate/shear-stress. As it is fairly improbable that in a log-log graph of rheological data a straight line can be obtained, it is better to determine n and K in the drill pipes (np and Kp) and in the annular space (na and Ka). The most accurate results will be obtained with the use of n and K in the range from 5 to 200 sec-1 in the annular space and from 200 to 100 sec-1 in the pipes (rpm = sec-1 ÷1.7).

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HYDRAULICS RHEOLOGICAL MODELS The reading of a normal instrument at six velocities, allows the determination of the power law constants. The formula for n and K are:

⎛τ 2 ⎞ log ⎜⎜ ⎟⎟ τ1 ⎠ ⎝ n= ⎛γ2 ⎞ log ⎜⎜ ⎟⎟ ⎝ γ1 ⎠

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HYDRAULIC RHEOLOGICAL MODELS Where:

τ2= reading shear stress at higher shear rate, τ1= reading shear stress at low shear rate, γ2 = high shear rate range (rpm), γ1 = low shear rate range (rpm). If readings from 600 to 300 rpm are carried out, the formula is simplified as follows:

⎛ θ 600 n = 3 . 32 log ⎜⎜ ⎝ θ 300

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⎞ ⎟⎟ ⎠

K

=

θ

300

511


n

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HYDRAULIC RHEOLOGICAL MODELS Herschel - Buckley (modified Power Law) It is a combination of the characteristics of the Newtonian model,Bingham Plastic and Power Law. The formula is:

τ = YP + K γ

n

This model using three parameters, reproduces the results of the three models described previously. Due to the difficulty to obtain this data, using the rheometer a γp = 3 RPM is assumed.

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HYDRAULIC RHEOLOGICAL MODELS

Guide to the hydraulics optimization

Where:

n = 3.32log [(θ 600 − YP ) ÷ (θ 300 − YP )] YP = θ3 (lb/100 ft2),

K

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θ 300 511

n


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HYDRAULIC RHEOLOGICAL MODELS REYNOLDS NUMBER The number of Reynolds (NRe) is a dimensionless number that is used to determine whether a fluid is in laminar or turbulent flow. According to the API rule, the number of Reynolds, lower or equal to 2100 indicates a laminar flow while if it is higher than 2.100 the regime is turbulent flow.

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HYDRAULIC RHEOLOGICAL MODELS •

The general formula for the number of Reynolds is:

N

Re

⎛ ρ ⎞ = VD ⎜⎜ ⎟⎟ ⎝µ ⎠

Where: V = Velocity D =Diameter ρ = Density µ =Viscosity

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HYDRAULIC RHEOLOGICAL MODELS The number of Reynolds in the pipes is:

N

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Rep

⎛ ρ ⎞ ⎟ = 15 . 467 V p D ⎜ ⎜ µ ep ⎟ ⎠ ⎝


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HYDRAULIC RHEOLOGICAL MODELS The number of Reynolds in the annular is:

N Rea

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⎛ ρ = 15 .467 Va (D 2 - D 1 )⎜⎜ ⎝ µ ea


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134

HYDRAULIC RHEOLOGICAL MODELS Where:

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D

= ID drill pipe or drill collars

D2

= ID hole or casing

D1

= OD drill pipe or drill collars

µep

= Effective viscosity (cP) pipe

µea

= Effective viscosity (cP) annulus


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FRICTION FACTORS FOR THE POWER-LAW MODULE

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HYDRAULIC EQUATIONS AVERAGE BULK VELOCITY The API calls in this way the velocity of a fluid in an annular space or inside the pipes, assuming that all the fluids flow at the same velocity. Average bulk velocity in the pipes (Vp):

V

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p

0 . 408 Q = 2 D


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HYDRAULIC EQUATIONS

Average bulk velocity in the annulus (Va):

Va

0 . 408 Q = 2 2 D 2 − D1

(

)

Where: V = Velocity (ft/sec) Q = Rate (gpm) D = Diameter (in.)

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HYDRAULIC EQUATIONS Pressure Losses In the drilling, the flow inside the pipes on the surface, their connections, drill pipe, drill collars and at the bit chokes is always turbulent.

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HYDRAULIC EQUATIONS The flow in the annulus can be both laminar and turbulent. However, the annular pressure losses are rather low compared to the total pressure fall. In the turbulent flow, the effect of the viscosity properties of the mud in the pressure losses are very limited. As a consequence, the calculations are usually made using programmes, rules, tables, etc based on turbulent flow.

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HYDRAULIC EQUATIONS In the indications of a hydraulic programme it is necessary to calculate the pressure losses in the one or more flow rates. The circulating pressure, Pc, corresponds to the sum of the following pressure losses: 1.

Surface connection (standpipe, rotary hose, kelly e swivel).

2.

Drillpipes.

3.

Heavy-weight drillpipes.

4.

Drill collars.

5.

Annulus between the drill collars and the open hole.

6.

Annulus between the drillpipes the open hole.

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ENGINEERING CALCULATIONS

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INTRODUCTION

This section is a guide for calculations regarding the drilling fluid such as capacity of mud, tanks volumes, tubulars and holes, circulation times, velocity of the mud in the annular and in the drill pipes and other important calculations. The capacity of calculating muds formulations and various situations by means of solids and liquids additives is necessary in fluids engineering.

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FIELD UNIT SYSTEM The unit of measure most commonly used is the Field Unit. However, the decimal measuring system is more and more used in drilling.

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FIELD UNIT SYSTEM

Agip KCO

Metric System

Field Unit

Mass

Kilogramme (kg)

Pounds (lb)

Length

Meters (m)

Feet (ft) and inches (in.)

Volume, Capacity and Displacement

Cubic meters (m3) and litres (l)

Barrels (bbl) and gallons (gal)

Density

grams/ cubic centimetres (g/cm3) e (kg/l) – Either equal at specific weight

Pounds/gallon (lb/gal) and pounds/foot3 (lb/ft3)

Pressure

kiloPascals (kPa), bar or atmospheres

pounds/inch2 (lb/in.2 o psi)

Concentration

kilogram/cubic meter (kg/m3)

pounds /barrel (lb/bbl)


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CONVERSION FACTORS Multiply

by

To obtain

Volume Barrels(bbl)

5.615

Feet3 (ft3)

Barrels(bbl)

0.159

Meters3 (m3)

Barrels(bbl)

42

Cubic feet (ft3)

0.0283

Cubic feet(ft3)

7.48

Gallons, U.S. (gal) Meters3 (m3) Gallons, U.S. (gal) Meters3 (m3)

Gallons, U.S. (gal)

0.00379

Gallons, U.S. (gal)

3.785

Litre(l)

Cubic Meters (m3)

6.289

Barrels(bbl)

Cubic Meters (m3)

1,000

Litres(l)

Pounds (lb)

453.6

Grams (g)

Pounds (lb)

0.4536

Kilograms(kg)

Kilograms (kg)

2.204

Pound (lb)

Metric tons (mt)

1,000

Kilograms (kg)

0.3048

Metres (m)

Mass or Weight

Length Feet (ft) Inches (in.)

2.54

Centimetre(cm)

Inches (in.)

25.4

Milliletre (mm)

Meters (m)

3.281

Feet (ft)

Miles (mi)

1.609

Kilometres (km)

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CONVERSION FACTORS Multiply

To obtain

by

Pressure lb/in.2 (psi)

6.8948

lb/in.2 (psi)

0.068948

lb/in.2 (psi)

0.0703

kg/cm2

kiloPascal (kPa)

0.145

lb/in.2 (psi)

bar (bar)

100

kiloPascal (kPa) bar (bar)

kiloPascal (kPa)

Concentration pounds/barrels (lb/bbl)

2.853

kg/m3

kilograms/cubic metre(kg/m3)

0.3505

lb/bbl

119.83

kg/m3

kilogram/cubic meter (kg/m3)

0.008345

lb/gal

pound/gallon (lb/gal)

0.11983

g/cm3, kg/l or SG

Density pounds /gallons (lb/gal)

pound/cubic feet (lb/ft3)

16.02

kg/m3 and g/l

g/cm3, kg/l or SG

8.345

lb/gal

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HOLE CALCULATIONS

¾ Capacity ¾ Volume ¾ Displacement

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HOLE CALCULATIONS

Mud pits and tanks – Capacity and Volume Rectangular Tanks V = Volume or Tank Capacity L = Length W = Width H = Height M = Fluid Level

Volume = L x W x H

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HOLE CALCULATIONS Vertical cylindrical tanks Vcyl = Cylindrical tank capacity D = Cylinder diameter Acqua distillata H = Cylinder height M = Fluid level π= 3.1416 To calculate the diameter, measure the circumference and divide by 3.1416: D = Circumference/π The formula to calculate the capacity of a vertical cylindrical tank is: Vcyl

=

2

π D

H

4

The volume of the fluid (VMud) in a vertical cylindircal tank is calculated as follows : π x D2 M 3 3 VMud (ft or m ) = 4

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HOLE CALCULATIONS Horizontal Cylindrical Tanks.

Figure 3

VCyl = Horizontal cylinder capacity D = Cylinder diameter L = Cylinder length M = Material level π = 3.1416

VCyl

Agip KCO

L = 2

⎡ ⎢ (2 M − D ) MD − M ⎢⎣

2

2 D2 ⎞ πD −1 ⎛ 2 M + − 1⎟ + sin ⎜ 2 4 ⎝ D ⎠


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151

HOLE CALCULATIONS Hole Volume The volume of each interval is calculated as follows:

V

Hole

=

section

2

π D

L

4

Where: DW = Internal diameter (ID) of the casing, liner or open hole L = Interval Length If the diameter (DW) is in inches:

VHole

D2 W ( in.) section (bbl / ft ) = 1029

On the contrary, with the decimal measuring system:

V Hole

Agip KCO

section

D 2 W ( in. ) m /m = 1974

(

3

)


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HOLE CALCULATIONS Drill Pipe or Drill Collars capacity (volume of the internal diameter) The volume, with the drill string in the hole, is the sum of the internal capacity of the drill pipes plus the volume of the annular space.

V

Pipe

(bbl

/ ft

)=

ID

)=

ID

2

P

(in . )

1029

With the metric system:

V Pipe

Agip KCO

(l

/ m

2

P

(in . )

1974


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HOLE CALCULATIONS Annular volume The volume of the annular space is determined subtracting the external volume of the drill string from the hole capacity(or casing).

V Annulus

ID 2 W (in .) − OD 2 P (in .) (bbl/f t ) = 1029

Where: IDW = Internal diameter hole or casing ODP = External diameter drill pipes or drill collars With the metric system:

ID 2 W (in.) − OD 2 P (in.) VAnnulus (l/m ) = 1974 Or: VAnnulus = CapacityWell – DisplacementDrillstring – CapacityDrillstring

Agip KCO


Annular Volume

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HOLE CALCULATIONS Displacement The drill string displacement can be estimated (VPipe Displ.) using the OD and the ID of the drill pipes and the drill collars.

V Pipe

Displaceme nt

(bbl/f t ) =

OD

2

P

(in .) − ID 2 P (in .) 1029

Where: ODP = External diameter drill pipe or drill collars IDP = Internal diameter drill pipe or drill collars With the metric system:

VPipe

Agip KCO

Displaceme nt

(l/m ) =

OD

2

P

(in .) − ID 2 P (in .) 1974


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PUMPS CAPACITY TRIPLEX PUMPS In the triplex pump, pistons are three and operate in the same direction. Generally, they are short stroke, (from 6-in. to 12-in.) and operate at rates from 60- to 120-stk/min. Triplex pump performance calculation formula:

V Pump

Agip KCO

Output

=

3 x 3 . 1416 xID

2

Liner

xLxP Eff

4


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PUMPS CAPACITY Where: VPump Output = Pumps Performance IDLiner = Liner Internal Diameter L = Stroke Length PEff = Volumetric efficiency (dimensionless)

V Pump

Output

(bbl

/ stk ) =

ID

2

Liner

(in .)xL (in .)xP Eff 4117 . 7


V Pump

Agip KCO

Output

(l

/ stk

)=

ID

2

Liner

(in . )xL (in . )xP Eff 25 . 90


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PUMPS CAPACITY

DUPLEX PUMPS The pistons of a duplex pump are two and operate in both directions. The difference between the calculations for a duplex and for a triplex is that the volume of the rod pistons must be subtracted form the volume of one of the cylinders plus the difference in number of cylinders, 4 for a duplex and 3 for a triplex. Generally, the duplex pumps have a longer stroke (from 10 to 18 in.) and operate at lower rate from 40 to 80-stk/min.

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PUMPS CAPACITY The general equation for a duplex pump is :

VPump Output

[

) ]

(

2π = x ID 2 Liner xL + ID 2 Liner − OD 2 Rod xL xPEff 4

Where:

IDLiner = Internal diameter liner VPump Output = Pump rate ODRod = Rode external diameter L = Stroke length PEff = Volumetric efficiency (dimensionless)

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PUMPS CAPACITY

Pump rate bbl/stroke for a duplex pump with ID liner, OD pipe and stroke length are in inches:

V Pump

Output

⎡ (bbl / stk ) = ⎢ 2 xID ⎢⎣

2

Liner

(in . ) − OD

2

Rod

6174

(in . ) ⎤ xL (in . )xP ⎥ ⎥⎦

Eff


VPump Output

Agip KCO

⎡ 2 xID 2 Liner (in.) − OD 2 Rod (in.) ⎤ (l / stk ) = ⎢ ⎥ xL (in.)xPEff 38.85 ⎦ ⎣


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ANNULAR VELOCITY The annular velocity (AV) is the average velocity of the fluid which flows in the annular space. For a correct well cleaning, a minimum velocity is needed. The latter depends on a certain number of factors such as ROP, cuttings size, hole inclination, mud density and rheology.

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ANNULAR VELOCITY •

Equations for the calculation of the anular velocity based on pumps rate and the annular volume:

AV

=

AV (ft/min

V

V

)=

AV (m/min ) =

Agip KCO

→ Output

Pump

Ann

V Pump

Output

(bbl

/ min

V Ann (bbl / ft )

)

VPump Output (l / min ) V Ann (l / m )


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CIRCULATION TIME The total circulation time is the time (or number of strokes) to make a complete loop, from the tank, following the path down to the bit, go up from the annular and come back in the circulating tank.

Total circulatio n time (min) =

Agip KCO

VSystem VPump Output


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CIRCULATION TIME Where: VSystem = Total attive volume (bbl o m3) V Pump Output = Pumps capacity (bbl/min o m3/min) Total circulation (strokes) = Total circulation time (min) x pumps velocity (stks/min)

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CIRCULATION TIME

Bottoms-up is the time (or number of strokes) the mud circulates from bottom hole throught the annular up on the surface. The bottoms-up is calculated as follows:

Bottom

− up time(min)

=

V Annulus V Pump

Output

Where: VAnnulus = Annular volume (bbl or m3) VPump Output = Pumps capacity (bbl/min or m3/min)

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CIRCULATION TIME

Bottoms-up (strokes) = Bottoms-up (min) x pumps velocity (stk/min) The total circulation time (or number of strokes) to make a complete circulation, starting from the tank down to the bit and up to the annular returning back in the tanks. The circulation total time is calculated as follows:

Hole cycle time(min)=

Agip KCO

VHole − VDrillsringDispl VPump Output


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CIRCULATION TIME Where: V Hole = Total well volume (bbl o m3) V Drillstring Displ = Pipes displacement (bbl o m3) V Pump Output = Pumps capacity (bbl/min o m3/min) Complete circulation (strokes) = time for a complete circulation (min) x pumps velocity(stk/min)

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HYDROSTATIC PRESSURE The hydrostatic pressure (Ph) is the pressure exerted by a column of liquid and depends on the density of the fluid and the vertical depth or True Vertical Depth (TVD). In a well, it is the pressure exerted on the wall to avoid cavings and controls the formation pressure as well.

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HYDROSTATIC PRESSURE Hydrostatic pressure is calculated as follows:

P H (bar

)=

Mud Weight

(kg/l )x TVD (m ) 10.2

Hydrostatic pressure = Mud density x TVD x conversion factor (0.0981)

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HYDROSTATIC PRESSURE Field unit system:

P H (lb/in.2) = Mud density (lb/gal) x TVD (ft) x 0.052

Conversion

Agip KCO

12 in./ft factor 0.052 = 231 in. 3 /gal


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EXAMPLES • Data: • Surface Casing: 1,600 m - 133/8-in. 48-lb/ft, (323-mm ID) • Bit diameter: 121/4 in • T.D.: 3,400 m • Drillstring: Drill Pipe: 5-in. 19.50-lb/ft, (127-mm OD, 108.6-mm ID), Drill collars 200 m of 71/4-in. x 23/4-in (185-mm OD x72-mm ID) • Surface systems: 2 tanks: depth 4-m, width 3-m, length 10-m. Both tanks have 2.5 m of mud with pipes in well. • Mud density: SG 1.50 o 1.5 kg/l • Mud pumps: Triplex: 6 in x 12 in (152.4 mm x 304.8 mm) 110 stk/min, with 90% efficiency

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EXAMPLES

13.3/8 in. casing 1600 m 12.1/4 in. Open Hole Drill Pipe 5 in. 19.5-lb/ft

DC 71/4-in. x 23/4-in 3400 m Well Diagram

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EXAMPLES

Part I: Determine the capacity of the surface system in m3, m3/m e m3/cm. VPit (m3) 1 tank = 4 m x 3 m x 10 m = 120 m3 VPit (m3) 2 tanks = 120 m3 x 2 = 240 m3 VPit (m3/m) 2 tanks = 240 m3 ÷ 4 = 60 m3/m VPit (m3/cm) 2 tanks = 60 m3/m ÷ 100 cm/m = 0.60 m3/cm Part II: Determine the total volume of the mud on the surface, in m3. VSurface (m3)= VMud (m3) 2 tanks = 60 m3/m x 2.5 m = 150 m3

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EXAMPLES Part III: Determine the total volume of the well without drill string. Calculate the volume of the mud in each interval and sum the volumes.

V Well

(m ) 3

( )

VCsgl m

3

ID 2 Well = 1,273,000

x L (m

)

323 2 mm 2 = x 1,600 (m ) = 131.1 m 3 1,273,000

( )

250.82 mm2 x1,800(m) = 88.9 m3 VOH m = 1,273,000 3

Total without the drill string: V System = V Csg + V OH = 131.1 m3 + 88.9 m3 = 220 m3

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EXAMPLES Part IV: Determine the total volume with the string in the well . Volume in the string: V Drillstrin

( )

VDP m

3

g

(m )

ID 2 DS (mm = 1,273,000

) x L (m )

108.6 2 (mm ) = x 3200 (m ) = 29.6 m 3 1,273,000

( )

V DC m

3

3

72 2 (mm = 1,273,000

)

x 200 (m ) = 0.8 m

3

Total volume in the string: V Drillstring = V DP + V DC = 29.6 m 3 + 0.8 m 3 = 30.4 m 3

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EXAMPLES Volume in the annulus:

VAnnulus

ID 2 Well (mm) − OD 2 DS (mm) x L (m ) m = 1,273,000

( ) 3

3232 (mm) − 1272 (mm) VAnn(Csg DP) m = x 1,600 (m) = 0.06927 x 1,600 = 110.8 m 3 1,273,000

( ) 3

250.82 (mm) − 127 2 (mm) VAnn(OH DP) m = x 1,600 (m ) = 0.03673 x 1,600 = 58.8 m 3 1,273,000

( ) 3

250.82 (mm) − 1852 (mm) VAnn (OH DC ) m = x 200 (m ) = 0.02252 x 200 = 4.5 m 3 1,273,000

( ) 3

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EXAMPLES

VAnnulus

(m ) = V 3

Total

Ann (Csg DP )

+ VAnn (OH DP ) + VAnn (OH DC )

V Annulus Total = 110.8 + 58.8 + 4.5 = 174.1 m3

Total volume of the hole with a string:

V Well/DS = V Annulus Total + V Drillstring = 174.1 + 30.4 = 204.5 m 3

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EXAMPLES Part V: Determine the circulation total volume. V Total = V Well/DS + V Surface = 204.5 m3 + 150 m3 = 354.5 m3

Part VI: Determine the time needed for a complete circulation and bottoms-up.

ID2 Liner (mm) x L (mm) x Eff (decimal) VP ump Output (l/stk) = 424,333

152.4 2 Liner (mm) x 304.8 (mm) x 0.9 (decimal) VP ump Output (l/stk ) = = 15.01 (l/stk ) 424,333

(

VP ump Output (l/min ) = 15.01 (l/stk ) x 110 stk/min = 1,651 (l/min ) = 1.651 m 3 / min

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)

EXAMPLES

354.5 m 3 Total circulatio n time (min ) = = 215 min 3 1.651 m /min

(

)

Total circulation (stk) = 215 min x 110 stk/min = 23,650 stk

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EXAMPLES Hole cycle time (min ) =

( )

V Well/DS m 3

(

V P ump Output m 3 / min

)

=

204 .5 = 124 min 1.651

Hole cycle (stk) = 124 min x 110 stk = 13,640 stk

Bottoms - up time (min ) =

V Annulus

Total

(

(m ) 3

V P ump Output m 3 / min

)

=

174 . 1 = 106 min 1.651

Bottoms-up (stk) = 106 min x 110 stk = 11,660 stk

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EXAMPLES Part VII: Determine the velocity in the annular per each single interval.

AV =

V Pump

AV(OH/DC) =

AV(OH/DP) =

AV(Csg/DP) =

Agip KCO

V

Output Ann

( (m

) = 1.651(m / min) = 73 m/min / m) 0.02252(m / m)

VPumpOutput m3 / min VAnn OH/DC

3

( (m

3

3

) = 1.651 (m / min) = 45 m/min 0.03673 (m / m ) / m)

VPump Output m 3 / min VAnn OH/DP

( (m

3

3

3

) = 1.651 (m / min) = 24 m/min / m) 0.06927 (m / m )

VPump Output m 3 / min VAnn Csg/DP

3

3

3


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EXAMPLES

Part VIII: Determine the hydrostatic pressure at the TD

P

P

H

H

Agip KCO

(bar ) =

(bar ) =

Mud

Weight

(kg/l

)x TVD

(m )

10.2

1.5

(kg/l ) x 10.2

3,400

(m ) =


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182

bars

DENSITY INCREASE THROUGH ADDITIVES

Weight increase (specific weight d = 4.2)

(

W kg/m

with:

3

( d 2 − d1 ) ) = 4200 (4.2 - d ) 2

W = weight of the barite to be added in kg/m3 d1 = mud initial denisty (specific weight) d2 = desired mud density (specific weight)

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DENSITY INCREASE THROUGH ADDITIVES

Weight increase using Calcium Carbonate

(

W M kg/m

with:

3

) = 2650

(d 2 − d 1 ) (2.65 - d 2 )

W = weight of the calcium carbonate to be added in kg/m3 d1 = mud initial density (specific weight) d2 = desired mud density (specific weight)

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DENSITY REDUCTION THROUGH WATER OR OIL

Water needed to reduce density (density H2O d = 1)

(

VWater liters/m

with:

3

) = 1000 x

(d1 − d 2 ) (d 2 − 1)

VWater = water volume (litres) to be added at 1 m3 of mud d1 = mud initial density (specific weight) d2 = desired mud density (specific weight)

Agip KCO


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DENSITY REDUCTION THROUGH WATER OR OIL

Oil needed to reduce the density (oil density d = 0.85)

(

Voil liters/m

3

) = 1000 x

(d 1 − d 2 )

(d 2 − 0 .85 )

with: VOil = oil volume (litres) to be added 1 m3 of mud d1 = mud initial density (specific weight) d2 = desired mud density (specific weight)

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VOLUME INCREASE WITH ADDITIVES

Final volume after the addition of additives

V

F

with:

(l ) =

V

I

+

M d

a a

VF = final volume (litres) VI = initial volume of 1,000 l (1 m3) Ma = additives weight da = additives specific weight

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MIXING OF LIQUIDS WITH DIFFERENT DENSITY Final volume after mixing VA + VB = VF (VA x dA)+ (VB x dB) = (VF x dF)

with: VA = fluid volume A (m3) VB = fluid volume B (m3) VF = final volume(m3) dA = fluid density A (kg/l) dB = fluid density B (kg/l) dF = final density of the mixed fluids (kg/l)

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MUD REPORT

Agip KCO


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MUD REPORT

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MUD REPORT

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2 Drilling Fluids

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