2002.Hydraulics

2.00.00

HYDRAULICS

2.01.01 2.01.02

Introduction, Circulating System Hydraulics Introduction, Study of Hydraulics

2.02.01 2.02.02 2.02.03 2.02.04 2.02.05 2.02.06 2.02.07 2.02.08

Hydraulic General Terms and Definitions Matter and Types of Matter Mass, Inertia, Density Specific Gravity Common Oilfield Volumes Area of shapes Weight Force & Pressure

2.03.01 2.03.02 2.03.03 2.03.04 2.03.05 2.03.06 2.03.07 2.03.08 2.03.09 2.03.10 2.03.11 2.03.12 2.04.01 2.04.02 2.04.03 2.04.04 2.04.05 2.04.06

Hydrostatic Fluids Hydrostatic Fluid Illustrations Single Liquid. Seeks own level Multiple Liquid. Seeks own level Hydrostatic Pressure Hydrostatic Pressure in Oilfield Hydrostatic Pressure Gradient Buoyancy & Archimedes Principle Buoyed Weight Buoyancy Factor Buoyancy Calculation Example. Hydrostatic Fluid Formulas. Hydraulic Power Transmission Pressure & Force Hydraulic Cylinders Work Hydraulic Power, horsepower Basic Hydraulic Pump

2.05.01 2.05.02 2.05.03 2.05.04 2.05.05 2.05.06 2.05.07 2.05.08 2.05.09 2.05.10

Circulating System & Fluid Flow Drilling Circulating System Hydraulic Horsepower Volumetric-Rate & Pumps Continuity of Flow Fluid Velocity Types of Flow, Turbulent & Laminar Fluid Flow Pressure Loss Pressure Drop Factors Annular Flow & Pressure Drop

2.06.01 2.06.02 2.06.03 2.06.04

Rheology - Study of Fluid Flow Terms Rheology - general terms and definitions Rheology - Fluid Flow Calculation Models Rheology - Fluid Flow Classifications

2.07.01 2.07.02

Bit - Mechanical Energy Bit - Hydraulic Energy

2.08.01 2.08.02

Rule of Thumb to Optimize Hydraulics Rule of Thumb to Optimize Hydraulics

2.09.01 2.09.02 2.09.03 2.09.04 2.09.06 2.09.08

Hydraulic & Related Formula Summary Hydraulic - Bit Hydraulic - Drill String & Annulus Hydraulic - Fluids & Well Control Hydraulic - Volumes & Misc Misc……….

E X I T

2.01.01 Hydraulics

Introduction

The Circulating System

2.01.02 Hydraulics

Stand Pipe

Introduction

Hydraulics is the study or application of liquids and their properties.

Rotary hose

PUMP

Hydraulics Hydraulicsmay maybe bedivided dividedinto intotwo twocategories: categories: Hydrostatic applies Hydrostatic appliestotoliquids liquidsatatrest rest

HOPPER

Hydrodynamic Hydrodynamic M

UD

PI T

RV SE RE

Note. Note. Some Somelaws lawsofofhydraulics hydraulicsalso alsoapply applytotogases gasesunder undercertain certain conditions, but unless specifically conditions, but unless specificallystated,only stated,onlyliquids liquids will willbebeused usedororconsidered. considered.Note Notealso, also,the theterm term“fluid” “fluid” will willapply applytotoliquids liquidsonly only- - unless unlessspecifically specificallynoted. noted.

E

OPEN HOLE

CASED HOLE

T PI

SHALE SHAKER

The circulating system is a large hydraulic system with the fluid serving multiple purposes, including the work of making hole, supporting hole walls, carrying cuttings from the hole, lubricating the drill string, etc.

applies appliestotoliquids liquidsininmotion motion

Drilling an oil or gas well involves extensive use of hydraulics in making and maintaining the hole, as well as, in the operation of some rig and down hole equipment. Sometimes only one fluid category is applicable to an operation, but in many circumstances principles of both types are in play at the same time. Many compromises must be made during the course of drilling a well between the “ideal” hydraulic parameters used and what is possible. It is necessary to understand each type of hydraulics, individually and the effect that one has on the other one.

2.02.01 Hydraulics General Terms and Definitions

Matter

Material Substance that occupies space and has weight

Mass

Property of matter that is a measure of its inertia.

Inertia

To remain at rest or in a straight line motion unless acted upon by a force.

Density

Weight per unit volume or weight-density is the common use. ( Lbs/Gal., etc.)

Specific Gravity

Ratio or density of a substance to density of a standard substance such as water.

Area

Measure of a surface equal to unit squares. ( sq.inches, sq. feet, etc )

Volume

Measure of amount of space an object occupies. ( Cu.Foot, Cu.Inch, Gallon, Barrel, etc.)

Weight

Measure of the downward force of object as a result of gravitational force and objects mass.

Force

Push or pull exerted on object to change its position or direction of movement.

Pressure

Amount of force exerted on substance per area over which force is applied.

Work

Measure of force through a distance.

Power

the time rate of doing work. ( Horsepower ) ( 33000 Ft - Lbs / Min. )


Matter: Any material substance that occupies space and has weight. It can be grouped as solids (rigid), or fluids (flow) with fluids being either a liquid or a gas.

Solids:

having a definite volume and shape independent of any container.

.

Liquids: having a definite volume with no shape of its own, assuming that of its container. Volume is affected only slightly by changes in temperature or pressure.

Gases having neither volume nor shape of its own, assuming that of its container. Gases are highly compressible with their volume dependant on temperature and pressure. Fluids….Liquids and gases are both fluids in that they have no shape of their own, constantly deforming with any application of force unless confined. However, fluids in this presentation will refer to a liquid only unless specifically noted as one of of the more common terms in the oilfield is “drilling fluids” which refers to drilling mud, a liquid.


Mass is the property of matter that is a measure of its inertia. Inertia is the property of matter by which it remains at rest or in a straight line motion unless acted upon by an outside force.


Specific Gravity .. is the relative density of a substance compared to the density of a standard substance. The most commonly used substance as a standard is water at the temperature of its maximum density, 39.20 F. Specific SpecificGravity Gravity Weight WeightofofSubstance Substanceper perUnit UnitVolume Volume Weight WeightofofWater Waterper perUnit UnitVolume Volume

A stationary object will remain stationary unless acted upon by an outside force Air Resistance

R E A S I I R S T A N C E

Gravity

A thrown ball will travel in a straight line were it not for the external forces of air friction and gravity.

Density is technically Mass per Unit Volume but is commonly used as Weight per Unit Volume. The term “density” will refer to weight-density or weight per unit volume unless specifically noted.

COMMON DENSITIES

RELATIVE

LBS PER LBS PER CU.FT. GALLON

DENSITY

GRAMS PER CU. CM

Water ( 39.20F )

1.000

62.4

8.34

1.000

Water ( 68.00F )

0.998

62.3

8.33

0.998

Sea Water

1.026

64.0

8.55

1.026

Steel

7.804

487.0

65.10

7.804

7.853

490.0

65.50

7.853

2.700

168.5

22.50

2.700

Iron Aluminum

Weight-Density Weight-Density Density Density ==Weight Weight / /Unit UnitVolume Volume

Because the weight of a liquid is expressed in terms of its volume, it is necessary to be familiar with and to be able to calculate volume.



Common Oilfield Volumes

Area: is equal to the unit squares of a surface.

Volume is the amount of space an object occupies. It is measured and calculated in cubic units such as cubic feet, cubic inches, cubic centimeters, etc. Volume may also be expressed as gallons, barrels, or other standards.

Units of measurements include square centimeters, square feet, square inches, etc. with square inches being among the most common, particularly when dealing with pressures. Calculations require that the unit of measurement be consistent in the equation and in the solution.

Columns 12 cubic inches

Barrel 12”

42 gal

Cubic Foot

Square Area ( Triangle ) Square Units = ( Base x Height ) / 2

1 gal

12” x 12” x 12”

Square Area ( Rectangle) Square Units = Length x Width 6” x 6” = 36 square inches

Volume Conversions

Cubic Inches

Gallons

Barrels

Cubic Feet

1728

7.48052

0.17811

1

Barrel

9702

42

1

5.61458

Gallon

231

1

0.02381

0.13368

Cubic Feet

Volume ( Cubic Units ) = Square Area x Length

( 6” x 6” ) / 2 = 18 sq. inches Square Area ( Circle ) Square Units = Diameter 2 x ( π / 4 ) 6 2 x ( 3.14 / 4 ) = 28.3 sq. inches Square Area ( Ring ) Sq. Units = ( DIA. 2 - dia. 2 ) x ( π / 4 )

Volumes must have consistent units of measurement in both calculations and solutions.

( 8 2 - 6 2 ) x .7854 = 22.0 sq. inches


Weight... is the measure of the downward force of an object as a result of gravitational force and the objects mass. The weight of an object may vary due to variations in the earth’s gravitational field or by its distance from the main body of the earth. Objects weigh more at sea level than far above it. This variation in weight is normally very small and will be ignored unless otherwise noted. While weight is used to measure individual objects, it is also used in hydraulics, with “ weight “ referring to a liquid as its density or weight per unit volume.

The weight of an object can be measured and identified as belonging to that single object. This is true mainly in relation to solid objects. The unit measured is that unique object. Unit Wt.

Wt / Gal

As liquid volumes may vary, a liquids weight or density refers to it as weight per a specified unit volume. This is also true of a gas and may be true of a solid if it noted as weight-density.

It is necessary to maintain consistency when referring to weight-densities of liquids as well as in calculations. Use the weight per the same unit volume in all.


Force… is a push or pull that is exerted on an object in order to change its position or the direction of its movement. This includes starting, stopping, changein speed, and direction of movement. Force is expressed in same terms as weight, such as grams, tons, dynes, etc. with pounds being the most common in the oilfield. Weight is a force which is directed downward, but force is not limited to any direction. Pressure is the amount of force exerted on an object or substance per area over which force is applied. Pressure may be expressed in various ways, such as newtons per square meter, dynes per square centimeter, etc. Most common measurement in the oilfield is Pounds per Square Inch ( PSI )

A 500 lb. force exerted on a 25 square inch surface equals a pressure of 20 PSI. Pressure of 20 PSI exerted on 25 square inches equals a 500 lb. force applied to the surface.

F O R C E

F O R C E

Pressure Pressure==Force Force/ /Area Area Pressure Pressure( (PSI PSI) )==Force Force( (Lbs. Lbs.) )/ /Area Area( (sq.in. sq.in.) ) Force Force==Pressure PressurexxArea Area Force Force( (Lbs.) Lbs.)==Pressure Pressure( (PSI PSI) )xxArea Area( (sq.in. sq.in.) )

2.03.01 Hydraulics Hydrostatic Fluid Properties

Hydrostatic Fluid : is a fluid at rest which having no shape of its own assumes the shape of its container.


Liquids Flow

A liquid assumes the shape of its container at lowest portion of the container equal to the liquids volume.

Gravity Liquids at rest, exert perpendicular forces on surfaces they touch as they cannot support tangential forces without flowing. Liquids are attracted by gravitational pull with each layer of liquid exerting its weight on the layers beneath it. Liquids are relatively incompressable making their density a constant. Liquids are only slightly affected by temperature changes.

Gravity’s constant pull downward on liquids creates pressure within the liquid called hydrostatic pressure.

Pressure in a fluid at rest: • Exists at every point within the liquid. • Is proportional to the depth below the suface • Is the same at all points at the same level within a single liquid. • Is of the same magnitude at any point regardless of the surface orientation that it touches.

Each layer of liquid exerts its weight on those below

Pressure is proportional to the depth in a liquid

At any level pressure is the same in single liquid

• Exerts a force which is everywhere perpendicular to the surface that it touches.

Pressure exists at all points in a liquid

At any point pressure is equal in all directions

Direction of force reacts perpendicular to surfaces

The well bore drilling fluids are subject to these same properties of a liquid.



Liquids seek their own level. Single Liquid

Liquids seek their own level. Two Liquids

As a liquid is pulled downward by gravity, filling its container, the liquid’s surface is on a flat horizontal plane parallel to the earth’s surface. The pressure at any point in the liquid is the same as all other points at the same depth and will always equalize despite the number of compartments in a container or its size, shape or orientation.

Liquids of two different densities in a container such as a u-tube equalize pressures from the point of separation and downward. Two distinct column heights result from the heavier liquid forcing the lighter upward until the pressure has been equalized at and below the point of separation.

Vertical depth

In a single liquid, equal pressures at the same depth provide equal support for the liquid producing equal vertical heights of the liquid.

LIGHT MUD

Liquids having different densities and that do not mix together will each seek their own level with the heavier of the liquids settling to the bottom of the container while the lighter liquid rises to the top.

Heavy Mud

Hydrostatic pressure with more than one liquid requires that each be calculated separately. The pressure to any point in the lower liquid will the sum of its calculation plus the total pressure of the liquids above.

HEAVY MUD

Hydrostatic Pressures are additive. Calculate each separately and add.

U- Tube Unequal Hydrostatic Pressures Point of Separation

L I G H T E R M U D

Equal Hydrostatic Pressures

In the oilfield, a common occurrence of the u-tube effect happens when cuttings weight-up the mud in the annulus. This “denser” mud having a greater hydrostatic pressure than the mud in the I.D. of the drill string seeks to balance the pressures by forcing the mud back up through the inside diameter of the drill string. This is seen during the make-up of connections when no float exists in the drill string or it does not work properly.


Hydrostatic Pressure is the pressure exerted by a column of fluid due to its own height and weight. Hydrostatic pressure in a fluid means the “ downward force per unit area “ equal to the weight of the column as defined by the area of the column and the fluids weight per unit volume. In the oilfield, hydrostatic pressure is measured in pounds per square inch ( PSI ). It equals the weight ( Lbs/Gal or PPG ) multiplied by the volume ( gallons ) of a column defined as 1 square inch in area and 1 foot ( 12 inches ) in depth which is then multiplied by the number of feet of the fluid column. A numerical constant of 0.05195, is frequently used in calculating pressures, which defines the number of gallons found in a column that is one foot tall with a cross-sectional area of 1 square inch. The 12 cubic inches in the column are divided by 231 cubic inches in a gallon to find the number of gallons in the one foot column in the format most commonly used. Pounds per Square Inch.

HYDROSTATIC HYDROSTATICPRESSURE PRESSURE H.P. H.P.( (PSI PSI) )==0.05195 0.05195xxMud MudWeight Weight xxDepth Depth ==0.05195 0.05195xxLbs/Gal Lbs/GalxxVertical VerticalFt Ft


Hydrostatic Pressure in Oilfield Hydrostatic pressure controls and promotes stability in the well bore, preventing cave-in and collapse. It is the primary means of well control used to prevent formation fluid flow into the well bore ( kicks ). It must be at least equal to the highest pressurized permeable zone of the well bore and yet not be excessive, as high pressures could lead to the break down of formations. Although the pressure is generated downward by the fluid’s weight, the pressure reacts perpendicular to the sides of the hole, providing support to them.

VERTICAL DEPTH

Note: Hydrostatic Pressures are based upon vertical heights of the fluid column only, irregardless of angle or shape of the column. In regards to wells, true vertical depth to point of interest, not measured depth, determines hydrostatic pressures.


Pressure Gradient is pressure change per foot of vertical depth in pounds per square inch (psi) due to hydraulic pressure.

Pressure PressureGradient Gradient PG PG==0.05195 0.05195xxMud MudWt. Wt.(Lbs/ (Lbs/Gal Gal) ) Pressure Gradient is simply a convenient number for calculations relating to hydrostatic pressures. It combines two of the three factors used to calculate hydrostatic pressure: • 0.05195 is a constant representing the number gallons equal to a column one foot tall and having an area of one square inch. It is 12 cubic inches divided by 231 cubic inches in a gallon.


Buoyancy is the power of a fluid to exert an upward force on a body placed in it. An object in a fluid is acted upon by hydrostatic pressures of the fluid on all its surfaces. Side pressures are balanced by pressures of opposing side. As the pressures are proportional to depth, upper and lower surfaces experience pressure differential with the bottom being greater than the top, generating a net upward force. Archimedes Principle states that “ a body, either wholly or partly submerged in a fluid experiences an upward force which is equal to the weight of the fluid being displaced. ” A solid which has less density than a fluid will float, sinking down to the point that:

• weight-density of a gallon of mud.

• Volume of fluid displaced equals the total weight of the object.

By establishing the pressure gradient, pressure at any point is found simply by multiplication of it by the current footage of interest. Examples: Find Pressure Gradient: water at 8.33 lbs per gallon Then: PG = 8.33 x 0.05195 = 0.433 psi / foot Find Hydrostatic Pressure: 9 lb/gal mud at 6000 feet. Then: PG = 0.05195 x 9 = .46755 psi / foot And: HP = 0.46755 x 6000 = 2805 psi Find Mud Weight: 3000 psi needed at 5000 feet. Then: PG = 3000 psi / 5000 feet = .600 psi/ft and: MW = .600 PG / 0.5195 = 11.55 lbs/gal

Net Pressure

• Hydrostatic pressure on the bottom surfaces of an object continually increase as the object sinks until it is sufficient to balance the total weight of the object.

Either of the two methods used to calculate buoyant forces are acceptable. They are the same principle stated differently and produce equal results. Use one best suited to data available.



Buoyancy Factor is a ratio of an object’s density to a fluid’s density. The factor when multiplied by the weight per unit volume of an object solves for the buoyed weight of the object in a fluid of a certain density ( weight per unit volume ).

Buoyed Weight is less than Air Weight A solid having more density than a fluid will sink into the fluid with its submerged weight then being less than its air weight by an amount equal to the: • weight of fluid displaced. • total pressure differential between that exerted on the bottom and top surfaces of the object.

BUOY WT AIR WT

Net Pressure

Buoyed Weight : Displacement Method Object. Wt. - ( Object Volume x Fluid Density ) Hydrostatic Pressure Differential ( HP x Lower Area ) - ( HP x Upper Area )

Buoyed Weight : Hydrostatic Press. Diff. Obj. Wt. - ( Sum of Hyd.Press.Diff.s )

Note: units of measurements must be consistent within formulas. Convert to common units as needed.

Buoyancy Factor ( Object Density - Fluid Density ) / Object Density Buoyed Weight : Buoyancy Factor Object Weight x Object Buoyancy Factor In drilling, the buoyancy factor is frequently used to predetermine the size and number of bottom hole assembly components to use in order to have the buoyed weight needed to do the job. Because the geometry and dimensions of some tools can be complex, it can be difficult to calculate the buoyancy effect of pressure differences. The use of the displacement or buoyancy factors may easier as the tool weight is often known or can easily be determined by calculation or rig equipment. Often the major variable is drilling fluid density.


Buoyancy Example Object: Densities:

10” x 10” x 5’ steel bar. Steel at 65.5 Lbs/Gal. Fluid at 9.0 Lbs./Gal. End Area (s) = 10” x 10” = 100 in2 Volume = 100 in2 x ( 5’ x 12” ) = 6000 in3 Volume = 6000 / 231 = 25.97 gal Air Wt. = 25.97 gal x 65.5 lbs = 1701 lbs

Using Displacement


Summary of formulas: Hydrostatic Fluids Density ( Weight-Density) = Weight / Unit Volume Specific Gravity =

Wt of Substance per Unit Vol. Wt of Water per Unit Vol.

Volume ( Cubic Units ) = Area x Length Circle Area = Diameter 2 x .7854 Ring Area = ( Dia2 - dia 2 ) x .7854

Eq.Fluid Wt. ( Vol) = 25.97gal x 9 lbs/gal = 234 lbs Buoyed Wt. = 1701 lbs - 234 lbs = 1467 lbs

Pressure ( PSI ) = Force ( Lbs) / Area ( Sq.In.)

Using Buoyancy Factor:

Force ( Lbs.) = Pressure ( PSI ) x Area ( sq.in. )

Buoyancy Factor = ( 65.5 - 9 ) / 65.5 = 0.863 Buoyed Weight = 0.863 x 1701 lbs = 1468 lbs Using Pressure Differential ( 1000 ft depth ) H.Press @ 1000 = 0.052 x 9 x 1000 = 467.55 psi H.Press @ 1005 = 0.052 x 9 x 1005 = 469.89 psi Force ( down ) = 100 sq.in. x 467.55 = 46755 lbs Force ( up ) = 100 sq.in. x 46989 = 46989 lbs. Total Diff. = 46989 - 46755 = 234 lbs Buoyed Wt. = 1701 lbs - 234 lbs = 1467 lbs.

Hydrostatic Pressure 0.05195 x Fluid (Lbs/Gal ) x Depth (Ft) Pressure Gradient = 0.05195 x Mud Wt. (Lbs/ Gal ) Buoyed Weight : Displacement Method Object. Wt. - ( Object Volume x Fluid Density ) Buoyancy Factor ( Object Density - Fluid Density ) / Object Density

2.04.01 Hydraulics Hydraulic Power Transmission


Pascal’s Law: If an external pressure is applied to a confined fluid, the pressure will be increased at every point in the fluid by the amount of external pressure. This is the basic principle upon which hydraulic power transmission systems are based. Pressure at any point in a fluid at rest is the same in all directions. Pressure applied to a confined fluid is transmitted undiminished throughout the fluid.

Force applied to a solid block is transmitted in a straight line through block. Force to confined liquid is transmitted: - in all directions - equally distributed - undiminished

Force exerted on a confined fluid results in a pressure increase which is equal to the amount of force applied divided by the area over which this force was applied.

Pressure Pressure( (PSI PSI) )==Force Force( (Lbs. Lbs.) )/ /Area Area( (sq.in. sq.in.) ) A pressure increase in a confined fluid is distributed equally throughout the fluid and against all sides of container. Force applied on any surface such as a piston is equal to the pressure applied multiplied by the area of the “piston”.

Force Force( (Lbs.) Lbs.)==Pressure Pressure( (PSI PSI) )xxArea Area( (sq.in. sq.in.) ) F O R C E

S O L I D

F O R C E

L I Q U I D

Force is directly related to pressure and pressure to force by the areas over which they act. This can be seen in hydraulic cylinders which are a common application of hydraulic power transmission systems. Single Cylinder FORCE

A 10 pound force exerted on a 10 square inch piston area of a confined fluid transmits a 1 PSI pressure to all surfaces of the confining container. The pressure transmitted is in addition to any existing pressures such as hydrostatic. Since these pressures were in a state of equilibrium, they can be ignored when considering these pressure transmissions.

FORCE FLUID

10 LBS

10 LBS

Input force equals output force if piston areas are the same. The force divided by the input piston area creates a pressure which multiplied by the identical area of the output piston creates a force which equals the input force.


2/04.04 Hydraulics Hydraulic Power Transmission

Force of an “Output” hydraulic cylinder is proportional to area of its piston to the area of the “Input” piston.

Work is the occurrence of a force moved through a distance. It is equal to the product of the force multiplied by the distance through which the force was applied. Common units are Pounds and Feet.

Force(out) = Force(in) x ( Area(out) / Area(in) ) Force(out) = Force(in) x ( Area(out) / Area(in) )

A 10 lb. force exerted on the 10 sq.in. of cylinder 1 transmits a pressure of 1 psi through-out the fluid, on all container sides. The 20 sq.in. piston area of Cylinder 2 then has a total upward force of 20 lbs.

1

2

Mechanical Work Work = Ft-Lbs Pounds x Feet

Length(out) Length(out)==Length(in) Length(in)xx( (Area(in) Area(in)/ /Area(out) Area(out)) ) Stroke Speed of an output cylinder is proportional to area of input piston to area of its output piston.

x Distance

1

2

=

Work

Work Work==Force ForcexxDistance Distance

Given that the piston of cylinder 1 travels downward, forcing fluid into cylinder 2 and raising its piston, the travel of piston 2 would be one-half that of piston 1 since the area of piston 1 is one-half piston 2. Travel of an output hydraulic cylinder is proportional to area of the input piston to area of its output piston.

Force

Hydraulic Work

Force ( psi )

a r e Distance a

p i s t o n

=

Work

Work Work(Inch-Lbs) (Inch-Lbs)==Force Force(Lbs) (Lbs)xxTravel Travel(inch) (inch) Work ( Foot-Lbs ) = Force ( Lbs ) / Travel ( Feet ) Work ( Foot-Lbs ) = Force ( Lbs ) / Travel ( Feet )

Speed(out) = Speed(in) x ( Area(in) / Area(out) ) Speed(out) = Speed(in) x ( Area(in) / Area(out) )

Note: Work ( Foot-Lbs ) = Work ( Inch-Lbs ) / 12


Hydraulic power systems are used to do work. Discounting losses from friction, the work which is input equals the work which is output. It is only adapted to meet the needs of a job to be done. Hydraulic Cylinders : Input and Output Work In

=

Pressure = Pressure Work Out =

Force

Distance

Power : is the rate of doing work. It is defined as an amount of work ( foot-pounds ) done in a given time. Power =

Work Work( (Ft./Lbs. Ft./Lbs.) ) Time Time( (minutes minutesororseconds seconds) )

Common power unit of measurement is Horse Power Horsepower =

33000 33000Ft-Lbs Ft-Lbs= 11minute minute

Energy: is used to do work or use power. The law of Conservation of Energy states that “ Energy cannot be created or destroyed, it can only be transformed.” Not all energy is used to perform work, some is expended, when doing work, to overcome the effects of friction. This energy is not lost, but changed to heat energy. Hydraulic pumps are used to convert electrical or other types of energy to hydraulic energy. Types of energy used in a basic hydraulic system include:

Distance

Force


550 550Ft-Lbs Ft-Lbs 11second second

The basic principles involved in transmitting power are readily seen using hydraulic cylinders as both in input and output, but not all power is input in this manner and not all work is done in this manner.

• Electrical • Hydraulic • Kinetic • Potential • Heat

to operate pump motor produced by the pump produced when hydraulic fluid moves a piston. produced when the piston has raised an object. produced by friction in pump,pipe, & fluid

Hydraulic pumps are used to create pressure increases used to do work. Basuc hydraulic systems are a closed piping circuit in which a fluid under controlled pressure is used to do work. Hydraulic pumps impart energy or power to the fluid which is transmitted to the work site where the work is done. The fluid is then returned to the pump to be energized again.

HYDRAULIC POWER SYSTEM FLUID RESERVOIR

PUMP

RETURN LINE

F ILTER PRESS REG

AIR FLUID

PRESSURE REGULATOR ACCUMULATOR

CHECK VALVES

GAGE

7

HAND PUMP

RELIEF VALVE CONTROL VALVE WORK SIDE

PRESSURE SIDE

Circulating hydraulic systems have fluids which flow which are called hydrodynamic fluids. While hydrostatic fluids can be described by relatively simple concepts of density and pressure, hydrodynamic fluids require new and more complex properties be considered. Hydrodynamics…... is the study or application of properties of liquids in motion. A liquid having no shape of its own, assumes that of its container as it cannot support a tangential force without loosing its shape or deforming. The continuous deformation of a liquid is known as “ Flow ”. The flow of a liquid always takes place in a conductor. A conductor is can be any shape or size, even a flat surface with the atmosphere serving as the sides and top. Flow conductors are often cylindrical shaped ( pipes ). Hydraulic Power is the power required to cause a fluid to flow; the product of flow rate and pressure drop. In drilling, two major conductors of flow are the drill string and annulus. Wells are often two to 4 miles deep, the pressures required to maintain the high flow rates required are substantial and together with the pressure used at the bit could be a limiting factor on flow rates. Additionally, the fluids or mud used are tailored to do different tasks associated with drilling the hole. These fluid characteristics impact fluid flow properties. It is essential that hydrostatic and hydrodynamic properties of fluid be understood as well as the impact that the mud properties may have.

2.05.02 Hydraulics Circulating System and Fluid Flow

The Drilling Circulating System

Stand Pipe

Rotary hose

PUMP HOPPER

M

UD

RE SE RV

E

PI T

PI T

SHALE SHAKER

A circulating system imparts energy or power to a fluid, transports it to the work site, does the work, and returns it to be energized again. Basic elements include geometry of the piping, fluid properties and flow rate with each of these influencing the total pressures realized. Major purposes of the fluid and its flow are: • transmit hydraulic horsepower to the bit to clean it and the bottom of hole. • cool and lubricate bit & drill string • transport cuttings produced out of hole. • support hole walls & prevent formation fluids from entering well bore.

CASED HOLE

The hydraulic system used to drill a well is called a circulation system. It, like the basic hydraulic power system, circulates a fluid under controlled pressure to do work.

OPEN HOLE



The circulating system has no pressure control valve as exists in a simple hydraulic system. This, along with the multiple duties the circulation system must perform, requires the complete hydraulic system and all of its elements be preplanned. Often, compromises between conflicting requirements must be done. Hydraulic Power… is the power required to cause a fluid to flow; the product of flow rate and pressure drop. • The product of low flow rate and high pressure may equal the product of high flow rate and low pressure. Hydraulic Horse Power is a measure of the energy delivered to the fluid being pumped. Hydraulic HydraulicHorse HorsePower Power Pressure PressureDrop Drop( (psi psi) )xxFlow FlowRate Rate( (gpm gpm) ) 1714 1714 Hydraulic Horse Power = Mechanical Horse Power. For Pressure in Lbs per Sq.In. and flow is in GPM. Then 231 cubic inches of a gallon divided by 12 equals 19.25 feet and 33,000 ft-lbs divided by the 19.25 feet equals the numerical constant 1714. Engine Horsepower required equals the Hydraulic Horsepower divided by pump efficiency. Note - newer pumps are usually 95 to 97% efficient.


Volumetric-Rate. Volume of liquid in units per a unit time as barrels per minute, gallons per minute, etc. The volumetric-rate output of a pump is found by the volume per stroke multiplied by the number of strokes per unit time at which pump is operated. Triplex: single acting with three cylinders NO ROD OD FLUID

P I S T O N

F ID

L U I D

Pumping action occurs on one side only.

Duplex: double acting with two cylinders FLUID ROD

OD

FLUID

P I S T O N

F ID

L U I D

Pumping action occurs on both sides of piston.

Volume Triplex Volume Triplex( (Bbl Bbl/ /stroke stroke) ) 2 33( (ID ID2/ /12353 12353) )xxStroke StrokeLength Length(inch) (inch) Volume ( (Bbl VolumeDuplex Duplex Bbl/stroke /stroke) ) 2 2 ( (44( (ID ID2/ /12353 12353) )- -22( (OD OD2/ /12353 12353) )) )xxStroke StrokeLength Length(inch) (inch) Constant: 12353 = 1cu.in. / ( π / ( 4 x 9702 cu.in.) ) Actual output per stroke of pump is found by multiplying above result by pump efficiency. Hydraulic Pumps are limited to a maximum volume and pressure. The maximums not only vary by manufacturer and type of pump but on the size of pump liners or cylinders used and stroke length.

2.05.05 Hydraulics Circulating System & Fluid Flow


Continuity of Flow. As liquids do not readily compress, volumetricrate input into a conductor equals volumetric-rate which is output. It is not affected by changes in inside area of the conductor. Volume-Rate in

= Volume-Rate out

Flow can be imagined as a cylinder having an area equal to crosssectional area of pipe and a distance of such length that would result in a volume equal to that which which is referenced.

Fluid Flow Velocity through a conductor is distance traveled by a fluid within a defined time. It is usually stated as feet per minute or feet per second. The speed of a flowing fluid is dependant on volume (GPM ) and area in square inches at a cross-section of the conductor. Fluid )) FluidFlow FlowVelocity Velocity( (Feet Feetper perMinute Minute 2 ( (24.51 24.51xxGPM GPM) )/ /Diameter Diameter2

Cross-section

D

SPEED

Fluid FluidFlow FlowVelocity Velocity( (Feet Feetper per2Second Second) ) ( (( (24.51 24.51xxGPM GPM) )/ /Diameter Diameter)2 )/ /60 60

A

Volume

Cross-section SPEED

Volume

D

A

Given that equal volumes per unit time flows through the pipes, it can be seen that the cross-sectional area influences the following: • Volume per a given length in proportion to Area. • Fluid Velocity per a given unit of time in inverse proportion to Area. Volumetric-Rate Volumetric-RateofofFlow Flow==Velocity Velocity( (ft/min ft/min) )xxArea Area Velocity Velocity( (ftft/min /min) )==Volumetric-Rate Volumetric-Rate/ /Area Area

Numerical constant 24.51 is derived from 231 cubic inches in a gallon equal to a cylinder of 1 square inch in area by 19.25 ( 231 / 12) feet long. Then the velocity ( ft / min ) would equal : 19.25 ft x GPM / .7854 x Diameter2 or 24.51 x GPM/Diameter 2 In hydraulic formulas related to drilling in the oilfield, it is common practice to express velocities of fluids related to the annulus in feet per minute. Fluid velocities related to the inside diameter of the drill string and to the bit are expressed in feet per second.


Fluid Flow: When external forces (pump) acting on a fluid are great enough to overcome viscous forces, fluid flows. The velocity of fluid particles at conduit wall is zero, increasing with distance from wall. PLUG FLOW

Plug Flow: occurs only at very slow rate where a thin layer of fluid slips at conduit wall with rest flowing as a unit. This flow regime to be given no further consideration. Not Considered.

Laminar Flow: can be viewed as relatively smooth, straight streamlines of flow having concentric layers of fluid beginning with a zero velocity at the conduit wall, with layers progressively faster, reaching maximum speed at the center. This flow pattern requires less energy. LOW VISCOSITY

CHAOTIC FLOW

HIGH VISCOSITY

The Velocity Differential between layers are greatest at wall and least in center. Low viscosity fluids have greater differentials than do the higher viscosity fluids.

Turbulent Flow is the “fast”, chaotic flow of fluid particles, moving in random loops except at wall of conduit where velocity is zero. Streamlines are irregular patterns with a flat profile. Maintaining the fast flow rate requires more energy versus the straighter streamlines of laminar flow.

While flow in the drill string is generally considered to be turbulent, annular flow may be turbulent or laminar. It is necessary to verify the flow pattern and to then use the formula applicable to the type. Each flow type has a different formulas to calculate the applicable pressure drop. Laminar flow uses less energy than does Turbulent flow.


Circulating pressure loss in a hydraulic system is energy or pressure required to force a liquid through a system to overcome the effects of friction of the fluid itself and friction of the fluid and the structure of the container. In a circulation system, not only is pressure required to do the work intended, but pressure is also required in getting the hydraulic fluid to the work site and returned to the starting point. The pressure that is required to move the fluid through the system is referred to as pressure drop or pressure loss as it is not available to do work. Circulating Pressure Loss:

Effects of Friction 5 4 3 2 1 0

P S I

Flow Direction Pipe with inserted glass tubes shows pressure losses as fluid flows. Pressure which is no longer available for additional flow or to do work.

In drilling, two major conductors of flow are the drill string and annulus. These have a large impact on flow rates and resulting pressures which may limit the flow rate. Wells are frequently 2 to 4 miles deep, and may reach 6 or more miles. Pressures required to generate fluid flow to the bit and back to surface are substantial and of prime importance. A well designed hydraulics program is one in which less than 50% of available hydraulic horse power is used for flow with 50% or more used by the bit in making hole.



Factors of Pressure Drop Fluid: Density and Viscosity Volume: Volumetric-Rate Dimensions: Length and ID

ID

I.D. I.D.Pressure PressureLoss Loss( (turbulent turbulentflow) flow) 0.18 0.82 x 1.82 0.0000765 0.0000765PV PV 0.18xxMW MW 0.82 xGPM GPM 1.82xxLL 4.82 ID ID 4.82

Analysis of the formula for turbulent flow in a conduit shows a complex relationship between components. The analysis below illustrates the degree that changes to a factor could result in. ANALYSIS OF PRESSURE DROP - PROPORTIONAL FACTORS Plastic Mud Length Gallons Viscosity Weight Feet per Min. 1 0.18 = 1.00 1 0.82 = 1.00 1 1 = 1.00 1 1.82 = 1.00 2 0.18 = 1.13 2 0.82 = 1.77 2 1 = 2.00 2 1.82 = 3.53 3 0.18 = 1.22 3 0.82 = 2.46 3 1 = 3.00 3 1.82 = 7.39 ANALYSIS OF PRESSURE DROP - INVERSELY PROPORTIONAL FACTORS DECREASE IN I.D.

DECREASE IN I.D.

DECREASE IN I.D.

4.82

4.82

4.82

1.1 = - 1.58 1.2 4.82 = - 2.41 1.3 4 82 = - 3.54

1.4 = - 5.06 1.5 4.82 = - 7.06 1.6 4.82 = - 9.64

1.7 = -12.91 1.8 4.82 = -17.00 1.9 4.82 = -22.06

Chart shows varying degrees of net change to pressure drop by changes in a component. Doubling Mud Weight increases PSI by 1.77 times. Doubling GPM increases PSI by 3.53 times. Increasing ID by 20% decreases PSI by 2.41 times.

Annular pressure drop does not typically affect the total system pressure to the same degree as do the drill string internal diameters, especially in the large, upper sections of the hole. However, as the hole becomes deeper and the annular space becomes smaller, its impact is larger. The pressure drop in the annulus is added to the hydrostatic pressure of the hole and as such plays a significant role in the open hole where it is exposed to the formation. Too little or to much pressure against the formation(s) can result in problems. Annular flow may be either turbulent or laminar depending on the velocity of the fluid flow. To calculate the annular pressure drop, the critical velocity must first be calculated. If critical velocity is below 2,000 the flow is laminar. If it is 2,000 or above the flow is turbulent. Annular AnnularCritical CriticalVelocity Velocity 2 2 0.5 H P 1.08PV PV 2• •9.3{ 9.3{DDH- -DDP}} •2 YP • YP• •MM) ) 0.5 1.08PV++1.08( 1.08(PV H - DP ) MM( (DD H - DP )

Annular AnnularPressure PressureLoss Loss( (Laminar LaminarFlow Flow) ) 4.82 H P AS ( {( {LL• •YP YP}}÷÷225 225{{DDH- -DDP}}) )++({({LL• •VVAS• •PV PV}}ID ID 4.82 H P ( (1500 1500{{DDH- -DD}P }) )

Annular AnnularPressure PressureLoss Loss( (turbulent turbulentflow) flow) 0.18 ( .0000765 • M 0.82 • G 1.82 • L ) ( .0000765PV PV 0.18 • M 0.82 • G 1.82 • L )

({ DH H- DP P} 3 3• { DH H+ DP }P 1.82 ({ D - D } • { D + D } 1.82) )

2.06.01 Hydraulics Rheology - Study of Fluid Flow Terms

Flow type of a liquid is affected by the cohesive internal attraction of the fluid to itself and the adhesive external attraction of the fluid to the conduit wall. Flow may be viewed as a series of parallel fluid layers. The first layer is held in place by fluid’s adhesive attraction to the conduit wall. The second layer rides on this fluid layer, gaining some velocity as it is retarded only by the fluids internal attraction. Each subsequent layer gains additional velocity as it rides on the previous layer which already has a Faster Velocity velocity of its own. This builds, Zero Velocity Fluid reaching its peak in the center. Conduit Wall

A bead of water on a window pane illustrates the cohesive internal attraction of a fluid within and to the fluid itself. The thin track of water on the glass as the bead of water slides down illustrates that a very thin layer of fluid at zero velocity exists at the fluid’s point of contact with a conductor surface due to the adhesive attraction of the fluid to a conduit surface.

Window Pane

Bead of water


Rheology is the study of the flow of fluids. Viscosity is descriptive of drilling mud in motion. The appearance (apparent viscosity) of high viscose mud is referred to as “thick” and low viscose mud called “thin”. Viscosity relates shear stress to shear rate or a resistance to flow. Plastic Viscosity is a measure of internal resistance to fluid flow. It is related to the type, amount, and size of solids present in the mud. It is an expression relating shear stress to shear rate or a resistance to flow. Shear Stress is the result from forces that tend to cause particles of fluid to slide relative to other particles in a direction parallel to the plane of plane of contact. It is the resistance or frictional drag to the sliding movement of two parallel fluid layer. Shear Rate is the force per unit of time or the velocity of fluid particles relative to their distance or separation. It is the difference in the velocities between two layers divided by the distance between them . Yield Point is a measure of the resistance to initial flow or stress required to start fluid movement which is caused by electrical forces on or near surfaces of the solid particles. Gel Strength is a measure of the same electrical forces on solid particles in mud considered by yield point, except it is measured at rest. A static mud solidifies or gels by arranging solid particles in a manner to best satisfy theses forces of attraction and repulsion. Gel strength indicates the strength of these forces

Fluid Flow Models are attempts to mathematically define behavior of fluids as they flow. Bingham Plastic Model - a finite stress must be applied to initiate flow. At greater stresses, the flow will be newtonian. Pressure losses are calculated using plastic viscosity (PV) and Yield Point (YP). This is a good model for clay muds having a high solids content. Power Law Model - Flow is initiated immediately as stress is applied. This is a good model for polymer muds having a low solids content. Pressure losses are calculated using a viscosity (k) and a flow-behavior index ( N).

Fluid Classifications Newtonian Fluids - Shear Stress is directly proportional to shear rate. ( water, oil, ) Non-Newtonian Fluids - Shear Stresses are not directly proportional to shear rates. Pseudo Plastic Fluids - the rate at which the viscous forces increase respective to shear rate decreases with the increasing shear rate. In other words, viscosity decreases with increasing shear rate ( drilling fluids ). Dilatent Fluids - the rate at which the viscous forces increase with shear rate increases with increased shear rate. In other words, viscosity increases with increasing shear rate. ( ink, blood)

Ideal Power Law

shear stress

ic Plast Plastic Viscosity

shear stress yield point

am Bingh


Dilatent


n

w Ne

a ni to

Pseudo-Plastic

nian

to New

Viscosity shear rate

shear rate

2.07.01 Hydraulics Bit - Mechanical Energy

Bits use both Mechanical and Hydraulic Energy in drilling the well bore. Bits are tailored to the formation characteristics. In general, the softer the rock to be drilled the larger the teeth. Bits are built to be rotated while weight is applied to supply the mechanical energy required.

TOOTH CHIP

2.07.02 Hydraulics Bit - Hydraulic Energy

Bit Hydraulic Energy

Bit Mechanical Energy

Hydraulic Energy is required to allow the bit to effectively drill by cleaning the bit and the hole bottom. As fluid is forced through the bit nozzles, its kinetic energy is greatly increased through high jet velocities having a high impact force. Hydraulic Horsepower at bit is measure of energy expended at bit. Impact Force is a measure of the force which the drilling fluid impinges upon the bore hole below the bit. Adequate jet velocity cleans the bit. A balled up bit acts as a cushion preventing effective drilling. There must be room between the bits teeth for new formation. Adequate fluid jet velocity and fluid volume cleans drilled chips from hole bottom. Re-drilling of chips is not efficient and generates added, unnecessary solids in the mud

Tooth penetrates formation, creating chip by fracture and shearing of rock. Higher Hydrostatic Pressure

Chip Weight on Bit controls Chip Size & Quantity RPM controls Fracture Rate.

Lower Formation Pressure

Adequate Jet Velocity and Impact Force releases differentially stuck chips. Chips can be held down when solids filtered from the mud seal cracks around them and where the hydrostatic pressure is greater than the formation pressure.

Jet Velocity and its Impact Force may “drill” some soft formations by its own hydraulic energy.

2.08.01 Hydraulics Rules of Thumb to Optimize Hydraulics

Factors to consider when planning hydraulics are: • Geology of well - Formations, pressures, Hole problems, etc. • Pumps - Volume & Pressure capabilities, limitations, etc. • Drill String Geometry. ID, OD, Length, Strength, etc. • Bit - Size, Type, Nozzles, Hydraulic Horsepower. • Mud - Type, Weight, Properties. Supply & availability. • Annulus: Pressure Loss, Flow Rate for cutting removal • Tool Needs - MWD, etc. Flow and Pressure requirements. • Drilling Rates - Expected and/or desired ROP. • Pressures with expected flow rates, hole sizes, and depths. Bit Pressure drop, Drill String Pressure Drop, Annular Pressure Drop, Bottom Hole Hydrostatic Pressure, etc. Rule of Thumb Guidelines for Optimal Hydraulics: Flow Rate: 30 to 60 GPM per inch of bit diameter. Maximize in soft formations, fast drilling, high angle holes for hole cleaning. Restrict only to the degree that hole wash out is a problem. Limit in slow drilling to rate needed. Limit in small and / or deep holes to reduce, annular friction, ECD, and potential for lost circulation, differential sticking, and hole instability. • Flow Rate too low - Inadequate hole cleaning. Hole could load with cuttings. - Bit may “ball“. • Flow Rate too high: - Increases Annular Friction, Bottom Hole Pressure, & ECD - Erodes soft, unconsolidated formations. Fast drilling & light mud weights need more flow. ( 50+ gpm/bit diameter ). Slow drilling needs less flow. Do not slow below minimum. High angle holes need higher flow to clean.

2.08.02 Hydraulics Rules of Thumb to Optimize Hydraulics

Rule of Thumb Guidelines for Optimal Hydraulics: Maintain 2.5 to 5 bit hydraulic horsepower per square inch of bit diameter. ( HHP/ Inch2 ) Bit hydraulic horsepower is based on ROP & Hole Size. • Large Bits require more HHP / Inch2 • Fast ROP requires maximum HHP / Inch2, even over normal maximum of 5. • Some rigs do not have pumps or horsepower to provide the needed hydraulic horsepower. • Do not use excessive pressure, costing unnecessary fuel and pump wear. Maintain Bit Jet Velocity between 350 to 450 feet per second. ( ft./sec.) Do not attempt to operate below 250 ft/sec. Jet velocity influences penetration rates, hole cleaning, & chip holddown. Impact force, the force exerted on the formation to assist in hole clearing is the product of mud weight and jet velocity and is directly proportional to jet velocity. • to improve penetration rates in a small hole of 9-1/2” or less consider running 2 larger jets rather than 3 of the same total flow area. Larger jets are less likely to plug. • asymmetrical jets of differing sizes may improve penetration rates versus 2 jets. • for long bit run which would force a lowering of the jet velocity, consider 3 jets with a diverting ball dropped in lower section to maintain the jet velocity. Bit Pressure Drop to be 50 % to 65 % of the total system pressure drop. Calculate total system losses and adjust if pressure drops through drill string and annulus exceed 50%. Do not adjust volume below 30 GPM/Inch of bit diameter. Consider drill string changes, nozzle adjustments, etc.

2.09.01 Hydraulics Hydraulic & Related Formulas

1. HYDROSTATIC FLUID FORMULAS WEIGHT DENSITY SPECIFIC GRAVITY HYDROSTATIC PRESSURE. ( PSI ) PRESSURE GRADIENT BUOYANCY FACTOR OF A FLUID Buoyed Weight Buoyancy Factor (Steel)

= Weight / Unit Volume = Weight of Substance per Unit Volume / Weight of Water per Unit Volume = 0.05195 • Mud Weight • Depth = 0.05195 • Mud Wt. (Lbs/ Gal ) = ( Object Density - Fluid Density ) / Object Density = Object Weight • Object Buoyancy Factor = ( 65.5 - M ) / 65.5 ( 65.5 lbs = steel in gallons )

2. POWER TRANSMISSION FORMULAS PRESSURE ( PSI ) FORCE ( LBS ) FORCE ( OUTPUT ) LENGTH ( OUTPUT) SPEED ( OUTPUT ) WORK WORK (INCH-L BS) WORK (FOOT-LBS) POWER HORSE POWER HYDRAULIC HORSEPOWER

= FORCE ( LBS ) / AREA ( SQUARE INCH ) = PRESSURE ( PSI ) • AREA ( SQUARE INCH ) = FORCE ( IN ) • ( AREA (OUT ) / AREA ( IN ) = LENGTH ( IN ) • (AREA ( IN ) / AREA (OUT ) ) = SPEED ( IN ) • (AREA ( IN ) / AREA (OUT ) ) = FORCE • DISTANCE = FORCE ( LBS ) • Travel ( INCH ) = FORCE ( LBS ) • Travel ( FEET ) = WORK(FTLBS) / TIME(MINUTES OR SECONDS) = 3300 FTL BS / 1 MINUTE OR 550 FTLBS / 1 SECOND = PRESSURE DROP (PSI) • FLOW R ATE (GPM )

3. CIRCULATION RELATED FORMULAS TRIPLEX PUMP (BBLS/STK ) DUPLEX PUMP (BBLS /STK) VOLUMETRIC-R ATE OF FLOW VELOCITY (FT/MIN) FLUID FLOW VELOCITY (FT/MIN) FLUID FLOW VELOCITY (FT/SEC)

= 3 ( ID2 / 12353 ) • STROKE LENGTH (IN.) = 4 ( ID2 / 12353 ) - 2 ( OD2 / 12353 ) ) • S.L.) = VELOCITY ( FPM ) • AREA = VOLUMETRIC-RATE / AREA = ( 24.51 • GPM ) / DIAMETER2 = ( 24.51 • GPM ) / ( DIAMETER2 • 60 )


Nomenclature CD CW DH DP G JV k L

Note: other nomenclature found in formulas themselves

= Chip Diameter, inch = Chip Weight, ppg = Diameter of Hole, inch = Diameter of Pipe, inch = Gallons Per Minute ( gpm ) = Jet Velocity (fps)(ft/sec) = Consistency Index = Length in feet

M NZ NA ∆PXX PV VAS VAM YP

= Mud Weight (ppg)(lbs/gal) = Nozzle Size ( 32nds of inch ) = Nozzle Area ( square inch ) = Pressure Loss ( psi )(lbs/sq.in.) = Plastic Viscosity, cps = Velocity, Annular Fluid ( fps ) = Velocity, Annular Fluid ( fpm ) = Yield Point (lbs / 100 ft.)

f n Re U Kb kf1

4. Hydraulic Formulas. Bit Related A. B. C. D. E. F. G. H. I. J.

K.

L. M.

Nozzle Pressure Loss ( psi ) Nozzle Volume ( gpm ) Nozzle Total Flow Area ( sq.in. ) Nozzle Area ( per size ) (sq.in. ) Nozzle Size ( 32nds inch ) Jet Velocity of Nozzles ( fps ) Impact Force of Nozzles ( psi ) “ “ “ ( psi ) Bit Hydraulic Horse Power, Total Bit Hydraulic Horse Power, per Sq.In. Drill Rate: Bit Weight - Rotational Speed a. Drill Rate, Soft Formation b. Drill Rate: Hard Formation Bit Size versus Penetration Rate a. Up to 17-1/2” b. 17-1/2” to 26” Minimum Flow Rate for PDC Bit Rules of thumb: Bit a. Bit Hydraulic HorsePower b. Flow Rate c. Max Bit Hydraulic HP d. WOB

∆PNZ = ( M • G 2 ) ÷ ( 10858 • NA 2 ) GNZ = ( { P NZ • 10858 • NA 2 } ÷ M ) 0.5 NA = ( { M • G 2 } ÷ { 10858 • PNZ } ) 0.5 NA = ( N Z ÷ 32) 2 • ( ¶ ÷ 4) or N Z 2 • 1303.8 NZ = 32 • (N A ÷ { .7854 • Qty }) 0.5 JV (F/S) = ( 0.32 • G ) ÷ N A IF = JV • 0.0173 • G • ( P NZ • M ) 0.5 IF = 0.000516 • JV• G • M BHHP (TOTAL) = PNZ • G ÷ 1713.6 HHP B / sq. in = BHHP ÷ ( BIT OD 2 • 0.7854 ) ROP = kf1 • W • R ROP = kf1 • W 1.2 • R 0.5 ROP2 = ROP 1 • ( Bit Dia 1 / Bit Dia 2 ) ROP2 = ROP 1 • ( Bit Dia 1 / Bit Dia 2 ) • 1.25 FRMIN = 12.72 • Bit Dia 1.42 • • • •

2-1/2 to 6 HP per Square Inch of Hole Diameter 30 to 50 GPM per Hole Diameter 50 to 75% of pump pressure across jet nozzles 3,000 to 5,000 per inch of Bit Diameter.

= fanning friction factor = Power Law Index = Reynolds Number, dimensionless = Viscosity, apparent, effective, cps = Formation Factor: Hard= 4 x 10 -5 Soft = 4 x 10 -4 = Apparent, corrected formation drillability factor


5. Drill String Bore Pressure Loss ( psi ) Turbulent flow ( sii ) Turbulent flow ( security ) Turbulent flow ( fanning )

∆PID = ( 0.0000765 PV 0.18 • M 0.82 • G 1.82 • L ) ÷ ID 4.82 ∆PID = ( 0.000061 • M • G 1.86 • L) ÷ ID 4.86 ∆PID = ( f • M • V ID (F/S) 2 • L ) ÷ 25.8 DP

6. Annulus Flow Annular Flow Velocity ( fpm ) Annular Flow Velocity ( fps) Annular Critical Velocity ( fps ) Optimum Annular Velocity (fpm ) Optimum Annular Flow ( gpm) Optimum Annular Flow ( gpm)

VAM = ( 24.51 • G ) ÷ ( DH 2 - DP 2 ) VAS = ( { 24.51 ÷ 60 } • G ) ÷ ( DH 2 - DP 2 ) VCA = 1.08PV + 1.08( PV 2 • 9.3{ DH - DP } 2 • YP • M ) 0.5 ÷ M • ( DH - DP ) VOA = 11800 ÷ ( M • DH ) Opt Flow ( Annulus ) = 482 ( DH 2 - DP 2 ) ÷ ( DH • M ) Opt Flow ( Open hole) = ( 265 DH + 10 DH 2 ) ÷ M

7. Annulus Pressure Loss ( psi ) Turbulent Flow ( sii ) Turbulent Flow ( security ) Turbulent Flow ( fanning ) Newtonian Laminar Flow ( hagan ) Plastic Laminar Flow ( beck, etc ) 8. Bottom Hole Pressure Bottom Hole Hydrostatic Pressure ( psi ) B. H. Circulating Pressure ( psi ) Equivalent Circulating Density ( ppg ) “ “ “ ( ppg )

∆PAN ∆PAN ∆PAN ∆PAN ∆PAN

= ( .0000765 PV 0.18 • M 0.82 • G 1.82 • L ) ÷ ({ DH - DP } 3 • { DH + DP } 1.82 ) = 0.00000014327 M • L • VA 2 ÷ ( DH - DP ) = ( f • M • VAN (fps) 2 • L ) ÷ 25.8 ( DH - DP ) Note: V = Apparent Viscosity (cps) = V • L • Va ÷ ( 1500 • {DH - DP } ) = ( { L • YP } ÷ 225 { DH - DP } ) + ({ L • Va • PV } ÷ ( 1500 { DH - DP } )

BHP = ( 0.5195 • M • L ) BHCP = BHP + ∆PAN ECD = BHCP ÷ ( 0.52 • L ) ECD = (∆PAN ÷ { 0.052 • L } ) + Mud Weight

9. Hole Cleaning Rock Chip Slip Velocity Lam. - Spherical Chips ( Stokes ) Lam. - Flat Chips ( Pigott ) Turb. - Spherical Chips ( Rittinger ) Turb. - Flat Chips ( Pigott ) Chip Rate ( ft / Min )

Note: use 21 as cutting density and 0.25 as cutting diameter if unknown V C = ( 8310 CD 2 { CW - M }) ÷ ( PV + ( 399 YP • { DH - DP }) ÷ VA ) V C = ( 3226 CD 2 { CW - M } ) ÷ ( PV + ( 399 YP • { DH - DP }) ÷ V A ) V C = 159 ( ( CD { CW - M } ÷ M ) 0.5 ) V C = 60.6 ( ( CD { CW - M } ÷ M ) 0.5 ) V S = VANN (fpm) - V C


e k SS U

2.718 Naperian Base Consistency Index Shear Stress ( dynes / cm2 ) Viscosity, apparent (cps)

Effective Viscosity A. Viscosity Definition B. Bingham Plastic C. Shear Stress D. Effective Viscosity E. Annular Shear Rate F. Consistency Index G. Power Law Index

f n Sr Ue

Fanning Friction Factor Power Law Index Shear Rate ( sec -1 ) Viscosity, effective ( cps )

U = SS / Sr U = ( PV + ( 399 YP • ( DH - DP)) / VA ) SS = k • Sr n ( Power Law Fluids ) Ue = k • Sr n-1 ( Power Law ) S r = 2.4 VA / ( DH - DP ) k = 511( YP + PV ) / 511n n = 3.32 log 10 ( YP + 2PV ) / ( YP + PV )

Drilling Fluid Property Change and its ROP effect: A. Plastic Viscosity ROP2 = R OP1 • 10 0.003 (PV1 - PV2) B. Bentonitic Clay ROP2 = ROP1 • e 0.051(VOL1% - V OL2%) C. Total Solids D. Water Loss

ROP2 = ROP1 • 10 0.0066 (V OL1% - VOL2%) ROP2 = ROP1 • ((WL2 + 35) / WL1 +35)) ROP2 = ROP1 • (sin(10.6 Vol%2 - 4.83) + 10.33)

E. Oil Content ( < 30%) (sin(10.6 Vol%1 - 4.83) + 10.33) F. Total Drilling Fluid effects, Density, Viscosity, Solids, Pressure Loss

ROP2 = ROP1 • e 0.382 (GPM1 - GPM2)

G. Density Effects (Fullerton)

Log10 kf2 = 0.000208 ( BHP 1 - BHP2 ) + log10 kf2 Results of Changes to M. Wt. or GPM A. B. C. D.


11. Well Control Related

10. Fluid Properties

HHP2 = ( M2 / M1 ) • HHP1 • ( G2 / G1 )3 HP2 = (M2 / M1 ) • HP1 • ( G2 / G1 )3 ∆P2 = (M2 / M1 ) • ∆P1 • ( G2 / G1 )2 T2 = (M2 / M1 ) • T 1 • ( G2 / G1 )2

A. Hydrostatic Pressure (psi) HP = M x 0.05195 x Depth B. Pressure Gradient PG = M x 0.05195 C. Differential Pressure M = Diff. Press. ( L x 0.5195 ) D. Shut In D.P. Pressure, new M2 = (SIDPP / ( 0.052 x L ) ) + M1’ E. Formation Pressure BHP = Hydrostatic Pressure + SIDPP F. Balance Weight Mud M2 = ( SIDPP / 0.052 / TVD ) + M1 G. Equivalent Mud Wt at Casing Seat EMW = SICP / 0.052 / TVDCS ) M1 H. Pressure Integrity Test at Casing Seat PIT = ( PumpIn(psi) / 0.052 / TVD ) + M I. Max. Surface Casing Pressure Allowable SCPMX = PIT - ( M x 0.052 x TVD(CS) J. Final Drill Pipe Circulating Pressure a. Friction Pressure = Initial Circ.Press. - Shut In Dp Press. B. Final DP Circ. Press = Friction (psi ) x Kill Wt. Mud Original Mud Wt. K. D.P. Circ. Press. Decrease per Point of Mud Wt. Increase. DPCPDEC ( Initial DPCP - Final DPCP ) total points of mud weight increase L. Barrels of Mud to bleed off : Stair Step Method. Bbls = ( Csng PSI Increase x Annular Capacity (bbls/ft) M x 0.052 M. Barite (sacks) to weight up mud (barrels) Sacks/Bbl. = 14.9 x (Kill Wt. Mud - Original Mud Wt.) 35.5 - Kill Wt Mud


12. Common Oilfield Volumes 1 gallon = 231 cu.inches = 0.13368 cu.feet = 0.02381 bbls 1 barrel = 42 gallons = 9702 cu.inches = 5.61458 cu.feet 1 cu.foot = 1728 cu.inches = 7.48052 gals. = 0.17811 bbls 13. Capacity of Open Hole or Pipe Barrels Per Foot Gallons Per Foot Cubic Feet / Foot

= ( Hole or Pipe ID )2 / 1029.4 = ( Hole or Pipe ID )2 / 24.51 = ( Hole or Pipe ID )2 / / 183.35

14. Capacity of Annulus : Barrels Per Foot Gallons Per Foot Cubic Feet / Foot

= ( Hole ID2 - Pipe ID2) / 1029.4 = ( Hole ID2 - Pipe ID2) / 24.51 = ( Hole ID2 - Pipe ID2) / 183.35

15. Displacement of Tubular Tools: (Pipe,Collars,Etc) Barrels Per Foot Gallons Per Foot Cubic Feet / Foot

= ( Tool OD2 - Tool ID2) / 1029.4 = ( Tool OD2 - Tool ID2) / 24.51 = ( Tool OD2 - Tool ID2) / 183.35

16. Displacement of Steel Tools based on weight) Barrels Per Foot Gallons Per Foot Cubic Feet / Foot

= Air Weight (lbs/ft) x 0.0003636 = Air Weight (lbs/ft) x 0.0043290 = Air Weight (lbs/ft) x 0.0005787

17. Weight of Steel per Unit Volume Gallon, Steel = 65.5 ppg Cubic Feet, Steel = 490 ppg Barrel, Steel = 2751 pounds 18. Buoyancy Factor of Steel in a Mud. BF = ( 65.5 - Mud Wt (ppg ) / Mud Wt (ppg)


19. Bit Press. Drop relationship to RPM & WOB COMMON DENSITIES

Water ( 39.20F ) Water ( 68.00F ) Sea Water Steel Iron Aluminum

RELATIVE DENSITY

1.000 0.998 1.026 7.804 7.853 2.700

LBS PER CU.FT.

LBS PER GALLON

GRAMS / CU. CM

62.4 62.3 64.0 487.0 490.0 168.5

8.34 8.33 8.55 65.10 65.50 22.50

1.000 0.998 1.026 7.804 7.853 2.700

20. Bit Press. Drop relationship to RPM & WOB ∆PBIT = a. 0.678•G•DH•(BWT • BRPM ) 0.5 for (BWT • BRPM ) <250 b. 0.678•G•DH•(BWT • BRPM ) 0.5 for 250<(B WT • BRPM) <350 c. 0.678•G•DH•(BWT • BRPM ) 0.5 for (BWT • BRPM ) >350

21. Hydraulic HorsePower HHP USED = (∆PSYSTEM • GPM) / 1714 HHP MAX = (∆PSYS. MAX • GPM) / 1714 HHP USED = (HHP USED / HHP MADE )

22. Mechanical Horsepower Created HrsPwr(Mech) = Torque x RPM/ 5252


Miscellaneous. Hook Load OverPull Maximum Neutral Point (STRAIGHT HOLE) Natural Frequency Excitation Frequency Excitation Frequency Frequency w/ Shock Mechanical Horsepower Created

HL = ( ( Pipe Wt/Ft x Feet ) + ( Collar Wt/Ft x Feet ) ) x Buoyancy Factor OP = ( Yield Strength of Pipe - Hook Load ) NP = Bit Weight / ( Weight/Foot * Buoyancy Factor ) FN = 4212 / Drill Collar Length (ft) FE = RPM / 20 Ncrit = FN x 20 FNS = P x ( shock spring rate (k) / Total Wt. DC’s (w) ) 0.5 HrsPwr(Mech) = Torque x RPM/ 5252

Maximum Tensile Loading w/ Collapse Pressure Applied Y = Minimum Yield Strength of Pipe ( PSI ) A = Cross-sectional area of Pipe Body P = Collapse Pressure on the Pipe ( PSI ) C = Collapse Rating w/ no load ( tables ) L= Y x A x (( 1-.75 x (P/C)2 ) 0.5 - ( .5 x (P/C) )

Expansion of Steel: Temperature Changes C L T

= Coefficient of Expansion for steel = 0.0000828” per foot, per degree Fo = Length in feet = Change in temperature, Fo

ET

= CxLxT

2002.Hydraulics

Recommend Documents