hydraulics presentation

2.05.01 Circulating System & Fluid Flow

2.05.02 Drilling Circulating System

2.05.03 Hydraulic Horsepower

2.05.04 Volumetric-Rate & Pumps

2.05.05 Continuity of Flow

2.05.06 Fluid Velocity

2.05.07 Types of Flow, Turbulent & Laminar

2.05.08 Fluid Flow Pressure Loss

2.05.09 Pressure Drop Factors

2.05.10 Annular Flow & Pressure Drop

2.06.01 Rheology - Study of Fluid Flow Terms

2.06.02 Rheology - general terms and definitions

2.06.03 Rheology - Fluid Flow Calculation Models

2.06.04 Rheology - Fluid Flow Classifications

2.07.01 Bit - Mechanical Energy

2.07.02 Bit - Hydraulic Energy

2.08.01 Rule of Thumb to Optimize Hydraulics

2.08.02 Rule of Thumb to Optimize Hydraulics

2.09.01 Hydraulic & Related Formula Summary




E X I T

2.01.01 Introduction, Circulating System Hydraulics

2.01.02 Introduction, Study of Hydraulics

2.02.01 Hydraulic General Terms and Definitions

2.02.02 Matter and Types of Matter

2.02.03 Mass, Inertia, Density

2.02.04 Specific Gravity

2.02.05 Common Oilfield Volumes

2.02.06 Area of shapes

2.02.07 Weight

2.02.08 Force & Pressure

2.03.01 Hydrostatic Fluids

2.03.02 Hydrostatic Fluid Illustrations

2.03.03 Single Liquid. Seeks own level

2.03.04 Multiple Liquid. Seeks own level

2.03.05 Hydrostatic Pressure

2.03.06 Hydrostatic Pressure in Oilfield

2.03.07 Hydrostatic Pressure Gradient

2.03.08 Buoyancy & Archimedes Principle

2.03.09 Buoyed Weight

2.03.10 Buoyancy Factor

2.03.11 Buoyancy Calculation Example.

2.03.12 Hydrostatic Fluid Formulas.

2.04.01 Hydraulic Power Transmission

2.04.02 Pressure & Force

2.04.03 Hydraulic Cylinders

2.04.04 Work

2.04.05 Hydraulic Power, horsepower

2.04.06 Basic Hydraulic Pump

DD Hydraulics 2.00.00 H Y D R A U L I C S

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

Hydraulics may be divided into two categories:

Hydrostatic applies to liquids at rest

Hydrodynamic applies to liquids in motion

Hydraulics may be divided into two categories:

Hydrostatic applies to liquids at rest

Hydrodynamic applies to liquids in motion

Note.

Some laws of hydraulics also apply to gases under certain conditions, but unless specifically stated,only liquids

will be used or considered. Note also, the term “fluid” will apply to liquids only - unless specifically noted.

Note.

Some laws of hydraulics also apply to gases under certain conditions, but unless specifically stated,only liquids

will be used or considered. Note also, the term “fluid” will apply to liquids only - unless specifically noted.

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 playat 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 typeof hydraulics, individually and the effect that one has on the other one.

DD Hydraulics 2.01.01 Introduction DD Hydraulics 2.01.02 Introduction

Stand Pipe

Rotary hose

The Circulating System

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.

RESERVE PIT

OPE

N H

OL

EC

ASE

D H

OL

E

PUMP

HOPPER

MUD PIT

SHALE SHAKER

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 Ratio or density of a substance to density of a Gravity 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 positionor 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.

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.

Gases

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

Solids:having a definite volume and shape independent of any container.

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.

.

DD Hydraulics 2.02.02 General Terms and DefinitionsDD Hydraulics 2.02.01 General Terms and Definitions

Air Resistance

A stationary object will remainstationary unless acted upon byan outside force

A thrown ball will travel in astraight line were it not for theexternal forces of air frictionand gravity.

Inertia is the property of matter by which it remains at restor in a straight line motion unless acted upon by an outsideforce.

Mass is the property of matter that is a measure of itsinertia.

Density is technically Mass per Unit Volume but iscommonly used as Weight per Unit Volume.

Weight-Density

Density = Weight / Unit Volume

Weight-Density

Density = Weight / Unit Volume

The term “density” will refer to weight-density or weightper unit volume unless specifically noted.

Specific Gravity

Weight of Substance per Unit Volume

Weight of Water per Unit Volume

Specific Gravity

Weight of Substance per Unit Volume

Weight of Water per Unit Volume

Water ( 39.20F )

Water ( 68.00F )

Sea Water

Steel

Iron

Aluminum

1.000

0.998

1.026

7.804

7.853

2.700

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

RELATIVE

DENSITY

LBS PERCU.FT.

LBS PERGALLON

GRAMSPER CU.

CM

COMMON

DENSITIES

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.

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


Gravity

AIR

RESISTANCE

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

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.

Square Area ( Rectangle)

Square Area ( Triangle )

Square Units = Diameter 2 x ( π / 4 )

Square Area ( Circle )

6” x 6” = 36 square inches

Square Units = Length x Width

Square Units = ( Base x Height ) / 2

( 6” x 6” ) / 2 = 18 sq. inches

6 2 x ( 3.14 / 4 ) = 28.3 sq. inches

Square Area ( Ring )

Sq. Units = ( DIA. 2 - dia. 2 ) x ( π / 4 )

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


Common Oilfield VolumesVolume 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.

Cubic Foot

1 gal12” x 12” x 12”

Columns

12 cubic inches

12”

Volume ( Cubic Units ) = Square Area x Length

Volumes must have consistent units of measurement in bothcalculations and solutions.

Barrel

42 gal

Volume Cubic Gallons Barrels CubicConversions Inches Feet

Cubic Feet 1728 7.48052 0.17811 1

Barrel 9702 42 1 5.61458

Gallon 231 1 0.02381 0.13368

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 )

FORCE

FORCE

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 ignoredunless 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.

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.

UnitWt.

GalWt /

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.

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.

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

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

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

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

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


LiquidsFlow

Gravity

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

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

Each layer of liquid exerts its weight on those below

Pressure exists at all points in a liquid

At any point pressure is equal in all directions

At any level pressure is the same in single liquid

Pressure is proportional to the depth in a liquid

Direction of force reacts perpendicular to surfaces

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

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

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.

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.

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

DD Hydraulics 2.03.02 Hydrostatic Fluid PropertiesDD Hydraulics 2.03.01 Hydrostatic Fluid Properties

Liquids seek their own level. Single Liquid

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 seek their own level. Two Liquids

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.

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 pressuresby 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.

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.

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.

Hydrostatic Pressures are additive. Calculate each separately and add.

Verticaldepth

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


HeavyMud

HEAVYMUD

LIGHT MUD

LIGHTER

MUD

UnequalHydrostaticPressures

Equal Hydrostatic Pressures

Point of Separation

U- Tube

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.

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.

HYDROSTATIC PRESSURE

H.P. ( PSI ) = 0.05195 x Mud Weight x Depth

= 0.05195 x Lbs/Gal x Vertical Ft

HYDROSTATIC PRESSURE

H.P. ( PSI ) = 0.05195 x Mud Weight x Depth

= 0.05195 x Lbs/Gal x Vertical Ft

VERTICALDEPTH

Note: Hydrostatic Pressures are basedupon vertical heights of the fluid columnonly, irregardless of angle or shape of thecolumn. In regards to wells, true verticaldepth to point of interest, not measureddepth, determines hydrostatic pressures.

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. Poundsper Square Inch.


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

Pressure Gradient

PG = 0.05195 x Mud Wt. (Lbs/ Gal )

Pressure Gradient

PG = 0.05195 x Mud Wt. (Lbs/ Gal )

Examples:

Find Pressure Gradient: water at 8.33 lbs per gallonThen: 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 / footAnd: HP = 0.46755 x 6000 = 2805 psi

Find Mud Weight: 3000 psi needed at 5000 feet.Then: PG = 3000 psi / 5000 feet = .600 psi/ftand: MW = .600 PG / 0.5195 = 11.55 lbs/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.

• weight-density of a gallon of mud.

By establishing the pressure gradient, pressure at any point is found simply by multiplication of it by the current footage of interest.

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 partlysubmerged 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:

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

• 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.

Net Pressure


AIRWT

NetPressure

BUOYWT

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.

Buoyed Weight : Displacement Method

Object. Wt. - ( Object Volume x Fluid Density )

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.

Hydrostatic Pressure Differential

( HP x Lower Area ) - ( HP x Upper Area )

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 ).

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 pressuredifferences. The use of the displacement orbuoyancy 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: 10” x 10” x 5’ steel bar. Densities: Steel at 65.5 Lbs/Gal.

Fluid at 9.0 Lbs./Gal.

Using Buoyancy Factor:

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 psiH.Press @ 1005 = 0.052 x 9 x 1005 = 469.89 psi Force ( down ) = 100 sq.in. x 467.55 = 46755 lbsForce ( up ) = 100 sq.in. x 46989 = 46989 lbs.Total Diff. = 46989 - 46755 = 234 lbs Buoyed Wt. = 1701 lbs - 234 lbs = 1467 lbs.

Using Displacement

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

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

Density ( Weight-Density) = Weight / Unit Volume

Wt of Substance per Unit Vol.

Wt of Water per Unit Vol.Specific Gravity =

Volume ( Cubic Units ) = Area x Length

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

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

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

Summary of formulas: Hydrostatic Fluids

Circle Area = Diameter 2 x .7854

Ring Area = ( Dia2 - dia 2 ) x .7854


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 to confined liquid is transmitted:

- in all directions- equally distributed- undiminished

Force applied to a solid block is transmitted in a straight line through block.

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.

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

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

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.

FORCE

10 LBS

FORCE

10 LBS

FLUID

Single Cylinder

A pressure increase in a confined fluid is distributed equally throughout the fluid and against all sides of container. Forceapplied on any surface such as a piston is equal to the pressure applied multiplied by the area of the “piston”.

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.

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.

F O R C E

F O R C E

SOLID

LIQUID

DD Hydraulics 2.04.02 Hydraulic Power TransmissionDD Hydraulics 2.04.01 Hydraulic Power Transmission

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

Travel of an output hydraulic cylinder is proportional to area of the input piston to area of its output piston.

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

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

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.

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

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.

Stroke Speed of an output cylinder is proportional to area of input piston to area of its output piston.

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

DistanceForce ( psi )

area

piston

Work=

Hydraulic Work

Work (Inch-Lbs) = Force (Lbs) x Travel (inch) Work (Inch-Lbs) = Force (Lbs) x Travel (inch)

Work ( Foot-Lbs ) = Force ( Lbs ) / Travel ( Feet )Work ( Foot-Lbs ) = Force ( Lbs ) / Travel ( Feet )

DD Hydraulics 2/04.04 Hydraulic Power TransmissionDD Hydraulics 2.04.03 Hydraulic Power Transmission

Work = Force x DistanceWork = Force x Distance

Mechanical Work Work = Ft-LbsPounds x Feet

Force Distance Workx =

1 2

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

DistanceForce

Force

Hydraulic Cylinders : Input and Output

Pressure = Pressure

Work In =

Work Out =

Distance

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.

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

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

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

Power =

33000 Ft-Lbs 550 Ft-Lbs 1 minute 1 second

33000 Ft-Lbs 550 Ft-Lbs 1 minute 1 second

Horsepower = =

Common power unit of measurement is Horse Power

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

DD Hydraulics 2.04.06 Hydraulic Power TransmissionDD Hydraulics 2.04.05 Hydraulic Power Transmission

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:

• Electrical to operate pump motor• Hydraulic produced by the pump• Kinetic produced when hydraulic fluid moves a piston. • Potential produced when the piston has raised an object. • Heat produced by friction in pump,pipe, & fluid

Hydraulic pumps areused to create pressureincreases used to dowork. Basuc hydraulicsystems are a closedpiping circuit in whicha fluid under controlledpressure is used to dowork. Hydraulic pumpsimpart energy or powerto the fluid which istransmitted to the worksite where the work isdone. The fluid is thenreturned to the pumpto be energized again.

7

FILTER

PRESS REG

RELIEF

VALVE

HANDPUMP

FLUID

PRESSURESIDE

AIR

WORK

SIDE

HYDRAULIC POWER SYSTEM

PRESSURE

REGULATOR

ACCUMULATOR

CONTROL

VALVE

GAGE

CHECK VALVES

RETURN LINE

PUMP

FLUID RESERVOIR

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.

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 withoutloosing its shape or deforming.

The continuous deformation of a liquid is known as “ Flow ”. Theflow 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 oftentwo to 4 miles deep, the pressures required to maintain the high flow rates required are substantial and together with the pressureused at the bit could be a limiting factor on flow rates.

Additionally, the fluids or mud used are tailored to do differenttasks associated with drilling the hole. These fluid characteristicsimpact fluid flow properties. It is essential that hydrostatic andhydrodynamic properties of fluid be understood as well as the impact that the mud properties may have.

Stand Pipe

Rotary hose

The DrillingCirculating System

RESERVE PIT

OPE

N H

OL

EC

ASE

D H

OL

E

PUMP

HOPPER

MUD PIT

SHALE SHAKER

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.

A circulating system imparts energy or power to a fluid, transports it to the work site, does thework, and returns it to be energized again.

Basic elements include geometry of the piping, fluid properties and flow rate with each of theseinfluencing the total pressures realized.

DD Hydraulics 2.05.02 Circulating System and Fluid FlowDD Hydraulics 2.05.01 Circulating System and Fluid Flow

DD Hydraulics 2.05.04 Circulating System and Fluid FlowDD Hydraulics 2.05.03 Circulating System and Fluid Flow

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.

Volumetric-Rate. Volume of liquid in units per a unit time asbarrels per minute, gallons per minute, etc.

The volumetric-rate output of a pump is found by the volume perstroke multiplied by the number of strokes per unit time at whichpump is operated.

Volume Triplex ( Bbl / stroke ) 3 ( ID2 / 12353 ) x Stroke Length (inch)

Volume Triplex ( Bbl / stroke ) 3 ( ID2 / 12353 ) x Stroke Length (inch)

Volume Duplex ( Bbl /stroke ) ( 4 ( ID2 / 12353 ) - 2 ( OD2 / 12353 ) ) x Stroke Length (inch)

Volume Duplex ( Bbl /stroke ) ( 4 ( ID2 / 12353 ) - 2 ( OD2 / 12353 ) ) x Stroke Length (inch)

Actual output per stroke of pump is found by multiplying aboveresult by pump efficiency.

Constant: 12353 = 1cu.in. / ( π / ( 4 x 9702 cu.in.) )

Pumping action occurs on one side only.

Pumping action occurs on both sides of piston.

Triplex: single acting with three cylinders

Duplex: double acting with two cylinders

FLUID

PISTON

OD ID ROD

NO

FLUID

PISTON

ROD

FLUID

FLUID

OD ID

FLUID

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.

Hydraulic Horse Power is a measure of the energy delivered to the fluid being pumped.

Hydraulic Horse Power = Mechanical Horse Power. For Pressure in Lbs per Sq.In. and flow is in GPM. Then 231 cubicinches 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.

Hydraulic Horse Power

Pressure Drop ( psi ) x Flow Rate ( gpm )

1714

Hydraulic Horse Power

Pressure Drop ( psi ) x Flow Rate ( gpm )

1714

DD Hydraulics 2.05.06 Circulating System & Fluid FlowDD Hydraulics 2.05.05 Circulating System & Fluid Flow

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

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

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 of Flow = Velocity ( ft/min ) x AreaVolumetric-Rate of Flow = Velocity ( ft/min ) x Area

Velocity ( ft /min ) = Volumetric-Rate / AreaVelocity ( ft /min ) = Volumetric-Rate / Area

A

Cross-section

DSPEED

A

Cross-section

DSPEED Volume

Volume

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.

Fluid Flow Velocity ( Feet per Minute )( 24.51 x GPM ) / Diameter2

Fluid Flow Velocity ( Feet per Minute )( 24.51 x GPM ) / Diameter2

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

The speed of a flowing fluid is dependant on volume (GPM ) and area in square inches at a cross-section of the conductor.

Fluid Flow Velocity ( Feet per Second )( ( 24.51 x GPM ) / Diameter2 ) / 60

Fluid Flow Velocity ( Feet per Second )( ( 24.51 x GPM ) / Diameter2 ) / 60

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 tothe inside diameter of the drill string and to the bit are expressed in feet per second.

Volume-Rate in = Volume-Rate out


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 workintended, 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

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

Flow Direction

PS

I

012345

In drilling, two major conductors of flow are the drillstring and annulus. These have a large impact on flowrates and resulting pressures which may limit the flowrate. Wells are frequently 2 to 4 miles deep, and mayreach 6 or more miles. Pressures required to generatefluid flow to the bit and back to surface are substantialand of prime importance. A well designed hydraulicsprogram is one in which less than 50% of availablehydraulic horse power is used for flow with 50% ormore used by the bit in making hole.

Plug Flow: occurs only at very slow rate wherea thin layer of fluid slips at conduit wall withrest flowing as a unit. This flow regime to begiven no further consideration. Not Considered.

Laminar Flow: can be viewed as relatively smooth, straight stream-lines of flow having concentric layers of fluid beginning with a zerovelocity at the conduit wall, with layers progressively faster, reachingmaximum speed at the center. This flow pattern requires less energy.

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

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 calculatethe applicable pressure drop. Laminar flow uses less energy than does Turbulent flow.

PLUGFLOW

Turbulent Flow is the “fast”, chaotic flow of fluidparticles, moving in random loops except at wallof conduit where velocity is zero. Streamlines areirregular patterns with a flat profile. Maintainingthe fast flow rate requires more energy versus thestraighter streamlines of laminar flow.

CHAOTICFLOW

The Velocity Differential betweenlayers are greatest at wall and leastin center. Low viscosity fluids havegreater differentials than do thehigher viscosity fluids.

LOWVISCOSITY

HIGHVISCOSITY

ID

Factors of Pressure DropFluid: Density and ViscosityVolume: Volumetric-Rate Dimensions: Length and ID

Analysis of the formula for turbulent flow in a conduit shows acomplex relationship between components. The analysis belowillustrates the degree that changes to a factor could result in.

I.D. Pressure Loss ( turbulent flow)

0.0000765 PV 0.18 x MW 0.82 x GPM 1.82 x L

ID 4.82

I.D. Pressure Loss ( turbulent flow)

0.0000765 PV 0.18 x MW 0.82 x GPM 1.82 x L

ID 4.82


ANALYSIS OF PRESSURE DROP - PROPORTIONAL FACTORS

Plastic Mud Length GallonsViscosity Weight Feet per Min.

1 0.18 = 1.00 1 0.82 = 1.00 1 1 = 1.00 1 1.82 = 1.002 0.18 = 1.13 2 0.82 = 1.77 2 1 = 2.00 2 1.82 = 3.533 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 DECREASE DECREASEIN I.D. IN I.D. IN I.D.

1.1 4.82 = - 1.58 1.4 4.82 = - 5.06 1.7 4.82 = -12.911.2 4.82 = - 2.41 1.5 4.82 = - 7.06 1.8 4.82 = -17.001.3 4 82 = - 3.54 1.6 4.82 = - 9.64 1.9 4.82 = -22.06

Chart shows varying degrees of net change to pressure drop by changes in acomponent. Doubling Mud Weight increases PSI by 1.77 times. DoublingGPM increases PSI by 3.53 times. Increasing ID by 20% decreases PSI by2.41 times.

Annular Pressure Loss ( Laminar Flow )

( { L • YP } ÷ 225 { DH - DP } ) + ({ L • VAS • PV } ID 4.82

( 1500 { DH - DP } )

Annular Pressure Loss ( Laminar Flow )

( { L • YP } ÷ 225 { DH - DP } ) + ({ L • VAS • PV } ID 4.82

( 1500 { DH - DP } )

Annular Pressure Loss ( turbulent flow)

( .0000765 PV 0.18 • M 0.82 • G 1.82 • L )

({ DH - DP } 3 • { DH + DP } 1.82 )

Annular Pressure Loss ( turbulent flow)

( .0000765 PV 0.18 • M 0.82 • G 1.82 • L )

({ DH - DP } 3 • { DH + DP } 1.82 )

Annular Critical Velocity

1.08PV + 1.08( PV 2 • 9.3{ DH - DP } 2 • YP • M ) 0.5

M ( DH - DP )

Annular Critical Velocity

1.08PV + 1.08( PV 2 • 9.3{ DH - DP } 2 • YP • M ) 0.5

M ( DH - DP )

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 isturbulent.

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 thehole becomes deeper and the annular space becomes smaller, itsimpact is larger. The pressure drop in the annulus is added to the hydrostatic pressure of the hole and as such plays a significant rolein the open hole where it is exposed to the formation. Too little or to much pressure against the formation(s) can result in problems.

DD Hydraulics 2.06.02 Rheology - Study of Fluid Flow TermsDD Hydraulics 2.06.01 Rheology - Study of Fluid Flow Terms

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

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.

Window Pane

Bead of water

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.

A bead of water on a window pane illustrates the cohesive internal attraction of a fluid within and to the fluid itself.

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 velocity of its own. This builds, reaching its peak in the center.

Conduit WallZero Velocity Fluid

Faster Velocity

shear rate

yiel

d po

int

shea

r st

ress

Bingham Plastic

Plastic Viscosity

Newtonian

Viscosity

IdealPowerLaw

shear ratesh

ear

stre

ss

Pseudo-Plastic

Dila

tent

Newto

nian

Bingham Plastic Model - a finite stress must be applied to initiateflow. 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)

Fluid Flow Models are attempts to mathematically define behavior of fluids as they flow.

DD Hydraulics 2.06.04 Rheology - Study of Fluid Flow TermsDD Hydraulics 2.06.03 Rheology - Study of Fluid Flow Terms

Tooth penetrates formation,

creating chip by fracture and

shearing of rock.

Weight on Bit controls

Chip Size & Quantity

RPM controls Fracture

Rate.

Bit Mechanical Energy

TOOTH

CHIP

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.

Bit Hydraulic Energy

Hydraulic Energy is required to allow the bit to effectively drill bycleaning the bit and the hole bottom. As fluid is forced through thebit nozzles, its kinetic energy is greatly increased through high jetvelocities 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 fluidimpinges upon the bore hole below the bit.

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

Chip

Higher Hydrostatic

Pressure

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.

Adequate jet velocity cleans the bit. A balled up bit acts as a cushion preventing effective drilling. There must be roombetween the bits teeth for new formation.

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

DD Hydraulics 2.07.02 Bit - Hydraulic EnergyDD Hydraulics 2.07.01 Bit - Mechanical Energy

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 hold-down. 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 tomaintain 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.

DD Hydraulics 2.08.02 Rules of Thumb to Optimize HydraulicsDD Hydraulics 2.08.01 Rules of Thumb to Optimize 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.

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:

CD = Chip Diameter, inchCW = Chip Weight, ppgDH = Diameter of Hole, inchDP = Diameter of Pipe, inchG = Gallons Per Minute ( gpm )JV = Jet Velocity ( fps )L = Length in feetM = Mud Weight (ppg )NZ = Nozzle Size ( 32nds of inch )NA = Nozzle Area ( square inch )∆PXX = Pressure Loss ( psi )

PV = Plastic Viscosity, cpsVAS = Velocity, Annular Fluid ( fps )VAM = Velocity, Annular Fluid ( fpm )YP = Yield Point (lbs / 100 ft.)f = fanning friction factorn = Consistency indexp = Numerical Constant, 3.14159Re = Reynolds Number, dimensionlessU = Viscosity, apparent, effective, cps

other nomenclature found in formulas themselves

NOMENCLATURE

Hydraulic Formulas. Bit RelatedHydraulic Formulas. Bit Related

Nozzle Pressure Loss ( psi ) ∆PNZ = ( M • G 2 ) ÷ ( 10858 • NA 2 )Nozzle Volume ( gpm ) GNZ = ( { P NZ • 10858 • NA 2 } ÷ M ) 0.5

Nozzle Total Flow Area ( sq.in. ) NA = ( { M • G 2 } ÷ { 10858 • PNZ } ) 0.5

Nozzle Area ( per size ) (sq.in. ) NA = ( N Z ÷ 32) 2 • ( ¶ ÷ 4)Nozzle Size ( 32nds inch ) N Z = 32 • (N A ÷ { .7854 • Qty } ) 0.5

Jet Velocity of Nozzles ( fps ) JV (F/S) = ( 0.32 • G ) ÷ N A

Impact Force of Nozzles ( psi ) IF = JV • 0.0173 • G • ( P NZ • M ) 0.5

“ “ “ ( psi ) IF = 0.000516 • JV• G • MBit Hydraulic Horse Power, Total BHHP (TOTAL) = PNZ • G ÷ 1713.6Bit Hydraulic Horse Power, per Sq.In. BHHP / sq. in = BHHP ÷ ( BIT OD • 0.7854 )

Drill String Bore Pressure Loss ( psi )

Turbulent flow ( sii ) ∆PID = ( 0.0000765 PV 0.18 • M 0.82 • G 1.82 • L ) ÷ ID 4.82

Turbulent flow ( security ) ∆PID = ( 0.000061 • M • G 1.86 • L) ÷ ID 4.86

Turbulent flow ( fanning ) ∆PID = ( f • M • V ID (F/S) 2 • L ) ÷ 25.8 DP

DD Hydraulics 2.09.01 Hydraulic & Related Formulas

∆PAN = Pressure Drop, Annulus ( psi )PSI = Pressure, Pounds per Square InchGPM = Gallons per MinuteFPS = Feet per SecondFPM = Feet per MinutePPG = Pounds per Gallon


Annulus Flow

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

Annulus Pressure Loss ( psi )

Turbulent Flow ( sii ) ∆PAN = ( .0000765 PV 0.18 • M 0.82 • G 1.82 • L ) ÷ ({ DH - DP } 3 • { DH + DP } 1.82 )Turbulent Flow ( security ) ∆PAN = 0.00000014327 M • L • VA 2 ÷ ( DH - DP )Turbulent Flow ( fanning ) ∆PAN = ( f • M • VAn (F/S) 2 • L ) ÷ 25.8 ( DH - DP )Newtonian Laminar Flow ( hagan ) ∆PAN = V • L • Va ÷ ( 1500 • {DH - DP } ) Note: V = Apparent Viscosity (cps)

Plastic Laminar Flow ( beck, etc ) ∆PAN = ( { L • YP } ÷ 225 { DH - DP } ) + ({ L • Va • PV } ÷ ( 1500 { DH - DP } )

Bottom Hole Pressure

Bottom Hole Hydrostatic Pressure ( psi ) BHP = ( 0.5195 • M • L )B. H. Circulating Pressure ( psi ) BHCP = BHP + ∆PAN

Equivalent Circulating Density ( ppg ) ECD = BHCP ÷ ( 0.52 • L ) “ “ “ ( ppg ) ECD = (∆PAN ÷ { 0.052 • L } ) + Mud Weight

Hole Cleaning

Rock Chip Slip Velocity Note: use 21 as cutting density and 0.25 as cutting diameter if unknown

Lam. - Spherical Chips ( Stokes ) V C = ( 8310 CD 2 { CW - M }) ÷ ( PV + ( 399 YP • { DH - DP }) ÷ VA )Lam. - Flat Chips ( Pigott ) V C = ( 3226 CD 2 { CW - M } ) ÷ ( PV + ( 399 YP • { DH - DP }) ÷ V A )Turb. - Spherical Chips ( Rittinger ) V C = 159 ( ( CD { CW - M } ÷ M ) 0.5 )Turb. - Flat Chips ( Pigott ) V C = 60.6 ( ( CD { CW - M } ÷ M ) 0.5 )Slip Velocity ( ft / Min ) V S = V C - V A

HYDOSTATIC FLUID FORMULASWEIGHT DENSITY = Weight / Unit Volume SPECIFIC GRAVITY = Weight of Substance per Unit Volume / Weight of Water per Unit Volume HYDROSTATIC PRESSURE. ( PSI ) = 0.05195 x Mud Weight x Depth PRESSURE GRADIENT = 0.05195 x Mud Wt. (Lbs/ Gal ) BUOYANCY FACTOR OF A FLUID = ( Object Density - Fluid Density ) / Object DensityBuoyed Weight = Object Weight x Object Buoyancy FactorPOWER TRANSMISSION FORMULASPRESSURE ( PSI ) = FORCE ( LBS ) / AREA ( SQUARE INCH )FORCE ( LBS ) = PRESSURE ( PSI ) x AREA ( SQUARE INCH )FORCE ( OUTPUT ) = FORCE ( IN ) x ( AREA (OUT ) / AREA ( IN )LENGTH ( OUTPUT) = LENGTH ( IN ) x (AREA ( IN ) / AREA (OUT ) )SPEED ( OUTPUT ) = SPEED ( IN ) x (AREA ( IN ) / AREA (OUT ) )WORK = FORCE x DISTANCE

WORK (INCH-LBS) = FORCE ( LBS ) x Travel ( INCH ) WORK (FOOT-LBS) = FORCE ( LBS ) x Travel ( FEET ) POWER = WORK(FTLBS) / TIME(MINUTES OR SECONDS)HORSEPOWER = 3300 FTLBS / 1 MINUTE OR 550 FTLBS / 1 SECOND

HYDRAULIC HORSEPOWER = PRESSURE DROP (PSI) x FLOW RATE (GPM)CIRCULATION RELATED FORMULASTRIPLEX PUMP (BBLS/STK) = 3 ( ID2 / 12353 ) x STROKE LENGTH (IN.)DUPLEX PUMP(BBLS/STK) = 4 ( ID

2 / 12353 ) - 2 ( OD2 / 12353 ) ) x S.L.)VOLUMETRIC-RATE OF FLOW = VELOCITY ( FPM ) x AREA

VELOCITY (FT/MIN) = VOLUMETRIC-RATE / AREA

FLUID FLOW VELOCITY (FT/MIN) = ( 24.51 x GPM ) / DIAMETER2 FLUID FLOW VELOCITY (FT/MIN) = ( 24.51 x GPM ) / ( DIAMETER2 x 60 )


Water ( 39.20F ) Water ( 68.00F ) Sea Water SteelIronAluminum

1.0000.9981.0267.8047.8532.700

62.462.364.0

487.0490.0168.5

8.348.338.55

65.1065.5022.50

1.0000.9981.0267.8047.8532.700

RELATIVEDENSITY

LBS PER CU.FT.

LBS PER GALLON

GRAMS / CU. CM

COMMONDENSITIES

Volume Cubic Gallons Barrels Cubic

Conversions Inches Feet

Cubic Feet 1728 7.48052 0.17811 1

Barrel 9702 42 1 5.61458

Gallon 231 1 0.02381 0.13368

New Press. (Change: GPM or MW ) P2 = P1 x ( M2 / M1 ) x ( G2 / G1 ) 2 x P1

Hook Load HL = ( ( Pipe Wt/Ft x Feet ) + ( Collar Wt/Ft x Feet ) ) x Buoyancy FactorOverPull Maximum OP = ( Yield Strength of Pipe - Hook Load ) Neutral Point (STRAIGHT HOLE) NP = Bit Weight / ( Weight/Foot * Buoyancy Factor ) Buoyancy Factor ( GALLONS ) BF = ( 65.5 - M ) / 65.5 ( 65.5 lbs = steel in gallons )BUOYED WEIGHT BW = AIR WEIGHT x BOUNCY FACTORNatural Frequency FN = 4212 / Drill Collar Length (ft)Excitation Frequency FE = RPM / 20Excitation Frequency Ncrit = FN x 20Frequency w/ Shock FNS = P x ( shock spring rate (k) / Total Wt. DC’s (w) ) 0.5 Mechanical Horsepower Created HrsPwr(Mech) = Torque x RPM/ 5252

Effective Viscosity

A. Viscosity Definition U = SS / Sr

B. Bingham Plastic U = ( PV + ( 399 YP x ( DH - DP)) / VA )

C. Shear Stress, Power Law Fluids SS = k x Sr nD. Effective Viscosity , Power Law Ue= k x Sr n-1

E. Annular Shear Rate S r = 2.4 VA / ( DH - DP ) F. Consistency Index k = 511( YP + PV ) / 511n

G. Power Law Index n = 3.32 log 10 ( YP + 2PV ) / ( YP + PV )


Expansion of Steel: Temperature Changes

C = Coefficient of Expansion for steel = 0.0000828” per foot, per degree Fo

L = Length in feetT = Change in temperature, Fo

ET = C x L x T

Maximum Tensile Loading w/ Collapse Pressure AppliedY = Minimum Yield Strength of Pipe ( PSI )A = Cross-sectional area of Pipe BodyP = 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) )


hydraulics presentation

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