Hydraulics Presentation

DD Hydraulics 2.00.00

HYDRAULICS

2.01.01

Introduction, Circulating System Hydraulics Hyd raulics

2.01.02

Introduction, Study of Hydraulics

2.02.01

Hydraulic General Terms and Definitions

2.02.02

2.05.01

Circulating System & Fluid Flow

2.05.02

Drilling Circulating System

2.05.03

Hydraulic Horsepower

Matter and Types of Matter

2.05.04

Volumetric-Rate & Pumps

2.02.03

Mass, Inertia, Density

2.05.05

Continuity of Flow

2.02.04

Specific Gravity

2.05.06

Fluid Velocity

2.02.05

Common Oilfield Volumes

2.05.07

Types of Flow, Turbulent & Laminar

2.02.06

Area of shapes

2.05.08

Fluid Flow Pressure Loss

2.02.07

Weight

2.05.09

Pressure Drop Factors

2.02.08

Force & Pressure

2.05.10

Annular Flow & Pressure Drop

2.03.01

Hydrostatic Fluids

2.06.01

Rheology - Study of Fluid Flow Terms Ter ms

2.03.02

Hydrostatic Fluid Illustrations

2.06.02

Rheology - general terms and definitions

2.03.03

Single Liquid. Seeks own level

2.06.03

Rheology - Fluid Flow Calculation Models

2.03.04

Multiple Liquid. Seeks own level

2.06.04

Rheology - Fluid Flow Classifications

2.03.05

Hydrostatic Pressure

2.03.06

Hydrostatic Pressure in Oilfield

2.07.01

Bit - Mechanical Energy

2.03.07

Hydrostatic Pressure Gradient

2.07.02

Bit - Hydraulic Energy

2.03.08

Buoyancy & Archimedes Principle

2.08.01

Rule of Thumb to Optimize Hydraulics

2.03.09

Buoyed Weight

2.08.02

Rule of Thumb to Optimize Hydraulics

2.03.10

Buoyancy Factor

2.09.01

Hydraulic & Related Formula Summary

2.03.11

Buoyancy Calculation Example.

2.09.02


2.03.12

Hydrostatic Fluid Formulas.

2.09.03


2.04.01

Hydraulic Power Transmission

2.09.04


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

E X I T


Introduction

The Circulating System

Stand Pipe


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

Rotary hose

P U M P

Hydraulics Hydraulicsmay maybe bedivided dividedinto intotwo twocategories: categories:

HOPPER

M U D P I T R E S E R V E P I T

Introduction

applies appliestotoliquids liquidsatatrest rest

Hydrodynamic Hydrodynamic

applies appliestotoliquids liquidsininmotion motion

Note. Note. Some laws of hydraulics also apply to gases under certain Some laws of hydraulics also apply to gases under certain conditions, but unless specifically stated,only liquids conditions, but unless specifically stated,only liquids will be used or considered. Note also, the term te rm “fluid” will be used or conside c onsidered. red. Note also, the term “fluid” will apply to liquids only - unless specifically noted. will apply to liquids only - unless specifically noted.

SHALE SHAKER

The circulating system is a large lar ge 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.

Hydrostatic Hydrostatic

E L O H D E S A C

E L O H N E P O

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 t he 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.


General Terms and Definitions

Matter

Material Substance that occupies space and has weight

Mass

Propert y of matter that i s a measure of its inertia.

Inertia

To remain at re rest or i n a strai ght li li ne motion unless acted upon by a force.

Density

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

Specifi ific Gravity

Ratio tio or dens ensity of a substanc ance to to density ity of a standard subst ance 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 di stance.

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 eithe r a liquid or a gas.

Solids:

having a definite volume volu me and shape independent of any container.

.

Liquids: having a definite volume volume with no shape 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 unless confined. However, fluids in this presentation 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 st raight 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, 39.2 0 F. Specific SpecificGravity Gravity

Weight WeightofofSubstance Substanceper perUnit UnitVolume Volume Weight UUnit nit Volume WeightofofWater Waterper perUnit Volume

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

AirER Resistance A I R

S I 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.

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

COMMON DENSITIES

RELATIVE DENSITY

LBS PER LBS PER GRAMS CU.FT. GALLON 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 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 a s 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 when dealing with pressures. pressures. Calculations require that the unit of measurement be consistent in the equation and in the solution.

Columns 12 cubic inches

Square Area ( Rectangle)

Barrel 12”

42 gal

Square Units = Length x Width 6” x 6” = 36 square inches

Cubic Foot

Square Area ( Triangle )

1 gal

Square Units = ( Base x Height ) / 2

12” x 12” x 12” Volume Conversions

Cubic

Cubic Feet

1728

7.48052

0.17811

1

Barrel

9702

42

1

5. 61458

Gallon

231

1

0.02381

0. 13368

Gallons

Barrels

Cubic

Inches

( 6” x 6” ) / 2 = 18 sq. sq. inches

Feet

Volume ( Cubic Units ) = Square Area x Length

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 m easurement in both calculations and solutions.

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



d ownward force of an object as a Weight... is the measure of the downward result of gravitational force and the objects mass. The weight of an object may m ay vary due to variations in the t he earth’s gravitational field or by its distance from the main body of the earth. Objects weigh more at sea level l evel than far above it. This variation in weight is normally very small and will be ignored i gnored unless otherwise noted. While weight is used to measure individual individual objects, it is also used in hydraulics, with 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 consist ency when referring to weight-densities of liquids as well as in calculations. c alculations. Use the weight per the same unit volume in all.



exerted on an object object in order to Force… is a push or pull that is exerted 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. object or substance Pressure is the amount of force exerted on an object per area over which force is applied. Pressure may be expressed in various ways, such as newtons per square meter, dynes per square square centimeter, etc. Most common measurement in the oilfield is Pounds Po unds 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 sq uare 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 ( PSI ) = Force ( L Lbs. bs. ) Pressure ( PSI ) = Force ( Lbs. )/ /Area Area( (sq.in. sq.in.) )

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


Hydrostatic Fluid Properties

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


Hydrostatic Hydrostatic Fluid Properties

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 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 weight on the layers beneath it. Liquids are relatively incompressable making their density a constant. Liquids are only slightly sli ghtly affected by temperature changes.

Gravity’s constant pull downward on liquids c reates 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 i t 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 e verywhere perpendicular to the surface that it i t touches.

Pressure At any point exists at all pressure is points in a equal in all liquid 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 separati on 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

LIGHT MUD

Heavy Mud HEAVY MUD

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

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 cal culated 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.

U- Tube Unequal Hydrostatic Pressures Point of Separation

Equal Hydrostatic Pressures

L I G H T E R M U D

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 connections when no float float exists in the drill string string or it does not work properly.



Hydrostatic Pressure is the pressure exerted by a column of fluid due to t o its i ts own height and weight.

Hydrostatic pressure in a fluid means the “ downward downward force per unit area “ equal to the weight w eight of the column as defined defi ned 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 wei ght ( 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 numbe r 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 f ormat most commonly used. Pounds per Square Inch.

HYDROSTATIC HYDROSTATICPRESSURE PRESSURE H.P. PPSI SI ) )==0.05195 H.P.( (PSI 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 shape of the column. In regards regards to wells, wells, true vertical depth to point of interest, not measured depth, determines hydrostatic pressures. 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 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.


power of a fluid to exert an an upward force on a Buoyancy is the power body placed in it. An object in a fluid is acted upon by hydrostatic pressures of the fluid on all its it s surfaces. Side pressures are balanced by pressures of opposing side. As the pressures pressures are are proportional 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 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 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 total weight of the 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 t han Air Weight

A solid having more density dens ity than a fluid will sink into the fluid f luid with its submerged weight then being less than its air weight by an amount equal to the:

Buoyancy Factor

• weight of fluid displaced.

BUOY WT

• total pressure differential between that exerted on the bottom and top surfaces of the object.


AIR WT

Net Pressure

( Object Density - Fluid Density ) / Object Density Buoyed Weight : Buoyancy Factor Object Weight x Object Buoyancy Factor

Buoyed Weight : Displacement Method Object. Wt. - ( Object Volume x Fluid Density )

Hydrostatic Pressure Differential

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

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

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

In drilling, the buoyancy factor is frequently used to predetermine the size siz e 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 displa cement 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: Dens Densiities ties::



Summary of formulas: Hydrostatic Fluids

10” x 10” x 5’ steel bar. Steel teel at 65.5 65.5 Lbs/ bs/Gal. Gal. Fluid at 9.0 Lbs./Gal. in2

End Area (s) = 10” x 10” = 100 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 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

Using Displacement

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 Factor = ( 65.5 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 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




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 t ransmitted 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 LLbs. bs. ) )/ /Area Pressure( (PSI PSI) )==Force Force( (Lbs. Area( (sq.in. sq.in.) ) A pressure increase in a confined fluid flui d is distributed equally throughout the fluid and against all sides of container. Force applied on any surface surface such as a piston is equal to the pressure 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 pre ssure and pressure to force by the areas over which they act. This can ca n be seen in hydraulic cylinders c ylinders which are a common application of hydraulic power transmission systems. Single Cylinder FORCE

A 10 pound force exerted on o n a 10 square inch piston pist on area of a confined fluid transmits a 1 PSI pressure to 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 e quilibrium, they can be ignored ig nored when considering these pressure transmissions.

FORCE FLUID

10 LBS

10 LBS

Input force equals output force if piston p iston 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.



DD Hydraulics 2/04.04

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

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

Force(out) = Force(in) x ( Area(out) A rea(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 pr essure 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

Force

x Distance

1

2

=

Work

Work = Force x Distance Dist ance Work = Force x Distance Dista nce

Given that the piston pist on 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.

Hydraulic Work

Force ( psi )

a r e Distance a

p i s t o n

=

Work

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

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

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 Foot-Lbs ) = Force 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 Inch-Lbs ) / 12



Hydraulic power systems are used to do work. Discounting losses lo sses from friction, the work which is input equals eq uals 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 ( Ft./Lbs. ) Work ( Ft./Lbs. ) Time ( minutes or seconds ) Time ( minutes or seconds )

Common power unit of measurement is Horse Power Horsepower =

33000 Ft Ft-Lbs 33000 Ft-Lbs= 1 minute 1 minute


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

Distance

Force


550 Ft-Lbs 550 Ft-Lbs 1 second 1 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 • Hyd Hydra raul uliic • Kin Kinet etic ic • Potential • Heat eat

to operate pump motor prod produc uced ed by by the the pum pump p prod produc uced ed when hen hydra hydrau ulic lic flu fluid id mov moves a pisto piston. n. produced when the piston has raised an object. produced by ffrricti ictio on in pump,pipe ipe, & flu 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

R ETURN ETURN LINE

PUMP FILTER EG PRESS R EG

AIR FLUID

PRESSURE EGULATOR R EGULATOR ACCUMULATOR

CHECK VALVES

GAGE

7 HAND

ELIEF R ELIEF VALVE

PUMP

CONTROL VALVE WORK SIDE

PRESSURE SIDE


Circulating System and Fluid Flow

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 f luids which flow which are called hydrodynamic fluids. While hydrostatic fluids can be described by relatively simple concepts c oncepts 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 it s 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 tak es place in a conductor. A conductor conduc tor is can be any shape or size, even a flat surface with the atmosphere at mosphere 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 toge ther 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 und erstood as well as the impact that the mud properties may have.



The Drilling Circulating System

Stand Pipe

Rotary hose

PUMP HOPPER

M U D P I T R E S E R V E P I T

SHALE SHAKER

E L O H D E S A C

A circulating system imparts 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 elements include geometry of the piping, fluid properties and flow rate with each of these influencing the total pressures pr essures realized. Major purposes of the fluid and its flow are: • transmit hydraulic hydraulic horsepower to to the bit to clean it and the bottom of hole. • cool and lubricate lubricate bit & drill string • transport cuttings cuttings produced out of hole. • support hole walls & prevent formation fluids from entering well bore.

E L O H N E P O



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 eleme nts 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. pressure.


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

to Hydraulic Horse Power is a measure of the energy delivered to the fluid being pumped. Hydraulic Horse Power Hydraulic Horse Power Pressure Drop ( psi ) x Flow Rate ( gpm ) Pressure Drop ( psi ) x Flow Rate ( gpm )

FLUID

Engine Horsepower required equals the Hydraulic Horsepower divided by pump efficiency. Note - newer pumps pumps are usually usually 95 to 97% efficient.

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 ( Bbl / stroke ) Volume Triplex ( Bbl / stroke ) 3 ( ID2 /2 12353 ) x Stroke Length (inch) 3 ( ID / 12353 ) x Stroke Length (inch)

1714 1714 Hydraulic Hydraulic Horse Power Power = Mechanical Mechanical Horse Power. Power. For Pressure 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.


Volume Duplex ( Bbl /stroke ) Volume Duplex ( Bbl /stroke ) ( 4 ( ID /2 12353 ) - 2 ( OD2 /2 12353 ) ) x Stroke St roke Length (inch) ( 4 ( ID / 12353 ) - 2 ( OD / 12353 ) ) x Stroke St roke Length (inch) 2

Constant: Constant: 12353 = 1cu.in. / ( π / ( 4 x 9702 cu.in.) ) Actual output per per stroke of pump pump is found by by multiplying above above result by pump efficiency. Hydraulic Hydraulic Pumps Pumps are limited to a maximum maximum volume volume and pressur pressure. e. The maximums not only vary by manufacturer and type of pump but on the size of pump liners or cylinders used and stroke length.



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 d istance of such length that would result in a volume equal equa l to that which which is referenced.



Fluid Flow Velocity through a conductor is distance traveled by a fluid within a defined time. ti me. 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 Flow Velocity ( Feet per Minute ) Fluid Flow Velocity ( Feet per Minute ) ( 24.51 x GPM ) / Diameter 2 2 ( 24.51 x GPM ) / Diameter

Cross-section

D

SPEED

Fluid Flow Velocity ( Feet per Second ) Fluid Flow Velocity ( Feet per2 Second ) ( ( 24.51 24.51 x GPM ) / Diameter ) / 60 ( ( 24.51 x GPM ) / Diameter 2 ) / 60

A

Volume

Cross-section SPEED

Volume

D

A

Given that equal volumes volu mes per unit time flows through the pipes, pi pes, it can be seen that the cross-sectional c ross-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 Flow = Velocity ( ft/min ft/min ) x Area Volumetric-Rate of Flow Flow = Velocity ( ft/min ) x Area Velocity ( ft /min ) = Volumetric-Rate / Area Velocity ( ft /min ) = Volumetric-Rate / Area

Numerical Numerical constant 24.51 24.51 is derived from from 231 cubic inches inches in a gallon equal to a cylinder of 1 square inch in area by 19.25 ( 231 / 12) feet long. long. Then the velocity velocity ( ft / min ) would equal : 19.25 ft x GPM / .7854 x Diameter 2 or 24.51 x GPM/Diameter 2

In hydraulic 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.


Rheology - Study of Fluid Flow Fl ow Terms

be havior Fluid Flow Models are attempts to mathematically define behavior of fluids as they flow. a pplied to initiate Bingham Plastic Model - a finite stress must be applied flow. At greater stresses, the flow will be newtonian. Pressure losses are calculated using plastic viscosity (PV) and Yield Point (YP). (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).

c i c a s t i P l a m a n g h Plastic Viscosity B i n s s e r t t n s i r o a p e d h l s e

Ideal Power Law


Rheology - Study of Fluid Flow Terms

Fluid Classifications Newtonian Fluids - Shear Stress is directly proportional to shear rate. ( water, oil, )

di rectly Non-Newtonian Fluids - Shear Stresses are not directly proportional to shear rates. the rate at which which the viscous forces increase increa se Pseudo Plastic Fluids - the respective respective to shear s hear rate decreases with the increasing shear rate. In other words, viscosity decreases with increasing increas ing 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)

s s e r t s r a e h s

t n e t a l i D

n i a n t o e w N Pseudo-Plastic

a n t o n i w e N

i y

Viscosity shear rate

shear rate


Bit - Mechanical Energy


Bit Hydraulic Energy

Bit Mechanical Energy Bits use both Mechanical and Hydraulic Energy in drilling the well bore. Bits are tailored to the formation characteristics.

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.

In general, the softer the rock to t o be

Hydraulic Horsepower at bit is measure of energy expended at bit.

drilled the larger the teeth. Bits are built

d rilling fluid Impact Force is a measure of the force which the drilling impinges upon the bore hole below the bit.

to be rotated while weight is applied to supply the mechanical energy required.

Adequate jet velocity cleans cle ans 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, TOOTH OOTH

CHIP

Bit - Hydraulic Energy

creating chip by fracture and shearing of rock. Higher Hydrostatic Pressure

Chip Weight on Bit controls Chip Size & Quantity

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.

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


Hydraulic & Related Formulas

NOMENCLATURE CD CW DH DP G JV L M NZ NA ∆PXX

= Chip Diameter, inch = Chip Weight, ppg = Diameter Diamete r of Hole, inch = Diameter of Pipe, inch = Gallons Per Minute ( gpm ) = Jet Velocity Veloci ty ( fps ) = Length in feet = Mud Weight (ppg ) = Nozzle Size ( 32nds of inch ) = Nozzle Area ( square inch ) = Pressure Loss Loss ( psi )

PV VAS VAM YP f n p R e U

= Plastic Viscosity, cps = Velocity, Velocity, Annular Fluid Fluid ( fps ) = Velocity, Annular Fluid ( fpm ) = Yield Point (lbs ( lbs / 100 ft.) = fanning friction factor = Consistency index = Numerical Constant, 3.14159 = Reynolds Number, dimensionless = Viscosity, apparent, effective, cps

∆PAN

= Pressure Drop, Annulus ( psi ) PSI = Pre Press ssur uree, Pou Pound ndss per per Squ Squaare Inc Inch h GPM = Gall Gallons ons per per Minu Minute te FPS = Feet eet pe per Seco Secon nd FPM = Fee Feett pe per Min Minute ute PPG = Pound oundss per per Gal Gallo lon n

other nomenclature found in formulas themselves

H ydraulic Formulas. Bit Related Hydraulic Nozzle Pressure Loss ( psi ) Nozzle Volume ( gpm ) Nozzle Total Flow Area ( sq.in. ) Nozzle Area ( per size ) (sq.in. (sq.in. ) Nozzle Size ( 32nds inch ) Jet Vel ocit y of Nozzl es ( fps ) Impact Force of Nozzles ( psi ) “ “ “ ( psi ) Bit Hydraulic Horse Power, Total Bit Hydraulic Horse Power, per S q.In.

= ( M • G 2 ) ÷ ( 10858 • NA 2 ) G NZ = ( { P NZ • 10858 • NA 2 } ÷ M ) 0.5 NA = ( { M • G 2 } ÷ { 10858 • P NZ } ) 0.5 NA = ( N Z ÷ 32) 2 • ( ¶ ÷ 4) N Z = 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) = P NZ • G ÷ 1713.6 BHHP / sq. in = BHHP ÷ ( BIT OD • 0.7854 ) ∆P NZ

Drill String Bore Pressure Loss ( psi ) Turbulent Turbulent flow ( sii si i ) Turbulent flow ( security ) Turbulent Turbulent flow ( fanning )

∆PID ∆PID ∆PID

= ( 0.0000 0.0000765 765 PV PV 0.18 • M 0.82 • G 1.82 • L ) ÷ ID 4.82 = ( 0.000061 0.0000 61 • M • G 1.86 • L) ÷ L) ÷ ID 4.86 = ( f • M • V ID (F/S) 2 • L ) ÷ 25.8 DP



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

Annulus Pressure Loss ( psi ) Turbulent Flow ( sii ) Turbulent Flow ( security ) Turbulent Flow ( fanning ) Newtonian Lamina Laminarr Flow ( hagan hagan ) Plasti c Laminar Flow ( beck, et c )

∆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 (F/S) 2 • L ) ÷ 25.8 ( DH - DP ) = Apparent Viscosity (cps) = V • L • Va ÷ ( 1500 • {DH - DP } ) Note: V = = ( { L • YP } ÷ 225 { DH - DP } ) + ({ L • Va • PV } ÷ ( 1500 { DH - DP } )

Bottom Hole Pressure

Bot otto tom m Ho Holle Hyd ydro rost staati ticc Pre ress ssur uree ( ps psii ) B. H. Circulating Pressure ( psi ) Equivale len nt Circula lattin ing g Den enssity ( ppg ) “ “ “ ( ppg )

BHP = ( 0. 0.51 5195 95 • M • L ) BHCP = BHP + ∆PAN ECD = BHCP ÷ ( 0.52 • L ) ECD = (∆PAN ÷ { 0.052 • L } ) + Mud Weight

Hole Cleaning

Rock Chip Sli p Velocity

Note: use 21 as cutt ing densi ty and 0.25 as cutt ing diamet er i f unknown

Lam. - Spherical Chips ( Stokes ) Lam . - Flat C hi ps ( Pi got t ) Turb. - Spheri cal Chips ( Ri ttinger ) Turb. - Fl at Chips ( Pigott ) Sl ip Vel ocity ( ft / Min )

V C = ( 8310 C D 2 { CW - M }) ÷ ( PV + ( 399 YP • { D H - DP }) ÷ VA ) V C = ( 3226 C D 2 { CW - M } ) ÷ ( PV + ( 399 YP • { D H - DP }) ÷ V A ) V C = 159 ( ( CD { CW - M } ÷ M ) 0.5 ) V C = 60.6 ( ( CD { CW - M } ÷ M ) 0.5 ) V S = V C - V A



HYDOSTATIC FLUID FORMULAS WEIGHT DENSITY = Weight / Unit Volume SPECIFIC GRAVITY = Weight of Substance per Unit Volume / W eight 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 Density Buoyed Weight = Object Weight x Object Buoyancy Factor POWER TRANSMISSION TRANSMISSION FORMULAS PRESSURE ( PSI ) = FORCE ( LBS ) / AREA ( SQUARE I NCH ) FORCE ( LBS ) = PRESSURE ( PSI ) x AREA ( SQUARE I NCH ) FORCE ( OUTPUT ) = FORCE ( I N ) x ( AREA (OUT ) / AREA ( I N ) LENGTH ( OUTPUT) = LENGTH ( I N ) x (AREA ( I N ) / AREA (OUT ) ) SPEED ( OUTPUT ) = SPEED ( I N ) x (AREA ( I N ) / AREA (OUT ) ) WORK = FORCE x DISTANCE WORK (I (I NCH-LBS) = FORCE ( LBS ) x Travel ( I NCH ) WORK (FOOT-LBS) = FORCE ( LBS ) x Travel ( FEET ) POWER = WORK (F (FTLBS) / TIME (MINUTES OR S SECONDS) HORSE POWER = 3300 FTLBS / 1 MINUTE OR 550 FTLBS / 1 SECOND HYDRAULIC HORSEPOWER = PRESSURE DROP (PSI) x FLOW R ATE ATE (GPM) CIRCULATION RELATED FORMULAS TRIPLEX PUMP (BBLS/STK ) = 3 ( ID2 / 12353 ) x STROKE LENGTH (IN.) DUPLEX PUMP(BBLS/ STK ) = 4 ( ID2 / 12353 ) - 2 ( OD2 / 12353 ) ) x S.L.) VOLUMETRIC-R ATE = VELOCITY ( FPM ) x AREA ATE OF F LOW VELOCITY (FT/MIN) = VOLUMETRIC -R ATE ATE / AREA FLUID FLOW VELOCITY (FT/MIN) = ( 24.51 x GPM ) / DIAMETER 2 FLUID FLOW VELOCITY (FT/MIN) = ( 24.51 x GPM ) / ( DIAMETER 2 x 60 ) Volume Conversions

Cubic

Cubic Feet

1728

7.48052

0. 17811

1

Barrel

9702

42

1

5.61458

Gallon

231

1

0. 02381

0.13368

Gallons

Barrels

Inches

Cubic Feet

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



Effective Viscosity

A. Viscosity Definition

U

B. Bingham Plastic

U

= SS / Sr

D. Effective Viscosity , Power Law

= ( PV + ( 399 YP x ( DH - DP)) / VA ) SS = k x Sr n U e = k x Sr n-1

E. Annular Shear Rate

S r =

F. Consistency Consiste ncy Index

k = 511( YP + PV ) / 511n

G. Power Law Index

n = 3.32 log 10 ( YP + 2PV ) / ( YP + PV )

C. Shear Stress, Power Law Fluids

New Press. (Change: GPM or MW ) Hook Load OverPull Maximum Neutral Point (STRAIGHT (STRAIGHT HOLE) Buoyancy Factor ( GALLONS GALLONS ) BUOYED WEIGHT Natural Frequency Excitation Frequency Excitation Frequency Frequency w/ Shock Mechanical Horsepower Created

2.4 VA / ( DH - DP )

P2 = P1 x ( M2 / M1 ) x ( G2 / G1 ) 2 x P1 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 Weight/Foot * Buoyancy Factor ) BF = ( 65.5 - M ) / 65.5 ( 65.5 lbs = steel in gallons ) BW = AIR WEIGHT x BOUNCY FACTOR F N = 4212 / Drill Collar Length (ft) FE = RPM / 20 Ncrit = FN x 20 F NS = P x ( shock spring rate (k) (k) / Total Wt. DC’s DC’s (w) ) 0.5 HrsPwr(Mech) = Torque x RPM/ 5252



Expansion of St eel: Temp eratur e Changes Changes C

= Coe Coeffic fficie ient nt of Expa xpansio nsion n for for stee steell = 0.0000828” per foot, per degree Fo L = Length in feet T = Change Change in temperature, temperature, Fo ET = C x L x T

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) )

Hydraulics Presentation

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