Drilling Training Course

Drilling Engineering Department

Drilling Training Manual

SAUDI ARAMCO

DRILLING ENGINEERING COURSE

September 2006 SEGMENT

DRILLING

CHAPTER

ROTARY DRILLING BITS

TABLE OF CONTENTS BIT TYPES DRAG BITS DIAMOND BITS

Page 1 1 3

- BIT CONSTRUCTION

3

POLYCRYSTALLINE DIAMOND COMPACT BITS

6

- PDC BIT WHIRL

11

ROLLING CUTTER BITS

13

- BIT DESIGN - CUTTERS - BEARINGS - BIT BODY

13 16 18 20

BIT ENHANCEMENTS

22

STANDARD CLASSIFICATION OF BITS

22

CLASSIFICATION OF FIXED-CUTTER BITS CLASSIFICATION OF ROLLER CONE BITS

22 24

ROCK FAILURE MECHANISM

28

BIT SELECTION

29

TYPES OF BITS USED IN SAUDI ARAMCO

30

DULL BIT GRADING

33

GRADING TOOTH WEAR

33

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TABLE OF CONTENTS FACTORS AFFECTING PENETRATION RATE BIT TYPE FORMATION CHARACTERISTICS DRILLING FLUID PROPERTIES TERMINATING A BIT RUN BIT WEIGHT AND ROTARY SPEED

OPERATING PROCEDURES OF ROLLING CONE BITS SURFACE HANDLING TRIPPING BIT IN THE HOLE

Page 40 40 41 41 43 46

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The purpose of this chapter is to introduce the engineer to

• • • •

the various types of bits criteria for bit selection evaluation of drill bits and optimization of bit weight and rotary speed

The process of drilling a well requires the use of drilling bits. The drilling engineer is responsible for the selection of the best drilling bit for a given situation and the optimization of the bit operating conditions. The performance of drilling bits has a direct impact on the total cost of drilling a well. It is, therefore, important for the drilling engineer to learn the fundamentals of bit design so that he can understand the differences among the various types of bits available.

BIT TYPES Rotary drilling bits are classified according to their design as either drag bits or rolling cutter bits. Drag bits consist of fixed cutter blades that are an integral part of the body of the bit and rotate as an unit with the drill string. Rolling cutter bits have two or more cones containing cutting elements, which rotate about the axis of the cone as the bit is rotated at the bottom of the hole. DRAG BITS The design features of the drag bit include the number, size and shape of the cutting blades or stones, the size and location of the water courses and the metallurgy of the bit and cutting elements. Drag bits drill by plowing cuttings from the bottom of the hole like a farmer’s plow cuts in the soil. There are two types of drag bits; diamond bits and polycrystalline diamond cutter (PDC) bits. An advantage of drag bits over rolling cutter bits is that they do not have any rolling or moving parts, which require strong and clean bearing surfaces. This feature is especially important in drilling small hole sizes, where space is not available for designing strong bit cutter elements and bearings needed for a rolling cutter. Also, since drag bits are made of one solid piece of steel, there is less chance of bit breakage and leaving junk in the hole.

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Fig. 1 Diamond cutter drag bit - design nomenclature

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DIAMOND BITS Diamond drilling bits are quite expensive and may cost three or four times as much as roller cone insert bits. They are used to drill hard and abrasive formations when they offer economic advantage over other types of bits. The most important advantage of a diamond bit is that it drills more hole than any other bit over its rotating life, thus fewer round trips are required. Diamond bits drill at lower penetration rates than other bits. Therefore, to be cost effective a diamond bit must drill at a reasonable rate of penetration; otherwise, the time lost in rotating would cancel out the savings in round trips. Bit Construction: Diamond bits have three major components: • the bit blank • the matrix crown • the shank Refer to the design nomenclature in Fig (1). The crown is made of tungsten carbide powder bonded together with nickel copper alloy binder. The use of the tungsten carbide alloy offers resistance to the corrosion and abrasion caused by the high pressure drops across the bit face, long bit runs and high solid content in the drilling mud. The shape of the crown is determined by the graphite mold in which it is furnaced. During this process, the nickel copper alloy also binds the crown to the steel bit blank. The blank is then machined, threaded and welded to a properly heat treated shank which has been machined for the appropriate API pin connection. The face or crown of the bit consists of many natural diamonds set in the tungsten carbide matrix. The diamonds come in various sizes and grades for a range of applications. Some natural diamonds are mechanically or chemically treated to provide smoother cutting surface for improved wear resistance. Small reclaimed natural diamonds are used for gauge protection. Under proper bit operation only the diamonds contact the bottom of the hole leaving a small clearance between the matrix and the hole bottom. Fluid courses (Fig 2) are provided in the matrix to direct the flow of drilling fluid over the face of the bit. These courses must be small enough so that some of the fluid is forced between the matrix and the hole bottom, thereby cleaning and cooling the diamonds. There are two types of hydraulic flow patterns; the radial flow and the feeder collector shown in Fig (2). The radial flow pattern is used in soft formation bits. The fluid flows from the bit's axis

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toward the shoulder through a series of parallel or expanding fluid courses keeping the cuttings off the bit's face in shale and soft formations. The feeder collector flow pattern is used in hard formation diamond bits where the fluid flows across diamond pads from high pressure feeders into low pressure zones.

Radial Flow

Feeder/Collector

Open Flow

Fig. 2 Natural Diamond Bit Hydraulic Flow Patterns

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Long Taper

Short Taper

Non-Taper

Fig. 3 Diamond Bit Shapes

• • • •

• • • •

Sharp-nosed crown profile Large diamond size Radial flow hydraulics Used in soft formations

Blunt-nosed crown profile Medium diamond size Feeder collector hydraulics Used in hard formations

Fig. 4 Examples of diamond bits used in soft and hard formations

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An important design feature of a diamond bit is its shape or crown profile (Fig 3). A bit with a long taper assists in drilling a straight hole and allows the use of higher bit weights. On the other hand, a short taper is easier to clean because the hydraulic energy is concentrated over a smaller surface area. A more concave bit face is used in directional drilling to assist in increasing the angle of deviation of the borehole from vertical. The size and number of diamonds used in a diamond bit depends on the hardness of the formation to be drilled. Bits for hard formations have many small (0.07 - 0.125 carat) stones, while bits for soft formations have a few large (0.75 - 2 carat) stones. Examples of diamond bits for soft and hard formations are shown in (Fig 4). If the diamonds are too large the unit loading on the diamond points will be excessive, resulting in localized heat generation and polishing of the cutting edge of the stones. The design of the water-course pattern cut in the face of the bit and the junk slots cut in the side of the bit face controls cuttings removal and diamond cooling. Diamond bits are designed to operate at a given flow rate and pressure drop across the face of the bit. Experiments by bit manufacturers indicated the need of 2.0-2.5 bhp/sq. in. of hole bottom with 500-1000 psi pressure drop across the face of the bit to clean and cool the diamond adequately. The pressure drop can be measured as the difference between the pump pressure with the bit off bottom and the pump pressure measured while drilling. The bit manufacturer usually provides the approximate circulating rate required to establish the needed pressure drop across the bit face. POLYCRYSTALLINE DIAMOND COMPACT BITS

Fig. 5 Diamond Compact Structure

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September 2006 SEGMENT CHAPTER

DRILLING ROTARY DRILLING BITS

In the mid-1970's a new type of drag bit had been made possible by the introduction polycrystalline diamond compact (PDC) as a bit cutter element. The compact consists of a thin layer of synthetic diamond about 0.5 mm thick bonded through high-pressure, high-temperature process to a tungsten carbide disc. The diamond layer consists of small diamond crystals which have random orientations for maximum strength and wear resistance. As shown in (Fig 5), the polycrystalline diamond compact is bonded to a tungsten carbide body matrix.

Short-Taper

Non-Taper Fig. 6 PDC Bit Crown Profiles

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Long Taper

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During drilling the diamond compact maintains a sharp cutting edge as it wears. The PDC cutter's self sharpening effect results in long bit life and high rates of penetration. PDC bits are evolving rapidly. They perform best in soft, firm and medium - hard nonabrasive formations that are not "gummy". Bit balling is a serious problem in very soft, gummy formations, and rapid cutter abrasion and breakage are serious problems in hard, abrasive formations. The bit shape or crown profile is an important design feature in PDC bits. There are three basic crown profiles for PDC bits: short taper, long taper and non-taper profiles (Fig 6). Short taper bits have a tapered crown with a relatively blunt nose. This design allows more cutters to be distributed toward the outside of the bit for a more even wear pattern on the cutters. The short taper bit also provides rotational and directional stability while drilling build curves in directional or horizontal wells. The high cutter density enhances bit life at higher rotational speeds, on down hole motors or in large hole sizes. The short parabolic shape provides a relatively small surface area for easy cleaning. Bits with non-taper have the least surface area which minimizes the number of cutters required to provide full coverage. The weight on bit is distributed evenly on the cutters. Because of the reduced surface area, the available hydraulic horsepower is more concentrated which improves hole cleaning. These features provide high penetration rates in soft to medium formations. Non-taper bits are usually used to drill 9" or smaller holes. Long taper bits provide a smooth load distribution across the entire profile of the bit. This reduces the potential for excessive point loading that can occur with other designs. This profile allows increased cutter density toward the shoulder and gauge of the bit, making this style well suited for high RPM drilling using down hole motors. The long taper profile, however, is more vulnerable to damage when hard stringers are encountered.

Fig. 7

Page 8

Nozzle Placement & Orientation

PDC bits use nozzles strategically positioned on the face of the bit to clean the bottom of the hole and cool the PDC cutters. The jetting action of the nozzles and the design of the bit face direct the cuttings toward junk slots located at the outside diameter of the bit (Fig 7). Nozzle placement and orientation depend on the individual bit style.

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d/4 d EXPOSURE

d

BACK RAKE ANGLE (negative)

SIDE RAKE ANGLE

Fig. 8 Cutter orientation expressed in terms of exposure, back rake and side rakes Page 9

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Typically, nozzles are positioned and oriented to clean a group of PDC cutters. In some designs each nozzle can be dedicated to a single PDC cutter. Other important features of a PDC bit include the size, shape and number of cutters used and the angle of attack between the cutter and the surface of the exposed formation. The number of cutters on a bit depends on the formation being drilled. Generally, the greater cutter concentration, the lower is the wear rate and slower the rate of penetration. High number of cutters are usually placed on hard formation bits to reduce the load per cutter and the cutter breakage. Fewer cutters are placed on soft formation bits to reduce chances of bit balling. Cutter orientation is defined in terms of back rake, side rake and chip clearance or cutter exposure (Fig 8). Back rake is the angle that the face of the cutter makes with the vertical. Back rake angle o o can be varied from 0 to 30 to match the drilling mechanics of the formation. Cutters with small or no back rake are suited for soft formations where the aggressive cutter orientation can improve ROP. Cutters with greater back rake are used in harder formations to shear rock more efficiency and resist impact damage. Increasing the back rake makes the bit less responsive to variations in weight on bit. This widens the operational range of the bit and makes it more versatile for directional and horizontal drilling. The side rake angle assists in directing the cuttings formed towards the junk slots and the annulus. The angle can be varied to enhance the bit's cleaning efficiency. Cutter exposure is the distance between the cutting edge and the bit face. The exposure of the cutter provides room for the cuttings to peel off the bottom without impacting against the bit body and packing in front of the cutter. Soft formation PDC bits have full exposure for maximum rate of penetration. In hard formations partial exposure may be desirable to increase cutter durability.

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PDC Bit Whirl: PDC bits have shown excellent performance in soft formations. In harder formations, however, accelerated cutter wear and short bit life severely limited their efficiency. Research conducted during the last few years has shown that cutter wear and damage resulted from the phenomenon known as "bit whirl". Basically, a whirling bit fails to rotate smoothly about its geometric center. Bit whirl occurs when the dynamic forces (vibrations) of the bottom hole drilling assembly on the bit cause its instantaneous center of rotation to move sideways as the bit rotates (Fig 9). The trajectory of the center of a whirling bit is shown in (Fig 10). Lab experiments have shown that whirling bits make many different star-shaped bottom hole patterns as shown in (Fig 11). By comparison a non-whirling bit makes a circular pattern. A whirling bit cuts an overgauge hole and subjects the cutters to very rapid accelerations backwards and sideways resulting in high impact loads. The severe impact loading causes accelerated cutter wear and chipping and lower rates of penetration.

Fig 9 “Bit Whirl” resulting from the failure of the bit to rotate smoothly about its true geometric center Page 11

Eastman Christensen has developed a PDC bit that resists whirl by directing the resultant cutter force vector into one direction thereby pushing the bit against the bore hole wall to provide maximum bit stability (Fig 12). Since the side of the bit in the direction of the resultant force will always be rubbing against the bore hole wall, the gauge of anti-whirl bits is protected with smooth low friction tungsten carbide or diamond pads to prevent eroding the bore hole wall. Anti-whirl bits drill at higher penetration rates, have greater bit life and are capable of penetrating harder formations than conventional PDC bits.

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Anti-whirl

Standard

Fig. 10 Bit Trajectory

Anti-whirl

Standard

Fig. 11 Bottom Hole Pattern

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Fig. 12 Cutter layout and orientation designed to create a net imbalance force to counter bit whirl forces and create a stable rotating condition

ROLLING CUTTER BITS The three-cone rolling cutter bit is the most common bit type currently used in rotary drilling. This bit is available with a large variety of tooth design and bearing types and thus is suited for a wide variety of formation characteristics. Bit Design: Rolling cutter bits consist of three major components: • the cutters • bearings and • the bit body Fig (13) shows a rolling cutter bit with the various parts labeled.

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Cutters: The cutters or teeth, which are placed or machined on the outer surfaces of the cones are the parts of the bit that break the rock. The cones are mounted on bearings which run on pins that are an integral part of the bit body. See (Fig 13). The drilling action of a rolling cutter bit depends to some extent on the offset of the cones. As shown in Fig (14), the offset of the bit is a measure of how much the cones are moved so that their axes do not intersect at a common point in the center of the hole. Offsetting causes the cones to slip as they Fig. 14 Offsetting cone centerlines from rotate and scrape the hole bottom the center of bit rotation to much like a drag bit. This action increase penetration rates in soft tends to increase penetration rate in formations soft formations. In hard formations where the rock must be fractured or broken, scraping contributes little to rock removal. In addition scraping against a hard formation is very abrasive and can wear the teeth down quickly. For these reasons hard formation bits are designed with little or no cone offset. Cone offset angle is expressed as the angle the cone axis would have to be rotated, to make it pass through the center line o of the hole. Cone offset angles vary from 4 for bits used in soft formations to zero for bits used for extremely hard formations. The cutting elements on the bit cones are either milled tooth cutters or tungsten carbide insert cutters. The milled tooth cutters are machined on the cones. The cones are made of forgings of nickel molybdenum alloy steel. The cones are hardened by special processing and heat treating to produce a 0.07” to 0.13” deep hard case on the teeth. All steel tooth cones have tungsten carbide hardfacing material applied to the gage surface. Tungsten carbide hardfacing is applied to the teeth as dictated by the intended use of the bit.

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The tungsten carbide cutting elements are made of sintered tungsten carbide teeth which are pressed into holes drilled in the cone surfaces. The positioning of the teeth on the cones of a roller cone bit is very important. The inner rows of teeth are positioned on the cones so that they intermesh. A relief ring is cut into the surface of one cone to provide space for the tooth rotation of an adjoining cone. This intermeshing allows more room for a stronger bit design, provides self cleaning action and prevents bit balling as the bit turns. The outer rows of teeth, the heel teeth, do not intermesh. These teeth do the hardest job because more rock must be removed from the outer most annular ring of the hole bottom. Because the heel teeth have a more difficult job, they may wear excessively causing the bit to drill an undersized hole. This causes a misalignment of the load on the bearings and premature bit failure. Premature failure of the next bit is likely if the hole remains undersized. Bit manufacturers offer different heel tooth designs to provide the gauge protection needed. Another important feature in the positioning of teeth on bit cones is the pitch. Pitch is the distance between adjacent teeth on a bit cone. If the pitch is the same for all teeth on a given row, then the teeth will impact the formation in the same location on each rotation and prevent the bit from making hole. To prevent this from happening, the pitch between teeth is varied. The shape and height of the teeth on a bit cone has a large effect on the drilling action of a rolling cone bit. Milled tooth bits used for soft formations have offset cones and long, widely spaced teeth. Long teeth give maximum penetration into the formation and generate large cuttings. The scraping action provided by the offset cones removes the drilled material. The wide spacing of the teeth promotes bit cleaning and prevents bit balling. Tooth wear is a problem in soft formation bits because of the scraping action of the offset cones. This problem is minimized by adding tungsten carbide hard facing to the teeth. When tungsten carbide inserts are used, abrasion is not a concern because of the exceptional wear resistance of tungsten carbide. Long and widely spaced insert usually chisel or conical in shape, are used for maximum penetration. In drilling medium hard formations, rolling cutter bits are designed to drill by a combination of crushing and scraping action. Milled tooth breakage becomes a problem because higher bit weights are required. So the teeth are shorter and less pointed. Wide tooth spacing is still required to permit adequate cleaning of the bit. The teeth on insert bits are more closely spaced to reduce load on each tooth. The inserts are more conically shaped and blunter.

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In hard formations the failure mechanism of the rock is primarily, by crushing. Drilling hard formations require heavy bit weights which cause severe bending forces on the teeth. Therefore the bit teeth are short, stubby and closely spaced to minimize breakage. Because there is little scraping action, hardfacing is only applied on the gauge row of teeth. In rolling cutter insert bits the inserts are spherical or elliptical and set deeply into the cone to reduce their tendency to pop out. Bearings: The purpose of the bit bearings is to allow the cones to turn on the pin with minimum friction. The most inexpensive bearing assembly consists of nonsealed roller-type outer bearing and a ball-type bearing. Refer to Fig (15). The roller bearing is the most heavily loaded member and tends to wear out first. The ball bearings carry some of the axial and thrust loads and serve to hold the cone in place on the bit. Since the bearings are neither sealed nor lubricated drilling mud is free to enter into the bearing area and erode the metal of the rollers and races and cause the cones to become Fig. 15 Non-lubricated ball & roller bearings loose. A loose bearing cannot used in the steel tooth of rock bit evenly distribute the load and finally become damaged. Nonsealed roller bearings are adequate to last as long or longer than the cutting structures. In some areas the bearings are not adequate to varying degrees. This type of bearing assembly is used in steel tooth bits for drilling shallow top hole sections. The intermediate cost bearing assembly in rolling cutter bits is the sealed bearing assembly which was introduced in carbide insert bits. In this type of bit the bearings are maintained in a grease environment to minimize wear and prolong life of bearings. In addition to the ball and roller bearing elements, this bearing requires a grease reservoir,

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pressure compensator, connecting passage and seal as shown in Fig (13a). The compensator allows the grease pressure to be maintained equal to the hydrostatic pressure at the bottom of the hole. The seal keeps the grease in place and prevents drilling fluid from entering the bearing area. As the bit wears, the seals eventually fail and allow drilling fluid to enter the bearings and accelerate bearing wear. The carbide tooth cutters still out lasted the lubricated ball and roller bearings. This led to the development of the journal bearing bit. Fig (13a). In this type of bit the roller bearings are removed and the cone rotates in contact with the journal bearing pin. Instead of a series of rollers, the journal bearing consists of two circular bearing surfaces which mate within very close tolerances of each other. A thin layer of grease must separate the two surfaces to prevent galling. This type bearing has the advantage of greatly increasing the contact area through which the weight on the bit is transmitted to the cone. Also, by eliminating one of the components (the rollers), additional space becomes available for strengthening the remaining components. Journal bearing bits require effective grease seals, special metallurgy, and extremely close tolerances during manufacture. Silver inlays in the journal help to minimize friction and prevent galling. While journal bearing bits are much more expensive than the standard or sealed bearing bits, much longer bit runs can be obtained, thus eliminating some of the rig time spent on tripping operations. High rotational speeds in motor drilling shorten the operating life of conventional elastomer seals. Hughes developed the metal-face seal (ATM) which enables the bit to run long hours at high rpm. The metal-face seal withstands high temperatures and provides greater bearing reliability even in abrasive environments.

Fig. 16 Metal-face Seal

Page 19

The metal-face seal consists of two hard metal alloy rings which are suspended between the cone and head bearing shaft of the bit as shown in Fig (16). Two elastomer rings, assembled under compression, position and energize the precision capped metal seal rings to create leak proof face contact. The advantage of the metal-face seal is that all relative rotary motion occurs between the lubricated hard metal surfaces, which greatly reduces friction generated heat; therefore the seal can withstand higher RPMs. The elastomer energizers do not rotate relative to the metal seals so they are not subject to wear. The metal-face seal withstands heat better and is more abrasion resistant than the elostomer seal. Bits with metal-face seal are normally used with downhole motors in drilling directional and horizontal wells.

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Bit body: Bit bodies consist of the threaded connection which attaches the bit to the drill stem, the bearing pins on which cones are mounted, the lubricant reservoirs which contain the lubricant supply for the bearings and the watercourses through which the drilling fluid flows to clean the cuttings from the hole.

Fig. 17 Nozzles in jet-type bits delivering high velocity streams of drilling fluid against the hole bottom

Page 20

One of the purposes of the bit body is to direct the drilling fluid where it will do the most effective job of cleaning. Most present-day drill bits are of the jet type which aim the fluid between cutters directly to the hole bottom as shown in Fig (17). Modern pumps provide adequate power to clean hole bottom and the cutters. In some softer formations the jets will remove material by their own forces. Fluid erosion in the bit body from high velocities is held to a minimum by use of tungsten carbide nozzles shown in Fig (17).

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Fig. 18 G/D Gage Enhancement

Fig. 20 Leg Inset Protection

Page 21

Fig. 19 Motor Hardfacing

Fig. 21 Wear / Stabilization Pads

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BIT ENHANCEMENTS Bit enhancements are made to increase the operating life of the rolling cone bit. Fig (18) shows an insert bit with gage enhancement. Ovoid shaped inserts are placed on the heel row and carbide button inserts on the gage row. The inserts are made from extremely wear resistant tungsten carbide grade and protect the gage of the bit from abrasive wear. In high-speed directional or abrasive applications, tungsten carbide particle hard facing and/or flat tungsten carbide inserts are applied along the shirtail to protect against excessive gage wear as shown in Figs (19) and (20). Wear pads can also be added to the outer diameter of a rolling cone bit to minimize wear on the bit leg and body. Flat tungsten carbide inserts are pressed into the pad to provide a wear resistant surface as shown in Fig (21).

STANDARD CLASSIFICATION OF BITS There is a large variety of bits available from several manufactures. The International Association of Drilling Contractors (IADC) approved the standard three-digit code classification system for identifying similar bit types from various manufacturers. CLASSIFICATION OF FIXED-CUTTER BITS The classification system for fixed cutter bits (PDC & diamond bits) consists of four characters. The first character describes the body material and it is either M for matrix body or S for steel body construction. The second digit designates the cutter density and ranges from 1 for soft formations to 4 for hard formation PDC bits, and from 6 to 8 for diamond bits. Numerals 0, 5 and 9 are reserved for future use. For PDC bits, 1 refers to 30 or fewer 1/2” cutters, 2 refers to 30 to 40 cutters, 3 indicates 40 to 50 cutters; and 4 refers to 50 or more cutters. For diamond bits, the number 6 represents diamond sizes larger than 3 stones per carat; 7 represents 3 stones to 7 stones per carat; and 8 represents sizes smaller than 7 stones per carat. Thus the diamond size becomes smaller as the digit increases from 6 (soft formation) to 8 (hard formation).

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Fishtail

Short

Fig. 22

Page 23

Long

Medium

Industry-Accepted Bit Profile Description

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The third digit designates the size or type of cutter. For PDC bits 1 indicates cutter larger than 24 mm in diameter, 2 represent 14-24 mm, 3 indicates 13.3 mm (1/2”), and 4 is used for the smaller 8 mm in diameter. For diamond bits the third digit represents diamond type, with 1 indicating natural diamonds; 2 to TSP material; 3 represents mixed natural diamonds and TSP, and 4 applies only to the highest density bit, indicating an impregnated diamond bit. The fourth digit gives the basic description of bit’s profile. The number 1 represents both fishtail PDC bit and ‘flat’ TSP and natural diamond bits. Numbers 2, 3 and 4 indicate increasingly longer bit profiles as shown in Fig (22). Classification systems for PDC and diamond bits are shown in Tables (1) and (2). CLASSIFICATION OF ROLLER CONE BITS The classification of roller cone bits consists of 4 characters, the first 3 characters are numeric and the fourth character is alphabetic. The first character designates the cutting structure series which describe the general formation characteristics. There are 8 series; the first 3 series refer to milled tooth bits; series 4 to 8 refer to insert bits. The formations become harder and more abrasive as the series number increases. The second character refers to cutting structure type. Each series is divided into 4 types or degrees of hardness. Type 1 for the softest formation in the series and type 4 for the hardest formation. The third character describes the bearing and gage protection. There are 7 categories of bearing and gage protection design as shown in Table (3). Categories 8 and 9 are reserved for future use. The fourth character is optional and describes special features available. There are 16 alphabetic characters as shown in Table (3).

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September 2006 CHAPTER:

DRILLING

SECTION:


BODY STYLE

CUTTERS DENSITY

SIZE

1 FISHTAIL EC

DBS

HYC

2 SHORT STC

R522(M) R573(M) R523(M)

SEC

EC

DBS

HYC

> 24

2

14-24

PD12(S)

DS40(S) S95(S) DS33(S)

B933(M)

3

< 14

R423(M) PD10(S) AR423(M) PD11(S)

S96(S) S93(M)

B923(M)

1

> 24

R525(M)

2

14-24

R526(M)

TD19L(M)

B925(M)

3

< 14

R426(M) Z426(M)

TD2A1(M) DS39(M) S93(M)

B935(M)

1

> 24

2

14-24

TD19M(M)

3

< 14

TD5A1(M)

1

> 24

2

14-24

3

< 14

(S) = STEEL BODY

3 MEDIUM STC

B943(M)

1

1

(M) = MATRIX BODY

SEC

EC

DBS

HYC

4 LONG STC

SEC

EC

DBS

HYC

STC

SEC

B17-4(M)

DS30(S)

S25(S)

B254(M)

R516(M)

B272(M)

Z528(M)

DS34(S)

2 R482(M)

PD1(S)

DS48(S) S10(S) HZ232(M) B2S(M)

LX201(M) DS26(S) S45(S) LX101(M) DS31(S) R535(M)

R535S(M) PD4(S)

3 B927(M)

TD19H(M)

AR435(M) TD268(M) DS23(S) TD260(M) DS49(M)

MX42(M)

R435(M)

PD5(S)

PD2(S)

S85(S) S43(S)

PD4HS(S)

4 TD290(M)

Table (1)

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HZ352(M) B352(M)

R437(M) Z437(M)

LX401(M) D247(M) S35(M) S292(S) LX301(M)

REVISED CLASSIFICATION SYSTEM TABLE (PDC)

R419(M) LX271(M) DS18(M) R429(M) TD115(M) DS19(M) Z429(M) LX291(M) DS20(M)

B102(M) B362(M)

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SECTION:


CUTTERS SIZE

6

BODY STYLE

ELEMENT

1 FLAT EC

1

NAT

2

TSP

3

COMB

1

NAT

D411

2

TSP

SST

3

COMB

1

NAT

2

TSP

3

COMB

4

IMP

DBS

HYC

2 SHORT STC

SEC

EC

DBS

HYC

3 MEDIUM STC

SEC

S725

EC

DBS

HYC

D262 D311

TB16

901 932

S225

TT16

211 241

TBT16

211ND 241ND

D262 D331 D311

TB601

S248 S226

4 LONG STC

SEC

EC

N37

D18

901 730 753 744

N39 N50

T51 T54

TT601

243 223

P341 P343

TBT601

243ND 223ND

DBS

HYC

N42

< 3 SPC

TB26

828

N4S

D41

TB521

7 828TSP

TT521

263

TBT521

263ND

P443

3 - 7 SPC

D24

525 585

N60

8 > 7 SPC

S279

TB5211

Table (2)

REVISED CLASSIFICATION SYSTEM TABLE (TSP / ND)

IADC CLASSIFICATION CHART FOR TSP & NATURAL DIAMONDS

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STC

TB593 TB703

TT593

TBT593 TBT703

901DT

SEC

SAUDI ARAMCO



DRILLING

SECTION:


Manufacturer :

Date:

FORMATIONS

S E R I E S

STEEL TOOTH

1

AND HIGH DRILLABILITY

2

HIGH COMPRESSIVE STRENGTH

3

FORMATIONS

4

SOFT TO MEDIUM

5

LOW COMPRESSIVE STRENGTH

BITS

6

EXTREMELY HARD

1) 2) 3)

…

2 3

A -

AIR APPLICATION

4

B -

SPECIAL BEARING SEAL

2

C -

CENTER JET

3

D -

DEVIATION CONTROL

E -

EXTENDED JETS (FULL LENGTH)

G -

GAGE / BODY PROTECTION ADDITIONAL

H -

HORIZONTAL / STEERING APPLICATION

4

J -

JET DEFLECTION

1

L -

LUG PADS

M -

MOTOR APPLICATION

S -

STANDARD STEEL TOOTH MODEL

2 3

2 3

2 3

2 3

8

2

T -

TWO CONE BIT

W -

ENHANCED CUTTING STRUCTURE

X -

PREDOMINANTLY CHISEL TOOTH INSERT

3 4

Y -

CONICAL TOOTH INSERT

1

Z -

OTHER SHAPE INSERT

2 3 4 th

Several features may be available on any particular bit – 4 character should describe predominant feature. All bit types are classified by relative hardness only and will drill effectively in other formations. Please check with the specific bit supplier for additional information.

Table (3) Page 27

AVAILABLE

1

7

FORMATIONS

FORMATIONS

≈

4

HARD SEMI-ABRASIVE

AND ABRASIVE

≡

FEATURES

1

HIGH COMPRESSIVE STRENGTH

AND ABRASIVE

≠

SEALED FRICTION BEARING GAGE PROTECTED

4

MEDIUM HARD FORMATIONS WITH

÷

SEALED FRICTION BEARING

1

AND HIGH DRILLABILITY

INSERT

•

SEALED ROLLER BEARING GAGE PROTECTED

4

SOFT FORMATIONS WITH

FORMATIONS WITH

SEALED ROLLER BEARING

1

HARD SEMI-ABRASIVE

LOW COMPRESSIVE STRENGTH

ROLLER BEARING GAGE PROTECTED

4

BITS AND ABRASIVE

∂

ROLLER BEARING AIR COOLED

1

MEDIUM TO MEDIUM HARD FORMATIONS WITH

STANDARD ROLLER BEARING

1

SOFT FORMATIONS WITH LOW COMPRESSIVE STRENGTH

BEARING / GAGE

T Y P E S

IADC CLASSIFICATION CHART

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ROCK FAILURE MECHANISM Drilling bits cut rock by five basic mechanisms

• • • • •

wedging scraping and grinding erosion by fluid jet action crushing torsion or twisting

While one mechanism may be dominant for a given bit design, more than one mechanism is usually present. PDC bits cut rock primarily by wedging, Fig (23), which is more efficient than crushing and grinding. In the right formations, the wedging action makes it possible for PDC bits to maintain high rates of penetration with less weight on bit. This reduces wear on bit while yielding high ROP. Fig. 23

PDC bits shear rock

Natural diamond bits are designed to drill with very small penetration into the formation. The diameter of sandstone grains may not be much smaller than the depth of penetration of the diamonds. The drilling action of diamond bits is primarily a grinding action in which the cementing material holding the individual sand grains together is broken by the diamonds Fig (24). Fig. 24 Natural diamond bits cut by indenting, plowing and grinding

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Fig. 25 Roller cone bits drill by complex modes that crush the rock

Rolling cutter bits designed with small or zero offset angle for drilling hard formations employ the crushing mechanism for rock removal Fig (25). This cutting action is inefficient and requires high weight on bit to deliver enough energy to the formation to achieve fast penetration. The drilling action of rolling cone bits designed with a large offset angle for drilling soft formations is more complex than the simple crushing action. Since each cone alternately rolls and drags, considerable wedging and twisting action is present in addition to the crushing action.

BIT SELECTION Unfortunately, the selection of the best bit to drill a given formation is made by trial and error. The criterion used for selecting a bit is based on the drilling cost per unit interval. The best bit for drilling a given formation is the bit that would give the least drilling cost. The drilling cost per foot. C is, C =

where, C: Cb: Cr : Tb: Tc: Tt: L:

Cb + C r ( T b + T c + T t ) .............………………………..............(1) L

drilling cost, $/ft cost of bit, $ rig cost, $/hr rotating time, hrs non-rotating time (connection time), hrs tripping time to change bit, hrs interval drilled, ft

The initial selection of bit type can be made on the basis of formation drillability and abrasiveness. Drillability is a measure of how easy the formation is drilled and it is inversely related to the compression strength of the rock. The abrasiveness of the formation is a measure of how rapidly the teeth of a milled tooth bit wear when drilling

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the formation. Normally, abrasiveness increases as the drillability decreases. In the absence of prior bit records, Table (3) can be used as a guide for initial bit selection. The drilling cost per foot is the final criterion that should be used for selecting a bit. Some of the rules of thumb used for bit selection are: 1. High-cost bits tend to be more applicable when the daily rig operation cost is high. 2. Rolling cone bits are the most versatile bits and are good initial choice. 3. When using a rolling cone bit: a) Use the longest tooth suitable for the application. b) When the rate of tooth wear is much less than the rate of bearing wear, select a longer tooth size, a better bearing design, or apply more bit weight. c) When the rate of bearing wear is much less than the rate of tooth wear, select a shorter tooth size, a more economical bearing design, or apply less bit weight. 4. Diamond drag bits perform best in nonbrittle formations having a plastic mode of failure, especially in the bottom portion of a deep well, where the high cost of tripping operations favors a long bit life, and a small hole size favors the simplicity of a drag bit design. 5. PDC drag bits perform best in uniform sections of carbonates or evaporites that are not broken up with hard shale stringers or other brittle rock types. 6. PDC drag bits should not be used in gummy formations, which have a strong tendency to stick to the bit cutters.

TYPES OF BITS USED BY SAUDI ARAMCO Most of the bits used by Saudi Aramco are roller cone type bits. The milled tooth bits are normally used to drill the shallow aquifers of Neogene, Khobar, Alat and UER. The deeper and harder formations are drilled by using the roller cone insert bits. The type of bit is selected based on the cost-per-foot criterion. New bits are purchased and used to drill the desired formation. The performance of the bit is evaluated on a cost-per-foot basis and compared with that of the existing bits used to drill the same formation. If the new bit is more cost effective it will replace the old bit and will be included in the SAMS (Saudi Aramco Material Supply System). The types of bits that are used to drill the varoius formations in Saudi Aramco are shown in Table (4) Diamond bits are used only in special situations to drill extremely hard and abrasive formations in deep exploratory wells. Diamond bits are not in the SAMS and are acquired when the need arises. PDC bits are being developed for use in horizontal and highly deviated wells. Since PDC bits have no moving parts they will last longer than the

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Table (4)

BIT TYPES USED IN DRILLING SAUDI ARAMCO FORMATIONS

FORMATION (MEMBER / RESERVOIR)

LITHOLOGY

BIT TYPE (IADC CODE)

DAMMAM

DOLOMITE & LIMESTONE

1-1-1 or 1-3-1

RUS

ANHYDRITE

1-1-1 or 1-3-1

UMM ER RADHUMA (UER)

DOLOMITIC LIMESTONE

1-1-1 or 1-3-1

ARUMA

LIMESTONE

1-1-1, 1-3-1 or 1-3-4

WASIA (MiSHRIF, RUMAILA, AHMADI, WARA & MAUDDUD)

LIMESTONE

5-1-7 or 5-3-7

SANDSTONE & SHALE

1-3-1, 5-1-7 or 5-3-7

SHU’AIBA

LIMESTONE & CALCARENITE

5-1-7 or 5-3-7

BIYADH

SANDSTONE & SHALE

5-1-7 or 5-3-7

BUWAIB

LIMESTONE & SHALE

5-1-7 or 5-3-7

YAMAMA

LIMESTONE

5-1-7 or 5-3-7

SULAIY

CALCARENITE & LIMESTONE

5-1-7 or 5-3-7

HITH

ANHYDRITE

5-1-7 or 5-3-7

ARAB-A, B, C & D

CALCARENITE & ANHYDRITE

5-1-7 or 5-3-7

JUBAILA


5-3-5, 5-3-7 or 6-1-7

HANIFA


5-3-5, 5-3-7 or 6-1-7

TUWAIQ, DHRUMA & MARRAT

LIMESTONE

5-3-5, 5-3-7 or 6-1-7 or PDC bit

MINJUR

SANDSTONE & SHALE

5-3-5, 5-3-7 or 6-1-7 or PDC bit

JILH

DOLOMITE & ANHYDRITE

5-3-5, 5-3-7 or 6-1-7 or PDC bit

SUDAIR

SHALE & SILTSTONE

5-3-5, 5-3-7 or 6-1-7 or PDC bit

KHUFF

DOLOMITE

6-1-7, 6-3-7 or PDC bit

UNAYZAH, BERWATH & JUBAH

SANDSTONE & SHALE

6-1-7, 6-3-7 or 7-3-7

JAUF

SANDSTONE & SILTSTONE

6-1-7, 6-3-7 or 7-3-7

TAWIL

SANDSTONE

6-1-7, 6-3-7 or 7-3-7

QALIBAH

SHALE & SILTSTONE

6-1-7, 6-3-7 or 7-3-7

SARAH

SANDSTONE, CONGLOMERATE & GRAVEL

6-1-7, 6-3-7 or 7-3-7

(SAFANIYA & KHAFJI)

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roller cone bits especially when run at high rpms with a downhole motor. One disadvantage of PDC bits is that they cannot be steered easily when drilling the curved sections of a deviated well in the sliding mode. The bit, if it is not properly designed, tends to take a large bite into the formation and stalls the bottom hole motor. When this happens, the bit is pulled out and replaced with a roller cone bit. For this reason, a considerable amount of time and effort is required to select the most suitable PDC bit for a particular formation. Example:

Rig PA-235 used a Smith 8-1/2” MF-2 rolling-cone insert bit type code 5-1-7 with a motor to drill the curved section across the Hith and Arab-A formations in HRDH-308. The bit drilled from 6987’ to 7582’ in 61.5 hours. The daily rig operating rate is $11,000 and the Saudi Aramco overhead is $6500 per day. Assuming a bit price of $5800 and tripping time of 1 hr per 1000 ft (round trip) what is the drilling cost per foot ? Solution:

Cb

=

$5800.

Tt

=

7582 1000

= 7.58 hrs

Tb + Tc =

61.5 hrs.

Cr

=

11,000 + 6500 24

L

=

7582 – 6987 = 595 ft.

From Eq (1) C

=

= $729 / hr

5800 + 729( 615 . + 7.58) 595

= $ 94.4 per foot

Problem:

Rig P-235 conducted a trial test to evaluate the performance of an 8-1/2” Reed EHP53A bit type code 5-3-7 in drilling the curved section across the Hith and the Arab-A

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formations in HRDH-314 which is located near HRDH-308. The regular price of the bit is $7000, however, since this was a trial test Saudi Aramco was able to purchase the bit at 50% discount. The bit drilled the interval from 6767’ to 7612’ in 67.5 hrs. Which bit should be used to drill the Hith and Arab-A in Haradh, MF-2 or EHP53A?

DULL BIT GRADING Since bit selection is made mainly by trial and error, evaluation of dull bits removed from the well is essential. Accurate records of bit performance should be kept in well files for future reference. The IADC Grading System is used to evaluate dull bits. Eight columns of information are used for reporting the condition of dull bits, TEETH

BEARING

CUTTING STRUCTURE Inner Rows (I)

Outer Rows (O)

Dull Char. (D)

Locatio n (L)

GAGE

REMARKS

B

G

Bearing Seal (B)

Gage 1/16 (G)

REMARKS Other Dull (O)

Reason Pulled (R)

GRADING TOOTH WEAR

Columns 1 and 2 are used to report the condition of the teeth on the inner and outer rows. The tooth wear of milled tooth bits is graded in terms of the fractional tooth height that has been worn away and it is reported to the nearest eighth. For example, if half the tooth height is worn then the bit is graded as T-4. If some teeth are worn more than others then the average wear of the row of teeth with the most severe wear is reported. The best way to obtain tooth wear is by measuring the tooth height before and after the bit is run in the hole. The first column is used to report the wear of the teeth on the inner rows that are not touching the wall of the hole. Column 2 is used to report the condition of the teeth on the outer row which touch the wall of the hole. The tooth wear of insert bits is a measure of the combined cutting structure reduction due to lost, worn and broken inserts. For example, if 50% of the inner inserts are broken and/or lost and the remaining inner inserts have no reduction in height, the bit is graded T-4 in column 1. If 50% of the outer inserts are lost and the remaining inserts on the outer rows have 50% reduction in height, the bit is graded T-6 in column 2. Page 33

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In column 3 a two-letter code is used to indicate the major dull characteristic of the cutting structure. Table (5) lists the two-letter codes to be used in this column. Table (5) DULL CHARACTERISTICS * BC BF BT BU * CC *CD CI CR CT ER FC HC JD * LC

-

Broken Cone Bond Failure Broken Teeth / Cutters Balled Up Bit Cracked Cone Cone Dragged Cone interference Cored Chipped Teeth / Cutters Erosion Flat Crested Wear Heat Checking Junk Damage Lost Cone

LN LT OC PB PN RG RO SD SS TR WO WT NO

-

Lost Nozzle Lost Teeth / Cutters Off Center Wear Pinched Bit Plugged Nozzle / Flow Passage Rounded Gage Ring Out Shirttail / Damage Self-Sharpening Wear Tracking Washed Out Bit Worn Teeth / Cutters No Dull Characteristic

* Show Cone # or #’s under location 4.

Some of the dull characteristics listed in Table (5) are self explanatory. More detailed description of the two-letter codes can be obtained from bit companies. Examples of some of the dull characteristics are shown in Figs (26) through (29).

A balled bit will show tooth wear. This is caused by cone(s) being unable to rotate because of formation cuttings being packed between cones.

Fig 26.

Page 34

Balled Bit (BU)

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The erosion is caused by abrasive cuttings in mud travelling at a high rate from right to left. The eddy effect caused cuttings to remove cone shell on the right side of the inserts.

Fig. 27

Erosion (ER)

Flat crested wear is an even reduction in height across the entire face of the cutters. It is often caused by reducing weight and increasing rpm.

Fig. 28 Flat Crested Wear (FC)

Heat checking occurs when a cutter overheats by being dragged on the formation and is later cooled by drilling fluid over many cycles. It can also occur when reaming undergage hole at high rpm using a motor.

Fig. 29 Heat Checking (HC) Page 35

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In column 4 a letter or number code is used to indicate the location on the face of the bit where the cutting structure drilling characteristic occurs. Table (6) lists the codes to be used for describing locations on roller cone bits. TABLE (6) LOCATION (ROLLER CONE BITS) N M G A

-

Nose Row Middle Row Gage Row All Rows

CONE # 1 2 3

Location is defined as follows: Gage Those cutting elements which touch the hole wall. Nose The centermost cutting element(s) of the bit. Middle Cutting elements between the nose and the gage All All Rows Cone numbers are identified as follows: • The number one cone contains the centermost cutting element. • Cones two and three follow in a clockwise orientation as viewed looking down at the cutting structure with the bit sitting on the pin. Column 5 (B-Bearing/Seals) uses a letter or a number code, depending on bearing types to indicate bearing condition of roller cone bits. For non-sealed bearing roller cone bits, a linear scale from 0-8 is used to indicate the amount of bearing life that has been used. A zero (0) indicates that no bearing life has been used (a new bearing) and an 8 indicates that all of the bearing life has been used (locked or lost). For sealed bearing (journal or roller) bits, a letter code is used to indicate the condition of the seal. An “E” indicates an effective seal, an “F” indicates a failed seal(s), and an “N” indicating “not able to grade” has been added to allow reporting when seal/bearing condition cannot be determined. Column 6 (G-Gage) is used to report on the gage of the bit. The letter “I” (IN) indicates no gage reduction. If the bit does have a reduction in gage it is to be recorded in 1/16th of an inch. The “Two Thirds Rule” is correct for three-cone bits.

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The Two Thirds Rule, as used for three-cone bits, requires that the gage ring be pulled so that it contacts two of the cones at their outermost points. Then the distance between the outermost point of the third cone and the gage ring is multiplied by 2/3’s and rounded to the nearest 1/16th of an inch to give the correct diameter reduction.

“TWO THIRDS RULE”

Column 7 (O-Other Dull Characteristics) is used to report any dulling characteristic of the bit, in addition to the cutting structure dulling characteristic listed in column 3 (D). Note that this column is not restricted to only cutting structure dulling characteristics. Table (5) lists the two-letter codes to be used in this column. Column 8 (R-Reason Pulled) is used to report the reason for terminating the bit run. Table (7) lists the two-letter or three-letter codes to be used in this column.

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ROTARY DRILLING BITS TABLE (7) REASON PULLED OR RUN TERMINATED BHA CM CP DMF DP DSF DST DTF FM HP HR LIH LOG PP PR RIG TD TQ TW WC

Page 38

-

Change Bottom Hole Assembly. Condition Mud Core Point Dwonhole Motor Failure Drill Plug Drill String Failure Drill Stem Testing Downhole Tool Failure Formation Change Hole Problems Hours on Bit Left in hole Run Logs Pump Pressure Penetration Rate Rig Repair Total Depth / Casing depth Torque Twist Off Weather Conditions

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ROTARY DRILLING BITS Example:

Describe the bit in Fig (30). The bit was pulled out of the hole because of low penetration rate. The seals were found effective and the bit in gauge. Solution:

Fig. 30

Fig. 31

Page 39

The bit looks to have been dulled by encountering a harder formation than the bit was designed for. This is indicated by the heavy tooth breakage on the inner rows, and by the fact the bit was pulled for penetration rate. The reduced penetration rate was caused by tooth breakage occurring when the bit encountered the hard formation. Excessive weight on bit also could cause the dull to have this appearance. Because of the heavy tooth breakage in the inner rows, the grade for column 1 would be 7. The inserts on the outer row have little tooth wear but none are broken, Therefore, the grade for column 2 is 1. Since the dull characteristics of the cutting structure are broken teeth, a ‘BT’ code is used in column 3. An ‘M’ code is used in column 4 because the broken teeth are on the middle rows. Since the seals are effective a letter ‘E’ is used in column 5. A letter ‘I’ is used in column 6 to indicate

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that the bit was in gauge. Another characteristic of the cutting structure in addition to broken teeth is the worn teeth on the outer row. Therefore, a ‘WT’ code is entered in column 7. Finally, since the bit was pulled out because of low ROP, a PR code is entered in column 8. So the bit description is 7, 1, BT, M, E, I, WT, RP. Problem:

Describe the dull bit shown in Fig (31). When pulled the bit was 2/16” under gauge and the seals were still in satisfactory condition. The bit was pulled when a specified number of hours had elapsed.

FACTORS AFFECTING PENETRATION RATE The penetration rate that can be achieved by a bit has an inverse effect on the drilling cost per foot. The main factors that affect penetration rate are • • • • •

bit type formation properties drilling fluid properties bit weight and rotary speed bit hydraulics

BIT TYPE

Penetration rate is highest when using bits with long teeth and large cone offset angle. These bits are practical only in soft formations. The lowest cost per foot drilled usually is obtained by using the longest teeth that are consistent with bearing life at optimum bit operating conditions. Drag bits such as diamond and PDC bits are designed to obtain a given penetration rate by the selection of the number and size of the diamonds or the PDC blanks. The penetration rate of PDC bits also depends on the back rake angle and the exposure of the blanks.

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FORMATION CHARACTERISTICS

The ultimate compressive strength of the rock is the most important rock property that affect penetration rate. The higher the compressive strength the lower the penetration rate. The mineral composition of the rock has some effect on penetration rate. Rocks that contain abrasive minerals can cause rapid dulling of the bit teeth. Rocks that contain gummy clays can cause the bit to ball up and drill inefficiently. The permeability of rock also affects penetration rate. In permeable rock, the drilling fluid filtrate can move into the rock ahead of the bit and equalize the pressure differential acting on the chips formed beneath each tooth and increase the penetration rate. DRILLING FLUID PROPERTIES

Drilling fluid properties that affect penetration rate are • • • •

Fig. 32 The effect of differential pressure on penetration rate in Berea sandstone

Page 41

density rheological properties filtration properties and solids content and size distribution

An increase in the drilling fluid density would decrease the penetration rate. Increasing the drilling fluid density would cause an increase in the bottom hole hydrostatic pressure beneath the bit and thus an increase in the differential pressure (overbalance) between the bore hole and the formation pore pressure. An increase in the differential pressure would increase the strength of the rock and, therefore, decrease the penetration. Also, a higher differential pressure prevents the ejection of the crushed fragments of rock formed beneath the teeth of the bit resulting in a lower penetration rate. The effect of differential pressure on penetration rate in Berea sandstone is shown Fig. (32). Note that the effect of overbalance on penetration rate is more pronounced at low values of overbalance than at high values of overbalance. If overbalance is quite large, additional increase in overbalance has little effect on penetration rate.

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Penetration rate tends to decrease with increasing viscosity and solids content and tends to increase with increasing filtration rate. Increasing the viscosity increases the frictional losses in the drillstring and thus decrease the hydraulic energy available at the bit for cleaning the bottom of the hole. The solid content of the mud controls the pressure differential across the zone of crushed rock beneath the bit. Increasing the solid content decreases the filtration rate and therefore increases the differential pressure. As mentioned earlier, an increase in the differential pressure results in a decrease in the penetration rate.

Fig. 33

Exponential relation between penetration rate and overbalance for rolling cutter bits

Many studies have been conducted on the effect of differential pressure (overbalance) on the penetration rate. Bourgoyne and Young observed that the relation between differential pressure and penetration rate can be represented by a straight line on a semi log paper as shown in Fig (33). The equation for the straight line is given by log ( where, Ro = R = Pbh = Pf = Page 42

R ) = 0.000666 (Pbh - Pf) .........................….…............(2) Ro

penetration rate at zero overbalance, ft/hr penetration rate, ft/hr bottom hole pressure, psi Formation pore pressure, psi

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Equation (2) can be expressed in a different form

where, ρ1 = R1 = D =

R2 R1

= e 80.6 × 10

−6

D ( ρ 1− ρ 2 )

…..............................….........….........(3)

equivalent circulating mud density, lb/gal penetration rate ft/hr for equivalent circulating mud density ρ1 depth, ft.

Example:

A 12000 ft deep well is being drilled at a penetration rate of 20 ft/hr using 12 lb/gal mud. Estimate the penetration rate if the mud density is increased to 13 lb/gal. Solution:

R1 = 20 ft/hr ρ 1 = 12 lb/gal ρ 2 = 13 lb/gal

From Eq (3), R 2 = R1 e80.6 ×10

−6

= 20 e 80.6 × 10 = 7.6 ft/hr

D ( ρ 1− ρ 2 )

−6

× 12000 (12−13)

Problem

The Jilh formation in HRDH-604 was drilled at a rate of penetration of 4.9 ft/hr using 105 pcf mud. The mud weight was decreased to 102 pcf at 9383 ft and drilling continued to 9538 ft. How long did it take to drill the interval 9383-9538 ft? Terminating A Bit Run

The decision to terminate a bit run is not always simple in drilling operations. A bit should be pulled out if the bearings are worn out or the cutters have worn out to the point that it is no longer economical to continue drilling with the bit. Badly worn bearings can be detected by monitoring the rotary table torque. When bearings are worn, one or more Page 43

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of the bit cones will lock and cause a sudden increase in the rotary torque needed to rotate the bit. When this happens the bit should be pulled out. When the penetration rate decreases rapidly as bit wear progresses, it may be advisable to pull the bit before it is completely worn. If the lithology is somewhat uniform, the total cost can be minimized by minimizing the cost of each bit run. In this case, the best time to terminate the bit run can be determined by keeping a current estimate of the cost-per-foot for the bit run, assuming that the bit would be pulled at the current depth. Even if significant bit life remains, the bit should be pulled when the computed cost-per-foot begins to increase. However, if the lithology is not uniform, this procedure will not always result in the minimum total well cost. In this case, an effective criterion for determining optimum bit life can be established only after enough wells are drilled in the area to define the lithologic variations. Example:

Determine the optimum bit life for the bit run described in the table below. The lithology is known to be essentially uniform in this area. The bit cost is $800, the rig cost is $600/hr, and the trip time is 10 hrs. Cumulative Footage Drilled ft

Cumulative Drilling Time (Tb + Tc) hrs

0 30 50 65 77 87 96 104 111

0 2 4 6 8 10 12 14 16

The cost per foot is calculated by using Eq (1). For the first interval the cost per foot is C =

Page 44

800 + 600 (2 + 10) = $ 266.66 / ft 30

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The cost per foot that would result if the bit were pulled at the various depths is as follows, Footage ft

Drilling Time, hrs

0 30 50 65 77 87 96 104 111

Drilling Cost $ / ft

0 2 4 6 8 10 12 14 16

0.00 266.66 184.00 160.00 150.65 147.12 145.83 146.15 147.75

The cost per foot is plotted versus the footage in Fig (34). Note that the cost per foot during the first 12 hours decreased from $266.66/ft to $145.83/ft. After 12 hrs the cost/ft increased to $147.75/ft. Therefore, the optimum time to pull out the bit is after 12 hours.

Drilling Cost ($/ft)

280 260 240 220 200 180 160 140 120 0

20

40

60

80

100

120

Footage (ft)

Fig. 34 The cost per foot when the bit is pulled at various depths

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Problem:

A 17” ATX-11 H bit was used to drill the Jilh in HWYH-200. Given the data below, determine the optimum time to terminate the bit run. Assume $24,500 bit cost, $790/hr rig cost and tripping speed of 2000 ft/hr. Bit Depth ft 8859 8900 8953 8990 9040 9090 9140 9200 9260 9340 9387 9430 9500 9550 9573

Time of Day, hr. 1300 2050 0700 1340 2110 0425 1125 1910 0240 1134 1950 0435 1720 0214 0600

Date 09-29-95 09-30-95

10-01-95

10-02-95

10-03-95 10-04-95

BIT WEIGHT AND ROTARY SPEED

In addition to selecting the best bit for the job, the drilling engineer is responsible for selecting the optimum weight on bit and rotary speed that will give the least cost drilling operation. The effect of bit weight and rotary speed on penetration rate has been studied extensively in the lab and in the field. A plot of penetration rate versus weight on bit at constant rotary speed has the typical shape shown in Fig (35). No significant penetration rate is obtained until the threshold bit weight is applied (Point a). Penetration rate then increases rapidly with increasing values of bit weight for moderate values of bit weight (Segment ab). A linear curve is often observed at moderate bit weights (Segment bc). However, at higher values of bit weight, subsequent increase in bit weight causes only slight improvements in penetration rate (Segment cd). In some cases, a decrease in penetration rate is observed at extremely high values of bit weight (Segment de). This type of behavior often is called bit floundering. The poor response of penetration rate at high values of bit weight usually is attributed to less efficient bottomhole cleaning at higher rates of cuttings generation or to a complete penetration of the cutting element into the hole bottom.

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Fig. 35 Typical response of penetration rate to increasing bit weight

Fig. 36 Typical response of penetration rate to increasing rotary speed

Page 47

A typical plot of penetration rate vs. rotary speed obtained with all other drilling variables held constant is shown in Fig (36). Penetration rate usually increases linearly with rotary speed at low values of rotary speed. At higher values of rotary speed, the penetration rate to increasing rotary speed diminishes. The poor response of penetration rate at high values of rotary speed usually is also attributed to less efficient bottomhole cleaning. The plots in Figs (35) and (36) can be obtained by measuring the rate of penetration at various bit weights and rotary speeds. However, frequent changes in lithology with depth make measurements very difficult because the lithology may change before the tests are completed. To overcome this problem, a drilloff test can be performed. A drilloff test consists of applying a large weight to the bit and then locking the brake and monitoring the decrease in bit weight with time while maintaining a constant rotary speed. Hook’s law of elasticity then can be applied to compute the amount the drillstring has stretched as the weight on the bit decreased and the hook load increased. In this manner, the penetration rate in ft/hr can be expressed as,

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R =

3420 L Δw EAs Δt

...............................................…..................(4)

where:

Δw = bit weight increment, lb E As Δt L

= = = =

modulus of elasticity = 30 x 106 psi wall cross-sectional area of drill pipe, inch2 time increment, sec. length of drill pipe, ft.

The following procedure can be used for conducting a drilloff test: 1. Select a depth where a uniform lithology is expected. 2. After proper break in, select a starting rpm which is less than the recommended maximum normal rpm. Measure rpm by counting revolutions of the rotary table. 3. Ensure that the weight indicator has been properly zeroed. Place on the bit the maximum weight recommended by the manufacturer. Make sure that the product of the selected bit weight in 1000 lbs times the rpm does not exceed the bearing capability or the WN number of the bit as recommended by the manufacturer. The WN number is the product of the weight on bit in 1000 lb times rpm recommended by manufacturer. 4. Lock the brake. By using a stop watch, measure the time in seconds required to drilloff 2000 lb increments. The average weight corresponding to the shortest time is the optimum bit weight for that rotary speed. 5. Repeat the test for various rotary speeds. Make sure that weight × rpm product does not exceed the WN number. Pick the optimum bit weight and rpm which corresponds to the shortest time. If the shortest time occurs at different weights and rpms, use the lowest weight or rpm. This will save bearing wear.

Page 48

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Example:

A drilloff test was conducted in UTMN-242 across the Arab-D at a constant speed of 72 rpm and 7300 ft using a Hughes 6-1/4” J-33 bit run on 3-1/2” drill pipe (ID= 2.764 in). Using the test data below, construct a table of penetration rate versus weight on bit. Bit weights 1000 lb

Time, sec

25 23 21 19 17 15

0 20 21 26 29 30

Solution:

The penetration rate can be calculated using Eq (4), R

=

3420 L Δw EAs Δt

For Δt = 20 sec, Δw = 2000 lb. As

=

R

=

π

(3.52 - 2.7642)

4 3420 × 7300 × 2000 30 × 10 6 × 3.62 × 20

= 3.62 in2 = 23 ft/hr

A tabulation of bit weight versus penetration rate is shown below: Average Bit Weight 1000 lb

24 22 20 18 16

Page 49

R, Ft/hr

23.0 21.9 17.7 15.8 15.3

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Problem:

The drilloff test of the previous example was repeated at 5 different rpms as shown below. What is the optimum bit weight and rpm? The WN number for 6-1/4” J-55 bit is 2550.

Bit Weight 1000 lbs.

63 rpm

25-23 23-21 21-19 19-17 19-17 17-15

20 21 27 27 27 32

72 rpm

79 rpm

87 rpm

Time to Drill-off, Seconds 20 21 26 29 29 30

14 19 26 27 27 29

15 25 29 27 27 29

A rotary-weight response test is usually conducted to confirm the results of the drilloff test. The response test is more reliable than the drilloff test and takes about one hour to run. The procedure for conducting a response test is as follows: 1. Pick the optimum weight obtained in the drilloff test and measure the time to drill one foot. Compute the rate of penetration in ft/hr. 2.

Keeping the weight on bit constant repeat step for various rpms. Make sure the weight rpm product does not exceed the WN number. Plot rate of penetration versus rpm. Pick the optimum rotary speed from the graph.

3. Keep the rpm constant at the optimum rotary speed obtained in step #2 and repeat the test for various bit weights. Calculate the rate of penetration. Plot rate of penetration versus bit weight. Pick the optimum bit weight from the plot.

Page 50

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Problem:

A rotary-weight response test was conducted to confirm the results of the drilloff test of the pervious example. The test data are tabulated below. Determine the optimum bit weight and rpm for drilling the Arab-D using 6-1/4” J-33 bit. ROTARY SPEED RESPONSE TEST FOR HUGHES 6-1/4” J-33 BIT Bit Weight Constant at 24,000 lbs

Page 51

Rotary Speed PPM

Time To Drill One Foot, Seconds

60 72 82 90 97

180 141 123 127.5 147

Rotary Speed Constant at 79 rpm Bit Weight 1000 lb 18 20 22 24 25

Time To Drill one ft 262 212 187 166 146

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OPERATING PROCEDURES OF ROLLING CONE BITS SURFACE HANDLING

New bits should be ordered without nozzles and stored in their boxes. Serial numbers and date of arrival should be recorded. Re-run bits should be cleaned and stored with pipe coating on the threads. Before a new bit is run in the hole, the drilling representative should check the bit and confirm it is of the correct type and size and has no missing inserts or protruding seals. Grease reservoir equalization ports should not be clogged. The drilling representative should witness the installation of new nozzles. Never force a nozzle into the bit. Measure nozzle size with a nozzle gauge. The bit should be made up using a properly sized bit breaker. The threads should be cleaned, greased and properly torqued. TRIPPING BIT IN THE HOLE

The bit should be run slowly in the hole to get through ledges or restrictions which may be present in blow out preventers, casing leaks, whipstocks, liner hangers, casing patches and casing shoes. Hitting obstructions at high running speed can break teeth and damage bearings. Bits should not be used to ream obstructions in casing and liner hangers or drill on junk. When in the open hole the bit should be run with care in areas which were tight in the previous bit run. Tight holes may be reamed at low bit weight and high RPM and pumping rate. Excessive reaming can damage teeth on the outer row where all the bit weight is applied during reaming. When the bit is on bottom, wash the last two joints and avoid running into fill which may plug the bit. Start the pump slowly and avoid pressure surges which could blow the nozzles out of the bit. Start drilling with low weight and rpm for few feet to enable the bit establish a bottomhole pattern. Do not exceed the rpm and weight recommended by bit manufacturer.

Page 52

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DRILL STRING DESIGN

TABLE OF CONTENTS DRILL PIPE DRILL PIPE DESCRIPTION SELECTION OF TOOL JOINTS BUOYANCY

DRILL STRING DESIGN MAXIMUM PULL SLIP CRUSHING COLLAPSE PRESSURE BURST PRESSURE TORSIONAL STRENGTH OF TOOL JOINTS TORSIONAL STRENGTH OF DRILL PIPE

DRILL COLLARS COLLAR SIZE SELECTION WEIGHT ON BIT CALCULATIONS THE PRESSURE-AREA METHOD

Page 1 2 7 8

11 14 15 17 21 22 26

29 30 32 41

HEAVY WEIGHT DRILL PIPE

44

BOTTOM HOLE ASSEMBLY

47

STABILIZERS REAMERS SHOCK ABSORBERS JARRING DEVICES MECHANICAL JARS

BOTTOM HOLE ASSEMBLY DESIGN

49 51 52 54 55

55

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DRILL STRING DESIGN

TABLE OF CONTENTS

Page

FATIGUE DAMAGE

59

DRILL STRING DESIGN FOR INCLINED AND HORIZONTAL WELLS

62

PREDICTION OF TORQUE AND DRAG CALCULATION OF DRAG CALCULATION OF DRAG WITH ROTAATION CRITICAL HOLE ANGLE CALCULATION OF TORQUE DETERMINATION OF FRICTION COEFFICIENT FACTORS AFFECTING TORQUE AND DRAG CALCULATION OF BUCKLING

62 66 74 75 76 80 81 84

- SINUSOIDAL BUCKLING IN STRAIGHT INCLINED HOLES - BUCKLING IN CURVED HOLES - HELICAL BUCKLING

84 87 89

FATIGUE DRILL STRING DESIGN FOR HIGH ANGLE AND HORIZONTAL WELLBORES MAXIMUM WEIGHT ON BIT BELOW TANGENT POINT MAXIMUM WEIGHT ON BIT ABOVE KICK-OFF POINT CALCULATION OF AXIAL MECHANICAL FORCES DRILL STRING DESIGN SUMMARY

90 91 92 93 95 96

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DRILL STRING DESIGN

DRILL STRING DESIGN The drill string is an important part of the rotary drilling process. It serves several purposes which include the following: • • • • •

provide a fluid conduit from rig to bit. impart rotary motion to the bit. provide weight on the bit. lower and raise the bit in the well. allow formation evaluation and testing.

The drill string consists primarily of the drill pipe and the bottom-hole assembly (BHA). The drill pipe can contain conventional drill pipe and heavy-weight drill pipe (HWDP). The BHA may contain the following major components: • • • • • •

drill collars stabilizers jars reamers shock subs bit

The BHA may also include downhole motors, non-magnetic drill collars for directional surveys, monitoring-while-drilling (MWD) tools, float subs, junk subs and crossovers. The major parts of the drill string are discussed in detail below. DRILL PIPE The longest section of the drill string is the drill pipe. Each joint of drill pipe consists of the tube body and the tool joint (connection). Drill pipe joints are available in three length ranges: Range 1 2 3

Range 2 is the most common.

Page 1

Length, ft 18-22 27-30 38-45

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DRILL STRING DESIGN

Drill Pipe Description Drill pipe is generally described by tube OD, nominal weight, pipe grade, type tool joint, thread connection and classification. For example, a 5” drill pipe can be described as, 5” 19.5 tube OD Nominal Weight

Grade ‘E’ Minimum Yield strength

XH NC50 Upset Connection Tool Joint thread

Premium Classification (wear)

Each description item is explained in more detail below. Tube OD is simply the plain end (tube) outside diameter. Common drill pipe sizes and other dimensional data can be found in Table (1) of API RP7G and are shown in Table (1). The average length of the tube is normally 29.4 ft. The nominal weight is the weight per foot including the weight of an API regular connection. The number serves no real purpose other than identification because drill pipe does not have API regular connection. The actual weight per foot or adjusted weight of the drill pipe is listed in Table (8) and (9) of API RP7G. The actual weight depends on the type of connection and pipe grade. As the grade (minimum yield strength) increases the actual weight increases because the upset has more metal in it. The actual weight of the 5” drill pipe in the above example is 20.89 lb/ft. The actual weight for grade ‘G’ pipe is 21.92 lb/ft. The pipe grade states the minimum yield strength of the metal which is the tensile stress that will result in 0.5% strain. Grade ‘E’ drill pipe has a yield strength of 75,000 psi. The minimum strength can be converted to a more usable strength in pounds by multiplying the minimum yield in psi by the cross-sectional area of the metal. From Table (1) the ID of 5” 19.5 # DP is 4.276 in and the cross-sectional area of the pipe wall is, A=

π (52 − 4.276 2 ) 4

= 5.271 in 2

Therefore, the working strength of the DP is, Working strength = 5.271 x 75000 = 395,394 lb

Page 2

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DRILL STRING DESIGN TABLE 1 DRILL PIPE TUBE DIMENSIONS

1 Nom. size (OD)

2 Nom. weight

3 Nom. ID

4 Nom. wall

5

(in) 2-3/8

(lb/ft) 4.85 6.65

(in) 1.995 1.815

(in) 0.190 0.280

Min. 2.344

Max 2.406

OD 4.430

2-7/8

6.85 10.40

2.441 2.151

0.217 0.362

2.844

2.906

6.492

4.679 3.634

1.812 2.858

1.121 1.602

3-1/2

9.50 13.30 15.50

2.992 2.764 2.602

0.254 0.368 0.449

3.469

3.531

9.621

7.031 6.000 5.317

2.590 3.621 4.304

1.962 2.572 2.923

4

11.85 14.00 15.70

3.476 3.340 3.240

0.262 0.330 0.380

9.480 8.762 8.244

3.078 3.805 4.322

2.700 3.229 3.579

4-1/2

13.75 16.60 20.00 22.82

3.958 3.826 3.640 3.500

0.271 0.337 0.430 0.500

4.478

4.545

15.904

12.303 11.497 10.406 9.621

3.600 4.407 5.498 6.283

3.592 4.272 5.116 5.673

5

16.25 19.50 25.60

4.408 4.276 4.000

0.296 0.362 0.500

4.975

5.050

19.635

15.261 14.364 12.566

4.374 5.275 7.069

4.859 5.708 7.245

5-1/2

19.20 21.90 24.70

4.892 4.778 4.670

0.304 0.361 0.415

5.473

5.555

23.758

18.796 17.930 17.128

4.962 5.828 6.629

6.111 7.031 7.844

25.20 5.965 27.72 5.901 (1) Z = (π/32){(OD4 - ID4)/OD}

0.330 0.362

6.592

6.691

34.472

27.945 27.349

6.526 7.123

9.786 10.578

6-5/8

6 OD. (in)

7 8 9 (Ai) (Ao) Section Area (in2) ID Wall 3.125 1.305 2.587 1.843

Common grades of drill pipe are listed in Table (2) below. Table (2) Common Grades of Drill Pipe Grade E X G S Page 3

Yield Strength, psi 75,000 95,000 105,000 135,000

10 (Z) 1 Section Modulus (in3) 0.661 0.867

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DRILL STRING DESIGN

The tool joint and type of upset are part of the drill pipe description. Tool joints are screw-type connectors that join the individual joints of drill pipe. Each joint of drill pipe is fitted with a pin (male thread) and box (female threads) tool joints or connectors. As was mentioned above, there are three types of tool joints which are widely used. 1 - IEU (Internal-External Upset) - The tool joint OD is larger than the OD of the drillpipe and the tool joint ID is less than the ID of the drillpipe. Generally, IEU tool joints are the strongest available. The large OD and small ID of the tool joint cause relatively high external and internal pressure losses. A schematic diagram of IEU connector is shown in Fig (1c). The dimensions of the tool joint can be obtained from API SPEC 5D Table 6.1. 2 - IF (External Upset) - The tool joint ID is the same as the ID of the drill pipe to minimize internal pressure losses. The upset is on the OD of the tool joint as shown in Fig (1b). 3 - IU (Internal Upset) - The tool joint ID is less than that of the drill pipe. The small ID causes relatively higher internal pressure losses. The tool joint OD is the same as the OD of the drill pipe. This type is called “slim-hole” drill pipe because of the small OD.

INTERNAL UPSET

EXTERNAL UPSET

(a)

(b)

INTERNAL-EXTERNAL UPSET

(c)

Types of Drill Pipe Tool Joints Fig (1)

Mechanical properties of new tool joints are listed in Table (8) of API RP7G. Tool joints for 4-1/2” 16.6 # grade ‘E” drill pipe are shown in Table (3).

Page 4

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DRILL STRING DESIGN

The type of connection is designated by the API NC number which is a two-digit number of the pitch diameter taken at the pin gauge point as shown in Fig (2). Gauge point pitch diameter is the distance at gauge point measured to imaginary lines that bisect the thread halfway between crest and root. For example, for pitch diameters of 5.042” and 4.628” the NC numbers are NC 50 and NC 46 respectively.

5/8" down from shoulder (gauge point)

Pitch diameter

Fig (2)

Table 3 Mechanical Properties of New Tool Joints and New Grade E Drill Pipe (1)

(2)

(3)

(4)

Drill Pipe Data

(5)

(6)

(7)

(8)

(9)

Drift Diam in.

Pipe

Tool Joint Data

Nom Size in.

Nom Wt lb/ft

Approx

4½

13.75

15.21 14.93 14.06 14.79

IU EU EU EU

6 6 3 /8 5¾ 6 1 /8

3¼ 3¾ 331/32 3 7 /8

3.125 3.625 3.770 3.750

16.60

18.14 17.81 17.98 17.10 16.79 18.37

IEU FH 6 IEU H90 6 EU NC50(IF) 63/8 EU OH 5 7 /8 IEU NC38(SH) 5 IEU NC46(XH) 6¼

3 3¼ 3¾ 3¾ 211/16 3¼

20.00

21.63 21.63 21.62 22.09

IEU FH IEU H90 EU NC50(IF) IEU NC46(XH)

6 6 6 3 /8 6¼

22.82

24.07 24.59

19.50

5

25.60 5½

Page 5

21.90 24.70

Wt* lb/ft

Type Upset

Tool Joint

Pipe

270034. 270034. 270034. 270034.

938984. 939095. 555131. 868775.

25907. 25907. 25907. 25907.

39021.p 37676.p 20965.p 34440.p

2.875 3.125 3.625 3.625 2.563 3.125

330558. 330558. 330558. 330558. 330558. 330558.

976156. 938984. 939095. 714267. 587308. 901164.

30807. 30807. 30807. 30807. 30807. 30807.

34780.p 39021.p 37676.p 27272.p 18346.p 33993.p

3 3 3 5 /8 3

2.875 2.875 3.452 2.875

412358. 412358. 412358. 412358.

976156. 1086246. 1025980. 1048426.

36901. 36901. 36901. 36901.

34780.p 45258.p 41235.p 39659.p

EU NC50(IF) 63/8 IEU NC46(XH) 6¼

3 5 /8 3

3.452 2.875

471239. 471239.

1025980. 1048426.

40912. 40912.

41235.p 39659.p

22.26 20.89 28.26 26.89

IEU 5½ FH IEU NC50(XH) IEU 5½ FH IEU NC50(XH)

7 6 3 /8 7 6 3 /8

3¾ 3¾ 3½ 3½

3.625 3.625 3.375 3.375

395595. 395595. 530144. 530144.

1448407. 939095. 1619231. 1109920.

41167. 41167. 52257. 52257.

62903.b 37676.p 62903.b 44673.p

23.77 26.33

IEU IEU

7 7

4 4

3.875 3.875

437116. 497222.

1265802. 1265802.

50710. 56574.

55933.p 55933.p

Conn H90 NC50(IF) OH WO

FH FH

OD in.

ID in.

(10) (11) (12) Mechanical Properties Tensile Torsional Yield, lb Yld, ft-lb Tool Joint

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DRILL STRING DESIGN

Another characteristic of the connection is the thread form or shape. Various types of the thread forms are shown in Fig (3). All NC connections have a V-038R thread form which has thread root whose radius of curvature is 0.038”. Thread forms which have rounded thread roots as in Fig (3a) and Fig (3b) are better than those which have sharp corners as shown in Fig (3d) and Fig (3e). The sharp corners tend to act as stress concentrators and make the thread more susceptible to failure than threads with rounded roots. There are two additional API connections which are in use. These are the API regular connection (Reg) and the API full hole connection (FH). These connections are inferior to the NC connection because they have a sharper root radius. There are other proprietary connections which were made obsolete such as Hughes XH, Hughes H-90, Reed open hole (OH), slim hole (SH), double streamline (DSL), PAC and external flush (EF). Table (4) shows some of the API, NC and obsolete connections that are interchangeable. Fig. (3) Thread Forms for Tool Joint Connections

Table 4 Rotary Shouldered Connection Nomenclature (Connections in the same column are interchangeable, though not identical)

(From T.H. Associates)

OBSOLETE API NAME Internal Flush (IF) Full Hole FH OTHER OBSOLETE NAME Extra Hole (XH) Double Streamline (DSL) Slim Hole (SH) External Flush (EF) Semi-Internal Flush

Page 6

CURRENT API NAME NC-38 NC-40 NC-46

NC-26

NC-31

-

NC-50

2-3/8 -

2-7/8 -

-

-

3-1/2

4

4 -

4-1/2 -

2-7/8 -

3-1/2 -

2-7/8 3-1/2 -

3-1/2 4 4-1/2 -

4-1/2 -

4-1/2 -

4-1/2 -

5 5-1/2 5

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DRILL STRING DESIGN

The classification of drill pipe is based on wear. As new drill pipe is rotated in the hole it wears and, therefore, it must be reclassified according to its wear. Exterior wear includes OD wear, dents and mashes, slip area crushing, cuts, pitting and corrosion. The working strength is reduced because of loss in cross-sectional area. The drill pipe is inspected periodically to detect cracks, pits, reduction in wall thickness and other defects. Inspection methods include electromagnetic inspection of pipe body to locate cracks and pits, sonic inspection to measure wall thickness, visual inspection to detect mashes and caliper measurements. The API has established guidelines for pipe classes in API RP7G. The classes are summarized as follows: New:Premium:Class 2:Class 3:-

No wear and has never been used. Uniform wear and a minimum wall thickness of 80% of original wall thickness. Allows drillpipe with a minimum wall thickness of 65% with all wear on one side so long as the cross-sectional area is the same as premium class, that is to say, based on not more than 20% uniform wall reduction. Allows drillpipe with a minimum wall thickness of 55% with all wear on one side.

Selection of Tool Joints

The factors that must be considered in the selection of tool joints are outside diameter, inside diameter, tensile and torsional ratings and cost. Tool joints with large OD and small ID will result in high pressure losses inside and outside the drill pipe. Large OD tool joints have better wear characteristics. The OD of the tool joint should be small enough to facilitate fishing the tool joint with standard fishing tools. One rule of thumb in the selection of tool joints is that when all other factors are equal the larger the tool joint, the better it will perform. In the selection of tool joints the operating engineer or supervisor must rely on the advice of the technical personal who represent the tool joint manufacturer. These people are knowledgeable in the design and operation of the tool joints. However, they may not be familiar with the conditions of the well. This may cause a problem, but it is the responsibility of the engineer to prevent such occurrence. To do this, the engineer should (1) be familiar with the tool joint design and operation, (2) be able to clearly communicate his needs to the sales representative, and (3) ensure that the people advising him are knowledgeable and clearly understand the problem.

Page 7

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DRILL STRING DESIGN

Buoyancy Drill strings and subsurface well equipment are subjected to forces created by the hydrostatic pressure of the fluid in the well. The effect of the hydrostatic pressure is called buoyancy. Buoyancy is understood most easily for a vertical solid bar of circular cross-section immersed in liquid as shown in Fig (4). Hydrostatic pressure acting on one side of the bar is balanced by an equal pressure acting on the opposite side. Thus the net force exerted by the fluid is the force F acting upward on the bottom end of the bar. The magnitude of the force is given by,

Surface of liquid

W

F Fig (4) (5) Fig

F = PA =

ρf 144

LA .........................................................................(1)

where, P = hydrostatic pressure, psi A = cross-sectional area of bar, in2 L = depth below surface of liquid, ft ρf = weight of fluid, 1b/ft3 The weight of the bar in the fluid is given by, Wf = W - F..................................................................................(2) where, Wf = weight in fluid, lb W = weight in air, lb F = buoyant force, lb Page 8

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DRILL STRING DESIGN

The weight of the steel bar in air can be expressed as the volume times density, or W= L

A ρ s ............................................................................(3) 144

Where, ρs is the density of steel in lb/ft3. If we substitute Equation (1) and (3) in Eq (2), Wf = L

Wf =

ρf A ρs − LA 144 144

LAρ s ⎛ ⎜1 − 144 ⎝

⎛ ρ ⎞ ρf ⎞ ⎟ = W ⎜1 − f ⎟ ρs ⎠ ⎝ ρs ⎠

⎛ ρs − ρf ⎞ ⎟ ..................................................(4) ⎝ ρs ⎠

Wf = W ⎜

Substituting the density of steel 490 lb/ft3 yields, ⎛ 490 − ρ f ⎞ ⎟ ...............................................(5) ⎝ 490 ⎠

Wf = W ⎜

The expression between the parenthesis is called the buoyancy factor.

Example (1)

A drill string which consists of 6000 ft of 5”,19.5 lb/ft drill pipe (ID = 4.276 in) and 500 ft of 9” OD x 3” ID drill collars is suspended in a well filled with 80 pcf mud. Calculate the weight of the string in the fluid (a) by using Eq (2), and (b) by using Eq (5). Solution

a) The hydrostatic forces acting on the drill string are shown in Fig (5). The force F1 is acting downward at the shoulder areas between the drill pipe and drill collars. The force F2 is acting upward on the wall cross-sectional area of the drill collars.

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DRILLING

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DRILL STRING DESIGN

The shoulder area between the DP and drill collars is, 314 314 . . 4.276 2 − 32 9 2 − 52 + A= 4 4

(

)

(

)

= 5124 . in 2

80 x 6000 = 3333 psi 144

Pressure at 6000 ft =

F1 = 3333 x 51.24 = 170,800 lb↓

5" DP 4.276" ID

Section area of DCs = 314 . 9 2 − 32 = 56.52 in 2 4

(

)

9" Drill Collars 3" ID

80 Pressure at 6500 ft = x 6500 = 3611 psi 144

F2 = 3611 x 56.52 = 204,100 lb↑

F1

6000

F2

6500

Fig (5)

The resultant force acting on drill string is, F = F2-F1 = 204,100 - 170,800 = 33,300 lb↑ Weight of drill string in air (disregard couplings) = weight of DP + weight of DC Weight DP = Vol. x density = Weight DC =

(

)

314 . 52 − 4.276 2 x 6000 x 490 = 107,635 lb 4 x 144

(

)

314 . 9 2 − 32 x 500 x 490 = 96,162 lb 4 x 144

Weight of string in air = 107,635 + 96,162 = 203,797 lb From Eq (2), Wf = W - F =203,797 - 33,300 = 170,497 lb b) Buoyancy factor =

490 − 80 = 0.836 490

From Eq (5), Wf = W x 0.836 = 203,797 x 0.836 = 170,524 lb Page 10

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DRILL STRING DESIGN

Example (2)

A 6000’ of 3-1/2” 14.69 lb/ft actual weight drill pipe (ID = 2.76 in) got stuck in the well while spotting a cement plug as shown in Fig (6). The weight of the mud in the well is 75 pcf. What is the weight of the drill pipe in the mud. Solution

Since the bottom end of the DP is sealed in hard cement there will be no hydraulic force acting on the bottom end of the DP. Therefore, from Eq (2),

3-1/2" DP 75 pcf mud

Wf = W - F Wf = W - 0 = W = 6000’ x 14.69 = 88,140 lb The weight of DP in fluid is the same as the weight in air. CEMENT

DRILL STRING DESIGN

Fig (6)

A drill string should be designed to deliver sufficient weight to the bit and provide sufficient torsional and tensile strength to withstand the vigorous and dynamic conditions of drilling. The drill string should also withstand burst and collapse pressure loads and be designed to minimize hole stability problems. There are many factors that must be considered in the design of the drill string. These factors are: • • • • • • • • • Page 11

Total depth Hole size Mud weight Over pull Bottom hole assembly Hole angle Pipe weights and grades Corrosive environment Ability to fish tools out of hole

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DRILL STRING DESIGN

The drill string is designed so that the uppermost joint of each section of drill pipe is loaded to no more that 80% of the minimum tensile yield strength of that particular weight and grade pipe for a single size and grade drill pipe. The total load exerted on the top joint of the drill pipe consists of the buoyed weight of the drill pipe plus the buoyed weight of the bottom hole assembly (heavy-weight drill pipe plus drill collars) and the margin of overpull (MOP). The MOP is the desired amount of load in excess of the buoyed weight of the drill string to account for hole drag and provide excess pull capacity in the event the drill string becomes stuck in the hole. The amount of overpull ranges from 50,000 to 100,000 lb. The design criterion can be expressed as, where,

Y x 0.8 = LWBf + LcWcBf + LHWHBf + MOP ..........................(6)

L = length of DP, ft W = Actual weight of drill pipe, lb/ft Lc = length of drill collars, ft Wc = weight of collars, lb/ft LH = length of heavy weight drill pipe, ft WH = weight of heavy weight drill pipe, lb/ft MOP = Margin of overpull, lb Bf = buoyancy factor Y = minimum yield strength, lb Solving Eq (6) for the maximum length of drill pipe that can be used, L=

0.8Y − MOP − LCWC Bf − L HW H Bf ...............................(7) WBf

If the drill string consists of two sections of drill pipe of different grade and weight, then the maximum length of the second drill pipe section (top section) is, L2 =

0.8Y2 − MOP − L1W1 B f − LcWc B f − L HWH B f W2 B f

...............(8)

where, L1 = length of first section of drill pipe (lower section), ft W1 = actual weight of first section of drill pipe, lb/ft L2 = length of second section of drill pipe, ft W2 = actual weight of second section of drill pipe, lb/ft Y2 = minimum tensile strength of second section of drill pipe (top section), lb.

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In drill string design the pipe of the lowest grade (weakest) is placed on bottom. Each section of drill pipe is designed starting with the BHA and working upwards. This design is checked at various depths, for often the most critical section of hole is not at TD, but further up the hole due to mud weight changes. Example

A vertical Khuff well is to be drilled to a total depth of 15000 ft by using 5” 19.5# grade ‘G’ and ‘S’ drill pipe with FH connections. The mud weight at 7000 ft is 70 pcf and increases to 90 pcf at 15000 ft. The BHA (heavy weight drill pipe plus drill collars) is 1200 ft long and weighs 150,000 lb in air. Calculate the length of each section of drill pipe assuming 100,000 lb overpull.

Solution

The actual weights and strengths of the drill pipe are obtained from API RP7G Table (9). 5”, 19.5# ‘G’, FH 5”, 19.5#, ‘S’, FH

Actual wt, lb/ft Tensile Strength, lb 80% Tensile Strength, lb 22.46 553,833 443,066 23.4 712,070 569,656

489 − 70 = 0.856 489 489 − 90 Buoyancy factor for 90 pcf mud = = 0.815 489

Buoyancy factor for 70 pcf mud =

Let us first design for a TD of 7000 ft. Using Eq (7), the length of the bottom section of drill pipe (grade G) is, L=

443,066 − 100,000 − 150,000 x 0.856 = 11162 ft 22.46 x 0.856

Since the length of the BHA and the ‘G’ drill pipe (1200 + 11162) is greater than 7,000 ft, there is no need to use grade ‘S’ drill pipe when drilling to 7000 ft TD. Now, for a TD of 15,000 ft, the length of the first drill pipe (grade G) is calculated by using Eq (7),

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L1 =

443,066 − 100,000 − 150,000 x 0.815 = 12,063 ft 22.46 x 0.815

The length of the second section of DP from Eq (8) is, L2 =

569,656 − 100,000 − 12063 x 22.46 x 0.815 − 150,000 x 0.815 = 6638 ft 23.4 x 0.815

We can use a maximum of 6638 ft of grade ‘S’ drill pipe, but all we need is, 15000 - 12,063 - 1200 = 1737 ft. Therefore, the string design at 15000’ is, BHA 5” 19.5#, grade ‘G’ 5” 19.5# grade ‘S’ Total

Length, ft 1,200 12,063 1,737 15,000

Maximum Pull The maximum pull that can be exerted on stuck drill pipe should not exceed 80% of the maximum tensile strength of the weakest grade drill pipe. The maximum pull for each crossover point must be calculated to determine the maximum pull that can be exerted on the drill string. The calculation procedure is illustrated in the following example. Example

If the drill string in the previous example becomes stuck at the bit at 14000 ft, what is the maximum pull that can be exerted on the DP at the surface? Solution

With the bit at 14000 ft the drilling assembly would consist of the following: Section BHA 5” Grade G DP 5” Grade S DP

Length, ft 1200 12,063 737 14,000 ft

Page 14

Section Air weight, lb 150,000 lb 270,934 lb 17,245 lb

80% Tensile Strength, lb 443,066 569,656

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First we check the maximum pull on the weakest DP. The maximum pull that can be exerted on the top of grade ‘G’ DP without exceeding 80% of Grade ‘G’ tensile strength is 443,066 lb. Now the maximum pull that can be exerted on top of grade ‘S’ DP is, Max pull = 443,066 + 17245 = 460,311 lb. Since 460,311 lb are less than 80% of the grade ‘S’ tensile strength (569,656 lb), then it is safe to pull 460,311 lb. So when we pull 460,311 lb at the surface the pull exerted on top of Grade ‘S’ DP is 460,311 lb. The pull exerted on top of Grade ‘G’ DP is 460,311 minus the weight of grade ‘S’ DP, or 460,311-17245 = 443,066 lb.

Slip Crushing Slips exert hoop compression on the drill pipe which can deform the pipe if conditions are unfavorable. A unit tensile stress St from hanging weight will result in a hoop stress Sh that is a function of many factors such as slip length, coefficient of friction between slips and bowl, pipe diameter and others. The slip crushing constant is defined for a given set of conditions as the ratio Sh/St. Slip crushing constants for a variety of conditions are listed in Table (6). Table (6) SLIP CRUSHING CONSTANTS Slip Length In 12

16

Coefficient of Friction 0.06 0.08 0.10 0.12 0.14 0.06 0.08 0.10 0.12 0.14

2-3/8

2-7/8

1.27 1.25 1.22 1.21 1.19 1.20 1.18 1.16 1.15 1.14

1.34 1.31 1.28 1.26 1.24 1.24 1.22 1.20 1.18 1.17

Pipe Size-Inches 3-1/2 4 4-1/2 Minimum Ratio Sh.St 1.58 1.50 1.43 1.52 1.45 1.39 1.47 1.41 1.35 1.43 1.38 1.32 1.40 1.34 1.30 1.41 1.36 1.30 1.37 1.32 1.28 1.34 1.29 1.25 1.31 1.27 1.23 1.28 1.25 1.21

5

5-1/2

1.66 1.59 1.54 1.49 1.45 1.47 1.42 1.38 1.35 1.32

1.73 1.66 1.60 1.55 1.50 1.52 1.47 1.43 1.39 1.36

A coefficient of friction of 0.08 between slips and bowl is normally used. If the pipe is not stuck, the maximum tension carried by the slips is the working load, Pw, which is the buoyed weight of the drill pipe and BHA. In order to prevent any deformation of the pipe, the working load Pw times the crushing constant should be less than 0.8Y, or

Page 15

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DRILL STRING DESIGN Sh x Pw ≤ 0.8Y ..................................................(9) St

Example

Will the slips cause any deformation of DP in the previous two examples. Assume coefficient of friction of 0.08 and slip length of 16 in. Solution

a) When drilling at 7000 ft the buoyed weight of the drill string is, Section

Length, ft

BHA 5” 19.5# ‘G’

1200 5800

Air Weight, lb

Buoyed Weight, ft

150,000 130,268

128,400 111,509 239,909

From Table (6), the crushing constant for 5” DP and 16” slip length is 1.42 Substituting in Eq (9), 0.8Y ≥ 239,909 x 1.42 443,066 ≥ 340,670 Therefore, there will be no deformation to the pipe. b) With the bit at 15,000 ft the buoyed weight of the drill string is as follows: Section BHA 5” 19.5 ‘G’ pipe 5” 19.5 ‘S’ pipe

Page 16

Buoyed Weight, lb

Length, ft

AirWeight, lb

1200 12,063 1737

150,000 270,934 40,645

122,250 220,812 33,126

15,000

461,579

376,188

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At 15000 ft the drill pipe that will be in the slips is the grade ‘S’ pipe whose yield strength, Y, is 712,070 lb. Substituting in Eq (9), 0.8(712,070) ≥ 1.42 x 376,188 569,656 ≥ 534,186 Therefore, there will be no damage to the 5” grade ‘S’ pipe. c) When the bit becomes stuck at 14000 ft, the grade ‘S’ drill pipe will be inside the slips. From the previous example, the maximum allowable pull was calculated to be 460,311 lb. Substituting in Eq (9), 0.8(712,070) ≥ 1.42 x 460,311 569,656 ≥ 653641 Since the above equality is not true, then the pipe will be deformed if it is set in the slips with 460,311 lb of tensile force exerted on it. Normally, when pull is applied to release stuck drill pipe, the pipe is not set in the slips and, therefore, no damage will occur to it. If it is required to have the DP set in the slips, how much maximum pull can be exerted without deforming the pipe? From Eq (9), 0.8 x 712,070 = 1.42 x P P=

569,656 , = 401166 lb 142 .

Collapse Pressure The drill pipe may at certain times be subjected to external pressure which is higher than the internal pressure. This condition usually occurs during drill stem testing and may collapse the drill pipe. The differential pressure (external pressure minus internal pressure) required to produce collapse is calculated for various sizes and grades of new and used drill pipe and is presented in API RP7G Tables (3), (5) and (7). Collapse pressure ratings for new drill pipe are presented in Table (7). The tabulated collapse pressure ratings must be divided by a safety factor in order to establish the allowable collapse pressure.

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Pc = Pac ...........................................................(10) S. F. where, Pc = theoretical collapse pressure rating from tables, psi Pac = allowable collapse pressure, psi S.F. = safety factor = 1.1 to 1.2 If the drill pipe is subjected to an axial tensile load, the collapse pressure ratings from the tables must be derated. The effective collapse corrected for tension load can be calculated from the equation, Effective Collapse =

where, Z =

Nominal Collapse ( 4 − 3Z 2 − Z ) ...............(11) 2

Tension Load, lb Area of Metal x Ave Yield Strength

The average yield strength of various grades of drill pipe are: Page 18

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DRILL STRING DESIGN Grade

Average Yield Strength, psi

E X G S

85,000 110,000 120,000 145,000

Table (7) New Drill Pipe Collapse and Internal Pressure Data (1) Size OD in. 2-3/8

(3)

(4)

(5)

(6)

Collapse Pressure Based on Minimum Values, psi E 95 105 135 11040. 13984. 15456. 19035. 15599. 19759. 21839. 28079.

(7)

(8)

(9)

(10)

Internal Pressure at Minimum Yield Strength. psi E 95 105 135 10500. 13300. 14700. 18900. 15474. 19600. 21663. 27853.

2-7/8

6.85 10.40

10467. 16509.

12940. 20911.

14020. 23112.

17034. 29716.

9907. 16526.

12548. 20933.

13869. 23137.

17832. 29747.

3-1/2

9.50 13.30 15.50

10001. 14113. 16774.

12077. 17877. 21247.

13055. 19758. 23484.

15748. 25404. 30194.

9525. 13800. 16838.

12065. 17480. 21328.

13335. 19320. 23573.

17145. 24840. 30308.

4

11.85 14.00 15.70

8381. 11354. 12896.

9978. 14382. 16335.

10708. 15896. 18055.

12618. 20141. 23213.

8597. 10828. 12469.

10889. 13716. 15794.

12036. 15159. 17456.

15474. 19491. 22444.

4-1/2

13.75 16.60 20.00 22.82

7173. 10392. 12964. 14815.

8412. 12765. 16421. 18765.

8956. 13825. 18149. 20741.

10283. 16773. 23335. 26667.

7904. 9829. 12542. 14583.

10012. 12450. 15886. 18472.

11066. 13761. 17558. 20417.

14228. 17693. 22575. 26250.

5

16.25 19.50 25.60

6938. 9962. 13500.

8108. 12026. 17100.

8616. 12999. 18900.

9831. 15672. 24300.

7770. 9503. 13125.

9842. 12037. 16625.

10878. 13304. 18375.

13986. 17105. 23625.

5-1/2

19.20 21.90 24.70

6039. 8413. 10464.

6942. 10019. 12933.

7313. 10753. 14013.

8093. 12679. 17023.

7255. 8615. 9903.

9189. 10912. 12544.

10156. 12061. 13865.

13058. 15507. 17826.

5500. 5321. 4788, 25.20 6538. 6036. 7103. 6755. 5894. 27.70 7172. 7813. Note: Calculations are based on formulas in API Bulletin 5C3.

8281. 9084.

9153. 10040.

11768. 12909.

6-5/8

Page 19

(2) Nom Weight Thds & Couplings lb. 4.85 6.65

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Example

A drill stem test is conducted using 5” 19.5# grade ‘G’ drill pipe and a packer set at 13000 ft. The DP-casing annulus has a surface pressure of 3000 psi and is filled with 90 pcf mud. The DP has 3000 ft of water cushion above the packer. The DP at 13000 ft has a tensile load of 50,000 lb. Will the DP collapse at 13000 ft? Use a S.F. of 3000 1.125. psi Solution

air

First calculate the differential pressure at 13000 ft.

90 pcf mud

10,000'

External Pressure at 13000 ft = Hydrostatic Pressure + Surface Pressure. External Pressure at 13000 ft =

water Packer at 13000'

90 x 13000 + 3000 = 11125 , psi 144

Internal Pressure at 13000 ft =

62.4 x (13000 − 10000) = 1299 psi 144

Differential Pressure (collapse load) at 13000 ft = 11125-1299 = 9826 psi. From Table (7), the collapse pressure rating of 5” 19.5 ‘G’ pipe = 12999 psi. Since the DP is under tension, then we have to derate the collapse pressure rating by using Eq (11). Z=

Tensile Load, lb Area of metal x Avg. Yield Strength, psi Area of Metal =

Ave Yield Strength = 120,000 psi Tensile load = 50,000 lb

Page 20

(

314 . 52 − 4.276 2 4

)

= 5.27in 2

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DRILL STRING DESIGN

Z=

50,000 = 0.079 5.27 x 120,000

From Eq (11), Nominal Collapse 4 − 3Z 2 − Z 2 12999 = 4 − 3 x 0.079 2 − 0.079 = 12,455 psi 2

Effective Collapse =

The allowable collapse =

12455 = 11071 psi 1125 .

Since the collapse load 9826 psi is less than the allowable collapse, then the drill pipe will not collapse. Burst Pressure

The differential pressure acting across the drill pipe wall due to an internal pressure greater than the external pressure is known as the burst load. The maximum burst load normally occurs at the depth where there is no external pressure, or backup. This normally occurs at the surface just below the wellhead where there is often no surface pressure. Burst pressure rating for various sizes and grades of new drill pipe are shown in Table (7). As in the case of the collapse pressure, a safety factor of 1.125 ± must be used to determine the maximum allowable burst load.

Example

A well is to be acidized by using 5” 19.5# ‘G’ drill pipe and bottom hole packer set at 13000’. The 15% HCl (67 pcf) will be pumped at maximum surface pressure of 8000 psi. The DP-casing annulus is filled with 90 pcf mud. a) Determine the worst burst load. b) Will the drill pipe burst?

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Solution

a) Burst load at 13000 ft = Internal pressure - External pressure Internal Pressure = External Pressure =

67 x 13000 + 8000 = 14,048 psi 144 90 x 13000 = 8125 psi 144 8000 psi

Burst load = 14048 - 8125 = 5923 psi Now let us calculate the burst load at the surface. Burst load = DP Surface Pressure - Surface annular pressure = 8000 - 0 = 8000 psi.

Acid 90 pcf mud

Since 8000 psi is greater than 5923 psi, the worst burst load is at the surface. b) The burst rating of 5” 19.5# ‘G’ drill pipe from Table (7) is 13304 psi The allowable burst pressure =

Packer

13000'

13304 = 11825 psi 1125 .

Since the burst load of 8000 psi is less than the allowable burst of 11825 psi, then the drill pipe will not burst.

Torsional Strength of Tool Joints Most standard tool joints are weaker in torsion than the drill pipe tubes to which they are welded. API sets tool joint torsional strength arbitrarily at 80 percent of tube torsional strength. This torsional strength ratio (TSR) of 0.8 is the basis for establishing tool joint’s ID’s and OD’s. As the tool joint OD wears the torsional strength decreases. Tool joints with TSR’s less then 0.8 are used successfully in low torsion drilling, and some high torsion drilling applications require tool joint torsional ratings higher than the standard TSR of 0.8. The torsional strengths of tool joints of various dimensions are given in Tables (8) and (9) of API RP7G. Tool joints for 4-1/2” and 5” grade ‘E’ drill pipe are shown in Table (3).

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DRILL STRING DESIGN

Tool joints are made up by applying sufficient makeup torque to force the pin and box shoulder tightly together and make a seal. This is accomplished when makeup stretches the pin and compresses the box shoulder as shown in Fig (7). Makeup torque is determined by the tool joint OD and ID and not by the properties and dimensions of the drill pipe tube. For a given pin ID the makeup torque increases as the tool joint OD increases. Similarly, for a given tool joint OD the makeup torque decreases as the pin ID increases. The standard makeup torque is the torque that would stress the weaker pin or box to 60% of its minimum yield strength of 72,000 psi. (Minimum yield strength of all tool joints is 120,000 psi regardless of the grade of the drill pipe). Recommended makeup torque of tool joints can be obtained from Figs 1-25 in API RP7G. Recommended torques for various selected sizes of NC -50 tool joints are shown in Table(5). Tool joints should be selected such that the makeup torque exceeds the maximum torsional load anticipated during drilling operations.

Fig. (7)

Page 23

Makeup puts elastic stretch in pin and box

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Table 5 Tool Joint Make-up Torque For NC-50 Connection, ft-1b 1

2

3

4

6- /8

6- /16

6-½

38040 36400 34680 33800 32890 31020 29090 27080 24990 22840

36170 36170 34590 33710 32810 30950 29020 27010 24930 22780

34190 34190 34190 33630 32720 30870 28940 26940 24870 22720

5

6

7

8

9

10

11

6-7/16

6-3/8

6-5/16

6-¼

6-3/16

6-1/16

6

32240 32240 32240 32240 32240 30790 28870 26870 24800 22660

30290 30290 30290 30290 30290 30290 28790 26800 24740 22610

28370 28370 28370 28370 28370 28370 28370 26740 24680 22550

26500 26500 26500 26500 26500 26500 26500 26500 24620 22490

24660 24660 24660 24660 24660 24660 24660 24660 24550 22430

21020 21020 21020 21020 21020 21020 21020 21020 21020 21020

OD(in.) 5

9

ID (in.) 2-3/4 2-7/8 3 3-1/16 3-1/8 3-1/4 3-3/8 3-1/2 3-5/8 3-3/4

Note: Box-weak connections are shown in bold type Example

A 5” 19.5 # grade ‘E’ drill pipe with 6-5/16” OD x 3-3/4” ID NC-50 tool joints will be used to drill a horizontal well where the anticipated torque is 20,000 ft.-lb. a-

Are the tool joint ID and OD adequate for the job?

b-

How much tool joint wear can be tolerated?

Solution

a) From Table (5) the recommended makeup torque of NC-50 6-5/16” OD x 3-3/4” ID tool joint is 22550 ft-lb. Since the anticipated torsion of 20,000 ft-lb is less than the makeup torque then the tool joint dimensions are adequate.

Page 24

19240 19240 19240 19240 19240 19240 19240 19240 19240 19240

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b) From Table (5) the makeup torque of 6-1/16” OD x 3-3/4” ID tool joint is 21020 ftlb which is higher than the anticipated torque of 20,000 ft-lb. If the OD of the tool joint wears down to 6” OD, the makeup torque drops to 19240 ft-lb which is unacceptable as it is less than the anticipated torque. Therefore, the smallest OD that can be tolerated is 6-1/16” or a wear of about ¼”. If the predicted torque is higher than the makeup torque, the torsional capacity of the tool joint can be increased by increasing the makeup torque. Higher makeup torque will increase the pin neck tensile stress and therefore reduce the pin neck capacity to carry external tensile load. The engineer should make sure that the pin tensile capacity is at least equal to the drill pipe tensile capacity. The relationship between the pin neck tensile capacity and makeup torque for NC50 connection for various pin IDs is shown in Fig (7a). The horizontal lines represent the tensile capacity of premium class drill pipe tube (under no torsion loads) being used with the tool joint. Example

Fig (7a) Failure of a rotary shouldered connection under static loads (top). Measurement points (center). Example problem (bottom)

Page 25

A 3-1/4” ID, 6.5” OD, NC50 tool joint is being used with 5” 19.5# grade S premium class drill pipe. You anticipate high operating torque and you wish to increase the makeup torque to 34,000 ftlb. Can you do this and still have a tool joint stronger for externally applied tension than the drill pipe tube?

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Solution

Tensile capacity of grade S DP Normal makeup torque of tool joint Connection tensile capacity at normal makeup torque New makeup torque New connection tensile capacity

= = = = =

560,000 (Fig 7a). 30870 ft-lb. 625, 000 lb (Fig 7a) 34,000 ft-lb 500,000 (Fig 7a)

So the new tensile capacity of tool joint is now weaker than that of the drill pipe. Now the tensile loads should be limited to less than 500,000 lb.

Torsional Strength of Drill Pipe The torsional strength of drill pipe becomes critical when drilling deviated or horizontal holes or when pipe is stuck. Torsional strength of various sizes and grades of new drill pipe are listed in Table (8). The actual torque applied to the drill pipe during drilling is difficult to measure, but may be approximated by the following equation, T=

HP x 5250 ...................................................(12) RPM

where, T = torque delivered to DP, ft-lb HP = horse power used to produce rotation of pipe, hp RPM = revolutions per minute The torque applied to DP while drilling should not exceed the tool joint make up torque listed in API RP7G Table (10). If a tensile load is exerted on the drill pipe, the torsional strength values in Table (8) must be derated. The torsional yield strength under tension may be calculated from the following equation, Q=

Page 26

0.09616 J Y2 − OD

P2 A2

...........................................(13)

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where. Q = minimum torsional yield strength, ft-lb J

= polar moment of inertia =

(

314 . OD 4 − ID 4 32

)

OD = outside diameter, in

ID = inside diameter, in

Y = minimum yield strength, psi

P = tensile load, lbs

A = metal cross section area, in2

Example

A well is being drilled using 12000 ft of 5” 19.5# ‘G’ drill pipe and BHA which weighs 150,000 lb in air. The weight on bit is 60,000 lb and the mud weight is 90 pcf. What is the maximum torque that can be applied to the drill pipe at surface? Use a safety factor of 1.2. Solution

First calculate the tensile load exerted on the drill pipe at surface. Buoyancy factor =

489 − 90 = 0.815 489

Buoyed weight of DP = 12000 x 22.46 x 0.815 = 219,915 lb Buoyed weight of BHA = 150,000 x 0.815 = 122,250 lb Tensile load at surface = 219,915 + 122,250 - 60,000 = 282,165 lb Now, calculate the minimum torsional strength.

( (

) )

314 . 54 − 4.276 4 = 28.5 32 314 . A= 52 − 4.276 2 = 5.27 in 2 4 J=

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DRILL STRING DESIGN Table (8) New Drill Pipe Torsional and Tensile Data

(1) Size OD in. 2-3/8

(2) Nom. Weight Thds & Couplings lb. 4.85 6.65

(3)

(4)

(5)

(6)

Torsional Data Torsional Yield Strength, ft-lb E 95 105 135 8574. 6668. 6033. 4763. 11251. 8751. 7917. 6250.

(7)

(8)

(9)

(10)

Tensile Data Based on Minimum Values Load at the Minimum Yield Strength, lb: E 95 105 135 97817. 123902. 136944. 176071. 138214. 175072. 193500. 248786.

2-7/8

6.85 10.40

8083. 11554.

10238. 14635.

11316. 16176.

14549. 20798.

135902. 214344.

172143. 271503.

190263. 300082.

244624. 385820.

3-1/2

9.50 13.30 15.50

14146. 18551. 21086.

17918. 23498. 26708.

19805. 25972. 29520.

25463. 33392. 37954.

194264. 271569. 322775.

246068. 343988. 408848.

271970. 380197. 451885.

349676. 488825. 580995.

4

11.85 14.00 15.70

19474. 23288. 25810.

24668. 29498. 32692.

27264. 32603. 36134.

35054. 41918. 46458.

230755. 285359. 324118.

292290. 361454. 410550.

323057. 399502. 453765.

415360. 513646. 583413.

4-1/2

13.75 16.60 20.00 22.82

25907. 30807. 36091. 40912.

32816. 39022. 46741. 51821.

36270. 43130. 51661. 57276.

46633. 55453. 66421. 73641.

270034. 330558. 412358. 471239.

342043. 418707. 522320. 596903

378047. 462781. 577301. 659734.

486061. 595004. 742244. 848230.

5

16.25 19.50 25.60

35044. 41167. 52257.

44389. 52144. 66192.

49062. 57633. 73159.

63079. 74100. 94062.

328073. 395595. 530144.

415559. 501087. 671515.

459302. 553833. 742201.

590531. 712070. 954259.

5-1/2

19.20 21.90 24.70

44074. 50710. 56574.

55826. 64233. 71660.

61703. 70944. 79204.

79332. 91278. 101833.

372181. 437116. 497222.

471429. 553681. 629814.

521053. 611963. 696111.

669925. 786809. 894999.

6-5/8

25.20 27.70

70580, 76295.

89402. 96640.

98812. 106813.

127044. 137330.

489464. 534199.

619988. 676651.

685250. 747877.

881035. 961556.

From Eq (13), 0.09616 x 28.5 282,1652 2 105,000 − = 49,507, ft - lb Q= 5 5.27 2 49507 Allowable torque = = 41,256 ft - lb 1.2 The combined load capacity of the tool joint should be checked, as it may be weaker than the drill pipe. Refer to API RP7G Tables 8,9 and 10. Page 28

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Fig (7b) Effect of Torsion on Tensile Strength

The effect of torsion on the tensile capacity of 5” 19.5 lb/ft premium class drill pipe of different grades is shown in Fig (7b). The graph is constructed by using Eq (13) using adjusted OD for premium class pipe.

DRILL COLLARS

Drill collars are large diameter-small bore steel pipes which posses’ great weight and great stiffness. Drill collars are run above the bit and make up the predominant component of the bottom hole assembly. Some of the functions of drill collars are: • • •

Page 29

provide weight on the bit minimize bit stability problems from vibrations, wobbling and jumping minimize directional control problems by providing stiffness to the BHA.

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Drill collar sizes range from 3” to 12” OD in increments of 1/4”. The inside diameter varies from 1” to 3-1/4”. The weight per foot can be obtained from published tables or calculated by using a steel density of 489 lb per ft3. The length of a drill collar joint is normally about 30 ft.

Example

Calculate the weight of a 9” OD x 3” ID x 30 ft long drill collar. Solution

The area of metal cross section =

(

)

314 . 9 2 − 32 ft 2 = 0.392 ft 2 4 x 144

The volume of 1 ft = area x length = 0.392 x 1 = 0.392 ft3 Weight of 1 ft = density x volume = 489 x 0.392 = 192 lb Weight of one joint = 192 lb/ft x 30 ft = 5760 lb Collar Size Selection

Woods and Lubinski pointed out that using an unstabilized bit and small OD drill collars can cause an undersized hole, making it difficult to run casing. They determined that the actual drift, or useful diameter, of the hole would be equal to the bit diameter plus the drill collar diameter, divided by two, Drift diameter =

Bit OD + Collar OD 2

The above equation can be rewritten to determine the minimum drill collar OD which would insure passage of casing with its larger coupling diameter. Substituting casing coupling diameter for drift diameter, casing coupling OD =

Bit OD + Collar OD 2

or, Minimum Collar OD = 2 (Casing Coupling OD) - Bit OD ....... (14)

Table (9) Page 30

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DRILL STRING DESIGN Minimum Drill Collar Sizes Bit Size, In.

Casing to be run Minimum drill Size, In. Coupling OD collar diam In. *3.875 5.000 4-1/2 6-1/8 *3.750 5.000 4-1/2 6-1/4 *3.250 5.000 4-1/2 6-3/4 7-7/8 4-1/2 5.000 *2.125 5-1/2 6.050 4.225 8-3/8 5-1/2 6.050 *3.725 6-5/8 7.390 6.405 8-1/2 6-5/8 7.390 6.280 7 7.656 6.812 8-3/4 6-5/8 7.390 6.030 7 7.656 6.562 9-1/2 7 7.656 5.812 7-5/8 8.500 7.500 9-7/8 7 7.656 5.437 7-5/8 8.500 7.125 10-5/8 7-5/8 8.500 6.375 8-5/8 9.625 8.625 8.250 9.625 8-5/8 11 9.000 10.625 9-5/8 12-1/4 11.250 11.750 10-3/4 12-1/4 9.750 11.750 10-3/4 13-3/4 10.750 12.750 11-3/4 14-3/4 11.250 14.375 13-3/8 17-1/2 14.000 17.000 16 20 15.500 19.750 18-5/8 24 16.000 21.000 20 26 *minimum drill collar size satisfies the equation but a larger size drill collar would be recommended

Table (9) lists the sizes of drill collars recommended for popular hole sizes. It should be noted that the collar sizes in Table (9) are the minimum sizes, and larger sizes are preferable. As a general rule, the largest drill collars that can be washed over and fished out should be selected. Table (10) lists the largest drill collar sizes that can be washed over and fished out. Large drill collars are preferred because fewer drill collars and less tripping time would be required. Also, large drill collars are stiffer and have less tendency to buckle or bend. This characteristic promotes even load distribution on the bit for better bit performance and lessens hole deviation problems. The stiffness of pipe is related to its moment of inertia, 314 . I= OD 4 − ID 4 .. ............................................(15) 64

(

)

It can be seen that the stiffness is proportional to the fourth power of the pipe diameter. Example Page 31

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Compare the stiffness of 9” x 3” and 6.5” x 2.25” drill collars Solution

Moment of inertia for the 9” x 3” drill collar is, I=

(

314 . 9 4 − 34 64

)

= 318

Moment of inertia for 6.5” x 2.25” drill collar is, I=

(

)

314 . 6.54 − 2.254 = 86 64

Note that by increasing the drill collar OD by a factor of 1.38, the stiffness increased by a factor of 3.69. In other words, if the 9 in diameter drill collar is deflected 1 in under a certain load, the 6.5 in diameter drill collar would deflect 3.69 in under the same load. Weight On Bit Calculations As was mentioned earlier, one of the functions of drill collars is to provide weight on the bit. The number of drill collars that are required to provide a desired weight on bit is usually based on buckling considerations in the BHA. There are two methods for calculating the required number of drill collars; the buoyancy factor method and the pressure area method.

The buoyancy factor method ensures that buckling is restricted to the drill collars and no buckling occurs in the heavy-weight drill pipe (HWDP) or drill pipe above the drill collars. Buckling is a problem that must be avoided at all times. Buckling in HWDP or DP induces stresses in the pipe which will cause premature pipe fatigue and pipe failure. The required collar length to provide a desired weight on bit can be calculated as follows, Lc =

WOB x SF .................................................(16) BF x Wc x COSθ

where, WOB = desired weight on bit, lb SF = safety factor (1.1-1.15) BF = buoyancy factor Wc = drill collar weight in air, lb/ft Θ = maximum hole angle at BHA, degrees

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DRILL STRING DESIGN Table (10) Minimum Drill Collar Sizes that Can Be Caught with Overshot and/or Washed Over with Washpipe Hole size, In.

Overshot Washpipe Maximum fish OD Size, In. Max. Size, In Max. fish to catch and/or catch, In OD, In washover, In. 6-1/8 *5-3/4 5-1/8 5-1/2 4-3/4 4-3/4 6-1/4 *5-3/4 5-1/8 5-3/4 4-7/8 4-7/8 6-3/4 *6-3/8 5-1/4 6 5-1/8 5-1/8 7-7/8 *7-3/8 6-1/4 7-3/8 6-1/2 6-1/4 8-3/8 *7-7/8 6-3/4 7-5/8 6-3/4 6-3/4 8-1/2 *8 6-7/8 7-5/8 6-3/4 6-3/4 8-3/4 *8-1/4 7-1/8 8-1/8 7-1/8 7-1/8 9-1/2 *9 7-7/8 9 8 7-7/8 9-7/8 *9-1/8 8 9 8 8 10-5/8 *9-3/4 8-5/8 9-5/8 8-1/2 8-1/2 11 10-1/2 8-7/8 10-3/4 9-5/8 8-7/8 12-1/4 11-3/4 10-1/8 11-3/4 10-1/2 10-1/8 13-3/4 12-3/4 11-1/4 12-3/4 11-1/2 11-1/4 14-3/4 13-3/4 12 13-3/8 12 12 17-1/2 15-1/8 13-3/8 26 14-1/2 13-3/8 20 16-3/4 14-3/4 18-5/8 17-3/8 14-3/4 24 20-1/4 16-3/4 21 19-1/2 16-3/4 26 24-3/4 22 21 19-1/2 19-1/2 *Overshots are not full strength and are limited in pulling, torsional and jarring strain. Note: Some sizes of Overshots and Washpipe may not be available.

As we will see later, the drill collar length calculated by Eq (16) is not enough to provide all the desired weight on the bit, and, therefore, the remainder of the weight will be provided by the HWDP or DP above the drill collars. Lubinski defined the buckling neutral point as the point in the drill collar string below which the pipe is buckled or will have a tendency to buckle, and above which no buckling will occur. The buckling neutral point should not be confused with the axial stress neutral point, the point where the axial stress is equal to zero (compressive and tensile stresses are zero). The buckling neutral point is calculated by the following equation, Buckling Neutral Point =

Page 33

WOB ...................................(17) BF x Wc

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Equation (17) states that no buckling occurs above the drill collars as long as the weight on the bit does not exceed the buoyed weight of the drill collars. It can be seen from Eq (16) that the buoyancy factor method considers only forces related to weight on bit and the weight of the drill collars and does not take into account the hydraulic forces acting on the bottom end of the drill collars and on the shoulder areas between the drill collar and the HWDP or DP as shown in Fig (10). The hydraulic forces are a result of the hydrostatic pressure of the mud and are computed by multiplying the hydrostatic pressure times the respective section area. In some cases it may be necessary to calculate the axial stress in the drill string or locate the axial stress neutral point. When axial stress must be determined, all forces acting on the BHA must be considered including the hydrostatic forces. F

HWDP

W

T

1

L P

F

F H

1

1

W

W

2

L

DC P

2

L C

C

2

WOB

F

2

Fig Fig(8) (8) To determine the axial stresses in the HWDP above the drill collars, consider the free body diagram in Fig (8). The BHA in Fig (8) consists of drill collars of length LC and HWDP of length LH. The hydraulic force acting on the cross sectional area between the DC and HWDP is denoted by F1. The hydraulic force acting on the bottom of the drill collars is denoted by F2. The weight on bit WOB acts on the formation, but since for every action there is a reaction equal in magnitude, there will be a reaction force equal to WOB acting upward on Page 34

T

F

WOB

2

Fig Fig (9) (9) Mud

DP F

3

HW DP

F1

DC F2 Fig (10) Fig (10)

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the bottom end of the drill collars. The total weights of the HWDP and DC are denoted by W1 and W2 respectively. The tensile force acting at the cut off point in the HWDP is denoted by FT. Now for the system to be in static equilibrium, the forces acting upward must be equal to the forces acting downward, or FT + WOB + F2 = W1 + W2 + F1 .......………...........................(18) If we define, W1 = LH WH W2 = LC WC F1 = P1 (A2-A1) F2 = P2 A2 Substituting in Eq (18) and solving for FT, FT = LH WH + LC WC + P1 (A2 - A1) - P2 A2 - WOB .................(19) where, WH = weight in air of HWDP, lb/ft WC = weight in air of drill collars, lb/ft LH = length of HWDP, ft LC = length of drill collars, ft P1 = hydrostatic pressure at top of drill collars, psi P2 = hydrostatic pressure at bottom of drill collars, psi A1 = steel cross sectional area for HWDP, in2 A2 = steel cross section area for drill collars, in2 FT = tensile force, lb It should be noted that FT is assumed to be a tensile force. If the magnitude of FT is negative then it is a compressive force. To determine the axial stresses in the drill collars consider the free body diagram in Fig (9). Note that in this case there is no hydrostatic force acting on the top of the drill collars because there is no change in diameter (no shoulder area). Adding the forces gives, FT + WOB + F2 = W2 FT + WOB + A2 P2 = LC WC Solving for FT, FT = LCWC - WOB - A2P2 .......…...................................(20) Page 35

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Example

A vertical Khuff well is drilled to 7600 ft with a mud weight of 103 pcf and WOB of 70,000 lb utilizing 10” OD x 3” ID drill collars and 5-1/2” OD x 3.375” ID 62.7 lb/ft HWDP. a) Calculate the length of drill collars required to provide 70,000 lbs WOB using the buoyancy factor method. Assume safety factor of 1.1. b) Determine the position of the buckling neutral point. c) Which parts of the BHA will buckle when weight is applied on the bit? d) Determine the axial stress 2 ft below the top of the drill collars. e) Determine the axial neutral point. Solution

a)

WOB = 70,000 lb 489 − 103 = 0.789 489 314 . 10 2 − 32 x 489 = 242.6 lb / ft WC = 4 x 144

BF =

(

)

SF = 1.1 cosθ = cos (0) = 1.0 From Eq (16), Length of drill collars = or b)

70,000 x 11 . = 402.3 ft 0.789 x 242.6 x 10 .

402.3 ≅ 14 drill collars 30

Buckling neutral point =

WOB 70,000 = = 365.7 ft BF x Wc 0.789 x 242.6

The buckling neutral point is 365.7 ft from bottom end of drill collars or 36.6 ft below the top of the drill collars.

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c)

All pipe below the buckling neutral point will buckle, i.e., the bottom 365 ft of the drill collars will buckle. The top 36.6 ft of the drill collars and all the HWDP will not buckle because they are above the buckling neutral point.

d)

Since we need to calculate the axial stress in the drill collars below the cross over between the drill collar and HWDP then Eq (20) must be used. LC = 402.3 - 2 = 400.3 ft WC = 242.6 lb/ft WOB = 70,000 lb A2 P2

(

)

314 . 10 2 − 32 = 7143 . in 2 4 103 = x 7600 = 5436 psi 144

=

Substituting in Eq (20), FT

= 400.3 x 242.6 - 70,000 - 71.43 x 5436 = - 361,180 lb

Since FT is negative then the axial force is a compressive force. This means that the drill collars are all in compression. e)

Since the drill collars are in compression, the point of zero axial stress must be in the HWDP. The axial neutral point can be calculated by using Eq (19) and setting the value of FT equal to zero, or 0 = LHWH + LCWC + P1(A2 - A1) - P2A2 - WOB Solving for LH, LH =

P2 A2 + WOB − LcWc WH

− P1 ( A2

Refer to Fig (11). P2 = 5436 psi P1 =

103 x 7197.7 = 5148 psi 144

LC = 402.3 ft WC = 242.6 lb/ft A2 = 71.43 in2

Page 37

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(

314 . 55 . 2 − 3375 . 2 4 WH = 62.7 lb/ft A1 =

) = 14.8 in

2

F1 = P1(A2-A1) = 5148(71.43 - 14.8) = 291,531 lb F2 = P2A2 = 5436 x 71.43 = 388,293 lb Substitution in the equation above, LH =

5436 x 71.43 + 70000 - 402.3 x 242.6 - 5148(71.43 - 14.8) = 1103 ft 62.7

This means that the axial neutral point is 1103 ft above the top of drill collars. In other words, the bottom 1103 ft of HWDP will be in compression when weight is applied on the bit. All HWDP and DP above the axial neutral point will be in tension. In the previous example we have seen that the drill collars and 1103 ft of HWDP are all in compression when weight is applied on the bit. It was also shown that the drill collars are buckled but the HWDP has no tendency to buckle. The function of the drill collars is to provide weight on the bit and, therefore, they are expected to be buckled and in compression. However, the questions that arise are if the WOB is provided by the drill collars, why 1103 ft of HWDP above the bit are in compression? and if they are in compression why aren’t they buckled?

F =0 T

LH P1

F1

7197

402.3

P2 F2

7600

WOB

Fig Fig(11) (11)

To answer the first question let us go back to the Buoyancy Factor method and Eq (16) which was used to calculate the length of the drill collars. This simple equation considers only the mechanical forces acting on the BHA, namely, the WOB and the weight of the drill collars, and does not take into account the hydraulic forces F1 and F2. In order to analyze the problem correctly, both the mechanical and the hydraulic forces should be taken into account. Referring to Fig (11), the forces that contribute weight to the bit are the forces that are acting downward, namely, F1 and the weight of drill collars LCWC. In the previous example F1 = 291,531 lb and the weight of drill collars is 402.3 x 242.6 or 97,598 lb for a total of 389,129 lb. The force F2(388,293 lb) acts upward and cancels out some of the downward force. So the net force acting downward is 389,129 - 388,293 or 836 lb. It is obvious that the available weight of 836 lb is much less than the desired weight on bit (70,000 lb).

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So when the driller slacks off 70,000 lb weight on the bit, where does the remaining 69,164 lb (70,000-836) come from? The answer is that they come from the HWDP or DP directly above the drill collars. The length of HWDP required to provide 69164 lb of weight is 69164 divided by 62.7 lb/ft or 1103 ft, which is equal to the length of HWDP that is in compression. Therefore, the answer to the first question is that the HWDP is in compression because it is providing some (or most) of the weight on the bit. The fraction of the total WOB that is provided by the HWDP depends on the total depth and mud weight. The fraction of the WOB contributed by the HWDP or DP above the drill collars increases as well depth and mud weight increase, and visa versa. Normally, if an axial compressive force is exerted at the end of a pipe, the pipe tends to buckle or bend. In the previous example the HWDP is under compression but we said it is not buckled because it is above the buckling neutral point. So how can we have a pipe under compression but not buckled? Lubinski has shown that buckling of drill strings can be induced only by mechanical compressive forces and that hydraulic compressive forces do not cause buckling (the reader is cautioned that this applies only to drill string assemblies and does not apply to tubing strings landed into bottom hole production packers). In the previous example if there were no hydraulic forces F1 and F2, the weight of the drill collars would have been more than enough to provide 70,000 lb on the bit, and the HWDP would have been in tension. What caused the HWDP to be in compression are the hydraulic forces F1 and F2, and these forces do not cause buckling. The drill collars are buckled because of the reaction force WOB acting upward on the bottom end of the drill collars. This force is a mechanical force and causes buckling.

Example 4000 psi

A 7000 ft string of open ended 5” 19.5# (ID = 4.276”) drill pipe is run in a cased well filled with 90 pcf mud. The BOP was closed on the DP at the surface and 4000 psi surface pressure was applied in the DP.

F2

a) Calculate the forces acting on the drill pipe. b) Calculate the axial neutral point. c) Does the drill pipe has a tendency to buckle?

Page 39

90 pcf mud

7000'

F1

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Solution

a) There are two hydraulic forces acting on the DP. A compressive force (F1) is acting upward on the bottom end and an upward tensile force (F2) acting on the top of the DP. F1 = P1 A1 P1 =

90 x 7000 + 4000 = 8375 psi 144

(

)

314 . 52 − 4.276 2 = 5.27 in 2 4 F1 = 8375 x 5.27 = 44,136 lb↑ F2 = P2 A2 314 . 4.276 2 = 14.35 in 2 A2 = 4 A1 =

(

)

P2 = 4000 psi F2 = 4000 x 14.35 = 57,412 lb↑ b) Take a free body diagram of the drill pipe as shown on the right. The forces acting on the system are the weight of the drill pipe W, the compressive force F1 and the tensile force FT. To be in equilibrium the upward forces must equal the downward forces, or FT + F1 = W Since the axial stress is zero at the axial neutral point, then FT = 0 or F1 = W = L x WH, Substituting for F1 and WH

Axial neutral point

FT

W

L

F1

44136 = L x 22.46 L = 1965 ft The axial neutral point is 1965 ft above the bottom end of the DP. This means that the bottom 1965 ft of the drill pipe are in compression and the remainder is in tension. c) The drill pipe will not buckle because the compressive force F1 acting on the bottom end is a hydraulic force which does not cause buckling. Page 40

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Most drilling engineers and drillers use the Buoyancy Factor method and Eq (16) to calculate the number of drill collars required to apply a desired weight on the bit. They erroneously believe that all the weight on bit is provided by the drill collars calculated from Eq (16) and that the DP or HWDP above the drill collars is always in tension. As we have shown above, all these concepts are incorrect. The sources of these misconceptions are errors and confusing statements made in some drilling books, service company literature and drilling seminars and courses. In summary, if the drilling engineer or driller decides to use the Buoyancy Factor method, he must remember the following: •

The number of drill collars calculated by the Buoyancy Factor method is not enough to provide all the WOB. Some of the WOB will be provided by the DP or HWDP directly above the drill collars. For this reason the DP or HWDP above the drill collars will be in compression but not buckled. It is an acceptable practice to use HWDP or DP in compression as long as it is not buckled.

•

The buckling neutral point is always near the top of the drill collars. The drill collars below the neutral point will have a tendency to buckle. The drill collars and HWDP above the neutral point will not buckle as long as the actual weight applied on the bit while drilling does not exceed the WOB used in the calculations. If the actual WOB exceeds the WOB used in the calculations then the number of drill collars must be increased, otherwise, the HWDP or DP above the drill collars will buckle. DP or HWDP should never be used in a buckled condition.

The Pressure-Area Method Unlike the Buoyancy Factor method, the Pressure-Area method takes into account all the forces acting on the BHA including the hydraulic forces. Consider the free body diagram in Fig (12). A force balance yields, F1 + W = F2 + WOB W = F2 + WOB - F1

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HWDP

where W is the weight of drill collars,

P1

F1

Substituting expressions for F1, F2 and W, LCWC = A2P2 + WOB - P1(A2-A1)

Lc

The length of drill collars required to provide weight on bit is, Lc =

A2 P2 + WOB − P1 ( A2 − A1 ) ....... 21 Wc

W P

2

WOB

F2

Fig (12)

Example

Using the same data in the previous example, a) Calculate the length of the drill collars required to provide WOB by using pressure area method. b) Determine the position of the buckling neutral point. c) Which parts of the BHA will buckle when weight is applied on the bit? d) Determine position of the axial neutral point. Solution

a) Using Eq (21), the length of the drill collars is, LC =

A2 P2 + WOB − P1 ( A2 − A1 ) WC

From the previous example, A2P2 = 388,293 lb

P1(A2-A1) = 291,531 lb

WOB = 70,000 lb

WC = 242.6 lb/ft

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LC =

388,293 + 70,000 − 291,531 = 687.4 ft 242.6

b) The buckling neutral point is calculated from Eq (17), Buckling Neutral Point =

WOB 70,000 = = 365.7 ft Wc x BF 242.6 x 0.789

c) All drill collars below the buckling neutral point will buckle. The top 321.7 ft of drill collars and HWDP will not buckle. d) Consider the free body diagram on the right F T + WOB + F2 = F 1 + W

FT F1

Neutral point

F T = 0 (neutral point) Substituting the values of the variables, 0 + 70,000 + 388,293 = 291,531 + L x 242.6 L W

Solve for L, L =

70,000 + 388,293 − 291,531 = 687.3 ft 242.6

The axial neutral point is at the top of drill collars. This means that the DP or HWDP on top of drill collars is in tension.

WOB

F

2

A comparison between the Buoyancy Factor method and the Pressure-Area method is shown in Fig (13). It can be seen that the drill collar length calculated by the PressureArea method is almost twice that calculated by the Buoyancy Factor method and, therefore, is enough to provide all the weight on the bit. For this reason, only the drill collars are in compression while the HWDP is in tension. The buckling neutral point is the same in both cases. Either of the two methods can be used to calculate the length of the drill collars. However, the Pressure-Area method has the following disadvantages: •

Requires more drill collars to keep the HWDP or DP in tension. This serves no useful purpose because whether the pipe above the buckling neutral point is in tension or compression is irrelevant to fatigue damage, if the pipe is not buckled.

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•

The need to procure, transport, maintain, inspect and handle the extra drill collars increase the cost of the drilling operation.

•

Adding more drill collars reduce the available overpull.

•

Adding more drill collars increases the weight of the drill string and the tensile stress in the drill pipe at all depths. The increase in stress will increase the rate of fatigue attack and reduce the life of the drill pipe. H WD P in T ension

Ax ia l Str e s s N e u tr a l Po in t

Ax ia l Str e s s N e u tr a l Po in t

1103'

6913'

C ollars in C ompression but not buckled

H WD P in C ompr ession ( N ot Buckled) 7198'

687' Bu c k lin g N e u tr a l Po in t

402'

365' C ollars in C ompr ession and Buckled 7600' 7600' WOB = 70,000 lb

WOB = 70,000 lb

Buoyancy F actor M ethod

Pr essure Area M ethod

Fig (13) F ig (13)

HEAVY WEIGHT DRILL PIPE In the past, the two main components of the drill string consisted of the drill pipe and the drill collars. The point where the relatively small OD and flexible drill pipe connects to the large stiff drill collars is called the transition zone. Field studies have shown that almost all of the drill pipe fatigue failures are the result of an accumulation of fatigue damage occurring when the drill pipe joints were run in the transition zone, or were stressed above the endurance limit in crooked holes. Downhole data has indicated that the large change in diameter at the transition zone caused accelerated fatigue damage as a result of the concentration of cyclic bending stress reversals in the bottom joints of the flexible drill pipe, since the stiff drill collars bend very little from these stress reversals. Field tests indicated that fatigue build up in the drill pipe in the transition zone is related to the relative stiffness of the drill collar and the adjacent drill pipe. The stiffness ratio of two sections of pipe in the drill string is expressed by the equation,

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SR = where,

(

( I / C) lower ( I / C) upper

314 . OD 4 − ID 4 64 OD = outside diameter, in

I = moment of inertia =

)

C = external radius =

OD 2

ID = inside diameter, in

The field tests showed that the higher the stiffness ratio at the transition zone the greater the fatigue build up. Criteria for permissible stiffness ratio varies between different operators and areas. The following maximums are typical:

•

For shallow or routine drilling or low failure rate experience, keep stiffness ratio below 5.5.

•

For more severe drilling or for significant failure rate experience, keep SR below 3.5.

In order to reduce the stiffness ratio and increase service life of the drill pipe, heavy weight drill pipe having the same OD as the drill pipe and a wall thickness of up to 1 in and weight up to 78 lb/ft is used between the drill pipe and the drill collars. The number of joints used varies between 15 to 21 joints. Use of heavy weight drill pipe offers the following advantages:

•

Reduces drilling cost by eliminating drill pipe failures in the transition zone.

•

Significantly increases performance of small rigs through the ease of handling

•

Provides substantial savings in directional drilling by replacing most of the drill collar string, reducing down hole drilling torque and drag.

•

reduces tendency to become differentially stuck. This is due to the fact that large diameters are easier to stick than small diameters.

Heavy weight drill pipe normally has the same external dimensions as the regular drill pipe. In some types of HWDP the tool joints are longer and, in some types, an extra mock tool joint is located in the center of the joint as shown in Fig (12). Dimensions and weights of HWDP vary for different manufacturers. Typical dimensions are shown in Table (11).

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Fig (14) Heavyweight Drill Pipe

Table (11) Page 46

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DRILL STRING DESIGN Heavy Weight Drill Pipe Dimensions and Weights 1 Nom. Size (in) 3-1/2 4-1/2 5 5-1/2 6-5/8

SMITH INTERNATIONAL (DRILCO) - STANDARD 2 3 5 6 7 8 9 10 (W) Appro Center Upset Tool Joint data x ID OD Length Type ID OD Length wt/ft (in) (in) (in) (in) (in) (in) (lbs) 26.7 2-1/4 4 24 NC38 2-3/8 4-3/4 25/23 45.0 2-3/4 5 24 NC46 2-7/8 6-1/4 25/23 53.7 3 5-1/2 24 NC50 3-1/16 6-1/2 25/23 62.7 3-3/8 6 24 5-1/2FH 3-1/2 7 25/23 76.3 4-1/2 7-1/8 24 6-5/8FH 4-1/4 8 25/23

BOTTOM HOLE ASSEMBLY The bottom hole assembly (BHA) is the part of the drill string below the drill pipe. It consists of several types of components or tools which provide different functions. The most common components of the BHA are drill collars, stabilizers, shock absorbing subs, jars, reamers, heavy weight drill pipe and the bit. The drill collars and HWDP were discussed in the previous sections, and the remaining components are described individually below. This section provides the reader with a general understanding of the concepts involved and how certain assemblies will react under average conditions. It is the responsibility of the drilling engineer to apply these concepts to his own particular case and gain practical experience to continue improving the selection of BHAs in his area of concern. The purpose of the BHA is to drill a usable hole economically. This objective is achieved with the proper selection of the drill bit and drill collars which are required to provide high bit weights to improve the penetration rate. Stabilizers are needed to minimize the rate of hole angle change and prevent the formation of doglegs and key seats. Shock absorber play an important part to prolong the service life of the bit and drill pipe. Down hole drilling jars may be needed to unstick the BHA.

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Fig (15)

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Stabilizers Stabilizers are used to centralize the drill collars in the hole and increase its rigidity or stiffness. This increases the ability of the drill collars to drill a smooth and straight hole and reduces the undesirable bit movement, such as bit wobble, which reduces bit life. Stabilizers also provide some reaming action and wipe the walls of the hole to ensure a full-gauge hole. There are several types of stabilizers which are described below. The most common type of stabilizers is the steel body spiral stabilizer. These stabilizers shown in Fig (15) have wings or blades that are an integral part of the stabilizer body. The blades make a 360o contact with the well bore. The outer surfaces of the blades are curved to fit the curvature of the well bore wall. This bearing surface provides bore hole wall contact and permits the stabilizers to hold the drill collar assembly centered in the hole. The outer surface of the blades are covered with hard metal such as tungsten carbide to resist erosion. The stabilization efficiency of the stabilizer increases as the OD of the stabilizer blades approaches bit diameter. For 6” through 12-1/4” holes sizes the blade diameter is equal to the bit diameter or 1/32” under gauge. For 13-3/4” through 171/2” hole sizes the blade diameter is equal to bit diameter or 1/16” under gauge. When the spiral blade stabilizer becomes worn, the bearing surfaces are built up by welding on a layer of hard metal that is then machined to the correct OD. The length and width of the blades depend on hole size and type of formation. For 14-3/4” to 26” hole sizes the blade length varies from 18” to 36”, whereas for 6” to 12-1/4” holes sizes the blade length is about 12”. Large blade areas are needed to provide adequate support in soft formations. Thicker blades increase the torque and are harder to mill over if the stabilizer becomes stuck. There are two types of spiral blade stabilizers: the integral blade stabilizer and the replaceable sleeve stabilizer. The blades of the integral blade stabilize are an integral part of the stabilizer body. Whenever the stabilizer has worn down to an unacceptable condition the entire stabilizer is sent to the shop for reconditioning. This stabilizer is sturdier than the replaceable sleeve stabilizer and is suitable for hard and abrasive formations. The replaceable sleeve stabilizer consists of the mandrel and the spiral sleeve. When the blades wear out, the sleeve can easily be detached from the mandrel at the rig and replaced with a reconditioned or new sleeve. Saudi Aramco uses both types of stabilizers; the integral blade stabilizer is used in the small hole sizes whereas the sleeve stabilizer is used in large holes.

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The replaceable wear pad (RWP) stabilizer shown in Fig (16) consists of four 3 ft long vertical (straight) replaceable pads. Large diameter, pressed-in tungsten carbide compacts on the surface of the pad prolong wear and keep the stabilizer in gauge. The long pads provide large contact area which make the stabilizer suitable for areas of extreme deviation tendencies. The wear pads can be changed easily at the rig. The stabilizer may be resized to a different hole size by replacing the pads with a set that has been manufactured to a different diameter.

Fig (16)

Fig (17)

The RWP stabilizer is good for deviation control. However, the RWP stabilizer is difficult to wash over because of the inserts used and it generates higher torque than the spiral stabilizer because of its longer pad length. The RWP stabilizer is not used in Saudi Aramco drilling operations.

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A third type of stabilizers is the non-rotating stabilizer shown in Fig (17). It consists of a mandrel and a polyurethane stabilizer sleeve which is free to rotate on the mandrel Since the polyurethane stabilizer sleeve does not rotate during drilling (only the mandrel rotates), the non-rotating stabilizer is used for reducing drilling torque by employing the ease of rotation between the polished mandrel and the polyurethane sleeve. Its use is most common in the very large hole sizes where the use of spiral steel stabilizers would generate excess torque. Non rotating stabilizers are rarely used as near bit stabilizers because their radial stiffness is no more than the drill bit itself. Rather, they are used two or three collars above the bit. Advantages of the non-rotating stabilizer are low cost and low torque and drag. The disadvantages include low radial stiffness, rapid wear and tearing of the urethane pads and inability to ream undergauge holes. The non-rotating stabilizers are used in Saudi Aramco to stabilize the drill string in large hole sizes.

Reamers The basic function of the reamer is to open an undergauge hole to its original full-gauge size. Most reamers today have roller cutters which are free to rotate on their own axes. A three-point reamer has three roller cutters spaced 120o apart on the reamer body as shown in Fig (18). A six-point reamer has two rows of roller cutters and are staggered such that the six rollers are spaced around the reamer body 60o apart. Various types of cutters are available, ranging from mill-toothed cutters, flat faced cutter from medium to hard formations and tungsten 3 PT. Reamer

6 PT. Reamer

Fig (18)

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Reamers are often run as a near bit stabilizer. Stabilizers may cause excessive torque due to the dragging action of the blades. Roller reamers prevent this action. If roller reamers are used as a stabilizer in a packed-hole assembly, hard cutters should be used to ensure that they maintain full gauge.

Shock Absorbers Shock absorbers are used between the bit and drill collars to reduce the vertical oscillation (bouncing) of the drill string. Field studies have shown that the frequency of the oscillations was consistently three oscillations per revolution with a three cone bit. It is believed that the bit does not drill the bottom of the hole evenly (completely flat) with a resulting condition of high and low places, the number of high and low places being the same as the number of cutters on the bit. The peak-to-peak amplitude is approximately 0.5 inch. This bottom hole condition could account for the three oscillations per revolution when drilling with a three cone bit.

Fig (19)

The studies show that the placement of a shock absorber between the bit and the drill collars can almost completely eliminate the vertical oscillations as shown in Fig (19). Vertical bouncing causes fluctuations in the bit load which has adverse effect on bit footage. Field tests have shown that the use of shock absorbers increases the bit footage (bit life). The use of the shock absorbers has also shown substantial reduction in drill collar connection failures in hard formations by dampening the high fluctuating buckling Page 52

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loads imposed by bit vibrations. Unfortunately, shock absorbers are prone to mechanical failure due to the complexity of the device and the fact that it cannot be made as strong as the drill collars. For maximum effectiveness, the shock tool should be placed immediately above the bit to minimize the amount of unsprung mass below the tool. It is important that the tool is placed in a position where it is exposed to minimum side loading or bending stress. Ideally, the tool should have the same degree of stabilization at both ends to minimize buckling. The principle of operation of a Griffith shock absorber can be explained with the aid of Fig (20). Vibrations from the bit cause the shaft to move down inside the mandrel and compress the spring (compression stroke). Some of the vibration force will not be transmitted to the drill string above the tool because some of that force has been stored as energy in the spring, thus reducing the shock or impact on the drill string. As the shaft moves up (expansion stroke) the oil below the spring support between the shaft and mandrel is forced to flow through a small restriction to the upper chamber. The flow of oil generates high pressure drop which is converted to heat, thus dissipating some of the vibration energy into heat. D r ill C o lla r

S h a ft S p ri n g S p rin g S u p p o r t O uter M an dr e l

O il O -r in g Seal attac hed to s haf t

D r i l l C o l l ar

(a )

(b ) E xpa nti on S tr oke

C om pr e s s i on s tr oke

F i Fig g (2(20) 0) S cSchematic he m a ti c of o faaGriffith G ri ffi th S ho Absorber c k A b s o rb e r Shock

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Jarring Devices Jars are placed in the BHA to generate upward or downward impact loads to free stuck pipe or release a fish. There are many factors which affect the decision of when to use jars and where to position them for maximum benefit. Jars should be used when:

• • • •

there are sloughing formations in a given area or at a particular depth. there are sensitive swelling shales. the mud system used does not have good suspension properties to suspend cuttings.. There is costly equipment in the bottom hole assembly, such as monel drill collars, downhole motor or MWD tools that need to be recovered.

Jars can be run in compression or tension according to manufacturer recommendation, but never at the axial neutral point. To be of most benefit, the jars should be positioned in the BHA a small distance above the point where sticking is likely to occur. If the jars are placed far above the stuck point, some of the jarring action is wasted in stretching the pipe between the jars and stuck point . Jars are rendered useless if they become stuck or are placed below the stuck point. Prediction of the location of the stuck point is not straight forward and depends on the types of formations drilled, wellbore conditions and driller’s experience in the area. For example, if a BHA with three stabilizers is used to drill sloughing shale, it is likely that the shale will slough and pack around all three stabilizers. In this case the jars are often placed above the top stabilizer. Usually, few drill collars are placed above the jars to achieve effective jarring force on the drill collars below the jars. If differential sticking is likely to occur, all drill collars could become stuck. In this case, it would be advantageous to place the jars in the HWDP above the drill collars. The most common type of hydraulic jars operates on a time delay sequence wherein hydraulic fluid is metered through a small opening for the initial extension of the mandrel. After moving a small distance over several minutes, the fluid opening size increases dramatically and the jar opens unrestrained. Finally, when the jar has reached the end of its stroke a tremendous jolt is achieved by rapidly decelerating the collars and drill pipe above the jars that had built speed during the unrestrained portion of the opening cycle. The magnitude of the jars impact depends on the tension applied to the jars when they are fired.

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The time required for a hydraulic jar to fire depends on the type of hydraulic fluid used, the size of the metering hole and the temperature of the hole. Repeated firing of jars generally increases the hydraulic fluid temperature and, therefore, lowers the viscosity of the fluid and the time required for firing to occur to a point where the jar becomes useless. The principle of operation of the jar can be explained with the aid of the schematic diagram in Fig (21). When the drill string is pulled, the piston moves up inside the housing and forces the hydraulic oil to move down through the small annular clearance between the piston and the housing which restrains the movement of the piston. It takes several minutes for the piston to move out of the small diameter housing. Once the piston moves into the larger ID area, the piston velocity increases dramatically, and when the hammer hits the anvil a tremendous jolt is achieved. The magnitude of the impact is directly proportional to the amount of tensile pull applied on the jars before firing.

Seal Anvil

oil

oil

oil

Hammer Mandrel Housing Piston

oil

Fig (21)

Mechanical Jars Another common type of jars are mechanically operated ones. These are preset at surface or in the shop to fire at a given tension. While they do not allow various jarring tensions, there is no time delay present as in the case of hydraulic jars. To increase or decrease the jarring tension, the jars must be tripped to surface, in other words the string must become unstuck.

BOTTOM HOLE ASSEMBLY DESIGN As was mentioned earlier, the purpose of the bottom hole assembly is to drill a useful vertical hole with full gauge, smooth bore and free of doglegs and ledges. The simplest bottom hole assembly is the slick assembly which consists of a bit and drill collars and no stabilizers as shown in Fig (22). In the vicinity of the bit, the string does not contact the wall of the hole. At some distance above the bit, the string contacts the wall. Above the point of contact the string lies on the low side of the hole. With no weight on the bit, the Page 55

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I Point of tangency

Active D.C. Length

F sin I

I F cos I

Drill collar axis

Hole axis

W2 W

W1

Fig (22)

F

only force acting on the bit is the force F which is the weight of the portion of the string between the bit and the point of contact. This force can be resolved into two components, F cost I along the centerline of the hole and F sin I perpendicular to the hole axis. The force component F sin I is a beneficial force because it tends to bring the hole toward vertical. The second force acting on the BHA is the weight on bit W, which can be resolved into components W1 and W2 . The force component W2 is responsible for hole deviation and its magnitude increases with increasing clearance between drill collars and hole and with increases in weight on bit. The magnitude and direction of the resultant hole deviation is dependent on the difference between W2 and F sin I.

The major source of natural hole deviation is formation characteristics. Laminated structures composed of alternating soft and hard bands can cause hole deviation. Practical experience has shown that laminated dipping formations cause hole deviation. When the dip angle is less than 45o, the bit tends to drill up dip. When the dip angle is greater than 45o, the bit tends to drill down dip. In general there are three types of bottom hole assemblies: slick, pendulum and packed BHA. The slick BHA consists of a bit and drill collars without stabilizers. This BHA is suitable for formations which have mild crooked hole tendencies. Slick bottom hole assemblies are seldom used. The pendulum BHA is used primarily to reduce or maintain hole deviation. The pendulum technique relies on the force of gravity to deflect the hole to vertical. The force of gravity, F sin I in Fig (22), is related to the length of drill collars between the drill bit and the point of tangency. The along-hole component of the force F is F cos I, which attempts to maintain the present hole direction. The pendulum BHA consists of a bit, several drill collars to provide the pendulum force and one or more stabilizers. Lubinski presented charts for determining the location of the first stabilizer from the bit and the weight that can be applied on the bit to maintain

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present hole deviation. It was found that the best position of the stabilizer is as high as possible provided the collars below stabilizer do not contact or barely contact the wall of the hole as shown in Fig (23). The optimum position of the stabilizer depends upon the size of the collars, size of hole, hole

Stabilizer

Drill Collars

One Stabilizer Two Stabilizers

Pendulum Assemblies (After Wilson, 1979)

Fig. (23)

inclination and weight on bit. If the weight on bit obtained from the graphs is maintained constant, the hole deviation will not change. If the weight on bit is increased, the hole angle will increase and the opposite is true. The main disadvantage of the pendulum BHA is that the stabilizer must be placed within a few feet of a definite position obtained from the charts. Also, if there is a need to change the weight on bit the BHA must be pulled out to change the position of the stabilizer.

The packed bottom hole assembly relies on the principle that two points will contact and follow a sharp curve, while three points will follow a straight line as shown in Fig (24). A three-point stabilization is obtained by placing three or more stabilizers in the portion of the hole immediately above the bit. There are three different types of packed BHA:

3 2 2

2

1 1 1

The packed BHA results from the 1. Mild Crooked Hole Packed BHA basic idea that three points cannot A typical BHA is shown in Fig (25a). This is contact and form a curved hole. used for formations which have mild crooked Fig. (24) hole tendency that produce little or no deviation such as hard and isotropic rocks. The three-point stabilization is provided at Zone-1, immediately above the bit, at Zone 2, immediately above a short, large OD drill collar; and at Zone 3, on top of a standard length drill collar. This type of BHA is commonly used in Saudi Aramco drilling operations. If a vibration dampener (shock absorber) is required, it should be placed at Zone-2 for maximum effectiveness.

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2. Medium Crooked Hole Packed BHA. This BHA is used for formations which have medium crooked hole tendency such as medium and soft formations. In this type of BHA a second stabilizer is included at Zone-1 in order to provide increased bit stabilization against deviation effects of the formation. 3. Severe Crooked Hole Packed BHA This type of BHA is used for drilling formations which have severe crooked hole tendency such as medium and soft formations which show a great degree of dipping, fracturing and variation in strength. In this type of BHA three stabilizers are included in Zone-1.

Stabilizer

Zone III

30 ft Drill Collar

Stabilizer

30 ft Drill Collar

Vibration Dampener, or Shock-Sub Stabilizer

Vibration Dampener, or Shock-Sub

Zone II

Short Drill Collar

Stabilizer Short Drill Collar

Zone I

Stabilizer

Stabilizer or Reamer

(a)

Stabilizer or Reamer

(b) Fig (25)

Packed Hole Assembly for (a) Mild, (b) Medium, (c) Severe, Crooked Hole Tendencies, (After Wilson, 1979)

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FATIGUE DAMAGE Most drill pipe failures are a result of fatigue damage. Drill pipe will suffer fatigue when it is rotated in a section of hole in which there is a change of hole angle and/or direction, commonly called a dogleg. The amount of fatigue damage depends upon the tensile load in the pipe at the dogleg, the severity of the dogleg and the dimension and properties of the pipe. Since tension in the pipe is critical, a shallow dogleg in a deep hole often becomes a source of difficulty. Rotating off bottom is not a good practice since additional tensile load results from the suspended drill collars. Lubinski and Nicholson have published methods of calculating forces on tool joints and conditions for fatigue to occur. Curves are published in API RP7G to determine the maximum permissible dogleg severity above which fatigue damage will occur for a given tensile load below the dogleg. The curves are based on the following equations, C=

432,000 σb tanh( KL ) .....................................(22) 314 . ED KL K=

T EI

where, C = maximum permissible dogleg severity, deg/100 ft E = Young’s modulus, psi (30 x 106 for steel). D = OD of pipe, in L = half the distance between tool joints, 180 in for 30 ft joint. T = buoyant weight suspended below the dogleg, lb σb = maximum permissible bending stress, psi 314 . I = moment of inertia = D4 − d4 64 d = drill pipe ID, in

(

)

Equation (22) holds true for range 2 pipe only. The maximum permissible bending stress, σb, is calculated from the buoyant tensile stress σt, T ...…….......................................................(23) σt = A where, Page 59

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A

= cross sectional area of pipe body, in2

For Grade E drill pipe, 10 0.6 σt − (σt − 33500) 2 .....................(24) 2 67 ( 670) Eq (24) holds true for values of σt up to 67,000 psi. For Grade S drill pipe, σt ⎞ ⎛ σb = 2000 ⎜1 − ⎟ .....................……….................(25) ⎝ 145000 ⎠ σb = 19500 −

If doglegs of sufficient magnitude are present, it is good practice to string ream the dogleg area. This reduces the severity of the hole angle change and reduces fatigue. Corrosive conditions have detrimental effect on the fatigue life of drill pipe. In corrosive environment, the dogleg severity calculated from Eq (22) should be reduced to a fraction of the calculated value (0.6 for very severe corrosive condition). The fatigue life of steel drill pipe may be increased by maintaining a mud pH of 9.5 or higher. Example

A 5” 19.5# Grade E, R-2 drill pipe and 600 ft of 9” x 2.5” drill collars are used to drill a well at 11,600 ft. A dogleg developed at 3,000 ft. What is the maximum dogleg severity that can be allowed at 3,000 ft to avoid fatigue damage? Drill pipe ID = 4.276 in Mud weight = 80 pcf Actual drill pipe weight = 22.6 lb/ft Weight on bit = 60,000 lb Weight of collars = 200 lb/ft Solution

489 − 80 = 0.836 489 The tensile load at the dogleg = [(11,000 - 3000) 22.6 + 600 x 200] 0.836 - 60,000

Buoyancy factor =

T = 191,468 lb A=

(

314 . 52 − 4.276 2 4

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σt =

191,468 = 36,318 psi 5.27

σb = 19,500 -

I=

0.6 10 2 (36,318) 2 (36,318 - 33,500) = 19,500 - 5420 - 10.6 = 14,069 psi 67 ( 670 )

(

314 . 54 − 4.276 4 64

K =

T = EI

) = 14.26

191,468 30 x 10 6 x 14.26

= 0.0211

KL = 0.0211 x 180 = 3.80 tanh KL =

e kl − e − kl e

kl

+e

− kl

=

44.7 − 0.0223 44.677 = = 0..9989 44.7 + 0.0223 44.722

From Eq (22), C=

432000 14069 0.9989 x = 3.4 deg / 100 ft 6 314 38 . . 30 x 10 x 5

Therefore, in order to avoid fatigue damage the dogleg severity at 3000 ft should be less than 3.4 deg/100 ft.

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DRILL STRING DESIGN FOR INCLINED AND HORIZONTAL WELLS As was discussed in the previous sections, the drill string in vertical wells is designed so that the mechanical weight on the bit is provided by the drill collars. The buckling neutral point should always be below the top of the drill collars to prevent any buckling to the DP and HWDP. The drill pipe in vertical hole should always be in tension because drill pipe has very low resistance to buckling. A small mechanical compressive force of few thousand pounds will buckle the drill pipe. It is common practice in drill string design that the drill pipe should not be rotated in a buckled condition as this will cause rapid fatigue failure. In drilling inclined and horizontal wells there are two additional factors, which are not present in vertical wells, that must be considered. These are (1) the frictional forces between the drill string and the hole and (2) the ability to use the drill pipe or HWDP to provide weight on the bit without buckling. Because of the hole geometry some or all of the weight of the drill string in inclined or horizontal wells is exerted on the low side of the hole. This will create a frictional force or drag between the drill string and the hole that will require additional force or pull to move the drill string up or down the hole. The frictional force will also increase the torque that is required to rotate the drill string. The second factor that must be considered in the design of the drill string is the fact that drill pipe can be used in compression to provide weight on the bit. In inclined and horizontal wells the drill pipe can tolerate significant levels of compression without buckling in small-diameter holes. The reason that the drill pipe in inclined holes is so resistant to buckling is that the hole is constraining and supporting the pipe throughout its length. The low side of the hole forms a trough that resists even a slight displacement of the pipe from its initial straight configuration. The effect of gravity and the curving sides of the hole form a restraint against buckling.

Prediction of Torque and Drag Ft

An object of weight W is resting on a horizontal plane as shown in Fig (26). In order to slide the object a force Ft must be exerted to overcome the force of friction Ff between the object and the plane surface.

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W

Fig. (26)

Ff

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The magnitude of the frictional force is Ff = µW ...................................................................................... (26) where, Ff = frictional force, lb µ = coefficient of friction, dimensionless W = normal force acting perpendicular to the surface, lb Therefore, the force Ft required to slide the object should be slightly greater than Ff or, Ft > Ff = µW.............................................................................. (27) The force Ft is called the drag force and acts in the opposite direction of the frictional force as shown in Fig (26). The coefficient of friction is a constant that depends on the roughness of the object and plane surface. The value of µ is determined experimentally. Now consider a section of drill pipe in a wellbore inclined at an angle θ as shown in Fig (27). The buoyed weight of the drill pipe W is acting vertically downward and can be resolved into two forces Fn and Fx. The force Fn is the normal force acting perpendicular to the hole and is equal to,

Ft

θ

Fx

Fn W

Ff

Fig. (27)

Fn = W sin θ ............................................................................... (28) The force Fx is the weight component acting parallel to the hole axis and is equal to, Fx = W cos θ ............................................................................. (29) The frictional or drag force Ff is the normal force times the coefficient of friction, or Ff = Fn µ = Wµ sin θ ............................................................... (30) The force Ft required to move the drill pipe is determined by making a force balance, forces acting up must equal forces acting down, or

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Ft = Fx + Ff Ft = W cos θ + Wµ sin θ ……………………… .....................(31) The frictional force or drag always acts in the opposite direction of the pulling force Ft. If the drill pipe is lowered into the hole, the frictional force will act in a direction opposite to that shown in Fig (27). Example

A 5”, 19.5# drill pipe is in a tangent section 2000 ft long inclined at an angle 60 deg from vertical. The mud weight is 90 pcf and the coefficient of friction is 0.25. (a) How much drag does the section of drill pipe contribute while tripping out of the hole? (b) What is the tension required to move the drill string up hole assuming no drag? (c) How much tension (pull) is required to move the drill string up hole? Solution

(a)

The normal force acting perpendicular to the hole is, Fn = W sin θ Buoyancy factor =

⎛ 490 − 90 ⎞ ⎜ ⎟ = 0.815 ⎝ 490 ⎠

W = 2000 x 19.5 x 0.815 = 31785 lb Fn = 31785 sin60 = 31785 x 0.866 = 27525 lb Ff = 27525 x 0.25 = 6881 lb (b) By doing a force balance along the wellbore axis in Fig (27), Ft = Fx + Ff Assuming there is no friction, Page 64

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Ff = 0,

and

Ft = Fx = W cos θ Ft = 2000 x 19.5 x 0.815 x cos60 = 15892 lb (c)

Ft = Fx + Ff Fx = W cos θ = 15892 lb From Eq (30) Ff = Fnµ = Wµ sin θ = 2000 x 19.5 x 0.815 x 0.25 x 0.866 = 6881 lb Ft = 15892 + 6881 = 22773 lb

The drag force expressed by Eq (30) is for straight inclined holes where the angle of inclination and azimuth are constant across the hole section. In a curved section of the wellbore where inclination and azimuth change across the section, the calculation of the drag force is more difficult. Fig (28) shows the forces acting on a short slightly curved element of the drill string. In addition to the normal force from the weight of the element, there are normal forces from the tension forces Ft and Ft + Δ Ft as a result of the change in the angle of inclination and the azimuth of the wellbore. The magnitude of the resultant normal force can be estimated by an equation presented Johancsik, Friesen and Dawson,

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Ft + ΔFt

NET SIDE LOAD, Fn Ff Wcosθ

W

Fig. (28) Force balance on drillstring element illustrating sources of normal force

Ft

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[(

Fn = Ft Δα sin θ

) + (F Δθ + W sin θ ) ]

1 2 2

2

t

................................... (32)

where Fn Ft

= resultant normal force, lb = axial tension on lower end of element, lb θ = average inclination angle, degrees Δ θ = change in inclination angle, radians (1 deg = 0.0174 radians) Δ α = change in azimuth, radians W = buoyed weight of element, lb

(

)

The first term in Eq (32) Ft Δα sin θ is the normal force due to change in azimuth, the second term Ft Δ θ is the normal force due to change in angle of inclination, and the third term W sin θ is the normal force due the weight of the element. If there is no change in azimuth and angle of inclination then Δ α and Δ θ are zero and Eq (32) will reduce to Fn = W sin θ which is the same as Eq (28).

Calculation of Drag Calculation of drag across a hole of constant inclination and azimuth (tangent section) is simple as was illustrated by the above example problem. Calculation of drag across a curved section is more complex. The drill string is divided into small elements 100 to 50 + ft long and the drag is calculated across each element. The total drag is the sum of the drag across all elements. The drag across an element is the coefficient of friction times the normal force, or Ff = µ Fn ..................................................................................... (33) Referring to Fig (28), if the tension at the bottom of the element is Ft, then the tension at the top of the element is (by force balance), or,

Ft + Δ Ft = Ft + Wcos θ + Fnµ.................................................... (34)

Δ Ft = Wcos θ + Fnµ .................................................................. (35)

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If the string is lowered in the hole, then

Δ Ft = Wcos θ - Fnµ ................................................................... (36) It should be noted here that θ is the average of the angles of inclination at the bottom and top of the element. The value of Fn is calculated by Eq (32) for each element. Δ θ in Eq (32) is the inclination at the bottom of the element minus the inclination at the top of the element. Δ α is calculated in a similar manner. The value of Ft + Δ Ft for the first element (bottom most element) becomes Ft for the second element. The calculation process is repeated for all elements to the surface. The value of Ft + Δ Ft for the last element at surface will be the total tension that must be applied to move the string up hole. The total tension is the sum of all drag forces and the weight components parallel to the hole axis, or

∑ (F ) + ∑W n

Total tension Ft =

n =1

n

f

n

n =1

n

cos θ n .......................................... (37)

Calculation of drag forces in an inclined wellbore with curved section is illustrated by the following example. Example

Given: Well bore 9-5/8” 40# casing at 5000 ft Total Depth = 10,000 ft Hole Size = 8-1/2” Kick off Point = 5000 ft Build Rate = 10 deg/100 ft End of Build at = 5600 ft Tangent Section = 5600 to 10,000 Inclination across tangent section = 60 deg Azimuth = 0 deg north (surface to 5000 ft) Azimuth is changed at the rate of 5 deg / 100 ft from 5000 ft to 5600 ft Azimuth = 30 deg from 5600 ft to 10,000 ft Mud Weight = 90 pcf Coefficient of friction = 0.25 in open hole and 0.2 in casing Drill String Page 67

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Drill pipe 5” OD, 5.276”ID, 19.5# normal weight 20.89# adjusted weight Drill Collars 7” OD, 2.25” ID, 117.42#, 400 ft long With the drill string at bottom, calculate: a) b) c)

The tension at surface required to move the drill string up hole Total drag force when moving string up hole The hook load when the string is off bottom and not moving

Solution

The first step is to divide the drill string into small elements from bottom to top as shown in Table (12). The drill collars are considered as one element since there is no change in inclination or azimuth across them. The drill pipe in the tangent section from 9600’ – 5600’ is also taken as one element for the same reason. The drill pipe in the curved section from 5600 to 5000 is divided into twelve 50-ft elements because the inclination and azimuth change across the curved section. The last element is the vertical section from 5000’ to surface. The entries in the columns are defined as follows: Column # 1:

element sequence number

Column # 2:

depth of the bottom of the element

Column # 3:

element length in feet

Column # 4:

angle of inclination in degrees at the bottom of the element

Column # 5:

average angle of inclination across the element, or angle at bottom of element plus angle at top of element divided by 2.

Column # 6:

change in angle of inclination across the element in radians, or angle at bottom of element minus angle at top of element. Multiply answer by 0.0174 to convert from degrees to radians.

Column # 7:

azimuth at bottom of element in degrees.

Column # 8:

change in azimuth in radians, or azimuth at bottom minus azimuth at top multiplied by 0.0174.

Column # 9:

buoyed weight of the element which is the weight in air times the buoyancy factor. For drill pipe use the actual or adjusted weight per foot.

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Column # 10:

tension force Ft at the bottom of the element. In this case since the drill string is off bottom Ft = 0 for element # 1. If the string is on bottom then Ft is a compressive force equal to the weight on bit.

Column # 11:

normal force Fn calculated by using Eq (32).

Column # 12:

Δ Ft calculated by Eq (35).

Column # 13:

tension at the top of the element and is equal to the value of Col# 11 plus the value of Col# 12.

Calculations for Element # 1 Column # 1:

Element #1 consists of the entire drill collars which lie across the tangent section.

Column # 2:

Depth at the bottom of drill collars is 10000 ft.

Column # 3:

Length of drill collars is 400 ft.

Column # 4:

Inclination at bottom of drill collars is 60 deg.

Column # 5:

Average inclination is 60 deg because it is across tangent section, i.e. inclination at bottom of drill collars is same as at top.

Column # 6:

Change in inclination in zero

Column # 7:

Azimuth at bottom of drill collars is 30 deg.

Column # 8:

Change in azimuth is zero because azimuths at top and bottom of drill collars are the same.

Column # 9:

buoyed weight of drill collars 490 − 90 Buoyancy factor = = 0.815 490 W = 400 x 117.42 x 0.815 = 38278 lb

Column # 10:

The tension at the bottom of the drill collars is zero because the drill string is off bottom.

Column # 11:

The normal force Fn is calculated by Eq (32) Fn =

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60-60=0

(F Δα sin θ ) + (F Δθ + W sin θ ) 2

t

2

t

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= 0 + (0 + 38278 × W sin 60 ) = 33149 lb

2

Column # 12:

Δ Ft is calculated by Eq (35) Δ Ft = Wcos θ + µFn = 38278cos60 + 0.25 x 33149 = 27426 lb

Column # 13:

Ft + Δ Ft = 0 + 27426 = 27426 lb This is the tension at the top of the drill collars which is equal to the tension Ft at the bottom of element # 2


This element is the bottom 4000 ft of 5” DP across the tangent section.

Column # 2:

Depth of bottom end is at 9600 ft.

Column # 3:

Length is 4000 ft.

Column # 4:

Inclination angle across tangent section is 60 deg.

Column # 5:

Average inclination angle is 60 deg.

Column # 6:

No change in inclination angle.

Column # 7:

Azimuth is constant at 60 deg.

Column # 8:

No change in azimuth, zero

Column # 9:

Weight of DP is, W = 4000 x 20.89 x 0.815 = 68101 lb

Column # 10:

This is the tension at the bottom of the element which is equal to the tension at the top of the previous element.

Column # 11:

The normal force is from Eq (32) Fn = 0 + (0 + 68101 sin 60) = 58977 lb

2

Column # 12:

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Δ Ft from Eq (35) Δ Ft = Wcos θ + µFn

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= 68101 x cos60 + 0.25 x 58977 = 48794 lb Column # 13:

Tension at top of element is Ft + Δ Ft = 27426 + 48794 = 76220 lb


This is a 50-ft element of 5” DP at the bottom of the build section. The inclination angle at the lower end is 60 deg and at the upper end is 55 deg (build rate of 10 deg/100 or 5 deg per 50 ft). The azimuth at lower end is 30 deg N and at upper end is 27.5 deg N (azimuth is changed at rate of 5 deg/100 ft or 2.5 deg per 50 ft.)

Column # 2:

Depth of lower end of element is 5600’ – end of tangent section.

Column # 3:

length of element is 50 ft.

Column # 4:

Inclination at bottom of element is 60 deg. Inclination at top of element is 55 deg.

Column # 5:

Average inclination is 55 + 60 = 57.5 deg 2

Column # 6:

Change in inclination angle is 60-55 = 5 degrees 5 x 0.01744 = 0.0872 radians

Column # 7:

Azimuth at lower end is 30 deg.

Column # 8: `

Change in azimuth is Azimuth at lower end – Azimuth at upper end 30 – 27.5 = 2.5 deg 2.5 x 0.01744 = 0.0436 radians

Column # 9:

W = 50 x 20.89 x 0.815 = 851 lb

Column # 10:

Tension at bottom of element is equal to tension at top of previous element 76220 lb.

Column # 11:

Fn = (76220 × 0.0436 sin 57.5) + (76220 × 0.0872 + 851 × sin 57.5) = 7879 lb

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Column # 12:

Δ Ft = 851cos57.5 + 0.25 x 7879 = 2427 lb

Column # 13:

Tension at top of element is 2427 + 76220 = 78647 lb

Calculations for elements 4 through 14 are done in the same manner. Calculations for Element # 15 Column # 1:

This is the last element. It is the DP in the vertical section from surface to the kick off point at 5000 ft.

Column # 2:

Depth of lower end is at 5000 ft.

Column # 3:

Length is 5000 ft.

Columns # 4 through 8: All values are zero because the pipe is vertical Column # 9:

W = 5000 x 20.89 x 0.815 = 85126 lb

Column # 10:

Tension at bottom end is 110427 lb

Column # 11:

Fn = (110406 × 0 × 0) + (110406 × 0 + 85126 × 0 ) =0 The normal force is zero because the pipe is vertical and is not lying on the hole.

Column # 12:

Δ Ft = 85126 x cos0 + 0.25 x 0 = 85126 x 1.0 + 0 = 85126 lb

Column # 13:

Tension at the surface is 85126 + 110427 = 195,553 lb

2

2

Therefore, it requires a pull of 195,553 lb to move the pipe uphole. The Landmark torque and drag software (WELL PLAN) gave a value of 193, 500 lb. b)

The total drag force is the sum of all drag forces generated by all elements, or 15

Total Drag Force =

∑F μ n

n =1

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This is the sum of all normal forces times the friction coefficient. The sum of all normal forces is obtained by adding all values of Fn in Col # 11 which is 195176 lb. Therefore, Total Drag = 195176 x 0.25 = 48794 lb c)

The hook load when the pipe is static is

195553 - 48794 = 146,759 lb

The hook load is also equal to 15

Hook load =

∑W Cosθ n =1

n

n

Calculation of Drag with Rotation If the drill string is moved up hole (or down hole) while being rotated as in back reaming, the drag force is calculated by using the following equation, Ff = Fn µ where, T = Fn = µ = V A

= = =

D

=

T V

............................................................................... (38)

Trip speed, in/sec Normal force, lb Coefficient of friction Resultant speed = T 2 + A2 Angular speed, in/sec rpm D x 3.14 x 60 Diameter, inch

Example In the previous example, what would be the approximate total drag if the drill string was pulled out while rotating it at 100 rpm? Assume a tripping speed of 2000 ft/hr

T = 2000

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ft 12in 1hr × × = 6.66 in/sec 3600 sec hr ft

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A= V=

5 × 3.14 × 100 = 26.16 in/sec 60 26.16 2 + 6.66 2 = 26.99 in/sec

Drag = Fnµ

T V

= 48794 x

6.66 = 12040 lb 26.99

This shows that rotation decreases drag force.

Critical Hole Angle When the weight component of the drill string in the direction of the hole axis is equal to the drag force resisting downhole movement, the drill string is not able to slide down hole by its own weight and when drilling in the sliding mode the drill string will require pushing with pipe higher in the hole. The angle at which down hole movement becomes impossible is called the critical hole angle, Referring to Fig (27), down hole movement becomes impossible when Ff = Wcos θ Note that in this case Ff is acting in the opposite direction of Wcos θ because the movement is downward Wµsin θ = Wcos θ sinθ 1 = cosθ μ tan θ =

1

μ

θcr = arctan

1

μ

................................................................. (39)

Where θcr is the critical hole angle in degrees.

Example Page 75

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At which hole angle the drill string cannot move downward by its own weight when drilling a tangent section where the coefficient of friction is 0.30 Solution

1 0.3 = arctan 3.33 = 73.3 degrees

θcr = arctan

Calculation of Torque The torque required to turn the drill string is calculated by the following equation, Torque = Fnrµ where T = Fn = r = µ A V

= = =

A ............................................................ (40) V

Torque, ft-lb normal force as defined by Eq (32) Radius of drill string component ft (for drill collars use outer radius of collar, for DP, HWDP and casing use outer radius of tool joint) Coefficient of friction Angular speed as defined in Eq (38) Resultant speed as defined in Eq (38) θ+Δ θ

If the drill string is not tripped while rotating, then T = 0 and A is equal to V and Equation (40) reduces to

M + ΔM Fn

Torque = Fnrµ ........................... (41) M θ, α

W

Fig. (29)

Page 76

Torque acting on a drillstring element

If the torque acting on the lower end of the drill string element is M, then the torque acting at the top of the element is M + Δ M as shown in Fig (29). The torque increment Δ M is

Δ M = Fnrµ ...............................(42)

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where Fn is the normal force exerted on the element. The torque required to turn the entire drill string is the sum of all torque increments for all the elements. Torque calculations are illustrated by the following example Example

In the last example, a) Calculate the surface torque required to turn the string off bottom with no tripping. b) Calculate the torque while drilling. Use a bit torque of 2000 ft-lb. Drill pipe tool joint OD is 6.625 in. Solution

The drill string is divided into small elements as was done for calculating drag. Calculations are tabulated in Table (13). The data in columns 10, 11 and 13 are calculated in the same manner as was done for the drag calculations in the previous example. For drag calculations the tension increment Δ Ft in column 12 is calculated by Eq (35)

Δ Ft = Wcos θ + µFn where µFn in the incremental drag caused by moving the element uphole. However, since in this example there is no tripping, the drag increment due to uphole or downhole movement is zero, and the tension increment Δ Ft is

Δ Ft = Wcos θ ............................................................... (43) Column 14 is the torque at the bottom of the element. If the string is off bottom the torque is zero. For the case of drilling the torque is the bit torque which have to be estimated. The torque increment Δ M in Column 15 is the torque resistance exerted by the element and is calculated by using Eq (42). Column 16 is the torque at the top of the element which is M + Δ M. The torque at the top of the element ( M + Δ M ) is the torque at the bottom of the next element. The calculations are repeated for all elements to the surface. Calculations for Element # 1 Columns 2 through 11 are calculated as in the previous example. Column # 12.

Page 77

is calculated by Eq (43) Δ Ft = 38278 x cos60 = 19139 lb


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Column # 14.

The torque at bottom of element in this case is zero because the drill string is off bottom (no bit torque)

Column # 15.

Torque increment Δ M

Δ M = Fnµr r= =

OD of DrillCollar 7 = =3.5 in 2 2 3.5 =0.2916 ft 12

Δ M = 33149 x 0.25 x 0.2916 = 2416 ft-lb Column # 16.

Torque at top of element #1 is M + Δ M = 0 + 2416 ft-lb = 2416 ft-lb

Calculation for element # 2 Columns # 10 and 11 are calculated in same manner as in the drag example Column # 12:

Δ Ft = 68101 cos60 = 34050 lb

Column # 14:

Torque at bottom of element # 2 is same as torque at top of element # 1 or 2416 ft-lb

Column # 15:

Δ M = Fnrµ r= =

tool joint OD 2

6.625 = 3.312 in 2

3.312 = .276 ft 12

Δ M = 58977 x 0.276 x 0.25 = 4069 ft - lb Column # 16:

Torque at top of element # 2 is Δ M + M = 4069 + 2416 = 6485 ft-lb

Calculations for elements 3 through 14 are done in the same manner as for element # 2. Page 79

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For element # 15 since the inclination angle is zero, the normal force Fn is zero and the torque increment, Δ M is also zero. In other words, drill pipe in vertical hole can be turned with no or very little torque. So the torque at top of element # 15 (the surface) is equal to torque at top of element #14 which is 11032 ft-lb b)

If the bit torque is 2000 ft-lb then the torque at surface is Torque = 11032 + 2000 = 13032 ft-lb

It can be seen from the above example problems that the calculation of torque and drag consumes large amount of time and is best done by using computer software.

Determination of Friction Coefficient The coefficient of friction determines how much of the normal force is transformed into drag or torque and it is an important factor in calculating torque and drag in a wellbore. Friction coefficients can be calculated from actual drilling situations for a particular well geometry using a computer program with drill string surface loads as input data. Input includes pickup weight, slack-off weight and torque readings, each of which can produce independent friction coefficient. Agreement among the three coefficients from one well lends credibility to the model and also provides confidence in the friction coefficient for its use in the prediction of torque and drag in subsequent wells. To calculate the friction coefficient during tripping out of hole, the pick up load is read from the weight indicator. The reading from the weight indicator includes the weight of the kelly and travelling equipment. In calculating drag forces, the tension at the top of the drill pipe below the kelly, is required. Thus, it is necessary to subtract the weights of the traveling equipment and kelly from weight indicator reading. The tension at the top of the drill pipe, the drill string and wellbore geometry data are entered in the computer program. The program calculates the tension required to pick up the drill pipe by using different coefficients of friction until the calculated tension is equal to the tension read by the indicator. The coefficient of friction at which the calculated tension is equal to the measured tension is the correct coefficient to use for that wellbore. To ensure accurate results, the weight indicator must be calibrated to give accurate readings. Also accurate weights of the traveling equipment and kelly must be known. The friction coefficient can also be determined by measuring the torque while turning the drill string off bottom. The torque must be in foot-pounds rather than amperes. The coefficient of friction where the measured torque is equal to the calculated torque is the correct coefficient for the wellbore. Most rigs are not equipped with calibrated torque Page 80

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indicators and, therefore, it is recommended to determine friction coefficients by measuring tension. Friction coefficients depend on the mud type and whether the hole is cased or open. Friction coefficients from a number of similar wells must be compared to verify useful values for prediction use. Typical ranges of friction coefficients are shown in Table (14). Table (14 ) Coefficients of Friction

Open Hole Cased Hole

Water Based Mud

Oil Based Mud

0.15 – 0.3

0.15 – 0.25

0.125 – 0.4

0.125 – 0.25

Torque and drag arise not only from friction, but also from the effects of hole tortuosity, cuttings accumulations, swelling shale, differential sticking, and other mechanical impediments to drill string movement. Thus, the “friction coefficient” in the torque and drag programs could more accurately be considered a “drag coefficient or coefficient factor,” that is, a composite coefficient that includes all factors affecting torque and drag. Furthermore, the drag coefficients for rotational movement, for axial movement or for a combination of the two, will often be different, and may also vary with the direction of movement. Finally, because these mechanical impediments to string movement frequently change with changing hole conditions, friction coefficient will also fluctuate, particularly for axial movement.

Factors that Affect Torque and Drag Kick-Off Point: The kick-off point is often the major factor influencing torque and drag in a well. This is due to the fact that shallow doglegs combined with pipe tension will cause a high normal force, thus a high drag and torque at that point. In deep directional wells, the kick-off point should be considered from a torque and drag standpoint, in addition to the other factors influencing the kick-off point. Dogleg Severity: As seen from Equation (32), the change in inclination and azimuth influence the normal force acting on the pipe. The larger the change in either, meaning increasing dogleg severity, the larger the normal force and the larger the increase in drag due to that point. Smaller doglegs lead to less torque and drag. Page 81

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Mud Lubricity: Mud lubricity is a term generated from a lab test (on a specific piece of equipment) that is intended to mimic the drill pipe/casing interaction. The lubricity is scaled opposite the friction factor, increasing lubricity reduces the friction factor. The measurement is a lab test, and the results do not always directly correlate with field observations. The mud lubricity, as measured in the lab, strongly influences the friction factor, or coefficient of friction, between the pipe and casing, or pipe and borehole, but does not always accurately reflect the influence of the specific material, be it pipe or formation. It is usually assumed, based on the mathematical model, that raising the mud lubricity will generally lower torque and drag by lowering the coefficient of friction. This is not always the case, because the source of the torque or drag may be generated by a factor not included in the mathematical model. These factors are discussed below. Cutting Beds: There is no provision in the mathematical model for the local influence of cutting beds. Cuttings beds will locally change the effective coefficient of friction, and can be modeled in that manner if the location, length and friction factor of the cuttings bed is known. Unfortunately, these parameters continually change in the presence of a cuttings bed. The typical technique is to deduce the presence of a cuttings bed when unexplained trends in torque and drag develop in well. This would be seen very practically on trip out of the hole when pick-up weight begins increasing during the trip, rather than decreasing, then suddenly returns to the baseline trend. Drilling Tools: Once again, there is no provision in the mathematical model for the impact of individual drilling tools. It is not hard to imagine that the influence of a stabilizer on drag would be much greater than that of a similar length of drill pipe or drill collar and that the interaction of the stabilizer with the wellbore may include additional components than does the simple model we have been working with. Drilling tools can greatly impact torque and drag are stabilizers, drill pipe protectors, bit type (roller cone vs. drag type), drill collars (flex vs. slick) and hard banding on drill pipe.

Minimizing Torque & Drag Several techniques have been developed to aid in the reduction of torque and drag in a wellbore. Many of these are direct consequences of the mathematical model we have studied, while others are based on field experience. Deep Kick-off Point: By lowering the kick-off point, the tension in the pipe is lowered and the normal force reduced. This will reduce the drag and torque generated at the point.

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Raise Mud Lubricity: As mentioned in prior section, raising the lubricity of the mud corresponds to lowering the coefficient of friction between the pipe and casing or pipe and bore hole. This can be accomplished by 1) changing mud type. Oil based mud generally has a lower lubricity than does water based mud. 2) Adding a lubricant. 3) Adding a mechanical aid to the mud, such as drilling beads or walnut hulls. Beads are typically very expensive, but have the benefit of being environmentally safe. Remove Cutting Beds: Quite often the presence of cuttings beds will dramatically increase the torque and drag observed in a well. The cuttings beds can be identified by various means, and the effect of a cutting bed can only be changed by eliminating the bed. Minimize DLS in Tangent Section: A smoothly drilled tangent section will help minimize the torque and drag on a well. Because the tangent section represents the major length of a well, cumulative DLS in this section can greatly impact the total drag and torque generated in the well. Soft Type Hard Banding: Tungsten carbide hard banding on drill pipe greatly increases the effective coefficient of friction, as well as causes casing wear. Current technology for hard banding is to apply a softer metal for hard banding, which minimizes casing wear and drag, while still protecting tool joints. Drill Pipe Protectors: Non-rotating drill pipe protectors have been demonstrated to dramatically reduce torque when placed in the build section of well. These protectors (Western Oil Tool) work by effectively reducing the normal force on the drill pipe through the build section by providing standoff from the casing and by creating a fluid bearing between the drill pipe and protector. Flex Drill Collars: Flex drill collars provide standoff from the bore hole, thus minimizing the differential sticking force on the collar. This in turn eliminates the need to use stabilizers, which create a large amount of drag, to prevent differential sticking. Tapered edges on the flex collars also reduce drag. Minimize use of Stabilizers: It has been found the use of conventional stabilization techniques are not necessary, and even detrimental, on horizontal, high angle, and extended reach wells. Stabilizers cause a large downward drag force which reduced or prevents transfer of weight-on-bit, thus affecting penetration rate. In addition, the movement of stabilizers is much more of a slip/stick motion, which leads to rapid application of weight-on-bit and subsequent stalling of a positive displacement motor. Page 83

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Calculation of Buckling Sinusoidal Buckling in Straight Inclined Holes

When a mechanical compressive load over a critical value is applied on drill pipe, the drill pipe will buckle. The drill pipe will first buckle into a sinusoidal wave shape as shown in Fig (30). As the compressive force increases it will ultimately buckle into a helix as in Fig(31).

Fig. (30)

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Sinusoidal “snake” buckling of pipe

Fig. (31)

Helical buckling of pipe

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The critical mechanical compressive load at which sinusoidal buckling of pipe is expected to occur in a straight inclined hole can be calculated from the following equation developed by Paslay, Fcrit = 2 Fcrit = E = I =

= OD = ID = W = = θ Kb = r =

EIWK b sin θ ............................................................. (44) 12r

Critical buckling force lb Young’s modulus, 30 × 10 6 , psi Moment of inertia, in4 3.14 ( OD 4 − ID 4 ) 64 Outside diameter of pipe, in inside diameter of pipe, in Adjusted (actual) weight of pipe in air, lb/ft Angle of inclination, degrees Buoyancy factor, unitless Radial clearance between pipe and hole, in (Some computer software use clearance between tool joint OD and hole which gives higher Fcrit)

Equation (44) is used to predict the onset of buckling of pipe in a straight inclined hole. For pipe in vertical hole the critical buckling force is, Fcrit = 1.94

3

EIW 2 K 2 b ............................................................ (45) 144

Example

Calculate the critical sinusoidal buckling force for 5” 19.5 # drill pipe in (a) 8-1/2” vertical hole and (b) in 8-1/2” deviated hole where the angle of inclination is 60 degrees. Mud Weight = Drill pipe ID = Actual air weight =

Page 85

90 pcf 4.276 in4 20.89 lb/ft

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Solution

a)

E

=

30 × 106 psi

I

=

3.14 4 ( 5 − 4.276 4 ) = 14.262 in4 64

EI

=

14.262 x 30 × 106 = 4.278 × 108

r

=

8.5 − 5 = 1.75 in 2

Kb

=

490 − 90 =0.816 490

From Eq (45), 4.278 × 108 × 20.89 2 × 0.816 2 3 = 1846 lb Fcrit = 194 144 b)

From Eq (44), 4.278 × 10 8 × 20.89 × 0.816 × sin60 12 × 1.75 = 34,677 lb

Fcrit = 2

It can be seen from the above example that drill pipe in vertical holes has small resistance to buckling. The critical buckling force is so small that it is assumed to be zero. This is the reason in vertical drilling the drill pipe should be kept in tension to prevent buckling. The critical buckling force in the inclined hole in the above example is large (34,677 lb). This means that it will require a compressive force of 34677 lb to buckle the drill pipe into a sinusoidal shape. The critical buckling force increases with the angle of inclination until it reaches a maximum at an angle of inclination of 90 deg. The reason that the drill pipe in an inclined hole is so resistant to buckling is that the low side of the hole forms a trough that resist even a slight displacement of the pipe from its initial straight configuration. The effect of gravity and the curving sides of the hole form a restraint that resists buckling. Because of its ability to withstand large mechanical compressive loads without buckling , drill pipe and HWDP are used in compression to provide weight on bit while drilling inclined and horizontal holes.

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Buckling in Curved Holes. The critical buckling load in a curved wellbore of increasing angle (build section) is always higher than that in a straight wellbore of the same inclination. Positive wellbore curvature has a stabilizing effect on pipe in mechanical compression because compression increases the side load of the pipe against the outside of the wellbore curve. This is added to the stabilizing effect of the pipe weight at any angle. Therefore, a drill string in a positive build section, being mechanically compressed from straight hole sections above and below must buckle first in one or the other of the straight sections. The amount of mechanical compression that a uniform string can carry without any where buckling will be limited by the critical buckling loads in the straight sections above and below a positive build section, not in the build section itself. In a dropping wellbore the critical buckling load of drill pipe can be more or less than that in a straight well bore of the same inclination angle. The critical buckling loads of pipe in a curved wellbore can be estimated by the following equation. In a build section of the well

2EIK Fcrit = +2 r

2

EIWK bsinθ ⎛ EIK ⎞ ............................... (46) ⎜ ⎟ + 12r ⎝ r ⎠

where K =

The wellbore curvature 1 = =BR in radians per inch R BR = Build rate in radians per inch R = Radius of build, inch 180 degrees = pi(3.14) radians 1 degree = 0.01744 radians In a drop section of the well, Ktest = If

K > Ktest

rWK bsinθ .................................................................... (47) 12 EI

then, 2

Fcrit =

Page 87

EIWK bsinθ 2EIK ⎛ EIK ⎞ −2 ⎜ ................................ (48) ⎟ − r 12r ⎝ r ⎠

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If

K < Ktest

then 2

EIWK bsinθ - 2EIK ⎛ EIK ⎞ Fcrit = +2 ⎜ .............................. (49) ⎟ + 12r r ⎝ r ⎠ Example

Calculate the critical buckling force for the 5” 19.5# drill pipe in the previous example in a build section drilled at a build rate of 5 degrees per 100 ft. Assume same angle of inclination of 60 degrees. Solution

Since this is a build section then Eq (46) will be used to calculate the critical buckling force. deg 5 deg K = BR = = 0.05 100 ft ft =

0.05 deg/inch 12

=

0.05 x 0.01744 radians / inch 12

= 0.0000727 radians/inch 2 × 4.278 × 108 × 0.0000727 Fcrit= +2 1.75 = 3.55 × 10 4 + 2

2

⎛ 4.278 × 108 × 0.0000727 ⎞ 4.278 × 108 × 20.89 × 0.816 × 0.866 ⎜⎜ ⎟⎟ + 1.75 12 × 1.75 ⎝ ⎠

3.15 × 108 + 3 × 108

= 85,128 lb It can be seen from the above example that the critical buckling force in the build section is 2.45 times (85128/34677) the critical force in the straight (tangent) section of the same angle of inclination of 60 degrees. Therefore, a drill string in a wellbore which has a build section will buckle first below the build section (below end of build) or above the build section (above the kickoff point). So the amount of mechanical compressive force that a drill string can carry without buckling is limited by the straight sections below and above the build section. Page 88

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Helical Buckling As the mechanical compressive force increases beyond the critical sinusoidal buckling force, the tubular will take the shape of a helix as shown in Fig (31). Helical buckling will occur at compressive loads which are 1.4 times the critical force for sinusoidal buckling. Once the tubular takes the shape of a helix it will be forced against the wall of the wellbore and additional drag forces will develop. The drag forces will ultimately prevent the tubular from sliding down the wellbore. This condition is called lock-up. Drill pipe should never be rotated when it is buckled. Example

What is mechanical compressive force required to helically buckle 5”, 19.5 # drill pipe in 8-1/2” wellbore inclined at 60 degrees? Solution

From the previous example, the critical sinusoidal buckling force for 5” DP was calculated to be 34,677 lb. Therefore, the force required to initiate helical buckling is 34677 x 1.4 = 48547 lb. Bending

Bending is a point load, that is the effect of bending occurs at the point of bending, and the stress it generates is localized. The force or stress generated by bending is additive to the existing tensile force or stress in the tubular at the point of bending, but does not increase the tensile force in the tubular at other points. Bending does not affect the hookload on the drill string. The maximum stress on the convex side of the pipe caused by bending is given by, z = 218BRd

and the equivalent force caused by bending is, Fb = 64 BRdW ............................................................... (50) Where Fb = BR = d = W = Page 89

Bending force, lb Dog leg severity or build rate, deg / 100 ft Outside diameter of tubular, in Weight of tubular in air, lb/ft

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The total tensile force in the pipe is the bending force plus the existing tensile force at that point, and the sum should remain below the tensile strength of the tubular. Bending forces become important at high build rates such as in drilling short radius horizontal wells. Example

A short radius sidetrack is performed in a well by using 2-7/8”, 6.85 lb/ft grade E drill pipe. The axial tensile force in the drill pipe in the build section as a result of applying weight on bit is 10,000 lb. The maximum build rate in the build section is expected to reach 110 degrees per 100 ft. Is it safe to use this drill pipe for drilling the short radius hole? Tensile strength for grade E pipe is 75,000 psi, drill pipe ID= 2.441 in. Solution

The force caused by bending in the build section is, Fb = 64 x 110 x 2.875 x 6.85 = 138644 lb Total axial force = 10,000 + 138644 = 148644 lb Tensile strength of drill pipe is 3.14 (2.8752 − 2.4412 ) 4 = 135,833 lb

Y = 75000 ×

Since the total axial force exceeds the tensile strength of the drill pipe, the drill pipe will fail and therefore, it is not safe to use it for drilling.

Fatigue Most drill pipe failures are related to fatigue, and result from the cyclical forces induced on the pipe during rotation. Fatigue implies a change in material properties with the total number of revolutions of the drill pipe. During rotation in a dogleg, the drill pipe experiences variable stress levels along a given circumference, and if the stress variation is severe enough, the metallurgy of the pipe will eventually be altered. Fatigue is not well understood, once fatigue occurs, the life of a given tubular is very difficult to predict. Most efforts are concentrated on operating below the conditions which cause fatigue. Page 90

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The essential factors in fatigue failure are tension in the pipe at the dogleg, dogleg severity, and number of revolutions (cycles) made by the pipe in a given DLS and under a given tensile load. The failure model is mathematically presented by Eq (22). The model does not guarantee elimination of failure but it provides an engineering basis in drill string design.

Drill String Design for High Angle and Horizontal Wellbores The design approach in high angle and horizontal wellbores differs from the approach in vertical wells in the following respects: a)

In high-angle holes, traditional BHA components are often eliminated. Bit weight is likely to be applied by running normal weight drill pipe in compression, a practice never recommended in vertical holes.

b)

For a given measured depth, surface tension load from hanging weight decreases in a high-angle hole due to wall support, but torque and drag required to move the drill string are higher compared to vertical holes. The load limit for the drill string will be its tensile capacity in a vertical hole, but is more likely to be its torsional capacity in horizontal and extended reach holes.

c)

In vertical wells, loads are calculated based on hanging weight. Friction effects are often small and are traditionally ignored. In horizontal wells, friction effects will probably be large enough that they cannot be ignored.

d)

Drill string design for vertical holes is a once-through calculation. In horizontal and ER wells, drill string design is an iterative process.

The objectives of drill string design in horizontal wells are: 1.

Provide adequate weight on bit without buckling the drill pipe or heavy weight drill pipe.

2.

Ensure that the components in the drilling assembly are not subjected to mechanical loads that exceeds their design limitation

Thus the problem is to determine where and at what bit weight the drill pipe will first begin to buckle. If this bit weight is sufficient to drill the well, buckling can be avoided by staying below it. If the bit weight needed causes the pipe to buckle, then heavy weight drill pipe should be run in the buckled sections.

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Fig. (32)

As was mentioned above, the critical buckling load in a curved section is greater than that across tangent and vertical sections. Therefore, buckling of drill pipe will initiate in the tangent section (below the tangent point) or above the kickoff point. The maximum weight that can be applied on the bit without buckling the drill pipe is determined in two steps: 1.

Calculate the maximum weight that can be applied on the bit without buckling the drill pipe below the tangent point refer to Fig (32)

2.

Calculate the maximum weight that can be applied on the bit without buckling the drill pipe above the kickoff point. The lower of the two weights is the correct weight to be used for drilling.

Maximum Weight on Bit Below Tangent Point If the tangent angle of a curved or build section is less than horizontal, the highest mechanical compression in the drill pipe will occur in the joint at the top of the bottom hole assembly. A free body diagram of the drilling assembly across a tangent section is shown in Fig (33). The force F1 is the effective weight of the HWDP and F2 is the effective weight (along the hole axis) of the BHA. Fcrit is the critical buckling force of the drill pipe at the top of HWDP which is equal to the maximum mechanical compressive force that can be applied (slacked off) without buckling the drill pipe. Taking a force balance along the hole axis,

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DF x Fcrit + F1 + F2 = Drag + WOB Substituting for F1 and F2 F1 F2

= =

Fcrit

LHWH BF cos θ LBHAWBHA BF cos θ HWDP

The maximum weight on bit that can be applied without buckling the drill pipe is

Drag

F1

WOB < DF x Fcrit + LHWH BF cosθ + LBHAWBHA BF cosθ – Drag ….(51) where Fcrit = DF LH WH BF θ LBHA WBHA WOB Drag

= = = = = = = = =

Critical sinusoidal buckling force, lb Design factor (0.9+) length of HWDP, ft Weight of HWDP in air, lb/ft Buoyancy factor, unitless Angle of inclination, degrees length of bottom hole assembly, ft Weight of BHA in air, lb/ft Weight on bit, lb Drag force, lb

(51) BHA F2

WOB

Fig. (33) Free Body Diagram of Drilling Assembly across a Tangent Section

If the hole below the tangent point is horizontal, then cos 90=0 and WOB < DF Fcrit – Drag............................................................. (52) For sliding mode drilling the drag force is calculated as was described previously. For rotary mode drilling, the drag force is small and is assumed to be zero. Maximum Weight on Bit Above the Kick-off Point

A free body diagram of the drilling assembly below the kick-off point is shown in Fig (34). The forces in the diagram are defined as follows: F2

= =

Page 93

Effective weight of BHA, lb WBHALBHA BF cosθ

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Fcrit Kickoff Point

F4

DP

Tangent Point DP

Drag

HWDP F3

BHA F1 F2 WOB

Fig. (34)

Free Body Diagram of Drilling Assembly below Kickoff Point

F1

= =

Effective weight of HWDP, lb WHLH BF cosθ

F3

= =

Effective weight of DP across tangent section, lb WDPLDP BF cosθ

F4

=

Effective weight of DP across build section, lb

Fcrit =

Critical buckling force of DP at the kickoff point, lb

Balancing the forces along the hole axis yields, WOB < Fcrit x DF + WBHALBHA BF cosθ + LHWH BF cosθ + WDPLDP BF cosθ + F4Drag….(53) where, LDP = WDP =

Page 94

Length of DP across tangent section, ft Weight of DP in air, lb/ft

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Since the inclination angle θ across the build section is not constant the value of F4 is computed by dividing the build section into small increments and adding the effective weights of all the increments. n

F4 = WDPBF

∑L n −1

nDP

cos θ n ....................................................................... (54)

A simpler way of calculating the effective weight of the drill pipe across the build section is by using the following equation, F4= where, BR = θt = α =

(

)

⎡ 5729.6 sinθ t − sinα ⎤ WDPBF ⎢ ⎥ ........................................................... (55) BR ⎥⎦ ⎣⎢ Build rate, degrees per 100 ft Inclination angle at tangent point (end of build), degrees. Inclination angle at kickoff point, degrees

Calculation of Axial Mechanical Forces In drill string design of horizontal wells, the engineer may want to check if the applied WOB will buckle the drill pipe at the top of BHA. This is accomplished by drawing a free body diagram of the BHA showing only the mechanical forces as shown in Fig (35). The mechanical force F at any point in the drill pipe is calculated by making a force balance. It should be noted here that the force F is the mechanical force trying to buckle the drill pipe and it is not the total or actual axial force. The actual axial force is calculated by including the hydrostatic pressure forces acting on the shoulder areas as was done in Fig (8) and Eq (19). The actual axial stress in the drill pipe is calculated by dividing the actual force by the metal area of the drill pipe. Taking a force balance, F + F1 + F2 = WOB + DRAG F1 = LHWH cos θ BF F2 = LBHAWBHA cos θ BF Page 95

F DP

HWDP F1

BHA

Drag F2

WOB

Fig. (35)

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Substituting for F1 and F2, F= - LHWH cos θ BF - LBHAWBHA cos θ BF + WOB + DRAG If drilling in the rotary mode, the drag force is zero. If the calculated value of F is less than the critical buckling force Fcrit of the drill pipe, then the drill pipe will not buckle. If F is greater than the Fcrit, then the drill pipe will buckle.

Drill String Design Summary 1.

Calculate the drag forces and torque by using a computer program such as Landmark Well Plan.

2.

Calculate the sinusoidal critical buckling force below the tangent point and above the kickoff point.

3.

Calculate the maximum WOB that can be applied without buckling the DP below the tangent point using Eq (51)

4.

Calculate the maximum WOB that can be applied without buckling the DP above the kickoff point by using Eq (53). Take the lower of the two values. If the WOB is not sufficient then add more HWDP in the drilling assembly.

5.

Calculate the actual forces on DP below the tangent point and above the kickoff point. Include bending forces and check for fatigue if the build rate is high. If the actual forces exceed 90% of the DP yield strength, then select a DP with higher yield strength.

6.

Check the actual torque at all depths and make sure it is less than the makeup torque of DP

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Example

You are drilling a build and hold wellbore with the following characteristics: Hole Geometry: Hole size Casing Kickoff point Angle at KP Build rate Tangent point TD TVD TVD at top of BHA Tangent angle Mud weight Drill String Drilling assembly HWDP DP Bit at 8000 ft Drag Forces BHA HWDP DP in tangent section DP in build section Buckling DF a) b) c) d) e)

= = = = = = = = = = =

8-1/2” @ 8000 ft 9-5/8” 40# @ 7000 ft 5500 ft 0 Deg 5 deg/100 ft 7000 ft 8000 ft 6865 ft 6841 ft 75 deg 75 pcf

= = =

90 ft, 82.68 lb/ft 6” OD, 2.25” ID 90 ft, 49.7 lb/ft, 3” ID, 5” OD 5”, 22.26#, grade E, 4.276” ID

= = = = =

2200 lb 1300 lb 5200 lb 5400 lb 0.9

What is the maximum rotary mode bit weight that can be applied without buckling the drill pipe? What is the maximum sliding mode bit weight that can be applied without buckling the drill pipe? What is the maximum rotary mode bit weight that can be applied if the DP is to remain in tension? Calculate the actual stress in the DP at top of HWDP when applying 27431 lb WOB in rotary mode. What is the mechanical force at top of HWDP that will cause buckling?

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Solution

a)

Let’s first calculate by using Eq (51) the maximum WOB that will not cause buckling above the drilling assembly. Calculate critical buckling force by using Eq (44) E = 30 × 10 6 psi I=

π

(5 64

4

− 4.276 4 )

= 14.26 in 4

EI = 30 × 10 6 × 14.26 = 4.278 × 108 W = 22.26 lb/ft r = KB =

8.5 − 5 = 1.75 in 2 490 − 75 = 0.846 490

sin θ = sin75 = 0.965 4.278 × 108 × 22.26 × 0.846 × 0.965 Fcrit = 2 12 × 1.75 = 38482 lb LHWHKBcos θ = 90 x 49.7 x 0.846 x 0.258 = 979.4 lb LBHAWBHAKBcos θ = 90 x 82.68 x 0.846 x 0.258 = 1624 lb Drag = 0 lb because of drilling in rotary made From Eq (51) 0.9 x 38482 + 979.4 + 1624 – 0 > WOB WOB < 37237 lb Now let’s calculate the maximum weight on bit for no DP buckling above the kickoff point by using Eq (53) From Eq (44) Fcrit = 0 lb (because θ = 0, sin0=0)

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LHWHKBcosθ = 979.4 lb LBHAWBHAKBcosθ = 1624 lb LDPWDPKBcosθ = (8000 – 90 – 90 – 7000) X 22.26 X 0.846 X 0.258 = 3984 lb F4 = effective weight of DP in build section From Eq (55) ⎛ 5729.6(sin 75 − sin 0) ⎞ F4 = 22.26 x 0.846 ( ⎜ ⎟ 5 ⎠ ⎝ = 20844 lb Drag = 0 lb (rotary mode) From Eq (53) WOB < 0 + 1624 + 979.4 + 3984 + 20844 – 0 < 27431 lb So the maximum WOB that can be applied without buckling the DP while drilling at 8000 ft in rotary mode is the smaller of the two values or 27,431 lb. b)

To calculate the maximum WOB in the sliding mode we repeat the calculations made in part (a) keeping in mind that the drag forces are not zero. Maximum WOB for no DP buckling above the drilling assembly using Eq (51) is, 0.9 x 38482 + 979.4 + 1624 – drag > WOB Drag = drag of BHA + drag of HWDP = 2200 + 1300 = 3500 lb WOB < 33,737 lb Maximum WOB for no DP buckling above the kickoff point is, WOB < 1624 + 979.4 + 3984 + 20844 - drag

Drag = drag of BHA+drag of HWDP+drag of DP in tangent section+drag in build section = 2200 + 1300 +5200 + 5400 = 14100 lb

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WOB < 13,331 lb The WOB is the smaller of the two values or 13,331 lb. c)

If the DP is to remain in tension, the WOB can be provided only by the effective weights of the BHA and the HWDP WOB < 1624 + 979.4 WOB < 2603.4 lb It can be seen from the above example that only 2603 lb can be applied on the bit if the DP is to remain in tension, whereas if the DP is used in compression (but below the critical buckling force) 27, 431 lb can be applied on the bit. Applying 27,431 lb of WOB with the DP in tension requires a very long HWDP which will increase the cost and the torque and drag that are associated with it. Therefore, the benefit of using DP in compression in horizontal drilling is that the BHA weight is kept low which in turn helps reduce torque and drag.

d)

Draw a free body diagram of the drilling assembly below the drill pipe. Show all mechanical and hydrostatic pressure forces acting on the drilling assembly as shown Fig (36). F DP F1 HWDP

F4 F2

BHA F5

F3 WOB Fig. (36)

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The forces are: F F1 F2 F3 F4 F5

= = = = = =

The axial force on DP, unknown hydrostatic pressure force acting at shoulder area between HWDP and DP Hydrostatic force acting at shoulder area between HWDP and BHA Hydrostatic force acting at bottom of BHA Effective air weight of HWDP Effective air weight of BHA

There are no drag forces because string is rotating. Making a force balance yields, F + F1 + F4 + F2 + F5 = F3 + WOB Now determine the TVD at top of HWDP and at top of BHA TVD at bottom

=

6865 ft

TVD at top of BHA

=

6865-90 x cos75 = 6841 ft

TVD at top of HWDP =

6865-180xcos75= 6818 ft

P1 =

75 × 6818 = 3551 psi 144

P2 =

75 × 6841 = 3563 psi 144

P3 =

75 × 6865 = 3575 psi 144

A1 = A2 =

π 4

(4.276

π

(6 4

2

2

− 32 ) = 7.28in 2

− 2.252 ) −

π

(5 4

2

− 32 )

= 24.28 – 12.56 = 11.72 in 2 A3 =

π

(6 4

2

− 2.252 ) = 24.28in 2

F1

=P1A1

F2

= P2A2

Page 101

= =

3551 x 7.28 = 25851 lb

3563 x 11.72 =41758 lb

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F3

= P3A3

F4

=

90 x 49.7 x cos75 = 1157 lb

F5

=

90 x 82.68 x cos75 = 1926 lb

WOB = F

=

3575 x 24.28 = 86801 lb

27431 lb

= F3 + WOB – F1 – F2 – F4 – F5 = 86801+27431-25851-41758-1157-1926 = 43540 lb

Actual axial stress = F/A A=

π 4

(5

2

− 4.276 2 ) = 5.27in 2

Actual axial stress =

43540 = 8261 psi 5.27

The stress is less than the minimum compressive strength of grade E drill pipe which is 75,000 psi. Note: The actual force F calculated above (43540 lb) is greater than the critical buckling force Fcrit (38482 lb). However, this does not mean that the DP will buckle. The reason is that the force F is a combination of mechanical forces and hydrostatic forces. Only the mechanical component of F will cause buckling. See part (e) below

e)

Do a force balance as in part (d) but do not include the hydrostatic forces. Use only the mechanical forces which are the effective buoyed weight of the string and WOB.

F + F4 + F5 = WOB F4 and F5 in this case are the effective buoyed weights of HWDP and BHA F4

=

90 x 49.7 x cos75 x 0.846 = 979 lb

F5

=

90 x 82.68 x cos75 x 0.846 = 1629 lb

F

= =

WOB – F4-F5 27431 – 979 – 1629 = 24823 lb

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The mechanical force F is less than Fcrit, therefore, the DP at top of HWDP will not buckle. In the previous discussion it was emphasized that drill pipe can be used in compression to provide weight on bit as long as the mechanical compressive force is less than the critical buckling force required to initiate sinusoidal buckling (Fcrit). Field practice has shown that drill pipe can tolerate sinusoidal buckling when there is no rotation, that is, when drilling in the sliding mode. This means that mechanical compressive forces up to 1.4 Fcrit, which is the force required to initiate helical buckling in the sliding mode, can be applied on drill pipe while drilling in the sliding mode. In rotary drilling mode the mechanical compressive force should not exceed Fcrit. The mechanical compressive force should never exceed 1.4 Fcrit in any drilling mode. In other words, helical buckling must be avoided at all times.

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TABLE OF CONTENTS

Page

HYDROSTATIC PRESSURE

1

RHEOLOGICAL FLUID MODELS

2

NEWTONIAN MODEL

2

- LAMINAR AND TURBULENT - PIPE FLOW OF NEWTONIAN LIQUIDS

4 4

REYNOLDS CRITERION

5

- LAMINAR FLOW - TURBULENT FLOW

6 7

BINGHAM PLASTIC FLUIDS

10

- PSEUDO PLASTIC AND DILATANT FLUIDS

12

THE POWER-LAW MODEL INITIATING CIRCULATION IN A WELL

13 18

HYDRAULIC POWER PRESSURE DROP ACROSS BIT NOZZLES HYDRAULIC IMPACT FORCE

20 20 21

CUTTINGS SLIP VELOCITY

23

OPTIMIZATION OF BIT HYDRAULICS

24

MAXIMUM BIT HYDRAULIC HORSEPOWER MAXIMUM JET IMPACT FORCE

24 25

- BOTTOM HOLE CLEANING NEEDS - GRAPHICAL SOLUTION

26 26

ON-SITE NOZZLE SELECTION

44

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Fluid mechanics is very important for the drilling engineer. Large fluid pressures are developed in the long wellbore and drill strings by the drilling fluid. The presence of these pressures must be considered in almost every well problem. In this chapter the relations to determine subsurface fluid pressures are presented for 1) the static condition in which the wellbore fluid and drill pipe are at rest, and 2) the circulating operation in which fluids are pumped down the drill pipe and up the drill pipe-hole annulus. Some of the important drilling applications of the fundamental concepts are also presented. These applications are 1) calculations of subsurface hydrostatic pressures, 2) pressure losses in circular pipe and annuli, and 3) bit nozzle size selection.

HYDROSTATIC PRESSURE The hydrostatic pressure of the drilling fluid is an essential feature in maintaining control of a well and preventing blowouts. It is defined as the static pressure of a column of fluid. The hydrostatic pressure of a mud column is a function of the mud weight and the true vertical depth of the well. Remember that the true vertical depth is used and not the measured depth. The formula to calculate hydrostatic pressure in the units common for Saudi Aramco is: PH =

ρ × D 144

.............…......……………….......... (1)

where, PH = ρ = D =

hydrostatic pressure, psi mud weight, pcf vertical depth, ft

Drilling operations often involve several fluid densities, pressures resulting from fluid circulating and induced surface pressures during kick control operations. For practicality these different pressures are put into a common descriptive system called “equivalent mud weight”. This provides the same pressures in a static system with no surface pressure. Page 1

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EMW = (total pressures × 144) / true vertical depth .....………........ (2) where, EMW is equivalent mud weight in pcf and 144 is the reciprocal of 0.0069456

RHEOLOGICAL FLUID MODELS The movement of drilling fluids in the drill string creates large frictional pressure losses which must be evaluated by the drilling engineer in many of the drilling engineering applications. A mathematical description of the viscous forces in a fluid is required for the development of the frictional pressure loss equations. The rheological fluid models used by drilling engineers to approximate fluid behavior are 1) the Newtonian model, 2) the Bingham plastic model, and 3) the power-law model. NEWTONIAN MODEL

The fluid property responsible for frictioned drag when one layer of fluid is caused to slide over another is called viscosity. Viscosity is defined as the ratio of the shear stress to the resulting shearing rate. In Fig (1), the shear stress acting on the moving upper plate is,

τ=

F ................……….…………............................. (3) A

where, F = force required to move plate at a velocity v A = area of plate. The shearing rate caused by the shear stress τ is equivalent to the velocity gradient

dv dy

The absolute viscosity of the fluid is given by µ =

Page 2

τ τ = ............……………............…...... (4) θ dv / dy

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where,

θ = shear rate =

dv dy

Fig. 1 Simple Planar Shearing of a Fluid

If the viscosity of a fluid is influenced only by temperature and pressure, the fluid is called Newtonian. Some of the Newtonian fluids are water, gases and thin oils. For a Newtonian fluid the ratio of shear stress to shear rate is constant, namely µ. A plot of shear stress versus shear rate for a Newtonian fluid is a straight line with a slope equal to µ as shown in Fig (2)

Fig. 2 Shear Stress v/s Shear Rate for Newtonian Fluid

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All fluids which do not have a direct proportionality between shear stress and shear rate at constant temperature and pressure are classified as non-Newtonian fluids. Examples of non-Newtonian fluids are, drilling mud, clay suspensions, cement slurries and viscous gelled fracturing fluids. The viscosity of these fluids will vary with the magnitude of applied shear stress. Laminar and Turbulent Flow

Fluid flows through a conduit according to either laminar or turbulent flow. When all the fluid particles move in straight lines parallel to the conduit axis, and adjacent layers of fluid slip past each other with no mixing between layers, the flow pattern is called laminar. Steady laminar flow in pipes can be visualized as a series of thin concentric cylinders each sliding past its neighbors like the tubes of a telescope. The cylinder in contact with the pipe wall remains stationary while the inner cylinders move progressively faster as their diameters become smaller. At higher average flow velocities when the fluid particles move down stream in a chaotic motion so that voticies and eddies are formed in the fluid, the flow is called turbulent. In turbulent flow there is no orderly shear between fluid layers but a random shearing and impact of the fluid masses caught up in the swirls and eddies of the flow. Pipe Flow of Newtonian Liquids

Theoretical and experimental evidence have established that within a pipe carrying a Newtonian liquid in laminar flow, the maximum velocity of the fluid occurs at the center of the pipe and decreases to zero at the pipe wall as shown in Fig (3). The average velocity defined as q/A is exactly half the maximum velocity. In turbulent flow, the velocity profile is flatter and the velocity gradient near the pipe wall is much larger than the laminar flow profile for a corresponding value of average velocity.

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Fig. 3 Velocity Profiles for Pipe Flow

Reynolds Criterion

The magnitude of the dimensionless Reynolds number indicates whether the pipe flow of a Newtonian fluid is laminar or turbulent. The Reynolds number is defined by, NR =

123.9 (dvρ )

μ

.................…………........…......... (5)

where, NR = Reynolds number, dimensionless. d = diameter of pipe, in. v = velocity, ft/sec. ρ = fluid density, pcf µ = viscosity, cp When the Reynolds number exceeds the critical value of approximately 2100 in round pipes, turbulent flow starts, while for lower values of NR the flow is laminar. Example

Water is circulated down 3.5” drill pipe (ID= 2.76”). Calculate the maximum pumping rate in bbl/min that will maintain laminar flow.

Solution Page 5

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d ρ

µ NR

= = = =

2.76 in 62 pcf 1 cp 2000

For laminar flow the Reynolds number should not exceed 2100, 2100 =

123 .9 × 2 .76 × 62 v 1

Solving for v, v=

Flow rate = Av = = =

2100 = 0 .099 ft / sec 123 .9 ( 2 .76 ) 62

π ( 2 . 76 ) 2 4 × 144

× 0 . 099

0.0041 ft3/sec 0.0041 × 60 = 0.04 bbl / min 5.61

Laminar Flow

The frictional losses (pressure loss) in laminar Newtonian flow in circular pipes can be calculated from the equation, ΔΡ =

μ Lv 1500 d 2

.......…………........…………........... (6)

Equation (4) is expressed to accept the practical engineering units: ΔP = L = µ

=

Page 6

pressure loss, psi. length, ft. viscosity, cp

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v d

= =

average velocity, ft/sec. diameter, in.

For annular flow, the pressure loss of Newtonian liquids is expressed in practical engineering units by the equation, ΔΡ =

μ Lv 1000 ( d 2 − d 1 ) 2

......………......................... (7)

where, d2 d1

= =

inside diameter of outer pipe, in outside diameter of inner pipe, in

Turbulent Flow

The pressure loss of Newtonian fluids in pipe is calculated by using the Fanning equation, ΔΡ

=

fL ρ v 2 193 d

..................………........…......... (8)

where f is the Fanning friction factor which depends on the Reynolds number and surface condition of the pipe. The value of f can be obtained from the plot shown in Fig (4). The pressure loss of a Newtonian fluid in an annulus is calculated by the equation ΔΡ =

fL ρ v 2 193 d e

where, de = equivalent diameter = 0.816 (d2-d1)

Page 7

.......................………......................... (9)


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The Reynolds number for an annulus is calculated from the equation NR =

123 .9 ρ v d e

μ

........….....................………........ (10)

The annular velocity is expressed v=

q 2.448( d 22 − d 12 )

.................................…....... (11)

Example

A workover rig is circulating brine in the tubing-casing annulus at a rate of 200 gpm. Calculate the pressure loss in 1000 ft of annulus using the following data: Brine density Brine viscosity Casing ID Tubing OD

= = = =

66 pcf 0.8 cp 4.892 in 2.375 in

Solution

de = 0.816(d2-d1) = (4.892 - 2.375) 0.816 = 2.053 in Annular velocity =

NR =

200 = 4 .4 6 ft / sec. 2 .4 4 8 ( 4 .8 9 2 2 − 2 .3 7 5 2 )

123.9(66)( 4.46)(2.05 3) = 93594 (turbulent ) 0.8

From Fig (4) (with de = 2.05 in) f = 0.005

From Eq.(9) Page 9

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ΔΡ =

0 . 0050 × 1000 × 66 × 4 . 46 2 = 16 . 5 psi 193 × 2 . 053

BINGHAM PLASTIC FLUIDS

The Bingham plastic flow model was proposed by E.C. Bingham and is described by the flow curve shown in Fig (5)

Fig. 5 Flow curves of Newtonian and Bingham Plastic Fluids

The defining equation of the curve is ,

(τ − τ y ) = μ pθ where τ y is the positive intercept on the shear stress axis and µp is the plastic viscosity. Unlike a Newtonian fluid, a Bingham plastic fluid will not deform (or flow) continuously

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until the applied shear stress

τ

exceeds a certain minimum value τ y which is known as the yield point. After the yield point value has been exceeded, equal increments of additional shear stress will produce equal increments of shear rate in proportion to the plastic viscosity µp. The plastic viscosity and the yield point specify completely the flow properties of a Bingham plastic fluid. At least two experimental determinations of shear rate at different values of applied shear stress are necessary to define the flow curve. The Bingham plastic model is used to approximate the behavior of drilling fluids and cement slurries. The rotational viscometer is used to measure the rheological properties of a Bingham plastic or drilling fluid. The fluid is sheared between an inner bob and a rotating sleeve. Six standard rotational speeds are available with a rotational viscometer. The plastic viscosity is computed using the equation.

μ p = θ 600 − θ 300 ...................................………........ (12) or

μp =

300(θ N2 − θ N1 ) N2 − N1

where θ 600 is the dial reading with the viscometer operating at 600 rpm and θ 300 is the dial reading at 300 rpm. The yield point Y in 1b/100 ft2 is computed using. Υ = θ 300 −

μ p ..............................………................ (13)

A third parameter called the gel strength, in units of lb/100 ft2 is obtained by noting the maximum dial reading when the viscometer is turned at a low rotor speed of 3 rpm. If the reading is obtained after the mud is allowed to remain static for 10 sec. the dial reading obtained is called the initial gel. If the mud is allowed to remain static for 10 minutes the maximum dial reading is called the 10 min gel strength. The frictional pressure loss of Bingham plastic fluids in laminar pipe flow can be calculated from the equation, ΔΡ =

μ p Lv 1500d

2

+

The pressure drop for annular laminar flow is, Page 11

YL 225d

...............…........................ (14)

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ΔΡ =

1000( d 2 − d 1 )

2

+

YL 220( d 2 − d 1 )

....................... (15)

The frictional pressure loss for turbulent pipe flow is, fL ρ v 2 ΔΡ = 193 d

....................……….......................... (16)

For pipe flow the Reynolds number is calculated by using µp instead of µ or NR =

123.9 ρ vd

..........…....………............................ (17)

μp

The Reynolds number for annular flow is, NR =

101.2 ρν ( d 2 − d 1 )

μp

............................…….....

(18) The frictional pressure loss for turbulent annular flow can be obtained from the equation, ΔP =

Lf ρ v 2 157 .8( d 2 − d 1 )

................................……....... (19)

Pseudo Plastic and Dilatant Fluids

Fluids which do not behave as Bingham plastic are assigned the class of generalized nonNewtonian. Fluids of this type do not have yield points but their viscosity is a nonlinear function of shear stress and possibly duration of shear. Within this general class are two major sub groups called pseudoplastic and dilatant fluids. The viscosity of a pseudo plastic fluid will decrease with increasing values of shear stress. A dilatant fluid displays rheological properties opposite to pseudoplastic in that its viscosity increases with increasing shear stress.

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THE POWER-LAW MODEL

The power-law model is represented by the relationship

τ

= Kθ n

........................................…………...... (20)

A plot of Eq. (20) is shown in Fig (7). The constants K and n characterize the flow behavior of the fluid. K is the consistency index which corresponds to the viscosity of a Newtonian fluid and n is the flow behavior index which indicates the degree of departure from Newtonian behavior. The value of n ranges between zero and 1.

Fig. 6 Flow Line for a Power-law Fluid

Fig. 7 Flow Curve for a Power-law Fluid

Equation (18) can be written as, log τ = log K + n log θ

....................…………........... (21)

A logarithmic plot of shear stress versus shear rate is linear as shown in Fig (6). The slope of the line is n and the intercept on the stress axis defines K at θ =1.

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The stress-shear relationship for a power-law fluid or a pseudoplastic fluid is nonlinear but approach linearity at high shear rates. Thus, if stress readings at high shear rates are extrapolated to the axis, there appears to be a yield point similar to that of a Bingham plastic fluid; hence the name pseudoplastic. Typical pseudoplastic fluids are suspensions of long-chain polymers such as XC polymer. The consistency curves of most drilling fluids are intermediate between the ideal Bingham plastic and power-law flow models. Low - solid, polymer fluids and oil-base muds tend towards power-law behavior, whereas high-solid muds and untreated and flocculated clay muds act more like Bingham plastic fluids. The frictional pressure loss in pipe for a power-law or pseudoplastic fluid in laminar flow is

ΔP =

LKv n 144000d 1+ n

1⎞ ⎛ ⎜3+ ⎟ n⎟ ⎜ . 0416 ⎜ ⎟ ⎝ ⎠

n

........................…....... (22)

where, n = K =

3.32 log 5 1 0θ

θ θ

600 300

300

5 1 1n

The frictional pressure drop in an annulus is 1 ⎞ ⎛ ⎜ 2+ ⎟ n ⎟ ....……..….. (23) ⎜ 0 . 0208 ⎜ ⎟ ⎝ ⎠ n

ΔP =

Page 14

Kv n L 144000 ( d 2 − d 1 ) 1+ n

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The Reynolds number for a power-law fluid flowing in a pipe is defined as,

NR =

11912 ρ v 2 − n K

(24)

⎛ ⎜ 0 . 0416 d ⎜ ⎜ 3+ 1 ⎜ n ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

n

.....……..............

A plot of the friction factor f versus the Reynolds number for a power-law fluid is shown in Fig (8). The critical Reynolds number, above which the flow is turbulent, is a function of the index n. For example, for an n value of 0.2 the critical Reynolds number is 4200. For power-law annular flow the Reynolds number is, n

N

R

14572 ρ v ( 2 − n ) = K

(25)

⎤ ⎡ ⎢ 0.0208 ( d 2 − d 1 ) ⎥ ⎥ ….. ⎢ 1 ⎥ ⎢ 2+ n ⎦ ⎣

The frictional pressure drop in a circular pipe for a power-law fluid in turbulent flow is, fρ v 2 L ΔP = 193 d

...........................……………......... (26)

where f is obtained from Fig (8). For turbulent annular flow the pressure drop is, ΔP =

Page 15

fρ v 2 L 157.8 ( d 2 − d 1 )

........….............……......... (27)


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Example

A 67 pcf bentonite clay drilling fluid is circulated down 5” open ended drill pipe which is inside a 9 58 ” casing at the rate of 5 BPM. Calculate the frictional pressure loss per 1000 ft of drill pipe and 1000 ft of annulus given the following: Casing ID Drill pipe ID Dial reading at 600 rpm Dial reading at 300 rpm

= = = =

8.9 in 4.27 in 34.5 24.5

Solution Pressure loss in drill pipe

Since the mud is bentonite clay, Bingham plastic model is used. v=

q 5 x 42 gpm = = 4 .7 ft / sec. 2 2.448 d 2.448 x 4 .27 2

μ = θ 600 − θ 300 = 34.5 − 24.5 = 10cp p

Y = θ 300 − μ p = 245 . − 10 = 145 . lb / 100 ft 2 NR =

123.9 ρvd

μp

=

123.9 × 67 × 4.7 × 4.27 = 16660 > 2100 10

Since NR is greater than 2100, the flow is turbulent. From Fig (4) the friction factor f is 0.007 From Eq (16) pressure loss per 1000 ft

=

fρ v 2 L 1 93 d

0.007 × 6 7 × 4.7 2 × 1000 = = 12.6 psi . 1 93 × 4.27 Page 17

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Pressure loss in annulus: v=

q 5 × 42 =× = 1 . 406 ft / sec . 2 2 2 . 448 ( d 2 − d 1 ) 2 . 448 ( 8 . 9 2 − 4 . 27 2 )

NR =

=

1 01 . 2 ρ v ( d 2 − d 1 )

μp

, Eq (18 )

101.2 × 67 × 1.4(8.9 − 4.27) = 4395 10

Since NR is greater than 2100, the flow is turbulent. From Fig (8) the friction factor f is 0.0095 From Eq(19) pressure loss in 1000 ft of annulus =

fρ v 2 L .0095 × 6 7 × 1.4 2 × 1000 = = 1.71 psi 157.8 ( d 2 − d 1 ) 157.8( 8 .9 − 4 .27 )

INITIATING CIRCULATION IN A WELL

Non-Newtonian fluids whose viscosity at a fixed shear rate do not remain constant, but change with the duration of shear, are classified as time-dependent. While subjected to a constant rate of shear, a thixotropic fluid exhibits a decrease in shear stress as time of shear is increased. Drilling fluids usually will exhibit a thixotropic behavior at the time circulation is started. If a drilling fluid is allowed to remain static it will develop a gel strength which will require higher pressure to initiate circulation. The circulation pressure decreases with time until a steady frictional pressure loss is observed.

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The frictional pressure loss equations presented in the previous sections do not take into account the thixotropic behavior of the mud and, therefore, should not be used to calculate the pressure required to initiate circulation. In some cases, the pressure required to initiate circulation is greater than the pressure required to sustain circulation at the desired rate. The pressure required to start circulation is ΔP =

Lτ g 300d

(For pipe) ...........………….................. (28)

and ΔP =

Lτ

g

300 ( d 2 − d 1 )

(For Annulus) .......….......... (29)

where τ g is the gel strength of the mud in lb/100 ft2 Example

Compute the pressure at the casing seat at 3000 ft when a mud having a density of 67 pcf and a gel strength of 50 lb/100 ft2 just begins to flow. The casing has an ID of 10 in and the drill pipe OD is 5 in. Solution

Using Eq (29) the pressure required to initiate circulation in the annulus is, ΔP =

50 × 3000 = 100 psi 300 (10 − 5 )

Pressure at casing seat is the hydrostatic mud pressure plus circulation pressure, P = 3000 ×

67 + 100 = 1495 psi 144

When the drilling fluid becomes severely gelled in an annulus of small clearance, excessive pressures may be required to break (start) circulation. In some cases, the pressure required to initiate circulation may exceed the fracture pressure of the exposed formation. To reduce the pressure requirements, the drill string can be rotated before the Page 19

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pump is started. In addition, the pump speed can be increased very slowly while the drill string is rotated.

HYDRAULIC POWER If the flow rate q is expressed in gpm and the pump pressure Δ P is expressed in psi, the hydraulic power output of the pump is, H =

ΔΡ × q 1714

.......................…………………........ (30)

where H is expressed in hydraulic horsepower . Example

A 70 pcf mud is circulated in a well at a rate of 500 gpm and 3000 psi surface pumping pressure. The pressure loss in the drill pipe is 1500 psi. Determine (a) the hydraulic horsepower developed by the pump and (b) the power lost due to viscous forces in the drill pipe. Solution

a) From Eq (30), the power output of the pump is. H =

3000 × 500 = 875 Hp 1714

b) Power consumed due to friction in drill pipe is, H =

1500 × 500 = 437 Hp 1714

PRESSURE DROP ACROSS BIT NOZZLES

The pressure drop across bit nozzles can be calculated from the equation,

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ΔΡ =

ρq 2 8 1000 A 2

.................................……..…....... (31)

where,

ΔΡ

ρ

A

= = =

pressure drop, psi mud density, ppg total cross sectional area of nozzles, in2

The velocity of flow through a nozzle is,

v =

q 3.1 1 7 A

..............................….…………......... (32)

where, A = the cross sectional area of a nozzle, in2 q = the flow rate through a nozzle, gpm The hydraulic horsepower across the bit is

H=

ΔΡ × q 1714

Substituting for ΔΡ from Eq (31),

H =

ρq 3 138.83 × 10 6 A 2

..……....…..….….............. (33)

HYDRAULIC IMPACT FORCE

The purpose of bit jet nozzles is to improve the cleaning action of drilling fluid at the bottom of the hole. Investigators have shown that the cleaning action is maximized by maximizing the hydraulic impact force of the jetted fluid against the hole bottom. The impact force developed by a bit is,

F = 0.00666 Cq

Page 21

ρ Δ Ρb

...........…......................... (34)

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where, q C

ρ Δ Pb

= = = =

flow rate, gpm discharge coefficient (0.95) mud density, pcf pressure drop across bit, psi.

Example

A 90 pcf mud is flowing through a bit having three 13/32 in nozzles at the rate of 400 gpm. Calculate

a) Pressure drop across bit b) Velocity of fluid through nozzles c) Impact force developed by bit

Solution

a)

b)

c) Page 22

π ( 13 32 ) 2

Total area of nozzles

= 0 . 129 in 2 4 = 3 × .129 = 0.388 in2

From Eq (31) pressure drop across bit

=

Flow rate through one nozzle

=

400 = 133.3gpm 3

From Eq (32) velocity through a nozzle

=

133 . 3 = 331 . 5 ft / sec 3 . 117 × . 129

Cross sectional area of one nozzle

From Eq (34) impact force

=

90 × 400 2 = 1181 psi 81000 ( 0 .388 ) 2

= .00666×0.95×400

9 0 × 1181

= 825 lb

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CUTTINGS SLIP VELOCITY Removal of drill cuttings is a primary function of the drilling fluid. Since the drill cuttings are heavier than the drilling fluid, they tend to fall through the mud or “slip” down the annulus. The settling rate of these cuttings is difficult to determine because the cuttings densities are not uniform and the size and diameter ds vary widely. It is important that the annular fluid velocity be greater than the slip velocity so that the drilled cuttings are circulated out of the hole properly. According to Chien the slip velocity for water based fluid is: ⎡ ⎤ ⎢ ⎥ ⎛ μ ⎞⎢ ⎥ ⎞ ⎛ 36 800 , d − ρ ρ f a ⎟ s s V = 0.0561⎜ + 1 − 1⎥ ..….. (35) ⎢ ⎜ ⎟ ⎜ρ d ⎟ s ⎥ ⎝ f S ⎠ ⎢ ⎛ ( μ a ) 7.48 ⎞ 2 ⎝ ρ f ⎠ ⎢ ⎜ ⎥ ⎟ ⎢⎣ ⎝ ρ f d s ⎠ ⎥⎦

Where μ a = PV for water based mud, for polymer based fluid, μ a = PV + 5 where,

YP × d s . v

ρ f = fluid density, pcf ρ s = cutting density, pcf PV = Plastic viscosity, cp YP = Yield point, lb/100 ft2 v = annular velocity, ft/sec µa = apparent viscosity, cp

Example:

A surface hole is to be drilled to 3,500 ft. Many wells in the same area have experienced loss circulation problems due to insufficient cuttings removal. If the chosen annular velocity is 60 ft/min. will the hole be cleaned adequately? Hole size Drillpipe: Mud: Fann viscometer:

Page 23

24 in. 4-1/2 in. 67 pcf (water based) θ 600 = 52, θ 300 = 31

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Cuttings:

0.25 in. (diameter) 157 pcf (density)

Solution:

Since the mud is water based, use the plastic viscosity as the apparent viscosity Using Eq.12,

PV = 52 - 31 = 21 cp

⎡ ⎢ 21 ⎛ ⎞⎢ Vs = 0 . 0561 ⎜ ⎟ ⎝ 67 × 0 . 25 ⎠ ⎢ ⎢ ⎣⎢

⎤ ⎥ 36 ,800 ( 0 . 25 ) ⎛ 157 − 67 ⎞ ⎟ + 1 − 1 ⎥⎥ = 0 . 76 ft / sec 2 ⎜ 67 ⎠ ⎛ ( 21 ) 7 . 48 ⎞ ⎝ ⎥ ⎜ ⎟ ⎝ 67 × 0 . 25 ⎠ ⎦⎥

Since the chosen annular velocity will be 60 ft/min = 1 ft/sec, is greater than the slip velocity the pump rate is sufficient to move the cuttings up hole and out of the annulus.

OPTIMIZATION OF BIT HYDRAULICS The selection of the proper jet bit nozzle sizes is important in the drilling operation. Significant increases in penetration rate can be achieved through the proper choice of bit nozzles. The penetration rate increase is due to mainly to improved cleaning action at the bottom of the hole. Wasteful regrinding of cuttings is prevented if fluid circulated through the bit removes the cuttings as rapidly as they are made. At present, there is still disagreement as to what hydraulic parameter should be used to indicate the level of the hydraulic cleaning action. The most commonly used hydraulic design parameters are (1) bit hydraulic horsepower and (2) jet impact force. Jet bit nozzle sizes are selected such that one of these parameters is a maximum. MAXIMUM BIT HYDRAULIC HORSEPOWER

It has been pointed out that the effectiveness of jet bits could be improved by increasing the hydraulic horsepower at the bit. It has been shown mathematically that the bit horsepower is a maximum when,

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Δ Ρd =

ΔΡ m+1

..........……………............................ (36)

where Δ P is the total pressure loss in the circulating system and Δ Pd is the parasitic pressure loss to and from the bit,

ΔΡ d = Δ Ps + ΔΡ dp + ΔΡ dc + Δ Pdca + Δ Pdpa ……...... (37) where,

Δ Ps Δ Pdp Δ Pdc Δ Pdca Δ Pdpa m

= = = = = =

Pressure loss in surface equipment Pressure loss in drill pipe Pressure loss in drill collars Pressure loss in drill collar annulus Pressure loss in drill pipe annulus constant approximately equal to 1.75

Since the total pressure loss is the pressure loss across the bit Δ Pb plus the parasitic pressure loss then for maximum horsepower at the bit the pressure drop across the bit should be, or

Δ Pb

=

Δ P - Δ Pd

Δ Pb

=

ΔP -

=

mΔP m+1

ΔP m+1 ...........…….................................. (38)

MAXIMUM JET IMPACT FORCE

Some operators prefer to select bit nozzle sizes so that the jet impact force is a maximum rather than bit hydraulic horsepower. It can be shown that maximum jet impact force is obtained when the parasitic pressure drop is, Δ Pd =

or the bit pressure drop is Page 25

2 Δ P .............….….......................... (39) m+ 2

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Δ Pb =

Δ P - Δ Pd 2ΔP m+2

=

ΔP -

=

m ΔP ..........................….......…........ (40) m+ 2

Bottom Hole Cleaning Needs

When high pressure pumps are available and the parasitic pressure loss is low because of large diameter drill string, it may be possible to achieve higher hydraulic bit horsepower or impact force than is needed to clean the bottom of the hole. The hole cleaning needs can be determined by measuring the penetration rate at various bit hydraulics. Once the cleaning needs are determined, it would be wasteful to provide higher bit hydraulics than needed. Under these conditions, the pumping rate should be reduced until the desired hydraulics are obtained. The rate should never be reduced below the rate required to lift the cuttings. As a rule of thumb bit power should be in the range of 2.5 to 5 HP per square inch of bit area. In hole sizes 12 ¼" and greater bit power of 5 to 6 HP/ in2 may be used. The engineer should not be overly concerned about which criterion, bit power or impact force is best. There is not a great difference in the application of the two procedures. If the bit horsepower is maximum the jet impact force will be within 90% of the maximum and vice versa. Graphical Solution

The selection of bit nozzle sizes can be simplified by graphical techniques. In the case of turbulent flow the parasitic pressure loss in the drill string can be represented by the equation Δ Pd = cqm ...................……..……......................... (41) Taking the logarithm of each side Page 26

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log Δ Pd= log c + m log q

................….….............. (42)

A plot of Δ Pd versus q on log-log paper is a straight line whose slope is m as shown in Fig (9). For turbulent flow, the value of m is close to 1.75.

Fig. 9 Use of log-log plot for selection of proper pump operation and bit nozzle sizes

Similarly, the hydraulic horsepower equation Hp =

P × q 1714

is represented by a straight line with a slope of -1.0 on a graph of log P versus q.

Shown in Fig (9) is a summary of the conditions for the selection of bit nozzle sizes using the various hydraulic parameters. The conditions for proper pump operations and bit Page 27

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nozzle sizes occur at the intersection of the parasitic pressure loss line and the path of the optimum hydraulics. The path of optimum hydraulics has three straight-line segments as shown in Fig(9). Segment 1, defined by q = qmax, corresponds to the shallow portion of the well where the pump is operated at maximum rate and pressure for the convenient pump liner size and horsepower rating. Segment 2, defined by constant parasitic pressure loss Δ Pd, corresponds to the intermediate portion of the well where the flow rate is reduced gradually to maintain Δ Pd / Pmax at the proper value for maximum bit hydraulic horsepower or impact force. Segment 3, defined by q = qmin, corresponds to the deep portion of the well where the rate is reduced to the minimum value that will efficiently lift the cuttings to the surface. In Fig (9), the intersection of the parasitic pressure loss line and the path of optimum hydraulics occurs in Segment 2. This corresponds to bit at intermediate depth. Since parasitic pressure loss increases with depth, a shallow bit run would intersect in Segment 1 and a deep bit run would intersect in Segment 3. Once the intersection point is obtained the proper flow rate, qopt is read from the graph. The proper pressure drop across the bit corresponds to Pmax - Δ Pd on the graph at the intersection point. The proper nozzle area is calculated from the equation Aopt

=

2 ρq opt

90000ΔPb ( opt ) C 2

.…............................. (43)

Example

Determine the proper pump operating conditions and bit nozzle sizes for maximum jet impact force for the next bit run given the following: Measured parasitic pressure loss @ 485 gpm Measured parasitic pressures loss @ 247 gpm Mud weight Pump horsepower Pump efficiency Minimum rate to lift cuttings Maximum allowable pumping pressure

Page 28

= = = = = = =

906 psi 409 psi 72 pcf 1250 hp 0.91 225 gpm 3000 psi

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Solution

The two parasitic pressure loss values are plotted on log-log paper to define the parasitic pressure loss line. The slope of the line is, m =

lo g 9 0 6 − lo g 4 0 9 0 .3 4 5 = = 1 .1 8 lo g 4 8 5 − lo g 2 4 7 0 .2 9 3

The path of optimum hydraulics is determined as follows: Segment 1

Hp

=

qmax =

qΡ 1714 E 1714 × H p E Pmax

=

1714 × 1250 × .91 = 650 gpm 3000

Segment 2

For maximum impact force the parasitic pressure loss should be ΔΡd =

2 2 × 3000 = 1886 psi Pm ax = m+2 1 . 18 + 2

Segment 3

qmin = 225 gpm The path of optimum hydraulics is determined by plotting the three segment lines as shown is Fig (10). It can be seen that the path of optimum hydraulic line intersects the parasitic pressure loss line in Segment 1 at qopt = 650 gpm

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Fig. 10 Application of graphical analysis techniques for selection of bit nozzle sizes

From the graph,

Δ Pd = 1300 psi

Thus ΔPb = 3000 - 1300 = 1700 psi The proper total nozzle area is, Aopt =

ρq 2 2

C Δ Pbopt 90000

Jet Impact Force = 0.00666 C q

ρ Δ Pb

= 0.00666 × 0.95 × 650

Page 30

= 0.47 in2

72 × 1700

= 1439 lb

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In the previous example the impact force criterion was not used to select the nozzle size because the segment 2 line which represents the conditions of maximum impact force did not intersect the parasitic pressure loss line. If the impact force criterion were used, the parasitic pressure loss would be 1886 psi. From the graph, the corresponding rate is 850 gpm. The pump pressure required to deliver 850 gpm is, P

=

1714 EH P q

=

1714 × .91 × 1250 = 2293 psi 850

Therefore, the bit pressure drop is

ΔPb = 2293 - 1886 = 407 psi The total nozzle area is

72 × 850 2 ( 0 .95 ) 2 407 × 90000

A

=

= 1.25 in2

F

= 0.00666 × 0.95 × 850 72 × 407

Jet impact force is, = 920 lb

The use of the impact force criterion in this example provides a higher pumping rate. However, a high rate does not always mean better cleaning. The high rate resulted in a larger nozzle area of 1.25 in2 and thus a lower bit pressure drop of 407 psi and a lower impact force of 920 lb compared to 1438 lb impact force obtained by using the qmax criterion. This shows that the qmax criterion is the correct method to use in this example. Example

Estimate proper pump operating conditions and bit nozzle sizes for maximum bit horsepower while drilling at 2000 ft and 5000 ft. The well plan calls for the following conditions:

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Pump

1000 psi 1500 psi 800 hp 0.90

maximum surface pressure at 1000 ft. maximum surface pressure at 5000 ft. maximum horsepower input. pump efficiency.

Drill string

5” 19.5 #/ft (4.276 in ID) drill pipe 600 ft of 9 in OD x 2.5 in ID drill collars for 17.5 in hole 600 ft of 6¼ in OD x 2.5 in ID for 8½” hole Surface Equipment

Equivalent to 580 ft of 5” DP Hole Sizes

17.5” at 2000 ft. 8.5” at 5000 ft. Minimum Annular Velocity

50 ft/min at 1000 ft. 120 ft/min at 5000 ft. Mud Properties

Depth 2000 5000

Mud Weight, pcf 64 75

Polymer mud is used at both depths.

Page 32

θ 600

Reading 29 38

θ 300

Reading 22 27

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Solution A) Drilling at 2000 ft.

Parasitic Pressure losses

Pressure losses can be calculated at any flow rate. Flow rates of 700 and 500 gpm will be used in this example. Since polymer mud is being used the pressure losses in each segment of the drill pipe will be calculated using the power-law model. n = 3.32 log θ

θ

= 3.32 log K =

600 300

29 = 0.398 22

510θ 300 510 x 22 = = 937 .6 n 511 511 0 .398

Pressure losses in drill pipe

v =

q 500 = = 11.17 ft / sec. 2 2 .448 d 2 .4 48 ( 4 .276 2 )

From Eq (24) NR =

11912 × 64 × 11.17 1.602 937.6

⎛ ⎜ .0416 × 4 .276 ⎜ ⎜ 3+ 1 ⎜ .398 ⎝

= 9900 (flow is turbulent) From Fig (8) f

= 0.004

From Eq (24), the pressure loss is Page 33

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

0 .398

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fρ v 2 L 1 93 d . 004 × 64 × 11 . 17 2 × 1400 = 193 × 4 . 276 = 55 psi

ΔΡ =

Pressure loss in drill collars

v

=

NR =

q 2 . 448 d

2

=

11912 × 64 × 32 . 68 1 .602 937 . 6

= 44642 f

500 = 32 . 68 ft / sec . 2 . 448 × 2 . 5 2 ⎞ ⎛ ⎜ . 0416 × 2 .5 ⎟ ⎟ ⎜ ⎟ ⎜ 3+ 1 ⎟ ⎜ .398 ⎠ ⎝

. 398

= 0.0027

Pressure loss =

0.0027 × 64 × 32.68 2 × 600 = 229 psi. 193 × 2.5

Pressure loss in drill collar-hole annulus

From Eq(9),

=

v q 500 = = 0 .9 0 6 ft / s e c . 2 2 2 .4 4 8 ( d 2 − d 1 ) 2 .4 4 8 (1 7 .5 2 − 9 2 )

From Eq (23) NR =

14572 × 6 4 × 0.906 937.6

1.602

= 237 (flow is laminar) From Eq (22), Page 34

⎡ ⎤ ⎢ . 0208 (17 . 5 − 9 ) ⎥ ⎢ ⎥ 1 ⎢ ⎥ 2+ . 398 ⎣ ⎦

. 398

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1 ⎛ ⎜2+ .398 937 .6 × .906 .398 ⎜ ⎜ .0208 ⎜ ⎝ Pressure loss = 144000 (17 .5 − 9 ) 1 .398 = 1.6 psi.

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

.398

× 600

Pressure loss in drill pipe-hole annulus

v =

500 = .7 2 6 ft / sec. (Flow is laminar) 2 .4 48 17 .5 2 − 5 2

(

)

1 ⎞ ⎛ 2+ ⎟ 937.6 × .726 .398 ⎜⎜ .398 ⎟ Pressure loss = 144000 (17.5 − 5)1.398 ⎜ .0208 ⎟ ⎟ ⎜ ⎠ ⎝ = 1.997 psi

.398

× 1400

Pressure loss in surface equipment is equivalent to pressure loss in 500 ft of drill pipe, Pressure loss =

53 . 8 × 580 = 22 . 2 psi 1400

Parasitic pressure loss at 500 gpm

= 53.8 + 228 + 1.6 +1.99 + 22.2 = 307.6 psi

The above calculations are repeated for a flow rate of 700 gpm The parasitic pressure loss at 700 gpm = 502 psi The parasitic pressure loss is plotted versus rate on log-log paper in Fig (11). The slope of the line is, m =

Page 35

log 502 − log 307 .6 = 1.455 log 700 − log 500

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Fig. 11 Hydraulic Plot for example problem

The path of optimum hydraulics is as follows: Segment 1

qmax =

Page 36

1714 × H p × E Pmax

=

1714 × 800 × .9 = 1234 gpm 1000

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Segment 2

From Eq (34) for optimum bit horsepower the parasitic pressure loss is, ΔPd =

1 1000 P m ax = = 407 psi 1 + 1.455 2 .455

Segment 3

For a minimum annular velocity of 55 ft/min, the minimum flow rate is,

(

where v is in ft/sec.

qmin = 2.448 × v d 22 − d 12

Therefore, qmin = 2.448 × = 630 gpm

)

55 17 .5 2 − 5 2 60

(

)

The path of optimum hydraulics is plotted in Fig (11). Note that the path intersects the parasitic pressure loss line at qmin = 630 gpm. Therefore, the optimum pumping rate should be 630 gpm. The parasitic pressure drop at 630 gpm is 425 psi (from graph). Therefore, the optimum bit pressure drop is, ΔPb = 1000 - 425 = 575 psi The total optimum nozzle area is, Aopt = Area of one nozzle =

Page 37

64 × 630 2 = 0 .737 in 2 0 .95 2 × 575 × 90000 0 .737 = .246 in 2 3

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Diameter of nozzle =

4A = 3.14

4(.246) = 0.56 in 3.14

= 1 8 3 2 in B) Drilling at 5000 ft.

Parasitic pressure losses are calculated at 500 and 700 gpm using power-law model. n = 3.32 log

θ 600 θ 300

= 3.32 log

K =

38 = 0.492 27

510 θ 300 510 x 27 = = 640.4 n 511 511 0 .492

Pressure Losses in Drill Pipe

v = 11.17 ft/sec (same as v at 2000 ft)

NR =

11912 × 75 × 11.17 1.508 640.4

= 10246

⎛ ⎜ .0416 × 4.276 ⎜ ⎜ 3+ 1 ⎜ .492 ⎝

f = 0.0048

Pressure loss = =

fρ v 2 L 193 d

0.0048 × 75 × 11 .17 2 × 4400 193 × 4.276

= 240 psi

Page 38

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

.492

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Pressure Loss in Drill Collars

v = 32.68 ft/sec. NR =

11912 × 75 × 32.68 640.4

1.508

= 39770

⎡ ⎢ .0416 × 2.5 ⎢ 1 ⎢ 3+ .492 ⎣

⎤ ⎥ ⎥ ⎥ ⎦

.492

f = 0.0032 .0032 × 75 × 32.68 2 × 600 193 × 2.5 = 319 psi

Pressure loss =

Pressure Loss in Drill Collar-Hole Annulus

v =

500 = 6.154 ft / sec. 2 .448 (8.5 2 − 6.25 2 )

From Eq (25) NR =

14572 × 75 × 6.154 1.508 640.4

= 2934 (turbulent flow)

⎡ ⎤ ⎢ .0208(8.5 − 6.25) ⎥ ⎢ ⎥ 1 ⎢ ⎥ 2+ .492 ⎣ ⎦

f = 0.007 From Eq (27), Pressure loss =

Page 39

.007 × 75 × 6.154 2 × 600 = 33.6 psi 157.8(8.5 − 6.25)

.492

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Pressure Loss in Drill Pipe-Hole Annulus

v

=

NR =

500 2 .4 4 8 ( 8 .5 2 − 5 2 )

= 4.32 ft/sec.

14572 × 75 × 4.32 1.508 640.4

⎡ ⎤ ⎢ .0 2 0 8 (8.5 − 5 ) ⎥ ⎢ ⎥ 1 ⎢ 2+ ⎥ 0 .4 9 2 ⎦ ⎣

= 2139 (flow is laminar)

.4 9 2

From Eq (23), .492

1 ⎞ ⎛ ⎟ ⎜2+ .492 ⎟ 640 .4 × ( 4 .32 ) .492 ⎜ ⎜ . 0208 ⎟ ⎟ ⎜ ⎠ ⎝ Pressure loss = 1 .492 144000 (8 .5 − 5 ) = 82 psi

× 4400

Pressure loss in surface pipe is equal to pressure loss in 580 ft of 5” drill pipe Presssure loss =

239 × 580 = 31 . 5 psi 4400

Parasitic pressure loss = 240 + 319 + 33.6 + 82 + 31.5 = 706 psi. The above calculations are repeated at 700 gpm The parasitic pressure losses at 700 gpm = 390+546+57+148+51 = 1192 psi The parasitic pressure losses at 500 and 700 gpm are plotted on log-log paper in Fig (12). The slope of the line is, m =

Page 40

log1192 − log706 log700 − log500

= 1.553

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Fig. 12 Hydraulic Plot for example problem

The path of optimum hydraulics is as follows: Segment 1

Pmax =

Page 41

1714 H p × E P max

=

1714 × 800 × .9 = 822 gpm 1500

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Segment 2

For optimum bit horsepower the parasitic pressure loss is, ΔΡd =

1 Pm ax = 587 psi 1 + 1.553

Segment 3

Minimum flow rate is, q min = 2 .448 ×

120 (8 .5 2 − 5 2 ) = 231 gpm 60

The path of optimum hydraulics is plotted in Fig (12). The path intersects the parasitic pressure loss line at qopt = 445 gpm and parasitic pressure loss Δ Pd of 587 psi.

Δ Pbit = Pmax - Δ Pd = 1500 - 587 = 913 psi Total optimum nozzle area is, Aopt

=

Area of one nozzle = Diameter of nozzle = =

Page 42

75 × 445 2 = 0 . 446 in 2 0 . 95 2 × 913 × 90000 0 .4 46 = 0 .14 9 in 2 3

4 × . 149 = 0 . 435 in 3 . 14 14 in 32

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Example

a) In the previous example calculate the bit horsepower per square inch while drilling at 5000 ft b) If the maximum pressure is maintained at 1500 psi, would increasing the pumping rate from 445 to 600 gpm increase the bit horsepower? Solution

a) The bit horsepower is calculated by using Eq (33), H =

Area of bit = Bit horsepower/in2 =

ρq 3 6

1 38 .83 ×10 A

π x 8 .5 2 4 265 56 .7

=

2

75 × 445

3

6

1 38 .83 ×10 ×.0446

2

= 265 H Ρ

= 56.7 in2

= 4.7 hp/in2

b) From the graph in Fig (12), the parasitic pressure loss at 600 gpm is 940 psi. The pressure loss at the bit

= 1500 - 940 = 560 psi.

The total nozzle area is,

A =

H

2

2

0 . 95 × 560 × 90000

= 0 . 77 in 2

75 × 600 3 = 218 hp = 138.83 × 10 6 ( .95 ) 2 ( .77 ) 2

Bit Horsepower/in2 =

Page 43

75 × 600

218 56.7

= 3.85 hp/in2

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Therefore, the bit horsepower would decrease from 4.2 to 3.46 hp/in2 if the pumping rate is increased from the optimum rate of 445 to 600 gpm. This shows that the maximum bit horsepower that can be obtained is 4.2 hp/in2 at an optimum pumping rate of 445 gpm. Increasing or decreasing the pumping rate (while maintaining maximum pressure of 1500 psi) would give less horsepower at the bit. On-Site Nozzle Selection

The calculation of the parasitic pressure loss in the drill string using the pressure loss equations is time consuming and subject to errors due to discontinuities in the drill strings (such as jars, tool joints), washouts in the open hole and variations in the mud properties. The easiest and the most accurate method for determining the total parasitic pressure loss at a given depth is by direct measurement of the pump pressure (stand pipe pressure) at the rig. The pump pressure can be measured at the rig for at least two pumping rates. Since the total nozzle area of the bit currently in use is known, the pressure loss across the bit can be computed at the given pumping rates by using Eq (31). The parasitic pressure loss then can be obtained at these pumping rates as the difference between the pump pressure and the pressure drop across the bit. The parasitic pressure loss is then plotted versus the pumping rates and bit nozzle selection can be performed as was shown in the previous example by using the maximum horsepower or the maximum impact force method.

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Example

A well is being drilled at 5000 ft using 10 ppg mud and three 14/32” (0.4375”) nozzles. The driller recorded that when the mud is pumped at 500 gpm a pump pressure of 1860 psi is observed, and when the rate is increased to 700 gpm a pump pressure of 3470 psi is observed. (a) Calculate the parasitic pressure loss at each pump rate, (b) Calculate the slope of the parasitic pressure loss line. Solution

a) Total area of nozzles

=

3.14 × .4375 2 × 3 4

= 0.450 in2

The pressure loss across the bit at 500 gpm is, ΔP = = =

ρq 2 8100 A 2 75 × 500 2 81000 × .45 2 1267 psi

Parasitic pressure at 500 gpm = 1860 - 1267 = 593 psi Pressure loss across bit at 700 gpm is, 75 × 700 2 ΔΡ = 81000 × .45 2

= 2482 psi

Parasitic pressure loss at 700 gpm = 3470 - 2482 = 988 psi b) The slope of the parasitic pressure loss line is, m

=

log 988 − log 593 log 700 − log 500

0 . 222 0 . 146 = 1.5

=

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TABLE OF CONTENTS

Page

INTRODUCTION

1

CASING DESCRIPTION

2

DIAMETER LENGTH RANGE API CASING & TUBING WEIGHT DESCRIPTION CASING GRADES

2 2 3 4

- PROPERTIES OF TUBULAR MATERIAL

4

• • • •

YIELD STRENGTH HARDNESS OF STEEL HEAT TREATMENTS CHEMICAL COMPOSITION

- API CASING GRADES - NON-API CASING GRADES - SAUDI ARAMCO NON-API CASING GRADES

CONNECTIONS

4 5 6 7

8 9 10

11

API CASING CONNECTIONS SAUDI ARAMCO API CASING CONNECTIONS

11 11

- API SHORT / LONG THREAD & COUPLING - API BUTTRESS THREAD & COUPLING

11 12

PROPRIETARY CONNECTIONS SAUDI ARAMCO PROPRIETARY CONNECTIONS

13 13

- VAM CONNECTION - NS-CC CONNECTION - VETCO LS, RL-4S, & DRIL-QUIP S-60 CONNECTIONS

13 14 14

CORROSION CORROSION MITIGATION SULFIDE STRESS CRACKING SAUDI ARAMCO APPLICATIONS IN H2S SERVICE

16 16 17 19

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TABLE OF CONTENTS

Page

CARE OF OILFIELD TUBULARS

19

CASING PERFORMANCE PROPERTIES

20

BURST COLLAPSE TENSILE FORCE BIAXIAL EFFECTS SAFETY FACTORS

20 21 23 25 27

TYPES OF CASING

28

CONDUCTOR CASING SURFACE CASING INTERMEDIATE CASING LINER PRODUCTION CASING

28 28 28 28 28

SELECTION OF CASING SETTING DEPTHS

30

SELECTION OF CASING & BIT SIZES

36

SELECTION OF WEIGHT, GRADE & COUPLINGS

39

SURFACE CASING INTERMEDIATE CASING INTERMEDIATE CASING WITH A LINER PRODUCTION CASING

GRAPHICAL METHOD, COLLAPSE & BURST DESIGN

40 42 47 47

49

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CASING CENTRALIZER SPACINGS

57

CASING LANDING PRACTICE

58

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CASING DESIGN INTRODUCTION From the earliest days of wells dug in the ground for various purposes, the need for some means of supporting the walls of the hole has been recognized. Many wells of ancient times were lined or “cased” with rock. Over the years casing technology has developed from rock to plaster to wood and then to steel. The functions of casing can be summarized as follows: 1. To keep the hole open and to provide a support for weak, or fractured formations. If the hole is left uncased, the hole may fall in and the redrilling of the hole may become necessary. 2. To isolate porous media with different fluid/pressure regimes from contaminating the pay zone. This is basically achieved through the combined presence of cement and casing, so production from a specific zone can be made. 3. To prevent contamination of near surface fresh water zones. 4. To provide a passage for oil and gas; most production operations are carried out through special tubing which is run inside the casing. 5. To provide a suitable connection with the wellhead equipment (e.g. christmas tree). The casing also serves to connect the blowout prevention equipment which is used to control the well while drilling. 6. To provide a hole of known diameter and depth to facilitate the running of testing and completion equipment. By the year 1900 the regular oil field products had been reasonably well standardized. The earliest American Petroleum Institute (API) specifications on oil well casing were issued in 1924. Beginning about 1930 and continuing for the next several years, specifications were published in several issues of API standards 5-A to cover lengths, sizes, weights, threads, joints, and grades of the steel. In the following years many improvements have been made in thread and coupling design as well as higher pipe grades.

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CASING DESCRIPTION Casing is described using the following parameters: 1. Diameter 2. Length range 3. Weight per unit length

4. Grade of steel 5. Type of coupling

Diameter There are three types of diameter designations. They are outer diameter, inner diameter and drift diameter. The outer diameter is the diameter of the casing measured from outer wall across to outer wall and is the diameter measurement casing is identified with. The inner diameter is the diameter of the casing measured from inner wall to inner wall. The third type of diameter is the drift diameter, which is the guaranteed minimum diameter of the casing, the drift diameter is important because it indicates whether the casing is large enough for a specified size of bit to pass through.

Length Range API has established three length ranges with limits and tolerances as shown below. API specifications for casing and tubing designate the length range of each joint. There are three length ranges for casing: Length Ranges for Casing Table 1 Range

Length

Maximum Length

1

Range, ft 16-25

Variation, ft 6

2

25-34

5

3

Over 34

6

Casing is mostly run in R-3 lengths. These longer lengths reduce the total number of threaded connections needed for the casing string. Since casing is usually run in single joints (instead of doubles or triples), the longer R-3 lengths are easier to handle.

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API Casing & Tubing Weight Designation Casing and tubing weights are expressed in lb/linear ft and are designated as either plainend weights or nominal weights. •

Plain-end weight per foot is the weight per foot of the pipe body excluding the threaded portion and coupling.

•

Nominal weight per foot is the weight per foot that is reflected in casing tables and is an approximate average weight per foot of the pipe with API connections, including upsets, threads, and couplings.

•

Average weight per foot is the total weight of an average joint of threaded pipe with one coupling divided by the total length of the average joint.

The plain end weight of casing can be calculated by knowing the outer and inner diameter of the pipe and the density of steel (489 lb/ft3):

W (lb/ft) =

(

)

π De 2 − Di 2 × 489 4 × 144

(

2

2

)

= D e − D i 2.67 . . . . . . . . . . . . .. . . . . . . . . . . .(1)

where, De: Di:

outside diameter, in. inner diameter, in.

The difference between nominal weight and average weight is generally small and most design calculations are performed by using nominal weight per foot.

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Casing Grades Steel pipe grades are identified by letters and numbers which indicate various characteristics of the pipe steel. It is a specification according to its yield stress, ultimate tensile strength, chemical composition, heat treatment or other characteristics. There are many grades of steel that make up oilfield tubulars. Properties of Tubular Material In order to understand strengths of tubular materials, it is important to understand the basic terminology and process of manufacture of these materials. Yield Strength The strength of a steel is usually indicated by its minimum yield strength or ultimate tensile strength. Casing and tubing are manufactured mostly from ductile steels. Whereas brittle steels fracture without appreciable deformation, ductile steels can withstand significant plastic deformation prior to fracture. Basic Stress-Strain Equations Stress and strain are common terms used in describing strengths of materials. If a tensile load (or force) is applied to a test sample cross-sectional area, then the tensile (or axial) stress is found by:

Stress = Force / Area Axial strain is defined as the ratio of the test sample axial elongation to the original length of the sample: Axial Strain = Axial Elongation / Original Length Hooke's Law defines stress as the product of the elastic constant or Young's modulus of elasticity (E) and strain: Stress = E x strain 6

Young's Modulus for steel is typically 30 x 10 psi. Figure 1 is a stress-strain diagram for a typical ductile steel. Point 'A' represents the yield strength or elastic limit of the steel. If the steel is stressed below the elastic limit, it will return to its original shape upon unstressing or unloading the test sample. Below the elastic limit, the stress-strain curve is linear. Page 4

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The API specifies that the yield stress (yield strength) is the tensile stress required to produce a total elongation of 0.5% of the tensile test sample length. This is shown by point 'B' in the diagram. Stresses greater than the elastic limit cause permanent deformation of the steel and the steel will not return to its original shape when the load is taken away. If a steel is stressed beyond its yield strength, it will deform plastically until its ultimate strength is reached as shown by point 'C'. The ultimate strength is the maximum stress that the steel can sustain before it begins to fail. Beyond this point the material will continue to deform plastically (with a reducing stress) until complete failure (breakage) occurs as shown by point 'D'. STRESS (psi)

A - Elastic Limit B - API Specified Minimum Yield

C - Ultimate Strength D - Failure

STRAIN (%) 0.5% STRESS-STRAIN DIAGRAM FOR DUCTILE STEEL FIGURE 1

Hardness of Steel Hardness is the measure of a steel's yield point in compression. When a material is required to resist wear, corrosion, erosion or plastic deformation, it may be necessary to specify a specific hardness. Hardness generally increases with increasing material ultimate tensile strength. Very hard materials are brittle and will crack or fracture easily. Hardness is determined by a test where a load is applied with a small ball or pointed object. The hardness of the material is then expressed by the depth of the indentation caused by the pointed object. The "Rockwell C", "Brinell", or “Charpy” hardness scales are used to quantify the degree of hardness of an oilfield tubular material. Hardness can be expressed by a Charpy Impact Test, where a weighted pendulum is dropped onto a sample and the amount of impact it takes to break the sample is measured. This amount of impact must exceed a minimum standard. Page 5

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Heat Treatments Mechanical properties of steel such as yield stress, ultimate tensile strength, ductility, or hardness can be achieved by controlling the heat treating portion of the manufacturing process and chemical composition of the steel. Heat treating affects changes in the microstructure, or grain structure of the steel which directly affects its mechanical properties. Heat treating is an operation involving heating and/or cooling the solid steel tubular to develop the desired steel microstructures.

The five basic heat treatments are: Quenched and Tempered The steel is heated to 1500-1600 oF. It is then rapidly quenched (or cooled) in water or oil to produce a desired microstructure. It is then tempered (or re-heated) at 10001300 oF to produce a desired combination of strength and ductility. This is the preferred method of producing high strength casing and tubing. Normalizing The steel is heated to 1600-1700 oF and then cooled in air to produce a uniform microstructure and to alter mechanical properties. Normalized and Tempered The steel is first normalized (as above) and then tempered and air cooled. This tempering process slightly lowers the strength from the normalized condition but improves ductility and helps to relieve residual stresses. Cold Drawn and Tempered The tubing or casing is shaped or rolled to the desired OD at room temperature. This process causes a high residual stresses in the tube and increases the hardness due to plastic deformation. The tubular is then tempered to reform the microstructure from the cold drawn state. Tempering reduces the hardness and relieves the residual stresses. Hot Rolled The tubing or casing is shaped or rolled to the desired OD at a very high temperature. Hot rolling does not cause changes in the microstructure as in the cold rolling process above. Hot rolling produces a steel similar to the normalized condition.

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Chemical Composition The chemical composition of steel directly affects all of its mechanical properties and corrosion resistance. Steels can be classified according to chemical composition as follows: Carbon Steels These steels are considered to be a mixture of iron and carbon with up to 2% carbon content. The high carbon steels contain up to 2% carbon, like J-55, while the low carbon steels, like L-80, contain as low as 0.25% carbon. Carbon steels can contain other elements such as manganese or silicon in small quantities. Most tubulars are made of carbon steel. Alloy Steels These steels contain significant quantities of alloying elements other than carbon. A steel is considered an alloy steel when the content of either manganese, silicon or copper exceeds 1.65%, 0.6% and 0.6% respectively. A steel is also considered an alloy if there is a minimum content specified for aluminum, boron, cobalt, chromium, niobium, molybdenum, or nickel. Alloy steels are less susceptible to corrosion and more expensive than carbon steel. High Alloy Steels High-alloy steels contain more than 5% alloy elements, in particular, high concentrations of chromium, molybdenum, and nickel are used for high-alloy tubulars. High-alloy steels which contain greater than 12% chromium are often called "stainless" steels. Low-Alloy Steels Low-alloy steels contain less than 5% metallic alloying elements. High Alloy Chrome-13 casing for Saudi Aramco GWI Wells The majority of the casing used in Saudi Aramco is J-55 and made of carbon steel. Alloy steel that contains 13% chrome is used in Abqaiq gravity water injection wells to combat corrosion caused by the Wasia water. Chrome-13 casing is also used across the corrosive UER and Aruma aquifers in the Safaniya field. The price of 13-chrome casing is 3 times that of carbon steel casing, however, its use is justified since it increases the useful life of the wells and reduces workover costs.

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API Casing Grades To understand API casing grades, it is important to understand the terms minimum yield stress, maximum yield stress, and minimum ultimate strength. To explain these terms, two popular grades of oilfield tubulars will be used as an example: L-80 and N-80.

The grade of steel is denoted by a letter of the alphabet followed by the minimum yield stress of the particular steel. For example, the API grade L-80, which is a common grade used by Saudi Aramco, has a minimum yield stress of 80,000 psi as shown by point "A" in Figure 2. In other words, it can support a stress of 80,000 psi with an elongation of 0.5%. STRESS (psi) B - API Specified Maximum Yield

95,000 80,000

C - API Minimum Ultimate Strength

A - API Specified Minimum Yield

STRAIN (%) 0.5% STRESS-STRAIN DIAGRAM FOR L-80 STEEL FIGURE 2

The 'L' is a distinguishing prefix to avoid confusion between different steel grades. The letter in conjunction with the number designates such parameters as the maximum yield strength and minimum ultimate yield strength. In L-80 the maximum yield strength is shown by point "B" as 95,000 psi which is 15,000 psi higher than the minimum yield stress. The minimum ultimate strength is shown by point "C" as 95,000 psi. Note that there is no maximum ultimate strength specified. N-80, another API grade (see Figure 3), also has a minimum yield stress of 80,000 psi, but is different from L-80 in that the former has a greater maximum yield stress of 110,000 psi (shown by point "B"). This is 30,000 psi higher than the minimum yield stress and twice the tolerance of L-80. The minimum ultimate strength of 100,000 psi is also higher as shown by point "C". Whereas N-80 has no hardness specification, L-80 has a hardness specification of 23 HRC. The tight tolerance on yield strength and hardness allow the L-80 to be more suitable for H2S service than N-80 grade tubulars.

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The following Table 2 lists the API casing grades and flags the ones common to Saudi Aramco.

B - API Specified Maximum Yield

110,000 100,000 80,000

C - API Minimum Ultimate Strength

A - API Specified Minimum Yield

STRAIN (%) 0.5% STRESS-STRAIN DIAGRAM FOR N-80 STEEL FIGURE 3

API CASING GRADES TABLE 2 Designation Casing Grades H-40 J-55 K-55 C-75 L-80 N-80 C-95 P-110

Min yield psi 40,000 55,000 55,000 75,000 80,000 80,000 95,000 110,000

Max yield psi 80,000 80,000 80,000 90,000 95,000 110,000 110,000 140,000

Min ultimate psi 60,000 75,000 95,000 95,000 95,000 100,000 105,000 125,000

Common to Aramco Yes Yes Yes Yes Yes

Casing sizes 24" and larger commonly have grade designations such as X-42, X-56, X60, and B. These are API designations specified under the Line Pipe Specifications. Non-API Casing Grades In addition to API grades, there are many proprietary steel grades which may not conform to the API specifications, but which are used in the industry. These extensively used special grades are often run for various applications requiring such properties as very high tensile strength, high collapse strength, or steels resistant to sulfide stress cracking. This pipe is manufactured to many, but not all of the API specifications with such variations as steel grade, wall thickness, OD, threaded connection, and related upset. As a result of these changes, the ratings of internal yield, collapse, and tension for both the pipe and the connection are non-API. Page 9

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The rating of these proprietary products are generally calculated using API formulas or are consistent with API methods. Also, such parameters as drift diameter, wall thickness tolerance, length range, and weight tolerance are kept the same as, or are consistent with API specifications. Saudi Aramco Non-API Casing Grades Several non-API casing grades are used in Saudi Aramco drilling and workover operations. These proprietary grades have different lettering designations than the familiar API standard. The two most common proprietary grades stocked by Saudi Aramco are SM (Sumotomo), NK (NKK), and NT (Nippon Steel).

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CONNECTIONS Oilfield tubulars may be equipped with plain ends (no threads), have API specified threaded connections or proprietary (non-API) threaded connections. API Casing Connections Oilfield casing conforming to API standards may be obtained with plain ends, but ends are usually threaded and furnished with couplings such as: • • • •

short thread and coupling (STC)* long thread and coupling (LTC) * buttress thread and coupling (BTC) Extreme-Line thread (X-line) for casing * with 8 round threads per inch (8 RD)

With the exception of Extreme-Line, male (or pin) threads are machined on plain-end pipe and later made up with a coupling. A reduced OD (special clearance) coupling is offered on some sizes and weights to allow for additional clearance between pipe and hole. While providing this additional clearance, special clearance couplings often reduce the rating of the connection, usually in tension or internal yield and test pressure. Saudi Aramco API Casing Connections Several API connections are used in Saudi Aramco drilling and workover operations. A brief description of the most popular connections used are as follows: API Short/Long Thread and Coupling The API Short Thread and Coupling (STC) and API Long Thread and Coupling (LTC) are used in pipe sizes of 4-1/2", 7” and 9-5/8”. Figure 4 shows the LTC design.

EXTERNALLY THREADED PIN INTERNALLY THREADED COUPLING (BOX)

ROUND CRESTS AND ROOTS

60 deg

API LONG THREAD & COUPLING (LTC)

FIGURE 4

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The STC design is the same except that the coupling and the threaded pins are shorter. This design is externally threaded on both ends of a non-upset pipe. The single lengths are joined with an internally threaded coupling. The thread profile has rounded threads and roots with a 60° angle between the thread flanks as shown in the figure. The thread density is 8 threads per inch (8 RD) on a 0.0625 inch per inch taper. When the coupling is made up, small voids exist at the roots of each thread. Thread compound must be used to fill these voids in order to obtain a seal. LTC is not made in casing sizes larger than 13-3/8” because of the possibility of joint pull-out that can occur with the heavier weight casing. API Buttress Thread and Coupling The API Buttress Thread and Coupling (BTC) is also a popular thread design used by Saudi Aramco in several casing sizes ranging from 9-5/8", 13-3/8”, and 18-5/8". This is used in conductor and surface casing applications, also where a higher joint strength is required.

Figure 5 shows the BTC design. This design is externally threaded on both ends of a nonupset pipe (as in the STC and EXTERNALLY THREADED PIN LTC). The single lengths are INTERNALLY THREADED COUPLING joined with an internally threaded coupling. The thread profile has flat crests and roots parallel to the taper cone. The thread density is 5 threads per inch on a 0.0625 inch per inch taper for sizes 13-3/8" and smaller, and 0.0833 inch per inch taper for sizes 16" and larger.

FLAT CRESTS AND ROOTS

API BUTTRESS THREAD & COUPLING (BTC)

FIGURE 5 The BTC thread has higher joint and bending strengths compared to LTC (or STC). As a result, this thread is used often in deeper wells where higher hook loads are experienced.

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Thread compound must also be used in order to obtain a seal with BTC. BTC is also run in horizontal wells where doglegs can cause high bending loads on the larger size casings. But BTC leak resistance is lower that that of LTC and STC.

Proprietary Connections Proprietary connections are available which offer premium features not available on API connections. Among the special features for proprietary connections are: • • • • • • • • • •

clearance OD of coupling for slimhole completions metal-to-metal seals for improved high pressure seal integrity high bending strength for deviated holes multiple shoulders for high torque strength a streamlined connection OD for easy running in multiple completions. recess-free bores through the connection ID for improved flow characteristics higher tensile strength for deep holes an integral connection to reduce the number of potential leak paths resilient seal rings for continuous corrosion protection high compressive strength for compressive loading situations

Saudi Aramco Proprietary Connections Several proprietary connections are used in Saudi Aramco drilling and workover operations. A brief description of the most popular connections used are as follows: FLUSH BORE DIAMETER

VAM Connection A proprietary connection which is very popular with Saudi Aramco is the VAM connection and is stocked in 4-1/2" and 7" sizes. This connection has a metal to metal seal for superior leak resistance. An internally threaded coupling with

TORQUE SHOULDER METAL TO METAL SEAL FACE

FLAT CRESTS AND ROOTS

PROPRIETARY VAM CONNECTION

internal shoulders provide positive

FIGURE 6

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make up torque and a non-turbulent bore (see Figure 6). It has become a standard completion tubing for the high pressure Khuff gas wells. Also, due to its superior joint and bending strength, it is used as the completion liner for the horizontal wells. NS-CC Connection The NS-CC (Nippon Steel Connection for Casing) is a proprietary connection used by Saudi Aramco in the Khuff Gas wells (see Figure 7). It is stocked in 7", 9-5/8" and 133/8" sizes. This connection is noteworthy for its gas leak tightness, low hoop stress, high joint strength (equivalent to API buttress thread), high collapse strength and easy stabbing design. Its two step pin nose which incorporate a primary and reserve torque shoulder and metal to metal seal make it a good candidate for the deep, high temperature, high pressure Khuff Gas service. TWO-STEP PIN NOSE DESIGN

Vetco LS, RL-4S, and Dril-quip S60 Connections for Large Casing Sizes The Vetco LS, RL-4S and Dril-quip S-60 connections are proprietary connections used by Saudi Aramco in the 24" casing size.

RESERVE SHOULDER METAL TO METAL SEAL

PRIMARY SHOULDER

API BUTTRESS

The Vetco LS connection is a high strength integral design which accommodates high internal operating pressures, bending PROPRIETARY NS-CC CONNECTION FIGURE 7 moments and tensile loads. The pin/box mating shoulder has a 30 degree taper (see Figure 8). This results in the open end of the box being captured by the tapered shoulder of the pin, and prevents the box from ballooning at the pin/box interface during periods of high internal pressure and large bending moments.

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REVERSE SHOULDER "O" RING SEAL ON PIN

COARSE THREE PITCH THREAD DESIGN

ELEVATOR SHOULDER

PROPRIETARY VETCO LS CONNECTION FOR LARGE CASING SIZES

FIGURE 8

The Vetco RL-4S connection features dual stabbing guides and a high stab angle for easy stabbing. Self locking, four start thread forms allow fast quarter-turn makeup. The Dril-quip S-60 connection features easy stabbing, no cross-threading, fast makeup, low torque and high pressure sealing. Both of these connections save rig time and are used in 24” and larger casing sizes.

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CORROSION

The presence of CO2 and H2S accompanied by water, can cause corrosion of the exposed tubulars. In addition, H2S can cause stress corrosion cracking. Corrosion Mitigation When CO2 or H2S are dissolved in water, they will create an acidic solution. These solutions react with the iron in the pipe causing local pitting which can eventually eat a hole in the pipe. Some of the ways of combating this corrosion are as follows: 1. Plastic Coatings Plastic coating on the pipe which is exposed to the produced fluids is one method of corrosion prevention. There are a variety of coating materials and thicknesses for the different chemical components and temperatures of the produced fluid. The application of a coating to the inside of the pipe can reduce its effective drift diameter. This will make it necessary to coordinate the plastic coating thickness with the proposed through tubing work.

Some disadvantages of internal plastic coating, IPC, is that it is difficult to apply to all exposed surfaces. This is particularly true of coupling recesses and accessories such as packers, seating nipples and safety valves. In order to maintain continuity of the plastic coating's corrosion barrier, some connections provide a Teflon ring on the ID between the pin end and the box recess. It is difficult to ensure that there are no holidays in the coating. Also IPC can be damaged by wireline tools. Saudi Aramco carries a stock of internally plastic coated (IPC) tubulars in the 41/2" and 7" sizes. 2. Fiberglass Lined Pipe Where the corrosion is very high, it is cost effective to have fiberglass lined tubing. Fiberglass lined tubing is constructed by inserting a fiberglass tube of ±0.1 in. wall thickness into the steel tubing. The small annulus between the fiberglass tube and steel tubing is filled with cement. This lining results in a longer lasting corrosion barrier that plastic coating does not provide. In addition, there are no holidays or interruptions in the lining, the fiberglass is continuous the entire length of the joint. Fiberglass is more resistant to wireline damage than plastic coatings. Page 16

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3. High Alloy Carbon, Stainless Steel or Chromium Tubulars Where plastic coating is impractical, corrosion control can be achieved through these alloy steels. This is not a common method since alloy steel tubulars usually cost much more than a conventional steel string.

Saudi Aramco maintains a stock of 4-1/2", 7" and 9-5/8" high alloy Chrome-13 tubulars for use in the gravity water injection wells (see section entitled "Properties of Tubular Materials" for more information about CR-13 casing). 3. Chemical Inhibition An inhibitor may periodically be pumped into a well to form a film on the pipe. This treatment is being performed in Wasia water supply wells. If there is no means to circulate down the inhibitor while producing the well, it will be necessary to shut in the well and pump down the tubing. In a gas lift installation, the inhibitor may be pumped into the gas system. Where wells are completed with concentric strings, the inhibitor can be continuously pumped down one string, with the produced fluid carrying the inhibitor into the other string. Sulfide Stress Cracking A type of corrosion caused by H2S can be a severe condition because it can lead to gross failure of steel equipment. Stress corrosion cracking attacks points subjected to a high tension stress. Once the stress crack is initiated, the tensile stress may increase due to the reduced area, thus leading to accelerated stress cracking. This process continues until the stress increases to the ultimate strength of the steel, at which point failure occurs.

In order to prevent stress corrosion cracking in tubulars due to the presence of H2S, certain design criteria can be applied. 1. Steel Properties One of the principal factors governing the resistance of tubulars to stress corrosion cracking is the physical properties of the steel. Through extensive testing it has been determined that the higher strength carbon steels are more susceptible to sulfide stress cracking.

The API Specification 5CT lists two steel grades, L-80, and C-95 which have a restricted yield strength range of 15,000 psi. This restricted range has the net effect of holding down the maximum strength of the steel while maintaining an

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adequate minimum yield strength. In addition to the narrower yield strength range, these grades have additional chemical and heat treatment controls not required on other API steel grades. These three have been widely used in H2S environments. With experimental work on the effect of the heat treatment methods on resistance to sulfide stress cracking, there has been an increased use of the quenched and tempered L-80 grade. In addition to the API grades, there are proprietary grades used in H2S service. Most of these have a minimum yield strength from 80,000 psi to 90,000 psi, with a controlled yield strength range of 15,000 psi. This is the same range as API restricted yield grades. 2. Temperature Susceptibility Another factor in susceptibility of tubulars to sulfide stress cracking is the temperature of the steel when it is exposed. It has been shown that at elevated temperatures, the higher strength steels are not susceptible to sulfide stress cracking. NACE Specification MR-10-75 refers to the use of API grade, P-110 and proprietary grades to a maximum 140,000 psi yield strength in an H2S environment where the temperature during exposure is not less than 175° F. The use of API grades N-80, C-95 and proprietary grades up to a maximum yield strength of 110,000 psi can be used in temperatures above 150° F. 3. Other Factors Other factors effecting sulfide stress cracking are the level of stress in the steel and the time of exposure. Lower stress levels reduce the chance of sulfide cracking. The steel chemical and mechanical properties, in addition to the time and temperature at exposure and the tensile stress level, determine the susceptibility of the steel to sulfide stress cracking. 4. Design Considerations In deep, high pressure gas wells where both internal pressure and tension would normally require high strength steels, design of casing and tubing strings becomes difficult with the restriction of the minimum yield strength to 90,000-95,000 psi in an H2S environment. Application of restricted yield strength steel grades dictates thicker-wall pipe in order to handle the high tension and internal pressure loads. A well with a high bottom hole temperature can use P-110 and/or X-125 casing and P-105 tubing in the lower section of the hole up to a point where the static

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temperature is no longer high enough. At this crossover temperature, it is then necessary to run the sulfide stress cracking resistant grades to the surface. By using high strength steel on the bottom, the wall thickness can generally be reduced, thus decreasing the total weight of the string. This is particularly important with the upper section of the string requiring lower strength steel, the reduced weight on the bottom sections will further reduce the weight required at the surface.

Saudi Aramco Applications in H2S Service Associated gas and non-associated (Khuff) gas can contain high levels of H2S. The L-80 grade has become a standard specification for several Saudi Aramco oil and gas fields which have high levels of H2S. Saudi Aramco stocks several sizes of L-80 tubing and casing such as 4-1/2", 7" and 9-5/8".

Proprietary (non-API) grades such as C-95VTS, and NT-90HSS are also used in Saudi Aramco high pressure sour Khuff gas applications where a high yield strength is required.

CARE OF OILFIELD TUBULARS With the large expense of tubular products to drill and complete an oil or gas well, it is important that the proper shipping, handling, storage, and running practices be followed to ensure that the investment made in tubulars yields its maximum benefit. Leaky joints are one cause of trouble which can be attributed to many forms of improper care. API Recommended Practice for Care and Use of Casing and Tubing (RP-5C1) lists common causes of trouble for casing and tubing. Of these, over half are related to poor shipping, handling, and running practices.

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CASING PERFORMANCE PROPERTIES Casing must have certain properties in order to achieve its functions in a well. The most important performance properties of casing include its rated values for axial tension, burst pressure, and collapse pressure. Axial tension loading results from the weight of the casing string suspended below the joint of interest. Body yield strength is the tensional force required to cause the pipe body to exceed its elastic limit. Joint strength is the minimal tensional force required to cause joint failure. Burst pressure rating is the calculated minimum internal pressure that will cause the casing to rupture in the absence of external pressure and axial loading. Collapse pressure rating is the minimum external pressure that will cause the casing walls to collapse in the absence of internal pressure and axial loading. Burst The burst loads on the casing must be evaluated to ensure the internal yield resistance of the pipe is not exceeded. The burst load is the force applied by the fluid inside the casing which acts to rupture the pipe in the absence of external pressure. The loads are normally caused by mud hydrostatic pressure inside the casing and perhaps some surface pressure. Fluids on the outside of the casing, called back-up fluids, supply a hydrostatic pressure that helps resist pipe burst. The resulting effective burst pressure is the internal pipe pressure minus the external pressure. Burst conditions are established and the least expensive pipe that will satisfy the burst pressure is tentatively selected. The API burst pressure rating of casing is given by: Y ×t Pi = 1.75 m (2) De

Where Ym is the minimum yield strength of the pipe, t is the wall thickness in inches and De is the outer diameter in inches. Example:

Calculate the burst rating for 7” 23# J-55 casing. Solution: From Table 2, Ym for J-55 casing is 55,000 psi. The ID of 7” 23# casing in ⎛ 7 − 6.366 ⎞ t =⎜ 6.377”, therefore, ⎟ = 0.317 in. 2 ⎝ ⎠ ⎛ 0.317 ⎞ Pi = 1.75 × 55,000 × ⎜ ⎟ = 4360psi ⎝ 7 ⎠

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Collapse The primary collapse loads are supplied by the column of fluid on the outside of the casing which act to collapse the pipe. These fluids are usually the mud and possibly the cement slurry in which the casing was set. Since the column of mud increases with depth, collapse pressure is the highest at the bottom of the hole section and is zero at the surface. The formula to calculate the hydrostatic pressure acting at a particular depth is:

Phyd, psi = (ρm /144) x h

(3)

where ρm is the density of the fluid in pcf, and h is depth in feet. Never allow the hydrostatic pressure to exceed the collapse rating of the casing. The worst case design conditions are when the casing is void of fluid and the external force (collapse load) is the maximum mud weight when the casing was run. In designing for collapse, the casing is assumed empty for surface casing, production casing and partially empty for intermediate casing. Once the casing is cemented and the cement is set the cement acts to help increase the collapse resistance. There are four formulas to calculate the collapse rating of casing (Pc) depending on the ratio of the pipe outer diameter to wall thickness. For: Table 3 Grade D/t Ratio H40 16.44 and less J & K55 14.8 and less C75 13.67 and less N80 13.38 and less C95 12.83 and less P105 12.56 and less P110 12.42 and less

Pc = 2Ym[((D/t)-1)/(D/t)²]

(4)

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For: Table 4 B’

Grade

A’

H40 J & K55 C75 N/L80 C95 P105 P110

2.95 2.99 3.06 3.07 3.125 3.162 3.18

0.0463 0.0541 0.0642 0.0667 0.0745 0.0795 0.082

C

D/t Ratio

755 1205 1805 1955 2405 2700 2855

16.44 to 26.62 14.8 to 24.99 13.67 to 23.09 13.38 to 22.46 12.83 to 21.21 12.56 to 20.66 12.42 to 20.29

Pc = Ym[A’/(D/t)-B’]-C

(5)

For: Table 5 A B

Grade H40 J & K55 C75 N80 C95 P105 P110

2.047 1.99 1.985 1.998 2.047 2.052 2.075

0.03125 0.036 0.0417 0.0434 0.049 0.0515 0.0535

D/t Ratio 26.62 to 42.7 24.99 to 37.2 23.09 to 32.05 22.46 to 31.05 21.21 to 28.25 20.66 to 26.88 20.29 to 26.2

Pc = Ym[A/(D/t)-B]

(6)

And for: Grade H40 J & K55 C75 N80 C95 P105 P110

Table 6 D/t Ratio 42.7 and greater 37.2 and greater 32.05 and greater 31.05 and greater 28.25 and greater 26.88 and greater 26.2 and greater

Pc = 46,950,000 / [(D/t) x ((D/t)-1)²]

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Example: A string of 9-5/8” 53.5# L-80 casing is to be set in 75 pcf mud at a depth of 6000 ft. Calculate the collapse rating for this casing, assume that the casing is empty. Then determine if the casing can safely be set to this depth in order to satisfy a safety factor for collapse of 1.125. Ym = 80,000 psi, ID = 8.535 in. Solution:

Phyd = (75/144) x 6000 ft = 3,125 psi;

Since: De/t = 9.625 / 0.545 = 17.6605 We use Pc = Ym[A’/(D/t)-B’]-C;

For N-80/L-80, A’ = 3.07; B’ = 0.0667; C = 1955

Pc = 80,000 [(3.07/17.6605) - 0.0667] - 1955 = 6616 psi

SF = 6616 / 3125 = 2.12

Allowable collapse = 6616 psi/1.125 = 5880 psi. Therefore since the collapse load at 6000’ (3125 psi) is less than the allowable collapse (5880 psi), then it is safe to run. Tensile Force The tensile load of the pipe is the weight of the casing which acts to pull the pipe apart. The tension is always the greatest at the surface and decreases with depth due to the casing weight below the point of interest. In designing a casing string the upper most joint of the string is considered to have the maximum load on it since it has to carry the total weight of the casing string. Tensile loads are used to select pipe couplings.

Tension loads are computed using the buoyant forces acting on the pipe and the pipe weight. The buoyancy force acts on the bottom joint of the casing and results in a reduction in the hanging weight of the casing. The buoyant forces are defined as the forces acting on submerged equipment due to hydrostatic pressure. The weight of the casing in fluid is given by: ⎛ MW ⎞ Wf = Wa ⎜1 − ⎟ 489 ⎠ ⎝ where, Wf = weight in fluid Wa = weight in air

MW = mud weight in pcf

The pipe body yield strength is calculated by: Page 23

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Ften (lb) = 0.7854 x Ym (De²- Di²)

(8)

where Di is the ID of the casing.

Example:

Calculate the body yield strength of 7 in, 26-lb/ft J-55 casing with long threads & couplings (LT&C). Solution:

The body yield strength is: Ften (lb) = 0.7854 x 55,000 [(7)²- (6.276)²] = 415,200 lb. To calculate the joint strength of a given thread depends on grade, size and weight of the casing and on the effective size of the threads. Formula (9) is for minimum strength of a joint failing for fracture and formula (10) is for minimum strength for a joint failing for thread pullout. The lesser of the two values govern. Fracture strength:

P j = 0.95A jp U p

(9)

Pullout strength:

⎡ 0.74D −0.59 U p ⎤ Yp Pj = 0.95A jp L ⎢ + ⎥ ⎣⎢ 0.5L + 0.14D L + 0.14D ⎦⎥

(10)

where: Pj Ajp D d L Yp Up

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= = = = = = = =

minimum joint strength, lb. cross sectional area of pipe under the last perfect thread at pin, sq. in. 0.7854 [(D-0.1425)2-d2] for eight round threads nominal outside diameter of pipe, in. nominal inside diameter of pipe, in. engaged thread length, in. minimum yield strength of pipe, psi. minimum ultimate strength of pipe, psi.

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Biaxial Effects The combination of stresses due to the weight of the casing and external pressures is referred to as ‘biaxial stresses’. Biaxial stresses can reduce collapse resistance of the casing and must be accounted for in designing deep wells. The collapse resistance, Pcc under tensile loading is given by the following formula: 2 ⎤ WPc ⎡ ⎛ (AYm) ⎞ ⎢ 4⎜ (11) Pcc = ⎟ − 3 − 1⎥ 2AYm ⎢ ⎝ W ⎠ ⎥ ⎣ ⎦

Where Pcc = minimum collapse pressure under axial tension stress (psi); Pc = collapse resistance with no tensile load (psi); W = weight supported by the casing (lb) ; Ym = average yield stress of steel (psi) with zero load, A = cross sectional area. Biaxial loading generates forces within the surfaces of the casing which reduce the casing collapse but increase its burst resistance. This equation can be represented in tabular form, showing the percentage reduction in collapse resistance for a given unit weight carried by the casing, see below. Table 8 Tensile ratio = weight carried yield strength 0 .01 .05 .1 .15 .2 .25 .3 .35 .4 .45 .5 .55 .6 .65 .7 .75 .8 .85 .90 .95

Remaining collapse resistance (%) 100.0 99.5 97.3 94.5 91.8 88.5 85.0 81.3 77.7 76.0 69.5 65.0 60.2 55.8 50.0 44.5 38.5 32.0 25.0 17.8 9.0

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0

To use Table 8, determine the ratio between the weight to be carried by the top joint of the weakest casing and the yield strength of the casing. Then from the table determine the corresponding reduction in collapse strength. Example:

A 13-3/8” 68# K-55 casing string with an average yield strength 1,069,000 lb, weighs 250,000 lb in air and is to be run in a well that contains 75 pcf mud. The ID of the casing is 12.415 in. What is the corrected weight of the casing and what is the collapse rating reduced due to biaxial loading? Solution:

Corrected weight (W) is: 250,000 lb [1 - (75/490)] = 211,735 lb The biaxial effect on the collapse rating of the pipe is: 211,735/1,069,000 = .198 and looking at Table 8 on page 41 for this ratio we can see that the collapse resistance needs to be multiplied by a 0.885 correction factor. So instead of a collapse rating of 1950 psi for this casing, it is actually 1726 psi once the biaxial effect is included. We can also solve this by using equation (11), where W = 211,735 lb; Pc = 1950 psi; and Ym = 55,000 psi, A = πt(OD-t) = 19.435 in2. 2 2 ⎡ ⎤ ⎤ WPc ⎡⎢ ⎛ (AYm) ⎞ 211,735 × 1950 ⎢ ⎛ (19.435 × 55,000) ⎞ ⎥ Pcc = 4⎜ 4⎜⎜ ⎟⎟ − 3 − 1⎥ = 1728 psi ⎟ − 3 −1 = ⎥ 2AYm ⎢ ⎝ W ⎠ 211,735 ⎥ 2 × 19.435 × 55,000 ⎢ ⎝ ⎠ ⎣ ⎦ ⎣ ⎦

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SAFETY FACTORS

Exact values of loading are difficult to predict through out the life of the well. For example, if mud of 75 pcf is on the outside of the casing during the running of the casing, this value cannot be expected to remain constant for the entire life of the well. The mud will become deteriorated with time and will reduce this value to perhaps a saltwater value of 64 pcf. Therefore, calculations of burst values assuming a column of mud at 75 pcf are not realistic throughout the life of the well. If the initial casing design is marginal, then over a period of time in the event of a gas leak the casing may burst. Since casing design is not an exact technique and because of the uncertainties in determining the actual loadings as well as the deterioration of the casing itself due to corrosion and wear, a safety factor is used to allow for such uncertainties in the casing design and to ensure that the rated performance of the casing is always greater than any expected loading. In other words the casing strength is always down rated by a chosen safety factor value. Usual safety factors are: Collapse: Burst:

1.125 1.1

Tension:

1.6

The safety factor is determined by the ratio of the body resistance to the magnitude of the applied pressure.

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TYPES OF CASING CONDUCTOR CASING This string of casing serves several purposes. It prevents erosion around the rigs foundation where the surface sand is weak, it makes the cellar more stable, provides a good start for subsequent drilling and to circulate drilling fluids through to the surface. The conductor casing setting depth is usually based on the amount required to prevent washout of the shallow borehole when drilling to the depth of the surface casing and to support the weight of the surface casing. The conductor also protects inner casings from corrosion. A diverter can be installed on the conductor to divert flow in case of a shallow kick. Normally when the surface sand is stable this string of casing is not necessary. SURFACE CASING This string of casing protects shallow fresh water sands from possible contamination, it prevents cave-in of unconsolidated, weaker, near surface sediments and in the event of a kick, it allows the flow to be contained by the BOP’s. It also supports and protects from corrosion any subsequent casing strings run in the well. INTERMEDIATE CASING Intermediate casing is similar to surface casing in that its function is to permit the final depth objective to be reached safely. This casing string is run to isolate problem zones, i.e. abnormal pressured, lost circulation, sloughing or caving zones between the surface casing depth and the production casing depth. More than one intermediate string can be set if necessary. LINER A liner is a different type of casing profile in that it does not extend from the bottom of the well to the surface but is suspended from the bottom of the next largest casing string. Normally it only extends a few hundred feet above the last casing shoe depth. A liner can be used for drilling purposes to isolate problem zones or production or both. The principal advantage of the liner is its lower cost. PRODUCTION CASING This string of casing is run across the production zone. Its purpose is to provide isolation between zones and in case a tubing leak occurs, it will contain the production fluid until remedial work can be performed to repair the leak. A typical casing configuration of an Arab-D well is shown in Fig (9). Figure 9

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GEOLOGIC HORIZONS

26" CONDUCTOR @ 100' 415 PNU 500

1000

725

KHOBAR

900

RUS

18-5/8" SURFACE CSG @ 950'' (50' INTO RUS)

1125 UER

1500

2000

2500

1950

ARUMA

2650

L.A.S.

3900

SHUAIBA BIYADH

13-3/8" INTERMEDIATE CSG @ 2700' (50' INTO LAS)

3000

3500

4000

4100

7" LINER HANGER @ 4200' 9-5/8" INTERMEDIATE CSG @ 4400' (300' INTO BIYADH)

4500

5000

5500

5760

SULAIY CEMENT

6000

6500

7000

7500

7300

ARAB-D

7575

TD

7" LINER @ 7295' (5' ABOVE ARAB-D) 6" OPEN HOLE @ TD 7575'

Typical Arab-D Casing Program

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SELECTION OF CASING SETTING DEPTHS The selection of the number of casing strings and their respective setting depths generally is based on a consideration of the pore pressure and fracture pressure gradients of the formations to be penetrated. The selection of casing depths is illustrated in Fig. (10). The pore pressure can be estimated from offset wells. For wildcat wells, the pore pressure is estimated by geophysicists using seismic data. The fracture gradient is defined as the bottom hole pressure required to keep the fracture open divided by the reservoir depth. The fracture gradient can be estimated from the Eaton equation, F=

S P × α + (1 − α ) × . . . . . . . . . . . . . . . . . . . . . . . . . .(12) D D

where: F = fracture gradient, psi/ft D = reservoir depth, ft S = overburden stress, psi P = bottom hole reservoir static pressure, psi α=

V 1− V

V = Poisson’s ratio, dimensionless. If the overburden stress gradient reduces to,

F = α + (1 − α ) ×

P D

S is assumed to be equal to 1.0 psi/ft, Eq. (12) D

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .(13)

The value of α varies between 0.3 and 0.5. The pore pressure and fracture pressure gradients can be expressed in terms of equivalent mud density in pcf, by using Eq. 14. Equivalent Density (pcf) = Pressure Gradient, psi/ft x 144 . . . . . .. . . . . . . . . . . . . .(14)

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The static reservoir pressure (pore pressure) at 8000 ft is 3700 psi. What is the equivalent mud density in lb/ft3 (pcf)? Solution: The pressure gradient is pressure divided by depth, or,

Pressure gradient =

3700 = 0.46 psi/ft; and Equivalent Density = 0.46 x 144 = 66 pcf 8000

The pore pressure and fracture pressure gradients expressed in equivalent mud density are plotted versus depth as shown in Fig. (10). A line representing the planned mud density is also plotted. The planned mud density is chosen to provide trip safety margins above the anticipated formation pore pressure, to allow for reductions in effective mud weight created by upward drill pipe movement (swabbing) during tripping operations. The safety margin allows for errors made in estimating the pore pressure. A commonly used margin of error is 4 pcf or one that will provide 200-500 psi of excess (overbalance) mud hydrostatic bottom hole pressure over the formation pore pressure. Similarly, a 4 pcf kick margin is subtracted from the true fracture gradient line to obtain a design fracture gradient line. If no kick margin is provided, it is impossible to take a kick at the casing setting depth without causing a fracture and a possible underground blowout. To reach the desired depth objective, the effective drilling fluid density shown at Point a is chosen to prevent the flow of formation fluid into the wellbore. To carry this drilling fluid density, without exceeding the fracture gradient of the weakest formation exposed within the wellbore, the protective intermediate casing string must be extended to at least a depth at Point b. This is, where the fracture gradient is equal to the mud density to drill to Point a. Similarly, to drill to Point b and set intermediate casing, the drilling fluid density shown at Point c will be needed and will require surface casing to be set at least to the depth at Point d. If possible, a kick margin is subtracted from the true fracturegradient line to obtain a design fracture-gradient line. If no kick margin is provided, it is impossible to take a kick at the casing-setting depth without causing a fracture and a possible underground blowout.

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Equivalent Mud Density

Conductor Fracture Gradient

D

6,000

Surface

Normal Pressure 9,000

Depth 12,000

Fracture Gradient Kick Margin

Pore Pressure Gradient

C

B

Intermediate

Mud Density (Pore Press + Trip Margin) 15,000

Depth Objective

A

Production

Sample relationship among casing setting depths, formation pore pressures gradient, and fracture gradient Figure 10

Other factors such as the protection of fresh ground water reservoirs, the presence of lost circulation zones, pressure depleted zones that tend to cause pipe sticking problems, and governmental regulations can also affect casing setting depths. Experience in some areas might determine where the best casing seat might be in order to get a good cement job. The conductor casing setting depth is based upon the amount require to prevent shallow washout of the shallow borehole when drilling to the depth the surface casing is set and to support the weight of the surface casing. The conductor casing must be able to sustain pressures that might be encountered during diverting operations without washing out around the outside of the conductor. The conductor is often driven into the ground with a big hammer, the resistance of the ground determines how much conductor is set. The casing driving operation is stopped when the number of blows per foot exceeds some specified number.

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Example:

A well is to be drilled to a depth of 15,000’. Determine the number of casing strings needed to reach this depth objective safely, and select the casing setting depth of each string. Pore pressure and fracture gradient data are given below. Allow a 4pcf trip margin, and a 4 pcf kick margin when making the casing seat selections. The minimum length of surface casing required to protect the freshwater aquifers is 2,000 ft. Approximately 180 ft. of casing is generally required to prevent washout on the outside of the conductor. Depth, ft 1,000 2,000 4,000 6,000 8,000 9,000 10,000 11,000 12,000 13,000 14,000 15,000

Pore Pressure, psi 457 914 1,828 2,742 3,656 4,114 4,800 6,643 9,235 10,883 11,930 12,950

Fracture Gradient, psi/ft 0.62 0.66 0.73 0.79 0.83 0.85 0.87 0.9 0.935 0.95 0.957 0.967

Solution:

1.

Calculate the equivalent mud density for the pore pressure gradient: Equivalent mud density @ 1000’ =

2.

PorePressure 457 × 144 = × 144 = 65.8 pcf Depth 1000

Calculate equivalent mud density for fracture gradient: Equivalent mud density for fracture gradient:

= Fracture Gradient × 144 = 0.62 × 144 = 89.3 pcf

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Equivalent mud densities for the remaining depths are tabulated below. The planned mud density is found by adding 4 pcf to the pore pressure equivalent mud density. Similarly, the design fracture equivalent mud density is obtained by subtracting 4 pcf from the fracture gradient equivalent mud density. Depth (ft) 1000 2000 4000 6000 8000 9000 10000 11000 12000 13000 14000 15000

Pore Pressure Gradient (pcf) 65.8 65.8 65.8 65.8 65.8 65.8 69.12 86.9 110.8 120.5 122.7 124.3

Equivalent Mud Densities, pcf Planned Mud Fracture Gradient Density (pcf) (pcf) 69.8 89.28 69.8 95.04 69.8 105.12 69.8 113.76 69.8 119.5 69.8 122.4 73.12 125.2 90.9 129.6 114.8 134.6 124.5 136.8 126.7 137.8 128.3 139.2

Design Fracture Gradient (pcf) 85.28 91.04 101.12 109.76 115.5 118.4 121.2 125.6 130.6 132.8 133.8 135.2

The pore-pressure equivalent mud density, the planned mud density, the fracture gradient equivalent density and the fracture design equivalent mud density are plotted in Fig (11). From the graph, it can be seen to drill to a depth of 15,000 ft, a 128.3 pcf mud will be required (Point A). This, in turn, requires intermediate casing to be set at 11,700 ft (Point B) to prevent fracture of the formation above 11,700 ft. Similarly, to drill safely to a depth of 11,700 ft to set intermediate casing, a mud density of 110 pcf is required (Point C). This requires surface casing to be set at 6,600 ft (Point D). Because the formation at 6,600 ft is normally pressured, the usual conductor casing depth of 180 ft is appropriate. Surface casing is set at 2000 ft to protect the freshwater aquifers.

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80

100

140

120

0

2,000

4,000

Design Fracture Gradient

6,000 D

Depth, ft 8,000

Fracture Gradient

Mud Density

10,000 Pore Pressure C

12,000

B

14,000

Equivalent Mud Density, pcf Figure 11

A

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SELECTION OF CASING AND BIT SIZES The design of the casing sizes is performed from the bottom to the top, starting with the production tubing. The tubing size is designed to maintain a specific wellhead flowing pressure or allow the well to produce at a specified flowrate. The size of the production casing or liner is based on the size of the tubing. The production casing should have an inside diameter such that there is adequate radial clearance between the tubing and casing to allow for fishing the tubing during workover operations. To enable the production casing to be placed in the well, the bit size used to drill the last interval of the well must be 0.375”-0.5” or preferably 0.75” larger than the OD of the production casing, see Table (9). The selected bit size should provide sufficient clearance between the borehole and the casing to allow for mud cake on the borehole and for installing centralizers and scratchers. Sufficient clearance is also necessary to prevent premature dehydration of the cement and the formation of cement bridges during cementing. The bit used to drill the hole for the production casing must fit inside the casing string above, see Table (10). This, in turn, determines the minimum size of the second deepest casing string. With similar considerations, the bit size and casing size of successively shallower well segments are selected. Table (9) provides commonly used bit sizes for drilling a hole which various API casing strings generally can be placed safely without getting casing stuck. In Table (10) are casing IDs and drift diameters for various standard casing sizes and wall thicknesses. The pipe manufacturers assures that a bit smaller than the drift diameter will pass through every joint of casing bought. In most instances, bits larger than the drift diameter but smaller than the ID will also pass, but this is not good practice. Commonly Used Bit Sizes for Running Casing Table 9 Casing Size (OD in) 4 1/2 5 5 1/2 6 5/8 7 8 5/8 9 5/8 10 3/4 13 3/8 18 5/8 20 24

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Coupling Size (OD in) 5.0 5.563 6.05 7.39 7.656 9.625 11.75 11.75 14.375 19.625 21.0 25.25 &, 25.5

Common Bit Sizes (in) 6, 6 1/8, 6 1/4 6 1/2, 6 3/4 7 7/8, 8 3/8 7 7/8, 8 3/8, 8 1/2 8 5/8, 8 3/4, 9 1/2 11, 12 1/4 12 1/4 15 17 1/2, 17 22 24,26 28

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Casing Size OD inches 4 1/2

Weight Per Ft, lb/ft

Internal Diameter, in

Drift Diameter, in

Common Bit Sizes inches

11.6

4.0

3.875

3 7/8

13.5

3.92

3.795

3 3/4

11.5

4.56

4.435

4 1/4

13

4.494

4.369

15

4.408

4.283

17

4.892

4.764

4 3/4

20

4.778

4.653

4 5/8

23

4.67

4.545

4 1/4

17

6.135

6.010

6

20

6.049

5.924

5 5/8

24

5.921

5.796

28

5.791

5.666

20

6.456

6.331

23

6.366

6.241

26

6.276

6.151

6 1/8

29

6.184

6.059

6

32

6.094

5.969

5 7/8

35

6.006

5.879

36

8.921

8.765

8 3/4, 8 1/2

40

8.835

8.679

8 1/2

43.5

8.755

8.599

47

8.681

8.525

53.5

8.535

8.379

58.4

8.435

8.375 (SD)

61

12.515

12.359

68

12.414

12.259

72

12.347

12.191

86

12.125

12.000 (SD)

12

18-5/8

87.5

17.755

17.567

17-1/2

20

94

19.124

18.936

17-1/2

5

5 1/2

6 5/8

7

9 5/8

13 3/8

6 1/4

8 3/8

12 1/4

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The most commonly used bit sizes are highlighted in Tables 9 and 10. Selection of casing sizes that permit the use of commonly used bits is advantageous because the bit manufacturers make readily available a much larger variety of bit types and features in these common sizes. Example: Using the data in the previous example, select casing sizes (ODs) for each casing string. A 4-1/2” tubing size is required to produce the well at optimum flow rate. Solution: A 5-7/8” overshot fishing tool is required to catch the coupling on 4-1/2” tubing. The overshot must be able to pass through the drift diameter of the production casing. Therefore from Table (10), the 7” OD casing has drift diameters larger than 5-7/8”. So a 7” OD production string is desired. From Table (9), a 8-1/2” bit is required to drill the hole for 7” casing. From Table (10), the 9-5/8” casing is the smallest OD casing that has a drift diameter larger than 8-1/2”. Therefore the size of the intermediate casing string at 11,800’ is 9-5/8”. From Table (10), a 12-1/4” bit will pass through the drift diameter of the 13-3/8” casing. A 17-1/2” bit is needed to drill the hole for the 13-3/8” casing. Finally, Table (10) shows that a 17-1/2” bit will pass through 18-5/8” casing to be set at 2000’.

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SELECTION OF WEIGHT, GRADE, AND COUPLINGS Once length and size of each casing string are established, the weight, grade and couplings used in each string can be determined. In general, each casing string is designed to withstand the most severe loading conditions anticipated during casing placement and the life of the well. The loading conditions that are always considered are burst, collapse, and tension. When appropriate, other loading conditions such as bending or buckling must also be considered. Because the loading conditions in a well tend to vary with depth, it is often possible to obtain a less expensive casing design with several different weights, grades, and couplings in a single casing string. It is often impossible to predict the various loading conditions that a casing string will be subjected to during the life of a well. The assumed design load must be severe enough that there is a very low probability of a more severe situation actually occurring and causing casing failure. When appropriate, the effects of casing wear and corrosion should be included in the design criteria. These effects tend to reduce the casing thickness and greatly increase the stresses where they occur. The casing design criteria used by various drilling companies differ significantly and are too numerous to include in this text. Instead, design criteria that are representative of current drilling engineering practice are presented. To achieve a minimum casing design, the most economical casing and coupling that will meet the design loading conditions must be used for all depths. Because casing prices change so frequently, a detailed list of prices in this text is not practical. In general, minimum cost is achieved when casing with the minimum weight per foot in the minimum grade that will meet the design criteria is selected. For this illustration, only API casing and couplings will be considered in the example applications. It will be assumed that the cost per foot increases with the burst strength and that the cost per connector increases with increasing joint strength. As stated before, casing strings serve several functions and therefore drilling conditions for surface casing are different from that for intermediate casing or liners. Thus each type of casing string will have different design criteria. General design criteria will be presented for surface casing, intermediate casing and production casing.

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Surface Casing Design loading conditions for surface casing are illustrated in Fig 10 for burst, collapse, and tension considerations. The high internal pressure loading condition used for the burst design is based on a well control condition assumed to occur while circulating out a large kick. The high external pressure loading condition used for the collapse design is based on a severe lost circulation problem. The high axial tension loading condition is based on assumption of stuck casing while the casing is run in the hole before cementing operations. COLLAPSE

BURST Normal Press. GAS

TENSION Empty Mud

Lost Circ. Z

Mud

Gas Kick GAS Lost Circulation

Fig. 10 Drilling casing design loads for burst, collapse, and tension

The burst design should insure that formation fracture pressure at the casing shoe will be exceeded before the casing burst pressure is reached. Therefore, this design uses formation fracture as a safety pressure release mechanism to assure that casing rupture will not occur at the surface and endanger lives. The design pressure at the casing seat is equal to the fracture pressure plus a safety margin to allow for an injection pressure that is slightly higher than the fracture pressure. If the fracture gradient is not known, a gradient of 1.0 psi/ ft may be safely assumed. The pressure inside the casing is calculated assuming that all of the drilling fluid in the casing is lost to the fractured formation, leaving only formation gas in the casing. The pressure at the surface is the bottom hole fracture pressure plus a safety margin, less the hydrostatic pressure of the gas column. If gas gradients are not known, it is practical to assume a minimum gas gradient of 0.10 psi/ft for pressures originally shallower than 10,000 ft and 0.15 psi/ft for pressure sources deeper than 10,000 ft. If the formations below the surface casing do not have any gas, then gradients of the formation fluids (oil or water) should be used. The external pressure, or back-up pressure outside the casing that helps resist burst, is assumed to be equal to the normal formation pore pressure. The beneficial effect of cement or higher density mud outside the casing is ignored because of the possibility of both a locally poor cement bond and mud degradation that occur over time. A safety factor of 1.1 is used to provide an

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additional safety margin during transportation and handling of the pipe. The burst load at the casing seat is the fracture pressure plus a safety margin minus formation porepressure (back-up pressure). The burst load at the surface is the surface pressure inside the casing. The burst load line is defined by two points; burst load at the casing seat and the burst load at the surface. Connecting the two points gives the burst load line in the casing from top to bottom. Multiplying the burst loads at the two points by a safety factor determines the burst design line. The collapse design is based on the most severe lost-circulation problem that is felt to be possible or on the most severe collapse loading anticipated when the casing is run. For both cases, the maximum possible external pressure that tends to cause casing collapse results from the fluid that is in the hole when the casing is placed and cemented. The beneficial effect of the cement and of possible mud degradation is ignored, but the detrimental effect of axial tension on collapse-pressure rating is considered. The collapse rating should be de-rated above the neutral point using Eq. (11). Below the neutral point the casing is in compression and adjustment of the collapse rating is not required. The depth of the neutral point of a casing string in mud can be calculated by the following formula: W ⎞ ⎛ Dn = Dt ⎜1 − ⎟ 489 ⎠ ⎝

where,

(14)

Dn = depth to neutral point, ft Dt = setting depth of casing string, ft W = mud weight, pcf

When correcting the collapse-pressure rating of the casing, it is recommended that the axial tension be computed as the hanging weight of the casing for the hydrostatic pressures present when the maximum collapse load is encountered plus any additional tension put in the pipe during and after landing. The beneficial effect of pressure inside casing can also be taken into account by the consideration of a maximum possible depression of the mud level inside the casing. A safety factor generally is applied to the design-loading condition to provide an additional safety margin. The minimum fluid level in the casing while it is placed in the well depends on field practices. The casing usually is filled with mud after each joint of casing is made up and run in the hole, and an internal casing pressure that is roughly equivalent to the external casing pressure is maintained. However, in some cases the casing is floated in or run in at least partially empty to reduce the maximum hook load before reaching bottom. If this practice is anticipated, the maximum depth of the mud level in the casing must be used in collapse calculations.

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Tension design requires a consideration of axial stress present when the casing is run, during cementing operations, when the casing is loaded in the slips, and during subsequent drilling and production operations. The design load is usually based on conditions that occur when the casing is run. It is assumed that the casing becomes stuck near the bottom of the hole and that a minimum amount of pull, in excess of the casing weight in mud, is required to pull the casing free. A minimum safety factor criterion is applied such that the design load will be dictated by the maximum load resulting from the use of either the safety factor or the overpull force whichever is greater. The minimum overpull force tends to control the design in the upper portion of the casing string, and the minimum safety factor tends to control the lower part of the casing string. Once the casing design is completed, maximum axial stress anticipated during cementing, casing landing, and subsequent drilling operations should also be checked to ensure that the design load is never exceeded.

When the selection of casing weight and grade in a combination string is determined by collapse, a simultaneous design for collapse and tension is best. The greatest depth at which the next most economical casing can be used depends on its corrected collapsepressure rating, which in turn depends on the axial tension at that depth. Therefore, the corrected collapse-pressure rating cannot be computed until the axial tension is calculated. It takes an iterative procedure, in which the depth of the bottom of the next most economical casing section is first selected on the basis of uncorrected table value of collapse resistance, to be used. The axial tension at this point is then calculated, and the collapse resistance is then corrected. This allows the depth of the bottom of the next casing section to be updated for a second iteration. Several iterations may be required to arrive at a solution.

Intermediate Casing Intermediate casing is similar to surface casing in that its function is to permit the final depth objective of the well to be reached safely. When possible, the general procedure outlined for surface casing is also used for intermediate casing strings. However, in some cases, the burst-design requirements in Fig (10) are extremely expensive to meet, especially when the resulting high working pressure is in excess of the working pressure of the surface BOP stacks and choke manifolds for the available rigs. In this case, the operator may accept a slightly larger risk of loosing the well and select a less severe design load in which the burst limitation is equivalent to the BOP stack rating. The design load remains based on an underground blowout situation assumed to occur while a gas kick is circulated out. However, the acceptable mud loss from the casing is limited to the

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maximum amount that will cause the working pressure of the surface BOP stack and choke manifold to be reached. If the existing surface equipment is to be retained, it is pointless to design the casing to have a higher working pressure than the surface equipment. When the surface burst-pressure load is based on the working pressure of the surface equipment, Pmax, internal pressure at intermediate depths should be determined, as shown in Fig 11. BURST Pmax Normal Press.

Dm

MUD

Dlc

Fracture

Gas GAS

Fig. 11 Modified burst design load for intermediate casing

It is assumed that the upper portion of the casing is filled with mud and the lower portion of the casing is filled with gas. The depth of the mud/gas interface, Dm, is determined with the following relationship. Dm =

144(Pi − Pmax ) ρ g × D lc − ρm − ρg ρm − ρg

(15)

where Pi , in psi, is the injection pressure opposite the fractured zone, ρm and ρg are the densities of the mud and gas in pcf, and Dlc is the depth of the fractured zone in ft. The density of the drilling mud is determined to be the maximum density anticipated while drilling to the depth of the next full-length casing string. This permits the calculation of the maximum intermediate pressures between the surface and the casing seat. The depth of the fractured zone is determined from the fracture gradient vs. depth plot to be the depth of the weakest exposed formation. The injection pressure is equal to the fracture pressure plus an assumed safety margin to account for a possible pressure drop within the hydraulic fracture.

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Example:

A gas exploration well has 13-3/8” casing set at 6200 ft. Design a 9-5/8” casing string to be set at 10,400 ft in 73 pcf mud that will be subjected, in the event of a kick, to a formation pressure gradient of 0.57 psi/ft from the next hole drilled to a TD of 13,900 ft. The 9-5/8” casing in stock at the Aramco pipe yard is: Grade

Weight (lb/ft)

C-75 L-80 C-95

43.5 47 53.5

Collapse Rating

Burst Rating

Tensile-1000 lb

3750 4750 8960

5930 6870 9410

776 905 1220

LT&C BUTT

1016 1161 1458

Collapse

At the surface collapse pressure = 0 At the casing shoe collapse pressure = (73 pcf x 10,400 ft)/144 = 5272 psi It is obvious that the 53.5# will have to be run on bottom. The question is how much 53.5# do we need to run before we can switch over to the less expensive lighter weight casing. Taking the collapse figure for the 47# we can calculate the deepest point to which this weight casing can be run and still satisfy the collapse requirements with the 1.125 SF. (4750 psi/1.125) ÷ (73 pcf/144 in2/ft2) = 8,330 ft Doing the same for the 43.5# casing we get: (3750psi/1.125)÷(73pcf/144in2/ft2)=6,575 ft. Therefore designing the casing string for collapse gives us: 10,400 ft - 8,330 ft 8,330 ft - 6,575 ft 6,575 ft - surface

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53.5# C-95 47# L-80 43.5# C-75

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Burst:

The 9-5/8 in casing will be subjected, in the event of a kick, to a formation pressure of: 0.57 psi/ft x 13,900 ft = 7,923 psi The burst pressure at the shoe is: internal pressure - external pressure or: Burst at shoe = [Pf - (TD - CSD) x G] - [CSD x (Mud Gradient)] Where: CSD is casing setting depth = 10,400 ft, and Mud Gradient is 73 pcf/144 in2/ft2 = 0.507 psi/ft A gas kick is expected for this well, since G is 0.1 psi/ft, we get: Burst at shoe = [7923 psi - ((13,900 ft-10,400 ft) x 0.1 psi/ft)]-(10,400 ft x 0.507 psi/ft)) = 2300 psi

Burst at the surface is: Pf - (TD x G) or: 7923 psi - (13,900 ft x 0.1 psi/ft) = 6533 psi x 1.1 S.F. = 7186 psi

By comparing 7186 psi with what the burst rating for 43.5# C-75 of 5930 psi, we can see that some heavier weight 47# L-80 and 53.5# C-95 casing is required for the top of the string instead of the 43.5#. We can calculate the amount required by each with: (Burst at surface - casing burst rating)/mud gradient = (7186 - 5930) / (73/144) = 2,478’

and since the burst rating for 47# L-80 is 6870 psi which is less than 7186 psi we then calculate how much 53.5# C-95 is required at the surface in order to satisfy the burst requirement. (7186 - 6870) ÷ (73/144) = 623 ft

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Therefore, from the previous collapse design we adjust the design for burst considerations and we can see that 47# and 53.5# casing is needed at the top of the string. We now have a casing design, in the order going in the hole, of: 10,400 ft - 8,330 ft 8,330 ft - 6,575 ft 6,575 ft - 2,478 ft 2,478 ft - 623 ft 623 ft - surface

53.5# C-95 47# L-80 43.5# C-75 47# L-80 53.5# C-95

Next we need to check the tensile strength of the design to ensure that this design will pass the tensile design criteria. Tension

The suitability of the selected design will be investigated by considering the total tensile load resulting from the buoyant weight of the string. The buoyancy factor for 73 pcf mud is: Fb = 1 - (73/490) = 0.849 Starting from the bottom, the weight carried by each section is as follows: Depth

Weight x 1000 lb

10,400 ft - 8,330 ft 8,330 ft - 6,575 ft 6,575 ft - 2,478 ft 2,478 ft - 623 ft 623 ft - surface

97.3 kips 82.5 kips 178.2 kips 87.2 kips 33.3 kips

Cum Weight in Air 97.3 kips 97.3 + 82.5 = 179.8 kips 179.8 + 178.2 = 358 kips 358 + 87.2 = 445.2 kips 445.2 + 33.3 = 478.5 kips

Weight in Mud 82.6 kips 152.6 kips 303.9 kips 378 kips 406.7 kips

To ensure that the selected design meets the safety factor for tension of 1.6 we divide the cumulative weight of each pipe section into the tensile rating of each grade. We then come up with a design factor: 53.5# C-95 47# L-80 43.5# C-75 47# L-80 53.5# C-95

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1220 kips / 82.6 kips = 14.8 905 kips / 152.6 kips = 5.9 776 kips / 303.9 kips = 2.5 905 kips / 378 kips = 2.4 1220 kips / 406.7 kips = 3.0

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From these results we can see that the design exceeds the tensile safety factor of 1.6. Biaxial Effects

Finally we need to check the weakest grade for biaxial collapse correction. Grade C-75 43.5# is the weakest grade, carrying a buoyant load of: 303,900 lb. By dividing this load by the yield strength of 942,000 lb we can see: 303.9 kips / 942 kips = 0.32 Looking at Table 8 on page 25 we see that the collapse reduces to 79% of its original value of 3750, or 2963 psi. Therefore, rechecking the collapse safety factor for that casing grade at that depth of 2478 ft we get: 2963 psi / [2478 ft x (73 pcf/144 in2/ft2 )] = 2.3, then since the safety factor is still greater than 1.125, the biaxial effect on collapse did not change the casing design.

Intermediate Casing With a Liner The burst design-load criteria for intermediate casing on which a drilling liner will be supported later must be based on the fracture gradient below the liner. The burst design considers the intermediate casing and liner as a unit. All other design criteria for the intermediate casing are identical to those previously presented.

Production Casing Example burst and collapse design loading conditions for production casing are illustrated in Fig. 12. The example burst-design loading condition assumes that a producing well has an initial shut-in BHP equal to the formation pore pressure and a gaseous produced fluid in the well. The production casing must be designed so that it will not fail if the tubing fails. A tubing leak is assumed to be possible at any depth. It generally is also assumed that the density of the completion fluid in the casing above the packer is equal to the density of the mud left outside the casing. If a tubing leak occurs near the surface, the effect of the hydrostatic pressure of the completion fluid in the casing would negate the effect of the external mud pressure on the casing. Mud degradation outside the casing is neglected because the formation pore pressure of any exposed formation would nearly equal the mud hydrostatic pressure.

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The collapse-design load shown in Fig. 12 is based on conditions late in the life of the reservoir, when reservoir pressure has been depleted to a very low (negligible) abandonment pressure. A leak in the tubing or packer could cause the loss of the completion fluid, so the low internal pressure is not restricted to just the portion of the casing below the packer. Thus, for design purposes, the entire casing is considered empty. As before, the fluid density outside the casing is assumed to be that of the mud in the well when the casing was run, and the beneficial effect of the cement is ignored. BURST GAS IN TUBING

COLLAPSE PRODUCTION CASING

COMPLETION FLUID

TUBING AT NEGLIGIBLE PRESSURE

FORMATION PRESSURE

MUD DENSITY CASING WAS RUN LEAK IN TUBING OR PACKER CAUSES LOSS OF COMPL FLUID DEPLETED FORMATION PRESSURE

Fig. 12 Production casing design loads for burst and collapse

In the absence of any unusual conditions, the tension design load criteria for production casing are the same as for surface and intermediate casing. When unusual conditions are present, maximum stresses associated with these conditions must be checked to determine whether they exceed the design load in any portion of the string.

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GRAPHICAL METHOD, COLLAPSE AND BURST DESIGN Graphical pipe selection is the most widely used method of picking proper weights, grades, and tension. Since the collapse and burst loads vary linearly with depth, a plot may be made using the previous example. The calculated values at the surface and setting depth for the collapse and burst pressures. A collapse design line is drawn on a graph of depth versus pressure by using the hydrostatic pressure of 73 pcf mud at 10,400 ft. of 5,272 psi and zero hydrostatic pressure at the surface. The appropriate design factor of 1.125 is applied to the hydrostatic pressure and a line is then drawn (see below). Similarly, the maximum burst load line is drawn on the same graph by connecting the burst load points of 2,300 psi at 10,400 ft and 6,533 psi at the surface. The burst design line is established by multiplying 2,300 and 6,533 psi by the burst design factor of 1.1 or 2,530 psi at 10,400 ft and 7,186 psi at the surface and drawing a line between these two points.

0

2000

4000

6000 1.1.25 S.F. Collapse

8000

10,000 Setting Depth

5931

12,000 0

1

2

3

4

5

6

7

8

Pressure, thousand psi

6533 psi

7186 psi

0

2000

4000 1.1 S.F. Burst Design Line

6000

8000

10,000

Setting Depth 2300 psi 2530 psi

12,000 0

1

2

3

4

5

6

7


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The first section of pipe is selected based on the collapse requirement at the setting depth. In this example 53.5# C-95 has a collapse rating of 8960 psi which is off the chart. The collapse rating of the next weaker section is plotted on the appropriate collapse design line and the changeover depth read at the intersection on the graph. A vertical line for the first section is drawn from the casing setting depth to the changeover depth and a horizontal line is drawn from the intersection of the second collapse rating plotted on the design line to the 6533 psi 7186 psi 53.5# C-95 0 collapse rating of the first section. Subsequent segments 2000 47# L-80 are similarly 1.1.25 S.F. Collapse Line determined. 4000 Concurrently burst 43.5# C-75 1.1 S.F. ratings are plotted and Burst Line vertical and horizontal 6000 lines are drawn. Burst Design Line 47# L-80

Above the cement top and when the casing is in tension, the collapse ratings are reduced by the effect of tension on collapse.

8000 Collapse Design Line 53.5# C-95

10,000

Setting Depth 2300 psi

2530 psi

5931 psi

12,000 0

1

2

3

4

5

6

7


At changeover depths above the cement top, the axial stress is calculated. Where the pipe is in tension, a percent of rated collapse is read from Table 8 based on the axial tension. Using the percent of rated collapse multiplied by the changeover depth adjusts the depth to the correct depth. The collapse design factor at the bottom of the weaker section then is calculated to determine if the collapse design requirements are sufficient. If the depth is not correct, the design factor calculated times the depth used will adjust the changeover point to the correct depth. By repetition the correct depth will finally be selected. If the pipe is not in tension, plot the collapse rating of the next weaker section in collapse on the design line and continue the design as before.

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As the design continues upward from the bottom a depth will be encountered where collapse no longer controls the design. Above this depth the design will be controlled by burst or tension. If burst controls the design, the burst ratings of the casing are plotted on the burst design line and the burst loads are read from the burst load line at the corresponding depth. Changeover depths are read directly from the graph. If tension is controlling the design, the changeover depth is calculated directly. The changeover depth is calculated by using the tension rating divided by 1.6 and subtracting the buoyed weight of the pipe below; from this remainder divide by the buoyed weight per foot of the pipe used to determine the footage of pipe to be used.

Example: Graphically design a 13-3/8” intermediate casing string to a depth of 6,250 ft. The mud weight is 67 pcf. For burst considerations, use an injection pressure gradient that is equivalent to a mud density of 2.5 pcf greater than the fracture gradient of 104 pcf, and a safety factor of 1.1. Assume any kick will be composed of gas with a 0.1 psi/ft gradient. The normal formation pore pressure for the area is 0.46 psi/ft. For collapse considerations, assume that a normal pressure, lost circulation zone could be encountered as deep as the next casing seat, that no permeable zones are present above the lost circulation zone, and use a safety factor of 1.125. Also assume that the axial tension results only from the hanging weight of the casing under prevailing borehole conditions. The next hole section will be drilled to a depth of 10,000 ft with 73 pcf mud. For tension considerations, use a minimum over-pull of 100,000 lb. or a safety factor of 1.6, whichever is greater. The 13-3/8” casing available is listed below. Grade

K-55 LT&C K-55 BT&C N-80 BT&C N-80 BT&C

Weight (lb./ft)

54.5 68 72 85

Collapse (psi)

1,130 1,950 2,670 3,870

Burst (psi)

2,730 3,450 5,380 6,360

Yield Strength (kips)

853 1,069 1,661 1,951

Solution: Page 51

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Burst: The fracture gradient at 6,250 ft. is equivalent to 104 pcf. For an injectionpressure gradient that is 2.5 pcf greater than the fracture pressure:

Pinj =

6,250(104 + 2.5) = 4,622 psig 144

Since the formation gas gradient is 0.1 psi/ft, the surface casing pressure for the design considerations is, 4,622 − (0.1 × 6,250 ) = 3,997 psig The external pressure at the surface is zero. For a normal pore pressure of 0.46 psi/ft, the external pressure at the casing seat is, 0.46 (6,250 ) = 2,875 psig

The pressure differential that tends to burst the casing is: 3,997 psig at the surface and 1,747 psig (4,622-2,875) at the casing seat. Multiplying each of these pressures by a safety factor of (1.1) yields a burst design load of 1.1 (3,997) = 4,397 psig at the surface, and 1.1 (1,747) = 1,921 psig at the casing seat. Graphically draw the burst line from 4,397 psig at the surface to 1,921 psig at 6,250 ft. in bold, as shown in Fig. (A). Plot the burst resistance values for the above grades of casing as shown in Fig. (A) also in bold. Collapse: The external pressure of the collapse-design load is controlled by the maximum loss in fluid level that could occur, if a severe lost circulation problem is encountered. The maximum depth of the mud level is calculated with:

Dm =

(ρ

max

− gp )

ρ max

D lc

Where, Dm is the depth to where the mud level will fall Dlc is the lost circulation depth gp is the pore pressure gradient of the lost circulation zone ρmax is the maximum mud density anticipated in drilling to Dlc

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It is assumed that a normal pressure, lost circulation zone unexpectedly is encountered near the depth of the next casing seat (10,000 ft.) while the planned 73 pcf mud is used, and if no permeability zones are exposed above this depth, then Dm =

(73 − (0.46 × 144)) 10,000 = 926 ft. 73

For these conditions the mud level could fall 926 ft. down the inside of the 13-3/8” casing. The internal pressure is assumed to be zero at 926 ft. and,

(6,250 − 926) 73

144

= 2,700 psig at the casing seat.

However, for this example lets consider the worse possible scenario, that the entire contents of the casing are lost to a lost circulation zone at 10,000 ft. Therefore, the internal pressure would be zero at 6,250 ft., the casing seat. The pressure differential that ⎛ 67 ⎞ tends to collapse the casing is zero at the surface and, 6,250⎜ ⎟ = 3,908 psig at 6,250 ft., ⎝ 144 ⎠ the casing seat. Multiplying this pressure by a safety factor of 1.125 yields a collapsedesign load of zero at the surface and 2,908 × 1.125 = 3,271 psig at 6,250 ft. Graphically draw the collapse line from 0 at the surface to 3,217 psig at 6,250 ft., as shown in Fig. (A). Plot the collapse resistance values for the above grades of casing as shown in Fig. (A). Pressure, psi

0 0 -250

1000 K-55 54.5#

2000 K-55 68#

3000 N-80 72#

K-55 54.5#

4000 K-55 68#

5000

6000 N-80 72#

Depth, Feet

-1250 N-80 85#

-2250 Collapse

-3250 Burst

-4250 -5250 -6250

Figure A

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The first section of pipe is selected based on the collapse requirement at the setting depth. In this example, 85# N-80 is the only casing type to satisfy the collapse requirement of 3,217 psig, having a collapse rating of 3,870 psig. A vertical line for the first section (85#) is drawn from the casing setting depth (6,250 ft.) to the changeover depth for the next lower weight casing, that lies to the right of the collapse design line, 72# at 5,100 ft. A horizontal line is drawn from the intersection of the second collapse rating plotted on the design line to the collapse rating of the first section. The collapse rating of the next weaker section is plotted on the appropriate line and the changeover depth read at the intersection on the graph. Subsequent segments are similarly determined. Concurrently burst ratings are plotted and vertical and horizontal lines are drawn, see Fig. (B). Based on the combined burst and collapse satisfaction, Figure B indicates the following casing selection. Depth

Grade & Weight L-80, 72 lb/ft K-55, 68 lb/ft L-80, 72 lb/ft N-80, 85 lb/ft

0-2,380 2,380-3,725 3,725-5,100 5,100-6,250

Weight in 67 pcf mud, kips 147.74 78.85 85.35 84.2 Total: 396.21

Pressure, psi

0 -250

0

1000 K-55 54.5#

2000 K-55 68#

3000 N-80 72#

K-55 54.5#

4000 K-55 68#

Depth, Feet

N-80 85#

Collapse

-3250 Burst

-4250 -5250 -6250 Starting

Figure B

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6000 N-80 72#

-1250 -2250

5000

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Tension: The third step in the casing design is to check the tension design requirements for the preliminary casing design found to satisfy the burst and collapse strength requirements. The design loading condition for tension was specified to be while the casing is run, when the well contains 67-pcf mud. If bending and shock loading is ignored, the casing design is obtained by comparing the buoyed tensions of each section multiplied by a safety factor of 1.6, with the buoyed tensions plus 100,000 lb. of each section, assuming the casing was stuck in the borehole near the bottom and 100,000 lb. of pull was imposed. Then select the larger of the two results (see table below) and insure that these results are below the lowest value of either the body or coupling yield strength of each selected casing section. Selected casing weights & grades 72#,L-80 68#,K-55 72#,L-80 85#,N-80

Weight in 67 pcf mud times 1.6, kips

236.4 126.2 136.6 134.8

Buoyant hanging weight of each section times 1.6, kips 634.1 397.7 271.4 134.8

Buoyant hanging weight of each section, plus 100M#, kips 496.2 348.5 269.6 184.3

Greater of 100M# overpull or 1.6 S.F., kips 634.1 397.7 271.4 184.3

Yield Strength kips 1,661 1,069 1,661 1,951

Satisfies design criteria, Y/N Y Y Y Y

Total: 634.1

The above table indicates the casing selection satisfies tension requirement. Biaxial Effects: The weakest grade of the selected casing should be checked for biaxial effects as follows: Buoyant Weight carried by weakest joint Tensile Ratio = Yield strength of body or coupling

Since the weakest grade is 68# K-55, the buoyant weight to be supported by the top joint of that section is 248.5 kips. 248.5 × 1,000 Tensile ratio = = 0.232 1,069,000 Table 8 shows that the collapse resistance of the casing is reduced to approximately 86% of its original value, PC = 1,950 for K-55, 68 lb/ft casing.

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CASING DESIGN Pca = 1,950 x 0.86 = 1,677 psi

Collapse pressure due to mud weight at 2,380 ft is. Pc = (67/144) x 2,380 = 1,107 psi

Therefore safety factor in collapse for top joint of K-55, 68 lb/ft casing is SF =

1,677 = 1.51 1,107

Since SF = 1.51 > 1.125, the selection is satisfactory. An extra two joints of the heaviest casing is placed on top for drift control. Therefore the following casing string meets the design criteria. Depth, ft. 0-80 80-2,380 2,380-3,725 3,725-5,100 5,100-6,250

Page 56

Grade & Weight N-80, 85 lb/ft L-80, 72 lb/ft K-55, 68 lb/ft L-80, 72 lb/ft N-80, 85 lb/ft

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CASING CENTRALIZER SPACINGS

The centralizers on a casing string are used to provide clearance between the casing and the wall of the hole. The clearance is called standoff. The major function of a centralizer is to centralize the casing in the hole and to prevent it from lying against the wall, thus providing a reasonable uniform cement layer around the casing. Centralizer spacing should be sufficiently close to keep the casing to wall clearance at some acceptable minimum distance. The casing couplings or various types of attachable stops control the vertical travel of the centralizer. The following equations apply to all pipe in normal oilfield service.

(WF)b

(

2

= Wcs + 0.0062963 ρ mi d i − ρ mo d o

(

2

2

)

2

)

T = 0.0062963 TVD ρ mi d i − ρ mo d o cosθo

CS

S

CS ⎞ ⎛ F = 2Tsin ⎜ DLS × ⎟ + (WF)b CSsinθ 2 ⎠ ⎝ CS =

F 0.0175T × DLS + (WF)b sinθ

Where: F = force on each centralizer if spaced Cs feet apart (lbf) Wcs = weight of casing (lb/ft) θ = average inclination angle (degrees) T = tension in the wall of the casing (lbf) TVD = true vertical depth S = distance from casing shoe to the centralizer in question ρmi and ρmo = the mud weights in and out of the casing, respectively DLS = dog leg severity do = outer casing diameter di = inner casing diameter

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CASING LANDING PRACTICE It is recommended that casing strings be landed as cemented without slacking off or picking up additional weight after the cement is set. In other words, the casing should be landed or suspended in the slips with the same weight as recorded when the cement plug reached the float collar. There are definite disadvantages to landing casing, by slacking off or picking up additional weight. Any slacking off of weight is particularly detrimental since it results in the string being put in compression and causes it to buckle to the extent the hole conditions will permit. The problem in attempting to pull additional weight after the cement is set, is that in most cases the string is stuck at some point far above the cement top. Another factor is that we can seldom anticipate all the conditions the casing string may be subjected to during its useful life. By landing the casing as cemented and using a design factor in tension of 1.6, based on weight in air, a considerable change in well conditions can occur without exceeding the joint strength of the casing or excessively increasing the amount of casing in compression. Keeping as much of the pipe in tension as possible is particularly critical for long, intermediate strings where subsequent drilling with heavier mud weights through higher temperature formations is anticipated. The increase in internal pressure due to heavier mud and the higher temperature both increase the tendency of the string to buckle, so landing the string in tension keeps this to a minimum. References 1. “Bulletin on Performance Properties of Casing, Tubing, and Drill Pipe,” Bull. 5C2, Dallas: API, (1970) 2. Adams, N., Drilling Engineering, Tulsa: Penn Well Publishing, (1985) 3. Bourgoyne, Chenevert, Milhelm, Young, Applied Drilling Engineering, Richardson, TX.: SPE, (1986) 4. Brantly, J.E., History of Oil Well Drilling, Gulf Publishing Co., (1971) 5. Craft, Holden, Graves, Well Design: Drilling and Production, New Jersey: Prentice-Hall, (1962) 6. Mian, M.A., Petroleum Engineering, Vol II, Tulsa: Penn Well Publishing, (1992) 7. Rabia, H., Oilwell Drilling Engineering, London: Graham & Trotman Ltd., (1985)

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TABLE OF CONTENTS HISTORY OF PORTLAND CEMENT MANUFACTURE OF PORTLAND CEMENT COMPONENTS OF PORTLAND CEMENT API CEMENT CLASSES CEMENT SETTING PROCESS

CEMENT TESTING DENSITY FREE WATER THICKENING TIME FLUID-LOSS RATE COMPRESSIVE STRENGTH RHEOLOGICAL PROPERTIES

CEMENT ADDITIVES ACCELERATORS RETARDERS FLUID-LOSS ADDITIVES ADDITIVES TO INCREASE DENSITY DISPERSANTS SILICA DEFORAMERS ADDITIVES TO DECREASE DENSITY

FACTORS THAT INFLUENCE SLURRY DESIGN CHEMICAL ENVIRONMENT BOTTOM-HOLE STATIC TEMPERATURE PORE PRESSURES FORMATION PERMEABILITY FORMATION INTEGRITY HOLE GEOMETRY

Page 1 2 2 3 4

5 5 6 6 9 10 12

12 14 15 16 17 18 18 19 19

21 21 22 23 24 24 25

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TABLE OF CONTENTS

Page

CEMENT PLACEMENT TECHNIQUES

25

SUBSURFACE CASING EQUIPMENT

26

- FLOATING EQUIPMENT - STAGE-CEMENTING TOOLS

PRIMARY CEMENTING PLANNING A CEMENT JOB TYPE AND VOLUME OF CEMENT CEMENT ADDITIVES CEMENT MIXING PREFLUSHING CEMENT PLACEMENT TECHNIQUES CEMENT DISPLACEMENT

LINER CEMENTING LINER EQUIPMENT CEMENTING TECHNIQUES RUNNING AND CEMENTING PROCEDURE

FLOW CALCULATIONS BINGHAM PLASTIC MODEL

26 29

32 32 32 39 41 43 44 50

52 53 55 56

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The purposes of this chapter are to present 1. 2. 3. 4. 5. 6.

The objectives of cementing The composition of cement Cement testing Cement additives Cement slurry design and Cement placement

In well drilling operations cement slurry is placed around the casing strings and liners by mixing powdered cement, water and additives on the surface and pumping it by hydraulic displacement into the annular space between the casing and the wellbore. When the cement slurry sets, it forms a rigid solid that exhibits favorable compressive strength characteristics. The primary objectives of well cementing are 1. 2. 3. 4.

to support the casing string protect the casing from corrosive fluids prevent fluid movement behind casing and plug an abandoned zone or well.

In designing a casing cementing job, the drilling engineer is responsible for selecting the cement composition and displacement techniques so that the cement slurry will fill the entire annular space behind the casing and achieve adequate compressive strength soon after it is placed at the desired location in the well. This minimizes the waiting time after cementing. The cement slurry must be designed such that it will remain pumpable until it is placed at the desired location. The density of the cement slurry must be adequate to control any movement of pore fluid while at the same time not cause any formation fracture.

HISTORY OF PORTLAND CEMENT Although cementatious materials have been used since ancient times, the invention of modern Portland cement is usually attributed to Joseph Aspdin, an Englishman, who filed a patent for Portland cement in 1824. He called it "Portland" cement because it resembled the limestone quarried in Portland, England.

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MANUFACTURE OF PORTLAND CEMENT Portland cement is manufactured with materials and methods that have changed little since Aspdin’s time. The material is prepared by sintering fixed proportions of calcium containing materials (limestone, chalk, seashells) with aluminosilicates (clays) in a kiln at 2600-2800 oF (1425-1535 oC). The resulting material, clinker, is then cooled and interground with gypsum which controls the setting time of the cement. Small percentages of other substances, such as sand, bauxite or iron ore are sometimes used in the kiln feed to adjust the properties of the clinker.

COMPONENTS OF PORTLAND CEMENT Portland cement consists primarily of the four chemical compounds shown in Table 1. All grades or classes of Portland cement contain these four compounds. However, the relative percentages of the compounds can vary, depending on the feed materials in the manufacturing process. The relative percentages of these compounds along with grind of the cement have been found to strongly affect the cement performance. Table 1 Principal Components of Portland Cement _________________________________________________________________________________________________________________________________________________________________________

Compound

Formula

Standard Designation

Typical % (Wt)

_________________________________________________________________________________________________________________________________________________________________________

Tricalcium Silicate Dicalcium Silicate Tricalcium Aluminate Tetracalcium Aluminoferrite Other Oxides

3CaO-SiO 2 2CaO-SiO2 3CaO-Al2O3 4CaO-Al2O3-Fe2O3

C3S C2S C3A C4AF

50% 25% 10% 10% 5%

_________________________________________________________________________________________________________________________________________________________________________

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API CEMENT CLASSES Specifications for cements used in oil-well applications have been written by the American Petroleum Institute (API). These specifications are found in “API Specifications for Materials and Testing for Well Cements”, (API Spec 10). There are eight API cement classes. Table 2 provides a summary of the chemical composition, grind and special properties of some of these API cements. Most oil-field operations use class A, C, G, or H. The different classes of API cement and their compositions are shown below and in Table 2. Class A:

Intended for use from surface to a depth of 6,000 ft when special properties are not required. Available only in Ordinary type (similar to ASTM C150, Type I).

Class B:

Intended for use from surface to a depth of 6,000 ft when conditions require moderate to high sulfate resistance. Available in both Moderate type (similar to ASTM C150, Type II) and High Sulfate Resistant types.

Class C:

Intended for use from surface to a depth of 6,000 ft when conditions require high early strength. Available in Ordinary type and in Moderate and High Sulfate Resistant types.

Class D:

Intended for use at depths from 6,000 to 10,000 ft and at moderately high temperatures and pressures. Available in both Moderate and High Sulfate Resistant types.

Class E:

Intended for use at depths from 10,000 to 14,000 ft and at high temperatures and pressures. Available in both Moderate and High Sulfate Resistant Types.

Class F:

Intended for use at depths from 10,000 to 16,000 ft and at extremely high temperatures and pressures. Available in High Sulfate Resistant types.

Class G:

Intended for use as a basic cement from the surface to a depth of 8,000 ft as manufactured. With accelerators and retarders it can be used at a wide range of depths and temperatures. It is specified that no additions except calcium sulfate or water, or both, shall be interground or blended with the clinker during the manufacture of Class ‘G’ cement. Available in Moderate and High Sulfate Resistant types.

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Class H:

Intended for use as a basic cement from the surface to a depth of 8,000 ft as manufactured. This cement can be used with accelerators and retarders at a wide range of depths and temperatures. It is specified that no additions except calcium sulfate or water, or both, shall be interground or blended with the clinker during the manufacture of class H cement. Available only in Moderate Sulfate Resistant type. Table 2 Typical Composition and Properties of API Classes of Portland Cement

API Class A B C D&E G&H

Compounds (percentage) C3S C2S C3A C4AF 53 47 58 26 50

24 32 16 54 30

8+ 58 2 5

8 12 8 12 12

Wagner Fineness (sq cm/gm) 1,600 1,600 1,800 1,200 1,600

to to to to to

1,800 1,800 2,200 1,500 1,800

Property High Early Strength Better retardation Low heat of hydration Resistance to sulfate attack

CEMENT SETTING PROCESS When water is added to Portland cement, a chemical reaction (hydration) takes place that eventually causes the cement particles to bond together to form an impermeable, hard, rock-like material. The strength and impermeability of the cement is due to the formation of a dense network of interlocking fibers. Two of the byproducts of cement hydration are calcium hydroxide [Ca(OH)2] crystals and heat. The [Ca(OH)2] crystals cause the cement to be very basic (high pH). Because of this, a cement sheath will provide corrosion protection for the steel casing. The heat given off during the hydration reaction is sometimes used to detect the top of cement by temperature logging. The time at which the slurry achieves its maximum temperature depends on the particular slurry and its curing conditions, but generally is between 3 and 12 hours.

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CEMENT TESTING Physical properties of drilling cement slurry are measured according to test procedures established by the API (American Petroleum Institute). These properties are used by drilling personnel to formulate the specifications of cement when designing a cementing job. Cement slurry testing is normally done by cementing service companies or by Saudi Aramco lab, however, it is important for the drilling engineer to understand the nature of these tests so that be can interpret cement specifications and test results properly. The basic properties of cement slurry are (1) (2) (3) (4) (5) (6)

density fluid loss thickening time free water compressive strength and rheological properties. DENSITY The density of a cement slurry is important for well control and the prevention of lost circulation while cementing. Density is also a useful field monitor of whether or not the slurry has been mixed with the designed water requirement. With the appropriate additives, cement slurries can be designed with densities ranging from about 60 pcf to about 150 pcf.

Fig. 1 Fluid Density Balance for weighing cement slurries

Page 5

Density of cement is measured by using either a unpressurized or pressurized mud balance. Because cement slurries often contain entrapped air, the pressurized mud balance, Fig (1), provides a more accurate

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measurement (at about 30 psi pressure). Errors of unpressurized balance.

7-15 pcf may occur using the

In the field, in-line radioactive densitometers are often used to monitor density as the cement slurry is pumped. FREE WATER The water added to the dry bulk cement is used both as a reactant in the hydration reaction and to provide fluidity to the slurry. When properly mixed, about 2/3 of the water is involved in the chemical reaction while 1/3 provides fluidity. All of the water in a properly mixed slurry, however, is either bound to the cement particles by chemical bonds or loosely attracted to the cement particles to form a stable suspension. If excess water is added, the cement particles will settle, leaving a layer of free water above the suspension. Excessive cement free water may lead to the formation of water pockets in a well, especially on the high side of deviated wells, Also, since excessive free water indicates solids settling, it may result in difficulty in mixing and displacing the slurry. Procedures for determining the free water content of a cement slurry have been specified by the API. There are two types of tests: a specification test conducted at 80 oF and a new (tentative) operating free water test conducted under downhole conditions. Under the API specification procedure, the maximum allowable free water is 1.4% (3.5 ml water from 250 ml of cement). THICKENING TIME Perhaps the most important property of a cement slurry for well applications is its thickening time. The thickening time provides an indication of the length of time the slurry will remain pumpable. A thickening time that is too short can result in the cement setting inside the casing, tubing or drill pipe with severe economic consequences. A thickening time that is too long, on the other hand, can necessitate an unduly long and costly delay waiting for the cement to set. The API defines the thickening time of a cement slurry to be the time required for the slurry to reach 100 Bearden units of consistency (Bc), using the methods of API Spec 10. One hundred Bearden units of consistency is roughly equivalent to a viscosity of 100 poise. Cement is considered to be unpumpable at this viscosity.

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The thickening time is measured in a device called a consistometer. Consistometers are designed so that the consistency of the cement slurry can be continually monitored while the cement is subjected to a temperature, shear, and pressure history that simulates what the cement will see as it is pumped downhole. A schematic diagram of a consistometer is shown in Fig (2). The apparatus consists of a rotating cylindrical slurry container equipped with a stationary paddle, all enclosed in a pressure chamber capable of withstanding pressures and temperatures encountered in cementing operations. As heat and pressure are applied on the slurry sample, the Fig. 2 Schematic of a High-Pressure Cement cylindrical slurry container is Consistometer rotated at 150 rpm. The consistency of the slurry is measured in terms of the torque exerted on the paddle which is recorded continuously on a strip chart. The limit of pumpability is reached when the paddle torque reaches 100 Bearden units. Since the thickening time depends not only on the slurry being tested, but also on the simulated downhole conditions, it is important to simulate these conditions as accurately as possible. The API has published a series of cementing schedules, based on field measurements, that can be used to simulate the downhole conditions for many wells. There are different API schedules, depending on the type of job (casing, liner, or squeeze), well depth, and the bottom hole static temperature. Schedule 6, designed to simulate average conditions encountered during cementing of casing at 10000 ft is shown in Table (3).

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TABLE 3 EXAMPLE CONSISTOMETER SCHEDULE (Schedule 6-10,000 ft (3050 m) casing cement specification test) Surface temperature, oF (oC) Surface pressure, psi (kg/cm2) Mud density lbm/ga (kg/l) lbm/cu ft psi/Mft (kg/cm3/m) Bottomhole temperature, oF(oC) Bottomhole pressure, psi (kg/cm2) Time to reach bottom, minutes

80(27) 1,250 (88) 12 (1.4) 89.8 623 (0.144) 144 (62) 7,480 (526) 36

_____________________________________ Time (minutes)

Pressure (psi) (kg/cm2)

Temperature (oF) (oC)

_____________________________________ 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36

1,250 1,600 1,900 2,300 2,600 3,000 3,300 3,700 4,000 4,400 4,700 5,100 5,400 5,700 6,100 6,400 6,800 7,100 7,480

88 113 134 162 183 211 232 260 281 309 330 359 380 401 429 451 478 499 526

80 84 87 91 94 98 101 105 108 112 116 119 123 126 130 133 137 140 144

27 29 31 33 34 37 38 41 42 44 47 48 51 52 54 56 58 60 62

_____________________________________ Final temperature and pressure should be held constant to completion o o 2 of test, within ± 2 F ( ± 1 C) and ± 100 psi ( ± 7 kg/cm ), respectively.

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The temperature and pressure of the slurry sample in the consistometer chamber is increased to the pressures and temperatures and time schedule in Table (3). When the final temperature and pressure are reached, they are held constant until the test is completed, that is when a consistency of 100 Bearden Units is reached. The API schedules have proven to be accurate and reliable over many years. However, there are certain situations where the API cementing schedules may not be appropriate. If unusual temperature conditions are encountered, such as geothermal gradients outside the 0.9-1.9 oF/100 ft API range, highly deviated wells or offshore cementing through long risers, it may be necessary to develop a cement testing schedule using computer simulation. The thickening time of a cement slurry is generally selected to be equal to the job time plus a safety factor. The job time is the estimated time required to mix the slurry and pump it into place. Usual practice is to employ a 50-100% safety factor, depending on the type of job and the experience in the area. Through the use of the appropriate additives, well cement slurries have been designed with thickening times as short as 60 minutes or as long as 12 hours.

FLUID-LOSS RATE The rate at which a cement slurry loses water through a permeable barrier when a differential pressure is imposed is referred to as filtration rate or fluid-loss rate. The water lost is the water that does not take part in the chemical reaction, that is, the water required for slurry fluidity. When this water is lost, the slurry viscosity increases, and the slurry loses its fluidity. In addition, as water is lost, the concentration of the cement particles increases. This may result in the formation of cement bridges which restrict or prevent flow in areas of narrow clearance. Thus, control of the fluid-loss rate of a slurry is necessary when: • Cementing past very permeable intervals • Cementing through narrow clearances (for example, liners) • Squeeze cementing perforation or channels

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Because the water lost is that used to maintain slurry fluidity, there is still sufficient water to complete the hydration reaction. In fact, because the cement particles are closer together, the strength of a slurry that has lost water is greater than the strength of the parent slurry (that is, the slurry that did not lose any water). Testing procedures for fluid loss rates are given in API Spec 10. There are two types of tests: 1) low temperature/low pressure (LT/LP) and 2) the well-simulation or high temperature/high pressure (HT/HP) The HT/HP fluid-loss rate of a neat cement slurry (i.e. just cement and water) is on the order of 1000-2000 cc/30 min. However, through the use of certain additives, the fluidloss rate can be adjusted to lower values. Table (4) presents some general fluid loss guidelines for different cementing operations. Table 4 Guidelines for Cement Slurry Fluid-Loss Rates _______________________________________________ Operation

HT/HP Fluid-Loss Rate (cc/30 min.)

_______________________________________________ Casing Cementing (past high permeablility formation) Liner Cementing Squeeze Perforation or Repair Channels

300-450 100 50

_______________________________________________ The API procedure for measuring fluid loss uses a 325-mesh screen as a filtration medium. A pressure of 1000 psi is applied on the slurry sample and the volume filtrate in 30 minutes is measured. COMPRESSIVE STRENGTH The compressive strength of set cement is the stress required to cause failure of the cement under a uniaxial compressive load. Fig (3) shows the compressive strength development for a class A cement. The rate of strength development depends on the type Page 10

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of cement, the type and concentration of additives and the curing temperature. However, 75-80% of the ultimate compressive strength is generally achieved within 3 days. Compressive strength data are used for • Establishing waiting on cement (WOC) time • Determining optimum time to perforate and • Monitoring the stability of the set cement. After cement has been pumped into the annulus, it must obtain sufficient strength so that further operations will not damage the cement sheath. Although the loadings placed on the cement downhole are not necessarily uniaxial compressive loads, the compressive strength has been found to be a convenient indirect measure of the ability of the cement to withstand these loads. The industry has generally accepted a value of 500 psi as the minimum required compressive strength before further drilling operations Fig. 3 Compressive Strength Development can commence. Tests have shown that a cement sheath with 500 psi can easily support the weight of the casing, even under rather poor bonding conditions. Similarly, laboratory experiments indicate that a well should not be perforated until the cement has achieved at least 2000 psi compressive strength. Above this value, the tests indicate that perforating does not damage the cement bond. The API testing procedures for determining compressive strength are in API Spec 10. These tests use conventional compressive strength testing equipment. An Ultrasonic Cement Analyzer (UCA) is also available for making non-destructive compressive strength measurements. The UCA is based on the measurement of the travel time of ultrasonic waves pulsed through a cement sample. While the UCA provides a useful time

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history of strength development, the actual values of compressive strength predicted by the UCA may not agree with conventional crush tests, especially for non-standard slurries. Therefore compressive strength values obtained from the UCA should be used with caution. RHEOLOGICAL PROPERTIES The rheological properties of cement slurry allows the drilling engineer to compute the frictional pressure losses in pipe and annulus from the flow of cement slurry and the annular velocity required to establish laminar, plug or turbulent flow. Cement slurry is a non-Newtonian fluid, that is it does not exhibit a direct proportionality between pressure loss and flow rate at constant temperature and pressure. The behavior of non-Newtonian fluids can be expressed by the Bingham plastic model or the powerlaw model. For cement slurries the power-law model is more accurate than the Binghamplastic model; therefore, the results are closer to the exact behavior of the cement slurry in the well. Cement slurry flow calculations are presented in the section on Flow Calculations.

CEMENT ADDITIVES Cement additives are solid or liquid chemicals that are mixed with cement slurry to change its properties so that it will meet cementing specifications of a particular job. Solid additives are free-flowing powders that either can be dry blended with the cement before transporting it to the well or can be dispersed in the mixing water at the job site. Liquid additives are mixed with the mixing water at the job site. By convention, the concentration of solid additives, except sodium and potassium chloride, is expressed as a percentage of the weight of dry cement used in mixing the slurry. Thus, a cement that contains 0.75% of additive A contains 0.75 lb of additive A for every 100 lb of dry cement used. The concentration of sodium chloride is usually expressed as a percent by weight of the mix water. The concentration of liquid additives is expressed in gals per sack of dry cement which weighs 94 lbs. The volume of slurry obtained by mixing one sack of cement with water and additives is called the yield, and is expressed in cu. ft/sack. Calculation of the cement yield is illustrated in the following example. Example Page 12

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It is desired to mix a slurry of Class ‘G’ cement containing 6.6% bentonite, 0.1% CFR-3 friction reducer and 0.9% Halad-22A fluid loss additive. Determine (a) (b) (c) (d)

the weights of bentonite, CFR-3 and Halad-22A to be mixed with one sack of cement volume of mix water slurry yield and slurry density.

Solution a) Weight of bentonite per sack of cement = 94 lb × 0.066 = 6.2 lb. Weight of CFR-3 = 94 lb × 0.001 = 0.094 lb Weight of Halad 22A = 94 lb × 0.009 = 0.846 lb b) The volume of mix water per sack is the sum of the water requirements for cement and each additive which can be obtained from the Halliburton Cementing Tables (Red Book) as follows: Water Requirement Class ‘G’ cement Halad 22A Bentonite (prehydrated) CFR-3

5 gal per 94 lb sk 0.4 gal / sk of cmt 0.43 gals / 2% in cmt

Water Volume, gals. 5.00 0.40 1.42 0.00 6.82 gals or 0.911 ft3 per sack

Total

c) The slurry yield is the sum of the volumes of cement, water and all the additives per sack of cement. The volumes are calculated by dividing the weight by the density of each additive from Halliburton Tables, Material Cement Halad 22A Bentonite CFR-3 Water

Specific Gravity

Density (lb/ft3)

Weight (lb)

Volume (ft3)

3.14 1.32 2.65 1.20 1.00

196.0 82.3 165.3 74.8 62.4

94.000 0.846 6.200 0.094 -

0.480 0.010 0.037 0.001 0.911 1.439 ft3/sack

d) Density is obtained by dividing the total weight by the total volume

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Material Cement Halad 22A Bentonite CFR-3 Water

Weight (lb) 94.000 0.846 6.200 0.094 56.840 157.980 lb

Density =

157.98 = 109.78 lb/ft3 1439 .

Problem

It is desired to mix a slurry of Class ‘G’ cement containing 35% silica flour, 0.3% Halad 413, 0.45% Halad 344 and 0.8 HR-15 retarder. Determine (a) the quantity of each additive per one sack of cement, (b) volume of mix water in gal, (c) slurry yield, and (d) slurry density. Water requirements and specific gravity of additives are as follows: Additive Silica Sand (Coarse) Halad 344 Halad 413 HR-15

Water Requirement

Specific Gravity

None None None None

2.63 1.22 1.48 1.57

ACCELERATORS

The additive most commonly used to accelerate the set of cement is calcium chloride (CaCl2). This compound is used in the concentration range 1 to 4%. The effect of CaCl2 concentration on thickening time is shown in Fig (4). Since CaCl2 is effective at relatively low concentrations, it is an economical additive. In addition, the accelerating effect of CaCl2 is predictable, and it has few adverse side effects. The presence of CaCl2, however, will decrease the effectiveness of some fluid loss additives.

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Fig. 4

Effect of CaCl2 on Thickening Time

Another additive sometimes used as an accelerator is sodium chloride (NaCl). At concentrations below 18% (by weight of mix water), NaCl accelerates the set of cement. At greater concentrations, however, NaCl acts as a retarder. Sodium chloride is not compatible with most fluid loss additives. In addition, it increases the tendency for slurry foaming.

RETARDERS

Retarders are additives that delay the set of cement. Most commercially available retarders are organic materials. Table 5 presents a summary of the generic types of organic retarders. Retarders are generally used in the concentration range of 0.1 to 1.0%. Since retarders are generally composed of heat-sensitive organic molecules, particular attention should be paid to the recommended temperature range for using the retarder. Information on specific retarders is available from cementing company literature. Another additive that will retard the set of cement at certain concentrations is sodium chloride (NaCl). At concentrations greater than about 18% (by weight of mix water), NaCl will act as retarder. Sodium chloride is incompatible with most fluid loss additives, has an increased tendency for slurry foaming, a limited extent of retardation, and has to be used in large concentrations to be effective as a retarder.

Table 5

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Application

Service Co. Equivalents

BHCT Range. oF

HB

D-S

Low Temperature Low Temperature/Dispersing

<200 <200

HR4 HR5 HR7

D13

Moderate Temperature Moderate Temperature

150-250 180-225

Diacel LWL

D120 D8

High Temperature

225-400

D28 D99

High Temperature

<300

HR12 HR15 HR20 Borax

D93

NOTES

-

CMHEC: Also acts as a fluid-loss additive, and viscosifies the slurry

Borax: Added to enhance the behavior of high temp. retarders. Not to be used alone

FLUID-LOSS ADDITIVES

Fluid-loss additives are used to reduce the rate of fluid loss from cement. There are two basic types of fluid loss additives: polymers and bentonite. Polymers function primarily by plugging the pore space in the cement filter cake. Polymeric fluid loss additives • • • •

are sensitive to temperature, seem to have a threshold concentration of about 0.8%, generally retard the slurry and tend to increase the viscosity of the slurry.

Bentonite functions as a fluid loss additive by decreasing the permeability of the cement filter cake, As a fluid loss agent, bentonite generally • will result in a lower slurry density Page 16

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Attapulgite clay is sometimes used as a fluid-loss additive in slurries containing salts because it is not sensitive to the salts. Attapulgite, however, does not have the same water-absorbing power as bentonite.

ADDITIVES TO INCREASE DENSITY

For purposes of well control, it is sometimes necessary to use additives that increase the slurry density. Table 6 presents a summary of the additives commonly used to increase slurry density. Table 6 Additives to Increase Density Additive Class ‘G’ Cement Barite Hematite Okla. # 1 Sand

Specific Gravity

Water Requirement

3.14 4.23 5.02 2.63

5.0 gal/ 94 lb 2.64 gal/ 100 lb 0.36 gal/ 100 lb 0

These additives generally increase the slurry density because they have a high specific gravity and/or a low water requirement in comparison to the cement. Hematite is more commonly used than barite because it has a higher specific gravity and a lower water requirement. A pumpable slurry with a density as high as 20 ppg can be achieved with hematite. Although Oklahoma #1 sand has a lower specific gravity than cement, it can increase slurry density (up to 17.5 ppg) because of its zero water requirement. Since these weighting additives “dilute” the cement particles, the final strength of the set cement will be lower than that of a neat cement. Reductions in compressive strength can be minimized by using a reduced water content in conjunction with a dispersant. This method is discussed further in the next section below.

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DISPERSANTS

Dispersants (also called thinners or turbulence inducers) are used to reduce slurry viscosity or increase slurry density. A reduction in slurry viscosity may sometimes be desirable to reduce friction pressures. This may occasionally be necessary when the cement column is long, the annulus is narrow or when the annulus might be partially obstructed. However, dispersants to thin a slurry should be used with care. Their misuse can lead to high free water breakout and they can affect the behavior of other additives. Dispersants are also helpful for increasing slurry density. Because dispersants thin the slurry, when a dispersant is present a pumpable slurry can be formulated with a lower water requirement than normally recommended for the neat cement. By adding a dispersant and reducing the water content, slurry densities as high as 17.5 ppg can be achieved without the addition of barite or hematite (Table 7). Table 7 Use of Dispersants to Increase Class G Slurry Density CFR-2 % (bwc)

Mix Water (gal/sack)

Density (lb/gal)

0.00 0.75 0.75 0.75

5.00 4.00 3.78 3.78

15.8 16.7 17.0 17.5

An advantage of using this technique for increasing slurry density is that cement particles are not diluted. In fact, since the concentration of cement particles is increased, the strength of the set cement will be higher than that of a neat cement.

SILICA

On being cured at temperatures in excess of 250 oF, one of the components of Portland cement (C2S) undergoes a change in structure that results in a significant loss in compressive strength and a significant increase in permeability. This phenomenon is called strength retrogression.

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It has been found that the addition of 35% or more of silica can prevent this degradation (see Fig 5). Any silica sand finer than 100 mesh can be used. Note that the use of less than 20% silica will intensify the problem, where as the maximum benefit is obtained at around 40% concentration.

Fig. 5

Effect of Silica Concentration on Strength Retrogression of Class A cement cured at 320 oF

DEFOAMERS

Excessive foam makes it difficult to maintain slurry density control and can cause other problems, such as “air locking” of the pumps. This is often a problem in slurries containing salt. Chemical foam inhibitors, which minimize air entrainment and foaming are available. These materials can be obtained in liquid or solid form. With the exception of foam cements, these compounds have no known detrimental effects on other cement properties.

ADDITIVES TO DECREASE DENSITY

There are a number of additives available to lower slurry density. Table 8 presents a summary of additives to lower slurry density. These additives lower the density of slurry because they have a lower specific gravity than the cement, and in most cases, have a higher water requirement than the cement. Perhaps the most widely used additive to decrease slurry density is bentonite. Bentonite in cement lowers density chiefly because of its high water requirement. Where as each pound of Class G cement requires 0.05 gal water, each pound of bentonite requires 0.69

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gal water. Thus, for example, the density of a Class G cement can be lowered from 15.8 ppg (neat) to 12.8 ppg by the addition of 12% bentonite. Because of loss of compressive strength, bentonite is generally not used at concentrations greater than 12%. When bentonite is dry blended with the cement, the high calcium content of the cement prevents full hydration of the bentonite. If bentonite is prehydrated, i.e. allowed to hydrate in fresh water before being added to the cement, it will have a greater capacity for water. One part by weight of bentonite prehydrated in the mix water has an effect that is essentially equivalent to 3.6 parts by weight of bentonite dry blended with the slurry. In other words, if the bentonite is to be prehydrated (usually 2-12 hours is sufficient) the amount of bentonite can be reduced by the factor 3.6. To obtain ultra light weight slurries, ceramic spheres, glass beads, or foam can be used. Although ceramic spheres and glass beads are relatively expensive, slurry densities as low as 8.3 ppg can be achieved while maintaining good compressive strength properties. However, because the spheres will crush at sufficiently high hydrostatic pressure (generally around 4000 psi), there are density and depth limitations associated with their use. Ultra lightweight slurries can also be achieved by incorporating air or nitrogen into the cement as a foam. Using foam, slurry densities as low as 9 ppg can be achieved while maintaining good strength properties in the cured cement.

Table 8 Additives to Lower Density Additive Bentonite Attapulgite Diatomaceous Earth Gilsonite Pozzolan Ceramic Spheres Glass Beads Sodium Meta-silicate

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Specific Gravity 2.65 2.89 2.10 1.07 2.50 0.72 0.39 2.40

Water Requirement 1.3 gal/ 2% sk cmt. 1.3 gal/ 2% sk cmt. 3.3-7.4 gal/ 10% sk cmt. 2.0 gal/ 50 lb 3.6-3.9 gal/ 74 lb 0.31 gal/ 2 lb 0.36 gal/ 2 lb 3.2-12.3 gal/ 2-3% sk cmt.

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FACTORS THAT INFLUENCE SLURRY DESIGN There are a number of factors that affect the behavior of the cement as it is pumped into place and as it solidifies. The important factors which must be considered by the drilling engineer in designing a cementing job are chemical environment, bottom-hole temperature and pressure, formation permeability, formation integrity and hole geometry.

CHEMICAL ENVIRONMENT

The parameters which make up the chemical environment of the cement include a) Mix water b) Wellbore fluids and c) Formation fluids Mix Water

The primary functions of water in a cement slurry is to carry the cement solids down the hole and react with the cement to form a rigid solid. Ideally, the water for mixing cement should be reasonably clean and free of soluble chemicals, organic matter and other contaminates. Inorganic materials such as chlorides, sulphates, hydroxide, carbonates will accelerate the setting of cement, the rate depends on the concentration of the material. Organic chemicals from decomposed plant life will retard the setting of cement. If potable water is not available, the purest water available may be useable but the slurry must be tested in the laboratory before it is pumped in the hole. In Saudi Aramco UER water from drilling water supply wells (TDS = 2-3 Mppm) is used for mixing cement on most onshore wells. Wellbore Fluids

Contamination or dilution of cement by drilling and workover fluids may damage the cementing system. The best way to combat the detrimental effects of drilling fluids is to use wiper plugs and spacers. Wiper plugs help eliminate contamination of the cement inside the casing, and flushes help to clean the annular space between the casing and the formation and minimize mixing of cement and drilling fluid in the annulus. Composition of spacers must be compatible with the drilling fluid and cement.

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Formation Fluids

Formation brines containing sodium sulphates, magnesium sulphate and magnesium chloride react with the hard cement and can cause eventual deterioration of the cement sheath behind the casing. The rate of attack depends on the concentration of the sulphate salts in the formation water. Sulphate attack is most pronounced at temperatures of 80 to 120 oF. Lowering the Tricalcium Aluminate (C3A) content in the cement increases the sulphate resistance of the cement.

BOTTOM-HOLE STATIC TEMPERATURE

The bottom-hole static temperature (BHST) is one of the most important parameters to establish when designing a cement job. It is important for two reasons:

Fig. 6 Effect of Temperature on Thickening Time of various API cements at atmospheric pressures Page 22

•

The bottom-hole static temperature is often used to help estimate the temperature history that the cement will see as it is pumped into place. The temperature history strongly affects the strength and thickening time of the cement. As the formation temperature increases, the cement slurry hydrates and sets faster and develops strength more rapidly. Also the thickening time is decreased as shown in Fig (6).

•

The bottom-hole static temperature is usually the maximum temperature the cement will see during its lifetime. This temperature affects the rate at which the cement gains compressive strength. Also, if this temperature exceeds 250 oF, silica should be added to the slurry to prevent long term strength retrogression.

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When designing a cement slurry, the bottom-hole circulating temperature (BHCT) is considered to be the temperature of an element of cement as it reaches the bottom of the hole. The BHCT will usually be less than the BHST because the inlet temperature of the cement at the surface is usually less than the BHST. For testing, the BHCT is taken to represent the highest temperature the slurry will see as it is pumped into place. The API thickening time testing procedure calls for holding the slurry at BHCT after it has been brought up to temperature according to the appropriate schedule. There are two methods for determining the slurry temperature history: • •

API Cementing Schedules Computer Simulation

To use the API Cementing Schedules, it is necessary to know only the type of job (casing, liner, or squeeze), the well depth, and BHST. The job type and well depth are used to select the appropriate schedule type. The BHST is used to calculate the temperature gradient from Eq (1). T. Grad. = (BHST - 80) ÷ (Depth/100 ft) ............................…....…...... (1) Once the temperature gradient is known, the particular schedule for that gradient can be identified. The BHCT is the highest (final) temperature for that gradient. As mentioned earlier, for unusual conditions such as highly deviated wells or offshore cementing through long risers, it may be necessary to use computer simulation to develop a cement testing schedule.

PORE PRESSURES

The pore pressures of the fluid-bearing formations also affect the design of the cement job. The density of the cement should be such that the hydrostatic pressure exceeds the pore pressure at all depths in the well. Generally, this will be the case if the cement density exceeds the drilling fluid density used to drill the well. However, in deep wells, such as Saudi Aramco’s Khuff wells, mud weights as high as 140 pcf are required to keep deep high-pressure formations under control. In this case additives should be added to the cement to being its density slightly above 140 pcf.

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FORMATION PERMEABILITY

Another factor to consider when designing a cement job is the formation permeability that the cement may see. Long intervals of high permeability formation increase the potential for fluid loss from the cement. This may cause high slurry viscosities leading to increased pumping pressures and lost circulation or perhaps total loss of slurry mobility. It should be recognized, however, that the drilling fluid filter cake or particle plugging from the drilling fluid may reduce the permeability that the cement sees. However, any broach in this shield (i.e. by fracturing or erosion) could lead to disaster without the appropriate cement fluid-loss control. In Saudi Aramco fluid loss additives are used in Arab-D wells when cementing 4-1/2” liners across the Arab-D or 7” liners across the permeable Biyadh, Sulaiy and ArabA,B,C formations.

FORMATION INTEGRITY

Another fundamental consideration in designing a cement job is formation integrity. The breakdown fracture pressure (often expressed as a fracture pressure gradient) will limit the density of the cement and/or the surface pumping pressure that can be used without losing returns. Losing returns while cementing is generally undesirable because •

Since some cement is lost to the formation, the top of cement (TOC) may not be high enough to cover all necessary zones.

•

A cement-filled fracture may adversely alter fluid flow in the reservoir.

•

Fracturing may cause undesired interzonal flow.

•

Fracturing could expose the cement to high permeability and lead to a costly bridge-off in the annulus.

•

Cement may plug up a naturally-fractured pay zone.

Information on formation integrity can often be obtained from the Daily Drilling Reports for the well. If returns were lost while drilling, the mud weight being used at the time provides some indication of the formation integrity. More direct information may be available from pressure integrity test (PITs).

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HOLE GEOMETRY

Hole geometry is another important factor in designing a cement job. The hole geometry can influence the cement job in a number of ways. For example: •

The hole size, casing size, and desired top of cement will affect the volume of cement to be pumped.

•

The amount of annular clearance may affect the amount of fluid loss control required to prevent bridging. It may also limit the pumping rate to prevent excessive friction pressures.

•

The angle of deviation may necessitate reducing the free water content of the slurry to prevent high-side water pockets. The deviation angle may also affect the placement of centralizers.

For many wells the hole geometry is obtained from caliper logs. In those wells where caliper logs are not run, the size of the annulus can be roughly estimated from a fluid caliper. In this method, the volume required to pump a marker pill down the casing and up the annulus is monitored. The annular volume is then obtained by subtracting the casing volume.

CEMENT PLACEMENT TECHNIQUES Different cementing equipment and placement techniques are used for • •

Cementing casing strings and Cementing liner strings

as illustrated in Fig. (7). A casing string differs from a liner in that the casing extends to the surface while the top of the liner is attached below the surface to a previously cemented casing. Fig. 7 Common Cement Placement Requirements

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SUBSURFACE CASING EQUIPMENT

Floating equipment, cementing plugs, stage tools, centralizers, cementing baskets and liner hangers are mechanical devices commonly used in running casing and liner strings and in placing cement around the pipe. This section provides a general description of these mechanical aids and discusses their functions and applications.

Floating Equipment

The guide shoe shown in Fig (8) is an open ended device attached at the bottom end of the casing to direct the casing away from ledges and minimize sidewall caving as the casing is lowered in deviated sections of the hole. The outer body of the guide shoe is made of steel which has the same strength as the casing while the nose is made of drillable concrete. Circulation is established down the casing through the open end of the guide shoe. Some types of guide shoes have side ports which allow circulation if the casing is set on bottom. Fig 8. Guide Shoe

The float shoe, Fig (9), is a guide shoe equipped with a springloaded back pressure valve which is enclosed in plastic and concert. The valve, which is closed by the spring or by the hydrostatic pressure of the fluid column around the casing, prevents fluids from entering the casing while pipe is lowered into the hole. Float shoes may also be used to reduce the load on the derrick by allowing the casing to be floated into the hole. This is done by lowering the casing empty or partially filled with water or drilling fluid. In Saudi Aramco float shoes are used on all casings string except the shallow conductors set at 100 ± ft from the surface.

Fig 9. Float Shoe

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shoe in the casing string and serve the same function as the float shoe. The float collar, Fig (10), serves as a back up to the float shoe in the event the back pressure valve in the float shoe fails to provide the necessary seal. The space between the float collar and the float shoe serves as a trap for contaminated cement or mud that may accumulate from the wiping action of the cementing plug. The contaminated cement is thus kept away from the shoe, where the best cement bond is required. When the cementing plug reaches (bumps) the float collar during cement displacement a pressure buildup is observed at the surface indicating that the cement displacement is complete.

Fig. 10 Float Collar

Fig. 11 Innerstring Cementer with latch-down plug

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For large casing strings, float shoes may be obtained with a special stab-in device that allows the cement slurry to be pumped through the drill pipe (this method of cementing is called inner-string cementing). This device, shown in Fig (11), eliminates the need for large cementing plugs and for leaving large amount of cement inside the casing above the shoe. The tool offers another advantage in cementing large casing across loss circulation zones which is discussed later in this chapter. In Saudi Aramco inner-string float shoes are used for cementing 18-5/8” conductors. A latch collar, shown in Fig (12), or a landing collar, is run one or two joints above the float collar. It is used only with liner strings to catch and seal the liner wiper plug. It keeps the wiper plug from moving up hole and seals pressure from below and above. It also prevents the wiper plug from turning while drilling out. The cementing plug, shown in Fig (13), is not a permanent part of the casing string but is placed in the casing after the cement slurry Fig. 12 Landing Collar is pumped. It separates the cement from the displacing fluid and minimizes Fig. 13 Five-wiper top contamination of the cement at the interface. When the cementing plug cementing plug reaches the float collar the surface pumping pressure increases abruptly indicating that the cement job is complete. The cementing plug has rubber wipers and drillable insert and constructed to withstand the force of the cement column and displacement fluid and provide a dependable seal. Since holes are rarely straight, the pipe will generally be in contact with the hole at several places. Centralizers are installed around the casing to center the casing in the hole and create a uniform annular flow area and minimize variations of resistive drag forces in this flow area. In addition, centralizers keep the pipe away from the wall and minimize differential sticking. Cementing casing without centralizers results in ineffective cement job, Fig (14), since the cement slurry tends to preferentially flow in the large annular space with the least flow resistance. Centralizers should be spaced to provide a minimum standoff of 70%, where,

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Standoff = 2 ×

casing / hole clearance hole diameter - casing diameter Service companies and Drilling Engineering Div. have computer programs which calculate the centralizer spacing to provide a given stand off. There are many types of centralizers which depend on the purpose and the manufacturer. The three major types of centralizers used by Saudi Aramco are 1) rigid or positive centralizer, 2) bow centralizer and 3) spiral rigid centralizer.

The rigid (positive) centralizer shown in Fig (15) is assembled without any welding to minimize breakage. The Fig. 14 Effect of Centralization on Uniformity of Cement Placement centralizer is normally used in cased hole to provide nearly 100% standoff. The non-weld bow centralizer, Fig (16), is used to centralize casing in the open hole. The spiral rigid centralizer (SRC) is made of high grade aluminum and provide fixed standoff regardless of the lateral loads. The SRC, shown Fig (17), allows the pipe to rotate freely inside the centralizer and the spiral blades restrict the flow area and create swirling motion which assists in preventing channeling during cementing. The SRC acts as a bearing surface which reduces torque necessary for rotating casing while cementing and reduces drag while running casing in the hole. The SRC is used in highly deviated and horizontal wells. Stage-cementing tools are used for cementing the casing in two or three stages. Stage

cementing is used to lessen the possibility of breaking the formation during cementing and to cement the casing above existing lost circulation zones.

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Fig. 15 Non-Weld Positive Centralizer

Fig. 16 Non-Weld Straight Centralizer

Fig. 17 Spiral Rigid Centralizer

The stage tool most commonly used in Saudi Aramco consists of 3-4 ft inflatable packer element and cementing side ports above the packer. The stage tool is installed at a specific point in the casing as the casing is being run in the hole. Cement slurry is placed around the casing from the casing shoe to 100-200 ft above the stage tool (first stage). Refer to Fig (18). Once the first stage shutoff plug seats in the shut-off baffle, a freefalling opening plug is dropped down the casing. After the plug seats in stage tool, pressure is applied on the plug to move a sleeve and expose inflation ports of the inflatable packer. A specified pressure is applied and the packer is inflated by pumping mud into the packer rubber element which will make a seal with the outer casing. Higher pressure is applied which will rupture a disk and open the side ports above the packer. Mud is circulated through the ports to remove any cement above the packer. Cement slurry is then placed through the ports from the packer to the surface followed by a closing plug which closes a sleeve over the side ports. After the cement sets, the opening and closing plugs are drilled out with a bit and the stage tool is pressure tested to ensure the side ports are shut off.

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Fig. 18 Multiple-stage cementing tool with free-falling plug for cementing a hole in two stages

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PRIMARY CEMENTING The success of a primary cement job cannot be over emphasized. A poor cementing job can result in a failure to isolate zones and can be very costly during the producing life of a well. Failure to isolate between zones can result in 1) 2) 3) 4) 5)

ineffective stimulation treatment, improper reservoir evaluation, production of unwanted fluids, accumulation of gas in the annulus and casing corrosion.

PLANNING A CEMENT JOB

For a cementing job to be successful it must be properly planned. The important factors that must be considered in the planning of a cement job are a) b) c) d) e) f)

Type and volume of cement Cement additives Cement mixing Preflushing Cement placement Cement displacement

TYPE AND VOLUME OF CEMENT

The type of cement used by Saudi Aramco is Class ‘G’ cement which is manufactured locally to API specifications. The properties of the cement slurry can be tailored by the use of additives to meet the requirements of shallow and deep wells. One sack of dry Class ‘G’ cement weighs 94 lbs and yields 1.15 cubic feet of 118 pcf neat slurry when mixed with 5 gals of water. Addition of additives changes the weight (density) and yield of the cement slurry. Calculation of cement weight and yield were discussed on page (12). The weight of the cement slurry must be little heavier than the mud weight and should exert a hydrostatic pressure which is greater than the pore pressures of the formations to be cemented. A cement slurry of 118 pcf is used in Saudi Aramco’s normal cementing

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jobs. In cases where there is a concern that the cement column may fracture a weak formation, then a lighter weight or multiple weight cement may be used. The light weight slurry (101 pcf is normally used by Saudi Aramco) is pumped first as the lead slurry followed by the heavier weight cement tail slurry. The volumes of the cement slurries are designed such that the heavy cement occupies the annulus from the casing shoe to the weak formation and the lighter slurry from the weak formation to the surface. If there is still danger that the light cement may break the formation, then the casing should be cemented in two stages with the stage tool placed above the weak formation. The volume of cement required to cement the casing must be calculated by the drilling engineer prior to the cementing job. The cement volume is usually based on past experience and regulatory requirements in the area. As little as 300 ft cement fill up has been used behind relatively deep casing strings. In Saudi Aramco the practice is to cement the casing from the shoe to the surface wherever it is economically possible. The volume of cement is based on the size of casing and the diameter of the open hole. It is usually necessary to include more slurry than theoretically required because of hole enlargement while drilling. The excess factor to be used is based on prior experience in the area. Hole volume can be calculated from an open hole caliper. If a caliper is not available, the open hole volume can be estimated by pumping a marker and circulating it back to the surface. The following guidelines may be used in calculating the cement volumes for cementing casing and liners: Casing:

a) An excess factor of 10-15% is used for cementing casing inside casing. b) If the openhole volume to be calculated from caliper, an excess factor of about 20-30% may be used. If the openhole volume is not known, then the excess factor depends on hole enlargement and is determined by experience. It may range from 50 to 100%. c) When a stage tool is used, the excess factor should result in 200-300 ft of cement rise above the stage tool. Liners:

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a) If the openhole is known from a caliper, enough cement should be used to obtain a cement rise of not more than 500 ft above the liner hanger.

3-1/2" DP

300' liner hanger @ 6700'

7" 26# casing @ 7000' 4-1/2" 11.6# liner cement 6" open hole

latch collar @ 7520' TD = 7600'

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Example No. 1 - Calculating Cement requirements

b) If a caliper is not available the excess factor is determined by experience. In this case the engineer should make sure that the thickening time of the lead cement is large enough to allow circulating out any excess cement above the hanger and prevent cementing the drill pipe and liner hanger setting tool in place.

Example No 1

Calculate the number of sacks of 118 pcf ‘G’ neat cement required to cement a 4-1/2” 11.6# liner across the Arab-D as shown in Fig (19). The volume of the open hole from the caliper log is 135 ft3. Given:

7” casing ID = 6.276 in 4-1/2” liner ID = 4.0 in Solution

Volume of cement inside liner below latch collar =

π 42 4 × 144

× (7600 − 7520) = 6.97 ft 3

Volume of annulus between open hole and liner = open hole volume - liner volume or Vol. of annulus

= 135 −

π 4.52 4 × 144

× (7600 − 7000)

= 135 - 66.23 = 68.77 ft3 Vol. of annulus between liner and 7” casing

=

π (6.276 2 − 4.52 )

4 × 144 = 31.3 ft3

Volume of 300 ft rise above hanger between 3-1/2” DP and 7” casing,

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× (7000 − 6700)

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=

π

× (6.276 2 − 35 . 2 ) × 300

Vol.

4 × 144 = 44.4 ft3

Total vol. of cement slurry

= 6.97 + 68.77 + 31.3 + 44.4 = 151.44 ft3

Yield of 118 pcf neat cement

= 1.15 ft3/sk

Number of sacks

=

15144 . 115 . = 132 sacks

5" DP ID = 4.276" 6700' long

9-5/8" 40# csg @ 4000'

9-5/8" 40# casing @ 4000'

8-1/2" hole

6" drill collars 2" ID 300' long float collar 6920' 7" 26# casing @ 7000'

TD = 7000'

(a) Page 36

(b)

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Example No. 2 - Estimating Lead & Tail Cement Volumes

Example No 2

A 7” casing string is to be cemented inside an 8-1/2” hole section which is known to have severe hole enlargements across shale sections. Refer to Fig (20 b). Before the drill string was pulled out of the hole, the foreman pumped a dyed mud pill and circulated it out to the surface by pumping a total of 560 bbls of mud as shown in Fig (20 a). The 7” casing is to be cemented with 101 pcf 1.68 ft3/sk yield lead ‘G’ cement from the 9-5/8” shoe to the surface and 118 pcf neat ‘G’ cement from TD to the 9-5/8” casing shoe. Calculate the lead and tail cement volumes. Given:

9-5/8” casing ID = 8.835 in 7” casing ID = 6.276 in Solution

First we have to calculate the volume of the open hole. Since it took 560 bbls of fluid to circulate the dyed marker to surface, we can say that, Vol. of DP + Vol. of drill collars + Vol. of DC-open hole annulus + Vol. of DP-open hole annulus + Vol. of DP-casing annulus

=

560 bbls

=

3141 ft3

667.8 + 6.54 + Vol. of DC-open hole + Vol. of DP-Open hole + 1156

=

3141 ft3

Vol. of DC-open hole annulus + Vol. of DP-openhole annulus

=

1310 ft3

or

π × 4.267 2

π × 22

× 300 + Vol. 4 × 144 4 × 144 of DC-openhole + Vol. of DP-open hole + π (8.8352 − 52 ) × 4000 4 × 144

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× 6700 +

=

3141 ft3

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Since the volume of the annulus between the DP and drill collars and the open hole is 1310 ft3, The volume of the open hole = 1310 + outside vol. of DP + outside vol. of DC π 52 π (6 2 x 300) × (3000 − 300) + = 1310 + 4 × 144 4 x144 = 1736 ft3 If there were no enlargements in the 8-1/2” open hole, the volume of the hole would be, π × 8.52 Vol. of gauge hole = × (7000 − 4000) 4 x144 = 1181 ft3 The volume of the washout is Vol. of washouts

= =

1736 - 1181 555 ft3

Volume of tail slurry

π 6.276 2 × 80

Vol. of cement inside 7” casing below float collar

=

Vol. of cement between 7” casing and open hole liner

= Vol. of open hole - Vol. of 7”

4 × 144 = 17.2 ft3

= 1736 −

π 72 4 × 144

× (7000 − 4000)

= 934 ft3 Using an excess factor of 25% for the open hole, Vol. of cement Vol. of tail slurry Number of sacks

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= 934 × 1.25 = 1167 ft3 = 1167 + 17.2 = 1184.2 ft3 1184 = 1030 sacks = 115 .

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Volume of lead slurry

Vol. between 7” and 9-5/8” casing

=

π (8.8352 − 7 2 )

4 × 144 = 633 ft3

× 4000

Using excess factor of 10%, Vol.

= 633 × 1.1 = 697 ft3

Number of sacks

=

697 168 . = 415 sacks

CEMENT ADDITIVES

In designing the cement composition the drilling engineer should ensure that the cement contains additives to provide the proper slurry properties. The most important properties are the thickening time and fluid loss. Cement retarders should be added to the cement to provide enough thickening time to mix, pump and displace the cement. The type and amount of retarders are determined by the service cementing Fig. 21 Circulating Temperature vs Depth companies or Saudi Aramco as a function of the temperature Cement Lab based on the volume gradient of cement, type of cementing job, the static bottom hole temperature or the circulating temperature which are provided by the drilling engineer. The bottom hole circulating temperature can be measured with a temperature gauge run in the bottom of drill string while conditioning the mud. If measurement of the temperature is not

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possible it can be estimated by using Fig (21). The drilling engineer is also responsible for providing the cementing lab with a sample of water to be used by the rig in mixing the cement. Fluid loss additives should be added to the cement to control the fluid loss of the slurry when cementing casing or liners across permeable formations. Failure to control the fluid loss will cause cement dehydration and bridging in the casing-hole annulus and may result in total failure of the cementing job. The fluid loss of the cementing slurry should be low enough to keep the slurry pumpable during the entire cement job. The value of the fluid loss is determined by experience and depends on the permeability of the formation and size of the casing-hole annulus. The higher the permeability and the smaller the annulus the lower the fluid loss. General guidelines for fluid loss values are shown in Table (4). It should be noted that fluid loss additives are fairly expensive and should be used prudently. Other additives discussed on pages 12-20 such as accelerators, silica flour, dispersants and gas blocking agents, should be added to the cement as required.

Example

The static bottom hole temperature of an Arab-D well is 210 oF at 7000 ft. Estimate the bottom hole circulating temperature at 7000 ft. Solution

Assume the static temperature at the surface is 80 oF, then, Bottom hole static temp (BHST)

= 80 + Temp Grad × Depth

Temp Grad

=

or BHST − 80 Depth 210 − 80 = 7000 = 0.0185 oF/ft = 1.85 oF/100 ft

From Fig (21), the circulating temperature at 7000’ is 138 oF.

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CEMENT MIXING

The purpose of the mixing system is to proportion and blend the dry cement and additives with the water and supply to the wellhead a cement slurry with predictable properties.

Fig. 22 (a)

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Schematic of Jet Mixer

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Typical Jet Mixing Operation

The drilling engineer is responsible for selecting the type of cement mixer. There are three types of mixers available. The most widely used mixer is the jet mixer shown in Figs (22a & 22b). It consists of funnel-shaped hopper, a mixer bowl and mixing tub. The mixer forces a stream of water through a jet and into the bowl where it mixes with cement from the hopper to form a slurry. The slurry is forced into the discharge line then into the mixing tub from which it is taken to the cementing pumps. The density of the slurry is checked by taking samples from the mixing tub. Slurry density can be changed by varying the water / cement ratio.

Fig. 23 Page 42

Halliburton’s RCM System Flow Schematic

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The recirculating cement mixer (RCM) mixes more uniform slurry with more accurate density. The Halliburton RCM shown Fig (23) consists of a two-compartment 8-bbl mixing tub equipped with a turbine agitator in each compartment. The rate of dry cement is controlled by the cement control value. The water rate is controlled by the mixing manifold on top of the tub. The slurry is blended by the agitator in the first compartment on the right, recirculated by a centrifugal pump and weighed by a densometer. Any variations in density are corrected by the operator. When the first compartment is full, the slurry flows over a weir to the second compartment. Flowing over the weir helps remove entrained air. The second compartment already contains same slurry at the desired weight. The combined slurries are blended further by the agitator to insure a uniform mixture. The slurry is then pumped into the well. The batch cement mixer gives the most accurate density and is used for critical cementing jobs were good isolation between zones is essential. In Saudi Aramco the batch mixer is used for cementing short liners across the producing zones and in cement squeezing operations. The batch mixer consists of a tank and paddle mixers. A measured volume of mix water is placed in the mixer and the required amount of cement is added. The slurry is mixed with the paddles until the desirable density is obtained.

PREFLUSHING

Preflushes or spacers are pumped ahead of the cement to minimize mixing and gellation in the annulus between the cement and the mud and also aid in the removal of mud cake. Preflushes have various characteristics, depending on the mud system, and various functions. Some contain additives to thin the mud and penetrate and loosen the wall cake, some contain abrasive material to scour the hole; and some have high viscosity to remove drilling mud by buoyancy. Spacer composition is normally provided by the service company. For simple water base muds, water is an excellent spacer since it is cheap, easy to put into turbulence and has little effect on the setting of cement. The volume of the spacer should be 300-500 ft of annular fill or enough volume to give a 10-minute contact time except when the hydrostatic head of the cement column in the annulus is reduced excessively. Studies indicate that when turbulent flow is attained, a contact time of 10 minutes or more provides excellent mud removal.

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CEMENT PLACEMENT TECHNIQUES

Most of the primary cement jobs are performed by the displacement method, that is by pumping the cement slurry down the casing and up the annulus. Fig.s (24 a, b, c & d) below illustrate the four cement placement methods used by Saudi Aramco.

1. Displacement method

The surface, intermediate and production casings are usually cemented using the normal single stage displacement method. The cement is pumped down the casing through the casing shoe using a top and bottom wiper plugs as shown in Fig (24a) (bottom plug is not used in Saudi Aramco cementing operations). The displacement fluid, mud or water, is pumped behind the top plug until the plug bumps the float collar. Fig. 24 (a)

Normal Displacement Method

2. Stage Cementing

Stage cementing is performed by the displacement method in two or three stages. Refer to Fig (24b). It is commonly used in wells that require long columns of cement and where weak formations cannot support the hydrostatic head of the cement. Stage cementing is also used to cement casing above loss circulation zones. One disadvantage of stage cementing is that the casing cannot be reciprocated or rotated after the first stage. For more information see Page 29. Fig. 24 (b)

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Two Stage Cementing

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3. Inner String Cementing

Inner string cementing is used in Saudi Aramco for cementing large-diameter (18-5/8” and 24”) conductor casing. Tubing or drill pipe is used as an inner string to place the cement. A sealing adapter (stinger) is run on the bottom of the drill pipe and stung into a special inner-string float shoe. Cement slurry is pumped down the drill pipe followed by a small-diameter cementing plug. The cement is displaced with drilling fluid or water until the plug bumps, latches and seals in the float shoe. The drill pipe is then unstung from the float shoe and pulled out.

Fig. 24 (c)

Inner String Cementing

The advantage of the inner-string cementer is that it reduces the cementing time and the amount of cement left inside the casing. It avoids having to drill out great amount of cement that a large casing could hold if it were cemented in the conventional manner. Use of the inner-string cementer avoids having a wet shoe when cementing across a loss circulation zone which has a static bottom hole pressure less than water gradient such as the Neogene aquifer. 4. Annulus Cementing

Cement is pumped through tubing or smalldiameter pipe run between the casings or between casing and hole to bring top of cement to the surface. This is done when the cement during a primary cementing job does not reach the surface because of a shallow loss circulation zone or other reasons. This types of cement placement technique is called top cement job. To bring the top of cement above a shallow loss circulation zone to the surface may require performing several top jobs. The cement must be given enough time to harden after each job. Usually cement baskets are installed around the casing above the loss circulation zone to help

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Fig. 24 (d)

Outside Cementing

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achieve cement fill up to the surface. Example

An 18-5/8” conductor is to be cemented at 600 ft across the Neogene aquifer which has a loss circulation zone at 400 ft. The static fluid level of the aquifer is at 300 ft. Refer to Fig (25). a) What type of cement placement technique would you use? Explain why. b) What is the volume in sacks of 118 pcf Class ‘G’ cement that should be used? ID of casing is 17.755 in. Solution

a) Let’s try to use the standard displacement technique. In this method the cement will be pumped down the casing followed by the cementing plug. The plug will be displaced with water from surface until it bumps the float collar at 520 ft. Since there is a L.C. zone at 400’, the cement will not be able to rise in the annulus above 400 ft. Now let’s make pressure balance calculations after the plug bumps the float collar. Refer to Fig (26). Pressure inside casing at float shoe at 600’ is,

Pc

62.4 ⎛ 118 ⎞ + ⎜ 80 × ⎟ 144 ⎝ 144 ⎠ = 225 + 65 = 290 psi

= 520 ×

Pressure in annulus at 600’ is,

Pa

62.4 118 (400 − 300) + (600 − 400) 144 144 = 206 psi =

Since the casing pressure is greater than the annulus pressure, the water level in the casing will drop and displace the cement below the float collar into the annulus until the pressures inside the casing and annulus are equal. This will result in a wet shoe (no cement around the shoe) which is not acceptable. Refer to Fig (27). Therefore, in order to use the displacement method and have a cemented shoe we have to underdisplace the cement so that pressures inside and outside the casing are equal.

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24" Hole

Water

24" Hole 300' Fluid Level

18-5/8" Casing

Water LC zone @ 400'

LC zone @ 400' Cement Float Collar @ 520'

Float Collar @ 520'

600'

600'

Fig. 25 Example Problem

Fig. 26 Example Solution: Cement Plug bumped into Float Collar

24" Hole Water

Cement Baskets

195' 300' Fluid Level

300' Fluid Level

Water

325'

Water

Water

LC zone @ 400'

LC zone @ 400' Cement Float Collar @ 520' Water

Water 600'

Fig. 27 Example Solution: Normal Displacement resulting in a Wet Shoe

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520'

600'

Fig. 28 Example Solution: Normal Displacement with under displaced cement

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The column of water to be pumped is L ×

62.4 118 118 62.4 + 80 × + 100 × = 200 × 144 144 144 144 L = 325 ft of water

So we displace the cement by 325 ft of water. The water level will drop inside the casing until the pressures inside and outside the casings are equal as shown in Fig (28). There is no need to use a float collar or a cementing plug.

Pc

= 325 ×

62.4 118 + 80 × 144 144

= 206 psi

Pa

62.4 118 + 200 × 144 144

= 100 × =

206 psi

The annulus above the basket is cemented by the top job technique. b) The volume of cement required is equal to the volume inside the casing plus the volume in the annulus from the shoe to the loss circulation zone, or Vol. inside casing =

π 17.7552

4 × 144 = 137 ft 3

Vol. of annulus

=

π (24 2 - 18.6252 ) × (600 - 400) 4 × 144

= 250 ft

3

Using excess factor of 50% Vol. of annulus Total volume

Page 48

= = = =

× 80

250 x 1.50 375 ft3 375 + 137 512 ft3

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Number of sacks = =

512 115 . 445 sacks

DP 24" Hole

Water

300' Fluid Level Water

LC zone @ 400'

Cement

600' Float Shoe

Plug

Fig. 29 Example Solution: Inner String Cementing

A better way to cement the casing is by using the inner-string technique as shown in Fig (29). The cement is pumped in the drill pipe and followed by the cementing plug. The plug is displaced with water until it reaches and seats in the float shoe where it makes a seal and stops the water from entering the float shoe. After the plug seats in the float shoe, the drill pipe is pulled out leaving the plug on bottom. This method provides good cement around the shoe with very little cement inside the casing. The volume of cement to be used is, Vol.

=

π (24 2 - 18.625 2 ) × (600 - 400) 4 × 144

= 250 ft

3

Using excess factor of 50%,

Vol.

=

250 × 1.5 = 375 ft 3 or

= 326 sacks

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375 1.15

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CEMENT DISPLACEMENT

The displacement of the mud in the casing-hole annulus by the cement slurry is the most important and critical phase of the primary cementing job. A successful cement job is one in which the cement slurry displaces all the mud from the casing-hole annulus. The predominant cause of cement job failure is channels of gelled mud remaining in the annulus after the cement is in place. Laboratory and field research show that the factors discussed below all contribute to the success of the cement job during the displacement period. Pipe centralization creates a uniform annular flow area and aids mud displacement in the

annulus. Centralizers do not provide perfect casing-hole concentricity. Flow pattern in an eccentric annulus is not uniform and the highest velocity occurs in the side of the hole with the largest clearance as shown in Fig (14). If the casing is close to the wall as in Fig (14), it may not be possible to pump the cement at a rate high enough to develop uniform flow throughout the annulus. Casing strings should be centralized to achieve at least 70% standoff. Pipe movement, either rotation or reciprocation, is a major driving force for mud removal

and should be used during primary cementing wherever possible. During rotation, (15-25 rpm) cement-casing drag forces are more effective than during reciprocation, as they tend to pull the cement into the bypassed mud column instead of along side it as shown in Fig (30). Centralizers that allow pipe rotation should be used if the casing is to be rotated. Reciprocating the casing in 20 ft strokes can cause changes in standoff as centralizers move across wellbore irregularities. Care must be taken not to swab in the well during casing reciprocation. This lateral movement alters the flow area and aids in the displacement of bypassed mud.

Fig. 30

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Rotational displacing drag forces aids in the removal of by-passed mud in the narrow side of an eccentric annulus

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Conditioning the mud before cementing to reduce the gel strength, plastic viscosity and

yield point greatly improves the displacement efficiency of the mud. It also reduces displacement drag forces to erode and remove bypassed mud by reducing casing to mud and wellbore to mud drag forces. The mud should be circulated and conditioned until the desired gel strength and viscosity are obtained. High displacement rates promote mud removal and improve displacement efficiency if

cement can be pumped in turbulent flow up the annulus. Conditions that may prevent turbulent flow include limited pumping capability and formation conditions that may limit downhole pumping pressures. Dispersants can be added to the cement to lower friction pressure and attain turbulent flow at lower displacement rates. If turbulent flow is difficult to obtain then the cement should be pumped in plug flow (Reynolds number of 100-200). Displacement rates required to induce turbulent or plug flow can be calculated by using flow equations presented in the Flow Calculations section.

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LINER CEMENTING A liner is a casing string that is usually used to case-off the open hole below an existing casing string, and which does not extend to the surface. In Saudi Aramco 4000-5000 ft 7” liners are set at the top of the producing Arab-D formation and extend to 200 ft above the 9-5/8” x 13-3/8” DV stage tool as shown in Fig (31). Sometimes a short 4-1/2” liner is also set across the Arab-D formation to shut off water producing zones and super permeability stringers. Liner cementing is one of the most difficult operations associated with drilling and completion. If a liner is not effectively cemented, the well’s capability to produce will likely be reduced and the advantages of liner installation will not be

26” CONDUCTOR

DV TOOL 18-5/8” CASING

DV TOOL 13-3/8” CASING

9-5/8” CASING

realized.

Page 52

oil

7” LINER

water

4-1/2” LINER

ARAB-D

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A liner is normally run on drill pipe that extends from the liner setting tool to the surface. Special tools are available to perform various running, setting and cementing operations. The following equipment is discussed from the float shoe to the cementing manifold. Equipment locations are shown in Fig (32). A float shoe is placed at the bottom of the liner. It contains a check valve to prevent back flow of the cement. A float collar can be installed 2 joints above the shoe to provide a backup check valve to assure that cement cannot re-enter the liner after displacement. A latch collar is run one joint above the float collar or two joints above the float shoe to provide space for mud contaminated cement inside the liner. The latch collar’s function is to latch and seal the liner wiper plug. It prevents the wiper plug from moving uphole if a check valve fails and also it prevents it from rotating, which aides in the drilling out operation.

Fig. 32 Typical Equipment used to Install and Cement a Liner

Page 53

The liner length is selected to extend across the open hole to overlap the existing casing or stage tool. The length of the overlap should provide good cement seal in the liner-casing annulus. A 300 ft overlap is normally used in Arab-D wells and 400-500 ft overlap is used in deep high pressured wells.

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An external casing inflatable packer is run to cement the liner in two stages, if required. The packer is run with a DV tool port collar set above super permeability or loss circulation stringer to allow placement of cement above the loss circulation zone. The cementing job is performed by pumping the first stage cement first and dropping the drill pipe plug. The cement and plug are displaced with drilling fluid until the DP plug latches in the liner wiper plug which is attached at the bottom of the liner setting tool. Pressure is applied to break the shear pins and release the wiper plug. Displacement is continued until the wiper plug and DP plug latch and seal in the latch collar. Pressure is applied to open the packer inflation ports. The packer is then inflated with mud or cement to provide a seal with the open hole above the loss circulation zone. Higher pressure is applied to open the DV ports and cement is pumped above the packer to the top of liner. The DV ports are closed by the DV shut off plug (not shown in Fig 32). The liner hanger is installed at the top of the liner. The function of the hanger is to hang and suspend the liner. There are mechanical and hydraulic set hangers. Mechanical hangers require manipulation of the drill pipe to engage the hanger slips with the casing. Slips of the hydraulic hanger engage the casing by applying pressure. Hydraulic hangers are used in deep or highly deviated wells where it is difficult to transmit pipe manipulation to the hanger because of excessive friction between drill pipe and casing. A mechanical liner hanger is shown in Fig (33). Liner hangers may be equipped with a liner packer installed at top of the hanger to seal between liner and casing after cement placement. Seal elements may be rubber, lead or both and are set by applying weight on the hanger. After the packer is set excess cement above the hanger can be circulated out without imposing high pressure on the formation below. Fig. 33

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Mechanical Liner Hanger

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The liner setting tool, which is a rental item furnished by liner hanger service company, provides the connection between the drill pipe and the liner hanger. A packoff bushing and slick joint are inserted into the liner to provide a seal between the setting tool and the liner. Once the hanger is set and the liner is cemented, the setting tool is released from the hanger and pulled out to surface. The liner wiper plug is attached at the bottom of the setting tool with shear pin arrangement. The function of the plug is to separate the cement from the displacing fluid. Once the plug latches in the latch collar it provides a pressure seal and allows pressure testing the liner above the latch collar or inflating the external casing packer. The drill pipe plug is dropped in the drill pipe after all the cement has been pumped. The plug is displaced by the drilling fluid until it latches with the liner wiper plug at the bottom of the setting tool. Higher pressure is applied to shear the wiper plug from the setting tool and both plugs are displaced with drilling fluid until they latch in the latch collar. Cementing Techniques

A good liner cement job is one that allows drilling to the next casing point without having to squeeze the liner shoe or the top of the liner. The cement job should also provide an effective seal (if required) between the open hole and liner such that remedial cement squeeze jobs are not required. Studies have shown that 1”-1.5” clearance between the liner and open hole resulted in successful liner cement jobs. A way to increase clearance would be to drill larger holes or run smaller liners. Another solution is to underream the open hole. Centralizing the liner is very essential for a successful cement job. The centralizers should be spaced out to provide at least a 70% stand off. Liner movement during cementing greatly enhances cement placement efficiency and

displacement of mud in the liner hole annulus. Special liner hanger equipment is now available which permit liner reciprocation or rotation while cementing. The maximum liner length that can be suspended and rotated below a rotating liner hanger should be confirmed with the manufacture. Short liners run across the Arab-D in Saudi Aramco are rotated while cementing. If it is not possible to rotate liner while cementing, the liner should be rotated while circulating and conditioning the mud. Wide temperature variations across a long liner require special cement formulation. It may be necessary to retard the cement to compensate for high temperatures at the bottom. But, at the same time, it is necessary that the cement set at the lower temperature at the top of the liner in a reasonable time. The cement thickening time should be long enough to allow reversing out the cement above the hanger if required.

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Fluid loss additives are usually required especially across permeable formations to prevent cement dehydration and bridging in the annulus. Fluid loss of 50-100 ml is recommended for cementing a production liner where isolation behind the liner is required. If isolation behind the liner is not essential, the amount of fluid loss should be low enough to prevent cement dehydration and bridging and is based on experience and economic considerations.

Cement volume for a liner cementing job should be based on the caliper log volume and an excess amount that will result in a rise of less than 500’ above the hanger. This excess serves to remove the mud from the overlap and allows high quality cement to take its place. Excessive cement columns above the hanger may prevent pulling out the setting tool out of the cement. If a caliper survey is not available the hole volume may be estimated by pumping a marker. As in the case of casing cementing, after the liner is run to bottom the mud should be circulated at least one hole volume and conditioned until low plastic viscosity and yield values are obtained. It is recommended that the cement be batch mixed to obtain the correct slurry density. Spacers should be used ahead of the cement and the cement should be pumped in turbulent or plug flow if possible. Laminar flow should be avoided. It is recommended that the cement be pumped by using the service company cement unit to insure displacement accuracy.

Running And Cementing Procedure

A step by step procedure for running and cementing a liner is outlined below. 1)

With the bit on bottom circulate and condition mud. Make short trip and circulate out any fill.

2)

Make up and run the liner with float shoe on bottom and latch collar 2-4 joints above shoe.

3)

Centralize the liner to provide at least 70% stand off. Fill liner with drilling fluid while running in hole.

4)

Make up hanger on top of liner. Install setting tool in hanger and run liner on drill pipe. Rabbit each stand of DP while RIH. Fill drill pipe before entering openhole.

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5)

In deep wells, stop and break circulation at least once half way in the open hole. Prior to circulating, reciprocate liner to break gel strength. Start slowly with low pump pressure. If hydraulic liner hanger is used, circulating pressure should be well below the setting pressure of the hanger.

6)

If rotating hanger is used, rotate liner prior entering open hole and record torque.

7)

Tag bottom and install cementing head. Reciprocate and break circulation at low rate and pressure. Circulate at higher rate at least one hole volume. Record weight of liner and drill pipe.

8)

Set hanger. Slack off weight on the hanger to ensure hanger is set.

9)

Release setting tool from hanger but keep setting tool stung in hanger. Set 10,000 lb weight on hanger while cementing.

10) Batch mix cement to the programmed weight. Use retarders, fluid loss and other additives as required. Measure slurry weight using pressurized mud balance. 11) Pump spacer followed by cement. 12) Drop drill pipe plug. Pump drilling fluid and displace plug and cement using service company pumps. Slow down pumping rate when DP plug approaches the liner wiper plug. Continue displacing cement until DP plug latches and shears liner wiper plug. Note increase in pressure on surface. Continue displacing cement until liner wiper plug and DP plug latch into latch collar. Reduce pumping rate before latching into latch collar and note increase in surface pressure when bumping the latch collar. 13) Bleed pump pressure and check for flow back. Pull out setting tool to top of hanger and reverse out excess cement above hanger. If reversing out is not required, pull out setting and observe U-tube effect. 14) Wait on cement until 500 psi compressive strength is developed. 15) Run with bit and drill cement to top of hanger.

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FLOW CALCULATIONS The flow properties of wellbore fluids are classified as Newtonian or non-Newtonian. Newtonian fluids are fluids such as oil or water which exhibit a direct and constant proportionality between shear rate (which related to velocity or flow rate) and shear stress (which is related to flowing pressure drop) as long as the regime is laminar. In a fluid of this type, viscosity is independent of shear rate at constant temperature and pressure. A Newtonian fluid will begin to flow immediately when pressure is applied. When pressure is released, the fluid returns to its previous state. See Fig (34).

Fig. 34 Flow-Rate / Shear-Stress Curves of Newtonian and non-Newtonian Fluids

Non-Newtonian fluids are fluids like mud, cement slurries and heavy asphaltic oil. These are rheologically complex and are described as Bingham plastics or power-law fluids. These fluids do not exhibit direct proportionality between shear rate and shear stress. Some types of non-Newtonian fluids, such as drilling fluid, do not start to move when a force is applied. The two mathematical models commonly used to describe the flow behavior of drilling fluids and cement slurries are the Bingham-plastic model and the power-law model.

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Bingham Plastic Model

This model assumes that all rheological calculations can be made from a linear relationship between shear rate and shear stress. This relationship can be obtained from a rotational viscometer or Fann VG meter. The cement slurry is sheared at a constant rate between an inner bob and an outer rotating sleeve. Six rotation speeds (600, 300, 200, 100, 6, and 3 rpm) are available. Two parameters are required to characterize fluids that follow the Bingham plastic model. These parameters are the plastic viscosity and the yield point. The plastic viscosity, µp, in centipoise is computed using Eq (2).

μ p = θ 600 - θ 300 ............…..................................................... (2) where θ 600 and θ 300 are the dial reading at 600 and 300 rpm. The apparent viscosity is,

μ a = 300 θ N ......................................................................... (3) N

The yield point, Y, in lb/100 sqft is computed using Eq (4).

τy =

θ 300 - μ p ....................................................................... (4)

The power-law model is based on the assumption that the cement slurry exhibits a proportionality between the logarithm of pressure loss and the logarithm of shear rate as shown in Fig (35). The equations of the power-law model are more complex but also more accurate than those of the Bingham plastic model. Fig. 35 Power-law Plot of non-Newtonian Slurry

The equation for the power-law

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is, S s = K ( S r ) n......................................................................... (5) where, K n Ss Sr

= = = =

intercept of lines (fluid consistency index) lb secn/ft2 slope of shear rate / shear stress line (flow behavior index), dimension less shear stress, lb/ft2 shear rate, 1/sec

Equation (5) can be written in the form, Log S s = log K + n log S r .......................................................... (6) The constant K is related to viscosity of the fluid, the larger K value the more viscous the fluid. The constant n characterizes the degree of the fluid’s non-Newtonian behavior. The valve of n for a non-Newtonian fluid is less than one and for a Newtonian fluid is one. In placing cement slurry down hole the preferred method, where hole conditions permit, is to thin the cement slurry and mud so that turbulent flow is induced. Turbulent flow will result in high displacement efficiency and increases the probability of moving mud from hole restrictions and reduces contamination of the cement sheath. Since the power-law model yields more accurate results than the Bingham plastic model, a summary of the frictional pressure loss equations for the power-law model in turbulent flow is presented below. The n and K constants are determined from the Fann viscometer as, n

=

⎛ θ 600 ⎞ 3.32 log ⎜ ⎟ ..........................................…............. (7) ⎝ θ 300 ⎠

K

=

510 θ 300 (511)n

.................................….................................

(8) where the unit of K is equivalent centipoise or dynes per secn per 100 cm2.

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For pipe flow, the Reynolds number is calculated from the equation, 89100 ρ v 2 − n K

NRe =

n

⎡ 0.0416d ⎤ ⎢⎣ 3 + 1 / n ⎥⎦ .............…............................(9)

where, v = average pipe velocity =

q 2.448d 2

ft / sec

d = inside diameter, in q = flow rate, gpm ρ = density, (lb/gal) The Reynolds number is plotted versus the Fanning friction factor in Fig (36). The critical Reynolds number, above which the flow is turbulent, is a function of the flowbehavior index n. It is recommended that the critical Reynolds number for a given n value be taken from Fig (36) as the starting point of the turbulent flow line for the given n value. For example, the critical Reynolds number for n = 2 is 4200. For annular flow, the Reynolds number is, 109000 ρ v 2 − n ⎡ 0.0208 (d2 - d1) ⎤ ⎢⎣ ⎥⎦ ..........…...............(10) K 2 +1/ n n

NRe = and the mean velocity is,

v

= average annular velocity q ..............................….......…..........(11) = 2.448 ( d2 2 - d12 )

where, d2 = inside diameter of hole, in d1 = outside diameter of pipe, in For laminar pipe flow, the frictional pressure loss is, n⎛ 3+1/

ΔP =

Page 61

n⎞ LKv ⎜ ⎟ ⎝ 0.0416 ⎠ 144000 d 1 + n

n

.....................................…..........(12)

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Fig. 36

Friction Factors for Power-law Fuild Model

SEGMENT

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For laminar annular flow, the frictional pressure loss is, n⎛ 2 +1/

ΔP =

n

n⎞ LKv ⎜ ⎟ ⎝ 0.0208 ⎠ ..................…...….......…..........(13) 144000 (d2 - d1) 1 + n

where, L = length of pipe or annulus, ft ΔP = pressure loss, psi For turbulent pipe flow, the frictional pressure loss is, ΔP =

L f ρ v2 ...............................….............................. (14) 2 5 .8 d

For turbulent annular flow, the frictional pressure loss is, ΔP =

L f v2ρ .....................................…..................... (15) 21.1 (d 2 - d 1)

where, f = fanning friction factor from Fig (36), dimensionless.

Example

A 600 ft 4-1/2” liner is to be cemented inside a 6-1/4” hole across the Arab-D reservoir to shut off production from the super permeability zone. A 118 pcf Class ‘G’ cement slurry with fluid loss additive will be used to cement the liner. In order to have good cement job that will provide isolation between the super permeability zone and the top of the ArabD, the liner will be rotated while cementing and the cement is to be pumped and displaced in turbulent flow. a) At what rate the cement should be pumped in order to have turbulent flow?, b) calculate the annular pressure loss across the liner. The Fann meter dial readings were measured by Halliburton as θ 600 = 209 and θ 300 = 122.

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Solution

a)

⎛ θ 600 ⎞ 3.32 log ⎜ ⎟ ⎝ θ 300 ⎠ 209 = 3.32 log 122 = 0.776

n

=

K

=

510 θ 300 511n

510 × 122 5110.776 = 492

=

From Fig (36), for n = 0.776 turbulent flow will start at NRe of 2000. Using Eq (10), we set the value of NRe = 2000 and solve for the average velocity v, or 2000 =

109000 ρ v 2 − n ⎡ 0.0208 (d2 - d1) ⎤ ⎥⎦ ⎢⎣ K 2 +1/ n

d1

= 4.5 in

d2

= 6.25 in

ρ

=

118 7.48 = 15.7 ppg

109000 × 15.7 v 2 − 0.776 2000 = 492 2000 = 3478 v1.224 × 0.0304 v1.224 = 18.91 v = 11.04 ft/sec

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n

⎡ 0.0208 (6.25 - 4.5) ⎤ ⎥⎦ ⎢⎣ 2 + 1 / 0.776

0.776

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From Eq (11),

v

=

q

2.448 ( d2 2 - d12 )

Substituting the value of v and solving for q, q

= = =

2.448 (6.252 - 4.52) 11.04 508 gpm 12.1 BPM

In order to have turbulent flow in the annulus the rate should be 12.1 BPM or greater. b) From Eq (15), the frictional pressure drop in the liner-hole annulus is, ΔP =

L f ρ v2 21.1 (d 2 - d 1 )

From Fig (36), for NRe = 2000 and n = 0.776, the valve f is 0.0105. Substituting in Eq (15), ΔP

600 × 0.0105 × 15.77 × 1104 . 2 = 211 . (6.25 - 4.5) = 328 psi

Problem

Using the same slurry in the previous example, calculate the minimum rate to achieve annular turbulent flow for cementing 7” liner inside 8.5” hole.

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TABLE OF CONTENTS

Page

INTRODUCTION

1

CAUSES OF KICKS

1

HOLE NOT FULL OF MUD SWABBING DURING A TRIP INSUFFICIENT MUD WEIGHT SPECIAL SITUATIONS THAT REQUIRE EXTRA CARE

2 2 3 3

- LOSS CIRCULATION - DRILL STEM TESTING - DRILLING INTO AN ADJACENT WELL - EXCESSIVE DRILLING RATE THROUGH A GAS SAND

3 3 3 4

SURFACE WARNING SIGNALS

4

VOLUME OF MUD TO KEEP HOLE FULL ON A TRIP IS LESS GAIN IN PIT VOLUME INCREASED FLOW FROM ANNULUS SUDDEN INCREASE IN PENETRATION RATE CHANGE IN PUMP SPEED OR PRESSURE GAS-CUT MUD

4 4 4 5 5 5

- DRILLED GAS - TRIP OR CONNECTION GAS - GAS FLOW

6 6 6

DRILLING A WATER SAND ACTION WHEN SURFACE WARNING SIGNAL IS RECOGNIZED

7 7

WELLBORE MECHANICS PRESSURE RELATIONSHIPS INFLUX BEHAVIOR

8 8 11

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TABLE OF CONTENTS UNSOUND WELL CONTROL TECHNIQUES CALCULATION OF KILL MUD WEIGHT CALCULATION OF DOWNHOLE FAILURE PRESSURE CIRCULATION RATE FOR KILL OPERATION

DRILL PIPE PRESSURE METHOD PUMP PRESSURE SCHEDULES FOR WELL CONTROL

Page 12 13 14 15

15 17

KICK IDENTIFICATION

20

ANNULAR PRESSURE PREDICTION

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INTRODUCTION Blowouts or the uncontrolled flow of oil and gas have long been a serious problem in drilling operations. If blowouts are not quickly brought under control they can cause casualties, environmental damage and economic losses. For these reasons, drilling personnel must be able to recognize the warning signs of potential blowouts, to plan effective well killing operations, and to take positive action to control the well. The catastrophic problems associated with uncontrolled blowouts can be avoided if the ”kick” is detected before a large volume of formation fluid enters the wellbore. The purpose of this chapter of Well Control is to explain why and how wells “kick” and to describe in detail the recommended method of well control.

CAUSES OF KICKS A kick is the entry or influx of formation fluids from a permeable formation into the wellbore. The goal of well control operations is to prevent a well kick from becoming a blowout (uncontrolled flow of formation fluid). An understanding of why wells kick coupled with the ability to recognize and evaluate the surface warning signs that indicate possible kick occurrence will substantially increase the probability of successfully controlling the well. The two conditions that must be present in the wellbore for a kick to occur are: 1) the pressure inside the wellbore at the face of the kicking formation must be less than the pore pressure of the formation, and 2) the kicking formation must have sufficient permeability to allow flow into the wellbore. Since formation permeability cannot be controlled, drilling personnel should utilize all techniques at their disposal to ensure that the pressure inside the wellbore is always greater than the formation pressure.

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Most kicks occur when one or more of the following conditions exist: 1. HOLE NOT FULL OF MUD Failure to keep the wellbore full of mud while pulling out of the hole on a trip has been the primary cause of 50-70% of all industry blowouts. As the drill string is pulled from the hole, the mud level will drop due to the volume of metal being removed. As the mud level drops, the hydrostatic pressure exerted by the mud column is reduced since the height of the column decreases. A decrease in mud column height can also be caused by down hole seepage losses. If no mud is added to the hole as the pipe is pulled, it is possible to reduce the hydrostatic pressure of the mud column to less than the formation pressure. When this happens, a kick can occur. To prevent this loss in hydrostatic pressure, it is only necessary to fill the hole on a regular schedule or continuously with a trip tank to replace the metal volume of the pipe being pulled, and to replace the mud lost through seepage. The metal volume of the pipe being pulled can be calculated, but the mud additions necessary to replace the seepage losses can be predicted only be comparison to the mud volume schedule required to keep the hole properly filled on previous trips. For this reason, it is imperative that a record of mud volume required versus the number of stands pulled on every trip be maintained on the rig. 2. SWABBING DURING A TRIP Even when the well is completely full of sufficiently heavy mud, the pressure in the wellbore opposite a permeable formation can be reduced by the swabbing action of the drill string. This wellbore pressure reduction could allow small volumes of formation fluid to feed-in during the time the pipe is in motion. Swabbing can cause the well to begin flowing since the hydrostatic pressure of the mud column is reduced by the formation fluid. Some loss in hydrostatic mud pressure due to swabbing cannot be avoided; however, the pressure reduction should not exceed the overbalance pressure of the mud column. Swabbing pressure is a function of pipe-pulling speed, mud properties, and annular clearances.

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3. INSUFFICIENT MUD WEIGHT The hydrostatic pressure exerted by the column of mud is the primary means of well control. If this hydrostatic pressure is equal to or greater than the pore pressure of all permeable formations exposed in the open hole, the well cannot flow. Kicks caused by insufficient mud weight are most prevalent when drilling exploration wells in abnormal pressure areas; however, this type of kick can also occur in development drilling because of “charged formations”. In a charged formation, the pore pressure has been increased by previous drilling or production operations and is not naturally occurring. Injection operations, casing leaks, poor cement jobs, improper abandonments, and previous underground blowouts can produce charged formations. 4. SPECIAL SITUATIONS THAT REQUIRE EXTRA CARE A) Loss of Circulation Loss of circulation can cause the mud level in the hole to drop. This reduction in height of the mud column reduces the hydrostatic well bore pressure and could result in a well kick if the wellbore pressure becomes less than the pore pressure of a permeable formation. B) Drill Stem Testing A drill stem test is performed by setting a packer above the formation to be tested, and allowing the formation to flow. Down hole chokes can be incorporated in the test string to limit surface pressures and flow rates to the capabilities of the surface equipment to handle or dispose of the produced fluid. During the course of the test, the borehole or casing below the packer and at least a portion of the drill pipe or tubing is filled with formation fluid. At the conclusion of the test, this fluid must be removed by proper well control techniques to return the well to a safe condition. Failure to follow the correct procedures to kill the well could lead to a blowout. C) Drilling into an Adjacent Well Frequently, a large number of directional wells are drilled from the same offshore platform or onshore drilling pad. If a drilling well penetrates the production string

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of a previously completed well, the formation fluid from the completed well will enter the wellbore of the drilling well, causing a kick. This is an extremely dangerous situation and could easily result in an uncontrolled blowout. D) Excessive Drilling Rate Through a Gas Sand Even if the mud weight in the hole is sufficient to control the gas sand pressure, gas from the drilled cuttings will mix with the mud. Excessive drilling rate through a gas sand can supply sufficient gas from the cuttings to reduce the hydrostatic pressure of the mud column through a combination of density reduction and mud loss from “belching” to the point that the formation will begin flowing into the wellbore resulting in a kick.

SURFACE WARNING SIGNALS The preceding discussion has shown how kicks can occur. The next subject for consideration is how to recognize and interpret the surface warning signs that give notice that a kick is or may be about to occur. Every effort must be made to recognize these signals and to evaluate down hole conditions promptly so that proper action can be taken. 1. VOLUME OF MUD TO KEEP THE HOLE FULL ON A TRIP IS LESS THAN CALCULATED OR LESS THAN TRIP BOOK RECORD This condition is usually caused by formation fluid entering the wellbore because of the swabbing action of the drill string. As soon as swabbing is detected, the drill string should be run back to bottom, bottoms up circulated, and the mud conditioned to minimize further swabbing. It may be necessary to increase the mud weight, but this should not be the first step considered because of the potential problems of lost returns or differential sticking. 2. GAIN IN PIT VOLUME A gain in the pit volume not caused at the surface is a positive sign that a kick is occurring. As the formation fluid feeds into the wellbore, it causes more mud to flow from the annulus than is pumped down the drill string, thus the pit volume increases.

3. INCREASED FLOW FROM ANNULUS Page 4

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If the pump rate is held constant, the flow from the annulus should be constant. If the annulus flow increases without a corresponding change in pump rate, the additional flow is caused by formation fluid feeding into the wellbore. 4. SUDDEN INCREASE IN PENETRATION RATE A sudden increase in penetration rate (drilling break) is usually caused by a change in the type of formation being drilled; however, it may also signal an increase in permeability. Faster penetration rates due to increasing pore pressures are usually not as abrupt as drilling breaks, but they can be. In order to be sure that the gradual increases in pore pressure are recognized, a penetration rate versus depth curve is usually required to highlight the trend of increasing pressures. 5. CHANGE IN PUMP SPEED OR PRESSURE Fluid entry into the wellbore causes a gradual decrease in pump pressure, accompanied by an increase in pump speed. Since the light formation fluid continues to flow into the wellbore, the hydrostatic pressure exerted by the annular column of fluid decreases, and the mud in the drill pipe tries to U-tube into the annulus. When this occurs, the pump pressure begins to drop and the pump speeds up. The lowering of pump pressure and increase in pump speed is also characteristic of a hole in the drill string commonly referred to as a washout. Until a determination can be made whether a washout or kick has occurred, a kick must be assumed. 6. GAS-CUT MUD Gas-cut mud often occurs during drilling operations and can be considered one of the early warning signs of a potential kick; however, it is not a definite indication that a kick has occurred or is impending. An essential part of analyzing this signal is being able to determine the downhole conditions causing the mud to be gas-cut. Gas-cut mud occurs as a result of one or more of the following downhole conditions:

A) Drilling a gas-bearing formation with the correct mud weight in the hole. B) Swabbing while making connections or making a trip. Page 5

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C) Feed-in of gas from a formation having a pore pressure greater than the hydrostatic pressure exerted by the mud. A) Drilled Gas When the hydrostatic pressure exerted by the mud is greater than the pore pressure of a gas-bearing formation being drilled, there will be no feed-in of gas from the formation. Nevertheless, gas from the drilled cuttings will mix with the mud causing the mud returns to be cut. As gas at bottom hole conditions is circulated up the annulus, it expands very slowly until just before reaching the surface. The gas then undergoes a rapid expansion that results in the mud weight being reduced considerably upon leaving the annulus. In some cases this reduction in mud weight can be quite extreme, but it may not mean that a kick is about to occur. Normally, only a small loss in bottom-hole pressure occurs because the majority of gas expansion and concurrent mud weight reduction occurs in the extreme top portion of the hole and an adequate mud weight is still maintained in most of the hole. Quite often when the drilled gas reaches the surface, the annular preventer must be closed and the mud circulated through the open choke manifold. This prevents the expending gas from “belching” mud through the bell nipple. If this “belching” is permitted, the hydrostatic head will be reduced due to loss of mud from the hole. B) Trip or Connection Gas After circulating “bottoms-up” following a trip or connection, a higher level of gas entrained in the mud returns may cause a short duration density reduction or gas unit increase. If the well did not flow when the pumps were stopped during the trip or connection, it can be reasonably assumed that the gas was swabbed into the wellbore by the pipe movement. These two values can indicate increasing formation pressure when compared with previous trips and connections.

C) Gas Flow

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Feed-in from a gas sand while drilling is a serious situation. The formation pore pressure must exceed the hydrostatic pressure of the mud plus the circulating friction losses in the annulus for gas from the formation to flow into the wellbore while drilling. Once feed-in begins, continued circulation without the proper control of surface pressures may induce additional flow, since the density of the hydrostatic column (annulus) is continually lessened by the flow of formation fluid. 7. DRILLING A WATER SAND When a permeable water sand having a pore pressure greater than the mud hydrostatic pressure is encountered, water will feed into the wellbore. Depending upon the pressure differential between the formation and the mud, feed-in may be detected by the appearance at the surface of lower density mud returns, by a gain in mud pit volume, and by a change in chlorides. Cut mud can also result from swabbing or drilling a water-bearing formation. When a permeable water sand is drilled, water, like gas and oil, will mix with the mud. Under these circumstances, the mud returns are cut by the solution gas in the water rather than by the water itself.

ACTION WHEN SURFACE WARNING SIGNAL IS RECOGNIZED If any of these Surface Warning Signals appear, an immediate Flow Check must be made to determine if a kick is occurring. FLOW CHECK PROCEDURE 1. 2. 3. 4.

Pick up the kelly until tool joint clears rotary table. Shut down the mud pump. Set the slips around the drill pipe and initiate rotation. Observe the flowline to see if mud stops flowing from the annulus.

Unless mud continues to flow from the annulus, drilling can usually be resumed without increasing the mud weight. If the mud continues to flow, even slightly, the well should be shut in and checked for pressure.

Early kick recognition, followed immediately by steps to kill the well, is the key to successful well control. Once feed in is detected and the well is shut in, the formation fluid must be properly circulated out before resuming normal operations. A complete understanding of the Page 7

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pressure relationships in the wellbore during killing operations as well as the correct operational procedures to kill the well are essential.

WELLBORE MECHANICS The primary objective in circulating out a kick is to establish and maintain a constant bottom hole pressure slightly greater than the pressure of the kicking formation during the entire well killing operation. This is necessary to prevent additional feed in from occurring while the kick is being circulated out and to minimize the pressure imposed on the casing or formations in the open hole. As a basis for considering well control methods, this section contains a review of the pressure relation ships that exist in the wellbore and the behavior of the different fluids that can be present.

520 PS I 735 PSI D R ILL PIPE

10# / G AL MUD

C ASIN G

10#/G AL. MUD

20 B B L. G AS EN TR Y

10,000 ft

B H P 5720 PSI FO R M ATIO N 5720 PSI

Fig. 1 Well shut in on a kick

A) Pressure Relationships

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After a kick is detected and the well shut in, no more formation fluid can flow into the wellbore after pressures stabilize (except in the case of an underground blowout). At this point, the well is in a pressure balanced condition. Figure 1 is a schematic diagram of a well shut in on a kick. In this well, a 20 barrel gas influx occurred while drilling at 10,000 feet with a 10 ppg mud. The stabilized shut in pressures were 520 psi on the drill pipe gage and 735 psi on the casing or annulus gage.

Fig. 2 Well Kick Static Pressure Diagram

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To understand how the various pressures interact in the wellbore, it is necessary to isolate and identify each one. Figure 2 is a representation of the pressures acting in the example problem. On the drill pipe, the gage pressure and hydrostatic pressure of the mud act in a downward direction. These pressures are exactly balanced by the formation pressure acting upward. Drill pipe gage pressure + Hydrostatic pressure of the mud

=

Formation pressure

Casing gage pressure + Hydrostatic pressure of the annulus mud + Hydrostatic pressure of the influx =

Formation pressure

This same pressure balance can be made for the annulus.

The casing pressure is greater than the drill pipe pressure because the gas influx does not exert as much hydrostatic pressure as the same height of mud in the drill pipe. The drill pipe provides a direct link to the bottom-hole pressure since the gage pressure can be read at the surface and the drill pipe should contain only mud of known weight. The drill pipe usually has a float (back pressure value) above the bit which will prevent the influx from entering the drill pipe. If we begin circulating this well under controlled conditions to maintain a constant bottom-hole pressure and do not increase the mud weight, the well can be shut in at any time, and the shut in drill pipe pressure will still be 520 psi. The drill pipe pressure which will be required to maintain a constant bottom hole pressure is dependent on the mud weight inside the drill pipe under stabilized shut in conditions. In the annulus, the pressure relationships are not predictable because of the presence of the formation fluid. The casing pressure that will be required to maintain a constant bottom hole pressure is dependent upon the type of formation fluid and the height or vertical length of formation fluid present in the annulus. Under actual conditions you really don’t know for sure either the type or the height of formation fluids. This is the reason we use the drill pipe pressure for well control.

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B) Influx Behavior Water, which is nearly incompressible, does not expand to any appreciable extent as the pressure is reduced. Because of this property, the pumping rate and returns rate will be equal as the water kick is circulated from the wellbore provided no further water influx is permitted. The casing pressure will change with a constant bottom hole pressure as the water height in the annulus changes with the hole geometry. The casing pressure will also change as the lighter water is displaced and the heavier mud replaces it in the annulus. Increases in mud weight during the kill operation will also cause changes in the casing pressure. Nearly all water influxes contain some solution gas which will make the surface pressures form the same pattern as seen during a gas kick, but to a lesser degree. Oil, like gas-charged salt water, behaves essentially like a gas influx. Gas, is a highly compressible fluid. The volume occupied by a certain amount of gas is a function of both temperature and pressure. For instance, consider a barrel of gas at the bottom of a 10,000 foot well. The bottom hole temperature is 240 oF and the well is full of 9.0 ppg mud which provides a hydrostatic pressure of 4680 psi on the gas. This same barrel of gas will expand to occupy a volume of 202 barrels under surface conditions of 14.7 psi and 80 oF. Temperature also affects the volume, but to a lesser extent (for the wellbore temperature range) and in the opposite direction. In the above example, the barrel of gas would have occupied 261 barrels at the surface if the temperature was held constant at 240 oF while the pressure was reduced from 4680 psi to 14.7 psi. If that barrel of gas were not allowed to expand in a controlled manner as it was circulated up the wellbore, it would maintain its initial pressure of 4680 psi as it moved up the annulus, and would create excessive wellbore pressures. The pressure relationships outlined in this section are the basis for effective well control. The Drill Pipe Pressure Method applies these principles and, in so doing, assures that a constant bottom-hole pressure is maintained throughout the killing circulation.

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UNSOUND WELL CONTROL TECHNIQUES In the past, it was standard practice to circulate an influx of formation fluid from the wellbore maintaining a constant pit level without regard to either the drill pipe or casing pressure. The assumption was made that no additional feed in would occur if the volume of returns from the annulus was held equal to the volume of mud pumped into the drill pipe. This is true, however, the correct objective of maintaining a constant bottom hole pressure during the well killing operation is ignored. This technique is unsound because it does not take into account the pressure density relationships in fluid columns and the laws governing the expansion of gas. In order to appreciate the dangers of using the constant pit level method to control the well, it is necessary to review the basic pressure relationships which are caused by the presence of a gas kick in a wellbore.

Ft x 1000 0

SURFACE PRESSURE = ZERO

PRESSURE AT TOP OF GAS = 4166 PSI GAS = 200 PSI

2

4 MUD = 75 PCF

MUD

6 P = 75 / 144 X 8000 = 4166 PSI PRESSURE AT TOP OF GAS BUBBLE IS 4166 PSI

8

10

GAS 0.1 X 2000 = 200 PSI BHP = 4366 PSI

Fig. 3 Wellbore Pressures

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BHP = 8532 PSI = 122.6 PCF

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In Figure 3, a gas influx has entered the wellbore, but no further flow is taking place because the hydrostatic pressure of the 75 pcf mud (4166 psi) plus the hydrostatic pressure of the gas (200 psi) exactly equal the bottom hole pressure of 4366 psi. If it were possible to insert a pressure gage at the top of the bubble, the gage would read 4166 psi. Using the constant pit level method, the gas is circulated to the top of the well. Since the volume of the gas bubble has not been allowed to change, the pressure at the top of the bubble has remained constant at 4166 psi. The hydrostatic pressure caused by the density of the gas and mud still exerts a combined pressure of 4366 psi on the bottom of the hole, but now the gas pressure of 4166 psi must be added so that the total pressure exerted on the bottom of the hole is 8532 psi. Had this situation occurred on a drilling well, lost returns could certainly have been expected. Excessive pressure buildups will result when the constant pit level method is used for well control. CALCULATION OF KILL MUD WEIGHT A positive shut-in drill pipe pressure indicates that the original mud weight is not adequate to balance the formation pressure. The mud weight increase required to balance the formation pressure is given by Δρ =

SID PP x 144 TD

…….………………..……… (1)

Normally, in kill operations a specified overbalance pressure is used to offset any reduction in hydrostatic pressure as a result of swabbing. The increase in mud required to provide a desired overbalance pressure is, Δ ρ ob =

Overbalance pressure x 144 TD

…...…… (2)

The mud weight required to kill the formation is ,

ρ = ρ i + Δ ρ + Δ ρ ob ............................……............. (3) where,

ρ

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= kill mud weight, pcf

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ρi Δρ

= = SIDPP = Δρob =

original mud weight prior to kick, pcf mud weight increase to balance formation, pcf shut-in drill pipe pressure, psi increase in mud weight required to provide desired overbalance pressure, pcf

CALCULATION OF DOWNHOLE FAILURE PRESSURE. This calculation is necessary to determine if the casing or the formation below the casing shoe will be the weak point. The equivalent mud weight that the casing shoe can hold is usually determined after the casing was cemented. The maximum allowable casing surface pressure that will not cause failure at the casing shoe is given by,

Pmax =

(ρ c − ρ i ) D 144

.......................…….......................

(4) where, Pmax = ρc = ρi = D =

maximum allowable surface casing pressure during kill operation, psi equivalent mud weight that the casing seat can hold, pcf original mud weight, pcf casing depth, ft

The maximum allowable surface casing pressure that will not burst the casing at shoe depth is,

Pmax =

Burst pressure ⎛ Mudwt . inCsg − M udwt . outsideCsg ⎞ − D⎜ ⎟ ... (5) ⎝ ⎠ S . F. 144

where, S.F. is a safety factor (1.1). The lower of the two values of Pmax from Eq (4) or Eq (5) must not be exceeded during the killing operation.

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CIRCULATION RATE FOR KILL OPERATION

The selection of the circulation rate is based upon the circulation rate before the kick, and the rig equipment. This rate should always be less than the normal circulation rate and should usually fall in the range of 1/2-2 1/2 barrels per minute (or no more than 50 ft./min. around drill collars). The lower rates are desirable since they allow more time to condition the mud while weighting up, give the choke operator more time to react, and simplify the handling of large volumes of gas at the surface. The low rates are also desirable to minimize lost returns.

DRILL PIPE PRESSURE METHOD The Drill Pipe Pressure Method is the simplest and most effective technique for controlling kicks and should be used any time a formation fluid influx is circulated out of a well. This method does not require that the volume or type of feed-in fluid be known and works equally well for gas, oil and water kicks. As long as the annulus choke is adjusted to maintain a constant drill pipe pumping pressure, gas, if present, will expand in the proper manner and the volume of mud returns will automatically be more than the constant volume pumped into the drill pipe until the influx is circulated out of the wellbore. The correct wellbore pressure relationships are maintained at all times by this method of control and the procedure is readily adaptable to field operations in all cases where the bit is near the bottom of the well. The drill pipe pressure method is the recommended procedure for killing any well if the well has been successfully shut-in without loss of returns. The kill procedure is summarized as follows: 1. Pick up kelly until tool joint is above rotary table. 2. Shut down pump and check for flow. 3. Close annular BOP. 4. Close adjustable choke. 5. Record shut-in drill pipe and casing pressures, pit volume gain, depth and original mud weight. The drill pipe float normally has a small port to transmit pressure to surface. If a non-ported drill pipe float is in use, the pressure to pump the float value open should be used as the drill pipe pressure.

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6. Determine the circulating drill pipe pressure. This can be done by holding the casing pressure constant at the shut in value and pumping down the drill pipe at constant and reduced rate of 0.5 to 2.5 BPM. As the pump starts, the casing pressure tends to increase. The choke on the annulus is manipulated to keep the annulus pressure constant at the shut-in value. By holding the annulus pressure constant at the shut-in value for the short time required to bring the pump to the desired speed, the bottom hole pressure remains essentially constant and no additional influx will enter the wellbore. With the pump running at the desired constant speed (rate) and the casing pressure maintained constant at the shut-in value, read and record the drill pipe circulating pressure The bottom hole pressure is common to both the annulus and drill pipe, and is being held constant by holding the casing pressure constant. (The drill pipe pumping pressure read at this point is that pressure necessary to maintain a constant bottom hole pressure). This is true as long as the pump rate and mud weight remain the same. The difference between the shut-in and pumping drill pipe pressure is the pressure necessary to cause the mud to circulate at the desired rate. Changes in drill pipe pressures due to choke manipulation will usually require approximately 2 seconds per 1000 feet of well depth to register on the stand pipe gauge; however, this lag in response time can be longer if the mud is contaminated or a large gas kick is present. 7. Once the drill pipe circulating pressure becomes stabilized at the desired casing pressure, hold the drill pipe pressure constant by adjusting the annulus choke. Continue to pump original mud at the same CONSTANT rate and circulate out the kick to surface. 8. Pump kill mud of heavier density to the bit, make downward correction in the drill pipe pressure as described in the following section. 9. Continue circulating at CONSTANT pumping rate until mud of the required density to kill the well occupies the wellbore. 10. Stop circulation and check for flow. In the above procedure the kick is circulated out in Step # 7 using the original mud weight and then kill mud is circulated to kill the well. This approach of weighting up the mud, called the Driller’s Method, offers the advantages of relative speed and simplicity, but

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generally results in a higher maximum surface pressure than other methods. If insufficient barite is on hand to weight up the mud, this method should be used rather than suspending operations until barite can be delivered. If sufficient barite is available on the well site, Step # 7 may be deleted and the well killed by circulating kill mud as in Step # 8. This method of weighting up the mud, called the Wait-and-Weight Method, will result in lower surface casing pressures and requires less time since the well is killed in one circulation. The drill pipe pumping pressure should be adjusted to correct for the increase in hydrostatic pressure and frictional pressure losses as a result of pumping the heavier kill mud. The procedure for adjusting the drill pipe circulating pressure is described in the following section. PUMP PRESSURE SCHEDULES FOR WELL CONTROL OPERATIONS

In the previous section a procedure was described to determine the circulating drill pipe pressure using the original mud weight. When kill mud of heavier density is pumped to kill the well, both the hydrostatic pressure and the frictional pressure losses in the drill string are altered. Thus, the circulating drill pipe pressure must be varied to offset the change in the hydrostatic and frictional pressure loss terms if the bottom hole pressure is to remain constant. The change in hydrostatic pressure due to a change in mud density is given by ⎛ ρ − ρi ⎞ ΔPh = ⎜ f ⎟ L .....................……...................... (6) ⎝ 144 ⎠ The change in pressure drop through the bit increases linearly with mud density. Also, the frictional pressure loss in the drill string for the usual case of turbulent flow can be assumed to increase linearly with mud density without introducing a large error. Since the annular pressure losses are small, the frictional pressure loss due to a change in mud density can be approximated using ⎛ ρ − ρi ⎞ ⎟⎟ .......................…..…................... (7) ΔPf = ΔPi ⎜⎜ f ρ i ⎠ ⎝

where, ΔPf = increase in frictional pressure, psi

ρf ρi

= final kill mud weight, pcf

= original mud weight, pcf Δ Pi = initial frictional pressure in drill string, psi

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The net decrease in circulating drill pipe pressure required to offset the increase in hydrostatic pressure and frictional pressure loss between the surface and the bit is given by

Δ P = Δ Ph − Δ Pf ⎛ L ΔP ⎞ = ( ρ f − ρ i )⎜ − i ⎟ .......................…............. (8) ⎝ 144 ρ i ⎠

Since this relation is linear with respect to mud density increase, it is convenient to calculate only the final circulating drill pipe pressure corresponding to the final kill mud density reaching the bit. Intermediate drill pipe pressures are determined by means of graphical interpolation. Example

A 20-bbl kick is taken at a depth of 10,000 ft. After the pressures stabilized, an initial drill pipe pressure of 520 psig and an initial casing pressure of 720 psig were recorded. The ID of the 9100-ft 5” drill pipe is 4.276” and the ID of the 900-ft drill collars is 2.5” Original mud weight of 72 pcf was pumped down the drill pipe at a constant rate of 20 strokes per minutes and a pressure of 1300 psi while holding the casing pressure constant at the initial shut-in value of 720 psi. The pump factor is 0.2 bbl/stroke. Calculate, a) the bottom hole circulating pressure that should be maintained constant during the kill operation, b) the kill mud weight that will provide 150 psi overbalance, c) the drill pipe pressure schedule required to keep the bottom hole pressure constant as the density of mud in drill pipe increases from the initial value of 72 pcf to the final kill mud density. Solution

a) The bottom hole circulating pressure is the initial shut-in drill pipe pressure plus the hydrostatic pressure of the mud. Bottom hole pressure Bottom hole pressure

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= 520 +

72 × 10000 = 5520 psi 144

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b) Mud weight increase to balance formation pressure is calculated from Eq (1), Δρ =

520 x144 = 7.5 pcf 10,000 = 72 + 7.5 = 79.5 pcf

Mud weight to balance formation pressure

Mud weight to overbalance formation pressure = 79.5 +

c) Capacity of drill pipe

Capacity of drill collars

=

=

314 . ( 4.276) 4 x144 x5.61 314 . ( 2.5)

150 x144 = 816 . pcf 10000

= 0.0177 bbl / ft

2

4 x144 x5.61

=.00607bbl / ft

The frictional pressure drop in drill pipe while pumping 72 pcf mud at 20 strokes/min is ΔPi = 1300 − 520 = 780 psi

The total drill pipe pressure change required to maintain the bottom hole pressure constant as the mud weight increases from 72 to 81.6 pcf is given by Eq (8) ⎛ 10000 780 ⎞ ΔΡ = ( 816 . − 72) ⎜ − ⎟ ⎝ 144 72 ⎠ = 562 psi

Thus, the final circulating drill pipe pumping pressure when kill weight mud reaches the bit is, 1300 − 562 = 738 psi It is usually convenient to determine the drill pipe pumping pressure as a function of the number of strokes or barrels pumped. The total number of strokes to pump kill mud to the bit is, 9 1 0 0 x 0 .01 7 7 + 9 0 0 x 0 .0 0 6 0 7 = 832 strokes 0 .2

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The initial circulating drill pipe pressure is 1300 psi and the final pressure when the kill mud reaches the bit after pumping 832 strokes is 738 psi. Assuming linear relation, intermediate values of drill pipe pressures can be obtained graphically as shown in Fig (4).

Fig. 4 Drill Pipe Pressure Schedule for the Wait-and-Weight Method

For example, after pumping 400 strokes (80 bbls of kill mud) the drill pipe circulating pressure should be 1030 psig in order to maintain constant bottom hole pressure.

KICK IDENTIFICATION The annular pressure profile that will be observed during well control operations depends to a large extent on the composition of the kick fluids. In general, a gas kick causes higher annular pressures than a liquid kick. This is true because a gas kick 1) has a lower density than a liquid kick and 2) must be allowed to expand as it is pumped to the surface.

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Both of these factors result in a lower hydrostatic pressure in the annulus. Thus, to maintain a constant bottom hole pressure, a higher surface annular pressure must be maintained using the adjustable choke. Kick composition must be specified for annular pressure calculations made for the purpose of well planning. Kick composition generally is not known during actual well control operations. However, the density of the kick fluid can be estimated from the observed drill pipe pressure, annular casing pressure, and pit gain. The density calculation often will determine if the kick is predominantly gas or liquid.

Fig. 5 Formation Fluid Influx Enters the Wellbore

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The density of the kick fluid is estimated most easily by assuming that the kick fluid entered the annulus as a slug. A schematic of initial well conditions after closing the blowout preventer on a kick is shown in Fig (5). The volume of kick fluid present must be ascertained from the volume of drilling fluid expelled from the annulus into the pit before closing the blowout preventer.

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The pit gain G usually is recorded by pit volume monitoring equipment. A pressure balance on the initial well system for a uniform mud density ρi , is given by, SIDPP +

ρ i TD 144

= SICP +

ρ i Li 144

+

ρ fLf 144

Rearranging the equation for the density of the kick yields,

ρf =

144 ( SIDPP − SICP) Lf

+ ρ i ...........................………. (9)

If the kick volume is smaller than the annular volume opposite the drill collars, the length of the kick can be expressed in terms of the kick volume and the annular capacity opposite the drill collar, Lf =

V f x 1029.4 2 2 − Ddc Doh

................……………….................. (10)

Substituting in Eq (9),

ρ =

(

2 144( SIDPP − SICP) Doh − Ddc2

f

V f x 10294 .

) +ρ

where, SIDPP SICP Li Lf ρ

= = = = =

Shut-in drill pipe pressure, psi Shut-in casing pressure, psi Length of mud column in annulus above kick, ft Length of kick, ft Density of mud, lb/ft3

TD Vf Ddc Doh

= = = =

Total depth, ft Volume of kick, bbl OD of drill collars, in Diameter of hole, in

i

Page 22

i

...............…...... (11)

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If the volume of the kick is larger than the annular volume opposite the drill collars, then the length of the kick becomes, L =L + f

dc

(V

f

)

− Vdc 1029.4 2 D − Ddp 2 oh

.............………................. (12)

where, Vdc = Annular volume opposite drill collars, bbl Ddp = OD of drill pipe, in Ldc = Length of drill collars, ft The density of the kick ρ f is obtained by substituting Eq (12) into Eq (9). If the calculated kick density is less than 22 pcf then the kick fluid is gas, and a kick density greater than 60 pcf indicates that the kick fluid is predominantly liquid. Several factors can cause large errors in the calculation of kick fluid density when the kick volume is small. Hole washout can make the determination of kick length difficult. In addition, the pressure gauges often do not read accurately at low pressures. Also, the effective annular mud density may be slightly greater than the mud density in the drill pipe because of entrained drilled solids. Furthermore, the kick fluid is mixed with a significant quantity of mud and often cannot be represented accurately as a slug. Thus, the kick density computed using Eq (9) should be viewed as only a rough estimate. Some improvement in the accuracy of the kick density calculation can be achieved if the volume of mud mixed with the formation fluids is known. The minimum mud volume that was mixed with the kick fluids can be estimated using the expression

Vm = qt d where q is the flow rate of the pumps, and td is the kick detection time before stopping the pump and closing the blowout preventer. The volume of the kick contaminated zone is, V mix = V f + V m ....................…………………................ (13) The mean density of the kick-contaminated zone ρ mix can be computed from Eq (11) by using Vmix instead of Vf . The density of the kick fluid can be calculated from the mixture equation,

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ρ = f

ρmixVmix − Vmρi Vf

..........................………………....... (14)

Example

A well is being drilled at a vertical depth of 10,000 ft while circulating a 9.6-lbm/gal mud at a rate of 8.5 bbl/min when the well begins to flow. Twenty barrels of mud are gained in the pit over a 3 minute period before the pump is stopped and the blowout preventers are closed. After the pressures stabilized, an initial drill pipe pressure of 520 psig and an initial casing pressure of 720 psig are recorded. Compute the density of the kick fluid. The total capacity of the drill string is 130 bbl. The size of the hole is 10.25 in. The drill pipe OD is 5 in and the OD of the 900 ft of drill collars is 8.125 in. Solution

The annular volume opposite 900 ft of drill collars is V dc =

(

)

. 10.25 2 − 8125 . 2 900 314 4 x 144 x 5.61

= 34.14 bbl

If it is assumed that kick fluid entered as a slug, then volume of kick is less than the annular volume opposite drill collars. Thus Eq (11) is used to calculate the density of kick fluid.

ρf =

(

. 2 144 ( 520 − 720) 10.252 − 8125 20 x1029.4

)

+ 9.6 x 7.48 = 17.18 pcf

The low density indicates that the kick fluid is gas. If it is assumed that the kick fluids are mixed with the mud pumped while the well was flowing, Vmix = 20 + Vm bbl x 3 min = 255 . bbl Vm = 8.5 min Vmix = 20 + 25.5 = 45.5 bbl

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Since the volume of kick is larger than the annular volume around the drill collars, the length of kick is calculated by using Eq (12), .4 Lmix = Ldc + (Vmix −2 Vdc ) 1029 2 Doh − Ddp

Lmix = 900 +

( 455. − 34.14) 1029.4 10.252 − 52

= 1046 ft

Using Eq (9), the density of contaminated kick (kick fluid + mud) is,

ρ

=

mix

144( 520 − 720) 1046

+ 9.6 x7.48 = 44.27 pcf

From Eq(14), the density of the kick fluid is,

ρf =

ρ = f

ρ mix Vmix − Vm ρ i Vf

44.27 x455 . − 255 . x9.6 x7.48 = 9.15 pcf 20

which also indicates that the kick fluid is a gas.

ANNULAR PRESSURE PREDICTION Although annular pressure calculations are not required for well control procedure, a prior knowledge of kick behavior helps in the preparation of the appropriate contingency plans. The same hydrostatic pressure balance approach used to identify the kick fluids also can be used to estimate the pressure at any point in the annulus for various well conditions. During well control operations, the bottom hole pressure will be maintained constant at a value slightly above the formation pressure through the operation of an adjustable choke. Thus, it is usually convenient to express the pressure at the desired point in the annulus in terms of the known bottom hole pressure. This requires a knowledge of only the length and density of each fluid region between the bottom of the hole and the point of interest. When a gas kick is involved, the length of the gas region must be determined using the real gas equation. For simplicity, it usually is assumed that the kick region remains as a continuous slug that does not slip relative to the mud.

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Fig. 6 Casing Pressure Calculations

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In Fig (6) the gas kick is shown to have moved up from the bottom of the hole to a new position. A pressure balance on the new well system yields, Pf = Pb + Pg + Pa + CP ......................………............ (15) where, Pf Pb Pg Pa CP

= = = = =

Since

formation pressure, psi hydrostatic pressure of mud column below gas bubble, psi pressure of the hydrostatic column of gas, psi hydrostatic pressure of mud column above gas bubble, psi casing surface pressure, psi Pf = Pb + Pgb Pgb = Pf - Pb

where Pgb is pressure at bottom of gas bubble. The volume of the gas kick at the new position is calculated from the real gas law, Vg =

V f Pf Tbg Z g Pgb T f Z f

............……………...................... (16)

where, Vg Vf Zf Zg Tf Tbg

= = = = = =

Gas volume at new position, ft3 Gas volume at initial position at TD, ft3 Gas compressibility factor at reservoir conditions Gas compressibility factor at the new position conditions Reservoir temperature, degree Rankine Temperature at bottom of gas bubble, degree Rankine

The density of the gas at the new position is also determined from the real gas law as, ρg =

Page 27

ρ f Pgb Tf Z f Pf Tg Zg

...................……………...................... (17)

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The length of the gas bubble is calculated from Eq (10). From the pressure balance Eq (15) the casing surface pressure is CP = Pf −

Lb ρ b L g ρ g La ρ a − − 144 144 144

.......…................... (18)

where, Lb = Length of mud column below gas bubble, ft Lg = Length of gas bubble, ft La = Length of mud column above gas bubble, ft ρ b = Density of mud below gas bubble, pcf ρ a = Density of mud above gas bubble, pcf ρ g = Density of gas at new position, pcf

Example

Consider the kick described in the previous example. Refer to Fig (7). a) Compute mud density required to kill the well. b) Compute the pressure at the casing seat located at 3,500 ft for initial well conditions. c) Compute pressure at casing seat after circulating 170 bbls of kill fluid in the annulus while maintaining the bottom hole pressure constant through the use of adjustable surface choke. d) Compute the new surface casing pressure after circulating 170 bbls kill mud in the annulus. Assume the following: Temperature at TD Temperature gradient Zf / Zg Casing ID Capacity of drillstring

Page 28

= = = = =

200 oF 0.1 oF / 100 ft 1.0 10.25 in 130 bbls

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Fig. 7 Well Schematic for Example Problem Solution

a) From Eq (2), the kill mud density with zero overbalance is

ρ =

Page 29

520 x144 + 9.6 x7.48 = 79.3 pcf 10,000

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b) Pressure at casing seat for initial conditions is P

9.6 x 7.48 = 2465 psi 144

= 720 + 3500 x

c) The fluids in the annulus from bottom to top are, i) ii) iii) iv)

170 bbls of kill fluid, 130 bbls of 9.6 ppf mud which were displaced from the drill string, the gas bubble, and initial mud

Annular volume around drill collars

=

(

900 x314 . 10.252 − 8125 . 2 4 x144 x5.61

(

314 . 10.252 − 52

) = 0.0778bbl / ft

Annular capacity around the drill pipe

=

Length of kill fluid around drill pipe

170 − 34.14 = 1746 ft .0778 =

4 x144 x5.61

) = 34.14bbls

Total length of kill fluid column in annulus = 900 + 1746 = 2646 ft The region above the kill mud in the annulus contains 130 bbls of 9.6 ppg mud which 130 = 1670 ft . was displaced from the drill pipe. The length of this column is .0778 The region above 9.6 ppg mud is the gas. The approximate pressure of the gas is needed to compute the gas volume and length. The pressure at the bottom of the gas bubble is Pgb = Pf - Pb Pf

= 520 +

9.6 x7.48 x10,000 = 5506 psi 144

Pb = 2646 ×

10.6 × 7.48 1670 × 9.6 × 7.48 + = 2290 psi 144 144

Pgb = 5506 - 2290 = 3216 psi Page 30

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This pressure occurs at a depth of 10000 - 2646 - 1670 = 5684 ft. The temperature of the gas at 5684 ft is Tbg = 200 - 0.1 x

(10000 − 5684) 100

= 195.6 oF = 195.6 + 460 = 655.6 oRankine. The gas volume at the new position calculated from Eq (16), 20 x5506 x655.6 = 34bbls 3216 x660

Vg =

The length of the gas region is

34 = 437 ft .0778

The density of the gas is given by Eq (17),

ρ = g

17.18 x3216 x660 . pcf = 101 5506 x655.6

The pressure at the casing seat is 5506 - (900+1746)

79.3 718 . 718 . 101 . - (5247 - 3500) - 1670 x - 437 x = 2316 psi 144 144 144 144

d) Casing surface pressure = 2316 - 3500 × = 570 psi

Page 31

718 . 144

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BLOWOUT PREVENTER EQUIPMENT

TABLE OF CONTENTS INTRODUCTION GENERAL REQUIREMENTS BLOW OUT PREVENTERS

ANNULAR PREVENTERS HYDRIL GK HYDRIL DIVERTER SYSTEMS

RAM PREVENTERS HYDRIL - PIPE RAM - BLIND RAM - SHEAR RAM

PRESSURE RATINGS AND SIZES PRESSURE RATINGS BOP SIZING

STACK ARRANGEMENTS ARRANGEMENTS STACKS WITH TWO RAM PREVENTERS ARRANGEMENT LOGIC STACKS WITH THREE RAM PREVENTERS ARRANGEMENT LOGIC BOP ARRANGEMENT - ONE PIPE SIZE BOP ARRANGEMENT - TWO PIPE SIZES "CLASS A" SAUDI ARAMCO 3000 PSI BOP STACK "CLASS A" SAUDI ARAMCO 10,000 PSI BOP STACK SUMMARY

Page 1 1 1

2 2 3 6

7 7 8 10 10

12 12 14

16 16 19 20 21 21 23 24 26 27 29

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TABLE OF CONTENTS ACCESSORY BLOWOUT PREVENTION EQUIPMENT

Page 31

PIT VOLUME TOTALIZERS MUD FLOW INDICATORS MUD GAS SEPARATORS

31 31 32

- DEGASSER - GAS BUSTER

32 32

SAFETY VALVES INSIDE BOP DRILLING CHOKES TRIP TANK STROKE COUNTERS GAS DETECTORS MUD LOGGING UNITS MUD WEIGHT RECORDERS DRILLING RATE RECORDERS DRILLPIPE FLOAT VALVES CHOKE MANIFOLD

33 33 34 34 35 36 36 36 37 37 37

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INTRODUCTION General Requirements The purpose of a blowout control system is to maintain control of the well when wellhead pressures develop, and this requires a means of closing the hole, controlled release of fluids, and a means of pumping into the hole. The first requirement implies having large valves (BOPS) attached to and supported by casing cemented in the well. In addition, the casing must be of sufficient burst strength and set at such a depth that the estimated formation strength is enough to resist rupture. There must be provisions for closing the well with and without pipe in the hole, a means of closing around drill pipe when it is present, and a means of stripping pipe in or out of the hole. Pipe in this case may be drill pipe, drill collars, tubing, or casing. Controlled release of fluids requires valves, chokes, and lines which allow mud, gas, oil, or water to bleed or flow at necessary rates under varying pressures, with further provision for directing these fluids to waste pit, separator, flow box, or mud tanks as desired. When drill pipe is in the hole, mud is usually pumped through the mud circulation system; but under excessive pressures or when pipe is out of the hole, additional pump-in connections become necessary. In addition, a power source and control system for quickly closing the BOPs and other control valves are needed for even the simplest low-pressure systems. Blow Out Preventers (BOP’s) There are two types of blow out preventers: Annular blowout preventer. Ram blowout preventer The annular BOP provides quick positive closing action with simplified controls to keep drilling fluids in the hole when a blowout threatens. The universal seal feature of the annular blow out preventer permits closing and sealing on virtually anything in the well bore; drill pipe, kelly, tool joints, collars, tubing or casing. The annular can strip (move pipe with pressure on the annular) any pipe, close the annulus or close the open hole if needed. The annular blowout preventer will be discussed in the annular section. The ram blowout preventer is essentially a specialized valve to close the wellbore. Similar in operation to a gate valve the ram BOP has gates called rams which meet at the center of the hole and close off the hole. These rams are designed to seal around pipe (pipe rams) or seal the wellbore completely without pipe in the hole (blind or shear rams). The ram blowout preventer will be discussed in the ram preventer section. Page 1

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Annular Preventers Annular preventers are composed of specially designed, steel-reinforced resilient elements that can seal around any cylindrical or nearly cylindrical objects that go through the BOPs. They will also seal over the open hole, and can pass drill-pipe tool joints without severe damage to the sealing element. Because of their flexibility, the annular preventers are often referred to as Universal Preventers.

Wear Plate Packing Unit Head Opening Chamber & Port Piston Closing Chamber & Port

Figure 1:

The annular preventer (Hydril GK)

Referring to Figure 1; 1. The wear plate serves as an upper non sealing wear surface for the movement of the packing unit. 2. The packing unit is the compressible element that closes around the pipe or closes on open hole to seal the wellbore . 3. The head is the part of the annular preventer that is removed when the packing unit needs to be replaced. The head is secured either by threads (Figure 2) or by latching (Figure 3). 4. The opening chamber accepts pressured hydraulic fluid to move the piston down. 5. The piston is the part that applies proper forces on the packing unit to effect a seal. 6. The closing chamber accepts pressured hydraulic fluid to move the piston up.

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Screwed Head Threads

Figure 2:

Annular with screwed head (Hydril)

Latched Head Latch

Figure 3:

Annular with latched head (Hydril)

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Annular preventers are actuated by applying hydraulic pressure to the closing chamber through the closing port. The pressured hydraulic fluid forces the wedge shaped piston upward causing the packing unit to move upward and inward thus squeezing it into the wellbore to seal on open hole or around whatever pipe is in the hole. To open the hole or annulus, hydraulic pressure is removed from the closing port and applied to the opening port. The pressured hydraulic fluid forces the piston down which allows the packing unit to relax. The resilient construction of the packing unit allows it to return to its original shape. Drill pipe can be rotated and tool joints stripped through the closed packing unit, while maintaining a full seal on the pipe. Stripping is the act of moving the pipe in or out of the hole through the closed packing unit without losing control of the well while there is pressure in the wellbore. Longest packing unit life is obtained by adjusting the closing chamber pressure just low enough to maintain a seal on the pipe with a slight amount of drilling fluid leakage as the tool joint passes through the packing unit. This leakage indicates the lowest usable closing pressure for minimum packing unit wear. Well fluid should not be the only lubrication during stripping through any annular preventer. Drilling mud or gel (bentonite) and water are excellent lubricants and will not react with rubber. The element should be lubricated continuously during stripping operations. Use of an annular preventer adds capabilities not possible with the ram type arrangement alone. The advantages are that: 1. 2. 3. 4. 5. 6. 7. 8.

Closure can be made on drill collars or casing. Closure can be made on tool joints or on the kelly. Closure can be made on any segment of a tapered drill string. Closure can be made on swab, logging and perforating lines and tools. Drill pipe can be reciprocated. Faster well closure is possible because the pipe does not need to be positioned. The string can be stripped in or out of the hole. A back-up for both blind rams and pipe rams is provided.

Life of the sealing element is shortened by repeated closure. For this reason, testing should be less frequent than normal for ram preventers. Closure on open hole tends to shorten the sealing element's life. To extend element life when testing the annular preventer, it should be closed with less than normal operating pressure, and always with pipe in the hole. If a kick occurs while running casing, ram-type preventers should be used for closure where available. In the absence of correctly sized rams, annular preventers must be used. Some casing sizes may be deformed and even crushed by annular preventers that are closed at pressures recommended for closure on drill pipe. Closing pressures to avoid casing damage vary with casing size, preventer size, preventer model and manufacturer. These recommendations are included in the operating and maintenance manual for each preventer and should be kept at the rig site. Page 4

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Old elements can be removed and new elements inserted with drill pipe in the hole by vertically cutting the elements in a groove between metal segments. The cut is opened to permit the element to be wrapped around the pipe. Cutting will not affect the sealing ability When the well pressure is high enough, the preventer will not leak well fluids even if the closing pressure were bled to zero. For example, a 13-5/8” GK closed on 4-1/2” drill pipe will leak at zero closing pressure if well pressure is 1700 psi or less. At higher well pressures it will not leak. High closing pressures, especially when combined with high well pressure and pipe movements, can destroy a GK element. This is particularly true of the larger models because of the high closing force caused by the larger piston area. There have been instances where 7-1/16” GKs have been closed at 3000 psi without severe seal damage; however, such high pressures can damage the element and are not required for a seal. Closing pressures of 3000 psi on 13-5/8” and larger will often blow the seal out of a GK preventer. If there is concern about an annular preventer's sealing around drill pipe, use a ram preventer. Generally, drilling contractors keep a GK Hydril adjusted at 700-900 psi when closing on pipe. If closing on wireline or open hole, closing pressure is adjusted to about 1100 psi. There is a relatively large selection of annular preventers available (see Table 3): Three types of resilient elements are available (Table 1): natural rubber, nitrile rubber, and neoprene rubber. Natural rubber should be used only when drilling with water-base drilling fluids without added oil. This material gives the longest life under such conditions, but always leaves a question about performance if closure is necessary on flowing oil. The synthetic resilient elements are for use with mud containing oil. Nitrile rubber is limited to use at temperatures above 20°F, and neoprene is used for temperatures down to -30°F. Table 1: Hydril packing unit types Packing Type

Natural Rubber

Color Code Black

Letter Code R

Recommended Use

Nitrile

Red

S

Oil muds with aniline points between 20°F and 190°F and for H2S service

Neoprene

Green

N

Oil based muds with operating temperatures between -30°F and 170°F

Water based muds -30°F to 225°F

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Diverter Systems A diverter system is a large, low-pressure annular preventer with large relief (discharge) lines to divert produced well fluids away from the rig while regaining control of a shallow gas flow. It may be used with drive pipe, conductor pipe, or with short surface strings that have not been set deeply enough for the well to be shut in, i.e. where broaching would occur, making a true shut in impossible. When properly designed, a diverter system allows the crew to work on the rig floor with less chance of fire or of being hit by debris blown from the well.

30" Annular Hydraulic Control Gate Valve

10-3/4" OD x 10.05" ID Diverter Line

Hydraulic Control Gate Valve Emergency

Valve Bore 6 1/8" 2" WECO Union Outlet for 2" Valve

Pump In Connection 30" 600 MSS Flanges

10-3/4" OD x 10.05" ID Diverter Line

Valve Bore 6 1/8"

Note: The same arrangement shall be used for 20 in. equipment when a diverter is required. Provisions shall be made to take pressures either flowing, shut-in or pumping

Figure 4: Saudi Aramco “Class D” Diverter BOP Stack Figure 4 is a Saudi Aramco diverter system. A true diverter system, by definition, closes the annular and opens the HCR (Hydraulically Controlled Remote) valve at the same time. The well cannot be shut in because it cannot hold the shut in pressure; the formation at the conductor shoe will fracture. This action would lead to an underground blowout which is more difficult to kill. The underground fracture could come to the surface thereby causing a blowout with no means of control and possible loss of the rig. On land rigs, the flowline is run to the reserve pit. These lines should be at least 100 ft long, discharging the effluent a safe distance from the rig. The minimum line ID should be 7 in.

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RAM PREVENTERS The ram preventer is the result of some eighty years of development. The first ram preventers were developed in the early 1900’s. The ram preventer will only seal on the specific condition for which the ram block is designed. Ram Body Seal Seat Bonnet Seal Ram Block Emergency Piston Rod Packing Bonnet & Bonnet Bolts

Fluid Connection Side Outlet for Choke/Kill Line

Piston Rod Mud Seal

Figure 5: Ram Preventer (Hydril) The major parts of the ram preventer consist of: 1. The ram body which houses all the parts that allow the ram preventer to operate. 2. The fluid connection which is the port for the hydraulic fluid connection that operates the piston. 3. the side outlet which is a connection built into the ram body that allows fluid to be circulated into or out of the wellbore. 4. The piston rod mud seal which is a secondary seal that keeps the drilling fluid from contaminating the inner seals on the piston rod. 5. The bonnet which is secured by the bonnet bolts allows access to the ram body to change the ram block. Page 7

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6. The emergency piston rod packing which is a grease injection port that allows a seal to be made on the piston rod should the primary seal begin to leak. 7. The bonnet seal which is an elastomeric seal that prevents wellbore fluid under pressure from escaping. 8. The seal seat which is a replaceable part that stops wear on the ram body by the operation of the ram block. It can be replaced in the field. 9. There are two types of ram blocks with packer seals: I. Ram blocks that seal on pipe (Figure 6) or wireline in the hole. The block is specifically designed for a particular size of drill pipe, tubing, casing or wireline. Block Body

Ram Packer

Figure 6: Ram Block for One Size Pipe (Hydril) Ram blocks designed for pipe should not be closed without pipe in them because the sealing elements may be damaged by extrusion. Ram Packer

Pipe

Figure 7: Open pipe rams Ram Packer

Pipe

Figure 8: Closed pipe rams Page 8

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BLOWOUT PREVENTER EQUIPMENT The pipe ram block can also be a variable bore block (Figure 9) which can seal on a size range. Ram Body

Ram Packer with Inserts

Figure 9: Varibore Bore Ram Block (Hydril) Variable rams became available by 1979. These rams extend the versatility of the ram-type preventer by sealing on pipe of various sizes. This flexibility could eliminate the need for changing rams when running a tapered string or when testing with tubing. Variable rams use steel fingers that move inward to seal around pipe smaller than the ram body. All manufacturers of ram preventers offer variable bore rams (Figure 9) which can close and seal on a range of pipe diameters. These rams can be especially useful when a tapered string is in use or when substructure space limitations restrict the addition of another ram preventer. Also if aluminum drill pipe is being used an effective seal cannot always be assured with regular pipe rams because the diameter of the pipe is larger near the tool joints than at the middle of the joint. Variable bore rams have limited hang off potential, depending on the size of pipe on which they are sealing. For example Cameron variable bore rams (13-5/8” U for 5” to 2-7/8”) will support 450,000 pounds of 5” drill pipe or 150,000 pounds of 3-1/2” drill pipe. Most variable bore rams are constructed in a similar fashion with the key element being a feedable rubber packer.

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II. Ram blocks that can be closed on open hole. There are two types of rams which do not have pipe openings and can be closed on open hole. The most widely used is the blind rams (Figure 10); Ram Block

Ram Packer

Figure 10:

Blind Ram (Hydril)

the other is the shear ram. The shear ram (Figures 11 & 12) is designed to cut through tubing, drillpipe or casing and it has the ability to seal on open hole. Shear rams are used mostly in a subsea stack.

Ram Body Packer Seal Lower Blade

Figure 11:

Lower blade on shear ram (Hydril)

Ram Body

Ram Packer Upper Blade

Figure 12:

Page 10

Upper blade on shear ram (Hydril)

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Lower Blade

Upper Blade Shear Rams Open

Shear Rams Closing

Shear Rams Closed

Figure 13:

Shear rams operation

Rams are interchangeable between ram-type preventers of the same design and pressure rating. Casing rams are usually substituted for pipe rams when casing is run. Most pipe rams can be locked in the closed position, operated manually and hydraulically and support the weight of the entire drill string. Pipe rams cannot be installed upside down because they are designed to hold pressure from one direction only.

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PRESSURE RATINGS & SIZES Pressure Ratings The assembly working pressure should be equal to the maximum possible surface pressure, and any choice will be arbitrary to some extent. The safest approach is to assume the maximum possible surface pressure to be the maximum anticipated bottom-hole pressure less gas gradient to the surface. When fracturing at the casing seat would limit pressure in the hole, the maximum possible surface pressure is the fracturing pressure less a gas gradient to the surface. For a very dry gas, 0. 1 psi/ft gradient may be used if the well is no deeper than 10,000 ft, and 0.15 psi/ft may be used for deeper wells. Casing burst strength should be designed accordingly, with wear factor included. Known conditions may allow these requirements to be reduced. If formations are known to contain only oil or water, the working pressure of the preventer need not be greater than anticipated bottom-hole pressure less 0.2-0.3 psi/ft. These conservative assumptions may lead to over design of the blowout prevention equipment in some cases; however, it is certainly safer than assuming a maximum surface pressure based on an arbitrary amount of mud left in the hole. The design principles stated cannot be followed in many deep, abnormally pressured wells, since probable pressures exceed the maximum working pressure of available equipment. Therefore, some risk must be taken if the well is to be drilled. Experienced and well-trained drilling crews decrease these risks. For production strings, the BOPs and casing should have ratings higher than surface tubing pressure. Rotating heads and strippers will have a lower rating than will the rest of the well-control system. It is generally accepted that annular preventers with a pressure rating 5000 psi lower than that of ram preventers can be used on stacks with 10,000 psi and higher ram preventers. This is accepted because the ram BOPs can be used if these lower pressure items are in use and well pressure approaches their limitation. Pressure ratings are given in the follwing tables: Table 2 Table 3 Table 4 Table 5

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API sizes and pressures for all preventers. Sizes, pressure ratings and types of Hydril annular preventers. Sizes and pressure ratings for Hydril ram preventers. Sizes and pressure ratings for Cameron U preventers.

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Table 2 BOP Sizes and Sealing Components Rated Working Pressure,

Flange or Hub Size,

Minimum Vertical Bore,

psi

in.

in.

RX

BX

500 (0.5 M) 2,000 (2 M)

29-1/2 16-3/4 21-1/4 26-3/4 7-1/16 9 11 13-5/8 20-3/4 26-3/4 7-1/16 11 13-5/8 16-3/4 18-3/4 21-1/4 7 1/16 9 11 13-5/8 16-3/4 18-3/4 21-1/4 7 1/16 9 11 13-5/8 18-3/4 7 1/16 9 11 13-5/8

29-1/2 16-3/4 21-1/4 26-3/4 7-1/16 9 11 13-5/8 20-3/4 26-3/4 7-1/16 11 13-5/8 16-3/4 18-3/4 21-1/4 7 1/16 9 11 13-5/8 16-3/4 18-3/4 21-1/4 7 1/16 9 11 13-5/8 18-3/4 7 1/16 9 11 13-5/8

65 73 45 49 53 57 74 46 54 -

160 162 163 165 156 157 158 159 162 164 166 156 157 158 159 164 156 157 158 159

3,000 (3 M)

5,000 (5 M)

10,000 (10 M)

15,000 (15 M)

20,000 (20 M)

Ring-Joint Gaskets

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BOP Sizing Table 2 lists the API recommended blow-out-preventer sizes and flanges for different working pressures. The preventers must be selected with a vertical bore capable of passing the casing and, in some cases, the casing hanger. Working pressure must be equal to or greater than the least of the anticipated maximum surface pressure, the burst-pressure rating of the casing, or the formationbreakdown pressure at the base of the casing. Many preventers have connections that can be used for choke and kill lines in lieu of a drilling spool. There is no standardization of component design except for those listed in Table 2. Each manufacturer selects and designs components he considers best. API Std 6A relates to working pressure, flange connections, plant test pressures, materials, throughbore dimensions, and marking of BOPS. Table 3 Sizes and Pressure Ratings for Hydril Annular Blowout Preventers

Bore Size 2-9/16 4-1/16 6-3/8 7-1/16 9 11 13-5/8 16-3/4 18-3/4 21-1/4 29-1/2 30

Page 14

500 MSP -

1000 MSP

Working Pressure Ratings (psi) 2000 3000 5000 10M GKS GKS GKS GS RS MSP GK/GKM GK GK MSP GK GK GK MSP GK GK GK GK GL/GK GK GK GK GL/GK GL GL MSP -

15M GS GK -

20M GK -

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BLOWOUT PREVENTER EQUIPMENT Table 4 Sizes and Pressure Ratings for Hydril Ram-Type Preventers Bore Size (in) 7-1/16 9· 11 13-5/8 16-3/4 18-3/4 20-3/4 21-3/4

Working Pressure (psi) 3,000; 5,000; 10,000; 15,000 3,000; 5,000 3,000; 5,000; 10,000 3,000; 5,000; 10,000 10,000 10,000 3,000 2,000

Table 5 Sizes and Pressure Ratings Available for Cameron Type U Rams

Size (in)

7-1/16 7-1/l6 7-1/16 71/16 11 11 11 11 13-5/8 13-5/8 13-5/8 13-5/8 Model B 16-3/4 Model B 16-3/4 Model B 16-3/4 18-3/4 21-1/4 20-3/4 21-1/4 26-3/4

Working Pressure (psi)

Vertical Bore

3,000 5,000 10,000 15,000 3,000 5,000 10,000 15,000 3,000 5,000 10,000 15,000 3,000 5,000 10,000 10,000 2,000 3,000 10,000 3,000

7-1/16 7-1/16 7-1/16 7-1/16 11 11 11 11 13-5/8 13-5/8 13-5/8 13-5/8 16-3/4 16-3/4 16-3/4 18-3/4 21-1/4 20-3/4 21-1/4 26-3/4 Page 15

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STACK ARRANGEMENTS A BOP stack is several blow out preventers connected together one on top of the other. The normal BOP stack consists of: 1. 2. 3. 4. 5. 6.

annular preventer ram preventer - blind ram preventer - pipe drilling spool (optional) This is also known as a drilling cross kill line, choke line and fill line bell nipple

A large number of stack arrangements are possible, including those using duplicate preventers, kill lines, choke lines, and relief lines. Duplicate ram preventers are necessary if tapered drill strings are used, and high-pressure hook-ups usually have an extra preventer, alternative choke and kill lines connected to the casing head, choke manifolds, and multiple chokes. When preventers are equipped with side outlets, the drilling spool may be omitted, and kill and choke lines attached to connections on the preventer. With the large number of ram arrangements possible, a convenient method of designation was needed. This was supplied by API Bulletin D13. For example,

5M- 13-5/8-RSRA indicates, in order, the working pressure (5000 psi), minimum bore (13-5/8”), and the arrangement of the stack from bottom to top (RSRA). In this system, R indicates a ram-type preventer (blind or pipe rams, not designated), S the drilling spool, and A an annular preventer. Other designations are Rd for a double ram-type preventer, and G for a rotary stripper head for gas, air, or aerated-fluid drilling. Attempts at standardization were made by API in Bulletin D13, while API RP53, which superseded the Bulletin, presented several arrangements that are acceptable without designating one as preferable. API RP53 has been discontinued. Although a large number of stack arrangements may be possible, these can be reduced to several arrangements that are preferred. Inability to standardize on the "most preferred" arrangement stems from the advantages and disadvantages of one arrangement of ram preventers not being outstanding over any of the others.

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The following discussions assume that the BOP stack has pressure and that there are no problems with the well such as lost circulation or a hole in the drill string. Activities that should be possible by stack operation:

1. Normal kill down drill pipe using either pipe ram (Figure 14). This is a prime requirement regardless of the geometry of the drill string.

ANNULAR

PIPE

RAM

1K SECONDARY KILL FLOWLINE CHECK VALVE HCR VALVE MANUAL VALVE

1C BLIND

2K

PRIMARY CHOKE FLOWLINE MANUAL VALVE HCR VALVE 2C

PIPE EMERGENCY KILL FLOWLINE CHECK VALVE HCR VALVE MANUAL VALVE

RAM

RAM

3K

3C

4K

4C

Figure 14.

SECONDARY CHOKE FLOWLINE MANUAL VALVE HCR VALVE

Normal kill down drill pipe

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2. Kill with blind ram closed.(Figure 15). This option is required normally for ram, annular or piping repair. The pipe is hung off on the lower pipe rams and the pipe is disconnected, the blind rams are closed and the kill circulation operation can be suspended or it can continue while repairs are made. In order to disconnect the drill pipe as shown, there must be a float in the drill string or kill weight mud must be in the drill pipe.

ANNULAR

PIPE

RAM

BLIND

RAM

1C


2K

2C PIPE

EMERGENCY KILL FLOWLINE CHECK VALVE HCR VALVE MANUAL VALVE

Figure 15.

Page 18

PRIMARY CHOKE FLOWLINE MANUAL VALVE HCR VALVE

RAM

3K

3C

4K

4C


Kill with blind or shear ram closed

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3. Ram to ram stripping (refer to figures 17 & 18 item C.). This option is required to be able to get the drillpipe in or out of the hole. Normally when a well kicks while tripping, the requirement is to get back to bottom or as close as possible to make the kill easier to accomplish. The stripping operation is called ram to ram stripping but if the annulus pressure is not too high the annular preventer and the top pipe ram would be used to strip back into the hole because if the lower pipe ram is damaged there is no method to repair it. One of the pipe ram preventers would need to be different size or variable rams if the drill string was tapered. Stacks with Two Ram-Type Preventers Reference is made to Figure 16, where these items are illustrated in a simple, two ram stack. Uses of the individual elements of BOP stacks are discussed next, and complete systems are discussed later in this section. Rotary Hose 5000 psi WP

4" Stand Pipe

Upper Kelly Cock Kelly Lower Kelly Cock

Pressure Gauge

Kelly Saver Sub 4 1/16" Stand Pipe Valve

Annular

Blind Rams 2 1/16" Kill Line 3 1/16" Min. ID Drilling Cross Check Valve

HCR

Manual

Manual

Pipe Rams

Casing Head

Figure 16.

Typical two-ram blowout-preventer stack arrangement. Saudi Aramco Class “B” 3000 psi BOP Stack

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The logic for the arrangement of a stack with two ram preventers is illustrated in the following discussion. Here, a spool was used, but using the side outlets on the preventer would serve the same purpose with fewer flanged connections. Where only two sets of rams are used, there are four arrangements for the blowout preventer stack as shown next. The discussion assumes the well is under pressure. Symbols are the API designations just discussed, but listed from top down. 1 R-Blind S-Spool R-Pipe

2 R-Blind R-Pipe S-Spool

3 R-Pipe R-Blind S-Spool

4 R-Pipe S-Spool R-Blind

Advantages: • • • • • • • • • • • •

Numbers 1 and 2: With drill pipe in the hole, the upper rams may be changed to pipe rams. When this is done, the drill pipe may be reciprocated through the upper rams while maintaining the lower rams as a reserve. Number 1: With two pipe rams and some pipe in the hole, the drill pipe can be stripped back to bottom. Number 1: If a leak develops above the rotary, the drill pipe may be suspended in the lower rams; and by closing the upper blind rams, the hole can be circulated. Also, this holds for Number 2, if there is sufficient distance between rams to place a tool joint box. Number 1: Since most kicks occur with some pipe in the hole, the lower pipe rams may be closed for repairs to the drilling spool or flowline. Numbers 1, 2, and 3: When the blind rams are closed, these permit use of the drilling spool flowlines and chokes. Numbers 2 and 3: With either of the rams closed, these permit the use of the choke flowlines and chokes. Numbers 2 and 3: By application of the double-type preventers, a low substructure height may be used. This is especially so if outlets of the preventer are used in lieu of the drilling spool. Numbers 3 and 4: Drill-pipe rams can be changed to casing rams while the drill pipe is out of the hole. Number 4: If any serious leaks develop in the stack, the drill pipe can be set on bottom or dropped and the well closed in as a last resort. Number 4: There are a minimum number of flanges exposed below the blind rams. Numbers 2, 3 and 4: When the pipe rams are closed, the choke line is available for use. Number 4: When the blind rams are closed, all of the previous connections can be stripped off or repaired.

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Primary Disadvantages: • • •

Numbers 1, 2, and 3: If the blind rams are closed, there would be no control if a leak occurred around the drilling spool. Numbers 2 and 3: There are more flanges exposed to well pressure. Numbers 1 and 4: With bottom rams closed, circulation involves secondary casing-head connections.

Flanges are considered weak points in any hookup; therefore, where side outlets are used, Numbers 2 and 3 are not considered the best arrangements by many because the drilling spool introduces an extra flange. However, the primary choke and kill line connections should be available when either blind or pipe rams are closed (Numbers 2 and 3). Overall, Number 2 is probably the most popular, however Saudi Aramco uses Number 1. Stacks with Three Ram-Type Preventers To obtain an appreciation for the evaluation of a stack with three ram preventers a generally accepted configuration will be developed for a string with one pipe size, and also for a tapered string. Then a second, yet just as popular, configuration will be presented. To experienced personnel, the “preferable” arrangement will be the one that most closely corresponds to his previous, personal experiences. Since experiences differ, preferences differ. Arrangement Logic The drilling business is often a series of compromises, both in equipment and practices. This is certainly true with BOP stack arrangements. Consider placement of blind rams in a three-ram surface BOP stack. If blind (or shear) rams are placed at the bottom of the stack, with no flowlines below, then the BOP stack has the advantage of a "master" valve for open hole situations, or a last resort valve if all else fails during a kick. But this placement also imposes limitations on stack utility. For example, drill pipe cannot be hung off on pipe rams below the blind ram and the well killed by circulating through the drill stem. This arrangement may also force placement of pipe rams so close together that adequate space is not available for ram-to-ram stripping. If blind rams are placed at the top of the ram BOP stack, they can be replaced with pipe rams for ram to ram stripping operations to either protect the lower pipe ram or, in the event of a tapered string, to furnish the pipe ram size that will fit the size of drill pipe being stripped. But this arrangement also presents a problem because it prevents utilizing the blind ram as a master valve in open hole situations for repair of items above it, or changing to casing rams. It also may force spacing of pipe rams so close that ram-to-ram stripping is impossible.

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The question arises as to how to best maximize advantages of both of these placements and minimize disadvantages. The two compromise arrangements illustrated herein place blind rams on top for tapered string drilling and in the middle when one size drill pipe is being used. This allows hanging off pipe in the pipe rams and circulating through the drill stem (with proper placement of kill and choke lines); adequate clearance for ram-to-ram stripping; and partial utilization of the blind ram as a master valve for equipment repair (top ram change to casing size obviously being safer with the blind ram in the middle). Arranging rams is important, but choke and kill flowline (wing valves) placement is equally important to fully utilize the BOPs. Again, compromises are made between the most conservative position of having no flowlines below the bottom ram and a middle road position of arranging the flowlines for maximum BOP usage. Figures 17 and 18 illustrate two BOP and wing valve arrangements. Activities possible with each of these two arrangements are summarized at the bottom of the figures and further illustrated in Figures 14 and 15. Before reviewing Figures 14 and 15, general observations can be made about both arrangements. 1. No spools are used. Choke and kill wing valves are connected directly to side outlets integral with the BOP ram body This reduces connections and chances of flange leaks. 2. A standard size 13-5/8 inch, flanged double ram is mounted on top of a single-ram unit. This provides sufficient space for shearing above a standard 5-inch NC50 connection hung in the bottom pipe ram, as illustrated in Figure 15. This is the best arrangement for use with a single drill-pipe size. Notes B2 and C1 on the bottom of Figure 18 (arrangement for tapered strings) indicate that space between the blind rams and small pipe rams limits certain activities. For tapered string application, this space problem could be eased by stacking the single-ram unit on top of the double-ram unit. However, Figure 18 shows the double on top. This illustrates another compromise, since in the field it would not be practical to rearrange the BOP stack before picking up a smaller drill string. Some contractors prefer to assemble the single on top so that the annular and single can be separated from the double for handling. Trade offs may be necessary in this matter. The primary aim here is not to debate each point, but to emphasize importance of critically reviewing BOP arrangements. Double-ram units can be specially ordered with enough room between rams to hang off and shear. This special height double-ram unit could be put on bottom, best satisfying both single and tapered string application. Here we consider standard height double and single BOP units only, with no spools or special stacks, so the most practical compromise is to place the doubleram unit on top (Figures 17 and 18).

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ANNULAR

PIPE

RAM

BLIND

RAM

PIPE

RAM

1C



2K

PRIMARY CHOKE FLOWLINE MANUAL VALVE HCR VALVE 2C

3K

3C

4K

4C


ACTIVITIES POSSIBLE A.

Normal kill down drill pipe using either pipe ram (Figure 14) 1. Choke flowlines 2C and 3C below each pipe ram.

B.

Kill with blind or shear ram closed (Figure 15) 1. Double ram unit must be on top of single ram to provide sufficient space for hang off and shear. 2. Kill flowline 2K and choke flowline 3C must be arranged as shown

C.

Ram to ram stripping (Figure 15) 1 Blind ram must be in middle to provide sufficient space with some models. 2 Kill flowline 2K to equalize pressure before opening bottom ram. 3 Choke flowlines 2C and 3C to bleed fluid and monitor pressures below each ram during stripping. 4 Kill flowline 3K to lubricate in fluid (volumetric method when bleeding gas) or kill below bottom ram. 5 Could also strip between annular and either ram and do items 2, 3 or 4 above.

D.

Location of blind ram in the middle 1. More room for ram to ram stripping as previously mentioned. 2. Safe out-of-hole top ram change, annular element change or repairs to the single ram unit or annular.

Note: Location of primary choke flowline 2C at alternate location 1C will allow all previously mentioned activities but is somewhat more exposed to mechanical damage.

Figure 17.

BOP arrangement for one pipe size.

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ANNULAR

SECONDARY KILL FLOWLINE CHECK VALVE HCR VALVE MANUAL VALVE

BLIND

RAM

Sm PIPE

RAM

PIPE

RAM

1C 1K


2K 2C


3K

3C

4K

4C


ACTIVITIES POSSIBLE A.

Normal kill down drill pipe using either, pipe ram (Figure 14) 1. Choke flowlines 2C and 3C below each pipe am.

B.

Kill with blind or shear ram closed (Figure 15) 1. Can hang off in large pipe (bottom) rams, shear, and kill. 2. Can hang off in small pipe (top) rams but normally cannot shear due to small space so must back off before closing blind rams. 3. Kill flowline K2 and choke flowlines 2C and 3C must be arranged as shown.

C.

Ram to ram stripping 1 Could change blind ram to large pipe size and strip ram to ram but the arrangement shown provides insufficient space to strip small pipe ram to ram in some cases. 2 Kill flowline 2K to equalize pressure before opening bottom rams. 3 Choke flowlines 2C to bleed fluid and 3C to monitor pressures below each ram during stripping. 4 Kill flowline 3K to lubricate in fluid (volumetric method when bleeding gas) or kill below bottom ram. 5 Could also strip between annular and either small or large ram and do items 2. 3 and 4 above.

Note:

Relocation of kill flowline 2K as shown required to accomplish kill procedures mentioned in items B3 and C2.

D. Location of blind ram on top 1. Can accomplish kill with either size pipe hung off. 2. Can change to large pipe size for ram to ram stripping. 3. Can change to either pipe size thereby minimizing wear on lower pipe rams, which inevitably occurs when pipe is worked with rams closed. 4. A disadvantage is open hole exposure while installing casing rams while out of hole. Note: If the single ram unit were arranged on top of double unit or enough space between top and middle ram provided some other way, then small pipe ram to ram stripping might be possible.

Figure 18.

Page 24

BOP arrangement for two pipe sizes.

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3.

Check (non-return) valves are located in each kill wing-valve assembly. Reasons: • To stop backflow in case the kill flowline ruptures while pumping into the well at high pressure. • Other valves between check valves and BOP can be left open during kicks for pumping into the well whenever desired without personnel having to open them. • Kill lines should not be used as fill-up lines. Constant use could result in erosion of lines and valves which would result in an unsuitable kill flowline. A separate line from the mud standpipe (independent of all choke and kill flowlines) is desirable for filling the hole during trips. Nevertheless, kill lines are sometimes used for fill up, so the check valve protects against a "blowout down the kelly" which might occur after picking up the kelly on a trip kick; assuming mud standpipe valves and various kill flowline valves have been left open for fill up.

4.

Inboard valves adjacent to the BOP stack on all flowlines are manually operated "master" valves to be used only for emergency Outboard valves should be used for normal killing operations. Hydraulic operators are generally installed on the primary (lines 2K and 2C in Figure 17 and lines 1K and 2C in Figure 18) choke and kill flowline outboard valves. This allows remote control during killing operations. During normal drilling, secondary (lines 3K and 3C in Figures 17 and 18) choke and kill flowline "master" valves should be left closed to prevent mud solids buildup. Conversely master values in the primary kill and choke flowlines must be open so that fluid passage to the choke manifold can be controlled remotely. Also, being able to operate the primary choke flowline valve remotely allows the .well to be closed "softly" on a kick. Hydraulic valve replacement is always possible.

5.

No choke or kill flowlines are connected to the casing head outlets, but valves and unions are installed. This provides: • Reserve outlets for emergency use only. • Relief opening to prevent pressuring the casing and open hole should a casing head plug tester leak during BOP testing. It is not good practice to flow into or from a casing head outlet. If this connection is ruptured or cut out, there is no control. Therefore, primary and secondary choke and kill flowlines should all be connected to heavy duty BOP outlets (or spool outlets), with wellhead outlets used only in an emergency.

6.

Figures 17 and 18 do not illustrate flowline variations such as relief lines direct from BOP to a gas/mud separator, or outlets for gauges or remote pump connections. These have merit, but were excluded to prevent diversion from the primary purpose of illustrating flowline locations that provide a high degree of utility during killing operations. Notations on Figures 17 and 18 illustrate various activities possible with the two subject arrangements. Certain limitations are also listed.

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BLOWOUT PREVENTER EQUIPMENT Rotary Hose 5000 psi WP

4" Stand Pipe Upper Kelly Cock

Mud Pump

Kelly

Pressure Gauge

Lowr Kelly Cock Kelly Saver Sub

4 1/16" Stand Pipe Valve

Mud Flow Line Rotating Head Annular

Check Valve Pipe Rams Blind Rams

2-1/16" 5M Kill Line

Drilling Cross 4 1/16" x 2 1/16" DSA Flange

Pipe Rams

4-1/16" ID 5M Choke Line

Manual HCR

Casing Head

Weco Union 1502 Emergency 2-1/16" Kill Line Laid To At Least 60 FT. From Wellbore. For Connection to Emergency Pump

Figure 19. “Class A” Saudi Aramco 3000 psi BOP stack Figure 19 is a suggested stack arrangement for the rigs working for Saudi Aramco while drilling production wells. In reviewing the previous discussion the arrangement is satisfactory however there needs to be another kill and choke flowline connected below the lower pipe ram. The valve arrangement should be reversed to make it complete.

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BLOWOUT PREVENTER EQUIPMENT Rotary Hose 5000 psi WP

4" Stand Pipe Upper Kelly Cock

Mud Pump

Kelly Lower Kelly Cock Pressure Gauge

Kelly Saver Sub

4 1/16" Stand Pipe Valve

Mud Flow Line Rotating Head Annular

Check Valve Drill Pipe Rams

4 1/16" x 2 1/16" DSA Flange

BlindRams

Drilling Cross 2-1/16" 10M Kill Line Blind Rams

4-1/16" 10M Choke Line

Master Pipe Rams

Manual HCR Casing Head Weco Union 1502

For Connection to Emergency Pump

Emergency 2-1/16" Kill Line Laid To At Least 60 FT. From Wellbore.

Figure 20. Class “A” 10,000 psi BOP Stack With similar logic, the stack configuration presented in Figures 20 and 21 can be developed.

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ANNULAR

PIPE

RAM

BLIND

RAM

1C

1K

2C


DRILLING 3K

3C BLIND

RAM

PIPE

RAM

4C

4K EMERGENCY KILL FLOWLINE CHECK VALVE HCR VALVE MANUAL VALVE


SPOOL


5K

5C

6K

6C

Figure 21. Figure 20 is a the Saudi Aramco arrangement as illustrated in the Saudi Aramco well control handbook. This handbook is received when the actual well control course is taken. Figure 21 is a simpler illustration that is used to describe some of the different scenarios that this four ram configuration can accomplish. Normal kill operation: 1. With the upper pipe ram closed the returns can be taken through connections 1C, 2C, 3C, 4C, 5C and 6C 2. With the lower or “master” pipe rams closed the returns can be taken through connections 5C and 6C Kill with blind rams closed: 1. With the upper blind ram closed the kill line connections can be made at 2K, 3K and 4K and the choke line connections can be made at 5C and 6C. 2. With the lower blind ram closed the kill line connection can only be made at 4K and the choke line connections can be made at 5C and 6C.

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Neither one of these options are good because 5C is an emergency connection and 6C is a connection of last resort. If the upper pipe ram and the upper blind ram were swapped then the kill could be accomplished while keeping the connection 5C in reserve. The lower blind rams are still in place to close on an open hole and the upper blinds or the upper pipe rams can be changed to casing rams. Ram to Ram stripping 1. This configuration can strip the way it set however the “master” ram must be used. 2. Pressure can be applied through 1K, 2K, 3K and 4K. Pressure can be bled off through 1C, 2C, 3C and 4C. Changing the lower blind ram to a pipe ram would allow stripping without using the “master” pipe ram. Changing the upper blind ram to a pipe ram would allow the stripping without using the “master” pipe ram. This can only be accomplished if there is enough room between the two rams to accommodate a tool joint. Normally this is not the case with double pipe rams unless they have been specifically designed for a stripping operation. For tapered string operation the lower blind ram becomes the small pipe ram for normal kill operation and for ram to ram stripping the upper pipe ram can be changed to fit the smaller pipe. The stack arrangement versatility changes by using varibore rams. In Saudi Aramco four ram arrangement the blind rams theoretically do not have to be removed if varibore rams are used. Or if put in place of the upper pipe ram and the lower blind ram, there would be emergency large pipe rams and stripping capability exists for all pipe sizes all the time; however, what are the disadvantages? Evaluating stack redundancy and flexibility becomes more tedious as the stack becomes more complicated. Also, the chances of standardization decrease. Summary The BOP stack must be designed so that all elements will pass the casing, casing hanger, and bit. The preventers should correspond as nearly as practical to the casing size. Adapter spools have extra flanges that may leak, and should be avoided if possible. Large preventers on smaller casing spools may have adequate pressure rating, but vibration of the large mass results in severe stress on the casing head and should be avoided. A separate fill-up line to the bell nipple should be used to fill the well during trips. Kill lines should not be used as fill-up lines because of increased erosion; if a check valve is not installed or if it plugs, the fill line would have to be shut in before the well could be shut in. Also, the kill line is more likely to plug if left filled with mud instead of clean water.

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When flowing into or from the well, only the heavy duty BOP outlets should be used, not the casing valves. In squeeze or kill operations, for example, some service companies recommend using the casing valve outlets. This is not a good practice because if the casing valves or outlets to the valves erode or rupture, a kick cannot be controlled. The stacks should be anchored with adjustable rods to derrick legs or other suitable support to minimize vibration and allow alignment. The tie rods should be as level as possible; otherwise, movement will still occur. On barges, jackups, and other platforms with long, unsupported casing lengths, the wellhead casing should be secured to the drive pipe. Locking screws should be provided with stem extensions, universal joints (if needed), and hand wheels. The hand wheels should be located outside the derrick substructure. When handling or moving stack elements, the flanges should be covered with steel or wooden plates, or mounted on a skid. Side outlets should be plugged and care exercised to avoid damaging control lines on hydraulic cylinders. Substructure height must be sufficient to accommodate the complete stack. Cellars are frequently dug and concreted to house casing heads, thereby lessening substructure height requirements. Drilling spools and preventers should not be below ground, as this requires bends in the choke flowline and results in poor access. Leaking gas or oil in a cellar is particularly dangerous for personnel working there because gas may displace all air. About three feet of space is required for the bell nipple to accept the flowline and prevent overflow. About 10 feet of substructure height is required for a double ram preventer, an annular preventer and flow stack, and the requirement may go as high as 30 feet when all elements discussed for a deep well are included in the stack.

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ACCESSORY BLOWOUT PREVENTION EQUIPMENT Various devices will indicate gain or loss of drilling fluids whether it is an indication of volume or flow rate. The indicating devices should write to charts near the driller’s position. The same read out should be installed in the tool pusher’s office and the drilling representative’s office. Warning devices (horns, lights, etc.) are necessary to alert the crew to a change in pit volume or flow rate. These should be installed and operate all the time while there is open hole on all rigs drilling in areas with hazardous or uncertain formation pressure. Pit Volume Totalizers The pit volume totalizer (Figure 22) totals the drilling fluid volume from all pits and records it on a chart. These devices may employ either air pressure or electric signals to monitor the pit volume.

Figure 22 Mud Flow Indicators Flowline monitoring devices detect early changes in the flow pattern of the drilling fluid system. Installed near the wellhead, they sense and respond before the pit volume devices do thus giving the driller early warning of a mud flow change so the prompt proper action can be taken. The flow indicators should be installed and kept operating continuously while there is open hole. There are two popular types of mud flow indicators on the market; electrical differential and flow sensor. The differential flow meter measures the difference between the fluid inflow and out flow for the well and records the difference on a chart. The flow sensor uses a paddle installed in the flow line which is deflected by increasing mud returns. Page 31

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Mud Gas Separators Gas Out

Degassers:

Gas cut mud

These very helpful units remove gas entrained in the drilling fluid during normal drilling operations, preventing circulation of gas cut mud. Circulating gas cut mud into the hole can lead to bottom hole hydrostatic pressure reduction and possibly a well kick. Compensating for entrained gas with weighting material unnecessarily increases costs. The degasser should be operated daily to keep it clean. Degassers are available in both atmospheric and vacuum models. Atmospheric models occupy less space and are generally easier to maintain but the vacuum models are generally more efficient. Gas Busters:

Mud out

Figure 23

Page 32

The typical gas buster (Figure 23) or “poorboy degasser” is constructed at the rig site or obtained from the local rental tool company. Gas busters are easy to use and maintain and should be cleaned periodically. Never circulate cement returns through a gas buster. Most gas busters are constructed from a length of large diameter pipe with a series of internal baffles to cause the gas to break free of the drilling fluid. A siphon arrangement at the bottom permits mud to flow to the shale shaker while maintaining a fluid head to hold the gas in the upper part. The gas vent pipe at the top should be large enough to permit gas to be vented at a safe distance away from the rig floor without much back pressure. Gas busters have a tendency to shake when gas cut mud is circulated through them so they should be well anchored. Short fat gas busters are generally preferred over the tall skinny variety.

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Safety Valves Safe operation requires that full opening safety valves (Figure 24) to fit each size of drill pipe and collar in use be kept in the open position on the rig floor. Then, should the well begin to flow when the kelly is not connected the correct size can be stabbed into the drill pipe tool joint and made up. Installing a valve as a precaution is good practice when the drill pipe is left in the slips for any length of time that requires the kelly to be disconnected. Care should be taken that all valves have the proper threads and they will go through the BOP’s and casing so that they could be stripped into the hole below a back pressure valve (inside BOP). Note that the term full opening does not mean that the ID of the valve is the same as the pipe but that the bore through the valve is not restricted.

Figure 24 Inside BOP This is a back pressure or float valve that allows stripping or running drill pipe into the hole without fluid flow upward through the valve. It can be stabbed and made up on the drill pipe only at very low flow rates. The best method is to stab and close the full opening safety valve then install the inside BOP if the decision is made to go back in the hole. The dart type inside BOP (Figure 25) is one of the more widely used tools. The dart is used to hold the tool open, making it possible to install the tool while the mud is flowing from the well. Releasing the dart permits the valve to close. The upper sub is then removed and additional drill pipe may be added as desired. Also available is a “drop in” inside BOP which can be pumped down the drillpipe. This tool lands and seats in a special sub installed in the bottom hole assembly.

Figure 25

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Drilling Chokes The prime function of a drilling choke (Figure 26) is to create a resistance to flow on the well which will increase the bottom hole pressure sufficiently to control formation flow while the well is circulated. Chokes are available in either positive or adjustable styles for flow control with a variety of sizes and pressure ranges. An adjustable choke regulates pressure better than a positive choke which has a fixed opening. Hydraulic chokes are easier to adjust and permit accurate regulation of the choke pressure. An important feature of most hydraulic drilling chokes is that the choke can be placed in the choke manifold but is controlled remotely from a panel which displays the casing and drill pipe pressure.

Figure 26 Trip Tank In order to keep the hole full on trips, measuring and monitoring the mud volume put in the hole is necessary. The measurement is most accurately determined through the use of a trip tank. The trip tank can be any tank or pit in which the mud volume can be measured to within ±1/2 barrel, with the measuring gauge visible to the driller on the floor.

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There are two types of trip tanks; the gravity type and the circulating type. The gravity trip tank is installed above the flow line and uses the head of fluid in the tank to fill the hole intermittently as pipe is pulled from the hole. The circulating trip tank, installed below the flow line, uses a centrifugal pump to pump mud into the well from the tank as pipe is pulled from the well. In this arrangement, returns from the well flow back into the trip tank. The pump should run continuously while tripping so that the hole fill can be monitored at all times. Intermittent filling only provides information once every several stands. Figures 27 and 28 are schematics of circulating trip tanks.

Figure 27 Stroke Counters In the absence of a trip tank, the pump stroke counter offers the driller an alternative means of measuring fluid volume used to fill the hole on trips. In order to use the stroke counter properly, the driller must know two things. First, the driller must know the fluid displacement for the particular pipe and hole size being used. Second, the volume discharged per stroke of the pump in operation must be known. This information gives the driller the ability to check for the correct fill volume required while tripping. Using the stroke counter to measure hole fill is less accurate than using the trip tank therefore it is not preferred. There is a tendency to use the kill line for filling the hole when the rig pumps and stroke counters are used. This action is never recommended. The kill line is an emergency piece of equipment that should not be used for routine trip hole fills. Stroke counters provide a means of correctly displacing special fluids or lost circulation pills. A stroke counter is especially useful to determine pumped volumes while executing well control problems.

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Figure 28 Gas Detectors Gas detectors are usually found in mud logging units to detect formation gas which can indicate abnormal pressure sections and hydrocarbon bearing formations. Rig supervisors should monitor trip gas, connection gas, and background gas for any significant change. Gas detectors can be misleading if absolute values of the gas unit rather than relative trends and magnitudes are used to interpret formation problems. Mud Logging Units Mud logging companies furnish personnel and equipment to analyze well cuttings, gas in the mud, gas type and drilling rate versus formation. They provide detailed mud analysis and gas analysis to predict hydrocarbon shows. These units should be used in unknown exploration areas. Mud Weight Recorders These devices periodically measure and record mud weight. The output is useful for detecting light or heavy streaks in the mud whatever the cause. The rig personnel should not depend wholly on these to provide absolute values. Any change should be checked manually on a routine basis by the mud personnel. These recorders should be checked frequently when using high mud weight. Page 36

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Drilling Rate recorders Drilling rate recorders are useful correlation tools when electric logs are available from other wells in the area. A particular formation that is used for setting casing can be picked quite accurately. A sudden increase in penetration rate can be the first signs of a well kick. Drillpipe Float Valves The drillpipe float valve provides instantaneous shutoff against pressures below the bit when the well is shut in thereby preventing back flow into the drill string. The drillpipe float valve is a one way valve that allows full flow through the drill string under normal circulating conditions. Allowing formation fluids along with drilling mud can be especially hazardous because the drill pipe can become evacuated very quickly or the bit can become plugged by formation cuttings. If the drillstring becomes contaminated by formation fluids when the well kicks, the kill weight mud cannot be accurately calculated. For these reasons Saudi Aramco recommends that a drillpipe float be installed above the bit. Choke Manifold The choke manifold (Figure 29) is the assembly of pipe, valves and chokes that allow the well to be controlled remotely or manually, to safely discharge the pressured formation fluid from the wellbore. The valves and the hydraulically actuated drilling choke are all open in the preferred fluid routing. The hydraulically actuated gate valve next to the wellhead is closed. When a kick is detected the hydraulically actuated gate valve is opened and the drilling choke is closed at the same time that the blowout preventers are closed. One of the manually actuated gate valves on the buffer tank will be connected to flowline that goes to the gas buster or to a degasser. A flowline to the flare pit is connected to one of the other valves. There is a pressure transmitter to send pressure information to the control console and a pressure gauge as a backup. The check valve is part of the kill line. Although this diagram shows two manual valves on the kill line, in most cases one of them is a remote controlled valve. Figure 30 is a schematic of a Saudi Aramco 5000 psi choke manifold. The lead target on the buffer is to stop erosion of the metal when the kick fluid is being circulated through the manifold.

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Pressure Transmitter

Pressure Gauge Hydraulically Actuated Drilling Choke

Buffer Tank

Manually Actuated Gate Valve Cross

Hydraulically Actuated Gate Valve

Manually Actuated Drilling Choke

Check Valve on Kill Line

Figure 29

Swaco Super Choke

Choke Manifold (Cameron)

2" Nom. - Manual Adjust. Choke Lead Target

To Flare

Weco Unions Emergency Flare Min. 3" Line To Mud Pit Instrument Flange 8" Spacer Spool 3 1/8" x 5M

* All Valves are Sized 3 1/8" x 5M

Figure 30 Page 38

To Gas Buster

Cameron 7" O.D. x 5 1/8" I.D. Buffer

Saudi Aramco 5000 psi Manifold Schematic

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FISHING OPERATIONS

TABLE OF CONTENTS

Page

INTRODUCTION

1

REASONS FOR FISHING

1

JUNK IN HOLE PARTED STRING STUCK PIPE AND THEIR CAUSES

2 2 3

- MECHANICAL STICKING - DIFFERENTIAL WALL STICKING

3 5

ECONOMICS OF FISHING

7

AVOIDING HAZARDS

8

PREPARATION FOR FISHING

9

CRITICAL INFORMATION PREPARATION OF HOLE AND DRILLING FLUID TYPE AND DETAILS OF FISH

9 10 10

DETERMINING STUCK POINT

11

MEASURING STRETCH FREE POINT INSTRUMENTS

11 13

PARTING THE PIPE STRING BACK-OFF JET CUT CHEMICAL CUT MECHANICAL CUT

14 14 15 15 19

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FISHING OPERATIONS

TABLE OF CONTENTS FISHING TOOLS RELEASING OVERSHOT RELEASING SPEAR OIL JARS MILLING TOOLS TAPER TAP JUNK BASKETS REVERSE CIRCULATING JUNK BASKETS FISHING MAGNETS PACKER RETRIEVING TOOLS

Page 21 21 22 23 24 26 27 28 29 30

WASHOVER OPERATIONS

32

WIRELINE FISHING TOOLS

34

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FISHING OPERATIONS

INTRODUCTION A fish is defined as any undesirable tool, piece of equipment, or other object found in the cased or uncased wellbore that stops or retards routine operational progress. Fishing can be defined as any operation required to remove undesirable objects (the fish) from the wellbore. Fishing requires the use of specialized procedures and equipment to remove, retrieve, or sidetrack a fish so that normal drilling or completion operations may continue. Almost every fishing job presents special problems requiring proper analysis, creative thinking, and the exercise of good judgement to successfully accomplish the objective. Often, fishing jobs require many tools and frequent trips with the work string, which may consume much rig time and can result in high operational costs.

REASONS FOR FISHING Tools and equipment are lost in the hole for a variety of reasons. In drilling operations, common causes of fishing are a result of • • • •

a twist-off or parted drill string, a stuck drill string, stuck wirline logging tools, and lost tools or junk

which inadvertently fall or are otherwise left in the wellbore. Each of these different types of 'fish' require special tools and techniques for retrieval. To explain and discuss all the tools and techniques as applied to the variety of fishing operations would require a large volume; therefore, this discussion of the fundamentals must be limited to the most common problems and the generally accepted methods of retrieval. The costs and inherent risks when fishing make it imperative that the operations and engineering personnel involved communicate freely. Predictions of the additional cost and risks associated with certain types of fishing operations may make it necessary to change the whole job plan and the final objective. For relatively simple, straightforward fishing jobs such as the recovery of pipe inadvertently dropped or left in the hole, an overshot can be used for a reasonably fast and inexpensive recovery. For a more complicated job such as the recovery of stuck or cemented pipe, or the recovery of several wireline tools - special fishing tools and skills will be required. When cases such as these arise, an experienced fishing tool operator should be considered.

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FISHING OPERATIONS

1. JUNK IN HOLE Junk lost in the hole is considered to be a frequent cause of fishing operations. There are many possible causes for loose junk to be lost in the hole, however, the most frequent causes are • the loss of a rock bit cone and bearings • hand tools or other miscellaneous objects that are inadvertently dropped from the surface • the loss of roller reamer parts • the loss of a hole opener arm and cone • the loss of an underreamer arm • the loss of a pilot bit during hole opening operations If at all possible, the first step in the recovery of loose junk lost in the hole is to identify what it is. This may be readily determined if something has been left in the hole on a trip or has been dropped into the hole accidentally. If the type and configuration of the junk is not known, an impression block should be considered. Once the type and size of the object is determined, a decision can be made if it can be recovered as a single whole piece or whether the object must be milled or otherwise broken-up. It is generally preferable to recover the junk whole rather than in pieces, however, this is not always possible. 2. PARTED STRING One frequent reason for a fishing job results when there is a twist-off and the drill string parts due to metal fatigue. Rough handling, scarring by tong dies, improper make-up torque, corrosion and erosion resulting in a washout in the tube body or cracks that form and enlarge under constant bending and torsional stresses during drilling operations. The most common place for this to occur is at the connection of a drill collar, at a crossover, or drillpipe tool joint where the higher stress level generates a crack. Connection fatigue is commonly found at the base or thread roots on the box or pin connection. When a box failure occurs, a dutchman (the box end thread) is left still threaded onto the pin end connection and recovered when the parted string is pulled. Although not as common, the drill pipe tube can sometimes fail in a long tear or split. Surface signs of a twist-off include loss of drill string weight, lack of penetration, reduced pump pressure, increased pump speed, reduced torque, and increased rotary speed.

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3. STUCK PIPE and THEIR CAUSES Quite often, circumstances during drilling operations will result in the drill string becoming stuck downhole. Both human error and mechanical failures can cause this. The recovery of stuck pipe can be a difficult, time-consuming and costly operation, therefore it is desirable to identify the type of sticking so that the most effective method of recovery may be initiated immediately. There are two general categories of drill string sticking - mechanical and differential. In mechanical sticking, the drill string is lodged in place by solid material, while in differential sticking, the drill string is held in place by the differential pressure exerted by the drilling fluid. Further discussion of these sticking causes is provided below 3.1. 3.1.1.

MECHANICAL STICKING SLOUGHING and SENSITIVE SHALES

There is a tendency for shale sections to absorb water from the mud causing the shales to swell and break off into the hole and lodge around tool joints, drill collars and stabilizers causing the drill string to become stuck. Otherwise, the shales or clays swell up from the adsorption of the free water becoming very sticky and when trying to pull the BHA back through this section the drill string may become stuck. 3.1.2.

TAPERED or UNDERGAUGE HOLE

- A tapered hole results from wear on the bit gauge when drilling hard and abrasive formations. When a new bit and BHA is run back into the hole, the full gauge bit could become stuck when the string is lowered into the tapered and undergauge hole. Extra care should be taken when running a PDC or diamond bit. Always ream through a suspect section of hole with caution if the previous bit was pulled undergauge. - Undergauge hole frequently occurs across shale formations. If the formation swells but does not slough off, the deformed layer may close in around the drill pipe, cutting off circulation and restricting clear passage of the tool joints, stabilizers and drill collars. A buildup of mud solids can have the same effect, especially in a permeable zone where water is lost to the formation. Periodic wiper trips up across the problem zone is necessary to prevent the formation from closing in around the pipe while drilling. 3.1.3.

Page 3

ABNORMAL PRESSURE and BLOWOUTS

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When drilling into a abnormally pressured shale or a plastic salt zone with insufficient mud density, the plastic flow into the wellbore can cause the string to pack-off around the pipe and become stuck.

- When a blowout occurs, a large volume of sand or shale is driven uphole by the formation fluids entering the wellbore packing off around the drillstring. 3.1.4.

INADEQUATE HOLE CLEANING AND MUD STICKING

This can occur in both cased and open hole. It is usually caused by the settling out of solids in the mud. - Inadequate hole cleaning will lead to an accumulation of drill cuttings and result in sticking. The fluid reology should be checked and adjusted to maintain desired gel strengths sufficient to support the cuttings when the pump is shut off. - Mud sticking can also be induced by high temperatures setting up the mud. 3.1.5.

LOST CIRCULATION

When fluid circulation to the surface is lost into a weak, low pressure, zone or a fractured and cavernous formation, the drill cuttings can accumulate at the lost circulation interval and pack-off around the pipe. 3.1.5.

JUNK IN THE HOLE

Metal fragments or broken-off or dropped equipment, may lodge between the hole wall and drill pipe tool joints, or drill collars. 3.1.7.

KEYSEATING

Keyseating occurs when drill pipe in tension wears an undergauge groove in the wall (low side) of a curved section, or dogleg, of the hole. When the drill string is raised or lowered, tool joints or drill collars may become lodged in the lower or upper end of the keyseat. 3.1.8.

CROOKED PIPE

Crooked pipe, often results from dropping the drill string or applying excessive weight to stuck pipe, may jam against the hole wall and become impossible to raise, lower, or rotate. 3.1.9.

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CEMENT STICKING

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FISHING OPERATIONS This can usually occur when spotting cement plugs or when cementing liners and is most prevalent in deep ‘hot’ holes where the cement can flash set due to insufficient quantity of chemical retarder that is required to prolong the setting time of the slurry.

3.2.

DIFFERENTIAL ‘WALL’ STICKING

This occurs in open hole when the drillstring comes in contact with a permeable formation. It is caused by a high hydrostatic pressure creating a differential force that holds the pipe in a thick filter cake across the permeable zone. It often occurs when rotation has stopped prior to making a connection. Differential sticking occurs only across a permeable zone, such as sand, and the friction resistance may be a function of the filter cake thickness. The extra force necessary to pull the pipe loose from the wall may be calculated by the formula below F = DP x Ac x Cf where F = DP = Ac = Cf =

force, lbs differential pressure, psi contact area, sq. in. coefficient of friction

It can readily be seen by calculating the forces in two hypothetical situations that the pull necessary to free the pipe frequently exceeds the tensile strength of any pipe available. EXAMPLE 1

Assume that drill pipe contacts the filter cake in a width of 3 inches along a 25 ft permeable sand interval with a differential pressure of 1600 psi and a friction coefficient of 0.2. F = [1600 psi x

[(25 ft x 12 in/ft) x 3 in.]] x 0.2

= 288,000 lbs

This force must be added to the normal hook load in order to pull the pipe free. In many cases, the total load would exceed the safe pull on the pipe.

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EXAMPLE 2

Assume that 8-1/4” drill collars are stuck in a 12” hole with a 14.5 ppg high waterloss mud. It is quite possible that shortly after the drill collars become stuck that one-third of the circumference of the drill collars is imbedded across a permeable sand in a thick filter cake leaving two-thirds of the drill collars exposed to the hydrostatic pressure of the mud column. At 11,000 ft, the estimated pore pressure of the permeable sand is 7000 psi and the mud density in use is 95 pcf. The calculated overpull to free the stuck drill collars is calculated as follows: F = [HP - PP (psi)] x [1 ft x 12 in x 1 C] x Cf ft 3 where HP = PP = C = Cf = HP =

95 144

hydrostatic pressure, psi pore pressure, psi drill collar circumference, in. coefficient of friction

x 11000 = 7257 psi

C=2x π x

8.25 2

F = [[7257 - 7000 (psi)] x [12 in x 1 (25.92 in)]] x 0.2 1 ft 3 F = 5329 lbs per ft of stuck drill collars Therefore, if one 30 ft drill collar is stuck, then an additional overpull force of 159,863 lbs is required to pull the string free. The causes of differential sticking may be summarized as follows: 1. 2. 3. 4. 5. Page 6

High ratio of wellbore hydrostatic pressure to formation pressure. Large, unstabilized, drill collar surface area. High mud filtration rate. High mud solids content. Excessive shutdown time opposite a permeable formation.

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Normally the sticking occurs when the drill pipe is not in motion, and usually full or partial circulation can be accomplished.

ECONOMICS of FISHING Some fishing jobs can go on for months before the fish is retrieved. After a certain period, however, the cost of fishing operations and lost drilling time become prohibitive. Therefore, a truly successful fishing job should not only be an operational success, but an economic success as well. Decisions made during fishing operations should address an economical solution to the problem in the well. Obviously, a shallow hole with little rig time and equipment invested can justify only the cheapest fishing operation. When the lost equipment and tools to be recovered represent a large capital investment, more time and expense can be justified. Generally, once these costs reach about one half of the cost of sidetracking and re-drilling, fishing operations should be abandoned. One approach used to calculate the number of rig days that should be allowed for fishing uses the following equation: D=

where D Vf Cs R Cd

Page 7

= = = = =

V

f

+Cs

R+Cd

number of days allowed for fishing replacement value of fish estimated cost of sidetracking daily cost of fishing tool rental and services daily rig operating cost

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AVOIDING HAZARDS The secret to a succesful drilling program is to avoid the hazards that historically have led to fishing jobs. • • • •

• • • •

Page 8

Take all precautions to ensure that objects are not dropped into the wellbore. Monitor the drilling torque carefully and learn the limitations of the drilling assembly . Begin with a full inspection (including electro-magnetic) of the drillstring and BHA. Continue a routine maintenance and inspection program of the BHA by monitoring operating hours in service. “If indoubt, leave it out”. Condition the hole and drilling mud prior to bit or BHA trips. Circulate a high- viscous sweep if required to unload the hole of excessive drill cuttings. If required, spot a high viscous pill on bottom to adequately suspend solids and contain loose, unconsolidated formation. Prior to drilling into a potential loss circulation zone, circulate a high viscous sweep to unload the hole of excessive drill cuttings and reduce the effective mud density. Make routine wiper conditioning trips to gauge the overall hole stability and the condition of possible water sensitive and tight shale sections. Carefully ream out any suspected undergauge hole. Formulate a plan to recover a downhole assembly should it become stuck or parted. Inventory and inspect all fishing tools that may be required immediately. Order missing or defective parts. • Can the BHA components be fished? What tools are required? • Can it be washed over? What tools are required?

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PREPARATION FOR FISHING Regardless of the cause, the success or failure of the fishing job greatly depends on the degree of preparedness taken prior to commencement of drilling operations. The following factors should be considered when preparing for a fishing job 1. CRITICAL INFORMATION The costs and inherent risks of fishing make it imperative that the following critical well information and BHA data be recorded in advance of drilling operations startup. Continue to update the information as required. •

A. CASING/LINER DATA

B. HOLE DATA

C. MUD DATA

-

-

-

Size (OD, ID) Weights Depths Grade Connection

Size Depth Angle Condition Lost Circulation Undergauge

Type Density Reology Water Loss

Record every dimension and the position of all drilling tools (as listed below) in the string before they are run below the rotary table. Continue to update the information as required. Maintaining a good record of all critical equipment dimensions is necessary if an economical fishing job is to be done. •

•

Page 9

DRILL PIPE DRILL COLLARS

-

Sizes (OD, ID) Tool Joint OD Weights Grade Quantity Length Connection Last Inspection?

STABILIZERS REAMERS DRILLING JARS SHOCK SUB

-

Size (OD, ID) Type/Serial No. Quantity Fish Neck Length

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-

Overall Length Connection Last Inspection?

2. PREPARATION of HOLE and DRILLING FLUID The hole should be conditioned and the drilling fluid should have the desired properties before starting in the hole with fishing tools. It may be necessary to make a trip with a bit to condition the hole and circulate out fill that has covered the fish. Always circulate and flush the hole first with a high viscous pill to unload drill cuttings and spot a balanced high viscous pill above the fish to suspend solids prior to trip out for fishing tools. 3. TYPE and DETAILS of FISH Once it is evident that fishing operations must be conducted, it is necessary to learn and acquire all relevant information about the fish that is to be recovered. Determine • • • • • • • •

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the probable cause of failure or reason for junk lost in hole. the size, length, amount, and type of fish (all dimensions are important). the location or top of the fish. the condition of the top of the fish. the mechanical condition of the wellbore casing. the condition and stability of the open hole interval down to the top of fish. the value or replacement cost of the fish if sidetracked and left in the hole. the probability of success, risks of failure, and estimated recovery time and cost.

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DETERMINING STUCK POINT 1. MEASURING STRETCH When pipe becomes stuck in the wellbore for any reason, one of the first steps is to determine at what depth the sticking has occurred. When retrieving production tubing from a well, it is often common to find that the tubing is stuck, with the seal unit seized, or locked, up in the production packer bore due to scale or an abundance of solids that have settled around the outside of the seal unit in the tubing- casing annulus. Stretch in pipe can be measured and a calculation made to estimate the depth to the top of the stuck pipe. All pipe is elastic and all formulae and charts are based on the modulus of elasticity of steel, which is approximately 30,000,000 lb/sq. in. If the length of stretch in the pipe with a given pull is measured, the amount of free pipe can be calculated or determined from a chart available in data books. Since all wellbores are crooked to some extent, there is friction between the pipe and the wellbore. Steps should be taken to reduce this friction to a minimum. The pipe should be worked for a period of time by pulling approximately 10%-15% more than the weight of the string and then slacking off an equal amount. There are certain techniques that reduce error in estimating stuck points from stretch data. It is also necessary to assume certain arbitrary conditions. Stretch charts and formulaes do not take into consideration drill collars or heavy weight drill pipe. First, pull tension on the pipe at least equal to the normal hook load (air weight) of the pipe prior to getting stuck. This should then be marked on the pipe as point "a". Next, pull additional tension which has been predetermined within the range of safe tensional limits on the pipe. Now slack off this weight back down to the hook load weight. Mark this point "b". It will be lower than point "a". This difference is accounted for by friction of the pipe in the wellbore. Next pull additional tension on the pipe to a predetermined amount within the safe working limits of the string. Mark this point as "c". Then pull additional tension on the pipe in the same amount used to determine points "a" and "b" and slack off to tension used to locate point "c". Mark this point "d". The mid-point between "a" and "b" and between "c" and "d" will be the marks used. Measure the distance between these average marks and use this number as the stretch in inches. The amount of free pipe can be determined by using the following formula:

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length of free pipe (ft) = 1,000,000 x (stretch - in.) K x (pull over string wt., lbs) where K = constant The constant in this formula can be determined by: K=

1.5 for drill pipe or pipe wt, lb/ft

K=

1.4 for tubing and casing. pipe wt, lb/ft

This method of estimating the stuck point of pipe is not completely reliable and accurate as there are many variables caused by friction, doglegs, hole angle, and pipe wear. However, it frequently indicates the cause of sticking such as possible areas of a key seat or differential sticking in open holes and collapsed tubing or tubing leaks in producing wells. In addition to the basic formula provided above for calculating the amount of free pipe, there are reference manuals available that provide stretch charts from which the length of free pipe can be read directly. The same procedures and precautions, as outlined above, should be followed to obtain the pipe stretch with a predetermined pull over the string weight. Accuracy of the charts and the formula is approximately the same, as both are affected by the same problems of hole friction, loss of material in used pipe, and the accuracy of weight indicators. Note, however, that the modulus of elasticity of all grades of steel is the same. The grade of the pipe does not affect its stretch. Also, when pipe is stuck, buoyancy forces are not effective. Immediately when the pipe is freed, the buoyant forces are again in effect and should be considered accordingly. Example A 13,000 feet of 23 #/ft casing is stuck near the bottom of the well. A differential load of 55,000lb was applied on the pipe which resulted in a stretch of 42in. What is the depth of the free point? Solution

K=

1.4 = 0.0608 23

Pull = 55000 lb Stretch = 42 in

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Length of free pipe =

1000000 x 42 0.0608 x 55000 = 12559 ft

2. Free-Point Instruments Electric wireline service companies run instruments on conductor lines inside the stuck drillpipe or tubing and are able to accurately determine the stuck point of pipe. The instruments are highly sensitive electronic devices which measure both stretch and torque movement in a string of pipe. This information is transmitted through the electric conductor cable to a surface panel in the control unit where the operator interprets the data. The basic free-point instrument consists of a mandrel which encompases a strain gauge or microcell. At the top and bottom of the instrument are friction springs, friction blocks, or magnets, which hold the tool rigidly in the pipe. When an upward pull or torque is applied at the surface, the pipe above the stuck point stretches or twists. The change in the current passing through the instrument is measured by the microcell and transmitted to the surface for interpretation. When the instrument is run in stuck pipe, there is no movement of the pipe, therefore there is no tension or torque transmitted to the instrument. In turn, the gauge at the surface shows no change in its reading. Free-point indicators are frequently run with collar locators and in combination with string shots, chemical cutters, and jet cutters. This combination run saves expensive rig time, and it will also maintain a continuous sequence in measuring so that there is less chance of a misrun in cutting or backing-off. Since fishing operations usually begin as soon as the pipe is parted following the free-point determination, it is a good practice to have the fishing tool supervisor or operator on the location during the free-point and back-off or cutting operations. Frequently there are suggestions that can be made to improve the fishing situation when the fishing operator is present to observe the free-point and parting operations.

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PARTING THE PIPE STRING After determining the stuck point in a pipe string, it is normal procedure to part the string so that fishing tools such as a jarring string or a washpipe string may be run. The cutting method for the particular job should be selected carefully. Only the back-off method, listed below, leaves threads looking up, and therefore should be selected if it is desired to screw back into the fish. There are four acceptable methods of parting the pipe string: 1.Back-Off - Unscrewing the pipe at a selective threaded joint above the stuck point using a prima cord explosive run on an electric wireline. Back-off is the procedure of applying left-hand torque to a pipe string and firing a shot of prima cord explosive across a tool joint which produces a concussion to effectively partially unscrew the threads. The back-off method of parting pipe is probably the most popular of all, particularlly in drill pipe. Tool joints on drill pipe, drill collars, and other drilling tools have coarse threads, large tapers and seal by the flat surfaces or faces. These characteristics make the back-off method attractive as it leaves a threaded connection looking up, making it possible to screw back into the fish with a jarring work string. Tubing or other coupled pipe does not lend itself to back-off in the same way as drill pipe. Tubing threads are usually fine, at least eight per inch; there is only a small taper, and the threads are commonly in tension with a high degree of thread interference. Furthermore, there is a high chance of damaging the fine threads such that cross-threading will likely occur if attempting to screw back into the fish. To prevent an accidental back-off in a loose connection up the hole, the pipe should first be tightened by applying right-hand torque and then reciprocating the pipe while holding the torque. Once the pipe is made up, left-hand torque is introduced in the string. This torque must also be "worked down" by reciprocating the pipe as the torque is increased. This action distributes the torque throughout the string and assures that there is left-hand torque at the point of back-off. Theoretically just prior to firing the string shot, the pipe at the back-off point should be in a neutral condition, with neither tension nor compression. Since this condition is very difficult to obtain, any choice should lead toward slight tension in the pipe. Since buoyancy is not effective in the stuck pipe, air weight of the string is used in calculations. However, the moment the pipe starts to spin free, buoyancy produces an upward lifting force against the free pipe string. The Page 14

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left-hand torque is held, and the determined weight of the string is picked up when the stringshot is fired. The concussion at the joint momentarily loosens the threads and the pipe begins to unscrew. It usually must be manually unscrewed completely and then the freed pipe can be removed from the well. When ordering a string-shot, the service company needs to know the size and weight of pipe to be backed off, the approximate depth of the stuck point, the weight of the mud or fluid in the hole, and the temperature of the well. This information will dictate the strength of the charge needed as well as the type of fuse. 2.Jet Cut

-

A cut made by an explosive shaped with a concave face and formed in a circle. It is also run and fired on an electric line.

The jet cutter is a shaped charge of explosive which is run on an electric wireline. The modified parabola face of the plastic explosive is formed in a circular shape to conform to the shape and size of the pipe to be cut. When an explosive such as this is used to cut pipe, the end of the pipe is flared, and it is necessary to mill over and remove this flare if the pipe is to be fished with an overshot from the outside. The jet cutter is often used when abandoning a well during salvage operations or when low fluid level, heavy mud, or cost would preclude the use of the chemical cutter. Jet cutters are available for practically all sizes of tubing, drill pipe, and casing. There is a possibility of damage to an adjacent string or to casing if the pipe to be cut is touching at the point where the cut is made. 3.Chemical Cut - An electric wireline tool and procedure that uses a propellant and a chemical, halogen fluoride, to burn a series of holes in the pipe thereby weakening it such that it is easily pulled apart. This method of cutting pipe is the most recent innovation. It was patented and for years was an exclusive process of one wireline company. Today it is available through most electric wireline service companies for practically all sizes of tubing and drill pipe and most popular sizes of casing. All wireline cuts are generally economical because rig time is reduced to a minimum. The big advantage of the chemical cut is that there is no flare, burr, or swelling of the pipe that is cut. Therefore, no dressing of the cut is necessary in order to catch it on the outside with an overshot or on the inside with a spear. The chemical cutting tool consists of a body having a series of chemical flow jets spaced around the lower part of the tool. The tool contains a propellant which forces the chemical reactant through the jets under high pressure and at high temperature to react with the metal of the pipe. Page 15

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Electric current ignites the propellant which forces the chemical, halogen fluoride or bromide trifluoride, through the reaction section which heats the chemical and forces it out the jets. The tool also contains pressure-actuated slips to prevent a vertical movement of the tool up the hole and insure a successful cut. The chemical cutting tool may also be explained as producing a series of perforations around the periphery of the pipe. The reaction of the chemical produces harmless salts which do not damage adjacent casing. The products of the chemical reaction are harmless and are rapidly dissipated in the well fluid. The chemical cutter will not operate successfully in dry pipe and requires at least one hundred feet of fluid above the tool when a cut is made. Since it is not necessary to apply torque to the pipe when chemically cutting as compared with the string shot back-off, it is a safer process for rig personnel.

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FISHING OPERATIONS Explosive Jet Cutter

Ex pl osi v e

Shaped Char ge

Pip e c ut w ith exp losive Jet Cutte r

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Ch e m i c a l Cut t e r

Pi p e cu t w i t h a Ch emi c al Cu t t er .

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4. Mechanical Cut -

A cut made with a set of knives installed in a tool and run on a small diameter work string. This is referred to as an internal or inside cut. Internal mechanical pipe cuts are most common when removing sections of casing and wellhead equipment during final well abandonment operations.

The internal cutter is made on a mandrel with a wickered sleeve or split nut fitted to threads on the mandrel. This allows the slips to be released and the tool to set at any specific depth desired. Friction blocks or drag springs are fitted to the mandrel to furnish back-up for this release operation. As weight is applied to the set tool, knives are fed out on tapered blocks, and as the tool is rotated, they engage the pipe and cut it in two. Upon reaching the desired cutting depth, the internal cutter is anchored by slowly rotating to the right while slowly lowering the work string. The wiper blocks resist rotation and lowering by maintaining friction on the pipe and continued lowering of the work string until the slips, which move upward and outward, engage and anchor the cutter to the pipe wall. The mandrel is free to travel downward under the knife blocks forcing the knives upward and outward to start the cut. Slight weight additions are applied while slowly rotating to the right. The main spring in the upper part of the cutter is partially compressed by the applied weight and assists in maintaining a uniform feed to the knives and to help absorb any shock that may accidentally be applied to the work string causing the knives to gouge or to break. Cutting is accomplished by slow rotation to the right with just enough weight being gradually applied to feed the knives into the metal. For best operation, the work string is lowered in 1/16" intervals (never more than 1/8") a total of 1-1/4" on the work string to complete the cut. Free rotation, with little or no reverse torque, indicates that the cut is completed. To prove the cut, increase the rotating speed; and if there is no increase in torque noted, it will indicate that the cut has been successfully completed. Care should be exercised not to hurry the cutting operation, as excess weight will cause the knives to dig into the pipe burning the knife points or possibly even breaking the knife blades. Fishing tool operators will usually run a bumper sub above the cutter so that excessive weight is not exerted on the knives causing them to break or dig into the pipe. To release the cutter, raise the work string one foot. This will cause the grip jaws to engage the wickered sleeve, and now the tool is ready to be raised or lowered, as desired.

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Internal Cutter Mandrel

Knife Knife Block

Main Spring

Slips Bowl

Wiper Block & Spring

Bottom Nut

Running In

MECHANICAL CASING CUTTER

Page 20

Cutting

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FISHING TOOLS Releasing Overshot

The Releasing Overshot is used to externally engage and retrieve all sizes of tubing, drill pipe, and casing. Top Sub

Packer

Bowl Spiral Grapple Grapple Control

Guide

The overshot is designed to assure positive external engagement over a large area of the fish and is ruggedly built to withstand severe jarring and pulling strains without damage or distortion to either tool or fish. Most overshots consist of a bowl, top sub, guide and the grapple or slip, a control, and packoff. The overshot bowl is turned with a taper on a helical spiral internally and then the grapple, which is turned with an identical spiral and taper, is fitted to it. Overshots are very versatile and may be fitted for a variety of problems. Mill controls may be used to dress the area that the grapple will catch in order to remove burrs and splinters on the pipe. When the pipe has been "shot off" or parted in such a way to heavily damage it, it may be necessary to fit a mill extension, or mill guide, to the overshot bowl so that extensive milling can be accomplished for the catch to be made on the same trip in the hole. These extensions, or guides, are "dressed" inside with tungsten carbide and can mill off a substantial amount of material so that the "fish" is trimmed down to the grapple size. Controls may also be designed with a pack-off, or packer, that seals off around the fish and allows the circulating fluid to be pumped through the fish to aid in freeing the stuck fish.

To properly engage an overshot on a fish, slowly rotate the overshot as it is lowered onto the fish. The pump may be engaged to help clean the fish and also to indicate when the overshot goes over the object. Once this has been indicated by an increase in pump pressure, stop the pump, as there may be a tendency to kick the overshot off the fish. Set the grapple with gradually increasing, light upward blows. An excessively hard upward impact may strip the grapple off the fish and cause the wickers to be dulled, resulting in a misrun and trip to replace the grapple. To release overshots, it is first necessary to free the two tapered surfaces, bowl, and grapple, from each other. This freeing of the grapple or "shucking" can be accomplished by jarring down with the fishing string. Usually a bumper sub is run just above the overshot and is used for this purpose. After bumping down on the overshot the grapple is usually free and the overshot can be rotated to the right and released from the fish. If a large amount of the fish has been swallowed, it may be necessary to free or " shuck" the grapple more than once. Page 21

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Releasing Spear

The Releasing Spear is used to internally engage and to retrieve all sizes of tubing, drill pipe, and casing as opposed to overshots which catch on the outside. It is designed to assure positive internal engagement with the fish and is ruggedly built to withstand severe jarring and pulling strains without distorting the fish.

Mandr el

Usually a spear is not the first choice, as the spear will have a smaller internal bore than an overshot which limits running of some tools and instruments through it for cutting, free-pointing, and in some cases, backing-off. Spears, however, are popular for use in pulling liners, picking up parted or stuck casing, or fishing any pipe that has become enlarged when parted due to explosive shots, fatigue, or splintering.

Gr appl e

Rel ease Ri ng

Nut

The most popular spears in use today are built on the same principles as overshots with a tapered helix on the mandrel and a matching surface on the inside of the grapple. The slip, or gripping surface is on the outside surface of the spear so that it will catch and grip the inside of the pipe that is being fished. Due to the design with the small bore in the mandrel, spears are usually very strong. The spear is run inside the fish and positioned. The slips are released by action of the J-slot by using left-hand torque, moving the drill string down a short distance, and then picking it back up slowly. This action releases the slips so they can slide up over a taper on the body of the spear as the spear is moved uphole. The slips move outward engaging the inner wall of the fish. In order to release a spear, it is rotated to the right. If the grapple is frozen to the mandrel, it may be necessary to bump down to free or 'shuck' the grapple. Usually a bumper sub is run just above the spear and this can be used to effectively jar down and free the grapple. The spear is a very versatile tool, in that it can be run in the string above an internal cutting tool if desired or in combination with other tools. Milling tools may be run below the spear to open up the pipe so that the spear can enter and catch the fish.

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Oil Jars

Jars are impact tools used to strike heavy blows either up or down upon a fish that is stuck. Jars fall into two categories as to use: drilling jars and fishing jars. Jars can further be classified as to the basic principle of operation; either hydraulic or mechanical. Most jarring strings used in conjunction with fishing operations consist of hydraulic "Oil" jars. Oil Jars are very effective in freeing stuck fish as the energy stored in the stretched drill pipe or tubing is converted to an impact force, which can be varied according to the pull exerted on the string. The oil jar is designed to strike a blow upward only, while an additional tool, the bumper sub is designed to strike a blow downward on the fish. The oil jar consists of a mandrel and piston operating within a hydraulic cylinder. When the oil jar is in the closed position, the piston is in the down position in the cylinder where it provides a very tight fit and restricts the movement of the piston within the cylinder. The piston is fitted with a set of packing which slows the passage of oil from the upper chamber to the lower chamber of the cylinder when the mandrel is pulled by picking up on the work string at surface. About half way through the stroke, the piston reaches an enlarged section of the cylinder and is no longer restricted so the piston moves up very quickly and strikes the mandrel body. The intensity of this impact can be varied by the amount of strain taken on the work string. This variable impact is the main advantage of the oil jar over the mechanical jar for fishing.

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Milling Tools

Sometimes a packer or fish cannot be removed from the wellbore intact. It is then necessary to reduce the fish to small pieces that can be circulated to the surface. Mills dressed on the bottom with tungsten carbide have been used extensively for this purpose and with good results. Flat bottom mills are often used to mill over the slip segments and packer element on permanent production packers. Milling tools are available in a number of sizes and design shapes for various applications. Some common types of mills are shown below.

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Pilot mills are used to mill casing inside the hole. New metal technology has improved the milling performance and increased the rate of penetration by four to twelve times compared to conventional tools. This performance represents a dramatic reduction in costs for milling tubing and casing strings. The conventional pilot mills have fabricated blades coated with crushed tungsten carbide for a cutting edge. The new design uses specially formulated tungsten carbide buttons brazed to the leading edge of the alloy blades. As the blades wear down, more buttons are exposed and form a continuous cutting edge until the blades are completely worn out. New technology pilot mills are more efficient and give the operator longer tool life, fewer trips and lower job cost. Mill manufacturers claim that one pilot mill can mill 600 ft of 7” casing at a penetration rate of 20-30 ft/hr.

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Taper Tap

Taper Tap

Fish

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The Taper Tap is used to engage a tubular fish internally. This tool screws into the fish and cuts threads as it goes. Cutting new threads is a more positive engagement than attempting simply to screw on or into existing threads on a fish that may be damaged, misaligned, or incomplete. New threads can also be cut on blank pipe. Frequently, the Taper Tap is used to retrieve a production packer after the slip segments and packer element have been milled with conventional mills.

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Junk Baskets

The Core-Type Junk Basket, as shown, was the old stand-by for years for fishing bit cones and similar junk from the open hole. It consists of the top sub, a barrel, a shoe, and usually two sets of fingertype catchers. This tool is still used quite often, and it is made to circulate out the fill and to cut a core in the formation. The two sets of fingers help to break the core off and retrieve it. Any junk that is in the bottom of the hole is retrieved on top of the core.

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Reverse Circulation Junk Baskets

The reversing action is extremely helpful in lifting junk into the barrel and catcher, that might otherwise be held away from the catcher by fluid flow. The reverse circulation junk basket design incorporates an inner barrel with the fluid flow between the outer and inner barrels when a ball is dropped and closes off the center flow through the seat. With this design, when the ball is circulated down, the flow is diverted between the two barrels and reverse circulation flow is created back up the inside of the junk catcher with the fluid exiting into the annulus through the upper ports near the top of the barrel.

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Fishing Magnets

Fishing magnets are either permanent magnets fitted into a body with circulating ports or electromagnets which are run on a conductor line. Permanent magnets, as shown, have circulating ports around the outer edge so that fill and cuttings can be washed away and contact made with the fish. Ordinarily the magnetic core is fitted with a brass sleeve between it and the outer body so that all of the magnetic field is contained and there is no drag on the pipe or casing. Permanent magnets have the advantage of the circulation washing away any fill so that the junk is exposed. Ordinarily, by rotation, one can detect when contact is made with the fish. The operator should then thoroughly circulate the hole, shut the pump off, and retrieve the fish without rotation

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Packer Retrieval Cutting over and retreving permanent production packers is a very common job during workover operations. A Packer Milling tool, such as the unit shown below, is designed to mill over and retrieve the production packer from the wellbore in one trip.

Junk Basket

Packer Milling Tool

Retainer Production Packer

Retainer Retrieved Production Portion Packer

Milling Shoe

of Packer

Catch Sleeve

Fig. 1 Drill String Make-Up

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Fig. 2 Milling Packer

Fig. 3 Retrieving Packer

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Normally only a small portion of a packer, the slips and packing element, needs to be milled up. The packer milling and retrieving tool consists of a carbide rotary mill shoe, an inner mandrel, with a catch sleeve on the lower end which extends through and below the packer body during milling operations. Retrieving tools are made in several designs, but most can be operated through a ‘J’ on the mandrel with springs to provide back-up for operation. The retriever is run in the retracted position and is small enough to go through the packer bore. It is then set so that the grapple or catch sleeve is extended so that it will not come back through the packer bore. After the slips and packing element have been milled up, the catch sleeve will catch the remaining packer body and it will be removed with the milling tool by pulling tension in the work string to dislodge the packer from the casing. When a seal bore extension and tail pipe assembly is run below the packer, a millout extension must also be included below the seal bore extension in order to accommodate the catch sleeve. The millout extension should be twice the length of the packer to fully accommodate the stinger and catch sleeve when the mill has cut through the entire packer. If the packer is small such as in 4-1/2” O.D. casing a rotary mill shoe should be selected rather than a milling and retrieving tool of this type, as a tool of this size would be weak. After the top packer slips are milled, the packer is retrieved with a taper tap or spear. A new technology packer milling and retrieving tool is shown in Fig (4). Instead of a mill shoe, the new design uses four alloy blades which have tungsten carbide buttons brazed on the leading face of the blades. As the blades wear down, more buttons are exposed and form a continuous cutting edge until the blades are completely worn out. The blades on this new tool are more than adequate to mill the entire packer. Fig (4) Smith Baron Packer Retrieving Tool

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WASHOVER OPERATIONS

Very often it is not enough merely to catch hold of the fish and pull. In those cases, washpipe and a rotary shoe can be used to rotate over the fish to remove annular material that may be causing it to stick and free up a section of stuck pipe so that it may be retrieved. The outside diameter of the washpipe must be small enough to run inside the casing, and its inside diameter must be large enough to fit over the fish. Washpipe is heavy N-80 grade casing cut into Range 2 lengths for ease of handling and equiped with special threads with good characteristics for torquing and strength. Washpipe is usually flush joint both inside and outside for maximum clearance. The length of run in the well must normally be limited to a few hundred feet. The rotary shoe is placed on the end of the washpipe to drill-up and circulate out any material around the fish.

Washover Pipe

Casing

Tubing Fish

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Rotary Shoe

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The rotary shoe run on the bottom of the washpipe string should be designed for the particular job. Tooth-type shoes are usually used if cuttings, fill, formation, or cement is to be cut. The teeth are shaped with a straight leading edge, and all the surfaces of the teeth are dressed with a wear material, usually tube borium, to prevent excessive wear and erosion from the fluids circulation. If steel, such as the tool joints tube or junk must be cut by the rotary shoe it is dressed with tungsten carbide in a configuration that is appropriate for the particular job. Care should be exercised in designing the shoe since it is necessary to have sufficient circulation to keep the carbide cool as well as wash away the cuttings. If the job is inside casing no cutting carbide should be allowed to remain on the outside as this will damage the casing. In some cases, smooth brass is applied on the outside diameter of the shoe to provide a bushing which reduces the friction and prevents damages to the casing. Tungsten carbide is applied to the bottom of the shoe and if possible to the inside diameter. Where it is possible to apply the carbide with a small shoulder inside the shoe, the chances of retrieving some or all of the fish inside the pipe are improved. This would save a trip with another tool to recover what has been washed over. The length of the washpipe string is most important. Realizing that the washpipe string is large stiff and smooth length becomes extremely important in preventing sticking. There is no rule or gauge for determining the maximum length but a judgment must be made based on careful consideration of the hole conditions. Long washpipe strings can be used for washing soft fill around stuck pipe. In washing over drill pipe cemented inside casing only 100-150 ft of washpipe could be run at one time. When longer washpipe is run, it may become stuck and increase the cost of the fishing operation. When the entire length of fish cannot be covered in one washover, it is necessary to part the string that has been freed from that which remains in the hole. This can be accomplished by one of several methods: 1. An overshot can be run after the washpipe has been removed and left-hand torque applied and the fish backed off with a string-shot. 2. An external or outside cutter can be run on the washpipe instead of the rotary shoe and the fish cut off above the lowest point that has been freed.

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Wireline Fishing Tools

Wireline Spears One of the most challenging of fishing operations may be the recovery of wireline and the tools or instruments run with it. Often times wireline has been parted. When this occurs, wireline slumps down the hole in a coil. The wireline center spear or the twopronged wire grab, shown at left, are used frequently to remove parted wireline from the wellbore.

TWO-PRONGED WIRELINE GRAB

WIRELINE CENTER SPEAR

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When the tools are used in casing, a guide should be run above the tool to prevent the wire from getting above the spear.

Drilling Training Course

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