## Pump To Surface Slip And Cut Program

A 10.5-lb/gal mud is being used to drill at 11,000 ft. A 600-ft, 7-in x 3-in. collar assembly is used with 4.5-in., 16.6-lb/ft drillpipe. Determine the number of ton-miles involved in a round trip. A 93-ft stand is pulled. The traveling block assembly weighs 28,000 lb.

### Solution:

1. The number of stands involved in the trips is;
2. The effective weight of the drillpipe, WM, is:
• 13.94 lb/ft
• 13.94 lb/ft

### 3. The effective weight of the collars, \Vr, is:

• 89,84 lb/ft
• 89,84 lb/ft

Therefore, C is 89.84 lb/ft minus 13.94 lb/ft, or 75.9 lb/ft.

The IA DC provides an extensive set of tables for ton-mile calculations. The tables are often available in the field and may be easier for some field people to use. Drilling engineers planning a well may prefer to use the equations.

In addition to fatigue wear from accumulated ton-miles of service, the wire rope will wear more at lap and pickup points. The pickup points arc on the top side of the crown block when the weight of the drillstring is lifted from its supports in the rotary table during tripping operations. The lap points on the drawworks drum occur when the line begins a new wrap.

Slip and cut programs are designed to avoid excessive wear at the lap and pickup points. Slipping involves loosening the dead-line anchor and placing a few more feet of line into service from the storage reel. Cutting requires that the line on the drawworks reel be loosened and a section cut and removed. Slipping changes the pickup points, and cutting changes the lap points. A line is usually slipped several times before it is cut. Care must be exercised to ensure the line is not slipped sufficiently to allow the pickup point to be moved from one sheave to another. Similar care must be given to the cutting program.

Derrick and Substructures. The derrick and substructure play an important role in drilling operations. The derrick provides the vertical height necessary for the hoisting system to raise and lower the pipe. The substructure provides the height required for the blowout preventer stack on the wellhead below the rig floor. The derrick and the substructure must have enough strength to support all loads, including the hook load, pipe set in the derrick, and wind loads.

The API has developed size classifications for the derrick shown in Fig. 16—25. The specifications arc summarized in Table 16—1. In addition. Table 16—1 contains data used in determining wind loading.

The derrick and substructure must be able to support the loads imposed by pipe weight on the block plus a portion of the drillstring standing in the derrick. Due to the manner in which the hook load is distributed over the derrick, the effective load may exceed the actual. When heavy casing strings arc run, it may be necessary to lay down some drillpipc initially so the dcrrick loading capacity is not exceeded.

The dcrrick load resulting from a hook load can be evaluated with Fig. 16-26, The force on the derrick (F„) includes the hook load, L; the tension in the fast line. TF; and the tension in the dead line, T^,:

The tension in the fast line in a nonideal environment, including friction, is shown in Eq. 16.9:

A - Vertical distance from the top of the base plate to the bottom of the crown block support beam

B - Distance between heel to heef of adjacent legs at the top ot the base plate

C - Window opening measure in the clear and parallel to the center line of the derrick side from top of base plate D - Smallest clear dimension at the top of the derrick that would restrict passage of crown block

E - Clearance between the horizontal header of the gin pole and the top of the crown support beam

Fig. 16-25 Derrick size classifications (Courtesy API)

Since the dead line does not move, the tension is shown in Eq. 16.12:

Eq. 16,11 can be rewritten as follows:

The total force on the derrick (FD) is not evenly distributed over each of the four legs (Fig. 16-27). The fast-line tension is distributed evenly between

Table 16-1 General Dimensions of Derrick Sizes

Nominal

Table 16-1 General Dimensions of Derrick Sizes

Nominal

 Base Wind Load Derrick Height (A) Square (B) Pipe Size, Total Length, Pipe Weight, Area, Size No. ft in. ft in. in. ft lb/ft ft£ 10 80 0 20 0 27/s 9,200' 6.5 264 [] 87 0 20 0 2% 9,200' 6.5 264 I2 94 0 24 0 27/s 9,200' 6.5 264 I6 122 0 24 0 414 4,500; 18.5 353 18 136 0 26 0 m 10,8003 18.5 510 18A 136 0 30 0 5 8,9004 22.5 510 19 146 0 30 0 5 15,0005 22.5 558 20 147 0 30 0 5 15,0005 22.5 558 25 189 0 37 6 5 20,000" 22.5 810

Courtesy American Petroleum Institute, Standard 4A

### See Fig. 16-25 for A and B

• 132 stands (12 stands by 11 stands)
• 48 stands (6 stands by S stands)

3110 stands 00 stands by 11 stands)

190 stands (9 stands by lü stands)

5160 stands

6148 stands

Courtesy American Petroleum Institute, Standard 4A

### See Fig. 16-25 for A and B

• 132 stands (12 stands by 11 stands)
• 48 stands (6 stands by S stands)

3110 stands 00 stands by 11 stands)

190 stands (9 stands by lü stands)

5160 stands

6148 stands

Fig. 16-26 Free body diagram of the block, fast, and dead lines

legs C and D since the drawworks is commonly positioned between the legs. The dead-line tension (TD) is applied almost exclusively to a single leg since the dead-line anchor is near a leg. The force on each leg can be summarized as follows:

Total

4 L 2EBN

4ErN

Derrick leg t

Fast line

 • • • • • • • • • • Lines to block

Fig. 16—27 Typical rig lloor layout for distribution of forces

The load on leg A is greater than any other leg if EB > 0.5. Therefore, the maximum derrick load can be defined as four times the strength of the weakest ¡eg:

Where:

The derrick wiil be exposed to loads created by wind acting horizontally on pipe set back in (he derrick. The wind load (Lw) is calculated from Bq. 16.15:

Where:

Lw = wind load, lb/ft" V — wind velocity, mph

The wind load (Lw) acts on the wind load area shown in Table 16-i, For example, a 50-mph wind acting on a type 18A derrick has the following load:

The total load is as follows:

Rotary System. The rotary system is responsible for imparting a rotating action to the drillstring and bit. The principal components are the kelly, rotary and drive bushings, swivel, and rotary hose. Fig, 15-1 illustrates some of the components of the rotary system.

The kelly is a square or hexagonal member that scrcws into the drillstring and provides a flat surface for applying torque to rotate the pipe. The standard kelly is 40 ft long with a 37-ft drive section. An optional 54-ft kelly is available. Kelly cock valves are located on cither end of the kelly. The upper kelly cock and all connections above it have left-handed threads. A saver sub usually is run on the lower end of the kelly. Too! joint wear is then confined to an inexpensive part.

The kelly and rotary bushings are responsible for turning the kelly (Fig. 16-28). The rotary bushing is driven by the prime movers via the compound or an electric motor. The kelly bushing sits in the rotary bushing and is held in place by four pins. The inner diameter of the kelly bushing is square or hex-agonally arrayed set of rollers to match the kelly.

The rotary swivel serves two important functions in the drilling process. It is a connecting point between the circulating system and the rotary system, [n addition, it provides a fluid seal that must absorb rotational wear while holding pressure (Fig. 16—29). The upper section of the swivel has a bail for connection to the elevator hook.

Master bushing-

Kelly square drive bushing removed from table

Cut-away showing master bushing

Fig, 16-28 Rotary table with square drive bushings (Courtesy API)

The rotary hose connects the standpipe to the swive! through the gooseneck The hose should not be used in temperatures above ]80°F. Oil muds with a high aromatic contcnt or an aniiinc point in excess of I50°F should be avoided {Fig, 16-30).

Rotary horsepower requirements depend on hole friction, angle, and straightness, Eq, 16.16 shows the basic procedure.

 Where: HP* = rotary horsepower T = rotary torque, ft-lb N — rotary speed, rpm

Although Fq. 16.16 is straightforward, it is difficult to estimate the rotary torque prior to actual drilling.

An empirical approach has been developed for estimating rotary horsepower requirements:

Fig. 16-29 Rotary swivel connections (Courtesy API)

where F is dependent on drillstring weight, weight oil bit, hole friction, and bit bearing condition. The torque factor (F) is generally acccpted by many industry personnel to be as follows;

1.5—shallow holes less than 10,000 ft with light drillstrings 1.75—10,000-15,000-ft wells with average conditions 2.0—deep holes with heavy drillstrings

Fig. 16-30 Layout for rotary hose (Courtesy API)

Although these empirical estimates are subject to many variables, they have provided reasonable estimates of rotary horsepower requirements.

### Example 16.7

A rig is expected to drill a 17,500-ft well with I7.6-lb/gal mud required at the bottom. If the rotary will operate at 90 rpm, determine the horsepower requirements.

### Solution:

1. For a 17,500-ft well, use a torque factor (F) of 2.0,
2. From Eq. 16.17.

Circulating System. The circulating system is a major component of the drilling operation, affecting its overall success. Its many purposes are similar to the purposes of the mud system. The fluid performs three essential and four supplemental functions and offers two desirable benefits.

### These jobs must be done on all wells:

• control subsurface pressures
• keep the hole walls from collapsing
• remove cuttings from the hole

### Fluid circulation should perform these beneficial functions:

• clean, cool, and lubricate the bit
• suspend cuttings when circulation is stopped
• protect potential pay zones from damage
• transmit hydraulic horsepower to the bit

### Extra benefits sometimes available include:

• provide information about the formation penetrated
• help buoy the weight of drillpipe and casing

The ability of the circulating system to do these jobs is controlled by fluid constituents (Chapter 8) and circulating hardware. The components of the circulating system arc shown in Fig. 16-31. The heart of the system, which is the mud pump and related equipment, will be discussed. Other components such as the drill string and kelly are discussed in other sections of the text.

Mud Pumps, The drilling fluid is circulated through the system shown in Fig. 16-31 with the mud pump. The primary components of the mud pump equipment include the following:

• suction line components
• valves
• liners
• pulsation dampeners

This equipment can be selected properly so the well to be drilled will be optimized with respcct to the available equipment.

Fig. 16-31 Circulating system (Courtesy !ADC)

Mud pumps are designed for pressure output. How rate, and horsepower requirements. High pressures arc required to circulate heavy muds in deep wells and to optimize hole cleaning below the bit. How rate considerations arc usually not a limiting criteria for most operations except when drilling large-diameter surface hole sections. Maximum available pump horsepower is sometimes used in surface holes or when operating downhole motors.

Pump suction requirements are an often-neglected consideration in mud pump planning that can seriously rcduce the efficiency of the pump. If the mud pump cannot gain access to the proper amounts of mud when needed, its output will be less than maximum for the particular pump rate, l ire two common types of suction systems are atmospheric and supercharged.

The pump suction system used during early drilling operations was a flooded or atmospheric suction. Atmospheric pressure, which is approximately 15 psia, is used in conjunction with hydrostatic pressure to force the mud into (he suction valves of the pump. This system was more effective as the height of the mud increased, as is the case when suction from a 2-ft earthen pit is compared to a steel mud pit with an 8-ft column of mud (Fig. 16—32). This "head" of pressure must overcome inertia forces arid friction pressures of the mud in the lines. Obviously, long suction lines with many bends significantly decrease the effectiveness of the system. In addition, gas-cut or high-viscosity muds impede the system's operation.

Fig. 16-33 illustrates the suction input pressure requirements for two pumps under several operating conditions. The optimum input requirements increase as the pump stroke rates increase. Under most conditions, a flooded or atmospheric suction system cannot meet the upper demands.

Improved pumping characteristics can be expected with a suction dam-pcncr. A dampener will not increase the horsepower of a pump, but it can help utilize most of the available horsepower. A dampener accomplishes this feature by increasing the speed at which the pump can run without the problems of knocking and accompanying pressure surges (Fig. 16—34). The extra speed advantage is the basic reason to use a dampener, but other advantages include the following:

• stabilizes pressures in the suction line
• allows the use of longer suction lines or smaller-diameter lines
• makes suction from deeper pits possible
• allows the use of heavier muds
• allows the use of higher temperature muds

The advantages of a dampener add up to more flexibility of present equipment and increased pump life.

The most successful method of correcting mud pump suction problems is the use of centrifugal pumps as boosters (superchargers or preehargers). The

Fig, 16-32 Suction characteristics from earthen and steel mud pits (Courtesy TRW Mission)

Fig. 16-33 Various pressure losses for two sizes of pistons (Courtesy TRW Mission)

addition of a supercharger offers many advantages: elimination of shock loads, smoother operation, increased bearing life, and higher-speed operation. The supercharger, shown in Fig. 16-35, also enables the mud pump to handle gascut or aerated mud, giving better filling characteristics with less chance of losing prime.

Most mud pumps currently used in the drilling industry are duplex or triplex positive displacement pumps. The duplex double-acting pump has two liners with valves on both ends of the liner. Fluid is displaced from the liner on the forward and backward strokes of the rod plunger (Fig. 16-36),

The triplex single-acting pump has three liners with valves on one end of the liner. Fluid is displaced on the forward stroke only. The triplex pump has a smooth action that can pump at higher stroke rates even though the volumetric output per stroke is less than the duplex pump.

The volumetric output of a pump on a per-stroke basis depends on the stroke length, the rod diameter, the liner size, and the volumetric efficiency of the pump. For a duplex double-acting pump, the volumetric output is calculated from Eq. 16.18:

Fig. 16-34 A typical shop-made suction dampener in use in the field (Courtesy TRW Mission)
Fig. 16-35 Centrifugal pump used as a supercharger (Courtesy TRW Mission)
 Where: vD = volumetric displacement, gal/stroke Ls = stroke length, in. dL = liner diameter, in. dR — rod diameter, in. Et = pump volumetric efficiency, %

Rod sizes are often assumed to be 2 in. in diameter. The volumetric output for a triplex pump is as follows:

Since the triplex pump does not displace fluid on the backward stroke, the rod diameter is not considered in Eq. 16.19.

+1 0

## The Productive Entrepreneur

Entrepreneurs and business owners. Discover 45 Insightful Tips To Motivate, Encourage And Energize You To Become A Successful Entrepreneur. These Tips Will Move You Forward Towards Your Goals As An Entrepreneur. Use It As A Handbook Whenever You Need To Get Motivated.

Get My Free Ebook

### Responses

• Ulrich
How to calculate DRILLING mud pump efficient?
9 years ago
• callan
What is supercharger for mud pump?
8 years ago
• Einojuhani Rautavaara
How slip and cut programme is executed in petroleum drilling?
4 years ago
• Andrea
How is a slip and cut program executed?
4 years ago
• Max
How is slip and cut program excuted?
3 years ago
• jean perry
How to calculate when to slip and cut drill line?
9 months ago
• SCARLETT
What are the procedure of slip and cut program?
6 months ago