What is a Drill Collar?

13 minutes read

Contents

General Description Weight Requirements: Vertical Hole Example 1:Solution:Weight Requirements: Inclined Hole Example 2:Solution:Drill Collar Dimensions and Strengths

General Description

In the early years of rotary drilling , a typical bottomhole assembly (BHA) consisted of nothing more than a short crossover sub, or collar, between the bit and the drill pipe. Over time, drillers observed that penetration rates tended to increase as more weight was applied to the bit,and so they began using BHAs with longer sections of heavy pipe (in keeping with the original terminology, these heavy joints of pipe have continued to be called drill collars).

But even as these heavier BHAs brought improved bit performance, it became clear that applying more weight to the bit involved a trade-off: a greater potential for drill pipe failure due to increased compressive force and resultant bending or buckling.

When drill pipe rotates in a compressed state, it experiences cyclic stress reversals. Each rotation of the string causes the drill pipe fibers to go from compression to tension to compression. This high cyclic stress bending causes metal fatigue, which is cumulative and eventually leads to drill pipe failure (Figure 1).

drill pipe failure, drill pipe, drill collars, cyclic stress reversals — **FIGURE 1**

Buckling tendencies are greatest in the pipe body between the end upsets, which is the thinnest and most flexible part of the drill string. The abrupt change in cross-sectional area between the pipe body and the upsets sets up stress concentrations just above and below the much stiffer tool joint.

Thus, a fundamental objective in BHA design and drill collar weight selection is to “eliminate the potential for axial and torsional buckling of the drill pipe” (Mitchell, 1995). This is illustrated in Figure 2.

Eliminate the potential for axial and torsional buckling of the drill pipe, drill pipe, drill pipe failure, drill collars — **FIGURE 2**

Weight Requirements: Vertical Hole

Figure 3 (after Mitchell, 1995) illustrates two points of transition for a drill string in a vertical well:

The neutral point of bending,or NPB,is the point of transition from a straight to a bent drill string.
The neutral point of tension and compression (NPT & C) is the point below which the drill string is in compression, and above which it is in tension.

The NPB should always be located in the bottomhole assembly, below the drill pipe. It is this quantity that is of practical interest in drill string design. The NPT&C, on the other hand, is normally located in the drill pipe and is not used in BHA design.

two points of transition for a drill string, drill string, drill pipe, drill collars — **FIGURE 3**

In practical terms, the buoyant weight of the drill collars should correspond to the desired weight-on-bit:

$\left ( W_{dc} \right )_{air}\cdot BF=WOB\cdot SF$ (1)

where:

(W_dc)_air= weight of drill collars in air
BF= buoyancy factor, which is equal to $1-\tfrac{\rho _{fluid}}{\rho _{solid}}$ . For steel drill collars in drilling fluid, ρ_solid is equal to 65.5 ppg (7.85 specific gravity), and ρ_fluid is the drilling fluid density or mud weight (ppg or specific gravity).
WOB= desired weight on bit, based on manufacturer’s recommendations
SF= predetermined safety factor. May range from 1.10 to 1.25, depending on local conditions. In any case, it must be high enough to ensure that the NPB will always be in the BHA and not in the drill pipe.

The number of drill collars needed to attain the required weight on bit can be determined from total weight of the collars in air, the average collar length and the unit weight (i.e., lb/ft) of each collar.

Example 1:

Given the following information:

Bit weight required = 60,000 lb
Mud weight = 12 lb/gal
Safety factor = 1.15
Drill collars available: Nine 8′′ OD⋅2 13/16′′ D⋅31′ length, 150 lb/ft; also unspecified number of 6 3/4′ OD⋅2 16/13′′ ID⋅30′ length, 100 lb/ft

If all nine 8′′ OD collars are used, how many 6 3/4′′ collars are needed?

Solution:

Buoyancy Factor $(BF) = 1 - \left( \dfrac{12}{65.5} \right ) = 0.82$

Required weight of collars in air = $WOB\cdot \dfrac{SF}{BF} = 60,000\cdot \dfrac{1.15}{0.82} = 84,146\ lb$

Weight of 8′′ collars = $9\cdot 31\cdot 150 = 41,850\ lb$

Weight of 6 3/4′′ OD collars required = $84,146 - 41,850 = 42,296\ lb$

Number of 6 3/4′′ OD collars = $\dfrac{42,296}{100\cdot 30}=13.6$

⇒ Pick up fourteen (14) 6 3/4′′ drill collars

Weight Requirements: Inclined Hole

In a wellbore with an inclination angle α from vertical, the amount of drill collar weight that can be transferred to the bit is reduced by a factor equal to cosα At the same time, a portion of the drill pipe weight can be transferred to the bit, because its helical buckling tendencies are reduced. The equation that accounts for both drill collar and drill pipe weight in an inclined wellbore (Mitchell, 1995) is :

$WOB=\left [ \left ( W_{dc} \right )_{air}\cdot BF\cdot \cos \alpha \right ]+1617\cdot \left [\dfrac{BF\cdot \left ( OD_{dp}^{2}-ID_{dp}^{2} \right )\cdot \left ( OD_{dp}^{4}-ID_{dp}^{4} \right )\cdot \sin \alpha }{D_{hole}-OD_{dp}} \right ]^{0.5}$

where:

WOB= desired weight on bit, lb
(W_dc)_air= weight of drill collars in air, lb/ft
BF buoyancy factor
OD_dp= outside diameter of drill pipe, inches
ID_dp= inside diameter of drill pipe, inches
D_hole= actual hole diameter, inches

Example 2:

Given the same information as in Example 1 above — except that in this case the average hole inclination in the BHA interval is 35 degrees — how many 6 3/4′′ collars are needed to attain the desired 60,000 lb weight-on-bit?

Drill pipe = 4.5 inch, 16.6 lb/ft (3.826 inch I.D.); average hole diameter = 12.5 inches.

Solution:

$60000=\left [ \left ( W_{dc} \right )_{air}\cdot 0.82\cdot \cos 35^{\circ} \right ]+1617\cdot \left [\dfrac{082\cdot \left ( 4.5^{2}-3.826^{2} \right )\cdot \left ( 4.5^{4}-3.826^{4} \right )\cdot \sin 35^{\circ} }{12.5-4.5} \right ]^{0.5}$

$\Rightarrow \quad \left ( W_{dc} \right )_{air}=69,977\ lb$

Weight of 8′′ collars = 9⋅31⋅150=41,850 lb

Weight of 6 3/4′′ OD collars required = 69,977−41,850=28,127 lb

Number of 6 3/4′′ OD collars = $\dfrac{28,127}{100\cdot 30} = 9.4$

⇒ Pick up ten (10) 6 3/4′′ drill collars

Drill Collar Dimensions and Strengths

Drill collars are available in both API and non-API sizes. Length ranges for API collars are Range 2 (30, 31 ft) and Range 3 (42, 43 1/2ft).

Below is listed a range of drill collar outside diameters and bores.

Drill Collar outside diameters, inches *	Inside bores, inches*
3.00	1.0	1.125
3.125, 3.25	1.0	1.125	1.25	1.5
3.375, 3.5, 3.75			1.25	1.5	1.75
3.875			1.25	1.5	1.75	2.00	2.25
4.00, 4.125, 4.25, 4.50				1.5	1.75	2.00	2.25
4.75, 5.00				1.5	1.75	2.00	2.25	2.50
5.25				1.5	1.75	2.00	2.25	2.50	2.8375
5.50, 5.75				1.5	1.75	2.00	2.25	2.50	2.8375	3.00
6.00, 6.25, 6.50, 6.75, 7.00, 7.25, 7.50, 7.75, 8, 8.25, 8.50, 8.75				1.5	1.75	2.00	2.25	2.50	2.8375	3.00	3.25
9.00, 9.25, 9.50, 9.75					1.75	2.00	2.25	2.50	2.8375	3.00	3.25
10.00, 10.25, 10.50, 10.75						2.00	2.25	2.50	2.8375	3.00	3.25
11.00, 11.25, 11.50, 11.75, 12.00								2.50	2.8375	3.00	3.25