Page 110 - Air and Gas Drilling Manual

3-38 Air and Gas Drilling Manual
where γ m is the specific weight of the drill fluid (lb/ft ),
γ s is the specific weight of steel (lb/ft ).
The maximum allowable design axial tension force, F a (lb), is
F = 09. F y 3 3 (3-3)
a
where F y is the drill pipe tension force to produce material (steel) yield (incipient
failure of the drill pipe) (lb). These tension failure values can be obtained from the
tables in Appendix B or Tables 3-3, 3-6, and 3-7.
For most drill string designs a factor of safety is used to insure there is a
margin-of-overpull (MOP) to allow for a stuck drill string. The drill string design
factor of safety, FS, is given by
F
FS = a (3-4)
F

The MOP is determined by the rotary drilling rig hoisting capacity. Thus, MOP is
MOP = F − F (3-5)
c
where F c is the hoisting capacity of the rotary drill rig (lb). But the total hoisting
axial force cannot exceed F a.

Illustrative Example 3.1 A section of a vertical well is to be rotary air drilled
from 7,000 ft to 10,000 ft with a 7 7/8 inch tri-cone roller cutter drill bit. This is to
be a direct circulation drilling operation. Above the drill bit the BHA is made up of
500 ft of 6 3/4 inch by 2 13/16 inch drill collars and similar diameter survey subs
and nonmagnetic drill collars. The drill pipe available from the drilling contractor is
API 4 1/2 inch, 16.60 lb/ft nominal, EU-S135, NC50(IF). Determine the FS and
the MOP associated with using this drill string to drill this section of the well.
From Appendix B, Table B-1 gives 100 lb/ft for the 6 3/4 inch by 2 13/16 inch
drill collar. Table B-5 gives 18.62 lb/ft for the drill pipe actual weight per unit
length (includes tool joints). Since the drilling fluid is air, then γ m = 0, and K b =
1.0. The maximum axial tension force in the top drill pipe element of the drill
string is when the depth is 10,000 ft and when the drill bit is lifted off the bottom
of the well (after the target depth has been reached). The maximum axial tension
force is determined from Equation 3-1. This is
F = ( [ 9 500 )(18 62. ) + (500 100 ) )
,
)( ] (1 0.

,
F = 226 890 lb
Table B-5 gives the tension force to produce yield in the drill pipe (in this case the
pipe body). This tension force to yield is 595,004 lb. Equation 3-3 can be used to