Drill String Design
WHAT THE FIELD MAN SHOULD KNOW ABOUT DRILL STRING DESIGN
A. Big Hole, Drill Collar, and Drill Pipe Design
1. Drill Collar Design
a. Formulas:
Bit weight (lbs)
(1)
Air Weight (D.C.) =
B. F. x S. F.
Where: B.F. = Buoyancy Factor
S.F. = Safety Factor (i.e., 0.90 = 10% S.F.; 0.70 – 30% S.F.)
(2) B.F. (P.C.F.)
(3) B.F. (lb/ gal) =
=
(489)- (Mud Weight Ib/cu.ft.)
489
(65.44) – (Mud weight lb/ gal)
65.44
(4) Correcting S.F. after Air Weight Design is complete =
100
-
(desired bit weight) x (100)
(Air Weight Calculated) x (B.F.)
(5) Rule of-Thumb for Tapered String:
•
Have at least one stand of transition per O.D. taper.
•
•
Have no more than 2" O.D. change.
Have crossover subs at least one foot for each one-inch change in taper.
Bottle neck X-0 less than 36" or better than the "big bevel".
Fifteen to twenty-one joints of Heavy- wahte will improve your tapered
assembly by spreading the transition zone out further and protecting the
d.p.
•
Remember that: Drill collars and Hevi wate connections are designed to be
run in compression, and drill pipe connection are designed
to be run in tension.
Remember that: A BAD KELLY SAVER SUB CAN DAMAGE A
COMPLETE STRING OF D.P.
1
b. Example of d.c. design
Find the correct d.c. Air Weight to pick up. if desired bit weight is 70,000 Ibs..
and the Mud Weight is 9.1 Ib/gal. (68 pcf). The hole size is 171/2", and a
tapered string with 15% S.F. is desired. Consider using;
11" x 3" (8970 Ib/single)
9-3/4" x 3" (6900 Ib/single)
8 1/2" x 3" (5070 Ib/single)
70,000 lbs
Air Weight =
=
95,800 lbs.
0.86 x 0.85
B.F.
= ( 65.44 – 9.1) / 65.44 = 0.86
*S.F. (15%) = 100 - 15 = 85% (0.85)
Try: (3) 11" d.c.
(3) 9 3/4" d.c.
(3) 8 1/2" d.C.
Total
Required
Remaining
= 26,910 Ibs,
= 20,700 Ibs,
= 15,210 Ibs,
= 62,820 Ibs,
= 95,800 Ibs,
= 32,980 Ibs,
Second Try: (3) Additional 9-3/4" = 20,700 Ibs.
(3) Add. 81/2"
= 15,210 Ibs.
Total
35,910 Ibs.
So With:
(3) 11" d.c.
(6) 9 3/4" d.c.
(6) 81/2" d.c.
S.F. Difference
Actual Air Weight
= 98.730 Ibs
Calculated Air Weight
Difference.
Corrected S.F.
= 26,910 Ibs
= 41,400 Ibs
= 30.420 |bs.
98,730 Ibs.
= 95,800 Ibs.
= 2,930 Ibs.
(70,000) x (100)
(98,730) x (0.86)
100 - 82.4 = 17.56% (New S.F.)
2
2. Drill Pipe Design
Formulas; (This material is from RP7G )
(1) Design
parameters
- It is intended to outline a step-by-step procedure to insure complete
consideration of factors, and to simplify calculations. The following design
criteria must be established.
• Anticipated total depth with this string.
•
•
•
•
•
•
Hole Size.
Expected mud weight.
Desired Factor of Safety in tension and/or Margin of Overpull.
Desired Factor of Safety in collapse.
Length of drill collars, O.D., I.D., and weight per foot.
Desired drill pipe sizes, and inspection class.
(2)Tension Loading
The design of the drill string for static tension loads requires sufficient
strength in the topmost joint of drill pipe to support the submerged weight of
all the drill pipe plus the sub merged weight of the collars, stabilizer and bit.
This load may be calculated as shown below. The bit and stabilizer weights
are either neglected or included with the drill collar weight.
P = ((Ldp x Wdp) + (Lc x Wc))Kb
Where P
submerged load hanging below this section of drill pipe,lb.
Ldp = length of drill pipe
Lc = length of drill collars, ft.
Wc = weight per foot of drill collars in air.
Kb = buoyancy factor
If the pipe is loaded to the extent shown in the tables, it is likely that some
permanent stretch will occur, and difficulty may be experienced in keeping the
pipe straight. To prevent this condition, a design factor of approximately 90% of
the tabulated tension value from the Hand book is sometimes used; however, a
better practice is to request a specific factor for the particular grade of pipe
involved from the drill pipe supplier .
3
Pa =
Pt x 0.9
Where:
Pa = maximum allowable design load in tension, lb.
Pt = theoretical tension load from hand book, Ib.
0.9
a constant relating proportional limit to yield strength.
The difference between the calculated load P and the maximum allowable tension
load represents the Margin of Overpull (M.O.P.).
M.O.P. = Pa / P
The selection of the proper safety factor and/or margin of over pull is of critical
importance and should be approached with caution. Failure to provide an
adequate safety factor can result in loss or damage to the drill pipe while an
overly conservative choice wilt result in an unnecessarily heavy and more
expensive drill string.
Normally, the designer will desire to determine the maximum length of a specific
size, grade and inspection class of drill pipe which can be used to drill a certain
well. By combining the two formulas given under, the following equation result;
Pt x 0.9
S.F. x Wdp x Kd
-
Wc Lc
Wdp
= Ldp
the string is to be a tapered string, i.e., to consist of more than one size, guide or
inspection class of drill pipe, the pipe having the lowest load capacity should be
placed just above the drill collars and the maximum length is calculated as
shown previously. The next stronger pipes placed next in the string, and the WL
term in equations is replaced by a term representing the weight in air of the drill
collars plus the drill pipe assembly in the lower string. The maximum length of
the next stronger pipe may then be calculated.
4
(3)Collapse Due to External Fluid
Pressure
Pp
=
Pac
S.F.
Pp
= Theoretical collapse pressure from tables, psi
S.F. = Safety factor
Pac = Allowable collapse pressure, psi
When the fluid level inside and outside the drill pipe are equal and provided
the density of the drilling fluid is constant, the collapse pressure which is
zero at any depth i.e., there is no fluid inside the pipe, the actual collapse
pressure may be calculated by the following equation.
Pc
= LWg
19.251
Or
Pc = LWf
144
Where
Pc = net collapse pressure, psi.
L = the depth at which Pc acts, ft.
Wg = weight of drilling fluid, Ib/gal.
W f = weight of drilling fluid, lb/cu.ft.
If there is fluid inside the drill pipe, but the fluid level is not as high inside as
outside, or if the fluid inside is not the same weight as the fluid outside, the
following equation may be used:
Pc = LW’g
- ( L- Y ) W’g
19.251
Or
Pc = LWg - ( L – Y) W’f
144
Where: Y = depth of fluid inside drill pipe, ft.
W’g = weight of drilling fluid inside pipe, Ib/gal.
W’f = weight of drilling fluid inside pipe, Ib/cu.ft.
5
(4) Torsional Strength
-
The torsional strength of drill pipe becomes critical when drilling
deviated holes, deep holes/ reaming, or when the pipe is stuck. The actual
torque applied to the pipe during drilling is difficult to measure, but may
be approximated by the following equation.
T
=
Where: T
HP x 5,250
RPM
= torque delivered to drill pipe, ft.-lbs.
HP = horse power used to produce rotation of pipe
RPM = revolutions per minute
b. Example
Anticipated T.D. – 12,000 ft.
Hole Size
- 121/4 inches.
Expected Mud Wt. - 13.4 Ib/gal. (100 pcf).
Desired S.F. or Safety (Margin of over pull) 50, 000 Ibs. (collapse S.F. 1.125).
Length and Size Drill Collars (3) 10" d.c. w/2 13/16" I.D. 2 13/16"
(12) 9" d.c. w/2-13/16: I.D. - 2-13/16"
Total Length - 450 ft.
Total Air Wt. - 92, 340 Ibs.
Hevi- wate Pipe - (21 joints) 5" x 3" x 50 Ib/ft.
Total Length 630 ft.
Total Air Wt. - 31, 500 Ibs.
Drill Pipe - 5", 19.5 Ib/ft., XH, Class-2. Adj. Wt. - 20.99 Ib/ft.
Grade "E" and/or Grade"95" (21.09 Ib/ft.)
Tensile Strength - Grade "E" API Class 2 = 311,540 Ibs.
6
Grade "95" API Class 2 = 394,600 Ibs.
Buoyancy Factor - 0.795 w/13.4 Ib/gal. mud.
c. Calculations:
(1) Length of d.p. (5", 19.5 Grade "E") to go above Hevi-wt. and d.c.'s.
Pt x 0.9 - M.O.P.
Wdp x Kb
Ldp =
– WcLc = Ldp
Wdp
( 311,540 x 0.9) - (50,000) _
20.99 x 0.795
123,840 lbs.
20.99
= 13,806 - 5900 = 7,906 ft. (will use 7900 ft.)
*123,840 - Total Air Weight of d.c.'s and Hevi Wt. d.p.
(2) Calculate Air Weight hanging below this top joint of Grade "E" d.p.
P = (LDP x Wdp ) + (Lc x Wc ) Kb
(not corrected for buoyancy)
P = [(7900 ft.) x (20.99) + (123,840)]
= 165,821 Ibs. + 123,840 Ibs. = 289,661 Ibs.
(3) Accumulate Length of Drilling Assembly
Tapered d.c.
450 ft.
Hevi-Wt. 5"
630 ft.
Grade "E" 5"
7900 ft.
Accum. Total
Require
8980 ft.
12,000 ft.
Remaining Grade
3,020 ft.
(4) Length of d.p. No.2 5” Grade "95”
Ldp No2 = ( 394,600 x 0.9) - (50,000) - 289,661 lbs.
(21.09 x 0.795)
21.09
18,199 - 13.735 = 4465 ft. (will use only 3020 ft.)
*•289,661 Ibs. = Total Air Wt.at D.C. Hevi Wt., Grade "E".
7
(5) Calculate Total Air Weight of string
P = {(3020 ft. x 21.09) + 289,661 Ibs.}
63.692 + 289,661 = 353,353 Ibs.
(6) Corrected for Buoyancy
Weight in 13.41 Ib/gal. mud = 353,353x 0.795= 280,916 Ibs.
( 7) Summary of Design Calculation
Weight in
Items
Length
13.4 Ib/gal.mud
D.C. 10" and 9" tapered
450 ft.
73,410 Ibs.
Hevi-Wt. d.p. (5" - 50 Ib/ft.)
630 ft.
25,043 Ibs.
D.P. Grade "E" (5". 19.5 - 20.99)
7,900 ft.
131,828 Ibs.
D.P. Grade '•95" (5". 19.5 - 21.09) 3,020 ft.
50,635 Ibs.
Totals.............................
12,000 ft.
280,916lbs.
(8) Check on Margin of Overpull
Grade “ E “
(311,540 x 0.9) - (230,281) = 50,105 Ibs
Grade "95" (394,600 x 0.9) - (280,916) = 74,224 Ibs.
(9) This design Could Be Used DownTo 13, 445 f t.
Note: Only 3020 ft. of Grade"95" was used and this allowed another 1466
ft. of Grade "95" for use.
(10) AIIowable Collapse Pressure
For 5", 19.5, Grade "E", API Class - 2 = 4760 psi. Btm 5" @10,920'
Pac = ( 4760) = 4231 psi
( 1.125)
(11) Net Collapse Pressure
Pc = (10,920 ft )(13.4 Ib/gal.) = 7600psi
19.251
8
(12) Tension Safety Factor
Grade “ E” =
Grade "95" =
355,140
280,386
355,140
280,916
= 1.22
= 1.26
3. What Happens to Margin of Overpull (M.O.P.) if Mud Weight is Increased?
Example: Mud weight increased to 18.0 Ib/gal.
B.F. = 0.725
M.O.P. Grade "É" (311,540 x 0.9) - (210,004) = 70,382
Therefore; M.O.P. increased by 20.000 Ibs.
M.O.P. Crade "95" ( 394. bOO x 0.9) - (256,181 Ibs.) =
= 98,959 Ibs.
Therefore: M.O.P. increased by 25,000 Ibs.
4. What happens To Margin Overpull (M.O_P.) If, Due To Rough Wear The
D.P. is Downgraded To API Class 3? (This example was designed based
on API Class 2.)
B.F.
= 0.795 (original 13.4 lb/gal.)
M.O.P. Grade "E" (240,300 x 0.9) — (230,281) = 216, 270 - 230, 281 =
= - 14,011
Therefore: M.Q.P. would be below safe design for tensile loading, and either a
newer pipe or more Grade '95"or higher grade would be required.
M.O.P. Grade "95" - (304,380 x 0.9) —(280,916) = - 6,974
Therefore; M.O.P. would be below safe design for tensile loading, and either a
newer (higher API Class) pipe or a higher grade pipe would be required.
B. Drill Pipe Design Using Two Different Grades of Pipe
Example:
Anticipated Total Depth with this String...................... 1 5, 000 feet
Hole Size (9-5/8" Casing (@ 12,000 feet) ....................... 81/2 inches
Expected Mud Weight.........................................
1 3. 4 Ib/gal.
9
Desired S.F. of Safety Margin - Collapse - 1.125, M.O.P. . 50, 000 Ibs.
Length and Size of Drill Collars 61/2 " x 2 13/16 x 950 ft.
wt. Ib/ft. 91
Total Wt. (air)
86,450 lbs.
Drill Pipe; 5", 19.5 x H, Grade "E" 20.99 Ib/ft. API Class No. 2
5” , 19.5, x H, Grade "95" 21.09 Ib/ft. APl Class No.2
Tensile Strength; Grade "E". API Class No. 2................ 311, 540 Ibs.
Grade "95". API Class No. 2 ............... 394,600 Ibs.
Buoyancy Factor w/13.4 Ib/gal................................ 0.795
CALCULATIONS
1. Length of d.p. (5" 19.5 Ib/ft., Grade "E") to go above d.c.
Ldp =
(311,540 x 0.9) - 50,000
20.99 x 0.795
_
86,450 Ibs.
20.99
= 13,806 - 4119 = 9688 ft.
2. Calculate Air Weight Hanging Below This Top Joint of Grade "E" d.p
p = (9688 x 20.99) + 86,450 Ibs. = 203,351 + 86,450 = 289,801 Ibs.
3. Accumulated Length of Drilling Assembly
D.C............................................
950 feet
D.P. Grade "E" .............................. 9,688 feet
Total........................................
Require......................................
10,638 feet
15, 000 feet
Remaining Crade "95"....................... 4, 362 feet
4. Length of d.p. No. 2 (Grade "95")
Ldp No2 =
(394,600 x 0.9) - 50,000 - 289,801
21.09 x 0.795
21.09
= 18,199 - 13,741 = 4,458 ft. (will use 4,362 ft.)
5. Calculate Total Air Weight of String
P = [4362 (21.09) + 289,801] = 91.995 + 289,801 = 381,796 Ibs.
10
6. Corrected For Buoyancy
Weight in 13.4 Ib/gal. mud = 0.795 x 381,796 Ibs. = 303, 528 Ibs.
7. Summary of Design Calculation
Item
Weight in
13.4 Ib/gal. mud
Length (ft.)
D.C. (61/2 x 2-13/16 x 9501)
950
68,728 Ibs.
D.P. (19.5 20.99 Grade "E")
9,688
1,6b4
D.P. (19.5 - 21.09 - Grade "95")
4,362
73,136
Total..................................
15,000 ft.
303,528 Ibs.
8. Check on Margin of Overpull (M.O.P.)
5", (19.5 - 20.99) Grade "E" -- (311,540 x 0.9) - 230.392
= 50.000 Ibs.
5". (19.5 21.09) Grade “95" - - (394,600 x 0.9) - 303.528 = 51,612 Ibs.
9. Tension Safety Factor
Grade -E"
= ( 311,540 x 0.9) = 1.22
230,392
Grade- “95” = (394,600 x 0.9) = 1.17
303,528
Many engineers feel that the design S.F. should be approximately 1.25 - 1.33
C. Tapered Drill Pipe String Down to 18,000 Feet
Example:
Anticipated Total Depth..........................................
18,000 feet
Hole Size (7" liner - 25,000 - 11, 500 ft.)........................ 6-1 / 8 inches
Expected Mud Weight (0.833 psi/ft.)..... ...................... 16. 0 Ib/gal.
Designed Safety Factor - M.O.P. ................................ 50,000 Ibs.
Length and Size of Drill Collars ................................. 950 ft.
Weight d.c. (Air) 43/4” x 2 x 950 ft.) 50 Ib/ft.. ................ 47, 500 Ibs.
Drill Pipe – 31/2. 13.30 - 14.06. Grade "E", API Class-2, XH
5", 19.50 - 20.99, Grade "E", API Class-2, XH
5", 19.50 - 21.09, Grade "95", API Class-2, XH
11
Tensile Strength
31/2", 13.30. Grade "E", API Class-2, 212,250 Ibs.
5", 19.50,
Grade "E", API Class 2, 311,500 Ibs.
5", 19.50,
Grade "95", API Class-2, 394,600 Ibs.
Buoyancy Factor w/16.0 Ib/gal. ................................. 0.756
1. D.P. Design (31/2, 13.30, Grade "E") for Below Top uf liner to T.D.
6600 ft. required.)
Ldp =
(212,250 x 0.9) - 50,000 14.06 x 0.756
47,500
14.06
= 13,267 - 3378 = 9889 ft. (will use 5650 feet) .
2. Air Weight of String Supported by Top Joints of 31/2" d.p.
P1
= (5650 ft.x 14.06) + (950 ft. x 50 Ib/ft.)
= 79,439 + 47,500 = 126,939 Ibs. (31/2" d.p. + 4-3/4" d.c.)
3. d.p. Design for 5" 19.5. Grade "E" (above 31/2", 13.30)
Ldp No.2 =
(311,500 x 0.9) - 50,000
20.99 x 0.756
- 126,939 = 8469 ft.
20.99
4. Air Weight of String Supported by Top Joint of 5", 19.5, "E"
P = (8469 x 20.99) + 126,939 = 177,764 + 126,939 =
= 304,703 Ibs. (5", 19.5, "E" +31/2 + 4-3/4d.c.)
5. d.p. Design for 5", 19.5, Grade "95", (above 5", Grade "E")
Ldp No.3 =
(394,600 x 0.9) - 50,000 - 304.703
21.09 x 0756
21.09
= 4690 feet (will use 2931 ft.)
6. Air Weight of String Supported by Top Joint of 5", 19.5, Grade
"95"
P3 =
(2931 ft. x 21.09) + 304,703 - 61,815 + 304,703 = 366, 518 Ibs.
(complete string weight in air)
12
7. Buoyancy Weight of String
366,518 x 0.756 = 277,087 lbs.
8. Summary of Drill Strinq Design
Item
Length (ft.)
Drill collars (4-3/4 x 2 x 950)
950
Drill Pipe (31/2, 13.30, Grade "E")
5,650
Weight in 16.0
Ib/gal. mud (Ibs.)
35,910
60,056
Drill Pipe (5", 19.5, Grade "E")
8,469
134,390
Drill Pipe (5", 19.5. Grade "95")
2,931
46,732
Totals. ............................
18,000 ft.
277,0881bs.
9. Margin of Overpull Check
31/2. 13.30, Grade "E" = (212,250 x 0.9) - 95,966 = 95,059 Ibs.
5", 19.50, Grade "E"
= (311,500 x 0.9) – 230,056 = 50,294 Ibs.
5", 19.50, Grade "95" = (394,600 x 0.9) - 277,088 = 78,052 Ibs
10. Torsional Yield
31/2", 13.30. Grade "E", API CIass No. 2 =11,710 ft. Ib.
5". 19.50, Grade "E", API Class No. 2
= 26,320 ft. Ib.
11. Collapse Pressure
31/2”, 13.30, Grade "E".............10,250 psi
Pc = 17,050 ( 16.0lb/gal.) = 14,170 psi
19.251
Therefore, this pipe must be protected against collapse and a D.S.T.
may result in failure.
13
CRITICAL ROTATION SPEEDS
Critical rotating speeds in drill pipe strings which cause vibrations are often the
cause of crooked drill pipe, excessive wear, and rapid deterioration and fatigue
failure. Critical speeds will vary with length and size of drill stem and collars and
hole size. Excessive power is required at the rotary to maintain a constant speed
at critical conditions. This power indicator, plus surface evidence of vibration,
should warn the crew that they are in the critical range.
Two types of vibration may occur. The pipe between each tool joint may vibrate
in nodes, as a violin string. This critical speed may be predicted approximately
by the formula:
4,760,000
RPM = L2
√ (D2 + d2)
Where
RPM = critical speed, revolutions per minute.
L
= length of one joint of pipe, inches.
D = outside diamefcer of pipe, inches.
d
= inside diamefcer of pipe, inches.
The critical speed predicted by this formula is probably accurate within 15 %. On this
basis the rotary speeds shown below should be avoided by plus or minu 15%.
Size:
RPM:
2-78 3-1/2 4-1/2
130
160
5
210 235
The second type of vibration is of the spring pendulum type, and may be
approximately predicted by the following formula:
RPM =
258,000
L
Where:
RPM = critical speed,revolutions per minute.
L = total length of string, feet.
14
Secondary and higher harmonic vibrations will occur at 4, 9, 16, 25, 36, etc., times
the speed determined from the equation above.
The vibrations of the spring pendulum type are probably less significant than the
node type. Each higher harmonic of the spring pendulum type vibration is also less
significant.
Particular care should be taken to avoid operating under those conditions which
would be the critical speed for both types of vibration because the combination is
particularly bad.
The depths and rotary speeds at which the two types of vibrations coincide are
shown below:
2 7/8”
130 RPM
3 ½”
160 RPM
4 ½”
210 RPM
1,200’
5”
235 RPM
1,960'
1,600'
1,100'
8,000'
6,600'
5,000'
18,000'
14,800'
11,200'
10,000'
32,000'
26,000'
19,700'
17, 500 '
4,500'
15