Design Procedure
A graphical approach to drillstring design is recommended. If one section of the string does not meet requirements it must be upgraded. The procedure is as follows:
1. Choose a weight and grade of pipe to satisfy the collapse conditions
2. Using the pipe chosen in 1. calculate the tension loading, including buoyancy effects. Draw the tension loading line and also the maximum allowable load line.
3. Modify the tension load as given in 2. by applying a design factor, MOP and Sh/St factor. Three design lines are thus generated.
4. If any of these design lines exceed the maximum allowable load, a higher rated drillpipe must be used for that section of pipe.
5. Calculate the new tension loading line for the new drill string and repeat steps 3. and 4.
Design a 5” 19.5 lb/ft drill string using new pipe to reach a TD of 12000 ft in a vertical hole. The BHA consists of 20 drill collars 6 1/4” x 2 13/14” (82.6 lb/ft) each 30 ft long. For design purposes assume the following:
MW = 10 ppg
MOP = 100000 lbs
Length of slips = 12”
Design factors = 1.125 (collapse)
= 85% (tension)
1. Collapse loading at 12000’ Pc = 0.052 x 10 x 12000 = 6240 psi
From Table 11
|
OD |
Grade |
Wt |
Collapse rating |
|
5” |
E |
19.5 lb/ft |
10000 = 8889 psi (x 1.125) |
|
5” |
E |
25.6 lb/ft |
13500 =12000 psi (x 1.125) |
choose: 19.5 lb/ft grade E drill pipe (ID = 4.276”)
2. Tension loading line (Figure 17)
at 12000’ Fj = (P x A)
where: P = 0.052 x 10 x 12000 = 6240 psi
A = п/4 (6.252 — 2.8122) = 24.47in2
Fj = (6240 x 24.47) = 152693 lbs
W11 = 20 x 30 x 82.6 = 49560 lbs
at 11400’ F2 = (P x A)
where: P = 0.052 x 10 x 11400 = 5928 psi
A = п/4 (6.252-52) + п/4 (4.2762-2.81252) = 19.19 in2
F2 = (5928 x 19.19) = 113758 lbs.
W2 = 11400 x 19.5 = 222300 lbs (Nominal weight used as approximation)
Calculating the tension at the top and bottom of each section:
at bottom of collars T = -152693 lbs
at top of collars T = -152693 + 49560 = -103133 lbs
at bottom of drill pipe T = -103133 + 113758 = 10625 lbs
at top of drillpipe T = 10625 + 222300 = 232925 lbs
Plot these figures on a graph, along with the maximum allowable load = 0.85×395000 = 335750 lbs
3. Construct Design loading lines:
a. multiply actual loads by 1.3 to obtain the design loads (Td) at surface T = 1.3 x 232925 = 302802 lbs
d
at 11400’ T = 1.3 x 10625 = 13812 lbs
d
b. add 100000 MOP to obtain Td
at surface Td = 232925 + 100000 = 332925 lbs at 11400’ Td = 10625 + 100000 = 110625 lbs
c. apply slip crushing factor
at surface T = 1.59 x 232925 = 370351 lbs
d
at 11400’ T = 1.59 x 10625 = 16894 lbs
d
Plot these 3 design lines on Figure 17
Choose 25.6 lb/ft grade E drill pipe.
5. Re-calculate tensile loading for new string and repeat 3. and 4. (Figure 18).
|
|
-150000 -50000 50000 150000 250000 350000 TENSION (LBS)
‘■ TENSION LOAD
O — TENSION LOAD X 1.3 (D. F.)
!• TENSION LOAD + 100000 LBS (MOP)
D — TENSION LOAD X 1.59 (Sh/St) — MAX. ALLOW. LOAD
Figure 17 Calculated Tension Loading on Drillpipe
|
Figure 18 Revised Tension Loading on Drillpipe |
