Evaluation of Axial Tension in Deviated Wells
The BFF obtained by the above method is not a measured value but rather is calculated from the hook load measurement. A major error can. therefore, arise if the axial load predicted from the mathematical model is incorrect: consequently, it is important to include all the factors in Eq. 4.65. Maidla (1987) has made a series of field investigations and reported that most values of borehole friction factor fall within the range of 0.25 to 0.45.
A deviated well with buildup, slant and dropoff sections will be considered here to study the effect of hole deviation on casing design. The well is kicked at 5.000 ft with a maximal inclination of 40° at an average buildup rate of 2°/100 ft. The well is then held at this inclination to 12.428 ft measured depth and Anally dropped off to a maximal inclination of 40° to the vertical axis at an average dropoff rate of 1.5°/100 ft. The drilling conditions, casing program and the well profile are presented in Table 4.1 and Fig. 4.11. The true vertical depth of the casing shoes, pore pressure gradient and drilling fluid program as shown in lable
1.1 remain the same in all the examples and. consequently, the collapse and burst load on each casing string will remain the same in all the examples.
In this section, tin1 suitability of the selected steel grades in tension will be in vest igate<l by considering the total tensile force resulting from casing buoyant weight, bending force, shock load, and frictional drag force. It is important to
note that although shock and drag forces occur when the pipe is in motion, it is unlikely that both can exist simultaneously, because the effect of drag force vanishes before the shock load is generated. Thus, the tension due to drag and shock load are calculated separately and only the maximal value is considered in the design of casing for tension (maximal design load concept).
Depth conversion will be made by projecting the actual well profile (measured depth) onto the vertical axis (true vertical depth). Vertical depth and inclination angle are calculated for all casing unit sections. The following formulas are used to calculate the depths and inclinations (refer to Fig. 4.11) :
1. For the vertical portion. 0 < ( < (kop-
Dx=lx (4.73)
2. For the buildup portion, (kop < t-i < (eob-
Let the buildup rate equal oh degrees per 100 ft. and the radius of curvature equal R ft. Thus:
bi 100
360 ~ 2 ж R _ 18,000 —> it — ;
7Г OL
The vertical projection of the measured depth ((2~ (кор) ‘п buildup section is:
D = R sin в where:
0 = {(-2 — (kop)
in the buildup section = di ((2 — (kop) x 10-2
Thus
D = 18 000 sin (di {(‘2 ~ (kop) x 10~2)
7Г C( ‘
= Rx sin (di(f2 — (kop) x 10~2)
True vertical depth for (kop < (г ^ (eob’-
18,0 . , . , г.-Л
‘kop d———— :— sln (Q1 ((‘2 — (kop) x 10 j
7Г Qi 4 ‘
3. For the first slant portion, £eob < (3 < (dop• the vertical projection of the measured depth, {£3 — (eob), is:
D — {l3 — (.eob) sin(90 — аг)
= (4 — (-eob) cose*!
True vertical depth for £eob < £3 < £dop■
D3 — Deop H — sin (qi (£eob — (kop) x 10~2)
7Г Q] 4 ‘
+ (^3 — £eob) COS Qi = DKOp + Iti sin ОЦ + (£3 — £E0B) cos an
= Deob + {£3 — £eob) cos Qi (4-75)
Thus, for a measured depth of 12,428 ft the TVD is:
Z)3 = 5,000 + 18,000 sin 40 + (12,428 — 7,000) cos 40 7Г 2
= 11,000ft
4. For the drop-off portion, £dop < £4 < (eod, the vertical projection of the measured depth, {(4 — £dop), is:
D = 18,000 sin (а2 (£4 — £dop) x 10~2)
TOC o "1-5" h z 7Г Q2 4 ‘
— ft? sin(d2 {(4 — £dop) x 10 2)
True vertical depth for £dop < £4 < (eod’-
D4 = ОкорЛ —sin (q’i (£eob — £kop) x 10-2) (4-76)
7Г Qj V ‘
-(-{(dop — £eob) cos Qi
H—— :— sin(a2 (£4 — (dop) x 10 2)
it q2
= Dkop + Ri sin qj — f (Ddop — Deob) (4-77)
+ Ri sin(o(2 (£4 — (dop) x 10 2)
= Deop + {Deob — Deop) + {Ddop — Deob)
+Л2 sin(a2 {(4 — (dop) x 10 2)
= Ddop + ^2 sin(d2 {(4 — (dop) x Ю 2) (4-78)
5. For the second slant portion, (eod < (5 < (t, the vertical projection of the
measured depth, {(ъ — (eod), is:
D = (4 — 1Eod) cos q2
True vertical depth for ^EOD < 4 < t-T’-
|
(4.79) (4.80) |
Ds = Dkop- — sin (di (i eob — ^ kop) x 10~2)
7Г £V[
+ (^DOP — VeOb) COS Oi
H ——- sin (oi-2 (?EOD — (dOp) X 10~2) + (4 — f EOD ) cos a2
it a 2
= Ddop + R2 sin(oi — a’2) + (4 — (eod) cos a2
= DeOD + (4 — @EOd) COS {>2
Thus, for a measured depth of 15,095 ft the TVD is:
18,000
Ds = 5,000 4——- —— sin 40 + (12,428 — 7,000) cos 40
7Г 2
18,0 . , ,
4——- sin(40— — 0)
7Г 1.5
= 13,455 ft
Similarly, for a measured depth of 20,638 ft the TVD is:
Ds = 13,455 + (20,638 — 15,095)
= 18,998 ft
|
where: D I Subscripts: DOP EOB EOD KOP |
|
true vertical depth, measured depth. dropoff point, end of build, end of dropoff, kickoff point. |
A friction factor of 0.35 will be used to calculate the drag associated tension on the casing. The effect of friction on axial load during downward movement is ignored.
For the buildup section, the tension load will be calculated by arbitrarily dividing this section into three equal parts: top, middle, and bottom. For the slant and dropoff sections, the tension load will be calculated by considering each of them as one section.
The approach to the buildup section is very arbitrary and not at all ideal, but this is an example of a hand calculation of a problem which can accurately be solved with a computer (Chapter 5 shows how).
In practice, the pipe will lie on the lower side. IT/? sin л > /v for most of the interval. At some point, probably quite near to the kick-off point. IT/? sin о = F, t and the pipe, like the ‘Grand Old Duke of York’s 10.000 Men’, is ‘neither up nor down’. Finally, near the top of the interval. IT/? sin о < Fa and the pipe will touch the top of the hole. Quite obviously, in the drop-off section of the hole. WR sin cv > Fa across the entire interval.
A two-dimensional model will be used to determine the drag-associated axial tension, because a numerical solution to the three-dimensional model is outside the scope of this section (refer to Chapter 5). For the purpose of casing design, a two-dimensional model has a strong practical appeal: it is simple to use.
The buoyant weight of the casing will be calculated using the true vertical depth of the well, because the horizontal component of the pipe is fully supported by the wall of the hole.
|
TVD MD Fig. 4.12: Example of well profile showing steel grade and weight for intermediate casing. |
The well profile with the steel grades and weights (based on collapse and burst loads) is presented in Fig. 4.1’2. Starting from the bottom, the tensional load due
to buoyant weight and frictional drag can be calculated as shown in the Example Calculations.
|
Table 4.2: Total tensile load in intermediate casing string.
|
|
(6) Bending force (63 d0WJ) в = 3°/100 ft (lbf) |
(?) Total tensional load = buoyant weight + frictional drag + bending force (lbf) |
(8) Total tensional load = buoyant weight + bending force + shock load (lbf) |
|
247,732 |
750.514 |
855.064 |
|
214.869 |
938.087 |
958.224 |
|
247,732 |
1.290.989 |
1.341.996 |
Example Calculation:
Tensional load due to the buoyant weight and frictional drag on pipe section P-110 (98 lb/ft) can be calculated as follows:
1. For the slant section, from 12.428 to 7.000 ft:
Fa = Fal + W (lx — 12) (fb sin o, + COSO])
= 430,395 lbf
where:
Fal = tensional load at 12.428 ft — 0 lbf
W = BF x 98 lb/ft
= 0.816 x 98 = 80 lb/ft
h-l2 = 12,428 — 7,000 = 5.428 ft
fb = 0.35 (assumed)
Ql = 40°
|
WR Г+7? |
|
Fa — Л В Fal + |
[(1 — fb) (I<B sinoi — sino2)
—2 fb (Kb cos qi — cos а2)]
= 459,578 + 43,204 = 502,782 lbf
where:
Fal = 430,395 lbf
ct-2 = 29.26° at 6,463 ft measured depth.
c*i = 40° at 7,000 ft measured depth.
KB = (=>2-0.!) = LQ678
R = 2,866 ft
The tensional load on the pipe section P-110 (85 lb/ft) from 6,463 ft to 5,000 ft can be calculated as follows:
3. For the bottom part of buildup section, from 29.26° to 26.66° inclination angle:
, WR Г 2
Fa = A’b + ——— (1 -/6[3]) (A’s sine*! — sina2)
1 + h
-2/b (KB COS Qj — cosq2)]
= 509,803 + 7,299 = 517,101 lbf
where:
Fai = 502,782 lbf W = 0.816 x 85 = 69.4 lb/ft Oil = 26.99°
Qi = 29.26°
К в = e~fb(ai2-ai) = 1.014 R = 2,866 ft
4. For the middle part of the buildup section, from 26.66° to 13.33° inclination angle:
Fa = Fai + WR (sin qj — sin q2)
= 517,101 + 43,386 = 560,487 lbf
where:
Fai = 517,101 lbf W = 69.4 lb/ft Oil = 13.33°
ai = 26.66°
R = 2,866 ft
5. For upper part of the buildup section, from 13.33° to 0° inclination angle, for P-110 (85 lb/ft) :
Fa = I< в Fal + — +—2 [(1 — fb) {KB sin аг — sin t>2)
‘ J b
+2 /ь (К в cos о i — cos o2)]
= 608,036 + 45,786 = 653,821 lbf
where:
Fai = 560,487 lbf
W = 69.4 lb/ft
= 0° ax = 13.33°
К в = е-Л(«2-Л1) = 1 0848
R = 2,866 ft
6. For vertical section, from 5,000 ft to 4,000 ft. for P-110 (85 lb/ft):
Fa = Fax + W (5,000 — 4.000)
= 732,217 lbf
where:
Fai = 653,821 lbf
W = 69.4 lb/ft
7. Tension load at the top of the casing section L-80 (98 lb/ft) is given by:
Fa = Fal + W (4,000)
= 1,043,257 lbf
where:
Fai = 732,217 lbf
W = 98 x 0.816 = 80 lb/ft.
Drilling Liner
Figure 4.13 presents the well profile and steel grades and weight selected based on the collapse and burst loads. Starting from the bottom, the tensional loads due to the buoyant weight are shown in Table 4.3.
Example Calculation:
Pipe section, L-50 (58.4 lb/ft), 14,128 ft to 15.638 ft.
|
TVD MD Fig. 4.13: Example of well profile showing steel grade and weight for a liner. |
1. For the vertical section from 15.638 ft to 15.095 ft:
Fa = Wn x BF x (15.638 — 15.095)
= 58.4 x 0.743 x 543 = 23,561 lbf
2. For the dropoff section from 15.095 ft to 14.128 ft. with inclination angle of 0° to 14.5°:
, WR r
Fa — KD Fal — j—— |(/t2 — 1) (sin q2 — KD sino-!)
+2 fb (cos q2 — К и cos «j)]
= 69,938 lbf
|
where: |
||
|
Fal = |
23,561 lbf |
|
|
W = |
58.4 x 0.743 |
= 43.39 lb/ft |
|
R = |
3,726 ft |
|
|
fb = |
0.35 |
|
|
«i = |
0° |
|
|
«2 = |
14.5° |
|
|
KD = |
ghi® 2—<>i) — |
1.0926 |
Pipe section P-110, (47 lb/ft), from 14,128 ft to 11.775 ft.
|
(1) |
(2) |
(3) |
(4) |
|
True vertical |
Grade, and |
Measured depth |
Angle of |
|
depth |
Weight |
(ft) |
inclination |
|
(ft) |
(lb/ft) |
(degrees) |
|
|
14,000 — 12.500 |
L-80, 58.4 |
15.638 — 14.128 |
0 — 14.5 |
|
12,500 — 10,500 |
P-110. 47 |
14.128 — 11.775 |
14.5 — 40 |
|
(5) |
(6) |
(") |
(8) |
|
Tensional load |
Bending force |
Total tensional load |
Total tensional load |
|
carried by the |
= 63 d0WJ |
= buoyant weight |
= buoyant weight |
|
top joint, Fa = |
II CO о О О *-*% |
+ frictional drag |
+ bending force |
|
Fbu + Fd (lbf) |
(lbf) |
+ bending force (lbf) |
4- shock load (lbf) |
|
69,938 |
106,237 |
176.175 |
358,222 |
|
169,587 |
85.499 |
255.086 |
370,865 |
3. For the dropoff section from 14.128 ft to 12.428 ft:
WR r ,
Fa = KD Fai — 1 + p [(fb — 1) (sin a2 — Fd sin o,)
+2 fb (cos a2 — KD cos oj)]
= 146,988 lbf
|
where: |
||
|
Fal = |
69,938 lbf |
|
|
W = |
47 x 0.743 |
= 34.92 lb/ft |
|
R = |
3,726 ft |
|
|
fb = |
0.35 |
|
|
Ql = |
14.5° |
|
|
«2 = |
40° |
|
|
I<D = |
eh(<*2-ai) — |
1.1686 |
4. For the slant section from 11.775 to 12.428 ft
Fa = Fal + W (l — l2) (fb sill O! + COSO])
= 169,587 lbf
where:
Fai = 146,988 lbf
W = 34.92 lb/ft
11 = 12,428 ft
12 = 11,775 ft
qj = 40°
|
TVD MD Fig. 4.14: Example of well profile showing steel grade and weight for production casing. |
Production Casing
The well profile, the steel grades and weight, selected on the basis of collapse and burst loads are presented in Fig. 4.14. Starting from the bottom, the tensional load based on the concept of frictional drag and shock load are shown in Table 4.4.
The values of drag-associated tensional load for intermediate, liner and production casings are found almost the same way as those of the tension due to the shock load. This suggests that for a well profile presented in Fig. 4.11 and an assumed value for friction factor of 0.35, shock load can be substituted for drag force for ease of calculation of the design load for tension.
Equations 4.30, 4.3‘2, 4.33, 4.35 and 4.38 are extremely useful to estimate the
drag force for casing in complicated well profiles, if the hole trajectory and other parameters such as bit walk, dog legs and bearing angles are known. It is equally important to know the exact value of the friction factor, because it is a major contributor to the frictional drag.
|
depth (ft) |
Weight (lb/ft) |
(ft) |
|
19,000 — 16,000 |
SOO-155, 46 |
20.638 — 17,638 |
|
16,000 — 8,000 |
V-150, 46 |
17.638 — 8,512 |
|
8,000 — 3,000 |
MW-155. 38 |
8.512 — 3.000 |
|
3,000 — 0 |
V-150, 38 |
3,000 — 0 |
|
(4) |
(5) |
(6) |
|
Angle of |
Tensional load |
Bending force |
|
inclination |
carried by the |
= 63 d0WJ |
|
(degrees) |
top joint |
0 = 3°/100ft |
|
Fa |
= Fbu + Fd (lbf) |
(lbf) |
|
0 |
100.212 |
60,858 |
|
0-40 |
468,226 |
360,858 |
|
40-0 |
711,846 |
50,274 |
|
0 |
794,630 |
50,274 |
|
Table 4.4: Total tensile load in production casing. |
|
(7) |
(8) |
|
Total tension |
Total tension |
|
= buoyant weight |
= buoyant weight |
|
+ frictional drag |
+ shock load |
|
+ bending force (lbf) |
+ bending force (lbf) |
|
161,070 |
308,308 |
|
529,084 |
575.648 |
|
762,120 |
677,494 |
|
844,904 |
760,304 |
|
fir m m True vertical Grade and Measured depth |
Using the planned trajectory data in Table 4.5, three production casing strings for a typical deviated well (Fig. 4.1). a single-build horizontal well (Fig. 4.15) and a double-build horizontal well (Fig. 4.15) were generated using the program introduced in Chapter 5.
To calculate the tensional load in the top joint of the strings, Eqs. 4.30, 4.3’2,
|
Table 4.5: Planned trajectories of (a) typical deviated well, (b) singlebuild horizontal well and, (c) double-build horizontal well.
|
was assumed that the casing rested on the upper-middle-bottom part of the hole for an equal third of the interval as in the previous example. In the case of the double build, like the dropoff, the casing was assumed to rest on the bottom of the hole for the entire section of the second build.
At this point it is worthwhile to reiterate what was said earlier about the validity of these assumptions and in particular the one concerning the buildup section. Namely, that they need bear little relation to what actually occurs in practice. Just how close they are to the computer generated solution is illustrated in Table 4.6, which summarizes the results for each case and records the error between the value calculated using this approach and that produced using the computer program. Even the ‘errors’ must be taken with a grain of salt because by changing the buildup assumption from upper-middle-bottom (each 1/3 of interval) to upper-bottom (each 1 /’2 of interval) the errors change to -7%. -0.3% and -12% for the typical, single-build and double-build wells, respectively.
|
Table 4.6: Production combination strings for: (a) typical deviated well, (b) single-build horizontal well, (c) double-build horizontal well.
|
|
“Computed value” is that generated in Examples 5.11 and 5.12 |