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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.

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.

w.

True vertical depth (ft)

(2) Grade and Weight (lb/ft)

(3)

Measured dept h

(ft)

(4)

Angle of inclination (degrees)

. ^

Tensional load

carried by the top joint, lbf:

Fa — Fhu + Fd

11,100 — 6,400

P-110, 98

12.428 — 6.463

40 — 29.26

502,782

6,400 — 4,000

P-110. 85

6.463 — 4.000

29.26 — 0

723.218

4,000 — 0

L-80, 98

4.000 — 0

0

1.043.257

(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) :

WR r

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 =

«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

Further Examples

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.

Typical Deviated

Single Build

Double Build

KickofF point (ft)

5.000

3,000

5.000

Buildup rate (dj) (°/100 ft)

2

2

2

End of buildup (ft)

7.000

7.500

7.000

Inclination angle (q[) (deg.)

40

90

40

Dropoff point (ft)

12.480

12,500

Second build (ft)

12.480

Dropoff rate (d2) (°/100 ft)

2

2

Buildup rate (d2) (°/100 ft)

2

End of dropoff (ft)

14.480

12,500

End of build (ft)

13.480

Inclination angle (q2) (deg.)

0

90

90

Total measured depth (ft)

16.720

12,500

16,720

Total vertical depth (ft)

15.120

5.865

11,449

Additional information common to all three examples:

Minimum casing interval = 2.000 ft

Design = minimum cost (see

Chapter 5)

Pseudo friction factor = 0.35

Design factor burst = 1.1

Design factor collapse = 1.12

5

Design factor yield =1.8

Specific weight of mud = 16.8 lb/gal

Design string = 7-in. production

4.33, 4.35 and 4.38 were used.

In all three cases.

for the top buildup section it

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.

Typical Deviated Well

(see Table 5.21 on page 307 ).

Casing interval, measured depth (ft)

Grade and Weight (lb/ft)

Tensional load on top joint

(lbf)

16,720 — 14,200 14,200 — 11,520 11,520 — 0

P-110, 38 P-110. 35 P-110, 32

73.612

167.449

490.333

Computed value = 484,623 lbf (Error = -1.29c)

Single Build Horizontal Well

(see Table 5.22 on page 308 ).

Casing interval, measured depth (ft)

Grade and Weight (lb/ft)

Tensional load on top joint (lbf)

12,500 — 0

S-95. 23

152.930

Computed value = 169,839 lbf (Error = 1091)

Double Build Horizontal Well

(see Table 5.23 on page 309 ).

Casing interval, measured depth (ft)

Grade and Weight (lb/ft)

Tensional load on top joint (lbf)

16,720 — 12,280 12,280 — 9,760 9,760 — 7,240 7,240 — 0

S-105. 32 P-110, 32 S-95. 29 S-95, 26

69.537

128.915

182.726

356.893

Computed value = 336,018 lbf (Error = -6.297)

“Computed value” is that generated in Examples 5.11 and 5.12

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