Buildup Section
Besides the frictional factor, borehole frictional drag is also controlled by the direction and the normal force. In a buildup section, the three positions of the casing as shown in Fig. 4.2 are possible: uppermost, middle and the lowermost position (Maidla, 1987). The normal and axial forces acting on each unit section are presented in Figs. 4.3 and 4.4. From the free-body diagram, the normal force Fn can be expressed as:
Fn = 2Facos^90-^j
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Fig. 4.3 : Determination of normal force in buildup section. |
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Fig. 4.4 : Forces acting on a small element within the buildup section. |
where:
Fa = axial force on the unit section, lbf.
Да = angle subtended by unit section at radius R.
Inasmuch as Да/2 is very small compared to R. sin (Да/2) % До/2. Hence, the Eq. 4.1 yields:
Fn=2Fa^ = FaAa (4.2)
Considering the buildup section in general, the resultant normal force while pulling out of the hole is the vector sum of the normal components of the weight and the axial force of the unit section (Fig. 4.4). Therefore:
Fn = Да WR sin а — Fa Да
= Aa(WRsm a — Fa) (4.3)
where:
W = weight of the unit section, lb/ft.
= WnBF R = radius of curvature, ft.
The magnitude of the drag force. Fj. which acts in a direction opposite to pipe movement is given by:
Fi = — ft, |F„| (4.4)
Fd = — /(, |Да И7/? sin а — Fa Да| (4.5)
where:
ft — borehole friction factor.
|Fn| = absolute value of the normal force, lbf.
The incremental axial force. ДFa (Fa, — Fai = ДFa > 0). over the incremental arc length (a2 — аг = Да < 0) when the casing is being pulled (indicated by negative Да W R cos a), is given by:
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(4.6) |
AFa = Fi — Да W R cos а
or,
<o
AFa — +fb [Aar VH/?sin a — Fa Aaj — Aq WR cos a (4-7)
Hence, at equilibrium, the following differential equation is obtained: dF
—~=—fbWRsma — Fa — WR cos a (4-8)
da
If the casing is in contact with the upper side of the hole, (WRsma — Fa) < 0 and, therefore, Eq. 4.8 can be rewritten as:
dF
—A = — fb(Fa — WRsma) — WRcosa (4.9)
da
Rearranging Eq. 4.9, yields: dF
—r^- + fb Fa = WR (/{, sin a — cos a) (4.10)
da
The value of Fa in Eq. 4.10 can be found by first considering the homogeneous solution and then the particular solution as follows:
К — Fahorno + Fapart (4.11)
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dFa da dFa da dFa Fa |
Considering first the homogeneous solution:
+ fb Fa — 0 = — fbFa
= ~fbda (4.12)
Integrating Eq. 4.12, yields:
In Fa = — fb Aq + С
г
ото
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(4.14) (4.15) (4.16) (4.17) (4.18) (4.19) (4.20) (4.21) (4.22) |
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apart |
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= -(1 + /Л |
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Ai |
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д, = |
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= 2Л WR |
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= WR(l-f>) |
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Аз — |
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Hence |
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Ai |
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— A sin a + В cos a + fb A cos a — f /ь В sin a = +/ь VE-Rsina — WR cos о Equating for the coefficients, yields: — A sin a + /ь В sin a — +fi, WR sin a -A +fb В = +fbWR +fb A cos a + В cos a = — W R cos a +fb A +B = — WR Now solving for A and В using a matrix solution yields: |
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where: С = constant of integration. Now, consider the particular solution of the form: — A cos a В sin a |
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Substituting Eqs. 4.14 and 4.15 into Eq. 4.10, yields: |
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— — — A sin a + В cos a |
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“1 +/f> +/ь 1 |
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2 fbWR |
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1 + Л2 |
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dF, |
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da |
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-1 fbWR h — WR |
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fbWR fb — WR 1 |
WR(l-f£)
Thus, the particular solution for Fa is obtained by substituting Eq. 4.22 and Eq.
4.23 into Eq. 4.14:
TOC o "1-5" h z n 2 fb WR WR(l-ff) .
Г a, =————— ГГ"—- COS———————— a Sill Ct 4.24 )
par’ 1 + fi 1 + Л2
Inasmuch as Fa = Fahomo + Fapart. the expression for Fa is obtained by combining Eqs. 4.13 and 4.24:
Fa(a) = С e~ha — ^ [2 fh cos a + sin q (1 — /2)] (4.25)
Applying the first boundary condition. Fa(rti) = F,, = constant, one obtains:
Fai = С e~hai — ^ Rf2 [‘2 fb coso, + sino, (1 — /2)] (4.26)
1 + h
Solving for C:
W R 1 r
с = e/i, ni F01 + + [2 /(, cos Qj + sin cq (1 — /*)] (4.27)
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F(a) = e-F(-m)Fai + ——————— ^—— [2 /6 cosq, + sina1(l -/62)] |
Substituting the expression for С into Eq. 4.25:
1+Я
IV ff г i
cosa + sino (1 — fZ) (4.28)
The second boundary condition is: Fa(«2) = Fa2 = constant
Substituting the second boundary condition into Eq. 4.28. the tensile force at the point of interest (position 2). where a = a2. is obtained:
Fa2 = e~h{a2~ai)Fai 4—————— 1 ^ j2 [‘2/6 cosq! + sinai (1 — Д2)]
[2/ь cosa-2 + sino2 (1 — fl)] (4.29)
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Representing е ail = Ig for buildup, one obtains the following expression for Fa2 in the uppermost section: |
1+Л2
Fa-i — I<B Fai + Y~-p [(! — fb) {Kb sin o, — sin a2)
+ 2 fb (Kg COS ax — COS О2 )] (4..10)
where:
R = ll~}’2 (4.31)
Q i — a2
l and l2 are the lengths of pipe in feet and the units of ax and o2 are radians.
Note that the angles ax and a2 are obtained from surveys taken during the drilling
of the well and, therefore, it is not true to say that:
R_ lx — l2 180 1 100
ax — Ot2 7Г Q 1
Whereas a and, therefore, R are constants in the planned well. R is not constant in an actual well.
The additional tension due to frictional drag for both the intermediate and upper sections can be obtained following the same procedure as illustrated for the upper section.
For the intermediate section:
Fa2 = Fai + WR (s’m ax — sin a2) (4.32)
For the lower section:
Fai = I<g Fai + WRf2 [(1 — /ь2) {Kb sin Qi — sin o2) 1 + h
—2 fb(Kg cos on — cosa2)j
Fa2 ■ ez
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Fa! ■ |
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Fig. 4.5 : Forces acting on a small element within the slant section. |