Total Effective Axial Force

19.10.2013 | Автор: admin

Total effective axial force, Fae, can be determined by summing all the forces:

Fae = Faw + Fap + A Faw + A Fap + Faj + Fas (2.204)

Fae = W„(D — x) — A0 {j GPo + GPcm (D — г)} + At(bPi D +GPt Daas(AUPi — AiOWl) + v {A,(AGPtx + 2 Apst)

-A0(AGPc x + 2 Apso)} +(AGPi DAAi + Apsi) {Aupi — Ai0Wl)

where:
F 1 aw	— weight of casing string carried by the joint above the cement
Fap	= piston force.
FaT	= force due to a change in temperature.
Fas	= force applied at the surface.
A’s	= changes in principle forces due to changes in
	fluid specific weight and surface pressures.
DA a.	= depth of change in pipe cross-section.

In Eq. 2.205, tension is considered as positive and compression as negative.

EXAMPLE 2-14:

Reconsider Example 2-13, assuming that a new section of hole has been drilled. The pressure gradient of the mud has been increased to 0.858 psi/ft and the average downhole temperature has increased by AT = 75° F. Is the string stable? Assume T = 6.9 x 10-6 in./in./°F and v — 0.3.

Solution:

The changes in mud weight and temperature will generate additional stresses in the string.

From the temperature change one can obtain from Eq. 2.201:

<JaT = — ЕГАТ = —15,525 psi

From the mud weight change (Eq. 2.192):

Aj(Gp, x + 2Apsl) — A0(AGPox + 2Apso)

Asx )

х + 0) — 9.625(0 4-0)4

Q3 M,-((0.858 — 0.702)

—А, )

0.0468 А{Х

From the change in piston effect (Eq. 2.195):

» (AGPtDAA, 4“ Ар5,)(.4ирд Aiow)

^——————-

_ ((0.156)ДдУ4,)(.Лцр1 — Ai0W1)

~ A

The final axial stress is:

‘Ti/.na! = (T,, + <Ta7 + Д<т

aw + A<rap

Table 2.10: EXAMPLE 2-14: Revised axial stress calculations.

Depth	Weight	A	A		&aT	А(Та1,	Arrap
(ft)	(lb/ft)	(sq. in.)	(sq. in.)	(psi)	(psi)	(psi)	(psi)
10,000	58.4	16.879	55.88	-7.323	-15,525	1,549	0
9,100	58.4	16.879	55.88	-4.209	-15.525	1,402	0
7,500	58.4	16.879	55.88	1.327	-15,525	1,162	0
7,500	47	13.572	59.19	2.935	-15,525	1,531	-285
2,000	47	13.572	59.19	21.984	-15.525	408	-285
2,000	58.4	16.879	55.88	17.398	-15,525	310	-224
0	58.4	16.879	55.88	24,318	-15,525	0	-224

Recalculating the radial and tangential stresses yields Table 2.11

Table 2.11: EXAMPLE 2-14: Recalculated radial and tangential

stresses.

Depth (ft)	Weight (lb/ft)	Ai (sq. in.)	Pi (psi)	Po (psi)	(a, + crr)/2 (psi)	(psi)
2,000	58.4	5-5.88	1,716	1,404	-371	1,958
2,000	47	59.19	1,716	1.404	-43	6,582
7,500	47	59.19	6.435	5.265	-162	-11.345
7,500	58.4	55.88	6,435	5.265	-1.392	-13,036
9,100	58.4	55.88	7.808	6.388	-1,687	-18,324
10,000	58.4	55.88	8,580	7.090	-2,158	-21,298

There are two options available to the designer to avoid buckling: (1) Place the cement top at the neutral point (Example 2-13); (2) application of overpull such that the neutral point ‘moves’ back to 9,100 ft (the current cement top) under these modified conditions.

The size of the overpull required to achieve this is:

(a> + <T(

—b = -1,687 -(-18,324)

V 2 /evaluated @ 9,100 ft

= 16,637 psi

As a pickup force this equates to:

Fas = 16,637 x 16.879 = 280,816 lbf

Stress (psi) stress (psi)

Fig. 2.30: Graphical solution of Examples 2-13 and 2-14.

Рубрика: Casing design theory and practice