Солнечная электростанция 30кВт - бизнес под ключ за 27000$

15.08.2018 Солнце в сеть




Производство оборудования и технологии
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Total Effective Axial Force

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

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

or

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)

+AsET(-&T) + Fas (2.205)

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.

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