Prevention of Casing Buckling
As discussed previously, if the axial load becomes less than the stability load. ( oy + cq ) / 2, a mechanical adjustment, such as application of surface pressure and/or mechanical pull or slackoff load, must be carried out to prevent casing buckling. Equation 2.207 shows that the amount of mechanical adjustment necessary to prevent buckling decreases as the depth of the cement top decreases. Thus, one or a combination of the following procedures can be used to adjust the axial load on the casing string:
1. Adjustment of cement height.
2. Application of surface pressure.
3. Alteration of mechanical or slackoff load.
The required depth of the cement top, Djoc, can be derived by applying the condition for no buckling in Eq. 2.209.
Substituting FbUc = 0 in Eq. 2.209, the distance from the surface to the top of the cement column, Djoc, is obtained:
n D(Wn — A0GPcm + ) — ASETAT + Fa.
TOC o "1-5" h z TOC ~ ш ИГЕ T’n ————————— continue -*•
Wn — (A0 GPo — A, GPt)
+ {AuPl — Ai0Wl)[DbA,(GPt + AGP) + ApSt] .
———- 71— / а л ^. … ,——————— continue —>
— (1 — i/)(A0AGPo — A, AGPt)
_> +11 -^^(АоАр^ — AjApSt) ^ •> 1 ■>i
+A0 (GPo — GPcm)
The distance between the top of the cement and casing head depth, Djoc, should be determined for the various surface pressures, fluid densities and temperatures to which the casing string is to be subjected bearing in mind that the specific weight of the drilling fluid and the bottomhole temperature will vary for subsequent drilling operations. To prevent buckling, an extra surface pressure and/or an overpull, equal in magnitude to the difference between the effective axial load and the stability load, must be applied.
In Example 2-14, it was determined that the string was liable to buckle in the interval 2,000 — 7,500 ft. Determine: (i) the depth of the cement top to avoid buckling, (ii) the magnitude of the internal pressure required if the casing cement cannot be cemented above 7,500 ft. Assume that GPcrn = 0.78, GPo = 0.702 psi/ft and in both cases that Fas = 0.
From Eq. 2.212, it is apparent that several ‘average weighted’ properties need to be determined before substituting to find Djoc-
Average weight, Wn:
ur _ (2,000 x 58.4)+ (5,500 x 47)+ (2,500 x 58.4) n ~ 10,000 = 52.13 lb/ft
Average internal area, A,-:
_ (4,500 x 55.88) + (5,500 x 59.19)
‘ ~ 10,000 = 57.70 sq. in.
Average steel cross-section is, therefore:
As = 72.76 — 57.70 = 15.06 sq. in.
(i) Substituting into Eq. 2.212 to obtain the depth of cement to avoid buckling yields:
_ D(Wn — A0GPcm + A{GPx) + (1 — 2i’)(A0ApS0 — A, ApS!) 4.
* “ Wn — (A0GPcm — AiGp,) + A0(GPo — GPcrn) con mue ^
+F/ — ETASAT + Fas -(l-u)(A0AGPa-AiAGp,)
104 (52.13 — 72.76 x.780 + 57.7 x.702) + 0 .
= ———————————————————— continue —►
52.13 — (72.76 x.78 — 57.7 x.702) + 72.76(-.078)
—285 x 15.06 — 15,525 x 15.06 + 0 -(.7)(-9.00)
120, 727.4
“ 36.51
= 3,307 ft
Thus the casing should be cemented to 3.307 ft to prevent buckling. Actually, the presence of buckling forces does not necessarily mean that the casing will buckle. Buckling will only occur if the buckling force exceeds the critical buckling force.
(ii) Essentially the same equation as used in (i) is employed in this case: only now the unknown is Apsl and the buoyancy force is calculated using the initial fluid densities inside and outside the casing, i. e.. the buoyancy term in Eq. 2.205. {A0(GPc(D — x) + GPox) — A, GPtD} is replaced by (A0GPoD — A, GPtD) and Eq. 2.212 becomes:
D(Wn — A0GP + AiGp) + (1 — 2r/)(A0ApS0 — A, Apsl) x = —— — у— —-— continue —>
— (AQGPo — A, GP:)
—Fp — ETASAT + Fas —(1 — v)(A0AGPo — A, AGPl)
Solving for Apsi yields:
|
7,500 |
_ 104 (41.558) + (0.4)(—57.7 Aps,) — (285 + 15.525) x 15.06 + 0 ~ 52.13 — (72.76 x.702 — 57.7 x.702) + 6.301
(7,500 x 47.859) — 415.579 + 238.099 ~ 23.08
= initial internal surface pressure — final internal surface pressure = 7, 862 psi
However, the burst rating for N-80, 9| in.. 47 lb/ft pipe is 6.870 psi (for aa = 0). Thus, to safely use this casing, a lower pressure needs to be applied in combination
!Fp — (Fp + AFap)
with the surface overpull, Fas. Assuming a maximum surface pressure of 60% of burst, i. e., 4,1’22 psi, the required surface overpull can be calculated as:
? 500 = 104 (41.56) + ,4(—57.7 x (-4,122)) — 238.099 + Fas ’ 52.13 — (72.76 x.702 — 57.7 x.702) + 6.301
Rearranging the above in terms of Fas yields
Fas = (7,500 x 47.86) — 415,579 + 238.099 — 95.136 = 86,326 lbf