CN200943449Y - Riser pipe for marine petroleum drilling well - Google Patents

Abstract

The utility model pertains to a marine riser for offshore petroleum drilling, characterized in that the marine riser is a tower-type tubular column of variable stiffness, and the upper tubular column is stiffer than the lower tubular column; after the concentrated load from the blowout preventer at the shaft mouth and the material used for the tubular column are determined, the length and radius of the upper/lower tubular column satisfies the following critical load formula for the elastic stability of the marine riser: P<cr>=ET<1>Pi/1<3>Alpha<1>+ET<2>Pi/1<3>Alpha<2>-(q<2>Beta<2>/21Pi<2>+q<1>Beta<1>/21Pi<2>). The above formula can be used to easily design the parameters for and monitor the engineering work. The utility model is very practical and may guide the offshore petroleum drilling work. The use of the marine riser here will save much steel, effectively lowers the cost and maintain a stable and reliable work. The utility model can be widely used in offshore petroleum drilling work.

Description

A kind of offshore oil well drilling used water

Technical field

The utility model relates to a kind of riser pipe, particularly about a kind of offshore oil well drilling used water.

Background technology

Along with a large amount of exploitations of marine marginal oil field, how to adopt most economical effective means, obtain the maximum profit in oil field, satisfy the demand that oilfield safety is effectively produced simultaneously again, be the problem of the applicant's primary study.When carrying out petroleum drilling operation at sea, usually need be in the external circumferential setting of bored shaft by a string 30 " * 28 " riser pipe formed of the tubing string of (OD * ID, external diameter * internal diameter).Owing at well head wellhead blowout preventor is installed, therefore, riser pipe need bear the effect of concentrated load P, and simultaneously, because the riser pipe caliber is big, it is also very big to conduct oneself with dignity, and therefore, needs also to consider that riser pipe is subjected to the effect of the uniformly distributed load q of own wt formation.In view of this, all be to adopt tubing string material to make usually as the waterproof casing of pressure-containing member, and the rigidity of these tubing strings all is identical from top to bottom usually with certain rigidity.Yet, by to critical pressure under riser pipe concentrated load and the uniformly distributed load effect and further investigation that operation is exerted an influence thereof, find that there is unreasonable place in this traditional design, the waste of material that causes therefrom also is huge, but, should use which type of rigidity material where, whether its security performance can be protected, when the well head blowout preventer set is installed, can carry on the riser pipe and should carry great weight and be advisable or the like, in existing literature and textbook, there is not ready-made answer to seek, these are the utility model problem that need solve just, also is one of key issue of controlling effectively in marginal field development development cost.

Summary of the invention

At the problems referred to above, the purpose of this utility model provides a kind ofly can bring into play well head blowout prevention effect to greatest extent, can save the offshore oil well drilling used water of steel again in a large number.

For achieving the above object, the utility model is taked following technical scheme: a kind of offshore oil well drilling used water, it is characterized in that: described riser pipe is tower change rigidity tubing string, top tubing string rigidity is little, bottom tubing string rigidity is big, after the material that concentrated load that wellhead blowout preventor forms and tubing string use was determined, the length of described upper and lower section tubing string and caliber satisfied the critical load formula of following riser pipe elastic stability:

$P_{\mathrm{cr}}=\frac{\mathrm{<figure-callout\; id="1"\; label="E\; I"\; filenames="Y20062011916300141.png"\; state="\{\{state\}\}">E</figure-callout>}{I}_{}{}_1\mathrm{\&pi;}}{{\mathrm{<figure-callout\; id="3"\; label="l"\; filenames="Y20062011916300152.png"\; state="\{\{state\}\}">l</figure-callout>}}^3}{\mathrm{\&alpha;}}_1+\frac{{\mathrm{<figure-callout\; id="2"\; label="EI"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">EI</figure-callout>}}_2\mathrm{\&pi;}}{{\mathrm{<figure-callout\; id="3"\; label="l"\; filenames="Y20062011916300152.png"\; state="\{\{state\}\}">l</figure-callout>}}^3}{\mathrm{\&alpha;}}_2-(\frac{{\mathrm{<figure-callout\; id="2"\; label="q"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">q</figure-callout>}}_2{\mathrm{\&beta;}}_2}{2l{\mathrm{\&pi;}}^2}+\frac{{\mathrm{<figure-callout\; id="1"\; label="q"\; filenames="Y20062011916300141.png"\; state="\{\{state\}\}">q</figure-callout>}}_1{\mathrm{\&beta;}}_1}{2l{\mathrm{\&pi;}}^2})$ In the formula:

${\mathrm{\&alpha;}}_1=\mathrm{\&pi;}(l-l_2)+\frac{\mathrm{<figure-callout\; id="2"\; label="l"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">l</figure-callout>}}{2}\sin \frac{2\mathrm{\&pi;}{l}_2}{l}$

${\mathrm{\&alpha;}}_2=\frac{\mathrm{<figure-callout\; id="2"\; label="l"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">l</figure-callout>}}{2}\sin \frac{2\mathrm{\&pi;}{l}_1}{l}-{\mathrm{\&pi;<figure-callout\; id="2"\; label="l"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">l</figure-callout>}}_2$

${\mathrm{\&beta;}}_1={\mathrm{\&pi;}}^2{{\mathrm{<figure-callout\; id="1"\; label="l"\; filenames="Y20062011916300141.png"\; state="\{\{state\}\}">l</figure-callout>}}_1}^2-{\mathrm{\&pi;<figure-callout\; id="1"\; label="ll"\; filenames="Y20062011916300141.png"\; state="\{\{state\}\}">ll</figure-callout>}}_1\sin \frac{2\mathrm{\&pi;}{l}_2}{l}-{\mathrm{<figure-callout\; id="2"\; label="l"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">l</figure-callout>}}^2\sin {\left(\frac{\mathrm{\&pi;}{l}_2}{l}\right)}^2$

${\mathrm{\&beta;}}_2={\mathrm{\&pi;}}^2{l}_2(l+l_1)+{\mathrm{\&pi;<figure-callout\; id="1"\; label="ll"\; filenames="Y20062011916300141.png"\; state="\{\{state\}\}">ll</figure-callout>}}_1\sin \frac{2\mathrm{\&pi;}{l}_2}{l}+{\mathrm{<figure-callout\; id="2"\; label="l"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">l</figure-callout>}}^2\sin {\left(\frac{\mathrm{\&pi;}{l}_2}{l}\right)}^2$

Wherein: P _CrBe critical axial load [N] that E is steel modulus of elasticity [N/m ²], generally: E=2.058 * 10 ⁵MPa; I is a polar moment of inertia, I=3.14 * (external diameter ⁴-internal diameter ⁴)/32[m ⁴]; L be riser pipe 1 from the seabed length [m] to the top; l ₁Length for the hypomere tubing string; l ₂Length for the hypomere tubing string; q ₁Unit weight [N/m] for the epimere tubing string; q ₂Unit weight [N/m] for the hypomere tubing string.

Work as l ₂=l q ₂During=q,

$P_{\mathrm{cr}}=\frac{{\mathrm{<figure-callout\; id="2"\; label="EI"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">EI</figure-callout>}}_2\mathrm{\&pi;}}{l^3}-\left(\mathrm{\&pi;l}\right)\left(\frac{{\mathrm{<figure-callout\; id="2"\; label="q"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">q</figure-callout>}}_2{\mathrm{\&pi;}}^2{\mathrm{<figure-callout\; id="2"\; label="l"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">l</figure-callout>}}^2}{{2\mathrm{l\&pi;}}^2}\right)=\frac{E{I}_2{\mathrm{\&pi;}}^2}{l^2}-\frac{q_2\mathrm{<figure-callout\; id="2"\; label="l"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">l</figure-callout>}}{2}.$

When ${\mathrm{<figure-callout\; id="1"\; label="l"\; filenames="Y20062011916300141.png"\; state="\{\{state\}\}">l</figure-callout>}}_1={\mathrm{<figure-callout\; id="2"\; label="l"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">l</figure-callout>}}_2=\frac{\mathrm{<figure-callout\; id="2"\; label="l"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">l</figure-callout>}}{2}$ The time:

$P_{\mathrm{cr}}=\frac{{\mathrm{<figure-callout\; id="1"\; label="EI"\; filenames="Y20062011916300141.png"\; state="\{\{state\}\}">EI</figure-callout>}}_1{\mathrm{\&pi;}}^2}{{2\mathrm{<figure-callout\; id="2"\; label="l"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">l</figure-callout>}}^2}+\frac{{\mathrm{<figure-callout\; id="2"\; label="EI"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">EI</figure-callout>}}_2{\mathrm{\&pi;}}^2}{{2l}^2}-\frac{l}{2}[q_2(\frac{3}{4}+\frac{1}{{\mathrm{\&pi;}}^2})+q_1(\frac{1}{4}-\frac{1}{{\mathrm{\&pi;}}^2})]$

The internal diameter of described upper and lower part tubing string is identical, and the external diameter of described top tubing string is less than the external diameter of described hypomere tubing string.

The external diameter of described upper and lower part tubing string is identical, and the internal diameter of described top tubing string is greater than the internal diameter of described bottom tubing string.

The utility model is owing to take above technical scheme, it has the following advantages: 1, the utility model is from the actual conditions of drilling well, use the theory of elastic stability, riser pipe under the drillng operation condition is carried out mechanical analysis, thereby obtained under the synergy condition that end concentrated load p and deadweight uniformly distributed load q are arranged, the elastic stability critical load mathematical formulae of riser pipe can select the utility model to become each section rigidity and the physical dimension of rigidity riser pipe according to this Mathematical Modeling convenient and reasonablely.2, the Mathematical Modeling of setting up according to the utility model, the length that can comprise each section of riser pipe by every definite riser pipe parameter, inside and outside footpath, weight, material stiffness coefficient etc., detect the elastic stability of riser pipe, simultaneously, according to safety factor require, load requirement such as sea water advanced, preventer weight, also can pass through the critical load formula of the elastic stability of riser pipe, in conjunction with the specification of existing riser pipe, parameters such as the length of selection upper and lower part riser pipe 1 and caliber.3, the utility model has high practicality and directive significance to the offshore oil drilling engineering, adopt riser pipe of the present utility model can save great deal of steel, the cost that effectively reduces cost guarantees the reliability stability of engineering, and it can be widely used in the various seawater petroleum drilling engineerings.

Description of drawings

Fig. 1 is the utility model structural representation

The general state schematic diagram of Fig. 2 offshore drilling pit shaft

Fig. 3 is reduced to typical stressed member deflection deformation schematic diagram with riser pipe

The specific embodiment

Below in conjunction with drawings and Examples the utility model is described in detail.

As shown in Figure 1, the utility model at use in the prior art from top to bottom with the riser pipe 1 of rigidity, a kind of riser pipe 1 of tower stiffness changing structure has been proposed, this riser pipe 1 is divided into top tubing string 11 and bottom tubing string 12 two parts, the material that upper and lower part tubing string 11,12 uses is generally identical, top tubing string 11 is thin, and rigidity is little, and bottom tubing string 12 is thick, rigidity is big; Perhaps the external diameter of upper and lower part tubing string 11,12 is identical, the internal diameter difference, and promptly the tube wall of top tubing string 11 is thin, and rigidity is little, the thickness of pipe wall of bottom tubing string 12, rigidity is big.Can save great deal of steel like this, reduce cost.Yet how the length and the caliber of upper and lower tubing string 11,12 are selected, guarantee safe requirement in the time of could both satisfying field operation, does economical with materials reduce cost to greatest extent again? this need provide the foundation of a selection, it is truly feasible that this foundation should have theoretical foundation and process actual verification, and does not have ready-made foundation to follow at present.

For this reason, the applicant has carried out following work:

At first the mud line of going into of riser pipe 1 is looked closely and is position of fixity 2, riser pipe 1 is considered as hinge support 3 in uphole end, the blowout preventer set that acts on riser pipe 1 top is considered as concentrated load P, the riser pipe own wt is considered as uniformly distributed load q, under the effect of concentrated load P and uniformly distributed load q, riser pipe 1 is a typical stressed member (as Fig. 2, shown in Figure 3).During this stressed member deflection deformation, establishing riser pipe 1 length is 1, is subjected to that the maximum displacement of rod member off-axis is δ behind the responsive to axial force, supposes that its deflection curve is a sine curve, and as general approximate, can obtain Flexural Equation to be:

$y=\mathrm{\&delta;}\sin \frac{\mathrm{\&pi;x}}{l}---\left(1\right)$

Ask first derivative to obtain to formula (1): $\mathrm{dy}/\mathrm{dx}=\frac{\mathrm{\&delta;\&pi;}}{l}\cos \frac{\mathrm{\&pi;x}}{l}---\left(2\right)$

Ask second dervative to obtain to formula (1): $\mathrm{<figure-callout\; id="2"\; label="d\; y"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">d</figure-callout>}{y}^{}{}^2/d{x}^2=-\frac{\mathrm{\&delta;}{\mathrm{\&pi;}}^2}{{\mathrm{<figure-callout\; id="2"\; label="l"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">l</figure-callout>}}^2}\sin \frac{\mathrm{\&pi;x}}{l}---\left(3\right)$

Hinged line of deflection is when bending, and the difference of the length of its deflection curve and chord length AB is:

$\mathrm{\&lambda;}=\frac{1}{2}{\&Integral;}_o^l{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2\mathrm{dx}=\frac{{\mathrm{\&delta;}}^2{\mathrm{\&pi;}}^2}{4l}---\left(4\right)$

Axially point load P acting is: ${\mathrm{<figure-callout\; id="1"\; label="w"\; filenames="Y20062011916300141.png"\; state="\{\{state\}\}">w</figure-callout>}}_1=P\&CenterDot;\mathrm{\&lambda;}=\frac{P{\mathrm{\&delta;}}^2{\mathrm{\&pi;}}^2}{4l}---\left(5\right)$

Consider that simultaneously the uniformly distributed load q acting that deadweight causes is:

${\mathrm{<figure-callout\; id="2"\; label="w"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">w</figure-callout>}}_2=\frac{1}{2}q{\&Integral;}_o^l(l-x){\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2\mathrm{dx}=\frac{q{\mathrm{\&delta;}}^2{\mathrm{\&pi;}}^2}{8}---\left(6\right)$

Its deformability is during member bending: $\mathrm{\&mu;}=\frac{1}{2}\mathrm{EI}{\&Integral;}_o^l{\left(\frac{{\mathrm{<figure-callout\; id="2"\; label="dy"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">dy</figure-callout>}}^2}{{\mathrm{dx}}^2}\right)}^2\mathrm{dx}=\frac{\mathrm{EI}{\mathrm{\&pi;}}^4{\mathrm{\&delta;}}^2}{{4l}^3}---\left(7\right)$

Wherein E is the rod member modulus of elasticity, and I is the polar moment of inertia of rod member, I=3.14 * (external diameter ⁴-internal diameter ⁴)/32[m ⁴].

When system reaches critical condition, must have: μ=w ₁+ w ₂(8)

Formula (5), (6), (7) substitution formula (8) can be got: $\frac{{\mathrm{<figure-callout\; id="2"\; label="EI"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">EI</figure-callout>}}^2{\mathrm{\&pi;}}^2}{{\mathrm{<figure-callout\; id="3"\; label="l"\; filenames="Y20062011916300152.png"\; state="\{\{state\}\}">l</figure-callout>}}^3}=\frac{P}{l}+\frac{q}{2}---\left(9\right)$

Can solve the critical load of riser pipe 1 according to formula (9), promptly the end when considering riser pipe 1 end concentrated load P and deadweight uniformly distributed load q props up the critical load with the hinged rod member of an end admittedly.Observation type (9) can be found:

Do not considering under the riser pipe 1 deadweight condition that 1, promptly (9) second of formulas are zero, then (9) formula becomes $P_{\mathrm{cr}}=\frac{\mathrm{EI}{\mathrm{\&pi;}}^2}{{\mathrm{<figure-callout\; id="2"\; label="l"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">l</figure-callout>}}^2},$ This is Euler's load of standard.

When 2, considering riser pipe 1 deadweight uniformly distributed load, its effect is to reduce critical load, and its critical load minimizing value is half of riser pipe gross weight ql.

As shown in Figure 1, in the utility model, when considering operation at the scene, under the precondition that guarantees safety, economical with materials as much as possible, therefore adopt tower stiffness changing structure:

${\mathrm{<figure-callout\; id="3"\; label="w"\; filenames="Y20062011916300152.png"\; state="\{\{state\}\}">w</figure-callout>}}_3=\frac{1}{2}{\mathrm{<figure-callout\; id="2"\; label="q"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">q</figure-callout>}}_2{\&Integral;}_o^{l_2}(l-x){\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2\mathrm{dx}$

${\mathrm{<figure-callout\; id="3"\; label="w"\; filenames="Y20062011916300152.png"\; state="\{\{state\}\}">w</figure-callout>}}_3=\frac{{\mathrm{<figure-callout\; id="2"\; label="q"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">q</figure-callout>}}_2{\mathrm{\&delta;}}^2}{8l^2}[\mathrm{l\&pi;}(l-l_2)\sin \frac{2{\mathrm{\&pi;l}}_2}{l}+{\mathrm{\&pi;}}^2{l}_2(2l-l_2)+l^2\sin {\left(\frac{{\mathrm{\&pi;l}}_2}{l}\right)}^2]---\left(10\right)$

$w_4=\frac{1}{2}{\mathrm{<figure-callout\; id="1"\; label="q"\; filenames="Y20062011916300141.png"\; state="\{\{state\}\}">q</figure-callout>}}_1{\&Integral;}_{l_2}^l(l-x){\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2\&CenterDot;\mathrm{dx}$

$w_4=\frac{{\mathrm{<figure-callout\; id="1"\; label="q"\; filenames="Y20062011916300141.png"\; state="\{\{state\}\}">q</figure-callout>}}_1{\mathrm{\&delta;}}^2}{{8l}^2}[{\mathrm{\&pi;}}^2{(l-l_2)}^2+\mathrm{\&pi;l}(l_2-l)\sin \frac{{2\mathrm{\&pi;l}}_2}{l}-{\mathrm{<figure-callout\; id="2"\; label="l"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">l</figure-callout>}}^2\sin {\left(\frac{{\mathrm{\&pi;l}}_2}{l}\right)}^2]---\left(11\right)$

Simultaneously, by formula (7) as can be known, its flexural deformation can for:

$\mathrm{\&mu;}={\mathrm{\&mu;}}_1+{\mathrm{\&mu;}}_2=\frac{{\mathrm{<figure-callout\; id="1"\; label="EI"\; filenames="Y20062011916300141.png"\; state="\{\{state\}\}">EI</figure-callout>}}_1}{2}{\&Integral;}_{l_2}^l\frac{{\mathrm{\&delta;}}^2}{l^4}\sin {\left(\frac{\mathrm{\&pi;x}}{l}\right)}^2\mathrm{dx}+\frac{{\mathrm{EI}}_2}{2}{\&Integral;}_o^{l_2}\frac{{\mathrm{\&delta;}}^2}{l^4}\sin {\left(\frac{\mathrm{\&pi;x}}{l}\right)}^2\mathrm{dx}$

$\mathrm{\&mu;}=\frac{{\mathrm{EI}}_1{\mathrm{\&pi;}}^3{\mathrm{\&delta;}}^2}{{4l}^4}[\mathrm{\&pi;}(l-l_2)+\frac{\mathrm{<figure-callout\; id="2"\; label="l"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">l</figure-callout>}}{2}\sin \frac{2\mathrm{\&pi;}{l}_2}{l}]+\frac{{\mathrm{EI}}_2{\mathrm{\&pi;}}^3{\mathrm{\&delta;}}^2}{{4l}^4}[\frac{l}{2}\sin \frac{{2\mathrm{\&pi;l}}_2}{l}-\mathrm{\&pi;}{l}_2]---\left(12\right)$

With formula (5), (10), (11) and (12) substitutions (8) Shi Kede:

$P_{\mathrm{cr}}=\frac{{\mathrm{<figure-callout\; id="1"\; label="EI"\; filenames="Y20062011916300141.png"\; state="\{\{state\}\}">EI</figure-callout>}}_1\mathrm{\&pi;}}{{\mathrm{<figure-callout\; id="3"\; label="l"\; filenames="Y20062011916300152.png"\; state="\{\{state\}\}">l</figure-callout>}}^3}{\mathrm{\&alpha;}}_1+\frac{{\mathrm{<figure-callout\; id="2"\; label="EI"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">EI</figure-callout>}}_2\mathrm{\&pi;}}{{\mathrm{<figure-callout\; id="3"\; label="l"\; filenames="Y20062011916300152.png"\; state="\{\{state\}\}">l</figure-callout>}}^3}{\mathrm{\&alpha;}}_2-(\frac{{\mathrm{<figure-callout\; id="2"\; label="q"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">q</figure-callout>}}_2{\mathrm{\&beta;}}_2}{2l{\mathrm{\&pi;}}^2}+\frac{{\mathrm{<figure-callout\; id="1"\; label="q"\; filenames="Y20062011916300141.png"\; state="\{\{state\}\}">q</figure-callout>}}_1{\mathrm{\&beta;}}_1}{2l{\mathrm{\&pi;}}^2})---\left(13\right)$

In the formula (13):

${\mathrm{\&alpha;}}_1=\mathrm{\&pi;}(l-l_2)+\frac{\mathrm{<figure-callout\; id="2"\; label="l"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">l</figure-callout>}}{2}\mathrm{<figure-callout\; id="2"\; label="sin"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">sin</figure-callout>}\frac{{2\mathrm{\&pi;l}}_2}{l}---\left(14\right)$

${\mathrm{\&alpha;}}_2=\frac{\mathrm{<figure-callout\; id="2"\; label="l"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">l</figure-callout>}}{2}\mathrm{<figure-callout\; id="2"\; label="sin"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">sin</figure-callout>}\frac{2\mathrm{\&pi;}{l}_1}{l}-{\mathrm{\&pi;l}}_2---\left(15\right)$

${\mathrm{\&beta;}}_1={\mathrm{\&pi;}}^2{{\mathrm{<figure-callout\; id="1"\; label="l"\; filenames="Y20062011916300141.png"\; state="\{\{state\}\}">l</figure-callout>}}_1}^2-{\mathrm{\&pi;<figure-callout\; id="1"\; label="ll"\; filenames="Y20062011916300141.png"\; state="\{\{state\}\}">ll</figure-callout>}}_1\sin \frac{2\mathrm{\&pi;}{l}_2}{l}-{\mathrm{<figure-callout\; id="2"\; label="l"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">l</figure-callout>}}^2\sin {\left(\frac{\mathrm{\&pi;}{l}_2}{l}\right)}^2---\left(16\right)$

${\mathrm{\&beta;}}_2={\mathrm{\&pi;}}^2{l}_2(l+l_1)+{\mathrm{\&pi;<figure-callout\; id="1"\; label="ll"\; filenames="Y20062011916300141.png"\; state="\{\{state\}\}">ll</figure-callout>}}_1\sin \frac{2\mathrm{\&pi;}{l}_2}{l}+{\mathrm{<figure-callout\; id="2"\; label="l"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">l</figure-callout>}}^2\sin {\left(\frac{\mathrm{\&pi;}{l}_2}{l}\right)}^2---\left(17\right)$

In the formula: P _CrBe critical axial load [N] that E is steel modulus of elasticity [N/m ²], generally: E=2.058 * 10 ⁵MPa; I is a polar moment of inertia, I=3.14 * (external diameter ⁴-internal diameter ⁴)/32[m ⁴]; L be riser pipe 1 from the seabed length [m] to the top; l ₁Length for the hypomere tubing string; l ₂Length for the hypomere tubing string; q ₁Unit weight [N/m] for the epimere tubing string; q ₂Unit weight [N/m] for the hypomere tubing string.

Now formula (13) is discussed:

1, works as l ₂=l q ₂=q I ₂=I (18)

Then

With α in the formula (19) ₂And β ₂Substitution formula (13) can get:

$P_{\mathrm{cr}}=\frac{{\mathrm{<figure-callout\; id="2"\; label="EI"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">EI</figure-callout>}}_2\mathrm{\&pi;}}{l^3}(-\mathrm{\&pi;l})-\left(\frac{{\mathrm{<figure-callout\; id="2"\; label="q"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">q</figure-callout>}}_2{\mathrm{\&pi;}}^2{\mathrm{<figure-callout\; id="2"\; label="l"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">l</figure-callout>}}^2}{2l{\mathrm{\&pi;}}^2}\right)=\frac{{\mathrm{<figure-callout\; id="2"\; label="EI"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">EI</figure-callout>}}_2{\mathrm{\&pi;}}^2}{l^2}-\frac{q_{2}l}{2}---\left(20\right)$

Observation type (20) can be found:

1) when considering riser pipe 1 for same rigidity tubing string, identical by the critical load that formula (20) draws with the critical load that draws by formula (9), this has illustrated two problems: the correctness of having verified formula (13) on the one hand, illustrated becoming under the rigidity condition, can adopt formula (13) to find the solution critical load; On the other hand, contact inherent between formula (13) and the formula (20) has been described.

2) formula (20) is a specific example of formula (13), illustrates to adopt general-purpose type (13), goes for finding the solution with rigidity condition lower critical load fully.

By (17) Shi Kede: ${\mathrm{\&beta;}}_2=[\frac{3}{4}{\mathrm{\&pi;}}^2{\mathrm{<figure-callout\; id="2"\; label="l"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">l</figure-callout>}}^2+l^2]$

With formula (21) substitution (13) Shi Kede:

$P_{\mathrm{cr}}=\frac{{\mathrm{<figure-callout\; id="1"\; label="EI"\; filenames="Y20062011916300141.png"\; state="\{\{state\}\}">EI</figure-callout>}}_1{\mathrm{\&pi;}}^2}{{2\mathrm{<figure-callout\; id="2"\; label="l"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">l</figure-callout>}}^2}+\frac{{\mathrm{<figure-callout\; id="2"\; label="EI"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">EI</figure-callout>}}_2{\mathrm{\&pi;}}^2}{{2l}^2}-\frac{l}{2}[q_2(\frac{3}{4}+\frac{1}{{\mathrm{\&pi;}}^2})+q_1(\frac{1}{4}-\frac{1}{{\mathrm{\&pi;}}^2})]---\left(22\right)$

Observation type (22) can be found:

1) works as I ₁=I ₂=I; q ₁=q ₂During=q, its substitution formula (22) can be drawn:

$P_{\mathrm{cr}}=\frac{{\mathrm{EI\&pi;}}^2}{l^2}-\frac{1}{2}\mathrm{ql}---\left(23\right)$

As seen: formula (23) is to work as l ₁=l ₂The time, q ₁-q ₂=q; I ₁-I ₂Special circumstances under the qualifications constraint of=I, this moment is identical with the situation of front formula (9) just, and this formula (13) is a general formula, and formula (22) is at l ₁=l ₂Special formula under the condition.

2) in formula (22), can find, in two of the fronts, work as I ₁=I ₂The time, $\frac{{\mathrm{<figure-callout\; id="1"\; label="EI"\; filenames="Y20062011916300141.png"\; state="\{\{state\}\}">EI</figure-callout>}}_1{\mathrm{\&pi;}}^2}{{2\mathrm{<figure-callout\; id="2"\; label="l"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">l</figure-callout>}}^2}+\frac{{\mathrm{<figure-callout\; id="2"\; label="EI"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">EI</figure-callout>}}_2{\mathrm{\&pi;}}^2}{{2\mathrm{<figure-callout\; id="2"\; label="l"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">l</figure-callout>}}^2}\&DoubleRightArrow;\frac{{\mathrm{EI\&pi;}}^2}{{\mathrm{<figure-callout\; id="2"\; label="l"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">l</figure-callout>}}^2}$

Being Euler's load of depression bar, and in the back in the 3rd, the 4th, then is that explanation deadweight uniformly distributed load is to critical influence, and at l ₁=l ₂Condition under, q ₂Influence than q ₁Influence is big.

Further specify the critical load formula application in practice that the utility model proposes below in conjunction with specific embodiment.

Embodiment 1

Adopt the utility model that Bohai Bay Oil one drilling riser critical load and stability are detected.

The opening of setting riser pipe 1 this moment is in the drilling well moon pool of offshore boring island, the top is in fixing state, known: oil field, Bohai Sea depth of water 25m, carry out drillng operation with self-elevating drilling platform, its air gap is 10m, and the conduit top is subjected to the constraint of temporary guide base, the top can be considered hinge support 3, mud 50m is gone in its bottom, can be considered solid end 2, and this moment, bored shaft mud was hung above pit shaft conduit and surface pipe data following (as shown in table 1):

Table 1

Sleeve pipe-grade of steel-#	External diameter "/mm	Internal diameter "/mm	Length m	Uniformly distributed load kg/m	Remarks
						30”-X52	30”/762	28”/711.2	35	461.3	Go into mud 50m
20”-K55	20”/508	18.73”/475.7	35	104	More than mud is hung
						13-3/8”-N80-68#	13-3/8”/339.7	12.915”/315.3	35	100	More than mud is hung

Wellhead blowout preventor group weight: 13-5/8 " * 10M (M=10 * 1000psi (pound/inch ²)), be combined as: omnipotent preventer * 5M+ double ram type preventer * 10M+ single ram type perventer * 10M+ drilling spool, gross weight is about: 44t, find the solution its elastic stability.

At first detect or gather the following technical data of riser pipe, getting the steel modulus of elasticity is E=2.1 * 10 ⁶Kg/cm ²

The riser pipe 1 of present embodiment is found the solution above-mentioned technical data substitution critical load formula (23) and is obtained (as shown in table 2), wherein P under the condition of the constant rigidity of identical caliber _Cr1Be the critical load value of not considering to conduct oneself with dignity and influence:

Table 2

Caliber " (inch)	External diameter cm	Internal diameter cm	Uniformly distributed load kg/m	The l length m	Critical load P crt	Critical load P cr1t	Load value
									Tubing string weight t	Tension
							30”	76.2			71.1 2	461.3	35	666.192	674.260	39.2	54.8
20”	50.8	47.5 7	198	35	124.155	127.620			16.8	57.3
							13-3/8 ”	33.97			31.5 3	100	35	26.719	28.469	8.5	83

Analyze the above-mentioned detection of present embodiment, result calculated as can be seen, after this drilling well is spudded in, be lowered to common 30 " riser pipe; install 30 at well head " * the ring-like preventer of 1M, its weight is 14.7t, and this moment, riser pipe 1 uniformly distributed load was 461.3kg/m, its critical load P _Cr=666.2t.30 " weight of ring-like preventer is 14.7t, only is 2.2% of critical load 666.2t.Show that its safety allowance of tubing string commonly used is very big, the safety factor of its critical load is $n_s=\frac{666}{14.7}=45.3.$

In subsequent job, with 13-5/8 " * after the weight of 10M blowout preventer set all was added on the riser pipe 1, its safety factor was $n_s=\frac{666}{44}=15.$

According to above parameter, the drilling riser of present embodiment exists according to critical load formula (22) when becoming the condition of rigidity $l=\frac{\mathrm{<figure-callout\; id="2"\; label="l"\; filenames="Y20062011916300151.png"\; state="\{\{state\}\}">l</figure-callout>}}{2}=17.5m$ The time find the solution the critical load (as shown in table 3) that obtains:

Table 3

The combination caliber

OD/ID cm

l 2 m

q2 kg/m

OD/ID cm

l 1 m

q1 kg/m

Critical load Pcr t

Tubing string deadweight t

The tension safety factor

30”×20”	76.2/71. 12	17.5	461	50.8/47 .5	17.5	194	393.566	34.6	27.9
										20” ×13-3/8”	50.8/47. 5	17.5	198	34/32	17.5	100	74.834	15.1	46.7
30”×30”	76.2/71. 12	17.5	461	76/73	17.5	292	430.668	36.2	37.5

Analyze the above-mentioned detection of present embodiment, result calculated as can be seen, adopted and become after the tower structure of rigidity, same outer diameter as size as shown in table 3, different inner diameters size 30 " * 30 " riser pipe, during the i.e. change rigidity combination of the thin two kinds of different inwalls of bottom wall thickness and top wall, its deadweight uniformly distributed load is 461kg/m and 292kg/m, its critical load is 430.7t, promptly show the deadweight of installing this moment be 14.7t 30 " * the ring-like preventer of 1M, the safety factor of its critical load is $n_s=\frac{430.7}{14.7}=29.3,$ Safety allowance also is flush with money.

In subsequent job with 13-5/8 " * after the weight of 10M blowout preventer set all was added on the riser pipe, its safety factor had also reached $n=\frac{461}{44}=10.4.$

When operating condition allows, adopt tower stiffness changing structure, promptly be to adopt large diameter pipe near sea bed one end, adopt narrow tube (as shown in Figure 1) near drilling well well head one end, be employing 30 as shown in table 3 " * 20 " the combination of tower change rigidity riser pipe, this moment, critical load was 393.56t, the 21-1/4 of installation " * the 5M preventer; its deadweight is 20t, the then safety factor of critical load $n_s=\frac{383.56}{20}=19.68;$ In subsequent job, with 13-5/8 " * total weight of 10M blowout preventer set adds that all its safety factor still is $n=\frac{393.56}{44}=8.9.$ As seen, allowance is very big at this moment, is enough to engineering demands.

Drillng operation reduces cost in consideration, when raising the efficiency, spud in and can adopt the tower structure (as shown in table 3) that becomes rigidity in the top layer, select 20 " * 13-3/8 " combination, this moment, the critical load of riser pipe 1 was 74.8t, install 211/4 " * preventer of 5M, its deadweight is 20t, the safety factor of its critical load is $n_s=\frac{74.8}{20}=3.74,$ As seen safety allowance still can engineering demands, promptly can be by the detection requirement of riser pipe 1 elastic stability.

Embodiment 2

Adopt the utility model that another drilling riser critical load of Bohai Bay Oil and stability are detected.

This drilling well depth of water 32m of living in, the technical data that all the other on-the-spot drilling conditions and institute detect and collect is identical with embodiment 1.According to detecting the technical data that is collected, the drilling riser of present embodiment is found the solution the critical load that obtains according to formula (23) under the condition of constant rigidity conduit, and the result is as shown in table 4.

Table 4

Caliber "/mm	External diameter cm	Internal diameter cm	Critical load kg/m	The l length m	Critical load P cr?t	Tubing string deadweight t	The tension safety factor
								30”	76.2	71.12	461.3	42	458.554	39.2	54.8
20”	50.8	47.57	198	42	89.625	16.8	57
								13-3/8”	33.97	31.53	100	42	17.670	8.5	83

According to above parameter, present embodiment according to formula (22), is found the solution the critical load (as shown in table 5) that obtains when using notch cuttype riser pipe 1 condition that becomes rigidity:

Table 5

The combination caliber	OD/ID cm	l 2 m	q 2 kg/m	OD/ID cm	l 1 m	q 1 kg/m	Critical load P cr t	Tubing string deadweight t	The tension safety factor
										30”×20”	76.2/71.1 2	21	461	50.84/47.5 7	21	194	79.777	13.7 6	69.8
20” ×13-3/8”	50.8/47.5	21	198	34/32	21	100	15.918	6.26	77.18
										30”×30”	76.2/71.1 2	21	461	76/73	21	292	295.21 8	15.8 7	85.75

Below the result of calculation of embodiment 1,2 is analyzed, as can be seen from Table 2:

If 1 is lowered to common 30 at spudding well " behind the riser pipe; suppose that we install 30 at well head " * the ring-like preventer of 1M, its weight is 14.7t, and our deadweight adopted is that the critical load (as shown in table 2) of 461.3kg/m is: 666.2t, that is this moment 30 usually " ring-like preventer weight only is 2.2% of critical load.As seen, the safety allowance of tubing string commonly used is big, and the safety factor of its critical load is $n_s=\frac{666}{14.7}=45.3,$ And in the follow-up operation, if 13-5/8 " * 10M blowout preventer set weight all is added on the riser pipe, and its safety factor also is $n_s=\frac{666}{44}=15.$

If 2 adopt the tower structure that becomes rigidity, as the same outside dimension in the table 3, but different inner diameters 30 " * 30 ", riser pipe, its deadweight be for 461kg/m and 292kg/m, and promptly bottom wall thickness and top wall are thin two kinds, when the change rigidity of different inwalls makes up, its critical load is 430.7t, that is install this moment deadweight for 14.7t 30 " * the ring-like preventer of 1M, the safety factor of its critical load is $n_s=\frac{430.7}{14.7}=29.3,$ As seen safety allowance is flush with money naturally, even consider in the subsequent job 13-5/8 " * 10M blowout preventer set weight all is added on the riser pipe 1, and its safety factor is $n=\frac{461}{44}=10.4.$

3, when operating condition allows, we adopt tower stiffness changing structure, promptly be to adopt large diameter pipe near sea bed one end, adopt the way (as shown in table 3) of narrow tube near drilling well well head one end, if adopt 30 " * 20 " the combination of tower change rigidity marine riser, this moment, critical load was 393.56t, and if 21-1/4 is installed this moment " * the 5M preventer; its deadweight is 20t, then this moment critical load safety factor $n_s=\frac{383.56}{20}=19.68,$ Consider in the subsequent job, even with 135/8 " * total weight of 10M blowout preventer set adds that all allowance is also very big, is enough to engineering demands, its safety factor still is $n=\frac{393.56}{44}=8.9.$

4, the demand of considering operation is wanted to reduce cost, raise the efficiency, spud in and adopt the tower structure (as shown in table 3) that becomes rigidity in the top layer, if adopt 20 " * 13-3/8 " combination, then this moment, its critical load was 74.8t, and we this moment if after only considering the top layer operation, 21-1/4 is installed " * preventer of 5M; its deadweight is 20t, then this moment critical load safety factor be $n_s=\frac{74.8}{20}=3.74.$

5, as seen by the contrast of table 4 and table 2: when the depth of water when 35m is increased to 42m, critical load reduces very fast, for 30 " deadweight is the riser pipe of 461kg/m, its critical load is reduced to 458.6t by 666.2t, promptly the depth of water increases by 16%, critical load reduction by 31%.

6, as seen: if the employing stiffness changing structure is 30 " * 20 " during the riser pipe structure by the contrast of table 5 and table 3, when the depth of water increases to 42m by 35m, its critical load is reduced to 79.8t by 393.6t, that is under the condition of depth of water increase by 16%, critical load reduces about 80%.

By above practice with the analysis showed that, the utility model is considered as an end build-in with the riser pipe on the self-elevating drilling platform 1, the mechanical model that one end hinged constraint is set up is found the solution the method for the critical load of its elastic stability, is feasible on principle and field practice.Operation on the sea top layer preventer 30 commonly used just at present " * 1M and 21-1/4 " * two kinds of patterns commonly used of 5M on, calculate with the utility model method, conventional 30 " and become rigidity 30 " * 20 " and 30 " * 30 " three kinds of various combination forms, its critical load all satisfies the operation needs.

In sum, as can be seen, become the critical load formula that the rigidity tower structure proposes according to the utility model, can comprise the length of riser pipe 1 each section by every definite riser pipe 1 parameter, inside and outside footpath, weight, stiffness factors etc. calculate the critical load value of riser pipe 1 elastic stability, act on gross weight on the riser pipe according to preventer etc., just can obtain safety factor, obtain the testing result of riser pipe elastic stability.Simultaneously, according to safety factor require, load requirement such as sea water advanced, preventer weight, also can pass through the critical load formula of the elastic stability of riser pipe, in conjunction with the specification of existing riser pipe, parameters such as the length of selection upper and lower part riser pipe 1 and caliber.

Claims (5)

1, a kind of offshore oil well drilling used water, it is characterized in that: described riser pipe is tower change rigidity tubing string, top tubing string rigidity is little, bottom tubing string rigidity is big, after the material that concentrated load that wellhead blowout preventor forms and tubing string use was determined, the length of described upper and lower section tubing string and caliber satisfied the critical load formula of following riser pipe elastic stability:

$P_{\mathrm{cr}}=\frac{{\mathrm{EI}}_1\mathrm{\&pi;}}{l^3}{\mathrm{\&alpha;}}_1+\frac{{\mathrm{EI}}_2\mathrm{\&pi;}}{l^3}{\mathrm{\&alpha;}}_2-(\frac{q_2{\mathrm{\&beta;}}_2}{{2\mathrm{l\&pi;}}^2}+\frac{q_1{\mathrm{\&beta;}}_1}{{2\mathrm{l\&pi;}}^2})$ In the formula:

${\mathrm{\&alpha;}}_1=\mathrm{\&pi;}\left({l-l}_2\right)+\frac{l}{2}\sin \frac{2\mathrm{\&pi;}{l}_2}{l}$

${\mathrm{\&alpha;}}_2=\frac{l}{2}\sin \frac{{2\mathrm{\&pi;l}}_1}{l}-{\mathrm{\&pi;l}}_2$

${\mathrm{\&beta;}}_1={\mathrm{\&pi;}}^2{l_1}^2-\mathrm{\&pi;}{\mathrm{ll}}_1\sin \frac{2{\mathrm{\&pi;l}}_2}{l}-l^2\sin {\left(\frac{\mathrm{\&pi;}{l}_2}{l}\right)}^2$

${\mathrm{\&beta;}}_2={\mathrm{\&pi;}}^2{l}_2\left({l+l}_1\right)+\mathrm{\&pi;}{\mathrm{ll}}_1\sin \frac{2{\mathrm{\&pi;l}}_2}{l}+l^2\sin {\left(\frac{\mathrm{\&pi;}{l}_2}{l}\right)}^2$

Wherein: P _CrBe critical axial load [N] that E is steel modulus of elasticity [N/m ²], generally: E=2.058 * 10 ⁵MPa; I is a polar moment of inertia, I=3.14 * (external diameter ⁴-internal diameter ⁴)/32[m ⁴]; 1 be riser pipe 1 from the seabed length [m] to the top; l ₁Length for the hypomere tubing string; l ₂Length for the hypomere tubing string; q ₁Unit weight [N/m] for the epimere tubing string; q ₂Unit weight [N/m] for the hypomere tubing string.

2, a kind of offshore oil well drilling used water as claimed in claim 1 is characterized in that: work as l ₂=1 q ₂During=q,

$P_{\mathrm{cr}}=\frac{{\mathrm{EI}}_2\mathrm{\&pi;}}{l^3}(-\mathrm{\&pi;l})-\left(\frac{q_2{\mathrm{\&pi;}}^2{l}^2}{{2\mathrm{l\&pi;}}^2}\right)=\frac{{\mathrm{EI}}_2{\mathrm{\&pi;}}^2}{l^2}-\frac{q_{2}l}{2}.$

3, a kind of offshore oil well drilling used water as claimed in claim 1 is characterized in that:

When $l_1=l_2=\frac{l}{2}$ The time:

$P_{\mathrm{cr}}=\frac{{\mathrm{EI}}_1{\mathrm{\&pi;}}^2}{{2l}^2}+\frac{{\mathrm{EI}}_2{\mathrm{\&pi;}}^2}{{2l}^2}-\frac{l}{2}[q_2(\frac{3}{4}+\frac{1}{{\mathrm{\&pi;}}^2})+q_1(\frac{1}{4}-\frac{1}{{\mathrm{\&pi;}}^2})]$

4, a kind of offshore oil well drilling used water as claimed in claim 1 is characterized in that: the internal diameter of described upper and lower part tubing string is identical, and the external diameter of described top tubing string is less than the external diameter of described hypomere tubing string.

5, a kind of offshore oil well drilling used water as claimed in claim 1 is characterized in that: the external diameter of described upper and lower part tubing string is identical, and the internal diameter of described top tubing string is greater than the internal diameter of described bottom tubing string.

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