CN101324174B - Method for preventing well drilling riser pipe and drill stem from generating resonance - Google Patents

Abstract

The invention relates to a method for preventing a drilling riser and a drilling string to generate resonance, which comprises the following steps: 1) supposing that the upper end of the drilling riser is a free end which is suspended above a floating drilling platform by a tensioner and limited by drifting value, and the lower end is an articulating supporting end which is connected with a blowout preventer and an underwater wellhead by a flexible joint, a floating drilling model is further formed; 2) the equation Omega (refer to the lower right formula) with natural frequency of the riser is established according to the floating drilling model; 3) the rotating speed N of the drilling string when the drilling is operated is substituted, the resonance rotating speed Nn (Nn is equal to 30 Omega/Pi) of the drilling string is calculated according to the natural frequency of the riser; 4) the resonance rotating speed Nn of the drilling string is avoided to select the rotating speed of the drilling string. The method reasonably selects the rotating speed of the drilling string by calculating the natural frequency of the drilling riser, thus effectively preventing the drilling riser and the drilling string from generating the resonance.

Description

A kind of drilling riser and drill string of preventing produces the method for resonance

Technical field

The present invention relates to a kind of method that prevents that resonance from producing, particularly produce the method for resonance about a kind of prevent drilling riser and drill string that is applied in the offshore oil drilling.

Background technology

When deepwater work, the deepwater drilling riser pipe is the unique passage that connects seabed and water surface drilling rig, has the seabed of the connection critical function such as preventer and water surface drilling rig control device under water, and the effect in whole deepwater drilling process is particularly important.The drill string that riser pipe is rotated drillng operation not only will for section within it provides closed environment, simultaneously or the circulation canal of drilling fluid.In the research of deepwater drilling problem, be a requisite link to the research of riser pipe.How reasonably to use riser pipe, be one of major issue in the drilling well work always.

Consider that from the motion state aspect motion state of riser pipe one end in the seabed is subjected to the constraint of underwater well head, and is subjected to the constraint of drilling rig at an end on sea.Riser pipe is subject to top tension force, the horizontal and axial thrust load that deadweight produces, the percussion of the Caspian Sea, operation waters stream; In addition, riser pipe also is subject to because the impact that is coupled motion that the drilling rig motion produces.Because in drillng operation, this under intrinsic frequency transverse vibration will occur drill string, riser pipe itself also has its intrinsic frequency simultaneously, and when these two frequencies reached consistent, whole riser pipe and drill string produced resonance.When resonance phenomena produced, the vibration amplitude of riser pipe and drill string reached maximum value, caused the contact between riser pipe and the drill string that violent friction occurs.Simultaneously, because the coating of carbide alloy is arranged on the drill string joint, therefore its hardness make the inside body generation excessive wear of riser pipe greater than the hardness of riser pipe in friction process.Excessive wear directly causes the bearing capacity of riser pipe greatly to descend, and shortens application life, when serious even can have influence on the safety of deepwater drilling.

Because the complexity of riser pipe and drill string motion in the deepwater drilling operation, resonance problems never is well solved, and more neither one is reasonably selected the method for drill string rotating speed.Conventional viewpoint is that riser pipe is considered as is the fixed form that upper and lower end is hinge support, and its defective is, it does not fully take into account in deepwater work, the drift impact that Current Movement produces drilling rig.The present invention is exactly the research for kinetic characteristic and the vibration characteristics of deep water riser pipe, has proposed the concrete solution to resonance problems.

Summary of the invention

For the problems referred to above, the purpose of this invention is to provide a kind of fully taking into account in deepwater work, Current Movement produces the method for resonance to prevent deepwater drilling riser pipe and the drill string of the drift impact that drilling rig produces.

For achieving the above object, the present invention takes following technical scheme: a kind of drilling riser and drill string of preventing produces the method for resonance, and it may further comprise the steps:

1) suppose that riser pipe upper end is to be suspended on the floating drilling platform and to be subjected to the free end of drift value restriction by tensioner, the lower end is to be connected to hinge support on preventer and the underwater well head by flexible joint, and then forms a floatation type drilling well model;

2) according to floatation type drilling well model, set up the equation ω of riser pipe intrinsic frequency:

$\mathrm{\ω}=1.414\sqrt{\frac{g[2\cos \mathrm{n\π}+(2n-1)\mathrm{\π}]}{a_n[2\cos \mathrm{n\π}(4-\sin \mathrm{n\π})+3(2n-1)\mathrm{\π}]}}$

In the formula, a _nThe panel indicator deflection coefficient, g represents acceleration of gravity, n represents natural number;

3) gone out the resonance rotational speed N of drill string by the calculation on Natural Frequency of riser pipe _n

$N_n=\frac{30\mathrm{\ω}}{\mathrm{\π}};$

4) avoid the resonance rotational speed N of drill string _nSelect the drill string rotating speed.

The equation ω of the riser pipe intrinsic frequency described step 2), take following steps to draw:

1. based on floatation type drilling well model, set up the riser pipe Flexural Equation:

$y=\underset{n=1}{\overset{\∞}{\mathrm{\Σ}}}{a}_n(1-\cos \frac{(2n-1)\mathrm{\πx}}{2l})$

Wherein, l is the length of riser pipe, a _nThe panel indicator deflection coefficient,

$a_n=\frac{{\mathrm{ql}}^4\sin \mathrm{\&alpha;}[2(\cos \mathrm{n\&pi;}+\mathrm{n\&pi;})-\mathrm{\&pi;}]}{{\mathrm{\&pi;}}^4\left\{\sin n\mathrm{\&pi;}\cos \mathrm{n\&pi;}\right[{\mathrm{<figure-callout\; id="1"\; label="TK"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">TK</figure-callout>}}_1+{\mathrm{EIK}}_2+\mathrm{ql}\cos \mathrm{\&alpha;}{K}_3]+\mathrm{\&pi;}[{\mathrm{TK}}_4+{\mathrm{EIK}}_5+\mathrm{ql}\cos \mathrm{\&alpha;}\&CenterDot;K_6\left]\right\}};$

Wherein, the deadweight of q-riser pipe, unit is Kg;

α-riser pipe inclination angle;

The T-axial force, unit is N;

E-modulus of elasticity, unit are N/m ²

The I-polar moment of inertia,

Unit is m ⁴

OD-riser pipe external diameter, unit are m;

ID-riser pipe internal diameter, unit are m;

G-acceleration of gravity;

K ₂＝-64n ⁴+128n ³-96n ²+32n-4

K ₃＝-32n ⁴+64n ³-48n ²+16n-2

K ₄＝32n ⁵-80n ⁴+80n ³-40n ²+10n-1

K ₅＝64n ⁵-160n ⁴+160n ³-80n ²+20n-2

K ₆＝32n ⁵-80n ⁴+80n ³-40n ²+10n-1；

2. according to the riser pipe Flexural Equation, calculate the riser pipe formula for natural frequency

Intrinsic frequency square be:

${\mathrm{\&omega;}}^2=\frac{g{\&Integral;}_0^l\mathrm{qydx}}{{\&Integral;}_0^{\mathrm{<figure-callout\; id="2"\; label="l\; qy"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">l</figure-callout>}}{\mathrm{qy}}^{}{}^2\mathrm{dx}}$

Intrinsic frequency ω as can be known then:

$\mathrm{\&omega;}=1.414\sqrt{\frac{g[2\cos \mathrm{n\&pi;}+(2n-1)\mathrm{\&pi;}]}{a_n[2\cos \mathrm{n\&pi;}(4-\sin \mathrm{n\&pi;})+3(2n-1)\mathrm{\&pi;}]}}.$

When n=1,

$y_{n=1}=\frac{32{\mathrm{ql}}^4\sin \mathrm{\&alpha;}(\mathrm{\&pi;}-2)}{\mathrm{EI}{\mathrm{\&pi;}}^5[2+\frac{1}{\mathrm{EI}{\mathrm{\&pi;}}^5}(T+\mathrm{ql}\cos \mathrm{\&alpha;})]}(1-\cos \frac{\mathrm{\&pi;x}}{2l}).$

Namely when x=l, drift value is:

$y_{\max }=\frac{32{\mathrm{ql}}^4\sin \mathrm{\&alpha;}(\mathrm{\&pi;}-2)}{\mathrm{EI}{\mathrm{\&pi;}}^5[2+\frac{1}{\mathrm{EI}{\mathrm{\&pi;}}^5}(T+\mathrm{ql}\cos \mathrm{\&alpha;})]}.$

The present invention is owing to take above technical scheme, it has the following advantages: 1, the present invention is from deepwater drilling operation actual conditions, consider simultaneously drilling rig because Current Movement produces the factor of drift, having set up the riser pipe lower end is hinge support, the upper end is one and is suspended on free-ended floatation type drilling well model on the floating drilling platform by tensioner, makes it be more suitable for actual condition in the deepwater drilling operation.2, the present invention's riser pipe Flexural Equation of trying to achieve by energy method, axial thrust load qcos α and the vertical component qsin α of the suffered axial force T of riser pipe, inclination alpha, deadweight q have been considered, drawn the design formulas of riser pipe intrinsic frequency by it, come the rotating speed of choose reasonable drill string with this, effectively avoided riser pipe and drill string to produce resonance.3, riser pipe calculation on Natural Frequency formula of the present invention has had theoretical guidance to the research of riser pipe motion state; has in actual applications great practice significance; the method is used in the deepwater drilling operation simultaneously; effectively protect riser pipe, thereby the safety that guarantees the deepwater drilling operation is carried out.

Description of drawings

Fig. 1 is floatation type drilling well model schematic diagram of the present invention

Fig. 2 is the stressed schematic diagram of riser pipe of the present invention

The specific embodiment

Be described in detail of the present invention below in conjunction with drawings and Examples.

In the process of research riser pipe problem, can find that its motion state has following characteristics: 1) riser pipe topmost links together with drilling rig usually, but consider in the actual condition of deepwater drilling, drilling rig on the sea is subject to action of ocean current will produce drift, therefore drives in the fixing riser pipe upper end of water surface well head and also produces related drift motion.2) the riser pipe lower end is connected on underwater well head and the preventer, obviously, is considered as fixing hinge support.3) top of riser pipe is subject to the effect of axial force T; Riser pipe itself also has deadweight q.

One, floatation type drilling well model

As shown in Figure 1, analysis based on above motion state, for the design that makes riser pipe 1 more realistic, floatation type drilling well model of the present invention is that fixing hypothesis is held by a hinge support in riser pipe 1 upper end, and displacement becomes one and is suspended on the floating drilling platform by tensioner 2 and is subjected to drift value y _MaxThe free-ended hypothesis of restriction.And the riser pipe lower end still is connected on preventer 4 and the underwater well head 5 by flexible joint 3, and the connected mode of flexible joint 3 is regarded as hinge support.Riser pipe 1 under the floatation type drilling well model is regarded as free end owing on it, so that have inclination alpha between the axis of riser pipe 1 and the vertical line like this.According to the actual mode of occupation of riser pipe 1, axially as the direction of abscissa x axle, with direction perpendicular to axial direction direction as ordinate y axle, the length of riser pipe 1 is l with it.Under the condition that produces inclination alpha, riser pipe 1 deadweight q will be created in axial component q _x=qcos α, and at the vertical component q perpendicular to axis direction _y=qsin α, these two orthogonal components all act on the riser pipe 1 simultaneously.

Two, riser pipe flexural equation

Based on the characteristics of floatation type drilling well model, riser pipe is set up Flexural Equation, the form of being write as trigonometrical number is:

$y=\underset{n=1}{\overset{\&infin;}{\mathrm{\&Sigma;}}}{a}_n(1-\cos \frac{(2n-1)\mathrm{\&pi;x}}{2l})---\left(1\right)$

Wherein, a _nThe panel indicator deflection coefficient.Then the deflection deformation of riser pipe can for:

$u=\frac{1}{2}\mathrm{EI}{\&Integral;}_0^l{\left(\frac{d^{2}y}{d{x}^2}\right)}^2\mathrm{dx}---\left(2\right)$

Wherein, E is the modulus of elasticity of riser pipe steel, generally: E=2.058 * 105MPa; I is moment of inertia, and I=3.14 (OD-ID)/32, OD are the external diameter of riser pipe, and ID is the internal diameter of riser pipe.Therefore, (1) formula carried out the second order differentiate after, substitution (2) formula integration obtains:

$u=\frac{1}{32l^3}\mathrm{EI}{a}_{\mathrm{<figure-callout\; id="2"\; label="n"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">n</figure-callout>}}^2{\mathrm{\&pi;}}^3\{2\sin n\mathrm{\&pi;}\cos \mathrm{n\&pi;}(-8n^3+12n^2-6n+2)---\left(3\right)$

$+\mathrm{\&pi;}[1+8(2n^4-4n^3+3n^2-2n)]\}$

Because riser pipe is when existing inclination alpha, the difference of its crooked radian and chord length is:

$\mathrm{\&lambda;}=\frac{1}{2}{\&Integral;}_0^l{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2\mathrm{dx}---\left(4\right)$

Therefore (1) formula is asked substitution (4) Shi Kede after the second dervative:

$\mathrm{\&lambda;}=\frac{1}{64l^3}{a}_{\mathrm{<figure-callout\; id="2"\; label="n"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">n</figure-callout>}}^2{\mathrm{\&pi;}}^2[2\sin n\mathrm{\&pi;}\cos \mathrm{n\&pi;}(-8n^3+12n^2-6n+1)---\left(5\right)$

$+\mathrm{\&pi;}({16n}^4-32n^3+24n^2-8n+1)]$

By λ to flexibility factor a _nDifferentiate, the displacement increment d λ that obtains is:

$\mathrm{d\&lambda;}=\frac{\&PartialD;\mathrm{\&lambda;}}{\&PartialD;a_n}d{a}_n$

$=\frac{1}{32l^3}{a}_n{\mathrm{\&pi;}}^3[2\sin n\mathrm{\&pi;}\cos \mathrm{n\&pi;}(1-6n+12n^2-8n^3)---\left(6\right)$

$+\mathrm{\&pi;}(1-8n+24n^2-32n^3+16n^4)]d{a}_n$

Because this moment, axial force T was axial tensile force, because it belongs to tensile force, what therefore do is negative work:

$\stackrel{.}{..}{W}_T=-T\&CenterDot;\mathrm{d\&lambda;}$

$=\frac{1}{32l^3}T{a}_n{\mathrm{\&pi;}}^3[2\sin n\mathrm{\&pi;}\cos \mathrm{n\&pi;}(8n^3-12{\mathrm{<figure-callout\; id="2"\; label="n"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">n</figure-callout>}}^2+6n-1)---\left(7\right)$

$-\mathrm{\&pi;}(16n^4+32n^3-24{\mathrm{<figure-callout\; id="2"\; label="n"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">n</figure-callout>}}^2+8n-1)]{\mathrm{da}}_n$

The vertical component q of the deadweight q of riser pipe _y=qsin α acting is:

$W_{\mathrm{qx}}=-{\&Integral;}_0^{l}q\cos \mathrm{\&alpha;dxd}{a}_n$

$=-\frac{1}{32l^2}q\cos \mathrm{\&alpha;}\&CenterDot;a_n{\mathrm{\&pi;}}^3[2\sin n\mathrm{\&pi;}\cos \mathrm{n\&pi;}(1-6n+12n-8n^3)---\left(8\right)$

$+(1-8n+24n^2-32n^3+16n^4)]d{a}_n$

The axial component q of the deadweight q of riser pipe _x=qcos α acting is: $W_{\mathrm{qy}}={\&Integral;}_0^{l}q\sin \mathrm{\&alpha;}[1-\cos \frac{(2n-1)\mathrm{\&pi;x}}{2l}]\mathrm{dx}{\mathrm{da}}_n$ (9)

$=\frac{\mathrm{ql}\sin \mathrm{\&alpha;}[2\cos \mathrm{n\&pi;}+2\mathrm{n\&pi;}-\mathrm{\&pi;}]}{(2n-1)\mathrm{\&pi;}}d{a}_n$

By the elastic and plastic properties mechanics as can be known, elastic system is done a micro-displacement for its equilbrium position, and the increment of this system potential energy just equals external force institute's work under this displacement.Therefore, the deflection deformation of riser pipe can equal axial force T acting, riser pipe axial component q by u _xDo work and vertical component q _yThat is done work closes.Be:

u＝W _T+W _qx+W _qy (10)

Above-mentioned (7) formula, (8) formula and (9) formula substitution (10) formula can be obtained:

$\frac{\mathrm{IE}{a}_{\mathrm{<figure-callout\; id="2"\; label="n"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">n</figure-callout>}}^2{\mathrm{\&pi;}}^3}{32l^3}\{2\sin n\mathrm{\&pi;}\cos \mathrm{n\&pi;}(-8n^3+12n^2-6n+2)+\mathrm{\&pi;}[8(2n^4-4n^3+3n^2-2n)+1\left]\right\}$

$=\frac{{\mathrm{Pa}}_n{\mathrm{\&pi;}}^3}{32l^3}[2\sin n\mathrm{\&pi;}\cos \mathrm{n\&pi;}(-8n^3-12{\mathrm{<figure-callout\; id="2"\; label="n"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">n</figure-callout>}}^2+6n-1)-\mathrm{\&pi;}(16n^4+32n^3-24{\mathrm{<figure-callout\; id="2"\; label="n"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">n</figure-callout>}}^2+8n-1)]{\mathrm{da}}_n$

$+\frac{\mathrm{ql}\cos \mathrm{\&alpha;}}{32l^2}{a}_n{\mathrm{\&pi;}}^3[2\sin n\mathrm{\&pi;}\cos \mathrm{n\&pi;}(8n^3-12{\mathrm{<figure-callout\; id="2"\; label="n"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">n</figure-callout>}}^2+6n-1)-({16n}^4+32n^3-24{\mathrm{<figure-callout\; id="2"\; label="n"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">n</figure-callout>}}^2+8n-1)]{\mathrm{da}}_n$

$+\frac{\mathrm{ql}\sin \mathrm{\&alpha;}}{(2n-1)\mathrm{\&pi;}}[2(\cos \mathrm{n\&pi;}+\mathrm{n\&pi;})-\mathrm{\&pi;}]d{a}_n---\left(11\right)$

Can solve thus:

$a_n=\frac{32{\mathrm{ql}}^4\sin \mathrm{\&alpha;}[2(\cos \mathrm{n\&pi;}+\mathrm{n\&pi;})-\mathrm{\&pi;}]}{{\mathrm{\&pi;}}^4\left\{\sin n\mathrm{\&pi;}\cos \mathrm{n\&pi;}\right[P(64n^3-48{\mathrm{<figure-callout\; id="2"\; label="n"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">n</figure-callout>}}^2+16n-2-32n^4)+\mathrm{EI}(128n^3-96{\mathrm{<figure-callout\; id="2"\; label="n"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">n</figure-callout>}}^2+32n-4-64n^4)}$

$\stackrel{\&OverBar;}{+\mathrm{ql}\cos \mathrm{\&alpha;}(64n^3-48{\mathrm{<figure-callout\; id="2"\; label="n"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">n</figure-callout>}}^2+16n-2-32n^4)]+\mathrm{\&pi;}[P(32n^5-80n^4+80n^3-40{\mathrm{<figure-callout\; id="2"\; label="n"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">n</figure-callout>}}^2+10n-1)}$

$\stackrel{\&OverBar;}{+\mathrm{EI}(64n^5-160n^4+160n^3-80{\mathrm{<figure-callout\; id="2"\; label="n"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">n</figure-callout>}}^2+20n-2)+\mathrm{ql}\cos \mathrm{\&alpha;}(32n^5-80n^4+80n^3-40{\mathrm{<figure-callout\; id="2"\; label="n"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">n</figure-callout>}}^2+10n-1)\left]\right\}}---\left(12\right)$

In order to simplify (12) formula, order:

K ₁＝64n ³-48n ²+16n-2-32n ⁴ (13)

K ₂＝-64n ⁴+128n ³-96n ²+32n-4 (14)

K ₃＝-32n ⁴+64n ³-48n ²+16n-2 (15)

K ₄＝32n ⁵-80n ⁴+80n ³-40n ²+10n-1 (16)

K ₅＝64n ⁵-160n ⁴+160n ³-80n ²+20n-2(17)

K ₆＝32n ⁵-80n ⁴+80n ³-40n ²+10n-1 (18)

Therefore, (13)～(18) formula substitutions (12) formula can be obtained:

$a_n=\frac{{\mathrm{ql}}^4\sin \mathrm{\&alpha;}[2(\cos \mathrm{n\&pi;}+\mathrm{n\&pi;})-\mathrm{\&pi;}]}{{\mathrm{\&pi;}}^4\left\{\sin n\mathrm{\&pi;}\cos \mathrm{n\&pi;}\right[{\mathrm{<figure-callout\; id="1"\; label="TK"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">TK</figure-callout>}}_1+{\mathrm{<figure-callout\; id="2"\; label="EIK"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">EIK</figure-callout>}}_2+{\mathrm{ql}\cos \mathrm{\&alpha;K}}_3]+\mathrm{\&pi;}[{\mathrm{TK}}_4+{\mathrm{<figure-callout\; id="5"\; label="EIK"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">EIK</figure-callout>}}_5+\mathrm{ql}{\cos \mathrm{\&alpha;}\&CenterDot;K}_6\left]\right\}}---\left(19\right)$

(19) formula entered the deflection equation that (1) formula draws riser pipe and was generation: $y=\underset{i=1}{\overset{\&infin;}{\mathrm{\&Sigma;}}}\frac{32{\mathrm{ql}}^4\sin \mathrm{\&alpha;}[2(\cos \mathrm{n\&pi;}+\mathrm{n\&pi;})-\mathrm{\&pi;}]}{{\mathrm{\&pi;}}^4\left\{\sin n\mathrm{\&pi;}\cos \mathrm{n\&pi;}\right[{\mathrm{<figure-callout\; id="1"\; label="TK"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">TK</figure-callout>}}_1+\mathrm{<figure-callout\; id="2"\; label="EI\; K"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">EI</figure-callout>}{K}_{}{}_2+\mathrm{ql}\cos \mathrm{\&alpha;}{K}_3]+\mathrm{\&pi;}[{\mathrm{TK}}_4+\mathrm{<figure-callout\; id="5"\; label="EI\; K"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">EI</figure-callout>}{K}_{}{}_5+\mathrm{ql}\cos \mathrm{\&alpha;}\&CenterDot;K_6\left]\right\}}$ (20)

$\&CenterDot;[1-\cos \frac{(2n-1)\mathrm{\&pi;}}{2l}x]$

Because series equation has well approximate, first the computational accuracy of getting progression just is enough to satisfy the needs of engineering calculation, so select n=1.(20) formula is write as:

$y_{n=1}=\frac{32{\mathrm{ql}}^4\sin \mathrm{\&alpha;}(\mathrm{\&pi;}-2)}{\mathrm{EI}{\mathrm{\&pi;}}^5[2+\frac{1}{\mathrm{EI}{\mathrm{\&pi;}}^5}(T+\mathrm{ql}\cos \mathrm{\&alpha;})]}(1-\cos \frac{\mathrm{\&pi;x}}{2l})---\left(21\right)$

(21) formula is analyzed the following characteristics of being not difficult to find to have:

1, the Flexural Equation of riser pipe is at the axial component q that considers deadweight q _x, vertical component q _yUnder axial force T effect, the characteristics of the amount of deflection y of riser pipe have been reflected accurately.By (21) formula as can be known, the amount of deflection y of riser pipe and riser pipe length l, vertical component q _yBe directly proportional, with the axial component q of axial force T, deadweight q _xBe inversely proportional to.This tells from another point of view, increases the axial force T on the tensioner 2 or increases riser pipe deadweight q and all can reduce amount of deflection y.

2, ask the amount of deflection y of riser pipe upper end, namely when x=l,

According to (21) formula, have maximum defluxion in the riser pipe upper end this moment is y _MaxSuch result tallies with the actual situation.

$y_{\max }=\frac{32{\mathrm{ql}}^4\sin \mathrm{\&alpha;}(\mathrm{\&pi;}-2)}{\mathrm{EI}{\mathrm{\&pi;}}^5[2+\frac{1}{\mathrm{EI}{\mathrm{\&pi;}}^5}(P+\mathrm{ql}\cos \mathrm{\&alpha;})]}---\left(22\right)$

Three, riser pipe formula for natural frequency

The intrinsic frequency of riser pipe is exactly the resonant frequency of drill string.Therefore, find the solution draw riser pipe Flexural Equation (21) after, what be not difficult to try to achieve its intrinsic frequency square is:

${\mathrm{\&omega;}}^2=\frac{g{\mathrm{\&beta;}}_1}{{\mathrm{\&beta;}}_2}=\frac{g{\&Integral;}_0^l\mathrm{qydx}}{{\&Integral;}_0^{\mathrm{<figure-callout\; id="2"\; label="l\; qy"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">l</figure-callout>}}{\mathrm{qy}}^{}{}^2\mathrm{dx}}---\left(23\right)$

Wherein, g is acceleration of gravity.Obviously, with equation substitution (23) formula of amount of deflection y, can be in the hope of parameter:

${\mathrm{\&beta;}}_1={\&Integral;}_0^l\mathrm{qydx}={\&Integral;}_0^{l}q{a}_n[1-\cos \frac{(2\mathrm{n\&pi;}-1)\mathrm{\&pi;x}}{2l}]\mathrm{dx}$ (24)

$=\frac{\mathrm{ql}{a}_n[2(\cos \mathrm{n\&pi;}+\mathrm{n\&pi;})-\mathrm{\&pi;}]}{(2n-1)\mathrm{\&pi;}}$

In like manner can get:

${\mathrm{\&beta;}}_2={\&Integral;}_0^{\mathrm{<figure-callout\; id="2"\; label="l\; qy"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">l</figure-callout>}}{\mathrm{qy}}^{}{}^2\mathrm{dx}={\&Integral;}_0^l{{\mathrm{qa}}_n}^2{[1-\cos \frac{(2\mathrm{n\&pi;}-1)\mathrm{\&pi;x}}{2l}]}^2\mathrm{dx}$ (25)

$=\frac{{{\mathrm{qla}}_n}^2[2\cos \mathrm{n\&pi;}(4-\sin \mathrm{n\&pi;})+3(2n-1)\mathrm{\&pi;}]}{2(2n-1)\mathrm{\&pi;}}$

In (24) formula and (25) formula substitution (23) formula, can draw intrinsic frequency:

$\mathrm{\&omega;}=1.414\sqrt{\frac{g[2\cos \mathrm{n\&pi;}+(2n-1)\mathrm{\&pi;}]}{a_n[2\cos \mathrm{n\&pi;}(4-\sin \mathrm{n\&pi;})+3(2n-1)\mathrm{\&pi;}]}}---\left(26\right)$

Flexibility factor a with (12) formula _nSubstitution (26) formula is got back:

$\mathrm{\&omega;}=0.25\sqrt{\frac{g{\mathrm{\&pi;}}^4[\sin n\mathrm{\&pi;}\cos \mathrm{n\&pi;}({\mathrm{<figure-callout\; id="1"\; label="TK"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">TK</figure-callout>}}_1+{\mathrm{<figure-callout\; id="2"\; label="EIK"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">EIK</figure-callout>}}_2+\mathrm{ql}\cos \mathrm{\&alpha;}{K}_3)+\mathrm{\&pi;}({\mathrm{TK}}_4+{\mathrm{<figure-callout\; id="5"\; label="EIK"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">EIK</figure-callout>}}_5+\mathrm{ql}\cos \mathrm{\&alpha;}{K}_6)][2\cos \mathrm{n\&pi;}+(2n-1)\mathrm{\&pi;}]}{{\mathrm{ql}}^4\sin \mathrm{\&alpha;}(2\cos \mathrm{n\&pi;}+2\mathrm{n\&pi;}-\mathrm{\&pi;})[2\cos \mathrm{n\&pi;}(4-\sin \mathrm{n\&pi;})+3(2n-1)\mathrm{\&pi;}]}}---\left(27\right)$

For numerous and diverse formula like this, although still feel inconvenient through simplifying to process to use.Because in actual applications, the rotating speed of drill string has a scope roughly, for the calculating that exceeds this scope all without practical significance.Therefore in following formula, only get the resonance rotational speed N that drill string is calculated on its front 5 rank _n, greatly simplified calculation procedure.

When n=1, be first natural frequency

${\mathrm{\&omega;}}_{n=1}=0.25\sqrt{\frac{g{\mathrm{\&pi;}}^5(P+2\mathrm{EI}+\mathrm{ql}\cos \mathrm{\&alpha;})}{{\mathrm{ql}}^4\sin \mathrm{\&alpha;}(3\mathrm{\&pi;}-8)}}---\left(28\right)$

When n=2, be the second order intrinsic frequency

${\mathrm{\&omega;}}_{n=2}=0.25\sqrt{\frac{243g{\mathrm{\&pi;}}^5(P+2\mathrm{EI}+\mathrm{ql}\cos \mathrm{\&alpha;})}{{\mathrm{ql}}^4\sin \mathrm{\&alpha;}(9\mathrm{\&pi;}+8)}}---\left(29\right)$

When n=3, be three rank intrinsic frequencies

${\mathrm{\&omega;}}_{n=3}=0.25\sqrt{\frac{3125g{\mathrm{\&pi;}}^5(P+2\mathrm{EI}+\mathrm{ql}\cos \mathrm{\&alpha;})}{{\mathrm{ql}}^4\sin \mathrm{\&alpha;}(15\mathrm{\&pi;}-8)}}---\left(30\right)$

When n=4, be the quadravalence intrinsic frequency

${\mathrm{\&omega;}}_{n=4}=0.25\sqrt{\frac{16807g{\mathrm{\&pi;}}^5(P+2\mathrm{EI}+\mathrm{ql}\cos \mathrm{\&alpha;})}{{\mathrm{ql}}^4\sin \mathrm{\&alpha;}(21\mathrm{\&pi;}+8)}}---\left(31\right)$

When n=5, be five rank intrinsic frequencies

${\mathrm{\&omega;}}_{n=5}=0.25\sqrt{\frac{59049g{\mathrm{\&pi;}}^5(P+2\mathrm{EI}+\mathrm{ql}\cos \mathrm{\&alpha;})}{{\mathrm{ql}}^4\sin \mathrm{\&alpha;}(27\mathrm{\&pi;}-8)}}---\left(32\right)$

So far, the present invention has drawn the design formulas of calculating intrinsic frequency ω, so just can by avoiding intrinsic frequency, determine the rotational speed N of drill string.(26) formula of observation, can find that its characteristics are as follows:

1, intrinsic frequency ω all is directly proportional with the elastic modulus E I of axial tensile force T riser pipe.

2, intrinsic frequency ω and l ⁴Be inversely proportional to, that is to say that in other words the intrinsic frequency ω of riser pipe reduces along with the increase of 4 powers of its length l.

3, along with the increase of vibration exponent number, front 5 rank intrinsic frequencies obviously increase, and amplification is more than 3 times.

Four, the analysis of the Flexural Equation of riser pipe and intrinsic frequency thereof

In being considered as riser pipe such as background technology, describe, during the fixed form of pin-ended support, then adopt riser pipe deflection deformation function to be:

$y^{\&prime;}=\underset{n=1}{\overset{\&infin;}{\mathrm{\&Sigma;}}}{a}_n\sin \frac{\mathrm{n\&pi;x}}{l}---\left(33\right)$

The Flexural Equation drawn with this function is:

$y^{\&prime;}=\frac{-2q{l}^4\sin \mathrm{\&alpha;}}{{\mathrm{EI\&pi;}}^5}\underset{m=1}{\overset{\&infin;}{\mathrm{\&Sigma;}}}\frac{(\cos \mathrm{n\&pi;}-1)}{n^3[{\mathrm{<figure-callout\; id="2"\; label="n"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">n</figure-callout>}}^2+\frac{{\mathrm{<figure-callout\; id="2"\; label="l"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">l</figure-callout>}}^2}{\mathrm{EI}{\mathrm{\&pi;}}^2}(T+\mathrm{ql}\cos \mathrm{\&alpha;})]}\sin \frac{\mathrm{n\&pi;}}{l}x---\left(34\right)$

Wherein, the implication of parameters is identical with (21) formula.When n=1, have:

$y^{\&prime;}=\frac{4q{l}^4\sin \mathrm{\&alpha;}}{\mathrm{EI}{\mathrm{\&pi;}}^5[1+\frac{{\mathrm{<figure-callout\; id="2"\; label="l"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">l</figure-callout>}}^2}{\mathrm{EI}{\mathrm{\&pi;}}^2}(T+\mathrm{ql})]}\sin \frac{\mathrm{\&pi;x}}{l}---\left(35\right)$

When

The time, reach y '=y _Max', then the maximum defluxion of riser pipe occurs in the middle part of beam as can be known, then has:

${y_{\max }}^{\&prime;}=\frac{4q{l}^4\sin \mathrm{\&alpha;}}{{\mathrm{EI\&pi;}}^5[1+\frac{{\mathrm{<figure-callout\; id="2"\; label="l"\; filenames="H2008101177569E00011.png"\; state="\{\{state\}\}">l</figure-callout>}}^2}{\mathrm{EI}{\mathrm{\&pi;}}^2}(P+\mathrm{ql})]}---\left(36\right)$

Intuitively, this is the actual conditions that do not meet the deepwater drilling operation.

Above analytic explanation:

1, riser pipe is considered as under the fixed form of pin-ended support, its maximum defluxion must be at the mid point of pin-ended support fixed point up and down

The place, and the situation of maximum value shown in (36) formula arranged.

2 and to adopt the lower end in floatation type drilling well model of the present invention be hinge support, the upper end is free-ended structure.Say intuitively the maximum defluxion y of riser pipe _MaxOccur in x=l place, upper end, be the free-ended drift value of restriction riser pipe.And shown in (22) formula, the conclusion that the present invention draws is consistent with actual conditions.

3, as drift value y _MaxWhen very little, the account form of getting Flexural Equation (34) formula and intrinsic frequency thereof is rational.If but riser pipe then should calculate according to Flexural Equation of the present invention (20) formula the amount of deflection y of riser pipe when larger drift value is arranged, its intrinsic frequency ω should be calculated as suitable according to (27) formula.

When 4, considering the deepwater drilling operation, need the rotational speed N of choose reasonable drill string, this moment the intrinsic frequency of riser pipe is converted into the resonance rotational speed N of drill string _n, then have

$N_n=\frac{30\mathrm{\&omega;}}{\mathrm{\&pi;}}---\left(37\right)$

So far, the present invention is with the resonance rotational speed N of intrinsic frequency ω and the drill string of riser pipe _nSet up inherent contact.Can instruct when selecting the rotational speed N of drill string with this, fully take into account the intrinsic frequency ω of riser pipe.Thereby avoid selecting the rotational speed N of post in the resonance rotational speed N as far as possible _nScope within, so just can effectively reduce and avoid wearing and tearing and the damage of riser pipe.Calculation on Natural Frequency formula of the present invention has had theoretical guidance to the research of riser pipe motion state, has in actual applications great practice significance, and the method is used in the deepwater drilling operation simultaneously, guarantees that effectively the safety of deepwater drilling operation is carried out.

Five, the specific embodiment

The below illustrates as follows to specific implementation process of the present invention:

Embodiment 1:

Known, deepwater drilling riser pipe external diameter OD=0.609m, inner diameter, ID=0.58m, the aerial weight of its unit length is q=232.4kg/m.Does examination ask the intrinsic frequency of its riser pipe when depth of water 500m-3000m what are?

Separate: be 1.3g/cm according to mud density in the pipe ³Calculate, consider simultaneously the buoyancy that the interior mud of conduit and density of sea water difference cause, can find the solution according to (28)-(32) formula, draw under the different water depth condition, the Changing Pattern of intrinsic frequency, detailed data sees Table 1.

Table 1:609.6mm (the front 5 rank calculation on Natural Frequency results of 24 ") riser pipe

Adding boldface type in the above table is the resonance rotational speed N _nScope should manage to be avoided.

Embodiment 2:

Known, deepwater drilling riser pipe external diameter OD=0.762m, inner diameter, ID=0.7239m, the aerial weight of its unit length is q=349kg/m.Does examination ask the intrinsic frequency of its riser pipe when depth of water 500m-3000m what are?

Separate: in like manner, the present invention can be 1.3g/cm according to mud density ³Calculate its weight and buoyancy, and find the solution according to (28)-(32) formula, obtain under the different water depth condition, the Changing Pattern of intrinsic frequency, detailed data sees Table 2.

The front 5 rank calculation on Natural Frequency results of table 2:762mm riser pipe

Adding boldface type in the above table is the resonance rotational speed N _nScope should manage to be avoided.

Observation table 1 and table 2 can be found:

1, under same length l and same conduit size OD, ID condition, the riser pipe inclination alpha is less, and intrinsic frequency ω is just higher.As in the table 1, the riser pipe depth of water under the condition of 2000m, when inclination alpha=3 °, its three rank intrinsic frequency ω _N=3Be 0.02sec ^-1, and during in inclination alpha=5 °, its three rank intrinsic frequency ω _N=3Be decreased to 0.01sec ^-1

2, under same length l and same inclination alpha condition, the conduit physical dimension increases, and intrinsic frequency ω increases.In table 1 and table 2, equally under depth of water 2000m inclination alpha=1 ° condition, when adopting φ 609.6mm riser pipe, its three rank intrinsic frequency ω _N=3Be 0.07sec ^-1, and this moment when employing φ 762mm riser pipe, its three rank intrinsic frequency ω _N=3But increase to 0.089sec ^-1

3, in same physical dimension, under the condition of the intrinsic frequency of same exponent number, along with the increase of length l, its intrinsic frequency ω reduces.Such as φ 609.6mm riser pipe inclination alpha in the table 1=1 °, under the three rank intrinsic frequency conditions, intrinsic frequency is 3.64sec during the 500m depth of water ^-1, and when the depth of water was 3000m, intrinsic frequency was reduced to 0.1sec ^-1φ 762mm riser pipe inclination angle is 1 ° in the table 2, intrinsic frequency ω during 500m length _N=3Be 4.484sec ^-1, and when the depth of water is 3000m, intrinsic frequency ω _N=3Be reduced to 0.126sec ^-1

Can be found by table 1 and table 2 that 4, under this official holiday fixed condition, drill string rotational speed N and the riser pipe that avoid commonly using resonate as far as possible.This moment in the riser pipe of φ 609mm and φ 762mm, mostly occur when 500m length, and in the conduit of φ 762mm resonance speed N _nWider.In drillng operation, along with the increase of riser pipe length, it avoids the situation of resonance also just more complicated.Therefore the scope that should as far as possible select the drill string rotating speed to avoid, otherwise be difficult to guarantee the safety of drill string and riser pipe.

The present invention studies the meaning of deep water riser pipe characteristic properties, the very important point is exactly for when carrying out drillng operation, selection to the drillng operation parameter provides decision-making foundation, should avoid near the rotating speed of selecting drill string of intrinsic frequency as far as possible, thereby avoided to greatest extent between drill string and the riser pipe resonance occuring, and the damage riser pipe, thereby guarantee as far as possible the safety of riser pipe and carrying out smoothly of drillng operation.

Claims (4)

1. one kind prevents that drilling riser and drill string from producing the method for resonance, and it may further comprise the steps:

1) suppose that riser pipe upper end is to be suspended on the floating drilling platform and to be subjected to the free end of drift value restriction by tensioner, the lower end is to be connected to hinge support on preventer and the underwater well head by flexible joint, and then forms a floatation type drilling well model;

2) according to floatation type drilling well model, set up the equation ω of riser pipe intrinsic frequency:

$\mathrm{\&omega;}=1.414\sqrt{\frac{g[2\cos \mathrm{n\&pi;}+(2n-1)\mathrm{\&pi;}]}{a_n[2\cos \mathrm{n\&pi;}(4-\sin \mathrm{n\&pi;})+3(2n-1)\mathrm{\&pi;}]}}$

In the formula, a _nThe panel indicator deflection coefficient, g represents acceleration of gravity, n represents natural number;

3) gone out the resonance rotational speed N of drill string by the calculation on Natural Frequency of riser pipe _n

$N_n=\frac{30\mathrm{\&omega;}}{\mathrm{\&pi;}};$

4) avoid the resonance rotational speed N of drill string _nSelect the drill string rotating speed.

2. a kind of drilling riser and drill string of preventing as claimed in claim 1 produces the method for resonance, it is characterized in that: the equation ω of the riser pipe intrinsic frequency described step 2), take following steps to draw:

1. based on floatation type drilling well model, set up the riser pipe Flexural Equation:

$y=\underset{n=1}{\overset{\&infin;}{\mathrm{\&Sigma;}}}{a}_n(1-\cos \frac{(2n-1)\mathrm{\&pi;x}}{2l})$

Wherein, l is the length of riser pipe, a _nThe panel indicator deflection coefficient,

$a_n=\frac{{\mathrm{ql}}^4\sin \mathrm{\&alpha;}[2(\cos \mathrm{n\&pi;}+\mathrm{n\&pi;})-\mathrm{\&pi;}]}{{\mathrm{\&pi;}}^4\left\{\sin n\mathrm{\&pi;}\cos \mathrm{n\&pi;}\right[{\mathrm{TK}}_1+{\mathrm{EIK}}_2+\mathrm{ql}\cos \mathrm{\&alpha;}{K}_3]+\mathrm{\&pi;}[{\mathrm{TK}}_4+{\mathrm{EIK}}_5+\mathrm{ql}\cos \mathrm{\&alpha;}\&CenterDot;K_6\left]\right\}};$

Wherein

The deadweight of q-riser pipe, unit is Kg;

α-riser pipe inclination angle;

The T-axial force, unit is N;

E-modulus of elasticity, unit are N/m ²

The I-polar moment of inertia,

Unit is m ⁴

OD-riser pipe external diameter, unit are m;

ID-riser pipe internal diameter, unit are m;

G-acceleration of gravity;

K ₂＝-64n ⁴+128n ³-96n ²+32n-4

K ₃＝-32n ⁴+64n ³-48n ²+16n-2

K ₄＝32n ⁵-80n ⁴+80n ³-40n ²+10n-1

K ₅＝64n ⁵-160n ⁴+160n ^3-80n ²+20n-2

K ₆＝32n ⁵-80n ⁴+80n ³-40n ²+10n-1；

2. according to the riser pipe Flexural Equation, calculate the riser pipe formula for natural frequency

Intrinsic frequency square be:

${\mathrm{\&omega;}}^2=\frac{g{\&Integral;}_0^l\mathrm{qydx}}{{\&Integral;}_0^l{\mathrm{qy}}^2\mathrm{dx}}$

Intrinsic frequency ω as can be known then:

$\mathrm{\&omega;}=1.414\sqrt{\frac{g[2\cos \mathrm{n\&pi;}+(2n-1)\mathrm{\&pi;}]}{a_n[2\cos \mathrm{n\&pi;}(4-\sin \mathrm{n\&pi;})+3(2n-1)\mathrm{\&pi;}]}}$

3. a kind of deepwater drilling riser pipe and drill string of preventing as claimed in claim 2 produces the method for resonance, it is characterized in that: when n=1,

$y_{n=1}=\frac{32{\mathrm{ql}}^4\sin \mathrm{\&alpha;}(\mathrm{\&pi;}-2)}{\mathrm{EI}{\mathrm{\&pi;}}^5[2+\frac{1}{\mathrm{EI}{\mathrm{\&pi;}}^5}(T+\mathrm{ql}\cos \mathrm{\&alpha;})]}(1-\cos \frac{\mathrm{\&pi;x}}{2l}).$

4. a kind of deepwater drilling riser pipe and drill string of preventing as claimed in claim 3 produces the method for resonance, and it is characterized in that: namely when x=l, drift value is:

$y_{\max }=\frac{32{\mathrm{ql}}^4\sin \mathrm{\&alpha;}(\mathrm{\&pi;}-2)}{\mathrm{EI}{\mathrm{\&pi;}}^5[2+\frac{1}{\mathrm{EI}{\mathrm{\&pi;}}^5}(T+\mathrm{ql}\cos \mathrm{\&alpha;})]}.$

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