Research Article Analyzing Axial Stress and Deformation...

Hindawi Publishing CorporationThe Scientific World JournalVolume 2013 Article ID 565891 9 pageshttpdxdoiorg1011552013565891

Research ArticleAnalyzing Axial Stress and Deformation ofTubular for Steam Injection Process inDeviated Wells Based on the Varied (119879 119875) Fields

Yunqiang Liu12 Jiuping Xu1 Shize Wang3 and Bin Qi3

1 Uncertainty Decision-Making Laboratory Sichuan University Chengdu 610064 China2 College of Economics amp Management Sichuan Agricultural University Chengdu 611130 China3 Research School of Engineering Technology The Southwest Petroleum and Gas Corp China Petroleum and Chemical CorpDeyang 618000 China

Correspondence should be addressed to Jiuping Xu xujiupingscueducn

Received 13 May 2013 Accepted 2 August 2013

Academic Editors G Carbone S Park S Torii and Q Yang

Copyright copy 2013 Yunqiang Liu et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

The axial stress and deformation of high temperature high pressure deviated gas wells are studied A new model is multiplenonlinear equation systems by comprehensive consideration of axial load of tubular string internal and external fluid pressurenormal pressure between the tubular and well wall and friction and viscous friction of fluid flowing The varied temperature andpressure fields were researched by the coupled differential equations concerning mass momentum and energy equations insteadof traditional methods The axial load the normal pressure the friction and four deformation lengths of tubular string are gotten by means of the dimensionless iterative interpolation algorithm The basic data of the X Well 1300 meters deep are used forcase history calculations The results and some useful conclusions can provide technical reliability in the process of designing welltesting in oil or gas wells

1 Introduction

The deviated wells had been wildly applicable for petroleumand natural gas industry Deviated wells have their distinctivecharacteristics which are distinguished from that of otherwells (1) High temperature high pressure the temperaturedistribution and pressure on the tubing are significantlydifferent when outputs are varied (flow velocity) but neitherhas a simple linear relationship because the fluid densityis not constant (2) Deep well the sensibility of force anddeformation influencing by the factors such as the temper-ature pressure density of fluid viscous friction and fluidvelocity and so forth will become high with the increase oftubing length The completion test of a deep well is a newproblem In the research of applied basic theory for deep welltesting tubular string mechanical analysis is very complexbut fluid temperature and tubing pressure affect the forceof the tubular string heavily Temperature pressure liquiddensity and fluid velocity within tubing may change with of

the hole depth time and operations so that the axial forcechanges constantly A large compression load at low end caninduce the tubing plastic deformation and make the packerdamaged A large tension load at the top end may unpack thepacker or cause the tubing to break If the tubing failed thewhole borehole can hardly maintain its integrity and safety[1]Therefore it is very important for deviatedwells to predictthe axial forces for the safety

Hammerlindl [2] had made a great contribution abouttubular mechanics He had put forth the four effects betweenthe packer forces and length change of tubing temperatureeffect ballooning effect axial load effect and the helicalbuckling effect There is a large amount of papers to researchthe effect of buckling behaviorTherefore it is considered thatinflexion is caused on its axial force under certain conditionsby which colliding on parts of the drill string with well bore isinduced When buckled of tubular beyond wellholersquos controlthe buckling configuration which will be transformed at thestate of stabilization sinusoidal buckling and helical buckling

2 The Scientific World Journal

with the increase of loadThe problem of buckling of the tubewas first studied and put into practice by Lubinski et al [3]They had done the emulation experiment for the bucklingbehavior of tube in deviated wells and found the computeformula on critical buckling load of tube in deviated wellsPaslay and Bogy [4] found that the number of sinusoids inthe buckling mode increases with the length of the tubeThe buckling behavior by inner and outer fluid pressure oftubing was analyzed and the mathematical relation betweenpitch and axial pressures was deduced based on the principleof minimum potential energy (see Hammerlindl [2]) Themptotic solution for sinusoidal buckling of an extremely longtube has been analyzed by Dawson and Paslay [5] based on asinusoidal buckling mode of constant amplitude Numericalsolutions were also sought by Mitchell [6] using the basicmechanics equations His solutions confirm the thought thatunder a general loading the deformed shape of the tube is acombination of helices and sinusoids while helical deforma-tion occurs only under special values of the applied loadTheformula about tubing forces had been put however which istoo simple for shallowwells to accommodate the complicatedstates of deep wells Up to now many researches are centeredonwater injection tubular but not on steam injection Amongthem the values of temperature and pressure are consideredas constant or lineal functions which will cause large errorson tubular deformation computing [7]

In fact the tubular string deformation includes trans-verse deformation and longitudinal deformation Becausethe transverse length (its order of magnitude is 10minus3m) ismuch and much smaller than the longitudinal length (itsorder of magnitude is 103m) we mainly consider the axial(longitudinal) deformation for the tubular string deformationanalysis in the paper In the paper the force states oftubular in the process of steam injection are analyzed Thevaried (119879 119875) fields are considered to compute the values ofseveral deformations The axial load and four deformationlengths of tubular string are obtained by the dimensionlessiterative interpolation algorithmThe basic data of the XWell(deviated well) 1300 meters deep in China are used for casehistory calculations Some useful suggestions are drawn

This paper is organized as follows Section 2 gives asystem model about tubular mechanics and deformationAnd the varied (119879 119875) fields were presented by model con-cerning mass momentum and energy balance Section 3gives the parameters initial condition and algorithm forsolving model In Section 4 we give an example from adeviated well at 1300 meters of depth in China and the resultanalysis are made Section 5 gives a conclusion

2 Model Building

21 Basic Assumption Before analyzing the force on themicroelement some assumptions are introduced as follows

(1) the curvature of the hole of the considered modularsection is constant

Pi

P0

D

d

o

r

z

Tube

Casing

Packer

Packerfluid

Figure 1 The physical figure of forces analysis on tube

(2) on the upper side or underside of the section whichis point of contact of the pipe and tube wall thecurvature is the same with the hole curvature

(3) the radius of steam injection string in contrast tocurvature of borehole is insignificant

(4) the string is at the state of linear elastic relationship

22 Forces Analysis of Tubular String The forces of tubularstring are shown in Figure 1 Consider the flow systemdepicted in Figure 1 a constant cross-sectional flow area 119860inner diameter 119889 outer diameter 119863 material density 120588

1

packer fluid density 120588

2

and a total length 119885 Through thistubing gas flows from the bottom to the top with a mass flowrate 119882 The distance coordinate in the flow direction alongthe tubing is denoted by 119911 The cylindrical coordinate system119903120579119911 origin of which is in wellhead and 119885 axis is down as theborehole axis is used

As shown fromFigure 1 the tubular string ismainly actedupon by the following forces at the process of steam injection

(1) Initial Axial Force The initial axial force of tubularshould include the deadweight buoyant weight andinitial pull force

(2) Thermal Stress On the process of steam injection thetemperature stress will act at the tubular with variedtemperature

(3) Axial Force by the Varied Internal and External Pres-sure Thanks to the varied pressure with internal andexternal pressure the tubular will be acted by thebending force piston force and other axial forces

(4) Friction Drag by Steam Injection On the process ofsteam injection the flow in tubular will produceviscous flow which will cause the friction drag

The Scientific World Journal 3

23 The Axial Load and Axial Stress of the Tubular

231 Initial Axial Load and Initial Axial Stress of SteamInjection Tubular

Initial Axial Load The section to which the distance from thewellhead is 119911(119898) was considered The axial static load by thedeadweight of tubular is as follows

119873119902119911

= int

119871

119911

119902 cos120572119889119911 =120587

41205881

119892 (1198632

minus 1198892

)int

119871

119911

cos120572119889119911 (1)

where 119873119902119911

is the deadweight of tubular 119902 is the average unitlength weight of tubing 119871 is the length of tubular 120588

1

is thedensity of tubular and 120572 is the inclination angle

The axial static load by the buoyant weight is as follows

119873119887119911

= minus1205882

1198921198602

int

119871

119911

cos120572119889119911 = minus1205882

119892119911120587(119863

2)

2

int

119871

119911

cos120572119889119911

(2)

where 119873119887119911

is the buoyant weight of tubular 1205882

is the densityof packer fluid

The axial load by the steam injection pressure

119873119901119911

=1198751199111

1205871198892

119911

4 (3)

where 1198751199111

represents the inner pressure at this sectionTherefore summing (1) (2) and (3) the axial forces in

the section are obtained as follows

119865119911

= 119873119902119911

+ 119873119887119911

+ 119873119901119911

(4)

Initial Axial Stress The axial stress can be derived from thefollowing equation

120590119911119894

=4119865119911

120587 (1198632

minus 1198892

) (5)

232 Axial Thermal Stress of Steam Injection Tubular In theprocess of steam injection the temperature of tubular willchange with time and depth which will make the tubulardeform as follows

120590119911119905

= 119864120573 (1198791199111

minus 1198791199110

) = 119864120573Δ119879 (6)

where 119864 represents the steel elastic modulus of tubular 120573 isthe warm balloon coefficient of the tubular string and Δ119879 isthe temperature change with before and after steam injection

233 Axial Stress of Steam Injection Tubular by the Changewith Pressure The effect acting the tubular with pressurechange which is called ballooning effect normally

Ballooning Stress Analysis The ballooning effect will beproduced from pressure acted in inner and outer of the tubeGenerally there are two kinds of tubular in oil wells One isthe tubulars whose outer diameter is 889mm inner diameter

Pz1

Pz2D

d

Figure 2 The radial and tangential stresses figure of tube

is 76mm and thickness of tubes is 65mm (120575(1198892) =

171 gt 5) the other is the tubular whose outer diameter is1143mm inner diameter is 1005mm and thickness of tubesis 69mm (120575(1198892) = 137 gt 5) Neither is the thin-wallproblemTherefore it should be solved by Lamersquos formula [8]

The radial and tangential stresses in the thick-wall cylin-der can be shown as Figure 2 The two can be calculated asfollows

120590119903119911

=1198892

1198751199111

minus 1198632

1198751199110

1198632

minus 1198892

minus(1198751199111

minus 1198751199110

)1198632

1198892

(1198632

minus 1198892

) 41199032

120590120579119911

=1198892

1198751199111

minus 1198632

1198751199110

1198632

minus 1198892

+(1198751199111

minus 1198751199110

)1198632

1198892

(1198632

minus 1198892

) 41199032

(7)

where 119903 is radial stress 120579 is tangential stress 119903 (119889 le 119903 le 119863) isradial coordinate 119875

1199111

is tube internal pressure at 119911 point and1198751199110

is tube external pressure at 119911 point

234 Axial Stress of Steam Injection Tubular by the FrictionLoss In fact the flow in the tubular should be multiflow Onthe process of steam injection the flow will be run and it willgive rise to friction effect to cause axial stress In our paperwe consider the flow gas-liquid mix flow and the liquid headloss is gotten by the Darcy-Weisbach formula [9] as follows

ℎ119891

=120582 (119885 minus 119911) ]2

119898

2119892119889 (8)

where ℎ119891

means heat loss of liquid flow 120582 is frictional headlosses coefficients and ]

119898

is the velocity of liquid flowThe friction drag in tubular is 119873

119891119911

= ℎ119891

120588119898

1198921205871198892 (120588119898

isdensity of liquid flow) The axial stress by fiction drag can beobtained as follows

120590119911119891

=

4119873119891119911

120587 (1198632

minus 1198892

) (9)

24 Analysis of Axial Deformation Based on the studiesand analyses mentioned above the axial deformation on thetubular is made up of the following parts


241TheAxial Deformation by the Axial Static Stress For themicroelement of the tubular 119889119911 the unit deformation by thestatic stress can be computed by generalized Hooke law

1205761

=1

119864[120590119911119894

minus 120583 (120590119903119911

+ 120590120579119911

)] (10)

where 120583 represents Poissonrsquos ratiosThe axial deformation at an element can be obtained

through integrating on the length of the element as follows

Δ1198711119894

= int

119885119894

119885119894minus1

1

119864[120590119911119894

minus 120583 (120590119903119911

+ 120590120579119911

)] 119889119911 (11)

Therefore the total axial deformation by the static stresscan be gotten accumulating each element as follows

Δ1198711

=

119873

sum

119894=1

Δ1198711119894

(12)

242The Axial Deformation with Temperature Changed Forthe microelement of the tubular 119889119911 the unit deformation bythe temperature change is as follows

Δ1198712119894

= int

119885119894

119885119894minus1

120590119911119905

119864119889119911 = 120573Δ119879

119894

Δ119871119894

(13)

The same principle is that the total axial deformation bythe varied temperature fields can be gotten accumulating eachelement as follows

Δ1198712

=

119873

sum

119894=1

Δ1198712119894

(14)

243 The Axial Deformation with the Friction Drag For themicroelement of the tubular 119889119911 the unit deformation by thefriction force is as follows

Δ1198713

= int

119885

0

120590119911119891

119864119889119911 =

120582120588119898

]2119898

1198891198852

119864 (1198632

minus 1198892

) (15)

244TheAxial Deformationwith the Tubular String BucklingResearchers in general call the buckling a bending effect Thetubular is freely suspended in the absence of fluid inside asshown in Figure 3(a) Because the force 119865 applied at the endof the tubular which is large enough the tubular will buckleas shown in Figure 3(b)

Lubinski et al [3] had done many researches on thephenomenon From theirworkwe can get the buckling effectDefine the virtual axial force of tubing as follows

119865119891

= 119860119901

(1198751

minus 1198750

) (16)

where 1198751

is the pressure inside the tubular at the packerlength 119875

0

is the pressure outside the tubular at the packerlength and 119860

119901

is the area corresponding to packer boreBy (16) whether the tubular will buckle or not can be

judged The string will buckle if 119865119891

is positive or remain

(a)

F

Neutral point

(b)

Figure 3 Buckling of tubular

straight if 119865119891

is negative or zero The axial deformation of thetubular string buckling is

Δ1198714119894

= minus

1199032

1198602

119901

(Δ1198751119894

minus Δ1198750119894

)2

8119864119868119882

(17)

where 119903means tubing-to-casing radial clearance 119868 ismomentof inertia of tubing cross-section with respect to its diameter(119868 = 120587(119863

4

minus 1198894

)64) Δ denotes change with before and afterinjection and119882 is the unit weight of tubing as

Δ1198714

=

119873

sum

119894=1

Δ1198714119894

(18)

In addition the position of the neutral point is needed Thelength (119899) from the packer to the point can be computed asfollows

119899 =

119865119891

119882 (19)

Generally the neutral point should be in tubular (119899 le 119885)However at the multipackers it will occur that the neutralpoint is outside the tubing between dual packers In thispaper we leave the latter phenomenon

To sum up the whole deformation length can be repre-sented as follows

Δ119871 = Δ1198711

+ Δ1198712

+ Δ1198713

+ Δ1198714

(20)

25 The Analysis of the Varied (119879 119875) Fields In the courseof dryness modeling we can find that the numerical valuesof deformation ((10) (13) and (17)) were affected by thetemperature and pressure In fact the two parameters variedaccording to the depth and time changing So the varied(119879 119875) fields need to be researched Under the China Sinopec


Group Hi-Tech Project ldquoStress analysis and optimum designof well completionrdquo in 2009 [6] undertaken by SichuanUniversity at early time The varied (119879 119875) fields had beendeduced strictly based on the mass momentum and energybalance The proof details can be shown in Xu et al [11] Thevaried (119879 119875) fields is

119889119875

119889119911=

minus (120591119894

119860) + 120588119898

119892 cos 120579 + (119898119860) 119877 (119889119909119889119911)

1 minus (119898119860) 119878

119889119879

119889119911= minus

]119898

119862119875119892

(119877119889119909

119889119911minus 119878

119889119875

119889119911) minus

119892 cos 120579119862119875119892

minus120587119891119903119905119894

120588119898

]3119898

4119862119875119892

+119886 (119879 minus 119879

119890

)

119862119875119892

119875 (1199110

) = 1198750

119879 (1199110

) = 1198790

119889119909 (1199110

) = 1198891199090

119909 (1199110

) = 1199110

(21)

3 Numerical Implementation

31 Calculation of Some Parameters In this section we willgive the calculating method of some parameters

(1) Each pointrsquos inclination

120572119895

= 120572119895minus1

+

(120572119896

minus 120572119896minus1

) Δ119904119895

Δ119904119896

(22)

where 119895 represents segment point of calculation Δ119904119896

represents measurement depth of inclination angle120572119896

and 120572119896minus1

Δ119904119895

is the step length of calculationTransient heat transfer function [12]

119891 (119905119863

) =

1128radic119905119863

(1 minus 03radic119905119863

) 119905119863

le 15

(04063 + 05 ln 119905119863

) (1 +06

119905119863

) 119905119863

gt 15

(23)

(2) The density of wet steam Since the flow of the watervapor in is the gas-liquid two-phase flow there aremany researches about this problem [13 14] In thepaper we adopt the M-B model to calculate theaverage density of the mixture

(3) The heat transfer coefficient 119880to from different posi-tions of the axis of the wellbore to the second surface

These resistances include the tubing wall possible insu-lation around the tubing annular space (possibly filled witha gas or liquid but is sometimes vacuum) casing wall andcementing behind the casing as follows

1

119880119905119900

= 119903119905119894

1

120582insln(

119903119888119894

119903119905119900

) +1

ℎ119888

+ ℎ119903

+ 119903119905119894

1

120582cemln(

119903cem119903119888119900

) (24)

120582ins and 120582cem are the heat conductivity of the heat insulatingmaterial and the cement sheath respectively ℎ

119888

and ℎ119903

are thecoefficients of the convection heat transfer and the radiationheat transfer

32 Initial Condition In order to solve model some definiteconditions and initial conditions should be addedThe initialconditions comprise the distribution of the pressure andtemperature at the well top In this paper we adopt thevalue at the initial time by actual measurement Before steaminjected the temperature of tubular just is initial temperatureof formation (119879

119911

= 1198790

+ 120574119911 cos120572 120574 is geothermal gradient)At the same time the pressure of inner tubular is assumed tobe equal to the outer tubular before steam injected

33 Steps of Algorithm To simplify the calculation wedivided the wells into several short segments of the samelength The length of a segment varies depending on varia-tions in wall thickness hole diameter fluid density inside andoutside the pipe and wells geometry The model begins withthe calculation at one particular position in the wells the topof the pipe

Step 1 Set step length of depth In addition we denotethe relatively tolerant error by 120576 The smaller ℎ 120576 is themore accurate the results are However it will lead to rapidincreasing calculating time In our paper we set ℎ = 1 (m)and 120576 = 5

Step 2 Give the initial conditions

Step 3 Compute each pointrsquos inclination

Step 4 Compute the parameters under the initial conditionsor the last depth variables

Step 5 Let 119879 = 119879119896

then we can get the 119879119890

by solving thefollowing equation

120597119879119890

120597119905119863

= (1205972

119879119890

1205971199032

119863

+1

119903119863

120597119879119890

120597119903119863

)

119879119890

1003816100381610038161003816119905119863=0= 1198790

+ 120574119911 cos 120579

120597119879119890

120597119903119863

10038161003816100381610038161003816100381610038161003816119903119863=1

= minus1

2120587120582119891

119889119902

119889119911

120597119879119890

120597119903119863

10038161003816100381610038161003816100381610038161003816119903119863rarrinfin

= 0

(25)

Let 119879119895119890119894

be the temperature at the injection time 119895 andradial 119894 at the depth 119911 We apply the finite different methodto discretize the equations as follows

119879119894+1

119890119895

minus 119879119894

119890119895

120593=

119879119894+1

119890119895+1

minus 2119879119895

119890119895+1

+ 119879119894minus1

119890119895+1

1205852

minus

119879119894+1

119890119895+1

minus 119879119894+1

119890119895

119903119863

120593 (26)

where 120593 is the interval of time and 120585 is the interval of radialrespectively It can be transformed into the standard form asfollows

minus(120593 +120593120585

119903119863

)119879119894+1

119890119895+1

+ (2120593 +120593120585

119903119863

)119879119894+1

119890119895

minus 120593119879119894+1

119890119895minus1

= 1205852

119879119894

119890119895

(27)


Table 1 Parameters of pipes

Diameter (m) Thickness (m) Weight (Kg) Expansion Elastic (Gpa) Poissonrsquos ratios Using length (m)00889 001295 2379 00000115 215 03 27000889 000953 1828 00000115 215 03 12000889 000734 1504 00000115 215 03 62000889 000645 1358 00000115 215 03 290

Table 2 Well parameters

Measured (m) Internal (m) External (m)3367 015478 017784226 01525 0177813000 010862 0127

Table 3 Parameters of azimuth inclination and vertical depth

Number Measured(m)

Inclination(∘)

Azimuth(∘)

Vertical depth(m)

1 135 263 24101 134722 278 123 23786 277913 364 143 21386 363824 393 217 2638 392535 422 185 4456 421286 450 082 19112 449627 486 293 26907 485478 514 103 29755 513839 543 358 32451 5417410 571 298 30305 5704311 600 203 20474 5994212 628 234 16433 6272813 660 185 19528 6595614 723 314 21484 7217015 782 098 21648 7813016 830 215 22931 8291217 860 267 24403 8597118 908 485 26662 9040819 928 672 25878 9214220 972 203 23688 9717121 1025 478 23927 10212522 1058 401 24459 10555823 1089 498 2282 10841724 1132 375 23388 11292825 1174 563 23514 11688726 1204 423 23438 12009927 1235 387 23499 12320828 1268 497 23257 12634529 1300 884 23328 128496

Then the different method is used to discretize theboundary condition For 119903

119863

= 1 we have

119879119890119894+1

1198902

minus (1 +119886120585

2120587120582119891

)119879119894+1

1198901

=119886119879119896

2120587120582119891

(28)

For 119903119863

= 119873 we have

119879119894+1

119890119899

minus 119879119894+1

119890119899minus1

= 0 (29)We can compute the symbolic solution of the temperature 119879

119890

of the stratum In this stepwewill get the discrete distributionof 119879119890

as the following matrix

[[[[[[[[[[[[

[

1198791

1198901

1198792

1198901

sdot sdot sdot T1198941198901

sdot sdot sdot

1198791

1198902

1198792

1198902

sdot sdot sdot 119879119894

1198902

sdot sdot sdot

sdot sdot sdot

1198791

119890119895

1198792

119890119895


119890119895

sdot sdot sdot

sdot sdot sdot

1198791

119890119899

1198792

119890119899


119890119899

sdot sdot sdot

]]]]]]]]]]]]

]

(30)

where 119894 represents the injection time and 119895 represents theradial

Step 6 Let the right parts of the coupled differential equa-tions be functions 119865

119896

where (119896 = 1 2) Then we can obtain asystem of coupled functions as follows

1198651

=minus (120591119894

119860) + 120588119898

119892 cos 120579 + (119898119860) 119877 (119889119909119889119911)

1 minus (119898119860) 119878

1198652

= minus]119898

119862119875119892

(119877119889119909

119889119911minus 119878

119889119875

119889119911) minus

119892 cos 120579119862119875119892

minus120587119891119903119905119894

120588119898

]3119898

4119862119875119892

+119886 (119879 minus 119879

119890

)

119862119875119892

(31)

where 119879119890

at 119903119863

= 1

Step 7 Assume that 119875 119879 are 119910119896

(119896 = 1 2) respectively Thenwe can obtain some basic parameters as follows

119886119896

= 119865119894

(1199101

1199102

)

119887119896

= 119865119894

(1199101

+ℎ1198861

2 1199102

+ℎ1198862

2)

119888119896

= 119865119894

(1199101

+ℎ1198871

2 1199102

+ℎ1198872

2)

119889119896

= 119865119894

(1199101

+ ℎ1198881

1199102

+ ℎ1198882

)

(32)


Table 4 The results of the axial force and various kinds of deformation lengths

Number Depth(m)

Axial force(N)

Displacement bytemperaturechanged (m)

Displacement bypressure

changed (m)

Axial deformation(m)

Bucklingdeformation (m)

Total deformation(m)

1 1 8952448 0 0 0 0 02 100 8547248 01201 000986 0024 0 015443 200 8142155 02362 0019392 0052 0 030724 300 7737177 03483 0028598 0082 0 04595 400 737970 04564 0037476 0115 minus0006 060296 500 7068773 05606 0046028 0152 minus0006 075237 600 6757639 06607 0054254 0192 minus0006 090068 700 6446023 07569 0062153 0235 minus0006 10489 800 6134372 0849 0069725 0283 minus0007 1194610 900 5822721 09371 0076968 0335 minus0007 1342211 1000 5511072 10212 0083883 0391 minus0007 1489612 1100 5199423 11014 0090471 0452 minus0007 1636713 1200 4887775 11775 0096731 0517 minus0009 1782214 1300 4576128 12496 0102662 0584 minus001 19261

Step 8 Calculate the pressure and temperature at point(119895 + 1)

119910119895+1

119896

= 119910119895

119896

+ℎ (119886119896

+ 2119887119896

+ 2119888119896

+ 119889119896

)

6

119896 = 1 2 119895 = 1 2 119899

(33)

Step 9 Calculate the deformation Δ1198711119895

Δ1198712119895

and Δ1198714119895

byprevious equations

Step 10 Repeat the third step to the tenth step until tubularlength 119885 is calculated

Step 11 Calculate the deformationΔ1198713

and total deformationlength as follows

Δ119871 =

119873

sum

119895=1

Δ1198711119895

+

119873

sum

119895=1

Δ1198712119895

+ Δ1198713

+

119873

sum

119895=1

Δ1198714119895

(34)

4 Numerical Simulation

41 Parameters To demonstrate the application of our the-ory we study a pipe in X well which is in Sichuan ProvinceChina All the basic parameters are given as follows depthof the well is 1300m ground thermal conductivity parameteris 206 ground temperature is 16

∘C ground temperaturegradient is 00218 (∘Cm) roughness of the inner surface ofthe well is 0000015 and parameters of pipes inclined wellinclination azimuth and vertical depth are given in Tables 12 and 3

42 Main Results and Results Analysis After calculation weobtain a series of results of this well as Table 4 The influenceof outputs on the axial deformation of tubingwas investigatedas shown by Figure 4

25

2

15

1

05

0

Tota

l axi

al d

efor

mat

ion

(m)

Depth (m)1 1200800400

700000m3d500000m3d300000m3d

Figure 4 The total axial deformation under varied outputs

From the results as shown in Figure 4 and Table 4 someuseful analysis can be drawn

(1) The amount of steam injected and inject pressureaffected the stretching force with special severity

(2) The results were as follows the length of tubulardeformation was risen with increased injected pres-sure or injected velocity

(3) The length of tubular deformation increases with theincreasing of outputs but more slowly

(4) The thermal stress is the main factor influencing thetubular deformation Therefore the temperature ofsteam injected should not be too high


(5) The lifting prestressed cementing technology hasimportant meanings to reduce the deformation oftubular

(6) The creeping displacement of downhole stings willproduce an upward contractility which causes packerdepressed or lapsedTherefore the effective measuresshould be adopted to control the companding oftubular

5 Conclusion

In this paper a total tubular deformation model aboutdeviated wells was given A coupled-system model of differ-ential equations concerning pressure and temperature in hightemperature-high pressure steam injection wells according tomassmomentum and energy balances which can reduce theerror of axial stress and axial deformation was given insteadof the average value or simple linear relationship in traditionalresearch The basic data of the Well (high temperature andhigh pressure gas well) 1300m deep in Sichuan China wereused for case history calculations The results can providetechnical reliance for the process of designing well tests indeviated gas wells and dynamic analysis of production

Nomenclature

119889 Inner diameter (m)119889119911

Microelement of the tubular119892 Acceleration of gravity (ms2)ℎ Depth of top tubular located at the packer

(m)119905 Time of down stroke (s)119905119863

Dimensionless time (dimensionless)V119894

Velocity of fluid in tubing (ms)V119904

Velocity of down stroke (ms)119911 Distance coordinate in the flow direction

(m) along the tubing119860 Constant cross-sectional flow area (m2)119860119888

Effective area (m2)119860119901

Area corresponding to packer bore (m2)119862119869

The Joule-Thomson coefficient(dimensionless)

119862119875

Heat capacity of fluids (JKg sdotK)119863 Outer diameter (m)119864 Steel elastic modulus of tubular (Mpa)119865119894

Axial forces in the section (N)119865119911

Axial tensile strength (N)119865119891

Friction force (N)119865119888

Piston force for supporting packerrsquos pressure(N)

119865119904

Pumping force (N)119871 Length of tubular (m)119873119887119911

Buoyant weight of tubular (Kg)119873119902119911

Dead weight of tubular (Kg)1198750

Pressure outside the tubular (Mpa)1198751

Pressure inside the tubular at the packerlength (Mpa)

119875119894

Pressure in tubing (Mpa)119879119894

Temperature in tubing (∘C)119879119890

Initial temperature of formation (∘C)119882 Mass flow rate (Kgs)119885 Total length (m)1205881

Material density (Kgm3)1205882

Packer fluid density (Kgm3)120572 Inclination angle (∘)120573 Warm balloon coefficient of the tubular

string (dimensionless)120575 Drop of any parameter120590119911119905

Axial thermal stress (119873)120588119894

Density of fluid in the tubing (Kgm3)Δ1198713

The tubular string buckling axialdeformation (m)

Δ119871119905

Total axial deformation by variedtemperature fields (m)

Δ119871119887

Total axial deformation by the variedpressure fields (m)

Δ119875start Differential pressure at startup (Mpa)Δ119875119894119894

Change in tubing pressure at the 119894 length(Mpa)

Δ119875119900119894

Change in annulus pressure at the 119894 length(Mpa)

Δ119875119888

Differential pressure from top to bottom(Mpa)

Δ119879 Temperature change with before and afterwell shut-in (∘C)

Δ120588119894119894

Change in density of liquid in the tubing atthe 119894 length (Kgm3)

Δ120588119900119894

Change in density of liquid in the casing atthe 119894 length (Kgm3)

AcknowledgmentsThis research was supported by the Key Program of NSFC(Grant no 70831005) and the Key Project of China Petroleumand Chemical Corporation (Grant no GJ-73-0706)

References

[1] D-L Gao and B-K Gao ldquoA method for calculating tubingbehavior in HPHT wellsrdquo Journal of Petroleum Science andEngineering vol 41 no 1ndash3 pp 183ndash188 2004

[2] D J Hammerlindl ldquoMovement forces and stresses associatedwith combination tubing strings sealed in packersrdquo Journal ofPetroleum Technology vol 29 pp 195ndash208 1977

[3] A Lubinski W S Althouse and J L Logan ldquoHelical bucklingof tubular sealed in packersrdquo Journal of Petroleum Technologyvol 14 no 6 pp 655ndash670 1962

[4] P R Paslay and D B Bogy ldquoThe stability of a circular rodlaterally constrained to be in contact with an inclined circularcylinderrdquo Journal of Applied Mechanics vol 31 pp 605ndash6101964

[5] R Dawson and P R Paslay ldquoDrillpipe buckling in inclinedholesrdquo Journal of PetroleumTechnology vol 36 no 10 pp 1734ndash1738 1984

[6] R FMitchell ldquoEffects ofwell deviation onhelical bucklingrdquo SPEDrilling and Completion vol 12 no 1 pp 63ndash68 1997


[7] P Ding and X Z Yan ldquoForce analysis of high pressure waterinjection stringrdquo PetroleumDring Techiques vol 36 no 5 p 232005

[8] Z F Li ldquoCasing cementing with half warm-up for thermalrecovery wellsrdquo Journal of Petroleum Science and Engineeringvol 61 no 2ndash4 pp 94ndash98 2008

[9] A M Sun ldquoThe analysis and computing of water injectiontubularrdquo Drilling and Production Technology vol 26 no 3 pp55ndash57 2003 (Chinese)

[10] J P Xu ldquoStress analysis and optimum design of well comple-tionrdquo Technical Report of Sinopec GJ-73-0706 2009

[11] J P Xu Y Q Liu S Z Wang and B Qi ldquoNumerical modellingof steam quality in deviated wells with variable (T P) fieldsrdquoChemical Engineering Science vol 84 pp 242ndash254 2012

[12] A R Hasan and C S Kabir ldquoTwo-phase flow in vertical andinclined annulirdquo International Journal of Multiphase Flow vol18 no 2 pp 279ndash293 1992

[13] HD Beggs and J R Brill ldquoA study of two-phase flow in inclinedpipesrdquo Journal of Petroleum Technology vol 25 no 5 pp 607ndash617 1973 paper 4007-PA

[14] H Mukherjee and J P Brill ldquoPressure drop correlations forinclined two-phase flowrdquo Journal of Energy Resources Technol-ogy vol 107 no 4 pp 549ndash554 1985

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Control Scienceand Engineering

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Hindawi Publishing Corporation httpwwwhindawicom

Journal ofEngineeringVolume 2014

Submit your manuscripts athttpwwwhindawicom

VLSI Design



Shock and Vibration


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Journal of

Advances inOptoElectronics


Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Modelling amp Simulation in EngineeringHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Chemical EngineeringInternational Journal of Antennas and

Propagation




Navigation and Observation



DistributedSensor Networks



with the increase of loadThe problem of buckling of the tubewas first studied and put into practice by Lubinski et al [3]They had done the emulation experiment for the bucklingbehavior of tube in deviated wells and found the computeformula on critical buckling load of tube in deviated wellsPaslay and Bogy [4] found that the number of sinusoids inthe buckling mode increases with the length of the tubeThe buckling behavior by inner and outer fluid pressure oftubing was analyzed and the mathematical relation betweenpitch and axial pressures was deduced based on the principleof minimum potential energy (see Hammerlindl [2]) Themptotic solution for sinusoidal buckling of an extremely longtube has been analyzed by Dawson and Paslay [5] based on asinusoidal buckling mode of constant amplitude Numericalsolutions were also sought by Mitchell [6] using the basicmechanics equations His solutions confirm the thought thatunder a general loading the deformed shape of the tube is acombination of helices and sinusoids while helical deforma-tion occurs only under special values of the applied loadTheformula about tubing forces had been put however which istoo simple for shallowwells to accommodate the complicatedstates of deep wells Up to now many researches are centeredonwater injection tubular but not on steam injection Amongthem the values of temperature and pressure are consideredas constant or lineal functions which will cause large errorson tubular deformation computing [7]

In fact the tubular string deformation includes trans-verse deformation and longitudinal deformation Becausethe transverse length (its order of magnitude is 10minus3m) ismuch and much smaller than the longitudinal length (itsorder of magnitude is 103m) we mainly consider the axial(longitudinal) deformation for the tubular string deformationanalysis in the paper In the paper the force states oftubular in the process of steam injection are analyzed Thevaried (119879 119875) fields are considered to compute the values ofseveral deformations The axial load and four deformationlengths of tubular string are obtained by the dimensionlessiterative interpolation algorithmThe basic data of the XWell(deviated well) 1300 meters deep in China are used for casehistory calculations Some useful suggestions are drawn

This paper is organized as follows Section 2 gives asystem model about tubular mechanics and deformationAnd the varied (119879 119875) fields were presented by model con-cerning mass momentum and energy balance Section 3gives the parameters initial condition and algorithm forsolving model In Section 4 we give an example from adeviated well at 1300 meters of depth in China and the resultanalysis are made Section 5 gives a conclusion

2 Model Building

21 Basic Assumption Before analyzing the force on themicroelement some assumptions are introduced as follows

(1) the curvature of the hole of the considered modularsection is constant

Pi

P0

D

d

o

r

z

Tube

Casing

Packer

Packerfluid

Figure 1 The physical figure of forces analysis on tube

(2) on the upper side or underside of the section whichis point of contact of the pipe and tube wall thecurvature is the same with the hole curvature

(3) the radius of steam injection string in contrast tocurvature of borehole is insignificant

(4) the string is at the state of linear elastic relationship

22 Forces Analysis of Tubular String The forces of tubularstring are shown in Figure 1 Consider the flow systemdepicted in Figure 1 a constant cross-sectional flow area 119860inner diameter 119889 outer diameter 119863 material density 120588

1

packer fluid density 120588

2

and a total length 119885 Through thistubing gas flows from the bottom to the top with a mass flowrate 119882 The distance coordinate in the flow direction alongthe tubing is denoted by 119911 The cylindrical coordinate system119903120579119911 origin of which is in wellhead and 119885 axis is down as theborehole axis is used

As shown fromFigure 1 the tubular string ismainly actedupon by the following forces at the process of steam injection

(1) Initial Axial Force The initial axial force of tubularshould include the deadweight buoyant weight andinitial pull force

(2) Thermal Stress On the process of steam injection thetemperature stress will act at the tubular with variedtemperature

(3) Axial Force by the Varied Internal and External Pres-sure Thanks to the varied pressure with internal andexternal pressure the tubular will be acted by thebending force piston force and other axial forces

(4) Friction Drag by Steam Injection On the process ofsteam injection the flow in tubular will produceviscous flow which will cause the friction drag





119873119902119911

= int

119871

119911

119902 cos120572119889119911 =120587

41205881

119892 (1198632

minus 1198892

)int

119871

119911

cos120572119889119911 (1)

where 119873119902119911


1



119873119887119911

= minus1205882

1198921198602

int

119871

119911

cos120572119889119911 = minus1205882

119892119911120587(119863

2)

2

int

119871

119911

cos120572119889119911

(2)

where 119873119887119911




119873119901119911

=1198751199111

1205871198892

119911

4 (3)

where 1198751199111



119865119911

= 119873119902119911

+ 119873119887119911

+ 119873119901119911

(4)


120590119911119894

=4119865119911

120587 (1198632

minus 1198892

) (5)


120590119911119905

= 119864120573 (1198791199111

minus 1198791199110

) = 119864120573Δ119879 (6)




Pz1

Pz2D

d





120590119903119911

=1198892

1198751199111

minus 1198632

1198751199110

1198632

minus 1198892

minus(1198751199111

minus 1198751199110

)1198632

1198892

(1198632

minus 1198892

) 41199032

120590120579119911

=1198892

1198751199111

minus 1198632

1198751199110

1198632

minus 1198892

+(1198751199111

minus 1198751199110

)1198632

1198892

(1198632

minus 1198892

) 41199032

(7)


1199111




ℎ119891

=120582 (119885 minus 119911) ]2

119898

2119892119889 (8)

where ℎ119891


119898


119891119911

= ℎ119891

120588119898

1198921205871198892 (120588119898


120590119911119891

=

4119873119891119911

120587 (1198632

minus 1198892

) (9)




1205761

=1

119864[120590119911119894

minus 120583 (120590119903119911

+ 120590120579119911

)] (10)



Δ1198711119894

= int

119885119894

119885119894minus1

1

119864[120590119911119894

minus 120583 (120590119903119911

+ 120590120579119911

)] 119889119911 (11)


Δ1198711

=

119873

sum

119894=1

Δ1198711119894

(12)


Δ1198712119894

= int

119885119894

119885119894minus1

120590119911119905

119864119889119911 = 120573Δ119879

119894

Δ119871119894

(13)


Δ1198712

=

119873

sum

119894=1

Δ1198712119894

(14)


Δ1198713

= int

119885

0

120590119911119891

119864119889119911 =

120582120588119898

]2119898

1198891198852

119864 (1198632

minus 1198892

) (15)



119865119891

= 119860119901

(1198751

minus 1198750

) (16)

where 1198751


0


119901




(a)

F

Neutral point

(b)




Δ1198714119894

= minus

1199032

1198602

119901

(Δ1198751119894

minus Δ1198750119894

)2

8119864119868119882

(17)


4

minus 1198894


Δ1198714

=

119873

sum

119894=1

Δ1198714119894

(18)


119899 =

119865119891

119882 (19)



Δ119871 = Δ1198711

+ Δ1198712

+ Δ1198713

+ Δ1198714

(20)




119889119875

119889119911=

minus (120591119894

119860) + 120588119898

119892 cos 120579 + (119898119860) 119877 (119889119909119889119911)

1 minus (119898119860) 119878

119889119879

119889119911= minus

]119898

119862119875119892

(119877119889119909

119889119911minus 119878

119889119875

119889119911) minus

119892 cos 120579119862119875119892

minus120587119891119903119905119894

120588119898

]3119898

4119862119875119892

+119886 (119879 minus 119879

119890

)

119862119875119892

119875 (1199110

) = 1198750

119879 (1199110

) = 1198790

119889119909 (1199110

) = 1198891199090

119909 (1199110

) = 1199110

(21)




120572119895

= 120572119895minus1

+

(120572119896


) Δ119904119895

Δ119904119896

(22)



and 120572119896minus1

Δ119904119895


119891 (119905119863

) =

1128radic119905119863

(1 minus 03radic119905119863

) 119905119863

le 15

(04063 + 05 ln 119905119863

) (1 +06

119905119863

) 119905119863

gt 15

(23)




1

119880119905119900

= 119903119905119894

1

120582insln(

119903119888119894

119903119905119900

) +1

ℎ119888

+ ℎ119903

+ 119903119905119894

1

120582cemln(

119903cem119903119888119900

) (24)


119888

and ℎ119903



119911

= 1198790







Step 5 Let 119879 = 119879119896



120597119879119890

120597119905119863

= (1205972

119879119890

1205971199032

119863

+1

119903119863

120597119879119890

120597119903119863

)

119879119890

1003816100381610038161003816119905119863=0= 1198790

+ 120574119911 cos 120579

120597119879119890

120597119903119863

10038161003816100381610038161003816100381610038161003816119903119863=1

= minus1

2120587120582119891

119889119902

119889119911

120597119879119890

120597119903119863

10038161003816100381610038161003816100381610038161003816119903119863rarrinfin

= 0

(25)

Let 119879119895119890119894


119879119894+1

119890119895

minus 119879119894

119890119895

120593=

119879119894+1

119890119895+1

minus 2119879119895

119890119895+1

+ 119879119894minus1

119890119895+1

1205852

minus

119879119894+1

119890119895+1

minus 119879119894+1

119890119895

119903119863

120593 (26)


minus(120593 +120593120585

119903119863

)119879119894+1

119890119895+1

+ (2120593 +120593120585

119903119863

)119879119894+1

119890119895

minus 120593119879119894+1

119890119895minus1

= 1205852

119879119894

119890119895

(27)







Number Measured(m)

Inclination(∘)

Azimuth(∘)

Vertical depth(m)

1 135 263 24101 134722 278 123 23786 277913 364 143 21386 363824 393 217 2638 392535 422 185 4456 421286 450 082 19112 449627 486 293 26907 485478 514 103 29755 513839 543 358 32451 5417410 571 298 30305 5704311 600 203 20474 5994212 628 234 16433 6272813 660 185 19528 6595614 723 314 21484 7217015 782 098 21648 7813016 830 215 22931 8291217 860 267 24403 8597118 908 485 26662 9040819 928 672 25878 9214220 972 203 23688 9717121 1025 478 23927 10212522 1058 401 24459 10555823 1089 498 2282 10841724 1132 375 23388 11292825 1174 563 23514 11688726 1204 423 23438 12009927 1235 387 23499 12320828 1268 497 23257 12634529 1300 884 23328 128496


119863

= 1 we have

119879119890119894+1

1198902

minus (1 +119886120585

2120587120582119891

)119879119894+1

1198901

=119886119879119896

2120587120582119891

(28)

For 119903119863

= 119873 we have

119879119894+1

119890119899

minus 119879119894+1

119890119899minus1


119890



[[[[[[[[[[[[

[

1198791

1198901

1198792

1198901


sdot sdot sdot

1198791

1198902

1198792

1198902


1198902

sdot sdot sdot

sdot sdot sdot

1198791

119890119895

1198792

119890119895


119890119895

sdot sdot sdot

sdot sdot sdot

1198791

119890119899

1198792

119890119899


119890119899

sdot sdot sdot

]]]]]]]]]]]]

]

(30)



119896


1198651

=minus (120591119894

119860) + 120588119898

119892 cos 120579 + (119898119860) 119877 (119889119909119889119911)

1 minus (119898119860) 119878

1198652

= minus]119898

119862119875119892

(119877119889119909

119889119911minus 119878

119889119875

119889119911) minus

119892 cos 120579119862119875119892

minus120587119891119903119905119894

120588119898

]3119898

4119862119875119892

+119886 (119879 minus 119879

119890

)

119862119875119892

(31)

where 119879119890

at 119903119863

= 1



119886119896

= 119865119894

(1199101

1199102

)

119887119896

= 119865119894

(1199101

+ℎ1198861

2 1199102

+ℎ1198862

2)

119888119896

= 119865119894

(1199101

+ℎ1198871

2 1199102

+ℎ1198872

2)

119889119896

= 119865119894

(1199101

+ ℎ1198881

1199102

+ ℎ1198882

)

(32)



Number Depth(m)

Axial force(N)



changed (m)






119910119895+1

119896

= 119910119895

119896

+ℎ (119886119896

+ 2119887119896

+ 2119888119896

+ 119889119896

)

6

119896 = 1 2 119895 = 1 2 119899

(33)


Δ1198712119895

and Δ1198714119895





Δ119871 =

119873

sum

119895=1

Δ1198711119895

+

119873

sum

119895=1

Δ1198712119895

+ Δ1198713

+

119873

sum

119895=1

Δ1198714119895

(34)





25

2

15

1

05

0

Tota

l axi

al d

efor

mat

ion

(m)

Depth (m)1 1200800400

700000m3d500000m3d300000m3d










5 Conclusion


Nomenclature

119889 Inner diameter (m)119889119911










119862119875






119865119904






119875119894










Δ119871119905


Δ119871119887




Δ119875119900119894


Δ119875119888



Δ120588119894119894


Δ120588119900119894



References


















RoboticsJournal of





Journal of



RotatingMachinery





VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation













119873119902119911

= int

119871

119911

119902 cos120572119889119911 =120587

41205881

119892 (1198632

minus 1198892

)int

119871

119911

cos120572119889119911 (1)

where 119873119902119911


1



119873119887119911

= minus1205882

1198921198602

int

119871

119911

cos120572119889119911 = minus1205882

119892119911120587(119863

2)

2

int

119871

119911

cos120572119889119911

(2)

where 119873119887119911




119873119901119911

=1198751199111

1205871198892

119911

4 (3)

where 1198751199111



119865119911

= 119873119902119911

+ 119873119887119911

+ 119873119901119911

(4)


120590119911119894

=4119865119911

120587 (1198632

minus 1198892

) (5)


120590119911119905

= 119864120573 (1198791199111

minus 1198791199110

) = 119864120573Δ119879 (6)




Pz1

Pz2D

d





120590119903119911

=1198892

1198751199111

minus 1198632

1198751199110

1198632

minus 1198892

minus(1198751199111

minus 1198751199110

)1198632

1198892

(1198632

minus 1198892

) 41199032

120590120579119911

=1198892

1198751199111

minus 1198632

1198751199110

1198632

minus 1198892

+(1198751199111

minus 1198751199110

)1198632

1198892

(1198632

minus 1198892

) 41199032

(7)


1199111




ℎ119891

=120582 (119885 minus 119911) ]2

119898

2119892119889 (8)

where ℎ119891


119898


119891119911

= ℎ119891

120588119898

1198921205871198892 (120588119898


120590119911119891

=

4119873119891119911

120587 (1198632

minus 1198892

) (9)




1205761

=1

119864[120590119911119894

minus 120583 (120590119903119911

+ 120590120579119911

)] (10)



Δ1198711119894

= int

119885119894

119885119894minus1

1

119864[120590119911119894

minus 120583 (120590119903119911

+ 120590120579119911

)] 119889119911 (11)


Δ1198711

=

119873

sum

119894=1

Δ1198711119894

(12)


Δ1198712119894

= int

119885119894

119885119894minus1

120590119911119905

119864119889119911 = 120573Δ119879

119894

Δ119871119894

(13)


Δ1198712

=

119873

sum

119894=1

Δ1198712119894

(14)


Δ1198713

= int

119885

0

120590119911119891

119864119889119911 =

120582120588119898

]2119898

1198891198852

119864 (1198632

minus 1198892

) (15)



119865119891

= 119860119901

(1198751

minus 1198750

) (16)

where 1198751


0


119901




(a)

F

Neutral point

(b)




Δ1198714119894

= minus

1199032

1198602

119901

(Δ1198751119894

minus Δ1198750119894

)2

8119864119868119882

(17)


4

minus 1198894


Δ1198714

=

119873

sum

119894=1

Δ1198714119894

(18)


119899 =

119865119891

119882 (19)



Δ119871 = Δ1198711

+ Δ1198712

+ Δ1198713

+ Δ1198714

(20)




119889119875

119889119911=

minus (120591119894

119860) + 120588119898

119892 cos 120579 + (119898119860) 119877 (119889119909119889119911)

1 minus (119898119860) 119878

119889119879

119889119911= minus

]119898

119862119875119892

(119877119889119909

119889119911minus 119878

119889119875

119889119911) minus

119892 cos 120579119862119875119892

minus120587119891119903119905119894

120588119898

]3119898

4119862119875119892

+119886 (119879 minus 119879

119890

)

119862119875119892

119875 (1199110

) = 1198750

119879 (1199110

) = 1198790

119889119909 (1199110

) = 1198891199090

119909 (1199110

) = 1199110

(21)




120572119895

= 120572119895minus1

+

(120572119896


) Δ119904119895

Δ119904119896

(22)



and 120572119896minus1

Δ119904119895


119891 (119905119863

) =

1128radic119905119863

(1 minus 03radic119905119863

) 119905119863

le 15

(04063 + 05 ln 119905119863

) (1 +06

119905119863

) 119905119863

gt 15

(23)




1

119880119905119900

= 119903119905119894

1

120582insln(

119903119888119894

119903119905119900

) +1

ℎ119888

+ ℎ119903

+ 119903119905119894

1

120582cemln(

119903cem119903119888119900

) (24)


119888

and ℎ119903



119911

= 1198790







Step 5 Let 119879 = 119879119896



120597119879119890

120597119905119863

= (1205972

119879119890

1205971199032

119863

+1

119903119863

120597119879119890

120597119903119863

)

119879119890

1003816100381610038161003816119905119863=0= 1198790

+ 120574119911 cos 120579

120597119879119890

120597119903119863

10038161003816100381610038161003816100381610038161003816119903119863=1

= minus1

2120587120582119891

119889119902

119889119911

120597119879119890

120597119903119863

10038161003816100381610038161003816100381610038161003816119903119863rarrinfin

= 0

(25)

Let 119879119895119890119894


119879119894+1

119890119895

minus 119879119894

119890119895

120593=

119879119894+1

119890119895+1

minus 2119879119895

119890119895+1

+ 119879119894minus1

119890119895+1

1205852

minus

119879119894+1

119890119895+1

minus 119879119894+1

119890119895

119903119863

120593 (26)


minus(120593 +120593120585

119903119863

)119879119894+1

119890119895+1

+ (2120593 +120593120585

119903119863

)119879119894+1

119890119895

minus 120593119879119894+1

119890119895minus1

= 1205852

119879119894

119890119895

(27)







Number Measured(m)

Inclination(∘)

Azimuth(∘)

Vertical depth(m)

1 135 263 24101 134722 278 123 23786 277913 364 143 21386 363824 393 217 2638 392535 422 185 4456 421286 450 082 19112 449627 486 293 26907 485478 514 103 29755 513839 543 358 32451 5417410 571 298 30305 5704311 600 203 20474 5994212 628 234 16433 6272813 660 185 19528 6595614 723 314 21484 7217015 782 098 21648 7813016 830 215 22931 8291217 860 267 24403 8597118 908 485 26662 9040819 928 672 25878 9214220 972 203 23688 9717121 1025 478 23927 10212522 1058 401 24459 10555823 1089 498 2282 10841724 1132 375 23388 11292825 1174 563 23514 11688726 1204 423 23438 12009927 1235 387 23499 12320828 1268 497 23257 12634529 1300 884 23328 128496


119863

= 1 we have

119879119890119894+1

1198902

minus (1 +119886120585

2120587120582119891

)119879119894+1

1198901

=119886119879119896

2120587120582119891

(28)

For 119903119863

= 119873 we have

119879119894+1

119890119899

minus 119879119894+1

119890119899minus1


119890



[[[[[[[[[[[[

[

1198791

1198901

1198792

1198901


sdot sdot sdot

1198791

1198902

1198792

1198902


1198902

sdot sdot sdot

sdot sdot sdot

1198791

119890119895

1198792

119890119895


119890119895

sdot sdot sdot

sdot sdot sdot

1198791

119890119899

1198792

119890119899


119890119899

sdot sdot sdot

]]]]]]]]]]]]

]

(30)



119896


1198651

=minus (120591119894

119860) + 120588119898

119892 cos 120579 + (119898119860) 119877 (119889119909119889119911)

1 minus (119898119860) 119878

1198652

= minus]119898

119862119875119892

(119877119889119909

119889119911minus 119878

119889119875

119889119911) minus

119892 cos 120579119862119875119892

minus120587119891119903119905119894

120588119898

]3119898

4119862119875119892

+119886 (119879 minus 119879

119890

)

119862119875119892

(31)

where 119879119890

at 119903119863

= 1



119886119896

= 119865119894

(1199101

1199102

)

119887119896

= 119865119894

(1199101

+ℎ1198861

2 1199102

+ℎ1198862

2)

119888119896

= 119865119894

(1199101

+ℎ1198871

2 1199102

+ℎ1198872

2)

119889119896

= 119865119894

(1199101

+ ℎ1198881

1199102

+ ℎ1198882

)

(32)



Number Depth(m)

Axial force(N)



changed (m)






119910119895+1

119896

= 119910119895

119896

+ℎ (119886119896

+ 2119887119896

+ 2119888119896

+ 119889119896

)

6

119896 = 1 2 119895 = 1 2 119899

(33)


Δ1198712119895

and Δ1198714119895





Δ119871 =

119873

sum

119895=1

Δ1198711119895

+

119873

sum

119895=1

Δ1198712119895

+ Δ1198713

+

119873

sum

119895=1

Δ1198714119895

(34)





25

2

15

1

05

0

Tota

l axi

al d

efor

mat

ion

(m)

Depth (m)1 1200800400

700000m3d500000m3d300000m3d










5 Conclusion


Nomenclature

119889 Inner diameter (m)119889119911










119862119875






119865119904






119875119894










Δ119871119905


Δ119871119887




Δ119875119900119894


Δ119875119888



Δ120588119894119894


Δ120588119900119894



References


















RoboticsJournal of





Journal of



RotatingMachinery





VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation











1205761

=1

119864[120590119911119894

minus 120583 (120590119903119911

+ 120590120579119911

)] (10)



Δ1198711119894

= int

119885119894

119885119894minus1

1

119864[120590119911119894

minus 120583 (120590119903119911

+ 120590120579119911

)] 119889119911 (11)


Δ1198711

=

119873

sum

119894=1

Δ1198711119894

(12)


Δ1198712119894

= int

119885119894

119885119894minus1

120590119911119905

119864119889119911 = 120573Δ119879

119894

Δ119871119894

(13)


Δ1198712

=

119873

sum

119894=1

Δ1198712119894

(14)


Δ1198713

= int

119885

0

120590119911119891

119864119889119911 =

120582120588119898

]2119898

1198891198852

119864 (1198632

minus 1198892

) (15)



119865119891

= 119860119901

(1198751

minus 1198750

) (16)

where 1198751


0


119901




(a)

F

Neutral point

(b)




Δ1198714119894

= minus

1199032

1198602

119901

(Δ1198751119894

minus Δ1198750119894

)2

8119864119868119882

(17)


4

minus 1198894


Δ1198714

=

119873

sum

119894=1

Δ1198714119894

(18)


119899 =

119865119891

119882 (19)



Δ119871 = Δ1198711

+ Δ1198712

+ Δ1198713

+ Δ1198714

(20)




119889119875

119889119911=

minus (120591119894

119860) + 120588119898

119892 cos 120579 + (119898119860) 119877 (119889119909119889119911)

1 minus (119898119860) 119878

119889119879

119889119911= minus

]119898

119862119875119892

(119877119889119909

119889119911minus 119878

119889119875

119889119911) minus

119892 cos 120579119862119875119892

minus120587119891119903119905119894

120588119898

]3119898

4119862119875119892

+119886 (119879 minus 119879

119890

)

119862119875119892

119875 (1199110

) = 1198750

119879 (1199110

) = 1198790

119889119909 (1199110

) = 1198891199090

119909 (1199110

) = 1199110

(21)




120572119895

= 120572119895minus1

+

(120572119896


) Δ119904119895

Δ119904119896

(22)



and 120572119896minus1

Δ119904119895


119891 (119905119863

) =

1128radic119905119863

(1 minus 03radic119905119863

) 119905119863

le 15

(04063 + 05 ln 119905119863

) (1 +06

119905119863

) 119905119863

gt 15

(23)




1

119880119905119900

= 119903119905119894

1

120582insln(

119903119888119894

119903119905119900

) +1

ℎ119888

+ ℎ119903

+ 119903119905119894

1

120582cemln(

119903cem119903119888119900

) (24)


119888

and ℎ119903



119911

= 1198790







Step 5 Let 119879 = 119879119896



120597119879119890

120597119905119863

= (1205972

119879119890

1205971199032

119863

+1

119903119863

120597119879119890

120597119903119863

)

119879119890

1003816100381610038161003816119905119863=0= 1198790

+ 120574119911 cos 120579

120597119879119890

120597119903119863

10038161003816100381610038161003816100381610038161003816119903119863=1

= minus1

2120587120582119891

119889119902

119889119911

120597119879119890

120597119903119863

10038161003816100381610038161003816100381610038161003816119903119863rarrinfin

= 0

(25)

Let 119879119895119890119894


119879119894+1

119890119895

minus 119879119894

119890119895

120593=

119879119894+1

119890119895+1

minus 2119879119895

119890119895+1

+ 119879119894minus1

119890119895+1

1205852

minus

119879119894+1

119890119895+1

minus 119879119894+1

119890119895

119903119863

120593 (26)


minus(120593 +120593120585

119903119863

)119879119894+1

119890119895+1

+ (2120593 +120593120585

119903119863

)119879119894+1

119890119895

minus 120593119879119894+1

119890119895minus1

= 1205852

119879119894

119890119895

(27)







Number Measured(m)

Inclination(∘)

Azimuth(∘)

Vertical depth(m)

1 135 263 24101 134722 278 123 23786 277913 364 143 21386 363824 393 217 2638 392535 422 185 4456 421286 450 082 19112 449627 486 293 26907 485478 514 103 29755 513839 543 358 32451 5417410 571 298 30305 5704311 600 203 20474 5994212 628 234 16433 6272813 660 185 19528 6595614 723 314 21484 7217015 782 098 21648 7813016 830 215 22931 8291217 860 267 24403 8597118 908 485 26662 9040819 928 672 25878 9214220 972 203 23688 9717121 1025 478 23927 10212522 1058 401 24459 10555823 1089 498 2282 10841724 1132 375 23388 11292825 1174 563 23514 11688726 1204 423 23438 12009927 1235 387 23499 12320828 1268 497 23257 12634529 1300 884 23328 128496


119863

= 1 we have

119879119890119894+1

1198902

minus (1 +119886120585

2120587120582119891

)119879119894+1

1198901

=119886119879119896

2120587120582119891

(28)

For 119903119863

= 119873 we have

119879119894+1

119890119899

minus 119879119894+1

119890119899minus1


119890



[[[[[[[[[[[[

[

1198791

1198901

1198792

1198901


sdot sdot sdot

1198791

1198902

1198792

1198902


1198902

sdot sdot sdot

sdot sdot sdot

1198791

119890119895

1198792

119890119895


119890119895

sdot sdot sdot

sdot sdot sdot

1198791

119890119899

1198792

119890119899


119890119899

sdot sdot sdot

]]]]]]]]]]]]

]

(30)



119896


1198651

=minus (120591119894

119860) + 120588119898

119892 cos 120579 + (119898119860) 119877 (119889119909119889119911)

1 minus (119898119860) 119878

1198652

= minus]119898

119862119875119892

(119877119889119909

119889119911minus 119878

119889119875

119889119911) minus

119892 cos 120579119862119875119892

minus120587119891119903119905119894

120588119898

]3119898

4119862119875119892

+119886 (119879 minus 119879

119890

)

119862119875119892

(31)

where 119879119890

at 119903119863

= 1



119886119896

= 119865119894

(1199101

1199102

)

119887119896

= 119865119894

(1199101

+ℎ1198861

2 1199102

+ℎ1198862

2)

119888119896

= 119865119894

(1199101

+ℎ1198871

2 1199102

+ℎ1198872

2)

119889119896

= 119865119894

(1199101

+ ℎ1198881

1199102

+ ℎ1198882

)

(32)



Number Depth(m)

Axial force(N)



changed (m)






119910119895+1

119896

= 119910119895

119896

+ℎ (119886119896

+ 2119887119896

+ 2119888119896

+ 119889119896

)

6

119896 = 1 2 119895 = 1 2 119899

(33)


Δ1198712119895

and Δ1198714119895





Δ119871 =

119873

sum

119895=1

Δ1198711119895

+

119873

sum

119895=1

Δ1198712119895

+ Δ1198713

+

119873

sum

119895=1

Δ1198714119895

(34)





25

2

15

1

05

0

Tota

l axi

al d

efor

mat

ion

(m)

Depth (m)1 1200800400

700000m3d500000m3d300000m3d










5 Conclusion


Nomenclature

119889 Inner diameter (m)119889119911










119862119875






119865119904






119875119894










Δ119871119905


Δ119871119887




Δ119875119900119894


Δ119875119888



Δ120588119894119894


Δ120588119900119894



References


















RoboticsJournal of





Journal of



RotatingMachinery





VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation











119889119875

119889119911=

minus (120591119894

119860) + 120588119898

119892 cos 120579 + (119898119860) 119877 (119889119909119889119911)

1 minus (119898119860) 119878

119889119879

119889119911= minus

]119898

119862119875119892

(119877119889119909

119889119911minus 119878

119889119875

119889119911) minus

119892 cos 120579119862119875119892

minus120587119891119903119905119894

120588119898

]3119898

4119862119875119892

+119886 (119879 minus 119879

119890

)

119862119875119892

119875 (1199110

) = 1198750

119879 (1199110

) = 1198790

119889119909 (1199110

) = 1198891199090

119909 (1199110

) = 1199110

(21)




120572119895

= 120572119895minus1

+

(120572119896


) Δ119904119895

Δ119904119896

(22)



and 120572119896minus1

Δ119904119895


119891 (119905119863

) =

1128radic119905119863

(1 minus 03radic119905119863

) 119905119863

le 15

(04063 + 05 ln 119905119863

) (1 +06

119905119863

) 119905119863

gt 15

(23)




1

119880119905119900

= 119903119905119894

1

120582insln(

119903119888119894

119903119905119900

) +1

ℎ119888

+ ℎ119903

+ 119903119905119894

1

120582cemln(

119903cem119903119888119900

) (24)


119888

and ℎ119903



119911

= 1198790







Step 5 Let 119879 = 119879119896



120597119879119890

120597119905119863

= (1205972

119879119890

1205971199032

119863

+1

119903119863

120597119879119890

120597119903119863

)

119879119890

1003816100381610038161003816119905119863=0= 1198790

+ 120574119911 cos 120579

120597119879119890

120597119903119863

10038161003816100381610038161003816100381610038161003816119903119863=1

= minus1

2120587120582119891

119889119902

119889119911

120597119879119890

120597119903119863

10038161003816100381610038161003816100381610038161003816119903119863rarrinfin

= 0

(25)

Let 119879119895119890119894


119879119894+1

119890119895

minus 119879119894

119890119895

120593=

119879119894+1

119890119895+1

minus 2119879119895

119890119895+1

+ 119879119894minus1

119890119895+1

1205852

minus

119879119894+1

119890119895+1

minus 119879119894+1

119890119895

119903119863

120593 (26)


minus(120593 +120593120585

119903119863

)119879119894+1

119890119895+1

+ (2120593 +120593120585

119903119863

)119879119894+1

119890119895

minus 120593119879119894+1

119890119895minus1

= 1205852

119879119894

119890119895

(27)







Number Measured(m)

Inclination(∘)

Azimuth(∘)

Vertical depth(m)

1 135 263 24101 134722 278 123 23786 277913 364 143 21386 363824 393 217 2638 392535 422 185 4456 421286 450 082 19112 449627 486 293 26907 485478 514 103 29755 513839 543 358 32451 5417410 571 298 30305 5704311 600 203 20474 5994212 628 234 16433 6272813 660 185 19528 6595614 723 314 21484 7217015 782 098 21648 7813016 830 215 22931 8291217 860 267 24403 8597118 908 485 26662 9040819 928 672 25878 9214220 972 203 23688 9717121 1025 478 23927 10212522 1058 401 24459 10555823 1089 498 2282 10841724 1132 375 23388 11292825 1174 563 23514 11688726 1204 423 23438 12009927 1235 387 23499 12320828 1268 497 23257 12634529 1300 884 23328 128496


119863

= 1 we have

119879119890119894+1

1198902

minus (1 +119886120585

2120587120582119891

)119879119894+1

1198901

=119886119879119896

2120587120582119891

(28)

For 119903119863

= 119873 we have

119879119894+1

119890119899

minus 119879119894+1

119890119899minus1


119890



[[[[[[[[[[[[

[

1198791

1198901

1198792

1198901


sdot sdot sdot

1198791

1198902

1198792

1198902


1198902

sdot sdot sdot

sdot sdot sdot

1198791

119890119895

1198792

119890119895


119890119895

sdot sdot sdot

sdot sdot sdot

1198791

119890119899

1198792

119890119899


119890119899

sdot sdot sdot

]]]]]]]]]]]]

]

(30)



119896


1198651

=minus (120591119894

119860) + 120588119898

119892 cos 120579 + (119898119860) 119877 (119889119909119889119911)

1 minus (119898119860) 119878

1198652

= minus]119898

119862119875119892

(119877119889119909

119889119911minus 119878

119889119875

119889119911) minus

119892 cos 120579119862119875119892

minus120587119891119903119905119894

120588119898

]3119898

4119862119875119892

+119886 (119879 minus 119879

119890

)

119862119875119892

(31)

where 119879119890

at 119903119863

= 1



119886119896

= 119865119894

(1199101

1199102

)

119887119896

= 119865119894

(1199101

+ℎ1198861

2 1199102

+ℎ1198862

2)

119888119896

= 119865119894

(1199101

+ℎ1198871

2 1199102

+ℎ1198872

2)

119889119896

= 119865119894

(1199101

+ ℎ1198881

1199102

+ ℎ1198882

)

(32)



Number Depth(m)

Axial force(N)



changed (m)






119910119895+1

119896

= 119910119895

119896

+ℎ (119886119896

+ 2119887119896

+ 2119888119896

+ 119889119896

)

6

119896 = 1 2 119895 = 1 2 119899

(33)


Δ1198712119895

and Δ1198714119895





Δ119871 =

119873

sum

119895=1

Δ1198711119895

+

119873

sum

119895=1

Δ1198712119895

+ Δ1198713

+

119873

sum

119895=1

Δ1198714119895

(34)





25

2

15

1

05

0

Tota

l axi

al d

efor

mat

ion

(m)

Depth (m)1 1200800400

700000m3d500000m3d300000m3d










5 Conclusion


Nomenclature

119889 Inner diameter (m)119889119911










119862119875






119865119904






119875119894










Δ119871119905


Δ119871119887




Δ119875119900119894


Δ119875119888



Δ120588119894119894


Δ120588119900119894



References


















RoboticsJournal of





Journal of



RotatingMachinery





VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation















Number Measured(m)

Inclination(∘)

Azimuth(∘)

Vertical depth(m)

1 135 263 24101 134722 278 123 23786 277913 364 143 21386 363824 393 217 2638 392535 422 185 4456 421286 450 082 19112 449627 486 293 26907 485478 514 103 29755 513839 543 358 32451 5417410 571 298 30305 5704311 600 203 20474 5994212 628 234 16433 6272813 660 185 19528 6595614 723 314 21484 7217015 782 098 21648 7813016 830 215 22931 8291217 860 267 24403 8597118 908 485 26662 9040819 928 672 25878 9214220 972 203 23688 9717121 1025 478 23927 10212522 1058 401 24459 10555823 1089 498 2282 10841724 1132 375 23388 11292825 1174 563 23514 11688726 1204 423 23438 12009927 1235 387 23499 12320828 1268 497 23257 12634529 1300 884 23328 128496


119863

= 1 we have

119879119890119894+1

1198902

minus (1 +119886120585

2120587120582119891

)119879119894+1

1198901

=119886119879119896

2120587120582119891

(28)

For 119903119863

= 119873 we have

119879119894+1

119890119899

minus 119879119894+1

119890119899minus1


119890



[[[[[[[[[[[[

[

1198791

1198901

1198792

1198901


sdot sdot sdot

1198791

1198902

1198792

1198902


1198902

sdot sdot sdot

sdot sdot sdot

1198791

119890119895

1198792

119890119895


119890119895

sdot sdot sdot

sdot sdot sdot

1198791

119890119899

1198792

119890119899


119890119899

sdot sdot sdot

]]]]]]]]]]]]

]

(30)



119896


1198651

=minus (120591119894

119860) + 120588119898

119892 cos 120579 + (119898119860) 119877 (119889119909119889119911)

1 minus (119898119860) 119878

1198652

= minus]119898

119862119875119892

(119877119889119909

119889119911minus 119878

119889119875

119889119911) minus

119892 cos 120579119862119875119892

minus120587119891119903119905119894

120588119898

]3119898

4119862119875119892

+119886 (119879 minus 119879

119890

)

119862119875119892

(31)

where 119879119890

at 119903119863

= 1



119886119896

= 119865119894

(1199101

1199102

)

119887119896

= 119865119894

(1199101

+ℎ1198861

2 1199102

+ℎ1198862

2)

119888119896

= 119865119894

(1199101

+ℎ1198871

2 1199102

+ℎ1198872

2)

119889119896

= 119865119894

(1199101

+ ℎ1198881

1199102

+ ℎ1198882

)

(32)



Number Depth(m)

Axial force(N)



changed (m)






119910119895+1

119896

= 119910119895

119896

+ℎ (119886119896

+ 2119887119896

+ 2119888119896

+ 119889119896

)

6

119896 = 1 2 119895 = 1 2 119899

(33)


Δ1198712119895

and Δ1198714119895





Δ119871 =

119873

sum

119895=1

Δ1198711119895

+

119873

sum

119895=1

Δ1198712119895

+ Δ1198713

+

119873

sum

119895=1

Δ1198714119895

(34)





25

2

15

1

05

0

Tota

l axi

al d

efor

mat

ion

(m)

Depth (m)1 1200800400

700000m3d500000m3d300000m3d










5 Conclusion


Nomenclature

119889 Inner diameter (m)119889119911










119862119875






119865119904






119875119894










Δ119871119905


Δ119871119887




Δ119875119900119894


Δ119875119888



Δ120588119894119894


Δ120588119900119894



References


















RoboticsJournal of





Journal of



RotatingMachinery





VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation











Number Depth(m)

Axial force(N)



changed (m)






119910119895+1

119896

= 119910119895

119896

+ℎ (119886119896

+ 2119887119896

+ 2119888119896

+ 119889119896

)

6

119896 = 1 2 119895 = 1 2 119899

(33)


Δ1198712119895

and Δ1198714119895





Δ119871 =

119873

sum

119895=1

Δ1198711119895

+

119873

sum

119895=1

Δ1198712119895

+ Δ1198713

+

119873

sum

119895=1

Δ1198714119895

(34)





25

2

15

1

05

0

Tota

l axi

al d

efor

mat

ion

(m)

Depth (m)1 1200800400

700000m3d500000m3d300000m3d










5 Conclusion


Nomenclature

119889 Inner diameter (m)119889119911










119862119875






119865119904






119875119894










Δ119871119905


Δ119871119887




Δ119875119900119894


Δ119875119888



Δ120588119894119894


Δ120588119900119894



References


















RoboticsJournal of





Journal of



RotatingMachinery





VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation












5 Conclusion


Nomenclature

119889 Inner diameter (m)119889119911










119862119875






119865119904






119875119894










Δ119871119905


Δ119871119887




Δ119875119900119894


Δ119875119888



Δ120588119894119894


Δ120588119900119894



References


















RoboticsJournal of





Journal of



RotatingMachinery





VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation




















RoboticsJournal of





Journal of



RotatingMachinery





VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation









Research Article Analyzing Axial Stress and Deformation...

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Transcript of Research Article Analyzing Axial Stress and Deformation...