Lecture 8: DIRECTIONAL DRILLINGLecture8:DIRECTIONALDRILLINGDrill String Design in Directional Wells

Arun S ChandelAssistant Professoraschandel@ddn.upes.ac.in

09997200339

1

NeutralPoint The neutral point in a drill string can be defined as the point

where the string changes from tension to compression

This point is a function of bit weight and buoyancy

The easiest way to conceptualize this is to imagine firstly that The easiest way to conceptualize this is to imagine, firstly, thatthe whole drillstring is suspended off bottom, in this case theentire string is in tension with neutral point right at the bit.Secondly, imagine the whole drillstring is set on bottom with noload being taken by the surface equipment, in this case the stringis in compression and the neutral point is at surface.

It i i t t t k th l ti f thi t iti i t It is important to know the location of this transition point orneutral point for several reasons. If the neutral point is at the jars,for example, then the drill string and jars could both be damaged.If the neutral point is allowed to move up into the drill pipe,t e eut a po t s a o ed to o e up to t e d p pe,buckling could occur.

The neutral point should be maintained in the stronger drill collarThe neutral point should be maintained in the stronger drill collarassembly for regular vertical and directional drilling operations whenpossible. There may be a problem in high angle and horizontal drillingin this respect because of the difficulty in maintaining bit weight.Damage at the neutral point may be strongly dependent upong p y g y p pdrillstring rotation, and consideration should be given to critical rotaryspeeds and their associated harmonics.

Theformulatocalculatethelengthfromthebittotheneutralpointinaverticalholeifonlydrillcollarsarebeingusedis:

Lnp = {BitWeight}/{WxBF}p

(Where W = Collar or pipe weight in lbs/ft & BF = Buoyancy Factor)

e.g. Determine the neutral point in:e.g. Determinetheneutralpoint in:8x213/16DCsifWOB=30klbsin11ppg mud

Lnp = {30,000} / {150 x 0.832} = 240Lnp {30,000}/{150x0.832} 240

SoNPis240ftupincollars

Butoftenwellhavetheneutralpointabovethecollars somewhereintheheviwate,lets

havealookatthis.

Lnphw = {BW (Wc xLc xBF)}/{Whw xBF}

Where:Lnphw = Distance from bottom of HWDP to NPnphwBW = Bit WeightWc= Weight per foot of collarsLc = Length of collarsc gWhw = Weight per foot of Hevi-WateBF = Buoyancy Factor

Check first though to see if the NP is within the collars

e g Determine the neutral point for: 300ft of 8 x 2 13/16e.g.Determinetheneutralpointfor: 300ftof8x213/16DCs&600ftof5HWDPifWOB=40k.lbsin13ppg mud

Is NP within collars?Lnp = Bit Weight / [W x BF] or:

40 000 / [150 x 0 801] 333 ft40,000 / [150 x 0.801] = 333 ftSo NP is above collars..

But where?

Lnphw = 40,000 [150 x 300 x 0.801] / [50 x 0.801]nphw , [ ] / [ ]Or: 3,955 / 40.05 = 98.75 ftSo - NP is ~ 99 ft into the HWDP

NeutralPointCalculations

in

Directional WellsDirectionalWells

(DrillCollars+HWDPs)

Directional Well, Neutral Point in the D ill Colla sDrill Collars

When the neutral point is in the drill collar section and the collars are all of the same diameter, the formula for neutral point is:

cosnpWOBL

W BF = Wh

cosDCW BF Where: = borehole inclinationWDC = weight per foot of the drill collars

Directional Well, Neutral Point in the HWDP,

When the neutral point is in the HWDP section and the drill collars are all of the same diameter, the formula for neutral point is:

{ ( )cos }cos

DC DCnphw

WOB W L BFLW BF

= Wh

coshwW BF Where: = borehole inclinationWhw = weight per foot of the HWDP

General formula for Directional WellsGeneralformulaforDirectionalWells

The last formula can be expanded in the case of a tapered BHA i h d ill ll f h di F l BHA with drill collars of more than one diameter. For example, if there were two sizes drill collars but the neutral point was in the hevi-wate the formula would become:

1 1 2 2{ cos ( }DC DC DC DCnphw

WOB BF W L W LL +=cosnphw hwW BF

= borehole inclinationWDC1 & WDC2 = weight per foot of first and second size of drill

collarscollars

PROBLEM3

Determine the neutral point in a 300 inclined well:300 of 6.5 x 2-1/4 DCs + 200 of 7-1/4x 2 DCs

+ 250 of 5 x 3 HWDP (50 lb/ft) if+ 250 of 5 x 3 HWDP (50 lb/ft), ifWOB = 45k-lbs in 12 ppg mud

Is NP within collars? Is NP within collars?

Lnp1 = Bit Weight / [W x BF] or:45 000 / [99 5 x 0 82 x cos30] = 636 86 ft> 300 ft45,000 / [99.5 x 0.82 x cos30] = 636.86 ft> 300 ftSo NP is above this collar..

L = Bit Weight / [W x BF] or:Lnp2 = Bit Weight / [W x BF] or:

45,000-(300x99.5x0.82xcos30)/[129.3 x 0.82 x cos30] = 259 2 ft> 200 ft= 259.2 ft> 200 ft

So NP is above this collar..

PROBLEM3

But where?

Lnphw = {45,000 [0.82x cos30x (99.5x 300+ 129.3x 200 ]} / [50 x 0.82x cos30]

Or: 523 / 41.0 = 153.15 ftSo - NP is ~ 153.15 ft into the HWDP

TENSIONDESIGN

OFOF

DRILLSTRINGDRILLSTRING

1. Static Load

The design of the drillstring for static tension loadsrequires sufficient strength in the topmost joint ofq g p jeach size, weight, grade and classification of drillpipe to support the submerged weight of all the drillpipe plus the submerged weight of the collarspipe plus the submerged weight of the collars,stabilizers, and bit.

Th bit d t bili i ht ith l t dThe bit and stabilizer weights are either neglectedor are included with the drill collar weight.

This load may be calculated as shown in thefollowing equation:

1.StaticLoad

TensileYieldStrengthg

inpoundscanbecalculatedforClassIdrillpipe(newdrillpipe)usingthefollowingformula:

TensileYieldStrength(lbs),Ym=

Min.YieldStrength(lb/in2)x/4(OD2 ID2)

If the pipe is loaded to the extent shown in the APIIf the pipe is loaded to the extent shown in the APIformula above it is likely that some permanent stretch willoccur and difficulty may be experienced in keeping thepipe straight.

To prevent this condition a design factor of approximatelyTo prevent this condition a design factor of approximately90% of the tabulated tension value is recommended.

2.Overpull If the drill string were to get stuck in the well bore,

the operator would want to know how much additionaltension o p ll can be applied to the st ing befo etension, or pull, can be applied to the string beforeexceeding the yield point of the drill pipe. This isknown as overpull since it is pull force over the weightof the stringof the string.

Maximum overpull is the difference between theyield strength and the hookload or margin ofyield strength and the hookload or margin ofoverpull (MOP) is normally applied)

MarginofOverpull(MOP)

The difference between the calculated load FTEN and the maximum allowable tension load represents the Margin of Over Pull (MOP):

The same values expressed as a ratio may be The same values expressed as a ratio may be called the Safety Factor (SF).

FinalDesignEquation

Example5:DrillStringDesignbasedonMOP

Design the drill string for the given wellDesign the drill string for the given welldata.Can the final well depth be reached with thisassembly?Finally make a table showing all the drillstring components with their air & buoyedstring components with their air & buoyedweight.

GivenGiven

1. The Yield Strength of grade E drill pipe=225,771 lb and weight/ft = 18.37 lb/ft.

2. The Yield Strength of grade X-95 drill pipe=g g p p329,542 lb and weight/ft = 18.88 lb/ft.

Solution