Lecture 8: DIRECTIONAL DRILLINGLecture 8: DIRECTIONAL DRILLINGDrill String Design in Directional Wells

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

09997200339

1

Neutral Point• 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.

The formula to calculate the length from the bit to the neutralpoint in a vertical hole if only drill collars are being used is: 

Lnp = {Bit Weight} / {W x BF}p

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

e.g. Determine the neutral point in:e.g. Determine the neutral point in:8” x 2‐13/16” DC’s if WOB = 30k‐lbs in 11 ppg mud

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

So NP is 240 ft up in collars

But often we’ll have the neutral point above the collars – somewhere in the hevi‐wate, let’s 

have a look at this….

Lnphw = {BW – (Wc x Lc x BF)} / {Whw x BF} 

Where:Lnphw = Distance from bottom of HWDP to NPnphw

BW = 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 13/16”e.g. Determine the neutral point for: 300ft of 8” x 2‐13/16” DC’s & 600 ft of 5” HWDP if WOB = 40k.lbs in 13 ppg 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

Neutral Point Calculations 

in 

Directional WellsDirectional Wells

(Drill Collars + 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 θ=

× ×

Wh

cosDCW BF θ× ×

Where:θ = 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

θθ

−=

× ×

Wh

coshwW BF θ× ×

Where:θ = borehole inclinationWhw = weight per foot of the HWDP

General formula for Directional WellsGeneral formula for Directional Wells

• 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 θ− +=

cosnphwhwW BF θ× ×

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

collarscollars

PROBLEM 3

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

+ 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…..

PROBLEM 3

• − 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

TENSION DESIGN

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. Static Load

Tensile Yield Strength…g

in pounds can be calculated for Class I drill pipe (new drill pipe) using the following formula:

Tensile Yield Strength (lbs), Ym= 

Min. Yield Strength (lb/in2) x π/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)

Margin of Overpull (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).

Final Design Equation

Example 5: Drill String Design based on MOP

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