Lecture 2 Directional Surveying

 

VI B Tech PE LECTURE 2 / 2015-16

DIRECTIONAL SURVEYING Well Path Correction and Well Monitoring BY Prof.A.K.Pathak Departmentt of Petroleum Engineering Departmen Indian School of Mines, Dhanbad – 826004 E-mail: [email protected]

CORRECTION OF DEFLECTED WELL PATH:

Tan DL. Sin TFS Δβ = Tan  [ -------------------------------------------------------------------------------------------------------- ] Sin α1  + Tan DL. Cos α1. Cos TFS -1

α2 = Cos-1 { Cos α1. Cos DL –Cos TFS. Sin α1. Sin DL}   Cos α1. Cos DL – Cos α2 -1 TFS = Cos  [ ------------------------------------------------------------------------------------------ ] Sin α1. Sin DL

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OBJECTIVES:

i.

To mon monititor or the pro progr gres esss of the well. well. Act Actua uall di dire rect ctio iona nall da data ta can be used used to plot plot th thee course of the well and than can be compared with the planned trajectory.

ii. To planned planned th thee well ppath ath co correcti rrection on strateg strategies ies for for the de deviat viated ed wells. wells. iii iii.. To prevent prevent coll collisio isionn of the prese present nt well with ne nearby arby exi existing sting we well. ll. iv. To determine determine the actual de deflection flection to tool ol orie orientation ntation in corre correct ct dire direction. ction. v. To determi determine ne ex exact act lo locatio cationn of th thee bit aass dri drillin llingg pro progres gresses. ses. vi. To plan an andd monito monitorr relie relieff well duri during ng blow ou outs, ts, Fire acc accide idents nts etc. vii. To calcula calculate te dogdog-leg leg sseveri everity. ty. METHODS OF WELL SURVEYING:

i.

Acid Acid Bott Bottle le Surv Survey eyin ing: g:

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 Acid Floating compass Gelatin compass

Drill pipe

This tool is based on simple principle that the free surface of a liquid always remains horizontal regardless of the position of container. In this instrument the container is a glass cylinder and the liquid is acid. The instrument is allowed to rest in an incline position for a certain period of time (30 min.). The acid will react with glass surface and leave a mark on the side of the cylinder indicating horizontal surface. The distance between the mark and the acid’s original position, when the cylinder was levelled, can be used to calculate the inclination angle. The strength of the acid should be chosen carefully so that it can leave distinct mark on the glass surface. To measure the direction an addition compartment was required containing gelatine and a magnetic compass needle nee dle.. The com compa pass ss nee needle dle was fre freee floating and aligned itself with magnetic north. It was held in this position by the gelatine. The direction of the deviated well could therefore be referenced to magnetic no nort rth. h. Th Thee ma majo jorr di disa sadv dvan anta tage ge of th this is method was that the acid did not react with the glass surface and failed to leave distinct mark. magnetic etic sing single le shot surveyin surveyingg instr instrumen umentt ii ii.. Magn Magnet etic ic Singl Single e Shot Shot Surve Surveyi ying ng:: The magn records simultaneously hole inclination and magnetic north direction of an uncased hole at a single measured depth. This instrument consists of –

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Angle indicating unit: It comprise of a magnetic compass  card pivoted

(a)

on a single point, divided in to 360 degrees in five degree increment and a pendulum assembly consists of a plumb bob or an inclinometer. A glass plate between the pendulum and compass having a series of co-centric rings in increment of one degree to represent the drift or the inclination angle. The entire unit is built in one case and is fluid filled to dampen movement and shock protection. Camera section: The camera section is placed above the angle unit and

(b)

comprises a lamp assembly, lens and film holder. The camera has no shutter and its exposure is controlled by time for which the light is on. Timing device: The timing device be either mechanical or electronic. It

(c)

holds the action of the camera until the single shot is located in non-magnetic collar  above the bit and it also controls the time for which the lamps are switched on for  proper exposure. Battery pack: The battery pack supplies electric power for the timer and

(d)

camera. OPERATION: The magnetic single shot is run on a wire line to hole bottom or dropped

from surface, when it sits on a baffle plate placed on top of the bit. The timer  activates the camera and a photographic record of the positions of plumb bob, magnetic compass and co-centric ring plate is taken. The instrument is then pulled out of the hole, the film is developed and placed in the chart recorder. It should be 22

 

noted that the magnetic recorder only records the direction with respect to the magnetic north. Therefore the correction for the magnetic north and the true north should be done to get the data with respect to the true north. For this, local magnetic declination declination angle should be added with sign for the data correction.

23

 

24

 

ii iii. i. Magn Magnet etic ic Mult Multii Shot Shots: s: The magnetic multi shot instruments are similar to the

single shot, except that a film magazine is used so that the survey data can be recorded at regular interval of time, also the camera is a modified movie camera. This instrument can be run on a wire line to land in a non-magnetic drill collar or  simply dropped from the surface. A continuous record is taken as the drill string is pulled out of the hole. Surveys are taken when each stand is broken off and the string is stationary. Magnetic multi shot instruments are useful for wells free from the magnetic materials or magnetic geological formations. A new generation of  multi-shot survey the well bore and store the data in digital form in down-hole memory for the later recovery.

iv iv.. Gyro Gyrosc scop opes es:: The magnetic survey results are liable to errors due to magnetic

distur dis turba bance ncess cau caused sed by cas casin ings gs of ne nearb arbyy wel wells ls or the mag magne netic tic geo geolo logic gical al formations. To overcome these errors, gyroscopes are used because of their ability to maintain pre-set direction irrespective of the earth magnetic fields or the factors disturbing it. Gyros are most widely used in orienting mud motors and bent sub assemblies.. The pprincipal assemblies rincipal element of gyro are as follows-

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Rapid Ra pidly ly spinni spinning ng weigh weighted ted whee wheel: l: It mai mainta ntain inss a preset preset dir direct ectio ionn by res resist istin ingg

changes in direction, owing to its inertia. Thus if the wheel is preset to maintain the direction parallel to the earth’s axis of rotation (N-S direction), then it will maintain this direction till drilling. Therefore it serves as a directiona directionall indicator.

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horizont zontal al and is mounted insi inside de two Supporting Gimbals:  The axis of the wheel is hori perpendicular gimbals to allow the gyro to maintain its direction irrespective of gyro casing movement. The inner gimbal is made of the spin motor and the t he housing. The motor spins at a rate of 18000 to 42000 rpm depending on gyro size and type of spin motor used. The outer gimbal is fitted with a compass card similar to that used in magnetic survey tools.

The compass card is set in known direction normally to the north, to serve as a reference for  survey directional data.   are similar to those of the magnetic The pendulum assembly, camera unit, timer  are survey single shot/ multi shot tools.

Operation: The orientation of the gyrocompass must be checked prior to running in the hole,

which is done by aligning the tool to a known direction by use of a telescope alignment kit. For single shot the gyro assembly is run on wire line to the bottom of the hole. A picture of  plumb bob, gyro compass compass and the co-centric ring ring plate is taken at the required depth. The tool if pulled out, the film is unloaded and developed to obtain the survey data by a chart recorder. For multishot, the tool is lowered on wire line through a section of cased hole. The tool is stopped and set depth intervals e.g. every 100 ft. and a survey is taken by multi shot camera. All gyros have tendency to drift from f rom preset direction and survey readings should be corrected for this effect. Drift is due to friction in the bearings of spinning wheel, gyro 27

 

imbalances imbalanc es and motor speed instability. During surveying, the tool is kept stationary at every 15 min. for 5 min. to monitor gyro drift.

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ANALYSIS OF DIRECTIONA SURVEY DATA:

i.

Tangential Method:

V

=

L Cos 2

H E

= =

L Sin 2 L Sin 2  . Sin 2

N

=

L Sin 2  . Cos 2 

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ii.

iii.

Balanced Tangential Method:

V

=

(L/2) (Cos 1 + Cos 2)

H

=

(L/2) (Sin 1 + Sin 2)

E

=

(L/2) (Sin 1 . Sin 1 + Sin 2 . Sin 2)

N

=

(L/2) (Sin 1 . Cos 1 + Sin 2 . Cos 2)

Average Angle Method:

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iv.

V

=

L Cos [(1 + 2) / 2]

H

=

L Sin [(1 + 2) / 2]

E

=

L Sin [(1 + 2) / 2] . Sin [( 1 + 2)/2]

N

=

L Sin [(1 + 2) / 2] . Cos [( 1 + 2)/2]

Radius of of Cu Curvature Me Method:

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RV 

=

180 . L / π (2 - 1)

V

=

RV Sin 2 - RV Sin 1 = RV ( Sin 2 - Sin 1)

V

=

360 . L . (Sin 2 - Sin 1) ---------------------------------------------2. . (2 - 1)

H

=

RV Cos 1 - RV Cos 2 = RV ( Cos 1 - Cos 2) 360 . L . (Cos 1 - Cos 2)

H

=

----------2. -----.--(---2- ------1)-------------------32

 

RH 

=

180 . H / π (2 - 1)

E

=

RH Cos 1 – RH Cos 2= RH (Cos 1 - Cos 2) 3602 . L . (Cos 1 - Cos 2). ) (Cos 1 - Cos 2)

E

=

---------------------------------------------------------------4. 2. (2 - 1). (2 - 1)

N

=

RH Sin 2 – RH Sin 1= RH (Sin 2 - Sin 1)

=

3602 . L . (Cos 1 - Cos 2). ) (Sin 2 - Sin 1) -------------------------------------------------------------4. 2. (2 - 1). (2 - 1)

N

v.

Minimum Curvature Method:

RF = (2/DL) (in radians) X Tan (DL/2) (in degrees) Where DL = dog leg angle Cos DL = Cos (2 - 1) - Sin 1 . Sin 2 [ 1 - Cos (ß 2 - ß1)] V

=

RF . (L/2) (Cos 1 + Cos 2)

H

=

RF . (L/2) (Sin 1 + Sin 2)

E

=

RF . (L/2) (Sin 1 . Sin 1 + Sin 2 . Sin 2)

N

=

RF . (L/2) (Sin 1 . Cos 1 + Sin 2 . Cos 2)

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TUTORIAL SHEET 1 Q.1: Δß = 250

TFS = 500

α1 = 80

Ø = 50 /100 ft. workout the correction

run for the deflected well. Q.2:

Calculate Calculate the stiffn stiffness ess of a steel steel drill drill collar collar OD 8 inch inch and and ID 3 inch. inch.

Q.3. Q.3.

A pendu pendulu lum m assem assembly bly is made made up of a numb number er of 8 inch inch drill drill collar collars s whose whose

weight is 147 ppf. The inclination of the hole is 200 and the tangent length is 40 ft. Calculate the side force on the bit.   Q.4.. Q.4

A correc correctio tion n run is is to be be made made using using a PDM PDM and a bent bent sub to to make make 2.5 2.50 / 100 ft.

build up rate. The present inclination and azimuth are 250 and N1600E respectively. If  the tool face is set at 600  to the right of high side, Calculate new inclination and azimuth after the correction run.

Q.5.

During a kick off maximum 50  / 10 100f 0ftt dog dog leg leg is pe perm rmit itte ted. d. The The pr pres esen entt

inclination in 80 and the hole must turned 250 to the left. The tool face is set at 50 0  to the left of high side, Calculate final inclination and length of the correction run.

Q.6.. Q.6

To corre correct ct the the path path of the the well well bore bore the the angl angle e must must drop drop from from 180 to 150 and the

direction must turn to the left from its present heading N700E. This must be done over  next 400 ft of correction run. The dog leg severity is 20 / 100ft. Calculate

tool face

setting and new direction allowing 300reactive torque due to motor.

Q.7.

At an an in inclination ion 80, the well must be turned to the right side with expected dog

leg 30. Find out tool face setting and new inclination for the right turn of the profile. Q. Q.8. 8.

At 4000 4000 ft. ft. dril drille led d de dept pth h pres presen entt incl inclin inat atio ion n an and d azim azimut uth h are are 80  and N700E

re respe specti ctivel vely. y. It was observed observed that that the well is deflec deflected ted to 100  left form presen presentt azimuth. Find out tool face setting to turn the well in next 100 ft of drilling with 1.5 0  / 100 ft build up rate.

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TUTORIAL SHEET 2

Q.1:

L = 200 ft.

α1 = 150

α2 = 200

ß1 = 3200

ß2 = 3050

Calculate the change in coordinates using Radius of Curvature Method And find out the new coordinate of the bit if the Kickoff point is at 2000 ft vertical depth.

Q.2:

L = 200 ft.

α1 = 80

α2 = 110

ß1 = 600

ß2 = 700

Calculate the change in coordinates using Radius of Curvature Method

Q.3:

L = 200 ft.

α1 = 80

α2 = 110

ß1 = 600

ß2 = 700

Calculate the change in coordinates using Balanced Tangential Method

Q.4:

L = 200 ft.

α1 = 80

α2 = 110

ß1 = 600

ß2 = 700

Calculate the change in coordinates using Minimum Curvature Method

Q.5.

KOP = 1000 ft.

NS = 2400 ft.

ES = 200 ft.

MD1 = 1200 ft.

α1 = 150

ß1 = 3200

MD2 = 1400 ft.

α2 = 190

ß2 = 3100

Using Radius of curvature method determine well path correction plan.

Q.6.

KOP = 1000 ft.

NS = 2400 ft.

ES = 200 ft.

MD1 = 1200 ft.

α1 = 150

ß1 = 3200

MD2 = 1400 ft.

α2 = 190

ß2 = 3100

Using Minimum curvature method determine well path correction plan.

Q.7: Calculate the trajectory for the well from 8000 to 8400 ft, where the kick off uis at 8000 ft and the rate of build up is 10 /100 ft, using a lead of 100 and a right hand walk rate of 10 / 100 ft. Direction to the bull’s eye is N300E. Assume that the first 200 ft is to set the lead, where the direction is held constant to 8200 ft. and then turns at a rate of  10 /100 ft.

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Q.8:: Using Q.8 Using follow following ing surve survey y data data compar compare e the three three dimens dimension ional al coordi coordinat nates es by Tangential method, Average angle method, Radius of curvature method –

Measured Depth (ft.)

3000 3300 3600 3700 5000 6000 7000 8000 9000 10000

Hole inclination in degrees

2 4 8 12 15 16 17 17 17 17

Hole Azimuth

N280E N100E N350E N250E N300E N280E N500E N200E N300E N250E

TUTORIAL 3 Q.1: It is desired to drill under the lake to the location designated for Well 2. For this well, a build and hold trajectory will be used. Horizontal departure to the target is 2655

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ft and at a TVD 9650 ft. The recommended rate of build up is 20 /100ft. The kickoff  depth is 1600 ft.. Determine(1) the radius of curvature (R1) (2) the maximum inclination angle (α1) (3) the measured depth to the end of build up. (4) the total measured depth. (5) Horizontal departure at the end of build up (6) the measure depth at TVD of 1915 ft. (7) the horizontal displacement at TVD of 1915 ft. (8) the measured depth at TVD of 7614 ft. (9) the horizontal departure at TVD of 7614 ft.

Q.2: What are the directions, in the alternative format of each of the following wells. Well A

N150E

Well B

2250

Well C

N00E

Solution: Well A is in the first quadrant and is at 150 from north Well B is in third quadrant and is at S450W Well C is in true north direction. -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------Q.3: Determine the new direction for a whipstock set at 705 meter with a tool face setting of 450 right of high side for a course length of 10 meter. The inclination is 70 and the direction is N150W OF 705 meter. The curve of whipstock will cause a total angle change of 30 /30 m over course course length.

TUTORIAL SHEET 1

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Q.1: Δß = 250

TFS = 500

α1 = 80

Ø = 50 /100 ft. workout the correction

run for the deflected well. Solution:

Tan DL . Sin TFS Tan Δß = ------------------------------------------------------------------------------------------------------------------------Sin α1 + Tan DL . Cos α1 . Cos TFS

Tan DL . Sin TFS Tan 25 = ------------------------------------------------------------------------------------------------------------------------Sin α1 + Tan DL . Cos α1 . Cos TFS

Tan DL X 0.7660 0.46631 = ------------------------------------------------------------------------------------------------------------------------0.1392 + Tan DL .X . X 0.9903 X 0.6428 0.06491 + 0.2969 Tan DL = 0.7660 Tan DL 0.4691 Tan DL = 0.06491 Tan DL = 0.13837 DL = 7.8780 DLS = 7.878 X 100 / 5 = 157.56 ft. α2 = Cos -1 [Cos α1 . Cos DL – Sin α 1 . Sin DL . Cos TFS] α2 = Cos -1 [Cos 8 . Cos 7.878 – Sin 8 . Sin 7.878 . Cos 50] α2 = Cos -1 [0.9903 X 0.9906 – 0.1392 X 0.1371 X 0.6428] α2 = 14.37 0 -----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------Q.2 Calculate the stiffness of a 8 inch x 3 inch steel drill collar. To accommodate an MWD tool has to be used of 3.25 inch id. By how how much will this reduce th the e stiffness of the collar? Solution:

S = E . I = E X (π/64) (de4 – di4) S1 = 5.913 X 109 psi S2 = 5.87 X 109 psi ΔS = 4.3 x 107 psi ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

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Q.3: A pendulum assembly is made up of a number of 8 inch drill collars whose weight is 147 ppf. The inclination of the hole is 200 and the tangent length is 40 ft. Calculate the side force on the bit. Solution:

F = L . WC . Sin α / 2 F = 40 X 147 X Sin 20 / 2 F = 1005.54 lbs. ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------Q.4: A correction run is to be made using a PDM and a bent sub which is expected to make a total change of 2.50 per 100 ft. drilled. The present inclination and azimuth are 250  and 1600  respectively. If the tool face is set at 600  to the right of high side, calculate the change in azimuth and the new inclination after drilling 100 ft. with this assembly. Solution:

Tan--------------------------------------} DL Sin TFS -------------------} Δß = Tan-1 {-----------------{------------------------------------Sin α1 + Tan α1 . Cos α1 . Cos TFS Tan 2.5 2.5 . Sin 60 Δß = Tan  {-----------------{--------------------------------------------------------------------------} -------------------} Sin 25 + Tan 25 . Cos 25 . Cos 60 -1

Δß = = 4.90 α2 = Cos -1 [Cos α1 . Cos DL – Sin α 1 . Sin DL . Cos TFS] α2 = Cos -1 [Cos 25 . Cos 2.5 – Sin 25 . Sin 2.5 . Cos 60] α2 = 26.330 ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Q.5: During a kick off using a jet deflection a maximum dog leg severity of 50 /100 ft. is permitted. The present inclination is 80 and the hole must turned 250 to the left. If the tool face is at 500  to the left left of high high si side de Calcu Calculat late e the expecte expected d length length of the correction run and final inclination after the job.

Solution: 39

 

Tan DL Sin TFS Δß = Tan  {-----------------{--------------------------------------------------------------------------} -------------------} Sin α1 + Tan α1 . Cos α1 . Cos TFS -1

Tan DL . Sin(-50) Δß = Tan  {-----------------{--------------------------------------------------------------------------} -------------------} Sin 8 + Tan 8 . Cos 8 . Cos (-50) -1

DL = = 7.90 The maximum DL is 50/100 ft. Length of correction correction run L = 7.9 x 1100 00 / 5 = 158 ft. α2 = Cos -1 [Cos α1 . Cos DL – Sin α 1 . Sin DL . Cos TFS] α2 = Cos -1 [Cos 8 . Cos 7.9 – Sin 8 . Sin 7.9 . Cos (-50)] α2 = 14.40 ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------Q.6: To correct the path of the well bore the angle must drop from 180 to 150, and the 0

direct dire ctio ion n must must run run to the the le left ft fr from om its its pr pres esen entt he head adin ing g of N70 N70 E. This must be accomplished over the next 400 ft and the dog leg severity is limited to 20 /100 ft. Calculate the required tool face setting allowing 300 reacting torque due to the motor  and final direction after correction run. Solution:

DL = 2 X 400 / 100 = 80  Cos α1. Cos DL – Cos α 2 TFS = Cos  [ ------------------------------------------------------------------------------------------ ] Sin α1. Sin DL -1

Cos 18. Cos----------------------------8 – Cos15 --------- ] TFS = Cos-1 [ ----------------------------------------------------Sin 18. Sin 8 0 TFS = 124.1  Actual Tool face setting setting = 124.1 – 30 30 = 94.10C Q.7: At an inclination of 80 the well must be turned to the right by setting a whip-stock. The expected dogleg is 30. Calculate the tool face and change in direction. Solution:

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TUTORIAL SHEET - 2 Q.1:

L = 200 ft.

α1 = 150

α2 = 200

ß1 = 3200

ß2 = 3050

Calculate the change in coordinates using Radius of Curvature Method And find out the new coordinate of the bit if the Kickoff point is at 2000 ft vertical depth. Solution:

ΔV = 180 L (Sin α2 – Sin α1) / π (α2 – α1) ΔV = 180 x 200 (Sin 20 – Sin 15) / π (20 – 15) ΔV = + 190.68 ft.   ΔH = 180 L (Cos α2 – Cos α1) / π (α2 – α1) ΔH = 180 x 200 (Cos 20 – Cos 15) / π (20 – 15) ΔH = + 60.12 ft. ΔN = (180 / π)2  L (Cos α1 – Cos α2) (Sin ß2 – Sin ß1) / [ (α2 – α1) (ß2 – ß1)] 2

ΔN = (180 / π)   200 (Co (Coss 15 – Cos 20) (Sin (Sin 305 – Sin 320) / [ ((20 20 – 15) (305 (305 – 320)] ΔN = + 40.51 ft. ΔE = (180 / π)2  L (Cos α1 – Cos α2) (Cos ß1 – Cos ß2) / [ (α2 – α1) (ß2 – ß1)] ΔE = (180 / π)2  200 (Co (Coss 15 – Cos 20) (Cos 320 – Cos 30 305) 5) / [ ((20 20 – 15) (305 – 320 320)] )] ΔE = - 44.20 ft. VB = 2000 ft. Ø = 1500 / 15 = 1 0 / 100 ft.

R = 18000 / πØ = 5729.58 ft.

V1 = VB + R Sin α1 = 2000 + 5729.58 Sin 15 = 3482.92 ft. H1 = R (1 – Cos α 1) = 195.23 ft. N1 = H1 Cos ß1 = 195.23 Cos 320 = 149.55 ft. E1 = H1 Sin ß1 = 195.23 Sin 320 = - 125.49 ft. V2 = V1 + ΔV = 3482.92 + 190.68 = 3673.6 ft. H2 = H1 + ΔH = 195.21 + 60.12 = 255.33 ft. N2 = N1 + ΔN = 2549.55 + 40.51 = 2590.06 ft. E2 = E1 + ΔE = 74.51 – 44.20 = + 30.31 ft. ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------  41

 

Q.2:

L = 200 ft.

α1 = 80

α2 = 110

ß1 = 600

ß2 = 700

Calculate the change in coordinates using Radius of Curvature Method

Solution: ΔV = 180 L (Sin α2 – Sin α1) / π (α2 – α1) ΔV = 180 x 200 (Sin 11 – Sin 8) / π (11 – 8) ΔV = + 197.23 ft.   ΔH = 180 L (Cos α1 – Cos α2) / π (α2 – α1) ΔH = 180 x 200 (Cos 8 – Cos 11) / π (11 – 8)  ΔH = 180 x 200 (8.641) / π x 3 ΔH = + 33.01 ft. ΔN = (180 / π)2  L (Cos α1 – Cos α2) (Sin ß2 – Sin ß1) / [ (α2 – α1) (ß2 – ß1)] ΔN = (180 / π)2  200 (Co (Coss 8 – Cos 11) (Sin 70 – Sin 60) 60) / [ (11 – 8) (70 – 60)] ΔN = + 13.93 ft. ΔE = (180 / π)2  L (Cos α1 – Cos α2) (Cos ß1 – Cos ß2) / [ (α2 – α1) (ß2 – ß1)] ΔE = (180 / π)2  200 (Cos 8 – Cos 11) (Cos 60 – Cos 70) / [ (11 – 8) (70 – 60)] ΔE = 29.88 ft. -----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------Q.3:

L = 200 ft.

α1 = 80

α2 = 110

ß1 = 600

ß2 = 700

Calculate the change in coordinates using Balanced Tangential Method

Solution: V V V

= = =

(L/2) (Cos 1 + Cos 2) (200 / 2) (Cos 81 + Cos 11) 197.19 ft.

H H H

= = =

(L/2) (Sin 1 + Sin 2) (200/2) (Sin 8 + Sin 11) 33.00 ft.

E E E

= = =

(L/2) (Sin 1 . Sin 1 + Sin 2 . Sin 2) (200/2) (Sin 8 . Sin 60 + Sin 11 . Sin 70) 29.98 ft.

N N N

= = =

(L/2) (Sin 1 . Cos 1 + Sin 2 . Cos 2) (200/2) (Sin 8 . Cos 60 + Sin 11 . Cos 70) 13.48 ft.

42

 

Q.4:

L = 200 ft.

α1 = 80

α2 = 110

ß1 = 600

ß2 = 700

Calculate the change in coordinates using Minimum Curvature Method

Solution: R RF F = (2/DL) (i(in radians)

.

Tan ((D DL/2) (i(in degrees)

Where DL = dog leg angle Cos DL = Cos (2 - 1) - Sin 1 . Sin 2 [ 1 - Cos (ß 2 - ß1)] Cos DL = Cos (11 - 8) - Sin 8 . Sin 11 [ 1 - Cos (70 - 50] Cos DL = Cos 3 – Sin 8 x Sin 11 [ 1 – Cos 10] Cos DL = 0.99863 – 0.026556 X [1 – 0.98481] Cos DL = 0.9982261 DL = 3.410 DL = 0.05952 RF = (2 / 0.05952) X Tan (3.41 / 2) = 1.000 1.0002234 2234 V V V

= = =

RF X (L/2) (Cos 1 + Cos 2) 1.0002234 X (200 / 2) (Cos 81 + Cos 11) 197.23 ft.

H H H

= = =

RF X (L/2) (Sin 1 + Sin 2) 1.0002234 X (200/2) (Sin 8 + Sin 11) 33.01 ft.

E E E

= = =

RF X (L/2) (Sin 1 . Sin 1 + Sin 2 . Sin 2) 1.0002234 X (200/2) (Sin 8 . Sin 60 + Sin 11 . Sin 70) 29.99 ft

= RF X (L/2) (Sin 1 . Cos 1 + Sin 2 . Cos 2) = 1.0002234 X (200/2) (Sin 8 . Cos 60 + Sin 11 . Cos 70) = 13.48 ft --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

N N N

43

 

Q.5.

KOP = 1000 ft.

NS = 2400 ft.

ES = 200 ft.

MD1 = 1200 ft.

α1 = 150

ß1 = 3200

MD2 = 1400 ft.

α2 = 190

ß2 = 3100

Using Radius of curvature method determine well path correction plan.

Solution: ΔV = 180 L (Sin α2 – Sin α1) / π (α2 – α1) ΔV = 180 x 200 (Sin 19 – Sin 15) / π (19 – 15) ΔV = + 191.22 ft.   ΔH = 180 L (Cos α1 – Cos α2) / π (α2 – α1) ΔH = 180 x 200 (Cos 15 – Cos 19) / π (19 – 15) ΔH = + 58.44 ft. ΔN = (180 / π)2  L (Cos α1 – Cos α2) (Sin ß2 – Sin ß1) / [ (α2 – α1) (ß2 – ß1)] ΔN = (180 / π)2  200 (Co (Coss 15 – Cos 19) (Sin (Sin 310 – Sin 320) / [ ((19 19 – 15) (320 (320 – 310)] ΔN = + 41.29 ft. ΔE = (180 / π)2  L (Cos α1 – Cos α2) (Cos ß1 – Cos ß2) / [ (α2 – α1) (ß2 – ß1)] ΔE = (180 / π)2  200 (Co (Coss 15 – Cos 19) (Cos 320 – Cos 31 310) 0) / [ ((19 19 – 15) (320 – 310 310)] )] ΔE = - 41.29 ft.  Azimuth between between station 1 & 2 Tan Δß(1-2) = (ΔE / ΔN) = (- 41.29 / 41.24) Δß = - 45.030 ß = 360 – 45.03 = N 314.97 0 E N2 = 2400 + 41.29 = 2441.29 ft. E2 = 200 – 41.29 = 158.71 ft. V1 = VB + R Sin α1 150 inclination achieved in at 1200 feet drilled depth after KOP that is 1000 feet depth. Ø = (100 α / 200) = 7.50 / 100 ft. R = 18000 18000 / π Ø = 76 763.944 3.944 ft. V1 = VB + R Sin α1 = 1000 + 763.944 Sin 15 V1 = 1197.73 ft. V2 = V1 + ΔV = 1197.73 + 191.22 = 1388.95 ft. H1 = R (1 – Cos α 1) = 763.944 X (1 – Cos 15) H1 = 26.031 ft. H2 = H1 + ΔH = 26.031 + 58.44 = 84.47 ft. 44

 

Q.6.

KOP = 1000 ft.

NS = 2400 ft.

ES = 200 ft.

MD1 = 1200 ft.

α1 = 150

ß1 = 3200

MD2 = 1400 ft.

α2 = 190

ß2 = 3100

Using Minimum curvature method determine well path correction plan.

Solution: RF = (2/DL) (in radians)

.

Tan (DL/2) (in degrees)

Where = dog  - angle Cos DLDL = Cos (2leg 1) - Sin 1 . Sin 2 [ 1 - Cos (ß 2 - ß1)] Cos DL = Cos (19 - 15) - Sin 15 . Sin 19 [ 1 - Cos (310 - 320] Cos DL = Cos 4 – Sin 15 x Sin 19 [ 1 – Cos (-10)] Cos DL = 0.9963 DL = 4.930 Ø = 2.460 / 100 ft. DL = 0.0859 RF = (2 / 0.00859) 0.00859) X Tan (4.93 / 22)) = 1.0003 V V V

= = =

RF X (L/2) (Cos 1 + Cos 2) 1.0003 X (200 / 2) (Cos 15 + Cos 19) 191.20 ft.

H H H

= = =

RF X (L/2) (Sin 1 + Sin 2) 1.0003 X (200/2) (Sin 15 + Sin 19) 58.46 ft.

E E E

= = =

RF X (L/2) (Sin 1 . Sin 1 + Sin 2 . Sin 2) 1.0003 X (200/2) (Sin 15 . Sin 320 + Sin 19 . Sin 310) - 41.59 ft

N N N

= = =

RF X (L/2) (Sin 1 . Cos 1 + Sin 2 . Cos 2) 1.0003 X (200/2) (Sin 15 . Cos 320 + Sin 19 . Cos 310) 40.77 ft

 Azimuth between between station 1 & 2 Tan Δß(1-2) = (ΔE / ΔN) = (- 41.29 / 41.24) Δß = - 45.030 ß = 360 – 45.03 = N 314.97 0 E N2 = 2400 + 40.77 = 2440.77 ft. E2 = 200 – 41.59 = 158.41 ft. V1 = VB + R Sin α1 150 inclination achieved in at 1200 feet drilled depth after KOP that is 1000 feet depth. Ø = (100 α / 200) = 7.50 / 100 ft. R = 18000 18000 / π Ø = 76 763.944 3.944 ft. V1 = VB + R Sin α1 = 1000 + 763.944 Sin 15 V1 = 1197.73 ft. V2 = V1 + ΔV = 1197.73 + 191.20 = 1388.93 ft. H1 = R (1 – Cos α 1) = 763.944 X (1 – Cos 15) H1 = 26.031 ft. H2 = H1 + ΔH = 26.031 + 58.46 = 84.49 ft.

45

 

Q.7: Calculate the trajectory for the well from 8000 to 8400 ft, where the kick off uis at 8000 ft and the rate of build up is 10 /100 ft, using a lead of 100 and a right hand walk rate of 10 / 100 ft. Direction to the bull’s eye is N300E. Assume that the first 200 ft is to set the lead, where the direction is held constant to 8200 ft. and then turns at a rate of  10 /100 ft. Solution:

Lead in right side walk is 100  The Bull’s Eye (TFS) is N300E So the direction changing tendency is 200 per 100 ft interval TVD of 8000 – 8100 ft interval V1 V1

V1 H1 H1 H1  N1 N1 N1 E1  1

E1

= = = = = = = = =

L Cos [(1 + 2) / 2] 100 Cos [[((0 + 10) / 2] = 99.996 ft. 8000 + 99.996 = 8099.996 ft. L Sin [(1 + 2) / 2] 100 Sin [(0 + 10) / 2] = 0.873 ft. 0.873 ft. L Sin [(1 + 2) / 2] . Cos [(1 + 2)/2] 100 Sin [(0 + 10) / 2] . Cos [(0 + 20)] first time direction direction is not averaged averaged 0.82 ft.

= =

2 1 ++20)] )/2] 1L0S0inS[in([1( + 0 +21)0/) 2] / 2]. Sin . Sin[([(0 0.30 ft.

TVD of 8100 – 8200 ft interval V2 V2

V2 H2 H2 H2  N2 N2

N2 E2 E2 E2

= = = = = =

L Cos [(1 + 2) / 2] 100 Cos [(10 + 20) / 2] = 99.966 ft. 8099.996 + 99.966 = 8199.962 ft. L Sin [(1 + 2) / 2] 100 Sin [(10 + 20) / 2] = 2.618 ft. 0.873 + 2.618 = 3.491 ft.

= = = = = =

L Sin [(1 + 2) / 2] . Cos [(1 + 2)/2] 100 Sin [(10 + 20) / 2] . Cos [(20 + 20)/2] = 2.460 ft 0.82 + 2.460 = 3.28 ft. L Sin [(1 + 2) / 2] . Sin [( 1 + 2)/2] 100 Sin [(10 + 20) / 2] . Sin [(20 + 20)/ 2] = 00.895 .895 ft 0.30 + 0.895 = 1.195 ft.

TVD of 8200 – 8300 ft interval the direction changes by 10 /100 ft. fro. N200E to N210E V3 V3

V3 H3 H3 H3  N3 N3 N3 E3 E3 E3

= = = = = = = = = = = =

L Cos [(1 + 2) / 2] 100 Cos [(20 + 30) / 2] = 99.905 ft. 8199.962 + 99.905 = 8299.867 ft. L Sin [(1 + 2) / 2] 100 Sin [(20 + 30) / 2] = 4.362 ft. 3.491 + 4.362 = 7.853 ft. L Sin [(1 + 2) / 2] . Cos [(1 + 2)/2] 100 Sin [(20 + 30) / 2] . Cos [(20 + 21)/2] = 4.08 4.0866 ft 3.28 + 4.086 = 7.366 ft. L Sin [(1 + 2) / 2] . Sin [( 1 + 2)/2] 100 Sin [(20 + 30) / 2] . Sin [(20 + 21)/ 2] = 11.528 .528 ft 1.195 + 1.528 = 2.723 ft. 46

 

TVD of 8300 – 8400 ft interval the direction changes by 10 /100 ft. fro. N210E to N220E V4 V4

= L Cos [(1 + 2) / 2] = 100 Cos [(30 + 40) / 2] = 99.8135 ft. V4 = 8299.867 + 99.8135 = 8399.6805 ft. H4 = L Sin [(1 + 2) / 2] H4 = 100 Sin [(30 + 40) / 2] = 6.105 ft. H4  = 7.853 + 6.105 = 13.958 ft. N4 = L Sin [(1 +0 2) /0 2] . Cos [(1 + 2)/2] N4 = 100 Sin [(3  + 4 ) / 2] . Cos [(21 + 22)/2] = 5.6801 ft N4 = 7.366 + 5.6801 = 13.0461 ft. = L Sin [(1 + 2) / 2] . Sin [( 1 + 2)/2] E4 E4 = 100 Sin [(30 + 40) / 2] . Sin [(21 + 22)/ 2] = 22.2374 .2374 ft E4 = 2.723 + 2.2374 = 4.9604 ft. Total departure = √ (total north2 + total east2) Total departure = √ (13.04612 + 4.96042) = 13.9573 ft Departure angle = tan-1 (total east / total t otal north) -1 Departure angle = tan  (4.9604 / 13.0461) = 20.820 Depth (X) 8000 8100

VX 8000 8099.996

HX 0 0.873

NX 0 0.82

EX 0 0.30

88230000 8400

88129999..986627 8399.6805

37..489513 13.958

37..23866 13.0461

21..712953 4.9604

 



20.1 2200..128 20.82

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------Q.8: Determine the dogleg severity following a jetting run where the inclination was changed from 4.30 to 7.10 and the direction from N890E to S800E  over a drilled interval of 85 ft. Solution: The total angle change is

Cos DL = Cos (2 - 1) - Sin 1 . Sin 2 [ 1 - Cos (ß 2 - ß1)] Cos DL == 0.99881 Cos (7.1 –– 0.07498 4.3) - Sinx0.1236 4.3 . Sin[ 17.1 [ 1 - Cos (11)] Cos DL – 0.98163] Cos DL = 0.99881 – 0.07498 x0.1236 x 0.018373 Cos DL = 0.99881 –0.00017 = 0.99864 DL = 2.990 Dog leg severity = 85 ft. Dog leg severity angle = 2.99 x 100 / 85 = 3.50/100 ft ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

47

 

TUTORIAL 3 Q.1: It is desired to drill under the lake to the location designated for Well 2. For this well, a build and hold trajectory will be used. Horizontal departure to the target is 2655 ft and at a TVD 9650 ft. The recommended rate of build up is 20 /100ft. The kickoff  depth is 1600 ft.. Determine(1) the radius of curvature (R1) (2) the maximum inclination angle (α1) (3) the measured depth to the end of build up. (4) the total measured depth. (5) Horizontal departure at the end of build up (6) the measure depth at TVD of 1915 ft. (7) the horizontal displacement at TVD of 1915 ft. (8) the measured depth at TVD of 7614 ft. (9) the horizontal departure at TVD of 7614 ft.

Solution:

(1) R = 18 18000/ 000/πΦ πΦ = 118000 8000 / π x 2 = 22864 864.79 .79 ft. (2 (2)) X = ta tann-1 [(Ht – R) / (Vt – Vb)] = tan-1 [(2655 – 2864.79) / (9650 – 1600)] = -1.49 0 Y = sin-1 [R cos x)/ (Vt – Vb)] – sin-1 [(2864.79 cos (-1.49)) / (9650 – 1600)] Y =20.83 α1 = 19.340

(3) MDC = 1600 +19.34X +19.34X 100 / 2 = 256 2567.50 7.50 ft. (4) MDt = MDc + (Vt – Vc) / cos α1  ,Vc = Vb + R Sin Sin α1  Vc = = 1600 + 2864.79 x Sin 19.34 = 2548.74 ft MDt = 2567.50 + (9650 – 2548.74) / Cos 19.34 = 10093.45 ft. (5) HC = R ( 1 – cos α 1) = 2864.69 x ( 1 – cos 19.34) = 161.66 ft. (6) MDX = MDB + 100 X αX / Φ VX – VB = 1915 -1 600 = 315 ft. = R sin αX αX = sin-1 (315 / 2864.79) = 6.31 0 MD1915 = 1600 + 100 X 6.31 / 2 = 1915.5 ft. (7) Hx = R ( 1 – cos α X) = 2864.69 x ( 1 – cos 6.31) = 17.36 ft. (8) MD7614 = MDc + (Vx – Vc) / cos α1 MD7614 = 2567.5 + (7614 – 2548.74) / cos 19.34 = 7845.96 ft (9) H7614 = Hc + (Vx – Vc) tan α1 = 161.66 + (7614 – 2548.74) tan 19.34 = 1939.46 ft. Q.2: What are the directions, in the alternative format of each of the following wells. Well A N150E 48

 

Well B

2250

Well C

N00E

Solution: Well A is in the first quadrant and is at 150 from north Well B is in third quadrant and is at S450W Well C is in true north direction. -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------Q.3: Determine the new direction for a whipstock set at 705 meter with a tool face setting of 450 right of high side for a course length of 10 meter. The inclination is 70 and the direction is N150W OF 705 meter. The curve of whipstock will cause a total angle change of 30 /30 m over course course length. Solution: The angle of inclination obtain by curvature is DL = 30 x 10 m / 30 m = 1 0 The change in azimuth is –

Tan DL . Sin TFS Tan Δß = ------------------------------------------------------------------------------------------------------------------------Sin α1 + Tan DL . Cos α1 . Cos TFS Tan (10) . Sin (450) Tan Δß = ------------------------------------------------------------------------------------------------------------------------0 0 0 Sin (7 )  + Tan (1 ) . Cos (7 )  . Cos (450) Δß = 5.30 The ß is in right of high side so ß2 = 3450 + 5.30 = N350.30E or N9.70W ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

49

 

Q. Q.4: 4: From From 1000 1000 ft KOP KOP Depth epth and and 20 /100 ft. Build up rate, following wells are progressing through curvature drilling. Draw the figures of progressing wells and determine final location (inclination, azimuth and three dimensional coordinates) after  completion of curvature drilling. Well A: progressing in E1150S direction for 100 inclination. Well B: progressing in S2250W direction for being horizontal. Well C: progressing on coordinates coordinates of N = - 234 ft an and d E = - 120 ft. Well D: progressing progressing on coordinates of S = - 130 ft and W = - 75 75 ft.

Solution: Well A: E1150S direction for 100 inclination

R = 2864.79 ft. V = 1000 + 2864.79 Sin 10 = 1497.47 ft. H = 2864.79 ( 1 – Cos 10) = 43.52 ft. W = 43 43.5 .522 Sin 25 = 18.39 8.39 ft ft.. S = 43. 3.225 Cos 25 = 16. 6.67 67 ft ft.. 0 α = 10 β = N2050E N = -16.67 ft. E = - 18.39 ft. Well B: S2250W direction for being horizontal

R = 2864.79 ft. V + 2864.79 3864.79ft.ft. H = 1000 2864.79 ( 1 – CosSin 90)90= =2864.79 N = 2864.79 Sin 45 = 2025.71 2025.71 ft. E = 2864.79 Cos 45 45 = 2025.71 ft. α = 900

β = N450E

Well C: N = - 234 ft and E = - 120 ft.

R = 2864.79 ft. H = √ (1202 + 2342) = 262.98 ft. α = 2244.740 R (1 – Cos α) = H = 262.98 V = 1000 + 2864.79 Sin 24.74 = 2198.92 ft. N = - 234 ft and E = - 120 ft.

α = 24.740 β = N207.150E

Well D: S = - 130 ft and W = - 75 ft.

R = 2864.79 ft. 2

2

H = H) == 1150.08 50.08 ft. α = 1188.630 R =(1√ –(130 Cos +α)75 V = 1000 + 2864.79 Sin 18.63 = 1915.17 ft. N = 130 ft and E = 75 ft.

α = 24.740

β = N29.980E

50