Well Control Preschool Exercises by Transocean Sedco Forex Jakarta Learning Centre

Transocean Sedco Forex

Jakarta Learning Centre

Pre-school exercises for Well Control

With Answers.

Pre-School Well Control exercises 25/01/00

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Contents.

Introduction_______________________________________________________________ 3

Section A

Well Control Equipment_____________________________________________________ 4

Section B

Pre-recorded information____________________________________________________ 14

Section C

Causes of kicks____________________________________________________________ 18

Section D

Indications of a kick________________________________________________________ 21

Section E

Shut-in Procedure_________________________________________________________ 25

Section F

Kick Data________________________________________________________________ 28

Answers__________________________________________________________________ 59

Formulae for Well Control__________________________________________________ 67


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Introduction.

Pre-School exercises for Well Control

This book of exercises is designed to help you prepare for well control school. The exerciseswere written to provide up to date questions for self-study either on the rig or at home.Answers are provided for all the questions at the back of the book.

Please bring this book with you to well control class if there is anything that you would like todiscuss.


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Section A

Well Control Equipment

A1 Blowout Preventers and Diverters.

1) Indicate the activities that may be carried out wih the BOP stack shown below.

a) With no drill pipe in the hole, shut in the well under pressure and repair the spool.b) With drill pipe in the hole, shut the well in and change pipe rams to blind rams.c) With drill pipe in the hole, circulate through the drill pipe.d) With drill pipe in the hole, shut in the well under pressure and repair the side outlets on

the spool .

2) What is the primary function of the weep hole (drain hole, vent hole) on a ram type BOP?

a) To show that ram body rubber is leaking.b) To show that the primary mud seal on the piston rod is leaking.c) To show that the Bonnet seals are leaking.d) To show that the closing chamber operating pressure is too high.

3) You only have one inside BOP with an NC 50 (41/2 IF) lower pin connection on your rigbut the drill string consist of 5 HWDP, and 8 collars. Which one of the following crossoverswould you have on the drill floor in case of kick while tripping?

a) 6-5/8 reg. Box X 7-5/8 reg. Pinb) NC50 (4-1/2 IF) Pin X 6-5/8 reg. Pinc) NC50 (4-1/2 IF) Box X 7-5/8 reg. Pind) NC50 (4-1/2 IF) Box X 6-5/8 reg. Pin

Annular

Blind

Spool

PipeChokeKill


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4) Two types of valves may be used in the drill string:

Type 1 Non return, stab in safety valve or inside BOPType 2 Fully opening stab in Kelly cock valve or fully opening safety valve

Indicate in the table which statement describes the valves.

Type 1 Type 2Requires the use of key to closeMust not run in the hole in the close positionHas to be pumped to read shut-in drill pipe pressureWill not allow wireline to be run inside the drill stringHas potential to leak through the open/close keyEasier to stab if strong flow is encountered up the string

5) A BOP stack is configured: Pipe ram / Blind-Shear ram / Pipe ram / Annular, kill andchoke lines are connected under the blind-shear rams. Is it possible to kill a well using theDriller's method if;

a) The upper pipe rams are closed?b) The blind shear rams are closed?c) The lower pipe rams are closed?

6) A BOP stack is configured: Pipe ram / pipe ram / Blind-Shear ram / Annular, kill and chokelines are connected under the blind-shear rams.

a) Can you repair the side outlets with pipe in the hole?b) Can you repair the outlets with no pipe in the hole?c) Is it possible to shut in with drill pipe in the hole andcirculate through the drill pipe?d) Can you change blind rams to pipe rams and kill the well?

7) A BOP stack is configured: Drilling spool / Pipe ram / Blind-Shear ram / Annular, kill andchoke lines are connected to the drilling spool.

a) With drill pipe in hole, can we repair the side outlets?b) With no drill pipe in the hole, can you shut in and repair the Drilling spool?c) With drill pipe in hole, can you circulate through the Drilling spool?

8) The kill line should enter a stack so that

a) The well can be circulated if the blind rams are in use.b) The well can be circulated if the pipe rams are being used.c) Both the above.

9) Which of the following statements are true concerning Ram Packing Elements?


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a) Reciprocating motion of the pipe increases the wear on seals.b) Closing pipe rams on open hole may damage the elements.c) The ram packer should normally be checked, and if worn, changed whenever the bonnet is

opened.d) All of above.

10) What do the term 6BX stamped on a flange represent?

a) serial numberb) pressure ratingc) typed) size

11) What is meant by the closing ratio for a ram type BOP?

a) Ratio between closing & opening volume.b) Ratio between closing & opening time.c) Ratio of the wellhead pressure to the pressure required to close the BOP.

12) Which option gives the advantage of using the kill line with static fluid to monitor wellhead pressure during a well kill operation?

a) Response on changes in well head pressure is quicker through the kill line.b) Effect of choke line friction is reduce to when monitoring on kill line gauge during the

kill operation.c) Effect of choke line friction is reduced to when monitoring on kill line gauge during the

kill operation.d) The kill line pressure can be kept constant while changing the pump speed, thus

eliminating the need to compensate for CLFL.

13) Study the two tables below which contain markings stamped on API flanges and ringgaskets. Each flange (1,2,3 and 4) mates with one of the ring gaskets (A,B,C or D). Write theappropriate flange number in the blanks.

Ring Gasket Marking FlangeA CI API BX154 S304-4B OES API R57 D-4C OES API RX66 S-4D CI API BX153 S316-4

Flange Marking

1. OES API 16-3/4 3M RX66 6A 89 300F PSL3 05/912. CI API 3-1/16 15M BX154 CRA 6A 89 250F PSL2 PRL2 08/923. OES API 2-9/16 20M BX153 CRA 6A 89 350F PSL4 PRL4 01/944. OES API 13-5/8 2M R57 6A 89 250F PSL1 PRL1 11/93

14) Write the pressure rating, bore diameter and temperature rating of each flange in theprevious question, in the blanks below.


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Flange#1___________psi __________ inches __________ deg.FFlange#2___________psi __________ inches __________ deg.FFlange#3___________psi __________ inches __________ deg.FFlange#4___________psi __________ inches __________ deg.F

15) Identify the one ram locking device from the list below that locks the ram in the sameposition regardless of wear.

a) Shaffer Ultralockb) Shaffer Poslockc) Hydril MPLd) Cooper(Cameron) Wedgelocke) Koomey Autolock

16) From the list below, identify the ring gaskets that are pressure energized. (Pick fouranswers)

a) Type RXb) Type BXc) Type AXd) Type R ovale) Type R octagonalf) Type CX

17) Which dimension from the list below is used to identify the Nominal Flange Size (Pickone answer).

a) Throughbore I.D.b) Flange O.D.c) Diameter of raised face.d) O.D. of ring groove.e) Bolt circle diameter.

18) What is the main function of a diverter?

a) To shut in a shallow kick.b) To direct fluid a safe distance away from the rig floor.c) To create a back pressure sufficient to stop formation fluids entering the wellbore.d) To act as a back up system if the annular preventer fails.


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19) In an area where local legislation requires that BOP equipment must be rated so thatmaximum anticipated formation pressures do not exceed 75% of BOP equipment pressureratings, what is the minimum acceptable rating for equipment to be used in drilling normallypressure formation to 16000 ft TVD?

a) 2000 psi BOP equipmentb) 3000 psi BOP equipmentc) 5000 psi BOP equipmentd) 10000 psi BOP equipmente) 15000 psi BOP equipment

20) What is normally considered the highest potential risk when diverting a shallow gasblowout through a long marine riser?

a) The marine riser may collapse.b) The marine riser may burst from the excess pressure exerted by the gas inside the riser.c) Buoyancy forces acting on the marine riser may require riser tension forces in excess of

situation where the riser is full of drilling fluid.

A2 BOP control systems

1) A BOP stack is configured Pipe Ram / Blind-Shear ram / Pipe Ram / Annular. Use the tablebelow to calculate the required accumulator volume if company policy is to provide sufficientvolume to close, open and close again all rams and the annular.

Component Volume to Open Volume to closeAnnular BOP 27 29

Ram BOP 13 15

2) The following statements relate to the drillers remote control BOP control panel located onthe rig floor. Decide if the statements are true or false.

a) If you operate a function without operating the master control valve that function will notwork.

b) The master control valve on an air operated panel allows air pressure to go to eachfunction in preparation for you operating the function.

c) The master control valve must be held depressed while BOP functions are operated.d) The master control valve must be depressed for five seconds then released before

operating a BOP function.

3) The API RP53 states that closing time should not exceed X seconds for annular BOPssmaller than 18-3/4". What is the value of X?

a) 30 sec.b) 60 sec.c) 2 min.d) 45 sec.

4) Which is the correct definition of the HPU reservoir volume according to API RP53?


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a) 2 times usable accumulator volume.b) 2 times accumulator volume.c) 5 times total accumulator volume .

5) Which two pressure readings decrease during normal operation of the pipe rams?

a) Manifold pressureb) Annular pressurec) Accumulator pressured) Precharge pressure

6) When closing the annular preventer from the remote panel, which two gauges show areduction in pressure?

a) Manifold pressureb) Annular pressurec) Accumulator pressured) Air pressuree) Bypass pressure

7) In each of the cases below, identify the most likely problem from the gauge readingsobserved on the remote control panel. The annular setting is 900 psi, the manifold setting is1,500 psi.

a) Everything is OK.b) Malfunction pressure regulating valve.c) Malfunction hydro-electric switchd) Leaking in hydraulic circuite) Precharge pressure is to low

Accumulatorpressure

Manifoldpressure

Annularpressure

Problem

(i) 2,900,increasing

1,500, steady 900 steady

(ii) 2,700increasing

1,800 steady 900 steady

(iii) 2,400increasing


(iv) 3,300increasing


8) A BOP operating unit has 8 accumulator bottles, each with a capacity of 10 gallons.Operating pressure is 3000 psi. Precharge pressure is 1000 psi. What is the total usable fluidvolume when the minimum BOP operating pressure is 1,200 psi?

9) On a 3000 psi accumulator system, what are the normal operating pressures seen on thefollowing gauges on the drillers remote control panel?


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(i) Air pressure, (ii) Accumulator pressure, (iii) Manifold pressure, (iv) Annular pressure

10) On which two gauges on the remote panel would you expect to see reduction in pressurewhen the annular preventer is being closed?

11) If the air pressure on the drillers panel reads 0 psi, which of the following statements istrue?

a) No stack function can be operated from the remote panel.b) All stack function can be operated from the remote panel.c) Choke and kill lines can still be operated from the remote panel.d) The annular preventer can still be operated from the remote panel.

12) Which of the problems below would not stop the BOP from closing?

a) Master control valve was not held down.b) Four-way valve did not shift position.c) Closing line in the BOP was blocked.d) Leak in the hydraulic line to the BOP or in the BOP closing chamber.e) Air pressure to the panel was lost.f) A bulb has blown on the remote panel.

13) When drilling, which may be the correct position of the 4-way valves on the BOPaccumulator unit?

a) openb) closec) neutrald) open or closed depending on BOP stack function

14) What is the normal precharge for the accumulator bottles on a 3000 psi accumulator unit?

a) 1000 psib) 3000 psic) 1200 psid) 200 psi

15) Name three indications that a function operated normally.


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16) A driller needs to close in a flowing well with drill pipe in a subsea BOP stack. Hepushes the Annular Close button and the pilot light changes, but all gauges and the flow-meter remain static. What is his best option?

a) Change pod and try again.b) Call and wait for the subsea engineer.c) Send assistant driller to manually operate the 4-ways valve on the Hydraulic Control

Manifold to close the annular.d) Close the lower annular preventer.

17) While drilling, an alarm goes off indicating low accumulator pressure and the flow meterindicates a rapid loss of fluid. The best course of action is:

a) Stop drilling and shut the well in.b) Stop drilling and call subsea engineer.c) Stop drilling and put all function in block one at a time until the flow stops.d) None of the above.

18) When a function is operated, which of the following is true?

a) SPM valve will operate in both pods.b) SPM valve will operate only on the active pod.c) The SPM valve will operate after the function is complete.

19) How much time is allowed for ram type preventers to close in API RP53?

20) Name two items on the stack that are supplied by fluid from the manifold regulator.

21) From which position in the hydraulic circuit is readback pressure taken?

a) Upstream of the regulator in the pod?b) The regulator itself?c) Down stream of the regulator in the pod?

22) What is the principal reason for fitting ram locking devices such as wedgelocks orposlocks to a subsea stack?

a) To give additional force when closing in, thus reducing delay times.b) To lock the ram in the closed position and maintain the shear rams locked during

disconnect.c) To lock the BOP stack to the well head and lock the lower Marine Riser Package to the

BOP stack.


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23) The subsea hydraulic BOP control system is divided into a Control System and a Pilotsystem. Which two statements are true with respect to the Pilot System?

a) The fluid in the Pilot System flows continuously while a function on the BOP takes place.b) The Pilot System dumps fluid to the sea at every operation of BOP functions.c) The Pilot System controls the position of all shuttle valves on the BOP stack directly.d) The Pilot system is a closed dead-end system.e) Pilot fluid consists of potable water, water soluble concentrate and glycol.

24) Which two statements are true with respect to shuttle valves on a subsea stack?

a) The shuttle valves automatically seal any hydraulic leaks in the selected pod.b) The shuttle valves prevent communication between the selected system and the redundant

system.c) The shuttle valves are pilot operated.d) The shuttle valves allow the retrieval of a malfunctioning pod without losing hydraulic

BOP control.

25) What is the purpose of the "Memory Function" on electric control panels?

a) Memory Function indicates a malfunction by giving permanent light on the alarm panelafter an alarm has been acknowledged and the audible alarm has stopped.

b) Memory Function reminds the driller to add anti-freeze fluid when the temperature dropsbelow a set level.

c) Memory Function indicates the previous position before Block position of three positionfunctions.

d) Memory Function reminds the driller to engage Wedge Locks before hanging off.

26) Mark the following statements true or false regarding to the use of manipulator type 4-ways valve used in subsea hydraulic BOP control systems.

a) If the valve is shifted to the center or block position, pressure will be vented from theline previously pressurized.

b) The center or block position can be used for troubleshooting hydraulic leaks.c) The pod selector valve on a subsea hydraulic BOP control system is of the manipulator

type.d) If the valve is shifted to the center or block position, pressure will be trapped in the line

previously pressurized.e) Manipulator type valves are the types typically installed inside the pod hose reels.

A4 Auxiliary Equipment

1) Mark the statements below "true" or "false" when drilling with a float valve in the string.

a) Surge pressure.is reduced.b) Reverse circulation is possible.c) Flowback through the drillstring often occurs after pumping a slug.d) Shut-in drillpipe pressure can be taken without starting the pumps.2) What is the primary function of the choke in the overall BOP system?


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a) To divert contaminant to burning pit.b) To hold back pressure while circulating up kick.c) To divert fluid to the mud tank.d) To prevent the loss of mud due to expansion of gas.e) To close the well in softly.

3) What is the reason for installing a riser fill-up valve in the marine riser of a subseaoperation?a) To relieve the diverter system on the rig when diverting a shallow gas kick.b) To prevent collapse of the marine riser in an emergency.c) To increase buoyancy on the marine riser in order to relieve the riser tensioning system on

the rig.d) To save time filling the hole when tripping out.

A5 BOP Testing

1) Identify the situations in which a BOP pressure test is required per API RP-53

a) After circulating out a gas kick.b) Prior to drilling into a known high pressure zone.c) After changing out BOP components or after maintenance.d) After setting a casing string.

2) Which tool would you use if you wanted to test the BOP stack, the casing head and uppercasing seals.

a) Plug type testerb) Cup type tester

3) While testing the BOP stack, it is noticed that hydraulic oil is leaking from the weep holeon the upper rams. Which one of the following best describes the proper action to be taken?

a) Energize plastic seal and repair BOP at next scheduled maintenance.b) A primary seal is leaking, secure the well and repair the seal.c) The rams packer is leaking due to wear. Change the worn packer.d) Do nothing. The seal requires a slight leak for lubrication purpose.

4) Why should the side outlet below a test plug be kept in the open position while testing asurface BOP stack?

a) Because of potential damage to casing/open hole.b) Because the test will create extreme hook load.c) Otherwise reverse circulation will be needed to release the plug


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Section B

Pre-recorded information.

B1 Slow Circulating Rates

1) Calculate the new pump pressure at the new pump speed for each of these situations:

Pump speed Pressure New speed New pressureA 40 200 80B 20 400 55C 30 600 40D 80 2,500 60E 70 1,800 65

2) Calculate the new pump pressure for different mud weights:

Mud weight(ppg)

Pressure New Mud weight New pressure

A 16 2,500 17.5B 10 1,700 14C 10 2,200 10.5D 9.5 1,800 9.8E 11.8 600 12.4

3) In which cases would you consider taking a new SCR?

a) Every shift.b) Mud weight changesc) Before and after a leak off testd) After each connection when drilling with top drives.e) Every 250 of open hole.f) After recharging pulsation dampeners on mud pump, discharge line.g) When returning to drilling after kick.

4) Why is the Choke Line Friction Loss (CLFL) recorded on rigs drilling with subsea BOPs?How is the CLFL measured?


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B2 Leak off test and MAASP.

1) Calculate the maximum allowable annulus surface pressure (MAASP) in each case.

MAMW MW TVDcsg MAASPA 14 10 6,000B 15.5 9 7,500

2) Calculate the hydrostatic pressure for each well.

MW or Gm MD TVD PhA 9.5 ppg 9,000 8,000B 15.5 ppg 21,000 18,000C 0.889 psi/ft 11,000 9,500

3) Change these pressures to an equivalent mud weight (ppg).

Ph TVD EMWA 3,500 7,000B 2,800 4,000C 5,250 9,750

4) Change the following Pressure Gradients to Mud Weights.

a) 0.56 psi/ftb) 0.81 psi/ft

5) Change the following Mud weights to Pressure Gradients:

a) 10.4 ppgb) 14 ppg

6) Change the following Circulating Densities to Bottom Hole Circulating Pressure:

E.D.C Depth T.V.D B.H.C.P.A 12.5 ppg 8000 ftB 10.2 ppg. 11400 ftC 9.4 ppg. 12500 ft


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7) Using the following data from Leak off test results. Calculate Maximum Allowable MudWeights:

L.O.T. Pressure Mud Wt. Shoe Depth. T.V.D.Max Mud Wt.A 1800 psi. 11.4 ppg. 9000 ftB 1560 psi. 10.6 ppg. 7400 ftC 1420 psi. 9.8 ppg. 6350 ft

8) Calculate new M.A.A.S.P. from the following Data:

Max Allow MudWt

Mud Wt. In use. Shoe Depth T.V.D.M.A.A.S.P.

A 19 ppg. 12 ppg. 8000 ftB 16.7 ppg 11.5 ppg 6800 ftC 15 ppg. 9.2 ppg. 5500 ft

9) Which three of the following conditions in the well increase the risk of exceeding theMAASP during the well kill operation?

a) Long open hole section.b) Large difference between formation breakdown pressure and mud hydrostatic pressure.c) Small influx.d) Short open hole section.e) Large influx.f) Small difference between formation breakdown pressure and mud hydrostatic pressure.

Questions 10-13 are base on the following information13 3/8 surface casing is set and cemented at 3126 ft. (TVD) The cement is drilled outtogether with 15 ft. of new hole, using a 10.2 ppg. mud. A Leak Off Pressure of 670 psi isdetermined.

10) What is the formation fracture gradient?

a) 0.619 psi/ftb) 0.837 psi/ftc) 0.745 psi/ftd) 0.530 psi/ft.


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11) What is the Maximum Allowable Annular Surface Pressure for 11.4 ppg mud used at6500 ft TVD.


12) How often should the MAASP be recalculated?

a) After every bit changeb) After a change in mud weightc) After every 500 ft. drilled

13) A gas kick is being circulated out. At the time the gas reaches the casing shoe (3126 ftTVD) the pressure at the top of the bubble is 2200 psi. If the original mud weight is 11.6 ppg,what is the casing pressure at surface.


14) The Fracture Gradient of an open hole formation at 3680 ft. is 0.618 psi/ft. The drillingmud currently in use is 9.8 ppg. Approximately how much Surface Casing Pressure can beapplied to the well before this formation breaks down?



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Section C

Causes of kicks.

C1 Normal and abnormal formation pressure.

1) What is primary well control?

a) The slow Circulating Rate Pressure used in the kill process.b) The used of Mud hydrostatic to balance fluid pressures in the formation.c) The use of Blow Out Preventers to close in a well that is flowing.d) The use of Pit Volume and Flow Rate measuring devices to recognize the kick.

2) What is meant by Abnormal High Pressure with regard to fluid pressure in the formation?

a) The excess pressure due to circulating mud at high rates.b) The excess pressure that needs to be applied to cause leak-off into a normally pressure

formation.c) High density mud used to create a large overbalance.d) Formation fluid pressure that exceeds normal water hydrostatic pressure.

3) Which factors most influence the rate at which shut in pressures stabilize after the well isshut in?

a) Gas migrationb) Friction lossesc) Permeabilityd) Type of influx

C3 Gas Cutting

1) When we are drilling through a gas zone, with the proper mud density, the mud hydrostaticpressure should be able to prevent the gas from coming into the well. However, if we still geta kick, which of the following reasons is the best explanation?

a) When a small volume of gas is circulated from the bottom of the hole,its pressuredecreases and volume increases. This may cause a sufficient reduction in hydrostatic tocause the well to flow.

b) The mud weight decreases due to the large splintered crescent-shaped cuttings that we getfrom a high pressured zone

c) The formation pressure increases suddenly as we drill into this zone since the gas inside isunder high pressure

d) The mud is leaking into the formation thereby reducing the effective hydrostatic head,causing an under balance

C4 Lost Circulation


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1) While running pipe back into the hole, it is noticed that the normal displacement of mudinto the trip tank is less than calculated. After reaching bottom and commencing circulation,the return flow meter is observed to reduce from 50% to 42%. A pit loss of 2 bbl. is noted.What is the most likely cause of these indications?

a) Partial lost circulation has occurred.b) Total lost circulation has occurred.c) A kick has been taken.d) The well has been swabbed.

2) If total losses occurred while drilling with water based mud what would you do?

a) Continue drilling blind.b) Stop drilling and fill the annulus up with water, from the top untill stabilized.c) Stop drilling, shut the well in and see what happens.

3) Lost circulation during a well control operation is usually detected by:

a) Monitoring the return flow with the flowshow.b) Monitoring the mud volume in the mud tanks.c) Monitoring the weight indicator.

4) A kick has been taken and it is known that a potential lost circulation zone exists in theopen hole. Select two correct actions which can be taken to minimize pressure in the annulusduring the kill operation.

a) Maintain extra back pressure on the choke for safety.b) Use the wait and weight method.c) Choose a lower circulating rate.d) Choose a higher circulating rate.

C5 Kicks as a Result of Surface Practices

1) Which of the following causes of well kicks is totally avoidable and is due to a lack ofalertness by the driller?

a) Lost circulation.b) Gas cut mud.c) Not keeping hole full.d) Abnormal Pressures.

2) Which two of the following cause swabbing?


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a) Pulling the pipe too fast.b) Insufficient trip margin.c) Improper circulating density.d) Going into the hole too fast.e) Failure to slug pipe prior to pulling out of hole.


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Section D

Indications of A Kick

D1 Kicks While Drilling

1) Which of the following is the First Reliable indication that you have taken a kick?

a) Increase in torque.b) Gas cut mud.c) Decrease in pump pressure.d) Increase in flow rate.

2) Why is a 20 barrel kick in a small annulus more significant than a 20 barrel kick in a largeannulus?

a) The kill weight mud cannot be calculated as easily.b) It result in higher annulus pressures, due to the height of the kick.c) The kicks are usually kick.d) The pipe usually get stuck.

3) Which one of the following is not an indication when a kick may be occurring?

a) Flow rate increase.b) Increase torque.c) Pit gain.d) Gas cut mud.

4) What should the driller do at a drilling break?

a) Circulate bottoms up.b) Flow checkc) Reduce weight on bit.d) Increase pump speed.

6) Which two practices are used to maintain primary well control as a precaution whenconnection gas is noticed?

a) Pumping a low viscosity pill around bit to assist in reduction of balled bit or stabilizers.b) Control drilling rate so that only one slug of connection gas is in the hole at any one time.c) Pulling out of the hole to change the bit.d) Raising Mud yield point.e) Minimizing the time during a connection when the pumps are off.


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7) Of all the following warning signs, which two signs would leave little room for doubt thatthe well is kicking?

a) flow line temperature increase.b) increased rotary torquec) flow rate increase.d) decrease drill string weighte) pit volume gainf) increased rate of penetration

8) Which of the following statements best describes formation porosity.

a) The ratio of the open spaces to the total volume of rock.b) The ability of fluid and gas to move within the rock.c) The presence of sufficient salt water volume to provide gas lift.d) All of the above

9) While drilling The active tank contained 200 bbls and the mud return line to the pitscontains 20 bbls. After having a kick the tank contains 240 bbls. What is the size of theinflux?.

a) 260 bblsb) 20 bblsc) 40 bblsd) 240 bbls.

10) If the cutting load in the annulus was high and the well had been shut in on kick. (AnswerYes or No to each question.)

a) Would the drill pipe pressure be higher than in a clean well? {Include a brief explanationof your answer.}

b) Would the casing pressure be higher than in a clean well?c) Would the casing pressure be lower than in a clean well?

11) Two early warning signs of kicks are an increase in flow rate and pit volume. For drillingon the floating rig these signs are difficult to detect due to the drilling vessel motion whichwill cause the fluctuation of the pit level. What is the equipment that we are using tocompensate and minimize these problems and explain roughly how it works?


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12) Which of the following is a problem when using oil base mud?

a) The oil base mud will contaminate the influx.b) In certain circumstances gas can dissolve in OBM.c) The gas will migrate in oil base mud faster than water base mud.d) It is difficult to detect the kick due to the gas dispersing in the oil base mud.

D2 Kick While Tripping

1) The driller is tripping pipe out of a 12 diameter hole. 25x92 ft. stand of 5 pipe havealready been pulled. There are 85 more stands to pull. The calculated metal displacement ofthe 9 collars is 0.08 bbls/ft. The capacity of the drill pipe is 0.01776 bbls/ft and the metaldisplacement 0.0075 bbls/ft. The trip tank volume has reduced from 27 barrels to 15 barrels.What action should be taken in this situation?

a) Flow check, if negative continue to pull out of hole.b) Shut the well in and circulate hole clean.c) Flow check, if negative displace a 100 ft. heavy slug into annulus and continue to pull out

of hole.d) Flow check, if negative run back to bottom and monitor returns.e) Pull remaining stands out of hole.

2) Prior to pulling out of the hole from 10485 ft. TVD, the pipe is full of 10.4 ppg. mud. Thepipe capacity is 0.01776 bbls/ft. A 25 bbls slug weighting 12.0 ppg is pumped into the drillpipe causing the level to drop some 216 ft. inside the drill pipe.What is the drop in bottom hole pressure due to pumping the slug into position?

a) 25 psib) 0 psi.c) 117 psid) 135 psi.

3) Which of the following possible indications suggest that mud hydrostatic pressure andformation pressure are almost equal?

a) A drilling break.b) Connection gas.c) Large, splintery cuttings.d) Trip gas.e) All of above.


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4) While pulling out of the hole it is noticed that mud required to fill the hole is less thancalculated. What action must be taken?

a) Flow check, if negative displace a 100 ft. heavy slug into annulus and continue to pull outof the hole.

b) Flow check, if negative run back to bottom circulate bottoms up and monitor returns.c) Pull remaining stands out of the hole.d) Flow check, if negative continue to pull out of the hole.e) Shut the well in and circulate the hole clean.

5) You are pulling out of hole. Two 93 ft. stands of 8 drill collars have been stood back in thederrick. The displacement is 0.0549 bbls/ft.According to your Assistant driller - 5.1 bbls should be pump into the well. It only takes 5bbls to fill the hole. (Answer Yes or No to each question.)

a) Are the calculations correct?b) Have you taken a 5 bbls influx?c) All OK, keeps going?

6) While tripping out of the hole a kick was taken and a full bore kelly cock was stabbed andclosed. A non return type safety valve was made up on top of the kelly cock prior to strippingin. (Answer Yes or No to each question.)

a) Should the kelly cock be closed?b) If the kelly cock is left in the open position, can a wire line be run inside the drill string?

7) You are planing to trip out of the hole. From the list below, circle six items that you wouldcheck before starting your trip.

a) Kelly-cock on drill floorb) Slow circulation rate recordedc) Sufficient power to drawworks.d) Choke and kill manifold lined up for drillinge) Make sure trip tank is half fullf) Trip sheets ready to record volumes displaced.g) Make up kick sheeth) Crossover sub on drill floor for kelly cock & drill collars


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Section E

Shut-in Procedure

1) From the list of practices shown below, choose the six most likely to lead to an increase inthe size of the influx.

a) Switch off the flow meter alarms.b) Regular briefing for the derrickman on his duties regarding the monitoring of pit levels.c) Drilling 20 ft further after a drilling brake, before flow checking.d) Running regular pit drills for drill crew.e) Maintaining stab in valves.f) Testing stab in valves during BOP tests.g) Excluding the drawworks from the SCR assignment.h) Keeping air pressure on choke control console at 10 psi.i) Calling toolpusher to floor prior to shutting in the well.j) Not holding down master air valve on remote BOP control panel while functioning a

preventer.

2) What is the reason for raising the kelly to bring the first tool joint above the rotary tablewhen shutting in a well?

a) Allow the free flow of mud around bit during kill operation.b) Allow access to the lower kelly cock and, if required, removal of the kelly.c) Extend closing time to give softest possible shut in.d) Allow annular to close around drillpipe because the annular is not designed to seal around

the kelly.

3) If flow through the drillpipe occurs while tripping, what should the first action be?a. Pick up and stab kelly.b. Run back into bottom.c. Close the annular preventer.d. Stab a full opening safety valve, close the valve.

4) Which list below (a, b, c or d) describes how the choke manifold will most likely be set upfor Hard Shut-in while drilling?

BOP Side Outlet HydraulicValve(HCR)

Auto Choke(Remote Adj.Choke)

Degasser Valves

A open closed closedB open open closedC closed open openD closed closed open


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5) While drilling along at a steady rate the derrickman asks to slow the mud pumps down sothat the shakers can handle the increase in cuttings coming back in the returns. Which one ofthe following would be the safest course of action.

a) Continue at the same rate allowing the excess to bypass the shakers and get caught in sandtraps which can be dumped later.

b) Pick up off bottom and check for flow, if there is not any then circulate bottoms up toreduce rate so shakers can handle cutting volume, flow check periodically duringcirculation.

c) Slow down the mud pump until the shakers can handle the volume of cuttings in thereturns as requested by derrickman.

d) Slow down the drilling rate and the pump rate until the shakers clear up then go back tothe original parameters.

6)

Choose from the following the list of valves that would normally be left in the open positionwhen lining the choke manifold up for a hard shut in procedure when drilling.

a) V1,V2,V3,V4,V5,V6,V8,R1b) V1,V2,V5,V6,V7c) V1,V2,V9,C1,V10,V11d) V1,V2,V4,V6,V8,R1

7) Which of the following would be the first action you would take if while circulating out akick the chicksans or hose connected to your drill string parted?

Manual ChokeC1

From BOP

V11V121

V2

V1

V9

V10

V5

V4

V3 V8V7

V6

Pressure Gauge OnChoke Control Panel

Remote Choke

R1

To PoorboyDegaser

CementPit

OverBoard

To StandPipe

To PoorboyDegaser


Page 27 of 769

a) Stop pump and close the full opening safety valve on the drill string. Close the choke.b) Close the shear rams. (Shear ram position above pipe rams being used).c) Drop the drill string and close blind/shear rams.

8) While circulating out the kick, No.1 mud pump fails. What is the first thing to do?

a) Immediately switch to No. 2 pump.b) Fix pump as soon as possiblec) Secure the well, isolate mud pump restart using No. 2 pump.d) Divert the well.


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Section F

Kick Data

F1 Pressure Observations

1) When a kick occurs, why is it important to get the well shut in as soon as possible? Pleaseanswer the following items True or False.

a) A larger pit gain will result in a higher SIDPP resulting in a heavier kill mud weightb) A larger pit gain will result in higher SIDPP and SICPc) A larger pit gain will result in higher SICP but SIDPP will stay the same

2) A flowing well is closed in. Which pressure gauge reading is normally used to determineformation pressure?

a) BOP manifold pressure gaugeb) Choke console drill pipe pressure gaugec) Drillers console drill pipe pressure gauged) Choke console casing pressure gauge

3) A flowing well is closed in. Which two pressure gauge readings might be used to determineformation pressure?

a) BOP manifold pressure gaugeb) Choke console drill pipe pressure gaugec) Drillers console drill pipe pressure gauged) Choke console casing pressure gauge

4) A kick is being circulated out at 30 SPM. The drill pipe pressure reads 550 psi, and casingpressure 970 psi. It is decided to slow the pumps to 20 SPM while maintaining 970 psi on thecasing gauge. How will this affect bottom hole pressure (exclude any Equivalent CirculatingDensity [ECD] effect)? Pick one answer.

a) Increaseb) Decreasec) Stay the samed) No way of knowing


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5) While killing a well, as pump speed is increased, what should happen to casing pressure inorder to keep bottom hole pressure steady?

a) Casing pressure should be held steady during SPM changeb) Casing pressure should be allowed to rise during SPM changec) Casing pressure should be allowed to fall during SPM change

6) The principle involved in Constant Bottom Hole Pressure methods of well control is tomaintain a bottom hole pressure that is :

a) Equal to the slow circulating rate pressureb) At least equal to the formation pressurec) Equal to the shut in drill pipe pressured) At least equal to the shut in casing pressure

7) At what point while correctly circulating out a gas kick is it likely that the pressure at thecasing shoe to be at its maximum?

a) At initial shut inb) When kill mud reaches the bitc) When kill mud reaches the shoed) When top of gas reaches the shoe

8) If Drill pipe Pressure is held constant while displacing the string with kill mud, what willhappen to Bottom Hole Pressure?

a) Increasesb) Remains the samec) Decreases

9) How is a choke wash-out recognized?

a) Rapid rise in casing pressure with no change in drill pipe pressureb) Increase in drill pipe pressure with no change in casing pressurec) Continually having to open choke to maintain drill pipe and casing pressured) Continually having to close choke to maintain drill pipe and casing pressure

10) The choke has to be gradually closed due to a string washout. What effect does thegradual closing of the choke have on the bottom hole pressure?

a) Decreasesb) Increasesc) Stays the same


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11) If Bottom Hole Pressure is held constant while circulating the influx out, the pressure onat the casing shoe will not increase after the influx passes, even though surface pressure on theannulus continues to rise.

a) Trueb) False

Questions 12-21 are based upon the following information :

A well is closed in having taken a 30 bbl gas kick, while drilling 8 hole at 11,000 ft. (TVD)with 5 drill pipe and 750 ft. of 6 drill collars

Annular capacities5" DP / 8 " Hole, 0.0459 bbls / ft. " Hole, 0.0292 bbls / ft

12) The mud weight is 12.3 ppg and the Shut in Drill Pipe Pressure is 350 psi. Assuming thegas Pressure Gradient to be 0.115 psi/ft, what will be the approximate Shut in Casing Pressure:


13) While preparing to circulate Kill Mud, the gas bubble begins to migrate. If no action istaken, what will happen to the pressure in the gas bubble as it rises:

a) Increaseb) Decreasec) Remain approximately the same

14) What will happen to Bottom hole Pressure?



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15) What will happen to Shut in Casing Pressure?


16) What will happen to the pressure on the Casing Seat?


17) If you decide to bleed enough mud to keep the Drill Pipe Pressure constant at 350 psi,what would the pressure in the bubble do as the gas rises?


18) What would happen to Bottom Hole Pressure?


19) What would happen to the Shut in Casing Pressure?


20) What would happen to the Pressure on the Casing Seat while the bubble is below theCasing Shoe?



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21) What would happen to the Pressure on the Casing Seat when the bubble is above theCasing Shoe?


22) A kick is being circulated from a well using the Drillers Method; Pumping pressurehaving been established as 1000 psi at 30 SPM. During the operation, pressure suddenlyincreases to 1350. You are reasonably sure that a Nozzle of the Bit is plugged. What shouldyou do?

a) Reduce pump pressure to 1000 psi by adjusting the chokeb) Shut the well in and re-establish the pumping pressurec) Hold casing pressure constant at the value recorded just before the bit pluggedd) (a) and (b) are acceptable courses of action

23) During the well kill operation, slowly but regularly you have had to reduce choke sizebecause the drill pipe and casing pressures keep dropping with constant pump strokes. What isthe likely cause of this?

a) A bit nozzle is washing outb) The choke is washing outc) You have a washed out pump swab

24) An influx is being circulated out using the Drillers Method and using 1100 psi at 30SPM. The operator increases pump speed to 35 SPM, while holding pump pressure constant.What happens to Bottom Hole Pressure?

a) Increasesb) Decreasesc) Remains approximately the same

25) Which of the following parameters can be affected by a drill string washout during a wellkill operation?

a) Bottom hole pressureb) Kick tolerancec) Formation fracture pressured) Slow circulating rate pressure


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26) You are killing a well using the Drillers Method, maintaining constant Drill pipe pressure.The drill pipe pressure begins to drift down, but the casing pressure remains unchanged. Thepump strokes remain constant. You close up your choke slightly, the drill pipe pressureremains unchanged but the casing pressure goes up. What is the probable cause for this?

a) Choke is plugging offb) Bit is plugging offc) Hole in drill piped) Choke is washing out

27) If regularly and rather slowly, you have to pinch in the choke to maintain drill pipe andchoke pressures while the pump strokes remain constant, you may have:

a) a washed out bit nozzleb) a washed out chokec) a pump failure

28) Problems that occur during a killing operation may affect the parameters you aremonitoring at the surface. These are: Drill pipe pressure, casing pressure, bottom holepressure. For each of the following problems state the immediate effect on each of the aboveparameters

For an increase draw + For a decrease draw - For no change draw =

Problem Drill Pipe PressureCasing Pressure Bottom Hole Pressure

Choke Washout

Hole in String

Nozzle blown out

Choke Plugging

Nozzle Plugging

29) How can a washout at the adjustable choke be recognized?

a) Drill pipe and casing pressures both fallingb) Drill pipe and casing pressures both risingc) Rapid rise in casing pressure with no change to drill pipe pressured) Increase in drill pipe pressure with no change to casing pressure


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F2 Shut in Pressure Interpretation

1) The reason shut in casing pressure is usually higher than the shut in drill pipe pressure is:

a) The cuttings in the annulus are lighter, therefore creating a lighter hydrostatic in theannulus.

b) The influx fluid is usually less dense than the existing mud weight.c) The casing pressure is not necessarily higher, it depends on whether it is an offshore or

land operation.d) The only difference is in the type of gauges used.

100

0

200

900800

700

600

500

400

300

TIME

2) In the diagram above, a well has been shut in and it is decided that the drilling engineer willplot the build up of drill pipe pressure against time as shown in the drawing above. WhatSIDPP would you use?

3) After shutting in on a kick, the SIDPP and SICP are observed to be stable for fifteenminutes. Both, then, start rising slowly by the same amount. Which one of the following is theprobable cause?

a) A further influx is occurringb) The influx is migrating up the well borec) The gauges are faultyd) The BOP stack is leaking

4) After a round trip at 9854 ft with 10.3 ppg mud, we kick the pump in and start to circulate.The well kicks and is closed in with 0 psi on the SIDPP and 150 psi on the SICP. There is nofloat in the drill string. What kill mud weight is required?

a) 10.3 ppgb) 11.3 ppgc) 10.7 ppgd) No way of knowing


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5) Shut in casing pressure is used to calculate

a) Kill weight mudb) Influx gradient and type when influx volume and well geometry are knownc) Maximum Allowable Annular Surface Pressured) Initial circulating pressure

F3 Kick Handling Methods

1) What should the driller do at a drilling break?

a) Circulate bottoms upb) Flow checkc) Reduce weight on bitd) Increase pump speed

2) A kicking well has been shut in. The drill pipe pressure is 0 because there is a non-returnvalve (float) in the string. To establish the SIDPP, what action should be taken?

a) Shearing the pipe and reading the SIDPP directly off the casing gaugeb) Pump at kill rate into the drill string with the well shut in. When casing pressure starts to

rise, read the pump pressure. This is the SIDPP.c) Pump very slowly into the drill pipe with the well shut in. When the pumping pressure

stabilizes, the float has opened. This pumping pressure is the SIDPP.d) Bring the pump up to the kill rate holding the casing pressure constant by opening the

choke. The pressure shown when the pump is at kill rate is the SIDPP.

3) After circulating out a kick using the drillers method (no weight up), are the SICP andSIDPP about the same?

4) A gas kick is being circulated up the well. What is the surface pit volume most likely to do?

a) Increaseb) Stay the samec) Decrease


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5) On a surface stack, what would happen if when bringing the pumps up to kill speed, thecasing pressure was allowed to fall below shut in casing pressure?

a) Formation would most probably break downb) More influx would be let into the well borec) It would have no effect on anything

6) For each of the following statements, note whether it relates to the Drillers Method or theWait and Weight Method.

a) Minimize pressures generated in the annulus due to gas expansion.b) Remove influx from well before pumping kill mudc) Pump kill mud while circulating influx up the annulusd) Maintain Drill Pipe pressure constant for 1st circulation

7) Which one of the following actions taken while stripping into the hole will help to maintainan acceptable bottom hole pressure?

a) Pumping a volume of mud into the well, equal to the drill pipe closed end displacement atregular intervals

b) Bleeding off the drill pipe steel displacement at regular intervalsc) Pumping a volume of mud into the well, equal to the drill pipe steel displacement, at

regular intervalsd) Bleeding off the drill pipe closed end displacement at regular intervals

8) Which of the following statements is true?

a) There is no difference between using the Drillers method and the Wait and Weight methodb) If the kill mud is being circulated up the annulus before the kick has reached the shoe then

Wait and Weight method will reduce the risk of breaking down the formation compared tousing the Drillers method

c) The Wait and Weight method should always be used because the pressure against the openhole will always be lower when using the Drillers method

9) Mud weight increase required to kill a kick should be based upon :

a) shut in drill pipe pressureb) shut in casing pressurec) original mud weight plus slow circulation rate pressure lossesd) shut in casing pressure minus shut in drill pipe pressure

10) How is the Initial Circulating Pressure found on a land rig or a jack-up, when the slowpump rate circulating pressure is not known but a kick has been taken?


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a) Circulate at desired strokes per minute to circulate out the kick, but hold 200 psi backpressure on drill pipe side with choke

b) Add 400 psi to casing pressure and bring pump up to kill rate while using the choke tokeep the casing pressure +400 constant

c) Bring pump strokes up to kill rate while keeping casing pressure constant by manipulatingthe choke, observed pump pressure is ICP

d) Add 1000 psi to shut in drill pipe pressure and circulate out the kick

11) Having completed the first circulation of the Drillers Method, the well is shut in. Shouldcasing pressure be:

a) Less than Shut in Drill Pipe Pressureb) Equal to Shut in Drill Pipe Pressurec) Greater than Shut in Drill Pipe Pressure

12) On the second circulation of the Drillers method, if the casing pressure was held constantuntil the kill mud reached Surface, what would happen to the bottom hole pressure?

a) Increaseb) Decreasec) Stay the same

13) Using Wait and Weight method, if the drill pipe pressure drops below the line of the graphas the kill mud goes down, what happens to the bottom hole pressure?

a) Increasesb) Decreasesc) Stays the same

14) You have taken a kick with a non-return valve (float) in the drill string. After shutting thewell in properly, it is best to :

a) Use the annulus pressure to calculate the kill weight mudb) Start raising the mud weight 1 ppg per circulation until the well is deadc) Use either the rig pump or cementing unit pump to increase pressure in 100 psi increments

until a change is seen on casing gauged) Pump slowly into the drill pipe. When the pump pressure stabilizes, the float is open. The

pumping pressure is the SIDPP used to calculate kill mud


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15) A well is being killed using the Drillers Method.Original shut-in drill pipe pressure= 500 psiOriginal shut-in casing pressure = 900 psiAfter the first circulation, the well is shut in and pressures allowed to stabilize. They thenread :Shut-in drill pipe pressure = 500 psiShut-in casing pressure = 650 psiIt is decided not to spend any more time cleaning the holeWhich one of the following actions should be taken

a) Prepare to use the Wait and Weight methodb) Bull-head the annulus until shut-in casing pressure is reduced to 500 psic) Reverse circulate until shut-in casing pressure is reduced to 500 psid) Continue with second circulation of Drillers Method (holding casing pressure constant

until mud reaches the bit)

16) If the slow pump circulating pressure was not known, and a kick has been taken with thewell closed in, how would you find the ICP?

a) Bring pump up to the desired rate, while holding the casing pressure 150 psi above theoriginal SICP

b) Bring pump up to desired rate, but hold 200 psi back pressure on the drill pipec) Bring pump up to the desired rate holding casing pressure constant by manipulating the

hydraulic choked) Circulate at desired kill rate but hold casing pressure 100 psi below MAASP

17) The correct gauge to use for calculating the kill weight mud is :

a) the gauge on the choke and kill manifoldb) the drill pipe pressure gauge on the drillers consolec) the casing gauge on the drillers consoled) the drill pipe gauge on the remote auto choke panele) the casing gauge on the remote auto choke panel

18) The following diagrams show the approximate changes in pressure at certain points in thewell during the first circulation of the Drillers Method. Match the following locations to theirrespective diagrams:

a) Surface casing pressureb) Casing shoe pressurec) Bottom hole pressured) Pump pressure


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(i)

TIME

PSI

(ii)

TIME

PSI

(iii)

TIME

PSI

(iv)

PSI

TIME

250

500

750

1000

1250

1500

(a) STATICPRESSURE

KILL MUD PUMPED

DRILLPIPEPRESSURE(PSI)

Drill pipe pressure graph of the one circulation method

(b)

(c)

(d)

DYNAMICPRESSURE

(e)

(x)


Page 40 of 769

19) From the figure above, match the following steps with correct explanation

a)b)c)d)e)x)

1) Drillstring displaced with kill mud2) ICP = SIDPP + kill rate pressure3) 0 psi (static)4) SIDPP (static)5) Drillstring volume pumped6) FCP= Kill rate pressure x kill mud

weight / original mud weight

250

500

750

1000

1250

1500

(e)STATICPRESSURE

DRILLPIPEPRESSURE(PSI)

Drill pipe pressure graph of the driller method

(c)

(d)

(a)

(b)

DYNAMICPRESSURE

(x) (y) (z)

20) From the figure above, match the following steps with correct explanation

a)b)c)d)e)x)y)z)

1) 0 psi (static)2) FCP = kill rate pressure x kill mud

weight / original mud weight3) Annular volume pumped4) Drillstring volume displaced5) ICP = SIDPP + kill rate pressure6) SIDPP (static)7) Annular volume pumped8) Drillstring volume pumped


Page 41 of 769

F4 Calculations

Exercise 1

Calculate the following

1)

Old Mud Wt.PPG

New Mud Wt.PPG

Old Pressurepsi.

New Pressurepsi.

9.68 10.0 185011.5 12.2 250011.0 12.6 300

2)

Old StrokesSPM

New StrokesSPM

Old Pressurepsi

New Pressurepsi

75 40 245030 60 40020 80 180

3)

Mud WtPPG

E.C.D.PPG

T.V.D.Ft.

Mud Hydrostaticpsi

B.H.C.P.psi

A.P.L.psi

10.0 10000 30011.2 5824 76

12.0 9800 600014.54 15.1 16000

11450 6200 3209.84 10.6 8700

4) Calculate the hydrostatic pressure for each well :

a) 9.5 PPG at 9,000 feet M.D. and 8,000 feet T.V.D.b) 15.5 PPG at 18,000 feet T.V.D. and 21,000 feet M.D.c) 0.889 PSI/FT at 11,000 feet M.D. and 9,500 feet T.V.D.

5) Change these pressures to an equivalent mud weight in ppg.

a) 3,500 psi at 7,000 feet.b) 4,000 feet with 2,800 psi.c) 12,000 feet M.D. / 10,500 T.V.D. with 9,000 psi.6) Calculate equivalent mud weights for these wells :


Page 42 of 769

a) BHP = 9,800 psi, Depth = 9,800 feetb) BHP = 4,580 psi, Depth = 10,000 ft TVD, 11,500 feet M.D.

7) Calculate kill mud weight for the following wells

a) SIDPP = 600 PSI, Depth = 10,000 ft, Mud wt. = 10 PPG, SICP = 900 PSIb) Mud wt = 9.5 PPG, SICP = 2,000 PSI, SIDPP = 1,200 PSI, Depth = 9,500 PSIc) SICP = 600 PSI, Depth = 15,000 feet T.V.D., 17,000 M.D. SIDPP = 300 PSI, Mud Weight

= 17 PPG

8) Calculate the new pump pressure at the new pump speed for each of these situations

Old strokes Old Pressure New Strokes New Pressurea) 30 spm 600 psi 40 spm psib) 80 spm 2500 psi 60 spm psic) 70 spm 1800 psi 20 spm psid) 40 spm 200 psi 80 spm psie) 20 spm 400 psi 55 spm psi

9) Calculate the new pump pressure for different mud weights :

a) 16 ppg mud at 40 spm = 2,500 psi. What is the pressure with 17.5 ppg mud at 40 spm?b) Present pump pressure = 1,700 psi at 50 spm with 10 ppg. What is the new pump pressure

at 50 spm with 14 ppg?

10) Calculate the maximum allowable annulus surface pressure (MAASP) for these wells:

a) Maximum allowable mud weight = 14 ppg, Mud weight = 10 ppg, Depth of casing =7,500 feet TVD

b) Maximum allowable mud weight = 15.5 ppg, Mud weight = 9.0 ppg, Depth of casing =7,500 feet TVD

11) Calculate the influx height for each situation :

a) 25 bbl gain, 900 feet of 6 " drill collars in an 8 " hole (Annular capacity =0.0292 bbl/ft)

b) 40 bbl gain, 500 feet of 6 " drill collars in an 8 " hole (0.0292 bbl/ft annularcapacity), 5" drill pipe precedes (.0459 bbl/ft annular capacity)


Page 43 of 769

12) Calculate the bottom hole circulating pressure of each well:

APL hydrostatic pressure Mud wt BHCPa) 400 psi 6500 psi 10 ftb) 250 psi 5000 psi 10.5 psi

13) Calculate the equivalent circulating density of each well :

APL hydrostatic pressureDepth TVD ECDa) 350 psi 5050 psi 9800 ftb) 40 psi 6800 psi 12000 psi

14) Calculate the annular pressure loss for each well:

BHCP Mud wt Depth TVD APLa) 6000 psi 11.6 psi 9450 ftb) 2600 psi 9.8 psi 5000 psi

15) Calculate the new pump pressure with the following new mud weights :

Old Mud wt Old pressure New Mud Wt New pressure

10 ppg 220 psi 10.5 ppg9.5 ppg 1800 psi 9.8 ppg11.8 ppg 600 psi 12.4 ppg

16) Change the following pressure gradients to mud weights :

Gradient Mud weighta) .56 psi/ftb) .81 psi/ft

17) Change the following mud weights to pressure gradients :

Mud weight Pressure Gradienta) 10.4 ppgb) ppg


Page 44 of 769

18) Change the following circulating densities to bottom hole circulating pressures :

ECD Depth TVD BHCPa) 12.5 ppg 8000 ftb) 10.2 ppg 11400 ftc) 9.4 ppg 12500 ft 19) Using the following data from the Leak Off test results, calculate maximum allowablemud weights :

LOT pressure Mud weight Shoe depth TVD Max mud weighta) 1800 psi 11.4 ppg 9000 ftb) 1560 psi 10.6 ppg 7400 ftc) 1420 psi 9.8 ppg 6350 ft

20) Using the same well data, calculate the height of each influx :

Drill Collar Length = 700 ftDC - OH Capacity = 0292 bbl/ftDP - OH Capacity = .0459 bbl/ftInflux volume = Height ft.

10 bbls = ft20 bbls = ft30 bbls = ft

21) Using the following well data, calculate Influx gradients :

SICP SIDPP MUD weight Ht of Influx Gradienta) 800 psi 720 psi 11.5 ppg 400 ftb) 950 psi 600 psi 10.6 ppg 840 ftc) 680 psi 550 psi 10.2 ppg 350 ft

22) Calculate new MAASP from the following data :

Max. allowancemud weight

Mud weightin use

Shoe depthTVD

MAASP

a) 19 ppg 12 ppg 8000 ftb) 16.7 ppg 11.5 ppg 6800 ftc) 15 ppg 9.2 ppg 5500 ft


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23) Using the following data, calculate kill mud weights :

SICP SIDPP Mud weight DepthTVD

DepthMD

Kill Mud

a) 600 psi 450 psi 10 ppg 9500 ft 10000 ftb) 850 psi 690 psi 11 ppg 12000 ft 12300 ftc) 780 psi 570 psi 10.5 ppg 11200 ft 11800 ftd) 700 psi 300 psi 14 ppg 13000 ft 13400 ft

24) A well is shut in with 500 psi on the casing pressure. The annulus contains 8,000 ft of10.0 ppg mud above 1,000 ft of 9.0 ppg saltwater. What is the equivalent mud weight at 8,000ft? At 9,000 ft?

a) 8,000 ft ; 11.2 ppgb) 8,000 ft ; 11.7 ppgc) 9,000 ft ; 11.0 ppgd) 9,000 ft ; 11.9 ppg

25) Prior to pulling out of the hole from 10,485 ft. TVD, the pipe is full of 10.4 ppg mud. Thepipe capacity is .01776 bbls/ft. A 25 bbls slug weighing 12.0 ppg is pumped into the drill pipecausing the level to drop some 216 ft. inside the drill pipe.What is the drop in bottom hole pressure due to pumping the slug into position?


26) Drill pipe capacity = 0.0178 bbls/ftDrill pipe metal displacement= 0.0082 bbls/ftAverage stand length = 93 ftCalculate :

a) Mud required to fill the hole per stand when pulled dry (bbls per stand)b) Mud required to fill the hole per stand when pulled wet (bbls per stand)

27) You are determining your kill rate pressure and bringing your pump rate up to a pre-determined 30 SPM by holding the shut in casing pressure constant. You have got a kick inthe well of 220 psi shut in drill pipe pressure. At 30 SPM your drill pipe circulating pressure is1060 psi. Calculate the slow circulating rate pressure loss.



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28) Calculate the equivalent circulating density in the following circumstances:

Circulating pressure = 3100 psi

Pressure lossesSurface equipment = 20 psiDrill string = 930 psiNozzles = 1800 psiAnnulus = 350 psi

Drilled depth 12,300 ft. (11,500 ft. TVD)Mud weight 11.4 ppgECD is

a) 10.8 ppgb) 12.0 ppgc) 11.4 ppgd) 12.3 ppg

29) A well is drilled to 13,000ft with 13.2 ppg mud. The formation pressure at that depth is8,700 psi. The intermediate casing is 43.5 lb/ft, 9 5/8 in. pipe set to 11,000 ft. The drill pipe is4 in.(ID 3.826) and the collars displacement is 0.04bbl/ft. The operator requires hole fillingafter 5 stands of drill pipe or collars are pulled. Will the well kick when pulling the drill pipedry? Drill collars dry? (Assume no swabbing effects with an average length of 90ft per stand)

a) Pulling pipe ; no kickPulling collars ; kick

b) Pulling pipe ; kickPulling collars ; no kick

c) Kicks for both drill pipes and drill collarsd) Does not kick for either drill pipe or drill collar


Page 47 of 769

F4 Calculations

Exercise 2

Questions 1-4 are based on the following information

13 3/8 surface casing is set and cemented at 4250 ft. (TVD). The cement is drilled outtogether with 15 ft. of new hole, using a 11 ppg mud. A leak off test pressure of 800 psi isdetermined. (Hole TVD 7000ft)

1) What is the formation fracture gradient?

a) 0.188 psi / ftb) 0.686 psi / ftc) 0.760 psi / ftd) 0.384 psi / ft

2) What is the maximum allowable annular surface pressure for 12.3 ppg mud in use at 7350ft. TVD :

a) 373 psia) 511 psib) 884 psic) 1982 psi

3) How often should the MAASP be recalculated?

a) After every bit changeb) After a change in mud weightc) After every 500 ft. drilled

4) A gas kick is being circulated out. At the time the gas reaches the casing shoe (4250 ft.TVD), the pressure at the top of the bubble is 3000 psi. If the original mud weight is 12 ppg,what is the casing pressure at the surface? (Hole TVD 7000ft)



Page 48 of 769

Questions 5-9 are based upon the following information

A deviated borehole has a measured depth of 12,320 ft. (TVD 10429 ft). 9 5/8, 47 lb/ft.casing in set at a measured depth of 9750 ft. (9200 ft. TVD). 11.4 ppg mud is in use when thewell kicks and is closed in.

Shut in Drill Pipe Pressure is 750 psiShut in Casing Pressure is 1050 psi and kick volume is 15 bbls.Pre- recorded information is as follows :

Fracture mud weight = 14.4 ppgCapacity of 19.5 lbs. Drill pipe= 0.01776 bbl/ft.Capacity of 9 5/8 J55 casing= 0.0732 bbl/ft.Slow Circulating Rate Pressure= 850 psi

5) The maximum allowable annular surface pressure is rounded off to :


6) The kill mud weight required to balance the formation pressure is:


7) The kill mud weight with a Safety Margin of 100 psi is:


8) The initial circulating pressure is:

a) 1400 psib) 1600 psic) 1900 psi


Page 49 of 769

9) The final circulating pressure (using kill mud weight with a 100 psi Safety Margin is) :a) 850 psib) 970 psic) 920 psid) 1050 psi

10) The Fracture Gradient of an open hole formation at 3680 ft. is 0.618 psi/ft. The drillingmud currently in use is 9.8 ppg. Approximately how much Surface Casing Pressure can beapplied to the well before this formation breaks down?


11) In the area where local legislation requires that BOP equipment must be rated so thatmaximum anticipated formation pressures do not exceed 75% of BOP equipment pressureratings, what is the Minimum Acceptable rating for equipment to be used in drilling NormallyPressured Formation to 16,000 ft. TVD?

a) 2,000 psi BOP Equipmentb) 3,000 psi BOP Equipmentc) 5,000 psi BOP Equipmentd) 10,000 psi BOP Equipmente) 15,000 psi BOP Equipment

Drilling 12 " hole at 7653 ft. TVD with 11.7 ppg mud, the well kicks and is closed in. Theshut in DPP is 430 psi. SICP is 600 psi. Pit gain 28 bbl.

Pre-Recorded Information

13 3/8" Casing Shoe at 3975 ft. TVDA Leak Off Test on the formation at shoe, using 11.2 ppg mud, showed a surface pressure of910 psi when Leak Off occurredDrill Collars 8 " OD x 2 " ID 546 ft. long. Capacity 0.0061 bbls / ftDrill Pipe 5" OD x 19.5 lbs / ft. Capacity 0.01776 bbls / ftOpen Hole - Drill Collar Annulus Capacity is 0.0796 bbls / ftOpen Hole - Drill Pipe Annulus Capacity is 0.1215 bbls / ft13 3/8" Casing - drill Pipe Annulus capacity is 0.1239Slow Pump Rate 30 SPM - Output 5 gal / stk at 730 psi

* For Subsea Candidates use* Subsea Information

Water Depth - 700 ftAir Gap - 75 ft


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1. Choke Line Capacity 0.0087 bbls / ft2. Riser Capacity 0.3892 bbls / ft3. Drill Pipe Displacement 0.0076 bbls / ft4. CLFL ( Choke Line Friction) 75 psi

Complete an IWCF kick sheet as required to assist in answering questions 12-21

12) What is the Fracture Gradient at the 3 3/8" casing shoe?

a) 0.792 psi / ftb) 0.811 psi / ftc) 0.834 psi / ftd) 0.861 psi / ft

13) What is the MAASP with 11.7 ppg mud in the hole?


14) The Kill Mud required is:

a) 11.7 ppgc) 12.8 ppgd) 13.3 ppgb) 13.9 ppg

15) The Surface to Bit strokes are :

a) 987b) 1016c) 1088d) 1164

16)

The Bit to Shoe strokes are Subsea:Strokes to displace riser to Kill Mud

a) 3562 3370b) 3257 2370c) 3752 2300d) 3604 2730


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17)

The Bit to Surface Strokes are : Subsea :shoe to BOP Strokes

a) 7100 4744b) 7300 3388c) 7500 3332d) 7700 3563

18) The Initial Circulating Pressure is :


19) The Final Circulating Pressure is :


20) The gradient of influx is:

a) 0.087 psi / ftb) 0.212 psi / ftc) 0.1225 psi / ftd) 0.327 psi /ft

21) The estimated time required to kill the well at 30 SPM is:

a) 3 hours approximatelyb) 5 hours approximatelyc) 7 hours approximatelyd) 11 hours approximately

22) You are tripping out of the hole :

Hole Depth 12,000 ftMud Weight 10 ppgDrill Pipe capacity 0.01776 bbls / ftSlug 12 ppg densityDisplaced 20 bbls


Page 52 of 769

The fluid level in the drill pipe has dropped 223 ft. How much bottom hole pressure has beenlost?


23) Given the following data, complete an IWCF kick sheet as far as required to completequestions 23(a) to 23(i)

Drill String Data Capacities5" Drill Pipe 0.01776 bbls /ft8 Drill Collars - 630 ft 0.0077 bbls / ft5" Heavy weight DP 450 ft 0.0087 bbls / ft

Hole DataCasing depth 4000 ft. 13 3/8" casingHole Depth 7591 ft. 12 "Mud weight 9.5 ppg

Annular Capacities8 DC 12 hole 0.0796 bbls / ft5" DP 12 " hole 0.1215 bbls / ft5" DP 13 3/8" casing 0.1238 bbls / ft

LOT 1070 psi test mud 9 ppg

Pump Data6" liners 97% efficiency = .102 bbls / stk

Subsea Information :Water Depth 300 ftAir Gap 90 ftChoke Line Capacity 0.0087 bbls / ftRiser Capacity 0.3892 bbls / ftDrill Pipe Steel Displacement 0.0076 bbls / ft

Questions :

a) What is the maximum mud weight?b) Total Drill string Capacityc)Total Annulus Capacityd) Bit to shoe volumee)Total system volumef) Surface to bit strokesg) Bit to shoe strokesh)Bit to surface strokes


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i) Total circulation strokes

24) You have taken a kick and shut the well in. The active tank while drilling contained 250bbls. And the mud return line to the pits contains 25 bbls.

The tank now contains 300 bbls. How many barrels of mud has been displaced from thewell?

a) 0 bblsb) 25 bblsc) 50 bblsd) 275 bbls

25) You are pulling out of hole. Two x 93 ft stands of 8" drill collars have been stood back inthe derrick. The displacement is 0.0538 bbls. / ft. According to your Assistant Driller, 10 bblsshould be pumped into the well. It only takes 10 bbls to fill the hole. (Answer yes or no toeach question)

a) Are the calculations correct?b) Have you taken a 5 bbls influx?c) All Ok, keep going

26) Use an IWCF kill sheet to assist you in answering questions 26(a) through 26(i)

Well DataHole size 8 inHole depth 11937 ft. TVD / MDCasing 9 5/8 in. casing set at 9474 ft.Drill pipe 5 in. capacity = 0.0178 bbls / ftHeavy weight pipe 5 in., 497 ft. long

Capacity = 0.0088 bbls / ftDrill Collars 6 in., 892 ft. long

Capacity = 0.006 bbls / ftMud density 14.3 ppgVolume open hole / collars 0.0322 bbls / ftVolume open hole / drill pipe / HWDP 0.0459 bbls / ftVolume casing / drill pipe 0.0493 bls / ftFracture mud wt. At the casing shoe 16.9 ppgMud pumps Output = 0.117 bbls / stkSlow circulating rate 820 psi at 30 spm / (Riser)

1180@30 spm ( Choke Line)The well has been shut in after a kick:Kick Data :

Shut-in Drill Pipe Pressure 580 psiShut-in Casing Pressure 755 psiPit Gain 12 bbls


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The well will be killed using the Drillers Method at 30 spm

* For Subsea use

Water Depth.........................................................................2120 ftAir Gap.................................................................................80 ftSteel Displacement...............................................................0.0076 bbls / ftRiser Capacity 19.75" ID......................................................0.3789 bbls / ftChoke Line Capacity 2 7/8" ID............................................0.008 bbls / ft

26a) Strokes to pump down inside drillstring from surface to bit

26b i). Strokes to pump from bit to shoe ii). Strokes to displace riser (subsea)

26c) Strokes to pump from bit to surface

26d) Kill mud weight (no safety factor)

26e i). Initial Circulating Pressure ii). Initial Dynamic Casing Pressure

26f) Final Circulating Pressure

26g) MAASP with current mud weight

26h) MAASP after circulation of kill mud

26i i). Time for complete circulation ii). Subsea excluding riser volume

15.0'

Inlet

Shakers

6.5'

27) From the diagram, calculate the pressure (psi) required to unload the MGS(M.W. = 12.5 ppg.)


Page 55 of 769

IWCF deviated well control exercise #1 (surface BOP stacks)

Prepare a deviated well control kick sheet (use attaches kick sheets) from the following welldata and answer the accompanies questions :

Well Data :Hole Size 8 "Hole Depth 14370 ft., MD (5250 ft., TVD)Kick-off Point 1640 ft., MD (1640 ft., TVD)End of Build 4265 ft., MD (3494 ft., TVD)Casing Shoe 9 5/8" x 47 lbs/ft @ 10600 ft., MD (4593 ft., TVD)

Capacities :Drill Pipe 5 OD x 19.5 # 0.01776 bbls/ftHeviwate 480 ft 5 OD x 3 ID 0.00874 bbls/ftDrill Collar 660 ft 6 OD x 2 13/16 ID 0.0077 bbls/ft

Drill Collar/Open Hole 0.0291 bbls/ftDrill Pipe / HWDP / Open Hole 0.0458 bbls/ftDrill Pipe / HWDP / Casing 0.0489 bbls/ft

Pump Data :Displacement 0.12 bbls/strokeActive Surface Volume 470 bblsSlow Circulating Rate 870 psi @ 30 spm

Formation Strength Data 1027 psi LOT using 10.5 ppg mud weight

Kick Data :Mud Weight 10.85 ppgKick depth 14370 ft., MD (5250 ft., TVD)Pit Gain 19 bblsSIDPP 725 psiSICP 785 psi


Page 56 of 769

Questions ( Surface BOP stack)

1. What is the pressure safety margin at the casing shoe with the well shut in?

2. How many strokes to pump from surface to bit?

3. What is the initial circulating pressure?

4. What is the final circulating pressure?

5. How many strokes to pump from surface to kick-off point depth?

6. What is the circulating pressure at the kick off point?

7. How many strokes to pump from surface to the end of build depth?

8. What is the circulating pressure at the end of build depth?

9. Calculate the pressure drop per 100 strokes of kill fluid pumped inside the string from theend of build depth to the bit.

10. Calculate the MAASP (Maximum Allowable Annular Surface Pressure) after circulation ofKill Mud.

11. If we neglected the directional nature of the well and decided to use a conventional VerticalWell Kill Sheet to remove the influx, calculate the pressure over balance at the end of builddepth.

12. What will be the consequences of this overbalance in the well bore?


Page 57 of 769

IWCF deviated well control exercise ( Subsea BOP stack)

Prepare a deviated well control kick sheet (use attached kick sheets) from the following welldata and answer the accompanied questions

Well Data :Hole Size 8 "Hole Depth 14370 ft., MD (5250 ft., TVD)Kick-off Point 1640 ft., MD (1640 ft., TVD)End of Build 4265 ft., MD (3494 ft., TVD)Casing Shoe 9 5/8" x 47 lbs/ft @ 10600 ft., MD (4593 ft., TVD)

Capacities :Drill Pipe 5 OD (NC50, S-135) 0.01776 bbls/ftDrill Pipe 5 Closed end displacement0.0254 bbls/ftHeviwate 480 ft 5 OD x 3 ID 0.00874 bbls/ftDrill Collar 660 ft 6 OD x 2 13/16 ID 0.0077 bbls/ftChoke Line 520 ft x 3 ID 0.0085 bbls/ftMarine Riser 505 ft 0.3892 bbl/ft

Drill Collar/Open Hole 0.0291 bbls/ftDrill Pipe / HWDP / Open Hole 0.0458 bbls/ftDrill Pipe / HWDP / Casing 0.0489 bbls/ft

Pump Data :Displacement 0.12 bbls/strokeActive Surface Volume 470 bblsSlow Circulating Rate Riser Circuit 870 psi @ 30 spmSlow Circulating Rate Choke Circuit960 psi @30 spm

Formation Strength Data 1027 psi LOT using 10.5 ppg mud weight

Kick Data :Mud Weight 10.85 ppgKick depth 14370 ft., MD (5250 ft., TVD)Pit Gain 19 bblsSIDPP 725 psiSICP 785 psi


Page 58 of 769

Questions ( Subsea BOP Stacks)

1. What is the pressure safety margin at the casing shoe with the well shut in?

2. How many strokes to pump from surface to bit?

3. What is the initial circulating pressure?

4. What is the final circulating pressure?

5. How many strokes to pump from surface to kick-off point depth?

6. What is the circulating pressure at the kick off point?

7. How many strokes to pump from surface to the end of build depth?

8. What is the circulating pressure at the end of build depth?

9. Calculate the pressure drop per 100 strokes of kill fluid pumped inside the string from theend of build depth to the bit.

10. Calculate the MAASP (Maximum Allowable Annular Surface Pressure) after circulationof Kill Mud.

11. If we neglected the directional nature of the well and decided to use a conventionalVertical Well Kill Sheet to remove the influx, calculate the pressure over balance at the endof build depth.

12. How many strokes to pump to displace Marine Riser to kill fluid before opening the BOP?


Page 59 of 769

Answers.

Section A

A1 Blowout Preventers and Diverters.

1 no, no, yes, yes2 b3 d4 A = 2, B = 2, C = 1, D = 1, E = 2, F = 25 Yes, yes, no, no6 Yes, no, yes, yes7 No, no, yes8 c9 d10 c11 c12 d13 2, 4, 1, 314 3000, 16 , 300; 15000, 3 1/16, 250; 20000, 2 9/16, 350; 2000, 13 5/8,

25015 b16 a, b c, f17 a18 b19 D20 A

A2 BOP control systems.

1 2142 T, T, T, F3 d4 a5 a, c6 b, c

7 (i) a7 (ii) b7 (iii) d7 (iv) c

8 409 120, 3000, 1500, 750-1500 psi10 2, 411 a12 113 d14 a


Page 60 of 769

15 a. light changes color, b. read back pressure drops, c. read backpressure recovers, d. flow meter stops at a proper reading.

16 a17 c18 a19 The BOP control system should be capable of closing each ram BOP in

45 seconds or less. API RP53 3rd edition March 1997 (13.3.5, p.31)20 pipe rams, connectors, failsafe valves, shear rams21 (b) The connection is to the regulator, the measurement is taken on the

downstream (regulated) side.22 b23 d & e24 b, d25 c26 a=T, b=T, c=T, d=F, e=F

A4 Auxiliary Equipment.

1 a=F, b=F, c=T, d=F2 b3 b

A5 BOP testing.

1 b, c, d2 b3 b4 a

Section B

B1 Slow Circulating Rates.

1 a. 800 psi, b. 3025 psi, c. 1067 psi, d. 1407 psi, e. 147 psi2 a. 2734 psi, b. 2380 psi, c. 2310 psi, d. 1857 psi, e. 631 psi3 a, b, g4 For deep water drilling we have significant pressure drop on the choke

line due to its length. This will affect the calculation of the initialcirculating pressure in a wll control situation.Three methods for measuring CLFL1) 1.1 Pump at SCR, taking return up the riser, read (SCRP)

1.2 Close BOP, open choke line failsafe valves 1.3 Circulate at SCR, Taking return through a wide openchoke.Record Drill Pipe pressure. 1.4 The difference between the two is CLFL2) 2.1 Pump down the choke line at SCR, taking returns up the riser 2.2 The pumping pressure record at SCR is CLFL.3) 3.1 Pump down the kill line and up the choke


Page 61 of 769

3.2 The pump pressure is twice the CLFL

B2 Leak Off Test & MAASP

1 a. 1248 psi, b. 2535 psi2 a. 3952 psi, b. 14508 psi, c. 8446 psi3 a. 9.62 ppg., b. 13.46 ppg. C. 16.48 ppg.4 a. 10.77 ppg., b. 15.58 ppg.5 a. 0.5408 psi/ft, b. 0.728 psi/ft6 a. 5200 psi, b. 6046 psi, c. 6110 psi.7 a. 15.25 ppg., b. 14.65 ppg, c. 14.1 ppg8 a. 2919psi, b. 1838.7 psi, c. 1658.8 psi.9 a, e, f10 c11 b12 b13 a14 d

Section C

C1 Normal and Abnormal Formation Pressure

1 b2 d3 c

C3 Gas Cutting

1 a

C4 Lost Circulation

1 a2 b3 b4 b-c

C5 Kicks As Result of Surface Practices

1 c2 a, b

Section D

D1 Kick While Drilling

1 d


Page 62 of 769

2 b3 b4 b5 b, e6 c, e7 a8 b9 a. = No, b. No, c. = Yes10 PVT (A pit volume totalizing system). It will report overall pit gain or loss

by using multiple pit monitors and resolving individual losses and gainsreported by each monitor into a single value.

11 b

D2 Kick While Tripping

1 d2 b3 e4 b5 a. = No (10.2), b. =Yes, c.= No.6 a.= No, b. = No7 a, b, c, e, f, h

Section E

E1 Shut-In Procedure

1 a, c, g, h, i, j2 b3 d4 d5 b6 a7 a8 a

Section F

F1 Pressure Observation

1 F, F, T2 b3 b, d4 c5 a6 b7 d8 a


Page 63 of 769

9 d10 b11 a12 d13 c14 a15 a16 a17 b18 c19 a20 a21 c22 b23 b24 b25 a, d26 c27 b28 - - -

- = =- = =+ + ++ = =

29 a30 c, e

F3 Shut In Pressure Interpretation

1 b2 700-7103 b4 a5 b

F3 Kick Handling Methods

1 b2 b3 Yes4 a5 b6 a. W&W

b. Drillerc. W&Wd. Driller

7 d8 b


Page 64 of 769

9 a10 c11 b12 a13 b14 d15 a16 c17 d18 d., b., a., c.19 a. 4)

b. 3)c. 2)d. 6)e. 1)x. 5)

20 a. 6)b. 1)c. 5)d. 2)e. 4)x. 3)y. 8)z. 7)

F4: Calculation Exercise 1

1)Old Mud Wt.

ppg.New Mud Wt.

ppg.Old Pressure

ppg.New Pressure

ppg.9.68 10 1850 191211.5 12.2 2500 265211 12.6 300 344

2)

Old Strokesspm.

New Strokesspm.

Old Pressurepsi

New Pressurepsi

75 40 2450 69730 60 400 160020 80 180 2880

3)

Mud Wtppg.

E.C.D.ppg.

T.V.D.ft

Hydrostaticpsi.

B.H.C.P.psi.

A.P.L.psi.

10 10.58 10000 5200 5500 30011.2 11.35 10000 5824 5900 76


Page 65 of 769

11.77 12 9800 6000 6115 11514.54 15.1 16000 12097 12563 46610.41 10.95 11450 6200 6520 3209.84 10.6 8700 4452 4795 343

4 a. 3952 psi., b. 14508 psi., c 8446 psi.5 a. 9.62 ppg., b. 13.46 ppg., c. 16.48 ppg.6 a. 19.23 ppg., b. 8.81 ppg.7 a. 11.16 ppg., b. 11.93 ppg., c. 17.39 ppg.8 a. 1067 psi., b. 1407 psi., c. 147 psi., d. 800 psi.,e. 3025 psi9 a. 2735 psi., b. 2308 psi10 a. 1560 psi., 2535 psi.11 a. 856.16 ft., b. 1053.38 ft.12 a. 6900 psi., b. 5250 psi.13 a. 10.6 ppg., b. 10.96 ppg.14 a. 300 psi., b. 52 psi.15 a. 231 psi., b. 1857 psi., c. 631 psi16 a. 10.77 ppg., b. 15.58 ppg.17 a. 0.5408 psi/ft., b. 0.728 psi/ft.18 a. 5200 psi., b. 6047 psi., c. 6110 psi.19 a. 15.24 ppg., b. 14.65 ppg., c. 14.10 ppg.20 a. 342.47 ft., b. 684.94 ft., c. 908.28 ft.21 a. 0.398 psi/ft., b. 0.1345 psi/ft., c. 0.159 psi/ft.22 a. 2912 psi., b. 1838 psi., c.1658 psi.23 a. 10.92 ppg., b. 12.11 ppg., c. 11.48 ppg., d. 14.45 ppg.24 a., c.25 b.26 a. 0.72-0.80, b. 2.3-2.527 d28 b29 a

F4 Calculation Exercise

1 c2 b3 b4 a5 c6 c7 b8 b9 b10 d11 d12 b13 d


Page 66 of 769

14 b15 c16 a ; // subsea // b17 d ; // subsea // c18 b19 a20 c21 b ; //subsea// b22 d23 a) 14.14 ppg, b) 124.4 bbls, c) 907.7 bbls , //subsea// 862.74, d) 412.5

bbls ,e) 1033 stks, //subsea// 988 stks ,f) 1221 stks g) 4040 stks, h) 8899stks, //subsea// 8460 stks, i) 10119 stks //subsea// 9680 stks

24 b25 a) Yes, b) No, c) Yes26 a) 1688, b) (i)862 +/- 5 stks, (ii) subsea 6647 stks, c) 4854 +/- 5 stks, *

subsea 4088 stks, d) 15.23 = 15.3 ppg, e) (i) 1400 psi, (ii) subsea 395 psi,f) 873, g) 1281, h) 822, i) (i)218(ii) subsea 192.6 mins

27 4.225 psi

F4 Surface BOP Stacks

1 158 psi2 2036 strokes3 1595 psi4 1084 psi5 243 strokes6 1393 psi7 632 strokes8 1175 psi9 6.53 psi / 100 strokes10 308 psi11 261 psi12 Breakdown at the shoe

F4 Subsea BOP Stacks1 158 psi2 2036 strokes3 1595 psi4 1084 psi5 243 strokes6 1393 psi7 632 strokes8 1175 psi9 6.53 psi / 100 strokes10 308 psi11 261 psi12 1532 strokes


Page 67 of 769

Formulae for Well Control.

1 Hydrostatic pressure: Ph (psi) P MW TVDh = ( . )0052

2 Pressure gradient: Gm (psi/ft) G MWm = ( . )0052

3 To convert pressure gradient to mudweight. MW

Gm=( . )0052

4 To convert pressure to mud weight.EMW

PTVD

h=( . ) ( )0052

5 Hydrostatic pressure using pressuregradient.

P G TVDh m=

6 Equivalent Circulating Density: ECD(ppg) ECD MW

APLTVD

= +( . ) ( )0052

7 Maximum Allowable Mud Weight fromleak off test: MAMW (ppg) MAMW MW

PTVDlot

lot

shoe

= +( ) ( . )0052

8 Maximum Allowable Annular SurfacePressure: MAASP (psi)

MAASP MAMW MW TVDshoe= - ( ) ( ) ( . )0052

9 Boyles gas law. P V P V1 1 2 2 =

10 Pressure change for change in strokerate. P P

SPMSPM

new oldnew

old=

2

11 Pressure change for a change of mudweight P P

MWMW

new oldnew

old=

12 Kill weight mud: MWkill (ppg)MW MW

SIDPPTVD

kill = +( ) ( . )0052

13 Shut In Casing Pressure (psi) SIDPPHiGiGmSICP +-= )(

14 Initial Circulating Pressure: ICP (ppg) ICP P SIDPPscr= +

15 Final Circulating Pressure: FCP (ppg)

Pscr = PL MWMW

PscrFCPkill

=

16 Barite required to increase mudweight: Barite (lbs/bbl)

)8.35())(1500

2

12

WWW

Barite-

-=


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17 Height of influx: Hi (ft)H

GainCapacityi

Pit

Annular

=

18 Trip Margin/Safety Factor (ppg)MW

TVDTM +

=

052.0MarginSafety

19 Estimated influx gradient (psi/ft).G MW

SICP SIDPPHInflux i

= --

( . )0052

20 Triplex pump output at 100%efficiency (bbl/stroke). (Liner diameterD and stroke in inches)

Output D Stroke= ( . ) ( )0000243 2

21 Percolation Rate (ft/hr)

052.0)/(

D

=MW

hrpsiPPR

22 Pipe capacity (BBL/ft).Capacity

IDPipe =

( ).

2

10294

23 Annular capacity (bbl/ft).Capacity

Dh DpAnnular=

-( ) ( ).

2 2

10294

24 Annular velocity (ft/min)

Outputpump= bbl/min Annular

Pump

ph

PumpAnnular Capacity

Output

DD

OutputVelocity =

-

=

22 )()(

4.1029

25 Formation Pressure (Pf) P SIDPP MW TVDf = + ( ( . ) )0052

26 Gas migration rate: Rm (ft/hrR

SICPG Gm m m

= =D Increase in

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