Well Control Exercises.pdf

Transocean Sedco Forex Jakarta Learning Centre

Pre-school exercises for Well Control With Answers.

Pre-School Well Control exercises 25/01/00

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 exercises were 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 to discuss.

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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) b) c) d)

With no drill pipe in the hole, shut in the well under pressure and repair the spool. With drill pipe in the hole, shut the well in and change pipe rams to blind rams. With drill pipe in the hole, circulate through the drill pipe. With drill pipe in the hole, shut in the well under pressure and repair the side outlets on the spool . Annular Blind Spool Kill

Pipe

Choke

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

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

3) You only have one inside BOP with an NC 50 (4”1/2 IF) lower pin connection on your rig but the drill string consist of 5” HWDP, and 8” collars. Which one of the following crossovers would you have on the drill floor in case of kick while tripping? a) b) c) d)

6-5/8” reg. Box X 7-5/8” reg. Pin NC50 (4-1/2” IF) Pin X 6-5/8” reg. Pin NC50 (4-1/2” IF) Box X 7-5/8” reg. Pin NC50 (4-1/2” IF) Box X 6-5/8” reg. Pin

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4) Two types of valves may be used in the drill string: Type 1 Type 2

Non return, stab in safety valve or inside BOP Fully opening stab in Kelly cock valve or fully opening safety valve

Indicate in the table which statement describes the valves. Type 1

Type 2

Requires the use of key to close Must not run in the hole in the close position Has to be pumped to read “shut-in drill pipe pressure” Will not allow wireline to be run inside the drill string Has potential to leak through the open/close key Easier 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 and choke lines are connected under the blind-shear rams. Is it possible to kill a well using the Driller'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 choke lines 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 and circulate 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 and choke 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? Page 5 of 769


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) b) c) d)

serial number pressure rating type 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 well head 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 ¼ when monitoring on kill line gauge during the kill operation. c) Effect of choke line friction is reduced to ½ 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 ring gaskets. Each flange (1,2,3 and 4) mates with one of the ring gaskets (A,B,C or D). Write the appropriate flange number in the blanks. Ring Gasket Marking Flange A CI API BX154 S304-4 B OES API R57 D-4 C OES API RX66 S-4 D CI API BX153 S316-4 Flange Marking 1. OES API 16-3/4 3M RX66 6A 89 300F PSL3 05/91 2. CI API 3-1/16 15M BX154 CRA 6A 89 250F PSL2 PRL2 08/92 3. OES API 2-9/16 20M BX153 CRA 6A 89 350F PSL4 PRL4 01/94 4. 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 the previous question, in the blanks below.

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Flange#1___________psi __________ inches __________ deg.F Flange#2___________psi __________ inches __________ deg.F Flange#3___________psi __________ inches __________ deg.F Flange#4___________psi __________ inches __________ deg.F 15) Identify the one ram locking device from the list below that locks the ram in the same position regardless of wear. a) b) c) d) e)

Shaffer “Ultralock” Shaffer “Poslock” Hydril “MPL” Cooper(Cameron) “Wedgelock” Koomey “Autolock”

16) From the list below, identify the ring gaskets that are pressure energized. (Pick four answers) a) b) c) d) e) f)

Type RX Type BX Type AX Type R oval Type R octagonal Type CX

17) Which dimension from the list below is used to identify the “Nominal Flange Size” (Pick one answer). a) b) c) d) e)

Throughbore I.D. Flange O.D. Diameter of raised face. O.D. of ring groove. Bolt circle diameter.

18) What is the main function of a diverter? a) b) c) d)

To shut in a shallow kick. To direct fluid a safe distance away from the rig floor. To create a back pressure sufficient to stop formation fluids entering the wellbore. 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 that maximum anticipated formation pressures do not exceed 75% of BOP equipment pressure ratings, what is the minimum acceptable rating for equipment to be used in drilling normally pressure formation to 16000 ft TVD? a) b) c) d) e)

2000 psi BOP equipment 3000 psi BOP equipment 5000 psi BOP equipment 10000 psi BOP equipment 15000 psi BOP equipment

20§) What is normally considered the highest potential risk when diverting a shallow gas blowout 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 table below to calculate the required accumulator volume if company policy is to provide sufficient volume to close, open and close again all rams and the annular. Component Annular BOP Ram BOP

Volume to Open 27 13

Volume to close 29 15

2) The following statements relate to the driller’s remote control BOP control panel located on the 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 not work. b) The master control valve on an air operated panel allows air pressure to go to each function 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 BOPs smaller than 18-3/4". What is the value of X? a) b) c) d)

30 sec. 60 sec. 2 min. 45 sec.

4) Which is the correct definition of the HPU reservoir volume according to API RP53? Page 8 of 769


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) b) c) d)

Manifold pressure Annular pressure Accumulator pressure Precharge pressure

6) When closing the annular preventer from the remote panel, which two gauges show a reduction in pressure? a) b) c) d) e)

Manifold pressure Annular pressure Accumulator pressure Air pressure Bypass pressure

7) In each of the cases below, identify the most likely problem from the gauge readings observed on the remote control panel. The annular setting is 900 psi, the manifold setting is 1,500 psi. a) b) c) d) e)

Everything is OK. Malfunction pressure regulating valve. Malfunction hydro-electric switch Leaking in hydraulic circuit Precharge pressure is to low

(i) (ii) (iii) (iv)

Accumulator pressure 2,900, increasing 2,700 increasing 2,400 increasing 3,300 increasing

Manifold pressure 1,500, steady

Annular pressure 900 steady

1,800 steady

900 steady

1,300 steady

900 steady

1,500 steady

900 steady

Problem

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 fluid volume 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 the following gauges on the drillers remote control panel? Page 9 of 769


(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 pressure when the annular preventer is being closed? 11) If the air pressure on the drillers panel reads 0 psi, which of the following statements is true? a) b) c) d)

No stack function can be operated from the remote panel. All stack function can be operated from the remote panel. Choke and kill lines can still be operated from the remote panel. 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) b) c) d) e) f)

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

13) When drilling, which may be the correct position of the 4-way valves on the BOP accumulator unit? a) b) c) d)

open close neutral 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) b) c) d)

1000 psi 3000 psi 1200 psi 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. He pushes the “Annular Close” button and the pilot light changes, but all gauges and the flowmeter 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 meter indicates a rapid loss of fluid. The best course of action is: a) b) c) d)

Stop drilling and shut the well in. Stop drilling and call subsea engineer. Stop drilling and put all function in block one at a time until the flow stops. 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 or poslocks 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 Pilot system. Which two statements are true with respect to the Pilot System? a) b) c) d) e)

The fluid in the Pilot System flows continuously while a function on the BOP takes place. The Pilot System dumps fluid to the sea at every operation of BOP functions. The Pilot System controls the position of all shuttle valves on the BOP stack directly. The Pilot system is a closed dead-end system. 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 panel after 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 drops below a set level. c) Memory Function indicates the previous position before “Block position” of three position functions. 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 4ways 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 the line 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) b) c) d) e)

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

3§) What is the reason for installing a riser fill-up valve in the marine riser of a subsea operation? 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) b) c) d)

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

2) Which tool would you use if you wanted to test the BOP stack, the casing head and upper casing seals. a) Plug type tester b) Cup type tester

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

Energize plastic seal and repair BOP at next scheduled maintenance. A primary seal is leaking, secure the well and repair the seal. The rams packer is leaking due to wear. Change the worn packer. 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 a surface 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:

A B C D E

Pump speed 40 20 30 80 70

Pressure 200 400 600 2,500 1,800

New speed 80 55 40 60 65

New pressure

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

A B C D E

Mud weight (ppg) 16 10 10 9.5 11.8

Pressure

New Mud weight

2,500 1,700 2,200 1,800 600

17.5 14 10.5 9.8 12.4

New pressure

3) In which cases would you consider taking a new SCR? a) b) c) d) e) f) g)

Every shift. Mud weight changes Before and after a leak off test After each connection when drilling with top drives. Every 250’ of open hole. After recharging pulsation dampeners on mud pump, discharge line. 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.

A B

MAMW 14 15.5

MW 10 9

TVDcsg 6,000 7,500

MAASP

TVD 8,000 18,000 9,500

Ph

2) Calculate the hydrostatic pressure for each well.

A B C

MW or Gm 9.5 ppg 15.5 ppg 0.889 psi/ft

MD 9,000 21,000 11,000

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

A B C

Ph 3,500 2,800 5,250

TVD 7,000 4,000 9,750

EMW

4) Change the following Pressure Gradients to Mud Weights. a) 0.56 psi/ft b) 0.81 psi/ft

5) Change the following Mud weights to Pressure Gradients: a) 10.4 ppg b) 14 ppg

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

A B C

E.D.C 12.5 ppg 10.2 ppg. 9.4 ppg.

Depth T.V.D 8000 ft 11400 ft 12500 ft

B.H.C.P.

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

A B C

L.O.T. Pressure 1800 psi. 1560 psi. 1420 psi.

Mud Wt. 11.4 ppg. 10.6 ppg. 9.8 ppg.

Shoe Depth. T.V.D. 9000 ft 7400 ft 6350 ft

Max Mud Wt.

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

A B C

Max Allow Mud Wt 19 ppg. 16.7 ppg 15 ppg.

Mud Wt. In use.

Shoe Depth T.V.D.

12 ppg. 11.5 ppg 9.2 ppg.

8000 ft 6800 ft 5500 ft

M.A.A.S.P.

9) Which three of the following conditions in the well increase the risk of exceeding the MAASP during the well kill operation? a) b) c) d) e) f)

Long open hole section. Large difference between formation breakdown pressure and mud hydrostatic pressure. Small influx. Short open hole section. Large influx. Small difference between formation breakdown pressure and mud hydrostatic pressure.

Questions 10-13 are base on the following information 13 3/8” surface casing is set and cemented at 3126 ft. (TVD) The cement is drilled out together with 15 ft. of new hole, using a 10.2 ppg. mud. A Leak Off Pressure of 670 psi is determined. 10) What is the formation fracture gradient? a) b) c) d)

0.619 psi/ft 0.837 psi/ft 0.745 psi/ft 0.530 psi/ft.

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11) What is the Maximum Allowable Annular Surface Pressure for 11.4 ppg mud used at 6500 ft TVD. a) b) c) d)

865 psi 474 psi 449 psi 563 psi

12) How often should the MAASP be recalculated? a) After every bit change b) After a change in mud weight c) After every 500 ft. drilled

13) A gas kick is being circulated out. At the time the gas reaches the casing shoe (3126 ft TVD) 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. a) b) c) d)

314 psi 442 psi 542 psi 506 psi

14) The Fracture Gradient of an open hole formation at 3680 ft. is 0.618 psi/ft. The drilling mud currently in use is 9.8 ppg. Approximately how much Surface Casing Pressure can be applied to the well before this formation breaks down? a) b) c) d)

350 psi 2275 psi 630 psi 400 psi

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Section C Causes of kicks. C1 Normal and abnormal formation pressure. 1) What is primary well control? a) b) c) d)

The slow Circulating Rate Pressure used in the kill process. The used of Mud hydrostatic to balance fluid pressures in the formation. The use of Blow Out Preventers to close in a well that is flowing. 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 is shut in? a) b) c) d)

Gas migration Friction losses Permeability Type of influx

C3 Gas Cutting 1) When we are drilling through a gas zone, with the proper mud density, the mud hydrostatic pressure should be able to prevent the gas from coming into the well. However, if we still get a 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 pressure decreases and volume increases. This may cause a sufficient reduction in hydrostatic to cause the well to flow. b) The mud weight decreases due to the large splintered crescent-shaped cuttings that we get from a high pressured zone c) The formation pressure increases suddenly as we drill into this zone since the gas inside is under 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 mud into 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) b) c) d)

Partial lost circulation has occurred. Total lost circulation has occurred. A kick has been taken. 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 the open hole. Select two correct actions which can be taken to minimize pressure in the annulus during the kill operation. a) b) c) d)

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

Lost circulation. Gas cut mud. Not keeping hole full. Abnormal Pressures.

2) Which two of the following cause swabbing?

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a) b) c) d) e)

Pulling the pipe too fast. Insufficient trip margin. Improper circulating density. Going into the hole too fast. 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) b) c) d)

Increase in torque. Gas cut mud. Decrease in pump pressure. Increase in flow rate.

2) Why is a 20 barrel kick in a small annulus more significant than a 20 barrel kick in a large annulus? a) b) c) d)

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

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

Flow rate increase. Increase torque. Pit gain. Gas cut mud.

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

Circulate bottoms up. Flow check Reduce weight on bit. Increase pump speed.

6) Which two practices are used to maintain primary well control as a precaution when connection gas is noticed? a) b) c) d) e)

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

flow line temperature increase. increased rotary torque flow rate increase. decrease drill string weight pit volume gain increased rate of penetration

8) Which of the following statements best describes formation porosity. a) b) c) d)

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

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

260 bbls 20 bbls 40 bbls 240 bbls.

10) If the cutting load in the annulus was high and the well had been shut in on kick. (Answer “Yes” or “No” to each question.) a) Would the drill pipe pressure be higher than in a clean well? {Include a brief explanation of 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 drilling on the floating rig these signs are difficult to detect due to the drilling vessel motion which will cause the fluctuation of the pit level. What is the equipment that we are using to compensate 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) b) c) d)

The oil base mud will contaminate the influx. In certain circumstances gas can dissolve in OBM. The gas will migrate in oil base mud faster than water base mud. 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 ¼” diameter hole. 25x92 ft. stand of 5” pipe have already been pulled. There are 85 more stands to pull. The calculated metal displacement of the 9 ½” collars is 0.08bbls/ft. The capacity of the drill pipe is 0.01776 bbls/ft and the metal displacement 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. The pipe capacity is 0.01776 bbls/ft. A 25 bbls slug weighting 12.0 ppg is pumped into the drill pipe 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) b) c) d)

25 psi 0 psi. 117 psi 135 psi.

3) Which of the following possible indications suggest that mud hydrostatic pressure and formation pressure are almost equal? a) b) c) d) e)

A drilling break. Connection gas. Large, splintery cuttings. Trip gas. 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 than calculated. What action must be taken? a) Flow check, if negative displace a 100 ft. heavy slug into annulus and continue to pull out of 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 the derrick. The displacement is 0.0549 bbls/ft. According to your Assistant driller - 5.1 bbls should be pump into the well. It only takes 5 bbls 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 and closed. A non return type safety valve was made up on top of the kelly cock prior to stripping in. (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 would check before starting your trip. a) b) c) d) e) f) g) h)

Kelly-cock on drill floor Slow circulation rate recorded Sufficient power to drawworks. Choke and kill manifold lined up for drilling Make sure trip tank is half full Trip sheets ready to record volumes displaced. Make up kick sheet 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 in the size of the influx. a) b) c) d) e) f) g) h) i) j)

Switch off the flow meter alarms. Regular briefing for the derrickman on his duties regarding the monitoring of pit levels. Drilling 20 ft further after a drilling brake, before flow checking. Running regular pit drills for drill crew. Maintaining stab in valves. Testing stab in valves during BOP tests. Excluding the drawworks from the SCR assignment. Keeping air pressure on choke control console at 10 psi. Calling toolpusher to floor prior to shutting in the well. 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 table when shutting in a well? a) b) c) d)

Allow the free flow of mud around bit during kill operation. Allow access to the lower kelly cock and, if required, removal of the kelly. Extend closing time to give softest possible shut in. 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 up for Hard Shut-in while drilling?

A B C D

BOP Side Outlet Hydraulic Valve(HCR) open open closed closed

Auto Choke(Remote Adj. Choke) closed open open closed

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Degasser Valves closed closed open open


5) While drilling along at a steady rate the derrickman asks to slow the mud pumps down so that the shakers can handle the increase in cuttings coming back in the returns. Which one of the 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 sand traps which can be dumped later. b) Pick up off bottom and check for flow, if there is not any then circulate bottoms up to reduce rate so shakers can handle cutting volume, flow check periodically during circulation. c) Slow down the mud pump until the shakers can handle the volume of cuttings in the returns as requested by derrickman. d) Slow down the drilling rate and the pump rate until the shakers clear up then go back to the original parameters.

6)

From BOP

Manual Choke

Remote Choke

V1

C1 V9

V2

R1

V4

Pressure Gauge On Choke Control Panel

V6

V10 V5 V12 1

V11

To Poorboy Degaser

V3

V8

V7

Over Board

To Stand Pipe To Poorboy Degaser

Cement Pit

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

V1,V2,V3,V4,V5,V6,V8,R1 V1,V2,V5,V6,V7 V1,V2,V9,C1,V10,V11 V1,V2,V4,V6,V8,R1

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

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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) b) c) d)

Immediately switch to No. 2 pump. Fix pump as soon as possible Secure the well, isolate mud pump restart using No. 2 pump. 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? Please answer the following items True or False. a) A larger pit gain will result in a higher SIDPP resulting in a heavier kill mud weight b) A larger pit gain will result in higher SIDPP and SICP c) 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 determine formation pressure? a) b) c) d)

BOP manifold pressure gauge Choke console drill pipe pressure gauge Driller’s console drill pipe pressure gauge Choke console casing pressure gauge

3) A flowing well is closed in. Which two pressure gauge readings might be used to determine formation pressure? a) b) c) d)

BOP manifold pressure gauge Choke console drill pipe pressure gauge Driller’s console drill pipe pressure gauge Choke console casing pressure gauge

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

Increase Decrease Stay the same No way of knowing

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5) While killing a well, as pump speed is increased, what should happen to casing pressure in order to keep bottom hole pressure steady? a) Casing pressure should be held steady during SPM change b) Casing pressure should be allowed to rise during SPM change c) Casing pressure should be allowed to fall during SPM change

6) The principle involved in Constant Bottom Hole Pressure methods of well control is to maintain a bottom hole pressure that is : a) b) c) d)

Equal to the slow circulating rate pressure At least equal to the formation pressure Equal to the shut in drill pipe pressure 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 the casing shoe to be at its maximum? a) b) c) d)

At initial shut in When kill mud reaches the bit When kill mud reaches the shoe When top of gas reaches the shoe

8) If Drill pipe Pressure is held constant while displacing the string with kill mud, what will happen to Bottom Hole Pressure? a) Increases b) Remains the same c) Decreases

9) How is a choke wash-out recognized? a) b) c) d)

Rapid rise in casing pressure with no change in drill pipe pressure Increase in drill pipe pressure with no change in casing pressure Continually having to open choke to maintain drill pipe and casing pressure 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 the gradual closing of the choke have on the bottom hole pressure? a) Decreases b) Increases c) Stays the same

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11) If Bottom Hole Pressure is held constant while circulating the influx out, the pressure on at the casing shoe will not increase after the influx passes, even though surface pressure on the annulus continues to rise. a) True b) 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 ½” hole at 11,000 ft. (TVD) with 5” drill pipe and 750 ft. of 6 ½” drill collars Annular capacities 5" DP / 8 ½" Hole, 0.0459bbls / ft. ½" Hole, 0.0292bbls / ft

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

480 psi 650 psi 975 psi 888 psi

13) While preparing to circulate Kill Mud, the gas bubble begins to migrate. If no action is taken, what will happen to the pressure in the gas bubble as it rises: a) Increase b) Decrease c) Remain approximately the same

14) What will happen to Bottom hole Pressure? a) Increase b) Decrease c) Remain approximately the same

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15) What will happen to Shut in Casing Pressure? a) Increase b) Decrease c) Remain approximately the same

16) What will happen to the pressure on the Casing Seat? a) Increase b) Decrease c) Remain approximately the same

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? a) Increase b) Decrease c) Remain approximately the same

18) What would happen to Bottom Hole Pressure? a) Increase b) Decrease c) Remain approximately the same

19) What would happen to the Shut in Casing Pressure? a) Increase b) Decrease c) Remain approximately the same

20) What would happen to the Pressure on the Casing Seat while the bubble is below the Casing Shoe? a) Increase b) Decrease c) Remain approximately the same

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21) What would happen to the Pressure on the Casing Seat when the bubble is above the Casing Shoe? a) Increase b) Decrease c) Remain approximately the same 22) A kick is being circulated from a well using the Driller’s Method; Pumping pressure having been established as 1000 psi at 30 SPM. During the operation, pressure suddenly increases to 1350. You are reasonably sure that a Nozzle of the Bit is plugged. What should you do? a) b) c) d)

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

23) During the well kill operation, slowly but regularly you have had to reduce choke size because the drill pipe and casing pressures keep dropping with constant pump strokes. What is the likely cause of this? a) A bit nozzle is washing out b) The choke is washing out c) You have a washed out pump swab

24) An influx is being circulated out using the Driller’s Method and using 1100 psi at 30 SPM. The operator increases pump speed to 35 SPM, while holding pump pressure constant. What happens to Bottom Hole Pressure? a) Increases b) Decreases c) Remains approximately the same

25) Which of the following parameters can be affected by a drill string washout during a well kill operation? a) b) c) d)

Bottom hole pressure Kick tolerance Formation fracture pressure 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. The pump strokes remain constant. You close up your choke slightly, the drill pipe pressure remains unchanged but the casing pressure goes up. What is the probable cause for this? a) b) c) d)

Choke is plugging off Bit is plugging off Hole in drill pipe Choke is washing out

27) If regularly and rather slowly, you have to pinch in the choke to maintain drill pipe and choke pressures while the pump strokes remain constant, you may have: a) a washed out bit nozzle b) a washed out choke c) a pump failure

28) Problems that occur during a killing operation may affect the parameters you are monitoring at the surface. These are: Drill pipe pressure, casing pressure, bottom hole pressure. For each of the following problems state the immediate effect on each of the above parameters For an increase draw + Problem

For a decrease draw -

Drill Pipe Pressure

For no change draw =

Casing 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) b) c) d)

Drill pipe and casing pressures both falling Drill pipe and casing pressures both rising Rapid rise in casing pressure with no change to drill pipe pressure 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 the annulus. 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.

900 800 700 600 500 400 300 200 100 0 TIME

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

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

A further influx is occurring The influx is migrating up the well bore The gauges are faulty 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 no float in the drill string. What kill mud weight is required? a) b) c) d)

10.3 ppg 11.3 ppg 10.7 ppg No way of knowing

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5) Shut in casing pressure is used to calculate a) b) c) d)

Kill weight mud Influx gradient and type when influx volume and well geometry are known Maximum Allowable Annular Surface Pressure Initial circulating pressure

F3 Kick Handling Methods 1) What should the driller do at a drilling break? a) b) c) d)

Circulate bottoms up Flow check Reduce weight on bit Increase pump speed

2) A kicking well has been shut in. The drill pipe pressure is ‘0’ because there is a non-return valve (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 gauge b) 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 driller’s method (no weight up), are the SICP and SIDPP about the same?

4) A gas kick is being circulated up the well. What is the surface pit volume most likely to do? a) Increase b) Stay the same c) Decrease

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5) On a surface stack, what would happen if when bringing the pumps up to kill speed, the casing pressure was allowed to fall below shut in casing pressure? a) Formation would most probably break down b) More influx would be let into the well bore c) It would have no effect on anything

6) For each of the following statements, note whether it relates to the Drillers Method or the Wait and Weight Method. a) b) c) d)

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

7) Which one of the following actions taken while stripping into the hole will help to maintain an acceptable bottom hole pressure? a) Pumping a volume of mud into the well, equal to the drill pipe closed end displacement at regular intervals b) Bleeding off the drill pipe steel displacement at regular intervals c) Pumping a volume of mud into the well, equal to the drill pipe steel displacement, at regular intervals d) 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 method b) 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 to using the Drillers method c) The Wait and Weight method should always be used because the pressure against the open hole will always be lower when using the Drillers method

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

shut in drill pipe pressure shut in casing pressure original mud weight plus slow circulation rate pressure losses 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 slow pump 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 back pressure on drill pipe side with choke b) Add 400 psi to casing pressure and bring pump up to kill rate while using the choke to keep the casing pressure +400 constant c) Bring pump strokes up to kill rate while keeping casing pressure constant by manipulating the 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 Driller’s Method, the well is shut in. Should casing pressure be: a) Less than Shut in Drill Pipe Pressure b) Equal to Shut in Drill Pipe Pressure c) Greater than Shut in Drill Pipe Pressure

12) On the second circulation of the Driller’s method, if the casing pressure was held constant until the kill mud reached Surface, what would happen to the bottom hole pressure? a) Increase b) Decrease c) Stay the same

13) Using Wait and Weight method, if the drill pipe pressure drops below the line of the graph as the kill mud goes down, what happens to the bottom hole pressure? a) Increases b) Decreases c) Stays the same

14) You have taken a kick with a non-return valve (float) in the drill string. After shutting the well in properly, it is best to : a) Use the annulus pressure to calculate the kill weight mud b) Start raising the mud weight 1 ppg per circulation until the well is dead c) Use either the rig pump or cementing unit pump to increase pressure in 100 psi increments until a change is seen on casing gauge d) 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 Driller’s Method. Original shut-in drill pipe pressure = 500 psi Original shut-in casing pressure = 900 psi After the first circulation, the well is shut in and pressures allowed to stabilize. They then read : Shut-in drill pipe pressure = 500 psi Shut-in casing pressure = 650 psi It is decided not to spend any more time cleaning the hole Which one of the following actions should be taken a) b) c) d)

Prepare to use the Wait and Weight method Bull-head the annulus until shut-in casing pressure is reduced to 500 psi Reverse circulate until shut-in casing pressure is reduced to 500 psi 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 the well closed in, how would you find the ICP? a) Bring pump up to the desired rate, while holding the casing pressure 150 psi above the original SICP b) Bring pump up to desired rate, but hold 200 psi back pressure on the drill pipe c) Bring pump up to the desired rate holding casing pressure constant by manipulating the hydraulic choke d) 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) b) c) d) e)

the gauge on the choke and kill manifold the drill pipe pressure gauge on the drillers console the casing gauge on the drillers console the drill pipe gauge on the remote auto choke panel the casing gauge on the remote auto choke panel

18) The following diagrams show the approximate changes in pressure at certain points in the well during the first circulation of the Driller’s Method. Match the following locations to their respective diagrams: a) b) c) d)

Surface casing pressure Casing shoe pressure Bottom hole pressure Pump pressure

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(i)

PSI TIME

(ii)

PSI TIME

(iii)

PSI TIME

(iv)

PSI TIME

(c) 1500

DYNAMIC PRESSURE

1250 D RI LL PI PE PR ES SU RE (P SI)

(d) 1000 750 (a) 500

(e)

STATIC PRESSURE

250 (b) (x) KILL MUD PUMPED

Drill pipe pressure graph of the one circulation method

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

1) Drillstring displaced with kill mud 2) ICP = SIDPP + kill rate pressure 3) 0 psi (static) 4) SIDPP (static) 5) Drillstring volume pumped 6) FCP= Kill rate pressure x kill mud weight / original mud weight

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

(c)

1500 1250

DYNAMIC PRESSUR E

1000 D RI LL PI PE PR ES SU RE (P SI)

(d) 750 (a)

500 250

(e) STATIC PRESSUR E

(b) (x)

(z)

(y)

Drill pipe pressure graph of the driller method

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 weight 3) Annular volume pumped 4) Drillstring volume displaced 5) ICP = SIDPP + kill rate pressure 6) SIDPP (static) 7) Annular volume pumped 8) Drillstring volume pumped

Page 40 of 769


F4 Calculations Exercise 1 Calculate the following 1) Old Mud Wt. PPG 9.68 11.5 11.0

New Mud Wt. PPG 10.0 12.2 12.6

Old Pressure psi. 1850 2500 300

New Pressure psi.

Old Strokes SPM 75 30 20

New Strokes SPM 40 60 80

Old Pressure psi 2450 400 180

New Pressure psi

2)

3) Mud Wt PPG 10.0 11.2

E.C.D. PPG

14.54

12.0 15.1

9.84

10.6

T.V.D. Ft. 10000

Mud Hydrostatic psi

9800 16000 11450 8700

5824 6000 6200

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 41 of 769

B.H.C.P. psi

A.P.L. psi 300 76

320


a) BHP = 9,800 psi, Depth = 9,800 feet b) 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 PSI b) Mud wt = 9.5 PPG, SICP = 2,000 PSI, SIDPP = 1,200 PSI, Depth = 9,500 PSI c) 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

a) b) c) d) e)

Old strokes 30 spm 80 spm 70 spm 40 spm 20 spm

Old Pressure 600 psi 2500 psi 1800 psi 200 psi 400 psi

New Strokes 40 spm 60 spm 20 spm 80 spm 55 spm

New Pressure psi psi psi psi 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 ½" drill collars in an 8 ½" hole (Annular capacity = 0.0292 bbl/ft) b) 40 bbl gain, 500 feet of 6 ½" drill collars in an 8 ½" hole (0.0292 bbl/ft annular capacity), 5" drill pipe precedes (.0459 bbl/ft annular capacity)

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

a) b)

APL 400 psi 250 psi

hydrostatic pressure 6500 psi 5000 psi

Mud wt 10 ft 10.5 psi

BHCP

13) Calculate the equivalent circulating density of each well :

a) b)

APL 350 psi 40 psi

hydrostatic pressure 5050 psi 6800 psi

Depth TVD 9800 ft 12000 psi

ECD

Depth TVD 9450 ft 5000 psi

APL

14) Calculate the annular pressure loss for each well:

a) b)

BHCP 6000 psi 2600 psi

Mud wt 11.6 psi 9.8 psi

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

Old pressure

New Mud Wt

10 ppg 9.5 ppg 11.8 ppg

220 psi 1800 psi 600 psi

10.5 ppg 9.8 ppg 12.4 ppg

16) Change the following pressure gradients to mud weights :

a) b)

Gradient .56 psi/ft .81 psi/ft

Mud weight

17) Change the following mud weights to pressure gradients :

a) b)

Mud weight 10.4 ppg ppg

Pressure Gradient

Page 43 of 769

New pressure


18) Change the following circulating densities to bottom hole circulating pressures : ECD 12.5 ppg 10.2 ppg 9.4 ppg

a) b) c)

Depth TVD 8000 ft 11400 ft 12500 ft

BHCP

19) Using the following data from the Leak Off test results, calculate maximum allowable mud weights :

a) b) c)

LOT pressure 1800 psi 1560 psi 1420 psi

Mud weight 11.4 ppg 10.6 ppg 9.8 ppg

Shoe depth TVD 9000 ft 7400 ft 6350 ft

Max mud weight

20) Using the same well data, calculate the height of each influx : Drill Collar Length DC - OH Capacity DP - OH Capacity Influx volume 10 bbls 20 bbls 30 bbls

= = = = = = =

700 ft 0292 bbl/ft .0459 bbl/ft Height ft. ft ft ft

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

a) b) c)

SICP 800 psi 950 psi 680 psi

SIDPP 720 psi 600 psi 550 psi

MUD weight 11.5 ppg 10.6 ppg 10.2 ppg

Ht of Influx 400 ft 840 ft 350 ft

22) Calculate new MAASP from the following data :

a) b) c)

Max. allowance mud weight 19 ppg 16.7 ppg 15 ppg

Mud weight in use 12 ppg 11.5 ppg 9.2 ppg

Shoe depth TVD 8000 ft 6800 ft 5500 ft

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MAASP

Gradient


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

a) b) c) d)

SICP

SIDPP

Mud weight

600 psi 850 psi 780 psi 700 psi

450 psi 690 psi 570 psi 300 psi

10 ppg 11 ppg 10.5 ppg 14 ppg

Depth TVD 9500 ft 12000 ft 11200 ft 13000 ft

Depth MD 10000 ft 12300 ft 11800 ft 13400 ft

Kill Mud

24) A well is shut in with 500 psi on the casing pressure. The annulus contains 8,000 ft of 10.0 ppg mud above 1,000 ft of 9.0 ppg saltwater. What is the equivalent mud weight at 8,000 ft? At 9,000 ft? a) b) c) d)

8,000 ft ; 11.2 ppg 8,000 ft ; 11.7 ppg 9,000 ft ; 11.0 ppg 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. The pipe capacity is .01776 bbls/ft. A 25 bbls slug weighing 12.0 ppg is pumped into the drill pipe 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) b) c) d)

25 psi 0 psi 117 psi 135 psi

26) Drill pipe capacity Drill pipe metal displacement Average stand length Calculate :

= = =

0.0178 bbls/ft 0.0082 bbls/ft 93 ft

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 predetermined 30 SPM by holding the shut in casing pressure constant. You have got a kick in the well of 220 psi shut in drill pipe pressure. At 30 SPM your drill pipe circulating pressure is 1060 psi. Calculate the slow circulating rate pressure loss. a) b) c) d)

700 psi 770 psi 800 psi 840 psi

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28) Calculate the equivalent circulating density in the following circumstances: Circulating pressure = 3100 psi Pressure losses Surface equipment Drill string Nozzles Annulus

= = = =

20 psi 930 psi 1800 psi 350 psi

Drilled depth 12,300 ft. (11,500 ft. TVD) Mud weight 11.4 ppg ECD is a) b) c) d)

10.8 ppg 12.0 ppg 11.4 ppg 12.3 ppg

29) A well is drilled to 13,000ft with 13.2 ppg mud. The formation pressure at that depth is 8,700 psi. The intermediate casing is 43.5 lb/ft, 9 5/8 in. pipe set to 11,000 ft. The drill pipe is 4 ½ in.(ID 3.826”) and the collars displacement is 0.04bbl/ft. The operator requires hole filling after 5 stands of drill pipe or collars are pulled. Will the well kick when pulling the drill pipe dry? Drill collars dry? (Assume no swabbing effects with an average length of 90ft per stand) a) Pulling pipe ; no kick Pulling collars ; kick b) Pulling pipe ; kick Pulling collars ; no kick c) Kicks for both drill pipes and drill collars d) Does not kick for either drill pipe or drill collar

Page 46 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 out together with 15 ft. of new hole, using a 11 ppg mud. A leak off test pressure of 800 psi is determined. (Hole TVD 7000ft) 1) What is the formation fracture gradient? a) b) c) d)

0.188 psi / ft 0.686 psi / ft 0.760 psi / ft 0.384 psi / ft

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

373 psi 511 psi 884 psi 1982 psi

3) How often should the MAASP be recalculated? a) After every bit change b) After a change in mud weight c) 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) a) b) c) d)

348 psi 442 psi 1368 psi 2625 psi

Page 47 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 the well kicks and is closed in. Shut in Drill Pipe Pressure is 750 psi Shut in Casing Pressure is 1050 psi and kick volume is 15 bbls. Pre- recorded information is as follows : Fracture mud weight Capacity of 19.5 lbs. Drill pipe Capacity of 9 5/8” J55 casing Slow Circulating Rate Pressure

= = = =

14.4 ppg 0.01776 bbl/ft. 0.0732 bbl/ft. 850 psi

5) The maximum allowable annular surface pressure is rounded off to : a) b) c) d)

1370 psi 1480 psi 1435 psi 1415 psi

6) The kill mud weight required to balance the formation pressure is: a) b) c) d)

13.1 ppg 12.6 ppg 12.8 ppg 12.2 ppg

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

13.4 ppg 13.0 ppg 12.4 ppg 11.8 ppg

8) The initial circulating pressure is: a) 1400 psi b) 1600 psi c) 1900 psi

Page 48 of 769


9) The final circulating pressure (using kill mud weight with a 100 psi Safety Margin is) : a) 850 psi b) 970 psi c) 920 psi d) 1050 psi

10) The Fracture Gradient of an open hole formation at 3680 ft. is 0.618 psi/ft. The drilling mud currently in use is 9.8 ppg. Approximately how much Surface Casing Pressure can be applied to the well before this formation breaks down? a) b) c) d)

350 psi 2275 psi 630 psi 400 psi

11) In the area where local legislation requires that BOP equipment must be rated so that maximum anticipated formation pressures do not exceed 75% of BOP equipment pressure ratings, what is the Minimum Acceptable rating for equipment to be used in drilling Normally Pressured Formation to 16,000 ft. TVD? a) b) c) d) e)

2,000 psi BOP Equipment 3,000 psi BOP Equipment 5,000 psi BOP Equipment 10,000 psi BOP Equipment 15,000 psi BOP Equipment

Drilling 12 ¼" hole at 7653 ft. TVD with 11.7 ppg mud, the well kicks and is closed in. The shut in DPP is 430 psi. SICP is 600 psi. Pit gain 28 bbl. Pre-Recorded Information 13 3/8" Casing Shoe at 3975 ft. TVD A Leak Off Test on the formation at shoe, using 11.2 ppg mud, showed a surface pressure of 910 psi when Leak Off occurred Drill Collars 8 ¼" OD x 2 ½" ID 546 ft. long. Capacity 0.0061bbls / ft Drill Pipe 5" OD x 19.5 lbs / ft. Capacity 0.01776 bbls / ft Open Hole - Drill Collar Annulus Capacity is 0.0796 bbls / ft Open Hole - Drill Pipe Annulus Capacity is 0.1215 bbls / ft 13 3/8" Casing - drill Pipe Annulus capacity is 0.1239 Slow Pump Rate 30 SPM - Output 5 gal / stk at 730 psi * For Subsea Candidates use * Subsea Information Water Depth - 700 ft Air Gap - 75 ft

Page 49 of 769


1. 2. 3. 4.

Choke Line Capacity Riser Capacity Drill Pipe Displacement CLFL ( Choke Line Friction)

0.0087 bbls / ft 0.3892 bbls / ft 0.0076 bbls / ft 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) b) c) d)

0.792 psi / ft 0.811 psi / ft 0.834 psi / ft 0.861 psi / ft

13) What is the MAASP with 11.7 ppg mud in the hole? a) b) c) d)

910 psi 730 psi 1004 psi 806 psi

14) The Kill Mud required is: a) c) d) b)

11.7 ppg 12.8 ppg 13.3 ppg 13.9 ppg

15) The Surface to Bit strokes are : a) b) c) d)

987 1016 1088 1164

16) The Bit to Shoe strokes are a) b) c) d)

3562 3257 3752 3604

§ Subsea: Strokes to displace riser to Kill Mud 3370 2370 2300 2730

Page 50 of 769


17) The Bit to Surface Strokes are : a) b) c) d)

7100 7300 7500 7700

§ Subsea : shoe to BOP Strokes 4744 3388 3332 3563

18) The Initial Circulating Pressure is : a) b) c) d)

1060 psi 1160 psi 1260 psi 1330 psi

19) The Final Circulating Pressure is : a) b) c) d)

800 psi 730 psi 1330 psi 1270 psi

20) The gradient of influx is: a) b) c) d)

0.087 psi / ft 0.212 psi / ft 0.1225 psi / ft 0.327 psi /ft

21) The estimated time required to kill the well at 30 SPM is: a) b) c) d)

3 hours approximately 5 hours approximately 7 hours approximately 11 hours approximately

22) You are tripping out of the hole : Hole Depth Mud Weight Drill Pipe capacity Slug Displaced

12,000 ft 10 ppg 0.01776 bbls / ft 12 ppg density 20 bbls

Page 51 of 769


The fluid level in the drill pipe has dropped 223 ft. How much bottom hole pressure has been lost? a) b) c) d)

702 psi 116 psi 70 psi 0 psi

23) Given the following data, complete an IWCF kick sheet as far as required to complete questions 23(a) to 23(i) Drill String Data 5" Drill Pipe 8 ¼ Drill Collars - 630 ft 5" Heavy weight DP 450 ft

Capacities 0.01776 bbls /ft 0.0077 bbls / ft 0.0087 bbls / ft

Hole Data Casing depth 4000 ft. Hole Depth 7591 ft. Mud weight 9.5 ppg

13 3/8" casing 12 ¼"

Annular Capacities 8 ¼” DC 12 ¼” hole 5" DP 12 ¼" hole 5" DP 13 3/8" casing

0.0796 bbls / ft 0.1215 bbls / ft 0.1238 bbls / ft

LOT 1070 psi test mud 9 ppg Pump Data 6" liners 97% efficiency = .102 bbls / stk § Subsea Information : Water Depth Air Gap Choke Line Capacity Riser Capacity Drill Pipe Steel Displacement

300 ft 90 ft 0.0087 bbls / ft 0.3892 bbls / ft 0.0076 bbls / ft

Questions : a) What is the maximum mud weight? b) Total Drill string Capacity c§) Total Annulus Capacity d) Bit to shoe volume e§) Total system volume f) Surface to bit strokes g) Bit to shoe strokes h§) Bit to surface strokes

Page 52 of 769


i§) Total circulation strokes

24) You have taken a kick and shut the well in. The active tank while drilling contained 250 bbls. 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 the well? a) b) c) d)

0 bbls 25 bbls 50 bbls 275 bbls

25) You are pulling out of hole. Two x 93 ft stands of 8" drill collars have been stood back in the derrick. The displacement is 0.0538 bbls. / ft. According to your Assistant Driller, 10 bbls should be pumped into the well. It only takes 10 bbls 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, keep going

26) Use an IWCF kill sheet to assist you in answering questions 26(a) through 26(i) Well Data Hole size Hole depth Casing Drill pipe Heavy weight pipe Drill Collars Mud density Volume open hole / collars Volume open hole / drill pipe / HWDP Volume casing / drill pipe Fracture mud wt. At the casing shoe Mud pumps Slow circulating rate

8 ½ in 11937 ft. TVD / MD 9 5/8 in. casing set at 9474 ft. 5 in. capacity = 0.0178 bbls / ft 5 in., 497 ft. long Capacity = 0.0088 bbls / ft 6 ¼ in., 892 ft. long Capacity = 0.006 bbls / ft 14.3 ppg 0.0322 bbls / ft 0.0459 bbls / ft 0.0493 bls / ft 16.9 ppg Output = 0.117 bbls / stk 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 Shut-in Casing Pressure Pit Gain

580 psi 755 psi 12 bbls

Page 53 of 769


The well will be killed using the Driller’s Method at 30 spm §* For Subsea use Water Depth .........................................................................2120 ft Air Gap.................................................................................80 ft Steel Displacement...............................................................0.0076 bbls / ft Riser Capacity 19.75" ID......................................................0.3789 bbls / ft Choke 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

Inlet 15.0'

6.5'

Shakers

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

Page 54 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 well data and answer the accompanies questions : Well Data : Hole Size Hole Depth Kick-off Point End of Build Casing Shoe

8 ½" 14370 ft., MD (5250 ft., TVD) 1640 ft., MD (1640 ft., TVD) 4265 ft., MD (3494 ft., TVD) 9 5/8" x 47 lbs/ft @ 10600 ft., MD (4593 ft., TVD)

Capacities : Drill Pipe 5” OD x 19.5 # Heviwate 480 ft 5” OD x 3” ID Drill Collar 660 ft 6 ½” OD x 213/16” ID

0.01776 bbls/ft 0.00874 bbls/ft 0.0077 bbls/ft

Drill Collar/Open Hole Drill Pipe / HWDP / Open Hole Drill Pipe / HWDP / Casing


Pump Data : Displacement Active Surface Volume Slow Circulating Rate

0.12 bbls/stroke 470 bbls 870 psi @ 30 spm

Formation Strength Data

1027 psi LOT using 10.5 ppg mud weight

Kick Data : Mud Weight Kick depth Pit Gain SIDPP SICP

10.85 ppg 14370 ft., MD (5250 ft., TVD) 19 bbls 725 psi 785 psi

Page 55 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 the end of build depth to the bit. 10. Calculate the MAASP (Maximum Allowable Annular Surface Pressure) after circulation of Kill Mud. 11. If we neglected the directional nature of the well and decided to use a conventional Vertical Well Kill Sheet to remove the influx, calculate the pressure over balance at the end of build depth. 12. What will be the consequences of this overbalance in the well bore?

Page 56 of 769


IWCF deviated well control exercise ( Subsea BOP stack) Prepare a deviated well control kick sheet (use attached kick sheets) from the following well data and answer the accompanied questions

Well Data : Hole Size Hole Depth Kick-off Point End of Build Casing Shoe

8 ½" 14370 ft., MD (5250 ft., TVD) 1640 ft., MD (1640 ft., TVD) 4265 ft., MD (3494 ft., TVD) 9 5/8" x 47 lbs/ft @ 10600 ft., MD (4593 ft., TVD)

Capacities : Drill Pipe 5” OD (NC50, S-135) Drill Pipe 5” Closed end displacement Heviwate 480 ft 5” OD x 3” ID Drill Collar 660 ft 6 ½” OD x 213/16” ID Choke Line 520 ft x 3” ID Marine Riser 505 ft

0.01776 bbls/ft 0.0254 bbls/ft 0.00874 bbls/ft 0.0077 bbls/ft 0.0085 bbls/ft 0.3892 bbl/ft

Drill Collar/Open Hole Drill Pipe / HWDP / Open Hole Drill Pipe / HWDP / Casing


Pump Data : Displacement Active Surface Volume Slow Circulating Rate Riser Circuit Slow Circulating Rate Choke Circuit

0.12 bbls/stroke 470 bbls 870 psi @ 30 spm 960 psi @30 spm

Formation Strength Data

1027 psi LOT using 10.5 ppg mud weight

Kick Data : Mud Weight Kick depth Pit Gain SIDPP SICP

10.85 ppg 14370 ft., MD (5250 ft., TVD) 19 bbls 725 psi 785 psi

Page 57 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 the end of build depth to the bit. 10. Calculate the MAASP (Maximum Allowable Annular Surface Pressure) after circulation of Kill Mud. 11. If we neglected the directional nature of the well and decided to use a conventional Vertical Well Kill Sheet to remove the influx, calculate the pressure over balance at the end of build depth. 12. How many strokes to pump to displace Marine Riser to kill fluid before opening the BOP?

Page 58 of 769


Answers. Section A A1 Blowout Preventers and Diverters. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

no, no, yes, yes b d A = 2, B = 2, C = 1, D = 1, E = 2, F = 2 Yes, yes, no, no Yes, no, yes, yes No, no, yes c d c c d 2, 4, 1, 3 3000, 16 ¾, 300; 15000, 3 1/16, 250; 20000, 2 9/16, 350; 2000, 13 5/8, 250 b a, b c, f a b D A

A2 BOP control systems. 1 2 3 4 5 6 7 (i) 7 (ii) 7 (iii) 7 (iv) 8 9 10 11 12 13 14

214 T, T, T, F d a a, c b, c a b d c 40 120, 3000, 1500, 750-1500 psi 2, 4 a 1 d a

Page 59 of 769


15 16 17 18 19 20 21 22 23 24 25 26

a. light changes color, b. read back pressure drops, c. read back pressure recovers, d. flow meter stops at a proper reading. a c a 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) pipe rams, connectors, failsafe valves, shear rams (b) The connection is to the regulator, the measurement is taken on the downstream (regulated) side. b d&e b, d c a=T, b=T, c=T, d=F, e=F

A4 Auxiliary Equipment. 1 2 3

a=F, b=F, c=T, d=F b b

A5 BOP testing. 1 2 3 4

b, c, d b b a

Section B B1 Slow Circulating Rates. 1 2 3 4

a. 800 psi, b. 3025 psi, c. 1067 psi, d. 1407 psi, e. 147 psi a. 2734 psi, b. 2380 psi, c. 2310 psi, d. 1857 psi, e. 631 psi a, b, g For deep water drilling we have significant pressure drop on the choke line due to its length. This will affect the calculation of the initial circulating pressure in a wll control situation. Three methods for measuring CLFL 1) 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 open choke.Record Drill Pipe pressure. 1.4 The difference between the two is CLFL 2) 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 60 of 769


3.2 The pump pressure is twice the CLFL B2 Leak Off Test & MAASP 1 2 3 4 5 6 7 8 9 10 11 12 13 14

a. 1248 psi, b. 2535 psi a. 3952 psi, b. 14508 psi, c. 8446 psi a. 9.62 ppg., b. 13.46 ppg. C. 16.48 ppg. a. 10.77 ppg., b. 15.58 ppg. a. 0.5408 psi/ft, b. 0.728 psi/ft a. 5200 psi, b. 6046 psi, c. 6110 psi. a. 15.25 ppg., b. 14.65 ppg, c. 14.1 ppg a. 2919psi, b. 1838.7 psi, c. 1658.8 psi. a, e, f c b b a d

Section C C1 Normal and Abnormal Formation Pressure 1 2 3

b d c

C3 Gas Cutting 1

a

C4 Lost Circulation 1 2 3 4

a b b b-c

C5 Kicks As Result of Surface Practices 1 2

c a, b

Section D D1 Kick While Drilling 1

d

Page 61 of 769


2 3 4 5 6 7 8 9 10

11

b b b b, e c, e a b a. = No, b. No, c. = Yes PVT (A pit volume totalizing system). It will report overall pit gain or loss by using multiple pit monitors and resolving individual losses and gains reported by each monitor into a single value. b

D2 Kick While Tripping 1 2 3 4 5 6 7

d b e b a. = No (10.2), b. =Yes, c.= No. a.= No, b. = No a, b, c, e, f, h

Section E E1 Shut-In Procedure 1 2 3 4 5 6 7 8

a, c, g, h, i, j b d d b a a a

Section F F1 Pressure Observation 1 2 3 4 5 6 7 8

F, F, T b b, d c a b d a Page 62 of 769


9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28

29 30

d b a d c a a a b c a a c b b b a, d c b ---== -== +++ +== a c, e

F3 Shut In Pressure Interpretation 1 2 3 4 5

b 700-710 b a b

F3 Kick Handling Methods 1 2 3 4 5 6

7 8

b b Yes a b a. W&W b. Driller c. W&W d. Driller d b Page 63 of 769


9 10 11 12 13 14 15 16 17 18 19

20

a c b a b d a c d d., b., a., c. a. 4) b. 3) c. 2) d. 6) e. 1) x. 5) 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. 9.68 11.5 11

New Mud Wt. ppg. 10 12.2 12.6

Old Pressure ppg. 1850 2500 300

New Pressure ppg. 1912 2652 344

Old Strokes spm. 75 30 20

New Strokes spm. 40 60 80

Old Pressure psi 2450 400 180

New Pressure psi 697 1600 2880

2)

3) Mud Wt ppg. 10 11.2

E.C.D. ppg. 10.58 11.35

T.V.D. ft 10000 10000

Hydrostatic psi. 5200 5824

Page 64 of 769

B.H.C.P. psi. 5500 5900

A.P.L. psi. 300 76


11.77 14.54 10.41 9.84 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29

12 15.1 10.95 10.6

9800 16000 11450 8700

6000 12097 6200 4452

a. 3952 psi., b. 14508 psi., c 8446 psi. a. 9.62 ppg., b. 13.46 ppg., c. 16.48 ppg. a. 19.23 ppg., b. 8.81 ppg. a. 11.16 ppg., b. 11.93 ppg., c. 17.39 ppg. a. 1067 psi., b. 1407 psi., c. 147 psi., d. 800 psi.,e. 3025 psi a. 2735 psi., b. 2308 psi a. 1560 psi., 2535 psi. a. 856.16 ft., b. 1053.38 ft. a. 6900 psi., b. 5250 psi. a. 10.6 ppg., b. 10.96 ppg. a. 300 psi., b. 52 psi. a. 231 psi., b. 1857 psi., c. 631 psi a. 10.77 ppg., b. 15.58 ppg. a. 0.5408 psi/ft., b. 0.728 psi/ft. a. 5200 psi., b. 6047 psi., c. 6110 psi. a. 15.24 ppg., b. 14.65 ppg., c. 14.10 ppg. a. 342.47 ft., b. 684.94 ft., c. 908.28 ft. a. 0.398 psi/ft., b. 0.1345 psi/ft., c. 0.159 psi/ft. a. 2912 psi., b. 1838 psi., c.1658 psi. a. 10.92 ppg., b. 12.11 ppg., c. 11.48 ppg., d. 14.45 ppg. a., c. b. a. 0.72-0.80, b. 2.3-2.5 d b a

F4 Calculation Exercise 1 2 3 4 5 6 7 8 9 10 11 12 13

6115 12563 6520 4795

c b b a c c b b b d d b d

Page 65 of 769

115 466 320 343


14 15 16 17 18 19 20 21 22 23

24 25 26

27

b c a ; // subsea // b d ; // subsea // c b a c b ; //subsea// b d 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) 8899 stks, //subsea// 8460 stks, i) 10119 stks //subsea// 9680 stks b a) Yes, b) No, c) Yes 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 4.225 psi

F4 Surface BOP Stacks 1 2 3 4 5 6 7 8 9 10 11 12

158 psi 2036 strokes 1595 psi 1084 psi 243 strokes 1393 psi 632 strokes 1175 psi 6.53 psi / 100 strokes 308 psi 261 psi Breakdown at the shoe

F4 Subsea BOP Stacks 1 158 psi 2 2036 strokes 3 1595 psi 4 1084 psi 5 243 strokes 6 1393 psi 7 632 strokes 8 1175 psi 9 6.53 psi / 100 strokes 10 308 psi 11 261 psi 12 1532 strokes

Page 66 of 769


Formulae for Well Control. 1

Hydrostatic pressure: Ph (psi)

Ph = MW × TVD × (0.052)

2

Pressure gradient: Gm (psi/ft)

Gm = MW × (0.052)

3

To convert pressure gradient to mud weight.

4

To convert pressure to mud weight.

5

Hydrostatic pressure using pressure gradient.

6

Equivalent Circulating Density: ECD (ppg)

7

Maximum Allowable Mud Weight from leak off test: MAMW (ppg)

8

Maximum Allowable Annular Surface Pressure: MAASP (psi)

9

Boyle’s gas law.

10

Pressure change for change in stroke rate.

 SPMnew  Pnew = Pold ×   SPMold 

11

Pressure change for a change of mud weight

 MWnew  Pnew = Pold ×   MWold 

12

Kill weight mud: MWkill (ppg)

13

Shut In Casing Pressure (psi)

14

Initial Circulating Pressure: ICP (ppg)

ICP = Pscr + SIDPP

15

Final Circulating Pressure: FCP (ppg)

FCP = Pscr ×

EMW =

Ph (0.052) × (TVD)

Ph = Gm × TVD ECD = MW +

APL (0.052) × (TVD)

MAMW = MWlot +

Plot (TVDshoe ) × (0.052)

MAASP = ( MAMW − MW ) × (TVDshoe) × (0.052) P1 × V 1 = P 2 × V 2

MWkill = MW +

2

SIDPP (TVD) × (0.052)

SICP = (Gm − Gi) × Hi + SIDPP

Pscr = PL 16

Gm (0.052)

MW =

Barite required to increase mud weight: Barite (lbs/bbl)

Barite =

Page 67 of 769

MWkill MW

1500(W 2 − W 1)) (35.8 − W 2)


17

Height of influx: Hi (ft)

18

Trip Margin/Safety Factor (ppg)

19

Estimated influx gradient (psi/ft).

20

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

21

Percolation Rate (ft/hr)

22

Pipe capacity (BBL/ft).

Hi =

Gain Pit Capacity Annular

Safety Margin + MW TVD × 0.052

TM =

 SICP − SIDPP  G Influx = MW × (0.052) −   Hi   Output = (0.000243) × ( D) 2 × Stroke

PR =

∆P( psi / hr ) MW × 0.052

Capacity Pipe 23

Annular capacity (bbl/ft).

24

Annular velocity (ft/min)

Capacity Annular = Velocity Annular =

Outputpump= bbl/min 25

Formation Pressure (Pf)

26

Gas migration rate: Rm (ft/hr

27

Height of slug.

28

29

( ID) 2 = 1029.4 ( Dh) 2 − ( Dp) 2 1029.4

1029.4 × Output Pump ( Dh ) 2 − ( D p ) 2

=

Output Pump Capacity Annular

Pf = SIDPP + ( MW × (0.052) × TVD)

Rm =

∆SICP Increase in pressure (psi / hr) = Gm Gm

H slug =

Height of dry pipe.

Volume slug Capacity Pipe

H Drypipe =

Slug weight.

MWslug =

=

Height Drypipe × MW ( MWslug − MW )

H Slug × ( MWslug − MW ) MW H Drypipe × MW H Slug

+ MW

30

Bottomhole pressure (BHP): Static case in drillstring.

BHP = SIDPP + Ph (Mud in drillstring)

31

Bottomhole pressure: Static case in annulus.

BHP = SICP + Ph (Mud and influx in annulus)

Page 68 of 769


32

Bottomhole pressure: Dynamic case.

BHP = Ph ( drillstring ) + APL

33

Cross sectional area (square inches).

A = Diameter 2 × 0.7854

34

Rectangular tank capacity.

35

Change in BHP when pulling dry pipe out of hole.

36

Capacity ( BBL) =

Change in BHP when pulling wet pipe out of hole.

∆BHP =

∆BHP =

Page 69 of 769

L × W × D (FT) 5.61

MW × 0.052 × LPipe × DisplacementMetal (Capacity Csg − DisplacementMetal )

MW × 0.052 × L Pipe × ( Displacement + Capacity Pipe ) ( Capacity Csg − Displacement − Capacity Pipe )

Well Control Exercises.pdf

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