Exercise Book V1.pdf

Well Control Training Manual

L3, L4 Exercises

1 Exercises 1.1 Chapter 1 questions Question 1.1. Which of the below elements are impacted by a well control event? SELECT THREE ANSWERS a) Personal wellbeing b) Environment c) Crew competence d) Reputation Question 1.2. What is the primary means used to prevent formation fluid entering the wellbore?

Question 1.3. What is the correct definition of primary well control during normal drilling operations?

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a) The BOP stacks. b) The annular preventer. c) Hydrostatic pressure of the mud. d) Monitoring trips.

a) Preventing the flow of formation fluid into the wellbore by maintaining the sum of hydro static pressure and dynamic pressure loss in the annulus equal to or greater than formation pressure. b) Preventing the flow of formation fluid into the wellbore by maintaining hydrostatic pressure equal to or greater than formation pressure. c) Preventing the flow of formation fluid into the wellbore by using BOP equipment when the hydrostatic pressure in the wellbore does not balance or exceed the formation pressure. d) Preventing the flow of formation fluid into the wellbore by maintaining the dynamic pressure loss in the annulus equal to or greater than formation pressure. Question 1.4. What is the secondary means used to control formation fluid pressure? a) The blowout preventers. b) Mud viscosity. c) Mud hydrostatic pressure. d) Cement plugs.

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Question 1.5. - API Calculate the hydrostatic pressure given the following data. a) TVD = 3000 ft. MD = 3000 ft, ρ = 10.5 ppg b) MD 8100 ft, TVD = 7750 ft. ρ = 10.7 ppg Convert the following densities into pressure gradients. c) ρ = 10.0 ppg d) ρ = 12.5 ppg Convert the following pressure gradients into densities. e) G = 0.462 psi/ft f) G = 0.544 psi/ft Question 1.6. - Metric Calculate the hydrostatic pressure given the following data. a) TVD = 1200 m, MD = 1200 m, p = 1.15 kg/I b) MD = 3200 m, TVD = 2930 m, p = 1.23 kg/I

Convert the following pressure gradients into densities. e) G = 0.135 bar/m f) G = 0.162 bar/m

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Convert the following densities into pressure gradients. c) ρ = 1.25 kg/I d) ρ = 1.50 kg/I

Question 1.7. Consider the following data: Formation pressure 5031 psi / 347 bar Depth 8330 ft TVD / 2515 m TVD Drilling fluid gradient 0.61 psi/ft / 0.14 bar/m What can you say about the state of the well? a) The well is overbalanced. b) The well is in balance. c) The well is underbalanced. The drilling fluid gradient is changed to 0.603 psi/ft / 0.137 bar/m. What can you say about the state of the well? d) The well is overbalanced.

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e) The well is in balance. f) The well is underbalanced. Question 1.8. API A well 9850 ft TVD is filled with 9.2 ppg brine. The plan is to run in the hole to 6200 ft TVD (6600 ft MD) and displace with 8.4 ppg drill water. What will the hydrostatic pressure be at 9850 ft when the drill water has been circulated back to the surface? Metric A well 3000 m (TVD) is filled with 1.10 kg/I brine. The plan Is to run in the hole to 1900 m TVD (2000 m MD) and displace with 1.00 kg/I drill water. What will the hydrostatic pressure be at 3000 m when the drill water has been circulated back to the surface? a) 4815 psi b) 4303 psi c) 4454 psi d) 5164 psi

/332 bar /294 bar /305 bar /356 bar Schlumberger Private

Question 1.9. API Consider a well with a depth of 8000 ft MD / 7500 ft TVD. Calculate the minimum drilling fluid density that is required to hold back a formation pressure of 4100 psi. Metric Consider a well with a depth of 2500 m MD I 2300 m TVD. Calculate the minimum drilling fluid density that is required to hold back a formation pressure of 283 bar. a) 9.8 ppg / 1.15 kg/l b) 9.9 ppg I 1.16 kg/I c) 10.5 ppg / 1.25 kg/I d) 10.6 ppg / 1.26 kg/I Question 1.10. Consider the following data: Formation pressure 4161 psi / 287 bar Depth 8484 ft TVD / 2586 m TVD a) Calculate the pressure gradient required to balance this pressure. b) The program calls for a trip margin of 200 psi / 15 bar. Calculate the required drilling fluid density. Question 1.11. Match the following statements to one of the answers below:

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1. Amount of space between the grains of rock. a) Permeability b) Porosity 2. Rock ability to allow flow of formation fluid through It. c) Permeability d) Porosity Question 1.12. Which pressure gradient value is used when calculating normal formation pressure? a) 0.433 psi/ft / 0.100 bar/m b) 0.465 psi/ft / 0.105 bar/m c) 10.0 ppg / 1.00 kg/I Question 1.13. API Calculate what the normal formation pressure would be at a depth of 10675 ft TVD. Metric Calculate what the normal formation pressure would be at a depth of 3254 m TVD.

a) Abnormal b) Normal c) Subnormal

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Question 1.14. API At a depth of 1315 ft the formation pressure is 653 psi. This formation pressure is: Metric At a depth of 400 m the formation pressure is 45 bar. This formation pressure is:

Question 1.15. What does abnormal pressure mean regarding fluid pressure in the formation? a) The excess pressure due to circulating drilling fluid at high rates. b) The excess pressure that needs to be applied to cause leak-off in a normally pressured formation. c) High density drilling fluid used to create a large overbalance. d) Formation pressure that exceeds the normal formation fluid pressure. Question 1.16. What is the most common cause of abnormal formation pressure worldwide? a) Depleted sands. b) Carbonate layers. c) Undercompacted shales. Question 1.17.

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Which statement is true for under compacted shales? a) The porosity Is lower than expected for a given depth. b) The porosity Is higher than expected for a given depth. Question 1.18. Undercompacted shales tend to drill slower than expected. a) True b) False Question 1.19. What is the reason for abnormal pressure in the crest of a dome shaped reservoir? a) This is due to the artesian effect. b) The highest point of the reservoir is more compacted. c) The reservoir gas Is less dense than the formation water. Question 1.20. Consider the following situation: Schlumberger Private

Gas/water contact in reservoir Top of reservoir Formation water gradient Gas gradient

5400 ft 4000 ft 0.465 psi/ft 0.1 psi/ft

1650 m 1200 m 0.105 bar/m 0.0226 bar/m

Calculate the pressure at top reservoir. a) 2511 psi / 173 bar b) 140 psi/ 10 bar c) 2371 psi/163 bar d) 2231 psi/ 153 bar Question 1.21. L4 A well is drilled into the top of a reservoir, the so-called crest. The top of the caprock will be encountered at 5675 ft / 1730 m. The caprock Is 30 ft I 10 m thick. The bottom of the reservoir, the so-called spill point, Is situated at 6463 ft I 1970 m and is in contact with the normal pressure gradient. Normal pressure gradient 0.465 psi/ft 0.105 bar/m Gas gradient 0.1 psi/ft 0.0226 bar/m Rev 1.0 [Nov 2017]

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a) Calculate the pressure at the top of the caprock. b) Calculate the drilling fluid weight to balance this pressure. c) Calculate the pressure at the crest of the reservoir at 5705 ft / 1740 m. d) Calculate the drilling fluid weight to balance this pressure.

Normal pressure gradient Gas gradient Oil gradient

0.465 psi/ft 0.1 psi/ft 0.347 psi/ft

0.105 bar/m 0.0226 bar/m 0.0785 bar/m

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Question 1.22. L4, Extra Question A well is drilled into the top of a reservoir, the so-called crest. The top of the caprock will be encountered at 5675 ft / 1730 m. The caprock is 30 ft / 10 m thick. The bottom of the reservoir, the so-called spill point, is situated at 6463 ft f 1970 m and is in contact with the normal pressure gradient.

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Question 1.23. A gas bearing formation is overpressured by an artesian effect. Which of the following has created the overpressure? a) Compaction of the formation by the overburden pressure b) A formation water source located at a higher level than the rig floor. c) The difference in density between the gas and the formation fluid. Question 1.24. When should a leak-off test be carried out? a) Before drilling out casing shoe. b) Before running casing. c) Immediately after running and cementing casing. d) After drilling out the casing shoe and 5 to 15 ft of new formation. Question 1.25. After setting casing, which of the following actions are normally taken prior to starting a leak-off test? SELECT THREE ANSWERS Schlumberger Private

a) Run the bit close to bottom. b) Circulate to get a uniform column of mud in the hole. c) To prevent damage to the formation, pull the bit inside the shoe. d) Drill out the casing shoe approximately 10-30 ft (3-10 m) into new formation. e) Line up the mud pump to do the leak-off test at the slow circulating rate. f) To minimize chance of stuck pipe, pull the bit inside the shoe. Question 1.26. Which of the following are needed for the calculation of accurate formation strength at shoe? SELECT THREE ANSWERS a) Calibrated pressure gauge. b) Accurate stroke counter. c) Accurate hole capacity. d) Exact vertical depth of casing shoe. e) Installation of retrievable packer approximately 1000 feet below the rig floor. f) Constant mud weight around the well.

Question 1.27. Which of the following parameters must be accurately recorded to ensure a correct leak-off test result? SELECT THREE ANSWERS

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a) b) c) d) e) f)

L3, L4 Exercises

Measured depth of the casing shoe. Mud volume in the casing. Pumping time till leak-off starts. True vertical depth of the casing shoe. Mud volume pumped till leak-off starts. Mud density in hole.

Question 1.28. Determine the leak-off point in the leak-off graph

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a) 800 psi

b) 700 psi

c) 680 psi

d) 740 psi

Question 1.29. Which of the following is a definition of MAASP? a) The pressure in excess of drilling fluid hydrostatic that, if exceeded, is likely to cause losses at the shoe. b) The maximum pressure allowed in the hole during a kill operation. c) The maximum pressure allowed on the drill pipe gauge during a kill operation. d) The total pressure applied at the shoe that is likely to cause losses Question 1.30. L4 Which of the following would give the highest MAASP? a) When formation fracture pressure is much higher than drilling fluid hydrostatic pressure. b) When the casing shoe is set deep. c) When formation pressure is close to drilling fluid hydrostatic pressure. d) When the casing shoe is set close to surface.

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Question 1.31. Which of the following affect MAASP? SELECT THREE ANSWERS a) The TVD of the last casing shoe. b) The maximum pump pressure. c) The mud weight in the hole. d) Viscosity and water loss of the mud. e) The fracture pressure of the formation at the shoe. f) The ID of the last casing string. Question 1.32. API A well is full of 11.0 ppg mud and a leak-off test has been performed at 6000 ft TVD. A leak-off pressure of 1000 psi has been recorded. Metric A well is full of 1.32 kg/I mud and a leak-off test has been performed at 2000 m TVD. A leak-off pressure of 70 bar has been recorded. Schlumberger Private

Calculate the maximum allowable mud density. a) 15.5 ppg / 1.85 kg/I b) 14.2 ppg / 1.67 kg/I c) 13.4 ppg / 1.50 kg/I d) 11.0 ppg / 1.32 kg/I Question 1.33. Calculate the MAASP using the following information: Casing shoe depth (TVD) Maximum allowable mud density Mud density in hole

6000 ft 15.4 ppg 12.0 ppg

1500 m 1.84 kg/I 1.44 kg/I

Question 1.34. Calculate the MAASP using the following information: Well depth Well depth Casing shoe depth Casing shoe depth Formation fracture gradient Mud density in hole

11200 ft TVD 13000 ft MD 3600 ft TV 3690 ft MD 0.590 psi/ft 9.6 ppg

3400 m TVD 3900 m MD 1100 m TVD 1160 m MD 0.132 bar/m 1.15 kg/I

Question 1.35.

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API A leak-off test is carried out and leak-off occurs at a surface pressure of 650 psi. The casing shoe is set at 3150 ft TVD. The drilling fluid weight is 10.4 ppg. Metric A leak-off test is carried out and leak-off occurs at a surface pressure of 45 bar. The casing shoe is set at 951 m TVD. The drilling fluid weight is 1.15 kg/I. a) b) c) d)

Calculate MAASP. Calculate the formation fracture pressure. Calculate the formation fracture gradient. Calculate the maximum allowable drilling fluid density.

Question 1.36. API 13.3/8-inch casing is set at 5210 ft TVD. Formation strength at the shoe is 3800 psi. Current mud density= 10.6 ppg. Metric 13.3/8-inch casing is set at 1590 m TVD. Formation strength at the shoe is 260 bar. Current mud density = 1.27 kg/I.

Question 1.37. API A leak-off test is carried out and leak-off occurs at a gradient of 0.737 psi/ft. The casing shoe is set at 8200 ft TVD. The drilling fluid weight is 11.9 ppg. Metric A leak-off test is carried out and leak-off occurs at a gradient of 0.167 bar/m. The casing shoe is set at 2400 m TVD. The drilling fluid weight is 1.43 kg/I. a) b) c)

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Calculate MAASP.

Calculate the formation fracture pressure. Calculate the maximum allowable drilling fluid density. Calculate the MAASP.

While drilling the hole section the drilling fluid weight is increased to 12.6 ppg / 1.52 kg/I d) e) f)

Calculate the formation fracture pressure. Calculate the maximum allowable drilling fluid density. Calculate the MAASP.

Drilling is continued and the drilling fluid weight is brought back to 12.1 ppg / 1.46 kg/I g) h)

Calculate the formation fracture pressure. Calculate the maximum allowable drilling fluid density.

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I) Calculate the MAASP. Question 1.38. What will happen to MAASP if drilling fluid weight is increased? a) b) c)

MAASP will stay the same. MAASP will increase. MAASP will decrease.

Question 1.39. Question 1.39 When should MAASP be recalculated? a) b) c) d)

At the beginning of each shift. Immediately before entering a reservoir. After each bit change. After changing the drilling fluid weight.

a) b) c) d)

MAASP has become irrelevant, a new leak-off test must be performed. MAASP hasn't changed. MAASP is now lower than before the well kill. MAASP is now higher than before the well kill.

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Question 1.40. Question 1.40 A well kill has just been completed and the full wellbore has been circulated to a heavier drilling fluid. What can be said about MAASP?

Question 1.41. Question 1.41 A well is being killed and part of the gas influx has just been circulated into the casing. What can be said about MAASP. a) b) c) d)

MAASP has become irrelevant. MAASP has not changed. MAASP is now lower than before. MAASP is now higher than before.

Question 1.42. Question 1.42 API While drilling, the drilling fluid weight is increased from 10.1to 10.7 ppg expecting little higher formation pressure below. The casing shoe is at 7060 ft TVD. Metric

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While drilling, the drilling fluid weight is increased from 1.21 to 1.30 kg/I expecting little higher formation pressure below. The casing shoe is at 2152 m TVD. a) Would this change of drilling fluid weight increase or decrease the MAASP? 0 b) Calculate the change in MAASP. Question 1.43. L4 Consider the following leak-off test data: Casing shoe depth Casing shoe depth Surface leak-off pressure Drilling fluid density at test

7700 ft MD 7001 ft TVD 493 psi 11.3 ppg

2347 m MD 2134 m TVD 34 bar 1.36 kg/I

While drilling the open hole section a kick was taken and the well was shut-in: Hole depth 9472 ft MD 2887 m MD Hole depth 7963 ft TVD 2427 m TVD SIDPP 174 psi 12 bar SICP 274 psi 19 bar Drilling fluid density 11.9 ppg 1.43 kg/I

a) b) c) d)

No margin 219 psi/ 15 bar 21 psi / 1.1 bar 100 psi / 7 bar

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Determine the margin between the MAASP and the SICP.

Question 1.44. L4 Assuming a gas kick, which of the following increases the risk of exceeding MAASP during a well kill operation? SELECT FOUR ANSWERS a) Small influx. b) Large influx. c) Small difference between formation fraction pressure and hydrostatic pressure. d) Large difference between formation fraction pressure and hydrostatic pressure. e) Short open hole section. f) Long open hole section. g) Small overbalance h) Large overbalance Question 1.45. Which statement about well barriers is true? a) Well barriers can be seen as an envelope of one or several dependent barrier elements.

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b) Well barriers can be seen as an envelope of one or several independent barrier elements. Question 1.46. Which statement about the minimum number of well barriers is true? a) When working with abnormally pressured formations with potential to flow to surface a minimum of one barrier needs to be in place at any time. b) When working with abnormally pressured formations with potential to flow to surface a minimum of two barriers needs to be in place at any time. Question 1.47. Which statement about the primary well barrier is true? a) It is the first object that prevents flow from a source. b) It is the second object that prevents flow from a source. Question 1.48. Which statement about the secondary well barrier is true?

Question 1.49. Select the statements about barriers that are true. SELECT THREE ANSWERS

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a) It is the first object that prevents flow from a source. b) It is the second object that prevents flow from a source. c) It is the BOP that closes the wellbore. d) It is the kill fluid that is pumped during a well kill.

a) All barrier pressure tests must be recorded; b) All well barriers can be directly pressure tested in the direction of flow. c) The drill string is not regarded as a well barrier. d) Barriers shall be tested in the direction of flow. e) All barrier pressure test documents shall be signed by an authorized person. f) Test charts do not require the signature of an authorized person. Question 1.50. An inflow test is carried out. On which side of the barrier is the hydrostatic head reduced? a) The upstream side. b) The downstream side. Question 1.51. Select the correct statement regarding an inflow test.

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a) An inflow test verifies the integrity of barriers that cannot be directly pressure tested in the direction of flow. b) During an inflow test the hydrostatic pressure above the barrier element is increased. c) During an inflow test the fluid expansion rate of the wellbore is tested. d) An inflow test is considered acceptable when the fluid returns from the wellbore at a constant rate. Question 1.52. There is pressure build up in annulus between two casing strings after the cement job of last casing. What would be the effect of this pressure build up? SELECT THREE ANSWERS a) The shoe of the inner casing may fracture. b) The shoe of the outer casing may fracture. c) The inner casing may collapse. d) The outer casing may collapse. e) The ECD while drilling will increase. f) The inner casing may burst. g) The outer casing may burst.

a) Severity of the hazard. b) Assessment of the risk. c) Probability of the hazard. d) Identification of hazards.

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Question 1.53. Risk is a factor of which two elements? SELECT TWO ANSWERS

Question 1.54. Select the correct risk statement regarding a blowout. a) The probability is high and the severity is very low. b) The probability is high and the severity is very high. c) The probability is low and the severity is very low. d) The probability is low and the severity is very high.

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1.2 Chapter 1 answers 1.1 1.2 1.3 1.4 1.5

1.6

1.7 1.8 1.9 1.10

1.22

1.23 1.24 1.25

1.26 1.27 1.28 1.29 1.30 1.31 1.32 1.33 1.34 1.35

1.36 1.37

1.38 1.39 1.40 1.41 1.42 1.43 1.44 1.45 1.46 1.47 1.48 1.49 1.50 1.51 1.52 1.53 1.54

A, D, F D, E, F C A A A,C,E B 1060 psi / 58 bar 318- 326 psi/ 20-21 bar a) 650 psi / 45 bar b) 2353 psi / 152 bar c) 0.747 psi/ft/ 0.160 bar/m d) 14.3 ppg / 1.63 kg/I 928 psi / 61 bar a) 6043 psi/ 400 bar b) 14.1 ppg / 1.70 kg/I c) 938-969 psi / 63- 64 bar d) 6043 psi / 400 bar e) 14.1 ppg / 1.70 kg/I f) 639-670 psi / 42 bar g) 6043 psi/ 400 bar h) 14.1 ppg / 1.70 kg/I i) 852-883 psi/ 56-57 bar C D C D a) MAASP decreases b) 220 psi/ 19 bar A B,C, F, H A B A B A,D,E B A B,C,G A,C D

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1.11 1.12 1.13 1.14 1.15 1.16 1.17 1.18 1.19 1.20 1.21

A,B,D C B A a) 1638 psi b) 4312 psi c) 0.520 psi/ft d) 0.650 psi/ft e) 8.9 ppg f) 10.5 ppg a) 135 bar b) 353 bar c) 0.123 bar/m d) 0.147 bar/m e) 1.38 kg/I f) 1.65 kg/I A, F C D a) 0.491 psi/ft/ 0.111 bar/m b) 9.9 ppg / 1.20 kg/I 1B;2C B 4964 psi / 342 bar A D C B B C C a) 2639 psi / 182 bar b) 9.0 ppg / 1.08 kg/I c) 2929 psi / 202 bar d) 9.9 ppg / 1.19 kg/I a) 2898 psi / 200 bar b) 2819 psi/ 194 bar c) 2789 psi/ 192 bar B D B, D, F

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1.3 Chapter 2 questions Question 2.1. Subsea The purpose of a riser booster line is to: a) Monitor flow from the riser. b) Assist in cleaning cuttings from the riser. c) Both of the above. d) None of the above. Question 2.2. Subsea What is the weakest link in the marine riser system while diverting a shallow gas kick? a) The upper flex joint at surface. b) The lower flex joint at the LMRP. c) The telescopic joint (slip joint) packer seal. d) The diverter adaptor spool.

a) The hydrostatic pressure of the drilling fluid in the standpipe causes the difference. b) The dynamic pressure loss from the pump to the standpipe affects the reading. c) The stand pipe pressure gauge is situated at a lower elevation than the gauge at the pump.

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Question 2.3. When pumping, the standpipe pressure will be slightly lower than the pressure at the pump. What is the most likely reason for this?

Question 2.4. Consider a well of 6500 ft TVD with a drilling fluid density of 10.8 ppg. When circulating the APL is estimated to be 215 psi. The formation pressure is 3700 psi. a) Calculate the BHP when circulating. b) Calculate the overbalance when circulating. c) Calculate the BHP with the pumps off. d) Calculate the overbalance with the pumps off. Question 2.5. Consider a well of 2000 m TVD with a drilling fluid density of 1.3 kg/I. When circulating the APL is estimated to be 15 bar. The formation pressure is 260 bar. a) Calculate the BHP when circulating. b) Calculate the overbalance when circulating. c) Calculate the BHP with the pumps off. d) Calculate the overbalance with the pumps off.

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Question 2.6. A well is flowing when in static condition (i.e. when not circulating) but is not flowing during circulation. What could be the possible reason? a) The dynamic pressure loss in the drill string causes sufficient overbalance while circulating. b) The annular pressure loss must be higher than the formation pressure while circulating. c) The hydrostatic head and the annular pressure loss in the annulus cause sufficient overbalance while circulating. Question 2.7. Which of the following contributes to the equivalent circulating density (ECD)? a) The pressure loss in the annulus. b) The pressure loss in the open hole section only. c) The pressure loss across the nozzles. d) The pressure loss in the drill string. e) The pressure loss in the surface system.

2815 m 1.27 kg/I 13 bar

a) Calculate the static BHP. b) Calculate the dynamic BHP. c) Calculate the ECD. Question 2.9. Consider the following data: Well TD Drilling fluid density Pump rate Surface lines pressure loss Drill string pressure loss Annulus pressure loss Bit pressure loss

9500 ft TVD 10.4 ppg 70 spm 100 psi 800 psi 87 psi 1015 psi

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Question 2.8. Consider the following data: Well TD 9236 ft Drilling fluid density 10.6 ppg Annular pressure loss 189 psi

2900 m TVD 1.23 kg/I 70 spm 7 bar 55 bar 6 bar 70 bar

a) Calculate the static BHP. b) Calculate the BHP when drilling fluid is circulated at 70 spm. c) Calculate the pump pressure when drilling fluid is circulated at 70 spm. d) Calculate the equivalent circulating density.

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Question 2.10. A vertical well is 5500 ft/ 1650 m deep and filled with 11.2 ppg / 1.34 kg/I drilling fluid. While circulating at 100 spm the pressure losses in the well system are as follows: Surface lines pressure loss Drill string pressure loss Bit pressure loss Annulus pressure loss

150 psi 900 psi 1700 psi 100 psi

10 bar 60 bar 113 bar 7 bar

Calculate the pump pressure when circulating at 90 spm. Question 2.11. Which of the following factors increase the circulating pressure? SELECT TWO ANSWERS a) Use of bigger size of bit nozzle. b) Increase in BHA length. c) Increase in drilled depth. d) Decrease in the drilling fluid density.

Question 2.13. Metric A well is being drilled with a fluid density of 1.33 kg/I and a pump pressure of 155 bar. The density is decreased to 1.20 kg/I. What will the new pump pressure be?

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Question 2.12. API A well is being drilled with a fluid density of 11.1 ppg and a pump pressure of 2240 psi. The density is decreased to 10.5 ppg. What will the new pump pressure be?

Question 2.14. API A well is being drilled with a pump rate of 12 bbl/min and a pump pressure of 2580 psi. The pump rate is increased to 13 bbl/min. What will the new pump pressure be? Question 2.15. Metric A well is being drilled with a pump rate of 2000 I/min and a pump pressure of 178 bar. The pump rate is increased to 2500 I/min. What will the new pump pressure be? Question 2.16. Consider the following data: Drilling fluid density Standpipe pressure@30 spm Drilled depth

10.0 ppg 390 psi 8400 ft

1.20 kg/I 27 bar 2922 m

Calculate the new standpipe pressure at 30 spm if the drilling fluid density is increased to 13.0 ppg / 1.56 kg/I.

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a) 507 psi / 35 bar b) 300 psi/ 21bar c) 659 psi/ 46 bar d) 390 psi/ 27 bar Question 2.17. L4 While circulating at 80 spm, the pump pressure is 130 bar with 1.20 kg/I drilling fluid density. What will be the new pump pressure, if the pump rate is reduced to 60 spm and the drilling fluid weight increased to 1.35 kg/I? While circulating at 80 spm, the pump pressure is 1885 psi with 10.0 ppg drilling fluid density. What will be the new pump pressure, if the pump rate is reduced to 60 spm and the drilling fluid weight increased to 11.3 ppg? a) 1598 psi/ 110bar b) 1198 psi/ 82 bar c) 2130 psi/ 146 bar d) 1060 psi/ 73 bar

Drilling fluid density Standpipe pressure Pump rate Pump efficiency

13.1 ppg 3103 psi 110spm 94%

1.57 kg/I 214 bar 110spm 94%

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Question 2.18. L4 The following data are recorded while drilling:

What will be the new standpipe pressure, if the pump rate is reduced to 90 spm and the drilling fluid density increased to 13.6 ppg / 1.63 kg/I? a) 2001 psi / 138 bar b) 2027 psi/ 140 bar c) 1881 psi / 30 bar d) 2156 psi / 149 bar Question 2.19. The Driller recorded the pump pressure at 3 different pump rates. Which pump rate/ pump pressure combination seems invalid? a) 30 spm: 320 psi / 22 bar b) 40 spm: 570 psi/ 40 bar c) 50 spm: 690 psi/ 52 bar Question 2.20.

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Why is the trip tank used to monitor volumes during trips? a) The reduced surface area of the trip tank makes monitoring of volumes more accurate. b) The trip tank has a small volume which makes monitoring of volumes more accurate. c) The large surface area of the trip tank makes monitoring of volumes more accurate. d) The trip tank has a large volume which makes monitoring of volumes more accurate. Question 2.21. Select the correct statement regarding pulling dry pipe. a) Pulling dry pipe means that only the pipe capacity is removed from the wellbore. b) Pulling dry pipe means that both the capacity and the pipe metal volume are removed from the wellbore. c) Pulling dry pipe means that only the pipe metal volume is removed from the wellbore. d) Pulling dry pipe means that the closed end displacement is removed from the wellbore.

Question 2.23. Pipe is pulled out of hole and the driller notices that the actual level drop in the trip tank is less than the predicted level drop. What can be cause of this? a) Drilling fluid is lost to the wellbore. b) Formation fluid is entering the wellbore. c) Dry pipe is being pulled instead of pipe wet. d) Drilling fluid is u-tubing from the drill string back into the annulus. Question 2.24. Consider the following data: Drill pipe capacity Drill pipe metal displacement Stand length

0.01776 bbl/ft 0.0083 bbl/ft 93 ft

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Question 2.22. When pulling out of hole et, drilling fluid from the pulled pipe returns back to the wellbore. a) True b) False

9.26 l/m 4.33 l/m 28 m

a) Calculate the volume of drilling fluid required to fill the hole per stand when pulled dry. b) Calculate the volume of drilling fluid required to fill the hole per stand when pulled wet. Question 2.25. Consider the following data: Drill pipe capacity Drill pipe metal displacement Stand length

0.01720 bbl/ft 0.00821 bbl/ft 90 ft

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Five stands of drill pipe are pulled from the well. a) b)

How much drilling fluid should return to the well when the stands are pulled dry? How much drilling fluid should return to the well when the stands are pulled wet?

Question 2.26. Consider the following data: Drill collar capacity Drill collar metal displacement Stand length

0.0073 bbl/ft 0.0370 bbl/ft 90 ft

3.81I/m 19.30 I/m 27 m

Two stands of drill collars are pulled from the well dry. How many barrels of drilling mud should be pumped into the well to keep the well full? a) 5.5 bbl/ 874 I b) 6.6 bbl/ 1042 I c) 1.3 bbl / 206 I d) 7.9 bbl/ 1248 I

11.0 ppg 0.0178 bbl/ft 0.0076 bbl/ft 0.0708 bbl/ft 98 ft

1.32 kg/I 9.28 l/m 3.96 I/m 36.93 l/m 30 m

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Question 2.27. Consider the following data: Drilling fluid density Drill pipe capacity Drill pipe metal displacement Casing capacity Stand length

I. Calculate the pressure drop when pulling 10 stands of drill pipe dry without filling the hole. a) 7 psi/ 0.5 bar b) 67 psi/ 4.7 bar c) 188 psi / 13.0 bar d) 314 psi/ 21.7 bar ii. Calculate the pressure drop when pulling 10 stands of drill pipe wet without filling the hole. a) 31 psi/ 2.2 bar b) 67 psi/ 4.7 bar c) 188 psi / 13.0 bar d) 314 psi/ 21.7 bar Question 2.28. Calculate the fluid level drop when pulling 2 stands of drill collars wet without filling the hole. Drill collar capacity Drill collar metal displacement

0.0073 bbl/ft 0.0370 bbl/ft

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Casing capacity Stand length a) b) c) d)

L3, L4 Exercises

0.0708 bbl/ft 90 ft

36.93 I/m 27 m

19 ft/ 5.6 m 56 ft/ 16.9 m 94 ft/ 28.2 m 113 ft/ 33.8 m

Question 2.29. Extra While tripping out the hole drill pipe the wellbore was kept full. The trip tank pump was stopped and the complete BHA was pulled out dry. Hole size BHA length BHA capacity BHA metal displacement Casing capacity Annular capacity SHA/casing Drilling fluid density

8.1/2 inch 600 ft 0.006 bbl/ft 0.030 bbl/ft 0.072 bbl/ft 0.035 bbl/ft 12.0 ppg

8.1/2 inch 183 m 3.13l/m 15.65 l/m 37.56l/m 18.26 l/m 1.44 kg/I

BHA length BHA capacity BHA metal displacement Casing capacity Annular capacity SHA/casing Drilling fluid density

580 ft 0.0061 bbl/ft 0.032 bbl/ft 0.0719 bbl/ft 0.0322 bbl/ft 11.0 ppg

177 m 3.18 l/m 16.69 l/m 37.50 l/m 16.80 l/m 1.32 kg/I

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Question 2.30. The BHA is pulled wet without filling the hole.

Calculate the reduction in bottom hole pressure. a) 148 psi/ 10.2 bar b) 307 psi/ 21.1 bar c) 258 psi/ 17.8 bar d) 176 psi/ 12.1 bar Question 2.31. Consider the following data: Overbalance Drilling fluid gradient Casing capacity Drill pipe metal displacement Drill pipe capacity Stand length

290 psi 0.539 psi/ft 0.07078 bbl/ft 0.00757 bbl/ft 0.01754 bbl/ft 98 ft

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Calculate the length of pipe to pull wet to lose overbalance. Question 2.32. A vertical well has been drilled to a depth of 7480 feet. Overbalance 179 psi 12 bar Drilling fluid gradient 0.624 psi/ft 0.141 bar/m Casing capacity 0.1571 bbl/ft 82.95 l/m Drill pipe metal displacement 0.008 bbl/ft 4.17 I/m Stand length 93 ft 28 m How many complete stands can be pulled dry before the overbalance is lost? a) b) c)

59 stands 58 stands 57 stands

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Question 2.33. Consider the following data: Overbalance 200 psi 14 bar Drilling fluid gradient 0.572 psi/ft 0.129 bar/m Casing capacity 0.157 bbl/ft 80.70 I/m Drill pipe metal displacement 0.0082 bbl/ft 4.28l/m Drill pipe capacity 0.0178 bbl/ft 9.28 I/m Stand length 90 ft 28 m A vertical well has been drilled to a depth of 9400 feet. How many complete stands can be pulled wet before the overbalance is lost? a) 19 stands b) 20 stands c) 21 stands Question 2.34. Extra Consider the following data: Well depth Casing shoe depth Drilling fluid gradient Formation pressure Casing capacity Open hole capacity Drill pipe metal displacement Drill pipe capacity Stand length

9629 ft TVD 4193 ft TVD 0.535 psi/ft 5003 psi 0.1497 bbl/ft 0.1458 bbl/ft 0.00757 bbl/ft 0.01754 bbl/ft 99 ft

2935 m TVD 1278 m TVD 0.121 bar/m 345 bar 78.08 I/m 76.04I/m 3.95 I/m 9.15 l/m 30 m

Calculate the number of stands that can be pulled dry before the well starts to flow.

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a) b) c) d) e)

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21 22 53 52 50

Question 2.35. Casing capacity Annular capacity Joint length 40 ft Drilling fluid density

L4 Consider the following data: 0.152 bbl/ft 79.3 l/m 0.1238 bbl/ft 64.6 l/m 12 m 12.6 ppg 1.50 kg/I

A 13.3/8" casing string is run into the wellbore without being filled up. After having run 14 joints of casing, the float valve fails and drilling fluid u-tubes back into the casing. Calculate the effect on bottom hole pressure.

Extra: Calculate the fluid level drop in the well due to the float failure. Question 2.36. After having reached TD and flow-checking the well, a 13.5 ppg / 1.6 kg/I slug is pumped to pull dry pipe. The slug causes a 295 ft/ 90 m level drop in the drill pipe. The drilling fluid density is 11.3 ppg / 1.35 kg/I. Calculate the BHP increase due to pumping the slug.

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a) BHP will decrease with 14 psi / 1.0 bar b) BHP will decrease with 165 psi/ 11.1 bar c) BHP will decrease with 202 psi/ 13.6 bar d) BHP will increase with 202 psi / 13.6 bar

a) 207 psi/ 14 bar b) 0 psi/ 0 bar c) 34 psi / 2.2 bar d) 173 psi/ 12 bar Question 2.37. A heavy slug is circulated into the well. At which point does BHP increase? a) As soon as the heavy slug is pumped into the drill string. b) Once the heavy slug enters the annulus. c) When the heavy slug has been fully displaced into the drill string. d) When the heavy slug has been fully displaced into the annulus. Question 2.38.

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Accidentally a light slug is pumped into the drill string. What happens to BHP once the top drive is disconnected? a) BHP remains the same. b) BHP increases. c) BHP decreases. Question 2.39. Consider the following data: Stand length Drill pipe capacity Slug density Drilling fluid density

98 ft 0.01735 bbl/ft 13.5 ppg 11.3 ppg

30 m 9.05 l/m 1.62 kg/I 1.26 kg/I

The driller likes to have 1 stand of dry pipe while POOH. Calculate the required slug volume.

Question 2.40. Consider the following data: Well depth TVD 8000 ft Drill pipe capacity 0.0176 bbl/ft Drilling fluid density 9.8 ppg Slug density 11.3 ppg Slug volume 20 bbl

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a) 1.7 bbl/ 271I b) 5.0 bbl/ 500 I c) 8.7 bbl/ 950 I d) 10.0 bbl/ 1000 I

2500 m 9.2I/m 1.17 kg/I 1.35 kg/I 3200 I

To pull dry pipe, a slug volume was displaced into the drill string. The trip tank was lined up to the well and the top drive was disconnected to allow the slug to fall. Calculate the volume increase in the trip tank due to slug u-tubing. a) 3 bbl/ 492 I b) 20 bbl / 3200 I c) 23 bbl / 3692 I Question 2.41. Prior to starting a trip out of hole, a heavy slug is pumped and followed by 10 bbl / 1500 I of regular drilling fluid. Well depth TVD Drill pipe capacity Annulus capacity DP/Casing

10200 ft 0.0177 bbl/ft 0.0514 bbl/ft

3100 m 9.23 I/m 26.21I/m

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Drilling fluid density 10.8 ppg 1.29 kg/I Slug density 13.3 ppg 1.59 kg/I Slug volume 15 bbl 2500 I How far will the drilling fluid level drop when the well has equalized? a) b) c) d)

196 ft/ 63 m 247 ft/ 75 m 597 ft/ 182 m 847 ft/ 271 m

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1.4 Chapter 2 answers 2.1 2.2 2.3 2.4

2.5

2.6 2.7 2.8

2.9

2.16

2.17 2.18 2.19 2.20 2.21 2.22 2.23 2.24

d) No 0/B - 5 bar U/B C

b) 11.4 bbl/ 1789 ltr 2.26 B

A a) 5091 psi / 351 bar b) 5280 psi/ 364 bar c) 11.0 ppg / 1.32 kg/I a) 5138 psi/ 350 bar b) 5225 psi/ 356 bar c) 2002 psi / 138 bar d) 10.6 ppg / 1.25 kg/I 2309 psi / 154 bar B,C 2119 psi 140 bar 3028 psi 278 bar

2.27 2.28 2.29 2.30 2.31 2.32 2.33 2.34 2.35 2.36 2.37 2.38 2.39 2.40

A

2.25

B D C A C B B a) 0.77 bbl/ 121I b) 2.42 bbl/ 380 I a) 3.7 bbl/ 578ltr

i) B; ii) D D C D 978.6 ft/ 298.1 m C A D C; level drop = 309 ft / 93 m B B C C A

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2.10 2.11 2.12 2.13 2.14 2.15

B C B a) 3865 psi b) 165 psi c) 3650 psi d) No 0/B - 50 psi U/B a) 270 bar b) 10 bar c) 255 bar

2.41 A

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1.5 Chapter 3 questions Question 3.1. A kick can only occur when primary well control has been lost and the formation has sufficient porosity. a) True b) False Question 3.2. Which one of the following causes of well kicks is totally avoidable and due to lack of alertness by the driller? a) Gas cut drilling fluid. b) Not keeping the hole full during a trip. c) Abnormal formation pressure. d) Reduction in hydrostatic pressure due to lost circulation.

a) There is a washout in the string. b) Partial lost circulation has occurred. c) A kick has been swabbed in. d) Total lost circulation has occurred.

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Question 3.3. While drilling ahead through a faulted formation, the flow meter drops from 60% to 35%. What is the most likely cause of this?

Question 3.4. While tripping into the hole the driller notices that the displaced volume from the well is less than calculated. What is the most likely cause? a) Gas has been swabbed into the wellbore. b) The well has kicked and gas is migrating up the wellbore. c) Total losses have occurred. d) Partial losses have occurred. Question 3.5. Does a kick always occur after a total loss of circulation? a) No, it depends on the drilling fluid level in the annulus and the formation pressure. b) Yes, losses will always occur above any potential kick zone. c) No, it depends on the reduction in drill string weight.

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Question 3.6. The flow sensor shows a total loss of returns and the drilling fluid level cannot be seen in the annulus. What immediate action should be taken? a) Shut the well in and pump lost circulation material. b) Fill the annulus with water (or lightest drilling fluid available) and record volume. c) Pump at reduced rate adding lost circulation material. d) Continue drilling ahead cautiously. Question 3.7. Severe losses occurred while drilling. The pumps were stopped and the drilling fluid in the well could not be seen. The well was then filled to the top with water. Drilling fluid density 12.0 ppg 1.44 kg/I Sea water density 8.6 ppg 1.03 kg/I Height of water column 150 ft 46 m What is the reduction in bottom hole pressure with the 150 feet/ 46 m of water?

Question 3.8. L4 Whilst drilling ahead, partial losses occur. Static loss rate is measured at 10 bbl/hour (1600 I/hr). A total power loss occurs. Drilling fluid density 10.2 ppg 1.22 kg/I Annular capacity 0.1512 bbl/ft 78.9 1/m If the hole cannot be filled, what will be the reduction in bottom hole pressure after 4 hours?

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a) 94 psi/ 6.5 bar b) 26 psi/ 1.8 bar c) 67 psi/ 4.6 bar d) 30 psi/ 2.5 bar

a) 250 psi/ 17.0 bar b) 560 psi / 38.8 bar c) 175 psi / 10.5 bar d) 140 psi/ 9.7 bar Question 3.9. Subsea A riser fill-up valve is generally installed on risers operating in deep water. What is the purpose of this device? a) To prevent burst pressure damaging the riser. b) To act as a diverter. c) To prevent riser collapse Question 3.10. Subsea From a floating rig, top hole is drilled with a marine riser in place

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Drilling fluid density Sea water density Current overbalance Well depth Water depth Riser length

L3, L4 Exercises

9.6 ppg 8.6 ppg 50 psi 1700 ft 1000 ft 1100 ft

1.15 kg/I 1.3 kg/I 5 bar 520 m 300 m 320 m

Calculate the minimum required drilling fluid density to maintain the current overbalance on bottom after disconnecting the riser. Question 3.11. Subsea From a floating rig, top hole is drilled with a marine riser in place. Drilling fluid density 11.6 ppg 1.39 kg/I Sea water density 8.6 ppg 1.03 kg/I Well depth 3000 ft 1000 m Air gap 100 ft 30 m Water depth 1000 ft 300 m

Question 3.12. You're drilling a sandstone that contains some gas. The well is overbalanced. What should you do?

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a) Determine the reduction in BHP when the riser is lost. b) Calculate the additional drilling fluid weight that is required to maintain the same BHP when the marine riser is lost. c) Determine the additional 0/B when drilling would continue with the new drilling fluid weight.

a) Remove the gas using the vacuum degasser. b) Close the BOP and route the return flow via the mud/gas separator Question 3.13. In a well with gas cut drilling fluid, when is the reduction in bottom hole pressure most? a) When the gas is at the casing shoe. b) When the gas is at bottom. c) When the gas reaches surface. Question 3.14. How will bottom hole pressure be affected by gas cut drilling fluid whilst drilling? a) There will be a small drop. b) There will be a large drop. c) There will be no change.

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Question 3.15. Extra - API Calculate the reduction in bottom hole pressure when circulating gas cut mud in the following well: Well TVD 5900 ft Surface to 650 ft 11.2 ppg 650 - 1300 ft 11.8 ppg 1300 ft to bottom 12.5 ppg Original drilling fluid density 12.5 ppg a) 88 psi b) 68 psi c) 24 psi

a) 6.2 bar b) 4.9 bar c) 1.8 bar

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Question 3.16. Extra - Metric Calculate the reduction in bottom hole pressure when circulating gas cut mud in the following well: Well TVD 1800 m Surface to 200 m 1.34 kg/I 200 - 400 m 1.41 kg/I 400 m to bottom 1.50 kg/I Original drilling fluid density 1.50 kg/I

Question 3.17. High volume swabbing causes the level of drilling fluid in the annulus to drop significantly. a) True b) False Question 3.18. Which of the following increase surge pressures when running in the hole? SELECT TWO ANSWERS a) Small annular clearance. b) Large bit nozzles. c) Running in slowly. d) High gel strength mud. e) Large annular clearance. f) Low gel strength mud.

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Question 3.19. Which of the following would be the immediate effect of swabbing? a) Reduction in bottom hole pressure. b) A kick. c) Losses. d) Increase in bottom hole pressure. Question 3.20. Which of the following are likely to increase the chance of swabbing? SELECT THREE ANSWERS a) Pulling through tight hole with the pump off. b) Balled up stabilizers. c) Pulling pipe too slowly. d) Pulling through tight hole with the pump on. e) Maintaining high mud viscosity.

a) Low viscosity in the drill string. b) Pulling out of the hole too fast. c) Large open hole/SHA capacity. d) Long bottom hole assembly. e) Low gel strength in the drilling mud.

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Question 3.21. Swabbing may occur when pulling the drill string out of the hole. Which of the following will increase the possibility of swabbing? SELECT TWO ANSWERS

Question 3.22. When pulling out of the hole from the top of the reservoir at 10000 ft/ 3050 m swab pressures are calculated to be 150 psi / 10 bar. Drilling fluid density 10.2 ppg 1.22 kg/I Formation pressure 5200 psi 360 bar Will the well flow? a) Yes b) No Question 3.23. L4, Extra Pressure recorders located below the drill stem test tools show that swab pressure when pulling a stand was 250 psi/ 17 bar.

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Drilling fluid density 10.0 ppg 1.20 kg/I Top of reservoir 9500 ft 2900 m If the well does not flow when the pipe is static, what would the reservoir pressure have to be at this swab pressure? a) 3800 psi/ 260 bar b) 5800 psi/ 400 bar c) 4690 psi/ 324 bar d) 4940 psi/ 341bar Question 3.24. What is the main function of the trip margin? a) The trip margin gives extra hydrostatic pressure and increases drilling rate. b) The trip margin provides an extra safety margin to prevent going underbalanced when swabbing takes place. c) The trip margin ensures swabbing can never take place.

a) Reduce weight on bit. b) Circulate bottoms up. c) Flow check the well. d) Reduce pump speed.

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Question 3.25. What action should a driller take after a positive drilling break?

Question 3.26. When picking up to check for flow the pumps are usually kept running, why? a) To take a slow circulating rate pressure. b) To check the pressure losses in the annulus. c) To clean the bottom of the hole of cuttings. d) To maximize the pressure on the bottom of the hole. Question 3.27. While drilling ahead at constant rate of penetration (ROP), the shale shakers cannot handle the number of cuttings. What would be the safest course of action? a) Slow down the mud pump until the shakers can handle the number of cuttings in the returns. b) Check for flow - if none, circulate bottoms up at a reduced rate so that shakers can handle cuttings volume, flow check periodically during circulation. c) Check for flow - if none, continue drilling at same ROP. d) Check for flow - if none, then continue at the same ROP. Allow half of the mud returns to bypass the shakers.

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Question 3.28. A well needs to be closed in immediately when a kick warning sign is observed. a) True b) False Question 3.29. During a trip, the driller notices that the well is swabbing. A flow check is negative. What should he do next? a) Because the well is not flowing, continue pulling pipe from the hole. b) Shut-in the well and check for pressures. c) Run to bottom and circulate bottoms up. d) Pull 5 stands and carry out another flow check.

a) Running into the swabbed fluids caused hydrostatic pressure in annulus to drop. b) In the hurry, it was forgotten to slug the pipe while preparing for running back. c) Derrick man mixed too light a slug prior to pulling out of the hole. d) Abnormal formation fluid pressure.

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Question 3.30. While POOH, swabbing is suspected. The flow check is negative and it was decided to run back to bottom (30 stands) checking for flow after lowering every 5 stands. The flow checks after 5, 10, 15 & 20 stands showed no flow. With stand number 25 back in the hole, the well was found to be flowing. What is the most probable cause of the well flow?

Question 3.31. Whilst pulling out of the hole, the driller observes that the hole is taking less than the calculated volume. Which is the safest action to follow? a) Flow check for at least 15 minutes. If the well is static, continue pulling out of the hole slowly and carry out a flow check after every 15 stands pulled. b) Flow check for at least 15 minutes. If the well is static, continue pulling out of the hole slowly to prevent swabbing. c) Flow check. If the well is static, circulate bottoms up and observe returns for any swabbed fluids. If negative, continue pulling out of the hole slowly to avoid any further swabbing. d) Flow check. If the well is static, run back to bottom, whilst monitoring the trip tank. Circulate bottoms up and observe returns for any swabbed fluids.

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Question 3.32. While pulling out of hole, the drilling fluid required to fill the hole is less than calculated. What action should be taken? a) Flow check, if negative continue to pull out of the hole. b) Flow check, if negative pump a heavy slug around the SHA then continue to pull out of hole. c) Flow check, if negative run back to bottom and monitor returns. d) Pump remaining pipe out of the hole. e) Shut the well in and circulate to clean the hole.

Question 3.33. After pulling 15 stands of pipe, the well has not taken the correct amount of mud. The driller takes a flow check, which is negative. What action should be taken?

Question 3.34. Consider the following data: Drill pipe capacity 0.01720 bbl/ft 8.97 I/m Drill pipe metal displacement 0.00821 bbl/ft 4.28I/m Drill collar capacity 0.00769 bbl/ft 4.011/m Drill collar metal displacement 0.05843 bbl/ft 30.4I/m Stand length 90 ft 27 m A drill string is pulled out of hole dry. After having pulled 8 stands of drill pipe the trip tank volume has reduced with 4 bbl/ 400 l. How should the driller proceed?

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a) Run or strip back to bottom and circulate bottoms up. b) Continue pulling pipe from the hole because there was no flow. c) Shut the well in and check for pressures. d) Pull another 15 stands and flow check.

a) Flow check the well. Strip back to bottom and monitor returns. b) Continue POOH, the volume reduction in the trip tank is correct. c) Run back to bottom at a reduced speed. d) Flow check the well. If negative continue POOH at a reduced speed.

Question 3.35. When the differential pressure between the hydrostatic drilling fluid pressure and the formation pressure reduces, the drilling rate decreases. a) True b) False

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Question 3.36. Which of the following drilling practices would be considered when connection gas is noticed? SELECT TWO ANSWERS a) Control drilling rate so that only one slug of connection gas is in the hole at any one time. b) Pulling out of the hole to change the bit. c) Raising the drilling fluid viscosity. d) Reduce the drilling fluid viscosity to minimize swabbing during trips. e) Minimizing the time during a connection when the pumps are switched off. Question 3.37. Which one of the following causes connection gas? a) Pulling pipe too fast during a trip. b) An increase in bottom hole pressure during a connection. c) Air in pipe when making a connection. d) A decrease in bottom hole pressure during a connection.

a) ROP meter b) Pump pressure gauge c) Flow line temperature gauge d) SPM counter e) Gas detector f) Return flow meter

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Question 3.38. Which of the following equipment may pick up a kick warning sign while drilling? SELECT THREE ANSWERS

Question 3.39. Which of the following are kick warning signs? SELECT TWO ANSWERS a) Increase in pump pressure. b) Reduction in rate of penetration. c) Change in cuttings size & shape. d) Increase in weight on bit. e) Increase in drilled gas percentage. Question 3.40. Which of the following is the first reliable indication that the well is flowing? a) Increase in torque. b) Gas cut drilling fluid.

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c) Decrease in pump pressure. d) Increase in return flow rate. Question 3.41. Which of the following is a kick indicator? SELECT TWO ANSWERS a) Increase in chloride content. b) Increase in return flow rate. c) Increase in temperature gradient. d) Increase in active tank volume. e) Increase in gas levels.

Question 3.42. When a kick indicator is observed the well should be flow checked immediately. a) True b) False

a) b) c) d)

Flow check the well immediately. Increase the drilling fluid density. Shut-in the well immediately. Close the diverter.

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Question 3.43. When an increase in return flow is observed the following should be done.

Question 3.44. While drilling, which of the following situations make kick detection with a PVT (pit volume totalizer) more difficult for the driller? a) Bypassing the solids control equipment at the pits. b) Reducing the range of the PVT high and low alarms settings. c) Allowing drilling fluid losses by overflowing the shakers. d) Keeping the active mud system transfers to a minimum when drilling ahead. Question 3.45. While drilling with water based drilling fluid a gas kick is taken and a total pit gain of 40 bbl. 6 m3 is observed. What would have been the pit gain in the case of oil based drilling fluid? a) Possibly larger than 40 bbl/ 6 m3. b) Also 40 bbl/ 6 m3. c) Possibly smaller than 40 bbl / 6 m3.

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Question 3.46. When a gas kick is dissolved in the oil based drilling fluid system it can migrate and expand freely. a) True b) False Question 3.47. Select True or False for each statement describing differences between drilling with water based mud and oil based mud.

Question 3.48. The solubility of gas in oil based or water based mud can make a difference to the shut-in well data following a kick under identical conditions. Which of the following statements is correct when using oil base mud? SELECT TWO ANSWERS

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a) Hydrocarbon gas is more soluble in water based mud than in oil based mud. True/ False b) Gas entering the wellbore whilst drilling with water based mud will reduce the hydrostatic pressure more than it would when drilling with oil based mud. True/ False c) Whilst circulating out a kick, gas expansion occurs in the annulus at the same rate in oil and water based mud. True/ False d) The solubility of gas in oil based drilling fluids generally increases with depth. True/ False e) Gas kicks in water based mud are normally harder to detect due to higher gas solubility. True/ False

a) The shut-in casing pressure will be higher. b) There will be no difference in pit gain compared with water based drilling mud. c) The initial pit gain will be lower. d)There will be no difference in shut in-casing pressure compared with water based drilling mud. e) The initial pit gain will be higher. f) The shut-in casing pressure will be lower.

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1.6 Chapter 3 answers 3.1 3.2 3.3 3.4 3.5

B B 8 D A 3.6 8 3.7 8 3.8 D 3.9 C 3.10 12.9 ppg / 1.45 kg/I 3.11 a) 216 psi/ 15 bar b) 2.2 ppg / 0.22 kg/I c) 343 psi / 22 bar A C A 8 B 8 A,D A A,B,E B,D A C

C D 8 B C A D C A

3.32 3.33 3.34 A 3.35 8 3.36 A, E 3.37 D 3.38 A,C, E 3.39 C,E 3.40 D 3.41 B,D 3.42 B 3.43 C 3.44 C 3.45 C 3.46 B 3.47 a) False; b) True; c) False; d) True; e) False 3.48 C, F

B

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3.12 3.13 3.14 3.15 3.16 3.17 3.18 3.19 3.20 3.21 3.22 3.23 3.24

3.25 3.26 3.27 3.28 3.29 3.30 3.31

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1.7 Chapter 4 questions Question 4.1. A rig is operating in an environmentally sensitive area and is in zero-discharge mode. One of the crew members notices that the shaker screens are blinding and overflowing. What should he do? a) He should inform the driller about the situation. b) He should inform the mud engineer about the situation. c) He should rig up a hose to direct the flow overboard. d) He should open the dump valve to the sand trap. Question 4.2. Why should the well be shut-in quickly after a kick has been detected? a) To minimize the shut-in drill pipe pressure (SIDPP). b) To reduce the migration speed of the influx. c) To minimize the size of the influx.

a) Running regular pit drills for the drill crew. b) Switching off the flow meter alarms. c) Regular briefing for the derrickman on his duties regarding the monitoring of pit level. d) Drilling a further 15 ft after a drilling break, before flow checking. e) Calling the toolpusher to the rig floor prior to shutting-in the well. f) Testing stab-in valves during BOP tests.

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Question 4.3. Which of the following practices would lead to a bigger influx when shutting the well in? SELECT THREE ANSWERS

Question 4.4. The valve setting for the hard-shut-in procedure is as follows: a) HCR open, choke closed, valve upstream of choke closed. b) HCR open, choke closed, valve upstream of choke open. c) HCR closed, choke open, valve upstream of choke closed. d) HCR closed, choke closed, valve near choke closed.' Question 4.5. Which shut-in procedure is followed with these steps? 1. Stop drilling; 2. Pick up the string and space out for the BOP; 3. Stop the pumps; 4. Close the annular or the pipe rams;

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5. Open HCR; a) Soft shut-in b) Hard shut-in Question 4.6. The valve setting for the soft shut-in procedure is as follows: a) HCR closed, choke open, valves upstream and downstream of choke open. b) HCR closed, choke closed, valve upstream of choke closed. c) HCR closed, choke open, valve upstream of choke closed. d) HCR closed, choke open, valve downstream choke closed. Question 4.7. Listed below are the two shut-in procedures for handling a kick: Procedure 1: With choke open, stop drilling, pick up off bottom, shut down pumps, open HCR, close BOP, close choke, close a valve near the choke Procedure 2: With choke already closed, stop drilling, pick up off bottom, shut down pumps, close BOP, open HCR.

Question 4.8. Subsea Which shut-in procedure is followed with these steps? 1. Stop drilling; 2. Pick up the string and space out for the BOP; 3. Stop the pumps; 4. Close the annular or the pipe rams; 5. Open choke line fail-safe valves; 6. Notify supervisor; 7. Start recording shut-in pressures.

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a) Which procedure refers to the soft shut-in? b) Which procedure refers to the hard shut-in?

a) Soft shut-in b) Hard shut-in Question 4.9. Subsea Why would a driller need to know the sea tides? a) This information is required to adjust tension in the riser tensioners. b) This information is required to enable correct drill string space-out during shut-in. c) This information is required to adjust the telescopic joint expansion. d) This information is required to adjust the upper flex joint angle.

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Question 4.10. The hard-shut-in procedure is much harder to execute and therefore takes longer than the soft shut-in procedure a) True b) False Question 4.11. The hard-shut-in procedure generally leads to a smaller influx volume. a) True b) False Question 4.12. When a kick indicator is observed during tripping the well is shut-in before the drill pipe safety valve is installed. a) True b) False

a) Stab a drill string safety valve and shut the well in. b) Pull back to shoe and shut-in. c) Run back to bottom as quickly as possible. d) Close the annular and make up the top drive or Kelly.

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Question 4.13. Whilst tripping out of the hole the well starts flowing. What is the first action that should be taken?

Question 4.14. Which of the following describes the procedure for a hard shut-in during tripping as per API RP 59? a) Stab full opening drill pipe safety valve. Close drill pipe safety valve. Close BOP. Open HCR. b) Stab full opening drill pipe safety valve. Close drill pipe safety valve. Open HCR. Close BOP. c)Make up top drive.

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Close BOP. Open HCR. Question 4.15. Which of the following describes the procedure for a soft shut-in during tripping as per API RP• 59?

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a) Stab a full opening drill pipe safety valve. Close the drill pipe safety valve. Open HCR. Close BOP. Close choke. Close a valve near the choke. b) Open HCR. Close BOP. Close choke. Close a valve near the choke. Stab full opening drill pipe safety valve. Close the drill pipe safety valve. c) Stab full opening drill pipe safety valve. Close the drill pipe safety valve. Close BOP. Open HCR. Close choke. Close a valve near the choke. Question 4.16. A well has been shut-in while tripping. The forward plan is to strip to bottom. Which additional equipment is required? a) Install a DPSV on top of the inside BOP and open the inside BOP. b) Install a DPSV on top of the inside BOP and close the inside BOP. c) Install an inside BOP on top of the DPSV and close the DPSV. d) Install an inside BOP on top of the DPSV and open the DPSV. Question 4.17. A shallow gas kick must be closed in with the soft shut-in procedure to prevent fracturing the formation a) True b) False

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Question 4.18. Which of the following are good operating practices in top hole when there is a risk of shallow gas? SELECT FOUR ANSWERS a) Circulate while pulling out of the hole on trips. b) Use high density drilling fluid to create maximum overbalance. c) Drill a pilot hole. d) Maintain a high fluid viscosity. e) Regularly pump fresh water pill to remove the cuttings from the hole. f) Drill at a controlled ROP. g) Install a non-return valve in the BHA. Question 4.19. During top hole drilling from a jack-up rig, the well starts to flow due to shallow gas. What will be the safest action to take to secure the safety of the rig and personnel? SELECT TWO ANSWERS

Question 4.20. At the first indication of shallow gas follow the following procedure:

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a) Line up to the mud/gas separator, close the diverter and send personnel to the lifeboats. b) Start pumping drilling fluid into the well at the highest possible rate. c) Shut-in the well and prepare to bullhead immediately. d) Shut-in the well using the blind/shear rams. e) Activate the diverter system and remove non-essential personnel from the rig floor.

a) Stop drilling - open downwind vent line - close the shaker valve - close the diverter - increase pump rate. b) Stop drilling - increase pump rate - open downwind vent line - close the shaker valve - close the diverter. c) Stop drilling - close the diverter - open downwind vent line - close the shaker valve - increase pump rate. d) Close the diverter - stop drilling - open downwind vent line - close the shaker valve - increase pump rate. Question 4.21. If a shallow gas flow is detected, which of the following actions should be taken at the same time as activating the diverter? a) Maintain the pump rate. b) Increase the pump rate. c) Decrease the pump rate.

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1.8 Chapter 4 answers 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8

A C B,D,E D B A a) Procedure 1 b) Procedure 2 B

4.9 B 4.10 B

4.11 4.12 4.13 4.14 4.15 4.16 4.17 4.18 4.19

A B A A A D B A,C,F,G B, E

4.20 A 4.21 B

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1.9 Chapter 5 questions Question 5.1. Consider a simple U-tube where both legs contain the same fluid. The U-tube is open at the top. One leg is displaced partly to a heavier fluid. What can you say about the pressure at the bottom of the U-tube. a) The BHP increases. b) The BHP decreases. c) The BHP stays the same Question 5.2. A gas kick is taken while drilling and the well is shut-in. Why is SICP usually higher than SIDPP? a) The annulus capacity is larger than the drill pipe capacity. b) The BHP on the casing side is smaller than on the drill pipe side so SICP is also smaller. c) The influx is usually less dense than drilling fluid. d) SICP is not necessarily higher, it depends on whether it is an offshore or land operation.

Question 5.4. Metric A gas kick is taken and the well is closed-in. The well is 3115 m MD/ 2330 m TVD deep, the formation pressure is 330 bar and the drilling fluid density is 1.33 kg/I. Determine the SIDPP.

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Question 5.3. API A gas kick is taken and the well is closed-in. The well is 10220 ft MD/ 7600 ft TVD deep, the formation pressure is 4800 psi and the drilling fluid density is 10.3 ppg. Determine the SIDPP.

Question 5.5. A gas kick is taken and the well is closed-in. The SIDPP is 300 psi and the SICP is 400 psi. The well is 4400 ft TVD deep and the formation pressure is 3000 psi. a) Calculate the BHP. b) Calculate the hydrostatic pressure of the drilling fluid. c) Calculate the drilling fluid density. Question 5.6. A gas kick is taken and the well is closed-in. The SIDPP is 30 bar and the SICP is 40 bar. The well is 1390 m deep and the formation pressure is 210 bar. a) Calculate the BHP. b) Calculate the hydrostatic pressure of the drilling fluid. c) Calculate the drilling fluid density.

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Question 5.7. A gas kick has been taken and the well is shut-in. The influx starts migrating up, what happens to the BHP? a) The BHP increases. b) The BHP remains constant. c) The BHP decreases. Question 5.8. A gas kick has been taken and the well is shut-in. The influx starts migrating up, what happens to the surface pressures? a) The surface pressure increases. b) The surface pressure remains constant. c) The surface pressure decreases. Question 5.9. Which of the following does not increase when the well is shut-in after taking a gas kick?

Question 5.10. Consider the following data: Kick volumes 12.6 bbl 2000 I Formation pressure 2900 psi 200 bar A well has been shut-in and due to gas migration, the surface pressure has increased with 435 psi/ 30 bar.

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a) Bottom hole pressure b) Casing shoe pressure c) Shut-in casing pressure d) Gas bubble pressure

How much volume must be bled off to restore BHP back to formation pressure? Question 5.11. A well was shut-in on a gas kick, SIDPP & SICP have stabilized. After some time both start rising slowly by the same amount. What is the probable cause? a) A further influx is entering the hole. b) The influx is migrating up the wellbore. c) The gauges are faulty. d) The BOP stack is leaking.

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Question 5.12. After shut-in & stabilization of surface pressures, which pressure is to be maintained constant to keep BHP constant? a) Shut-in casing pressure b) Casing shoe pressure c) Formation pressure d) Shut-in drill pipe pressure Question 5.13. What could happen if gas migrates after a well is shut-in and pressures have stabilized? a) Only the drill pipe pressure will increase. b) Shut-in pressures will remain constant. c) Both drill pipe and annulus pressures will increase. d) Only the annulus pressure will increase.

a) The influx is gas and is migrating up the annulus and drill string. b) The influx is gas and is migrating up the annulus. c) The influx is gas and is migrating up the drill string. d) The influx is gas and is expanding rapidly.

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Question 5.14. A well is shut-in following a kick whilst drilling. After pressures have been stabilized, both drill pipe and casing pressures are observed to be gradually increasing by the same amount. What is the most likely reason for this?

Question 5.15. Which of the following parameters affect the value of SIDPP? SELECT TWO ANSWERS a) Drilling fluid density in drill string. b) Volume of influx c) Annular volume from bit to shoe. d) Influx gradient e) Formation pressure Question 5.16. Which of the following parameters affect the value of SICP? SELECT THREE ANSWERS a) Formation pressure b) Drill string capacity c) Annulus capacity d) Bottom hole temperature e) Volume of influx

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Question 5.17. Which of the following parameters is influenced by the permeability of the reservoir? SELECT THREE ANSWERS a) SICP b) Time for pressures to stabilize c) ICP d) Kill fluid density e) Volume of the influx f) SIDPP Question 5.18. SICP is usually higher than SIDPP. If there was a large volume of cuttings in the annulus how would this affect the pressure readings? a) SICP would be higher than expected. b) It would not affect pressure readings. c) SICP would be lower than expected.

a) SICP is lower when compared to drilling a short reservoir section. b) SICP is equal when compared to drilling a short reservoir section. c) SICP is higher when compared to drilling a short reservoir section.

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Question 5.19. A well is drilled into a reservoir and due to high ROP a long reservoir section is drilled before the well is shut-in. Which of the following statements is correct?

Question 5.20. L4 A well is shut-in and following data are recorded: SIDDP 500 psi 35 bar SICP 600 psi 41 bar Drilling fluid density 11.5 ppg 1.36 kg/I The gas migrates 500 ft/ 150 m up the well. What would the shut-in pressures be if no action is taken? a) SIDPP = 500 psi, SICP = 900 psi b) SIDPP = 800 psi, SICP = 600 psi c) SIDPP = 600 psi, SICP = 900 psi d) SIDPP = 800 psi, SICP = 900 psi

SIDPP = 35 bar, SICP = 61 bar SIDPP = 55 bar, SICP = 41 bar SIDPP = 41 bar, SICP = 61 bar SIDPP = 55 bar, SICP = 61 bar

Question 5.21. A well is shut-in on a gas kick.

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SIDPP 400 psi 30 bar SICP 600 psi 40 bar Drilling fluid density 13.0 ppg 1.56 kg/I Influx gradient 0.1 psi/ft 0.23 bar/m After 30 minutes both pressures have risen by 150 psi/ 10 bar due to gas migration. Calculate the gas migration rate in ft/hr or m/hr. a) 325 ft/hr/ 75 m/hr b) 444 ft/hr/ 131m/hr c) 590 ft/hr/ 190 m/hr d) 200 ft/hr/ 50 m/hr Question 5.22. Extra A well is shut-in on a gas kick. Drilling fluid density 10.5 ppg TVD 9200 ft SIDPP 348 psi

1.26 kg/I 2800 m 24 bar

Question 5.23. A gas kick is taken while drilling and the well is shut-in. The SIDPP reads zero pressure. What's going on? a) The influx must be below the bit. b) The influx is already above the BHA. c) There is a float valve in the string.

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After pressure stabilization, both pressures are found to be rising 29 psi/ 2 bar every 10 min. Calculate the gas migration rate in ft/hr or m/hr.

Question 5.24. On shutting in a flowing well, the drill pipe pressure is zero because there is a float in the string. To determine SIDPP, what action should be taken? a) Bring the pump up to kill rate holding the casing pressure constant. The pressure shown when the pump is at kill rate is SIDPP. b) Shear the pipe and read the SIDPP directly off the casing gauge. c) Pump very slowly into the drill string until the float valve opens and SICP starts rising. Read off SIDPP and stop pumping. d) Pump at 2 bbl/min/ 300 I/min into the annulus with the well shut-in. When the pressure is equalized the float will open. This pump pressure is the SIDPP.

Question 5.25. L4 Consider the following data:

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SIDPP Drilling fluid density Influx height Influx gradient

500 psi 9.6 ppg 850 ft 0.1 psi/ft

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35 bar 1.20 kg/I 250 m TVD 0.023 bar/m

Calculate SICP. Question 5.26. A gas kick is swabbed in while tripping and the well is shut-in. The SIDPP and SICP are equal. What can be said about the influx? a) The influx must be above the bit. b) The influx must be below the bit. c) Impossible to say. Question 5.27. A well is shut-in with a gas kick. The bit is 90 ft/ 30 m off bottom and the influx is on bottom and 30 ft/ 10 m long (all the influx is below the bit). What is the SICP likely to be?

Question 5.28. When drilling a horizontal section of a well a gas kick is taken and the well shut-in. If the influx is in the horizontal section what would you expect the SIDPP and SICP to be?

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a) Lower than the SIDPP because of the effect of the annular pressure loss. b) Higher than the SIDPP. c) The same as the SIDPP. d) Will depend on the gradient of the influx.

a) SIDPP equal to SICP. b) SICP is much greater than the SIDPP. c) SIDPP is much greater than the SICP. d) SICP will be zero. Question 5.29. If a kick occurs while drilling a horizontal well, why is there little or no difference between SIDPP and SICP? a) In horizontal wells, the influx is not able to migrate. b) Drilling fluid density and formation fluid density are the same when drilling horizontal wells. c) Volume inside the string is the same as volume in the annulus. d) The vertical height of the influx has little effect on the hydrostatic pressure in the annulus.

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5.1 C 5.2 C 5.3 729 psi 5.4 26 bar 5.5 a) 3000 psi b) 2700 psi c) 11.8 ppg 5.6 a) 210 bar b) 180 bar c) 1.32 kg/I 5.7 A 5.8 A 5.9 D 5.10 2.2 bbl/ 353 I 5.11 B 5.12 D 5.13 C

5.14 5.15 5.16 5.17 5.18 5.19 5.20 5.21 5.22 5.23 5.24 5.25 5.26 5.27 5.28 5.29

B A, E A,C,E A, B,E C C D B 319 ft/hr/ 97 m/hr C C 840 psi/ 59 bar 8 C A D

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Chapter 6 questions

Question 6.1. What is the fundamental principle of all well kill methods? a) Never exceed MAASP. b) Keep BHP constant and equal to formation pressure. c) Always use the Driller’s Method. d) Never exceed a circulation rate of 30 spm. Question 6.2. Which of the following should be considered for selecting a slow circulating rate? SELECT TWO ANSWERS a) Rated working pressure of BOP. b) Casing capacity. c) Barite mixing capability. d) Gas handling capacity of mud/gas separator.

a) Allow choke operator time to make the necessary choke adjustments. b) To reduce damage to the pump. c) Minimize pressures in the wellbore. d) To allow kick fluids to be handled at surface. e) To reduce gas migration. f) To reduce the chance of overloading the mud/gas separator.

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Question 6.3. Why is the influx displaced from the hole at a pump rate slower than used when drilling? SELECT FOUR ANSWERS

Question 6.4. Which 2 important simplifications that can be made when circulating at a reduced rate (SCR) using a surface BOP. SELECT TWO ANSWERS a) There is no APL and associated BHP increase. b) The APL increase is equal to the DPL decrease; hence BHP will stay constant. c) All dynamic pressure losses occur in the drill string and across the bit. d) The gas influx will remain a single bubble when pumping at a reduced rate.

Question 6.5. Which of the following situations require recording of slow circulating rate pressures? SELECT FIVE ANSWERS

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a) b) c) d) e) f) g) h)

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Every shift. Before and after a leak-off test. Drilling fluid property changes. Drilling fluid density changes. After each connection when drilling with top drives. After recharging mud pump pulsation dampener. When long sections of hole are drilled rapidly. When returning to drilling after a kick.

Question 6.6. Which of the following statements about slow circulating rates (SCR) are correct? SELECT THREE ANSWERS a) SCRs should be taken through the choke manifold. b) SCRs are needed to calculate formation pressure. c) SCRs should be taken when drilling fluid properties are changed. d) SCRs should be read on the drill pipe pressure gauge at the remote choke panel. e) SCRs should be taken with the bit near the bottom.

a) Using larger bit nozzles. b) Drilling deeper. c) A decrease of the drilling fluid weight. d) An increase of the drilling fluid viscosity.

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Question 6.7. Which of the following would cause the slow circulation rate pressure to increase if SPM is constant? SELECT TWO ANSWERS

Question 6.8. A well kill is started and after reaching the desired pump rate it is observed that the ICP is higher than the calculated ICP. Which of the below statements is correct? a) Stop the well kill and take new SCR readings. b) Continue the well kill using the observed ICP and adjust the kill graph. c) Open the choke and bleed drill pipe pressure back to the calculated ICP. d) Lower the pump rate so drill pipe pressure drops back to the calculated ICP. Question 6.9. Subsea During a subsea BOP well kill, gas is starting to displace the drilling fluid in the choke line. What must be done to keep BHP constant? a) Open the choke some more. b) Close the choke some more. c) Keep the choke position constant.

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Question 6.10. Subsea A well kill is about to be started. What can be done to keep BHP constant during the start-up. SELECT TWO ANSWERS a) Keep the static kill line pressure constant during start-up. b) Keep the casing pressure constant during start-up. c) Reduce the kill line pressure with the CLPL during start-up. d) Increase the kill line pressure with the CLPL during start-up. e) Reduce the casing pressure with the CLPL during start-up. f) Increase the casing pressure with the CLPL during start-up. Question 6.11. When can the SIDPP be used to calculate the required kill fluid density. a) Take the SIDPP directly after shut-in of the well. b) Wait for 5 to 7 minutes, then take the SIDPP. c) Wait until surface pressure is no longer increasing, then take the SIDPP. d) Wait until the initial build-up is over, then take the SIDPP.

a) Shoe pressure b) Formation pressure c) Hydrostatic pressure d) Fracture pressure e) Kill mud density

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Question 6.12. If the drill string is plugged and SIDPP cannot be read, which of the following parameters cannot be calculated? SELECT TWO ANSWERS

Question 6.13. A kick has been taken and the well is shut-in. SIDPP Drilling fluid density Well depth (MD) Well depth (TV)

326 psi 9.7 ppg 5610 ft MD 5085 ft TVD

23 bar 1.17 kg/I 1710 m MD 1550 m TVD

Calculate the required kill fluid density. Question 6.14. Which gauge is affected by the transmission time?

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a) The drill pipe pressure gauge. b) The casing pressure gauge. Question 6.15. Calculate the transmission time in a well with a measured depth of 5000 ft/ 1500 m a) 5 seconds b) 10 seconds c) 20 seconds Question 6.16. A kick has been taken and the well is shut-in. SIDPP 300 psi 20 bar SICP 700 psi 50 bar SCR @ 40 spm 400 psi 30 bar Circulation is started with the original mud. Which pressure must be held constant while the pump is being brought up to 40 spm?

Question 6.17. Subsea Which actions should be considered before opening the subsea BOP, after circulating out a gas kick? SELECT FIVE ANSWERS

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a) 1000 psi / 70 bar at the casing gauge. b) 700 psi/ 50 bar at the casing gauge. c) 700 psi/ 50 bar at the drill pipe gauge. d) 1100 psi / 80 bar at the drill pipe gauge.

a) Displace the marine riser to kill fluid. b) Reduce the pressure of the trapped gas below the BOP as much as possible by circulating sea water through the upper kill and choke lines under controlled conditions. c) Close the bottom set of rams. d) Open the annular BOP slowly by reducing the hydraulic regulated pressure. e) Open the diverter element. f) Close the diverter element. g) Disconnect the riser. h) Switch pods. Question 6.18. A well kill is about to be started. Both SIDPP and SICP have increased significantly from their initial stabilized values. What should be done? a) Start the well kill using the increased SIDPP and SICP value. b) Bleed off at the choke until SIDPP returns to its initial stabilized value. c) Bleed off at the choke until SICP returns to its initial stabilized value.

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d) Bleed off at the choke until both SIDPP and SICP return to their initial stabilized value.

Question 6.19. A kick has been taken and the well is closed in. The stabilized SIDPP is 609 psi/ 42 bar. The well kill is started and an ICP of 826 psi/ 57 bar is recorded. Calculate the DPL. Question 6.20. A well kill is started. The DPL is determined at 188 psi / 13 bar. The original fluid density is 9.4 ppg / 1.13 kg/I and kill fluid density is 11.9 ppg / 1.43 kg/I. Calculate the FCP. Question 6.21. Gas is circulated up the well during well kill operations.

Question 6.22. What will happen to the bottom hole pressure if a gas kick is not allowed to expand as it is circulated up the hole?

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a) What happens to the pressure inside the gas bubble? Increases/ Decreases/ Stays the same/ Increases, then stays the same b) What happens to the pressure at the choke? Increases/ Decreases/ Stays the same/ Increases, then stays the same c) What happens to the pressure at the casing shoe? Increases/ Decreases/ Stays the same/ Increases, then stays the same d) What happens to the pressure at the bottom of the well? Increases/ Decreases/ Stays the same/ Increases, then stays the same

a) Increases b) Decreases c) Remains constant Question 6.23. In a well with a long open hole section, which of the following kill methods will minimize the risk of losses? a) Bullheading Method. b) Driller's Method. c) Wait & Weight Method. d) Volumetric Method. Question 6.24. A kick is shut-in on a surface BOP stack and the slow circulating rate pressures is not known. What procedure should be used to obtain the correct Initial Circulating Pressure?

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a) Only use the Driller's Method. As the drill pipe pressure does not change during the entire circulation it is only required to observe that the drill pipe pressure remains constant and equal to the Shut in Drill Pipe Pressure. b) When starting to kill the well, keep the casing pressure as close as possible to the Shut in Casing Pressure. When the selected kill pump rate has been reached, read the drill pipe pressure and use that as the Initial Circulating Pressure. c) Check the records and choose the kill rate circulating pressure taken with the last BHA in the hole nearest to the depth where the kick took place. Add 100 psi to this value as a precautionary measure. d) Contact the mud logger and request the best possible calculation of the Initial Circulating Pressure intended for use during the well kill. Question 6.25. While circulating out a gas kick, when is it possible for the pressure at the casing shoe to be at its maximum? SELECT TWO ANSWERS

Question 6.26. Whilst drilling, a 20 bbl gas kick is taken and the well shut-in. The pressures at surface stabilize after a 15 minute. Due to a delay in mixing kill mud the kill operation cannot start. The pressures at surface begin to increase due to gas migration. What action should be taken to keep bottom hole pressure constant (no float in the string)?

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a) When kill mud reaches the casing shoe. b) At initial shut-in. c) When kill mud reaches the bit. d) When top of gas reaches the casing shoe.

a) Start bleeding off mud and let the casing pressure decrease according to volumetric calculations. b) Bleed mud from the choke, keeping the drill pipe pressure constant. c) Bleed off mud keeping the casing pressure constant. d) Leave it as it is. Gas migration will not affect the bottom hole pressure. Question 6.27. During the first phase of the Wait & Weight Method kill fluid is pumped down the drill string. How can BHP be kept constant? a) Keep the drill pipe pressure constant at ICP. b) Keep the drill pipe pressure constant at SIDPP. c) Keep the drill pipe pressure constant at FCP. d) Reduce the drill pipe pressure from ICP to FCP.

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Question 6.28. Final Circulating Pressure (FCP) is maintained constant: a) Once influx is out of the well. b) While pumping bottoms up after kill mud has passed the bit. c) Once kill mud reaches the casing shoe. d) When pumping kill mud down to the bit. Question 6.29. When does the Wait & Weight Method result in lower pressures at the shoe? a) When the kill fluid enters the annulus after the top of the influx reaches the shoe. b) The Wait & Weight Method never results in lower pressures at the shoe. c) When the drill string volume is less than the open hole volume. d) When the drill string volume is larger than the open hole volume.

a) Ignore it, this mistake will only have a small effect on the BHP. b) Increase the back pressure by the equivalent difference in drilling fluid hydrostatic. c) Stop the pump and shut the well in. Evaluate pressures, reset stroke counter and restart circulation.

Schlumberger Private

Question 6.30. A well is being killed using the Wait & Weight Method. Unfortunately, a quantity of original drilling fluid is pumped down the well instead of kill fluid. The quantity pumped is equivalent to a volume of 650 ft/ 200 m of drill pipe. Which one of the following actions should be taken?

Question 6.31. A kill operation is ready to start using the Wait & Weight Method. The kill mud is ready to be pumped. It takes 100 strokes to displace surface lines. What is the correct procedure? a) Ignore the 100 strokes. It will not affect the drill pipe pressure schedule, and may give an added safety factor. b) Subtract the 100 strokes from the surface to bit strokes and re-calculate the drill pipe pressure schedule. c) Re-Zero stroke counter after 100 strokes are pumped. Question 6.32. What will happen if we ignore a large surface line volume when preparing a kill sheet for the Wait & Weight Method? SELECT TWO ANSWERS a) The drill pipe pressure schedule will result in a bottom hole pressure that is too low. b) The total time to kill the well will be shorter than calculated. c) The total time to kill the well will be longer than calculated.

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d) The drill pipe pressure schedule will result in a bottom hole pressure that is too high. e) There will be no effect on the bottom hole pressure. Question 6.33. L4 A well is shut-in with the following data: SIDPP 480 psi 33 bar SICP 570 psi 39 bar Drilling fluid density 10.0 ppg 1.20 kg/I Kill fluid density 11.0 ppg 1.40 kg/I DPL@ 30 spm 520 psi 36 bar String volume 1400 strokes 1400 strokes The well is killed using the Wait & Weight Method. At 650 strokes, the pump rate is increased from 30 spm to 35 spm keeping the casing pressure constant. Determine the drill pipe pressure at 700 strokes

Schlumberger Private

Question 6.34. L4, Extra A well is shut-in with the following data: Depth (TVD) 10000 ft 3000 m SIDPP 500 psi 35 bar SICP 850 psi 60 bar Drilling fluid density 9.6 ppg 1.15 kg/I DPL@40 spm 1000 psi 65 bar String volume 1200 strokes 1200 strokes The well is killed using the Wait & Weight Method. At 600 strokes, the pump rate is decreased from 40 spm to 30 spm, keeping the casing pressure constant. Determine the drill pipe pressure at 800 strokes. Question 6.35. During the first circulation of the Driller's Method original drilling fluid is pumped. How can BHP be kept constant? a) Keep the drill pipe pressure constant at ICP. b) Keep the drill pipe pressure constant at SIDPP. c) Keep the drill pipe pressure constant at FCP. d) Reduce the drill pipe pressure from ICP to FCP. Question 6.36. A well is killed using the Driller's Method. The first circulation has been completed and the well is shut-in. What can be said about the SIDPP and SICP? a) SIDPP = 0 but SICP is still equal to the stabilized value. b) SICP = 0 but SIDPP is still equal to the stabilized value. c) SIDPP is still equal to the stabilized value but SICP is now equal to SIDPP. d) SIDPP and SICP are still equal to their stabilized values.

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Question 6.37. During first phase of the second circulation of the Driller's Method kill fluid is pumped down the drill string. How can BHP be kept constant? a) Keep the drill pipe pressure constant at ICP. b) Keep the drill pipe pressure constant at SIDPP. c) Keep the drill pipe pressure constant at FCP. d) Reduce the drill pipe pressure from ICP to FCP. Question 6.38. During the second phase of the second circulation of the Driller's Method kill fluid is pumped up the annulus. How can BHP be kept constant? a) Keep the drill pipe pressure constant at ICP. b) Keep the drill pipe pressure constant at SIDPP. c) Keep the drill pipe pressure constant at FCP. d) Reduce the drill pipe pressure from ICP to FCP.

Interval

Drill pipe pressure

Casing pressure

Shoe pressure

Schlumberger Private

Question 6.39. Indicate how the drill pipe pressure, casing pressure, shoe pressure and bottom hole pressure will change during the following intervals. Assume the Driller’s Method is used. Use the following indicators: + for pressure increases - for pressure decreases = for pressure remains constant

BHP

a) Pumping influx from bit to shoe b) Pumping influx from shoe to surface c) Pumping kill fluid from surface to bit d) Pumping kill fluid from bit to shoe e) Pumping kill fluid from shoe to surface

Question 6.40. L4 A salt water kick is circulated out using the Driller's Method. When will the surface pressure be at its maximum value? a) Immediately after shutting-in the well. b) When the salt water kick has reached the choke. c) When the salt water kick has been circulated to the shoe. d) When the kill fluid is entering the annulus.

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Question 6.41. During a well kill the pump rate is reduced whilst holding casing pressure constant. How will this affect bottom hole pressure? a) There is no way of knowing the effect on bottom hole pressure. b) There will be a large drop in bottom hole pressure. c) Bottom hole pressure will increase and may cause formation breakdown. d) Bottom hole pressure will stay constant. Question 6.42. The pump rate is increased during a well kill. The procedure is executed correctly. What can you say about the new drill pipe pressure? a) The new drill pipe pressure is the same as before. b) The new drill pipe pressure is lower than before. c) The new drill pipe pressure is higher than before.

Question 6.44. During the second circulation of the Driller's Method the drill pipe pressure is kept constant until kill fluid is at the bit. What happens to bottom hole pressure?

Schlumberger Private

Question 6.43. While bringing the pumps up to kill rate the casing pressure can increase above SICP. What happens to bottom hole pressure? a) Increases b) Decreases c) Remains constant

a) Increases b) Decreases c) Remains constant Question 6.45. During the first circulation of the Driller's Method casing pressure is kept constant. What happens to bottom hole pressure? a) Increases b) Decreases c) Remains constant Question 6.46. During the second circulation of the Driller's Method casing pressure is kept constant until kill fluid is at the bit. What happens to bottom hole pressure? a) Increases

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b) Decreases c) Remains constant Question 6.47. During the first circulation of the Driller's Method the mud pump develops a leak. The choke operator is unaware of the situation and maintains a constant drill pipe pressure. What happens to bottom hole pressure? a) Increases b) Decreases c) Remains constant Question 6.48. During the first circulation of the Driller's Method the pump rate is increased from 30 spm to 40 spm. The drill pipe pressure is kept at the original ICP while increasing the pump rate. What happens to bottom hole pressure?

Question 6.49. L4 You're in the first circulation of the Driller's Method. SIDPP is 800 psi and ICP is 1020 psi. The pump rate is increased from 30 spm to 40 spm while keeping the drill pipe pressure constant at 1020 psi. What happens to bottom hole pressure?

Schlumberger Private

a) Increases b) Decreases c) Remains constant

a) Increases with 171 psi b) Decreases with 171 psi c) Increases with 391 psi d) Decreases with 391 psi e) Remains constant

Question 6.50. L4 You're in the first circulation of the Driller's Method. SIDPP is 45 bar and ICP is 63 bar. The pump rate is increased from 30 spm to 40 spm while keeping the drill pipe pressure constant at 63 bar. What happens to bottom hole pressure? a) Increases with 14 bar b) Decreases with 14 bar c) Increases with 32 bar d) Decreases with 32 bar e) Remains constant

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Question 6.51. A well is shut in on a gas kick. SIDPP = 400 psi/ 25 bar and SICP = 500 psi/ 35 bar. There is a power failure on the rig and the influx starts to migrate. Drill pipe and casing pressures are increasing. If the choke was used to keep casing pressure constant at 500 psi I 35 bar, what would happen to bottom hole pressure? a) Increases b) Decreases c) Remains constant Question 6.52. During the first circulation of the Driller's Method the drill pipe pressure decreases slowly. The choke man responds by closing the choke to keep the drill pipe pressure at the planned value. Indicate what happens to the pressure at the following locations.

Question 6.53. A well is being killed and the first circulation of the Driller's Method has just been completed. With the well shut-in, it is noticed that SICP is higher than SIDPP. How should the well kill continue?

Schlumberger Private

a) Casing pressure Increases/ Decreases/ Remains constant b) Formation pressure Increases/ Decreases/ Remains constant c) Casing shoe pressure Increases/ Decreases/ Remains constant d) Bottom hole pressure Increases/ Decreases/ Remains constant

a) Continue the well kill using the Wait & Weight Method. b) Continue with the second circulation of the Driller's Method. Keep casing pressure constant until kill fluid is at the bit. c) Continue the well kill using the Bullheading option. d) Continue the well kill using the reverse circulation technique. Question 6.54. L4 While drilling a horizontal section of a well, flow was observed and the well was shut-in. The influx is still in the horizontal section. Which of the following is true? a) SIDPP will be more than SICP. b) SICP & SIDPP will be approximately same. c) SICP will be more than SIDPP.

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Question 6.55. L4 While drilling a horizontal well the return flow alarm is activated. The well is shut-in but both SIDPP and SICP read zero. Drilling can safely resume. a) True b) False Question 6.56. L4 Kill mud is being circulated down the drill string in a horizontal well. The pumps are stopped and the well is shut-in as the kill mud reaches the start of the horizontal section. What would you expect the SIDPP to be? a) Zero. b) The same as the slow circulation rate. c) The same as the original SIDPP. d) The same as the SICP. e) Original SIDPP less the hydrostatic of the horizontal section.

a) ECD is much greater in the horizontal section. b) This is because kill mud was not pumped from the start of the kill. c) This is normal for any well when the Driller’s Method is used. d) There is no loss of hydrostatic head until the gas arrives at the inclined or vertical section of the well.

Schlumberger Private

Question 6.57. L4 A kick is taken in a horizontal well and SIDPP = SICP. The influx is circulated out using Driller's Method. When the influx is circulated out of the horizontal section and into the more vertical section, the casing pressure increase quite rapidly, why?

Question 6.58. Under which circumstances can the Volumetric Method be used? SELECT TWO ANSWERS a) In case of a non-migrating influx. b) When the string must be stripped back to bottom. c) When it's not possible to circulate the well. d) In case of a migrating influx. Question 6.59. Under which circumstances can the Combined Volumetric and Stripping Method be used? SELECT TWO ANSWERS a) When the influx is above the bit and circulation is possible. b) When the influx is below the bit and drill string is blocked. c) When the influx is below the bit and the influx is non-migrating. d) When the drill string must be brought to bottom to kill the well.

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Question 6.60. During a trip out, the well kicks due to swabbing. The pipe is stripped back to bottom. Will the original mud density be sufficient to control the well? a) Yes b) No Question 6.61. L4 When stripping into the hole which of the following will maintain a constant bottom hole pressure? (Assume there is no influx migration). a) Bleed off the drill pipe closed end displacement while stripping each stand. b) Pump a volume of mud into the well equal to the drill pipe closed end displacement while stripping each stand. c) Bleed off the drill pipe steel displacement while stripping each stand. d) Pump a volume of mud into the well equal to the drill pipe steel displacement while strip- ping each stand.

Schlumberger Private

Question 6.62. L4 5” drill pipes is stripped into the hole through the annular preventer. Stand length 93 ft 28 m DP capacity 0.0178 bbl/ft 9.27 1/m DP closed end displ. 0.0254 bbl/ft 13.231/m DP metal displ. 0.0076 bbl/ft 3.96 1/m For each stand of pipe stripped, how many barrels/ liter of drilling fluid must be bled off from the annulus? a) 0.16 bbl/ 260 ltr b) 2.36 bbl/ 370 ltr c) 0.71 bbl/ 111ltr Question 6.63. During a well kill the drill pipe pressure increases suddenly but the casing pressure remains the same. What could be going on? a) Most likely a washed-out choke. b) Most likely a plugged choke. c) Most likely a plugged nozzle. d) Most likely a pump failure. Question 6.64. During the first circulation of the Driller's Method the drill pipe pressure suddenly increases from 740 psi to 1150 psi. All other parameters have remained constant. What procedure should be followed?

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a) Open the choke to bring the drill pipe pressure back to 740 psi and maintain the same pump rate. b) Reduce the pump rate to bring the drill pipe pressure back to 740 psi. c) Continue as per plan but use the higher drill pipe pressure to re-calculate the kill fluid density for second circulation of the Driller’s Method. d) Stop the well kill and evaluate the situation. Commence the well kill, read off the new ICP and draw a line parallel to the old line and continue the well kill. Question 6.65. During a well kill the casing pressure drops and shortly thereafter the drill pipe pressure as well. What could be going on? a) Most likely a washed-out choke. b) Most likely a plugged choke. c) Most likely a plugged nozzle. d) Most likely a pump failure.

a) Rapid rise in casing pressure and no change in drill pipe pressure. b) Continually closing the choke to maintain correct circulating pressure. c) Increase in drill pipe pressure and no change to casing pressure. d) Continually opening the choke to maintain correct circulating pressure.

Schlumberger Private

Question 6.66. How would a choke washout be recognized?

Question 6.67. During a well kill operation, the choke operator notices that both drill pipe and casing pressures are slowly decreasing. He reacts by adjusting the choke to maintain the original pump pressure. What effect does this choke adjustment have on the bottom hole pressure? a) Bottom hole pressure decreases. b) Bottom hole pressure is returned to correct value. c) Bottom hole pressure is not affected by choke adjustment. Question 6.68. During a well kill the casing pressure increases and shortly thereafter the drill pipe pressure as well. What could be going on? a) Most likely a washed-out choke. b) Most likely a plugged choke. c) Most likely a plugged nozzle. d) Most likely a pump failure.

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Question 6.69. When killing a well using the Driller's Method the choke pressure suddenly increases by 200 psi. Shortly after, the choke operator observes the same pressure increase on the drill pipe pressure gauge. What would be the correct response to this problem? a) Reduce the pump rate to reduce both pressures by 200 psi. b) No action required. c) Shut the well in and change to another pump. d) Shut the well in and change to another choke.

Question 6.70. During a kill operation, which one of the following problems require the pump to be shut down quickly to prevent over-pressuring the formation?

Question 6.71. During a well kill both the drill pipe pressure and casing pressure drop suddenly. What could be going on?

Schlumberger Private

a) A washout in the choke. b) A plugged choke. c) A washout in the drill pipe. d) A plugged standpipe. e) A washed-out bit nozzle. f) A plugged bit nozzle.

a) Most likely a washed-out choke. b) Most likely a plugged choke. c) Most likely a plugged nozzle. d) Most likely a pump failure.

Question 6.72. Whilst circulating out a kick, the mud pump fails. What is the first action to take? a) Change over to mud pump#2 b) Shut the well in. c) Divert the well. d) Fix the mud pump as soon as possible.

Question 6.73.

Question 6. 73

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Indicate how the drill pipe pressure, casing pressure and bottom hole pressure would respond to the following problems. Use the following indicators: + for pressure increases - for pressure decreases = for pressure remains constant Problem

Drill pipe pressure

Casing pressure

BHP

a) Plugged choke b) Washed out choke c) Plugged bit nozzle d) Pump failure

Question 6.74. During a well kill the drill pipe pressure decreases suddenly but the casing pressure remains the same. What could be going on?

Question 6.75. Whilst killing a well a sudden loss in standpipe pressure is observed. If the choke was closed to compensate for the reduction in pressure, what would happen to the bottom hole pressure? a) It would increase. b) It would decrease. c) It would remain constant.

Schlumberger Private

a) Most likely a washed-out choke. b) Most likely a lost bit nozzle. c) Most likely a washout in the drill string. d) Most likely a plugged nozzle.

Question 6.76. When killing a well what is the correct action to take if a sudden loss in drill pipe pressure is observed? a) Increase the pump rate to return drill pipe pressure to the correct value. b) Continue at the same pump rate holding the new drill pipe pressure. c) Shut the well in and investigate the pressure loss. d) Close the choke to compensate for the pressure loss. Question 6.77. During a well kill the drill pipe pressure decreases slowly but the casing pressure remains the same. What could be going on? a) Most likely a washed-out choke. b) Most likely a lost bit nozzle. Rev 1.0 [Nov 2017]

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c) Most likely a washout in the drill string. d) Most likely a plugged nozzle. Question 6.78. During a kill, a washout develops in the drill string above the influx. What is the safest way to handle this situation? a) Stop pumping. Use the Volumetric Method until the influx is above the washout. Then resume pumping. b) Stop pumping. Strip out of hole to locate the washout in the drill string, then replace the washed out joint and strip back to bottom. c) Keep pumping until the influx is above the washout, then use the Volumetric Method to remove the influx. Question 6.79. A well is being killed and the gas influx is still in open hole. The casing pressure is approaching MAASP. Which of the following reduces the risk of fracturing the formation?

Question 6.80. Whilst drilling ahead, the well kicks and is shut-in. Drill pipe pressure and casing pressure both start to build up, but before stabilizing both start to drop quite rapidly. Which of the following might have occurred?

Schlumberger Private

a) Reduce the pump rate as much as possible to minimize any extra pressure in the annulus without allowing BHP to fall below formation pressure. b) Keep the casing pressure at or just below MAASP by opening the choke. c) Stop the pumps and allow the gas to migrate into the casing shoe. Once in the shoe, continue circulating out the gas kick. d) Increase the kill fluid density to reduce the pressure on the casing shoe.

a) Gas is migrating up the well. b) The drill string has washed out. c) The bottom hole assembly has packed off. d) Both gauges have malfunctioned. e) A weak formation has broken down.

Question 6.81. A well was shut-in after a kick and following pressures were recorded: Time 06:00 hrs 06:05 hrs 06:10 hrs 06:20 hrs 06:25 hrs 06:30 hrs

SIDPP 300 psi 350 psi 400 psi 350 psi 200 psi 100 psi

SICP 480 psi 540 psi 600 psi 500 psi 350 psi 180 psi

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120 psi

What has probably happened in the well? a) he influx is migrating up in the well. b) The drill string has parted. c) The hole is packed around drill collar. d) A weak formation in the open hole got fractured. Question 6.82. The well is shut-in with the following pressure readings: SIDPP 435 psi 30 bar SICP 0 psi 0 bar The annulus is observed through the choke, and there is no flow. What is the probable cause? a) The well was swabbed in. b) The hole is packed off around the BHA. c) The formation at the shoe has been fractured. d) The drill string has twisted off. e) The choke gauge failed.

a) By a decrease in the mud volume in the pits. b) By a decrease in the weight indicator. c) By monitoring the return flow with the flow-show. d) By an increase in pump speed.

Schlumberger Private

Question 6.83. How is lost circulation usually detected during a well control operation?

Question 6.84. Why is it important to monitor the pit volume during a well control operation? SELECT TWO ANSWERS a) To check for mud losses. b) Tells you when to adjust drill pipe pressure. c) To monitor the gas expansion. d) To maintain bottom hole pressure constant. e) Tells you when to adjust pump speed. Question 6.85. What action should be taken if the choke line parts while circulating out a kick? a) Stop the pump and close the choke. b) Stop the pump and close the HCR. c) Stop the pump and close the top drive or Kelly valve.

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d) Stop the pump and close the shear rams. Question 6.86. Whilst circulating out a kick the rotary (Kelly) hose in the derrick bursts, what is the first action to take? a) Hang tool joint on pipe ram and close the shear rams. b) Stop the pump, close the upper Kelly valve, then close the choke. c) Drop the drill string and close the blind/shear rams. d) Stop the pump, close the choke and record pressures. Question 6.87. How can hydrates be inhibited? a) By adding antifreeze material such as salt, glycols and glycerin. b) By adding methanol. c) By reducing the temperature. d) by increasing the pressure.

a) To close the BOP when primary well control has been lost b) To maintain hydrostatic pressure above formation pressure c) To isolate the trouble zone by creating some sort of plug

Schlumberger Private

Question 6.88. What is the objective of tertiary well control?

Question 6.89. Which of the below are tertiary well control methods. SELECT THREE ANSWERS a) Spot a barite plug b) Spot a viscous pill c) Set a cement plug d) Spot a gunk squeeze e) Pump kill fluid

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Question 6.90. Fill out a vertical kill sheet using the data below 9520 ft 9410 ft 5580 ft 5520 ft 11.0 ppg

2900 m 2870 m 1700 m 1680 m 1.32 kg/I

Surface leak-off pressure LOT mud weight

740 psi 10.6 ppg

51 bar 1.27 kg/I

Pump displacement DPL@ 30 spm

0.115 bbl/stk 490 psi

18.3 I/stk 34 bar

HWDP length DC length DP capacity HWDP capacity DC capacity DC X Open Hole DP/HWDP x Open Hole DP X Casing

540 ft 360 ft 0.0174 bbl/ft 0.0088 bbl/ft 0.0077 bbl/ft 0.0226 bbl/ft 0.0449 bbl/ft 0.0510 bbl/ft

150 m 120 m 9.05 I/m 4.61I/m 4.011/m 11.8 1/m 23.4 I/m 26.3 I/m

SIDPP SICP Pit Gain

360 psi 490 psi 15 bbl

25 bar 34 bar 2000 liter

a) Calculate the maximum allowable drilling fluid density. b) Calculate the initial MAASP. c) How many strokes are required to pump from surface to the bit? d) How many strokes are required to pump from the bit to the casing shoe? e) Calculate the total annular volume. f) How many minutes does it take to do one full circulation at 30 spm? g) Calculate the required kill fluid density. h) Calculate the initial circulating pressure. i) Calculate the final circulating pressure. j) Calculate the new MAASP after circulating the well to kill fluid.

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Question 6.91. Fill out a vertical kill sheet using the data below 12250 ft 11200 ft 7500 ft 12.2 ppg

3750 m 3400 m 2300 m 1.46 kg/I

Fracture fluid density at test

16.0 ppg

1.90 kg/I

Pump displacement DPL@ 35 spm

0.2 bbl/stk 1100 psi

28 I/stk 75 bar

DC length DP capacity DC capacity DC X Open Hole DP X Open Hole DP X Casing

800 ft 0.01776 bbl/ft 0.005 bbl/ft 0.031 bbl/ft 0.0418 bbl/ft 0.0529 bbl/ft

240 m 9.261/m 2.611/m 16.17 1/m 21.81I/m 27.60 I/m

SIDPP SICP Pit Gain

750 psi 1000 psi 20 bbl

50 bar 70 bar 3000 liter

a) Calculate the required kill fluid density. b) Calculate the MAASP at shut-in. c) How many strokes are required to pump from surface to bit? d) What is the time required to pump from surface to bit? e) How many strokes are required to pump from the casing shoe to surface? f) What is the time required for one complete circulation? g) Calculate the initial circulating pressure. h) Calculate the final circulating pressure. i) What is the drill pipe pressure reduction per 100 strokes as kill fluid is pumped to the bit? j) Calculate the new MAASP after circulating the well to kill fluid.

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Question 6.92. Fill out a vertical kill sheet using the data below 9800 ft 6500 ft 10.8 ppg

2900 m 2000 m 1.29 kg/I

Fracture gradient at shoe

0.806 psi/ft

0.182 bar/m

Pump displacement DPL@40 spm

0.15 bbl/stk 950 psi

23.8 I/stk 66 bar

DC length DP capacity DC capacity DC X Open Hole DP X Open Hole DP X Casing

680 ft 0.01776 bbl/ft 0.007 bbl/ft 0.0756 bbl/ft 0.1215 bbl/ft 0.1279 bbl/ft

210 m 9.26 I/m 3.65 I/m 39.44I/m 63.38 I/m 66.72 1/m

SIDPP SICP Pit Gain

800 psi 1000 psi 30 bbl

55 bar 70 bar 3500 liter

a) Calculate the required kill fluid density. b) How many strokes are required to pump from surface to bit? c) How many strokes are required to pump from bit to the casing shoe? d) How many strokes are required to pump from bit to surface? e) Calculate the MAASP at shut-in. f) Calculate the new MAASP after circulating the well to kill fluid. g) How many strokes are required for one complete circulation? h) Calculate the initial circulating pressure. i) Calculate the final circulating pressure. j) What is the drill pipe pressure reduction per 100 strokes as kill fluid is pumped to the bit?

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Question 6.93. Fill out a vertical kill sheet using the data below 10285 ft 10131 ft 5171 ft 5112 ft 10.3 ppg

3135 m 3088 m 1576 m 1558 m 1.23 kg/I

Surface leak-off pressure LOT mud weight

870 psi 10.1 ppg

60 bar 1.21 kg/I

Pump displacement DPL@ 30 spm

0.102 bbl/stk 348 psi

16.2 I/stk 24 bar

HWDP length DC length DP capacity HWDP capacity DC capacity DC X Open Hole DP/HWDP X Open Hole DP X Casing

295 ft 394 ft 0.01735 0.00884 0.00769 0.02053 0.04283 0.05051

90 m 120 m 9.05 1/m 4.611/m 4.011/m 10.71I/m 22.34I/m 26.35 I/m

SIDPP SICP Pit Gain

450 psi 624 psi 19.5 bbl

bbl/ft bbl/ft bbl/ft bbl/ft bbl/ft bbl/ft

31 bar 43 bar 3100 liter

a) Calculate the maximum allowable drilling fluid density. b) Calculate the initial MAASP. c) How many strokes are required to pump from surface to the bit? d) How many strokes are required to pump from the bit to the casing shoe? e) Calculate the total well system volume? f) How many minutes does it take to do the first circulation of the Driller's Method at 30 spm g) Calculate the required kill fluid density. h) Calculate the initial circulating pressure. i) Calculate the final circulating pressure. j) Calculate the new MAASP after circulating the well to kill fluid.

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Question 6.94. Subsea Fill out a vertical subsea BOP kill sheet using the data below 12000 ft 8500 ft

3700 m 2500 m

Marine riser Choke line Mud Weight in use

2000 ft 2050 ft 11.9 ppg

610 m 625 m 1.38 kg/I

Surface leak-off pressure LOT mud weight

1000 psi 11.3 ppg

60 bar 1.32 kg/I

Pump displacement PL riser @ 30 spm PL choke line @ 30 spm

0.11 bbl/stk 650 psi 900 psi

16.3 I/stk 45 bar 60 bar

HWDP length DC length DP capacity HWDP capacity DC capacity DC X Open Hole DP/HWDP x Open Hole DP X Casing Choke line capacity DP X Riser

600 ft 450 ft 0.0172 bbl/ft 0.0088 bbl/ft 0.0077 bbl/ft 0.0226 bbl/ft 0.0448 bbl/ft 0.0454 bbl/ft 0.009 bbl/ft 0.325 bbl/ft

180 m 120 m 8.97 1/m 4.61I/m 4.011/m 11.78 I/m 23.36 I/m 23.67 I/m 4.56 I/m 169.67 1/m

SIDPP SICP Pit Gain

350 psi 430 psi 10 bbl

24 bar 31 bar 2000 liter

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a) What is the maximum allowable mud weight? b) How many strokes are required to pump from surface to the bit? c) How many strokes are required to pump from bit to the shoe? d) Calculate the total well system volume. e) Calculate the required kill fluid density. f) Calculate the initial circulating pressure. g) Calculate the final circulating pressure. h) Calculate the initial dynamic casing pressure. i) Calculate the new MAASP after circulating the well to kill fluid.

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Question 6.95. Subsea Fill out a vertical subsea BOP kill sheet using the data below 9331 ft 9131 ft 6211 ft 6119 ft

2844 m 2783 m 1893 m 1865 m

Marine riser Choke line Mud Weight in use

820 ft 840 ft 12.0 ppg

250 m 256 m 1.44 kg/I

Surface leak-off pressure LOT mud weight

1566 psi 11.0 ppg

108 bar 1.32 kg/I

Pump displacement PL riser @ 40 spm PL choke line @ 40 spm

0.119 bbl/stk 827 psi 914 psi

18.92 1/stk 57 bar 63 bar

HW length DC length DP capacity HW capacity DC capacity DC X Open Hole DP/HWDP x Open Hole DP X Casing Choke line capacity DP X Riser

476 ft 650 ft 0.01781 bbl/ft 0.00880 bbl/ft 0.00780 bbl/ft 0.08360 bbl/ft 0.12019 bbl/ft 0.12420 bbl/ft 0.00861 bbl/ft 0.36359 bbl/ft

145 m 198 m 9.29 1/m 4.591/m 4.07 1/m 43.611/m 62.70 1/m 64.791/m 4.49 1/m 189.67 1/m

SIDPP SICP Pit Gain

798 psi 914 psi 30 bbl

55 bar 63 bar 4770 liter

Schlumberger Private

Hole - MD Hole -TVD Casing shoe - MD Casing shoe - TVD

a) Calculate the initial MAASP. b) How many strokes are required to pump from surface to the bit? c) Calculate the annulus volume. d) How many minutes takes a full well system volume circulation? e) Calculate the required kill fluid density. f) Calculate the initial circulating pressure. g) Calculate the final circulating pressure. h) Calculate the initial dynamic casing pressure. i) Calculate the new MAASP after circulating the well to kill fluid. j) What is the drill pipe pressure reduction per 100 strokes as kill fluid is pumped to the bit?

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Question 6.96. Subsea Question 6.96 Fill out a vertical subsea BOP kill sheet using the data below Hole - MD Hole -TVD Casing shoe - MD Casing shoe - TVD

8750 8630 5280 5200

Marine riser Choke line Mud Weight in use

1200 ft 1300ft 11.3 ppg

410 m 450 m 1.35 kg/I

Fracture gradient at shoe

0.859 psi/ft

0.194 bar/m

Pump displacement PL riser @ 30 spm PL choke line @ 30 spm

0.103 bbl/stk 730 psi 1150 psi

16.38 I/stk 50 bar 79 bar

HW length DC length DP capacity HW capacity DC capacity DC x Open Hole DP/HWDP x Open Hole DP X Casing Choke line capacity DP X Riser

450 ft 300 ft 0.0176 bbl/ft 0.0087 bbl/ft 0.0077 bbl/ft 0.0210 bbl/ft 0.0430 bbl/ft 0.0510 bbl/ft 0.0061 bbl/ft 0.3650 bbl/ft

180 m 90 m 9.05 I/m 4.60 I/m 4.101/m 10.70 I/m 22.34I/m 26.35 I/m 3.10 I/m 189.5 I/m

SIDPP SICP Pit Gain

980 psi 1130 psi 15 bbl

67 bar 78 bar 2500 liter

ft ft ft ft

2700 m 2630 m 1610 m 1585 m

Schlumberger Private

a) Calculate the initial MAASP. b) How many strokes are required to pump from surface to the bit? c) Calculate the annulus open hole volume. d) How many minutes is a bottom up circulation? e) Calculate the required kill fluid density. f) Calculate the initial circulating pressure. g) Calculate the final circulating pressure. h) Calculate the initial dynamic casing pressure. i) Calculate the new MAASP after circulating the well to kill fluid.

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Question 6.97. L4 - Metric Use the completed deviated kill sheet to answer the following questions. The well is killed using the Driller’s Method.

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a) The first circulation is started using the original drilling fluid. Which pressure is held constant while bringing the pumps up to speed? 1) 41 bar drill pipe pressure 2) 49 bar casing pressure 3) 78 bar drill pipe pressure 4) 44 bar drill pipe pressure b) After a couple of minutes, the following readings are recorded: • Drill pipe pressure: 88 bar • Casing pressure: 58 bar • Pump rate: 33 spm • Stroke counter: 41 strokes Which of the following is required?

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1) Continue, everything is okay. 2) Open the choke more. 3) Close the choke more. 4) Increase the pump rate. 5) Decrease the pump rate. c) After 30 minutes, the following readings are recorded: • Drill pipe pressure: 78 bar • Casing pressure: 51 bar • Pump rate: 30 spm • Stroke counter: 883 strokes Which of the following is required? 1) Continue, everything is okay. 2) Open the choke more. 3) Close the choke more. 4) Increase the pump rate. 5) Decrease the pump rate.

Schlumberger Private

d) At the end of the first circulation, the well was shut-in and the following readings are recorded: • SIDPP: 41 bar • SICP: 41 bar The second circulation is started using the kill fluid. The stroke counter is reset after having pumped the surface line volume. After 600 strokes, the following readings are recorded: • Drill pipe pressure: 64 bar • Casing pressure: 48 bar • Pump rate: 30 spm Which of the following is required? 1) Continue, everything is okay. 2) Open the choke more. 3) Close the choke more. 4) Increase the pump rate. 5) Decrease the pump rate. e) About 200 strokes later the casing pressure starts decreasing quickly followed by the drill pipe pressure. • Drill pipe pressure: 45 bar • Casing pressure: 34 bar • Pump rate: 30 spm • Stroke counter: 800 strokes Which of the following is the problem? 1) A plugged choke

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2) A washed-out choke 3) A plugged bit nozzle. 4) A lost bit nozzle. 5) A pump problem f) The problem is solved and the well kill is continued. The following readings are recorded: • Drill pipe pressure: 57 bar • Casing pressure: 49 bar • Pump rate: 30 spm • Stroke counter: 1220 strokes Which of the following is required? 1) Continue, everything is okay. 2) Open the choke more. 3) Close the choke more. 4) Increase the pump rate. 5) Decrease the pump rate.

Schlumberger Private

g) After some time, the following readings are recorded: • Drill pipe pressure: 40 bar • Casing pressure: 35 bar • Pump rate: 30 spm • Stroke counter: 1870 strokes Which of the following is required? 1) Continue, everything is okay. 2) Open the choke more. 3) Close the choke more. 4) Increase the pump rate. 5) Decrease the pump rate. h) Again, some later, the following readings are recorded: • Drill pipe pressure: 44 bar • Casing pressure: 31 bar • Pump rate: 30 spm • Stroke counter: 2800 strokes Which of the following is required? 1) Continue, everything is okay. 2) Open the choke more. 3) Close the choke more. 4) Increase the pump rate. 5) Decrease the pump rate. i)

A few strokes later the drill pipe suddenly increases from 44 bar to 64 bar and then remained constant. The casing pressure did not change. • Drill pipe pressure: 64 bar • Casing pressure: 30 bar • Pump rate: 30 spm • Stroke counter: 2840 strokes Which of the following is the problem?

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1) A plugged choke 2) A washed-out choke 3) A plugged bit nozzle. 4) A lost bit nozzle. 5) A pump problem. 6) A washout in the drill string. j) Which of the following is required? 1) Open the choke more. 2) Close the choke more. 3) Increase the pump rate. 4) Shutdown the well kill and analyses the problem.

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k) Kill fluid is coming back to surface and the choke is fully open. The following readings are recorded: • Drill pipe pressure: 64 bar • Casing pressure: 0 bar • Pump rate: 30 spm • Stroke counter: 6180 strokes Which of the following is required? 1) Open the choke more. 2) Close the choke more. 3) Reduce the pump rate. 4) Stop pumping and open the well 5) Stop pumping and close the choke.

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Question 6.98. L4 - API Use the completed deviated kill sheet to answer the following questions. The well is killed using the Driller’s Method

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a) The first circulation is started using the original drilling fluid. Which pressure is held constant while bringing the pumps up to speed? 1) 595 psi drill pipe pressure 2) 711 psi casing pressure 3) 1132 psi drill pipe pressure 4) 624 psi drill pipe pressure b) After a couple of minutes, the following readings are recorded: • Drill pipe pressure: 1290 psi • Casing pressure: 869 psi • Pump rate: 33 spm •Stroke counter: 41 strokes Which of the following is required? Rev 1.0 [Nov 2017]

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1) Continue, everything is okay. 2) Open the choke more. 3) Close the choke more. 4) Increase the pump rate. 5) Decrease the pump rate. c) After 30 minutes, the following readings are recorded: • Drill pipe pressure: 1132 psi • Casing pressure: 739 psi • Pump rate: 30 spm • Stroke counter: 883 strokes Which of the following is required? 1) Continue, everything is okay. 2) Open the choke more. 3) Close the choke more. 4) Increase the pump rate. 5) Decrease the pump rate.

Schlumberger Private

d) At the end of the first circulation, the well was shut-in and the following readings are recorded: • SIDPP: 595 psi • SICP: 595 psi The second circulation is started using the kill fluid. The stroke counter is reset after having pumped the surface line volume. After 600 strokes, the following readings are recorded: • Drill pipe pressure: 923 psi • Casing pressure: 693 psi • Pump rate: 30 spm 0 Which of the following is required? 1) Continue, everything is okay. 2) Open the choke more. 3) Close the choke more. 4) Increase the pump rate. 5) Decrease the pump rate. e) About 200 strokes later the casing pressure starts decreasing quickly followed by the drill pipe pressure. • Drill pipe pressure: 648 psi • Casing pressure: 493 psi • Pump rate: 30 spm • Stroke counter: 800 strokes Which of the following is the problem? 1) A plugged choke 2) A washed-out choke 3) A plugged bit nozzle.

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4) A lost bit nozzle. 5) A pump problem f) The problem is solved and the well kill is continued. The following readings are recorded: • Drill pipe pressure: 805 psi • Casing pressure: 711 psi • Pump rate: 30 spm • Stroke counter: 1220 strokes Which of the following is required? 1) Continue, everything is okay. 2) Open the choke more. 3) Close the choke more. 4) Increase the pump rate. 5) Decrease the pump rate.

h) Again, some later, the following readings are recorded: • Drill pipe pressure: 624 psi • Casing pressure: 450 psi • Pump rate: 30 spm • Stroke counter: 2800 strokes Which of the following is required? 1) Continue, everything is okay. 2) Open the choke more. 3) Close the choke more. 4) Increase the pump rate. 5) Decrease the pump rate.

Schlumberger Private

g) After some time, the following readings are recorded: • Drill pipe pressure: 574 psi • Casing pressure: 507 psi • Pump rate: 30 spm • Stroke counter: 1870 strokes Which of the following is required? 1) Continue, everything is okay. 2) Open the choke more. 3) Close the choke more. 4) Increase the pump rate. 5) Decrease the pump rate.

i) A few strokes later the drill pipe suddenly increases from 624 psi to 928 psi and then remained constant. The casing pressure did not change. • Drill pipe pressure: 928 psi • Casing pressure: 435 psi • Pump rate: 30 spm • Stroke counter: 2840 strokes Which of the following is the problem? 1) A plugged choke 2) A washed-out choke

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3) A plugged bit nozzle. 4) A lost bit nozzle. 5) A pump problem. 6) A washout in the drill string. j) Which of the following is required? 1) Open the choke more. 2) Close the choke more. 3) Increase the pump rate. 4) Shutdown the well kill and analyses the problem.

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k) Kill fluid is coming back to surface and the choke is fully open. The following readings are recorded: • Drill pipe pressure: 928 psi • Casing pressure: 0 psi • Pump rate: 30 spm • Stroke counter: 6180 strokes Which of the following is required? 1) Open the choke more. 2) Close the choke more. 3) Reduce the pump rate. 4) Stop pumping and open the well 5) Stop pumping and close the choke.

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Question 6.99. L4, Extra Fill out a deviated kill sheet using the data below and draw a kill graph for the first phase of the Wait & Weight Method 9843 ft 5427 ft 4265 ft 3517 ft 1640 ft 3937 ft 3406 ft 10.7 ppg

3000 m 1654 m 1300m 1072 m 500 m 1200 m 1038 m 1.28 kg/I

Surface leak-off pressure LOT mud weight

870 psi 9.60 ppg

60 bar 1.15 kg/I

Pump displacement DPL@ 30 spm

0.119 bbl/stk 493 psi

18.92 I/stk 34 bar

DP capacity DC length DC capacity DC X Open Hole DP X Open Hole DP X Casing

0.01754 bbl/ft 328 ft 0.00491 bbl/ft 0.03221 bbl/ft 0.04467 0.04524

9.15 I/m 100 m 2.561/m 16.8 I/m 23.3 I/m 23.6 I/m

SIDPP SICP Pit Gain

392 psi 508 psi 10 bbl

27 bar 35 bar 1500 liter

Schlumberger Private

Hole - MD Hole -TVD Casing shoe - MD Casing shoe - TVD KOP - MD/TVD EOB - MD EOB -TVD Mud Weight in use

a) Calculate the maximum allowable drilling fluid density. b) Calculate the initial MAASP. c) How many strokes are required to pump from bit to the shoe? d) Calculate the total well system volume. e) Calculate the required kill fluid density. f) Calculate the initial circulating pressure. g) Calculate the final circulating pressure. h) Calculate the circulating pressure at KOP. i) Calculate the circulating pressure at EOB.

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Question 6.100. L4, Subsea extra Fill out a deviated subsea BOP kill sheet using the data below 10000 ft 5892 ft 5000 ft 4182 ft 3000 ft 4167 ft 3897 ft

3048 m 1796 m 1524 m 1275 m 914 m 1270 m 1188 m

Marine riser Choke line Mud Weight in use

2000 ft 2050 ft 11.5 ppg

610 m 625 m 1.38 kg/I

Surface leak-off pressure LOT mud weight

1200 psi 11.0 ppg

83 bar 1.32 kg/I

Pump displacement PL riser @ 30 spm PL choke line @ 30 spm

0.103 bbl/stk 1020 psi 1200 psi

16.38 I/stk 70 bar 83 bar

HW length DC length DP capacity HW capacity DC capacity DC X Open Hole DP/HW X Open Hole DP X Casing Choke line capacity DP X Riser

300 ft 400 ft 0.01720 bbl/ft 0.00884 bbl/ft 0.00769 bbl/ft 0.02259 bbl/ft 0.04478 bbl/ft 0.04537 bbl/ft 0.00874 bbl/ft 0.32527 bbl/ft

91 m 122 m 8.97 I/m 4.611/m 4.011/m 11.78 I/m 23.36 1/m 23.67 I/m 4.56 I/m 169.68 I/m

SIDPP SICP Pit Gain

450 psi 500 psi 20 bbl

30 bar 36 bar 3180 liter

Schlumberger Private

Hole - MD Hole -TVD Casing shoe - MD Casing shoe - TVD KOP- MD/TVD EOB - MD EOB -TVD

a) What is the initial MAASP? b) How many strokes are required to pump from surface to the bit? c) How many minutes does a bottom up circulation take? d) Calculate the required kill fluid density. e) Calculate the initial circulating pressure. f) Calculate the final circulating pressure. g) Calculate the circulation pressure at EOB. h) Calculate the initial dynamic casing pressure. i) Calculate the new MAASP after circulating the well to kill fluid.

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Chapter 6 answers 6.42 C 6.43 A 6.44 A 6.45 B 6.46 C 6.47 A 6.48 B 6.49 B 6.50 B 6.51 B 6.52 a) Increases; b) Remains constant; c) Increases; d) Increases 6.53 A 6.54 B 6.55 B 6.56 A 6.57 D 6.58 C, D 6.59 C, D 6.60 A 6.61 A 6.62 B 6.63 C 6.64 D 6.65 A 6.66 B 6.67 B 6.68 B 6.69 D 6.70 B 6.71 D 6.72 B 6.73 a) +,+,+ b) -,-,c) +,=,= d) -,-,6.74 B 6.75 A 6.76 C 6.77 C 6.78 A 6.79 A 6.80 E 6.81 D 6.82 B 6.83 A 6.84 A, C

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6.1 B 6.2 C,D 6.3 A, C, D, F 6.4 A, C 6.5 A, C, D, G, H 6.6 C, D, E 6.7 B, D 6.8 B 6.9 B 6.10 A, E 6.11 D 6.12 B, E 6.13 11.0 ppg / 1.33 kg/I 6.14 A 6.15 B 6.16 B 6.17 A, B, C, D, F 6.18 B 6.19 217 psi/ 15 bar 6.20 238 psi / 17 bar 6.21 a) Decreases; b) Increases; c) Increases, then stays the same; d) Stays the same 6.22 A 6.23 C 6.24 B 6.25 B, D 6.26 B 6.27 D 6.28 B 6.29 C 6.30 C 6.31 C 6.32 A, C 6.33 984 psi / 70 bar 6.34 769 psi / 51 bar 6.35 A 6.36 C 6.37 D 6.38 C 6.39 a) =,+,+,= b) =,+,=,= c) -,=,=,= d) =,-,-,= e) =,-,=,= 6.40 A 6.41 D

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c) 1335 strokes/ 1635 strokes d) 657 bbl/ 106038 liter e) 12.5 ppg / 1.45 kg/I f) 1000 psi/ 69 bar g) 683 psi/ 48 bar h) 180 psi / 16 bar i) 442 psi/ 26 bar 6.95 a) 1240 psi/ 85 bar b) 1306 strokes/ 1306 strokes c) 1028 bbl/ 163447 liter d) 249 min / 249 min e) 13. 7 ppg / 1.65 kg/I f) 1625 psi / 112 bar g) 945 psi/ 66 bar h) 827 psi/ 57 bar i) 700 psi/ 47 bar j) 52 psi/100 stk / 3.52 bar/100 stk 6.96 a) 1406 psi/ 96 bar b) 1427 strokes/ 1416 strokes c) 143 bbl / 23303 liter d) 116 min / 115 min e) 13.5 ppg / 1.61 kg/I f) 1710 psi/ 117 bar g) 873 psi / 60 bar h) 710 psi / 49 bar i) 811 psi/ 55 bar 6.97 a) 2; b) 5; c) 1; d) 2; e) 2; f) 2; g) 3; h) 1; i) 3; j) 4; k) 5; 6.98 a) 2; b) 5; c) 1; d) 2; e) 2; f) 2; g) 3; h) 1; i) 3; j) 4; k) 5; 6.99 a) 14.3 ppg / 1.72 kg/I b) 658 psi / 46 bar c) 2059 strokes / 2059 strokes d) 607 bbl / 96431 liter e) 12.1 ppg / 1.45 kg/I f) 885 psi / 61 bar g) 558 psi / 39 bar h) 777 psi / 54 bar i) 664 psi / 46 bar 6.100 a) 1087 psi/ 75 bar b) 1609 strokes/ 1608 strokes c)119 min/ 119 min d) 13.0 ppg / 1.56 kg/I e) 14 70 psi / 100 bar f) 1154 psi / 80 bar g) 1222 psi / 84 bar h) 320 psi/ 23 bar i) 761 psi / 52 bar

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6.85 B 6.86 B 6.87 A 6.88 C 6.89 A, C, D 6.90 a) 13.1 ppg / 1.57 kg/I b) 602 psi / 41 bar c) 1370 strokes / 1365 strokes d) 1469 strokes / 1458 strokes e) 453 bbl / 71398 liter f) 177 min / 176 min g) 11.8 ppg / 1.41 kg/I h) 850 psi / 59 bar i) 526 psi / 37 bar j) 373 psi / 26 bar 6.91 a) 13.5 ppg / 1.61 kg/I b) 1482 psi / 99 bar c) 1037 strokes / 1183 strokes d) 30 min / 34 min e) 1984 strokes / 2267 strokes f) 113 min / 129 min g) 1850 psi / 125 bar h) 1218 psi / 83 bar i) 60 psi/100 stk / 3.54 bar/100 stk j) 975 psi / 65 bar 6.92 a) 12.4 ppg / 1.49 kg/I b) 1112 strokes / 1079 strokes c) 2465 strokes / 2185 strokes d) 8007 strokes / 7792 strokes e) 1588 psi / 109 bar f) 1047 psi / 70 bar g) 9119 strokes / 8871 strokes h) 1750 psi / 121 bar i) 1091 psi / 77 bar j) 59 psi/100 stk / 4.07 bar/100 stk 6.93 a) 13.3 ppg / 1.60 kg/I b) 797 psi / 56 bar c) 1688 strokes / 1689 strokes d) 2061 strokes / 2064 strokes e) 644 bbl / 102328 liter f) 154 min / 154 min g) 11.2 ppg / 1.34 kg/I h) 798 psi / 55 bar i) 379 psi / 27 bar j) 558 psi / 39 bar 6.94 a) 13.5 ppg / 1.56 kg/I b) 1792 strokes/ 1951 strokes

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Chapter 7 questions

Question 7.1. Using the BOP configuration shown below, answer the following questions.

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a) With drill pipe in the hole, is it possible to shut the well in under pressure and repair the side outlets on the drilling spool? Yes I No b) With no drill pipe in the hole, is it possible to shut the well in under pressure and repair the drilling spool? Yes I No c) Is it possible to shut the well in with drill pipe in the hole and circulate through the drill pipe? Yes I No d) With drill pipe in the hole and the well shut in under pressure with the annular preventer, is it possible to circulate through the kill line and choke line? Yes I No e) With no drill pipe in the hole, is it possible to shut the well in under pressure using the annular preventer and change pipe rams to blind rams? Yes I No f) While replacing the ring gasket on the drilling spool choke line flange the well starts to flow. There is no drill pipe in the hole. Can the well be shut in under pressure? Yes I No

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Question 7.2. Using the BOP configuration shown below, answer the following questions.

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a) With no drill pipe in the hole, is it possible to shut the well in under pressure and repair the side outlets on the drilling spool? Yes I No b) With no drill pipe in the hole, is it possible to shut the well in under pressure and repair the drilling spool? Yes/ No c) Is it possible to shut the well in with drill pipe in the hole and circulate through the drill pipe? Yes/ No d) While changing blind rams to pipe rams with drill pipe in the hole the well start to flow. Can the well be shut in? Yes I No e) With no drill pipe in the hole, is it possible to shut the well in under pressure and change the pipe rams? Yes I No f) With drill pipe in the hole, is it possible to shut the well in under pressure and change blind rams to pipe rams? Yes I No

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Question 7.3. Using the BOP configuration shown below, answer the following questions.

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a) With the well shut in under pressure on 5” drill pipe in the hole, is it possible to repair the side outlets on the drilling spool? Yes/ No b) With no drill pipe in the hole, is it possible to shut the well in under pressure and change the 3.1/2” rams to 5” rams? Yes/ No c) With the well shut in on 3.1/2” rams (on 3.1/2“pipe) under pressure and with a safety valve in the string, is it possible to change 5” rams to variable bore rams? Yes/ No d) With the well shut in on 5” pipe rams under pressure, is it possible to change blind rams to 5” pipe rams? Yes/ No e) With the well shut in on 5” pipe rams under pressure, can the annular element be replaced? Yes/ No f) With the well shut in on 5” pipe rams under pressure, can the manual valve on the choke line be replaced? Yes/ No

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Question 7.4. Using the BOP configuration shown below answer the following questions.

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a) With drill string in the hole and the well shut-in on the upper pipe ram, can the well be circulated while making repairs on the annular? Yes/ No b) Can the well be circulated and killed with the lower pipe rams closed when the drill string is in the hole? (i.e. circulate via the casing head valves) Yes I No c) Can the casing head valves be repaired with the string in the hole and the well closed on the annular? Yes I No

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Question 7.5. Using the BOP configuration shown below, answer the following questions.

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a) With the drill string in the hole and the well shut-in on 5" pipe ram, can the HCR be repaired? Yes/ No b) With no drill string in the hole and the well shut-in on the blind shear ram, can the HCR be repaired? Yes/ No c) With the drill string in the hole and the well shut-in on the 5" pipe rams, can the blind shear rams be changed to pipe rams? Yes/ No

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Question 7.6. The well is shut-in on the upper pipe ram. It is planned to circulate using mud pump 1 through the kill line into the annulus and bleed off mud or gas through the manual choke to the mud/gas separator.

a) b) c) d) e)

Schlumberger Private

Which if the following groups of valves must be open to kill the well safely and monitor the operation? Valves no. 2, 4, 5, 7, 8, 10, 14, 16, 25 Valves no. 1, 4, 5, 6, 8, 9, 10, 11, 12, 19, 25 Valves no. 2, 4, 7, 9, 10, 12, 15, 18, 25 Valves no. 1, 3, 10, 11, 14, 19, 25 Valves no. 1, 4, 9, 10, 11, 12, 14

Question 7.7. In which order should the valves for the choke line be installed on a surface BOP stack with a rated working pressure of 10000 psi according to best practice? (inside means: close to the BOP) a) Inside: a hydraulically operated valve; Middle: a manual valve; Outside: a hydraulically operated valve. b) Inside: a hydraulically operated valve; Outside: a manual valve. c) Inside: a manual valve; Outside: a hydraulically operated valve. d) Inside: a manual valve; Middle: a check valve; Outside: a hydraulically operated valve. e) Inside: a check valve; Middle: a hydraulically operated valve; Outside: a hydraulically operated valve.

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Question 7.8. The well is shut-in on the upper pipe rams. It is planned to circulate using mud pump 2 down the drill string, through the remote choke and mud/gas separator.

a) b) c) d) e) f)

Valves no. 2, 7, 8, 9, 16, 17, 18, 19, 25 Valves no. 2, 3, 7, 8, 10, 11, 14, 19 Valves no. 2, 7, 9, 11, 12, 15, 18 Valves no. 1, 3, 7, 8, 10, 11, 13, 14, 19 Valves no. 2, 7, 8, 9, 10, 11, 12, 14, 20 Valves no. 2, 3, 7, 8, 10, 13, 16, 17, 25

Schlumberger Private

Which if the following groups of valves must be open to kill the well safely and monitor the operation?

Question 7.9. On a surface BOP stack, in which position must the valves on the kill line and choke line be placed during drilling? a) Both type of valves closed on the kill line and opened on the choke line. b) Manual valves closed and hydraulic valves opened. c) Hydraulic valves closed on manual valves opened. d) All valves must be closed. e) All valves must be opened.

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Question 7.10. The well is shut-in on the annular BOP. It is planned to circulate from the cement pump down the drill string and bleed off through the manual choke to the mud/gas separator.

a) b) c) d) e) f)

Valves no. 2, 3, 5, 8, 9, 10, 11, 14, 19 Valves no. 1, 3, 4, 6, 7, 8, 10, 11, 13, 18, 25 Valves no. 3, 7, 8, 9, 10, 11, 12, 19, 25 Valves no. 2, 3, 5, 8, 9, 10, 11, 12, 15, 17 Valves no. 3, 5, 8, 9, 10, 11, 14, 16, 19, 25 Valves no. 2, 3, 4, 6, 7, 8, 9, 10, 11, 14, 16

Schlumberger Private

Which if the following groups of valves must be open to kill the well safely and monitor the operation?

Question 7.11. A 13.5/8’’ BOP is installed on the wellhead. What does the 13.5/8" refer to? a) b) c) d)

The height of the BOP measured from the bottom flange to the top flange. The size of the flange. The outer diameter of the BOP. The through bore size (ID) of the BOP.

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Question 7.12. Subsea Identify the LMRP components in the drawing below. a) b) c) d) e) f) g)

Control Hose Bundles = ...... Riser Connector= ...... Subsea Accumulators Bottles = ...... Flex (Ball) joint = ...... Flexible loop = ...... Annular Preventer = ...... Control Pod = ......

Question 7.13. Which of the following determines the rated working pressure of the BOP? The reservoir pressures. The MAASP. The maximum anticipated surface pressure. The dynamic bottom hole pressure.

Question 7.14. Select True or False for each of the below statements. a) Pipe rams have a seal at the bottom. True/ False b) Pipe rams are wellbore pressure assisted. True/ False c) Pipe rams can be used to close on tool joints. True/ False

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a) b) c) d)

Question 7.15. Ram type preventers are designed to hold pressure: a) b) c)

Only from above Only from below. From both above and below.

Question 7.16. What is the standard operating pressure of pipe rams?

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a) 3000 psi b) 1500 psi c) 500 - 1500 psi d) 75 - 125 psi Question 7.17. Due to well bore conditions, there is a need to change the top pipe rams to casing rams. When must the casing rams be installed? a) The casing rams must be installed while pulling out of hole with the bit at the casing shoe. b) The casing rams must be installed when rigging up for logging operations. c) The casing rams must be installed once the casing shoetrack has been tested and run past the BOP. d) The casing rams must be installed prior to rigging up for running casing. Question 7.18. A long series of E-line logging operations is to be carried out with only a small overbalance on bottom. Select the safest course of action from the statements below. All should be okay, continue operations. Spot a heavy pill on bottom prior to starting logging operations. Change and pressure test the BOP rams to allow installation of a riser & lubricator.

Question 7.19. A rig crew is about to start rigging up for running 7" casing. The BOP contains a 5” to 7" variable bore ram. Is it required to change over to 7" casing rams? a) b)

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a) b) c)

Yes No

Question 7.20. Can all ram type BOPs close on rated working pressure in the wellbore with a hydraulic operating pressure of 1500 psi? a) b)

Yes No

Question 7.21. Can all ram type BOPs open where rated working pressure is contained below the ram and hydrostatic pressure to the flow line is above the ram? a) b)

Yes No

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Question 7.22. When a ram type surface BOP is operated, the hydraulic fluid on the opposite side of the operating piston is being displaced. Indicate what happens to the fluid. a) The fluid leaves the operating cylinder and drains off in the borehole through a check valve. b) The fluid leaves the operating cylinder and returns to the opposite side of the piston to enforce the closing pressure. c) The fluid leaves the operating cylinder and returns to the fluid reservoir. Question 7.23. Select the correct statements about fixed bore ram type BOPs. SELECT THREE ANSWERS a) Ram type BOPs are designed to contain and seal rated working pressure from above the closed rams as well as from below. b) Ram type BOPs should be equipped with a mechanical locking system. c) Fixed bore ram type BOPs can close and seal on various pipe sizes. d) Fixed bore ram type BOPs can be used to hang off the drill string. e) Ram type BOPs are designed to contain and seal rated working pressure only from below the closed rams.

a) To shear tubulars like drill pipe and simultaneously seal the wellbore. b) To shear tubulars like drill pipe without sealing the wellbore. c) To affect a seal with drill collars in the hole.

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Question 7.24. What is the main purpose of the blind shear ram?

Question 7.25. Which pressure would be used to shear and seal 5", grade G drill pipe? a) 1000 psi b) 1500 psi c) 3000 psi d) Drill pipe cannot be sheared Question 7.26. Why are the front packers of pipe ram preventers enclosed between the steel plates? SELECT TWO ANSWERS a) To support the weight of drill string during hang-off. b) To prevent the rubber extruding top and bottom when the rams are closed. c) To feed new rubber into sealing contact with pipe when sealing face becomes worn. d) To prevent any swelling when used during elevated temperature operations.

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Question 7.27. What is the meaning of closing ratio for a ram type BOP? a) The ratio between opening and closing volume. b) The ratio of the wellhead pressure and the ram type BOP closing pressure. c) The ratio between opening and closing time. d) The ratio between the BOP rated working pressure and the hydraulic control unit working pressure. Question 7.28. A 15K BOP has a closing ratio of 6.8 and an opening ratio of 3. The accumulator operating pressure is 3000 psi. Determine the minimum closing pressure to close the BOP at maximum wellbore pressure. a) 1500 psi b) 2206 psi c) 3000 psi d) 5000 psi

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

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Question 7.29. What is the primary function of a weep hole on a ram type BOP?

Question 7.30. What are the main functions of a weep hole on a ram type BOP? SELECT TWO ANSWERS a) To show that the bonnet seal is leaking. b) To show that the primary mud seal on the piston rod is leaking. c) To release any overpressure that may occur during testing. d) To prevent damage to the opening chamber. Question 7.31. In an emergency, it is possible to activate a secondary seal on a ram type preventer. Which of the following pressures will it seal against? a) Wellbore pressure b) Closing chamber pressure c) Opening chamber pressure

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Question 7.32. What is the correct meaning of the term primary seal and secondary seal when used in relation with ram type BOPs? a) Primary seal is shutting in the well using the annular BOP. Secondary seal is shutting in the well using the rams after the annular BOP has already been closed. b) Primary seal is well control utilizing only mud hydrostatic pressure. Secondary seal is well control utilizing both mud hydrostatic pressure and the BOPs. c) Primary seal is the mechanical ram shaft packing. Secondary seal is injected plastic pack-in intended to active an extra seal on the ram shaft in an emergency if the primary seal is leaking. d) Primary seal is a seal between the ring gasket and the connection on the side or end outlets. Secondary seal is a seal established by ring gasket wound with Teflon tape. Question 7.33. During a routine test, it is noticed that the weep hole on one of the BOP bonnets is leaking fluid. What action should be taken?

Question 7.34. Subsea Which locking system stays in a fixed position, regardless of packer wear?

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a) The weep hole only checks closing chamber seals. Leave it till next maintenance schedule. b) Energies emergency plastic packing. If leak stops, leave it till next maintenance schedule. c) A leak is normal because the metal to metal sealing face in the bonnet needs some lubrication to minimize damage. d) Ram packing elements on ram body are worn out, replace immediately. e) The primary mud seal is leaking, secure the well and replace immediately.

a) MPL b) Poslock c) Ultralock d) Wedgelock e) Bakerlock Question 7.35. Annular preventers can seal the wellbore with all pipe removed. a) True b) False Question 7.36. Which type of rubber is suited for an annular preventer being used in extreme cold condition? SELECT TWO ANSWERS a) Natural rubber

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b) Nitrile rubber c) Neoprene rubber Question 7.37. Which one of the following statements defines 'well pressure assistance' for annular preventers? a) The well pressure acting on the piston that produces an increasing pressure in the closing chamber. b) The pressure exerted by the well on the exposed surface of the piston gives a result force that is added to the force produced by the pressure in the closing chamber. c) The pressure exerted by the well on the exposed surface of the piston gives a result force that is subtracted from the force produced by the pressure in the closing chamber. Question 7.38. What pressure must be kept in the annular BOP closing chamber during stripping operations?

Question 7.39. Why is it important to reduce the regulated hydraulic pressure for annular BOP before running a generous sized casing?

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a) The minimum pressure of BOP closure that ensures proper sealing. b) The minimum pressure that allows the tool joint to go through the packing. c) 500 psi. d) 300 psi less than the closing pressure of the ram operation.

a) To prepare for the soft shut-in procedure. b) To reduce the closing time. c) To avoid collapsing the casing during closing. d) To enable the packing unit to fit uniformly around the casing body without damaging the steel segments. Question 7.40. Which of the below statements about annular preventer are true? SELECT THREE ANSWERS a) Can be used as a means of secondary well control. b) Is designed to seal around any object in the wellbore. c) Cannot seal on a square or hexagonal Kelly. d) Will not allow tool joints to pass through. e) Will allow reciprocating or rotating the drill string while maintaining a seal against wellbore pressure. f) Can require a variable closing pressure according to the task carried out. Question 7.41.

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What should be checked before the installation of any annular packing element? SELECT TWO ANSWERS a) Temperature rating of the element. b) Type of mud to be used. c) Desired hydraulic closing pressure. d) Maximum pipe outside diameter

Question 7.42. L4, Extra When annular BOPs are pressure tested, it often happens that the test pressure cannot be kept steady during the first attempt. Test pressure may have to be charged up two or more before an acceptable test is obtained. Why is this?

Question 7.43. API A well is closed in 5" drill pipe using the annular preventer. The stabilized SICP = 1000 psi. The graph below indicates the recommended annular closing pressure as a function of well pressure.

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a) Annular BOPs always leak until the packing element finds its new shape. This motion can take several minutes. b) The compressibility of the hydraulic fluid from the hydraulic control unit below the closing piston causes the pressure drop. c) The packing unit elastomer is flowing into a new shape because the rate of flow is influenced by the applied pressure.

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100 - 200 psi 200 - 300 psi 300 - 400 psi 1000 psi

Question 7.44. Metric A well is closed in 5" drill pipe using the annular preventer. The stabilized SICP = 69 bar. The graph below indicates the recommended annular closing pressure as a function of well pressure.

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Which annular closing pressure is recommended? a) 7 - 14 bar b) 14 - 21 bar c) 21 - 28 bar d) 69 bar Question 7.45. What is the main function of a diverter? a) To divert formation and wellbore fluids away from the rig floor. b) To shut-in a shallow gas kick. c) To act as a backup in case the BOP fails to close. d) To direct fluid into the mud/gas separator. Question 7.46. Select True or False for each of the below statements. a) Diverter lines must have a small diameter to create sufficient backpressure.

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True/False b) Diverter systems are designed to shut-in the wellbore. True/False c) The main purpose of a diverter is to divert shallow gas. True/False d) The requirements of a diverter system are a low pressure annular preventer and an overboard line to the mud/gas separator. True/False Question 7.47. Which of the below statements is 'good practice’ is relation to diverter systems? a) Open the diverter line before closing the diverter. b) For safety, the diverter should only be operated some distance away from the rig floor. c) The diverter is only used occasionally and is therefore not included in the rig maintenance program. Question 7.48. Which of the below statements is 'good practice' is relation to diverter systems? Schlumberger Private

a) Diverter vent line are not designed for high pressure and can therefore contain bends when needed. b) If a rig can only install one vent line, then that line should discharge down-wind of the rig. c) Full bore rubber hoses are acceptable on diverter vent line bends.

Question 7.49. What are the components of an annular type diverter? SELECT TWO ANSWERS a) A low pressure annular preventer with a large internal diameter. b) A vent line of sufficient diameter to permit safe venting using the mud/gas separator. c) A high pressure ram type preventer with a large internal diameter. d) A vent line with a manually operated full opening valve. e) A vent line of sufficient diameter to permit safe venting and proper return flow disposal. Question 7.50. Which of the following factors would limit the success of diverter operations in case of a shallow gas blowout? SELECT THREE ANSWERS a) Rig air pressure of zero (0) psi. b) The formation strength at the conductor/casing shoe. c) Diverter lockdown dogs unlocked. d) Diverter lockdown dogs locked. e) Rig air pressure of 125 psi.

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f) Mud pumps running and pumping drilling fluid to the bottom of the well. Question 7.51. What is the main function of the choke in the overall BOP system? a) To direct hydrocarbons to the flare. b) To direct wellbore fluids to the mud/gas separator. c) To shut-in the well softly. d) To hold backpressure while circulating out a kick. Question 7.52. Why are some choke manifolds equipped with a glycol or methanol injection system? a) To minimize the effect of hot climates. b) To help prevent hydrate formation while circulating a kick. c) To help fluids flow better during well testing. d) To protect rubber goods in high temperature wells.

a) To directs returns to the separator. b) To direct returns to the pit. c) To direct returns to the flare. d) To minimize backpressure when circulating through the manifold. e) To provide backup if a problem occurs with the active choke.

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Question 7.53. Why are two chokes fitted on most choke manifolds?

Question 7.54. What is the recommended diameter for the choke manifold bleed line bypassing the chokes according to API Std 53? a) The same diameter as the other line on the choke manifold. b) At least equal to the diameter of the choke line. c) At least 5 inches. Question 7.55. What is the purpose of the choke manifold vent/bleed line that bypasses the chokes? a) To connect the choke manifold to the mud/gas separator. b) To bypass the chokes and the connect the choke manifold to the kill line. c) To bypass the chokes and bleed off high volumes of fluid/gas.

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Question 7.56. Which method is used to operate the remotely operated valves in the choke and kill line? a) Hydraulic fluid b) Air c) Nitrogen d) Wires Question 7.57. The hydraulically operated valves in the choke and kill line are designed to close automatically when operating pressure is lost. a) True b) False Question 7.58. Which of the following parameters determines the operating pressure of the mud/gas separator?

Question 7.59. Which of the following parameters determines the maximum operating pressure of the mud/gas separator?

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a) The height of the mud leg. b) The diameter and height of the mud/gas separator. c) The length of the vent line. d) The diameter and length of the vent line.

a) The length of the vent line. b) The height of the mud leg. c) The diameter and length of the vent line. d) The height of the mud/gas separator. e) The diameter of the mud/gas separator. Question 7.60. Why does pressure build up in the mud/gas separator dangerous whilst circulating out a kick? a) Pressure build up will increase the risk of lost circulation. b) Pressure build up may allow gas to be blown up the derrick vent line. c) Pressure build up may allow gas to enter the shale shaker area. d) Pressure build up will make choke adjustment difficult. Question 7.61. Determine the maximum operating pressure of the mud/gas separator in the figure below.

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Drilling fluid density is 10.9 ppg / 1.30 kg/I. Hl = 30 ft/ 9.0 m H2 = 15 ft/ 5.0 m H3 = 12 ft/ 4.0 m

a) 6.8 psi/ 0.51 bar b) 8.5 psi/ 0.64 bar c) 14.5 psi/ 1.0 bar d) 17.0 psi/ 1.15 bar Question 7.62. During a well kill gas-cut mud with a 10.0 ppg density is coming back to surface and flowing through the mud/gas separator. The original drilling fluid density is 12.5 ppg. Determine the maximum operating pressure of the mud/gas separator. The mud leg height is 15 ft.

Question 7.63. During a well kill gas-cut mud with a 1.20 kg/I density is coming back to surface and flowing through the mud/gas separator. The original drilling fluid density is 1.50 kg/I. Determine the maximum operating pressure of the mud/gas separator. The mud leg height is 4.5 m.

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a) 9.75 psi b) 17.6 psi c) 1.95 psi d) 7.8 psi

a) 0.66 bar b) 1.19 bar c) 0.1364 bar d) 0.53 bar Question 7.64. During a well kill the hot fill line may be used to circulate fresh drilling fluid across the Li-tube. From which tank should the fresh drilling fluid be circulated? a) The active tank b) The trip tank Question 7.65. What is the main purpose of the vacuum degasser? a) It is only used when circulating out a kick.

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b) It is mainly used to remove gas from the drilling fluid while drilling. c) It is mainly used to separate gas from liquids while testing. d) It is a standby in the event of the mud/gas separator failing. Question 7.66. Do the mud/gas separator and the vacuum degasser share the same purpose. a) Yes b) No Question 7.67. Where should the suction line to the vacuum degasser be connected according to best practice? a) Upstream of the mud/gas separator. b) From the mud/gas separator vent line. c) Inside the mud/gas separator. d) Downstream of the mud/gas separator

a) Because it has capacity limitations. b) Because it is not sited in an explosion proof area. c) Because cuttings must be removed first.

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Question 7.68. Why can vacuum degasser not be used in place of a mud/gas separator while circulating out a kick?

Question 7.69. Which of the following statements are correct regarding the use of a drill pipe safety valve in the string. SELECT THREE ANSWERS a) Easier to stab if strong flow is encountered up the drill string. b) Must not be run in the hole in closed position. c) Has to be pumped open to read SIDPP. d) Will not allow wireline to be run inside the drill string. e) Is kept in open position by a rod secured by a T-handle. f) Requires the use of a key to close. Question 7.70. A slip and cut operation is carried out with the bit at the casing shoe. What piece of equipment is installed on the drill pipe to make the operation safe? a) Circulating head

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b) Make up top drive/Kelly c) Full opening safety valve d) Inside BOP e) Full opening safety valve with an inside BOP on top Question 7.71. The upper top drive DPSV or upper Kelly valve is installed to isolate the surface installation from well pressure. When should this valve be closed? a) When making up connections to prevent spilling of drilling fluid. b) In well control situations when the surface pressure may exceed the rated working pressure of the rotary Kelly hose and the standpipe manifold. c) Only when the swivel packing is being replaced. d) Always when the rotary Kelly hose is being replaced Question 7.72. What is an inside BOP?

Question 7.73. Which of the following statements are correct regarding the use of a non-return valve in the string. SELECT THREE ANSWERS

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a) An element inserted into the annular preventer to reduce the inner diameter. b) A ball valve installed immediately above the bit. c) A device that can be installed in the drill string to act as a non-return valve.

a) Easier to stab if strong flow is encountered up the drill string. b) Must not be run in the hole in closed position. c) Has to be pumped open to read SIDPP. d) Will not allow wireline to be run inside the drill string. e) Has potential to leak through the open/close key. f) Is kept in open position by a rod secured by a T-handle. Question 7.74. A well kicks with the bit off bottom and is shut-in. The DPSV has been used to close off the drill string. The decision is made to strip back to bottom. What equipment should be made up onto the string to perform the stripping operation safely? a) Only the DPSV in closed position. b) Only the inside BOP c) The DPSV in open position with an inside BOP installed on top. d) The inside BOP with a DPSV in closed position installed on top. e) Only the DPSV in open position.

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Question 7.75. With a float valve installed in the drill string a kick is taken with the bit off bottom. Is it necessary to install an inside BOP above the DPSV to be able to strip back to bottom? a) Yes b) No Question 7.76. In which of the following situations is it an advantage to use a full closing float valve in the drill string? a) To avoid flow back while tripping or during a connection. b) To read the shut-in drill pressure value following a well kick. c) To allow reverse circulation. d) To reduce surge pressure.

a) It increases the risk of hydraulic collapse of the drill pipe if not filled. b) It increases tripping time. c) It increases flow back through the drill string. d) It reduces surge pressure on the formation. e) It reduces flow back in the flow line. f) It allows reverse circulation at any time.

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Question 7.77. A conventional flapper type float valve is installed in the bit sub in closed position. What effect does the float valve have on the drill string when tripping into the well? SELECT TWO ANSWERS

Question 7.78. While pulling out of hole a kick is taken. The Hydril drop in back-pressure valve is dropped and pumped down and the well is shut-in. After a while it is observed that the pressure on the drill pipe gauge continues to increase. Which of the following are the causes of this pressure increase? SELECT TWO ANSWERS a) There is an obstruction in the annulus. b) The special ring to stop the drop in back-pressure valve has not been inserted. c) The stabilizers are balled up. d) The special seat has not been inserted in the drill string. Question 7.79. While pulling out of hole a kick is taken. The Hydril drop in back-pressure valve is dropped and pumped down and the well is shut-in. After a while it is observed that the pressure on the drill pipe gauge continues to increase. Which of the following are the causes of this pressure increase? SELECT TWO ANSWERS

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a) The bit nozzles are plugged. b) The drop-in check valve is not yet seated. c) The indented surface inner seat is washed out by the mud flow. d) The stabilizers are balled up. Question 7.80. Indicate whether the following operations can or cannot be performed with a float valve in the string. a) Can the correct drill pipe pressure be read on the gauge after the pumps stopped? Yes I No b) Is it possible to get drill pipe back flow while tripping? Yes I No c) Is it possible to do forward circulation? Yes/ No d) Is it possible to do reverse circulation? Yes/ No Question 7.81. Select True or False for each of the below statements. Schlumberger Private

a) Full opening safety valves are easier to stab than non-return valve if back flow occurs. True/ False b) Non-return valves require the use of a key to close. True/ False c) Full opening safety valves have to be pumped open to read SIDPP. True/ False d) Full opening safety valves must not be run into the hole in closed position. True/ False Question 7.82. Select True or False for each of the below statements. a) Floats allow reverse circulation. True/ False b) Floats increase surge pressure. True/ False c) Floats prevent back flow up the drill string. True/ False d) Floats protect the bit from plugging. True/ False e) Floats should be run if shallow gas is anticipated. True/ False f) Floats allow SIDPP to be read without further action. True/ False Question 7.83. The following drill string is in the well:

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• 5” drill pipe with NC50 connections. • 5” HW drill pipe with NC50 connections. • 8" drill collars with 6.5/8” REG connections. The inside BOP has a NC50 pin connection. Which one of the following crossovers must be on the floor while tripping? a) NC50 box x 7.5/8" REG pin b) NC50 box x 6.5/8" REG box c) 6.5/8" REG. Box 7.5/8" REG pin d) NC50 box x 6.5/8" REG pin Question 7.84. What is the meaning of '6BX' when referring to a flange? a) The flange type b) The flange serial number c) The flange dimension d) The flange trademark

a) It is designed for RX ring joint gasket type. b) It has a 10000 psi test pressure and 5000 psi working pressure. c) It has a 10000 psi working pressure and a 7.1/16 “ID. d) It has a 7.1/16 “OD and a 10000 psi working pressure.

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Question 7.85. What is a 7.1/16 “, 10000 psi flange?

Question 7.86. What would be the effect of fitting a 7.1/16” x 5000 psi flange to a BOP stack with a rated working pressure of 10000 psi? a) The rating would remain 10000 psi. b) The rating would become 5000 psi. c) The rating would become 7500 psi. Question 7.87. The figure below illustrates an API type 68 flange.

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Which dimension indicates the nominal flange diameter? a) Dimension 1 b) Dimension 2 c) Dimension 3 d) Dimension 4 Question 7.88. What is meant with the expression 'stand-off between flanges' when installed and made up?

Question 7.89. Select the correct answers for each of the following questions a) R and RX ring joints are interchangeable. True/ False b) RX and BX ring joint are interchangeable. True/ False c) Which type flange is not designed for face-to-face make-up? 6B / 6BX d) Which type flange is designed for face-to-face make-up? 6B / 6BX e) Which type of ring joint gaskets are used on the 6B flange? R / RX / R or RX / BX f) Which type of ring joint gaskets are used on the 6BX flange? R / RX / R or RX / BX g) Which ring joint gasket types are energized by well pressure? SELECT TWO ANSWERS R /RX/ BX h) Which ring joint gasket type always contains a pressure balance hole? R /RX/ BX i) Which type flange must be used when the required rated working pressure is 10000 psi? 6B / 6BX j) The nominal size of a flange is the diameter of the ring joint gasket. True/ False

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a) The external diameter of the flange. b) The internal diameter of the flange. c) The distance between 2 flanges when installed with the studs and nuts tightened. d) The total height of the 2 flanges assembly.

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k) All ring joint gasket can be re-used. True/ False Question 7.90. A 15000 psi WP BOP will be subjected to a body test after manufacturing in the manufacturer's facility. To what pressure should it be tested? a) 30000 psi b) 22500 psi c) 15000 psi d) 20000 psi Question 7.91. How often should BOP equipment be function tested? a) Only after the installation of the BOP stack. b) At least once a week. c) Once per shift.

a) Drain the accumulator cylinders and check the nitrogen precharge pressure. b) Function test all items on the stack. c) Place all functions in neutral position and start pressure testing the BOP stack. d) Perform the accumulator unit pump capacity test.

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Question 7.92. What is the first action that should be taken after connecting the open and close hydraulic lines to a surface BOP stack?

Question 7.93. When should BOP equipment be pressure tested? SELECT THREE ANSWERS a) After the disconnection or repair of BOP equipment. b) Prior to a known high-pressure zone. c) Once per 21 days. d) Prior to spud or upon installation. e) After each new casing string. Question 7.94. Which of the following determines the rated working pressure selection of BOP equipment? a) Maximum anticipated bottom hole pressure. b) Maximum anticipated pore pressure. c) Maximum anticipated wellhead pressure. d) Maximum anticipated hydrostatic drilling fluid pressure.

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e) Maximum anticipated dynamic choke pressure. Question 7.95. Drill pipe safety valve (DPSV) are the to be tested: SELECT TWO ANSWERS a) Less often than the BOP. b) Each time the BOP is tested. c) To the same pressure as the ram BOPs. d) To the same rated working pressure as the Kelly/top drive.

Question 7.96. Components exposed to well pressure should be pressure tested. The first test is called the "initial pressure test". The following test are called ‘subsequent pressure tests'. To which pressure should the DPSV and inside BOPs be tested at the subsequent tests? a) Always use a pressure test equal to 10000 psi. b) Twice the rated working pressure of the tool used (up to 5000 psi). c) Test to a pressure at least equal to the maximum anticipated wellhead/surface pressure. d) One and half times the rated working pressure of the tool used.

a) From the wellbore side; with the check valve installed. b) From the pump side; with the check valve removed so that the pressure can bled off. c) From the wellbore side; with the check valve removed and the kill line vented. d) From the pump side; because the check valve on the outside of the valves prevents the detection of a faulty valve if they are pressure tested from the wellbore side.

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Question 7.97. How should the manually operated and hydraulically operated kill line valves on the BOP be pressure tested?

Question 7.98. A surface BOP stack is tested using a plug type tester. Why should the side outlet valves below the test plug be kept in the open position? SELECT TWO ANSWERS a) Because the test will create extreme hook loads. b) Otherwise reverse circulation would be needed to release test plug. c) To check for a leaking test plug. d) Because of the potential damage to casing/open hole. Question 7.99. When testing the BOP stack with a test plug or cup type tester in place, why is a means to communication established from below the tool to atmosphere?

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a) To avoid creation of extreme hook load. b) To avoid potential damage to the casing/open hole. c) Otherwise reverse circulation will be needed to release the tool. d) To avoid swabbing a kick during the test. Question 7.100. Under what circumstances would a cup type tester be preferred to the plug type tester when testing a surface BOP stack? a) To test the BOP stack without applying excess pressure to the wellhead and casing. b) When you require to test the casing head, outlets and casing to wellhead seals. c) There is no difference as they are interchangeable. Question 7.101. A test cup for 9.5/8” casing is used to test a BOP stack to a pressure of 10000 psi/ 690 bar using 5” drill pipe. The area of the test cup subjected to pressure is 42.4 square inches / 273.55 square centimeter. What is the minimum grade of drill pipe to use (exclude any safety margin).

Question 7.102. Extra API A cup type tester is run on 5" drill pipe inside a 13.3/8" casing. The drill pipe hangs in the elevator. • Casing ID = 12.35" • Casing burst pressure = 5380 psi • Drill pipe tensile yield = 436205 lbf a) Explain what happens to the hookload when test pressure is applied. b) Calculate the area that is exposed to the test pressure in in2 c) According to the drilling program, the casing needs to be tested to 90% of the burst pressure. Calculate the tension in the drill pipe in lbf. d) What would be the consequence of applying this test pressure? e) What is the maximum test pressure taking the tensile yield strength of the drill pipe into account?

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a) Grade E-75 premium pipe, tensile strength = 311200 lbs/ 141160 kgf b) Grade X-95 premium pipe, tensile strength = 394200 lbs/ 178800 kgf c) Grade G-105 premium pipe, tensile strength = 436150 lbs/ 197840 kgf d) Grade 5-135 premium pipe, tensile strength = 560100 lbs/ 254060 kgf e) Any grade will withstand the stress of the test.

Metric A cup type tester is run on 5" (12.7 cm) drill pipe inside a 13.3/8" casing. The drill pipe hangs in the elevator. • Casing ID = 31.37 cm • Casing burst pressure = 371 bar • Drill pipe tensile yield = 194.1 x 103 daN.

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a) Explain what happens to the hookload when test pressure is applied. b) Calculate the area that is exposed to the test pressure in cm2 and m2. c) According to the drilling program, the casing needs to be tested to 90% of the burst pressure. Calculate the tension in the drill pipe in Newton, kgf and metric tons. d) What would be the consequence of applying this test pressure? e) What is the maximum test pressure taking the tensile yield strength of the drill pipe into account?

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1.14 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 7.11 7.12

Chapter 7 answers a) Yes; b} No; c) Yes; d) Yes; e) No; f) No a) No; b) No; c) Yes; d) Yes; e) No; f) Yes a) No; b) No; c) Yes; d) Yes; e) Yes; f) No a) Yes; b) No; c) No a) Yes; b} Yes; c) Yes B C A C C D a) 3; b) 7; c) 4; d) 1; e) 2; f) 5; g) 6 C a) False; b) True; c) False B B D C B B B C B,D,E A C B,C B B C B,D A C E B A A,C B A C A, E, F A,B C C

7.44 7.45 7.46 7.47 7.48 7.49 7.50 7.51 7.52 7.53 7.54 7.55 7.56 7.57 7.58 7.59 7.60 7.61 7.62 7.63 7.64 7.65 7.66 7.67 7.68 7.69 7.70 7.71 7.72 7.73 7.74 7.75 7.76 7.77 7.78 7.79 7.80 7.81 7.82 7.83 7.84 7.85 7.86 7.87 7.88 7.89

C A a) False; b) False; c) True; d) False A B A, E A,B,C D B E B C A B D B C A D D A B B D A A, B, F C B C C,D,F C A A A,B B,D B,C a) No; b) No; c) Yes; d) No a) True; b) False; c) False; d) True a) False; b) True; c) True; d) True; e) True; f) False D A C B D C a) True; b) False; c) 68; d) 6BX;

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e) R or RX; f) BX; g) RX and BX; h) BX; i) 6BX; j) False; k) False B B B A,C,D C B,C C C C,D B

7.100 B 7.101 C 7.102 a) Pressure acts on cup area, hookload increases. b) 100.16 in2 646.2 cm 2, 0.06462 m2 c) 484975 lbf 2157662 N, 219945 kgf, 219.9 MT d) Tension > drill pipe tensile yield; drill pipe breaks. e) 4355 psi / 300 bar

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Chapter 8 questions

Question 8.1. The figure below illustrates a hydraulic control schematic for a BOP control system.

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i. Select the list below that indicates the valves that should be open while drilling. a) Valves: 2, 3, 7, 8, 11, 13, 14, 16, 17, 18 b) Valves: 1, 3, 7, 8, 10, 11, 14, 15, 17, 18 c) Valves: 2, 3, 4, 7, 9, 10, 12, 13, 15, 16, 18 d) Valves: 1, 2, 4, 7, 8, 9, 11, 12, 14, 17 ii. Select the list below that indicates the valves that should be closed while drilling. a) Valves: 1, 4, 9, 10, 12, 15 b) Valves: 2, 4, 8, 10, 11, 15, 17 c) Valves: 3, 7, 9, 13, 16, 18 d) Valves: 3, 4, 9, 11, 16, 17 Question 8.2. What should be the usable capacity of the hydraulic fluid reservoir? a) At least twice the stored hydraulic fluid capacity of the accumulator system. b) At least twice the usable hydraulic fluid capacity of the accumulator system.

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Question 8.3. Which gas is used to pre-charge accumulator bottles? a) Air b) Oxygen (0 2) c) Nitrogen (N2) d) Hydrogen e) Carbon dioxide (CO2) Question 8.4. A BOP consists of: 1 annular preventer 3 ram preventers 2 HCR valves

22 gals to close and 20 gals to open Each ram: 16 gals to close and 13 gals to open Each valve: 1.5 gals to close and 1.5 gals to open

The operator requires the full BOP to close, open and close with the pumps isolated. How many gallons are required to perform these actions?

Question 8.5. Which of the following functions on a BOP stack are supplied from the manifold pressure? SELECT TWO ANSWERS

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a) 203.5 gals b) 135 gals c) 170 gals d) 208 gals

a) Ram preventers b) Hydraulically operated choke and kill line valves c) Annular preventer d) All BOP stack functions Question 8.6. Which function on a BOP stack is supplied from the annular pressure? a) Ram preventers and hydraulically operated choke and kill line valves b) Annular preventer only c) Annular preventer and hydraulically operated choke and kill line valves d) Ram preventer, annular preventer and hydraulically operated choke and kill line valves e) No function is supplied with this pressure; the value on the gauge only indicates the maximum allowable working pressure for the annular preventer in use.

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Question 8.7. What is the correct description of a hydraulic regulator on the hydraulic BOP control unit manifold? a) A hydraulic device that reduces upstream supply pressure to a desired (regulated) pressure. b) A hydraulic device that maintains supply pressure to a desired (regulated) pressure. c) A device that is built into a hydraulic or pneumatic system to relieve any excess pressure. d) A device that automatically converts hydraulic pressure signals into electric signals or into pneumatic pressure signals. Question 8.8. On the remote panel the high/low bypass button allows you to put full accumulator pressure on which of the following? a) Rams only b) Annular only c) All functions d) Rams and HCR valve only

a) To increase hydraulic b) To increase hydraulic c) To increase hydraulic d) To increase hydraulic

annular pressure to existing accumulator pressure. accumulator pressure to 3000 psi. manifold pressure to 2000 psi. manifold pressure to existing accumulator pressure.

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Question 8.9. What is the purpose of the bypass button on the driller's remote control panel for a surface BOP installation?

Question 8.10. In what position should the 3-position/4-way valves be kept on the accumulator unit while drilling? a) All closed. b) All open. c) Some open and some closed. d) All in neutral. Question 8.11. Select the correct pressure for each of the below statements. a) What is the normal manifold pressure on a 3000 psi surface BOP control system? 500 1500 psi/ 1500 psi / 3000 psi b) What is the normal annular pressure on a 3000 psi surface BOP control system? 500 - 1500 psi / 1500 psi / 3000 psi

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c) What is the normal accumulator pressure on a 3000 psi surface BOP control system? 500 1500 psi / 1500 psi / 3000 psi

Question 8.12. What is the maximum available hydraulic pressure to close a pipe ram? a) 1000 psi b) 1200 psi c) 1500 psi d) 3000 psi Question 8.13. What would be the pressure on the ram opening lines between the BOP control system and the BOP stack during drilling?

Question 8.14. What is the maximum closing time for an 13.5/8" ram type preventer? a) Less than 30 seconds b) Less than 45 seconds c) Less than 60 seconds

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a) Zero b) 500 psi c) 1500 psi d) 3000 psi

Question 8.15. What is the maximum closing time for an 18.3/4” ram type preventer? a) Less than 30 seconds b) Less than 45 seconds c) Less than 60 seconds Question 8.16. What is the maximum closing time for a 18.3/4" annular preventer? a) Less than 30 seconds b) Less than 45 seconds c) Less than 60 seconds Question 8.17. What is the maximum closing time for a 21.1/4” annular type preventer?

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a) Less than 30 seconds b) Less than 45 seconds c) Less than 60 seconds Question 8.18. During drilling operations, the hydraulic lines between the BOP control system and the BOP stack are not pressurized. a) True b) False Question 8.19. What pressure rating should the valves and fittings have between the closing unit and a 10K BOP stack? a) 1500 psi b) 3000 psi c) 10000 psi

a) The regulated power fluid inlet will be blocked. Both outlets will be vented back into the fluid reservoir tank. b) The inlet and both outlets will be blocked. c) The inlet and both outlets will be vented back into the fluid reservoir tank.

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Question 8.20. On a surface BOP control unit, the 3-position/4-way selector valve is in the upright or center position. Which of the below statements is correct?

Question 8.21. On a control unit 3-position/4-way valves are used. Which of the following statements about these valves are true? SELECT TWO ANSWERS a) b) c) d)

They can be manually operated. They cannot be remotely operated. They can be placed in 4 positions. They have 4 active connections.

Question 8.22. On a surface BOP control unit, many selector valves are installed. Which of the following statements correctly describes a selector valve? a) A selector valve is a 3-position/4-way directional control valve that has the pressure inlet port blocked, the operator ports blocked and the pressure trapped in the center position.

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b) A selector valve has two or more supply pressure ports and only one outlet port. When fluid is flowing through one of the supply ports the internal shuttle seals off the other inlet port and allows fluid to the outlet port only. c) A selector valve is a 3-position directional control valve that has the pressure inlet port blocked and the operator ports vented in the center position. Question 8.23. Select True or False for each of the below statements.

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a) On the BOP remote control panel, the master control valve button must be depressed for 5 seconds then released before operating a BOP function. True/ False b) On the BOP remote control panel, the master control valve button must be held depressed while BOP functions are operated. True/ False c) On the BOP remote control panel, the master control valve button allows air pressure to go to each function in preparation for operating the handle/button. True/ False d) On the BOP remote control panel, if a function is activated without operating the master control valve button, that function will work. True/ False e) If the upper ram close light on the panel illuminates you know that the ram is closing. True/ False f) When the close light on the panel illuminates then you know that the 3-position/4-way valve on the accumulator has moved to the close position. True/ False g) Operating the master control valve button only will illuminate all lights on the panel. True/ False Question 8.24. What is the reason of operating the master control valve button for at least 5 seconds on an air operated remote BOP panel? a) To check if rig air pressure is correct. b) To allow buildup of air pressure to operate the 3-position/4-way valve. c) To bleed air from the system. d) To give the operator time to think about what he is doing. Question 8.25. If the air pressure gauge on the: remote control panel reads zero which of the following statements is true? a) The annular preventer can still be operated from the remote panel. b) The choke and kill lines can still be operated from the remote panel. c) No stack function can be operated from the remote panel.

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d) All function on the remote panel will operate normally. Question 8.26. What happens when the handle on the BOP remote control panel is activated to close the upper pipe ram preventer? Note: The master valve has been operated. a) The handle opens the hydraulic valve in the back of the remote-control panel and hydraulic fluid flows to the preventer. b) The handle operates an electric switch in the back of the remote panel. The electric current operates the hydraulic valve at the accumulator unit and this enables the hydraulic fluid to flow to the preventer. c) The handle operates an air valve in the back of the remote panel. The air activates a piston at the accumulator unit that operates the 3-position/4-way valve enabling flow of hydraulic fluid to the preventer. Question 8.27. Where are the electric (activating) switches for the BOP remote control panel lights located?

Question 8.28. On a driller's remote-control panel, the annular preventer is closed and the close light illuminates. What is the light indicating?

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a) On the pressure gauge mounted on the remote-control panel. b) On the BOP hydraulic control unit. c) Inside the BOP operating chambers d) On the remote-control panel operating handles.

a) A hydraulic signal has been sent to the accumulator unit. b) The annular has closed. c) The annular 3-position/4-way valve on the accumulator unit has functioned and fluid should be going to the annular. d) A micro switch in the back of the driller's remote-control panel has been activated, indicating that you have pushed the lever to the close position. Question 8.29. If the manifold gauge on the remote-control panel reads zero and the other gauges read normal values, which of the following statements is true? a) Everything is correct. b) The annular preventer can still be operated from the remote-control panel. c) No stack functions can be operated from the remote-control panel. d) All stack functions can be operated from the remote-control panel.

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Question 8.30. A ram preventer is closed from the remote panel. Which gauges would show a reduction in pressure? SELECT TWO ANSWERS a) Accumulator pressure gauge b) Manifold pressure gauge c) Annular pressure gauge d) Air pressure gauge Question 8.31. The annular preventer is closed from the remote panel. Which gauges would show a reduction in pressure? SELECT TWO ANSWERS

Question 8.32. Some BOP control systems have a selector valve labelled "Unit/Remote". What is the purpose of this selector valve? a) To shift the capability of adjusting the annular pressure reducing and regulating valve from the remote panel to the main BOP control unit and vice versa. b) To operate the bypass valve from the remote panel c) To operate the hydraulic manifold reducing and regulating valve.

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a) Accumulator pressure gauge b) Manifold pressure gauge c) Annular pressure gauge d) Air pressure gauge

Question 8.33. The "Unit/Remote" selector valve on a surface BOP control unit is placed in "Unit" position, which of the following is true? SELECT TWO ANSWERS a) All BOP functions can be operated from the Driller’s remote-control panel. b) No BOP functions can be operated from the Driller's remote-control panel. c) The hydraulic annular pressure regulator cannot be adjusted from the Driller's remote-control panel. d) The air-operated pumps are isolated. Question 8.34. On the driller's air, operated panel for a surface BOP the ram close function was activated and the following was observed: • Green light went out.

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• Red light came on. • Annular pressure did not change. • Manifold pressure decreased and later returned to the original position. • Accumulator pressure decreased to 2500 psi and remained steady. What is the most probable cause of the problem? a) There is a blockage in the hydraulic line connecting the BOP to the BOP control unit. b) There is a leak in the hydraulic line connecting the BOP to the BOP control unit. c) The selector valve (3-position/4-way valve) is stuck in the open position. d) The pressure switch or the pumps on the BOP control unit did not work. e) The electric position switches are malfunctioning.

What is the most probable cause of the problem? a) The selector valve (3-position/4-way valve) is stuck in the open position. b) There is a leak in the hydraulic line connecting the BOP to the BOP control unit. c) The electric position switches are malfunctioning. d) The pumps on the BOP control unit are malfunctioning. e) There is a blockage in the hydraulic line connecting the BOP to the BOP control unit.

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Question 8.35. On the driller's air, operated panel for a surface BOP the ram close function was activated and the following was observed: • Green light remained on. • Red light remained off. • Annular pressure did not change. • Manifold pressure did not change. • Accumulator pressure did not change.

Question 8.36. When shutting in the well from the remote-control panel the normal sequence may not occur. What has happened if the pressure gauge drops but does not rise back up? a) The 3-position/4-way valve on the hydraulic closing unit failed to shift. b) The hydraulic closing line to the BOP is plugged. c) There is a leak in the hydraulic line to the BOP. d) The bulb has blown.

Question 8.37. When shutting in the well from the remote-control panel, a problem may occur that causes doubt about whether the selected function has operated.

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What has happened if the light illuminates but the pressure gauge does not drop? a) The 3-position/4-way valve on the hydraulic closing unit failed to shift. b) The hydraulic closing line to the BOP is plugged. c) There is a leak in the hydraulic line to the BOP. d) The bulb has blown. Question 8.38. When closing the upper rams from the remote-control panel on the rig floor the green light indicator goes out but the red-light indicator does not come on. The accumulator pressure and the manifold pressure readings decrease and then return to normal What could be the reason for this? a) The 3-position/4-way valve on the hydraulic closing unit did not move. b) There is a leakage in the hydraulic circuit. c) The rams did not close. d) The is an electrical fault with the lights.

a) Try and close another ram BOP. b) Open the bypass. c) Instruct the assistant driller to repair the BOP control system. d) Manually close the ram preventer.

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Question 8.39. During a well kill, the annular preventer starts leaking badly. The driller closes a ram pre- venter but manifold pressure is lost. What should now be the first action of the driller to regain control of the situation and shut-in the BOP?

Question 8.40. You are drilling ahead and the gauges on the BOP control unit show: (The BOP has not been operated and the charge pump is not running) ACCUMULATOR

2900 psi decreasing

MANIFOLD

1500 psi constant

ANNULAR

700 psi decreasing

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b) There is a leak in the hydraulic system. c) There is a malfunction in the pressure transducer assembly. d) There is a malfunction in the hydraulic regulators. e) Faulty pump start/stop switch. Question 8.41. You are drilling ahead and the gauges on the BOP control unit show: (The BOP has not been operated and the charge pump is not running) ACCUMULATOR

3000 psi constant

MANIFOLD

1000 psi decreasing

ANNULAR

1500 psi constant

Select the best answer. Schlumberger Private

a) Everything is OK. b) There is a leak in the hydraulic system. c) There is a malfunction in the pressure transducer assembly. d) There is a malfunction in the manifold regulator. e) Faulty pump start/stop switch.

Question 8.42. You are drilling ahead and the gauges on the BOP control unit show: (The BOP has not been operated and the charge pump is not running ACCUMULATOR

2900 psi constant

MANIFOLD

ANNULAR

1500 psi constant

1500 psi constant

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d) There is a malfunction in the hydraulic regulators. e) Faulty pump start/stop switch. Question 8.43. After stopping drilling and closing the annular, annular pressure dropped and recovered. The gauges on the BOP control unit now show: ACCUMULATOR

2500 psi constant

MANIFOLD

1500 psi constant

ANNULAR

1500 psi constant

Select the best answer.

Question 8.44. You are drilling ahead and the gauges on the BOP control unit show: (The BOP has not been operated and the charge pump is not running) ACCUMULATOR

3000 psi constant

MANIFOLD

1500 psi constant

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a) Everything is OK. b) There is a leak in the hydraulic system. c) There is a malfunction in the pressure transducer assembly. d) There is a malfunction in the hydraulic regulators. e) Faulty pump start/stop switch.

ANNULAR

600 psi decreasing

Select the best answer. a) Everything is OK. b) There is a leak at one of the accumulator bottles. c) There is a malfunction in the remote panel master switch/handle. d) There is a malfunction in the annular regulator. e) Electric motor start/stop switch is faulty.

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Question 8.45. You are drilling ahead and the gauges on the BOP control unit show: (The BOP has not been operated) ACCUMULATOR

3200 psi increasing

MANIFOLD

1500 psi constant

ANNULAR

900 psi constant

Select the best answer.

Question 8.46. You are drilling ahead and the gauges on the BOP control unit show: (The BOP has not been operated and the charge pump is not running) ACCUMULATOR

3000 psi constant

MANIFOLD

1500 psi constant

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a) Everything is OK. b) There is a leak in the hydraulic system. c) There is a malfunction in the remote panel master switch/handle. d) There is a malfunction in the manifold hydraulic regulator. e) Electric motor start/stop switch is faulty.

ANNULAR

900 psi constant

Select the best answer. a) Everything is OK. b) There is a leak in the hydraulic system. c) There is a malfunction in the pressure transducer assembly. d) There is a malfunction in the hydraulic regulators. e) Electric motor start/stop switch is faulty. f) D and E are correct.

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Question 8.47. You are drilling ahead and the gauges on the BOP control unit show: (The BOP has not been operated and the charge pump is not running) ACCUMULATOR

3000 psi constant

MANIFOLD

3000 psi constant

ANNULAR

900 psi constant

What is the likely reason for the change?

Question 8.48. You are drilling ahead and the gauges on the BOP control unit show: (The BOP has not been operated and the charge pump is not running) ACCUMULATOR

2800 psi decreasing

MANIFOLD

1700 psi increasing

ANNULAR

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a) Everything is OK. b) Leak in the hydraulic circuit. c) Problem with the automatic hydro-electric pressure switch. d) Leaking bypass valve.

900 psi constant

What is the likely reason for the change? a) Fault on the annular regulator. b) Leakage in the hydraulic circuit. c) Problem with the automatic hydro-electric pressure switch. d) Fault on the manifold hydraulic regulator.

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Question 8.49. You are drilling ahead and the gauges on the BOP control unit show: (The BOP has not been operated and the charge pump is not running) ACCUMULATOR

2800 psi decreasing

MANIFOLD

1300 psi decreasing

ANNULAR

900 psi constant

What is the likely reason for the change? a) Problem with the bypass valve. b) Everything is OK. c) Problem with the charging pump. d) Leak in the hydraulic circuit. Schlumberger Private

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Chapter 8 answers i) A; ii) A A C D A, B B A D D C a) 1500 psi b) 500 - 1500 psi c) 3000 psi D C A A B B B B B A,D A a) False; b) True; c) True; d) False; e) False; f) True; g) False

8.24 8.25 8.26 8.27 8.28 8.29 8.30 8.31 8.32 8.33 8.34 8.35 8.36 8.37 8.38 8.39 8.40 8.41 8.42 8.43 8.44 8.45 8.46 8.47 8.48 8.49

B C C B C B A,B A,C A A,C D A C B D B B D A E D E A D D D

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