Basic Pressure and Its Calculation In Well Control
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In this well control quiz, there are a total of 25 questions related to basic pressure and its calculation in well control covered from the basic to some harder calculations. Additionally, this content is based on both IWCF and IADC well control. Each question contains 3 possible answers and there are full explanations for more understanding. As a minimum, you will need a pen or pencil, paper, well control formula and calculator.
Please take your time and think carefully before checking each answer in order to maximize your learning.
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Question 1 of 25
1. Question
Well information: TD at 8,000’MD/7,500′ TVD. Current mud weight is 9.8 ppg.
What is the hydrostatic pressure?
Correct
Hydrostatic pressure = 0.052 x 7,500 x 9.8 =3,822 psi
Incorrect
Hydrostatic pressure = 0.052 x 7,500 x 9.8 =3,822 psi
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Question 2 of 25
2. Question
What is Equivalent Mud Weight in ppg for a formation with 0.62 psi/ft?
Correct
Equivalent Mud Weight in ppg = 0.62÷0.052 =11.9 ppg
Incorrect
Equivalent Mud Weight in ppg = 0.62÷0.052 =11.9 ppg
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Question 3 of 25
3. Question
What is the pressure gradient (psi/ft) for 13.0 ppg mud?
Correct
Pressure gradient (psi/ft) = 13.0 x 0.052 = 0.676 psi/ft
Ref: Calculate Pressure Gradient and Convert Pressure Gradient
Incorrect
Pressure gradient (psi/ft) = 13.0 x 0.052 = 0.676 psi/ft
Ref: Calculate Pressure Gradient and Convert Pressure Gradient
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Question 4 of 25
4. Question
Hole depth = 9,000’MD/8,500’TVD.
Current mud weight is 9.2 ppg.
Annular pressure loss = 400 psi.What is the Equivalent Circulating Density (ECD) at the bottom of the well?
Correct
ECD = Current Mud Weight + (Annular pressure loss ÷ 0.052 ÷ TVD)
ECD = 9.2 + (400÷0.052÷8500) = 10.1 ppg
Incorrect
ECD = Current Mud Weight + (Annular pressure loss ÷ 0.052 ÷ TVD)
ECD = 9.2 + (400÷0.052÷8500) = 10.1 ppg
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Question 5 of 25
5. Question
Mud gradient = 0.83 psi/ft.
What is the bottom hole pressure at 9,800′ MD/9,000′ TVD?
Correct
Hydrostatic pressure at the bottom hole = 0.83 psi/ft x 9,000′ TVD = 7,470 psi.
Incorrect
Hydrostatic pressure at the bottom hole = 0.83 psi/ft x 9,000′ TVD = 7,470 psi.
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Question 6 of 25
6. Question
Formation pressure is 5,500 psi and the formation is at 8,000’MD/7,500′ TVD.
What is the pressure in term of mud weight (ppg)?
Correct
Mud weight (ppg) = 5,500 ÷0.052÷7,500 = 14.1 ppg
Ref: Convert Pressure into Equivalent Mud Weight
Incorrect
Mud weight (ppg) = 5,500 ÷0.052÷7,500 = 14.1 ppg
Ref: Convert Pressure into Equivalent Mud Weight
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Question 7 of 25
7. Question
Formation top = 12,000’MD / 11,000 TVD.
Reservoir pressure is 6,000 psi.
What is minimum mud weight for drilling without allowing influx into the wellbore?Correct
Mud weight should be at least balance formation pressure.
Mud weight = 6,000÷0.052÷11,000 = 10.5 ppg.Incorrect
Mud weight should be at least balance formation pressure.
Mud weight = 6,000÷0.052÷11,000 = 10.5 ppg. -
Question 8 of 25
8. Question
What are factors contributing to bottom hole pressure when the well is shut in?
Correct
Hydrostatic pressure and shut in pressure will affect the bottom hole pressure.
Bottom Hole Pressure = Hydrostatic Pressure + Shut in Pressure
Incorrect
Hydrostatic pressure and shut in pressure will affect the bottom hole pressure.
Bottom Hole Pressure = Hydrostatic Pressure + Shut in Pressure
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Question 9 of 25
9. Question
The well is shut in with 450 psi.
Current mud weight = 11.5 ppg
Well depth = 8,000’MD/7,500’TVD
What is the formation pressure, psi?
Correct
Formation pressure = Hydrostatic Pressure + Shut in Pressure
Formation pressure = (0.052×11.5×7500) + 450 =4,935 psi
Ref: Formation Pressure from Kick Analysis
Incorrect
Formation pressure = Hydrostatic Pressure + Shut in Pressure
Formation pressure = (0.052×11.5×7500) + 450 =4,935 psi
Ref: Formation Pressure from Kick Analysis
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Question 10 of 25
10. Question
The well is shut in with 450 psi.
Current mud weight = 11.5 ppg
Well depth = 8,000’MD/7,500’TVD
What is the Equivalent Mud Weight of formation pressure?
Correct
Equivalent Mud Weight = Current Mud Weight + (Shut in casing pressure ÷ 0.052 ÷TVD)
Equivalent Mud Weight = 11.5 + (450 ÷ 0.052 ÷7,500) = 12.65 ppg
Incorrect
Equivalent Mud Weight = Current Mud Weight + (Shut in casing pressure ÷ 0.052 ÷TVD)
Equivalent Mud Weight = 11.5 + (450 ÷ 0.052 ÷7,500) = 12.65 ppg
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Question 11 of 25
11. Question
Formation pressure = 4,200 psi.
Formation depth = 6,000’MD/5,500′ TVD
Drill the well with 13.0 ppg.
What is the shut in drill pipe pressure at this depth?
Correct
Hydrostatic pressure = 0.052x13x5,500 = 3718 psi.
Shut in drill pipe pressure = Formation Pressure – Hydrostatic Pressure
Shut in drill pipe pressure = 4,200 – 3,718 = 482 psi.Incorrect
Hydrostatic pressure = 0.052x13x5,500 = 3718 psi.
Shut in drill pipe pressure = Formation Pressure – Hydrostatic Pressure
Shut in drill pipe pressure = 4,200 – 3,718 = 482 psi. -
Question 12 of 25
12. Question
15 bbl of gas bubble migrates up 2,500′ TVD in 9,000′ TVD well while the well is shut in. Hole section is 12-1/4″ and mud weight is 10.5 ppg.
What is the approximate increase in casing pressure?
Correct
Increase in casing pressure = 0.052 x 2,500 x 10.5 = 1,365 psi
Incorrect
Increase in casing pressure = 0.052 x 2,500 x 10.5 = 1,365 psi
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Question 13 of 25
13. Question
Pump pressure at 35 spm is 1,500 psi.
What is the pump pressure at 30 spm?
Correct
New pump pressure = Old pump pressure x (New SPM ÷ Old SPM)2
New pump pressure = 1500 x (30 ÷ 35)2 = 1,102 psi
Ref: Pump pressure and pump stroke relationship
Incorrect
New pump pressure = Old pump pressure x (New SPM ÷ Old SPM)2
New pump pressure = 1500 x (30 ÷ 35)2 = 1,102 psi
Ref: Pump pressure and pump stroke relationship
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Question 14 of 25
14. Question
Pump pressure at 25 spm is 1,000 psi.
What is the pump pressure at 50 spm?
Correct
New pump pressure = Old pump pressure x (New SPM ÷ Old SPM)2
New pump pressure = 1000 x (50 ÷ 25)2 = 4,000 psi
Incorrect
New pump pressure = Old pump pressure x (New SPM ÷ Old SPM)2
New pump pressure = 1000 x (50 ÷ 25)2 = 4,000 psi
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Question 15 of 25
15. Question
With 13.0 ppg mud weight, pumping pressure is 3,000 psi.
What is the pumping pressure with 14.5 ppg mud?
Correct
New pump pressure = Old pump pressure x (New MW ÷ Old MW)
New pump pressure = 3,000 x (14.5 ÷ 13.0) = 3,346 psi
Incorrect
New pump pressure = Old pump pressure x (New MW ÷ Old MW)
New pump pressure = 3,000 x (14.5 ÷ 13.0) = 3,346 psi
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Question 16 of 25
16. Question
If a 15 bbl of gas bubble at 4,000 psi is allowed to expand to 30 bbl, what will be the bubble pressure?
Note: Neglect temperature
Correct
Use Boyle’s Gas Law: P1 x V1 = P2 x V2
P2 = (4000 x 15) ÷30 = 2,000 psi
Ref: Boyle’s Gas Law and Its Application in Drilling
Incorrect
Use Boyle’s Gas Law: P1 x V1 = P2 x V2
P2 = (4000 x 15) ÷30 = 2,000 psi
Ref: Boyle’s Gas Law and Its Application in Drilling
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Question 17 of 25
17. Question
A 15 bbl kick in larger hole creates less Shut In Casing Pressure (SICP) than a 15 bbl kick in smaller hole.
What is the reason for this case?
Correct
Height of influx in the annulus is less in a larger hole therefore there is less reduction in hydrostatic pressure which result in lower SICP.
Incorrect
Height of influx in the annulus is less in a larger hole therefore there is less reduction in hydrostatic pressure which result in lower SICP.
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Question 18 of 25
18. Question
The well is planned to TD at 9,500’MD/9,000′ TVD.
Expected formation pressure at TD is 4,500 psi.
Plan to have over balance 300 psi over formation pressure.How much mud weight in ppg should be?
Correct
Desired hydrostatic pressure at the bottom = 4500 +300 = 4800 psi
Equivalent Mud Weight = 4,800÷0.052÷9,000 = 10.26 ppg.
This figure is rounded up to 10.3 ppg.
Ref: Convert Pressure into Equivalent Mud Weight
Incorrect
Desired hydrostatic pressure at the bottom = 4500 +300 = 4800 psi
Equivalent Mud Weight = 4,800÷0.052÷9,000 = 10.26 ppg.
This figure is rounded up to 10.3 ppg.
Ref: Convert Pressure into Equivalent Mud Weight
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Question 19 of 25
19. Question
Which two of the following do not increase with gas migration in shut in well?
Correct
Gas bubble pressure and pit volume will not increase in shut in well.
Ref: Gas Behavior and Bottom Hole Pressure in a Shut in well
Incorrect
Gas bubble pressure and pit volume will not increase in shut in well.
Ref: Gas Behavior and Bottom Hole Pressure in a Shut in well
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Question 20 of 25
20. Question
The well is shut in with 6-3/4″ drill collar (3.34″ ID) and shut in casing pressure is 350 psi.
What is the force pushing the drill collar upward?
Correct
Force = Pressure x Area
Force = 350 x 0.7845 x 6.752 = 12,525 lb
Ref: Pressure and force relationship and applications
Incorrect
Force = Pressure x Area
Force = 350 x 0.7845 x 6.752 = 12,525 lb
Ref: Pressure and force relationship and applications
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Question 21 of 25
21. Question
The well is shut in and drill pipe pressure is 350 psi.
Current mud weight in hole is 9.2 ppg and well depth is 8,200’MD/7,500’TVD.
What is the kill weight mud?
Correct
Kill Weight Mud = Current Mud Weight + (SIDPP ÷0.052÷TVD)
Kill Weight Mud = 9.2 + (350 ÷0.052÷ 7,500) = 10.1 ppg
Ref: Kill Weight Mud
Incorrect
Kill Weight Mud = Current Mud Weight + (SIDPP ÷0.052÷TVD)
Kill Weight Mud = 9.2 + (350 ÷0.052÷ 7,500) = 10.1 ppg
Ref: Kill Weight Mud
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Question 22 of 25
22. Question
The plan is to cement 7″casing in 9.5″ hole and 35 bbl of sea water is pumped ahead as spacer.
Shoe depth = 8,000’MD/8,000’TVD
Mud weight = 10.0 ppg
Cement weight = 14.0 ppg
Water weight = 8.6 ppgHow much bottom hole pressure will be reduced when all of sea water is out of the shoe?
Correct
Annular capacity = (9.52 – 72) ÷ 1029.4 = 0.04007 bbl/ft
Length of seawater in the annulus = 35 ÷ 0.04007 = 874 ft
Hydrostatic pressure reduction = (10.0 – 8.6) x 0.052 x 874 = 64 psiIncorrect
Annular capacity = (9.52 – 72) ÷ 1029.4 = 0.04007 bbl/ft
Length of seawater in the annulus = 35 ÷ 0.04007 = 874 ft
Hydrostatic pressure reduction = (10.0 – 8.6) x 0.052 x 874 = 64 psi -
Question 23 of 25
23. Question
Fracture gradient at shoe = 0.65 psi/ft. Casing shoe depth is 8,500’MD/8,500′ TVD.
Current mud weight is 10.0 ppg.
Well depth is 12,000’MD/12,000’TVD.What is the maximum annular pressure loss before fracturing formation at shoe?
Correct
Fracture gradient in ppg = 0.65 ÷ 0.052 = 12.5 ppg
Annular pressure loss = (Fracture Pressure – Mud Weight) x 0.052 x TVD
Annular pressure loss = (12.5-10.0) x 0.052 x 8,500 = 1,105 psi
Incorrect
Fracture gradient in ppg = 0.65 ÷ 0.052 = 12.5 ppg
Annular pressure loss = (Fracture Pressure – Mud Weight) x 0.052 x TVD
Annular pressure loss = (12.5-10.0) x 0.052 x 8,500 = 1,105 psi
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Question 24 of 25
24. Question
Current depth = 11,000′ MD/10,000′ TVD.
Formation pressure gradient = 0.52 psi/ftWhat is surface pressure if the well is full of gas (0.1 psi/ft)?
Correct
Formation pressure = 0.52 x 10,000 = 5,200 psi
Hydrostatic pressure of gas in the well bore = 0.1 x 10,000 = 1,000 psi
Surface pressure = Formation Pressure – Hydrostatic Pressure
Surface pressure = 5,200 – 1,000 = 4,200 psiRef: Formation Pressure from Kick Analysis
Incorrect
Formation pressure = 0.52 x 10,000 = 5,200 psi
Hydrostatic pressure of gas in the well bore = 0.1 x 10,000 = 1,000 psi
Surface pressure = Formation Pressure – Hydrostatic Pressure
Surface pressure = 5,200 – 1,000 = 4,200 psiRef: Formation Pressure from Kick Analysis
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Question 25 of 25
25. Question
Gas cut mud reduces mud density from 13.0 ppg to 12.8 ppg.
Well depth = 12,000’MD/10,000′ TVD
What is the reduction of hydrostatic pressure at the bottom hole?
Correct
Reduction in hydrostatic pressure = (13.0-12.8) x 0.052 x 10,000 = 104 psi
Ref: Understand Hydrostatic Pressure
Incorrect
Reduction in hydrostatic pressure = (13.0-12.8) x 0.052 x 10,000 = 104 psi
Ref: Understand Hydrostatic Pressure
Leaderboard: Basic Pressure and Its Calculation In Well Control
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There is something wrong in here I marked the right answer but program take it as incorrect while showing the wrong answer as correct one.
Usama,
Please let me know which questions have this issue. I will fix it.
Regards
shyne
dear Shyne,
The error question is number 18. and also check the answer n0. 21.
Thanks..
Regards,
deni m
Yes, I found the same – on question 18 of 25
18. Question = The well is planned to TD at 9,500′MD/9,000′TVD. Expected formation pressure at TD is 4,500 psi. Plan to have over balance 300 psi over formation pressure. How much mud weight in ppg should be?
9.7 ppg
10.3 ppg (answer selected)
11.0 ppg
System responded – Incorrect
Desired hydrostatic pressure at the bottom = 4500 +300 = 4800 psi
Equivalent Mud Weight = 4,800÷0.052÷9,000 = 10.3 ppg
Ref: Convert Pressure into Equivalent Mud Weight
Dave,
There is an error in the system on No 18 and right now the issue is solve.
Thanks
Shyne
Hy Shyne,
I agree with Usama, there is something error at this Quiz Program, I marked the right answer, but the program take it is incorrect (Question 18).
And please check the answer no 21.
Great Quiz Program Actually.. 🙂
Best Regards,
Deni M
Deni,
Both 18 and 21 are fixed now. Sorry for the error.
Regards,
Shyne.
Thanks Shyne.. 🙂
Cheers,
Deni M
kindly review question number 18 & 21. The Answers are not right.
Aftab,
The no 18 is fixed and the no 21 is correct.
Regards,
Shyne
Number 18 is incorrect. The explanation and the answer shown there is correct, however the answer shown in the selections is incorrect.
Rick,
I’ve fixed the issue with No. 18. It works well now.
Regards,
Shyne.
THX………! When do you update the leaderboard…..?
Comment on question 25. Although the question and the answer are correct as stated in the quiz, in reality the mud is usually only gas cut at the flow line and will not be gas cut once it reaches its solution gas pressure or bubble point of the system. This is a complex physical calculation not usually available to field personnel and involves the knowledge of the solubility of hydrocarbon gases in water/oil drilling fluids. This condition is more dramatic in oil base fluids and can and does lead to the double kick (inital and bubble point) condition observed with these fluids. Craveat: If in doubt do a flow check, etc.
George,
Thanks for your comment. In this question, we would like people to simply understand of the gas cut mud in a very basic concept. Therefore, people can apply well control concept on this question.
Regards,
Shyne
Many thanks
THANKS ALLOT ADMIN
HI; Regarding Q 11 ( IT SHOULD BE SIDPP NOT SICP )
Fahmi,
Thanks for your comment.
Regards,
Shyne.
Good Web site
Sir ,
Shyne Hi,
This website is very good and through these questions revision of IWCF Exm has been made . I Suggest that more questions include in it where as in your book have all these questions but this’s the v.simple and easy way.
Sir are you have a easy solution to solve the NEW kill sheets
Regards and thanks.
Thanks
I am trying to get the quiz started but it is not coming up
Can you try again?
I fix this issue already.