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

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

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

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

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.

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

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

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

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

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 121/4″ 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

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

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

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

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

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.

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

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

Question 20 of 25
20. Question
The well is shut in with 63/4″ 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 6.75^{2 }= 12,525 lb
Ref: Pressure and force relationship and applications
Incorrect
Force = Pressure x Area
Force = 350 x 0.7845 x 6.75^{2 }= 12,525 lb
Ref: Pressure and force relationship and applications

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

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 = (9.5^{2} – 7^{2}) ÷ 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 = (9.5^{2} – 7^{2}) ÷ 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 = (12.510.0) 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 = (12.510.0) x 0.052 x 8,500 = 1,105 psi

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

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 = (13.012.8) x 0.052 x 10,000 = 104 psi
Ref: Understand Hydrostatic Pressure
Incorrect
Reduction in hydrostatic pressure = (13.012.8) x 0.052 x 10,000 = 104 psi
Ref: Understand Hydrostatic Pressure
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