Basic Well Control Calculations
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This interactive well control quiz contains 25 questions related to well control calculations. Additionally, this content is based on both IWCF and IADC well control. Each question contains 3 possible answers and there is full explanation for each one.
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
The well is shut in and Shut In Drill Pipe Pressure (SIDPP) = 900 psi. Current depth of the well is 12,500′ MD/12,500′ TVD.
Mud weight in hole is 13.0 ppg.
What is the kill weight mud for this well (round up figure)?Correct
Kill Weight Mud = Mud Weight +(SIDPP÷0.052÷TVD)
Kill Weight Mud = 13 + (900 ÷0.052÷12,500) = 14.4 ppg
Ref: kill weight mud
Incorrect
Kill Weight Mud = Mud Weight +(SIDPP÷0.052÷TVD)
Kill Weight Mud = 13 + (900 ÷0.052÷12,500) = 14.4 ppg
Ref: kill weight mud

Question 2 of 25
2. Question
The well is shut in and pressures are observed as follows: Shut In Casing Pressure (SICP) = 500 psi, Shut In Drill Pipe Pressure (SIDPP) = 400 psi. Current depth of the well is 9,500′ MD/9,000′ TVD. Mud weight in hole is 10.4 ppg.
What is the kill weight mud for this well (round up figure)?
Correct
Kill Weight Mud = Mud Weight +(SIDPP÷0.052÷TVD)
Kill Weight Mud = 10.4 + (400 ÷0.052÷9000) = 11.3 ppg
Ref: kill weight mud
Incorrect
Kill Weight Mud = Mud Weight +(SIDPP÷0.052÷TVD)
Kill Weight Mud = 10.4 + (400 ÷0.052÷9000) = 11.3 ppg
Ref: kill weight mud

Question 3 of 25
3. Question
Well information: LOT = 16.0 ppg, mud weight = 10.0 ppg and well depth = 6,000’MD/5,000’TVD.
What is the MAASP (Maximum Allowable Surface Pressure)?
Correct
MAASP = (LOT Current Mud Weight) x 0.052 x TVD
MAASP = (1610) x 0.052 x 5,000 = 1,560 psiRef: Maximum Initial ShutIn Casing Pressure (MISICP)
Incorrect
MAASP = (LOT Current Mud Weight) x 0.052 x TVD
MAASP = (1610) x 0.052 x 5,000 = 1,560 psiRef: Maximum Initial ShutIn Casing Pressure (MISICP)

Question 4 of 25
4. Question
Well information: LOT = 15.5 ppg, mud weight = 11.0 ppg and well depth = 7,000’MD/6,500’TVD.
What is the MAASP (Maximum Allowable Surface Pressure)?
Correct
MAASP = (LOT Current Mud Weight) x 0.052 x TVD
MAASP = (15.511.0) x 0.052 x 6,500 = 1521 psiIncorrect
MAASP = (LOT Current Mud Weight) x 0.052 x TVD
MAASP = (15.511.0) x 0.052 x 6,500 = 1521 psi 
Question 5 of 25
5. Question
Drill pipe capacity is 0.0188 bbl/ft.
Drill collar capacity is 0.0078 bbl/ft.Well depth is at 12,000’MD/11,000’TVD.
Drill collar length is 1,000 ft.What is the string volume for this case?
Correct
String volume = volume in drill pipe + volume in drill collar
String volume = (11,000 x 0.0188) + (1,000 x 0.0078) = 214.6 bblIncorrect
String volume = volume in drill pipe + volume in drill collar
String volume = (11,000 x 0.0188) + (1,000 x 0.0078) = 214.6 bbl 
Question 6 of 25
6. Question
The total string volume is 250 bbl and pump output is 0.102 bbl/stroke.
How many strokes are required to pump mud to the bit?
Correct
Total strokes pump to the bit = string volume ÷ pump output
Total strokes pump to the bit = 250 ÷ 0.102 = 2,541 strokesIncorrect
Total strokes pump to the bit = string volume ÷ pump output
Total strokes pump to the bit = 250 ÷ 0.102 = 2,541 strokes 
Question 7 of 25
7. Question
Well depth is 12,000’MD/10,000’TVD.
Drill collar length is 1,000 ft.
Casing shoe depth is 6,000 ft.
Annular capacity between drill pipe and open hole = 0.0559 bbl/ft.
Annular capacity between drill pipe and casing = 0.0680 bbl/ft
Annular capacity between drill collar and open hole = 0.031 bbl/ft
Pump output = 0.131 bbl/strokeHow many strokes are required to circulate from bottom up to surface?
Correct
Total strokes pump to surface = Annular volume ÷ Pump output
Annular volume = (5000 x 0.0559) + (1000 x 0.031) + (6000 x 0.068) = 718.5 bbl
Total strokes pump to surface = 718.5 ÷ 0.131 = 5485 strokesRef: Calculate Annular Capacity
Pump Output Calculation for Duplex Pump and Triplex Pump
Incorrect
Total strokes pump to surface = Annular volume ÷ Pump output
Annular volume = (5000 x 0.0559) + (1000 x 0.031) + (6000 x 0.068) = 718.5 bbl
Total strokes pump to surface = 718.5 ÷ 0.131 = 5485 strokesRef: Calculate Annular Capacity
Pump Output Calculation for Duplex Pump and Triplex Pump

Question 8 of 25
8. Question
7″ casing shoe is set at 6,500’MD/6,000’TVD.
Leak Off Test (LOT) is performed with 9.2 ppg mud and leak off pressure is 1,600 psi.What is the equivalent mud weight for LOT at shoe?
Correct
LOT = (Leak off pressure ÷0.052 ÷shoe TVD) +current mud weight
LOT = (1,600 ÷0.052 ÷ 6,000) + 9.2 = 14.3 ppgRef: Leak Off Test (Procedures and Calculation)
Incorrect
LOT = (Leak off pressure ÷0.052 ÷shoe TVD) +current mud weight
LOT = (1,600 ÷0.052 ÷ 6,000) + 9.2 = 14.3 ppgRef: Leak Off Test (Procedures and Calculation)

Question 9 of 25
9. Question
7″ casing shoe is set at 7,500’MD/7,000’TVD.
Leak Off Test (LOT) is performed with 10.0 ppg mud and leak off pressure is 1,200 psi.What is the leak off pressure at 7″ shoe?
Correct
Pressure at shoe = leak off pressure + (0.052 x MW x Shoe TVD)
Pressure at shoe = 1,200 + (0.052 x 10.0 x 7,000) = 4,840 psi.Ref: Bottom Hole Pressure Relationship
Incorrect
Pressure at shoe = leak off pressure + (0.052 x MW x Shoe TVD)
Pressure at shoe = 1,200 + (0.052 x 10.0 x 7,000) = 4,840 psi.Ref: Bottom Hole Pressure Relationship

Question 10 of 25
10. Question
Current mud weight = 10.5 ppg. Casing shoe = 8000’MD/7,500’TVD. Perform leak off test with 9.5 ppg mud. Leak off pressure = 2,500 psi.
What is MAASP?
Correct
LOT = (Leak off pressure ÷0.052 ÷ Shoe TVD) + MW used in LOT
MAASP = (LOT – Current MW) x 0.052 x Shoe TVD
LOT = (2,500 ÷0.052 ÷ 7,500) + 9.5 = 15.9 ppg
MAASP = (15.9 – 10.5) x 0.052 x 7,500 = 2,110 psiIncorrect
LOT = (Leak off pressure ÷0.052 ÷ Shoe TVD) + MW used in LOT
MAASP = (LOT – Current MW) x 0.052 x Shoe TVD
LOT = (2,500 ÷0.052 ÷ 7,500) + 9.5 = 15.9 ppg
MAASP = (15.9 – 10.5) x 0.052 x 7,500 = 2,110 psi 
Question 11 of 25
11. Question
When should personnel recalculate MAASP?
Correct
MAASP = (LOT – Current MW) x 0.052 x Shoe TVD
Based on the equation above, the MAASP should be recalculated every time that mud weight in the hole is changed.
Incorrect
MAASP = (LOT – Current MW) x 0.052 x Shoe TVD
Based on the equation above, the MAASP should be recalculated every time that mud weight in the hole is changed.

Question 12 of 25
12. Question
String volume is 65 bbl. Pump output is 0.102 bbl/stroke.
If the circulation is performed at 30 SPM, how long will it take to circulate kill mud to the bit?Correct
String volume (stokes) = string volume(bbl) ÷ pump output (bbl/stroke)
String volume (stokes) = 65 bbl ÷ 0.102 bbl/strokes = 637 strokes
Time required to pump kill mud to bit = String volume (stokes) ÷ SPM
Time required to pump kill mud to bit = 637 ÷ 30 = 21 minutes
Incorrect
String volume (stokes) = string volume(bbl) ÷ pump output (bbl/stroke)
String volume (stokes) = 65 bbl ÷ 0.102 bbl/strokes = 637 strokes
Time required to pump kill mud to bit = String volume (stokes) ÷ SPM
Time required to pump kill mud to bit = 637 ÷ 30 = 21 minutes

Question 13 of 25
13. Question
Fracture gradient at shoe is 0.71 psi/ft.
Casing shoe = 8,000’MD/7,000’TVD.
Current mud weight is 11.5 ppg.How much pressure can be shut in before breaking formation at shoe?
Correct
Mud gradient (psi/ft) = MW (ppg) x 0.052
Mud gradient (psi/ft) = 11.5 x 0.052 = 0.598 psi/ft
Maximum pressure = (Fracture gradient – Mud gradient) x Shoe TVD
Maximum pressure = (0.71 – 0.598) x 7,000 = 784 psi.
Ref: Calculate Pressure Gradient and Convert Pressure Gradient
Incorrect
Mud gradient (psi/ft) = MW (ppg) x 0.052
Mud gradient (psi/ft) = 11.5 x 0.052 = 0.598 psi/ft
Maximum pressure = (Fracture gradient – Mud gradient) x Shoe TVD
Maximum pressure = (0.71 – 0.598) x 7,000 = 784 psi.
Ref: Calculate Pressure Gradient and Convert Pressure Gradient

Question 14 of 25
14. Question
Annulus volume is 320 bbl. Pump output is 0.155 bbl/stroke.
If the circulation is performed at 40 SPM, how long will it take to circulate bottom up?
Correct
Annulus volume (stokes) = annulus volume (bbl) ÷ pump output (bbl/stroke)
Annulus volume (stokes) = 320 bbl ÷ 0.155 bbl/strokes = 2065 strokes
Time to circulate bottom up = Annulus volume (stokes) ÷ SPM
Time to circulate bottom up = 2065 ÷ 40 = 52 minutes
Incorrect
Annulus volume (stokes) = annulus volume (bbl) ÷ pump output (bbl/stroke)
Annulus volume (stokes) = 320 bbl ÷ 0.155 bbl/strokes = 2065 strokes
Time to circulate bottom up = Annulus volume (stokes) ÷ SPM
Time to circulate bottom up = 2065 ÷ 40 = 52 minutes

Question 15 of 25
15. Question
7″ casing shoe is set at 7,500’MD/6,500′ TVD. LOT performed using 9.5 ppg mud is equivalent to 16.5 ppg at 7″ casing shoe.
Current depth of the well is 12,000’MD/10,000′ TVD and mud weight while drilling is 10.5 ppg.What is the MAASP?
Correct
MAASP = (LOT – Current MW) x 0.052 x Shoe TVD
MASSP = (16.5 – 10.5) x 0.052 x 6,500 = 2,028 psi.Incorrect
MAASP = (LOT – Current MW) x 0.052 x Shoe TVD
MASSP = (16.5 – 10.5) x 0.052 x 6,500 = 2,028 psi. 
Question 16 of 25
16. Question
Current depth of the well is 9,800’MD/9,000’TVD. The well takes 20 bbl kick while drilling with 13.0 ppg mud.
SICP = 1500 psi and SIDP = 600 psiWhat is the kill weight mud?
Correct
Kill weight mud = (SIDPP ÷0.052÷TVD)+current mud weight
Kill weight mud = (600÷0.052÷9,000)+13.0 = 14.3 ppg
Ref: Kill Weight Mud
Incorrect
Kill weight mud = (SIDPP ÷0.052÷TVD)+current mud weight
Kill weight mud = (600÷0.052÷9,000)+13.0 = 14.3 ppg
Ref: Kill Weight Mud

Question 17 of 25
17. Question
The well is shut in and 15 bbl of pit gain observed. There is 800 ft of drill collar as BHA.
Annular capacity DC and OH = 0.0766 bbl/ft.
What is the influx height?
Correct
Influx height,ft = Pit gain, bbl ÷ Annular capacity DC and OH, bbl/ft
Influx height, ft = 15 ÷ 0.0766 =196 ftIncorrect
Influx height,ft = Pit gain, bbl ÷ Annular capacity DC and OH, bbl/ft
Influx height, ft = 15 ÷ 0.0766 =196 ft 
Question 18 of 25
18. Question
The well is shut in while drilling at 10,000’MD/9,000’TVD. Pit gain is 60 bbl. BHA consists of drill collar whose length is 600 ft.
Annular capacity DC and OH = 0.0376 bbl/ft.
Annular capacity DP and OH = 0.0680 bbl/ft.What is the influx height?
Correct
Volume between open hole and drill collar = BHA length x Annular capacity DC and OH
Volume between open hole and drill collar = 600 x 0.0376 = 22.56 bbl.
For this case, there is some influx above the BHA because total pit gain is 60 bbl.
Influx volume above BHA = 60 – 22.56 = 37.44 bbl
Length of influx between drill pipe and open hole = Influx volume above BHA ÷ Annular capacity DP and OH
Length of influx between drill pipe and open hole = 37.44 ÷ 0.0680 = 551 ft
Total influx height = 600 + 551 = 1,151 ft
Incorrect
Volume between open hole and drill collar = BHA length x Annular capacity DC and OH
Volume between open hole and drill collar = 600 x 0.0376 = 22.56 bbl.
For this case, there is some influx above the BHA because total pit gain is 60 bbl.
Influx volume above BHA = 60 – 22.56 = 37.44 bbl
Length of influx between drill pipe and open hole = Influx volume above BHA ÷ Annular capacity DP and OH
Length of influx between drill pipe and open hole = 37.44 ÷ 0.0680 = 551 ft
Total influx height = 600 + 551 = 1,151 ft

Question 19 of 25
19. Question
The well is shut in but SIDPP is “0” because there is a float sub in the drill string?
What is the proper way to determine SIDPP?
Correct
With the float sub in the string, the proper way to determine the shut in drill pipe pressure is to slowly bring the pump up till you see “lull” on drill pipe pressure gauge.
Incorrect
With the float sub in the string, the proper way to determine the shut in drill pipe pressure is to slowly bring the pump up till you see “lull” on drill pipe pressure gauge.

Question 20 of 25
20. Question
The well is shut in due to gas kick and the current mud weight = 13.0 ppg.
Well information: Well depth = 12,000’MD/11,000’TVD, SIPP = 900, SICP = 1200, pit gain 20 bbl and annular capacity = 0.0350 bbl/ft.
What is the influx gradient?
Correct
Influx height = pit gain ÷ annular capacity
Influx height = 20 ÷ 0.0350 = 571 ftInflux gradient (psi/ft) = (MW x 0.052) – ((SICP – SIDPP) ÷ Influx height)
Influx gradient (psi/ft) = (13 x 0.052) – ((1200900) ÷571) = 0.151 psi/ftIncorrect
Influx height = pit gain ÷ annular capacity
Influx height = 20 ÷ 0.0350 = 571 ftInflux gradient (psi/ft) = (MW x 0.052) – ((SICP – SIDPP) ÷ Influx height)
Influx gradient (psi/ft) = (13 x 0.052) – ((1200900) ÷571) = 0.151 psi/ft 
Question 21 of 25
21. Question
Well information: SIDPP = 900 psi, SICP = 1200 psi, MAASP = 2,100 psi, Slow Circulating Rate (SCR) = 750 psi, Annular pressure loss = 300 psi.
What is Initial Circulating Pressure (ICP)?
Correct
Initial Circulating Pressure (ICP) = Shut In Drill Pipe Pressure (SIDPP) + Slow Circulating Rate (SCR)
Initial Circulating Pressure (ICP) = 900 + 750 = 1,650 psi
Incorrect
Initial Circulating Pressure (ICP) = Shut In Drill Pipe Pressure (SIDPP) + Slow Circulating Rate (SCR)
Initial Circulating Pressure (ICP) = 900 + 750 = 1,650 psi

Question 22 of 25
22. Question
Well information: SIDPP = 900 psi, SICP = 1200 psi, MAASP = 2,100 psi, Slow Circulating Rate (SCR) = 750 psi, Annular pressure loss = 300 psi.
Current depth = 13,000’MD/11,000’TVD. Mud weight in hole = 13.0 ppg.What is the kill weight mud with 100 psi safety margin?
Correct
Formation pressure = (current mud weight x 0.052 x TVD) + SIDPP
Formation pressure = (13 x 0.052 x 11,000) + 900 = 8,336 psi
With 100 psi safety margin, the pressure required is 8,436 psi.
Convert 8,436 psi in to Equivalent Mud Weight.
Equivalent Mud Weight = 8,436 ÷ 0.052 ÷11,000 = 14.8 ppg
Ref: Convert Pressure into Equivalent Mud Weight
Formation Pressure from Kick Analysis
Incorrect
Formation pressure = (current mud weight x 0.052 x TVD) + SIDPP
Formation pressure = (13 x 0.052 x 11,000) + 900 = 8,336 psi
With 100 psi safety margin, the pressure required is 8,436 psi.
Convert 8,436 psi in to Equivalent Mud Weight.
Equivalent Mud Weight = 8,436 ÷ 0.052 ÷11,000 = 14.8 ppg
Ref: Convert Pressure into Equivalent Mud Weight
Formation Pressure from Kick Analysis

Question 23 of 25
23. Question
At 35 SPM with 12.0 ppg, Slow Circulating Pressure (SCR) is 350 psi. Kill weight mud require is 13.5 ppg.
What is the Final Circulating Pressure (FCP)?
Correct
Final Circulating Pressure (FCP) = Slow Circulating Pressure (SCR) x (Kill Weight Mud ÷ Current Mud Weight)
Final Circulating Pressure (FCP) = 350 x (13.5 ÷ 12.0) = 394 psi
Incorrect
Final Circulating Pressure (FCP) = Slow Circulating Pressure (SCR) x (Kill Weight Mud ÷ Current Mud Weight)
Final Circulating Pressure (FCP) = 350 x (13.5 ÷ 12.0) = 394 psi

Question 24 of 25
24. Question
The well is shut in due to kick into the wellbore. Current detph is 9,500′ MD/8,500′ TVD. SICP = 450 psi and SIDPP = 300 psi.
Bring the pump up to speed at 30 spm and drill pipe pressure reading is 1,200 psi.What is the Slow Circulating Pressure at 30 spm?
Correct
Slow Circulating Rate (SCR) = Initial Circulating Pressure (ICP) – Shut In Drip Pipe Pressure (SIDPP)
Slow Circulating Rate (SCR) = 1,200 – 300 = 900 psi.
Incorrect
Slow Circulating Rate (SCR) = Initial Circulating Pressure (ICP) – Shut In Drip Pipe Pressure (SIDPP)
Slow Circulating Rate (SCR) = 1,200 – 300 = 900 psi.

Question 25 of 25
25. Question
The well is shut in due to kick into the wellbore. Current detph is 9,500′ MD/8,500′ TVD and mud weight is 10.0 ppg.
SICP = 450 psi and SIDPP = 300 psi.
Bring the pump up to speed at 30 spm and drill pipe pressure reading is 1,200 psi.What is the Final Circulating Pressure at 30 spm?
Correct
Firstly, determine kill weight mud
Kill Weight Mud = Current Mud Weight + (Shut In Drill Pipe Pressure ÷ 0.052÷TVD)
Kill Weight Mud = 10 + (300 ÷ 0.052÷ 8,500) = 10.7 ppg
Secondly, determine Slow Circulating Rate (SCR)
Slow Circulating Rate (SCR) = Initial Circulating Pressure (ICP) – Shut In Drip Pipe Pressure (SIDPP)
Slow Circulating Rate (SCR) = 1,200 – 300 = 900 psi.
Finally, determine the Final Circulating Pressure (FCP)
Final Circulating Pressure (FCP) = Slow Circulating Pressure (SCR)x (Kill Weight Mud ÷ Current Mud Weight)
Final Circulating Pressure (FCP) = 900 x (10.7 ÷ 10) = 963 psi
Incorrect
Firstly, determine kill weight mud
Kill Weight Mud = Current Mud Weight + (Shut In Drill Pipe Pressure ÷ 0.052÷TVD)
Kill Weight Mud = 10 + (300 ÷ 0.052÷ 8,500) = 10.7 ppg
Secondly, determine Slow Circulating Rate (SCR)
Slow Circulating Rate (SCR) = Initial Circulating Pressure (ICP) – Shut In Drip Pipe Pressure (SIDPP)
Slow Circulating Rate (SCR) = 1,200 – 300 = 900 psi.
Finally, determine the Final Circulating Pressure (FCP)
Final Circulating Pressure (FCP) = Slow Circulating Pressure (SCR)x (Kill Weight Mud ÷ Current Mud Weight)
Final Circulating Pressure (FCP) = 900 x (10.7 ÷ 10) = 963 psi
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THANK YOU FOR YOUR HELP
Questions 6 and 7 have the wrong dimensions, it should be Bbls/stk not Bbls/ft.
George,
Thanks for checking this. I’ve changed the unit already.
Why do you use TVD for volume calculation in question 5?
Raquele,
I use MD which consists of DP (11,000 ft) and DC (1,000 ft).
Regards,
Shyne.
please check the question no 13 the constant should be Maximum pressure = (Fracture gradient – Mud gradient) x 0.052 x Shoe TVD
Maximum pressure = (0.71 – 0.598) x 0.052 x 7,000 = 784 psi .rong answer
right answer is=40.768
Sajjad,
It should be Maximum pressure = (0.71 – 0.598) x 7,000 = 784 psi.
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Work at SLB
Hello, I have a question on #9
the question gave information over 7″ casing, but it’s asking the pressure of the shoe at 9 5/8″ casing, so is the 7″ casing not cemented?
Thanks,
Donovan
It is 7″ casing shoe. Sorry for typo mistake. It is fixed.
Question 19 what is the lull?
The position shows the position of the float opening. You will see that the increasing trend of drill pipe is slower that typical trend.