Wait and Weight Well Control Quiz
Quiz-summary
0 of 20 questions completed
Questions:
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- 11
- 12
- 13
- 14
- 15
- 16
- 17
- 18
- 19
- 20
Information
Wait and Weight or Engineer’s Method is one of well control methods which are widely utilized to control the well. This quiz will quiz consists of 20 questions which cover both theory and calculation. You need to use a calculator in order to figure out answers in some questions. If you are not clear about this subject, you may need to read some well control books. Please feel free to leave questions if you don’t understand.
Enjoy the quiz.
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Results
0 of 20 questions answered correctly
Your time:
Time has elapsed
You have reached 0 of 0 points, (0)
Categories
- Not categorized 0%
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- 11
- 12
- 13
- 14
- 15
- 16
- 17
- 18
- 19
- 20
- Answered
- Review
-
Question 1 of 20
1. Question
Fill in the blank: Wait and Weight (Engineer’s method) is a well killing method that requires only ____ completion circulation.
Correct
Incorrect
-
Question 2 of 20
2. Question
Who is responsible for actually closing the blow out preventers and shutting the well in?
Correct
Incorrect
-
Question 3 of 20
3. Question
Which one is NOT the correct statement for wait and weight well control method?
Correct
Incorrect
-
Question 4 of 20
4. Question
If a drillpipe float is installed in the drillstring, the pressure gauge on the drillpipe will probably show zero. In order to get shut-in drillpipe pressure, what action should be performed?
Correct
Incorrect
-
Question 5 of 20
5. Question
What is the advantage of Wait and Weight well control method?
Correct
Incorrect
-
Question 6 of 20
6. Question
Fill in the blank: As the gas in the mud is circulated to the surface, the gas will begin to ___ resulting in ____ both the pressure and pit volume.
Correct
Incorrect
-
Question 7 of 20
7. Question
The following table shows sample drillpipe pressure schedule for the wait and weight method:
From the following, which is the correct statement about the above table?
Correct
Incorrect
-
Question 8 of 20
8. Question
From the following list, which parameter is not known prior to the kick while a well is being drilled?
Correct
Incorrect
-
Question 9 of 20
9. Question
What is the disadvantage of Wait and Weight well control method?
Correct
Incorrect
-
Question 10 of 20
10. Question
Well is drilled to 10,000’MD/9,000’TVD. The well is shut in due to well control. The float is bump and SIDP is 500 psi. The current mud weight is 9.5 ppg.
What is kill weight mud?
Correct
KWM = 9.5 + (500 ÷ (0.052 x 9,000)) = 10.6 ppg (round up figure)
Incorrect
KWM = 9.5 + (500 ÷ (0.052 x 9,000)) = 10.6 ppg (round up figure)
-
Question 11 of 20
11. Question
Initial circulating pressure at 30 spm is 1,500 psi. Shut in drill pipe pressure is 500 psi with 9.5 ppg current mud weight. The required kill weight mud is 11.0 ppg. What is the Final Circulating Pressure (FCP)?
Correct
SCR = ICP – SIDPP
SCR = 1,500 – 500 = 1,000 psi
FCP = SCR x (KWM÷OMW)
FCP = 1,000 x (11 ÷ 9.5)
FCP = 1,158 psi
Incorrect
SCR = ICP – SIDPP
SCR = 1,500 – 500 = 1,000 psi
FCP = SCR x (KWM÷OMW)
FCP = 1,000 x (11 ÷ 9.5)
FCP = 1,158 psi
-
Question 12 of 20
12. Question
When you bring the pump up to speed for wait and weight well control, which pressure do you need to maintain constant?
Correct
Incorrect
-
Question 13 of 20
13. Question
If the pre-recorded slow circulating pressure at various rates are unavailable for some reasons, the initial circulating pressure can be determined by proceeding the following action:
Correct
Incorrect
-
Question 14 of 20
14. Question
The casing shoe has quite low Leak off pressure. Which well control method should you use for kill the well safety and minimize the broken u-tube?
Correct
Incorrect
-
Question 15 of 20
15. Question
If the initial circulating pressure = 600 psi, Final circulating pressure is 400 psi and total internal capacity is 800 strokes (surface to bit) then what will be drillpipe pressure drop (per stroke) ?
Correct
Drillpipe pressure drop (per stroke) = (600 – 400) ÷ 800 = 0.25 psi/stroke
Incorrect
Drillpipe pressure drop (per stroke) = (600 – 400) ÷ 800 = 0.25 psi/stroke
-
Question 16 of 20
16. Question
Drill pipe pressure schedule is quite straight forward for straight hole with non-taper string. However, if the well is a deviated well and/or taper string, the drill pipe pressure schedule is quite complex. It is quite difficult to figure it out by hand for this case. Is this statement correct?
Correct
Incorrect
-
Question 17 of 20
17. Question
The well is shut in and SIDP is 600 psi. The current mud weight is 13.0 ppg and the present depth is 9,500’MD/9,000’TVD. What is the reservoir pressure?
Correct
Formation pressure = 600 + (0.052 x 13 x 9,000) = 6,684 psi
Incorrect
Formation pressure = 600 + (0.052 x 13 x 9,000) = 6,684 psi
-
Question 18 of 20
18. Question
The current mud weight is 9.2 ppg and the kill weight mud required is 10.0 ppg. The overall volume to weight up is 500 bbl. How much barite do you require to weight up the system?
Correct
Sacks of Barite per 100 bbl of mud = 1470 x (W2 – W1) ÷ (35 – W2)
Where; W1 = current mud weight
W2 = new mud weight
Sacks of Barite per 100 bbl of mud = 1470 x (10.0 – 9.2) ÷ (35 – 10) = 47
Total sack required = 47 x 5 = 235 sack
Incorrect
Sacks of Barite per 100 bbl of mud = 1470 x (W2 – W1) ÷ (35 – W2)
Where; W1 = current mud weight
W2 = new mud weight
Sacks of Barite per 100 bbl of mud = 1470 x (10.0 – 9.2) ÷ (35 – 10) = 47
Total sack required = 47 x 5 = 235 sack
-
Question 19 of 20
19. Question
The current mud weight is 9.2 ppg and the kill weight mud required is 10.0 ppg. The overall volume to weight up is 500 bbl. How much is the volume increase in the system?
Correct
Volume increase per 100 bbl of mud = 100 x (W2 – W1) ÷ (35 – W2)
Where;
W1 = current mud weight
W2 = new mud weight
Volume increase per 100 bbl of mud = 100 x (10.0 – 9.2) ÷ (35 – 10.0) = 3.2 bbl
Total volume increase = 3.2 x 5 = 16 bbl
Incorrect
Volume increase per 100 bbl of mud = 100 x (W2 – W1) ÷ (35 – W2)
Where;
W1 = current mud weight
W2 = new mud weight
Volume increase per 100 bbl of mud = 100 x (10.0 – 9.2) ÷ (35 – 10.0) = 3.2 bbl
Total volume increase = 3.2 x 5 = 16 bbl
-
Question 20 of 20
20. Question
Initial circulating pressure = 900 psi and final circulating pressure is 400 psi. Total internal capacity from surface to bit = 900 strokes. What is the drillpipe pressure drop per 200 stroke increment?
Correct
Drillpipe pressure drop (per stroke) = (900 – 400) ÷ 900 = 0.56 psi/stroke
Drillpipe pressure drop per 200 strokes = 0.56 x 200 = 112 psi
Incorrect
Drillpipe pressure drop (per stroke) = (900 – 400) ÷ 900 = 0.56 psi/stroke
Drillpipe pressure drop per 200 strokes = 0.56 x 200 = 112 psi
Leaderboard: Wait and Weight Well Control Quiz
Pos. | Name | Entered on | Points | Result |
---|---|---|---|---|
Table is loading | ||||
No data available | ||||