تم الحل ✓
categoryهندسة النفط
schoolبكالوريوس
event_available2026-07-15
السؤال
Transcribed Image Text:
A well has the following profile:
Casing
Total depth 16,400' (vertical well)
Drill string:
16" Open Hole, Surface casing 13 3/8" OD and 12 3/7" ID of 68.00 lb/ft set at 2000'.
12.5" Open Hole, Intermediate Casing 9 5/8" OD and ID of 8 2/3", 47 lb/ft, set at 8,458'
8.5" Open Hole, Drilling liner 7" (6" ID) (Grade N-80) interval between 8,400' to 14,200'
Open Hole, Bit diameter 5.5"
5" drill pipe 19.50 lb/ft nominal weight, (4.276" ID) surface to 8300'.
3 1/2 drill pipe 13.30 lb/ft nominal weight between 8,300' to 15,200'.
(For the ID on the 3 1/2" drill pipe remember that density of steel is 65.5 lb/gal)
Drill collars, 4 3/4" OD, x 2 1/4" ID, 46.70 ppf, 15200' to TD.
Surface mud system:
3 mud tanks, each 9' high, 7' wide, 36' long.
With the entire drill string out of the hole:
Mud level in tanks No. 1 & 2 is 72" of mud and tank No. 3 has 60" of mud (M.W: 11 ppg)
For the following questions 2 to 9, do calculation by hand using equations from the lectures.
2 Calculate the total capacity of the surface mud system in bbl.
bbls
3 Calculate total mud volume in the three tanks in bbls.
bbls
4 Calculate the capacity in bbl/ft for each well section.
bbl/ft
5 Calculate total hole volume with the entire drill string out of the hole.
bbls
6 Calculate the displacement of the 5" drill pipe.
bbl/ft
7 Calculate the displacement of the complete drill string.
bbls
8 Calculate the total hole volume with the bit on the bottom.
bbls
9 Calculate the hydrostatic pressure at bottom of hole.
psi
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