What is the MD of the Problem #3 well if at a TVD of 6,000 ft drops at 2 deg/100 ft until vertical?
This is easy since in #3 build was 895 ft (MD) at 3 deg/100 ft. If the well then drops at the same rate, we merely add this MD to the sequence: 6,198 + 895 = 7,093 ft (C).
Showing posts with label Directional Drilling. Show all posts
Showing posts with label Directional Drilling. Show all posts
Wednesday, September 23, 2020
Friday, September 18, 2020
Directional Drilling: 2016 #3
What is the MD at 6,000 ft TVD for a build, hold, and drop well (KOP, TVD & HD at end of hold, and build/drop rates will all be given).
This is a standard directional drilling problem; see Guidebook on 2 DRL 9 or TS2 (note TS12 doesn't have a good section here). First calculate r1 & r2 (2,865 ft); next calculate Theta1 & Theta2 (17.9 deg). Finally, use the MD equation and find the MD of each component of the vertical, build, hold sections:
MD = 1,500 + 895 + 3,803 = 6,198 ft or (C).
Note these equations are all in the GB or TS2; just plug-n-chug through them.
This is a standard directional drilling problem; see Guidebook on 2 DRL 9 or TS2 (note TS12 doesn't have a good section here). First calculate r1 & r2 (2,865 ft); next calculate Theta1 & Theta2 (17.9 deg). Finally, use the MD equation and find the MD of each component of the vertical, build, hold sections:
MD = 1,500 + 895 + 3,803 = 6,198 ft or (C).
Note these equations are all in the GB or TS2; just plug-n-chug through them.
Monday, March 9, 2020
MD from TVD: 2017 #3 & #4
Guidebook 2 DRL 9 has a quick and simple method of calculating directional drilling problems.
This problem asks for MD given TVD, as well as MD if the well turns vertical. The GB shows both the radius of curvature calculation (2,865 ft) as well as the max inc (17.9 degrees).
The MD calculation is as follows:
1,500 + 17.9/(2/100) + [8,200 - 1,500 -2,865sin(17.9)]/[cos(17.9)]
1,500 - 895 + 6,119 = 8,514 ft (C).
In the second part of the problem, if the well turns vertical from 8,200 ft TVD using the same DLS, the 895 ft is merely inverted. So just take the build calculation (inc/build) already done and use it on the drop section. If we start at 8,514 ft, that's 8,514 + 895 = 9,408 ft (C).
This problem asks for MD given TVD, as well as MD if the well turns vertical. The GB shows both the radius of curvature calculation (2,865 ft) as well as the max inc (17.9 degrees).
The MD calculation is as follows:
1,500 + 17.9/(2/100) + [8,200 - 1,500 -2,865sin(17.9)]/[cos(17.9)]
1,500 - 895 + 6,119 = 8,514 ft (C).
In the second part of the problem, if the well turns vertical from 8,200 ft TVD using the same DLS, the 895 ft is merely inverted. So just take the build calculation (inc/build) already done and use it on the drop section. If we start at 8,514 ft, that's 8,514 + 895 = 9,408 ft (C).
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