Directional Drilling no.1
After a survey in a deviated well it is apparent that the target will be missed. A correction run must be made but it is uncertain whether the target can be reached within the along hole distance remaining.
Check whether the correction run will allow the borehole to pass through the centre of the target and the length of the correction run.
The hole angle = 44 deg which is acceptable therefore requiring only azimuth change.
DL = 3 deg/100'
6450 AH = N 90 deg E
Survey coord :
N = 134 700'
E = 187 100'
Target coord :
N = 134 260'
E = 187 960'
Answers Directional Drilling 1
Step 1 is to calculate true length Lt from survey point to target point
Lt = (((134700 - 134260)^2 + (187960 - 187100)^2)^0.5) /sin 44 deg= 1390'
Step 2 is to determine the magnitude of the azimuth correction , where is the included angle between survey pint and target
= inv tan d/D = inv tan 440/860 = 27.1 deg
Step 3 is to determine the minimum rate of azimuth change
DL = 2 * 27.1 * 100 / 1390 = 3.9 deg/100'
Directional Drilling no. 2
In a deviated well at 1980 m AH a correction run must be made to correct the wellpath. In order to hit the proposed target, the present hole inclination of 18 deg must drop to 15 deg and the hole direction should simultaneously turn to the left from its present position of N 70 deg E. To enable the correction to take place in the presently medium soft formation before entering a hard formation, the correction must take place within the next 120 m.
The dogleg severity is limited to 0.66 deg/10m by the DTU which is in the hole.
Establish :
The required toolface setting for the first 10 m drilled, allowing for 70 deg of reactive torque from the DTU motor and the new direction at the end of the run and after drilling the first 10 m.
Answers Directional Drilling 2
Since the correction run must take plave over a 120 m drilled hole section, the change in inclination required per 10 m becomes
3 / 120 = 0.025 deg/m = 0.25 deg/10m
The new inclination after 10 m is then 18 - 0.25 = 17.75 deg
Toolface setting = inv cos ((cos 18 * cos 0.66 - cos 17.75) / (sin 18 * sin 0.66)) = 113.2
TFmag = N 70 deg E - 113.2 = 316.8 + 70 deg for reactive torque = N 26.8 deg E
Azimuth = inv tan ((tan 0.66 * sin 113.2) / (sin 18 + tan 0.66 * cos 18 * cos 113.2)
= 1.99 deg/10m
After drilling 120m the azimuth will then turn through 120 / 10 * 1.99 = 23.88 deg and the inclination will be 15 deg
Directional Drilling no. 3
A vertical well has blown out and is on fire.
Details of the blowing well :
N 1262372.4
E 0120831.6
The producing reservoir lies TVD 11250' and P0 = 8438 psi.
A relief well has to be drilled and it's coordinates are :
N 1251902.8
E 0117630.7
It is planned to kick off at 600' BDF.
Build up angle at 3 deg/100' to 50 deg and maintain inclination constant until the intersection point with the blowing well is reached.
Establish :
1) What is the azimuth of the relief well.
2) What is the required mud wt at the time of intersection.
3) What is the AHD of the relief well at the intersection.
Answers Directional Drilling 3
1) Surface coord :
N 1251902.8
E 0117630.7
Target coord :
N 1262372.4
E 0120831.6
N = 10469.6
E = 3200.9
Azimuth = inv tan (E / N) = inv tan (3200.9 / 10469.6) = N 17deg E
2) Horizontal displacement : d = (E^2 + N^2)^0.5 = 10948'
D = TVDtarget - TVDkop
R = 360/2 * characteristic length/BUR = 360/ 2* 100/3 = 1909.9'
For BU section, TVD = R * sin I = 1909.9 * sin 50 = 1463.1'
d = R * (1 - cos I) = 682.2'
For tangent section, TVD = d/ tan I = 10948.0 - 682.2 / tan 50= 8614'
TVD at time of intersection = 600 + 1463.1 + 8614 = 10677.1'
Assuming gas gradient = 0.1 ppf;
BHP10677' = 8438 - (11250 - 10677) * 0.1 = 8381 psi
Mud wt = 8381 / 10677 = 0.785 ppf
3) For BU section, AHD = I * 2 * R/ 360 = 50 * 2 * 1909.9 / 360 = 1666.7'
For tangent section, AHD = TVD/cos I = 8614 / cos 50 = 13401'
Total AHD = 600 + 1666.7 + 13401 = 15668'