# Reactive Torque

A clockwise rotating downhole motor like the Navi-drill applies right-hand torque to the bit. An equal but reverse torque is therefore created and given by to the stator housing, which equals the torque consumed by the bit. Called 'reactive torque', it can easily be controlled by the operator, by controlling weight on bit.

During directional drilling, this reactive torque must be taken into consideration, because it tends to force the drill string to the left. The actual amount of twist angle created by reactive torque depens upon:

1. Weight on bit
2. Type and length of drill pipes
3. Inclination of hole
4. Type and length of drill collars and heavy weight drill pipes
5. If installed, number and placement of stabilizers.
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### The occuring torsional angle of a drill string with a mud motor may be determined roughly as follows:

• Measure the standpipe pressure with the bit on bottom, when flowrate and weight on bit are adjusted to drilling conditions.
• Measure the standpipe pressure when the bit is lifted off bottom with the flow rate being kept constant.
• Calculate difference in standpipe pressure.
• When a diamond bit is run, reduce above value by the bit's pressure drop.
• Read reactive torque values for calculated differential pressure out of tables.
• Read torsional angle for 1000 ft. (1000 m) length of drill pipe from Fig. 8a or

The effective angle of torsion is calculated by:

Torsional angle - (Angle for 1000ft) + (Length of DP/ft) * 1000

Note: Calculation does not consider wall friction.

Torque Itl/lb) 1S00 3000 4500

Torque Itl/lb) 1S00 3000 4500

Figure 8a: Torsional angle of drill pipe lor 1000ft length

Figure 8a: Torsional angle of drill pipe lor 1000ft length

Figure Bb: Torsional angle of drill pipe lor 1000ft length

Figure Bb: Torsional angle of drill pipe lor 1000ft length

After orientation by single shot measurement, the string has to be adjusted to the required borehole direction. To do so, the above calculated reactive torque angle is considered as a right-hand angle in addition to the direction change. Having applied the accumulated angle of the string with the rotary table, the string should be raised and lowered several times over a 30 ft. interval.

Following a survey in a deviated well it is apparent that the target will be missed if no action is taken.

A correction run must be made, but first of all a calculation will be required to determine if it is still possible to reach the target from the present depth

The dogleg severity as used in this calculation is an example only and can be different in various area's.

No units for length (depth) are used in this example to make it usable for both field and S.I. units

Last survey results:

6450 TVBDF 100 DEGRTN 44DEGR North : 134700 East : 187100

Target coordinates: North East

134260 187960

Target depth: 7500

J DEGR

Maximum DOGLEG severity standard length

In this example the standard length is : 30.

Note

The method shown in this example is only to calculate the possibility to reach the target but is not suitable to design the new well path as the line between the present hole depth and the target is assumed to be a continues curve. A more realistic or efficient wellpath might be chosen once it has been established that the target can be reached from the present depth.

SOLUTION

As the coordinates of both survey point A and target E are known the length of GF and GD can be calculated

GF = 134700 - 134260 = 440 GD = 187960 - 187100 = 860

Distance AG can be calculated as the TVD of survey point A and target are known AG = 7500 - 6450 = 1050

The angle BGC can be calculated from the azimuth reading of survey point A. In this case: angle BGC = 100 - 90 = 10 DEGR.

The rest of the calculation continues as follows; GB = AG tan angle BAG = 1050 x tan 44 = 1014 AB = VAG2 + GB2 =1460 GE=ÏGF2 + FE2 = 966

1427

BC = GB sin angle BGC = 1014* sin 10 = 176 = HD EH = ED-HD = 264.. ..

GC = GB cos angle BGC = 1014" cos 10 = 999 DC = GC - GD = 999 - 860 = 139 = HB EB - (EH^THB2 = 298

ae2=aj2+ej:

survey point A

JB? = AB2 - 2AJ x AB + AJ 2 AE2 = ^i8f EB AB + 2 x AJ x AB - ^'' 2 x AJ x AB = AE2-EB2+AB2 _AE2-EB2^ AB2

E (TARGET)

2 x 1460

1397

The length of the curve AE will be:

2x7txRx-2a =1436 360

Dogleg severity will be :

2 x a x standard length for DLS curve AE

0.48 DEG/30

This means that the target can be hit by making a correction run without exceeding the maximum dogleg severity.

NOTE: The radius R, point X are in the same plain as the points A, B and E