DETERMINATION OF THE LENGTH OF FREE PIPE IN A STUCK STRING

Find out the length of free pipe by the following equations, 2.1 x 103 x A x e

L = ———————— P2 - P1 Where :

L = Length of free pipe (m)

A = Cross sectional area of drill-pipe (cm2₎

e = Differential stretch (m) P2 - P1 = Differential pull (T)

Considering a correction factor of 1.05 for tool joint the equivalent length (Leq) of free drill pipe is given by:

Leq = 1.05 x L

Note: This method is fairly accurate in straight wells. 13.6.1 Method of Application of this Technique

Prior to measuring the differential stretch the string should be thoroughly worked so as to minimize the effect of residual stress in the string.

A pull (P1) of 10-15T greater than the air weight of the drill string is applied to the stuck string and a mark is made on the kelly or pipe as the case my be.

This pull is released and equal pull (P1) is applied once again. Another mark is made on the kelly. The two marks do not coincide due to the friction in the hole. The mid point between the two marks is taken as the upper reference mark A.

A pull P2 (P1+ 10-15T) is applied and a lower reference mark B is made following the steps mentioned above.

The distance between the two marks A and B is measured as ‘e’.

Note : The pull must be within the safe limits of the margin of over-pull of the string. Example

String stuck at 4200m in 8-1/2" hole with mud density 1.4.

Drilling String:

1500m of 5", 19.5 ppf# (NC50) E grade class 2 drill pipe weight per meter 31.02 Kg/m. 1500m of 5", 19.5 ppf# (NC50) G grade class 2 drill pipe weight per meter 32.63 Kg/m. 1000m of 5", 19.5 ppf# (NC50) S grade class 2 drill pipe weight per meter 33.64 Kg/m. 200m of 6-1/2" x 2-13/16" drill collars weight per meter 136.4 Kg/m.

Step I :

Calculate the margin of over pull (MOP).

Weight of the drill collars in air = 200 x 136.4 = 27.280T

Weight of E grade drill pipe in air = 1500 x 31.02 = 46.53T

Weight of G grade drill pipe in air = 1500 x 32.63 = 48.945T

Weight of S grade drill pipe in air = 1000 x 33.64 = 33.64T

Buoyancy factor in mud of 1.4 SG is 1 - 1.4 x 8.33 = 0.822

MOP of the drill string is given by

Tensile strength of the pipe x factor of safety (0.9) minus the buoyant weight of the string. MOP of E grade drill pipe (class 2) with tensile strength (122.67T) is:

122.67T x 0.9 - 0.822 x (27.280 + 46.53) = 49.73T

MOP of G grade drill pipe (class 2) with tensile strength (171.73T) is: 171.73 x 0.9 - 0.822 x (27.280 + 46.53 + 48.945) = 53.65T MOP of S grade drill pipe (class 2) with tensile strength (220.8T) is:

220.8 x 0.9 - 0.822 x (27.280 + 46.53 + 48.945 + 33.64) = 70.16T Hence the MOP of the string is the minimum of the three i.e. = 49.73T Buoyant wt. of drill string is:

0.822 x (27.280 + 46.53 + 48.945 + 33.64) = 128.56T. 128.56

Air of wt. of the string = ———— = 156.4T 0.822

Hence while working on pipe or during calculation of the free point the hook load of the string should not exceed 128.56 + 49.73 = 178.3T

Step II:

Apply a pull (P1) of 160T and make a mark of the string. Release the string weight to 128T and again pull to 160T. Make another mark of the string. Mid point between the two marks is point A. Apply a pull (P2) of 175T and make a mark on the string. Release the weight to 128 T and again pull to 175T. Mid point between these marks is point B.

The distance between A and B is the stretch for a differential pull of (P2-P1) i.e. 175-160 = 15T. Let us assume this distance is 62.8 cm i.e. 0.628m

Step III:

Apply the formulae with A = 34.03 cm2, e = 0.8m, P2-P1 = 15T 2.1 x 1.05 x 103 x 34.03 x 0.8

Length of free pipe = —————————————— = 4001.93m 15

Back Off of a stuck pipe

1. Before back off first determine the free length of drill string by either the differential stretch method or more accurately by electro-logging technique.

2. Make up the string to the maximum of 80% of the torsional limit. 3. Put the neutral point on a level with the joint to be backed off.

The weight indicator tension is given by: P x A

T = W + ———— 1000 where

T = Weight indicator tension in tons.

W = Buoyant weight of free drill string plus weight of kelly, traveling block, hook, etc. (tons) P = Hydrostatic pressure at the point of back off (kg/cm2₎

4. Work 80% of left hand torsion of that applied in step 2 to the point of back off. 5. Run in the string shot to point of back off desired and detonate the string shot.

6. If the string is not fully opened, count the number of turns released. If it is less than that applied in point 4, then the joint is partially opened and by working the torque down the joint may get fully opened.

Note: If back off is done without the kelly connected, then in this case the weight of kelly & swivel

etc. is not be considered.

13.6.2 Back-off Procedure in Highly Deviated (ERD) Wells

When backing off in a deviated hole it is difficult to calculate the loss of weight due to buoyancy, friction in the hole & weight of pipes lying on the low side of the hole.

Example

Assume that following drill string is in the hole :

1500m x 31.24 Kg/m drill pipe = 46.86 T

192m x 75 Kg/m HWDP = 14.4 T

118m x 219 Kg/m of 8” D/C, stab, reamers = 25.842 T 118m x 219 Kg/m of 8” D/C, stab, reamers = 25.842 T

TOTAL WEIGHT IN AIR = 87.102 T

In this case assume that the weight indicator reading before pipe got stuck was 50 T. (this should be recorded in the driller’s daily log book.)

The wt. of pipe in the hole would be = 50T

Minus the wt. of block = 11 T

TOTAL WT. IN THE HOLE = 39 T

Wt. in air = 87.102 T

Wt. in the hole = 39 T

LOSS OF WT. = 48.102 T

Loss of wt. due to buoyancy, friction & pipe lying on the low side of the hole is 48.102 Kg. or 52.22 % .

To back off at the top of the D/C. we have

1500m x 31.24 Kg/m drill pipe = 46.86 T

192m x 75 Kg/m HWDP = 14.4 T

Total Wt. in air = 61.26 T

Minus 52.22% buoyancy & friction etc. = 31.99 T

Total wt. in the hole = 29.27 T

Plus wt. of the block = 11 T

Plus 5000 Kg. pull = 5 T

WEIGHT TO BACK OFF FROM TOP OF 8” D/C = 45.27 T

Note : Another method that could be used, starting the calculation with wt. of what is left behind in

the hole. In that case use rotational weight.

13.6.3 Precautions While Backing Off with String Shots (Deviated Holes)

z When taking the free point with a logging unit, the tool pusher and logger should be in the logging cabin with the logging engineer to study the behavior of the tensionand torque indicators and to get an idea of the string reaction to the pull and torque required to overcome friction.

z Always record the tension reading first, prior to taking torque readings. Especially in deviated holes, residual torque can give false indications of the pipe being free when applying tension.

z After every torque reading, rotate the string a few turns to the left to get rid of any residual torque.

z Check any reading taken on the indicators with the logging unit monograph to see if the meter point readings confirm. After the free point is determined, the left hand torque has to be worked down the hole before the string shot.

z· Sometimes in deviated holes it is necessary to use the kelly to work down the torque, even if it is required to back off high up the hole first to be able to install the kelly.

Note : If a fish is left in the hole after a back-off, then the actual weight of the fishing string is

known. When this fishing string has been latched onto the fish, then the total string weight to the next lower back off point can be calculated as :

The neutral weight of the fishing string PLUS the weight of the stuck string down to the new back off point MINUS the percentage buoyancy and friction loss previously calculated for the original string. Counter check this figure again with the remainder of the fish in the hole and the neutral point of the string when it was still free.

When running back in the hole with a fishing string, recheck all connections and make them up to maximum tightness. Remove the rubber protectors because they can account for approximately 10% of the total friction.

Transmitting Torque Down the Hole

The following information is required

1. What is the equivalent in amperes or ft-lbs or m-kgs of the drill string make up torque? This value should not be exceeded when applying left hand torque, and preferably 60 - 80% of this value should be used.

2. What is the string weight while rotating?

This is considered the neutral weight. Do not confuse this weight with the free weight of the string, which is the weight of the string above the stuck point.

3. What is the weight going down? 4. What is the weight pulling up?

For an estimate, assume that the rotating weight is slightly less than 50% of the difference between weights pulling up and going down, added to the weight going down.

5. What is the free weight of the string to the back-off point?

For accurate determination, see the calculation in example above.

Procedures