5-28-1999 E.S.P. tapered string.doc 1

Est. Stuck Pt. = Tapered StringMeasure stretch in inches with the following overpulls:

> 2-7/8 D.P. = 25,000 lbs.

> 3-1/2 D.P. = 30,000 lbs.

> 4-1/2 D.P. = 35,000 lbs.

> 5 D.P. = 40,000 lbs.

5-28-1999 E.S.P. tapered string.doc 2

Formula for Tapered String

E.S.P. = 735,294 x e x W + L [1 - W2 ] P W1

L = Length of big pipe ( above liner )

L = Length of small pipe ( in or below liner)

W = Plain end weight of big dp ( wo/tool Jt.)

W = Plain end weight of small dp ( wo/T.Jt. )

e = Stretch in inches

P = Over pull

5-28-1999 E.S.P. tapered string.doc 3

Example Estimated Stuck PointWhat is the estimated stuck point for this tapered string?

Given: L = 10,000 & L = 2,200

Plain end W = 17.93 lbs/ft ( 5x 19.5# )

Plain end W = 12.31 lbs/ft (3-1/2 x 13.30)

Stretch = 41 Over pull = 40,000 lbs.

5-28-1999 E.S.P. tapered string.doc 4

Observation> If E.S.P.>L Stuck point is in or below

L

> If E.S.P.<L Stuck point is in the upper string and the formula ( method 2 ) should be used and the estimated

stuck point recalculated.

5-28-1999 E.S.P. tapered string.doc 5

Results of FormulaE.S.P.=735,294 x 41 x 12.31 + 10,000 [ 1- 12.31 ] = 12,412 40,000 17.93

> Since E.S.P.( 12,412 ) is > L (10,000 ), stuck point is at or below L.

> The number 735,294 is a fixed constant.

5-28-1999 E.S.P. tapered string.doc 6

Method # 2 E.S.P. = 735,294 x e x Wdp

P

Where:

> e = Stretch

> Wdp = Plain end weight of dp( wo/T.Jt.)

> P = Over pull

5-28-1999 E.S.P. tapered string.doc 7

ExampleWhat is the estimated stuck point for this non-tapered drill string?

Given: > D.P. = 12,500 of 5, 19.5 lb/ft plain

end (tube) wt. = 17.93 lb/ft

> Stretch = 38

> Over pull = 40,000 lbs.

5-28-1999 E.S.P. tapered string.doc 8

Results of FormulaE.S.P.= 735,294 x 38 x 17.93 = 12,525 40,000

This number indicates that the pipe is stuck in the B.H.A.

5-28-1999 E.S.P. tapered string.doc 9