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STANDARD HANDBOOK OF PETROLEUM & NATURAL GAS ...

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768 Drilling and Well Completions<br />

From offset wells, it is known that six joints of heavy-weight drill pipe are<br />

desirable<br />

Assume vertical hole.<br />

Solution<br />

Selection drill collar size, Table 4-73, 7q x 2% in., unit weight = 139 lb/ft.<br />

Such drill collars can be caught with overshot or washed over with washpipe.<br />

(60,000)(1’15) = 608ft<br />

Length of drill collars = (139)( 0.816)<br />

Note: Buoyant factor = 0.816.<br />

Select 21 joints of 2 x 2# in. drill collars that give the length of 630 ft.<br />

Section modulus of drill collars calculate to be 89.6 in.3.<br />

Determine size of heavy-weight drill pipe.<br />

To maintain BSR of less than 5.5, selection 5-in. heavy-weight drill pipe with<br />

unit weight of 49.3 lb/ft (see Table 491) and section modulus of 21.4 in3.<br />

Length of heavy-weight drill pipe hw = (5)(30) = 180 ft.<br />

Selection 5 in. IEU new drill pipe with unit nominal weight 19.5 lb/ft (see<br />

Table 4-79), steel grade X-95, with NC 50 tool joint (see Table 4-89).<br />

Unit weight of drill pipe corrected for tool joint is 21.34 lb/ft. Section<br />

modulus of this pipe can be calculated to be 5.7 in3.<br />

From Table 4-80, the minimum tensile load capacity of selected drill pipe<br />

PI = 501,090 lb.<br />

From Table 4-89, the recommended makeup torque T = 26,000 ft/lb.<br />

The tensile load capacity of drill pipe corrected for the effect of the maximum<br />

allowable torque, according to Equation 458 is<br />

Determine the maximum allowable length of the selected drill pipe from<br />

Equation 4-72:<br />

L, = 403,271 - (630)(139) - (180)(49.3) = 12,022ft<br />

(1.4)(0.816)(21.34) 21.34 21.34<br />

Required length of drill pipe L,.p = 10000 - (630+180) = 9,190 ft.<br />

Since the required length of drill pipe (9,190 ft) is less than the maximum<br />

allowable length (12,022 ft), it is apparent that the selected drill pipe satisfies<br />

tensile load requirements.<br />

Obtained margin of overpull:<br />

MOP = (0.9)(501090) - [(630)( 139) + (180)(49.3) + (9190)(21.34)](0.816)<br />

= 212,253 lb (greater than required 100,000 lb).<br />

In the above example, the cost of drill string is not considered. From a practical<br />

standpoint, the calculations outlined above should be performed for various drill

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