A cased and cemented hole with a measured depth of 7112 ft is perforated from 6967 ft to 6992 ft. the customer wants tubing to be at least 90 ft above the top of the perforations but wants the tubing to be at the rig floor located 18 ft above ground level. John is using 33 foot joints of tubing. how many joints should he use to satisfy the customers request?

Question

Diana Waller

Mathematics

27 November, 22:55

A cased and cemented hole with a measured depth of 7112 ft is perforated from 6967 ft to 6992 ft. the customer wants tubing to be at least 90 ft above the top of the perforations but wants the tubing to be at the rig floor located 18 ft above ground level. John is using 33 foot joints of tubing. how many joints should he use to satisfy the customers request?

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Answers (1)

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Ray Brock · Answer 1

209 joints

Step-by-step explanation:

Depth of hole = 7112_ft

Perforated depth = 6967_ft to 6992_ft

Tubing set depth = 90_ft above top of perforation

and 18_ft above ground

Length of tubing required = (6967-90) + 18 = 6895_ft of tubing

One joint of tubing = 33_ft

6895_ft requires =

6895_ft/33_ft = 208.93 ~~209 joints