18 Casing Design Example

# W 2 8000 6382 53 5 86 563 lbf 86 563 lbf s2

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Unformatted text preview: pipe to the new depth of 6,382 ft. W 2 = ( 8,000 − 6,382 ) * 53 . 5 = 86 ,563 lbf 86 ,563 lbf S2 = = 6,378 psi 2 13 . 572 in 24 Casing Design Interpolating again, S − S1 1 P1 − Pc1 = S − S (P1 − P2 ) D.F. 1 2 1 6,378 − 5000 pcc2 = * (4,680 − 4,600) = 4,140 psi 4,680 − 1.125 5000 This is the pressure at a depth of 4,140 = 6,369 ft h3 = 0 .052 * 12 .5 25 Casing Design This is within 13 ft of the assumed value. If more accuracy is desired (generally not needed), proceed with the: Third Iteration h 3 = 6,369 ' W 3 = ( 8,000 − 6,369 ) * 53 . 5 = 87 ,259 lbf 87 ,259 S3 = = 6,429 psi 13 . 572 Pcc3 = ? 26 Casing Design Third Iteration, cont’d 1 6,429 − 5,000 thus Pcc3 = * (4,680 − 4,600) 4,680 − 5,000 1.125 = 4,140 psi = Pcc 2 27 Casing Design Third Iteration, cont’d This is the answer we are looking for, i.e., we can run 47 #/ft N-80 pipe to a depth of 6,369 ft, and 53.5 #/ft pipe between 6,369 and 8,000 ft. Perhaps this string will run all the way to the surface (check tension), or perhaps an eve...
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## This note was uploaded on 12/03/2012 for the course PMRE PMRE6004 taught by Professor M.tamim during the Summer '12 term at Bangladesh University of Eng and Tech.

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