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# Formation pressure160 ppg gas gradient012 psift

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Unformatted text preview: 4 ½” D.P. (16.6 #/ft), drilling up to 15.0 PPG mud. Formation pressure=16.0 PPG. Gas gradient=0.12 psi/ft. , Temp.Grad.=0.014 ° F /ft, TATM=60 ° F, Gain=10 bbls. Solutions: (1) From tables, Pipe Cap.=0.01422bbl/ft=15000ft×0.01422bbl/ft=213 bbls 15 PPG×0.052=0.78 psi/ft BHP=16.0PPG ×0.052psi.ft/PPG × 15000 ft =12,480 psi (2) HydroStatic Pressure= 15.0 PPG×0.052×15000=11,700psi This is the shut in drill pipe pressure. (3) KWM=15.0+780psi/15000ft ×19.23 PPG/(Psi/ft)=16.0 PPG From Tables, 8 ½” hole with 4 ½” D.P.=0.05 bbl/ft 15,000 ft ×0.05(bbl/ft)=750 bbls Copyright(c) 2005 by Ali Ghalambor &amp; Boyun Guo 39 79 4 ½“ 8 ½“ F.P.=16 PPG 15,000 ft 10 BBLs Gain 80 Copyright(c) 2005 by Ali Ghalambor &amp; Boyun Guo 40 (1) GAS column HT: 10 bbl×20 ft/bbl=200 ft (Height of Gas) (2) Gas Pressure: 200ft×0.12psi/ft=24 psi (3) MID-PT Gas Pressure (U-Tube Analog): 12480-24/2=12468 psi 12468+15=12483 psi (4) Temperature 15000ft×0.014 °F /ft+60 °F =270 °F MID-PT of U-Tube For 12,483 psi. And 270 °F, z=1.54 15000ft – 200ft =14800 psi 14800ft...
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## This document was uploaded on 03/28/2014.

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