钻井工程师lecture6

PE3043-FALL/2004-Page1SandContent•200-meshsieve•Funneltofitsieve•Aglasstubecalibratedtoreaddirectlythe%ofsandbyvolumePE3043-FALL/2004-Page2TestProcedure•Fillglassmeasuringtubetomarkwithmud•Addwatertonextmark•Shakethetubevigorously•Pourcontentsoftubeontoscreen,discardingliquid•AddMorewatertotube,shakeandpourthroughscreen•Repeatuntiltubeisclean•Washsolidsonthescreenuntiltheyareclean•Putfunnelupsidedownovertopofsieve•Insertfunneltipintomouthofglasstube•Washsandoffscreenintoglasstubewithwaterspray•Allowsandtosettleandreadvol%“sand”•Report“Sand”asvol%onAPIMudReportPE3043-FALL/2004-Page3•WhatImportantInformationIsObtainedwithThisTest?–Amountofcoarsesolidsinmud,thatmaybeabrasive–Drillingcontractorsapplythistesttowarnofpossiblewearonpumps,etc•WhatIsThisTestUsedintheAnalysisOf?–Excesspumpwear–FailureofunconsolidatedsandsPE3043-FALL/2004-Page4Errors•Incompletewashingofsolidsonscreen•Usingwronggraduationmarksontube•Screenplugs,sosmallerparticlesareretained•Can’tuseoil-basemudinthistestPE3043-FALL/2004-Page5MudRetort•Todeterminethevolumefractionofoil,water,andsolids•Distilledwater•DistilledoilPE3043-FALL/2004-Page6SolidFraction•fsisthevolumefractionofsolids•fwisthevolumefractionofdistilledwatercollectedinthegraduatedcylinder•foisthevolumefractionofdistilledoil•Cfisthevolumeincreasefactor(tables2.3&2.4)ofwsfCff1PE3043-FALL/2004-Page7A12-lbm/galsaltwatermudisretortedandfoundtocontain6%oiland74%distilledwater.Ifthechloridetestshowsthemudtohavechloridecontentof79000mgCl-/L,whatisthesolidsfractionofmud?Assumethemudisasodiumchloridemud.Solution:MgNaCl/L=79000/35.457*(23+35.457)=130350mg/LFromTable2.5wt%=12%fromTable2.3Cf=1.045Example2.8167.006.0045.1*74.011ofwsfCffPE3043-FALL/2004-Page8PilotTests•Evaluationofmixturesofgivenconcentrationsanddensities•Idealmixturing–Vtotal=V1+V2+…..+Vn–•Themixturedensity:•Typicaldensities:Table2.7iiimVnnVVVmmm2121PE3043-FALL/2004-Page9Computethevolumeanddensityofamudcomposedof25lbmofbentoniteclay,60lbmofAPIbariteand1bbloffreshwater.Solution:MaterialsDensityClay910lbm/bblBentonite1470lbm/bblVtotal=V1+V2+V3=V1+m1/1+m2/2=1.0+25/910+60/1470=1.0683bblsDensityofwater:350lbm/bblMixturedensity=(350+25+60)/1.0683=407lbm/bblExample2.10PE3043-FALL/2004-Page10Non-IdealMixturing•UseTables2.3&2.4todeterminethefinaltotalvolume•Idealassumptionisinvalid–VtotalV1+V2+…..+VnPE3043-FALL/2004-Page11Determinethevolumeanddensityofabrinecomposedof110.5lbmofNaCland1bbloffreshwaterat68F.Solution:Totalvolumeis1.113bbls(Table2.3)densityis:9.85lbm/gal(Table2.3)Ex:trythiswithidealmixturingassumptionandcomparetheresults.Example2.11PE3043-FALL/2004-Page12EffectofClayonDrillingFluidDensity•Thedensityofclayis21.7lbm/gal•Thedensityofwateris8.33lbm/gal•Densityofmud:(basedon100lbmofmud)33.8%)1(*1007.21%*1001002121wtwtVVmmPE3043-FALL/2004-Page13SolidsVolumeofRockFragmentsEnteringtheMud–Vs=Thesolidsvolumeenteringthemud–=porosity–D=thebitdiameter–dD/dt=thepenetrationrateofthebitdtdDdVs4)1(2PE3043-FALL/2004-Page14DensityControlAdditives•BariumSulfateistheprimaryadditiveusedtoincreasedensity•Specificgravityofpurebariumsulfateis4.5•Commercialgrade(APIbarite)hasanaveragespecificgravityof4.2•Alternativedensity-controlagents:–Hematite(Fe2O3)specificgravity=4.9~5.3–Ilmenite(FeO•TiO2)specificgravity=4.5~5.11–Galena(PbS)specificgravity=7.5PE3043-FALL/2004-Page15DensityControlAdditivesWhenthefinalvolumeisnotlimited:Problem:largeamountofAPIbaritecancausethedrillingfluidtobecomequiteviscousSolution:Addingwaterwiththeweightmaterial.Butadditionalweightmaterialrequiredbecausewaterloweredthedensity.BBBBmVVmVVVV1122112BBBBVVmVV1212212112BBVVPE3043-FALL/2004-Page16Itisdesiredtoincreasethedensityof200bblof11-lbm/galmudto11.5-lbm/galusingAPIbarite.Thefinalvolumeisnotlimited.ComputetheweightofAPIbariterequired.Solution:densityofAPIbariteis35.0lbm/gal(Table2.7)Example2.16lbmVVmbblVVBBBB62554235)200255.204(255.2045.110.350.110.35200122112PE3043-FALL/2004-Page17DensityControlAdditivesVwB=requiredwatervolumeperunitmassofbaritewBBWBwBBBBWBVmmVVVmmVVVVV112211212122111111VVVmVVVVVVwBBBBwBBwBwBwBBwBwBPE3043-FALL/2004-Page18Itisdesiredtoincreasethedensityof800bblof12-lbm/galmudto14-lbm/gal.Onegallonofwaterwillbeaddedwitheach100lbmsackofAPIbaritetopreventexcessivethickeningofthemud.Afinalmudvolumeof800bblisdesired.ComputethevolumeofoldmudthatshouldbediscardedandthemassofAPIbaritetobeadded.Example2.17PE3043-FALL/2004-Page19Solution:VwB=1/100=0.01gal/lbm800-700.53=99.47bblofoldmudshouldbediscarded.Volumeofwatertobeaddedwiththebariteis:0.01mB=0.01*108312=1083galor25.79bbllbmVVVmbblVVVVVVwBBBBwBBwBwBwBBwBwB10831242*)53.700800(01.0*0.3510.35153.7000.1201.0*35101.0*33.81350.1401.0*35101.0*33.81358001111121221PE3043-FALL/2004-Page20DensityControlAdditivesfc2=volumefractionoflow-gravity-solids(desirednewvolume)fc1=volumefraction