Fdp Assignment 1

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Reservoir thickness 16.4 ft Connate water saturation 0.2 Oil saturation 0.8 Water compressibility 3.00E- 06 psi -1 Oil compressibility 5.40E- 05 psi -1 Rock compressibility 4.00E- 06 psi -1 Average Porosity 19.20% Oil viscosity 0.22 cP Wellbore radius 0.354 Ft Oil formation volume factor 1.6 bbl/ STB Oil Production Rate in STB/DAY 6774 STB/D Table 1. Reservoir 1 Properties Data q o = ( StudentID 2600 ) = ( 17612793 2600 ) Time Oil Rate Pressure Time Oil Rate Pressure (hr) (STB/D) (psi) (hr) (STB/D) (psi) 0.00100 6774 3372.8 12.00783 0 3429.9 0.00138 6774 3367.5 12.00962 0 3433.2 0.00191 6774 3362.2 12.01182 0 3436.6 0.00264 6774 3356.9 12.01452 0 3440 0.00365 6774 3351.7 12.01784 0 3443.3 0.00505 6774 3346.4 12.02192 0 3446.7 0.00698 6774 3341.1 12.02693 0 3450.1 0.00965 6774 3335.8 12.03309 0 3453.4 0.01334 6774 3330.5 12.04065 0 3456.8 0.01845 6774 3325.2 12.04994 0 3460.1 0.02550 6774 3319.8 12.06135 0 3463.5 0.03500 6774 3314.5 12.07538 0 3466.8 0.04900 6774 3309.2 12.09260 0 3470.2 0.06700 6774 3303.9 12.11377 0 3473.5 0.09300 6774 3298.6 12.13977 0 3476.9 0.12900 6774 3293.3 12.17171 0 3480.2 0.17800 6774 3288 12.21100 0 3483.5 0.24600 6774 3282.7 12.25900 0 3486.8 0.34000 6774 3277.4 12.31800 0 3490.1 0.47100 6774 3272.1 12.39100 0 3493.4 0.65000 6774 3266.8 12.48100 0 3496.6

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FDP Assignment

Transcript of Fdp Assignment 1

Reservoir thickness 16.4 ftConnate water saturation 0.2Oil saturation 0.8Water compressibilit !.00"#06 psi#1Oil compressibilit $.40"#0$ psi#1Rock compressibilit 4.00"#06 psi#1%vera&e 'orosit 1(.20)Oil viscosit 0.22 c'Wellbore ra*ius 0.!$4 +tOil formation volume factor 1.6 bbl,-./Oil 'ro*uction Rate in -./,0%1 6774 -./,0Table 1. Reservoir 1 Properties Dataqo=( Student ID2600)=(176127932600).ime Oil Rate 'ressure .ime Oil Rate 'ressure2hr3 2-./,03 2psi3 2hr3 2-./,03 2psi30.00100 6444 !!42.8 12.0048! 0 !42(.(0.001!8 6444 !!64.$12.00(620 !4!!.20.001(1 6444 !!62.212.011820 !4!6.60.00264 6444 !!$6.(12.014$20 !4400.00!6$ 6444 !!$1.412.014840 !44!.!0.00$0$ 6444 !!46.412.021(20 !446.40.006(8 6444 !!41.112.026(!0 !4$0.10.00(6$ 6444 !!!$.812.0!!0(0 !4$!.40.01!!4 6444 !!!0.$12.0406$0 !4$6.80.0184$ 6444 !!2$.212.04((40 !460.10.02$$0 6444 !!1(.812.061!$0 !46!.$0.0!$00 6444 !!14.$12.04$!80 !466.80.04(00 6444 !!0(.212.0(2600 !440.20.06400 6444 !!0!.(12.11!440 !44!.$0.0(!00 6444 !2(8.612.1!(440 !446.(0.12(00 6444 !2(!.!12.141410 !480.20.14800 6444 !28812.211000 !48!.$0.24600 6444 !282.412.2$(000 !486.80.!4000 6444 !244.412.!18000 !4(0.10.44100 6444 !242.112.!(1000 !4(!.40.6$000 6444 !266.812.481000 !4(6.60.8((00 6444 !261.$12.$(0000 !4((.(1.24!00 6444 !2$6.212.42$000 !$0!.11.41(00 6444 !2$0.(12.8(1000 !$06.22.!4600 6444 !24$.61!.0($000 !$0(.!!.28$00 6444 !240.!1!.!4$000 !$12.44.$4100 6444 !2!$1!.6$2000 !$1$.46.24(00 6444 !22(.414.0!0000 !$18.!8.68000 6444 !224.!14.4(4000 !$21.212.00000 6444 !21(1$.064000 !$2!.(12.001000 !!(6.21$.464000 !$26.$12.0012!0 !!((.616.62$000 !$2(12.001$10 !40!14.682000 !$!1.412.0018$0 !406.!18.(80000 !$!!.612.002280 !40(.420.$4$000 !$!$.412.002800 !41!22.$!$000 !$!4.$12.00!440 !416.424.(4!000 !$!(.!12.004220 !41(.824.(01000 !$40.812.00$1(0 !42!.1!1.$!$000 !$42.212.006!80 !426.$!6.000000 !$4!.4Table 2. Reservoir 1 Pressure Drawdown and Build-Up DataQuestion 12a3 %nalsethewell test*ata&ivenbelowusin&sprea*sheetan*"crin-oftwaretocalculate followin& parameters. 540 points67 8nitial reservoir pressure Reservoir 'ermeabilit -kin factor 'ressure *rop *ue to skin e9ect 8n or*er to calculate the above parameter: pressure buil*#up *ata was manipulate* intoa ;ornersin& the followin& formula: hei&ht an* reservoir hei&ht are calculate* usin& capillarpressure *ata in Table ,7h=pc( wo)hres="#h"Fample for Table / calculation: 'olu*n 17h=20( 0.4340.269)=121.40 fthres=8314.80121.40=8193.4 ft-w'c#ow 2psi3 h 2ft3 h#res 2ft30.2 20 121.40 81(!.400.2 1(.28 114.0! 81(4.440.2 16.!( ((.4( 821$.!10.2 1!.$1 82.01 82!2.4(0.2 12.12 4!.$4 8241.2!0.2 (.$! $4.8$ 82$6.($0.2 4.4$ 44.04 8264.460.2 $.82 !$.!! 824(.440.2 4.(4 !0.14 8284.6!0.2 !.22 1(.$$ 82($.2$0.2 2.6 1$.48 82((.020.22 1.8 10.(! 8!0!.840.24 1.4 8.$0 8!06.!00.! 0.8 4.86 8!0(.(40.!$ 0.6 !.64 8!11.160.4 0.$ !.04 8!11.460.$ 0.!$ 2.12 8!12.680.6 0.2$ 1.$2 8!1!.280.4 0.2 1.21 8!1!.$(0.8 0.1$ 0.(1 8!1!.8(0.( 0.1 0.61 8!14.1(1 0 0.00 8!14.80Table /. +apillar- Pressure %or Di2erent Reservoir 3nterval.he water saturation for each reservoir interval was then interpolate* as tabulate* below.-wh#res 2ft3 -wh#res 2ft3 -wh#res 2ft30.200 820!.28 0.216 8!02.(( 0.!80 8!11.$20.2008216.400.21(8!0!.6$0.40(8!11.8$0.20082!2.800.2248!04.!00.4468!12.180.200824(.200.22(8!04.(60.4818!12.$00.20082$(.040.2!48!0$.620.$2$8!12.8!0.2008260.$$0.2!48!0$.(40.$808!1!.160.2008260.680.2408!06.240.6688!1!.4(0.200826$.600.2$08!06.(!0.4448!1!.820.2008282.000.2618!04.$80.88!8!14.140.20082(8.400.2428!08.240.(468!14.440.20082((.060.28!8!08.(01.0008!14.800.20!82((.410.2(48!0(.$$0.2068!00.!40.2((8!0(.880.2088!01.020.!118!10.210.2118!01.680.!!88!10.860.2148!02.!40.!$!8!11.1(Table 1.. 3nterpolated +apillar- Pressure Data %or Reservoir 2/ulk volume of each contour interval is calculate* usin& .rapeEoi*al Rule7"Fample for Table 11 calculation: 12.3.21 4 1216.4. interval7$%=h(A1+A2)2$%= ( 8216.408203.28) ( 4.73E06+4.32E07)2=3.15E08 fth 2ft3%rea2ft23Gol. 2ft!3 h 2ft3%rea2ft23Gol. 2ft!3 h 2ft3%rea2ft23Gol. 2ft!3820!.284.4!"D06 0.00"D008!02.!44.42"D082.(0"D088!10.86$.!$"D08!.4$"D088216.404.!2"D04!.1$"D088!02.((4.48"D082.(0"D088!11.1($.!("D081.44"D0882!2.808.4!"D041.04"D0(8!0!.6$4.$$"D082.(8"D088!11.$2$.42"D081.48"D08824(.201.!$"D081.82"D0(8!04.!04.61"D082.(8"D088!11.8$$.46"D081.80"D0882$(.041.68"D081.4("D0(8!04.(64.64"D08!.06"D088!12.18$.4("D081.81"D088260.$$1.44"D082.$8"D088!0$.624.4$"D08!.11"D088!12.$0$.$!"D081.46"D088260.681.44"D082.26"D048!0$.(44.4("D081.$!"D088!12.8!$.$6"D081.8!"D08826$.601.(2"D08(.01"D088!06.244.8!"D081.$("D088!1!.16$.60"D081.84"D088282.002.80"D08!.84"D0(8!06.(!4.(0"D08!.21"D088!1!.4($.6!"D081.8$"D0882(8.404.06"D08$.6!"D0(8!04.$84.(8"D08!.21"D088!1!.82$.64"D081.86"D0882((.064.12"D082.40"D088!08.24$.06"D08!.!1"D088!14.14$.40"D081.82"D0882((.414.18"D082.40"D088!08.(0$.1!"D08!.!6"D088!14.44$.44"D081.8("D088!00.!44.24"D082.48"D088!0(.$$$.20"D08!.!6"D088!14.80$.44"D081.(0"D088!01.024.!0"D082.48"D088!0(.88$.24"D081.42"D088!01.684.!6"D082.86"D088!10.21$.28"D081.44"D08Table 11. Bul5 6olu*e %or &a') Reservoir 3nterval-ince COC is at 8260.$$ ft. Cumulative bulk volume can be estimate* from Table 11an* then usin& the followin& formula to calculate &as cap siEe7=$%(8203.288260.55) hneth'ross (1Sw)%ssumin&hnetHh&rossan*sincethewatersaturationinthisinterval hasaconstantvalue of 0.2: therefore7=4.96E0910.192(10.2)=761 (()#!-.O88' calculation is more compleF since the water saturation within oil bearin&formation is chan&in& an* therefore it is much better to calculate the -.O88' for eachinterval an* then summin& up all the -.O88'. .he volume which nee*s to be calculate*is ran&in& from8260.$$ to 8!14.80 ft which are the value for COC an* WOCrespectivel.STII*=$% hneth'ross (1Sw) 1Boi"Fample for Table 12 calculation: 'olu*n 1 interval7STII*=4.02E0610.192( 10.2) 11.6=386,329 STBh 2ft3 %rea 2ft23 Gol. 2bbl3 -w-.O88' 2-./3 Cumu. -. O88' 2-./38260.68 2.26"D04 4.02"D06 0.200 !86:!2( !86:!2(826$.60(.01"D08 1.60"D08 0.200 1$:!(8:04( 1$:484:4088282.00!.84"D0( 6.(0"D08 0.200 66:2!!:648 82:018:0$682(8.40$.6!"D0( 1.00"D0( 0.200 (6:20$:468 148:22!:$2482((.062.40"D08 4.81"D04 0.200 4:612:41( 182:8!6:24282((.412.40"D08 4.80"D04 0.20! 4:$(4:820 184:4!1:0628!00.!42.48"D08 4.($"D04 0.206 4:414:$(8 1(2:148:6608!01.022.48"D08 4.(4"D04 0.208 4:6(4:204 1(6:84$:86!8!01.682.86"D08 $.0("D04 0.211 4:820:62( 201:666:4(!8!02.!42.(0"D08 $.16"D04 0.214 4:841:(21 206:$!8:4148!02.((2.(0"D08 $.16"D04 0.216 4:848:((8 211:!84:4128!0!.6$2.(8"D08 $.!1"D04 0.21( 4:(44:$4$ 216:!61:($48!04.!02.(8"D08 $.!0"D04 0.224 4:(!4:(4( 221:2((:(!68!04.(6!.06"D08 $.4$"D04 0.22( $:046:010 226:!4$:(448!0$.62!.11"D08 $.$4"D04 0.2!4 $:08$:604 2!1:4!1:$$48!0$.(41.$!"D08 2.42"D04 0.2!4 2:484:$46 2!!:(1(:12(8!06.241.$("D08 2.82"D04 0.240 2:$46:(41 2!6:4(6:0408!06.(!!.21"D08 $.42"D04 0.2$0 $:14!:!82 241:6!(:4$!8!04.$8 !.21"D08 $.42"D04 0.261 $:042:61$ 246:412:0688!08.24 !.!1"D08 $.(0"D04 0.242 $:1$4:422 2$1:866:4(08!08.(0 !.!6"D08 $.(("D04 0.28! $:1$!:444 2$4:020:2!48!0(.$$ !.!6"D08 $.(8"D04 0.2(4 $:041:88( 262:0(2:12!8!0(.88 1.42"D08 !.04"D04 0.2(( 2:$82:!$4 264:644:4448!10.21 1.44"D08 !.0("D04 0.!11 2:$$$:(!4 264:2!0:4118!10.86 !.4$"D08 6.1$"D04 0.!!8 4:88(:4$0 242:11(:8618!11.1( 1.44"D08 !.16"D04 0.!$! 2:4$1:$6$ 244:$41:42$8!11.$2 1.48"D08 !.18"D04 0.!80 2:!64:642 246:(!6:0(48!11.8$ 1.80"D08 !.20"D04 0.40( 2:266:($4 24(:20!:0$48!12.18 1.81"D08 !.22"D04 0.446 2:141:!$( 281:!44:4148!12.$0 1.46"D08 !.14"D04 0.481 1:($4:106 28!:!01:$1(8!12.8! 1.8!"D08 !.26"D04 0.$2$ 1:8$$:(40 28$:1$4:48(8!1!.161.84"D08 !.28"D04 0.$80 1:6$!:$4( 286:811:0688!1!.4(1.8$"D08 !.!0"D04 0.668 1:!1!:1$$ 288:124:22!8!1!.821.86"D08 !.!2"D04 0.444 888:0!8 28(:012:2618!14.141.82"D08 !.24"D04 0.88! 4$6:$64 28(:468:82$8!14.441.8("D08 !.!6"D04 0.(46 21(:24$ 28(:688:0418!14.801.(0"D08 !.!8"D04 1.000 0 28(:688:041Table 12. Reservoir 2 +u*ulative 7il Produ'tion.he -.O88' is the last value in the cumulative -.O88' column shown inTable 12 whichis 2/. (TB.2ii3 >sin& the concept of /uckle an* =everette theor an* +ractional +low concept performthe followin& tasks7 540 points6 Construct oil pro*uction rate: water pro*uction rate an* water oil ratio plot withtimebeforean*after breakthrou&husin&atotal pro*uctionrateof67074.(TB8D. 510 points6 Compute cumulative oil an* water pro*uction before an* after breakthrou&h an*plot them as a function of time. 510 points6 Calculate the recover factors before an* after breakthrou&h. 5$ 'oints6/ase* on relative permeabilit *ata in Table 6: the value of fwis calculate* an* thenplot a&ainst *epth as shown below: where7f w=11+ wokrokrw-wkrwkrofw0.20 0.000 1.000 0.0000.!0 0.01$ 0.266 0.0!20.40 0.0$( 0.0$2 0.!(60.4$ 0.0(! 0.01( 0.4!(0.$0 0.1!4 0.006 0.(280.$$ 0.182 0.002 0.(810.60 0.2!8 !"#04 0.((80.6$ 0.!01 !"#0$ 1.0000.40 0.!42 ("#04 1.0000.44 0.4!! 4"#0( 1.0000.48 0.$00 0 1.000Table 13. %w values %or Reservoir 2IeFt: Wi* is calculate* usin& the followin& formulas7"id=1( + fw+Sw )Whereas: Ip* after is calculate* as follow:,pd=(Sw-Swc )+(1fw-)where:Sw-(.)=(Sw(.)+Sw(.1))/ 2f w-=(fw( .)+fw(.1))/ 2 Sw=Sw( .)Sw(.1) fw=fw(.)fw( .1).he result is tabulate* in Table 14 below.-wfw -wfwWi*-Jw=fJw=Ip*0.20 0.0000.!0 0.0!20.40 0.!(60.4$ 0.4!( 2bt30.$0 0.(28 0.0$ 0.18(0 0.264481 0.44$ 0.8!4 0.!1(0.$$ 0.(81 0.0$ 0.0$!2 0.(40$40 0.$2$ 0.($$ 0.!680.60 0.((8 0.0$ 0.016$ !.0!8600 0.$4$ 0.((0 0.4040.6$ 1.000 0.0$ 0.0020 24.(($24 0.62$ 0.((( 0.4$40.40 1.000 0.0$ 0.0002 2(4.4166 0.64$ 1.000 0.$010.44 1.000 0.04 0.0000 (6!6.!61 0.42 1.000 0.$400.48 1.000 0.04 0.0000 14!2481 0.46 1.000 0.$80Table 14. 9id and :pd values %or di2erent (w!i"ure 3. !ra'tional !low +urve+rom fractional Kow curve in !i"ure 3: we can see that the Sw H 0.$24: S%t H 0.$:an* Swc H 0.2..ime to breakthrou&h can be calculate* b usin& the followin& formula7t%t= (SwSwc) *$qwiBw365where: inLection rate eAuals to7qwi=qoBo=67,7401.6=108,384 STB/ D'G is calculate* from the summation of bulk volume in oil bearin& formulas in the thir*column of Table 12: multiplie* b porosit. 'G H 6!0 MMbbl..he value of /w is assume* to be 1: since water is hi&hl incompressible..herefore:t%t= ( 0.5240.2) 630E06108,3841365 =5.16 /e0rs.he calculate* Ip* values shown in Table 14 is onl vali* for after breakthrou&h watersaturation. /efore breakthrou&h:,pd="idan* at breakthrou&h,pd(%t )="id (%t)=SwSwc+rom Ip* values: Ip can be calculate* as7,p= ,pd*$BoOil pro*uction rate can be calculate* as7qo=,p/ tWater pro*uction rate can be calculate* as7qw=qo")where:")=BofwBw( 1f w)where: the f w values are interpolate* from Table 13.t 2ears3Wi*Ip*Ip 2-./3Ao 2-./,03fwWORAw 2-./,03Wp 2-./3At 2-./,03Iote0 0.000 0.000 0 0 0 0 00000 000010.0624$020.0624$0224:42$:10064:440 0 064:44020.12$$0040.12$$0044(:4$0:20064:440 0 064:440!0.1882$060.1882$0644:14$:!0064:440 0 064:44040.2$100080.2$10008(8:(00:40064:440 0 064:440$0.!1!4$100.!1!4$1012!:62$:$0064:440 0 064:440$.160.!2400000.!240000124:66!:8!(64:440 0 064:440bt60.!46$0120.!240284128:8$4:112 !:(040.(!402022!.80$(!:01428:40$:826(6:(2440.4!(2$1$0.!!1$!!21!0:6!2:088 4:86!0.(41($442$.(6$126:26444:4(2:2821!1:12480.$0200140.!!60!4(1!2:404:064 4:86!0.(46888428.$2$1!8:41812$:124:22014!:$81(0.$644$1(0.!40$4261!4:182:0!( 4:86!0.($182!0!1.6111$!:422181:2!2:4!!1$8:$8$100.624$0210.!4$04441!$:($4:01$ 4:86!0.($64$4!!$.400142:1$0244:064:648144:01!Table 1,. ;o0 ;w0 97R0 :p and 9p values %or Reservoir 2 .he recover factor can be calculate* as7)!=1)STII* 100 /efore breakthrou&h: the ultimate recover value is the value of Ipin the ear$.16 shown in Table 1,.)!=127,663,839290E06100=44.07 +or after breakthrou&h: our %O+ is limite* to 81:288 -./,0 which means that ourAtmust not eFcee* this value. .herefore: this value lies between ear $.16 an*ear 6 in Table 1,. -imple interpolation of Ip usin& At H 81:288 -./,0 results in:Ip H 128:214:48( -./. .herefore: R+ after breakthrou&h7)!=128,217,789290E06100=44.26 .he &raph of annual oil pro*uction rate: water pro*uction rate an* water#oil#ratiois plotte* in !i"ure 4. .he &raph of annual cumulative pro*uce* oil an* watervolume is shown in !i"ure ,.0 1 2 ! 4 $ 6 4 8 ( 10020:00040:00060:00080:000100:000120:000140:000160:000180:0002022242628!0!2!4!6!8Oil Rate Water Rate WOR1ear+low Rate 2-./,03 WOR !i"ure 4. ;o0 ;w and 97R vs Ti*e0 1 2 ! 4 $ 6 4 8 ( 100$0:000:000100:000:0001$0:000:000200:000:0002$0:000:000Oil Cumu. Gol Water Cumu. Gol.1earCumulative Golume 2-./3 !i"ure ,. +u*ulative Produ'ed 7il 6olu*e