Reservoir Formation Damage Episode 3 Part 10 ppsx

Reservoir Formation Damage Episode 3 Part 11 ppt

Reservoir Formation Damage Episode 3 Part 11 ppt

... assisted analysis, 2 13 Experimental set-up, formation damage testing, 459, 564 Expert system, formation damage, 702 Evaluation, formation, 1 03 Evaluation, formation damage, laboratory, 456, ... 27 thickness-averaged, 30 4 transient analysis, near- wellbore permeability alteration, 694 Problems, formation damage, 4 Program, laboratory formation damage test...

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Reservoir Formation Damage Episode 1 Part 4 ppsx

Reservoir Formation Damage Episode 1 Part 4 ppsx

... wells nos. 1- 9, 11 , and 12 (after Hohn et al., 19 94; reprinted by permission of the U.S. Depart- ment of Energy). 74 Reservoir Formation Damage P c CcosG <r,cose, (4 -10 ) for ... Dekker Inc., New York, Ch. 9, 19 91, pp. 319 -373. 76 Reservoir Formation Damage 1. 0 (a) 10 0% WW 66% WW 50% WW 25% WW — 0%WW 1. 0 (b) .* 0 .4 0.0 Figure...

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Reservoir Formation Damage Episode 1 Part 10 potx

Reservoir Formation Damage Episode 1 Part 10 potx

... Civan (19 95): (10 -12 9) (10 -13 0) subject to =0, 0<jc<L, f = (10 -13 1) (10 -13 2) k is a particle exchange rate coefficient. A solution of Eqs. 10 -12 9 through 13 2 along with the particle ... are presented in Figure 10 -25. Next, they have solved their model equations, Eqs. 10 -12 1, 11 1, 11 2, 11 4, 11 5, 11 6, numerically by assuming t...

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Reservoir Formation Damage Episode 2 Part 1 ppsx

Reservoir Formation Damage Episode 2 Part 1 ppsx

... fraction is given by: (11 - 12 ) (11 -13 ) where (() is porosity and S } is the saturation of phase J. Thus, substituting Eqs. 11 - 12 and 11 -13 into Eq. 11 -11 yields the following ... core samples 24 6 Reservoir Formation Damage and MS w 1 + (M -1) S W (11 -26 ) Consequently, Eq. 11 -19 can be simplified significantly by substituting Eqs....

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Reservoir Formation Damage Episode 2 Part 2 ppsx

Reservoir Formation Damage Episode 2 Part 2 ppsx

... Modeling 27 1 K c L f )aK f ( 12- 22) aK ( 12- 23) Alternatively, eliminating (p c -p e ] between Eqs. 12- 18 and 12- 19 and then solving for 8 yields: aK t 5 = i- ( 12- 24) aK, Notice that Eq. 12- 24 ... Eqs. 12- 2, 12- 9, 12- 46, and 12- 48 into Eq. 12- 45 results in (1999a) dt r w -8 ' where ( 12- 49) ^yp^ ( 12 ~ 50) and B is given by Eq. 12- 13. The...

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Reservoir Formation Damage Episode 2 Part 6 ppsx

Reservoir Formation Damage Episode 2 Part 6 ppsx

... 1177-1185. 3 72 Reservoir Formation Damage f + Fe +2 + 16H + + Ue~ t=> FeS 2 + 8// 2 0 Fe^O 23 + 38/T + 30e~ & 2FeS- + 19HQ + 6H + + 2e~ d 2Fe +2 + 3// 2 0 from ... \*"J—\jj) Fe 2 O 3 +6H + +6H + + 2e~ <=>2Fe +2 + 3H 2 O Fe 2 O 3 +2HCO 3 +4H + +2e FeCO 3 +H + <=> Fe +2 + HCO 3 + 5H + + 2e~ ^ 3FeCO 3 + 4...

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Reservoir Formation Damage Episode 2 Part 10 pot

Reservoir Formation Damage Episode 2 Part 10 pot

... ambient laboratory conditions. 470 Reservoir Formation Damage Guidelines and Program for Laboratory Formation Damage Testing Recommended Practice for Laboratory Formation Damage Tests* Introduction The ... the wellbore. 458 Reservoir Formation Damage effect for clays and fines in the reservoir, an effect that would be removed by extraction." Fu...

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Reservoir Formation Damage Episode 3 Part 1 pps

Reservoir Formation Damage Episode 3 Part 1 pps

... % BC Q Kf Pf Cc D Farrar 1 McCreery, Dale Consolidated Field — AP1 12 055 234 56 3, 19 0.5 3, 19 1.5 3, 19 2.7 3, 19 4.0 3, 19 5.7 3, 19 7.9 3, 19 9.4 3, 2 01. 0 3, 2 03. 3 3, 205.8 3, 207 .1 3, 208.4 3, 209.9 3, 212 .7 3, 216 .7 3, 219 .2 49.0 ... 20.4 3. 2 1. 4 0.7 81. 5 23. 4 2.8 1. 5 0.7 10 6.0 25.6 2.5 1. 1 0.6 11 6.0 23. 6 3. 4 2.5 0.9 55....

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Reservoir Formation Damage Episode 3 Part 2 pptx

Reservoir Formation Damage Episode 3 Part 2 pptx

... McCreery 120 55 23 4 56 Zeigler 2 Mack 120 55 23 7 50 Boyd 9 Sanders 120 81 026 28 Depth ft 2 ,39 3.5 2 ,38 8 .3 2 ,39 1.1 2 ,39 0 2 ,38 8 2 ,39 3 2 ,39 2. 8 3, 198.7 3, 20 0.6 2, 611 2, 627 2, 1 63 SEM/EDX Yes Yes Yes No No Yes Yes Yes No Yes Yes No XRD No Yes No Yes No No Yes No No Yes Yes No Thin section No Yes No Yes No No Yes Yes No Yes Yes No After ... Bud...

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Reservoir Formation Damage Episode 3 Part 3 ppt

Reservoir Formation Damage Episode 3 Part 3 ppt

... 542 Reservoir Formation Damage 2Ax (16 -36 ) The central difference formula is obtained as, by substituting Eqs. 16 -33 and 34 for b { ,b 2 at x = x f =0 into Eq. 16 -35 : dx 2Ax (16 -37 ) The ... substituting Eqs. 16 -33 and 34 for b { ,b 2 at x = x i _ l =-A;c into Eq. 16 -35 : 546 Reservoir Formation Damage The outlet boundary condition given by...

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Reservoir Formation Damage Episode 3 Part 4 pps

Reservoir Formation Damage Episode 3 Part 4 pps

... no. 01120 0 031 2 2 34 65 2 34 91 0 039 2 2 34 56 237 44 237 50 237 53 237 68 237 69 01070 015 43 2 34 99 00078 01862 01972 01950 01 935 0 049 6 0 048 8 0 049 0 00090 0 045 9 0 130 1 01 131 00795 0 131 9 01169 00291 0 133 0 00788 0 032 5 02187 00551 01266 01812 01165 Well ... no. EOR-B115 EOR-B92 EOR-B17 EOR-B70 EOR-B51 EOR-B 93 EOR-B52 EOR-B107 EOR-B101 EOR-B99 EOR-B-60 EOR-B9...

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Reservoir Formation Damage Episode 3 Part 5 pdf

Reservoir Formation Damage Episode 3 Part 5 pdf

... change 8 47 03 3 93 699 0 0 29 0 1 65 86 80 47 03 3 93 7 35 0 0 29 0 161 1 15 0.2 90 47 03 3 93 1 056 109 90 (muscovite) 87 0 0.2 37 0 .3 0 0 -36 1 - 154 -112 +3 -17 +1 65 +86 +80 -31 0 +16.9 25. 7 0 0 -32 5 - 154 -112 +3 -17 +161 +1 15 +0.2 +90 - 239 + 13 24.9 0 0 -4 - 45 d ... Plagioclase Calcite (%) -31 90 .5 -31 91 .5 -31 92.7 -31 94.0 -31 95. 7 -31...

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Reservoir Formation Damage Episode 3 Part 6 pptx

Reservoir Formation Damage Episode 3 Part 6 pptx

... by: ,9£ dr (18 -3) 62 4 Reservoir Formation Damage U 1.0- 0.8- 0 .6- 0.4- 0.2- 0.0- _,_,., C r Sr Sc So M * r w ã 0.25 - 0.258 r " 0 .3 (max) 3 ã 0.2 - 100 mm |iii (a) 1.0- 0.8- 0 .6- 0.4- 0.0- 200 jL 30 0 Radius, ... are required: (19-10) 62 6 Reservoir Formation Damage International Symposium on Oilfield Chemistry, February 28-March 3, 19 93,...

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Reservoir Formation Damage Episode 3 Part 7 potx

Reservoir Formation Damage Episode 3 Part 7 potx

... /,(*./*,- 1 ) (19 -36 ) (19 - 37 ) Thus, they calculate the harmonic average permeability of the damaged portion of the core by: K(t\- Xf { dx K(t) -T 0 n^t (19 -38 ) Substituting Eqs. 19 -33 , 34 , and 36 ... Study," SPE Paper 38 180, Proceedings of the 19 97 SPE European Formation Damage Conference held in the Hague, The Netherlands, June 2 -3, 19 97,...

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Reservoir Formation Damage Episode 3 Part 10 ppsx

Reservoir Formation Damage Episode 3 Part 10 ppsx

... of, 6 93 Damage by formation fines migration, 251 Damage by mud filtration, 255 Damage by particle invasion, 2 53 Damage ration, 687 Deposition calcite, model, 674 Formation Damage ... Deposition," SPE 238 10 paper, SPE International Symposium on Formation Damage Control, Lafayette, Louisiana, February 26-27, 1992, pp. 38 3 -39 5. Table 23- 2 (...

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