exploatare secundara

8/18/2019 exploatare secundara

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Proiectarea exploatari unei sonde prin injectie ciclica

de abur

n 1:=

Sa se proiecteze o operatie de injectie ciclica de abur la o sonda la

care se cunosc urmatoarele date:

C 273.15K :=in 0.0254:=

n 1:=kcal 414:=

! 400 20 n⋅+" #:= ! 420= $adincimea pina la capatul per%oraturilor

& 7 0.4 n⋅+" #:= & 7.4= $'rosimea inter(alului per%orat

de 0.0)03:=d*+ 2

3

+ i:= $diametrul te(ilor de e,tractie

di 0.0503:=

k r 2.25 414⋅:= - $conducti(itatea termica a rocii

k r .414 103×= -

/ 0.003:= $coe%icientul de di%uzibilitate termica

/ 3. 103−×=

13.25 414⋅:= - $coe%icientul 'lobal de transmitere a caldurii de la

interiorul te(ilor de e,tractie la e,teriorul coloanei

5.543 104×=

/e 0.137:=/C+ 5

1

2+

i:= $diametrul coloanei de e,ploatare/i 0.1273:=

a 0.032:= b 10:= bar 105:=

ts a !⋅ b+:= $'radient 'eotermic

ts 23.44=

m 23.5:= $porozitate

Sa 23 n+" #:= Sa 0.24= $saturatia in apa



ρa 1000:= $densitatea apei

ρr 24)5:= $densitatea rocii

ρt 25 ) n⋅−" #:= ρt 1= $densitatea titeiului

cr 0.2355 414×:= $caldura speci%ica a rocii

$caldura speci%ica a apeica 1 414⋅:= -

ct 0.71 414×:= ct 2.70)4 103×= $caldura speci%ica a titeiului

,0 0.:= $titlul aburului

pinj )0:= pentru aceasta presiune aburul are urmatoarele propietati:

*s 274.2:=

iabur 2.74033) 10)⋅:=

ia 1.20)2472 10)⋅:=

5500 110 n⋅+" #:= $debit masic de abur 5)10=



Metodologia de proiectare pentru un proces de injectie

ciclica de abur

1.Se calculeaza entalpia aburului la supra%ata

i ,0 iabur ⋅ 1 ,0−( ) ia⋅+:=i 2.4)47) 10

)×=

i 2.47012 10)⋅:=

2.Se calculeaza cantitatea totala de caldura injectata in sonda la supra%ata

t 24:=

os i⋅ ⋅:= os 3.3257 1011×=

$calculul se (a %ace pe zile

k 0 20..:=tk 24 k ⋅:= t0 1:= tk

1

24

48

72

96

120

144

168

192

216

240

264

288

312

336

360

384

408

432

456

480

=



3.Se determina (aloarea %unctiei %"t#

r 1

di

2:= r 2

/e

2:=

τ/ t⋅

r 22

:=r 1 ⋅k r

0.1410)=

lo'% τk

0.35−0.20

0.25

0.30

0.34

0.37

0.3

0.41

0.44

0.4)

0.4

0.50

0.52

0.54

0.5)

0.57

0.5

0.5

0.)0

0.)1

0.)2

:=% τ

k 10

lo'% τk :=

% τ

k

0.446684

1.584893

1.778279

1.995262

2.187762

2.344229

2.454709

2.570396

2.754229

2.884032

3.019952

3.162278

3.311311

3.467369

3.630781

3.715352

3.801894

3.890451

3.981072

4.073803

4.168694

=

τ

0

0

1

2

3

4

5

6

7

8

9

10

11

12

13

1415

16

17

18

19

20

0.79934

19.184171

38.368341

57.552512

76.736682

95.920853

115.105024

134.289194

153.473365

172.657535

191.841706

211.025877

230.210047

249.394218

268.578388287.762559

306.94673

326.1309

345.315071

364.499241

383.683412

= lo' τ( )

0

0

1

2

3

4

5

6

7

8

9

10

11

12

13

1415

16

17

18

19

20

-0.097268

1.282943

1.583973

1.760064

1.885003

1.981913

2.061094

2.128041

2.186033

2.237186

2.282943

2.324336

2.362124

2.396886

2.4290712.459034

2.487063

2.513392

2.538216

2.561697

2.583973

=



4.Se calculeaza cantitatea de caldura cedata stratelor tra(ersate de sonda6la

un anumit moment de timp considerat de la inceputul procesului de injectie

ck

2 π⋅ r 1⋅ ⋅ k r ⋅

k r r 1 ⋅+ % τk

+*s b−( ) !⋅

a !2⋅

2−

⋅:=

7ck

88.254109·10

88.25324·10

88.253093·10

88.252927·10

88.25278·10

88.25266·10

88.252576·10

88.252488·10

88.252347·10

88.252248·10

88.252145·10

88.252036·10

88.251922·1088.251803·10

88.251678·10

88.251614·10

88.251548·10

88.25148·10

88.251411·10

88.25134·10

88.251268·10

=



Se calculeaza cantitatea de caldura cedata stratelor tra(ersate

c1

c0

c1

+

223⋅:=

c1

1.345 1010×=

k 2 20..:=

ck

ck 1− ck +2

24⋅:=

k 1 20..:=

3ck

101.898345·10

101.98076·10

101.980722·10

101.980685·10

101.980653·10

101.980628·10

101.980608·10

101.98058·10

101.980552·10

101.980527·10

101.980502·10

101.980475·10

101.980447·10

101.980418·10

101.980395·10

101.980379·10

101.980363·10

101.980347·10

101.98033·10

101.980313·10

=



5.Se determina cantitatea de caldura ima'azinata in abur cind

acesta ajun'e la intrarea in strat

isk

os ck

−:=3

isk 113.135946·10

113.127704·10

113.127708·10

113.127712·10

113.127715·10

113.127718·10

113.12772·10

113.127722·10

113.127725·10

113.127728·10

113.12773·10

113.127733·10

113.127736·10

113.127739·10

113.127741·10

113.127742·10

113.127744·10

113.127746·10

113.127747·10

113.127749·10

=



).Se determina titlul aburului la intrarea in strat

Con%orm punctului doi entalpia aburului la intrarea in strat este:

k 1 20..:=

,isk

1

k

i

isk ∑

= tk ⋅ iabur ia−( )⋅

ia

iabur ia−−:=

,isk

0.711685

0.707805

0.707807

0.707809

0.70781

0.707811

0.707812

0.707814

0.707815

0.707816

0.707817

0.707819

0.707820.707821

0.707822

0.707823

0.707824

0.707825

0.707826

0.707826

=

0 100 200 300 400 5000.70)

0.70

0.71

0.712

,isk

tk



7.Se calculeaza cantitatea de caldura ima'azinata de un metru cub de

roca impreuna cu %luidele continute in pori

St 1 Sa−:=St 0.7)=

8 1 m−( ) ρr cr ⋅ m ρa⋅ ca⋅ Sa⋅+ m ρ t⋅ ct⋅ S⋅+:=

8 2.51)24 10)×=

.Se determina (aloarea parametrului adimensional 9

92 k r ⋅ t⋅

8 &⋅ /⋅:=

9

0.015781

0.077313

0.109337

0.13391

0.154626

0.172877

0.189378

0.204551

0.218674

0.231939

0.244485

0.256418

0.26782

0.278756

0.289279

0.299432

0.309252

0.31877

0.328011

0.337

0.345754

=



.Se determina e%icienta termica a procesului

k 0 20..:=

+tk

1

9k ( )2

e9k ( ) 2

er%c 9k ( )⋅2 9k ⋅

π+ 1−

:=

+t

0.988252

0.944696

0.923357

0.907559

0.894612

0.883478

0.873624

0.864735

0.856607

0.849097

0.842103

0.835546

0.829368

0.823519

0.817961

0.812662

0.807596

0.802741

0.798077

0.793588

0.789259

=



10.Se determina caldura cedata stratelor adiacente

cak

1 +tk

−( ) isk

⋅:=

3cak

0

101.734292·10

102.397155·10

102.891276·10

103.296218·10

103.644481·10

103.952694·10

104.230712·10

104.484944·10

104.719829·10

104.938592·10

105.143662·10

105.336923·10

105.519871·10

10

5.693713·10105.859444·10

106.017891·10

106.169757·10

106.315638·10

106.456049·10

106.591436·10

=



11.Se calculeaza raza sondei incalzite ca raza interioara a zonei in(adata

de apa calda

isk

1

k

i

isk ∑

=tk

:= isk

iabur ,isk

⋅ 1 ,isk

−( ) ia⋅+⋅:=

3is

0

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

0

101.306644·10

101.30321·10

101.303212·10

101.303213·10

101.303215·10

101.303216·10

101.303216·10

101.303218·10

101.303219·10

101.30322·10

101.303221·10

101.303222·10

101.303223·10

101.303224·10

101.303225·10

101.303226·10

101.303227·10

101.303227·10

101.303228·10

101.303229·10

= :is

0

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

96.767047·10

101.306644·10

101.30321·10

101.303212·10

101.303213·10

101.303215·10

101.303216·10

101.303216·10

101.303218·10

101.303219·10

101.30322·10

101.303221·10

101.303222·10

101.303223·10

101.303224·10

101.303225·10

101.303226·10

101.303227·10

101.303227·10

101.303228·10

101.303229·10

=



Supra%ata zonei incalzite se determina cu relatia

;k

isk

8⋅ /⋅ &⋅

4 k r 2⋅ *s ts−( )⋅

e9k ( ) 2

er%c 9k ( )⋅2 9k ⋅

π+ 1−

:=

;

0

61.819

120.52765

177.698453

233.551608

288.306232

342.108886

395.066336

447.260782

498.757699

549.610803

599.865335

649.559857

698.727758

747.398291

795.597136

843.34734

890.669833

937.583392

984.105098

31.030251·10

=



Se determina raza zonei incalzite

r k ;k

π:=

r

0

4.435944

6.19396

7.520849

8.622168

9.579704

10.435355

11.213988

11.931786

12.59998

13.226736

13.818215

14.379198

14.913482

15.424146

15.913718

16.384316

16.837726

17.275476

17.698881

18.109085

=



12.Se determina raza zonei ocupate de abur la %iecare moment de timp

r iabur ia−:= $caldura latenta de condensare

isk

,isk

⋅ r⋅:=

3isk

0

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

0

96.299394·10

96.265055·10

96.26507·10

96.265086·10

96.265099·10

96.265109·10

96.265118·10

96.26513·10

96.265141·10

96.265152·10

96.265162·10

96.265173·10

96.265185·10

96.265197·10

96.265207·10

96.265213·10

96.26522·10

96.265227·10

96.265234·10

96.265241·10

=



;k

isk

8⋅ /⋅ &⋅

4 k r 2⋅ *s ts−( )⋅

e9k ( ) 2

er%c 9k ( )⋅2 9k ⋅

π+ 1−

:=

rk ;k

π:=

;k

0

29.803238

57.942483

85.426895

112.277931

138.600899

164.466231

189.925256

215.017558

239.774586

264.222114

288.381935

312.272588

335.91013

359.308626

382.480321

405.43627

428.18663

450.740426

473.105873

495.290479

= rk

0

3.080043

4.294609

5.214617

5.978225

6.642141

7.235415

7.775287

8.272981

8.736282

9.170851

9.580961

9.969927

10.340383

10.694461

11.033914

11.36021

11.674589

11.978111

12.271686

12.556108

=

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