3. Water Conning and Fingering

7/22/2019 3. Water Conning and Fingering

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WATER CONNING AND FINGERING-TEKRES

WATER CONNING AND FINGERING

Definisi water Conning

yaitu terjadinya pengerucutan water akibatpressure drop didekat well terlalu besar, yaitumelebihi differensial gravity pressure, akibatnya

produksi migas berkurang.P = Pavg Pwf > 0.433 (w - o) hcP: Pressure drawdoen at the well, psi; w: SG water,

hc: vertical distance, ft

Definisi Fingering tonguing:

terproduksinya water yang terlalu cepat akibat

mobilitas horizontal dari water yang terlalu besardibanding mobilitas oil, dan terjadi pada arahhorizontal.

Historis:

1. dijaga agar tidak terjadi water connig andfingering.

2. Conning akan terjadi jika laju produksimelebihi laju kritis terjadinya conning (vc).

3. Menentukan kapan terjadinya Breaktrogh(tembus air)

4. Kinerja water Conning5. Air dibiarkan terproduksi asal surface

facilities mendukung

Metode Penentuan Water Conning:

1. Meyer and Garder2. Chaney et al.3. Sobinski and Cornelius4. Bournazel and Jeansen5. Kuo and Desbrisay.

Meyer and Garder Methode:

Assumsi untuk menentukan critical oil rate1. The flow of oil and or gas to the wellbore is

strictly radial2. The flow of water from the water/ oil contact

to the bottom of the wellbore is strictlyvertical

3. The pressure drawdown controlling the flowof oil or gas is restricted to the pressuredifference

( ) (( )rw/relnBoo

Dhkow001535.0qc

22

=

qc : critical oil rate (max oil rate that precludesconning water, STB/D); k: formation

permeability (md); h : oil zone thickness, ft; D :

completion interval thickness , f; re : externaldrainage radius, ft; rw : wellbore radius, ft

o ; formation oil density, gr/cc

Chaney et al

tidak hanya water conning tapi juga gasconning

1. qCURVE = figure 13-1 - 13-5

dg input : perforated intervaldistance from top perforation

to top of GOCoutput : qCURVENBqCURVE dihitung pada : k=1000

mD

o = 1 cp

= 0.3

if completion interval is at the top of theoil zone maka distance from topperforation to top of GOC = 0

2. qCURVE alternativesyarat : completion interval is at the topof the oil zone

( ) 3.0h

D250

44

h5734Dh1313.0q

2

22CURVE

+

+=

3.

( )

Bo*

k*q*00333.0qc

o

owCURVE

=

Sobinski and Cornelius:

Determine the producing time until waterbreaktrough occurs due to conning

1. tentukan

for M < 1 ; = 0.5

for M 1 ; = 0.6

M : mobility ratio2. tentukan permeability ratio (Fk)

Fk = kh/kvkh : horizontal permeabilitykv : vertical permeability

3. tentukan cone height (hc)hc = pay thickness perforated interval(at top of pay)

4. tentukan dimensionless cone height

( Z )( )

oo

chow

qBo

hhk00307.0Z

=

5. from figure 13-7 hal: 13-15 tracy; td =0.5

6. tentukan time ( t )

( ) ( )+

=M1k00137.0

Fhtt

how

kod

days7. Kuo and DesBrisay develop an aquation

S&C (no.4)

( )( )Z274

Z3Z716Zt

2

dBT +=

untuk Z = 3.5 pers akan infinite

1 dari 4


2/5


shg diperoleh qo = qc

( )

oo

chow

qBo

hhk00307.0Z

=

menjadi

( )Bo

hhk000877.0

qco

chow

=

(STB/day)

Bournazel and Jeansen

1. Dimensionless Cone height ( Z ) from S & C)

( )

oo

chow

qBo

hhk00307.0Z

=

2. Dimensionless time to Breaktrough

Z7.00.3

ZtdBT

=

3. untuk : 0.14 M 7.3

= 0.7

4. actual breaktrough time

( ) ( )+

=M1k00137.0

Fhtt

how

koBTd

days5. pd pers no.2 jika : Z = 4.28 maka akan

infiniteshg diperoleh qo = qc

( )

oo

chow

qBo

hhk00307.0

Z

=

menjadi

( )Bo

hhk000717.0qc

o

chow

=

(STB/day)

Kuo and Desbrisay

1.

( ) ( )Bo2049

DhkwA

o

22

o

=

2. ( )rw/reln1416.3

432.0B +=

3. ( ) 14.0re/hC =

day/STBAxBxCqc =

Contoh Kasus

Ho : 42 Fk : 10 & 1 O : 1.44

cp

Hw : 60 D : 21 W : 1.095

g/cc

hc : 21 M : 3.27 O : 0.861

g/cc

Sor : 0.337 re : 1053 Bo : 1.102Swc : 0.288 rw : 0.29 qT : 100

STB/day

: 0.25 kh : 90 mD : 0.05

Ramalkan kinerja bottom water drive (kerucutair)

Solusi :1. IOIP

( ) ( )( )STB

Bo615.5

Swc1.hrwreN

22 =

N = 4.21 x 106 STB2. Laju Produksi kritis

( )( )day/STB

Bo

h.Hk000717.0qc

o

coowh

=

day/STB4.8qc =

karena qT > qO maka terjadi Conningsebelum tembus air : qO = qT3. Dimensionless cone Height

( )qoBo

h.Hk00307.0Z

o

cohow

=

Z = 0.3594. Bournazel & Jeanson breaktrough time

untuk : 0.14 M 7.3 maka = 0.7

* Dimensionless breaktrough time

Z7.00.3

ZtdBT

=

131.0tdBT =* Breaktrough time

( ) ( )+

=M1k00137.0

tFkHt

how

dBTooBT

(day)

day15.208tBT =5. Kuo and desbrisay Breaktrough time

BTBT t5.0*t = (day)day07.104*tBT =

6. Produksi cumulative sampai breaktrough

)STB(*tqN BTTpBT =

STB10407NpBT =ITERASI I

Assumsi : NP = 5000 STB dan t = 40

day

7.N

NNn

ppBT

1j

+=+

00367.0n 1j =+8. penentuan hw dan ho

a.

+= +SorSwc1

Swc1nHoHwhw 1j

hw = 60.29

2 dari 4


3/5


b.

= +SorSwc1

Swc1n1Hoho 1j

ho = 41.71

9.hohwM

hwMWC lim

+=

825.0WClim =

10

*BT

t*BT

d

t

tt

+=

td = 1.38411 Dimensionless WC

for : td < 0.5 WCd = 0.0for : 0.5 td 5.7

29.0)log(94.0 += dd tWCfor : td > 5.7 WCd = 1.0

WCd = 0.42312 dlim WCxWCfwWC ==` WC = fw = 0.349

13 T1j1j q.fw1qo ++ =day/STB1.65qo 1j =+

14. ( )day/STB2

qoqqo

1jT ++=

day/STB55.82qo =

15.qo

Nt

p

hit

=

day57.60thit =16.

t

tt hit

=

diygerror514.0 = minta

ITERASI II

Assumsi : t = t hit = 60.57 day dan Np =

5000 STB

17

*BT

t*BT

d

t

tt

+=

582.1td =18 29.0)tlog(94.0WC dd +=

477.0WCd =

19 dlim WCxWCfwWC ==

(WClim selalu sama)

394.0fwWC ==20 T1j1j q.fw1qo ++ =

day/STB61.60qo 1j =+

21 ( )day/STB2

qoqqo

1jT ++=

day/STB304.80qo =

22qo

Nt

p

hit

=

day26.62thit =

23t

tt hit

=

ygerror028.0 1

OWC stabil jika < 0; G>M-1; M>1

tidak stabil jika : =0; GM-1

o

*kro

w

*krwM

=

3 dari 4


4/5


wT

4

q

sin.A.*krw.k10.9.4G

=

( )1Mq

qG

T

critic =

)1M(10.0133.1

singA*krwkq

w6critic

=

Contoh Kasus:Pendesakan Torak (pendesakan linear)

Data :krw* = 0.3 h =

40 ft w = 1.04

kro* = 0.8 W = 625 ft o =

0.81

Swc = 0.2 QT = 1000 B/D =

250

Sor = 0.2 L = 2000 ft W =

0.5 cp

= 0.18 k = 2 D = 2000 mD O = 5

cp

Tentukan : NpD & WiDSolusi :1. Mobili ty Ratio (M)

o

w

/*kro

*krwM

=

75.35/8.0

5.013.0M ==

2. Laju alir air kritis (qC)

( )1M.

si..A.krw.k10x9.4Q

w

4

critical

=

(( 75.3x5.0

.1xAx13.0x2000x10x9.4Q

4

critical =

= 520 rb water /day3. Pendesakan stabil :

0 shg : G > M 1

Assumsi : pendesakan stabil untuk : Q =90% QCritical

a. criticalTstabil Q.9.0Q =

D/rb468520.9.0QTstabil ==

b.Tstabil

critical

Q

QG = ( M 1 )

( ) 056.3175.3468

520G ==

c.

== tanG

G1Mtan

dx

dy

025tan056.3

056.3175.3tan

dx

dy ==

0467.0tandx

dy==

= acr tan (0.0467) = 2.6730

4. Moveable Oil Vollume (MOV)

( )SwcSor1WhLMOV =MOV = 625 * 40 * 2000 * 0.18 * ( 1 0.2

0.2 )= 5.4 x 106 cuft

MOV = 9.617 x 105 bbl

Kondisi saat Breakthroughyc = 0

5. NpD saat breaktrough

( )

=

tanhL2

ych1N

2

pD

saat Breakthrough :

( ))0467.0)(40)(2000(2

020001N

2

pD

=

= 0.786 MOV= 0.786 ( 1 Swc Sor ) PV= 0.786 ( 1 0.2 0.2 ) PV= 0.472 PV

6. WiD saat breakthrough

+=

tanhL2

ycNpWi

2

DD

)0467.0)(40)(2000(2

0786.0Wi

2

D +=

= 0.786 MOV= 0.768 ( 1 Swc Sor ) PV

= 0.472 PV7. Waktu terjadinya breakthrough

D

Tstabil

T Wi*39.4xQ

Qt =

DWi*39.4x468

1000t =

DWi38.9t = yearst @ bt = 9.38 * 0.472 = 4.423 PV years

8. produksi saat breakthrough (Np @ bt)

MOVNpNp D=Np @ Bt = ( 0.786 ) ( 9.617 x 105 ) =

0.755 x 106 bbl

Berikut disajikan untuk data saat yc = 10 ft

4 dari 4


5/5


A.( )

=

tanhL2

ych1N

2

pD

=( )

)0467.0)(40)(2000(2

1020001

2

= 0.880 MOV

B. MOV880.0NpD == 0.880 ( 1 0.2 0.2 ) = 0.526 PV

C.

+=tanhL2

ycNpWi

2

DD

=

)0467.0)(40)(2000(2

10880.0

2

+

= 0.893 MOV

D. MOV893.0WiD =

= 0.893 ( 1 0.2 0.2 ) = 0.536 PV

E. DWi38.9t = = 9.38 * 0.536 PV = 5.028 PV

years

F. MOVNpNp D== 0.880 MOV= 0.880 x 9.617 x 105 = 8.46 x 105bbl

Berikut table selengkapnya

yc

(ft)

NpD

(MOV

)

NpD

(PV)

WiD

(MO

V)

WiD

(PV)

t (PV

years)

Np (x

105)

0=BT 0.786 0.472 0.786 0.472 4.423 7.552

10 0.880 0.528 0.893 0.536 5.026 8.452

20 0.946 0.568 1.000 0.600 5.628 9.096

30 0.987 0.592 1.107 0.664 6.231 9.481

40 1.000 0.600 1.214 0.728 6.833 9.610

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3. Water Conning and Fingering