Teknik Produksi i

Materi Kuliah

TEKNIK PRODUKSI I

Wibowo – JTM UPN “Veteran” o!"a#arta



_{PRODUKTIVITAS FORMASI}



VERTICAL LIFT PERFORMANCE



CHOKE PERFORMANCE



HORIZONTAL FLOW

PERFORMANCE



_{NATURAL FLOW WELL}



_{NODAL ANALYSIS}

I$ PRODUKTIVIT%S

&ORM%SI

Ter'iri 'ari (



)$ %liran &lui'a Melalui Me'ia Pori

*$ Pro'u+ti,it"

%liran &lui'a 'ala1 Me'ia

5er3ori



'i3ela6ari oleh 7enr" Dar+" "an! 1en!e1u#a#an hubun!an e13iri2 'ala1 bentu# 'i00erential

L P k A q v δ δ µ

−

=

=

q 8 9a6u aliran 0lui'a: ++;2e+

A 8 9ua2 1e'ia 3ena13an! 1e'ia ber3ori: +1*

v 8 Ke+e3atan aliran 0lui'a: +1;2e+

k 8 Per1eabilita2: Dar+"

%2u12i Dar+"



< %liran 1anta3 /

steady state

4 = 'P;'t 8>

<&lui'a "an! 1en!alir 2atu 0a2a 'an

incompressible

<Vi2#o2ita2 0lui'a "an! 1en!alir #on2tan

<Kon'i2i aliran i2other1al

Per2a1aan %liran Ra'ial Min"a#



qo 8 la6u aliran 1in"a# 'i3er1u#aan: ST5;D

ko 8 3er1eabilita2 e0e#ti0 1in"a#: 1D

h 8 #etebalan la3i2an: 0t

µ o 8 ,i2+o2ita2 1in"a#: +3

Bo 8 0a#tor ,olu1e 0or1a2i 1in"a#: 5bl;ST5

Pe 8 Te#anan re2er,oir 3a'a 6ari?6ari re: 32i

Pwf 8 Te#anan alir 'a2ar 2u1ur: 32i

re 8 6ari?6ari 3en!ura2an: 0t rw 8 6ari?6ari 2u1ur: 0t

)

rw

/

re

ln(

)

Pwf

Pe

(

B

.

h

.

k

,

qo

o o o

−

=

µ

00708

0

(teady tate!

Pen!e1ban!an

Per2a1aan %liran Ra'ial Min"a#



S

rw

re

Pwf

Pe

B

h

k

qo

o o o

+

−

=

)

/

ln(

)

(

.

.

00708

,

0

µ

( ) dP B k function pressure Pseudo s wf P P o o ro

∫

=

µ ψ ) / ln( ) ( . . . 00708 , 0 rw re Pwf Pe B h k k qo o o abs ro

−

=

µ

(

)

B dP k rw re h k qo s wf P P o o ro abs

∫

=

µ / ln . 00708 , 0 (Steady State) (Steady State)

Pen!e1ban!an

Per2a1aan %liran Ra'ial Min"a#



S

rw

re

Pwf

Pe

B

h

k

qo

o o o

+

−

=

)

/

ln(

)

(

.

.

00708

,

0

µ

(Steady State) (Steady State)

S

rw

re

Pwf

Pe

B

h

k

qo

o o o

+

−

−

=

5

.

0

)

/

ln(

)

(

.

.

00708

,

0

µ

(Semi/Pseudo Steady State) (Semi/Pseudo Steady State)

Per2a1aan %liran Ra'ial @a2



q" 8 la6u aliran !a2 'i3er1u#aan: SA&;D

µ " 8 Vi2#o2ita2 !a2: +3

k" 8 3er1eabilita2 e0e#ti0 !a2: 1D

# 8 te13eratur re2er,oir:

°

&

$ 8 0a#tor #o13re2ibilita2 !a2

(

)

(

re

/

rw

)

ln

Pwf

Pe

T

h

.

k

,

q

! ! ! 2 2

000703

0

−

=

µ



Prod%ctivity &nde'

< In'e#2 "an! 1en"ata#an #e1a13uan 0or1a2i untu# ber3ro'u#2i 3a'a 2uatu #on'i2i te#anan tertentu

< Meru3a#an 3erban'in!an la6u 3ro'u#2i "an! 'iha2il#an 0or1a2i 3ro'u#ti0 3a'a te#anan 'raw 'own /P2?Pw04 tertentu

P Pwf wf )>>> *>>> )>> *>> B 1a-*C> q q P2 > θ > tanθ 8 PI w dP dq J PI

=

=

Per2a1aan Pro'u+ti,it"

In'e-

Pwf)

#

(Ps

q

$

P%

₌

₌

'i1ana (

P& 8  8 Pro'u#ti,it" In'e-: bbl;hari;32i

q 8 la6u 3ro'u#2i aliran total: bbl;hari

Ps 8 Te#anan 2tati2 re2er,oir: 32i

Pwf 8 Te#anan 'a2ar 2u1ur wa#tu a'a aliran: 32i

)

rw

/

re

ln(

)

Pwf

Pe

(

B

.

h

.

k

,

qo

o o o

−

=

µ

00708

0

= J ~ PI = J ~ PI

Contoh Soal Productivity

Contoh Soal Productivity

Index

Index



<Di#etahui 'ata la3an!an 2eba!ai beri#ut (

P2 8 *>>> 32i

Pw08 )F>> 32i

Bo 8

CG b3'

•

_{Pertanyaan :}

_{Pertanyaan :}

5era3a#ah

Prod%ctivity &nde'

?n"a H

•

_{Jawab :}

_{Jawab :}

(

2000-1400

)

65

Soal Productivity Index

Soal Productivity Index



<Di#etahui 'ata la3an!an 2eba!ai beri#ut (

P2

8 **>> 32i

Pw0 8 )GC> 32i

Bo

8  b3'

<

Pertanyaan :

Pertanyaan :

5era3a#ah

Prod%ctivity &nde'

?n"a H

(

2200

-

1560

)

Inflow Performance

Inflow Performance

Relationship (IPR)

Relationship (IPR)

@ra0i# "an! 1en"ata#an 3erila#u aliran 0lui'a 'ari

re2er,oir 1enu6u 2u1ur: 2e2uai nilai

Prod%ktivitas (P&!

0or1a2in"a$

@ra0i# ini 1eru3a#an hubun!an antara Te#anan alir /Pw04 terha'a3 9a6u Pro'u#2i /L4

Dibe'a#an 2e2uai 6u1lah 0a2a 0lui'a "an! 1en!alir (

)$ IPR Satu &a2a

*$ IPR Dua &a2a

?

P2  Pb

?

P2  Pb 'an Pw0  Pb

?

P2  Pb 'an Pw0  Pb

3.

Prosedur umum : Prosedur umum :

Prosedur umum : Prosedur umum :

< Tentu#an be2ar nilai PI 0or1a2i /#hu2u2 2atu 0a2a4

< Tentu#an be2ar Lo

1a-< %2u12i#an bebera3a har!a Pw0: #e1u'ian 'en!an

1en!!una#an 3er2a1aan "an! 2e2uai 'a3at 'itentu#an 1a2in!?1a2in! har!a Lo 2e2uai har!a Pw0 a2u12i

< Pa'a #erta2 !ra0i# #arte2ian /1ili1eter !ra0i#4 2ia3#an 2u1bu

ab2i2 untu# har!a Lo 'an 3a'a 2u1bu Or'inat untu# har!a Pw0

< 5uat hubun!an antara Pw0 terha'a3 L 2e2uai ha2il

1.

1. IPR

IPR Satu

Satu Fasa

Fasa

1.

1. IPR

IPR Satu

Satu Fasa

Fasa



Da2ar Per2a1aan (

Pwf)

-(Ps

q

J

PI

=

=

)o = P&

/

Ps * Pwf

4

7ar!a Lo 1a- 'i+a3ai bila Pw0 8 >

Sehin!!a 'en!an 1e1,aria2i#an har!a Pw0 'a3at

'itentu#an be2ar Lo

Contoh Soal IPR Satu Fasa

Contoh Soal IPR Satu Fasa



< Di#etahui 'ata la3an!an 2eba!ai beri#ut ( P2 8 *>>> 32i Lo 8 CG b3' Pw0 8 )F>> 32i < 5a!ai1ana#ah IPRn"a H Men+ari PI( = 0,1083 !d/!si = 0,1083 !d/!si )o = P& /Ps * Pwf 4 )o 8 +,+-./+++*0+ 4 8-1,0- bpd %2u12i#an Pw0 *G>( )o 8 +,+-./+++*+++ 4 8 +-,10- bpd %2u12i#an Pw0 )>>>( Pwf) # (Ps q $ P%

=

=

( #&') *+ P%

=

P

Pw

wff

q

qo

o

0

0



1

1!

!""#

#



$

$0

0

1

1%

%&

&""!

!

$

$0

00

0

1

1!

!

""$

$

1

10

00

00

0

1

10

0%

%""'

'

1

1$

$0

00

0

$

$

""





0

00

00

0

0

0



*$ IPR Dua &a2a

*$ IPR Dua &a2a

S0 * F+1 S0 * F+1

< Dar+" /P2eu'o Pre22ure &un+tion4

< Vo!el

S,0

S,0 * * F+-1 F+-1   F+/1F+/1

< Stan'in! /Vo!el Mo'i0ie'44

< 7arri2on /Stan'in! Mo'i0ie'4

< Aouto /Stan'in! Mani3ulate'4

< Pu'6o Su#arno /Vo!el ba2e'  Si1ulate'4

urulensi 2an S,0 urulensi 2an S,0

< Jone2: 5lount  @lae /Per0orate' Well4

< &et#o,i+h /@a2 Well Mo'i0ie'4

3eto2e : 3eto2e :

3eto2e : 3eto2e :

*$ IPR Dua &a2a

< Vo!el 1en!e1ban!#an 3er2a1aan ha2il re!re2i 2e'erhana 'an 1u'ah 3e1a#aiann"a: 'en!an an!!a3an (

? Re2er,oar 5er3en'oron! @a2 Terlarut ? Te#anan Re2er,oar bera'a 'i 5awahTe#anan 5ubble Point$ ? &a#tor S#in 2a1a 'en!an >

2 max

8

.

0

2

.

0

1









−









−

=

Ps

Pwf

Ps

Pwf





_o

a.

a. IPR

IPR 4ua

4ua Fasa"

Fasa"

Ps - P

Ps - P

Ps - P

Ps - P



Per2a1aan Vo!el (

2 8 0 2 0 1

_













−















−

=

s wf s wf o -a. P P , P P , q q

































−

















−

=

2 max 1 0,2 0,8 s wf s wf o P P P P q q

Contoh

Contoh Soal

Soal IPR

IPR 4ua

4ua Fasa"

Fasa" Ps

Ps -

- P

P



Di#etahui 'ata la3an!an 2eba!ai beri#ut ( P2 8 *>>> 32i Pw0 8 )G>> 32i Pb 8 *)>> 32i Bo 8 CG b3' Pertanyaan : Pertanyaan : 5a!ai1ana#ah IPRn"a H

5a

5awaan

waan Contoh

Contoh IPR

IPR 4ua

4ua Fasa"

Fasa" Ps

Ps -

- P

P



Men+ari B_1a-( 2

8

0

2

0

1

_

_











−















−

=

s wf s wf o -a.

P

P

,

P

P

,

q

q

2 2000 1500 8 0 2000 1500 2 0 1 65













−













−

=

,

,

q

_-a.

_{8 )C*:G b3'}

5a

5awaan

waan Contoh

Contoh IPR

IPR 4ua

4ua Fasa"

Fasa" Ps

Ps -

- P

P



Men+ari B_o 3a'a Pw0 8 G>> (

8 )FC:*G b3'





























−













−

=

2 8 0 2 0 1 _s wf s wf -a o _P P , P P , q q





























−













−

=

2 2000 500 8 0 2000 500 2 0 1 5 162 , , , q_o

5a

5awaan

waan Contoh

Contoh IPR

IPR 4ua

4ua Fasa"

Fasa" Ps

Ps -

- P

P

Pwf Pwf

qo

qo

0

0

1

1!

!

""$

$0

0

$

$0

00

0

1

1

!

!""

$

$

1

10

00

00

0

11

1

1'

'""#

#$

$

1

1$

$0

00

0

!

!$

$



0

00

00

0

0

0



P2  Pb /H4

< IPR ter'iri 'ari 'ua ba!ian (

? 5a!ian 9inier /Pw0  Pb4: #ur,a 1en!i#uti 3er2 (

? 5a!ian Non?9inier /Pw0  Pb4: 1en!i#uti 3er2 (

< Proble1 (

P2  Pb  Pw0 te2t  Pb

P2  Pb  Pw0 te2t  Pb

(

Ps-Pwf

)

J o q

=

(

)

































_











−













−

−

+

=

max 1 0,2 0,8 2 b wf b wf b b o P P P P q q q q 8 . 1  P J  q

=

.

. IPR

IPR 4ua

4ua Fasa

Fasa

Ps

Ps /

/ P

P 2an

2an Pwf

Pwf /

/ P

P

Ps

Ps /

/ P

P 2an

2an Pwf

Pwf /

/ P

P



Per2a1aan "an! 'i!una#an

(

)





_











−

+

=

wf s b o b -a.

P

P

,

P

q

q

q

8

1

Pb)

#

(Ps

P%

q

_b

=

(

)

_













































−















−

−

+

=

 b wf b wf b -a b o P P  ,  P P  ,  & q q q q

Contoh Soal IPR 4ua Fasa

Contoh Soal IPR 4ua Fasa

Ps / P 2an Pwf / P

Ps / P 2an Pwf / P



Di#etahui 'ata la3an!an 2eba!ai beri#ut ( P2 8 *.G> 32i Pw0 8 )>> 32i Pb 8 )>> 32i Bo 8 C>> b3' Pertanyaan : Pertanyaan : 5a!ai1ana#ah IPRn"a H Men+ari PI (

8 ):...

Pwf)

#

(Ps

q

$

P%

=

=

(

1+

#

&0

)

*

P%

=

5awaan Contoh IPR 4ua Fasa

5awaan Contoh IPR 4ua Fasa

Ps / P 2an Pwf / P

Ps / P 2an Pwf / P



Men+ari B_b (

q

_b

=

P%

(Ps

#

Pb)

&2)

#

(1+

&,11111

q

_b

=

8 CC:CC

Men+ari B_1a-(

8 *)*G:*. b3'

&,

Pb

P%

q

-a

q

=

_b

+

×

&,

&2

&,11111

**,***2

-a

q

=

+

×

5awaan Contoh IPR 4ua Fasa

5awaan Contoh IPR 4ua Fasa

Ps / P 2an Pwf / P

Ps / P 2an Pwf / P



Men+ari B_o 3a'a Pw0 8 F>> (

8 *>)>:. b3'

(

)













































−













−

−

+

=

2 1700 400 8 0 1700 400 2 0 1 6667 866 923 2125 6667 866 , , , , , q_o

(

)







































−













−

−

+

=

2 8 0 2 0 1 b wf b wf b -a. b o P P , P P , q q q q

5awaan Contoh IPR 4ua Fasa

5awaan Contoh IPR 4ua Fasa

Ps / P 2an Pwf / P

Ps / P 2an Pwf / P

P Pwwff qqoo > *)*G:*C F>> *>)>:. >> )F:.)F )*>> )FFC:) )C>> C:G)F )>> CC:CC )>> C>>:>>> *.G> >



c.

c. IPR

IPR 4ua

4ua Fasa

Fasa

Ps

Ps /

/ P

P 2an

2an Pwf

Pwf -

- P

P



Per2a1aan "an! 'i!una#an

 (Pwf/Pb) , # (Pwf/Pb) , # & A

=

(Pb/&,)A

Pb

#

Ps

qo

P%

₌

₊

Pb)

#

(Ps

P%

qb

=

_{( )}

8

1

,

Pb

P%

q

=

q

q

-a

q

=

_b

+

(

)

_































 











−

 











−

−

+

=

) b wf b wf . -a. b o

P

P

/

,

(

P

P

)

,

(

&

q

q

q

q

Contoh Soal IPR 4ua Fasa

Contoh Soal IPR 4ua Fasa

Ps / P 2an Pwf - P

Ps / P 2an Pwf - P



Di#etahui 'ata la3an!an 2eba!ai beri#ut ( P2 8 )G> 32i Pw0 8 >> 32i Pb 8 )*>> 32i Bo 8 C>> b3' Pertanyaan : Pertanyaan : 5a!ai1ana#ah IPRn"a H

5awaan Contoh IPR 4ua Fasa

5awaan Contoh IPR 4ua Fasa

Ps / P 2an Pwf - P

Ps / P 2an Pwf - P



Men+ari % (

8 >:F

Men+ari PI (

8 >:.FC

2 8 0 2 0 1













−













−

=

Pb Pwf , Pb Pwf , A 2 1200 900 8 , 0 1200 900 2 , 0 1













−













−

=

A (Pb/&,)A Pb # Ps qo P%

+

=

,' (&/&,) & # &2+ * P%

+

=

5awaan Contoh IPR 4ua Fasa

5awaan Contoh IPR 4ua Fasa

Ps / P 2an Pwf - P

Ps / P 2an Pwf - P



Men+ari Bb ( 8 F>F:>)C b3' Men+ari B- ( 8 F:G b3'

Pb)

#

(Ps

P%

qb

=

&) # (&2+ ,21'*0' qb

=

( )

8 1 , Pb P% q

=

(

)

8 1 1200 734694 0 , , q

=

Men+ari B1a- ( q-a

=

q_b

+

q

, '',&*

-a.

q

=

+

4897959

5awaan Contoh IPR 4ua Fasa

5awaan Contoh IPR 4ua Fasa

Ps / P 2an Pwf - P

Ps / P 2an Pwf - P



Men+ari B_o 3a'a Pw0 8 .>> ( 8 C:GG) b3'

(

)

_











































−













−

−

+

=

2 8 0 2 0 1 b wf b wf . -a. b o _P P , P P , q q q q

(

)













































−













−

−

+

=

2 1200 900 8 0 1200 900 2 0 1 7959 489 8776 893 0816 404 , , , , , q_o

5awaan Contoh IPR 4ua Fasa

5awaan Contoh IPR 4ua Fasa

Ps / P 2an Pwf -

Ps / P 2an Pwf - P

P

P Pwwff qqoo > *:.G* .>> .G:*F C>> F):)C >> C>>:>>> )*>> F)):CG )G> >



S Q > = &E Q )

< Stan'in! 1e1o'i0i#a2i 3er2a1aan Vo!el

ber'a2ar#an #en"ataan bahwa 0or1a2i "an!

1en!ala1i #eru2a#an /'a1a!e4 a#an ter6a'i

ta1bahan #ehilan!an te#anan

< Pw0 i'eal /ti'a# 'i3en!aruhi 2#in 0a+tor4 

P

P

wf wf ’

< Pw0 a+tual /'i3en!aruhi 2#in 0a+tor4 

P

P

wf wf

< &E /&low E00i+ien+"4 (

) s (P ) ! s (P "# wf P wf P − − = 34 wf P s P s P ( ) ! wf P = − − 2 ! ! 1 max 8 . 0 2 . 0 1 _      −       − = = _Ps Pwf Ps Pwf   34 o

Aontoh Meto'e Stan'in!

Di#etahui 'ata la3an!an 2eba!ai beri#ut (

P2 8 *C>> 32i

Bo 8 G>> b3' 3a'a Pw0 8 )>> 32i

&E 8 >$C

Pertanyaan :

Pertanyaan :

5a!ai1ana#ah IPRn"a H

Jawaban Meto'e Stan'in!

2120 6 . 0 ) 1800 2600 ( 2600 ! wf P

=

−

−

=

1639 2600 2120 8 . 0 2600 2120 2 . 0 1 500 2 1 max

=









−









−

=

= 34 

L1a- 3a'a &E8>$C Pw08>

1040 6 . 0 ) 0 2600 ( 2600 ! wf P

=

−

−

=

1298 2600 1040 8 . 0 2600 1040 2 . 0 1 1639 2 6 . 0 max

=

























−









−

=

= 34 

Jawaban Meto'e Stan'in!



Dari bebera3a har!a Pw0 a2u12i 'i'a3at (

P

P

_{wf wf}

P

P

_{wf wf} ’’

₆

₆

o o

0

0

10

1

0

0

0

1

1

&

&%

%

$

$0

00

0

1'

1

'

0

0

1

11

1

&

&

1

10

00

00

0

1!

1

!

0

0

&

&1

11

1

1

1$

$0

00

0

1&

1

&

0

0

!

!!

!





0

00

00

0







0

0

'

'%

%'

'



!

!0

00

0

!



!0

00

0

0

0

S Q > = &E Q )

< 7arri2on 1e1o'i0i#a2i 3er2a1aan Stan'in!

#arena 3a'a har!a &E "an! 2an!at #e+il atau &E

3o2iti0 be2ar /Pw0 ne!ati04 1en!ha2il#an bentu#

IPR "an! ti'a# 2e1e2tin"a

< Kon2e3 &E teta3 'i!una#an untu# #on'i2i 22atu

0a2a

< Per2a1aan 7arri2on (

34 wf P s P s P ( ) ! wf P = − −         =

=

−

Ps Pwf 34 o e   ! 792 . 1 1 max 2 . 0 2 . 1

Jawaban Meto'e 7arri2on

2120 6 . 0 ) 1800 2600 ( 2600 ! wf P

=

−

−

=

L1a- 3a'a &E8>$C Pw08>

1340 6 . 0 ) 500 2600 ( 2600 ! wf P

=

−

−

=

47 . 859 2 . 0 2 . 1 16 . 1480 2600 1940 792 . 1 6 . 0 max

=

















−

=

      = e  34       =

=

−

2600 2120 792 . 1 1 max

2

.

0

2

.

1

500

e



34 16 . 1480 1 max

=

=

34 

Aontoh Meto'e 7arri2on

Di#etahui 'ata la3an!an 2eba!ai beri#ut (

P2 8 *C>> 32i

Bo 8 G>> b3' 3a'a Pw0 8 )>> 32i

&E 8 *$C

Pertanyaan :

Pertanyaan :

5a!ai1ana#ah IPRn"a H

Jawaban Meto'e 7arri2on

2120 6 . 0 ) 1800 2600 ( 2600 ! wf P

=

−

−

=

L1a- 3a'a &E8>$C Pw08>

1040 6 . 0 ) 0 2600 ( 2600 ! wf P

=

−

−

=

96 . 1169 2 . 0 2 . 1 16 . 1480 2600 1040 792 . 1 6 . 0 max

=

















−

=

      = e  34       =

=

−

2600 2120 792 . 1 1 max

2

.

0

2

.

1

500

e



34 1480.16 1 max

=

=

34 

Jawaban Meto'e 7arri2on



Dari bebera3a har!a Pw0 a2u12i 'i'a3at (

P

P

_{wf wf}

P

P

_{wf wf} ’’

₆

₆

o o

0

0

10

1

0

0

0

1

11

1#

#0

0

$

$0

00

0

1'

1

'

0

0

1

10

0'

'1

1

1

10

00

00

0

1!

1

!

0

0

%

%!

!0

0

1

1$

$0

00

0

1&

1

&

0

0

!

!

&

&



0

00

00

0







0

0

'

'&

&0

0



S Q > = &E Q )

< Aouto 1e1ani3ula2i 3er2a1aan Stan'in! 'en!an 1en!!abun!#an #on2e3 PI < Per2a1aan 7arri2on (

(

)

B . P

(

34

)

. A k rw re h  r o o o o































=

µ 472 . 0 ln 00419 . 0

( )

(

( )( )

)

(

5

34

5

)

A

=

1

−

1.8

−

0.8 1

−

r wf P P 5 =

Di1an

a:

Kon2tanta A

₎

: A

_*

: A

_.

: 'an A

_F a ann cc11 cc cc'' cc a a11 0..101%%&& 7700..''!!''%% 00..%%11$$11 7700..00$$$$%%##'' a a 7711..##!!&&$$00 7700..$$!!!!'' 11..!!!!!! 7700..''00!! a

a'' 77..11&&## 7700..11&&$$&&##!! ..%%&& 7700..00''''''

a

a 7700..0011##%%'' 00..00%%%%%%!! 7700..!!00''%%$$ 7700..1100%%0011

a

Aontoh Meto'e Pu'6o Su#arno /*

Ø

₄

Di#etahui 'ata la3an!an 2eba!ai

beri#ut (

P2 8 )G> 32i

Bo 8 *F b3' 3a'a Pw0 8 *F>32i

S 8 *$F. /&E 8 >$>4

Pertanyaan :

Pertanyaan :

5a!ai1ana#ah IPRn"a H

Jawaban Meto'e Pu'6o Su#arno

/*

Ø

₄

< 7itun! #on2tanta a) 2;' aG < a* 8 >$>G>F a . 8 >$>>G** aF 8 ?>$). a G 8 ?>$) < Pw0;P2 8 *F>;)G> 8 >$)G>F < Lo1a- 8 *F;>$C.C 8 )*>$F bbl;hari

< %2u12i#an bebera3a har!a Pw0 untu# 1enentu#an har!a Lo

78658 . 0 814541 . 0 182922 . 0 ( 0.364438 2.43) ( 0.055873 2.43) 1

=

−

+

−

=

. . e e a

(

)

( )

(

2

)

2 0 max 1 0.07504 0.15094 0.183 0.15094 15094 . 0 7819 . 0 15094 . 0 00522 . 0 78658 . 0 924 − + − + = = S o 

Tabula2i ha2il 3erhitun!an

P

Pw

wf

f 6

6o

o P

Pw

wff 6

6o

o

1

1

&

&0

0

0

0..0

0

%

%0

00

0

#0.

#0.

&'

&'

100

100

'!.

'!.

#1

#1

!00

!00

%1#.

%1#.

0$

0$

100

100

'!.

'!.

'%

'%

00

00

%%!.

%%!.

&

&

1000

1000

$&$.

$&$.

'!

'!

0.0

0.0

&$1.

&$1.

1&

1&

.$ %liran &lui'a Ti!a &a2a

< 8paila flui2a yan9 men9alir 2ari formasi e luan98paila flui2a yan9 men9alir 2ari formasi e luan9

sumur ter2iri 2ari ti9a fasa"

sumur ter2iri 2ari ti9a fasa" yaitu minya" air 2anyaitu minya" air 2an 9as" maa 2i9unaan 3eto2e Pu2;o

9as" maa 2i9unaan 3eto2e Pu2;o Suarno.Suarno.

(

) (

)

2 2 1 max , r wf r wf o t o

_A

_A

_P

_P

_A

_P

_P

q

q

+

+

=

• An = konstanta persamaan (n = 0, 1 dan 2), yang

harganya berbeda untuk water cut yang

berbeda.

• Hubungan antara konstanta tersebut dengan

water-cut dtentukan pu!a dengan ana!ss regres, dan

dpero!eh persamaan berkut "

 _{An 8 #on2tanta untu# 1a2in!?1a2in! har!a %n}

'itun6u##an 'ala1 Tabel beri#ut ini

( )

( )

A_n C₀ C₁ C₂ $ 0 0.980321

−

0.115661

_×

10 -1 0.179050

_×

10 -4 $1

−

0.414360 0.392799

×

10 -2 0.237075

×

10-5 $2

−

0.564870 0.762080

×

10 -2

₋

0.202079

×

10-4

Tabel Kon2tanta An Untu# Ma2in!?Ma2in! har!a %n

A_n C₀ C₁ C₂ $ 0 0.980321

−

0.115661

_×

10 -1 0.179050

_×

10 -4 $1

−

0.414360 0.392799

×

10 -2 0.237075

×

10-5 $2

−

0.564870 0.762080

×

10 -2

₋

0.202079

×

10-4 A_n C₀ C₁ C₂ $ 0 0.980321

−

0.115661

×

10 -1 0.179050

×

10 -4 $1

−

0.414360 0.392799

×

10 -2 0.237075

×

10-5 $2

−

0.564870 0.762080

×

10 -2

₋

0.202079

×

10-4 A_n C₀ C₁ C₂ $ 0 0.980321

−

0.115661

×

10 -1 0.179050

×

10 -4 $1

−

0.414360 0.392799

×

10 -2 0.237075

×

10-5 $2

−

0.564870 0.762080

×

10 -2

₋

0.202079

×

10-4

< Se'an!#an hubun!an antara te#anan alir 'a2ar 2u1ur

terha'a3 water?+ut 'a3at 'in"ata#an 2eba!ai Pw0;Pr

terha'a3 WA;/WA  Pw0 =Pr4:

'i1ana /WA  Pw0 =Pr4 telah 'itentu#an 'en!an

anali2i2 re!re2i 'an 1en!ha2il#an 3er2a1aan beri#ut (

[

_{wf r}

]

r wf

P

P

P

4.p

P

P

P

76

76

/

%

≈

=

1

×

2

Aatatan (

234P

_wf5

P

_{r =}

236

< 'i1ana har!a P) 'an P* ter!antun! 'ari

har!a water?+ut 3en!u#uran: 'i1ana(

)

(

130447

.

0

606207

.

1

1

Ln

76

P

=

−

×

)

(

110604

.

0

517792

.

0

2

Ln

76

P

=

−

+

×

'i1ana ( water?+ut 'in"ata#an 'ala1 3er2en /4 'an 1eru3a#an 'ata u6i 3ro'u#2i

Pro2e'ur Perhitun!an

< 5er'a2ar#an har!a WA 3en!u#uran tentu#an

WAPw0=Pr: P

)

'an P

*

< 7itun! har!a?har!a %

>

: %

)

: 'an %

*

2e2uai har!a %

n

2e3erti tertera 'ala1 tabel

< 7itun! L

t

1a-< %2u12i#an bebera3a har!a Pw0 'an hitun! Lo 2e2uai

har!a Pw0 a2u12i

< 5uat hubun!an /3lot4 antara Pw0 'an Lo 3a'a #erta2

#arte2ian untu# 1en'a3at#an #ur,a IPR

PER%M%9%N IPR M%S% %K%N

PER%M%9%N IPR M%S% %K%N

D%T%N@

D%T%N@

<

S 0 :

S 0 :

3eto2e Stan2in9 (<o9el ase) :

3eto2e Stan2in9 (<o9el ase) :

( 5

( 5

pp

))

==

 PI searan9

 PI searan9

( 5

( 5

ff

))

==

 PI masa aan 2atan9

 PI masa aan 2atan9

( )

(

)

s o p

P



$

&

=

8

1

.

max ( ) ( ) ( ) ( ro o o) p f o o ro p f B k B k $ $ µ µ / / & &

₌

(

)

( )

( )

s _f f f o

P

$



& max

8

.

1

=

PER%M%9%N IPR M%S% %K%N D%T%N@

PER%M%9%N IPR M%S% %K%N D%T%N@

< S Q> (

Meto'e E+#1ier ( 3en"e'erhanaan 'ari 3er2a1aan &et#o,i+h

'en!an an!!a3an n 8 ): 2ehin!!a 3erban'in!an Lo1a- 3a'a wa#tu 3ro'u#2i t) 'an t* 'in"ata#an (

3 max max       =       sp sf p o f o P P , ,

L_{o1a- 3} 3a'a P_{2 3} 'itentu#an ber'a2ar#an u6i te#anan 'an 3ro'u#2i 3a'a t) 1en!!una#an 3er2a1aan Vo!el

2e'an!#an

L_{o1a- 0} 3a'a P_{2 0} 'itentu#an ber'a2ar#an 3er2a1aan 'iata2

Selan6utn"a untu# 1e1buat IPR 'i!una#an 3er2a1aan Vo!el ber'a2ar#an har!a Lo1a- 0 _{'an P}2 0

OPTIM%SI PRODUKSI

'en!an

%N%9ISIS SISTEM

NOD%9

 wibowo?t1 u3n”,eteran”"o!"a#arta

OVERVIEW

TUJU%N 'an S%R%T

?5?8@

?5?8@

• Mendapatkan laju produksi optimum sumur

dengan melakukan evaluasi secara lengkap

dan terintegrasi pada sistem produksi sumur

SA8R8

SA8R8

• Tersedia Infow Perormance (IPR)

• Tersedia Outfow Perormance (

!P"#P"$%P"

&P)

METODO9O@I

< Me1aha1i #o13onen In0low Per0or1an+e

< Me1aha1i #o13onen Out0low Per0or1an+e: "an! ter'iri 'ari #iner6a (

 Verti+al 9i0t Per0or1an+e  Aho#e Per0or1an+e

 7oriontal &low Per0or1an+e  Se3arator

< Me1aha1i hubun!an in0low 'an out0low 3er0or1an+e

< Me1aha1i 'i2#ri32i hubun!an Te#anan ,er2u2 Ke'ala1an 3a'a berba!ai 1eto'e 3ro'u#2i /li0tin! 1etho'24

Pe  Pr Pw02 Pw0 P'r Pur Pu2, Pwh P'2+ P2e3

DP) 8 Pr ? Pw02 8 9o22 in Porou2 Me'iu1

DP* 8 Pw02 ? Pw0 8 9o22 a+ro22 Ao13letion

DP. 8 Pur ? P'r 8 9o22 a+ro22 Re2tri+tion

DPF _{8 P}u2, _{? P}'2, _{8 9o22 a+ro22 Sa0et" Val,e}

DPG 8 Pwh ? P'2+ 8 9o22 a+ro22 Sur0a+e Aho#e

DPC 8 P'2+ ? P2e3 8 9o22 in &lowline

DP 8 Pw0 ? Pwh 8 Total 9o22 in Tubin!

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5otto1 7ole Re2tri+tion Sa0et" Val,e Sur0a+e Aho#e Se3arator

M%N&%%T %N%9ISIS SISTEM

NOD%9

• Optimasi laju produksi

• Menentukan laju produksi 'ang dapat

diperole secara semur alam

• Meramalkan kapan sumur akan *mati+

• Memeriksa setiap komponen dalam sistem

produksi untuk mementukan adan'a

amatan aliran

• Menentukan saat 'ang teraik untuk

mengua sumur semur alam menjadi

semur uatan atau metode produksi satu

ke metode produksi lainn'a

%9IR%N ME9%9UI PIP%

I$ VERTA%9 9I&T PER&ORM%NAE /TU5IN@4

II$ 7ORIONT%9 &9OW PER&ORM%NE /PIPE9INE4

Pen'ahuluan

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Ke1a13uan re2er,oir 'a3at 'i3ro'u#2i#an #e

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ter!antun! 3a'a te#anan 'an #on0i!ura2i 2i2te1

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3er3i3aan: 2ehin!!a 'a3at 'ituli2

<

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Untu# 1e1entu#an #e1a13uan 2i2te1 2e+ara

total 3erlu 1en!hitun! #ehilan!an te#anan

total 3erlu 1en!hitun! #ehilan!an te#anan

1a2in!?1a2in! #o13onen

1a2in!?1a2in! #o13onen

rts tb ch fl sep wf

P

P

P

P

P

P

=

+

∆

+

∆

+

∆

∆

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< Den!an 1en!!una#an 3rin2i3 ter1o'ina1i#a:

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c c s c c

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m

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:

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2bb

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v

;

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1

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4

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p

p

p

p

k

k

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d

p

6

q

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1

γ

Pen!!unaan #ur,a 3re22ure

tra,er2e

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Pen!!unaan #ur,a 3re22ure

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