Mata Kuliah
Teknik Reservoir I Dr. Ir. Wahju Wibowo Wib owo
Daftar Pustaka 4 1 0 2
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@ G N U D N A B S N I A S N A D I G O L O N K E T T U T I T S N I B S
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•
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Craft, B. C. and Hawkins, M. F.: Applied Petroleum Reservoir e dition by R. E. E . Terry err y, Prentice-Hall, Prentice -Hall, Inc., In c., Engineering , Revised edition Englewood Cliffs, NJ, 1991. L . P.: Fundamentals of Reservoir Engineering , Elsevier Dake, L. Science Publisher B. V., Amsterdam, the Netherlands, 1978.
Dake, L. P.: The Practice of Reservoir Engineering , Revised Edition, Elsevier Science Scienc e Publisher Publis her B. V., V., Amsterdam, Amsterd am, the Netherlands, Nether lands, 2001. 2001. T.: Reservoir Engineering Handbook , Fourth Edition, Gulf Ahmed, T. Professional Publishing, United States of America, 2010.
Ekspektasi 4 1 0 2 @ G N U D N A B S N I A S N A D I G O L O N K E T T U T I T S N I B S
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Hadir disetiap jam perkuliahan
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Menyimak dan memahami materi perkuliahan
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Bertanya Bertanya jika tidak mengerti
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Mengerjakan tugas atau pekerjaan rumah
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Mengerjakan ujian yang dilaksanakan (Quiz, UTS, UAS)
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Komunikasi dilakukan melalui Ketua Ketua Kelas
Topik Bahasan 4 1 0 2
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Pendahuluan
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Review Sifat Fisik Fluida
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Review Sifat Fisik Batuan
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Cadangan Migas
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Mekanisme Pendorong Reservoir
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Persamaan Aliran dan Aplikasinya
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Konsep Produktivitas
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Analisis Decline Curve
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Persamaan Kesetimbangan Materi
@ G N U D N A B S N I A S N A D I G O L O N K E T T U T I T S N I B S
Mata Kuliah Teknik Reservoir I
Analisis Kesetimbangan Materi (Material Balance Analysis)
Dr. Ir. Wahju Wibowo
Pendahuluan 4 1 0 2
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@ G N U D N A B S N I A S N A D I G O L O N K E T T U T I T S N I
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Salah satu cara atau alat yang digunakan oleh seorang reservoir engineer untuk melakukan interpretasi dan prediksi perilaku suatu reservoir hidrokarbon. Konsep material balance pertama kali dikembangkan oleh Schilthuis di tahun 1941 berdasarkan konsep kesetimbangan materi dalam suatu volume tertentu. Volume awal = Volume tersisa + Volume terproduksi
Volume Terproduksi Volume Awal
Volume Tersisa
Kondisi Awal
Kondisi saat pengamatan
B S
Material Balance: Definitions of Variables 4 1 0 2 @ G N U D N A B S N I A S N A D I G O L O N K E T T U T I T S N I B S
Production data G p W p
= cumulative gas produced (scf) = Cumulative water produced (stb)
Reservoir Data p i p S wc c r
= Initial mean pressure in the reservoir (psi) = current mean pressure in the reser voir, (psi) = connate water saturation, (fraction) = Compressibility of reservoir rock (psi -1)
Fluid PVT Data B gi B g c w B w Z Z i
= Initial gas volume factor at p i (ft3 /scf) = Gas volume factor at current pressure p (ft 3 /scf) = Compressibility of water (psi -1) = Formation volume factor of water at current pressure p (rb/stb) = compressibility factor, fraction = initial compressibility factor, fraction
Material Balance: Gas Reservoir 4 1 0 2 @
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G N U D N A B S N I A S N A D I G O L O N K E T T U T I T S N I B S
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Aplikasi MB pada reservoir gas biasanya digunakan untuk memperkirakan atau menghitung: –
Volume gas in place
–
Gas reserves
–
Gas recovery factor
–
Identifikasi adanya mekanisme pendorong lain selain ekspansi gas itu sendiri
Data yang diperlukan: –
Data produksi dan tekanan reservoir
–
Data sifat fisik batuan dan fluida reservoir
Akurasi perhitungan tergantung kualitas data dan waktu saat evaluasi dilakukan (sebaiknya setelah tekanan reservoir turun > 10% dari tekanan awal atau telah berproduksi > 20% dari volume awal)
Material Balance: Gas Reservoir 4 1 0 2 @ G N U D N A B S N I A S N A D I G O L O N K E T T U T I T S N I B S
Pada kuliah ini, akan dibahas secara mendalam perhitungan MB untuk sistem reservoir: 1.
Volumetric dry gas reservoir: berlaku pada reservoir gas yang tertutup tanpa adanya energi/influx dari luar reservoir. Mekanisme pendorong reservoir adalah ekspansi dari fluida (gas dan connate water) dan batuan yg ada didalam reservoir tersebut.Karena ekspansi connate water dan batuan relatif sangat kecil dibanding gas, maka biasanya hal ini diabaikan.
2.
Dry gas reservoir dengan water influx: terjadi jika ada sumber energi lain masuk ke dalam reservoir gas (air dari akuifer). Volume reservoir gas berubah sebagai akibat adanya sebagian volume reservoir gas ditempati air akuifer yang bergerak masuk ke dalam reservoir. Selain itu, adanya air akuifer yang masuk ke reservoir akan memberikan tambahan energi dan harus diperhitungkan dalam perhitungan material balance.
Material Balance: Volumetric Dry Gas Reservoir 4 1 0 2
G p Volume Terproduksi
@
N A D I G O L O N K E T T U T I T S N I B S
= G − G p )⋅ Bg G p ⋅ Bg = G ⋅ Bg − G ⋅ Bgi G ⋅ Bgi
G N U D N A B S N I A S
G p
G p G
G ⋅ Bgi
G − G p ⋅ Bg
Volume Awal
Volume Tersisa
Kondisi Awal Pi
Kondisi setelah berproduksi P(t)
G p
G p
Bgi = G − G ⋅ Bg B = G ⋅ 1 − gi Bg
Z i 0.0282 ⋅ T ⋅ Z i P P i = G ⋅ 1 − i = G ⋅ 1 − P Z i = G ⋅ 1 − Z T ⋅ Z Z Pi 0 . 0282 ⋅ P P = 1−
P Z i Z Pi
P Z i Z Pi
= 1−
Gp G
P Z
=
Pi Z i
−
Pi G p Z i G
Bg
RF =
P Z
=
Pi Z i
−
G p G
= 0.0282 ⋅
P Z i = 1 − Z P i
Pi G p Z i G
T ⋅ Z P
,
ft 3 scf
Material Balance: Volumetric Dry Gas Reservoir 4 1 0 2
Persamaan MB untuk volumetric dry gas reservoir
@ G N U D N A B S N I A S N A D I G O L O N K E T T U T I T S N I B S
P Z P Z
=
Pi
−
Z i
=−
Pi G p
Pi
Z i G
Z i
Pi G p Z i G
+
Pi Z i
Y = − m ⋅ X + C Y = m=
P
slope
P
=
−
m
=
−
Pi Z i G ⋅
Z
Z Pi Z i ⋅ G
X = G p C =
Pi
G
Z i G p
= Gas In Place
Material Balance: Volumetric Dry Gas Reservoir 4 1 0 2
Persamaan MB untuk volumetric dry gas reservoir
@
P
G N U D N A B S N I A S
Z
N A D I G O L O N K E T T U T I T S N I B S
=−
Pi G p Z i G
+
Pi Z i
Pi Z i
slope
=
−
m
=
−
Pi Z i G ⋅
P
• Kondisi awal, G p = 0, memberikan
Pi Z i
• Gas in place dapat dihitung dengan mengetahui: •
Perpotongan pada sumbu Gp pada saat
P Z
Z
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Slope atau garis miring - m
G = Gas In Place
G p
Note…. Gas In Place = G
= 43560 ⋅
A ⋅ h ⋅ φ ⋅ (1 − S wi ) B
=0
Material Balance: Volumetric Dry Gas Reservoir Step by step 4 1 0 2 @ G N U D N A B S N I A S N A D I G O L O N K E T T U T I T S N I B S
1.
Pengumpulan data yang diperlukan: –
Data tekanan reservoir
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Data produksi sumur / reservoir
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Hitung z-factor sebagai fungsi dari tekanan
3.
Hitung P/z dan plot vs. Gp
4.
Tarik garis lurus melalui data point yang diplot
5.
Ekstrapolasi garis lurus tersebut sampai pada P/z = 0 (memotong sumbu x)
6.
Tentukan harga G atau gas in place, yaitu harga Gp @ P/z=0
8.
Pi
Z
Z i
slope
Data sifat fisik dan batuan reservoir
2.
7.
P
Perkirakan harga ultimate recovery @ tekanan abandonment
=
−
m
=
Pi
−
Z i G ⋅
G = Gas In Place
Pa Z a
remaining reserves cum. prod. gas
ultimate recovery
Hitung parameter lain yang diinginkan Note: Pa = Tekanan abandonment Za = z-factor pada tekanan abandonment
G p
Material Balance: Volumetric Dry Gas Reservoir Contoh Soal – 1 4 1 0 2 @ G N U D N A B S N I A S N A D I G O L O N K E T T U T I T S N I B S
Dari peta isopach suatu lapangan gas A, secara volumetrik memberikan volume gas in place sebesar 44 mmscf. Berikut adalah data produksi dan tekanan yang diambil dari sumur yang berproduksi dari lapangan gas A. •
Reservoir Pressure (Pr), psi
Cum. Gas Prod. (Gp), mmscf
Z
4000
0.00
0.80
3480
2.46
0.73
2970
4.92
0.66
2500
7.88
0.60
2100
11.20
0.55
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Dengan menggunakan metode P/Z, hitung besarnya gas in place lapangan gas A. (bandingkan dengan hasil volumetrik… Mana yang lebih valid?....) Jika diasumsikan sumur tidak mampu lagi mengalir pada tekanan abandonment (di reservoir) sebesar 1000 psi (Z = 0.52), hitung kumulatif produksi gas dan faktor perolehan maksimum lapangan A
Material Balance: Volumetric Dry Gas Reservoir Contoh Soal – 1 4 1 0 2
Hitung P/Z berdasarkan data produksi yang diketahui dan plot P/Z vs. Gp
@ G N U D N A B S N I A S N A D I G O L O N K E T T U T I T S N I
Reservoir Cum. Gas Pressure (Pr), Prod. (Gp), psi mmscf
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•
B S
•
6000
Z
P/Z, psi
4000
0.00
0.80 5000.0
3480
2.46
0.73 4767.1
2970
4.92
0.66 4500.0
2500
7.88
0.60 4166.7
2100
11.20
0.55 3818.2
Dari plot P/Z, didapat gas in place sebesar 47 mmscf. Pada Pr = 1000 psi, maka P/Z = 1000/0.52 = 1923 psi. dari plot P/Z perkiraan kumulatif produksi gas adalah sebesar = 29 mmscf. RF = 29 / 47 = 0.617 atau 61.7%
5000
4000 i s p , Z / P
3000
2000
P/Z = 1923 psi
1000
Gp = 29 mmscf
0 0
10
20
30
40
50
Cum. Gas Production (Gp), mmscf G = 47 mmscf
Material Balance: Volumetric Dry Gas Reservoir Contoh Soal – 2 4 1 0 2
A volumetric gas reservoir has the following production history.
@ G N U D N A B S N I A S N A D I G O L O N K E T T U T I T S N I B S
The following data are also available: φ = 13%
Swi = 0.52
h = 54 ft
T = 164oF
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A = 1060 Acres
Calculate the gas initially in place volumetrically and from the MBE
Material Balance: Volumetric Dry Gas Reservoir Contoh Soal – 2 4 1 0 2
Step – 1. Calculate
@ G N U D N A B S N I A S N A D I G O L O N K E T T U T I T S N I B S
Bgi
= 0.02827 ⋅
(164 + 460) ⋅ 0.869
ft 3
= 0.00853
1798
scf
Step – 2. Calculate the gas initially in place volumetrically A ⋅ h ⋅ φ ⋅ (1 − S wi ) 1060 ⋅ 54 ⋅ 0.13 ⋅ (1 − 0.52) = 43560 ⋅ G = 43560 ⋅ Bgi 0.00853 G = 18239925270.8 scf = 18.24 bcf Step – 3. Plot P/z vs. G p
2500
2000
Reservoir Cum. Gas Time, Pressure (Pr), Prod. (Gp), years psi bscf
Z
P/Z, psi
0.00
1798
0.00
0.87 2069.0
0.50
1680
0.96
0.87 1931.0
1.00
1540
2.12
0.88 1750.0
1.50
1428
3.21
0.89 1604.5
2.00
1335
3.92
0.90 1483.3
1500
i s p , Z / 1000 P
G = 14.2 bscf
500
0 0
10
Cum. Gas Production (Gp), bscf
20
Material Balance: Volumetric Dry Gas Reservoir dengan Water Influx 4 1 0 2
Volume Terproduksi
G p & W p
@ G N U D N A B S N I A S N A D I G O L O N K E T T U T I T S N I B S
Volume gas Tersisa
G − G p )⋅ Bg
G ⋅ Bgi
)
i g
B −
g
B
Volume Awal
)
w
∆V p
W e
Kondisi setelah berproduksi P(t)
Kondisi Awal Pi
G ⋅ Bgi
=
G ⋅ Bgi
= (G − G p )⋅ Bg + W e − W p ⋅ Bw
G ⋅ Bgi
+ W p ⋅ Bw = G ⋅ Bg − G p ⋅ Bg + W e
G p ⋅ Bg
+ W p ⋅ Bw = G ⋅ Bg − G ⋅ Bgi + W e
G p ⋅ Bg
+ W p ⋅ Bw = G ⋅ ( Bg − Bgi ) + W e
G p ⋅ Bg
+ W p ⋅ Bw
( B
G − G p ⋅ Bg
B
)
B ⋅ p
W +
g
B ⋅ p
G
G = Gas In Place
+ ∆V p → ∆V p = W e − W p ⋅ Bw
=G+
W e
( B
B
)
W e Bg
− Bgi )
Material Balance: Oil Reservoir 4 1 0 2
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@ G N U D N A B S N I A S N A D I G O L O N K E T T U T I T S N I B S
Aplikasi MB pada reservoir oil biasanya digunakan untuk memperkirakan atau menghitung / memperkirakan: –
Volume oil in place
–
Oil reserves & oil recovery factor
–
•
•
Perilaku reservoir dimasa mendatang pada berbagai macam mekanisme pendorong reservoir
Data yang diperlukan: –
Data produksi dan tekanan reservoir
–
Data sifat fisik batuan dan fluida reservoir
Akurasi perhitungan tergantung kualitas dan banyaknya data dan waktu saat evaluasi dilakukan (sebaiknya setelah tekanan reservoir turun > 30% dari tekanan awal atau telah berproduksi > 15% dari volume awal)
Material Balance: Definitions of Variables 4 1 0 2 @ G N U D N A B S N I A S N A D I G O L O N K E T T U T I T S N I B S
Production data N p N G p G i W p W i W e R p
= Cumulative oil produced (stb) = initial oil in place (stb) = cumulative gas produced (scf) = cumulative gas injected (scf) = Cumulative water produced (stb) = Cumulative water injected (stb) = Cumulative water influx (stb) = G p /N p = Cumulative produced gas-oil ratio (scf/stb)
Reservoir Data p i p ∆P S wc c r c w V b V p S o S g S w m φ
= Initial mean pressure in the reser voir (psi) = current mean pressure in the reservoir, (psi) = P2-P1 = connate water saturation, (fraction) = Compressibility of formation (psi -1) = Compressibility of water (psi -1) = bulk volume (bbl) = pore volume (bbl) = oil saturation, fraction = gas saturation, fraction = water saturation, fraction = ratio initial gas cap vol to vol. of oil zone = porosity, fraction
Fluid PVT Data B gi B g B oi B o B wi B w c w c r R si R s Z
= Initial gas volume factor at p i (ft3 /scf) = Gas volume factor at current pressure p (ft 3 /scf) = Initial oil volume factor at p i (rb/stb) = Oil volume factor at current pressure p (rb/stb) = initial formation volume factor of water (rb/stb) = Formation volume factor of water at current pressure p (rb/stb) = Compressibility of water (psi -1) = Compressibility of reser voir rock (psi -1) = solution gas-oil ratio at initial pressure p i (scf/stb) = solution gas-oil ratio at current pressure p (scf/stb) = compressibility factor, fraction
Sifat Fisik Minyak (Black Oil) 4 1 0 2
Bo
Rs
Bw
Bg
@ G N U D N A B S N I A S N A D I G O L O N K E T T U T I T S N I B S
Material Balance: Oil Reservoir 4 1 0 2
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@ G N U D N A B S N I A S N A D I G O L O N K E T T U T I T S N I B S
Aplikasi MB pada reservoir oil biasanya digunakan untuk memperkirakan atau menghitung / memperkirakan: –
Volume oil in place
–
Oil reserves & oil recovery factor
–
•
•
Perilaku reservoir dimasa mendatang pada berbagai macam mekanisme pendorong reservoir
Data yang diperlukan: –
Data produksi dan tekanan reservoir
–
Data sifat fisik batuan dan fluida reservoir
Akurasi perhitungan tergantung kualitas dan banyaknya data dan waktu saat evaluasi dilakukan (sebaiknya setelah tekanan reservoir turun > 30% dari tekanan awal atau telah berproduksi > 5% dari volume awal)
Bentuk Umum Persamaan Material Balance dari Reservoir Minyak 4 1 0 2
Kondisi awal, Pi Gascap Gas
@
mNBoi
G N U D N A B S N I A S
(r-bbl)
N A D I G O L O N K E T T U T I T S N I B S
Kondisi setelah terjadi penurunan tekanan, P = P i – P
B
Oil + Originally dissolved gas
C NBoi
A
(r-bbl)
Perubahan volume di reservoir akibat penurunan tekanan reservoir, DP
A = ekspansi volume minyak dan gas yang terlarut didalamnya B = ekspansi volume gascap gas C = Perubahan volume pori-pori HC karena ekspansi connate water dan perubahan volume pori karena adanya kompresibilitas formasi Underground withdrawal (r-bbl)
=
ekspansi volume minyak dan gas yang terlarut didalamnya (r-bbl)
+
ekspansi gascap gas (r-bbl)
+
pengurangan volume pori-pori HC karena ekspansi connate water dan penurunan volume pori (r-bbl)
Bentuk Umum Persamaan Material Balance dari Reservoir Minyak 4 1 0 2 @ G N U D N A B S N I A S N A D I G O L O N K E T T U T I T S N I B S
Kondisiawal, Pi
Kondisi setelah terjadi penurunan tekanan, P = Pi – P
Gascap Gas
mNBoi
B
IOIP = N =
V ⋅ φ ⋅ (1 − S wc )
Oil + Originally dissolved gas
C NBoi
A
(r-bbl)
( stb)
Boi
(r-bbl)
IOIP = N ⋅ Boi (r - bbl) m=
volume gascap awal (r - bbl)
=
volume minyak awal (r - bbl) G ⋅ Bgi
= m ⋅ N ⋅ Boi
G ⋅ Bgi N ⋅ Boi
(r - bbl)
A = ekspansi volume minyak dan gas yang terlarut didalamnya •
Ekspansi minyak = N ⋅ ( Bo − Boi ) (r - bbl)
•
Ekspansi gas yang terbebas dari minyak = N ⋅ ( Rsi − Rs ) ⋅ Bg (r - bbl) A = N ⋅ ( Bo − Boi ) + N ⋅ ( Rsi − Rs ) ⋅ Bg (r - bbl)
Bo
Rs
Bo1
Rs1
Bo2
Rs2
, maka
Bentuk Umum Persamaan Material Balance dari Reservoir Minyak 4 1 0 2 @ G N U D N A B S N I A S N A D I G O L O N K E T T U T I T S N I
Kondisiawal, Pi
Kondisi setelah terjadi penurunan tekanan, P = Pi – P
IOIP = N =
Gascap Gas
mNBoi
B
V ⋅ φ ⋅ (1 − S wc )
Oil + Originally dissolved gas
C NBoi
( stb)
Boi
(r-bbl)
IOIP = N ⋅ Boi (r - bbl)
A
(r-bbl)
m=
volume gascap awal (r - bbl) volume minyak awal (r - bbl)
G ⋅ Bgi
= m ⋅ N ⋅ Boi
=
G ⋅ Bgi
, maka
N ⋅ Boi
(r - bbl)
B = ekspansi volume gascap gas •
Volume gascap awal = G =
m ⋅ N ⋅ Boi
(scf)
atau
Bgi •
G ⋅ Bgi
= m ⋅ N ⋅ Boi
Volume gascap karena penurunan tekanan ∆P = G ⋅ Bg = m ⋅ N ⋅ Boi B = m ⋅ N ⋅ Boi
Bg Bgi
Bg − m ⋅ N ⋅ Boi = m ⋅ N ⋅ Boi − 1 B gi
Bg (r - bbl)
-
Bg2
B S
Bg1
(r - bbl)
Bg Bgi
(r - bbl)
Bentuk Umum Persamaan Material Balance dari Reservoir Minyak 4 1 0 2 @ G N U D N A B S N I A S N A D I G O L O N K E T T U T I T S N I B S
Kondisiawal, Pi
Kondisi setelah terjadi penurunan tekanan, P = Pi – P
IOIP = N ⋅ Boi (r - bbl)
Gascap Gas
mNBoi
B
(r-bbl)
IGIP = G ⋅ Bgi
Oil + Originally dissolved gas
C NBoi
A
= m ⋅ N ⋅ Boi
(r - bbl)
HCPV = V ⋅ φ ⋅ (1 − S wc ) = IOIP + IGIP
(r-bbl)
HCPV = N ⋅ Boi
+ m ⋅ N ⋅ Boi
(r - bbl)
(r - bbl)
HCPV = N ⋅ Boi ⋅ (1 + m ) (r - bbl)
C = Perubahan/pengurangan volume pori-pori HC •
Ekspansi volume connate water ( ∆Vw)
•
Perubahan/penurunan volume pori akibat compressibilitas formasi ( ∆Vf) C = ∆( HCPV ) = − ∆V w + ∆V f
INGAT !!!
C = ∆( HCPV ) = + cw ⋅V w ⋅ ∆P + c f ⋅ V f ⋅ ∆P
C = ∆( HCPV ) = cw ⋅
HCPV
(1 − S wc )
cw ⋅ S wc + c f ∆P ) (1 S )
C = N ⋅ Boi ⋅ 1 + m
(
⋅ S wc ⋅ ∆P + c f ⋅
HCPV = V ⋅ φ ⋅ (1 − S wc )
∆V V ∂P V ∆P V ⋅ φ ⋅ (1 − S wc ) HCPV = V ⋅ φ = = (1 − S wc ) (1 − S wc )
c=−
⋅ ∆P (1 − S wc ) HCPV
(r - bbl)
V f
V w
1 ∂V
=−
= V f ⋅ S wc =
1
HCPV
⋅ S wc
Bentuk Umum Persamaan Material Balance dari Reservoir Minyak 4 1 0 2 @ G N U D N A B S N I A S N A D I G O L O N K E T T U T I T S N I
Kondisiawal, Pi
Kondisi setelah terjadi penurunan tekanan, P = Pi – P
Gascap Gas
mNBoi
B
(r-bbl) Oil + Originally dissolved gas
C NBoi
= N p + G p = N p + N p ⋅ R p
free gas production = ( N p ⋅ R p ) − ( N p ⋅ Rs ) (scf)
(r-bbl)
free gas production = N p ( R p − Rs )⋅ Bg (r - bbl)
D = Underground withdrawal (volume pengurasan reservoir) •
Volume karena produksi minyak dan gas terlarut dalam satuan reservoir (r-bbl)
•
Volume karena produksi gas bebas (free gas) dalam satuan reservoir (r-bbl) Volume karena produksi minyak dan gas terlarut dalam satuan reser voir (r-bbl)
= N p ⋅ Bo
Volume karena produksi gas bebas (free gas) dalam satuan reservoir (r-bbl)
= N p R p − Rs ⋅ Bg
D = N p ⋅ Bo + N p R p − Rs ⋅ Bg
dimana R p
=
G p N p
free gas production = total gas production - solution gas production
A
B S
Total produksi minyak dan gas di permukaan
(r - bbl)
(r - bbl)
(r - bbl)
Bentuk Umum Persamaan Material Balance dari Reservoir Minyak 4 1 0 2 @ G N U D N A B S N I A S N A D I G O L O N K E T T U T I T S N I B S
Kondisiawal, Pi
Kondisi setelah terjadi penurunan tekanan, P = Pi – P
A = N ⋅ ( Bo − Boi ) + N ⋅ ( Rsi − Rs ) ⋅ Bg (r - bbl)
Gascap Gas
mNBoi
B
(r-bbl) Oil + Originally dissolved gas
C NBoi
B = m ⋅ N ⋅ Boi
A
Bg Bgi
(r-bbl)
B − m ⋅ N ⋅ Boi = m ⋅ N ⋅ Boi g − 1 B gi cw ⋅ S wc + c f ∆P (1 − S wc )
C = N ⋅ Boi ⋅ 1 + m
(
)
D = N p ⋅ Bo + N p R p − Rs ⋅ Bg Underground withdrawal (r-bbl)
ekspansi volume minyak dan gas yang terlarut didalamnya (r-bbl)
=
+
ekspansi gascap gas (r-bbl)
+
(r - bbl)
(r - bbl)
(r - bbl)
pengurangan volume pori-pori HC karena ekspansi connate water dan penurunan volume pori (r-bbl)
D = A + B + C
(
)
N p ⋅ Bo + N p ⋅ R p − Rs ⋅ Bg
Bg cw ⋅ S wc + c f ∆P = N ⋅ ( Bo − Boi ) + N ⋅ ( Rsi − Rs ) ⋅ Bg + m ⋅ N ⋅ Boi − 1 + N ⋅ Boi ⋅ (1 + m) B (1 − S ) wc gi
( B − B ) + ( R − R ) ⋅ B Bg cw ⋅ S wc + c f o oi si s g ∆P N p ⋅ ( Bo + ( R p − Rs )⋅ Bg ) = N ⋅ Boi + m⋅ − 1 + (1 + m) Boi (1 − S wc ) Bgi
Bentuk Umum Persamaan Material Balance dari Reservoir Minyak 4 1 0 2 @ G N U D N A B S N I A S N A D I G O L O N K E T T U T I T S N I B S
Kondisiawal, Pi
Kondisi setelah terjadi penurunan tekanan, P = Pi – P
A = N ⋅ ( Bo − Boi ) + N ⋅ ( Rsi − Rs ) ⋅ Bg (r - bbl)
Gascap Gas
mNBoi
B
(r-bbl) Oil + Originally dissolved gas
C NBoi
B = m ⋅ N ⋅ Boi
A
Bg Bgi
(r-bbl)
B − m ⋅ N ⋅ Boi = m ⋅ N ⋅ Boi g − 1 B gi cw ⋅ S wc + c f ∆P (1 − S wc )
C = N ⋅ Boi ⋅ 1 + m
(
Underground withdrawal (r-bbl)
=
ekspansi volume minyak dan gas yang terlarut didalamnya (r-bbl)
+
)
D = N p ⋅ Bo + N p R p − Rs ⋅ Bg
(r - bbl)
E = W e ⋅ Bw − W p ⋅ Bw
⋅ Bw
ekspansi gascap gas (r-bbl)
+
=
W e − W p
(r - bbl)
(r - bbl)
(r - bbl)
pengurangan volume poripori HC karena ekspansi connate water dan penurunan volume pori (r-bbl)
+
Net water influx(r-bbl)
( B − B ) + ( R − R ) ⋅ B Bg cw ⋅ S wc + c f o oi si s g ∆P + (W e − W p ) ⋅ Bw N p ⋅ ( Bo + ( R p − Rs )⋅ Bg ) = N ⋅ Boi + m⋅ − 1 + (1 + m) (1 − S ) Boi wc Bgi
Linierisasi Bentuk Persamaan Umum MB 4 1 0 2
F = N p ⋅ Bo
@ G N U D N A B S N I A S N A D I G O L O N K E T T U T I T S N I
+ R p − Rs ⋅ Bg + W p ⋅ Bw
E o
= ( Bo − Boi ) + ( Rsi − Rs ) ⋅ Bg
E g
Bg = Boi − 1 B gi
E f , w
Underground
r - bbl stb
cw ⋅ S wc + c f ∆P oi ⋅ (1 − S ) wc
= (1 + m )⋅ B
withdrawal
ekspansi
volume minyak dan gas yang terlarut didalamnya
r - bbl stb
ekspansi
r - bbl stb
gascap gas
pengurangan
volume pori-pori HC karena ekspansi connate water dan penurunan volume pori
Bentuk umum persamaan MB: ( B − B ) + ( R − R ) ⋅ B Bg cw ⋅ S wc + c f o oi si s g ∆P + (W e − W p ) ⋅ Bw N p ⋅ ( Bo + ( R p − Rs )⋅ Bg ) = N ⋅ Boi + m⋅ − 1 + (1 + m) Boi (1 − S wc ) Bgi Menjadi: Net F = N ⋅ E o
+ m ⋅ E g + E f ,w + W e ⋅ Bw
B S
(r - bbl)
water
water influx
Pers. MB untuk Solution Gas Drive Reservoir 4 1 0 2 @ G N U D N A B S N I A S
Bentuk umum persamaan MB: ( B − B ) + ( R − R ) ⋅ B Bg cw ⋅ S wc + c f o oi si s g ∆P + W e ⋅ Bw N p ⋅ ( Bo + ( R p − Rs )⋅ Bg ) + W p ⋅ Bw = N ⋅ Boi + m⋅ − 1 + (1 + m) B (1 − S ) Boi wc gi oil + gas production
water production
oil + dissolved gas expansion
gas cap expansion
change of HC pore volume
Eo
Eg
Ef,w
F
N A D I G O L O N K E T T U T I T S N I
Above bubble point (P > Pb)
-
( B − B ) cw ⋅ S wc + c f oi ∆P + N p ⋅ Bo = N ⋅ Boi o ( )
B S
•
•
•
water influx
Karena P > Pb, maka no gas cap, m = 0 Karena P > Pb, maka no free gas prod., R p − Rs ⋅ Bg
=0
Karena P > Pb, maka no dissolved gas escape from oil solution, ( Rsi
Pers. MB menjadi:
− Rs ) = 0 F
( B − B ) cw ⋅ S wc + c f oi ∆P + W e ⋅ Bw N p ⋅ Bo + W p ⋅ Bw = N ⋅ Boi o + Boi (1 − S wc )
N
For the case no water influx F = N ⋅ E o
+ E w, f
E o+E w,f
Pers. MB untuk Solution Gas Drive Reservoir 4 1 0 2 @ G N U D N A B S N I A S N A D I G O L O N K E T T U T I T S N I B S
Pers. MB menjadi: N p ⋅ Bo
F
( B − B ) c ⋅ S + c + W p ⋅ Bw = N ⋅ Boi o oi + w wc f ∆P + W e ⋅ Bw Boi (1 − S wc )
N
For the case no water influx
( B − B ) cw ⋅ S wc + c f oi ∆P N p ⋅ Bo = N ⋅ Boi o + Boi (1 − S wc )
F = N ⋅ E o
E o+E w,f
+ E w, f
Dalam bentuk lain: co
1 ∂ Bo
=−
Bo
∂P
=
1 Bo
∆ Bo 1 ( Bo − Boi ) = ∆P Boi ∆P
cw ⋅ S wc + c f ∆P N p ⋅ Bo = N ⋅ Boi C o ⋅ ∆P + (1 − S wc ) N p ⋅ Bo
Karena
S o
= 1 − S wc
maka,
C ⋅ S + c ⋅ S + c = N ⋅ Boi o o w wc f ⋅ ∆P (1 − S wc )
N p ⋅ Bo
= N ⋅ Boi ⋅ ce ⋅ ∆P
dimana,
co ⋅ S o + cw ⋅ S wc + c f ce = (1 − S wc )
Effective saturation weighted compressibility of the
Pers. MB untuk Solution Gas Drive Reservoir 4 1 0 2 @ G N U D N A B S N I A S N A D I G O L O N K E T T U T I T S N I B S
Bentuk umum persamaan MB: ( B − B ) + ( R − R ) ⋅ B Bg cw ⋅ S wc + c f o oi si s g ∆P + W e ⋅ Bw N p ⋅ ( Bo + ( R p − Rs )⋅ Bg ) + W p ⋅ Bw = N ⋅ Boi + m⋅ − 1 + (1 + m) B (1 − S ) Boi wc gi oil + gas production
water production
F
oil + dissolved gas expansion
gas cap expansion
change of HC pore volume
Eo
Eg
Ef,w
Below bubble point (P < Pb) •
•
Karena P < Pb, maka gas escapes from solution and free gas saturation formed in the reservoir Free gas compressibility is way higher than formation or water (50-100X), so that c w,f usually neglected
Pers. MB menjadi:
( B − B ) + ( R − R ) ⋅ B Bg o oi si s g N p ⋅ ( Bo + ( R p − Rs )⋅ Bg ) + W p ⋅ Bw = N ⋅ Boi + m ⋅ − 1 + W e ⋅ Bw Boi Bgi
water influx
Pers. MB untuk Solution Gas Drive Reservoir 4 1 0 2 @ G N U D N A B S N I A S N A D I G O L O N K E T T U T I T S N I B S
For the case no water influx and no initial gas cap gas: Pers. MB menjadi: N p ⋅ Bo N p
=
N
Jadi
+ R p − Rs ⋅ Bg = N ⋅ ( Bo − Boi ) + ( Rsi − Rs ) ⋅ Bg
( Bo − Boi ) + ( Rsi − Rs ) ⋅ Bg Bo + ( R p − Rs )⋅ Bg N p N
RF
= RF ≈
1 R p
N p
dimana
dimana
= RF = Recovery Factor
N R p
=
G p
= cumulative prod. gas oil ratio
N p
Recovery factor (RF) berbanding terbalik terhadap cumulative production gas oil ratio. Ini berarti bahwa semakin banyak gas yang terproduksi akan memperkecil RF. Oleh karena itu, unruk mendapatkan RF yang optimum sebaiknya menjaga gas sebanyak mungkin tetap berada di reservoir.
Pers. MB untuk Gas Cap Drive Reservoir 4 1 0 2 @ G N U D N A B S N I A S N A D I G O L O N K E T T U T I T S N I B S
Bentuk umum persamaan MB: ( B − B ) + ( R − R ) ⋅ B Bg cw ⋅ S wc + c f o oi si s g ∆P + W e ⋅ Bw N p ⋅ ( Bo + ( R p − Rs )⋅ Bg ) + W p ⋅ Bw = N ⋅ Boi + m⋅ − 1 + (1 + m) B (1 − S ) Boi wc gi oil + gas production
water production
F
oil + dissolved gas expansion
gas cap expansion
change of HC pore volume
Eo
Eg
Ef,w
Assumptions •
Gas cap gas compressibility is way higher than formation or water (50-100X), so that c w,f usually neglected
For the case no water influx : Pers. MB menjadi:
( B − B ) + ( R − R ) ⋅ B Bg o oi si s g N p ⋅ ( Bo + ( R p − Rs )⋅ Bg ) = N ⋅ Boi + m ⋅ − 1 B Boi gi oil + gas production (withdrawal rate)
Oil & dissolved gas expansion
Gascap gas expansion
water influx
Pers. MB untuk Gas Cap Drive Reservoir 4 1 0 2 @ G N U D N A B S N I A S N A D I G O L O N K E T T U T I T S N I B S
For the case no water influx : Pers. MB menjadi:
( B − B ) + ( R − R ) ⋅ B Bg o oi si s g N p ⋅ ( Bo + ( R p − Rs )⋅ Bg ) = N ⋅ Boi + m ⋅ − 1 B Boi gi Pers. di atas dalam bentuk lain: F = N ⋅ E o
F
+ m ⋅ E g
N E o+m.E g