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Mata Kuliah
Teknik Reservoir I Dr. Ir. Wahju Wibowo
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Daftar Pustaka • Craft, B. C. and Hawkins, M. F.: Applied Petroleum Reservoir Engineering, Revised edition by R. E. Terry, Prentice-Hall, Inc., Englewood Cliffs, NJ, 1991. • Dake, L. P.: Fundamentals of Reservoir Engineering, Elsevier Science Publisher B. V., Amsterdam, the Netherlands, 1978. • Dake, L. P.: The Practice of Reservoir Engineering, Revised Edition, Elsevier Science Publisher B. V., Amsterdam, the Netherlands, 2001. • Ahmed, T.: Reservoir Engineering Handbook, Fourth Edition, Gulf Professional Publishing, United States of America, 2010.
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Ekspektasi • • • • • •
Hadir disetiap jam perkuliahan Menyimak dan memahami materi perkuliahan Bertanya jika tidak mengerti Mengerjakan tugas atau pekerjaan rumah Mengerjakan ujian yang dilaksanakan (Quiz, UTS, UAS) Komunikasi dilakukan melalui Ketua Kelas
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Topik Bahasan • Pendahuluan • Review Sifat Fisik Fluida • Review Sifat Fisik Batuan • Cadangan Migas • Mekanisme Pendorong Reservoir • Persamaan Aliran dan Aplikasinya • Konsep Produktivitas • Analisis Decline Curve • Persamaan Kesetimbangan Materi
Mata Kuliah Teknik Reservoir I
Analisis Kesetimbangan Materi (Material Balance Analysis) Dr. Ir. Wahju Wibowo
• 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 Pi
Kondisi saat pengamatan P(t)
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Pendahuluan
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Material Balance: Definitions of Variables Production data Gp Wp
= cumulative gas produced (scf) = Cumulative water produced (stb)
Reservoir Data pi p Swc cr
= Initial mean pressure in the reservoir (psi) = current mean pressure in the reservoir, (psi) = connate water saturation, (fraction) = Compressibility of reservoir rock (psi-1)
Fluid PVT Data Bgi Bg cw Bw Z Zi
= Initial gas volume factor at pi (ft3/scf) = Gas volume factor at current pressure p (ft3/scf) = Compressibility of water (psi-1) = Formation volume factor of water at current pressure p (rb/stb) = compressibility factor, fraction = initial compressibility factor, fraction
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Material Balance: Gas Reservoir •
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)
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Material Balance: Gas Reservoir 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.
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Material Balance: Volumetric Dry Gas Reservoir G p Volume Terproduksi
G ⋅ Bgi = (G − G p )⋅ Bg
G ⋅ Bgi
(G − G )⋅ B
Volume Awal
Volume Tersisa
Kondisi Awal Pi
Kondisi setelah berproduksi P(t)
T ⋅ Zi 0.0282 ⋅ Pi G p = G ⋅ 1 − T ⋅Z 0.0282 ⋅ P
Gp G
G p ⋅ Bg = G ⋅ Bg − G ⋅ Bgi
= 1−
p
= G ⋅ 1 −
g
Zi Pi Z P
Bgi Gp = G − G ⋅ B g Bgi G p = G ⋅ 1 − B g = G ⋅ 1 − P Z i Z P i
Gp P Zi P Zi P Pi Pi G p = 1 − = − Z Pi Z Pi G Z Zi Zi G
RF =
ft 3 T ⋅Z , Bg = 0.0282 ⋅ scf P
P Zi = 1 − G Z Pi
Gp
P Pi Pi G p = − Z Zi Zi G
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Material Balance: Volumetric Dry Gas Reservoir Persamaan MB untuk volumetric dry gas reservoir P Pi Pi G p = − Z Zi Zi G P G p Pi P =− i + Z Zi G Zi
Y = −m ⋅ X + C P Y= Z P m= i Zi ⋅ G
Pi Zi
slope = −m = −
P Z
Pi Zi ⋅ G
X = Gp C=
Pi Zi
G Gp
= Gas In Place
Persamaan MB untuk volumetric dry gas reservoir P G p Pi P =− i + Z Zi G Zi
Pi Zi slope = −m = −
Pi Zi ⋅ G
• Kondisi awal, Gp = 0, memberikan
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Pi Zi
• Gas in place dapat dihitung dengan mengetahui: • Perpotongan pada sumbu Gp pada saat
P Z
P =0 Z
• Slope atau garis miring - m
G = Gas In Place
Gp Note….
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Material Balance: Volumetric Dry Gas Reservoir
Gas In Place = G = 43560 ⋅
A ⋅ h ⋅ φ ⋅ (1 − S wi ) Bgi
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Material Balance: Volumetric Dry Gas Reservoir Step by step 1.
Pengumpulan data yang diperlukan: –
Data tekanan reservoir
–
Data produksi sumur / reservoir
–
Data sifat fisik dan batuan reservoir
2.
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
7.
Perkirakan harga ultimate recovery @ tekanan abandonment
8.
Hitung parameter lain yang diinginkan
P Z
Pi Zi slope = −m = −
Pi Zi ⋅ G
G = Gas In Place
Pa Za remaining reserves cum. prod. gas
ultimate recovery
Note: Pa = Tekanan abandonment Za = z-factor pada tekanan abandonment
Gp
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Material Balance: Volumetric Dry Gas Reservoir Contoh Soal – 1 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
•
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
Hitung P/Z berdasarkan data produksi yang diketahui dan plot P/Z vs. Gp
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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
P/Z, psi
Reservoir Cum. Gas Pressure (Pr), Prod. (Gp), psi mmscf
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Material Balance: Volumetric Dry Gas Reservoir Contoh Soal – 1
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
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Material Balance: Volumetric Dry Gas Reservoir Contoh Soal – 2 A volumetric gas reservoir has the following production history.
The following data are also available: φ = 13%
Swi = 0.52
h = 54 ft
T = 164oF
•
A = 1060 Acres
Calculate the gas initially in place volumetrically and from the MBE
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Step – 1. Calculate
Bgi
(164 + 460) ⋅ 0.869 = 0.00853 = 0.02827 ⋅
ft 3
1798
scf
Step – 2. Calculate the gas initially in place volumetrically A ⋅ h ⋅ φ ⋅ (1 − S wi ) 1060 ⋅ 54 ⋅ 0.13 ⋅ (1 − 0.52 ) G = 43560 ⋅ = 43560 ⋅ Bgi 0.00853
G = 18239925270.8 scf = 18.24 bcf Step – 3. Plot P/z vs. Gp
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
P/Z, psi
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Material Balance: Volumetric Dry Gas Reservoir Contoh Soal – 2
1000
G = 14.2 bscf
500
0 0
10
Cum. Gas Production (Gp), bscf
20
Material Balance: Volumetric Dry Gas Reservoir dengan Water Influx
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(G − G )⋅ B
G ⋅ Bgi
p
Volume Awal Kondisi Awal Pi
∆V p
We
Kondisi setelah berproduksi P(t)
G ⋅ Bgi = (G − G p )⋅ Bg + ∆V p
G ⋅ Bgi + W p ⋅ Bw = G ⋅ Bg − G p ⋅ Bg + We G p ⋅ Bg + W p ⋅ Bw = G ⋅ Bg − G ⋅ Bgi + We G p ⋅ Bg + W p ⋅ Bw = G ⋅ (Bg − Bgi ) + We
(B
g
− Bgi )
=G+
(B
We g
G = Gas In Place
→ ∆V p = We − W p ⋅ Bw
G ⋅ Bgi = (G − G p )⋅ Bg + We − W p ⋅ Bw
G p ⋅ Bg + W p ⋅ Bw
g
p
Volume gas Tersisa
⋅ Bg + W p ⋅ Bw ) (Bg − Bgi )
G p & Wp
(G
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Volume Terproduksi
− Bgi )
We (Bg − Bgi )
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Material Balance: Oil Reservoir •
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)
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Material Balance: Definitions of Variables Production data Np N Gp Gi Wp Wi We Rp
= 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) = Gp/Np = Cumulative produced gas-oil ratio (scf/stb)
Reservoir Data pi p ∆P Swc cr cw Vb Vp So Sg Sw m
φ
= Initial mean pressure in the reservoir (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 Bgi Bg Boi Bo Bwi Bw
cw cr Rsi Rs Z
= Initial gas volume factor at pi (ft3/scf) = Gas volume factor at current pressure p (ft3/scf) = Initial oil volume factor at pi (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 reservoir rock (psi-1) = solution gas-oil ratio at initial pressure pi (scf/stb) = solution gas-oil ratio at current pressure p (scf/stb) = compressibility factor, fraction
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Sifat Fisik Minyak (Black Oil) Bo Rs
Bw Bg
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Material Balance: Oil Reservoir •
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)
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(r-bbl)
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Kondisi awal, Pi
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Bentuk Umum Persamaan Material Balance dari Reservoir Minyak Kondisi setelah terjadi penurunan tekanan, P = Pi – ∆P
Gascap Gas
B
mNBoi 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)
Kondisi awal, Pi
Kondisi setelah terjadi penurunan tekanan, P = Pi – ∆P
Gascap Gas
B
mNBoi (r-bbl)
Oil + Originally dissolved gas
C
NBoi
A
(r-bbl)
IOIP = N =
V ⋅ φ ⋅ (1 − S wc )
( stb)
Boi IOIP = N ⋅ Boi (r - bbl)
m=
volume gascap awal (r - bbl)
=
volume minyak awal (r - bbl)
G ⋅ Bgi
, maka
N ⋅ Boi
G ⋅ Bgi = m ⋅ 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
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Bentuk Umum Persamaan Material Balance dari Reservoir Minyak
P2
P1
P
P2
P1
P
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Bentuk Umum Persamaan Material Balance dari Reservoir Minyak Kondisi awal, Pi
Kondisi setelah terjadi penurunan tekanan, P = Pi – ∆P
IOIP = N =
Gascap Gas
B
mNBoi (r-bbl)
C
( stb)
Boi
Oil + Originally dissolved gas
NBoi
V ⋅ φ ⋅ (1 − S wc )
IOIP = N ⋅ Boi (r - bbl)
A
(r-bbl)
m=
volume gascap awal (r - bbl)
=
volume minyak awal (r - bbl)
G ⋅ Bgi
, maka
N ⋅ Boi
G ⋅ Bgi = m ⋅ N ⋅ Boi (r - bbl)
B = ekspansi volume gascap gas •
Volume gascap awal = G =
m ⋅ N ⋅ Boi
(scf)
atau
Bgi
•
G ⋅ Bgi = m ⋅ N ⋅ Boi (r - bbl)
Volume gascap karena penurunan tekanan ∆P = G ⋅ Bg = m ⋅ N ⋅ Boi B g B = m ⋅ N ⋅ Boi − m ⋅ N ⋅ Boi = m ⋅ N ⋅ Boi − 1 (r - bbl) Bgi Bgi Bg
Bg
Bg
(r - bbl)
Bgi
Bg2 Bg1 P2
P1
P
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INSTITUT TEKNOLOGI DAN SAINS BANDUNG
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Bentuk Umum Persamaan Material Balance dari Reservoir Minyak Kondisi awal, Pi
Kondisi setelah terjadi penurunan tekanan, P = Pi – ∆P
IOIP = N ⋅ Boi (r - bbl)
Gascap Gas
B
mNBoi (r-bbl)
IGIP = G ⋅ Bgi = m ⋅ N ⋅ Boi (r - bbl)
Oil + Originally dissolved gas
C
NBoi
HCPV = V ⋅ φ ⋅ (1 − S wc ) = IOIP + IGIP (r - bbl)
A
(r-bbl)
HCPV = N ⋅ Boi + m ⋅ N ⋅ Boi
(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 ) = −(∆Vw + ∆V f )
INGAT !!!
C = ∆(HCPV ) = +(cw ⋅Vw ⋅ ∆P + c f ⋅ V f ⋅ ∆P )
c=−
HCPV HCPV C = ∆(HCPV ) = cw ⋅ ⋅ S wc ⋅ ∆P + c f ⋅ ⋅ ∆P (1 − S wc ) (1 − S wc ) cw ⋅ S wc + c f C = N ⋅ Boi ⋅ (1 + m ) (1 − S ) wc
∆P
HCPV = V ⋅ φ ⋅ (1 − S wc )
(r - bbl)
1 ∂V V ∂P
V f = V ⋅φ =
=−
1 ∆V V ∆P
V ⋅ φ ⋅ (1 − S wc )
(1 − S wc )
Vw = V f ⋅ S wc =
HCPV
(1 − S wc )
=
HCPV
(1 − S wc )
⋅ S wc
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INSTITUT TEKNOLOGI DAN SAINS BANDUNG
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Bentuk Umum Persamaan Material Balance dari Reservoir Minyak Kondisi awal, Pi
Kondisi setelah terjadi penurunan tekanan, P = Pi – ∆P
Gascap Gas
B
mNBoi (r-bbl)
Oil + Originally dissolved gas
C
NBoi
Total produksi minyak = N + G = N + N ⋅ R dimana R = G p p p p p p p dan gas di permukaan Np
free gas production = total gas production - solution gas production
A
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 reservoir (r-bbl)
= N p ⋅ Bo (r - bbl)
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
(r - bbl)
(r - bbl)
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Bentuk Umum Persamaan Material Balance dari Reservoir Minyak Kondisi awal, Pi
Kondisi setelah terjadi penurunan tekanan, P = Pi – ∆P
A = N ⋅ (Bo − Boi ) + N ⋅ (Rsi − Rs ) ⋅ Bg (r - bbl)
Gascap Gas
B
mNBoi (r-bbl)
B g B = m ⋅ N ⋅ Boi − m ⋅ N ⋅ Boi = m ⋅ N ⋅ Boi − 1 (r - bbl) Bgi Bgi cw ⋅ S wc + c f ∆P (r - bbl) C = N ⋅ Boi ⋅ (1 + m ) (1 − S ) wc
Oil + Originally dissolved gas
C
NBoi
Bg
A
(r-bbl)
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)
pengurangan volume pori-pori HC karena ekspansi connate water dan penurunan volume pori (r-bbl)
D = A+ B +C B cw ⋅ S wc + c f g N p ⋅ Bo + N p ⋅ (R p − Rs )⋅ Bg = 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 o oi si s g N p ⋅ (Bo + (R p − Rs )⋅ Bg ) = N ⋅ Boi Boi
+ m ⋅ Bg − 1 + (1 + m ) cw ⋅ S wc + c f B (1 − S ) wc gi
∆P
∆P
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Bentuk Umum Persamaan Material Balance dari Reservoir Minyak Kondisi setelah terjadi penurunan tekanan, P = Pi – ∆P
Kondisi awal, Pi Gascap Gas
B
mNBoi (r-bbl)
B g B = m ⋅ N ⋅ Boi − m ⋅ N ⋅ Boi = m ⋅ N ⋅ Boi − 1 (r - bbl) Bgi Bgi cw ⋅ S wc + c f ∆P (r - bbl) C = N ⋅ Boi ⋅ (1 + m ) (1 − S ) wc
Oil + Originally dissolved gas
C
NBoi
A = N ⋅ (Bo − Boi ) + N ⋅ (Rsi − Rs ) ⋅ Bg (r - bbl) Bg
A
(r-bbl)
D = N p ⋅ Bo + N p (R p − Rs )⋅ Bg
(r - bbl)
E = We ⋅ Bw − W p ⋅ Bw = (We − W p ) ⋅ Bw (r - bbl) Underground withdrawal (r-bbl)
=
ekspansi volume minyak dan gas yang terlarut didalamnya (r-bbl)
+
ekspansi gascap gas (r-bbl)
(B − B ) + (R − R ) ⋅ B o oi si s g N p ⋅ (Bo + (R p − Rs )⋅ Bg ) = N ⋅ Boi Boi
+
pengurangan volume poripori HC karena ekspansi connate water dan penurunan volume pori (r-bbl)
B c S c + m ⋅ g − 1 + (1 + m ) w ⋅ wc + f B (1 − S ) wc gi
+
Net water influx(r-bbl)
∆P + (W − W ) ⋅ B e p w
F = N p ⋅ (Bo + (R p − Rs )⋅ Bg ) + W p ⋅ Bw Eo = (Bo − Boi ) + (Rsi − Rs ) ⋅ Bg B g E g = Boi − 1 B gi
ITSB
Underground withdrawal
(r - bbl)
r - bbl stb
ekspansi volume minyak dan gas yang terlarut didalamnya
r - bbl stb
cw ⋅ S wc + c f E f , w = (1 + m ) ⋅ Boi ⋅ (1 − S ) wc
ekspansi gascap gas ∆P
r - bbl stb
pengurangan volume pori-pori HC karena ekspansi connate water dan penurunan volume pori
Bentuk umum persamaan MB: (B − B ) + (R − R ) ⋅ B o oi si s g N p ⋅ (Bo + (R p − Rs )⋅ Bg ) = N ⋅ Boi Boi
Menjadi: F = N ⋅ (Eo + m ⋅ E g + E f , w ) + We ⋅ Bw
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INSTITUT TEKNOLOGI DAN SAINS BANDUNG
@ 2014
Linierisasi Bentuk Persamaan Umum MB
water influx
B c S c + m ⋅ g − 1 + (1 + m ) w ⋅ wc + f B (1 − S ) wc gi
∆P + (W − W ) ⋅ B e p w Net water influx
ITSB
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INSTITUT TEKNOLOGI DAN SAINS BANDUNG
@ 2014
Pers. MB untuk Solution Gas Drive Reservoir Bentuk umum persamaan MB: (B − B ) + (R − R ) ⋅ B o oi si s g N p ⋅ (Bo + (R p − Rs )⋅ Bg ) + W p ⋅ Bw = N ⋅ Boi Boi oil + gas production
water production
B c S c + m ⋅ g − 1 + (1 + m ) w ⋅ wc + f B (1 − S ) wc gi
oil + dissolved gas expansion
gas cap expansion
change of HC pore volume
Eo
Eg
Ef,w
F
∆P + W ⋅ B e w water influx
Above bubble point (P > Pb) •
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 − Rs ) = 0
Pers. MB menjadi:
(B − B ) cw ⋅ S wc + c f oi N p ⋅ Bo + W p ⋅ Bw = N ⋅ Boi o + (1 − S ) Boi wc
∆P + W ⋅ B e w
N
For the case no water influx
(B − B ) cw ⋅ S wc + c f oi + N p ⋅ Bo = N ⋅ Boi o (1 − S ) Boi wc
F
∆P
F = N ⋅ (Eo + Ew, f )
Eo+Ew,f
ITSB
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INSTITUT TEKNOLOGI DAN SAINS BANDUNG
@ 2014
Pers. MB untuk Solution Gas Drive Reservoir Pers. MB menjadi:
(B − B ) cw ⋅ S wc + c f oi N p ⋅ Bo + W p ⋅ Bw = N ⋅ Boi o + Boi (1 − S wc )
F
∆P + W ⋅ B e w
N
For the case no water influx
(B − B ) cw ⋅ S wc + c f oi + N p ⋅ Bo = N ⋅ Boi o (1 − S ) Boi wc
∆P
F = N ⋅ (Eo + Ew, f )
Eo+Ew,f
Dalam bentuk lain:
co = −
1 ∂Bo Bo ∂P
=
1 ∆Bo Bo ∆P
=
1 (Bo − Boi ) Boi
∆P
cw ⋅ S wc + c f N p ⋅ Bo = N ⋅ Boi Co ⋅ ∆P + (1 − S ) wc Co ⋅ S o + cw ⋅ S wc + c f N p ⋅ Bo = N ⋅ Boi (1 − S wc )
N p ⋅ Bo = N ⋅ Boi ⋅ ce ⋅ ∆P
∆P
Karena
S o = 1 − S wc
maka,
⋅ ∆P
dimana,
co ⋅ S o + cw ⋅ S wc + c f ce = (1 − S wc )
Effective saturation weighted compressibility of the system
ITSB
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INSTITUT TEKNOLOGI DAN SAINS BANDUNG
@ 2014
Pers. MB untuk Solution Gas Drive Reservoir Bentuk umum persamaan MB: (B − B ) + (R − R ) ⋅ B o oi si s g N p ⋅ (Bo + (R p − Rs )⋅ Bg ) + W p ⋅ Bw = N ⋅ Boi Boi oil + gas production
water production
F
B c S c + m ⋅ g − 1 + (1 + m ) w ⋅ wc + f B (1 − S ) wc gi
oil + dissolved gas expansion
gas cap expansion
change of HC pore volume
Eo
Eg
Ef,w
∆P + W ⋅ B e 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 cw,f usually neglected
Pers. MB menjadi:
(B − B ) + (R − R ) ⋅ B o oi si s g N p ⋅ (Bo + (R p − Rs )⋅ Bg ) + W p ⋅ Bw = N ⋅ Boi Boi
+ m ⋅ B g − 1 + W ⋅ B e w B gi
water influx
For the case no water influx and no initial gas cap gas: Pers. MB menjadi:
N p ⋅ (Bo + (R p − Rs )⋅ Bg ) = N ⋅ ((Bo − Boi ) + (Rsi − Rs ) ⋅ Bg ) Np
=
N Jadi
(Bo − Boi ) + (Rsi − Rs ) ⋅ Bg Bo + (R p − Rs )⋅ Bg
Np N
= RF ≈
1 Rp
ITSB
Np
dimana
= RF = Recovery Factor
N dimana
RF
Rp =
Gp
= cumulative prod. gas oil ratio
Np
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.
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INSTITUT TEKNOLOGI DAN SAINS BANDUNG
@ 2014
Pers. MB untuk Solution Gas Drive Reservoir
Rp
ITSB
-
INSTITUT TEKNOLOGI DAN SAINS BANDUNG
@ 2014
Pers. MB untuk Gas Cap Drive Reservoir Bentuk umum persamaan MB: (B − B ) + (R − R ) ⋅ B o oi si s g N p ⋅ (Bo + (R p − Rs )⋅ Bg ) + W p ⋅ Bw = N ⋅ Boi Boi oil + gas production
water production
F
B c S c + m ⋅ g − 1 + (1 + m ) w ⋅ wc + f B (1 − S ) wc gi
oil + dissolved gas expansion
gas cap expansion
change of HC pore volume
Eo
Eg
Ef,w
∆P + W ⋅ B e w
Assumptions •
Gas cap gas compressibility is way higher than formation or water (50-100X), so that cw,f usually neglected
For the case no water influx : Pers. MB menjadi:
(B − B ) + (R − R ) ⋅ B o oi si s g N p ⋅ (Bo + (R p − Rs )⋅ Bg ) = N ⋅ Boi Boi oil + gas production (withdrawal rate)
Oil & dissolved gas expansion
B g + m⋅ − 1 B gi Gascap gas expansion
water influx
ITSB
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INSTITUT TEKNOLOGI DAN SAINS BANDUNG
@ 2014
Pers. MB untuk Gas Cap Drive Reservoir For the case no water influx : Pers. MB menjadi:
(B − B ) + (R − R ) ⋅ B o oi si s g N p ⋅ (Bo + (R p − Rs )⋅ Bg ) = N ⋅ Boi Boi Pers. di atas dalam bentuk lain:
F = N ⋅ (Eo + m ⋅ E g )
+ m ⋅ B g − 1 B gi
F
N Eo+m.Eg
Bentuk umum persamaan MB: (B − B ) + (R − R ) ⋅ B o oi si s g N p ⋅ (Bo + (R p − Rs )⋅ Bg ) + W p ⋅ Bw = N ⋅ Boi Boi Assumptions •
We = (cw + c f )⋅ Wi ⋅ ∆P
•
∆P + W ⋅ B e w
dimana Wi = Volume air akuifer
Karena adanya water influx, tekanan reservoir biasanya terjaga dengan baik, sehingga ∆P akan menjadi kecil. Oleh karena itu biasanya Ew,f diabaikan. Maka persamaan MB Menjadi:
F = N ⋅ (Eo + m ⋅ E g ) + We
assume Bw = 1
Untuk case no initial gas cap gas, m = 0, dan pers MB menjadi
F = N ⋅ Eo + We
-
•
B c S c + m ⋅ g − 1 + (1 + m ) w ⋅ wc + f B (1 − S ) wc gi
Pergerakan air akuifer ke dalam zona minyak diakibatkan oleh adanya sifat kompresibilitas air dan formasi batuan akuifer
F
ITSB
INSTITUT TEKNOLOGI DAN SAINS BANDUNG
@ 2014
Pers. MB untuk Natural Water Drive Reservoir
Eo
=N+
We Eo
F Eo
45o N
We Eo