analisis data

1. KERJA ULANG SUMUR
TUGAS (SOAL DAN JAWABAN)
DOSEN PEMBIMBING:
IDHAM KHALID, ST.MT
OLEH:
ILHAM RABIKHI
123210575
KELAS B
JURUSAN PERMINYAKAN
FAKULTAS TEKNIK
UNIVERSITAS ISLAM RIAU
PEKANBARU
2015

2. SOAL DAN JAWABAN
1. Dari data di bawahini, buatlahkurva IPR menggunakanrumus Vogel,
asumsikanpenurunanPwf setiap 500 psi:
Porosity ∅ = 0.19
Effective horizontal permeability k = 8.2 md
Pay zone thickness h = 53 ft
Reservoir pressure P = 5,651 psia
Bubble point pressure Pb = 5,651 psia
Fluid formation volume factor Bo = 1.1
Fluyid viscosity μo = 1.7 cp
Total compressibility ct = 0.0000129 psi−1
Drainage area A = 640 acres, re = 2,980 ft
Wellbore radius rw = 0.328 ft
Skin factor S = 0
Penyelesaian:
Rumus Vogel:
Qo
Qmaks
= 1 − 0.2 (
Pwf
Ps
) − 0.8 (
Pwf
Ps
)
2
Pwf Pwf/Pr Qo/Qmaks
5661 1.017982377 -0.032626972
5161 0.928070491 0.125334033
4661 0.838158605 0.270360402
4161 0.748246718 0.402452135
3661 0.658334832 0.521609233
3161 0.568422946 0.627831695
2661 0.478511059 0.721119521
2161 0.388599173 0.801472712
1661 0.298687286 0.868891267
1161 0.2087754 0.923375186
661 0.118863514 0.964924469
161 0.028951627 0.993539117
0 0 1

3. 2. Berapakahkemampuanlajualirmaksimum(Qmax) dari well KB#059 (5 Juli 2005):
Total fluid proud = 3000BFPD 300 BOPD⁄
SFL = 400 ft
WFL = 1500 ft
Perforation depth = 4800 ft
Oil = 33°API
SGwater = 1.02
Penyelesaian:
𝑆𝐺 𝐹𝑙𝑢𝑖𝑑 = (1 − 𝑤𝑐) 𝑥𝑆𝐺 𝑂𝑖𝑙 + 𝑤𝑐 𝑥 𝑆𝐺 𝑤𝑎𝑡𝑒𝑟
𝑆𝐺 𝐹𝑙𝑢𝑖𝑑 = (1 − 0) 𝑥 33 + 0 𝑥 1.02
= 33
𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑓𝑙𝑢𝑖𝑑 = 0,433 𝑥 33
0
0.2
0.4
0.6
0.8
1
1.2
-0.2 0 0.2 0.4 0.6 0.8 1 1.2
Pwf/Pr
Qo/Qmaks
IPR VOGEL
IPR

4. = 0,143
𝑃𝑟 = (ℎ – 𝑆𝐹𝐿) 𝑥 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑓𝑙𝑢𝑖𝑑
= ( 4800 – 400 ) 𝑥 0,143 = 6,29 𝑥 102
= 629
𝑃𝑤𝑓 = (ℎ − 𝑊𝐹𝐿) 𝑥 𝑔𝑓
= ( 4800 – 1500 ) 𝑥 0,143 = 471
𝑃𝐼 = 𝑄 / (𝑃𝑟 – 𝑃𝑤𝑓)
=
3000
( 629−471 )
= 18,89
𝑄𝑚𝑎𝑥 = 𝑃𝐼 𝑥 ( 𝑃𝑟 – 0 )
= 18,98 𝑥 ( 629 – 0 )
= 11, 93 𝐵𝑃𝐹𝐷
3. Turunkan persamaan rumus PI di bawah ini dari persamaan awalnya, PI =
Q (Pr − Pwf)⁄ !
PI = Q (WFL − SFL)⁄
PI = BPD psi⁄
Penyelesaian:
Penurunan Rumus 𝑃𝐼 =
𝑄
(Pr −𝑃𝑤𝑓)
𝐵𝑃𝐷
𝐹𝑡⁄
𝑃𝐼 =
𝑄
(Pr −𝑃𝑤𝑓)
𝑃𝐼 =
𝐵𝑃𝐷
𝑃𝑠𝑖
𝑃𝐼 =
𝑄
𝑊𝐹𝐿−𝑆𝐹𝐿
𝑃𝐼 =
𝑄
𝑊𝐹𝐿 − 𝑆𝐹𝐿
=
𝐵𝑃𝐷
𝐹𝑡
Makasatuan PI adalahBPD ft⁄
𝑄 𝐵𝑃𝐷
Pr −𝑃𝑤𝑓 𝑃𝑠𝑖
𝑃𝑟 = ( ℎ – 𝑆𝐹𝐿 ) × 𝐺𝑟
= ( 𝐹𝑡 – 𝐹𝑡 ) ×
𝑝𝑠𝑖
𝑓𝑡
= 𝐹𝑡 ×
𝑝𝑠𝑖
𝑓𝑡
𝑃𝑟 = 𝑃𝑠𝑖
𝑃𝑟 = ( ℎ – 𝑆𝐹𝐿 ) × 𝑔𝑟
( ℎ – 𝑆𝐹𝐿 ) =
𝑃𝑟
𝑔𝑓
𝑓𝑡 =
𝑝𝑠𝑖
𝑝𝑠𝑖
𝑓𝑡⁄
𝑓𝑡 = 𝑓𝑡

5. 4. Berapakahkemampuanlajualirmaksimumdari well KB#059?
Total fluid proud = 2742BFPD 137 BOPD⁄
WC = 95%
SFL = 403 ft
WFL = 1452 ft
Perforation depth = 4807 ft
Oil = 33°API
SGwater = 1.02
Penyelesaian:
𝑆𝐺 𝐹𝑙𝑢𝑖𝑑 = (1 − 𝑤𝑐) 𝑥𝑆𝐺 𝑂𝑖𝑙 + 𝑤𝑐 𝑥 𝑆𝐺 𝑤𝑎𝑡𝑒𝑟
𝑆𝐺 𝐹𝑙𝑢𝑖𝑑 = (1 − 0,95 ) 𝑥 33+ 0,93 𝑥 1.02
= 2,6
𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑓𝑙𝑢𝑖𝑑 = 0,433 𝑥 𝑆𝐺 𝑓𝑙𝑢𝑖𝑑
= 0,433 𝑥 2,6
= 1,125
𝑝𝑠𝑖
𝑓𝑡⁄
𝑃𝑟 = (ℎ – 𝑆𝐹𝐿) 𝑥 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑓𝑙𝑢𝑖𝑑
= ( 4807 – 403 )𝑥 1,125 = 6,29 𝑥 102
= 4954 𝑝𝑠𝑖
𝑃𝑤𝑓 = (ℎ − 𝑊𝐹𝐿) 𝑥 𝑔𝑓
= ( 4807 – 1452 )𝑥 1,125 = 4193 𝑝𝑠𝑖
𝑃𝐼 = 𝑄 / (𝑃𝑟 – 𝑃𝑤𝑓)
=
2742
( 4954 −4193 )
= 3,6
𝑄𝑚𝑎𝑥 = 𝑃𝐼 𝑥 ( 𝑃𝑟 – 0 )
= 3,6 𝑥 ( 4954 – 0 )
= 17,834 𝐵𝑃𝐹𝐷
PI =
Q
Pr − Pwf
=
2742
1929.83 − 1470.1608
= 5.9651
Qmaks = PI × (Pr − Pwf)
= 5.9651 × (1929.83− 1470.1608)
= 2742 BPD