Page 90 - Petroleum Production Engineering, A Computer-Assisted Approach

Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap06 Final Proof page 82 3.1.2007 8:40pm Compositor Name: SJoearun

6/82 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
At the junction points, Solution Example Problem 6.9 is solved with the
spreadsheet program MultilateralGasWellDeliverability
: (6:40)
(Radial-FlowIPR).xls. Table 6.9 shows the appearance
p kf i ¼ p hf i 1
Equations (6.32), (6.33), (6.38), and (6.40) contain (4n 1) of the spreadsheet for the Input data and Result sections.
at the top of It indicates that the expected total gas flow rate is
equations. For a given flowing pressure p hf n
lateral n, the following (4n 1) unknowns can be solved 4,280 Mscf/d from the four laterals. Lateral 3 will steals
from the (4n 1) equations: 6,305 Mscf/d.
q g1 , q g2 , .. . q gn
6.3.2 Oil well
The inflow performance function for oil wells can be ex-
p wf 1 , p wf 2 , .. . p wf n
pressed as
p kf 1 , p kf 2 , .. . p kf n
p ), (6:44)
q o i ¼ J i ( p i p wf i
p hf 1 , p hf 2 , .. . p hf n 1
where J i ¼ productivity index of lateral i.
Then the gas production rate of the multilateral well can The fluid flow in the curvic sections can be approxi-
be determined by mated as
X p wf i ¼ p kf i þr R i R i , (6:45)
n
q g ¼ q gi : (6:41)
i¼1 where r R i ¼ vertical pressure gradient in the curvic sec-
tion of lateral i.
Thus, the composite IPR,
The pressure gradient r R i may be estimated by the
, (6:42) Poettmann–Carpenter method:
q g ¼ fp hf n

2
k
can be established implicitly. The solution procedure has r r þ k i
i
been coded in the spreadsheet program MultilateralGas r R i ¼ 144 i , (6:46)
r
r
WellDeliverability(C-nIPR).xls. It has been found that
the program does not allow cross-flow to be computed where
because of difficulty of computing roof of negative number M F
with Eq. (6.32). Therefore, another spreadsheet was dev- r r i ¼ , (6:47)
V m i
eloped to solve the problem. The second spreadsheet
ð
is MultilateralGasWellDeliverability(Radial-FlowIPR).xls M F ¼ 350:17 g o þ WOR g w Þ þ 0:0765GOR g g , (6:48)
and it employs the following IPR model for individual V m i ¼ 5:615 B o þ WOR B w Þ þ z GOR R s Þ
zð
ð
laterals:
2 3 29:4 T i , (6:49)
520
2
2
þ p xf i
p wf i
p
k h h( p p ) 6 1 7
wf
q g ¼ 6 7 (6:43)
2
1424mZT 4 0:472r eh 5 f 2F q M F
ln k k i ¼ oi : (6:50)
10 5
L=4 7:4137 10 d
i
The fluid flow in the vertical sections may be expressed as
Example Problem 6.9 For the data given in the
H i i ¼ 1, 2, .. . , n, (6:51)
þr h i
¼ p hf i
p kf i
following table, predict gas production rate against
1,000 psia wellhead pressure and 100 8F wellhead where r h i ¼ pressure gradient in the vertical section of
temperature: lateral i.
Horizontal sections
Lateral no.: 1 2 3 4
Length of horizontal section (L) 500 600 700 400 ft
Horizontal permeability (k) 1 2 3 4 md
Net pay thickness (h) 20 20 20 20 ft
Reservoir pressure (p-bar) 3700 3500 1,800 2,800 psia
Radius of drainage (r eh ) 2,000 2,500 1,700 2,100 ft
Gas viscosity (m g ) 0.02 0.02 0.02 0.02 cp
Wellbore diameter (Di) 8.00 8.00 8.00 8.00 in.
Bottom-hole temperature (T) 270 260 250 230 8F
Gas compressibility factor (z) 0.85 0.90 0.95 0.98
Gas-specific gravity (g g ) 0.85 0.83 0.80 0.75 air ¼ 1
Curvic sections
Lateral no.: 1 2 3 4
Radius of curve (R) 250 300 200 270 ft
Average inclination angle (u) 45 45 45 45 8F
Tubing diameter (di) 3 3 3 3 in.
Pipe roughness (e) 0.0018 0.0018 0.0018 0.0018 in.
Vertical sections
Lateral no.: 1 2 3 4
Interval length (H) 250 300 200 8,000 ft
Tubing diameter (d i ) 3 3 3 3 in.
Pipe roughness (e) 0.0018 0.0018 0.0018 0.0018 in.