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Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap18 Final Proof page 273 4.1.2007 10:04pm Compositor Name: SJoearun
PRODUCTION OPTIMIZATION 18/273
L
p 1 q D 1 p 2 D 2 p 3 D 3 p 4
L 1 L 2 L 3
(a)
p 2 L
q 1 D 1
q t p 1 q 2 D 2 p 2 q t
q 3 D 3
(b)
Figure 18.9 Sketch of (a) a series pipeline and (b) a parallel pipeline.
L 1 L 3
q 1 D 1
q t p 1 p 3 D 3 q t p 2
q 2 D 2
L
Figure 18.10 Sketch of a looped pipeline.
q t ¼ q 1 þ q 2 q t ¼ 18:062T b
s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi q ffiffiffiffiffiffiffiffiffiffiffi q ffiffiffiffiffiffiffiffiffiffiffi p b
2
2
( p p ) v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
T b 1 2 16=3 16=3
¼ 18:062 D 1 þ D 2 (18:37) u p p 2
2
u
p b g g TzL u 0 1 2 1:
u
or u B L 3 C
u
C
B
2 u g g Tz q ffiffiffiffiffiffiffiffiffiffiffi L 1 q þ C
B
u
2 2 g g TzL 1 q t p b t @ ffiffiffiffiffiffiffiffiffiffiffi 2 D 16=3 A
p p ¼ q ffiffiffiffiffiffiffiffiffiffiffi q ffiffiffiffiffiffiffiffiffiffiffi 2 : (18:38) D 16=3 þ D 16=3 3
3
1
1
2
D 16=3 þ D 16=3 18:062T b
2
1
(18:41)
Applying the Weymouth equation to the third segment Capacity of a single-diameter (D 3 ) pipeline is expressed as
v
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u
(with diameter D 3 ) yields 18:062T b u p p 2
2
q 3 ¼ u 1 2 !: (18:42)
2 p b u L
g g TzL 3 q t p b t g g Tz
2
2
p p ¼ : (18:39) 16=3
3 2 16=3 D
D 3 18:062T b 3
Dividing Eq. (18.41) by Eq. (18.42) yields
Adding Eqs. (18.38) and (18.39) results in v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
!
2 u L
u
2
2
p p ¼ g g Tz q t p b u 16=3
u
2
1
u
18:062T b q t ¼ u 0 D 3 1: (18:43)
0 1 u
q 3 u
C
C
B L 3 C u B L 1 L 3 C
B
u
B
C
B q ffiffiffiffiffiffiffiffiffiffiffi L 1 q þ C (18:40) u B q ffiffiffiffiffiffiffiffiffiffiffi q ffiffiffiffiffiffiffiffiffiffiffi 2 þ 16=3 A
t
@
@ ffiffiffiffiffiffiffiffiffiffiffi 2 D 16=3 A 16=3 16=3 D 3
D 16=3 þ D 16=3 3 D 1 þ D 2
2
1
Let Y be the fraction of looped pipeline and X be the
or increase in gas capacity, that is,
