Page 272 - Petroleum Production Engineering, A Computer-Assisted Approach

Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap18 Final Proof page 272 4.1.2007 10:04pm Compositor Name: SJoearun

18/272 PRODUCTION ENHANCEMENT
where
Gas composition
V Vsc ¼ specific volume of vapor phase under
standard condition, scf/mol-lb Compound Mole fraction
3
R ¼ gas constant, 10:73 ft -psia/lb mol-R
C 1 0.6599
T sc ¼ standard temperature, 520 8R
C 2 0.0869
p sc ¼ standard pressure, 14.7 psia 0.0591
3
V L ¼ specific volume of liquid phase, ft /mol-lb. C 3
i-C 4 0.0239
Finally, the gas–oil ratio (GOR) in the separator can be n-C 4 0.0278
calculated by i-C 5 0.0157
n-C 5 0.0112
V Vsc 0.0181
GOR ¼ : (18:24) C 6
V L C 7þ 0.0601
Specific gravity and the American Petroleum Institute N 2 0.0194
(API) gravity of oil at the separation pressure can be CO 2 0.0121
calculated based on liquid density from Eq. (18.21). The H 2 S 0.0058
lower the GOR, the higher the API gravity, and the higher
the liquid production rate. For gas condensates, there
exists an optimum separation pressure that yields the
lower GOR at a given temperature. Capacity of a single-diameter (D 1 ) pipeline is expressed as
v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u
2
Example Problem 18.2 Perform flash calculation under q 1 ¼ 18:062T b u p p 2 4 !: (18:30)
1
u
the following separator conditions: p b u L
t
Pressure: 600 psia g g Tz 16=3
Temperature: 200 8F D 1
Specific gravity of stock-tank oil: 0.90 water ¼ 1
Specific gravity of solution gas: 0.70 air ¼ 1 Dividing Eq. (18.29) by Eq. (18.30) yields
!
Gas solubility (R s ): 500 scf/stb v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u
u L
u
Solution The flash calculation can be carried out using q t u D 16=3
u
1
the spreadsheet program LP-Flash.xls. The results are ¼ u !: (18:31)
u
q 1 t L 1 L 2 L 3
shown in Table 18.1. þ þ
D 16=3 D 16=3 D 16=3
3
1
2
18.6 Pipeline Network 18.6.2 Pipelines in Parallel
Optimization of pipelines mainly focuses on de-bottle- Consider a three-segment gas pipeline in parallel as
necking of the pipeline network, that is, finding the most depicted in Fig. 18.9b. Applying the Weymouth equation
restrictive pipeline segments and replacing/adding larger to each of the three segments gives
segments to remove the restriction effect. This requires the v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u
u
2
2
knowledge of flow of fluids in the pipe. This section pre- T b t ( p p )D 16=3
2
1
1
sents mathematical models for gas pipelines. The same q 1 ¼ 18:062 g g T zL (18:32)
p b
principle applies to oil flow. Equations for oil flow are v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
presented in Chapter 11. u 2 2 16=3
u
T b t ( p p )D
q 2 ¼ 18:062 1 2 2 (18:33)
p b g g T zL
18.6.1 Pipelines in Series v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
Consider a three-segment gas pipeline in a series of total u ( p p )D 16=3
u
2
2
length L depicted in Fig. 18.9a. Applying the Weymouth q 3 ¼ 18:062 T b t 1 2 3 : (18:34)
equation to each of the three segments gives p b g g T zL
2
g g T zL 1 q h p b Adding these three equations gives
2
2
p p ¼ (18:25)
2
1
D 16=3 18:062T b
1
2 q t ¼ q 1 þ q 2 þ q 3
g g T zL 2 q h p b s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2
p p ¼ (18:26) ( p p )
2
2
2
3
D 16=3 18:062T b ¼ 18:062 T b 1 2
2
2 p b g g TzL
g g T zL 3 q h p b
2
2
p p ¼ : (18:27) q ffiffiffiffiffiffiffiffiffiffiffi q ffiffiffiffiffiffiffiffiffiffiffi q ffiffiffiffiffiffiffiffiffiffiffi
4
3
D 16=3 18:062T b D 16=3 þ D 16=3 þ D 16=3 : (18:35)
3
2
3
1
Adding these three equations gives
! Dividing Eq. (18.35) by Eq. (18.32) yields
2
2
p p ¼ g g T z L 1 þ L 2 þ L 3 q ffiffiffiffiffiffiffiffiffiffiffi q ffiffiffiffiffiffiffiffiffiffiffi q ffiffiffiffiffiffiffiffiffiffiffi
1
16=3
4
16=3
16=3
D 16=3 D 16=3 D 16=3 q t D 1 þ D 2 þ D 3
2
3
1
2 ¼ q ffiffiffiffiffiffiffiffiffiffiffi : (18:36)
q 1 16=3
q h p b D
(18:28) 1
18:062T b
or
v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u
2
18:062T b u p p 2
q h ¼ u 1 4 !: (18:29) 18.6.3 Looped Pipelines
p b u Consider a three-segment looped gas pipeline depicted in
t L 1 L 2 L 3
g g Tz þ þ Fig. 18.10. Applying Eq. (18.35) to the first two (parallel)
D 16=3 D 16=3 D 16=3 segments gives
1
3
2