Ref.: Brill & Beggs, Two Phase Flow in Pipes, 6 th Edition, 1991. Chapter 3.
7
Two Phase Pipeline Example Ref.: Brill & Beggs, Two Phase Flow in Pipes, 6 th Edition, 1991. Chapter 3.
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Transcript of Ref.: Brill & Beggs, Two Phase Flow in Pipes, 6 th Edition, 1991. Chapter 3.
Two Phase PipelineExample
Ref.: Brill & Beggs, Two Phase Flow in Pipes, 6th Edition, 1991. Chapter 3.
Two Phase Flow Example 1: Description
P2
P1
d1 , L1
DZ1
For the above two-phase pipeline, calculate the exit pressure (P2) based on the Beggs and Brill equation. Compare the manual calculation results with Hysys software results (using Pipe Segment and Pipesys) .
Feed specifications: T1 = 60 oF, P1 = 1000 psia, qgtsc= 50 MMscfd, qosc =100000 bbl/day, γgt = 0.7896, γo = 31.3 oAPI,
Pipeline specifications:L1 = 900 ft, ΔZ1 = 10 ft, d1= 12 in nominal (Mild Steel, Schedule 40), T2 = 60 oF 2
Two Phase Flow Example 1: Description
For using compositional model (in Hysys), the following analysis of produced oil and gas at standard conditions are exist:
3
Produced Gas at S. C. Light Ends Analysis Distillation Curve and Density (oAPI=31.3)
Mole, % Components Liquid Vol. % Components Density, oAPI Cut Temp, oF Liquid Vol. %
6.8311 N2 0.0013 N2 ---- 145 2.85
1.3467 H2S 0.014 H2S 57.7 167 3.83
0.9777 CO2 0.0037 CO2 56.4 212 5.63
72.8749 C1 0.0733 C1 55.4 257 7.33
7.5713 C2 0.0803 C2 52.3 302 10.03
2.9463 C3 0.1307 C3 47.8 347 13.43
2.9119 i-C4 0.4177 i-C4 42.6 392 18.53
1.8989 n-C4 0.3947 n-C4 39.2 437 24.43
1.2841 i-C5 0.9029 i-C5 35.8 482 34.23
0.8183 n-C5 0.7487 n-C5 33.0 527 44.93
0.5388 C6+ Properties of C6+ : Molecular Weight = 89.53, Density = 55.49 oAPI, N. Boiling point = 210 oF
Two Phase Flow Example 1: solution
In hand calculation, a single segment with Pav= P1 was used (for first iteration). I didn’t find any reliable correlation for calculation of dissolved gas specific gravity, and therefore the specific gravity of free gas was selected from HYSYS software at input conditions: γgf = 0.6397
Calculation of Rs and γgd needs a try and error, based on these equations:
Therefore:
4
83.0/1
)(00091.0
)(0125.0
10
10
18:eq. Standing
T
API
gds
o
PR
s
gfspgtPgd R
RRR
)(:eq. balance mass Gas
8988.0,3.289 gdsR
3m3
ft
lb355.51sec,/ft 31.7bbl/day 112500
125.1)standing(MMscfd,07.21)(
oooo
oospgf
sc
scsc
qBq
BqRRq
Two Phase Flow Example 1: solution
Volume flow rate of free gas at inlet conditions can be calculated with using Bg:
Viscosity of oil and gas and surface tension between them can be calculated as:
5
42.1,49.1R,1.367psia, 672:method alet Brown o prprpcpc TPTP
cp56.5cp835.16:method Standing oOD
01207.082.0:method Katz and Standing gBz
3m
3 /ftlb049.4sec,/ft943.2 gfgfggf scqBq
cp0.0127gr/cm06496.0:method alet Lee 3 gfgf
dynes/cm 153.31:method Swerdloff andBaker o OD
Two Phase Flow Example 1: solution
Therefore, the two-phase no slip properties can be calculated as follows:
Flow regimes:
Liquid Hold and ρs:
6
713.0ft/sec, 185.13ft/sec, 785.3,ft/sec 4.9,ft7776.0 2 LmsgsL vvvA
3f
3m)()0( /ftlb46.0
d
d/ftlb374.41789.0
elsLL L
PHH
cp97.3,/ftlb78.37 3mn nggfLo
8.24,885.4,163.0,002.0,3.285,43.5 4321 LvFr NLLLLN
)( ,4.0 4 ntIntermittedDistributeLNFrL
Two Phase Flow Example 1: solution
Pressure gradient due to friction:
Outlet pressure:
7
0155.0)smooth(1086.11488 5Re n
n
nm fvd
N
0205.02774.0)2.12.2ln(145.1 Sntp effySy
3f /ftlb103.2
d
d
fL
P
3f
4 /ftlb564.21
dd
dd
d
d105.4
k
felk E
LP
LP
L
PE
984PipeSys982,:PipeSegpsia,984144/900564.212 PP
