Ref.: Brill & Beggs, Two Phase Flow in Pipes, 6th Edition, 1991.
Chapter 3.
Two Phase Flow
Example 1: Description
P2
P1
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:
Distillation Curve and Density (oAPI=31.3)
Light Ends Analysis
Produced Gas at S. C.
Liquid Vol. %
Cut Temp, oF
Density, oAPI
Components
Liquid Vol. %
Components
Mole, %
2.85
145
----
N2
0.0013
N2
6.8311
3.83
167
57.7
H2S
0.014
H2S
1.3467
5.63
212
56.4
CO2
0.0037
CO2
0.9777
7.33
257
55.4
C1
0.0733
C1
72.8749
10.03
302
52.3
C2
0.0803
C2
7.5713
13.43
347
47.8
C3
0.1307
C3
2.9463
18.53
392
42.6
i-C4
0.4177
i-C4
2.9119
24.43
437
39.2
n-C4
0.3947
n-C4
1.8989
34.23
482
35.8
i-C5
0.9029
i-C5
1.2841
44.93
527
33.0
n-C5
0.7487
n-C5
0.8183
C 6+
0.5388
Properties of C6+ : Molecular Weight = 89.53, Density = 55.49 oAPI, N. Boiling point = 210 oF
3
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:
 P 10 0 .0125 ( API ) 

0 . 00091 ( T ) 
18
10


o
Standing
eq. : R s   gd
Gas mass balance eq. :  gd 
Therefore:
1 / 0 . 83
 R s  289 . 3 ,  gd  0 . 8988
R P  gt  ( R p  R s ) gf
Rs
q gf sc  ( R p  R s ) q o sc  21 . 07 MMscfd,
B o ( standing )  1 . 125
 q o  B o q o sc  112500 bbl/day  7 . 31 ft / sec,  o  51 . 355
3
lb m
ft
3
4
Two Phase Flow
Example 1: solution
Volume flow rate of free gas at inlet conditions can be calculated with using Bg:
Brown et al method : P pc  672 psia, T pc  367 . 1 R, P pr  1 . 49 , T pr  1 . 42
o
Standing
and Katz method : z  0 . 82  B g  0 . 01207
 q gf  B g q gf sc  2 . 943 ft / sec,  gf  4 . 049 lb m /ft
3
3
Viscosity of oil and gas and surface tension between them can be calculated as:
Standing
method :  OD  16 . 835 cp   o  5 . 56 cp
Lee et al method :  gf  0 . 06496 gr/cm
Baker and Swerdloff
3
  gf  0.0127 cp
method :  OD  31 . 3   o  15 dynes/cm
5
Two Phase Flow
Example 1: solution
A  0 . 7776 ft , v sL  9 . 4 ft/sec , v sg  3 . 785 ft/sec, v m  13 . 185 ft/sec,  L  0 . 713
2
Therefore, the two-phase no slip properties can be calculated as follows:
 n   o  L   gf  g  37 . 78 lb m /ft ,  n  3 . 97 cp
3
Flow regimes:
N Fr  5 . 43 , L1  285 . 3, L 2  0 . 002 , L3  0 . 163 , L 4  4 . 885 , N Lv  24 . 8
 L  0 . 4 , N Fr  L 4  Distribute d ( Intermitte nt )
Liquid Hold and ρs:
H L ( 0 )  H L (  )  0 . 789   s  41 . 374 lb m /ft
3
 dP 
 
  0 . 46 lb f /ft
 d L  el
3
6
Two Phase Flow
Example 1: solution
Pressure gradient due to friction:
N Re  1488
d vm  n
n
 1 . 86  10  f n ( smooth )  0 . 0155
5
y  1 . 145  S  ln( 2 . 2 y  1 . 2 )  0 . 2774  f tp  f n e  0 . 0205
S
 dP 
 
  2 . 103 lb f /ft
 dL  f
3
Outlet pressure:
E k  4 . 5  10
4
 dP 
 

 dL 
 dP 
 dP 

  

 d L  el  d L  f
1  Ek
 2 . 564 lb f /ft
 P2  P1  2 . 564  900 / 144  984 psia, PipeSeg : 982, PipeSys
3
 984
7