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Beggs and Brill method
The Beggs and Brill method works for horizontal or vertical flow and everything in between. It
also takes into account the different horizontal flow regimes. This method uses the general
mechanical energy balance and the average in-situ density to calculate the pressure gradient. The
following parameters are used in the calculations.
gD
uN mFR
2
(2-38)
m
ll
u
u
302.01 316 lL 4684.2
2 0009252. lL (2-39, 40)
4516.13 10. lL
738.6
4 5. lL (2-41, 42)
Determining flow regimes
Segregated if
l < .01 and NFR < L1 or l >= .01 and NFR < L2
Transition if
l >= .01 and L2 < NFR = l.
8.1sin333.8.1sin1 3 C gFRfvlell NNdC ln1 (2-45,46)
Where a, b, c, d, e, f and g depend on flow regimes and are given in the following table
For transition flow, the liquid holdup is calculated using both the segregated & intermittent
equations and interpolating using the following:
ntIntermitteBySegregatedAyy lll (2-47)
23
3
LL
NLA FR
AB 1 (2-48,49)
ggll yy _
144
sin_
cPE g
g
dl
dp
(2-50,51)
The frictional pressure gradient is calculated using:
Dg
uf
dl
dp
c
mmtp
F
22
(2-52)
ggllm n
tp
ntpf
fff (2-53,54)
The no slip friction factor fn is based on smooth pipe (D =0) and the Reynolds number,
mmmm
DuN
1488Re where ggllm (2-55,56)
ftp the two phase friction factor is
Sntp eff (2-57)
where
42 )ln(01853.0)ln(8725.0)ln(182.30523.0
)ln(
xxx
xS
(2-58)
and
2
l
l
yx
(2-59)
Since S is unbounded in the interval 1 < x < 1.2, for this interval
)2.12.2ln( xS (2-60)
Using Beggs & Brill (Same data is Duklar example)
First find the flow regime, calculate NFR, l, L1, L2, L3, and L4.
NFR = 18.4, l = .35, L1=230, L2=.0124, L3= .456, L4= 590.
So .01 < l < .4 and L3 < NFR < L1 so flow is intermittent.
Using the table to get a, b and c:
454.06.29
35.*845.0173.0
5351.0
0 cFR
b
ll
N
ay
Find C and d, e, f and g from table:
0351.06.29*28.10*35.*96.2ln35.1ln1 0978.04473.0305.0 gFRfvlell NNdC
01.1)90*8.1(sin333.)90*8.1sin(0351.18.1sin333.8.1sin1 33 C
Find yl 459.01.1*454.0 ll yy
The in-situ average density is
3_
/29.246.2*)459.1(9.49*459. ftlbyy ggll
Potential gradient is
ftpsig
g
dl
dp
cPE
/169.144
1*29.24
144
sin_
For friction gradient
First find the mixture density and viscosity
3/1.1965.*6.235.*9.49 ftlbggllm
cpggllm 709.65.*0131.35.*2
The Reynolds Number
109184709.
1488*203.*39.13*1.191488Re
m
mmm
DuN
From Moody plot fn is .0045, solve for S
66.1459.
35.2
l
l
yx
42 )ln(01853.0)ln(8725.0)ln(182.30523.0
)ln(
xxx
xS
379.)66.1ln(01853.0)66.1ln(8725.0)66.1ln(182.30523.0
)66.1ln(42
S
Solve for ftp
0066.0045. 379. eeff Sntp
Find the friction gradient
ftpsiftlbDg
uf
dl
dp
c
mmtp
F
/032./62.4203*17.32
94.10*1.19*0066.*22 322
1)Using the Beggs and Brill method find the length of pipe between the points at 1000psi and 500
psi with the following data. Both vertical and horizontal cases.
d = 1.995 g = .65 oil 22o API qo = 400 stb/day
qw = 600 bpd g = .013 cp o = 30 dynes/cm w = 70 dynes/cm GLR = 500 scf/stb
@ average conditions
Rs = 92 scf/stb o = 17 cp w = .63 z = .91
Pipe Fittings in Horizontal flow
To find the pressure drop through pipe fitting such as elbows, tees and valves an equivalent length
is add to the flow line. This will account for the additional turbulence and secondary flows which
cause the additional pressure drop.
These equivalent lengths have been determined experimentally for the most of the fittings. These
are found in the following tables. They are given in pipe diameters, which are in feet.
So to find the equivalent length for a 45o elbow in 2 inch pipe, find the equivalent length for the
elbow in the table, 16, and multiply it by .166 feet, which gives 2.66 feet. This is added to the
length of the flow line, the pressure drop for the system is then calculated using one of the
methods for horizontal flow.