Transcript
7/23/2019 Beggs - Brill Method.pdf
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522 Production
16 05'
500(
144)-
21.22
x 32.2
[0.0051(480)5(462)*]
[
7.41
x
10 (0.1662)* 21.221
AL=
21.22(1)+
= (72,000- 84.88)/(21.22 0.005774) = 3,388 ft
r
-=--
500 - 0.1476
psi/ft
AI. 3,388
The Beggs Br i l l Method [20 25]
The parameters studied in this method and their range of variation were
as follows:
gas flowrate 0 to 300 Mscfd
liquid flowrate
0 to
5 gal/min
average system pressure 35 to 95 psia
pipe diameter 1 and
1.5
in.
liquid holdup 0 to 0.870
pressure gradient 0 to 0.8 psi/ft
inclination angle -90 to
90°
also horizontal flow patterns
A
flow diagram for calculating a pressure traverse in a vertical well is shown
in Figure
6-75.
The depth increment equation for
AI.
is
(6-130)
where y = two-phase specific weight in lb/ft3
v, = twephase superficial velocity v, = v , ~ v ~ )n ft/s
f, = two-phase friction factor
G, = two-phase weight flux rate lb/s ftg)
detailed procedure for the calculation of a pressure traverse is following:
1. Calculate the average pressure and average depth between the two points:
p = p, P V2
14.7
2. Determine the average temperature T at the average depth. This value
3. From P-V-T analysis or appropriate correlations, calculate Rs o B
po,
must be known from a temperature versus depth survey.
pw, , bo
ow
and Z at T and p.
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Flow of Fluids 5 3
alculate
Ah
Flgure
6 75.
Flow
diagram for
the Beggs Brill method [19]
4 Calculate the specific gravity
of
the oil SG :
141 5
131 5
PI
SG
5
Calculate the liquid and
gas
densities at the average conditions of pressure
and temperatures:
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5 Production
350SG0
0
0764RSGg
5.615
o
=
350SGw
Yw
=
5.615BW
0 0764SGgp(520)
Yg = (14.'7)(T 460)Z
6. Calculate the
in s tu
gas and liquid flowrates.
3 27
x
10-'Zqo(R- R,)(T 460)
P
g
=
6.49 x 10-5(q0B0 qwBw)
7.
Calculate the
in s tu
superficial gas, liquid and mixture velocities:
V L
= e/
sg =
q
v,
=
V
v1
G,
=
YLVL
Gg
Y A g
8 Calculate the liquid, gas and total weight flux rates:
G, = G, Gg
9. Calculate the input liquid content (neslip holdup):
10 Calculate the Froude number N the liquid viscosity, pL, the mixture
viscosity
pm
and the liquid surface tension oL:
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Flow of Fluids
545
pt
=
p h ~ 1
X)(6.72
x
lo4
aL
oofo awfw
11.
Calculate the no-slip Reynolds number and the liquid velocity number:
a 5
N, =
1. 8v,( )
12. To determine the flow pattern that would exist if flow were horizontal,
calculate the correlating parameters,
L , L3
and
L4:
= 316hO.0302
1
= 0.1oh-1.4516
L, = 0.0009252h-'.4684
L =
0.5h 6.738
4
13. Determine flow pattern using the following limits:
Segregated
C 0.01 and
N,
L,
or
0.01 and
N,
L
Transition:
2
0.01
and
L
N,
Ls
Intermittent:
0.01
0.4
and Ls
N,
L,
or
0.4 and
L, N, .5 L4
Distributed:
0.4
and
N, 2
L,
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546 Production
or
0.4 and N,
L,
14. Calculate the horizontal holdup H,(O):
where a, b and c re determined for each flow pattern from the follow-
ing table:
~~~
Flow pattern
a
b
~~ ~
Segregated
Intermittent
Distributed
~~
0.98
0.845
1.065
~
0.4846 0.0868
0.5351 0.01 73
0.5824 0.0609
15. Calculate the inclination correction factor coefficient:
C
=
(1 - h)ln(&Ni Jk)
where d, e,
f,
and g are determined for each flow condition from the
following table:
Flow
attern d e
t
g
Segregated uphill 0.011 3.768 3.539 1.614
Intermittent uphill
2.96 0.305 4.4473 0.0978
Distribu ted uphill
No
correction
c=o
All flow
patterns downhill 4.70 0.3692 0.1244 4.5056
16.
Calculate the liquid holdup inclination correction factor:
1 Crsin(1.80) 0.333 sin3(1.80)]
=
1 0.3C
for vertical well
17. Calculate the liquid holdup and the two-phase density:
H, 0) =
HLWW
P =
PLHL
p, l -
HL)
18.
Calculate the friction factor ratio:
f /f..
=
el
where
S =
[ln(y)l/{-0.0523 3.182 ln(y) 0.8725
[ln y)I4
0.01853 [ln(y)I4)
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Flow of Fluids
527
=
W H L ( ~ ) l 4
S becomes unbounded at a point in the interval
1
y 1.2; and for
y
in
this interval, the function S s calculated from
s
= h 2 .2y 1.2)
19. Calculate the no-slip friction factor:
f,
=
1/{2 log [NJ(4.5223 log Nkm - 3.8215)])
or
f,
= 0.0056 0.5/(N,)0.34
20
Calculate the two-phase friction factor:
f, = f,/ f,/fJ
21.
Calculate AL. If the estimated and calculated values for AL are not
sufficiently close, the calculated value is taken as the new estimated value
and the procedure is repeated until the values agree. A new pressure
increment is then chosen and the process is continued until the sum of
the
AL s
is equal to the well depth.
Example
Solve the problem in Example
2
using the Beggs-Brill method.
Solution
1.
p =
1,719.7
psia
2. T
=
90°F
3. R, = 947.3
scf/stb
o = 1.495
bbl/stb
4. SG,
=
0.736,
y
= 8.823
lb/ft5
5.
yo =
38.32 lb/fpts (from Example 3)
6.
q, =
0.08855
ft5/s
qL=
0.0466
ft3/s
7.
A
= 0.0217
ft
vsL=
e / A , =
2.147
ft/s,
vs
=
4.081
ft/s
8.
Calculate the liquid, gas and total weight flux rates:
pw
=
0.5
cp, 6
= 28
dyn/cm, Z
= 0.72
GL
= YLvT.L,G =
Ygvq
G
=
GL G,
=
38.32 x 2.147 8.823) 4.081
= 118.3 lb/(s ftp)
9.
Calculate the input liquid (no-slip holdup):
k = q L 0m0466 = 0.3448= 0.345
qL+q,
0.0466+0.08855
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5 8
Production
10. The Froude number, viscosity and surface tension
=
7.26
m = < =
6.23'
gd 32.174~0.1662
=
pofo pwfw
0.5(1.0) ~ ~ 0 . 0 )0.5
p,
=
(6.72 x 104)[0.5 x 0.345 0.02(1 0.345)]
=
1.164 x
= 0.0001164
lbm/(ft/s)
0
=
oofo owfw 28 x 1.0
=
28 dyn/cm
11. Calculate the no-slip Reynolds number and the liquid velocity number:
N, =
1.938
x
2.147(38.32/28)0.*5
=
4.5
12. Determine the flow pattern that would exist if flow were horizontal:
L
=
316h0.30* 316 x (0.345)0.3
=
229.14
=
0.0009252(0.345)-*.46&41.2796 x lo-*
L
= 0.10h-'.*16
=
0.10(.345)-1. 16
=
0.4687
L4
=
0.5h-6.ns = 0.5(0.345)-6.7s8 650.3
13.
Determine flow pattern:
0.4 > > 0.01
and p
< N, <
L
The flow pattern is intermittent.
14.
Calculate the horizontal holdup:
HJO) =
0.845 0.345)0~5551/7.260~0175
0.462
15. Calculate the inclination correction factor coefficient:
C
=
1 - 0.345) ln(2.96
x
0.345°~5054.5-0~~737.260~0g78)
=
0.18452
16. Calculate the liquid holdup inclination correction factor:
\v
=
1 C[sin(l.8 x
90)
0.333 ~ i n ~ 1 . 9
90)
1
C(0.309 0.009826) =
1 0 3C
= 1 0.3(0.18452) = 1.055
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Flow of Fluids 5 9
17 Calculate the liquid holdup and the two-phase density:
HL 90)=
H L 0 ) ~
0.462 1.055 = 0.4876
r
=
y,H,
r
1
HL)
=
38.32 0.4876) 8.823
1
0.4876)
= 23.2 Ib/ft3
18. Calculate the friction factor ratio:
y = [0.345/ 0.4876)2]= 1.451, In 1.451 = 0.3723
fc/fns= exp[0.3723/ -0.0523 3.182 0.3723) - 0.8725 0.3723p
0.1853 0.37234)]
= exp 0.3723/1.0118) = 3°.367961.4447
19. Calculate the no-slip friction factor:
f = 1/{2 l0g[NRcns/ 4.5223og Nk .3.8215)])*
= 1/36.84
=
0.0271
20. Calculate the two-phase friction factor:
t
= fnJft/fns) 0.0271 1.4447) = 0.0391
21. Determine the distance AL for Ap = 500 psi from Equation 6-130
AL=
23.2 6.23)4.081
144
50011 32.174 1.719.7
23.2 1.0)+ 0.0391 118.3)6.23
2 32.174)O. 1662
and
_ 5
=
0.18 psi/ft
AL 2,750
xample
Solve example 1 using the Beggs-Brill method
q = 40 MMscf/d, p = 2,000 psia
q
= 40,000 stb/d,
T
= 80°F
ID = 9 in.
SGp
=
0.75 at p
=
14.7 psia in
T
= 60°F
Rp
= 990 scf/bbl
=
2,750 ft
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530
Production
Solution
1. SG = 141.5/131.5
+
API = 0.86
2.
Calculate
R
Bo,
potp Zg
at
pav
and
T
Z, = 0.685
pg
=
0.0184 cp
R,
= 477
scf/stb
Bo
= 1.233
rb/stb
po = 2.96 cp
3. Calculate yo and Y at average parameters:
350(0.86) +0.0764(477)(0.75) =
47.42
I w f t 3
5.614( 1.233)
o
=
0.07Wgp(520)
- 0.0764 0.73) 2,000) 520) =
I w f t 3
Yg
=
(14.?)(T 460)Zg (14.7)(80 x 460)(0.685)
4.
Calculate the
n situ
gas and liquid flowrates,
3.27 X lo9 gqo
R
-
R
T
460)
P
=
-
3.27~10 (0.685)(40,000)(990- 477)(80+ 460)
2,000
=
1.241ft3/s
qL
=
6.49
x
10 (qoBo qwBw)
=
6.49
x
10 [40,000(1.233) 01
=
3.201
ft3/s
5. Calculate
A :
6. Calculate the in situ superficial gas, liquid and mixture velocities:
vL
= q /A =
3.201/0.4418
=
7.25 ft/s
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low
of Fluids 531
vsg
=
q Ap
=
1.241/0.4418
=
2.81 ft/s
v,,, - V~ v ~
7.25 2.81
=
10.06
ft/s
Calculate the liquid, gas and total mass flux rates:
G = pLvsL= (4'7.42)('1.25)= 343.6 lb/(s ft)
G
=
p v
=
(10.96)(2.81)
=
30.79
Ib/(s ft)
G = G, G = 343.6 30.8
=
374.4 lb/(s ft)
Calculate the no-slip holdup:
z0.72
qL+qs 32+1.241
Calculate the Froude number
N
the mixture viscosity pm and surface
tension oL:
pm = 6.27
x
10-4 [p h pg(i
- h ]
= 6.27 x
=
1.44 lb/(ft/s)
[2.96(0.72) 0.0184(0.28)]
oL=
37.5
-
0.257(API) = 37.5
-
0.257(33)
=
29.0 dyn/cm
10. Calculate the non-slip Reynolds number and the liquid velocity number:
0 25
N
= 1.938vsL( ]
= 1.938 7.25) 47.42/29)0.*5
= 15.88
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534 Production
12. Determine flow pattern:
Since 0.721 0.4 and L, < N,
I
L4
Flow
is intermittent.
13. Calculate the horizontal holdup H, O):
H, O) = ahb/Nc, = 0.845
x
0.721°~555 /7.1860~01m0.692
14. Calculate \~r
nd
HL (0) and two-phase specific weight:
Since e = O , = 1 = 1
H,(O )
=
H, O)h
= 0.692
7
=
yLH, r,Cl HL)= 47.42(0.692) 10.96(1 -0.692)
=
36.19 lb/ft3
15. Calculate the friction factor ratio:
ln(y) = 0.4092
S
=
ln(y)/[-0.0523 3,182 In
y
- 0.8725(1n
y)*
0.01853(1n
y)
= 0.3706
fJf,
=
e =
eoJ706
1.449
16. Calculate the non-slip friction factor f,:
f, = 1/{2 log[ReJ4.5223 log
ReN
3,8215)]1
= 1/(2 1og[195,000/(4.5223 log 195,000 - 3.8215)])2
=
0.01573
17. Calculate the two-phase friction factor:
f, = f,(fJf,)
=
0.01573(1.449) = 0.0227
18. Calculate the pressure gradient:
7 t V m V . p (36.19)(10.06)(2.81)
l 1
(32.2)2,000(
144
-- - gp
L
*~(144) yPsin e
0.0227(374.4)(10.06)
t mvm 36.19( 1)(0
2(32.2)(9/12)
gd
= 0.5646
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Natural Flow Performance
533
L AP
P
L
81.3 or 1.23 x lo- psi/ft
because pressure is decreasing in flow direction to proper value of
Ap/AL
-1.23
x
lO-?si/ft.
Summary
In this work attention was paid only to five methods. These are flow regime
maps, the Duns-Ros method, the Orkiszewski method, the Hagedorn-Brown
method and the Beggs-Brill method. They are the most often used. However, it
is necessary to point out that in literature [19 20 25 26 27] it is possible to find
a lot of other methods. Large numbers of correlations indicate that this problem
has not been properly solved s far. Pressure loss in pipe is a function of a few
parameters. The most important are sort of fluid mixture, working temperature
and pressure, pipe diameters and inclination. In practice the best way
to
evaluate
methods is to make measurement of pressure drop distribution in wells or pipes,
and, next, adjust a proper correlation. It means that for various oil-gas fields
different methods could satisfy the above requirements.
For production purposes pressure gradient is often evaluated based on
Gilbert’s type curves. This method is not accurate, but still is used.
NATURAL FLOW PERFORMANCE
The most important parameters that are used to evaluate performance or
behavior of petroleum fluids flowing from an upstream point in reservoir) to
a downstream point at surface) are pressure and flowrate. According to basic
fluid flow through reservoir, production rate is a function of flowing pressure
at the bottomhole of the well for a specified reservoir pressure and the fluid
and reservoir properties. The flowing bottomhole pressure required to lift the
fluids up to the surface may be influenced by size of the tubing string, choke
installed at downhole or surface and pressure loss along the pipeline.
In oil and gas fields, the flowing systems may be divided into at least four
components, as follow:
1 reservoir
2. wellbore
3.
chokes and valves
4
surface flowline
Each individual component, through which reservoir fluids flow, has its own
performance and, of course, affects each other. A good understanding of the flow
performances is very important
in
production engineering. The combined perfor-
m nces re often used
s
a tool for optimizing well production and sizing equipment.
Futhermore, engineering and economic judgments can depend on good infor-
mation on the well and reasonable prediction of the future performances.
As has been discussed in previous sections, hydrocarbon fluids produced can
be either single phase oil or gas) or two phases. Natural flow performance
of
oil, gas and the mixture will therefore be discussed separately. Some illustrative
examples are given at the end of each subsection.
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