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.



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:



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:



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,



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)



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



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



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



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



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



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



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.

beggs - brill method.pdf

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