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.