2 Phase Fluid

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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 u

N mFR 2

= (2-38)

m

l l u

u =λ

(2-39, 40) 302.01 316 lL λ=

4684.2 2 0009252.

−= lL λ (2-41, 42) 4516.13 10.

−= lL λ 738.6

4 5. −= lL λ

Determining flow regimes Segregated if λl < .01 and NFR < L1 or λl >= .01 and NFR < L2 Transition if λl >= .01 and L2 < NFR <= L3 Intermittent if .01 <= λl <.4 and L3 < NFR <= L1 or λl >= .4 and L3 < NFR <= L4 Distributed if λl < .4 and NFR >= L1 or λl >= .4 and NFR > L4 For segregated, intermittent and distributed flow regimes use the following:

ψ0ll yy = c FR

b l

l N a

y λ

=0 (2-43, 44)

with the constraint of that yl0 >= λl.

( ) ( )[ ]θθψ 8.1sin333.8.1sin1 3−+= C ( ) ( )gFRfvlell NNdC λλ 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 NL

A FR −

− = AB −= 1 (2-48,49)

ggll yy ρρρ += _

144 sin

_ θρ

cPE g g

dl dp

=⎟ ⎠ ⎞

⎜ ⎝ ⎛ (2-50,51)

The frictional pressure gradient is calculated using:

Dg uf

dl dp

c

mmtp

F

22 ρ =⎟

⎠ ⎞

⎜ ⎝ ⎛ (2-52)

ggllm λρλρρ += n

tp ntp f f

ff = (2-53,54)

The no slip friction factor fn is based on smooth pipe (ε/D =0) and the Reynolds number,

m

mm m

Du N

μ ρ 1488

Re = where ggllm λμλμμ += (2-55,56)

ftp the two phase friction factor is (2-57) Sntp eff = where

[ ] [ ]( )42 )ln(01853.0)ln(8725.0)ln(182.30523.0 )ln(

xxx x

S +−+−

= (2-58)

and 2

l

l

y x

λ = (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 qo = 2000 bpd qg = 1 mmcfpd Temp = 175

oF μo = 2 cp μg = .0131 Pipe = 2.5” ρo = 49.9 lb/ft

3 ρg = 2.6 lb/ft 3 Pressure = 800 psi

First find the flow regime, calculate NFR, λl, L1, L2, L3, and L4. NFR = 18.4, λl = .35, L1=230, L2=.0124, L3= .456, L4= 590. So .01 < λl < .4 and L3 < NFR < L1 so flow is intermittent. Using the table to get a, b and c:

454.0 6.29 35.*845.

0173.0

5351.0

0 === c FR

b l

l N a

y λ

Find C and ψ, d, e, f and g from table:

( ) ( ) ( ) ( ) 0351.06.29*28.10*35.*96.2ln35.1ln1 0978.04473.0305.0 =−=−= −gFRfvlell NNdC λλ

[ ] [ ] 01.1)8.1(sin333.)8.1sin(0351.1)8.1(sin333.)8.1sin(1 33 =−+=−+= θθθθψ C

Find yl 459.01.1*454.0 === ψll yy

The in-situ average density is 3

_ /29.246.2*)459.1(9.49*459. ftlbyy ggll =−+=+= ρρρ

Potential gradient is

ftpsi g g

dl dp

cPE

/169. 144

1*29.24 144 sin

_

===⎟ ⎠ ⎞

⎜ ⎝ ⎛ θρ

For friction gradient First find the mixture density and viscosity 3/1.1965.*6.235.*9.49 ftlbggllm =+=+= λρλρρ cpggllm 709.65.*0131.35.*2 =+=+= λμλμμ

The Reynolds Number 109184

709. 1488*203.*39.13*1.191488

Re === m

mm m

Du N

μ ρ

From Moody plot fn is .0045, solve for S 66.1

459. 35.

2 === l

l

y x

λ

[ ] [ ]( )42 )ln(01853.0)ln(8725.0)ln(182.30523.0 )ln(

xxx x

S +−+−

=

[ ] [ ]( ) 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

ftpsiftlb Dg uf

dl dp

c

mmtp

F

/032./62.4 203.*17.32

94.10*1.19*0066.*22 3 22

====⎟ ⎠ ⎞

⎜ ⎝ ⎛ ρ

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” γg = .65 oil 22

o API qo = 400 stb/day qw = 600 bpd μg = .013 cp σo = 30 dynes/cm σw = 70 dynes/cm GLR = 500 scf/stb @ average conditions βο = 1.063 Rs = 92 scf/stb μo = 17 cp μw = .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.

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 u

N mFR 2

= (2-38)

m

l l u

u =λ

(2-39, 40) 302.01 316 lL λ=

4684.2 2 0009252.

−= lL λ (2-41, 42) 4516.13 10.

−= lL λ 738.6

4 5. −= lL λ

Determining flow regimes Segregated if λl < .01 and NFR < L1 or λl >= .01 and NFR < L2 Transition if λl >= .01 and L2 < NFR <= L3 Intermittent if .01 <= λl <.4 and L3 < NFR <= L1 or λl >= .4 and L3 < NFR <= L4 Distributed if λl < .4 and NFR >= L1 or λl >= .4 and NFR > L4 For segregated, intermittent and distributed flow regimes use the following:

ψ0ll yy = c FR

b l

l N a

y λ

=0 (2-43, 44)

with the constraint of that yl0 >= λl.

( ) ( )[ ]θθψ 8.1sin333.8.1sin1 3−+= C ( ) ( )gFRfvlell NNdC λλ 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 NL

A FR −

− = AB −= 1 (2-48,49)

ggll yy ρρρ += _

144 sin

_ θρ

cPE g g

dl dp

=⎟ ⎠ ⎞

⎜ ⎝ ⎛ (2-50,51)

The frictional pressure gradient is calculated using:

Dg uf

dl dp

c

mmtp

F

22 ρ =⎟

⎠ ⎞

⎜ ⎝ ⎛ (2-52)

ggllm λρλρρ += n

tp ntp f f

ff = (2-53,54)

The no slip friction factor fn is based on smooth pipe (ε/D =0) and the Reynolds number,

m

mm m

Du N

μ ρ 1488

Re = where ggllm λμλμμ += (2-55,56)

ftp the two phase friction factor is (2-57) Sntp eff = where

[ ] [ ]( )42 )ln(01853.0)ln(8725.0)ln(182.30523.0 )ln(

xxx x

S +−+−

= (2-58)

and 2

l

l

y x

λ = (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 qo = 2000 bpd qg = 1 mmcfpd Temp = 175

oF μo = 2 cp μg = .0131 Pipe = 2.5” ρo = 49.9 lb/ft

3 ρg = 2.6 lb/ft 3 Pressure = 800 psi

First find the flow regime, calculate NFR, λl, L1, L2, L3, and L4. NFR = 18.4, λl = .35, L1=230, L2=.0124, L3= .456, L4= 590. So .01 < λl < .4 and L3 < NFR < L1 so flow is intermittent. Using the table to get a, b and c:

454.0 6.29 35.*845.

0173.0

5351.0

0 === c FR

b l

l N a

y λ

Find C and ψ, d, e, f and g from table:

( ) ( ) ( ) ( ) 0351.06.29*28.10*35.*96.2ln35.1ln1 0978.04473.0305.0 =−=−= −gFRfvlell NNdC λλ

[ ] [ ] 01.1)8.1(sin333.)8.1sin(0351.1)8.1(sin333.)8.1sin(1 33 =−+=−+= θθθθψ C

Find yl 459.01.1*454.0 === ψll yy

The in-situ average density is 3

_ /29.246.2*)459.1(9.49*459. ftlbyy ggll =−+=+= ρρρ

Potential gradient is

ftpsi g g

dl dp

cPE

/169. 144

1*29.24 144 sin

_

===⎟ ⎠ ⎞

⎜ ⎝ ⎛ θρ

For friction gradient First find the mixture density and viscosity 3/1.1965.*6.235.*9.49 ftlbggllm =+=+= λρλρρ cpggllm 709.65.*0131.35.*2 =+=+= λμλμμ

The Reynolds Number 109184

709. 1488*203.*39.13*1.191488

Re === m

mm m

Du N

μ ρ

From Moody plot fn is .0045, solve for S 66.1

459. 35.

2 === l

l

y x

λ

[ ] [ ]( )42 )ln(01853.0)ln(8725.0)ln(182.30523.0 )ln(

xxx x

S +−+−

=

[ ] [ ]( ) 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

ftpsiftlb Dg uf

dl dp

c

mmtp

F

/032./62.4 203.*17.32

94.10*1.19*0066.*22 3 22

====⎟ ⎠ ⎞

⎜ ⎝ ⎛ ρ

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” γg = .65 oil 22

o API qo = 400 stb/day qw = 600 bpd μg = .013 cp σo = 30 dynes/cm σw = 70 dynes/cm GLR = 500 scf/stb @ average conditions βο = 1.063 Rs = 92 scf/stb μo = 17 cp μw = .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.