ByBy

Oba Fred AjubolakaOba Fred AjubolakaG2011/MENG/PNG/FT/846G2011/MENG/PNG/FT/846

Abiodun Benjamin OdunayoAbiodun Benjamin OdunayoG2010/MENG/PNG/FT/804G2010/MENG/PNG/FT/804

MULTIPHASE FLOW IN PIPESMULTIPHASE FLOW IN PIPES

DEPARTMENT OF PETROLEUM & GAS ENGINEERINGDEPARTMENT OF PETROLEUM & GAS ENGINEERINGUNIVERSITY OF PORT HARCOURTUNIVERSITY OF PORT HARCOURT

AUGUST, 2012AUGUST, 2012

&

Project Data:

D/STBq 71400

D/mmcf.qg 725

700.g API. 408300 .ind 12

RFT.,Ave 55090

psiaP 425

Divide Pipeline into two sectionsDivide Pipeline into two sections

Section 1: Rises 300ft., in one mile(5280 ft.)Section 1: Rises 300ft., in one mile(5280 ft.)Section 2: Drops 300ft., in 3000ft., Section 2: Drops 300ft., in 3000ft.,

SolutionSolutionSection 1

1.0 Estimate Average Pressure,let,

2.0 Determine fluid properties from relevant correlation at 410psia and 90°F

Determine from:

psiP 30P

2

PPP

psia/ 410230425

sR

20481

000910

01250

10

10

18

.

t.

API.gs

PR

Where, t=90°F , °API = 40, P =410 psia

.STB/scf.).(.Rs 70961468138700

21

25100012097590

.

t.sRoo

g..B

Where, Rs =96.70scf/STB, t= 90°F

Bo =1.0457Bbl/STB

Determine Bo from the correlation below:

Z-Factor determinationusing standing & katz correlation

gg ..pcP 537015667

gg .pcT 512325168

psia.pcP 75503

R.pcT 28394

pcP

Ppr

P 810

75503

410.

.

pcT

Tpr

T 401

28394

550.

.

But, z = ƒ(Ppr, Tpr) = 0.93

Determination of oil viscosity, µo

(d) Using the following correlations:

cp.

R.b

R.

tx

o

o

g

od

.s

bod

.s

.

..

x

573

150445

10071510

10

110

3380

5150

1631

0202303243

Where, t = 90°F, Rs =96.70scf/stb and 700.g

Determine gas viscosity,µg

(e) First we determine ρg

Using Lee et al Correlation

3521550930

4107007272

ft/Ibm..

..

zT

. Pg

g

cp.g

g

g

.g

.g

Kg

.X..Y

.M.T

.X

.TM

TM..K

molIbm/Ibm.Mg

Y

.g

X

Exp

011550

51

9728

10

3012042

505010986

53

5311019209

02049

2820

4624

Determine surface tension σo and Bg

(f) From the plot of Baker and Swerdloff at 40°API

Correction factor of 70% , σo will be:

Bg is determined from z-factor, average pressure and

temperature:

cm/dynesod

28

cm/dynes.%dyneso 19700287028

scf/ft...

p

zT.Bg

303530410

5509300283002830

3.0 Determine flowrates & Densities(g) Gas density,

(h)Oil density,

(i)Oil flowrate,

(j)Gas flowrate,

3521550930

4107007272ft/Ibm.

.

..

zT

. Pgg

336506155

0760350ft/Ibm,

B.

R.

o

gsoo

s/ft..Bq

q ooo

34852086400

6155

s/ft.Bqq

qgRsog

g321810

86400

4.0 Determine superficial velocities

s/ft.

.

d

q

A

qV o

po

sl 61801

4852044

22

s/ft.

.

d

q

A

qV

g

p

gsg 00913

1

2181044

22

Superficial velocity of liquid:

Superficial velocity of gas:

s/ft...VVV sgslm 62713009136180

Mixture velocity:

5.0 Determine flow patternNo slip liquid holdup,

Froude Number,

Liquid Velocity Number,

Modified Flow pattern equations:

0454062713

6180.

.

.

V

V

msl

l

775117432

62713 22.

.

.

gd

VN m

Fr

5215161619

3650618093819381

250250..

.

...V.N

..

oo

sllv

2012404540316316 302030201

..L ..l

873864

45213

4682

1059550

918100

91100092502

..L

..L

..L

.l

.l

.l

Flow PatternSince, ,Flow pattern is Transition: transition pattern is

between segregated and intermittent flow pattern

010.l 32 LNLFr

6.0 Determine HL for Segregated & Intermittent flow patterns

Segregated Pattern Calculation:

1880575

045409800

08680

484600 .

.

..

N

aH

.

.bl

FrL

300ft

5280ft 3000ftϴ1

ϴ2

Pipeline uphillPipeline downhill

7153000

300

2535280

300

12

11

.tan

.tan

Fig. 1.1 Schematic of the Pipeline as it rises and dropsFig. 1.1 Schematic of the Pipeline as it rises and drops

Segregated HL Determination (contd)

01 LL

HH

2935056111880

5611

5235

0110

01

8133308101

01

13

1

...HH

.

.c

.e

NNeln.C

.sin..sinc.

LsegL

hFr

glv

fll

76803.f 53903.g 61401.h

Determine HL for Intermittent flow patterns

159001241115670

01241

12160

9602

01

15670775

045408450

01

01

0 01730

53510

....HH

.

.c

.e

NNeln.C

HH

..

..

N

aH

LerintL

LermittentintL

L

hFr

gLv

fLL

l

.

.

cFr

bLl

30500.f 44730.g 09780.h

HHLL for Transitional flow pattern for Transitional flow pattern

Determine, A from:

21930

1590448612935044860

1

44860

911918

775918

1

1

111

233

.H

....H

HAAHH

.A

..

..

NN

NNA

transL

transL

.IntL.segLtransL

Fr

7.0 Determine Actual & Non slip denties

.ft/Ibm.

....

.ft/Ibm.

....

HH

ns

ns

LgLLns

s

s

LgLLs

3

3

743

045401521045403650

1

2312

219301521219303650

1

8.0 Determine the friction factor, ƒtp.

Determine Nre., µn, and ƒn:

)chartmoody(.Nff

)assumption(oooo.od/e

...

N

cp.

....

dVN

d/e,

.

gn

Ren

Re

n

n

LLLn

mnsRe

020

6

10384162713743

1488

1730

04540101155004540573

1

1488

51730

Determine the friction factor, ƒtp.(Contd)

Determine y, s and ƒtp:

025480

02027412741

2741

2420

01853087250182305230

944021930

04540

2420

42

22

1

.f

..fn.f

.eef

f

.s

yln.yln.yln..

ylns

..

.

H

y

tp

tp

n

tp

L

L

.s

Trans

9.0 Determine Pressure Gradient

.ft/psi.dL

dP

.

.sin..

.

...

dL

dP

g

sing

dg

Vftp

dL

dP

,dL

dP

dL

dP

dL

dP

dL

dP

dL

dP

T

T

T

.acc

.accelfT

c

s

c

mns

006720

17432

253174322312

1174322

62713743025480

2

0

2

12

10.0 Determine uphill Pressure Drop, ∆PFor an horizontal distance of 0ne mile(5280ft.)Pressure drop will be:

psi.P

.ft.ft

psi.L

dL

dPP

T4835

5280006720

Section 2 of Pipeline Fluid properties have been determined at averagePressure of 410psia and Temperature of 90°F. So we continue to with these properties for downhill

calculations. We shall proceed to determining the Liquid holdups

forsegregated and intermittent flow patterns and there

after determine the transition holdup. From that point we can determine the two phaseFriction factor and move on to compute the pressure

dropfor the downhill segment.

Determine Liquid Holdup for Segregated flow pattern

12920

687401880

68740

8133308101

77061

7004

1

1880

7153000

300

2

02

0

02

23

2

12

.H

..HH

.

.sin..sinc.

.c

.e

NNelnc

.H

.tan

HH

segL

LsegL

L

LsegL

hFr

gLv

fLL

36920.f 12440.g 50560.h

Determine Liquid holdup for Intermittent flow pattern.

We have already determined Holdup for Intermittent flow

pattern, HL(0) = 0.1567 in the uphill segment of pipeline

and is the same for both segregated and

intermittent flow patterns in downhill. Hence,

77061.c

68740.

11730

107704486011292044860

1

107706874015670

2

2

222

2

02

.H

....H

HAAHH

...H

HH

TransL

TransL

.intLsegLTransL

.intL

L.intL

Determine actual density of fluid on downhill

3257

117301521117303650

1

ft/Ibm.

....

HH

s

s

LgLLs

Determine 2-phase friction factor

30311730

04540

020

000060

10384

222

5

..

.

H

Ly

.d/e,Nffn

pipesmooth.d/e

edminerdetalready.N

.TrL

Re

Re

0320020601601

601

46990

01853087250182305230

46990

42

...f.f

.eef

f

.s

yln.yln.yln..

ylns

ntp

.sn

tp

Determine Pressure gradient

ft/psi.ft/psf...dL

dP

.

.sin..

.

...

dL

dP

g

.sing

dg

Vf

dL

dP

dL

dP

dL

dP

dL

dP

dL

dP

T

T

cc

m

T

.accelfT

snstp

0026110376072103450

17432

71517432257

1174322

627137430320

715

2

2

2

Determine Pressure DropFor an horizontal distance, L=3,000ft.

psi.P

psi.P

ft,ft

psi.P

LdL

dPP

T

837

30000026110

00030026110

Total Pressure drop for both uphill & downhill

psiP

psi..P

PPP

T

T

downhilluphillT

43

438374835

The Beggs and Brill Correlation is iterative. The calculated pressure drop is not equal to the estimated pressure drop, hence, the calculated pressure becomes our new estimated pressure drop and process is repeated to achieved the condition where, estimated pressure drop equals calculated pressure drop.