Calculation of Pressure Traverse Using Beggs and Brill

Calculation of Pressure Drop for a hilly terrain pipeline using Beggs & Brill correlation By

Oba Fred Ajubolaka G2011/MENG/PNG/FT/846

& Abiodun Benjamin Odunayo G2010/MENG/PNG/FT/804

MULTIPHASE FLOW IN PIPES

DEPARTMENT OF PETROLEUM & GAS ENGINEERING UNIVERSITY OF PORT HARCOURT AUGUST, 2012

Project Data: q

0

 7140STB/ D

qg  25.7mmcf / D

 g  0.70   0.83  40 API 0

P  425psia Ave.,T  90 F  550 R

d  12in.

Divide Pipeline into two sections Section 1: Rises 300ft., in one mile(5280 ft.) Section 2: Drops 300ft., in 3000ft.,

Solution 1.0

Section 1 Estimate Average Pressure,P let, P  30 psi P  P  P  425 30 / 2  410psia 2

2.0

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



 P 100.0125  API  Rs   g   18 100.00091t 

 1.2048  

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

Rs  0.70( 138.1468)  96.70scf / STB. Determine Bo from the correlation below:    Bo  0.9759 0.00012 R s    

Where, Rs =96.70scf/STB, t= 90°F

Bo =1.0457Bbl/STB

1.2

    g  1.25t    o   

Z-Factor determination using standing & katz correlation Ppc66715.0 37.5 g g Ppc503.75 psia

T pc168325 12.5 g g

T pc394.28R P

pr 

T

pr 

But,

P Ppc

T T pc





410  0.81 503.75

550  1 .4 0 3 9 4.2 8

z = ƒ(Ppr, Tpr) = 0.93

Determination of oil viscosity, µo (d) Using the following correlations: 

od

x



 1 0x  1

3.0324.02023  g

10

t1.163

   1 0.7 1 5Rs  1 0 00.515   b o   od

b  5.4 4Rs  1 5 00.338



o

 3.5 7cp

Where, t = 90°F, Rs =96.70scf/stb and  g  0.70

Determine gas viscosity,µg (e) First we determine ρg 2.7

g 

g

zT

P



2.7  0.70 410  1.52Ibm / ft 3 0.93 550

Using Lee et al Correlation 

g 104 K

 Y    g    X  62.4       

Exp

M g  28.97

g

 2 0.2 8Ib m / Ib m  mo l

 9.4  0.0 2M g T 1.5 K 2 0 9 1 9M g  T   1 1 0.5 3

 9 8 6 X  3 .5     0.0 1M g  5.5 0 T   Y  2.4  0.2 X  1.3 0

 g 0.01155cp

Determine surface tension σo and Bg (f) From the plot of Baker and Swerdloff at 40°API 

od

 28dynes/ cm

Correction factor of 70% , σo will be:  o  2 8d yn es 7 0%  2 8 0.7 0  1 9d yn es/ cm

Bg is determined from z-factor, average pressure and temperature: 0.0283zT 0.0283 0.93 550 Bg    0.0353ft3 / scf p 410

3.0 Determine flowrates & Densities (g)

Gas density, g 

(h)

gP

zT

2.7  0.70 410   1.52Ibm / ft3 0.93 550

Oil density,   o

(i)

2.7

350 o  0.076Rs g 5.615Bo

Oil flowrate, q B  5.615

q  o o o 86400

(j)

 50,36Ibm / ft3

 0.4852ft 3 / s

Gas flowrate,

q  q B  g R o s  g  q   10.218ft 3 / s g 86400

4.0 Determine superficial velocities Superficial velocity of liquid:

Vsl 

qo 4qo 4  0.4852    0.618ft / s Ap   12 d 2

Superficial velocity of gas:

Vsg 

qg Ap



4q g

d 2



4 1 0.2 1 8

  12

 1 3.0 0 9ft / s

Mixture velocity:

Vm  Vsl  Vsg  0.61813.009 13.627ft / s

5.0 Determine flow pattern No slip liquid holdup, Froude Number,

V 0.618   sl   0.0454 l

Vm

13.627

2 Vm 1 3.6 2 72 N    5.7 7 Fr 3 2.1 7 41 gd

Liquid Velocity Number,  N  1.938V  o lv sl    o

   

0.25

 50.36    1.9380.618 19 . 6  

Modified Flow pattern equations:

0.25

L  316l 0.302  3160.04540.302  124.20 1

L  0.000925 2.468  1.91

l L3  0.101.452  8.91 l L4  0.56.738  5.59108 l 2

 1.516  1.52

Flow Pattern Since, l  0.0 1 L , 2  N Fr  L3 Flow pattern is Transition: transition pattern is between segregated and intermittent flow pattern

6.0 Determine HL for Segregated & Intermittent flow patterns Segregated Pattern Calculation: ab

 0.9800.04540.4846 l H 0    0.188 L 0 . 0868 N 5.57 Fr

300ft ϴ1

5280ft

ϴ2 3000ft

Fig. 1.1 Schematic of the Pipeline as it rises and drops  300    3.25 5280  

1  tan1

 300    5.71 3000  

 2  tan1

Segregated HL Determination (contd) H

L1

H

L0



  3       1.0  c  sin 1.81  0.3 3 3sin 1.81     f g h  C  1.0  l ln e N N  Fr l lv   e  0.0 1 1 f  3.7680 g  3.5390 h  1.6140  c  5.5 2 3   1.5 6 1 H

L 1 seg

H

L 0 

  0.1 8 8 1.5 6 1 0.2 9 3 5

Determine HL for Intermittent flow patterns 

ab

L Nc Fr

 0.8 450.0 45 40.5351   0.1 56 7 5.7 70.0173

Hl

L 0 

Hl

L 1 int ermitte nt

H

L 0 



 f g  h C  1.0  L lne N N  L Lv Fr   e  2.9 60 f  0.3050 g  0.4473 h  0.0978  c  0.1 21 6   1.0 12 4 H H   0.1 56 7 1.1.0 12 4 0.1 59 L 1 int er

L 0 

HL for Transitional flow pattern Determine, A from: N 3  N Fr 8.9 1 5.7 7 A  N3  N 2 8.9 1 1.9 1 A  0 .4 4 8 6 H

L 1 trans

H

L 1 trans

H

L 1 trans

 AH

L 1 se g .

 1  AH

L 1 Int.

 0.4 4 8 60.2 9 3 5  1  .4 4 8 60.1 5 9  0 .2 1 9 3

7.0 Determine Actual & Non slip denties s  LH

L



g

1  H  L

 s  5 0.3 60.2 1 9 3  1.5 21  0.2 1 9 3  s  1 2.2 3Ib m / ft 3 .



 ns   L  L   g 1   L



 ns  5 0.3 60.0 4 5 4  1.5 21  0.0 4 5 4  ns  3.7 4Ib m / ft 3 .

8.0 Determine the friction factor, ƒtp. Determine Nre., µn, and ƒn: N

Re

 1488

n   LL

 ns V m d

n  g 1  L





1  0.0 4 5 4  n  3.5 70.0 4 5 4  0.0 1 1 5 5  n  0.1 7 3cp N

3.7 41 3.6 2 71

Re

 1488

 4.3 8 1 05

0.173

e / d  o .o o o o6( a ssu mp tio n ) f

n



 f N

R e ,e



/ d  0.0 2( mo o d y

ch a rt)

Determine the friction factor, ƒtp.(Contd) Determine y, s and ƒtp: y



L

H   L 1 Trans 

2



0.0 4 5 4

0.2 1 9 32

 0.9 4 4

ln y

s

 0.0 5 2 3 3.1 8 2ln y  0.8 7 2 5ln y 2  0.0 1 8 5 3ln y 4 s  0.2 4 2 f f f f

tp

 e s  e0.242  1.2 7 4

n

tp tp

 1.2 7 4fn  1.2 7 4 0.0 2  0.0 2 5 4 8

9.0 Determine Pressure Gradient  dP   dP   dP   dP              d L T  d L  f  d L el  d L  acc.  dP     0,  d L  acc.  dP     d L  T

 ftp ns V m 2 2g d c

  g sin1  s

g

c

 dP   1 2.2 33 2.1 7 4sin3.2 5 0.0 2 5 4 83.7 41 3.6 2 72     23 2.1 7 41 3 2.1 7 4  d L T  dP     0.0 0 6 7 2p si / ft .  d L T

10.0 Determine uphill Pressure Drop, ∆P For an horizontal distance of 0ne mile(5280ft.) Pressure drop will be:  dP  psi P     L  0.00672  5280ft . ft .  dL T P  35.48psi

Section 2 of Pipeline Fluid properties have been determined at average Pressure 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 for segregated and intermittent flow patterns and there after determine the transition holdup.  From that point we can determine the two phase Friction factor and move on to compute the pressure drop 

Determine Liquid Holdup for Segregated flow pattern H

L  2 se g

 H

L 0 



  300   2  ta n1    5.7 1 H

L 0 

 3000  0.1 8 8

 f g Nh  c  1   L ln e N  L Lv Fr   e  4.7 0 0f  0.3692 g  0.1 2 4 4 h  0.5056 c  1.7 7 0 6     1.0  c  sin1.8 2   0.3 3 3sin3 1.8 2      0 .6 8 7 4 H H

L  2 se g L  2 se g

 H

L 0 

   0.1 8 8 0.6 8 7 4

 0.1 2 9 2

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 c  1.7706 and is the same for both segregated and intermittent flow patterns in downhill. Hence,   0.6874 H

L  2 int .

H

H

L  2 int .

 0.1567 0.6874 0.1077

L 0 



     2 H L Trans  0.1173 2

 

 H L  Trans  AH L  seg  1  AH L  int . 2 2 2 H L  Trans  0.44860.1292  1  0.44860.1077

Determine actual density of fluid on downhill   H s

L

L



g

1  H  L

  50.360.1173  1.521  0.1173 s

  7.25Ibm / ft 3 s

Determine 2-phase friction factor    4.38105  already det er min ed  Re ln y   s  0.0523 3.182 ln y  0.8725ln y     e / d  0.00006 smooth pipe s  0.4699   ftp fn  f N ,e / d  0.02 s 0.4699 N

 



y

Re

L



H L 2Tr .2



0.0454 0.11732

 3.30

fn

e e

 2  0.01853ln y 4

 1.60

ftp  1.60 f n  1.60 0.02  0.032

Determine Pressure gradient  dP   dP   dP   dP              dL T  dL  f  dL el  dL acc.

2 f  Vm

 dP  tp ns     2 gc d  dL T

 g sin 5.71

 s

gc

 dP   0.0323.7413.6272 7.2532.174 sin5.71      232.1741 32.174  dL T  dP     0.345 0.721 0.376psf / ft  0.002611psi / ft  dL T

Determine Pressure Drop For an horizontal distance, L=3,000ft.  dP    L P    d L T p si P  0.0 0 2 6 1 1  3,0 0 0ft ft P  0.0 0 2 6 1 1  3 0 0 0p si P  7.8 3p si

Total Pressure drop for both uphill & downhill

P  P T

uphill

 P

downhill

P  35.48 7.83  43psi T

P  43psi T

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.