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Calculation of Pressure Traverse Using Beggs and Brill

Evaluation of Models to Predict Liquid Loading in Gas wells - NTNU

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tion of Pressure Traverse Using Beggs and Brill

Description

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,

Project Data: q

7mmcf / D

P  425psia Ave

,T  90 F  550 R

Divide Pipeline into two sections Section 1: Rises 300ft

Solution 1

Section 1 Estimate Average Pressure,P let,

P  30 psi P  P  P  425 30 / 2  410psia 2

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

°API = 40,

P =410 psia

Rs  0

70( 138

Determine Bo from the correlation below:    Bo  0

9759 0

70scf/STB,

Z-Factor determination using standing & katz correlation Ppc66715

0 37

75 psia

T pc168325 12

5 g g

T pc394

28R P

410  0

550  1

4 0 3 9 4

Tpr) = 0

Determination of oil viscosity,

µo (d) Using the following correlations: 

0324

Determine gas viscosity,µg (e) First we determine ρg 2

7  0

93 550

Using Lee et al Correlation 

 Y    g    X  62

M g  28

4  0

4  0

2 X  1

01155cp

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

Correction factor of 70% ,

σo will be:  o  2 8d yn es 7 0%  2 8 0

Bg is determined from z-factor,

average pressure and temperature: 0

0283zT 0

0283 0

Gas density,

7  0

93 550

Oil density,

Oil flowrate,

Gas flowrate,

q  q B  g R o s' g  q   10

Vsl 

Superficial velocity of gas:

Vsg 

4 1 0

Mixture velocity:

Vm  Vsl  Vsg  0

61813

009 13

627ft / s

Froude Number,

2 Vm 1 3

Liquid Velocity Number,

9380

6  

Modified Flow pattern equations:

L  316l 0

04540

L  0

468  1

101

452  8

56

738  5

516  1

Flow Pattern Since,

0 1 L',

9800

04540

188 L'0

57 Fr

300ft ϴ1

ϴ2 3000ft

Segregated HL Determination (contd) H

L1

L0

  3       1

3 3 3sin 1

5 6 1 H

L 1 seg

L 0 

1 8 8 1

5 6 1 0

2 9 3 5

Determine HL for Intermittent flow patterns 

L Nc Fr

0 45 40

7 70

L 0 

L 1 int ermitte nt

L 0 

L 0 

HL for Transitional flow pattern Determine,

A from: N 3  N Fr 8

9 1 5

9 1 1

4 4 8 6 H

L 1 trans

L 1 trans

L 1 trans

L 1 se g

L 1 Int

2 1 9 3

3 60

 ns   L' L'  g 1   L

3 60

Determine Nre

5 70

1 7 3cp N

Determine the friction factor,

(Contd) Determine y,

H   L'1 Trans 

0 4 5 4

242  1

2 7 4 0

0 2  0

0 2 5 4 8

 dP     0,

 dP     d'L' T

 ftp ns V m 2 2g d'c

∆P For an horizontal distance of 0ne mile(5280ft

) Pressure drop will be:  dP  psi P     L' 0

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

L 0 

  300   2  ta n1    5

L 0 

 f g Nh  c' 1   L'ln e N  L'Lv Fr   e  4

L  2 se g L' 2 se g

L 0 

1 8 8 0

6 8 7 4

1 2 9 2

Determine Liquid holdup for Intermittent flow pattern

We have already determined Holdup for Intermittent flow pattern,

HL(0) = 0

L  2 int

L  2 int

1567 0

6874 0

L 0 

     2 H L Trans  0

 H L' Trans  AH L' seg  1  AH L' int

1077

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

360

1173 s

Determine 2-phase friction factor    4

0523 3

60 0

02  0

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

71

0323

7413

2532

345 0

721 0

Determine Pressure Drop For an horizontal distance,

L=3,000ft

 dP    L'P    d'L'T p si P  0

8 3p si

Total Pressure drop for both uphill & downhill

48 7

The Beggs and Brill Correlation is iterative

The calculated pressure drop is not equal to the estimated pressure drop,

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