Thickness(1)

7/30/2019 Thickness(1)

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Calculate the Velocity of Gas in 6" Pipeline (Pipeline Rules of Thumb Handbook)

Min Max

v = 1440 * Z * Q * T where : v = velocity of gas, ft/sec 5 30

d2

* P Z = Compressibility 0.94 0.94

Q = Volume, MMSCFH

T = Operating Temperatur, R 660 660

d = Inside diameter of pipe , inches 6.193 6.193

P = Operating Pressure, psia 600 600

Q = 0.128792073 MMSCFH

3.088538921 MMSCFD Minimum

Q = 0.772752438 MMSCFH

18.53123352 MMSCFD Maksimum

maksimum velocity should be limited to acceptable noise level (60 to 80 ft/s)

minimum velocity should be limited to about 3 - 10 ft/s refer to API RP 14 E if there are solid particles

Hasil perhitungan diatas menunjukan bahwa pipa 6" dapat mengalirkan gas dengan kapasitas maksimum sebesar

37 MMSCFD dan minimum sebesar 3 MMSCFD

Sumur Makmur #25 berada pada satu area dengan sumur-sumur makmur 1, 2, 4, 6, 17 dan 23. Sumur-sumur tersebu

mempunyai zona G-50 series yang kandungan gasnya sangat potensial dan apabila ke-6 sumur tersebut zona gasny

produksikan, maka fasilitas 6" trunkline yang diajukan akan dapat menampung gas dari ke-6 sumur tersebut.


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t

di


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LINE NO : 121-6-GR-CCD-XXX

LEGENT : Trunkline Calculation

AFE NO : 08-1318

TABLE OF CONTENTS

1 GENERAL DATA

2 Pipe Thickness Calculation


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0



AFE NO : 08-1318

1. GENERAL DATA

Design Code =

Design Pressure =

Corrosion Allowance =

Design Temperatur =

MATERIAL =

Basic Quality Factor For longitudinal weld joints =

Radiography =

Diameter =


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AFE NO : 08-1318

2 Thickness Calculation

tm = t + c

1 - u t = Pi do

2 Smys E F T

where :

tm = minimum required thickness

t = pressure design thickness, mm

c = the sum of the mechanical allowance

Pi = internal design pressure, Psi

do = outside diameter of pipe, mm

Smys = Basic allowable Stres forValue

F = Design Factor

E' = longitudinal weld joint factor

T = temperature derating factor

Pi = 1350 Psi

do = 6.625 inch

c = 0.063 inch

= 0.0362 inch (Accumalation corrosion for 5 years operation) NORSOK-M506

S' = 35000 Psi

E' = 1

F = 0.72

u = 12.5%

T = 1 (Less than 250 F)

Pipe thickness caused by pressure

t = 0.177 inch

Minimum required thickness

tm = 0.316 inch

Sch 40 for NPS 6 = 0.280 inch


Conclusion

We choose Sch 80 NPS 6


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LINE NO :

LEGENT : Wellhead Flowline Calculation

AFE NO : AFE 09-1333

TABLE OF CONTENTS

1 GENERAL DATA



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LINE NO :



1. GENERAL DATA

Design Code =

Design Pressure =


Design Temperatur =

MATERIAL =


Radiography =

Diameter =


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LINE NO :




tm = t + c

1 - u t = Pi do

2 (S . E + Pi . Y)

where :




Pi = internal design pressure, Psi


S' = Basic allowable Stres forValue


Y' = coefficient having values ferritic steels

as follow 0.4 up to and including 480C

0.5 for 510C

0.7 for 540C and above

u = mill tolerance

Pi = 1350 Psi

do = 3.5 inch

c = 0.063 inch

= 0.066 inch 2% of CO2 contentS' = 20000 Psi

E' = 1

Y = 0.4

u = 12.5%

Pipe thickness caused by pressure

t = 0.115 inch

Minimum required thickness

tm = 0.279 inch



Conclusion

We choose Sch 80 NPS 3


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1 GENERAL DATA

Design Code = ASME B31.3

Operating Pressure = 1137 Psi

Working Allowable Pressure = 1350 Psi (ANSI 600)

Corrosion Allowance = 0.063 inch = 1.6 mm

Spesific Gravity = 0.74

Design Temperatur = 200 F = 93.33 C

MATERIAL = API 5L Gr 5 Sch 40

Basic Quality Factor For longitudinal weld joints = 1

Radiography = Spot

Diameter = NPS 4


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LINE NO :



TABLE OF CONTENTS

1 GENERAL DATA



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LINE NO :



1. GENERAL DATA

Design Code =

Design Pressure =


Design Temperatur =

MATERIAL =


Radiography =

Diameter =


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LINE NO :




tm = t + c Pi do where :

1-u 2 Smys E 0.72

tm = minimum requ

u = mill tolerances

c = the sum of the

Pi = internal design

do = outside diamet

S' = Specific minim

E' = longitudinal we

P = 1350 Psi

do = 4.5 inch

c = 0.063 inch

S' = 35000 Psi (ASME B31.4 hal 12)

E' = 1 (ASME B31.4 hal 12)

Y' = 0.40

u = 12.50%

t = 0.120535714 inch

tm = 0.20976 inch

Thickness 4" Sch 40 = 0.237 inch

tm < tSch40The 4" Sch 40 thickness is acceptable

To determine allowable internal working pressure for this wellhead flowline in accordance with ANSI B31.4, Code for pipline transportatio

t =


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Pi do

2 (S . E + Pi . Y)

where :




Pi = internal design pressure, kPa


S' = Basic allowable Stres forValue


Y' = coefficient having values ferritic steels

as follow 0.4 up to and including 480C

0.5 for 510C

0.7 for 540C and above

Pi = 203 Psi

do = 4 inch

c = 0.12 inchS' = 20000 Psi 1.77

E' = 0.85

Y' = 0.4

u = 12.5% t

Pi do = 0.024 inch

2 (S . E + Pi . Y)

tm = 0.147 inch 3.72719

sch 40 for NPS 4 = 0.237

and the calculation = 0.242 it shows that sch 40 is not adequate

Design thickness calculation indicates that Schedule 80 is appropriate

t =

To determine allowable internal working pressure for this wellhead flowline in accordance with ANSI B31.3, Co

transportation systems, use the following :

tm = t + c + u t =


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203

for pipline


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tm = t + ca Pi do where :

1-a 2 Smys E 0.72


t = pressure design thickness, m

a = mill tolerancese = erosion allowance

c = the sum of the mechanical all



S' = Specific minimum yield stren


Y' = coefficient having values ferri

as follow

0.5 for 510C

0.7 for 540C and

Pi = 1350 Psi

do = 3.5 inch

c = 0.063 inch

e = 0.02 inch

S' = 35000 Psi (ASME B31.4 hal 12)

E' = 1 (ASME B31.4 hal 12)

Y' = 0.40

a = 12.50%

Pi do = 0.09375 inch

2 Smys E 0.72

tm = 0.202 inch



Thickness calculation = 0.184 inch


To determine allowable internal working pressure for this wellhead flowline in accordance with ANSI B31.4, Code for pi

transportation systems, use the following :

0.4 up to and incl

t =

t =


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wance

th

tic steels

above

line

uding 480C


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tm = t + c Pi do where :

1-u 2 Smys E F T


t = pressure design thickness, m

a = mill tolerancese = erosion allowance

c = the sum of the mechanical all



S' = Specific minimum yield stren


F = design factor

T = temperature derating factor

Pi = 1350 Psi

do = 6.25 inch

c = 0.063 inch

= 0.036 inch (Accumalation corrosion for 5 years operation) NORSOK-M506

e = inch

S' = 20000 Psi

E' = 1

F = 0.72

T = 1

a = 12.50%

Pi

do

= 0.29296875 inch

2 Smys E 0.72

tm = 0.448192857 inch



Thickness calculation = 0.184 inch


To determine allowable internal working pressure for this wellhead flowline in accordance with ANSI B31.8, Code for G

and distribution piping systems, use the following :

t =

t =


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wance

th

s Trnamission


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Gas Flowrate :35.9 Mscfd = 35900 Mscfd

Operating Pressure : 260 Psi

S1 : 0.858

Sg : 0.65

Fluid Flowrate : 200000 BFPD

Ve = c = A =

Where :

Ve : Fluid erosional velocity, feet/second

c : empirical constant, Solid-free Fluids 100 for continuous service,

125 for intermittent service

S1 : liquid specific gravity

Sg : gas specific gravity

: Gas or liquid mixture density at flowing pressure and temperature, lbs/ft3

A : minimum pipe cross sectional flow area

R : gas/liquid ratio, ft/barrel at standard conditions. 0.1795 scfd/bfpd

T : operating temperatur, R

Z : gas compressibility factor

Ve = 100 A = 9.367

7.314 13.673

= 13.67 = 0.685

= 137.0200476 inch2

D = 13.21165513 inch

This calculation describe that NPS 14 is appropriate

To determine internal diameter pipe for Walio pipeline in accordance with API RP 14 E (Multiphase Flow)

12409SlP + 2.7 RSgP

198.7P + RTZ

9.35 + (RTZ/21.25P)

Ve


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Thickness(1)

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