HW5 2016 15512006

8/15/2019 HW5 2016 15512006

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Tugas 5 KL- 4220 PIPA BAWAH LAUTDosen : Prof. Dr. Ir. Ricky Lukman Tawekal, MSE/ Eko Charnius Ilman, ST, MT

Diberikan : Se lasa, 16 Februari 2016Dikumpul : Senin, 29 Februari 2016 (Jam 7.00 WIB Pagi sebelum kuliah dimulai)

Nama : ______________________________________

NIM : ______________________________________

BAGIAN A – FREE SPAN ANALYSIS (BOBOT 100%)

Didapatkan data-data dua buah pipa penyalur bawah laut yang bersebelahan dengan jarak 100meter pada kedalaman 20 s.d. 100m dengan lokasi sekitar 450m dari Platform, seperti berikut:

Data proses: Design temperature 1000C Berat jenis gas 35 kg/m3 or 2.185

pcf

Pressure design 1230 psi Production life 25 thn

Corrosion Rate 0.06 mm/yearData pipa :

Material API 5L X-65 Berat jenis baja 490 pcf Pipa pertama: NPS 60”, Wall thickness 1

inch. Tebal concrete coating 3 inch. Pipa kedua: NPS 16”, Wall Thickness 0.5

inch. Tebal concrete coating 2 inch.

Berat jenis beton 190 pcf

Young’s Modulus baja 3 x 107 psi

Poisson’s Ratio baja 0.3Data lingkungan:

Seawater Density 64 pcf Kinematic Viscosity of Seawater = 1.05E-6 m2sec-1

Hydrodynamic coeff:

Cd = 0.9Cm = 3.29CL = 0.9

Astronomical Tide:o HAT: 1.8mo LAT: 1.4m

Kecepatan arus:1 Year (m/s) 100 Year (m/s)

0% WD 0.74 0.93

20% WD 0.69 0.8140% WD 0.57 0.68

90% WD 0.51 0.63100% WD 0.47 0.57

Gelombang:

Return Period 1-yrs 100-yrs

Significant wave height (m) 2 3.7

Significant wave period (s) 4.2 5.5

Maximum Individual height (m) 4 6

Note: Tp = 1.05*TsTugas Anda:

Untuk kedua pipa di atas, hitung Allowable free span secaraa. Static danb. Dynamic (3 kondisi: hydrotest, operasi, instalasi)

15512006 REEZALI RAHARJAYA 1 of 84.

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NPS 60 --> 60 inch

REEZALI RAHARJAYA

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BAGIAN B – STRESS ANALYSIS AND GLOBAL BUCKLING (BOBOT 100%)

Pipa bawah laut dengan NPS 20 Schedule 40 sepanjang 40 km, dengan material baja API 5L X52memiliki lapisan pelindung anti korosi tipe 3LPP (tebal 3 mm), lapisan pemberat beton dengan tebal1 in. Service berupa gas dengan densitas 80 pcf. Pipa didesain dengan tekanan desain 15 MPa dantemperature desain 100C ini berada di atas seabed dengan kedalaman 30 meter. Dari hasil soil

investigation, diketahui tipe tanah dasar laut adalah pasir dengan sudut gesert tanah 35 dengan

densitas 18835.6 N/m3 dan undrained shear strength 14000 Pa. Kondisi perairan memilikitemperature ambient 25C. Diketahui factor friksi pipa dengan tanah sebesar 0.3. Koefisienekspansi thermal sebesar 1.1710-5 /C. Jarak antara ujung bebas 62 m.

Tugas Anda:

1. (40%) Lakukan analisis upheaval buckling mengikuti metode pada paper AC Palmer OTC

6335. Jika diketahui tinggi imperfection sebesar 1 meter,a. cek apakah diperlukan tambahan gravel dumping untuk mengubur pipa sebagai tambahan

pemberat untuk kestabilan;b. bandingkan regangan yang terjadi terhadap regangan izin berdasarkan kriteria strain based

design.

2. (40%) Untuk pipa yang sama, jika diketahui batimetri dan profil temperature sebagaimanaGambar 1. Lakukan analisis expansi thermal. Kemudian hitunglah besar total ekspansi thermaldan virtual anchor length pada hot-end dan cold-end!

Gambar 1 Profil kedalaman pipa (kiri) dan profil temperatur (kanan) sepanjang pipa

Catatan: Untuk data yang Anda anggap kurang, dapat digunakan asumsi namun ingat untuk mencantumkan sumbernya.

3. (15%) Dengan kata-kata Anda sendiri, berikanlah penjelasan berikut sketsanya mengenai

perilaku global buckling pada suatu potongan pipa dengan kapasitas buckling yang beragamsepanjang pipa, hingga terjadi buckling pertama dan kedua. Anda harus menjelaskanbagaimana prilaku sebelum buckling dan mekanisme yang terjadi saat buckling pertama dankedua. Berikan sketsa yang jelas dan mudah dimengerti untuk setiap penjelasan Anda. Di akhirpenjelasan Anda, tambahkan pula penjelasan apa yang dimaksud dengan dengan globalbuckling sebagai fenomena beban, bukan fenomena kegagalan.

4. (5%) Buatlah sebuah tabel yang memberikan perbedaan antara teori kegagalan : (1) vonmises, (2) tresca, dan (3) rankine. Penjelasan yang komprehensif berbobot lebih.

Catatan:

1. Semua proses perhitungan tahap demi tahap beserta rumus dan keterangan rumus yang dipakai

WAJIB Anda tuliskan! TIDAK BOLEH hanya melampirkan angka-angka tanpa rumus yang dipakai atau

bahkan hanya hasil akhir saja! Pekerjaan Anda harus bisa ditelusuri dengan mudah. Sumber setiap

rumus yang dipakai harus dicantumkan. Misal: “Sumber: Equation 2.3 DnV RP E305 page 20”. Jangan

mencantumkan referensi misalnya:”Slide kuliah balapan atau contoh spreadsheet dst..”. Cek di

standar/codenya langsung.

2. Perhitungan yang Anda lakukan untuk soal ini harus menggunakan EXCEL, Mathcad atau MATLAB.

Tidak boleh menggunakan tulis tangan.--Soal Selesai—

“Perilaku Anti korupsi dimulai sejak dini.. Dilarang mencontek!”

0

50

100

150

0 10 20 30 40 T e m p e r a t u r e ( C )

KP


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BAGIAN A

Untuk pipa pertama:

NPS 60. Wall thickness= 1 in. Concrete thickness= 4.5 in (dibesarkan karena dengan tebal 3 inch, pipa

tidak stabil)

Dari perhitungan pada halaman selanjutnya ini, dapat disimpulkan panjang free span yang diijinkan:

Condition Allowable Free Span (ft)

Installation Hydrotest Operation

Static 437.992 306.759 355.971

Dynamic 372.89 343.262 333.69

Keterangan: Perhitungan On-Bottom Stability (OBS) dapat diperoleh di bagian lampiran.


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Free Span Analysis - Hydrotest NPS 60

1. INPUT PARAMETER pcf lb ft

3-:= C K

1.1 Pipeline Design Parameter

Outer Diameter Ds 60in:=

Wall Thickness ts 1in:=

Internal Diameter ID Ds 2ts-:=

Concrete Coang tcc 4.5in:=

Modulus Elascity E 30000ksi:=

Design Pressure Po 1845psi:=

Concrete Coang Density ρcc 190pcf :=

Content Density ρcont 62.4pcf :=

Seawater Density ρsw 64pcf :=

Steel Density ρs 490pcf :=

Design Temperature Td 373K :=

Poisson Rao υ 0.3:=

SMYS API 5L X-65 SMYS 65ksi:=

Corrosion Coang THK tcorr 0in:=

Thermal Coe cient α 1.17 10 5-

1

K :=

Structural Damping δ 0.126:= DNV RP F105 para. 6.2.11

Corrosion Coang Density ρcorr 60pcf :=

Seabed Temperature Tsw 296K :=

1.2 Environmental Parameter

Signi cant Wave Height Hs 2m:=


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Signi cant Wave Period Ts 4.2s:= T p 1.05 Ts 4.41 s=:=

Water Depth d 100m:=

Current 10% WD Above Seabed Uc 0.51m s 1-:= Zr 10m:=

Kinemac Viscosity of Seawater ν 1.05 10 6-

m2

s 1-

:=

1.3 Soil Parameter

Soil Type 1 = Sand, 2 = Clay soil 2:=

From table A.1 DNV RP

E305Undrained Shear Strength Su 2kPa:= z0 5.21 10

6-m:=

2. CALCULATION

2.1 Submerged Weight Calculation

Outside Diameter Dtot Ds 2tcorr + 2tcc+:= Dtot 1.753m=

Inera I π

64Ds

4ID

4-:= I 0.034m

4=

Elasc Modulus EI E I:= EI 6.946 109

m

3kg

s2

=

Weight of Pipe

Steel Weight Wstπ

4Ds

2ID

2- ρs:= Wst 938.612

kg

m=

Corr. Coat. Weight Wcorr π

4Ds 2tcorr +( )

2Ds

2- ρcorr := Wcorr 0

kg

m=

Concrete Weight Wccπ

4Dtot

2Ds 2tcorr +( )

2- ρcc:= Wcc 1.79 10

3

kg

m=

Content Weight Wcontπ

4ID

2ρcont:= Wcont 1.704 10

3

kg

m=

Added Mass Waddπ

4Dtot

2ρsw:= Wadd 2.473 10

3

kg

m=

Total E ecve Weight Weff Wst Wcorr + Wcont+ Wcc+ Wadd+:= Weff 6.906 103

kg

m=

External Pressure Pe ρsw g d:= Pe 1.005 10

6

Pa=


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Pressure Di erence ΔP Po Pe-:= ΔP 1.172 107

Pa=

2.2 Wave Induced Velocity Calculation

Angle between wave direcon and pipeline direcon ϕwave 90deg:=

Calculaon of Wave Length

Lg T p

2

2π:= L 30.354 m=

Wave Length

λ L d,( )g T p

2

2π

tanh 2π d

L

:=

Given

λ L d,( ) L=

L d( ) Find L( ):=

L d( ) 30.354m=

Horizontal Water Parcle Velovity

k d( ) 2π

L d( ):= k d( ) 0.207

1

m=

Phase Angle Range: i 0 90..:= θi

i deg:=

u d θ,( )Hs

2

g

d

d

L d( )

1

20<if

π Hs

T p

ek d( ) Dtot d-( )

d

L d( )

1

2

>if

Hs

2

g T p

L d( )

cosh k d( ) Dtot( )cosh k d( ) d( )

otherwise

cos θ( ):= Shallow approximaon

Deep approximaon

Intermediate depth

approximaon

uw max u d θ,( )( ):= uw 2.097 10 9-

m

s=

Reducon factor (direconality) R wave sin ϕwave( ):= R wave 1=


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Signi cant wave vel. perpendicular

to pipeUw d θ,( ) uw R wave

ln0.5 Ds

z0

lnZr

z0

:=

Horizontal Water Parcle Acceleraon

Uw d θ,( ) 1.724 10 9-

m

s=

Aw d θ,( )Hs π

T p

g

d

d

L d( )

1

20<if

2Hsπ

T p

2

ek d( ) Dtot d-( )

d

L d( )

1

2>if

Hsg π

L d( )


otherwise

sin θ( ):=

Aw max Aw d θ,( )( ):= Aw 2.988 10 9- m

s2

=

Signi cant wave acceleraon

perpendicular to pipeAw d θ,( ) Aw R wave:= Aw d θ,( ) 2.988 10

9-

m

s2

=

2.3 Free Span - Dynamic

Stability Number K s

2 Weff δ

ρsw Dtot2

:= K s 0.553=

Formula from DNV 1981 paragraph A.2.1.4

Reduced Velocity Vr 1.62:= From Fig A.3 DNV 1981

Note :

con1 "In-line Oscillation":=

con2 "Cross-flow Oscillation":=

Type of Oscillaon Otype con1 1 Vr < 3.5< K s 1.8<if

con2 otherwise

:=

Otype "In-line Oscillation"=

Strouhal Number St 0.24:= From Fig. A.2 DNV 1981


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Vortex Shedding Frequency f v

St Uw d θ,( ) Uc+( )

Dtot

:= f v 0.07 Hz=

Crical Pipespan Lcr 9.87

2π

EI

Weff Wadd-

DtotVr

Uw d θ,( ) Uc+

:=

Lcr 104.626 m=

2.4 Free Span - Static

Yield requirement

i 0 200..:= Li

i m:=

Longitudinal Stress σx L( )

Wst L2

0.5 Ds g

10I

Ds Po Pe-( )

4ts

+

:=

0 200 400 600 800

0

5 104

1 105

1.5 105

σx L( )

psi

L

ft

Liming Longitudinal Pressure σxa L( ) 0.8SMYS:= σxa L( ) 5.2 104

psi=


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Lcrit L( ) L 0.5m

Lcrit L

L L 1m+

σx L( ) σxa L( )<while

Lcrit

:=

Lcrit L( ) 93.5 m=

Longitudinal Stress σx Lcrit L( )( ) 5.197 104

psi=

Hoop Stress σh

Ds Po Pe-( )

2ts

5.098 104

psi=:=

Von Mises Stress σe σx Lcrit L( )( )2 σh2+ 7.28 104 psi=:=

0.9SMYS 5.85 104

psi=

3. SUMMARY RESULTS

Crical pipespan - Dynamic Lcr 343.262 ft=

Crical pipespan - Stac Lcrit L( ) 306.759 ft=


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Free Span Analysis - Instalasi NPS 60


3-:= C K









Content Density ρcont 0pcf :=








1

K :=







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Pa=




Lg T p

2

2π:= L 30.354 m=

Wave Length

λ L d,( )g T p

2

2π

tanh 2π d

L

:=

Given

λ L d,( ) L=

L d( ) Find L( ):=

L d( ) 30.354m=


k d( ) 2π

L d( ):= k d( ) 0.207

1

m=


i deg:=

u d θ,( )Hs

2

g

d

d

L d( )

1

20<if

π Hs

T p

ek d( ) Dtot d-( )

d

L d( )

1

2

>if

Hs

2

g T p

L d( )


otherwise


Deep approximaon

Intermediate depth

approximaon

uw max u d θ,( )( ):= uw 2.097 10 9-

m

s=



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ln0.5 Ds

z0

lnZr

z0

:=


Uw d θ,( ) 1.724 10 9-

m

s=

Aw d θ,( )Hs π

T p

g

d

d

L d( )

1

20<if

2Hsπ

T p

2

ek d( ) Dtot d-( )

d

L d( )

1

2>if

Hsg π

L d( )


otherwise

sin θ( ):=

Aw max Aw d θ,( )( ):= Aw 2.988 10 9- m

s2

=



9-

m

s2

=



2 Weff δ

ρsw Dtot2

:= K s 0.416=



Note :




con2 otherwise

:=




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Dtot

:= f v 0.07 Hz=


2π

EI

Weff Wadd-

DtotVr

Uw d θ,( ) Uc+

:=

Lcr 113.657 m=


Yield requirement

i 0 200..:= Li

i m:=


Wst L2

0.5 Ds g

10I

Ds Po Pe-( )

4ts

+

:=

0 200 400 600 800

5- 104

0

5 104

1 105

1.5 105

σx L( )

psi

L

ft


psi=


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Lcrit L( ) L 0.5m

Lcrit L

L L 1m+


Lcrit

:=

Lcrit L( ) 133.5 m=


psi=

Hoop Stress σh

Ds Po Pe-( )

2ts

4.374- 103

psi=:=


0.9SMYS 5.85 104

psi=

3. SUMMARY RESULTS




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Free Span Analysis - Operaon NPS 60


3-:= C K

















1

K :=





Signi cant Wave Height Hs 3.7m:=


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Signi cant Wave Period Ts 5.5s:= T p 1.05 Ts 5.775s=:=




m2

s 1-

:=

1.3 Soil Parameter




6-m:=

2. CALCULATION



Inera I π

64Ds

4ID

4-:= I 0.034m

4=


m

3kg

s2

=

Weight of Pipe

Steel Weight Wstπ

4Ds

2ID

2- ρs:= Wst 938.612

kg

m=


4Ds 2tcorr +( )

2Ds


kg

m=


4Dtot

2Ds 2tcorr +( )

2- ρcc:= Wcc 1.79 10

3

kg

m=


4ID

2ρcont:= Wcont 59.66

kg

m=

Added Mass Waddπ

4Dtot

2ρsw:= Wadd 2.473 10

3

kg

m=

Total E ecve Weight Weff Wst Wcorr + Wcont+ Wcc+ Wadd+:= Weff 5.262 103

kg

m=


6

Pa=


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Pa=




Lg T p

2

2π:= L 52.053 m=

Wave Length

λ L d,( )g T p

2

2π

tanh 2π d

L

:=

Given

λ L d,( ) L=

L d( ) Find L( ):=

L d( ) 52.053m=


k d( ) 2π

L d( ):= k d( ) 0.121

1

m=


i deg:=

u d θ,( )Hs

2

g

d

d

L d( )

1

20<if

π Hs

T p

ek d( ) Dtot d-( )

d

L d( )

1

2

>if

Hs

2

g T p

L d( )


otherwise


Deep approximaon

Intermediate depth

approximaon

uw max u d θ,( )( ):= uw 1.424 10 5-

m

s=



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ln0.5 Ds

z0

lnZr

z0

:=


Uw d θ,( ) 1.17 10 5-

m

s=

Aw d θ,( )Hs π

T p

g

d

d

L d( )

1

20<if

2Hsπ

T p

2

ek d( ) Dtot d-( )

d

L d( )

1

2>if

Hsg π

L d( )


otherwise

sin θ( ):=

Aw max Aw d θ,( )( ):= Aw 1.549 10 5- m

s2

=



5-

m

s2

=



2 Weff δ

ρsw Dtot2

:= K s 0.421=



Note :




con2 otherwise

:=




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Dtot

:= f v 0.086 Hz=


2π

EI

Weff Wadd-

DtotVr

Uw d θ,( ) Uc+

:=

Lcr 101.709 m=


Yield requirement

i 0 200..:= Li

i m:=


Wst L2

0.5 Ds g

10I

Ds Po Pe-( )

4ts

+

:=

0 200 400 600 800

0

5 104

1 105

1.5 105

σx L( )

psi

L

ft


psi=


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Lcrit L( ) L 0.5m

Lcrit L

L L 1m+


Lcrit

:=

Lcrit L( ) 108.5 m=


psi=

Hoop Stress σh

Ds Po Pe-( )

2ts

3.253 104

psi=:=


0.9SMYS 5.85 104

psi=

3. SUMMARY RESULTS




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Untuk pipa kedua:

NPS 16. Wall thicknes=0.5 in. Concrete Thickness=2 in

Dari perhitungan pada halaman selanjutnya ini, dapat disimpulkan panjang free span yang diijinkan:

Condition Allowable Free Span (ft)

Installation Hydrotest Operation

Static 106.616 102.981 96.254

Dynamic 218.176 185.367 198.491

Keterangan: Perhitungan On-Bottom Stability (OBS) dapat diperoleh di bagian lampiran.


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Free Span Analysis - Hydrotest NPS 16


3-:= C K



Wall Thickness ts 0.5in:=


Concrete Coang tcc 2in:=












1

K :=







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m2

s 1-

:=

1.3 Soil Parameter




6-m:=

2. CALCULATION



Inera I π

64Ds

4ID

4-:= I 3.047 10

4- m

4=


kg m

3

s2

=

Weight of Pipe

Steel Weight Wstπ

4Ds

2ID

2- ρs:= Wst 123.292

kg

m=


4Ds 2tcorr +( )

2Ds


kg

m=


4Dtot

2Ds 2tcorr +( )

2- ρcc:= Wcc 222.072

kg

m=


4ID


kg

m=

Added Mass Waddπ

4Dtot

2ρsw:= Wadd 207.787

kg

m=

Total E ecve Weight Weff Wst Wcorr + Wcont+ Wcc+ Wadd+:= Weff 667.11 kg

m=


6

Pa=


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Pa=




Lg T p

2

2π:= L 30.354 m=

Wave Length

λ L d,( )g T p

2

2π

tanh 2π d

L

:=

Given

λ L d,( ) L=

L d( ) Find L( ):=

L d( ) 30.354m=


k d( ) 2π

L d( ):= k d( ) 0.207

1

m=


i deg:=

u d θ,( )Hs

2

g

d

d

L d( )

1

20<if

π Hs

T p

ek d( ) Dtot d-( )

d

L d( )

1

2

>if

Hs

2

g T p

L d( )


otherwise


Deep approximaon

Intermediate depth

approximaon

uw max u d θ,( )( ):= uw 1.621 10 9-

m

s=



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ln0.5 Ds

z0

lnZr

z0

:=


Uw d θ,( ) 1.184 10 9-

m

s=

Aw d θ,( )Hs π

T p

g

d

d

L d( )

1

20<if

2Hsπ

T p

2

ek d( ) Dtot d-( )

d

L d( )

1

2>if

Hsg π

L d( )


otherwise

sin θ( ):=

Aw max Aw d θ,( )( ):= Aw 2.309 10 9- m

s2

=



9-

m

s2

=



2 Weff δ

ρsw Dtot2

:= K s 0.635=



Note :




con2 otherwise

:=




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Dtot

:= f v 0.241 Hz=


2π

EI

Weff Wadd-

DtotVr

Uw d θ,( ) Uc+

:=

Lcr 31.389 m=


Yield requirement

i 0 200..:= Li

i m:=


Wst L2

0.5 Ds g

10I

Ds Po Pe-( )

4ts

+

:=

0 200 400 600 800

0

1 105

2 105

3 105

4 105

5 105

σx L( )

psi

L

ft


psi=


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Lcrit L( ) L 0.5m

Lcrit L

L L 1m+


Lcrit

:=

Lcrit L( ) 56.5 m=


psi=

Hoop Stress σh

Ds Po Pe-( )

2ts

2.719 104

psi=:=


0.9SMYS 5.85 104

psi=

3. SUMMARY RESULTS




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Free Span Analysis - Instalasi NPS 16


3-:= C K









Content Density ρcont 0pcf :=








1

K :=







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m2

s 1-

:=

1.3 Soil Parameter




6-m:=

2. CALCULATION



Inera I π

64Ds

4ID

4-:= I 3.047 10

4- m

4=


kg m

3

s2

=

Weight of Pipe

Steel Weight Wstπ

4Ds

2ID

2- ρs:= Wst 123.292

kg

m=


4Ds 2tcorr +( )

2Ds


kg

m=


4Dtot

2Ds 2tcorr +( )

2- ρcc:= Wcc 222.072

kg

m=


4ID

2ρcont:= Wcont 0=

Added Mass Waddπ

4Dtot

2ρsw:= Wadd 207.787

kg

m=


m=


6

Pa=


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Pa=




Lg T p

2

2π:= L 30.354 m=

Wave Length

λ L d,( )g T p

2

2π

tanh 2π d

L

:=

Given

λ L d,( ) L=

L d( ) Find L( ):=

L d( ) 30.354m=


k d( ) 2π

L d( ):= k d( ) 0.207

1

m=


i deg:=

u d θ,( )Hs

2

g

d

d

L d( )

1

20<if

π Hs

T p

ek d( ) Dtot d-( )

d

L d( )

1

2

>if

Hs

2

g T p

L d( )


otherwise


Deep approximaon

Intermediate depth

approximaon

uw max u d θ,( )( ):= uw 1.621 10 9-

m

s=



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ln0.5 Ds

z0

lnZr

z0

:=


Uw d θ,( ) 1.184 10 9-

m

s=

Aw d θ,( )Hs π

T p

g

d

d

L d( )

1

20<if

2Hsπ

T p

2

ek d( ) Dtot d-( )

d

L d( )

1

2>if

Hsg π

L d( )


otherwise

sin θ( ):=

Aw max Aw d θ,( )( ):= Aw 2.309 10 9- m

s2

=



9-

m

s2

=



2 Weff δ

ρsw Dtot2

:= K s 0.527=



Note :




con2 otherwise

:=




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Lcrit L( ) L 0.5m

Lcrit L

L L 1m+


Lcrit

:=

Lcrit L( ) 66.5 m=


psi=

Hoop Stress σh

Ds Po Pe-( )

2ts

2.333- 103

psi=:=


0.9SMYS 5.85 104

psi=

3. SUMMARY RESULTS




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Free Span Analysis - Operaon NPS 16


3-:= C K

















1

K :=





Signi cant Wave Height Hs 3.7m:=


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Signi cant Wave Period Ts 5.5s:= T p 1.05 Ts 5.775s=:=




m2

s 1-

:=

1.3 Soil Parameter




6-m:=

2. CALCULATION



Inera I π

64Ds

4ID

4-:= I 3.047 10

4- m

4=


kg m

3

s2

=

Weight of Pipe

Steel Weight Wstπ

4Ds

2ID

2- ρs:= Wst 123.292

kg

m=


4Ds 2tcorr +( )

2Ds


kg

m=


4Dtot

2Ds 2tcorr +( )

2- ρcc:= Wcc 222.072

kg

m=


4ID


kg

m=

Added Mass Waddπ

4Dtot

2ρsw:= Wadd 207.787

kg

m=


m=


6

Pa=


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Pa=




Lg T p

2

2π:= L 52.053 m=

Wave Length

λ L d,( )g T p

2

2π

tanh 2π d

L

:=

Given

λ L d,( ) L=

L d( ) Find L( ):=

L d( ) 52.053m=


k d( ) 2π

L d( ):= k d( ) 0.121

1

m=


i deg:=

u d θ,( )Hs

2

g

d

d

L d( )

1

20<if

π Hs

T p

ek d( ) Dtot d-( )

d

L d( )

1

2

>if

Hs

2

g T p

L d( )


otherwise


Deep approximaon

Intermediate depth

approximaon

uw max u d θ,( )( ):= uw 1.225 10 5-

m

s=



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ln0.5 Ds

z0

lnZr

z0

:=


Uw d θ,( ) 8.952 10 6-

m

s=

Aw d θ,( )Hs π

T p

g

d

d

L d( )

1

20<if

2Hsπ

T p

2

ek d( ) Dtot d-( )

d

L d( )

1

2>if

Hsg π

L d( )


otherwise

sin θ( ):=

Aw max Aw d θ,( )( ):= Aw 1.333 10 5- m

s2

=



5-

m

s2

=



2 Weff δ

ρsw Dtot2

:= K s 0.531=



Note :




con2 otherwise

:=




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Dtot

:= f v 0.298 Hz=


2π

EI

Weff Wadd-

DtotVr

Uw d θ,( ) Uc+

:=

Lcr 29.338 m=


Yield requirement

i 0 200..:= Li

i m:=


Wst L2

0.5 Ds g

10I

Ds Po Pe-( )

4ts

+

:=

0 200 400 600 800

0

1 105

2 105

3 105

4 105

5 105

σx L( )

psi

L

ft


psi=


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Lcrit L( ) L 0.5m

Lcrit L

L L 1m+


Lcrit

:=

Lcrit L( ) 60.5 m=


psi=

Hoop Stress σh

Ds Po Pe-( )

2ts

1.735 104

psi=:=


0.9SMYS 5.85 104

psi=

3. SUMMARY RESULTS




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BAGIAN B

JAWABAN NO 1 – UPHEAVEL BUCKLING ANALYSIS

a)

Tidak diperlukan tambahan gravel dumping, karena dari hasil perhitungan pada halaman berikut

ini diperoleh bahwa

b)

Allowable bending strain pada kasus ini adalah = 1 .73 × 10−

Namun, bending strain yang terjadi adalah = 9.212 × 10−4

Dari hasil di atas, bending strain yang terjadi masih diperbolehkan (aman).


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Upheavel Buckling Analysis1. Input Parameter

pcf lb

ft3

1.1 Pipeline Data

Outside diameter OD 20in:=

Wall thickness WT 0.594in:= Schedule 40

Corrosion allowance CA 0in:=

Steel density ρs 490lb ft 3-

:=

Corrosion coating thickness tcorr 3mm:=

Concrete coating thickness tcc 1 in:=

Corrosion coating density ρcorr 60lb ft 3-

:= density of 3LPP

Product density ρcont 80lb ft 3-

:=

Modulus Young E 207GPa 3.002 107

psi=:=

Poisson ratio v 0.3:=

Coeff. of linear expansion α 1.17 10

5-

1 °C:=

Imperfection Height δ 1 m=

1.2 Operating Data

Max. operating pressure Po 15MPa:=

Max. operating temp. To 100 °C:=

Installation temp. Ti 25°C:=

1.3 Soil Data

Soil type Sand (Pasir) soilt "Sand":=

Soil specific weight γs 18835.6N m 3-

:=

Undrained shear strength Su 14000Pa:=

Distance between free end Lf 62m:=


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Soil friction f r 0.3:=

Internal friction angle of soil ϕ 35deg:=

Coef. of lateral soil stress rest K o 1 sin ϕ( )-:= K o 0.426=

2. Calculation

2.1 Pipe Properties

Total outside diameter Dtot OD 2 tcc+ 2 tcorr +:= Dtot 0.565m=

Internal diameter of pipe ID OD 2 WT-:= ID 0.478m=

Moment of inertia I π

64

OD4

ID4

-( ):= I 7.102 10 4-

m4

=

Weight of steel pipe Wstπ

4OD

2ID

2-( ) ρs g:= Wst 1.798 10

3

N

m=

Steel pipe cross section area Asπ

4OD

2ID

2-( ):= As 0.023m

2=

Corroded Condition

Internal diameter corroded IDc OD 2 WT 0.5 CA-( )-:= IDc 0.478m=

Moment of inertia corroded Ic

π

64 OD

4

IDc

4

-:= Ic 7.102 10

4-

m

4

=

Weight of steel pipe Wstcπ

4OD

2IDc

2- ρs g:= Wstc 1.798 10

3

N

m=

Internal cross section area Aicπ

4IDc

2:= Aic 0.179m

2=

Steel pipe cross section area Ascπ

4OD

2IDc

2-:= Asc 0.023m

2=

Weight of corrosion coating Wcorr

π

4 OD 2 tcorr +( )2

OD

2

- ρcorr g:=

Wcorr 45.393 N

m=

Weight of internal content Wcontπ

4ID

2ρcont g:= Wcont 2.254 10

3

N

m=

Operating condition weight Wsop Wst Wcont+ Wcorr +:=

Wsop 4.097 103

N

m=


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Installation condition weight Win Wsop Wcont-:= Win 1.844 103

N

m=

Cover Depth H 30m Dtot- 29.435m=:= distance of top of pipe to

water surface Axial Load

Hoop stress σy

Po OD

2 WT:= σy 3.663 10

4 psi=

Compression force due to

temperatureσT α E To Ti-( ) Asc:= σT 1.548 10

4 N=

Tension due to poisson's

effectσv v-

Po OD 2 WT-( )

2WT

Asc:=

σv 1.665- 106

N=

Compression force due to

end cap effectσe

Po OD Asc

4 WT:= σe 2.95 10

6 N=

Total axial force Fax σT σv+ σe+:= Fax 1.301 106

N=

Friction force

f f r π γs OD

2 H

OD

2+

1 K o+( ) 4

32 K o+( )

Wsop

π

+ 4

32 K o+( ) γs

OD

2

2

-

:=

f 1.91 10

5

N

m=

Axial load p min f Lf

2 Fax,

:= p 1.301 10

6 N=

Required downward force for pipe stability

Wreq 0.0646 δ p

2

E I

δ

4.494

E I Win

72 p2

<if

0.67 p

δ W

in

E I 1.23 Win-

4.494

E I W

in

72 p2

δ<

8.064

E I W

in

72 p2

<if

1.16pδ Win

E I 4.76 Win-

otherwise

:=

Wreq 818.108 N

m=

Uplift resistance of soil, q


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q Su Dtot min 3 H

Dtot

,

soilt 1=if

γs H Dtot 1 0.01 H

Dtot

+

otherwise

:=

q 4.763 105

N

m=

3. Check for Upheaval Buckling

Imperfection height δ 1m:= q Wsop+ 4.804 105

N m 1-

=

q Wsop+ Wreq> Wreq 818.108 N m 1-

=

Extra weight Wanc Wreq q Wsop+( )- q Wsop+ Wreq<if

0 N

m

otherwise

:=

Wanc 12.2 m 0 N=

Allowable imperfection length,

relevant to span lengthLrel

72E I δ

W

in

1

4

:= Lrel 48.95m=

Pipe strain P buckleπ

2E I

Lrel2

6.056 105

N=:=

ΔL p P buckle-( ) 2 Lrel

Asc E 2

p P buckle-( )2

2 Asc E f +:=

ΔL 0.015m=

Total uplift buckle amplitude a δΔL 2 Lrel

π2

4

+:=

a 1.761m=

Total pipe strain ε OD π

2 a

2 Lrel( )2

:= ε 9.212 10 4-

=

Allowable bending strain

εallow

εallow 1.73 10 3-

:=

4. Executive Summary


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Imperfection height δ 1m

Required downward load Wreq 818.108 N

m=

Uplift soil resistance q 4.763 105

N

m=

Operation weight Wsop 4.097 103

N

m=

Actual downward force Wactual Wsop q+ 4.804 105

N

m=:=

Check for stability

if q Wsop+ Wreq< "Not OK", "OK",( ) "OK"= if it is OK, no required

extra weight needed

Safety factor q Wsop+

Wreq

587.253=

Extra weight needed per joint Wanc 12.2 m 0 N=

Allowable bending strain, allow

if Re ε( ) εallow< "OK", "Not OK",( ) "OK"=


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JAWABAN NO 2 – EXPANSION OPERATION

Dari hasil perhitungan pada halaman selanjutnya ini diperloeh hasil sebagai berikut


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Thermal Expansion

Loadcase : Operation - Unburied

K C kN 1000N bar 105Pa MPa 106Pa ksi 103 psi pcf lb ft 3-

1. INPUT DATA

1.1 Design Data

Pipeline Parameters Environmental Parameters

External pipeline diameter De 20in:= Density of sea water ρsw 1025 kg

m3

:=

Specified Minumum Yield

StrengthSMYS 52000psi:=

Installation temperature To 25C:=

Corrosion allowance being

used upTc 0.125in:=

Marine growth tmar 0m:=

Young's Modulus of pipe

materialE 207000MPa:=

Marine growth density ρmar 0 kg

m3

:=

Steel density ρst 490pcf := Axial friction

factor(?) ax 0.1:=

Wall thicknessWT

15MPa De

2 SMYS 1( ) 1( ) 0.72( ):=

Residual lay tension Nlay 0kN:=

Coefficient of thermal

expansionα 1.17 10

5- C

1-:=

Poisson's ratio ν 0.3:=

Coating Parameters Operating Parameters

Corrosion coating density ρcor 900kg m 3-

:= Content density ρcont 80pcf :=

Concrete coating density ρcc 190pcf := Design pressure Pin 15MPa:=

Corrosion coating thickness tcor 3mm:= Inlet reference elevation Elv 0:=

Concrete coating thickness tcc 1in:= Thermal insulation density ρth 0pcf :=

Soil parameter

μlong 0.3:=Longitudinal friction factor

1.2 Input Data Variables with KP


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Input KP step size KPstep 1km:=

Input temperature profile Number of temperature input point n 5:=

Product temperature Corresponding KP point i 0 n 1-..:=

Ti

100C

90C

80C

70C

60C

:= KPi

0km

10km

20km

30km

40km

:=

0 2 104

4 104

60

70

80

90

100

Ti

KPi

ΔTi

Ti

To-:=

Temp ΔT:=

Temp 126.984 C=


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Input water depth profile (LAT)

Number of Input points

Input corrosion coating thickness/density with KP

nw 31:= Number of Points ncr 2:=

iw 0 nw 1-..:=icr 0 ncr 1-..:=

Water Depth Corresponding

KP Point Thickness Density KP Points

WDiw

0m

1- m

2- m

3- m

4.01- m

5.01- m

6- m7- m

8- m

9.01- m

10- m

12- m

13- m

14.7- m

20- m

23- m

22.3- m

13.5- m

13- m

12- m

10- m

9.01- m

8- m

7- m

6- m

5- m

4- m

3- m

2- m

1- m

0m

:= KPdiw

0km

1.49km

1.6km

1.91km

2.52km

3.12km

3.48km3.88km

4.19km

4.59km

5.07km

5.29km

6.18km

6.73km

8km

13.44km

19.5km

22.5km

32.18km

32.38km

33.07km

33.76km

34.03km

34.36km

34.67km

34.97km

35.3km

35.82km

36.32km

36.86km

36.96km

:=Tcor

icr

tcor

tcor

:= ρcor icr

ρcor

ρcor

:= KPcr icr

0km

KPn 1-

:=

0 1 104

2 104

3 104

4 104

30-

20-

10-

0

WDiw

KPdiw


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Wall thickness and OD profi le with KP Marine growth thickness

Number of Points nwt 2:= Number of points nm 2:=

iwt 0 nwt 1-..:= im 0 nm 1-..:=

Wall Thickness Corresponding KP Marine Growth Corresponding KP Points

WTiwt

WT 10% Tc( )-

WT 10% Tc( )-

:= KPwtiwt

0km

KPn 1-

:= Tmar im

tmar

tmar

:= KPmim

0km

KPn 1-

:=

Input concrete coating thickness with KP Input concrete coating thickness with KP

Number of points ncc 2:= Number of points nlf 11:=

icc 0 ncc 1-..:= ilf 0 nlf 1-..:=

Concrete

Thickness

Corresponding

Kp Points

KPcf ilf

0km

1.91km

5.07km

6.78km

8km

13.44km

19.5km

22.5km32km

34.97km

36.96km

:=μlong

ilf

μlong

μlong

μlong

μlong

μlong

μlong

μlong

μlong

μlong

μlong

μlong

:=

Tconicc

tcc

tcc

:= KPccicc

0km

KPn 1-( )

:=

Thermal Insulation with KP

Number of Points nti 2:= Corroded WT with KP

Number of Points nwtc 2:=iti 0 nti 1-..:=

iwtc 0 nwtc 1-..:=

Additional Weight Corresponding KP Corresponding KPCorroded WT

Tthiti

0in

0in

:= KPthiti

0km

KPn 1-

:=WTc

iwtc WT

iwtc Tc-:= KP

iwtc

0km

KPn 1-

:=


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Jacket Material with KP

Number of Points nti 2:=

iti 0 nti 1-..:=

Additional Weight Corresponding KP

Tjiti

0in

0in

:= KPthiti

KP0

KPn 1-

:=

2.0 Calculation

2.1 Calculated parameters constant with KP

Define range of KP variable x KP0

KPstep, KPn 1-

..:=

Total outside diameter Do De 2. Tcor Tth+ Tj+ Tcon+ Tmar +( )+:=

Internal diameter Di De 2WT-:=

Corroded Internal diameter Dic Di 2 10 % Tc+:=

Nominal Steel Area Ast π

4De

2Di

2-( ):=

Astc π

4De

2Dic

2-( ):=

Corroded steel area

Mst Astc ρst:=Steel mass

Acorr π

4De 2 Tcor +( )

2De

2-:=

Corrosion coating area

Mcorr Acorr ρcor :=Corrosion coating mass

Ath π

4De 2 Tcor + 2Tth+( )

2De 2 Tcor +( )

2-:=

Thermal Insulation area

Mth Ath ρth:=Thermal Insulation mass

Aj π

4De 2 Tcor + 2Tth+ 2.Tj+( )

2De 2 Tcor + 2Tth+( )( )

2-:=

Jacket area

Jacket mass M j Aj ρst:=

Concrete coating area Acc π

4Do 2 Tmar +( )

2

De 2 Tcor Tth+ Tj+( )+[ ]2

-+

...

:=

Concrete coating Mcc Acc ρcc:=


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Marine growth areaAmar

π

4Do( )

2De 2 Tcor Tth+ Tj+ Tcon+( )+[ ][ ]

2-:=

Marine growth mass Mmar Amar ρmar :=

Content mass Mcontπ

4Dic

2 ρcont:=

Buoyancy force Fbπ

4ρsw Do

2 g:=

Submerged Weight Ws Mst Mcorr + Mth+

M j Mcc+ Mmar + Mcont+( )-+

...

g Fb-:=

Dry Weight Wdry Ws Fb+:=

0 1 104

2 104

3 104

4 104

6- 103

4- 103

2- 103

0

2 103

4 103

Wdry

Ws

Fb

KP

2.2 Effective Axial Force Derivation - Restrained Pipeline

Define function with KP x KP0

KPstep, KPn 1-

..:=

Define External pressure with KP Po ρsw g WD-:=

Define internal pressure with KP Pin Pin ρcont g WD-+:=

Define Pressure difference with KP ΔP Pin Po-:=

Thermal expansion force with KP Ft E- Astc α Temp:=

Poissons Force with KP Fp ν ΔP Astc De 2 WTc-

2WTc:=

Feπ

4Pin Dic( )

2Po De

2-:=

Apparent Force with KP

Fully restrained axial force with KP Pr Fe- Fp+ Ft+ Nlay+:=


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Summary of Results

At cold end Total Pipeline Expansion Expand1 0.106=

Anchor Length Lanchor1 1.98 104

m=

At hot end Total Pipeline Expension Expand2 0.106=

Anchor Length Lanchor2 1.98 104

m=


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JAWABAN NO 3 – GLOBAL BUCKLING

Perilaku pipeline sebelum terjadi buckling dan mekanisme buckling saat buckling pertama dan

kedua

Efek temperature dan tekanan menghasilkan gaya ekspansi (expansion effective forces), yang

merupakan kombinasi pipe wall force (N) dan tekanan internal dan eksternal, memberikanpotensi untuk terjadinya global buckling.

Garis titik-titik biru menunjukkan kapasitas bucklingpada bagian pipa tertentu.

a.

Buckling capacity along the pipeline

Sepanjang pipa yang terinstall akan memiliki

kapasitas buckling yang berbeda-beda pada tiap

bagian-bagiannya.

Dari gambar di samping, ditunjukkan bahwa pipa

tersebut memiliki imperfection. Pada titik ini

kapasitas global buckling, Sinit adalah di titik

terendahnya.

b.

Sebelum terjadi global buckling

Meningkatnya temperature dan tekanan juga akan

meningkatkan compressive effective force kepada

nilai maksimum S0.

c.

First buckle

Ketika effective axial force S0 mencapai kapasitas

global buckling pipa, Sinit, pipeline akan mengalami

buckle (tekuk) dan effective axial force pada puncak

akan turun pada Spost. Nilai maksimum akan

berubah sebanding dengan gaya tahan antara pipa

dengan tanah, memberikan lembah pada diagram

efektif. Effective axial force direpresentasikan

dalam garis solid. Buckle yang terjadi akan

sebanding dengan area yang diarsir antara garis

solid dan garis potential effective force.

d.

Second buckle

Jika tekanan dan temperature terus menaik,imperfection sekitarnya bisa mengalami buckle dan

mengubah diagram gaya. Pada titik ini gaya akan

tetap konstan tapi tekuk akan bertambah,

sebanding dengan area yang diarsir. Lembah yang

diarsir menunjukkan kea rah mana pipa bergerak

dan tekuk seperti apa yang terjadi.


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JAWABAN NO 4 – TEORI KEGAGALAN

Teori Kegagalan Konsep Dasar Kondisi Kegagalan Sifat Teori Failure Criteria

Relationship

Von Mises Energi distorsi Kegagalan diprediksi terjadi

pada keadaan tegangan

multiaksial jika energydistorsi per satuan volume

sama atau lebih besar

dibandingkan energy

distorsi per satuan volume

pada saat terjadinya

kegagalan dalam pengujian

tegangan uniaksial.

The most

representative

for ductilematerials

=

Tresca Tegangan geser

maksimum

Kegagalan terjadi jika

tegangan geser maksimum

pada suatu titik di pipa

sama atau lebih besar daritegangan geser maksimum

pada saat material leleh

(yield) pada tes beban Tarik

uniaksial.

The most

conservative for

all materials

= 0.5

Rankine Tegangan

normal

maksimum

Kegagalan terjadi jika

tegangan Tarik maksimum

pada suatu titik di pipa

sama atau lebih besar dari

tegangan Tarik maksimum

pada saat material leleh

(yield) pada tes beban tarik

uniaksial.

The best fit for

brittle materials

= 0.577


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LAMPIRAN


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DNV RP305 (Installation) OBS - NPS 16

1. Input Parameter

1.1 Pipeline Design Parameter Outer Diameter D_s 16in 0.406m=:=

Pipe Wall Thickness t_s 0.5 in:= ID D_s 2 t_s- 0.381m=:=

Corrosion Coating Thickness t_corr 0mm 0 in=:=

Concrete Thickness t_cc 2in 0.051m=:=

Pipe Joint Length L 12.2m:=

Steel Pipe Density ρs 490 lb

ft3

:=

Concrete Coating Density ρcc 190 lb

ft3

:=

Corrosion Coating Density ρcorr 940 kg

m3

58.682 lb

ft3

=:=

Density of Sea Water ρsw 1025 kg

m3

63.989 lb

ft3

=:=

Density of Pipeline Content ρcont 0 kg

m3

:=


Significant Wave Height Hs 2m 6.562 ft=:=

Significant Wave Period Ts 4.2s:=

Spectral Peak Period Tp 1.05 Ts 4.41 s=:=

Water Depth d 100m 328.084 ft=:=

Current velocity Ur 0.51 m

s1.673

ft

s=:= Zr 10m:=

Kinematic Viscosity of Seawater υ 1.076 10 5-

ft

2

s:=

Corrosion Allowance ca 3mm:=

Marine Growth Thickness t_mg 0mm:=

Marine Growth Density ρmg 1400 kg

m3

:=


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1.3 Soil Parameter

Soil Type [1 = sand, 2 = clay] soil 2:= ρsoil 326.309 kg

m3

:=

Undrained Shear Stress Su 2kPa:=

2. Calculation2.1 Vertical Stability 2.1.1 Submerged Weight CalculationTotal Outer Diameter D_out D_s 2t_corr + 2 t_mg+ 2 t_cc+ 0.508m=:=

Internal Diameter D_in D_s 2t_s- 0.381m=:=

D_corr D_s 2 t_corr + 0.406m=:= D_mg D_s 2 t_mg+ 0.406m=:=

Ws π4

D_s2 ID2-( ) ρs D_corr 2 D_s2-( ) ρcorr + D_out2 D_corr 2-( )ρcc+

D_in2

ρcont D_out2

ρsw-+

...

:=

Ws 137.614 kg

m=

Buoyancy B π

4D_out

2ρsw 207.75

kg

m=:=

2.1.2 Determining Stability

Specific Gravity SG Ws B+

B1.662=:=

if SG 1.1 "UNSTABLE", "STABLE",( ) "STABLE"=

Specific Gravity of Product SG_prod ρcont

ρsw0=:=

Specific Gravity of Soil SG_soil ρsoil

ρsw0.318=:=

2.2 Lateral Stablity 2.2.1 Water Particle Velocities

Tn d

g3.193s=:=

Tn

Tp0.724=

ϕ Tp

Hs3.118

s

m=:=


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Peakedness Parameter γ 5 ϕ 3.6 s

mif

1 ϕ 5 s

m

if

3.3 otherwise

5=:=

From Fig. 2-1 DNV RP E305

Tn

Tp0.724=

Us 0.000001 Hs

Tn6.263 10

7-

m

s=:= Us:kecepatan partikel air signifikan akibat gelombang


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Zero Up-Crossing Period Tu 1.4 Tp 6.174s=:=

Table A.1 Soil parameter can be found as

Grain Size d50 0.125mm:=

Roughness Zo 1.04 10

5-m:=

Nikuradse's equivalent sand roughness parameter

Kb 2.5 d50 0.313 mm=:=

Zob Kb

301.042 10

5- m=:=

Using Graph, Fig. 2.2

D_out

Zo4.885 10

4=

Zr Zo

9.615 105=

Ratio average velocity to reference velocity

U_D 1

ln Zr

Zo1+

1 Zo

D_out+

ln D_out

Zo1+

1-

Ur 0.363 m

s=:=

2.2.2 Hydrodinamic Coefficient


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Wave Particle Acceleration As 2 π Us

Tu6.374 10

7-

m

s2

=:=

Current to Wave Velocity M U_D

Us5.791 10

5=:=

KC Number K Us Tu

D_out7.612 10

6-=:=

Reynold's Number Re U_D Us+( ) D_out

υ1.843 10

5=:=

Hydrodynamics Coefficient

Cd 1.2 Re 3 105

< M 0.8if

0.7 otherwise

1.2=:=

CL 0.9:=

CM 3.29:=

Soil Coefficient

Clay Soil

Ratio = 1/S Ratio D_out Su

Ws g 0.753=:=

From Figure 5.11 (Friction Factor = μc)

μc 0.24:=

Sand Soil

μs 0:=

Friction Factor Coefficient

μ μs soil 1=if

μc otherwise

:=

μ 0.24=

Calibration Factor

M 5.791 105

=

K 7.612 10 6-

=

Fw 1:=From Fig. 5.12

2.2.3 Lateral Stability Calculation

Hydrodynamic Forces and Required Submerged Weight


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Phase Angle Range i 0 360..:= θi

i deg:=

Lift Force FL 1

2

ρsw

g D_out CL Us cos θ( ) U_D+( )

2:=

Drag Force FD 1

2

ρsw

g D_out Cd Us cos θ( ) U_D+( )

2:=

Inertia Force FI π D_out

2

4

ρsw

g CM As sin θ( ):=

Required Submerged Weight W_s FD FI+ μ FL+

μ

Fw:=

Result of Calculation

0 100 200 300 40020.6023

20.6024

20.6025

20.6026

20.6027

20.6028

W_s

θ

deg

Change the concrete thickness until its ok.

Concrete thickness t_cc 50.8 mm=

Wso Ws g 1.35 103

N

m=:=

Wreq max W_s( ) 20.603 kg

m

=:= Ws 137.614 kg

m

=

if Ws Wreq "Change Concrete Thickness", "OK",( ) "OK"=

Safety Factor for submerged weight due to requirement weight SFw Ws

Wreq6.679=:=

Specific Gravity SG 1.662=

Submerged Weight Ws 137.614 kg

m=


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DNV RP305 (Operation) OBS - NPS 16

1. Input Parameter


Pipe Wall Thickness t_s 0.5 in:= ID D_s 2 t_s- 0.381m=:=


Concrete Thickness t_cc 2in 0.051m=:=



ft3

:=


ft3

:=


m3

58.682 lb

ft3

=:=


m3

63.989 lb

ft3

=:=


m3

:=


Significant Wave Height Hs 3.7m 12.139 ft=:=


Spectral Peak Period Tp 1.05 Ts 5.775s=:=



s2.067

ft

s=:= Zr 10m:=


ft

2

s:=




m3

:=


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1.3 Soil Parameter


m3

:=





Ws π4


D_in2

ρcont D_out2

ρsw-+

...

:=

Ws 141.605 kg

m=

Buoyancy B π

4D_out

2ρsw 207.75

kg

m=:=



B1.682=:=



ρsw0.034=:=


ρsw0.318=:=


Tn d

g3.193s=:=

Tn

Tp0.553=

ϕ Tp

Hs3.002

s

m=:=


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mif

1 ϕ 5 s

m

if

3.3 otherwise

5=:=


Tn

Tp0.553=

Us 0.000001 Hs

Tn1.159 10

6-

m



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5-m:=


Kb 2.5 d50 0.313 mm=:=

Zob Kb

301.042 10

5- m=:=


D_out

Zo4.885 10

4=

Zr Zo

9.615 105=


U_D 1

ln Zr

Zo1+

1 Zo

D_out+

ln D_out

Zo1+

1-

Ur 0.448 m

s=:=



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Tu9.005 10

7-

m

s2

=:=


Us3.867 10

5=:=

KC Number K Us Tu

D_out1.844 10

5-=:=


υ2.277 10

5=:=


Cd 1.2 Re 3 105

< M 0.8if

0.7 otherwise

1.2=:=

CL 0.9:=

CM 3.29:=

Soil Coefficient

Clay Soil


Ws g 0.732=:=


μc 0.24:=

Sand Soil

μs 0:=


μ μs soil 1=if

μc otherwise

:=

μ 0.24=

Calibration Factor

M 3.867 105

=

K 1.844 10 5-

=





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i deg:=

Lift Force FL 1

2

ρsw


2:=

Drag Force FD 1

2

ρsw


2:=


2

4

ρsw

g CM As sin θ( ):=


μ

Fw:=


0 100 200 300 40031.4382

31.4384

31.4386

31.4388

31.439

W_s

θ

deg



Wso Ws g 1.389 103

N

m=:=


m

=:= Ws 141.605 kg

m

=



Wreq4.504=:=



m=


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DNV RP305 (Installation) OBS - NPS 60

1. Input Parameter


Pipe Wall Thickness t_s 1 in:= ID D_s 2 t_s- 1.473m=:=


Concrete Thickness t_cc 4.5in 0.114m=:=



ft3

:=


ft3

:=


m3

58.682 lb

ft3

=:=


m3

63.989 lb

ft3

=:=


m3

:=


Significant Wave Height Hs 2m 6.562 ft=:=


Spectral Peak Period Tp 1.05 Ts 4.41 s=:=



s1.673

ft

s=:= Zr 10m:=


ft

2

s:=




m3

:=


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1.3 Soil Parameter


m3

:=





Ws π4


D_in2ρcont D_out

2ρsw-+

...

:=

Ws 256.325 kg

m=

Buoyancy B π

4D_out

2ρsw 2.473 10

3

kg

m=:=



B1.104=:=



ρsw0=:=


ρsw0.318=:=


Tn d

g3.193s=:=

Tn

Tp0.724=

ϕ Tp

Hs3.118

s

m=:=


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mif

1 ϕ 5 s

m

if

3.3 otherwise

5=:=


Tn

Tp0.724=

Us 0.000001 Hs

Tn6.263 10

7-

m



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5-m:=


Kb 2.5 d50 0.313 mm=:=

Zob Kb

301.042 10

5- m=:=


D_out

Zo1.685 10

5=

Zr Zo

9.615 105=


U_D 1

ln Zr

Zo1+

1 Zo

D_out+

ln D_out

Zo1+

1-

Ur 0.409 m

s=:=



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i deg:=

Lift Force FL 1

2

ρsw


2:=

Drag Force FD 1

2

ρsw


2:=

Inertia Force FI πD_out

2

4

ρsw

g CM As sin θ( ):=


μ

Fw:=


0 100 200 300 40058.335

58.336

58.337

58.338

58.339

58.34

58.341

W_s

θ

deg



Wso Ws g 2.514 103

N

m=:=


m

=:= Ws 256.325 kg

m

=



Wreq4.394=:=



m=


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DNV RP305 (Operation) OBS - NPS 60

1. Input Parameter


Pipe Wall Thickness t_s 1 in:= ID D_s 2 t_s- 1.473m=:=


Concrete Thickness t_cc 4.5in 0.114m=:=



ft3

:=


ft3

:=


m3

58.682 lb

ft3

=:=


m3

63.989 lb

ft3

=:=


m3

:=


Significant Wave Height Hs 3.7m 12.139 ft=:=


Spectral Peak Period Tp 1.05 Ts 5.775s=:=



s2.067

ft

s=:= Zr 10m:=


ft

2

s:=




m3

:=


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1.3 Soil Parameter


m3

:=





Ws π4


D_in2

ρcont D_out2

ρsw-+

...

:=

Ws 315.984 kg

m=

Buoyancy B π

4D_out

2ρsw 2.473 10

3

kg

m=:=



B1.128=:=



ρsw0.034=:=


ρsw0.318=:=


Tn d

g3.193s=:=

Tn

Tp0.553=

ϕ Tp

Hs3.002

s

m=:=


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mif

1 ϕ 5 s

m

if

3.3 otherwise

5=:=


Tn

Tp0.553=

Us 0.000001 Hs

Tn1.159 10

6-

m



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5-m:=


Kb 2.5 d50 0.313 mm=:=

Zob Kb

301.042 10

5- m=:=


D_out

Zo1.685 10

5=

Zr Zo

9.615 105=


U_D 1

ln Zr

Zo1+

1 Zo

D_out+

ln D_out

Zo1+

1-

Ur 0.505 m

s=:=



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Tu9.005 10

7-

m

s2

=:=


Us4.355 10

5=:=

KC Number K Us Tu

D_out5.345 10

6-=:=


υ8.847 10

5=:=


Cd 1.2 Re 3 105

< M 0.8if

0.7 otherwise

0.7=:=

CL 0.9:=

CM 3.29:=

Soil Coefficient

Clay Soil


Ws g 1.131=:=


μc 0.24:=

Sand Soil

μs 0:=


μ μs soil 1=if

μc otherwise

:=

μ 0.24=

Calibration Factor

M 4.355 105

=

K 5.345 10 6-

=





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i deg:=

Lift Force FL 1

2

ρsw


2:=

Drag Force FD 1

2

ρsw


2:=


2

4

ρsw

g CM As sin θ( ):=


μ

Fw:=


0 100 200 300 40089.016

89.018

89.02

89.022

89.024

89.026

W_s

θ

deg



Wso Ws g 3.099 103

N

m=:=


m

=:= Ws 315.984 kg

m

=

if Ws Wreq "Change Concrete Thickness" "OK"( ) "OK"


Documents

HW5 2016 15512006