test calc - bj abz 20 inch lp gas pipeline

Upload: mubariz841

Post on 09-Mar-2016

216 views

Category:

Documents


0 download

DESCRIPTION

Subsea pigging calculation

TRANSCRIPT

  • SYSTEM & OPERATOR DETAILS

    Page 1

    INTRODUCTION

    PROJECT: Socar Gas Pipelines PrecommissioningJOB No.: TBACLIENT: MCCILOCATION: Oil RocksSYSTEM: 20" LP Gas Pipeline

    PIPELINE DETAILSLENGTH: 18500.000 mOD: 508.000 mmWT: 15.900 mmID (nominal): 476.200 mmTEST PRESSURE: 125.000 BarCOMP OF FLUID: 0.00004125 BAR -1 at 15C & 3.5% salt Figure for Temperature CalcsEXP OF FLUID: 0.0002325 DEG C at 15C & 3.5% salt Figure for Temperature CalcsCOMP OF FLUID: 0.00004235 BAR -1 at 15C & 3.5% salt Figure for Pressurising CalcsPOISSONS RATIO: 0.300 C (at 50% of test pressure)YOUNGS MODULUS: 2070000 BAR

    CHEMICAL INJECTIONCHEMICAL 1CHEMICAL: SeadyePPM REQUIRED: 100.000

    CHEMICAL 2CHEMICAL:PPM REQUIRED:

    CHEMICAL 3CHEMICAL:PPM REQUIRED:

    CHEMICAL 4CHEMICAL:PPM REQUIRED:

    ACCEPTANCE:BJPPS Rep Name: TK AnandClient Rep Name:Company Rep Name:

  • CALCULATIONS

    Page 2 of 5

    20" LP Gas Pipeline

    1.1 System Volume Calculation

    V= ? m^3 Pipeline VolumeL= 18500.000 m Lengthd= 0.4762 m Internal Diameter

    V= pi x d^2 x L4

    V= 3.142 0.2268 18500.0004

    V= 3294.886 m^3

    1.2 Volume Per Metre Calculation

    Vm= ? m^3 Volume Per MetreV= 3294.886 m^3 System VolumeL= 18500.000 m System Length

    Vm= VL

    Vm= 3294.88618500.000

    Vm= 0.178 m^3

  • CALCULATIONS

    Page 3 of 5

    20" LP Gas Pipeline

    1.3 Pressure/Volume Formulae

    K= ? Bar 1 / CompressibilityC= 0.00004235 BAR -1 Compressibility of fluidV= ? m^3 Change in line volumeP= 125.000 Bar Change in pressurev= 3294.886 m^3 Line volumeu= 0.300 C Poissons ratio

    ID= 0.4762 m Internal diameterE= 2070000 Bar Young's modulust= 0.0159 m Wall thickness

    K Factor:

    K= 1Compressibility of fluid

    K= 10.00004235

    K= 23612.75 BAR

    Restrained Pipe:

    V= P x ( v / K ) + ((v x ( 1 - u^2) x ID))E x t

    V= 125.00 3294.8860 23612.75 3294.8860 1 0.09 0.4762070000 0.0159

    V= 125.00 x ( 0.1395 + 0.0434 )

    V= 22.8650 m^3

    Unrestrained Pipe:

    V= P x ( v / K ) + ((v x (5 - 4 x u) x ID))4 x E x t

    V= 125.00 3294.8860 23612.75 3294.8860 5 1.2 0.4768280000 0.0159

    V= 125.00 x ( 0.1395 + 0.0453 )

    V= 23.1033 m^3

    Terminated Pipe:

    V= P x [( v / K ) + ((v x (5 - 4 x u) x ID))4 x E x t

    V= 125.00 3294.8860 23612.75 3294.8860 2 0.30 0.4764140000 0.0159

    V= 125.00 x ( 0.1395 + 0.0405 )

    V= 22.5074 m^3

    1.4 Pressure/Volume Formulae

    Pump Sizing (1 BAR/MINUTE):

    Restrained Pipe: 182.9198 Litres/BARUnrestrained Pipe: 184.8267 Litres/BARTerminated Pipe: 180.0595 Litres/BAR

  • CALCULATIONS

    Page 4 of 5

    20" LP Gas Pipeline

    1.5 Temperature Effect

    Since changes in temperature are likely to be relatively small, the following approxiamte formula shall be utilised:

    P= ? Bar Resulting change In pressureE= 2070000 Bar Young's modulust= 0.0159 m Wall thickness

    B= 0.000233 DEG C Coefficient expansion fluidA= 0.00001116 PER DEG C Coefficient exp of materialC= 0.0000413 BAR -1 Compressibility of fluidu= 0.30 C Poissons ratio

    OD= 0.5080 m Outside diameterT= 0.1 DEG C Change in temperature

    Restrained Pipe:

    P= E x t x (B- 2 x A) x TE x t x C + ( 1 - u^2) x OD

    P= 2070000 0.0159 0.000233 2 0.000001116 0.102070000 0.0159 0.000041 1 0.09 0.5080

    P= 0.416 Bar

    Unrestrained Pipe:

    P= 4 xE x t x (B- 3 x A) x T4 x E x t x C + ( 5 - 4 x u) x OD

    P= 8280000 0.0159 0.000233 3 0.00000112 0.108280000 0.0159 0.000041 5 1.2 0.5080

    P= 0.410 Bar

    Terminated Pipe:

    P= 2 xE x t x (B- 3 x A) x T2 x E x t x C + ( 2 - u) x OD

    P= 4140000 0.0159 0.000233 3 0.00000112 0.104140000 0.0159 0.000041 2 0.30 0.5080

    P= 0.421 Bar

  • CALCULATIONS

    Page 5 of 5

    20" LP Gas Pipeline

    1.6 Chemical injection requirements (Unrestrained)

    Total volume of chemicals required (Chemical 1):

    V= Seadye m^3 ?P.P.M.= Parts per million 100

    FV= Volume to pressurise m^3 23.1033

    V= P.P.M.1000000 x FV(m^3)

    V= 1001000000 x 23.1033

    V= 0.0023 m^3 2.31 litres

    V= 0.0028 m^3(+20%) 2.77 litres(+20%)

    Total volume of chemicals required (Chemical 2):

    V= 0 m^3 ?P.P.M.= Parts per million 0

    FV= Volume to pressurise m^3 23.1033

    V= P.P.M.1000000 x FV(m^3)

    V= 01000000 x 23.1033

    V= 0.0000 m^3 0.00 litres

    V= 0.0000 m^3(+20%) 0.00 litres(+20%)

    Total volume of chemicals required (Chemical 3):

    V= 0 m^3 ?P.P.M.= Parts per million 0

    FV= Volume to pressurise m^3 23.1033

    V= P.P.M.1000000 x FV(m^3)

    V= 01000000 x 23.1033

    V= 0.0000 m^3 0.00 litres

    V= 0.0000 m^3(+20%) 0.00 litres(+20%)

    Total volume of chemicals required (Chemical 4):

    V= 0 m^3 ?P.P.M.= Parts per million 0

    FV= Volume to pressurise m^3 23.1033

    V= P.P.M.1000000 x FV(m^3)

    V= 01000000 x 23.1033

    V= 0.0000 m^3 0.00 litres

    V= 0.0000 m^3(+20%) 0.00 litres(+20%)

    SYSTEM & OPERATOR DETAILSCALCULATIONS