test calc - bj abz 20 inch lp gas pipeline
DESCRIPTION
Subsea pigging calculationTRANSCRIPT
-
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