Cairn Energy India PTY Ltd BSPL Project - Barmer to Salaya Pipeline CALCULATION
PROJECT NO.
Stability Calculation 24" Oil Pipeline (Empty)
52893.03 REF
03-P-2-003
No OF SHEETS
1 of 4 DOCUMENT No
01 REV
OFFICE CODE
PROJECT No
AREA
DIS
TYPE
NUMBER
05
5893
03
P
2
003
PD
JC
PD
BS
CEIL
BY
CHK
ENG
PM
Client
21.05.07 Issued For IDC DATE
DESCRIPTION
Cairn Energy India Limited PIPELINE STABILITY CALCULATION JP Kenny CLIENT
Cairn Energy India Limited
SUBJECT
24" Crude Oil Pipeline
DOC NO.
05 - 2893 - 03 - P - 2 - 003
PRPD BY
Paul Docherty
DATE
5-May-07
CHKD BY DATE
Jateen Chohan 21-May-07
JOB NO.
52893.03
.
REV NO.
0
.
APP'D BY
P Docherty
DATE
21-May-07
1.0 INTRODUCTION The aim of this calculation is to determine the stability of the pipeline through water course crossings.
2.0 METHOD The method used to determine the stability of the pipeline through water courses is to calculate the total weight of the pipeline and compare it with the weight of the external fluid (water) it would displace. In order for the pipeline to be considered stable, the pipeline weight must be 10% greater than the weight of the water displaced. The above method is overly conservative as it considers a pipeline free floating in water. In reality the pipeline lays in a trench covered by soil which will provide a hold down force to prevent floatation. This force is calculated by determining the submerged weight of soil above the pipeline. 3.0 REFERENCE DATA The reference data for this calculation has come from: Reference 1: ASME B 31.4 Design and OISD - STD - 141 2001 X65
API 5L
Reference 2: Crude Pipeline Design Refer to Wall thickness Report 052893 - 03 - P - 3 - 020 and range of Calculations from 002 to 010
4.0 CALCULATIONS 4.1 Pipe Details Nom. Pipe Outside Diameter Specified Wall Thickness Concrete Coating Thickness
OD
=
0.610
m
=
24
t tc
=
0.018
m
=
17.5
in mm
=
0.102
m
=
102
mm
tp
=
0.008
m
=
8.0
mm
=
996
3
4.2 Coating and Insulation Details Thickness of the outer coating Density of the outer coating
Dp
=
9771
Thickness of the Insulation
ti
=
0.100
Density of the insulation
3
=
100.0
kg/m mm
3
=
60
kg/m
3
3
=
1000
kg/m
3
N/m m
Di
=
589
N/m
De
=
9810
N/m
Density Pipe Contents (Empty) Density Pipe Steel Density Concrete
Di Ds Dc
= = =
0 76420 29921
N/m N/m3 3 N/m
3
= = =
0 7790 3050
kg/m kg/m3 3 kg/m
Density Soil
=
14862
N/m
3
=
1515
kg/m
3
Density of Liquified Soil
Dg DL
=
12753
3
=
1300
kg/m
3
Stability Requirement
Sm
=
1.1
H
=
1.0
4.3 Density and Stability Factor Density External Fluid (Water)
Depth of Cover
N/m m
03-P-2-003 Rev 01 - 24 inch Stability Empty with insulation.xls 21/05/2007
3
Page 2 of 4
Self Check........................... Date...........................
Cairn Energy India Limited PIPELINE STABILITY CALCULATION JP Kenny 4.4 Weight of Pipeline Cross-sectional area Steel
Weight of steel
Cross-sectional area - internal
As
Ws
Ai
= pi()*(OD^2-(OD-2*t)^2) 4 2
=
0.033
=
As*Ds
=
2488
=
pi()*(OD-2*t)^2
=
0.259
=
Ai*Di
=
0
m
N/m
4
Weight of pipe contents
Wi
2
m
N/m
4.5 Insulation and Protective Coating CSA Insulation
Ain
=
pi()*(((2*ti)+OD)^2-(OD)^2) 4
=
0.223 Ain*Di
Weight of Insulation
Win
=
CSA Polyurathane Coating
Ap
=
=
2
m
131 N/m pi()*(((2*(tp+ti))+OD)^2-((2*ti)+OD)^2) 4
Weight of Coating
Wp
=
0.021
=
Ap*Dp
=
201
2
m
N/m
4.6 Concrete Coating ( Note: no consideration has been made for reinforcement) OD + 2*(tp+ti+tc) ODT = Total Pipeline Outside Diam.
Cross-sectional area - coating
Ac
=
1.0296
m
=
pi()*(ODT^2 -(2*tp+2*ti+OD)^2) 4
Weight of concrete coating
Total Weight of Pipeline
Wc
Wt
=
0.297
=
Ac*Dc
=
8894
2
m
N/m
= Wp+Win+Ws+Wi+Wc =
11713
N/m
4.7 Pipeline Buoyancy in Water CSA Pipe+ Ins + coatings
At
=
pi()*ODT^2 4
Weight of displaced fluid
Required Buoyancy
We
Br
=
0.833
=
At*De
=
8168
=
Sm*We
=
8984
2
m
N/m
N/m
03-P-2-003 Rev 01 - 24 inch Stability Empty with insulation.xls 21/05/2007
Page 3 of 4
Self Check........................... Date...........................
Cairn Energy India Limited PIPELINE STABILITY CALCULATION JP Kenny Pipeline Buoyancy, open trench
Cross-sectional area of trench
Bp
Ad
=
Wt-Br
=
2729
= =
Submerged weight of soil
Wg
Bs
(OD+2*ti+2tp)*H 0.826
m2
= Ad*(Dg-De) =
Pipeline Buoyancy, buried
N/m
4171
N/m
= (Wg+Wt)-Br =
6900
=
At*DL
=
10618
=
Sm*Wm
=
11680
=
Wt-Bm
=
34
N/m
4.8 Pipeline Buoyancy in Liquified Soils Weight of displaced fluid
Required Buoyancy
Pipeline Buoyancy, liquified soil
Wm
Bm
BL
N/m
N/m
N/m
5.0 Summary of Results The results of the stability calculations are detailed in the following table: Variable
Value
Required
Pipeline Buoyancy, open (N/m)
2728.9
0.0
OK
Pipeline Buoyancy, buried (N/m)
6899.9
0.0
OK
Pipeline Buoyancy, liq soil (N/m)
33.6
0.0
OK
The pipeline in an open trench will be stable through flooded water courses. The buried pipeline will be stable through flooded water courses. The pipeline will be stable through flooded water courses where soil liquification has occurred.
03-P-2-003 Rev 01 - 24 inch Stability Empty with insulation.xls 21/05/2007
Page 4 of 4
Self Check........................... Date...........................