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Tishreen University
mechanical and electrical engineering faculty
mechanical engineering department
Rehabilitation of pipeline between Banias-Homs
Study in hydraulic and pipeline integrity to Hold
B.Sc. in mechanical engineering prepared
by student
Mohammed Badr Mustafa Alzeer
Study steps :
1- Basic Informations .
2- Maximum allowable operation pressure(MAOP)
3- Hydraulic calculation
4-Analysis of corrosion inspection results
5- Modified MAOP calculation
1-Basic Informations :
1-1- Available Pipelines :
Pipe GradeWall
inches
Distance of
pipe
Date of
Installation
Pipe
Diameter
inches
5l GRB0.375106 Km195026
5L X520.37534 Km199224
5L X520.31172 Km197518
Schedule-1-
1-Basic Informations :
1-2- oil specifications :
Oil density : 0.879 Kg/m3 at 60F°
Oil Viscosity : 23 Centi Stokes .
API Quality degree : 34
Red Vapor Pressure : 5 psi .
2- MAOP Calculation :
P= 2StEFT/D (Barlow’s Formula)
Where :
S : specified minimum yield strength (psi) .
t : wall thickness ( inches ) .
F : Design Factor,( according the area crossing ) .
E : longitudinal joint factor .
T : temperature factor
D : nominal outside diameter (inches) .
2- MAOP Calculation :
62% SMYS
Bar
Psi
100% SMYS
Bar
psi
Pipe GradeWall Thickness
inches
Pipe Diameter
inches
43.16
625.88
69.62
1009
5L GRB0.37526
69.44
1006
112
1625
5L X520.37524
76.8
1113
123.9
1796
5L X520.31118
Schedule -2-
3-Hydraulic calculation :
3-1-Static Pressure calculation :
The maximum static pressure that generated by the filling
of pipeline in crude oil from Banias Pumps uptil the highest
point in pipe section .
P=ρ.g.h where
ρ : fluid density (Kg/m3 ) .
g : earth gravity constant ( 9.81 Kg.m/s2)
P = 855 *9.8066*(585-15 ) =4780903 N/m2
P = 47.8 Bars
3-Hydraulic calculation :
3-2- Pressures & Stresses by friction ( Linear Losses ) :
When Q= 100,000 barrel/day = 662 m3/h =0.184 m3/s .
HL=λ.L/Di .V2/2.g (Darcy-Weisbach Equation) .
Where : HL= friction Losses (m)
Di= internal Diameter (m)
V= liquid velocity (m/s2)
λ= Hydraulic Resistance Factor .
3-Hydraulic calculation :
3-2-1- Pressures & Stresses by friction ( Linear Losses ) :
Reynolds number, Hydraulic Resistance Factor and propotional roughness
are showed in schedule below :
λǨKReV
M/S
Q
M3/s
Di
inches
t
inches
D
Inches
0.028180.0000704μ5158900.56990.18425.25
0.6413
0.37526
0.027600.0000764μ5172580.672210.18423.25
0.590
0.37524
0.025660.0001024μ5231041.2050.18417.37
0.441
0.31118
Schedule-3-
3-Hydraulic calculation :
3-3- minor losses (specified losses)
Both of pipelines contain 7 (gate valves), 120
and 4 (no return valve).
Φ1,2,3=K1,2,3 *
𝑽 𝟐
𝟐.𝒈
Φ4 =
𝑪𝒄 .(𝑽𝟏−𝑽𝟐 ) 𝟐
𝟐.𝒈
3-Hydraulic calculation :
3-3-1- minor losses (specified losses)
Reducer Φ4No return
valve Φ3
Elbow Φ2Gate valve
Φ1
Pipeline
inches
-0.041380.00165530.003310726
-0.05710.0022870.004575924
0.001458840.1850.00740.014818
Schedule-4-
3-Hydraulic calculation :
3-4- Total Losses calculation :
For line 26’’- 106 Km :
H =(120* CL1 ) + (7* CL2 ) + (4 * CL3 ) + HL
H=𝟕𝟔.𝟒𝟏(𝒎 )
For line /24’’-18’’/-106 Km :
H =(120* CL1 ) + (7* CL2 ) + (4 * CL3 ) + HL
H=314.7 (m)
Hydraulic calculation :
3-5- Operation pressure calculation at beginning of line :
According to Pernoli formula :
𝑷𝟏
𝜸
+
𝜶. 𝑽𝟏 𝟐
𝟐 . 𝒈
+ 𝒁𝟏 =
𝑷𝟐
𝜸
+
𝜶. 𝑽𝟐 𝟐
𝟐 . 𝒈
+ 𝒁𝟐
For line 26’’ -106 Km :
V= constant
α=1
𝑷𝟏
𝜸
+ 𝒁𝟏 = 𝐇𝒕𝒐𝒕𝒂𝒍 + 𝒁𝟐
P1 = 51.9 (Bar ) .
3-Hydraulic calculation :
3-5-1 Operation pressure calculation at beginning of line :
For line 24’’- 18’’ 106 Km :
𝑷𝟏
𝜸
=
𝑽𝟐 𝟐−𝑽𝟏 𝟐
𝟐.𝒈
+∆z +H total
P1 =74.1 (bar ) .
3-Hydraulic calculation :
resultOperation
pressure Bar
62%
SMYS
Bar
GradeDiameter
inches
Line number
rejected51.943.16Gr B261
rejected74.1
69.44X5224
2
76.8X5218
Schedule-5-
3-Hydraulic calculation :
According to schedule-5- the operation pressure of the pipe>MAOP at flow
rate 100000 barrel/day, so we will reduce the flow rate to 500 M3/h .
Due that the pipeline pass through many cities like Banias city ,Tartous City,
and many highways (Damascus –Homs – Tartous international highway ) and
pass near many lakes and river like ( Tal Hosh lake ), The pipeline with
Diameter 26’’ which its Grade – 5L GrB - is not suitable to pumping .
3-Hydraulic calculation :
At flow rate 500 M3/h in line 24’’-18’’ is P1= 63 Bar .
3-Hydraulic calculation :
3-6-Pressure on-site at every point calculation :
We calculate the pressure at the location of each point of the
defining points selected at the peaks and
valleys characteristic and the 24 points along
the track, generated by pumping the required quantity, and this
is a very significant pressure to choose the right way of reform
Has been selected points of topographic planned within the
attached tables below
3-Hydraulic calculation :
3-6-2Pressure on-site at every point calculation :
where we take the distance between these points with determining linear
losses generated within these distances, and as we define topographic
planned height between each two points in a row to calculate the static
pressure including teams .. In order to pressure account location of each
proceed from the starting point being a known pressure from previous
calculations that we collect algebraic for losses (friction + local) between the
first point and the second from the pressure of the starting point as well as
the collection of algebraic values elevations between each of the two points,
planned topographic in the event of a landing, we get after that the
in the second point. Thus, we calculate the following points based on the
pressure points before and after the completion of these accounts, we have
calculated the pressure at each point of the selected points and in which
can draw a curved dynamic pressure, or the pressure generated at the site
shown below
3-Hydraulic calculation :
Pressure on
the site Bar
Total losses
+ static
height
Total losses
(m)
Length
between
the two
sites Km
Static
height
difference
(m)
Attributed
above sea
level
(m)
site
63.7515554/10
42.2168.21612.46334
60.2149546/16
-36.9367.06410.716-44
63.305439/25
-5.7070.7071.073-5
63.7870539/11
3-Hydraulic calculation :
Pressure on
the site Bar
Total losses
+ static
height
Total losses
(m)
Length
between
the two
sites Km
Static
height
difference
(m)
Attributed
above sea
level
(m)
site
12.28996.2899.5406
62.7576533/13
18.45450.45450.69418
61.2024532/25
to Homs
38.6252.62513618 inches
57.964601/634
3-Hydraulic calculation :
Pressure on
the site
Bar
Total losses
+ static
height
Total losses
(m)
Length
between
the two
sites Km
Static
height
difference
(m)
Attributed
above sea
level
(m)
site
33.9383.931.530
55.122902/632.5
-2.9350176.5-20
55.369703/626
-14.2415.7516-30
56.56404/620
70.5010.500460
50.651005/616
3-Hydraulic calculation :
Pressure
on the site
Bar
Total losses
+ static
height
Total losses
(m)
Length
between
the two
sites Km
Static
height
difference
(m)
Attributed
above sea
level
(m)
site
47.8767.876340
46.6351406/613
-14218-35
47.801057/605
-14.4910.504-25
49.02808/601
175.74815.7486160
34.282409/595
3-Hydraulic calculation :
Pressure
on the site
Bar
Total losses
+ static
height
Total losses
(m)
Length
between
the two
sites Km
Static
height
difference
(m)
Attributed
above sea
level
(m)
site
123.12613.1265110
23.9535010/590
37.357.352.830
23.938011/587.2
-94.2245.7752.2-100
28.72728012/585
24.85117.8516.87
26.6428713/578.2
3-Hydraulic calculation :
Pressure
on the site
Bar
Total losses
+ static
height
Total losses
(m)
Length
between
the two
sites Km
Static
height
difference
(m)
Attributed
above sea
level
(m)
site
161.48.403.2153440
13.114/575
155.5010.5014145
0.0658515/571
summit
point
-8.4366.5632.5-15
0.7756016/568.5
3-Hydraulic calculation :
554/10 539/11
Topographi…
SUMMIT…
Operation pressure
diagram at…
-5
15
35
55
75
95
0
100
200
300
400
500
600
700
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
‫الجغرافية‬ ‫اإلرتفاعات‬ ‫مخطط‬ ‫الضغوط‬ ‫مخطط‬
MAOP 72% SMYS MAOP 72% SMYS
MAOP 62% SMYS
MAOP 62% SMYS
Pressure diagram
4-Analysis of corrosion
inspection results
The 2008 inspection identified a total of 107026 metal loss
features. Of these, 102765 were reported as Internal Corrosion
features & the remaining 4261 were classified as External
Corrosion features.
The inspection also identified 1063 Dents, 112 Metal Objects,
22 Eccentric Casings, 20 Weld
Anomalies & 107 repair shells.
4-Analysis of corrosion
inspection results
The 2008 MFL inspection identified a total of 1063 dents. A
description of these dents is provided in schedule-7- :
TotalBottomTopDent description
1022585437Plain dents
211Dents associated by corrosion
871Dent Associated with Girth Welds
31526Dent Associated with Seam Welds
1063598465Total
Schedule-7-
5- Modified MAOP Calculation :
- According to the American Standard ANSI/ASME B31G the continuing corrosion
Area which has a depth of 10% up to 80% of the nominal thickness of the pipe
should not be through longitudinal extension greater than the calculated length L
of the pipeline through the pipeline .
L=1.12 *A* 𝑫.𝒕
Where : L : the length of corrosion Area
D: Nominal outer Diameter ( inch )
A: Calculated by the formula
A = (
𝒅
𝒕
𝟏.𝟏∗
𝒅
𝒕
−𝟎.𝟏𝟓
) 𝟐−𝟏
5-Modified MAOP Calculation :
IF the Length L is greater than the calculated value of the previous
relationship The value of A is given by the following relationship :
A = 0.893 *
𝑳
𝑫.𝒕)
5-Modified MAOP Calculation :
The Modified MAOP calculated by the formula below :
P* = 1.1*
𝟏−
𝟐
𝟑
∗
𝒅
𝒕
𝟏−
𝟐
𝟑∗𝒅/𝒕
𝑨+𝒂
For values for A less than 4 A<4 , The Modified MAOP calculated by :
P* = 1.1 *P * (1-d/t )
Where P is the design pressure .
For 50%Wt = 0.5*0.311=0.1555 inch
L=1000 mm=39.370 inch A= 14.8>4
P* = 1.1* 76.8*(1-0.1555/0.311) = 42.24 Bar
5-Modified MAOP Calculation :
554/10 539/11
Topographi…
SUMMIT…
Operation pressure…
-5
15
35
55
75
95
0
100
200
300
400
500
600
700
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
‫الجغرافية‬ ‫اإلرتفاعات‬ ‫مخطط‬ ‫الضغوط‬ ‫مخطط‬
MAOP 72% SMYS MAOP 72% SMYS
MAOP 62% SMYS MAOP 62% SMYS
Modified MAOP = 42 Bar
Rehabilitation of  pipeline between Banias-Homs

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Rehabilitation of pipeline between Banias-Homs

  • 1. Tishreen University mechanical and electrical engineering faculty mechanical engineering department Rehabilitation of pipeline between Banias-Homs Study in hydraulic and pipeline integrity to Hold B.Sc. in mechanical engineering prepared by student Mohammed Badr Mustafa Alzeer
  • 2. Study steps : 1- Basic Informations . 2- Maximum allowable operation pressure(MAOP) 3- Hydraulic calculation 4-Analysis of corrosion inspection results 5- Modified MAOP calculation
  • 3. 1-Basic Informations : 1-1- Available Pipelines : Pipe GradeWall inches Distance of pipe Date of Installation Pipe Diameter inches 5l GRB0.375106 Km195026 5L X520.37534 Km199224 5L X520.31172 Km197518 Schedule-1-
  • 4. 1-Basic Informations : 1-2- oil specifications : Oil density : 0.879 Kg/m3 at 60F° Oil Viscosity : 23 Centi Stokes . API Quality degree : 34 Red Vapor Pressure : 5 psi .
  • 5. 2- MAOP Calculation : P= 2StEFT/D (Barlow’s Formula) Where : S : specified minimum yield strength (psi) . t : wall thickness ( inches ) . F : Design Factor,( according the area crossing ) . E : longitudinal joint factor . T : temperature factor D : nominal outside diameter (inches) .
  • 6. 2- MAOP Calculation : 62% SMYS Bar Psi 100% SMYS Bar psi Pipe GradeWall Thickness inches Pipe Diameter inches 43.16 625.88 69.62 1009 5L GRB0.37526 69.44 1006 112 1625 5L X520.37524 76.8 1113 123.9 1796 5L X520.31118 Schedule -2-
  • 7. 3-Hydraulic calculation : 3-1-Static Pressure calculation : The maximum static pressure that generated by the filling of pipeline in crude oil from Banias Pumps uptil the highest point in pipe section . P=ρ.g.h where ρ : fluid density (Kg/m3 ) . g : earth gravity constant ( 9.81 Kg.m/s2) P = 855 *9.8066*(585-15 ) =4780903 N/m2 P = 47.8 Bars
  • 8. 3-Hydraulic calculation : 3-2- Pressures & Stresses by friction ( Linear Losses ) : When Q= 100,000 barrel/day = 662 m3/h =0.184 m3/s . HL=λ.L/Di .V2/2.g (Darcy-Weisbach Equation) . Where : HL= friction Losses (m) Di= internal Diameter (m) V= liquid velocity (m/s2) λ= Hydraulic Resistance Factor .
  • 9. 3-Hydraulic calculation : 3-2-1- Pressures & Stresses by friction ( Linear Losses ) : Reynolds number, Hydraulic Resistance Factor and propotional roughness are showed in schedule below : λǨKReV M/S Q M3/s Di inches t inches D Inches 0.028180.0000704μ5158900.56990.18425.25 0.6413 0.37526 0.027600.0000764μ5172580.672210.18423.25 0.590 0.37524 0.025660.0001024μ5231041.2050.18417.37 0.441 0.31118 Schedule-3-
  • 10. 3-Hydraulic calculation : 3-3- minor losses (specified losses) Both of pipelines contain 7 (gate valves), 120 and 4 (no return valve). Φ1,2,3=K1,2,3 * 𝑽 𝟐 𝟐.𝒈 Φ4 = 𝑪𝒄 .(𝑽𝟏−𝑽𝟐 ) 𝟐 𝟐.𝒈
  • 11. 3-Hydraulic calculation : 3-3-1- minor losses (specified losses) Reducer Φ4No return valve Φ3 Elbow Φ2Gate valve Φ1 Pipeline inches -0.041380.00165530.003310726 -0.05710.0022870.004575924 0.001458840.1850.00740.014818 Schedule-4-
  • 12. 3-Hydraulic calculation : 3-4- Total Losses calculation : For line 26’’- 106 Km : H =(120* CL1 ) + (7* CL2 ) + (4 * CL3 ) + HL H=𝟕𝟔.𝟒𝟏(𝒎 ) For line /24’’-18’’/-106 Km : H =(120* CL1 ) + (7* CL2 ) + (4 * CL3 ) + HL H=314.7 (m)
  • 13. Hydraulic calculation : 3-5- Operation pressure calculation at beginning of line : According to Pernoli formula : 𝑷𝟏 𝜸 + 𝜶. 𝑽𝟏 𝟐 𝟐 . 𝒈 + 𝒁𝟏 = 𝑷𝟐 𝜸 + 𝜶. 𝑽𝟐 𝟐 𝟐 . 𝒈 + 𝒁𝟐 For line 26’’ -106 Km : V= constant α=1 𝑷𝟏 𝜸 + 𝒁𝟏 = 𝐇𝒕𝒐𝒕𝒂𝒍 + 𝒁𝟐 P1 = 51.9 (Bar ) .
  • 14. 3-Hydraulic calculation : 3-5-1 Operation pressure calculation at beginning of line : For line 24’’- 18’’ 106 Km : 𝑷𝟏 𝜸 = 𝑽𝟐 𝟐−𝑽𝟏 𝟐 𝟐.𝒈 +∆z +H total P1 =74.1 (bar ) .
  • 15. 3-Hydraulic calculation : resultOperation pressure Bar 62% SMYS Bar GradeDiameter inches Line number rejected51.943.16Gr B261 rejected74.1 69.44X5224 2 76.8X5218 Schedule-5-
  • 16. 3-Hydraulic calculation : According to schedule-5- the operation pressure of the pipe>MAOP at flow rate 100000 barrel/day, so we will reduce the flow rate to 500 M3/h . Due that the pipeline pass through many cities like Banias city ,Tartous City, and many highways (Damascus –Homs – Tartous international highway ) and pass near many lakes and river like ( Tal Hosh lake ), The pipeline with Diameter 26’’ which its Grade – 5L GrB - is not suitable to pumping .
  • 17. 3-Hydraulic calculation : At flow rate 500 M3/h in line 24’’-18’’ is P1= 63 Bar .
  • 18. 3-Hydraulic calculation : 3-6-Pressure on-site at every point calculation : We calculate the pressure at the location of each point of the defining points selected at the peaks and valleys characteristic and the 24 points along the track, generated by pumping the required quantity, and this is a very significant pressure to choose the right way of reform Has been selected points of topographic planned within the attached tables below
  • 19. 3-Hydraulic calculation : 3-6-2Pressure on-site at every point calculation : where we take the distance between these points with determining linear losses generated within these distances, and as we define topographic planned height between each two points in a row to calculate the static pressure including teams .. In order to pressure account location of each proceed from the starting point being a known pressure from previous calculations that we collect algebraic for losses (friction + local) between the first point and the second from the pressure of the starting point as well as the collection of algebraic values elevations between each of the two points, planned topographic in the event of a landing, we get after that the in the second point. Thus, we calculate the following points based on the pressure points before and after the completion of these accounts, we have calculated the pressure at each point of the selected points and in which can draw a curved dynamic pressure, or the pressure generated at the site shown below
  • 20. 3-Hydraulic calculation : Pressure on the site Bar Total losses + static height Total losses (m) Length between the two sites Km Static height difference (m) Attributed above sea level (m) site 63.7515554/10 42.2168.21612.46334 60.2149546/16 -36.9367.06410.716-44 63.305439/25 -5.7070.7071.073-5 63.7870539/11
  • 21. 3-Hydraulic calculation : Pressure on the site Bar Total losses + static height Total losses (m) Length between the two sites Km Static height difference (m) Attributed above sea level (m) site 12.28996.2899.5406 62.7576533/13 18.45450.45450.69418 61.2024532/25 to Homs 38.6252.62513618 inches 57.964601/634
  • 22. 3-Hydraulic calculation : Pressure on the site Bar Total losses + static height Total losses (m) Length between the two sites Km Static height difference (m) Attributed above sea level (m) site 33.9383.931.530 55.122902/632.5 -2.9350176.5-20 55.369703/626 -14.2415.7516-30 56.56404/620 70.5010.500460 50.651005/616
  • 23. 3-Hydraulic calculation : Pressure on the site Bar Total losses + static height Total losses (m) Length between the two sites Km Static height difference (m) Attributed above sea level (m) site 47.8767.876340 46.6351406/613 -14218-35 47.801057/605 -14.4910.504-25 49.02808/601 175.74815.7486160 34.282409/595
  • 24. 3-Hydraulic calculation : Pressure on the site Bar Total losses + static height Total losses (m) Length between the two sites Km Static height difference (m) Attributed above sea level (m) site 123.12613.1265110 23.9535010/590 37.357.352.830 23.938011/587.2 -94.2245.7752.2-100 28.72728012/585 24.85117.8516.87 26.6428713/578.2
  • 25. 3-Hydraulic calculation : Pressure on the site Bar Total losses + static height Total losses (m) Length between the two sites Km Static height difference (m) Attributed above sea level (m) site 161.48.403.2153440 13.114/575 155.5010.5014145 0.0658515/571 summit point -8.4366.5632.5-15 0.7756016/568.5
  • 26. 3-Hydraulic calculation : 554/10 539/11 Topographi… SUMMIT… Operation pressure diagram at… -5 15 35 55 75 95 0 100 200 300 400 500 600 700 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 ‫الجغرافية‬ ‫اإلرتفاعات‬ ‫مخطط‬ ‫الضغوط‬ ‫مخطط‬ MAOP 72% SMYS MAOP 72% SMYS MAOP 62% SMYS MAOP 62% SMYS Pressure diagram
  • 27. 4-Analysis of corrosion inspection results The 2008 inspection identified a total of 107026 metal loss features. Of these, 102765 were reported as Internal Corrosion features & the remaining 4261 were classified as External Corrosion features. The inspection also identified 1063 Dents, 112 Metal Objects, 22 Eccentric Casings, 20 Weld Anomalies & 107 repair shells.
  • 28. 4-Analysis of corrosion inspection results The 2008 MFL inspection identified a total of 1063 dents. A description of these dents is provided in schedule-7- : TotalBottomTopDent description 1022585437Plain dents 211Dents associated by corrosion 871Dent Associated with Girth Welds 31526Dent Associated with Seam Welds 1063598465Total Schedule-7-
  • 29. 5- Modified MAOP Calculation : - According to the American Standard ANSI/ASME B31G the continuing corrosion Area which has a depth of 10% up to 80% of the nominal thickness of the pipe should not be through longitudinal extension greater than the calculated length L of the pipeline through the pipeline . L=1.12 *A* 𝑫.𝒕 Where : L : the length of corrosion Area D: Nominal outer Diameter ( inch ) A: Calculated by the formula A = ( 𝒅 𝒕 𝟏.𝟏∗ 𝒅 𝒕 −𝟎.𝟏𝟓 ) 𝟐−𝟏
  • 30. 5-Modified MAOP Calculation : IF the Length L is greater than the calculated value of the previous relationship The value of A is given by the following relationship : A = 0.893 * 𝑳 𝑫.𝒕)
  • 31. 5-Modified MAOP Calculation : The Modified MAOP calculated by the formula below : P* = 1.1* 𝟏− 𝟐 𝟑 ∗ 𝒅 𝒕 𝟏− 𝟐 𝟑∗𝒅/𝒕 𝑨+𝒂 For values for A less than 4 A<4 , The Modified MAOP calculated by : P* = 1.1 *P * (1-d/t ) Where P is the design pressure . For 50%Wt = 0.5*0.311=0.1555 inch L=1000 mm=39.370 inch A= 14.8>4 P* = 1.1* 76.8*(1-0.1555/0.311) = 42.24 Bar
  • 32. 5-Modified MAOP Calculation : 554/10 539/11 Topographi… SUMMIT… Operation pressure… -5 15 35 55 75 95 0 100 200 300 400 500 600 700 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 ‫الجغرافية‬ ‫اإلرتفاعات‬ ‫مخطط‬ ‫الضغوط‬ ‫مخطط‬ MAOP 72% SMYS MAOP 72% SMYS MAOP 62% SMYS MAOP 62% SMYS Modified MAOP = 42 Bar