Rehabilitation of pipeline between Banias-Homs

Tishreen Universitymechanical and electrical engineering faculty

mechanical engineering department

Rehabilitation of pipeline between Banias-HomsStudy 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 Grade Wall

thickness inches

Distance of pipe

Date of Installatio

n

Pipe Diameter

inches5l GRB 0.375 106 Km 1950 265L X52 0.375 34 Km 1992 245L X52 0.311 72 Km 1975 18

Schedule-1-

1-Basic Informations:

1-2 -oil specifications: Oil density : 0.879 Kg/m3 at 60FOil Viscosity : 23 Centi Stokes.

API Quality degree : 34Red 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 factorD : nominal outside diameter (inches).

2 -MAOP Calculation:

62% SMYSBarPsi

100% SMYSBarpsi

Pipe Grade Wall Thickness

inches

Pipe Diameter

inches43.16

625.8869.621009

5L GRB 0.375 26

69.441006

1121625

5L X52 0.375 24

76.81113

123.91796

5L X52 0.311 18

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/m2P = 47.8 Bars


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.


3-2-1 -Pressures & Stresses by friction ( Linear Losses ) : Reynolds number, Hydraulic Resistance Factor and propotional roughness are showed in schedule below:

λ Ǩ K Re VM/S

QM3/s

Diinches

tinches

DInches

0.02818 0.000070 4μ5 15890 0.5699 0.184 25.250.6413

0.375 26

0.02760 0.000076 4μ5 17258 0.67221 0.184 23.250.590

0.375 24

0.02566 0.000102 4μ5 23104 1.205 0.184 17.370.441

0.311 18

Schedule-3-


3-3 -minor losses (specified losses)Both of pipelines contain 7 (gate valves), 120 (elbow) , and 4 (no return valve).

Φ1,2,3=K1,2,3* Φ4=


3-3-1 -minor losses (specified losses)

Reducer Φ4

No return valve Φ3

Elbow Φ2 Gate valve Φ1

Pipelineinches

- 0.04138 0.0016553 0.0033107 26- 0.0571 0.002287 0.0045759 24

0.00145884 0.185 0.0074 0.0148 18

Schedule-4-


3-4 -Total Losses calculation: For line 26’’- 106 Km :

H =(120* CL1 ) + (7* CL2 ) + (4 * CL3 ) + HLH=For line /24’’-18’’/-106 Km:

H =(120* CL1 ) + (7* CL2 ) + (4 * CL3 ) + HLH=314.7 (m)

Hydraulic calculation:

3-5 -Operation pressure calculation at beginning of line: According to Pernoli formula:

For line 26’’ -106 Km: V= constantα=1

P1 = 51.9 (Bar ).


3-5-1 Operation pressure calculation at beginning of line: For line 24’’- 18’’ 106 Km:

+ ∆z +H totalP1 =74.1 (bar ).


result Operation pressure

Bar

62%SMYSBar

Grade Diameterinches

Line number

rejected 51.9 43.16

Gr B 26 1

rejected 74.1

69.44

X52 242

76.8

X52 18

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 .


At flow rate 500 M3/h in line 24’’-18’’ is P1= 63 Bar .


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 reformHas been selected points of topographic planned within the attached tables below


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 point 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, if planned topographic in the event of a landing, we get after that the pressure 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 they can draw a curved dynamic pressure, or the pressure generated at the site shown below


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

63.75 15 554/1042.216 8.216 12.463 34

60.21 49 546/16-36.936 7.064 10.716 -44

63.30 5 439/25-5.707 0.707 1.073 -5

63.787 0 539/11


Pressure on the

site Bar

Total losses +

static height

Total losses

(m)


Static height

difference (m)


( m)

site

12.2899 6.289 9.540 662.757 6 533/13

18.4545 0.4545 0.694 18

61.20 24 532/25to Homs

38.625 2.625 1 36 18 inches

57.964 60 1/634


Pressure on the

site Bar

Total losses +

static height

Total losses

(m)


Static height

difference (m)


( m)

site

33.938 3.93 1.5 3055.122 90 2/632.5

-2.9350 17 6.5 -2055.369 70 3/626

-14.24 15.751 6 -3056.56 40 4/620

70.50 10.500 4 6050.65 100 5/616


Pressure on the

site Bar

Total losses + static

height

Total losses

(m)


Static height

difference (m)


( m)

site

47.876 7.876 3 4046.635 140 6/613

-14 21 8 -3547.80 105 7/605

-14.49 10.50 4 -2549.02 80 8/601

175.748 15.748 6 16034.28 240 9/595


Pressure on the

site Bar

Total losses + static height

Total losses

(m)


Static height

difference (m)


( m)

site

123.126 13.126 5 11023.95 350 10/590

37.35 7.35 2.8 3023.9 380 11/587.2

-94.224 5.775 2.2 -10028.727 280 12/585

24.851 17.851 6.8 726.64 287 13/578.2


Pressure on the

site Bar

Total losses + static

height

Total losses

(m)


Static height

difference (m)


( m)

site

161.4 8.40 3.2 153 44013.1 14/575

155.50 10.501 4 1450.06 585 15/571

summit point

-8.436 6.563 2.5 -150.77 560 16/568.5


1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 240

100

200

300

400

500

600

700

-5

15

35

55

75

95

554/10 539/11

Topo-graphic diagram

SUMMIT POINT

Operation pressure dia-

gram at flowarate 500

m3/h

الOجغOرOافOيOة اإلرOتفاOعاOت مخطط الOضغOوط مخOطط

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 WeldAnomalies & 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: - Total Bottom Top Dent description1022 585 437 Plain dents

2 1 1 Dents associated by corrosion8 7 1 Dent Associated with Girth Welds

31 5 26 Dent Associated with Seam Welds

1063 598 465 TotalSchedule-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 *

Where : L : the length of corrosion AreaD: Nominal outer Diameter ( inch )A: Calculated by the formula

A =

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 *


The Modified MAOP calculated by the formula below: P* = 1.1 *

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 inchL=1000 mm=39.370 inch A= 14.8>4P* = 1.1* 76.8*(1-0.1555/0.311) = 42.24 Bar


1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 240

100

200

300

400

500

600

700

-5

15

35

55

75

95

554/10 539/11

Topo-graphic diagram

SUMMIT POINT

Operation pressure

diagram at flowarate 500 m3/h

الOجغOرOافOيOة اإلرOتفاOعاOت مخOطط الOضغOوط مخطط

MAOP 72% SMYS MAOP 72% SMYSMAOP 62% SMYS MAOP 62% SMYS

Modified MAOP = 42 Bar

Rehabilitation of pipeline between Banias-Homs

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