2009 Solution

8/10/2019 2009 Solution

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8/10/2019 2009 Solution


The smaller buckling pressure is because of the imperfect roundness of pipe and deflection of pipesurface during manufacturing or installation process. Furthermore, the damage on pipe surface fromdropped objects will results in the smaller critical pressure. The mechanical stress resulting in sag

bending causes internal stress in the pipe which cause lower in buckling pressure.

There are several hydrodynamics forces acting on the pipeline.

1. The inertia and drag force can be estimated by Morrison equation

2 1( ) (1 )

4 2m d F t C D v C D v v

The wave speed and acceleration can be estimated by linear wave, non linear wave theory andspectral wave analysis

The C m and C d value are related with Reynolds’ numbers, Keulegan-Carpenter number and

relative roughness The current speed can be given from the ADCP measurement

2. Weight and Buoyancy force –using the Archimedes’ s law 3. Friction force - The friction coefficient is about 0.3 if pipe laid in the clay and 0.7 if laid on in

sand

1. Concrete coating – to increase the unit weight of pipeline which enhances the stability of pipe2. Mattress – The mattress will cover pipeline in appropriate intervals will provide sufficient weight

to the pipe.3. Trenching – Seabed was trenched along the pipeline to make pipe completely laid on the seabed

and avoid the sag bending and scour.4. Rock dumpling – rock was damped on top of the pipeline to provide additional weight

5. Buried in soil

8/10/2019 2009 Solution


The lateral buckling mainly causes by the excessive compressive stress due to thermal expansion in the

pipeline during operations. The hot reservoir fluid will cause extension of pipeline in the longitudinaldirection causing excessive compressive stress in the pipeline. Pipeline will try to find the easiestmitigate ways to relief this stress which will cause the lateral buckling and upheaval buckling. Theupheaval buckling is the commonly found if the pipe is laid cross the dune where top of the dune iscritical point that where it is easiest to buckling.

1. At given flow rate,2. Guess diameter (used empirical formula), calculate the average flow velocity3. Calculate Reynolds’ numbers 4. Calculate the friction loss (Re, roughness) by using Moody’s chart 5. Estimate the friction loss – if pressure drop is too high, assume larger diameter and recalculate

the pressure loss again until obtain reasonable pressure loss

212 2

P f v x D

8/10/2019 2009 Solution


Assume diameter is 7”, 0.1778 m

2

80,000*0.15924*3600 5.929 /

*(0.1778 )

4

QV m s

A

4850*5.929*0.17788.9*10

0.01eVD

R

( , ) 0.019e f R D

2 20.019*850*5.929* 20,000 5.38

2 2*0.1778 f V

P L MPa D

The sour corrosion condition is when the partial pressure of hydrogen sulphide is more than 0.0035 Bar.The sour corrosion results in fast cracking failure of the structure.

The domain diagram is the diagram to indicate the recommended steel in API5L using for pipelineoperated in the sour condition. Since the H 2S will attack at the molecule of steel especially high strengthsteel subjecting the high tensile loads. Therefore the plain steel used in the H 2S environment will be

limited by their hardness to ensure the integrity of structure over the service life.

8/10/2019 2009 Solution


The first corrosion mechanism of sour corrosion is sulphide stress cracking where the hydrogenmolecules are in the solution and they locally attack at the molecule of ferrous which will lead to the

embrittlement in the steel. This can happen in the high strength steel subjected under high tensile load.

Another type of sour corrosion is hydrogen induce cracking where the hydrogen in the gaseous phasediffuses in to the steel accumulated around non-metallic inclusion such as MnS. The hydrogen gasaccumulated around in the steel will cause high pressure area. The hydrogen gas will cause swelling ,

cracking along the metal surface and breaking steel in the lowest strength plain.

The insulation layer is wrapped and pipe was buried in soil because it is to prevent the fluid temperaturegetting lower than the critical temperature in which may cause operational problems such as hydration ,

wax deposition, flow insurance due to multiphase flow.

,2 ,1T T T

21

1

2 lncosh ( )1

2 2 2o

total soil layer i i

r H r D

R Lk Lk Lr h

8/10/2019 2009 Solution


8/10/2019 2009 Solution


The frictional pressure gradient in single phase flow is the pressure lost according to the friction force at

the pipe wall. This results from a homogeneous phase of fluid flowing in the pipe. The equation ofpressure gradient can be written as

2

2dP f V dx D

However, the multiphase flow model has to take account for multiple fluid phases flowing in thepipeline. Therefore, the frictional pressure gradient has to account for frictional force due to multiphaseflow. The liquid hold up determined by the average cross section area of liquid volume to the totalvolume will be introduced to estimate the proportion of liquid phase and gas phase in particular pipe

section. So, the friction pressure drop can be estimate.

Homogeneous flow model

2

(1 )

(1 )

2

L

T

m L g

m L g

m m m

V H

V

H H

H H

f V dP dx D

Separated flow model

2

2

e

2

2

R >2300 will be the turbulent flow regime

SL L SL

SL

SG G SG

SG

f vdP dx D

f vdP dx D

where

1/2

SL

SG

dP dx

X dP dx

8/10/2019 2009 Solution


22

2 2

13

2

2 2

11

1

1

L

G

L L

L GSL SG

C X X CX X

H

dP dP dP dx dx dx

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2009 Solution