Technical catalogue submarine installations of pipes

Technical Catalogue
for Submarine
Installations
of Polyethylene Pipes

Table of content :
0.0 INTRODUCTION.......................................................................................................... 4
0.1 DIFFERENT TYPES OF SUBMARINE PIPELINES ............................................................... 4
0.1.1 Intake pipeline......................................................................................................... 4
0.1.2 Transit pipeline........................................................................................................ 5
0.1.3 Outfall pipeline ........................................................................................................ 6
0.2 SINKING OF SUBMARINE PE-PIPE, EXAMPLE FROM A REAL PROJECT. (SEE ALSO
SECTION A.5) ........................................................................................................................ 8
0.2.1 Introduction ............................................................................................................. 8
0.2.2 Sinking of the pipeline............................................................................................. 8
0.2.3 Installation of diffuser............................................................................................ 12
0.2.4 Weather conditions ............................................................................................... 13
0.2.5 Summary .............................................................................................................. 13
A. HYDRAULIC AND TECHNICAL DESIGN ......................................................................... 15
A.1 TECHNICAL DATA FOR DESIGN OF PE-PIPELINES........................................................ 15
A.2 HYDRAULIC DESIGN ................................................................................................. 18
A.2.1 Coefficient of friction ............................................................................................. 18
A.2.2 Coefficient for singular head losses ...................................................................... 20
A.2.3 Density head loss.................................................................................................. 22
A.2.4 Hydraulic capacity................................................................................................. 22
A.2.5 Self cleaning velocity ............................................................................................ 25
A.2.6 Air transport .......................................................................................................... 25
A.3 STATIC DESIGN ....................................................................................................... 28
A.3.1 Internal pressure................................................................................................... 28
A.3.1.1 HOOP DIRECTION................................................................................................................................28
A.3.1.2 LONGITUDINAL DIRECTION ................................................................................................................... 29
A.3.2 External loads / buckling....................................................................................... 31
A.3.2.1 BUCKLING OF UNSUPPORTED PIPE........................................................................................................ 32
A.3.2.2 BUCKLING OF PIPE IN TRENCH / SOIL PRESSURE .................................................................................... 35
A.3.3 Water hammer ...................................................................................................... 36
A.3.4 Temperature stresses........................................................................................... 38
A.3.5 Bending stresses .................................................................................................. 40
A.3.5.1 BUCKLING OF PE PIPE DURING BENDING............................................................................................... 41
A.3.6 Other stresses....................................................................................................... 43
A.3.6.1 CURRENT AND WAVE FORCES .............................................................................................................. 44
A.3.6.2 HOVERING PIPELINE............................................................................................................................ 45
A.3.6.3 CONCENTRATED LOADS ...................................................................................................................... 45
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A.3.7 Combined loads.................................................................................................... 46
A.4 DESIGN OF LOADING BY CONCRETE WEIGHTS ............................................................ 48
A.4.1 Degree of loading.................................................................................................. 48
A.4.2 Types of concrete weights .................................................................................... 50
A.4.3 Stability of PE-pipeline on the seabed .................................................................. 51
A.4.4 Recommended “air filling rate” for subwater pipelines .......................................... 54
A.4.5 Current forces ....................................................................................................... 55
A.4.6 Wave forces.......................................................................................................... 58
A.5 DESIGN OF PARAMETERS FOR THE SINKING PROCESS................................................. 66
A.5.1 Internal air pressure.............................................................................................. 67
A.5.2 Pulling force .......................................................................................................... 67
A.5.3 Sinking velocity ..................................................................................................... 71
B. INSTALLATION......................................................................................................... 76
B.1 JOINTING OF PE PIPES ............................................................................................ 76
B.2 BUTT FUSION OF PE PIPES....................................................................................... 77
B.2.1 Welding parameters.............................................................................................. 77
B.2.2 Welding capacity................................................................................................... 78
B.3 INSTALLATION......................................................................................................... 79
B.3.1 Buried PE pipes .................................................................................................... 79
B.3.2 Pipe laying on seabed........................................................................................... 81
AUTHOR : TOM A. KARLSEN, INTERCONSULT ASA............................................................... 84
LIST OF REFERENCES : ......................................................................................................... 84
REFERENCE PROJECTS…..………………………………………………..………………...85
Side 3 av 88

0.0 Introduction
Description of different types of submarine applications for polyethylene pipes.
Submarine PE-pipes have been used for transport of drinking water and sewage water since 1960.
The pipes were then produced in length of 12 m, welded together by butt fusion, weighted by
concrete loads and sunk to the sea bottom by entering water at one end and releasing air at the
other.
The method is nearly the same today. However there is more emphasis on design and calculations
to secure a safe installation and avoid damages.
Another innovation is use of long length (up to 500 m) pipes continuously extruded at the factory,
towed by boat to the site and jointed by flange connections.
This solution has been used successfully in overseas projects.
Since 1960 there has also been a significant improvement in the development of raw materials and
methods of production.
Therefore PE-pipes are today the most common pipe material in submarine applications.
The combination of flexibility and strength makes it superior to other materials.
In Norway, for instance, more than 95% of submarine pipelines are PE-pipes. The diameters vary
within the range Ø 50 mm - Ø 1600 mm, and the water depth can in special cases reach 250 m.
Damages happen very rarely.
This is due to :
- Excellent materials
- Proper design
- Experienced contractors
- Well educated supervisors
The consecutive technical catalogue deals with the design subject.
Here you will find theory and formulas that will enable you to calculate and solve the most common
problems occurring in submarine pipeline projects.
However, as an introduction, we first will mention the different types of submarine installations and
briefly describe a typical project example regarding the sinking of a pipeline.
0.1 Different types of submarine pipelines
If we follow the natural transport direction for consumer water, we can divide the installation into
3 categories :
- Intake pipeline
- Transit pipeline
- Outfall pipeline
0.1.1 Intake pipeline
Intake pipelines serve both civil and industrial applications.
The sources can be rivers, lakes and fiords. The intake depths vary from 2 m to 250 m.
The water is normally transported in the pipeline by gravity to an intake chamber.
In some cases, the intake pipeline is connected directly to the pump in a pumping station.
An intake pipeline is always exposed to negative pressure.
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Special problems to be aware of :
- Under-pressure
- Fouling
- Air release
- Current
- Waves
The intake end of the pipeline is normally supplied with a screen.
Fig. 0.1.1.1 shows an example from a river water intake. The figure shows a new water intake
in Glomma river. The 1200 mm diameter pipeline in 3 km long. The pipe material is PE PN80
SDR17.
The hydraulic capacity is 1.5 m3
/sec. The whole pipeline lies in a ditch 2-3 m deep for protection
against current, erosion, ice and floating timber. PE-pipes were chosen because of their flexibility,
strength and ease of installation.
Fig. 0.1.1.1 River water intake
0.1.2 Transit pipeline
In many cases it can be suitable to cross lakes and fiords by subwater pipelines instead of using
a longer route along the waterside.
In other situations it is necessary to cross rivers and seas to supply cities and islands with water,
or to remove wastewater.
The water can be transported by gravity or by pumping. During operation there is always an
overpressure in the pipe except in case of pressure surge.
It is normal to install a manhole/shaft on each waterside to establish an interface between the under
water pipeline. The equipment in the shafts depends on the service level. It is normal to install shut-
off valves.
Special problems to be aware of for transit pipelines are:
- Pressure
- Air transport
- Current
- Waves
- Fishing equipment
- Anchoring
Fig. 0.1.2.1 indicate a river crossing. The figure shows a profile of a PE-pipeline, a sewerage
crossing of the Glomma, the longest river in Norway. The diameter of the pipeline is 600 mm and
its wall thickness is 55 mm (PN10). The line length is 450 m. A five-metre deep pipeline-trench at
the river bottom was required to avoid damages to the pipeline from boat anchors. A PE-pipe was
chosen because of its flexibility, which permitted producing the whole length in one piece at
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the factory, towing it to the site and submerging it into the trench at the river bottom.
After submersion the trench was filled with gravel.
Fig. 0.1.2.1 Sewerage river crossing.
0.1.3 Outfall pipeline
Treated sewage water will normally be conveyed into the recipient discharge area at a certain depth
and distance from coast. A depth water outlet will provide excellent dilution of the wastewater.
Outlet deep will vary in the range 10-60 m dependent of the recipient’s self-purification capacity.
The recipient can be river, lake, fiord or sea.
The outfall usually starts from an outfall chamber at the waterfront to which the wastewater is lead by
gravity or pumping.
Use of pumping directly on the outfall pipeline is rather rare and not recommended. If pumping is
necessary, the best solution is to pump the sewage water into the outfall chamber and conduct it with
gravity into the recipient.
The main task for the outfall chamber is to prevent air from entering the pipeline.
Air can cause floatation of the pipe due to buoyancy.
It is also necessary to take into account the variations in low tide and high tide when designing
an outfall chamber.
Special problems to take into consideration regarding outfall pipelines are :
- Air entrainment in pipe flow
- Bio fouling
- Current and wave induced forces
- Sediment transport
Fig. 0.1.3.1 represent an industrial outfall. The figure shows the outfall system to the sea from
a steel plant in northern Norway. The main components in the outfall system are :
- 430 m pre-stressed concrete pipes with a diameter of 1.800 mm buried in the seabed at a water
depth of 4 m. The sea end of the concrete pipeline is connected to concrete anchor block.
The land end is connected to an outfall chamber.
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- 90 m PE-pipes PN3.2 with a diameter of 1.600 mm on the steep seabed from the anchor block
on to a depth of 30 m.
The PE-pipe was produced, transported 1.200 km by rail and submerged in one piece.
PE was selected over other pipe materials, because of its flexibility and because it required very little
construction work under water.
Fig. 0.1.3.1 Outfall system to the sea from an iron plant.
The example above is not very characteristic of an outfall. Usually the PE-pipe starts from the outfall
chamber.
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0.2 Sinking of submarine PE-pipe, example from a real project. (See
also section A.5)
In the following sequence we will introduce a typical example regarding sinking of a PE-pipeline
produced in long length. The example deals with an outfall pipeline.
0.2.1 Introduction
The project has the following characteristics:
· Pipe material: Ø1200 mm PE100 SDR26
· Length of pipeline: 4600 m
· Length of diffuser: 400 m
· Maximum depth: 61 m
· Loading percentage: 20 %
The consecutive description deals with the sinking process and the necessary precautions to be
taken to secure a safe installation at the bottom.
There are two different methods to be used, one for the pipeline itself and another for the diffuser.
Sinking of the pipeline is mainly carried out by Nature’s own forces, i.e. gravity, buoyancy and air
pressure, while sinking of the diffuser involved use of cranes.
This note is only a rough description of the main elements in the sinking phase. There must be
prepared a detailed sinking procedure prior to the real installation.
0.2.2 Sinking of the pipeline
The pipes will be towed from the production plan in Norway by tugboats to the installation site.
The pipeline will be delivered in sections of 400-600m. At arrival the pipes will be stored in surface
position as shown in fig. 0.2.2.1 below.
Fig.0.2.2.1 Storing the pipeline sections
It is important to find an assembly site sheltered from waves and currents.
Every section remains filled with air and is equipped with stub ends and blind flanges on each end.
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Next phase of work is to install the concrete weights. They are fixed to the pipeline at a certain centre
distance. This distance can vary along the pipeline dependent of the calculated forces to act at a
special depth. The weights can be installed on shore or off shore. Fig. 0.2.2.2 shows an installation
where the concrete weights are fixed to the pipe on shore and floated out on the water using cranes
or excavators. Usually the weights have rectangular shape and not round.
Fig.0.2.2.2 Concrete weights are fixed to the pipeline
When all sections are weighted they have to be fitted together by flanges or support sleeves.
This work is usually done off shore supported by barges and cranes. Fig. 0.2.2.3 shows a typical
installation.
Fig.0.2.2.3 Two pipe sections are flanged together
When all pipe sections are fitted together, the pipeline is ready for the sinking process. The pipeline
is equipped with blind flanges in each end. At the outmost end the blind flange is also equipped with
pipes and valves for air evacuation and air filling.
Before start of the sinking, the route has to be marked properly by buoys floating at the sea surface.
It is also very important to listen to the local weather forecast. There should be very little wind and
waves during the sinking process.
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The total pipeline is positioned in the correct route by boats, barges and small boats. The inmost end
is connected to the flange in the Outfall Shaft. There must be a pipe through the wall in
the shaft, so that seawater can enter the shaft during the sinking. A valve can be fitted to regulate the
flow.
Before the flange connecting take place, the inside air pressure in the pipeline has to be adjusted to
the pressure at connecting depth (for instance +0,3 bar if the start depth is 3m). A compressor does
this adjustment. The reason is to prevent the pipeline to “run away”.
It is also important to apply a pulling force in the outmost end of the pipe before the sinking starts.
This force can vary during the sinking operation and will be specially calculated beforehand.
Preliminary calculations show that the maximum pulling force will be approx. 40 tons.
The sinking starts by opening the air valve in the outmost end carefully and controlling the inside
pressure by a manometer if required to charge the pipe with compressed air. Beforehand there will
be calculated a curve showing the necessary air pressure as a function of the sinking depth. By
regulating the inside pressure according to this curve, we will get a controlled sinking with a nearly
constant speed. The sinking velocity may be approx. 0.3m/s.
The S-bend configuration expresses a balance between the forces acting downward (i.e. concrete
weights) and the forces acting upward (i.e. buoyancy of air filled section). This situation
is illustrated in fig.0.2.2.4.
Fig.0.2.2.4 PE-pipeline during sinking process.
The critical factor is the radius of curvature at the sea surface. If this radius is less than approx.
50 m in this case, the pipeline runs the risk for buckling (safety factor =2).
It is necessary to carry out the sinking operation as a continuously process. If the sinking stops,
the E-modulus for the PE material will decrease by time and the minimum radius of curvature will be
reduced analogously. This can cause buckling of pipe. If, for any reason, it should be necessary
to interrupt the installation, it is important to start the compressor and reverse the sinking process.
This action must take place within 15 minutes. The compressor must be able to work at 7 bars.
As we can imagine the S-configuration will be transformed to a J-configuration when the sinking
reaches the outmost end of the pipe. In this position we have to apply a correct pulling force and
a correct sinking speed to prevent dynamic acceleration forces when the last volume of air leaves
the pipe. The length of the pulling wire must also be in accordance to the maximum depth to secure
a safe “landing” of the pipe end at the bottom. The “landing” takes place when the pulling force is
gradually reduced to zero.
Side 10 av 88

Fig.0.2.2.5 and 0.2.2.6 show the pipeline during the sinking process. Observe the assistance boat
and the pulling wire from the tugboat at the outmost end.
Fig.0.2.2.5 The submerging process has started
Fig.0.2.2.6 Shortly before the end of the pipeline is leaving the surface.
It should also be mentioned that the concrete weights have to be fixed properly to the pipeline to
prevent sliding during installation. To increase the coefficient of friction and to avoid scratches
in the surface of the pipe, we install an EPDM rubber gasket between the pipe and the concrete
weights. An example of a concrete weight system is shown in fig.0.2.2.7.
Side 11 av 88

Fig.0.2.2.7 Concrete weight system
The torque moment for the bolts will be calculated to secure a sufficient bolt force. Sometimes
it is also adequate to use rubber cushions on the bolts.
0.2.3 Installation of diffuser
Sinking of the diffuser has to be carried out in a different way than the pipeline.
The diffuser will be produced or assembled in one piece, 406m long, and towed to the site in
the same way as the pipeline sections. The pipe material is PE 100 SDR26 and diameter is
staggered from Ø1200mm to Ø500mm. The contractor will drill the holes in the diffuser on site.
Concrete weights and buoyancy elements will be fixed on the pipe before submerging.
The capacity of the buoyancy elements must be greater than the weight of the pipe including
the fixed weights.
The way of doing the submersion is to lower the pipe as a beam from barges. Fig.0.2.3.1 on next
page shows the installation in principle.
The diffuser section must not be lifted out of the water. In such a case the stresses will be too high in
the PE 100 material and the diffuser will suffer damage.
There must be carried out a proper calculation of the static system during submerging.
This calculation includes how many fix points and hook points are needed to get a safe installation.
For the moment we assume 3 or 4 hooking points. This means that we need 4 boats/barges with
cranes if the diffuser shall be submerged in one piece. There is an alternative to divide the diffuser in
4 pieces and submerge them separately. In this case they will be “mated” together on sea bottom or
some distance above by flange connections.
Choice of method will depend on resources available and on costs / risks assessments.
Side 12 av 88

Fig.0.2.3.1 Principle of sinking a diffuser as a beam
If the ratio between the radius of curvature and diameter of the pipe (R/D) =20, there will be
a collapse or buckling of the pipe. Maximum allowable stress in the pipe material in the sinking
phase should not exceed 10 Mpa.
Preliminary calculations show that the sinking cannot be done without support from buoyancy
bodies. It means that only a part of the installed buoyancy bodies, from the operation process at
the water surface can be removed before submerging by the crane.
In the calculations of necessary support from such bodies the safety factor against buckling shall not
be less than 3, taking into account the sinking process will be influenced also by waves and current.
Safety factor against buckling = 3.0 gives R/D min. = 60
The modules of elasticity for the PE material are assumed to be 300 Mpa. Such a value corresponds
to a 1.5% strain in the material during approximately 24 hours at a temperature of 30o
C. If the sinking
takes more time, the situation will be more unfavourable because of a decrease in the modules of
elasticity.
The buoyancy bodies must stand the water pressure at the water depth of 60m. They are not
allowed to slide along the pipeline during the sinking.
As indicated in fig.0.2.3.1, the cranes working simultaneously will lower the diffuser. This method
requires a safe communication system among the human operators.
0.2.4 Weather conditions
Expected timeframe for the whole operation with the main pipe, including joining of the different
sections and the sinking process, is expected to be approx. 3-5 days. The sinking process should
require a weather window of 12 hours.
Expected timeframe for sinking of the diffuser is assumed to be 12 hours. Including the preparation
for the sinking, the timeframe is expected to be 1-2 days.
Weather/wave forecast data is essential in the preparation for the sinking processes. The wave
height should not exceed 1m during the submersion of the pipeline. It will raise the safety factor
against damage to the pipes if the wave action is as small as possible.
0.2.5 Summary
During sinking of the outfall pipeline in this example one had to consider the following factors:
Side 13 av 88

· Detailed sinking procedure must be worked out including technical parameters, necessary
resources, communication systems and emergency procedures
· Detailed calculations of the sinking curvatures must be carried out by computer programs
· The pulling force in the end shall be approximately 40 tons
· The sinking speed shall not exceed 0.3 m/s
· The compressor shall work at a pressure up to 7 bar
· Air pressure curve as a function of depth shall be calculated
· The critical radius of curvature is approximately 50m
· The sinking shall be carried out in an continuous process
· Concrete weights must be fixed securely
· The weather conditions must be satisfactory
· The diffuser must be installed as a beam system by use of cranes
· The static system during lowering of the diffuser must be calculated
· The diffuser must be “mated” to the main pipeline at sea bottom
· The sinking shall be carried out under assistance from a supervisor with experience in this field
Generally it is recommended to do as much as possible of the installation work from sea surface
position. Use of divers shall be minimized. It is also favourable to do all butt welding at the
manufacturer’s plant if possible.
We hope this introduction has given the reader an idea of how PE-pipes can be applied in
subwater applications.
In the following sequences we shall deal with the design problems.
Side 14 av 88

A. Hydraulic and technical design
A.1 Technical data for design of PE-pipelines
To carry out calculations we need figures for mechanical properties.
The essential mechanical properties are described in terms of :
EO = modulus of elasticity at zero loading time and low load (Mpa)
EC = creep modulus, time > 0, stress s > 0 and constant (Mpa)
ER = relaxation modulus, time > 0, strain e > 0 and constant (Mpa)
sO = burst strength at time zero (Mpa)
sC = creep strength at time > 0 (Mpa) (also called burst stress)
e
n = Poisson’s ratio = l
er
el = strain in the axial direction
er = strain in the ring direction
a = thermal expansion (º C -1
)
For practical purposes the relaxation modulus (ER) and the creep modulus (EC) are assumed to be
equal.
ER = EC = E (E-modulus) as being function of load and loading time
The mechanical properties for a PE-pipe are also dependent on the temperature. Normally the
properties are given at 20ºC or 23ºC .
Fig. A.1.1 and A.1.2 show examples of how the E-modulus and the creep strength (burst stress)
vary as a function of time and stress. For the creep strength the influence of the temperature is also
indicated.
The curves are taken from the Borealis book “Plastics Pipes for Water Supply and Sewage
Disposal” written by Lars-Eric Janson [1].
= E
Fig. A.1.1 The relationship between creep modulus E and tensile stress with time as parameter for
HDPE Type bars HE2467 (full lines) and HDPE Type 2 bars HE2467-BL (dotted lines) at 23ºC
[1].
Side 15 av 88

= sc
Fig. A.1.2 Principal stress/time curves for PE80 and PE100 pipes at 20ºC and 80ºC
The standard curve for HDPE Type 2 at 80ºC (acc. to DIN8075) is shown for
comparison. The minimum required strength (MRS) at 20ºC and 50 years is
10 Mpa for PE100 and 8 Mpa for PE80 giving the design stress 8 Mpa
and 6.3 Mpa, respectively.
For PE-pipes, 50 years operation time is usually chosen as service life.
The design stress (sd) is introduced by the formula :
sC,50year
s = A.1-1)d
C
sC,50year = burst stress (creep stress) for the PE material for a constant load in 50 years
C = design factor (safety factor)
The safety factor varies from country to country dependent on the national standards.
Normal values are C = 1.25 or C = 1.6.
Today we are mainly talking about the material qualities PE80 and PE100.
These materials have burst stress of 8Mpa and 10Mpa respectively for a constant stress
in 50 years at 20ºC.
The design stresses are shown in table A.1.1:
Material Design stress
C = 1.6
Design stress
C = 1.25
PE80
PE100
5.0 Mpa
6.3 Mpa
6.4 Mpa
8.0 Mpa
Table A.1.1 Design stress
The client must assess the risks in his project when deciding the design factor.
For submarine applications, we normally use a design factor of 1.6.
Side 16 av 88

In table A.1.2, we have listed guiding mechanical properties for PE-materials to be used
in calculations (T = 20ºC ).
Property Unit PE80 PE100
Density
Design stress 50years sd,50
Design stress at time zero sd,0
Modulus of elasticity
at time zero E0
Modulus of elasticity
after 50 years E50
Poisson’s ratio n
Average coefficient of
thermal expansion a
kg/m
3
Mpa
Mpa
Mpa
Mpa
-
-1
ºC
950
5.0/6.4 *
8.0/10.4 *
800
150
0.4-0.5
0.2×10
-3
960
8.0/6.3 *
9.4/12.0 *
1050
200
0.4-0.5
0.2× 10
-3
* Safety factors are 1.6 and 1.25 respectively
Table A.1.2 Mechanical properties for PE-pipes.
There is a continuous improvement and development of PE materials. In the particular case we
recommend you to contact the pipe producer or the raw material manufacturer to get exact figures
for the properties.
Another important factor is the roughness according to Nikuradse regarding calculation of the
hydraulic capacity for the pipeline.
A new pipe will have a low roughness, but fouling may occur as a function of time and increase the
roughness factor.
The quality of the water running through the pipe is important for development of the roughness.
Normally we distinguish between potable water and waste water.
For a new pipe the roughness value can be as low as 0.05 mm but this is only of theoretical
interest.
In table A.1.3 we have proposed design values for equivalent roughness based on experience in
Norway.
Type of water Type of PE-pipeline
Intake Transit Outlet
Potable
Sewage
2 mm
-
0.25 mm
0.50 mm
-
1 mm
Table A.1.3 Design values for equivalent roughness (e)
If the pipes are regularly flushed supported by a cleaning pig, the values in table A.1.3 may be
reduced.
Side 17 av 88

R
=
v =
=
A.2 Hydraulic design
The pressure (Dh) drop in a pipeline can generally be described by the formula :
2 2
Dh = f
L
×
v
+ åk ×
v
+
Dr
× y A.2-1)
D 2
× g
2 × g ro
f = coefficient of friction (see diagram fig. A.2.1.1)
L = length of pipe (m)
D = internal diameter (m)
v = velocity in pipe (m/s)
g = acceleration of gravity (= 9.81 m/s2
)
k = sum of coefficients for singular head losses
Dr = density difference between water inside the pipe and water in recipient (kg/m3
)
ro = density of water inside the pipe (kg/m3
)
y = water depth at outlet point in recipient
A.2.1 Coefficient of friction
The friction coefficient (f) is dependent of Reynolds number (Re) :
v × D
e
n
A.2-2)
v = velocity
n = viscosity of water (m2
/s)
The viscosity of water depends on the temperature.
T = 20ºC n = 1.0 × 10 –6
m2
/s
T = 10ºC n = 1.3 × 10 –6
m2
/s
We recommend applying the value for 10ºC.
The velocity (v) can be calculated by the formula :
Q = flow (m3
/s)
v =
4 × Q
pD2
A.2-3)
As we see, the Reynolds number can be calculated if we know the flow and the internal diameter.
Example 1
Destine Reynolds number for a flow of 100 l/s in a pipe with internal diameter 327.2 mm. T = 10ºC
Solution :
First we calculate the velocity, v, from A.2-3)
4 × 0,100
m / s =1,19 m / s
p × 0,32722
Reynolds number is found from A.2-2)
1,19 × 0,3272
Re
1,31×10-6
= 2,09×105
When we know the Reynolds number, the friction coefficient can be found from the Moody chart, fig.
A.2.1.1.
Side 18 av 88

Fig. A.2.1.1 The Moody chart for pipe friction with smooth and rough walls
The entrance parameter on the horizontal axis (x-axis) is the Reynolds number.
To find the right curve, we need to decide the relative roughness (rr) for the pipe wall.
e
rr = A.2-4)
D
e = absolute roughness, taken from table A.1.3 (mm)
D = internal diameter (mm)
On the right hand side in the Moody chart you will find figures for relative roughness representing
different curves.
The intersection point between Reynolds number and the relative roughness curve gives the
coefficient of friction (f). The value for (f) is found on the vertical axis (y-axis) on the left hand side in
Moody's chart.
Example 2
Assume that example 1 represent a pipeline for transport of potable water crossing a fiord.
Find the coefficient of friction (f).
Solution :
5
We have already calculated the Reynolds number in example 1 Re = 2.97 × 10
Now we need to find the relative roughness (rr) : e = 0.25 mm
Side 19 av 88

=
p =
v
2
Hence :
0,25
rr = 0,0008
327,2
Knowing Re and rr we take f from fig. A.2.1.1 as indicated in the diagram with dotted lines and
arrows.
The result is : f » 0.02
For rough estimates without any Moody chart in hand, it is often usual to use f = 0.02 as an average
value.
Knowing f, we can calculate the friction pressure drop (Dhf) for the pipeline from part one in
formula A.2-1)
Dh = f ×
L
×
v
A.2-5)f
D 2 × g
Example 3
Calculate the friction pressure drop for the pipeline described in example 1 and 2 if the length
is 2500 m.
Solution :
Formula A.2-5) gives the result in the unit mwc (meter water column) :
2500 1,192
Dh = 0,02× × mwc = 11,03 mwcf
0,3272 2 ×9,81
To convert this unit to Pa (N/m2
) we introduce the relationship :
p = r × g × h A.2-6)
p = pressure (N/m2
= Pa)
r = density of water (1000 kg/m3
)
g = acceleration of gravity (9.81 m/s2
)
This gives : p = 1000 × 9.81× 11.03 Pa = 108204 Pa
If we divide this figure with 105
we get the unit (bar), and if we divide it with 106
we have
the unit Mpa.
108204
bar = 1.08 bar
100000
108204
p =
1000000
MPa = 0.108 MPa
A.2.2 Coefficient for singular head losses
Part two of formula A.2-1) represent the singular pressure drops (Dhs) :
2
Dhs = åk ×
2 × g
A.2-7)
The expression k means a sum of discrete head losses.
Head losses arise for instance in bends, in diameter changes, in inlet and outlet of pipe, in beads,
in valves, in screens, in water meters and in diffusers.
Side 20 av 88

Table A.2.1.1 gives guiding values for singular coefficients.
Singular headloss k-factor
k-factor
V Inlet 1
V
Inlet 2
V Outlet
q Elbow
q Smooth bend
k = 1,0
k = 0,5
k = 1,0
2
k=1,1.( q
)90o
k= 0,2 . sin q (rough)
k= 0,1 . sin q (smooth)
V
Diffuser k = 16
V Intake
screen k = 0,03
Bead
Gate valve (open)
Non return valve
k = 0,03
k= 0,2
k= 10
Table A.2.1.1 Guiding coefficients for singular head losses.
Example 4
The pipe described in example 1 is equipped with 3x90º elbow, 25 beads and has an outlet
in a elevated reservoir. Calculate the total head loss.
Solution :
From table A.2.1.1 we find the coefficients :
90º elbow Þ k = 1,1× (
90
)2
90
= 1.1
Bead Þ k = 0.03
Outlet Þ k = 1.0
Total sum of coefficients comes to : Sk = 3 × 1.1 + 25 × 0.03 + 1.0 = 5.05
Total singular head loss: Dhs
1.192
= 5.05× mwc = 0.36 mwc
2 × 9.81
Side 21 av 88

=
=
A.2.3 Density head loss
Term 3 in formula A.2-1) describe the density head loss (called saltwater resistance) when
water is flowing into a recipient where the density of the water (for instance seawater) is higher.
Dhr
Dr
× y
ro
A.2-8)
This term normally comes into account only when dealing with outfall pipelines if difference in density
between wastewater and recipient water.
Difference in density can be due to content of salt in water or difference in temperature.
Example 5
Calculate the saltwater resistance for an outlet pipeline installed to 50 depth in the sea.
Density for wastewater is 1000 kg/m3
while density for seawater is 1025 kg/m3
.
Solution :
Formula A.2-8) gives the result: Dhr
1025 -1000
×50 mwc = 1.25 mwc
1000
As we see the saltwater resistance reaches a significant value and must always be taken into
consideration for outfall pipelines in saltwater recipients.
A.2.4 Hydraulic capacity
In previous chapters we have calculated the pressure drops for a given pipe diameter and a given
design flow.
Sometimes the case is opposite. We know the available pressure and flow and want to decide
the actual diameter.
We therefore have to calculate the diameter from the formulas A.2-1) and A.2-3).
This gives the equation :
g×(Dh -
Dr
× y)×p2
×D5
- åk ×8×Q2
×D -8×f ×Q2
×L = 0
ro
A.2-9)
The equation of degree 5 for the diameter, D, can not be solved explicitly.
We therefore have to make a simplification.
Since the singular head loss normally is small compared to the friction loss, we neglect term 2 in A.2-
9) and find an approximately diameter:
1
5
é ù
ê 2 ú
D = ê= 8× f × Q × L ú A.2-10)
ê Dr ú
êg × p2
(Dh - × y)ú
ë ro û
The factor f is chosen to be 0.02.
After we have decided the theoretical diameter from A.2-10), we pick the nearest standard diameter
above in the manufacturers programme.
This diameter is put into formula A.2-1 to check that the total pressure drop is less than
the allowable.
Side 22 av 88

× y) × D × g
×
00
ê
û
Another approach to the problem is to solve the flow (Q) from equation A.2-9)
1
é Dr ù 2
2 ê2 × (Dh -
r ú
Q =
pD ê o ú A.2-11)
4 ê f × L + Sk × D ú
ê ú
ë û
If we choose the value f = 0.02, only the diameter D is unknown on the right side in the equation A.2-
11)
By choosing values for D in steps, it is possible to solve the problem by iteration.
The diameter (D) that gives the correct flow (Q) is the solution in the equation.
Equation A.2-10) is applied to find the “start value” in the iteration process.
Knowing the flow and diameter it can be useful to control the friction coefficient from Moody’s chart,
fig. A.2.1.1.
If necessary the value is corrected and a new iteration carried out.
Example 6
Find the optimal diameter, D, for the pressure drops given in example 3, 4 and 5 for a requested flow
Q = 100 l/s. SDR = 11.
Solution :
We find the approximately diameter from A.2-10)
1
é 8× 0.02 × 0.12
× 2500 ù
D = ê ú m = 0.325 m = 325 mm
êë9.81× p2
(11.03 + 0.36 +1.25 -1.25)úû
The nearest standard diameter above for SDR11 is 327.2 mm (Ø 400 mm).
This diameter value is inserted in A.2-11).
Hence:
p × 0.32722
é2 × (11.03 + 0.36 +1.25 -1.25) × 0.3272 ×9.81
Q = ×
1
2
= 0.1m3
/ s =100 l /
s
q.e.d.
4 ë 0.02 × 2500 + 5.05× 0.3272 ú
By the system of formulas previous described in chapter A.2, we can do exact hydraulic calculations
for subwater pipelines.
In cases where a roughness estimate is required, we can use diagrams based on work carried out
of Colebrook-Prandtl-Nikuradse.
In fig. A.2.4.1 is shown a chart for absolute roughness k = 1.0 mm [3].
If we know the friction drop available in 0
/ (=
Dh
×1000)
L
we can find the necessary diameter
when the flow is given.
Generally we can solve one of the quantities Q, Dh, D when 2 of them are known. In the chart you
also can read the velocity.
Side 23 av 88

Dr =
oo
00
o
Q = Flow (l/s)
Fig. A.2.4.1 Hydraulic capacity, e = 1 mm
Example 7
An outfall pipeline is 2500 m long and ends at 50 m depth. The design flow is 100 l/s and available
pressure drop is 13 mwc. Density in the sea water is 1025 kg/m3
.
Estimate the necessary pipe diameter when neglecting the singular head losses.
Solution :
First we calculate the density loss :
1025 -1000
× 50 mwc = 1.25 mwc
1000
Total available friction drop is : Dhf = (13 -1.25) mwc = 11.75 mwc
We find the incline of
the friction drop line (I) : I =
11.75
×1000
2500
/oo
=
4.7 o
/
We enter chart A.2.4.1 with the quantities
Q = 100 l/s and I = 4.7 0
/ .
The intersection point gives : D = 340 mm
We choose the nearest standard diameter above to include the singular head losses. For SDR11
this gives Ø 450 mm, di = 368.2 mm.
Example 6 is similar to example 7. In the last case we got a one step bigger diameter.
The approximate cost difference between the two results for a 2500 m long pipeline amount
to 70.000 Euro.
This example can be a motivation to carry out proper hydraulic calculations.
Side 24 av 88

A.2.5 Self cleaning velocity
Another important factor for subwater pipelines is to prevent deposits inside the pipe and to prevent
accumulation of air/gas.
To check the pipelines capacity for self-cleaning, we introduce the flow’s shear stress (t) :
t = rg ×
D
× I
4
r = density of water (kg/m3
)
g = acceleration of gravity (= 9.81 m/s2
)
A.2.12)
Dh
I = incline of friction drop line
L
To be self-cleaning the shear stress shall be ³ 4 N/m2
Example 8
Check if the pipeline Ø 400 mm PE SDR11 in example 6 is self-cleaning?
Solution:
We must find the incline of the friction drop line: I =
11,03
= 0.0044
2500
Hence using A.2.12): t =1000 × 9.81×
0.3272
× 0.0044 N/m = 3.5 N/m2
4
As we see the shear stress is < 4.0. We therefore must expect some deposits in the pipeline.
In such a case it can be useful to install equipment for flushing and use of cleaning pig.
A.2.6 Air transport
Air and gas accumulations are the ”worst enemies” for subwater pipelines.
To handle the problems there are 2 possible solutions:
a) Prevent air from entering the pipeline
b) Provide a sufficient velocity in the pipe to transport air/gas through the pipeline
Air/gas accumulations in a pipeline will/can bring :
- reduce the hydraulic capacity
- entail flotation or vertical displacement
If possible, we recommend method a) to be the safest solution.
For an outfall pipeline the outlet chamber must be constructed in a way that air can not enter the
pipeline. It means that you have to take into account :
- Lowest low water level in recipient / source (LLW)
- Vortex
- Fluctuations in water level due to sudden change in flow
In most cases this means that top of the outfall pipeline in the point where it leaves the chamber shall
be in the range 0.5-1.5 m below LLW.
For inlet pipelines the maximum under-pressure shall be less than 4 mwc to avoid air release from
the water. Siphon constructions are normally not recommended.
For both outfall pipelines and intake pipelines, we recommend avoiding high points in the trace.
Side 25 av 88

For transit pipelines it must be possible to remove air in the manholes at the shoreline when starting
up the water transport during general operation and in case of repair work.
For sewage transport the retention period must not exceed the time limit for H2S emission.
As an indicator, 4 hours retention period should not be exceeded (depends however on operating
temperature).
In solution b) the critical speed, Uc, must be obtained by the flow to remove air bubbles present
in the pipe.
The critical speed of water, Uc, is given by : Uc = f (Di sin a) A.2.13)
Di = pipe internal diameter (m)
a = pipe gradient
A simplified expression gives Uc as a function of gDi
Uc = k
×
gDi A.2.14)
g = acceleration of gravity (9.81 m/s2
)
The factor k is displayed in Fig. A.2.6.1 as a function of sin a
The curve of k in Fig. A.2.6.1 is applicable for a= 0º ®90º.
1,2
1,0
0,5
a £ 5o
a > 5o
0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8
0,9 1,0
1o
2o
5o
10o
20o
30o
40o
50o
60o
90o
Ö sin a
(a)
Fig. A.2.6.1 Critical velocity for transport of air in a pipeline.
Example 9
Calculate the critical velocity for transport of air in a pipeline with slope a = 10º and internal
diameter Di = 500 mm.
Solution :
From fig. A.2.6.1 gives : k = 0.75
If we insert this value in A.1-13 we got : Uc = 0.75 × 9.81× 0.5 m/s = 1.66 m/s
As we see, the system requires quite a high velocity to transport air.
Side 26 av 88

If the velocity in the pipe is higher than 1.66 m/s, air bubbles are carried away with the water.
If the speed is less than 1.66 m/s, air bubbles will move backward to be released onshore provided
there are no high points in the trace.
This is a theoretical consideration. In the real case there is a diffuse transition for Uc.
Formula A.2-13 gives however an qualified indication.
Side 27 av 88

P
A.3 Static design
In this chapter we will present formulas to decide the wall thickness of the pipe taking into account
internal and external forces acting on the pipeline.
The internal diameter of the pipe is decided by the formulas in chapter A.2.
We will underline that quite a few of these calculations are necessary to carry out in a real
project. It is important to sort out the significant factors with respect to the pipe’s service life.
A.3.1 Internal pressure
The internal pressure will create stress in the pipe wall both in the hoop direction and the longitudinal
direction. The stress in the longitudinal direction is dependent on the way the pipeline is able to
move (fixed or free movement).
A.3.1.1 Hoop direction
Fig. A.3.1.1.1 indicate the static system.
Fig. A.3.1.1.1 Static system for
internal pressure, cut pipe.
p
N N
S
sr Dm sr
No shear stress will occur due to the internal pressure. There will only be a tensile force (N) in
the ring direction.
If we integrate the pressure components we find the following result based on equilibrium of
forces :
2 × N = p × Dm A.3-1)
N = tensile force (N)
p = pressure (N/m2
=
Pa) Dm = mean diameter (m)
Introducing the ring stress (sr) and the wall thickness (s), we can develop the following formulas :
N= sr ×S A.3-2)
s =
p × Dm
r
2 ×S
A.3-3)
S =
p × Dm
2 × sr
p × D
A.3-4)
Since Dm = D – s s = A.3-5)
(2 × sr + p)
sr = design stress (see table A.1.2)
D = external diameter
Side 28 av 88

sl
s =
=
=
C =
Example 1
Find the wall thickness for a Ø 200 mm PE80 pipe exposed to a design pressure of 1Mpa (10 bar)
Design safety factor = 1.6.
Solution :
The wall thickness (s) is found from formula A.3-5). s is taken from table A.1.2
1× 0.2
m = 0.0182 m = 18.2 mm
(2 ×5 +1)
The stress (s) for a given pipe in the hoop direction exposed to a pressure (p) can be calculated from
the formula :
p
sr (SDR -1)
2
where SDR=
D
s
A.3-6)
Example 2
Given a PE100 pipe SDR 17.6 exposed for a pressure of 0.8Mpa (8bar). Calculate the stress in the
pipe wall and the safety factor against burst after 50 years of loading ?
Solution :
Formula A.3-6) gives the hoop stress :
0.8
sr (17.6 -1) MPa
2
= 6.64 MPa
Safety factor confer A.1-1) :
10
=1.50
6.64
A pipe is always exposed to additional forces besides the internal pressure, for instance temperature
forces, forces in bends and reducers, depth of backfill in trenches, water hammer, forces from
current and waves, installation forces etc.
You have to consider the factor of safety (design factor) taking into account these other forces.
The method process is to calculate all acting forces and find the maximum combined stress.
This is the method adopted in the consecutive chapters.
A.3.1.2 Longitudinal direction
Fig. A.3.1.2.1 below shows the stress and strains for a pipe exposed to internal pressure.
er
el
DL sr
Q p Q
L
Fig. A.3.1.2.1 Pipe exposed to internal pressure
The internal pressure will give a deformation in the longitudinal direction if the pipe is free to move.
The tube will try to be shorter due to contraction :
Side 29 av 88

=
DL =
DL =
=
el = -n × er A.3-7)
el = strain in longitudinal direction
er = strain in ring direction
n = Poisson’s figure (0.4-0.5)
If there is no friction force acting against the movement, there will be no permanent stress in the
longitudinal direction and the shortening (DL) will be fully developed as indicated by formula A.3-8).
This is the case for a pipeline floating free in surface position :
DL = -n × L × er
L = length of pipe
To estimate er we have to introduce Hook’s law :
A.3-8)
s
er =r
E
A.3-9)
sr = stress in ring direction (ref. formula A.3-6)
E = modulus of elasticity (creep modulus) (ref. table A.1.2)
This gives :
er
p
(SDR -1)
2 × E
A.3-10)
n × L × p
(SDR -1)
2 × E
A.3-11)
Example 3
Calculate the shortening of a PE80 pipe SDR11 exposed to an internal pressure p=1.2Mpa and able
to move free. The length of pipe is 100 m. Short time E-modulus can be set to 800 Mpa and
Poisson’s figure is 0.5.
Solution :
The task is solved applying formula A.3-11)
-0.5×100 ×1.2
(11-1) m = - 0.375 m
2 ×800
As we see the shortening can be significant. If the end coupling for such a pipeline is not tensile,
leakage will occur. We also see that the result is independent of the diameter.
In most cases the movement of the pipe is prevented due to anchor blocks, soil cover etc.
It means that stresses will occur in the longitudinal direction.
The maximum stress appears when the strain is zero :
slmax = n × sr A.3-12)
slmax
n × p
(SDR -1)
2
A.3-13)
The stress will be a tension.
Side 30 av 88

=
=
e = =
reach
Example 4
Calculate the maximum longitudinal stress for the data given in example 3.
Solution :
By use of formula A.3-13) we get : slmax
0.5×1.2
(11-1) MPa = 3 MPa
2
As we see the longitudinal stress can reach the half of the hoop stress.
The stress in the longitudinal direction will decrease by time due to relaxation in the PE-material.
This is due to a permanent strain while the E-modulus is reduced by time. This fact can be seen
from Hook’s law :
s = E × e A.3-14)
Constant
Decreasing
Example 5
Discover the long-term stress in longitudinal direction for a fixed pipe exposed to a constant
pressure 1Mpa. Assume SDR = 11, short time E-modulus = 800 Mpa, long time E-modulus =
150 Mpa and n = 0.5.
Solution :
First we calculate the stress from A.3-13) :
0.5×1
sl (11-1) MPa = 2.5 MPa
2
The corresponding strain from A.3-14) :
s 2.5
×100% = 0.31%
E 800
Long term stress for this constant fictive
strain can also be found from A.3-14) : sl,long term =150 × 0.0031MPa = 0.465 MPa
As we see the long-term stress is
0.465
×100% =18.6% of the short-term stress in
2.5
longitudinal direction. The relaxation is significant.
Compared to the hoop stress, which is constant over time, the stress in longitudinal direction
0.465
×100% = 9.3%
5
after about 50 years of operation.
A.3.2 External loads / buckling
In this chapter we shall study the risk for buckling of a PE pipe exposed to external loads.
These loads for subwater pipes can be :
- Under-pressure
- Soil cover in trench
The under-pressure can be created in several ways :
- Friction and singular losses in intake pipes
- Pressure surge
- Under-pressure during sinking of pipe
- External water pressure on air filled pipes used as buoyancy elements
Side 31 av 88

= 3
Buckling occurs when compressive forces in the pipe’s hoop direction exceed the stability of the
material.
Fig. A.3.2.1 shows “buckling pictures” for a pipe in firm soil trench and in loose soil/air/water.
Firm soil Loose soil
water or air
Fig. A.3.2.1 Different types of buckling
There is a significant difference for a pipe’s
resistance against buckling if it is installed
in a trench or installed at sea bed.
n > 2 n = 2
A.3.2.1 Buckling of unsupported pipe
A pipe during sinking or laying on the seabed can be considered as unsupported for normal
distances between the concrete weights.
The buckling pressure for an unsupported pipe can be calculated by the formula :
pbuc
2 × E s
× ( )× k A.3-15)
1- n2 Dm
Pbuc = buckling pressure (Mpa)
E = modulus of elasticity (For long lasting loads the creep modulus shall be applied.
For pressure surge we apply the short term modulus of elasticity)
n = Poisons figure (0.4-0.5)
s = wall thickness (m)
Dm = mean diameter (m)
k = correction factor due to ovaling, ref. fig. A.3.2.1.1
Fig. A.3.2.1.1 Correction factor due
to ovaling.
k 1,0
0,9
0,8
0,7
0,65
0,6
0,5
0,4
0,3
0,2
0,1
1 2 3 4 5 6
%
Degree of ovaling
Side 32 av 88

=
=
buc
=
=
= 0,5
Formula A.3-15) can be transformed by introducing the SDR ratio (SDR =
D
):
s
pbuc
2 × E
×
1- n2
k
(SDR -1)3
A.3-16)
From fig. A.3.2.1.1 we realize that the ovaling of the installed pipe is of significant importance
regarding the capacity against buckling. For a standard pipe an ovaling corresponding
1-1,5% is acceptable. This gives a reduction factor k = 0.65.
Example 6
Calculate the buckling pressure capacity pbuc for an unsupported pipe Ø 900 mm PE100, SDR26
exposed to pressure surge. Short time E modulus is 1050 Mpa. Assume ovaling 1% and n = 0.4.
Solution :
By use of formula A.3-16) and fig. A.3.2.1.1 we get :
2 ×1050 0,65
p × MPa = 0,099 MPa =10 mwcbuc
1- 0,42
(26 -1)3
In practice this means that the pipe for a short period of time can withstand full vacuum.
It is, however, usual to introduce a safety factor, F=2.0, for such calculations.
We will not recommend to expose the actual pipe for an under-pressure greater than
p 10
mwc = 5 mwc
F 2
Example 7
Calculate the safety factor against buckling for an unsupported PE100 pipe, SDR33 used as an
intake pipeline. The pipeline is exposed to a constant under-pressure 2 mwc in the most critical
point. Long term E-modulus can be set to 200 Mpa. Ovaling is 1% and n= 0.4.
Solution :
Using formula A.3-16) and. fig. A.3.2.1.1 we obtain :
2 × 200 0,65
p × MPa = 0,0094 MPa =1mwcbuc
1- 0,42
(33 -1)3
Factor of safety : F =
pbuc
=
1
pappear 2
The pipe will buckle due to under-pressure before it reaches a lifetime of 50 years.
Theoretically the pipe will buckle when the E modulus equals 400 Mpa. This will happen already
after 1-2 years of operating (ref. fig. A.1.1).
Underwater pipelines exposed to under-pressure can be supported by concrete weights if the
distance between the weights is small enough. Hence the capacity against buckling will increase.
If the distance (l) between the supports (weights or rings) is in the range :
4 ×
s × Dm
< l £
1.56 × s
0.5
A.3-17)
2 (s/Dm )
Side 33 av 88

=
s =
=
=
F=
The buckling pressure pbucl can be written :
pbucl
2.2 ×s × E
×
l
k
pbuc ×
F
A.3-18)
l = length between the supports (centre distance – width of support)
pbuc = buckling pressure for unsupported pipe (ref. formula A.3.16), k=1.0)
s = wall thickness
k = reduction factor due to ovaling, ref. fig. A.3.2.1.1
F = safety factor (2.0)
Example 8
The pipe described in example 7 is equipped with concrete weights with a centre distance of 3 m.
The width of the blocks is 0.4 m and the diameter of the pipe is Ø 600 mm.
Calculate the safety factor against buckling.
Solution :
The wall thickness is :
600
mm =18,2 mm
33
Distance between the supports : l = (3-0.4) m = 2.6 m
Buckling pressure for unsupported pipe, assumed k= 1.0 in formula A.3-16) :
2 × 200 1
p × MPa = 0.0145MPa »1.5 mwcbuc
1- 0.42
(33 -1)3
We apply formula A.3-18) with F = 1.0 :
pbucl
2.2 ×18.2 × 200
×
2600
0.0145 ×0.65 MPa = 0.017
MPa
»1.7 mwc
The safety factor against buckling is :
1.7
= 0.85
2.0
As we see the safety factor has increased from 0.5 to 0.85, but the pipe will still buckle.
Buckling will happen when the creep modulus. E, is approximately 275 Mpa. This happens after
about 10 years of operation (ref. fig. A.1.1).
To reach a safety factor of 2.0 in this specific case, the following solutions can be considered :
- Shorter distance between the concrete weights
- Support of steel rings
- Installation of the pipe in a trench
- Increase of pipe diameter to reduce the under-pressure caused by friction
- Increase the wall thickness to improve the capacity against buckling
The choice of solution must be based on a technical/economical assessment. For more advanced
calculations see [12].
In next chapter we will consider buckling of a pipe installed in a trench.
Side 34 av 88

t
A.3.2.2 Buckling of pipe in trench / soil pressure
A pipe installed in a trench has a significant better capacity against buckling than an unsupported
pipe. The most important factors are :
- Ring stiffness of the pipe
- Modulus of elasticity for the soil (tangential modulus)
The buckling pressure (q) can be estimated by formula A.3.19) [8]:
q = 5.63
× SR
F
a =
1- 3×
d
× E 1
× a
A.3-19)
A.3-20)
D
SR = ring stiffness, short term
SR =
1
E
12 × (SDR -1)3
1
A.3-21)
Et = 2 Es
1
= tangential modulus for the soil
Es = secant modulus for the soil (ref. fig. A.3.2.2.1)
d
D = ovaling ( » 0.05)
F = safety factor (should never be less than 2.0)
Secant Modulus
E`s MN/m2
Filling height H m
Fig. A.3.2.2.1 Secant modulus for granular soil versus filling height
in submarine trenches.
For a pipe installed in a trench we have to add the pressure caused by soil cover to the under-
pressure caused by hydraulic flow.
The soil pressure (qs) around a PE-pipe is considered to be uniformly distributed along
the perimeter.
qs = (g-gw)×h A.3-22)
g = specific gravity of soil
gw = specific gravity of water
h = height of soil cover
Side 35 av 88

s
=
s t
q =
F = =
Es
Example 9
Let us return to example 7 and 8.
We choose to dig down the pipe in a trench with 1 m soil cover. Decide the safety factor against
buckling in this case. Assume short time E-modulus for pipe = 1000 Mpa, = 0.05, g = 20
kN/m3
D
and mod. Proctor for soil = 80%.
Solution :
First we decide the pipe’s ring stiffness from A.3-21) :
1000 ×1000
S kPa = 2.54 kPaR
12 × (33 -1)3
Correction factor, s, due to ovaling is taken from A.3-20) : a = 1 - 3× 0.05 = 0.85
1
is found from fig. A.3.2.2.1 : E 1
= 600 kPa Þ E 1
= 2× 600 kPa = 1200 kPa
Using formula A.3-19) assuming F= 1.0 we get the buckling pressure :
5.63
×
1
2.54 ×1200 × 0.85 kPa = 264 kPa = 0.264 MPa » 27 mwc
We realize that the pipe now can withstand an external pressure corresponding to approximately
27 mwc.
Compared to example 7 and 8 there is an external soil pressure caused by the cover, ref. formula
A.3-22)
qt = (20-10) × 1 kN/m2
= 10 kN/m2
= 0.01 Mpa » 1 mwc
Total external pressure is : qt = under-pressure + soil pressure = (2 mwc + 1 mwc) = 3 mwc
Safety factor against buckling :
q 27
» 9
qt 3
By installing the pipe in a trench with soil cover the safety factor has increased from 0.85 to » 12.
This indicates that we ought to install pipes in trenches if they are exposed to significant external
forces and the SDR class is high.
Regarding subwater pipelines it may be economically favourable to reduce the SDR-class compared
to installing the pipe in a trench.
A.3.3 Water hammer
Water hammer (pressure surge) occurs in a pipeline when there is a sudden change in the flow.
The result is a pressure wave going backwards and forwards in the system.
The most common reason for pressure surge is sudden start and stop of pumps or closing/opening
of valves. Even if there is installed frequency converter on pumps the electric power supply can fail.
Exact calculations of water hammers are complicated and must be carried out by computer
programs.
However there is a simplified method that gives an indication of the maximum and minimum
amplitude of the pressure wave. This method will be presented below.
For intake and outfall pipelines water hammer is normally not a problem if the pipes are not directly
connected to the pump, but suddenly closing of gates must be avoided.
Change in flow will be damped in the intake and outfall chambers.
The area of the chamber should be designed for the expected variations in flow.
Side 36 av 88

× ê
In such cases the amplitude of the fluctuations will be in the range of ± 1 m above maximum and
below minimum operating level respectively.
For transit pipelines and intake and outfall pipelines connected directly to pumps, the water hammer
can imply damage to the pipe if the pressure class is too low.
The most critical is normally the under-pressure which can reach values >10 mwc if there are
significant high points in the trace.
To reduce the pressure surge, we can install flywheel mass on the pumps or connect pressure
vessels.
Such solutions are most often economic favourable compared to reduction of the SDR ratio for the
pipe, but depend on the length of the pipeline and the diameter.
It shall also be mentioned that pressure surge can occur during sinking of PE-pipes [12].
The size of the water hammer is derived from the general relationship of surge
Dp =
Dv × c
g
A.3-23)
The surge pressure/water hammer is said to linearly depend on the pressure wave speed, c, in water
inside a pipe. Dv is the change of water flow speed (acceleration/retardation) and
g = 9.81 m/s2
. The pressure wave speed, c, is given by :
é ù
1/ 2
Eo s
c = ú A.3-24)
(1- n2
) ×r ëDm û
Eo = short time modulus of elasticity, ref. table A.1.2.
n = 0.4-0.5 = Poisson ratio
r = density of water
s = wall thickness
Dm = Do-e
With rewriting, we get an equation of c as a function of the pipe SDR class :
c =
Eo
×
(1- n2
) ×r
1
(SDR -1)1/ 2
A.3-25)
Surge is a short time condition (few seconds) under which a PE pipe, applied to a constant long term
stress, returns to its initial E-modulus at time zero.
In table A.3.3.1, we have calculated the pressure wave speed for PE100 and PE80 materials as
a function of SDR class.
Polyethylene Pipe
Pressure wave speed in water inside a PE-pipe c m/sec
SDR33 (PN322) SDR26 (PN4) SDR17.6 (PN6) SDR11 (PN10)
PE100
2
Eo = 1050 N/mm
PE80
2
Eo = 800 N/mm
203
180
230
200
282
250
263
320
n =
0.45
Table A.3.3.1 Pressure wave speed for PE
Side 37 av 88

Dp =
In practice, Dv in A.3-23) may be positive or negative :
positive, as caused by shutting a valve at the end of a transmission line,
or by starting a pump
negative, as caused by a pump failure or by a sudden change of hydraulic
conditions that reduce the flow-rate and speed.
Example 10
Find the size of the water hammer for a PE100 pipe SDR 17.6 if the change in water velocity
= 0.15 m/s (reduction).
Solution :
From table A.3.3.1 we get c = 282 m/s
Hence using formula A.3-23) :
- 0.15×282
= - 4,3 mwh =
9.81
- 0.44 bar
The pressure is an under-pressure.
This result must be added to other external loads to check the risk for buckling.
Assuming the required space of time to shut a valve to be from one to two minutes, when operated
properly, the maximum surge pressure should be in the range :
Dpmax = 10 – 15 % times the pipes pressure rating PN (bar)
If water hammer repeats regularly over a pipe’s service life, it may cause fatigue failure.
As a rule of thumb, a PE-pipe can sustain 107
oscillations of amplitude + 0.5 x nominal pressure
without diminishing its service lifetime.
Under-pressure will never lead to fatigue, only ovalisation.
A.3.4 Temperature stresses
If a pipe is exposed to a change in temperature, it will try to adjust its length if it can move freely.
The change in length DL can be expressed :
DL = a × Lo × DT A.3-26)
a = thermal expansion coefficient (» 0.2×10-3
ºC –1
)
Lo = initial length at installation
DT = change in temperature
As we see the change in length is independent of the diameter and the wall thickness.
Example 11
How much shorter will a PE-pipe be if it is installed in sea water at 4ºC when it had a length of
3000 m at 20ºC in the production factory ?
Solution :
We apply formula A.3-26) and get : DL = 0.2 × 10-3
× 3000 × (4-20) m = -9.6 m
There have been some real examples where subwater pipelines have been too short due to change
in temperature.
Side 38 av 88

If this is not discovered in time, it can cause conflicts and extra costs.
When estimating the length of a pipe, we always have to take into consideration temperature
changes before placing an order.
If the movement of the pipe is prevented, stress in the pipe wall will be the result.
Concrete weights, anchor blocks or cover in trenches can prevent the pipe’s movement.
If the pipe is totally fixed, the stress (sT) can be expressed :
sT = -E ×s × DT
E = modulus of elasticity (creep modulus) (Mpa)
A.3-27)
A positive value is regarded as a tension stress. As A.3-27) indicated, the stress is independent
of the pipe length and the diameter. The stress will be reduced by time as the E-modulus decreases
due to relaxation in the PE-material.
Example 12
A submarine pipeline is installed in the winter when the sea temperature is 4ºC. In the summer the
temperature can reach 20 ºC. The pipe is a PE100 Ø 315 mm SDR11 and can be considered to be
totally fixed by the concrete weights.
Calculate the stress caused by change in temperature the first summer assuming E=500 Mpa.
What happens after 50 years ?
Solution :
Formula A.3-27) gives : sT = - 500 × 0.2 × 10-3
× (20-4) Mpa = -1.6 Mpa
A compression stress will occur since the sign is negative.
After 50 years the E-modulus is reduced to 200 Mpa, ref. table A.1.2. This gives :
sT,50 = - 200 × 0.2 × 10-3
× (20-4) Mpa = -0.64 Mpa
The stresses are acting in the longitudinal directions of the pipe and must be added/subtracted to
other stresses caused by internal pressure, water hammer and soil cover.
So far we have considered a homogenous temperature change over the whole pipeline.
Another situation can be described as a temperature difference over the pipe wall.
There can be one temperature in the water flowing through the pipe and another in the surrounding
water outside the pipeline.
In this case both extra compression and extra tension stresses can occur.
The stresses will act in the ring direction.
The maximum stresses can be calculated from formula A.3-28) :
s =
E × a × (Toutside - Tinside )
t
2
A.3-28)
A negative sign means compression stress, while a positive sign indicates tension stress.
These stresses will also undergo a relaxation as the time passes.
Example 13
Calculate the maximum stress in the hoop direction if the temperature in the water inside the pipe is
20 ºC and the ambient water has a temperature of 4 ºC
? Assume E=800 Mpa and a = 0.2 × 10-3
ºC -1
.
Solution :
We apply formula A.3-28) and get :
The stress’ nature is compression.
800 × 0.2 ×10-3
(4 - 20)
st = MPa = -1.28 MPa
2
Side 39 av 88

4
A.3.5 Bending stresses
A PE-pipe can, due to its flexibility, be bent to a certain curvature. However, there is a minimum
radius that can not be ”exceeded” if buckling should be avoided.
During such bending it will occur stress and strains both in longitudinal and radial direction of
the pipe.
When the bending radius is too little, the pipe will buckle.
Especially during sinking of a subwater pipeline it is necessary to ensure that the bending radius
is greater than the critical buckling radius.
During installation the balance between forces; weight of the concrete block, forces from boats,
buoyancy forces, forces from currents and waves or other man made forces defines the
configuration and the maximum curvature.
When a pipe is bent to a curvature with radius R in axial direction there will occur a strain, ea, in
the pipe wall. This strain can be expressed :
r D
ea = =
R 2 × R
A.3-29)
r = pipe radius
R = bending radius
D = pipe’s!outside diameter
ea
The case is shown in fig. A.3.5.1.
r
Fig. A.3.5.1 PE-pipe under pure bending
To bend a pipe to this radius, R, it must be subject to an external moment caused by the forces
mentioned earlier. The moment (M) can be expressed :
M =
E × I
R
A.3-30)
E = modulus of elasticity (creep modulus)
I = p
× (D
64
- d4
) (moment of inertia) A.3-31)
D = outside diameter
d = inside diameter
The maximum stress in the pipe wall can be estimated from Hook’s law (ref.A.3-14) :
sa = E
× ea
= E ×
r
= E
×
R
D
2 × R
A.3-32)
Side 40 av 88

=
=
The stress is a tension in the outer curve and a compression in the inner curve.
The value of the stress will decrease by time due to relaxation in the PE material.
We often introduce the ratio
R
= a .
D
Formula A.3-29) and A.3-32) can be rewritten :
1
ea
2 × a
A.3-33)
E
sa
2 × a
A.3-34)
Note that the stress and strain in longitudinal direction are independent of the pipe’s SDR class.
Example 14
Estimate the maximum bending stress in a Ø 1200 mm PE100 pipe bent to a radius 30 × D
during sinking. Assume E-modulus = 700 Mpa.
Solution :
First we decide the radius of curvature : R = 30 × 1.2 m = 36 m
The stress is, for instance, calculated from formula A.3-32) : s = 700 ×
1.2
MPa =11.71MPaa
2 × 36
If we look back to table A.1.2, we can find the burst stress for short-term loads to be 15 Mpa.
The safety factor against rupture is F =
15
=1.3
11,7
For practical purposes a bending radius of 30 × D can be considered to be minimum radius for
a PE-pipe during sinking (SDR < 26).
As we have seen the bending stresses can be significant.
When a pipe is permanent installed in a curve over lifetime, these stresses can contribute to
a reduction in allowable pressure.
As a rule of thumb, in situations with combined loads e.g. pressure, temperature loads, waves etc.,
we recommend :
R min = 60 × D
As mentioned earlier the relaxation in the PE material will reduce the stresses due to bending
more than the reduction in the burst stress for the material. Hence the factor of safety will increase
as time passes.
A.3.5.1 Buckling of PE pipe during bending
When a pipe is bent continuously it will sooner or later buckle. Theoretically there are 2 possible
cases :
- Axial buckling
- Radial buckling
Side 41 av 88

×
ç ÷
For subwater pipelines the radial buckling will be critical unless the internal pressure is significant
[12].
The critical strain for radial buckling in the state of pure bending can be written :
ecrit,r
=
0.28
æ s
ö
A.3-35)
è Dm ø
The relationship between the axial and radial strain is given by Poisson’s figure :
er
= n × ea
A.3-36)
If we choose n= 0.50 and put A.3-36) into A.3-35) we can find the critical strain
in axial direction ecrit,a :
0.28 s
ç ÷ =
0.56
e = ×
æ ö A.3-37)
crit,a
D
SDR =
s
n è Dm ø SDR -1
Dm = mean diameter
s = wall thickness
If we now combine A.3-37) and A.3-33), we can determine the critical bending ratio for a PE-pipe
in axial direction :
=
SDR -1
acrit
1.12
= 0.89 (SDR –1) A.3-38)
It is normal to introduce a safety factor, F = 1.5 for such calculations.
R
Hence allowable bending ratio : aallowable,F=1.5 = =1.34×(SDR -1)
D
A.3-39)
Example 15
Make a table showing allowable bending ratio (R/D) for the SDR classes 33, 26, 22, 17, 11 and 9,
assuming a safety factor of 1.5.
Solution :
We use formula A.3-39) and get table A.3.5.1.1 below :
SDR-class
Allowable bending
R
ratio F=
1.5
D
33
26
22
17
11
9
44
34
28
21
13
11
Table A.3.5.1.1 Allowable bending ratio during sinking.
If the pipe is exposed to an internal pressure during bending, the ovaling will be reduced and the
critical strain will increase.
Side 42 av 88

=
Fig. A.3.5.1.2 shows the effect on a pipe with an internal overpressure of 1 bar for SDR-classes 26,
17.6 and 11.
Fig. 3.3.1.2 Increase in allowable strain due to internal pressure of 1 bar
Fig. A.3.5.1.2 indicates that the internal pressure has a significant stabilizing effect on
SDR-class 26 (27 %).
For PE-pipe SDR11 or lower, the stabilizing effect of an internal overpressure is more or less
insignificant.
Example 16
What will the allowable bending ratio (R/D) be for a SDR26 pipe if it is subject to an internal pressure
of 1 bar during sinking? Assume a safety factor of 1.5.
Solution :
From table A.3.5.1.1 in example 15 we find the bending ratio without any internal pressure. a = 34
Since the bending ratio is in inverse ratio to the allowable strain (ref. formula A.3-37) and A.3-38))
we get by use of fig. 3.5.1.1 :
k = 1.27 ap=1bar
34
= 27
1.27
As we see the bending ratio has decreased from 35 to 28.
If the pipe has been a Ø 1000 mm, the bending radius would have been reduced from 35 m
to 28 m.
For low pressure pipes (£ PN4) internal pressure will increase the safety factor against buckling.
A.3.6 Other stresses
So far we have discussed stresses caused by :
- Internal pressure
- External pressure (water and soil)
- Water hammer
- Temperature changes
- Bending
Side 43 av 88

More or less there can be other forces acting on a subwater pipeline, for instance :
- Concentrated load where the pipe is resting on rock or stone
- Weight of hovering pipeline
- Current forces
- Wave forces
A.3.6.1 Current and wave forces
Current and wave forces will be studied in next chapter in accordance to design of concrete weights.
There will be both drag and lift forces caused by these elements. For a pipeline lying stable on the
seabed, the forces can be considered to be uniformly distributed along the pipe section between the
supports (concrete weights), but limited by the crest length of the waves.
The magnitude of these forces can be described roughly by the formula :
v2
f = C × D × r ×
2
A.3-40)
f = force pr. unit m pipe
C = coefficient
D = external diameter
r = density of surrounding water
v = speed of surrounding water vertical to the pipe axis
For wave forces we also have to consider the inertia forces, especially for large diameters (see
chapter A.4.6).
Example 17
Find a rough estimate for the magnitude of current and wave forces assuming combined maximum
speed 3 m/s and coefficient C = 1.0. Diameter of pipe is 1.0 m and r = 1025 kg/m3
Solution :
We apply formula A.3-40) :
32
f =1×1×1025 × N / m = 4612 N/m = 4.6 kN/m
2
This indicates that these forces can be significant and must be taken into consideration when
deciding the design factor for the project.
If the pipe in example 17 has been SDR-class 22 the unit mass is 140 kg/m » 1.4 KN/m in air.
The current and wave forces are in this case approx. 3.3 times the pipe’s unit weight.
For high SDR-classes this ratio can be about 6 and for low SDR-classes it can reach about 2.5.
We have to underline that the example above only is an indication of the maximum magnitude of the
forces from current and waves. For proper design, comprehensive calculations must be carried
out. We will also mention that the wave forces are significantly reduced as the water depth
increases.
When we know the acting uniform force pr. unit length of the pipe, the stresses can be calculated by
well-known formulas from static beam design.
If we for instance choose the case fixed beam, we get :
4 × f × l2
× D
smax =
3× p × (D4
- d4
)
A.3-41)
f = force pr. unit length
l = distance between supports
d = inside diameter
If we go back to example 17 and assume SDR = 22 and l =10 m we get :
Side 44 av 88

=
3 3
=
=
-3 2
smax
4 × 4.6 ×10 × l0 ×1
MPa = 0.60 MPa
3×3.14 × (14
- 0.9094
)
Compared to a design stress of for instance 5 Mpa this stress amount to 12 %.
If the pipe has been SDR 33 the corresponding percentage would reach 17.5 %.
A.3.6.2 Hovering pipeline
If we now return to the case of a hovering pipeline, the situation is quite similar to what we have seen
regarding a uniformly distributed load from current and wave forces. In this case we get an extra
force component from the concrete weights over the length, l, between the supports. This means
that the stresses in the pipe wall also increase. Such situations can be the fact in very rough under
water terrain. There are several examples from Norway.
If we look to example 17 and assume a loading percentage equal 30 % of the displacement, the
weight pr. unit length from the concrete weights will amount to 2.4 KN/m. This is about 50 % of
the current and wave forces.
However the current component will mainly act in the horizontal direction for a hovering pipeline,
while the concrete weights will act in the vertical direction.
The wave components will act in all directions as the wave passes.
If we assume the wave component to be 2
/
maximum force including concrete weights :
and the current component to be 1
/ , we get the
fmax
=
(4.6 ×1/ 3)2
+ (0.5× 4.6 + 4.6 ×1/ 32
kN/m = 5.6 kN/m
If we put this result into formula A.3.41) we get a maximum stress of 0.73 Mpa for a span of 10 m.
If we can accept a stress in longitudinal direction for instance equal 2 Mpa, the maximum span in this
case can reach :
lmax
2
×10 m = 27 m
0.73
This example shows that it is important to put effort into the work finding an optimal trace and
location for a subwater pipeline.
A.3.6.3 Concentrated loads
Where the pipeline is resting on a rock or a stone extra stresses will occur. The magnitude of
the stresses depends mainly on :
- number of concrete weights hovering on both sides of the attack point
- surface area of attack point
It is a good idea to repair all concentrated loads by putting extra protections material between pipe
and stone/rock.
The magnitude of the stress caused by concentrated load can be estimated roughly from the
formula :
scon
3× P
2 × p ×s2
A.3-42)
P = total concentrated load
s = wall thickness
We generally recommend avoidance of contact with stones. In a lot of cases however, we
experience that this ideal situation is impossible without including enormous costs.
Example 18
Side 45 av 88

s =
s 2 2
A Ø 1000 mm PE80 SDR 17.6 is resting on a stone in such a way that 2 concrete weights on each
side of the stone are hovering. The weight in water for each concrete weight is 14 KN.
Estimate the maximum stress in the pipe wall due to the concentrated load.
Solution :
First we find the wall thickness :
1000
mm = 56.8 mm
17.6
We apply formula A.3.42) and get, assuming that 2 of the weights is contributing to the concentrated
load :
scon
3× 2 ×14 ×10-3
= MPa = 4.2 MPa
2 × 0.05682
× 3.14
As we see this stress will be significant and can reduce the lifetime of the pipe.
The pipeline must be moved sideways to a better position or a protection material, with sufficient
thickness, must be placed between the pipeline and the stone.
A.3.7 Combined loads
In chapter A.3 we have considered different types of forces that can act on a subwater pipeline
in operation.
These forces create stresses and strains in the pipe wall.
Some stresses are compressive and some are tensile. Some are acting in the longitudinal
direction and some are acting in the hoop direction. In some situations there can also be shear
stresses, but we will not deal with them in this technical catalogue.
For a subwater pipeline, shear stresses will not be critical.
When we have calculated all actual stresses (ref. A.3.1-A.3.6), we sum them up in the hoop direction
and in the longitudinal directions.
Tensile stresses are positive and compressive stresses are negative.
n
sh = å si, h A.3-43)
i=1
n
sl = å si, l A.3-44)
i=1
sh = total stress in hoop direction
si,,h = stress no.i in hoop direction
sl = total stress in longitudinal direction
si,l = stress no.i in longitudinal direction
To find a combination/comparison (scomp) stress, one often use Von Mises criteria:
scomp
=
h + sl - sh
× sl
A.3-45)
As the formula express a combination of compressive stress in one direction and tensile stress in the
other, is more critical than only compressive stress or tensile stress in both directions.
Side 46 av 88

Example 19
Calculate the comparison stress for a situation where the total stress in hoop direction sh = 4 Mpa
and in longitudinal direction is sl = - 2.5 Mpa (compressive) for a PE80 pipe ?
Solution :
We apply formula A.3-45) : scomp
=
42
+ (-2.5)2
- 4 × (-2.5)MPa = 5.7 MPa
This comparison stress should be compared to the allowable stress for the PE-material
(ref. table A.1.1)
We see that even if the values for sl and sh is less than the design stress 5.0 Mpa,
the comparison stress exceeds 5.0 Mpa.
This is a motivation to include all relevant stresses in a proper design, especially when dealing with
low design factors (e.g. C = 1.25).
As we have mentioned earlier in this catalogue (for instance chapter A.1, A.3.1.2) the PE
material will undergo creep and relaxation. This means that the stresses and strains due to a
certain load situation will be a function of time. We therefore have to check both short term
and log term situations.
Side 47 av 88

3
A.4 Design of loading by concrete weights
Submarine pipelines of PE material will float due to buoyancy if they are not loaded by concrete
weights, since the specific gravity of PE material is less than the surrounding water.
The purpose of the weights is also to provide stability against :
- Air and gas accumulation (although preferably this is not “solved” by weights)
- Current forces
- Wave forces
A.4.1 Degree of loading
Dependent of the project’s technical specifications, we have to calculate the amount of loading.
This degree of loading is often related to the pipe’s displacement :
ad
=
wcw
D2
p × × gw
4
×100% A.4.1)
wcw = weight of concrete weights in water distributed pr. m pipe
gW = specific gravity of surrounding water
Another way to describe the degree of loading is to compare it to the buoyancy of internal volume of
the pipe. This is called the air fill rate, and is nearly always used in Norway to describe the degree of
loading:
wcw + wpipe w
aa = ×100%
d2
p × × gw
4
A.4.2)
wpipe w = weight of pipeline in water (negative)
d = inside diameter
The degree of air filling tells us which degree of the internal pipe volume has to be filled with air to
make the pipe buoyant. This definition also includes the weight of the pipe. We have to underline
that an air fill rate of for instance 30% doesn’t mean that we expect 30% of the internal volume
to be filled with air during operation, but is simply a practical way to describe the degree of
loading.
The difference between ad and aa is not so big. Fig. A.4.1 on next page gives an indication based on
the assumptions :
rPE = 950 kg/m
3
(density PE)
rC = 2400
kg/m
(density concrete)
rw,sea = 1025 kg/m3
(density sea water)
rw = 1000 kg/m
3
(density fresh water)
Side 48 av 88

Degree of air filling
aa (%)
SDR 11
110
100
90
80
70
60
50
40
30
20
10
52%
45%
30%
SDR 26
SDR 41
Sea water
Fresh water
SDR 17,6
0
-10 10 20 30 40 50 60 70 80 90 100 Degree of displacement
ad (%)
Fig. A.4.1 Relationship between degree of displacement and
degree of air filling for concrete weights.
Normally we are speaking about air filling rates in the range 10-60 %.
If a pipe is loaded in accordance to an air-filling rate of 30%, it mean that 30% of the pipe’s internal
volume must be filled with air to obtain equilibrium in the system.
It can often be useful to know the relationship between the weight of a body in air and in water.
This can be written :
w r - rw
= w
A.4.3)
wa r
w w = weight in water
w a = weight in air
r = density of body
rw = density of water
Example 1
A Ø 500 mm PE80 SDR22 pipe is loaded with concrete weights with centre distance 5 m.
The weight in air pr. concrete weight is 5.6 kN. The weight of the pipe in air is 0.35
kN/m. Assume rPE = 950 kg/m3
, rw = 1025 kg/m3
and rc = 2400 kg/m3
Calculate the air filling rate aa.
Side 49 av 88

=
= 6.5×
= 0.35×
=
Solution :
First we find the weight of the concrete weights and the pipe in water by use of A.4.3) :
wcw
wcw
w
2400 -1025
kN= 3.2 kN
2400
3.2
kN / m= 0.64 kN/m
5
950 -1025
kN / m = - 0,28 kN/m
(a piece)
(pr. m pipe)
pipe w
950
The internal diameter of the pipe is :
d = (500 -
2 × 500
)mm = 454.6 mm
22
We apply formula A.4.2) to calculate the degree of air filling :
(0.64 - 0.028) ×100
a % = 37.5%a
0.45462
3.14 × ×1025×9.81
4
The corresponding degree of displacement can be found from fig. A.4.1 by interpolating between
SDR 17.6 and SDR 26 for sea water.
This gives : ad » 32 %
For all practical situations will aa > ad
A.4.2 Types of concrete weights
There are 3 types of concrete weights by reference to the shape :
- Rectangular
- Circular
- Starred
These are schematic shown in fig. A.4.2.1.
Rectangular Circular Starred
Rectangular Circular Starred
Fig. A.4.2.1 Different types of concrete weights
All weights are to be bolted on the pipe. The fix force should be sufficiently to avoid sliding during
sinking and rotation on seabed.
As a rule of thumb, the bolt force shall be in the range 2-3 times the weight of the concrete weight in
air.
Between the concrete weight and the pipe wall there shall be a rubber band, type EPDM or
equivalent. In most cases we also recommend rubber compensators in the bolts to reduce local
stresses in the pipe wall caused by internal pressure.
Side 50 av 88

W
g
It is obvious that the weights shown in fig. A.4.2.1 have different grip on the seabed when they are
subject to a wave or current force.
The rectangular weight is the classic shape. It has an overall good performance and can be utilised
in most cases.
Circular shaped weights are used in trenches, in smooth water and in places where fishing and
anchoring take place.
Star shaped weight may conveniently be applied in cases where the impact from waves and currents
is significant. The special shape gives increased stability.
Below are listed approximate friction coefficients for the 3 types of concrete weights :
Type Friction
coefficient
Rectangular
Circular
Starred
0.5
0.2
0.8
Table A.4.2.1 Guiding friction coefficient for concrete weights
A.4.3 Stability of PE-pipeline on the seabed
We will now establish formulas to check the stability of an underwater pipeline subject to
air/gas accumulation and external forces from currents and waves. The situation is shown
in fig. A.4.3.1.
FB FL
gsea
n ga FD
gw gP
p
Ff
Wc c
Fig. A.4.3.1 Stability of PE-
pipe on seabed
Wcw m
FN
Ww
Wp
Wa
We suppose that the forces from current and waves can be decomposed in a drag force, FD, in
horizontal direction and a lift force, FL, in vertical direction acting simultaneously on the pipeline.
To avoid sliding, these two forces must be overcame by the weight of the system and the friction
force between the concrete weights and the sea-bottom.
Side 51 av 88

Force balance in vertical direction gives pr. unit m of pipe:
FN = wCw + ww + wP + wa – FB - FL A.4.4)
FN = normal force against seabed
wcw = submerged weight pr. m pipe of concrete weights
w w = weight in water pr. m inside pipe
wp = weight of pipe pr. m in air
w a = weight of air/gas pr. m inside pipe
FB = buoyancy of pipe pr. m
FL = lift force
Force balance in horizontal direction gives : Ff ³ FD A.4.5)
Since Ff = m×FN we get the criteria for stability : m ³
FD
FN
A.4.6)
The coefficient of friction should be greater than the ratio drag force/normal force.
The elements in formula A.4.4) can be expressed more exactly as follows :
wcw = w
ca
g - g
× c sea
g
A.4.7)
c
Wca = weight of concrete weights in air pr. m pipeline
gc = specific weight of concrete
gsea = specific weight of seawater
p × d2
w w = (1- n) × × gw
4
A.4.8)
n = amount of air filled section, e.g. 30%, h = 0.3
d = internal diameter
gw = specific gravity of water inside the pipe
p × d2
wa = n × × ga
4
ga = specific gravity of air inside the pipe
wa can in most cases be neglected.
A.4.9)
p × D2
FB = × gsea
4
D = external diameter of pipe
A.4.10)
We have now a complete set of formulas to check the pipeline’s stability on the seabed when the
drag force, FD, and lift force, FL, are known. For calculation of FD and FL, see chapter A.4.5 and
A.4.6.
Example 2
A Ø 500 mm PE100 pipe SDR22 is laying on the sea bottom and is attacked by waves
and currents.
Design drag force is FD = 0.4 kN and design lift force is FL = 0.2 kN.
Degree of air accumulation is assessed to be n = 0.15. On the pipeline there are installed
concrete weight for every 3 m. The concrete weight has a weight of 5.6 kN in air.
The weight of the pipeline is 0.345 kN/m.
Assume specific gravity of concrete to be 23.5 kN/m3
, specific gravity of seawater to be
10.05 kN/m3
and specific gravity of sewage water to be 10 kN/m3
Specific gravity of air/gas can be neglected. Is the pipeline stable on the seabed ?
Side 52 av 88

=
=
=
=
=
Solution :
We use formulas A.4.4) – A.4.10) to solve the problem.
First we calculate the weight of concrete weights pr. m pipeline in seawater by formula A.4.7) :
5.6 23.5 -10.05
w × kN / m =1.068 kN/mcw
3 23.5
Then we apply A.4.8) to find the weight of water inside the pipe pr. m :
p × 0.45462
w w = (1- 0.15) × ×10 kN/m = 1.379 kN/m
4
The buoyancy is given by formula A.4.9) :
p × 0.52
FB = ×10.05 kN/m = 1.972 kN/m
4
The normal force, FN, is decided by putting values into A.4.4) :
FN = (1.068+1.379+0.345+0-1.972-0.2) kN = 0.62 kN
The minimum friction coefficient is calculated from A.4.6)
mmin
0.4
= 0.65
0.62
If the pipe shall avoid sliding, the friction coefficient between concrete weights and sea bottom
must be greater than 0.65.
If we return to table A.4.2.1., we see than only the starred weight can perform this friction coefficient.
The conclusion is that the pipe is stable only if the concrete weights have a starred shape.
Else it will slide sideways.
To get it stable by rectangular or circular weights, we have to increase the weight of the concrete
weights to 6.54 kN and 9.34 kN respectively.
It is also possible to adjust the centre distance to 2.57 m and 1.8 m and keep the original weight.
The corresponding air filling rate is given by formula A.4.2). This gives :
Starred weight :
Rectangular weight :
Circular weight :
1.068 - 0.027
a ×100 % = 64.2 %a
0.45462
p × ×10
4
1.247 - 0.027
a ×100 % = 75.2 %a
0.45462
p × ×10
4
1.781- 0.027
a ×100 % = 108.1%a
0.45462
p × ×10
4
In reality it is not possible to use circular weights without introducing buoyancy elements temporary
during sinking/installation of the pipe.
Side 53 av 88

A.4.4 Recommended “air filling rate” for subwater pipelines
As mentioned earlier the loading by concrete weights on a subwater pipeline depends on :
i) Buoyancy of PE-material
ii) Air/gas accumulation
iii) Current forces
iv) Wave forces
v) Fishing equipment
i) Buoyancy of the PE-pipeline is dependent of the diameter and the SDR-class, but will
normally be within the range 0.3-2.5 % measured as “air filling rate” aa
ii) How much air/gas that will be accumulated in a subwater pipeline, depends on the project
design and must normally be calculated accurately.
Especially for outfall pipelines air/gas accumulation can be a problem. The topography in
the trace is critical.
Generally we will advice “air filling rate” as shown in table A.4.4.1 below :
Type of pipeline
Type of transport/topography
Gravitation Pumping Big highpoints
Potable water
Sewerage water
10 %
25 %
15 %
30 %
20 %
50 %
Table A.4.4.1 Guiding ”air filling rates” for subwater pipelines
with respect to air/gas accumulation.
iii) Current forces can be significant for subwater pipelines installed directly on the bottom,
especially in rivers.
If the forces are too big, the pipeline has to be buried in a trench.
Stability calculations (ref. A.4.3) must be carried out for each project.
Generally, in rivers, pipelines must be buried, especially when crossing the stream direction.
In the sea and in lakes, it will often be sufficient to increase the “air filling rate” with 10%
to obtain stability. This extra amount shall be added to the values in i) and ii)
iv) Wave forces must be calculated separately. Generally we advise to bury the pipeline to
a water depth where the wave is breaking. This will normally mean 10-15 m water depth
in exposed areas.
Further we recommend a total “air filling rate” in the range 70-30% dependent on
the projects characteristics.
This loading is kept to a depth corresponding to half the wavelength of the design wave.
On deeper water the general rules given in i), ii) and iii) are applied.
It can be accepted that the pipeline moves a little on the seabed when the waves are
passing.
Experiences show that the tube will move back and forth within a limited area if the “air-
filling rate” is properly calculated. In such cases starred weights are always applied. The
movements of the pipe are actually both rotation and sliding. Comprehensive computer
programmes must be used for such calculations.
v) It is not normal to design the “air filling rate” to include influence from fishing equipments like
fishing nets and trawls.
However, the concrete weights can be shaped to avoid the equipment to be stuck in the
pipeline. In these cases we apply circular weights.
As a summary, we can say that the degree of loading for a PE-subwater pipeline will correspond to
an “air filling rate” in the range : 15-60 %.
Side 54 av 88

F
In some cases, it can be economically favourable to secure the pipeline against under-pressure (ref.
A.3.2.1) by reducing the centre distance for the concrete weights. The “air filling rate” in such cases
will be higher than indicated above.
Example 3
A gravitation sewerage pipeline Ø 500 mm PE80 SDR22 shall be installed as an outfall pipeline
in a lake. There is a significant high point in the trace. We can neglect forces from currents and
waves.
Can you advice a design “air filling rate” ?
Solution :
We go through the points i), ii), iii), iv) and v) :
i) aai = 2.5 % (assume the highest value, can eventually be calculated)
ii) We use table A.4.4.1 by input values :
- sewerage water
- big highpoints
Þ aaii) = 50%
iii)
iv)
v)
give no extra contribution
Overall result : aa = 52.5 %
We shall underline that this “air-filling rate” is only representative in the high point area.
Generally aa = 27.5 % will be more suitable.
A.4.5 Current forces
Calculations of current forces acting on a pipeline can be complicated. In the following chapter we
shall deal with a simplified method to estimate the forces roughly. For concise calculations experts
in the field must be contacted.
When a current attacks a pipeline, it will be subject to a force. The force can be split into two
elements, a drag force, FD, and a lift force, FL, ref. fig. A.4.5.1.
FL
Fig. A.4.5.1 Current forces v
acting on a pipeline
D
f
The amount of the forces is mainly dependent on :
Side 55 av 88

=
- Current velocity (v)
- Pipe's diameter (D)
- Density of streaming water (r)
- Pipe's distance above seabed (f)
The forces can mathematically be expressed as follows :
FD =
FL =
C ×
1
×r × v2
× DD
2
C ×
1
×r × v2
× DL
2
A.4-10)
A.4-11)
CD = drag coefficient
CL = lift coefficient
r = density of streaming water (kg/m3
)
v = current velocity (m/s)
D = external diameter of pipe (m)
FD = drag force (N/m)
FL = lift force (N/m)
The coefficients FD and FL are in principle dependent of Reynolds number and roughness of the
bottom.
Reynolds number (ref. A.2-1) can be expressed :
v × D
Re
n
A.4-12)
n = viscosity of water » 1.3 ×10-6
(m2
/s)
The coefficients will normally vary within the range 0.5-1.2. Values for a pipeline laying on the
seabed can be taken from fig. A.4.5.2 and A.4.5.3 .
Re
Fig. A.4.5.2 Drag coefficient, CD
Side 56 av 88

Re
Fig. A.4.5.3 Lift coefficient, CL
The lifting force will be reduced as the distance (f) between pipe and seabed increases.
If f = 0.5 ×D the lifting force will be approximately 10% of the lifting force for a pipeline laying directly
on the seabed. This is a vital detail in design of concrete weights.
Example 4
A current is attacking a pipeline at an angle 45º
from the centreline as shown beside.
The pipe's diameter is 500 mm.
Density of water is 1000 kg/m3
.
Assume that CL = 0.20 and CD = 1.0.
Calculate the drag force and the lift force.
V= 1 m/s
Solution :
The velocity component perpendicular to the pipeline
can be written :
a = 45o
D= 500mm
vN = v ×sin a A.4-13)
Inserting this expression into A.4-10) and A.4-11) we get :
1
FD = C × ×r × v2
× sin2
a × DD
2
1
A.4-14)
FL = CL × ×r × v2
×sin2
a × D
2
A.4-15)
Putting in values from example 4 gives :
1
FD =
FL =
1.0 × ×1000 ×12
× sin2
45× 0.5 N/m = 125 N
2
0.2 ×
1
×1000 ×12
× sin2
45× 0.5 N/m = 25 N
2
Side 57 av 88

d
As we can see, the forces are reduced significantly if the attack angle, a, is small. It is therefore
a good idea to avoid the current to run perpendicular to the pipe.
Finally we should mention that formulas A.4-14) and A.4-15) must be corrected taking into account
the shape and area of the concrete weights, if comprehensive calculations are done. In this case we
can introduce a "shadow coefficient" k. Usually k will be in the range 1.0-1.5.
This means that the forces calculated in example 4 can be 50 % higher if the concrete weights are
taken into account, dependent on shape, dimension and centre distance.
A.4.6 Wave forces
Waves will apply big forces on a subwater pipeline installed directly on seabed.
The main factors are :
- Wave height
- Wave period
- Pipe diameter
- Distance between pipe and sea bottom
- Angle between pipeline and the wave's moving direction
- Depth of water
- Condition of seabed
Waves approaching the shore will be influenced by the bottom conditions and soon or later they will
reach a depth where they are breaking.
A breaking wave will release a big amount of energy that eventually can damage the pipe structure.
A good rule is therefore : "Burry the pipeline to a depth equal or greater then the depth
where the design wave is breaking"
Practically speaking this mean a depth in the range 5-15 m dependent on the site's local conditions.
Description of waves and wave forces involves a complicated basis of formulas.
There are several theories, but a common feature is the dividing of the force components into
3 elements :
- Drag force
- Lift force
- Inertial force
The movement of water particles in a wave take place in circular or elliptic orbits as shown in
fig. A.4.6.1.
L
Shallow water Semi deep water Deep water
h
£
1
1 < <
h
>
L 20 h 1 1
20 L 2 L 2
Fig. A.4.6.1 Movement of particles in a wave
Side 58 av 88

L
As shown in fig. A.4.6.1, the orbits for the particles at deep water are circles. The deep is so great
that the movement of the wave does not "touch the bottom"
Deep water is defined as the water depth (h) deeper than half the wavelength (h >
L
)
2
Wave forces will never influence a pipeline installed at deep water.
At semi deep water (
L
< h <
L
)
the forces can be significant while they can reach
20 2
extreme values as shallow water (h >
L
)
20
Since the wave particles are moving continually by time, the wave forces will change both direction
and magnitude.
At a fixed moment, forces acting in one direction will influence a section of the pipeline
while another section will be exposed to forces acting in opposite direction.
To check the stability of the pipe it is sufficient to know the extreme values of the forces.
These can be calculated by the following formulas :
pD2
H
Fi = p × Ci × f × g × × o
A.4-16)
4 Lo
pD2
H H
FD = CD × f 2
× g × × o
×o A.4-17)
4
pD2
Lo D
H H
FL = C × f 2
× g × × o
×o A.4-18)
4
Fi = inertial force
FD = drag force
FL = lift force
Lo D
f = refraction factor
Ci = inertial coefficient
CD = drag coefficient
CL = lift coefficient
g = specific gravity of water (N/m3
)
D = external diameter of pipe (m)
Ho =
wave height on deep water (m)
(vertical distance from wave bottom to wave crest)
Lo = wave length on deep water (m)
There is a phase angle between Fi, FD and FL, which indicates that they never occur
simultaneously. For instance the Fi is 90º out of phase with the FL force.
If the wave hits the pipeline under an angle a, the forces must be corrected by the factor sin a.
As the formulas A.4-16), -17) and –18) indicate there are several values which must be known to
calculate wave forces.
Subsequently we shall discuss the most important factors.
Force coefficients
The coefficients Ci, CD and CL are decided experimentally. The coefficients are dependent of the
distance between the pipeline and the seabed (ref. fig. A.4.5.1).
If there is a passage for the water under the pipeline, the coefficients will be reduced.
Table A.4.6.1 below gives some practical values for calculations.
Coefficient Distance to bottom = 0
D
Distance to bottom ³
4
Ci
CD
CL
3.3
1
2
2
0.7
0
Table A.4.6.1 Force coefficients for waves Side 59 av 88

Wave height and wave length
If there are no measurements of the wave heights, the heights can be decided on the basis of wind
statistics and "fetch length" for the wind.
The diagrams in fig. A.4.6.2 and A.4.6.3 give the significant wave height, H1/3 and the
corresponding wave period, knowing the wind speed and "fetch length".
In calculations we apply the maximum wave height (Ho) on deep water, which is :
Ho = 1.8 × H1/3 A.4-19)
The wave period for Ho is assumed to be the same as for H1/3 and can be taken directly from
fig. A.4.6.3.
To = T1/3 A.4-20)
80
70
60
50
40
30
20
10
0.1 0.2 0.3 0.4 0.5 0.7 1.0 2.0 3.0 4.0 5.0 7.0 10 20 30 40
Fetch length F
(miles)
Fig. 4.6.2 Wave height H1/3 as function of wind speed and "fetch length"
80
70
60
50
40
30
20
10
0.1 0.2 0.3 0.4 0.5 0.7 1.0 2.0 3.0 4.0 5.0 7.0 10 20 30 40
Fetch length
F(miles)
Fig. 4.6.3 Wave period as function of wind speed and "fetch length"
Side 60 av 88

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