Pipelife Norge AS · PDF filePE CATALOGUE-SUBMARINE APPLICATIONS, PIPELIFE NORGE AS, ... The outfall usually starts from an outfall chamber at the waterfront to which the

Pipelife Norge AS

Technical Catalogue for Submarine Installations

of Polyethylene Pipes

Pipelife Norge AS

PE CATALOGUE-SUBMARINE APPLICATIONS, PIPELIFE NORGE AS, December 2002.

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

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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

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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. PLEASE NOTE: This catalogue is presented only as information and without any legal obligation or responsibility on Pipelife side.

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.

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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.

0.1

Fig. 0.1.1.1 River water intake

.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.

0.1.3 OTaOT TgUng TA Ia S−−−− Fa −

Fig. 0.1.2.1 Sewerage river crossing.

utfall pipeline reated sewage water will normally be conveyed into the recipient discharge area at a certain depth nd distance from coast. A depth water outlet will provide excellent dilution of the wastewater. utlet deep will vary in the range 10-60 m dependent of the recipient’s self-purification capacity. he recipient can be river, lake, fiord or sea.

he outfall usually starts from an outfall chamber at the waterfront to which the wastewater is lead by ravity or pumping. se of pumping directly on the outfall pipeline is rather rare and not recommended. If pumping is ecessary, the best solution is to pump the sewage water into the outfall chamber and conduct it with ravity into the recipient.

he main task for the outfall chamber is to prevent air from entering the pipeline. ir can cause floatation of the pipe due to buoyancy.

t is also necessary to take into account the variations in low tide and high tide when designing n outfall chamber.

pecial problems to take into consideration regarding outfall pipelines are : Air entrainment in pipe flow Bio fouling Current and wave induced forces Sediment transport

ig. 0.1.3.1 represent an industrial outfall. The figure shows the outfall system to the sea from 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.

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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.

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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.

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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.

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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 30oC. 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

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During sinking of the outfall pipeline in this example one had to consider the following factors:

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• 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.

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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 σ > 0 and constant (Mpa) ER = relaxation modulus, time > 0, strain ε > 0 and constant (Mpa) σO = burst strength at time zero (Mpa) σC = creep strength at time > 0 (Mpa) (also called burst stress) ν

= Poisson’s ratio =

r

l

εε

εl = strain in the axial direction εr = strain in the ring direction α = 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].

Fig. A.1.1 THDPE Type[1].

= E

he relationship between creep modulus E and tensile stress with time as parameter for bars HE2467 (full lines) and HDPE Type 2 bars HE2467-BL (dotted lines) at 23ºC

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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 (σd) is introduced by the formula :

C

yearCd

50,σσ = A.1-1)

σC,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.

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= σc

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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 σd,50

Design stress at time zero σd,0 Modulus of elasticity at time zero E0 Modulus of elasticity after 50 years E50 Poisson’s ratio ν Average coefficient of thermal expansion α

kg/m3

Mpa Mpa

Mpa

Mpa -

ºC -1

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 PE-pipeline Type of water Intake Transit Outlet

Potable Sewage

2 mm

-

0.25 mm

0.50 mm

-

1 mm

Table A.1.3 Design values for equivalent roughness (ε) If the pipes are regularly flushed supported by a cleaning pig, the values in table A.1.3 may be reduced.

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A.2 Hydraulic design The pressure (∆h) drop in a pipeline can generally be described by the formula :

yg2

vkg2

vDLfh

o

22⋅

ρρ∆

+⋅

⋅+⋅

⋅= ∑∆ A.2-1)

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) Σk = sum of coefficients for singular head losses ∆ρ = density difference between water inside the pipe and water in recipient (kg/m3) ρo = 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) : ν⋅

=DvR e A.2-2)

v = velocity D = internal diameter (m) ν = viscosity of water (m2/s)

The viscosity of water depends on the temperature. T = 20ºC ν = 1.0 ⋅ 10 –6 m2/s T = 10ºC ν = 1.3 ⋅ 10 –6 m2/s

We recommend applying the value for 10ºC.

The velocity (v) can be calculated by the formula : 2DQ4v

π

⋅= A.2-3)

Q = flow (m3/s)

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) s/m 19,1 s/m3272,0100,04v 2 =

⋅π

⋅=

Reynolds number is found from A.2-2) 56e 10 2,09

1031,13272,019,1R ⋅=

⋅

⋅=

−

When we know the Reynolds number, the friction coefficient can be found from the Moody chart, fig. A.2.1.1.

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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.

D

rrε

= A.2-4)

ε = 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 : We have already calculated the Reynolds number in example 1 Re = 2.97 ⋅ 105

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Hence : 0,0008 2,327

25,0rr ==

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 (∆hf) for the pipeline from part one in formula A.2-1)

g2v

DLfh

2

f ⋅⋅⋅=∆ A.2-5)

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) :

mwc 11,03 mwc81,92

19,13272,0

250002,0h2

f =⋅

⋅⋅=∆

To convert this unit to Pa (N/m2) we introduce the relationship : p = ρ ⋅ g ⋅ h A.2-6) p = pressure (N/m2 = Pa) ρ = 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.

bar 1.08 bar 100000108204p == MPa 0.108 MPa

1000000108204

==p

A.2.2 Coefficient for singular head losses Part two of formula A.2-1) represent the singular pressure drops (∆hs) :

∑ ⋅⋅=∆

g2vkh

2

s A.2-7)

The expression Σk 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.

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Table A.2.1.1 gives guiding values for singular coefficients.

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 ⇒ k = 2)9090(1, ⋅1 = 1.1

Bead ⇒ k = 0.03 Outlet ⇒ k = 1.0

Total sum of coefficients comes to : Σk = 3 ⋅ 1.1 + 25 ⋅ 0.03 + 1.0 = 5.05

Total singular head loss: mwc 0.36 mwc81.92

19.105.5h2

s =⋅

⋅=∆

Side 21 av 84

��

Singular headloss k-factor

k-factor

Inlet 1

Inlet 2

Outlet

k = 1,0

k = 0,5

k = 1,0

Elbow

Smooth bend

k=1,1

k= 0,2 sin (rough) k= 0,1

Diffuser k = 16

k = 0,03Intakescreen

Bead k = 0,03Gate valve (open) k= 0,2 Non return valve k= 10

.( )θ

θ

θ90o

2

.

.θθsin (smooth)

��

V

V

V

V

V

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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.

yho

⋅ρ

ρ∆=∆ ρ 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: mwc 1.25 mwc 501000

10001025h =⋅−

=∆ ρ

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 :

0LQf8DQ8kD)yh(g 2252

o=⋅⋅⋅−⋅⋅⋅∑−⋅π⋅⋅

ρρ∆

−∆⋅ 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:

51

)yh(g

LQf8D

o

2

2

⋅ρ

ρ∆−∆π⋅

⋅⋅⋅= A.2-10)

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 84

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Another approach to the problem is to solve the flow (Q) from equation A.2-9)

21

o2

DkLf

gD)yh(2

4DQ

⋅Σ+⋅

⋅⋅⋅ρ

ρ∆−∆⋅

⋅π

= A.2-11)

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)

mm 325 m 0.325 m)25.125.136.003.11(81.9

25001.002.08D51

2

2==

−++π⋅

⋅⋅⋅=

The nearest standard diameter above for SDR11 is 327.2 mm (Ø 400 mm). This diameter value is inserted in A.2-11).

Hence: s/l 100s/m 0.1 3272.005.5250002.0

81.93272.0)25.125.136.003.11(243272.0Q 32

12

==

⋅+⋅⋅⋅−++⋅

⋅⋅π

= q.e.d.

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 0/00 (= )1000Lh

⋅∆ we can find the necessary diameter

when the flow is given. Generally we can solve one of the quantities Q, ∆h, D when 2 of them are known. In the chart you also can read the velocity.

Side 23 av 84

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Fig. A.2.4.1 Hydraulic capacity, ε = 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 : mwc 1.25 mwc 501000

10001025=⋅

−=ρ∆

Total available friction drop is : mwc 11.75 mwc )25.113(h f =−=∆ We find the incline of

the friction drop line (I) : ooo

ooo / 4.7 / 1000

250075.11I =⋅=

We enter chart A.2.4.1 with the quantities Q = 100 l/s and I = 4.7 0/00. 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 84

(D)

(D)

(v)

(v)

I = ∆

h ·

1000

= F

rictio

n lo

ss (%

)

L

Q = Flow (l/s)

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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 (τ) :

I4Dg ⋅⋅ρ=τ A.2.12)

ρ = density of water (kg/m3 ) g = acceleration of gravity (= 9.81 m/s2) D = internal diameter (m)

I = incline of friction drop line Lh∆

To be self-cleaning the shear stress shall be ≥ 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: 0044.02500

03,11I ==

Hence using A.2.12): 2N/m 3.5 N/m 0044.04

3272.081.91000 =⋅⋅⋅=τ

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.

Side 25 av 84

For both outfall pipelines and intake pipelines, we recommend avoiding high points in the trace.

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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 α) A.2.13)

Di = pipe internal diameter (m) α = pipe gradient

A simplified expression gives Uc as a function of igD

ic gDkU ⋅= 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 αsin The curve of k in Fig. A.2.6.1 is applicable for α= 0º →90º.

Example 9Calculate thdiameter D Solution : From fig. A If we insert As we see,

√ sin α

(α)1o 2o 5o 10o 20o 30o 40o 50o 60o 90o

0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1,0

α ≤ 5o α > 5o

√g

⋅ Di

Uc

K =

1,2

1,0

0,5

Fig. A.2.6.1 Critical velocity for transport of air in a pipeline.

e critical velocity for transport of air in a pipeline with slope α = 10º and internal

i = 500 mm.

.2.6.1 gives : k = 0.75

this value in A.1-13 we got : Uc = 0.75 ⋅ 5.081.9 ⋅ m/s = 1.66 m/s

the system requires quite a high velocity to transport air.

Side 26 av 84

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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.

Side 27 av 84

Formula A.2-13 gives however an qualified indication.

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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.

N N SDm

P

σr σr

NotheIf wfor

Int

p

Fig. A.3.1.1.1 Static system for internal pressure, cut pipe.

shear stress will occur due to the internal pre ring direction. e integrate the pressure components we find

ces : 2 ⋅ N = p ⋅ Dm N = tensile force (N) p = pressure (N/m2 = Pa) Dm = mean diameter (m) roducing the ring stress (σr) and the wall thick

SN r ⋅σ=

S2

Dp mr ⋅

⋅=σ

r

m

2Dp

Sσ⋅

⋅=

Since Dm = D – s sp2(

Dp

r +σ⋅⋅

=

σr = design stress (see table A.1.2) D = external diameter

ssure. There will only be a tensile force (N) in

the following result based on equilibrium of

A.3-1)

ness (s), we can develop the following formulas :

A.3-2)

A.3-3)

A.3-4)

) A.3-5)

Side 28 av 84

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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). σ is taken from table A.1.2

mm 18.2 m 0.0182 m)152(

2.01==

+⋅⋅

=s

The stress (σ) for a given pipe in the hoop direction exposed to a pressure (p) can be calculated from the formula :

sD SDR where)1SDR(

2p

r =−=σ 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 : MPa 6.64 MPa )16.17(28.0

r =−=σ

Safety factor confer A.1-1) : 50.164.6

10C ==

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.

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 84

∆Lεlσl

εrσr

QQ p

L

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A.3-7) rl ε⋅ν−=ε εl = strain in longitudinal direction εr = strain in ring direction ν = 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 (∆L) will be fully developed as indicated by formula A.3-8). This is the case for a pipeline floating free in surface position :

A.3-8) rLL ε⋅⋅ν−=∆ L = length of pipe

To estimate εr we have to introduce Hook’s law :

E

rr

σ=ε A.3-9)

σr = stress in ring direction (ref. formula A.3-6) E = modulus of elasticity (creep modulus) (ref. table A.1.2)

This gives :

)1SDR(E2

pr −

⋅=ε A.3-10)

)1SDR(E2

pLL −⋅

⋅⋅ν=∆ 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) m 0.375- m )111(8002

2.11005.0L =−⋅

⋅⋅−=∆

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 :

A.3-12) rlmax σ⋅ν=σ

)1SDR(2

plmax −

⋅ν=σ A.3-13)

The stress will be a tension.

Side 30 av 84

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Example 4 Calculate the maximum longitudinal stress for the data given in example 3. Solution :

By use of formula A.3-13) we get : MPa 3 MPa )111(2

2.15.0lmax =−

⋅=σ

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 : σ = E ⋅ ε 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 ν = 0.5. Solution :

First we calculate the stress from A.3-13) : MPa 2.5 MPa )111(2

15.0l =−

⋅=σ

The corresponding strain from A.3-14) : % 0.31 %100800

5.2E

=⋅=σ

=ε

Long term stress for this constant fictive strain can also be found from A.3-14) : MPa 0.465 MPa 0031.0150term long,l =⋅=σ

As we see the long-term stress is %6.18%1005.2

465.0=⋅ of the short-term stress in

longitudinal direction. The relaxation is significant. Compared to the hoop stress, which is constant over time, the stress in longitudinal direction

reach %3.9%1005465.0

=⋅ 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 84

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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.

Theresiin a

A.3.2.1 BucA pidistThe

PE

νs Dk

Firm soil Loose soil water or air

Fig. A.3.2.1 Different types of buckling

re is a significant difference for a pipe’s stance against buckling if it is installed trench or installed at sea bed.

n > 2 n = 2

kling of unsupported pipe pe during sinking or laying on the seabed can be considered as unsupported for normal ances between the concrete weights. buckling pressure for an unsupported pipe can be calculated by the formula :

k)D

s(1

E2p 3

m2buc ⋅⋅

ν−

⋅= A.3-15)

buc = buckling pressure (Mpa) = modulus of elasticity (For long lasting loads the creep modulus shall be applied.

For pressure surge we apply the short term modulus of elasticity)

= Poisons figure (0.4-0.5) = wall thickness (m)

m = mean diameter (m) = correction factor due to ovaling, ref. fig. A.3.2.1.1

e

k 1,0

0,9

0,8

0,7

0,650,6

0,5

0,4

0,3

0,2
Fig. A.3.2.1.1 Correction factor du to ovaling.
Side 32 av 84

%

Degree of ovaling1 2 3 4 5 6

0,1

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Formula A.3-15) can be transformed by introducing the SDR ratio :)sDSDR =(

32buc)1SDR(

k1

E2p−

⋅ν−

⋅= 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 ν = 0.4. Solution : By use of formula A.3-16) and fig. A.3.2.1.1 we get :

mwc 10 MPa 099,0MPa)126(

65,04,01

10502p 32buc ==−

⋅−

⋅=

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

mwc 5mwc2

10F

pbuc ==

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 ν= 0.4. Solution : Using formula A.3-16) and. fig. A.3.2.1.1 we obtain :

mwc 1 MPa 0094,0MPa)133(

65,04,01

2002p 32buc ==−

⋅−

⋅=

Factor of safety : 5,021

pp

Fappear

buc ===

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 :

1.5m

m

)(s/Ds1.56 l

2Ds

4 ⋅≤<

⋅⋅ A.3-17)

Side 33 av 84

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The buckling pressure pbucl can be written :

Fkp

lEs2.2p bucbucl ⋅⋅

⋅⋅= 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 : mm 2,18mm33

600s ==

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) :

mwc 1.5 MPa 0.0145 MPa)133(

14.01

2002p 32buc ≈=−

⋅−

⋅=

We apply formula A.3-18) with F = 1.0 :

mwc 1.7 MPa 0.017 MPa 0.65 0145.02600

2002.182.2pbucl ≈=⋅⋅⋅⋅

=

The safety factor against buckling is : 85.00.27.1F ==

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 84

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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 = α⋅⋅⋅ 1

tR ESF63.5 A.3-19)

α = D

31 δ⋅− A.3-20)

SR = ring stiffness, short term SR =

3)1SDR(12E

−⋅ A.3-21)

Et1 = 2 Es

1 = tangential modulus for the soil Es

1 = secant modulus for the soil (ref. fig. A.3.2.2.1)

Dδ

=

ovaling ( ≈ 0.05)

F = safety factor (should never be less than 2.0)

Filling height H m

Secant ModulusE`s MN/m2

For a pipe inpressure cau The soil presthe perimete

qs = γ = γw = h =

stallesed

surer.

(γ-spespehe

Fig. A.3.2.2.1 Secant modulus for granular soil versus filling heightin submarine trenches.

d in a trench we have to add the pressure caused by soil cover to the under-by hydraulic flow.

(qs) around a PE-pipe is considered to be uniformly distributed along

γw)⋅h A.3-22) cific gravity of soil cific gravity of water

ight of soil cover

Side 35 av 84

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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, ,05.0D

=σ γ = 20 kN/m3

and mod. Proctor for soil = 80%. Solution :

First we decide the pipe’s ring stiffness from A.3-21) : kPa 54.2kPa)133(12

10001000S 3R =−⋅

⋅=

Correction factor, σ, due to ovaling is taken from A.3-20) : α = 1 - 3⋅ 0.05 = 0.85 Es

1 is found from fig. A.3.2.2.1 : Es1 = 600 kPa ⇒ Et

1 = 2⋅ 600 kPa = 1200 kPa

Using formula A.3-19) assuming F= 1.0 we get the buckling pressure :

mwc 27 MPa 0.264 kPa 264 kPa 85.0 120054.2163.5q ≈==⋅⋅⋅=

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 : 9 3

27qqF

t≈==

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.

Side 36 av 84

The area of the chamber should be designed for the expected variations in flow.

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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

g

cvp ⋅∆=∆ A.3-23)

The surge pressure/water hammer is said to linearly depend on the pressure wave speed, c, in water inside a pipe. ∆v is the change of water flow speed (acceleration/retardation) and g = 9.81 m/s2. The pressure wave speed, c, is given by :

2/1

m2o

Ds

)1(E

c

⋅

ρ⋅ν−= A.3-24)

Eo = short time modulus of elasticity, ref. table A.1.2. ν = 0.4-0.5 = Poisson ratio ρ = 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 :

2/12o

)1SDR(1

)1(E

c−

⋅ρ⋅ν−

= 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.

Pressure wave speed in water inside a PE-pipe c m/sec Polyethylene Pipe SDR33 (PN322) SDR26 (PN4) SDR17.6 (PN6) SDR11 (PN10) PE100 Eo = 1050 N/mm2 PE80 Eo = 800 N/mm2

203

180

230

200

282

250

263

320

ν = 0.45

Table A.3.3.1 Pressure wave speed for PE

Side 37 av 84

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In practice, ∆v 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) : bar 0.44 - mwh 3,4 81.9

28215.0=−=

⋅−=∆p

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 : ∆pmax = 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 ∆L can be expressed :

A.3-26) TLL o ∆⋅⋅α=∆ α = thermal expansion coefficient (≈ 0.2⋅10-3 ºC –1) Lo = initial length at installation ∆T = 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 : ∆L = 0.2 ⋅ 10-3 ⋅ 3000 ⋅ (4-20) m = -9.6 m

Side 38 av 84

There have been some real examples where subwater pipelines have been too short due to change in temperature.

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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 (σT) can be expressed :

A.3-27) TET ∆⋅σ⋅−=σ E = modulus of elasticity (creep modulus) (Mpa)

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 : σT = - 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 :

σT,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) :

2

)TT(E insideoutside −⋅α⋅=στ 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 α = 0.2 ⋅ 10-3 ºC -1. Solution :

We apply formula A.3-28) and get : MPa 1.28 - MPa2

)204(102.0800 3=

−⋅⋅=σ

−

τ

Side 39 av 84

The stress’ nature is compression.

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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, εa, in the pipe wall. This strain can be expressed :

R2

DRr

a ⋅==ε A.3-29)

r = pipe radius R = bending radius D = pipe’s outside diameter

The case is shown in fig. A.3.5.1. To bend a pipe to this radius, R, it must mentioned earlier. The moment (M) can

R

IEM ⋅=

E = modulus of elasticity (creepI =

)dD(64

44 −⋅π (moment

D = outside diameter d = inside diameter

The maximum stress in the pipe wall can

R2

DERrEE aa ⋅

⋅=⋅=ε⋅=σ

εa

Fig. A.3.5.1 PE-pipe under pur

be subject to an external moment caused by the forces be expressed :

A.3-30)

modulus)

of inertia) A.3-31)

be estimated from Hook’s law (ref.A.3-14) :

A.3-32)

Side 40 av 84

r

R

D

e bending

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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 . aDR

=

Formula A.3-29) and A.3-32) can be rewritten :

A.3-33) a2

1a ⋅

=ε

A.3-34) a2

Ea ⋅

=σ

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) :

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

MPa 11.71 MPa 3622.1700a =

⋅⋅=σ

3.17,11

15F ==

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 : D60R min ⋅= 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

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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 :

⋅=

mrcrit D

s28.0,ε A.3-35)

The relationship between the axial and radial strain is given by Poisson’s figure : A.3-36) ar ε⋅ν=ε If we choose ν= 0.50 and put A.3-36) into A.3-35) we can find the critical strain in axial direction εcrit,a :

1

56.028.0, −

=

⋅=

SDRDs

macrit ν

ε A.3-37)

SDR = sD

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 :

12.1

1−=

SDRcrita = 0.89 (SDR –1) A.3-38)

It is normal to introduce a safety factor, F = 1.5 for such calculations.

Hence allowable bending ratio : )1(34.15.1, −⋅=== SDRDRa Fallowable 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

ratio DR F= 1.5

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.

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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

Fig. A.3.5.1.2 iSDR-class 26 For PE-pipe Sinsignificant. Example 16 What will the aof 1 bar during Solution : From table A.3 Since the bendwe get by use

k = 1

As we see the If the pipe hasto 28 m. For low pressu

A.3.6 Other stress

So far we have− Internal pr− External p− Water ham− Temperatu− Bending

.3.1.2 Increase in allowable strain due to internal pressure of 1 bar

ndicates that the internal pressure has a significant stabilizing effect on (27 %). DR11 or lower, the stabilizing effect of an internal overpressure is more or less

llowable bending ratio (R/D) be for a SDR26 pipe if it is subject to an internal pressure sinking? Assume a safety factor of 1.5.

.5.1.1 in example 15 we find the bending ratio without any internal pressure. a = 34

ing ratio is in inverse ratio to the allowable strain (ref. formula A.3-37) and A.3-38)) of fig. 3.5.1.1 :

.27 2727.1

341 === barpa

bending ratio has decreased from 35 to 28.

been a Ø 1000 mm, the bending radius would have been reduced from 35 m

re pipes (≤ PN4) internal pressure will increase the safety factor against buckling.

es

discussed stresses caused by : essure ressure (water and soil) mer re changes

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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 :

2

vDCf2

⋅ρ⋅⋅= A.3-40)

f = force pr. unit m pipe C = coefficient D = external diameter ρ = 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 ρ = 1025 kg/m3 Solution :

We apply formula A.3-40) : kN/m 4.6 N/m 4612m/N2

3102511f2

==⋅⋅⋅=

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 :

)dD(3

Dlf444

2

max−⋅π⋅

⋅⋅⋅=σ A.3-41)

f = force pr. unit length l = distance between supports D = outside diameter d = inside diameter

Side 44 av 84

If we go back to example 17 and assume SDR = 22 and l =10 m we get :

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MPa 60.0MPa)909.01(14.33

10l106.4444

23

max =−⋅⋅

⋅⋅⋅⋅=σ

−

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 2/3 and the current component to be 1/3 , we get the maximum force including concrete weights :

kN/m 5.6 kN/m 3/16.46.45.0()3/16.4(f 22max =⋅+⋅+⋅=

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 :

m 27 m 1073.02lmax =⋅=

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 :

2con s2P3⋅π⋅

⋅=σ 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.

Side 45 av 84

Example 18

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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 : mm 56.8 mm6.17

1000s ==

We apply formula A.3.42) and get, assuming that 2 of the weights is contributing to the concentrated load :

MPa 2.4MPa14.30568.02

1014232

3

con =⋅⋅

⋅⋅⋅=σ

−

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 σh = ∑ σi, h A.3-43) i=1

n σl = ∑ σi, l A.3-44) i=1

σh = total stress in hoop direction σi,,h = stress no.i in hoop direction σl = total stress in longitudinal direction σi,l = stress no.i in longitudinal direction

To find a combination/comparison (σcomp) stress, one often use Von Mises criteria:

lh2

l2

hcomp σ⋅σ−σ+σ=σ 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.

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Example 19 Calculate the comparison stress for a situation where the total stress in hoop direction σh = 4 Mpa and in longitudinal direction is σl = - 2.5 Mpa (compressive) for a PE80 pipe ? Solution : We apply formula A.3-45) : MPa 7.5MPa)5.2(4)5.2(4 22

comp =−⋅−−+=σ 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 σl and σh 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).

Side 47 av 84

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.

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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 :

%100

4Dw

a

w

2cw

d ⋅

γ⋅⋅π

= A.4.1)

wcw = weight of concrete weights in water distributed pr. m pipe D = outside diameter γW = 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:

%100

4d

wwa

w

2w pipecw

a ⋅

γ⋅⋅π

+= 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 : ρPE = 950 kg/m3 (density PE) ρC = 2400 kg/m3 (density concrete) ρw,sea = 1025 kg/m3 (density sea water) ρw = 1000 kg/m3 (density fresh water)

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110

100

90

80

70

60

50

40

30

20

10

0-10 10 20 30 40 50 60 70 80 90 100 Degree of displacement

ad (%)

52%

45%

30%

Degree of air fillingaa (%)

SDR 11

SDR 17,6

SDR 26

SDR 41

Sea water

Fresh water

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 :

ρρ−ρ

= w

a

w

ww A.4.3)

w w = weight in water w a = weight in air ρ = density of body ρw = 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 ρPE = 950 kg/m3, ρw = 1025 kg/m3 and ρc = 2400 kg/m3 Calculate the air filling rate aa.

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Solution : First we find the weight of the concrete weights and the pipe in water by use of A.4.3) :

kN 2.3kN2400

102524005.6w cw =−

⋅= (a piece)

kN/m 64.0m/kN52.3w cw == (pr. m pipe)

kN/m 28,0 m/kN950

102595035.0w w pipe −=−

⋅=

The internal diameter of the pipe is :

mm 6.454mm)225002500(d =

⋅−=

We apply formula A.4.2) to calculate the degree of air filling :

%5.37%81.91025

44546.014.3

100)028.064.0(a 2a =

⋅⋅⋅

⋅−=

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.

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 84

Rectangular Circular Starred


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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


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. We suppose that horizontal directioTo avoid sliding, tforce between the

Fig. A.4.3.1

Sp

the forces from current and waves can be decomposed in a drag force, FD, in n and a lift force, FL, in vertical direction acting simultaneously on the pipeline. hese two forces must be overcame by the weight of the system and the friction concrete weights and the sea-bottom.

Side 51 av 84

��

��

��

��

��

��

��

��

��

��

γcWc

n γa

γw γP

Ff

FD

Wp

γsea

FB FL

FN

Wcw

Ww

Wp

Wa

µ

tability of PE- ipe on seabed

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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 ≥ FD A.4.5)

Since Ff = µ⋅FN we get the criteria for stability : N

D

FF

≥µ 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 :

c

seaccacw ww

γγ−γ

⋅= A.4.7)

Wca = weight of concrete weights in air pr. m pipeline γc = specific weight of concrete γsea = specific weight of seawater

w

2

w 4d)n1(w γ⋅

⋅π⋅−= A.4.8)

n = amount of air filled section, e.g. 30%, η = 0.3 d = internal diameter γw = specific gravity of water inside the pipe

a

2

a 4dnw γ⋅

⋅π⋅= A.4.9)

γa = specific gravity of air inside the pipe wa can in most cases be neglected.

sea

2

B 4DF γ⋅

⋅π= A.4.10)

D = external diameter of pipe 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

Side 52 av 84

Specific gravity of air/gas can be neglected. Is the pipeline stable on the seabed ?

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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) :

kN/m 1.068 m/kN5.23

05.105.2336.5w cw =

−⋅=

Then we apply A.4.8) to find the weight of water inside the pipe pr. m :

kN/m 1.379 kN/m 1044546.0)15.01(w

2

w =⋅⋅π

⋅−=

The buoyancy is given by formula A.4.9) :

kN/m 1.972 kN/m 05.104

5.0F2

B =⋅⋅π

=

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)

65.062.04.0

min ==µ

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 : % 64.2 % 10010

44546.0

027.0068.1a2a =⋅

⋅⋅π

−=

Rectangular weight : % 75.2 % 10010

44546.0

027.0247.1a 2a =⋅

⋅⋅π

−=

Circular weight : % 108.1 % 10010

44546.0

027.0781.1a 2a =⋅

⋅⋅π

−=

In reality it is not possible to use circular weights without introducing buoyancy elements temporary during sinking/installation of the pipe.

Side 53 av 84

Pipelife Norge AS


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 transport/topography Type of pipeline 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.

Side 54 av 84

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 %.

Pipelife Norge AS


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 ⇒ 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. The amount of the forces is mainly depe

Fig. A.4.5.1 Current forces acting on a pipeline

ndent on :

Side 55 av 84

D

FL

FD

v

f

Pipelife Norge AS


- Current velocity (v) - Pipe's diameter (D) - Density of streaming water (ρ) - Pipe's distance above seabed (f)

The forces can mathematically be expressed as follows :

FD = Dv21C 2

D ⋅⋅ρ⋅⋅ A.4-10)

FL = Dv21C 2

L ⋅⋅ρ⋅⋅ A.4-11)

CD = drag coefficient CL = lift coefficient ρ = 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 :

ν⋅

=DvR e A.4-12)

ν = 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 .

Fig. A.4.5.2 Drag coefficient, CD

Side 56 av 84

CD

Re

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CL

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. Solution : The velocity component perpendicular to the pipeline can be written :

vN = α⋅ sinv A.4-13) Inserting this expression into A.4-10) and A.4-11) we get :

FD = Dsinv21C 22

D ⋅α⋅⋅ρ⋅⋅ A.4-14)

FL = Dsinv21C 22

L ⋅α⋅⋅ρ⋅⋅ A.4-15)

Putting in values from example 4 gives :

FD = N 125 N/m 5.045sin11000210.1 22 =⋅⋅⋅⋅⋅

FL = N 25 N/m 5.045sin11000212.0 22 =⋅⋅⋅⋅⋅

Side 57 av 84

D= 500mm

V= 1 m/s

α = 45o

Pipelife Norge AS


As we can see, the forces are reduced significantly if the attack angle, α, 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.

Shallow water Semi deep water Deep water

L

d

≤ < < >hL

120

120

hL

12

hL

12

Fig. A.4.6.1 Movement of particles in a wave

Side 58 av 84

Pipelife Norge AS


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 > )2L

Wave forces will never influence a pipeline installed at deep water.

At semi deep water 20L( < h < )

2L the forces can be significant while they can reach

extreme values as shallow water (h > )20L

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 :

Fi = o

o2

i LH

4DfC ⋅

π⋅γ⋅⋅⋅π A.4-16)

FD = D

HLH

4DfC o

o

o2

2D ⋅⋅

π⋅γ⋅⋅ A.4-17)

FL = D

HLH

4DfC o

o

o2

2L ⋅⋅

π⋅γ⋅⋅ A.4-18)

Fi = inertial force FD = drag force FL = lift force f = refraction factor Ci = inertial coefficient CD = drag coefficient CL = lift coefficient γ = 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 α, the forces must be corrected by the factor sin α. 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 Distance to bottom ≥ 4D

Ci CD CL

3.3 1 2

2

0.7 0

Side 59 av 84

Table A.4.6.1 Force coefficients for waves

Pipelife Norge AS


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

Fetch length F (miles)

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

0,2

0,3

0,4

0,5

0,6

2.2

0,70,8

0,91.0

1.21.4

1.61.8

2.0

2.62.8

3.02.4

4.0

5.0

6.07.0

8.09.0

10.0

12.0

14.0

10090

8070

60

50

40

30

25

20

15

10

3

4

56

78

9

Win

d sp

eed

(mile

s/ho

ur)

Fig. 4.6.2 Wave height H1/3 as function of wind speed and "fetch length"

80

Fig. 4.6.3 Wave period as function of wind speed and "fetch length" Side 60 av 84

70

60

50

40

30

20

10

Fetch length F(miles)

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

1.0

1.2

1.41.6

1.8

2.02.2

2.4

3.0

3.4

3.84.2

4.6

5.0

6.0

7.09.0

Win

d sp

eed

(mile

s/ho

ur)

Pipelife Norge AS


Note that the wave height, H1/3, is given in the unit feet when the wind speed is in miles/hour and the "fetch length" is in miles ( 1 mile = 1609 m, 1 foot ≈ 0.3 m) If statistic information is not available, one can apply the following formulas for a roughly calculation of Ho and Lo [1].

Ho = 0.045 ⋅ u ⋅ F A.4-21)

Lo = 0.56 ⋅ u ⋅ F A.4-22)

u = wind speed (m/s) F = fetch length (km)

This means that the abruptness of the wave, o

o

LH is approximately 8 %.

Variations within the range 7-9 % are usually. Formulas A.4-21) and A.4-22) tend to give a little to high values and should therefore be on the safe side. We recommend to use wave statistics if available for a given project. When the wave period is known, the wavelength on deep water can be calculated by formula : Lo = 1.56 ⋅ T1/3

2 A.4-23) Refraction factor This factor tries to describe how the waves are influenced by the bottom conditions when the waves are approaching the shore. Mathematically this factor can be expressed :

α⋅⋅

= sinH

a2fo

A.4-24)

a = wave particles amplitude in orbit at the bottom (ref. fig. A.4.6.1) α = angle between wave's speed direction and pipeline

There exist diagrams for the refraction factor based on certain conditions. These assume that the contour intervals on the bottom are straight lines and parallel to the shoreline. The input values in the diagrams are :

αo = angle between wave's speed direction and perpendicular to the shoreline β = angle between pipeline and perpendicular to the shoreline

Side 61 av 84

Fig. A.4.6.4, A.4.6.5 and A.4.6.6 give possibilities to decide a value for f [4].

Pipelife Norge AS


Wave direction

Shore line

Pipe

Fig. A.4.6.4 Refraction factor for β = 0

Fig. A.4.6.5 Refraction factor for β = 15º

Side 62 av 84

Fig. A.4.6.6 Refraction factor for β = 30º

Pipelife Norge AS


The breaking depth (hb) for a wave can roughly be calculated by formula A.4-25)

hb = 0.05 ⋅ Lo A.4-25) Lo = wave length at deep water (ref. A.4-22) and A.4-23))

On the following pages we shall deal with some examples where we apply the formulas for calculation of wave properties and wave forces. Example 5 Find the wave height, Ho, and wave length, Lo, at deep water for a wind speed u = 30 m/s (strong storm) and a fetch length, F = 10 km. Apply both the method with diagrams and the method with formulas. Compare the results. At which depth will the wave break ? Solution : a) Use of wind statistic : First we find the wind speed in miles/hour and the fetch length in miles.

h/miles 67h/miles 1609

606030s/m 30u =⋅⋅

==

miles 2.6miles 1609

10000mk 10F ===

By use of diagrams in fig. A.4.6.2 and fig. A.4.6.3 we get :

m 1.2feet 7H 3/1 == s 3.5 T 3/1 = Maximum wave height is given by formula A.4-19) : Ho = 1.8 ⋅ 2.1 m = 3.8 m Wavelength can be calculated by formula A.4-23) : Lo = 1.56 ⋅ 5.32 m = 44 m b) Use of formulas : We apply formulas A.4-21) and –22) : Ho = 0.045 ⋅ 30 ⋅ m10 = 4.2 m Lo = 0.56 ⋅ 30 ⋅ m10 = 53 m

c) Comparison of results : If we compare the results from a) and b), we find that the formula method gives higher

values than the method applying wind statistics. The differences expressed in percentage are :

Wave height % 10 %1008.3

8.32.4=⋅

−

Wavelength % 20 %10044

4453=⋅

−

Side 63 av 84

Pipelife Norge AS


d) Breaking depth : Formula A.4-25) gives an estimate of the breaking depth. a) ⇒ hb = 0.05 ⋅ 44 m = 2.2 m b) ⇒ hb = 0.05 ⋅ 53 m = 2.6 m

The pipeline has to be buried to a depth minimum equal the breaking depth. Normally we bury the pipe to a depth equal the maximum wave height, H. In example 5 this recommendation gives a trench to approximately 4 m water depth. Example 6 Consider the wave data calculated in example 5 by the wind statistic method (Ho = 3.8 m Lo = 44 m). Fig. A.4.6.7 shows that the waves speed direction hits the perpendicular to the contour intervals under an angle αo = 45º . An outfall pipeline Ø 500 mm PE is installed perpendicular to the contour lines, it means β = 0º The pipeline is buried to 5 m depth. Decide the refraction factor f and calculate the forces attacking the pipeline at 20 m depth. Assume that the pipeline is laying directly at the seabed (no space between pipe and bottom) Assume α = 10000 N/m3

Fig. A.4.6.7 Outfall pipeline attacked by waves Solution :

We apply fig. A.4.6.4, α = 45º, 45.04420

Lh

o==

This gives a refraction factor : f = 0.1 The wave forces maximum values can be found by the formulas A.4-16), -17) and –18). Force coefficients are taken from table A.4.6.1.

Inertial force : N/m 180 N/m 44

8.34

5.0100001.03.32

=⋅⋅

⋅⋅⋅⋅=π

πiF

Drag force : N/m 15 N/m 0.53.8

448.3

45.0100001.01

22 =⋅⋅

⋅⋅⋅⋅=π

DF

Side 64 av 84

Trench

Outfall Chamber

β= 0o

-20-15-10

-50

αo= 45o

Wave front

Pipelife Norge AS


Lift force : N/m 25 N/m 0.53.8

448.3

45.0100001.02

22 =⋅⋅

⋅⋅⋅⋅=π

LF

As we see the inertial force is the dominating force. It is out of phase with the lift force and the drag force. For stability calculations we can assume FD to be zero when Fi = Fi,max and vice versa. Thus we get the following critical combination : Horizontal 180 N Vertical 25 N

Side 65 av 84

For stability calculations see point A.4.3.

Pipelife Norge AS


A.5 Design of parameters for the sinking process

The critical phase for installations of a PE-pipe is the sinking. Fig. A.5.1 below shows the situation during sinking of a PE-pipeline.

To(qanwe TothToWth If FoThthTo Wdu- - - In If of

Fig. A.5.1 PE-pipeline during sinking

carry out a safe installation we should consider the balance between the forces acting downward 1) d the forces acting upwards (q2). The downward forces are created mainly by the concrete ights on the pipeline and the upward forces are due to buoyancy of the air filled section.

start and to continue the sinking process, the downward forces must be a little greater than e upward forces. control this difference is essential and the main challenge during sinking. e must try to avoid acceleration forces on the system. This can be controlled by recording e sinking speed (v) and regulating the internal pressure (p).

the speed is increasing, we can increase the air pressure and vice versa. r regulation of the air pressure we apply valves and compressor. e most critical situation for the pipeline regarding damage, is buckling at the sea surface or at

e bottom due to "exceeded" buckling radius, ref. A.3.5.1. secure a sufficient radius it is necessary to apply a pulling force (P) in the end of the pipe.

ith reference to fig. A.5.1, we have the following parameters available for control and regulations ring sinking :

Air pressure (p) Pulling force (P) Sinking velocity (v)

the next chapters we shall give some recommendations for calculation of these parameters.

we can carry out a slowly sinking, the installation will be successful from the pipeline's point view.

Side 66 av 84

V P

H

aa·H

(1-aa)·H

Pipelife Norge AS


A.5.1 Internal air pressure The internal pressure is dependent of loading from the concrete weights. Calculation of the loading is described in A.4. The important parameter is the air filling rate, aa (ref. A.4-2). To obtain an air filled section in fig. A.5.1, which will balance the weight of the concrete collars, we must apply an internal pressure (p) in the pipelines.

p = aa ⋅ H A.5-1) p = internal pressure (mwc) aa = degree of air filling H = water depth (m)

As we see from formula A.5-1) the internal pressure is dependent of the water depth. This means that we have to increase the pressure as the depth increases. The compressor must have capacity to produce sufficient air against a pressure corresponding to maximum depth including pressure drop in the transmission pipelines. If we know the longitudinal section for a pipeline, we can calculate the balance pressure in each point. This curve or table will be the basis for a successful installation. We have to use the pipe sections as reference regarding the length. For instance every 50 m of the pipe must be marked. Example 1 Calculate the internal balance pressure for a Ø 500 mm PE80 SDR22 which has a loading degree of concrete weights corresponding to an air filling rate aa = 30 %. Assume the depths to be 10 m, 20 m and 30 m. Solution : We apply formula A.5-1) and get : P10m = 0.3 ⋅ 10 m = 3 mwc = 0.29 bar P20m = 0.3 ⋅ 20 m = 6 mwc = 0.59 bar P30m = 0.3 ⋅ 30 m = 9 mwc = 0.88 bar Notice that the internal pressure is independent of the pipe diameter.

A.5.2 Pulling force The pulling force in the end of the pipe is applied to control the pipe's position and to increase the bending radius in the S configurations (ref. fig. A.5.2) If the loading percentage is less than 50 %, which normally is the case, the critical radius will be at the sea surface. Otherwise it will be at the seabed. To carry out a proper calculation of the sinking process is complicated and must be done by computer programs. However, there is a simple method to find an estimate for the pulling force. This method is based on the chain link theory and is valid for deep water. Fig. A.5.2 shows the situation.

Side 67 av 84

Pipelife Norge AS


D Inthw F

Itin If

water level inside pipewater into pipe

air under pressure

Pulling forceP

controlled air ventilation

P

Fig. A.5.2 Process, shape and technical parameters for a PE-pipe during sinking

eep water is defined as : H > 12 ⋅ D A.5-2)

shallow water (start phase of the sinking) it is impossible to apply a force in the end before e pipe is connected to a fixed installation. When submerging the end for connecting the pipe, e have to check that the bending radius is greater than the buckling radius (ref. table A.3.5.1.1).

ormula A.5.3) can be applied for this purpose :

H2

LR2

⋅= A.5-3)

R

=

bending radius

L = submerged length of pipe ("cantilever length") H = connection depth

can be necessary to use several attack points for submerging the pipe during connection (not only the end)

we return to fig. A.5.2, we have the following parameters :

H = depth (m) h = internal water height (m) w2 = net buoyancy in air filled section (N/m) w1 = net weight of water filled section (N/m) P = pulling force (N) T = tension force in turning point (N) α = angle between pipe axis and horizontal in turning point (º )

Side 68 av 84

h

H

P

R1minR1

EJW1

R2R2min

W2

Tα

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R1 = bending radius in water filled section (m) R2 = bending radius in air filled section (m) EJ = pipe stiffness Following mathematical relations can be developed :

R1 min =

1wP A.5-4)

R2 min = 2w

P A.5-5)

T = P + w1 ⋅ h A.5-6)

cosα = hwP

P

1 ⋅+ A.5-7)

As mentioned earlier w2 > w1 if the design air filling rate is less than 50 %. The radius R2 at surface will be critical in this case, ref. A.5-4) and A.5-5). The relationship between w1 and w2 can approximately be written :

1a

a2 w

aa1

w ⋅−

= A.5-8)

aa = air filling rate (a > 20%) From formula A.5-4) and A.5-5) we can find the necessary pulling force using the critical radius, Rmin, from table A.3.5.1.1.

P1 = w1 ⋅ Rmin A.5-9)

P2 = w2 ⋅ Rmin A.5-10)

The greatest force of P1 and P2 will be the pulling force to be applied in the project.

Example 2 A Ø 500 mm PE80 SDR26 pipe is to be installed to 50 m water depth. The pipeline has a loading degree corresponding to 25% air filling. The sinking shall be carried out using a safety factor 2.0 against buckling. Density of seawater can be set to 1025 kg/m3 Decide the following factors : a) Minimum bending radius b) Necessary pulling force in the end of the pipe c) Maximum tension stress in pipe wall d) The angle, α, at the return point in the S-curve Solution : a) Minimum bending ratio is taken from table A.3.5.1.1 :

SDR = 26 ⇒ DR = 34 Rmin = 35 ⋅ 0.5 m = 17 m

Side 69 av 84

Pipelife Norge AS


b) To decide the puling force we have to know w1 and w2. We apply formula A.4-2) to estimate w1 (w1 = wcw + wpipw w)

w

2

a1 4daw γ⋅⋅π⋅= N 420 N/m 81.91025

44618.025.0w

2

1 =⋅⋅⋅π⋅=

w2 can be estimated from A.5-8) : N 1260 N 25.0

25.01420w 2 =−

⋅=

The maximum pulling force is given by formula A.5-10) : P = 1260 ⋅ 17 N = 21.4 kN c) Maximum tension force in the pipeline will appear in the return point. Formula A.5-6) gives :

T = P + w1 (1-aa) ⋅ H T = (21400 + 420 (1-0.25) ⋅ 50 ) N = 37.2 kN The corresponding stress in the pipe wall can be decided to :

MPa 1.3 N/m )4618.05.0(

4

37200

)dD(4

T 2

2222=

−⋅π

=−⋅

π=σ

In addition there will be a stress in longitudinal direction due to internal pressure and Poisson’s number, ref. A.3.1.2. Formula A.3-13) gives :

MPa 0.78 MPa )126(2

125.05.0max l =−⋅

⋅=σ

Maximum tension stress is the sum of σ and σmax :

σmax = (1.3+0.78) Mpa = 2.08 Mpa

From table A.1.2 we see that short time burst stress is 8 ⋅ 1.6 Mpa = 12.8 Mpa for PE80.

Safety factor regarding tension stress becomes: 15.608.28.12

==C

This result underlines that buckling is the most critical damage that can occur during sinking (C = 2.0) d) The angle, α, at the return point is given by formula A.5-7) :

54 58.050)25.01(42021400

21400cos ==⋅−⋅+

= αα º

As shown in example 2, we can calculate the pulling force, P, to secure a safe installation combined with internal pressure. Experiences indicate that the pulling force calculated by the formulas in this chapter, give higher values than more advanced methods.

Side 70 av 84

Pipelife Norge AS


A.5.3 Sinking velocity To avoid acceleration forces on the pipeline the sinking speed shall be kept as constant as possible during installation. Since there will always be some variations in the velocity during a practical installation, it is also important to keep the speed at a low level. If we look at Newton’s law;

tvmK

∆∆

⋅= A.5-10)

K = acceleration force m = mass in movement ∆v = change in speed ∆t = change in time

We see that a big change in tv

∆∆ will create a big force K to act on the water string and on

the pipeline. If v is kept low, we will secure that ∆v also is low for a given time ∆t. As a rule of thumb, it has often been recommended that the sinking velocity should not exceed 0.3 m/s ≈ 1 km/h. There are, however, several examples from successfully projects where the sinking speed has been greater than 0.3 m/s. The sinking speed is governed by the flow, Q, entering the pipe. This flow is again dependent on the available driving pressure.

A.5-11) ia pHah −⋅=∆ ∆h = available pressure drop (mwc) H = depth (m) pi = internal pressure (mwc) aa = design air filling rate

The pressure drop can be expressed, ref. A.2-1) :

g2

vkg2

vDLfh

2

s

2

⋅⋅+

⋅⋅⋅=∆ A.5-12)

f = friction factor (≈ 0.02) L = length of water filled section (m) D = internal diameter (m) v = velocity (m/s) g = gravity acceleration (≈ 9.81 m/s2) ks = singular loss coefficient

If we combine A.5-11) and A.5-12) we can express the sinking velocity as :

21

s

ia

DkLf)pHa(Dg2

v

⋅+⋅

−⋅⋅⋅⋅= A.5-13)

From A.5-13) we see that v is dependent of the length of water filled section (L), the depth (H) and the internal pressure (pi). Other parameters are nearly constant. To keep a constant speed the internal pressure (pi) must be regulated in accordance to the changes in L and H.

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Since the relation between L and H is known for a given project, it is possible to calculate a "sinking curve" for pi as a function of L. This curve is essential during installation. ∆h will appear as an under-pressure in the pipe and must be controlled against risk of buckling, (ref. A.3-16). To make the regulation of the sinking speed easier in the initial phase of the submerging procedure, it is an advantage to have a high value for the singular loss coefficient, ks. This can be done by using a reduced inlet diameter (T-pipe) compared to the main pipe. A suitable diameter can be in the range 1/3 D to 1/20 D (under-pressure must be checked). We recommend to keep this opening constant during sinking. There is a maximum sinking speed at which the sinking PE-pipe has a risk to oscillate. This speed can be estimated roughly from the formula :

21

2

2)

SDR2E(

nHS2Dkv

ρ⋅⋅⋅

⋅⋅⋅

⋅π⋅= A.5-14)

k = support factor (k = 1.0 for freely supported, k = 2.25 for a fixed situation) S = Strouhals number (≈ 0.2) D = external diameter (m) H = maximum installation depth (m) n = safety factor (assume n = 2.0) E = modulus of elasticity (short time) (kN/m2)

ρ = mass of pipe, content (water) and oscillating water pr. unit of pipe volume (ρ ≈ 3.0 t/m3 /m)

If we assume E = 8⋅105 kN/m2 , k = 2.0, and maximum depth = 50 m, formula A.5-10) can be transformed to :

21 2 )SDR(D2.1v

−⋅⋅= A.5-15)

Formula A.5-15) gives an indication about the maximum sinking speed, but the buckling risk must also be considered. For small diameters, the formula is a bit conservative compared to experiences. Oscillations during sinking will normally not harm the pipe. Example 3 Calculate the maximum sinking speed for a PE-pipeline as a function of diameter SDR classes 26 and 33. Assume maximum depth to be 50 m and D ≥ 600 mm. Solution : We use formula A.5-15) and plot the result graphically as shown in fig. A.5.3. Fig. A.5.3 Maximum sinking speed to avoid oscillations

V

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600 800 1000 1200 1400 1600 D(mm)

(m/s 0,9

0,8

0,7

0,6

0,5

0,4

0,3

0,2

0,1

SDR = 26

SDR = 33

Pipelife Norge AS


Fig. A.5.3 indicates that a speed 0.3 m/s can be applied for pipe diameters ≥ 1200 mm without risks for oscillations. For Ø 1600 mm the maximum sinking speed with respect to oscillations is 0.6 m/s for SDR26. For practical purposes the sinking speed will be in the range 0.1 – 0.6 m/s for a controlled submerging. Example 4 Assume a Ø 1200 mm PE80 SDR26 to be sunk to 50 m depth. The longitudinal section of the bottom is given by table A.5.1 below :

L (m) H (m) L (m) H (m) 0

100 200 300

5 15 20 30

400 500 600 700

35 40 45 50

Table A.5.1 Longitudinal section

The intention is to sink the pipe by a constant velocity of v = 0.3 m/s. The loading is equal to an air filling rate, aa = 30 %. Assume the inlet opening for water during sinking to be 1/4 of the pipes internal diameter. Calculate the balance pressure inside the pipe and the sinking pressure. Find the maximum speed regarding buckling. Solution : Formula A.5-1) is used to calculate the balance overpressure : pb = aa ⋅ H aa = 0.3 This gives :

L (m) H (m) ρb (mwc) L (m) H (m) ρb (mwc) 0

100 200 300

5 15 20 30

1.5 4.5 6.0 9.0

400 500 600 700

35 40 45 50

10.5 12.0 13.5 15.0

Table A.5.2 Balance pressure Formula A.5-13) must be used to estimate pi. We can rewrite this formula :

Dg2

)DkLf(v - Ha s

2

ai ⋅⋅⋅+⋅

⋅=ρ A.5-16)

Input values : D = 1107.6 mm f = 0.02 ks = 0.5 ⋅ (4)2 = 8 g = 9.81 m/s2 v = 0.3 m/s

The result is given in table A.5.3 :

L (m) H (m) pb (mwc) pi (mwc) ai (%) 0

100 200 300 400 500 600 700

5 15 20 30 35 40 45 50

1.5 4.5 6.0 9.0

10.5 12.0 13.5 15.0

1.46 4.46 5.95 8.94

10.43 11.92 13.42 14.91

29.2 29.8 29.8 29.8 29.8 29.8 29.8 29.8

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Table A.5.3 Internal sinking pressure.

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As we see there is a very low under-pressure in the pipe (0.04-0.09 mws) during sinking. It can be difficult to regulate the pressure with sufficient accuracy. The manometer must have a scale designed for the purpose. In this case it can be favourable to reduce the inlet area for water in order to increase ks (singular loss coefficient). If, for instance, a valve is applied and the inlet diameter is equivalent to 1/20 ⋅ D, the pressure drop will add to 0.83 mwc. From formula A.3-16) we can find the buckling pressure :

mwc 8.0 MPa 0.079 MPa )126(

65.04.01

8002p32

==−

⋅−

⋅=

If we introduce a safety factor 2.0, available pressure drop is approximately 4.0 mwc. If we put this value for (aa ⋅ H – pi) into formula A.5-13) we get :

m/s 3.13 m/s 107.18002.0

0.4107.181.92v21

o,max =

⋅+⋅⋅⋅⋅

=

m/s 1.95 m/s 107.1870002.0

0.4107.181.92v21

700,max =

⋅+⋅⋅⋅⋅

=

Critical speed in the start point is 3.13 m/s and 1.95 m/s in the end point regarding buckling. The maximum speed will be limited by the drag force due to the current appearing when the pipe moves through the water, ref. A.4-10). As we see there will be no risk for buckling of the pipe if a controlled sinking is carried out. To control the sinking speed in reality, it can be adequate to record the time between, for instance, 3 consecutive concrete weights disappearing from the water surface. If the centre distance is known, the speed is :

tc3v ⋅

= A.5-17)

c = centre distance between concrete weights t = recorded time for 3 times c

The recorded/calculated speed should be compared to the design speed and necessary actions carried out if the speed is too high (e.g. close valve, start compressor) As you have noticed there is no exact theory to decide the maximum sinking speed for a pipeline. A good idea is to keep the speed low in case of waves, current, failure in pulling force equipment, failure in internal air pressure, regulation equipment etc. As an information we can tell you that two successful sinking operations were carried out as follows i) Ø 1600 mm PE80 SDR26 Length = 2500 m Max. depth = 50 m Max. pulling force = 500 kN Weighting, aa = 28 % vmax = 0.46 m/s

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ii) Ø 800 mm PE8 SDR17.6 Length = 500 m Max. depth = 200 m Max. pulling force = 40 kN Weighting, aa = 15 % vmax = 1.0 m/s If we derive a formula based on a simple energy balance where kinetic energy is transformed to elastic energy in the pipe, and only consider the axial direction, we get the expression :

21

2

2m

max ))Dd1(

E2(

Fv −⋅

ρ⋅⋅

σ= A.5-18)

vmax = critical sinking speed (m/s) σm = σa - σT σm = stress (Mpa) σa = design stress (5.0 Mpa) σT = stress created by pulling force and weighting (≈ 2.5 Mpa) F = factor of safety / correction factor E = Modulus of Elasticity (800 Mpa) d = internal diameter D = external diameter ρ = density of water (kg/m3 )

If we put in values and assume the pipe to be thin-walled, we get the formula :

SDRF9.7vmax

⋅= A.5-19)

If we now compare this formula to the real projects i) and ii) we get the safety factors / correction factors : Fi) = 3.4 Fii) = 1.9 If we, for instance, use the mean value of the two factors we get the result (F = 2.65) :

SDR

0.3vctitical = A.5-20)

This formula gives an indication of the maximum sinking speed during a controlled sinking (pulling force in the end) before the pipe is damaged. We will, however, not recommend to use these speeds without risk analysis. Table A.5.4 gives the critical sinking speeds :

SDR-class Critical sinking speed (m/s)

SDR-class Critical sinking speed (m/s)

41 33

27.6 26 22

0.47 0.52 0.57 0.59 0.64

17.6 17

13.6 11 9

0.72 0.73 0.81 0.90 1.00

Table A.5.4 Critical sinking velocity for PE-pipes

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Pipelife Norge AS


B. Installation This chapter gives some practical information to the contractor regarding installation of subwater pipelines. We also refer to chapter 0.2.

B.1 Jointing of PE pipes Various jointing methods have been developed since polyethylene pipes first came on the market in the early 1960s. To day there are many jointing methods suitable for all sizes of PE pipes : - butt fusion welding - electro-fusion fittings - stub end with steel backing ring - mechanical couplers

Fusion welded joints are by far the most common. Butt fusion can be used on all sizes of PE pipes, but is mainly used on pipes from 110 mm to 2500 mm in diameter. Electro-fusion couplers are now available in sizes up to 500 mm and in the future even bigger diameters will be available. A stub end with backing ring is mainly used for jointing longer sections of pipes together, for connections to valves or manholes or to pipes made of other materials. Butt fusion welding of pipe strings combined with stub end / backing rings are usual for the larger PE pipes for installation on land or in marine environments. Butt fusion and electro fusion fittings are mainly used for the smaller pipe sizes. Mechanical fittings for all sizes of PE pipes are now available in various metal and plastic designs. They are preferable under conditions of : - extreme short-term bending stress as during submerging and laying - difficult or impossible welding conditions - under water jointing of repair of pipes in general

In case of using long prefabricated sections towed by boat to the site, it can be favourable to join them together with support sleeves and stub ends as shown in fig. B.1.1. Support sleeves are also used in connecting PE pipes to grouted flanges in manholes. They may be delivered to fit all sizes of pipes and can be adjusted to suit customer preferences for length and corrosion protection.

Fig. B.1.1 Support sleeve

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B.2 Butt fusion of PE pipes The standard butt fusion cycle, according to DS-INF-70/INSTA 2072, is shown in fig. B.2.1. Fig. B.2.1 Butt fusion cycle. P1 = heating pressure, high (Mpa) P2 = heating pressure, low (Mpa) P3 * = welding pressure (Mpa) P4 = cooling pressure (Mpa) ≈ P3 t1 = heating time (s) with high pressure t2 = heating time (s) with low pressure ∆t = change-over time (s) t3,1 = pressure build-up time (s) t3,2 = cooling time (s), relative to a cooling pressure P3 *) The welding pressure may differ from that stipulated in DS-INF-70/INSTA 2072, as it depends on the welding criteria stated in the standard

In next chapter there is given some guidelines regarding the welding parameters.

B.2.1 Welding parameters The welding parameters listed below are average of guideline values. Wall thickness (e) and diameter (de) are stated in millimetres. 1. Welding Temperature – T The welding temperature, T, shall be in the range of T = 210ºC ± 10ºC and shall be measured continuously and verified for each weld using a thermo stick. 2. Heating-up Pressure (high) – P1 The heating-up pressure shall be P = 0.18 N/mm2 ± 0.01 N/mm2 3. Heating-up Time - t1 This is the bead formation time in seconds. It shall be recorded. See item 4.

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4. Bead Width - A The bead width A, at the end of the heating-up time, is a function of the wall thickness, e : A = 0.5 mm + 0.1 x e 5. Heating Soak Pressure (low) – P2 The heating soak pressure is normally zero and shall be no more than 0.01 N/mm2 6. Heating Soak Time – t2 The heating soak time is a function of the wall thickness, e : t2 = 15e ± e (sec.) 7. Change-over Time - ∆t The change-over time, is a function of pipe diameter, de : ∆t ≤ 3 sec. + 0.01 de (sec.) 8. Pressure Build-up Time – t3,1 The pressure build-up time, is a function of pipe diameter, de : t3,1 ≤ 3 sec. + 0.03 de (sec.) 9. Welding Pressure – P3 The welding pressure shall be P3 = 0.18 N/mm2 ± 0.01 N/mm2 10. Cooling Pressure – P4 The cooling pressure shall be P4 = 0.18 N/mm2 11. Cooling Time Under Pressure – t3,2 The minimum cooling time shall be t3,2 = 10 + 0.5e (minutes) During the cooling time, the pipe and welding structure shall rest completely and not be subjected to any movement, in any direction. For bigger projects we recommend to carry out a welding procedure and confirm the parameters by destructive tensile tests.

B.2.2 Welding capacity These are guideline figures based on eight-hours working days.

Dimension mm

Number of welds per day,

> SDR26

Number of welds per day, < SDR22

Dimension mm

Number of welds per day, > SDR26

Number of welds per day, < SDR22

1600 1400 1200 1000 900 800 710 630 560 500 450 400

2 2 3 3 4 4 5 6 7 7 8

10

- - 3 3 4 4 5 6 7 7 8

10

355 280 250 225 200 180 160 140 125 110 90 75

10 14 16 18 20 22 22 22 25 25 25 26

10 14 16 17 18 18 20 20 22 25 25 25

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Pipelife Norge AS


Fig. B.2.2 shows a welding machine for big diameters. Fig. B.2.2 BW 1600 butt fusion welding machine.

B.3 Installation As described in previous chapters a submarine pipeline will normally be installed as a combination of trench installation and directly laying on the seabed installation.

B.3.1 Buried PE pipes Whenever the water tables is higher than the centre of a PE pipe, the pipe may be subjected to buoyancy forces when it is partly filled with water, as illustrated in fig. B.3.1. The buoyancy forces must be overcame by the backfilling and the concrete weights. Fig. B.3.1 PE pipe in underwater trench The backfill materials on top of pipe combined with the concrete weights provide the weight that counterbalances the uplift due to buoyancy, preferable with a safety factor not less than 2. Note that the specific gravity of soil is diminished when it is submerged in water :

γsea = γair - γW γsea = specific gravity in sea γair = specific gravity in air γW = specific gravity of water

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Compaction under water is not possible. If a compaction of more than DPR = 85 % is required, it can be attained by using gravel with a natural degree of compaction in the range of : DPR = 85 – 90 %. If a pipe is laid in ground that has a constant water table close to the surface, which is the fact for a sub marine pipeline, it can not be covered to counteract uplift. So it should be weighted using ordinary concrete weights (2 half-pipes, mounted with bolts), ref. A.4. Concrete weights are fixed to the pipe, which is filled with air and floated into position on water filled in an open-cut trench, as shown in fig. B.3.2. When in position, the pipe is filled with water and will sink to bottom of trench. The backfilling can now start. If the water is deep and the trench can not be seen from surface position, the route has to be marked with buoys. Fig. B.3.2 Installation in trench Trenching in soft soil under water may be done using air or water jets to remove material, which is then sucked up while the trench is flushed. Excavators on barge is however more efficient and has greater capacity. The trench depth depends on pipe diameter. Recommended values of H, for normal conditions (see fig. B.3.3) : Do ≤ 500 mm H = 1.75 → 2.0 m Do ≤ 1000 mm H = 2.25 → 2.5 m Do > 1000 mm H = 2.5 m

Fig. B.3.3 Trench depth in soil

Sea bed material (deposits) or gravel should beit should be restored to its original condition. Ochanged profile. In areas where the seabed is exposed to erosioprotection. Fig. B.3.4 illustrates trenching in rock. Rock tretrenching in soft soil.

used for backfilling. After the pipe is laid, bed above therwise, waves and ocean currents will erode the

n, gabions filled with gravel should be used for

nching is considerably more expensive than

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Rock is often located in the surf zone close to shore. To protect the pipe, it is recommended that the top of a trench shall be covered with a layer of concrete cast under water. The layer of concrete should be reinforced and anchored as illustrated in fig. B.3.4. Otherwise the lifting forces generated by wave action might remove the concrete. As an alternative, gabions filled with gravel can be u The pipe should have a backfill of gravel or crushedsay d = 22 → 32 mm.

B.3.2 Pipe laying on seabed This procedure has briefly been described and calcBefore sinking there has to be worked out a sinkingconditions that can occur during installation. A submarine pipeline is built by welding individual 1"section" or by continuous extension of long lengthsA string or section should be as long as possible, bavailable at the site. In general, lengths practically L (section) = 500 → 700 m When produced in long lengths, each string or sect L (section) = 500 → 700 m depending on pipe diameter and the towing conditio The pipes should be weighted with concrete weightis launched into the sea, or on a barge if it is deliverPipes towed to a job site should be stored in a floatwaves, and the sections should be securely anchorSkilled Marine Contractors can fix concrete weights Fig. B.3.5 and B.3.6 shows schematic how a subma

Fig. B.3.4 Trenching in rock

sed for the same purpose.

rock, with a high degree of self compaction,

ulated in chapter 0.2 and A.5. procedure taking into account all relevant

0 m to 25 m lengths of pipe into a string or at the factory.

ut its overall length depends on the space possible to handle are :

ion could have an overall length of :

ns (open sea, weather conditions, etc).

s. The weights can be attached before the pipe ed in long lengths. ing position, at a location protected from wind and ed. onto the floating length of pipeline.

rine pipeline is installed on seabed.

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F

PipeNormfromcan Aftethe sectbuck ThecontdeeShosinkbaseDuri 1. 2. 3.

1

Fig. B.3.5 Installation of a transit pipeline in principal

Fig. B.3.6 Installation of an intake or an outfall pipeline in principal.

line systems are sunk in the direction from shore to the outer end, or other side of water. ally the whole pipe is sunk in one operation. If the ambient conditions are adverse and vary

calm to rough sea, the pipe could be installed in sections, step-by-step. For practical reasons it also be convenient to install the pipe in sections.

r one section is sunk, its sealed flange end rests on the sea bed. In a period with calm weather, pipe is filled with air to lift its end to the surface, and the sinking proceeds, as soon as the next ion in connected to the flange. During connection there must be applied a pulling force to avoid ling.

re are also other methods to connect pipe sections in a step-by-step sinking. Dependent on the ractor’s resources over and below sea-level, the connection can take place at seabed, semi-p or in surface position. rt pipe pieces may be needed for connection between section ends on seabed dependent on the ing method. Divers install those pieces after the sinking has taken place and the lengths are d on exact surveying.

ng the sinking of pipe, water can be filled in one of the following ways :

Blind flange with a valve located onshore, and water delivered by a pump or from a water main.

Blind flange with a valve located offshore, at a depth of 2 to 5 meters, with water filled directly by opening the valve

Blind flange with valve in the inlet/outlet chamber, pipe connected to separate flange in chamber wall.

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The sinking speed, v, should be regulated and should not be greater than the calculated vmax (ref. A.5.3). In the outmost end there must be a valve arrangement for air release and pressure control and a compressor for air filling. The pulling force, F1, and the internal air pressure should be applied according to the calculations in chapter A.5.1 and A.5.2. The sinking speed shall be checked and recorded during sinking. If the speed is too high the internal pressure must be adjusted. The pulling force can be constant or adjusted to the water depth. Before sinking, ensure that : - all bolts are re-tightened to their final torque. This applies to bolts for the concrete weights and

the bolts for flange connections. - all concrete weights are at their correct locations/positions ; verify by measurement - all ancillary devices needed are on hand, including :

- air pressure gauge, increments of 0.01 bar - water valves of suitable diameter - blind flanges fitted with air valves, 1" to 2" (in/out) - water valve (in/out) - air compressor of sufficient capacity and pressure - eventually water pump of sufficient capacity and pressure

A tug boat or other vessel should be available to supply the necessary pulling force. Its engine power at full speed must be known accurately, to within 10 %. Small tug boats or other vessels may be used to provide transverse control of the floating pipe as it is positioned along the route. Experience shows that sinking of a PE pipeline is normally "a piece of cake" if the planning is good, the resources sufficient and weather conditions taken into consideration.

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Pipelife Norge AS


Author : Tom A. Karlsen, Interconsult ASA

List of references : [1] Lars-Eric Janson and Borealis Plastics Pipes for Water Supply and Sewage Disposal, 1995 [2] Svensk Byggetjänst 1983 ISBN 91-7732-186-9 Rörbok yttre rörledningar [3] Frank M.White Fluid Mechanics, 1986 ISBN 0-07-069673-X [4] Lars-Eric Janson Plaströr i VA-tekniken, 1971 [5] Aksel Lydersen Kjemiteknikk, 1972 ISBN 8251900085 [6] PRA 5.1, Vassdrags- og havnelaboratoriet Ytre krefter på utslippsledninger, 1975 [7] AS Byggemästarens Förlag Bygg, huvuddel 1A og 1B Allmänna grunder, 1971 [8] Lars-Eric Janson and Jan Molin Design and installation of Buried Plastic Pipes, ISBN 87-983636-0-3 [9] Thor Meland Dimensjonering av fleksible rør i senkefasen og under drift NIF-kurs 1979 [10] NOU 1974:40 Rørledninger på dypt vann [11] Mabo, Einar Gann-Meyer Polyethylene Pipe Systems Handbook, 1997 [12] Ian Larsen: Marine Waste Water Discharges 2002 Design of marine PE pipes for transient and long-term under-pressure [13] Ian Larsen: Marine Waste Water Discharges 2000 Controlling and installation of marine outfalls of large diameters PE pipes [14] Ian Larsen: Senking av HDPE ledningar. Dykarseminarium 1999, Bergen.

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[15] Torstein Langgård: Installation of offshore pipelines, sinking procedure, Sogusku – Kumkøy, 1998

PLEASE NOTE: This catalogue is presented only as information, with no any obligation or responsibilityon Pipelife side.

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Pipelife Norge AS · PDF filePE CATALOGUE-SUBMARINE APPLICATIONS, PIPELIFE NORGE AS, ... The outfall usually starts from an outfall chamber at the waterfront to which the