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PE Pipe Technical Catalogue (PDF) - Pipelife Norge AS

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<strong>Pipe</strong>life <strong>Norge</strong> <strong>AS</strong><strong>PE</strong> CATALOGUE-SUBMARINE APPLICATIONS, PI<strong>PE</strong>LIFE NORGE <strong>AS</strong>, December 2002.d) Breaking depth :Formula A.4-25) gives an estimate of the breaking depth.a) ⇒ h b = 0.05 ⋅ 44 m = 2.2 mb) ⇒ h b = 0.05 ⋅ 53 m = 2.6 mThe 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 6Consider the wave data calculated in example 5 by the wind statistic method (H o = 3.8 mL o = 44 m).Fig. A.4.6.7 shows that the waves speed direction hits the perpendicular to the contour intervalsunder an angle α o = 45º .An outfall pipeline Ø 500 mm <strong>PE</strong> is installed perpendicular to the contour lines, it means β = 0ºThe 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/m 3α o = 45 oWave frontTrench-20-15-10β= 0 o -50Outfall ChamberFig. A.4.6.7 Outfall pipeline attacked by wavesSolution :h 20We apply fig. A.4.6.4, α = 45º, = = 0. 45L 44This gives a refraction factor : f = 0.1oThe 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 :Fiπ ⋅0.5= π ⋅3.3⋅0.1⋅10000⋅423.8⋅ N/m44=180 N/mDrag force :FD22 π ⋅0.5= 1⋅0.1⋅10000⋅43.8 3.8⋅ ⋅ N/m44 0.5=15 N/mSide 64 av 84

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