tutorial tlp offshore

8/11/2019 Tutorial TLP Offshore

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By Mazlan Muslim, MEng

UniKL MIMET


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KC No. The Keulegan Carpenter no. is a measure for the

viscous effects. An increasing KC value means thatflow separation becomes important.

KC = UmT/D

where Um the maximum velocity, T the wave period &D the cylinder diameter.

For the wave experiment the diffraction number isalways smaller than 0.2. The cylinder can beconsidered a small body with respect to the wavelength. The KC no. ranges from 3 < KC < 25.


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KC Value The KC value corresponds with velocities determined

by nonlinear stream function wave models. The valuesof the KC no. mean that both the drag & inertiacomponents are important for the horizontal waveforce in the wave experiment.

The max. wave force range from 16 to 166 kN. The min.

forces work in opposite wave direction. The min. forcesrange from -9 to -69 kN.

The wave loads on the cylinder are calculated usingMorrison equation.


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Morisons Equation The wave loads are calculated by Morisons equation in

conjunction with linear Airy wave theory. Morisons equationexpressed the wave force as a summation of inertia force & dragforce:

F = Fi + Fd F = (CmD**2/4)u + (CdD/2)IuIu Where F is the wave force; Fi inertia force Fd drag force Cm, Cd inertia & drag coefficients IuI and uare water particle velocity normal to the cylinder uwater particle acceleration seawater density, D member diameter


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Water Particle The water particle kinematics are determined by the

following equations: Horizontal water particle velocity:

u = ((H coshks)/(T sinhkd))cos Horizontal water particle acceleration: u = ((2**2H coshks)/(T sinhkd))sin where s = z + d = kx t kwave no. (2/L); Twave period zheight of the point of evaluation of wave particle kinematics


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Contd x is point of evaluation of water particle kinematics

from the origin in the horizontal direction;

t is time instant at which water particle kinematics isevaluated;

L is wave length;

H is wave height;

d water depth.


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Problem 1A TLP is moored over a well in 120 m of water offshore.

It has a vertical rigid marine riser (diameter of 0.70 m)that is subjected to wave loads. For a wave 90 m long &4.5 m high, determine which horizontal wave loadcomponent dominates at each of the following depths:z= 0, -5, & -10 m. Base the comparison on a unitlength of riser at each depth. Assume that the drag &added mass coefficients are both 1. State anyassumptions you make & show all work to justify youranswer.


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SolutionWe recognize that the max. drag force will occur under

the crest (at a phase angle of /2) & the max. inertiaforce will occur at a phase angle of .

Wave no.: k = 2/L = 2/90 = 0.06981

Calculate the wave frequency note that the tanhterms approaches 1 because the water is actually deep,

as indicated by our rule of thumb guide that says thewater is not deep if d/L < 0.5. Here we have d/L =120/90. We can still use the intermediate water depthequations, but they do simplify:


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Contd **2 = kgtanhkd= 0.06981 x 9.806 x 1 = 0.6845

= 0.8274

Calculate the wave no. k = 2/L = 2/90 = 0.06981

Calculate the Keulegan-Carpenter no. Kc= H/D

= x 4.5/0.70 = 20.2

It tells us the drag & inertia loads will be about

compatible. We simply have to calculate the loadcomponents at each depth & compare them we needonly consider the max. loads for each component & weknow where these occur.


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Contd Max. drag occurs at = /2 where max. horizontal

wave particle velocity occurs.

Max. inertia occurs at = where max. horizontalwave particle acceleration occurs.

Plug into the equations & solve for velocities &accelerations at each depth


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Contd max u max du/dt dF drag dF inertia

1.862 1.540 1243 1215

1.313 1.086 619 857 0.926 0.766 308 605

The load components are of similar magnitude: drag isslightly higher at the top, but as it decays more quicklythan the inertia force, the latter becomes the greater atthe other 2 depths.

tutorial tlp offshore

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