Offshore pipeline riser calculation

Universidad de los Andes

Introducción a la ingeniería offshore – Subsea Infrastructure

Sergio Antonio Muñoz Pinzón

November, 2015

1.    The drillship Transocean Deepwater Discovery has a drilling riser with the following properties:

Length: 2000 m

External diameter: 21,25”

Wall width: 0,812”

Young’s modulus: 2,07e+11 Pa

Axial force: 9e+6 N

Assuming the whole riser is immersed in fresh water in a water depth of 2000. A constant current (same direction and velocity) of 2 knots is loading the whole riser. It is necessary to take in account the axial force. However, the effects of    gravity, internal and external hydrostatic pressure can be neglected.

Plot the graphs of (i) deflection, (ii) angle of rotation, (iii) shear force and (iv) bending moment along the whole drilling riser for the following cases:

a.       Riser with both ends fixed

b.      Riser with both ends equipped with ball joints

For both cases a. and b. the riser can be assumed to be a tensioned beam,

Where E is the elasticity modulus, I the area moment of inertia, v or x is the deflection of the riser measured from the neutral axis, T is the axial pretension load, and q an external load applied on the riser by the current.

Yamamoto has solved the differential equation. 

In this formula there are four constants name C₁, C₂, C₃, C₄ that should be determined with the boundary conditions.

Before going to solve the cases it is necessary to find the parameters of the equations, the external load q can be determined with the Morison’s equation

To solve this equation we assume no inertial forces (C_m=0), and C_d can be determined from diagrams as a function of Reynolds number Re.

Current velocity, U	1,029	m/s (2 knots)
Density, rho	1.000,0	kg/m3 Fresh water
Viscosity, nu	1,31E-06	m2/s Fresh water 10°C
Re	4,25E+05	Re=UD/ν
Cd	1,0	From Diagram
q	285,7	N/m
E	2,07E+11	N/m2
D	5,40E-01	m
w(thickness)	2,06E-02	m
I	1,13E-03	m4
T	9,00E+06	N
Length	2000	m
Lambda	0,195731448	1/m

a)      Riser with both ends fixed;

In this paper the system of equations 6 was solve with Engineering Equation Solver EES® by F-Chart Software. LLC.

C1=-0,00121743208

C2=0,00618813269

C3=0,162353970

C4=1,58594097E-171

Now it is possible to plot the main variables of the riser as a beam

Deflection

Angle of rotation

Moment of bending

Shear Force

a)      Riser with both ends equipped with ball joints.

The equations 9 were solved with EES to find:

C1=-0,00121743208

C2=-4,87559171E-22

C3=0,000829473096

C4=8,10263751E-174

Now it is possible to plot the main variables of the riser as a beam

Deflection

Angle of Rotation

Bending Moment

Shear Force

c.       Conclusion

For both cases, the deflection and angle are very similar; however, the bending moment and shear force are significantly lower in case b, which is the riser with ball joints who bears less stresses, especially at the ends.

    References

Yamamoto, M. A study about the dynamic behavior of flexible tubes including internal flow. 2001. Doctoral Thesis. Graduate School of Environment and Information Sciences, Yokohama              National University, Yokohama, Japan, 2011.