Effects of nonlinear wave loads on large monopile offshore wind turbines with and without ice-breaking cone configuration

Abstract

In the present paper, the computational fluid dynamics (CFD) method is used to investigate the variation of linear and nonlinear wave loads on a 10-MW large-scale monopile offshore wind turbine under typical sea conditions in the eastern seas of China. The effect of adding a structural ice-breaking cone configuration close to the mean water level on the monopile’s hydrodynamic response is studied further. Results are derived with the use of the CFD model and are compared with the relevant results that are calculated using the Morison equation and the potential flow theory based on the high-order boundary element method. The fifth-order Stokes’ theorem is used to model the incoming wave kinematics, and the volume of fluid (VOF) method is used to capture the free surface of waves and to accurately calculate the wave run-up on the monopile and cone configuration. The influence of different water depths and wave heights on the wave maximum vertical extent of wave uprush on the structure, pressure and horizontal wave forces on the monopile is investigated for both with and without the use of the cone configuration. Up–downward cone configuration results in better performance compared to the inverted cone configuration in terms of reduction of hydrodynamic nonlinear excitation loads and wave maximum vertical extent of wave uprush on the structure.

Appendix A

Equations related to fifth-order stokes waves

$$s = \sinh \left( {kd} \right),$$

$$c = \cosh \left( {kd} \right),$$

$$A_{11} = {\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 s}}\right.\kern-0pt} \!\lower0.7ex\hbox{$s$}},$$

$$A_{13} = \frac{{ - c^{2} \left( {5c^{2} + 1} \right)}}{{8s^{5} }},$$

$$A_{15} = \frac{{ - \left( {1184c^{10} - 1440c^{8} - 1992c^{6} + 2641c^{4} - 249c^{2} + 18} \right)}}{{1536s^{11} }},$$

$$A_{22} = \frac{3}{{8s^{4} }},$$

$$A_{24} = \frac{{\left( {192c^{8} - 424c^{6} - 312c^{4} + 480c^{2} - 17} \right)}}{{768s^{10} }},$$

$$A_{33} = \frac{{\left( {13 - 4c^{2} } \right)}}{{64s^{7} }},$$

$$A_{35} = \frac{{\left( {512c^{12} + 4224c^{10} - 6800c^{8} - 12808c^{6} + 16704c^{4} - 3154c^{2} + 107} \right)}}{{4096s^{13} \left( {6c^{2} - 1} \right)}},$$

$$A_{44} = \frac{{\left( {80c^{6} - 816c^{4} + 1338c^{2} - 197} \right)}}{{1536s^{10} \left( {6c^{2} - 1} \right)}},$$

$$A_{55} = \frac{{ - \left( {2880c^{10} - 72480c^{8} + 324000c^{6} - 432000c^{4} + 163470c^{2} - 16245} \right)}}{{61440s^{11} \left( {6c^{2} - 1} \right)\left( {8c^{4} - 11c^{2} + 3} \right)}},$$

$$B_{22} = \frac{{\left( {2c^{2} + 1} \right)}}{{4s^{3} }}c,$$

$$B_{24} = \frac{{c\left( {272c^{8} - 504c^{6} - 192c^{4} + 322c^{2} + 21} \right)}}{{384s^{9} }},$$

$$B_{33} = \frac{{3\left( {8c^{6} + 1} \right)}}{{64s^{6} }},$$

$$B_{35} = \frac{{\left( {88128c^{14} - 208224c^{12} + 70848c^{10} + 54000c^{8} - 21816c^{6} + 6264c^{4} - 54c^{2} - 81} \right)}}{{12288s^{12} \left( {6c^{2} - 1} \right)}},$$

$$B_{44} = \frac{{c\left( {768c^{10} - 448c^{8} - 48c^{6} + 48c^{4} + 106c^{2} - 21} \right)}}{{384s^{9} \left( {6c^{2} - 1} \right)}},$$

$$B_{55} = \frac{{\left( {192000c^{16} - 262720c^{14} + 83680c^{12} + 20160c^{10} - 7280c^{8} + 7160c^{6} - 1800c^{4} - 1050c^{2} + 225} \right)}}{{12288s^{10} \left( {6c^{2} - 1} \right)\left( {8c^{4} - 11c^{2} + 3} \right)}},$$

$$C_{1} = \frac{{8c^{4} - 8c^{2} + 9}}{{8s^{4} }},$$

$$C_{2} = \frac{{3840c^{12} - 4096c^{10} - 2592c^{8} - 1008c^{6} + 5944c^{4} - 1830c^{2} + 147}}{{512s^{10} \left( {6c^{2} - 1} \right)}}.$$