﻿ 随机波浪下ROV缆索动力响应无因次分析
 舰船科学技术  2016, Vol. 38 Issue (8): 43-46 PDF

Non-dimensional dynamic response of ROV system under excitation of random wave
HUANG Zhu-lin, YAO Bao-heng, ZENG Zheng
State Key Laboratory of Ocean Engineering, Shanghai Jiaotong University, Shanghai 200240, China
Abstract: Develop a single-degree-of-freedom model for the cable-body system to analyze the dynamic response. By making the differential equation non-dimensional, the relation between excitation and drag coefficient with the response is analyzed. Non-dimensional PM Spectrum with dual parameter is used as the excitation. The differential equation is solved by using Runge-Kutta iterative algorithm to get relation between the largest tension and parameters above. These results are compared with the response under harmonic excitation. Results show that the largest tension has multiple peak values, and maintains at a high level for a wide range, while the peak tension of cable is smaller than the response under harmonic excitation.
Key words: cable     dynamic response     non-dimensional     random wave
0 引言

1 缆索动力学模型 1.1 ROV作业系统运动模型

ROV作业系统，由母船、缆索和ROV三部分组成，如图 1所示。考虑ROV作业时缆索的动态响应，建立一维垂向运动方程，在缆索张紧阶段，ROV受到缆索的张力与非线性的流体阻力，在缆索松弛阶段，缆索张力为0。缆索上端与母船相连，受到波浪提供的激励项f，为便于分析，忽略缆索张力对母船运动的影响并假定母船运动与波浪运动一致。可以得到ROV作业时的运动方程：

 图 1 ROV-母船作业系统示意图 Fig. 1 Sketch of cable-body system
 $m\ddot x + c\dot x|\dot x| + kx = w + kf\text{，}$ (1)

 $k = \left\{ {\begin{array}{*{20}{c}} {\frac{{ES}}{L}\;\;\;x-f > 0}\text{，} \\ {0\;\;\;x-f \leqslant 0} \text{。} \end{array}} \right.$

1.2 运动方程无因次化

 $\begin{array}{*{20}{c}} {y = \frac{Y}{{0.5H}},\;\;{y_s} = \frac{w}{{0.5HES/L}},\;\;\tau = \frac{{2\pi t}}{T},}\\ {{\omega _n} = \sqrt {\frac{{ES/L}}{m}} ,\;\;\eta = \frac{{2\pi }}{{T{\omega _n}}},\;\;\zeta = \frac{{ca\pi }}{{T\sqrt {mES/L} }}。} \end{array}$

 \begin{aligned} \frac{{{{\rm d}^2}y}}{{{\rm d}{\tau ^2}}} + & \frac{{2\zeta }}{\eta }\left| {\frac{{{\rm d}y}}{{{\rm {\rm d}}\tau }} + \frac{1}{{0.5H}}\frac{{{\rm d}f}}{{d\tau }}} \right|(\frac{{{\rm d}y}}{{{\rm d}\tau }} + \frac{1}{{0.5H}}\frac{{df}}{{{\rm d}\tau }}) + \\ & \frac{1}{{{\eta ^2}}}y + \frac{{{{\rm d}^2}f}}{{0.5Hd{\tau ^2}}} = \frac{1}{{{\eta ^2}}}{y_s}\;\;\;\;\;\;y > 0\text{；} \end{aligned} (2)
 \begin{aligned} \frac{{{{\rm d}^2}y}}{{{\rm d}{\tau ^2}}} + & \frac{{2\zeta }}{\eta }\left| {\frac{{{\rm d}y}}{{{\rm d}\tau }} + \frac{1}{{0.5H}}\frac{{{\rm d}f}}{{{\rm d}\tau }}} \right|(\frac{{dy}}{{{\rm d}\tau }} + \frac{1}{{0.5H}}\frac{{{\rm d}f}}{{{\rm d}\tau }}) + \\ & \frac{{{{\rm d}^2}f}}{{0.5H{\rm d}{\tau ^2}}} = \frac{1}{{{\eta ^2}}}{y_s}\;\;\;\;\;\;y \leqslant 0 \text{。} \end{aligned} (3)

2 随机波浪数值模拟

 $f(t) = \sum\limits_{i = 1}^\infty {{a_i}\cos ({\omega _i}t + {\varepsilon _i})} \text{。}$ (4)

 ${a_i} = \sqrt {2S({\varpi _i})\Delta {\omega _i}} \text{。}$ (5)

 $S(\omega ) = \frac{{173H_{1/3}^2}}{{{\omega ^5}T_p^4}}\exp \left( {-\frac{{691}}{{{\omega ^4}T_p^4}}} \right)\text{。}$ (6)

 ${\omega _i} = {k_i}\frac{{2{\rm{\pi }}}}{{{T_P}}}。$ (7)

 $\bar f = \frac{f}{{0.5H}} = \sum\limits_{i = 1}^\infty {\sqrt {\frac{{173}}{{2k_i^5{{\rm{\pi }}^4}}}\exp \left( {\frac{{ - 691}}{{{{(2{\rm{\pi }})}^4}k_i^4}}} \right)\Delta \eta } \cos ({k_i}\tau + {\varepsilon _i})} 。$ (8)

 图 2 无因次随机波浪时域历程图和波浪谱 Fig. 2 Time history of non-dimensional random wave and wave spectrum
3 数值仿真 3.1 时域分析

 图 3 ζ=0.1，η=0.3，ys=10时，缆索伸缩量和ROV位移时域历程图 Fig. 3 Time history of cable's tension and ROV's displacement(ζ=0.1, η=0.3, ys=10)
3.2 频域分析

 图 4 ys=1和ys=4时，随机激励下缆索最大、最小张力与阻尼比、频率比的关系 Fig. 4 Tension of cable under random excitation for various η and ζ with ys=1 and ys=4

 图 5 ys=1和ys=4时，谐波激励下缆索最大、最小张力与阻尼比、频率比的关系 Fig. 5 Tension of cable under harmonic excitation for various η and ζ with ys=1 and ys=4

4 结语

1）缆索的最大张力随阻尼的增大而减小，在随机波浪作用下，频域内最大张力有多个峰值，其中不管缆索多长，在固有频率附近的峰值十分明显；频率较大时，由于随机波浪中包含了不同频率的谐波成分，共同作用下，在很大的频率范围内最大张力保持较大的数值。

2）相比谐波激励，随机激励下缆索的最大张力在频域的分布更为分散，虽然峰值较小，但不安全的频率范围大得多，因此在真实海浪中作业时，为避开缆索张力大的频率增加了难度，为ROV作业系统设置补偿系统减小张力十分必要。

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