海洋科学  2018, Vol. 42 Issue (8): 62-70   PDF    
http://dx.doi.org/10.11759/hykx20171228001

文章信息

翟振峰, 黄华, 詹杰民, 张敖. 2018.
ZHAI Zhen-feng, HUANG Hua, ZHAN Jie-min, ZHANG Ao. 2018.
双层直立圆弧型防波堤上绕射波浪力的解析计算
Analysis of diffracted wave force on a double-layered vertical arc-shaped breakwater
海洋科学, 42(8): 62-70
Marina Sciences, 42(8): 62-70.
http://dx.doi.org/10.11759/hykx20171228001

文章历史

收稿日期:2017-12-28
修回日期:2018-03-27
双层直立圆弧型防波堤上绕射波浪力的解析计算
翟振峰, 黄华, 詹杰民, 张敖     
中山大学 工学院应用力学与工程系, 广东 广州 510275
摘要:基于微幅波绕射理论, 应用特征函数展开法, 推导了双层直立圆弧型透空防波堤的波浪绕射解析解, 从而将已有的比例边界有限元法拓展为解析算法, 并据此对外层与内层防波堤所受波浪载荷进行了解析计算。计算结果表明:应用本文方法对直立透空圆环柱的绕射波浪载荷进行验证计算, 所得结果与现有的解析解完全吻合, 说明方法可靠。双层堤较单层堤能更有效地减弱波浪作用。波浪的入射角度和特征参数、防波堤张角与半径、防波堤透空系数以及水深等因素的相对变化对双层堤的波浪作用均存在一定影响。
关键词微幅波理论    特征函数展开法    双层圆弧型透空防波堤    波浪力    波浪绕射    
Analysis of diffracted wave force on a double-layered vertical arc-shaped breakwater
ZHAI Zhen-feng, HUANG Hua, ZHAN Jie-min, ZHANG Ao     
Department of Applied Mechanics and Engineering, College of Engineering, Sun Yat-sen University, Guangzhou 510275, China
Abstract: Based on the small amplitude wave diffraction theory, we derive an analytical solution for the wave diffraction of a double-layered vertical arc-shaped porous breakwater. To do so, we apply the eigenfunction expansion approach, in which we extend the existing scaled-boundary finite element method to an analytical algorithm and calculate the wave loads on the outer and inner breakwaters. The evaluation results reveal that the obtained wave force on a vertical concentric porous cylinder is in good agreement with the existing analytical solution. This indicates that the method we present in this paper is reliable and that a double-layered breakwater can weaken the wave effect more effectively than a single-layer breakwater. Variations in the incident wave angle and wave characteristic parameters, breakwater angle and radius, porous coefficient, and water depth may have some influence on the wave effects.
Key words: small amplitude wave    eigenfunction expansion approach    double-layered arc-shaped porous breakwater    wave force    wave diffraction    

防波堤是用于抵抗波浪作用以保护海工设施和近岸作业水域的重要水工结构之一, 其中圆弧形防波堤为常见的设计与建造形式。Wiegel[1]研究了半无限长薄壁直立防波堤的波浪绕射的相关问题。黄培基等[2]给出了波浪与直立堤的作用下的立波压力的计算方法。Losada等[3]解析研究了波浪与位于有限水深中的垂直薄板的相互作用问题。Abul-azm等[4]以及Mciver[5]分别应用解析法和积分方程法计算了无限长分段式等距直立防波堤的波浪绕射作用。李雪艳等[6]研究了斜防波堤典型胸墙波浪力的影响因素。程建生等[7]和楚玉川等[8]对单层圆弧型贯底式薄壁防波堤的防浪效果和绕射波浪力进行了解析研究。张敖等[9-10]分别引入椭圆余弦波模型和孤立波模型推导了浅水波对单层圆弧型防波堤绕射问题的解析解。

研究表明, 对防波堤表面布孔可以降低波浪对防波堤的荷载作用。Chwang[11]研究了透空造波机的理论, 给出了波浪与透空薄板的相互作用关系。Yu[12]计算了波浪对半无限长直线型薄壁防波堤的绕射作用。Darwiche等[13]与Williams等[14]分别解析求解了部分透空圆柱型防波堤对波浪绕射的波势解。Duan等[15]对单层圆弧型浮式透空防波堤的防浪效果进行了解析研究。林皋和刘俊[16]应用比例边界有限元法对双层圆弧型贯底式开孔介质防波堤的水波绕射进行了计算。

基于微幅波的水波绕射理论, 运用特征函数展开法, 本文给出了波浪对双层直立圆弧型透空防波堤绕射问题的解析解, 从而将现有的双层堤比例边界有限元法以及现有的单层堤波浪绕射解析法同时加以了拓展。此外, 双层圆弧堤的波浪绕射解是对仅有的双圆筒(即圆环柱)波浪绕射解的重要拓展, 其所涉及的问题和解法更为复杂, 适用的结构类型更为广泛。

1 双层防波堤的绕射波浪场求解

图 1所示, 在水深为d的均匀海中, 设置有外径和内径分别为ab且具有同圆心角的双层直立圆弧型透空防波堤。外层堤透空系数为G1, 内层堤透空系数为G2。建立坐标系Oxyz (即坐标系Orθz), 令Oxy平面位于水底, 原点位于圆弧型防波堤的圆心处, 设置Ox轴使内外层防波堤对称位于θ=αθ=2π–α间, 外层和内层防波堤相应的圆心角为γ= 2π–2α, Oz轴竖直向上。设入射波为Airy微幅波, 波高为H, 入射角(与Ox轴正向夹角)为β, 频率为ω。分别以ab为半径划一同心圆环柱面, 将流场划分为圆柱外流区Ω0和圆柱内流区Ω1Ω2

图 1 双层圆弧型透空防波堤 Fig. 1 Double-layered arc-shaped porous breakwater

对于大尺度海工结构, 海水可视为无粘性的不可压缩流体且运动无旋。设对应流域Ωj的Airy微幅波波总波势为${\varPhi ^{(j)}} = {\varPhi _i} + \varPhi _s^{(j)} = {\mathop{\rm Re}\nolimits} ({\varphi ^{(j)}})$$(j = 0, 1, 2)$, 则水波绕射速度势的边值问题为:

$ {\nabla ^2}{\varphi ^{(j)}} = {\nabla ^2}\varphi _s^{(j)} = 0\, \, \, \, (j = 0, 1, 2) $ (1)
$\frac{{\partial {\varphi ^{(j)}}}}{{\partial z}} = \frac{{\partial \varphi _s^{(j)}}}{{\partial z}} = 0\, \, \, \, \, (z = 0)$ (2)
$ g\frac{{\partial {\varphi ^{(j)}}}}{{\partial z}} - {\omega ^2}{\varphi ^{(j)}} = g\frac{{\partial \varphi _s^{(j)}}}{{\partial z}} - {\omega ^2}\varphi _s^{(j)} = 0\, \, \, \, \, (z = d) $ (3)
$ \mathop {\lim }\limits_{r \to \infty } {r^{\frac{1}{2}}}\left( {\frac{{\partial \varphi _s^{(0)}}}{{\partial r}} - ik\varphi _s^{(0)}} \right) = 0 $ (4)

式中, ${\varphi _i}$$\varphi _s^{(j)}$分别为入射波势和散射波势; 公式(4)为无穷远处的辐射条件。另外${\varphi ^{(j)}}$需要满足内外防波堤透空表面条件。

线性平面入射波势可表示为:

$ \begin{array}{l} {\varPhi _i} = {\mathop{\rm Re}\nolimits} ({\varphi _i}), \\ \begin{array}{*{20}{c}} {{\varphi _i} = } \end{array} - \frac{{igH}}{{2\omega }}{{\rm{e}}^{ - i\omega t}}\frac{{\cosh kz}}{{\cosh kd}}[\sum\limits_{n = 0}^\infty {{\beta _n}} {J_n}(kr)\cos n\theta \cos n\beta \\ \, \, \, \, \, \, \, \, \, + \sum\limits_{n = 0}^\infty {{\beta _n}} {J_n}(kr)\sin n\theta \sin n\beta ] \end{array} $ (5)

基于波浪绕射理论, 应用特征函数展开法, 可取:

$ {\varphi ^{(1)}} = - \frac{{igH}}{{2\omega }}\frac{{\cosh kz}}{{\cosh kd}}{e^{ - i\omega t}} $ (6)
$ \begin{array}{c} \times \sum\limits_{n = 0}^\infty {\{ [A_n^{(1)}} \cos n\theta + B_n^{(1)}\sin n\theta ]{J_n}(kr)\\ \;\; + [C_n^{(1)}\cos n\theta + D_n^{(1)}\sin n\theta ]H_n^{(1)}(kr)\} \end{array} $ (7)
$ \begin{array}{l} {\varphi ^{(2)}} = - \frac{{igH}}{{2\omega }}\frac{{\cosh kz}}{{\cosh kd}}{e^{ - i\omega t}}\\ \, \, \, \, \, \, \, \, \, \, \;\;\; \times \sum\limits_{n = 0}^\infty {[A_n^{(2)}} \cos n\theta + B_n^{(2)}\sin n\theta ]{J_n}(kr) \end{array} $ (8)

式中, $A_n^{(0)}$$B_n^{(0)}$$A_n^{(1)}$$B_n^{(1)}$$A_n^{(2)}$$B_n^{(2)}$$C_n^{(1)}$$D_n^{(1)}$均为待求系数, Jn(x)和$H_n^{{\rm{(1)}}}(x)$分别为第一类Bessel函数和第一类Hankel函数, 其中取$B_n^{(0)} = $ $B_n^{(1)} = D_n^{(1)} = B_n^{(2)} = 0$

首先考虑外层堤物面和界面条件的应用。外层堤前后侧两流域径向分界面条件以及透空物面条件为:

$ {\varphi ^{(0)}} = {\varphi ^{(1)}}\, \, \, \, \, \, \, \, \, \, (r = a, 0 \le \theta \le \alpha , 2{\rm{ \mathit{ π} }} - \alpha \le \theta \le 2{\rm{ \mathit{ π} }}) $ (9)
$ \frac{{\partial {\varphi ^{(0)}}}}{{\partial r}} = \frac{{\partial {\varphi ^{(1)}}}}{{\partial r}}\, \, \, \, \, \, \, \, \, \, (r = a, 0 \le \theta \le 2{\rm{ \mathit{ π} }}) $ (10)
$ \begin{array}{l} \frac{{\partial \varphi _{}^{(0)}}}{{\partial r}} = \frac{{\partial \varphi _{}^{(1)}}}{{\partial r}} = \frac{{{\gamma _1}\rho }}{\mu }\left( {\frac{{\partial \varphi _{}^{(0)}}}{{\partial t}} - \frac{{\partial \varphi _{}^{(1)}}}{{\partial t}}} \right){\kern 1pt} \, \\ \, \, \, \, \, \, \, \, (r = a, \alpha \le \theta \le 2{\rm{ \mathit{ π} }} - \alpha ) \end{array} $ (11)

式中, ργ1μ分别为海水密度、具有长度量纲的外层防波堤材料系数及粘性系数。由公式(10)可得:

$ \begin{align} & {{\beta }_{m}}\cos m\beta {{{{J}'}}_{m}}(ka)+A_{m}^{(0)}H{{_{m}^{(1)}}^{\prime }}(ka) \\ & =A_{m}^{(1)}{{{{J}'}}_{m}}(ka)+C_{m}^{(1)}H{{_{m}^{(1)}}^{\prime }}(ka)\,\,\,\,\,\,\,\,\,\,(m=0,1,\cdots ) \\ \end{align} $ (12)
$ \begin{align} & {{\beta }_{m}}\sin m\beta {{{{J}'}}_{m}}(ka)+B_{m}^{(0)}H{{_{m}^{(1)}}^{\prime }}(ka) \\ & =B_{m}^{(1)}{{{{J}'}}_{m}}(ka)+D_{m}^{(1)}H{{_{m}^{(1)}}^{\prime }}(ka)\,\,\,\,\,\,\,\,\,\,(m=1,2,\cdots ) \\ \end{align} $ (13)

定义函数:

$ f(\theta ) = \left\{ \begin{array}{l} {f_1}(\theta ), 0 \le \theta < \alpha , 2{\rm{ \mathit{ π} }} - \alpha < \theta \le 2{\rm{ \mathit{ π} }}\\ {f_2}(\theta ), \alpha \le \theta \le 2{\rm{ \mathit{ π} }} - \alpha \end{array} \right. = 0 $ (14)

其中, 由式(9)和式(11)分别得到:

$ \begin{array}{c} {f_1}(\theta ) = \sum\limits_{n = 0}^\infty {\{ [{\beta _n}\cos n\beta {J_n}(ka) + A_n^{(0)}H_n^{(1)}(ka)]\cos n\theta } \\ + [{\beta _n}\sin n\beta {J_n}(ka) + B_n^{(0)}H_n^{(1)}(ka)]\sin n\theta \} \\ - \sum\limits_{n = 0}^\infty {\{ [A_n^{(1)}} \cos n\theta + B_n^{(1)}\sin n\theta ]{J_n}(ka)\\ + [C_n^{(1)}\cos n\theta + D_n^{(1)}\sin n\theta ]H_n^{(1)}(ka)\} \\ = 0\, \, \, \, \, \, \, \, \, \, (0 \le \theta \le \alpha , 2{\rm{ \mathit{ π} }} - \alpha \le \theta \le 2{\rm{ \mathit{ π} }}) \end{array} $ (15)
$ \begin{align} & {{f}_{2}}(\theta )=\sum\limits_{n=0}^{\infty }{\{[A_{n}^{(1)}}\cos n\theta +B_{n}^{(1)}\sin n\theta ]k{{{{J}'}}_{n}}(ka) \\ & \,\,\,\,\,\,\,\,\,\,\,\,+[C_{n}^{(1)}\cos n\theta +D_{n}^{(1)}\sin n\theta ]kH{{_{n}^{(1)}}^{\prime }}(ka)\} \\ & \,\,\,\,\,\,\,\,\,\,\,\,-i\begin{matrix} {{G}_{1}} \\ \end{matrix}\sum\limits_{n=0}^{\infty }{\{[A_{n}^{(1)}}\cos n\theta +B_{n}^{(1)}\sin n\theta ]k{{J}_{n}}(ka) \\ & \,\,\,\,\,\,\,\,\,\,\,\,+[C_{n}^{(1)}\cos n\theta +D_{n}^{(1)}\sin n\theta ]kH_{n}^{(1)}(ka)\} \\ & \,\,\,\,\,\,\,\,\,\,\,\,+i\begin{matrix} {{G}_{1}} \\ \end{matrix}\sum\limits_{n=0}^{\infty }{\{[{{\beta }_{n}}\cos n\beta k{{J}_{n}}(ka)+A_{n}^{(0)}kH_{n}^{(1)}(ka)]\cos n\theta } \\ & \,\,\,\,\,\,\,\,\,\,\,\,+[{{\beta }_{n}}\sin n\beta k{{J}_{n}}(ka)+B_{n}^{(0)}kH_{n}^{(1)}(ka)]\sin n\theta \} \\ & \,\,\,\,\,\,\,\,\,\,\,\,=0\,\,\,\,\,\,\,\,\,\,(\alpha \le \theta \le 2\text{ }\!\!\pi\!\!\text{ }-\alpha ) \\ \end{align} $ (16)

其中, ${G_1} = \frac{{{\gamma _1}\rho \omega }}{{\mu k}}$为外层堤径向表面的透空系数。

$f(\theta )$$0 \sim 2{\rm{ \mathit{ π} }}$上展开为傅氏级数, 可得:

$ f(\theta ) = \frac{{{a_0}}}{2} + \sum\limits_{m = 1}^\infty {({a_m}\cos m\theta + {b_m}\sin m\theta ) = 0} $ (17)

${a_m} = {b_m} = 0$有:

$ \begin{array}{l} \int_{\;0}^{\;\alpha } {{f_1}(\theta )\cos m\theta d\theta + \int_{\;2{\rm{ \mathit{ π} }} - \alpha }^{\;2{\rm{ \mathit{ π} }}} {{f_1}(\theta )\cos m\theta } } d\theta + \\ \int_{\;\alpha }^{\;2{\rm{ \mathit{ π} }} - \alpha } {{f_2}} (\theta )\cos m\theta d\theta = 0\;\;(m = 0, 1, \cdots ) \end{array} $ (18)
$ \begin{array}{l} \int_{\;0}^{\;\alpha } {{f_1}(\theta )\sin m\theta d\theta + \int_{\;2{\rm{ \mathit{ π} }} - \alpha }^{\;2{\rm{ \mathit{ π} }}} {{f_1}(\theta )\sin m\theta } } d\theta + \\ \int_{\;\alpha }^{\;2{\rm{ \mathit{ π} }} - \alpha } {{f_2}} (\theta )\sin m\theta d\theta = 0\;\;(m = 1, 2, \cdots ) \end{array} $ (19)

相应有:

$ \begin{align} & \sum\limits_{n=0}^{\infty }{{{\beta }_{n}}\cos n\beta {{J}_{n}}(ka)[I_{nm}^{(1)}+i\begin{matrix} {{G}_{1}} \\ \end{matrix}kI_{nm}^{(2)}]} \\ & +\sum\limits_{n=0}^{\infty }{A_{n}^{(0)}}H_{n}^{(1)}(ka)[I_{nm}^{(1)}+i\begin{matrix} {{G}_{1}} \\ \end{matrix}kI_{nm}^{(2)}] \\ & +\sum\limits_{n=0}^{\infty }{A_{n}^{(1)}}[k{{{{J}'}}_{n}}(ka)I_{nm}^{(2)}-{{J}_{n}}(ka)(I_{nm}^{(1)}+i\begin{matrix} {{G}_{1}} \\ \end{matrix}kI_{nm}^{(2)})] \\ & +\sum\limits_{n=0}^{\infty }{C_{n}^{(1)}}[kH{{_{n}^{(1)}}^{\prime }}(ka)I_{nm}^{(2)} \\ & -H_{n}^{(1)}(ka)(I_{nm}^{(1)}+i\begin{matrix} {{G}_{1}} \\ \end{matrix}kI_{nm}^{(2)})]=0\,\,\,\,\,\,\,\,\,\,(m=0,1,\cdots ) \\ \end{align} $ (20)
$ \begin{align} & \sum\limits_{n=1}^{\infty }{{{\beta }_{n}}\sin n\beta {{J}_{n}}(ka)[I_{nm}^{(3)}+i\begin{matrix} {{G}_{1}} \\ \end{matrix}kI_{nm}^{(4)}]} \\ & +\sum\limits_{n=1}^{\infty }{B_{n}^{(0)}}H_{n}^{(1)}(ka)[I_{nm}^{(3)}+i\begin{matrix} {{G}_{1}} \\ \end{matrix}kI_{nm}^{(4)}] \\ & +\sum\limits_{n=1}^{\infty }{B_{n}^{(1)}}[k{{{{J}'}}_{n}}(ka)I_{nm}^{(4)}-{{J}_{n}}(ka)(I_{nm}^{(3)}+i\begin{matrix} {{G}_{1}} \\ \end{matrix}kI_{nm}^{(4)})] \\ & +\sum\limits_{n=0}^{\infty }{D_{n}^{(1)}}[kH{{_{n}^{(1)}}^{\prime }}(ka)I_{nm}^{(4)} \\ & -H_{n}^{(1)}(ka)(I_{nm}^{(3)}+i\begin{matrix} {{G}_{1}} \\ \end{matrix}kI_{nm}^{(4)})]=0\,\,\,\,\,\,\,\,\,\,(m=1,2,\cdots ) \\ \end{align} $ (21)

其中:

$ \left\{ \begin{array}{l} I_{nm}^{(1)} = \int_{\; - \alpha }^{\;\alpha } {\cos m\theta } \cos n\theta d\theta , I_{nm}^{(2)} = \int_{\;\alpha }^{\;2{\rm{ \mathit{ π} }} - \alpha } {\cos m\theta } \cos n\theta d\theta \\ I_{nm}^{(3)} = \int_{\; - \alpha }^{\;\alpha } {\sin m\theta \sin n\theta d\theta } , I_{nm}^{(4)} = \int_{\;\alpha }^{\;2{\rm{ \mathit{ π} }} - \alpha } {\sin m\theta \sin n\theta d\theta } \end{array} \right. $ (22)

公式(12)、(13)、(20)和(21)为待定系数完备代数方程组的组成部分, 其中取$B_0^{(0)} = B_0^{(1)} = D_0^{(1)} = 0$

再考虑内层堤物面和界面条件的应用。内层堤前后侧两流域径向分界面以及透空物面条件为:

$ {\varphi ^{(1)}} = {\varphi ^{(2)}}\, \, \, \, \, \, \, \, \, \, (r = b, 0 \le \theta \le \alpha , 2{\rm{ \mathit{ π} }} - \alpha \le \theta \le 2{\rm{ \mathit{ π} }}) $ (23)
$ \frac{{\partial {\varphi ^{(1)}}}}{{\partial r}} = \frac{{\partial {\varphi ^{(2)}}}}{{\partial r}}\, \, \, \, \, \, \, \, \, \, (r = b, 0 \le \theta \le 2{\rm{ \mathit{ π} }}) $ (24)
$ \begin{array}{l} \frac{{\partial \varphi _{}^{(1)}}}{{\partial r}} = \frac{{\partial \varphi _{}^{(2)}}}{{\partial r}} = \frac{{{\gamma _2}\rho }}{\mu }(\frac{{\partial \varphi _{}^{(1)}}}{{\partial t}} - \frac{{\partial \varphi _{}^{(2)}}}{{\partial t}}){\kern 1pt} \\ \, \, \, \, \, \, \, \, \, \, (r = b, \alpha \le \theta \le 2{\rm{ \mathit{ π} }} - \alpha ) \end{array} $ (25)

式中, ${\gamma _2}$为内层防波堤材料系数。由公式(24), 可得:

$ \begin{align} & A_{m}^{(1)}{{{{J}'}}_{m}}(kb)+C_{m}^{(1)}H{{_{m}^{(1)}}^{\prime }}(kb)=\,A_{m}^{(2)}{{{{J}'}}_{m}}(kb) \\ & \,\,\,\,\,\,\,\,\,\,(m=0,1,\cdots ) \\ \end{align} $ (26)
$ \begin{align} & B_{m}^{(1)}{{{{J}'}}_{m}}(kb)+D_{m}^{(1)}H{{_{m}^{(1)}}^{\prime }}(kb)=\begin{matrix} B_{m}^{(2)}{{{{J}'}}_{m}}(kb) \\ \end{matrix} \\ & \,\,\,\,\,\,\,\,\,\,(m=1,2,\cdots ) \\ \end{align} $ (27)

定义函数:

$ g(\theta ) = \left\{ \begin{array}{l} {g_1}(\theta ), 0 \le \theta < \alpha , 2{\rm{ \mathit{ π} }} - \alpha < \theta \le 2{\rm{ \mathit{ π} }}\\ {g_2}(\theta ), \alpha \le \theta \le 2{\rm{ \mathit{ π} }} - \alpha \end{array} \right. = 0 $ (28)

其中, 由公式(23)和公式(25)得到:

$ \begin{array}{c} {g_1}(\theta ) = \sum\limits_{n = 0}^\infty {\{ [A_n^{(1)}} \cos n\theta + B_n^{(1)}\sin n\theta ]{J_n}(kb)\\ + [C_n^{(1)}\cos n\theta + D_n^{(1)}\sin n\theta ]H_n^{(1)}(kb)\} \\ - \sum\limits_{n = 0}^\infty {[A_n^{(2)}} \cos n\theta + B_n^{(2)}\sin n\theta ]{J_n}(kb)\\ = 0\, \, \, \, \, \, \, \, \, \, (0 \le \theta \le \alpha , 2{\rm{ \mathit{ π} }} - \alpha \le \theta \le 2{\rm{ \mathit{ π} }}) \end{array} $ (29)
$ \begin{array}{c} {g_2}(\theta ) = \sum\limits_{n = 0}^\infty {[A_n^{(2)}} \cos n\theta + B_n^{(2)}\sin n\theta ]k{{J'}_n}(kb)\\ - i{G_2}\sum\limits_{n = 0}^\infty {[A_n^{(2)}} \cos n\theta + B_n^{(2)}\sin n\theta ]k{J_n}(kb)\\ + i{G_2}\sum\limits_{n = 0}^\infty {\{ [A_n^{(1)}} \cos n\theta + B_n^{(1)}\sin n\theta ]k{J_n}(kb)\\ + [C_n^{(1)}\cos n\theta + D_n^{(1)}\sin n\theta ]kH_n^{(1)}(kb)\} \\ = 0\, \, \, \, \, \, \, \, \, \, (\alpha \le \theta \le 2{\rm{ \mathit{ π} }} - \alpha ) \end{array} $ (30)

其中, ${G_2} = \frac{{{\gamma _2}\rho \omega }}{{\mu k}}$为内层堤径向表面的透空系数。

g(θ)在0~2π上展开为傅氏级数, 即有:

$ g(\theta ) = \frac{{{A_0}}}{2} + \sum\limits_{m = 1}^\infty {({A_m}\cos m\theta + {B_m}\sin m\theta ) = 0} $ (31)

Am=Bm=0可得:

$ \begin{array}{l} \int_{\;0}^{\;\alpha } {{g_1}(\theta )\cos m\theta d\theta + \int_{\;2{\rm{ \mathit{ π} }} - \alpha }^{\;2{\rm{ \mathit{ π} }}} {{g_1}(\theta )\cos m\theta } } d\theta + \\ \int_{\;\alpha }^{\;2{\rm{ \mathit{ π} }} - \alpha } {{g_2}} (\theta )\cos m\theta d\theta = 0\;\;(m = 0, 1, \cdots ) \end{array} $ (32)
$ \begin{array}{l} \int_{\;0}^{\;\alpha } {{g_1}(\theta )\sin m\theta d\theta + \int_{\;2{\rm{ \mathit{ π} }} - \alpha }^{\;2{\rm{ \mathit{ π} }}} {{g_1}(\theta )\sin m\theta } } d\theta + \\ \int_{\;\alpha }^{\;2{\rm{ \mathit{ π} }} - \alpha } {{g_2}} (\theta )\sin m\theta d\theta = 0\;\;(m = 1, 2, \cdots ) \end{array} $ (33)

相应有:

$ \begin{array}{l} \sum\limits_{n = 0}^\infty {A_n^{(1)}[{J_n}(kb)I_{nm}^{(1)} + i{G_2}k{J_n}(kb)I_{nm}^{(2)}]} \\ + \sum\limits_{n = 0}^\infty {A_n^{(2)}} \{ [k{{J'}_n}(kb) - i{G_2}k{J_n}(kb)]I_{nm}^{(2)} - {J_n}(kb)I_{nm}^{(1)}\} \\ + \sum\limits_{n = 0}^\infty {C_n^{(1)}[H_n^{(1)}(kb)I_{nm}^{(1)} + i{G_2}kH_n^{(1)}(kb)I_{nm}^{(2)}} ] = 0\, \\ \begin{array}{*{20}{c}} {}&{}&{} \end{array}(m = 0, 1, \cdots ) \end{array} $ (34)
$ \begin{array}{l} \sum\limits_{n = 0}^\infty {B_n^{(1)}[{J_n}(kb)I_{nm}^{(3)} + i{G_2}k{J_n}(kb)I_{nm}^{(4)}]} \\ + \sum\limits_{n = 0}^\infty {B_n^{(2)}} \{ [k{{J'}_n}(kb) - i{G_2}k{J_n}(kb)]I_{nm}^{(4)} - {J_n}(kb)I_{nm}^{(3)}\} \\ + \sum\limits_{n = 0}^\infty {D_n^{(1)}[} H_n^{(1)}(kb)I_{nm}^{(3)} + i{G_2}kH_n^{(1)}(kb)I_{nm}^{(4)}] = 0\\ \, \begin{array}{*{20}{c}} {}&{}&{} \end{array}(m = 1, 2, \cdots ) \end{array} $ (35)

公式(26)、(27)、(34)和(35)为待定系数完备代数方程组的另一组成部分, 其中取$B_0^{(1)} = B_0^{(2)} = D_0^{(1)} = 0$。公式(12)、(13)、(20)、(21)、(26)、(27)、(34)和(35)构成了全部待定系数的联立方程组。在实算中进行适度有限维运算即可满足计算精度要求(具体实算取m=60)。

进一步由下式可计算绕射波浪场压力分布:

$ {P^{(j)}} = - \rho \frac{{\partial {\varPhi ^{(j)}}}}{{\partial t}} = Re({p^{(j)}}), (j = 0, 1, 2) $ (36)

${F_{x1}}$${F_{y1}}$分别为外层堤上沿OxOy轴的水平波浪力分量, ${F_{x2}}$${F_{y2}}$分别为内层堤上的对应值, 则有:

$ \left\{ \begin{array}{l} {F_{x1}} = - \int_{\;0}^{\;d} {{\rm{d}}z} \int_{\;\alpha }^{\;2\pi - \alpha } {[{P^{(0)}} - {P^{(1)}}]\left| {_{r = a}} \right.a\cos \theta {\rm{d}}\theta } \\ {F_{y1}} = - \int_{\;0}^{\;d} {{\rm{d}}z} \int_{\;\alpha }^{\;2\pi - \alpha } {[{P^{(0)}} - {P^{(1)}}]\left| {_{r = a}} \right.a\sin \theta {\rm{d}}\theta } \end{array} \right. $ (37)
$ \left\{ \begin{array}{l} {F_{x2}} = - \int_0^d {dz} \int_\alpha ^{2\pi - \alpha } {[{P^{(1)}} - {P^{(2)}}]\left| {_{r = b}} \right.b\cos \theta d\theta } \\ {F_{y2}} = - \int_0^d {dz} \int_\alpha ^{2\pi - \alpha } {[{P^{(1)}} - {P^{(2)}}]\left| {_{r = b}} \right.b\sin \theta d\theta } \end{array} \right. $ (38)

相应作用于外、内层防波堤的总波浪力为:

$ {F_1} = \sqrt {{F_{1x}} + {F_{1y}}} , {F_2} = \sqrt {{F_{2x}} + {F_{2y}}} $ (39)
2 算例与分析

为方便计, 以符号$\frac{{{F_1}}}{{\rho g(H/2)ad}}$$\frac{{{F_2}}}{{\rho g(H/2)bd}}$分别表示外、内层堤的无量纲波浪力幅值。为验证文中结果的可靠性, 取双层圆弧防波堤圆心角γ=360°, 透空系数G1≠0和G2=0, 则结构化为外层透空的圆环柱结构(即双圆筒), 参照文献[13]取相同参数, 即G1=1, G2=0, b/a=1/5, d/a=3/2, 计算圆环柱所受的最大无量纲波浪力, 其中内外圆柱的无量纲波浪力幅值分别记为${f_1} = \frac{{{F_1}}}{{\rho gHad}}$${f_2} = \frac{{{F_2}}}{{\rho gHbd}}$。如图 2所示, 计算结果与文献[13]所算结果完全吻合, 说明所给解析方法是正确可靠的。

图 2 双圆筒柱的最大无量纲波浪力随kd的变化(${G_1} = 1, $ ${G_2} = 0, d/a = 3/2, b/a = 1/5, \gamma = {360^ \circ }$) Fig. 2 Variation of maximum dimensionless wave force on a concentric porous cylinder versus $kd$

图 3为不同波浪入射角β下双层透空防波堤的最大无量纲总波浪力随参数kd的变化趋势。由计算结果可知:内外层防波堤最大波浪力均随kd先增大后减小。对于外层堤, 当kd < 1时, 波浪力随入射角的增大而减小, 当kd > 1时波浪力呈交错变化趋势, 而对于内层堤波浪力保持随入射角的增大而减小的趋势。其中平均而言, 波浪正入射(β=0)情形对应最大无量纲波浪力幅值。

图 3 不同入射角下最大无量纲波浪力$\left( {{G_1} = 1, {G_2} = 1, d/a = 1/5, b/a = 1/2, \gamma = {{120}^ \circ }} \right)$ Fig. 3 Maximum dimensionless wave force at different incident wave angles

图 4为波浪正入射情况下, 单层圆弧型透空防波堤与双层圆弧透空防波堤之内层堤所受最大无量纲波浪力的比较。选取单层堤与内层堤在相同位置, 其他相关参数取值相同。计算结果表明:双层堤中内层堤的最大无量纲波浪力幅值明显低于相同位置上单层堤的对应值。由此可知, 外层堤的构建大为减弱了波浪对内层堤的直接作用, 而外层堤表面透空又可同时减弱波浪对外堤的直接作用。

图 4 单层堤和内层堤无量纲波浪力的比较 Fig. 4 Comparison of the maximum dimensionless wave force on single breakwater and internal breakwater of double-layered breakwater

图 5为不同防波堤张角下双层堤的最大无量纲波浪力随$kd$的变化趋势。由图可知:外层堤与内层堤的最大无量纲波浪力随防波堤张角γ的变化趋势相似。当kd < 1时, 无量纲波浪力幅值随防波堤张角的增大而增大, 特别当防波堤处于对于相对浅水区时(kd < 0.5), 这种变化态势更为明显, 而内层堤相对外层堤的变化差别更为明显一些。当kd > 1时, 内层堤与外层堤的无量纲波浪力幅值均呈现交错变化态势, 且变化幅度不大。

图 5 不同防波堤张角下最大无量纲波浪力$\left( {{G_1} = {G_2} = 1, \beta = 0, d/a = 1/5, b/a = 1/2} \right)$ Fig. 5 Maximum dimensionless wave force at different central breakwater angles

图 6为不同透空系数下双层防波堤的最大无量纲波浪力随kd的变化趋势, 其中内外层堤取相同透空系数, 即取G1=G2。由图可知:内外层防波堤表面同时布孔可明显减小外层堤的波浪承载, 且有效降低内外层堤波浪力幅值的振荡变化, 尤其可避免内层防波堤在特定kd取值下波浪力幅值出现的较为急剧的峰值突增现象。此外, 当防波堤透空系数增大到一定值时, 波浪力幅值下降现象不再明显, 说明对防波堤表面的透空适度进行即可, 防波堤表面过大的透空度对结构的稳定可能产生负面影响。

图 6 相同透空系数下的最大无量纲波浪力$\left( {\beta = 0, d/a = 1/5, b/a = 1/2, \gamma = {{120}^ \circ }} \right)$ Fig. 6 Maximum dimensionless wave force with the same porous coefficients of outer and inner breakwaters

图 7为四种不同透空形式下双层防波堤的最大无量纲波浪力随kd的变化趋势。其中透空形式包括内外层堤均密实(G1=G2=0)、外层堤密实内层堤透空(G1=0, G2=1)、外层堤透空内层堤密实(G1=1, G2=0)以及内外层堤均透空(G1=G2=1)四种。图 7a表明:外层堤表面透空能有效降低外层堤的无量纲波浪力幅值。当内外层堤表面均密实时, 外层堤在特定的kd取值处可能出现陡增的波浪力峰值, 而内层堤表面透空可以有效避免这些陡增峰值的出现。此外, 内层堤表面透空与否基本不影响外层堤波浪力幅值的大小。图 7b表明:当内外层堤均密实时, 内层堤在特定的kd取值处也会出现陡增的波浪力峰值, 且较外层堤陡增现象更为急剧。当外层堤表面透空时可以避免这些峰值的出现, 当然会略为增加内层堤的波浪力幅值, 此时可让内层堤表面同时透空以减弱内层堤的波浪承载。综合而言, 采用内外层堤均透空的结构形式既可避免内外层堤最大波载陡增峰值的出现, 又可同时降低双层堤的波浪力幅值大小, 因而为均衡考虑后相对合理的优化方案。内外层堤均密实所引起的波浪力幅值的陡增现象可以归结为波浪在内外层堤间径向方向的透射受到限制, 导致波浪力幅值整体大幅提高, 该处波浪呈现多重反射的复杂作用机制。

图 7 不同透空形式下的最大无量纲波浪力$\left( {\beta = 0, d/a = 1/5, b/a = 1/2, \gamma = {{120}^ \circ }} \right)$ Fig. 7 Maximum dimensionless wave force with different porous structures

图 8为不同内外半径比下双层密实防波堤的最大无量纲波浪力随kd的变化趋势。结果表明:外层堤的无量纲波浪力幅值整体随内外半径比的变化不大, 当b/a较小时外堤波浪力峰值陡增现象明显趋弱, 说明较小的密实内堤结构对波浪在内域径向方向传播的影响趋弱。内外半径比b/a的变化对密实内层堤的最大无量纲波浪力变化具有一定影响。在内外层堤均密实的情形, 内层堤的无量纲波浪力幅值的陡增现象始终存在, 当b/a较小时(即内外层堤间距增大), 陡增的峰值有所降低。

图 8 不同内外半径比下的最大无量纲波浪力$\left( {{G_1} = {G_2} = 0, \beta = 0, d/a = 1/5, \gamma = {{120}^ \circ }} \right)$ Fig. 8 Maximum dimensionless wave force at different ratios of internal and external breakwater radii

图 9为不同内外半径比下双层透空防波堤的最大无量纲波浪力随$kd$的变化趋势。图中结果表明:外层堤的无量纲波浪力幅值随内外半径比$b/a$的变化呈现交错变化态势, 且数值大小变化不大, 说明外圆弧堤的波浪力受内外半径比影响较小, 另外无波浪力幅值陡增现象。不过$b/a$变化对内堤的最大无量纲波浪力影响较大。内层堤无量纲波浪力幅值随着$b/a$增大(即内外层堤间距减小)而较为明显地减小。同样在防波堤表面透空条件下, 无波浪力幅值陡增现象。

图 9 不同内外半径比下的最大无量纲波浪力$\left( {{G_1} = {G_2} = 1, \beta = 0, d/a = 1/5, \gamma = {{120}^ \circ }} \right)$ Fig. 9 Maximum dimensionless wave force at different ratios of internal and external breakwater radii

图 10为不同外径水深比下(因为b/a为定值, a/d等价于b/d)双层防波堤的最大无量纲波浪力随kd的变化趋势。由图可见, 整体而言内外层堤的无量纲波浪力幅值均随着外径水深比的增大而减小。

图 10 不同外径水深比下的最大无量纲波浪力$\left( {{G_1} = {G_2} = 1, \beta = 0, b/a = 1/2, \gamma = {{120}^ \circ }} \right)$ Fig. 10 Maximum dimensionless wave force at different ratios of outer breakwater radius and water depth
3 结论

基于Airy微幅波绕射理论, 推导了双层直立圆弧型透空防波堤波浪绕射问题的解析解, 从而将原有的单层堤问题的解析解进行了拓展, 同时将原有的双圆筒波浪绕射问题的理论解进行了拓展。按本文方法取双层圆弧防波堤圆心角γ=360°, 使结构化为圆环透空柱体(即双圆筒), 对其所受到的最大无量纲波浪力进行计算, 所算结果与已有的双圆筒解析解结果完全吻合, 说明本文方法正确可靠。与单层堤波浪力幅值计算结果的对比说明, 双层堤因外层堤的引入将明显减小内堤承受的波浪荷载。此外, 内层堤表面透空与否虽对外层堤的波浪力幅值大小影响不大, 但内层堤透空可以有效避免外层堤密实时可能出现的波浪力峰值陡增的现象。同样外层堤表面透空可以避免内层堤密实时可能出现的更为急剧的波浪力峰值陡增现象。综合考虑采用内外层堤均透空的结构形式既可避免内外堤波载峰值陡增现象的出现, 又可分别各自降低内外堤的波浪力大小, 因而为相对优化的结构透空形式, 当然外堤的透空将略为增加内堤的波浪承载。进一步波浪的入射角、防波堤张角与半径以及水深等因素的相对变化均一定程度地影响双层堤的绕射波浪作用力。

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