﻿ 管路结构振动能量传递规律研究
 舰船科学技术  2019, Vol. 41 Issue (2): 56-60 PDF

Vibrational energy attenuation between two pipelines
LI Zheng-yang, CHE Chi-dong, HU Fan
State Key Laboratory of Ocean Engineering, School of Naval Architecture, Ocean and Civil Engineering, Collaborative Innovation Center for Advanced Ship and Deep-Sea Exploration, Shanghai Jiaotong University, Shanghai 200240, China
Abstract: A longitudinal-bending coupled vibration model of two semi-infinite long pipelines is put out in this paper. Vibration energy transmission of two semi-infinite long pipelines is indicated by wave approach under different kinds of boundary conditions. Several non-dimensional coefficients are introduced to simplify the equations, and the reflection and transmission coefficients are derived. Through numerical analysis, the parameter change on the vibration energy transfer is obtained, under the bending wave incidence. The results show that the vibration energy transmission is mainly affected by connecting rigidity at high frequency, and mainly affected by the angle at low frequency, and the rotational rigidity will cause vibrational energy transmission curves to generate troughs and has influence at low frequency under low rotational rigidity, which can be useful to the selection of transmission shafts parameters.
Key words: wave approach     pipelines     longitudinal-bending coupled vibration
0 引　言

1 理论分析 1.1 模型描述

 图 1 半无限长管道局部坐标及受力示意图 Fig. 1 Local coordinate and force diagram of semi-infinite long pipelines
1.2 位移表达式

 ${v_{y1}}({x_1}) = {e^{ - j{k_{B1}}{x_1}}} + {r_b}{e^{ + j{k_{B1}}{x_1}}} + {r_j}{e^{ + {k_{B1}}{x_1}}},{x_1} < 0\text{，}$ (1)
 ${v_{y2}}({x_2}) = {\tilde v_{y1}}({t_b}{e^{ - j{k_{B2}}{x_2}}} + {t_j}{e^{ - {k_{B2}}{x_2}}}),{x_2} > 0\text{，}$ (2)
 ${v_{x1}}({x_1}) = {r_L}{e^{ + j{k_{L1}}{x_1}}},{x_1} < 0\text{，}$ (3)
 ${v_{x2}}({x_2}) = {t_L}{e^{ - j{k_{L2}}{x_2}}},{x_2} > 0\text{。}$ (4)

1.3 边界条件

1）力平衡

 ${F_{x1}} = {k_x} \cdot \frac{1}{{j\omega }}({v_{x2}} \cdot \cos \theta - {v_{y2}} \cdot \sin \theta - {v_{x1}})\text{，}$ (5)
 ${F_{y1}} = {k_y} \cdot \frac{1}{{j\omega }}({v_{y2}} \cdot \cos \theta + {v_{x2}} \cdot \sin \theta - {v_{y1}})\text{，}$ (6)
 ${F_{x1}} = {F_{x2}} \cdot \cos \theta + {F_{y2}} \cdot \sin \theta \text{，}$ (7)
 ${F_{y1}} + {F_{x2}} \cdot \sin \theta = {F_{y2}} \cdot \cos \theta \text{；}$ (8)

2）力矩平衡

 ${M_1} = {k_r} \cdot ({\alpha _1} - {\alpha _2})\text{，}$ (9)
 ${M_1} = {M_2}\text{。}$ (10)

1.4 透反射系数控制方程

 $\small\begin{split} &\left[ {\begin{array}{*{20}{c}} \!\! \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!0\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!&\!\!\!\!\!\!\!\!\!\!0\!\!\!\!\!\!\!\!&\!\!\!\!\!\!\!\!{1 + {\eta _1}}\!\!\!\!\!\!\!\!\!&\!\!\!\!\!\!\!\!\!{{\eta _1}\sin \theta }\!\!\!\!\!\!\!\!\!&\!\!\!\!\!\!\!\!\!{{\eta _1}\sin \theta }\!\!\!\!\!\!\!\!&\!\!\!\!\!\!\!\!{ - {\eta _1}\cos \theta }\!\!\!\!\!\!\!\!\!\\ {1 - {\eta _2}}&{j - {\eta _2}}&0&{{\eta _2}\cos \theta }&{{\eta _2}\cos \theta }&{{\eta _2}\sin \theta } \\ {j{\eta _3}{\rm{ + 1}}}&{{\eta _3}{\rm{ - 1}}}&0&{j\chi {\eta _3}}&{\chi {\eta _3}}&0 \\ { - 1}&1&0&\psi &{ - \psi }&0 \\ 0&0&1&{ - {\beta _{\rm{1}}}\sin \theta }&{j \cdot {\beta _{\rm{1}}}\sin \theta }&{\alpha \cos \theta } \\ { - {\beta _{\rm{2}}}}&{ - j \cdot {\beta _{\rm{2}}}}&0&{ - {\gamma _{\rm{2}}}\cos \theta }&{ - j \cdot {\gamma _{\rm{2}}}\cos \theta }&{ - \sin \theta } \end{array}} \right] \times\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \\ &\quad \left[ {\begin{array}{*{20}{c}} {{r_b}} \\ {{r_j}} \\ {{r_l}} \\ {{t_b}} \\ {{t_j}} \\ {{t_l}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0 \\ {1 + {\eta _2}} \\ {j{\eta _3}{\rm{ - 1}}} \\ 1 \\ 0 \\ { - {\beta _{\rm{2}}}} \end{array}} \right]\text{，} \end{split}$ (11)
 $\small \begin{split} &\left[ {\begin{array}{*{20}{c}} \!\!\!\! 0\!\!\!\!&\!\!\!\!0\!\!\!\!&\!\!\!\!{1 + {\eta _1}}\!\!\!\!&\!\!\!\!{{\eta _1}\sin \theta }\!\!\!\!&\!\!\!\!{{\eta _1}\sin \theta }\!\!\!\!&\!\!\!\!{ - {\eta _1}\cos \theta } \!\!\!\!\\ {1 - {\eta _2}}&{j - {\eta _2}}&0&{{\eta _2}\cos \theta }&{{\eta _2}\cos \theta }&{{\eta _2}\sin \theta } \\ {\rm{1}}&{{\rm{ - }}1}&0&{\rm{0}}&{\rm{0}}&0 \\ {{\rm{ - 1}}}&{\rm{1}}&0&\psi &{{\rm{ - }}\psi }&0 \\ 0&0&1&{ - {\beta _{\rm{1}}}\sin \theta }&{j \cdot {\beta _{\rm{1}}}\sin \theta }&{\alpha \cos \theta } \\ { - {\beta _{\rm{2}}}}&{ - j \cdot {\beta _{\rm{2}}}}&0&{ - {\gamma _{\rm{2}}}\cos \theta }&{ - j \cdot {\gamma _{\rm{2}}}\cos \theta }&{ - \sin \theta } \end{array}} \right]\times \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\\ &\quad\left[ {\begin{array}{*{20}{c}} {{r_b}} \\ {{r_j}} \\ {{r_l}} \\ {{t_b}} \\ {{t_j}} \\ {{t_l}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0 \\ {1 + {\eta _2}} \\ {{\rm{ - 1}}} \\ {\rm{1}} \\ 0 \\ { - {\beta _{\rm{2}}}} \end{array}} \right]\text{，} \end{split}$ (12)
 $\small \begin{split} &\left[ {\begin{array}{*{20}{c}} 0&0&{\rm{1}}&{\sin \theta }&{\sin \theta }&{ - \cos \theta } \\ { - {\rm{1}}}&{ - {\rm{1}}}&0&{\cos \theta }&{\cos \theta }&{\sin \theta } \\ j&{\rm{1}}&0&{j\chi }&\chi &0 \\ { - 1}&1&0&\psi &{ - \psi }&0 \\ 0&0&1&{ - {\beta _{\rm{1}}}\sin \theta }&{j \cdot {\beta _{\rm{1}}}\sin \theta }&{\alpha \cos \theta } \\ { - {\beta _{\rm{2}}}}&{ - j \cdot {\beta _{\rm{2}}}}&0&{ - {\gamma _{\rm{2}}}\cos \theta }&{ - j \cdot {\gamma _{\rm{2}}}\cos \theta }&{ - \sin \theta } \end{array}} \right]\times\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \\ &\quad\left[ {\begin{array}{*{20}{c}} {{r_b}} \\ {{r_j}} \\ {{r_l}} \\ {{t_b}} \\ {{t_j}} \\ {{t_l}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0 \\ 1 \\ j \\ 1 \\ 0 \\ { - {\beta _{\rm{2}}}} \end{array}} \right]\text{。} \end{split}$ (13)

1.5 能量传递效率

 ${\tau _{BB}} = \frac{{{P_{BT}}}}{{{P_{BI}}}} = \frac{{{\beta _{\rm{1}}}}}{{{\gamma _{\rm{1}}}}}{\left| {{t_b}} \right|^2}{\rm{ = }}\frac{{{c_{B2}}{\rho _2}{A_2}}}{{{c_{B1}}{\rho _1}{A_1}}}{\left| {{t_b}} \right|^2}\text{，}$ (14)

 ${\rho _{BB}} = \frac{{{P_{BR}}}}{{{P_{BI}}}} = {\left| {{r_b}} \right|^2}\text{，}$ (15)

 ${\tau _{BL}} = \frac{{{P_{LT}}}}{{{P_{BI}}}} = \frac{1}{{2{\beta _2}}}{\left| {{t_l}} \right|^2}{\rm{ = }}\frac{{{c_{L2}}{\rho _2}{A_2}}}{{{\rm{2}} \cdot {c_{B1}}{\rho _1}{A_1}}}{\left| {{t_l}} \right|^2}\text{，}$ (16)

 ${\rho _{BL}} = \frac{{{P_{LR}}}}{{{P_{BI}}}} = \frac{1}{{2{\gamma _1}}}{\left| {{r_l}} \right|^2}{\rm{ = }}\frac{{{c_{L1}}}}{{2 \cdot {c_{B1}}}}{\left| {{r_l}} \right|^2}\text{。}$ (17)

 ${\rho _{BB}} + {\rho _{BL}} + {\tau _{BB}} + {\tau _{BL}} = 1\text{。}$ (18)
2 数值计算

 图 2 能量透射与反射效率曲线 Fig. 2 Energy transmission and reflection efficiencies

 图 3 忽略扭转刚度时能量透射与反射效率曲线 Fig. 3 The energy efficiencies without rotation rigidity

1）在计算频率范围内，相同连接方式下，4种能量系数和为1，符合式（18），即能量守恒定律；

2）2根管道间存在扭转刚度时，在计算范围内，透射波总能量曲线随频率的增加既可能产生波峰又可能产生波谷；

3）弯曲波能量（ ${\tau _{BB}}{\rm{ + }}{\rho _{BB}}$ ）远大于纵波能量（ ${\tau _{BL}}{\rm{ + }}{\rho _{BL}}$ ）；

4）2根管道之间不存在扭转刚度时，在计算范围内，透射波总能量，即 ${\tau _{BB}}{\rm{ + }}{\tau _{BL}}$ ，随频率增大先升高后降低。

 $TL = 10\log \left(\frac{1}{{{\tau _{BB}}}}\right)\text{。}$ (19)

1）连接刚度一定（1E8 N/m），扭转刚度为0；

2）连接刚度一定（1E8 N/m），扭转刚度为无穷大；

3）连接刚度和扭转刚度均为无穷大，即两管焊接。

 图 4 特殊边界下的弯曲波传递损失系数变化曲线 Fig. 4 TL under special boundary condition

 图 5 弯曲波传递损失系数随扭转刚度变化曲线 Fig. 5 TL curves with the change of rotation rigidity

 图 6 弯曲波传递损失系数随扭转刚度变化曲线 Fig. 6 Bend waves TL with the change of rotation rigidity

 图 7 弯曲波传递损失系数随连接刚度变化曲线 Fig. 7 Bend waves TL with the change of connecting rigidity

 图 8 弯曲波传递损失系数随连接刚度变化曲线 Fig. 8 Bend waves TL with the change of connecting rigidity

 图 9 扭转刚度极小时TL随两管夹角变化曲线 Fig. 9 The angle change curve of the non-rotating rigidity

 图 10 带有扭转刚度时TL随两管夹角变化曲线 Fig. 10 The angle change curve of the rotating rigidity

3 结　语

1）2根管道连接会产生一定的连接和扭转刚度，其刚度值会影响振动能量的传递。

2）两管间能量的传递弯曲波能量占主导地位，在工程中，更关注透射弯曲波对下游管道的工作状态影响，通过改变传动参数可以改善管道工作状态。

3）扭转刚度会使弯曲波能量传递效率曲线产生波谷，当扭转刚度较低时，扭转刚度对弯曲波能量透射传递效率低频段影响较大；当扭转刚度较高时，增加到一定数值之后，对弯曲波能量透射效率的影响将极小。

4）连接刚度主要影响高频弯曲波透射能量传递效率。

5）忽略扭转刚度时，两管夹角的变化对高频弯曲波的传递损失影响较小，对低频段影响较大；存在扭转刚度时，角度变化影响效率曲线波峰附近的频段。

6）本文理论分析中引入的无量纲数，将管道振动透射和反射系数与管道特征阻抗联系在一起，简化了公式表达，有利于进一步研究管道物性参数对振动能量传递的影响。

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