vault backup: 2025-07-21 17:00:32
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{"id":"82708a439812fdc7","type":"text","text":"# 7月已完成\n\nP1 工况点稳态变形量求解,F=kx\n- 文献调研,初步确定思路 done\n- 推导方程 done\n- 编写组建增广矩阵,求解广义坐标代码 done\n- 测试广义坐标到叶片变形量功能 可以变形,气动Cp会改变\n\nP1 职称评审系统填写,材料梳理上传 盖章","x":-220,"y":134,"width":440,"height":560},
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{"id":"505acb3e6b119076","type":"text","text":"# 6月已完成\n\n\nP1 结果对比\n- Herowind 带3.5气动与fast3.5对比 相同\n- Herowind 带4.0气动与fast4.0对比 相同\n- Herowind 带hrl气动与fast对比 需气动支持15MW\n- 叶根坐标系转换 \n\t- 叶尖变形量 - 变形向量 dot product 叶根坐标系方向\n\t- 叶片载荷输入量呢 载荷传递在blade mesh.force moment,mesh.orientation = coord_sys.n\n\nP1 Bladed交流问题汇总\n\nP1 模型线性化原理 done\n- Bladed 线性化理论手册 仔细阅读\n- multibody blade transform\n- fast线性化理论\n- 梳理Bladed线性化方法框架\n\n\nP1 编写线性化理论手册 done\nP1 上手Bladed \\ fast 线性化功能,研究OpenFAST线性化实现原理 done","x":-700,"y":134,"width":440,"height":560},
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{"id":"30cb7486dc4e224c","type":"text","text":"# 8月已完成\n\n","x":260,"y":134,"width":440,"height":560},
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{"id":"c18d25521d773705","type":"text","text":"# 计划\n这周要做的3~5件重要的事情,这些事情能有效推进实现OKR。\n\nP1 必须做。P2 应该做\n\n\nP1 柔性部件 叶片、塔架主动力惯性力算法 主线\n- 变形体动力学 简略看看ing\n- 柔性梁弯曲变形振动学习,主线 \n\t- 广义质量 刚度矩阵及含义\n- 如何静力学求解 \n\t- 基于本构方程 读孟的论文\n\t- normal mode shape 能否使用?\n\t\n- 梳理bladed动力学框架 this week\n\t- 子结构文献阅读\n\t- 叶片模型建模 done\n\nP1 工况点稳态变形量求解,F=kx\n\n- 连接气动测试,完成。存在一个问题,气动是否要用稳态模型\n- 直接迭代到变形量收敛 思路确定了\n- x.qt x.qdt数据如何从dxdt.qdt拿来/更新,预估校正方法\n\n\nP1 数值扰动+回归的线性化方法原理探究\n\n\n\nP2 如何优雅的存储、输出结果。\nP2 yaw 自由度再bug确认 已知原理了\n","x":-597,"y":-693,"width":453,"height":347}
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{"id":"c18d25521d773705","type":"text","text":"# 计划\n这周要做的3~5件重要的事情,这些事情能有效推进实现OKR。\n\nP1 必须做。P2 应该做\n\n\nP1 柔性部件 叶片、塔架主动力惯性力算法 主线\n- 变形体动力学 简略看看ing\n- 柔性梁弯曲变形振动学习,主线 \n\t- 广义质量 刚度矩阵及含义\n\t\n- 梳理bladed动力学框架 this week\n\t- 子结构文献阅读\n\t- 叶片模型建模 done\n\nP1 工况点稳态变形量求解,F=kx\n\n- 连接气动测试,完成。存在一个问题,气动是否要用稳态模型\n- 直接迭代到变形量收敛 思路确定了\n- x.qt x.qdt数据如何从dxdt.qdt拿来/更新,预估校正方法 steady中预估矫正方法去掉了\n\n\nP1 控制信号到多体\n\nP1 数值扰动+回归的线性化方法原理探究\n\nP1 产出的报告 线性化理论手册编写\n\nP2 如何优雅的存储、输出结果。\nP2 yaw 自由度再bug确认 已知原理了\n","x":-597,"y":-693,"width":453,"height":347}
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@ -184,7 +184,7 @@ where $\mathbf{u}=\mathbf{u}(s,\,t)=[u_{1}(s,\,t)$ , $\nu_{1}(s,\,t),\,\theta_{1
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The spatial derivatives in the linear equations of aeroelastic motion (equation (11)) are approximated by the f inite difference scheme (Table I) with N discretization points. The f inite difference implementation includes the spatial boundary conditions (equations (4) and (5)). The second-order differential equation is then rewritten into f irst-order form by introducing the f irst-order time derivatives as states and combining it with the unsteady aerodynamic model. The spatial discretized f irst-order equation of aeroelastic motion becomes
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# $\dot{\mathbf{x}}=\mathbf{A}\mathbf{x}+\mathbf{B}\mathbf{f}$
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$$\dot{\mathbf{x}}=\mathbf{A}\mathbf{x}+\mathbf{B}\mathbf{f}$$
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where $\dot{\mathbf{x}}$ includes the linear deformation around the linearization point, speed and the aerodynamic states for each discretization point, giving $3N+3N+4N=10N$ degrees of freedom, A is the linear coeff icients, $\mathbf{B}$ is the linear gains on the external inf luences and f is the linear variation of the external inf luences. The unforced version of equation (12) forms a differential eigenvalue problem.29 The differential eigenvalue problem is casted into an algebraic eigenvalue problem by assuming a complex exponential solution. The eigenvalues and corresponding eigenvectors can be grouped into two sets: real valued and complex valued eigenvalues. Generally, the real valued eigenvalues are related to the aerodynamic states and correspond to the aerodynamic time lags. However, overdamped aeroelastic modes will also have real valued eigenvalues. The complex valued eigenvalues are related to the aeroelastic states and give the aeroelastic frequencies and damping. The corresponding eigenvectors give the aeroelastic mode shapes of the particular mode.
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@ -134,6 +134,17 @@ $$
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图 1:示例线性回归计算单元 ${\bf A}_{7,4}$,其值为 -1.315,相关系数为 0.9982。平衡点以绿色显示。
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# 平衡点计算
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Bladed中,为了进行线性分析,Bladed依次取每个工作点,并找到风电机组的稳态条件(如同时域运行中的初始条件)。这意味着风轮没有加速,模态变形使得弹性载荷平衡外部载荷。这定义了$\mathbf{x_{0}},\,\mathbf{y_{0}}$和$\mathbf{u_{0}}$的值,即所有扰动围绕的主要平衡点。
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对每个工况计算工况点,工况点加速度为0,以叶片为例,每只叶片有三个自由度,1阶挥舞、1阶摆振,2阶挥舞,则每个自由度的加速度都为0。
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考虑到移除方位角依赖性,叶片上仅考虑气动力作用,忽略重力。刚度上,由于加速度为0,忽略叶片的阻尼刚度,仅考虑叠加离心刚度的广义刚度。动力学方程变为。
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$$
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F_{Aero} = K_{generalized} q
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$$
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# 2. 多叶片坐标系转换
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