vault backup: 2025-05-29 08:15:40
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@ -1253,7 +1253,7 @@ $$
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The lateral deflection of the tower is assumed to be small so that:
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塔的横向挠度被假设为很小,以使:
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$$
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\frac{\partial u(h,t)}{\partial h}\!<\!<\!I
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\frac{\partial u(h,t)}{\partial h}\!<\!<\!1
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$$
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Applying Eq. (3.61) to Eq. (3.60) results in two simplified solutions to the gradient of the axial deflection of the tower:
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@ -1286,7 +1286,7 @@ where $h^{\,\prime}$ is a dummy variable representing the elevation along the fl
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Employing the shape functions through the use of Eq. (3.39), Eq. (3.65) can be rewritten:
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利用形函数,通过公式 (3.39),公式 (3.65) 可以改写为:
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$$
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\nu(h,t)\!=\!\frac{1}{2}\sum_{i=p}^{N+p-1}\!\sum_{j=p}^{N+p-1}\!\left(\int_{\theta}^{h}\!\frac{d\varphi_{i}\bigl(h^{\prime}\bigr)}{d h^{\prime}}\frac{d\varphi_{j}\bigl(h^{\prime}\bigr)}{d h^{\prime}}\!d h^{\prime}\right)\!\!c_{i}(t)c_{j}(t)
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\nu(h,t)\!=\!\frac{1}{2}\sum_{i=p}^{N+p-1}\!\sum_{j=p}^{N+p-1}\!\left(\int_{\theta}^{h}\!\frac{d\varphi_{i}\bigl(h^{\prime}\bigr)}{d h^{\prime}}\frac{d\varphi_{j}\bigl(h^{\prime}\bigr)}{d h^{\prime}}\!d h^{\prime}\right)c_{i}(t)c_{j}(t)
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$$
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Substituting Eq. (3.66) into Eq. (3.56) results in the following equation for the potential energy of the tower caused by gravity:
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@ -1298,7 +1298,7 @@ $$
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Equations (3.44), (3.54), (3.55), and (3.67) show that the generalized stiffness of the tower is:
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$$
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\begin{array}{l}{{k_{i j}=\int_{\boldsymbol{\theta}}^{H}E I_{T}(h)\frac{d^{\,2}\varphi_{i}\,\left(h\right)}{d h^{\,2}}\frac{d^{\,2}\varphi_{j}\,\left(h\right)}{d h^{\,2}}d h-}}\\ {{g\Bigg[M_{T o p}\int_{o}^{H}\frac{d\varphi_{i}\,\left(h\right)}{d h}\frac{d\varphi_{j}\,\left(h\right)}{d h}d h+\int_{o}^{H}\mu_{T}\,\bigl(h\biggr)\bigg(\int_{o}^{h}\frac{d\varphi_{i}\,\left(h^{\prime}\right)}{d h^{\prime}}\frac{d\varphi_{j}\,\left(h^{\prime}\right)}{d h^{\prime}}d h^{\prime}\bigg)d h\Bigg]}}\end{array}
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\begin{array}{l}{{k_{i j}=\int_{\boldsymbol{0}}^{H}E I_{T}(h)\frac{d^{\,2}\varphi_{i}\,\left(h\right)}{d h^{\,2}}\frac{d^{\,2}\varphi_{j}\,\left(h\right)}{d h^{\,2}}d h-}}\\ {{g\Bigg[M_{T o p}\int_{0}^{H}\frac{d\varphi_{i}\,(h)}{d h}\frac{d\varphi_{j}\,\left(h\right)}{d h}d h+\int_{o}^{H}\mu_{T}\,(h)\bigg(\int_{0}^{h}\frac{d\varphi_{i}\,\left(h^{\prime}\right)}{d h^{\prime}}\frac{d\varphi_{j}\,\left(h^{\prime}\right)}{d h^{\prime}}d h^{\prime}\bigg)d h\Bigg]}}\end{array}
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$$
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After integrating by parts and simplifying, Eq. (3.68) is equivalent to14:
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@ -1317,9 +1317,9 @@ m_{i j}=M_{T i p}+\int_{0}^{R-R_{H}}\mu_{B}(r)\varphi_{i}(r)\varphi_{j}(r)d r
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$$
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where $\mu_{B}(r)$ is the distributed lineal density of the beam (blades). Equation (3.70) is valid for both the flapwise and edgewise directions.
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其中 $\mu_{B}(r)$ 为梁(叶片)的分布线密度。方程 (3.70) 对挥舞方向和摆振方向均有效。
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Neglecting gravity, the potential energy of each blade has a component associated with the distributed stiffness of the beam and a component associated with centrifugal stiffening as a result of rotor rotation:
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其中 $\mu_{B}(r)$ 为梁(叶片)的分布线密度。方程 (3.70) 对挥舞方向和摆振方向均有效。
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忽略重力,每个叶片的势能包含两部分:一部分与梁的分布刚度相关,另一部分是由于风轮旋转引起的离心刚度。
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$$
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@ -1333,23 +1333,29 @@ V_{R o t a t i o n}=\varOmega^{2}\biggl[M_{T i p}R\nu\bigl(R-R_{H}\,,t\bigr)+\in
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$$
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The centrifugal forces tend to increase the stiffness of each blade.
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离心力倾向于增加每一叶片的刚度。
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**The potential energy caused by rotor rotation is essentially the product of the centrifugal forces and the distances upon which they act**. For example, the centrifugal force acting on a blade’s tip brake as used in Eq. (3.72) is $\Omega^{2}M_{T i p}R$ , and the distance upon which this force acts is $\nu(R{-}R_{H},t)$ . This expression assumes centrifugal forces acting on a blade do not change when the blade deflects. To be technically correct, the centrifugal forces should be regarded as functions of the blade deflection since, for example, the centrifugal force acting on a blade’s tip brake would be slightly less than $\varOmega^{2}M_{T i p}R$ if the tip was deflected substantially since the distance to the tip from the center of rotation would be shorter than $R$ by a distance $\nu(R{-}R_{H},t)$ . If this correlation between centrifugal force and deflection were accounted for, however, terms involving $\nu(r,t)^{2}$ would appear in Eq. (3.72). This action would prevent the centrifugal forces from being derivable from a conservative force field and would beget considerable difficulties in the ensuing analysis. Therefore, FAST_AD and Modes assume that the centrifugal forces are not functions of the blade deflection. If the deflections of the blades are assumed to be small (the blades are relatively stiff), this action introduces negligible errors.
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**风轮转动产生的势能本质上是离心力及其作用距离的乘积**。例如,在公式(3.72)中使用的叶片尖端所受的离心力为$\Omega^{2}M_{T i p}R$,作用在该力上的距离为$\nu(R{-}R_{H},t)$。该表达式假设叶片挠曲时,作用在叶片上的离心力不发生变化。从技术上讲,应将离心力视为叶片挠曲的函数,因为例如,如果叶片尖端发生大幅挠曲,则作用在叶片尖端上的离心力将略小于$\varOmega^{2}M_{T i p}R$,因为从旋转中心到叶片尖端的距离将比$R$短,缩短的距离为$\nu(R{-}R_{H},t)$。然而,如果考虑离心力与挠曲之间的这种相关性,则公式(3.72)中将出现涉及$\nu(r,t)^{2}$的项。这将导致离心力无法从保守力场导出,并会产生后续分析中的相当大的困难。因此,FAST_AD和Modes假设离心力不是叶片挠曲的函数。如果假设叶片的挠曲是较小的(叶片相对较硬),则此操作引入的误差可以忽略不计。
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The potential energy caused by rotor rotation is essentially the product of the centrifugal forces and the distances upon which they act. For example, the centrifugal force acting on a blade’s tip brake as used in Eq. (3.72) is $\itOmega^{2}M_{T i p}R$ , and the distance upon which this force acts is $\nu(R{-}R_{H},t)$ . This expression assumes centrifugal forces acting on a blade do not change when the blade deflects. To be technically correct, the centrifugal forces should be regarded as functions of the blade deflection since, for example, the centrifugal force acting on a blade’s tip brake would be slightly less than $\varOmega^{2}M_{T i p}R$ if the tip was deflected substantially since the distance to the tip from the center of rotation would be shorter than $R$ by a distance $\nu(R{-}R_{H},t)$ . If this correlation between centrifugal force and deflection were accounted for, however, terms involving $\nu(r,t)^{2}$ would appear in Eq. (3.72). This action would prevent the centrifugal forces from being derivable from a conservative force field and would beget considerable difficulties in the ensuing analysis. Therefore, FAST_AD and Modes assume that the centrifugal forces are not functions of the blade deflection. If the deflections of the blades are assumed to be small (the blades are relatively stiff), this action introduces negligible errors.
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As for the tower, the axial (radial or span-wise) deflection of the flexible cantilever beam at time $t$ and an original span $r$ , $\nu(r,t)$ , is a direct result of the assumption that the flexible beam remains fixed in length (measured along the beam’s central axis) when deflecting. Thus, assuming small deflections, the axial deflection of each blade is directly related to its lateral deflection [see Eqs. (3.65) and (3.66) for towers]:
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关于塔架,柔性悬臂梁在时间 $t$ 时的轴向(径向或展向)挠曲 $\nu(r,t)$,是基于柔性悬臂梁在挠曲时长度保持不变(沿梁的中心轴测量)这一假设的直接结果。因此,假设挠曲较小,每个叶片的轴向挠曲直接与它的摆振挠曲相关 [见 Eqs. (3.65) 和 (3.66) 关于塔架]。
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$$
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\nu(r,t)\!=\!\frac{I}{2}\!\int_{0}^{r}\!\left[\frac{\partial u(r^{\prime},t)}{\partial r^{\prime}}\right]^{2}\!d r^{\prime}
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\nu(r,t)\!=\!\frac{1}{2}\!\int_{0}^{r}\!\left[\frac{\partial u(r^{\prime},t)}{\partial r^{\prime}}\right]^{2}\!d r^{\prime}
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$$
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$$
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\nu(r,t)\!=\!\frac{I}{2}\sum_{i=p}^{N+p-I}\sum_{j=p}^{N+p-I}\!\left(\int_{\theta}^{r}\!\frac{d\varphi_{i}(r^{\prime})}{d r^{\prime}}\frac{d\varphi_{j}(r^{\prime})}{d r^{\prime}}\!d r^{\prime}\right)\!\!c_{i}(t)c_{j}(t)
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\nu(r,t)\!=\!\frac{1}{2}\sum_{i=p}^{N+p-1}\sum_{j=p}^{N+p-1}\left(\int_{\theta}^{r}\frac{d\varphi_{i}(r^{\prime})}{d r^{\prime}}\frac{d\varphi_{j}(r^{\prime})}{d r^{\prime}}d r^{\prime}\right)c_{i}(t)c_{j}(t)
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$$
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where $\acute{r}$ is a dummy variable representing the span along the flexible part of each blade.
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where $r'$ is a dummy variable representing the span along the flexible part of each blade.
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Substituting Eq. (3.74) into Eq. (3.72) results in Eq. 3.75 for the potential energy of each blade caused by rotor rotation:
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其中 $r'$ 是代表风轮柔性部分叶片展向的虚拟变量。
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将公式 (3.74) 代入公式 (3.72) 得到公式 3.75,用于表示因风轮旋转而导致每个叶片的势能:
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$$
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\begin{array}{l}{{\displaystyle V_{R o t a t i o n}=\frac{I}{2}\varOmega^{2}\sum_{i=p}^{N+p-I}\sum_{j=p}^{N+p-I}\Biggl[M_{T i p}R\!\int_{\partial}^{R-R_{H}}\!\frac{d\varphi_{i}(r)}{d r}\frac{d\varphi_{j}(r)}{d r}d r\mathop{+}}}\\ {{\displaystyle\int_{0}^{R-R_{H}}\mu_{B}\bigl(r\bigr)\bigl(R_{H}+r\bigr)\biggl(\int_{0}^{r}\!\frac{d\varphi_{i}(r^{\prime})}{d r^{\prime}}\frac{d\varphi_{j}(r^{\prime})}{d r^{\prime}}d r^{\prime}\biggr)d r^{\prime}\Biggr]c_{i}(t)c_{j}(t)}}\end{array}
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$$
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73
学术讲座-交流-面试/2025.5.28 DNV年度用户会议.md
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73
学术讲座-交流-面试/2025.5.28 DNV年度用户会议.md
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@ -0,0 +1,73 @@
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## bladed开发计划与进展
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### turbine challenge
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- growth
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- challege in structures aerodynamics stabilities
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- 固定式 - 行架式 - floating
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- 计算数增加,时间增长、工况复杂
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- 浮式平台多种形式
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### bladed4
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100+ industral customers
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long flexible blades
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deep stall
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叶片的高频振动,instability
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SIV-1 stall induced vibration
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攻角在-150-150,也会引起vibration SIV-2 ,存在两个问题 :1没有气动数据,没有unsteady 模型
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IAG dynamic stall model
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IAG在Cl Cd Cm上都更准确 13%,21%, 24%厚度
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IEA 15MW DLC 6.3
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BEM + no DS BEM + BL model 都出现叶片抖动
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BEM + IAG model 不抖动,但是也出现停顿
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风场是否有负切变仿真能力
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bladed5
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flexible turbine assembly tree
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superelement ()
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model imblence
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- add 模块之间的offset
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better support of early blade design process
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independent meshes between aerodynamic and sur
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基于api前处理和后处理
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bladed 5 new features
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yaw bear model
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add sensors to simulation
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