obsidian_backup/书籍/力学书籍/OpenFast/FASTKinetics/auto/Drivetrain.md

The inertia of the drivetrain brings about generalized inertia forces and the load in the generator, high-speed shaft brake, gearbox (friction forces resulting from nonzero GBoxEff ) and the flexibility of the low speed shaft bring about generalized active forces. Note that all of these equations assume that the rotor is spinning about the positive c_{1} axis (they assume that the rotor can’t be forced to rotate in the opposite direction). This model works for any gearbox arrangement (including no gearbox, single stage, or multi-stage) as long as the generator rotates about the shaft axis (it may not be skewed relative to the shaft, even though it may rotate in the opposite direction of the low-speed shaft due to the gearbox stages). If there is no gearbox, simply set G B R a t i o\,=\,G B o x E f f=G e n D i r=1 (GB R e\nu e r s e={\mathrm{False}} ).
传动系惯性带来广义惯性力和发电机负载、高速轴制动器、齿轮箱（由于GBoxEff非零导致的摩擦力）以及低速轴挠曲性，都会产生广义主动力。需要注意的是，这些方程都假设转子绕正$c_{1}$轴旋转（即假设转子不能被强制旋转到相反方向）。该模型适用于任何齿轮箱布置（包括无齿轮箱、单级或多级），只要发电机绕轴旋转即可（即使它可能由于齿轮箱级数而以低速轴相反的方向旋转，但不能相对于轴倾斜）。如果不存在齿轮箱，只需将G B R a t i o\,=\,G B o x E f f=G e n D i r=1 (GB Rev e r s e={\mathrm{False}} )。

The mechanical torque within the generator is applied to the high speed shaft and equally and oppositely to the structure that furls with the rotor as follows:
发电机内部的机械扭矩施加到高速轴上，并以相等且相反的方式作用于与转子一起摆动的结构，具体如下：

\left.F_{r}\right|_{G e n}=\left(^{E}\omega_{r}^{G}-{^{E}\!\omega_{r}^{R}}\right)\cdot M_{G e n}^{G}\quad\left(r=l,2,...,22\right)\tag{1}

Thus,

\left.F_{r}\right|_{G e n}=\left\{\begin{array}{l l}{{^{E}\pmb{\omega}_{G e A z}^{G}\cdot M_{G e n}^{G}}}&{{f o r~~r=G e A z}}\\ {{0}}&{{o t h e r w i s e}}\end{array}\ \right.\tag{2}

where

\boldsymbol{M_{G e n}^{G}}=-G e n D i\boldsymbol{r}\cdot\boldsymbol{T^{G e n}}\left(G B R a t i o\cdot\dot{q}_{G e d z},t\right)\boldsymbol{c}_{1}\tag{3}

Note that a positive T^{G e n} represents a load (positive power extracted) whereas a negative T^{G e n} represents a motoring-up situation (negative power extracted, or power input). Thus,
请注意，正的 T^{G e n} 表示负载（提取正功率），而负的 T^{G e n} 表示电动运行情况（提取负功率，或功率输入）。因此，

F_{r}\Big|_{G e n}=\left\{\begin{array}{l l}{\left(G e n D i r\cdot G B R a t i o\cdot c_{1}\right)\cdot\left[-G e n D i r\cdot T^{G e n}\left(G B R a t i o\cdot\dot{q}_{G e d z},t\right)c_{1}\right]}&{f o r\;\;\;r=G e Az}\\ {0}&{o t h e r w i s e}\end{array}\right.\tag{4}

Or since G e n D i r^{2}=1 ,

F_r \Big|_{Gen} = \begin{cases}
-GRRatio \cdot T^{Gen} (GBRatio \cdot \dot{q}_{GeAz}, t) & \text{for } r = GeAz \\
0 & \text{otherwise}
\end{cases}\tag{5}

Thus,

[C(q,t)]|_{\text{Gen}} = 0\tag{6}

\{-f(\dot{q},q,t)\}|_{\text{Gen}} = \left\{
\begin{array}{l}
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
-GBRatio \cdot T^{\text{Gen}} (GBRatio \cdot \dot{q}_{\text{GeAz}}, t) \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots
\end{array}
\right\}\tag{7}

Similarly, the mechanical torque applied to the high-speed shaft from the high-speed shaft brake is applied equally and oppositely to the structure that furls with the rotor. Thus,
同样地，从高速轴制动器施加到高速轴上的机械扭矩，也会以相等且相反的方式作用于与转子一同回转的结构上。因此，

F_{r}\big|_{B r a k e}=\left\{\begin{array}{l l}{{^{E}\pmb{\omega}_{G e A z}^{G}\cdot M_{B r a k e}^{G}}}&{{f o r\;\;\;r=G e A z}}\\ {{0}}&{{o t h e r w i s e}}\end{array}\!\!\right.\tag{8}

where $$M_{B r a k e}^{G}=-G e n D i r\cdot T^{B r a k e}\left(t\right)c_{1}\qquad\mathrm{andwhere}\qquad T^{B r a k e}\left(t\right)=H S S B r k T\left(t\right)\tag{9}

which is assumed to be positive in value always.  Thus,  

F_{r}\bigr|_{B r a k e}=\left{\begin{array}{l l}{-G B R a t i o\cdot T^{B r a k e}\left(t\right)}&{f o r;;;r=G e A z}\ {O}&{o t h e r w i s e}\end{array}\right.\tag{10}

  

Thus,  

[C(q,t)]|_{\text{Brake}} = 0\tag{11}

{-f(\dot{q},q,t)}|_{\text{Brake}} = \left{ \begin{array}{l} \ldots \ \ldots \ \ldots \ \ldots \ \ldots \ \ldots \ -GBRatio \cdot T^{\text{Brake}}(t) \ \ldots \ \ldots \ \ldots \ \ldots \ \ldots \ \ldots \end{array} \right}\tag{12}

If the translational inertia of the drivetrain is assumed to be incorporated into that of the structure that furls with the rotor, then the high-speed shaft generator inertia generalized force is as follows: 
如果假设动能惯性由驱动系传递至随转子摆动的结构部分，则高速轴发电机惯性概括力如下： 

F_r^*|{\text{G}} = {}^E \omega_r^{\text{G}} \cdot (-\overline{\overline{{I}}}^{\text{G}} \cdot {}^E {a}^{\text{G}} - {}^E \boldsymbol{\omega}^{\text{G}} \times \overline{\overline{{I}}}^{\text{G}} \cdot {}^E \boldsymbol{\omega}^{\text{G}}) \quad \text{where} \quad \overline{\overline{{I}}}^{\text{G}} = \text{GenInerc}{1}c_1 \quad\left(r=1,2,...,22\right)\tag{13}

or, 

F_r^*|_{\text{G}} = {}^E \boldsymbol{\omega}r^{\text{G}} \cdot \left{ -\overline{\overline{{I}}}^{\text{G}} \cdot \left{ \left( \sum{i=4}^{13} {}^E \omega_i^{\text{G}} \ddot{q}i \right) + \left[ \sum{i=7}^{13} \frac{d}{dt} \left( {}^E \boldsymbol{\omega}_i^{\text{G}} \right) \dot{q}_i \right] \right} - {}^E \boldsymbol{\omega}^{\text{G}} \times \overline{\overline{{I}}}^{\text{G}} \cdot {}^E \boldsymbol{\omega}^{\text{G}} \right} \quad\left(r=1,2,...,22\right)\tag{14}

However, since $c_{1}\cdot\frac{d}{d t}(^{E}\omega_{G e A z}^{G}\Big)\propto c_{1}\cdot(^{E}\omega^{R}\times c_{1}\Big)={^{E}\omega^{R}}\cdot\Big(c_{1}\times c_{1}\Big)=0$ (the first $c_{1}$ coming from $\overline{\overline{{I}}}^{\text{G}}$ ), this simplifies as follows:  

F_r^*|_{\text{G}} = {}^E \boldsymbol{\omega}r^{\text{G}} \cdot \left{ -\overline{\overline{{I}}}^{\text{G}} \cdot \left{ \left( \sum{i=4}^{13} {}^E \boldsymbol{\omega}_i^{\text{G}} \ddot{q}i \right) + \left[ \sum{i=7}^{12} \frac{d}{dt} \left( {}^E \boldsymbol{\omega}_i^{\text{G}} \right) \dot{q}_i \right] \right} - {}^E \boldsymbol{\omega}^{\text{G}} \times \overline{\overline{{I}}}^{\text{G}} \cdot {}^E \boldsymbol{\omega}^{\text{G}} \right} \quad\left(r=1,2,...,22\right)\tag{15} $$Or,

F_r^*|_{G} = \begin{cases}
- {}^E \boldsymbol{\omega}_r^{\text{R}} \cdot \overline{\overline{{I}}}^{\text{G}} \cdot \left\{ \left( \sum_{i=4}^{13} {}^E \boldsymbol{\omega}_i^{\text{G}} \ddot{q}_i \right) + \left[ \sum_{i=7}^{12} \frac{d}{dt} \left( {}^E \boldsymbol{\omega}_i^{\text{R}} \right) \dot{q}_i \right] \right\} - ({}^E \boldsymbol{\omega}_r^{\text{R}} \cdot ({}^E \boldsymbol{\omega}^{\text{G}} \times \overline{\overline{{I}}}^{\text{G}} \cdot {}^E \boldsymbol{\omega}^{\text{G}})) & \text{for } r = 4,5,...,12 \\
- \text{GenDir} \cdot \text{GenIner} \cdot \text{GBRatio}  \left\{ \left( \sum_{i=4}^{12} {}^E \boldsymbol{\omega}_i^{\text{R}} \ddot{q}_i \right) + \left[ \sum_{i=7}^{12} \frac{d}{dt} \left( {}^E \boldsymbol{\omega}_i^{\text{R}} \right) \dot{q}_i \right] \right\} \cdot c_1 - \text{GenIner} \cdot \text{GBRatio}^2 \cdot \ddot{q}_{\text{GeAz}}  - \text{GenDir} \cdot \text{GBRatioc}_1 \cdot ({}^E \boldsymbol{\omega}^{\text{G}} \times \overline{\overline{{I}}}^{\text{G}} \cdot {}^E \boldsymbol{\omega}^{\text{G}}) & \text{for } r = \text{GeAz} \\
0 & \text{otherwise}
\end{cases}\tag{16}

However since c_{1}\cdot\left({\,^{E}}{\omega^{G}}\times c_{1}\,\right)={^{E}}{\omega^{G}}\cdot\left(c_{1}\times c_{1}\right)=0 (the first c_{1} coming from \overline{{\overline{{I}}}}^{G} ), this simplifies again as follows:
然而由于 c_{1}\cdot\left({\,^{E}}{\omega^{G}}\times c_{1}\,\right)={^{E}}{\omega^{G}}\cdot\left(c_{1}\times c_{1}\right)=0 （第一个 c_{1} 来自 $\overline{{\overline{{I}}}}^{G}$），这再次简化为：

F_r^*|_{G} = \begin{cases}
- {}^E \boldsymbol{\omega}_r^{\text{R}} \cdot \overline{\overline{{I}}}^{\text{G}} \cdot \left\{ \left( \sum_{i=4}^{13} {}^E \boldsymbol{\omega}_i^{\text{G}} \ddot{q}_i \right) + \left[ \sum_{i=7}^{12} \frac{d}{dt} \left( {}^E \boldsymbol{\omega}_i^{\text{R}} \right) \dot{q}_i \right] \right\} - ({}^E \boldsymbol{\omega}_r^{\text{R}} \cdot ({}^E \boldsymbol{\omega}^{\text{G}} \times \overline{\overline{{I}}}^{\text{G}} \cdot {}^E \boldsymbol{\omega}^{\text{G}})) & \text{for } r = 4,5,...,12 \\
- \text{GenDir} \cdot \text{GenIner} \cdot \text{GBRatio}  \left\{ \left( \sum_{i=4}^{12} {}^E \boldsymbol{\omega}_i^{\text{R}} \ddot{q}_i \right) + \left[ \sum_{i=7}^{12} \frac{d}{dt} \left( {}^E \boldsymbol{\omega}_i^{\text{R}} \right) \dot{q}_i \right] \right\} \cdot c_1 - \text{GenIner} \cdot \text{GBRatio}^2 \cdot \ddot{q}_{\text{GeAz}} & \text{for } r = \text{GeAz} \\
0 & \text{otherwise}
\end{cases}\tag{17}

The terms associated with DOFs 4,5,…,12 represent the fact that the rate of change of angular momentum of the generator can be considered as an additional torque on the structure that furls with the rotor (i.e., in addition to the torques on the structure transmitted directly from the low-speed shaft).
与DOF 4、5、…、12相关的项，表示发电机角动量变化率可以被视为一个附加于随转子展开的结构上的扭矩（即，除了直接从低速轴传递到结构上的扭矩之外）。

[C(q,t)]|_G = \begin{cases}
{}^E \boldsymbol{\omega}_{\text{Row}}^{\text{G}} \cdot \overline{\overline{{I}}}^{\text{G}} \cdot {}^E \boldsymbol{\omega}_{\text{Col}}^{\text{G}} & \text{for (Row, Col = 4,5,...,13)} \\
0 & \text{otherwise}
\end{cases}\tag{18}

\{-f(\dot{q},q,t)\}|_G = \begin{cases}
- {}^E \boldsymbol{\omega}_{\text{Row}}^{\text{G}} \cdot \left\{ \overline{\overline{{I}}}^{\text{G}} \cdot \left[ \sum_{i=7}^{12} \frac{d}{dt} \left( {}^E \boldsymbol{\omega}_i^{\text{R}} \right) \dot{q}_i \right] + {}^E \boldsymbol{\omega}^{\text{G}} \times \overline{\overline{{I}}}^{\text{G}} \cdot {}^E \boldsymbol{\omega}^{\text{G}} \right\} & \text{for (Row = 4,5,...,12)} \\
- {}^E \boldsymbol{\omega}_{\text{Row}}^{\text{G}} \cdot \overline{\overline{{I}}}^{\text{G}} \cdot \left[ \sum_{i=7}^{12} \frac{d}{dt} \left( {}^E \boldsymbol{\omega}_i^{\text{R}} \right) \dot{q}_i \right] & \text{for (Row = 13)} \\
0 & \text{otherwise}
\end{cases}\tag{19}

F_r |_{\text{ElasticDrive}} = -\frac{\partial V_{\text{Drive}}}{\partial q_r} \quad (r = 1,2,...,22) \quad \text{where} \quad V_{\text{Drive}} = \frac{1}{2} \text{DTTorSpr} \cdot q_{\text{DrTr}}^2\tag{20}

So,

F_r |_{\text{ElasticDrive}} = \begin{cases}
-DTTorSpr \cdot q_{DrTr} & \text{for } r = DrTr \\
0 & \text{otherwise}
\end{cases}\tag{21}

and likewise

F_r |_{\text{DampDrive}} = \begin{cases}
-DTTorDmp \cdot \dot{q}_{DrTr} & \text{for } r = DrTr \\
0 & \text{otherwise}
\end{cases}\tag{22}

Thus,

[C(q,t)]|_{\text{ElasticDrive}} = 0\tag{23}

\{-f(\dot{q},q,t)\}|_{\text{ElasticDrive}} = \begin{Bmatrix}
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
-DTTorSpr \cdot q_{DrTr} \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\end{Bmatrix}\tag{24}

[C(q,t)]|_{\text{DampDrive}} = 0\tag{25}

\{-f(\dot{q},q,t)\}|_{\text{DampDrive}} = \begin{Bmatrix}
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
-DTTorDmp \cdot \dot{q}_{DrTr} \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\end{Bmatrix}\tag{26}

Similar to the generator and HSS brake, the mechanical friction torque applied to the high speed shaft is applied equally and oppositely to the structure that furls with the rotor. Thus,
类似于发电机和HSS制动器，施加到高速轴上的机械摩擦扭矩，以相等且相反的方式作用于与转子一同旋转的结构上。因此，

F_r |_{\text{GBFric}} = \begin{cases}
{}^E\omega_{\text{GeAz}}^G \cdot M_{\text{GBFric}}^G & \text{for } r = \text{GeAz} \\
0 & \text{otherwise}
\end{cases}
\quad \text{where} \quad
M_{\text{GBFric}}^G = \frac{T_{\text{GBFric}}^G(\ddot{q}, \dot{q}, q, t) c_1}{\text{GBRatio} \cdot \text{GenDir}}\tag{27}

where, from a free-body diagram of the high and low-speed shafts, it is easily seen that the friction torque applied on the LSS upon the gearbox entrance, T^{G B F r i c}\left(\ddot{q},\dot{q},q,t\right) , is always positive in value and equal to: 从高速轴和低速轴的受力图上可以清楚地看到，作用于低速轴进入变速箱处的摩擦转矩，$T^{G B F r i c}\left(\ddot{q},\dot{q},q,t\right)$，其值始终为正，且等于：

T^{\text{GBFric}}(\ddot{q}, \dot{q}, q, t) = \left[ \text{GenIner} \cdot \text{GBRatio}^2 \cdot \ddot{q}_{\text{GeAz}} + \text{GenDir} \cdot \text{GenIner} \cdot \text{GBRatio} \cdot {}^Ea^R \cdot c_1 + \text{GBRatio} \cdot T^{\text{Gen}}(\text{GBRatio} \cdot \dot{q}_{\text{GeAz}}, t) + \text{GBRatio} \cdot T^{\text{Brake}}(t) \right] \cdot \left[ \frac{1}{\text{GBBoxEff}^{\text{SIGN(LSShftTq)}}} - 1 \right]\tag{28}

Thus,

F_r |_{\text{GBFric}} = \begin{cases}
-T^{\text{GBFric}}(\ddot{q}, \dot{q}, q, t) & \text{for } r = \text{GeAz} \\
0 & \text{otherwise}
\end{cases}\tag{29}

F_r |_{\text{GBFric}} = \begin{cases}
- \left[ \text{GenIner} \cdot \text{GBRatio}^2 \cdot \ddot{q}_{\text{GeAz}} + \text{GenDir} \cdot \text{GenIner} \cdot \text{GBRatio}^E a^R \cdot c_1 + \text{GBRatio} \cdot T^{\text{Gen}}(\text{GBRatio} \cdot \dot{q}_{\text{GeAz}}, t) + \text{GBRatio} \cdot T^{\text{Brake}}(t) \right] \cdot \left[ \frac{1}{\text{GBoxEff}^{\text{SIGN(LSShftTq)}}} - 1 \right] & \text{for } r = \text{GeAz} \\
0 & \text{otherwise}
\end{cases}\tag{30}

Or,

F_r |_{\text{GBFric}} = \begin{cases}
- \left[ \text{GenIner} \cdot \text{GBRatio}^2 \cdot \ddot{q}_{\text{GeAz}} \right. \\
\quad \left. + \text{GenDir} \cdot \text{GenIner} \cdot \text{GBRatio} \left\{ \sum_{i=4}^{12} {^E \omega_i^R \ddot{q}_i} + \left[ \sum_{i=7}^{12} \frac{d}{dt} ({^E \omega_i^R}) \dot{q}_i \right] \right\} \cdot c_1 \right. \\
\quad \left. + \text{GBRatio} \cdot T^{\text{Gen}}(\text{GBRatio} \cdot \dot{q}_{\text{GeAz}}, t) + \text{GBRatio} \cdot T^{\text{Brake}}(t) \right] \cdot \left[ \frac{1}{\text{GBoxEff}^{\text{SIGN(LSShftTq)}}} - 1 \right] & \text{for } r = \text{GeAz} \\
0 & \text{otherwise}
\end{cases}\tag{31}

Thus,

[C(q,t)]_{\text{GBFric}} = \begin{cases}
{^E\omega_{\text{Row}}^G \cdot \overline{\overline{{I}}}^{\text{G}} \cdot {^E\omega_{\text{Col}}^G}} \left[ \frac{1}{\text{GBoxEff}^{\text{SIGN(LSShftTq)}}} - 1 \right] & \text{for } (\text{Row}=13, \text{Col}=4,5,...,13) \\
0 & \text{otherwise}
\end{cases}\tag{32}

\{-f(\dot{q},q,t)\}_{\text{GBFric}} = \begin{cases}
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
- \left\{ {^E\omega_{\text{GeAz}}^G \cdot \overline{\overline{{I}}}^{\text{G}} \cdot \left[ \sum_{i=7}^{12} \frac{d}{dt} ({^E\omega_i^R}) \dot{q}_i \right]} + \text{GBRatio} \cdot T^{\text{Gen}}(\text{GBRatio} \cdot \dot{q}_{\text{GeAz}}, t) + \text{GBRatio} \cdot T^{\text{Brake}}(t) \right\} \cdot \left[ \frac{1}{\text{GBoxEff}^{\text{SIGN(LSShftTq)}}} - 1 \right] \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\end{cases}\tag{33}

It is noted that the gearbox friction generalized active force effects both the mass matrix and the forcing function. Its effect on the mass matrix can be thought of as an apparent mass in the system (i.e., an active friction force which is a function of the generator acceleration). The gearbox friction causes the mass matrix to become unsymmetric. Note that the equation for DOF GeAz is an implicit equation (since the gearbox friction is a function of DOF GeAz), which has no closed-form solution. To avoid the complications resulting from this implicit characterization, the value of the LSShftTq from the previous time step is used in the SIGN() function in place of the value of the LSShftTq in the current time step. This may be done since it may be assumed that LSShftTq will not change much between time steps if the time step is considered small enough. Note that gearbox friction is the only term in the equations of motion that cause the mass matrix to be unsymmetrical. The mass matrix will only be unsymmetric if GBoxEff is not 100\% —if GBoxEff is 100\% , then the mass matrix is completely symmetric.

值得注意的是，齿轮箱摩擦广义主动力同时影响质量矩阵和激励函数。它对质量矩阵的影响可以被视为系统中的一个表观质量（即，一个主动摩擦力，它是发电机加速度的函数）。齿轮箱摩擦导致质量矩阵变为非对称。请注意，自由度GeAz的方程是一个隐式方程（因为齿轮箱摩擦是自由度GeAz的函数），它没有闭式解。为了避免这种隐式特性带来的复杂性，在SIGN()函数中使用了上一时间步的LSShftTq值，而不是当前时间步的LSShftTq值。这样做是可行的，因为如果时间步长足够小，可以假设LSShftTq在时间步之间不会有太大变化。请注意，齿轮箱摩擦是运动方程中唯一导致质量矩阵不对称的项。质量矩阵只有在GBoxEff不是100%时才不对称——如果GBoxEff是100%，那么质量矩阵是完全对称的。