# Wind Turbines Around Cut-In Speed: Startup Optimization and Behavior Analysis Reported to MPP
## 5 Mathematical Model of Dual Powered Asynchronous/Synchronous Motor
### Visualization of the Process of Bringing It to the Optimal Rotational Speed  
Bringing the turbine to the point of maximum power by switching the electric generator to engine mode is a dynamic process. For this purpose, the kinetic momentum equation is used in the following form:  
\[J\cdot\frac{d\omega}{dt}=M_{WT}+M_{ENGINE} \tag{90}\]  
where \(\omega\) is \(MAS\), of the shaft of the electric motor, \(J\) is the equivalent moment of inertia, \(\mathrm{du}/\mathrm{dt}\) is derived in relation to time of \(MAS\), \(M_{WT}\) is the moment, in relation to the shaft of the electric motor, given by the \(WT\), \(M_{ENGINE}\) is the electromagnetic moment of the electric motor. Multiplying the equation of the kinetic moment by \(\omega\), at an equivalent moment of inertia of value \(J=136\) kg\(\cdot\)m\({}^{2}\), we obtain the equation of powers:  
\[136\cdot\frac{d\omega}{dt}\cdot\omega=P_{WT}+P_{ENGINE} \tag{91}\]  
where \(P_{WT}\) is the useful power given by \(WT\) in relation to the shaft of the electric motor, \(P_{ENGINE}\) is the electromagnetic power to the shaft of the electric motor. At a wind speed of 3 m/s, the power output of the turbine is:  
\[P_{WT}(\omega,3)=5.692\cdot 10^{5}\cdot\left(\frac{3}{\omega}-1.7488\cdot 10^{-2 }\right)\cdot e^{-41.495\cdot\frac{3}{\omega}}\cdot 3^{3} \tag{92}\]  
resulting:  
\[136\cdot\frac{d\omega}{dt}\cdot\omega=5.692\cdot 10^{5}\cdot\left(\frac{3}{ \omega}-1.7488\cdot 10^{-2}\right)\cdot e^{-41.495\cdot\frac{3}{\omega}} \cdot 3^{3}+P_{ENGINE} \tag{93}\]  
By summing up the stator and rotor active powers, \(P_{S}\) and \(P_{R}\), the shaft power of the motor is \(P_{ENGINE}=P_{S}+P_{R}=U_{S}\cdot\)8695.6\(\cdot\)sin \(\theta\) at \(U_{S}\) voltage and rotational speed \(n\), \(n=f=U_{S}/13.8\). The maximum power point (\(MPP\)) at a wind speed of 3 m/s is achieved at the optimal mechanical angular velocity, \(\omega_{OPTIM}\):  
\[\omega_{OPTIM}=24.046\cdot V=24.046\cdot 3=72.138\ [\mathrm{rad./s.}] \tag{94}\]  
The ideal mechanical angular speed, \(\omega_{OPTIM}\), should be attained within a time frame of one minute, leading to a change in frequency over that duration:  
\[f=\frac{\omega_{OPTIM}}{2\cdot\pi\cdot 60}\cdot t=\frac{72.138}{2\cdot\pi \cdot 60}\cdot t=0.19135\cdot t \tag{95}\]  
The power developed by the motor (83) does not depend on the \(MAS\), \(\omega\) being dependent only on the \(\mathrm{U_{S}}\) voltage and the load angle \(\theta\). In this case, the power equations are:  
\[136\cdot\frac{d\omega}{dt}\cdot\omega=5.692\cdot 10^{5}\cdot\left(\frac{3}{ \omega}-1.7488\cdot 10^{-2}\right)\cdot e^{-41.495\cdot\frac{3}{\omega}} \cdot 3^{3}+U_{S}\cdot 8695.6\cdot\sin\theta \tag{96}\]  
At the time moment \(t=60\) s, the system operates at the optimal mechanical angular speed, \(\omega_{OPTIM}\), in the \(MPP\) of the turbine, at the wind speed of 3 m/s:  
\[\omega_{OPTIM}=24.046\cdot V=24.046\cdot 3=72.138\ [\text{rad./s.}] \tag{105}\]  
Over the time interval \(\Delta t=60\) \(-\) 15.83 = 44.17 s, the double-powered synchronous generator outputs power to the grid, at variable frequency and voltage:  
\[U_{S}(t)=41.801+2.6406\cdot t \tag{106}\]  
In this case, the equation of motion becomes:  
\[ \begin{split} 136\frac{d\omega}{dt}\cdot\omega=5.692\cdot 10^{5}[(3/ \omega)-1.7488\cdot 10^{-2}]\cdot e^{-41.495(3/\omega)}\cdot 3^{3}-(41.801+2.2406 \cdot t)\cdot 8695.6\cdot\sin\vartheta\\ \frac{d\theta}{dt}-\omega+(19.032+1.2023\cdot t)=0\\ \omega(0)=19.032\\ \vartheta(0)=0.01\end{split}  \tag{107}\]  
by solving it, obtaining the variations in the load angle \(\vartheta\) and the mechanical angular speed \(\omega\), Figure 10.  
The power output from the generator:  
\[P_{EG}(72.138)=(41.801+2.6406\cdot t)\cdot 8695.6\cdot\sin\vartheta \tag{108}\]  
being dependent on the load angle that shows pronounced oscillations, it will also oscillate pronouncedly (Figure 11a). Figure 11: Variations in generator power: (**a**) On reaching the optimum speed and switching to generator operation; (**b**) In stationary operation.