Wind Turbines Around Cut-In Speed: Startup Optimization and Behavior Analysis Reported to MPP

Abstract

The conversion of air currents through wind turbine technology stands as one of the most significant and effective means of generating green electricity. Wind turbines featuring a horizontal axis exhibit the greatest installed capacity. The study establishes a mathematical model for large wind turbines, categorized by megawatt output, utilizing measured data for key parameters, including wind speed, power output from the generator, and rotational speed. The analysis of the system’s behavior on startup—the cut-in wind speed, is conducted by transitioning the electric generator into motor mode. A mathematical model has been established for the dual-powered motor configuration, wherein both the stator and rotor are connected to a common frequency network, facilitating a shift to synchronous motor functionality. The equation that describes the kinetic moment highlights the importance of attaining optimal velocity, while simultaneously accounting for variations in the load angle. These fluctuations are observable in both the power output and the electrical currents. The simulations that have been processed are derived from experimental data, specifically inputs obtained from a 1.5 MW wind turbine located in the Oravita region of southwestern Romania. The paper thus outlines essential elements concerning the functionality of high-power wind turbines that utilize wound rotor induction generators, aiming to guarantee optimal performance from the moment the wind speed reaches the cut-in threshold.

1. Introduction

The fulfillment of the Sustainable Development Goals also envisages the development of renewable energy production systems, so that it is accessible to all socio-economic sectors, and the price is convenient for consumers [1].

Wind energy is one of several renewable energy sources used in the production of electricity due to rising electricity consumption and environmental pollution. In choosing the type of wind turbine, the control system must consider the characteristics of the area, wind speed, climatic conditions, local and regional regulations as well as the length of the investment, which for a wind farm is about 10 years [2,3]. Currently, wind systems are also used together with other electricity production and/or storage systems. One of the most widely used storage systems is the pumped storage plant, which has a large storage capacity and contributes to the stability of the power supply system [4,5].

The overall efficiency of the wind energy system is significantly affected by the aerodynamic characteristics of the turbine, which can be enhanced through advanced aerodynamic modeling and refined design techniques. In the realm of design, certain researchers advocate for the integration of Industry 5.0 principles alongside optimization methods utilizing genetic algorithms. This innovative strategy has resulted in production improvements of as much as 10% [6]. Most wind turbines are designed to provide power output depending on wind speed, not being able to quickly change their power to sudden changes in load [7]. In order to ensure maximum capture of wind energy, it is necessary to identify the dependence of the optimal mechanical angular speed (MAS) on wind speed. This dependence can be determined by calculating the maximum power provided by the wind turbine, WT, at different wind speeds, also considering the energy efficiency of the wind system [8,9]. In the case of offshore wind systems, it has been found that the tracking of the maximum power point (MPP) and the control of the total output power of the wind system can be performed by controlling the current of the d-axis of the active rectifier related to the system [10]. Recent studies examine and outline various control strategies, including maximum power point tracking, control of turbine blade pitch angles, and stationary wind turbine control [11].

A challenge for variable renewable energy sources (RES) in isolated grids is to ensure control of the frequency of supply voltage and system stability [12]. Thus, in the situation where the share of RES is 100%, and the production system is hybrid, consisting of a hydroelectric unit, a pumping plant and a wind farm, it is also necessary to introduce energy storage system [13,14]. This will significantly improve frequency control and ensure continuity in supply, even if the wind speed is not constant.

Power oscillations affect the structural system of turbines. To dampen these oscillations, power oscillation damping (POD) regulators are used, whose effect on the wind tur-bines is comparable to that determined by variations in wind speed and much lower than that determined by transient effects in the case of three-phase short circuit [15]. In cases of the use of a double-powered induction generator (DFIG) with an adequate controller in wind power systems, following determinations based on numerical simulations, it was found that it works properly in dynamic regimes determined by fluctuations in wind speed and changes in grid energy values [16,17].

In order to analyze the operation of the wind system, it is necessary to know the wind speed [18]. This is usually measured with the sensor located on the nacelle, but there are situations in which the measured value is different from the actual wind value. In [19], to determine the effective wind speed it is proposed to reverse the aerodynamic model of the turbine after estimating the related torque and the speed of the turbine rotor. This method proved to be accurate under laboratory conditions and can be used in regulation and control structures related to wind power systems. Another method used to measure wind speed is to determine the speed with a low-power turbine, which operates without load at mechanical angular speed [20]. Previous research on wind speed forecasting, essential for modeling, employs artificial intelligence that utilizes statistical models grounded in machine learning [21]. In all cases, the control functions of the entire system play an essential role in achieving increased wind turbine efficiency [9,22,23,24].

The use of applications for modeling and simulating the operation of wind systems is important both in the design and operation optimization phases, as they allow, through the mathematical models used, the realization of different configurations and different operating scenarios for the steady state and transient regime [25,26,27,28]. Also based on simulations and experimental data regarding the value of the wind turbines provided power, it is possible to establish whether their operation is in the optimal zone in terms of captured energy [29,30,31,32].

In this paper, the mathematical models of the wind turbine and the double powered asynchronous motor are determined by switching the electric generator into motor mode, so that the turbine reaches the optimal speed in the shortest possible time interval and operates in MPP [8,9]. One of the main contributions consists of visualizing the commissioning process of high-power wind turbines equipped with dual-powered induction generators. In the specialized literature of the approached field, for the commissioning of high-power wind turbines to operate at the point of maximum power, simulations with the dual motor powered by the stator and rotor connected to the same frequency network have not been identified [2,5,20].

Wind turbines reach the MPP at the optimal mechanical angular speed, in a time interval of tens of seconds. This article addresses precisely this problem: the optimal operation of high-power wind power plants by driving the wind turbine with the dual motor powered by the stator and rotor at the same frequency. At wind speeds that vary over time, the wind turbine operates permanently in dynamic mode at variable speeds [10,13].

The process being dynamic, we use the equation of the kinetic momentum [7,14,16] of the form:

$$J\cdot {\displaystyle \frac{d\omega }{dt}}\cdot \omega =P_{WT}+P_{ENGINE}$$

(1)

where ω—mechanical angular speed, MAS, at the shaft of the electric motor/generator, J—the equivalent moment of inertia, dω—the derivative with respect to time of MAS, P_WT—the useful power given by WT relative to the shaft of the electric motor/generator, P_ENGINE—the electromagnetic power at the shaft of the induction motor.

The double-powered induction motor, coming from the double-powered induction generator, having the same frequency in the stator and rotor, becomes the synchronous motor. The synchronous motor operates at variable frequency and controlled stator flow.

This paper examines the challenges associated with the integration of wind turbines into the energy system, particularly as wind speed increases from zero to the cut-in speed. The power generated by the turbine is contingent upon the wind speed, v, and the rotational speed, n. The wind energy system operates at the maximum power point within the optimal range at the ideal rotational speed [29], as exemplified in Figure 1.

The rotational speeds of wind turbines equipped with a gearbox are referenced to the shaft of the electric generator. At the turbine shaft, the rotational speed values are k_T times lower, where k_T represents the gear ratio of the transmission. The wind speed is a fundamental parameter that determines the optimal rotational speed, n_OPTIM, ensuring that the turbine operates at its maximum power point.

It is evident that at low rotational speeds, specifically below 5 revolutions per second, the power output generated by the turbine is minimal. Consequently, the time required to reach optimal rotational speed can be significantly prolonged, particularly when wind speeds are low and the equivalent moment of inertia, J, is on the order of hundreds of kgm². Therefore, during this range, wind energy is not effectively harnessed.

2. The Mathematical Model of the Double-Powered Induction Generator

The wind turbine operates at the point of maximum power by changing the load on the double-powered induction generator.

The wind turbine on which the analysis is based is Fuhrlander FL MD 70, equipped with double-powered induction generators, with the following nominal data [31]:

-

Nominal power P_N = P_EG = 1.5 MW;

-

Nominal voltage, U_N—690 V;

-

Nominal current, I_N = P_EG/3U_N = 1,500,000/3 · 690 = 724.64 A;

-

Nominal rotation, n_N—1500 rpm.;

-

Maximum rotation, n_max—1800 rpm.;

-

Equivalent moment of inertia, J—136 [kgm²].

The generator’s parameters are established based on the nominal data P_N, U_N, I_N. The short-circuit impedance, Z_sc, is calculated from the ratio between the nominal voltage, U_N and the short-circuit current I_sc, considering I_sc ≅ 9 · I_N.

$$Z_{sc}={\displaystyle \frac{U_N}{9\cdot I_N}}={\displaystyle \frac{690}{9\cdot 724.64}}=0.1058 [\Omega ]$$

(2)

$$Z_{sc}=\sqrt{R_{sc}^2+X_{sc}^2}=0.1058 [\Omega ]$$

(3)

Short-circuit resistance, R_sc, is calculated from short-circuit power [33]:

$$P_{sc}\cong 0.05\cdot P_N$$

(4)

$$3\cdot R_{sc}\cdot I_N^2=0.05\cdot P_N=0.05\cdot \mathrm{1,500,000}=\mathrm{75,000} [\mathrm{W}]$$

(5)

Obtaining R_sc = 4.7610·10⁻².

The short-circuit reactance, X_sc, is calculated from the short-circuit impedance:

$$X_{sc}^2=Z_{sc}^2-R_{sc}^2={0.1058}^2-{(2.3805\cdot {10}^{-2})}^2=1.0627\cdot {10}^{-2} [\Omega ]$$

(6)

$$X_{sc}=X_1+X_2=\sqrt{1.0627\cdot {10}^{-2}}=0.10309 [\Omega ]$$

(7)

$$X_1=\omega \cdot L_1=X_2=\omega \cdot L_2={\displaystyle \frac{0.10309}{2}}=5.1545\cdot {10}^{-2}=314 \mathrm{L}$$

(8)

Short-circuit inductances: L₁ = L₂ = L = 1.6416∙10⁻⁶ [H].

The magnetization reactance, X_M, is calculated using the current from idle operation: I_no-load = 0.05∙I_N.

$$X_M\cong {\displaystyle \frac{U_N}{I_{no-load}}}\cong \cdot {\displaystyle \frac{690}{0.05\cdot 724.64}}=19.044 [\Omega ]$$

(9)

$$X_M=\omega \cdot L_M=314\cdot L_M=19.044 [\Omega ]$$

(10)

Magnetization inductance: L_M = 6.0650∙10⁻² H.

With these parameters of the generator, the stator and rotor equations are:

$$U_S=\left(R_1+j\cdot X_1\right)\cdot I_S+j\cdot X_M\cdot \left(I_R+I_S\right)$$

(11)

$$U_R=\left({\displaystyle \frac{R_2}{s}}+j\cdot X_2\right)\cdot I_S+j\cdot X_M\cdot I_R$$

(12)

where s is the slip, s = (n₁ − n)/n₁, n₁ is the speed of the rotating field, and n—the rotational speed of the rotor.

The rotor parameters R₂ and X₂ are reduced to the stator in the form of

$$R_2=k^2\cdot R_{2-REAL} X_2=k^2\cdot X_{2-REAL}$$

(13)

where k = N₁/N₂, N₁ is the number of turns per phase in the stator, and N₂ the number of turns per phase in the rotor.

Rotor tension U_R is obtained from the real rotor tension U_R––REAL by multiplying by k and dividing by sliding s:

$$U_R={\displaystyle \frac{k\cdot U_{R-REAL}}{s}}$$

(14)

Since the rotor is wound, a rotor tension can be imposed U_R to obtain the desired rotational speed.

3. Mathematical Model of the Wind Turbine

The mathematical models of WT are relatively well presented in the literature [4,8,19]. In order to achieve, at a given wind speed, a maximum capture of wind energy, the WT must operate at the optimal MAS, ω_OPTIM, in the maximum power point, MPP [22,27,34].

The mathematical model of WT, in the authors’ conception, is the characteristic of WT power, the function P_WT(ω, V, β), with V—wind speed [m/s] and β—the angle of inclination of the blades [28,35,36]:

$$P_{WT}\left(\omega ,V,\beta \right)=\rho \cdot \pi \cdot R_p^2\cdot C_p\left(\lambda ,\beta \right)\cdot V^3$$

(15)

where ρ is the density of the air in the operating location of the WT, R_p blades radius, C_p(λ) is the power conversion coefficient, and λ = ω∙R_p/V.

The power conversion coefficient, C_p(λ), at WT with three blades and the inclination angle of the blades β, fixed, is determined with the relationship [19,20,22,23]:

$$C_p\left(\lambda \right)=c_1\cdot \left({\displaystyle \frac{c_2}{\Lambda }}-c_3\right)\cdot e^{-\left({\displaystyle \frac{c_4}{\Lambda }}\right)}$$

(16)

where c₁, c₂, c₃, c₄ are constructive constants, given in the catalog.

$${\displaystyle \frac{1}{\Lambda }}={\displaystyle \frac{1}{\lambda }}-0.035={\displaystyle \frac{V}{R_p\cdot \omega }}-0.035={\displaystyle \frac{V}{1.5\cdot \omega }}-0.035$$

(17)

In wind turbines with blades having an adjustable angle of inclination, the power conversion coefficient, C_p(λ, β) is determined by the relation [23,28]:

$$C_p\left(\lambda ,\beta \right)=c_1\cdot \left({\displaystyle \frac{c_2}{\Lambda }}-c_3\right)\cdot e^{-\left({\displaystyle \frac{c_4}{\Lambda }}\right)}$$

(18)

where the parameter Λ has the form:

$${\displaystyle \frac{1}{\Lambda }}={\displaystyle \frac{1}{\lambda }}\exp \left(-d\cdot \beta \right)-0.035={\displaystyle \frac{V}{R_p\cdot \omega }}\exp \left(-d\cdot \beta \right)-0.035={\displaystyle \frac{V}{1.5\cdot \omega }}\exp \left(-d\cdot \beta \right)-0.035$$

(19)

By substituting Λ, the power conversion coefficient, C_p(λ, β), is obtained in the form:

$$\begin{array}{l}{C}_p\left(\lambda ,\beta \right)=c_1\cdot \left({\displaystyle \frac{c_2}{\Lambda }}-c_3\right)\cdot e^{-\left({\displaystyle \frac{c_4}{\Lambda }}\right)}=c_1\left(c_2\cdot {\displaystyle \frac{V}{1.5\cdot \omega }}\exp \left(-d,\beta \right)-0.035-c_3\right)\cdot e^{-c_4\left({\displaystyle \frac{V}{1.5\cdot \omega }}-0.035\right)}=\\ a_1\cdot \left[{\displaystyle \frac{V}{\omega }}exp\left(-d,\beta \right)-b\right]\cdot e^{-c\cdot {\displaystyle \frac{V}{\omega }}}\end{array}$$

(20)

where the values of the parameters a, b, c and d are determined from the experimental data.

In this paper, we present a mathematical model of WT, MM-WT, deduced by measuring the wind speed, V, the power output by the generator, P_EG and the MAS at the generator, or WT, n/ω.

As atmospheric conditions, mechanical stresses and even turbine geometry change over time, and the initial MM-WT changes over time. Therefore, a periodic recalculation of it is required based on measurements of the fundamental quantities: wind speed, power and MAS, at the electric generator.

In most of the works [11,37,38] the existing wind systems are treated on the basis of inadequate MM-WT and without efficient processing of experimental data. The experimental data can be analyzed through various algorithms, methodologies, and specified conditions [39,40]. On the basis of MM-WT, the dependence of the optimal mechanical angular speed, ω_OPTIM(V), on the wind speed, V—reference quantity in the r control algorithms, is determined for wind systems operating at time-varying wind speeds.

A maximum capture of wind energy, at wind speeds that vary significantly over time, can only be achieved on the basis of a valid, constantly updated MM-WT. It is preferable that the determination of the MM-WT is carried out during the operation of the wind system. To ensure the operation of the wind turbine at maximum power, it is essential to know the mathematical model of WT as accurately as possible.

The determination of MM-WT is based on experimental data from the 1.5 MW WT of type Fuhrländer FL MD 70, within the Oravita wind farm. The mathematical model of WT defined by the experimental power characteristics, P_WT(ω, V), [4,16,19,30] was initially deduced, at a fixed blade tilt angle, β = 0.

The value of the maximum power is obtained at ω_OPTIM, by canceling the power derivative:

$${\displaystyle \frac{d{P}_{WT}}{d\omega }}=0$$

(21)

results in:

$$\omega -V\cdot c+b\cdot c\cdot \omega =0$$

(22)

and from here [29,37,38]:

$${\omega }_{OPTIM}={\displaystyle \frac{c}{1+b\cdot c}}\cdot V=k\cdot V$$

(23)

The period was defined as an interval, during which the wind speed remains steady, allowing the wind turbine to function at its maximum efficiency, the maximum power point. The measured values for wind speed, output power, P_EG and MAS at the generator, ω, the experimental data were collected [30]: ω_OPTIM = 204.39 rad/s, V = 8.5 m/s, P_WT-MAX = 1500 kW.

From the experimental data, we obtained ω_OPTIM/V = 24.046, or c/(1 + b·c) = 24.046. This is the first equation of the system in the unknowns a, b, and c.

The value of the maximum power of the WT corresponds to the optimal MAS, ω_OPTIM:

$${\displaystyle \frac{V}{{\omega }_{OPTIM}}}={\displaystyle \frac{1+b\cdot c}{c}}$$

(24)

and it is:

$$P_{WT-MAX}\left(V\right)=a\cdot \left({\displaystyle \frac{1+b\cdot c}{c}}-b\right)\cdot e^{-c\left({\displaystyle \frac{1+b\cdot c}{c}}\right)}\cdot V^3={\displaystyle \frac{a}{c}}\cdot e^{-1-c\cdot b}\cdot V^3$$

(25)

Results:

$$k_p={\displaystyle \frac{a}{c}}\cdot e^{-1-c\cdot b}$$

(26)

Analogous to ω_OPTIM, from the experimental data, the proportionality factor k_p has the value:

$$k_p={\displaystyle \frac{P_{WT-MAX}}{V^3}}={\displaystyle \frac{1500}{{8.5}^3}}=2.4425$$

(27)

Thus, the second equation of the system results in the unknowns a, b and c:

$${\displaystyle \frac{a}{c}}\cdot e^{-1-c\cdot b}=2.4425$$

(28)

These two equations are essential in determining MM-WT and in establishing MPP, the ω_OPTIM and P_WT-MAX coordinates. To determine the unknown variables a, b, and c, an extra equation is necessary. However, this equation is not critical for establishing the coordinates of the Maximum Power Point (MPP), which can instead be derived from the ratio: ω_OPTIMAL/ω_MAXIM, where ω_MAXIM is the maximum mechanical angular speed, at the idle operation of the WT. The value of the ω_OPTIM/ω_MAX ratio generates the third equation. Since the WT power is zero at idle:

$$P_{WT}=a\cdot \left({\displaystyle \frac{V}{\omega }}-b\right)\cdot e^{-c\cdot {\displaystyle \frac{V}{\omega }}}\cdot V^3=0$$

(29)

at MAS ω = ω_MAXIM results:

$${\omega }_{MAXIM}={\displaystyle \frac{V}{b}}$$

(30)

Substituting in the value of the ω_OPTIMAL/ω_MAXIM ratio, we obtain:

$${\displaystyle \frac{{\omega }_{OPTIM}}{{\omega }_{MAXIM}}}={\displaystyle \frac{\frac{c}{1+b\cdot c}}{\frac{1}{b}}}={\displaystyle \frac{b\cdot c}{b\cdot c+1}}$$

(31)

The value of the ratio ω_OPTIM/ω_MAXIM generates the 3. equation. Since the WT power is zero at idle, from the power conversion coefficient, C_p(λ) for a three-bladed WT [16,19,20], we obtain e^{−1−0.725 66} = 0.17806, ω_OPTIM/ω_MAXIM = 0.42051.

The following illustrates how the system of equations led to the determination of the unknown variables a, b, and c:

$$\left\{\begin{array}{l}{\displaystyle \frac{c}{1+(b\cdot c)}}=24.046\\ {\displaystyle \frac{a}{c}}\cdot e^{-1-c\cdot b}=2.4425\\ {\displaystyle \frac{b\cdot c}{(b\cdot c)+1}}=0.42051\end{array}\right.$$

(32)

with the solution: a = 569.2, b = 1.7488·10⁻², c = 41.495.

With the experimental data of the WT from the Oravita wind farm, the wind turbine mathematical model was determined as follows, in the form:

$$P_{WT}\left(\omega ,V\right)=569.2\left({\displaystyle \frac{V}{\omega }}-1.7488\cdot {10}^{-2}\right)\cdot e^{-41.495\cdot {\displaystyle \frac{V}{\omega }}}\cdot V^3 [\mathrm{kW}]$$

(33)

or:

$$P_{WT}\left(\omega ,V\right)=5.692\cdot {10}^5\left({\displaystyle \frac{V}{\omega }}-1.7488\cdot {10}^{-2}\right)\cdot e^{-41.495\cdot {\displaystyle \frac{V}{\omega }}}\cdot V^3 [\mathrm{W}]$$

(34)

The maximum power point is obtained at the optimal MAS, ω_OPTIM, by canceling the power derivative:

$${\displaystyle \frac{d{P}_{WT}\left(\omega ,V\right)}{d\omega }}={\displaystyle \frac{d\left[5.692\cdot {10}^5\cdot \left(\frac{V}{\omega }–1.7488 · 10^{-2}\right){\mathrm{e}}^{-41.495\cdot \frac{V}{\omega } }·{\mathrm{V}}^3\right]}{d\omega }}$$

(35)

or:

$$3.0695\cdot {10}^{10}\cdot \omega -7.3809\cdot {10}^{11}\cdot V=0$$

(36)

with the solution ω = 24.046·V.

4. Operating the Turbine by Switching the Generator to Engine Mode

When wind speeds exceed V = 3 m/s, wind systems become operational [20,26,33]. Bringing the system into MPP requires enhancing its speed to achieve the optimal MAS. This can be accomplished within a brief timeframe through two approaches:

-

disconnecting the EG from the grid (slower method);

-

switching the EG to engine mode (faster method).

WT is operated only by the action of the wind, but the time interval for the speed to reach the optimal value is, from an economic perspective, it is deemed unacceptable due to the significant loss of potential wind energy. It is essential to position the wind turbine at the point of maximum power output within the shortest time frame feasible. For the cut-in wind speed, this can be achieved by switching the electric generator to engine mode and absorbing power from the grid, P.

The mains power, P, is injected into the stator, P₁ and the rotor of the machine, P₂, resulting:

$$P=P_1+P_2=P_{ENGINE}$$

(37)

Thus, the turbine rotor accelerates according to the equation of motion:

$$J{\displaystyle \frac{d\omega }{dt}}\cdot \omega =P_{WT}+P_{ENGINE}$$

(38)

where ω is the MAS at the shaft of the electric generator, J is the equivalent moment of inertia, dω/dt is the derivative with respect to time of MAS, P_WT is the useful power given by the wind turbine relative to the shaft of the electric generator, P_ENGINE is the electromagnetic power at the shaft of the induction motor.

4.1. Case Study 1

For the equivalent moment of inertia J = 136 kg·m², the equation of motion results:

$$136{\displaystyle \frac{d\omega }{dt}}\cdot \omega =5.692\cdot {10}^5\cdot \left({\displaystyle \frac{3}{\omega }}-1.7488\cdot {10}^{-2}\right)\cdot e^{-41.495\cdot {\displaystyle \frac{3}{\omega }}}\cdot 3^3+P_{ENGINE}$$

(39)

At the wind speed of 3 m/s, the MPP is obtained at the optimal mechanical angular speed, ω_OPTIM:

$${\omega }_{OPTIM}=24.046\cdot V=24.046\cdot 3=72.138 [\mathrm{rad}./\mathrm{s}.]$$

(40)

as can be seen in Figure 1.

The achievement of the value of the optimal mechanical angular speed is analyzed in two variants:

-

only with the wind turbine and without power absorption from the network;

-

with wind turbine and motor, with power absorption from the grid.

4.1.1. Variant 1—Only with the Wind Turbine and Without Power Absorption from the Grid

In this instance, the equations governing power are as follows:

$$\begin{array}{c}136{\displaystyle \frac{d\omega }{dt}}\cdot \omega =5.692\cdot {10}^5\left({\displaystyle \frac{3}{\omega }}-1.7488\cdot {10}^{-2}\right)\cdot e^{-41.495\cdot {\displaystyle \frac{3}{\omega }}}\cdot 3^3\\ \omega (0)=11\end{array}$$

(41)

At ω(0) = 11 rad/s the optimal mechanical angular speed is reached after 58.5 s, ω(58.5) = 72.151 rad/s.

$$\begin{array}{c}136{\displaystyle \frac{d\omega }{dt}}\cdot \omega =5.692\cdot {10}^5\cdot \left({\displaystyle \frac{3}{\omega }}-1.7488\cdot {10}^{-2}\right)\cdot e^{-41.495\cdot {\displaystyle \frac{3}{\omega }}}\cdot 3^3\\ \omega (0)=8\end{array}$$

(42)

At ω(0) = 8 rad/s the optimal mechanical angular speed is reached after 829.85 s, ω(829.85) = 72.248 rad/s. By solving them, the time variations in the MAS ω are shown in Figure 2.

The duration necessary for the wind turbine (WT) to achieve its maximum power point (MPP) is contingent upon the initial mechanical angular velocities, denoted as ω(0). Specifically, when the initial mechanical angular velocities are low, below cut-in speed, the time required for the turbine to attain the optimal speed is significantly prolonged, often extending to several minutes.

At a wind speed of 3 m/s, with ω(0) = 11 rad/s, at the time moment t = 58.5 s, the values of the optimal mechanical angular speeds, ω_OPTIM, and real, ω, being equal, the system operates at the maximum power point of the turbine. For ω(0) = 8 rad/s, the system gets to work in WT’s MPP at t = 829.85 s, the values of the optimal mechanical angular speeds, ω_OPTIM, and real angular speeds, ω, are then equal.

It is evident from Figure 2 that at the initial MAS, ω(0), of low values, specifically below 8 rad/s, the wind system attains its maximum power point after 14 min. This time interval in which the turbine reaches the point of maximum power, becomes even higher, for example, at ω(0) = 7 rad/s it reaches 70 min, and at ω(0) = 6 rad/s it exceeds the order of hours: 11 h, as Figure 3.

4.1.2. Variant 2—With Wind Turbine and Motor with Power Absorption from the Network

The turbine is brought to the MPP in the least amount of time, by converting the electric generator to operate in asynchronous motor mode and supplying it with power from the grid, utilizing variable frequency and regulated stator flow (see Figure 4).

At the nominal voltage U_N = 690 V and f = 50 Hz, the stator flux has the value:

$${\Psi }_S={\displaystyle \frac{U_N}{\omega }}={\displaystyle \frac{U_N}{2\cdot \pi \cdot f}}={\displaystyle \frac{690}{2\cdot \pi \cdot 50}}=2.1963 [\mathrm{Wb}]$$

(43)

At controlled stator flow, the U/f ratio is:

$${\displaystyle \frac{U}{f}}=2.1963\cdot 2\cdot \pi =13.8 [\mathrm{Vs}]$$

(44)

4.2. Operation at Frequency Equality f₁ = f₂ = f

The stator and rotor are connected to the same frequency network f₁. From the slippage definition relationship:

$$s={\displaystyle \frac{n_1-n}{n_1}}$$

(45)

or:

$$s=\pm {\displaystyle \frac{f_2}{f_1}}$$

(46)

results:

$$n=n_1\left(1\mp {\displaystyle \frac{f_2}{f_1}}\right)$$

(47)

Take the sign—at the same sequence of the phases in the stator as the phases in the rotor, and at an inverse sequence of phases, we take the + sign. The result is a synchronous motor regime with n = 0 at the same phase sequence and n = 2n₁ at an inverse phase sequence.

At the reverse sequence of phases, the rotor speed is n = 2n₁ = 2f/p₁ = 2f/2 = f; therefore, the MAS ω, at the shaft of the electric motor results in the form:

$$\omega =2\cdot \pi \cdot n=2\cdot \pi \cdot f$$

(48)

The relationship between the voltage time variation U(t) = k·t and the supply frequency of the double-powered asynchronous motor must be aligned with the temporal changes in rotational speed to prevent the occurrence of saturation phenomena.

Considering that the process takes place at a constant stator flux of value Ψ_S = 2.1963 Wb and at a load angle, ϑ, also constant, ϑ = 30°, the power to be generated by the asynchronous motor, P_ENGINE, is given by the equation of motion:

$$136{\displaystyle \frac{d\omega }{dt}}\cdot \omega =5.692\cdot {10}^5\left[{\displaystyle \frac{3}{\omega }}-1.7488\cdot {10}^{-2}\right]\cdot e^{-41.495\cdot {\displaystyle \frac{3}{\omega }}}\cdot 3^3+P_{ENGINE}$$

(49)

It is required that at the wind speed of 3 m/s, the optimal MAS ω_OPTIM = 72,138 rad/s has to be obtained within a time interval of one minute, resulting in:

$${\displaystyle \frac{d\omega }{dt}}={\displaystyle \frac{72.138}{60}}=1.2023 [\mathrm{rad}./{\mathrm{s}}^2]$$

(50)

and the time variation in the MAS, ω(t) = 1.2023 t.

In this case, the equation of motion becomes:

$$136\cdot 1.2023\cdot 1.2023\cdot t=5.692\cdot {10}^5\left[{\displaystyle \frac{3}{1.2023\cdot t}}-1.7488\cdot {10}^{-2}\right]\cdot e^{-41.495\left({\displaystyle \frac{3}{1.2023\cdot t}}\right)}\cdot 3^3+P_{ENGINE}$$

(51)

resulting in the necessary power that the double-powered asynchronous motor must develop:

$$P_{ENGINE}\left(t\right)=136\cdot 1.2023\cdot 1.2023\cdot t-5.692\cdot {10}^5\left[{\displaystyle \frac{3}{1.2023\cdot t}}-1.7488\cdot {10}^{-2}\right]\cdot e^{-41.495\left({\displaystyle \frac{3}{1.2023\cdot t}}\right)}\cdot 3^3$$

(52)

At the increasing slope of the mechanical angular speed, ω, of 1.2023 rad/s² the motor is energized up to t_M = 15.83 s, at which point the power developed by the asynchronous motor becomes zero, as seen from Figure 5. This result is also confirmed by solving the equation:

$$136\cdot 1.2023\cdot 1.2023\cdot t-5.692\cdot {10}^5\left[{\displaystyle \frac{3}{1.2023\cdot t}}-1.7488\cdot {10}^{-2}\right]\cdot e^{-41.495\left({\displaystyle \frac{3}{1.2023\cdot t}}\right)}\cdot 3^3=0$$

(53)

The variations over time of the engine and turbine power outputs are illustrated in Figure 5. At the time moment t_M, the MAS ω has the value ω(t) = 1.2023∙15.83 = 19.032 rad/s and, further, the turbine rotor is accelerated only by the action of the wind. The representation of the process is derived from equations characterized by powers of the following form:

$$J\cdot {\displaystyle \frac{d\omega }{dt}}\cdot \omega =P_{WT}\left(\omega ,t\right)=5.692\cdot {10}^5\cdot \left({\displaystyle \frac{3}{\omega }}-1.7488\cdot {10}^{-2}\right)\cdot e^{-41.495\cdot {\displaystyle \frac{3}{\omega }}}\cdot 3^3$$

(54)

or:

$$\begin{array}{c}136{\displaystyle \frac{d\omega }{dt}}\cdot \omega =5.692\cdot {10}^5\cdot \left({\displaystyle \frac{3}{\omega }}-1.7488\cdot {10}^{-2}\right)\cdot e^{-41.495\cdot {\displaystyle \frac{3}{\omega }}}\cdot 3^3\\ \omega (0)=19.032\end{array}$$

(55)

Given that ω(10.12) = 72.148 rad/s, it can be concluded that the optimal mechanical angular speed, denoted as ω_OPTIM, is attained at 10.12 s, as illustrated in Figure 6.

At a wind speed of 3 m/s, the optimal mechanical angular speed, ω_OPTIM, is obtained in 25.95 s, lower than initially prescribed.

At low values of mechanical angular velocities, the contribution of the power developed by the electric motor is preponderant, the turbine developing a low power, contributing very little to the acceleration of the masses in the rotational movement. The necessity of transitioning the electric generator to engine mode is warranted, next to the cut-in wind speed, to enable the system to reach its maximum power point in the shortest possible time.

5. Mathematical Model of Dual Powered Asynchronous/Synchronous Motor

At cut-in wind speed, the turbine reaches its peak power output in the least amount of time by converting the electric generator into either an asynchronous or synchronous motor mode, utilizing a mains-powered motor with variable frequency and regulated stator flow.

The values of the inductances are derived from the parameters of the generator established based on the nominal data: stator inductance L_S = L₁ + L_M = 7.5812∙10⁻² H, rotor inductance L_R = L₂ + L_M = 7.5812∙10⁻² H, magnetization inductance L_M = 6.0650∙10⁻² H, global dispersion factor σ = [(L_S∙L_R) – L_M²]/L_S∙L_R.

$$\sigma ={\displaystyle \frac{\left[7.5812\cdot {10}^{-2}\cdot 7.5812\cdot {10}^{-2}-{\left(6.065\cdot {10}^{-2}\right)}^2\right]}{7.5812\cdot {10}^{-2}\cdot 7.5812\cdot {10}^{-2}}}=0.36$$

(56)

When neglecting losses, the stator and motor rotor equations are:

$$\begin{array}{c}\underset{¯}{U_S}=j\cdot \omega \cdot \underset{¯}{{\Psi }_S}\\ \underset{¯}{U_R}=j\cdot \omega \cdot \underset{¯}{{\Psi }_R}\end{array}$$

(57)

where Ψ_S is the stator flow, Ψ_R is the rotor flow, ω is the pulsation of U_S and U_R tensions, and f is the frequency of the supply voltage.

$$\begin{array}{c}\underset{¯}{{\Psi }_S}=L_S\cdot \underset{¯}{I_S}+M\cdot \underset{¯}{I_R}\\ \underset{¯}{{\Psi }_R}=L_R\cdot \underset{¯}{I_R}+M\cdot \underset{¯}{I_S}\end{array}$$

(58)

ϑ is the phase shift in the supply voltages U_S and U_R, Figure 7, and is the same as the phase shift between flows, Ψ_S and Ψ_R.

Considering the phasor of the U_S supply voltage in the real axis, it follows:

$$\underset{¯}{U_R}=U_R\cdot e^{j\vartheta }$$

(59)

or:

$$\underset{¯}{U_R}=U_R\cdot {\displaystyle \frac{e^{j\vartheta }}{k}}$$

(60)

where k = N₁/N₂, N₁ is the number of turns per phase in the stator, N₂ is the number of turns per phase in the rotor. From the system:

$$\begin{array}{c}\underset{¯}{U_S}=j\cdot \omega \cdot (L_S\cdot \underset{¯}{I_S}+L_M\cdot \underset{¯}{I_R})\\ \underset{¯}{U_R}=j\cdot \omega \cdot (L_R\cdot \underset{¯}{I_R}+L_M\cdot \underset{¯}{I_S})\end{array}$$

(61)

stator I_S and rotoric I_R currents are obtained:

$$\underset{¯}{I_S}=U_S\cdot {\displaystyle \frac{L_R+L_M\cdot e^{j\cdot \vartheta }/k}{j\cdot \omega \cdot \beta }}$$

(62)

$$\underset{¯}{I_S}=\left|U_S\cdot {\displaystyle \frac{L_R+M\cdot e^{j\cdot \vartheta }/k}{j\cdot \omega \cdot \beta }}\right|=\left({\displaystyle \frac{{\Psi }_S}{\beta }}\right)\sqrt{L_R^2-2\cdot L_R\cdot L_M\cdot (\cos \vartheta /k)+{(L_M/k)}^2}$$

(63)

$$\underset{¯}{I_R}=U_S\cdot {\displaystyle \frac{L_M+L_S\cdot e^{j\cdot \vartheta }/k}{j\cdot \omega \cdot \beta }}$$

(64)

$$\underset{¯}{I_R}=\left|U_S\cdot {\displaystyle \frac{L_M+L_S\cdot e^{j\cdot \vartheta }/k}{j\cdot \omega \cdot \beta }}\right|=\left({\displaystyle \frac{{\Psi }_S}{\beta }}\right)\sqrt{{L_M}^2-2\cdot L_S\cdot L_M\cdot (\cos \vartheta /k)+{(L_S/k)}^2}$$

(65)

where β = (L_S∙L_R) – L_M²:

$$\sigma =7.5812\cdot {10}^{-2}\cdot 7.5812\cdot {10}^{-2}-{\left(6.065\cdot {10}^{-2}\right)}^2=2.069\cdot {10}^{-3}$$

(66)

Knowing the currents and voltages, the stator P_S and P_R rotor active power can be performed as follows:

$$\begin{array}{c}{P}_S=3\cdot \underset{¯}{U_S}\cdot {\underset{¯}{I_S}}_{REAL}\\ P_R=3\cdot \underset{¯}{U_R}\cdot \underset{¯}{I_R}\end{array}$$

(67)

or:

$$\begin{array}{c}{P}_S=3\cdot U_S^2\cdot {\displaystyle \frac{L_M\cdot \sin \vartheta }{\omega \cdot \beta \cdot k}}=3\cdot U_S\cdot {\Psi }_S\cdot {\displaystyle \frac{L_M\cdot \sin \vartheta }{\beta \cdot k}}=3\cdot U_S\cdot 2.1963\cdot {\displaystyle \frac{L_M\cdot \sin \vartheta }{\beta \cdot k}}\\ P_R=3\cdot U_S^2\cdot {\displaystyle \frac{L_M\cdot \sin \vartheta }{\omega \cdot \beta \cdot k}}=3\cdot U_S\cdot 2.1963\cdot {\displaystyle \frac{L_M\cdot \sin \vartheta }{\beta \cdot k}}\end{array}$$

(68)

5.1. Stator Current Under Load and No-Load

The nominal curent, I_N, is obtained at a load angle ϑ of value ϑ_N = 30° and has the value:

$$I_N={\displaystyle \frac{P_{EG}}{3\cdot U_N}}={\displaystyle \frac{\mathrm{1,500,000}}{3\cdot 690}}=724.64 [\mathrm{A}]$$

(69)

Thus, it is obtained:

$${\displaystyle \frac{2.1963}{\beta }}\cdot \sqrt{L_R^2-2\cdot L_R\cdot L_M\cdot \cos \vartheta /k+{(L_M/k)}^2}=724.64$$

(70)

or:

$${\displaystyle \frac{2.1963}{2.069\cdot {10}^{-3}}}\cdot \sqrt{L_R^2-L_R\cdot L_M\cdot \sqrt{3}/k+{(L_M/k)}^2}=724.64$$

(71)

The no-load current, I_Sno-load, is obtained at the zero value of the charge angle ϑ and is approximately 5% of the nominal current I_N:

$$I_{Sno-load}={\displaystyle \frac{5}{100}}\cdot 724.64=36.232 [\mathrm{A}]$$

(72)

resulting:

$${\displaystyle \frac{2.1963}{\beta }}\cdot \sqrt{L_R^2-2\cdot L_R\cdot L_M/k+{(L_M/k)}^2}=36.232$$

(73)

or:

$$\left({\displaystyle \frac{2.1963}{2.069\cdot {10}^{-3}}}\right)\cdot \left(L_R-{\displaystyle \frac{L_M}{k}}\right)=36.232$$

(74)

With the two values of the stator current, I_N and I_Sno-load, we obtain the system of equations in the unknown L_R and L_M/k:

$$\left\{\begin{array}{c}{\displaystyle \frac{2.1963}{2.069\cdot {10}^{-3}}}\cdot \sqrt{L_R^2-L_R\cdot L_M\cdot {\displaystyle \frac{\sqrt{3}}{k}}+{\left({\displaystyle \frac{L_M}{k}}\right)}^2}=724.64\\ {\displaystyle \frac{2.1963}{2.069\cdot {10}^{-3}}}\cdot \left(L_R-{\displaystyle \frac{L_M}{k}}\right)=36.232\end{array}\right.$$

(75)

with the solution L_R = 1·1.3343 [H], L_M/k = 1.3 [H].

Considering that the process takes place at a constant stator flux of value:

$${\Psi }_S={\displaystyle \frac{U_S}{\omega }}=2.1963 [\mathrm{Wb}]$$

(76)

and at a load angle $\vartheta$, the value of the stator current is:

$$I_S=\left|U_S\cdot {\displaystyle \frac{L_R+M\cdot e^{j\cdot \vartheta }/k}{j\cdot \omega \cdot \beta }}\right|=\left({\displaystyle \frac{2.1963}{2.069\cdot {10}^{-3}}}\right)\sqrt{{1.3343}^2-2\cdot 1.3343\cdot 1.3\cdot \cos \vartheta +{(1.3)}^2}$$

(77)

$$I_S=1061.5\sqrt{3.4704-3.4692\cdot \cos \vartheta }$$

(78)

Since between the inductances L_R, L_S and L_M are the bond relations:

$$L_S=k^2\cdot L_R=k\cdot L_M$$

(79)

with k = N₁/N₂, the obtained rotor current value is:

$$I_R=\left|U_S\cdot {\displaystyle \frac{L_M+L_S\cdot e^{j\cdot \vartheta }/k}{j\cdot \omega \cdot \beta }}\right|=k\left({\displaystyle \frac{2.1963}{2.069\cdot {10}^{-3}}}\right)\sqrt{{1.3}^2-2\cdot 1.3343\cdot 1.3\cdot \cos \vartheta +{1.3343}^2}$$

(80)

$$I_R=1061.5\cdot k\cdot \sqrt{3.4704-3.4692\cdot \cos \vartheta }$$

(81)

The examination indicates that the current in the stator is k times less than that in the rotor; conversely, with respect to voltage, the rotor voltage is k times less than the stator voltage.

5.2. Power to Synchronous Motor Shaft

The shaft power of the motor, P_ENGINE, can be determined by summing the active powers of the stator and rotor, P_S and P_R:

$$P_{ENGINE}=P_S+P_R=U_S\cdot 8695.6\cdot \sin \vartheta$$

(82)

at U_S voltage and rotational speed n:

$$n=f={\displaystyle \frac{U_S}{13.8}}$$

(83)

Since the stator flux has constant value,

$${\psi }_S={\displaystyle \frac{U_S}{\omega }}={\displaystyle \frac{U_S}{2\cdot \pi \cdot f}}=2.1963 [\mathrm{Wb}]$$

(84)

results:

$$f={\displaystyle \frac{U_S}{2\cdot \pi \cdot 2.1963}}={\displaystyle \frac{U_S}{13.8}}$$

(85)

When neglecting the resistance, at a constant stator flux of value 2.1963 Wb and at the load angle ϑ, the power value at the motor shaft is:

$$P_{arbore}=P_{ENGINE}=P_S+P_R=6\cdot U_S\cdot {\displaystyle \frac{L_M\cdot \sin \vartheta }{\omega \cdot \beta \cdot k}}=6\cdot U_S\cdot {\psi }_S\cdot 659.87\cdot \sin \vartheta =8695.6\cdot U_S\cdot \sin \vartheta$$

(86)

In conclusion, at the U_S voltage, the power developed by the motor is:

$$P_{ENGINE}=U_S\cdot 8695.6\cdot \sin \vartheta$$

(87)

at rotational speed:

$$n={\displaystyle \frac{U_S}{13.8}}$$

(88)

Figure 8 illustrates the changes associated with the load angle of both currents and power in the context of asynchronous and synchronous motors, which are supplied through the stator and rotor under variable frequency conditions and regulated stator flow.

Since the stator and rotor are connected to the same frequency network f1, the motor speed depends on the frequency value of the supply voltage. Given that the stator flux is kept constant, and the operating regime is synchronous motor, the succession of the phases in the stator with those in the rotor is reversed, resulting in:

$$n=2\cdot n_1=f={\displaystyle \frac{U_S}{13.8}} [\mathrm{rps}]$$

(89)

Also, the motor being synchronous, implies that its speed is determined solely by the frequency of the supply voltage, and not the magnitude of the power supplied to the shaft. In the synchronous motor, the phenomenon of pendulum occurs.

5.3. 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:

$$J\cdot {\displaystyle \frac{d\omega }{dt}}=M_{WT}+M_{ENGINE}$$

(90)

where ω is MAS, of the shaft of the electric motor, J is the equivalent moment of inertia, du/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 ω, at an equivalent moment of inertia of value J = 136 kg∙m², we obtain the equation of powers:

$$136\cdot {\displaystyle \frac{d\omega }{dt}}\cdot \omega =P_{WT}+P_{ENGINE}$$

(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:

$$P_{WT}\left(\omega ,3\right)=5.692\cdot {10}^5\cdot \left({\displaystyle \frac{3}{\omega }}-1.7488\cdot {10}^{-2}\right)\cdot e^{-41.495\cdot {\displaystyle \frac{3}{\omega }}}\cdot 3^3$$

(92)

resulting:

$$136\cdot {\displaystyle \frac{d\omega }{dt}}\cdot \omega =5.692\cdot {10}^5\cdot \left({\displaystyle \frac{3}{\omega }}-1.7488\cdot {10}^{-2}\right)\cdot e^{-41.495\cdot {\displaystyle \frac{3}{\omega }}}\cdot 3^3+P_{ENGINE}$$

(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∙8695.6∙sin ϑ 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, ω_OPTIM:

$${\omega }_{OPTIM}=24.046\cdot V=24.046\cdot 3=72.138 [\mathrm{rad}./\mathrm{s}.]$$

(94)

The ideal mechanical angular speed, ω_OPTIM, should be attained within a time frame of one minute, leading to a change in frequency over that duration:

$$f={\displaystyle \frac{{\omega }_{OPTIM}}{2\cdot \pi \cdot 60}}\cdot t={\displaystyle \frac{72.138}{2\cdot \pi \cdot 60}}\cdot t=0.19135\cdot t$$

(95)

The power developed by the motor (83) does not depend on the MAS, ω being dependent only on the U_S voltage and the load angle ϑ. In this case, the power equations are:

$$136\cdot {\displaystyle \frac{d\omega }{dt}}\cdot \omega =5.692\cdot {10}^5\cdot \left({\displaystyle \frac{3}{\omega }}-1.7488\cdot {10}^{-2}\right)\cdot e^{-41.495\cdot {\displaystyle \frac{3}{\omega }}}\cdot 3^3+U_S\cdot 8695.6\cdot \sin \vartheta$$

(96)

The change in time of the voltage and frequency of the motor supply is made at controlled stator flow, the U/ω ratio being:

$${\displaystyle \frac{U_S}{\omega }}={\psi }_S={\displaystyle \frac{U_N}{\omega }}={\displaystyle \frac{690}{2\cdot \pi \cdot 50}}=2.1963 [\mathrm{Wb}]$$

(97)

resulting:

$$U_S=2.1963\cdot \omega =2.1963\cdot 2\cdot \pi \cdot f=2.1963\cdot 2\cdot \pi \cdot 0.19135\cdot t=2.6406\cdot t$$

(98)

At the mechanical angular speed ω(0) of value ω(0) = 0.1 rad/s, we obtain:

$$\begin{array}{c}136{\displaystyle \frac{d\omega }{dt}}\cdot \omega =5.692\cdot {10}^5[(3/\omega )-1.7488\cdot {10}^{-2}]\cdot e^{-41.495(3/\omega )}\cdot 3^3+2.1963\cdot \omega \cdot 8695.6\cdot \sin \vartheta \\ \omega (0)=0.1\end{array}$$

(99)

Between the angular speed of the spinning field ω = 2 ∙ π ∙ f and the mechanical angular speed of the electric motor shaft, ω, there is the bonding relationship:

$${\displaystyle \frac{d\vartheta }{dt}}+\omega -\omega =0$$

(100)

where dϑ/dt is derived from the load angle ϑ. The angular speed of the spinning field, ω, is imposed by the frequency of the motor supply voltage, f, resulting:

$$\omega =2\cdot \pi \cdot f=2\cdot \pi \cdot 0.19135\cdot t=1.20232\cdot t$$

(101)

Consequently, the equation of motion is expressed as follows:

$$\begin{array}{c}136{\displaystyle \frac{d\omega }{dt}}\cdot \omega =5.692\cdot {10}^5\left({\displaystyle \frac{3}{\omega }}-1.7488\cdot {10}^{-2}\right)\cdot e^{-41.495\cdot {\displaystyle \frac{3}{\omega }}}\cdot 3^3+2.1963\cdot 1.2023\cdot t\cdot 8695.6\cdot \sin \vartheta \\ {\displaystyle \frac{d\vartheta }{dt}}+\omega -1.2023\cdot t=0\\ \omega \left(0\right)=0.1\\ \vartheta \left(0\right)=0\end{array}$$

(102)

By solving it, the time variations in the mechanical angular speed ω and of the load angle ϑ, shown in Figure 9, are obtained.

5.4. Achieving Optimal Speed and Switching to Generator Mode

At the time t = 15.83 s, the power developed by the synchronous motor becomes zero and the load angle, ϑ, becomes negative, as can be seen in Figure 9—the synchronous motor switching to generator mode and delivering power to the grid, at MAS:

$$\omega \left(15.83\right)=1.2023\cdot 15.83=19.032 [\mathrm{rad}./\mathrm{s}.]$$

(103)

and voltage:

$$U_S\left(15.83\right)=2.6406\cdot 15.83=41.801 [\mathrm{V}]$$

(104)

At the time moment t = 60 s, the system operates at the optimal mechanical angular speed, ω_OPTIM, in the MPP of the turbine, at the wind speed of 3 m/s:

$${\omega }_{OPTIM}=24.046\cdot V=24.046\cdot 3=72.138 [\mathrm{rad}./\mathrm{s}.]$$

(105)

Over the time interval ∆t = 60 − 15.83 = 44.17 s, the double-powered synchronous generator outputs power to the grid, at variable frequency and voltage:

$$U_S\left(t\right)=41.801+2.6406\cdot t$$

(106)

In this case, the equation of motion becomes:

$$\begin{array}{c}136{\displaystyle \frac{d\omega }{dt}}\cdot \omega =5.692\cdot {10}^5[(3/\omega )-1.7488\cdot {10}^{-2}]\cdot e^{-41.495(3/\omega )}\cdot 3^3-(41.801+2.2406\cdot t)\cdot 8695.6\cdot \sin \vartheta \\ {\displaystyle \frac{d\vartheta }{dt}}-\omega +(19.032+1.2023\cdot t)=0\\ \omega \left(0\right)=19.032\\ \vartheta \left(0\right)=0.01\end{array}$$

(107)

by solving it, obtaining the variations in the load angle ϑ and the mechanical angular speed ω, Figure 10.

The power output from the generator:

$$P_{EG}\left(72.138\right)=\left(41.801+2.6406\cdot t\right)\cdot 8695.6\cdot \sin \vartheta$$

(108)

being dependent on the load angle that shows pronounced oscillations, it will also oscillate pronouncedly (Figure 11a).

The intense oscillations of the power output by the generator are at the end of the transient process when the power to the generator is:

$${\displaystyle \frac{P_{EG}}{P_{WT}}}={\displaystyle \frac{\mathrm{1,200,000}}{\mathrm{65,946}}}=18.197$$

(109)

higher than that of the turbine.

5.5. Stationary Mode

Upon reaching the optimal mechanical angular speed ω_OPTIM = 72,138 rad/s, at the time moment t = 60 s, the system operates in the MPP of the turbine at the wind speed of 3 m/s. The frequency and voltage being constant, they stabilize at the values:

$$f={\displaystyle \frac{72.138}{2\cdot \pi }}=11.481[\mathrm{Hz}]; \omega =2\cdot \pi \cdot f=2\cdot \pi \cdot 11.481=72.137 [\mathrm{Hz}]$$

(110)

$$U_S=13.8\cdot f=13.8\cdot 11.481=158.44 [\mathrm{V}]$$

(111)

the regime becomes stationary. In this case, at constant tension, the equation of motion becomes:

$$\begin{array}{c}\begin{array}{c}136{\displaystyle \frac{d\omega }{dt}}\cdot \omega =5.692\cdot {10}^5[(3/\omega )-1.7488\cdot {10}^{-2}]\cdot e^{-41.495(3/\omega )}\cdot 3^3-158.44\cdot 8695.6\cdot \sin \vartheta \\ {\displaystyle \frac{d\vartheta }{dt}}+\omega +72.137=0\\ \omega (0)=72.1371\\ \vartheta (0)=0.01\end{array}\end{array},$$

(112)

obtaining the variation in the power at the generator, Figure 11b:

$$P_{EG-b}=158.44\cdot 8695.6\cdot \sin \vartheta$$

(113)

The fluctuations in value are significantly more evident during the process of ramping up the turbine to achieve maximum power, in contrast to the variations observed in a steady state (see Figure 11a,b).

The oscillations are, in this case, smaller compared to the previous case and the power to the generator is:

$${\displaystyle \frac{P_{EG-2}}{P_{WT-2}}}={\displaystyle \frac{\mathrm{79,135}}{\mathrm{65,946}}}=1.2$$

(114)

higher than that of the turbine.

The oscillations are pronounced in both regimes due to the fact that the electrical losses in the winding resistances and in the iron have been neglected. From the power characteristic WT, P_WT(V, ω), at ω_OPTIM:

$$\begin{array}{c}P=5.692\cdot {10}^5[(3/\omega )-1.7488\cdot {10}^{-2}]\cdot e^{-41.495(3/\omega )}\cdot 3^3\\ \omega =72.138\end{array}$$

(115)

the power developed by the WT results:

$$P_{WT}\left(3,72.138\right)=\mathrm{65,946}$$

(116)

The power developed by the WT is during the process of bringing the WT to the optimal constant speed, because the wind speed is constant, at 3 m/s. The power output of the generator oscillates around the power developed by WT. However, the power developed by the synchronous motor is variable during the process.

The variations over time of the mechanical angular speed and load angle, as illustrated in Figure 9, reveal several key aspects:

• At the beginning of the transient process, the mechanical angular speed, ω, and the charge angle, ϑ, vary linearly over time;
• At the end of the transient process, the mechanical angular speed, ω, and the charge angle, ϑ, vary oscillating over time;
• The oscillations of the load angle, ϑ, are more pronounced in cases where the motor is powered at variable voltage and frequency;
• The turbine enhances the fluctuations of the load angle, which are reflected in the variations in power output and current.

The fluctuations in the load angle, and consequently in the power output, are further intensified by the temporal variations in voltage and frequency, which are influenced by the regulated stator flow. In conclusion, by changing the voltage and frequency of the generator, in synchronous mode, there is a significant disturbance in the Electrical Power System. Consequently, the asynchronous generator regime is used to mitigate these disturbances.

The modification of the voltage and frequency of the generator is imposed by the wind speed variable in time and the capture of a maximum wind energy at these wind speeds can only be achieved when the WT operates at the point of maximum power, at the optimal mechanical angular speed.

The wind turbine consistently functions in dynamic mode at variable speeds, adapting to the fluctuations in wind speeds that occur over time.

6. Discussion

At low wind speeds, specifically above the connection threshold of 3 m/s, the duration required to attain optimal speed solely through wind input is significantly prolonged, resulting in considerable wastage of wind energy. In power converters that facilitate bidirectional power flow, the electric generator can be transitioned into engine mode, thereby reducing the time needed to achieve optimal speed.

Analyzing the operation of these wind turbines, it was possible to:

• Develop a mathematical model for the wind turbine (WT) and the double-fed asynchronous/synchronous motor, derived from the induction generator with a wound rotor, utilizing experimental data as the foundation;
• Follow the transition of the asynchronous generator to a double-powered synchronous motor, with the stator and rotor connected to the same frequency network, so that the turbine operates at the point of maximum power in the shortest possible time interval;
• Visualize the process of bringing it to the optimal speed and interpreted the oscillations of the load angle and power at the double powered synchronous motor;
• Determine the oscillation of the load angle, the power when switching to generator mode as well as the operation at the maximum power point.

The results obtained reveal several key aspects that can be emphasized:

• When switching the asynchronous generator to the mode of a double-powered synchronous motor, at the same frequency, oscillations of the load angle and power occur;
• The most pronounced oscillations are in the stator and rotor power supply at variable frequency.

Consequently, the power supply of the motor obtained from the induction generator with the wound rotor, having the same frequency in stator and rotor, is new and of interest in high-performance electric drives, with direct applicability in starting wind turbines at cut-in wind speeds. The work demonstrated the importance of modeling the double synchronous motor powered from the same variable frequency source.

7. Conclusions

This paper examines the challenges associated with the integration of wind turbines into the energy system, particularly at low wind speeds. To minimize the time required to reach optimal rotational speed, the advantage of operating the electric generator in motor mode has been demonstrated. High-capacity wind systems exhibit significant moments of inertia, which can pose challenges in optimizing the time required to reach optimal rotational speed.

In the present paper, the mathematical models of the wind turbine and the asynchronous/synchronous double powered motor, obtained by switching the asynchronous generator to motor mode, have been deduced, analyzing the oscillations of the load angle and power at the asynchronous/synchronous motor, double powered, at variable frequency, in the process of bringing it to the optimal speed. The solution presented—feeding the motor with the same frequency in the stator and in the rotor—is a technical novelty, based on the processing of experimental data: wind speed, generator power and mechanical angular speed from a WT type FL MD 70 1.5.