Learning Machine Based on Optimized Dimensionality Reduction Algorithm for Fault Diagnosis of Rotor Broken Bars in Induction Machine

To be able to focus on the study of the simulation of bar break defects, a model of the rotor is established in the form of electrically connected and magnetically coupled meshes, where a mesh consists of two bars and the two portions of rings that connect them. Each bar and ring segment is characterized by resistance and inductance (Figure 2). The model assumptions are:

• Negligible saturation and skin effect;
• Uniform air-gap;
• Sinusoidal mmf of stator windings in air-gap;
• Rotor bars are insulated from the rotor, thus no inter-bar current flows through the laminations;
• Relative permeability of machine armatures is assumed infinite.

Although a sinusoidal mmf of the stator winding is assumed, other winding distributions could also be analyzed by simply using superposition. This is justified by the fact that different space harmonic components do not interact [9, 10].

2.jpg

Figure 2. Rotor cage equivalent circuit

2.1 Equation model

By using the Clark transformation to go from three-phase stator quantities (a, b, c) to two-phase quantities (α, β). The simulation can be performed with two separate frames for the stator and the rotor. To reduce the calculation time, we eliminate the angle "ɵ" of the coupling matrix by choosing the most appropriate reference and which is that of the rotor.

In this frame, all the quantities have a pulsation "gω_s" in steady-state. This characteristic can be used for the analysis of the rupture of rotor bars in the machine by observing the current "I_as" [2].

We are therefore looking for the set of independent differential equations defining the model of the induction machine.

2.1.1 Stator voltage equation

All the stator phases voltages in matrix form and flux equations are deduced [11]:

$\left[V_{a b c s}\right]=\left[R_s\right]\left[I_{a b c s}\right]+\frac{d}{d t}\left[\emptyset_{a b c s}\right]$          (1)

$\left[\emptyset_{a b c s}\right]=\left[L_s\right]\left[I_{a b c s}\right]+\left[M_{s r}\right]\left[I_{r k}\right]$          (2)

where, [V_abcs]=[V_as V_bs V_cs]^T is the stator voltage vector, [I_abcs]=[I_as I_bs I_cs]^T is the stator current vector, [I_rk]=[I_r0 I_r1 … I_rk… I_r(N_r-1)]^T is the vector of the currents in the rotor mesh and $\left[\emptyset_{a b c s}\right]=\left[\emptyset_{a s} \emptyset_{b s} \emptyset_{c s}\right]^T$ is the stator flux vector.

We, therefore, write the matrices of the resistances, the inductances, and the stator mutuals respectively:

$\left[R_s\right]=\left[\begin{array}{ccc}R_s & 0 & 0 \\ 0 & R_s & 0 \\ 0 & 0 & R_s\end{array}\right] ; \quad\left[L_s\right]=\left[\begin{array}{ccc}L_s & M_s & M_s \\ M_s & L_s & M_s \\ M_s & M_s & L_s\end{array}\right] ;$

$\left[M_{s r}\right]=\left[\begin{array}{ccc}\cdots & -M_{s r} \cos (\theta+k \alpha) & \cdots \\ \cdots & -M_{s r} \cos \left(\theta+k \alpha-\frac{2 \pi}{3}\right) & \cdots \\ \cdots & -M_{s r} \cos \left(\theta+k \alpha-\frac{4 \pi}{3}\right) & \cdots\end{array}\right]$

With:

$K=0,1,2, \ldots,\left(N_r-1\right)$.

$M_{s r}=\frac{4 \mu_0 N_s R l}{e p^2 \pi} \sin \left(\frac{\alpha}{2}\right) ;[H]$.

$\alpha=\frac{2 \pi}{N_r} ;[\mathrm{rad}]$.

where, N_r is the number of rotor bars, M_sr is the mutual inductance between stator phase and rotor mesh and α is the angle between the rotor bars.

The transition to two-phase components of the stator components is carried out using the Park transformation matrix, knowing that the homopolar component is zero.

$\left[X_{\alpha \beta s}\right]=[P(\theta)]\left[X_{d q s}\right]$          (3)

where, $P(\theta)=\left[\begin{array}{cc}\cos (\theta) & -\sin (\theta) \\ \sin (\theta) & \cos (\theta)\end{array}\right]$, P(θ): Park's rotation matrix is as follows.

$\emptyset_{\alpha \beta s}=\left[\begin{array}{cc}L_{s c} & 0 \\ 0 & L_{s c}\end{array}\right] \cdot\left[I_{\alpha \beta s}\right]-M_{s r} \cdot\left[\begin{array}{cc}\ldots \cos (\theta+k \alpha) \ldots \\ \ldots \sin (\theta+k \alpha) \ldots\end{array}\right] \cdot\left[I_{r k}\right]$          (4)

With: L_sc=L_sp-M_s+L_sf; [H]; $L_{s p}=\frac{4 \mu_0 N_s^2 R l}{e \pi p^2} ;[H]$; $M_s=-\frac{L_{s p}}{2} ;[H]$; $\mu_0=4 \pi 10^{-7} ;\left[H . m^{-1}\right]$.

where, L_sc is the inductance cyclique statorique, L_sp is the main inductance of a stator phase, M_s is the mutual between two stator windings and μ₀ is the magnetic air permeability.

So, the flux and the stator voltage in the Park reference (d, q) are written respectively:

$\emptyset_{d q s}=\left[\begin{array}{cc}L_{s c} & 0 \\ 0 & L_{s c}\end{array}\right] \cdot\left[I_{d q s}\right]-M_{s r} \cdot\left[\begin{array}{ll}\ldots \cos (k \alpha)\ldots \\ \ldots \sin(k \alpha)\ldots\end{array}\right] \cdot\left[I_{r k}\right]$          (5)

$V_{d q s}=R_s \cdot I_{d q s}+\omega \cdot P\left(\frac{\pi}{2}\right) \cdot \emptyset_{d q s}+\frac{d}{d t} \emptyset_{d q s}$          (6)

After transformation and rotation, the electric equations in the rotor reference are written:

$\left\{\begin{array}{l}V_{d s}=R_s \cdot I_{d s}-\omega \cdot \emptyset_{q s}+\frac{d}{d t} \emptyset_{d s} \\ V_{q s}=R_s \cdot I_{q s}+\omega \cdot \emptyset_{d s}+\frac{d}{d t} \emptyset_{q s}\end{array}\right.$          (7)

Finally, the electrical equation of the stator in the rotor frame is written in matrix form:

$\left[\begin{array}{c}V_{d s} \\ V_{q s}\end{array}\right]=\left[\begin{array}{cccc}R_s & -\omega \cdot L_{s c} & \vdots \cdots+\omega \cdot M_{s r} \cdot \sin (k \alpha) \ldots \\ \omega \cdot L_{s c} & R_s & \vdots \cdots-\omega \cdot M_{s r} \cdot \cos (k \alpha) \ldots\end{array}\right] \cdot\left[\begin{array}{c}I_{d q s} \\ I_{r k}\end{array}\right]$

$+\left[\begin{array}{ccc}L_{s c} & 0 & \vdots \cdots-M_{s r} \cdot \cos (k \alpha) \ldots \\ 0 & L_{s c} & \vdots \cdots-M_{s r} \cdot \sin (k \alpha) \ldots\end{array}\right] \cdot \frac{d}{d t}\left[\begin{array}{c}I_{d q s} \\ I_{r k}\end{array}\right]$          (8)

2.1.2 Rotor voltage equation

Figure 3 represents the equivalent electric circuit of a mesh of the rotor cage, where the rotor bars and the portions of short-circuit rings are represented by their resistances and corresponding leakage inductances [12].

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Figure 3. Electric diagram equivalent of a rotor mesh

Knowing that:

$\left\{\begin{array}{c}I_{e k}=I_{r k}-I_e \\ I_{b k}=I_{r k}-I_{r(k+1)}\end{array}\right.$          (9)

The voltage equation for a mesh "k" of the rotor cage is given by:

$-R_{b(k-1)} I_{r(k-1)}+R_{b k} I_{b k}+\frac{R_e}{N_r} I_{e k}+\frac{R_e}{N_r} I_{r k}+\frac{d}{d t} \emptyset_{r k}=0$           (10)

The total flux "$\emptyset_{r k}$" for an elementary circuit of index "k" is composed of the sum of the following terms:

Main flux: L_rp.I_rk;

Mutual flux with the other circuits of the rotor:

$M_{r r} \cdot \sum_{\substack{j=0 \\ j \neq k}}^{N_r-1} I_{r j}$

Mutual flux with the stator, given after transformation:

$-\frac{3}{2} M_{s r} \cdot\left[\begin{array}{ccc}\vdots & \vdots & \vdots \\ \cos (k \alpha) & \vdots & \sin (k \alpha) \\ \vdots & \vdots & \vdots\end{array}\right] \cdot\left[I_{d q s}\right]$

The flux induced in the rotor mesh is given by:

$\emptyset_{r k}=\left(L_{r p}+2 L_b+2 \frac{L_e}{N_r}\right) I_{r k}+M_{r r} \cdot \sum_{\substack{j=0 \\ j \neq k}}^{N_r-1} I_{r j}-L_b\left(I_{r(k-1)}+I_{r(k+1)}\right)$

$-\frac{3}{2} M_{s r}\left(I_{d s} \cos (k \alpha)+I_{q s} \sin (k \alpha)\right)-\frac{L_e}{N_r} I_e$          (11)

It is finally necessary to complete the system of the equation of the circuits of the rotor by that of the ring of short-circuits; we then have:

$\frac{R_e}{N_r} \sum_{k=0}^{N_r-1} I_{r k}+\frac{L_e}{N_r} \sum_{k=0}^{N_r-1} \frac{d I_{r k}}{d t}-R_e I_e-L_e \frac{d I_e}{d t}=0$          (12)

with: $L_{r p}=\frac{2 \pi\left(N_r-1\right) \mu_0 R l}{e N_r^2} ;[H]$; $M_{r r}=-\frac{2 \pi \mu_0 R l}{e N_r^2} ;[H]$.

where, L_rp is the main inductance of a rotor mesh, M_rr is the mutual inductance between adjacent, I_rk is the mesh current "k” and I_e is the short-circuit ring current.

The complete system is:

$\left[\begin{array}{c}V_{d s} \\ V_{q s} \\ 0 \\ \vdots \\ 0\end{array}\right]=[L] \cdot \frac{d}{d t}\left[\begin{array}{c}I_{d s} \\ I_{q s} \\ \vdots \\ I_{r k} \\ \vdots\end{array}\right]+[R]\left[\begin{array}{c}I_{d s} \\ I_{q s} \\ \vdots \\ I_{r k} \\ \vdots\end{array}\right]$          (13)

So, becomes:

The representation of the state shows a very high order system because it consists of the number of phases of the stator, the number of phases of the rotor, and the electromechanical equations. The rank of the system is, therefore "N_r+3". It is therefore necessary to reduce the size of the matrices to reduce the simulation time [13].

To do this, we use the generalized Clarke matrix extended to the rotor system. This makes it possible to move from "n-phases" modeling to equivalent two-phase modeling written as follows [14]:

The electromagnetic torque is obtained by derivation of the co-energy:

$C_{e m}=\frac{3}{2} p\left[I_{d q s}\right]^T \frac{\delta}{\delta \theta}\left[\begin{array}{ccc}\cdots & -M_{s r} \cos (\theta+k \alpha) & \cdots \\ \cdots & -M_{s r} \sin (\theta+k \alpha) & \cdots\end{array}\right]\left[\begin{array}{c}\vdots \\ I_{r k} \\ \vdots\end{array}\right]$          (15)

$C_{e m}=\frac{3}{2} p M_{s r}\left\{I_{d s} \sum_{k=0}^{N_r-1} I_{r k} \sin (k \alpha)-I_{q s} \sum_{k=0}^{N_r-1} I_{r k} \cos (k \alpha)\right\}$          (16)

We add the mechanical equation to have the mechanical speed "Ω=ω/P”.

$\frac{d \Omega}{d t}=\frac{1}{J} p\left(C_{e m}-C_r-\frac{f}{p} \omega\right)$          (17)

with: $\frac{d \theta}{d t}=\omega$.

$\left[T_{3 n}\left(\theta_r\right)\right]=\frac{2}{n} \cdot \left[\begin{array}{ccccc}\frac{1}{2} & \cdots & \cdots & \cdots & \frac{1}{2} \\ \cos \left(\theta_r\right) & \cdots & \cos \left(\theta_r-k p \frac{2 \pi}{n}\right) & \cdots & \cos \left(\theta_r-(n-1) p \frac{2 \pi}{n}\right) \\ -\sin \left(\theta_r\right) & \cdots & -\sin \left(\theta_r-k p \frac{2 \pi}{n}\right) & \cdots & -\sin \left(\theta_r-(n-1) p \frac{2 \pi}{n}\right)\end{array}\right]$           (18)

The inverse matrix is given by:

$\left[T_{3 n}\left(\theta_r\right)\right]^{-1}=\left[T_{n 3}\left(\theta_r\right)\right]=\left[\begin{array}{ccc}1 & \cos \left(\theta_r\right) & -\sin \left(\theta_r\right) \\ \vdots & \vdots & \vdots \\ \vdots & \cos \left(\theta_r-k p \frac{2 \pi}{n}\right) & -\sin \left(\theta_r-k p \frac{2 \pi}{n}\right) \\ 1 & \vdots & \vdots \\ & \cos \left(\theta_r-(n-1) p \frac{2 \pi}{n}\right) & -\sin \left(\theta_r-(n-1) p \frac{2 \pi}{n}\right)\end{array}\right]$          (19)

After applying this transformation matrix, a state vector [X] is defined, will give:

$\left\{\begin{array}{c}{\left[X_{0 d q s}\right]=\left[T\left(\theta_s\right)\right] \cdot\left[X_{a b c s}\right] \Leftrightarrow} \\ {\left[X_{a b c s}\right]=\left[T\left(\theta_s\right)\right]^{-1} \cdot\left[X_{0 d q s}\right]} \\ {\left[X_{0 d q r}\right]=\left[T_{3 N_r}\left(\theta_r\right)\right] \cdot\left[X_{r k}\right] \Leftrightarrow} \\ {\left[X_{r k}\right]=\left[T_{3 N_r}\left(\theta_r\right)\right]^{-1} \cdot\left[X_{0 d q r}\right]}\end{array}\right.$          (20)

• For the following part of the stator voltages:

$\left[V_{a b c s}\right]=\left[R_s\right]\left[I_{a b c s}\right]+\frac{d}{d t}\left\{\left[L_s\right] \cdot\left[I_{a b c s}\right]\right\}+\frac{d}{d t}\left\{\left[M_{s r}\right] .\left[I_{r k}\right]\right\}$           (21)

Applying the generalized equation to equation (21) gives:

$\left[V_{0 d q s}\right]=\left\{\left[T\left(\theta_s\right)\right] \cdot\left[R_s\right] \cdot\left[T\left(\theta_s\right)\right]^{-1}\right\} \cdot\left[I_{0 d q s}\right]+$

$\left\{\left[T\left(\theta_s\right)\right] .\left[L_s\right] .\left[T\left(\theta_s\right)\right]^{-1}\right\} \cdot \frac{d}{d t}\left[I_{0 d q s}\right]+$

$\left\{\left[T\left(\theta_s\right)\right] \cdot\left[M_{s r}\right] \cdot\left[T_{3 N_r}\left(\theta_r\right)\right]^{-1}\right\} \cdot \frac{d}{d t}\left[I_{0 d q r}\right]$           (22)

• For the rotor part:

$\left[V_{a b c r}\right]=\left[R_r\right]\left[I_{r k}\right]+\frac{d}{d t}\left\{\left[L_r\right] .\left[I_{r k}\right]\right\}+\frac{d}{d t}\left\{\left[M_{s r}\right] .\left[I_{a b c s}\right]\right\}$          (23)

$\left[V_{0 d q r}\right]=\left\{\left[T\left(\theta_r\right)\right] \cdot\left[R_r\right] \cdot\left[T_{3 N_r}\left(\theta_r\right)\right]^{-1}\right\} \cdot\left[I_{0 d q r}\right]+$

$\left\{\left[T\left(\theta_r\right)\right] \cdot\left[L_r\right] \cdot\left[T_{3 N_r}\left(\theta_r\right)\right]^{-1}\right\} \cdot \frac{d}{d t}\left[I_{0 d q r}\right]+$

$\left\{\left[T\left(\theta_r\right)\right] \cdot\left[M_{s r}\right] \cdot\left[T\left(\theta_r\right)\right]^{-1}\right\} \cdot \frac{d}{d t}\left[I_{0 d q s}\right]$          (24)

By choosing a referential linked to the rotor such that "θ_s=θ et θ_r=0". This change of landmark makes it possible to obtain, after simplification, a reduced-size model of the asynchronous machine [15]:

$\left[\begin{array}{ccccc}L_{s c} & 0 & -\frac{N_r}{2} M_{s r} & 0 & 0 \\ 0 & L_{s c} & 0 & -\frac{N_r}{2} M_{s r} & 0 \\ -\frac{3}{2} M_{s r} & 0 & L_{r c} & 0 & 0 \\ 0 & -\frac{3}{2} M_{s r} & 0 & L_{r c} & 0 \\ 0 & 0 & 0 & 0 & L_e\end{array}\right] \cdot \frac{d}{d t}\left[\begin{array}{c}I_{d s} \\ I_{q s} \\ I_{d r} \\ I_{q r} \\ I_e\end{array}\right] =\left[\begin{array}{c}V_{d s} \\ V_{q s} \\ V_{d r} \\ V_{q r} \\ V_e\end{array}\right]-\left[\begin{array}{ccccc}R_s & -L_{s c} \omega & 0 & \frac{N_r}{2} M_{s r} \omega & 0 \\ L_{s c} \omega & R_s & -\frac{N_r}{2} M_{s r} \omega & 0 & 0 \\ 0 & 0 & R_r & 0 & 0 \\ 0 & 0 & 0 & R_r & 0 \\ 0 & 0 & 0 & 0 & R_e\end{array}\right] \cdot\left[\begin{array}{c}I_{d s} \\ I_{q s} \\ I_{d r} \\ I_{q r} \\ I_e\end{array}\right]$           (25)

with: $\left\{\begin{array}{c}L_{r c}=L_{r p}-M_{r r}+\frac{2 \cdot L_e}{N_r}+2 \cdot L_b(1-\cos (\alpha)) \\ \quad R_r=2 \frac{R_e}{N_r}+2 \cdot R_b(1-\cos (\alpha))\end{array}\right.$.

After establishing the model of the squirrel-cage asynchronous machine (Table 1) considering the structure of the rotor without fault, we now proceed to the modeling of the machine taking into account the rotor fault of the bar break type.

Modeling this type of failure can be done using two different methods to cancel the current passing through the faulty bar. The first modeling method is to completely reconstruct the electrical circuit of the rotor. In this type of approach, the failed rotor bar is removed from the electrical circuit, which requires recalculation of the [R_r] resistance and inductance [L_r] matrices of the machine. Indeed, removing a bar from the cage gives us a matrix [R_r] and [L_r] of lower rank than that developed for the healthy machine. The change in the order of the rotor matrices forces the electrical and magnetic laws of the 'k' loop to be recalculated [13, 16].

The second possible approach is to artificially increase the value of the resistance of the failed bar by a factor sufficient for the current that passes through it to be as close as possible to zero in a steady-state. Compared to the first method, the structure of the rotor electrical circuit is not modified because it is considered, in this type of modeling, that a bar break does not modify the own and mutual inductances of the rotor cage.

Therefore, modeling the partial break of the bars is possible in the latter approach, as the matrix [R_r] needs to be modified.

The rotor defect matrix is therefore written as follows:

$\left[R_{r f}\right]=\left[R_r\right]+\left[\begin{array}{ccccccc}0 & \cdots & \cdots & 0 & 0 & 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & \cdots & 0 & 0 & 0 & 0 & \cdots \\ 0 & \cdots & 0 & R_{b k} & -R_{b k} & 0 & \cdots \\ 0 & \cdots & 0 & -R_{b k} & R_{b k} & 0 & \cdots \\ 0 & \cdots & 0 & 0 & 0 & 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots\end{array}\right]$          (26)

The new rotor resistance matrix, after transformations, becomes:

$\left[R_{r f d q}\right]=\left[T\left(\theta_r\right)\right] \cdot\left[R_{r f}\right] \cdot\left[T\left(\theta_r\right)\right]^{-1}$          (27)

So, the reduced-size model (5X5) of the induction machine with rotor bar breakage defects becomes:

$\left[\begin{array}{ccccc}L_{s c} & 0 & -\frac{N_r}{2} M_{s r} & 0 & 0 \\ 0 & L_{s c} & 0 & -\frac{N_r}{2} M_{s r} & 0 \\ -\frac{3}{2} M_{s r} & 0 & L_{r c} & 0 & 0 \\ 0 & -\frac{3}{2} M_{s r} & 0 & L_{r c} & 0 \\ 0 & 0 & 0 & 0 & L_e\end{array}\right] \cdot \frac{d}{d t}\left[\begin{array}{c}I_{d s} \\ I_{q s} \\ I_{d r} \\ I_{q r} \\ I_e\end{array}\right]= \left[\begin{array}{c}V_{d s} \\ V_{q s} \\ V_{d r} \\ V_{q r} \\ V_e\end{array}\right]-\left[\begin{array}{ccccc}R_s & -L_{s c} \omega & 0 & \frac{N_r}{2} M_{s r} \omega & 0 \\ L_{s c} \omega & R_s & -\frac{N_r}{2} M_{s r} \omega & 0 & 0 \\ 0 & 0 & R_{r d d} & R_{r d q} & 0 \\ 0 & 0 & R_{r q d} & R_{r q q} & 0 \\ 0 & 0 & 0 & 0 & R_e\end{array}\right] \cdot\left[\begin{array}{c}I_{d s} \\ I_{q s} \\ I_{d r} \\ I_{q r} \\ I_e\end{array}\right]$             (28)

with:

$R_{r d d}=2 \cdot R_b(1-\cos (\alpha))+\frac{R_e}{N_r}+\frac{2}{N_r}(1-\cos (\alpha))$.

$\sum_k R_{b f k}(1-\cos (2 k-1) \cdot \alpha)$

$R_{r d q}=-\frac{2}{N_r}(1-\cos (\alpha)) \sum_k R_{b f k} \sin (2 k-1) \cdot \alpha$

$R_{r q d}=-\frac{2}{N_r}(1-\cos (\alpha)) \sum_k R_{b f k} \sin (2 k-1) \cdot \alpha$

$R_{r q q}=2 \cdot R_b(1-\cos (\alpha))+2 \cdot \frac{R_e}{N_r}+\frac{2}{N_r}(1-\cos \alpha) \cdot \sum_k R_{b f k}(1+\cos (2 k-1) \cdot \alpha)$          (29)

The index "k" indicates the broken bar.

For the mechanical part, after application of the generalized transformation on the expression of the torque (16) is obtained:

$C_{e m}=\frac{3}{2} \cdot p \cdot \frac{N_r}{2} \cdot M_{s r} \cdot\left(I_{d s} \cdot I_{q r}-I_{q s} \cdot I_{d r}\right)$          (30)

Table 1. Induction machine parameters

2.3 Simulation results and discussion

Once the model of the induction machine is established. The simulation aspect can be addressed in the MATLAB/Simulink environment, which provides the ability to observe and interpret visualized phenomena and quantities in real-time. We will represent the induction machine in different states (Figure 4), healthy and defective. The results of the simulation in these cases are as follows:

4.jpg

Figure 4. Simulation model of an induction machine with healthy and all cases of broken bars state

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Figure 5. Three-phase supply voltage Vabcs

Initially, all statistical and wavelet packet features were evaluated without performing feature screening; the results are shown in Table 4. ANN and RF using hybrid features (statistical and wavelet packet parameters) for detection of healthy and broken rotor bar faults states produced higher accuracy than the other combinations. The RF classifier with PCA achieved the highest accuracy (98.333%) among all classifiers with different types of characteristics.

Table 4. Results of ANN and RF classifiers without and with optimization algorithms