**2. The equations of the three-phase induction machine in phase coordinates**

The structure of the analyzed induction machine contains: 3 identical phase windings placed on the stator in an 120 electric degrees angle of phase difference configuration; 3 identical phase windings placed on the rotor with a similar difference of phase; a constant air-gap (close slots in an ideal approach); an unsaturated (linear) magnetic circuit that allow to each winding to be characterized by a main and a leakage inductance. Each phase winding has *Ws* turns on stator and *WR* turns on rotor and a harmonic distribution. All inductances are considered constant. The schematic view of the machine is presented in Fig. 1a.

The voltage equations that describe the 3+3 circuits are:

$$
\mu\_{as} = R\_s \dot{i}\_{as} + \frac{d\nu\_{as}}{dt}, \quad \mu\_{bs} = R\_s \dot{i}\_{bs} + \frac{d\nu\_{bs}}{dt}, \quad \mu\_{cs} = R\_s \dot{i}\_{cs} + \frac{d\nu\_{cs}}{dt} \tag{1}
$$

$$
\mu\_{AR} = R\_R \dot{i}\_{AR} + \frac{d\nu\_{AR}}{dt}, \quad \mu\_{BR} = R\_R \dot{i}\_{BR} + \frac{d\nu\_{BR}}{dt}, \quad \mu\_{CR} = R\_R \dot{i}\_{CR} + \frac{d\nu\_{CR}}{dt} \tag{2}
$$

In a matrix form, the equations become:

Mathematical Model of the Three-Phase Induction Machine

for the Study of Steady-State and Transient Duty Under Balanced and Unbalanced States 5

$$
\left[\left.\boldsymbol{\mu}\_{\rm abss}\right\|\right] = \left[\left.\boldsymbol{R}\_{s}\right]\right] \left[\left.\boldsymbol{i}\_{\rm abss}\right] + \frac{d\left\lfloor\boldsymbol{\mu}\right\rfloor\_{\rm abss}}{dt} \tag{3}
$$

$$
\left[\left.u\right|\_{ABCD}\right] = \left[\left.R\_R\right]\right] \left[\left.i\_{ABCD}\right] + \frac{d\left[\left.\nu\right|\_{ABCD}\right]}{dt} \tag{4}
$$

**Figure 1.** Schematic model of three-phase induction machine: a. real; b. reduced rotor

4 Induction Motors – Modelling and Control

**coordinates** 

2002; Marino et al., 2010; Ong, 1998; Sul, 2011; Wach, 2011).

dynamic processes (Bose, 2006; Marino et al., 2010; Wach, 2011).

*phasor* concept (Boldea & Tutelea, 2010; Marino et al., 2010; Sul, 2011).

electric, magnetic and mechanical quantities under transient operation.

**2. The equations of the three-phase induction machine in phase** 

considered constant. The schematic view of the machine is presented in Fig. 1a.

The voltage equations that describe the 3+3 circuits are:

In a matrix form, the equations become:

The family of mathematical models with concentrated parameters comprises different approaches but two of them are more popular: *the phase coordinate* model and the *orthogonal (dq)* model (Ahmad, 2010; Bose, 2006; Chiasson, 2005; De Doncker et al., 2011; Krause et al.,

The first category works with the real machine. The equations include, among other parameters, the mutual stator-rotor inductances with variable values according to the rotor position. As consequence, the model becomes non-linear and complicates the study of

The orthogonal (dq) model has begun with Park's theory nine decades ago. These models use parameters that are often independent to rotor position. The result is a significant simplification of the calculus, which became more convenient with the defining of the *space* 

Starting with the ″classic″ theory we deduce in this contribution a mathematical model that exclude the presence of the currents and angular velocity in voltage equations and uses total fluxes alone. Based on this approach, we take into discussion two control strategies of induction motor by principle of constant total flux of the stator and rotor, respectively.

The most consistent part of this work is dedicated to the study of unbalanced duties generated by supply asymmetries. It is presented a comparative analysis, which confronts a balanced duty with two unbalanced duties of different unbalance degrees. The study uses as working tool the Matlab-Simulink environment and provides variation characteristics of the

The structure of the analyzed induction machine contains: 3 identical phase windings placed on the stator in an 120 electric degrees angle of phase difference configuration; 3 identical phase windings placed on the rotor with a similar difference of phase; a constant air-gap (close slots in an ideal approach); an unsaturated (linear) magnetic circuit that allow to each winding to be characterized by a main and a leakage inductance. Each phase winding has *Ws* turns on stator and *WR* turns on rotor and a harmonic distribution. All inductances are

> , , *as bs cs as s as bs s bs cs s cs ddd u Ri u Ri u Ri*

*AR R AR BR R BR CR R CR d d <sup>d</sup> u Ri u Ri u Ri*

*dt dt dt*

 

, , *AR BR CR*

(2)

*dt dt dt*

(1)

 

The quantities in brackets represent the matrices of voltages, currents, resistances and total flux linkages for the stator and rotor. Obviously, the total fluxes include both main and mutual components. Further, we define the self-phase inductances, which have a leakage and a main component: Ljj=Lσs+Lhs for stator and LJJ=LΣR+LHR for rotor. The mutual inductances of two phases placed on the same part (stator or rotor) have negative values, which are equal to half of the maximum mutual inductances and with the main self-phase component: Mjk=Ljk=Lhj=Lhk. The expressions in matrix form are:

$$
\begin{bmatrix} L\_{\sigma s} \ \end{bmatrix} = \begin{bmatrix} L\_{\sigma s} + L\_{\text{hs}} & -(1/2)L\_{\text{hs}} & -(1/2)L\_{\text{hs}} \\ -(1/2)L\_{\text{hs}} & L\_{\sigma s} + L\_{\text{hs}} & -(1/2)L\_{\text{hs}} \\ -(1/2)L\_{\text{hs}} & -(1/2)L\_{\text{hs}} & L\_{\sigma s} + L\_{\text{hs}} \end{bmatrix} \tag{5-1}
$$

$$\begin{aligned} \left[L\_{\Sigma R}\right] = \begin{bmatrix} L\_{\Sigma R} + L\_{HR} & -(1/2)L\_{HR} & -(1/2)L\_{HR} \\ -(1/2)L\_{HR} & L\_{\Sigma R} + L\_{HR} & -(1/2)L\_{HR} \\ -(1/2)L\_{HR} & -(1/2)L\_{HR} & L\_{\Sigma R} + L\_{HR} \end{bmatrix} \end{aligned} \tag{5-2}$$

$$\begin{bmatrix} \begin{bmatrix} L\_{sR} \end{bmatrix} = \begin{bmatrix} L\_{Rs} \end{bmatrix}\_{\boldsymbol{\theta}} = L\_{sR} \cdot \begin{bmatrix} \cos \theta\_R & \cos \left(\theta\_R + u\right) & \cos \left(\theta\_R + 2u\right) \\ \cos \left(\theta\_R + 2u\right) & \cos \theta\_R & \cos \left(\theta\_R + u\right) \\ \cos \left(\theta\_R + u\right) & \cos \left(\theta\_R + 2u\right) & \cos \theta\_R \end{bmatrix} \tag{5-3}$$

where *u* denotes the angle of 1200 (or 2π/3 rad).

The analysis of the induction machine usually reduces the rotor circuit to the stator one. This operation requires the alteration of the rotor quantities with the coefficient k=Ws/WR by complying with the conservation rules. The new values are:

$$
\mu\_{\rm absr} = k \cdot \mu\_{\rm ABCR}; \quad \mathcal{W}\_{\rm absr} = k \cdot \mathcal{W}\_{\rm ABCR}; \quad i\_{\rm absr} = \left(1 \, k\right) \cdot i\_{\rm ABCR};
$$

$$
R\_r = k^2 \cdot R\_R; \quad L\_{\rm hr} = k^2 \cdot L\_{\rm HR} = \left(\frac{\mathcal{W}\_s}{\mathcal{W}\_R}\right)^2 \cdot \frac{\mathcal{W}\_R^2}{\mathcal{R}\_h} = \frac{\mathcal{W}\_s^2}{\mathcal{R}\_h} = L\_{\rm hs};\tag{6}
$$

Mathematical Model of the Three-Phase Induction Machine

cos cos cos 2

cos cos cos 2

cos cos cos 2

*r*

*bs cs RR R*

*r*

*cs as RR R*

*i ui u*

n sin 2 sin

 

*i ui u*

n sin 2 sin

 

 

*L u*

*L*

 

*L u*

*L*

 

*L u*

*L*

 

*<sup>r</sup> as*

 

*<sup>r</sup> bs*

 

*<sup>r</sup> cs*

 

 

for the Study of Steady-State and Transient Duty Under Balanced and Unbalanced States 7

sin sin sin 2

 

 

 

sin sin sin 2

*i i ui u*

sin sin sin 2

*i i ui u*

*i i ui u*

<sup>2</sup>

*R bs cs as bs cs s*

<sup>2</sup>

*s s s hs r r hs bs cs as br RR R cr ar*

 

*R L RL L R L s i i i i i ui u*

<sup>2</sup>

*R cs as as bs cs s*

<sup>2</sup>

*s s s hs r r hs cs as bs cr RR R ar br*

 

*R L RL L R L s i i i i i ui u*

<sup>2</sup>

*R as bs as bs cs s*

<sup>2</sup>

 

*hs cr RR R ar br*

cos cos 2 cos

 

*R L L R L s i i i ui u i i LL L L*

*L LL*

*R L L R L s i i i ui u i i LL L L*

3 2 2,6

*u u uu u*

cos cos 2 cos

 

<sup>2</sup> 2 cos cos 2 cos 2,6

*r r hs s hs br bs RR R R cs as cr ar*

<sup>2</sup> 2 cos cos 2 cos 2,6

*r r hs s hs ar as RR R R bs cs br cr*

*L L L L i i uuu*

*hs r hs hs*

cos cos cos 2 ,

*<sup>L</sup> L L u u uu u uuu <sup>L</sup> L L*

*hs s hs bs RR R cs as ar br cr*

 

> 

*<sup>L</sup> L L u u uu u uuu <sup>L</sup> L L*

*hs s hs as RR R bs cs ar br cr*

 

> 

*R cr RR R ar br*

 

*hs br RR R cr ar*

2

3 2 2,6

*u u uu u*

*<sup>L</sup> L L <sup>L</sup> i i uuu*

*hs r hs hs*

cos cos cos 2 ,

 

*R br RR R cr ar*

 

*hs ar RR R br cr*

2

*L LL*

*L LL*

*s s s hs r r hs as bs cs ar RR R br cr*

 

*R L RL L R L s i i i i i ui u*

3 2 2,6

*u u uu u*

*<sup>L</sup> L L <sup>L</sup> i i uuu*

*hs r hs hs*

cos cos cos 2 ,

 

*R ar RR R br cr*

2

 

*i i*

*s hs r cr ar*

23 23 si

,

*LL LL uL i L L*

*s hs s hs br bs R hs*

*i i*

*s hs r br cr*

23 23 si

,

*LL LL uL i L L*

*s hs s hs ar as R hs*

3 2

*LL L*

*hs hs r*

*L*

3 2

*LL L*

*hs hs r*

*L*

3 2

*LL L*

*LL LL L*

 

*L*

*L*

 

*L*

*L*

 

*L*

*L*

2

*L LR*

*r*

2

*r*

*L L* 

*L LR*

*L L* 

*LL LL L*

*LL LL L*

*hs hs r*

*L*

*r*

*r*

*s s*

*s s*

*s s*

$$L\_{\sigma r} = k^2 L\_{\Sigma R} = \left(\frac{W\_s}{W\_R}\right)^2 \frac{W\_R^2}{\Re\_{\sigma R}} = \frac{W\_s^2}{\Re\_{\sigma r}} \approx L\_{\sigma s};\ L\_{sr} = kL\_{sR} = \left(\frac{W\_s}{W\_R}\right)\frac{W\_s W\_R}{\Re\_h} = L\_{hs}$$

where the reluctances of the flux paths have been used. The new matrices, with rotor quantities denoted with lowercase letters are:

$$\begin{aligned} \left[\begin{array}{c} \left[L\_{rr}\right] = k^2 \left[L\_{RR}\right] = \begin{bmatrix} L\_{\sigma r} + L\_{\text{hs}} & -(1/2)L\_{\text{hs}} & -(1/2)L\_{\text{hs}} \\ -(1/2)L\_{\text{hs}} & L\_{\sigma r} + L\_{\text{hs}} & -(1/2)L\_{\text{hs}} \\ -(1/2)L\_{\text{hs}} & -(1/2)L\_{\text{hs}} & L\_{\sigma r} + L\_{\text{hs}} \end{bmatrix} \end{aligned} \tag{7-1}$$

$$\begin{bmatrix} L\_{sr} \\ \end{bmatrix} = k \begin{bmatrix} L\_{sR} \\ \end{bmatrix} = \begin{bmatrix} L\_{rs} \\ \end{bmatrix}\_{l} = L\_{hs} \cdot \begin{bmatrix} \cos \theta\_R & \cos \left(\theta\_R + u\right) & \cos \left(\theta\_R + 2u\right) \\ \cos \left(\theta\_R + 2u\right) & \cos \theta\_R & \cos \left(\theta\_R + u\right) \\ \cos \left(\theta\_R + u\right) & \cos \left(\theta\_R + 2u\right) & \cos \theta\_R \\ \end{bmatrix} \tag{7-2}$$

By virtue of these transformations, the voltage equations become:

$$\begin{cases} \left[\boldsymbol{u}\_{abcs}\right] = \left[\boldsymbol{R}\_s\right] \left[\boldsymbol{i}\_{abcs}\right] + \frac{d\left[\boldsymbol{\nu}\_{abcs}\right]}{dt} = \left[\boldsymbol{R}\_s\right] \left[\boldsymbol{i}\_{abcs}\right] + \left[\boldsymbol{L}\_{ss}\right] \frac{d\left[\boldsymbol{i}\_{abcs}\right]}{dt} + \frac{d\left\{\left[\boldsymbol{L}\_{sr}\right] \left[\boldsymbol{i}\_{abcr}\right]\right\}}{dt} \\\\ \left[\boldsymbol{u}\_{abcr}\right] = \left[\boldsymbol{R}\_r\right] \left[\boldsymbol{i}\_{abcr}\right] + \frac{d\left[\boldsymbol{\nu}\_{abcr}\right]}{dt} = \left[\boldsymbol{R}\_r\right] \left[\boldsymbol{i}\_{abcr}\right] + \left[\boldsymbol{L}\_{rr}\right] \frac{d\left[\boldsymbol{i}\_{abcr}\right]}{dt} + \frac{d\left\{\left[\boldsymbol{L}\_{sr}\right] \left[\boldsymbol{i}\_{abcs}\right]\right\}}{dt} \end{cases} \tag{8}$$

By using the notations:

$$\begin{aligned} \left(\Sigma L\_{\Pi}\right) &= L\_{\sigma r} \left(\Im L\_{\text{hs}} + L\_{\sigma s}\right) + L\_{\sigma s} \left(\Im L\_{\text{hs}} + L\_{\sigma r}\right) \\ \left(\Pi L\_{s}\right) &= L\_{\sigma r} \left(L\_{\text{hs}} + L\_{\sigma s}\right) + L\_{\sigma s} \left(\Im L\_{\text{hs}} + L\_{\sigma r}\right) \\ \left(\Pi L\_{r}\right) &= L\_{\sigma s} \left(L\_{\text{hs}} + L\_{\sigma r}\right) + L\_{\sigma r} \left(\Im L\_{\text{hs}} + L\_{\sigma s}\right) \end{aligned} \tag{9}$$

and after the separation of the currents derivatives, (8) can be written under operational form as follows:

Mathematical Model of the Three-Phase Induction Machine

### for the Study of Steady-State and Transient Duty Under Balanced and Unbalanced States 7

6 Induction Motors – Modelling and Control

form as follows:

By using the notations:

quantities denoted with lowercase letters are:

where *u* denotes the angle of 1200 (or 2π/3 rad).

complying with the conservation rules. The new values are:

2 2 2

2

By virtue of these transformations, the voltage equations become:

*abcs s abcs s abcs ss*

*u Ri Ri L*

*abcr r abcr r abcr rr*

*u Ri Ri L*

The analysis of the induction machine usually reduces the rotor circuit to the stator one. This operation requires the alteration of the rotor quantities with the coefficient k=Ws/WR by

; ; 1/ ; *abcr ABCR abcr ABCR abcr ABCR u ku*

*W W <sup>W</sup> R kR L kL <sup>L</sup>*

<sup>2</sup> ; *<sup>s</sup> <sup>R</sup> <sup>s</sup> s sR r R s sr sR hs RRr R h*

where the reluctances of the flux paths have been used. The new matrices, with rotor

(1 / 2) (1 / 2)

*rr RR hs r hs hs*

*L kL L L L L*

*sr sR rs hs R t R R*

 

 

*L L LL L LL L LL L L L L L LL L L L L*

 

 

 

*s r hs s s hs r r s hs r r hs s*

and after the separation of the currents derivatives, (8) can be written under operational

3 3

*r hs s s hs r*

*L kL L L u u*

 

 

*W W <sup>W</sup> W WW L kL L L kL <sup>L</sup> W W*

 

2 2 ; ; *s s <sup>R</sup> r R hr HR hs*

*W*

*k i ki*

2 2 2

(6)

(7-1)

(7-2)

(8)

*R hh*

(1 / 2) (1 / 2)

*hs hs r hs*

*RR R*

 

 

 

cos cos cos 2

*RR R*

*u u*

 

> 

(9)

*L LLL*

(1 / 2) (1 / 2)

*d d i dL i*

*dt dt dt d d i dL i*

*dt dt dt*

3 3

cos 2 cos cos cos cos 2 cos

*abcs abcs sr abcr*

*sr abcs abcr abcr t*

*u u*

*LL L L*

*r hs hs hs*

 

 <sup>2</sup> 2 cos cos cos 2 3 2 sin sin sin 2 3 2 2,6 *s s s hs r r hs as bs cs ar RR R br cr s s hs hs r R ar RR R br cr hs r hs hs R bs cs as bs cs s R L RL L R L s i i i i i ui u LL LL L LL L i i ui u L <sup>L</sup> L L <sup>L</sup> i i uuu L LL* <sup>2</sup> cos cos cos 2 , *<sup>r</sup> as hs ar RR R br cr L u L L u u uu u L* <sup>2</sup> 2 cos cos cos 2 3 2 sin sin sin 2 3 2 2,6 *s s s hs r r hs bs cs as br RR R cr ar s s hs hs r R br RR R cr ar hs r hs hs R cs as as bs cs s R L RL L R L s i i i i i ui u LL LL L LL L i i ui u L <sup>L</sup> L L <sup>L</sup> i i uuu L LL* <sup>2</sup> cos cos cos 2 , *<sup>r</sup> bs hs br RR R cr ar L u L L u u uu u L* <sup>2</sup> 2 cos cos cos 2 3 2 sin sin sin 2 3 2 2,6 *s s s hs r r hs cs as bs cr RR R ar br s s hs hs r R cr RR R ar br hs r hs hs R as bs as bs cs s R L RL L R L s i i i i i ui u LL LL L LL L i i ui u L L L L L i i uuu L LL* <sup>2</sup> cos cos cos 2 , *<sup>r</sup> cs hs cr RR R ar br L u L L u u uu u L* <sup>2</sup> 2 cos cos 2 cos 2,6 *r r hs s hs ar as RR R R bs cs br cr R L L R L s i i i ui u i i* 

 2 cos cos 2 cos 23 23 si *r hs s hs as RR R bs cs ar br cr r s hs s hs ar as R hs LL L L <sup>L</sup> L L u u uu u uuu <sup>L</sup> L L LL LL uL i L L* n sin 2 sin , *bs cs RR R s hs r br cr r i ui u L LR i i L L* 

 <sup>2</sup> 2 cos cos 2 cos 2,6 2 cos cos 2 cos 23 23 si *r r hs s hs br bs RR R R cs as cr ar r hs s hs bs RR R cs as ar br cr r s hs s hs br bs R hs R L L R L s i i i ui u i i LL L L <sup>L</sup> L L u u uu u uuu <sup>L</sup> L L LL LL uL i L L* n sin 2 sin , *cs as RR R s hs r cr ar r i ui u L LR i i L L* 

$$\begin{split} \left[\tilde{s} + \frac{R\_{s}\left(\boldsymbol{\Sigma}\boldsymbol{L}\_{r}\right)}{L\_{\sigma\sigma}\left(\boldsymbol{\Sigma}\boldsymbol{L}\_{\Pi}\right)}\right]\tilde{i}\_{lr} &= \frac{2L\_{bs}R\_{s}\left[\tilde{i}\_{ls}\cos\theta\_{R} + \tilde{i}\_{ls}\cos\left(\theta\_{R} + 2u\right) + \tilde{i}\_{ls}\cos\left(\theta\_{R} + u\right)\right] + 2,6\dot{\theta}\_{R}\left(\tilde{i}\_{lr} + \tilde{i}\_{lr}\right) - \\ &- \frac{2L\_{bs}}{\left(\boldsymbol{\Sigma}\boldsymbol{L}\_{\Pi}\right)}\left[\tilde{i}\_{ls}\cos\theta\_{R} + \tilde{i}\_{ls}\cos\left(\theta\_{R} + 2u\right) + \tilde{i}\_{ls}\cos\left(\theta\_{R} + u\right)\right] + \frac{L\_{qs}L\_{bs}}{L\_{\sigma\sigma}\left(\boldsymbol{\Sigma}\boldsymbol{L}\_{\Pi}\right)}\left(\tilde{i}\_{lr} + \tilde{i}\_{lr} + \tilde{i}\_{lr}\right) + \\ &+ \frac{2L\_{qs} + 3L\_{bs}}{\left(\boldsymbol{\Sigma}\boldsymbol{L}\_{\Pi}\right)}\boldsymbol{i\_{l}}\_{llr} + \dot{\theta}\_{R}L\_{bs}\frac{2L\_{cs} + 3L\_{bs}}{\left(2\boldsymbol{L}\_{\Pi}\right)}\left[\tilde{i}\_{ls}\sin\theta\_{R} + \tilde{i}\_{ls}\sin\left(\theta\_{R} + 2u\right) + \tilde{i}\_{ls}\sin\left(\theta\_{R} + u\right)\right] - \\ &- \frac{L\_{qs}L\_{bs}R\_{r}}{L\_{\sigma\sigma}\left(\boldsymbol{\Sigma}\_{\Pi}\right)}\left(\tilde{i}\_{sr} + \tilde{i}\_{lr}\right),\end{split}$$

Mathematical Model of the Three-Phase Induction Machine

*sR R R*

*abcabc*

*dt* 

*d*

(17)

 

> 

(14)

 

 

> 

> >

(18)

2 2

 

> 

2 2

 

cos cos cos 2

2 2

 

*R hs s r hs s*

*u LL L LL*

*u u L LL LL*

cos 2 cos cos cos cos 2 cos

for the Study of Steady-State and Transient Duty Under Balanced and Unbalanced States 9

*s R RR*

*l uu l uu*

1 1 / 2 1 / 2 cos cos cos 2 1/2 1 1 / 2 cos 2 cos cos 1 / 2 1 / 2 1 cos cos 2 cos cos cos 2 cos 1 1/2 1/2 cos cos cos 2 1 / 2 1 1/2

 

*s RR R*

*l u u*

 os 1/2 1/2 1 *R r l*

(15)

It is well known that the total fluxes have a self-component and a mutual one. Taking into consideration the rules of reducing the rotor circuit to the stator one, the matrix of

 

*u ul u L u l*

*R hs R R r*

*abcabc s r abcabc abcabc abcabc abcabc*

1 1 <sup>1</sup> , *abcabc abcabc abcabc abcabc abcabc abcabc abcabc abcabc <sup>L</sup>*

1

This is an expression that connects the voltages to the total fluxes with no currents involvement. Now, practically the reciprocal matrix must be found. To this effect, we suppose that the reciprocal matrix has a similar form with the direct matrix. If we use the condition: <sup>1</sup> 1 , *abcabc abcabc L L* than through term by term identification is obtained:

> *s hs r hs r R R R hs r s hs r R R R hs r hs r s R R R R R R r hs s hs s*

*R R R hs s hs s r*

*u u LL LL L*

 ( )3 3 2 ; 2 ;

*s hs r hs s r s r r hs s hs r r s s*

*LD L L L L L L L L LL L L LL LL L L L L LL LL L L L*

 

 

32 ; 32 *hs s hs r r s r s hs s r*

 

 

(19)

*L LL LL u u L L L LL u u*

*L L i or i L* (16)

*u Ri where L i dt* 

*RR R r*

, , : *abcabc*

inductances can be written as follows:

*LD*

1

2 2

*R R*

 *u*

1

*abcabc*

*L*

*abcabc hs*

*L L*

than (15) becomes:

*R R*

Now, the equation system (8) can be written shortly as:

By using the multiplication with the reciprocal matrix:

cos 2 cos c

*u u*

  *d*

,

 

 

*LL LL L u u*

 

os 2

 

2 2

2 2

 

 

cos cos 2 cos cos cos c

cos 2 cos cos

where the following notations have been used:

*u RL*

*abcabc s r abcabc abcabc*

Besides (10), the equations concerning mechanical quantities must be added. To this end, the electromagnetic torque has to be calculated. To this effect, we start from the coenergy expression, *Wm* , of the 6 circuits (3 are placed on stator and the other 3 on rotor) and we take into consideration that the leakage fluxes, which are independent of rotation angle of the rotor, do not generate electromagnetic torque, that is:

$$\boldsymbol{W}'\_{m} = \frac{1}{2} \left[ \boldsymbol{i}\_{\mathrm{abs}} \right]\_{\boldsymbol{t}} \left( \left[ \boldsymbol{L}\_{\mathrm{ss}} \right] - \boldsymbol{L}\_{\sigma \boldsymbol{\varsigma}} \left[ \boldsymbol{1} \right] \right) \left[ \boldsymbol{i}\_{\mathrm{abs}} \right] + \frac{1}{2} \left[ \boldsymbol{i}\_{\mathrm{abs}} \right]\_{\boldsymbol{t}} \left( \left[ \boldsymbol{L}\_{\mathrm{rr}} \right] - \boldsymbol{L}\_{\sigma \boldsymbol{\tau}} \left[ \boldsymbol{1} \right] \right) \left[ \boldsymbol{i}\_{\mathrm{abs}} \right] + \left[ \boldsymbol{i}\_{\mathrm{abs}} \right]\_{\boldsymbol{t}} \left[ \boldsymbol{L}\_{\mathrm{sr}} \left( \boldsymbol{\theta}\_{\mathrm{R}} \right) \right] \left[ \boldsymbol{i}\_{\mathrm{abs}} \right] \tag{11}$$

The magnetic energy of the stator and the rotor does not depend on the rotation angle and consequently, for the electromagnetic torque calculus nothing but the last term of (11) is used. One obtains:

$$\begin{split} T\_c &= \frac{1}{2} p \left[ i\_{abs} \right]\_t \frac{d \left[ L\_{sr} \left( \theta\_R \right) \right]}{d \theta\_R} \Big| i\_{abs} \Big] = \\ &= \frac{1}{2} p L\_{bs} \sin \theta\_R \Big[ i\_{as} \left( -2i\_{ar} + i\_{br} + i\_{cr} \right) + i\_{bs} \left( +i\_{ar} - 2i\_{br} + i\_{cr} \right) + i\_{cs} \left( +i\_{ar} + i\_{br} - 2i\_{cr} \right) \Big] + \\ &+ \frac{\sqrt{3}}{2} p L\_{bs} \cos \theta\_R \Big[ i\_{as} \left( i\_{cr} - i\_{br} \right) + i\_{bs} \left( i\_{ar} - i\_{cr} \right) + i\_{cs} \left( i\_{br} - i\_{ar} \right) \Big] \end{split} \tag{12}$$

The equation of torque equilibrium can now be written under operational form as:

$$\overline{\rho}\overline{\rho}\_{\rm R}\left(\frac{\overline{\rho}+k\_{z}}{p}\right) = \frac{1}{2}pL\_{\rm hs}\left(\sin\theta\_{\rm R}\cdot\left[i\_{\rm as}\left(-2i\_{ar}+i\_{br}+i\_{cr}\right)+i\_{bs}\left(i\_{ar}-2i\_{br}+i\_{cr}\right)+\right.\right.\tag{12.4}$$

$$+i\_{cs}\left(i\_{ar}+i\_{br}-2i\_{cr}\right)] + \sqrt{3}\cos\theta\_{\rm R}\cdot\left[i\_{as}\left(i\_{cr}-i\_{br}\right)+i\_{bs}\left(i\_{ar}-i\_{cr}\right)+i\_{cs}\left(i\_{br}-i\_{ar}\right)\right] \quad \left\{-T\_{\rm ss}\right\}$$

$$\overline{s}\theta\_{\rm R}=o\rho\_{\rm R}=\dot{\theta}\_{\rm R}\tag{13}$$

where ωR represents the *rotational pulsatance* (or *rotational pulsation*).

The simulation of the induction machine operation in Matlab-Simulink environment on the basis of the above equations system is rather complicated. Moreover, since all equations depend on the angular speed than the precision of the results could be questionable mainly for the study of rapid transients. Consequently, the use of other variables is understandable. Further, we shall use the *total fluxes* of the windings (3 motionless windings on stator and other rotating 3 windings on rotor).

It is well known that the total fluxes have a self-component and a mutual one. Taking into consideration the rules of reducing the rotor circuit to the stator one, the matrix of inductances can be written as follows:

$$
\begin{bmatrix} L\_{\text{abcabc}} \end{bmatrix} = L\_{\text{ks}} \cdot \begin{vmatrix} 1+l\_{\text{cs}} & -(1/2) & -(1/2) & \cos\theta\_{\text{R}} & \cos(\theta\_{\text{R}}+u) & \cos(\theta\_{\text{R}}+2u) \\ -(1/2) & 1+l\_{\text{cs}} & -(1/2) & \cos(\theta\_{\text{R}}+2u) & \cos\theta\_{\text{R}} & \cos(\theta\_{\text{R}}+u) \\ -(1/2) & -(1/2) & 1+l\_{\text{cs}} & \cos(\theta\_{\text{R}}+u) & \cos(\theta\_{\text{R}}+2u) & \cos\theta\_{\text{R}} \\ \cos\theta\_{\text{R}} & \cos(\theta\_{\text{R}}+2u) & \cos(\theta\_{\text{R}}+u) & 1+l\_{\text{cr}} & -(1/2) & -(1/2) \\ \cos(\theta\_{\text{R}}+u) & L\_{\text{ks}}\cos\theta\_{\text{R}} & \cos(\theta\_{\text{R}}+2u) & -(1/2) & 1+l\_{\text{cr}} & -(1/2) \\ \cos(\theta\_{\text{R}}+2u) & \cos(\theta\_{\text{R}}+u) & \cos\theta\_{\text{R}} & -(1/2) & -(1/2) & 1+l\_{\text{cr}} \end{vmatrix} \tag{14}$$

Now, the equation system (8) can be written shortly as:

$$
\begin{bmatrix} \boldsymbol{\mu}\_{\text{abcabc}} \end{bmatrix} = \begin{bmatrix} \boldsymbol{R}\_{s,r} \end{bmatrix} \begin{bmatrix} \boldsymbol{i}\_{\text{abcabc}} \end{bmatrix} + \frac{d\begin{bmatrix} \boldsymbol{\nu}\_{\text{abcabc}} \end{bmatrix}}{dt}, \quad \text{where} \; : \; \begin{bmatrix} \boldsymbol{\nu}\_{\text{abcabc}} \end{bmatrix} = \begin{bmatrix} \boldsymbol{L}\_{\text{abcabc}} \end{bmatrix} \begin{bmatrix} \boldsymbol{i}\_{\text{abcabc}} \end{bmatrix} \tag{15}
$$

By using the multiplication with the reciprocal matrix:

$$\left[\boldsymbol{L}\_{\text{abcabc}}\right]^{-1}\left[\boldsymbol{\nu}\_{\text{abcabc}}\right] = \left[\boldsymbol{L}\_{\text{abcabc}}\right]^{-1}\left[\boldsymbol{L}\_{\text{abcabc}}\right]\left[\boldsymbol{i}\_{\text{abcabc}}\right]\_{\text{\textdegree}} \text{ or } \left[\boldsymbol{i}\_{\text{abcabc}}\right] = \left[\boldsymbol{L}\_{\text{abcabc}}\right]^{-1}\left[\boldsymbol{\nu}\_{\text{abcabc}}\right] \tag{16}$$

than (15) becomes:

8 Induction Motors – Modelling and Control

2

*r*

*L L L*

*L L L*

Besides (10), the equations concerning mechanical quantities must be added. To this end, the electromagnetic torque has to be calculated. To this effect, we start from the coenergy

take into consideration that the leakage fluxes, which are independent of rotation angle of

 1 1 1 1 2 2 *m abcs ss s abcs abcr rr r abcr abcs sr R abcr t tt W i LL i i LL i i L i*

The magnetic energy of the stator and the rotor does not depend on the rotation angle and consequently, for the electromagnetic torque calculus nothing but the last term of (11) is

*hs R as cr br bs ar cr cs br ar*

The equation of torque equilibrium can now be written under operational form as:

*R RR s*

The simulation of the induction machine operation in Matlab-Simulink environment on the basis of the above equations system is rather complicated. Moreover, since all equations depend on the angular speed than the precision of the results could be questionable mainly for the study of rapid transients. Consequently, the use of other variables is understandable. Further, we shall use the *total fluxes* of the windings (3 motionless windings on stator and

 

*pL i i i i i i i i i*

<sup>1</sup> sin 2 <sup>2</sup>

2 3 cos

where ωR represents the *rotational pulsatance* (or *rotational pulsation*).

*R hs R as ar br cr bs ar br cr*

*Js k pL i i i i i i i i*

*hs R as ar br cr bs ar br cr cs ar br cr*

<sup>1</sup> sin 2 <sup>2</sup> <sup>2</sup>

*pL i i i i i i i i i i i i*

cos cos 2 cos

 

*R L L R <sup>L</sup> s i i i ui u i i LL L L*

<sup>2</sup> 2 cos cos 2 cos 2,6

*r r hs s hs cr cs RR R R as bs ar br*

*s hs s hs cr R hs cs*

,

*LL LL uL i L L*

23 23

*s hs r ar br*

*L LR i i L L* 

the rotor, do not generate electromagnetic torque, that is:

*sr R*

 

*e abcs abcr <sup>t</sup> <sup>R</sup>*

*d L T pi i d*

 

, of the 6 circuits (3 are placed on stator and the other 3 on rotor) and we

 

(11)

*ii i i ii i ii i ii i T*

*cs ar br cr R as cr br bs ar cr cs br ar st*

*hs s hs cs RR R as bs ar br cr*

*u u uu u uuu*

 

 

*r*

*RR R as bs*

*i ui u*

 

(13)

(12)

(10)

n sin 2 sin

 

si

*r*

expression, *Wm*

used. One obtains:

1 2

2

2

other rotating 3 windings on rotor).

*z*

*p*

<sup>3</sup> cos <sup>2</sup>

$$
\left[\begin{array}{c} \boldsymbol{\mu}\_{\boldsymbol{abc}\boldsymbol{abc}} \end{array}\right] = \left[\begin{array}{c} \boldsymbol{R}\_{s,r} \end{array}\right] \left[\begin{array}{c} \boldsymbol{L}\_{\boldsymbol{abc}\boldsymbol{abc}} \end{array}\right]^{-1} \left[\begin{array}{c} \boldsymbol{\nu}\_{\boldsymbol{abc}\boldsymbol{abc}} \end{array}\right] + \frac{d\left[\begin{array}{c} \boldsymbol{\nu}\_{\boldsymbol{abc}\boldsymbol{abc}} \end{array}\right]}{dt} \tag{17}
$$

This is an expression that connects the voltages to the total fluxes with no currents involvement. Now, practically the reciprocal matrix must be found. To this effect, we suppose that the reciprocal matrix has a similar form with the direct matrix. If we use the condition: <sup>1</sup> 1 , *abcabc abcabc L L* than through term by term identification is obtained:

 1 2 2 2 2 2 2 2 2 1 cos cos cos 2 cos 2 cos cos cos cos 2 cos cos cos 2 cos cos cos c *abcabc s hs r hs r R R R hs r s hs r R R R hs r hs r s R R R R R R r hs s hs s R R L LD L LL LL u u L L L LL u u LL LL L u u u u L LL LL u* 2 2 2 2 os 2 cos 2 cos cos *R hs s r hs s R R R hs s hs s r u LL L LL u u LL LL L* (18)

where the following notations have been used:

$$\begin{aligned} \text{(\tiny\,\text{II}LD)} &= \left( \text{3L}\_{\text{hs}} \text{L}\_{\sigma s} + \text{3L}\_{\text{hs}} \text{L}\_{\sigma r} + \text{2L}\_{\sigma r} \text{L}\_{\sigma s} \right) \text{L}\_{\sigma r} \text{L}\_{\sigma s}; \qquad \Gamma = -2 \text{L}\_{\text{hs}} \text{L}\_{\sigma s} \text{L}\_{\sigma r};\\ \text{(\tiny\,\text{II}L}\newline \text{L}\_{s\sigma} = \left( \text{L}\_{\text{hs}} \text{L}\_{\sigma r} + \text{3L}\_{\text{hs}} \text{L}\_{\sigma s} + \text{2L}\_{\sigma r} \text{L}\_{\sigma s} \right) \text{L}\_{\sigma r}; \quad \text{(\tiny \,\text{II}L)} = \left( \text{L}\_{\text{hs}} \text{L}\_{\sigma s} + \text{3L}\_{\text{hs}} \text{L}\_{\sigma r} + \text{2L}\_{\sigma r} \text{L}\_{\sigma s} \right) \text{L}\_{\sigma s} \end{aligned} \tag{19}$$

Further, the matrix product is calculated: <sup>1</sup> *R L s r abcabc abcabc* , , which is used in (17). After a convenient grouping, the system becomes:

$$\begin{split} \frac{d\boldsymbol{\nu}\_{\rm as}}{dt} + \frac{\boldsymbol{\Pi} \boldsymbol{L}\_{\rm s\sigma} \boldsymbol{R}\_{\rm s}}{\left(\boldsymbol{\Pi} \boldsymbol{L} \boldsymbol{D}\right)} \boldsymbol{\nu}\_{\rm as} &= \boldsymbol{\nu}\_{\rm as} - \frac{\boldsymbol{L}\_{\rm bs} \boldsymbol{L}\_{\sigma\tau}^{2} \boldsymbol{R}\_{\rm s}}{\left(\boldsymbol{\Pi} \boldsymbol{L} \boldsymbol{D}\right)} \left(\boldsymbol{\nu}\_{\rm bs} + \boldsymbol{\nu}\_{\rm cs}\right) + \frac{\boldsymbol{L}\_{\rm bs} \boldsymbol{L}\_{\sigma\sigma} \boldsymbol{L}\_{\sigma\sigma} \boldsymbol{R}\_{\rm s}}{\left(\boldsymbol{\Pi} \boldsymbol{L} \boldsymbol{D}\right)} \times \\ &\times \left[ \left(2\boldsymbol{\nu}\_{\rm ar} - \boldsymbol{\nu}\_{\rm br} - \boldsymbol{\nu}\_{\rm cr}\right) \cos \theta\_{\rm R} + \sqrt{3} \left(\boldsymbol{\nu}\_{\rm cr} - \boldsymbol{\nu}\_{\rm br}\right) \sin \theta\_{\rm R} \right] \end{split} \tag{20-1}$$

Mathematical Model of the Three-Phase Induction Machine

*u u*

(23)

1 1

   

> 

 

> 

*RR R R RR RR R*

*u u*

*u u*

 

> 

*u u*

3

*RR R R RR RR R*

1 1 2 2

1 2

*T p <sup>d</sup>*

2

2

*s s hs r s hs s r s bs bs cs as*

*s s hs r s hs s r s as as bs cs*

*as br cr bs cr ar cs ar br R*

3 cos

 

where the following notation has been used:

*u u*

1

*abcabc*

should be replaced, that is:

torque equation *in fluxes* alone:

 

*s u*

*s u*

*T p*

*R*

in total fluxes.

*<sup>d</sup> <sup>L</sup> d*

3

*u u*

 

> 

(22)

(24)

(25)

(26-1)

(26-2)

for the Study of Steady-State and Transient Duty Under Balanced and Unbalanced States 11

0 0 0 sin sin sin 2 0 0 0 sin 2 sin sin 0 0 0 sin sin 2 sin sin sin 2 sin 0 0 0 sin sin sin 2 0 0 0 sin 2 sin sin 0 0 0

1 (3 / 2)( ) / *s r r s hs L L LL L*

 

1

 

 

 

 

> 

 

*abcabc*

 

This expression defines the permeance of a three-phase machine for the mathematical model

*Observation:* One can use the general expression of the electromagnetic torque where the direct and reciprocal matrices of the inductances (which link the currents with the fluxes)

*abcabc abcabc e abcabc abcabc abcabc abcabc <sup>t</sup> t t R R*

*d L*

*d L d L T pi ip L L*

*<sup>e</sup> abcabc abcabc <sup>t</sup> <sup>R</sup>*

A more convenient expression that depends on sinθR and cosθR, leads to the electromagnetic

*e as ar br cr bs br cr ar cs cr ar br R*

<sup>3</sup> 1/2 2 2 2 sin

Ultimately, by getting together the equations of the 6 electric circuits and the movement equations we obtain an 8 equation system, which can be written under operational form:

*LD LD LD*

*LR LL R LL L R*

 

*LD LD LD*

*LR LL R LL L R*

 

 

 

2 cos 3 sin

 

*ar br cr R R cr br*

2 cos 3 sin

*ar br cr R R ar cr*

*d d*

$$\begin{split} \frac{d\boldsymbol{\nu}\_{\rm bs}}{dt} + \frac{\boldsymbol{\Pi} \boldsymbol{L}\_{\rm s\sigma} \boldsymbol{R}\_{\rm s}}{\left(\overline{\boldsymbol{\Pi} \boldsymbol{L} \boldsymbol{D}}\right)} \boldsymbol{\nu}\_{\rm bs} &= \boldsymbol{\mu}\_{\rm bs} - \frac{\boldsymbol{L}\_{\rm hs} \boldsymbol{L}\_{\sigma\tau}^{2} \boldsymbol{R}\_{\rm s}}{\left(\overline{\boldsymbol{\Pi} \boldsymbol{L} \boldsymbol{D}}\right)} \left(\boldsymbol{\nu}\_{\rm cs} + \boldsymbol{\nu}\_{\rm as}\right) + \frac{\boldsymbol{L}\_{\rm hs} \boldsymbol{L}\_{\sigma\tau} \boldsymbol{L}\_{\sigma\tau} \boldsymbol{R}\_{\rm s}}{\left(\overline{\boldsymbol{\Pi} \boldsymbol{L} \boldsymbol{D}}\right)} \times \\ &\times \left[ \left(-\boldsymbol{\nu}\_{\rm ar} + 2\boldsymbol{\nu}\_{\rm br} - \boldsymbol{\nu}\_{\rm cr}\right) \cos \theta\_{\rm R} + \sqrt{3} \left(\boldsymbol{\nu}\_{\rm ar} - \boldsymbol{\nu}\_{\rm cr}\right) \sin \theta\_{\rm R} \right] \end{split} \tag{20-2}$$

$$\begin{split} \frac{d\boldsymbol{\nu}\_{\rm cs}}{dt} + \frac{\boldsymbol{\Pi} \boldsymbol{L}\_{\rm s\sigma} \boldsymbol{R}\_{\rm s}}{\left(\boldsymbol{\Pi} \boldsymbol{L} \boldsymbol{D}\right)} \boldsymbol{\nu}\_{\rm cs} &= \boldsymbol{\nu}\_{\rm cs} - \frac{\boldsymbol{L}\_{\rm hs} \boldsymbol{L}\_{\sigma\sigma}^{2} \boldsymbol{R}\_{\rm s}}{\left(\boldsymbol{\Pi} \boldsymbol{L} \boldsymbol{D}\right)} \left(\boldsymbol{\nu}\_{\rm as} + \boldsymbol{\nu}\_{\rm bs}\right) + \frac{\boldsymbol{L}\_{\rm hs} \boldsymbol{L}\_{\sigma\sigma} \boldsymbol{L}\_{\sigma\sigma} \boldsymbol{R}\_{\rm s}}{\left(\boldsymbol{\Pi} \boldsymbol{L} \boldsymbol{D}\right)} \times \\ &\times \left[ \left(-\boldsymbol{\nu}\_{\rm ar} - \boldsymbol{\nu}\_{\rm br} + 2\boldsymbol{\nu}\_{\rm cr}\right) \cos \theta\_{\rm R} + \sqrt{3} \left(\boldsymbol{\nu}\_{\rm br} - \boldsymbol{\nu}\_{\rm ar}\right) \sin \theta\_{\rm R} \right] \end{split} \tag{20-3}$$

$$\begin{split} \frac{d\boldsymbol{\nu}\_{\rm ar}}{dt} + \frac{\boldsymbol{\Gamma} \boldsymbol{\Pi}\_{r\sigma} \boldsymbol{\mathcal{R}}\_{r}}{\left(\boldsymbol{\Gamma} \boldsymbol{\Pi} \boldsymbol{D}\right)} \boldsymbol{\nu}\_{\rm ar} &= \boldsymbol{u}\_{\rm ar} - \frac{\boldsymbol{L}\_{\rm bs} \boldsymbol{L}\_{\sigma\boldsymbol{s}}^{2} \boldsymbol{R}\_{r}}{\left(\boldsymbol{\Gamma} \boldsymbol{\Pi} \boldsymbol{D}\right)} \left(\boldsymbol{\nu}\_{\rm br} + \boldsymbol{\nu}\_{\rm cr}\right) + \frac{\boldsymbol{L}\_{\rm bs} \boldsymbol{L}\_{\sigma\boldsymbol{s}} \boldsymbol{L}\_{\sigma\boldsymbol{r}} \boldsymbol{R}\_{r}}{\left(\boldsymbol{\Gamma} \boldsymbol{\Pi} \boldsymbol{D}\right)} \times \\ &\times \left[ \left(2\boldsymbol{\nu}\_{\rm as} - \boldsymbol{\nu}\_{\rm bs} - \boldsymbol{\nu}\_{\rm cs}\right) \cos \theta\_{\rm R} + \sqrt{3} \left(\boldsymbol{\nu}\_{\rm bs} - \boldsymbol{\nu}\_{\rm cs}\right) \sin \theta\_{\rm R} \right] \end{split} \tag{20-4}$$

$$\begin{split} \frac{d\boldsymbol{\eta}\_{br}}{dt} + \frac{\boldsymbol{\Pi} \boldsymbol{L}\_{r\sigma} \boldsymbol{R}\_{r}}{\left(\boldsymbol{\Pi} \boldsymbol{L} \boldsymbol{D}\right)} \boldsymbol{\nu}\_{br} &= \boldsymbol{u}\_{br} - \frac{\boldsymbol{L}\_{\rm hs} \boldsymbol{L}\_{\sigma\rm cs}^{2} \boldsymbol{R}\_{r}}{\left(\boldsymbol{\Pi} \boldsymbol{L} \boldsymbol{D}\right)} \left(\boldsymbol{\nu}\_{cr} + \boldsymbol{\nu}\_{ar}\right) + \frac{\boldsymbol{L}\_{\rm hs} \boldsymbol{L}\_{\sigma\rm cs} \boldsymbol{L}\_{\sigma\rm cr} \boldsymbol{R}\_{r}}{\left(\boldsymbol{\Pi} \boldsymbol{L} \boldsymbol{D}\right)} \times \\ &\times \left[ \left(-\boldsymbol{\nu}\_{as} + 2\boldsymbol{\nu}\_{bs} - \boldsymbol{\nu}\_{cs}\right) \cos \theta\_{\rm R} + \sqrt{3} \left(\boldsymbol{\nu}\_{cs} - \boldsymbol{\nu}\_{as}\right) \sin \theta\_{\rm R} \right] \end{split} \tag{20-5}$$

$$\begin{split} \frac{d\boldsymbol{\nu}\_{cr}}{dt} + \frac{\boldsymbol{\Pi} \boldsymbol{L}\_{r\sigma} \boldsymbol{R}\_{r}}{\left(\boldsymbol{\Pi} \boldsymbol{L} \boldsymbol{D}\right)} \boldsymbol{\nu}\_{cr} &= \boldsymbol{u}\_{cr} - \frac{\boldsymbol{L}\_{\rm hs} \boldsymbol{L}\_{\sigma s}^{2} \boldsymbol{R}\_{r}}{\left(\boldsymbol{\Pi} \boldsymbol{L} \boldsymbol{D}\right)} \left(\boldsymbol{\nu}\_{ar} + \boldsymbol{\nu}\_{br}\right) + \frac{\boldsymbol{L}\_{\rm hs} \boldsymbol{L}\_{\sigma s} \boldsymbol{L}\_{\sigma r} \boldsymbol{R}\_{r}}{\left(\boldsymbol{\Pi} \boldsymbol{L} \boldsymbol{D}\right)} \times \\ &\times \left[ \left(-\boldsymbol{\nu}\_{as} - \boldsymbol{\nu}\_{bs} + 2\boldsymbol{\nu}\_{cs}\right) \cos \theta\_{R} + \sqrt{3} \left(\boldsymbol{\nu}\_{as} - \boldsymbol{\nu}\_{bs}\right) \sin \theta\_{R} \right] \end{split} \tag{20.6-6}$$

For the calculation of the *electromagnetic torque* we can use the principle of energy conservation or the expression of stored magnetic energy. The expression of the electromagnetic torque corresponding to a multipolar machine (*p* is the number of pole pairs) can be written in a matrix form as follows:

$$T\_e = -\frac{p}{2} \cdot \left\{ \left[ \nu\_{\text{abcabc}} \right]\_t \cdot \frac{d \left[ \begin{matrix} L\_{\text{abcabc}} \end{matrix} \right]^{-1}}{d \theta\_R} \cdot \left\lbrack \nu\_{\text{abcabc}} \right\rbrack \right\} \tag{21}$$

To demonstrate the validity of (21), one uses the expression of the matrix <sup>1</sup> *abcabc <sup>L</sup>* , (18), in order to calculate its derivative:

$$\begin{aligned} &\frac{d}{d\theta\_R} \Big[L\_{\text{cch}\text{ch}}\right]^{-1} = \Lambda\_3 \cdot \\ &\begin{bmatrix} 0 & 0 & 0 & \sin\theta\_R & \sin\left(\theta\_R + u\right) & \sin\left(\theta\_R + 2u\right) \\ 0 & 0 & 0 & \sin\left(\theta\_R + 2u\right) & \sin\theta\_R & \sin\left(\theta\_R + u\right) \\ 0 & 0 & 0 & \sin\left(\theta\_R + u\right) & \sin\left(\theta\_R + 2u\right) & \sin\theta\_R \\ \sin\theta\_R & \sin\left(\theta\_R + 2u\right) & \sin\left(\theta\_R + u\right) & 0 & 0 \\ \sin\left(\theta\_R + u\right) & \sin\theta\_R & \sin\left(\theta\_R + 2u\right) & 0 & 0 \\ \sin\left(\theta\_R + 2u\right) & \sin\left(\theta\_R + u\right) & \sin\theta\_R & 0 & 0 \end{bmatrix} \end{aligned} \tag{22}$$

where the following notation has been used:

10 Induction Motors – Modelling and Control

Further, the matrix product is calculated: <sup>1</sup> *R L s r abcabc abcabc* ,

*u*

*u*

*u*

*u*

*u*

*u*

2

 

 

 

 

2

*d L R LLR LL L R*

*dt LD LD LD*

*p d L*

*cr r r hs s r hs s r r cr cr ar br*

 

2

*d L R LLR LL L R*

*dt LD LD LD*

*br r r hs s r hs s r r br br cr ar*

2

*d L R LLR LL L R*

*dt LD LD LD*

*ar r r hs s r hs s r r ar ar br cr*

2

*d LR LL R LL L R*

*dt LD LD LD*

*cs s s hs r s hs s r s cs cs as bs*

 

2

*d LR LL R LL L R*

*dt LD LD LD*

*bs s s hs r s hs s r s bs bs cs as*

  

 

 

 

> 

 

 

 

For the calculation of the *electromagnetic torque* we can use the principle of energy conservation or the expression of stored magnetic energy. The expression of the electromagnetic torque corresponding to a multipolar machine (*p* is the number of pole

*<sup>e</sup> abcabc abcabc <sup>t</sup> <sup>R</sup>*

To demonstrate the validity of (21), one uses the expression of the matrix <sup>1</sup>

2

*d LR LL R LL L R*

*dt LD LD LD*

*as s s hs r s hs s r s as as bs cs*

 

1

 

*abcabc*

 

*d*

 

2 cos 3 sin

 

*as bs cs R bs cs R*

2 cos 3 sin

*as bs cs R cs as R*

*as bs cs R as bs R*

 

 

2 cos 3 sin

 

*ar br cr R cr br R*

2 cos 3 sin

*ar br cr R ar cr R*

*ar br cr R br ar R*

After a convenient grouping, the system becomes:

pairs) can be written in a matrix form as follows:

*T*

in order to calculate its derivative:

  

 

 

> 

 

> 

 

 

 

 

 

 

2 cos 3 sin

2 cos 3 sin

, which is used in (17).

 

> 

 

 

> 

 

(21)

*abcabc <sup>L</sup>* , (18),

(20-1)

(20-2)

(20-3)

(20-4)

(20-5)

(20-6)

$$\Lambda\_3 = \frac{1}{(\Im/\Im)(L\_{\sigma s} + L\_{\sigma r}) + L\_{\sigma r}L\_{\sigma s}/L\_{\text{hs}}} \tag{23}$$

This expression defines the permeance of a three-phase machine for the mathematical model in total fluxes.

*Observation:* One can use the general expression of the electromagnetic torque where the direct and reciprocal matrices of the inductances (which link the currents with the fluxes) should be replaced, that is:

$$T\_c = \frac{1}{2} p \left[ \overline{\boldsymbol{i}}\_{\text{abcabc}} \right]\_t \frac{d \left[ \overline{\boldsymbol{L}\_{\text{abcabc}}} \right]}{d \boldsymbol{\theta}\_R} \left[ \overline{\boldsymbol{i}}\_{\text{abcabc}} \right] = \frac{1}{2} p \cdot \left[ \boldsymbol{\nu}\_{\text{abcabc}} \right]\_t \left[ \boldsymbol{L} \right]\_t^{-1} \cdot \frac{d \left[ \overline{\boldsymbol{L}\_{\text{abcabc}}} \right]}{d \boldsymbol{\theta}\_R} \cdot \left[ \boldsymbol{L} \right]^{-1} \left[ \boldsymbol{\nu}\_{\text{abcabc}} \right]$$

$$T\_c = -\frac{1}{2} p \left[ \boldsymbol{\nu}\_{\text{abcabc}} \right]\_t \frac{d \left[ \overline{\boldsymbol{L}\_{\text{abcabc}}} \right]^{-1}}{d \boldsymbol{\theta}\_R} \left[ \boldsymbol{\nu}\_{\text{abcabc}} \right]$$

A more convenient expression that depends on sinθR and cosθR, leads to the electromagnetic torque equation *in fluxes* alone:

$$\begin{split} T\_{\varepsilon} &= -\Big(1/2\Big)p\Lambda\_{3}\Big[\Big(\nu\_{as}\Big(2\boldsymbol{\nu}\_{ar} - \boldsymbol{\nu}\_{br} - \boldsymbol{\nu}\_{cr}\Big) + \boldsymbol{\nu}\_{bs}\Big(2\boldsymbol{\nu}\_{br} - \boldsymbol{\nu}\_{cr} - \boldsymbol{\nu}\_{ar}\Big) + \boldsymbol{\nu}\_{cs}\Big(2\boldsymbol{\nu}\_{cr} - \boldsymbol{\nu}\_{ar} - \boldsymbol{\nu}\_{br}\Big)\Big]\sin\theta\_{\mathbb{R}} + \\ &+ \sqrt{3}\Big[\boldsymbol{\nu}\_{as}\left(\boldsymbol{\nu}\_{br} - \boldsymbol{\nu}\_{cr}\right) + \boldsymbol{\nu}\_{bs}\Big(\boldsymbol{\nu}\_{cr} - \boldsymbol{\nu}\_{ar}\Big) + \boldsymbol{\nu}\_{cs}\Big(\boldsymbol{\nu}\_{ar} - \boldsymbol{\nu}\_{br}\Big)\Big]\cos\theta\_{\mathbb{R}}\Big] \end{split} \tag{25}$$

Ultimately, by getting together the equations of the 6 electric circuits and the movement equations we obtain an 8 equation system, which can be written under operational form:

 2 2 cos 3 sin *s s hs r s hs s r s as as bs cs ar br cr R R cr br LR LL R LL L R s u LD LD LD* (26-1) 2 2 cos 3 sin *s s hs r s hs s r s bs bs cs as ar br cr R R ar cr LR LL R LL L R s u LD LD LD* (26-2)

$$\begin{split} \overline{\boldsymbol{\Psi}}\_{cs} \left( \overline{\boldsymbol{s}} + \frac{\boldsymbol{\Pi} \boldsymbol{L}\_{s\sigma} \boldsymbol{R}\_{s}}{\left( \overline{\boldsymbol{\Pi} \boldsymbol{L} \boldsymbol{D}} \right)} \right) &= \overline{\boldsymbol{u}\_{cs}} - \frac{\boldsymbol{L}\_{\rm hs} \boldsymbol{L}\_{\sigma\tau}^{2} \boldsymbol{R}\_{s}}{\left( \overline{\boldsymbol{\Pi} \boldsymbol{L} \boldsymbol{D}} \right)} \left( \overline{\boldsymbol{\nu}\_{as}} + \overline{\boldsymbol{\nu}\_{bs}} \right) + \frac{\boldsymbol{L}\_{\rm hs} \boldsymbol{L}\_{\sigma\sigma} \boldsymbol{L}\_{\sigma\tau} \boldsymbol{R}\_{s}}{\left( \overline{\boldsymbol{\Pi} \boldsymbol{L} \boldsymbol{D}} \right)} \times \\ &\times \left[ \left( -\overline{\boldsymbol{\nu}\_{ar}} - \overline{\boldsymbol{\nu}\_{br}} + 2 \overline{\boldsymbol{\nu}\_{cr}} \right) \cos \theta\_{\rm R} + \sqrt{3} \left( \overline{\boldsymbol{\nu}\_{br}} - \overline{\boldsymbol{\nu}\_{ar}} \right) \sin \theta\_{\rm R} \right] \end{split} \tag{26-3}$$

Mathematical Model of the Three-Phase Induction Machine

for the Study of Steady-State and Transient Duty Under Balanced and Unbalanced States 13

the equation system (26-1...8) or on the basis of symmetric components theory with three distinct mathematical models for each component (positive sequence, negative sequence

The vast majority of electric drives uses however the 3 wires connection (no neutral). Consequently, there is no homopolar current component, the homopolar fluxes are zero as well and the sum of the 3 phase total fluxes is null. This is an asymmetric condition with *single unbalance,* which can be studied by using the direct and inverse sequence components when the transformation from 3 to 2 axes is mandatory. This approach practically replaces the three-phase machine with unbalanced supply with two symmetric three-phase machines. One of them produces the positive torque and the other provides the negative torque. The resultant torque comes out through

The operation of the machine with 2 unbalances can be analyzed by considering certain expressions for the instantaneous values of the stator and rotor quantities (voltages, total fluxes and currents eventually, which can be transformed from (a, b, c) to (α, β, 0) reference

1 1/2 1/2

3 2 2 ; ( ) <sup>332</sup> <sup>3</sup>

*hs r hs s r s s s*

 

 

*hs s hs r r s s s r hs s*

 

3 2 2 ; ( ) <sup>332</sup> <sup>3</sup>

*hs r hs s r s r r*

 

 

*hs s hs r r s r r s hs r*

 

*L L R*

*L L R*

 

 

1

1 2

 

*s s*

*L L* 

1

1 2

 

*r r*

*L L* 

 

*R*

 

> 

*R*

 

*r*

*s*

 

  (27)

(28-1)

(28-2)

*st*

*sr*

*s*

*rt*

*rs*

*r*

<sup>2</sup> 0 3/2 3/2 <sup>3</sup> 2/2 2/2 2/2

 

 

> 

 

*r s r hs*

*r r hs s hs r r s r r*

 

*L R LL LL L L R R LD LL LL L L L L LLR LLR R LD LL LL L L L L*

*LL LL*

 

 

 

*s r s hs*

*LL LL*

*hs s r hs s r r*

 

 

 

32 / 3 / 32 /

*hs r s hs r s s*

 

 

32 / 3 / 32 /

; ( ) <sup>332</sup> <sup>6</sup>

 

; ( ) <sup>332</sup> <sup>6</sup>

 

*s s hs r hs s r s s s*

 

*L R LL LL L L R R LD LL LL L L L L LL R LL R R LD LL LL L L L L*

*s as s bs cs s*

and homopolar).

superposition of the effects.

**3.1. The abc-αβ0 model in total fluxes** 

frames in accordance with the following procedure :

We define the following notations:

2

*st sr*

2

*rt rs*

 

  0

$$\begin{split} \overline{\boldsymbol{\nu}}\_{ar} \left( \overline{\boldsymbol{s}} + \frac{\boldsymbol{\Pi} \boldsymbol{L}\_{r\sigma} \boldsymbol{R}\_{r}}{\left( \overline{\boldsymbol{\Pi} \boldsymbol{L} \boldsymbol{D}} \right)} \right) &= \overline{\boldsymbol{u}}\_{ar} - \frac{\boldsymbol{L}\_{bs} \boldsymbol{L}\_{cs}^{2} \boldsymbol{R}\_{r}}{\left( \overline{\boldsymbol{\Pi} \boldsymbol{L} \boldsymbol{D}} \right)} \left( \overline{\boldsymbol{\nu}}\_{br} + \overline{\boldsymbol{\nu}}\_{cr} \right) + \frac{\boldsymbol{L}\_{bs} \boldsymbol{L}\_{cs} \boldsymbol{L}\_{\sigma\sigma} \boldsymbol{R}\_{r}}{\left( \overline{\boldsymbol{\Pi} \boldsymbol{L} \boldsymbol{D}} \right)} \times \\ &\times \left[ \left( 2 \overline{\boldsymbol{\nu}}\_{as} - \overline{\boldsymbol{\nu}}\_{bs} - \overline{\boldsymbol{\nu}}\_{cs} \right) \cos \theta\_{\mathbb{R}} + \sqrt{3} \left( \overline{\boldsymbol{\nu}}\_{bs} - \overline{\boldsymbol{\nu}}\_{cs} \right) \sin \theta\_{\mathbb{R}} \right] \end{split} \tag{26-4}$$

$$\begin{split} \overline{\boldsymbol{\Psi}}\_{br} \overline{\boldsymbol{\Psi}}\_{br} \left( \overline{\boldsymbol{s}} + \frac{\boldsymbol{\Pi} \boldsymbol{L}\_{r\sigma} \boldsymbol{R}\_{r}}{\left( \overline{\boldsymbol{\Pi} \boldsymbol{L} \boldsymbol{D}} \right)} \right) &= \overline{\boldsymbol{u}}\_{br} - \frac{\boldsymbol{L}\_{\text{hs}} \boldsymbol{L}\_{\sigma\text{s}}^{2} \boldsymbol{R}\_{r}}{\left( \overline{\boldsymbol{\Pi} \boldsymbol{L} \boldsymbol{D}} \right)} \overline{\boldsymbol{\psi}}\_{cr} + \overline{\boldsymbol{\psi}}\_{ar} \overline{\boldsymbol{\psi}}\_{ar} + \frac{\boldsymbol{L}\_{\text{hs}} \boldsymbol{L}\_{\sigma\text{s}} \boldsymbol{L}\_{\sigma\text{r}} \boldsymbol{R}\_{r}}{\left( \overline{\boldsymbol{\Pi} \boldsymbol{L} \boldsymbol{D}} \right)} \times \\ &\times \left[ \left( -\overline{\boldsymbol{\nu}}\_{\text{as}} + 2 \overline{\boldsymbol{\nu}}\_{\text{bs}} - \overline{\boldsymbol{\nu}}\_{\text{cs}} \right) \cos \theta\_{\text{R}} + \sqrt{3} \left( \overline{\boldsymbol{\nu}}\_{\text{cs}} - \overline{\boldsymbol{\nu}}\_{\text{as}} \right) \sin \theta\_{\text{R}} \right] \end{split} \tag{26-5}$$

$$\begin{split} \overline{\boldsymbol{\nu}}\_{cr} \left( \overline{\boldsymbol{s}} + \frac{\boldsymbol{\Pi} \boldsymbol{L}\_{r\sigma} \boldsymbol{R}\_{r}}{\left( \overline{\boldsymbol{\Pi} \boldsymbol{L} \boldsymbol{D}} \right)} \right) &= \overline{\boldsymbol{u}\_{cr}} - \frac{\boldsymbol{L}\_{\mathrm{ls}} \boldsymbol{L}\_{\sigma\circ s}^{2} \boldsymbol{R}\_{r}}{\left( \overline{\boldsymbol{\Pi} \boldsymbol{L} \boldsymbol{D}} \right)} \overline{\boldsymbol{\nu}}\_{ar} + \overline{\boldsymbol{\nu}}\_{br} + \frac{\boldsymbol{L}\_{\mathrm{ls}} \boldsymbol{L}\_{\sigma\circ s} \boldsymbol{L}\_{\sigma\circ r} \boldsymbol{R}\_{r}}{\left( \overline{\boldsymbol{\Pi} \boldsymbol{L} \boldsymbol{D}} \right)} \times \\ &\times \left[ \overline{\boldsymbol{\nu}}\_{\mathrm{as}} - \overline{\boldsymbol{\nu}}\_{\mathrm{bs}} + 2 \overline{\boldsymbol{\nu}}\_{\mathrm{cs}} \right) \cos \theta\_{\mathrm{R}} + \sqrt{3} \left( \overline{\boldsymbol{\nu}}\_{\mathrm{as}} - \overline{\boldsymbol{\nu}}\_{\mathrm{bs}} \right) \sin \theta\_{\mathrm{R}} \right] \end{split} \tag{26-6}$$

$$\begin{split} \dot{\theta}\_{R} \Big( \overline{\hat{s}} + \boldsymbol{k}\_{z} \, \big/ \, \mathrm{J} \Big) &= \left( p \, \middle/ \mathrm{J} \right) \Big( - \left( 1 \, \mathrm{2} \right) p \Lambda\_{3} \Big( \sin \theta\_{R} \Big[ \overline{\nu}\_{\mathrm{as}} \left( 2 \overline{\nu}\_{\mathrm{ar}} - \overline{\nu}\_{\mathrm{br}} - \overline{\nu}\_{\mathrm{cr}} \right) + \\ &\quad + \overline{\nu}\_{\mathrm{bs}} \Big( 2 \overline{\nu}\_{\mathrm{br}} - \overline{\nu}\_{\mathrm{cr}} - \overline{\nu}\_{\mathrm{ar}} \Big) + \overline{\nu}\_{\mathrm{cs}} \Big( 2 \overline{\nu}\_{\mathrm{cr}} - \overline{\nu}\_{\mathrm{ar}} - \overline{\nu}\_{\mathrm{br}} \Big) \Big] + \sqrt{3} \cos \theta\_{R} \end{split} \tag{26-7}$$
 
$$\begin{split} \overline{\pi}\_{\mathrm{as}} \left( \overline{\nu}\_{\mathrm{br}} - \overline{\nu}\_{\mathrm{cr}} \right) + \overline{\nu}\_{\mathrm{bs}} \Big( \overline{\nu}\_{\mathrm{cr}} - \overline{\nu}\_{\mathrm{ar}} \Big) + \overline{\nu}\_{\mathrm{cs}} \Big( \overline{\nu}\_{\mathrm{ar}} - \overline{\nu}\_{\mathrm{br}} \Big) \Big] \ \underline{\phantom{3}} - T\_{\mathrm{si}} \end{split} \tag{26-7}$$

$$\frac{d\theta\_R}{dt} = \dot{\theta}\_R = \alpha\_R \tag{26-8}$$

This equation system, (26-1)-(26-8) allows the study of any operation duty of the three-phase induction machine: steady state or transients under balanced or unbalanced condition, with simple or double feeding.
