**5. Study of the unbalanced duties**

24 Induction Motors – Modelling and Control

**Figure 6.** Resultant stator voltage vs. pulsatance, *UsR=f(ωs)* at *ΨrR=const. (1.3Wb)*

0

100

200

300

400

500

600

700

ΨrRk=0,5Wb (short-circuit) and ΨrR0=1,78Wb (synchronism).

Voltage UsR [V]

2 2

2 2 3

*rR*

When the applied voltage and pulsation are two times smaller regarding the rated values then the operation points lie between B1 and B2 since the rotor flux varies within

Rotational pulsatance ω<sup>s</sup> [rad/s]

0 78.5 157 235.5 314

s=0.001 s=0.1 s=0.3

2

3

1

The control based on constant rotor flux strategy ensures parallel mechanical characteristics. This is an important advantage since the induction machine behaves like shunt D.C. motor. A second aspect is also favorable in the behavior under this strategy. The mechanical

The modification of the flux value (generally with decrease) leads to a different slope of the characteristics, which means a significant decrease of the torque for a certain angular speed. The question is ″what variation rule of *UsR/ωs* must be used in order to have constant rotor

characteristic has no sector of unstable operation as the usual induction machine has.

flux″? The expression of the modulus of the resultant rotor flux can be written as:

*s r s r*

 

 

 

3 2 2 3 3

<sup>2</sup> <sup>2</sup> *sR sR rR*

*U U As Bs C As Bs C*

2 2 2 22 2 2 : ; ; ; ( )/ *with A B C*

Fig. 6 presents the variation of the stator voltage with pulsatance at constant resultant rotor flux (1,3 Wb), which are called the *control characteristics* of the static converter connected to the induction machine. The presented characteristics correspond to three constant slip values, *s*=0,001 (no-load)-curve 1, *s*=0,1 (rated duty)-curve 2 and *s*=0,3 (close to pull-out point)-curve 3. It can be seen that the operation with high slip values (high loads) require an increased stator voltage for a certain pulsation. As a matter of fact, the ratio *UsR3/ωs* must be increased with the load when the pulsatance (pulsation) and the angular speed rise as well. Such a strategy is indicated for fans, pumps or load machines with speed-dependent torque.

 

 

*s s s r r tt tt s r s r s*

(51)

 

   . The unbalanced duties (generated by supply asymmetries) are generally analyzed by using the theory of symmetric components, according to which any asymmetric three-phase system with *single unbalance* (the sum of the applied instantaneous voltages is always zero) can be equated with two symmetric systems of opposite sequences: positive (+) (or direct) and negative (-) (or inverse) respectively. There are two possible ways for the analysis of this problem.

a. When the amplitudes of the phase voltages are different and/or the angles of phase difference are not equal to 2π/3 then the *unbalanced three-phase* system can be replaced with an equivalent *unbalanced two-phase* system, which further is taken apart in two systems, one of *direct sequence* with higher two-phase amplitude voltages and the other of *inverse sequence* with lower two-phase amplitude voltages. Usually, this equivalence process is obtained by using an orthogonal transformation. Not only voltages but also the total fluxes and eventually the currents must be established for the two resulted systems. The quantities of the unbalanced two-phase system can be written as follows:

$$
\begin{bmatrix}
\underline{\underline{U}}\_{\alpha s} \\
\underline{\underline{U}}\_{\beta s} \\
\underline{\underline{U}}\_{\alpha s}
\end{bmatrix} = \sqrt{\frac{2}{3}} \begin{bmatrix}
1 & -1/2 & -1/2 \\
0 & \sqrt{3}/2 & -\sqrt{3}/2 \\
1/\sqrt{2} & 1/\sqrt{2} & 1/\sqrt{2}
\end{bmatrix} \underbrace{\underline{\underline{U}}\_{\alpha s}}\_{\text{LS}} \leftrightarrow \begin{cases}
\underline{\underline{U}}\_{\alpha s} = \sqrt{\frac{3}{2}} \underline{\underline{U}}\_{\alpha s}; \underline{\underline{U}}\_{\beta s} = \sqrt{\frac{3}{2}} \frac{\underline{\underline{U}}\_{\beta s} - \underline{\underline{U}}\_{\alpha s}}{\sqrt{3}} \\
\underline{\underline{U}}\_{\alpha s} = 0; \quad \underline{\underline{U}}\_{\alpha s} + \underline{\underline{U}}\_{\beta s} + \underline{\underline{U}}\_{\alpha s} = 0
\end{cases} \tag{52}
$$

Further, the unbalanced quantities are transformed to balanced quantities and we obtain:

$$
\begin{bmatrix}
\underline{\underline{\underline{L}}}\_{\alpha s(+)} \\
\underline{\underline{\underline{L}}}\_{\alpha s(-)}
\end{bmatrix} = \frac{1}{2} \begin{bmatrix} 1 & j \\ 1 & -j \end{bmatrix} \begin{bmatrix} \underline{\underline{L}}\_{\alpha s} \\ \underline{\underline{L}}\_{\beta s} \end{bmatrix}, \text{or: } \begin{cases} \underline{\underline{L}}\_{\alpha s(+)} = \left( \underline{\underline{L}}\_{\alpha s} e^{j\pi/6} + j \underline{\underline{L}}\_{\beta s} \right) / \sqrt{2}; \\\underline{\underline{L}}\_{\alpha s(-)} = \left( \underline{\underline{L}}\_{\alpha s} e^{-j\pi/6} - j \underline{\underline{L}}\_{\beta s} \right) / \sqrt{2} \end{cases} \tag{53}
$$

The quantities of the three-phase system with *single unbalance* can be written as follows:

$$
\Delta u\_{as} = \text{LI}\sqrt{2}\cos\alpha t \Leftrightarrow \underline{\text{LI}}\_{as} = \text{LI}e^{j\beta}; \underline{\text{LI}}\_{bs} = k\text{LI}e^{-j\beta}; \underline{\text{LI}}\_{cs} = -\text{LI}(1 + ke^{-j\beta})\tag{54}$$

and further:

$$
\underline{\mathbf{U}}\_{\alpha s(+)} = \mathbf{U} (e^{i\pi/6} + ke^{i(\pi/2 - \beta)}) / \sqrt{2} ; \underline{\mathbf{U}}\_{\alpha s(-)} = \mathbf{U} (e^{-i\pi/6} - ke^{i(\pi/2 - \beta)}) / \sqrt{2} \tag{55}
$$

Modulus of these components can be determined at once with:

$$\mathcal{U}\_{\alpha s(+)} = \mathcal{U}\sqrt{1 + k^2 + 2k\sin(\beta + \pi/6)} / \sqrt{2};\\ \mathcal{U}\_{\alpha s(-)} = \mathcal{U}\sqrt{1 + k^2 - 2k\sin(\beta - \pi/6)} / \sqrt{2} \tag{56}$$

For the transformation of the unbalanced two-phase quantities in balanced two-phase components (53) must be used:

$$\begin{cases} \underline{\underline{\mathbf{U}}}\_{\alpha s} = \underline{\underline{\mathbf{U}}}\_{\alpha s(+)} + \underline{\underline{\mathbf{U}}}\_{\alpha s(-)}\\ \underline{\underline{\mathbf{U}}}\_{\beta s} = -j\underline{\underline{\mathbf{U}}}\_{\alpha s(+)} + j\underline{\underline{\mathbf{U}}}\_{\alpha s(-)} \end{cases} \tag{57}$$

Mathematical Model of the Three-Phase Induction Machine

*j js*

*j js*

2 3 ( )

*A s B sC*

2 3 ( )

 

> 

*B C* and

   

(69)

 

 

> 

 

 

> 

*s r*

 

> 

(65)

(66)

(67)

for the Study of Steady-State and Transient Duty Under Balanced and Unbalanced States 27

 

> > *s*

*s r*

*s*

 2 2 3 ( ) ( ) 2 2

*As Bs C A s B s C*

(68)

*As Bs C A s B s C*

 

 

 

(63)

 

  (62)

( ) ( )

( ) () () ( ) ( ) ( ) ; 0 (2 ) *U j <sup>s</sup> ss s <sup>s</sup> xr rrs <sup>s</sup> xr j s*

( ) ( )

To determine the electromagnetic torque developed under unbalanced supply condition we use the symmetric components and the superposition effect. The *mean electromagnetic torque* 

2 2 2

3 3 2 2

2 2 2 22

The *mean resultant torque*, as a difference of the torques produced by M2D and M2I, can be

2 2 (2 ) 2 (2 ) *s s r*

2 2 ( ) ( ) 2 1 2 sin( / 6); 2 1 2 sin( / 6) *U Ukk s s U U kk*

2 2 <sup>3</sup> 1 2 sin( / 6) 1 2 sin( / 6) <sup>2</sup> 2 2 (2 ) 2 (2 )

*p U k k k k T ss*

The influence of the supply unbalances upon *Te=f(s)* characteristic are presented in Fig. 7. To this effect, let us take again into discussion the machine with the following parameters: supply voltages with the amplitude of 490 V (Uas=346.5V) and 2π/3 rad. shifted in phase; Rs=Rr=2; Lhs=0,09; Lσs= Lσr=0,01; J=0,05; p=2; kz=0,02; ω1=314,1 (SI units). The characteristic corresponding to the three-phase symmetric machine is the curve A (the motoring pull-out

3 2 22

2 2 2

 

where we have defined the notations: 2 2 2 2 ;; ; *s s <sup>A</sup> s r r tt* 

> 

Finally, the expression of the mean resultant torque with the slip is:

*p s U s U*

*p p U s*

*p p U s T j As Bs C*

(64)

;; 2 *r s <sup>s</sup> <sup>r</sup> <sup>s</sup> s xr s sr s sr*

 

() 3 () () 2 3 3 2

() 3 () () 2

 

*s xr s s r s sr*

( ) ( ) ( ) ( ) ( ) ; ; *r s s s <sup>r</sup>*

 

*M2D* results from (25) but transformed in simplified complex quantities:

2Re

Similarly, the expression of the *mean electromagnetic torque M2D* is:

2Re

*s*

 

*e s xr*

*T j*

*erez*

3

*s*

*r*

 

*T*

written by using (65) and (66):

*erez*

*e s xr*

 

( ) ( ) ( ) ( ) ( )

 

*jsU U*

 Similarly, for M2I we obtain:

  

 

2

*js U U*

The matrix equation of the two-phase model is written in a convenient way hereinafter:

$$
\begin{bmatrix}
\underline{\mathbf{U}}\_{\alpha s} \\
\underline{\mathbf{U}}\_{\beta s} \\
0 \\
0 \\
\end{bmatrix} = \begin{bmatrix}
\nu\_s + jo\_s & 0 & 0 & -\nu\_{\sigma s} \\
0 & \nu\_s + jo\_s & -\nu\_{\sigma s} & 0 \\
0 & \nu\_{\sigma r} & -(\nu\_r + jo\_s) & o\_R \\
\nu\_{\sigma r} & 0 & -o\_R & -(\nu\_r + jo\_s)
\end{bmatrix} \times \begin{bmatrix}
\underline{\mathbf{V}}\_{\alpha s} \\
\underline{\mathbf{V}}\_{\beta s} \\
\underline{\mathbf{V}}\_{\beta r} \\
\underline{\mathbf{V}}\_{\alpha r}
\end{bmatrix} \tag{58}
$$

Using elementary math (multiplications with constants, addition and subtraction of different equations) we can obtain the equations of the two-phase *direct* (M2D) and *inverse* (M2I) models:

$$\begin{array}{ll} \text{(M2D)} & \begin{bmatrix} \underline{\mathbf{U}}\_{\alpha s(+)} \\ 0 \end{bmatrix} = \begin{bmatrix} \nu\_s + jo\_s & -\nu\_{\sigma s} \\ \nu\_{\sigma r} & -\left(\nu\_r + js\_d o\_s\right) \end{bmatrix} \times \begin{bmatrix} \underline{\mathbf{u}}\_{\alpha s(+)} \\ \underline{\mathbf{u}}\_{\alpha r(+)} \end{bmatrix} \\ & & \tag{59} \end{array} \tag{59}$$

$$\begin{array}{cc} \text{(M2I)} & \begin{bmatrix} \underline{\mathbf{U}}\_{\alpha s \text{(-)}} \\ \mathbf{0} \end{bmatrix} = \begin{bmatrix} \nu\_s + jo\_s & -\nu\_{\sigma s} \\ \nu\_{\sigma r} & -\left(\nu\_r + js\_i o\_s\right) \end{bmatrix} \times \begin{bmatrix} \underline{\Psi}\_{\alpha s \text{(-)}} \\ \underline{\Psi}\_{\alpha r \text{(-)}} \end{bmatrix} \end{array} \tag{60}$$

We have defined the slip values for the direct (+) and respectively inverse (-) machines: ; *sR sR d i s s ss s* with the interrelation expression: 2 *is s* .

The two machine-models create self-contained torques, which act simultaneously upon rotor. The resultant torque emerges from superposition effects procedure (Simion et al., 2009; Simion & Livadaru, 2010). The equation set (59), for M2D, gives two equations:

$$
\underline{\mathbf{U}}\_{\rm av\text{(+)}} = (\nu\_s + j\alpha\_s)\underline{\Psi}\_{\rm av\text{(+)}} - \nu\_{\sigma s}\underline{\Psi}\_{\rm ar\text{(+)}}; \quad \mathbf{0} = \nu\_{\sigma r}\underline{\Psi}\_{\rm av\text{(+)}} - \left(\nu\_r + j\alpha\alpha\_s\right)\underline{\Psi}\_{\rm ar\text{(+)}}\tag{61}
$$

which give further

Mathematical Model of the Three-Phase Induction Machine for the Study of Steady-State and Transient Duty Under Balanced and Unbalanced States 27

$$\underline{\mathbf{V}}\_{\alpha s(+)} = \frac{\left(\nu\_r + j \text{s}o\_s\right)\underline{\mathbf{U}}\_{\alpha s(+)}}{\underline{\Delta}\_{(+)}};\\\underline{\mathbf{V}}\_{\omega r(+)} = \frac{\nu\_{\sigma r}\underline{\mathbf{U}}\_{\alpha s(+)}}{\underline{\Delta}\_{(+)}};\\\underline{\Delta}\_{(+)} = \left(\nu\_s + j o\_s\right)\left(\nu\_r + j \text{s}o\_s\right) - \nu\_{\sigma s}\nu\_{\sigma r} \tag{62}$$

Similarly, for M2I we obtain:

26 Induction Motors – Modelling and Control

components (53) must be used:

*U j*

 

; *sR sR*

 

> 

*s s*

*U j*

  

Modulus of these components can be determined at once with:

 

0 0 ()

(M2D)

(M2I)

We have defined the slip values for the direct (+) and respectively inverse (-) machines:

The two machine-models create self-contained torques, which act simultaneously upon rotor. The resultant torque emerges from superposition effects procedure (Simion et al.,

( ) () () ( ) ( ) ( ) ; 0 *U <sup>s</sup> ss s <sup>s</sup> xr r rs <sup>s</sup> xr j js*

2009; Simion & Livadaru, 2010). The equation set (59), for M2D, gives two equations:

  *U j*

*U j*

with the interrelation expression: 2 *is s* .

/6 ( /2 ) /6 ( /2 ) ( ) ( ) ( )/ 2; ( )/ 2 *j j j j U U e ke s s U U e ke*

 (55)

 

> 

() () () ()

 

 

> 

*r rs R yr*

 

*j*

 

*r R rs xr*

( ) ( ) 0 ( ) *s ss s s*

 

( ) ( ) 0 ( ) *s ss s s*

 

 

 

*r r ds xr*

*r r is xr*

 

*js*

 

(61)

   

 

  *js*

 

*j*

For the transformation of the unbalanced two-phase quantities in balanced two-phase

*ss s ss s*

*UU U U jU jU*

The matrix equation of the two-phase model is written in a convenient way hereinafter:

0 0 0 0

*s s s s s s s ss s*

 

 

Using elementary math (multiplications with constants, addition and subtraction of different equations) we can obtain the equations of the two-phase *direct* (M2D) and *inverse*

 

0 0 ()

(56)

2 2 ( ) ( ) 1 2 sin( / 6) / 2; 1 2 sin( / 6) / 2 *U Ukk s s U Ukk*

 

> 

(57)

(58)

(59)

(60)

and further:

(M2I) models:

*d i*

which give further

*ss s* 

$$
\underline{\mathbf{U}}\_{\alpha s(-)} = (\nu\_s + jo\_s)\underline{\Psi}\_{\alpha s(-)} - \nu\_{\sigma s}\underline{\Psi}\_{\alpha r(-)}; \quad \mathbf{0} = \nu\_{\sigma r}\underline{\Psi}\_{\alpha s(-)} - \left[\nu\_r + j(\mathbf{2} - s)o\_s\right]\underline{\Psi}\_{\alpha r(-)}\tag{63}
$$

$$\underline{\boldsymbol{\Psi}}\_{\text{acs(-)}} = \frac{\underline{\boldsymbol{\nu}}\_r + j(\mathbf{2} - s)\boldsymbol{\alpha}\_s \underline{\boldsymbol{\Pi}}\_{\text{acs(-)}}}{\underline{\boldsymbol{\Delta}}\_{\text{(-)}}};\\ \underline{\boldsymbol{\Psi}}\_{\text{arr(-)}} = \frac{\boldsymbol{\nu}\_{\sigma r} \underline{\boldsymbol{\underline{\Omega}}}\_{\text{acs(-)}}}{\underline{\boldsymbol{\Delta}}\_{\text{(-)}}};\\ \underline{\boldsymbol{\Delta}}\_{\text{(-)}} = \left(\boldsymbol{\nu}\_s + j\boldsymbol{\alpha}\_s\right) \left[\boldsymbol{\nu}\_r + j(\mathbf{2} - s)\boldsymbol{\alpha}\_s\right] - \boldsymbol{\nu}\_{\sigma s} \underline{\boldsymbol{\nu}}\_{\sigma r} \tag{64}$$

To determine the electromagnetic torque developed under unbalanced supply condition we use the symmetric components and the superposition effect. The *mean electromagnetic torque M2D* results from (25) but transformed in simplified complex quantities:

$$T\_{\epsilon(+)} = -\frac{3p}{2} \Lambda\_3 \cdot 2 \operatorname{Re} \left( j \underbrace{\mathcal{V}}\_{\alpha s(+)} \cdot \overleftarrow{\mathcal{W}}\_{xr(+)}^{\bullet} \right) = \frac{3p \nu\_{\sigma r} \Lambda\_3}{2\alpha \iota\_s} \cdot \frac{2 \mathcal{U}^2\_{\alpha s(+)} s}{A s^2 + 2Bs + C} \tag{65}$$

Similarly, the expression of the *mean electromagnetic torque M2D* is:

$$T\_{\epsilon(-)} = -\frac{3p}{2} \Lambda\_3 \cdot 2 \operatorname{Re} \left( j \underbrace{\underline{\boldsymbol{\nu}} \underline{\boldsymbol{\nu}}\_{\alpha s(-)} \cdot \underline{\boldsymbol{\underline{\boldsymbol{\nu}}}}\_{\alpha r(-)}^\* \right) = \frac{3p \nu\_{\sigma r} \Lambda\_3}{2 \alpha\_s} \cdot \frac{2 \boldsymbol{\mathcal{U}}\_{\alpha s(-)}^2 \left( 2 - s \right)}{A \left( 2 - s \right)^2 + 2B \left( 2 - s \right) + C} \tag{66}$$

The *mean resultant torque*, as a difference of the torques produced by M2D and M2I, can be written by using (65) and (66):

$$T\_{rev} = \frac{3p\nu\_{\sigma\tau}\Lambda\_3}{2o\_s} \left[ \frac{s \cdot 2U\_{\alpha s(+)}^2}{As^2 + 2Bs + C} - \frac{\left(2 - s\right) \cdot 2U\_{\alpha s(-)}^2}{A(2 - s)^2 + 2B(2 - s) + C} \right] \tag{67}$$

where we have defined the notations: 2 2 2 2 ;; ; *s s <sup>A</sup> s r r tt B C* and

$$\sqrt{2}\mathcal{U}\_{as(+)} = \mathcal{U}\sqrt{1 + k^2 + 2k\sin(\beta + \pi/6)};\\\sqrt{2}\mathcal{U}\_{as(-)} = \mathcal{U}\sqrt{1 + k^2 - 2k\sin(\beta - \pi/6)}\tag{68}$$

Finally, the expression of the mean resultant torque with the slip is:

$$T\_{err} = \frac{3\mathcal{p}\nu\_{\sigma\tau}\Lambda\_3\mathcal{U}^2}{2o\_s} \left[ \frac{1+k^2+2k\sin(\beta+\pi/6)}{As^2+2Bs+C}s-\frac{1+k^2-2k\sin(\beta-\pi/6)}{A(2-s)^2+2B(2-s)+C}(2-s) \right] \tag{69}$$

The influence of the supply unbalances upon *Te=f(s)* characteristic are presented in Fig. 7. To this effect, let us take again into discussion the machine with the following parameters: supply voltages with the amplitude of 490 V (Uas=346.5V) and 2π/3 rad. shifted in phase; Rs=Rr=2; Lhs=0,09; Lσs= Lσr=0,01; J=0,05; p=2; kz=0,02; ω1=314,1 (SI units). The characteristic corresponding to the three-phase symmetric machine is the curve A (the motoring pull-out

torque is equal to 124 Nm and obviously *Uas(-)* = 0). If the voltage on phase *b* keeps the same amplitude as the voltage in phase *a*, for example, but the angle of phase difference changes with *π/24=7,5* degrees (from *2π/3=16π/24* to *17π/24* rad.) then the new characteristic is the B curve. The pull-out torque value decreases with approx. 12% but the pull-out slip keeps its value. Other two unbalance degrees are presented in Fig. 7 as well.

Mathematical Model of the Three-Phase Induction Machine

(72)

for the Study of Steady-State and Transient Duty Under Balanced and Unbalanced States 29

(71)

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

*As Bs C A s B s C*

which is replaced by the *direct* and *inverse* symmetric systems. The mean resultant torque is the difference between the torques developed by the two symmetric machinemodels. Taking into consideration their slip values (*sd = s* and *si = 2-s*) we can deduce the

3 / 2 [3Re <sup>3</sup> 1 1 3Re 2 2 ] *T pj j erez as ar as ar*

*p sU s U*

2 2 (2 ) 2 (2 ) *r as as*

The mathematical model described by the equation system (26-1…8) allows a complete simulation study of the operation of the three-phase induction machine, which include startup, any sudden change of the load and braking to stop eventually. To this end, the machine parameters (resistances, main and leakage phase inductances, moments of inertia corresponding to the rotor and the load, coefficients that characterize the variable speed and torque, etc.) have to be calculated or experimentally deduced. At the same time, the values of the load torque and the expressions of the instantaneous voltages applied to each stator phase winding are known, as well. The rotor winding is considered short-circuited. Using the above mentioned equation system, the structural diagram in the Matlab-Simulink environment can be carried out. Additionally, for a complete evaluation, virtual oscillographs for the visualization of the main physical parameters such as voltage, current, magnetic flux, torque, speed, rotation angle and current or specific characteristics (mechanical characteristic, angular characteristic or flux hodographs) fill out the structural

The study of the *symmetric three-phase* condition in the Matlab-Simulink environment takes into consideration the following parameter values: three identical supply voltages with the amplitude of 490 V (Uas=346.5V) and 2π/3 rad. shifted in phase; uar=ubr=ucr=0 since the rotor winding is short-circuited; Rs=Rr=2; Lhs=0,09; Lσs= Lσr=0,01; J=0,05; p=2; kz=0,02; ω1=314,1 (SI

135,71 32,14 *as as bs cs* 32,14 2 *ar br cr* cos 55,67( )sin *R R cr br s u*

135,71 *bs bs* 32,14 *cs as* 32,14 2 *br cr ar* cos 55,67( )sin *R R ar cr s u*

 

 

 

   

 

3 3 3 2

**6. Simulation study upon some transient duties of the three-phase** 

torque expression:

**induction machine** 

diagram.

**6.1. Symmetric supply system** 

units). The equation system becomes:

 *erez*

and this is the same with (69) as we expected.

*T*

*s*

 

**Figure 7.** Te=f(s) characteristic for different unbalance degrees

Usually, the *unbalance degree* of the supply voltage is defined as the ratio of inverse and direct components:

$$u\_n = \frac{\mathcal{U}\_{\alpha s(-)}}{\mathcal{U}\_{\alpha s(+)}} = \frac{\sqrt{1 + k^2 - 2k\sin(\beta - \pi/6)}}{\sqrt{1 + k^2 + 2k\sin(\beta + \pi/6)}} \cdot 100 \,\mathrm{[\,\%]}\tag{70}$$

The curves A, B, C, and D from Fig. 7 correspond to the following values of the unbalance degree: *un*= 0; 8%; 16% and 27%. The highest unbalance degree (27% - curve D) causes a decrease of the pull-out torque by 40%.

b. The second approach takes into consideration the following reasoning. When the amplitudes of the three-phase supply system and/or the angles of the phase difference are not equal to 2π/3 then the *unbalanced system* can be replaced by two *balanced threephase* systems that act in opposition. One of them is the *direct sequence* system and has higher voltages and the other is the *inverse sequence* system and has lower voltages. A transformation of the unbalanced voltages and total fluxes into two symmetric systems is again necessary. In other words, there is an unbalanced voltage system (*Uas, Ubs, Ucs*), which is replaced by the *direct* and *inverse* symmetric systems. The mean resultant torque is the difference between the torques developed by the two symmetric machinemodels. Taking into consideration their slip values (*sd = s* and *si = 2-s*) we can deduce the torque expression:

$$T\_{errz} = -\left(\Im / 2\right) p \Lambda\_3 \cdot \left[ 3 \text{Re}\left( j \underline{\Psi}\_{as1} \underline{\Psi}\_{ar1}^{\prime \ast} \right) - 3 \text{Re}\left( j \underline{\Psi}\_{as2} \underline{\Psi}\_{ar2}^{\prime \ast} \right) \right] \tag{71}$$

$$T\_{rev} = \frac{3p\nu\_{\sigma\tau}\Lambda\_3}{2o\_s} \left[ \frac{3s\mathcal{U}\_{as1}^2}{As^2 + 2Bs + C} - \frac{3(2-s)\mathcal{U}\_{as2}^2}{A(2-s)^2 + 2B(2-s) + C} \right] \tag{72}$$

and this is the same with (69) as we expected.
