**3. Mathematical models used for the study of steady-state under balanced and unbalanced conditions**

Generally, the symmetrical three-phase squirrel cage induction machine has the stator windings connected to a supply system, which provides variable voltages according to certain laws but have the same pulsation. Practically, this is the case with 4 wires connection, 3 phases and the neutral. The sum of the phase currents gives the current along neutral and the homopolar component can be immediately defined. The analysis of such a machine can use the symmetric components theory. This is the case of the machine with *two unbalances* as concerns the supply. The study can be done either using 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 and homopolar).

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 superposition of the effects.

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

12 Induction Motors – Modelling and Control

simple or double feeding.

*s u*

*s u*

*s u*

*s u*

*s k J pJ p*

**balanced and unbalanced conditions**

*LD LD LD*

*LD LD LD*

*LD LD LD*

*L R LLR LL L R*

 

 

*L R LLR LL L R*

 

*LD LD LD*

*LR LL R LL L R*

 

> 

*L R LLR LL L R*

 

2

2

2

2

*R z R as ar br cr*

 

*R*

*d dt* 

**3. Mathematical models used for the study of steady-state under** 

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

 

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

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

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

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

 

 

*R R*

 

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

Generally, the symmetrical three-phase squirrel cage induction machine has the stator windings connected to a supply system, which provides variable voltages according to certain laws but have the same pulsation. Practically, this is the case with 4 wires connection, 3 phases and the neutral. The sum of the phase currents gives the current along neutral and the homopolar component can be immediately defined. The analysis of such a machine can use the symmetric components theory. This is the case of the machine with *two unbalances* as concerns the supply. The study can be done either using

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

 

2 2 3 cos

*bs br cr ar cs cr ar br R*

 

2 cos 3 sin

 

*as bs cs R R bs cs*

2 cos 3 sin

*as bs cs R R cs as*

*as bs cs R R as bs*

 

 

 

2 cos 3 sin

 

> 

 

 

 

 

 

 

> 

2 cos 3 sin

 

 

> 

 

*T*

(26-8)

 

(26-3)

(26-4)

(26-5)

(26-6)

(26-7)

*ar br cr R R br ar*

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 frames in accordance with the following procedure :

$$
\begin{bmatrix}
\boldsymbol{\nu}\_{\alpha s} \\
\boldsymbol{\nu}\_{\beta s} \\
\boldsymbol{\nu}\_{\alpha s}
\end{bmatrix} = \sqrt{\frac{2}{3}} \cdot \begin{bmatrix}
1 & -1/2 & -1/2 \\
0 & \sqrt{3}/2 & -\sqrt{3}/2 \\
\sqrt{2}/2 & \sqrt{2}/2 & \sqrt{2}/2
\end{bmatrix} \cdot \begin{bmatrix}
\boldsymbol{\nu}\_{as} \\
\boldsymbol{\nu}\_{bs} \\
\boldsymbol{\nu}\_{cs}
\end{bmatrix} \tag{27}
$$

We define the following notations:

$$\frac{\Pi L\_{\rm s\sigma} R\_{\rm s}}{\langle \Pi L \boldsymbol{D} \rangle} = \frac{\left(L\_{\rm hs} L\_{\sigma\tau} + 3L\_{\rm hs} L\_{\sigma\sigma} + 2L\_{\sigma\tau} L\_{\sigma\sigma}\right)}{\left(3L\_{\rm hs} L\_{\sigma\tau} + 3L\_{\rm hs} L\_{\sigma\sigma} + 2L\_{\sigma\tau} L\_{\sigma\sigma}\right)} \left(\frac{R\_{\rm s}}{L\_{\sigma\sigma}}\right) \equiv \frac{2}{3} \left(\frac{R\_{\rm s}}{L\_{\sigma\sigma}}\right) = \nu\_{\rm st};$$

$$\frac{L\_{\rm hs} L\_{\sigma\tau}^2 R\_{\rm s}}{\langle \Pi L \boldsymbol{D} \rangle} = \frac{L\_{\rm hs} L\_{\sigma\tau} R\_{\rm s}}{\left(3L\_{\rm hs} L\_{\sigma\tau} + 3L\_{\rm hs} L\_{\sigma\tau} + 2L\_{\sigma\tau} L\_{\sigma\sigma}\right) L\_{\sigma\sigma}} \equiv \frac{1}{6} \left(\frac{R\_{\rm s}}{L\_{\sigma\sigma}}\right) = \nu\_{\rm s\tau};\tag{28-1}$$

$$\nu\_{\rm st} - \nu\_{\rm sr} = \frac{3 + 2\left(L\_{\sigma\tau} \mid L\_{\rm hs}\right)}{3\left(L\_{\sigma\tau} \mid L\_{\sigma\sigma}\right) + 3 + 2\left(L\_{\sigma\tau} \mid L\_{\rm s\sigma}\right)} \left(\frac{R\_{\rm s}}{L\_{\sigma\sigma}}\right) \equiv \frac{1}{2} \left(\frac{R\_{\rm s}}{L\_{\sigma\sigma}}\right) = \nu\_{\rm s}$$

$$\frac{\Pi L\_{r\sigma}R\_r}{\langle \Pi L \boldsymbol{D} \rangle} = \frac{\left(L\_{\rm hs}L\_{\sigma\sigma} + 3L\_{\rm hs}L\_{\sigma\tau} + 2L\_{\sigma\tau}L\_{\sigma\sigma}\right)}{\left(3L\_{\rm hs}L\_{\sigma\tau} + 3L\_{\rm hs}L\_{\sigma\sigma} + 2L\_{\sigma\tau}L\_{\sigma\sigma}\right)} \left(\frac{R\_r}{L\_{\sigma\tau}}\right) \equiv \frac{2}{3} \left(\frac{R\_r}{L\_{\sigma\tau}}\right) = \nu\_{rt};$$

$$\frac{L\_{\rm hs}L\_{\sigma\sigma}^2 R\_r}{\left(\Pi L\boldsymbol{D}\right)} = \frac{L\_{\rm hs}L\_{\sigma\sigma}R\_r}{\left(3L\_{\rm hs}L\_{\sigma\sigma} + 3L\_{\rm hs}L\_{\sigma\tau} + 2L\_{\sigma\tau}L\_{\sigma\sigma}\right)L\_{\sigma\tau}} \equiv \frac{1}{6} \left(\frac{R\_r}{L\_{\sigma\tau}}\right) = \nu\_{rs};\tag{28-2}$$

$$\nu\_{rt} - \nu\_{rs} = \frac{3 + 2\left(L\_{\sigma\sigma} \neq L\_{\rm hs}\right)}{3\left(L\_{\sigma\sigma} \neq L\_{\sigma\tau}\right) + 3 + 2\left(L\_{\sigma\sigma} \neq L\_{\rm hs}\right)} \left(\frac{R\_r}{L\_{\sigma\tau}}\right) \equiv \frac{1}{2} \left(\frac{R\_r}{L\_{\sigma\tau}}\right) = \nu\_r$$

$$\frac{\Im L\_{\rm hs} L\_{\sigma s} L\_{\sigma r} R\_s}{\{\Im L D\}} = \frac{\Im L\_{\sigma s}}{\left(\Im L\_{\sigma s} + \Im L\_{\sigma r} + 2L\_{\sigma r} L\_{\sigma s} / L\_{\rm hs}\right) \left(\frac{R\_s}{L\_{\sigma s}}\right)} \stackrel{\scriptstyle \triangleq}{=} \frac{1}{2} \left(\frac{R\_s}{L\_{\sigma s}}\right) = \nu\_{\sigma s};\tag{28-3}$$

Mathematical Model of the Three-Phase Induction Machine

 

> 

 

 

> 

 

*s u* (32-3)

 

> 

*s u* (32-6)

 

 

*r r* cos sin (32-1)

*r r* sin cos (32-2)

*s s* cos sin (32-4)

*s s* sin cos (32-5)

 

(32-8)

 

 

 

> 

> >

 

  (30-1, 2, 3)

(30-4, 5, 6)

for the Study of Steady-State and Transient Duty Under Balanced and Unbalanced States 15

 

 

cos sin

 

 

 

> 

cos sin

sin cos

 

 

sin cos

<sup>0</sup>

 

<sup>0</sup>

 

Ultimately, the 8 equation system under operational form is:

 

 

 

 

*R z s k J pJ p* / 3/2 <sup>3</sup>

2

 

*rt rs r r*

*s*

 

 

*dt d*

*dt d*

*dt*

*d*

 

 

*d*

 

 

convenient transformations:

*s*

*s*

*r*

*dt d*

*dt d*

*dt*

*r*

*r*

2

 

*u*

 

*u*

 

*u*

 

 

 

*u*

 

*st sr s s*

 

 

0 0

0 0

*u*

Further, the movement equation has to be attached. It is necessary to establish the detailed expression of the electromagnetic torque in *fluxes* alone starting with (25) and using

3/2 <sup>3</sup> sin cos *<sup>e</sup> sr sr R sr sr R T p*

 

*s sR R*

*s sR R*

*r rR R*

*r rRR*

*sr sr* cos *R st <sup>T</sup>* (32-7)

*R R*

<sup>0</sup> <sup>0</sup> 2 *<sup>s</sup> <sup>s</sup> st sr*

 

<sup>0</sup> <sup>0</sup> 2 *<sup>r</sup> <sup>r</sup> rt rs*

 

 *sr sr* sin *<sup>R</sup>*

> 

(31)

 

 

 

 

> 

 

*<sup>s</sup> s u*

*<sup>s</sup> s u*

 

*<sup>r</sup> s u*

*<sup>r</sup> s u*

 

 

 

 

 

 

*s* 

*s* 

*r* 

*r* 

*d dt* 

*R*

*u*

*ss s s r R r R*

*ss s s r R r R*

*rr r r s R s R*

*rr r r s R s R*

 

 

$$\frac{\Im \mathcal{L}\_{\text{lis}} L\_{\sigma s} L\_{\sigma r} R\_r}{\text{(IILD)}} = \frac{\Im L\_{\sigma r}}{\left(\Im L\_{\sigma s} + \Im L\_{\sigma r} + 2L\_{\sigma r} L\_{\sigma s} / L\_{\text{lis}}\right) \left(\frac{R\_r}{L\_{\sigma r}}\right) \equiv \frac{1}{2} \left(\frac{R\_r}{L\_{\sigma r}}\right) = \nu\_{\sigma r};\tag{28-4}$$

By using these notations in (17) and after convenient groupings we obtain:

$$\begin{split} \frac{d\boldsymbol{\nu}\_{\rm as}}{dt} + \boldsymbol{\nu}\_{\rm st} \boldsymbol{\nu}\_{\rm as} &= \boldsymbol{u}\_{\rm as} - \boldsymbol{\nu}\_{\rm sr} \left( \boldsymbol{\nu}\_{\rm bs} + \boldsymbol{\nu}\_{\rm cs} \right) + \frac{1}{3} \boldsymbol{\nu}\_{\rm crs} \times \\ &\times \Big[ \Big( 2\boldsymbol{\nu}\_{\rm ar} - \boldsymbol{\nu}\_{\rm br} - \boldsymbol{\nu}\_{\rm cr} \Big) \cos \theta\_{\rm R} + \sqrt{3} \Big( \boldsymbol{\nu}\_{\rm cr} - \boldsymbol{\nu}\_{\rm br} \Big) \sin \theta\_{\rm R} \Big] \end{split} \tag{29-1}$$

$$\begin{split} \frac{d\boldsymbol{\nu}\_{\rm bs}}{dt} + \boldsymbol{\nu}\_{\rm st} \boldsymbol{\nu}\_{\rm bs} &= \boldsymbol{u}\_{\rm bs} - \boldsymbol{\nu}\_{\rm sr} \left( \boldsymbol{\nu}\_{\rm cs} + \boldsymbol{\nu}\_{\rm as} \right) + \frac{1}{3} \boldsymbol{\nu}\_{\rm cs} \times \\ &\times \Big[ \Big( -\boldsymbol{\nu}\_{\rm ar} + 2\boldsymbol{\nu}\_{\rm br} - \boldsymbol{\nu}\_{\rm cr} \Big) \cos \theta\_{\rm R} + \sqrt{3} \left( \boldsymbol{\nu}\_{\rm ar} - \boldsymbol{\nu}\_{\rm cr} \right) \sin \theta\_{\rm R} \Big] \end{split} \tag{29-2}$$

$$\begin{split} \frac{d\boldsymbol{\nu}\_{\rm cs}}{dt} + \boldsymbol{\nu}\_{\rm st} \boldsymbol{\nu}\_{\rm cs} &= \boldsymbol{u}\_{\rm cs} - \boldsymbol{\nu}\_{\rm sr} \left( \boldsymbol{\nu}\_{\rm as} + \boldsymbol{\nu}\_{\rm bs} \right) + \frac{1}{3} \boldsymbol{\nu}\_{\rm cs} \times \\ &\times \Big[ \Big( -\boldsymbol{\nu}\_{\rm ar} - \boldsymbol{\nu}\_{\rm br} + 2\boldsymbol{\nu}\_{\rm cr} \Big) \cos \theta\_{\rm R} + \sqrt{3} \Big( \boldsymbol{\nu}\_{\rm br} - \boldsymbol{\nu}\_{\rm ar} \Big) \sin \theta\_{\rm R} \Big] \end{split} \tag{29-3}$$

$$\begin{split} \frac{d\boldsymbol{\nu}\_{ar}}{dt} + \boldsymbol{\nu}\_{rl}\boldsymbol{\nu}\_{ar} &= \boldsymbol{u}\_{ar} - \boldsymbol{\nu}\_{rs} \left(\boldsymbol{\nu}\_{br} + \boldsymbol{\nu}\_{cr}\right) + \frac{1}{3} \boldsymbol{\nu}\_{cr} \times \\ &\times \left[ \left(2\boldsymbol{\nu}\_{as} - \boldsymbol{\nu}\_{bs} - \boldsymbol{\nu}\_{cs}\right) \cos\theta\_{\mathbb{R}} + \sqrt{3} \left(\boldsymbol{\nu}\_{bs} - \boldsymbol{\nu}\_{cs}\right) \sin\theta\_{\mathbb{R}} \right] \end{split} \tag{29-4}$$

$$\begin{split} \frac{d\boldsymbol{\upmu}\_{br}}{dt} + \boldsymbol{\upnu}\_{r} \boldsymbol{\upmu}\_{br} &= \boldsymbol{\upmu}\_{br} - \boldsymbol{\upnu}\_{rs} \left( \boldsymbol{\upnu}\_{cr} + \boldsymbol{\upnu}\_{ar} \right) + \frac{1}{3} \boldsymbol{\upnu}\_{cr} \times \\ &\times \left[ \left( -\boldsymbol{\upnu}\_{as} + 2\boldsymbol{\upnu}\_{bs} - \boldsymbol{\upnu}\_{cs} \right) \cos \boldsymbol{\uptheta}\_{\mathbb{R}} + \sqrt{3} \left( \boldsymbol{\upnu}\_{cs} - \boldsymbol{\upnu}\_{as} \right) \sin \boldsymbol{\uptheta}\_{\mathbb{R}} \right] \end{split} \tag{29-5}$$

$$\begin{split} \frac{d\boldsymbol{\nu}\_{cr}}{dt} + \boldsymbol{\nu}\_{rl}\boldsymbol{\nu}\_{cr} &= \boldsymbol{\nu}\_{cr} - \boldsymbol{\nu}\_{rs} \left(\boldsymbol{\nu}\_{ar} + \boldsymbol{\nu}\_{br}\right) + \frac{1}{3}\boldsymbol{\nu}\_{cr} \times \\ &\times \left[ \left( -\boldsymbol{\nu}\_{bs} + 2\boldsymbol{\nu}\_{cs} - \boldsymbol{\nu}\_{as} \right) \cos\theta\_{\rm R} + \sqrt{3} \left( \boldsymbol{\nu}\_{as} - \boldsymbol{\nu}\_{bs} \right) \sin\theta\_{\rm R} \right] \end{split} \tag{29-6}$$

Typical for the cage machine or even for the wound rotor after the starting rheostat is shortcircuited is the fact that the *rotor voltages become zero*. The equations of the six circuits get different as a result of certain convenient math operations. (29-2) and (29-3) are multiplied by (-1/2) and afterwards added to (29-1); (29-3) is subtracted from (29-2); (29-1), (29-2) and (29-3) are added together. We obtain three equations that describe the stator. Similarly, (29- 4), (29-5) and (29-6) are used for the rotor equations. The new equation system is:

Mathematical Model of the Three-Phase Induction Machine for the Study of Steady-State and Transient Duty Under Balanced and Unbalanced States 15

$$\begin{cases} \frac{d\boldsymbol{\nu}\_{\rm{as}}}{dt} + \boldsymbol{\nu}\_{s}\boldsymbol{\nu}\_{\rm{as}} = \boldsymbol{u}\_{\rm{as}} + \boldsymbol{\nu}\_{\rm{es}} \left(\boldsymbol{\nu}\_{\rm{ar}}\cos\theta\_{\rm{R}} - \boldsymbol{\nu}\_{\boldsymbol{\beta}r}\sin\theta\_{\rm{R}}\right) \\ \frac{d\boldsymbol{\nu}\_{\boldsymbol{\beta}s}}{dt} + \boldsymbol{\nu}\_{s}\boldsymbol{\nu}\_{\boldsymbol{\beta}s} = \boldsymbol{u}\_{\boldsymbol{\beta}s} + \boldsymbol{\nu}\_{\boldsymbol{\sigma}s} \left(\boldsymbol{\nu}\_{\rm{ar}}\sin\theta\_{\rm{R}} + \boldsymbol{\nu}\_{\boldsymbol{\beta}r}\cos\theta\_{\rm{R}}\right) \\ \frac{d\boldsymbol{\nu}\_{\rm{os}}}{dt} + \left(\boldsymbol{\nu}\_{\rm{st}} + \boldsymbol{2}\boldsymbol{\nu}\_{\boldsymbol{\sigma}r}\right)\boldsymbol{\nu}\_{\rm{os}} = \boldsymbol{u}\_{\rm{os}} \end{cases} \tag{30-1,2,3}$$

$$\begin{cases} \frac{d\boldsymbol{\varmu}\_{\alpha r}}{dt} + \boldsymbol{\varmu}\_{r} \boldsymbol{\upmu}\_{\alpha r} = \boldsymbol{u}\_{\alpha r} + \boldsymbol{\upnu}\_{\sigma r} \left( \boldsymbol{\upnu}\_{\alpha s} \cos \theta\_{\mathbb{R}} + \boldsymbol{\upnu}\_{\boldsymbol{\upbeta}s} \sin \theta\_{\mathbb{R}} \right) \\ \frac{d\boldsymbol{\upmu}\_{\boldsymbol{\upbeta}r}}{dt} + \boldsymbol{\upnu}\_{r} \boldsymbol{\upmu}\_{\boldsymbol{\upbeta}r} = \boldsymbol{u}\_{\boldsymbol{\upbeta}r} + \boldsymbol{\upnu}\_{\sigma r} \left( -\boldsymbol{\upnu}\_{\alpha s} \sin \theta\_{\mathbb{R}} + \boldsymbol{\upnu}\_{\boldsymbol{\upbeta}s} \cos \theta\_{\mathbb{R}} \right) \\ \frac{d\boldsymbol{\upnu}\_{0r}}{dt} + \left( \boldsymbol{\upnu}\_{rt} + 2\boldsymbol{\upnu}\_{rs} \right) \boldsymbol{\upnu}\_{0r} = \boldsymbol{u}\_{0r} \end{cases} \tag{30-4,5,6}$$

Further, the movement equation has to be attached. It is necessary to establish the detailed expression of the electromagnetic torque in *fluxes* alone starting with (25) and using convenient transformations:

$$T\_e = -\left(3/2\right) p\Lambda\_3 \left[ \left( \left\nu\_{as}\nu\_{ar} + \nu\_{\beta s}\nu\_{\beta r} \right) \sin\theta\_R + \left( \nu\_{as}\nu\_{\beta r} - \nu\_{\beta s}\nu\_{ar} \right) \cos\theta\_R \right] \tag{31}$$

Ultimately, the 8 equation system under operational form is:

14 Induction Motors – Modelling and Control

 

 

*as*

*dt*

*bs*

*cs*

*ar*

*dt*

*br*

*cr*

*dt*

*d*

*dt*

*d*

*d*

*dt*

*d*

*dt*

*d*

*d*

 3 3 1

 3 3 1

 

4), (29-5) and (29-6) are used for the rotor equations. The new equation system is:

Typical for the cage machine or even for the wound rotor after the starting rheostat is shortcircuited is the fact that the *rotor voltages become zero*. The equations of the six circuits get different as a result of certain convenient math operations. (29-2) and (29-3) are multiplied by (-1/2) and afterwards added to (29-1); (29-3) is subtracted from (29-2); (29-1), (29-2) and (29-3) are added together. We obtain three equations that describe the stator. Similarly, (29-

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

 

 

*st as as sr bs cs s*

*st bs bs sr cs as s*

*st cs cs sr as bs s*

*rt ar ar rs br cr r*

*rt br br rs cr ar r*

*rt cr cr rs ar br r*

By using these notations in (17) and after convenient groupings we obtain:

 

> 

> 

> 

> 

> 

*u*

*u*

*u*

*u*

*u*

*u*

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

; ( ) 332 / <sup>2</sup> *hs s r s s s s*

; ( ) 332 / <sup>2</sup> *hs s r r r r r*

 

 

*s r r s hs s s*

*s r r s hs r r*

1 3

 

1 3

 

 

 

 

1 3 2 cos 3 sin

 

1 3

 

1 3

 

1 3 2 cos 3 sin

 

 

 

 

(28-3)

(28-4)

*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*

*as bs cs R bs cs R*

2 cos 3 sin

2 cos 3 sin

*bs cs as R as bs R*

*as bs cs R cs as R*

 

 

2 cos 3 sin

 

 

 

  *s*

*r*

(29-1)

(29-2)

(29-3)

(29-4)

(29-5)

(29-6)

 

> 

 

 

> 

   

 

$$
\overline{\boldsymbol{\nu}}\_{\alpha s} \left( \overline{\boldsymbol{s}} + \boldsymbol{\nu}\_s \right) = \overline{\boldsymbol{\mu}}\_{\alpha s} + \boldsymbol{\nu}\_{\sigma s} \left( \overline{\boldsymbol{\nu}}\_{\alpha r} \cos \theta\_R - \overline{\boldsymbol{\nu}}\_{\beta r} \sin \theta\_R \right) \tag{32-1}
$$

$$
\overline{\boldsymbol{\nu}}\_{\beta\ast} \left( \overline{\boldsymbol{s}} + \boldsymbol{\nu}\_{\boldsymbol{s}} \right) = \overline{\boldsymbol{\mu}}\_{\beta\ast} + \boldsymbol{\nu}\_{\sigma\ast} \left( \overline{\boldsymbol{\nu}}\_{ar} \sin \theta\_{\mathbb{R}} + \overline{\boldsymbol{\nu}}\_{\beta\boldsymbol{r}} \cos \theta\_{\mathbb{R}} \right) \tag{32-2}
$$

$$
\overline{\boldsymbol{\nu}}\_{0s} \left( \overline{\boldsymbol{s}} + \boldsymbol{\nu}\_{st} + \boldsymbol{\mathcal{D}} \boldsymbol{\nu}\_{sr} \right) = \overline{\boldsymbol{\nu}}\_{0s} \tag{32-3}
$$

$$
\overline{\boldsymbol{\nu}}\_{ar} \left( \overline{\boldsymbol{s}} + \boldsymbol{\nu}\_r \right) = \overline{\boldsymbol{u}}\_{ar} + \boldsymbol{\nu}\_{\sigma r} \left( \overline{\boldsymbol{\nu}}\_{as} \cos \theta\_{\mathbb{R}} + \overline{\boldsymbol{\nu}}\_{\rho s} \sin \theta\_{\mathbb{R}} \right) \tag{32-4}
$$

$$
\overline{\boldsymbol{\Psi}}\_{\beta r} \left( \overline{\mathbf{s}} + \boldsymbol{\nu}\_r \right) = \overline{\boldsymbol{\mu}}\_{\beta r} + \boldsymbol{\nu}\_{\sigma r} \left( -\overline{\boldsymbol{\nu}}\_{\alpha s} \sin \theta\_{\mathbb{R}} + \overline{\boldsymbol{\nu}}\_{\beta s} \cos \theta\_{\mathbb{R}} \right) \tag{32-5}
$$

$$
\overline{\boldsymbol{\nu}}\_{0r} \left( \overline{\mathbf{s}} + \boldsymbol{\nu}\_{rt} + \mathfrak{D} \boldsymbol{\nu}\_{rs} \right) = \overline{\boldsymbol{\mu}}\_{0r} \tag{32-6}
$$

$$
\partial\_R \left( \overline{s} + k\_z \mid I \right) = \begin{pmatrix} p/I \end{pmatrix} \cdot \begin{pmatrix} -\left( 3 \ / \ 2 \end{pmatrix} p \Lambda\_3 \left[ \left( \overline{\nu}\_{as} \overline{\nu}\_{ar} + \overline{\nu}\_{\beta s} \overline{\nu}\_{\beta r} \right) \sin \theta\_R + \left( \overline{\nu}\_{as} \overline{\nu}\_{\beta r} - \overline{\nu}\_{\beta s} \overline{\nu}\_{ar} \right) \cos \theta\_R \right] - T\_{st} \end{pmatrix} \tag{32-7}
$$

$$\frac{d\theta\_R}{dt} = \dot{\theta}\_R = \alpha\_R \tag{32-8}$$

These equations allow the study of three-phase induction machine for any duty. It has to be mentioned that the electromagnetic torque expression has no homopolar components of the total fluxes.

### **3.2. The abc-dq model in total fluxes**

For the study of the single unbalance condition is necessary to consider expressions of the instantaneous values of the stator and rotor quantities (voltages, total fluxes and eventually currents in *a,b,c* reference frame) whose sum is null. The real quantities can be transformed to (d,q) reference frame (Simion et al., 2011). By using the notations (28-1), (28-2), (28-3) and (28-4) then after convenient grouping we obtain (Simion, 2010):

$$\begin{aligned} \frac{d\boldsymbol{\nu}\_{\alpha s}}{dt} + \boldsymbol{\nu}\_{s}\boldsymbol{\nu}\_{\alpha s} &= \boldsymbol{u}\_{\alpha s} + \boldsymbol{\nu}\_{\sigma s} \left(\boldsymbol{\nu}\_{\alpha r} \cos \theta\_{R} - \boldsymbol{\nu}\_{\beta r} \sin \theta\_{R}\right) \\\\ \frac{d\boldsymbol{\nu}\_{\beta s}}{dt} + \boldsymbol{\nu}\_{s}\boldsymbol{\nu}\_{\beta s} &= \boldsymbol{u}\_{\beta s} + \boldsymbol{\nu}\_{\sigma s} \left(\boldsymbol{\nu}\_{\alpha r} \sin \theta\_{R} + \boldsymbol{\nu}\_{\beta r} \cos \theta\_{R}\right) \\\\ \frac{d\boldsymbol{\nu}\_{\alpha r}}{dt} + \boldsymbol{\nu}\_{r}\boldsymbol{\nu}\_{\alpha r} &= \boldsymbol{\nu}\_{\sigma r} \left(\boldsymbol{\nu}\_{\alpha s} \cos \theta\_{R} + \boldsymbol{\nu}\_{\beta s} \sin \theta\_{R}\right) \\\\ \frac{d\boldsymbol{\nu}\_{\beta r}}{dt} + \boldsymbol{\nu}\_{r}\boldsymbol{\nu}\_{\beta r} &= \boldsymbol{\nu}\_{\sigma r} \left(-\boldsymbol{\nu}\_{\alpha s} \sin \theta\_{R} + \boldsymbol{\nu}\_{\beta s} \cos \theta\_{R}\right) \end{aligned} \tag{33-3,4}$$

Mathematical Model of the Three-Phase Induction Machine

for the Study of Steady-State and Transient Duty Under Balanced and Unbalanced States 17

(*two-phase* mathematical model). Its parameters can be deduced by linear transformations of

**Figure 2.** Induction machine schematic view: a.Two-phase model; b. Simplified view of the total fluxes

The windings of two-phase model are denoted with (*αs, βs*) and (*αr, βr*) in order to trace a correspondence with the real two-phase machine, whose subscripts are (*as, bs*) and (*ar, br*) respectively. We shall use the subscripts *xs* and *ys* for the quantities that corresponds to the three-phase machine but transformed in its two-phase model. This is a rightful assumption since (*αs, βs*) axes are collinear with (*x, y*) axes, which are commonly used in analytic geometry. Further, new notations (35) for the flux linkages of the right member of the



> *xr r R r R yr r R r R XS s R s R YS s R s R*

Some aspects have to be pointed out. When the machine operates under motoring duty, the pulsation of the stator flux linkages from (*αs, βs*) axes is equal to *ωs.* Since the rotational pulsation is *ωR* then the pulsation of the rotor quantities from (*αr, βr*) axes is equal to *ωr=sωs=ωs –ωR*. The pulsation of the rotor quantities projected along the stator axes with the subscripts *xr* and *yr* is equal to *ωs*. The pulsation of the stator quantities projected along the rotor axes with the subscripts *XS* and *YS* is equal to *ωr*. The equations (33-1...4) become:

 

 

cos sin , sin cos

 

 

 

   

> 

(35)

the original parameters including the supply voltages (Fig. 2a).

in stator reference frame; c. Idem, but in rotor reference frame

equations (33-1...4) will be defined by following the next rules:

from the right member of the first two equations, Fig. 2b.

cos sin , sin cos

 

> 

from the last two equations, Fig. 2c.

 

> 

 

 

Further, the movement equation (31) must be attached. The operational form of the equation system (4 electric circuits and 2 movement equations) is:

$$
\overline{\boldsymbol{\nu}}\_{\alpha s} \left( \overline{\boldsymbol{s}} + \boldsymbol{\nu}\_s \right) = \overline{\boldsymbol{\mu}}\_{\alpha s} + \boldsymbol{\nu}\_{\sigma s} \left( \overline{\boldsymbol{\nu}}\_{\alpha r} \cos \theta\_{\mathbb{R}} - \overline{\boldsymbol{\nu}}\_{\beta r} \sin \theta\_{\mathbb{R}} \right) \tag{34-1}
$$

$$
\overline{\boldsymbol{\nu}}\_{\beta\circ} \left( \overline{\boldsymbol{s}} + \boldsymbol{\nu}\_{\boldsymbol{s}} \right) = \overline{\boldsymbol{\mu}}\_{\beta\circ} + \boldsymbol{\nu}\_{\sigma\circ} \left( \overline{\boldsymbol{\nu}}\_{\alpha r} \sin \theta\_{\boldsymbol{R}} + \overline{\boldsymbol{\nu}}\_{\beta r} \cos \theta\_{\boldsymbol{R}} \right) \tag{34-2}
$$

$$
\overline{\boldsymbol{\nu}}\_{\alpha r} \left( \overline{\mathbf{s}} + \boldsymbol{\nu}\_r \right) = \boldsymbol{\nu}\_{\sigma r} \left( \overline{\boldsymbol{\nu}}\_{\alpha s} \cos \theta\_{\mathbb{R}} + \overline{\boldsymbol{\nu}}\_{\beta s} \sin \theta\_{\mathbb{R}} \right) \tag{34-3}
$$

$$
\overline{\boldsymbol{\nu}}\_{\boldsymbol{\beta}r} \left( \overline{\boldsymbol{s}} + \boldsymbol{\nu}\_r \right) = \boldsymbol{\nu}\_{\sigma r} \left( -\overline{\boldsymbol{\nu}}\_{\boldsymbol{\alpha}s} \sin \theta\_{\boldsymbol{\mathcal{R}}} + \overline{\boldsymbol{\nu}}\_{\boldsymbol{\beta}s} \cos \theta\_{\boldsymbol{\mathcal{R}}} \right) \tag{34-4}
$$

$$\partial\_{R}\left(\overline{\mathbf{s}} + \mathbf{k}\_{z} \times \mathbf{J}\right) = \left(p/l\right) \cdot \left\{-\left(\mathbf{3} \times \mathbf{2}\right) p \Lambda\_{3} \left[\left(\overline{\boldsymbol{\nu}}\_{as}\overline{\boldsymbol{\nu}}\_{ar} + \overline{\boldsymbol{\nu}}\_{\rho\_{\overline{s}}}\overline{\boldsymbol{\nu}}\_{\rho\_{\overline{r}}}\right) \sin\theta\_{\overline{R}} + \left(\overline{\boldsymbol{\nu}}\_{as}\overline{\boldsymbol{\nu}}\_{\rho\_{\overline{r}}} - \overline{\boldsymbol{\nu}}\_{\rho\_{\overline{s}}}\overline{\boldsymbol{\nu}}\_{ar}\right) \cos\theta\_{\overline{R}}\right] - T\_{st}\right\},\tag{34-5}$$

$$\frac{d\theta\_R}{dt} = \dot{\theta}\_R = \alpha\_R \tag{34-6}$$

The equation sets (33-1...4) and (34-1...6) prove that a three-phase induction machine connected to the supply system by 3 wires can be studied similarly to a two-phase machine (*two-phase* mathematical model). Its parameters can be deduced by linear transformations of the original parameters including the supply voltages (Fig. 2a).

16 Induction Motors – Modelling and Control

**3.2. The abc-dq model in total fluxes** 

(28-4) then after convenient grouping we obtain (Simion, 2010):

 

 

*r*

*r*

system (4 electric circuits and 2 movement equations) is:

 

 

> 

 

*R z s k J pJ p* / 3/2 <sup>3</sup>

 

 

 

*u*

 

 

 

*u*

 

 

 

 

*s*

*dt d*

*dt*

*d dt d dt*

*d*

*s*

total fluxes.

These equations allow the study of three-phase induction machine for any duty. It has to be mentioned that the electromagnetic torque expression has no homopolar components of the

For the study of the single unbalance condition is necessary to consider expressions of the instantaneous values of the stator and rotor quantities (voltages, total fluxes and eventually currents in *a,b,c* reference frame) whose sum is null. The real quantities can be transformed to (d,q) reference frame (Simion et al., 2011). By using the notations (28-1), (28-2), (28-3) and

*ss s s r R r R*

*ss s s r R r R*

*rr r s R s R*

 

> 

> >

 

 

*rr R R*

 

*rr R R*

*s sR R*

*s sR R*

*rr r s R s R*

 

Further, the movement equation (31) must be attached. The operational form of the equation

 

*<sup>s</sup> s u*

*<sup>s</sup> s u*

 

 *sr sr* sin *<sup>R</sup>*

*R R*

*sr sr* cos *R st <sup>T</sup>*

(34-5)

 

The equation sets (33-1...4) and (34-1...6) prove that a three-phase induction machine connected to the supply system by 3 wires can be studied similarly to a two-phase machine

*r ss s*

*r ss s*

 

 

*R*

*d dt* 

 

*s* 

*s* 

 

 

cos sin

sin cos

 

> 

 

 

 

> 

 

cos sin

   

(33-1, 2)

(33-3, 4)

 

 

> 

> >

> > >

 

cos sin (34-3)

sin cos (34-4)

*r r* cos sin (34-1)

*r r* sin cos (34-2)

 

(34-6)

   

sin cos

**Figure 2.** Induction machine schematic view: a.Two-phase model; b. Simplified view of the total fluxes in stator reference frame; c. Idem, but in rotor reference frame

The windings of two-phase model are denoted with (*αs, βs*) and (*αr, βr*) in order to trace a correspondence with the real two-phase machine, whose subscripts are (*as, bs*) and (*ar, br*) respectively. We shall use the subscripts *xs* and *ys* for the quantities that corresponds to the three-phase machine but transformed in its two-phase model. This is a rightful assumption since (*αs, βs*) axes are collinear with (*x, y*) axes, which are commonly used in analytic geometry. Further, new notations (35) for the flux linkages of the right member of the equations (33-1...4) will be defined by following the next rules:


$$\begin{cases} \boldsymbol{\nu}\_{xr} = \boldsymbol{\nu}\_{ar} \cos \theta\_{\mathbb{R}} - \boldsymbol{\nu}\_{\beta r} \sin \theta\_{\mathbb{R}'} & \boldsymbol{\nu}\_{yr} = \boldsymbol{\nu}\_{ar} \sin \theta\_{\mathbb{R}} + \boldsymbol{\nu}\_{\beta r} \cos \theta\_{\mathbb{R}} \\ \boldsymbol{\nu}\_{XS} = \boldsymbol{\nu}\_{\alpha s} \cos \theta\_{\mathbb{R}} + \boldsymbol{\nu}\_{\beta s} \sin \theta\_{\mathbb{R}'} & \boldsymbol{\nu}\_{YS} = -\boldsymbol{\nu}\_{\alpha s} \sin \theta\_{\mathbb{R}} + \boldsymbol{\nu}\_{\beta s} \cos \theta\_{\mathbb{R}} \end{cases} \tag{35}$$

Some aspects have to be pointed out. When the machine operates under motoring duty, the pulsation of the stator flux linkages from (*αs, βs*) axes is equal to *ωs.* Since the rotational pulsation is *ωR* then the pulsation of the rotor quantities from (*αr, βr*) axes is equal to *ωr=sωs=ωs –ωR*. The pulsation of the rotor quantities projected along the stator axes with the subscripts *xr* and *yr* is equal to *ωs*. The pulsation of the stator quantities projected along the rotor axes with the subscripts *XS* and *YS* is equal to *ωr*. The equations (33-1...4) become:

$$
\overline{\boldsymbol{\nu}}\_{\alpha s} (\overline{\mathbf{s}} + \boldsymbol{\nu}\_s) = \overline{\boldsymbol{\mu}}\_{\alpha s} + \boldsymbol{\nu}\_{\sigma s} \overline{\boldsymbol{\nu}}\_{\boldsymbol{x} \boldsymbol{r}} \tag{36-1}
$$

Mathematical Model of the Three-Phase Induction Machine

*s r*

(40)

(41)

*j j*

 

*s r*

 

*j j*

 

for the Study of Steady-State and Transient Duty Under Balanced and Unbalanced States 19

(*αs*) axis and has a 3 times higher modulus. In a similar way can be obtained the space phasors of the stator voltages and rotor fluxes and the system equation (39-1...6) that

*sR s s sR s rR s s sR s rR*

*U j j ej e*

 

 3 3 3 3 3 3 3 3 3 0

 

 

*js j e js e*

*r sR r s rR r sR r s rR*

 

<sup>333</sup> 3/2 sin *<sup>e</sup> sR rR s r T p*

When the speed regulation of the cage induction machine is employed by means of voltage and/or frequency variation then the simultaneous control of the two total flux space vectors is difficult. As consequence, new strategies more convenient can be chosen. To this effect, we shall deduce expressions of the electromagnetic torque that include only one of the total flux

One of the methods used for the control of induction machine consists in the operation with

*r rr s sR*

*<sup>s</sup> <sup>j</sup> js ss s*

*r s srsr sr r sR j s r*

*s s s*

where *θ* is the angle between stator and rotor total flux space vectors. This angle has the

The expression of the magnetic torque that depends with the stator total flux space vector

3 3 3 3 <sup>3</sup> <sup>3</sup> 22 2

*cr s r e sR*

*<sup>s</sup> s r s s r r s tt r*

 

 

3 3 Re Re (cos sin ) 2 2

Assuming the ideal hypothesis of maintaining constant the stator flux, for example equal to

<sup>3</sup> 2sin cos 1 , ; <sup>4</sup>

max <sup>3</sup> <sup>2</sup> <sup>2</sup> 2 2

*sR r*

2 /2 / 2

*s and T p*

 

22 2 22 2 22 2 ;( );sin ;cos

*s*

*r*

*s r*

 

max 3 3

*r*

 

*r*

*s*

 

 

 

(43)

(42)

 

2

 

*s r sr sr*

3 3 3 22 2 22 2 22 2

 

**4.1. Variation of the torque with the stator total flux space vector** 

*constant stator total flux space vector*. From (40), the rotor total flux space vector is:

*s r*

*e sR rR sR sR*

 

 

 

*T pj p j j*

the no-load value, then the pull-out torque, Temax , corresponds to sin2θ = 1 that is:

2 3

> 

 

 

 

  describe the steady state becomes:

space vectors either from stator or rotor.

3

meaning of an *internal angle of the machine*.

2 4

*r sr r*

 

 

*s p p s*

*r r s r*

3

 

*<sup>U</sup> or T p*

*e r*

becomes:

 *e*

 

2 2 3 3 22 2 3 3

*sR sR*

3 3 sin 2 .

*rR sR*

 

> 

$$
\overline{\boldsymbol{\nu}}\_{\beta\boldsymbol{s}}(\overline{\boldsymbol{s}} + \boldsymbol{\nu}\_{\boldsymbol{s}}) = \overline{\boldsymbol{\nu}}\_{\beta\boldsymbol{s}} + \boldsymbol{\nu}\_{\sigma\boldsymbol{s}} \overline{\boldsymbol{\nu}}\_{\boldsymbol{y}\boldsymbol{r}} \tag{36-2}
$$

$$
\overline{\boldsymbol{\nu}}\_{\alpha r} (\overline{\boldsymbol{s}} + \boldsymbol{\nu}\_r) = \boldsymbol{\nu}\_{\sigma r} \overline{\boldsymbol{\nu}}\_{XS} \tag{36-3}
$$

$$
\overline{\boldsymbol{\nu}}\_{\beta r}(\overline{\mathbf{s}} + \boldsymbol{\nu}\_r) = \boldsymbol{\nu}\_{\sigma r} \overline{\boldsymbol{\nu}}\_{\text{YS}} \tag{36-4}
$$

The first two equations join the quantities with the pulsation *ωs* and the other two, the quantities with the pulsation *ω<sup>r</sup> = sωs*. The expression of the magnetic torque, in *total fluxes* and *rotor position angle* becomes:

$$T\_e = -\left(\Im / \Im\right) p \Lambda\_3 \left(\overline{\nu}\_{\alpha s} \overline{\nu}\_{\,\,yr} - \overline{\nu}\_{\,\,\beta s} \overline{\nu}\_{\,\,xr}\right) \tag{37}$$

or a second equivalent expression:

$$T\_e = (\Im / \Im) p \Lambda\_3 \left( \overline{\nu}\_{ar} \overline{\nu}\_{YS} - \overline{\nu}\_{\beta r} \overline{\nu}\_{XS} \right) \tag{38}$$

which shows the ″total symmetry″ of the two-phase model of the three-phase machine regarding both stator and rotor. The equations of the four circuits together with the movement equation (37) under operational form give:

$$
\overline{\boldsymbol{\nu}}\_{\alpha s} \left( \overline{\mathbf{s}} + \boldsymbol{\nu}\_{s} \right) = \overline{\boldsymbol{\mu}}\_{\alpha s} + \boldsymbol{\nu}\_{\sigma s} \overline{\boldsymbol{\nu}}\_{\boldsymbol{x} \boldsymbol{r}} \tag{39-1}
$$

$$
\overline{\overline{\nu}}\_{\beta\overline{s}} \left( \overline{\overline{s}} + \nu\_{\overline{s}} \right) = \overline{\overline{\nu}}\_{\beta\overline{s}} + \nu\_{\sigma\overline{s}} \overline{\overline{\nu}}\_{yr} \tag{39-2}
$$

$$
\overline{\boldsymbol{\nu}}\_{\alpha r} \left( \overline{\mathbf{s}} + \boldsymbol{\nu}\_r \right) = \boldsymbol{\nu}\_{\sigma r} \overline{\boldsymbol{\nu}}\_{\text{XS}} \tag{39.3}
$$

$$
\overline{\boldsymbol{\nu}}\_{\beta r} \left( \overline{\mathbf{s}} + \boldsymbol{\nu}\_r \right) = \boldsymbol{\nu}\_{\sigma r} \overline{\boldsymbol{\nu}}\_{\mathrm{YS}} \tag{39-4}
$$

$$
\partial\_R \left( \overline{\mathbf{s}} + k\_z \;/\; J \right) = \left( p/J \right) \cdot \left\{ \begin{array}{c} \left( \mathbf{3} \;/\; \mathbf{2} \right) p \Lambda\_3 \left( \overline{\boldsymbol{\nu}}\_{\beta s} \overline{\boldsymbol{\nu}}\_{xr} - \overline{\boldsymbol{\nu}}\_{\alpha s} \overline{\boldsymbol{\nu}}\_{yr} \right) - T\_{st} \right\} \tag{39-5}
$$

$$\frac{d\theta\_{\mathbb{R}}}{dt} = \dot{\theta}\_{\mathbb{R}} = \alpha\_{\mathbb{R}}\tag{39-6}$$

This last equation system allows the study of transients under single unbalance condition. It is similar with the frequently used equations (Park) but contains as variables only total fluxes and the rotation angle. There are no currents or angular speed in the voltage equations.

### **4. Expressions of electromagnetic torque**

For the steady state analysis of the symmetric three-phase induction machine, one can define the simplified space phasor of the stator flux, which is collinear to the total flux of the (*αs*) axis and has a 3 times higher modulus. In a similar way can be obtained the space phasors of the stator voltages and rotor fluxes and the system equation (39-1...6) that describe the steady state becomes:

$$\begin{aligned} \mathbf{U}\_{s\mathbf{R}3} &= \left(\nu\_s + j\alpha\_s\right) \underline{\Psi}\_{s\mathbf{R}3} - \nu\_{\sigma s} \underbrace{\Psi}\_{rR3} = \left(\alpha\_s - j\nu\_s\right) \Psi\_{s\mathbf{R}3} e^{j\alpha\_s} + j\nu\_{\sigma s} \Psi\_{rR3} e^{j\alpha\_r} \\ \mathbf{0} &= \nu\_{\sigma r} \underline{\Psi}\_{s\mathbf{R}3} - \left(\nu\_r + j\alpha s\_s\right) \underline{\Psi}\_{rR3} = -j\nu\_{\sigma r} \Psi\_{s\mathbf{R}3} e^{j\alpha\_s} + \left(j\nu\_r - s\alpha\_s\right) \Psi\_{rR3} e^{j\alpha\_r} \end{aligned} \tag{40}$$
 
$$T\_e = \left(\mathbf{3} \,\mathrm{/} \mathbf{2}\right) p \Lambda\_3 \Psi\_{s\mathbf{R}3} \Psi\_{rR3} \sin\left(\alpha\_s - \alpha\_r\right)$$

When the speed regulation of the cage induction machine is employed by means of voltage and/or frequency variation then the simultaneous control of the two total flux space vectors is difficult. As consequence, new strategies more convenient can be chosen. To this effect, we shall deduce expressions of the electromagnetic torque that include only one of the total flux space vectors either from stator or rotor.

### **4.1. Variation of the torque with the stator total flux space vector**

18 Induction Motors – Modelling and Control

and *rotor position angle* becomes:

or a second equivalent expression:

**4. Expressions of electromagnetic torque** 

movement equation (37) under operational form give:

( ) *<sup>s</sup> <sup>s</sup> s s xr*

( ) *<sup>s</sup> <sup>s</sup> s s yr*

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

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

which shows the ″total symmetry″ of the two-phase model of the three-phase machine regarding both stator and rotor. The equations of the four circuits together with the

*<sup>s</sup> <sup>s</sup> s s xr*

*<sup>s</sup> <sup>s</sup> s s yr*

*<sup>r</sup> r r XS s*

*<sup>r</sup> r r YS s*

*R z s k J pJ p* / 3/2 <sup>3</sup>

*R*

*d dt* 

 

 

 

 

 

 

 

 

*R R*

 

This last equation system allows the study of transients under single unbalance condition. It is similar with the frequently used equations (Park) but contains as variables only total fluxes and the rotation angle. There are no currents or angular speed in the voltage equations.

For the steady state analysis of the symmetric three-phase induction machine, one can define the simplified space phasor of the stator flux, which is collinear to the total flux of the

*s xr s yr Tst* (39-5)

 

 

 

 

The first two equations join the quantities with the pulsation *ωs* and the other two, the quantities with the pulsation *ω<sup>r</sup> = sωs*. The expression of the magnetic torque, in *total fluxes*

 

 

(36-1)

(36-2)

(36-4)

*s yr s xr* (37)

*r YS r XS* (38)

(39-1)

(39-2)

(39-3)

(39-4)

(39-6)

(36-3)

 

 

 

 

 *s u*

*s u*

*T p <sup>e</sup>* 3/2 <sup>3</sup>

*T p <sup>e</sup>* (3 / 2) <sup>3</sup>

*s u*

*s u*

 

 

  

> One of the methods used for the control of induction machine consists in the operation with *constant stator total flux space vector*. From (40), the rotor total flux space vector is:

$$\begin{aligned} \underline{\Psi\_{rR3}} &= \frac{\nu\_{\sigma r}}{\nu\_r + j \text{s}o\_s} \underline{\Psi\_{sR3}} = \frac{\nu\_{\sigma r}}{\sqrt{o\_s^2 s^2} + \nu\_r^2} \left( \frac{\nu\_r}{\sqrt{o\_s^2 s^2} + \nu\_r^2} - j \frac{s o\_s}{\sqrt{o\_s^2 s^2} + \nu\_r^2} \right) = \\\ \frac{\nu\_{\sigma r} \underline{\Psi\_{sR3}}}{\sqrt{o\_s^2 s^2} + \nu\_r^2} e^{-j\theta}; (\theta = \alpha\_s - \alpha\_r); \sin\theta = \frac{s o\_s}{\sqrt{o\_s^2 s^2} + \nu\_r^2}; \cos\theta = \frac{\nu\_r}{\sqrt{o\_s^2 s^2} + \nu\_r^2} \end{aligned} \tag{41}$$

where *θ* is the angle between stator and rotor total flux space vectors. This angle has the meaning of an *internal angle of the machine*.

The expression of the magnetic torque that depends with the stator total flux space vector becomes:

$$\begin{split} T\_{\varepsilon} &= -\left(\frac{3}{2}\right) p\Lambda\_{3} \operatorname{Re}\left(j\underbrace{\boldsymbol{\Psi}\_{s\mathcal{R}3}\boldsymbol{\Psi}\_{r\mathcal{R}3}^{\*}}\_{\mathcal{R}3}\right) = -\left(\frac{3}{2}\right) p\Lambda\_{3} \operatorname{Re}\left(j\underbrace{\boldsymbol{\Psi}\_{s\mathcal{R}3}}\_{\mathcal{R}3} \frac{\boldsymbol{\nu}\_{\sigma\tau}}{\sqrt{\boldsymbol{\alpha}\_{s}^{2}\boldsymbol{s}^{2} + \boldsymbol{\nu}\_{r}^{2}}} \cdot \underbrace{\boldsymbol{\Psi}^{\*}}\_{\mathcal{R}3} \mathbf{(\cos\theta + j\sin\theta)}\right) = \\ &= \frac{3}{2} \frac{\boldsymbol{\nu}\_{\sigma\tau}}{\boldsymbol{\nu}\_{r}} p\Lambda\_{3} \boldsymbol{\Psi}\_{s\mathcal{R}3}^{2} \frac{\boldsymbol{\nu}\_{\sigma\mathcal{R}3} \boldsymbol{\nu}\_{r}}{\boldsymbol{\alpha}\_{s}^{2}\boldsymbol{s}^{2} + \boldsymbol{\nu}\_{r}^{2}} = \frac{3}{4} \frac{\boldsymbol{\nu}\_{\sigma\tau}}{\boldsymbol{\nu}\_{r}} p\Lambda\_{3} \boldsymbol{\Psi}\_{s\mathcal{R}3}^{2} \sin 2\theta. \end{split} \tag{42}$$

Assuming the ideal hypothesis of maintaining constant the stator flux, for example equal to the no-load value, then the pull-out torque, Temax , corresponds to sin2θ = 1 that is:

$$\begin{aligned} 2\sin\theta\cos\theta &= 1 \leftrightarrow s\_{cr}\alpha\_s = \nu\_r, \text{and } T\_{e\text{max}} = \frac{3}{4} \frac{\nu\_{\sigma r}}{\nu\_r} p\Lambda\_3 \Psi\_{s\text{R3}}^2; \\ \text{or } T\_{e\text{max}} &= \frac{3}{2} \nu\_{\sigma r} p\Lambda\_3 \left(\frac{\text{II}\_{s\text{R3}}}{\alpha\_s}\right)^2 \frac{\nu\_r}{\nu\_s^2 \left(\nu\_r / \alpha\_s\right)^2 + 2\nu\_{\sigma s}\nu\_{\sigma r}\nu\_r / \alpha\_s + \nu\_{\text{th}}^2 + 2\nu\_r^2} \end{aligned} \tag{43}$$

Now an observation can be formulated. Let us suppose an ideal static converter that operates with a U*/f=constant=k1* strategy. For low supply frequencies, the pull-out torque decreases in value since the denominator increases with the pulsatance decrease, *ωs* (Fig. 3). Within certain limits at low frequencies, an increase of the supply voltage is necessary in order to maintain the pull-out torque value. In other words, *U/f = k2,* and *k2>k1.*

Mathematical Model of the Three-Phase Induction Machine

22 2

 

() ()

(44)

(45)

for the Study of Steady-State and Transient Duty Under Balanced and Unbalanced States 21

A proper control of the induction machine requires a strategy based on *U/f = variable*. More precisely, for low frequency values it is necessary to increase the supply voltage with respect to the values that result from *U/f = const.* strategy. At a pinch, when the frequency becomes zero, the supply voltage must have a value capable to compensate the voltage drops upon the equivalent resistance of the windings. Lately, the modern static converters can be parameterized on the basis of the catalog parameters of the induction machine or on the

 

3 3 2 2 3 3 22 2

 

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

 

 

 

22 2 2

: () ; () 2

fluxes after which the stator flux must keep its prescribed value.

means a significant rise of the voltage and current.

 

slip (and load as well), we can point out the following observations.

 

2 2 2 2

*where F s s G s s*

3 3 2 3 3 22 2 <sup>2</sup> ( ) *sR s sR s r sR sR*

*j U s js <sup>s</sup>*

*s r s sr*

*U ss U F s*

 

 

if the term *νtt* was neglected. By inspecting the square root term, which is variable with the




If the machine parameters are established, then a variation rule of the supply voltage can be settled in order to have a constant stator flux (equal, for example, to its no-load value) both

Fig. 4 presents (for a machine with predetermined parameters: supply voltage with the amplitude of 490 V (Uas=346.5V); Rs=Rr=2; Lhs=0,09; Lσs= Lσr=0,01; J=0,05; p=2; kz=0,02;

*s s tt r s s r*

*U s j As Bs C*

<sup>2</sup> ( ) <sup>1</sup>

*s F s sG s*

 

 

basis of some laboratory tests results.

From (40) we can deduce:

against G, (45).

for frequency and load variation.

and further:

**Figure 3.** Mechanical characteristics, *Me=f(ΩR)* at *ΨsR3=const.* 

**Figure 4.** Resultant stator voltage vs. pulsatance *UsR3=f(ωs) at ΨsR3=const. (1,91Wb)*

A proper control of the induction machine requires a strategy based on *U/f = variable*. More precisely, for low frequency values it is necessary to increase the supply voltage with respect to the values that result from *U/f = const.* strategy. At a pinch, when the frequency becomes zero, the supply voltage must have a value capable to compensate the voltage drops upon the equivalent resistance of the windings. Lately, the modern static converters can be parameterized on the basis of the catalog parameters of the induction machine or on the basis of some laboratory tests results.

From (40) we can deduce:

$$\underline{\Psi}\_{s\text{R3}} = -j \frac{\underline{\mathcal{U}}\_{s\text{R3}}}{\alpha\_s} \frac{\left(s\alpha\_s - j\nu\_r\right)}{\left(s\alpha\_s - \nu\_H\right) - j\left(\nu\_r + s\nu\_s\right)} \leftrightarrow \mathcal{U}\_{s\text{R3}}^2 = \frac{\Psi\_{s\text{R3}}^2 \alpha\_s^2 \left(s\alpha^2 + 2Bs + \mathcal{C}\right)}{\alpha\_s^2 s^2 + \nu\_r^2} \tag{44}$$

and further:

20 Induction Motors – Modelling and Control

Now an observation can be formulated. Let us suppose an ideal static converter that operates with a U*/f=constant=k1* strategy. For low supply frequencies, the pull-out torque decreases in value since the denominator increases with the pulsatance decrease, *ωs* (Fig. 3). Within certain limits at low frequencies, an increase of the supply voltage is necessary in

Angular velocity ΩR [rad/s]

ωs/ωsN

0 0.2 0.4 0.6 0.8 1

s=0

s=0.1 s=0.3

0 20 40 60 80 100 120 140 160

 UN fN

0.79UN

0.75fN 0.58UN

 0.5fN 0.36UN 0.25fN

order to maintain the pull-out torque value. In other words, *U/f = k2,* and *k2>k1.*

**Figure 3.** Mechanical characteristics, *Me=f(ΩR)* at *ΨsR3=const.* 

Voltage UsR3 [V]

0

100

200

300

400

500

600

700

Electromagnetic torque Te [Nm]

**Figure 4.** Resultant stator voltage vs. pulsatance *UsR3=f(ωs) at ΨsR3=const. (1,91Wb)*

$$\begin{aligned} \frac{\mathbf{U}\_{sR3}^2}{\alpha\_s^2} &= \Psi\_{sR3}^2 \left[ 1 + \frac{\nu\_s^2 \mathbf{s}^2 + 2\nu\_{\sigma s} \nu\_{\sigma r} \mathbf{s} + \nu\_{tt}^2}{\alpha\_s^2 \mathbf{s}^2 + \nu\_r^2} \right] \leftrightarrow \Psi\_{sR3} = \frac{\mathbf{U}\_{sR3}}{\alpha\_s} \sqrt{\frac{F(\mathbf{s})}{F(\mathbf{s}) + \mathbf{s}G(\mathbf{s})}} \\ \text{where:} \quad F(\mathbf{s}) &= \alpha\_s^2 \mathbf{s}^2 + \nu\_r^2; \quad G(\mathbf{s}) \approx \nu\_s^2 \mathbf{s} + 2\nu\_{\sigma s} \nu\_{\sigma r} \end{aligned} \tag{45}$$

if the term *νtt* was neglected. By inspecting the square root term, which is variable with the slip (and load as well), we can point out the following observations.


If the machine parameters are established, then a variation rule of the supply voltage can be settled in order to have a constant stator flux (equal, for example, to its no-load value) both for frequency and load variation.

Fig. 4 presents (for a machine with predetermined parameters: supply voltage with the amplitude of 490 V (Uas=346.5V); Rs=Rr=2; Lhs=0,09; Lσs= Lσr=0,01; J=0,05; p=2; kz=0,02;

ω1=314,1 (SI units)) the variation of the resultant stator voltage with the pulsatance (in per unit description) for three constant slip values. The variation is a straight line for reduced loads and has a certain inflection for low frequency values (a few Hz). For under-load operation, a significant increase of the voltage with the frequency is necessary. This fact is more visible at high slip values, close to pull-out value (in our example the pull-out slip is of 0,33).

Mathematical Model of the Three-Phase Induction Machine

for the Study of Steady-State and Transient Duty Under Balanced and Unbalanced States 23

 (50)

> *ΨrR0=1.78Wb ωs=314 s-1*

> > A1

A2

which is a straight line, A1 in Fig. 5. The two intersection points with the axes correspond to

The pull-out torque is extremely high and acts at start-up. This behavior is caused by the hypothesis of maintaing constant the rotor flux at a value that corresponds to no-load operation (when the rotor reaction is null) no matter the load is. The compensation of the magnetic reaction of the rotor under load is *hypothetical* possible through an unreasonable

Another unreasonable possibility is the maintaining of the rotor flux to a value that corresponds to start-up (*s = 1*) and the supply voltage has its rated value. In this case the expression of the mechanical characteristic is (50) and the intersection points with the axes (line A2, Fig. 5) correspond to synchronism (Te=0, ΩR=ωs/2=157) and start-up (Te=78 Nm,

> 3 2 32,14 <sup>2</sup> 2 0,25 2 2 96,43 *<sup>e</sup> rRk s R s R <sup>T</sup>*

The supply of the stator winding with constant voltage and rated pulsation determines a variation of the resultant rotor flux within the short-circuit value (ΨrRk=0,5Wb) and the synchronism value (ΨrR0=1,78Wb). The operation points lie between the two lines, A1 and A2, on a position that depends on the load torque. When the supply pulsation is two times smaller (and the voltage itself is two times smaller as well) and the resultant rotor flux is maintained constant to the value ΨrR0=1,78Wb, then the mechanical characteristic is described by the straight line B1, which is parallel to the line A1. Similarly, for ΨrRk=0,5Wb,

> *ΨrR0=1.78Wb ωs=*157s*-1*

B1

Angular velocity ΩR [rad/s]

0 20 40 60 80 100 120 140 160 <sup>0</sup>

*ΨrRk=0.5Wb ωs=157s-1*

B2

*ΨrRk=0.5Wb ωs=314s-1*

synchronism (Te=0, ΩR=ωs/2=157) and start-up (Te=995 Nm, ΩR=0) respectively.

increase of the supply voltage. Practically, the pull-out torque is much lower.

the mechanical characteristic become the line B2, which is parallel to A2.

**Figure 5.** Mechanical characteristics *Te=f(ΩR), ΨrR=const.* 

Electromagnetic torque Te[Nm]

ΩR=0) respectively.

The variation rule based on *UsR=f(ωs)* strategy (applied to the upper curve from Fig. 4) provide an operation of the motor within a large range of angular speeds (from start-up to rated point) under a developed torque, whose value is close to the pull-out one. Obviously, the input current is rather high (4-5 *I1N*) and has to be reduced. Practically, the operation points must be placed within the upper and the lower curves, Fig. 4. It is also easy to notice that the operation with higher frequency values than the rated one does not generally require an increase of the supply voltage but the developed torque is lower and lower. In this case, the output power keeps the rated value.

### **4.2. Variation of the torque with the rotor total flux space vector**

Usually, the electric drives that demand high value starting torque use *constant rotor total flux space vector* strategy. The stator total flux space vector can be written from (41) as:

$$\begin{aligned} \underline{\Psi}\_{sR3} &= \frac{\nu\_r + j \text{so}\_s}{\nu\_{\sigma r}} \underline{\Psi}\_{rR3} \Leftrightarrow \Psi\_{sR3} = \Psi\_{rR3} \frac{\sqrt{\alpha\_s^2 \,^2 + \nu\_r^2}}{\nu\_{\sigma r}}; \underline{\Psi}\_{sR3} = \frac{\sqrt{\alpha\_s^2 \,^2 + \nu\_r^2}}{\nu\_{\sigma r}} \underline{\Psi}\_{rR3} e^{j\theta}; \\\ \underline{\Psi} (\theta = \alpha\_s - \alpha\_r); \sin \theta &= \frac{\text{so}\_s}{\sqrt{\alpha\_s^2 \,^2 + \nu\_r^2}}; \cos \theta = \frac{\nu\_r}{\sqrt{\alpha\_s^2 \,^2 + \nu\_r^2}} \end{aligned} \tag{46}$$

and the expression of the electromagnetic torque on the basis of rotor flux alone becomes:

$$T\_e = -\left(\frac{3}{2}\right) p\Lambda\_3 \operatorname{Re}\left(j\underline{\Psi\_{sR3}}\underline{\Psi}\underline{\Psi}\_{rR3}^\*\right) = \frac{3}{2} \frac{p\Lambda\_3}{\nu\_{\sigma r}} \underline{\Psi}\_{rR3}^2 \operatorname{sao}\_s\tag{47}$$

Assuming the ideal hypothesis of maintaining constant the rotor flux, for example equal to the no-load value, then the electromagnetic torque expression is:

$$T\_e = \frac{3}{2} \frac{p\Lambda\_3}{\nu\_{\sigma r}} \Psi\_{rR30}^2 \text{so} \boldsymbol{\rho}\_s \approx \frac{3}{2} \frac{p\Lambda\_3}{\nu\_{\sigma r}} \left(\frac{\nu\_{\sigma r}\mathcal{U}\_{sR3}}{\nu\_r \alpha\_s}\right)^2 \text{so} \boldsymbol{\rho}\_s = \frac{3}{2} \frac{p\Lambda\_3 \nu\_{\sigma r}}{\nu\_r^2} \left(\frac{\mathcal{U}\_{sR3}^2}{\alpha\_s^2}\right) \left(\alpha\_s - p\Omega\_R\right) \tag{48}$$

where the voltage and pulsation is supposed to have rated values. Taking into discussion a machine with predetermined parameters (supply voltage with the amplitude of 490 V (Uas=346.5V); Rs=Rr=2; Lhs=0,09; Lσs= Lσr=0,01; J=0,05; p=2; kz=0,02; ω1=314,1 (SI units)) then the expression of the mechanical characteristic is:

$$T\_e = \frac{3}{2} \frac{2 \cdot 32, 14 \cdot 96, 43}{103, 57^2} \left(\frac{\mathcal{U}\_{sR3N}^2}{\alpha\_{sN}^2}\right) \left(\alpha\_s - 2\Omega\_R\right) = 3,17\left(\alpha\_s - 2\Omega\_R\right) \tag{49}$$

which is a straight line, A1 in Fig. 5. The two intersection points with the axes correspond to synchronism (Te=0, ΩR=ωs/2=157) and start-up (Te=995 Nm, ΩR=0) respectively.

22 Induction Motors – Modelling and Control

this case, the output power keeps the rated value.

( );sin ;cos

  *s*

the no-load value, then the electromagnetic torque expression is:

*<sup>U</sup> <sup>T</sup>*

33 3 22 2

 

> 2 3

2 2

 

*s r*

 

expression of the mechanical characteristic is:

 

**4.2. Variation of the torque with the rotor total flux space vector** 

0,33).

ω1=314,1 (SI units)) the variation of the resultant stator voltage with the pulsatance (in per unit description) for three constant slip values. The variation is a straight line for reduced loads and has a certain inflection for low frequency values (a few Hz). For under-load operation, a significant increase of the voltage with the frequency is necessary. This fact is more visible at high slip values, close to pull-out value (in our example the pull-out slip is of

The variation rule based on *UsR=f(ωs)* strategy (applied to the upper curve from Fig. 4) provide an operation of the motor within a large range of angular speeds (from start-up to rated point) under a developed torque, whose value is close to the pull-out one. Obviously, the input current is rather high (4-5 *I1N*) and has to be reduced. Practically, the operation points must be placed within the upper and the lower curves, Fig. 4. It is also easy to notice that the operation with higher frequency values than the rated one does not generally require an increase of the supply voltage but the developed torque is lower and lower. In

Usually, the electric drives that demand high value starting torque use *constant rotor total* 

3 3 3 3 3 3

*js s s*

 

*r s sr sr j sR rR sR rR sR rR r r r s r*

 

 

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

22 2 22 2

*sr sr*

and the expression of the electromagnetic torque on the basis of rotor flux alone becomes:

*<sup>p</sup> T pj <sup>s</sup>*

Assuming the ideal hypothesis of maintaining constant the rotor flux, for example equal to

3 2 33 33 30 2 2

 

*e rR s s s R r r rs r s p pU pU Ts s <sup>p</sup>* 

where the voltage and pulsation is supposed to have rated values. Taking into discussion a machine with predetermined parameters (supply voltage with the amplitude of 490 V (Uas=346.5V); Rs=Rr=2; Lhs=0,09; Lσs= Lσr=0,01; J=0,05; p=2; kz=0,02; ω1=314,1 (SI units)) then the

> 3 2 32,14 96,43 2 3,17 2 <sup>2</sup> 103,57 *sR N e sR sR sN*

 

3 3 Re 2 2 *<sup>e</sup> sR rR rR s*

*s s*

22 2 22 2

*r*

(47)

 

 

 

 

2 2

*r sR r sR*

; ;

 

> 

*e*

(46)

(48)

(49)

*flux space vector* strategy. The stator total flux space vector can be written from (41) as:

The pull-out torque is extremely high and acts at start-up. This behavior is caused by the hypothesis of maintaing constant the rotor flux at a value that corresponds to no-load operation (when the rotor reaction is null) no matter the load is. The compensation of the magnetic reaction of the rotor under load is *hypothetical* possible through an unreasonable increase of the supply voltage. Practically, the pull-out torque is much lower.

Another unreasonable possibility is the maintaining of the rotor flux to a value that corresponds to start-up (*s = 1*) and the supply voltage has its rated value. In this case the expression of the mechanical characteristic is (50) and the intersection points with the axes (line A2, Fig. 5) correspond to synchronism (Te=0, ΩR=ωs/2=157) and start-up (Te=78 Nm, ΩR=0) respectively.

$$T\_e = \frac{3}{2} \frac{2 \cdot 32.14 \cdot}{96 \,\mu\text{A}} \Psi\_{r\text{Rk}}^2 \left(\alpha\_s - 2\Omega\_R \right) = 0, 25 \left(\alpha\_s - 2\Omega\_R \right) \tag{50}$$

The supply of the stator winding with constant voltage and rated pulsation determines a variation of the resultant rotor flux within the short-circuit value (ΨrRk=0,5Wb) and the synchronism value (ΨrR0=1,78Wb). The operation points lie between the two lines, A1 and A2, on a position that depends on the load torque. When the supply pulsation is two times smaller (and the voltage itself is two times smaller as well) and the resultant rotor flux is maintained constant to the value ΨrR0=1,78Wb, then the mechanical characteristic is described by the straight line B1, which is parallel to the line A1. Similarly, for ΨrRk=0,5Wb, the mechanical characteristic become the line B2, which is parallel to A2.

**Figure 5.** Mechanical characteristics *Te=f(ΩR), ΨrR=const.* 

Mathematical Model of the Three-Phase Induction Machine

for the Study of Steady-State and Transient Duty Under Balanced and Unbalanced States 25

When the pulsation of the stator voltage is low (small angular velocities) then the torque that has to be overcame is small too, but it will rise with the speed and the frequency along a parabolic variation. Since the upper limit of the torque is given by the limited power of the machine (thermal considerations) then this strategy requires additional precautions as concern the safety devices that protect both the static converter and the supply source itself. The analysis of the square root term from (51) generates similar remarks as in the above

Finally is important to say that a control characteristic must be prescribed for the static converter. This characteristic should be simplified and generally reduced to a straight line

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

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:

↔

( )

<sup>0</sup> 2 cos ; ; (1 ) *jj j*

( )

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

, or:

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

*as as bs cs u U t U Ue U kUe U U ke*

0

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

*U U U UU*

 

/ 2;

/ 2

*s as bs cs*

*s as s*

/6

*j s as bs j s as bs*

*U U e jU*

(54)

*U U e jU*

/6

*U UUU*

*bs cs*

(53)

(52)

1 1/2 1/2 <sup>2</sup> 0 3/2 3/2 <sup>3</sup> 1/ 2 1/ 2 1/ 2

*s as s bs cs s*

*U U U U U U*

> ( ) ( )

 

1 1 2 1 *s s s s*

*U j U U j U*

discussed control strategy.

problem.

0

placed between the curves 1 and 2 from Fig. 6.

**5. Study of the unbalanced duties** 

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

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

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

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

$$\frac{\text{U}\_{sR3}^2}{\text{o}\_s^2} = \Psi\_{rR3}^2 \frac{As^2 + 2Bs + C}{\nu\_{\sigma r}^2} \leftrightarrow \frac{\text{U}\_{sR3}}{\text{o}\_s} = \frac{\Psi\_{rR3}}{\nu\_{\sigma r}} \sqrt{As^2 + 2Bs + C} \tag{51}$$

$$\text{with:}\\\quad A = \text{o}\_s^2 + \nu\_s^2\\\therefore B = \nu\_{\sigma s} \nu\_{\sigma r};\\\quad C = \nu\_r^2 + \nu\_{tt}^2 \nu\_{tt}^2 = \left(\nu\_s \nu\_r - \nu\_{\sigma s} \nu\_{\sigma r}\right)^2 / \text{o}\_s^2.$$

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.

When the pulsation of the stator voltage is low (small angular velocities) then the torque that has to be overcame is small too, but it will rise with the speed and the frequency along a parabolic variation. Since the upper limit of the torque is given by the limited power of the machine (thermal considerations) then this strategy requires additional precautions as concern the safety devices that protect both the static converter and the supply source itself.

The analysis of the square root term from (51) generates similar remarks as in the above discussed control strategy.

Finally is important to say that a control characteristic must be prescribed for the static converter. This characteristic should be simplified and generally reduced to a straight line placed between the curves 1 and 2 from Fig. 6.
