**4. Delta connected generator–delta connected load**

A delta-connected generator feeding a delta-connected load is shown in Figure 7, where the elements of the delta-connected load may be defined as follows,

$$\mathbf{Z}\_{ab} = \mathbf{Z}\_{\mathbf{L}\_{ab}} \wedge \mathbf{\slash} - \mathbf{j} \mathbf{X}\_{\mathbf{C}\_{ab}} \tag{39}$$

It is known that for a Delta connected load, *VabL0 = 0*, hence, from Eq. (43)

This equation yields:

Substituting Eq. (45) in Eqs. (46) and (47), yields

Substituting in Eqs. (48) and (49), yields:

From Eq. (43)

However,

0 0 02112 0 *V Z I Z I Z I abL abL abL abL* == + + (44)

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151

Steady State Modeling of Three Phase Self–Excited Induction Generator Under Unbalanced/Balanced Conditions

*abL*0 2 1 12 0 *abL abL I - (Z I Z I ) / Z* = + (45)

*V Z I Z I Z I abL abL abL abL* 1100 12 2 =++ (46)

*V Z I Z I Z I abL abL abL abL* 220110 2 = ++ (47)

(48)

(49)

(50)

(51)

2

2

1 2 1 10 12 2 0 0 *abL abL abL Z Z <sup>Z</sup> VZ IZ I*

*Z Z* æ ö æ ö = - ç ÷ +ç - ÷ ç ÷ è ø è ø

2 1 2 2 1 10 2 0 0 *abL abL abL <sup>Z</sup> Z Z VZ IZ I*

*Z Z* æ ö æ ö =ç - ÷ + - ç ÷ ç ÷ è ø è ø

*VabG*<sup>1</sup> = *VabL* <sup>1</sup>and *VabG*<sup>2</sup> = *VabL* <sup>2</sup>

*VabG*<sup>1</sup> = − *IabG*1*ZG*<sup>1</sup> and *VabG*<sup>2</sup> = *IabG*2*ZG*<sup>2</sup>

1 2 1 11 0 1 2 2 0 0

*Z Z*

2 1 2 22 1 1 0 2 0 0

*Z Z* æ ö æ ö =ç - ÷ + - ç ÷ ç ÷ è ø è ø

*abG G abL abL Z Z <sup>Z</sup> IZ Z I Z I*

*abG G abL abL <sup>Z</sup> Z Z IZ Z I Z I*

æ ö æ ö - =- ç ÷ +ç - ÷ ç ÷ è ø è ø

2

2

Since both the generator and the load are Delta connected, hence:

$$\mathbf{Z}\_{\rm bc} = \mathbf{Z}\_{\rm L\_{bc}} \wedge \mathfrak{h} - \mathrm{j}\mathbf{X}\_{\rm C\_{bc}} \tag{40}$$

$$\mathbf{Z}\_{\rm ca} = \mathbf{Z}\_{\rm L\_{ca}} \wedge \mathbf{\slash} - \mathbf{j} \mathbf{X}\_{\rm C\_{ca}} \tag{41}$$

The symmetrical components for this type of load connection are as follows:

$$
\begin{bmatrix} Z\_0 \\ Z\_1 \\ Z\_2 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 1 & 1 & 1 \\ 1 & \gamma & \gamma^2 \\ 1 & \gamma^2 & \gamma \end{bmatrix} \begin{bmatrix} Z\_{ab} \\ Z\_{bc} \\ Z\_{ca} \end{bmatrix} \tag{42}
$$

**Figure 7.** Delta connected generator - delta connected load.

Since the load as well as the SEIG is connected in delta, hence, the phase (line) voltage of both the generator (*VabG*) and the load (*VabL*) is equal. The symmetrical components of the phase voltage (*Vab*) at the load side are as follows:

$$
\begin{bmatrix} V\_{abL0} \\ V\_{abL1} \\ V\_{abL2} \end{bmatrix} = \begin{bmatrix} Z\_0 & Z\_2 & Z\_1 \\ Z\_1 & Z\_0 & Z\_2 \\ Z\_2 & Z\_1 & Z\_0 \end{bmatrix} \begin{bmatrix} I\_{abL0} \\ I\_{abL1} \\ I\_{abL2} \end{bmatrix} \tag{43}
$$

It is known that for a Delta connected load, *VabL0 = 0*, hence, from Eq. (43)

$$V\_{abL0} = 0 = Z\_0 \, I\_{abL0} + Z\_2 \, I\_{abL1} + Z\_1 \, I\_{abL2} \tag{44}$$

This equation yields:

$$I\_{abL0} = - (Z\_2 I\_{abL1} + Z\_1 I\_{abL2}) / Z\_0 \tag{45}$$

From Eq. (43)

**4. Delta connected generator–delta connected load**

150 New Developments in Renewable Energy

elements of the delta-connected load may be defined as follows,

A delta-connected generator feeding a delta-connected load is shown in Figure 7, where the

*ab ab*

*bc bc*

*ca ca*

2

*Ia*

*Ib*

Since the load as well as the SEIG is connected in delta, hence, the phase (line) voltage of both the generator (*VabG*) and the load (*VabL*) is equal. The symmetrical components of the phase

> 0 021 0 1 102 1 2 210 2

*abL abL abL abL abL abL*

ë û ë ûë û

é ù é ùé ù ê ú ê úê ú <sup>=</sup>

*V ZZZ I V ZZZ I V ZZZ I*

*ab bc ca*

2

g g

é ù é ù é ù ê ú ê ú ê ú <sup>=</sup> ê ú ê ú ê ú ê ú ê ú ê ú ë û ë û ë û

11 1

*Z Z Z Z Z Z* g g

The symmetrical components for this type of load connection are as follows:

0

1

2

*VabG*

*Vab*

*IbcG*

*Vbc*

c *b*

a

*IabG*

**Figure 7.** Delta connected generator - delta connected load.

voltage (*Vab*) at the load side are as follows:

*Vca*

*IcaG* *Z Z / / jX ab L <sup>C</sup>* = - (39)

*Z Z / / jX bc L <sup>C</sup>* = - (40)

*Z Z / / jX ca L <sup>C</sup>* = - (41)

*Ic Zbc*

*Zab*

*VabL*

> *IabL*

(42)

(43)

*Zca*

*IbcL*

*IcaL*

$$\mathbf{V}\_{abL1} = \mathbf{Z}\_1 \mathbf{I}\_{abL0} + \mathbf{Z}\_0 \mathbf{I}\_{abL1} + \mathbf{Z}\_2 \mathbf{I}\_{abL2} \tag{46}$$

$$\mathbf{V}\_{abL2} = \mathbf{Z}\_2 \mathbf{I}\_{abL0} + \mathbf{Z}\_1 \mathbf{I}\_{abL1} + \mathbf{Z}\_0 \mathbf{I}\_{abL2} \tag{47}$$

Substituting Eq. (45) in Eqs. (46) and (47), yields

$$V\_{abL1} = \left(Z\_0 - \frac{Z\_1 Z\_2}{Z\_0}\right) I\_{abL1} + \left(Z\_2 - \frac{Z\_1^2}{Z\_0}\right) I\_{abL2} \tag{48}$$

$$V\_{abL2} = \left(Z\_1 - \frac{Z\_2}{Z\_0}\right) I\_{abL1} + \left(Z\_0 - \frac{Z\_1 Z\_2}{Z\_0}\right) I\_{abL2} \tag{49}$$

Since both the generator and the load are Delta connected, hence:

$$\begin{array}{rcl} V\_{\text{abG1}} = V\_{\text{abL}\\_1} \text{and } V\_{\text{abG2}} = \begin{array}{rcl} V\_{\text{abL}\\_2} \end{array} \end{array}$$

However,

$$\boldsymbol{V}\_{abG1} = \, -\, I\_{abG1} \boldsymbol{Z}\_{G1} \quad \text{and} \, \boldsymbol{V}\_{abG2} = \, \boldsymbol{I}\_{abG2} \boldsymbol{Z}\_{G2}$$

Substituting in Eqs. (48) and (49), yields:

$$-I\_{abG1}Z\_{G1} = \left(Z\_0 - \frac{Z\_1 Z\_2}{Z\_0}\right)I\_{abL1} + \left(Z\_2 - \frac{Z\_1^2}{Z\_0}\right)I\_{abL2} \tag{50}$$

$$I\_{dbG2}Z\_{G2} = \left(Z\_1 - \frac{Z\_2}{Z\_0}\right)I\_{dbL1} + \left(Z\_0 - \frac{Z\_1 Z\_2}{Z\_0}\right)I\_{dbL2} \tag{51}$$

It can be shown using symmetrical components technique that the sequence components of phase and line currents are related as follows:

$$I\_{a1} = \left(1 - \gamma\right) I\_{ab1} \tag{52}$$

2 2 (1 ) *V V abL ak* = g

*Ia*

*Ib*

*Ic*

Substituting Eq. (59), Eq. (60), Eq. (61), and Eq. (62) into Eq. (63) and Eq. (64), yields:

1 11 (1 ) *ak abG G* - =

2 22 (1 ) *ak abG G* - =

2 1 1 01 22 (1 ) *abG G a a - I Z (Z I Z I )* =- + g

2

g

Substituting Eq. (57) and Eq.(58) into Eq. (65) and Eq. (66), yields:

g

impedances *ZG1* and *ZG2* are;

*Vca*

**a**

*Vab*

*Vbc*

**Figure 8.** Delta connected generator - star connected load

These voltages equal to the load line voltages, as;

c b

The positive and negative sequence components of generator voltage in terms of input

Steady State Modeling of Three Phase Self–Excited Induction Generator Under Unbalanced/Balanced Conditions

(60)

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153

*Zb* *Zc* k

*Za*

*V V abG abL* 1 1 = (63)

*V V abG abL* 2 2 = (64)

 *V -I Z* (65)

 *V I Z* (66)

(67)

*V -I Z abG abG G* 1 11 = (61)

*V I Z abG abG G* 2 22 = (62)

$$I\_{a2} = \left(1 \cdot \gamma^2\right) I\_{ab2} \tag{53}$$

Substituting Eqs. (52) and (53) into Eqs. (50) and (51) ), yields

$$-\frac{I\_{a1}}{(1-\gamma)}\ Z\_{\rm G1} = \left(Z\_0 - \frac{Z\_1 Z\_2}{Z\_0}\right)\frac{I\_{a1}}{(1-\gamma)} + \left(Z\_2 - \frac{Z\_1^2}{Z\_0}\right)\frac{I\_{a2}}{(1-\gamma^2)}\tag{54}$$

$$\frac{I\_{a2}}{(1-\gamma^2)}\ Z\_{G2} = \left(Z\_1 - \frac{Z\_2}{Z\_0}\right)\frac{I\_{a1}}{(1-\gamma)} + \left(Z\_0 - \frac{Z\_1 Z\_2}{Z\_0}\right)\frac{I\_{a2}}{(1-\gamma^2)}\tag{55}$$

Since excitation is assumed to occur and rearranging yields,

$$
\left(\frac{Z\_1 Z\_2}{Z\_0} - Z\_0 - Z\_{G1}\right) \left(Z\_0 - \frac{Z\_1 Z\_2}{Z\_0} - Z\_{G2}\right) - \left(Z\_2 - \frac{Z\_1^2}{Z\_0}\right) \left(\frac{Z\_2^2}{Z\_0} - Z\_1\right) = 0\tag{56}
$$

#### **5. Delta connected generator–star connected load**

The connection for this case is shown in Figure 8. It is known that *Ia0 = 0* for a star connected load, substituting in Eq. (5) and expanding *Vak1*, *Vak2*;

$$V\_{ak1} = Z\_0 I\_{a1} + Z\_2 I\_{a2} \tag{57}$$

$$V\_{ak2} = Z\_1 I\_{a1} + Z\_0 I\_{a2} \tag{58}$$

The load sequence components of line-to-line voltage may be expressed in terms of the sequence components of line to neutral voltage as;

$$V\_{abL1} = \left(1 - \gamma^2\right) V\_{ak1} \tag{59}$$

Steady State Modeling of Three Phase Self–Excited Induction Generator Under Unbalanced/Balanced Conditions http://dx.doi.org/10.5772/53914 153

$$\left.V\_{abL2}\right. = \left(1-\gamma\right)V\_{ak2}\tag{60}$$

The positive and negative sequence components of generator voltage in terms of input impedances *ZG1* and *ZG2* are;

$$V\_{ab\gets1} = -I\_{ab\gets1}Z\_{\gets1} \tag{61}$$

$$V\_{abG2} = I\_{abG2} Z\_{G2} \tag{62}$$

**Figure 8.** Delta connected generator - star connected load

It can be shown using symmetrical components technique that the sequence components of

(52)

(53)

g

 g (54)

(55)

(56)

2

*( )*

*( )*

2 2

*V Z I ZI ak*1 0 1 22 *a a* = + (57)

*V Z I Z I ak*2 11 02 *a a* = + (58)

(59)

( ) 1 1 1 *<sup>a</sup> ab I I* = g

( ) <sup>2</sup> 2 2 1- *<sup>a</sup> ab I I* = g

1 12 1 2 1 1 0 2 2

2 12 2 1 2 2 2 2 1 0 0 0 1 1 1 *a aa*

g

0 10 2 2 1

The connection for this case is shown in Figure 8. It is known that *Ia0 = 0* for a star connected

The load sequence components of line-to-line voltage may be expressed in terms of the

2 1 1 (1 ) *V V abL ak* = g

0 *G G*

*I II <sup>Z</sup> Z Z Z Z <sup>Z</sup>*

æ ö æ ö =ç - ÷ + - ç ÷ - - ç ÷ - è ø è ø

 g

*Z( ) Z*

0 0 1 1 1 *a aa*

æ ö æ ö - =- ç ÷ +ç - ÷ - - ç ÷ è ø - è ø

*( ) Z( ) Z*

2

1 2 1 2 1 2

0 0 0 0

*Z Z Z Z Z Z ZZ Z Z Z <sup>Z</sup> Z Z Z Z* <sup>æ</sup> ö æ <sup>ö</sup> æ öæ ö <sup>ç</sup> - - - - -ç - ÷ç - ÷= ÷ ç <sup>÷</sup> <sup>ç</sup> ÷ ç <sup>÷</sup> ç ÷ç ÷ <sup>è</sup> ø è <sup>ø</sup> è øè ø

*I II Z Z <sup>Z</sup> Z Z <sup>Z</sup>*

phase and line currents are related as follows:

152 New Developments in Renewable Energy

Substituting Eqs. (52) and (53) into Eqs. (50) and (51) ), yields

*G*

*G*

Since excitation is assumed to occur and rearranging yields,

**5. Delta connected generator–star connected load**

load, substituting in Eq. (5) and expanding *Vak1*, *Vak2*;

sequence components of line to neutral voltage as;

g

*( )* g

These voltages equal to the load line voltages, as;

$$V\_{abG1} = V\_{abL1} \tag{63}$$

$$V\_{abG2} = V\_{abL2} \tag{64}$$

Substituting Eq. (59), Eq. (60), Eq. (61), and Eq. (62) into Eq. (63) and Eq. (64), yields:

$$\left(\left(1-\gamma^{2}\right)V\_{ak1} = -I\_{abG1}Z\_{G1}\tag{65}$$

$$\left(1 - \gamma\right) V\_{\text{ak2}} = I\_{\text{abG2}} Z\_{\text{G2}} \tag{66}$$

Substituting Eq. (57) and Eq.(58) into Eq. (65) and Eq. (66), yields:

$$-I\_{abG1}Z\_{G1} = \left(1 - \gamma^2\right) \left(Z\_0 \, I\_{a1} + Z\_2 \, I\_{a2}\right) \tag{67}$$

$$I\_{abG2} Z\_{G2} = (1 - \gamma) \left( Z\_1 \, I\_{a1} + Z\_0 \, I\_{a2} \right) \tag{68}$$

The symmetrical components of line current are related to the symmetrical components of phase current in a delta connected generator as follows;

$$I\_{a1} = \left(1 - \gamma\right) I\_{ab \gets 1} \tag{69}$$

$$I\_{a2} = \left(1 - \gamma^2\right) I\_{abG2} \tag{70}$$

*V Z I Z I Z I abL abL abL abL* 220110 2 = ++ (76)

*Ic Zbc*

*Zab* *Zca*

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155

2

(79)

(80)

V I Z an1 G1 G1 = - (81)

V I Z an2 G2 G2 = (82)

(77)

(78)

1 2 1 10 12 2 0 0 *abL abL abL Z Z <sup>Z</sup> VZ IZ I*

*Ia*

Steady State Modeling of Three Phase Self–Excited Induction Generator Under Unbalanced/Balanced Conditions

*Ib*

*Z Z* æ ö æ ö = - ç ÷ +ç - ÷ ç ÷ è ø è ø

2 1 2 2 1 10 2 0 0 *abL abL abL <sup>Z</sup> Z Z VZ IZ I*

*Z Z* æ ö æ ö =ç - ÷ + - ç ÷ ç ÷ è ø è ø

> <sup>2</sup> V (1- ) V abG1 an1 <sup>=</sup> g

V (1- ) V abG2 an2 = g

Substituting Eq. (81) and Eq. (82) into Eq.(79) and Eq. (80), gives,

2

Substituting Eq. (74) in Eqs. (75) and (76), yields:

**Figure 9.** Star connected generator - delta connected load

*Van*

a

b **c**

n

*V*

*cn*

*V*

*bn*

at the generator side,

It is known that

Since excitation is assumed to occur, by substituting Eq. (69) and Eq. (70) in Eqs. (67) and (68), and rearranging, yields:

$$(\mathcal{G}Z\_0 + Z\_{G1})(\mathcal{G}Z\_0 - Z\_{G2}) - \mathcal{G}Z\_1Z\_2 = 0\tag{71}$$

### **6. Star connected generator–delta connected load**

At the load side of Figure 9, the symmetrical components of the load are as follows:

$$
\begin{bmatrix} Z\_0 \\ Z\_1 \\ Z\_2 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 1 & 1 & 1 \\ 1 & \gamma & \gamma^2 \\ 1 & \gamma^2 & \gamma \end{bmatrix} \begin{bmatrix} Z\_{ab} \\ Z\_{bc} \\ Z\_{ca} \end{bmatrix} \tag{72}
$$

It is known that for a delta connected load, *VabL0 = 0*. Hence, from Eq. (43);

$$V\_{abL0} = 0 \ = Z\_0 \ I\_{abL0} + Z\_2 \ I\_{abL1} + Z\_1 \ I\_{abL2} \tag{73}$$

from this equation, it can be shown that;

$$I\_{abL0} = - (\mathbf{Z\_2} I\_{abL1} + \mathbf{Z\_1} I\_{abL2}) / \mathbf{Z\_0} \tag{74}$$

From Eq. (43);

$$\mathbf{V}\_{abL1} = \mathbf{Z}\_1 \mathbf{I}\_{abL0} + \mathbf{Z}\_0 \mathbf{I}\_{abL1} + \mathbf{Z}\_2 \mathbf{I}\_{abL2} \tag{75}$$

Steady State Modeling of Three Phase Self–Excited Induction Generator Under Unbalanced/Balanced Conditions http://dx.doi.org/10.5772/53914 155

**Figure 9.** Star connected generator - delta connected load

$$\mathbf{V}\_{abL2} = \mathbf{Z}\_2 \ I\_{abL0} + \mathbf{Z}\_1 \ I\_{abL1} + \mathbf{Z}\_0 \ I\_{abL2} \tag{76}$$

Substituting Eq. (74) in Eqs. (75) and (76), yields:

$$V\_{abL1} = \left(Z\_0 - \frac{Z\_1 Z\_2}{Z\_0}\right) I\_{abL1} + \left(Z\_2 - \frac{Z\_1^2}{Z\_0}\right) I\_{abL2} \tag{77}$$

$$V\_{abL2} = \left(Z\_1 - \frac{Z\_2}{Z\_0}\right) I\_{abL1} + \left(Z\_0 - \frac{Z\_1 Z\_2}{Z\_0}\right) I\_{abL2} \tag{78}$$

at the generator side,

2 2 11 02 (1 ) *abG G a a I Z (Z I Z I )* =- + g

> 1 1 (1 ) *<sup>a</sup> abG I I* = g

At the load side of Figure 9, the symmetrical components of the load are as follows:

It is known that for a delta connected load, *VabL0 = 0*. Hence, from Eq. (43);

2

g g

é ù é ù é ù ê ú ê ú ê ú <sup>=</sup> ê ú ê ú ê ú ê ú ê ú ê ú ë û ë û ë û

11 1

*Z Z Z Z Z Z* g g

2

*ab bc ca*

0 0 02112 0 *V Z I Z I Z I abL abL abL abL* == + + (73)

*abL*0 2 1 12 0 *abL abL I - (Z I Z I ) / Z* = + (74)

*V Z I Z I Z I abL abL abL abL* 1100 12 2 =++ (75)

2 2 2 (1 ) *<sup>a</sup> abG I I* = g

Since excitation is assumed to occur, by substituting Eq. (69) and Eq. (70) in Eqs. (67) and (68),

phase current in a delta connected generator as follows;

**6. Star connected generator–delta connected load**

0

1

2

from this equation, it can be shown that;

From Eq. (43);

and rearranging, yields:

154 New Developments in Renewable Energy

The symmetrical components of line current are related to the symmetrical components of

(68)

(69)

(70)

(72)

0 1 0 2 12 3 3 90 *G G ( Z Z ) ( Z - Z ) - Z Z* + = (71)

$$\mathbf{V}\_{\text{abG1}} = \left(\mathbf{1}\text{-}\boldsymbol{\mathcal{V}}^{2}\right) \mathbf{V}\_{\text{an1}} \tag{79}$$

$$\mathbf{V}\_{\text{abG2}} = \begin{pmatrix} \mathbf{1} \text{-} \boldsymbol{\gamma} \end{pmatrix} \mathbf{V}\_{\text{an2}} \tag{80}$$

It is known that

$$\mathbf{V}\_{\rm an1} = \begin{array}{c} -\mathbf{I}\_{\rm G1} \; \mathbf{Z}\_{\rm G1} \end{array} \tag{81}$$

$$\mathbf{V\_{ar2}} = \mathbf{I\_{G2}} \,\mathbf{Z\_{G2}} \tag{82}$$

Substituting Eq. (81) and Eq. (82) into Eq.(79) and Eq. (80), gives,

$$\mathbf{V}\_{\text{abG1}} = \text{ -(1 - \chi^2)} \mathbf{I}\_{\text{GI}} \mathbf{Z}\_{\text{G1}} \tag{83}$$

2 2

Steady State Modeling of Three Phase Self–Excited Induction Generator Under Unbalanced/Balanced Conditions

where for each connection the appropriate value of β parameters are given in table 1 below.

SEIG-Load <sup>β</sup><sup>1</sup> <sup>β</sup><sup>2</sup> <sup>β</sup><sup>3</sup> <sup>β</sup><sup>4</sup> <sup>β</sup><sup>5</sup> <sup>β</sup><sup>6</sup> <sup>β</sup><sup>7</sup> <sup>β</sup><sup>8</sup> <sup>β</sup><sup>9</sup> <sup>β</sup><sup>10</sup>

Δ - Δ -1 -1 1 -1 1 -1 -1 1 1 -1 Δ - Y 1 3 0 -1 3 0 3 0 3 0 Y - Δ 3 1 -1 -3 1 -1 1 -1 1 -1 Y - Y 1 1 0 -1 1 0 1 0 1 0

A mathematical model based on the sequence equivalent circuits of the SEIG and the sequence components of the three-phase load was developed to study the performance of the SEIG in the steady state condition. Core loss resistance is included in the model as a function of *Xm*. Furthermore, the magnetizing reactance *Xm* is taken as a variable in the negative sequence equivalent circuit. The performance of the SEIG was determined for no-load, balanced and unbalanced load and/or excitation for different SEIG and load connections. The operating conditions were found by solving the proposed model iteratively using MATHCAD® software as described in section 1. Self-excited induction generator uses excitation capacitors at the terminals. For a given speed, and load situation, there is a specific value of the excitation capacitor that ensures voltage build up. Residual magnetism is a must in SEIG to initiate excitation. In balanced mode of operation, per phase equivalent circuit is solved to find *F* and *Xm*. Unbalanced operation of SEIG can be analyzed through the method of symmetrical

An experimental setup has been built to verify the results obtained from the theoretical model. It is found that the theoretical results are in good agreement with those recorded experimen‐ tally. The model is generalized to cover all connection types of SEIG and/or load. The charac‐ teristic equation of each type may be found by substituting the appropriate parameters, i.e., *β*<sup>1</sup>

<sup>2</sup> + *β*3*Z*1*Z*2)(*β*4*Z*0*Zn* + *β*5*Z*<sup>0</sup>

2)=0

2)(*β*9*Z*0*Z*<sup>2</sup> + *β*10*Z*<sup>1</sup>

<sup>2</sup> + *β*6*Z*1*Z*2)

components and the phase voltages may differ from each other notably.

 bb


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157


2 2 10 1 20 312 40 2 50 612

( )( )

*ZZ Z ZZ ZZ Z ZZ G G*

+ + + +

 b

> b

7 0 1 8 2 9 0 2 10 1

*ZZ Z ZZ Z*

 bb

 bb

b

Connection

Y - Y with neutral

**8. Conclusions**

b

**Table 1.** Values of Parameter β of characteristics equation

up to *β*10 from table 1, in the general model.

(*β*1*Z*0*Zp*

+ *β*2*Z*<sup>0</sup>

−(*β*7*Z*0*Z*<sup>1</sup> + *β*8*Z*<sup>2</sup>

( )( ) 0

$$\mathbf{V}\_{\text{abG2}} = \begin{array}{c} \begin{pmatrix} 1 \ - \ \end{pmatrix} \end{array} \begin{pmatrix} \ I\_{\text{G2}} \mathbf{Z}\_{\text{G2}} \end{pmatrix} \tag{84}$$

Since the line voltages at the generator and the load side are equal, hence,

$$-(1-\chi^2)I\_{G1}Z\_{G1} = \left(Z\_0 - \frac{Z\_1 Z\_2}{Z\_0}\right)I\_{abL1} + \left(Z\_2 - \frac{Z\_1^2}{Z\_0}\right)I\_{abL2} \tag{85}$$

$$(1 - \gamma)I\_{G2}Z\_{G2} = \left(Z\_1 - \frac{Z\_2}{Z\_0}\right)I\_{abL1} + \left(Z\_0 - \frac{Z\_1 Z\_2}{Z\_0}\right)I\_{abL2} \tag{86}$$

It is known that,

$$I\_{G1} = I\_{a1} \tag{87}$$

$$I\_{G2} = I\_{a2} \tag{88}$$

$$I\_{a1} = (1 - \gamma) I\_{abL1} \tag{89}$$

$$I\_{a2} = \left(1 - \gamma^2\right) I\_{abL2} \tag{90}$$

Since excitation is assumed to occur, by substituting Eqs. (87), (88), (89), and (90) into Eqs. (85) and (86), and rearranging yield,

$$\left(Z\_0 + 3Z\_{G1} - \frac{Z\_1 Z\_2}{Z\_0}\right)\left(Z\_0 - 3Z\_{G2} - \frac{Z\_1 Z\_2}{Z\_0}\right) - \left(Z\_1 - \frac{Z\_2}{Z\_0}\right)\left(Z\_2 - \frac{Z\_1^2}{Z\_0}\right) = 0\tag{91}$$

### **7. Generalization of steady state model**

The characteristics equations derived in the previous sections can be represented by one single general equation of the following form:

Steady State Modeling of Three Phase Self–Excited Induction Generator Under Unbalanced/Balanced Conditions http://dx.doi.org/10.5772/53914 157

$$\begin{aligned} & \left( \beta\_1 \mathbf{Z}\_0 \mathbf{Z}\_{G1} + \beta\_2 \mathbf{Z}\_0^{\ \ \ \ \ \beta} + \beta\_3 \mathbf{Z}\_1 \mathbf{Z}\_2 \right) (\beta\_4 \mathbf{Z}\_0 \mathbf{Z}\_{G2} + \beta\_5 \mathbf{Z}\_0^{\ \ \ \ \ \beta} + \beta\_6 \mathbf{Z}\_1 \mathbf{Z}\_2) \\ & - (\beta\_7 \mathbf{Z}\_0 \mathbf{Z}\_1 + \beta\_8 \mathbf{Z}\_2^{\ \ \ \ \ \ \ ) (\beta\_9 \mathbf{Z}\_0 \mathbf{Z}\_2 + \beta\_{10} \mathbf{Z}\_1^{\ \ \ \ \ \end{aligned})} = 0 \end{aligned} \tag{92}$$

where for each connection the appropriate value of β parameters are given in table 1 below.


**Table 1.** Values of Parameter β of characteristics equation
