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

Written By

A. Alsalloum and A. I. Alolah

Submitted: April 4th, 2012 Published: March 13th, 2013

DOI: 10.5772/53914

From the Edited Volume

## New Developments in Renewable Energy

Edited by Hasan Arman and Ibrahim Yuksel

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## 2. Star connected generator–star connected load without a neutral connection

The equivalent load impedance shown in Figure 1 may be described as follows;

Za = ZLa // jXCa E1
Zb = ZLb // jXCb E2
Zc = ZLc // jXCc E3

where,

ZLa,b,c =   Load impedance at base frequency connected across phase a, b and c, respectively.

XCa,b,c =   Excitation capacitor reactance at base frequency connected across phase a, b and c, respectively.

Za,b,c =   Equivalent impedance of load and excitation capacitor at base frequency connected across phase a, b and c respectively.

At the load side, the phase voltages are:

[VakVbkVck]=[Za000Zb000Zc][IaIbIc]E4

Since the load and/or the excitation capacitors are expected to be unbalanced, it is more appropriate to describe the different quantities involved in Eq. (4) in terms of their symmetrical components. Using the symmetrical components technique, the following is found:

[Vak0Vak1Vak2]=[Z0Z2Z1Z1Z0Z2Z2Z1Z0][Ia0Ia1Ia2]E5

Where the subscripts 0, 1, and 2 stands for zero, positive and negative sequence components, respectively. The symmetrical components of the load phase voltages may be found from the three-phase values as follows:

[Vak0Vak1Vak2]=[1111γγ21γ2γ][VakVbkVck]E6

where γ=1120

On the other hand, the three phase voltages may be found in terms of their symmetrical components by using the following transformation:

[VakVbkVck]=[1111γ2γ1γγ2][Vak0Vak1Vak2]E7

The transformation matrix shown in Eq. (6) can be used to find the symmetrical components of currents, namely, Ia0, Ia1, and Ia2. The symmetrical components of the three phase impedances Za, Zb, and Zc are as follows;

[Z0Z1Z2]=13[1111γγ21γ2γ][ZaZbZc]E8

Since in an isolated-neutral star-connected load, the zero sequence component of line current (phase current) equals to zero, substituting Ia0=0 in Eq. (5) and expanding yields:

Vak1 = Z0Ia1 + Z2Ia2E9
Vak2 = Z1Ia1 + Z0 Ia2E10

It can be shown using the symmetrical components technique that the relation between the positive and negative sequence components of both the line and the phase voltages are as follows:

VLL1 = (1 - γ2) VLN1E11
VLL2 = (1γ)VLN1E12

Hence,

Vab1 = (1 - γ2) Vak1E13
Vab2 = (1γ)Vak2E14

Now looking at the generator side, the following positive and negative sequence circuits of Figure 2 can be used to model the generator. As can be seen in Figure 2, the core loss resistance is taken into consideration in the positive-sequence equivalent circuit of the SEIG. Figure 2.Induction generator equivalent circuits: (a) +ve seq. (b) –ve seq.

As the core loss is variable according to saturation, the core loss resistance is expressed as a function of the magnetizing reactance (Xm) as shown in Eq. (15). Figure 3 shows the variation of the core resistance as a function of magnetizing reactance. Although this will increase the complexity of the model but the model will be closer to the actual case.

Rc(Xm)=a2Xm2+a1Xm+a0p.u.where,a2=17.159,a1=33.372,a0=35.699E15

Furthermore, the air gap voltage may be approximated over the saturated region as a function of Xm by the following sixth order polynomial. Figure 4 shows this variation.

Vg/F=b6Xm6+b5Xm5+b4Xm4+b3Xm3+b2Xm2+b1Xm+b0p.u.where,b6=0.95,b5=8.28,b4=28.91,b3=51.78,b2=50.18,b1=25.07,b0=6.21E16 Figure 3.Variation of core resistance as a function of magnetizing reactance. Figure 4.Variation of air gap voltage as a function of magnetizing reactance.

The terminal voltage of the positive and negative sequence equivalent circuits is given by;

Van1  =  IG1 ZG1E17
Van2  =  IG2 ZG2E18

Eq. (11), Eq. (12), Eq. (17), and Eq. (18) yield the following,

Vab1 = -(1 - γ2) IG1ZG1E19
Vab2 =(1γ)IG2ZG2E20

where

ZG1, ZG2 Input impedance of positive and negative sequence equivalent circuits, respectively.

Equating symmetrical components of line-to-line voltages yields:

- IG1 ZG1 = Z0 Ia1 + Z2 Ia2 E21
IG2 ZG2 = Z1 Ia1 + Z0 Ia2 E22

Since the phase current in a star connected generator is the same as the line current, hence,

Ia1 = IG1 E23
Ia2 = IG2 E24

Substituting Eq. (23) and Eq. (24) into Eq. (21) and into Eq. (22) and rearranging, yields the following:

(ZG1+Z0) Ia1+ Z2Ia2 = 0E25
Z1Ia1 + (Z0-ZG2)Ia2 = 0E26

Solving these two equations simultaneously yields,

( Z0+ ZG1) ( Z0- ZG2) - Z1Z2 = 0E27

This is the characteristic equation of an isolated-neutral star connected induction generator. It consists of two parts, namely, the real part and the imaginary part. Ia1 and Ia2 does not equal zero because self-excitation is assumed to occur, hence, the real and imaginary parts of Eq. (27) must equal to zero. By substituting the machine parameters, speed, excitation capacitor values, a nonlinear equation with constant coefficients in F and Xm can be found. Solving iteratively to find the real roots of the equation that satisfies the constraints, the values of F and Xm are found; hence, the performance of the generator under these conditions can be determined. This procedure is carried out using MATHCAD® software. The flow chart describing the performance evaluation is shown in Figure 5.

## 3. Star connected generator–star connected load with a neutral connection

The connection for this case is shown in Figure 6. In this type of connection the zero sequence component of line currents is present (i.e. Ia0 ≠ 0) while the zero sequence component of phase voltages (Vak0 = 0) equals zero.

Expanding Eq. (5) yields:

Vak0 = Z0 Ia0 + Z2 Ia1 + Z1 Ia2E28
Vak1 = Z1 Ia0 + Z0 Ia1 + Z2 Ia2E29
Vak2 = Z2 Ia0 + Z1 Ia1 + Z0 Ia2E30

Substituting Vak0 = 0 in Eq. (28) and rearranging, yield:

Ia0 = - (Z2 Ia1 + Z1 Ia2) / Z0 E31

Substituting this result in Eqs. (29) and (30), yields:

Vak1=(Z0Z1Z2Z0)Ia1+(Z2Z12Z0)Ia2E32
Vak2=(Z1Z22Z0)Ia1+(Z0Z1Z2Z0)Ia2E33

Since the phase voltage of both the generator and the load are equal, hence,

Vak1 = Van1 = - IG1 ZG1E34
Vak2 = Van2 = IG2 ZG2E35

Substituting Eqs. (34) and (35) into Eqs. (32) and (33), and taking into consideration that IG1 = Ia1 and IG2 = Ia2, yield,

Ia1ZG1=(Z0Z1Z2Z0)Ia1+(Z2Z12Z0)Ia2E36
Ia2ZG2=(Z1Z22Z0)Ia1+(Z0Z1Z2Z0)Ia2E37

Since sequence currents does not equal to zero, hence, the characteristics equation of this system equals to zero;

(Z0Z1Z2Z0+ZG1)(Z0Z1Z2Z0ZG2)(Z1Z22Z0)(Z2Z12Z0)=0E38

## 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,

Zab = ZLab // jXCab E39
Zbc = ZLbc // jXCbc E40
Zca = ZLca // jXCca E41

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

[Z0Z1Z2]=13[1111γγ21γ2γ][ZabZbcZca]E42

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:

[VabL0VabL1VabL2]=[Z0Z2Z1Z1Z0Z2Z2Z1Z0][IabL0IabL1IabL2]E43

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

VabL0 = 0 = Z0 IabL0 + Z2 IabL1 + Z1 IabL2E44

This equation yields:

IabL0 = - (Z2 IabL1 + Z1IabL2) / Z0E45

From Eq. (43)

VabL1 = Z1 IabL0 + Z0 IabL1 + Z2 IabL2E46
VabL2 = Z2 IabL0 + Z1 IabL1 + Z0 IabL2E47

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

VabL1=(Z0Z1Z2Z0)IabL1+(Z2Z12Z0)IabL2E48
VabL2=(Z1Z22Z0)IabL1+(Z0Z1Z2Z0)IabL2E49

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

VabG1= VabL1and VabG2 = VabL2

However,

VabG1=  IabG1ZG1  and VabG2 = IabG2ZG2

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

IabG1ZG1=(Z0Z1Z2Z0)IabL1+(Z2Z12Z0)IabL2E50
IabG2ZG2=(Z1Z22Z0)IabL1+(Z0Z1Z2Z0)IabL2E51

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

Ia1 = (1γ)Iab1E52
Ia2=(1-γ2) Iab2E53

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

Ia1(1γ)ZG1=(Z0Z1Z2Z0)Ia1(1γ)+(Z2Z12Z0)Ia2(1γ2)E54
Ia2(1γ2)ZG2=(Z1Z22Z0)Ia1(1γ)+(Z0Z1Z2Z0)Ia2(1γ2)E55

Since excitation is assumed to occur and rearranging yields,

(Z1Z2Z0Z0ZG1)(Z0Z1Z2Z0ZG2)(Z2Z12Z0)(Z22Z0Z1)=0E56

## 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;

Vak1 = Z0 Ia1 + Z2Ia2E57
Vak2 = Z1 Ia1 + Z0 Ia2E58

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

VabL1 = (1γ2) Vak1E59
VabL2 = (1γ) Vak2E60

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

VabG1 = -IabG1 ZG1E61
VabG2 = IabG2 ZG2E62

These voltages equal to the load line voltages, as;

VabG1 = VabL1E63
VabG2 = VabL2E64

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

(1γ2) Vak1 = -IabG1 ZG1E65
(1γ) Vak2 = IabG2 ZG2E66

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

- IabG1 ZG1 = (1γ2)(Z0 Ia1 + Z2 Ia2)E67
IabG2 ZG2 = (1γ) (Z1 Ia1 + Z0 Ia2)E68

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

Ia1 = (1γ) IabG1E69
Ia2 = (1γ2) IabG2E70

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

(3 Z0 + ZG1) (3 Z0 - ZG2) - 9 Z1 Z2 = 0E71

## 6. Star connected generator–delta connected load

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

[Z0Z1Z2]=13[1111γγ21γ2γ][ZabZbcZca]E72

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

VabL0 = 0 = Z0 IabL0 + Z2 IabL1 + Z1 IabL2E73

from this equation, it can be shown that;

IabL0 = - (Z2 IabL1 + Z1IabL2) / Z0E74

From Eq. (43);

VabL1 = Z1 IabL0 + Z0 IabL1 + Z2 IabL2E75
VabL2 = Z2 IabL0 + Z1 IabL1 + Z0 IabL2E76

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

VabL1=(Z0Z1Z2Z0)IabL1+(Z2Z12Z0)IabL2E77
VabL2=(Z1Z22Z0)IabL1+(Z0Z1Z2Z0)IabL2E78

at the generator side,

VabG1  = (1-γ2) Van1 E79
VabG2  = (1-γ) Van2 E80

It is known that

Van1  =  IG1 ZG1E81
Van2  =  IG2 ZG2E82

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

VabG1  =  -(1 - γ2) IG1ZG1E83
VabG2  =  (1 - γ) IG2ZG2E84

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

(1γ2)IG1ZG1=(Z0Z1Z2Z0)IabL1+(Z2Z12Z0)IabL2E85
(1γ)IG2ZG2=(Z1Z22Z0)IabL1+(Z0Z1Z2Z0)IabL2E86

It is known that,

IG1=Ia1E87
IG2=Ia2E88
Ia1=(1γ)IabL1 E89
Ia2=(1γ2) IabL2 E90

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

(Z0+3ZG1Z1Z2Z0)(Z03ZG2Z1Z2Z0)(Z1Z22Z0)(Z2Z12Z0)=0E91

## 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:

(β1Z0ZG1+β2Z02+β3Z1Z2)(β4Z0ZG2+β5Z02+β6Z1Z2)(β7Z0Z1+β8Z22)(β9Z0Z2+β10Z12)=0E92

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

 ConnectionSEIG-Load β1 β2 β3 β4 β5 β6 β7 β8 β9 β10 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 Y - Ywith neutral -1 -1 1 -1 1 -1 -1 1 1 -1

### Table 1.

Values of Parameter β of characteristics equation

## 8. Conclusions

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

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 experimentally. The model is generalized to cover all connection types of SEIG and/or load. The characteristic equation of each type may be found by substituting the appropriate parameters, i.e., β1 up to β10 from table 1, in the general model.

(β1Z0Zp+β2Z02+β3Z1Z2)(β4Z0Zn+β5Z02+β6Z1Z2)(β7Z0Z1+β8Z22)(β9Z0Z2+β10Z12)=0

Written By

A. Alsalloum and A. I. Alolah

Submitted: April 4th, 2012 Published: March 13th, 2013