Phase-Shifting and Zig-Zag Transformers

155

7Phase-Shifting and Zig-Zag Transformers

7.1 Introduction

Phase-shifting transformers are used in power systems to help control power flow and line losses. They shift the input voltages and currents by an angle that can be adjusted using a tap changer. They operate by adding a 90 volt-age to the input voltage, that is, in quadrature. For three-phase transformers, the quadrature voltage to be added to a given phase voltage can be derived by interconnecting the other phases, which can be done in many ways; this gives rise to a large number of configurations for these transformers. We will deal with only a few common types of configurations in this chapter. The phase-shifting capability can be combined with voltage magnitude control in the same transformer. This results in a more complex unit involving two sets of tap changers; however, we will not discuss this in this chapter.

The use of phase-shift transformers can be illustrated by considering the feeding of a common load from two voltage sources that are out of phase with each other, as shown in Figure 7.1. Solving for the currents, we get

I V V I V V1 12

2 2 21

1=( )

=+( )

Z Z

KZK

Z Z

KZK

+ L L L L, (7.1)

where K = Z1Z2 + Z1ZL + Z2ZL. The current into the load, IL, is

I I I V VL = + = +1 2 1 2 2 1ZK

ZK

(7.2)

and the voltage across the load, VL, is

V V I V VLL= = +( )1 1 1 1 2 2 1Z ZK Z Z (7.3)

Thus, the complex power delivered to the load is

V I V VL LL* = +Z

KZ Z2 1 2 2 1

2 (7.4)

where * denotes complex conjugation.

2010 by Taylor and Francis Group, LLC

156 Transformer Design Principles

Consider a case in which Z1 = Z2 = Z, V1 = V, V2 = Vej. Then, Equation 7.4 becomes

V IL LL* ( cos )= +2 1

2 2

2

Z Z VK

(7.5)

Thus, the maximum power is transferred when = 0. In this case, a phase-shifting transformer can be used to adjust the phase of V2 so that it equals that of V1.

In modern interconnected power systems the need for these devices has increased. Other methods for introducing a quadrature voltage are being developed, for instance, by means of power electronic circuits in conjunc-tion with ACDC converters, which can act much faster than on-load tap changers. However they are presently more costly than phase-shifting transformers and are used primarily when response time is important. (Electronic on-load tap changers are also being developed for fast response-time applications.)

In this chapter, we will develop a circuit model description for three com-mon types of phase-shifting transformers. This is useful to understand the regulation behavior of such devices, that is, how much the output voltage magnitude and phase change are when the unit is loaded as compared with that of the no-load voltage output. In the process, we will also find that the phase angle depends on the tap position, a relationship that can be nonlinear. In addition to the positive-sequence circuit model, which describes normal operations, we will also determine the negative- and zero-sequence circuit models for use in short-circuit fault current analysis, which is carried out for the standard types of faults for two phase-shifting transformers.

Very little has been published on phase-shifting transformers in the open literature beyond general interconnection diagrams and how they are used in specific power grids. References that emphasize the basic principles include [Hob39, Cle39, Kra98, Wes64].

V1 V2

Z2 Z1

ZL

I2 I1

Figure 7.1Two possibly out-of-phase sources feeding a common load.


Phase-Shifting and Zig-Zag Transformers 157

7.2 Basic Principles

The basic principles have been discussed in Chapters 3 through 6, but we will repeat some of them here for easier reference. We neglect the exciting current and model the individual phases in terms of their leakage impedances. For a two-winding phase, we use the circuit model shown in Figure 7.2. Z12 is the two-winding leakage impedance, referred to as side 1. Most of the develop-ment described here is carried out in terms of impedances in Ohms. Because of the differences in per-unit bases for the input and winding quantities, per-unit quantities are not as convenient in this analysis. We indicate the appro-priate places where the per-unit quantities might prove useful. We assume that positive currents flow into their respective windings. With N1 equal to the number of turns on side 1 and N2, the number of turns on side 2, the ideal transformer voltages satisfy

E12

1

2E= N

N (7.6)

and the currents satisfy

II

1

2

2

1

= NN

(7.7)

For three windings per phase, we use the model shown in Figure 7.3. In this case, the ideal transformer voltages satisfy

E11

2

2

3

3N N N= =E E (7.8)

and the currents satisfy

N N N1 1 2 2 3 3 0I I+ + =I (7.9)

V1

Z12

E1 E2

I2I1 N1:N2

Figure 7.2Model of a two-winding transformer phase.



The single-winding impedances are given in terms of the two-winding impedances by

Z Z ZNN

Z

ZNN

1 12 131

2

2

23

22

12

12

= +

=11

2

121

2

2

23 13

31

+

=

ZNN

Z Z

Z22

3

1

2

131

2

2

23 12NN

ZNN

Z Z

+

(7.10)

Here, the first subscript refers to the winding for which the two-winding impedance is measured and the second subscript refers to the winding that will be shorted when measuring the impedance. To refer impedances to the opposite winding, we use

V1

Z1

E1

I1 N1

V2

Z2

E2

I2 N2

V3

Z1

E3

I1 N3

Figure 7.3Model of a three-winding transformer phase. The vertical lines connecting the circuits repre-sent the common core.



ZNN

Ziji

jji=

2

(7.11)

For a three-phase system, the positive-sequence quantities correspond to the order of the unit phasors shown in Figure 7.4. We let

= = +e j j120 12

32

(7.12)

Then, the ordering is 1, 2, and . This corre-sponds to the phase ordering a, b, and c for the windings and 1, 2, and 3 for the terminals. Negative-sequence ordering is 1, , and 2, which corre-sponds to the phase ordering a, c, and b or 1, 3, and 2 for the windings or terminals, respectively. This is obtained by interchanging two phases. Note that

2 240 120 12

32

= = = = e ej j j* (7.13)

The zero-sequence quantities are all in phase with each other.We will choose interconnections to produce a positive phase shift at a posi-

tive tap setting. By interchanging two phases, we can produce a negative phase shift at a positive tap setting. Interchanging two phases is equivalent to imputing a negative-sequence set of voltages. This implies that negative-sequence circuits have the opposite phase shift in relation to positive-sequence circuits. Zero-sequence circuits have zero phase shifts. By positive phase shift, we mean that output voltages and currents lead the input voltages and currents, that is, output quantities are rotated counterclockwise on a phasor diagram relative to the input quantities.

7.3 Squashed Delta Phase-Shifting Transformer

One of the simplest phase shifters to analyze is the squashed delta configu-ration shown in Figure 7.5, where S denotes the source or input quantities and L indicates the load or output quantities. The input and output set of voltages and the corresponding currents form a balanced positive-sequence set. In this study, we take the S currents as positive into the terminals and the L currents as positive out of the terminals. The input and output voltage

b phase

c phase

1

2

j

a phase

Figure 7.4Positive-sequence unit phasors.



phasor diagram is shown in Figure 7.5b. These are the voltages to ground; the currents form a similar set but are not shown. Similarly, the internal volt-ages and corresponding currents form a positive-sequence set, as shown in Figure 7.5c for the voltages. Note that Figures 7.5b and 7.5c can be rotated relative to each other if shown on a common phasor diagram. Information on their relative orientation is not contained in the figure although the phase order is consistent between the figures. This applies to the corresponding current phasors as well. a, a, and so on refer to windings on the same leg, with the prime labeling the tapped winding. The two-winding impedances, that is, Zaa, and so on, will be referred to the unprimed coil. Because all these impedances are the same by symmetry, we need only one symbol. This transformer can be designed with a single three-phase core and is generi-cally referred to as a single-core design. It is also best adapted to phase-angle shifts in one direction.

Using the two-winding phase model described in the Section 7.2, adapting it to the present labeling scheme, and concentrating on one inputoutput pair, we can write

VS L1 a S1 L c aa c

S1 c a L1,1 2 = = += +

V E V V I E

I I I I

, Z

== +I Ia b (7.14)

VS1VL1

VS3

VL3

VS2

VL2

Ea

Eb

Ec

Eb

Ec

Ea

IS1 IL1

IS3

IL3IS2

IL2

Ia

Ib

Ia

IbIc

Ic

(a)

VS1

VL1

VL2

VS2

VS3

VL3

(b)

Ea

Ea

Ec

Ec

Eb Eb

(c)

Figure 7.5Squashed delta configuration, with the phase quantities labeled: (a) circuit, (b) input and output phasors, and (c) internal phasors. The ideal transformer voltages, the Es, increase in the opposite direction of the assumed current flow. is a positive phase-shift angle.



Note that, in Equation 7.14, the two-winding leakage impedance is assigned to the unprimed side of the transformer. We also have the following trans-former relations:

E E E E

I I I

= = =

= =

aa

aa a c

aa

aa

eNN n nNN

n

j1 120

aa ce= n j120 I (7.15)

where we define n = Na/Na, the turns ratio. We also have the following relationships:

V V V V

V V E

S S2 S1 S1

S2 L2 b

e

e

12 30

1

1 3 = ( ) =

= =

j

j 220120

=E Ea c

e j

n

(7.16)

A detailed solution of these equations is essential because we will need some of the intermediate results later. We assume that VS1 and IS1 are given. From Equations 7.14 and 7.15, we get

I I I Ic S1 a See

e=

=

11 1120

120

120nn

njj

j, 11 120=

nn je S1

I (7.17)

We also have

I I I I IL1 a b a ceee

= + = + =

j

j

j

nn

120120

120

=

=+

I IS1 S1e

tan

j

n

2 32 1

1

where

I I I I IL1 a b a ceee

= + = + =

j

j

j

nn

120120

120

=

=+

I IS1 S1e

tan

j

n

2 32 1

1 (7.18)

Because n is determined by the tap position, this shows that is a nonlinear function of the tap position, assuming the taps are evenly spaced.

Adding Equation 7.16 and using Equation 7.14, we get

V V V E I ES1 L2 S1 c c aa cee = + = +

3 30120

jj

nZ

Solving for Ec and using Equation 7.17, we get

E Vc S1aae

ee

e=

+

(

3 30120

120

120

nn

n Z

n

j

j

j

j ))2IS1 (7.19)



Combining this equation with the first expressions of Equations 7.14 and 7.15, we get

V V IL1 S1aa

S1 e= + +

Zn n

j2 1

(7.20)

where is given by Equation 7.18. Although the current shifts by in all cases, in general, the voltage shifts by only under no load conditions.

The circuit model suggested by Equation 7.20 is shown in Figure 7.6. By symmetry, this applies to all three phases with appropriate labeling. The equivalent impedance shown in Figure 7.6 is given by

ZZ

n neqaa=

+ +

2 1 (7.21)

Figure 7.6 is a positive-sequence circuit model. The negative-sequence model is obtained simply by changing to , with Zeq unchanged. Note that Zeq depends on the tap setting.

Although the input power per phase is Pin = VS1IS1*, the transformed or winding power per phase Pwdg is relatively less. Ignoring impedance drops, from Equations 7.17 and 7.19, we have

P jn

n nPwdg c c in= = + +

E I*3

12 (7.22)

where the factor j shows that the transformed power is in quadrature with the input power. As a numerical example, let = 30. Then, from Equations 7.18 and 7.22, we have

tan wdg in2

32 1

2 732 0 4227 = +

= =n

n P j P. . (7.23)

VS1

Zeq

VL1

IL1IS1 : 1 ej

Figure 7.6Circuit model of one phase of a phase-shifting transformer for a positive sequence. For a nega-tive sequence, change to .



We can express Equation 7.20 in per-unit terms; however, because of the difference between the input power and voltage base and that of the wind-ings, we must be careful to specify the base used. As usual, we take the power base to be the terminal power base. This is the terminal power per phase. Thus, the rated input power base is Pb,in; the rated input voltage base is Vb,in; and the rated input current base that can be derived from the power and voltage base is Ib,in = Pb,in/Vb,in. Similarly, we can derive the input imped-ance base from the power and voltage base, Zb,in = Vb,in/Ib,in = (Vb,in)2/Pb,in. Because the transformation ratio is 1:1 in terms of magnitude, the output base values are the same as the input base values. Thus, Equation 7.20 can be written as follows:

V V IL1b,in

S1

b,ineq

S1

b,in

b,in

b,inV VZ

IIV

=

=

e S1

b,in

eq

b,in

S1

b,in

j

V

Z

Z I V I

=

e

eL1 S1 eq S

j

jz

v v i 1

(7.24)

or

V V IL1b,in

S1

b,ineq

S1

b,in

b,in

b,inV VZ

IIV

=

=

e S1

b,in

eq

b,in

S1

b,in

j

V

Z

Z I V I

=

e

eL1 S1 eq S

j

jz

v v i 1

where lowercase letters are used to denote the per-unit quantities.Assuming the rated input power per phase to be the common power base,

the winding and input impedance bases are proportional to the square of the respective voltage magnitudes. Using Equation 7.19 and ignoring impedance drops, we have

Z

Zn

n nb,wdg

b,in

c

S1

= =+ +

EV

22

2

31

(7.25)

Thus, using Equation 7.21, zeq can be expressed as

zZ

Z n n

Z

ZZ

Zeqeq

b,in

b,wdg

b,in

aa

b

= =+ +

1

12 ,,wdgaa

= + +( )

31

2

2 2

n

n nz (7.26)

where zeq is on an input base and zaa is on a winding base. Note that the winding base depends on the turns ratio, as indicated in Equation 7.25. Thus, this base is perhaps most useful for design purposes when n refers to the maximum phase angle. At the other extreme, with = 0, we have n = , so that Zeq = zeq = 0. In this case, the input is directly connected to the output, bypassing the coils.

7.3.1 Zero-Sequence Circuit Model

The zero-sequence circuit model may be derived with reference to Figure 7.5, assuming all quantities are of zero sequence. Thus, Figures 7.5b and 7.5c



should be replaced by diagrams with all phasors in parallel. Using a zero subscript for zero-sequence quantities, we can write

V V E V V IS1,0 L1,0 a ,0 S L2,0 c,0 aa , c = = + , ,1 0 0Z E ,,0

S1,0 c,0 a ,0 L1,0 a ,0 b,0I I I I I I= + = + , (7.27)

and

E E E E

I I

= = =

=

aa

aa a c

aa

aa

, , , ,

,

0 0 0 0

0

1 1NN n nNN ,, , ,0 0 0

= = n nI Ia c (7.28)

Because Ib,0 = Ic,0, Equation 7.27 shows that

I IL1,0 S= 1 0, (7.29)

so the current is not phase-shifted. We also have

V V V V E E ES S S L b ,0 b c1 0 2 0 2 0 2 0 001 1

, , , , ,, = = = = n n ,,0 (7.30)

Solving the above zero-sequence equations, we get

I E I Ic,0S1,0

c,0 c,0 aa ,0 S= =

=

( )I

nn

nZ

nn1 1 1 2

, 11,0 aa ,Z 0 (7.31)

and

V V IL1,0 S1,0aa ,0

S1,0= ( )Z

n1 2 (7.32)

The circuit model for Equation 7.32 is shown in Figure 7.7, where we define

ZZ

neq,0

aa ,0=( )

1 2 (7.33)

Zeq,0 has a different dependence on n compared to the positive-sequence circuit and becomes infinite when n = 1. This is reasonable because, from Equation 7.28, Ia,0 = Ic,0 when n = 1; therefore, from Equation 7.27, IS1,0 = 0. Thus, no zero-sequence current can flow into the squashed delta transformer when n = 1. Internal current can, however, circulate around the delta. From Equation 7.18, = 60 when n = 1.



Equation 7.32 can be rewritten in per-unit terms referring to the same input base as used for the positive sequence:

v v iL1,0 S1,0 eq S1,0= z ,0 (7.34)

Using the positive-sequence winding base for the two-winding zero-sequence impedance, Zaa,0, we can write

zn

n n nzeq,0 aa ,0= + +

31 1

2

2 2( )( ) (7.35)

This expression is primarily useful in designs in which the winding base is used for the maximum phase-shift angle.

We will postpone both a discussion of the regulation effects and calcula-tions of the short-circuit current until other types of phase-shifting trans-formers are treated because the positive-, negative-, and zero-sequence circuit diagrams for all the cases treated will have the same appearance, although Zeq or zeq and differ among the various types.

7.4 Standard Delta Phase-Shifting Transformer

As opposed to the squashed delta design, the standard delta design uses an unsquashed delta winding but is still a single-core design. The connec-tion diagram is given in Figure 7.8. The tapped windings are on the same core as the corresponding parallel windings on the delta. The taps are sym-metrically placed with respect to the point of contact at the delta vertex. This ensures that there is no change in the current or no-load voltage magnitude from the input to the output. It also means that Ea = Ea, and so on, for the other phases. Each phase really consists of three windings: the two tap windings and the winding opposite and parallel in Figure 7.8, where primes

VS1,0

Zeq,0

VL1,0

IS1,0 IL1,0

Figure 7.7Zero-sequence circuit model of one phase of a squashed delta phase-shifting transformer.



and double primes are used to distinguish them. Thus, a three-winding per-phase model is needed. Za, Za, and Za will be used to label the single-winding impedances for phase a. Because all the phases have equivalent impedances by symmetry, the same designations will be used for the other phases as well. The same remarks apply to the phasor diagrams as in the squashed delta case. Although not shown, the current phasor diagrams have the same sequence order as their corresponding voltage-phasor diagrams; however, the two diagrams can be rotated relative to each other.

(b)

(a)

(c)

VS1 VL1

VS3

V2

VL2

VL3

VL1VL3

VL2

EaEa

EcEb

Ec

IS1

VS2

VS3

V3

VS1

IS2

VS2

IL1V1

IS3IL2

IL3

Ia Ia

Ib

IaEa

IcIb

Ib

Eb

Ec

Eb

Ea

Ea

Eb

Ec

Eb

EaEc

Eb

EcIc

Ic

Figure 7.8Standard delta configuration, with the phase quantities labeled: (a) circuit, (b) input and output phasors, and (c) internal phasors. The ideal transformer voltages, the Es, increase in the oppo-site direction of the assumed current flow. The taps are symmetrically positioned.



In Figure 7.8a, V1, V2, and V3 designate the phasor voltages to ground at the delta vertices. Thus, using Ea = Ea and so on, and Ia = IS1; Ia = IL1, and so on, we have

V V I E V V I E

V V

S1 S a a L L a a = + = +

= 1 1 1 1 1

2 3

Z Z,

(

21 1

12

3 = = +

= =

)

( )

V V I E

I I I I I

j Za a a

S1 L c b aa a3= j I

(7.36)

In addition, we have the following transformer relations:

EE

I I I I I

a

a

a

a

a a a S a L a Sor

= =

+ + = + +

NN

n

N N N n1 1 10 IIL1 0=

(7.37)

where we define the turns ratio n as the ratio between the turns in one of the delta windings to the turns in one of the tap windings. Both tap wind-ings have the same number of turns. Solving for Ia in the last equation in Equation 7.36 and inserting it into the last equation in Equation 7.37, we get

I IL1 S1e with tan= =

j

n 2 31 (7.38)

We also have

I Ia S1=

2

13

n jn

(7.39)

From Equations 7.36 through 7.39, we get

E Va S a

a=

+

j

jn

j

jn

ZjZ3

13

3

13

21

33 13

1

n jn

IS (7.40)

and

V V IL1 S1 a aa

S1 e= + + +

Z ZZ

nj4

32 (7.41)



with as given in Equation 7.38. This is the no-load phase-angle shift. This conforms to the circuit model shown in Figure 7.6 with

Z Z ZZ

neq a aa= + ++

432

(7.42)

or in terms of two-winding impedances, using Equation 7.10:

Z Zn

Z Z n Zeq a a aa aa a a= + +

+ ( )

232

2 (7.43)

This is the positive-sequence circuit. Again, the negative-sequence circuit is found by changing to without any change in Zeq.

We again determine the winding power per phase Pwdg in terms of the input power per phase Pin = VS1IS1*. Similarly, ignoring the impedance drops, we get

P jn

nPwdg a a* in= = +

E I2 3

32 (7.44)

where j indicates that the transformed power is in quadrature with the input power. Using trigonometric identities, we can show that

P j Pwdg insin= (7.45)

with as given in Equation 7.38. (This relation does not hold for the squashed delta design.) Thus, for = 30, |Pwdg| = 0.5 |Pin|.

In per-unit terms, based on input quantities, Equation 7.41 can be cast in the form of Equation 7.24. Again, we need to determine the relationship between the winding bases and the input base for impedances. The two-winding impedances given above are referred to by either the a or a winding. Their bases are related to the input base by maintaining the power base the same as the input power per phase:

ZEV

nn

Z Zb,a wdga

S1b,in b,a wd= = +

22

2

33

, gga

S1b,in= = +

EV n

Z2

2

33

(7.46)

Thus, in per-unit terms, Zeq in Equation 7.43 becomes

zn

nzn

z z zeq

a a aa aa a=+

+

+ 33

222 2

aa( )+( )

n2 3

(7.47)

Again, this is primarily useful for design purposes when n refers to the maximum phase angle. At zero phase shift, n = , Equation 7.47 indicates that zeq = 0. At this tap position, the tap windings are effectively out of the circuit, and the input is directly connected to the output.




The zero-sequence circuit model can be derived with reference to Figure 7.8 but with all phasors taken to be zero sequence. Rewriting Equations 7.36 and 7.37 accordingly and appending a zero subscript, we get

V V I E V V IS1,0 S1,0 a ,0 a ,0 L L1, = + = 1 0 1 0 1 0, , ,,Z 00 a ,0 a ,0

a, a,0 a, S1

Z

Z +

= = +E

V V I E I2 0 3 0 0 00, , , ,,0 L1,0 c,0 b,0

a

a ,0

a

aa,

= =

= =

I I I

EE

I

0

0, ,NN

n n 00 0+ + =I IS1,0 L1,0

(7.48)

Solving, we find that

I IL1,0 S1,0= (7.49)

thus, the zero-sequence current undergoes no phase shift. We also have

E I I Ia,0 a,0 a,0 a, S1,0,= = Z n02 (7.50)

and

V V IL1,0 S1,0 a ,0 a ,0 a,0 S1,0= + +

Z Z n

Z4

2 (7.51)

Thus, the voltage undergoes no phase shift at no-load. The circuit model of Figure 7.7 applies here, with

Z Z Zn

Zeq a ,0 a ,0 a,0= + + 4

2 (7.52)

or, in terms of two winding impedances

Z Zn

Z Z n Zeq,0 a a ,0 aa ,0 aa ,0 a a ,0= + + 2

22(( ) (7.53)

On a per-unit basis, using rated input quantities, Equation 7.51 has the same appearance as Equation 7.34. The equivalent per-unit impedance is given by

zn

z z zeq,0 aa ,0 aa ,0 a a ,0= +

+( )

33

22

(7.54)

Again, this is primarily useful when n refers to the maximum phase angle shift. At zero phase shift, the output is directly connected to the input as was the case for the positive sequence.



7.5 Two-Core Phase-Shifting Transformer

Phase shifters for large power applications are often designed as two unitsthe series unit and the excitor unit, each with its own core and associated coils. Depending on their size, the two units can be housed inside the same tank or in separate tanks. This construction is largely dictated by tap-changer limitations. A commonly used circuit diagram is shown in Figure 7.9. The phasor diagrams refer to the positive-sequence quantities and, although the phase ordering is consistent among the dia-grams, their relative orientation is not specified. Currents have the same phase ordering as their associated voltages. The series unit uses a three-winding model, whereas the excitor unit uses a two-winding model. We have used two different labeling schemes for the series and excitor units. Subscript letters are used for the series quantities, and subscript numbers are used for the excitor quantities. Primes and double primes are used to distinguish different windings associated with the same phase. Note that the inputoutput coils in Figure 7.9 are a part of the series unit but are attached to the excitor unit at their midpoints. The input and output voltages are voltages to ground. We assume the input voltage and current phasors are given.

Following an analysis similar to that of the Section 7.4 for the series unit, and using Ea = Ea, and so on, we can write

V V I E V V I E

E E

S S Z Z1 1 1 1 1 1

2 3

= + = +

a a L L a a ,

== ( ) = = += + =

2 1 11 1 1 1 1

3E E I EV I E I

j Z

Z

a a a1 , II I I I IS1 1 2 = L c a,

(7.55)

We also have the following transformer relations:

EE

EE

II

a

a

a

as

1e, ,

= = = = = =N

Nn

NN

nNN

1

1 1

1

1

1

1

+ + = + + =

n

N N N n

e

a a a S1 a L1 s a S1 L1orI I I I I I0 0

(7.56)

where we define the turns ratio of the series unit, ns, as the ratio of the turns in a coil of the delta to the turns in the first or the second half of the inputoutput winding, that is, from the input to the midpoint or from the output to the midpoint. We have also defined the excitor-winding ratio, ne, as the ratio of the turns in the winding connected to the midpoint of the inputoutput winding to the turns in the tapped winding. The latter ratio will depend on the tap position. From Equations 7.55 and 7.56 and the phasor diagrams, we get



I I I I I I = = = = = 2 120 1 120 1 1e e c a aj e jn ( ) 33

3

30

1

e a

ea

=

j

jn

I

I I

(7.57)

which implies

I I I I I I = = = = = 2 120 1 120 1 1e e c a aj e jn ( ) 33

3

30

1

e a

ea

=

j

jn

I

I I

(a)

(b)

(c)

(d)

Series unit

Excitor unit

Ec

Ea

Ea EaEc

Ec

Eb

E3

E2

E1

E1

E2

E3

Eb Eb

Ec Ib

Ic

Ia

V1

E1

E3

E3

E1

E2VL2

IL2Eb

V2IS2

EbVS2 VL3

IL3

V3

IS3Ec

VS3

Ec

E2

I1

I3

I3

I1

I2

I2

IL1VL1VS1

ISI

Eb

Ea

EaEa

VS3VL1

VS1

VL2

VS2

VL3

Figure 7.9Circuit diagram of a two-core phase-shifting transformer: (a) circuit, (b) input and output pha-sors, (c) internal phasors for series, and (d) internal phasors for excitor. The ideal transformer voltages, the Es, increase in the opposite direction of the assumed current flow. The inputoutput coil is equally divided.



Substituting I1 from Equation 7.55 and Ia from Equation 7.56 into Equation 7.57, we get

ILe s

e s

1

13

13

=+

jn n

jn n

= =

I IS1 S1e s

e where tanjn n

2 31 (7.58)

Thus, in terms of the known input current, using Equations 7.55 and 7.56, we can write

I I1 1

23

13

=

jn n

jn n

e s

e s

S , II Ia

se s

S1=

2

13

n jn n

(7.59)

From the initial set of equations, we also get

E I E1 3= +( )j n Ze a a a (7.60)

Substituting Equations 7.57 and 7.60 into the expression for V1 in Equation 7.55, we find that

V I E1 112

13 3= +

+Z

nZ j

nea

ea (7.61)

Substituting this into the first relationship of Equation 7.55, and using Equation 7.56 and the first equation of Equation 7.59, we get

E Vae

e s

S1

s

=

+j

n

jn n

n3

13

6nn n

jn n

Zn

Ze s

e s

ea

( )

+ 2

2 11

2

13 3

+

jn

jn n

Z

3

13

e

e s

a

IS1

(7.62)

Adding the first two expressions in Equation 7.55 and using Equation 7.62, we get

V VL1 S1 a ae s

ea= + + ( ) +

+

Z Zn n

Zn

Z12

3 3211

2

IS1 e j (7.63)



where is given in Equation 7.58. Equation 7.63 can be represented with the same circuit model as Figure 7.6 by applying the following relation:

Z Z Zn n

Zn

Zeq a ae s

e

3= + +

( ) +

+ 12

3211

2

aa

(7.64)

or, in terms of the two-winding impedances, we can express it as

Z Zn n

Zn

Zeq a ae s

e2

aa6= +

( ) +

+ +

1232

11 ZZ n Zaa s a a ( )

2 (7.65)

Using Equations 7.59 and 7.60 and ignoring impedance drops, we can show that the winding power per phase is the same for the series and excitor units, that is,

P jn n

n nPwdg a a*

e s

e si= = = ( ) +

E I E I1 1 22 3

3*

nn (7.66)

where Pin = VS1IS1*. In terms of the phase shift given in Equation 7.58, Equation 7.66 can be written as

P j Pwdg insin= (7.67)

similar to the case for the standard delta phase shifter.In per-unit terms, keeping the power base constant at the rated input

power, Equation 7.63 can be cast in the form of Equation 7.24. To do this, we need to find the ratio of the impedance bases for both the series and excitor windings to the input or terminal impedance base. Using the above formu-las, we find that

Zn

n nZ

Z

b,awdga

S1

s2

e sb,in

b

= =( ) +

EV

2

2

3

3

,,a wdga

S1 e sb,in

b,1

= =

( ) +

EV

2

2

3

3n nZ

Z wwdg1

S1

e s

e sb,in= =

( )( ) +

EV

2 2

23

n n

n nZ

(7.68)



Applying these relations, the per-unit equivalent impedance can be written as

zn n

zn n

n neq

e s

a ae s

e s

=( ) +

+ ( )( ) +

3

3

4

32

2

2

+ + ( )

z z z z11

12 aa aa a a

(7.69)

The negative-sequence circuit has the same equivalent impedance, but it has a phase-angle shift in the opposite direction to the positive-sequence circuit.


The zero-sequence circuit model is derived with reference to Figure 7.9 by assuming that all quantities are zero sequence. Thus, we simply rewrite the basic equations, appending a zero subscript:

V V I E V V IS1,0 S1,0 a ,0 a ,0 L L1,, = + = 1 0 1 0 1 0, , ,Z 00 a ,0 a ,0

a,0 a,0 a,0

Z

Z

+ = = +

E

E E I E

V2 0 3 0

1

0, ,,, , , , , ,0 1 0 11 0 1 0 1 0 2 0= + = = I E I I I IZ , ,S1,0 L1,0 II Ic a,0,0 0 =

(7.70)

and

EE

EE

Iaa

a

as e

,

,

,

,

,, ,00

1 0

1 0

1

1

1 0

= = = =NN

nNN

nII

I I I1 0

1

1

0 1 0 1 0 0,

, , ,

= =

+ + =

NN

n

N N N

e

a a a S a L orr ns a S LI I I, , ,0 1 0 1 0 0+ + =

(7.71)

Solving, we find that

I I I I I

E I

1 0 1 0 1 002

2

, , ,= = =

=

, ,S L a,0s

S1

a,0s

S1,

n

n 00 a,0,Z V1 0 1 0, ,= E

(7.72)

Notice that, even if both the Y windings of the excitor are grounded, no zero-sequence current flows into the excitor because the secondary current from the tap winding would have to flow into the closed delta of the series unit, and this is not possible for zero-sequence currents.

From the above formulas, we get

V V IL1,0 S1,0 a ,0 a ,0s

a,0 S1,= + +

Z Zn

Z4

2 00 (7.73)



Thus, we see that the zero-sequence current and no-load voltage have no phase angle shift and the circuit of Figure 7.7 applies with

Z Z Zn

Zeq,0 a ,0 a ,0s

a,0= + + 4

2 (7.74)

or, in terms of two-winding impedances

Z Zn

Z Z n Zeq,0 a a ,0s

aa aa s a a= + + ( ) 22 2 (7.75)

Using the same basis as for the positive sequence, the per-unit version of this equation is

zn n

z z zeq,0e s

aa aa a a= ( ) + +( )

3

322 (7.76)

7.6 Regulation Effects

Because all the phase-shifting transformers examined here have the same basic positive-sequence (as well as negative- and zero-sequence) circuits (Figure 7.6), with different expressions for Zeq or zeq and , the effect of a load on the output can be studied in common. The relevant circuit model is shown in Figure 7.10.

We assume a balanced positive-sequence system. Since all phases are identical, we exclude the phase subscript. We use per-unit quantities for this development, which can be accomplished by simply changing from upper- to lowercase letters and using a unit turns ratio. The ideal transformer is included in the figure to account for the phase shift. From the figure, we see that

v v i i iL S eq S L L S Le e= ( ) = = z z zj j (7.77)

iLiS 1:e j

vS

zeq

zLvL

Figure 7.10One phase of a phase-shifting transformer under load, represented using per-unit quantities.



where zL is the per-unit load impedance. Solving for iS, we find that

iv

SS

eq L

=+( )z z (7.78)

so that

v v vL Seq

eq L eq

L

e= +( )

=+

1

1

1

z

z z z

z

jS

e j (7.79)

Thus, any shift in the phase angle or magnitude from the no-load conditions is due to a nonzero zeq/zL. Because zeq is almost entirely inductive, a purely inductive or capacitive load will not affect but will result only in a magni-tude change in the voltage. On the contrary, a resistive or complex load will lead to shifts in both magnitude and phase angle.

Under no-load (NL) conditions (zL = ), Equation 7.79 shows that

v vL,NL Se= j (7.80)

Thus, taking ratios, we see that

v

vL

L,NL eq

L

=+

1

1z

z

(7.81)

Therefore, the presence of a load will lower the magnitude and shift the phase, depending on the phase of the above ratio of impedances. This analy-sis is similar to that given in Chapter 3, except there we ignored the shift in phase. Because these transformers are designed to shift the phase by a given amount under no-load conditions, finding the additional shift in this phase caused by transformer loading is important. Let zeq/zL = pej. Then, Equation 7.81 becomes

v

v p p

jppL

L,NL

tansincose=

+ +

+

11 2 2

11

cos

(7.82)

As an example, let zeq be a 10% inductive leakage impedance and zL be a 100% resistive load. Then, zeq/zL = 0.1ej90 and vL/vL,NL = 0.995 5.71, where indi-cates the angular dependence of the complex number. This angle of nearly 6 will be subtracted from the phase angle of the transformer output, so the behavior of the phase under load is quite important. In the above example,



the magnitude will only be lowered by 0.5%. Because the transformer imped-ance varies with the tap setting (or phase shift), the angle shift due to loading also varies with the tap setting.

7.7 Fault Current Analysis

The phase-shifting transformers discussed in this section are all two- terminal transformers that have a single reactance, Zeq or Zeq,0, for positive- or negative- and zero-sequence circuits, respectively. We assume that the positive- and negative-sequence reactances are the same and do not distin-guish them with subscripts. Thus, the theory developed in Chapter 6 can be applied to these phase-shifting transformers with the recognition that neg-ative-sequence currents have a phase shift opposite of the positive-sequence currents. We retain the notations source, S, and load, L, used above instead of H and X. We assume the fault is on the L or output terminal. The fault cur-rents at the fault position are the same for the various fault types, as shown in Chapter 6.

The faults of interest are as listed below:

1. Three-phase line-to-ground fault 2. Single-phase line-to-ground fault 3. Line-to-line fault 4. Double line-to-ground fault

We will summarize the fault currents for the above types here. We will use per-unit notation.

For fault type 1, we have

iv

zi ia

pfa2 a0,1

1

0= = = (7.83)

For fault type 2, we assume that the a-phase line is shorted, so that

iv

z z zi ia1

pfa2 a0= + +( ) = =0 1 2 (7.84)

For fault type 3, assuming that the b and c lines are shorted together, we have

iv

z zi ia1

pfa2 a,= +( ) = =1 2 0

0 (7.85)



For fault type 4, assuming that the b and c lines are shorted to ground, we get

i vz z

z z z z z z

i vz

z

a1 pf

a2 pf

= ++ +

=

0 2

0 1 0 2 1 2

0

00 1 0 2 1 2

2

0 1 0 2 1

z z z z z

i vz

z z z z z z

+ +

= + +a0 pf 22

(7.86)

Assuming equal positive and negative reactances for the transformer and the systems, we have the following expressions for the Thevenin impedances:

z zz z z

z z zz

z z1 2 0

0 0= =+( )

+ +=SL eq SS

eq SS SL

SL eq,

, , ++( )+ +

z

z z zSS

eq SS SL

,

, , ,

0

0 0 0

(7.87)

The per-unit currents feeding the fault from the transformer side are given by

i iz

z zi i

zz zeq a eq SS

eq2 a2eq SS

,1 11 1=+

= +

=+

i i

zz zeq0 a0 eq,0 SS,0

0 (7.88)

Figure 7.11 shows the terminal circuit diagrams for the three sequences, cov-ering all the fault types. For simplicity, we assume that no current flows in the prefault condition. These can always be added later.

As shown in Figure 7.11, the per-unit currents flowing out of the trans-former into the fault are on the load side of the transformer and are given by Equation 7.88 in terms of the fault sequence currents. Thus, in terms of finding the fault currents in the individual windings, we need to find their expres-sions in terms of the load currents. However, in a phase-shifting transformer, positive- and negative-sequence currents experience opposite phase shifts. This makes the analysis of fault currents within these transformers quite dif-ferent from the fault currents in a standard two-winding transformer.

We have already considered the currents in the windings of a phase-shifting transformer when we derived their equivalent impedance for positive-, negative-, and zero-sequence circuits. Now, we use those results and the fact that negative-sequence currents within a phase-shifting trans-former can be obtained from the positive-sequence currents by changing their phase shift to its negative. Note that ieq in Equation 7.88 equals iS1 in per-unit terms. We now apply the fault formulas from Equation 7.88 to the phase-shifting transformers described in Sections 7.3 through 7.5. Because the per-unit notation is used, we need to express the formulas for the currents



in per-unit terms. Considering the lengths of the formulas, we only present the results for the winding-sequence currents.

7.7.1 Squashed Delta Fault Currents

From Equation 7.17, we see that

I I I =

=

a S1 L1e ee

nn

nnj j

j120 120

(7.89)

where

= +

23

2 11tan

n (7.90)

Using Equation 7.90, Equation 7.89 can be manipulated to the following form:

I I=

+ +a L1e

nn n

j

22

1

(7.91)

(a)

(b)

(c)

ZSS

ZSS

ZSS,0

Zeq

Zeq

Zeq,0

ZSL

ZSL,0

ZSL

eSLia1

ia2

ia0

iL1

iSL2

iSL0

ieq1

ieq2

ieq0

Va1

Va2

Va0

eSS

Figure 7.11Sequence circuits showing system impedances and transformer equivalent impedances. The fault is on the load-side terminal: (a) positive sequence, (b) negative sequence, and (c) zero sequence.



To convert this expression to the per-unit form, we need the ratios of the cur-rent bases. These are the reciprocals of the ratios of the voltage bases, assum-ing the same power level for both bases. This is given by

II

VV n n

b,L1

b,a

b,a

b,L1

= =+ +3

12 (7.92)

Using this relation, we can convert Equation 7.91 to the per-unit form:

i i=

+ +a L1e

nn n

j312

2

(7.93)

Because this is a two-winding transformer, ia = ia.For the zero-sequence currents, from Equations 7.28 and 7.31, we have

I I I = = a ,0 c,0 L1n

nn 1

(7.94)

Note that the zero-sequence current is the same in all the phases. Using the base transformation given in Equation 7.92, this can be converted to per-unit form:

i i = + +a ,0 L1n

n n n3

1 12( ) (7.95)

We also have ia0 = ia0.Using Equation 7.88 and the fact that negative-sequence currents have

an opposite phase shift from positive-sequence currents, we can find the sequence currents in the windings as follows:

i = + +

+

a ,1 a

eq SS

ein

n nz

z zj

1 2 213

1

=+ +

+

ia ,2 a

eq SS

ein

n nz

z zj

2 2 213

1

= + +

+

ia ,0 aeq,0 SS

in

n n nz

z z0 203

1 1( )

,,0

(7.96)

Here, ia1, ia2, and ia0 are the fault sequence currents for the various fault types described in Chapter 6. These can then be transformed to the phase or wind-ing currents by using Equation 6.1.



7.7.2 Standard Delta Fault Currents

This transformer has three windings. From Equation 7.39, we have

I I Ia S1 L1e=

=

=2

13

2

13

n jn

n jn

j +

232

2

nj

e L1I (7.97)

where

I IL1 S1e and tan= =

j

n 2 31 (7.98)

Converting to the per-unit system, we find for the ratio of current bases:

II

EV

nn

b,L1

b,a

a

L1

= =+3

32 (7.99)

Using this ratio, Equation 7.97 becomes (in per-unit terms)

i ia L1e= +2 3

322

nn

j

(7.100)

We also have

I I I I = =a L1 a L1e ,j (7.101)

The ratio of current bases is the same for the currents in Equation 7.101 and is given by

II

EV n

b,L1

b,a'

a

L1

= =+

332

(7.102)

Using this, Equation 7.101 becomes (in per-unit terms),

i i i i = +=

+a L1 a L1e ,

33

332 2n n

j (7.103)

Using trigonometric identities and the formula for from Equation 7.98, we can show that ia + ia + ia = 0.



For the zero-sequence currents, from Equations 7.49 and 7.50, we get

I I I Ia,0 L1,0 a ,0 L1,0 a ,0 L1,0, ,= = = 2n

I I (7.104)

Putting these on a per-unit basis, using the same current bases as for the positive-sequence case, we get

i i i i ia,0 L1,0 a ,0 L1,0 a ,0, ,= +=

+=

2 33

33

32 2n n nn2 3+

iL1,0 (7.105)

These zero-sequence per-unit currents also sum to zero, as expected.Using Equation 7.88 and noting that ieq corresponds to iL1, we get the

following expressions for the winding sequence currents:

i ia,1 a1eq SS

a,2e ,= + +

=

i

nn

zz z

j2 332

2 1

iin

nz

z z

in

ja2

eq SS

a,0 a0

e2 3

3

2 3

22 1

2

+ +

=

i++ +

=+

3

33

0

2

1

zz z

in

zz

j

eq,0 SS,0

a ,1 a1 ei eeq SS

a ,2 a2eq SS

, e+

= + +

z

in

zz z

ji3

321

=+ +

i

i

a ,0 a0eq,0 SS,0

a ,1

in

zz z

332

0

==+ +

= +

in

zz z

in

za1

eq SSa ,2 a2,

33

332

1

2

1izz z

in

zz z

eq SS

a ,0 a0eq,0 SS,0

+

=+ +

i

332

0

(7.106)

The per-unit currents for the various fault types can be inserted for ia1, ia2, and ia0, as shown in Chapter 6. Accordingly, one can then get the winding phase currents in per-unit terms by applying Equation 6.1 for the various fault types.

7.7.3 Two-Core Phase-Shifting Transformer Fault Currents

We will quote the results for this transformer based on the formulas in Section 7.5. This transformer has five windings, labeled 1, 1, a, a, and a. The per-unit winding currents, expressed in terms of the output L1 current, are given by



i i i i1 2 2 1 12 3

32 3=

( ) += = j

n n

n n

nje s

e sL a

ee

, nn

n n

n n

j

j

s

e sL

a

e s

L

e

e

( ) +

=( ) +

22 1

2 1

3

3

3

i

i i , ii i

i i

=( ) +

= = =

a

e s

L

a

e

3

3

02 3

2 1

1 0 1 0 0

n n

in

, , ,, nn

n n

s

L

a

e s

L a

( ) +

=( ) +

=

2 1 0

0 2 1 0 0

3

3

3

i

i i i

,

, , ,

(7.107)

The winding sequence currents are given by

i i i1 1 2 2 1 1 1 22 3

32

, , ,, = ( ) += j

n n

n ne j

je s

e sL

333

0

2

22 1 2

1 0 1 0

1

n n

n ne

i

jL

e s

e s

a

( ) += =

=

i

i

i

,

, ,

,33

32 3

22 1 1 2

n n

n ne

n n

n nje s

e sL a

e s

e s( ) +=

i i, ,, (( ) +

= ( ) +

=

22 1 2

0 2 1 0

1

3

2 3

3

e

n n

ji

i i

i

L

a

e s

L

a

,

, ,

,33

3

3

321 1 2 2 1n n

en n

ej j

e s

L a

e s

L( ) +

=( ) +

i i i, ,, ,,

, ,

,

2

0 2 1 0

1 2

3

3

3

3

i i

i i

=( ) +

=( ) +

a

e s

L

a

e s

L

n n

n n 11 1 2 0 2 1 0

33, , , ,

,( )

= =+

i i ia ae s

L n n

(7.108)

Substituting the ieq currents from Equation 7.88 for iL1 into the formulas in Equation 7.108, we get the sequence currents in all the windings. Note that j in Equation 7.108 is a rotation by 90 in the complex plane, so in the negative-sequence current formula, it has to be replaced by j for a 90 rotation in the opposite direction.



7.8 Zig-Zag Transformer

A delta HV (H) winding and zig-zag low-voltage (X) winding transformer is analyzed. The complex currents are calculated, as well as the terminal positive- or negative- and zero-sequence impedances. Although explicit expressions for these impedances are given in terms of the two-winding impedances, they can also be obtained directly from a program that calcu-lates the magnetic field associated with complex currents. These impedances can then be used to obtain fault currents for a short-circuit analysis.

The three-phase connections for the delta primary and zig-zag second-ary windings are given schematically in Figure 7.12. Parallel lines denote the windings on the same phase. The delta winding phases are denoted a, b, and c. The zig-zag windings consist of windings from the two phases connected in series. Thus, the zig-zag X1 terminal is attached to a series combination of windings from the b and c phases. The orientation of the zig-zag windings in Figure 7.12 is such that the terminal voltage of the X terminals to ground is in phase with the terminal voltage of the delta ter-minals V to ground. This requires the two separate windings of the zig-zag to have the same number of turns. A 180 phase shift between X and V can be achieved by winding the X windings in the opposite sense of the delta windings.

Figure 7.13 shows the positive-sequence phasors associated with the wind-ing and terminal voltages of the delta winding. Ea, Eb, and Ec are a positive-sequence set of winding voltages, and V1, V2, and V3 are a positive-sequence set of terminal-to-ground voltages. The Es are no-load voltages due to the core excitation; the diagram in Figure 7.13 does not take into account the voltage drops caused by the currents and impedances. For the voltage drops,

a

b cc

b

ab c

a

V1

V2V3

X1

X2X3

Figure 7.12Schematic three-phase connection diagram of the zig-zag transformer.



we use a three-winding T-equivalent circuit model. The single- winding impedances are the same for all the three phases. Concentrating on the a phase, the winding voltages are given by

V E I

V E Ia a a a

a a a a

Delta HVZig-Y LV

= += +

Z

Z

( )( ))( )V E I = +a a a a Zag-Y LVZ

(7.109)

where the currents are taken to be positive into the winding terminals. The Es are proportional to the winding turns and increase in the opposite sense to the current direction.

The connection diagram of the winding for all the three phases is shown in Figure 7.14. The subscripts a, a, and a refer to the delta, zig, and zag wind-ings, respectively, of the a phase. The b and c subscripts have the same sig-nificance for the b and c phases. The zig and zag windings are connected in such a way that their phase voltages are subtractive. This gives the maxi-mum net voltage across both windings.

7.8.1 Calculation of electrical Characteristics

Like the voltages, the currents I1, I2, and I3 and Ia, Ib, and Ic form three-phase sets, as shown in Figure 7.14. From this figure, we have

I I I

I I I

I I I

1

2

3

= = =

c b

a c

b a

(7.110)

Ea

Eb

Ec

V2

V1

V3

Figure 7.13Phasor diagram of the delta terminal and winding voltages.



From Figure 7.14, we also have

V V E I

V V E I

V V E I

1 2

3 1

2 3

= + = + = +

c c a

b b a

a a a

Z

Z

Z

(7.111)

Note that Za = Zb = Zc; similar is the case for Za and Za. Therefore, we will only use the subscripts a, a, and a for the Zs

For the X1 terminal, from Figure 7.14, we see that

V E I E IX1 c c a b b a= + Z Z (7.112)

We let N1 = the number of turns in the HV delta winding and N2 = the num-ber of turns in the zig winding = the number of turns in the zag winding; thus, we have

EEEE

a

a

a

a

= =

=

NN

n12

1

(7.113)

where we define n as the ratio of turns in the delta winding to turns in the zig or zag winding. Because of the phase relationship among the constitu-ents of a positive-sequence set, we can write

V1

V2V3

VX1

VX3

VX2

Ib

Ea+ Ia

Eb+

Ic

Ec+

IaIa

EaEa

IcIc

EcEcIb

Ib

EbEb

I3 I2

I1

Figure 7.14Detailed connection diagram of the three-phase zig-zag transformer, with the electrical quan-tities labeled.



E E E E E = ( ) = =c b a a ae ej j j j n120 120 3

3 (7.114)

From Equation 7.111, we have

E V V I V Ia a a a ae e= ( ) = ( ) = 2 3 120 120 1 3Z Z jj j VV I1 a aZ (7.115)

Substituting from Equations 7.114 and 7.115 into Equation 7.112, we get

V V I I IX1 a a c a b a= + 3 3

1nj

nZ Z Z (7.116)

Due to the continuity of the current in the zig and zag windings, we must have

I I

I I

I I

= = =

c b

a c

b a

(7.117)

so that Equation 7.116 can be rewritten as

V V I IX1 a a c a a= + +( ) 3 31n j n Z Z Z (7.118)

Since the currents form a balanced set, we have

I I =c ae j120 (7.119)

From amp-turn balance, we have

N N Nn

1 2 2 00

I I I

I I Ia a a

a a a

+ + =+ + =

(7.120)


I I Ia b ae = = j120 (7.121)

Thus, from Equations 7.119 through 7.121, we get

I I = c ajn3

(7.122)

Using this result, Equation 7.118 becomes

V V IX1 a a a a= + +( )

3 331

2

nj

nZ

nZ Z (7.123)




I I I1 120 120 3= ( ) = e e a aj j j (7.124)

Substituting this into Equation 7.123, we get

V VI

X1 a a a= + +( )

331

12

n nZ

nZ Z (7.125)

Thus, under no-load conditions, we see from Equation 7.125 that the terminal voltages are related by

V VX1 1=3n

(7.126)

This shows that the no-load terminal voltages H and X are in phase, as desired. From Equations 7.122 and 7.124, the current into the X1 terminal is

I I IX1 c 1= = n3

(7.127)

Thus, we see that the power into the X1 terminal, ignoring the impedance drop, is

V I V IX1 X1 1 1* *= (7.128)

where * indicates complex conjugation. This equation shows that the power flows out of the X1 terminal and that it equals the power into the V1 terminal. From Equations 7.115 and 7.124, ignoring the impedance drop, we also get

E I V Ia a* *= 1 1 (7.129)

so that the delta winding power equals the input power. The a-phase wind-ing currents can be obtained from Equations 7.120 and 7.121.

I I

I I

= +

=

a a

a a

nj

nj

12 2 3

12 2 3

(7.130)

Multiplying Equation 7.125 by n/3, we get

n Zn

Z Z3 3 31

12

V VI

X1 a a a= + +( )

(7.131)



Thus, as seen from the high-voltage (HV) side, the effective impedance, Zeff, is given by

Z Zn

Z Zeff a a a= + +( )

13 3

2

(7.132)

7.8.2 Per-unit Formulas

At this point, it is desirable to represent everything on per-unit bases. We will use the input power per phase as the power base. Using Equation 7.129, this also equals the delta winding power. Thus, Pb = V1I1 = EaIa. The aster-isks are not needed because these quantities are real. Thus, V1 and I1 are the voltage and current bases for the H terminal. We label these as Vb,1 and Ib,1. In terms of these and from Equations 7.126 and 7.127, the base voltage and current at the X1 terminal are

Vn

V

In

I

b,X1 b,1

b,X1 b,1

=

=

3

3

(7.133)

Dividing Equation 7.131 by Vb,1 and using Equation 7.133, we find that

v v iX1 eff= 1 1z (7.134)

where lowercase letters have been used for per-unit quantities, and

zZZeff

eff

b,1

= (7.135)

where

ZVI

VPb,1

b,1

b,1

b,12

b

= = (7.136)

Since the H winding power equals the terminal power, we can resort to a winding base to calculate zeff. From Equations 7.115 and 7.124, ignoring impedance reductions, we get

V V

I I

ZVI

VP

b,a b,1

b,a b,1

b,ab,a

b,a

b,a2

b

=

=

= = =

3

13

33Zb,1

(7.137)



For the a and a windings, using Equation 7.113 and maintaining the same power base, we have

V V

Vn

ZVP

Z

b,a b,ab,a

b,a b,ab,a

b

b,

= =

= = =Z2

aa

n2

(7.138)

From Equations 7.135 and 7.132, we can write

zZZ

ZZ

ZZ

n Zeff

eff

b

b a

b

a

b a

b= =

+

,

,

, ,

1 1

213 3

,,

,

,

, ,

ab

b a

b a

a a

bZZZ

Z ZZ1

+( )

aa

(7.139)

Using Equations 7.137 and 7.138, we can put each winding on its own base to get

z z z zeff a a a= + +( ) 13 (7.140)

In terms of two-winding per-unit impedances, we have the standard formulas

zz z z

zz z z

aaa aa a a

aaa a a aa

= +

= +

2

2

zzz z z

= + a aa a a aa2

(7.141)

Substituting these into Equation 7.140, we get

zz z z

effaa aa a a= +

2 6 (7.142)

Thus, the effective terminal impedance can be derived from the two- winding impedances.

7.8.3 Zero-Sequence impedance

We can use a similar analysis to get the zero-sequence impedance, using a zero subscript to label the zero-sequence quantities. Thus, instead of Equation 7.112, we have

V E I E IX1,0 c ,0 c ,0 a ,0 b ,0 b ,0 a ,0= + Z Z (7.143)



However, Ec,0 and Eb,0 are now in phase and have the same magnitude, and therefore, they cancel each other in Equation 7.143. In addition, using Equation 7.117, which also applies to zero-sequence currents, Equation 7.143 becomes

V IX c a a1 0 0 0 0, , , ,= +( ) Z Z (7.144)From Equation 7.117 and the equality of the zero-phase currents in the three phases, we have

I I I = = a b a, , ,0 0 0 (7.145)

Hence, applying Equation 7.120 to zero-phase currents, we get

Ia,0 = 0 (7.146)

Thus, no zero-sequence current flows in the delta HV winding. Since IX1,0 = Ic,0, Equation 7.144 becomes

V IX1,0 X1,0 a ,0 a ,0= +( ) Z Z (7.147)Reverting to per-unit quantities, we divide Equation 7.147 by Vb,X1 to get

v iX1,0 X1,0 eff,0= z (7.148)

where

zZ Z

Zeff,0a ,0 a ,0

b,X1

=+

(7.149)

and

ZV

PVV

VP

Zb,X1b,X12

b

b,X1

b,a

b,a2

b

= =

=

2

3 bb,a (7.150)

Substituting this into Equation 7.149 and using Equation 7.141, we get

z z z zeff,0 a ,0 a ,0 a a ,0= +( ) = 1313

(7.151)

Thus, we can find the effective zero-sequence impedance from the per-unit two-winding zero-sequence impedance between the zig and zag windings. Normally, we calculate the positive-sequence impedance and adjust it, via an empirical factor, to get the zero-sequence impedance.



7.8.4 Fault Current analysis

The phase-sequence diagrams are shown in Figure 7.15. We will calculate the sequence currents in the three windings in terms of the fault-sequence currents, which depend on the fault type. The phase currents can then be obtained by the transformation given in Equation 6.1. We will consider equal positive- and negative-sequence reactances for the transformer and the sys-tems and will not distinguish them with subscripts. In addition, we will con-sider an X-terminal fault, which is fed from the HV line.

The Thevenin impedances, in terms of those given in Figure 7.15, are

z zz z zz z z

zz z

1 2 00= =

+( )+ +

=SX eff SHeff SH SX

SX ef, , ff SH

eff SH SX

, ,

, , ,

0 0

0 0 0

+( )+ +

z

z z z (7.152)

iX,2

iX,0

ia2

ia0

ZSX

ZSX,0

ZSH

eSH eSX

iX,1

ia1

ZSXZef f

Zef f

Zef f,0

ZSH

ZSH,0

(c)

(b)

(a)

Figure 7.15Sequence circuits for the zig-zag transformer: (a) positive sequence, (b) negative sequence, and (c) zero sequence.



zSH,0 must be set equal to zero in the formulas. The per-unit sequence cur-rents feeding the fault from the transformer side are given by

i iz

z zi i

zz zX,1 a1 eff SH

X,2 a2eff S

,=+

=+

1 1HH

X,0 a0eff,0 SH,0

=+

i i

zz z

0 (7.153)

We need to find the per-unit fault-sequence currents in the windings in terms of the fault-sequence currents out of the X1 terminal. The number 1 preceded by a comma in Equation 7.153 refers to a positive sequence, and so forth, whereas X1 refers to the X terminal number 1. From Equations 7.124, 7.127, and 7.130, we find that

I I I I I Ia X1 a a a a, ,= = +

= j nn j n

3 12

12 3

12

j1

2 3 (7.154)

From Equations 7.133, 7.137, and 7.138, the current bases are related by

II

n II

II

b,X1

b,a

b,X1

b,a'

b,X1

b,a

,= = =3

13

(7.155)

Using these base-current relations, Equation 7.154 can be written in per-unit terms as

i i i i i ia X1 a X1 a X1, ,= = +

= j j j j12

12 3

12

j1

2 3 (7.156)

The three per-unit currents sum to zero as expected.Replacing X1 by X,1, X,2, and X,0 for the sequence currents obtained from

Equation 7.153 for the different fault types, we can obtain, from Equation 7.156, the fault-sequence currents in the various windings. From these, we can then obtain the phase currents from Equation 6.1.


Chapter 7 Phase-Shifting and Zig-Zag Transformers7.1 Introduction7.2 Basic Principles7.3 Squashed Delta Phase- Shifting Transformer7.3.1 Zero- Sequence Circuit Model

7.4 Standard Delta Phase- Shifting Transformer7.4.1 Zero- Sequence Circuit Model

7.5 Two- Core Phase- Shifting Transformer7.5.1 Zero- Sequence Circuit Model

7.6 Regulation Effects7.7 Fault Current Analysis7.7.1 Squashed Delta Fault Currents7.7.2 Standard Delta Fault Currents7.7.3 Two- Core Phase- Shifting Transformer Fault Currents

7.8 Zig- Zag Transformer7.8.1 Calculation of Electrical Characteristics7.8.2 Per- Unit Formulas7.8.3 Zero- Sequence Impedance7.8.4 Fault Current Analysis

Documents

Phase-Shifting and Zig-Zag Transformers