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Appendices

A. A Review of Transformation and Symmetrical Components

A.l The Transformation Matrix A 385

A.l The Transformation Matrix A

The transformation matrix A is given by

Its inverse is

The complex conjugate is

( 1 1

A*= 3A- 1 = 1 a 1 a2

The product AA - 1 gives

~)

Some characteristics of the transformation matrix A are:

• A is complex • A is symmetric => AT = A • The magnitude of each term is 1, iaii = 1 • The determinant of A is det(A) = 3 (a- a2 ) = J3 J3 = Jffi = J5.1962 • The determinant of A - 1 is det(A - 1) = 1/det(A) = (a2 -a)/9 = -)0.19245 • a= 1L120° • a2 = a* = 1L - 120° • a3 = 1 • a +a2 = -1 • a- a2 = JJ3 • a2 - a= -JJ3 • 1 +a+a2 = 0 • a+ 1 = -a2

• a2 + 1 =-a • a - 1 = J3L150° • a2 - 1 = J3L - 150°.

386 A. A Review of Transformation and Symmetrical Components

A.2 Symmetrical Components

• Phase voltages in terms of sequence voltages: V abc = A Vo12

• Sequence voltages in terms of phase voltages: Vo12 = A - 1 V abc

• Phase currents in terms of sequence currents: labc = A l012

• Sequence currents in terms of phase currents: lo12 = A - 1 labc

( Io ) 1 ( 1 1 1 ) ( Ia ) h = 3 1 a a2 h h 1 a2 a Ic

• Phase impedance matrix in terms of the sequence impedance matrix

Vabc =A Vo12

=A Zo12 l012

= A Zo12 A -l labc

= Zabc labc

=:? Zabc =A Zo12 A - 1

• Sequence impedance matrix in terms of the phase impedance matrix

Vo12 =A - 1 Vabc

=A - 1 Zabc labc

= A - 1 Zabc A lo12

= Zo12 lo12

"* Zo12 =A - 1 Zabc A

A.2 Symmetrical Components 387

• Phase admittance matrix in terms of the sequence admittance matrix

Iabc =A lo12

=A Yo12 Vo12

=A Yo12 A - 1 Vabc

= Yabc Vabc

=? Yabc =A Yo12 A - 1

• Sequence admittance matrix in terms of the phase admittance matrix

lo12 = A - 1 Iabc

=A - 1 Yabc Vabc

=A - 1 Yabc A Vo12

= Yo12 Vo12

=? Yo12 =A - 1 Yabc A

• In = I a + h + Ic = 3 Io =? The absence of a neutral wire means that zero sequence currents are not present.

• For a balanced Y-connected system (Ia + h + Ic = 0) or in the absence of a ground connection, zero sequence currents are not present (Io = 0), so that

( Ia ) ( 1 1 h 1 a2

Ic 1 a

which can be re-written as

One can now solve for Ia 1 and Ia2 as

or

1 ( 1L- 30° = J3 1L30°

) ( ~: )

) ( ~:)

388 A. A Review of Transformation and Symmetrical Components

or

1 ( 1L150° = J3 1L -150°

1L -150° 1L150°

• Vng = Znln = Zn(3lo) = 3Znlo => Zn in the neutral appears as 3Zn in the zero sequence.

• Va + Vb+ Ve = 3 Va0 => Phase voltages contain no zero sequence components if Va + Vb + v;, = 0.

• The sum of line voltages is always zero regardless of unbalance, that is

Vab + Vbe + V ca = 3 Vab0 = 0 => Vab0 = 0.

This implies that zero sequence components are never present in line voltages.

• A balanced system (Va + Vb + Ve = 0) contains only positive sequence components.

Vabe= ( VL~120o) => Vo12=A-1 Vabe= ( ~O) V L120°

• If a single phase voltage source is connected to all three phases, then only zero sequence voltage components are present.

• For aLl, - Zero sequence currents circulate inside a Ll but are never present in the

line. - Positive sequence phase currents appear shifted in the line by -30°. - Negative sequence phase currents appear shifted in the line by 30°.

la =lab- lea = lab0 + lab1 + lab2 - (lca0 + lea1 + lea2 )

= labo + lab1 + lab2 - (!abo + lab1 L120° + lab2 L - 120°)

= J3 (Iab1 L- 30° + lab2 L30°)

= lao + lal + la2

which implies that

lao= 0,

A.2 Symmetrical Components 389

• For a Y, - Zero sequence voltages are never present in the line. - Positive sequence phase voltages appear shifted in the line by 30°. - Negative sequence phase voltages appear shifted in the line by -30°.

Vab = Va- Vb = Vao + Va 1 + Va 2 - (Vbo + Vb1 + Vb2) = Vao + 'Va 1 + Va2 - (Va0 + Va 1 L- 120° + Va2 L120°)

= /3 (Va1L30o + Va2L- 30°)

= Vabo + Vabl + Vab2

which implies that

Vabo = 0,

B. Phase and Sequence Admittance Matrices for Three-Phase Transformers

In this Appendix, we provide derivations for the sequence and phase admittance matrices of the different vector groups of Dy and Y d transformers. We will designate the primary by ABC and the secondary by abc.

Let a be the transformation ratio, that is the ratio of secondary to primary line voltages, and N be the turns ratio, that is the ratio of secondary to primary phase voltages, then

For a Y -.1 transformer,

In pu,

For a .1-Y transformer,

In pu,

a=l

N=-1 v'3

(B.l)

(B.2)

(B.3)

(B.4)

(B.5)

(B.6)

(B.7)

(B.8)

392 B. Phase and Sequence Admittance Matrices for Three-Phase Transformers

For a Y-Y transformer,

In pu,

a=l

N=l

For a L1 - L1 transformer,

In pu,

a=l

N=l

(B.9)

(B.lO)

(B.ll) (B.l2)

(B.l3)

(B.l4)

(B.15) (B.16)

B.l Y 9 - <:1 Transformers 393

B.l Y g - L1 Thansformers

The different vector groups for Y -Ll transformers are shown in Fig. B.l.

Ydl Yd3

Ydll Yd7

J. -~ j_ ·~ < • vb < • V'

' . Yd3 Yd9

Fig. B.l. Vector groups for Y -.1 transformers


B.l.l Y 9 dl Transformers

Consider the Y9 d1 transformer of Fig. B.2. The phase admittance matrix relates the currents and voltages as I = Y · V. With y being the transformer admittance in pu, one can write

!A= y (VA- Vab/v/3) = y[VA- (Va- Vb)jv/3]

IB = y (VB - Vbcfv/3) = y[VB - (Vb- \/;:)jv/3]

Ic = Y (Vc- Vca/v/3) = y[Vc- (Vc- Va)/v/3]

Ia = ha- lac= (/A- fc)/v/3

= y [(VA- Vc)fv/3- (2 Va- Vb- Vc)/3]

h =feb- ha = (/B- !A)/v/3

= y [(VB- VA)/v/3- (2Vb- Vc- Va)/3]

fc =lac- feb= (Ic- IB)/v/3

= y [(Vc- VB)/v/3- (2 Vc- Va- Vb)/3]

In matrix form,

Ic =y

-h

A

1 0 0 -1 v'3

0 1 0 0

0

-1 v'3

1 v'3

0 1

0 )J

1 v'3

2 3

~ 0 -t

1 v'3

-1 v'3

0

1 -3

2 3

0

1 v'3

-1 v'3

1 -3

1 -3

2 3

VA

VB

Vc

Va

vb

Vc

A -lA

B -Ia

- a I,

~' c - C B

Ic

Fig. B.2. A Y 9dl transformer

(B.17a)

(B.17b)

(B.17c)

(B.17d)

(B.17e)

(B.17f)

(B.18)

a

b

B.l Y 9 - .1 Transformers 395

The sequence admittance matrix Yo12 can now be calculated as Y 012 = A - 1 Y A and relates the currents and voltages as

!Po 1 0 0 0 0 0 VPo

IPl 0 1 0 0 e 0 VPl

IP2 0 0 1 0 0 f VP2

I so =y (B.19)

0 0 0 0 0 0 Vso

Is1 0 f 0 0 1 0 Vs1

ls2 0 0 e 0 0 1 VS2

where e = -1L30°, f = -1L- 30° and

1 1 1 0 0 0

1 a2 a 0 0 0

1 a a2 0 0 0 a= 1L120° A= '

(B.20) 0 0 0 1 1 1 a2 = a* = 1L - 120°

0 0 0 1 a2 a

0 0 0 1 a a2

From the matrix equation for Yo12, the primary and secondary sequence currents are given by

fpo = Y. Vpo

JP1 = Y ' (Vp1 - Vs 1 L30°)

lp2 = Y · (Vp2 - Vs 2 L- 30°)

180 = 0

fs 1 = Y · (Vs 1 - Vp1L- 30°)

ls2 = y · (Vs2 - Vp2 L30°)

(B.21a)

(B.21b)

(B.21c)

(B.21d)

(B.21e)

(B.21f)

This is in agreement with Vp = V8 L30° in the positive sequence and VP = V8 L - 30° in the negative sequence for a Y d1 transformer.


B.1.2 Y9 d3 Transformers

Consider the Y 9 d3 transformer of Fig. B.3. The phase admittance matrix relates the currents and voltages as I = Y · V. With y being the transformer admittance in pu, one can write

lA = Y (VA - Vcb/J3) = y[VA - (Vc- Vb)/J3]

lB = y (VB - Vac/J3) = y[VB - (Va - Vc)/J3]

lc = y (Vc - Vba/J3) = y[Vc- (Vb- Va)/J3]

la =lea- lab= (IB- lc)/J3

= y [(VB- Vc)/J3- (2Va- Vb- Vc)/3]

h =lab- lbc = (lc- 1A)/J3

= y [(Vc- VA)/J3- (2 Vi,- Vc- Va)/3]

lc = lbc- lea= (IA- 1B)/J3

= y [(VA- VB)/J3- (2Vc- Va- Vb)/3]

In matrix form,

lA 1 o o oJaJa

lB

Ic

-la

-h

-lc

0 1 0 -1 v'3 0 ta

0 0 1 1 -1 v'3 v'3

=y 0 -1 1 2

v'3 v'3 3 1 -3

1 0 -1 1 v'3 v'3 -3

2 3

t:.J3

0

1 -3

1 -3

2 3

VA

VB

Vc

Va

vb

Vc

(B.22a)

(B.22b)

(B.22c)

(B.22d)

(B.22e)

(B.22f)

(B.23)

A - - e lA I.

c b - -Ic Ib

B - ----+ a

,~.·~· Ia I, b

Fig. B.3. A Y9 d3 transformer


The sequence admittance matrix Y 012 can now be calculated as Y 012 = A - 1 Y A and relates the currents and voltages as

lpo 1 0 0 0 0 0 Vpo

JPt 0 1 0 0 e 0 VPl

IP2 0 0 1 0 0 f VP2

I so

=y 0 0 0 Vso

(B.24) 0 0 0

fs 1 0 f 0 0 1 0 Vst

182 0 0 e 0 0 1 Vs2

where e = -1Lgoo = -J and f = -1L- goo= J. From the matrix equation for Yo12, the primary and secondary sequence currents are given by

lpo = Y · VPo

!Pt = Y. (VPt- VstLgoo) = Y · (VPt- JVst)

]P2 = Y · (VP2 - Vs2L- goo) = Y · (VP2 + iVs2)

lso = 0

1st = Y · (Yst - VPt L- goo) = Y · (Vst + ]Vpt)

ls2 = y · (Ys2 - Vp2Lgoo) = y · (Ys2 - ]Vp2)

(B.25a)

(B.25b)

(B.25c)

(B.25d)

(B.25e)

(B.25f)


B.1.3 Y9 d5 Transformers

Consider the Y9 d5 transformer of Fig. B.4. The phase admittance matrix relates the currents and voltages as I = Y · V. With y being the transformer admittance in pu, one can write

lA = y (VA- Vea/v'3) = y[VA- (Ve- Va)/J3]

lB = y (VB- Vab/J3) = y[VB- (Va- Vb)/J3]

lc = y (Vc- Vbe/v'3) = y[Vc- (Vb- Ve)/v'3]

la =lea- lab= (IB- lA)/J3

= y [(VB- VA)/J3- (2 Va- Vb- Ve)/3]

h =lab- he= (lc- lB)/v'3

= y [(Vc- VB)/J3- (2 Vb- Ve- Va)/3]

le =he- lea= (IA- lc)/J3

= y [(VA- Vc)/J3- (2 Ve- Va- Vb)/3]

In matrix form,

Ic

-h

A -JA

B -Is

c -Ic

=y

1

0

0

1 y'3

0

-1 y'3

0

1

0

-1 y'3

1 y'3

0 ~

0 -1 y'3

1

0

0

2 3

0 ~

- c I,

- a I.

- b Ih

Fig. B.4. A Y 9d5 transformer

0

1 y'3

-1 y'3

1 -3

2 3

-1 y'3

0

1 v'3

1 -3

1 -3

2 3

VA

VB

Vc

Va

Vb

Vc

(B.26a)

(B.26b)

(B.26c)

(B.26d)

(B.26e)

(B.26f)

(B.27)

a



lpo 1 0 0 0 0 0 Vpo

IPl 0 1 0 0 e 0 VPl

IP2 0 0 1 0 0 f VP2

I so

=y 0 0 0 0 0 0 Vs0

(B.28)

lst 0 f 0 0 1 0 Vs 1

ls2 0 0 e 0 0 1 Vs2

where e = -1L:150° and f = -1L- 150°. From the matrix equation for Y 012 , the primary and secondary sequence currents are given by

lpo = Y. Vpo

JPt = Y . (VPl - Vst L150o)

lp2 = Y · (Vp2 - Vs2L- 150°)

lso = 0

ls 1 = Y · (V. 1 - Vp1 L- 150°)

ls2 = Y · (V82 - Vp2 L150°)

(B.29a)

(B.29b)

(B.29c)

(B.29d)

(B.29e)

(B.29f)


B.1.4 Y 9 d7 Transformers

Consider the Y 9 d7 transformer of Fig. B.5. The phase admittance matrix relates the currents and voltages as I = Y · V. With y being the transformer admittance in pu, one can write

IA = y (VA- Vba/J3) = y[VA- (VI,- Va)/J3]

IB = Y (VB - Vcb/J3) = y[VB - (Vc- Vl,)/J3]

Ic = Y (Vc- Vac/J3) = y[Vc- (Va- Vc)/J3]

Ia = ha- lac= (Ic- IA)/J3

= y[(Vc- VA)/J3- (2Va- Vb- Vc)/3]

h =feb- ha = (IA - IB)/J3

= y [(VA- VB)/J3- (2 Vb- Vc- Va)/3]

Ic =lac- feb= (IB- Ic)/J3

= y [(VB- Vc)/J3- (2 Vc- Va- Vl,)/3]

In matrix form,

Ic =y

-h

A

1 0 0 ~

0 1 0 0

0 0

1 V3

-1 V3

0

1 V3

1 -1

V3

-1 2 V3 3

0 1 -3

-1 0 V3

1 -1 V3 V3

0 1 V3

1 1 -3 -3

2 1 3 -3

VA

VB

Vc

Va

vb

Vc

(B.30a)

(B.30b)

(B.30c)

(B.30d)

(B.30e)

(B.30f)

(B.31)

-lA

c -lc

B ---+ a

I, ~·

c • ~' IB a

Fig. B.S. A Y 9 d7 transformer

B.1 Y 9 - Ll Transformers 401

The sequence admittance matrix Y 012 can now be calculated as Y o12 = A - 1 Y A and relates the currents and voltages as

lpo 1 0 0 0 0 0 Vpo

JPl 0 1 0 0 e 0 VPl

JP2 0 0 1 0 0 f v;,2

I so =y

0 0 0 0 0 0 Yso (B.32)

ls1 0 f 0 0 1 0 Val

ls2 0 0 e 0 0 1 Va2

where e = -1L210° and f = -1L - 210°. From the matrix equation for Y 012 , the primary and secondary sequence currents are given by

lpo = Y. Vpo

lp1 = y · (Vp1 - V81 L210°)

JP2 = y. (VP2 - \'s2L- 210o)

180 = 0

ls1 = Y · (Vs 1 - Vp 1 L - 210°)

ls2 = Y · (l's2 - Vp2L210°)

(B.33a)

(B.33b)

(B.33c)

(B.33d)

(B.33e)

(B.33f)


B.1.5 Y 9 d9 Transformers

Consider the Y 9 d9 transformer of Fig. B.6. The phase admittance matrix relates the currents and voltages as I= Y · V. Withy being the transformer admittance in pu, one can write

lA = y (VA - Vbc/J3) = y[VA- (\i- Vc)/J3]

IB = y (VB- Vca/J3) = y[VB- (Vc- Va)/J3]

Ic = Y (Vc - Vab/J3) = y[Vc- (Va- li)/v'3]

Ia = Iba- lac= (Ic- IB)/J3

= y [(Vc- VB)/J3- (2 Va- \i- Vc)/3]

lb =feb- lba = (IA- Ic)/J3

= y [(VA- Vc)/J3- (2\i- Vc- Va)/3]

lc =lac- feb= (IB - JA)/J3

= y [(VB- VA)/J3- (2Vc- Va -\i)/3]

In matrix form,

IA

IB

Ic =y

-fa

-Ib

-Ic

A

1 0 0 0 -1 1 73 v'3

01 O}a 01a

0 0 1 -1 1 v'3 7a

0 1 -1 2 v'3 v'3 3

1 -3

2 3

0

1 -3

1 -3

2 3

VA

VB

Vc

Va

li

Vc

A

(B.34a)

(B.34b)

(B.34c)

(B.34d)

(B.34e)

(B.34f)

(B.35)

-lA

B -Is ~.~4 b

c -Ic

Fig. B.6. A Y 9 d9 transformer

-+a I,

C B c

B.l Y 9 - Ll Transformers 403


lpo 1 0 0 0 0 0 Vpo

/Pl 0 1 0 0 e 0 VPl

IP2 0 0 1 0 0 f v,2 I so

=y 0 0 0 Vso

(B.36) 0 0 0

ls1 0 f 0 0 1 0 Vs 1

ls2 0 0 e 0 0 1 Vs2

where e = -1L270° = J and f = -1L-270° = -J. From the matrix equation for Y 012 , the primary and secondary sequence currents are given by

Ipo = Y. Vpo

/Pl = Y. (VPl - Vsl L270o) = Y. (VPl + JV.l)

IP2 = y. (VP2 - V,2L- 270°) = y. (V,2 - JV.2)

lso = 0

181 = Y · {V,1 - Vp1 L- 270°) = Y · {V,1 - ]Vp1 )

ls2 = Y · {Ys2 - Vp2L270°) = Y · {Vs2 + JVp2)

{B.37a)

(B.37b)

(B.37c)

(B.37d)

{B.37e)

(B.37f)


B.1.6 Y 9 dll Transformers

Consider the Y9dll transformer of Fig. B.7. The phase admittance matrix relates the currents and voltages as I = Y · V. With y being the transformer admittance in pu, one can write

lA = y (VA - Vae/vJ) = y[VA - (Va- Ve)/vJ]

lB = y (VB - Vba/vJ) = y[VB - (Vb- Va)/vJ]

lc = y (Vc - Veb/vJ) = y[Vc - (Vc- Vb)/vJ]

la =lea -lab= (/A - lB}/vJ

= y [(VA- VB)/vJ- (2Va- Vb- Ve}/3]

lb =lab- he= (/B- lc}/vJ

= y [(VB- Vc}/vJ- (2 Vb- Ve- Va}/3]

le =he- lea = (lc- lA}/vJ

= y [(Vc- VA)/vJ- (2 Vc- Va - ltb}/3]

In matrix form,

1 0 0 -1 0 1 lA v'3 v'3

0 1 0 1 -1 0 lB v'3 v'3

lc 0 0 1 0 1 -1 7a v'3

=y -fa -1 1 0 2 1 1

v'3 v'3 3 -3 -3

-h 0 -1 1 1 2 1 v'3 v'3 -3 3 -3

-lc 1 0 -1 1 1 2 v'3 v'3 -3 -3 3

1:--/3

VA

VB

Vc

Va

vb

Vc

(B.38a}

(B.38b}

(B.38c}

(B.38d)

(B.38e)

(B.38f)

(B.39)

A - - a IA I.

c c - -Ic I,

B - - b

c~.·~b Is r., c

Fig. B.7. A Y9 dll transformer

B.l Y 9 - L1 Transformers 405


/Po 1 0 0 0 0 0 l'vo IP1 0 1 0 0 e 0 v;,l

IP2 0 0 1 0 0 I VP2

I so

=y 0 0 0 0 0 0 Vso

(B.40)

ls1 0 I 0 0 1 0 Val ls2 0 0 e 0 0 1 Va2

where e = -1L330° and I = -1L - 330°. From the matrix equation for Yo12, the primary and secondary sequence currents are given by

Ipo = Y. Vpo

IP1 = Y · (Vp1 - Vs1 L330°)

IP2 = y. {VP2 - Vs2L- 330°)

Iso = 0 Isl = Y · (Ys 1 - Vp1 L - 330°)

ls2 = Y · (Va2 - Vp2L330°)

(B.41a)

(B.41b)

(B.41c)

(B.41d)

(B.41e)

(B.41f)


B.2 Y -..d Transformers

B.2.1 Y dl Transformers

Consider the Y dl transformer of Fig. B.S. With z being the transformer impedance in pu andy the transformer admittance in pu, one can write two loop equations at the Y side as

VAc = zIt + Vab/.J3 + z {It - !2) + Vac/V3

VcB = z (!2- It)+ Vca/V3 + z !2 + Vcb/.J3

Multiplying by two and adding, one can solve for h and /2 as

Y [ 2 Vab 2 Vac Vca Vcb] h = 3 2VAc + VcB- .J3 - .J3 - y'3- y'3

= y [(2 VA- VB- Vc)/3 + (V,- Va)/.J3]

Y [ 2 Vcb 2 Vca Vac Vab] !2 = 3 2 VcB + VAc - y'3 - .J3 - y'3 - y'3

= y [( -2 VB+ Vc + VA)/3 + (Vb- Vc)/J3] Primary and secondary currents can now be expressed as

IA = h = y [(2VA- VB- Vc)/3 + (Vb- Va)/.J3]

IB = -!2 = y [(2 VB- Vc- VA)/3 +We- Vb)/Va]

Ic = !2- It = Y [(2 Vc -VA - VB)/3 + (Va- Vc)/.J3]

Ia = Iba- lac= (IA- Ic)/.J3

= y [(VA- Vc)/.J3 + (-2Va + V, + Vc)/3]

Ib = Icb- Iba = (IB - IA)/.J3

= y ((VB - VA)/.J3 + ( -2 V, + Vc + Va)/3]

Ic =lac- Icb = (Ic- IB)/.J3

= y [(Vc- VB)/Va + ( -2 Vc + Va + Vb)/3]

(B.42a)

(B.42b)

(B.42c)

(B.42d)

(B.42e)

(B.42f)

A -lA

B -Is

-+ a I. ~,~·

c('0~ c - - e Ic I.: b

Fig. B.S. A Y dl transformer

B.2 Y -.:1 Transformers 407

In matrix form,

2 1 1 -1 1 0 IA 3 -3 -3 v'3 v'3 VA

1 2 1 0 -1 1 Is -3 3 -3 v'3 v'3 Vs

Ic 1 1 2 1 0 -1

Vc -3 -3 3 v'3 v'3 =y (B.43)

-fa -1 0 1 2 1 1 Va v'3 v'3 3 -3 -3

-h 1 -1 0 1 2 1 Vb v'3 v'3 -3 3 -3

-fc 0 1 -1 1 1 2 Vc v'3 v'3 -3 -3 3

The sequence admittance matrix Yo12 can now be calculated as Y 012

A - 1 Y A and relates the currents and voltages as

fpo 0 0 0 0 0 0 Vpo

IPl 0 1 0 0 e 0 VPl

IP2 0 0 1 0 0 f VP2

I so

=y 0 0 0 0 0 0 Vs0

(B.44)

fsl 0 f 0 0 1 0 Vs1

fs2 0 0 e 0 0 1 Vs2

where e = -1L30° and f = -1L- 30°. From the above matrix equation for Y 012 , the primary and secondary sequence currents are given by

lp0 = 0

IPl = y . (VPl - VSl L30°)

JP2 = Y · (Vp2 - Vs2L- 30°)

180 = 0

fs 1 = Y · (V81 - Vp1L- 30°)

fs 2 = Y · (V82 - Vp2L30°)

(B.45a)

(B.45b)

(B.45c)

(B.45d)

(B.45e)

(B.45f)


B.2.2 Yd3 Transformers

Consider the Y d3 transformer of Fig. B.9. With z being the transformer impedance in pu and y the transformer admittance in pu, one can write two

loop equations at the Y side as

VAs= z h + Veb/J3 + z (h- h)+ Vea/J3

Vsc = z (!2- h)+ Vae/J3 + z !2 + Vab/J3

Multiplying by two and adding, one can solve for h and h as

h = y [(2 VA- Vs- Vc)/3 + (Vb- Ve)/J3]

h = y [( -2 Vc +VA+ Vs)/3 + (Vb- Va)/J3]

Primary and secondary currents can now be expressed as

A

c

8

IA = h = y [(2 VA- Vs- Vc)/3 + (Vb- Ve)/J3]

Is= !2- h = Y [(2 Vs- Vc- VA)/3 + (Ve- Va)/J3]

Ic = -h = y [(2 Vc- VA- Vs)/3 + Wa- V,)/J3]

Ia =lea- lab= (Is- Ic)/J3

= y [(VB- Vc)/.J3 + ( -2 Va + Vb + Vc)/3]

h =lab- he= (Ic- IA)/.J3

= y [(Vc- VA)/J3 + ( -2 Vb + Ve + Va)/3]

Ie =he- lea= (IA- fs)/J3

= y [(VA- Vs)/J3 + ( -2 Ve + Va + V,)/3]

tdJ --IA

-Ic

- -- a I a I,

Fig. B.9. A Yd3 transformer

(B.46a)

(B.46b)

(B.46c)

(B.46d)

(B.46e)

(B.46f)

B.2 Y-Ll Transformers 409

In matrix form,

Ic =y

-h

0

1 v'3

-1 v'3

-1

v'3

0

1 v'3

0

-1

v'3

1 v'3

0

2 1 -1 3 v'3 v'3

1 v'3

0 -i

-1 v'3

1

v'3

0

2 3

Vc (B.47)


=y

0 0 0 0 0 0

0 1 0 0 e 0

0 0 1 0 0 f

0 0 0 0 0 0

0 f 0 0 1 0

0 0 e 0 0 1

(B.48)

where e = -1L90° = -J and f = -1L- 90° = J· From the above matrix equation for Y 012 , the primary and secondary sequence currents are given by

lp0 = 0

JPl = Y · (VPl- Vs1L90°} = Y · (VPl- J'Va1)

JP2 = Y. (VP2 - Vs2L- 90o) = Y. (VP2 + J'Va2)

lso = 0

ls 1 = Y · ('Va1 - Vp1 L- 90°) = Y · (Vs1 + J'Vp1)

ls2 = y · (V82 - Vp2L90°) = y · ('Va2 - JVp2)

(B.49a)

(B.49b)

(B.49c)

(B.49d)

(B.49e)

(B.49f)



Consider the Yd5 transformer of Fig. B.lO. With z being the transformer impedance in pu andy the transformer admittance in pu, one can write two loop equations at the Y side as

VAc = z l1 + Vca/h + z (h- l2) + Vcb/h

VcB = z (l2- h)+ Vbc/h + z l2 + Vba/h

Multiplying by two and adding, one can solve for l1 and l2 as

ft = y [(2 VA- VB- Vc)/3 + (Va- Vc)/v'J]

l2 = y [( -2 VB+ Vc + VA)/3 + (Va- Vb)/h]


A

B

c

lA = ft = y[(2VA- VB- Vc)/3+ (Va- Vc)/v'J]

lB = -12 = y [(2 VB- Vc- VA)/3 + (Vb- Va)/v'J]

lc = l2 - h = y [(2 Vc -VA - VB)/3 + (Vc- Vb)/h]

la =lea- lab= (lB - lA)/h

= y [(VB- VA)/J3 + (-2Va + Vb + Vc)/3]

h =lab- he= (lc- lB)/h

= y [(Vc - VB)/h + ( -2 Vb + Vc + Va)/3]

lc = lbc- lea= (IA- lc)/h

= y [(VA- Vc)/h + (-2Vc + Va + Vb)/3]

1=--/3 - - c A

JA lc

~· a - -Is Ia

b C~B - -Ic Ih

Fig. B.lO. A Yd5 transformer

(B.50a)

(B.50b)

(B.50c)

(B.50d)

(B.50e)

(B.50f)

c

a

B.2 Y -Ll Transformers 411

In matrix form,

2 1 1 1 0 -1

IA 3 -3 -3 V3 V3 VA

1 2 1 -1 1 0 lB -3 3 -3 V3 V3 VB

Ic 1 1 2 0 -1 1

Vc -3 -3 3 V3 V3 =y (B.51)

-fa 1 -1 0 2 1 1 Va V3 V3 3 -3 -3

-h 0 1 -1 1 2 1 Vb V3 V3 -3 3 -3

-Ic -1 0 1 1 1 2 Vc V3 V3 -3 -3 3

The sequence admittance matrix Yo12 can now be calculated as Yo12

A -l Y A and relates the currents and voltages as

!Po 0 0 0 0 0 0 Vpo

JP1 0 1 0 0 e 0 VPl

JP2 0 0 1 0 0 f VP2

fso =y

0 0 0 0 0 Vso (B.52)

0

fs1 0 f 0 0 1 0 Vs 1

fs2 0 0 e 0 0 1 Vs2

where e = -1L:150° and f = -1L- 150°. From the above matrix equation for Y 012 , the primary and secondary sequence currents are given by

lp0 = 0

JPl = Y. (VPl - Vsl L:150o)

lp2 = y · (Vp2 - V82 L- 150°)

fso = 0

fs 1 = Y · (Vs1 - Vp 1 L - 150°)

fs2 = Y · (Vs2 - Vp2L150°)

(B.53a)

(B.53b)

(B.53c)

(B.53d)

(B.53e)

(B.53f)



Consider the Yd7 transformer of Fig. B.ll. With z being the transformer impedance in pu and y the transformer admittance in pu, one can write two loop equations at the Y side as

VAB = z ft + Vba/J3 + z (ft- h)+ Vbc/J3

VBc = Z (12 -h)+ Vcb/J3 + Z l2 + Vca/J3

Multiplying by two and adding, one can solve for h and l2 as

fr = y [(2 VA- VB- Vc)/3 + (Va- 'Vb)/J3]

l2 = y [( -2 Vc +VA+ VB)/3 + (Va- Vc)/J3]


A

c

B

lA = ft = y [(2 VA- VB- Vc)/3 + (Va- Vb)/J3]

lB = h- fr = y [(2 VB- Vc- VA)/3 + (Vb- Vc)/J3]

lc = -f2 = y [(2 Vc- VA- VB)/3 + (Vc- Va)/J3]

la = ha- lac= (lc- 1A)/J3

= y [(Vc- VA)/J3 + ( -2 Va + Vb + Vc)/3]

h =feb- ha = (IA- 1B)/v'3

= y [(VA- VB)/v'3 + ( -2 Vb + V, + Va)/3]

fc =lac- feb= (IB- lc)/v'3

= y [(VB- Vc)/v'3 + ( -2 Vc + Va + Vb)/3]

-IA

-Ic

-I a

--+ a I,

Fig. B.ll. A Yd7 transformer

a

(B.54a)

(B.54b)

(B.54c)

(B.54d)

(B.54e)

(B.54f)

c

B.2 Y -.1 Transformers 413

In matrix form,

2 1 1 1 -1 0 IA 3 -3 -3 y'3 y'3 VA

1 2 1 0 1 -1 IB -3 3 -3 y'3 y'3 VB

Ic 1 1 2 -1 0 1

Vc -3 -3 3 y'3 y'3 =y (B.55)

-I a 1 0 -1 2 1 1 Va y'3 y'3 3 -3 -3

-h -1 1 0 1 2 1 % y'3 y'3 -3 3 -3

-lc 0 -1 1 1 1 2 Vc y'3 y'3 -3 -3 3



lpo 0 0 0 0 0 0 Vpo

JPl 0 1 0 0 e 0 VPl

IP2 0 0 1 0 0 f VP2

I so =y

0 0 Vso (B.56)

0 0 0 0

lsl 0 f 0 0 1 0 Vs1

ls2 0 0 e 0 0 1 Vs2

where e = -1L210° and f = -1L- 210°. From the above matrix equation

for Yo12, the primary and secondary sequence currents are given by

lp0 = 0

JPl = y . (Vpl - VSl L210°)

JP2 = Y · (Vp2 - Vs 2 L- 210°)

180 = 0

ls 1 = Y · (Vs 1 - Vp1 L - 210°)

182 = y . (Vs2 - VP2 L210°)

(B.57a)

(B.57b)

(B.57c)

(B.57d)

(B.57e)

(B.57f)


B.2.5 Y d9 Transformers

Consider the Yd9 transformer of Fig. B.l2. With z being the transformer impedance in pu and y the transformer admittance in pu, one can write two loop equations at the Y side as

VAc = z h + Vbc/J3 + z (h- /2) + Vba/V3

VcB = Z (!2- h)+ Vab/J3 + Z !2 + Vac/V3"

Multiplying by two and adding, one can solve for h and / 2 as

h = y [(2 VA- VB- Vc)/3 + (Vc- Vb)/J3]

/2 = y [( -2 VB+ Vc + VA)/3 + (Vc- Va)/J3]


!A= ft = y [(2VA- VB- Vc)/3 + (Vc- Vb)/J3]

lB = -!2 = y [(2 VB- Vc- VA)/3 + (Va- Vc)/J3]

lc =!2-ft= y [(2 Vc- VA- VB)/3 + (Vb- Va)/J3]

la = ha- lac= (Ic- IB)/J3

= y [(Vc- VB)/J3 + ( -2 Va + Vb + Vc)/3]

h =feb- ha = (IA- Ic)/v'3

= y [(VA- Vc)/v'3 + ( -2 Vb + Vc + Va)/3]

fc =lac- feb= (IB- fA)/J3

= y [(VB- VA)/J3 + ( -2 Vc + Va + %)/3]

(B.58a)

(B.58b)

(B.58c)

(B.58d)

(B.58e)

(B.58f)

~.~' c~,~

A -IA

B --Is c -Ic

Fig. B.12. A Yd9 transformer

- a I. c


In matrix form,

Is

Ic =y

-h

0

-1

V3

1 V3

1 V3

0

-1 V3

0

1 V3

-1

V3

0

2 -1 1 3 V3 V3

-1

V3

1 V3

-1 V3

0

2 3

Vc (B. 59)



=y

0 0 0 0 0 0

0 1 0 0 e 0

0 0 1 0 0 f 0 0 0 0 0 0

0 f 0 0 1 0

0 0 e 0 0 1

(B.60)

where e = -1L270° = J and f = -1L - 270° = - J. From the above matrix equation for Y 012 , the primary and secondary sequence currents are given by

/Po= 0

lp1 = y · (Vp1 - V81 L270°) = Y · (Vp 1 + ]Vs 1 )

lp2 = Y · (Vp2 - Vs 2 L- 270°) = Y · (Vp2 - JVs 2 )

fso = 0

fs 1 = Y · (Vs 1 - Vp 1 L- 270°) = Y · (Vs 1 - ]Vp1 )

fs 2 = Y · (Vs2 - Vp2 L270°) = Y · (Vs 2 + ]Vp2 )

(B.61a)

(B.61b)

(B.61c)

(B.61d)

(B.61e)

(B.61f)


B.2.6 Y dll Transformers

Consider the Y dll transformer of Fig. B.l3. With z being the transformer impedance in pu and y the transformer admittance in pu, one can write two loop equations at the Y side as

VAs= Z ft + Vae/J3 + Z (ft- !2) + Vab/J3

Vsc = z (h -h)+ Vba/J3 + z h + Vbe/J3

Multiplying by two and adding, one can solve for It and h as

it = y [(2 VA- Vs- Vc)/3 + (Ve- Va)/J3]

!2 = y [( -2 Vc +VA+ Vs)/3 + (Ve- Vb)/J3]


A

c

B

!A= it= y [(2VA- Vs- Vc)/3 + (Ve- Va)/J3]

Is= h-it= y [(2Vs- Vc- VA)/3 + (Va- Vb)/J3]

Ic = -!2 = y [(2 Vc- VA- Vs)/3 + (Vb- Ve)/J3]

Ia =lea- lab= (IA- ls)/J3

= y [(VA- Vs)/.J3 + ( -2 Va + Vb + Ve)/3]

h =lab- he= (Is- Ic)/J3

= y [(Vs- Vc)/J3 + ( -2 Vb + Ve + Va)/3]

Ie =he- lea= (Ic- IA)/J3

= y [(Vc- VA)/J3 + ( -2 Ve + Va + Vb)/3]

t:'/3 - - a IA I,

- -Ic I,

- - b

Io Ih

Fig. B.13. A Ydll transformer

c

(B.62a)

(B.62b)

(B.62c)

(B.62d)

(B.62e)

(B.62f)


In matrix form,

2 1 1 -1 0 1

!A 3 -3 -3 v'3 v'3 VA

1 2 1 1 -1 0 IB -3 3 -3 v'3 v'3 VB

Ic 1 1 2 0 1 -1

Vc -3 -3 3 v'3 v'3 =y (B.63)

-fa -1 1 0 2 1 1 Va v'3 v'3 3 -3 -3

-h 0 -1 1 1 2 1 Vb v'3 v'3 -3 3 -3

-fc 1 0 -1 1 1 2 Vc v'3 v'3 -3 -3 3



fpo 0 0 0 0 0 0 Vpo

JPl 0 1 0 0 e 0 VPl

IP2 0 0 1 0 0 f VP2

fsn =y

0 0 0 0 Vso (B.64)

0 0

fs 1 0 f 0 0 1 0 Vs1

fs 2 0 0 e 0 0 1 Vs2

where e = -1.L330° and f = -1.L - 330°. From the above matrix equation for Y 012 , the primary and secondary sequence currents are given by

lp0 = 0

JPl = Y . (VPl - Vsl .L330o)

IP2 = Y · (Vp2 - Vs2L- 330°)

fso = 0

fs 1 = Y · (V81 - Vp1 L- 330°)

fs 2 = Y · (V82 - Vp2L330°)

(B.65a)

(B.65b)

(B.65c)

(B.65d)

(B.65e)

(B.65f)


B.3 .L1-Y 9 Transformers

The different vector groups for ..1-Y transformers are shown in Fig. B.14.

Dyl Dy5

Dyll Dy7

Dy3 Dy9

Lh \j -. Lh·- V' c ./ c .\

h '

Fig. B.14. Vector groups for Ll-Y transformers

B.3 ..:1-Y 9 Transformers 419

B.3.1 Dy9 1 'fransformers

Consider the Dy9 1 transformer of Fig. B.15.

-./3:1

- a I.

c-Ic

Fig. B.15. A Dy9 1 transformer

With y being the transformer admittance in pu, one can write

la = y (VAcf../3- Va) = y[(VA- Vc)f../3- Val

h = y (VBA/../3- Vb) = y[(VB- VA)/../3- Vbl

Ic = y (VcB/../3- Vc) = y[(Vc- VB)/../3- Vel

fA= lAc- fBA =(fa- h)/../3

= y [(2VA- VB- Vc)/3 + (Vb- Va)f../31

IB = IBA- feB= (h- Ic)f../3

= y [{2 VB - Vc - VA)/3 + (Vc- Vb)/v'3l

Ic =feB -lAc = {Ic- Ia)/../3

= y [{2 Vc- VA- VB)/3 + (Va- Vc)/../31

In matrix form,

fA

IB

Ic =y

0

-fa -1 0 1 V3 V3 1 0 0

-h 1 -1 0 V3 V3 0 1 0

-Ic 0 1 -1

V3 V3 0 0 1

VA

VB

Vc

Va

vb

Vc

(B.66a)

(B.66b)

(B.66c)

(B.66d)

(B.66e)

(B.66f)

(B.67)


This matrix is similar to that for the Y 9 d1 transformer except that the upper left and lower right sub-matrices are interchanged.


lpo 0 0 0 0 0 0 Vpo

/Pl 0 1 0 0 e 0 VPl

IP2 0 0 1 0 0 f VP2

I so

=y 0 0 0 1 0 0 Vso

(B.68)

ls1 0 f 0 0 1 0 Vs1

fs2 0 0 e 0 0 1 Vs 2

where e = -1L30° and f = -1L- 30°. From the matrix equation for Yo12, the primary and secondary sequence currents are given by

/Po= 0

fp 1 = Y · (Vp1 - V81 L30°)

IP2 = y. (VP2 - Vs2L- 30°)

fso = Y · Vso

ls1 = Y · (V81 - Vp1L- 30°)

fs 2 = Y · (Vs2 - Vp2 L30°)

(B.69a)

(B.69b)

(B.69c)

(B.69d)

(B.69e)

(B.69f)

B.4 .:1-Y Transformers 421

B .4 ...::1-Y Transformers

B.4.1 Dyl Transformers

Consider the Dyl transformer of Fig. B.16.

-./3:1 A - - a

JA r. a

c - - c Ic Ic

8 - - b Io I~ b

Fig. B.16. A Dyl transformer

With z being the transformer impedance in pu and y the transformer admittance in pu, one can write two loop equations at they (secondary) side as

Vac = z h + VAc/v'a + z (h- !2) + VBc/v'a

Vcb = z (12- h)+ VcB/v'a + z !2 + VAB/v'a

Multiplying by two and adding, one can solve for h and !2 as

It= y [(2 Va- Vb- Vc)/3 + (Vc- VA)/vta]

h = y [( -2 Vb + Vc + Va)/3 +(VB- VA)/J3]


Ia = -!1 = y [(VA- Vc)/VJ + ( -2 Va + Vb + Vc)/3]

Ib = !2 = y [(VB - VA)/VJ + ( -2 Vb + Vc + Va)/3]

Ic = h- !2 = y [(Vc- VB)/vta + ( -2 Vc + Va + Vb)/3]

!A= lAc- lBA = (Ia- h)jvta

= y [(2 VA- VB- Vc)/3 + (Vb- Va)/vta]

IB = lBA- lcB = (Ib- Ic)/v'a

= y [(2VB- Vc- VA)/3 +We- Vb)/vta]

Ic =feB -lAc = (Ic- Ia)/v'a

= Y [{2 Vc- VA- VB)/3 + (Va- Vc)/v'3]

(B.70a)

(B.70b)

(B.70c)

(B.70d)

(B.70e)

(B.70f)


In matrix form,

Ic =y

-h

-1 V3 0

1 -1 V3 V3

0 1 V3

1 V3

-1 V3

0

1 V3

0 -!

1 V3

-1 V3

0

0

1 V3

-1 V3

2 3

Vc (B.71)

which is the same as the phase admittance matrix for a Yd1 transformer. The sequence admittance matrix Yo12 can now be calculated as Y 012 =


=y

0 0 0 0 0 0

0 1 0 0 e 0

0 0 1 0 0 f 0 0 0 0 0 0

0 f 0 0 1 0

0 0 e 0 0 1

(B.72)

where e = -1L30° and f = -1L- 30°. From the above matrix equation for Yo12 , the primary and secondary sequence currents are given by

fpo = 0

lp1 = Y · (Vp1 - V81 L30°)

JP2 = Y · (VP2 - Vs2L- 30°)

180 = 0

fs 1 = Y · (Vs 1 - Vp 1 L- 30°)

fs 2 = Y · (V82 - Vp2 L30°)

(B.73a)

(B.73b)

(B.73c)

(B.73d)

(B.73e)

(B.73f)

B.5 Y 9 - Y 9 Transformers 423

B.5 Y 9 - Y 9 Transformers

The different vector groups for Y-Y transformers are shown in Fig. B.l7.

Yy4

YylD Yy8

Fig. B.17. Vector groups for Y-Y transformers


B.5.1 Y 9 y9 0 Transformers

Consider the Y 9 y90 transformer of Fig. B.18.

1: 1 A - - a A a

lA I,

~ ~ 8 - - b

Ig Ib

c - - c c 8 c b

Ic I.

Fig. B.18. A Y 9 y90 transformer


lA = y(VA- Va)

lB = y(VB- Vb)

Ic = y(Vc- Vc)

la = !A = y (VA - Va)

Ib = IB = y(VB- V,)

Ic = Ic = y (Vc - Vc)

In matrix form,

IA 1 0 0

IB 0 1 0

Ic 0 0 1

-I a =y

-1 0 0

-h 0 -1 0

-Ic 0 0 -1

-1 0

0 -1

0 0

1 0

0 1

0 0

0

0

-1

0

0

1

VA

VB

Vc

Va

v;,

Vc

(B.74a)

(B.74b)

(B.74c)

(B.74d)

(B.74e)

(B.74f)

(B.75)


B.5 Y 9 -Y9 Transformers 425

JPO 1 0 0 -1 0 0 Vpo

Ivt 0 1 0 0 -1 0 VPl

IP2 0 0 1 0 0 -1 VP2 =y

Vso (B.76)

I so -1 0 0 1 0 0

ls1 0 -1 0 0 1 0 Val

ls2 0 0 -1 0 0 1 Vs2


fpo = Y · {Vpo- Vso)

/Pl = Y · (VPl - VsJ

IP2 = y. (VP2 - Vs2) fso = Y. (Vso- Vpo)

fsl = Y · (Vsl- Vpl)

182 = y. (Vs2 - VP2)

(B.77a)

(B.77b)

(B.77c) (B.77d)

(B.77e) (B.77f)



Consider the Y 9 y92 transformer of Fig. B.19.

1:1 A - - b A

lA Ih

c~·'*' B - - c

Ia I,

c - - a Ic I,

b

Fig. B.19. A Y 9y 9 2 transformer


lA = y(VA + Vt,) (B.78a)

IB = y(VB + Vc) (B.78b)

Ic = y(Vc + Va) (B.78c)

Ia = -Ic = -y(Vc + Va) (B.78d)

Ib = -IA = -y (VA+ Vt,) (B.78e)

lc = -lB = -y(VB + Vc) (B.78f)

In matrix form,

fA 1 0 0 0 1 0 VA

fB 0 1 0 0 0 1 VB

Ic 0 0 1 1 0 0 Vc

-fa =y

0 1 0 Va (B.79)

0 1 0

-h 1 0 0 0 1 0 vb -fc 0 1 0 0 0 1 Vc

The sequence admittance matrix Y 012 can now be calculated as Y 012 = A -l Y A and relates the currents and voltages as


fpo 1 0 0 1 0 0 Vpo

JPl 0 1 0 0 e 0 VPl

IP2 0 0 1 0 0 f VP2

I so =y

0 0 0 Vso (Bo80)

1 0 1

fsl 0 f 0 0 1 0 Vs1

fs2 0 0 e 0 0 1 Vs2

where e = -1L60° and f = -1L- 60°0 From the matrix equation for Y012, the primary and secondary sequence currents are given by

lp0 = Y 0 (Vpo + Vs0 )

lp1 = y 0 (Vp1 - V 81 L60°)

fp2 = Y 0 (Vp2 - Vs2L- 60°)

fso = Y 0 (Vso + Vpo)

ls1 = y 0 (Vs 1 - Vp1L- 60°)

182 = Y 0 (V82 - Vp2L60°)

(Bo81a)

(Bo81b)

(Bo81c)

(Bo81d)

(B.81e)

(Bo81f)


B.5.3 Y 9 y9 4 'fransformers

Consider the Y 9 y9 4 transformer of Fig. B.20.

1:1 A - - c A c

lA 1,

cA. A 8 - - a Ia I.

c - b b a -Ic Ih

Fig. B.20. A Y 9 y 94 transformer


!A= y(VA- ll;,)

IB = y (VB - Va)

Ic = y(Vc- ~)

la = IB = y (VB- Va)

h = Ic = y (Vc - lib)

lc = lA = y (VA - Vc)

In matrix form,

lA 1 0 0

IB 0 1 0

Ic 0 0 1

-fa =y

0 -1 0

-h 0 0 -1

-Ic -1 0 0

0 0

-1 0

0 -1

1 0

0 1

0 0

-1

0

0

0

0

1

VA

VB

Vc

Va

~

Vc

(B.82a)

(B.82b)

{B.82c)

(B.82d)

(B.82e)

(B.82f)

{B.83)

The sequence admittance matrix Y 012 can now be calculated as Y 012 = A -l Y A and relates the currents and voltages as

B.5 Y 9 -Y9 Transformers 429

1po 1 0 0 -1 0 0 Vpo

1Pl 0 1 0 0 e 0 VPl

1P2 0 0 1 0 0 f VP2

1so =y

-1 0 0 1 0 0 Vso (B.84)

1s1 0 f 0 0 1 0 Vsl

182 0 0 e 0 0 1 Vs2

where e = -1L120° and f = -1L - 120°. From the matrix equation for Y 012 , the primary and secondary sequence currents are given by

1p0 = Y · {Vpo- Vs0 )

1Pl = Y. {VPl - Vsl L120o)

1P2 = Y. {VP2 - Vs2L- 120o)

1so = Y · {Vso- Vpo)

1s1 = Y · {Vs1 - Vp 1 L- 120°)

182 = y. (V:,2 - VP2L120°)

(B.85a)

(B.85b)

{B.85c)

(B.85d)

(B.85e)

(B.85f)




1 : 1 A - - a A

JA I, A b I ' B b - -Is lh

C B 1 c - - c lc I,

a

Fig. B.21. A Y 9 y 96 transformer


IA = y(VA + Va) (B.86a)

Is= y(Vs + Vb) (B.86b)

Ic = y(Vc + Vc) (B.86c)

Ia = -IA = -y(VA + Va) (B.86d)

Ib = -IB = -y(VB + Vb) (B.86e)

Ic = -Ic = -y(Vc + Vc) (B.86f)

In matrix form,

IA 1 0 0 1 0 0 VA

Is 0 1 0 0 1 0 Vs

Ic 0 0 1 0 0 1 Vc

-I a =y

0 Va (B.87)

1 0 0 1 0

-h 0 1 0 0 1 0 Vb -Ic 0 0 1 0 0 1 Vc

The sequence admittance matrix Yo12 can now be calculated as Yo12 = A -l Y A and relates the currents and voltages as


/Po 1 0 0 1 0 0 Vpo

IPl 0 1 0 0 1 0 VPl

IP2 0 0 1 0 0 1 VP2

I so =y

Vs0

(B.88) 1 0 0 1 0 0

lsl 0 1 0 0 1 0 Vs1

ls2 0 0 1 0 0 1 Vs2


lp0 = Y · (Vpo + Vs0 )

JPl = Y · (VPl + Vsl)

JP2 = Y · (VP2 + Vs2)

180 = Y · (V80 + Vp0 )

ls1 = Y · (Vsl + VPl)

182 = y. (VS2 + VP2)

(B.89a)

(B.89b)

(B.89c)

(B.89d)

(B.89e)

(B.89f)



Consider the Y9 y9 8 transformer of Fig. B.22.

1 : 1 A - - b A b

IA Ib

~ ~ B - - c

Ia I,

c - a C B a c -Ic I,

Fig. B.22. A Y 9 y 9 8 transformer


IA = y(VA- Vb)

JB = y(VB- Vc)

Ic = y(Vc- Va)

Ia = Ic = y (Vc - Va)

h =!A= Y (VA- Vb)

fc = JB = Y (VB- Vc)

In matrix form,

fA 1 0

IB 0 1

Ic 0 0

-fa =y

0 0

-h -1 0

-Ic 0 -1

0 0 -1

0 0 0

1 -1 0

-1 1 0

0 0 1

0 0 0

0

-1

0

0

0

1

VA

VB

Vc

Va

vb

Vc

(B.90a) (B.90b) (B.90c) (B.90d) (B.90e)

(B.90f)

(B.91)

The sequence admittance matrix Yo12 can now be calculated as Yo12 A -l Y A and relates the currents and voltages as


Jpo 1 0 0 -1 0 0 Vpo

JPl 0 1 0 0 e 0 VPl

JP2 0 0 1 0 0 f VP2

I so

=y -1 0 0 Vso

(B.92) 1 0 0

fsl 0 f 0 0 1 0 v.,l

182 0 0 e 0 0 1 Vs2

where e = -1L240° and f = -1L - 240°. From the matrix equation for Yo12, the primary and secondary sequence currents are given by

lp0 = Y · (VPo - Yso) JPl = y . {VPl - v.,l L2400)

JP2 = y . (VP2 - v.,2 L - 240°)

fso = Y · {V.,o - VPo)

ls1 = Y · {V.,1 - Vp1 L- 240°)

ls2 = Y · (Vs2 - Vp2 L240°)

(B.93a)

(B.93b)

(B.93c)

(B.93d)

(B.93e)

(B.93f)




1:1 A - - c A

JA I,

CA.''f' B - --+ a Is I,

c b - -Ic Ih

c

Fig. B.23. A Y 9 y 9 10 transformer


IA = Y (VA + Vc) (B.94a)

IB = y (VB + Va) (B.94b)

Ic = y(Vc + Vb) (B.94c)

Ia = -IB = -y(VB + Va) (B.94d)

h = -Ic = -y(Vc +VI,) (B.94e)

Ic =-!A= -y(VA + Vc) (B.94f)

In matrix form,

IA 1 0 0 0 0 1 VA

IB 0 1 0 1 0 0 VB

Ic 0 0 1 0 1 0 Vc

-I a =y

0 1 0 1 0 0 Va (B.95)

-Ib 0 0 1 0 1 0 Vb

-Ic 1 0 0 0 0 1 Vc


B .5 Y 9 - Y 9 Transformers 435

fpo 1 0 0 1 0 0 Vpo

IPl 0 1 0 0 e 0 VPl

IP2 0 0 1 0 0 f VP2

fs 0

=y Vso

(B.96) 1 0 0 1 0 0

fst 0 f 0 0 1 0 Vs1

Is2 0 0 e 0 0 1 Vs2

where e = -1L300° and f = -1L - 300°. From the matrix equation for Yo12, the primary and secondary sequence currents are given by

fp0 = Y · (Vpo + Vs 0 )

JPl = Y · (Vp1 - Vs 1 L300°)

IP2 = Y · (Vp2 - Vs2L- 300°)

fso = Y · (Vso + Vp0 )

fs 1 = Y · (Vs 1 - Vp1 L - 300°)

ls2 = y · (Vs 2 - Vv2L300°)

(B.97a)

(B.97b)

(B.97c)

(B.97d)

(B.97e)

(B.97f)


B.6 Yg- Y Transformers

B.6.1 YgyO Transformers

For a Y 9 y0 transformer, the primary and secondary sequence currents are expressed as

Jpo = Y. Vpo

JPl = Y · (VPl- Vsl)

JP2 = Y · (VP2 - Vs2)

fso = 0

fsl = Y · (Vsl- VPl)

fs2 = Y · (Vs2 - Vp2)

In matrix form,

!Po 1 0 0

JP1 0 1 0

JP2 0 0 1

I so =y

0 0 0

fsl 0 -1 0

fs 2 0 0 -1

0 0

0 -1

0 0

0 0

0 1

0 0

0

0

-1

0

0

1

Vpo

VPl

VP2

Vso

Vs 1

Vs2

(B.98a)

(B.98b)

(B.98c)

(B.98d)

(B.98e)

(B.98f)

(B.99)

The phase admittance matrix Y can now be calculated as Y =A Y012 A - 1

and relates the currents and voltages as

IA 1 0 0 2 1 1 VA -3 3 3

IB 0 1 0 1 2 1 VB 3 -3 3

Ic 0 0 1 1 1 2 Vc 3 3 -3 =y (B.100)

-fa 2 1 1 2 1 1 Va -3 3 3 3 -3 -3

-h 1 2 1 1 2 1 Vb 3 -3 3 -3 3 -3

-fc 1 1 2 1 1 2 Vc 3 3 -3 -3 -3 3

From the above matrix equation for Y, primary and secondary currents are given by

B.6 Y 9 -Y Transformers 437

lA = y [VA+ (-2Va + Vb + Vc)/3] IB = y [VB+ (-2"\tb + Vc + Va)/3] Ic = y [Vc + (-2Vc + Va + Vb)/3] Ia = y [(2VA- VB- Vc)/3 + (-2Va + Vb + Vc)/3]

Ib = y [(2 VB - Vc- VA)/3 + ( -2 Vb + Vc + Va)/3]

lc = y [(2 Vc- VA - VB)/3 + ( -2 Vc + Va + Vb)/3]

(B. lOla)

(B.lOlb)

(B.lOlc)

(B.lOld)

(B.lOle)

(B.lOlf)


B.6.2 Y 9 y2 Transformers

For a Y9y2 transformer, the primary and secondary sequence currents are expressed as

fpo = Y. Vpo

fp1 = Y · (Vp1 - V81 L60°)

JP2 = y. (VP2 - Vs2L- 60o)

fso = 0

ls1 = y · (Vs1 - Vv1L- 60°)

fs 2 = Y · (V82 - Vp2 L60°)

In matrix form,

/Po 1 0 0 0 0 0

JP1 0 1 0 0 e 0

IP2 0 0 1 0 0 f I so

=y 0 0 0 0 0 0

fst 0 f 0 0 1 0

ls2 0 0 e 0 0 1

where e = -1L60° and f = -lL- 60°.

Vpo

VPl

VP2

Vso

Vs1

Vs2

(B.102a)

(B.l02b)

(B.102c)

(B.l02d)

(B.102e)

(B.102f)

(B.103)

The phase admittance matrix Y can now be calculated as Y = A Y 012 A - 1


!A 1 0 0 1 2 1 VA -3 3 -3

IB 0 1 0 1 1 2 VB -3 -3 3

Ic 0 0 1 2 1 1 Vc 3 -3 -3 =y (B.104)

-fa 1 1 2 2 1 1 Va -3 -3 3 3 -3 -3

-h 2 1 1 1 2 1 Vb 3 -3 -3 -3 3 -3

-fc 1 2 1 1 1 2 Vc -3 3 -3 -3 -3 3


B.6 Y 9 - Y Transformers 439

IA = y [VA+ (2 Vb- \'c- Va)/3]

IB = y [VB+ (2 Vc- Va- Vb)/3]

Ic = y[Vc + (2Va-%- Vc)/3]

Ia = y [(-2Vc +VA+ VB)/3 + (-2Va + vb + Vc)/3]

h = y[(-2VA +VB+ Vc)/3 + (-2% + Vc + Va)/3]

Ic = y[(-2VB + Vc + VA)/3 + (-2Vc + Va + %)/3]

(B.105a)

(B.l05b)

(B.l05c)

(B.105d)

(B.105e)

(B.l05f)



For a Y 9y4 transformer, the primary and secondary sequence currents are expressed as

/Po= Y · Vpo

/Pl = Y . (VPl - V8l L120o)

lp2 = Y · (Vp2 - V82 L- 120°)

/8o = 0

/81 = y . (Val - VPl L - 120°)

/82 = y. (Vs2 - VP2L120°)

In matrix form,

/Po 1 0 0 0

Ivl 0 1 0 0

JP2 0 0 1 0

/8o =y

0 0 0 0

fsl 0 f 0 0

/82 0 0 e 0

0 0

e 0

0 f 0 0

1 0

0 1

where e = -1L120° and f = -1L - 120°.

Vpo

VPl

VP2

V8o

V8l

Vs2

(B.106a)

(B.106b)

(B.106c)

(B.106d)

(B.106e)

(B.106f)

(B.107)



fA 1 0 0 1 1 2 VA 3 3 -3

IB 0 1 0 2 1 1 VB -3 3 3

Ic 0 0 1 1 2 1 Vc 3 -3 3 =y (B.108)

-fa 1 2 1 2 1 1 Va 3 -3 3 3 -3 -3

-Ib 1 1 2 1 2 1 vb 3 3 -3 -3 3 -3

-Ic 2 1 1 1 1 2 Vc -3 3 3 -3 -3 3


B.6 Y 9 -Y Transformers 441

fA= Y [VA+ ( -2 Vc + Va + Vb)/3]

IB = y [VB+ ( -2 Va + Vb + Vc)/3]

Ic = y [Vc + ( -2 Vb + Vc + Va)/3]

Ia = y [(2 VB- Vc- VA)/3 + ( -2 Va +% + Vc)/3]

h = y [(2 Vc- VA- VB)/3 + ( -2% + Vc + Va)/3]

lc = y [(2 VA- VB- Vc)/3 + ( -2 Vc + Va + %)/3]

(B.l09a)

(B.l09b)

(B.l09c)

(B.109d)

(B.109e)

(B.l09f)




/Po= Y · VPo

lp1 = Y · (Vp1 - Va 1 L180°) = Y • (Vp1 + Va 1 )

lp2 = Y · (Vp2 - Va2L- 180°) = Y · {Vp2 + Va2)

lao= 0

Ia 1 = y · (Vs1 - Vv1 L -180°) = y · (Vs1 + Vv1 )

Ia2 = y. (Vs2 - VP2L180°) = y. (Vs2 + Vp2)

In matrix form,

!Po 1 0 0 0 0 0 Vpo

JP1 0 1 0 0 e 0 VPl

IP2 0 0 1 0 0 f VP2

I so

=y 0 0 0 0 0 0 Vao

Ial 0 f 0 0 1 0 Val

182 0 0 e 0 0 1 va2

where e = -1L180° = 1 and f = -1L- 180° = 1.

(B.llOa)

(B.llOb)

(B.llOc)

{B.llOd)

(B.llOe)

{B.llOf)

(B.111)

The phase admittance matrix Y can now be calculated as Y = A Yo12 A - 1


IA 1 0 0 2 1 1 VA 3 -3 -3

IB 0 1 0 1 2 1 VB -3 3 -3

Ic 0 0 1 1 1 2 Vc -3 -3 3 =y {B.112)

-I a 2 1 1 2 1 1 Va 3 -3 -3 3 -3 -3

-h 1 2 1 1 2 1 Vb -3 3 -3 -3 3 -3

-Ic 1 1 2 1 1 2 Vc -3 -3 3 -3 -3 3



IA = y [VA+ (2 Va- Vb- Vc)/3]

IB = y [VB+ (2 vb- Vc- Va)/3]

Ic = y [Vc + (2 Vc- Va- Vb)/3]

la = y [(-2VA +VB+ Vc)/3 + (-2Va + Vb + Vc)/3]

h = y [( -2 VB+ Vc + VA)/3 + ( -2 Vb + Vc + Va)/3]

Ic = y [( -2 Vc +VA+ VB)/3 + ( -2 Vc + Va + Vb)/3]

(B.l13a)

(B.l13b)

(B.l13c)

(B.l13d)

(B.113e)

(B.l13f)


B.6.5 Y9 y8 Transformers


Ipo = Y. Vpo (B.l14a}

Ip1 = Y · {Vp1 - Vs 1 L240°} (B.l14b}

lp2 = y · {Vp2 - Vs 2 L- 240°} (B.114c}

Iso = 0 (B.l14d}

Isl = Y · (Vs 1 - Vp1 L - 240°} (B.114e}

ls2 = Y · (Vs 2 - Vp2 L240°} (B.114f}

In matrix form,

Ipo 1 0 0 0 0 0 Vpo

JP1 0 1 0 0 e 0 VPl

JP2 0 0 1 0 0 f VP2

I so =y

0 0 0 0 0 0 Vso (B.115}

Isl 0 f 0 0 1 0 Vsl

Is2 0 0 e 0 0 1 Vs2

where e = -1L240° and f = -1L- 240°. The phase admittance matrix Y can now be calculated as Y =A Y 012 A - 1


IA 1 0 0 1 2 1 VA 3 -3 3

IB 0 1 0 1 1 2 VB 3 3 -3

Ic 0 0 1 2 1 1 Vc -3 3 3 =y (B.l16}

-I a 1 1 2 2 1 1 Va 3 3 -3 3 -3 -3

-h 2 1 1 1 2 1 Vb -3 3 3 -3 3 -3

-Ic 1 2 1 1 1 2 Vc 3 -3 3 -3 -3 3



IA = y [VA+ ( -2 vb + Vc + Va)/3]

IB = y [VB+ (-2Vc + Va + %)/3]

Ic = y [Vc + ( -2 Va + Vb + Vc)/3]

Ia = y[(2Vc- VA- VB)/3+ (-2Va + Vb + Vc)/3]

h = y [(2VA- VB- Vc)/3 + (-2% + Vc + Va)/3]

Ic = y [(2VB- Vc- VA)/3 + (-2Vc + Va + %)/3]

(B.l17a)

(B.l17b)

(B.l17c)

(B.l17d)

(B.l17e)

(B.l17f)




Jpo = Y. Vpo (B.l18a)

fp 1 = Y · {Vp1 - V 81 L300°) (B.l18b)

fp2 = y · (Vp2 - V 82 L- 300°) (B.118c)

180 = 0 (B.l18d)

181 = Y · (V8l - Vp1 L - 300°) (B.l18e)

182 = y · (V82 - Vp2L300°) (B.l18f)

In matrix form,

Ipo 1 0 0 0 0 0 Vpo

JP1 0 1 0 0 e 0 VPl

JP2 0 0 1 0 0 f VP2 =y

V8o (B.119)

J8o 0 0 0 0 0 0

J81 0 f 0 0 1 0 v81

/82 0 0 e 0 0 1 v82

where e = -1L300° and f = -1L- 300°. The phase admittance matrix Y can now be calculated as Y = A Y 012 A -l


iA 1 0 0 1 1 2 VA -3 -3 3

IB 0 1 0 2 1 1 VB 3 -3 -3

Ic 0 0 1 1 2 1 Vc -3 3 -3

=y (B.120) -fa 1 2 1 2 1 1

Va -3 3 -3 3 -3 -3

-h 1 1 2 1 2 1 vb -3 -3 3 -3 3 -3

-Ic 2 1 1 1 1 2 Vc 3 -3 -3 -3 -3 3


B.6 Y 9 - Y Transformers 44 7

JA = Y [VA+ (2 Vc- Va- Vb)/3]

IB = y [VB+ (2 Va- Vb- Vc)/3]

Ic = y [Vc + (2 Vb- Vc- Va)/3]

Ia = y [( -2 VB+ Vc + VA)/3 + ( -2 Va + Vb + Vc)/3]

h = y [( -2 Vc +VA+ VB)/3 + ( -2 Vb + Vc + Va)/3]

Ic = y[(-2VA +VB+ Vc)/3+ (-2Vc + Va + Vb)/3]

(B.l21a)

(B.l2lb)

(B.l21c)

(B.121d)

(B.l21e)

(B.l21f)


B. 7 Y-Y 9 Transformers

B. 7.1 Y y 9 0 Transformers

For a Yy9 0 transformer, the primary and secondary sequence currents are expressed as

lp0 = 0

JPt = Y · (VPI - Vst)

JP2 = Y · (VP2 - Vs2)

fso = Y · V.o

lsi = Y · (V,l- Vpl)

ls2 = Y · (V,2 - VP2)

In matrix form,

lpo 0 0 0

JPt 0 1 0

IP2 0 0 1

I so =y

0 0 0

fst 0 -1 0

182 0 0 -1

0 0

0 -1

0 0

1 0

0 1

0 0

0

0

-1

0

0

1

Vpo

VPl

VP2

V.o

Vst

Vs2

(B.122a)

(B.122b)

(B.122c)

(B.122d)

(B.122e)

(B.122f)

(B.123)

The phase admittance matrix Y can now be calculated as Y =A Yo12 A - 1


fA 2 1 1 2 1 1 VA 3 -3 -3 -3 3 3

fB 1 2 1 1 2 1 VB -3 3 -3 3 -3 3

Ic 1 1 2 1 1 2 Vc -3 -3 3 3 3 -3 =y (B.124)

-fa 2 1 1 1 0 0 Va -3 3 3

-h 1 2 1 0 1 0 vb 3 -3 3

-fc 1 1 2 0 0 1 Vc 3 3 -3

This matrix is similar to that for the Y9 y0 transformer except that the upper left and lower right sub-matrices are interchanged.


B. 7 Y-Y 9 Transformers 449

lA = y [(2 VA- VB- Vc)/3 + ( -2 Va + Vb + Vc)/3]

IB = y [(2 VB - Vc - VA)/3 + ( -2 Vb + Vc + Va)/3]

lc = y [(2Vc- VA- VB)/3 + (-2Vc + Va + Vb)/3]

la = y [(2VA- VB- Vc)/3- Val

Ib = y [(2 VB - Vc - VA)/3- Vb]

lc = y [(2Vc- VA- VB)/3- Vel

(B.125a)

(B.125b)

(B.l25c)

(B.125d)

(B.l25e)

(B.l25f)


B.8 Y-Y 'fransformers

B.8.1 YyO Transformers

For a YyO transformer, the primary and secondary sequence currents are expressed as

lp0 = 0 (B.126a)

JPI = Y · (V,l - Vst) (B.126b)

JP2 = Y · (VP2 - Vs2) (B.126c) 180 = 0 (B.126d)

fst = Y ·(Val - VP1) (B.126e)

ls2 = Y · (Vs2 - Vv2) (B.126f)

In matrix form,

lpo 0 0 0 0 0 0 Vpo

IPl 0 1 0 0 -1 0 VPl

IP2 0 0 1 0 0 -1 VP2

fs 0

=y 0 Vso

(B.127) 0 0 0 0 0

lsi 0 -1 0 0 1 0 Vs1

ls2 0 0 -1 0 0 1 Vso

The phase admittance matrix Y can now be calculated as Y =A Y012 A - 1


!A 2 1 1 2 1 1 VA 3 -3 -3 -3 3 3

IB 1 2 1 1 2 1 VB -3 3 -3 3 -3 3

Ic 1 1 2 1 1 2 Vc -3 -3 3 3 3 -3 =y (B.128)

-fa 2 1 1 2 1 1 Va -3 3 3 3 -3 -3

-h 1 2 1 1 2 1 \tb 3 -3 3 -3 3 -3

-Ic 1 1 2 1 1 2 Vc 3 3 -3 -3 -3 3


B.8 Y-Y Transformers 451

fA= y [(2VA- VB- Vc)/3 + (-2Va + Vb + Vc)/3] IB = y[(2VB- Vc- VA)/3+ (-2% + Vc + Va)/3] Ic = y [(2 Vc- VA- VB)/3 + ( -2 Vc + Va + Vb)/3]

Ia = IA

h =IB

Ic = Ic

(B.129a)

(B.l29b)

(B.l29c)

(B.l29d)

(B.l29e)

(B.l29f)


B.8.2 Yy2 Transformers

For a Yy2 transformer, the primary and secondary sequence currents are expressed as

lp0 = 0

/P1 = y . (VP1 - Vs1 L60o)

lp2 = y · (Vp2 - V82 L- 60°)

/8o = 0

/81 = y. (V,1- Vp1L- 60o)

Is2 = y. (V,2 - Vp2L60o)

In matrix form,

lpo 0 0 0 0 0 0

JP1 0 1 0 0 e 0

/p2 0 0 1 0 0 I

I8o

=y 0 0 0 0 0 0

/81 0 f 0 0 1 0

/82 0 0 e 0 0 1

where e = -1L60° and I= -1L- 60°.

Vpo

VP1

VP2

Yso

v,1

vs2

(B.130a)

(B.130b)

(B.130c)

(B.130d)

(B.130e)

(B.130f)

(B.131)

The phase admittance matrix Y can now be calculated as Y =A Y 012 A - 1


IA 2 1 1 1 2 1 VA 3 -3 -3 -3 3 -3

IB 1 2 1 1 1 2 VB -3 3 -3 -3 -3 3

Ic 1 1 2 2 1 1 Vc -3 -3 3 3 -3 -3 =y (B.132)

-I a 1 1 2 2 1 1 Va -3 -3 3 3 -3 -3

-Ib 2 1 1 1 2 1 Vb 3 -3 -3 -3 3 -3

-Ic 1 2 1 1 1 2 Vc -3 3 -3 -3 -3 3



IA = y [(2 VA- VB- Vc)/3 + (2 Vb- Vc- Va)/3] IB = Y [(2 VB- Vc- VA)/3 + (2 Vc- Va- V/,)/3] Ic = y [(2 Vc- VA- VB)/3 + (2 Va- VI,- ll;,)/3]

Ia = -Ic

h = -IA

Ic = -IB

(B.133a)

(B.l33b)

(B.l33c)

(B.133d)

(B.l33e)

(B.l33f)




lpo = 0

lPl = y. (VPl -Val L120°)

lP2 = y. (VP2- Va2L- 120°)

lao= 0

lal = y . ('llal - VPl L - 120°)

la2 = y. (V,2 - VP2L120°)

In matrix form,

lpo 0 0 0 0 0 0

lpl 0 1 0 0 e 0

lp2 0 0 1 0 0 f lao

=y 0 0 0 0 0 0

lsl 0 f 0 0 1 0

la2 0 0 e 0 0 1

where e = -1L120° and f = -1L - 120°.

Vpo

VPl

VP2

Vao

Val

Vs2

(B.134a)

(B.134b)

(B.134c)

(B.134d)

(B.134e)

(B.134f)

(B.135)



lA 2 1 1 1 1 2 VA 3 -3 -3 3 3 -3

lB 1 2 1 2 1 1 VB -3 3 -3 -3 3 3

lc 1 1 2 1 2 1 Vc -3 -3 3 3 -3 3 =y (B.136)

-la 1 2 1 2 1 1 Va 3 -3 3 3 -3 -3

-lb 1 1 2 1 2 1 vb 3 3 -3 -3 3 -3

-lc 2 1 1 1 1 2 Vc -3 3 3 -3 -3 3



IA = y [(2VA- VB- Vc)/3 + ( -2 V, + Va + Vb)/3] IB = y [(2 VB- Vc- VA)/3 + ( -2 Va + vb + Vc)/3] Ic = y [(2Vc- VA- VB)/3 + (-2Vb + Vc + Va)/3]

Ia =lB

Ib = Ic

Ic = JA

(B.137a)

(B.137b)

(B.l37c)

(B.l37d)

(B.137e)

(B.l37f)



With -1L:180° = -1L - 180° = 1, the primary and secondary sequence currents, for a Yy6 transformer, are expressed as

fp 0 = 0

JPt = Y · (VPt + Vst)

IP2 = Y · (VP2 + Vs2)

180 = 0

fst = Y · (Vst + Vpt)

ls2 = Y · (Vs2 + VP2)

In matrix form,

lpo 0 0

JPt 0 1

lp2 0 0

I so

=y 0 0

fst 0 1

182 0 0

0 0 0

0 0 1

1 0 0

0 0 0

0 0 1

1 0 0

0

0

1

0

0

1

Vpo

VPl

VP2

Vs 0

Vst

Vs2

(B.138a)

(B.138b)

(B.138c)

(B.138d)

(B.138e)

(B.138f)

(B.139)



!A 2 1 1 2 1 1 VA 3 -3 -3 3 -3 -3

Is 1 2 1 1 2 1 Vs -3 3 -3 -3 3 -3

Ic 1 1 2 1 1 2 Vc -3 -3 3 -3 -3 3 =y (B.140)

-fa 2 1 1 2 1 1 Va 3 -3 -3 3 -3 -3

-h 1 2 1 1 2 1 vb -3 3 -3 -3 3 -3

-fc 1 1 2 1 1 2 Vc -3 -3 3 -3 -3 3



IA = y [(2VA- VB- Vc)/3 + (2Va- V&- Vc)/3]

lB = y [(2 VB- Vc- VA)/3 + (2 V&- Vc- Va)/3]

lc = y [(2 Vc- VA- VB)/3 + (2 Vc- Va- V&)/3]

la = -IA

l& = -lB

Ic = -Ic

(B.l41a}

(B.141b}

(B.l41c}

(B.141d}

(B.141e}

(B.l41f)




lp0 = 0

lp1 = y · (Vp1 - V 81 L240°)

lp2 = Y · (Vp2 - V 82 L- 240°)

/80 = 0

181 = Y · (Vs1 - Vp1 L- 240°)

/82 = y . (Vs2 - VP2 L240°)

In matrix form,

lpo 0 0 0 0 0 0

/Pl 0 1 0 0 e 0

Jp2 0 0 1 0 0 f I so

=y 0 0 0 0 0 0

/81 0 f 0 0 1 0

182 0 0 e 0 0 1

where e = -1L240° and f = -1L - 240°.

Vpo

VPl

VP2

Vso

V81

Vs2

(B.142a)

(B.142b)

(B.142c)

(B.l42d)

(B.142e)

(B.142f)

(B.143)



IA 2 1 1 1 2 1 VA 3 -3 -3 3 -3 3

IB 1 2 1 1 1 2 VB -3 3 -3 3 3 -3

Ic 1 1 2 2 1 1 Vc -3 -3 3 -3 3 3 =y (B.I44)

-I a 1 1 2 2 1 1 Va 3 3 -3 3 -3 -3

-h 2 1 1 1 2 1 vb -3 3 3 -3 3 -3

-Ic 1 2 1 1 1 2 Vc 3 -3 3 -3 -3 3


B.B Y-Y Transformers 459

fA= y [(2VA- VB- Vc)/3- (2Vb- Vc- Va)/3]

fB = y [(2 VB - Vc - VA)/3- (2 Vc- Va - Vb)/3]

fc = y[(2Vc- VA- VB)/3- (2Va- Vb- Vc)/3]

fa= fc

fb =fA

fc = fB

(B.l45a)

(B.l45b)

(B.145c)

(B.145d)

(B.l45e)

(B.l45f)


B.8.6 YylO Transformers


lp0 = 0

lp1 = y · (Vp1 - Vs1 L300°)

JP2 = Y · (Vp2 - Vs2L- 300°)

lso = 0

fs 1 = Y · (Vs 1 - Vp1 L - 300°)

ls2 = y · (Vs2 - Vp2 L300°)

In matrix form,

[Po 0 0 0 0 0 0

IP1 0 1 0 0 e 0

IP2 0 0 1 0 0 f I so

=y 0 0 0 0 0 0

fs1 0 f 0 0 1 0

ls2 0 0 e 0 0 1

where e = -1L300° and f = -1L - 300°.

Vpo

VP1

VP2

Vso

Vs1

Vs2

(B.146a)

(B.146b)

(B.146c)

(B.146d)

(B.146e)

(B.146f)

(B.147)

The phase admittance matrix Y can now be calculated as Y = A Y o12 A - 1


IA 2 1 1 1 1 2 VA 3 -3 -3 -3 -3 3

Is 1 2 1 2 1 1 Vs -3 3 -3 3 -3 -3

Ic 1 1 2 1 2 1 Vc -3 -3 3 -3 3 -3 =y (B.148)

-fa 1 2 1 2 1 1 Va -3 3 -3 3 -3 -3

-h 1 1 2 1 2 1 Vb -3 -3 3 -3 3 -3

-lc 2 1 1 1 1 2 Vc 3 -3 -3 -3 -3 3



IA = y [(2 VA- VB- Vc)/3 + (2Vc- Va -li)/3]

IB = y [(2 VB - Vc- VA)/3 + (2 Va- Vb- Vc)/3]

Ic = y [(2 Vc- VA- Vs)/3 + (2 Vb- Vc- Va)/3]

Ia = -IB

Ib = -Ic

Ic = -IA

(B.l49a)

(B.149b)

(B.l49c)

(B.l49d)

(B.l49e)

(B.149f)


B. 9 ..d - ..d Transformers

The different vector groups for L1- L1 transformers are shown in Fig. B.24.

DdO Dd6

"~-~. Dd4

DeliO Dd8

~ -~· ~ It\ c • \V c •• ~,

Fig. B.24. Vector groups for Ll - Ll transformers

B.9 Ll- Ll Transformers 463

B.9.1 DdO Transformers

Consider the DdO transformer of Fig. B.25.

1: 1 A - A a

JA

B -Is

c -

- a I.

- c c~• .~.

lc I.,

Fig. B.25. A DdO transformer


IA =lAB -leA = Y [(VAB - Vab) - (VeA - Vca)]

= y[(2VA- VB- Ve) + (-2Va + Vb + Vc)] IB =I Be -lAB = Y [(VBe- Vbc) - (VAB - Vab)]

= y [(2 VB- Ve- VA)+ ( -2 vb + Vc + Va)]

Ie =leA -!Be = Y [(VeA- Vca)- (VBe- Vbc)]

= y[(2Ve- VA- VB)+ (-2Yc + Va + Vb)]

la = lba -lac = lAB - leA = IA

lb =feb- ha =I Be -lAB = IB

lc =lac- lcb =leA - lBe = lc

In matrix form,

IA 2 -1 -1 -2 1 1

IB -1 2 -1 1 -2 1

Ie -1 -1 2 1 1 -2

-la =y

-2 1 1 2 -1 -1

-h 1 -2 1 -1 2 -1

-lc 1 1 -2 -1 -1 2

(B.150a)

(B.150b)

(B.150c)

(B.150d)

(B.150e)

(B.150f)

(B.151)

The sequence admittance matrix Y 012 can now be calculated as Y 012



fpo 0 0 0 0 0 0 Vpo

JPl 0 3 0 0 -3 0 VPl

IP2 0 0 3 0 0 -3 VP2

I so =y

Vso (B.l52)

0 0 0 0 0 0

JSl 0 -3 0 0 3 0 Vs1

fs2 0 0 -3 0 0 3 Vs2

From the matrix equation for Yo12 , the primary and secondary sequence currents are given by

lp0 = 0

JPl = 3y · {VPl - VsJ

JP2 = 3y · (VP2- Vs2)

180 = 0

ls1 = 3y · {Vs1 - Vp1)

fs2 = 3y · (Vs2 - VP2)

(B.l53a)

{B.153b)

(B.l53c)

(B.153d)

(B.153e)

(B.l53f)

B.9 .1- .1 Transformers 465

B.9.2 Dd2 Transformers

Consider the Dd2 transformer of Fig. B.26.

1 : 1 A - A

[A

- c B - I,

Is

- a c - I,

Ic C~B 'W'

L.:..----+--~- b b

r.

Fig. B.26. A Dd2 transformer


IA =lAB -leA = Y [(VAs - Vcb) - (VeA - Vba)]

= y[(2VA- Vs- Ve) + (2Vt,- Vc- Va)]

Is= lse- lAB= Y [(Vse- Vac)- (VAs- Vcb)]

= y [(2 Vs - Ve -VA)+ (2 Vc- Va - Vb)]

Ie =leA - Ise = y [(VeA - Vba) - (Vse- Vac)]

= y [(2Ve- VA- Vs) + (2Va- Vb- Vc)]

Ia = ha- lac= lse- leA= -Ie

h =feb- ha =leA- JAB= -JA

Ic =lac- Icb =lAB- lse =-Is

In matrix form,

fA 2 -1 -1 -1

Is -1 2 -1 -1

Ie -1 -1 2 2

-I a =y

-1 -1 2 2

-h 2 -1 -1 -1

-Ic -1 2 -1 -1

2 -1

-1 2

-1 -1

-1 -1

2 -1

-1 2

(B.154a)

(B.154b)

(B.154c)

(B.154d)

(B.154e)

(B.154f)

(B.155)

The sequence admittance matrix Y 012 can now be calculated as Yo12



/Po 0 0 0 0 0 0 Vpo

JPl 0 3 0 0 e 0 VPl

IP2 0 0 3 0 0 f VP2

I so

=y 0 0 0 0 0 0 Vso

(B.156)

fsl 0 f 0 0 3 0 Vsl

fs2 0 0 e 0 0 3 Vs2

where e = -3L60° and f = -3L - 60°. From the matrix equation for Y 012 , the primary and secondary sequence

currents are given by

lpo = 0

fp 1 = 3 Y · (Vp 1 - V 81 L60°)

IP2 = 3y · (Vp2 - Vs 2L- 60°)

180 = 0

ls 1 = 3y · (Vs 1 - Vp 1 L- 60°)

fs 2 = 3y · (Vs2 - Vp2L60°)

(B.l57a)

(B.l57b)

(B.157c)

(B.l57d)

(B.l57e)

(B.157f)




1:1 A --+ --+ c A c

JA I,

8 - - a

Ia '· ~~ C 8 b a c - - b

Ic Ih



IA =lAB -leA = Y [(VAB - Vca) - (VeA - Vbc)]

= y [(2VA- VB- Ve) + (-2"Vc + Va + Vb)]

IB =!Be- JAB= Y [(VBe- Vab)- (VAB- Vca)]

= y [(2 VB - Ve -VA) + ( -2 Va + vb + Vc)] Ie =leA -!Be = Y [(VeA - Vbc) - (VBe - Vab)]

= y [(2Ve- VA- VB)+ (-2Vb + Vc + Va)]

Ia = ha -lac = !Be -JAB = IB

h =feb- ha =leA- !Be= Ie

Ic =lac- feb =JAB -leA = IA

In matrix form,

fA 2 -1 -1 1 1 -2

IB -1 2 -1 -2 1 1

Ie -1 -1 2 1 -2 1

-fa =y

1 -2 1 2 -1 -1

-h 1 1 -2 -1 2 -1

-Ic -2 1 1 -1 -1 2

(B.158a)

(B.158b)

(B.158c)

(B.158d)

(B.158e)

(B.l58f)

(B.159)

The sequence admittance matrix Y 012 can now be calculated as Y 012

A -l Y A and relates the currents and voltages as


fpo 0 0 0 0 0 0 Vpo

IPl 0 3 0 0 e 0 VPl

IP2 0 0 3 0 0 f VP2

fso =y

0 0 0 0 0 Vso (B.160)

0

fsl 0 f 0 0 3 0 Vs1

fs2 0 0 e 0 0 3 Vs2

where e = -3L120° and f = -3L - 120°. From the matrix equation for Yo12, the primary and secondary sequence


Ipo = 0

Iv1 = 3 y · (Vv1 - "Vs 1 L120°)

IP2 = 3y · (Vp2 - Vs2L -120°)

180 = 0

fs 1 = 3y · (V81 - Vp1L -120°)

fs 2 = 3 Y · (V82 - Vp2 L120°)

(B.161a)

(B.161b)

(B.161c)

(B.161d)

(B.161e)

(B.161f)




1 : 1

A

cIc

-- b I, ~-~·

C B \V -- a

'· Fig. B.28. A Dd6 transformer


IA =lAB -leA = Y [(VAB - Vba) - (VeA - Vac)]

= y [(2VA- VB- Ve) + (2Va- Vb- Vc)]

IB =I Be- JAB= Y [(VBe- Vcb)- (VAB- Vba)]

= y [(2 VB - Ve -VA)+ (2 Vb - Vc- Va)]

Ie =leA -!Be = Y [(VeA - Vac) - (VBe- Vcb)]

= y [(2Ve- VA- VB)+ (2Vc- Va- Vb)]

Ia = ha- lac= leA- JAB= -IA

h =feb- ha =JAB- fBe = -JB

Ic =lac- feb= I Be- leA= -Ie

In matrix form,

h 2 -1 -1 2 -1 -1

IB -1 2 -1 -1 2 -1

Ie -1 -1 2 -1 -1 2

-fa =y

2 -1 -1 2 -1 -1

-h -1 2 -1 -1 2 -1

-Ic -1 -1 2 -1 -1 2

a

(B.162a)

(B.162b)

(B.162c)

(B.162d)

(B.162e)

(B.162f)

(B.163)

The sequence admittance matrix Yo12 can now be calculated as Yo12 A -l Y A and relates the currents and voltages as


fpo 0 0 0 0 0 0 Vpo

JPI 0 3 0 0 3 0 VPl

IP2 0 0 3 0 0 3 VP2

fso =y

Vso (B.l64)

0 0 0 0 0 0

fsl 0 3 0 0 3 0 Vsl

fs2 0 0 3 0 0 3 Vs2


lpo = 0

JPI = 3y · (VPl + Vst)

JP2 = 3y · (VP2 + Vs2) 180 = 0

181 = 3y · (Vs 1 + Vp1 )

182 = 3y · (Vs 2 + Vp2 )

The above result is expected as -1L180° = -lL - 180° = 1.

(B.l65a)

(B.l65b)

(B.l65c)

(B.l65d)

(B.l65e)

(B.165f)

B.9 L1- L1 Transformers 471



1:1 A - A b

~~ C B a e

JA

B -Is - e I.

c - - a Ic I,



IA = lAB -leA = Y [(VAB - Vbc) - (VeA - Vab)]

= y [(2VA- VB- Ve) + (-2Vb + Vc + Va)]

IB =!Be- lAB = Y [(VBe- Vca) - (VAB - Vbc)] = y [(2 VB - Ve - VA) + ( -2 Vc + Va + Vb)]

Ic =leA -!Be = Y [(VeA - Vab) - (VBe- Vca)]

= y [(2 Ve- VA- VB)+ ( -2 Va + Vb + Vc)]

Ia = ha -lac =leA -!Be = Ie

h = Icb - Iba = lAB -leA = IA

lc =lac- feb =!Be- lAB = IB

In matrix form,

IA 2 -1 -1 1 -2 1

IB -1 2 -1 1 1 -2

Ie -1 -1 2 -2 1 1

-Ia =y

1 1 -2 2 -1 -1

-Ib -2 1 1 -1 2 -1

-Ic 1 -2 1 -1 -1 2

(B.166a)

(B.166b)

(B.166c)

(B.166d)

(B.166e)

(B.166f)

(B.167)



!Po 0 0 0 0 0 0 Vpo

JPl 0 3 0 0 e 0 VPl

JP2 0 0 3 0 0 f VP2

fso =y

0 0 0 0 Vso (B.l68)

0 0

fsl 0 f 0 0 3 0 Vsl

182 0 0 e 0 0 3 vs2

where e = -3.L240° and f = -3L- 240°. From the matrix equation for Y 012 , the primary and secondary sequence


lp0 = 0

lp1 = 3 Y · (Vp1 - Vs1 .L240°)

lp2 = 3y · (Vp2 - V82 L- 240°)

180 = 0

ls1 = 3y · (V81 - Vp 1L- 240°)

ls2 = 3y · (V82 - Vp2L240°)

(B.l69a)

(B.l69b)

(B.l69c)

(B.169d)

(B.l69e)

(B.l69f)


B.9.6 DdlO Transformers


A -lA

B -Ia

c -lc

1:1

- a I,

b -•• - c

Ic

Fig. B.30. A DdlO transformer

A

c~• a


lA =lAB -leA = Y [(VAB - Vac) - (VeA - 'Vcb)] = y [(2VA- VB- Ve) + (2Vc- Va- Vb)]

lB =!Be -lAB = Y [(VBe- %a) - (VAB - Vac)]

= y [(2 VB - Ve- VA)+ (2 Va - Vb- Vc)]

Ie =leA -!Be = Y [(VeA- Vcb) - (VBe- Vba)]

= y[(2Ve- VA- VB)+ (2Vb- Vc- Va)]

la = ha- lac= lAB- !Be= -IB

lb =feb- ha =!Be -leA = -Ie

lc =lac- feb= leA- lAB= -IA

In matrix form,

2 -1 -1 -1 -1 2

-1 2 -1 2 -1 -1

-1 -1 2 -1 2 -1 =y

-1 2 -1 2 -1 -1

-1 -1 2 -1 2 -1

2 -1 -1 -1 -1 2

c

b

(B.170a)

(B.170b)

(B.170c)

(B.170d)

(B.170e) {B.170f)

(B.171)



1po 0 0 0 0 0 0 Vpo

1Pl 0 3 0 0 e 0 VPl

1P2 0 0 3 0 0 f VP2

18o =y

0 0 0 0 Yso (B.l72)

0 0

181 0 f 0 0 3 0 v;,l 182 0 0 e 0 0 3 v82

where e = -3L300° and f = -3L - 300°. From the matrix equation for Y 012 , the primary and secondary sequence


1p0 = 0

1Pl = 3 y. (VPl - v;,l L300°)

1p2 = 3y · (Vp2 - li;,2L- 300°)

180 = 0

181 = 3y. (V81- VPlL- 300°)

182 = 3y · (V82 - Vp2 L300°)

(B.l73a)

(B.173b)

(B.173c)

(B.173d)

(B.l73e)

(B.l73f)

B.lO Summary 475

B.lO Summary

The phase admittance matrix Y _ = _A _Yo12 A -l of a three-phase transformer relates the primary (p or ABC) and secondary (s or abc) currents and voltages as

(B.l74)

where I is the vector of nodal current injections. The self and mutual admittance submatrices of the phase admittance

matrix of a three-phase two-winding transformer are summarised in Table B.l for nine different connections. The different connections and vector groups of three-phase two-winding transformers can be represented by two matrices, namely Y u and Yv given by Equations 8. 71 and 8. 72. The two other matrices Y w and Y x given by Equations 8. 73 and 8. 75 can be expressed in terms of Y u and Y v as shown next. Y k and Y1 have been defined as Y k = Yv + Yr and Yz = Y w - Y~ = (Y v - Yr) I v'3 for simplicity.

Table B.l. Admittance matrices for three-phase transformers

Ypp Yps Ysp = Y~s Yss

Y9 dl Yu Yw yT w yk

Dy9 1 yk Yw yT w Yu

Ydl yk Yw yT w yk

Dyl yk Yw yT w yk

DdO 3Yk -3Yk -3Yf 3Yk

Y 9 y9 0 Yu -Yu _yT u Yu

Y 9 y0 Yu -Yk -Yf yk

Yy9 0 yk -Yk yT - k Yu

YyO yk -Yk yT - k yk

Yk = Yv + Yr Y w = -(Yv + 2Yr)/v'3

Yf=Yk Y~=Yu


where

(B.175)

(B.176)

(B.177)

-1 -1) 2 -1

-1 2 (B.178)

and y is the transformer admittance in pu. As the vector group varies, the self admittance matrices Y PP and Y 88 re

main the same and only the mutual admittance matrix Yp8 varies. Table B.2 shows the mutual admittance matrix Y p8 for the different vector groups of Dy, Yd, Yy and Dd transformers.

Table B.2. The mutual admittance matrix Yps for different vector groups

n Y9 d, Yd

n YgYg Yyy

Dd Dy9 , Dy Yy

1 Yw 0 -Yu -Yk -3Yk

3 Yz 2 Yx _yT v -3YT v

5 _yT w 4 _yT

X Yv 3Yv

7 -Yw 6 Yu yk 3Yk

9 -Yz 8 -Yx yT v 3YT v 11 yT w 10 yT

X -Yv -3Yv

Y w = -(Yv + 2Y'[)/J3

Yz = Yw- Y~ = (Yv- Y'[)jJ?,

Yk=Yv+Y'[

Yx = Yu + J3Yw = Yu- (Yv + 2Y'[)

with Y x and Yz given by

1 -1 ) 0 1

-1 0

and y being the transformer admittance in pu. Noting that

1 1 ) -2 1 1 -2

it is concluded that

T -1 T Y v + Y v = J3 (Y w + Y w)

Y v - Y~ = J3 (Y w - Y~)

B.lO Summary 477

(B.179)

(B.180)

(B.181)

(B.182)

(B.183)

(B.184)

Solving the above two equations simultaneously for Y v and Y w gives

1 T Y v = J3 (Y w - 2 Y w)

-1 T Y w = J3 (Yv + 2 Y v )

Finally, one can see that

Yx =Yu+J3Yw

= Yu- (Yv + 2Y~).

(B.185)

(B.186)

(B.187a)

(B.187b)

To summarise, the phase admittance matrix of a three-phase two-winding transformer can be constructed with the aid of Tables B.1 and B.2. The procedure is:

• For the desired transformer, pick Ypp and Yss from Table B.l. These matrices are not affected by the vector group.

• For any specific vector group of the chosen transformer, pick Y ps from Table B.2. Ysp = Y~s·

• Then, the phase admittance matrix is

y = ( Ypp Yps ) Ysp Yss


Having found the phase admittance matrix, the sequence admittance matrix is computed as Yo12 =A -l Y A.

General expressions can then be given for the primary and secondary sequence currents as

lp0 = Y · (Vp0 =F Vs0 )

I so = Y · (V:,o =F Vpo) JPl = k y . (VPl - v:,l LO)

lp2 = k y · (Vp2 - V:, 2 L- 0) 181 = ky · (V:,1 - Vp1 L- 0)

182 = k Y · (V:,2 - Vp2 LO).

where

• k = 3 for Ll - Ll transformers and 1 otherwise • 0 = 30° · n • n is the vector group • For the zero sequence currents,

- - is for n = 0, 4, 8 and + for n = 2, 6, 10 - Vp0 and V:,0 only exist for a Y 9 winding - V Po = Vso = 0 for Y and Ll windings.

(B.188a)

(B.188b)

(B.188c)

(B.188d)

(B.188e)

(B.188f)

C. Transmission Matrices for Three-Phase Transformers

In this Appendix, matrices for the transmission of voltage and current from primary to secondary and vice versa are derived for all vector groups of twowinding Dy, Y d, Dd and Yy transformers.

C.l .1-Y and Y- .1 Transformers 481

C.l .Ll-Y andY -.Ll Transformers

Let a be the transformation ratio, that is the ratio of secondary to primary line voltages, and N be the turns ratio, that is the ratio of secondary to primary phase voltages, then

For a Y-Ll transformer,

In pu,

For aLl-Y transformer,

In pu,

n:=l

N=-1 J3

With k defined as

k={ ~, J3'

Ll - Y transformers

Y - Ll transformers

the transformation ratio becomes

a=kN

(C.l)

(C.2)

(C.3)

(C.4)

(C.5)

(C.6)

(C.7)

(C.8)

(C.9)

(C.lO)

482 C. Transmission Matrices for Three-Phase Transformers

C.l.l Y -L1 Transformers

The different vector groups for Y -Ll transformers were presented in Fig. B.l.

C.l.l.l Ydl Transformers

Secondary currents can be expressed as

( fa ) ( 1 0

Iabc = lb = -1 1 fc 0 -1

~ laa ( ~ laa (

1 0 -1 1

0 -1

1 0 -1 1

0 -1

= !:_ I ( 117 ~ 13s0;o )

o: 1L90°

=> Iabc = !:_lABeL - 30° 0:

Primary voltages can be expressed as

(C.ll)

(C.12)

C.l ..1-Y andY -..1 Transformers 483

C.1.1.2 Yd3 Transformers



V ABC ~ ( ~~ ) ~ )a 0 ( ~= ) ~lao ( j -~ -D ( ~) ~Jao(j -~ -i)(~)v = .!_ V ( l~L!~~o )

a lL -150°

=? V ABC = .!_ V abcL90° a

(C.l3)

(C.14)




( la ) ( 1 -1 0 ) ( lea )

Iabe = h = 0 1 -1 lab le -1 0 1 he

0) ( -lA ) -1 -lB 1 -lc

_!)(~)I


V ABC = ( r~ ) = ~ 0 ( r~ ) =~OCi -: j)(~) = ~ ° Ci -: j )( ~ ) v

= ~ v ( 11~135~0 )---Q 1L- 90°

=? V ABC = ~ V abeL150° Q

(C.15)

(C.16)

C.l Ll-Y and Y-Ll Transformers 485



I.oc~ 0) ~ ( -~ 0 -1 )( Ioo) 1 0 feb

-1 1 lac

1 ( 1 0

-1 )( -IA) =v'3a -~ 1 0 -IB

-1 1 -Ic

c1 0

j)(~)I 1 1 -1 =v'3a 0 1

( 1L150° ) = ]:_I 1L30° a 1L- 90°

::::} Iabc = ]:_ IABcL150° (C.17) a Primary voltages can be expressed as

V ABC = ( ~~ ) = ~a ( ~~ ) Vc Vac

- 1 ( -~ -~ -v'3a 1 0

~ laa c~ -~ = ]:_ V ( 1~~9~~oo )

a 1L- 30°

1 v 0 ::::} V ABC = - abeL - 150 a

(C.18)




( la ) ( 1 0

labc = h = -1 1 lc ,0 -1

(C.19)


(C.20)

C.l Ll-Y and Y-Ll Transformers 487

C.1.1.6 Ydll Transformers


(C.21)


(C.22)


C.1.2 ...:1-Y Transformers

The different vector groups for~-Y transformers were presented in Fig. B.14.

C.1.2.1 Dyl Transformers

Secondary voltages can be expressed as

c·) cAC) Vabc = vb = ~ vBA Vc VeB

~:a ( -i 0

-~ )( ~) 1 -1

~:a( 1 0 -n ( ~) v -1 1 0 -1

= a V ( 11j ~ 13s~o )

1L90°

:::} Vabc =a V ABeL- 30° (C.23)

Primary currents can be expressed as

IABC ~ ( ~; ) ~ ( ~ -1 0) CAC) 1 -1 fBA Ie -1 0 1 feB

( I -1 -OU) - a 0 1

- J3 -1 0

~:a( 1 -1 -n (~)I 0 1

-1 0

~a! ( 1L30° ) 1L- 90° 1L150°

:::} I ABe = a IabcL30° (C.24)


C.1.2.2 Dy3 Transformers


v.k ~ ( ~ ) ~ ~ ( ~n ~ ~ ( -: j -i )( ~ ) ~ ~ ( -: j -i )( ~ ) v

= a V ( 1fL~5~~o ) 1L30°

=> Vabc = aV ABeL- 90°


( lA ) ( 1 0 -1 ) ( lAB )

IABe = IB = -1 1 0 !Be Ie 0 -1 1 leA

-D 0:) -i )(~)I

(C.25)

(C.26)




=> Vabc = aVABCL -150° (C.27)


( lA ) ( 1 0

IABC = IB = -1 1 Ic 0 -1

-~) ( ~~~) 1 leA

0 -1) ( -fa ) 1 0 -Ib

-1 1 -Jc

j)(~)J

(C.28)

C.l .1-Y andY -.1 Transformers 491



V abc = ( ~ ) = ~ ( ~~; ) Vc VBc

- a ( -~ -~ -v'3 0 1

- ~ ( -~ -~ -v'3 0 1

( 1L150° )

= aV 1L30° 1L- 90°

=> V abc = a V ABcL150° (C.29)


-~ 0 ( 1

- v'3 -1

( -1

= Js ~

( 1L -150° )

=a! 1L90° 1L- 30°

(C.30)




- ~ ( ~ -~ - v'3 -1 1

-~ 1 0 ( 0 -1

- v'3 -1 1

= aV ( 1~L~~~o ) 1L -150°

=> Vabc = a V ABcL90° (C.31)


(C.32)


C.1.2.6 Dyll Transformers


:::} V abc = a V ABC L30° (C.33)


(C.34)


C.1.3 Summary

Summarising the results of our detailed derivation, transmission matrices, T, for phase voltages and line currents across three-phase ..1-Y, Y -..1, Z-Y and Y -Z transformers are presented in Table C.1 and Fig. C.l. In general, primary (ABC) and secondary (abc) voltages and currents are related as

Yabc =a· V ABeL-()

1 V ABC = - · V abeL()

a

1 labc = - ·lABeL-()

a

() = n · 30°

where

(C.35)

(C.36)

(C.37)

(C.38)

(C.39)

a is the transformation ratio, that is the ratio of secondary to primary line voltages

() is the phase shift n is the vector group number, n = 1, 3, 5, 7, 9 and 11.

In terms of the transmission matrices, primary (ABC) and secondary (abc) voltages as well as primary and secondary currents are related as

v abc = a . T . v ABC

. 1 . V ABC = - · T · V abc

a

. 1 . labc = - · T · IABC

a

iABC = a · T · iabc

(C.40)

(C.41)

(C.42)

(C.43)

C.l .1-Y andY -.1 Transformers 495

Table C.l. Transmission matrices for .1-Y, Y -.1, Z-Y andY -Z transformers

n P-tS S-tP

1 T TT

3 T-TT TT -T

5 -TT -T

7 -T -TT

9 TT-T T-TT

11 TT T

T=)J( -: 0

-~) 1

-1

PRIMARY to secondary . . . Vabc=a·T·VASe labc = 1/a · T · lASe

1 5

T =_1_[ -~ 0 ·n [·1 0

-~] 1 1 1 -1 3

1 ·1] -./3 0 :j3 0 9 -1 1

1 [ 0 1 [ 0 -1 ·i] T3 -~ 0 1 11 7 - 1 0 -1 0 .! ]"3 ·1

1

[ 1 ·1 0] [ ·1 1 9 _L 0 1 -1 1 0 -1 3

" 3 -1 0 1 ..J3 1 0

1 5 .

VASe = 1/a · T · Vabc lASe = a . T . labc

secondary to PRIMARY

Fig. C.l. Transmission matrices for phase voltages and line currents across .1-Y, Y- .1, Z-Y and Y- Z transformers


C.2 Y-Y, L1 - L1, Z-Z, L1-Z and Z-L1 Transformers

Vector groups for Y-Y and L1- L1 transformers were depicted in Figs. B.l7 and B.24, respectively. Let o: be the transformation ratio, that is the ratio of secondary to primary line voltages, and N be the turns ratio, that is the ratio of secondary to primary phase voltages, then

For a Y-Y transformer,

In pu,

o:=l

N=l

For a L1- L1 transformer,

In pu,

o:=l

N= 1

(C.44)

(C.45)

(C.46)

(C.47)

(C.48)

(C.49)

(C.50)

(C.51)

C.2 Y-Y, .1- .1, Z-Z, .1-Z and Z-.1 Transformers 497

With Ia + h + Ic = 0, one can write

2 I a - Ib - Ic = 3 I a

2Ib-Ic-Ia=3h

2Ic-Ia-h=3lc

In matrix form, this can be expressed as

( fa ) 1 ( 2 -1 -1 ) ( fa ) h = 3 -1 2 -1 Ib fc -1 -1 2 fc

Applying similarity transformation, we get

which confirms the absence of zero-sequence currents. Line-to-line voltages are expressed as

Vab=Va-Vb

Vbc=Vb-Vc

Vca=Vc-Va

Multiplying the first equation by two and subtracting the second and third equations, we get

or

Similar expressions can be obtained for Vbc and Vca- In matrix form, this can be written as

( Vab ) 1 ( 2 -1 -1 ) ( Vab ) Vbc = 3 -1 2 -1 Vbc Vca -1 -1 2 Yea


C.2.1 YyO and DdO Transformers

Secondary voltages are expressed as

( Vab ) 1 ( 2 Vbc =- -1

3 Vca -1

~~( On the primary,

2 -1 -1 2 -1 -1

( VAB ) 1 ( 2 -1 -1 ) ( VAB ) Vsc = 3 -1 2 -1 Vsc VcA -1 -1 2 VcA

= 2._ -1 2 -1 ~c ( 2 -1 -1 ) ( V, b )

3 a -1 -1 2 Vca

(C.52)

(C.53)

C.2 Y-Y, Ll- Ll, Z-Z, Ll-Z and Z-Ll Transformers 499

C.2.2 Yy2 and Dd2 Transformers


c®) I ( 2 -1 -1 )( V®) Vbc =- -1 2 -1 Vbc 3 -1 2 'Vca Vca -1

~~ ( =: -1 -1) ( VAc) 2 -1 VBA -1 2 VcB

~~ ( -~ 1

-2) c··) 1 1 VBc (C.54) -2 1 VcA

On the primary,

( v •• ) I ( 2 -I -I ) c·· ) VBc = '3 -1 2 -1 VBc VcA -1 -1 2 VcA

~_!_( -~ -1 =I) c~) 2 1 Vac 3a _ 1 -1 2 Vba

1 ( I -2 l)(v"') -- 1 1 -2 Vbc (C.55) 3a _ 2 1 1 Vca


C.2.3 Yy4 and Dd4 Transformers


( Vab ) 1 ( 2 Vbc =- -1

3 Vca -1

On the primary,

( VAB ) 1 ( 2 -1 -1 ) ( VAB ) VBc = 3' -1 2 -1 VBc VcA -1 -1 2 VcA

~~ ( _; -1 -1) cro) 2 -1 Vab 3a _ 1 -1 2 Vbc cl -1 2 )( v.,)

= 31a -i -1 -1 Vbc 2 -1 Vca

(C.56)

(C.57)

C.2 Y-Y, .1- .1, Z-Z, .1-Z and Z-.1 Transformers 501

C.2.4 Yy6 and Dd6 Dz6 Transformers


c·') 1( 2 -1 ~1 )( v.,) %c =- -1 2 -1 Vbc Vca 3 -1 -1 2 Vca

= i ( =l -1 ~1 )( VBA)

2 -1 VcB -1 2 VAc

c2 1 1) CAB) = ~ 1 -2 1 VBc (C.58) 3 1 1 -2 VcA

On the primary,

CAB) 1 ( 2 ~1 ~1) CAB) VBc = 3 -1 2 -1 VBc VcA -1 -1 2 VcA

= __1_ ( ~i -1 ~1 ) ( Voo ) 2 -1 Vcb 3a _ 1 -1 2 Vac

1 c2 1 1 )( v.,) -- 1 -2 1 Vbc (C.59)

3a 1 1 -2 Vca


C.2.5 Yy8 and Dd8 Dz8 'fransformers


( v~) 1 ( 2 -1 -1) c··) Vbc =- -1 2 -1 Vbc Vca 3 -1 -1 2 Vca

~i ( =: -1 -1) CCA) 2 -1 VAB -1 2 VBc cl -1 2) CAB) = ~ 2 -1 -1 VBc

3 -1 2 -1 VcA (C.60)

On the primary,

( VAB ) 1 ( 2 -1 -1 ) ( VAB ) VBc = 3 -1 2 -1 VBc VcA -1 -1 2 VcA

~ 1_ ( -~ -1 =: )( ~~) 2 3a _ 1 -1 2 Vab cl 2 -1) c~)

= 31a -~ -1 2 \'be

-1 -1 Vca (C.61)

C.2 Y-Y, ..1- ..1, Z-Z, ..1-Z and Z-..1 Transformers 503

C.2.6 YylO and DdlO Transformers


c··) 1( 2 -1 -1) c··) Vbc = 3 -1 2 -1 Vbc Vca -1 -1 2 Vea

~ ~ ( ::: -1 -1) ccs) 2 -1 VAc -1 2 VBA

~~ ( : -2 1 )(VAs) 1 -2 VBc (C.62) 3 -2 1 1 VcA

On the primary,

(VAs) 1 ( 2 -1 -1 )(VAs) VBc = 3 -1 2 -1 VBc VcA -1 -1 2 VcA

~ 1_ ( _; -1 -1)(v"') 2 -1 vba

3a _ 1 -1 2 Vcb

~ 1_ ( -~ 1 -2) c®) 1 1 Vbc (C.63) 3a 1 -2 1 Vca


C.2. 7 Summary

Based on the presented derivation, transmission matrices, Te, for line voltages and currents across three-phase Y-Y, L1 - L1, Z-Z, L1-Z and Z-L1 transformers are summarised in Table C.2 and Fig. C.2. a is the transformation ratio, that is the ratio of line voltages. n = 0, 2, 4, 6, 8, and 10 represents the vector group, with the phase shift (} equal to n · 30°. In general, primary (ABC) and secondary (abc) voltages and currents are related through the transmission matrices as

Vabc = a . T e . v ABC

. 1 . V ABC = - · Te · V abc

a

. 1 . Iabc = - · T e · I ABC a

iABC = a . T e . iabc

(C.64)

(C.65)

(C.66)

(C.67)

Table C.2. Transmission matrices for Y-Y, Ll- Ll, Z-Z, Ll-Z and Z-Ll transformers

n P--tS S--tP

0 Te+Tr Te+Tr

2 Te TT e 4 -TT e -Te

6 -T -TT e e -T -TT e e

8 -Te -TT e

10 TT e Te

T.= j ( 1 1 -:) -2 1

1 -2

C.2 Y-Y, ..:1- ..:1, Z-Z, ..:1-Z and Z-..:1 Transformers 505

PRIMARY to secondary . . labc = 1/a · T8 • IAac

2 4 -~ 2 -1 ·1

....6-. 0 T,= ~ p ~ -~] 1 [ ~1] :_:_1 3 -1 2 -1 10

-1 -1 2 -

.1[ ~ -~ -~]

[·1 -1 2] -1 2 -1

-8 -[-1 2 ·1]

1 [·2 1 1 ] 3 1 -2 1 1 1 -2

3 -2 1 1 ~ -1 ·1 2

2 -1 ·1 - -2 4 ... . . .

VABc = 1/a. T •• Vabc IABc =a. T. ·labc

secondary to PRIMARY

Fig. C.2. Transmission matrices for Y-Y, ..:1- ..:1, Z-Z, ..:1-Z and Z-..:1 transformers


C.3 The Transmission Matrices T and Te

Following the detailed discussion on the transmission matrices T and T e, let's try to see how they are related.

1 ( I 0 -n T, ~ ~ ( -~

1 -n T=y3 -~ 1 1 -1 -2

TT=_I ( ~ -1 -n ~~~( : -2 -D 1 1 v'3 -1 0

e 3 1 -2

1 ( 2 -1 -1) 1 ( 2 -1 -1) T+TT = y3 =~ 2 -1 Te+Tr= 3 -1 2 -1

-1 2 -1 -1 2

1 ( 0 1 -n T,-T;~ ( -:

1 -n T-TT = y3 -~ 0 0 -1 -1

The above equations show that

T - TT = __!__ (T - TT) y3 e e (C.68)

T+TT = v'3(Te +T;) (C.69)

Solving simultaneously, we get

1 T = y3 (2Te + Tr) (C.70)

1 T Te= y3(2T-T ). (C.71)

bibliography978-3-662-04343... · 2019-06-12 · trol in electrical power systems. standard, ieee...

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