Enrolamentos

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Enrolamentos dos Transformadores de Potência

Enroulements des Transformateurs de Puissance

Power Transformers Windings

Descritores

Descripteurs

Descriptors

Borne

Borne

Terminal

Borne de Linha

Borne de Ligne

Line Terminal

Borne Neutro

Borne Neutre

Neutral Terminal

Bornes Homólogos

Bornes Homologues

Corresponding Terminals

Enrolamento

Enroulement

Winding

Enrolamento de Alta Tensão

Enroulement Haute Tension

High-Voltage Winding

Enrolamento de Baixa Tensão

Enroulement Basse Tension

Low-Voltage Winding

Enrolamento de Fase

Enroulement de Phase

Phase Winding

Grupos de Ligação

Groupes de Couplage

Groups Connection

Índices Horários

Indices Horaires

Clock Hour Figures

Ponto Neutro

Point Neutre

Neutral Point

Símbolos de Ligações

Symboles des Couplages

Connection Symbols

Transformadores de Potência

Transformateurs de Puissance

Power Transformers

Introdução No seguimento dos artigos que publiquei (em www.pdfcoke.com): “Índices Horários dos Transformadores de Potência” e “Grupos de Ligação dos Transformadores de Potência”, entendo ser útil abordar o tema “Enrolamentos dos Transformadores de Potência”.

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Embora alguns dos enrolamentos, da BT, não estejam normalizados pela C. E. I., julgo ser didáctico apresentá-los. “Enrolamento y0 ”: u y0 = u na = u ∠ 0?⇔ u y0 = u na = u ∠ 0h . “Enrolamento y2 ”: u y2 = u cn = u ∠ − 60?⇔ u y2 = u cn = u ∠ + 2h . “Enrolamento y4 ”: u y4 = u nb = u ∠ − 120?⇔ u y4 = u nb = u ∠ + 4h . “Enrolamento y6 ”: u y6 = u an = u ∠ − 180?⇔ u y6 = u an = u ∠ + 6h . “Enrolamento y8 ”: u y8 = u nc = u ∠ - 240º ⇔ u y8 = u nc = u ∠ + 8h . “Enrolamento y10 ”: u y10 = u bn = u ∠ − 300?⇔ u y10 = u bn = u ∠ + 10h . “Enrolamento d1 ”: u d1 = u ca = 3 ⋅ u ∠ − 30?⇔ u d1 = u ca = 3 ⋅ u ∠ + 1h . “Enrolamento d3 ”: u d3 = u cb = 3 ⋅ u ∠ − 90?⇔ u d3 = u cb = 3 ⋅ u ∠ + 3h . “Enrolamento d5 ”: u d5 = u ab = 3 ⋅ u ∠ − 150?⇔ u d5 = u ab = 3 ⋅ u ∠ + 5h .

“Enrolamento d7 ”: u d7 = u ac = 3 ⋅ u ∠ - 210?⇔ u d7 = u ac = 3 ⋅ u ∠ + 7h . “Enrolamento d9 ”: u d9 = u bc = 3 ⋅ u ∠ − 270?⇔ u d9 = u bc = 3 ⋅ u ∠ + 9h . “Enrolamento d11 ”: u d11 = u ba = 3 ⋅ u ∠ − 330?⇔ u d11 = u ba = 3 ⋅ u ∠ + 11h . “Enrolamento z1 ”: u z1 =

1 1 ⋅ u ∠ − 30?⇔ u z1 = ⋅ u ∠ + 1h . 3 3

“Enrolamento z3 ”: u z3 =

1 1 ⋅ u ∠ − 90?⇔ u z3 = ⋅ u ∠ + 3h . 3 3

“Enrolamento z5 ”: u z5 =

1 1 ⋅ u ∠ − 150?⇔ u z5 = ⋅ u ∠ + 5h . 3 3

“Enrolamento z7 ”: u z7 =

1 1 ⋅ u ∠ − 210º ⇔ u z7 = ⋅ u ∠ + 7h . 3 3

“Enrolamento z9 ”: u z9 =

1 1 ⋅ u ∠ − 270?⇔ u z9 = ⋅ u ∠ + 9h . 3 3

“Enrolamento z11 ”: u z11 =

1 1 ⋅ u ∠ − 330?⇔ u z11 = ⋅ u ∠ + 11h . 3 3

Faço votos para que este documento seja útil e agradeço, antecipadamente, os vossos comentários. A. A. A. C. Barrias

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Algumas definições

Enrolamento

Conjunto das espiras que constituem o circuito eléctrico associado a uma das tensões estipuladas para as quais o transformador foi concebido. Nota – Para um transformador polifásico, o “enrolamento” é o “conjunto dos enrolamentos de fase”.

Enrolamento de alta tensão

Enrolamento de tensão estipulada mais elevada.

Enrolamento de baixa tensão

Enrolamento de tensão estipulada mais baixa.

Enrolamento de fase

Conjunto das espiras que constituem uma fase de um enrolamento polifásico. Nota – O termo “enrolamento de fase” não deve ser utilizado para designar o conjunto de bobinas dispostas numa dada coluna (perna) do núcleo.

Enrolamento primário

Enrolamento que, em serviço, recebe da rede de alimentação a potência activa, real ou efectiva.

Enrolamento secundário

Enrolamento que, em serviço, fornece a potência activa, real ou efectiva ao circuito de utilização.

Terminais homólogos

Terminais dos diferentes enrolamentos de um transformador, identificados com as mesmas letras ou com símbolos correspondentes.

Enrolamento à Direita (Horário)

Se (para percorrer integralmente o enrolamento desde o "terminal de entrada" até ao "terminal de saída") tivermos de caminhar, rodando, no sentido horário.

Enrolamento à Esquerda (Trigonométrico)

Se (para percorrer integralmente o enrolamento desde o "terminal de entrada" até ao "terminal de saída") tivermos de caminhar, rodando, no sentido trigonométrico.

Número de terminais de um enrolamento Neste estudo, apenas consideramos enrolamentos com dois terminais: o "terminal 1" e o "terminal 2".

Designação dos terminais de um enrolamento

Os "terminais 1 e 2 de um enrolamento" são designados, arbitrariamente, por "entrada" e "saída", respectivamente.

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Outras designações comummente utilizadas na literatura técnica são:

Terminal 1

Borne 1

Terminal 1

Entrada

Entrée

Entrance

Início

Début

Beginning

Origem

Origine

Origin

Partida

Départ

Start

Terminal 2

Borne 2

Terminal 2

Chegada

Arrivée

Arrival

Extremidade

Extrémité

Extremity

Fim

Fin

Finish

Saída

Sortie

Exit

Sentidos dos enrolamentos:

Sentido Horário (Espacial)

Anti-trigonométrico

Anti-trigonométrique

Counter-trigonometric

Horaire

Clockwise

Inverse

Inverse

Negativo

Négatif

Negative

Retrógrado

Rétrograde

Retrograde

Dextrorso

Dextrorsu

Horário Inverso

Inversu

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Sentido Trigonométrico (Temporal)

Anti-horário Directo

Directu

Positivo Sinistrorso

Trigonométrico

Anti-horaire

Counter-clockwise

Direct

Direct

Positif

Positive

Trigonométrique

Trigonometric

Sinistrorsu

Definições dos dicionários Movimento Directo: Movimento de um corpo celeste no mesmo sentido que a terra gira em torno do sol. Movimento Retrógrado: Movimento de um corpo celeste no sentido contrário ao que a terra efectua em torno do sol. Dextrorso: Adjectivo com o significado “que se desenvolve, gira, torce, da esquerda para a direita, no sentido em que observamos o movimento dos ponteiros de um relógio. (lat. Dextrorsu). Sinistrorso: Adjectivo com o significado “que se enrola ou move em sentido contrário aos ponteiros de um relógio”. (lat. Sinistrorsu).

Nas figuras seguintes, apresentamos exemplos de “enrolamentos à esquerda e à direita”:

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Enrolamentos à Esquerda

Φ

I

Φ

I

A

A

UAN

UAN II

I

I

I





Enrolamentos à Direita

I

I A

A UAN

UAN I I

I 

 Φ

Φ

Usualmente, os fabricantes de transformadores constroem os enrolamentos com o mesmo sentido de bobinagem: ou à esquerda ou à direita. Nas figuras seguintes apresentamos as duas construções utilizadas.

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Transformador Monofásico com os Enrolamentos à Esquerda Φ

A I +

UAN

Receptor

I -

Φ

N a

Φ

i

+

uan

-

i

Gerador

n Φ

UAN A

I N

i

a

uan n

Transformador Monofásico com os Enrolamentos à Direita Φ

A I +

Receptor

UAN -

N a uan

Φ

Φ

i

+

Gerador -

n Φ

UAN A

I N

a

i

uan n

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Cálculos Justificativos dos Símbolos de Ligação dos Transformadores de Potência Permutações Circulares no Sentido Horário e no Sentido Trigonométrico Dd0 ⇔ D5 - d5 1 FD5 N D N N U= ⋅ ⋅ u ∠Θ D5 − θ d5 ⇔ U = 3 ⋅ D ⋅ u ∠ + 150?−150º ⇔ U = D ⋅ u ∠ 0? 1 f d5 n d nd nd 3

Ou: U D5 u d5

=

3 U ∠ + 5h N U U N N ⋅ ⇔ D = ∠ 0h ⇔ = D ∠ 0h ⇔ U = D ⋅ u ∠ 0h nd u u nd nd 3 u ∠ + 5h

Dd4 ⇔ D5 - d1 1 FD5 N D N N U= ⋅ ⋅ u ∠Θ D5 − θ d1 ⇔ U = 3 ⋅ D ⋅ u ∠ + 150?−30º ⇔ U = D ⋅ u ∠ + 12 0? 1 nd f d1 n d nd 3

Ou: U D5 u d1

=

3 U ∠ + 5h N U U N N ⋅ ⇔ D = ∠ + 4h ⇔ = D ∠ − 4h ⇔ U = D ⋅ u ∠ − 4h nd u u nd nd 3 u ∠ + 1h Dd8 ⇔ D5 (17) - d9

1 FD17 N D N N U= ⋅ ⋅ u ∠Θ D17 − θ d9 ⇔ U = 3 ⋅ D ⋅ u ∠ + 510?−270º ⇔ U = D ⋅ u ∠ + 2 40? 1 f d9 n d nd nd 3

Ou: U D17 3 U ∠ + 17 h N U U N N = ⋅ ⇔ D = ∠ + 8h ⇔ = D ∠ − 8h ⇔ U = D ⋅ u ∠ − 8h u d9 nd u u nd nd 3 u ∠ + 9h

Na página seguinte, apresentamos os “Símbolos de Ligação” Dd0, Dd4 e Dd8.

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Dd0

D5-d5

D5

d5

UD

0 A

A

ud B

b

1

1

2U

B

c

2

UD

8 C

C

a

Dd4

A

ud B

a

1

1

b

2

UD

8 C

C

c

Dd8

A

c

1

2

C

C

3

b

0

c

4

d

C

a

UD

8

8

b

2u

D

B

B

b

3

ud B

1 2U

4

a

d9

UD A

4

D5-d9

D5

0

a

c

ud A

3

0

d

C

2

B

c

c

2u

D

B

8

d1

2U

4

c

3

UD A

b

D5-d1

D5

0

4

a

ud A

3

b

d

C

2

B

0

2u

D

4

a

2

b

c

ud A

b

3

a

8 a

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Yy0 ⇔ Y6 - y6

U=

FY6 N Y 1 N N ⋅ ⋅ u ∠Θ Y6 − θ y6 ⇔ U = ⋅ Y ⋅ u ∠ + 180?−180? ⇔ U = Y ⋅ u ∠ 0? f y6 n y 1 ny ny

Ou: U Y6 1 U ∠ + 6h N U U N N = ⋅ ⇔ Y = ∠ 0h ⇔ = Y ∠ 0h ⇔ U = Y ⋅ u ∠ 0h u y6 1 u ∠ + 6h ny u u ny ny

Yy4 ⇔ Y6 - y2

U=

FY6 N Y 1 N N ⋅ ⋅ u ∠Θ Y6 − θ y2 ⇔ U = ⋅ Y ⋅ u ∠ + 180?−60? ⇔ U = Y ⋅ u ∠ + 12 0? f y2 n y 1 ny ny

Ou: U Y6 1 U ∠ + 6h N U U N N = ⋅ ⇔ Y = ∠ + 4h ⇔ = Y ∠ − 4h ⇔ U = Y ⋅ u ∠ − 4h u y2 1 u ∠ + 2h ny u u ny ny Yy8 ⇔ Y6 (18) - y10

U=

FY18 N Y 1 N N ⋅ ⋅ u ∠Θ Y18 − θ y10 ⇔ U = ⋅ Y ⋅ u ∠ + 540?−300? ⇔ U = Y ⋅ u ∠ + 24 0? f y10 n y 1 ny ny

Ou: U Y18 1 U ∠ + 18h N N U U NY = ⋅ ⇔ Y = ∠ + 8h ⇔ = ∠ − 8h ⇔ U = Y ⋅ u ∠ − 8h u y10 1 u ∠ + 10h ny u u ny ny

Na página seguinte, apresentamos os “Símbolos de Ligação” Yy0, Yy4 e Yy8.

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Yy0

Y6-y6

Y6

y6

UY A

0

B

4

uy

0 1

n

N

2U Y

2

2u

y

n

N

UY C

8 3

n

N

Y6-y2 y2

UY

uy

0

0 1

n

N

y

n

N

n

N

a

b

3

Y6-y10

Y6

y10

UY

uy

0

0 1

N

n

2U Y

b

1 2u

y

4 2

N

n

UY C

4

8

8

Yy8

B

2 uy

UY

A

c

1 2u

4

3

c

3

Y6

2

C

b

uy

2U Y

B

4 2

8

Yy4

A

a

1

4 2

c

uy

8

8 3

N

n

3

a

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Dz0 ⇔ D5 - z5

1 F N U = 2 ⋅ D5 ⋅ D ⋅ u ∠Θ D5 − θ z5 f z5 n z

2 N N ⇔ U = 2 ⋅ 3 ⋅ D ⋅ u ∠ + 150?−150? ⇔ U = ⋅ D ⋅ u ∠ 0? nz

3

3 nz

Ou: U D5 3 U ∠ + 5h N U U 2 ND 2 N = ⋅ ⇔ D = 3 ⋅ ∠ 0h ⇔ = ⋅ ∠ 0h ⇔ U = ⋅ D ⋅ u ∠ 0h 1 u ∠ + 5h nz u z5 u u 3 nz 3 nz 2 3

Dz4 ⇔ D5 - z1

1 F N U = 2 ⋅ D5 ⋅ D ⋅ u ∠Θ D5 − θ z1 f z1 n z

2 N N ⇔ U = 2 ⋅ 3 ⋅ D ⋅ u ∠ + 150?−30? ⇔ U = ⋅ D ⋅ u ∠ + 12 0? nz

3

3 nz

Ou: U D5 3 U ∠ + 5h N U U 2 ND 2 N = ⋅ ⇔ D = 3 ⋅ ∠ + 4h ⇔ = ⋅ ∠ − 4h ⇔ U = ⋅ D ⋅ u ∠ − 4h 1 n u z1 u ∠ + 1h u u 3 nz 3 nz z 2 3

Dz8 ⇔ D5 (17) - z9

1

U = 2⋅

FD17 N D ⋅ ⋅ u ∠Θ D17 − θ z9 f z9 n z

N 2 N ⇔ U = 2 ⋅ 3 ⋅ D ⋅ u ∠ + 510?−270? ⇔ U = ⋅ D ⋅ u ∠ + 240? 3

nz

3 nz

Ou: U D17 3 U ∠ + 17h N U U 2 ND 2 N = ⋅ ⇔ D = 3 ⋅ ∠ + 8h ⇔ = ⋅ ∠ − 8h ⇔ U = ⋅ D ⋅ u ∠ − 8h 1 u ∠ + 9h nz u z9 u u 3 nz 3 nz 2 3

Na página seguinte, apresentamos os “Símbolos de Ligação” Dz0, Dz4 e Dz8.

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Dz0

D5-z5

D5

z5

UD

0 A

A

1

n

2U

4

C

2

n

C

C

n

Dz4

A

uz n

C

n

C

C

n

Dz8

A

C

C

3

4 2

a

8 b

3

uz

2

D

C

n

y

1

x

z

2

y

uz A

n

3

b uz

4 2

uz x

z

0

1 2

uz

UD

8

z

uz n

2

uz

z9

1

B

B

x

3

c

uz

B

2U

4

y

0

D5-z9

UD A

z

2

D5

0

c

1 2

uz A

3

x

uz

UD

8

b

8

3

uz y

1 2

D

2

4 2

z1

1

B

B

z

B

2U

4

uz

D5-z1

UD A

y

a

uz x

3

D5

0

z

2

0

1 2

uz A

3

x

uz

UD

8

uz y

1 2

D

B

B

uz B

3

c

8 a

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Dy1 ⇔ D7 - y6 1 F N U = D7 ⋅ D ⋅ u ∠Θ D7 − θ y6 f y6 n y

N 1 ND ⇔ U = 3 ⋅ D ⋅ u ∠ + 210?−180? ⇔ U = ⋅ ⋅ u ∠ + 30? 1

ny

ny

3

Ou: U D7 3 U ∠ + 7h N U U 1 ND 1 ND = ⋅ ⇔ D = 3 ⋅ ∠ + 1h ⇔ = ⋅ ∠ − 1h ⇔ U = ⋅ ⋅ u ∠ − 1h u y6 1 u ∠ + 6h ny u u 3 ny 3 ny

Dy5 ⇔ D7 - y2 1 F N U = D7 ⋅ D ⋅ u ∠Θ D7 − θ y2 f y2 n y

N 1 ND ⇔ U = 3 ⋅ D ⋅ u ∠ + 210?−60? ⇔ U = ⋅ ⋅ u ∠ + 150? 1

ny

ny

3

Ou: U D7 u y2

=

3 U ∠ + 7h N U U 1 ND 1 ND ⋅ ⇔ D = 3 ⋅ ∠ + 5h ⇔ = ⋅ ∠ − 5h ⇔ U = ⋅ ⋅ u ∠ − 5h 1 u ∠ + 2h ny u u 3 ny 3 ny

Dy9 ⇔ D7 (19) - y10 1

F N U = D19 ⋅ D ⋅ u ∠Θ D19 − θ y10 f y10 n y

N 1 ND ⇔ U = 3 ⋅ D ⋅ u ∠ + 570?−300? ⇔ U = ⋅ ⋅ u ∠ + 270? 1

ny

3

ny

Ou: U D19 u y10

=

3 U ∠ + 19h N U U 1 ND 1 ND ⋅ ⇔ D = 3 ⋅ ∠ + 9h ⇔ = ⋅ ∠ − 9h ⇔ U = ⋅ ⋅ u ∠ − 9h 1 u ∠ + 10h ny u u 3 ny 3 ny

Na página seguinte, apresentamos os “Símbolos de Ligação” Dy1, Dy5 e Dy9.

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Dy1

D7-y6

D7

y6

UD

0

uy

A

1

C

1

A

2U

4

n

2u y

D

2

A

n

UD

9 B

n

Dy5

y2

UD

uy

A

1

C

1

A

2U

n

2u y

D

2

A

n

UD

9 B

n

Dy9

y10 uy

UD A

1

C n

1

A

2U

2u y

D

2

b

1

B

B

c

3

D7-y10

D7

A

n

UD C

b

uy

3

8

5 2

C

C

4

c

1

B

B

0

c

3

D7-y2

D7

8

b

uy

3

4

2

C

C

0

5

B

B

8

a

1

2

5 c

uy

9

C

3

B

n

3

a

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Yd1 ⇔ Y6 - d5

U=

FY6 N Y ⋅ ⋅ u ∠Θ Y6 − θ d5 f d5 n d

⇔ U=

1 NY ⋅ ⋅ u ∠ + 180?−150º 1 nd 3

⇔ U= 3⋅

NY ⋅ u ∠ + 30? nd

Ou: U Y6 N N N 1 U ∠ + 6h 1 U U = ⋅ ⇔ Y = ⋅ ∠ + 1h ⇔ = 3 ⋅ Y ∠ − 1h ⇔ U = 3 ⋅ Y ⋅ u ∠ − 1h u d5 u ∠ + 5h n u u n nd 3 3 d d

Yd5 ⇔ Y6 - d1

U=

FY6 N Y ⋅ ⋅ u ∠Θ Y6 − θ d1 f d1 n d

N ⇔ U = 1 ⋅ N Y ⋅ u ∠ + 180?−30º ⇔ U = 3 ⋅ Y ⋅ u ∠ + 150? 1

nd

nd

3

Ou: U Y6 u d1

=

1

U ∠ + 6h N 1 U U N N ⋅ ⇔ Y = ⋅ ∠ + 5h ⇔ = 3 ⋅ Y ∠ − 5h ⇔ U = 3 ⋅ Y ⋅ u ∠ − 5h nd u nd nd 3 u ∠ + 1h 3 u

Yd9 ⇔ Y6 (18) - d9

U=

FY18 N Y ⋅ ⋅ u ∠Θ Y18 − θ d9 f d9 n d

⇔ U=

1 NY ⋅ ⋅ u ∠ + 540?−270º 1 nd 3

⇔ U= 3⋅

NY ⋅ u ∠ + 270? nd

Ou: U Y18 u d9

=

1

U ∠ + 18h N 1 U U N N ⇔ Y = ⋅ ∠ + 9h ⇔ = 3 ⋅ Y ∠ − 9h ⇔ U = 3 ⋅ Y ⋅ u ∠ − 9h u ∠ + 9h n u u n nd 3 3 d d ⋅

Na página seguinte, apresentamos os “Símbolos de Ligação” Yd1, Yd5 e Yd9.

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Yd1

Y6-d5

Y6

d5

UY

ud

0

b

1

A

1

N

2U Y

2u

4 2

B

c

UY

3

a

UY

1

N

2U Y

2u

4

b

UY

3

1

N

2U Y

2u

a N

UY

b

b

1

c

5

d

2

b

c

ud

8 3

9

ud c

4

C

b

Y6-d9

UY

2

5

d9

0

B

a

c

a

3

N

Y6

1

1

ud c

Yd9

A

c

d

2

N

8 C

c

ud a

2

9

d1

0

B

c

Y6-d1

Y6

1

5 b

3

N

Yd5

A

b

a

ud

8 C

1

d

2

N

a

b N

3

a

9 a

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Yz1 ⇔ Y6 - z5 U = 2⋅

FY6 N Y 1 NY 2 NY ⋅ ⋅ u ∠ + 180?−150º ⇔ U = ⋅ ⋅ u ∠ + 30? ⋅ ⋅ u ∠Θ Y6 − θ z5 ⇔ U = 2 ⋅ f z5 n z 3 nz 3 nz

Ou: U Y6 N 1 U ∠ + 6h U U 2 NY 2 NY = ⋅ ⇔ Y = 3 ⋅ ∠ + 1h ⇔ = ⋅ ∠ − 1h ⇔ U = ⋅ ⋅ u ∠ − 1h 1 u ∠ + 5h nz u z5 u u n 3 3 nz z 2 3

Yz5 ⇔ Y6 - z1 U = 2⋅

FY6 N Y 1 NY 2 NY ⋅ ⋅ u ∠ + 180?−30º ⇔ U = ⋅ ⋅ u ∠ + 150? ⋅ ⋅ u ∠Θ Y6 − θ z1 ⇔ U = 2 ⋅ f z1 n z 3 nz 3 nz

Ou: U Y6 u z1

=

1 U ∠ + 6h N U U 2 NY 2 NY ⋅ ⇔ Y = 3 ⋅ ∠ + 5h ⇔ = ⋅ ∠ − 5h ⇔ U = ⋅ ⋅ u ∠ − 5h 1 u ∠ + 1h nz u u n 3 3 nz z 2 3

Yz9 ⇔ Y6 (18) - z9

U = 2⋅

FY18 N Y 1 NY 2 NY ⋅ ⋅ u ∠Θ Y18 − θ z9 ⇔ U = 2 ⋅ ⋅ ⋅ u ∠ + 540?−270º ⇔ U = ⋅ ⋅ u ∠ + 270? f z9 n z n 3 3 nz z

Ou: U Y18 u z9

=

1 U ∠ + 18h N U U 2 NY 2 NY ⋅ ⇔ Y = 3 ⋅ ∠ + 9h ⇔ = ⋅ ∠ − 9h ⇔ U = ⋅ ⋅ u ∠ − 9h 1 u ∠ + 9h nz u u 3 nz 3 nz 2 3

Na página seguinte, apresentamos os “Símbolos de Ligação” Yz1, Yz5 e Yz9.

A. A. A. C. Barrias

19 / 59

Yz1

Y6-z5

Y6

z5

UY A

0

B

4

1

uz n

N

2

2U Y

2

n

N

uz n

N

Yz5

B

4

1

n

N

2

n

N

n

N

3

N

n

y

2

x

z

3

uz

a

9 b

N

n

uz y

1 2

4

x

2

z

y

x

z

uz N

n

3

1 b

1 2

uz

UY

8 3

uz

5

z

uz

2U Y

C

c

z9

0

2

1

Y6-z9

UY

B

c

1 2

uz

2

Y6

1

x

uz

Yz9

A

b

9

3

uz y

1

8 3

z

uz

UY C

5 2

z1

2U Y

2

uz

Y6-z1

UY A

a

uz x

3

Y6 0

y

1

1 2

z

2

8 3

x

uz

UY C

uz y

1

uz

5 2

uz 3

c

9 a

A. A. A. C. Barrias

20 / 59

Dd2 ⇔ D7 - d5 1 FD7 N D N N U= ⋅ ⋅ u ∠Θ D7 − θ d5 ⇔ U = 3 ⋅ D ⋅ u ∠ + 210?−150? ⇔ U = D ⋅ u ∠ + 60? 1 nd f d5 n d nd 3

Ou: U D7 = u d5

3 U ∠ + 7h N U U ND N ⋅ ⇔ D = ∠ + 2h ⇔ = ∠ − 2h ⇔ U = D ⋅ u ∠ − 2h nd u u nd nd 3 u ∠ + 5h

Dd6 ⇔ D7 - d1 1 FD7 N D N N U= ⋅ ⋅ u ∠Θ D7 − θ d1 ⇔ U = 3 ⋅ D ⋅ u ∠ + 210?−30º ⇔ U = D ⋅ u ∠ + 180? 1 f d1 n d nd nd 3

Ou: U D7 u d1

=

3 U ∠ + 7h N U U N N ⋅ ⇔ D = ∠ + 6h ⇔ = D ∠ − 6h ⇔ U = D ⋅ u ∠ − 6h u ∠ + 1h n u u n nd 3 d d

Dd10 ⇔ D7 (19) - d9 1 FD19 N D N N U= ⋅ ⋅ u ∠Θ D19 − θ d9 ⇔ U = 3 ⋅ D ⋅ u ∠ + 570?−270? ⇔ U = D ⋅ u ∠ + 300? 1 f d9 n d nd nd 3

Ou: U D19 u d9

=

3 U ∠ + 19 h N U U N N ⋅ ⇔ D = ∠ + 10 h ⇔ = D ∠ − 10 h ⇔ U = D ⋅ u ∠ − 10 h nd u u nd nd 3 u ∠ + 9h

Na página seguinte, apresentamos os “Símbolos de Ligação” Dd2, Dd6 e Dd10.

A. A. A. C. Barrias

21 / 59

Dd2

D7-d5

d5

D7 UD

0

ud

A

b

C

1

1

A

2U

2u

D

4

B

c

2

B

UD

8

a

3

C

a

1

1

2U

2u

D

B

2

b

A

3

c

B

Dd10

1

2U

2u

D

2

A

a

UD

8 C

10 b

b

2

c

6

d

2

b

c

ud

C

3

b

3

c

B

B

a

ud C

1

4

6

d9

UD A

A

a

c

D7-d9

D7 0

2

ud

C

C

c

d

2

UD

8

c

ud

A

B

10

d1

UD

4

c

D7-d1

D7

A

6 b

3

B

Dd6

0

b

a

ud

C

C

2

d

2

A

a

B

b

3

a

10 a

A. A. A. C. Barrias

22 / 59

Dz2 ⇔ D7 - z5 1 F N U = 2 ⋅ D7 ⋅ D ⋅ u ∠Θ D7 − θ z5 ⇔ U = 2 ⋅ f z5 n z

3 ⋅ N D ⋅ u ∠ + 210?−150? ⇔ U = 2 ⋅ N D ⋅ u ∠ + 60? 3 nz 3 nz

Ou: U D7 3 U ∠ + 7h N U U 2 ND 2 N = ⋅ ⇔ D = 3 ⋅ ∠ + 2h ⇔ = ⋅ ∠ − 2h ⇔ U = ⋅ D ⋅ u ∠ − 2h 1 u ∠ + 5h nz u z5 u u 3 nz 3 nz 2 3

Dz6 ⇔ D7 - z1 1 F N U = 2 ⋅ D7 ⋅ D ⋅ u ∠Θ D7 − θ z1 ⇔ U = 2 ⋅ f z1 n z

3 ⋅ N D ⋅ u ∠ + 210?−30? ⇔ U = 2 ⋅ N D ⋅ u ∠ + 180? 3 nz 3 nz

Ou: U D7 3 U ∠ + 7h N U U 2 ND 2 N = ⋅ ⇔ D = 3 ⋅ ∠ + 6h ⇔ = ⋅ ∠ − 6h ⇔ U = ⋅ D ⋅ u ∠ − 6h 1 n u z1 u ∠ + 1h u u 3 nz 3 nz z 2 3

Dz10 ⇔ D7 (19) - z9 1 F N U = 2 ⋅ D19 ⋅ D ⋅ u ∠Θ D19 − θ z9 ⇔ U = 2 ⋅ f z9 n z

3 ⋅ N D ⋅ u ∠ + 570?−270? ⇔ U = 2 ⋅ N D ⋅ u ∠ + 300? 3 nz 3 nz

Ou: U D19 u z9

=

3 U ∠ + 19h N U U 2 N 2 N ⋅ ⇔ D = 3 ⋅ ∠ + 10h ⇔ = ⋅ D ∠ − 10h ⇔ U = ⋅ D ⋅ u ∠ − 10h 1 u ∠ + 9h nz u u 3 nz 3 nz 2 3

Na página seguinte, apresentamos os “Símbolos de Ligação” Dz2, Dz6 e Dz10.

A. A. A. C. Barrias

23 / 59

Dz2

D7-z5

z5

D7 UD

0

uz

A

C

n

1

A

2U

4

2

B

n

A

uz

3

n

B

Dz6

n

2

n

A

n

B

Dz10

uz

6 2

n

1

uz

2

A

n

y

1 2

D

B

B

b

C

2U

x

2

uz z

2

y

uz

UD B

n

3

b uz

6 2

uz x

z

2

1

C

3

10

3

z9 uz

A

A

z

a

D7-z9

UD

C

y

c

uz x

3

D7

8

z

2

2

1 2

uz

3

4

c

C

C

0

x

uz

UD

8

b

10

3

uz y

1 2

D

B

B

z

uz C

2U

4

6 2

z1

1

A

uz

D7-z1

UD A

a

uz x

3

D7 0

y

2

1 2

z

2

C

C

x

uz

UD

8

y

1 2

D

B

uz

3

c

10 a

A. A. A. C. Barrias

24 / 59

Dd4 ⇔ D5 - d1 1 FD5 N D N N U= ⋅ ⋅ u ∠Θ D5 − θ d1 ⇔ U = 3 ⋅ D ⋅ u ∠ + 150?−30? ⇔ U = D ⋅ u ∠ + 120? 1 nd f d1 n d nd 3

Ou: U D5 = u d1

3 U ∠ + 5h N U U ND N ⋅ ⇔ D = ∠ + 4h ⇔ = ∠ − 4h ⇔ U = D ⋅ u ∠ − 4h nd u u nd nd 3 u ∠ + 1h

Dd8 ⇔ D5 (17) - d9 1 FD17 N D N N U= ⋅ ⋅ u ∠Θ D17 − θ d9 ⇔ U = 3 ⋅ D ⋅ u ∠ + 510?−270? ⇔ U = D ⋅ u ∠ + 240? 1 nd f d9 n d nd 3

Ou: U D17 u d9

3 U ∠ + 17h N U U N N ⋅ ⇔ D = ∠ + 8h ⇔ = D ∠ − 8h ⇔ U = D ⋅ u ∠ − 8h nd u u nd nd 3 u ∠ + 9h

=

Dd0 ⇔ D5 - d5 1 FD5 N D N N U= ⋅ ⋅ u ∠Θ D5 − θ d5 ⇔ U = 3 ⋅ D ⋅ u ∠ + 150?−150? ⇔ U = D ⋅ u ∠ 0? 1 f d5 n d nd nd 3

Ou: U D5 u d5

=

3 U ∠ + 5h N U U N N ⋅ ⇔ D = ∠ 0h ⇔ = D ∠ 0h ⇔ U = D ⋅ u ∠ 0h nd u u nd nd 3 u ∠ + 5h

Na página seguinte, apresentamos os “Símbolos de Ligação” Dd4, Dd8 e Dd0.

A. A. A. C. Barrias

25 / 59

Dd4

D5 - d1

D5 0 A

ud

UD A

1

1 2u

D

B

2

UD

8 C

C

3

Dd8

A

1

1

2

C

C

c

A

a

2

3

3

A

a

1

1

2u

D

2

C

b

C

C

3

0

b

4

c

8

A

c

a

0

d

2

b

b

c

ud

UD

8

b

a

ud B

B

B

8

d5

UD

2U

4

a

c

D5 – d5

D5

A

4

c

ud

Dd0

0

c

d

C

UD

8

0

2u

D

B

B

c

b

ud b

B

2U

4

8

d9

UD A

b

a

D5-d9

D5

0

4

ud b

A

3

a

d

a

C

2

B

c

B

2U

4

d1

3

a

A. A. A. C. Barrias

26 / 59

Dz4 ⇔ D5 - z1 1 F N U = 2 ⋅ D5 ⋅ D ⋅ u ∠Θ D5 − θ z1 ⇔ U = 2 ⋅ f z1 n z

3 ⋅ N D ⋅ u ∠ + 150?−30? ⇔ U = 2 ⋅ N D ⋅ u ∠ + 120º 3 nz 3 nz

Ou: U D5 3 U ∠ + 5h N U U 2 ND 2 N = ⋅ ⇔ D = 3 ⋅ ∠ + 4h ⇔ = ⋅ ∠ − 4h ⇔ U = ⋅ D ⋅ u ∠ − 4h 1 u ∠ + 1h nz u z1 u u 3 nz 3 nz 2 3

Dz8 ⇔ D5 (17) - z9 1 F N U = 2 ⋅ D17 ⋅ D ⋅ u ∠Θ D17 − θ z9 ⇔ U = 2 ⋅ f z9 n z

3 ⋅ N D ⋅ u ∠ + 510?−270? ⇔ U = 2 ⋅ N D ⋅ u ∠ + 240º 3 nz 3 nz

Ou: U D17 u z9

=

3 U ∠ + 17h N U U 2 N 2 N ⋅ ⇔ D = 3 ⋅ ∠ + 8h ⇔ = ⋅ D ∠ − 8h ⇔ U = ⋅ D ⋅ u ∠ − 8h 1 u ∠ + 9h nz u u 3 nz 3 nz 2 3

Dz0 ⇔ D5 - z5 1 F N U = 2 ⋅ D5 ⋅ D ⋅ u ∠Θ D5 − θ z5 ⇔ U = 2 ⋅ f z5 n z

3 ⋅ N D ⋅ u ∠ + 150?−150? ⇔ U = 2 ⋅ N D ⋅ u ∠ 0º 3 nz 3 nz

Ou: U D5 u z5

=

3 U ∠ + 5h N U U 2 N 2 N ⋅ ⇔ D = 3 ⋅ ∠ 0h ⇔ = ⋅ D ∠ 0h ⇔ U = ⋅ D ⋅ u ∠ 0h 1 u ∠ + 5h nz u u 3 nz 3 nz 2 3

Na página seguinte, apresentamos os “Símbolos de Ligação” Dz4, Dz8 e Dz0.

A. A. A. C. Barrias

27 / 59

Dz4

D5-z1

D5

z1

UD

0 A

A

1

n

2U

4

B

B

uz

B

C

C

2

n

3

n

C

n

2

C

n

3

1

n

C

C

3

c uz

8 2

uz y

3

n

z

a

0 b

3

2

C

n

uz z

1

x

x

2

y

uz A

n

3

b uz

8 2

uz y

z

4

1 2

uz

UD

8

y

4

1

D5-z5

D

2

B

x

uz

B

c

z5 B

2U

4

b

0

3

2

uz A

UD A

x

uz

2

D5

A

z

uz z

1

D

Dz0

0

y

3

UD C

8 2

z9

1

2

8

uz

uz

uz

B

B

a

D5-z9

B

2U

4

y

uz

UD A

x

2

D5

A

2

uz

4

1

A

Dz8

0

x

C

UD

8

z

1 2

D

uz

3

c

0 a

A. A. A. C. Barrias

28 / 59

Dy5 ⇔ D5 - y0 1 FD5 N D N 1 ND U= ⋅ ⋅ u ∠Θ D5 − θ y0 ⇔ U = 3 ⋅ D ⋅ u ∠ + 150?−0? ⇔ U = ⋅ ⋅ u ∠ + 150? f y0 n y 1 ny 3 ny

Ou: U D5 3 U ∠ + 5h N U U 1 ND 1 ND = ⋅ ⇔ D = 3 ⋅ ∠ + 5h ⇔ = ⋅ ∠ − 5h ⇔ U = ⋅ ⋅ u ∠ − 5h u y0 1 u ∠ 0h ny u u 3 ny 3 ny

Dy9 ⇔ D5 (17) - y8 1 FD17 N D N 1 ND U= ⋅ ⋅ u ∠Θ D17 − θ y8 ⇔ U = 3 ⋅ D ⋅ u ∠ + 510?−240? ⇔ U = ⋅ ⋅ u ∠ + 270? f y8 n y 1 ny 3 ny

Ou: U D17 u y8

=

3 U ∠ + 17h N U U 1 ND 1 ND ⋅ ⇔ D = 3 ⋅ ∠ + 9h ⇔ = ⋅ ∠ − 9h ⇔ U = ⋅ ⋅ u ∠ − 9h 1 u ∠ + 8h ny u u 3 ny 3 ny

Dy1 ⇔ D5 - y4 1 FD5 N D N 1 ND U= ⋅ ⋅ u ∠Θ D5 − θ y4 ⇔ U = 3 ⋅ D ⋅ u ∠ + 150?−120º ⇔ U = ⋅ ⋅ u ∠ + 30? f y4 n y 1 ny 3 ny

Ou: U D5 u y4

=

3 U ∠ + 5h N U U 1 ND 1 ND ⋅ ⇔ D = 3 ⋅ ∠ + 1h ⇔ = ⋅ ∠ − 1h ⇔ U = ⋅ ⋅ u ∠ − 1h 1 u ∠ + 4h ny u u 3 ny 3 ny

Na página seguinte, apresentamos os “Símbolos de Ligação” Dy5, Dy9 e Dy1.

A. A. A. C. Barrias

29 / 59

Dy5

D5 - y0

D5

y0 uy

UD

0 A

A

B

B

n

1 2U

4

5

B

2u

D

C

2

n

C

C

n

Dy9

A

uy

4

n

C

2

n

C

C

n

Dy1

A

uy B

1 2U

4

B

B

5 n

2

C

n

C

C

3

b

1 2u

D

y

9 2

c

uy

UD

8

b

3

y4

UD A

a

D5 – y4

D5 0

9

2

1

A

3

y

uy

UD

8

c

1 2u

D

B

B

5

B

1 2U

c

3

y8

UD A

b

D5 – y8

D5 0

9

2

1

A

3

y

uy

UD

8

a

1

A

1 n

3

a

A. A. A. C. Barrias

30 / 59

Yd5 ⇔ Y6 - d1

U=

FY6 N Y N 1 NY ⋅ ⋅ u ∠Θ Y6 − θ d1 ⇔ U = ⋅ ⋅ u ∠ + 180?−30? ⇔ U = 3 ⋅ Y ⋅ u ∠ + 150? 1 nd f d1 n d nd 3

Ou: U Y6 N N N 1 U ∠ + 6h 1 U U = ⋅ ⇔ Y = ⋅ ∠ + 5h ⇔ = 3 ⋅ Y ∠ − 5h ⇔ U = 3 ⋅ Y ⋅ u ∠ − 5h u d1 u ∠ + 1h n u u n nd 3 3 d d

Yd9 ⇔ Y6 (18) - d9

U=

FY18 N Y N 1 NY ⋅ ⋅ u ∠Θ Y18 − θ d9 ⇔ U = ⋅ ⋅ u ∠ + 540?−270? ⇔ U = 3 ⋅ Y ⋅ u ∠ + 270? 1 nd f d9 n d nd 3

Ou: U Y18 u d9

=

1

U ∠ + 18h N 1 U U N N ⇔ Y = ⋅ ∠ + 9h ⇔ = 3 ⋅ Y ∠ − 9h ⇔ U = 3 ⋅ Y ⋅ u ∠ − 9h nd u nd nd 3 u ∠ + 9h 3 u ⋅

Yd1 ⇔ Y6 - d5

U=

FY6 N Y N 1 NY ⋅ ⋅ u ∠Θ Y6 − θ d5 ⇔ U = ⋅ ⋅ u ∠ + 180?−150? ⇔ U = 3 ⋅ Y ⋅ u ∠ + 30? 1 nd f d5 n d nd 3

Ou: U Y6 u d5

=

1

U ∠ + 6h N 1 U U N N ⇔ Y = ⋅ ∠ + 1h ⇔ = 3 ⋅ Y ∠ − 1h ⇔ U = 3 ⋅ Y ⋅ u ∠ − 1h u ∠ + 5h n u u n nd 3 3 d d ⋅

Na página seguinte, apresentamos os “Símbolos de Ligação” Yd5, Yd9 e Yd1.

A. A. A. C. Barrias

31 / 59

Yd5

Y6 - d1

Y6

d1

UY

ud

0

c

1

A

2U

1

N

2u

Y

4

2

N

UY b

3

3

N

Yd9

UY

1

N

2u

Y

4 B

2

N

UY a

3

3

N

Yd1

UY

2U

a

1 2u

Y

2

N

b

2

UY

3

1

b

5

c

9

a

1

b

b

d

c

ud

8 C

b

a

ud N

4 B

9

d5

0 1

a

c

Y6 – d5

Y6

A

5

ud

8 C

c

c

d

c

2

1

ud b

2U

c

b

d9

0 1

9

Y6 – d9

Y6

A

b

a

ud

8 C

5

d

a

2

B

a

N

c

3

a

A. A. A. C. Barrias

32 / 59

Yz5 ⇔ Y6 - z1 FY6 N Y 1 NY 2 NY ⋅ ⋅ u ∠Θ Y6 − θ z1 ⇔ U = 2 ⋅ ⋅ ⋅ u ∠ + 180?−30? ⇔ U = ⋅ ⋅ u ∠ + 150? f z1 n z 3 nz 3 nz

U = 2⋅

Ou: U Y6 N 1 U ∠ + 6h U U 2 NY 2 NY = ⋅ ⇔ Y = 3 ⋅ ∠ + 5h ⇔ = ⋅ ∠ − 5h ⇔ U = ⋅ ⋅ u ∠ − 5h 1 u ∠ + 1h nz u z1 u u n 3 3 nz z 2 3

Yz9 ⇔ Y6 (18) - z9

U = 2⋅

FY18 N Y 1 NY 2 NY ⋅ ⋅ u ∠Θ Y18 − θ z9 ⇔ U = 2 ⋅ ⋅ ⋅ u ∠ + 540?−270? ⇔ U = ⋅ ⋅ u ∠ + 270? f z9 n z 3 nz 3 nz

Ou: U Y18 u z9

=

1 U ∠ + 18h N U U 2 NY 2 NY ⋅ ⇔ Y = 3 ⋅ ∠ + 9h ⇔ = ⋅ ∠ − 9h ⇔ U = ⋅ ⋅ u ∠ − 9h 1 u ∠ + 9h nz u u 3 nz 3 nz 2 3

Yz1 ⇔ Y6 - z5

U = 2⋅

FY6 N Y 1 NY 2 NY ⋅ ⋅ u ∠Θ Y6 − θ z5 ⇔ U = 2 ⋅ ⋅ ⋅ u ∠ + 180?−150? ⇔ U = ⋅ ⋅ u ∠ + 150? f z5 n z 3 nz 3 nz

Ou: U Y6 u z5

=

1 U ∠ + 6h N U U 2 NY 2 NY ⋅ ⇔ Y = 3 ⋅ ∠ + 1h ⇔ = ⋅ ∠ − 1h ⇔ U = ⋅ ⋅ u ∠ − 1h 1 u ∠ + 5h nz u u n 3 3 nz z 2 3

Na página seguinte, apresentamos os “Símbolos de Ligação” Yz5, Yz9 e Yz1.

A. A. A. C. Barrias

33 / 59

Yz5

Y6-z1 z1

Y6 UY

uz

0 A

1 2U

n

N

2

B

n

N

2

n

N

Yz9

3

UY

2U

n

N

n

N

3

n

N

N

n

c uz

9 2

uz z

a

1 b

3

2

Y

N

n

uz z

1

x

2

N

n

3

b uz

9

x

y

2

y

z

3

uz

UY

5

1 2

uz

8 3

y

5

1 2

uz

4

C

x

z5

UY

2

1

Y6-z5

0 1

b

c

uz

y

3

Y6

B

z

uz

uz

Yz1

2U

y

3

x

2

UY

A

2

uz

8 C

y

z

1 2

Y

2

9

x

uz

4 B

uz

z9

0 1

a

Y6-z9

Y6

A

5

1 2

uz

UY

3

x

uz

8 C

z

1 2

Y

4

uz

uz

c

1 a

A. A. A. C. Barrias

34 / 59

Dd6 ⇔ D5(D17) - d11 1 FD17 N D N N U= ⋅ ⋅ u ∠Θ D17 − θ d11 ⇔ U = 3 ⋅ D ⋅ u ∠ + 510?−330? ⇔ U = D ⋅ u ∠ + 180? 1 f d11 n d nd nd 3

Ou: U D17 = u d11

3 U ∠ + 17h N U U ND N ⋅ ⇔ D = ∠ + 6h ⇔ = ∠ − 6h ⇔ U = D ⋅ u ∠ − 6h u ∠ + 11h n u u n nd 3 d d

Dd10 ⇔ D5(D17) - d7 1 FD17 N D N N U= ⋅ ⋅ u ∠Θ D17 − θ d7 ⇔ U = 3 ⋅ D ⋅ u ∠ + 510?−210? ⇔ U = D ⋅ u ∠ + 300? 1 f d7 n d nd nd 3

Ou: U D17 3 U ∠ + 17h N U U N N = ⋅ ⇔ D = ∠ + 10h ⇔ = D ∠ − 10h ⇔ U = D ⋅ u ∠ − 10h u d7 u ∠ + 7h n u u n nd 3 d d

Dd2 ⇔ D5 - d3 1 U=

FD5 N D N N ⋅ ⋅ u ∠Θ D5 − θ d3 ⇔ U = 3 ⋅ D ⋅ u ∠ + 150?−90? ⇔ U = D ⋅ u ∠ + 60? 1 nd f d3 n d nd 3

Ou: U D5 u d3

=

3 U ∠ + 5h N U U N N ⋅ ⇔ D = ∠ + 2h ⇔ = D ∠ − 2h ⇔ U = D ⋅ u ∠ − 2h nd u u nd nd 3 u ∠ + 3h

Na página seguinte, apresentamos os “Símbolos de Ligação” Dd6, Dd10 e Dd2.

A. A. A. C. Barrias

35 / 59

Dd6

D5-d11

D5

d11

UD

0 A

A

ud B

b

1

1

2U

B

B

c

2

UD

8 C

C

a

Dd10

A

a

1

1

B

2

C

C

b

2

c

Dd2

A

c

1

1

B

C

C

2

3

b

6

c

10

d

C

a

UD

8

2

b

2u

D

B

b

3

ud B

2U

4

a

d3

UD A

10

D5-d3

D5 0

a

c

ud A

3

6

d

C

UD

8

c

c

2u

D

B

2

ud B

2U

4

c

3

d7

UD A

b

D5-d7

D5 0

10

a

ud A

3

b

d

C

2

6

2u

D

4

a

2

b

c

ud A

b

3

a

2 a

A. A. A. C. Barrias

36 / 59

Yy6 ⇔ Y6 - y0 FY6 N Y 1 N N ⋅ ⋅ u ∠Θ Y6 − θ y0 ⇔ U = ⋅ Y ⋅ u ∠ + 180?−0? ⇔ U = Y ⋅ u ∠ + 180? f y0 n y 1 ny ny

U=

Ou: U Y6 1 U ∠ + 6h N N U U NY = ⋅ ⇔ Y = ∠ + 6h ⇔ = ∠ − 6h ⇔ U = Y ⋅ u ∠ − 6h u y0 1 u ∠ 0h ny u u ny ny

Yy10 ⇔ Y6 (18) - y8

U=

FY18 N Y 1 N N ⋅ ⋅ u ∠Θ Y18 − θ y8 ⇔ U = ⋅ Y ⋅ u ∠ + 540?−240? ⇔ U = Y ⋅ u ∠ + 300? f y8 n y 1 ny ny

Ou: U Y18 u y8

1 U ∠ + 18h N U U N N = ⋅ ⇔ Y = ∠ + 10h ⇔ = Y ∠ − 10h ⇔ U = Y ⋅ u ∠ − 10h 1 u ∠ + 8h ny u u ny ny

Yy2 ⇔ Y6 - y4

U=

FY6 N Y 1 N N ⋅ ⋅ u ∠Θ Y6 − θ y4 ⇔ U = ⋅ Y ⋅ u ∠ + 180?−120? ⇔ U = Y ⋅ u ∠ + 60? f y4 n y 1 ny ny

Ou: U Y6 u y4

1 U ∠ + 6h N U U N N = ⋅ ⇔ Y = ∠ + 2h ⇔ = Y ∠ − 2h ⇔ U = Y ⋅ u ∠ − 2h 1 u ∠ + 4h ny u u ny ny

Na página seguinte, apresentamos os “Símbolos de Ligação” Yy6, Yy10 e Yy2.

A. A. A. C. Barrias

37 / 59

Yy6

Y6 - y0

Y6

y0 uy

UY

6

0 1

A

2U

n

N

2u y

Y

4 2

B

n

N

2

n

N

Yy10

y8 uy

UY

0

6 1 2U

4

n

N

2u y

Y

2

n

N

UY

10 2 uy

n

N

Yy2

y4 uy

UY

0

6 1 2U

N

n

2

b

1 2u y

Y

4 N

n

UY

10 2

c

uy

8 C

b

3

Y6 – y4

Y6

B

a

2 3

A

c

1

8 C

c

3

Y6 – y8

Y6

B

b

2 3

A

10

uy

UY

8 C

a

1

2 3

N

n

3

a

A. A. A. C. Barrias

38 / 59

Dz6 ⇔ D5(D17) - z11 1 F N U = 2 ⋅ D17 ⋅ D ⋅ u ∠Θ D17 − θ z11 ⇔ U = 2 ⋅ f z11 n z

3 ⋅ N D ⋅ u ∠ + 510?−330? ⇔ U = 2 ⋅ N D ⋅ u ∠ + 180? 3 nz 3 nz

Ou: U D17 3 U ∠ + 17h N U U 2 ND 2 N = ⋅ ⇔ D = 3 ⋅ ∠ + 6h ⇔ = ⋅ ∠ − 6h ⇔ U = ⋅ D ⋅ u ∠ − 6h 1 n u z11 u ∠ + 11h u u 3 nz 3 nz z 2 3

Dz10 ⇔ D5(D17) - z7 1 F N U = 2 ⋅ D17 ⋅ D ⋅ u ∠Θ D17 − θ z7 ⇔ U = 2 ⋅ f z7 n z

3 ⋅ N D ⋅ u ∠ + 510?−210? ⇔ U = 2 ⋅ N D ⋅ u ∠ + 300? 3 nz 3 nz

Ou: U D17 u z7

3 U ∠ + 17h N U U 2 N 2 N ⋅ ⇔ D = 3 ⋅ ∠ + 10h ⇔ = ⋅ D ∠ − 10h ⇔ U = ⋅ D ⋅ u ∠ − 10h 1 u ∠ + 7h nz u u 3 nz 3 nz 2 3

=

Dz2 ⇔ D5 - z3 1 U = 2⋅

FD5 N D ⋅ ⋅ u ∠Θ D5 − θ z3 ⇔ U = 2 ⋅ f z3 n z

3 ⋅ N D ⋅ u ∠ + 150?−90? ⇔ U = 2 ⋅ N D ⋅ u ∠ + 60? 3 nz 3 nz

Ou: U D5 u z3

=

3 U ∠ + 5h N U U 2 N 2 N ⋅ ⇔ D = 3 ⋅ ∠ + 2h ⇔ = ⋅ D ∠ − 2h ⇔ U = ⋅ D ⋅ u ∠ − 2h 1 u ∠ + 3h nz u u 3 nz 3 nz 2 3

Na página seguinte, apresentamos os “Símbolos de Ligação” Dz6, Dz10 e Dz2.

A. A. A. C. Barrias

39 / 59

Dz6

D5-z11

D5

z11

UD

0 A

A

1 2U

4

uz n

C

2

n

C

C

n

Dz10

3

A

1 2

D

C

2

B

C

C

n

2

3

n

3

n

1

C

c uz

10

z

y

2

x

z

3

uz

3

a

2 b

uz

C

n

x

n

b uz

10 z

2

y

uz A

6

1 2

uz

UD C

6

1 2

y

1 2

D

2

8

uz

uz

B

B

c

B

2U

4

2

3

z3

UD A

x

b

D5-z3

D5

A

y

uz A

Dz2

0

z

uz

UD

8

x

uz n

B

10 2

uz

B

1 2U

4

uz

z7

UD A

y

a

D5-z7

D5 0

z

2

6

1 2

uz A

3

x

uz

UD

8

y

1 2

D

B

B

uz

B

3

2

uz x

z

3

c

2 a

A. A. A. C. Barrias

40 / 59

Dy7 ⇔ D7 - y0 1 FD7 N D N 1 ND U= ⋅ ⋅ u ∠Θ D7 − θ y0 ⇔ U = 3 ⋅ D ⋅ u ∠ + 210?−0? ⇔ U = ⋅ ⋅ u ∠ + 210? f y0 n y 1 ny 3 ny

Ou: U D7 3 U ∠ + 7h N U U 1 ND 1 ND = ⋅ ⇔ D = 3 ⋅ ∠ + 7h ⇔ = ⋅ ∠ − 7h ⇔ U = ⋅ ⋅ u ∠ − 7h u y0 1 u ∠ 0h ny u u 3 ny 3 ny

Dy11 ⇔ D7 (19) - y8 1 FD7 N D N 1 ND U= ⋅ ⋅ u ∠Θ D7 − θ y0 ⇔ U = 3 ⋅ D ⋅ u ∠ + 210?−0? ⇔ U = ⋅ ⋅ u ∠ + 210? f y0 n y 1 ny 3 ny

Ou: U D19 u y8

3 U ∠ + 19h N U U 1 ND 1 ND ⋅ ⇔ D = 3 ⋅ ∠ + 11h ⇔ = ⋅ ∠ − 11h ⇔ U = ⋅ ⋅ u ∠ − 11h 1 u ∠ + 8h ny u u 3 ny 3 ny

=

Dy3 ⇔ D7 - y4 1 FD7 N D N 1 ND U= ⋅ ⋅ u ∠Θ D7 − θ y0 ⇔ U = 3 ⋅ D ⋅ u ∠ + 210?−0? ⇔ U = ⋅ ⋅ u ∠ + 210? f y0 n y 1 ny 3 ny

Ou: U D7 u y4

=

3 U ∠ + 7h N U U 1 ND 1 ND ⋅ ⇔ D = 3 ⋅ ∠ + 3h ⇔ = ⋅ ∠ − 3h ⇔ U = ⋅ ⋅ u ∠ − 3h 1 u ∠ + 4h ny u u 3 ny 3 ny

Na página seguinte, apresentamos os “Símbolos de Ligação” Dy7, Dy11 e Dy3.

A. A. A. C. Barrias

41 / 59

Dy7

D7 - y0

D7

y0

UD

0

uy

A

7

C

1

A

n

2U

2u y

D

4

2

A

n

UD

8

3

B

n

Dy11

y8

UD

uy

A

7

C

1

n

2U

2u y

D

2

A

n

UD

8

n

y4

UD

uy

A

7

C

1

n

2U

2u y

D

2

A

n

UD

8 C

11 2

c

uy

3

C

3

b

1

B

B

b

3

D7 – y4

D7

4

a

3

B

Dy3

A

11

uy

3

0

2

C

C

c

1

B

B

c

3

D7 – y8

D7

A

b

uy

3

4

2

C

C

0

11

B

B

a

1

B

n

3

a

A. A. A. C. Barrias

42 / 59

Yd7 ⇔ Y6(Y18) - d11 FY18 N Y N 1 NY ⋅ ⋅ u ∠Θ Y18 − θ d11 ⇔ U = ⋅ ⋅ u ∠ + 540?−330? ⇔ U = 3 ⋅ Y ⋅ u ∠ + 210? 1 nd f d11 n d nd 3

U=

Ou: U Y18 N N N 1 U ∠ + 18h 1 U U = ⋅ ⇔ Y = ⋅ ∠ + 7h ⇔ = 3 ⋅ Y ∠ − 7h ⇔ U = 3 ⋅ Y ⋅ u ∠ − 7h u d11 nd u nd nd 3 u ∠ + 11h 3 u

Yd11 ⇔ Y6(Y18) - d7 FY18 N Y N 1 NY ⋅ ⋅ u ∠Θ Y18 − θ d11 ⇔ U = ⋅ ⋅ u ∠ + 540?−330? ⇔ U = 3 ⋅ Y ⋅ u ∠ + 210? 1 f d11 n d nd nd 3

U=

Ou: U Y18 u d7

1

U ∠ + 18h N 1 U U N N ⋅ ⇔ Y = ⋅ ∠ + 11h ⇔ = 3 ⋅ Y ∠ − 11h ⇔ U = 3 ⋅ Y ⋅ u ∠ − 11h nd u nd nd 3 u ∠ + 7h 3 u

=

Yd3 ⇔ Y6 - d3

U=

FY18 N Y N 1 NY ⋅ ⋅ u ∠Θ Y18 − θ d11 ⇔ U = ⋅ ⋅ u ∠ + 540?−330? ⇔ U = 3 ⋅ Y ⋅ u ∠ + 210? 1 nd f d11 n d nd 3

Ou: U Y6 u d3

=

1

U ∠ + 6h N 1 U U N N ⋅ ⇔ Y = ⋅ ∠ + 3h ⇔ = 3 ⋅ Y ∠ − 3h ⇔ U = 3 ⋅ Y ⋅ u ∠ − 3h nd u nd nd 3 u ∠ + 3h 3 u

Na página seguinte, apresentamos os “Símbolos de Ligação” Yd7, Yd11 e Yd3.

A. A. A. C. Barrias

43 / 59

Yd7

Y6-d11

Y6

d11

UY

ud

0

b

1

A

1

N

2U Y

2

N

2

UY

3

N

UY N

1

2U Y

2

N

UY N

1

b

b

7

c

11

2u

d

N

a

UY

2

b

c

ud

8 3

3

ud

2U Y

C

b

3

c

4 2

a

d3

0

B

11

Y6-d3

Y6

1

a

ç

ud c

Yd3

A

7

d

N

UY

3

c

c

2u

b

8 C

3

ud a

4 2

c

3

d7

0

B

b

Y6-d7

Y6

1

11

a

ud a

Yd11

A

b

d

c

8 C

7

2u

4 B

a

N

b

3

a

3 a

A. A. A. C. Barrias

44 / 59

Yz7 ⇔ Y6 (Y18) - z11

U = 2⋅

FY18 N Y 1 NY 2 NY ⋅ ⋅ u ∠Θ Y18 −θ z11 ⇔ U = 2 ⋅ ⋅ ⋅ u ∠ + 540?−330? ⇔ U = ⋅ ⋅ u ∠ + 210? f z11 n z n 3 nz 3 z

Ou: U Y18 N 1 U ∠ + 18h U U 2 NY 2 NY = ⋅ ⇔ Y = 3 ⋅ ∠ + 7h ⇔ = ⋅ ∠ − 7h ⇔ U = ⋅ ⋅ u ∠ − 7h 1 u ∠ + 11h nz u z11 u u 3 nz 3 nz 2 3

Yz11 ⇔ Y6 (Y18) - z7

U = 2⋅

FY18 N Y 1 NY 2 NY ⋅ ⋅ u ∠Θ Y18 −θ z7 ⇔ U = 2 ⋅ ⋅ ⋅ u ∠ + 540?−210? ⇔ U = ⋅ ⋅ u ∠ + 330? f z7 n z 3 nz 3 nz

Ou: U Y18 u z7

1 U ∠ + 18h N U U 2 NY 2 NY ⋅ ⇔ Y = 3 ⋅ ∠ + 11h ⇔ = ⋅ ∠ − 11h ⇔ U = ⋅ ⋅ u ∠ − 11h 1 u ∠ + 7h nz u u 3 nz 3 nz 2 3

=

Yz3 ⇔ Y6 - z3

U = 2⋅

FY6 N Y 1 NY 2 NY ⋅ ⋅ u ∠Θ Y6 −θ z3 ⇔ U = 2 ⋅ ⋅ ⋅ u ∠ + 180?−90? ⇔ U = ⋅ ⋅ u ∠ + 90? f z3 n z 3 nz 3 nz

Ou: U Y6 u z3

=

1 U ∠ + 6h N U U 2 NY 2 NY ⋅ ⇔ Y = 3 ⋅ ∠ + 3h ⇔ = ⋅ ∠ − 3h ⇔ U = ⋅ ⋅ u ∠ − 3h 1 u ∠ + 3h nz u u 3 nz 3 nz 2 3

Na página seguinte, apresentamos os “Símbolos de Ligação” Yz7, Yz11 e Yz3.

A. A. A. C. Barrias

45 / 59

Yz7

Y6-z11

Y6

z11

UY

uz

0 1

A

2U

4 B

N

n

2

Y

2

N

n

2

UY

3

N

n

3

UY

2U

B

n

2

Y

N

n

N

n

n

y

z

7 c

1

uz

11 2

uz

a

3 b

3

N

n

uz y

1

x

z

2

UY

y

uz N

n

3

b uz

11 2

uz x

z

7

1 2

uz

8 3

x

2

x

3

2

Y

4

C

3 c

uz

uz N

2

uz

z3

UY

B

3

b

Y6-z3

0 2U

z

z

2

Y6

1

x

uz

Yz3

A

2

uz

UY

3

y

y

1

8 C

11

z

uz N

2

uz

z7

0

4

a

Y6-z7

Y6

1

7

1 2

uz

Yz11

A

x

uz

8 C

uz y

1

3

c

3 a

A. A. A. C. Barrias

46 / 59

Dd8 ⇔ D7(D19) - d11 1 FD19 N D N N U= ⋅ ⋅ u ∠Θ D19 − θ d11 ⇔ U = 3 ⋅ D ⋅ u ∠ + 570?−330? ⇔ U = D ⋅ u ∠ + 240? 1 f d11 n d nd nd 3

Ou: U D19 = u d11

3 U ∠ + 19h N U U ND N ⋅ ⇔ D = ∠ + 8h ⇔ = ∠ − 8h ⇔ U = D ⋅ u ∠ − 8h u ∠ + 11h n u u n nd 3 d d

Dd0 ⇔ D7 - d7 1 FD7 N D N N U= ⋅ ⋅ u ∠Θ D7 − θ d7 ⇔ U = 3 ⋅ D ⋅ u ∠ + 210º −210º ⇔ U = D ⋅ u ∠ 0h 1 nd f d7 n d nd 3

Ou: U D7 u d7

=

3 U ∠ + 7h N U U N N ⋅ ⇔ D = ∠ 0h ⇔ = D ∠ 0h ⇔ U = D ⋅ u ∠ 0h nd u u nd nd 3 u ∠ + 7h

Dd4 ⇔ D7 - d3 1 FD7 N D N N U= ⋅ ⋅ u ∠Θ D7 − θ d3 ⇔ U = 3 ⋅ D ⋅ u ∠ + 210?−90? ⇔ U = D ⋅ u ∠ + 120? 1 f d3 n d nd nd 3

Ou: U D7 u d3

=

3 U ∠ + 7h N U U N N ⋅ ⇔ D = ∠ + 4h ⇔ = D ∠ − 4h ⇔ U = D ⋅ u ∠ − 4h nd u u nd nd 3 u ∠ + 3h

Na página seguinte, apresentamos os “Símbolos de Ligação” Dd8, Dd0 e Dd4.

A. A. A. C. Barrias

47 / 59

Dd8

D7-d11

d11

D7

ud

UD

0

A

b

C

1

1

A

2U

4

2u

D

B

c

2

B

UD

8

a

3

a

C

1

1

2u

D

B

b

2

B

c

3

c

C

1

1

2u

D

B

a

2

B

A

UD

8 C

b

b

8

c

0

d

2

b

c

ud

C

b

3

4

ud

A

2U

b

d3

UD

4

0

D7-d3

D7

A

a

c

a

3

B

Dd4

0

8

ud

C

C

c

d

2

A

UD

8

c

ud

A

2U

4

d7

UD

4

c

D7-d7

D7

A

0 b

3

B

Dd0

0

b

a

ud

C

C

8

d

2

A

a

B

3

a

4 a

A. A. A. C. Barrias

48 / 59

Dz8 ⇔ D7(D19) - z11 1 F N U = 2 ⋅ D19 ⋅ D ⋅ u ∠Θ D19 − θ z11 ⇔ U = 2 ⋅ f z11 n z

3 ⋅ N D ⋅ u ∠ + 570?−330? ⇔ U = 2 ⋅ N D ⋅ u ∠ + 240? 3 nz 3 nz

Ou: U D19 3 U ∠ + 19h N U U 2 ND 2 N = ⋅ ⇔ D = 3 ⋅ ∠ + 8h ⇔ = ⋅ ∠ − 8h ⇔ U = ⋅ D ⋅ u ∠ − 8h 1 n u z11 u ∠ + 11h u u 3 nz 3 nz z 2 3

Dz0 ⇔ D7 - z7 1 F N U = 2 ⋅ D19 ⋅ D ⋅ u ∠Θ D19 − θ z11 ⇔ U = 2 ⋅ f z11 n z

3 ⋅ N D ⋅ u ∠ + 570?−330? ⇔ U = 2 ⋅ N D ⋅ u ∠ + 240? 3 nz 3 nz

Ou: U D7 u z7

=

3 U ∠ + 7h N U U 2 N 2 N ⋅ ⇔ D = 3 ⋅ ∠ 0h ⇔ = ⋅ D ∠ 0h ⇔ U = ⋅ D ⋅ u ∠ 0h 1 u ∠ + 7h nz u u 3 nz 3 nz 2 3

Dz4 ⇔ D7 - z3 1 F N U = 2 ⋅ D7 ⋅ D ⋅ u ∠Θ D7 − θ z3 ⇔ U = 2 ⋅ f z3 n z

3 ⋅ N D ⋅ u ∠ + 210?−90? ⇔ U = 2 ⋅ N D ⋅ u ∠ + 120? 3 nz 3 nz

Ou: U D7 3 U ∠ + 7h N U U 2 ND 2 N = ⋅ ⇔ D = 3 ⋅ ∠ + 4h ⇔ = ⋅ ∠ − 4h ⇔ U = ⋅ D ⋅ u ∠ − 4h 1 u ∠ + 3h nz u z3 u u 3 nz 3 nz 2 3

Na página seguinte, apresentamos os “Símbolos de Ligação” Dz8, Dz0 e Dz4.

A. A. A. C. Barrias

49 / 59

Dz8

D7-z11

D7

z11

UD

0

A

uz

C

1

A

2U

4

n

2

D

2

A

n

2

B

n

Dz0

3

A

2U

n

1 2

D

3

y

x

uz

2

A

n

2

z

y

x

z

y

x

uz

3

B

n

Dz4

3

c uz

0

uz

uz

C

1 2U

2

n

1 2

D

2

b

uz

A

n

2

UD

b uz

0

z

y

2

x

z

3

uz B

n

3

8

1 2

uz

uz

C

3

4

3

B

B

a

z3

UD A

8

D7-z3

D7 A

4

1 2

C

C

b

c

uz

uz

UD

C

z

B

B

8

x

uz

C

1

A

4

2

z7

UD

0

uz

D7-z7

D7

8

y

uz

3

C

4

a

0

z

C

8

1 2

uz

UD

0

x

B

B

8

uz y

1

c

4 a

A. A. A. C. Barrias

50 / 59

Dd10 ⇔ D5(D17) - d7 1 FD17 N D N N U= ⋅ ⋅ u ∠Θ D17 − θ d7 ⇔ U = 3 ⋅ D ⋅ u ∠ + 510?−210? ⇔ U = D ⋅ u ∠ + 300? 1 f d7 n d nd nd 3

Ou: U D17 = u d7

3 U ∠ + 17h N U U ND N ⋅ ⇔ D = ∠ + 10h ⇔ = ∠ − 1 0h ⇔ U = D ⋅ u ∠ − 10h u ∠ + 7h n u u n nd 3 d d

Dd2 ⇔ D5 - d3 1 FD5 N D N N U= ⋅ ⋅ u ∠Θ D5 − θ d3 ⇔ U = 3 ⋅ D ⋅ u ∠ + 150?−90? ⇔ U = D ⋅ u ∠ + 60? 1 nd f d3 n d nd 3

Ou: U D5 u d3

=

3 U ∠ + 5h N U U N N ⋅ ⇔ D = ∠ + 2h ⇔ = D ∠ − 2h ⇔ U = D ⋅ u ∠ − 2h nd u u nd nd 3 u ∠ + 3h

Dd6 ⇔ D5(D17) - d11 1 FD17 N D N N U= ⋅ ⋅ u ∠Θ D17 − θ d11 ⇔ U = 3 ⋅ D ⋅ u ∠ + 510?−330? ⇔ U = D ⋅ u ∠ + 180º 1 nd f d11 n d nd 3

Ou: U D17 u d11

=

3 U ∠ + 17h N U U N N ⋅ ⇔ D = ∠ + 6h ⇔ = D ∠ − 6h ⇔ U = D ⋅ u ∠ − 6h nd u u nd nd 3 u ∠ + 11h

Na página seguinte, apresentamos os “Símbolos de Ligação” Dd10, Dd2 e Dd6.

A. A. A. C. Barrias

51 / 59

Dd10

D5-d7

D5

d7 ud

UD

0

A

4

B

A

c

B

1

1 2u

2U

D

2

B

2

C

C

b

A

3

3

Dd2

4

B

A

1

1 2u

D

2

C

a

A

3

3

Dd6

4

B

A

1

a

1 2u

D

2

C

C

C

3

6

b

10

c

2

a

6

d

b

2

b

b

c

ud

UD

8

b

a

ud B

2U

B

2

d11

UD A

a

c

D5 (17)-d11

D5

0

10

c

ud

UD C

c

d

c

C

2

8

6

ud b

B

2U

B

c

b

d3

UD A

2

D5-d3

D5

0

b

a

ud

UD

8

10

d

a

C

a

A

c

3

a

A. A. A. C. Barrias

52 / 59

Dz10 ⇔ D5(D17) - z7 1 F N U = 2 ⋅ D17 ⋅ D ⋅ u ∠Θ D17 − θ z7 ⇔ U = 2 ⋅ f z7 n z

3 ⋅ N D ⋅ u ∠ + 510?−210? ⇔ U = 2 ⋅ N D ⋅ u ∠ + 300? 3 nz 3 nz

Ou: U D17 3 U ∠ + 17h N U U 2 ND 2 N = ⋅ ⇔ D = 3 ⋅ ∠ + 10h ⇔ = ⋅ ∠ − 10h ⇔ U = ⋅ D ⋅ u ∠ − 10h 1 n u z7 u ∠ + 7h u u 3 nz 3 nz z 2 3

Dz2 ⇔ D5 - z3 1 F N U = 2 ⋅ D5 ⋅ D ⋅ u ∠Θ D5 − θ z3 ⇔ U = 2 ⋅ f z3 n z

3 ⋅ N D ⋅ u ∠ + 150?−90? ⇔ U = 2 ⋅ N D ⋅ u ∠ + 60? 3 nz 3 nz

Ou: U D5 u z3

=

3 U ∠ + 5h N U U 2 N 2 N ⋅ ⇔ D = 3 ⋅ ∠ + 2h ⇔ = ⋅ D ∠ − 2h ⇔ U = ⋅ D ⋅ u ∠ − 2h 1 u ∠ + 3h nz u u 3 nz 3 nz 2 3

Dz6 ⇔ D5(D17) - z11 1 U = 2⋅

FD17 N D ⋅ ⋅ u ∠Θ D17 − θ z11 ⇔ U = 2 ⋅ f z11 n z

3 ⋅ N D ⋅ u ∠ + 510?−330? ⇔ U = 2 ⋅ N D ⋅ u ∠ + 180? 3 nz 3 nz

Ou: U D17 u z11

=

3 U ∠ + 17h N U U 2 N 2 N ⋅ ⇔ D = 3 ⋅ ∠ + 6h ⇔ = ⋅ D ∠ − 6h ⇔ U = ⋅ D ⋅ u ∠ − 6h 1 u ∠ + 11h nz u u 3 nz 3 nz 2 3

Na página seguinte, apresentamos os “Símbolos de Ligação” Dz10, Dz2 e Dz6.

A. A. A. C. Barrias

53 / 59

Dz10

D5-z7

D5

z7

UD

0 A

A

C

2

n

C

C

n

Dz2

A

n

1

C

n

C

C

n

Dz6

A

1

n

2

C

C

3

z

C

n

x

a

6 b

3

x

2

y

uz A

n

3

b uz

2 2

uz y

z

10

1 2

uz

UD

8

2 2

uz z

1 2

D

B

B

y

3

uz

uz

uz B

2U

4

y

c

z11

UD A

x

2

10

D5-z11

D5 0

c

1 2

uz A

3

x

uz

UD

8

b

6

3

uz z

1 2

D

2

B

z

uz

B

2 2

z3 B

2U

4

uz

D5-z3

UD A

y

a

uz y

3

D5 0

x

2

10

1 2

uz A

3

x

uz

UD

8

uz z

1 2

D

B

B

n

1 2U

4

uz B

3

c

6 a

A. A. A. C. Barrias

54 / 59

Dy11 ⇔ D5 (D17) - y6

1 FD17 N D N 1 ND U= ⋅ ⋅ u ∠Θ D17 − θ y6 ⇔ U = 3 ⋅ D ⋅ u ∠ + 510?−180? ⇔ U = ⋅ ⋅ u ∠ + 330? f y6 n y 1 ny 3 ny

Ou: U D17 3 U ∠ + 17h N U U 1 ND 1 ND = ⋅ ⇔ D = 3 ⋅ ∠ + 11h ⇔ = ⋅ ∠ − 11h ⇔ U = ⋅ ⋅ u ∠ − 11h u y6 1 u ∠ + 6h ny u u n 3 3 ny y

Dy3 ⇔ D5 - y2

1 FD5 N D N 1 ND U= ⋅ ⋅ u ∠Θ D5 − θ y2 ⇔ U = 3 ⋅ D ⋅ u ∠ + 150?−60? ⇔ U = ⋅ ⋅ u ∠ + 90? f y2 n y 1 ny 3 ny

Ou: U D5 3 U ∠ + 5h N U U 1 ND 1 ND = ⋅ ⇔ D = 3 ⋅ ∠ + 3h ⇔ = ⋅ ∠ − 3h ⇔ U = ⋅ ⋅ u ∠ − 3h u y2 1 u ∠ + 2h ny u u 3 ny 3 ny

Dy7 ⇔ D5 (D17) - y10

1 U=

FD17 N D N 1 ND ⋅ ⋅ u ∠Θ D17 − θ y10 ⇔ U = 3 ⋅ D ⋅ u ∠ + 510?−300? ⇔ U = ⋅ ⋅ u ∠ + 210? f y10 n y 1 ny 3 ny

Ou: U D17 3 U ∠ + 17h N U U 1 ND 1 ND = ⋅ ⇔ D = 3 ⋅ ∠ + 7h ⇔ = ⋅ ∠ − 7h ⇔ U = ⋅ ⋅ u ∠ − 7h u y10 1 u ∠ + 10 h ny u u 3 ny 3 ny

Na página seguinte, apresentamos os “Símbolos de Ligação” Dy11, Dy3 e Dy7.

A. A. A. C. Barrias

55 / 59

Dy11

D5-y6

D5

y6 uy

UD

0 A

A

C

2

n

C

C

n

Dy3

A

uy n

1

C

2

n

C

C

n

Dy7

A

uy B

1 2U

4

11 n

2

C

n

C

C

3

y

3 2

c

uy

UD

8

b

1 2u

D

B

B

b

3

y10

UD A

a

D5-y10

D5 0

3 2

7

A

3

y

uy

UD

8

c

1 2u

D

B

B

11

B

2U

4

c

3

y2

UD A

b

D5-y2

D5 0

3 2

7

A

3

y

uy

UD

8

a

1 2u

D

B

B

n

1 2U

4

11

B

A

7 n

3

a

A. A. A. C. Barrias

56 / 59

Yd11 ⇔ Y6(Y18) - d7

U=

FY18 N Y N 1 NY ⋅ ⋅ u ∠Θ Y18 − θ d7 ⇔ U = ⋅ ⋅ u ∠ + 540?−210? ⇔ U = 3 ⋅ Y ⋅ u ∠ + 330? 1 nd f d7 n d nd 3

Ou: U Y18 N N N 1 U ∠ + 18h 1 U U = ⋅ ⇔ Y = ⋅ ∠ + 11h ⇔ = 3 ⋅ Y ∠ − 11h ⇔ U = 3 ⋅ Y ⋅ u ∠ − 11h u d7 u ∠ + 7h n u u n nd 3 3 d d

Yd3 ⇔ Y6 - d3

U=

FY6 N Y N 1 NY ⋅ ⋅ u ∠Θ Y6 − θ d3 ⇔ U = ⋅ ⋅ u ∠ + 180º −90º ⇔ U = 3 ⋅ Y ⋅ u ∠ + 90? 1 f d3 n d nd nd 3

Ou: U Y6 N N N 1 U ∠ + 6h 1 U U = ⋅ ⇔ Y = ⋅ ∠ + 3h ⇔ = 3 ⋅ Y ∠ − 3h ⇔ U = 3 ⋅ Y ⋅ u ∠ − 3h u d3 nd u nd nd 3 u ∠ + 3h 3 u

Yd7 ⇔ Y6(Y18) - d11

U=

FY18 N Y N 1 NY ⋅ ⋅ u ∠Θ Y18 − θ d11 ⇔ U = ⋅ ⋅ u ∠ + 540?−330? ⇔ U = 3 ⋅ Y ⋅ u ∠ + 210? 1 nd f d11 n d nd 3

Ou: U Y18 N N N 1 U ∠ + 18h 1 U U = ⋅ ⇔ Y = ⋅ ∠ + 7h ⇔ = 3 ⋅ Y ∠ − 7h ⇔ U = 3 ⋅ Y ⋅ u ∠ − 7h u d11 nd u nd nd 3 u ∠ + 11h 3 u

Na página seguinte, apresentamos os “Símbolos de Ligação” Yd11, Yd3 e Yd7.

A. A. A. C. Barrias

57 / 59

Yd11

Y6-d7

d7

Y6 UY

ud

0

c

1

A

1

N

2U

2

N

UY b

3

3

N

Yd3

UY

1

N

2

N

UY a

3

3

N

Yd7

UY

1

1

N

11

c

3

a

7

d

b

2

N

UY

b

b

c

ud

8

c

3

b

2u

Y

4

C

7

ud a

2

b

a

d11

0

B

3

Y6-d11

Y6

2U

a

c

ud

8

A

11

d

c

2

c

2u

Y

4

C

c

ud b

1

B

7

d3

0 2U

c

b

Y6-d3

Y6

A

3

a

ud

8 C

b

d

a

2

11

2u

Y

4 B

a

N

3

a

A. A. A. C. Barrias

58 / 59

Yz11 ⇔ Y6(Y18) - z7

U = 2⋅

FY18 N Y 1 NY 2 NY ⋅ ⋅ u ∠Θ Y18 −θ z7 ⇔ U = 2 ⋅ ⋅ ⋅ u ∠ + 540?−210? ⇔ U = ⋅ ⋅ u ∠ + 330? f z7 n z n 3 3 nz z

Ou: U Y18 u z7

1 U ∠ + 18h N U U 2 NY 2 NY ⋅ ⇔ Y = 3 ⋅ ∠ + 11h ⇔ = ⋅ ∠ − 11h ⇔ U = ⋅ ⋅ u ∠ − 11h 1 u ∠ + 7h nz u u 3 nz 3 nz 2 3

=

Yz3 ⇔ Y6 - z3

U = 2⋅

FY6 N Y 1 NY 2 NY ⋅ ⋅ u ∠Θ Y6 −θ z3 ⇔ U = 2 ⋅ ⋅ ⋅ u ∠ + 180?−90? ⇔ U = ⋅ ⋅ u ∠ + 90? f z3 n z 3 nz 3 nz

Ou: U Y6 u z3

=

1 U ∠ + 6h N U U 2 NY 2 NY ⋅ ⇔ Y = 3 ⋅ ∠ + 3h ⇔ = ⋅ ∠ − 3h ⇔ U = ⋅ ⋅ u ∠ − 3h 1 u ∠ + 3h nz u u 3 nz 3 nz 2 3

Yz7 ⇔ Y6(Y18) - z11

U = 2⋅

FY18 N Y 1 NY 2 NY ⋅ ⋅ u ∠Θ Y18 −θ z11 ⇔ U = 2 ⋅ ⋅ ⋅ u ∠ + 540?−330? ⇔ U = ⋅ ⋅ u ∠ + 210? f z11 n z 3 nz 3 nz

Ou: U Y18 u z11

=

1 U ∠ + 18h N U U 2 NY 2 NY ⋅ ⇔ Y = 3 ⋅ ∠ + 7h ⇔ = ⋅ ∠ − 7h ⇔ U = ⋅ ⋅ u ∠ − 7h 1 u ∠ + 11h nz u u 3 nz 3 nz 2 3

Na página seguinte, apresentamos os “Símbolos de Ligação” Yz11, Yz3 e Yz7.

A. A. A. C. Barrias

59 / 59

Yz11

Y6-z7 z7

Y6 UY

uz

0 A

1 2U

4 B

n

N

2

Y

2

N

n

uz

3

N

n

Yz3

UY

uz

0 n

N

2

Y

2

n

N

N

n

Yz7

UY

2

N

n

2

Y

z

N

n

x

3 2

a

7 b

3

x

2

y

uz N

n

3

b uz

3 2

uz y

z

11

1 2

uz

UY

3

uz

uz z

1

8 C

y

3

c

uz

uz

1

B

y

11

1

z11

0

2U

c

Y6-z11

Y6

4

x

uz

3

b

7

3

2

x

2

UY

A

z

uz

8 C

3 2

uz z

1

4 B

uz

Y6-z3 z3

1

a

uz y

3

Y6

2U

y

11

1 2

x

2

UY

A

x

uz

8 C

uz z

1

3

c

7 a

A. A. A. C. Barrias