aula9

Circuitos Trifásicos

Aula 9

Trafos Trifásicos

Engenharia Elétrica Universidade Federal de Juiz de Fora

tinyurl.com/profvariz

Transformadores Trifásicos

Existemduas maneiraspara se obter umtransformador trifásico:

Conectando3 transformadores monofásicos idênticos: banco de

transformadores;

Enrolando-se as bobinas em um mesmo núcleo: transformador trifásico

Para a mesma potência aparente (kVA),o transformador trifásicoé sempremenor e mais baratose comparado com3transformadores monofásicos.

Existem4 maneiras básicasde se conectar os transformadores trifásicos:

Y–Y ∆ − ∆ Y−∆ ∆−Y

Transformadores Trifásicos

Independentemente da conexão ST= √ 3VLIL (1) PT= STcos(θ) = √ 3VLILcos(θ) (2) QT = STsen(θ) = √ 3VLILsen(θ) (3)

Em que VLe ILsão, respectivamente, a tensão de linha e a corrente de

linha.

Podem ser utilizadas tanto as grandezas do primário (VLpe ILp) quanto as

do secundário (VLse ILs);

Uma vez que a potência deve ser conservada no transformador ideal.

Conexão Y–Y

Relação entre as tensões e correntes

VLs= nVLp (4)

ILs=

ILp

n (5)

CHAPTER 13

Magnetically Coupled Circuits

557

For the - connection (Fig. 13.47), Eq. (13.70) also applies for

the line voltages and line currents. This connection is unique in the sense

that if one of the transformers is removed for repair or maintenance, the

other two form an open delta, which can provide three-phase voltages at

a reduced level of the original three-phase transformer.

+

−

V

_{L p}

+

−

V

_{L s}

= nV

_{L p}

I

_{L p}

I

L s

=

I

_{L p}

n

1:n

Figure13.46

Y-Y three-phase transformer connection.

+

−

V

_{L p}

+

−

V

_{L s}

= nV

_{L p}

I

_{L p}

I

L s

=

I

_{L p}

n

1:n

Figure13.47



- three-phase transformer connection.

For the Y- connection (Fig. 13.48), there is a factor of

√

3 arising

from the line-phase values in addition to the transformer per phase turns

ratio n. Thus,

V

Ls

=

nV

Lp

√

3

(13.71a)

I

Ls

=

√

3I

Lp

n

(13.71b)

Similarly, for the -Y connection (Fig. 13.49),

V

Ls

= n

√

3V

Lp

(13.72a)

I

Ls

=

I

Lp

n

√

3

(13.72b)

+

−

V

_{L p}

+

−

I

_{L p}

1:n

I

_{L s}

=

3 I

_{L p}

n

V

_{L s}

=

nV

_{L p}

3

Figure13.48

Y- three-phase transformer connection.

+

−

V

_{L s}

= n V

_{L p}

I

_{L p}

1:n

I

_{L s}

=

I

_{L p}

3

n

3

+

−

V

_Lp

Figure13.49



-Y three-phase transformer connection.

Figura 1:Conexão Y–Y de transformadores trifásicos.

Conexão ∆ − ∆

Relação entre as tensões e correntes

VLs= nVLp (6)

ILs=

ILp

n (7)

CHAPTER 13

Magnetically Coupled Circuits

557

For the - connection (Fig. 13.47), Eq. (13.70) also applies for

the line voltages and line currents. This connection is unique in the sense

that if one of the transformers is removed for repair or maintenance, the

other two form an open delta, which can provide three-phase voltages at

a reduced level of the original three-phase transformer.

+

−

V

_{L p}

+

−

V

_{L s}

= nV

_{L p}

I

_{L p}

I

L s

=

I

_{L p}

n

1:n

Figure13.46

Y-Y three-phase transformer connection.

+

−

V

_{L p}

+

−

V

_{L s}

= nV

_{L p}

I

_{L p}

I

L s

=

I

_{L p}

n

1:n

Figure13.47



- three-phase transformer connection.

For the Y- connection (Fig. 13.48), there is a factor of

√

3 arising

from the line-phase values in addition to the transformer per phase turns

ratio n. Thus,

V

Ls

=

nV

Lp

√

3

(13.71a)

I

Ls

=

√

3I

Lp

n

(13.71b)

Similarly, for the -Y connection (Fig. 13.49),

V

Ls

= n

√

3V

Lp

(13.72a)

I

Ls

=

I

Lp

n

√

3

(13.72b)

+

−

V

_{L p}

+

−

I

_{L p}

1:n

I

_{L s}

=

3 I

_n

L p

V

_{L s}

=

nV

_{L p}

3

Figure13.48

Y- three-phase transformer connection.

+

−

V

_{L s}

= n V

_{L p}

I

_{L p}

1:n

I

_{L s}

=

I

_{L p}

3

n

3

+

−

V

_Lp

Figure13.49



-Y three-phase transformer connection.

Figura 2:Conexão ∆ − ∆ de transformadores trifásicos.

Conexão ∆−Y

Relação entre as tensões e correntes, sequência abc VLs= √ 3nVLp 30◦ (8) ILs= ILp √ 3n 30 ◦ ₍₉₎

CHAPTER 13

Magnetically Coupled Circuits

557

For the - connection (Fig. 13.47), Eq. (13.70) also applies for

the line voltages and line currents. This connection is unique in the sense

that if one of the transformers is removed for repair or maintenance, the

other two form an open delta, which can provide three-phase voltages at

a reduced level of the original three-phase transformer.

+ − V_{L p} + − V_{L s} = nV_{L p} I_{L p} IL s = I_{L p} n 1:n

Figure13.46

Y-Y three-phase transformer connection.

+ − V_{L p} + − V_{L s} = nV_{L p} I_{L p} I_{L s} = I_{L p} n 1:n

Figure13.47

- three-phase transformer connection.

For the Y- connection (Fig. 13.48), there is a factor of

√

3 arising

from the line-phase values in addition to the transformer per phase turns

ratio n. Thus,

V

Ls

=

nV

Lp

√

3

(13.71a)

I

Ls

=

√

3I

Lp

n

(13.71b)

Similarly, for the -Y connection (Fig. 13.49),

V

Ls

= n

√

3V

Lp (13.72a)

I

Ls

=

I

Lp

n

√

3

(13.72b) + − V_{L p} + − I_{L p} 1:n I_{L s} = 3 IL p n V_{L s} = nV_{L p} 3

Figure13.48

Y- three-phase transformer connection.

+ − V_{L s} = n V_{L p} I_{L p} 1:n I_{L s} = I_{L p} 3 n 3 + − V_Lp

Figure13.49

-Y three-phase transformer connection. Figura 3:Conexão ∆-Y de transformadores trifásicos.

Conexão Y−∆

Relação entre as tensões e correntes, sequência abc VLs= nVLp √ 3 −30 ◦ ₍₁₀₎ ILs= √ 3ILp n −30 ◦ ₍₁₁₎

CHAPTER 13

Magnetically Coupled Circuits

557

For the - connection (Fig. 13.47), Eq. (13.70) also applies for the line voltages and line currents. This connection is unique in the sense that if one of the transformers is removed for repair or maintenance, the other two form an open delta, which can provide three-phase voltages at a reduced level of the original three-phase transformer.

+ − VL p + − VL s = nVL p IL p IL s = IL p n 1:n

Figure13.46

Y-Y three-phase transformer connection.

+ − VL p + − VL s = nVL p IL p IL s = IL p n 1:n

Figure13.47

- three-phase transformer connection.

For the Y- connection (Fig. 13.48), there is a factor of√3 arising from the line-phase values in addition to the transformer per phase turns ratio n. Thus, VLs = nVLp √ 3 (13.71a) ILs = √ 3ILp n (13.71b)

Similarly, for the -Y connection (Fig. 13.49),

VLs= n √ 3VLp (13.72a) ILs = ILp n√3 (13.72b) + − VL p + − IL p 1:n IL s = 3 IL p n VL s = nVL p 3

Figure13.48

Y- three-phase transformer connection.

+ − V_{L s} = n V_{L p} IL p 1:n IL s = IL p 3 n 3 + − VLp

Figure13.49

-Y three-phase transformer connection. Figura 4:Conexão Y−∆ de transformadores trifásicos.

Ex. 1

Um transformador Y−∆ é conectado a uma carga que consome 60 kVA com fator de potência 0,85 (adiantado) através de um alimentador cuja impedância é 0,05 + j0,1Ω, por fase, como mostrado na figura a seguir. Determine a magnitude de:

(a) A corrente de linha na carga

(b) A tensão de linha no secundário do transformador

(c) A corrente de linha no primário do transformador

580

PART 2

AC Circuits

13.57 A Y- three-phase transformer is connected to a 60-kVA load with 0.85 power factor (leading) through a feeder whose impedance is 0.05+ j0.1  per phase, as shown in Fig. 13.122 below. Find the magnitude of:

(a) the line current at the load,

(b) the line voltage at the secondary side of the transformer,

(c) the line current at the primary side of the transformer.

13.58 The three-phase system of a town distributes power with a line voltage of 13.2 kV. A pole transformer connected to single wire and ground steps down the high-voltage wire to 120 V rms and serves a house as shown in Fig. 13.123.

(a) Calculate the turns ratio of the pole transformer to get 120 V.

(b) Determine how much current a 100-W lamp connected to the 120-V hot line draws from the high-voltage line.

13.2 kV 120 V

; ;;

;;;;;;;;;;;;;;;;;

Figure13.123

For Prob. 13.58.

Section 13.8 PSpice Analysis of Magnetically Coupled Circuits

13.59 Rework Prob. 13.14 using PSpice.

13.60 Use PSpice to find I1, I2, and I3in the circuit of Fig.

13.124. j15 Ω j80 Ω j0 Ω j100 Ω j10 Ω + − V 60 0° j50 Ω – j20 Ω 20 90° + − V I1 I3 I2 40 Ω 80 Ω

Figure13.124

For Prob. 13.60.

13.61 Rework Prob. 13.15 using PSpice.

13.62 Use PSpice to find I1, I2, and I3in the circuit of Fig.

13.125. I1 I2 I3 100 Ω 8 H 2 H 1 H 4 H 3 H 2 H 60 mF 200 Ω 70 Ω 50 mF + − V 120 0° f = 100

Figure13.125

For Prob. 13.62.

1:n 0.05 Ω j 0.1 Ω 0.05 Ω j 0.1 Ω 0.05 Ω j 0.1 Ω Balanced load 60 kVA 0.85 pf leading 2640 V 240 V

Figure13.122

For Prob. 13.57.

Figura 5:Circuito do exemplo 1.

Ex. 1

Resposta

(a) 144,34 A

(b) VLs= 238,7 V

(c) ILp= 13,05 A