3-phase transformer report 2

Group 7- Three-phase transformer 1

Group 7 :

1. Võ Quốc Cường (20900328)

2. Trương Hoàng Trí (80902940)

3. Nguyễn Tiến Khoa (40901248)

4. Lê Nguyễn Anh Nga (60901653)


Contents

1. Introduction .......................................................................................... 3

- Usage ..................................................................................................... 5

- Structure ................................................................................................ 6

2. Three-phase transformer bank .......................................................... 8

- Introduction

- Advantages and Defects

3. Three Phase Transformer Winding Configurations ....................... 9

4. Problems .............................................................................................. 11


1. Introduction

Three-phase power

The power company generators produce electricity by rotating 3 coils or windings through a magnetic

field within the generator . These coils or windings are spaced 120 degrees apart. As they rotate

through the magnetic field they generate power which is then sent out on three lines as in three-

phase power.

Three–phase transformer

- 3 coils or windings connected in the proper sequence.

- to match the incoming power.

- transform the power company voltage to the level of voltage we need and maintain the proper

phasing or polarity.


Usage

- Electricity transmission and distribution systems.

- Industrial electricity network.

Structure

- Three-legged iron core.

- Each leg has a respective primary and secondary winding.

Steel core: 3 cylinder to wrap wire and yoke to close the magnetic circuit.

Steel core is made of steel foil, both sides painted for electricity-insulation and merge them

into a cylinder.

Wires: windings (copper) are insulated, wrapped around steel core.

(AX, BY,CZ):primary windings.

(ax, by, cz):secondary windings.


2. THREE-PHASE TRANSFORMER BANK

- Connect 3 similar single-phase transformer.

- The primary and secondary windings may be connected in either star or delta configurations.


Advantages

- Ease of transportation.

- 1 phase of the transformer at fault, the other 2 are not affected.

Defects

- Inefficient magnetic circuit.

- Higher capital cost than a single one.

2. Three Phase Transformer Winding Configurations

Transformer design concept


Two ways to configure

Delta

(∆)

Wye (Y)


(positive pole of coil is marked by a dot)


Combining the Winding Configurations

Wye-Delta

Y/∆

Wye-Wye with neutral

Y/Y0


Delta-Delta

∆ /∆

Delta-Wye

with neutral

∆/Y0


The voltage transformation ratio K

� Phase transformation ratio:

� Winding transformation ratio:


a. Case 1: Transformer Wye-Wye with neutral (Y/Yo)

Primary is Wye configuration: Ud1 = √3Up1

Secondary is Wye configuration : Ud2 =√3Up2

So:

Kd =��

��=

√��

√��=

��

��=Kp

b. Case 2: Transformer Wye-Delta with neutral (Y/∆∆∆∆)

Primary is Wye configuration: Ud1 = √3Up1

Secondary is Wye configuration : Ud2 =Up2

So:

Kd =��

��=

√��

��=

√��

��=√3Kp

c. Case 3: Transformer Delta-Delta (∆∆∆∆/∆∆∆∆)

Primary is Wye configuration: Ud1 = Up1

Secondary is Wye configuration : Ud2 =Up2

So:

Kd =��

��=

��

��=Kp

d. Case 4: Transformer Delta-Wye with neutral (∆∆∆∆/Yo)

Primary is Wye configuration: Ud1 = Up1

Secondary is Wye configuration : Ud2 =√3Up2

So:

Kd =��

��=

��

√��=

√�Kp


4.Problem:

1) Three one-phase transformers 10kVA, 2300/460V, connected together,

make a three-phase transformer .It supplies 18kW to a three-phase load,

balanced 460V. Power factor 0,8 (lagging).

�� , ��, ��, ��, ��, ��?

Each one-phase transformer has apparent power is 10kVA, primary voltage 2300V, secondary 460V.

So we have:

��=2300V; ��=460V

Primary connection is Y, and secondary connection is ∆. So:

��=��√3 � 3988�; ��=��=460V

��=��=√3��

⇒ ��="#

√��$%&'#=

()))

√�.+,).),(=18,1A

• ��=-��

√�=

(,

√�=10,5A (connection ∆)

We have: ��=�� ⇒ ��=��-��

��=2,1A

��=��=2,10 (connection Y)


2) A three-phased motor ,50hp, 440 V, having

performance 0,88 and power factor 0,82 is

supplied by a three-phase transformer 6600/440

V. Connection ∆-Y

2) Caculate ��?

3)Caculate ��, ��, ��, ��?

2) When we write a three-phase motor 50hp,440V. It means the power motor supplies to Load is

�2=50hp, and voltage source supplies to motor is 440V

��="3

4=

5)67+,8

),((=42386W

3) Similar to exercise 1.

��=√3��

Suy ra ��="�

√��$%&'#=

+��(,

√�6++)6),(�=67,83A

��=��

√�=254V; ��=��= 67,83A

��=��=6600V

��=�� ⇒ ��=�5+6,7,(�

,,))=2,61A


4) A three-phase transformer is combinated from three ideal one-phase transformers, supplied by

three-phase power –supply 2400V .Connection Y-Y. It supplies 600kVA to a three-phase load,

balanced at 240V. Caculate ��?

We

have

9�=9�=√3��

⇒ ��=:�

√��=

,)))

√�.�+)=1443A

��=��=1443A (connect Y)

��=��

√�=139V; ��=2400V; ��=1385,6V

��=�� ⇒ ��=�;6++�

�(5,,=144,76A

5) a combinating three-phase transformer, connection Y-∆, supplies 500kW to to a three-phase load,

balanced at 1100V, power factor 0,85 (late). ��= 11000V. Caculate ��? (Circuit is the same as circuit

in exercise 1)

We have: �<�=1100V

��="#

√�.��.$%&'#=

5)))))

√�6))6),(5=223A

��= -��

√� =129A

��=��

√�=6351V

�� =))6�;

,5�=22,3A


With this data: a combinating three-phase transformer, connection ∆-Y, decreases voltage from

12600V to 660V and supplies 55kVA to load, having power factor 0,866 (late)

6) Calculate the transformer ratio of each one-phase transformer (=�) ?

7) Caculate apparent power (9�> (kVA) and effective power (��) (kW) of each one-phase transformer?

8)Caculate ��?

• 9�=55kVA; cos �=0,866

We have :��=12600V; ��=660V

��=��=12600V; ��=��/√3=381V

6) =�=��/��=12600/381=33

7) 9� � 55/3=18,3kVA

��=9�.cos �=18,3x0,866=15,9kW

8) 9�=√3x��x��

So ��=55000/(√3x660)=48,1A=��

�D � E381x48,1)/12600=1,45A

�< � √3�D=1,92A

3-phase transformer report 2

Documents