Group 7 : 1. Võ Quốc Cường (20900328) 2. Trương Hoàng Trí (80902940) 3. Nguyễn Tiến Khoa (40901248) 4. Lê Nguyễn Anh Nga (60901653)
Group 7- Three-phase transformer
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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
Group 7- Three-phase transformer
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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 threephase 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.
Group 7- Three-phase transformer
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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.
Group 7- Three-phase transformer
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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.
Group 7- Three-phase transformer
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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
Group 7- Three-phase transformer
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Two ways to configure
Delta (∆)
Wye (Y)
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(positive pole of coil is marked by a dot)
Group 7- Three-phase transformer
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(positive pole of coil is marked by a dot)
Group 7- Three-phase transformer
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Combining the Winding Configurations
Wye-Wye with neutral Y/Y0
Wye-Delta Y/∆
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Delta-Delta ∆ /∆
Delta-Wye with neutral ∆/Y0
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The voltage transformation ratio K
Phase transformation ratio:
Winding transformation ratio:
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a. Case 1: Transformer Wye-Wye with neutral (Y/Yo) Primary is Wye configuration: Ud1 = √3Up1 Secondary is Wye configuration : Ud2 =√3Up2 So:
√ Kd = = = =K √ p 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 = = = K √ √ p Group 7- Three-phase transformer
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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)
Group 7- Three-phase transformer
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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 "
= 43=
5)67+,8 ),((
=42386W
3) Similar to exercise 1. =√3 Suy ra = =
√
"
=
+(,
=67,83A
√ $%&'# √6++)6),(
=254V; = = 67,83A
= =6600V = ⇒ =
5+6,7,( ,,))
=2,61A
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4) A three-phase transformer is combinated from three ideal one-phase transformers, supplied by three-phase power –supply 2400V .Connection Y-Y. It supplies 600kVA to a three-phase load, balanced at 240V. Caculate ?
We have
9 =9 =√3 ⇒ =
:
=
,)))
=1443A
√ √.+)
= =1443A (connect Y) =
√
=139V; =2400V; =1385,6V
= ⇒ =
;6 ++
(5,,
=144,76A
5) a combinating three-phase transformer, connection Y-∆, supplies 500kW to to a three-phase load, balanced at 1100V, power factor 0,85 (late). = 11000V. Caculate ? (Circuit is the same as circuit in exercise 1) We have: < =1100V = =
"#
5)))))
=
=223A
√. .$%&'# √6
))6),(5 - √
=
=129A
√
=6351V
=
))6 ; ,5
=22,3A
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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 < √3D =1,92A
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