1. Electrical Engineering Principle DNT 235/3
CHAPTER 10
THREE PHASE TRANSFORMER

2. Three Phase Transformer

3. The knowledge that students can obtain from this chapter are:
Basic principle and structure of a three phase transformer.
Description of winding connections, delta and star configurations and autotransformer.

4. Construction
A three-phase transformer is constructed by winding three single-phase transformers on a single core.
Three-phase transformers are connected in either wye or delta configurations.

5. Three-phase transformer construction

6. The basic principle of a 3 phase transformer.
The figure shows which primary windings have been shown inter connected in star output across 3-phase supply.
The three cores are 120o apart and their empty legs are shown in contact with each other.
The centre leg, formed by these three, carries flux produced by the three-phase currents IR, IY and IB. As any instant IR+ IY + IB = 0, hence the sum of three flux is also zero.
Therefore, it will make no difference if the common legs will act as the return for the third just as in a three phase of system any two conductors act as the return fore the current in the third conductor.

7. Basic Structure
Three similar limbs are connected by the top and bottom yokes, and each limb having primary and secondary windings are arranged concentrically.
The primary is shown by star-connected and the secondary by delta connected.

8. Based on STRUCTURE OF COILS
1. air core
-- inside area of coil is empty
-- inefficient induction
2. open core
-- ferromagnetic material inside coils
-- B = induction

9. 3. closed core 4. shell -- ferromagnetic ends connected
(cont.)
3. closed core
-- ferromagnetic ends connected
-- ­B = ­ induction
4. shell
-- lazy square eight
-- overlap of coils on same ferromagnetic core

10. Transformer Type

11. Three phase transformer connections.
Labeling of transformer terminal
Terminal on the HV side of each phase will be labeled as capital letters A, B, C .
Terminal on the LV side will be labeled as small letters a, b, c.
Terminal polarities are indicated by suffixes 1 and 2 with 1’s indicating similar polarity ends and so do 2’s.
Assuming the transformer to be ideal, VA2A1 (voltage of terminal A2 with respect to A1) is in phase with Va2a1 and IA is in phase with Ia.

12. There are only 4 possible transformer combinations:
Delta to Delta - use: industrial applications
Delta to Wye - use : most common; commercial and industrial
Wye to Delta - use : high voltage transmissions
Wye to Wye - use : rare, don't use causes harmonics and balancing problems.

13. Star/ Delta or Y- ∆ Connection.
The equation for Y - ∆ connection are ;
Primary side
I1phase = I1line.
V1phase =
Secondary side
I2phase =
V2phase = V2line
Output Power = √3. V2Line.I2Line. cos Ф
Transformation ratio. K =

14. Star/delta -30o connection
From the phasor diagram that phase a to neutral on the delta side lags by -30o to the phase to neutral voltage on the star side; this is also the phase relationship between the respective line to line voltages.
This connection, therefore, is known as – 30o connection.

15. Star/delta +30o connection

16. Star/ Star or Y – Y Connection
This connection is most economical for small, high voltage transformers because the number of turns/phase and the amount of insulation are minimum (as phase voltage is only 1/√3 of line voltage).

17. Star/star 0o-connection

18. Delta / Delta or ∆ - ∆ Connection.
This connection is economical for large, low voltage transformers in which insulation problem is not so urgent, because it increases the number of turns/phase.

19. Delta/delta 0o-connection
The sum of voltages around the secondary delta must be zero; otherwise delta, being a closed circuit, means a short circuit.
With polarities indicated on primary and secondary side, voltages Va2a1, Vb2b1 and Vc2c1 add to zero as per phasor diagram if the delta is formed by connecting a1b2, b1c2 and c1a2.
It is easily seen from the phasor diagram that the primary and secondary line voltages are in phase so it is the 0o – connection.
The phase transformation ratio is x: 1, the transformation ratio for line quantities is also x: 1.

20. Auto transformer
An auto transformer is an electrical transformer with only one winding.
This single coil has one or more extra taps or electrical connections in various positions along the winding.
Each tap corresponds to a different voltage so that effectively a portion of the same inductor acts as part of both the primary and secondary winding.
I1and I2 = Primary and secondary currents respectively

21. Step-down auto transformer
A single phase auto transformer having N1 turns primary with N2 turns tapped for a lower voltage secondary. The winding section BC of N2 turns is common to both primary and secondary circuits.
In fact it is nothing but a conventional two winding transformer is connected in a special way. The winding section AC must be provided with extra insulation, being a higher voltage.
Current flows in the winding section BC is Ic = I2 – I1
= K =

22. Step-up auto transformer
A single phase autotransformer having N1 turns primary with N2 turns tapped for a higher voltage secondary.
Current flows in the winding section AB is Ic = I1 – I2

23. Example 4.1
A Three phase transformer bank consisting of three 1 phase transformers is used to step down the voltage of a three phase, 6,600 V transmission line. If the primary line current is 10 A, calculate the secondary line voltage, line current and output kVA for the following connections: (a) Y/∆ and (b) ∆/Y. The turn ratio is 12. Neglect ratio.

24. Solution

26. Example 4.2
Determine the core area, the number of turns and the position of the tapping point for a 500 kVA, 50 Hz, single phase, 6,600/5,000 V autotransformer, assuming the following approximate values : e.m.f. per turn 8 B. Maximum flux density 1.3 Wb/m2.

27. Solution

28. Example 4.3
The 2000/200-V, 20-kVA transformer is connected as a step-up auto transformer as in Figure 4.17 in which AB is 200 V winding and BC is 2000-V winding. The 200-V winding has enough insulation to withstand 2200-V to ground. If Core loss of transformer is 120 W and full load copper loss is 300 W. Calculate (a) The LV and HV side voltage ratings of the autotransformer;
(b) Its KVA rating;
(c) Its efficiency at full-load 0.8 pf.

29. Solution

30. Cont’