Three-Phase Transformers and Autotransformers
Three-phase transformer connections and phase shifts, vector groups, autotransformer copper saving and power transfer, parallel operation and load sharing, and per-unit impedance.
Power systems use three-phase transformers, autotransformers for small voltage changes, and transformers in parallel to share load.
Three-phase connections
| Connection | Line voltage ratio | Notes |
|---|---|---|
| Y–Y | N₁/N₂ | Third-harmonic problems without a neutral or tertiary |
| Δ–Δ | N₁/N₂ | Can run as open delta (V–V) at 57.7% of the rating |
| Y–Δ | N₁/(√3 N₂) | Step-down at the receiving end; 30° shift |
| Δ–Y | √3 N₁/N₂ | Step-up at the generating end; 30° shift |
Vector groups (for example Dyn11) show the phase shift between HV and LV in clock positions of 30° each. Only transformers of the same group can be paralleled.
Open delta (V–V): capacity = √3 × VA of one transformer = 57.7% of the original three-unit delta bank.
Autotransformer
One winding with a tap; part of the power is transferred conductively.
- With ratio a = V_LV/V_HV (< 1): copper saved = a times that of a two-winding transformer (copper used = 1 − a).
- Power transferred inductively = (1 − a) × total; conductively = a × total.
- Rating as an autotransformer = two-winding rating/(1 − a).
- Disadvantages: no isolation between HV and LV; higher short-circuit current.
Parallel operation
Conditions: same voltage ratio, same polarity, same phase sequence and vector group, and (for good load sharing) similar per-unit impedances and X/R ratios.
Load sharing: each transformer takes a share inversely proportional to its impedance: S₁ = S Z₂/(Z₁ + Z₂) (with impedances in ohms on a common base). Transformers with equal per-unit impedance share load in proportion to their ratings.
Per-unit values
Z_pu = Z_actual/Z_base, with Z_base = V_base²/S_base. A transformer has the same per-unit impedance on both sides. Change of base: Z_new = Z_old (S_new/S_old)(V_old/V_new)².
Worked example
A 230/115 V two-winding transformer rated 10 kVA is connected as a 345/230 V autotransformer.
- a = 230/345 = 2/3; autotransformer rating = 10/(1 − 2/3) = 30 kVA.
Practice
An autotransformer steps 400 V down to 300 V. What fraction of the copper of an equivalent two-winding transformer does it save?
Show answer and explanation
Answer: 0.75Copper saved = a = V_LV/V_HV = 0.75.
Three 100 kVA transformers in delta lose one unit and run in open delta. What load (kVA) can the bank supply? (Give 1 decimal place.)
Show answer and explanation
Answer: 173.2V–V capacity = √3 × 100 = 173.2 kVA (57.7% of 300 kVA).
Two transformers in parallel have impedances 2 Ω and 3 Ω (same angle). They share a 500 kVA load. How much (kVA) does the 2 Ω unit carry?
Show answer and explanation
Answer: 300S₁ = S Z₂/(Z₁ + Z₂) = 500 × 3/5 = 300 kVA.
A star-delta transformer introduces a phase shift between primary and secondary line voltages of:
Show answer and explanation
Answer: BY–Δ and Δ–Y connections shift the line voltages by 30°.