Three Phase Transformer
Three-phase transformers are used throughout the industry to change values of three-phase voltage and current. Since three-phase power is the most common way in which power is produced, transmitted, and used, an understanding of how three-phase transformer connections are made is essential.
- A three-phase transformer is constructed by winding three single-phase transformers on a single core.
- These transformers are put into an enclosure which is then filled with dielectric materials such as air, plastic or oil.
- The dielectric material performs several functions.
- Since it is a dielectric, a non-conductor of electricity, it provides electrical insulation between the windings and the case.
- It is also used to help provide cooling and to prevent the formation of moisture, which can deteriorate the winding insulation.
Three-Phase Transformer Connections
- Three-phase transformers can consist of either three separate single-phase transformers, or three windings on a three-legged, four-legged, or five-legged core.
- The high-voltage and low-voltage sides can be connected independently in either wye or delta.
- As a result, the ratio of the 3‑phase input voltage to the 3‑phase output voltage depends not only upon the turn ratio of the transformers but also on how they are connected.
There are only 4 possible transformer combinations:
- Delta-to-Delta Connection: It is used for industrial applications.
- Delta to Wye Connection: It is popular for stepping up transmission lines to four-wire services when neutrals are needed.
- Wye to Delta Connection: It is used to step-down utilities' high line voltages.
- Wye-to-Wye Connection: It is commonly used for interior wiring systems
Y/Y Connection: A Y/Y connection for the primary and secondary windings of a three-phase transformer is shown in the figure below.
Y/Y connected a three-phase transformer
- The line-to-line voltage on each side of the three-phase transformer is √3 times the nominal voltage of the single-phase transformer.
- The main advantage of the Y/Y connection is that we have access to the neutral terminal on each side and it can be grounded if desired.
- Without grounding the neutral terminals, the Y/Y operation is satisfactory only when the three-phase load is balanced.
- The electrical insulation is stressed only to about 58% of the line voltage in a Y-connected transformer.
Y/Δ Connection: This connection as shown in the figure below is very suitable for step-down applications.
Y/Δ connected three-phase transformer secondary winding current is about 58% of the load current.
- On the primary side, the voltages are from line to neutral, whereas the voltages are from line to line on the secondary side.
- Therefore, the voltage and the current in the primary are out of phase with the voltage and the current in the secondary.
- In a Y/Δ connection, the distortion in the waveform of the induced voltages is not as drastic as it is in a Y/Y-connected transformer when the neutral is not connected to the ground the reason is that the distorted currents in the primary give rise to a circulating current in the Δ-connected secondary.
Δ/Y Connection: This connection as shown in the figure below is proper for a step-up application.
Δ/Y connected three-phase transformer
- However, this connection is now being exploited to satisfy the requirements of both three-phase and single-phase loads.
- In this case, we use a four-wire secondary.
- The single-phase loads are taken care of by the three line-to-neutral circuits.
- An attempt is invariably made to distribute the single-phase loads almost equally among the three phases.
Δ/Δ Connection: as shown below the three transformers with the primary and secondary windings are connected as Δ/Δ.
Δ/Δ connected three-phase transformer
- The line-to-line voltage on either side is equal to the corresponding phase voltage. Therefore, this arrangement is useful when the voltages are not very high.
- The advantage of this connection is that even under unbalanced loads the three-phase load voltages remain substantially equal.
- This disadvantage of Δ/Δ connection is the absence of a neutral terminal on either side.
- Another drawback is that the electrical insulation is stressed to the line voltage.
- Therefore, a Δ-connection winding requires more expensive insulation than a Y-connected winding for the same power rating.
Example-1: Three single‑phase transformers are connected in delta‑delta to step down a line voltage of 138 kV to 4160 V to supply power to a manufacturing plant. The plant draws 21 MW at a lagging power factor of 86 per cent. Calculate the following.
- (i) Apparent power is drawn by the plant,
- (ii) Apparent power furnished by the HV line,
- (iii) The current in the HV lines,
- (iv) The current in the LV lines
- (v) The currents in the primary windings of each transformer
- (vi) The currents in the secondary windings of each transformer
- (vii) The load carried by each transformer
Solution: (i) The apparent power drawn by the plant is:
S= P Cos φ = 21/0.86 = 24.4 MVA
(ii) The transformer bank itself absorbs a negligible amount of active and reactive power because the losses and the reactive power associated with the mutual flux and the leakage fluxes are small. So, the apparent power furnished by the HV line is also 24.4 MVA.
(iii) The current in each HV line is :
(iv) The current in LV lines is :
(v) The current in each primary winding is:
(vi) The current in each secondary winding is:
(vii) The load carried by each transformer
- Plant load is balanced, hence each transformer carries one‑third of the total load: 24.4/3 = 8.13 MVA
OR
- The individual transformer load can also be obtained by multiplying the primary voltage times the primary current:
Example-2:: Determine the interrupting capacity, in amperes, of a circuit breaker or fuse required for a 75 kVA, three-phase transformer, with a primary of 480 volts delta and secondary of 208Y/120 volts. The transformer impedance (Z) = 5%.
Solution:
The primary circuit of the transformer is the only winding being interrupted. Hence the secondary voltage is not used in the calculation.
If the secondary is short-circuited (faulted),
- Normal Full Load Current = Volts Amps / (Square root (3) * Line volts) = 75,000 VA / (Square root (3) * 480 V) = 90 Amps
- Maximum Short Circuit Line Current = (Full load amps / 5%) = (90 Amps / 5%) = 1800 Amps
Therefore, the breaker or fuse would have a minimum interrupting rating of 1,800 amps at 480 volts.
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