The series impedance of a transformer consists of a resistance that accounts for the load losses and a reactance that represents the leakage reactance. This impedance has a very low power factor, consisting almost entirely of leakage reactance with only a small resistance.

As discussed earlier, the transformer design engineer can control the leakage reactance by varying the spacing between the windings. Increasing the spacing ‘‘decouples’’ the windings and allows more leakage flux to circulate between the windings, increasing the leakage reactance.

While leakage reactance can be considered a transformer loss because it consumes reactive power, some leakage reactance is necessary to limit fault currents. On the other hand, excessive leakage reactance can cause problems with regulation.

Regulation is often defined as the drop in secondary voltage when a load is applied, but regulation is more correctly defined as the increase in secondary output voltage when the load is removed. The reason that regulation is defined this way is that transformers are considered to be ‘‘fully loaded’’ when the secondary output voltage is at the rated secondary voltage.

This requires the primary voltage to be greater than the rated primary voltage at full load.

Let Ep equal the primary voltage and let Es equal the secondary voltage when the transformer is fully loaded. Using per-unit values instead of primary and secondary voltage values, the per-unit secondary voltage will equal Ep with the load removed. Therefore, the definition of regulation can be expressed by the following equations.

Regulation = (Ep - Es)/ Es

Since Es = 1 by definition,
Regulation Ep - 1 (3.8.2)
Regulation depends on the power factor of the load. For a near-unity power
factor, the regulation is much smaller than the regulation for an inductive load
with a small lagging power factor.

Example 3.4
A three-phase 1500 KVA 12470Y-208Y transformer has a 4.7% impedance. Calculate the three-phase fault current at the secondary output with the primary connected to a 12,470 V infinite bus. Calculate the regulation for a power factor of 90% at full load.

The three-phase fault is a balanced fault, so the positive-sequence equivalent circuit applies. The full-load secondary current is calculated as follows:

I 1.732 500,000 VA per phase/208 V 4167 A per phase
The per-unit fault current is the primary voltage divided by the series impedance:
1/0.047 = 21.27 per unit

The secondary fault current is equal to the per-unit fault current times the fullload current:
If 21.27 per unit 4167 A per phase 88,632 A per phase To calculate regulation, the secondary voltage is 1∠0° per unit by definition.

Applying a 1 per unit load at a 90% lagging power factor, I 1.0∠ 25.8°. Since the series impedance is mainly inductive, the primary voltage at full load Ep can be calculated as follows:

Ep 1∠0° + 1.0∠ 25.8° X 0.047∠90°
1.02 + j0.042 = 1.021 per unit
Regulation = Ep - 1 = 0.021 = 2.1%

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