5.1 Explain the principle of mutual induction with reference to transformers - NSC Electrical Technology Power Systems - Question 5 - 2022 - Paper 1

1. Transformer ratio: The primary to secondary voltage ratio must be the same.
2. Current rating: The rated current for both phases should match.
3. Power factor: The power factor rating of the transformers should be consistent.

The ‘Star’ (Y) connection on the secondary side of a three-phase transformer creates a neutral point, which allows for balanced load distribution. This configuration is essential for providing a reference point for voltages in the system.

Copper losses in transformers are primarily caused by the internal resistance of the copper conductors. When current flows through the windings, the resistance leads to heat generation, which is proportional to the square of the current ($I^2R$ losses).

Iron losses, also known as core losses, are mainly caused by eddy currents and hysteresis effects within the transformer core. Eddy currents are generated due to the changing magnetic field, which induces circulating currents, while hysteresis losses occur due to the periodic magnetization and demagnetization of the core materials as the magnetic flux reverses.

In dry-type transformers, insulation failure is typically controlled through the use of tubular radiators that facilitate airflow to cool the windings. This helps prevent overheating and the breakdown of insulation materials, ensuring reliable performance and safety.

A core-type transformer has limbs where the coils are wound around three limbs, whereas a shell-type transformer incorporates coils wound around the central section of the core. The core arrangement impacts the transformer's efficiency and magnetic performance.

A balanced earth-fault relay operates by monitoring the three-phase currents. Under normal conditions, the vector sum of the three phase currents equals zero. If there is an earth fault on one of the phases, the imbalance will cause the relay to detect a difference in current, which will then isolate the transformer from the supply to prevent damage.

Using the formula: $I_{L2} = \frac{P}{\sqrt{3} V_{L2} \cos \theta}$ For a load of 200 kW, $V_{L2} = 400 \text{ V}$, and power factor $\cos \theta = 0.8$: $I_{L2} = \frac{200000}{\sqrt{3} \times 400 \times 0.8} \\ = 360.84 \, A$

In a star configuration, the phase current is equal to the line current: $I_{ph2} = I_{L2} = 360.84 \, A$

Apparent power is given by: $S = \frac{P}{\cos \theta} \\ = \frac{200000}{0.8} \\ = 250000 \, VA \, (or 250 \, kVA)$

Using the formula: $P = \sqrt{3} V_{L1} I_{L1} \cos \theta$ For the primary side: $200000 = \sqrt{3} \times 6000 \times I_{L1} \times 0.8 \\ I_{L1} = 24.06 \, A$