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 turns ratio needs to be equal for optimal performance across all phases.
2. Voltage rating: All transformers must have the same voltage rating to ensure compatibility and safety.
3. Current rating: All transformers must have identical current ratings to maintain balance during operation.

The secondary side connection that will create a neutral point is the Star (Y) connection.

Copper losses refer to the energy lost in the form of heat due to the resistance of the copper conductors in the transformer windings when current flows. These losses are proportional to the square of the current flowing, mathematically represented as:

$P_{copper} = I^2 R$ where $P_{copper}$ is the power loss, $I$ is the current, and $R$ is the resistance.

Iron losses, also known as core losses, occur due to the eddy currents and the changing magnetic field within the core material. These losses are influenced by the frequency of the electrical supply and are typically made up of hysteresis losses and eddy current losses, both resulting in energy dissipation as heat.

Insulation failure in dry-type transformers is controlled by using materials that are resistant to degradation from heat and environmental factors. Furthermore, proper design includes strategies such as installing ventilation systems to maintain an optimal thermal environment and regular maintenance checks to inspect and replace aging insulation materials.

In a shell-type transformer, the core surrounds the coils and the coils are wound around three limbs of the core. Conversely, in a core-type transformer, the core is structured with three limbs, and the coils are wound around the central part of the core, allowing for better magnetic flux utilization.

A balanced earth-fault relay monitors the current flowing through the three phases. Under normal conditions, the three-phase currents sum to zero. However, if there is an earth fault in one of the phases, the difference in current will trigger the relay. This action isolates the transformer from the power supply, thus protecting it from damage.

To calculate the secondary line current ($I_L2$), we use the formula:

$I_L2 = \frac{P_{load}}{\sqrt{3} \cdot V_{L2} \cdot \cos \theta}$ Plugging in the values: $I_L2 = \frac{200000}{\sqrt{3} \cdot 400 \cdot 0.8} = 360.84\, A$

In a star connection, the line current equals the phase current, so:

$I_{p2} = I_{L2} = 360.84\, A$

The apparent power ($S$) can be calculated using the formula:

$S = \frac{P_{load}}{\cos \theta}$ Substituting the given values: $S = \frac{200000}{0.8} = 250000\, VA$

To find the primary line current ($I_{L1}$), we use the formula:

$I_{L1} = \frac{P}{\sqrt{3} \cdot V_{L1} \cdot \cos \theta}$ Substituting the values: $I_{L1} = \frac{200000}{\sqrt{3} \cdot 6000 \cdot 0.8} = 24.06\, A$