5.1 State whether this is a core-type or shell-type three-phase transformer - NSC Electrical Technology Power Systems - Question 5 - 2024 - Paper 1

A shell-type transformer is less prone to magnetic stray losses compared to a core-type transformer.

The configuration of the primary and secondary windings is a star (delta) configuration.

When an alternating voltage is applied to the primary windings, an alternating flux is generated which links with the secondary windings, inducing an e.m.f. of the same frequency. When the load is connected to the secondary winding, a current flows through it. Power is transferred magnetically from the primary to the secondary windings.

An increase in the load will increase the secondary current, resulting in a more significant primary current drawn from the supply.

Dielectric oil provides electrical insulation between the windings and the case.

Hysteresis losses and eddy current losses are two types of losses that occur in transformers.

1. Air Natural: Transformers are cooled naturally by the air flowing across them.
2. Air Forced: Transformers are cooled by being equipped with a fan that enhances the airflow.

The conductor size for a single-phase transformer is larger than that for a three-phase transformer for the same power rating.

The line current can be calculated using the formula: $I_L = \frac{P_{tr}}{\sqrt{3} \cdot V_{L1} \cdot \cos \phi}$. Substituting the values: $I_L = \frac{300000}{\sqrt{3} \cdot 600 \cdot 0.87} = 331.82 \text{ A}$.

Using the formula: $I_{ph} = \frac{I_L}{\sqrt{3}}$, we find: $I_{ph} = \frac{331.82}{\sqrt{3}} = 191.58 \text{ A}$.

Using the formula: $I_{L1} = \frac{P}{\sqrt{3} \cdot V_{L1} \cdot \cos \phi}$, we can calculate: $I_{L1} = \frac{300000}{\sqrt{3} \cdot 8 \times 1000 \cdot 0.87} = 24.89 \text{ A}$.

1. Buchholz relay (B)
2. Restricted earth fault relay (REF)