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 loss than 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 set up in the core 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 will flow through it. Power is therefore transferred magnetically from the primary to the secondary windings.

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

It provides electrical insulation between the windings and the case.

1. Hysteresis losses
2. Eddy current losses

1. Air Natural: transformers are cooled by natural air flowing across the transformer windings.
2. Air Force: transformers are cooled by being air forced cooled with a fan above the windings of a transformer.

A single-phase transformer requires a larger conductor size compared to a three-phase transformer due to the need for a larger size to handle the same power.

The line current of the load can be calculated using the formula: $I_L = \frac{P_T}{\sqrt{3} \times V_{L1} \times \text{cos} \phi}$ Substituting in the values: $I_L = \frac{300 \times 10^3}{\sqrt{3} \times 600 \times 0.87} = 331.82 \text{ A}$

The phase current can be calculated using the line current: $I_{ph} = \frac{I_L}{\sqrt{3}}$ Thus: $I_{ph} = \frac{331.82}{\sqrt{3}} = 191.58 \text{ A}$

The primary line current can be calculated using: $I_{L1} = \frac{P_T}{\sqrt{3} \times V_{L1} \times \text{cos} \phi}$ Plugging in the values: $I_{L1} = \frac{300 \times 10^3}{\sqrt{3} \times 8 \times 10^3 \times 0.87} = 24.89 \text{ A}$

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