4.1 Explain the purpose of a step-up transformer - NSC Electrical Technology Power Systems - Question 4 - 2020 - Paper 1

1. High load current: An increase in load can lead to higher current flow, causing increased heat generation.
2. Poor ventilation: Insufficient airflow around the transformer can trap heat, elevating temperature.
3. Insulation breakdown: Over time, insulation materials can degrade, leading to short circuits and increased heating.

1. Oil Natural Air Natural (ONAN): The heat is dissipated through the natural circulation of oil and air.
2. Oil Forced Air Forced (OFAF): This method employs a fan to circulate air and oil to enhance cooling efficiency.

Cooling failure can lead to overheating of the transformer, which may cause insulation damage, reduced efficiency, and ultimately failure of the transformer. This could lead to costly repairs, loss of service, and potential hazards such as fire.

An increase in load will cause the primary current to rise, as the transformer will draw more current from the supply to meet the increased demand. This is due to the conservation of power, where the apparent power on the primary side is equal to that on the secondary side.

1. Star (Y) connection: This connection ties one terminal of each winding together, forming a neutral point.
2. Delta (Δ) connection: In this setup, each winding is connected end-to-end in a closed loop.

1. Circuit breakers: Protect against overcurrent conditions.
2. Fuses: Provide overcurrent protection by melting and disconnecting the circuit.

The diagram should depict the core construction with three limbs where the windings are placed. Label the primary and secondary windings, and show the magnetic flux paths in the core.

Ensure the supply is switched off before wiring the transformer to avoid electric shock and accidents.

Given: $P_{out} = 200,000 imes 0.85 = 170,000 \text{ W}$ Calculate the input power using: $P_{in} = 171,800 \text{ W}$
Efficiency $\eta = \frac{P_{out}}{P_{in}} \times 100 = \frac{170,000}{171,800} \times 100 \approx 98.95\%$.

Use the formula: $TR = \frac{V_{l(3)}}{V_{l(1)}} = \frac{380\,V}{11\,kV} = 50:1$.

The secondary line voltage can be calculated from the turns ratio and primary line voltage, confirmed through the transformer equations derived from the basic transformer principles.