5.1 Name TWO parts of the cooling system in three-phase transformers - NSC Electrical Technology Power Systems - Question 5 - 2023 - Paper 1

The type of loss that contributes the most to heat in three-phase transformers is copper losses, also known as I²R losses.

One application of a star-delta transformer in a transmission network is to allow the connection of high voltage supplies in step-down transformers for residential or commercial use.

Eddy current losses are reduced by constructing the core out of thin laminations instead of solid material. This increases resistance to the flow of eddy currents, thus minimizing losses.

Heat should be dissipated in transformers to prevent the temperature from rising to levels that would cause degradation of the insulation system. Prolonged high temperatures can lead to insulation failure and ultimately result in transformer damage.

The preferred cooling method for very large transformers, typically those rated at several hundred MVA, is oil forced/water forced cooling.

The Buchholz relay monitors the gas formation inside the oil of a transformer. It sounds an alarm when gas is formed, isolates the transformer from the supply, and protects the transformer when gas formation exceeds a certain level.

To calculate the primary line current ($I_{L1}$):

$I_{L1} = \frac{S}{\sqrt{3}V_{L1}}$

Substituting the values:

$I_{L1} = \frac{100000}{\sqrt{3} \times 11000} = 5.25 \text{ A}$

To find the secondary phase voltage ($V_{ph2}$):

Using the transformer ratio:

$\frac{N_1}{N_2} = \frac{V_{ph1}}{V_{ph2}}$

We have:

$V_{ph2} = \frac{V_{L1}}{\sqrt{3}} = \frac{11000}{\sqrt{3}} \approx 2297.17 ext{ V}$

The active power (P) can be calculated as:

$P = S \times ext{pf}$

Substituting the values:

$P = 100000 \times 0.9 = 90000 \text{ W} = 90 ext{ kW}$

The diagram should include:

1. Three primary coils labelled L1, L2, L3.
2. Three secondary coils labelled L1', L2', L3'.
3. Labels for the transformer ratio of 48:1 between the primary and secondary sides.

This is a step-down transformer because the winding ratio is 48:1, meaning the primary voltage is reduced to a lower secondary voltage.

In a step-down transformer, the voltage is decreased while the current increases proportionally. According to the conservation of power principle, assuming ideal conditions, we have:

$P_{primary} = P_{secondary}$

Thus:

$V_1I_1 = V_2I_2$

This shows that as the voltage decreases (step-down), the current must increase to maintain the power level.