5.1 Name TWO types of transformer core constructions used in three-phase transformers - NSC Electrical Technology Power Systems - Question 5 - 2022 - Paper 1

Dielectric oil serves two primary purposes in a transformer. Firstly, it acts as a non-conductor of electricity, providing essential electrical insulation between the winding and the transformer's metallic parts. Secondly, it provides cooling support, helping to maintain optimal operating temperatures by dissipating heat generated during the transformation process.

The Buchholz relay is strategically placed in-line between the conservator and the transformer housing. This positioning allows it to monitor gas accumulation within the transformer, functioning as an important safety device that detects abnormal conditions such as overheating or faults.

To represent a three-phase delta-star step-down transformer, you can draw three identical single-phase transformers arranged in a delta configuration on the primary side and a star configuration on the secondary side. Ensure to label each winding and indicate the common neutral connection on the secondary side.

The output power can be calculated using the formula:

$P = S imes ext{p.f.}$

Substituting the given values:

$P = 10,000 ext{ VA} imes 0.8 = 8,000 ext{ W}$

Thus, the output power of the transformer is 8,000 W.

The efficiency of the transformer can be calculated using the formula:

$ext{Efficiency} ( ext{η}) = \frac{P_{out}}{P_{out} + ext{losses}} \times 100$

Where the losses are the sum of the copper loss and core loss: $\text{losses} = 300 ext{ W} + 50 ext{ W} = 350 ext{ W}$ Substituting the values:

$η = \frac{8,000}{8,000 + 350} \times 100 \approx 95.8 \%$

Therefore, the efficiency is approximately 95.8%.

To find the apparent power (kVA), we use:

$S = \sqrt{3} \times V_L \times I_L$

Where:

• $V_L = 6,000 ext{ V}$
• $I_L = 2 ext{ A}$

Calculating:

$S = \sqrt{3} \times 6,000 \times 2 = 20,784.61 \text{ VA} \approx 20.78 \text{ kVA}$

Thus, the rating of the transformer is approximately 20.78 kVA.

The power factor (pf) can be calculated using:

$\text{pf} = \frac{P}{S}$

Where: $P = 18,000 ext{ W}$ $S = 20,784.61 \text{ VA}$ Substituting the values:

$\text{pf} = \frac{18,000}{20,784.61} \approx 0.87$

Hence, the power factor of the load is approximately 0.87.

In a delta configuration, the line voltage ($V_L$) is equal to the phase voltage ($V_{ph}$) related by:

$V_{L} = V_{ph}$

For this transformer:

• Given: $V_L = 6,000 ext{ V}$, therefore,
• $V_{ph} = 6,000 ext{ V}$.

The turns ratio can be calculated using:

$\text{TR} = \frac{V_{ph(1)}}{V_{ph(2)}}$

For the primary and secondary voltage values:

• Given: $V_{ph(1)} = 6,000 ext{ V}$ and $V_{ph(2)} = 240 ext{ V}$,

Calculating: $\text{TR} = \frac{6,000}{240} = 25$

Thus, the turns ratio is 25.