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 multiple purposes in a transformer:

• Insulation: It acts as a non-conductor, providing electrical isolation between the internal components, thus preventing short circuits.
• Cooling: The oil dissipates heat generated during operation, maintaining optimal temperature levels for efficient performance.

The Buchholz relay is situated in-line between the conservator and the transformer housing. It operates as a safety device to detect gas accumulation or sudden oil flow changes, indicating potential faults within the transformer.

To construct the delta-star step-down transformer unit, draw three identical single-phase transformers connected in a delta configuration on the primary side and in a star configuration on the secondary side. Make sure to label the terminals appropriately, indicating primary and secondary sides along with load connections.

To calculate the output power (P_out), use the formula:

$P_{out} = S imes ext{p.f.}$

Substituting the given values:

$P_{out} = 10,000 imes 0.8 = 8,000 ext{ W}$

The efficiency ($\eta$) of the transformer can be calculated with the formula:

$\eta = \frac{P_{out}}{P_{out} + \text{losses}} \times 100$

Where losses are the sum of copper and core losses:

Losses = 300 W (copper) + 50 W (core) = 350 W

Substituting the values:

$\eta = \frac{8000}{8000 + 350} \times 100 = 95.8\%$

The apparent power (S) can be calculated using the formula:

$S(P_{app}) = \sqrt{3} \times V_{L} \times I_{L}$

Given:

• $V_L = 6 kV$
• $I_L = 2 A$

Substituting the values:

$S(P_{app}) = \sqrt{3} \times 6000 \times 2 = 20,784.61 VA \approx 20.78 kVA$

The power factor (cos φ) can be calculated using:

$\cos φ = \frac{P}{S}$

Where:

• $P = 18,000 W$ (given)
• $S = 20,784.61 VA$

Substituting the values:

$\cos φ = \frac{18,000}{20,784.61} = 0.87$

For a delta connection, the phase voltage ($V_{ph}$) can be calculated using:

$V_{ph} = \frac{V_{L}}{\sqrt{3}}$

Given:

• $V_L = 6 kV$

Substituting the values:

$V_{ph} = \frac{6000}{\sqrt{3}} = 3464.1 V$

The turns ratio (TR) can be calculated as:

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

Substituting the values:

$TR = \frac{V_{Ph(1)}}{240} = \frac{6,000}{240} = 25$

Thus, the turns ratio is 25:1.