• The heating of oil during transformer operation can lead to the generation of oil vapors, which are flammable and pose a significant fire risk if exposed to ignition sources.
• The cooling oil can degrade due to continuous agitation. This deterioration can lead to overheating of the transformer.
• The cooling oil can degrade in marine environments due to continuous agitation and exposure to seawater. This deterioration can lead to overheating of the transformer.
• The cooling oil requires periodic replacement, necessitating transformer isolation. This may not always be feasible, causing operational disruptions.

(b) Losses in a Transformer

1. Core Loss or Iron Loss: Core loss occurs in the transformer's magnetic core and consists of eddy current loss and hysteresis loss.

• Eddy Current Loss: When AC is supplied to the primary winding, it produces an alternating magnetizing flux in the transformer. While most of this flux links with the secondary winding to induce emf, some flux links with other conducting parts such as the steel core or transformer body. This induces small circulating currents in those parts, called eddy currents, which dissipate energy as heat.
• Hysteresis Loss: This loss arises due to the repeated reversal of magnetization in the transformer core. It depends on:
  • Volume and grade of the iron used
  • Frequency of magnetic reversals
  • Magnitude of flux density

2. Copper Loss (I²R Loss): Copper loss occurs due to the ohmic resistance of the transformer windings. It can be expressed as:

• Primary winding: ( I_1^2 R_1 )
• Secondary winding: ( I_2^2 R_2 )

Where:

• ( I_1 ) and ( I_2 ) are currents in the primary and secondary windings
• ( R_1 ) and ( R_2 ) are resistances of the primary and secondary windings

Key points:

• Copper loss is proportional to the square of the current.
• Since current depends on the load, copper loss varies with load.

3. Stray Losses: Stray losses occur due to the leakage flux linking with metallic parts of the transformer.

Note: Stray losses are small compared to copper and iron losses and are often negligible in calculations.

4. Dielectric Loss: Dielectric loss is caused by the transformer oil, which serves as an insulating material. If the insulating oil deteriorates, it leads to energy loss and affects the efficiency of the transformer.

$$KVA \space = \space 25$$

$$\cos \phi \space = \space 0.8$$

$$W_{iron \space FL} \space = \space 350W \space = \space 0.35kW$$

$$W_{cu \space FL} \space = \space 400W \space = \space 0.4kW$$

$$3300/233 \space = \space step \space down \space transformer$$

To find half load efficiency η

$$Loading \space factor \space (x) \space = \space {{1} \over 2}$$

∴ Half load copper loss = $$x^2 \space W_{cu \space FL}$$

Iron losses remain same

$$= \space \left(1 \over 2 \right)^2 \times 0.4 \space = \space {{0.41} \over 4} \space = \space 0.1 kW$$

$$%η \space = \space {{x \space KVA \space \cos \phi} \over x \space KVA \cos \phi + W_{iron} + x^2 \space W_{cu}} \times 100$$

$$= \space {{(1/2) \times 25 \times 0.8} \over (1/2) \times 25 \times 0.8 + 0.35 + 0.1} \times 100$$

$$%η \space = \space 95.69%$$