Question

A 60 kVA, single-phase transformer has copper losses and core losses at full load as 34.78 kW and 17.39 kW, respectively. Determine its efficiency at 60% of full load at 0.8 p.f. lag. Also, determine the maximum efficiency of the transformer.

Answer 1

The efficiency of the 60 kVA single-phase transformer at 60% of full load with a 0.8 power factor is approximately 57.0%. The maximum efficiency occurs when copper losses equal core losses, which happens at around 70.7% of full load.

Step-by-step explanation:

Efficiency at 60% of Full Load

To calculate the efficiency of the transformer at 60% of full load with a power factor of 0.8 lagging, we first determine the output power. The transformer's rated power is 60 kVA, so at 60% load:

• Output power (real) = 60 kVA × 0.6 × 0.8 (p.f.) = 28.8 kW

Copper loss (Pcopper) at full load is given as 34.78 kW. At 60% load (and assuming it varies with the square of the current), copper loss can be calculated as:

• Pcopper60% = 34.78 kW × (0.6)2 = 12.52 kW

Core loss (Pcore) does not change with the load and remains 17.39 kW.

Now, we can use the formula for efficiency η:

• η = Output Power / (Output Power + Copper Loss + Core Loss)
• η = 28.8 kW / (28.8 kW + 12.52 kW + 17.39 kW)
• η ≈ 0.570 or 57.0%

Maximum Efficiency of the Transformer

For a transformer, maximum efficiency occurs when the copper losses equal the core losses.

• Pcopper at maximum efficiency = Pcore = 17.39 kW

Since copper losses at full load are 34.78 kW:

• ηmax Load Factor = (Core Loss / Full Load Copper Loss)1/2 = (17.39 / 34.78)1/2
• ηmax Load Factor ≈ 0.707 or 70.7%

Thus, maximum efficiency occurs at approximately 70.7% of full load.