Estimating Losses and Efficiency of a Transformer

Given:
- Transformer rating: 100 kVA
- Full load efficiency: 98%
- Maximum efficiency occurs at (3/4) x full load
- Power factor: 0.8 lagging at all loads

1. Estimating Copper Loss at Full Load
- Copper loss is the power loss due to the resistance of the transformer windings.
- Copper loss can be calculated using the formula: I^2R, where I is the current and R is the resistance.
- At full load, the transformer output power is 100 kVA, and the efficiency is 98%, so the input power is:
Input power = Output power / Efficiency
= 100 kVA / 0.98
= 102.04 kVA
- The input current can be calculated using the formula: Input power / (sqrt(3) x Input voltage x Power factor)
- At full load, the input voltage is assumed to be equal to the rated voltage of the transformer, which is 100 kV / sqrt(3) = 57.74 kV.
- Therefore, the input current at full load is:
Input current = 102.04 kVA / (sqrt(3) x 57.74 kV x 0.8)
= 1,293.9 A
- Let's assume that the resistance of the transformer windings is 0.1 ohm.
- Therefore, the copper loss at full load is:
Copper loss = I^2R
= (1,293.9 A)^2 x 0.1 ohm
= 167,351.3 W or 167.4 kW

2. Estimating Iron Loss
- Iron loss is the power loss due to the hysteresis and eddy currents in the transformer core.
- Iron loss is usually expressed as a percentage of the transformer rating.
- Let's assume that the iron loss is 2% of the transformer rating.
- Therefore, the iron loss is:
Iron loss = 2% x 100 kVA
= 2,000 W or 2 kW

3. Estimating Percentage Efficiency at Half Load
- At (3/4) x full load, the efficiency of the transformer is maximum.
- Therefore, the efficiency at (3/4) x full load is:
Efficiency at (3/4) x full load = 98%
- Let's assume that the load on the transformer is halved, i.e., 50 kVA.
- The input current at half load can be calculated using the same formula as before:
Input current at half load = 50 kVA / (sqrt(3) x 57.74 kV x 0.8)
= 647.0 A
- The copper loss at half load can be calculated using the formula: I^2R
Copper loss at half load = (647.0 A)^2 x 0.1 ohm
= 42,024.9 W or 42.0 kW
- The total losses at half load are:
Total losses at half load = Copper loss + Iron loss
= 42,024.9 W +