Evaluates active electrical output power against total losses (no-load core losses plus load-dependent I²R copper losses scaled by the square of the loading factor) to determine precision efficiency curves per international standards.
Governing Formulas & Standards
Standards Basis: IEC 60076-1 / IS 2026 / IEEE C57.12.00
\eta = \frac{x \cdot S \cdot \cos\phi}{x \cdot S \cdot \cos\phi + P_0 + x^2 \cdot P_k} \times 100\%
Multiplies loading fraction x by capacity and power factor, divided by input power which includes fixed iron loss P0 and load copper loss scaled by x squared.
Worked Engineering Example: 1600 kVA Transformer Efficiency at 75% Load & 0.85 PF
- Active Output Power: P_out = 0.75 × 1600 × 0.85 = 1020.0 kW → 1020.0 kW
- Total Losses at 75% Load: P_loss = 2.4 + (0.75² × 16.8) = 2.4 + 9.45 = 11.85 kW → 11.85 kW
- Percentage Efficiency: η = 1020 / (1020 + 11.85) × 100 → 98.85%
Final Solution: Transformer Efficiency at 75% Load: 98.85% (Losses: 11.85 kW)
Frequently Asked Questions
- Why does transformer efficiency vary with loading?
- Core (iron) losses are constant regardless of load, dominating at light loads. Copper losses increase with the square of current (I²R), dominating at high loads. The efficiency curve therefore peaks where variable copper losses equal fixed core losses.
Interactive calculation engine and real-time CAD solver available online at https://amithvijayan.in/tools/transformer-efficiency.