Computes symmetrical short-circuit current density (J in A/mm²) and adiabatic final winding temperature reached during a through-fault of rated duration (typically 2.0 seconds) to ensure mechanical insulation integrity per IEC 60076-5.
Governing Formulas & Standards
Standards Basis: IEC 60076-5 / IEEE C57.12.00 / IS 2026-5
\theta_1 = \theta_0 + \frac{2(\theta_0 + 235)}{\frac{106000}{J^2 \cdot t} - 1} \le 250^\circ\text{C (Copper)}
Calculates temperature rise assuming zero heat dissipation into oil during the 2-second short-circuit pulse.
Worked Engineering Example: 2.5 MVA Transformer Winding Thermal Withstand (2.0s Fault)
- Current Density J: J = 5570 / 95 = 58.63 A/mm² → 58.6 A/mm²
- Adiabatic Temperature Rise: Denom = 106000 / (58.63² × 2.0) - 1 = 15.42 ; Δθ = 2(95 + 235) / 14.42 → 45.8°C Rise
- Final Temperature: θ1 = 95.0 + 45.8 = 140.8°C → 140.8°C (Pass: < 250°C)
Final Solution: Final Winding Temperature: 140.8°C (PASS: Safely below 250°C IEC threshold)
Frequently Asked Questions
- Why is the short-circuit temperature calculated adiabatically?
- Because short circuits clear in 0.5 to 2.0 seconds. Oil thermal time constants are typically 2 to 3 hours, meaning virtually zero heat is transferred into the surrounding oil during the fault pulse.
Interactive calculation engine and real-time CAD solver available online at https://amithvijayan.in/tools/transformer-short-circuit-withstand.