Primary/Secondary winding of transformer

This tool calculates the maximum symmetrical short-circuit current at the output of a transformer substation, based on IEC 60865 and IEEE C37.100 standards. Results are essential for selecting circuit breakers, fuses, busbars, and cables, as well as verifying equipment’s short-circuit withstand capability. Input Parameters Power Net Fault (MVA): Short-circuit power of upstream network, indicating source strength. Higher values lead to higher fault currents. Primary Voltage (kV): Rated voltage on the high-voltage side of the transformer (e.g., 10 kV, 20 kV, 35 kV). Secondary Voltage (V): Rated voltage on the low-voltage side (typically 400 V or 220 V). Transformer Power (kVA): Apparent power rating of the transformer. Voltage Fault (%): Short-circuit impedance percentage (U k %), provided by manufacturer. Key factor in determining fault current. Joule Effect Losses (%): Load loss as a percentage of rated power (P c %), used to estimate equivalent resistance. Medium Voltage Line Length: Length of MV feeder from transformer to load (in m, ft, or yd), affecting line impedance. Line Type: Choose conductor configuration: Overhead line Unipolar cable Multipolar cable Medium Voltage Wire Size: Conductor cross-section, selectable in mm² or AWG, with Copper or Aluminum material options. Medium Voltage Conductors in Parallel: Number of identical conductors connected in parallel; reduces total impedance. Conductor Material: Copper or Aluminum, affecting resistivity. Low Voltage Line Length: Length of LV circuit (m/ft/yd), usually short but significant. Low Voltage Wire Size: Cross-sectional area of LV conductor (mm² or AWG). Low Voltage Conductors in Parallel: Number of parallel conductors on the LV side. Output Results Three-phase short-circuit current (Isc, kA) Single-phase short-circuit current (Isc1, kA) Peak short-circuit current (Ip, kA) Equivalent impedance (Zeq, Ω) Short-circuit power (Ssc, MVA) Reference Standards: IEC 60865, IEEE C37.100 Designed for electrical engineers, power system designers, and safety assessors conducting short-circuit analysis and equipment selection in low-voltage distribution systems.

This tool calculates the required reactive power compensation for a distribution transformer to improve the system power factor and enhance efficiency. Power factor correction reduces line current, minimizes copper and iron losses, increases equipment utilization, and avoids utility penalties. Input Parameters Transformer Rated Power: The rated apparent power of the transformer (in kVA), typically found on the nameplate No-Load Current (%): The no-load current as a percentage of the rated current, provided by the transformer manufacturer. This value represents magnetizing current and core losses, which are key inputs for reactive power calculation Calculation Principle When operating under no-load conditions, a transformer consumes reactive power to establish the magnetic field in the core. This reactive power lowers the overall power factor of the system. By installing capacitors in parallel at the low-voltage side, part of this inductive reactive power can be compensated, thereby improving the power factor to a target value (e.g., 0.95 or higher). Output Results Required capacitor capacity (kvar) Comparison of power factor before and after correction Estimated energy savings and payback period Reference Standards: IEC 60076, IEEE 141 Ideal for electrical engineers, energy managers, and facility operators to evaluate capacitor bank sizing and optimize power system performance.

What Is Economic Transformer Capacity? In power system design, the economic transformer capacity refers to the rated capacity that minimizes the total cost—balancing initial investment and long-term operational losses —while meeting load requirements. Oversizing increases equipment cost and no-load losses; undersizing risks overloading, reduced efficiency, or even equipment failure. Therefore, accurately calculating the economic capacity is essential for a safe, efficient, and cost-effective distribution system. Calculation Principle and Core Formula This calculator implements a widely accepted engineering model aligned with standards such as GB 50052 Code for Design of Power Supply and Distribution Systems : Core Formula: S e = (A × α) / (cosφ × T 1 ) × √(K × T / 8760) Where: S e : Transformer economic capacity (kVA) A: Annual electricity consumption (kWh) α: Load development factor (typically 1.1–1.3) cosφ: Annual average power factor (usually 0.85–0.9) T 1 : Load operating time per year (h) K: Loss ratio (typically 1.05–1.2) T: Total power connection time per year (h), usually 8760 h Note: This formula accounts for load growth, power factor, operating hours, and transformer losses, making it suitable for practical engineering applications. How to Use This Calculator Enter annual energy consumption (kWh) — from utility bills or historical data Set load growth factor (default: 1.2; ≥1.2 recommended for new projects) Select load factor (e.g., 0.75 for industrial, 0.65 for commercial, 0.6 for residential) Input annual operating hours (e.g., 8760 for 24/7 operation) Click [Calculate] to instantly get the recommended economic transformer size (kVA) Real-World Examples Example 1: Industrial Park Power Design Annual consumption (A): 5,000,000 kWh Load development factor (α): 1.25 Average power factor (cosφ): 0.85 Annual load operating time (T₁): 7200 h Loss ratio (K): 1.10 Annual supply time (T): 8760 h Result: S e = (5,000,000 × 1.25) / (0.85 × 7200) × √(1.10 × 8760 / 8760) = 6,250,000 / 6120 × √1.10 ≈ 1021.24 × 1.0488 ≈ 1071 kVA → Recommended: 1250 kVA standard transformer Example 2: Commercial Complex Annual consumption (A): 1,200,000 kWh Load development factor (α): 1.15 Average power factor (cosφ): 0.85 Annual load operating time (T₁): 4000 h Loss ratio (K): 1.10 Annual supply time (T): 8760 h Result: S e = (1,200,000 × 1.15) / (0.85 × 4000) × √(1.10 × 8760 / 8760) = 1,380,000 / 3400 × √1.10 ≈ 405.88 × 1.0488 ≈ 426 kVA → Recommended: 500 kVA standard transformer Typical Applications Power planning for new factories or plants Retrofitting commercial building distribution systems Capacity assessment for data centers Transformer sizing for renewable energy projects (solar + storage) Substation design for residential communities Engineering consulting and feasibility studies Why Use Our Online Calculator? Free to use—no registration required Works on desktop, tablet, and mobile devices All calculations run locally—your data never leaves your device Results include standard size recommendations Built-in explanations for students, engineers, and designers Frequently Asked Questions (FAQ) Q: How do I determine the load factor? A: Load factor = Average Load / Peak Load. If unknown, typical values are: Industrial 0.7–0.85, Commercial 0.6–0.7, Residential 0.5–0.6. Q: Is annual operating time always 8760 hours? A: No. For non-24/7 facilities (e.g., malls open 10 hours/day), estimate actual high-load hours (e.g., 300 days × 10 h = 3000 h). Q: Does this work for dry-type and oil-immersed transformers? A: Yes. The method applies to all distribution transformers since economic sizing depends on load profile, not cooling type. References & Standards GB 50052-2009 Code for Design of Power Supply and Distribution Systems DL/T 572-2021 Operation Code for Power Transformers Industrial and Civil Power Distribution Design Manual (4th Edition)

The cross-sectional area of the iron core is the result of a combination of power, material characteristics, and multi-physical field effects (thermal, mechanical, coupled fields), according to IEC 60076.