Short-Circuit Current in Cables Calculator – IEC

Accurately calculating short-circuit current in cables is critical for electrical system safety and reliability. This process ensures cables withstand fault conditions without damage.

This article explores the IEC standards for short-circuit current calculations, providing formulas, tables, and practical examples. Learn to apply these methods confidently.

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  • Calculate short-circuit current for a 3-phase 400 V cable, 50 m length, copper conductor.
  • Determine short-circuit withstand current for 95 mm² aluminum cable, 100 m, 11 kV system.
  • Find short-circuit current for 3-core XLPE cable, 35 mm², 0.4 kV, 30 m length.
  • Evaluate short-circuit current for 150 mm² copper cable, 20 m, 6.6 kV system voltage.

Common Values for Short-Circuit Current in Cables According to IEC Standards

Cable TypeConductor MaterialCross-Sectional Area (mm²)Rated Voltage (kV)Short-Circuit Current (kA)Short-Circuit Duration (s)
XLPE InsulatedCopper160.4101
PVC InsulatedAluminum350.47.51
EPR InsulatedCopper5011203
XLPE InsulatedAluminum956.6253
PVC InsulatedCopper1500.4301
EPR InsulatedAluminum18511403

Fundamental Formulas for Short-Circuit Current Calculation in Cables (IEC)

Short-circuit current calculation in cables is governed by IEC standards such as IEC 60949 and IEC 60287. The key objective is to determine the maximum current a cable can safely carry during a fault without exceeding its thermal limits.

1. Basic Short-Circuit Current Formula

The short-circuit current (Isc) can be calculated using the following formula derived from the adiabatic heating equation:

Isc = k × √(A / t)
  • Isc: Short-circuit current in amperes (A)
  • k: Material constant (A·s0.5/mm²0.5)
  • A: Cross-sectional area of the conductor (mm²)
  • t: Duration of the short-circuit current (seconds)

The constant k depends on the conductor material and initial and final temperatures:

  • Copper: k ≈ 115 for initial temperature 30°C and final temperature 250°C
  • Aluminum: k ≈ 95 for initial temperature 30°C and final temperature 250°C

2. Derivation of the Constant k

The constant k is derived from the adiabatic equation:

k = √(2 × ρ × c × (Tf – Ti))
  • ρ: Resistivity of the conductor (Ω·mm²/m)
  • c: Specific heat capacity of the conductor (J/kg·K)
  • Tf: Final allowable temperature (°C)
  • Ti: Initial temperature before fault (°C)

Typical values for copper and aluminum are:

MaterialResistivity ρ (Ω·mm²/m)Specific Heat c (J/kg·K)Density (kg/m³)Typical k Value (A·s0.5/mm²0.5)
Copper0.01753858960115
Aluminum0.0282900270095

3. Calculating Short-Circuit Current with Cable Length and Impedance

In practical systems, the short-circuit current is also influenced by the cable’s impedance. The total impedance (Z) includes resistance (R) and reactance (X):

Z = √(R² + X²)

The short-circuit current at the cable end can be approximated by:

Isc = V / (√3 × Z)
  • V: Line-to-line voltage (Volts)
  • Z: Cable impedance (Ohms)

Resistance and reactance depend on cable construction, length, and frequency. IEC 60287 provides methods to calculate these parameters.

4. Resistance Calculation

The resistance of the cable conductor at operating temperature is:

R = R20 × [1 + α × (T – 20)]
  • R20: Resistance at 20°C (Ω)
  • α: Temperature coefficient of resistivity (Copper ≈ 0.00393 /°C, Aluminum ≈ 0.00403 /°C)
  • T: Operating temperature (°C)

Resistance at 20°C is calculated by:

R20 = (ρ × L) / A
  • ρ: Resistivity at 20°C (Ω·mm²/m)
  • L: Cable length (m)
  • A: Cross-sectional area (mm²)

5. Reactance Calculation

Reactance is mainly inductive and can be approximated by:

X = 2π × f × L × Lind
  • f: Frequency (Hz)
  • L: Cable length (m)
  • Lind: Inductance per unit length (H/m)

Typical inductance values for power cables range from 0.3 to 1 mH/km depending on construction.

Real-World Application Examples

Example 1: Short-Circuit Current Calculation for a 3-Phase 400 V Copper Cable

A 50 m long copper cable with a cross-sectional area of 35 mm² is installed in a 400 V, 50 Hz system. Calculate the maximum short-circuit current the cable can withstand for 1 second.

  • Given: A = 35 mm², t = 1 s, k (copper) = 115

Using the formula:

Isc = k × √(A / t) = 115 × √(35 / 1) = 115 × 5.916 = 680.34 A

The cable can safely carry a short-circuit current of approximately 680 A for 1 second without damage.

Example 2: Calculating Short-Circuit Current Considering Cable Impedance

Calculate the prospective short-circuit current at the end of a 100 m, 95 mm² aluminum cable in an 11 kV system. Assume the following:

  • Resistivity ρ = 0.0282 Ω·mm²/m
  • Operating temperature = 90°C
  • Temperature coefficient α = 0.00403 /°C
  • Inductance per unit length Lind = 0.7 mH/km
  • Frequency f = 50 Hz

Step 1: Calculate resistance at 20°C

R20 = (ρ × L) / A = (0.0282 × 100) / 95 = 0.0297 Ω

Step 2: Adjust resistance for operating temperature 90°C

R = R20 × [1 + α × (T – 20)] = 0.0297 × [1 + 0.00403 × (90 – 20)] = 0.0297 × 1.2821 = 0.0381 Ω

Step 3: Calculate reactance

X = 2π × f × L × Lind = 2 × 3.1416 × 50 × 0.1 × 0.0007 = 0.022 Ω

Step 4: Calculate total impedance

Z = √(R² + X²) = √(0.0381² + 0.022²) = √(0.00145 + 0.00048) = √0.00193 = 0.0439 Ω

Step 5: Calculate short-circuit current

Isc = V / (√3 × Z) = 11,000 / (1.732 × 0.0439) = 11,000 / 0.076 = 144,736 A

This value is theoretical and represents the prospective short-circuit current limited by cable impedance. Protection devices must be rated accordingly.

Additional Technical Considerations

  • Thermal Limits: IEC 60949 specifies maximum permissible conductor temperatures during short-circuit conditions, typically 250°C for copper and aluminum.
  • Duration of Fault: Short-circuit duration is critical; longer faults require cables with higher thermal withstand capability.
  • Ambient Conditions: Installation environment affects cable temperature and resistance; corrections may be necessary.
  • Multiple Cables: Parallel cables reduce impedance and increase short-circuit current capacity.
  • Standards Compliance: Always verify calculations against IEC 60287 and IEC 60949 for compliance and safety.

Understanding and applying these principles ensures electrical installations are safe, reliable, and compliant with international standards.

References and Further Reading