Electro-Dynamic Forces in Conductors During Short Circuits

When a short-circuit current flows through two parallel conductors, it creates significant mechanical forces between them. These forces are known as electro-dynamic forces and are a result of the interaction between the magnetic fields generated by the currents.

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Key Formula
The force per unit length (N/m) between two conductors is calculated using the following formula:

• F = (0.2 / a) × Icc_peak²
  

Where:

• F: Force between conductors (N/m).
  
• a: Axis-to-axis spacing between conductors (m).
  
• Icc_peak: Peak short-circuit current (kA), which is:
     
  • Icc_peak = 2.5 × Icc_RMS
       
• Icc_RMS: RMS short-circuit current (kA).
  

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Understanding the Formula

1. Relationship Between Current and Force:
   - The force increases with the square of the short-circuit current (Icc_peak²).
   - Higher short-circuit currents lead to exponentially larger forces.

2. Effect of Spacing (\(a\)):
   - The force is inversely proportional to the distance between the conductors (1/a).
   - As the spacing increases, the mechanical forces decrease.

3. Units:
   - Force (F): Measured in Newtons per meter (N/m).
   - Distance (a): Measured in meters (m).
   - Current (Icc_peak): Measured in kiloamperes (kA).

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Physical Implications of Electro-Dynamic Forces

1. Compression or Repulsion:
- If currents flow in the same direction, the conductors attract each other.
- If currents flow in opposite directions, the conductors repel each other.

2. Mechanical Stress:
- High forces can lead to mechanical deformation of conductors.
- Improperly secured conductors can cause equipment damage or even failure.

3. Design Considerations:
- Proper spacing (a) must be maintained to minimize forces.
- Conductors and their supports must withstand the maximum expected forces during a fault.

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Practical Applications

The formula is widely used in the design and validation of:

• Busbars: Calculating mechanical forces to ensure proper supports for busbar systems in electrical panels.
  
• Overhead Transmission Lines: Ensuring that conductor spacers and structures can handle the forces during faults.
  
• Underground Cables: Verifying that conductors within a cable remain stable under fault conditions.
  
• Switchgear and Substations: Ensuring all conductors, busbars, and connectors are mechanically secure.
  

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Example Calculation

**Given:**

• Icc_RMS = 20 kA
  
• a = 0.1 m
  

**Steps:**

• Calculate the peak current:

Icc_peak = 2.5 × Icc_RMS = 2.5 × 20 = 50 kA

• Substitute into the formula:

F = (0.2 / a) × Icc_peak²
   F = (0.2 / 0.1) × 50²
   F = 2 × 2500 = 5000 N/m
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**Result:** The force between the conductors is 5000 N/m.

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Conclusion

Understanding and calculating electro-dynamic forces are critical for the safe design of electrical systems. High fault currents can generate substantial forces that, if unaccounted for, may lead to equipment failure or safety hazards. The formula provided ensures designers can predict these forces and implement appropriate measures to secure conductors and maintain system integrity.

Electro-Dynamic Force Formulae

The calculation of electro-dynamic forces follows these formulas:

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1. Force Between Two Conductors

• F = (0.2 / a) × Icc_peak²
  

Where:

• F: Force between the conductors (N/m)
  
• a: Axis-to-axis spacing between conductors (m)
  
• Icc_peak: Peak short-circuit current (kA)
  

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2. Peak Short-Circuit Current

• Icc_peak = 2.5 × Icc_RMS
  

Where:

• Icc_RMS: RMS short-circuit current (kA)
  

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3. Heat Generated in the Conductor

• Q = I² × R × t
  

Where:

• Q: Heat energy (Joules)
  
• I: Short-circuit current (Amperes)
  
• R: Resistance of the conductor (Ohms)
  
• t: Duration of the short circuit (Seconds)
  

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4. Temperature Rise in the Conductor

• ΔT = Q / (m × c)
  

Where:

• ΔT: Temperature rise (Kelvin)
  
• m: Mass of the conductor (kg)
  
• c: Specific heat capacity (J/kg·K)
  

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5. Resistance at Temperature T

• R_T = R_0 × (1 + α × (T - T_0))
  

Where:

• R_T: Resistance at temperature T (Ohms)
  
• R_0: Resistance at reference temperature T₀ (Ohms)
  
• α: Temperature coefficient of resistance (1/K)
  
• T: Final temperature (Kelvin)
  
• T₀: Initial temperature (Kelvin)
  

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6. Maximum Short-Circuit Current

• I_max = √((A × ρ × c × ΔT) / t)
  

Where:

• I_max: Maximum short-circuit current (Amperes)
  
• A: Cross-sectional area of the conductor (m²)
  
• ρ: Resistivity of the conductor material (Ohm·m)
  
• c: Specific heat capacity (J/kg·K)
  
• ΔT: Temperature rise (Kelvin)
  
• t: Duration of the fault (Seconds)
  

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Conclusion
These formulas are used to ensure safe design and operation of conductors under fault conditions. They are particularly critical for short-circuit analysis and system design in compliance with standards like **IEC 60909** and **IEEE Std 605**.