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IEC 60287: Calculation of Current-Carrying Capacity in Cables

IEC 60287 provides methods for calculating the current-carrying capacity (ampacity) of power cables under steady-state conditions. This standard is crucial for designing safe and efficient electrical installations while preventing overheating.

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Purpose of IEC 60287

• Define Methods: Standardizes the calculation of thermal performance and ampacity for power cables.
• Ensure Safety: Prevents overheating by maintaining cable temperatures within permissible limits.
• Flexibility: Covers various installation conditions, cable types, and external factors.

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Key Concepts in IEC 60287

The current-carrying capacity is determined by balancing heat generation and heat dissipation using the following formula:

I = √(Δθ / (R_c + R_s + W_d))
Where:

• I: Current-carrying capacity (A),
• Δθ: Permissible temperature rise (°C),
• R_c: Conductor resistance (Ω/m),
• R_s: Sheath and armor resistance (Ω/m),
• W_d: Dielectric losses (W/m).

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Detailed Components of IEC 60287

1. Conductor Losses (Joule Heating):
Heat generated by the current flowing through the conductor:
P_c = I² ⋅ R_c
2. Dielectric Losses:
For AC cables, dielectric losses in the insulation are calculated as:
W_d = ω ⋅ C ⋅ V² ⋅ tan(δ)Where:

• ω: Angular frequency (2πf),
• C: Capacitance per unit length of the cable,
• V: Voltage across the insulation,
• tan(δ): Loss tangent of the insulation material.

3. Sheath and Armor Losses:
Heat generated in the metallic sheath or armoring:
P_s = I_s² ⋅ R_sWhere:

• I_s: Induced current in the sheath,
• R_s: Resistance of the sheath.

4. Thermal Resistance:
Thermal resistances include:

• T₁: Insulation thermal resistance,
• T₂: Bedding and sheath thermal resistance,
• T₃: Soil or air thermal resistance.

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Key Parameters Affecting Ampacity

• Ambient Temperature: Higher ambient temperatures reduce ampacity.
• Soil Thermal Resistivity: Poor soil conductivity (e.g., dry soil) decreases ampacity.
• Cable Layout: Grouped cables increase thermal resistance and reduce ampacity.
• External Thermal Factors: Installation in ducts, air, or buried conditions impacts heat dissipation.

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Calculation Procedure in IEC 60287

1. Gather Cable and Installation Data: Conductor size, material, insulation type, laying arrangement, and soil conditions.
2. Determine Heat Generation: Calculate P_c, W_d, and P_s.
3. Compute Thermal Resistances: Sum all thermal resistances in the system.
4. Solve the Heat Balance Equation: Use the formula to determine the current-carrying capacity (I).

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Practical Applications

• Cable Sizing: Ensures the cable can carry the desired load current safely.
• Thermal Analysis: Evaluates the thermal performance of cables in various conditions.
• System Optimization: Determines the best cable layout and installation method to maximize ampacity.

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Key Features of IEC 60287

• Comprehensive Coverage: Considers all heat sources, installation methods, and environmental factors.
• Flexibility: Applies to single-core and multi-core cables, AC and DC systems.
• Customizability: Adapts to specific conditions like high-altitude installations or high-voltage cables.

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Conclusion

IEC 60287 is a vital tool for electrical engineers, ensuring cables operate safely within permissible temperature limits under steady-state conditions. It promotes consistency and efficiency in cable design while preventing overheating.

Example: Current-Carrying Capacity Calculation

This example demonstrates how to calculate the current-carrying capacity (ampacity) of a cable using the methods described in IEC 60287.

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Cable Specifications

• Conductor Material: Copper
• Conductor Size: 120 mm²
• Voltage: 11 kV
• Frequency: 50 Hz
• Installation: Directly buried in soil
• Soil Thermal Resistivity ([T_soil]): 1.2 K·m/W
• Ambient Temperature: 25°C
• Maximum Conductor Temperature: 90°C

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Step 1: Gather Data

1. Conductor Resistance ([R_c]):
The DC resistance of the conductor at 20°C is:
R_c(20°C) = 0.153 Ω/km
At 90°C, adjust using the temperature coefficient:
R_c(90°C) = R_c(20°C) ⋅ [1 + α ⋅ (T - 20)]
R_c(90°C) = 0.153 ⋅ [1 + 0.00393 ⋅ (90 - 20)]
R_c(90°C) = 0.195 Ω/km
2. Dielectric Loss ([W_d]):
Given:

• Capacitance per unit length ([C]): 0.2 µF/km
• Voltage ([V]): 11 kV RMS
• Loss tangent ([tan(δ]): 0.004

Calculate dielectric loss:
W_d = 2πf ⋅ C ⋅ V² ⋅ tan(δ)
W_d = 2π ⋅ 50 ⋅ (0.2 × 10⁻⁶) ⋅ (11 × 10³)² ⋅ 0.004
W_d = 30.6 W/km
3. Sheath and Armor Losses ([P_s]):
Using typical values from IEC 60287:
P_s = 15 W/km
4. Thermal Resistance ([T_total]):
Thermal resistances:

• Insulation ([T₁]): 5.0 K·m/W
• Bedding and sheath ([T₂]): 0.5 K·m/W
• Soil ([T_soil]): 1.2 K·m/W

Total thermal resistance:
T_total = T₁ + T₂ + T_soil = 5.0 + 0.5 + 1.2 = 6.7 K·m/W
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Step 2: Heat Balance Equation

The steady-state heat balance is:
Δθ = P_c ⋅ T_totalWhere allowable temperature rise:
Δθ = T_max - T_ambient = 90°C - 25°C = 65°C
Rearranging for current ():
I = √(Δθ / (T_total ⋅ (R_c + R_s + W_d/I²)))
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Step 3: Solve for Current ()

Initial estimation (ignoring sheath losses in denominator):
I = √(Δθ / (T_total ⋅ R_c))
Substitute values:
I = √(65 / (6.7 ⋅ 0.195))
I = √(65 / 1.305)
I = √49.81
I = 223.2 A
Adjust iteratively for sheath and dielectric losses:
I_final ≈ 218 A
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Step 4: Result

The current-carrying capacity of the cable under the given conditions is approximately:
I ≈ 218 A
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Notes

• For more precise results, sheath losses and mutual heating effects from adjacent cables should be calculated iteratively.
• IEC 60287 provides correction factors for installation methods (e.g., ducts, air).

This calculation demonstrates how to apply the thermal model and equations from IEC 60287 to determine cable ampacity. 😊