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Three-Phase • 400V • Aluminum Cable

Calculate Wire Cross-Section for Three-Phase 400V

Calculate the required wire cross-section for an aluminum cable based on power, cable length, cos φ and permissible voltage drop.

Calculation Values

Note: This calculation is a technically sound simplification typical of online calculators. For practical sizing, standards, installation method, temperature, grouping, protective devices, cable type and manufacturer specifications must also be checked.

Formula Used

For three-phase systems:

I = P / (√3 × U × cos φ)

A = (√3 × L × I × cos φ) / (κ × ΔU)

Here, A is the conductor cross-section in mm², L the one-way cable length in meters, I the current in amps, κ the conductivity of aluminum, defaulting to 37, and ΔU the permissible voltage drop in volts. The power factor cos φ is properly accounted for in the current and voltage drop calculation.

FAQ

Frequently Asked Questions about Calculate Wire Cross-Section for Three-Phase Aluminum Cable

Briefly explained: formula, use case and key limits of the calculation.

Which conductivity is used for aluminum? +

For aluminum, the calculator defaults to κ = 37. The value can be adjusted if a different material figure should be used.

Why does aluminum often need a larger cross-section than copper? +

Aluminum has lower electrical conductivity than copper. For the same power, cable length and voltage drop, a larger conductor cross-section is usually required.

Which formula does the calculator use? +

For three-phase systems the calculator uses I = P / (√3 × U × cos φ) followed by A = (√3 × L × I × cos φ) / (κ × ΔU).

What else must be checked besides the calculation? +

Current-carrying capacity, installation method, ambient temperature, grouping, protective devices, contact points and manufacturer specifications must also be checked.

Practical guidance

Assess an aluminium feeder for a three-phase load

The calculator first derives current from active power, then finds the aluminium cross-section required by the voltage-drop target.

Input values explained

Power in kilowatts
Enter the electrical active power in kW. One kW equals 1,000 W; motor output power and electrical input power are not necessarily the same.
Cable length
Enter the one-way distance from supply point to load. The required return path is already represented by the selected formula.
Voltage
Enter the line-to-line voltage between two phases. This is commonly 400 V in low-voltage systems; 230 V is not the correct input here.
Power factor cos φ
cos φ is the ratio of active to apparent power. It is close to 1 for resistive heating loads; use the data-sheet value for motors and transformers.
Permitted voltage drop
This percentage limits the calculated voltage loss. It is a design assumption and must suit the circuit and applicable requirements.
Conductivity κ
κ is the calculation value for the conductor material. Common approximations are 58 for copper and 37 for aluminium; heating increases actual resistance.

Example: 11 kW over 25 m

Assume 400 V, 11 kW, cos φ 0.90, 25 m one-way length and 3% permitted voltage drop.

I ≈ 17.64 A; A = (√3 × 25 × 17.64 × 0.90) ÷ (37 × 12) ≈ 1.55 mm²

The next listed size is 2.5 mm² on voltage drop. Terminations, minimum sizes and thermal ampacity still require verification.

Account for aluminium terminations

Aluminium has lower conductivity than copper and requires suitable terminals, oxide control and workmanship. The numeric result cannot verify those conditions.

Calculation limits

  • Thermal ampacity and correction factors are not calculated.
  • Joint resistance and copper-to-aluminium transitions are outside the model.