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Single-Phase • 230V • Aluminum Cable

Calculate Wire Cross-Section for Single-Phase 230V

Calculate the required wire cross-section for an aluminum cable on a single-phase system 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 single-phase systems:

I = P / (U × cos φ)

A = (2 × 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 Single-Phase Aluminum Cable

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

Which formula does the single-phase aluminum calculator use? +

The calculator uses I = P / (U × cos φ) and A = (2 × L × I × cos φ) / (κ × ΔU). For aluminum, κ defaults to 37.

Why is the cross-section larger for aluminum? +

Aluminum conducts less well than copper. Under the same conditions this increases the required cross-section.

Why is cos φ taken into account? +

cos φ describes the power factor. Motors, pumps and other inductive loads draw more current than a purely resistive load with cos φ = 1.

What does the permissible voltage drop mean? +

The permissible voltage drop is the maximum allowed voltage loss on the cable, for example 3 percent of the mains voltage.

Practical guidance

Size a single-phase aluminium conductor from load power

High current at 230 V combines with aluminium's lower conductivity, increasing the voltage-drop cross-section.

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 voltage between line and neutral. This is commonly 230 V in low-voltage systems.
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: 3.5 kW over 25 m

At 230 V, 3.5 kW, cos φ 0.90 and 3%, current is approximately 16.91 A.

A = (2 × 25 × 16.91 × 0.90) ÷ (37 × 6.9) ≈ 2.98 mm²

The next standard size is 4 mm². Suitable aluminium terminals and thermal limits remain essential.

Use material-compatible connections

A numerically adequate conductor is only safe with approved terminals and correct aluminium preparation.

Calculation limits

  • Copper-to-aluminium transitions and corrosion risk are excluded.
  • The calculation does not establish a permitted installation method.