FISEKON GmbH – Fischer Elektrokonstruktion

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04 — Calculators

Calculate power loss in a cable

Every cable consumes part of what it carries. Over short runs that is irrelevant; over long routes at high current it becomes a figure with two faces: running cost over the lifetime, and heat that has to go somewhere.

Input

One shift is about 2000 h, three shifts about 6000 h, continuous 8760 h.

Result

Power loss
—W
Energy lost per year
—kWh
Cost per year
—€
Heat per metreDecides whether the cable becomes a problem in a duct or an enclosure.
—W/m

A guide, not a design to standard. We give no warranty for the correctness of the results, the applicable standards and case-by-case verification govern.

Worked example

The default case the calculator starts with, worked through once. Change the values above and it recalculates immediately.

Inputs

System
Three-phase (3~)
Operating current
32 A
One-way cable length
45 m
Conductor cross-section
6 mm²
Conductor material
Copper (κ = 56 m/Ω·mm²)
Conductor temperature
70 °C
Operating hours per year
4,000 h
Electricity price
25 ct/kWh

Result

Power loss
492 W
Energy lost per year
1,969 kWh
Cost per year
492 €
Heat per metre
10.9 W/m

Formula

  • Three-phase: P_v = 3 · I² · R
  • Single-phase and DC: P_v = 2 · I² · R
  • R = L / (κ · A) · [1 + α · (ϑ − 20 °C)]
  • W = P_v · t / 1000

Assumptions and standards

  • The calculation covers the resistance of the live conductors, not the protective conductor.
  • For balanced three-phase load no current flows in the neutral; three line conductors are counted.
  • Harmonics are not included. A strong third-harmonic content loads the neutral in addition.

Frequently asked

Is a larger cross-section worth it for the losses alone?

On long, heavily loaded routes, often yes. Losses fall in inverse proportion to cross-section: one size up saves roughly a third. Weigh the saving over the planned lifetime against the extra cost of cable and installation.

Why is heat per metre shown?

Because it decides whether the cable causes trouble elsewhere. In an open tray it hardly matters. In a filled duct, in thermal insulation or inside an enclosure it is part of the heat balance.

Does this hold at partial load?

It holds for the current you enter. Because losses go with the square of the current, you must not use an average when the load varies strongly — the result would be too low. Calculate section by section instead.

Calculating is the easy part.

A formula gives you a number. Designing a plant also demands installation method, grouping, discrimination, the standards in force and a look at the installed base. That is what we take on.

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