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Bereken uw besparing bij upgrade naar hogere IE-klasse

Minimum rendement volgens IEC 60034-30-1 en IEC TS 60034-30-2.

Binnen 2 jaar terugverdiend, elke dag winst

Bij continu-werkende motoren in industrie- en proces-omgevingen is de meerprijs van een IE4- of IE5-upgrade gemiddeld binnen 18-24 maanden terugverdiend.

18-24 mnd Gemiddelde terugverdientijd
20+ Jaren nettowinst na terugverdientijd
Uit voorraad Snelle realisatie IE4/IE5/IE6 upgrade
1

Wat heeft u nu?

kW
2

Naar welke klasse wilt u upgraden?

3

Uw gebruik en stroomprijs

uur
/ kWh
Meer precisie (optioneel)

De berekening werkt ook zonder deze instellingen. Vul alleen in wat u weet.

Technische details van de motor

%

Een specifieke motor uit onze productlijst kiezen

Optioneel: koppel de berekening aan een motor en uw korting. Handig voor offerte-aanvraag.

%

Twee motoren met elkaar vergelijken

Uw besparing per jaar IE2 → IE4
Kostenbesparing per jaar -
Energiebesparing per jaar - kWh
CO₂-reductie per jaar - kg
Rendement: -% -%

Bronnen: IEC 60034-30-1:2014 (IE1-IE4), IEC TS 60034-30-2:2016 (IE5), CO2-factor 0,355 kg/kWh (Europees mix).

Calculation method & example

How we calculate the savings

Annual energy consumption follows from power × load factor × operating hours ÷ efficiency. The efficiency difference between your current IE class and the target class determines the annual savings in kWh; multiplied by your electricity price, that gives the savings in euros. The payback time is the price premium of the more efficient motor divided by that annual saving. The efficiency values follow IEC 60034-30-1 (IE1–IE4) and IEC TS 60034-30-2 (IE5).

Worked example

A 22 kW motor running 6,000 hours per year at 75% load. At €0.30/kWh an IE3 motor costs about €15,970 per year in electricity; an IE5 motor (SynRM) about €15,500. The saving is then roughly €470 per year. The premium of IE5 over IE3 is typically between €800 and €1,500, so the upgrade pays for itself in 2 to 3 years and then delivers net profit for well over twenty years.

Frequently asked questions about energy savings

When does an IE upgrade pay for itself?

In continuous operation (4,000+ running hours per year) an upgrade to IE4 or IE5 usually pays back within 18 to 24 months. The more running hours and the higher the electricity price, the faster.

How much does IE5 save compared to IE3 or IE4?

IE4 reduces losses by roughly 2 to 4 percent compared to IE3. IE5 (usually SynRM, such as ABB SynRM) cuts losses by a further 20 percent or so compared to IE4.

Which data do I need for the calculation?

The motor power (kW), the current IE class, the desired target class, the number of running hours per year and your electricity price per kWh. Load and number of poles can optionally be refined for more precision.

Does the saving also apply at partial load or variable speed?

Yes. The calculator works with a load factor. With strongly varying loads, a variable frequency drive on top of the IE upgrade yields additional savings; ask us for a combined calculation.

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