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Calculate TCO in the drivetrain

Alexander Olenberger Alexander Olenberger | May 8, 2026 | 9 min. reading time |
Zuletzt geprüft: 8. Mai 2026 durch Alexander Olenberger

Purchase price does not equal total cost

Anyone procuring an electric drive usually compares catalog prices: Motor, gearbox, frequency inverter. However, with continuously operated industrial drives, these acquisition costs are only 5-15 % of the total life cycle costs out. By far the biggest cost driver is energy - and this is hardly calculated when purchasing.

For an 11 kW drive with 6,000 operating hours per year and a service life of 15 years, the energy costs alone amount to over €190,000 - the purchase price of the motor is between €800 and €1,200. Total Cost of Ownership (TCO) makes this imbalance visible and provides the basis for well-founded purchasing decisions.

Key message

At 6,000 h/year of operation, over 80 % of the TCO are energy costs. An improvement in overall efficiency of just 5 percentage points saves more over 15 years than the motor actually costs.

TCO components in the drive context

The total costs of a drive system are divided into five blocks. The following table shows typical proportions for continuous industrial operation at 6,000 h/year over a service life of 15 years:

TCO block Typical contents Share of TCO
CAPEX components Motor, gearbox, frequency inverter, clutch 3–8 %
CAPEX installation Installation, cabling, commissioning 1–4 %
Energy costs Electrical energy over the entire operating time 70–90 %
Maintenance costs Lubrication, bearing change, seals, oil change 3–10 %
Downtime costs Production downtime, emergency repair, express delivery 1–8 %
Waste disposal Waste oil disposal, metal recycling, dismantling < 1 %

The shares vary depending on the operating profile: drives with a few hundred operating hours per year have relatively higher CAPEX shares; continuous operation systems (pumps, compressors, fans) are at the upper end of the energy range.

Energy cost model with sample calculation

Basic formula

The annual energy costs of a drive are calculated as follows:

E_year = (P_mech / η_total) × t_operation × c_current

  • P_mech - Mechanical effective power at the output shaft [kW]
  • η_total - Overall efficiency motor × gearbox (dimensionless)
  • t_Operation - annual operating hours [h/year]
  • c_current - Industrial electricity price [€/kWh]

Example calculation IE3 vs. IE4 over 15 years

Boundary conditions: 11 kW mechanical output power, 6,000 h/year, electricity price 0.18 €/kWh, planetary gearbox η = 0.96, motor efficiency according to IEC 60034-30-1.

Parameters IE3 (η = 0.915) IE4 (η = 0.932) Difference
η Motor 0.915 0.932 +1.7 %
η total (motor × gearbox 0.96) 0.879 0.895
Electrical power consumption P_el 12.51 kW 12.29 kW 0.22 kW
Energy consumption / year 75,065 kWh 73,741 kWh 1,324 kWh
Energy costs / year 13,512 € 13,273 € 239 €
Energy costs over 15 years 202,672 € 199,096 € 3,576 €

The table shows the overall system comparison including the gearbox. Isolated to the motor (without gearbox), the difference in efficiency IE3→IE4 is even clearer: IE3 consumes 12.02 kW, IE4 only 11.80 kW - a difference of 1,320 kWh/year or €238/year.

Maintenance Costs: Lubrication, Bearings, Wear Parts

Lubrication intervals

Transmission mineral oil should be changed after 10,000-15,000 operating hours or every two years at the latest - regardless of the operating time. Synthetic gear oils (PAO, PAG) last up to 30,000 hours and reduce maintenance costs despite their higher cost price. For a complete method of calculating intervals, see the guide to Set lubrication intervals correctly.

Bearing replacement after L10 service life

Rolling bearings are wear parts with a calculable service life. The dynamic load rating and the load profile provide the L10 service life (10 % probability of failure). Typical guide values for rolling bearings in gearboxes:

  • Lightly loaded bearings (deep groove ball bearings, n < 1,500 1/min): L10 = 20,000-50,000 h
  • Medium-loaded bearings (tapered roller bearings, worm gear bearings): L10 = 10,000-25,000 h
  • Highly loaded bearings (planetary gears, hollow shaft): L10 = 15,000-30,000 h with good lubrication

Preventive bearing replacement after reaching 80 % of the L10 service life typically costs € 200-600 (bearing + labor). Reactive replacement after bearing damage causes additional consequential damage to shafts and housings as well as downtime costs - often five to ten times higher.

Other wearing parts

  • Rotary shaft seals (RWDR): every 15,000-20,000 h or in the event of visible oil leakage - neglected leakage leads to bearing damage
  • Coupling elements: Check elastomer inserts every 5-10 years depending on the load spectrum
  • Brake pads (for holding brakes): according to manufacturer's specifications; typically every 3-5 years in continuous operation

Rule of thumb: Preventive maintenance costs amount to approx. 1-2% of the purchase price per year. For a gearbox worth €1,500, this amounts to €15-30 per year - a minimal TCO item, but one that prevents downtime costs in the four-digit range.

Comparative Calculation: Worm Gearbox vs. Planetary Gearbox

The difference in efficiency between worm gearboxes (η ≈ 70 %) and planetary gearboxes (η ≈ 96 %) has a massive impact on the TCO over the service life. The following calculation shows the pure energy cost comparison with the same effective mechanical power.

Boundary conditions: 11 kW mechanical output power, IE3 motor (η = 0.915), 6,000 h/year, 0.18 €/kWh, 10-year observation period.

Parameters Worm gearbox
η = 70 %
Planetary gearbox
η = 96 %
Difference
Gearbox efficiency 70 % 96 % 26 %
η total (motor × gearbox) 0.641 0.879
Electrical power consumption P_el 17.16 kW 12.51 kW 4.65 kW
Energy consumption / year 102,971 kWh 75,065 kWh 27,906 kWh
Energy costs / year 18,535 € 13,512 € 5,023 €
Energy costs over 10 years 185,347 € 135,115 € 50,232 €
CO₂ over 10 years (0.38 kg/kWh) 391 t CO₂ 285 t CO₂ 106 t CO₂

Result: The worm gearbox costs around €50,000 more in energy over 10 years. A planetary gearbox of this size costs around €500-1,500 more to purchase - amortization takes less than 4 months. Further efficiency considerations can be found in the guides Calculate gearbox efficiency and Worm gearbox vs. planetary gearbox.

IE3 vs. IE4: Motor classes and amortization

The IEC 60034-30-1 standard defines the efficiency classes IE1 to IE5 for three-phase motors. IE4 (Super Premium Efficiency) exceeds IE3 by approx. 1.5-2.5 percentage points in efficiency. The additional price for 11 kW motors is 15-20% (approx. €150).

Parameters IE3 IE4
Motor efficiency (11 kW, 4-pole) 91.5 % 93.2 %
Electrical power consumption 12.02 kW 11.80 kW
Energy consumption / year (6,000 h) 72,131 kWh 70,813 kWh
Annual energy savings (0.18 €/kWh) 238 €/year
Typical purchase price approx. 800 € approx. 950 € (Δ +150 €)
Amortization of the additional price approx. 8 months

Calculation: Δ = €150 surcharge / €238/year Savings ≈ 0.63 years (approx. 8 months). For any drive with more than 2,000 operating hours per year, IE4 is clearly the economically superior choice. An overview of all efficiency classes IE1-IE5 can be found in the guide IE Efficiency Classes: IE1 to IE5 Explained.

CO₂ balance Scope 2: Emissions from operating electricity

Scope 2 emissions are caused by purchased electricity. According to data from the German Federal Environment Agency (UBA), the emission factor for the German electricity mix in 2025 is:

CO₂ [kg] = E_Consumption [kWh] × 0.38 kg CO₂/kWh

Source: Federal Environment Agency, emission factor electricity mix DE 2025. For green electricity contracts, the factor is reduced to 0-0.10 kg CO₂/kWh.

Based on the comparative gearbox calculation (11 kW, 6,000 h/year, 10 years), the following Scope 2 emissions result:

Drive Energy consumption 10 years CO₂ emissions
IE3 + worm gear unit (η=70 %) 1,029,706 kWh 391 t CO₂
IE4 + planetary gearbox (η=96 %) 708,127 kWh 269 t CO₂
Savings through optimal selection 321,579 kWh 122 t CO₂

122 tons of CO₂ correspond to the mileage of a mid-range car of around 750,000 kilometers. For companies with ESG reporting and CO₂ reduction targets, gearbox selection is therefore a directly measurable lever - without changing the process, just by selecting better components.

Influence of the frequency inverter: Cubic characteristic curve

For applications with variable load requirements - especially pumps, fans and compressors - a frequency inverter (FI) offers the greatest energy-saving potential. The reason for this is the cubic dependence of the power on the speed (affinity laws):

P₂/P₁ = (n₂/n₁)³

With speed reduction to 80 %: P₂ = (0.8)³ × P₁ = 0.512 × P₁ - i.e. only 51.2 % of the rated output, a saving of 48.8 %.

Example: Pump Drive 11 kW

A pump runs continuously at full load without a frequency converter. With frequency converter and demand-based control at an average speed of 80 %:

  • Without FI: 11 kW × 6,000 h/year × 0.18 €/kWh / 0.879 = 13,512 €/year
  • With FI (80 % speed → 51.2 % load): 0.512 × 13,512 = 6,918 €/year
  • Savings: 6,594 €/year - a VFD (approx. €800–1,500) pays back in less than 3 months

This cubic characteristic curve does not apply to all load profiles. Conveyor belts, strokes and constant-torque-loaded drives follow a linear P ~ n dependency - the savings through speed control are lower in these cases. More on this in the guide Frequency inverters: When is it worth using them?

10-point checklist for purchasing (B2B)

This checklist helps purchasers and designers to anchor TCO-relevant requirements in specifications and bid comparisons:

1

Specify efficiency class

Require motor class IE3 as minimum, IE4 for >2,000 h/year. Specify minimum gearbox efficiency in the specification (e.g., η ≥ 94%).

2

Select gearbox type based on operating hours

Worm gearbox only for <2,000 h/year or when self-locking is absolutely required. Specify planetary gearbox for continuous operation.

3

Request L10 bearing service life

Require minimum L10 for all load-bearing bearings in the offer (e.g., L10 ≥ 20,000 h at n = 1,500 rpm).

4

Have lubrication intervals documented

Oil/grease change intervals and recommended lubricant grades must be included in the delivery scope.

5

Ensure spare parts availability

Manufacturer commitment for bearings, rotary shaft seals, and gearbox assemblies for at least 10 years after delivery.

6

Check variable frequency drive compatibility

Motor must be designed for VFD operation (insulation class F minimum, bearing protection for VFD use).

7

Operating hour counter / condition monitoring

For drives >15 kW or critical machines, plan operating hour counting or vibration/temperature monitoring.

8

Include TCO calculation in bid comparison

Calculate not only the purchase price, but energy costs over 5–10 years as a comparison metric (use the formula from this article).

9

Request manufacturer warranty and MTBF

Mean Time Between Failures (MTBF) and warranty scope (incl. consequential damages) must be explicitly requested and compared.

10

Disposal plan for operating fluids

Waste oil disposal certification and recyclability of gearbox housings (aluminum/gray cast iron) for ESG documentation.

TCO analysis for your drive project?

Our application engineers will calculate the total cost of ownership for your application - free of charge and without obligation. Bring along your operating profile (performance, operating hours, electricity price) and receive a well-founded recommendation.

Request a consultation now →

Further guide articles

FAQ: TCO Calculation in the Drivetrain

TCO (Total Cost of Ownership) encompasses all costs over the entire service life of a drive system: acquisition (motor, gearbox, variable frequency drive, installation), ongoing energy costs, maintenance (lubrication, bearing replacement, wear parts), downtime costs in the event of failure, and disposal costs at end of life. The purchase price typically accounts for only 5–15% of the TCO.

For continuously operated drives (pumps, fans, conveyor technology, 6,000–8,000 h/year), typically 70–90% of the total costs over a 15-year service life are attributable to energy. Only for drives used infrequently with few operating hours per year does the ratio shift in favor of acquisition costs.

Formula: E_year = (P_mech / η_total) × operating hours × electricity price. Example: 11 kW mechanical output, overall efficiency 0.915 (IE3 motor), 6,000 h/year, 0.18 €/kWh → E_year = (11 / 0.915) × 6,000 × 0.18 = 12.02 kW × 6,000 h × 0.18 €/kWh = 12,984 €/year.

For drives with more than 2,000 operating hours per year, an IE4 motor pays back compared to IE3 typically within one year. The price premium of approx. 15–20% (typically €150 for 11 kW motors) is quickly offset by the annual energy savings of approx. €238/year at 6,000 h/year and 0.18 €/kWh. IEC 60034-30-1 defines the efficiency classes IE1–IE5.

Example: Worm gearbox (η=70%) vs. planetary gearbox (η=96%) at 11 kW, 6,000 h/year, 10 years of operation, 0.18 €/kWh. The worm gearbox incurs €169,714 in energy costs, the planetary gearbox only €123,750 — a difference of €45,964. In addition, the worm gearbox generates 97 t more CO₂.

For variable-speed applications with a cubic characteristic (pumps, fans), power consumption decreases with the cube of the speed: P ~ n³. A speed reduction to 80% lowers power consumption to 0.8³ = 0.512 × P_rated — by approximately 49%. For an 11 kW drive with 6,000 h/year, this results in savings of approximately 35,000 kWh/year.

For Scope 2 emissions (purchased electricity, DE grid): CO₂ [kg] = energy consumption [kWh] × 0.38 kg CO₂/kWh (DE electricity mix 2025, source: UBA — German Federal Environment Agency). For a worm gearbox with 942,857 kWh over 10 years, this amounts to 358 t CO₂; a planetary gearbox requires only 261 t — a saving of 97 t.

The greatest impact comes from: 1) Adhering to oil change intervals (gearbox mineral oil every 10,000–15,000 h or every 2 years), 2) Replenishing grease on bearings as specified by the manufacturer, 3) Replacing seals and rotary shaft seals in time (prevents costly bearing damage from oil leakage), 4) Planning bearing replacement according to L10 service life rather than reactively. Preventive maintenance reduces downtime costs by 60–80% compared to reactive maintenance.

Industry figures quote downtime costs of €1,000–10,000 per hour for unplanned plant shutdowns (depending on the sector and production value). Even at a moderate €500/h and a mean repair duration of 4 hours, this amounts to €2,000 in downtime costs — often more than the drive component itself costs. Preventive maintenance and spare parts availability are therefore not cost factors but risk reduction.

For a drive with 6,000 h/year and 11 kW of mechanical output power, the annual energy saving from the efficiency difference (η 70% → 96%) is approximately €4,596/year. A planetary gearbox of this size typically costs €800–2,000 more than a comparable worm gearbox. The payback period is therefore 0.2–0.5 years — the decision is almost always clear.

Alexander Olenberger

About the Author

Alexander Olenberger

Senior Application Engineer · Technische Antriebselemente GmbH

Alexander Olenberger advises design engineers and purchasers on the energetic evaluation and design of drive systems. His focus is on the economic optimization of drive life cycle costs — from gearbox selection and motor efficiency classes to operating cost calculation.

Reviewed on May 8, 2026

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