VFD Energy Savings Calculator
Estimate the input power, yearly kWh, cost and CO₂ saved by running a fan, pump or other motor load on a variable frequency drive, and the simple payback on your installed cost.
VFD Energy Savings Calculator
How to Use the VFD Energy Savings Calculator
- 1Enter the motor rating and running time. Rated power is the nameplate shaft output (kW). Hours per day and days per year set the annual energy.
- 2Enter the shaft load before and after. Use measured or pump/fan-curve values as a percentage of rated power. Or enter the reduced speed and let the calculator apply the affinity law (fans and pumps) or a linear law (constant torque).
- 3Enter efficiencies. Motor efficiency converts shaft power to electrical input; drive efficiency adds the VFD losses to the "after" case only.
- 4Enter tariff, VFD cost and grid factor. Payback is shown only when you enter your installed cost.
How VFD Energy Savings Are Calculated
Pafter = Prated × Loadafter / (ηmotor × ηVFD)
Loadafter (from speed) = Loadbefore × (N2/N1)³ for fans and pumps, × (N2/N1) for constant torque
Energy saved = (Pbefore − Pafter) × hours/day × days/year
Cost saved = energy saved × tariff; payback = installed VFD cost / cost saved
CO₂ saved (t) = energy saved (kWh) × grid factor (kg/kWh) / 1000
Example 1: 30 kW Fan, Measured Load
30 kW fan, 100 % load before, 70 % after, 12 h/day, 300 days/year, $0.15/kWh, motor 93.6 %, VFD 96 %.
Pafter = 30 × 0.70 / (0.936 × 0.96) = 23.37 kW
Energy saved = (32.05 − 23.37) × 12 × 300 = 31,250 kWh/year
Cost saved = 31,250 × 0.15 = $4,687.50 per year (shown as $4,687)
Example 2: Same Fan Slowed to 80 % Speed
Pafter = 30 × 0.512 / (0.936 × 0.96) = 17.09 kW
Energy saved = (32.05 − 17.09) × 3,600 h = 53,846 kWh/year → $8,077 per year
With an installed cost of $6,000 (your quote), payback = 6,000 / 8,077 = 0.7 years
For a 30 kW conveyor at 90 % load slowed to 80 % speed, power falls only in proportion to speed: 28.85 kW to 24.04 kW, saving 17,308 kWh per year.
Affinity Laws and Their Limits
For a centrifugal fan or pump on a system with no static head, flow is proportional to speed, pressure to speed², and power to speed³. The table shows the ideal values.
| Speed (%) | Flow (%) | Pressure / head (%) | Shaft power, ideal (%) | Constant-torque power (%) |
|---|---|---|---|---|
| 100 | 100 | 100 | 100 | 100 |
| 90 | 90 | 81 | 72.9 | 90 |
| 80 | 80 | 64 | 51.2 | 80 |
| 70 | 70 | 49 | 34.3 | 70 |
| 60 | 60 | 36 | 21.6 | 60 |
| 50 | 50 | 25 | 12.5 | 50 |
- Static head: a pump lifting water or holding a fixed pressure must keep a minimum speed, and its power falls much less than the cube law suggests. Use the pump curve and system curve.
- Throttled baseline: if flow was reduced with a damper or valve, the throttled power is already below full-speed power, so applying the cube law to it overstates savings. Enter the after-VFD load from the fan or pump curve instead.
- Efficiencies change: motor and drive efficiency fall at low load and speed, and the pump or fan efficiency moves along its curve.
- Running hours: savings scale directly with hours at reduced speed; a fan that runs at full speed most of the time saves little and loses the drive's 2–5 % losses.
Efficiency Standards That Affect the Result
| Item | IEC | US / North America |
|---|---|---|
| Motor efficiency classes | IEC 60034-30-1: IE efficiency classes (IE1, IE2, IE3, IE4 and higher) | NEMA MG 1 (NEMA Premium); DOE 10 CFR 431 minimum efficiencies |
| Motor efficiency test method | IEC 60034-2-1 | IEEE 112 |
| Drive and drive-system losses | IEC 61800-9-2: IE0–IE2 for drives, IES0–IES2 for motor + drive | Manufacturer data |
| Variable-speed requirements in buildings | National building codes | ASHRAE 90.1 requires variable-speed control for some fan and pump systems |
| Energy management and verification | ISO 50001 (energy management systems) | ISO 50001; IPMVP for measuring savings |
A drive on a motor used near full speed most of the time adds losses rather than saving energy. VFDs on motors that were not designed for converter supply may also need checking for insulation stress (NEMA MG 1 Part 31, IEC TS 60034-25).
VFD Energy Savings FAQs
It depends on the load and how long it runs at reduced speed. A centrifugal fan run at 80 % speed ideally needs about 51 % of full-speed shaft power; a constant-torque conveyor at 80 % speed needs about 80 %. A motor that runs at full speed saves nothing and loses the drive's losses.
The cube law assumes all of the pump head is friction. When part of the head is static (lift or a fixed pressure), the pump cannot slow as much for the same flow reduction, so its power falls by less.
Yes, at lower load the drive's input current falls with the power drawn. The motor-side current depends on load torque, so a constant-torque load at low speed still draws close to its full-torque current on the motor side.
Simple payback is the installed cost (drive, panel, cabling, commissioning) divided by the annual cost saving. Add maintenance and any utility incentive for a fuller comparison.
A drive with a diode rectifier has a high displacement power factor (close to 1), but it draws harmonic current, so the true power factor is lower; a line reactor or DC choke improves it. Power factor correction capacitors should not be connected on the drive output.
