Motor & Operating Parameters
Operating Speed
80%
40%60%80%100%
Energy Savings Results
Full-Speed kW Input
kW
Reduced-Speed kW
kW
Annual kWh Saved
kWh/yr
Annual $ Saved
$/yr
Simple Payback
years
CO₂ Avoided
lbs/yr
Speed vs. Power (Affinity Law Cube Relationship)
Speed% of Full PowerkW at SpeedkWh/yr$/yr
Power scales with the cube of speed (P ∝ N³). At 80% speed: power = 51% of full; at 60% speed: power = 22% of full. Actual savings depend on system curve, VFD efficiency, and motor loading at reduced speed.
About This Calculator

This calculator quantifies the energy a variable frequency drive (VFD) saves on a fan or pump motor. Enter the motor horsepower, motor and drive efficiencies, electricity rate, operating hours, and installed drive cost, then set the average operating speed. The tool returns full-speed and reduced-speed input kW, annual kWh and dollars saved, a simple payback, and the avoided CO₂.

The speed-versus-power table shows the steep, cubic shape of the affinity law, which is why even modest speed reductions on variable-volume systems produce outsized savings. It is a fast screening tool for justifying a drive retrofit before detailed analysis.

Formula & Method
Full-speed input kWkW = HP × 0.7457 ÷ (motor eff. × drive eff.)
Reduced-speed kWkWred = kWfull × (speed)³
Annual savings$ = (kWfull − kWred) × hours × rate
Simple paybackyears = drive cost ÷ annual $ saved
CO₂ avoidedlb = kWh saved × 0.92 lb/kWh (grid average)

The method follows the fan and pump affinity laws (power ∝ speed³) and the exact 0.7457 kW-per-horsepower conversion. The 0.92 lb CO₂ per kWh is a representative US grid-average emission factor. The cube law assumes a friction-dominated system curve; fixed static pressure and low-speed efficiency losses make real savings somewhat lower, so treat the result as an upper bound.

Frequently Asked Questions
How does a VFD save energy?
On a fan or pump, shaft power follows the affinity law and varies with the cube of speed. Slowing a motor to 80 percent speed drops power to about 0.8 cubed, or roughly 51 percent of full power; 60 percent speed drops it to about 22 percent. Because the relationship is cubic, small speed reductions yield large energy savings, which is why variable frequency drives pay back quickly on variable-flow systems.
How is full-speed motor input kW calculated?
The tool converts shaft horsepower to kilowatts (1 HP = 0.7457 kW), then divides by both the motor efficiency and the drive efficiency to get the electrical input kW. Reduced-speed kW is the full-speed input kW multiplied by the speed ratio cubed. Annual kWh saved is the difference between full and reduced kW times the operating hours.
How is simple payback computed?
Simple payback equals the installed drive cost divided by the annual dollar savings, where annual savings is kWh saved times the electricity rate. The calculator reports payback in years. It does not discount future savings, so for a longer-horizon view that accounts for the time value of money, use a payback and ROI tool with NPV.
Is the cube law always exact in practice?
No. The cube law assumes a system curve dominated by friction and ignores fixed static pressure, minimum-speed limits, and motor efficiency falling off at low load. Real savings are usually a little less than the ideal cube relationship predicts. Treat the result as an upper-bound estimate and confirm with the actual system curve and drive data.
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