Energy Saving Concepts · Variable-speed drives
Slow the load.
Cube the saving.
On suitable variable-torque pumps, fans and blowers, reducing speed can reduce required shaft power approximately with the cube of speed. Drag the frequency control and watch flow, pressure, power and motor current respond differently.
VFD control panel● OUTPUT ENABLED
5 HzDrag to adjust50 Hz
The cube law
P₂ / P₁ = (n₂ / n₁)³(40 / 50)³ = 51.2%Flow follows speed. Pressure follows speed squared. Power follows speed cubed.
Frequency40.0Hz
Motor speed1,176estimated loaded rpm
Airflow80.0%of full-speed flow
Pressure64.0%of full-speed pressure
SPEED
PRESSURE
LOAD POWER
MOTOR CURRENT
The motor still needs flux-producing current. The current gauge therefore does not follow the cube-law power bar.
At 40 Hz the motor runs at about 80% speed. The centrifugal fan needs about 51.2% of its full-speed shaft power—an ideal reduction of 48.8%.
Practical comparison · Water transfer
Fill the same tank: DOL versus VFD
A 10 m³ process tank is filled from empty. The DOL pump runs at 50 Hz. Drag the VFD control before or during the test: the water level and accumulated energy will not reset.
5 HzDrag to adjust — works while running50 Hz
DOL pump 50 Hz · full speed
0%
Flow20.0 m³/h
Fill time30.0 min
Shaft power15.00 kW
Energy used0.00 kWh
Simulation clock0.0 min
StatusReady
VFD pump 40.0 Hz · 80.0% speed
0%
Flow16.0 m³/h
Fill time37.5 min
Shaft power7.68 kW
Energy used0.00 kWh
Simulation clock0.0 min
StatusReady
At 40 Hz the VFD takes 7.5 minutes longer but uses 36.0% less ideal shaft energy.
Illustrative assumptions: 10 m³ tank, 20 m³/h full-speed flow, 15 kW full-speed shaft power and a friction-dominated system where flow follows speed and power follows speed cubed. The animation is accelerated. For a real tank with significant or changing static head, use the manufacturer’s pump curve and the actual system curve.
Theoretical study · Five fills per day
What could a £350 drive save?
This worked example compares the 15 kW DOL pump with an assumed Siemens SINAMICS V20 drive. It applies the ideal affinity laws to the same 10 m³ tank and shows how the longer fill time affects the energy used for each completed tank.
Assumed drive cost£350
Electricity tariff25p/kWh
Tank fills5 per day
DOL baseline7.50 kWh/fill
Why energy per tank follows speed squared:
power ∝ speed³, but fill time ∝ 1/speed. Therefore energy per tank = 7.5 × (Hz ÷ 50)² kWh.| VFD speed | Fill time | Energy per fill | Energy saved per fill | Saving per day | Saving per year | £350 payback |
|---|---|---|---|---|---|---|
| 50 Hz | 30.0 min | 7.50 kWh | 0.00 kWh | £0.00 | £0 | — |
| 45 Hz | 33.3 min | 6.08 kWh | 1.43 kWh | £1.78 | £650 | 197 days |
| 40 Hz | 37.5 min | 4.80 kWh | 2.70 kWh | £3.38 | £1,232 | 104 days |
| 35 Hz | 42.9 min | 3.68 kWh | 3.83 kWh | £4.78 | £1,745 | 73 days |
| 30 Hz | 50.0 min | 2.70 kWh | 4.80 kWh | £6.00 | £2,190 | 58 days |
| 25 Hz | 60.0 min | 1.88 kWh | 5.63 kWh | £7.03 | £2,566 | 50 days |
| 20 Hz | 75.0 min | 1.20 kWh | 6.30 kWh | £7.88 | £2,874 | 44 days |
Theoretical figures only. Annual savings assume five fills every day for 365 days. Payback covers the assumed £350 drive price only and excludes installation, controls, commissioning and maintenance. Actual savings will be affected by drive and motor losses, pump efficiency, minimum safe speed and especially static head. A real project must be checked against the pump and system curves.
For the same suitable pump, fan or blower, flow is approximately proportional to rotational speed: Q₂/Q₁ = n₂/n₁.
Pressure or pump head is approximately proportional to speed squared: H₂/H₁ = (n₂/n₁)².
Shaft power is approximately proportional to speed cubed: P₂/P₁ = (n₂/n₁)³. A small speed reduction can therefore produce a much larger power reduction.
Why current remains: while a VFD output is enabled and maintaining flux in an induction motor, the motor requires magnetising current even when the mechanical load is light. That current produces the magnetic field; it is not all converted into useful shaft power. A motor isolated by an open contactor draws no motor current. If the VFD run output is disabled, motor current also stops, although the drive electronics still consume a small standby power. The illustrated 28% current floor is educational only—actual no-load current depends on the motor, drive control method and settings.
Where the cube law applies: the simple relationship applies to suitable variable-torque fluid-moving machinery following the affinity-law assumptions and operating mainly against quadratic system resistance. This includes centrifugal, axial and mixed-flow pumps and fans, plus applicable centrifugal or axial blowers and compressors. It does not describe positive-displacement pumps or constant-torque loads such as conveyors, extruders, mixers and loaded hoists. High static head, changing efficiency, drive losses, minimum safe speed and the actual operating point can make the simple cube calculation overstate real savings.
Technical basis: ABB guide to variable-speed drives and motors · Danfoss Facts Worth Knowing about AC Drives. Values on this page are illustrative, not a drive-selection calculation.