# How VFDs save energy

## Analysis and examples of power conversion by variable-frequency drives.

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Variable frequency drives (VFD) are becoming more common place and more widely used in applications. They are capable of varying the output speed of a motor without the need for mechanical pulleys, thus reducing the number of mechanical components and overall maintenance. But the biggest advantage that a VFD has is the ability to save the user money through its inherit nature to save energy by consuming only the power that’s needed. The main question now is, How does a VFD accomplish this? The simple answer to this question is power conversion.

A VFD is similar to the motor to which it’s attached, they both convert power to a usable form. In the case of an induction motor, the electrical power supplied to it is converted to mechanical power through the rotation of the motor’s rotor and the torque that it produces through motor slip. A VFD, on the other hand, will convert its incoming power, a fixed voltage and frequency, to a variable voltage and frequency. This same concept is also the basis to vary the speed of the motor without the need of adjustable pulleys or gearing changes.

#### Electrical

Electrical power is defined as the following:

Power (P) = √3 x Voltage (V) x Current (I) x Power Factor (PF)

In an ideal VFD, the following would hold true:

Powerin = Powerout

But because a VFD has inefficiencies and requires a small amount of power consumption to power the brains of the drive, the input power will be slightly greater than the output power. For this, we will assume that this extra power draw is negligible.

With these two equations, we can then define the relationship between the VFD’s input and output:

Vin x Iin x PFin = Vout x Iout x PFout

Taking these equations into account, let’s use a 100-hp motor as an example with the following properties:

Power = 100 hp
Speed = 1,785 rpm
Voltage = 460 V
FLA = 115 A
Power factor = 0.86

A VFD will convert its incoming power, a fixed voltage and frequency, to a variable voltage and frequency.

– Edward Tom, Yaskawa America, Inc.

Assume that the motor is running at 60 Hz on a VFD, drawing a no-load current of 40 A on the output of the VFD. With this, one would assume that the input current would also be the same, 40 A. However, using an ammeter on the drive’s input, a person is reading nearly zero amps! How is this possible? Is the drive creating power somehow? The answer simply is no, the drive is not creating power. The power factor causes this “discrepancy” in current When a motor is running at no load, the motor’s power factor can be assumed to be zero, not 0.86 86 as stated on the nameplate. The reason the power factor isn’t at 0.86 is because this is the motor power factor at full load. Alternatively, mechanical (friction) and electrical (resistive) losses in the motor prevent the power factor from being zero when running no load, but we’ll assume these losses to be zero just like we did for the VFD. Therefore, you would have the following:

Pout = 460V x 40A x 0
Pout = 0

Because the output power is zero, the input power also will be zero. With a fixed input voltage, the two variables would be current and power factor. Because current is needed for a power factor to exist, both current and power factor are zero, which means the low input current reading is indeed correct.

This explains why the input current to the VFD is so low when the motor is operating under no load conditions. But what about under load? The same concept still applies when the motor is under load. For example, assume the same motor is now operating at half speed, 30 Hz and producing full motor rated torque and drawing the motor’s full-load amps (FLA). This means that the electrical power that the motor is drawing is:

Pout = √3 x 230V x 115A x 0.86 = 39.4kW

Because the VFD is a power converter, this means that the input current is (assuming a 0.89 input power factor from a 3% line impedance):

Iin = (39.4kW) / (√3 x 460V x 0.89) = 55.6A

Because the VFD is operating at half speed and under full load, the input current is less than half of the output current.

 Power (kW) Voltage (V) Current (A) Frequency (Hz) Power Factor Input 39.4 460 55.6 60 0.89 Output 39.4 230 115 30 0.86

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