Often times those using a Variable Frequency Drive (VFD) may find they need to connect a higher horsepower VFD to a single phase input power source. Since most high horsepower VFDs only accept three phase input as a power source, they are left with few options or alternatives. Don’t fret, there is a solution.
If you are using a Variable Frequency Drive (VFD) rated for three phase input and the only power source you have available to you is single phase input, then you can derate the Variable Frequency Drive (VFD) to accept the single phase input power source. You can almost always use a VFD rated for three phase input with a single phase input power source. When only a three phase input VFD is available, it is acceptable and common practice to derate the VFD to work with a single phase input power source.
Before you derate your VFD, it is most important to ensure the VFD you are using is properly suited for your application. The following are some basic guidelines to help you in determining whether or not your Variable Frequency Drive (VFD) is suitable for your application: Gather motor nameplate data including horsepower (HP), current (Amps), motor voltage, input line voltage and power source phase.
Determine which type of VFD your application will require. The type will fall under the category of Volts per Hertz (V/Hz), closed-loop vector, or open-loop vector (Sensor-less Vector). The internal components of the three phase input VFD are rated for the appropriate current expected when three phase input power is applied. When using single phase input the line side current from the single phase is always higher. To “derate” is the process of ensuring that these components are rated for the higher current that will flow from the single phase input instead of the three phase input.
You can derate a VFD by:
Determining the Horsepower of the Motor the VFD will be connected too, then choosing VFD with a Horsepower higher than the Horsepower of the motor to compensate for the additional input current from the single phase power source. The simplest formula used for these types of applications is:
VFD Input Current > Motor Current Rating * 1.73
The VFD input current must be equal to or greater than the Motor Current Rating * 1.73
When installing most three phase input Variable Frequency Drives (VFDs) on an application where single phase input power is used, you will almost always connect the input line leads to L1 and L2 of the VFD. L3 will be left open with nothing connected. Consult with the VFD manufacturer or knowledgeable integrator to be sure.
An application has a 230 VAC single phase input power source and needs to connect it to a conveyor that has a Variable Frequency Drive (VFD) connected to a 10 Horsepower 230 VAC 3 phase induction motor. Let us assume it has been determined that this application will operate well with a simple Volts per Hertz (V/Hz) VFD. The issue is, since there are no VFD manufacturers that offer a 10 Horsepower (HP) single phase input Variable Frequency Drive (VFD), we will need to de-rate a VFD with a three phase input for single phase input. Most manufacturers of VFDs only offer products up to 3 Horsepower (HP) for single phase input. The 10 Horsepower (HP) AC motor nameplate reveals that the motor is rated for approximately 27 amps at 230 VAC. We must use the equation above:
VFD Input Current > Motor Current Rating * 1.73
VFD Input Current > 27 Amps * 1.73
VFD Input Current > 46.71
This application will need a 230 VAC 3 phase Volts per Hertz (V/Hz) Variable Frequency Drive (VFD) with an input current rated at or above 47.0 amps.
Motors connected to VFDs receive power that includes a changeable fundamental frequency, a carrier frequency, and very rapid voltage buildup. These factors can have negative impacts, especially when existing motors are used.
There are a number of potential problems that can become real when a variable frequency drive (VFD) is used to power an existing induction motor. As such, you should carry out a careful study to determine if these problems could be sufficiently bad to cause reconsideration of such an installation. With a VFD, an existing motor normally having a number of useful years left in it could abruptly fail.
Existing motors are designed for 60 Hz only, 50 Hz only, or 60/50 Hz service. As such, you have to question whether or not a new VFD can be matched to your existing motor and still have the motor perform reasonably well. In other words, will the motor be able to handle additional factors that may cause greater vibration, heat rise, etc., and a possible increase in audible noise?
High Frequencies Can Cause Problems
You should be aware of possible side effects caused by high pulsing frequency when applying a VFD to an existing motor. These negative effects include additional heat, audible noise, and vibration. Also, pulse width modulation (PWM) circuitry, which causes a high rate of voltage rise of the carrier frequency, can cause insulation breakdown of the end turns of motor windings as well as feeder cable insulation.
The carrier frequency, a by-product of obtaining current at a variable fundamental frequency, is the cause for having additional watts in the motor; this power is essentially wasted energy that adds heat to the motor. The amount of such loss varies, depending upon the motor’s stator and rotor designs and frequency of the carrier wave.
With frequencies other than the fundamental, a motor runs at very high slippage and, therefore, is running somewhat inefficiently. (Slip is the difference between the rotational speed of the stator’s magnetic field [the synchronous speed of the induction motor] and the speed of the rotor.) Also, numerous lines of magnetic flux are being cut by the rotor; this phenomenon produces additional watts and additional heat. (Note that the high-frequency ripples in the current are at low magnitude, and the additional heat is in the order of 5% to 10% above that produced by a pure sine wave).
The synchronous speed of a four-pole motor served by 60-Hz power is 1800 rpm. This same motor, when considering “overtones” or ripples in the current’s fundamental frequency caused by a voltage carrier frequency of 4 kHz, will have current flowing through it based on that high frequency. Thus, the rotor of a four-pole, 60-Hz designed motor (with a rated full-load speed of 1750 rpm) being supplied power by a VFD adjusted to 10-Hz output will be turning at 1/6 rated speed. If the load’s torque requirements are constant at low- through full-rated speeds, slip rpm remains constant. For the motor above, which is operating at 10 Hz, the shaft will be turning at 250 rpm.
The rotor, while turning at 250 rpm and crossing lines of flux (the magnetic field) based on the 10-Hz fundamental frequency and the synchronous speed of 300 rpm (1/6 of 1800 rpm), is also crossing lines of flux due to the carrier frequency voltage of 4 kHz. The synchronous speed at 4 kHz is 120,000 rpm ([120 x 4000] [divided by] 4).
Based on a synchronous speed of 120,000 rpm and a shaft speed of 250 rpm, you can see that the magnetic lines of flux being cut due to the carrier frequency (4 kHz) are substantial compared with a synchronous speed of 300 rpm caused by the 10-Hz frequency. This additional current, which is transmitted to the rotor bars by the cutting of additional magnetic flux caused by carrier frequency, produces very little useful power. Most of this current is dissipated as heat, adding to the temperature rise of the motor. This additional heat represents about another 5% to 10% thermal buildup in the motor and can place an additional thermal strain on the motor’s rotor bars and stator windings, if it is running at full load. This high frequency power is an inefficient producer of torque.
Because of these and other conditions mentioned, you may wish to derate an existing motor when it’s connected to a VFD. The amount of energy from the carrier frequency that’s dissipated by the motor depends upon the amplitude and the frequency of the voltage, and the reactance and resistance of the motor at the resultant frequency. The amplitude of the current is determined by the ratio of the voltage over the impedance, while the watts lost is a product of the current squared times the resistance.
Other Undesired Side Effects
You also should be aware of other potential side effects caused by high frequency. These include undesirable audible noise, harmful vibration, and bearing problems.
Vibration and noise problems. To avoid noise and vibration problems, it’s recommended that the motor being used not have components that can resonate at the frequencies the motor (and its load) will generate. This is possible on systems where the frequency of the power is known, such as with 60 Hz. However, today’s VFDs have no standard carrier frequency, and the fundamental frequency can range from less than 10% of 60 Hz to 100% of 60 Hz, and beyond. Depending on which brand and model number of VFD is mated to the existing motor, and other factors such as the characteristics of the electrical system at the site, resonances in certain components may or may not be excited.
You must also consider that when a 60-Hz designed motor is operating at a different electrical frequency, various components of the motor might go into mechanical resonance, such as the fan or shaft. Each component has its own natural mechanical frequency, and an electrical frequency going through the coils and rotor bars can cause mechanical vibrations that are different from the initial design parameters. When an electrical frequency matches the natural frequency of a mechanical component, serious problems may occur. This may include the disintegration of a component.
Bearing problems. Another possible problem, which still isn’t fully understood, is the slow disintegration of the roller/ball (anti friction) bearings that support the shaft. It appears this is caused by bearing current and static discharge. What happens is that pitting occurs on the roller/ball surface and, when accumulated, causes the bearing to make noise. If not addressed, vibration will begin to develop.
Air flow problems. An additional factor you should consider when operating a standard 60-Hz motor at very low speed is that the fan, which is fixed and attached to the rotor, may not create enough air flow to effectively cool the motor. This is true because air flow is proportional to shaft speed. Thus, at half shaft speed, the air flow is half normal flow. To compensate for low-volume air flow at low motor speeds, if installation is possible, the attachment of a constant velocity air blower package to the back of the motor will usually provide adequate cooling.
Conductor Insulation Breakdown
As mentioned, PWM circuitry, which causes the high rate of voltage rise at the carrier frequency, can cause insulation breakdown of the end turns of the motor windings, as well as possible breakdown of the feeder cable insulation. This relates to the very high rate of rise of the voltage (rate of voltage change with respect to time) in combination with the very rapidly repeating voltage pulse caused by the VFD. Conductor insulation failures in motors have occurred because of this phenomenon. This subject is not completely understood and is presently being researched. The known facts about the matter are summarized as follows.
Switches in the inverter section of VFDs used today cause instantaneous turn-to-turn voltage inside a motor’s windings to be significantly higher than what an equivalent normal sine wave supply produces.
Each cycle of the fundamental voltage consists of numerous pulses of voltage.
Long distance between a motor and its VFD causes the turn-to-turn voltage to get even higher.
There are different approaches in explaining why there’s an increase of voltage at the motor terminals. Some explain it in terms of resonant capacitance/inductance (LC) circuits; others explain it in terms of standing wave theory. Both approaches end up with a similar result. When the distance between a motor and its VFD exceeds a critical distance (which may be as low as 30 ft), there is a voltage overshoot that may exceed twice the amplitude of the voltage pulse originally delivered at the VFD output terminals.
This higher voltage comes at the motor at such a high rate of change for each of the PWM pulses, from zero volts to its peak value, that it’s unevenly distributed across the winding, causing high turn-to-turn voltages in the turns connected closest to the power leads. The result places very high stress on the conductor insulation, which can cause early breakdown of the insulation.
Special inverter duty motors are available that are designed to meet or exceed the voltage amplitudes and rise times defined in NEMA Standard MG1, Motors and Generators, Section .220.127.116.11., Voltage Spikes. When connecting existing motors to VFDs with long cable lengths, you should consider using a filter to reduce the effects caused by the long cable.
Skin effect contributes to losses
In addition to the problems described above, there’s yet another loss component you should be aware of: skin effect. Skin effect induces the current in an AC system to crowd to the outside surface of a conductor. This phenomenon causes resistance to be directly related to the square root of the frequency of the current. In other words, the greater the frequency, the greater the resistance due to skin effect. The carrier frequencies are usually between 800 Hz to 15 kHz, and currents at these high frequencies will cause [I.sup.2]R losses. While the high frequency currents are relatively nominal, the loss relates to the current’s square power. And the carrier frequency, even at its square root, can be somewhat effective because of its basic high value. The geometry of the rotor bars also determines the degree to which the skin effect impacts rotor losses.
Motor application is very important
You should remember that a motor is a constant torque machine. In other words, at rated speed and rated torque, it will produce a certain horsepower. When speed is reduced through frequency and voltage reduction, the motor, by consuming more current, will try maintaining constant horsepower, if called for by the load. This can be done to a limited extent. As more current flows, more heat is produced, and it will not take long for the motor to overheat.
For situations where, throughout the speed range being used, there is a constant horsepower requirement, it’s critical that the motor be sized to match the horsepower requirement at the lowest shaft speed anticipated. For example, if the required speed range is from 50% to 100% of rated speed and the load’s horsepower requirement is 100 hp, then the motor must still be able to produce 100 hp at 50% speed. This also means that at 100% speed, the motor’s horsepower output, as called for by its load, also will be 100 hp; however, the load’s torque requirement will be reduced by 50%. At full-rated speed, the motor will be capable of producing 200 hp, meaning the motor will be larger than normal.
Using a VFD, with fundamental frequency being reduced to achieve lower speed, the voltage also is reduced in direct proportion to the speed reduction. As mentioned earlier, a 460V motor at half rotor speed will have 230V across its lines. Thus, if the motor’s rating is 100 hp at full speed, its output would only be 50 hp at half speed.
Certain loads, like lathes and grinders, require constant horsepower throughout their operating speed range. Let’s assume a VFD is serving a 20-hp lathe motor that’s operating at a 25% reduction in speed (3/4 rated speed). The lathe’s rotating chuck, which holds some material being worked by a cutting tool, will need constant horsepower over the entire speed range being used. If speed is reduced by 25%, voltage will be reduced by 25%. For the motor to maintain constant horsepower output, it will draw 33% more current (4/3 of normal amperage). Because current produces heat (primarily [I.sup.2]R losses), the motor will have to have sufficient thermal capacity to handle the extra amperage.
Some motors can withstand a certain amount of excess thermal load based on the motor’s service factor (SF). Usually an SF ranges from 1.0 to 1.15; beyond that point, motor damage will occur. Because voltage is reduced using a VFD, a motor’s horsepower rating must be increased to match the load requirement at the lowest speed used, should constant horsepower be required. Of course, this means the motor is overbuilt when used at higher speeds and will have higher losses and lower power factor (PF) at the higher speeds when operating at less than full load. However, the lower PF is compensated for by the VFD. This is a condition that must be accepted. Otherwise, you’re asking for trouble.
When working with motors, you’ll find it helpful to remember the following relationships:
1 hp = 0.746kW = [3 ft-lb x 1750 rpm] [divided by] 5250
Any of these numbers can be changed. When doing so, however, the equality of both sides of the equation must be maintained. Torque is ft-lb. If horsepower remains constant and speed (rpm) is reduced, obviously torque must be increased. Thus, in the above motor application (where there is a 25% reduction in speed), the motor’s torque output must be increased by 33%. And if kW remains constant and the voltage is reduced (which will happen using a VFD to reduce speed), current must be increased. This could lead to overheating. Incorrect application of motors is one of the main reasons why they fail.
If someone recommends getting a VFD for your existing motor, with the idea of making adjustments that would cause the output voltage to be set to any value (with a limit up to the VFD’s incoming voltage) for any particular fundamental frequency, use caution. Such an adjustment can be made; for example, you can adjust a VFD to produce 460V at 30 Hz. If 460V is the line voltage (thus the maximum voltage), then as the fundamental frequency increases beyond the set point, the voltage going to the motor remains constant.
Let’s look at one of the above examples again. Say 100 hp is required at half speed and the VFD is adjusted for delivering 460V at 30 Hz. If you use an existing motor rated for 100 hp, what will happen? Well, the motor will try to deliver 100 hp at half speed and will continue to try should the fundamental frequency be increased while the voltage remains constant at 460V. (Note that when the fundamental frequency gets below the set value [say 15 Hz], the voltage will be reduced proportionally, in this case to 230V.) At 30 Hz and 460V, the iron in the stator of this existing motor is magnetically saturated, which causes more current to flow and the motor to become excessively hot. This condition may destroy conductor insulation as well as negatively affect other motor components. Motors usually have enough iron in their stators to handle a certain ratio of volts to frequency (V/Hz). But when the ratio increases extensively, more iron is needed; otherwise, there will be overheating.
Still, using 30 Hz at 460V is an effective way of getting adjustable speed at constant horsepower, providing the iron in the motor’s stator is designed to take a higher V/Hz ratio. This means that more iron has to be placed in the motor’s stator. There are certain motors built today that have extra iron in their stator for operation at high V/Hz ratios. You’ll have to pay a premium for them. But for certain type applications, such as above, such motors can be cost effective when compared with using an existing motor of twice the capacity. This is because the premium motor can operate at 30 Hz, 460V, and normal current, whereas the high-capacity existing motor, operating at 30 Hz, 230V, will have to use double the current, and will experience the losses associated with high-current operation.