# AC Motors and Variable Frequency Drives

by JonasS
Tags: drives, frequency, motors, variable
 P: 16 Hello all. I have a question about AC motors that has been driving me crazy. I understand how variable frequency drives work, by adjusting the power frequency and thus changing the rpm of the motor based on the equation: rpm = 120 x f / p where f = frequency (Hz) and p = number of motor poles. What I don't understand is how this increase in motor speed relates to voltage and current. From the equation it doesn't look like they matter. For example: 3-phase 4-pole AC motor connected to a 15A 110VAC 60Hz power source (US utility power). With no load to cause slip this motor will spin at 1800rpm. Now put a variable frequency drive in between the motor and the outlet. Then crank the frequency up to 120Hz. The motor will reach 3600rpm. But what effect will this boost in rpm have on the power being drawn from the line? From what I have read the only thing that changes is the frequency, so you should be drawing no more power than when it was spinning at 1800rpm. But that doesn't make sense, since there is more energy available in the faster spinning motor. Can anyone help me understand this issue?
 Mentor P: 22,312 The power being drawn from the line depends on the load, so how much the power changes depends on how much the load changes when you change the frequency. I'm an HVAC engineer and I often use vfds on fans. Fan horsepower is a cube function of rpm, so in my applications, a small increase in rpm makes for a big increase in amperage/wattage.
 P: 16 Well let us say it is the same load...a fan...for both instances. If the load, which I assume means the torque needed to spin the fan, stays the same at both 60Hz and 120Hz then how can it drain more power?
P: 4,777
AC Motors and Variable Frequency Drives

Power = Torque x what?

Once you look in your book and find out what 'what' is, then look up for me the relationship between frequency and 'what', and you will have your answer.

Generally speaking, the load is circuit element placed in your network. So the fan is the load, not the torque. The load of the fan will change as the fan draws more or less power depending on the speed of the fan blades, as said by Russ below.

 Quote by Russ Fan horsepower is a cube function of rpm, so in my applications, a small increase in rpm makes for a big increase in amperage/wattage.
If you have the same load, I don't care what it is or what you call it, its going to use the same power. You can make the load a black box. As long as the value is the same, its effect on the network is going to be the same, period.

I want you to tell me the answer, because you already know it but are not seeing it.
 P: 16 Power = Torque x rpm I agree that I am just not getting something. Doesn't the speed at which the fan turns dependent on the speed of the rotor...which is dependent on the AC frequency? Power just determines how quickly the fan will accelerate, right? Okay, here is a real world example. A battery pack feeding DC current into a pulse width modulation inverter, which then feeds three-phase AC into a three-phase four-pole AC induction motor. I set the inverter to an output of 12V at 60Hz. The rotor will accelerate until it reaches synchronous speed, which at 60Hz with four poles is 1800rpm. Now I add a fan onto the motor shaft. Once again I set the inverter to 12V @ 60Hz. The rotor accelerates but does not reach synchronous speed because of the torque needed to spin the fan, which causes slip. Lets say it reaches 1750rpm. Same setup, but now I set the inverter to 12V @ 120Hz. The rotor will accelerate up to, say, 3500rpm. In both instances, at 60Hz and 120Hz, the voltage is the same so the battery pack should provide the same amount of running time. Only in the 120Hz setup the fan will be spinning twice as fast. Is that right? Am I missing something?
 Sci Advisor P: 5,095 I think your assumption that at the two frequencies that your running time will be equal is false. The speed is controlled by the frequency of the rotating field. The current draw is going to dependent on the motor construction. Every motor should have some sort of load curve. The equation you reference that is a function of frequency and # of poles assumes the motor is running at rated load.
Mentor
P: 22,312
 Quote by JonasS Power = Torque x rpm I agree that I am just not getting something. [snip] Power just determines how quickly the fan will accelerate, right?
Where in that equation does it say anything about acceleration? If you disconnect a fan from a motor it does slow down and stop.... there is a load on it.

Take a step back and think about it. Its a fan! You know how simple a fan is, you're just thinking too hard. If a fan is spinning slowly, there is almost no air resistance because there is no air movement over the fan. Ie, at low rpm, there is almost no torque required to spin a fan. At higher (but constant) rpm, the air provides a higher, constant torque on the fan.
 Doesn't the speed at which the fan turns dependent on the speed of the rotor...which is dependent on the AC frequency?
Right. If you want, you can use different-sized pulleys to make the fan rotor spin at the same rpm while the motor rpm doubles. What happens to your load if your fan rpm hasn't changed...? Has the rpm and torque at the fan changed? What happens to your motor if you gear it down? Has the rpm and torque changed there?
 P: 16 My question isn't about the fan. My question isn't about gear ratios. My question is about AC motors, and I still do not understand what voltage and current have to do with anything if the speed of the motor is based solely on the frequency of the AC power. Let's see if I can make the question even more simple. You have two identical setups: a battery pack connected to a PWM inverter, connected to a 4-pole AC motor, connected to a fan. In one setup you set the inverter to output 12V at 60Hz. In the other setup you set the inverter to output 12V at 120Hz. In the first setup, the fan will spin at 1800rpm minus slip (120x60/4 - slip)...so figure 1750rpm. In the second setup the fan will spin at 3800rpm minus slip (120x120/4 - slip)...so figure 3500rpm. The second setup will move more air since the fan in spinning faster, but both setups should drain the battery in the exact same time because they are both using 12V. Is this correct?
 P: 2,535 I would venture to say that both JonasS and those who have posted after have missed something. I can see where JonasS is coming from since he ASSUMES that the only thing changing when a motor is sped up to double speed is the frequency using 120 hertz instead of 60 hertz. - Cyrus: He already knows that power = torque x speed. That's his whole point. He is ASSUMING that the only thing changing is the frequency which is NOT included in the formula. This is why he is confused. - Russ: You've fallen into the same thing Cyrus has. Although I won't argue the whole relationship to air movement vs. power drawn. - Fred: While the current drawn is dependent upon motor construction, the main thing that the current drawn depends on is the load. - JonasS: I'm not sure why you're figuring in acceleration. - Here's the deal: The voltage fed into the motor must increase as frequency increases. This is due to the inductance of the motor. In order to force the same current through the motor at double the frequency the voltage has to increase as well because the inductive reactance goes up impeding current flow. The same thing happens in reverse. If we cut the frequency down to 15 hertz the voltage applied to the motor will have to be reduced in order to keep the motor from drawing too much current due to the lowered inductive reactance. You should now see where the extra power comes from when the speed of the motor is doubled.
Mentor
P: 22,312
 Quote by JonasS My question isn't about the fan. My question isn't about gear ratios. My question is about AC motors, and I still do not understand what voltage and current have to do with anything if the speed of the motor is based solely on the frequency of the AC power.
But in order to know how the motor reacts, you need to know what the load is doing....
 The second setup will move more air since the fan in spinning faster, but both setups should drain the battery in the exact same time because they are both using 12V. Is this correct?
No, it isn't correct. The fact that they are connected to a 12V battery does not tell you anything about how fast they will drain that battery. How fast they drain the battery depends on the amperage they draw from the battery and the amperage depends on impedance of the induction field and the impedance of the induction field depends on the torque that the load is applying to it.
Mentor
P: 22,312
 Quote by Averagesupernova I would venture to say that both JonasS and those who have posted after have missed something. I can see where JonasS is coming from since he ASSUMES that the only thing changing when a motor is sped up to double speed is the frequency using 120 hertz instead of 60 hertz. - Cyrus: He already knows that power = torque x speed. That's his whole point. He is ASSUMING that the only thing changing is the frequency which is NOT included in the formula. This is why he is confused.
I think you are missing something: speed = frequency. The speed of the motor is a multiple of the frequency in an a/c motor. So saying that the frequency is increasing is equivalent to saying the rpm is increasing. Hopefully, JonasS knows that....
 Here's the deal: The voltage fed into the motor must increase as frequency increases. This is due to the inductance of the motor. In order to force the same current through the motor at double the frequency the voltage has to increase as well because the inductive reactance goes up impeding current flow. The same thing happens in reverse. If we cut the frequency down to 15 hertz the voltage applied to the motor will have to be reduced in order to keep the motor from drawing too much current due to the lowered inductive reactance. You should now see where the extra power comes from when the speed of the motor is doubled.
NO!! The voltage is supplied by a power source and it is constant. When the frequency is increased, if the load increases, the amperage increases.

I think you - maybe both of you - need to review how an ac motor works: http://en.wikipedia.org/wiki/Electric_motor#AC_motors
 In operation, the squirrel cage motor may be viewed as a transformer with a rotating secondary - when the rotor is not rotating in sync with the magnetic field, large rotor currents are induced; the large rotor currents magnetize the rotor and interact with the stator's magnetic fields to bring the rotor into synchronization with the stator's field. An unloaded squirrel cage motor at synchronous speed will consume electrical power only to maintain rotor speed against friction and resistance losses; as the mechanical load increases, so will the electrical load - the electrical load is inherently related to the mechanical load. This is similar to a transformer, where the primary's electrical load is related to the secondary's electrical load.
 P: 2,535 It depends on what you're calling the power source russ. I know that the power source the AC drive is connected to feeds it a constant voltage. I don't have time read the link right now but give me about 12 hours and I will. I doubt that the link can tell me alot I don't know about induction motors already. The amperage cannot increase past the design limit at ANY frequency. So where is the extra power coming from russ? I would bet it's voltage which is reflected in more current draw from the source feeding the drive. - Do a little searching on this forum russ. I have at least once explained correctly how induction motors work.
 P: 16 Trust me, I have read just about everything there is to read about AC motors recently, and I have a very good understanding about how they work. Problem is, Super seems to be right. I have just found several articles that talk about keeping the voltage/frequency ratio constant as the frequency is ramped up and the motor speed is increased. While I usually don't trust Wikipedia as a sole source, I even found a few mentions of this that I must have missed in my first pass: From http://en.wikipedia.org/wiki/Variable_frequency_drive: "When a VFD starts, the applied frequency and voltage are increased at a controlled rate or ramped up to accelerate the load without drawing excessive current." I have seen this on several sites, so it seems that while the frequency of the power is used to determine the speed of the rotor, you must increase voltage as you increase frequency. But is this a "virtual" voltage increase just from the controller? I have read a few articles that make it seem that the controller simulates a higher voltage, and that the actual draw from the line (or battery, in this case) stays the same. I don't know this for sure. If you do need to increase the voltage as you increase the frequency, how do you calculate what that increase needs to be for a given frequency? I would assume that information would come from the motor, but is it the controller instead? In my previous example, how do I know how much power the motor is pulling from the battery at a given frequency?
Mentor
P: 22,312
 Quote by Averagesupernova It depends on what you're calling the power source russ. I know that the power source the AC drive is connected to feeds it a constant voltage.
Huh? If you know that the power source is at constant voltage, then what is there that depends? A typical vfd is fed from a 480v circuit and always feeds 480v to the motor.
 I don't have time read the link right now but give me about 12 hours and I will. I doubt that the link can tell me alot I don't know about induction motors already.
Well....
 The amperage cannot increase past the design limit at ANY frequency. So where is the extra power coming from russ?
Huh? No, again. As load is added to the motor (regardless of whether the load is increasing due to increaing the frequency - maybe the pump is just getting clogged with mud), the amperage will increase until the motor physically burns up unless you have safety devices to prevent it. That's what causes motors to fail! [see below]
 Do a little searching on this forum russ. I have at least once explained correctly how induction motors work.
Eeek. Then I must have missed it and I'm sorry. You're really, really wrong.

I spent my day today doing startup services on a central chilled and warm water plant at a manufacturing plant. The controls are not in place, so we had to start up the plant manually, opening and closing valves and adjusting the vfds to get the water flow we needed. The VFDs are fed from and feed 480v, 3phase power - the most common type of power for such applications.

The VFDs have data points for the specifications of the motor they are connected to. They monitor the amperage of the motor and will protect it from overamping should something happen to the motor (ie, if the pump clogs and overloads the motor). If that information is entered wrong, you can manually drive the vfd up past 60hz (120 is the typical limit of a drive's capability) and the amerage will increase as a cube function of frequency/rpm in a fan or pump.
 P: 2,535
P: 2,535
 Quote by russ_watters Huh? If you know that the power source is at constant voltage, then what is there that depends? A typical vfd is fed from a 480v circuit and always feeds 480v to the motor.
My point was defining what you call the source. I think we are in agreement that the 480V 3 phase line feeding the VFD is the source. No arguement there. However, my argument is the "...always feeds 480v to the motor..." part. It cannot do that at lower frequencies because the lowered inductive reactance will cause excessive current draw even at light to no load and the way it is remedied is to lower the voltage. I'm pretty sure that the voltage has to change at higher frequencies where the higher inductive reactance impedes current flow to the point that there isn't enough of a magnetic field to turn the rotor. However, I'm NOT saying that you can get 2 HP out of a 1 HP motor simply by doubling the frequency and voltage.

 Quote by russ_watters Huh? No, again. As load is added to the motor (regardless of whether the load is increasing due to increaing the frequency - maybe the pump is just getting clogged with mud), the amperage will increase until the motor physically burns up unless you have safety devices to prevent it. That's what causes motors to fail! [see below]
You have no argument from me that as the load increases the current will go up. Please do not imply that since in my first post I pointed this out specifically to fred. I probably did not correctly phrase what you quoted. The current can certainly go up but should not be allowed to run past the design limit for any length of time at all. It is not uncommon at all to have motors develop several times their rated horsepower intermittently. In order to do this at frequencies above 60 hertz I don't see how the voltage can remain the same as it is when the motor is running at 60 hertz. It may vary from motor to motor.
P: 5,095
 Quote by Averagesupernova Fred: While the current drawn is dependent upon motor construction, the main thing that the current drawn depends on is the load.
Obviously. My point was that each motor with the same load will draw differet currents based on their construction.
 P: 2,535 http://en.wikipedia.org/wiki/Variable_frequency_drive Specifically the part that is about half way down the page and around the schematic. They talk about volts per hertz. On a 460 volt motor designed to run at 60 hertz this would be 7.67. For every hertz that the frequency is adjusted the voltage is also adjusted by 7.67 volts. Voltage is reduced to half when the motor is running at 30 hertz for instance. They don't specifically mention increasing the voltage above 460 volts when running above 60 hertz. I suppose it isn't necessary if you simply want to spin something up faster that doesn't require the torque that would otherwise stall the motor at higher frequencies.

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