Induction motor running at FLA at reduced speed on VFD 2

[eeprom]

Hello,
I have a fan on a 200 Hp motor (460V / 1750 RPM) on a 200 Hp VFD. The fan has to run at 1430 RPM. The fan will draw 180 bHp at that speed. The questions I have are:

1. Can the VFD produce 180 Hp at 48 Hz? For that matter, can the VFD produce full power at all speeds?
2. If I can get the VFD to produce 180 Hp at 48 Hz, this will require full voltage and full current at 80% rated speed of the motor. This will affect the torque speed curve. Will this damage the motor? Can an induction motor produce rated power at a reduced speed?

Thanks,
EE

 

[davidbeach]

Yes, the right VFD can produce full nameplate torque at zero speed. That's the easy part. The hard part is getting the motors cooled. If you're running a load that looks for near full torque at well below full speed the built in fan won't be providing enough airflow over the motor. Reduced motor life will result. That type of application needs a externally cooled motor so that it can get full cooling at less than full speed.

 

[FreddyNurk]

A somewhat left of field suggestion, does the motor nameplate happen to have 415V/50Hz ratings on it? If it does, and its still within the nameplate at 50Hz (i.e. 180HP) then chances are cooling will be ok, as the motor would effectively be operating within 50HZ specification (albeit a bit lower). Conversely, if it has 50HZ ratings and its below your horsepower requirements, its pretty certain that its not going to be good for the motor.

If not, then what davidbeach has stated is quite prevalent.

I don't get to see too much that isn't IEC specification, so may be way off on what might be on the nameplate, or the size of the motor for that matter.

 

[crshears]

If I can get the VFD to produce 180 Hp at 48 Hz, this will require full voltage and full current at 80% rated speed of the motor.

and

I have a fan on a 200 Hp motor (460V / 1750 RPM) on a 200 Hp VFD.

Perhaps I am misunderstanding the OP...

Is the motor a 60 Hz one being operated @ ~80% of rated speed? If so, will not the V/Hz ratio you are proposing as being the required one be way off spec, leading to the motor maybe letting smoke out?

Plus the behaviour described does not seem to comport well with the load affinity laws, unless the planned fan operating speed has led to its being deliberately oversized so as to match the expected output of the motor [again, if I understand it correctly, the OP states that the plan is to load the VFD @ > 90% of its rating...]

CR

"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]

 

[ScottyUK]

Your motor can produce rated torque at any speed right down to zero. It can't produce rated power at reduced speeds because power is proportional to speed if torque is a constant, or conversely the torque would have to increase as speed drops to maintain constant power. Your motor isn't big enough to do what you require.

The VFD almost certainly can be reconfigured for 50Hz base speed in which case it will do what you need if you sort out the motor problem.

 

[itsmoked]

No 180HP at 48Hz for a 200HP 60Hz motor as Scotty sez.
You would need a:

x_60HzHP = (180*60)/48
= 225HP

Keith Cress
kcress -

http://www.flaminsystems.com

 

[eeprom]

Using Power = Torque x Speed makes it pretty easy to see that you cannot run the motor at full power for all speeds. That's a pretty simple check. Thank you ScottyUK. Otherwise at 1 Hz the motor would be operating at 60 x rated torque.

Thanks
EE

 

[Compositepro]

Also, in order to run full torque at low speed for any length of time will require a separately powered cooling fan for the motor.

 

[ScottyUK]

eeprom,

Another analogy which you might find useful is the gearbox as a mechanical transformer. You're familiar with V_prim x I_prim = V_sec x I_sec. Substitute rpm and torque for voltage and current for the mechanical equivalent. Granted, the gearbox is less efficient than the transformer but it's otherwise accurate.

 

[waross]

Have you considered taking advantage of the fan laws and running slightly below 1430 RPM?

I have always been under the impression that similar rotors were used in squirrel cage induction motors in the speed range from 1200 RPM to 3600 RPM. It has been my understanding that the reduced air flow at slower speed was offset to some extent be the greater heat transfer from the slightly hotter motor. Many integral rotor fans are not very efficient and may not move that much more air at higher speeds.
I would only anticipate serious cooling problems below 1000 RPM.
However, this case notwithstanding, most motor applications demand somewhat less than full rated power and the safety margin is generally sufficient to allow adequate cooling at reasonable speeds below rated speed.
I am open to correction here.

Comments?

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[eeprom]

I'm not convinced that I understand, nor that this cannot work.

If the VFD were set to produce max voltage at 48 Hz, the motor would have to generate some additional torque in order to match the fan load. I don't think cooling would be an issue at 48 Hz.

Exact values are ---
Motor: 200 Hp
Volts: 460V
FLA: 240A
RPM: 1750
Eff: 0.90
PF: .84

Fan load: 181 Hp
Fan speed: 1434 RPM

Torque needed by fan: 181*746/(1434*pi/30)=899 Nm
Rated torque at rated speed of motor: 200*746/(1750*pi/30) = 814 Nm

Disregarding the speed for the moment, if the motor saw full voltage at 48 Hz, the motor would match the fan load at 224A
Power = 1.732*V*I*eff*pf

181*746/(1.732*460*.84*.90) = 224A

So the motor would have to produce 899 Nm, which is only 4% higher than the rated 814 Nm.

It seems to me that the motor would be able to do this work. It would just operate at a higher slip. I don't have the torque speed curve, but I doubt that 4% extra torque would push the motor near the breakdown torque.

I don't see why this won't work. Please note, I am not suggesting this is a preferred design. I'm pretty sure we are going to use a larger motor. I am just trying to understand if this will work.

Thanks
EE

 

[itsmoked]

You are maybe neglecting that when you reduce the frequency the magnetic domains are now subjected to the field longer and hence more domains reach a static state which, in your motor's reality, means the magnetic circuit will find itself saturated. That's why at lower frequencies -everything else remaining the same- the voltage must come down to prevent saturation.

Your drive will either fry or trip on over-current due to saturation.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[FreddyNurk]

waross' understanding is effectively what I was thinking in terms of lower speed and cooling of the motor. I've heard various rules of thumb thrown around for lowest speed of a motor in order to allow proper cooling, obviously its application dependent, and if the motor is directly coupled to the fan shaft (as opposed to some sort of pulley or gearbox arrangement, which would in itself change the torque requirements), then one would expect that the motor cooling would be ok down to a certain speed regardless, particularly if its driving a fan.

I've also heard discussions on whether its better to remove the fan for TEFC motors in fan applications as the impeller is the wrong orientation for the air flow from the driven fan, but never really worried about the details.

That said, 180HP on reflection still appears too high for the motor, although I was thinking the same as waross in terms of fan laws with previous responses.

 

[eeprom]

Itsmoked,
I am certain that the magnetic circuit of the motor will be affected, but it's hard to believe it will be affected enough to smoke the motor. The stator current will be about 90% of rated, the stator voltage will be at 100% of rated, and the magnetizing impedance will be at 80% of what it would be at 60 Hz. Therefore there will be more magnetizing current at 48 Hz than there would be for the same voltage and current at 60 Hz. The magnetizing current will be approximately 120% of what it would be at 60 Hz. Is that enough to saturate the core? I am confident that the motor would be sufficiently cooled at 48 Hz, but I can't say with certainty that a 20% increase of magnetizing current will cause saturation.

Again, I am not advocating this as a good design. I'm just trying to understand why this would or wouldn't work.

EE

 

[LionelHutz]

No, it won't work without overloading the motor. No need to do the torque calculations either since HP is proportional to percentage of motor speed.

Ignoring any loss of cooling, the motor is rated 200hp x 1434rpm / 1750rpm = 164hp @ 1434rpm.

You also can't apply full voltage to the motor at 43Hz or you'll saturate the magnetic circuit which will cause a high current and burn the motor out. The V/Hz ratio applied to the motor has to be maintained. You should apply 460V x 1434rpm / 17650rpm = 377V @ 1434rpm.

 

[waross]

The motor is too small for the intended load. No amount of fun with figures will correct this basic issue.
Running the motor on 480 Volts at 43 Hz will increase the V/Hz ratio by over 20%
Past 10% over voltage or over V/Hz ratio and magnetic saturation is an issue. When a motor's magnetic circuit is completely saturated, burnout may occur in less than a minute, even with no load on the motor.
A VFD maintains a safe V/Hz ratio as the frequency varies.


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[crshears]

Thanks Bill, you stated in much clearer terms what I was trying to get at above.

CR

"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]

 

[waross]

It may be useful to consider overvoltages and saturation in induction coils. Lets consider an induction coil with an impedance of 10 Ohms.
This coil has resistance of 1 Ohm and inductive reactance of 9.95 Ohms.
As the voltage is increased, the current increases proportionately. For each 10 Volts increase the current will increase by 1 Ampere.
Until the magnetic circuit saturates.
When the magnetic circuit reaches saturation, inductive reactance no longer limits increases in current. Further increases in current are limited only by the 1 Ohm resistance. In full saturation, a 10 volt increase in applied voltage will cause a 10 Ampere increase in current.
Another way of saying this is that in this example, once the magnetic circuit reaches saturation, the ratio of current increase to voltage increase increases ten fold.
Transformers are subject to similar limits, but the calculations of the actual current is somewhat complicated by the addition of the load current, which itself may be complicated by the characteristics of the loads when subjected to over voltages.
In motors the calculations are further complicated by the slip frequency, and other factors.
But the bottom line is no amount of calculation will change the color of the smoke when a motor is run into magnetic saturation by the application of over voltage.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[eeprom]

Waross,
This was a good, instructive discussion. As I stated earlier, a larger motor had already been decided upon. The goal of the discussion was not to try to justify the use an undersized motor. The goal was to understand why it would or wouldn't work. I'm certain you are correct about the V/f relationship, and that increasing that ratio would lead to magnetic saturation. A review of the equivalent circuit makes this clear. Thanks again for your thoughts

EE

 

[waross]

"The goal was to understand why it would or wouldn't work."
"This was a good, instructive discussion."
That was my intent. Try to explain the basics as simply as possible. Once you have a grasp of the very basics, the rest comes so much easier.
Thank you for the kind feed back.
Yours
Bill

Bill
--------------------
"Why not the best?"
Jimmy Carter