Motor larger than VFD? 2

[cokeguy]

Somebody here at my plant wants to change a 250 HP motor (driven by a 250 HP VFD) to 300 HP, mainly because we are running the motor to its FLA and sometimes a bit more right now, and since we have an unused 300 HP motor sitting at the shop they reason that if we make the change at least the motor will be running at a safer operating point. I have stated that the 300 HP motor driving a 250 HP load will probably demand more current than the 250 HP motor with an equal load, and therefore the VFD will not be able to supply the current and we´ll end up worse than how we stand right now. Am I right, or do they have a point? We have motor amperage trends and alarms on our HMI, so we can keep a close look at motor amps, but I still think it is a lousy idea. Any comments and experiences most welcome. By the way, this is an I.D. fan controlled by a PID loop.

 

[Warpspeed]

If the 300 Hp motor is only delivering 250Hp, then it should not demand any more full load current than the original motor (once up to speed). It certainly should run a lot cooler.

The larger motor may show a higher starting current, and that could be a potential problem for the VFD. It should be possible to program in a suitable acceleration ramp up rate, and overload current limit settings to keep the VFD within it's maximum ratings. The motor will take care of itself, just be sure the peak currents during acceleration are not excessive for the drive electronics.

 

[jraef]

Well stated by Warpspeed. That really is the only major concern. The VFD will only deliver as much current as it can (max amp rating), after which it will current limit which may mean it will slow down. You also will likely not be able to set the motor FLA to higher than the VFD max amp rating, so the overload Protection will trip earlier than you would expect from a 300HP motor, but that certainly will not hurt the motor.

Also, you didn't mention what kind of load you have and what kind of VFD. Many Constant Torque rated VFDs can be used on larger motors if the load is Variable Torque, such as a centrifugal pump or fan. Of course, if your VFD is already sized asd a VT drive on a VT load, you will not have any headroom in it.

Hey, you already have the VFD, doesn't hurt to try!

Eng-Tips: Help for your job, not for your homework Read faq731-376

 

[cokeguy]

Thanks Warpspeed, we can certainly limit the ramp up rate to avoid problems at startup if we do change the motor, but I still worry because of the following reasoning, an extreme example perhaps but maybe valid:

A 100 HP motor can perhaps demand 110 amps or so when delivering 100 HP, but will demand probably 50 amps or so when delivering 20 HP. Now, a 200 HP motor will demand about 220 amps when delivering 200 HP, but will demand probably 60 or 70 amps or so when delivering 20 HP, more than the 100 HP motor at 20 HP (I´m guessing, just to get my point through, maybe it doesn´t work that way).

Assuming I´m correct, and extrapolating to my case, I would think that if the 250 HP motor is right now demanding about 290-300 peak amps with a certain load, the 300 HP motor will demand perhaps 300-310 or more amps for the same load, and now the VFD could be the one that struggles or even shuts down. COuld this be the case?

 

[Marke]

Hello cokeguy

Provided that your current is well above the magnetising current of the motor, there whould be little difference between the current drawn by the two motors, infact the larger motor could even be less due to better power factor and efficiency.
Typically, the magnetising current will be in the order of 20 - 25% of the motor rating so there should be no problems.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[DickDV]

cokeguy, you are basically correct that the larger motor will draw more current to drive the same load. This is because it takes more current to magnetize the larger motor than the smaller one.

Since this is a fan load and short term overloads for things like fast accelleration are not needed, you need to ask the drive manufacturer for the continuous rated drive output amps at 40 degrees C ambient. Make them understand that you want maximum continuous amps with no overload capability. The same drive will have a continuous amp rating at no overload, a lower continuous amp rating for 10% short-term overload (commonly called variable torque rating altho it has nothing at all to do with load torque behavior), and an even lower continuous rating for 50% short-term overload.

Compare this no-overload amp rating with the FLA on the 300hp nameplate. This will give you a pretty good indication of how close to full load you can get on the 300hp motor. It is likely that you will not be able to get to 300hp FLA and you don't want to.

My calculator tells me that if magnetizing amps is equal to 25% of FLA, then you will reach 250hp on a 300hp motor at .807 times FLA. That's the number you should use to compare to the drive continuous amps. If the drive amps is larger than this .807 number, you will be able to get more hp out of the 300hp motor. If less than the .807 number, you are better off pushing the 250hp motor a bit.

If pushing the 250hp motor is your only option, make some attempt to keep the ambient temperature under 40 degrees C and you will reduce the tendency to shorten the life of the motor.

 

[Warpspeed]

Dick has a point. The drive itself will have at least two different maximum current ratings, the short term maximum set by the semiconductors themselves, and the long term steady maximum available load current, determined by both heat sinking capacity and ambient temperature.

These days with microprocessor control, there will most likely be an instantaneous shutdown due to any excessive (fault) current, and also a thermal sensor fitted to the main heatsink which will shut down the drive in the event of any overheating. These will be inbuilt limits and non user adjustable to protect the drive itself.

The programmable current settings in the user menu will be lower, and intended to protect the motor. I still cannot really see any problems with going to a larger motor, provided startup conditions are not too violent.

If the VFD does spit the dummy and shut down, I am sure there will be a very specific error message to guide you in the right direction.

 

[waross]

Hello DickDV
I'm having some trouble verifying your figures.
Using a magnetizing current of 25% and Pythagoras' theorem I calculate the in phase current to be 96.8% of line current.
Adjusting this from 300 hp to 250 hp I get 80.7% of full load current. But this is the in-phase component. Don't we have to still calculate the line current at this loading? I get 84.5% That is a 300 hp motor with 25% magnetizing current should draw .845 of full load current with a 250 hp. load.

Hello cokeguy;
In regards to your observation that a lightly loaded motor draws a disproportionate ammount of current, your observation is valid. However the results cannot be extrapolated as a straight line relationship.
This can be best explained if you sketch a few triangles on a sheet of paper. Accuracy is not essential to understand the relationships.
There are two components of current in an induction motor.
One component is the magnetizing current. This acts at a 90 degree angle to the applied voltage. This depends on the motor charicteristics and on the applied voltage. For this explanation we will consider it to be unchanging. It cannot be measured directly with a normal ammeter.
The other component is the real power component of the current. This current is in phase with the applied voltage. It cannot be measured directly with a normal ammeter.
The vector addition of these two current results in the apparent power component. This is what we measure with an ammeter and this is the current that heats the wires and this is the current that the VFD must supply.

Draw a vertical line 25 units high to represent the magnetizing current. (To represent 25%).
Remember, this does not change when the other two values change.
The hypotenuse will represent the apparent power. This is the value that we can measure with a normal ammeter.
The hypotenuse will be 100 units long.

The base will represent the real power. This is closely proportional to the motor load but cannot be measured directly with an ammeter. The base will be 96.8 units long. You can see that there is less than 4% difference between apparent power and real power at full load.

Now consider the motor with a 10% load. The altitude of the triangle is still 25 units. The base line has shrunk to 10% of 96.8 units or 9.68 units. The hypotenuse is 26.8 units long.
At full load the current measured with an ammeter is only about 3.3% greater than the real power component of the current.
At 10% load the current measured with an ammeter is 177% greater than the real power component of the current.
I have neglected to consider losses in this explanation. You are probably more interested in the general relationship between loading and current than a precise calculation. And if you consider the losses to be part of the load, as the Watt-Hr. meter does, the explanation becomes much more accurate.
Hope this helps your understanding of why the current will be greater but still reasonable with a 250 hp. load on a 300 hp. motor but will be disproportionate on a lightly loaded motor.
respectfully.

 

[Marke]

By my calculation, if both motors have a magnetising current of 25%, then the current of the 300HP would be higher than the 250HP by 1.3% - a very small increase. I would expect that the magnetising current would not be exactly the same percentage and would lie in the range of 20 - 25%. It is quite possible for the 300HP motor to deliver 250HP at a lower current than the 250HP, the efficiency of the 300HP operating at 250HP would inevitably be higher than the 250HP at 250HP load if they are a similar design. This would likely offset any increase due to a higher magnetising current.
The bottom line is that there is probably going to be negligable difference between the current draw of the two motors at 250HP load if the two motors exhibit similar characteristics. If there are variations in the efficiency and magnetising currents of the tow motors, this will have a far greater influence than the change in size.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[UKpete]

cokeguy, having read all the above posts I can't help thinking - if it works (and specifically if your ambient temperature isn't at 40°C or above) it seems pointless changing the motor. The new motor will have a bit more windage loss (and some extra iron loss) and will have a bit more inertia to accelerate, but I don't suppose it's significant.

You'll just have a spare 250hp motor in the shop instead of a 300hp one.

 

[DickDV]

Waross, you are correct. The number is .847. I forgot to put it back in the triangle and solve for the hypotenuse.

 

[waross]

UKpete has a good point.

we are running the motor to its FLA and sometimes a bit more right now,

Another option is to tweak your control loop settings so that the pid controller does not run the motor over its FLA rating.
respectfully