Motor Curves

[timm33333]

I have question about torque curves for a low voltage VFD controlled motor. This motor has a 250% overload for 4 seconds after every 10 minutes. Is there some software available by which I can draw the torque-speed and torque-current curves for this motor? Thanks

 

[waross]

You may want to look at The section of the Cowern papers that deals with RMS loading.

http://www.baldor.com/pdf/manuals/PR2525.pdf

.pdf page 49

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

 

[jraef]

But torque-speed curves are based on rated voltage AND frequency only, they don't really apply to VFD driven motors.


"You measure the size of the accomplishment by the obstacles you had to overcome to reach your goals" -- Booker T. Washington

 

[timm33333]

Thanks, the RMS HP method as mentioned in Baldor's document looks very good !

I think there are situations when torque-speed curves are drawn for the VFD operated motors as well, for example there is a torque speed curve for a VFD operated motor on PDF page 4 at the following link:

http://www.cemanet.org/wp-content/uploads/2011/09/CEMAScrewConvStd-351-2007.pdf

On the same PDF page 4 in bullet 4, it says that VFD operated motors can withstand overload up to 150%.

So does it mean that we should apply the RMS HP method for VFD operated motors only when the overload is greater than 150%?

 

[timm33333]

Is the rms HP method only for variable torque applications, or can it be applied to constant torque applications as well?

 

[waross]

On servomotors and other adjustable speed applications, similar calculations are frequently made. In
these cases armature amperes or required torques are substituted in place of horsepower. The resulting
RMS amperes or RMS torque requirement is then compared to the motor’s continuous and peak ratings
to determine adequacy.

Torque loading and torque overloading are related to slip frequency or slip speed and current in the rotor.
Eg: A 1760 RPM motor has a slip of 1800 RPM - 1760 RPM = 40 RPM, or a frequency of 1.33 Hz.
If the volts per Hertz ratio is maintained, the motor will develop full torque safely at any speed fast enough to provide adequate cooling.

Select a motor which has enough breakdown torque to handle the torque overload. The common industrial induction motor is a design B. The breakdown torque is around 175%. If the 250% is based on a 10 HP motor then you will need 25 HP. The 10 HP motor will only develop about 17 HP or 18 HP. A 15 HP motor will develop about 26 HP at breakdown torque. You may want to use a 20 HP motor for the extra safety margin.
But, based on 15 HP, the needed 250% will now be about 167% of normal rated torque. (Note; This selection has almost no safety factor for low voltage situations.)
Now look at the speed/torque curve for the current at 167% torque. If you convert the curve from percent to actual RPM, You may slide the upper portion of the curve down to your operating point.
Example:
Rated speed = 1760 RPM
Normal slip speed = 40 RPM
Slip speed at 167% torque = about 70 RPM.

Slow speed operation @ 50% speed:
New synchronous speed = 900 RPM
Normal rated speed at 50% frequency = 900 RPM - 40 RPM slip = 860 RPM.
Speed at 50% frequency and 167% torque loading = 900 RPM - 70 RPM slip = 830 RPM.
In the example above, a 20 HP motor will only be overloaded 25HP / 20HP = 125%. This would be my choice. You may scale these numbers to your actual HP needs.
To your original question: Convert your speed torque curve into RPM and use synchronous speed and the slip speed to find your operating point on the curve.
Consider cooling for very slow applications.

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

 

[timm33333]

I used current (instead of torque) and replaced HP with current in the formula at the top of PDF page 50, and so calculated Irms.

In the example in the Cowern paper, they calculated HPrms and then selected the next HP size motor; and also made sure that the pull out torque is satisfied. In my case I have calculated the Irms (instead of HPrms). But in my case the motor is VFD operated, so the “drive” means VFD+motor.

So I have several options: should I compare the calculated Irms with full load current of motor (or VFD), or should I compare the calculated Irms with the maximum current of motor (or VFD)?

 

[mikekilroy]

So I have several options: should I compare the calculated Irms with full load current of motor (or VFD), or should I compare the calculated Irms with the maximum current of motor (or VFD)?

Easy answer is: rms is the heating equiv of something, so use rms, not some maximum value.

now for the not so easy answer..... whew.... careful here. You do not seem to realize current in an ac induction motor is not linear with load; not sure how you are using current to calculate.... you want to use torque or hp but not current if you want meaningful answer.

Waross already told you how slip relates to load and current. The fact remains you want to pick the correct size motor. You seem to know you need 250% torque(?) for some time every 10 minutes. So you can calculate rms equiv to pick the right motor.

Once picked, then look at waross reply again and make sure as part B of your answer you also have a motor & drive that do the maximum (250%) peak you require. Two parts to your answer.

http://www.KilroyWasHere<dot>com

 

[LionelHutz]

What exactly are you trying to do?

The motor supplier generally supplies the curves you asked about when the motor has line power applied.

To generalize, the torque vs speed curve when the motor is applied on a VFD will be a horizontal line. A torque vs current curve could be determined by using the torque vs speed and current vs speed curves.

To find the maximum torque possible, start at the motor operating point on the motor curves. Then, apply more load, which requires more torque, which means the motor slip increases and the motor current increases. As you apply this extra load, the maximum toque available from the VFD and motor is either the torque available when the VFD hits it's current limit or the breakdown torque of the motor.

The curves on page 4 of the CEMA document do NOT represent the possible speed vs torque curves you can get when those motors are applied on a VFD. They simply represent the speed vs torque curves when rated voltage and frequency is applied (ie line power is applied).

 

[timm33333]

I now did it using torque: Found rated motor torque and breakdown motor torque from datasheet. Calculated maximum permissible load torque from curve (which is 0.9 times rated motor torque.) Took the actual load-torque, and plugged it in the equation of Cowern paper to find Trms. Calculated torque at 250% overload (2.5 times load torque).

Results: Trms is less than maximum permissible load torque, and torque at 250% overload is less than breakdown motor torque: so the motor is acceptable.

 

[mikekilroy]

Excellent! Now that you know the MOTOR can do it, final step will be to verify your SYSTEM can do it; most generically applied VFDs will be picked to provide motor nameplate RATED (rms) torque. At that, most generic VFDs will provide 150-200% MAX torque over that nameplate value. So verify your 250% required torque (NOW use the amps for this value) will be available from your VFD for your required 4 seconds - probably requires upsizing 1 rating.

http://www.KilroyWasHere<dot>com