Where is the truth ??

leitmotif

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Join Date
Nov 2004
Location
Seattle Wa. USA
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With respect to motors controlled by a VFD

Between O RPM and Baseline
Torque "capacity" motor is equal to that of rated torque.
Voltage is proportional to frequency ie 240/60 or 480/60
RPM is proportional to frequency.

ABOVE baseline -
(let us assume we double baseline speed to keep math simple)
Voltage is held constant
RPM doubles
Torque drops as the inverse of speed change squared ie is now at 1/4 of rated.

IF you crank the data thru HP = (T x RPM) /5252 then HP should halve (2 x 0.25 = 0.5). However manufacturers state that this is constant HP range.

I dont understand this disparity. Can someone explain this to me please?


Dan Bentler
 
imagine torque as the motor current and use KW to calculate motor power. HP and KW is 2 unit to represent power, 1hp= 746w or so...

Let say you ask for a 100kw 60hz 1800 rpm nominal. with around 390lbs/f of torque.
at 120hz if your ,manufacturer was right, you still get 100kw but with 195lbs.......
Let see if the sailesman got the truth!!!

Let check this out:
P = Square-V / R or V2/R Everyone will agree with that basic electric law!
At 60HZ nominal 480x480 / 100000w = aprox 2.304ohms
At 120 hz, V doesn't change but R is the sum of resistive and inductive impedance.
resistive part remain the same and is is usually way lower than inductive resistance.
IMP = 2''PI''F L OR in other word, motor R alsmost double at 120hz meaning the amperage will drop at the same speed as frequency rise.
Assuming in the 2.3 ohm, the 0.3 is resistive so we will get a 4.3 ohms at 120hz.

(480x480)/4.3 = 53.6 kw which is close to 50% of nominal power and it will give the same results with HP units
 
Dan, your problem with the numbers is the formula you are using. Continuous torque drops off above base speed by the inverse of the overspeed, not the square of the inverse ratio. Thus the available hp stays constant. Of course, at some point typically around 90-100 hz, the torque starts to fall off even faster causing the hp to fall off as well. For this reason, unless you have a motor specifically designed for extended overspeed, you should probably not deliberately design a system for more than about 90hz max.

The short-term overload torque does drop off by the square of the overspend ratio.
 
Went back to the publication that has started this confusion. It is not the best written for student understanding. In his case I am going to call it an error.

Checked factory web sites and literature. They do not call out for a formula but everyone says constant HP over baseline for 3 phase motors.

No one else including my source makes any distinction between torque at 100 or 150 or >150 baseline. Thank you Dick. If you have any referances for future or more in depth study that would be great.

What I am going to do is trust multiple resources and believe
1. torque falls off as inverse of speed increase over baseline above 100 to 150 percent baseline.
2. Assume torque falls off more at 150% and more of baseline maybe be conservative and use inverse square of speed increase ratio.
3. Assume overload torque falls off above baseline proportinal to square of inverse speed increase. Probably too conservative I know.

Never was too wild about operating motors above baseline just because of the diminishing torque issue. Someone will always try to "hog" and create problems.

Dan Bentler
 
I'd say your assumptions are all good for commodity and ordinary inverter-duty motors. The exception would be motors designed for extended overspeed which will hold their constant hp rating out to double speed.

Check out the specs or nameplate data for a Marathon Black Max motor for an example of extended rage design
 

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