Calculating Motor Shaft Torque/HP when using VFD

[toygasm4u]

I may or may not have a delimma of sorts. I have a large rotating mass coupled to a gear reducer, hanging vertically from a ceiling. In the mounting structure, I have a 12,000 in.lb. torque cell. Assuming that the torque cell is properly calibrated (2 point scaling method used @ 150 ft.lbs max), I use the measured counter torque on the mounting structure to calculate the Motor HP required to drive the load. I've been using the following formula:

Motor HP= (((Output in.lbs.)/(Ratio*Eff.))*(Output RPM*Ratio))/63,000

Using this formula in an example with the following variables:
Motor NP HP = 1.5
Motor FLA = 2.8
Motor NP RPM = 1750
Reducer = 21.67 : 1
Reducer Eff. = 0.97

Measured Output Counter-torque = 389.1 in.lbs
Measured Output RPM = 83.5
VFD Output Frequency = 62.0 Hz
Measured Motor Current = 2.34A

When I plug the measured torque and speed into the above formula, I get a Motor Shaft HP of 0.532, but I also see that at 60Hz, my motor is loaded at 84% based on FLA.

For a 1.5HP motor loaded at 84%, my calculated value of 0.532 seems to be too low. It's making me feel funny about making assumptions for future testing based on these numbers. Would you guys disagree with my method?

Is there an alternate "electrical" method of calculating actual motor HP output, when Current, RPM, NP eff. are known, but voltage is not? I would naturally want to assume there isn't, since there's no good way to calculate power consumption on the motor leads because of PWM voltages.

I have a scope. I have a fluke 43B PQ analyzer, but I question accuracy because of the PWM content.

Help? Me? Thanks in advance guys (and gals)

 

[CJCPE]

If the proper voltage is not applied to the motor, the manufacturer’s data would not be accurate. Operation with the VFD also introduces the problem of not knowing what the optimum voltage should be for the applied frequency. The proper voltage is generally proportional to the applied frequency (V/Hz=constant=460/60=7.67) but some “voltage boost” above the constant V/Hz is required at low frequencies. At 62 Hz, you probably don’t have the “proper” voltage because the VFD puts out a constant voltage above 60 Hz.

Another way to estimate torque is to assume torque is proportional to slip. Since nameplate RPM is 1750 and synchronous RPM is 1800, slip at full load is 50 RPM. At 62 Hz, synchronous RPM = 120 x 62 /4 = 1860 RPM. Operating RPM = 83.5 x 21.67 = 1809 RPM and slip = 1860 – 1809 = 51 RPM. That would indicate the motor is operating near full load. Above 60 Hz, “full load” would be considered to be constant Hp and decreasing torque up to perhaps 90 Hz. However, slip will increase as frequency increases.

At 1-1/2 Hp, I would assume the VFD is a relatively simple design. More sophisticated VFDs can provide a pretty good load indication based on internal measurements.

There are power measuring devices that you can put in the motor lines like: Load Controls Inc. and Emotron . I believe that both of these devices give good results with VFD waveforms.

See also thread237-167751

 

[CJCPE]

Oops. I wrote my reply off line and didn’t paste the first paragraph:

An AC motor draws a significant amount of current at no load. You can probably get manufacturer’s data for your motor current at no load, 1/4, 1/2, 3/4 and full load. You can estimate the load by comparing your measured current with the manufacturer’s data and interpolating. A 1-1/2 Hp motor that I looked up had more than 50% current at no load. Using that data, your 84% current figure would indicate 78% load and at 35% load (.53 Hp), it would draw 60% current.

 

[LionelHutz]

I got 0.532hp.

389.1 * 83.5 / 63000 = hp to turn load

Then, divide that by 0.97 to get input hp.

Your formula has a "Ratio" that will cancel out.