Monitoring VFD Power

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211214-0850 EST

Output HP = output Torque ( in #-ft ) * RPM ( in rev per minute ) / 5252

Is torque proportional to input current? Possibly roughly. Sometimes fairly close, and in other configurations grossly in error.

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MOT RPM@HZREV /HZTOTAL GEAR REDRQD TORQUE IN FT/LBRQD HPRQD KW
1765.0​
60.0​
29.4​
1​
74.39​
25.00​
18.5​
1588.5​
54.0​
74.39​
22.50​
16.65​
1500.3​
51.0​
74.39​
21.25​
15.72​
1470.8​
50.0​
74.39​
20.83​
15.42​
1176.7​
40.0​
74.39​
16.67​
12.33​

Given line one is exact requirements. (We know it's not.)

The speed reductions via the VFD would give me the required HP at that RPM if torque is constant?
 
MOT RPM@HZREV /HZTOTAL GEAR REDRQD TORQUE IN FT/LBRQD HPRQD KW
1765.0​
60.0​
29.4​
1​
74.39​
25.00​
18.5​
1588.5​
54.0​
74.39​
22.50​
16.65​
1500.3​
51.0​
74.39​
21.25​
15.72​
1470.8​
50.0​
74.39​
20.83​
15.42​
1176.7​
40.0​
74.39​
16.67​
12.33​

Given line one is exact requirements. (We know it's not.)

The speed reductions via the VFD would give me the required HP at that RPM if torque is constant?
Correct.

If torque is constant, then motor output power is simply proportional to speed.

Jon
 
Tried explaining why I'm monitoring power and not amps to manager. Because I want to wasn't good enough.
Current will vary based on load but can also vary with voltage too. The VFD is controlling both so generally current will follow torque, but only up to the full voltage available at the line side. If that dips, it can change things. So by using power instead of current, it is always a true representation of the actual load on the motor shaft. By reading power, and you will know the RPM from the drive, you can always calculate torque if you want it. But really, you are after the information on what the leg is capable of under the different product variations. Shift power tells you exactly that. Current is more indirect.
 
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