Hey Aiph5u, thanks for the reply.
I understand your calculations in the examples perfectly. In fact, we used almost exactly the same rule-of-thumb numbers for power factor and motor efficiency. In this example, you used a Power Factor of 0.82, and a motor efficiency of 85%. Your terminology was much more precise than mine, to be sure, but if you multiply your 0.82 by 85%, you get 0.697 – almost exactly my 70% “system efficency” (loosely speaking).
I didn’t bother with calculating the motor phase current, because it’s not relevant to the system efficiency computation. But I do agree with your calculations in the example.
I think the difference in our conclusions about running a 1.5 kW (continuous mechanical output) spindle on 120V may come down to the efficiency of the VFD, which I had implicitly approximated as 100%.
So I think we agree that for a spindle to make 1500W of mechanical output power, it needs about 2143 W of electrical input power. A 120V, 20A supply can provide up to 2400 W continuous. The three-phase IGBT brushless servomotor drives that I once designed were about 96% efficient, IIRC, and so would need about 2143 / 0.96 = 2232 W of electrical input power to deliver 2143 W of power to the motor. That leaves just enough overhead for a 2232 / 2400 = 0.93 power factor at the input of the VFD. I honestly don’t know if that’s achievable, and it almost certainly doesn’t leave room for even a 5% supply-voltage sag.
So bottom line, I would have to agree that a 1.5kW spindle / VFD system that runs on 120 V does need a supply current of more than 20A in order to deliver its rated continuous output power. Or else maybe a few drops of snake-oil lubrication in the bearings. 