Excellent information John! I learned something today. I may have to go with a larger motor and VFD for my mill if I decide to remove the variable sheave belt drive.
On 1/7/2016 4:10 PM, John Kasunich wrote: > Actually, the domestic (input) power is not an issue. The only factor is > the cost of the oversized motor and VFD. Anyone lucky enough to > find a really good deal on a motor/VFD rated significantly more than > the nominal HP of their machine can take advantage of it to get more > torque at low speeds. > > Think about what a VFD does: > > The rectifier section of the VFD converts fixed frequency and voltage > input power into fixed voltage DC power and stores it in a capacitor bank. > > The inverter section of the VFD converts fixed voltage DC power from > the capacitor bank into variable voltage variable frequency AC power > and sends it to the motor. > > The power delivered to the motor contains two components - real and > reactive power. The real power is converted to mechanical power at > the shaft. The reactive power provides the magnetic fields needed to > make the motor work. The reactive power sloshes back and forth > between the motor phases and the power line (if across-the-line) or > between motor phases and the cap bank (if VFD), with no net power > transfer. > > Lets say the original motor on the machine was 2HP, three phase. > Connected across-the-line on a big industrial 3-phase line, and loaded > to 2HP, it might draw 6A - this reflects both real and reactive power. > Uncouple the shaft, and the no load current reflects only the reactive > power. Typically 30-50% of full load current - let's say it is 40% for this > motor, so 2.4A. So the reactive power required or this motor is > 2.4A*240V*sqrt(3) = 1kVA. This is constant, regardless of the load > on the shaft. > > Put the 10HP motor across that same big industrial line, and load it > down to 10HP. The full load phase current might be 30A. No-load > current might be 40% of that, or 12A. The reactive power requirement > is 12*240*sqrt(3) = 5kVA. Again, it is constant regardless of load. > > If you had a three phase 6A power line capable of running the 2HP > motor without a VFD at full load, it couldn't even run the 10HP motor > at NO load, because it couldn't supply the 12A of reactive power. > > But with the VFD things change. Since the inverter provides all the > reactive power, the line never sees it. The line and the rectifier section > only have to handle the real power. So the 10HP VFD can spin up > the unloaded 10HP motor, while drawing next to no power from the > 2HP power line. > > If you are running at rated speed (say 1500 RPM) and start to apply > load torque, the line will have to supply the real power. When the > torque reaches 7 ft-lbs, the shaft power will be 2HP. At that point > you need to stop increasing the torque, or you will overload the line. > > But the motor and drive were both designed to deliver 10HP. That > means the motor COULD deliver 35 ft-lbs of torque at any speed > from zero (with adequate cooling) to 1500 RPM. > > If you drop the speed from 1500 RPM to 300 RPM, you can increase > the torque from 7 ft-lbs to 35 ft-lbs and the real power is still only 2HP. > Since the real power is 2HP, the power line can deliver it. An ammeter > on the single phase input line might show 8-9 amps, but the same > ammeter on the motor connections could show 30 amps. > > This is rarely ever done in industry because industry doesn't want > to pay for the extra iron, copper, silicon, and capacitors needed > to build a 10HP motor and VFD when they're only going to get > 2HP from the shaft. But for a one-off where you find the motor > and/or VFD cheap on eBay or a scrap-heap, it can work very well. > ------------------------------------------------------------------------------ _______________________________________________ Emc-users mailing list Emc-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/emc-users
