On Thu, Jan 7, 2016, at 03:56 PM, andy pugh wrote: > On 7 January 2016 at 20:49, Stephen Dubovsky <[email protected]> wrote: > > The lack of torque multiplication by removing a variator/gearbox can be > > partially solved by just using a bigger motor. > > Not really an option on 240V domestic power. > 3HP is almost certainly enough, with the reduction gear for big work. >
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. -- John Kasunich [email protected] ------------------------------------------------------------------------------ _______________________________________________ Emc-users mailing list [email protected] https://lists.sourceforge.net/lists/listinfo/emc-users
