3phase industrial motors, anybody have experience with them? - Page 1
Forum section: electronics / beginners

author: V_King
author_rank: Regular Contributor
timestamp: September 12, 2016, 12:19:38 pm
content: Hello, I will be buying a 2.2kW 3 phase electric motor with variable frequency drive to run some tests for the project.I have two options of either 2pole or 4 pole electric motor which with correct worm gearbox produces me the same output RPM and torque (theoretically). 4 pole motor configuration is bulkier and costs more.The torque requirements from el motor will be varying as well as the frequency it will be run at (mostly between 15-40Hz).I wonder if anybody could tell mefrom experience, what are the benefits and drawbacks of 2pole vs 4pole motor for my application?I called a few local suppliers and it seems to be hard to get an objective answer from the tech/sales people.

author: Berni
author_rank: Super Contributor
timestamp: September 12, 2016, 01:12:44 pm
content: The number of poles on a AC motor determines the speed it runs at. For 50 Hz a 2 pole motor would run at 3000 RPM max while a 4 pole one would only run at 1500 RPM maxThe whole thing is the same as a gearbox. You are trading off speed versus torque. So while you only get half the speed out of a 4 pole motor you do get twice as much torque. This means when you VFD your motor to the speed you want the 4 pole motor will be only half as stressed as the alternative when used to drive a given machine.

author: tatus1969
author_rank: Super Contributor
timestamp: September 12, 2016, 02:12:55 pm
content: you could approach this from this side: A faster running motor / gearbox combination is likely to have a lower efficiency and lifetime. If that is acceptable for you go for 2 poles, otherwise spend the additional cost for 4 pole.p.s. when comparing electric motors of a specific construction type, a rule of thumb is (very roughly, of course ignoring VFD and mechanical losses): motors of equivalent volume or weight have roughly the same output powerand efficiencywhen operated at the same speed. The idea behind this is that they roughly have the same total winding cross section. As the efficiency mainly relates to the produced torque (idea: copper losses dominate), you can add another estimation: if you double output power by doubling RPM (less poles and/or higher VFD frequency) you double output power (rated torque is constant, power = torque * rpm), while basically preserving the same efficiency. In other words: the higher RPM the better. For example: the tiny 1000 grams BLDC that I use in my 1:8 RC monster truck develops > 3kW (!) peak power, because it runs at ridiculously high RPM. Sounds like a dentist drill. Power to weight ratio of this beast is almost that of a Formula 1 carFor your situation this means: if you have both 2 and 4 pole motors of same size, and let the VFD run them at the same speed, then they behave roughly the same in terms of efficiency.It also means: for same motor sizes, the 2 pole motor is better because it can run at twice speed. And to use all of the frequency capabilities of your VFD. Could be even that you come out using a much smaller motor then.

author: JS
author_rank: Frequent Contributor
timestamp: September 12, 2016, 02:41:51 pm
content: If you are changing the gearbox for each case, so both would run at the same frequency from the VFD, you may be better going for the 2 pole, as the fan of the 4 pole would run too slow to cool it properly at 15Hz. External cooling is the other option, of course. As rule of thumb you don't want to drive a motor close to full load lower than half it's intended speed without external cooling, with the 3000RPM ones you can get away with a bit more. If you end up lowering the speed to half because you started with a 2 pole, you are better off with the 4 poles for start.Here we only have 4-pole with 30:1 gearbox and then adjust the final speed with the ratios for the chain drive. Except for 2 machines we need to run at 3krpm, where we have 2 pole motors. We also have VFD for the feeding machines (4 poles, 30:1 gear, 2:1 chain drive), vectorial for the bigger which is running all day, that helps. With this we get away having just a few spares, as most the motors are alike.For production you may be better with the 4 pole overall, as if you need slower speeds better start with a slower motor. For your tests the 2 pole may be more practical, but will depend on what you are testing. If you are particularly interested in how the motor behaves at different frequencies you are better with the final one. If you are interested in get up an running ASAP to test the machine being driven the 2 pole may bring less headaches as is smaller, lighter, cheaper, may run cooler, etc.JS

author: tatus1969
author_rank: Super Contributor
timestamp: September 12, 2016, 04:16:41 pm
content: another aspect: if your load requires significant torque at low speed, or (worse) standstill, and if you dont want extra expenses in form of velocity/rotor angle sensors, choose 2 pole motor because this extends the low-velocity range that the VFD can handle. That is a difficult task for it because there is no direct way to measure/control torque for AC induction motors.

author: V_King
author_rank: Regular Contributor
timestamp: September 13, 2016, 06:38:51 pm
content: Thanks for all the answers. All the calculations and logic thinking was leading towards 2pole motor with an added benefit of lower price. I just wanted to see what would be a reason to pay more for 4pole motor as I might have missed on something. But it seems that 2pole motor is the way to go

author: tautech
author_rank: Super Contributor
timestamp: September 13, 2016, 07:32:47 pm
content: One thing that has not been offered is noise, IME the lower the RPM of a motor the lower the noise.VFD's do offer soft start, be sure to match it well to the expected start currents of any motor you select.

author: R005T3r
author_rank: Frequent Contributor
timestamp: September 14, 2016, 07:37:30 am
content: 4 pole costs more because there's more than double the work you have to do for a 2 pole motor: did one myself at school (took me 2 months) and lot of copper. Regarding the speeds: lower you go more it costs...Actually if you are running a motor to 15hz be aware that you will need to cool the motor otherwise you will set it on fire. Also, In my opinion, you shouldn't make it work at such speeds: asynchronous motor are not designed to work in these conditions: they are designed to be reliable and cheap, but not to be precise. The worm gearbox is one of the most inefficient items out there, you should consider an inverter: works better. There are other options available, such as DC motors: you can fully control their speeds and they are also pretty linear.If you want a really on-time motor, you can always use a synchronous motor + inverter: the torque can also be fully adjusted at any given frequency. Not to mention that can be used as a generator if you want..

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 14, 2016, 08:56:38 am
content: Note that you only get nameplate power at nameplate voltage, frequency and current.Voltage and frequency, current and torque must all change proportionally. So in general, the power drops quadratically.For the same reason that a 240V 2A 50Hz transformer is only good for 120V 1A 25Hz: if you try using it at the higher current level, the voltage drop (or for an induction motor, the slip) will be undesirably high.High slip might not be a mechanical issue, but the RPM loss comes with extra heating in the rotor -- as R005T3r said, trying to load it heavily at low speed isn't recommended!Because of the steep power reduction curve, VFDs are best suited to proportional loads like fans. Or if you're running a machine tool, remember to take very light cuts with low motor speeds!If you need a lot of power at any RPM, your best bet is a transmission. The VFD can be used to interpolate between gears, say in +/-30% steps, so that the power reduction is minimal.(By the same token, you canincreasemotor power, by raising voltage and frequency above nameplate! But this increases core losses, so again the real potential isn't proportional, but goes more like sqrt(x).)Tim