2 kw vfd on 5.5 kw motor
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  1. #1
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    Default 2 kw vfd on 5.5 kw motor

    hi there! I am trying to run a three phase motor that as the title says is 5.5 kw but i only have a 2 kw vfd.So my question.. will it start? i dont want to run at its full potential i would like to run as a 2 kw motor till this power i mean..i previously have run tis vfd with a one phase motor without connecting the third phase and on my surprise it run and it run for around a year without anything bad happening...so please help me there..

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    Probably the biggest problem is starting. The motor requires 5-6x full load amps to start. The VFD will not be able to supply this. The soft start MIGHTget it turning if the motor is totally unloaded but I would not count on it. Same thing with running. A 5.5 KW motor has different characteristics from a 2 KW. I seriously doubt the motor would supply any usable torque.

    Tom

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    Program the vfd for 60% of the motors nameplate volts per hertz.

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    TDegenhart thanks for your reply i will try it though cause it is very difficult to get a motor or a 5.5 vfd...i think my power supply cant even handle it..
    johansen the motor from my lathe has no label on it as regarding volts kw etc. i learned from a similar lathe that it must be a 5.5 kw...but in any case will it fry anything?i mean destroying the vfd or the motor?

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    I don't think anything will be damaged, the drive has built in protections.

    Tom

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    I would just adjust the ramp up on start until the VFD goes OL then back off (increase) the ramp time by about a third. How to look at it is a 1kw vfd could run a 10kw motor at no load if it drew less than 1kw. Just ramp it up and down slow enough to stay within the drives current ratings.

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    Are you trying to use the VFD to convert single phase to 3 phase? I ask because 2.2KW is the typical cutoff for every manufacturer. However many drives like the Hitachi SJ200 will run on single phase power if oversized 200%. this compensates for the much larger ripple current fro single phase and the fact that only 2/3 of the DC bridge rectifier is carrying all the load.

    We run a 3KW motor continuously driving a pump from a 2.2 VFD because the pump only requires 1 hp or 0.75Kw most of the time. Once a month we run it at 2.2 KW for 4 hours or so during a cleaning cycle.

    A lathe is just about worse case for an undersized drive because of very high load inertia both accelerating and decelerating. So your acceleration and decel both will have to be very slow. On accel you will easily trip overcurrent and on decel you will trip on excessive DC bus voltage. A braking resistor will help a lot but you lathe is going to be a slug.

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    I have two cheap Hung yang style VFDs. they were sold as having terminals for braking resistors. One manual admits the terminals are not connected internally the other manual seems to say they are brake terminals. A closer look shows that neither one actually has any wire or circuit traces to the terminals. So there is no way to attach a brake resister without opening up the circuit board and adding wires to somewhere if that is even possible.
    Bill D.

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    KW is the wrong question. Look at current capability.

    The motor will have a full load amps rating, the VFD should have a maximum current rating, plus a rating at 150% and maybe 200% rating, both of which are short term, the 150% is often a 1 minute or so rating, and the 200% may be as short as just a couple seconds.. Those are typical rating points, but the limit is a "current-time" limit that can be tripped by any prolonged current over the continuous limit.

    You are not likely to get a VFD rated at less than about 40% of the motor full loaf amps to really even spin it, because the motor takes about that much at idle (at a very low power factor so low watts), and there is bound to be some drive system friction that needs more current/power, plus you do have to speed it up, drawing more current.

    If you mess around with the accel time, you may get it going. The VFD usually does NOT have to supply the 5x current at start, because it brings up the speed slowly, and does not supply full voltage to start.

    There is a balance point between a too slow start, and a too fast one. Fast does pull extra current, in fact ALL acceleration does. But the 150% short term can carry you through that if not too fast.

    If the accel is too slow, you may time out by being within the current limit, but drawing it for too long. So sometimes a faster accel will not go over current, and will last a short enough time to avoid a timeout problem.

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    First of all i would like to thank aain everyone for taking the time to help me out! I need to run a three phase motor but in my workshop we have a single phase power supply. I dont know if the motor is exactly this cause it doesnt have a label on it...but i have seen a similar lathe same model so i guess they have the same motor.. you can see this lathe i am talking about in this thread. UMA 17 lathe i will try to get more info onthe motor and maybe more photos and stuff during the weekend and i will come back! i may also try to start it with the vfd and see if it does really spin...i think this lathe has an oversized motor for my uses and a little less power wouldnt hurt..

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    JST is right and that's why i said to program the vfd for 60% of nameplate volts per hz. your motor is probably a 240v 60hz motor but there is a small chance its a 200 volt motor, or a 200v 50hz motor

    anyhow, program the vfd for the square root of the ratio of hp.

    2kw/5.5kw is 0.36 square root of .36 is .60 so that's why i said 60%.

    you can go a bit higher than that because most vfds can deliver more current than the line amps of the motor its rated for.

    end result is you can get 2KW shaft power from the 5.5 kw motor and the motor will be more efficient (because its larger) than if you had replaced the 5.5kw motor with a 2KW motor.

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    Quote Originally Posted by johansen View Post
    JST is right and that's why i said to program the vfd for 60% of nameplate volts per hz. your motor is probably a 240v 60hz motor but there is a small chance its a 200 volt motor, or a 200v 50hz motor

    anyhow, program the vfd for the square root of the ratio of hp.

    2kw/5.5kw is 0.36 square root of .36 is .60 so that's why i said 60%.

    you can go a bit higher than that because most vfds can deliver more current than the line amps of the motor its rated for.

    end result is you can get 2KW shaft power from the 5.5 kw motor and the motor will be more efficient (because its larger) than if you had replaced the 5.5kw motor with a 2KW motor.
    wow! tis info was really well explained!! thanks alot again! I will try it and reply here with the results!

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    I would look at it a little differently. You have magnetizing current and torque producing current in a motor. You can determine magnetizing current pretty close by running a motor unloaded and see what the amps are. 40% of full load amps +/- is fairly typical for magnetizing current.

    I believe that running a much larger motor on a VFD that you use up a lot of the available amps producing magnetizing current and leave little left for torque producing current. That would certainly be the case if it was across the line and I don't believe it will be different on a drive.

    This is why amps is not a good indicator of motor hp until you get up close to full load amps.

    As far as starting torque, you control the starting amps and torque by the amps out of the drive. The 600% starting current does not apply on a VFD. As stated before, you can extend start time if you have enough torque to get it moving.

    You actually can get more starting torque from a motor on a vfd than across the line. On a VFD the motor is always running to the right of the speed/torque curve so you can get close to breakdown torque for starting versus locked rotor (starting) torque - breakdown torque can be substantially higher than starting torque. But you need to have adequate amps out of the drive to take advantage of the higher starting torque capability.

    Bottom line, I seriously doubt you well get much usable torque out of the motor on a 2.2 kw limited drive - certainly not 2 kw.

    PS - I re-read the posts and see that JST covered much of what I said.

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    I ran this by my buddy who is a NEMA sized product manager for a leading motor manufacturer - best motor product manager I know.

    This question made him stop and think - but at the end of the conversation he agreed with me. He does not think you will have much, if any, torque available out of the the motor running it on a 2 kw drive. He checked and the range of magnetizing current was 30 - 50% of full load amps for his motors. The smaller motors (like 5 kw) were on the high side up to 50%.

    I will be very interested to see what your results are actually running the motor to see if my conclusion is correct or not.

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    Running at a lower V/Hz as suggested by johansen can lower the magnetizing current (which is voltage dependent), but will not really hurt the amps much, particularly since the ":driving voltage" remains the same high bus voltage of the VFD

    May let you do better.

    If you run at full nameplate volts, you will not have much if anything left to supply shaft power.

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    in this case, somewhere around 60-70% of nameplate volts per hz is probably optimal to get the most torque out of the motor before reaching the vfd's current limit.


    but for a smaller vfd such as a 1kw drive, and a 4kw motor, wire the motor for 480 volts instead of lowering the vfd voltage. you should still be able to get 1kw out of a 4kw motor in such a case.

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    Hi
    I have a Yaskawa V1000 ba0018 3.8kW 1ph to 3ph VFD connected to a 4kW motor. This is the largest purpose built 1ph VFD I could find. It works very well.


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