Lathe motor replacement

A good point but I am struggling to see how to update the diagram without making it over-complicated (especially considering Ian's comment). I guess as a starting point I should state that the isolator should also contain a fuse/protection if there is not adequate elsewhere. That would cover the situation of a fused plug, fused connection unit or switched outlet with seperate MCBs etc.

Perhaps I should also remove the 230V supply to other equipment as where that it taken from depends on what that equipment is, what fuse it needs and whether it needs a separate isolator.

I knew I was going to regret posting that

Hywel, Reading your first post and looking at your pictures - I will take a guess that your lathe used to work by selecting forwards or reverse direction with the black knob and then pressing the green button started it up and the red button turned it off?

If that is the case then you are half way to having everything you need for an inverter. As you mention it is single phase I will also guess that it plugs into the wall socket?

If that is the case, then take the supply from the plug directly into the inverter supply side. Then, rewire the original direction, start and stop switches on the low voltage control side of the inverter. You will need to follow the instructions (there will be a wiring diagram in the manual) for the correct method of linking the two switches which will be "momentary" push button switches and need to be wired to latch (done internally in the inverter by electrickary). You then simply connect the motor directly to the inverter output and program the inverter with the correct parameters - another manual reading exercise unfortunately...

I hope that helps, I have two Boxfords wired like this (although mine do not plug in, they are wired permanently to switched spurs supplied from dedicated MCB's in the distribution board). Take care with the air flow requirements when you box in the inverter!

Mark

Edited By Mark C on 16/01/2017 10:32:40

Just read your manual, you need to read the page about three wire control wiring on page 10-5 (advanced section). I did not read it completely, you will need to check the inputs to confirm that is start, stop and direction switches but it certainly looks like it...

Edited By Mark C on 16/01/2017 10:38:37

Hywel, Rather than another thousand words, here are some pictures of my lathe installations. The VSL has an old inverter on the brick wall at the back but the other Boxford has the same sort of setup that you are doing (I think) with the inverter located in the lathe cabinet behind the control panel. The VSL is different in that it has the original rotary switch wired to the inverter to set direction and then start (the switch is 5 positions, centre off, detent in first position either side of off with a momentary start position if you keep turning the switch which then springs back to the first position - this is how they originally worked but it is all done by the inverter logic now). All my machines have a separate E-Stop fitted (wired as required by the inverters).

The lathe is permenantly wired as mentioned earlier and you can see the switched spur on the wall at the back on this picture

VSL looks like this (the white knob is the multifunction switch)

Hope this helps a bit

Mark

Thanks for the pictures Mark, very useful. I have now done away with all the existing switchgear because I was not confident enough to start fiddling with it. The motor is in place and wired. Now I need to wire in the inverter and mount in the cabinet. I will use it by controlling the the motor from the inverter box at first then see about using a pendant or maybe mounting remote controls in the original switchgear enclosure and make use of the original Viceroy faceplate so it all looks tidy . . . maybe. When I get around to that bit, the information here will be very useful, but for now I just need to get the lathe working again to finish the project which is currently still mounted in the 4-jaw chuck since the motor went boom.

Many thanks

Hywel

Dear All,

Returning to this thread, I have now installed a 0.75 kW 3 phase motor with 1P to 3P inverter on my Viceroy lathe. The motor is 4 pole so runs at ~1400 rpm normally. I have set it up so that it has intermediate gearing on the pulleys (geared down from 1400 to 930 running at 50 Hz), intending to run it between 20 to 100 Hz to vary the speed over the most convenient range, between about 350-1800 rpm, without have to change the gearing - clearly this is a compromise, but I am only using it for home projects and I also have a back-gear to run it at slow speeds when necessary.

The full-load current rating for the motor is 3.1 A at 50Hz/230 V. The inverter is set to trip at 5.6 A (180% of full-load current), so when I approach 100 Hz the motor starts to pull about 5.6 A and the inverted trips. I called the supplier and they said that this is likely due to the chuck acting like a flywheel, so that as it spins faster the motor needs more current. A few questions have arisen which would benefit from the combined knowledge of the forum as follows:

1. Does this sound like a reasonable analysis?

2. Will it damage the motor to run it at 100 Hz puling 5.6 A for any length of time provided it does not overheat?

3. If the answer to 2 is 'no' how high can I set the trip?

4. How fast could I reasonably run it without losing too much torque in any case? In other words, does my compromise set up seem reasonable for the use to which I want to put the lathe.

Many thanks in advance

Hywel

Hywel

I'm not an expert on VFDs but I can pass on my relevant experience.

Firstly the range of frequency and gearing you have mentions are totally suitable. I actually use a wider frequency range (4Hz to 150Hz) and although I can change the belt ratio I have actually never needed to. taking a smaller cut occasionally to machine something is more time and fuss free than moving belt position.

I have VFDs and motors on several machines, none with any extra cooling and no setup has ever got remotely hot. 'Warm' is the most I have ever seen and on my lathe I have converted the VFD internal fan to thermostatically operated (the fan noise annoyed me) and the fan now only runs very very infrequently. In fact if you asked me now, I would not even know if the fan was working OK as its a long time since it came on!

Regarding the current measurements. How and where are you monitoring the amperage?

The current for a three phase motor is 'per winding' so on full load each phase is 3.1A If you measure on the input to the VFD it will be a lot higher. I'm not sure where you overload trip is situated but as far as I know it is bad practice (even terrible) to put tit on the VFD output. The chuck acting as flywheel statement is irrelevant and will not affect the steady current drawn. If the inertia of the whole drive train is very high then the most likely effect is for it to act as a generator when you press the stop button and the VFD might throw up an error message. All you need to do is ensure the decelleration time is set long enough.

Before you worry about overheating, make use of the actual machine and see if it is really a problem.

Ian P

I sort of agree with Ian but with one proviso.....

As Ian says, the chuck will effect the startup current but not the steady state running current. If it is the startup current that is high then perhaps increase the startup ramp time?

I assume from what you say that this current is what is reported by the VFD and hence phase currents averaged. Plus it is the VFD that is "tripping" and shutting the motor down rather than an overload trip somewhere else?

You also don't say if the 5.6A is with the lathe fully loaded. If it is then it doesn't sound unreasonable even if it is a bit high. Personally I would only run it like that for very short periods. If I was machining something that was going to need a bit of grunt I would change the belts to bring the motor speed closer to its design speed (ie 50hz) it will be more efficient and less stressed then.

fwiw, my meddings drill has a FLA of 3A @230V. I have yet to get it past 2A on the VFD display but then I am currently only running at 50Hz and have only drilled up to 10mm holes so far. Unloaded it starts and runs at about 1.7A if I remember correctly.

Thanks Toby & Ian

To clarify some points:

1. The 5.6 A is the value measured by the VFD itself and it is the internal trip on the VFD which is tripping - it is set at 5.6 A

2. the VFD is programmed to start and stop slowly so there does not seem to be a problem with tripping during these operations

3. by fully loaded I mean just with the chuck in place and nothing else.

I have successfully bored out some Al bar running, at 80 Hz for an extended period, without any discernable heating and without it tripping, it's only when I get above this that the current ramps up - presumably the relationship is non-linear, the supplier who was quite helpful said it was quadratic.

Hywel

hmm... with just the chuck fitted that is no-load, not full load. I would expect less than 2/3 FLA unloaded. I could be wrong but I don't think that what you are seeing is right. I am also concerned the supplier appears to have suggested that having a chuck fitted constitutes a load. That is is not right......

Does the inverter have the option to display the current when it is running. I suspect so (most do) and that would tell you what it is doing at 50Hz. Once you are happy about that I would then move on to other speeds.

fwiw, I would expect less than 2A no load at 50Hz (ie just chuck) on a 3A FLA motor.

Like I said above, my drill is at about 1.7A without load, grabbing the chuck and trying to slow it down (don't try this at home kids....) only gets me up to about 1.8A before my hand/rag starts burning. My point being that it should take very significant load before it gets anywhere near full load current.

What make/model inverter is this?

I had one of the Chinese inverters on my Eagle 30 mill with a 1.1kw motor....drawing less than 2 amps (current indicated by the inverter readout ) while cutting steel with a 16mm end mill and a decent doc....

Without any load at all - belt removed - the current reads at about 2 A on the inverter. With the belt on the pulley and the chuck on this increases to between 2 to 3 A - can't remember without checking.

it's this package from Inverterdrive Supermarket

https://inverterdrive.com/group/AC-Inverter-and-Motor-Kits/Variable-Speed-Inverter-Motor-230V-1HP-2800RPM-5/

That no load current doesn't look too bad although it is a bit high, especially compared to John's figures.

But I am guessing it is working ok at 50Hz.

I am not sure what the supplier meant by "quadratic". That is a bit of a vague term, it implies a square law but in practice the function could be basically linear over the range we are looking at! I think fans and pumps can create loads that are roughly the square of the speed but that is not what we are looking at here.......

Unfortunately (as is typical) the motor spec is for 50Hz and no mention of FLA or efficiency at other speeds so this is hard to really judge.

Given it was bought as a kit, surely the supplier should be able to say what the FLA is at 100Hz?. They obviously think it is capable of this (and I think it does vary depending on motor design), so they should know what setup is required on the inverter to run reliably like this.

This is an armchair offering as I know more about electronics than motors!

It strikes me there are a couple of things that might explain why excessive current is being drawn at 100Hz, whilst all is well below that.

If your motor is a 50Hz type not designed to work with a VFD, taking it up to 100Hz could upset it in various ways.

One possibility is that high voltages generated by the pulse action of the VFD are causing something in the motor to breakdown at higher frequencies. This could be a capacitor, the windings, the cabling, an EMC filter, or some less obvious route to ground. You should be able to detect flashover by putting a LW/MW radio tuned between stations near the motor. (FM and digital radios aren't suitable.) A harsh ripping noise is typical of breakdown; an unpleasant whine is more likely to be the VFD in good health. Also, sparking may be visible if you run the lathe in the dark (but take care) If this is the cause, the cure is to either replace the motor or to keep the frequency below 100Hz.

A second possibility is that the windings are resonating electrically. It's a complete guess because I don't know what the inductance of a motor is. Same cure as above.

As high frequency AC behaves in odd ways the cable between VFD and motor should be as short as possible; reflected volts could confuse the VFD into miscalculating the current.

Dave

that is quite a drop off in current from 50Hz to 100Hz at no load, I am starting to guess here but I suspect that is because the inverter stops increasing the voltage at 50Hz (ie V/f becomes flat). I think this leads to flux weakening and the consequence will be a drop in torque as the speed increases. (hopefully I am not mixing my induction and syncronous motors up....)

Anyway, that doesn't explain the large current when connected to the lathe.

I'd do another test. If you can, change the belts/pulleys to get the same spindle speed as you currently get at 100Hz but now running the motor at 50Hz. So the load will be the same but motor speed will be the motor design speed for the 230V input.

What does the current do then? If it is still high then it might well be that you have excessive drag in the lathe, so maybe you need to find out why? Or if it reduces significantly then it means you have a problem running the motor at 100Hz, in which case I would be talking to the supplier again.

I was curious about this so I checked my lathe (boxford VSL, 1.5hp motor, running from cold) to see what current it takes at various speeds.

First I should point out that I appear to have it set to allow speeds between 0 and 50Hz. This also means the V/f curve is probably different to yours. So not quite the same but anyway I used speeds of 25Hz and 50Hz as a comparison.

This inverter doesn't give a current reading, rather a percentage of max rated load of the inverter. Unfortunately I cannot remember what that is and I am not removing it from its housing to check the rating plate. Anyway.......

No load (but driving the spindle/chuck) I get 28% at 25Hz and 40% at 50Hz. So that is a 43% increase in current for a doubling of input frequency. You are getting a 66% increase (taking the warmed up 100Hz figure).

Not sure if that helps, I would still suggest doing the test mentioned in my last post if you can, just to see if it is the motor causing the problems or just load from the lathe.

Hi Toby

Thanks for the info. I can't do the test at the moment because I only have one pulley wheel on the new motor and the old 4-wheel pulley is seized on the old motor shaft. I am going to do some more tests this weekend on a warmed up lathe.

I have done some internet research but it can all be very contradictory depending on whether there is a load or not and so forth. Given that the voltage is constant above 50 Hz the V/f ratio will decrease and so I understand the torque must also decrease. If that is the case then the current should also decrease, in a no load situation, which it does from 1.9 to 0.6 A on going from 50 to 100 Hz. However, when hooked to the lathe it increases, regardless of whether the chuck is on or not, so the gear train is obviously having an effect but doesn't seem to have anything amiss with it. Interestingly, when the backgear is engaged the current only increases from 2.1 to 2.8 A which seems OK to me.

Hywel

Hywel

Please could you explain what you mean by 'when the chuck is on of off'?

Removing the chuck from the spindle should not make a visible difference to the current consumption, There might be an increased radial load on the bearings or a windage loss but I would be surprised if the VFD monitoring electronics would indicate such an insignificant amount.

Two points. I suspect that there is nothing wrong withe your system that would not be cured by setting the VFD parameters (particularly relating to the motor data) and secondly the supplier should be involved if it was bought as a complete package.

Ian P

That certainly makes sense. The implication of the backgear results is that the main resistance in the gear train is after the backgear, ie spindle?