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Vfd For Woodturning Lathe


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#1 GrahamW

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Posted 09 June 2007 - 12:29 AM

Excellent site, I have read almost every post here on variable speed drives but as an amateur would be grateful for some clarification if someone would be so kind. Apologies for the length of this post!

I have a large wood turning lathe which ran on a 0,75 kw 1 phase 6 pole 900 rpm motor with variable speed provided by contracting pulleys. This gave a speed range of 300 to 2700 rpm. A good although now elderly machine except that the motor pulley slides directly on the long motor shaft which is worn so the pulley rattles. This motor is not readily available so rather than repair it I chose to go with a new 3 ph 4 pole TEFC motor, upsized 50% to 1.1 kw and a secondhand 1.5 kw Mitsubishi FR-520S EC inverter single phase in and 3 phase 230 v out.

The lathe will be used between about 2000 to 2700 max rpm for spindle turning, say 3" or smaller diameter pieces, to ideally 200 rpm for a bowl of say 28" diameter. In all cases the tool is presented and fed into the work manually so the cutting load is limited. The machine had enough power with its 750 watt motor to turn large objects but of course had the advantage at low speed of running through the 3:1 reduction ratio provided by the pulley system.

The maximum supply current shown on the inverter plate is 17.4 amps which suggests to me that the maximum load on the supply could be 4kw. If the inverter output is 1.5 kw where does all the rest of the electrical energy go? I read in the manual that the inverter can reach 150 degrees C so some of it is heat but 2.5 kw dissipated in a box that size makes me think it should be glowing red hot! The inverter has a large finned heat sink and its own cooling fan.

The motor has class F insulation and the rating plate says 4.9 amps at 230 volts which I make just over 1.1 kw.

Is 4.9 amps the maximum current drawn by the motor, the average current or something else? If it is 4.9 amps for the whole motor what is the current per phase?

I assume 4.9 amps is the current that should be entered into the inverter when setting parameters

What size supply and mcb should I use? I think Oneway suggest a 20amp breaker with their 1.5 hp lathes to avoid nuisance tripping. Elsewhere on the web I have seen advice that one can run up to 3 hp from an inverter powered from a 230 volt 13 amp socket outlet.

If 50 Hz gives standard motor speed of about 1400 rpm what is the lowest motor speed I should use for good service life of the motor? I saw 15Hz mentioned as a minimum is this ok? I think this would give 420 rpm with this motor, have I understood this correctly?

From posts on this and other sites it seems heat is the problem at low motor speeds. Below what motor speed/frequency should I consider fitting an independent cooling fan to the motor?

Is it acceptable to set a current in the inverter parameters that limits the load on the supply to a desired value?

Finally, I anticipate that to get reasonable performance over the required speed range a pair of 2 or 3 step pulleys may still be required on the motor and headstock spindle. Any thoughts on this?

Thanks in anticipation for any advice.

Graham

#2 marke

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Posted 09 June 2007 - 08:14 AM

Hello Graham
Welcome to the forum.

A mechanical means that you use to reduce the speed, such as belt drives, chain drives or gear boxes, will increase the torque as the speed reduces such that the torque times the speed equals the power. If you have the speed, you double the torque.
When you use a VFD to reduce the speed, the torque remains the same and the power reduces.
This can be a problem for a wood lathe because when you use a larger diameter of product in the chuck, you need to run at a lower speed. The larger diameter requires a higher shaft torque to produce the same force at the cutter.
I expect that you will still need to have some mechanical speed changing to step the torque up when operating at lower speed. Trial and error may be the best way to find out.

As you have an oversized motor, you could organise the gearing such that the full speed of the lathe is achieved when the motor is operating at 75 Hz. This will give you a higher torque output a a given low speed than if the gearing was for a full speed at 50Hz. - you can only do this if the motor is oversized.

Running a motor at reduced speed will reduce the cooling available and if the load on the motor is high, this can cause the motor to over heat. With small motors, the iron loss is high and this alone can cause the motor to overheat at low speeds. The lowest speed that you can operate at is dependant on the characteristics of the motor, (losses and cooling) and the shaft load on the motor at that lower speed. I would expect that with a lathe, at low speed, the torque will be high so the minimum speed will be restricted.

When power is initially applied to the inverter, there will be a high current for a short time to charge up the internal capacitors. I suspect that it is this charging current that is being referred to by the 17.4 amps.
The maximum continuous input current should be in the order of 5 amps at full load, but the current waveform would be very distorted so you would need to have a circuit breaker rated at greater than 10 amps for running, and possibly greater than 15 amps for the initial power up current, depending on the trip curves of the breaker.
If the motor was direct on line started, (no VFD) you would need to size the breaker to withstand the high start current and this could be the reason for the recommendation of 20 Amps. In your case, with the VFD, the start current is no more than the run current.

Hope this answers some of your questions.

Best regards,



#3 kens

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Posted 10 June 2007 - 10:30 PM

Hi Graham, I would like to clarify a couple of points for you. I'm not sure how experienced you are with electrical calculations so I will try to keep things simple. I apologise if it is too simple though.

Your motor is a 3 phase motor with name plate rating of 230V 4.9A. The 4.9A is the phase current. This equals 1.95 kVA (kiloVoltAmps)

(Volts x amps x √3) I dant know if you are familiar with power factor or not but simply put it s the difference between apparant power (kVA) and real power (kW). I will guess that the motor will have a power factor of 0.8 so the kW input is 1.95 x 0.8 = 1.56 kW. There is then an efficiency rating of the motor. The 1.1 kW is the output at the shaft of the motor. This would indicate that the motor has an efficiency of around 72%.

What I am trying to get to is that the inverter would have a three phase output of 1.95kVA which equals a single phase input of 8.3A + losses. As marke has mentioned that the 17.3A is most likely the capacitor charge at start up.

You cannot really set a limit on the input current to the drive as it is uncontrolled. The drive only controls the output current and although the two are related there is not any physical control on the input.

You should set the nameplate rating in the drive parameters.

I hope that this helps,

Ken
An expert is one who knows more and more about less and less until he knows absolutely everything about nothing

#4 GrahamW

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Posted 11 June 2007 - 02:46 PM

Marke and Ken, thank you both for your advice. No need for apologies, an explanation can never be too simple for me! Good estimate on the figures, the manufacturer's data says 79% efficiency and 0.77 power factor.

I knew very little about electrical calculations for motors and hadn't followed up power factor and the relationship between KW and KVA but know more now. The bottom line is that I ought to have a 20 A supply to be safe for the reasons you made clear.

Can you suggest please what might be the lowest speed that I might run this TEFC motor without cooking it? The motor was sold to me as Lafert T 90S A4 although the plate says ICME another company in that group.

Using Marke's suggestion of full speed at 75Hz I think the motor would need to run at 8Hz to match the lowest speed and range of the existing mechanical drive. I think from what I have seen elsewhere that is unacceptably slow. I have this figure of 15Hz as a minimum stuck in my mind from somewhere.

As mentioned before I will add at least a 2 step pulley and I think additional cooling from an inline fan.

Thanks very much for your help

Graham

#5 marke

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Posted 11 June 2007 - 02:58 PM

Hello Graham

The minimum speed is very much a function of the inherent losses in the particular motor you are using, plus the cooling of that motor, plus the load required at the reduced speed.
Generally, I would avoid running the motor at less than 20Hz, and if the low speed torque is high, even higher speed. Some small motors can cook at 15 Hz even with no shaft load due to the high iron losses associated with small motors.
One solution is to add an external fan to cool the motor so that it is always cooled properly irrespective of the speed that it is going.
If you apply external cooling, you could reliably run down to around 15 Hz. Below this, I expect that the torque output may well reduce and the motor may show a tendency to stall.

Best regards,
Mark.




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