WM250 inverter drive conversion.

Iirc, there was an article in MEW about adding a 3 ph motor and vfd to a lathe, cant remember if it was the WM250......

I'm converting my SPG lathe ( I think it may be the equivalent of the 290 but with a brushed motor...) to 3 phase with a 1.5kw motor ( going for a 2 pole job, the 4 pole 1400 rpm version only comes in a 90 frame size...tooooooo big enough ) and vfd, but undecided on choice of inverter....

I chose a 2 pole motor for 2 reasons, at low spindle speed, the motor will still be turning resonably quick to avoid any overheating issue, and secondly, the 80 frame size motor has the same footprint...I've turned up a replacement pulley to suit the motor shaft...14mm rather than mod the orig...if at some point I want to revert back, I can...

Edited By John Rudd on 27/09/2015 16:41:24

Motor speed is one thing you will have to watch if sticking with the same belt/pully ratios as the DC ones run at about 4600rpm at full lick.

Set of brushes would be a lot easier, the replacements I fitted have lasted far longer than the originals

Which is why I went with the 2 pole job. Run the inverter at 85hz should be around the same rpm....

John, I have found a 1.5Hp "80"frame 4 pole motor which I'm hoping should go in the hole vacated by the old motor.

Mark P.

Mark,

Good find....but if you want to stick with the same speed ranges, you'll need to change at least the motor pulley size........

Even running the inverter at 100hz will give a motor shaft speed of 2800.....-and I wouldnt want to push the motor much beyond that...

Is there a risk of loosing torque by gearing up a slower reving motor to run at higher spindle speeds?

Edited By JasonB on 28/09/2015 12:41:58

No Jason, I will be gearing down a high revving motor so it should increase torque?

Mark, I thought the 4 pole motors run at 1400rpm, 2 pole at 2800rpm (both at 50hz) but the DC one that is on the lathe at 4600 so you will need higher gearing/pullies to get the same spindle speeds you have now

Top speed on the lathe is 1000 or 2000 rpm depending on the belt/pulley position.

So for a motor spindle speed of 4600, the pulley ratio is 2.3:1 by my reckoning.....

If you maintain the same ratio, then for a motor speed of 1400 rpm, top speed now falls by a factor of 0.3 or in revs it falls to 600 ish....(608....)

This assumes you run the motor at 50 hz....which is why I went for the 2 pole variant .....by running the motor at 85 hz, I can get near to that same speed...

So what you thinking now Mark?

Someone tell me I got this wrong.....(I rather hope Im not wrong else Ive got the wrong motor again...already bought a previous one.... - )

Mark,

You're correct, the ratios will be the same.

If you use a 2 pole motor, its "native" speed is 2800rpm at 50Hz. If the VFD is set to 82Hz the motor will run at 4600rpm.

Above 50hz, the motor will give full torque (same as at 50hz), but below 50Hz the power will be the same as at 50Hz, but the torque will be reduced (power = torquexRPM).

Edited By Gary Wooding on 28/09/2015 15:41:26

I hope your motor designed for 2850 rpm has a good margin of strength built in to withstand the increase in the stresses of running at 4600. If it won't I don't think I'd want to be in the same room as it when it lets go.

Let's get this straight, assuming we're dealing with an induction motor with a base speed at a frequency of 50Hz. Base speed generally means where the motor is running at it's rated voltage, and hence rated current and up to rated power.

Above 50Hz, at the rated voltage, there is less current, as the motor is turning faster and therefore there is more back emf opposing the current. So the current, and torque, go down but since the angular velocity has gone up the power stays roughly constant. So, above base speed you get constant power.

Below 50Hz more current can be delivered at the rated voltage. However, to protect the motor the current is normally limited to the rated current. Since the current is constant the torque is constant, but the power decreases as the angular velocity reduces. So below base speed you get constant torque.

To be precise the above equation should be:

power (Watts) = torque (Nm) x angular velocity (radians/second)

Andrew

My comment was not what torque the motor would give at a given frequency but more what loss of torque there would be due to having to alter pully ratios to get the same spindle speed as the existing motor.

Lets take the slow speed of 50-950rpm, DC motor is running at approx a 5:1 reduction ratio on the small pinion and large spindle pully. Motor running at 250-4600rpm.

Using the same gear ratios the 4 pole would be giving speeds of 15-300rpm, and on the high ratio 30-600. So not a very fast top speed and do you want to run the motor as slow as 75rpm with little effect from the fan?

To get nearer the same spindle speeds rather than a 5:1 reduction you will have to probably go for 1:1 and overrun the motor to get the 1900rpm top spindle speed. Low range would be 2:1 but you would then need to be running the motor at 100rpm to equal the current 50rpm slowest speed which could still give overheating issues .So assuming both motors have a similar torque you will only get 20% the mechanical advantage on the high speed and 40% on low

Edited By JasonB on 28/09/2015 17:08:56

I agree with Andrew. If Gary's statement was correct it would imply infinite torque at zero speed.

Les.

Same here, only used it once when new to see what it was like and never again.

Still not sure that running the motor at 50rpm with the 1:1 ratio would not give overheating problems