On Thu, Jan 7, 2016, at 11:02 AM, andy pugh wrote:
> On 7 January 2016 at 15:37, John Kasunich <[email protected]> wrote:
> 
> Answering the only question that I can answer immediately:
> 
> > Second question: what is the nameplate amp rating of the motor?
> > If this is a 30A motor drawing 18A, then the motor and load is
> > probably fine and you need a VFD that can deliver 30A without
> > tripping.  If the motor nameplate says 15A, then the VFD is doing
> > its job and protecting the motor from an overcurrent that might
> > otherwise burn it up.
> 
> The motor has no nameplate. However was supplied as a unit with the Variator
> (Model M18, the same as a modern M318)
> www.allspeeds.co.uk/wp-content/files_mf/brochure.pdf

Well that is disappointing.  Did it fall off or something?  Or was there never
a nameplate?  I've never seen an industrial motor without a nameplate,
They are required by code so the installer can select the appropriate wire
size and set the overcurrent protection.

> The manual states 5A / Phase @400V but I have rewired the motor for 240V.
>
> (Maybe I should try wiring it back to 440V?)

You wrote "5A @ 400V", and then wrote "back to 440V".  Is it 400 or 440?

In the states, dual voltage motors are 460/230 (older motors might be labeled
440/220).  Either way it is a 2:1 ratio and it is done by having dual windings 
that
are connected in either series or parallel.  There are usually nine leads in the
terminal box.

My understanding is that in EU and UK, dual voltage motors are 400/230.
That is a sqrt(3):1 ratio and it is done by reconnecting from wye to delta.
There are six leads in the terminal box.

Which one applies here?  Was there a wiring diagram for the reconnection
or did you have to guess?  How confident are you that the reconnection is
correct?

If the nameplate data is truly 5A at 400V, and you converted it from wye 
to delta to run it on 230V, then the current rating should be 8.66A at 230V.

5A * 400V * sqrt(3) = 3.46kVA, and 8.66A at 230V is the same.

3HP is 2.24kW.  2.24kW / 3.46kVA = 0.65 power factor, which is maybe
a bit lower than I'm used to but not at all out of line.

There is no way you should be seeing 18A, and no reason to switch back
to 400 (or 440)V.  Doing so would hurt motor performance.  The only reason
you might want to _temporarily_ switch back is if there is uncertainty about
the connections.  Restoring it to the "as received" configuration might help
troubleshooting - although there is no guarantee that it wasn't already 
wrong.
 
> If I was to use a clamp ammeter, would I look at VFD input current or
> an individual phase current? I am not 100% clear which the VFD is
> displaying.

Look at motor current.  I would be utterly astonished if the VFD was 
displaying input current.  The vast majority don't have any form of sensor
on the input, it is the output current that matters for both motor control
and protection.

More questions:  when does the trip happen?  During acceleration,
or some time later after it has reached normal speed?  If during 
acceleration, does it happen right away (say less than 10% of 
full speed)?  Or later, at 50% speed or higher?

If it is happening during acceleration, my first guess would be that
you are trying to accelerate too fast.  It looks like your two speed
ranges have either a 6:1 (based on top speed) or 8:1 (based on 
bottom speed) ratio.  That means the spindle inertia reflected back
to the motor is either 36 or 64 times higher in high range than in
low range.

When the lathe was operated across-the-line, the motor and the
line would just grunt a little.  Current could hit 600-800% of rated,
torque might be 300-500% of nominal, and up she comes.  The 
overload relays have a time/current profile that is specifically 
designed to pass such short-term overloads, and the fuses are
also selected to deal with them.

With the VFD it is another story completely.  IGBTs don't have the
thermal mass of copper and iron, and can't handle such large 
overloads.  Most have instantaneous trip at about 200% of rating,
with an inverse time/current trip curve for lower currents.  A typical
"normal duty" drive might trip in 3 seconds at 150% and one minute
at 110%.  A "heavy duty" drive could typically give 200% for 3 seconds
and 150% for a minute.

You want to keep the current during acceleration below 150%,
by slowing down the acceleration rate.  For a load like a lathe 
where you won't be cutting during accel I'd actually aim to keep
the accel current at 100% or less - why beat things up to save
a couple seconds?

It looks like parameters 00-14 and 00-16 are acceleration times.
What do you have them set to?  If they are short, make them
longer.  The default 10 seconds should be OK, so I'm a little
perplexed - did you change the defaults?  For experimentation
purposes don't hesitate to set them to 60 seconds or so, that
will let you watch the meters as it slowly accelerates and get
a better idea of what is going on.  If a 60 second accel still
results in current of more than 8A we will need to dig deeper.

There are additional parameters telling the drive how to deal
with overloads - look at 08-00 thru 08-06.

To be honest, for something like this I strongly believe that 
the very first step is to tell the drive everything you know about
the motor.  The highest priority are parameters  01-00, 01-01,
and 02-01.  Those are absolutely necessary.  You probably 
want "1", "230", and "8.66" respectively.

If you had a nameplate it would give you 02-03.  Keep 02-02
zero since you don't know the proper value for 02-03.  If you 
can decouple the motor from all mechanical loads you could 
run it up to 50Hz to determine 02-00,   Otherwise, set 02-00 
to about 35-40% of 02-03, and keep 02-02 zero.



> 
> -- 
> atp
> If you can't fix it, you don't own it.
> http://www.ifixit.com/Manifesto
> 
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-- 
  John Kasunich
  [email protected]

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