Re: Adequate torque from VFD driven motors?

Your first idea is likely the correct path, if the machine does not
have a DC drive now
then the newer electronics are a good choice for a complete retro
job. You trade motor
cost and complexity for a simple motor and more complex controller.

The VFD or inverter is in practice an AC to DC converter coupled to
an inverter section
to provide variable frequency and voltage output. Better drives will
control both voltage
and frequency to provide current management and better control. Some
simple inverters
simply spit out a 3 phase sign wave and have far less control. You
can even use a VFD on
single phase applications however the 3 phase motor has many
advantages for any
significant size so you don't often see single phase units used.

When using a VFD on single phase lines there is usually a derating.
This is due primarily
to the nature of the power. Remember the VFD is an AC to DC converter
and an inverter.
3 phase power rectifies to DC with a higher effective power due to
the overlapping AC
cycles. To sustain the same capabilities an equivalent single phase
conversion requires
higher storage capacity and higher peak current. This is easy up to a
point simply buy a
single phase to 3phase unit with the proper rating or you need a
larger rated 3 phase unit
to get the input capacity. Example the Baldor book I am looking at
gives a derating of
40% for single phase operation so for a 5 H.P. 3 phase motor I need
an 8.3 or larger rated
control for single phase operation. This derating is a function of
the input section when
operated from single phase vs. 3phase .

The question of performance is dependent on many variables and
intended use but a few
general concepts may help. First in real general terms horse power is
the ability to do
work and can be expressed in units terms as 550 foot Lbs./second. or
746 watts electrical
power.

Here is where component selection comes in. In very general terms
the VFD will deliver
constant torque from the low speed up to 60 Hz. or the motor rated
speed. Note the first
units, the motor is rated as say 5 H.P. at a nominal speed, in rough
terms this equates to
maximum torque available and time applied by the RPM. The speed of
the AC motor is
determined by the frequency of the drive current minus a small
slippage but the key is, it
is frequency not load that sets the AC motor speed. Again very
general here but this
factor gives one guideline for motor size. Given a motor rating you
can loosely equate
this to a motor torque. example.

If I say I deliver 1 H.P. at 3500 RPM I am also saying that I
delivered a total of 550 foot
pounds / second for 60 seconds or 33000 foot-pound / minute. I can
also then say I
deliver 9.43 foot pound/ revolution. An odd term but handy for
comparison.

Now take the same 1 H.P. rating at 1750 RPM , since time remains the
same I conclude
that the motor delivers 18.86 foot-pound / revolution. Both provide
the same total work
but the lower rated motor does so by applying more torque.

There are many more factors that can apply such as motor design peak
vs. continuous and
etc.. but in a dull pencil approach you can use this type analysis
then fill in where actual
data is available such as motor torque curves and such.

Now let's take the 1.5 horse case with a 5:1 total reduction and
a
1750 RPM motor. This
gives a spindle speed of 350 RPM. This is (1.5HP * 550 foot-pound/sec
* 60 seconds)/
350 RPM or 141.43 foot-pound / revolution expressed in our bizarre
units. Yoiks! this
may be difficult . Now to be fair we could likely whack 20-30% off
this just for gear train
losses but lets work the numbers at 100%.

Given the constraint that the VFD delivers roughly constant torque
from the low end up
to rated speed we can work out approximate motor requirements. We
know this is big so
select a 2:1 direct drive ratio. This drops our torque requirement to
a rounded 71
foot-pound per revolution. Remember the VFD sets the speed so what we
need is the
number at motor rated speed.

So 71foot-pound/rev * a 1750 rpm motor = 124250 foot-pound / minute
divide this by 60
seconds then by 550 foot-pound/sec = 3.76 Hp. motor rating.

At a 1:1 drive ratio we need (142 *1750)/ 33000 = 7.5 Hp. rating

At 1:1 drive with an 1150 rpm motor (142*1150)/33000 = 4.98 Hp.
rating

And so on....

As you see as the speed drops the problem gets larger for torque so
keeping a back gear
really helps if large work is needed. This is very off the cuff but
if you go back and figure
the actual requirements for the large work you can find what rough
requirements the
motor and drive need to meet. Most good drives will provide good over
torque capacity
and frequency over 200Hz. these days . It's late so I must quit
for
now , hope this helps
some.

The DC approach is an option as you can find motors with large low
speed torque but
they are costly . In either case most good controls can derate a
motor so extra capacity
does not hurt to a point.

Brian
BSP