Re: Static phase converter

On 2009-02-06, pentagrid@xxxxxxxxx <pentagrid@xxxxxxxxx> wrote:
If the run capacitor is selected for current balance (more
accurate than voltage balance) at the nameplate motor load the
short answer is yes. Under these conditions the supply to the
motor is as well balanced or better than the same motor operating
from the more elaborate rotary converter/idler setup.

Pentagrid, this is very interesting indeed. Thank you.

I have some questions in this regard.

This is for a compressor duty application. The compressor is a Quincy
QR-25, model 340 pump based compressor, 80 gallon tank, that I bought
the other day for $200.

I have several options for running it.

1. Use a 5 HP, 1725 single phase RPM motor that I have. The plus is
that it is easy, the minus is that it is only 5 HP, 13 CFM at 175
PSI. Maybe I can get more CFM is I run it slightly faster, but
regulate to be between 110-140 PSI. I will call Quincy today about it.

2. Try to make a 10HP VFD which I have, run a 10HP 3ph motor that I
have, but I would need to try to supply its DC bus so that the drive
does not "phase fault".

3. Run from phase converter. (will work fine but painful!)

4. Try to develop some sort of starting and running circuitry for the
10 HP motor that we are discussing in this thread.

5. Buy a 7.5 or 10 HP single phase motor (very expensive! for a real motor)

For this particular compressor, the motor starts, and spins up the
pulley. Then unloader valves make the pump start pumping, only when
oil pressure reaches a certain number. This way, the motor has a
chance to spin up the pump unloaded.

After that, depending on initial tank pressure, and air use, the motor
load would gradually increase until the tank is almost full and the
motor reaches its maximum power output.

So, here are the q u e s t i o n s for option 4.

1. Suppose that I found out the capacitance needed to run the motor at
full nameplate horsepower. How would the motor behave, if it was
connected to the same capacitance, but at lower power output (and
therefore slightly higher RPM).

2. Practically, what would be a range of capacitance for a 10 HP 1725
RPM motor. I would need to buy a set of run caps to make up the
capacitance I need, and it would be good to know the cost beforehand.

3. How would I find this capacitance. Is that a function of hertz and
industance? If so, it would seem that I would need to find out the
inductance of the motor at full load first. So how would I do it? If I
run the motor from real three phase, I can connect a capacitor between
two legs and compute inductance from knowing the current through the
capacitor, frequency (60 Hz), and the capacitance. Right?

4. Under the less favorable assumption that capacitance needs to be
adjusted as tank pressure rises, would it be possible, as it seems, to
use pressure switches based on tank pressure, and zero crossing SSRs.

I will make some comments to your post below.

The following repeat of an earlier post gives more info.


It's true that static converters (start and run capacitor systems
with no idler) can deliver the full rated power of the motor for
surprisingly long periods but that is not the whole story.

A converter of this type is basically a capacitor/inductor
phase shift system which produces an open vee 3 phase system.
This phase shifter is a series resonant circuit and when it is
set up to give the 60 deg phase shift it is working a long way
below its natural resonant frequency. 60 deg is of course the
correct phase angle between the two legs of an open vee system.

The motor(s) is the inductor in the system and
unfortunately the apparent inductance of the motor changes with
rotor speed. For any particular rotor speed greater than about
90% of synchronous speed (the lower limit varies a bit with motor
type) it is possible to choose a capacitor combination which
produces a pretty close approximation to balanced 3 phase at the
motor terminals.

This means that if, say, the motor is 1740 RPM, as is my 10 HP motor,
then it runs at 96.6% of synchronous speed at highest output power,
meaning that it is even faster (closer to 100%) at lower outputs. So,
it would appear that it is a good case for giving it a phase shifted
third leg with a capacitor.

For near the full load rated speed of the motor, large run
capacitance

Do you have any idea how large, in uF per HP?

is needed with most or all of it as a single capacitor feeding the
phantom phase from supply live. At light load the impedance of the
rotor rises and if the capacitor value is chosen to achieve the
right phase angle the phantom phase voltage will be excessive. This
could be corrected by feeding the capacitor from a lower voltage
single phase source but this would mean feeding it from an auto
transformer across the supply.

You are referring to feeding the capacitor with voltage, but I thought
that it would be simply plugged between legs 1 and 3, without any
voltage connected to leg 3?

It is much simpler (and of course everybody does this) to use two
capacitors arranged as a voltage divider to simultaneously achieve
the correct phase angle and phase voltage.


The effective capacitanceof the two capacitors connected in
series across the supply is the sum of the capacitances because
the source impedance of the supply is zero and this effectively
parallels the two capacitors.

Because the they also act as a voltage divider, this sum
capacitance is effectively fed from a voltage of supply voltage
times C1/(C1+C2) where C1 is the top capacitor and C2 is
connected phantom phase to neutral.

Because it looks nicely symmetrical there seems to be a
tendency to believe that C1 and C2 should be equal and any
inequality in their optimum value must result from some strange
second order effect. This is NOT true. There is nothing magic
about equal C1 and C2. It simply results in a capacitor of value
C1+C2 fed from half the supply voltage. At this low effective
supply voltage it is only possible to get close to balanced
operation at no load or light loads which enable the rotor to
operate close to synchronous speed. As the load increases with
consequent slowing of the rotor speed the total capacitance needs
to increase with both more in C1 and less in C2. By the time full
load is reached the optimum value for C2 is usually zero.

How fine would these changes need to be? Would one change be enough,
for example, as the motor increases its output due to pressure rise?

These effects are very noticeable if you're using a
single motor on a variable load up to near rated full load power
and some compromise necessary. The saving grace is that
industrial motors are surprisingly tolerant of reasonable
overvoltage when operating at light loads. The trick is to size
the capacitors for at or near full load and to accept some
overvoltaqe at light loads. This increases the motor losses at
light load but the total motor losses still remain below the
losses at rated full load so temperature rise is acceptable.

This sounds good. The 10 HP motor is a great Reliance motor that
should be as good as any motor.

Summing up - if you need to cope with heavy loads on a
static converter throw away the bottom capacitor and be sure to
choose C1 for operation near full load.

Sounds easy enough.

None of this helps with starting torque - this is
inherently poor with the static converter arrangement however
large the starting capacitor. This is because correct low speed
phasing requires the capacitor to be fed fed from a voltage many
times the supply voltage.

Starting is not a problem, I am not worried, this is something that
can be solved very easily with A/C starting caps. The motor would
start at low torque anyway.

I appreciate your very thoughtful reply. I bought and read some motor
books a while ago, and I believe that one was yours.
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