Awl--

What's more important when adjusting capacitor values:
Getting equal voltage between each pair of legs, or balanced current in
each leg? Seems like you can't have both.

At equalized voltage, the current through the generated leg is through the
roof.
At equalized current, if line L1-L2 is 240, then L1-L3 and L2-L3 are about
230-233, which is not so bad.

On a 10 hp idler, it takes only 50 uF per leg to equalize the current in the
legs, but much more to equalize the voltage--with the resulting excessive
current in the third leg.

I also noticed that two 5 hp idlers draw much more current than the sum of
either operating by themselves! And much more than my 10 hp idler.

Idears?

--
DT

On 2008-08-06, DrollTroll wrote:
Awl--

What's more important when adjusting capacitor values:
Getting equal voltage between each pair of legs, or balanced current in
each leg? Seems like you can't have both.

Equal voltage between legs, means that your RPC has phases that are
exactly 120 degrees apart.

At equalized voltage, the current through the generated leg is through the
roof.

What exactly current are you measuring, from the generated leg to
load?

At equalized current, if line L1-L2 is 240, then L1-L3 and L2-L3 are about
230-233, which is not so bad.

On a 10 hp idler, it takes only 50 uF per leg to equalize the current in the
legs, but much more to equalize the voltage--with the resulting excessive
current in the third leg.

I also noticed that two 5 hp idlers draw much more current than the sum of
either operating by themselves! And much more than my 10 hp idler.

Idears?


What is your schematic , you can skip contactors and stuff and jyst
post how it is wired when running.

--
Due to extreme spam originating from Google Groups, and their inattention
to spammers, I and many others block all articles originating
from Google Groups. If you want your postings to be seen by
more readers you will need to find a different means of
posting on Usenet.
http://improve-usenet.org/

"Ignoramus11522" wrote in message
...
On 2008-08-06, DrollTroll wrote:
Awl--

What's more important when adjusting capacitor values:
Getting equal voltage between each pair of legs, or balanced current in
each leg? Seems like you can't have both.

Equal voltage between legs, means that your RPC has phases that are
exactly 120 degrees apart.

At equalized voltage, the current through the generated leg is through
the
roof.

What exactly current are you measuring, from the generated leg to
load?

Right now, just the idler(s) itself.


At equalized current, if line L1-L2 is 240, then L1-L3 and L2-L3 are
about
230-233, which is not so bad.

On a 10 hp idler, it takes only 50 uF per leg to equalize the current in
the
legs, but much more to equalize the voltage--with the resulting excessive
current in the third leg.

I also noticed that two 5 hp idlers draw much more current than the sum
of
either operating by themselves! And much more than my 10 hp idler.

Idears?


What is your schematic , you can skip contactors and stuff and jyst
post how it is wired when running.

Start caps disconnected when running.
With 240 V coming in to L1, L2, 50 uF (or whatever) connected between
L1-L3, 50 uF between L2-L3.

I wired up a switchable bank of capacitors, dual rows of 100, 80, 60, 25,
25, that can be switched in in any combination to any combination of idlers.
L1-L3 and L2-L3 can have different caps, and usually do for equal voltage.
But as mentioned above, for the equal current state, the caps are much
lower, and equal.

I will be wiring up heater elements (about a 5 A draw), in both delta and
wye configs, to test-load the idler/cap combo's. Hopefully today/tomorrow.
--
DT






--
Due to extreme spam originating from Google Groups, and their
inattention
to spammers, I and many others block all articles originating
from Google Groups. If you want your postings to be seen by
more readers you will need to find a different means of
posting on Usenet.
http://improve-usenet.org/

You can't adjust voltage across the L1 - L3, L2 - L3 legs of a RPC by observing current flow with a
clamp-on ammeter. Balance is achieved by measuring voltage across the respective legs with a
voltmeter. Suggest you go to Metal Web News, scroll down to Electric Phase Connverter Information
and do some reading on RPCs.

Bob Swinney




"DrollTroll" wrote in message ...
Awl--

What's more important when adjusting capacitor values:
Getting equal voltage between each pair of legs, or balanced current in
each leg? Seems like you can't have both.

At equalized voltage, the current through the generated leg is through the
roof.
At equalized current, if line L1-L2 is 240, then L1-L3 and L2-L3 are about
230-233, which is not so bad.

On a 10 hp idler, it takes only 50 uF per leg to equalize the current in the
legs, but much more to equalize the voltage--with the resulting excessive
current in the third leg.

I also noticed that two 5 hp idlers draw much more current than the sum of
either operating by themselves! And much more than my 10 hp idler.

Idears?

--
DT


** Posted from http://www.teranews.com **

"Robert Swinney" wrote in message
...
You can't adjust voltage across the L1 - L3, L2 - L3 legs of a RPC by
observing current flow with a
clamp-on ammeter. Balance is achieved by measuring voltage across the
respective legs with a
voltmeter. Suggest you go to Metal Web News, scroll down to Electric
Phase Connverter Information
and do some reading on RPCs.

I read your article--not bad, but perhaps a little heady for the diy-er.

Strange, tho, that someone who apparently knows so much about rpc's
repeatedly demonstrates that he can't properly comprehend simple g-d Qs
about rpc's on a ng.

I *did not* say I was trying to adjust voltage via current readings.

I was simply asking what the story was, that in achieving voltage balance
(which I can do readily, with a switchable capacitor bank), the 3rd leg
current was so disproportionately high, to the point that it will trip
breakers.

In fact, with no caps at all, current in L1 and L2 is also very high, much
higher than no-load should be, and L3 is near-zero. Adding caps shifts
this, seemingly linearly.

My Q was basically on the relative importance of the capacitor-induced
voltages and currents, and if perhaps there was a best compromise.

I'm sure I'm not the only one with this problem.

--
DT




Bob Swinney




"DrollTroll" wrote in message
...
Awl--

What's more important when adjusting capacitor values:
Getting equal voltage between each pair of legs, or balanced current in
each leg? Seems like you can't have both.

At equalized voltage, the current through the generated leg is through the
roof.
At equalized current, if line L1-L2 is 240, then L1-L3 and L2-L3 are about
230-233, which is not so bad.

On a 10 hp idler, it takes only 50 uF per leg to equalize the current in
the
legs, but much more to equalize the voltage--with the resulting excessive
current in the third leg.

I also noticed that two 5 hp idlers draw much more current than the sum of
either operating by themselves! And much more than my 10 hp idler.

Idears?

--
DT


** Posted from http://www.teranews.com **

Droll sez:

"Strange, tho, that someone who apparently knows so much about rpc's
repeatedly demonstrates that he can't properly comprehend simple g-d Qs
about rpc's on a ng.

I *did not* say I was trying to adjust voltage via current readings."

Very well, then. If you are as familiar with current measurement as you seem to imply, it should
come as no surprise that it is possible to overload your source circuit. You should seek the help
of a qualified electrician to install a heavier circuit breaker.

As a final step and after you have voltage balanced your idler - load(s) network you should install
capacitance between L1 and L2 (the input line) and adjust it until the current through that
capacitance is minimum. This is the only effective measurement you can get with a clamp-on ammeter
in a RPC. Doing this will minimize the load as seen at the breaker.

Bob Swinney





** Posted from http://www.teranews.com **

On Wed, 6 Aug 2008 02:28:01 -0400, "DrollTroll"
wrote:

Awl--

What's more important when adjusting capacitor values:
Getting equal voltage between each pair of legs, or balanced current in
each leg? Seems like you can't have both.

At equalized voltage, the current through the generated leg is through the
roof.
At equalized current, if line L1-L2 is 240, then L1-L3 and L2-L3 are about
230-233, which is not so bad.

On a 10 hp idler, it takes only 50 uF per leg to equalize the current in the
legs, but much more to equalize the voltage--with the resulting excessive
current in the third leg.

I also noticed that two 5 hp idlers draw much more current than the sum of
either operating by themselves! And much more than my 10 hp idler.

Idears?


Pretty well all the discussions you find in the many
posts on this subject are primarily based on voltage
measurements. If you haven't got a clip on ammmeter, voltage
measurements are so much more convenient and, if the setup is
near balanced, they give a reasonably good indication on which
way to change things.

It is true that, with a pure sinusoid input EQUAL
voltages on all three legs truly indicates a balanced system.
A correctly wound 3 phase motor connected to this system will of
course draw equal currents

However, in a system with substantial unbalance, voltage
measurements can only give a rough indication of what is
happening.

As far as the motor is concerned it it is always the
current balance that is important because output torque is
directly proportional to current* and (within the normal working
range) independent of voltage.

A clear example of the limitations of voltage measurement
is an RPC operating without a correction capacitor. Voltage
measurements will show a large but apparently measurable
unbalance ratio. Current measurement will correctly indicate the
true 100% unbalance.

* Excludes the wattless reactive component of motor current.


Jim

DrollTroll wrote:
Awl--

What's more important when adjusting capacitor values:
Getting equal voltage between each pair of legs, or balanced current in
each leg? Seems like you can't have both.

Balanced current in the MACHINE TOOL motor, or in the RPC? You
don't care about balanced current in the RPC, and it won't be
balanced, because each winding is doing a different job. (The
240 V line windings are running the RPC, the 3rd leg is
providing the shifted phase.)

What you are concerned about is getting roughly equal current in
the machine tool motor. And, equal voltages on the line DO NOT
guarantee that you have the proper phase shift of the generated
leg. The voltage from the 240 V mains ceter tap (neutral) to
the generated leg should be at 90 degrees to the mains, and
about 207 V. Without an oscilloscope, however, it is pretty
hard to measure phase shifts. If the machine tool motor draws
equal current on all 3 legs, with roughly balanced L-L voltages,
then your phase shift is right.

As long as the current in the generated leg of the RPC motor is
not excessive, I'd ignore it. If it IS much above the nameplate
rated amps, then you need to reduce the caps to bring the
current down.
At equalized voltage, the current through the generated leg is through the
roof.
I'm not too surprised, but what is "through the roof"? twice
rated current? More?
At equalized current, if line L1-L2 is 240, then L1-L3 and L2-L3 are about
230-233, which is not so bad.

What are you complaining about? That is quite good for an RPC.

Jon

"Jon Elson" wrote in message
...
DrollTroll wrote:
Awl--

What's more important when adjusting capacitor values:
Getting equal voltage between each pair of legs, or balanced current in
each leg? Seems like you can't have both.

Balanced current in the MACHINE TOOL motor, or in the RPC? You don't care
about balanced current in the RPC, and it won't be balanced, because each
winding is doing a different job. (The 240 V line windings are running
the RPC, the 3rd leg is providing the shifted phase.)

What you are concerned about is getting roughly equal current in the
machine tool motor. And, equal voltages on the line DO NOT guarantee that
you have the proper phase shift of the generated leg. The voltage from
the 240 V mains ceter tap (neutral) to the generated leg should be at 90
degrees to the mains, and about 207 V. Without an oscilloscope, however,
it is pretty hard to measure phase shifts. If the machine tool motor
draws equal current on all 3 legs, with roughly balanced L-L voltages,
then your phase shift is right.

As long as the current in the generated leg of the RPC motor is not
excessive, I'd ignore it. If it IS much above the nameplate rated amps,
then you need to reduce the caps to bring the current down.
At equalized voltage, the current through the generated leg is through
the roof.
I'm not too surprised, but what is "through the roof"? twice rated
current? More?
At equalized current, if line L1-L2 is 240, then L1-L3 and L2-L3 are
about 230-233, which is not so bad.

What are you complaining about? That is quite good for an RPC.

Well, I don't know enough to know if I *should* be complaining or not.... so
I just complain reflexively....

I bought a bunch of cheapie analog VOMs (identical), and now have them
permanently wired among the three legs--convenient. I have clamp-on
ammeters now, but I also plan to get 3 cheapie digital clampons (about $20
each, if I'm lucky), and have them installed permantly as well.

A 10 hp baldor Super E will read over 20 A in the 3rd leg, with closely
matched voltages (more capacitance)--idler only.
With matched currents (much less capacitance), it's as above, 240, 230-233.

I will wire up 3 ph resistive loads today (5 A heating elements), in both
delta and wye, and adjust the caps see how the currents/voltages respond *in
the resistive load*. Is this a good test method? Will the delta/wye
config of the load affect things?

Also, what do you think of the comment to add capacitance between L1 and L2,
and that the current through this capacitor "is the only effective
measurement you can get with a clamp-on ammeter in a rpc"?

I did notice that 25 uF between L1 and L2 drops the current in L1 and L2 by
about 2 amps, without affecting the current in L3 (no load).

--
DT






Jon

DrollTroll wrote:

A 10 hp baldor Super E will read over 20 A in the 3rd leg, with closely
matched voltages (more capacitance)--idler only.
With matched currents (much less capacitance), it's as above, 240, 230-233.

You keep saying " idler only " Youre missing the point.
There is nothing to measure CURRENT when you don't have
the LOAD (machine) connected. THAT is where you measure
current, to the load.
...lew...

From reading previous RCM discussions of this topic, it seems to
sometimes generate as much heated discussion as politics or religion.
For that reason, I’ll disclaim being an expert and “tread lightly.”

There was mention of running two 5 hp idlers and why they behaved so
differently from one 10 hp idler. I assume that the two 5 hp idlers
were running freely and not connected to each other. I believe that
would be like trying to hook up two independent gas-powered 120 VAC
generators and trying to get 240 VAC (single phase). I don’t think
that it will happen because the generators would not likely be in
phase with each other. Same goes with trying to run two free-wheeling
5 hp idlers. The currents and voltages would not add in phase and
would probably bounce back thru the mains or otherwise dissipate by
heat in the windings. More properly done, the motors shafts should be
connected to each other with the proper angle between them so that
they are acting in concert (like a single idler).

As far as using a resistive load to test the RPC, I don’t think that
it will help. What you want to drive with an RPC is a motor which is
a reactive load. A resistive load always has a power factor of 1,
whereas un-tuned reactive circuits generally have a PF 1. The motor
reacts with the idler causing phase shifts between the voltage and
current curves and this affect how much real power can be used by the
motor. If the PF=0.7, then you need to have 1.4 x the current and
double the input power to get the rated output power.

If you had the right tools (multi-channel oscilloscope, current &
voltage taps, etc or LabView data acquisition setup) you might be
better equipped to see what is happening in the circuit. If you had
lots of cash, I’d get one of these:
http://www.duncaninstr.com/power-quality-analyzer.htm

I bought an old watt-hr meter at a flea market and have used it to
verify that I needed to buy a new refrigerator. Maybe you can find
one and monitor your RPC performance. Since you probably can’t afford
a dynamometer, you would need to make sure that you set up your 3
phase motor with a constant workload (like a circulating water pump
and reservoir). If you measure your electrical use over time vs. the
amount of work done, you could use this as a tool to adjust the
capacitance and get some idea of improving the efficiency and PF of
your RPC. Even without a watt-hr meter, you can still make
measurements on the legs of the RPC, but I think that you need to set
up the motor with a constant mechanical load to see the effects of
changing capacitance values. (Measuring voltage and current (RMS
scalar values) will only tell you the “apparent power” used in each
leg and not the real power delivered in producing work. (see
references). I think that even using the watt-hr meter will just give
you an idea of the “apparent power” used in the system, but it is
roughly analogous to the measurement of efficiency, so it could be
thought of as measuring “input power”.)

Here are more links (if you are academically-inclined or just want
torture):

Wattmeter and power factor:
http://en.wikipedia.org/wiki/Power_factor
http://en.wikipedia.org/wiki/Wattmeter

3-Phase power measurements:
http://www.6mgb.com/E-News/08-04-01/...-493_short.pdf
http://www.manumesure.com/fiches/pdf...ical_power.pdf
https://www.ohiosemitronics.com/pdf/...-method(E).pdf

Wattmeter – construction and principles of operation:
http://www.usbr.gov/power/data/fist/...10/vol3-10.pdf

I rate the replies to this post as A+.

This is what newsgroups are all about!

Pete Stanaitis
-------------------

"Denis G." wrote in message
...
From reading previous RCM discussions of this topic, it seems to
sometimes generate as much heated discussion as politics or religion.
For that reason, I’ll disclaim being an expert and “tread lightly.”

There was mention of running two 5 hp idlers and why they behaved so
differently from one 10 hp idler. I assume that the two 5 hp idlers
were running freely and not connected to each other. I believe that
would be like trying to hook up two independent gas-powered 120 VAC
generators and trying to get 240 VAC (single phase). I don’t think
that it will happen because the generators would not likely be in
phase with each other. Same goes with trying to run two free-wheeling
5 hp idlers. The currents and voltages would not add in phase and
would probably bounce back thru the mains or otherwise dissipate by
heat in the windings. More properly done, the motors shafts should be
connected to each other with the proper angle between them so that
they are acting in concert (like a single idler).

As far as using a resistive load to test the RPC, I don’t think that
it will help. What you want to drive with an RPC is a motor which is
a reactive load. A resistive load always has a power factor of 1,
whereas un-tuned reactive circuits generally have a PF 1. The motor
reacts with the idler causing phase shifts between the voltage and
current curves and this affect how much real power can be used by the
motor. If the PF=0.7, then you need to have 1.4 x the current and
double the input power to get the rated output power.

================================================== =

See my data post.
I used purely resistive loads, bec I thought this would *simplify* the
problem!
No power factors, phase shifts, no bouncing back and forth between two
"co-generators"--which is what 3 ph motors in parallel do.

========================================


If you had the right tools (multi-channel oscilloscope, current &
voltage taps, etc or LabView data acquisition setup) you might be
better equipped to see what is happening in the circuit. If you had
lots of cash, I’d get one of these:
http://www.duncaninstr.com/power-quality-analyzer.htm

I bought an old watt-hr meter at a flea market and have used it to
verify that I needed to buy a new refrigerator. Maybe you can find
one and monitor your RPC performance. Since you probably can’t afford
a dynamometer, you would need to make sure that you set up your 3
phase motor with a constant workload (like a circulating water pump
and reservoir). If you measure your electrical use over time vs. the
amount of work done, you could use this as a tool to adjust the
capacitance and get some idea of improving the efficiency and PF of
your RPC. Even without a watt-hr meter, you can still make
measurements on the legs of the RPC, but I think that you need to set
up the motor with a constant mechanical load to see the effects of
changing capacitance values. (Measuring voltage and current (RMS
scalar values) will only tell you the “apparent power” used in each
leg and not the real power delivered in producing work. (see
references). I think that even using the watt-hr meter will just give
you an idea of the “apparent power” used in the system, but it is
roughly analogous to the measurement of efficiency, so it could be
thought of as measuring “input power”.)

========================================

There was a guy here about a year ago I think who gave a spectacular and
detailed description of how watt-hour meters work. A real keeper, which of
course I didn't keep. It was in, iirc, partial response to one of my
queries, made under the nym Proctologically Violated. No foolin.... (ahm
tryna clean up my act a little under Droll Troll. g).

I think he was involved with building them or designing them, and he said
that regardless of power factor, they do in fact measure *true* power by
inducing eddy currents or hysteresis or sumpn in the metal rotor, which
somehow made phase angles irrelevant. A bit over my head, but worth
googling, if you can wade through all the spam.

I haven't yet checked out your last link--hopefully it will be consistent
with what I've cited. g

--
DT


Here are more links (if you are academically-inclined or just want
torture):

Wattmeter and power factor:
http://en.wikipedia.org/wiki/Power_factor
http://en.wikipedia.org/wiki/Wattmeter

3-Phase power measurements:
http://www.6mgb.com/E-News/08-04-01/...-493_short.pdf
http://www.manumesure.com/fiches/pdf...ical_power.pdf
https://www.ohiosemitronics.com/pdf/...-method(E).pdf

Wattmeter – construction and principles of operation:
http://www.usbr.gov/power/data/fist/...10/vol3-10.pdf

"spaco" wrote in message
.. .


I rate the replies to this post as A+.

This is what newsgroups are all about!

Pete Stanaitis
-------------------

Indeed.

Here's some data, with *purely resistive loads*.

Resistance: 5 A heating elements (120V).
Set 1: hooked up 6 of these elements in "delta".
Set 2: hooked up 3 of these elements in "wye".

Why both?

To see if rpc "behavior" changed much. Didn't seem to, but keep in mind
that delta also consumes much more power than wye, but see the data.

I also figgered pure resistance load would prove less ambiguous ito power
factors, motor-motor interaction, etc. True, real application would be
motor-motor, or, in my case, rpc-CNC

Idler: 10 hp Baldor Super E motor.

L1-L2: 240V

Capacitance:
Strangely, best results were achieved with just one value: 160 uF, between
L1 and L3!! nothing on L2-L3!

Well, actually, I'm not sure what "best results" really are, as it is hard
to independently adjust voltages, even with individually switched caps on
each leg.


====== V: L1-L2 L2-L3 L1-L3 Amps: Leg1 Leg2 Leg3

Load:

delta 233 248 228 10.9 12.4 10.9

wye 237 252 238 7.6 8.2 8.1

delta+wye 231 246 208 17.6 20. 18.2

Hopefully the table is readable.

Notice that in the delta resistance, the current in Legs 1 and 3 are
identical.
In wye, legs 2 and 3 are identical.
But in neither case is the current in Leg1 and Leg2 (line V) identical.

In delta + wye (heavy load), it may be that the low 208 V reflects exceeding
the capacity of the idler. But despite this, the currents in each leg
seemed proportionately close as in the lighter loads.

I also put 25 uF and 60 uF caps between L1 and L2 as per Swinney's advice
(with the wye resistance load), but saw no sig. effect, even at no load at
all.
Current between L1 and L2 (through the cap) seemed to increase linearly with
capacitance, so the minimization he described does not seem to apply here.

Seems like with some idlers you can tune the system "by ear", because
unbalanced caps create a racket in the motor, presumably from the unbalanced
torques mentioned by others. The baldor, however, displays much less of a
racket than the two 5 hp idlers.

It may also prove useful to keep a light load on the rpc (mebbe three 100 W
bulbs in wye), to reduce no-load spikes.

I am still wondering if a mixture of delta- and wye-wound idlers would offer
some inherent stability in an rpc system.
That said, with 5 idlers totaling 25 hp, I seem to get as stable a result
with the one 10 hp idler as with combinations of the others.

Let me tell you, it sure got hot in that little motor room with 9 600W
heating elements blasting away....

If I had $5K to burn, I'd certainly spring for phaseperfects digital
converter, which they say is true 120 deg sine wave power at 1% up to 20 hp,
iirc. goodgawd.....

Comments on the data? Seem reasonable?
Reasonable that just one capacitance value seemed to best accommodate all
three loads?
Sure is a lot better than the 275 V from the 3rd leg that I used to have,
off a phaseamatic.

--
DT

I looked at your data, but….

Resistive loads and reactive loads act differently in AC circuits.
http://en.wikipedia.org/wiki/Load_bank

Resistive circuits don’t cause phase shifts. That’s the problem in
trying to using resistors as a substitute for your induction motor.
In this type of circuit with a motor, you HAVE phase shifts and the
process of tuning is to try and minimize them by adjusting
capacitance. It’s just the nature of the beast.

You can adjust a resistive load to make it draw the same as a motor,
but it will only be a good substitute for a motor circuit with a PF=1.
It won’t be a good substitute for your motor and it won’t help you to
tune your RPC.

With reactive loads (such as motors), you get phase shifts between the
current and voltage curves (PF1) and when you tune the circuit by
adding capacitance to balance the phase shift caused by inductance in
the motor. In essence, you try to shift the voltage and current
curves back together and get as close to PF=1 as possible. This
optimizes power transfer to the motor.

I know that you are just trying to get this thing to work and all this
probably sounds like “hand waving”, but you are better off trying to
balance your RPC with your real-live motor in place and under some
mechanical load.

I hope that you get your project working, and my apologies if this was
not helpful or I’ve erred.

On Thu, 7 Aug 2008 08:27:58 -0400, "DrollTroll"
wrote:


"Jon Elson" wrote in message
m...
DrollTroll wrote:
Awl--

What's more important when adjusting capacitor values:
Getting equal voltage between each pair of legs, or balanced current in
each leg? Seems like you can't have both.

Balanced current in the MACHINE TOOL motor, or in the RPC? You don't care
about balanced current in the RPC, and it won't be balanced, because each
winding is doing a different job. (The 240 V line windings are running
the RPC, the 3rd leg is providing the shifted phase.)

What you are concerned about is getting roughly equal current in the
machine tool motor. And, equal voltages on the line DO NOT guarantee that
you have the proper phase shift of the generated leg. The voltage from
the 240 V mains ceter tap (neutral) to the generated leg should be at 90
degrees to the mains, and about 207 V. Without an oscilloscope, however,
it is pretty hard to measure phase shifts. If the machine tool motor
draws equal current on all 3 legs, with roughly balanced L-L voltages,
then your phase shift is right.

As long as the current in the generated leg of the RPC motor is not
excessive, I'd ignore it. If it IS much above the nameplate rated amps,
then you need to reduce the caps to bring the current down.
At equalized voltage, the current through the generated leg is through
the roof.
I'm not too surprised, but what is "through the roof"? twice rated
current? More?
At equalized current, if line L1-L2 is 240, then L1-L3 and L2-L3 are
about 230-233, which is not so bad.

What are you complaining about? That is quite good for an RPC.

Well, I don't know enough to know if I *should* be complaining or not.... so
I just complain reflexively....

I bought a bunch of cheapie analog VOMs (identical), and now have them
permanently wired among the three legs--convenient. I have clamp-on
ammeters now, but I also plan to get 3 cheapie digital clampons (about $20
each, if I'm lucky), and have them installed permantly as well.

A 10 hp baldor Super E will read over 20 A in the 3rd leg, with closely
matched voltages (more capacitance)--idler only.
With matched currents (much less capacitance), it's as above, 240, 230-233.

I will wire up 3 ph resistive loads today (5 A heating elements), in both
delta and wye, and adjust the caps see how the currents/voltages respond *in
the resistive load*. Is this a good test method? Will the delta/wye
config of the load affect things?

Also, what do you think of the comment to add capacitance between L1 and L2,
and that the current through this capacitor "is the only effective
measurement you can get with a clamp-on ammeter in a rpc"?

I did notice that 25 uF between L1 and L2 drops the current in L1 and L2 by
about 2 amps, without affecting the current in L3 (no load).



At the moment, as you have described it, you are running
the idler alone. Matched current setup is right for most
efficient running of an idler alone running light. Matched
current is when the current through the correction capacitance to
the third motor terminal is roughly equal to the average of the
other two motor currents.


This is NOT optimum when it is then connected to one one
or more loaded or unloaded additional motors.

Each added motor even if running light needs the
capacitance to be increased and every increase in mechanical load
on any of the motors requires further increase.

This means that your current balance is best setup with
your load motor(s) connected and preferably working into their
worst case full mechanical load.

Measurements driving a resistive dummy load are of no
help because of the very different characteristics of a motor
load.

While ideally the correction measurements should be made
with the load motors under full load, correction made with partly
or completely unloaded motors is usually acceptable.

This is where the size of your idler becomes important. If
your idler is only the same size as the load motor, addition of
the load motor running light DOUBLES the optimum capacitor size
with a further increase needed for full load operation. If your
idler HP is 5 times the load motor the optimum change drops to
about 20% and this makes the setup adjustment correspondingly
less sensitive.

Your operating setup now consists of two or more motors
all directly connected in parallel.Because of this, as far as
correction is concerned,there's no distinction between idler and
load motor, single motor or multiple motors. It's the total HP in
the system that matters.

With your load motor(s) switched off and the idler running
light the idler will be substantially overcorrected with
excessive phantom phase current. However this is mainly reactive
wattless current with only a small I squared R (i.e. real power
dissipation) component. Provided it remains below idler rated
full load current this is OK. It can be avoided by splitting the
correction capacitance and giving each load motor its own
correction capacitance but it's rarely worth the additional
complication.

Because of the non unity power factor of the motor load
the current drawn from the single phase input is unnecessarily
high and can be reduced by adding power factor correction
capacitance directly across the single phase line input. Although
this connects to L1 & L2 this connection distinction is important
because current balance measurements must only measure the
current into the motor terminals - it must NOT include the power
factor correction current which is drawn by this capacitor.
directly from the supply.



Jim

"Denis G." wrote in message
...
I looked at your data, but….

Resistive loads and reactive loads act differently in AC circuits.
http://en.wikipedia.org/wiki/Load_bank

Resistive circuits don’t cause phase shifts. That’s the problem in
trying to using resistors as a substitute for your induction motor.
In this type of circuit with a motor, you HAVE phase shifts and the
process of tuning is to try and minimize them by adjusting
capacitance. It’s just the nature of the beast.

You can adjust a resistive load to make it draw the same as a motor,
but it will only be a good substitute for a motor circuit with a PF=1.
It won’t be a good substitute for your motor and it won’t help you to
tune your RPC.

With reactive loads (such as motors), you get phase shifts between the
current and voltage curves (PF1) and when you tune the circuit by
adding capacitance to balance the phase shift caused by inductance in
the motor. In essence, you try to shift the voltage and current
curves back together and get as close to PF=1 as possible. This
optimizes power transfer to the motor.

I know that you are just trying to get this thing to work and all this
probably sounds like “hand waving”, but you are better off trying to
balance your RPC with your real-live motor in place and under some
mechanical load.

I hope that you get your project working, and my apologies if this was
not helpful or I’ve erred.
=============================================

Well, I can do exactly the same thing using a 5 hp motor as the load.

Maybe your point that pure resistive loads don't cause phase shifts is why I
never had to change the capacitance value!

As you pointed out earlier, it would not be a motor under load, unless it
were, well, loaded!
Your pump idea would be very good, if I had one, or even a compressor head.
Or an auxiliary generator head.
Or mebbe just find a motor with really bad bearings!!

Actually, I do have one possible solution:
A kalamazoo hydraulic cutoff saw, while cutting. The hydraulic motor should
be a constant load (for the hydraulic vise), and the blade motor, while
cutting a thick material, should offer a pretty constant load, as well.

RPC's can be tuned only approximately anyway (iiuc), as what is right for a
given motor under one load won't be quite right for that same motor under
another load. Or other motors. Correct?

So you sort of have to anticipate a load in advance, for optimal tuning.

I'll keep an eye out for some kind of generator head. That would be a very
cool way of doing this, as a resistive load (and the current thereof) on a
generator is accurately measurable and reproducible.

--
DT

"Denis G." wrote in message
...
From reading previous RCM discussions of this topic, it seems to
sometimes generate as much heated discussion as politics or religion.
For that reason, I’ll disclaim being an expert and “tread lightly.”

There was mention of running two 5 hp idlers and why they behaved so
differently from one 10 hp idler. I assume that the two 5 hp idlers
were running freely and not connected to each other. I believe that
would be like trying to hook up two independent gas-powered 120 VAC
generators and trying to get 240 VAC (single phase). I don’t think
that it will happen because the generators would not likely be in
phase with each other. Same goes with trying to run two free-wheeling
5 hp idlers. The currents and voltages would not add in phase and
would probably bounce back thru the mains or otherwise dissipate by
heat in the windings. More properly done, the motors shafts should be
connected to each other with the proper angle between them so that
they are acting in concert (like a single idler).

As far as using a resistive load to test the RPC, I don’t think that
it will help. What you want to drive with an RPC is a motor which is
a reactive load. A resistive load always has a power factor of 1,
whereas un-tuned reactive circuits generally have a PF 1. The motor
reacts with the idler causing phase shifts between the voltage and
current curves and this affect how much real power can be used by the
motor. If the PF=0.7, then you need to have 1.4 x the current and
double the input power to get the rated output power.

If you had the right tools (multi-channel oscilloscope, current &
voltage taps, etc or LabView data acquisition setup) you might be
better equipped to see what is happening in the circuit. If you had
lots of cash, I’d get one of these:
http://www.duncaninstr.com/power-quality-analyzer.htm

I bought an old watt-hr meter at a flea market and have used it to
verify that I needed to buy a new refrigerator. Maybe you can find
one and monitor your RPC performance. Since you probably can’t afford
a dynamometer, you would need to make sure that you set up your 3
phase motor with a constant workload (like a circulating water pump
and reservoir). If you measure your electrical use over time vs. the
amount of work done, you could use this as a tool to adjust the
capacitance and get some idea of improving the efficiency and PF of
your RPC. Even without a watt-hr meter, you can still make
measurements on the legs of the RPC, but I think that you need to set
up the motor with a constant mechanical load to see the effects of
changing capacitance values. (Measuring voltage and current (RMS
scalar values) will only tell you the “apparent power” used in each
leg and not the real power delivered in producing work. (see
references). I think that even using the watt-hr meter will just give
you an idea of the “apparent power” used in the system, but it is
roughly analogous to the measurement of efficiency, so it could be
thought of as measuring “input power”.)
============================

I think the notion of using a wattmeter for tuning an rpc is very elegant
and also fundamental.
Unfortunately not so easy to set up!

The ultimate wattmeter would be the fuel consumption rate of the engine used
to drive the primary generator. The efficiency of all conversions
"downline" from this would be reflected in this consumption rate.

Talk about impractical!
--
DT






Here are more links (if you are academically-inclined or just want
torture):

Wattmeter and power factor:
http://en.wikipedia.org/wiki/Power_factor
http://en.wikipedia.org/wiki/Wattmeter

3-Phase power measurements:
http://www.6mgb.com/E-News/08-04-01/...-493_short.pdf
http://www.manumesure.com/fiches/pdf...ical_power.pdf
https://www.ohiosemitronics.com/pdf/...-method(E).pdf

Wattmeter – construction and principles of operation:
http://www.usbr.gov/power/data/fist/...10/vol3-10.pdf

DrollTroll wrote:
"Jon Elson" wrote in message
...

DrollTroll wrote:

Awl--

What's more important when adjusting capacitor values:
Getting equal voltage between each pair of legs, or balanced current in
each leg? Seems like you can't have both.


Balanced current in the MACHINE TOOL motor, or in the RPC? You don't care
about balanced current in the RPC, and it won't be balanced, because each
winding is doing a different job. (The 240 V line windings are running
the RPC, the 3rd leg is providing the shifted phase.)

What you are concerned about is getting roughly equal current in the
machine tool motor. And, equal voltages on the line DO NOT guarantee that
you have the proper phase shift of the generated leg. The voltage from
the 240 V mains ceter tap (neutral) to the generated leg should be at 90
degrees to the mains, and about 207 V. Without an oscilloscope, however,
it is pretty hard to measure phase shifts. If the machine tool motor
draws equal current on all 3 legs, with roughly balanced L-L voltages,
then your phase shift is right.

As long as the current in the generated leg of the RPC motor is not
excessive, I'd ignore it. If it IS much above the nameplate rated amps,
then you need to reduce the caps to bring the current down.

At equalized voltage, the current through the generated leg is through
the roof.

I'm not too surprised, but what is "through the roof"? twice rated
current? More?

At equalized current, if line L1-L2 is 240, then L1-L3 and L2-L3 are
about 230-233, which is not so bad.


What are you complaining about? That is quite good for an RPC.


Well, I don't know enough to know if I *should* be complaining or not.... so
I just complain reflexively....

I bought a bunch of cheapie analog VOMs (identical), and now have them
permanently wired among the three legs--convenient. I have clamp-on
ammeters now, but I also plan to get 3 cheapie digital clampons (about $20
each, if I'm lucky), and have them installed permantly as well.

A 10 hp baldor Super E will read over 20 A in the 3rd leg, with closely
matched voltages (more capacitance)--idler only.
With matched currents (much less capacitance), it's as above, 240, 230-233.

A 10 Hp 3-phase, 240 V motor would normally have a full-load
line current around 33 A. So, 20 doesn't sound like much of a
problem. When the RPC is idling (no load motor connected) then
ALL that current is going through the balancing caps, make sure
they can handle that much current.
I will wire up 3 ph resistive loads today (5 A heating elements), in both
delta and wye, and adjust the caps see how the currents/voltages respond *in
the resistive load*. Is this a good test method? Will the delta/wye
config of the load affect things?

I "think" you are overthinking this! Make sure the idler motor
and cap bank are not overstressed when idling, and make sure the
load motor currents are roughly balanced (+/- 10%) and that the
machine tool starts smoothly in all speed ranges. If it meets
all these needs, you are set!
Also, what do you think of the comment to add capacitance between L1 and L2,
and that the current through this capacitor "is the only effective
measurement you can get with a clamp-on ammeter in a rpc"?

I did notice that 25 uF between L1 and L2 drops the current in L1 and L2 by
about 2 amps, without affecting the current in L3 (no load).


Not really sure, but I don't think it matters much.

Jon