3-Phase Converter Question
adam smith
is not suitable for a VFD (motor starts with hard load, and must start
*real* quickly). I'm noticing that all the rotary converter companies
are actually passing the 2 wires of 1-phase service directly though to
the load. The converter is thus only manufacturing the "3rd leg".
My question is: is there any problem arising from the fact that the
votlages in the 2 utility-supplied legs are still 180 degrees apart,
rather than 120 as with true 3-phase power? It seems to me that if you
visually represent the converter's "fake" 3-phase they way they do
with real 3-phase (voltage phasor diagram), you would not have a "Y"
shape diagram, but rather a "T" that is basically single phase power
with an extra leg somewhere roughly perpendicular to the utility
lines.
Grant Erwin
lines as L1 and L2. You are thinking of a voltage between L1 and some
reference, and a voltage between L2 and some reference, and you are
confused because you don't see how there could be a voltage between
a generated L3 and that same reference such that the 3 voltages make
up 3 phase power.
Instead, consider the voltage *between* L1 and L2. Now add a generated
L3, and then think of the voltage between L1 and L3 as another leg, and
the voltage between L2 and L3 as the 3rd leg.
Does that make it easier? Anyway, the 3 phase power I get from my
homemade phase converter is really really 3 phase power. Really.
So will yours be.
I'm worried about the hard loading fast starting requirement, though.
That's why I converted my big air compressor to single phase ..
Grant Erwin
Kirkland, Washington
Adam
Grant, thank you. Put that way, it makes perfect sense to me. I was
indeed thinking of both legs of 1-phase with respect to the neutral.
-Adam
adam,airraidsirens,com
Bob Swinney
terminal to terminal, to the load and the converter does nothing more than
provide a shifted 3rd phase for starting purposes. After starting, the
3-phase load motor will continue to run on single phase power and the static
converter is out of circuit.
If the 3-phase load motor is loaded at startup, it can still operate from a
static converter. In this case you will have to select a so-called "heavy
duty" static phase converter. Check with Phase-a-Matic or any of the phase
converter manufacturers and tell them how much load you are starting. They
can advise an appropriate "heavy duty" static converter.
A rotary phase converter might be a good choice for you. Suitable rotaries
can be built (or bought) for loaded start-up as well.
A minor point:
You wrote: "...My question is: is there any problem arising from the fact
that the voltages in the 2 utility-supplied legs are still 180 degrees
apart, rather than 120 as with true 3-phase power?..."
A 3-phase motor operating from a static phase converter is running on single
phase current. Essentially, 2 of the "phases" are applied to 2 "phases" of
the load motor with the 3rd phase windings held open. That constitutes a
single-phase machine.
In 3-phase machines the phase winding sets are separated by 120 mechanical
degrees, whereas the phase voltages and currents within the machine are
separated by 60 electrical degrees. The math would be: Phase A = sine
theta. Phase B = sine (theta + 60) and Phase C = sine (theta + 240 - 180).
The strange expression above is on account of the "middle" winding is
reversed in polarity ref. the other 2 windings. Any 3 winding sets A, B and
C in a 3-phase machine are displaced around the stator by 120 (mechanical)
degrees. Because of the middle winding "turnover" the voltages generated or
induced in these 3 windings are 60 degrees apart.
Bob Swinney
"adam smith" <ad...@airraidsirens.com> wrote in message
news:2d8f1684.03032...@posting.google.com...
> I'm looking at rotary phase converters for an application I have which
> is not suitable for a VFD (motor starts with hard load, and must start
> *real* quickly). I'm noticing that all the rotary converter companies
> are actually passing the 2 wires of 1-phase service directly though to
> the load. The converter is thus only manufacturing the "3rd leg".
>
Bob Powell
news:LaOga.234294$F1.26065@sccrnsc04...
<snip>
> You wrote: "...My question is: is there any problem arising from the fact
> that the voltages in the 2 utility-supplied legs are still 180 degrees
> apart, rather than 120 as with true 3-phase power?..."
>
> A 3-phase motor operating from a static phase converter is running on
single
> phase current. Essentially, 2 of the "phases" are applied to 2 "phases"
of
> the load motor with the 3rd phase windings held open. That constitutes a
> single-phase machine.
>
> In 3-phase machines the phase winding sets are separated by 120 mechanical
> degrees, whereas the phase voltages and currents within the machine are
> separated by 60 electrical degrees. The math would be: Phase A = sine
> theta. Phase B = sine (theta + 60) and Phase C = sine (theta + 240 -
180).
> The strange expression above is on account of the "middle" winding is
> reversed in polarity ref. the other 2 windings. Any 3 winding sets A, B
and
> C in a 3-phase machine are displaced around the stator by 120 (mechanical)
> degrees. Because of the middle winding "turnover" the voltages generated
or
> induced in these 3 windings are 60 degrees apart.
I'll try a different way of responding to:
"...My question is: is there any problem arising from the fact
that the voltages in the 2 utility-supplied legs are still 180
degrees
apart, rather than 120 as with true 3-phase power?..."
The answer is they are *not* 180 degrees apart. The peaks of the 2 single
phase utility-supplied legs only appear 180 degrees apart when you measure
them relative to neutral.
But the 3 phase motor does not see the neutral, in fact the "center" of the
3 phase power provided by the rotary phase converter is offset from the
single phase neutral by 30 AC volts or so for a 240V rotary converter. It
is a different neutral and if you were to generate it using transformers or
resistors to divide down the converter's 3 phase output you would then see
the 2 single phase legs appear to be 120 degrees apart not 180.
Bob Powell
C What I Mean
> is not suitable for a VFD (motor starts with hard load, and must start
> *real* quickly). I'm noticing that all the rotary converter companies
> are actually passing the 2 wires of 1-phase service directly though to
> the load. The converter is thus only manufacturing the "3rd leg".
>
VFD's can start pretty fast. Some about .1sec. I doubt your motor could
get to speed any faster on a rotary phase converter.
Adam
> I'm worried about the hard loading fast starting requirement, though.
> That's why I converted my big air compressor to single phase ..
The motor I'm starting is a 3HP piston-driven horn. The mechanism is
similar to a 1-stage air compressor piston, except that the displacement
per stroke is very small and the reciprocation rate is very high
(140Hz). The motor is a 2-pole 60Hz type with a 2.5:1 speed increaser
gearbox. In order to make a responsive horn blast, the motor has to come
up to speed as close to intantaneously as possible.
The phase converter companies I called are all recommending that I
simply oversize my converter. I have a surplus 15HP 215T motor here in
my shop that I'm going to try before I shop around for anything else. I
have already made a capacitor starter that brings this motor up to speed
fine, by roughly following George Carlson's static converter design.
-Adam
adam at airraidsirens dot com
s falke
"Adam" <please_look@my_signature.com> wrote...
>
> > I'm worried about the hard loading fast starting requirement, though.
> > That's why I converted my big air compressor to single phase ..
>
To me, a second or two spin-up for a siren is somewhat charming, and reminds
me of hearing them as a kid, growing up across the street from a fire station.
Aside from the converter, for 3HP in the home shop, another cause of
less-than-snappy acceleration may be voltage drop during motor inrush. If you
can get it running, the 15HP motor will probably give the best voltage profile
for starting the 3ř 3HP motor. There will be a degree of stored electrical
energy in the 15HP unit.
Given the short duty [I'd hope :/ ] of planned tests, for future research, you
may stumble across a nice surplus ą25kW engine-generator set. For the typical
12-lead genset stator, that would also give you the capability for multiple
output voltages. A trailer-mounted rig would also permit you to have a
greater selection of "test sites."
--s falke
PH
have no problems starting your motor. They use them on rock crushers all
the time!!
"adam smith" <ad...@airraidsirens.com> wrote in message
news:2d8f1684.03032...@posting.google.com...
Adam
> "adam smith" <ad...@airraidsirens.com> wrote in message
> news:2d8f1684.03032...@posting.google.com...
> > I'm looking at rotary phase converters for an application I have which
> > is not suitable for a VFD (motor starts with hard load, and must start
> > *real* quickly). I'm noticing that all the rotary converter companies
> > are actually passing the 2 wires of 1-phase service directly though to
> > the load. The converter is thus only manufacturing the "3rd leg".
>
> VFD's can start pretty fast. Some about .1sec. I doubt your motor could
> get to speed any faster on a rotary phase converter.
There are a few other factors that turned me away from considering a VFD
as a phase converter, primarily price relative to a used 3-phase motor
from a local shop and some run caps from ebay ($180 total). But there
were also these considerations:
1. The motor I need to run is 460V-only, 3-wires (not 6 or 9 or anything
that can be converted to 240 or 277). I have a 240-to-480V 1-phase
transformer, but I have not found any VFD that takes 460V 1-phase input
(most 460V VFDs are 3-phase in and out). One may very well exist, but at
what cost?
2. The horn needs to start in well under 1-second, 0.1 seconds not being
unreasonable. The motor load is quite heavy, consisting of a 1:2.5 speed
increaser (output shaft at 8400 rpm) driving an air compressor type
piston at 140Hz.
3. All of the VFD manufacturers I've spoken with have indicated that for
a heavy-load, fast-start application I need to oversize the VFD so that
it does not need to "soft start" the motor to prevent it from exceeding
its output current rating. This seems to translate into more $$$.
So what I have now to work with are a 5kVA 240/480 transformer, a used
215T (15HP) motor and a bank of run caps (Ten 52uF 370V caps configured
as a bank rated for 130uF @ 740V). Under $300 spent total, and the only
downside seems to be the weight of the converter (about 200 pounds)
which I can live with, as nice as a light solid-state solution would be.
Gary Coffman
No. First a couple of definitions. A voltage is the potential difference
between two points. So a single wire doesn't have a particular voltage.
In other words, it takes at least two wires to make a circuit, one voltage
appears between any given pair, and one current flows through them.
That's one phase.
At any given instant, current is flowing out of one wire, through the
load, and into the other wire. It is the same current, flowing in the
same direction at each and every point of the circuit (Kirchhoff's law).
Only if you shuffle your point of view back and forth can you call
one wire 180 degrees with respect to the other.
In other words, 180 degrees would indicate a reversal of direction
of current flow in the circuit, either toward or away from the point of
observation. If you pick any point on any wire carrying current, you'll
see current flowing toward that point if you look in one direction
(0 *physical* degrees), and current flowing away from that point if
you look in the other direction (180 *physical* degrees). But that's
*cheating*. You're required by vector math to consistently observe
vectors from one orientation, conventionally that's from tail to head.
Phase is defined as the time difference between identical points on
two *separate* repetitive waveforms, or the angle on a phasor diagram
between two vectors when both are properly viewed tail to head. For
sinewaves, it is conventional to pick the positive going zero crossing
points as the points of reference.
Note that since one cycle represents a fixed time increment for any
given frequency, it is conventional to express the time difference of 2
sinewaves of the same frequency in terms of degrees, ie 1 degree is
1/360th of a cycle. (Radian measure can also be used, and often is
in certain calculations.)
Now obviously, since 2 wires are required for one phase, polyphase
power requires more than two wires.
In 2 phase power, the two voltages and currents are in quadrature, ie at
90 degrees to each other, or put another way, one waveform is delayed
by 1/4 of a cycle from the other waveform. Since 360/4=90, but 360/2=180,
if we were to try to carry it on 2 wires, we'd have a situation where the phase
voltages would fight each other, so normally 2 ph power requires 4 wires.
(There's a way to do it with 3 wires, but the phase center is displaced.)
In 3 phase power, there are three voltages and three currents offset from
each other by 120 degrees. This has the nice property that, since 360/120=3
and 360/3 =120, we can combine one wire of each pair with a wire from the
adjacent pair without any voltage disturbance. (If this isn't clear, the voltages
on adjacent wires of each pair of pairs are identical at every point in time.)
So we have 3 phases, A-B, B-C, and C-A, with only 3 wires instead of 6.
Now the idler motor has 3 sets of coils (for a 3600 RPM motor) that are
spaced equally around the armature, physically 360/3=120. When a
current is induced into the armature (transformer action), it induces
currents into each of the 3 windings in turn as it rotates past them (rotary
transformer action) once per cycle (3600 RPM/60 seconds in a minute =
60 cycles per second).
These 3 currents (and their associated voltages), are separated from each
other by 120 electrical degrees (time delay). They can't do anything else,
the physical system demands it, and they *must* sum to 360 degrees over
a complete cycle or you'd have discontinuous waveforms instead of sinewaves.
The input line voltage and current L1-L2, is *one* of these 3 phases. The other
two are generated by the rotary transformer action of the idler.
Note, for an 1800 RPM motor, there are 6 coils paired in staggered fashion,
1-4, 2-5, 3-6, which also works out because one revolution of the motor takes
2 cycles, and 6/2=3, so we're back to the same situation.
Note also that physically for either idler motor, each coil is divided in half,
with one half placed on either side of the armature, but wired in series so
that each pair forms a single electrically complete coil. This is just an
implementation detail, necessary because there are no such things as
magnetic monopoles, and is of no consequence to the rotary transformer
action being described.
It sometimes leads to a confused view which claims there are 60 degrees
between phases instead of 120 degrees. But that view is wrong since 60 x 3
= 180 instead of 360. In other words, it only describes half a cycle instead of
a full cycle. When we properly treat the coil halves as a unit (the way they're
actually wired to produce an electromagnet with *2* poles) we're back to
360/3=120, which is what you'd measure with an oscilloscope if you looked
at the phases.
Gary
Randal O'Brian
of about 5X the horn motor KVA rating. This will insure that their
impedance(s) are not going to seriously limit the inrush current(and also
the motor acceleration rate) when you energize the horn. The rotary
converter is problem enough, so don't handicap yourself with undersized
transformers.
Randy
"Adam" <please_look@my_signature.com> wrote in message
news:please_look-EF29...@reader1.news.rcn.net...
Adam
AC/DCdude17 <Je...@prontoREMOVETHISmail.com> wrote:
> X-No-Archive: Yes
>
> Is there anyway you can convert it to universal motor?
No, because the piston has to hit 140 Hz on the nose in order to
resonate the horn. The good thing about the induction motor is that it
snaps right to its slip speed, which is consistant since the load is not
variable. A universal motor would require external speed regulation
(probably a tach and feedback PWM controller) and I don't think I could
make that modification affordably, if at all. The motor in question is
also custom and built right into the horn. It does not come apart from
the horn or have standard NEMA or IEC frame.
If this horn were to be designed from scratch for single phase, I think
a repulsion-induction motor would be the obvious choice (starting torque
like a universal, fixed top speed of an induction motor).
> For high speed, relatively low torque application that requires fast start
> up, I think universal motor is better suited. Induction motor rotor is
> heavy meaning it has a lot of rotational inertia. Because acceleration =
> force/mass and force is dependent on current, it will take a lot of inrush
> to accelerate quickly. I don't think you can get it to accelerate as fast
> as it would when operated on low impedance 3phase utility when you run it on
> a small rotary converter.
I don't have true 3-phase to compare to, but I fired up my oversized
rotary (15HP) yesterday and it works much better than I had hoped for.
s falke
> > "adam smith" <ad...@airraidsirens.com> wrote...
> So what I have now to work with are a 5kVA 240/480 transformer, a used
> 215T (15HP) motor and a bank of run caps (Ten 52uF 370V caps configured
> as a bank rated for 130uF @ 740V). Under $300 spent total, and the only
> downside seems to be the weight of the converter (about 200 pounds)
> which I can live with, as nice as a light solid-state solution would be.
Keep in mind that there may be significant momentary voltage drop during motor
spin-up, {higher stator current; rotten power factor} through your dry-type
and in the circuit to and including the utility transformer in the alley.
I don't want to discourage your experimentation, but if it doesn't perform
exactly as planned, understand what the potential limitations may be.
Fast-starting, higher-torque motors like to see a "stiff source."
--s falke
Bob Swinney
toward the end. They are:
(Much snippage) and the statement:
"> Note, for an 1800 RPM motor, there are 6 coils paired in staggered
fashion,
> 1-4, 2-5, 3-6, which also works out because one revolution of the motor
takes
> 2 cycles, and 6/2=3, so we're back to the same situation."
If by "coils paired in staggered fashion" is meant 6 poles, then that would
define a 1200 rpm machine.
And then:
"> Note also that physically for either idler motor, each coil is divided in
half,
> with one half placed on either side of the armature, but wired in series
so
> that each pair forms a single electrically complete coil. "
Each coil divided in half (really coil group) with one half placed 180
*electrical* degrees from the other, and in series, defines a pair of
poles. Schematically, each coil half (each pole) of a single electrically
complete coil appears to be 180 degrees apart, but in reality the 180
degrees are electrical degrees, not mechanical degrees.
And finally:
"> It sometimes leads to a confused view which claims there are 60 degrees
> between phases instead of 120 degrees. But that view is wrong since 60 x 3
> = 180 instead of 360. In other words, it only describes half a cycle
instead of
> a full cycle. When we properly treat the coil halves as a unit (the way
they're
> actually wired to produce an electromagnet with *2* poles) we're back to
> 360/3=120, which is what you'd measure with an oscilloscope if you looked
> at the phases."
Correct, there are 120 degrees between phases - as stated, it has to be that
way. But from one phase group (of coils) to the next there are 60
electrical degrees. The arithmetic would be: 3 X 60 = 180. And there are
180 electrical degrees in each half of the armature thus each phase group of
coils on each side of the armature occupies 60 electrical degrees.
This may be more clear with a quote from Electrical Engineers Handbook,
McGraw-Hill, sect 7 Alternating-Current Generators and Motors:
"Electrical and Mechanical Degrees. It is customary to refer to a cycle as
360 elec deg because the voltage wave is sinusoidal and one cycle of a sine
wave is 360 deg. Thus one pair of poles always represents 360 elec deg,
although the number of mechanical degrees a pair of poles occupies on the
periphery of the rotor is only 360 deg divided by the number of pairs of
poles.
Synchronous Speed: Two poles must pass a given point on the stator every
cycle, so that:
Rpm = cycles per minute / pairs of poles; which becomes 7,200 / # of poles
for 60 cycles, etc."
Unquote and back to Bob: Visualize 3 coils (3 phases, if you will) on one
half of a 2 pole motor circle diagram. Naturally there would be three
similar coils on the other half. The coils are separated by 60 electrical
degrees but because of the way they alternate in polarity, they produce emfs
that are separated by 120 degrees. The alternation in polarity is
accomplished externally to the coils by jumper connections. The coils are
usu. wound all in the same direction with the jumpers between coils
accounting for the polarity reversal from coil to coil. If 6 wires were
brought out instead of 3 and there were no coil to coil jumpers, then it
would be a 3-phase, 6 wire system.
Bob Swinney
">
Gary Coffman
>Nice dissertation from Gary, however, I am unclear on a couple of points
>toward the end. They are:
>
>(Much snippage) and the statement:
>
>"> Note, for an 1800 RPM motor, there are 6 coils paired in staggered
>fashion,
>> 1-4, 2-5, 3-6, which also works out because one revolution of the motor
>takes
>> 2 cycles, and 6/2=3, so we're back to the same situation."
>
>If by "coils paired in staggered fashion" is meant 6 poles, then that would
>define a 1200 rpm machine.
No. Every electromagnet has 2 poles, a North and a South, remember that
there is no such thing as a magnetic monopole, so 6 coils means12 poles.
It also means 1800 RPM when the coils are paired in staggered fashion.
>And then:
>"> Note also that physically for either idler motor, each coil is divided in
>half,
>> with one half placed on either side of the armature, but wired in series
>so
>> that each pair forms a single electrically complete coil. "
>
>Each coil divided in half (really coil group) with one half placed 180
>*electrical* degrees from the other, and in series, defines a pair of
>poles. Schematically, each coil half (each pole) of a single electrically
>complete coil appears to be 180 degrees apart, but in reality the 180
>degrees are electrical degrees, not mechanical degrees.
Again no. Because the coils are in series, they have to be the same
electrical phase, 1 current=1 phase. The halves are arranged 180
*physical* degrees from each other, each half is on opposite sides
of the armature. That's also physically necessary because their
magnetic poles have to be on opposite sides.
>And finally:
>"> It sometimes leads to a confused view which claims there are 60 degrees
>> between phases instead of 120 degrees. But that view is wrong since 60 x 3
>> = 180 instead of 360. In other words, it only describes half a cycle
>instead of
>> a full cycle. When we properly treat the coil halves as a unit (the way
>they're
>> actually wired to produce an electromagnet with *2* poles) we're back to
>> 360/3=120, which is what you'd measure with an oscilloscope if you looked
>> at the phases."
>
>Correct, there are 120 degrees between phases - as stated, it has to be that
>way. But from one phase group (of coils) to the next there are 60
>electrical degrees. The arithmetic would be: 3 X 60 = 180. And there are
>180 electrical degrees in each half of the armature thus each phase group of
>coils on each side of the armature occupies 60 electrical degrees.
Monopoles again. Don't exist. Against the laws of physics.
>This may be more clear with a quote from Electrical Engineers Handbook,
>McGraw-Hill, sect 7 Alternating-Current Generators and Motors:
>
>"Electrical and Mechanical Degrees. It is customary to refer to a cycle as
>360 elec deg because the voltage wave is sinusoidal and one cycle of a sine
>wave is 360 deg. Thus one pair of poles always represents 360 elec deg,
>although the number of mechanical degrees a pair of poles occupies on the
>periphery of the rotor is only 360 deg divided by the number of pairs of
>poles.
Yes.
>Synchronous Speed: Two poles must pass a given point on the stator every
>cycle, so that:
>Rpm = cycles per minute / pairs of poles; which becomes 7,200 / # of poles
>for 60 cycles, etc."
That expression is correct *per phase*. Of course a 3 ph motor has 3 phases,
so the number of poles is times 3 also. That's the mistake you made up above.
>Unquote and back to Bob: Visualize 3 coils (3 phases, if you will) on one
>half of a 2 pole motor circle diagram. Naturally there would be three
>similar coils on the other half. The coils are separated by 60 electrical
>degrees but because of the way they alternate in polarity, they produce emfs
>that are separated by 120 degrees. The alternation in polarity is
>accomplished externally to the coils by jumper connections. The coils are
>usu. wound all in the same direction with the jumpers between coils
>accounting for the polarity reversal from coil to coil. If 6 wires were
>brought out instead of 3 and there were no coil to coil jumpers, then it
>would be a 3-phase, 6 wire system.
Those half coils are in phase with their other halfs on the other side
of the armature. The full coils are 120 degrees apart electrically, and
in the case of a 3600 RPM motor, physically too. That gives you the
full 360 degrees per revolution that is a necessary condition for the
motor to operate. In other words, the armature is locked to the rotation
of the phase vector. That rotation is continuous (120+120+120=360),
there is no discontinuity as would be the case if your 60 degree separation
were true.
Gary
Bob Swinney
Gary sez:
> No. Every electromagnet has 2 poles, a North and a South, remember that
> there is no such thing as a magnetic monopole, so 6 coils means12 poles.
> It also means 1800 RPM when the coils are paired in staggered fashion.
The internationally accepted dictum has it that when one describes a machine
according to the number of poles, it is understood that he is describing the
number of poles per phase. If your statement, "so 6 coils means 12 poles"
really implies a 4-pole, 3-phase machine then I concur, it must have a
synchronous speed of 1800 RPM.
Gary sez:
Again no. Because the coils are in series, they have to be the same
> electrical phase, 1 current=1 phase. The halves are arranged 180
> *physical* degrees from each other, each half is on opposite sides
> of the armature. That's also physically necessary because their
> magnetic poles have to be on opposite sides.
So what's to argue here? Of course each end of the coil/phase group is
diametrically opposite. That would mean 180 mechanical degrees apart, as
you said, a physical necessity.
Gary sez:
> That expression is correct *per phase*. Of course a 3 ph motor has 3
phases,
> so the number of poles is times 3 also. That's the mistake you made up
above.
Not a mistake. Read above, the part about "internationally accepted
nomenclature."
Gary sez:
> Those half coils are in phase with their other halfs on the other side
> of the armature. The full coils are 120 degrees apart electrically, and
> in the case of a 3600 RPM motor, physically too. That gives you the
> full 360 degrees per revolution that is a necessary condition for the
> motor to operate. In other words, the armature is locked to the rotation
> of the phase vector. That rotation is continuous (120+120+120=360),
> there is no discontinuity as would be the case if your 60 degree
separation
> were true.
Of course, each end of the phase coil groups are diametrically opposite each
other. It makes no difference whether it is a 3600 RPM machine or not. No
one can argue that the phases (full coils ?) are separated by 120 physical
degrees in a 3-phase machine. And I think most of us accept the "rotor
locked to the revolving field on the stator" notion. Belaboring the
obvious, aren't we? Currents and voltages, however, in each phase of a
3-phase machine are separated by 60 electrical degrees. External
connections between phase coil groups, call them jumpers if you will,
account for the familiar 120 degree phase separation.
Bob Swinney
PS: You want some arithmetic? Visualize a rudimentary 3-phase, 2-pole
machine with 6 coils physical coils spread around the stator. Count 'em:
1, 2, 3, 4, 5, 6. Right? Now count 'em again:. 60 degrees, 120 degrees,
180 degrees, 240 degrees, 300 degrees, 360 degrees.
Gary Coffman
>Gary sez:
> Again no. Because the coils are in series, they have to be the same
>> electrical phase, 1 current=1 phase. The halves are arranged 180
>> *physical* degrees from each other, each half is on opposite sides
>> of the armature. That's also physically necessary because their
>> magnetic poles have to be on opposite sides.
>
>So what's to argue here? Of course each end of the coil/phase group is
>diametrically opposite. That would mean 180 mechanical degrees apart, as
>you said, a physical necessity.
What's to argue is that you said previously that the coil halves are 180
degrees *electrical* out of phase. They aren't, and they can't be because
that would result in a net *zero* magnetic field.
>Gary sez:
>> Those half coils are in phase with their other halfs on the other side
>> of the armature. The full coils are 120 degrees apart electrically, and
>> in the case of a 3600 RPM motor, physically too. That gives you the
>> full 360 degrees per revolution that is a necessary condition for the
>> motor to operate. In other words, the armature is locked to the rotation
>> of the phase vector. That rotation is continuous (120+120+120=360),
>> there is no discontinuity as would be the case if your 60 degree
>separation
>> were true.
>
>Of course, each end of the phase coil groups are diametrically opposite each
>other. It makes no difference whether it is a 3600 RPM machine or not. No
>one can argue that the phases (full coils ?) are separated by 120 physical
>degrees in a 3-phase machine. And I think most of us accept the "rotor
>locked to the revolving field on the stator" notion. Belaboring the
>obvious, aren't we? Currents and voltages, however, in each phase of a
>3-phase machine are separated by 60 electrical degrees. External
>connections between phase coil groups, call them jumpers if you will,
>account for the familiar 120 degree phase separation.
Now this is the source of disagreement. How are you getting the 60 degree
electrical phase shifts? The driving voltages are separated by 120 electrical
degrees, and hence their currents must also be separated by 120 electrical
degrees. You can't pull the -180 trick because we've agreed the coil halves
must be in phase, otherwise the net magnetic field would be zero. So where
do these 60 degree phase shifts come from?
Gary
Bob Swinney
> Now this is the source of disagreement. How are you getting the 60 degree
> electrical phase shifts? The driving voltages are separated by 120
electrical
> degrees, and hence their currents must also be separated by 120 electrical
> degrees. You can't pull the -180 trick because we've agreed the coil
halves
> must be in phase, otherwise the net magnetic field would be zero. So where
> do these 60 degree phase shifts come from?
I suppose the best explanation I've seen is in the 1933 ed. of "Standard
Handbook for Electrical Engineers". I will quote passages here and attempt
to describe the accompanying diagrams. Here goes:
4. Electrical and magnetic degrees. It is customary to refer to a cycle as
360 electrical degrees (see Fig. 2), or one electrical revolution, and to
the circumferential distance from the center of one pole to that of the next
pole of the same polarity, as 360 magnetic degrees. Thus two active
armature conductors displaced from each other by B magnetic degrees will
experience e.m.fs. differing in phase by B electrical degrees. Two active
conductors 180 magnetic degrees apart would naturally be connected in series
to form a loop or coil, since their two e.m.fs. will then be always in the
same direction around the circuit of the loop (see coils I and II, Fig.3).
{Fig. 2 is merely a drawing of a single cycle of a sine wave, entitled, "A
graph of a cycle of e.m.f."}
{Fig. 3 shows 2 adjacent coils on the outside of an armature labeled I and
II - and entitled, "Fig.3 - Part of the armature and field of a
quarter-phase alternator"}
5. The two-phase (quarter-phase) alternator. Referring to Fig. 3, coils I
and II are displaced from each other by 90 magnetic degrees. The e.m.fs.
induced in these two coils therefore differ in phase by 90 deg. If all
coils in the same phase as I, are connected in series, and all those in the
same phase as II in another series, the two circuits being separate and
distinct electrically, the machine is called a two-phase machine. The circu
its themselves are referred to as the two phases or phase windings of the
machine. Sometimes this type of machine is called a quarter-phase machine,
since its two e.m.fs. differ in phase by 90 deg. or one-quarter cycle. The
electromotive force in coil I is zero when that in coil II is a maximum, and
vice versa. If the value of the electromotive force from instant to instant
is a sine function of the relative angular position of the armature coils
and revolving field, the curves of the electromotive forces in coils I and
II may be plotted as indicated in Fig. 4, in which the ordinates represent
the instantaneous values of the electromotive force, and the abscissas
represent time. The four terminals of the two windings of a two-phase
alternator are usually brought out to the terminal board. The load may be
fed by three wires, one the common return for the other two, or four wires
may be employed as two independent circuits.
{Fig 4 shows 2 sine waves labeled I and II, with the positive going zero
crossings of each displaced by 90 degrees. It is entitled, "Graph of
e.m.fs. in a quarter-phase alternator, plotted against time.}
6. The three-phase alternator. If three sets of overlapping coils
displaced by 60 magnetic degrees (Fig. 5) be connected to three separate
circuits, their e.m.fs. will differ in phase by 60 deg. (see Fig. 6) By
reversing the terminal connections of phase II we reverse the phase of its
e.m.fs. with respect to the load and obtain the three voltages shown in Fig.
7. These voltages differ in phase by 120 deg. The difference between the
three e.m.fs. of Fig. 6 and those of Fig. 7 is obviously external to the
machine in which they are generated, which is called a three-phase
alternator. In many three-phase alternators the phases are internally
connected, with only three terminals brought out. There are two principal
methods of making these connections, Y and Delta (see Figs. 8 and 9) also
called "star" and "mesh".
{Fig. 5 shows 3 adjacent coils on the outside of an armature labeled I, II
and III and entitled "Part of the armature and field of a three-phase
alternator"}
{Fig. 6 shows three sine waves, labeled left to right as I, II and III, with
the positive going zero crossings of each displaced by 60 degrees}
{Fig. 7 shows three sine waves, labeled left to right as I, III and II with
I beginning at 0 degrees. I and III are displaced by 120 degrees as
indicated by their positive going zero crossings. Wave II is drawn as a
dotted line with its positive going zero crossing at 240 degrees. Wave II,
dotted line, has its beginning as a negative going zero crossing at 60
degrees, the same point as where wave II in Fig. 6 has its beginning as a
positive going zero crossing.}
{Figs. 8 and 9 are Y and Delta diagrams entitled "Internal and external
e.m.fs. and currents in three-phase alternators with Y and Delta
connections"}
7. Note on phase classification. There is a slight inconsistency in the
usual classification of alternators and systems. In the ordinary
three-phase alternator (Fig. 5) there are, in each 360 magnetic degrees of
circumference, six coil groups or belts of conductors. In these groups are
being generated six different e.m.fs. differing in phase progressively by 60
deg. or one-sixth of a cycle. If the six terminals of the three phases be
connected to three equal loads, the six currents in the six leads, when
counted positive in the same direction along the line, will differ in phase
progressively by 60 deg. If, however the three phases be connected in Y or
Delta, with three leads connected to the load, the three currents, counted
positive in the same direction along the line, will differ in phase by 120
deg. or one-third of a cycle. To be consistent these two systems should be
referred to as "six-phase" and "three-phase" respectively, although the
alternator is the same. Moreover when a closed-coil armature is used for
alternating e.m.f. generation, as in the case of a synchronous converter, it
is called "three-phase" when it has three 120-deg. taps and three
phase-belts of 120-deg. span, and "six-phase" when it has six 60-deg. taps
and six phase-belts each of 60-deg. span. Thus, to be consistent, the
ordinary three-phase alternator should be called a six-phase alternator,
although mostly used to supply three-phase circuits.
Regards,
Bob
daestrom
"Gary Coffman" <ke...@bellsouth.net> wrote in message
news:0a5h8v03sq3k1ue1u...@4ax.com...
Having re-wound many a three-phase motor (and a few generators as well as DC
machinery), I'll jump in here with my $0.35 worth (inflation on $0.02 don't
you know ;-)
If you 'travel' around a 2 pole, 3-phase motor's stator (3600 sync speed on
60 hz), you will find the coil-phase groups like this...
A, C', B, A', C, B'
The ' are the second half of a coil-phase winding. After a positive peak on
the A phase, the next phase to reach maximum is actually C. It is negative
however. Because of this, the C' half is connected *backwards* from the C
coils. That is to say, when the current flows 'Positively' through the A
coils and creates a magnetic pole of a given polarity, a current flowing
'negatively' through the C' coils will create a magnetic pole of the same
polarity. So after the A current peaks positively and then the C current
peaks in the negative direction, there is a smooth transition of magnetic
field from A to the C' coil faces.
So with this in mind, you will find the C' coil 60 mechanical degrees along
the stator from the A coil, in the direction of rotation. The B coil is
another 60 degrees (120 from the original A coil). The A' another 60
degrees (180 from orig. A) and the C coil 210 degrees from the original A
coil.
For a four-pole 3-phase motor (1800 sync speed on 60 hz)....
A, C', B, A', C, B', A, C', B, A', C, B'
Four poles total for each phase. The individual coil-phase groups *may* be
wired in series, but often they are wired in a series-parallel combination.
Sometimes, all four coil-phase groups may even be wired in parallel. But
the 'prime' coils will be connected "backward" from the non-prime coils.
This gives the effect as if they were physically wound around the iron in
the opposite direction to give opposite magnetic polarity.
In the four pole unit, the coil-phase groups are spaced 30 mechanical
degrees apart around the stator.
The A and A' are said to be 180 electrical degrees out of phase by virtue of
the reversed connections, even though they are only 90 mechanical degrees
apart on a four-pole machine.
daestrom
And don't *even* get me started on 'consequent-pole' configurations :-)
s falke
"Gary Coffman" <ke...@bellsouth.net> wrote
> Now this is the source of disagreement. How are you getting the 60 degree
> electrical phase shifts? The driving voltages are separated by 120
electrical
> degrees, and hence their currents must also be separated by 120 electrical
> degrees. You can't pull the -180 trick because we've agreed the coil halves
> must be in phase, otherwise the net magnetic field would be zero. So where
> do these 60 degree phase shifts come from?
Er, connect a phase-angle meter to 3ø 4-wire Aø-Bø and Bø-Cø voltages -- for
the default {balanced} conditions, it reads 60°. Do the same for Aø-N and
Cø-N -- the meter will read 120°.
--s falke
Bob Swinney
Bob Swinney
"daestrom" <daes...@twcny.rr.com> wrote in message
news:cH3ia.1060$YQ2...@twister.nyroc.rr.com...
Adam
<Hgkha.31617$ja4.2...@bgtnsc05-news.ops.worldnet.att.net>,
"Randal O'Brian" <ra...@worldnet.att.net> wrote:
> Bear in mind that any transformers in the circuit should have a KVA rating
> of about 5X the horn motor KVA rating. This will insure that their
> impedance(s) are not going to seriously limit the inrush current(and also
> the motor acceleration rate) when you energize the horn. The rotary
> converter is problem enough, so don't handicap yourself with undersized
> transformers.
Randy, your points and those from s_falke are duly noted. I do
understand that my transformer is small for the amount of inrush I'm
expecting, and for the idler, and my converter setup is never going to
be ideal. There are two limiting factors though with my experimentation
that are not electrical in nature: budget and weight. The surplus 5kVA
transformer, for example, is the biggest one I found that I could afford
AND lift to my 3rd floor attic. The 10kVA xfmr I found failed on both
counts.
I'm going to hook it all up to the horn today and just see what I get.
if it honks, great! If not, I probably won't sink any more money into
the converter, and I'll instead try to fit a new 240V motor to the
mounts.
-Adam
daestrom
"s falke" <bus...@pacbell.net> wrote in message
news:OH4ia.368$9n7...@newssvr19.news.prodigy.com...
Ah... but remember, connecting the voltage between A and B is not the same
phase angle as A-N (nor is B-C the same as C-N). Many a delta-wye
transformer connection has been blown by not considering the phase-shift
caused by connecting two transformer banks fed from same primary line ('just
*has* to be in phase, right???'). One wye-delta and the other delta-delta
and *BOOM*.
Measured A-B to B-C at 60 degrees, B-C to C-A is 60 degrees, and C-A to A-B
is 60 degrees. That only adds up to 180. This is because measured this
way, you're only measuring the interior angles of an equilateral triangle
formed by the three phase vectors. That is not correct.
To measure the angle between phases, you must measure them with the correct
'polarity' for the meter. That means if you measure A-B to C-B (reverse the
leads of the second voltage) you get 120 (180 - 60). This requires a
four-wire meter with isolated inputs. If you get only 60 it's because you
hooked your meter up wrong.
daestrom
Romy Singh
> In article
> <Hgkha.31617$ja4.2...@bgtnsc05-news.ops.worldnet.att.net>,
> "Randal O'Brian" <ra...@worldnet.att.net> wrote:
>
> > Bear in mind that any transformers in the circuit should have a KVA
rating
> > of about 5X the horn motor KVA rating. This will insure that their
> > impedance(s) are not going to seriously limit the inrush current(and
also
> > the motor acceleration rate) when you energize the horn. The rotary
> > converter is problem enough, so don't handicap yourself with undersized
> > transformers.
>
> Randy, your points and those from s_falke are duly noted. I do
> understand that my transformer is small for the amount of inrush I'm
> expecting, and for the idler, and my converter setup is never going to
> be ideal. There are two limiting factors though with my experimentation
> that are not electrical in nature: budget and weight. The surplus 5kVA
> transformer, for example, is the biggest one I found that I could afford
> AND lift to my 3rd floor attic. The 10kVA xfmr I found failed on both
> counts.
>
I really have to admire that even though people have limited budgets, they
still do experiments.
Guys like you are admired by people like me. "Doing it for the love of the
game".
Romy
Gary Coffman
>"Gary Coffman" <ke...@bellsouth.net> wrote in message
>news:0a5h8v03sq3k1ue1u...@4ax.com...
>> On Mon, 31 Mar 2003 18:11:26 GMT, "Bob Swinney" <jud...@attbi.com> wrote:
>> >
>> >Of course, each end of the phase coil groups are diametrically opposite
>each
>> >other. It makes no difference whether it is a 3600 RPM machine or not.
>No
>> >one can argue that the phases (full coils ?) are separated by 120
>physical
>> >degrees in a 3-phase machine. And I think most of us accept the "rotor
>> >locked to the revolving field on the stator" notion. Belaboring the
>> >obvious, aren't we? Currents and voltages, however, in each phase of a
>> >3-phase machine are separated by 60 electrical degrees. External
>> >connections between phase coil groups, call them jumpers if you will,
>> >account for the familiar 120 degree phase separation.
>>
>> Now this is the source of disagreement. How are you getting the 60 degree
>> electrical phase shifts? The driving voltages are separated by 120
>electrical
>> degrees, and hence their currents must also be separated by 120 electrical
>> degrees. You can't pull the -180 trick because we've agreed the coil
>halves
>> must be in phase, otherwise the net magnetic field would be zero. So where
>> do these 60 degree phase shifts come from?
>>
>
>60 hz), you will find the coil-phase groups like this...
>
>A, C', B, A', C, B'
>The ' are the second half of a coil-phase winding. After a positive peak on
>the A phase, the next phase to reach maximum is actually C. It is negative
>however. Because of this, the C' half is connected *backwards* from the C
>coils. That is to say, when the current flows 'Positively' through the A
>coils and creates a magnetic pole of a given polarity, a current flowing
>'negatively' through the C' coils will create a magnetic pole of the same
>polarity. So after the A current peaks positively and then the C current
>peaks in the negative direction, there is a smooth transition of magnetic
>field from A to the C' coil faces.
>
>So with this in mind, you will find the C' coil 60 mechanical degrees along
>the stator from the A coil, in the direction of rotation. The B coil is
>another 60 degrees (120 from the original A coil). The A' another 60
>degrees (180 from orig. A) and the C coil 210 degrees from the original A
>coil.
Yes, *physical* degrees. They aren't separated by 60 *electrical* degrees,
however. That's the dispute I'm having with Bob.
>In the four pole unit, the coil-phase groups are spaced 30 mechanical
>degrees apart around the stator.
Yes, no argument there either.
>The A and A' are said to be 180 electrical degrees out of phase by virtue of
>the reversed connections, even though they are only 90 mechanical degrees
>apart on a four-pole machine.
Now that's an issue. A and A' are not in phase opposition. Remember that
every electromagnet must have both a N and a S pole, there are no magnetic
monopoles. Thus A and A' are series wired parts of the same coil, carrying
one current, producing a single N-S pair of poles (which are simultaneously
reacting with the corresponding N and S poles of the armature).
Gary
daestrom
> On Mon, 31 Mar 2003 22:55:04 GMT, "daestrom" <daes...@twcny.rr.com>
wrote:
> >"Gary Coffman" <ke...@bellsouth.net> wrote in message
>
of
> >the reversed connections, even though they are only 90 mechanical degrees
> >apart on a four-pole machine.
>
> Now that's an issue. A and A' are not in phase opposition. Remember that
> every electromagnet must have both a N and a S pole, there are no magnetic
> monopoles. Thus A and A' are series wired parts of the same coil, carrying
> one current, producing a single N-S pair of poles (which are
simultaneously
> reacting with the corresponding N and S poles of the armature).
>
If we have only one coil, and leave the opposite side of the stator empty,
that one coil will create a magnetic pole pair. Perhaps N right in front of
it inside the stator for the positive half-cycle if wound a certain way.
And the matching S pole will be on the other side of the coil, deep in the
iron of the stator. The stator iron will 'conduct' the magnetic flux around
the coil on each side where it will enter the rotor on each side, and then
leave the rotor and link up with the N pole of the stator. If you have some
sort of stator design such that there is a large air-gap preventing the flux
from entering the rotor nearby, most of it will enter the rotor long before
it gets even halfway around to the opposite side.
So I would say that one coil creates a pole pair, but that pair is somewhat
limited to only one side of the rotor. When we add the second coil to the
opposite side, how it is connected has important implications. If it is
connected so the pole formed right inside the stator in front of it forms a
S pole when the same current flows through it as the first coil, then the
two fields 'aid' or 'combine' to form a flux that passes from one side of
the rotor to the other. But it can be connected to form a pole that is N
the same time as the first coil does. Now we have four poles. Two N right
in front of the coils and two S that come out from the iron halfway between
the coils. This is what is sometimes called a 'consequent pole' setup. The
two S poles aren't explicitly wound, they form as a consequence of having
two N coils on opposite sides of the stator. Of course, this setup can be
done with pole/coil counts other than just two.
Many two-speed motors, where one speed is half the other are simply
consequent pole setups where the internal connections between the two
'coil's are brought out to an external controller that reverses the
connections to the second coil independent of the first coil. If you have a
two speed motor with nine-wires, it's a good bet that it is this sort of
setup. Depending on how the connections are made, you can have two-speed
motor that is constant-torque or constant-horsepower.
But as far as calling it 180 out of phase or not, as Einstien said, it's all
relative. Stepping back from the motor and looking at the total flux, you
can say the two coils are 'in-phase' since they are oriented to have their
mmf aligned in the same direction through the rotor. Others may say they
are 180 out (as I did) because when winding the durn thing with your head
halfway into the stator iron, the 'current' enters one coil and travels
clockwise up one slot, over to the right a few lots and down another slot.
Whereas the other coil has the current traveling CCW, it enters the coil of
wire and travels up one slot, over the the *left* a few slots, and down
another. The jumper from the first coil to the second connects the
conductor in the right-hand slot of the first to the conductor entering the
right-hand slot of the opposite coil. So from one side of the line
(terminal connection), the current flows into the left-hand side of the
first coil and goes CW (as viewed from inside the stator's center) several
times around that coil, then passes across to the opposite side of the
stator, enters that coil on the right hand side and makes several passes
around that coil CCW (as viewed again from inside the stator), and then goes
to either the center tie if it's a 'star' or to the next phase terminal if
it's a delta connection.
Of course when I'm looking at the second coil, I've turned the stator over
180 degrees so I can work on the dang thing. In larger machines that can't
be turned over, we lay on our back to work the second coil, but the effect
is the same, we're looking at the stator upside down. So I guess two
"180's" (turning the stator over, and connecting the coil up reversed) make
for 360 and you could be right.
But when you're doing this stuff (or teaching it as I have also done), it's
much easier to visualize it as 180 out of phase. Keeping track of how the
stator iron is oriented as you're looking at it is harder to grasp. So we
think of it this way. If you visualize cutting the stator along it's axis
and unrolling it out to just a long row of iron with slots across it, my
method works as well. We often draw it out on paper this way and verify our
work by later rolling the paper up and placing it inside the stator. Of
course there's a whole bit of topology there, going from 2D flat plane to
circular iron, but it works.
Whew.... Didn't intend to write so much, but sometimes I get to rambling...
I'd say in the end, you're right, but to a re-wind guy, it's not the
'conventional' 'view' point.
daestrom
Gary Coffman
>I'd say in the end, you're right, but to a re-wind guy, it's not the
>'conventional' 'view' point.
Yeah, the thing is with respect to phase, the only viewpoint that matters is the
viewpoint of the current. Shifting your physical viewpoint halfway through can
make it look backwards to you, but to the current it is still all the same direction.
This is the same thing as when people call single phase house wiring two
phase. They're shifting the direction they're looking at each end from the
center tap. But the current doesn't actually shift direction. You haven't actually
split one phase into two.
In the case of a 3 ph motor, you haven't actually split the 120 electrical degree
separated phases into two to give twice as many 60 electrical degree phases.
There's nothing in the circuit which can electrically do that. You're just changing
the way you're looking at it halfway through. In other words, you're being
inconsistent.
This is important only if you want to understand precisely how the motor works.
If you shift viewpoints in the middle like that, a rigorous analysis says you get a
result that violates one or more electrical laws. But of course you can't actually
violate electrical law, so this is telling you that you've done something wrong in
the way you're viewing the system.
What I'm trying to say here is that the rotating field that makes the motor work
is actually occupying the space where the armature sits. It is *between* the coil
halves. So you have to analyze it that way. If you try to analyze each coil half
separately, implicitly shifting the direction you look at it with each half, you don't
wind up with that rotating field vector. The signs come out wrong, and you wind
up with two vectors split between quadrants instead of a single smoothly rotating
vector.
Since you know that the motor actually does work, you then have to fudge the
equations by introducing artifical phase factors (the -180 in Bob's system of
equations) to make your 6 ph model come out with 3 ph answers. The current
in the coil halves isn't actually flowing in two different directions at the same time,
but that's what you've had to make your equations say in order to get the single
smoothly rotating field vector that you know has to be there.
I've made a big issue of this because I believe it is important that the conceptual
mathematical model of the system should actually describe the physical system,
and should not describe a different system which has to be fudged with artificial
factors in order to come out with the correct answers.
Gary
Fitch R. Williams
>So what I have now to work with are a 5kVA 240/480 transformer, a used
>215T (15HP) motor and a bank of run caps (Ten 52uF 370V caps configured
>as a bank rated for 130uF @ 740V).
If you are going to have run caps connected in series to get the higher
voltage rating, you need to put some equalizing resistors across them to
keep the voltage balanced (and discharge them some reasonable time after
the machine is shut off. Run caps, unlike starting caps, have extremely
low leakage. Otherwise the center connection to the end can creep to
full voltage in one direction or the other, vastly exceeding the rating
of the cap seeing the high voltage, which may cause its smoke seal to
leak, a lot.
Fitch
Bob Swinney
have to fudge the
> equations by introducing artifical phase factors (the -180 in Bob's system
of
> equations) to make your 6 ph model come out with 3 ph answers. The current
> in the coil halves isn't actually flowing in two different directions at
the same time,
> but that's what you've had to make your equations say in order to get the
single
I will try to validate the equations again (I didn't intend them to be a
system BTW). I only intended the show how 3 sine waves are formed on three
coil ends of an induction motor.
Visualize 3 sets of coils separated by 60 mechanical, or physical, degrees.
In motor winder's parlance these coil sets are labeled A, C', and B. These
are the first 3 sets of coils in line from the beginning of the stator, say,
at 12 o'clock. It follows the other side of the stator, 180 mechanical
degrees away, has connections to these phase coil sets and they would be
labeled A', C and B'. Equations describing the sine waves on the stator
would be: 1. sine theta 2. sine (theta [240 degrees minus 180 degrees] )
and 3. sine (theta + 120 degrees).
The 3 phase-coil sets are driven, from the line, by 3 phase power, commonly
referred to as Phases, A, B, and C. These phases, on the line, are
separated by 120 degrees. Phase A drives the first coil set at A on the
stator, so call that phase A at time zero. Phase A's waveform on the stator
would be described as sine theta. Next is Phase C, which, on the line, is
120 degrees later. However the winding polarity of coil set C is reversed,
turned over, start to end and end to start, bottom to top and top to bottom,
however you would like to describe it. Accordingly Phase C, at 240 degrees
on the line, is turned over or reversed at coil set C' on the stator. This
is accounted for by the prime sign at the second coil set on the stator, or
by the equation "sine (theta [240 degrees minus 180 degrees] )".
Line phase B at 120 degrees drives coil set B. Therefore the 3 driving
phases, which are separated by 120 degrees on the line, develop currents in
the 3 coil sets that are separated by 60
degrees.
The opposite ends of the phase coil sets are wound 180 degrees with respect
to the input sides we are discussing here; so the currents developed there
form opposite magnetic poles. In a Y configuration, the opposite ends, A',
C, and B' go to the star point, or to next phase set in a Delta
configuration.
I stand by the equations above, describing "3 Phases on a Stator". They are
consistent with motor winders' terminology, motor circle diagrams, fully
developed motor schematics, and any other methods used to describe the
operation of induction motors. The algebraic sum of the 3 sets of
waveforms (actually integrated in the flux field across the stator) is what
forms the stator's rotating magnetic field.
Respectfully,
Bob Swinney
daestrom
"Bob Swinney" <jud...@attbi.com> wrote in message
news:9B4ka.393342$sf5.7...@rwcrnsc52.ops.asp.att.net...
But you're comparing A, B and C' currents. Not A, B and C. The currents
through C are sine(theta+240). The currents of A, B and C are 120 degrees
apart. Certainly, the currents through the B, A' and C coils are 60 degrees
apart as well by your method, but that isn't the normal way to refer to the
three phases.
One can say that A, C', B, A', C and B' are all spaced 60 degrees apart, but
you may as well say you have six phases, 3 that are space 120 degrees apart
and 3 others that are exactly 180 degrees opposite the first three. But
each pair that are exactly 180 out from each other are connected together to
be additive. Very cumbersome way of saying you have 3 phases.
>
> The opposite ends of the phase coil sets are wound 180 degrees with
respect
> to the input sides we are discussing here; so the currents developed there
> form opposite magnetic poles. In a Y configuration, the opposite ends,
A',
> C, and B' go to the star point, or to next phase set in a Delta
> configuration.
Now, sometimes we connect the line to the B' and C' coil group because it's
closer to the terminal connections and we don't have enough room or don't
want to run the leads nearly halfway around. If you connect C' and B'
phases to the line and C and B to the star, you're reversing the current
flow through those phases. This means you connect the opposite side of the
coil from the side you used in A. Then on the star, you use the opposite
side of coils C, B to connect to A' to form the star. I've seen this
technique used in *large* machinery so the 'lead wire' (which is actually
heavy copper bars in that large a machine) doesn't need to be supported as
far around the circumference.
If you run leads 120 degrees around the stator to connect the line to A, B
and C, then the current flowing into A at peak will flow reversed through
the B and C windings. This means you connect the same side of coils A, B,
and C to A, B, and C line respectivly. And the same side of A', B', and C'
to the star. Although this method uses a little more lead wire, it is a lot
less prone to f**ing up. You always connect the left side of a coil to
either line or star, and the right side to another coil (or vice versa if
you want).
But in the first method, you're only reversing the coil leads because you've
reversed the ends of C and C' (and B and B') by connecting C' to the line
instead of C. Two reversals put you back to rights. By making C' the end
tied to the line, you have to reverse which side of the coil you bring out
(compared to the side you bring out of A). If you bring out A B C (not
their primes) you do *not* have to reverse which side of the coil you bring
out, it is always the same side.
Wiring it delta is A' to B, B' to C and C' to A. In this case, the lines
are connected to C', B and A' and that spans half-way around the stator (for
two pole machines anyway). Electrically, this is the same as A, B, C, so
you always bring out the same side of the coils. I don't think I've ever
seen delta done any other way. Current through A is sine(theta+60), B is
sine(theta+180) and C is sine(theta+300)
daestrom
Bob Swinney
and C. The currents
> through C are sine(theta+240). The currents of A, B and C are 120 degrees
> apart. Certainly, the currents through the B, A' and C coils are 60
degrees
> apart as well by your method, but that isn't the normal way to refer to
the
> three phases.
>
> One can say that A, C', B, A', C and B' are all spaced 60 degrees apart,
but
> you may as well say you have six phases, 3 that are space 120 degrees
apart
> and 3 others that are exactly 180 degrees opposite the first three. But
> each pair that are exactly 180 out from each other are connected together
to
Thanks for the confirmation. I was not referring to the "3 phases" but to
the currents forming the 3 phase pole sets on the stator.
I referred to the currents as I did because of a diagram which showed 3 sine
waves, 60 degrees apart (only 1 end of three phases) The caption on the
diagram was "Three Phases on a Stator". The diagram was from an article I
wrote on 3-phase induction motors and rotary phase converters. It was
intended to illustrate the way 3-phase motors utilize 120 degree separated
"phases" to form poles that are separated by 60 degrees. It was one diagram
out of several used to build the concept. Like many technical descriptions,
it could lose something when taken out of context.
Bob Swinney
Gary Coffman
>Gary sez: "...> Since you know that the motor actually does work, you then
>have to fudge the
>> equations by introducing artifical phase factors (the -180 in Bob's system
>of
>> equations) to make your 6 ph model come out with 3 ph answers. The current
>> in the coil halves isn't actually flowing in two different directions at
>the same time,
>> but that's what you've had to make your equations say in order to get the
>single
>> smoothly rotating field vector that you know has to be there."...
>
>I will try to validate the equations again (I didn't intend them to be a
>system BTW). I only intended the show how 3 sine waves are formed on three
>coil ends of an induction motor.
You do have to form the equations into a system. The three phases
have to satisfy the vector sum
A + B + C = R
with the constraints that R must have constant magnitude and it must
increment in phase smoothly and unidirectionally all the way 360 degrees
around the cycle. Otherwise the motor wouldn't work as it does.
Try it with your 60 degree separation equations (all four quadrants,
mind the signs), and see if R retains constant magnitude and that its
phase increment is smooth and unidirectional all the way around the
cycle. Hint: it won't be. Now try it with 120 degree separations.
Gary
Bob Swinney
> You do have to form the equations into a system. The three phases
> have to satisfy the vector sum
>
> A + B + C = R
>
> with the constraints that R must have constant magnitude and it must
> increment in phase smoothly and unidirectionally all the way 360 degrees
> around the cycle. Otherwise the motor wouldn't work as it does.
>
> Try it with your 60 degree separation equations (all four quadrants,
> mind the signs), and see if R retains constant magnitude and that its
> phase increment is smooth and unidirectional all the way around the
> cycle. Hint: it won't be. Now try it with 120 degree separations.
To begin with if R is a magnetmotive force vector launched into the rotor at
all loci around the stator it cannot have constant magnitude. As a
combination of 3 sinusoids, it must be sinusoidal itself.
Presuming A, B and C to be the familiar depiction of a star diagram with 3
equal vectors separated by 120 degrees. Lets assign a magnitude of say 10
units and do the vector arithmetic for A = (10 angle 0), B = (10 angle 120),
and C = (10 angle 240). The result is R = (0 angle 0)
Converting to rectangular notation and adding we have: (10 +j0) + (-5
+j8.66) + (-5 -j8.66) and the result R = (0 +j0). R is still = 0.
Now visualize the same 3 equal vectors, only this time they are separated by
60 degrees. The vectorial combination would be: (10 angle 0) + (10 angle
60) + (10 angle 120) and the result is (20 angle 60). The rectangular
notation would yield (10 +j17.32) which is, of course equivalent to the
polar expression, (20 angle 60).
Assuming the same magnitude of 10 units each, and plotting all loci around
the stator yields a sine wave swinging between a maximum of 17.32 and a
minimum of -17.32. This resultant sine wave would be the combination of 3
vectors, each spaced 60 degrees apart and oscillating in space. The
resultant wave would appear to rotate around the stator.
Respectfully,
Bob Swinney
"Gary Coffman" <ke...@bellsouth.net> wrote in message
news:4il49vgp2pd3afs2u...@4ax.com...
> On Mon, 07 Apr 2003 01:35:01 GMT, "Bob Swinney" <jud...@attbi.com> wrote:
> >Gary sez: "...> Since you know that the motor actually does work, you
then
> >have to fudge the
> >> equations by introducing artifical phase factors (the -180 in Bob's
system
> >of
> >> equations) to make your 6 ph model come out with 3 ph answers. The
current
> >> in the coil halves isn't actually flowing in two different directions
at
> >the same time,
> >> but that's what you've had to make your equations say in order to get
the
> >single
> >> smoothly rotating field vector that you know has to be there."...
> >
> >I will try to validate the equations again (I didn't intend them to be a
> >system BTW). I only intended the show how 3 sine waves are formed on
three
> >coil ends of an induction motor.
>
>
> Gary
>
Gary Coffman
>Gary sez:
>> You do have to form the equations into a system. The three phases
>> have to satisfy the vector sum
>>
>> A + B + C = R
>>
>> with the constraints that R must have constant magnitude and it must
>> increment in phase smoothly and unidirectionally all the way 360 degrees
>> around the cycle. Otherwise the motor wouldn't work as it does.
>>
>> Try it with your 60 degree separation equations (all four quadrants,
>> mind the signs), and see if R retains constant magnitude and that its
>> phase increment is smooth and unidirectional all the way around the
>> cycle. Hint: it won't be. Now try it with 120 degree separations.
>
>To begin with if R is a magnetmotive force vector launched into the rotor at
>all loci around the stator it cannot have constant magnitude. As a
>combination of 3 sinusoids, it must be sinusoidal itself.
No, R is the resultant of 3 *electro*motive forces, voltages. It has to be
of constant magnitude, or a 3 ph motor wouldn't deliver uniform torque
throughout each rotation, a primary benefit of 3 ph motors.
>Presuming A, B and C to be the familiar depiction of a star diagram with 3
>equal vectors separated by 120 degrees. Lets assign a magnitude of say 10
>units and do the vector arithmetic for A = (10 angle 0), B = (10 angle 120),
>and C = (10 angle 240). The result is R = (0 angle 0)
>Converting to rectangular notation and adding we have: (10 +j0) + (-5
>+j8.66) + (-5 -j8.66) and the result R = (0 +j0). R is still = 0.
Yes, they're symmetric AC waveforms, arranged so the contant resultant they
sum to is zero. That's why the neutral point of a wye connected 3 ph system
isn't offset from zero.
>Now visualize the same 3 equal vectors, only this time they are separated by
>60 degrees. The vectorial combination would be: (10 angle 0) + (10 angle
>60) + (10 angle 120) and the result is (20 angle 60). The rectangular
>notation would yield (10 +j17.32) which is, of course equivalent to the
>polar expression, (20 angle 60).
>
>Assuming the same magnitude of 10 units each, and plotting all loci around
>the stator yields a sine wave swinging between a maximum of 17.32 and a
>minimum of -17.32. This resultant sine wave would be the combination of 3
>vectors, each spaced 60 degrees apart and oscillating in space. The
>resultant wave would appear to rotate around the stator.
Halfway around the stator. You've forgotten that the coils are paired,
each separated from its mate around the stator by 180 mechanical
degrees. The latter have a negative sign in front of their magnitudes
to account for that physical location. Now when you do the sums,
you get constant zero again.
Gary
Dave Martindale
>Yes, they're symmetric AC waveforms, arranged so the contant resultant they
>sum to is zero. That's why the neutral point of a wye connected 3 ph system
>isn't offset from zero.
It's also why you can connect three phases into a delta and not have any
currents flowing around the triangle. And that's why you only need 3
wires to transmit 3-phase, compared with 4 wires for 2-phase. (Well,
you could use 3 wires with one larger than the other two, but that
wouldn't be so elegant).
Dave
Bob Swinney
" No, R is the resultant of 3 *electro*motive forces, voltages. It has to be
of constant magnitude, or a 3 ph motor wouldn't deliver uniform torque
throughout each rotation, a primary benefit of 3 ph motors."
Good. Now you're getting there. Torque is a result of uniform rotation of
the mmf wave through the rotor. Go back and read the part:
"To begin with if R is a magnetmotive force vector launched into the rotor
at
all loci around the stator it cannot have constant magnitude. As a
combination of 3 sinusoids, it must be sinusoidal itself."
The operative phrase here is "launched". The mmf vector, which is what this
discussion boils down to, cannot have constant magnitude - it begins on one
side of the stator and passes through the rotor to the other side. I will
grant that an opposite pole exists on the other side.
Gary sez: (after this statement - The resultant wave would appear to
rotate around the stator.)
"Halfway around the stator. You've forgotten that the coils are paired,
each separated from its mate around the stator by 180 mechanical
degrees. The latter have a negative sign in front of their magnitudes
to account for that physical location. Now when you do the sums,
you get constant zero again."
Haven't forgot squat! Of course the coils are paired, and the magnetic
poles are 180 degrees apart. What's to argue, here? Three coil sets, call
'em phases, on one side occupy 180 mechanical degrees (3 x 60). At any
given instant, the resultant of those coil sets is a magnetic pole; that
pole has it's "mate" of opposite polarity on the other side of the stator,
also occupying (3 x 60) mechanical degrees. Expand "given instant" to all
loci on the stator and you will have a smoothly rotating (around the stator)
magnetic field that induces current into the rotor.
Bob
>
daestrom
> Gary sez:
> " No, R is the resultant of 3 *electro*motive forces, voltages. It has to
be
> of constant magnitude, or a 3 ph motor wouldn't deliver uniform torque
> throughout each rotation, a primary benefit of 3 ph motors."
>
> Good. Now you're getting there. Torque is a result of uniform rotation
of
> the mmf wave through the rotor. Go back and read the part:
>
> "To begin with if R is a magnetmotive force vector launched into the rotor
> at
> all loci around the stator it cannot have constant magnitude. As a
> combination of 3 sinusoids, it must be sinusoidal itself."
If you are looking at the mmf established by the phase coils in 'cartesion
coordinates', the X and Y components of the MMF *are* sinusoidal. But the
absolute magnitude of their vector sum is constant. The magnitude sum of
the mmf created by the three phases is a constant. The vector's angle is
what rotates around the stator.
>
> The operative phrase here is "launched". The mmf vector, which is what
this
> discussion boils down to, cannot have constant magnitude - it begins on
one
> side of the stator and passes through the rotor to the other side. I will
> grant that an opposite pole exists on the other side.
I do not understand your special use of the word 'launched'. The magnitude
of the mmf from a single phase coil is sinusoidal, but the summation of the
mmf from all the phase-coils is a nearly constant magnitude vector whose
angle rotates around the frame. Perhaps the misunderstanding is you're
talking about just the mmf from a single phase-coil? But when combined with
the others it is constant magnitude (as one phase-coil's mmf drops, the next
phase-coil's mmf rises, etc...)
>
> Gary sez: (after this statement - The resultant wave would appear to
> rotate around the stator.)
> "Halfway around the stator. You've forgotten that the coils are paired,
> each separated from its mate around the stator by 180 mechanical
> degrees. The latter have a negative sign in front of their magnitudes
> to account for that physical location. Now when you do the sums,
> you get constant zero again."
>
> Haven't forgot squat! Of course the coils are paired, and the magnetic
> poles are 180 degrees apart. What's to argue, here? Three coil sets,
call
> 'em phases, on one side occupy 180 mechanical degrees (3 x 60). At any
> given instant, the resultant of those coil sets is a magnetic pole; that
> pole has it's "mate" of opposite polarity on the other side of the stator,
> also occupying (3 x 60) mechanical degrees. Expand "given instant" to all
> loci on the stator and you will have a smoothly rotating (around the
stator)
> magnetic field that induces current into the rotor.
>
I think we all agree on that. I think it's just a misunderstanding
discussing the mmf from one phase-coil, or the total combined mmf of all
three phase-coils.
But you said in an earlier post, "...As a combination of three sinusoids, it
[the resulting mmf] must be sinusoidal itself." The magnitude of the
resulting mmf is *not* sinusoidal, but nearly constant. The three sinusoids
being out of phase simply makes the resultant mmf rotate and have a small
amount of 'ripple' to its magnitude.
daestrom
Bob Swinney
"> I do not understand your special use of the word 'launched'. The
magnitude
> of the mmf from a single phase coil is sinusoidal, but the summation of
the
> mmf from all the phase-coils is a nearly constant magnitude vector whose
> angle rotates around the frame. Perhaps the misunderstanding is you're
> talking about just the mmf from a single phase-coil? But when combined
with
> the others it is constant magnitude (as one phase-coil's mmf drops, the
next
> phase-coil's mmf rises, etc...)"
Sorry about "launched". Somehow it sounded better than "a bunch of flux
lines between 2 poles". Anyway that is what it is - flux between opposite
stator poles - through the rotor. That flux, even though "rotating" around
the stator, is pulsating at line frequency through the rotor. It is this
flux, varying sinusoidally, that induces current into the rotor, and
ultimately is responsible for rotation.
Bob
daestrom
> Daestrom sez;
>
> Sorry about "launched". Somehow it sounded better than "a bunch of flux
> lines between 2 poles". Anyway that is what it is - flux between opposite
> stator poles - through the rotor. That flux, even though "rotating"
around
> the stator, is pulsating at line frequency through the rotor. It is this
> flux, varying sinusoidally, that induces current into the rotor, and
> ultimately is responsible for rotation.
>
You persist in the idea that the flux magnitude is varying sinusoidally. I
disagree completely. In a properly designed stator, the summation of the
three phase currents will be generating an mmf of equal strength at every
moment throughout the cycle. If you calculate the total mmf generated at
*any* point in time throughout the cycle, you will find it always the same
magnitude.
When you calculate the summation of the mmf generated by each phase coil,
for each moment through the cycle, you will find the magnitude the same,
only the angle differs.
In a single phase machine, the mmf actually pulsates. This pulsation does
not by itself create any net torque. The currents it induces into the rotor
are symetrical and forces are balanced about the rotor's center. This is
why such a single phase machine requires additional features for creating a
starting torque. Only after the rotor has begun to rotate and the fact of
inductance in the rotor winding can the voltage that is induced, and the
resulting current that lags that voltage produce a net torque on the shaft
allowing the rotor to continue rotating.
But although the mmf from a given phase pulsates in a 3-phase machine, the
combined mmf from all three is nearly steady. The relative motion between
this constantly shifting/rotating flux and the rotor (presumably not
rotating at the same speed) that induces currents into the rotor. If this
were not so, then 3-phase machines would not be self-starting and they would
need additional starting devices.
I suppose if you wanted to treat each phase group separately, then you would
have to consider the inductance in the rotor. Then a 'pulse' of mmf
increasing/decreasing in one plane would induce currents that would lag in
time from the pulse that created them. Thus, the 'peak' rotor current
generated from the pulse of one phase would occur later in time when another
phase's mmf was greatest. Thus, the rotor current would not be distributed
symetrically across the mmf of the second phase and a torque would develop.
This would imply, however, that a rotor of high resistance and low
inductance would have the rotor currents very nearly in-phase with the
'pulse' that generates them and thus the forces on each side would very
nearly cancel.
My experience shows that high R and low L rotors actually develop *higher*
starting torques. Up to the point when R == Xl whereupon the torque is at
maximum (pullout torque is often equated to when reactance and resistance
are equal).
Consider further, the synchronous motor, even one with a simple, reluctance
rotor. Once started, the poles of the rotor have no currents induced into
them by 'pulsating' mmf of the different phases. The torque generated is
very smooth and does not pulsate. This is because the salient poles of the
rotor are within a flux of nearly constant magnitude. The slight phase
angle between the centerline of the pole and the centerline of the mmf
creates a steady torque as the two rotate around in locked-step.
daestrom
Gary Coffman
>"Bob Swinney" <jud...@attbi.com> wrote in message
>news:IDola.434245$S_4.495757@rwcrnsc53...
>> Daestrom sez;
>>
>> Sorry about "launched". Somehow it sounded better than "a bunch of flux
>> lines between 2 poles". Anyway that is what it is - flux between opposite
>> stator poles - through the rotor. That flux, even though "rotating" around
>> the stator, is pulsating at line frequency through the rotor. It is this
>> flux, varying sinusoidally, that induces current into the rotor, and
>> ultimately is responsible for rotation.
>>
>
>You persist in the idea that the flux magnitude is varying sinusoidally. I
>disagree completely. In a properly designed stator, the summation of the
>three phase currents will be generating an mmf of equal strength at every
>moment throughout the cycle. If you calculate the total mmf generated at
>*any* point in time throughout the cycle, you will find it always the same
>magnitude.
>
>When you calculate the summation of the mmf generated by each phase coil,
>for each moment through the cycle, you will find the magnitude the same,
>only the angle differs.
Yes. My Mathcad simulation shows a 2.6E-14 "ripple" using Bob's magnitude
10 sets of values. That's probably just numeric noise in the simulation algorithm,
though. If you measure the ripple in an actual motor, you find that it is not at the
AC line frequency. It is actually harmonic currents due to the nonlinearity of the
B-H curves of the magnetics.
>In a single phase machine, the mmf actually pulsates. This pulsation does
>not by itself create any net torque. The currents it induces into the rotor
>are symetrical and forces are balanced about the rotor's center. This is
>why such a single phase machine requires additional features for creating a
>starting torque. Only after the rotor has begun to rotate and the fact of
>inductance in the rotor winding can the voltage that is induced, and the
>resulting current that lags that voltage produce a net torque on the shaft
>allowing the rotor to continue rotating.
>
>But although the mmf from a given phase pulsates in a 3-phase machine, the
>combined mmf from all three is nearly steady. The relative motion between
>this constantly shifting/rotating flux and the rotor (presumably not
>rotating at the same speed) that induces currents into the rotor. If this
>were not so, then 3-phase machines would not be self-starting and they would
>need additional starting devices.
Yes, yes, yes.
>I suppose if you wanted to treat each phase group separately, then you would
>have to consider the inductance in the rotor. Then a 'pulse' of mmf
>increasing/decreasing in one plane would induce currents that would lag in
>time from the pulse that created them. Thus, the 'peak' rotor current
>generated from the pulse of one phase would occur later in time when another
>phase's mmf was greatest. Thus, the rotor current would not be distributed
>symetrically across the mmf of the second phase and a torque would develop.
>This would imply, however, that a rotor of high resistance and low
>inductance would have the rotor currents very nearly in-phase with the
>'pulse' that generates them and thus the forces on each side would very
>nearly cancel.
>
>My experience shows that high R and low L rotors actually develop *higher*
>starting torques. Up to the point when R == Xl whereupon the torque is at
>maximum (pullout torque is often equated to when reactance and resistance
>are equal).
Yes they do. There's even an electrical theorem which proves that must be
the case.
>Consider further, the synchronous motor, even one with a simple, reluctance
>rotor. Once started, the poles of the rotor have no currents induced into
>them by 'pulsating' mmf of the different phases. The torque generated is
>very smooth and does not pulsate. This is because the salient poles of the
>rotor are within a flux of nearly constant magnitude. The slight phase
>angle between the centerline of the pole and the centerline of the mmf
>creates a steady torque as the two rotate around in locked-step.
>
>daestrom
You've done a much better job summarizing what I've been trying to tell
Bob than I have.
Gary
Gary Coffman
Yes, a much better example.
Gary
Bob Swinney
sinusoidally. I
> disagree completely. In a properly designed stator, the summation of the
> three phase currents will be generating an mmf of equal strength at every
> moment throughout the cycle. If you calculate the total mmf generated at
> *any* point in time throughout the cycle, you will find it always the same
> magnitude.
How can you disagree completely when you are, or have been, a motor winder?
I mean no disrespect by that question, but I wonder if we are talking about
the same elements in motors.
Before, you described the partial winding scheme for a motor. Remember? We
were discussing 6 coil sets for 3 phases, and you suggested the standard
winding notation of A, C', B on one side of the stator and A',C,B' on the
other side, 180 mechanical degrees away. It was understood, or at least I
think it was understood, that this is a method whereby poles are formed on
each side of the stator.
The purpose of opposite poles on either side of the stator is to cause flux
to line up between the poles. That flux is linked through the coil bars of
the rotor. (we aren't considering a rotating magnetic field for the moment,
just flux from one pole to the other pole, threaded through the rotor)
The coil sets, forming the pair of poles are driven by alternating current.
Alternately, then one pole is N and then S, and vs, in step with the AC
driving the coils and forming the poles. The magnetic lines of flux go back
and forth between the poles in this fashion and change direction every 180
degrees with the driving current. That flux is caused by sinusoidal current
and thus it changes direction in step with the current driving it. Current
is induced in the coils of the rotor because of the changing flux. Steady
flux can not induce any current flow - the lines of flux have to "cut" coils
in order to cause current flow. Call it transformer action, if you will.
In fact an induction motor is very similar to a transformer with a moveable
secondary, except the secondary (rotor) is caused to spin around rather than
being pushed away. It is necessary for the flux to be changing for
transformer action to take place. So far we've been talking about a
stationery rotor (secondary). The effect of flux cutting the rotor coils is
augmented by the rotating flux field as well.
When running, the rotor continuously cuts, or is cut by, the lines of flux.
Compared to the stator, a very heavy current flows in the bars (coils) of
the rotor because of the large step-down, or dynamic impedance difference
between stator and rotor. The rotor will revolve at nearly the synchronous
speed of the rotating magnetic field on the stator, differing in a few
percentage points, known as slip. Rotor speed is essentially constant in an
induction motor, with speed being regulated, or governed, over a narrow
range by the slip effect. If the magnetism in the rotor (remember it is
induced) was replaced by permanent magnets, the rotor would be synchronous
and revolve at the same speed as the revolving field on the stator.
I have appreciated your explanations in this overlong thread and hope that
in some way I may have been able to help you as well.
Best regards,
Bob
Bob Swinney
Bob Swinney
In this very long thread, much has been written re. opinions of those that
supposedly know what they are talking about. I am guilty of this to the
extent that I may inadvertently use terminology that doesn't always "fit"
well. For example, I have used "magnetmotive force (mmf) when I really
should have said, "flux". In several of my posts, I have quoted material
from engineering hand books in an attempt to state (only) substantiated
facts and remove the stigma of "opinion". Here is another quote that is
quite descriptive of the operation of 3-phase machines. The quote is from a
book by P.L. Alger entitled, "The Nature of Polyphase Induction Machines"
John Wiley & Sons, New York, 1951.
"Principle of Operation: An induction motor is simply an electric
transformer whose magnetic circuit is separated by an air gap into two
relatively movable portions, one carrying the primary and the other the
secondary winding. Alternating current supplied to the primary winding from
an electric power system induces an opposing current in the secondary
winding, when the latter is short-circuited or closed through an external
impedance. Relative motion between the primary and secondary structures is
produced by the electromagnetic forces corresponding to the power thus
transferred across the air gap by induction. The essential feature which
distinguishes the induction machine from other types of electric motors is
that the secondary currents are created solely by induction, as in a
transformer or other external power source, as in sychronous and DC
machines.
The following paragraph is from a later electrical engineering handbook:
"Induction motors are classified as squirrel-cage and wound-rotor motors.
The secondary windings on the rotors of squirrel-cage motors are assembled
from conductor bars short-circuited by end rings or are cast in place from
aluminum or another conductive alloy. The secondary windings of wound-rotor
motors are wound with discrete conductors with the same number of poles as
the primary windings on the stator. The rotor windings art terminated on
slip rings on the motor shaft. The windings can be short-circuited by
brushes bearing on the slip rings, or they can be connected to resistors or
solid-state converters for starting and speed control."
Bob Swinney
daestrom
> daestrom sez: >You persist in the idea that the flux magnitude is varying
> sinusoidally. I
> > disagree completely. In a properly designed stator, the summation of the
> > three phase currents will be generating an mmf of equal strength at every
> > moment throughout the cycle. If you calculate the total mmf generated at
> > *any* point in time throughout the cycle, you will find it always the same
> > magnitude.
>
> How can you disagree completely when you are, or have been, a motor winder?
> I mean no disrespect by that question, but I wonder if we are talking about
> the same elements in motors.
Because what I stated doesn't disagree with motor theory, just your
statements about sinusoidal variations in the flux developed by the
stator :-)
> Before, you described the partial winding scheme for a motor. Remember? We
> were discussing 6 coil sets for 3 phases, and you suggested the standard
> winding notation of A, C', B on one side of the stator and A',C,B' on the
> other side, 180 mechanical degrees away. It was understood, or at least I
> think it was understood, that this is a method whereby poles are formed on
> each side of the stator.
Yes...
>
> The purpose of opposite poles on either side of the stator is to cause flux
> to line up between the poles. That flux is linked through the coil bars of
> the rotor. (we aren't considering a rotating magnetic field for the moment,
> just flux from one pole to the other pole, threaded through the rotor)
Sure, pretend the flux travels straight across the stator from one
coil to it's opposite 'mate'.
>
> The coil sets, forming the pair of poles are driven by alternating current.
> Alternately, then one pole is N and then S, and vs, in step with the AC
> driving the coils and forming the poles. The magnetic lines of flux go back
> and forth between the poles in this fashion and change direction every 180
> degrees with the driving current.
Not exactly. The MMF of the coil under discussion certainly
alternates, and so the flux strength exiting one coil face and
entering the opposite coil face also alternates.
BUT THAT IS *NOT* the total flux!!!
Consider carefully, as the flux from coil A and A' are declining, the
flux from C' and C are rising. The flux from coil B and B' actually
drop to zero and reverse during the same time that A is declining and
C' is rising. It is the vector SUMMATION of the MMF from all three
sets of coils that develops the total flux which actually crosses the
air gap to the rotor, crosses the rotor and exits the other side.
The MMF of all three phases must be added to find the effective total.
> That flux is caused by sinusoidal current
> and thus it changes direction in step with the current driving it. Current
> is induced in the coils of the rotor because of the changing flux. Steady
> flux can not induce any current flow - the lines of flux have to "cut" coils
> in order to cause current flow.
Yes, it can if there is relative motion. Either mechanical movement
of the conductors, or movement of the magnetic field. Field movement
can be caused by mechanical rotation (AC generator), variations in
total strength (think of it as expanding/contracting from its
centerline position), -OR- variation the magnetic field orientation
(i.e. rotating stator field).
> Call it transformer action, if you will.
> In fact an induction motor is very similar to a transformer with a moveable
> secondary, except the secondary (rotor) is caused to spin around rather than
> being pushed away.
And this is where the transformer/motor analogy breaks down. This is
exactly what happens in a single phase motor with no start winding.
Yes, there are currents induced into the rotor by the
expanding/collapsing field, but *zero* torque is created if the rotor
is at rest. If you look at the currents induced in the rotor, they
are exactly symetrical about the centerline of the magnetic poles.
This symetry means the torques produced by the rotor currents on one
side of the shaft are exactly canceled by torques developed in the
opposite direction on the other side of the shaft. Net result, single
phase are not self-starting without additional features.
In two, three or more phase motor, as the orientation of the field
shifts, it 'moves' in relation to the conductors, inducing currents.
Since the currents are out of phase with the flux that induced them
(inductance of rotor winding), they are not symetrical around the
centerline of the flux that induced them. Thus, the torques developed
on each side of rotor are *not* exactly balanced and the thing starts
spinning.
If the rotor windings were *perfectly* resistive, the currents would
always be exactly balanced about the centerline of the flux that
induces them and such a three phase motor would not be self-starting
(although, like a stalled single phase motor, it would draw a lot of
current and overheat).
> It is necessary for the flux to be changing for
> transformer action to take place.
Not so. If there is a steady flux and just relative motion, then
currents can be induced that way as well. Simple proof, look at the
operation of a synch. AC generator. The rotor is a source of steady
flux levels. The spinning of the rotor provides relative motion of
the flux from the rotor with the stationary conductors of the stator.
In fact, to induce a current, the 'more correct' description is
'relative motion between the magnetic flux and conductor'. This
motion can be achieved mechanically (as in a generator), or by
changing the flux magnitude (simple transformer), or by having the
flux shift its position (three-phase stator).
> So far we've been talking about a
> stationery rotor (secondary). The effect of flux cutting the rotor coils is
> augmented by the rotating flux field as well.
>
> When running, the rotor continuously cuts, or is cut by, the lines of flux.
> Compared to the stator, a very heavy current flows in the bars (coils) of
> the rotor because of the large step-down, or dynamic impedance difference
> between stator and rotor. The rotor will revolve at nearly the synchronous
> speed of the rotating magnetic field on the stator, differing in a few
> percentage points, known as slip. Rotor speed is essentially constant in an
> induction motor, with speed being regulated, or governed, over a narrow
> range by the slip effect.
Yes, yes, I agree with all that. In fact
If the magnetism in the rotor (remember it is
> induced) was replaced by permanent magnets, the rotor would be synchronous
> and revolve at the same speed as the revolving field on the stator.
Now for a real 'zinger' for you to ponder...
In induction motor, at what speed does the magnetic field of the rotor
rotate with respect to the frame? Synchronous!!! The currents
induced in the rotor are of a low frequency (1-slip) such that the
magnetic field *advances* around the rotor. The magnetic field lags
behind the stator's field by a few degrees (somewhat like a torque
angle), but remains in locked step with it.
daestrom
Bob Swinney
I won't take up each point that you made in your rather lengthy reply, but
it is clear that you may understand most of the key points that have been
set forth in this thread.
One thing you don't seem to grasp is the flux we have been talking about is
not 3 individual "fluxes" but a composite flux, basically sinusoidal in
nature, caused by the three 60 degree coils sets on each side of the stator.
It may not be perfectly clear to you, but the seemingly rotating magnetic
field around the stator derives its rotation from the three (OK 6 ) phase
coil sets. You touched upon this in one of your paragraphs where you sought
to explain the ebb and flow of currents in the phase coil sets.
And you challenged the transformer analogy of the induction motor - as if it
was my idea. I believe the passages quoted from experts in the field
adequately qualifies the transformer analogy.
You left me with a "zinger" to ponder. I agree with your reasoning which is
roughly: "The flux field across the rotor rotates at synchronous speed. It
must do so because it is a product of the magnetic field rotating around the
stator. The rotor can never quite "catch up" to the rotating field that
induces current, resulting in torque, in it. The few degrees of lag is
known as slip as you say.
Regards,
Bob
daestrom
> Hi daestrom, and others with the patience to be still following this
thread.
>
> I won't take up each point that you made in your rather lengthy reply, but
> it is clear that you may understand most of the key points that have been
> set forth in this thread.
How gracious...
Perhaps some of the points I made don't fit well with your idea's about how
things work.
> It may not be perfectly clear to you, but the seemingly rotating magnetic
> field around the stator derives its rotation from the three (OK 6 ) phase
> coil sets.
On the contrary, the concepts have been quite clear to me for 25 years...
As well as DC machinery, synchronous machinery, 'consequent poles', and a
host of other electrical machinery concepts.
So you still maintain that the magnitude of flux through the rotor varies
sinusoidally???
If that is so, at what point during the cycle does the sum of all three
fluxes reach zero?? If you believe the flux magnitude varies sinusoidally
(as you've stated on several occasions), then you should be able to point
out a particular moment in the cycle when all three mmf's added together sum
to zero. When is it? (hint, never)
>
> And you challenged the transformer analogy of the induction motor - as if
it
> was my idea. I believe the passages quoted from experts in the field
> adequately qualifies the transformer analogy.
>
I challenge it because the analogy has limits. And the analogy does not
address the summation of all three phases. It describes only a single
phase. The experts in the field use this analogy within its limitations.
The experts don't apply it where its limitations prevent proper results.
And I am perfectly aware the idea is not your own. It is used in many texts
to calculate rotor currents. Usually preceded by a phrase such as, "If we
assume the three phases are symetrical, we can simplify the analysis to a
single phase." But beyond the limits, the analogy is useless. The analogy
works well with single-phase. But it does not explain how torque is
developed (the secondary *must* have some reactance). Nor does it describe
the sumation of the three circuits. And *that* is the crux of the
discussion.
The single phase, simplified circuit transformer analogy works well with
calculating the current induced into a rotor. But it is *simplified* to
only calculate a single phase. It shows the flux contribution from a single
phase varies sinusoidally, but that is not the 'air-gap' flux that passes
through to the rotor of a 3 phase machine.
The flux *at any fixed spot* on the stator varies sinusoidally, but that is
not what we've been talking about. The air-gap flux is a constant magnitude
vector that rotates at synch. speed around the stator. Follow the
'rotating' flux and you'll see it is nearly constant magnitude as it rotates
around. As it passes a fixed point, the mmf at that single point varies,
but the vector's magnitude does not.
There is *no* point in the cycle of a 3 phase stator when the magnitude of
the flux through the rotor is zero. When you sit down and calculate the
exact point when you *think* it's zero, add the flux from the other two
phases and you will find the total is about the same magnitude as always,
just oriented at a different angle. But the *magnitude* is never zero. If
you can show a single moment in the cycle when the summation of all three
(six) mmf's is zero, then I'll eat my hat.
To be equally condescending, it is apparent you've read a couple of books on
the subject. Yet it doesn't seem that you've actually done many
calculations to test your understanding.
daestrom
C What I Mean
I have to shake your hands (cyber world of course).. You three have this
long discussion going and nothing has slipped into name calling and
profanity! WOW!!! I can't remember a thread lasting this long with only
real discussion in a very long time! It is refreshing to see.... I do hope
at some point you all agree with each other ;-) or agree to disagree.. But
in any case, this is what the new group should be all about! Please keep it
up!
Craig
Brian Lawson
I've TRIED to follow along, and have learned lots (I think).
Brian Lawson,
Windsor, Ontario.
XXXXXXXXXXXXXXXXXXXXXXXXXX
On Sat, 12 Apr 2003 14:15:45 GMT, "C What I Mean" <Nos...@yahoo.com>
wrote:
Brian Lawson
XXXXXXXXXXXXXXXXXXXXX
On Sat, 12 Apr 2003 14:15:45 GMT, "C What I Mean" <Nos...@yahoo.com>
wrote:
>To Daestrom, Gary Coffman, and Bob Swinney...
daestrom
"C What I Mean" <Nos...@yahoo.com> wrote in message
news:lcVla.156770$OV.238821@rwcrnsc54...
> To Daestrom, Gary Coffman, and Bob Swinney...
>
> I have to shake your hands (cyber world of course).. You three have this
> long discussion going and nothing has slipped into name calling and
> profanity! WOW!!! I can't remember a thread lasting this long with only
> real discussion in a very long time! It is refreshing to see.... I do
hope
> at some point you all agree with each other ;-)
> or agree to disagree..
But that is the next best thing...
> But
> in any case, this is what the new group should be all about! Please keep
it
> up!
New Group???
daestrom
P.S. Thank you.
Gary Coffman
>Hi daestrom, and others with the patience to be still following this thread.
>
>I won't take up each point that you made in your rather lengthy reply, but
>it is clear that you may understand most of the key points that have been
>set forth in this thread.
>
>One thing you don't seem to grasp is the flux we have been talking about is
>not 3 individual "fluxes" but a composite flux, basically sinusoidal in
>nature, caused by the three 60 degree coils sets on each side of the stator.
>
>It may not be perfectly clear to you, but the seemingly rotating magnetic
>field around the stator derives its rotation from the three (OK 6 ) phase
>coil sets. You touched upon this in one of your paragraphs where you sought
>to explain the ebb and flow of currents in the phase coil sets.
I expect daestrom will explain this better than I can, but your previous calculations
neglected some factors that must be accounted for to describe the resultant MMF
vector. The first thing is that the 3 driving electrical phases are always separated
by 120 *electrical* degrees, and that the electrical vectors are in the same direction
in both halves of each coil pair.
The second thing is that each of the coil half pairs are separated by 120 *mechanical*
degrees in a "2 pole" motor. (60 degrees only if you count each coil pair twice as you
go around the circle.) Both the electrical and mechanical displacements must show up
in your calculations of the resultant vector.
Thirdly, the magnetic fields are dipolar (no magnetic monopoles), which affects the
signs of the magnitudes and the value of the resultant. When you resolve all these
various factors, the resultant flux vector is constant in polar magnitude, and rotates
about the center at line frequency.
Now if you attempt to *unroll* the motor, you will get what appears to be a sinewave
resultant for the MMF. But that's only because you're viewing it in a way that *doesn't
physically exist* in the motor for an observer stationary with respect to the stator (which
you must do, because your equations described the vectors in space with respect to
the stator). Daestrom did a good job of explaining what this does to the transformer
analogy, so I won't repeat that here.
Gary
Bob Swinney
always separated
> by 120 *electrical* degrees, and that the electrical vectors are in the
same direction
> in both halves of each coil pair."
Sorry, if you were unclear on the quotes from several electrical engineering
handbooks. In those quotes, I believe, it was explained that while
the driving "phases" are truly 120 degrees apart, because of the middle
winding reversal, the voltages developed in the phase coil sets are
separated by 60 degrees. And of course, those windings are diametrically
opposite across the stator, from one side to the other. Daestrom, as an
experienced motor winder showed this very well in his description of a phase
coil set, i.e, on one side, (A C' B) and on the other side, (A' C B').
And Gary wrote:
"Now if you attempt to *unroll* the motor, you will get what appears to be
a sinewave resultant for the MMF. But that's only because you're viewing it
in a way
that *doesn't physically exist* in the motor for an observer stationary with
respect to
the stator (which you must do, because your equations described the vectors
in space with
respect to the stator). Daestrom did a good job of explaining what this does
to the
transformer analogy, so I won't repeat that here."
Sorry, I choose not to unroll my motor, it is complex enough when *rolled*
the way it has been for 100 years or so. I know you are striving for
clarity here, but I feel we are only whipping the poor dead horse
needlessly. Please, if you will, approach this from a proper engineering
perspective and give us the appropriate equations. Can you do that?
And again it was written re. flux across the rotor:
..." When you resolve all these various factors, the resultant flux vector
is constant in polar magnitude, and rotates about the center at line
frequency"
Let's talk a bit about flux:
I want you to go stick your finger in a light socket. Ouch!! Right? But,
Hey! Remember for the briefest instant between jolts, it errr, well, felt
pretty good. Actually there was no shock at all for a tiny interval between
those massive shocks. What's going on here??
Here's whass happenin: Look at the big picture and visualize an alternator
on the grid. I think, or hope, you realize that the stator of a synchronous
alternator and an induction motor are practically the same. Three-phase
current from the grid ultimately comes to a transformer near you and 1 of
the 3 phases is parceled out to your light socket. Now, the comfortable
interval between shocks must be caused by ... Oh! Could it be there was
zero current doing that particular tiny instant? Hmmnnn, "zero current =
zero pain" you say? Yep! At that particular instant there was zero flux in
the great alternator that sent you the current. That tiny interval of "zero
flux" corresponds to a zero crossing of the AC waveform from the alternator.
And so it is with the resultant, or composite, flux in an induction motor.
It is not, can not, be constant. If flux was a constant quantity, there
would be no induced current in the rotor.
Respectfully,
Bob
BUSHBADEE
two converters.
.
.
I DO NOT FOLLOW MANY OF THESE NEWS GROUPS
To answere me address mail to
Bush...@aol.com
Gunner
3ph 208/220-440/480 motor. It was only removed to install a varidrive
system.
This would make a really nice and powerful rotary phase converter.
Located in the Van Nuys, California area. $100 obo.
Gunner
"Confronting Liberals with the facts of reality is very much akin to
clubbing baby seals. It gets boring after a while, but because Liberals are
so stupid it is easy work." Steven M. Barry
--------------------------------------------------------------------------------
Dave Martindale
>Here's whass happenin: Look at the big picture and visualize an alternator
>on the grid. I think, or hope, you realize that the stator of a synchronous
>alternator and an induction motor are practically the same. Three-phase
>current from the grid ultimately comes to a transformer near you and 1 of
>the 3 phases is parceled out to your light socket. Now, the comfortable
>interval between shocks must be caused by ... Oh! Could it be there was
>zero current doing that particular tiny instant? Hmmnnn, "zero current =
>zero pain" you say? Yep! At that particular instant there was zero flux in
>the great alternator that sent you the current. That tiny interval of "zero
>flux" corresponds to a zero crossing of the AC waveform from the alternator.
Here you are talking about single-phase power. The voltage and current
go to zero 120 times a second. At that point, *one* phase of the
3-phase alternator had zero voltage across one of its three sets of
coils.
>And so it is with the resultant, or composite, flux in an induction motor.
>It is not, can not, be constant. If flux was a constant quantity, there
>would be no induced current in the rotor.
In a single-phase single-coil induction motor (e.g. split phase after
the start winding has kicked out, or a shaded-pole motor, or some
synchronous designs), you're right. The flux increases and decreases,
and goes entirely to zero in some of these examples.
But I thought the subject was 3-phase motors. In 3-phase motors each
individual phase has a sinusoidal flux. But if you add the three sets
of flux together, this vector sum of magnetic flux never goes to zero
amplitude, and ideally it is completely constant. Ideally, the net flux
just rotates.
If the rotor was stationary with respect to the flux vector (i.e. the
motor was somehow already rotating at synchronous speed, then indeed
there would be no induced current in the rotor. But the rotor never
reaches synchronous speed. So the net flux is always rotating with
respect to the rotor, or the rotor is always rotating with respect to
the net flux, so there is current induced in the rotor.
Imagine building a device with the armature of a 3-phase induction
motor, but the field magnets and frame of a permanent magnet DC motor.
Now mount the whole device in a pair of bearings, and spin the housing
(and thus the permanent magnet field) using some external source of
power. The rotor of the device will rotate, and provide power to the
device's output shaft. Yet the field is constant amplitude - it has to
be because it's produced by a pair of permenent magnets.
In a 3-phase AC motor, the 3 sets of field coils produce the same
(ideally) constant-amplitude rotating field, without needing to spin any
magnets. But the rotor can't tell the difference.
Dave
Bob Swinney
Bob Swinney
> Here you are talking about single-phase power. The voltage and current
> go to zero 120 times a second. At that point, *one* phase of the
> 3-phase alternator had zero voltage across one of its three sets of
> coils.
Yes, and as far as flux through the rotor goes, I am also talking about the
composite flux formed by phase coil sets in a 3-phase motor.
And he sez: "> But I thought the subject was 3-phase motors. In 3-phase
motors each
> individual phase has a sinusoidal flux. But if you add the three sets
> of flux together, this vector sum of magnetic flux never goes to zero
> amplitude, and ideally it is completely constant. Ideally, the net flux
> just rotates."
This is not quite correct: I would amend that statement to read, "...In
3-phase motors each individual phase develops sinusoidal MMF. If you add
the three sets of sinusoidal currents together, a vector sum of MMF results
which is sinusoidal in its own right. That MMF goes to zero at each
crossing and accordingly the lines of flux from that MMF, through the rotor,
reverses direction every 1/2 cycle." That MMF is caused to seemingly rotate
around the peripher of the stator by virtue of the separation between phase
coil sets. Two things are at work here - 1) rotating magnetic field. 2)
flux through the rotor that induces current therein.
Dave sez: "... In a 3-phase AC motor, the 3 sets of field coils produce the
same
> (ideally) constant-amplitude rotating field, without needing to spin any
> magnets. But the rotor can't tell the difference."
True about the rotating field but we are discussing flux, a related but
different entity, here. Magnetic lines of force (flux) interact and react
with magnetism in the rotor, whether or not that magnetism was caused by
induction or permanent magnets.
Bob
Bob Swinney
capacitance required. I wish I was within range!
Bob
"Gunner" <gun...@lightspeed.net> wrote in message
news:bfbj9vcbeqc2ffbt6...@4ax.com...
Gary Coffman
>Dave sez: "... In a 3-phase AC motor, the 3 sets of field coils produce the
>same
>> (ideally) constant-amplitude rotating field, without needing to spin any
>> magnets. But the rotor can't tell the difference."
>
>True about the rotating field but we are discussing flux, a related but
>different entity, here.
phi (webers) = B (tesla) * A (m^2)
Since the area A enclosed by the stator is fixed for any given motor,
flux = kB.
Field strength H is related to B by
u = B/H
and u is a constant, so
B = k/H
Substituting,
phi = k*k/H
In words, flux is the inverse of field strength. If one varies in magnitude,
the other varies inversely in exact proportion. If one is unvarying in
magnitude, so is the other.
Gary
Gary Coffman
>On Mon, 14 Apr 2003 02:00:21 GMT, "Bob Swinney" <jud...@attbi.com> wrote:
>>Dave sez: "... In a 3-phase AC motor, the 3 sets of field coils produce the
>>same
>>> (ideally) constant-amplitude rotating field, without needing to spin any
>>> magnets. But the rotor can't tell the difference."
>>
>>True about the rotating field but we are discussing flux, a related but
>>different entity, here.
>
>phi (webers) = B (tesla) * A (m^2)
>
>Since the area A enclosed by the stator is fixed for any given motor,
>
>flux = kB.
>
>Field strength H is related to B by
>
>u = B/H
>
>and u is a constant, so
>
>B = k/H
Arg! I should really proofread better.
B= kH
Substituting,
phi = k*kH
In words, flux is directly proportional to field strength. If one varies in
magnitude, the other varies in exact proportion. If one is unvarying in
Bob Swinney
Gary. We all make mistakes.
I'm happy to see you are posting some math, though. BTW, in an earlier post
I asked you to give us *your* equations supporting the combination of three
120 degree separated vectors on a stator.
Here it is again:
["Sorry, I choose not to unroll my motor, it is complex enough when *rolled*
the way it has been for 100 years or so. I know you are striving for
clarity here, but I feel we are only whipping the poor dead horse
needlessly. Please, if you will, approach this from a proper engineering
perspective and give us the appropriate equations. Can you do that?"]
So, how about it Gary - please show us *your* equations.
Regards,
Bob
daestrom
> Dave sez:
>
> > Here you are talking about single-phase power. The voltage and current
> > go to zero 120 times a second. At that point, *one* phase of the
> > 3-phase alternator had zero voltage across one of its three sets of
> > coils.
>
> Yes, and as far as flux through the rotor goes, I am also talking about
the
> composite flux formed by phase coil sets in a 3-phase motor.
>
> And he sez: "> But I thought the subject was 3-phase motors. In 3-phase
> motors each
> > individual phase has a sinusoidal flux. But if you add the three sets
> > of flux together, this vector sum of magnetic flux never goes to zero
> > amplitude, and ideally it is completely constant. Ideally, the net flux
> > just rotates."
>
> This is not quite correct: I would amend that statement to read, "...In
> 3-phase motors each individual phase develops sinusoidal MMF. If you add
> the three sets of sinusoidal currents together, a vector sum of MMF
results
> which is sinusoidal in its own right. That MMF goes to zero at each
> crossing and accordingly the lines of flux from that MMF, through the
rotor,
> reverses direction every 1/2 cycle."
Hi Bob,
Now lets see, the mmf from A and A' don't cancel because they are wound in
such a way as to combine their mmf in producing a flux through the rotor.
Similar, with B and B' and the pair C and C'.
So, whenever A and A' current crosses zero (and their mmf equals zero), the
current through B and B' are not zero. Nor is the current through C and C'
for that same instant. So the only way the composite mmf can be zero is if
the mmf generated by B&B' somehow exactly cancels the mmf from C and C'.
Let's look at how those two coil sets are arranged. Using the stator
windings that we've considered before, the B coil is in the 4 o'clock
position with B' in the 10 o'clock position. Similarly, C is in the 8
o'clock position while C' is in 2 o'clock.
The currents through B and C are opposite. So if B is wound to create a N
with it's current, then C must be creating a S with it's current flowing in
the opposite direction. Similarly, B' would be creating an S pole, while C'
would be creating an N pole.
So, we have a S pole at 10 o'clock and 8 o'clock for an 'average' or
centerline of the resultant mmf at 9 o'clock. By similar addition, we have
a N pole at 2 o'clock and another at 4 o'clock for a centerline of the
resultant mmf at 3 o'clock.
So at the moment that A and A' are generating zero mmf between the 12
o'clock and 6 o'clock position, we find the resultant of B, B', C and C'
creating an mmf running from 3 o'clock to 9 o'clock (exactly perpendicular
to the A and A' positions).
Similar results can be found at every moment that one phase is at zero.
When plotting the three phase vectors, consider the coordinates of the
graph. If you are plotting the electrical displacement, the three vectors
are equal magnitude and 120 degrees displaced. Their sum is zero. As
another pointed out, this allows one to 'close the delta' of delta connected
transformers without large currents. The time displacement between phases
is shown by the different angles on the plot, but there is no way to show
the mechanical displacement.
But if you are plotting the mmf from the three phases and wish to calculate
the summation, then the coordinates are mechanical placement. When plotted
this way, the three vectors (representing the three mmf's) are not all the
same length. Their length varies with time and that is hard to plot when
the coordinates are not time related. As one phase's mmf is increasing in
length, another is decreasing. When a phase's current crosses zero and goes
negative, and its mmf reverses, the vector shrinks to a point then begins
growing in length, but in the opposite direction from before.
Try looking at it this way...
Plot the A vector in an arbitrary direction, with length equal to the mmf
generated by peak current. At the same moment in time, the B vector is only
50% of that value and opposite polarity. So although the B coils are 120
degrees displaced from A in the CW direction, the mmf from the B coils (at
this particular moment) is actually 50% of the peak value, but 60 degrees
CCW from A. And the C coils are mechanically 120 degrees from A CCW, but
its current is also reversed from A and only 50% of the value. So it's mmf
is oriented 60 CW from A.
A
|
B | C
\ | /
\|/
When these are added together (head to toe), we get a vector that is
centered on the A phase, that is 1.5 times longer than the A vector alone.
Meaning, the mmf of the combined is 150% of that of the A alone, but still
in line with it.
Another mechanical plot can be done at another moment in the cycle, say when
A crosses zero. At that moment, B would be 86.6% of it's maximum and C is
86.6% of it's maximum, but opposite polarity. When added, you get a vector
that is 1.5 times the maximum that occurs, oriented exactly perpendicular to
A.
There is *no* moment in the entire cycle when the mmf of the three phases,
when added together, sums to zero.
daestrom
Bob Swinney
formed by three 60 deg. separated vectors. For some unexplained reason, he
said the values of these (normalized) vectors were A = 1, B = .5 and C = .5.
Actually, the value of each vector would have to be = 1, if they were all
generated by equal currents. The vector sum would be, in polar form ( 2 at
angle 0 ) or in rectangular form, ( 1 + j1.732 ) and in x & y coordinates,
( x = 1, y = 1.732 ).
Nothing in Daestrom's post was compelling in the argument that flux reverses
direction every 180 electrical degrees. In an ordinary transformer,
magnetic flux is sinusoidal (barring saturation distortion) An induction
motor is a special case of the transformer, and it follows that its flux
through the rotor, will also reverse direction every 180 electrical degrees.
I believe the point of confusion here stems from the fact that in an AC
machine, motor action is continuous as flux reverses direction. This is in
accordance with Fleming's left-hand rule, describing motor action.
Regards,
Bob Swinney
daestrom
> Daestrom wrote a long narrative basically describing the resulting flux
> formed by three 60 deg. separated vectors. For some unexplained reason, he
> said the values of these (normalized) vectors were A = 1, B = .5 and C = .5.
> Actually, the value of each vector would have to be = 1, if they were all
> generated by equal currents. The vector sum would be, in polar form ( 2 at
> angle 0 ) or in rectangular form, ( 1 + j1.732 ) and in x & y coordinates,
> ( x = 1, y = 1.732 ).
Sheesh Bob, quite thinking in RMS or averaged over time or whatever.
Think about the mmf created by each phase at the exact instant that
A's instantaneous current has reached its maximum. Look at a plot of
three sine waves that are displaced by 120 degrees. Notice carefully
that at the exact moment that A is peaked, B and C are exactly -.5 and
-.5 respectively. Now remember that the coils on each side of A are
B' and C'. So the mmf in these coils *AT THAT EXACT MOMENT* are +.5
and +.5 while the mmf in A is +1.0
Add *THESE* vectors together and you get +1.5. Do it Bob, get out
some graph paper and draw it out!!!
>
> Nothing in Daestrom's post was compelling in the argument that flux reverses
> direction every 180 electrical degrees. In an ordinary transformer,
> magnetic flux is sinusoidal (barring saturation distortion) An induction
> motor is a special case of the transformer, and it follows that its flux
> through the rotor, will also reverse direction every 180 electrical degrees.
>
You *STILL* haven't tried adding the mmf of all three phases together
for a given instant in time!!! Do it. Find the exact magnitude of
each phasor at any given instant in the cycle and add the
instantaneous mmf's together.
You *CANNOT* find any point in the cycle when all three mmf's are zero
at the same instant. They *NEVER* cancel each other out and result in
zero either!!!
For the love of pete, sit down and try to calculate this out.
The flux is 180 displaced every half cycle, but the magnitude of the
total passing through the rotor *never* is zero. It travels from
vertical (up) to horizontal (say, right), then vertical (down) in a
half-cycle. But it's strength is constant as it rotates.
> I believe the point of confusion here stems from the fact that in an AC
> machine, motor action is continuous as flux reverses direction. This is in
> accordance with Fleming's left-hand rule, describing motor action.
>
Horse hockey.
The confusion here is that you think all three currents vary in exact
step or don't vary at all. When you plot the currents (or their
resulting mmf's) for any moment in the cycle, the are *NOT* the same
magnitude.
You're looking at a plot of the three RMS values that you find in some
book that shows three equal magnitude vectors displaced by 120
degrees. But that graph is *not* the right coordinates. It is rms
magnitude versus phase angle.
You want to plot the instantaneous mmf created by each set of coils
versus mechanical angle. This requires a different graph for each
moment in time. So pick *ONE* moment in time and examine the
instantaneous magnitude of all three vectors. If you think they are
always equal in magnitude, you don't understand the phase displacement
or you're using rms values, not instantaneous.
Ask yourself this very simple question. The moment the instantaneous
value of A is at zero, are B and C also at zero?? No, they are not,
if they were, then they would not have 120 degrees of phase
displacement. They are in different parts of their sinusoid, one at
+0.86 and the other at -0.86. So all three are *NOT* at zero the same
time!!!
Then ask yourself another simple question. When the instantaneous
value of B is at +0.86 and C is at -0.86, and the two coils are
arranged 120 mechanical degrees apart, do their mmf's somehow cancel
out?? No, they do not. They create a combined vector that is 1.5 in
magnitude, exactly perpendicular to the A coils. The field has
rotated a 90 degrees from the peak of 1.5 at A. In another quarter
cycle, the field will be 180 from the initial peak at A, and it's
magnitude will still be 1.5.
If you break down the total mmf into vertical and horizontal
components, you'll find the vertical component varies sinusoidally in
phase with A. And the horizontal component varies sinusoidally, but
displaced 90 degrees from the vertical component. The *SUM* of the
two sinusoids is what passes through the rotor and it never varies in
magnitude, only orientation. But the orientation changing is enough
to induce rotor currents and make it go.
Mathematically, combining a vertical component that varies
sinusoidally with a horizontal one that varies sinusoidally but
displaced in time by 90 degrees is the same as plotting sine vs
cosine. The result is a constant length vector that rotates around
the origin (i.e. the basis for sine and cosine in the first place).
The radius of the unit circle used to derive sine and cosine in from
the X and Y legs of the right-triangle created by the radius and the
X-Y coordinate system.
daestrom
Bob Swinney
daestrom
>
> "daestrom" <daes...@twcny.rr.com> wrote in message
> news:2862e26f.03041...@posting.google.com...
> > "Bob Swinney" <jud...@attbi.com> wrote in message
> news:<JUVma.462878$L1.133128@sccrnsc02>...
> > > Daestrom wrote a long narrative basically describing the resulting
flux
> > > formed by three 60 deg. separated vectors. For some unexplained
reason,
> he
> > > said the values of these (normalized) vectors were A = 1, B = .5 and C
=
> .5.
> > > Actually, the value of each vector would have to be = 1, if they were
> all
> > > generated by equal currents. The vector sum would be, in polar form
( 2
> at
> > > angle 0 ) or in rectangular form, ( 1 + j1.732 ) and in x & y
> coordinates,
> > > ( x = 1, y = 1.732 ).
> >
Hi Bob,
Yeah, the NG's sometimes 'loose' some of my posts too. I wonder what it has
against me, but then I just shrug it off.
You still are adding the vectors on the wrong coordinate axes to 'see' what
I'm trying to say here.
When you plot three vectors, 120 apart and of equal length, consider what
are you actually plotting. The 'magnitude' of the vectors is not
instantaneous voltage levels. It is an RMS value. Always same polarity.
And the angle is a representation of the time displacement.
So you have
A = sin(wt)
B = sin(wt - 2/3 pi)
C = sin(wt - 4/3 pi)
where w is supposed to be omega, the frequency in radians/sec and t is the
time.
Looked at this way, the magnitude of all three are constant and when you add
the three vectors, you (quite rightly) get a sum of zero.
But that is not the only way to represent things. With your method of
plotting, how do you represent the A voltage/current/mmf varying?
Instead of plotting the three phases as vectors of magnitude and angle,
consider how the magnitude varies with time through a cycle. Using the
above formula, we can easily calculate the instantaneous values of each
phase at any point in the cycle. Let's pick when t=0.
A=sin(wt) = sin(w*0) = 0.0
B=sin(wt - 2/3 pi) = sin(w*0 - 2/3 pi) = sin(-2/3 pi) = -0.866
C=sin(wt - 4/3 pi) = sin(w*0 - 4/3 pi) = sin(-4/3 pi) = 0.866
These are the instantaneous values of the three phases. I think you can
agree with this. Notice, they are *scalar* at this point, there is no
'direction' other than the polarity sign. One isn't 'leading' or 'lagging'
another, or displaced by some number of degrees, these are the magnitudes of
all three phases at the same instant in time. Agree? If you think they
should be vector's consider this, what is the reference that the angles are
measured from? In a typical plot of three phase, the 'reference' is taken
to be one of the phases, and the angle is a measure of displacement from
that phase. But the angle is the displacement in *time* and that is already
fixed because we choose an arbitrary moment in time. If we picked a moment
in time, we can't then say that one voltage is displaced in time from the
moment we picked. That's the whole point of talking 'instantaneous' values.
Another way to see this is to plot three sine waves, displaced by 120
degrees on a common set of axes. The Y axes would be voltage and the X axis
would be time. As time progresses from left to right, each voltage varies
in magnitude sinusoidally. But they do *not* all peak at the same moment in
time. On such a plot, draw a vertical line through the plot at any point on
the X axis. This line will intersect each of the three sine waves and the X
axis. This line represents one instant in time (as read on the X axis) and
shows the magnitude of each of the three voltages. They are never all the
same magnitude *at the same instant*(although when one sine wave is at its
peak, the other two are equal in magnitude at -1/2 the magnitude of the one
at its peak).
And I think we both agree, that for this sort of application, barring
saturation or some sort of asymmetry in the iron, the mmf and resulting
fluxes are directly proportional to these voltage magnitudes.
But now, this is where I seem to be loosing you. Having calculated the
magnitude of each of the three sine waves for some arbitrary instant in
time, how to combine the mmf's to find the 'total'.
Although the magnitudes of the three voltages are scalars at this instant,
the mmf's created by them *have* to be vectors since mmf is a vector
quantity. So we must now consider what 'direction' the mmf is generated in.
This is determined by exactly two things. The direction of current through
the coil of wire, and the orientation of that coil of wire. Unless we start
using exotic shapes of the iron to 'turn' the flux, those are the only two
items determining direction of the mmf.
The direction of the current is determined solely by the polarity of the
impressed voltage, so that is easy, we already have the polarity in our
scalar.
The orientation of the coils is determined by the physical geometry of the
stator and how its wound. For the case that we've been discussing, I'll
quickly reiterate my 'layout'. As viewed from one end of the
shaft/rotor....
A 12 o'clock position (+90 degrees on Cartesion coordinates)
C' 2 o'clock position (+30 degrees on Cartesion coordinates)
B 4 o'clock position (-30 degrees on Cartesion coordinates)
A' 6 o'clock position (-90 degrees on Cartesion coordinates)
C 8 o'clock position (-150 degrees on Cartesion coordinates)
B' 10 o'clock position (+150 degrees on Cartesion coordinates)
So, the mmf created by each phase can be calculated using the magnitude of
the impressed voltage and the coil orientation angle....
Ammf = sin(wt) at angle +90
Bmmf = sin(wt - 2/3 pi) at angle -30
Cmmf= sin(wt - 4/3 pi) at angle -150
Because of the connections to the 'prime' windings, their mmf's are inverted
by 180.
A'mmf = win(wt) at angle (-90 + 180) = sin(wt) at angle +90
B'mmf = sin(wt + 2/3 pi) at angle (+150 - 180) = sin(wt + 2/3 pi) at
angle -30
C'mmf = sin(wt + 4/3 pi) at angle (+30 - 180) = sin(wt + 4/3 pi) at
angle -150
Notice the 'prime' mmf's are the same exact direction/angle as the
non-primes. This is consistent with the other discussion in that the two
coils in a pair 'aid' each other in creating the total mmf. Also notice
that the angle for each phase is fixed. The stator doesn't rotate, so the
orientation of each coil is a constant :-)
Picking t=0 again (although you can do the calculation at any time 't'), we
have six mmf's to add up. Breaking them up into X and Y components, we
get.....
A = sin(0) = 0.0 at angle +90
Obviously, with magnitude zero, both X and Y components are 0.0 But for the
sake of completeness...
X = 0.0 * cos(+90) = 0.0 * 0.0 = 0.0
Y = 0.0 * sin(+90) = 0.0 * 1.0 = 0.0
B=sin(-2/3 pi) at -30 = -0.86 at angle -30
X = -0.86 * cos(-30) = -0.86 * 0.86 = -0.75
Y = -0.86 * sin(-30) = -0.86 * (-.5) = 0.433
C=sin(-4/3 pi) at -150 = 0.86 at angle -150
X = 0.86 * cos(-150) = 0.86 * (-0.86) = -0.75
Y = 0.86 * sin(-150) = 0.86 * (-0.5) = -0.433
Summing, we have...
X = 0.0 + (-0.75) + (-0.75) = -1.5
Y = 0.0 + 0.433 + (-0.433) = 0.0
Converting to polar form...
magnitude = sqrt(X^2 + Y^2) = sqrt((-1.5)^2 + 0) = 1.5
angle = atan ( Y/X) = atan(0.0) = 180 degrees. (neg. X implies between third
and forth quadrant)
So at time t=0, we have a total mmf that is magnitude 1.5, at angle 180.
Similar calculations can be made for any instant in time. The magnitude of
the resulting mmf is always 1.5, the angle is the only thing that changes
with time. The only 'trick' here is to be sure to 'inspect' the signs of X
and Y in order to place the angle in the correct quandrant.
Quickly, for 't' such that wt = 30 degrees or one-sixth radians (one
twelfth of a cycle later than above)...
A = sin(pi / 6) at +90 = 0.5 at angle +90
X = 0.5 * cos(+90) = 0.5 * 0.0 = 0.0
Y = 0.5 * sin(+90) = 0.5 * 1.0 = 0.5
B=sin(pi/6 - 2/3 pi) at -30 = -1.0 at angle -30
X = -1.0 * cos(-30) = -1.0 * 0.86 = -0.86
Y = -1.0 * sin(-30) = -1.0 * (-.5) = 0.5
C=sin(pi/6 - 4/3 pi) at -150 = 0.5 at angle -150
X = 0.5 * cos(-150) = 0.5 * (-0.86) = -0.433
Y = 0.5 * sin(-150) = 0.5 * (-0.5) = -0.25
Summing, we have...
X = 0.0 + (-0.86) + (-0.433) = -1.293
Y = 0.5 + 0.5 + (-0.25) = 0.75
Converting to polar form...
magnitude = sqrt( (-1.293)^2 + (0.75)^2) = 1.49... (rounding error, but
exact value would be 1.5)
angle = atan( (0.75)/(-1.293)) = +150 degrees (pos Y and neg X implies 2nd
quadrant)
The magnitude is the same, but the orientation has shifted clockwise by 30
degrees (i.e. the field has rotated, but not varied in its overall
magnitude).
daestrom