
[email protected] wrote:
>
> 
[email protected] wrote:
> >
> > A very common power distribution transformer uses a DELTA wired primary
> > and a WYE wired secondary (thus including a neutral connection).
>  (snip)
> > The interesting thing about this is that while the hottoneutral lines
> > provide power in the same 3 phases that the primary windings have, the
> > hottohot lines are off by 30 degrees.
> 
>  Actually, you have that backwards, the line to neutral phases
>  correspond to the line to line primary phases. This is because each
>  branch of the Y shares a core with one of the sides of the delta.
>  (snip)

> That's actually what I said, but in different terms (hot instead of line)
> and probably not clear enough. I should have added "compared to the
> hottohot lines of the primary" on the last sentence of that paragraph.

 I hope I didn't imply by my answers to your post that this was a
 trivial problem that I could understand at a glance. I think it took
 me three tries before I could get to the end of my review without
 contradicting myself. I think this is one of those contradictions
 that I forgot to go back and fix. Your question was a good
 opportunity for me to have to think about 3 phase power and I should
 have started of a little more humble.
I'm still working on even getting the terms right on all this.
 (snip)

> > At the load the phase relationship between the
> > voltage and current will be 0 degrees. But what about the relationship
> > between the voltage and current in the two windings that are feeding
> > this power? Wouldn't they be off by +30 degrees and 30 degrees?
> >
> > So wouldn't this appear to them to be a partially reactive load?
> > And what about the current on the primary side. Wouldn't that reflect the phase
> > difference there, too?
> 
>  The different phase voltages in the Y branches add up vectorially to a
>  new voltage, and that voltage produces a resistive current that is in
>  phase with that voltage. That current passes through two branches of
>  the Y and it is, as you say, 30 degrees out of phase with each of
>  those branch voltages.
> 
>  However, since each Y branch is driven by one side of the primary
>  delta, those currents must also be 30 degrees out of phase with two of
>  the delta sides. but this primary current is carried by the voltage
>  between two primary line phases, with that voltage and current exactly
>  in phase. So the resistive load across two branches of the Y appears
>  to the primary lines as a resistive load between two of the line
>  phases with no phase shift.
>
> I don't see how that happens. But the picturing below might be a better
> reference.

 Whoops. Another point I later contradicted and did not go back and
 correct. My first (non graphic) analysis somehow converted a
 horizontal (vectorially speaking) load into a vertical one (I reversed
 the sign of one of the delta currents). Glad to see you actually read
 what I wrote. Sorry I didn't catch this correction, myself.

 (snip)

>  If the loads of all phases are balanced, the 30 degree phase shifts
>  all cancel out. If the load is completely unbalanced, as you
>  specified, then there is a reduction in the total power the
>  transformer can carry do to the power factor.
>
> Yes, one would want to have balance if possible (usually is). But I want
> to explore how it all works out when not in balance (worst case scenario).
>
> Later I want to work it out for Delta to 6Star configuration, as well
> as a 2 core TT configuration.

 My head hurts, already. Phasor diagrams will help.
Yes, they do help. I should drag out shome graphics plotting tools. I can
visualize them in my head, but I find it hard to draw them in ASCII.
 (snip)

>  Pictorially:
> 
>  The instantaneous secondary load voltage and current is represented by
>  a horizontal vector (say, to the right).
> 
>  RRR>
>  A \ /B
>  \ /
>  Y n
>  
>  C
> 
>  So the winding currents are A down to n and n up to B.
> 
> 
>  The primary current that carries the load passes through c down to b
>  (the current A to n) and c up to a (the current n to B), since those
>  windings share magnetic flux with branches A (to n) and (n to) B,
>  above. If branches A and B are at rated current, so must be windings
>  c to a and c to b.
>  a
>  / 
>  / 
>  c< 
>  \ 
>  \ 
>  b
> 
>  So all the load current passes through c, but splits and exits equally
>  out of a and b. There is no current in a to b winding, since there is
>  no vertical component in the load current. So the 30 degree power
>  factor is also reflected in 2 of the primary windings. These windings
>  are carrying only 87% of their power capability at rated current,
>  because of the power factor.
>
> But isn't the current in winding ca and in winding cb, as well line c,
> at a horizontal phase angle here? But there's no voltage at that angle.

 There is no voltage at precisely that phase but there is a voltage
 between c and a (and between c and b) that can be resolved into two
 components, 1 of which is in that direction.
Which means the current will have to be seen as 30 degrees reactive?
>  If you picture the line wave as three symmetrical vectors with respect
>  to a floating common point in the middle of the delta (that is
>  rotating, and just happened to be frozen, momentarily), with the ends
>  of those vectors pointing to the corners of the delta, you find that
>  the line current in the c phase is in phase with the voltage, but the
>  line currents in phases a and b are 30 degrees shifted with respect to
>  their voltage. So a three phase transformer connected this way is
>  inefficient (2/3 * .87 of its rated capacity because only 2/3s of the
>  windings are being used, and those operate at .87 of power capacity)
>  driving a single phase load. Is this a surprise, when you could have
>  just used a single phase transformer? But the three phase transformer
>  still has the capacity to drive three loads that add up to 173% more
>  than this single load that hit the winding current limit, if those
>  loads are balanced.
>
> I'm not surprised at all regarding the reduced capacity, and that a
> single phase transformer would be better.
>
> Of course if the power is fed in one phase only, that's drawing off
> some 3 phase distribution point ... but it's likely to only be a tiny
> portion of the total load at that point.

 I was analyzing the case where the single phase load reached the
 current capacity of the transformer, as the most extreme example.
I can see where that would be pushing things. That 1 phase would be
more than its contribution in a fully loaded 3 phase scenario.
I do know if I had a 208Y/120 secondary and put 100 amps load on each
of the 3 ways to get 208 volts, I'm really looking at 173.2 amps on
each of the hot lines (some of the current crosses between the phases
but most of it combines in the lines back to the transformer secondary).
With a WYE secondary, that 173.2 amps is the current through each of
the windings. But with a DELTA secondary, the windings end up with
just 100 amps current, same as the loads. But this is right because
the 173.2 amps in the windings of a WYE secondary are really over the
120 volt potential, while the 100 amps in the windings of a DELTA
secondary are really over the 208 volt potential. So the KVAs work
out to be the same.
Europeans can work this out with 400 and 230 volts.
>  So the transformer is only 58% utilized for this single phase load.
>
> But I could get 3x the power at 100% utilization if the load were in
> proper balance.

 Not quite. This single load uses all of the current capacity of two
 of the 3 sets of windings. Balancing the loads will eliminate the
 power factor current on those two sets, and also make use of the third
 set, but this first load has to be lowered before two other equal
 loads can be added and still keep the winding currents at the same
 peak they had, originally.
The scenario I am thinking of is where 3 phase is the power source, and
the available loads, equally divided over all three phases on the secondary
of the transformer, will add up to the total transformer capacity. Then
the loads are selectively turned on so that all of the loads on ONE of
the phases are on, and none of the loads on the other 2 phases on.
Then a later scenario I will explore is 2 of the 3 load sets are on.
I want to understand what capacity every part of the system might need to
have for the worst case scenarios. And I haven't even gotten into the
harmonic and pulsed loads that can really throw three phase out of whack.
For example I read that it might be necessary to double the current capacity
of the neutral, but it looks to me like the very worst case would require a
triple capacity, or a DELTA secondary with 15.5% extra current capacity
(each leg carrying the full 200 amps, instead of 173.2 amps, from the above
scenario, if the current pulse times are so narrow that there is no way to
"share current" between phases).
With 1 load set on, I can see where the transformer is at 58% of capacity
( 100% / sqrt(3) = 57.735% ). That's not overloading the total capacity,
but I want to be certain there are not spot overloads anywhere.
> But if the load needed to be 208
> volts, and the secondary is WYE, then we have the 30 degree reactive case.
> A delta secondary might be a better choice if the load is going to be
> frequency or regularly out of balance, but all three phases are still
> needed at times ... or maybe even paralleling 3 phase and 1 phase.

 There is no real advantage to one over the other, as far as I know.
 One case puts all the load current on two primary lines with a 30
 degree phase shift (remember that the primary phase voltages do not
 match any of the voltages across any delta winding. The delta winding
 voltages are the vectorial difference of two line voltages. So a pure
 resistive current in one delta winding puts a 30 degree power factor
 load on two of the lines. The Y loaded case puts zero phase shift on
 one line and a lower current 30 degree shifted load on the other two.
 The only way to have a single phase load not produce power factor
 current on a 3 phase line is if it is a 4 wire setup, and the load is
 connected one line to neutral. Any single phase line to line load
 with or without a transformer will produce power factor current
 somewhere.
 (I think.)
I believe I have a full understanding of adding vectors to get the voltages.
What I'm working on right now is understanding the currents. But what I
see is that when there is only current drawn in 1 phase angle, it stays in
that angle all the way back to the power source. If voltage were supplied
to the 3 phase system in only 1 phase (1 leg opens in a delta primary for
example), then all the voltage presented everywhere will be only in that 1
phase angle, although in different voltages in various parts because with a
delta primary, losing the B leg means windings AB and BC are now in series
with the remaining voltage source between lines A and C, resulting in the
two corresponding secondary windings to become "flattened" at half voltage.
The scenario with 1 load set appears to me to be simple in transposing the
current back to the power source. But with 2 load sets on, you have to add
those current vectors and ultimately it will be their sum; what I want to
understand are the intermediate steps to get there.
The TT transformer will be interesting to study this on, too:
* *** * *
   
   
 <=>  OR  <=> **
   
   
*** * *** *
If you do this with 240/120 center tapped on the phase with the tap in the
middle, you get 240 volts deltalike from it, or 208 volts in one phase
along the untapped winding. The interesting thing about TT is it uses
only 2 cores.
The 6star would be nothing more than a special case of WYE, I believe:
* *
* \ /
/ \ \ /
/ \ <=> ***
/ \ / \
** / \
* *