# transformer phasing question

Discussion in 'Electronic Basics' started by [email protected], Jan 22, 2004.

1. ### Guest

A very common power distribution transformer uses a DELTA wired primary
and a WYE wired secondary (thus including a neutral connection). In the
WYE secondary, you then can get a lower voltage from the hot-to-neutral
connections, or a higher voltage from the hot-to-hot connections. Some
common voltages:
120 volts hot-to-neutral and 208 volts hot-to-hot in US, Canada
230 volts hot-to-neutral and 400 volts hot-to-hot in Europe
277 volts hot-to-neutral and 480 volts hot-to-hot in US
346 volts hot-to-neutral and 600 volts hot-to-hot in Canada
provide power in the same 3 phases that the primary windings have, the
hot-to-hot lines are off by 30 degrees. Normally that is no big deal
since 3 phase machines only care about the phases on what is hooked up.

Now here is my question. Suppose I have a 1 phase device hooked up to
just one pair of hot-to-hot lines. Further suppose this is a resistive
load (power factor 1). 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?

If the above is true, and it would seem to me to be the case (but I am
asking to have that confirmed since there might be some physics that are
involved that I am not aware of), then if one were expecting to have
loads that are routinely out of balance on a 3 phase line, that the best
choice would be to configure the secondary so that the exact voltage one
needs is provided on exactly one winding (whether the secondary is wired
WYE or DELTA)?

In many commercial buildings, power is provided in three phase due to
the higher capacity, at the 120/208 volt level in the US. This is the
right voltage for the lights and most appliances at the hot-to-neutral
lines. Then the hot-to-hot 208 volts is expected to serve the higher
voltage needs like larger 1 phase air conditioners, electric stoves,
and whatever else needs voltage in that range. But doesn't this kind
of setup potentially put an unbalance demand on the 3 phase connection?

2. ### John PopelishGuest

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)
So the current in that load involves current in two branches of the Y
and so, current in two sides of the delta.
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.
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.
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.

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

So the transformer is only 58% utilized for this single phase load.

Of course, if you had hooked a single phase load from neutral to one
branch of the Y, it would have been only 33.3% utilized (since that
load would have used exactly 1/3 of the windings running at a power
factor of 1). Phew!

3. ### Guest

| wrote:
|>
|> A very common power distribution transformer uses a DELTA wired primary
|> and a WYE wired secondary (thus including a neutral connection).
| (snip)
|> provide power in the same 3 phases that the primary windings have, the
|> hot-to-hot 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
hot-to-hot lines of the primary" on the last sentence of that paragraph.

|> Suppose I have a 1 phase device hooked up to
|> just one pair of hot-to-hot lines. Further suppose this is a resistive
|
| So the current in that load involves current in two branches of the Y
| and so, current in two sides of the delta.

That would seem to be the case. I can't think of a reason why it wouldn't.

|> 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.

|> If the above is true, and it would seem to me to be the case (but I am
|> asking to have that confirmed since there might be some physics that are
|> involved that I am not aware of), then if one were expecting to have
|> loads that are routinely out of balance on a 3 phase line, that the best
|> choice would be to configure the secondary so that the exact voltage one
|> needs is provided on exactly one winding (whether the secondary is wired
|> WYE or DELTA)?
|
| 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 6-Star configuration, as well
as a 2 core T-T configuration.

|> In many commercial buildings, power is provided in three phase due to
|> the higher capacity, at the 120/208 volt level in the US. This is the
|> right voltage for the lights and most appliances at the hot-to-neutral
|> lines. Then the hot-to-hot 208 volts is expected to serve the higher
|> voltage needs like larger 1 phase air conditioners, electric stoves,
|> and whatever else needs voltage in that range. But doesn't this kind
|> of setup potentially put an unbalance demand on the 3 phase connection?
|
| 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 c-a and in winding c-b, as well line c,
at a horizontal phase angle here? But there's no voltage at that angle.

| 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

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.

| 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.

| Of course, if you had hooked a single phase load from neutral to one
| branch of the Y, it would have been only 33.3% utilized (since that
| load would have used exactly 1/3 of the windings running at a power
| factor of 1). Phew!

If the secondary was 208Y/120 and all the load were on a single 120 volt
line, that would be as you describe. 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.

5. ### Guest

| OH Dear !!
| I hope you are not employed as an engineer somewhere.

Not an electrical engineer. My area of expertise is computer network and
systems administration and programming. Electronics is a hobby, and hence
the questions to acquire the knowledge. So do you have any further answers
to contribute?

6. ### John PopelishGuest

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
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.

(snip)
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)
(snip)
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.
I was analyzing the case where the single phase load reached the
current capacity of the transformer, as the most extreme example.
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
Yes, that was what I was describing.
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.)

7. ### Guest

| wrote:
|>
|> | wrote:
|> |>
|> |> A very common power distribution transformer uses a DELTA wired primary
|> |> and a WYE wired secondary (thus including a neutral connection).
|> | (snip)
|> |> provide power in the same 3 phases that the primary windings have, the
|> |> hot-to-hot 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
|> hot-to-hot 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
| 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 6-Star configuration, as well
|> as a 2 core T-T configuration.
|

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 c-a and in winding c-b, 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
|>
|> 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

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 A-B and B-C 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 T-T 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 delta-like from it, or 208 volts in one phase
along the untapped winding. The interesting thing about T-T is it uses
only 2 cores.

The 6-star would be nothing more than a special case of WYE, I believe:

* *
* \ /
/ \ \ /
/ \ <=> *----*----*
/ \ / \
*-------* / \
* *