How to convert 2 phase to 3 phase 1

[eeprom]

Hello,
I am trying to understand a common rural utility transformer circuit, wherein two transformers are used to convert 2 primary phases of 7200V into 3 phase 480V.

We have two incoming phases (A and C) and two single phase transformers. On the primary side, the transformers are wired in series; the connection point between the two transformers is tied to neutral, phases A and C are on the other two connections. On the secondary we somehow end up with 3 phase 480V. I know that this circuit is common, but I don't understand it. I've drawn it out a few times and my calculations results in a very unbalanced voltage on the secondary.

Can anyone refer me to a book or some calculations which can mathematically describe how this works?

thanks

 

[Overvoltage]

Try looking for a Scott-T connection, I believe.

 

[Overvoltage]

Here's a link to Wikipedia.

 

[davidbeach]

Nope, not a Scott-T connection, but rather an open wye-open delta connection.

Look at

http://tinyurl.com/r6nnls.

A couple of pages in is a diagram of the connection.

 

[Overvoltage]

Erf...apparently the second half of the post didn't make it to my brain when I read it. My mistake...

 

[jghrist]

Also see page 12 of

http://www.cooperpower.com/Library/pdf/R201902.pdf

 

[jraef]

2 phases out of a 3 phase system is NOT 2 phase. 2 phase power is characterized by the phases being separated by 180 degrees. 3 phase power has the phases separated by 120 degrees. In the scenario you describe, the 2 phases would be 120 degrees apart. We call that single phase.

2 phase power systems are an anachronism that was relatively common in the earliest days of electical distribution in the US, but was supplanted by 3 phase long ago. yet some isolated pockets of true 2 phase power still exist in places like Philadelphia. A Scott-T transformer is designed to create 3 phases from a true 2 phase supply and vice versa for places like that. It is not related to what you are talking about.


"If I had eight hours to chop down a tree, I'd spend six sharpening my axe." -- Abraham Lincoln
For the best use of Eng-Tips, please click here -> faq731-376

 

[jraef]

I should have added however that in other English speaking countries (and non-English translated documents), 2 out of 3 phases is often referred to as "2 phase", but that is because they do not have to contend with the existence of true 2 phase systems as we do. So I guess it depends on where you are. but being that you referred to 7200V and 480V, I thought it a safe assumption that you were in the US.


"If I had eight hours to chop down a tree, I'd spend six sharpening my axe." -- Abraham Lincoln
For the best use of Eng-Tips, please click here -> faq731-376

 

[stevenal]

jraef,

I think you meant 90 degrees, not 180.

faq238-777

 

[jraef]

Oops, yes. 4 wires, each 90 degrees from the next.


"If I had eight hours to chop down a tree, I'd spend six sharpening my axe." -- Abraham Lincoln
For the best use of Eng-Tips, please click here -> faq731-376

 

[busbar]

A slang term for the service feeding open-wye primary banks is “V-phase.” Another form of open-delta secondary bank uses an open-delta primary. An easy way to differentiate the two is ‘two cutouts on the pole—open-wye’ hi-side; ‘three cutouts—open delta’ high-side.

Voltage balance [and subsequent stator heating] can be somewhat of a problem with 3ø 2-transformer banks, so many utilities limit load to 20 or 30 horsepower.

 

[waross]

Look up a wye primary, delta secondary connection. Work out the voltages. Draw a vector sketch of the delta secondary voltages. Now remove one of the transformers. (For the sake of this explanation, your wye primary must have the wye point connected to the system neutral. Although this neutral connection is not much used in practice, it is valid if the primary voltages are well balanced.)
You will see that when you remove any one of the three transformers, the vector sum of the two remaining transformer secondaries equals vector of the transformer that was removed. This may help to understand the relationships.
This connection is called an open delta in North America but may be called a "V" connection in IEC land.
The three transformer delta secondary connection with one corner unconnected is called a broken delta in North America (NEMA land) and an open delta in IEC land. Beware of confusion on an international forum.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[eeprom]

Okay, let's work this out verbally. On the primary, I have a 7200V line. I have two single phase transformers (7200 - 480/240) wired in series. The common point of the two transformers is tied to neutral. Across the primaries, I have phases A and C. Let's make some assumptions. Primary phase voltages for a positive sequence are:

Van = 4157 angle 0
Vcn = 4157 angle 120
from this...
Vac = Van-Vcn = 7200 angle -30


Primary connections at 7200V


Phase A -----\/\/\/\------/\/\/\/\------Phase C
|
|
N

Secondary connections

Phase A -----\/\/\/\------/\/\/\/\------Phase C
|
|
Phase B

So, across both transformers on the primary, I have Vac which is equal to 7200 angle -30. Across the individual windings, I have Van (from A to neutral tie), and Vcn (from C to neutral tie).

So if I treat the primary connections as a three phase system, and I call the neutral point B, I can calculate voltages Vab, Vbc, and Vca, and their sum should be zero. I already know Vab, it is the same as Van. And Vca is -Vac, and I know that. And Vbc is the same as -Vcn. So,

Vab = 4157 angle 0
Vbc = 4157 angle -60
Vca = 7200 angle 150

Ignoring the rounding errors, this system does sum to zero. Before I start wrestling with the secondary voltages, can anyone see anything wrong with my statements so far?

thanks

 

[davidbeach]

Don't introduce B in the primary, it isn't there. Say that phase A (V_AN) is 4157[∠]0 and phase C (V_CN) is 4157[∠]240. (Side note, 7200V as a line-neutral voltage is far more common than 7200V as a line-line voltage.)

Given that, one of the secondary windings will be V_ca = 480[∠]240 and the other will be V_ab = 480[∠]0. The missing voltage, V_bc will calculate to be 480[∠]120.

 

[trosepe]

eeprom,
See if the attached helps.

 

[waross]

The transformers are NOT wired in series. They are wired to different phases.
With a wye delta transformer bank, it is a compromise as to whether or not to connect the neutral. When the neutral is connected, any one of the three transformers may be removed and the bank will still produce three phase power at reduced capacity.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[eeprom]

I don't see it. Here is the wiring
Transformers are 7200 - 480/240
Primary
T1h T2h
Phase A -----\/\/\/\------/\/\/\/\------Phase C
|
|
N

Secondary
T1L T2L
Phase A -----\/\/\/\------/\/\/\/\------Phase C
|
|
Phase B

Here is what is assumed (I did have my voltages incorrect earlier).
Primary:
Phase A to neutral is 7200 angle 0.
Phase C to neutral is 7200 angle 120
V_AC = 12470 angle -30

Secondary:
The voltage Vab (across the winding T1L) will be the reflected voltage from the primary winding T1h. The primary voltage is V_AN and is equal to 7200 angle 0, therefore secondary Vab will be 480 angle 0.

The voltage Vcb (across the winding T2L) will be the reflected voltage from the primary winding T2h. That primary voltage is V_CN and is equal to 7200 angle 120, so the secondary voltage Vcb will be equal to 480 angle 120.

Conclusion(??) - Secondary voltages are:
Vab = 480 angle 0
Vbc = -Vcb = 480 angle -60
Vca then becomes 831V

I am obviously making some improper assumptions somewhere here. Can someone please explain where I am going astray?

 

[jghrist]

You need to reverse the polarity of one of the primary windings. See connections on p. 12 of

http://www.cooperpower.com/Library/pdf/R201902.pdf

 

[NAZ55]

Please refer to trosepe's diagram. The answer is in the polarity and how they are connected for primary and secondary.

 

[davidbeach]

Did you look at my post? It is all there. I've corrected the primary voltages below based on your correction.


Say that phase A (V_AN) is 7200?0 and phase C (V_CN) is 7200?120.

Given that, one of the secondary windings will be V_ca = 480?120 and the other will be V_ab = 480?0. The missing voltage, V_bc = -(V_ca+V_ab) will calculate to be 480?240.

In each case, V_ab will have the same phase angle as V_AN, V_bc will have the same phase angle as V_BN, and V_ca will have the same phase angle as V_CN. If you are getting different angles you have polarity wrong.

Draw it out, line up the primary windings with the secondary windings, use all three windings and then delete the missing winding if necessary.

Something like:

C                               c 
 \                               \
  \                            .  \
   \                          .    \
    \                        .      \
     \                      .        \
      --------- A        b ------------ a
     .
    .
   .
  .
 .
B

Make sense?