Another High Leg question

[dicklaxt]

Okay getting the so called white jumpers out of the picture,now I see two windings. I also see where A-C has a 240v potential,I see where C-B has a 240v potential and that B to ground is 208v and of course the center tap and the 2-120v potentials. The kicker here is not readily apparent that A-B is also 240v,,,,,,,,,is that derived from the vector sums of A-C and C-B?

Please say it is,,,,,,,,,,, then I'm fat dumb and happy!!!!!.

dick

 

[charlie]

Open Delta and a little Wye

Open Delta and a little Wye

You may have had a problem with the first link since the "jumpers" are actually connections of both halves of the secondary windings of a 120/240 volt 1? transformer. Where the transformer with the neutral connection is made (white jumper in diagram) a jumper is shown for simplicity. The other transformer is identical to the first except there is no connection made to that point.

On smaller overhead transformers, the coil windings are brought out to three bushings. The two ends are taken to the two outside bushings and the two center ends are taken to the center bushing. This is all done on the inside of the tank and the coils are connected in series, just like a couple of flashlight batteries.

Where larger transformers are used, four bushings are brought out so there is actually an external jumper between the two center bushings to connect the coils in series.

---------------------------------------------------------------------
Slight change of subject​

When 120/240 volt pole mounted transformers are to be used to make a 208Y/120 volt bank, the secondary coils are connected in parallel and three transformers are required.

The two coils on the secondary side are connected in parallel by reconnecting the leads on the bushings on the inside of the tank. One bushing is not used and is normally taped up. Again, this is just like a couple of flashlight batteries but this time you would get 1.5 volts instead of 3 volts from the two batteries.

Where larger transformers are used, the external jumper must now be two jumpers and will connect the two outside bushings on each side to connect the coils in parallel. This works because the two center coils were crossed when they were connected to the center bushings at the factory.

If this is confusing, forget the second part. :smile:

 

[dicklaxt]

Yep,Charlie I got all that and no problem with any of it.

How about the question of the 240v potential question in my last post ,have you got an answer to that.

dick

 

[charlie]

. . . The kicker here is not readily apparent that A-B is also 240v,,,,,,,,,is that derived from the vector sums of A-C and C-B? . . .

You need to understand that I am not an engineer so I do not have an in depth knowledge of a lot of theory. To answer your question, I believe that you are correct.

I know that you can draw diagrams of the transformer configurations and then use simple geometry or trigonometry to calculate the missing potentials that may be confirmed with meters.

I know that 100% of the 1? load goes on the lighter (IPL slang for the transformer with the grounded center tap) and 57.7% of the 3? load goes on both the lighter and kicker (IPL slang for the transformer that "kicks" the high leg up). :smile:

 

[gar]

080727-1144 EST

dicklaxt:

The vector sum from C-A plus B-C = C-A.

The sum of sine waves of the same frequency is a sine wave of the same frequency and a phase angle of a value relative to one component sine wave. That angle could vary from 0 to 360 minus an infinitesimal amount modulo 360.

If you take the example and draw the sine waves for C-A and B-C and point by point add these you will get the sine wave for B-A. The sum of the instantaneous voltages around a loop has to equal zero.

.

 

[dicklaxt]

Okay then we still have the question outstanding,where is that 240v coming from if its not a vector sum which Gar says is "O".

I'm sure glad I don't work with this system(never have) give me the striaght old delta-wye or delta-delta.

Where are the engrs when you need them ???? Having coffee I guess LOL:grin: :grin: :grin:

dick

 

[charlie]

I'm sure glad I don't work with this system(never have) give me the straight old delta-wye or delta-delta.

Where are the engrs when you need them?

If the serving electric utility gave you a 124/240 V, 3?, 4w service, you would not be able to tell if it came from an open delta or a closed delta bank. In other words, you may have already worked with an open delta bank.

By the way, GAR is an engineer.

 

[dicklaxt]

No Charlie,I get to specify most of the time from client specs,,,I do only inplant industrial type stuff Chem ,Petro Chem,LNG etc .

Gar's an Engineer huh ,hey Gar where 'd that 240V come from???:grin:

dick

 

[gar]

080727-1328 EST

dicklaxt:

Draw two sine waves of the same amplitude with one shifted 120 degrees relative to the other. Now add these together. The result is another sine wave shifted 240 degrees from the 0 degree sinewave and the 240 degree sine wave is the same amplitude as the first two waveforms.

It is like adding two batteries together as has been previously mentioned, but here we need to also consider phase angle.

View the two secondaries as a single secondary and because of the voltages and phase angles of these components the output from this equivalent circuit is 240 V. You can put a voltmeter between the terminals and read 240 V and you can put a load across the terminals and draw power from the source.

There is a closed path for current to flow.

.

 

[dicklaxt]

Okay I think I got it,,,,,,a little review on wave forms will get me there I think ,thanks.

dick

 

[rattus]

Typo:

Typo:

080727-1144 EST

dicklaxt:

The vector sum from C-A plus B-C = C-A.

The sum of sine waves of the same frequency is a sine wave of the same frequency and a phase angle of a value relative to one component sine wave. That angle could vary from 0 to 360 minus an infinitesimal amount modulo 360.

If you take the example and draw the sine waves for C-A and B-C and point by point add these you will get the sine wave for B-A. The sum of the instantaneous voltages around a loop has to equal zero.

.

gar, I think your first equation has a typo in it.

 

[rattus]

If it makes you happy:

If it makes you happy:

Okay getting the so called white jumpers out of the picture,now I see two windings. I also see where A-C has a 240v potential,I see where C-B has a 240v potential and that B to ground is 208v and of course the center tap and the 2-120v potentials. The kicker here is not readily apparent that A-B is also 240v,,,,,,,,,is that derived from the vector sums of A-C and C-B?

Please say it is,,,,,,,,,,, then I'm fat dumb and happy!!!!!.

dick

Yes, you are fat, dumb, and happy.

In a delta, open or closed,

Vab + Vca + Vbc = 0 (corrected typo)

Then,

|Vab| = |Vca + Vbc| (phasorially that is)

The preferred term is "phasor" because these values are not true vectors even if they behave that way.

Edited by Charlie (please let me know if I did it wrong)

 

[gar]

080727-1449 EST

rattus:

I think there is a typo also.
So C-A + B-C should equal B-A

I universally like to use the name vector for something with magnitude and direction, and here I consider phase angle a direction.

.

 

[rattus]

080727-1449 EST

rattus:

I think there is a typo also.
So C-A + B-C should equal B-A

I universally like to use the name vector for something with magnitude and direction, and here I consider phase angle a direction.

.

Yup, I made a typo, I left off the "= 0", and for some reason I can't edit it.

Yes, remove the magnitude bars and,

Vca + Vbc = -Vab = Vba

I would argue that a phase angle is a measure of time, not direction. That is the reason that "phasor" is preferred. However, I do not expect that my opinion will change many minds on this matter.

 

[gar]

080727-1614 EST

dicklaxt:

Here is another way to intuitively view the open delta.

Consider a closed delta. If there was an unbalanced voltage around the delta loop, then there would be a circulating current in the secondaries with no load. Assume no circulating current, then the voltages must sum to zero or there would be a current.

Next at one corner of the delta, B, break the connection at B between A-B and C-B. Call these two points B1 and B2. Let B1 be at the end of A-B and B2 at the end of C-B. When this node is open what is the voltage difference between B1 and B2? It is zero based on the conditions when B1 was connected to B2.

For the voltage between B1 and B2 to be 0 means that the voltage from B1 to A is identical to the voltage between B2 and C in both magnitude and phase. Thus, there is no difference if the load is between B1 and A or between B2 and C assuming perfect voltage sources and zero internal impedances.

.

 

[roger]

Can't edit my own post!

Rattus, you have a ten minute window to edit a post, after that you have to make corrections in another post.

Roger

 

[gar]

080727-1630 EST

rattus:

I saw your typo, but thought it was self-evident and did not comment.

It is real tough to type these posts accurately even after looking at them a couple times before posting. 10 minutes is short, but the philosophy has some merit.

.

 

[charlie]

Yup, I made a typo, I left off the "= 0", and for some reason I can't edit it. . .

Make sure that I edited it the way you wanted it done, I have deleted your subsequent post. :smile:

 

[jrannis]

080727-1614 EST


dicklaxt:

Here is another way to intuitively view the open delta.

Consider a closed delta. If there was an unbalanced voltage around the delta loop, then there would be a circulating current in the secondaries with no load. Assume no circulating current, then the voltages must sum to zero or there would be a current.

Next at one corner of the delta, B, break the connection at B between A-B and C-B. Call these two points B1 and B2. Let B1 be at the end of A-B and B2 at the end of C-B. When this node is open what is the voltage difference between B1 and B2? It is zero based on the conditions when B1 was connected to B2.

For the voltage between B1 and B2 to be 0 means that the voltage from B1 to A is identical to the voltage between B2 and C in both magnitude and phase. Thus, there is no difference if the load is between B1 and A or between B2 and C assuming perfect voltage sources and zero internal impedances.

.

http://forums.mikeholt.com/attachment.php?attachmentid=1301&d=1204377498
Ok here it is, AGAIN. The "Open Delta". I hope this helps.:grin:

 

[charlie]

http://forums.mikeholt.com/attachment.php?attachmentid=1301&d=1204377498
OK here it is, AGAIN. The "Open Delta". I hope this helps.:grin:

There is just something about the picture that doesn't seem like its a transformer!?! :grin: