Transformer Loading:

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Re: Transformer Loading:

Sorry to be redundant, but this is an interesting question and I wonder if someone can bring the answer down to non-PE level.

The statement is (as I understand it) with a 10 amp load connected to A-phase and B-phase you have a VA load of (208*10) 2080. With a 10 amp load connected between A-phase and neutral and another 10 amp load connected between B-phase and neutral, you have a connected load of (120*10+120*10) 2400.

The question is why is there a difference in the VA when you have the same current going through the two phases of the transformer?

Is the agreed upon answer that when connecting at 120 volts you don't have current out of phase with voltage and when you connect at 208 volts you have a phase shift between current and voltage that causes the lower VA at the same current draw?
 
Re: Transformer Loading:

Originally posted by hardworkingstiff:
Sorry to be redundant, but this is an interesting question and I wonder if someone can bring the answer down to non-PE level.

The statement is (as I understand it) with a 10 amp load connected to A-phase and B-phase you have a VA load of (208*10) 2080. With a 10 amp load connected between A-phase and neutral and another 10 amp load connected between B-phase and neutral, you have a connected load of (120*10+120*10) 2400.

The question is why is there a difference in the VA when you have the same current going through the two phases of the transformer?

Is the agreed upon answer that when connecting at 120 volts you don't have current out of phase with voltage and when you connect at 208 volts you have a phase shift between current and voltage that causes the lower VA at the same current draw?
Click to expand...
Stiff, there is no agreed upon answer, but your take is basically right. Some even claim that 2400VA has no place in this discussion, but that is what it is all about. Here is my explanation.

1) My contention is that apparent power in VA computed from Vphase x Iphase does not always equate to apparent power computed from Vload x I load. This argument is proven by this example. Therefore, there is significant error in computing transformer loading from the load itself. This though is a special case, and in a reasonably balanced system there would be no problem in using conventional methods to compute the load.

However, if you throw in a power factor, real power in watts does equate.

2) I also contend that power, real or apparent, adds arithmetically, not vectorially. Power is not a vector quantity. You cannot say that 1200 + 1200 = 2080 when added vectorially because these are scalar quantities.

3) The reason for the discrepancy is that the line to line voltage, obtained vectorially, is 208V, not 240V. It produces the current, but this current is 30 degrees displaced from the phase voltages. This means that the PF in the windings is 87% and 100% in the load. Some of my critics have a hard time accepting that fact though.

It is also my contention that the "exact" method to determine transformer loading is Vphase x Iphase. In this case, that is 1200VA per transformer. This is of no great importance though unless you are dealing with a severely unbalanced system.
 
Re: Transformer Loading:

Ed,

Your take on this matter is correct. It does boil down to the current. This is tantamount to saying that one should use Vphase x Iphase to compute the load if they should ever encounter a situation like this which is unlikely. This is really a brain teaser which should help in the understanding of 3-phase loading if we can ever get the point across.

Rattus
 
Re: Transformer Loading:

rattus
I have removed the word "balanced" from grlsounds
post. The grlsound report says
This example is of a purely resistive single phase load, DELTA connected to a WYE source. (ie: transformer) I am told that when this connection happens, the current and voltage become out of phase by 30 degrees.
Click to expand...
The report says that when the load is connected in delta that the current and voltage is out of phase by 30 degrees. This is not correct.
You then go on to say
grlsound reports that three, count them, EEs agree that the wye phase currents and wye phase voltages are out of phase by 30 degrees. This is should be obvious from the vector diagram.
Click to expand...
The report did not say anything about the wye phase current and wye phase voltage being out of phase. The current in the delta is out of phase with the line current which is connected wye. It appears that you have misquoted the report.
 
Re: Transformer Loading:

Bob,

You are sniping again. The quote is rather vague, but it means that Vphase and Iphase are out of phase by 30 degrees. Obviously Vload and Iload cannot be out of phase.

Why don't you put down your sniper's rifle and work out these angles yourself. It only takes a few minutes. It is obvious that you have not done so. If you had, you would not be posting these baseless criticisms.

I would also suggest to Jim Dungar and anyone else who has doubts that they work out the vector diagram for this problem. It is not hard.

[ March 10, 2005, 10:05 AM: Message edited by: rattus ]
 
Re: Transformer Loading:

rattus,

I believe the quote you keep using is related to the reduced loading required for open-delta and open-wye transformer banks. I am not sure if the quote is being used out of context. I would be interested in seeing the entire discussion.

You seem to be saying there are special/different rules for calculations involving real power versus apparent power. This can not be possible.

There is no such thing as a delta connected single phase load connected to a wye transformer. A delta connection means all three phases are used. In your examples all we have is a single phase load connected phase to phase and two single phase loads connected phase to neutral.
 
Re: Transformer Loading:

Ed,

Excellent "layman's" summary - I genuinely appreciated it.

Most of the time, day-to-day, we facilites design types try to avoid vectors, scalar and vector products, phasors, etc. "VA" and "RMS" values usually permit us to do it. Occassionally, especially for certain metering or relay protection schemes, paralleling sources, uncommon load characteristics, etc. we have to bite the bullet and at least consider them.
 
Re: Transformer Loading:

Originally posted by jim dungar:
rattus,

I believe the quote you keep using is related to the reduced loading required for open-delta and open-wye transformer banks. I am not sure if the quote is being used out of context. I would be interested in seeing the entire discussion.


You seem to be saying there are special/different rules for calculations involving real power versus apparent power. This can not be possible.

Reply: All I am saying is that in this case, the "per phase" apparent power does not equate to the apparent power in the load, and in general it will not, but the error is acceptable under normal circumstances. I think rbalex has said as much.

There is no such thing as a delta connected single phase load connected to a wye transformer. A delta connection means all three phases are used. In your examples all we have is a single phase load connected phase to phase and two single phase loads connected phase to neutral.

Reply: Jim, the quote is poorly worded. It is obvious that a single line to line load is meant.
Click to expand...
Now, put aside your formulas which work very well with balanced loads and work out the vector diagram for this problem. Until you do so, you will not grasp what I am saying.
 
Re: Transformer Loading:

Bob,

I have to zing you on this one:

"The report did not say anything about the wye phase current and wye phase voltage being out of phase. The current in the delta is out of phase with the line current which is connected wye. It appears that you have misquoted the report."

There is only one current, Iphase = Iline = Iload. You are saying that the current is out of phase with itself. This cannot be! Now who needs to be educated?

Again, I challenge you to work out the angles yourself. Until you do, you cannot make any serious criticisms.

Rattus
 
Re: Transformer Loading:

Rattus
You are using the the grlsound report to try and make your case. In the report it says
This example is of a purely resistive single phase load, DELTA connected to a WYE source. (ie: transformer) I am told that when this connection happens, the current and voltage become out of phase by 30 degrees
Click to expand...
If the load is delta connected then you must be using all 3 phases. If the load is connected delta then the phase current is not the same as the line current.

You continue to say the following
grlsound reports that three, count them, EEs agree that the wye phase currents and wye phase voltages are out of phase by 30 degrees. This is should be obvious from the vector diagram.
Click to expand...
That is not what the report said. It did not reference the line current and voltage.
You continue bounce all over the problem. My comment is to the report and how you misquoted the results. If the load is delta, to infer that there is only one current is a complete misunderstanding. Each phase has a phase current and it is 30 degrees out of phase with the line current.
There is only one current, Iphase = Iline = Iload. You are saying that the current is out of phase with itself. This cannot be! Now who needs to be educated?
Click to expand...
In a delta connect load the Iphase is the Iload.
It is not the same as the Iline.
 
Re: Transformer Loading:

Originally posted by bob:
Rattus
You are using the the grlsound report to try and make your case. In the report it says
This example is of a purely resistive single phase load, DELTA connected to a WYE source. (ie: transformer) I am told that when this connection happens, the current and voltage become out of phase by 30 degrees
Click to expand...
If the load is delta connected then you must be using all 3 phases. If the load is connected delta then the phase current is not the same as the line current.

Reply: As I pointed out to Jim Dungar, the quote is poorly worded. We all know this issue is about a single line to line load. You are obfuscating now.

You continue to say the following
grlsound reports that three, count them, EEs agree that the wye phase currents and wye phase voltages are out of phase by 30 degrees. This is should be obvious from the vector diagram.

Reply: Again, the quote is poorly worded. grl is not specific about which voltages are meant, but it cannot be the load voltage because the phase angle is zero there. That leaves only the phase voltages. More obfuscation.
Click to expand...
That is not what the report said. It did not reference the line current and voltage.
You continue bounce all over the problem. My comment is to the report and how you misquoted the results. If the load is delta, to infer that there is only one current is a complete misunderstanding. Each phase has a phase current and it is 30 degrees out of phase with the line current.

Reply: grl meant line to line when he said delta connected, not a full delta. This has always been the case.

There is only one current, Iphase = Iline = Iload. You are saying that the current is out of phase with itself. This cannot be! Now who needs to be educated?
Click to expand...
In a delta connect load the Iphase is the Iload.
It is not the same as the Iline.

Reply: Iphase = Iline always with a wye transformer, and in this case it is also equal to Iload. To say that a current is out of phase with itself is nonsense. Get it right!

Click to expand...
Why don't you work out the phase angles yourself? Until you do, your criticisms are merely sniping and obfuscation. If you can't or won't do this, then your criticisms are hollow.
 
Re: Transformer Loading:

rattus

Please show me your calculations. My formulas simply determine each individual winding VA which are then added vectorially due to the time domain difference between Van and Vnc.

average power P=E*I*cos(α)
reactive power Q=E*I*sin(α)
total power VA=((P^2)+Q^2))^-2

given your values of:
Van =120, Ian=10, αan=30
Vcn =120, Icn=10, αcn=-30

I calculate:
Pan = 120*10*.866 = 1040W
Qan = 120*10*.5 = 600VAR
VAan = 1200VA
Pcn = 120*10*.866 = 1040W
Qcn = 120*10*-.5 = -600VAR
VAcn = 1200VA

But because this is a wye connected transformer Vcn is 120? out of phase with Van so VAan and VAcn must be added vectorially not arithmetically giving a total VA = 2080.

A detailed method for handling unbalanced or unsymmetrical vectors, especially if the phase angles are not 120?, is to resolve them into their zero, positive, and negative sequence components.
 
Re: Transformer Loading:

WOW! There is someone more stubborn then myself! :D

Rattus are you sure you don't want to open your mind to all these other peoples thoughts? :eek:
 
Re: Transformer Loading:

Whew,

What topic, learned something and got to use www.dictionary.com.

Rattus,

I would request that you post the calculations for phase angles that you are requesting others who question you to work out. That is the true way to prove your statement, show the work for everyone to see.

Tony
 
Re: Transformer Loading:

First and foremost, I'm not an engineer, so let that be known.

This is how I understand the argument that Rattus is making: If we connect a single phase purely resistive load to a 3 phase transformer, it will inherently "create" a power factor of 87%, due to the phase relationship of the voltage and current.

To make sure I am clear about this, these assumptions must be made: The load is 10 amps single phase, purely resistive, connected to two active lines of a 3 phase wye system.

The apparent power, as seen by the transformer, is 2400 VA . The true power, as consumed by the load, is 2080 watts .

If we had to go buy a transformer to drive this load, and assuming we actually had the choice between a 2080 VA transformer, and a 2400 VA transformer, we must choose the 2400 VA transformer or we will let the smoke out of the 2080 VA one.

As has been said, this is a rare and special case indeed, and won't be encountered very often in the world of faily balanced 3 phase systems.
 
Re: Transformer Loading:

Tony, I have done this before, and no one has paid me any mind. But I will do it again.

All you have to do is draw the vectors for the two 120V phase voltages, then draw in the 208V line to line voltage. By inspection you can see that the line to line voltage is 30 degrees out from either phase voltage.

Since the load is resistive, the current is in phase with the line to line voltage.

The same current flows through the secondary windings, therefore it it 30 degrees out of phase with the phase voltages. That being the case, the PF is 87% in the secondary windings. Then 2400VA x 0.87 = 2080W. This is the way it all adds up.

At least 5 EEs, including myself, have agreed that this is true, and the loading on the transformers is 1200VA per winding. For a given current, the heat generated is the same irrespective of the load connections.

ONE MORE TIME: Power does not add vectorially!
Power is not a vector quantity! Jim, you are a P.E., you should know this.

Iwire, I have heard what they have to say. Bob, the consultant, has merely sniped and made one erroneous comment after another plus harshly questioning my credibility.

Jim is still stuck on adding powers vectorially. You would not add real power vectorially, you cannot add apparent power vectorially either. Power is a scalar quantity, therefore it must be added arithmetically. I can't seem to get this point across!

Rattus
 
Re: Transformer Loading:

Rattus
Tony had this to say :
I would request that you post the calculations for phase angles that you are requesting others who question you to work out. That is the true way to prove your statement, show the work for everyone to see.
Click to expand...
Why don't you do this so we can see what you are talking about. You might ask Ed to add a drawing to the post showing the vectors.

[ March 10, 2005, 07:38 PM: Message edited by: bob ]
 
Re: Transformer Loading:

Bob, I have done this more than once, and no one took note. I am asking you and Jim to do it yourselves. Anyone proficient in AC system design can do it blindfolded.

Also, I responded to Tony with a graphical approach which you can sketch out yourself in a few seconds. This proves the angles. The computation is routine.

Ed posts some nice diagrams, but his vectors don't have a direction, so they are not vectors. He himself admits he is no mathematician or expert in vector analysis.

Also, read the post by peter d. He is not an engineer, but he expressed the argument very well.
 
Re: Transformer Loading:

peter d, you seem to have your mind wrapped around this problem and its solution. Do you now think you understand three phase just a little bit better?
 
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