Four-wire delta, phasors, and Kirchoff:

Status
Not open for further replies.

Rick Christopherson

Senior Member
winnie said:
Directed sum means 'add while cognizant of direction".
This is exactly my point. Rattus has artificially changed the direction, so the only way to account for that is to subtract. The direction is backwards, and that is why you are suddenly forced to subtract. Put the minus sign back into the phasor (polar coord) and you now have a phasor diagram that meets Kirchoff's Law.
 
Last edited:

coulter

Senior Member
mivey said:
Carl,

If you don't agree with the labels for graphs #1-#6, please list the labels you would like to see so the terminology you are referencing will be clear.

I'm not saying I have the best terminology but I think my understanding of how the waves propagate is correct (disclaimer: barring any stupid errors).

carl said:
no, its fine.

mivey said:
I really would like your labels for my curves as a reference. I may change the way I label things.
mivey -
I really can't help here or add anything useful. Anything I could say would just be repetitiuos. Like this next statement:

The model I propose has a 240D with the arrows all pointing head to tail. Changing to 240/120D does not magically reverse the sense of one half of one of the 240 arrows. The model you propose requires this flip.

As I see this discussion, you, rattus, (maybe winnie - can't tell for sure), and a few others, want to use a convention that that I see as unneccessarily limiting.

You (plural) understand the math model well enough to handle the occasional need for an additional minus sign or two that it takes to make this model work - so, really, its okay with me.

carl
 

rattus

Senior Member
Typo:

Typo:

Rick Christopherson said:
Which 3 are rises? The Blue, Violet, and Red polar coordinates all have the same Real component, but you are suggesting that only the Red should be subtracted?

I put these into polar coordinates so it becomes straight mathematics--no vectors anymore. If you can't add them up as Kirchoff states, that means that one of them is improperly defined.

Kirchoff doesn't say that you can selectively subtract when you get to a voltage source that doesn't fit his law. If you need to subtract, then it means that the polarity of the voltage source is wrong.

Kirchoff states that you sum the voltages as you go around a closed loop:
G to A, A to B, B to C, C to G

You are trying to jump from C to G and then go back to C:
G to A, A to B, B to C, G to C

That's not a closed path. You are ending at C instead of G, where you started.

Typo: C to G is a drop
 

Rick Christopherson

Senior Member
rattus said:
Typo: C to G is a drop
So is it your assertion that when you move from C to G you are decreasing in voltage, but when you move from G to A you are increasing in voltage? (given the positive half cycle of the sine wave)?
 

mivey

Senior Member
homerun

homerun

Just found I can get this on my Blackberry...that is just so wrong

I think winnie hit the nail on the head-homerun
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
rattus said:
Think DC, say a battery driving a lamp. Summing the voltages around the loop, we see a voltage rise and a voltage drop. One is positive, one is negative, therefore we have to subtract.
Here's where (semantics abound!) we look at it differently. To me, to obtain a difference, you either add a negative number to a positive, or you subtract one positive number from another.

To subtract a negative number from a positive number is like using a double negative in grammar, isn't it? :-?
 

rattus

Senior Member
Still waiting:

Still waiting:

Rick Christopherson said:
So is it your assertion that when you move from C to G you are decreasing in voltage, but when you move from G to A you are increasing in voltage? (given the positive half cycle of the sine wave)?

With phasors there is no need to consider instantaneous values, although in this case it is true because voltage drops from 120V to zero in the first case, and rises from zero to 120V in the second.

Go back and study my example of the opposing phasors in a wye.

And provide some references telling me that phasors must be connected head to tail. Until you do, your arguments cannot be accepted.
 

Rick Christopherson

Senior Member
rattus said:
Rick,

How about a reference of the "proper rules of vector analysis"?
This scan was taken from "Introduction to Vector Analysis", Fourth Edition, by Davis and Snider:

This passage clearly states that when you are adding vectors, you place them tip-to-tail.
rattus said:
Can't do that Rick. You have changed the reference on Vcg so it adds instead of subtracts. This is absolutely wrong! Both 120V phasors must be drawn tail to tail! In other words, you screwed up my perfectly good diagram.
This passage also explains why I did not screw up your drawing. Vectors are commutative, and that means it does not matter which order you place them in the diagram.

When I rearranged the vectors in your diagram, I was following the rules of commutation to show that the answer resulted in a non-zero sum of the vectors--hence, a violation of Kirchoff's Law.

VectorAdd.jpg
 

Rick Christopherson

Senior Member
You sidestepped my previous question without answering it. Forget about it, I will move to the next question. I do have a reason for asking these questions, so please answer them.

Looking at this from a nodal aspect of the system, and given the stipulation that we will use node G as our zero reference, do you agree with the voltages (in polar coordinates) I have assigned to each of the nodes in the diagram below.

Nodes.jpg
 

rattus

Senior Member
Yes, but?

Yes, but?

Rick Christopherson said:
This scan was taken from "Introduction to Vector Analysis", Fourth Edition, by Davis and Snider:

This passage clearly states that when you are adding vectors, you place them tip-to-tail.
This passage also explains why I did not screw up your drawing. Vectors are commutative, and that means it does not matter which order you place them in the diagram.

When I rearranged the vectors in your diagram, I was following the rules of commutation to show that the answer resulted in a non-zero sum of the vectors--hence, a violation of Kirchoff's Law.

VectorAdd.jpg

That is true for addition, but what about subtraction? You did not provide the rules for subtraction! Give us the rest of that section. I know what it will say. You have given us a half-truth.

Since Vcg requires subraction, you would have to arrange the phasor head to head at point A or point B, but that makes a clumsy diagram.
 

rattus

Senior Member
Rectangular or polar?

Rectangular or polar?

Rick Christopherson said:
You sidestepped my previous question without answering it. Forget about it, I will move to the next question. I do have a reason for asking these questions, so please answer them.

Looking at this from a nodal aspect of the system, and given the stipulation that we will use node G as our zero reference, do you agree with the voltages (in polar coordinates) I have assigned to each of the nodes in the diagram below.

Nodes.jpg

I would agree that the values are correct in rectangular, not polar cooriidnates.

And, my calculator says that the polar equivalent of -120, j0 is 120 @ 180.
 

mivey

Senior Member
mixed loops

mixed loops

I'm kind of thinking out loud so don't bust my chops too hard...

Aren't we mixing single phase and three phase loops when we add in the neutral reference?

For the three phase, we have the a-b-c loop. For single phase, we have a-load-g, c-load-g, b-load-g, or a-load-c?
 

mivey

Senior Member
two diagrams

two diagrams

How about one diagram for the three-phase phasors and one diagram for the single-phase phasors?

[added] yuck and yumm as you see fit.
 

Rick Christopherson

Senior Member
rattus said:
That is true for addition, but what about subtraction? You did not provide the rules for subtraction! Give us the rest of that section. I know what it will say. You have given us a half-truth.
First off, I didn't care about subtraction because Kirchoff's law is "summation". Secondly, in the passage below from the same book referenced above, you will see that for subtraction, you invert the vector and add it. Viola, there is the missing minus sign I have been talking about that lead you to perform subtraction. As I said way back at the beginning, you still got the correct answer, but you violated Kirchoff's law by subtracting, and the reason is because you redefined the voltage source instead of simply changing your point of reference.

VectorSub.jpg


Kirchoff's Law doesn't care if you sum your voltages clockwise or counterclockwise, but you can't do both at the same time, and that is what you tried to do in the original document in post #1 of this thread. Instead of simply saying that you changed your point of reference, you redefined the voltage source itself.

In your table in post #1, all of your voltages that you were summing were being reference counterclockwise until you got to Vcg, where you suddenly switched to clockwise. You hid this fact by putting in the equal signs between each operation. Had you not put in those equal signs, the mistake would have been much clearer to everyone that you changed directions from counterclockwise to clockwise.

Your phasor diagram makes no reference to the fact you changed how you define each phasor, and that is why it is wrong, and that is why the magnitude of the phasor between C and G needs a negative value to properly solve Kirchoff's Law.
 

mivey

Senior Member
Who's been violated?

Who's been violated?

Rick Christopherson said:
Kirchoff's law is "summation"...violated Kirchoff's law by subtracting

Maybe that's the reason I'm not happy with the whole thing yet. I don't see subtraction as a "violation". I think it falls right in line with Kirchoff's concept because you are still going around the loop. I don't see the word "subtraction" as a deal killer.

I don't think Kirchoff mentioned phasors in his law either but I'm not going to say we are "violating" anything because we are talking his law and applying it to phasors.
 

rattus

Senior Member
Rick Christopherson said:
First off, I didn't care about subtraction because Kirchoff's law is "summation". Secondly, in the passage below from the same book referenced above, you will see that for subtraction, you invert the vector and add it. Viola, there is the missing minus sign I have been talking about that lead you to perform subtraction. As I said way back at the beginning, you still got the correct answer, but you violated Kirchoff's law by subtracting, and the reason is because you redefined the voltage source instead of simply changing your point of reference.

VectorSub.jpg


Kirchoff's Law doesn't care if you sum your voltages clockwise or counterclockwise, but you can't do both at the same time, and that is what you tried to do in the original document in post #1 of this thread. Instead of simply saying that you changed your point of reference, you redefined the voltage source itself.

In your table in post #1, all of your voltages that you were summing were being reference counterclockwise until you got to Vcg, where you suddenly switched to clockwise. You hid this fact by putting in the equal signs between each operation. Had you not put in those equal signs, the mistake would have been much clearer to everyone that you changed directions from counterclockwise to clockwise.

Your phasor diagram makes no reference to the fact you changed how you define each phasor, and that is why it is wrong, and that is why the magnitude of the phasor between C and G needs a negative value to properly solve Kirchoff's Law.

1. There is nothing in KVL that says you can't subtract. In fact, it is required in order for the sum to be zero. "Summation" is algebraic, you must include negative values and/or subration. Give us a reference to prove that claim.

2. The direction of summation is clearly stated as CCW, period. No clockwise to it. Where do you get this idea?

3. Since Vcg is a drop, we subtract it. No different from adding Vgc. We must do this because we are summing from head to tail instead of tail to head. Quite meet, right, and proper, and the solution is right.

4. "Magnitude" in this case is a positive constant. It cannot change. Quit saying it has a negative value.

5. "Drop" in DC means a less positive value. With phasors it means summing that phasor from head to tail which is not necessarily a decrease in voltage.
 

rattus

Senior Member
Something different:

Something different:

Please open the attachment which is a partial phasor diagram of a 120/208 wye.

Now, we already know that Van = 120 @ 0 and Vbn = 120 @ -120. How would one use KVL and phasors to determine the magnitude and phase of Vab?
 

Rick Christopherson

Senior Member
rattus said:
2. The direction of summation is clearly stated as CCW, period. No clockwise to it. Where do you get this idea?
I hate to say this because the discussion has been going moderately well, but your understanding of the principles you are arguing is so light that you don't even realize that you have just grossly contradicted your founding premise.

You just asserted that the "Summation is clearly stated as CCW", yet you defined your voltage between points C and G in a clockwise fashion.

This is the single most critical point I have been making from the very outset. You have defined all of the voltages in your phasor diagram in a counterclockwise direction, with the exception of the voltage between points C and G. You defined this voltage in the clockwise direction.

Here is a hint: Take a look at the nodal analysis diagram that you have already agreed was correct.

I will address the rest of your posting at a later time, but I do not want to distract you from this single, rather critical, point.
 

mivey

Senior Member
for argument's sake

for argument's sake

Rick Christopherson said:
but your understanding of the principles...you don't even realize

Please don't do that. Until you can get into someone's mindset, you don't know what they are thinking or what they understand. That is the whole point of having some of these discussions. Debate what is said or the concepts.

I like to try to gather information about someone's point of view so I can try to see where they are coming from. If I can get enough info, I may be able to determine if the difference is about actual concepts or terminology.

If it is about concept differences like "do electrons fly through the air" vs " electrons don't jump the gap but are pushed away", that is a concept difference that can be debate (from "AC Current Flow Question" thread)

If it is a terminology difference, just use the other person's terminology (even if you don't like it) and see if you have the same understanding of the concept. Don't get too caught up in a standard terminology. Like someone posted elsewhere: "the best thing about standards is that there are so many to choose from."

I try, the best I can, to see if I can get enough info to "look" through someone else's eyes. Pretend you are critiquing a book's content, not the author.
 
Status
Not open for further replies.
Top