120v 277v insame 3ph box

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Re: 120v 277v insame 3ph box

Thank all of you for your time and answer, and i like tshea answer. They put 277v on lug A, 120v on lug B, and 120v on lug c. and one neutral do not know from witch transformer. It is a leased building left it alone after I found this problem.

[ December 28, 2005, 10:39 AM: Message edited by: terrytc ]
 
Re: 120v 277v insame 3ph box

Originally posted by don_resqcapt19: Charlie, You have a right triangle with a hypotenuse of 240 and one side of 120. Solve for the third side. Don
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Come on Don, I know how to do that. What I did not know for sure is whether to do that. You have two vectors (sides of a triangle). One has a length of 120 and the other has a length of 240. How do we know that the 240 is the longest side? How do we know that the result of adding 120 volts to 240 volts is going to be less than 240 volts? Or should it have been subtracting, and not adding?

The angle between the two sides is either 60 degrees or 120 degrees, depending on how you draw it, but only one way to draw it is right. You can add the two or subtract the two, and only one is right. There are two possible triangles that you wind up having to solve, and only one of them is right. You described one of the two triangles. The other triangle has sides of 120 and 240, and the angle between them is 120 degrees. You are solving for the long side (it's not a "hypotenuse," since this is not a right triangle). You use the Law of Cosines to find the length of that long side: X^2 = (120)^2 + (240)^2 ? 2(120)(240) x cosine (120 degrees). That is how I came up with X = 317.

The problem is setting up the problem. The trigonometry is the easy part. I'm still not certain why one triangle is the correct one to model this situation. Can anyone explain this to me?
 
Re: 120v 277v insame 3ph box

Charlie,
Come on Don, I know how to do that.
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I thought you would know how to do that. The way it was taught to me is you solve it just as it looks when you draw the center tapped delta.
Don
 
Re: 120v 277v insame 3ph box

Don,
I apologize for jumping in, but is this what you are talking about?
;) and I am wondering if what I have drawn is what you are talking about.

If I were able to tap this delta system where ever I want, could I find the voltage between any two points by re-drawing line NB to the tapped points?

[ December 28, 2005, 01:15 PM: Message edited by: jbwhite ]
 
Re: 120v 277v insame 3ph box

Charlie,

There is only one possible triangle.

The system is a 3 phase 240V delta with one leg center tapped. This arrangement can only be drawn as an equilateral triangle where each long leg is 240V and the middle point of one leg is identified.

In any three phase system, if all line-line voltages are equal then the resulting voltage triangle (a-b, b-c, c-a), must be an isosceles one regardless if the voltages are additive or subtractive.
 
Re: 120v 277v insame 3ph box

Thay have came in with 2 power supplies 120/240 and 277/480V. put 277 on A lug 120 on B lug 120 on C lug and only came in with one neutral.
 
Re: 120v 277v insame 3ph box

Originally posted by jim dungar: In any three phase system, if all line-line voltages are equal then the resulting voltage triangle (a-b, b-c, c-a), must be an isosceles. . . .
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They are not only isosceles, but they are also equilateral. Not the point.
Originally posted by jim dungar: This arrangement can only be drawn as an equilateral triangle where each long leg is 240V and the middle point of one leg is identified.
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I'm not talking about drawing a physical model of the transformer connections. I am talking about a phasor diagram of the voltages.
Originally posted by jim dungar: Charlie, there is only one possible triangle.
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No, there are two, and only one is correct for this application. The other is nonsense. But which is which? I did not explain clearly how the two come into existence. What I am looking for is a reason to believe it should be the right triangle, and not the obtuse triangle, as explained below.

Take a look at jbwhite's sketch (good job, by the way, jb). Ignore the arrows, as they only serve as pointers, and do not represent phasors or vectors.
</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">Consider the voltage from N to B. Represent it as an arrow with its origin at N and its tip at B. Give it the name "Vnb."</font>
<font size="2" face="Verdana, Helvetica, sans-serif"></font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">Now consider the voltage from B to C. Represent it as an arrow with its origin at B and its tip at C. Give it the name "Vbc."</font>
<font size="2" face="Verdana, Helvetica, sans-serif"></font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">Give the name Vnc to the voltage from N to C.</font>
<font size="2" face="Verdana, Helvetica, sans-serif">Question:
</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">Is Vnc = Vnb + Vbc?</font>
<font size="2" face="Verdana, Helvetica, sans-serif"> or
</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">Is Vnc = Vnb ? Vbc?</font>
<font size="2" face="Verdana, Helvetica, sans-serif">These two give different triangles. From the first equation, we get a right triangle, with side BC as its hypotenuse. From the second equation, we get an obtuse triangle, with a 120 degree angle between side NB and side "-BC."

Which is the correct representation of this set of voltages, and why? The part of phasor math I was never really good at was applying the correct subscript notation (Vnc versus Vcn). I was a Navy nuke, as you might have read in other posts. The problem with us Navy nukes is that if you give us a 50/50 guess, we will guess wrong 90% of the time. So we are not encouraged to do much guessing.
 
Re: 120v 277v insame 3ph box

Originally posted by jbwhite:. . . I am wondering if what I have drawn is what you are talking about.
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Yes it is.
Originally posted by jbwhite: If I were able to tap this delta system where ever I want, could I find the voltage between any two points by re-drawing line NB to the tapped points?
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You could measure the voltage between any to points in that manner. But you might have a hard time calculating in advance what the voltage should be. It is not a simple matter of drawing a line between two points on the picture and measuring the length of that line.
 
Re: 120v 277v insame 3ph box

Charlie,

My point is to not get confused by starting with a neutral point. The line to line voltages must be connected into an equilateral transformer first (after all it is a delta system), this establishes all of the phase relationships. After the phases are connected it becomes a simple matter to chose the midpoint of one leg and then derive the remaining relationships.

Neutral should never be used as the starting refence point because of the probabilty the center tap voltages will be listed as Van and Vcn when they are really Van and Vnc. this inaccurate method of voltage identification is what often leads people to try to argue the case for "two phase" 120/240.
 
Re: 120v 277v insame 3ph box

Originally posted by jim dungar: After the phases are connected it becomes a simple matter to chose the midpoint of one leg and then derive the remaining relationships.
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I am trying to, in your words, "derive the remaining relationships." Specifically, I am trying to derive the voltage from the high point (Phase C) to ground (or equivalently, to Neutral). What I am trying to get is an easy way to explain to the non-engineer why 240 plus 120 is not 360, and why it is indeed not even as high as 240.

To get the voltage from C to Neutral, you have to go from C to B, and then from B to Neutral. That is 240 plus 120. The sum is not 360 because the 240 and the 120 are not in phase with each other. So if you choose Vnc as being at an angle of zero, is Vcb at +120 or at -60? And why?
 
Re: 120v 277v insame 3ph box

You are trying to explain it too simply and not simply enough.

The answer is that a high leg system is an equilateral triangle cut in half. That results in the 60 degree angle being bisected into two 30 degree angles. So in your example, Vnb = Vbc*sin(30) and Vcn=Vbc*cos(30) or Vbn = Vcb*cos(60) and Vnc=Vcb*sin(60). But the simple way is Vbc^2 = SQRT(Vbn^2 + Vcn^2).

How do you explain 208Y/120?
 
Re: 120v 277v insame 3ph box

terrytc's original question re. 120 and 277 in same panel reminded me of a job I did 10 yrs. ago, a 120/208 3Ph. panel replacement in a commercial parking garage. Garage lighting was 277V from a panel fed by 480/277 4 wire Y from main service.
Panel I was working on was fed from a transformer, 480V 3Ph. Delta primary, 120/208 3Ph. Y secondary. Having opened transformer primary disconnect to work on 208 panel, I was surprised when I got an arc when I lifted a #10 from the neut. bus. That neut. was 277V to ground because it was the neutral of a hot lighting circuit being fed from the 480 panel. A previous "electrician" had used the neutral bus in the 208 panel to repair a lost
277 neutral.
 
Re: 120v 277v insame 3ph box

Terry's question seems to be getting lost in the 3-phase discussion, but a couple of us have caught on.

He's asking if there is a legal way to connect one phase of a 480/277 system into a 3-phase panel, with 208/120 or 240/120 on the other two phases.

The asnwer is: NO!
 
Re: 120v 277v insame 3ph box

Originally posted by LarryFine: Terry's question seems to be getting lost in the 3-phase discussion. . . .
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Spoil sport! :)
Originally posted by jim dungar:How do you explain 208Y/120?
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By recognizing that the voltages are all 120 degrees apart. That is an easy set of phasors to draw.
 
Re: 120v 277v insame 3ph box

Hey, we're only at three pages. That's hardly a thread jack most of the time.

So, Charlie you use phasors to show how simple it is to get Vbn in a wye sytem if all you know is Vcn and Vcb? Or are you really showing how Vcn and Vbn combine to create Vcb?

What phasors do you use to show how a closed delta works? How about an open delta? Use the same phasors just make one of them 1/2 the length.

In a high leg delta Vbn is 90 degrees out of phase with Vnc and Van. This is one reason we need one method to easily explain wye systems and a different method for deltas.
 
Re: 120v 277v insame 3ph box

Originally posted by charlie b:
By recognizing that the voltages are all 120 degrees apart. That is an easy set of phasors to draw.
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Looking at jb's drawing, and thinking of how a "high leg" service is connected at the pole (the 120/240-volt part on one transformer), aren't a-phase and b-phase 180 degrees out from each other, not 120 degrees?
 
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