What's the formula for assymetric phaset-to-phase voltage?

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Electric-Light

Electric-Light

Senior Member
Suppose you got a wye tranny wired as:

neutral to:

A = 100v
B= 120v
C = 127v

If N to A,B,C = 120, it is easy because its 120 * sqrt(3) = 208..

120+120 x sin(120deg) = 208


So, would A to C = 100 + 127 x sin(120) =196.7v ?
 
Electric-Light said:
Suppose you got a wye tranny wired as:

neutral to:

A = 100v
B= 120v
C = 127v

If N to A,B,C = 120, it is easy because its 120 * sqrt(3) = 208..

120+120 x sin(120deg) = 208
Click to expand...

sin(120) is 0.866 so your calculation would not give 208V

You'd get 1.866*120 = 224V
 
Law of Cosines:
Assuming 120? displacements (probably not a good assumption with voltages that are way off), the law of cosines yields:
sqrt(100^2 +127^2 -2*100*127*cos(120?)) = 197.05V
 
Electric-Light...

As Mivey pointed out you must determine the relative angles usin the Law of Cosines.

Thus the three Phase-neutral phasors (or vectors) for an A-B-C sequence with Van as the reference are:

Van=100.0<0 deg; Vbn=120.0<-110 deg; Vcn=127.0< 117.5deg.

The corresponding Phase-Phase phasors, using vector-subtraction are:

Vab=Van-Vbn=180.8<38.6 deg; Vbc=Vbn-Vcn=226.0<-86.6 deg; Vca=Vcn-Van=194.6<144.6 deg.

Regards, Phil Corso
 
Phil Corso said:
Electric-Light...

As Mivey pointed out you must determine the relative angles usin the Law of Cosines.
Click to expand...
Actually, you don't.
The cosine formula gives:

cosinerule01.jpg


You have a triangle with two sides, b and c of known length, 100 and 127 in this case and their included angle A, 120deg.
The one unknown is a and a little bit of arithmetic gets you that without having to calculate any of the other angles.
It is as Mivey posted, 197.05V.

Oops, the formula turned out to be sized larger than I expected. It isn't altogether to write equations here so I copied it from one of my word documents into Paint and posted from there.
Simple and I like simple.
 
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Besoker...

Always a PtP to respond if one angle is adequate. Note, I said angles!

Now your smart enough to know that a phasor (or vector) is desribed as having both magnitude and direction!

But, keep up the good work!

Phil
 
Phil Corso said:
Besoker...

Always a PtP to respond if one angle is adequate. Note, I said angles!
Click to expand...
Your comment "you must determine the relative angles" ............
You don't need to calculate any angles to answer the question posed in the opening post.
Both Mivey and myself provided the correct answer without calculating any angles.
Do you want me to provide you a graphical representation? Would that help?
 
Besoeker said:
sin(120) is 0.866 so your calculation would not give 208V

You'd get 1.866*120 = 224V
Click to expand...

I meant (120+120) [sum of two voltages in question] x sin(120).
240 x sin(120) does give me 208.

But this isn't the right way to do it?

In the example I used in first pot, the voltage for each phase is different, because of different coil voltage or due to voltage drop. So assume they're 120deg apart.
 
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I have lost the ability to do trigonometry (if I ever had it) since I learned to do AutoCAD. Here is my graphical solution. If there is no neutral shift, the voltages make a triangle;

Drawing1 Rename_This_Layout (1).jpg
 
The voltages vectors are;

A-N = 100<0?
B-N = 127<110?
C-N = 127<242?

B-A = 181<141?
C-B = 226<266?
A-C = 195<325?

The theoretical voltage vectors would be;

A-N = 120<0?
B-N = 120<120?
C-N = 120<240?

B-A = 208<150?
C-B = 208<270?
A-C = 208<30?

This shows that the phases have unequal loads and power factors.
 
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Besoeker said:
Actually, you don't.
The cosine formula gives:

cosinerule01.jpg


You have a triangle with two sides, b and c of known length, 100 and 127 in this case and their included angle A, 120deg.
The one unknown is a and a little bit of arithmetic gets you that without having to calculate any of the other angles.
It is as Mivey posted, 197.05V.

Oops, the formula turned out to be sized larger than I expected. It isn't altogether to write equations here so I copied it from one of my word documents into Paint and posted from there.
Simple and I like simple.
Click to expand...

I don't think you can assume the anble between b and c is 120?. The triangle is defined as having three sides; 100, 120 and 127.

Drawing1 Rename_This_Layout (2) .jpg
 
JoeStillman said:
I don't think you can assume the anble between b and c is 120?. The triangle is defined as having three sides; 100, 120 and 127.
Click to expand...
It's wye connected. It is thus likely that the open circuit voltages are equal and mutually displaced by 120deg being equal secondary windings of a D/Y or Y/Y transformer. On load, the secondary voltage vectors don't have to form a closed triangle in the way that you have shown.
 
JoeStillman said:
I don't think you can assume the anble between b and c is 120?.
Click to expand...

Sure you can. Take three single phase transformers with turns ratios of 4.8:1, 4:1 and 3.8:1. Connect the transformer in a delta-wye (Dy1) configuration to a primary where the voltages are Vab=480<60, Vbc=480<-60 and Vca=480<180. On the wye side you will get Van=100<0, Vbn=120<-120 and Vcn=127<120. That gives you Vab=190.8<33.0, Vbc=213.9<-89.1 and Vca=197.1<146.1

This is what the OP was suggesting.

Electric-Light said:
In the example I used in first pot, the voltage for each phase is different, because of different coil voltage or due to voltage drop. So assume they're 120deg apart.
Click to expand...
 
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