nisri said:
From your msg, did you mean that the symmetrical component of the line to line voltages cannot be obtained directly using the symmetrical components equation of the LN voltages.
After thinking about it, you can certainly use the transformation on the L-L values. It should have been clear to me before but I just had a brain clog. It will give you the zero sequence value if the L-L values do not form a closed delta. You will also get a number for V1 and V2 but these will be different from the ones you get if you use the L-N values. The value for the L-L will be higher than the L-N transformation by Sqrt(3). They will also have a 30 degree phase shift, one positive, one negative. For example, with your data:
With the L-L transformation, you get:
V0=0.009905033<115.3098335
V1=0.011182217<-82.40535524
V2=0.580811945<114.0455805
With the L-N transformation, you get:
V0=0.009905033<115.3098335
V1=0.006456056<-112.4053552
V2=0.335331933<144.0455805
I guess the best thing to say is that if you have L-N data, use it because it will also give you the zero sequence data. Converting to delta makes you lose the zero sequence info. Getting readings from a closed delta won't show you the zero sequence information. Converting from L-L to L-N may make it easier to see the sequence order. If you build a symmetrical component network, I think you will need the L-N sequences.
nisri said:
We need to transform the L-L voltages to equivalent phases, and then use the equation below (got from your previous message) to get the positive and negative sequance.
equation: Vab=(1-a^2)V1 + (1-a)V2
Vbc= (a^2-a)V1 + (a-a^2)V2
No. These equations are for finding Vab and Vbc once you already have the zero, positive and negative sequence. It is just an equation crunch of these (you will find that the V0 term goes away for the L-L result):
Van=V0+V1+V2
Vbn=V0+a^2*V1+a*V2
Vcn=V0+a*V1+a^2*V2
Vab=Van-Vbn
Vbc=Vbn-Vcn
Vca=Vcn-Van
Th following steps are the method to use when you start with L-L voltages and want to use L-N inputs & results. If you start with L-L values, it is easier to use the sqrt(3) factor and the 30 degree shift if you are sure of your sequence order:
Step #1: use the sum of the L-L values to find V0
Step #2: Find the equivalent L-N values from the L-L values by using (Va=1/3*Vcb+2/3*Vac, Vb=1/3*Vac+2/3*Vba, Vc=1/3*Vba+2/3*Vcb).
Step #3: If the rotation of the equivalent L-N set is CW then change to CCW (Vab=-original Vca, Vbc=-original Vbc, Vca=-original Vab) and go back to #1.
Step #4: Transform the equivalent L-N values to get V1 & V2
nisri said:
We dont have to calculate the zero sequence value since it equal to 0 (is it)?
We don't know if it is zero in the system. With a closed delta, we don't know if there are zero sequence values in the rest of the system because a closed delta will not allow you to see the zero sequence. If we have a set of L-L values that are not closed (a "broken delta") we can sum the L-L values to find the zero sequence (the sum is 3*V0). The L-N sum gives us the same result.
nisri said:
since my data is acb reversed phase, is the equation to get the positive and negative sequence will be the same with the abc phase sequence?
You can use the transformation matrix but you would need to adjust your end results if you want an a-b-c representation (counter-clockwise rotation). This is easy to see when you find the equivalent L-N values.
If it is a-c-b then you realize that Vab is really Vac, Vbc is really Vcb, and Vca is really Vba (swap the "b" and "c"). If we call our original, a-c-b reverse-order L-L vectors Vab', Vbc', and Vca' then we can find the a-b-c rotation vectors by:
Vab = -Vca'
Vbc = -Vbc'
Vca = -Vab'
You could use the transformation matrix on the a-c-b set but you would have to take the end result and add 180 degrees to the zero sequence, and exchange V1 & V2 (I think). I would have to think about the value of transforming set of phasors that are sequenced in a clockwise rotation.
I think this is all correct but I reserve the right to change my mind.
[edit: typo]
