I have another doubt on the symmetrical component for line-to-line voltage again...As being stated in a book that i've found in the library, the value of the zero sequence voltage for line to line voltage should be zero. My question is, what if we get a nonzero value for the zero sequence voltage for this line to line volatge? Do we need to make it zero? How can it be done?
zero sequence voltage for line to line voltage
- Thread starter nisri
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mivey
Senior Member
Are you talking about a line-line fault?
[edit: see if this thread helps you http://forums.mikeholt.com/showthread.php?t=89943&highlight=symmetrical]
[edit: see if this thread helps you http://forums.mikeholt.com/showthread.php?t=89943&highlight=symmetrical]
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Not the line to line fault. It is the line to line volatages (Vab, Vbc, Vca). The symmetrical component for the line-to-line voltages (Vab, Vbc, Vca) can be calculated using the same formulas for the star connected phase voltages (Va, Vb, Vc). But in the book, it is mentioned that the zero sequence of the line to line voltages value should be 0 since Vab + Vbc + Vca =0.
I'm developing a coding to calculate the symmetrical component of this line-to-line voltages (Vab, Vbc, Vca) using the standard equation for the symmetrical component. After executing the code, it give a nonzero value for the zero sequence voltage of the line-to-line voltages. I've tried to figure out what makes the result be like that, but it ends up with a headache :grin: .
So, is there anyone out there know why the zero sequence voltages for the line to line voltages is not equal to zero, and how to make it become zero?
FYI: The data that I used is a signal with a disturbance (contain sag and swell ).
I'm developing a coding to calculate the symmetrical component of this line-to-line voltages (Vab, Vbc, Vca) using the standard equation for the symmetrical component. After executing the code, it give a nonzero value for the zero sequence voltage of the line-to-line voltages. I've tried to figure out what makes the result be like that, but it ends up with a headache :grin: .
So, is there anyone out there know why the zero sequence voltages for the line to line voltages is not equal to zero, and how to make it become zero?
FYI: The data that I used is a signal with a disturbance (contain sag and swell ).
Perhaps for an ideal power supply. However, in real world power supplies this is seldom true.nisri said:
For real world values, I don't believe you should zero the result, provided the calculation is accurate.
Are you using simulated data from an ideal and stiff source, or real world data from an otherwise imperfect source?
I'm developing a software that can calculate the symmetrical component of the three phase voltage data obtained from IEEE1159.2 webpage.Its a real data obtained from the distribution line somewhere in this world (I'm not sure where :grin: ). The data are the three phase voltage (Va, Vb , Vc or Vab, Vbc, Vca) that contains power quality disturbances, such as sag, swell, transient,etc. I'm using Matlab as the software to develope my system. I need to obtain the symmetrical component of the three phase voltages for sag characterization.
The formulas that I used to get the symmetrical value of the line to line voltages(Vab, Vbc, Vca) are as follow:
[Vab0; Vab1; Vab2]=(1/3)*[1 1 1; 1 a a^2; 1 a^2 a]*[Vab; Vbc; Vca]
(the semicolon indicate that the matrix is in the next row).
From the formulas, the equation for the zero sequence voltage is:
Vab0 = (1/3)*(Vab + Vbc + Vca);
The negative and positive sequence voltages are given as:
Vab1 = (1/3)*(Vab + a*Vbc + a^2*Vca);
Vab2 = (1/3)*(Vab + a^2*Vbc + a*Vca);
The book state that For the line-to-line voltages, the zero sequence should equal to zero. But I got a nonzero value for the zero sequence voltage. Maybe Smart are right. Maybe in the real world the value is not equal to zero, but at least the value should be almost zero. The value that I get for the zero sequence component is quite big. Bigger than the value of the positive sequence. So I thought, maybe I'm using a wrong formulas, but it seem very impossible as this equations is stated in a book. Or it is true that actually this formulas is not the right formulas that can be used to determine the symmetrical component of the line to line voltages? Or actually the data is imperfect? hmm....:-?
The formulas that I used to get the symmetrical value of the line to line voltages(Vab, Vbc, Vca) are as follow:
[Vab0; Vab1; Vab2]=(1/3)*[1 1 1; 1 a a^2; 1 a^2 a]*[Vab; Vbc; Vca]
(the semicolon indicate that the matrix is in the next row).
From the formulas, the equation for the zero sequence voltage is:
Vab0 = (1/3)*(Vab + Vbc + Vca);
The negative and positive sequence voltages are given as:
Vab1 = (1/3)*(Vab + a*Vbc + a^2*Vca);
Vab2 = (1/3)*(Vab + a^2*Vbc + a*Vca);
The book state that For the line-to-line voltages, the zero sequence should equal to zero. But I got a nonzero value for the zero sequence voltage. Maybe Smart are right. Maybe in the real world the value is not equal to zero, but at least the value should be almost zero. The value that I get for the zero sequence component is quite big. Bigger than the value of the positive sequence. So I thought, maybe I'm using a wrong formulas, but it seem very impossible as this equations is stated in a book. Or it is true that actually this formulas is not the right formulas that can be used to determine the symmetrical component of the line to line voltages? Or actually the data is imperfect? hmm....:-?
There must be a measurement error, time difference between phase measurements, or some filtering of the data after measurements. If you measure the voltage between point A and point B (VAB), between point B and point C (VBC), and between point C and point A (VCA), the sum of these (VAB+VBC+VCA) is going to be zero if the measurements are accurate and made at the same time.
If you have a sag or a swell, your voltage won't be a pure sine wave. It will have some harmonic distortion. Some of the harmonics will be in phase for all three phases. So the zero sequence voltage isn't going to be zero.
The zero sequence voltage will only be zero for 3 balanced, steady sine waves.
Steve
The zero sequence voltage will only be zero for 3 balanced, steady sine waves.
Steve
mivey
Senior Member
I'm not sure about this L-L sequence transformation. I guess you could derive it but I just don't remember seeing like that. I only remember the transformation with the L-N values like this:
VA = (V0 + V1 + V2);
VB = (V0 + a^2*V1 + a*V2);
VC = (V0 + a*V1 + a^2*V2);
which leads to:
V0 = (1/3)*(VA + VB + VC);
V1 = (1/3)*(VA + a*VB + a^2*VC);
V2 = (1/3)*(VA + a^2*VB + a*VC);
if you take VA,VB,VC and derive VAB,VBC,VCA you would get (insert standard disclaimer here for stupid math errors):
VAB = (1-a^2)*V1 + (1-a)*V2;
VBC = (a^2-a)*V1 + (a-a^2)*V2;
VCA = (a-1)*V1 + (a^2-1)*V2);
and there are no zero sequence voltages because it is a set of delta voltages.
VA = (V0 + V1 + V2);
VB = (V0 + a^2*V1 + a*V2);
VC = (V0 + a*V1 + a^2*V2);
which leads to:
V0 = (1/3)*(VA + VB + VC);
V1 = (1/3)*(VA + a*VB + a^2*VC);
V2 = (1/3)*(VA + a^2*VB + a*VC);
if you take VA,VB,VC and derive VAB,VBC,VCA you would get (insert standard disclaimer here for stupid math errors):
VAB = (1-a^2)*V1 + (1-a)*V2;
VBC = (a^2-a)*V1 + (a-a^2)*V2;
VCA = (a-1)*V1 + (a^2-1)*V2);
and there are no zero sequence voltages because it is a set of delta voltages.
mivey
Senior Member
Consider these examples:
If you transform a 120<0, 120<-120, 120<120 set of wye voltages, you get:
V0=0, V1=120<0, V2=0
If you take the delta voltages for these wye values and transform them you get:
VAB=208<30,VBC=208<-90,VCA=208<150
with a transform of
V0=0, V1=208<30, V2=0
Now if you transform an unbalanced 120<0, 125<120, 120<240 set of wye voltages, you get:
V0=1.7<-120, V1=121.7<0, V2=1.7<120
If you take the delta voltages for these wye values and transform them you get:
VAB=212.2<30.7,VBC=212.2<-90.7,VCA=208<150
with a transform of
V0=0, V1=210.7<30, V2=2.9<90
And for one more: transform these 120<0, 120<125, 120<240 set of wye voltages, you get:
V0=3.5<147.5, V1=120<0, V2=3.5<27.5
If you take the delta voltages for these wye values and transform them you get:
VAB=212.9<2.75,VBC=202.4<-92.5.7,VCA=208<150
with a transform of
V0=0, V1=208<28.3, V2=6<-2.5
And now it's supper time. Playtime is over for now.
If you transform a 120<0, 120<-120, 120<120 set of wye voltages, you get:
V0=0, V1=120<0, V2=0
If you take the delta voltages for these wye values and transform them you get:
VAB=208<30,VBC=208<-90,VCA=208<150
with a transform of
V0=0, V1=208<30, V2=0
Now if you transform an unbalanced 120<0, 125<120, 120<240 set of wye voltages, you get:
V0=1.7<-120, V1=121.7<0, V2=1.7<120
If you take the delta voltages for these wye values and transform them you get:
VAB=212.2<30.7,VBC=212.2<-90.7,VCA=208<150
with a transform of
V0=0, V1=210.7<30, V2=2.9<90
And for one more: transform these 120<0, 120<125, 120<240 set of wye voltages, you get:
V0=3.5<147.5, V1=120<0, V2=3.5<27.5
If you take the delta voltages for these wye values and transform them you get:
VAB=212.9<2.75,VBC=202.4<-92.5.7,VCA=208<150
with a transform of
V0=0, V1=208<28.3, V2=6<-2.5
And now it's supper time. Playtime is over for now.
mivey
Senior Member
Supper's over and the dishes washed, nothing left but a piece of squash.
Now let's pick some wacky delta with 120,140, & 250 volt legs that looks like:
c.\
....\..\
......\....\
........\.....\
..........\.......\
..........b.______ a
[edit: this was supposed to look like a triangle but the spaces were being stripped out
]
If VB'=0, VA'=120<0, VC'=140<130.76, then VN'=250<17.254
If we use VN' as the reference point we get
VA=140<-148, VB=250<-162.75, VC=331.7<174.48 which transforms to
V0=232.48<-170.39, V1=86.29<5.8, V2=50.48<-60.8
If you take the delta voltages for these values and transform them you get:
VAB=120<0, VBC=140<-49.24,VCA=236.51<153.36
with a transform of
V0=0, V1=149.45<35.8, V2=87.43<-90.8
So you can see, even with a wacky delta, as long as the delta is closed, we are getting no zero sequence voltage.
Now let's take our nice delta from before of
VAB=207.85<30,VBC=207.85<-90,VCA=207.85<150
which transformed to:
V0=0, V1=207.85<30, V2=0
and open the delta like this:
VAB=210<30,VBC=207.85<-90,VCA=207.85<150
gives us a transform of:
V0=120.62<0.17, V1=120<-0.34, V2=120.62<120.17
The reverse is also true. If we take our sequence vectors of:
V0=0, V1=207.85<30, V2=0
and insert a zero sequence like this:
V0=1<1, V1=207.85<30, V2=0
We get:
VAB=208.72<29.87,VBC=207.83<-89.724,VCA=206.99<149.86
which does not close, showing an additional path and yields a net of
VAB+VBC+VCA=3<1
So, if you are getting a zero sequence, the delta voltages you are using can't be summed to zero. The legs can be different, they just must sum to zero in order to not have a zero sequence using the L-L transformation.
You can always draw the line-to-line voltages from the wye and the delta will close because the legs of the wye, by definition, will have a common reference point.
BONUS INFORMATION FOR THE 120/240 DELTA NEUTRAL:
For those that are paying attention, think about this common reference point on the delta and you will see why the center tap ground on the lighting pot of a 120/240 delta bank CAN NOT BE THE NEUTRAL POINT for the 3-phase delta system, but CAN ONLY BE the neutral point for the single phase portion of the delta system.
Now let's pick some wacky delta with 120,140, & 250 volt legs that looks like:
c.\
....\..\
......\....\
........\.....\
..........\.......\
..........b.______ a
[edit: this was supposed to look like a triangle but the spaces were being stripped out
]If VB'=0, VA'=120<0, VC'=140<130.76, then VN'=250<17.254
If we use VN' as the reference point we get
VA=140<-148, VB=250<-162.75, VC=331.7<174.48 which transforms to
V0=232.48<-170.39, V1=86.29<5.8, V2=50.48<-60.8
If you take the delta voltages for these values and transform them you get:
VAB=120<0, VBC=140<-49.24,VCA=236.51<153.36
with a transform of
V0=0, V1=149.45<35.8, V2=87.43<-90.8
So you can see, even with a wacky delta, as long as the delta is closed, we are getting no zero sequence voltage.
Now let's take our nice delta from before of
VAB=207.85<30,VBC=207.85<-90,VCA=207.85<150
which transformed to:
V0=0, V1=207.85<30, V2=0
and open the delta like this:
VAB=210<30,VBC=207.85<-90,VCA=207.85<150
gives us a transform of:
V0=120.62<0.17, V1=120<-0.34, V2=120.62<120.17
The reverse is also true. If we take our sequence vectors of:
V0=0, V1=207.85<30, V2=0
and insert a zero sequence like this:
V0=1<1, V1=207.85<30, V2=0
We get:
VAB=208.72<29.87,VBC=207.83<-89.724,VCA=206.99<149.86
which does not close, showing an additional path and yields a net of
VAB+VBC+VCA=3<1
So, if you are getting a zero sequence, the delta voltages you are using can't be summed to zero. The legs can be different, they just must sum to zero in order to not have a zero sequence using the L-L transformation.
You can always draw the line-to-line voltages from the wye and the delta will close because the legs of the wye, by definition, will have a common reference point.
BONUS INFORMATION FOR THE 120/240 DELTA NEUTRAL:
For those that are paying attention, think about this common reference point on the delta and you will see why the center tap ground on the lighting pot of a 120/240 delta bank CAN NOT BE THE NEUTRAL POINT for the 3-phase delta system, but CAN ONLY BE the neutral point for the single phase portion of the delta system.
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jghrist said:
Yes, but one post mentioned he was similating a sag or a swell.
I reacall very little of sequence voltages and currents. However, I suspect all the equations mivey just posted are only valid for pure sine wave inputs at the same frequency. As strange as it sounds, a sine wave that changes in amplitude is not a sine wave.
I think the reason the OP is getting a nonzero zero sequence voltage might be because of the sag or swell he is simulating.
mivey
Senior Member
I'm not so sure about the amplitude thing. Maybe it is a transient thing. As for the frequency, I can believe that.steve66 said:
As for amplitude, consider how the current waveform can change in amplitude based on the load. We are only looking at the instantaneous values.
To the best of my recollection, the symmetrical components were only good for waves with the same frequency, but I may be wrong. I do remember seeing a paper on harmonics and symmetrical components but I'll have to see if I can find it.
I don't recall a lot of this stuff either but can always dig through Wagner & Evans if I'm feeling extra energetic
Probably right. That's why I asked for the input he was using so we could see if we could tell what was going on.steve66 said:
wirenut1980
Senior Member
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With a delta source, there should not be any zero sequence path. It is often shown modeled as an "open," which mivey looks to have proven (made my head spin!). So if you are getting a large zero sequence voltage, I would suspect the model is a wye source, and if you have anything other than a 3 phase fault, there will be voltage imbalance and plenty of zero sequence current over the zero sequence impedance.
Or there is a calculation error somewhere in the coding.
Or there is a calculation error somewhere in the coding.
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wirenut1980 said:So if you are getting a large zero sequence voltage, I would suspect the model is a wye source, and if you have anything other than a 3 phase fault, there will be voltage imbalance and plenty of zero sequence current over the zero sequence impedance.
Or there is a calculation error somewhere in the coding.
I agree. While there are math equations that let you convert a wye system into an equivalent delta, only a physical delta will have no zero sequence voltage. Computers let us do all sorts of things, Mother Nature does not. We always need to confirm that our models simulate the actual world.
It doesn't matter whether the source is sinusoidal, wye, delta, or anything else. For example:
Van = 345?sin(2?pi?f+20?)+34
Vbn = 22?cos(24?pi?f)
Vcn = 57 (dc)
Vab = Van-Vbn = 345?sin(2?pi?f+20?)+34-22?cos(24?pi?f)
Vbc = Vbn-Vcn = 22?cos(24?pi?f)-57
Vca = Vcn-Van = 57-345?sin(2?pi?f+20?)-34
Vab + Vbc + Vca = Van-Vbn+Vbn-Vcn+Vcn-Van = 0
= 345?sin(2?pi?f+20?)+34-22?cos(24?pi?f)+22?cos(24?pi?f)-57+57-345?sin(2?pi?f+20?)-34 = 0
V0 = (Vab + Vbc + Vca)/3 = 0
Van = 345?sin(2?pi?f+20?)+34
Vbn = 22?cos(24?pi?f)
Vcn = 57 (dc)
Vab = Van-Vbn = 345?sin(2?pi?f+20?)+34-22?cos(24?pi?f)
Vbc = Vbn-Vcn = 22?cos(24?pi?f)-57
Vca = Vcn-Van = 57-345?sin(2?pi?f+20?)-34
Vab + Vbc + Vca = Van-Vbn+Vbn-Vcn+Vcn-Van = 0
= 345?sin(2?pi?f+20?)+34-22?cos(24?pi?f)+22?cos(24?pi?f)-57+57-345?sin(2?pi?f+20?)-34 = 0
V0 = (Vab + Vbc + Vca)/3 = 0
jghrist said:
So you're saying the zero sequence voltage is always zero, no matter what?
If so, then what would be the point in even having something called a "zero sequence voltage". It would just be another name for "zero".
Again, I remember almost nothing of sequence voltages, but I still wonder if the equations you are using aren't limited to sine waves that are all the same frequency.
At the very least, I'm betting those equations can't be used for waveforms which are changing in frequency.
Steve
mivey
Senior Member
In the physical world it matters. As far as drawing a closed triangle between 3 points, as I showed in several examples, you can call this a "closed delta" and the transformation will have to show no zero sequence voltage. The problem is, like Jim stated earlier, just because you can draw this "closed delta" on paper doesn't mean it is represented by an actual delta in the physical world.jghrist said:
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