Ignore that request... I discovered an error or three :roll:
Calculating Phase Currents for Unbalanced Loads
- Thread starter mea03wjb
- Start date
- Status
New formulas. Verified myself on one example (CAD, graphical). Could use additional verification from others
It works with your numerical example in Posts 53-56. How was it derived?New formulas. Verified myself on one example (CAD, graphical). Could use additional verification from others
Using a 100 samples/cycle cosine filter and using the va angle as a reference, I get the line currents to be:
ILa = 21.84 @ -39.66?
ILb = 23.15 @ -157.32?
ILc = 23.19 @ 78.73?
Using sequence components and assuming no circulating load current (no zero-sequence), I get the delta load currents to be:
IDab = 12.83 @ -7.48?
IDbc = 13.64 @ -129.26?
IDca = 12.89 @ 108.5?
You have to have at least 1-1/4 cycles of data to use a cosine filter, so I duplicated the values to get 2 identical cycles.
Using the equations for 4 samples per cycle on the 75th and 100th samples, I get slightly different values of line current (again referenced to Va angle):
ILa = 21.62 @ -40.08?
ILb = 23.15 @ -159.25?
ILc = 23.22 @ 78.71?
Derived geometrically...
Given three delta currents of any magnitude and angle, the only requirement (KCL) is the currents add to zero.
The Va vector is not necessary. It is only there for visual referrence.
Now that we have the three delta current vectors, we arrange them in pairs, tail to head (reversing direction as needed) to add them to get line current vectors.
In doing the above, I noticed there is a gemoetric equivalent vector across the heads of the base delta vectors. Note the parallelograms.
So we can actually draw it up as such...
Here I accidentally closed my graphics program and lost the drawing, so I'll try to expain the rest without visual aid
In the last image, if we extended the delta current vectors' tails, they fall on the midpoints of the line current vectors. Through analysis it can be determined the required extension is equal to 1/2 the delta leg's current magnitude. This creates multiple triangles with common sides and angles with three known magnitudes. This is enough info to determine the unknowns using the law of cosines to solve for each delta leg's current magnitude.
W, it can be used for instantaneous values but there are some considerations. One, the accuracy is dependent in part on the three values adding to zero. I noticed your sample data consistently does not, but resonably close to a zero total. Two, the result is always positive. A plot of the result will appear as if an ABS function was applied to sine waveform data (plotted appears similar to rectified sinewave current). Third, for some imbalanced data near zero crossings, the value under the square root function will be negative, yielding an indeterminate result.
Your solution uses the magnitude of the current vectors. Instantaneous values are not the magnitudes of the vectors, but are the values of points along the sinusoid. Using your Post #53 example instantaneous values in your equation would give:
IAB = 4.413, IBC = 1.085, ICA = 5.498
instead of the actual
IAB = 4.267, IBC = 5.511, ICA = 4.648
See my Post #44.
See http://en.wikipedia.org/wiki/Symmetrical_components for methods of computing the sequence components.
Instantaneous values are magnitudes... and they have an angle too. Consider the difference between phasors and vectors... the former being dynamic versus the latter being static.
That said, it amounts to instantaneous input to the formulas yields instantaneous output... rms input yields rms output, and magnitude only in both cases.
Here's some waveforms of the instantaneous. In this first one, the top chart is of the values as calculated using ideal input (pure sinusoids, equal magnitudes, equal power factors). In the lower chart I added a polarity factor to the equation: Ia-Ib / |Ia-Ib| for Iab polarity, and shifted appropriately for the other two.
Below is what happens when I slightly imbalance the magnitudes and power factors...
...and if I use W's data, first let's look at the line current plot...
...and here's the plot of delta currents... and remember, the line current values do not add up to zero...
The magnitudes all have to be divided by sqrt(2) to get rms values.
I have implemented the equations (supplied by Smart) in MATLAB - they are attached since I don't know how to put them in the text
The code I used is given below:
------------------------------
% WyeDelta.m
% Transforms the measured line currents into
% phase currents for delta connected load
function [ia ib ic] = WyeDelta(i1,i2,i3)
ia = abs(sqrt( 2*(i1).^2 + 2*(i2).^2 - (i3).^2) / 3);
ib = abs(sqrt( 2*(i2).^2 + 2*(i3).^2 - (i1).^2) / 3);
ic = abs(sqrt( 2*(i3).^2 + 2*(i1).^2 - (i2).^2) / 3);
% Calculate the polarity
ia_pol = (i1-i2)./abs(i1 - i2);
ib_pol = (i2-i3)./abs(i2 - i3);
ic_pol = (i3-i1)./abs(i3 - i1);
% Correct the polarity
ia = ia.*ia_pol;
ib = ib.*ib_pol;
ic = ic.*ic_pol;
end
---------------------
I added the 'abs' function to remove the imaginary components introduced due to the -ve square roots.
They seem to be fairly good except at the point of polarity change.
Thanks for posting your plots - it helped with my understanding.
I have another question: I want to analyse the negative sequence current in the delta load, at the moment I am applying the symmetrical component transform to the instantaneous data by applying a phase shifts in MATLAB. Would it be okay just to use a set of calculated phasors then apply the symmetrical components transform to the phasors?
The code I used is given below:
------------------------------
% WyeDelta.m
% Transforms the measured line currents into
% phase currents for delta connected load
function [ia ib ic] = WyeDelta(i1,i2,i3)
ia = abs(sqrt( 2*(i1).^2 + 2*(i2).^2 - (i3).^2) / 3);
ib = abs(sqrt( 2*(i2).^2 + 2*(i3).^2 - (i1).^2) / 3);
ic = abs(sqrt( 2*(i3).^2 + 2*(i1).^2 - (i2).^2) / 3);
% Calculate the polarity
ia_pol = (i1-i2)./abs(i1 - i2);
ib_pol = (i2-i3)./abs(i2 - i3);
ic_pol = (i3-i1)./abs(i3 - i1);
% Correct the polarity
ia = ia.*ia_pol;
ib = ib.*ib_pol;
ic = ic.*ic_pol;
end
---------------------
I added the 'abs' function to remove the imaginary components introduced due to the -ve square roots.
They seem to be fairly good except at the point of polarity change.
Thanks for posting your plots - it helped with my understanding.
I have another question: I want to analyse the negative sequence current in the delta load, at the moment I am applying the symmetrical component transform to the instantaneous data by applying a phase shifts in MATLAB. Would it be okay just to use a set of calculated phasors then apply the symmetrical components transform to the phasors?
Hmm... in the grand picture of conventional electrical math, that is of course correct... but you're missing the context of my statement.
You have instant values of Ia, Ib, and Ic. They occur at a point in the time of a cycle, which has an angle associated with it. Now even with that true, let's set it aside for a moment...
We have three lines feeding a 3? delta load. In the same manner in which the loads are connected, the instant values are connected. Though they are just values, they must sum to zero.
In graphical terms, that means a line connected to another line at one of their endpoints and their other endpoints are bridged by connecting to a third line's endpoints. That forms a triangle. As such they form angles relative to each other, but they are angles nonetheless. And because this is planar geometry, the lines also have direction relative to each other. If one continues to pursue the issue, I'm fairly certain correltaion to other angles and directions regarding the matter are possible.
So when it is all said and done, you can stick with the conventional sense... or not. You can be right and wrong either way... or from my viewpoint, totally right
...which brings to mind something I was once told and recall when pertinent: The difference between right and wrong is merely what we think...
Food for thought, what is the difference between instant values and instantaneous values...???
If everyone was conventional, there'd be no discoveries. I am knowledgeable about conventional ways, but I do not subscribe to those ways incessantly... and if a person thinks I don't know what I'm talking about, it is not I they should be analyzing. I seldom talk about anything which I do not know, at least to some degree, and seldom stretch that degree beyond my belief that I know what I'm talking about... and that is to say I may not know what I'm talking about but only to the degree I believe I know what I'm talking about.
With that said, why would I care whether someone I don't know or care about thinks I don't know what I'm talking about?
Do you know what I'm talking about?
So, are you going to back that statement up with the not roundabout way?
How about by both, math and reference... Note the equation at the top of the image below, then click here and refer to equation (4).
I can see from your reference and math that your equations for the delta load currents are correct if the neutral point of the delta currents is at the centroid of the triangle formed by the line currents.
Is there any proof that the neutral point is at the centroid? It is in at least one example, but is it always true?
Is there any proof that the neutral point is at the centroid? It is in at least one example, but is it always true?
Last edited:
- Location
- Connecticut
- Occupation
- Engineer
How about by both, math and reference... Note the equation at the top of the image below, then click here and refer to equation (4).
Smart$, this was really well done, but I don't think that it works. In post 66 you mention that the accuracy depends on the "values adding to zero." While this should be true for the line currents Ia+Ib+Ic=0, using the triangle centroid also means that is necessarily true for the phase currents too Iab+Ibc+Ica=0. I don't believe that this should be the case.
For instance, imagine a 208V, 3ph system with phase voltages of Vab=208<-60, Vbc=208<180 and Vca=208<60. If you connect a 2080W (resistive) load between A-B and a 1560W (resistive) load between B-C, and connect no load between C-A, then you would have phase currents of Iab=10<-60 and Ibc=7.5<180 and Ica=0
We know that Ia=Iab-Ica (etc) so, Ia=10<0, Ib=15.2<145.3 and Ic=7.5<0. If you add Ia+Ib+Ic, you will find that the sum is zero. However, if you add Iab+Ibc+Ica, you will find that the sum is 9<-106.1
If you took the line vectors and graphed them, and then used the centroid lines for the phase currents, you'd have a value for Ica which was not zero. But in the example, there is no load between C-A, so Ica must be zero. I don't think the triangle centroid method will provide the correct phase currents in all cases.
- Status