Using Symmetrical Componets to calculate an unbalanced load 3

[mcspipe]

I understand the concept of calculating unblanaced faults using symmetrical components, however I'm having a hard time calculating currents/volatges of a three-phase unbalanced load with no faults.

A seemingly simple example: an ungrounded 3-phase Y voltage source connected to a Y impedance bank where the A-phase has a capacitor and the B and C phases have resistors. The neutrals of the source and the load are NOT connected. I'm fairly sure the zero sequence currents are zero (no neutrals or grounds), but I don't know how to calculate the positive and negative sequence impendances for each phase.

Help ???

 

[jghrist]

The Westinghouse T&D Reference Book has sequence network connections for unbalanced loads. Chapter 2, Fig. 21 in the 1984 Edition. Fig. 21(l) seems to fit your case, with Z_a connected from ØA, and Z_b connected from ØB and ØC in wye, with no neutral connection. It's a little difficult to describe the connections without a diagram. Try this:
[tt]
+-----------+-------------------------+
| | |
Zb Zb |
| | |
+---------+ +---------+ +---------+ |
| oF1 | | oF2 | | oFo | |
|P | |N | |Z oNo | (1/3)*(Za-Zb)
| oN1 | | oN2 | | oGo | |
+---------+ +---------+ +---------+ |
| | |
+-----------+-------------------------+
[/tt]

F1 and F2 are the points in the sequence networks where the load is connected. There is no zero-sequence current because the zero-sequence network is not connected.

I have found that it is just as easy to analyze load situations without sequence networks (and a lot easier to understand). Draw the circuit diagrams with the three phase voltage sources (with proper phase angles) and use classical network analysis (mesh equations or node equations) to calculate the currents and voltages. If you need sequence values, calculate them after calculating the phase values.

 

[nbucska]

The B and C with resistors can be converted with one
x phase with a calculated amplitude, phase angle and'
series resistance. Now add the C to phase A.


<nbucska@pcperipherals DOT com> subj: eng-tips
read FAQ240-1032

 

[ijl]

Basically, the method of symmetrical components deals only with unsymmetrical currents and voltages, not with unsymmetrical networks. The network itself _must_ be symmetrical: identical components in all phases. Thus, the problem of the unsymmetrical load cannot directly be solved using the symmetrical components. But a trick will help. This trick demonstrates (beautifully, I think the basic idea of the symmetrical components.

The load in the example mcspipe mentions has a capacitor in phase a and a resistor in phases b and c, thus, the load is clearly unsymmetrical. A symmetrical load would have a capacitor and a resistor in series in each phase. The unnecessary components (the resistor in phase a and the capacitors in phases b and c) can be eliminated by the mentioned trick.

To eliminate a component, inject a current in one end of the component, and drain the same current from the other end of the component. Adjust the current in such a way, that the voltage over the component becomes zero, and (voila’) the component has been eliminated!

The method of symmetrical components can be used to solve the injection currents that are needed to eliminate the unnecessary components. For simplicity, the elimination of only one component, the capacitor is described here.

First, calculate the nominal voltage over the capacitor by solving the original, symmetrical network in the usual way (by using the positive sequence equations!). Let us call this voltage Uo. Next, set the voltage of all sources to zero, inject and drain a unit current to and from the ends of the capacitor, and solve the network in order to get the voltage over the capacitor. The method of symmetrical components is needed here.

That is, inject a current of 1A to phase a, current 0A to phase b, and a current 0A to phase c at one end of the capacitor. Similarly, inject a current –1A , 0A, 0A to the phases a,b,c, respectively, at the other end of the capacitor. Calculate the positive, negative and zero sequence components of these injection currents (which are [1,0,0]’ and [-1,0,0]’ in vector form) by multiplying them with the transformation matrix. Construct the positive, negative and zero sequence networks of the whole network (with zero voltages at all sources). Note, you will need also the zero sequence network, because the injected currents have a zero sequence component. (Often, the zero sequence network is very simple).

Inject the positive sequence currents to the ends of the capacitor in the positive sequence network and calculate the positive sequence current through the capacitor by solving the positive sequence network. Do the same for the negative and zero sequence. Transform the sequence currents through the capacitor to phase currents by multiplying them with the inverse transformation matrix. Calculate the voltage over the capacitor by multiplying the current and impedance of the capacitor. Let us call this voltage Ui. Because the injected current was equal to 1 (Ampere), and the network is liner (an assumption) the voltage over the capacitor for an arbitrary injection current Ii is equal to U = Uo + Ui*Ii. This voltage must be zero. Thus, the current to be injected (and drained) is Ii = -Uo/Ui.

If there are several components to be eliminated, several injection currents must be solved simultaneously. The method is the same as described above, but the currents must be solved from a system of linear equations.

So what is the benefit of using the symmetrical components? The currents can be calculated by solving four single phase (or actually single sequence) equations, when the symmetrical components are used. Without them, the currents must be calculated by solving the currents in all three phases simultaneously. In this example, the benefit of the symmetrical components is small (if any), but in a larger network the difference in computational work may much larger. In addition, if there are generators in the network, the method of symmetrical components should actually be used (not necessarily in the form described here, see the answer of jghrist). The reason is that the negative and zero sequence impedances of a generator are much different from the positive sequence impedance.

This was a lengthy explanation, but hopefully it helps?

 

[cuky2000]

The enclose solution intend to calculate the symmetrical sequence current of the unbalanced system.

I hope this help to clarify the questions.

http://cuky2000.250free.com/Symmetrical_Component.pdf

 

[jghrist]

cuky2000,

I think you have some misnomenclatures in your diagrams. If you letter your phases a-b-c clockwise with a at the top, then the a source voltage would be V_an, the lower-right would be V_bn, and the lower-left would be V_ac. Only two loop currents would be needed, I_ab and I_bc, drawn as you have them.

The Z11 impedance in your matrix would be Z_ao-Z_co. The first voltage of your voltage vector would be V_an-V_bn, or -V_ab. The second voltage of your voltage vector would be V_bn-V_cn, or -V_bc.

With these changes, using V_an=277V, Z_ao= -j·0.767 ohm (100 kvar), and Z_bo=Z_co=0.767 ohm (100W), I calculate:

I₀ = 0A
I₁ = (288.8 + j216.6)A
I₂ = (-72.2 + j216.6)A

Using the sequence network connection in my previous post, I calculate the exact same values.

 

[cuky2000]

Jghrist, thanks for the correction of the nomenclature and the verification of your calc.

The good news is that the zero sequence current is zero as expected.

Even thought this is an academic exercise, I guess that this prove that the symmetrical component techniques could be used for unbalance load.