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Notes 5 Symmetrical Components

By Jill Evans,2014-05-28 14:12
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Notes 5 Symmetrical Components

    Symmetrical Components 2

    Sequence impedances

    Although the following focuses on loads, the results apply equally well to lines, or lines and loads. Read these notes together with sections 12.6 and 12.9 of text.

    Consider the Y-connected balanced load, Fig. 1. It is grounded through an impedance Z. Because of this, the neutral point may n

    not be at the same potential as the ground.

     Ia

    Vag

    I c

    ZY ZY

    VI cg n

    Zn ZY I b

    Vbg

    Fig. 1

     1

Let’s use KVL to write the voltage

    equations for the three phase-to-ground voltages V, V, and V as a function of agbgcg

    the line currents and the load impedances. Assume no mutual coupling between phases.

    VZIZIagYann

    VZIZIbgYbnn (1)

    VZIZIcgYcnn

    We can replace I if we apply KCL at the n

    junction node at the center of the Y.

    IIIInabc (2)

    Substitution of (2) into (1) yields:

    ;;VZIZIIIagYanabc

    ;;VZIZIIIbgYbnabc (3)

    ;;VZIZIIIcgYcnabc

    Expanding the Z through and then n

    collecting terms with like currents yields:

    ;;VZZIZIZIagYnananb

    ;;VZIZZIZIbgnaYnbnc (4)

    ;;VZIZIZZIcgnanbYnc

     2

    Now let’s write these voltage equations in matrix form.

    (VZZZZI((agYnnna

    VZZZZIbgnYnnb (5) VZZZZIcgnnYnc~~~

In compact notation, eq. (5) is:

    VZIabcabcabc (6)

Now I pose the following question….

    If we can somehow find a way to transform eq. (5) into an equation that relates sequence voltages V on the LHS to sequence currents S

    I on the RHS, what will the impedance S

    matrix look like?

    To answer this question, we need to derive eq. (7).

    VZISSS (7)

     3

We refer to the impedance matrix, Z, that S

    relates sequence voltages to sequence currents, as the sequence impedance matrix.

To derive eq. (7), consider what we have: eq.

    (6).

Recall that V=AV and I=AI. abcSabcS

    Substituting into eq. (6) yields:

    AVZAIabcSS (8)

    -1

    Now pre-multiply both sides by A. This is:

    11

    AAVAZAIabcSS (9)

    The left hand side is just V. S

    1

    VAZAIabcSS (10)

    Comparison of (10) with (7) indicates that the sequence impedance matrix, Z, is given S

    by

    1

    ZAZASabc (11)

     4

So what does Z look like? We know all S

    three elements of eq. (11) so why don’t we do the matrix math and find out…

    1ZAZASabc

    111ZZZZ111(((Ynnn

    1221ZZZZ1nYnn3221ZZZZ1nnYn~~~

    Multiplying the two right-hand matrices:

    0(Z111Z3ZZZ((SYnYY

    122ZZ1Z3ZZZSSYnYY3 22Z1Z3ZZZSYnYY~~~

    Now multiply the remaining matrices:

    0(Z3Z9Z00Z3Z00((SYnYn

    1ZZ03Z00Z0SSYY3Z003Z00ZSYY~~~

    Plugging this expression into eq. (10)…

    1

    VAZAIZIabcSSSS

     5

    00

    ((VZ3Z00I(agYna

    

    V0Z0IagYa

    (12) V00ZIagYa~~~

Now this is an amazing thing…

     all off-diagonal terms are zero!

What does this mean?

It means that

    ; the only current that determines the zero

    sequence voltage is the zero sequence

    current.

    ; the only current that determines the

    positive sequence voltage is the positive

    sequence current.

    ; the only current that determines the

    negative sequence voltage is the negative

    sequence current.

    This is the case whenever the impedance matrix is diagonal, with off-diagonals all 0.

     6

    We say that the three equations represented by the matrix relation are uncoupled in that

    no variable (current) appears in more than one equation.

    So these 3 uncoupled equations are:

    00

    VZZI;;3agYna

    

    VZIagYa

     (13)

    VZIagYa

    The really nice thing about these 3 equations is that they represent 3 separate and distinct SINGLE PHASE CIRCUITS!!!!

    Therefore we can just apply EE 303 per-phase analysis to analyze them. Fig. 2 illustrates the single phase circuits.

     7

0I aZ Y0=Z+3Z Z0YnV a

    3Z n

    Zero sequence network

    I aV aZ ZZY SYPositive sequence network

    I aZ YVZZ SY a

    Negative sequence network

    Fig. 2

     8

Some questions:

    1.Why doesn’t the neutral impedance appear

    in the positive & negative sequence

    networks?

    Because the positive and negative

    sequence networks contain balanced

    currents only, and balanced currents sum

    to 0 and therefore do not contribute to

    flow in the neutral.

2.Why do we have 3Z in the zero sequence n

    network instead of just Z? n

    10IIanRecall I+I+I=I. We defined . ABCn3

    0I3ISo actually flows in the Z. But our nan

    0Izero sequence network has only flowing. a

    Therefore, to obtain the correct voltage

    drop seen in the neutral conductor with a

    0Iflow of only , we model the zero-a

    sequence impedance as 3Z. Then the n

    03IZvoltage drop is , as it should be .an

     9

    3.What do these three networks look like if

    the neutral is solidly grounded (no neutral

    impedance)?

    Positive and negative sequence networks

    are the same. Zero sequence is the same

    except Z=0. n

    4.What do these three networks look like if

    the neutral is ungrounded (floating)?

    Positive and negative sequence networks

    are the same. Zero sequence has an open

    0II0circuit, which means . na

    5.What is benefit of the SC transformation?

    Answer: If the load (or line, or load and line) is symmetric (so that Z is diagonal), then S

    the three networks will decouple and we can analyze an unbalanced situation with three separate per-phase analyses.

     10

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