Lineas

SYMMETRICAL COMPONENTS: LINE TRANSPOSITION
Stanley E. Zocholl Schweitzer Engineering Laboratories, Inc. Pullman, WA USA

LINE IMPEDANCE MATRIX
Symmetrical components are derived from the impedance matrix of a transposed transmission line. The impedance matrix is obtained by writing equations for the voltage drop due to current in each phase:
Va = Z aa I a + Z ab I b + Z ac I c Vb = Z ba I a+ Z bb I b + Z bc I c Vc + Z ca I a + Z cb I b + Z cc I c

(1)

When using matrix notation, the impedance is seen as a 3-by-3 matrix of self and mutual impedance terms, and the voltage and current are seen as 1-by-3 column vectors:
æ Va ö æ Z aa ç ÷ ç ç Vb ÷ = ç Z ba çV ÷ çZ è c ø è ca Z ab Z bb Z cb Z ac ö æ I a ö ÷ç ÷ Z bc ÷ ç I b ÷ Z cc ÷ ç I c ÷ øè ø

(2)

Equation (2) can be writtenin more compact notation where bold letters indicate matrices: V=ZI (3)

SYMMETRICAL COMPONENT MATRIX
The symmetrical component equations for voltage are:
Va 0 = Va1 = Va 2 = 1 ( Va + Vb + Vc ) 3 1 ( Va + aVb + a 2 Vc ) 3 1 ( Va + a 2 Vb + aVc ) 3

(4)

Which when written in matrix notation appear as:
æ Va 0 ö æ1 1 ç ÷ 1ç ç Va1 ÷ = ç1 a ç V ÷ 3 ç1 a 2 è a2 ø è 1 ö æ Va ö ÷ç ÷ a 2 ÷ ç Vb ÷a ÷ ç Vc ÷ øè ø

(5)

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Using the more compact notation: Vsym = A V Matrix A contains the vector 1 and the plus and minus 120° rotation unity vectors a and a2. Similarly, the symmetrical component equations for current are:
I a0 = I a1 = I a2 = 1 ( Ia + Ib + Ic ) 3 1 ( I a + aI b + a 2 I c ) 3 1 ( I a + a 2 I b + aI c ) 3

(6)

(7)

Which can be written in matrix form as:
æ I a0 öæ1 1 ç ÷ 1ç ç I a1 ÷ = ç 1 a ç I ÷ 3 ç1 a 2 è a2 ø è 1 ö æ Ia ö ÷ç ÷ a 2 ÷ çIb ÷ a ÷ ç Ic ÷ øè ø

(8)

and in more compact notation: Isym = A I Matrix equations (6) and (9) are the 1-by-3 voltage and current column vectors in terms of the symmetrical component operator A. The relation between the symmetrical components and the impedance matrix is obtained by substituting these equations for Vand I in (3): A Vsym = A Z Isym Vsym = A Z A-1 Isym Observing the position of terms in (10), we can write the symmetrical component impedance matrix as: Vsym = Zsyn Isym where the symmetrical component impedance matrix is: Zsyn = A Z A-1 and where Z is the transmission line impedance matrix: Z = ç Zba
ç è Zca æ Zaa ç Zab Z bb Zcb Zac ö ÷ Zbc ÷ ÷ Zcc ø

(9)

(10)

(11)

(12)

(13)

Inthe Z matrix, the self-impedance terms are equal such that: Zs = Zaa = Zbb = Zcc (14)

2

For a transposed line, the off-diagonal terms, which are the mutual impedances between conductors, are equal such that: Zm = Zab = Zbc = Zca The impedance matrix can then be written as:
æ Zs ç Z = ç Zm çZ è m Zm Zs Zm Zm ö ÷ Zm ÷ Zs ÷ ø

(15)

(16)

and the component matrix for the transposedtransmission line is:
æ Zs + 2Z m ç Zsyn = A Z A-1 = ç 0 ç 0 è 0 Zs − Zm 0 ö ÷ 0 ÷ ÷ Zs − Zm ø 0

(17)

In the symmetrical component matrix, the diagonal terms are the zero-, positive-, and negative-sequence impedance of the transmission line. Note that the off-diagonal terms are zero and indicate that there is no coupling between the zero-, positive-, and negativesequence networks for thetransposed line.
Z0 = Zs + 2Z m Z1 = Z s − Z m Z2 = Zs − Z m

(18)

The impedance matrix for a transposed 100-mile, 500 kV, flat construction line is given in (19). The impedance is given in per unit on a 500 kV, 1000 MVA base:
æ 0.048 + j0.432 ç Z = ç 0.036 + j0.17 ç è 0.036 + j0.17 . 0.036 + j017 0.048 + j0.432 . 0.036 + j017 0.036 + j0.17 ö ÷ 0.036 + j0.17 ÷ ÷ 0.048 + j0.432ø

(19)

Thesymmetrical component matrix is obtained by applying equation (12):
æ 0.12 + j0.771 ç Zsyn = A Z A = ç 0 ç 0 è
-1

0 0.012 + j0.262 0

ö ÷ ÷ 0 ÷ 0.012 + j0.262ø 0

(20)

Where the zero-, positive-, and negative-sequence impedance terms are:
. Z 0 = 012 + j0.771 Z1 = 0.012 + j0.262 Z 2 = 0.012 + j0.262

(21)

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TRANSPOSED LINES
The three-phase fault calculations for a transposed...