3.2.1.1 Calculation of lumped RLC parameters

The computation of three-phase transmission line parameters becomes cumbersome by the

existence of inductive and capacitive couplings between conductors, and between

conductors and ground (Anderson, 1973). Moreover, resistances and self and mutual

inductances vary nonlinearly with frequency and, together with the capacitive effects, vary

nonlinearly with the electrical distance of the line (Acha and Madrigal, 2001; Arrillaga et al.,

1997).

In fundamental frequency power system applications, it is normal practice to calculate the

inductive and capacitive effects of the transmission line independently and then to combine

46 MODELLING OF CONVENTIONAL POWER PLANT

them together to give the ¬nal transmission-line representation. Once the resistances,

inductances, and capacitances associated with a particular transmission line con¬guration

have been determined, a transmission-line model in the form of a p-circuit, or any other

alternative representation, become feasible.

The series impedance matrix Zseries of a multiconductor transmission line, which takes

account of geometric imbalances and frequency dependency but not long-line effects, may

be assumed to consist of the following components:

Zseries ¼ Zinternal þ Zgeometric þ Zground : °3:2Þ

In Equation (3.2), Zinternal is the impedance inside the conductors, Zground is the impedance

contribution of the ground return path, and Zgeometric is the impedance contribution from

the magnetic ¬‚uxes in the air surrounding the conductors. For most practical purposes, the

parameters Zgeometric may be taken to be linear functions of the potential coef¬cients P.

Unlike Zinternal, the parameters Zground, Zgeometric, and P are a function of the physical

geometry of the conductor™s arrangement in the tower. The capacitive effects are

incorporated in the shunt admittance matrix Yshunt, which is a linear function of P. Shunt

parameters are addressed in Section 3.2.1.2.

If the surfaces of the conductors and the earth beneath the conductors may be assumed to

be equipotential surfaces then the standard method of images may be used to calculate the

potential coef¬cients P.

The method of images allows the conducting plane to be replaced by a ¬ctitious

conductor located at the mirror image of the actual conductors. Figure 3.3 shows the case

when phase conductors a, b, and c above ground have been replaced by three equivalent

conductors and their images (Anderson, 1973).

dac

dab dbc

c

a b

ha hb hc

Dac′

Dab′

hc′

hb′

ha′

a′ b′ c′

Figure 3.3 Line geometry and its image

47

TRANSMISSION LINE MODELLING

The self-potential coef¬cient of an overhead conductor is solely a function of the height

of the conductor above ground, say h, and the external radius of the conductor, say rext. In

contrast, the mutual potential coef¬cient between two conductors is a function of the

separation between the two conductors, d, and the separation between one conductor and the

image of the second conductor, D. For the three conductors in the transmission line shown in

Figure 3.3, the matrix of potential coef¬cients is

2 3

2ha Dab

ln Dac

ln rext a ln dab ac

6 7

d

6 7

6 7

P ¼ 6 ln Dba ln Dbc 7: °3:3Þ

ln r2hbb

ba bc

6 7

d d

ext

4 5

Dca Dcb

ln r2hcc

ln dca ln dcb ext

It should be noted that potential coef¬cients are dimensionless and reciprocal.

The geometric impedance matrix for the circuit of Figure 3.3 is

!0

P kmÀ1 ;

Zgeometric ¼ j °3:4Þ

4p

where Zgeometric varies linearly with the base frequency f ; ! ¼ 2pf, and the permeability of

free space is 0 ¼ 4p ‚ 10À4 H kmÀ1 .

3.2.1.2 Shunt admittances

Shunt admittance parameters vary linearly with frequency and are completely de¬ned by the

inverse potential coef¬cients (Anderson, 1973). The matrix of shunt admittance parameters

for the circuit of Figure 3.3 is

Yshunt ¼ j!2p"0 PÀ1 S kmÀ1 ; °3:5Þ

where "0 , equal to 8:85 ‚ 10À9 F kmÀ1 ; is the permittivity of free space.

3.2.1.3 Internal impedances

It has long been recognised that the internal resistance and inductance of conductors vary

with frequency in a nonlinear manner. The reason for this effect is attributed mainly to the

nonuniform distribution of current ¬‚ow over the full area available, with current tending to

¬‚ow on the surface. This trend increases with frequency and is termed the ˜skin effect™. The

overall effect is an increase in resistance and a decrease in internal inductance (Arrillaga

et al., 1997).

In power systems applications, the established formula for evaluating the impedance of an

annular conductor, at a given frequency, uses the Bessel functions of zero order, ¬rst kind,

and second kind and their derivatives, which are solved, within speci¬ed accuracy, using

their associated in¬nite series. However, at power frequencies, the skin effect is negligibly

small and there is little error in calculating the internal impedance of conductors by

assuming that the magnetic ¬eld inside the conductor is con¬ned to an area lying between

the external radius, rext, of the conductor and the geometric mean radius (gmr), rgmr, of the

conductor (Grainger and Stevenson, 1994).

48 MODELLING OF CONVENTIONAL POWER PLANT

Cross-section of a power conductor

Figure 3.4

As illustrated in Figure 3.4, the gmr lies between the external and internal radii of the

conductor. The gmr is normally measured and made available by the manufacturer. An

approximated, frequency-independent relationship is given by

rgmr ¼ eÀ1=4 rext : °3:6Þ

If the frequency of interest is low enough for the skin effect to be of no consequence, then

the concept of potential coef¬cients can be applied to calculate the internal impedance of

conductors. For the three conductors in Figure 3.3, the matrix of internal potential

coef¬cients is,

2 3

rext

ln rgmr1 0 0

6 7

1

6 7

Pinternal ¼ 6 7:

rext 2

°3:7Þ

0 ln rgmr 0

6 7

4 5

2

rext

0 0 ln rgmr3

3

Hence, the conductor impedance matrix for this circuit is

!0

Pinternal kmÀ1 ;

Zinternal ¼ Rac þ j °3:8Þ

4p

where Rac is a diagonal matrix with entries corresponding to the AC power frequency

resistances (50 or 60 Hz) of the various conductors in the transmission circuit.

3.2.1.4 Ground return impedances

The impedance of the ground return path varies nonlinearly with frequency and exhibits an

effect similar to that of the skin effect in conductors, where the effective area available for

the current to ¬‚ow reduces with frequency.

The problem of current-carrying wires above a ¬‚at earth of homogeneous conductivity,

and the related issue of transmission-line parameter calculation, received a great deal of

research attention almost a century ago. It was J.R. Carson who in 1926 published a

comprehensive solution to the problem. The solution involves an in¬nite integral that cannot

be solved analytically or in closed form. However, the integral is conveniently expressed by

a set of in¬nite series that show good convergence characteristics for most problems

encountered in the areas of electromagnetic ¬elds, propagation characteristics, and magnetic

49

TRANSMISSION LINE MODELLING

induction effects caused by power lines. Ever since, and perhaps because of the existence of

the in¬nite series, the solution published by Carson has been extensively used by power

engineers worldwide.

As the need arises to calculate ground impedances for a wide spectrum of frequencies,

and also because of the uncertainty in the available data, in particular regarding ground

conductivity, the tendency is to go for simpler formulations aiming at a reduction in

computing time while keeping the accuracy at a reasonable level. Recent formulations use

the concept of a complex mirroring surface beneath the ground. Rigorous mathematical

analyses have shown these formulations to be good physical and mathematical approxima-

tions to Carson™s solution.

The most popular equations in power system applications are those attributed to C.

Dubanton (Deri et al., 1981). The reason is their simplicity and good accuracy for the whole

span of frequencies for which Carson™s equations are valid. With reference to Figure 3.5,

the equations for calculating the self-impedance of conductor l and the mutual impedance

between conductors l and m take the following form:

!

!0 2°hl þ pÞ

kmÀ1 ;

Zll ¼ j °3:9Þ

ln

2p rext l

( )

2 2 1=2

!0 ½°hl þ hm þ 2pÞ þ dlm

kmÀ1 ;

Zlm ¼ j °3:10Þ

ln 2 2 1=2

½ ° hl À hm Þ þ d

2p

lm

where p, equal to °j!0 g Þ1=2 , is the complex depth beneath the ground at which the

mirroring surface is located.

dac

a b c

dab dbc

ha hb hc

Earth surface

p

Fictitious surface

p′ Dac′

Dab′

Mirroring image

of earth surface

ha′ hb′ hc′

a′ b′ c′

Line geometry showing the complex depth

Figure 3.5

50 MODELLING OF CONVENTIONAL POWER PLANT

It should be noted that the use of Equations (3.9) and (3.10) yields combined information

of Zgeometric þ Zground . Moreover, if the transmission-line parameters are intended for power

frequency applications (50 or 60 Hz) then the skin effect inside the conductors can be

ignored and Equation (3.9) can be combined with Equation (3.8) to take account of the

impedance contribution from Zinternal :

!

!0 2° hl þ pÞ

kmÀ1 :

Zll ¼ Rac l þ j °3:11Þ

ln

2p rgmr l

In summary, for the purpose of low-frequency power applications, Equations (3.10) and

(3.11) may be used to calculate the individual elements of Equation (3.1). The impedance

parameters include geometric imbalances and ground return effects but no full frequency

dependency.

3.2.2 Ground Wires

Using the same notation as in Section 3.2.1, we may express the voltage-drop equation of a

three-phase transmission line with two ground wires, w and v, as follows (Anderson, 1973):

232 32 3 2 0 3

ZaaÀg ZabÀg ZacÀg ZawÀg ZavÀg Va

Va Ia

6 7 6Z 76 7 6 0 7

6 Vb 7 6 baÀg ZbbÀg ZbcÀg ZbwÀg ZbvÀg 76 Ib 7 6 Vb 7

676 76 7 6 7

6 Vc 7 ¼ 6 ZcaÀg ZcbÀg ZccÀg ZcwÀg ZcvÀg 76 Ic 7 þ 6 V 0 7: °3:12Þ

676 76 7 6 c 7

676 76 7 6 0 7

4 Vw 5 4 ZwaÀg ZwbÀg ZwcÀg ZwwÀg ZwvÀg 54 Iw 5 4 Vw 5

0

ZvaÀg ZvbÀg ZvcÀg ZvwÀg ZvvÀg Vv

Vv Iv

It is assumed that the individual impedance elements are calculated by using Equa-

tions (3.10) and (3.11). In compact notation, we have,

ÁVabc ¼ AIabc þ BIwv ; °3:13Þ

ÁVwv ¼ CIabc þ DIwv ; °3:14Þ

where

9

2 3

2 3>

0

Va À Va

Ia > >

>

6 7

6 7> >

6 07

¼ 4 Ib 5; >

>

ÁVabc ¼ 6 Vb À Vb 7 ; >

Iabc

>

4 5 =

Ic °3:15Þ

0

Vc À Vc >

>

2 3

!> >

0

Iw >

Vw À Vw >

>

;>

ÁVwv ¼ 4 5; ¼ >

Iwv

Iv > ;

0

V v À Vv

2 3 2 3

ZaaÀg ZabÀg ZacÀg ZawÀg ZavÀg

6 7 6 7

A ¼ 4 ZbaÀg ZbcÀg 5; B ¼ 4 ZbwÀg ZbvÀg 5;

ZbbÀg

°3:16Þ

ZcaÀg ZcbÀg ZccÀg Zcw--g ZcvÀg

! !

ZwaÀg ZwbÀg ZwcÀg ZwwÀg ZwvÀg

C¼ ; D¼ :

ZvaÀg ZvbÀg ZvcÀg ZvwÀg ZvvÀg

51

TRANSMISSION LINE MODELLING

As it is normal practice to connect ground wires to earth at both ends of every transmission

span, ÁVwv ¼ 0, and it is possible to simplify Equation (3.12) to

ÁVabc ¼ AIabc þ BIwv ; °3:17Þ

0 ¼ CIabc þ DIwv: °3:18Þ

Solving Equation (3.18) for Iwv ,

Iwv ¼ ÀDÀ1 CIabc ; °3:19Þ

and substitution of Equation (3.19) into Equation (3.17) yields

‚ Ã

ÁVabc ¼ A À BDÀ1 C Iabc ¼ ZabcÀwvÀg Iabc ; °3:20Þ

where

ZabcÀwvÀg ¼ A À BDÀ1 C: °3:21Þ

Equation (3.20) can be written in expanded form as

2 32 32 3

ÁVa ZaaÀwvÀg ZabÀwvÀg ZacÀwvÀg Ia

4 ÁVb 5 ¼ 4 ZbaÀwvÀg ZbbÀwvÀg ZbcÀwvÀg 54 Ib 5: °3:22Þ

ÁVc Ic

ZcaÀwvÀg ZcbÀwvÀg ZccÀwvÀg

The reduced equivalent matrix Equation (3.22) is fully equivalent to matrix Equation (3.12),

where the ground wires have been mathematically eliminated. For most system analysis

purposes, Equation (3.22) provides a suitable representation for transmission lines with

ground wires. Symmetrical components can be applied to Equation (3.22), and it is therefore

preferred over Equation (3.12).

3.2.3 Bundle Conductors

The use of more than one conductor per phase (i.e. bundle conductors) reduces the

equivalent transmission-line impedance and allows for an increase in power transmission. It

also allows for a reduction in corona loss and radio interference owing to a reduction in

conductor-surface voltage gradients.

For cases of transmission lines of 400 kV and above it is standard practice to have four

bundle conductors per phase, whereas for 230 kV lines, only three or two bundle conductors

per phase are required. These arrangements are shown in Figures 3.6(a), 3.6(b), and 3.6(c),

respectively (Grainger and Stevenson, 1994).

d

d

d

d d

(b) (c)

(a)

Figure 3.6 Typical arrangements of bundle conductors: (a) four, (b) three, and (c) two bundle

conductors (open circles)

52 MODELLING OF CONVENTIONAL POWER PLANT

In power system studies the interest is rarely on individual conductors but, rather, on the

individual phases. Hence, steps are taken to ¬nd reduced equivalents that involve only one

conductor per phase. The equivalent conductors correctly account for the original

con¬guration but keep essential information only.

The bundle-conductor reduction may be achieved in a number of ways. Use of the

concept of an equivalent geometric mean radius (GMR), Rgmr is one of them, and, although

it is a frequency-independent method, it yields reasonably accurate solutions, particularly at

power frequencies.

For cases of two, three, and four bundle conductors per phase, the following relations are

used to calculate the equivalent GMR:

Rgmr ¼ °rgmr ‚ dÞ1=2 ; °3:23Þ

Rgmr ¼ °rgmr ‚ d ‚ dÞ1=3 ; °3:24Þ

p¬¬¬

Rgmr ¼ °rgmr ‚ d ‚ d ‚ 2dÞ1=4 : °3:25Þ

In this case, the equivalent phase resistance is simply obtained by dividing the resistance of

one of the original phase conductors in the bundle by the number of conductors in the

bundle. This simple approach takes the very practical view that all conductors in the bundle

are equal and that they are at the same potential.

A more rigorous approach, which includes frequency dependency for the reduction

of the series impedance matrix, involves matrix reduction using Kron™s method

(Anderson, 1973). In this situation, all conductor impedances are calculated explicitly

and, after a suitable manipulation of terms in the impedance matrix, the mathematical

elimination of bundle conductors is carried out. The actual elimination is the same process

as that for the matrix reduction given by Equations (3.17)“(3.21) in the elimination of

ground wires.

To illustrate the elimination procedure used when bundle conductors are present, take the

case of a three-phase transmission line (a, b, c) with two conductors per phase (1, 2) and no

ground wires. Using similar notation as in Sections 3.2.1 and 3.2.2, the matrix of series

impedance parameters representing such a transmission line would be:

2 32 32 3

ÁVa1 Za1a1Àg Za1b1 À g Za1c1Àg Za1a2Àg Za1b2Àg Za1c2Àg Ia1

6 ÁVb1 7 6 Zb1a1Àg Zb1b1Àg Zb1c1Àg Zb1a2Àg Zb1b2Àg Zb1c2Àg 76 Ib1 7

6 76 76 7

6 ÁV 7 6 Z Zc1b1Àg Zc1c1Àg Zc1a2Àg Zc1b2Àg Zc1c2Àg 76 Ic1 7

6 c1 7 6 c1a1Àg 76 7

7¼6 76 7: °3:26Þ

6

6 ÁVa2 7 6 Za2a1Àg Za2b1Àg Za2c1Àg Za2a2Àg Za2b2Àg Za2c2Àg 76 Ia2 7

6 76 76 7

4 ÁVb2 5 4 Zb2a1Àg Zb2b1Àg Zb2c1Àg Zb2a2Àg Zb2b2Àg Zb2c2Àg 54 Ib2 5

ÁVc2 Zc2a1Àg Zc2b1Àg Zc2c1Àg Zc2a2Àg Zc2b2Àg Zc2c2Àg Ic2

The individual elements of Equation (3.26) are calculated by using Equations (3.10) and

(3.11). In compact notation, we have,

! ! !

ÁVabc1 Zabc11 Zabc12 Iabc1

¼ : °3:27Þ

ÁVabc2 Zabc21 Zabc22 Iabc2

If it is assumed that the two conductors in the bundle are at equal potential, then the row and

column corresponding to one of the conductors in the bundle, say 2, is mathematically

eliminated. There are three main steps involved in the elimination process:

53

TRANSMISSION LINE MODELLING

Step 1: The voltage equality constraint ÁVabc2 À ÁVabc1 ¼ 0 is incorporated into

Equation (3.27):

! ! !

ÁVabc1 Zabc11 Zabc12 Iabc1

¼ : °3:28Þ

Zabc21 ÀZabc11 Zabc22 ÀZabc12 I abc2

ÁVabc2 ÀÁVabc1 ¼ 0

Step 2: matrix symmetry is restored. This is achieved by adding and subtracting the terms

Zabc11 Iabc2 and °Zabc21 ÀZabc11 ÞIabc2 in rows 1 and 2, respectively:

! ! !

Zabc12 À Zabc11

ÁVabc1 Iabc1 þ Iabc2

Zabc11

¼ :

Zabc21 À Zabc11 °Zabc22 þ Zabc11 À Zabc12 À Zabc21 Þ Iabc2

0

°3:29Þ

Step 3: the actual matrix reduction is carried out. This is fully equivalent to that given by

Equations (3.17)“(3.21) used for the mathematical elimination of ground wires:

ÁVabc--b ¼ Zabc--b--g Iabc--b ; °3:30Þ

where

9

Zabc--b--g ¼ A À BDÀ1 C; >

>

>

>

>

A ¼ Zabc11 ; >

>

>

>

>

B ¼ Zabc12 À Zabc11 ; >

=

°3:31Þ

C ¼ Zabc21 À Zabc11 ;

>

>

D ¼ Zabc22 þ Zabc11 À Zabc12 À Zabc21 ; >

>

>

>

>

>

>

ÁVabc--b ¼ ÁVabc1 ; >

>

;

Iabc--b ¼ Iabc1 þ Iabc2 :

As illustrated in Figure 3.7, the current Iabc--b may be interpreted as the phase current just

before it splits into the individual currents Iabc1 and Iabc2 in the bundle.

L

R

Iabc 2

Iabc’b

L

Iabc 1 R

Currents in a two-bundle conductor

Figure 3.7

An alternative, more elegant approach than that carried out in steps 1 and 2 is achieved by

applying the following set of transformation matrices:

! !

1 À1 10

Tb ¼ and Ttb ¼ °3:32Þ

À1 1

01

to the impedance matrix in Equation (3.27), that is,

! ! !

À1

1 0 Zabc11 Zabc12 1

: °3:33Þ

À1 1 Zabc21 Zabc22 0 1

54 MODELLING OF CONVENTIONAL POWER PLANT

This operation yields the same result as Equation (3.30). In Equations (3.32) and (3.33), 1 is

a 3‚3 unit matrix, and the superscript t indicates the transpose of the matrix.

In the more general case, when there are n conductors per phase, the transformation

matrix Tb may have the following form:

2 3

1 À1 Á Á Á À1

60 1 ÁÁÁ 0 7

Tb ¼ 6 . . 7: °3:34Þ

. ..

4. .5

.

. . . .

ÁÁÁ

0 0 1

Independent of the procedure used to determine the reduced equivalent matrix Equation

(3.30), this can be written in expanded form as

2 32 32 3

ÁVaÀb ZaaÀb ZabÀb ZacÀb IaÀb

4 ÁVbÀb 5 ¼ 4 ZbaÀb ZbbÀb ZbcÀb 54 IbÀb 5: °3:35Þ

ÁVcÀb ZcaÀb ZcbÀb ZccÀb IcÀb

3.2.4 Double Circuit Transmission Lines

Often two or more three-phase transmission lines are operated in parallel (Anderson, 1973;

Grainger and Stevenson, 1994). A common arrangement is to place two three-phase circuits

in the same tower, as shown in Figure 3.8. In this case, the magnetic interaction between the

phase conductors of both three-phase circuits can be represented by the following

Ground wire 1 Ground wire 2

Phase c

Phase A

Phase b

Phase B

Phase a

Phase C

Double circuit, three-phase transmission line

Figure 3.8

55

TRANSMISSION LINE MODELLING

impedance matrix equation:

2 32 32 3

ÁVa Ia

ZaaÀg ZabÀg ZacÀg ZaAÀg ZaBÀg ZaCÀg

6 ÁVb 7 6 Z 76 Ib 7

6 7 6 baÀg ZbCÀg 76 7

ZbbÀg ZbcÀg ZbAÀg ZbBÀg

6 ÁV 7 6 76 7

6 c7 6 ZcaÀg ZcCÀg 76 Ic 7

ZcbÀg ZccÀg ZcAÀg ZcBÀg

7¼6 76 7: °3:36Þ

6

6 ÁVA 7 6 ZAaÀg ZACÀg 76 IA 7

ZAbÀg ZAcÀg ZAAÀg ZABÀg

6 76 76 7

4 ÁVB 5 4 ZBaÀg ZBCÀg 54 IB 5

ZBbÀg ZBcÀg ZBAÀg ZBBÀg

ÁVC ZCaÀg ZCbÀg ZCcÀg ZCAÀg ZCBÀg ZCCÀg IC

It is assumed here that neither ground wires nor bundle conductors are present in the double

circuit transmission line; alternatively, it is assumed they have been mathematically

eliminated by using the methods discussed in Sections 3.2.2 and 3.2.3, respectively. It is also

assumed that the individual elements of Equation (3.36) were calculated by using Equa-

tion (3.11) and (3.10).

3.2.5 The Per-unit System

Transmission-line parameters as calculated by Equations (3.10) and (3.11) are given in

ohms per kilometre. However, when dealing with transmission lines at the system level,

there are several advantages to be gained by expressing the line parameters in a uniform

units system, termed the per-unit system.

Moreover, equipment manufacturers also specify the equipment characteristics in either

percentage or per-unit values with respect to their nominal values. This is a simple

mechanism that enables the electrical power network to be analysed as a single entity

regardless of the voltage level at which the equipment operates.

The following electrical parameters are handled in per-unit values: voltage, current,

power, and impedance. In each case, the corresponding per-unit value is the ratio of the

actual value to a base value; that is,

actual value

per-unit value ¼ : °3:37Þ

base value

The per-unit (p.u.) value is dimensionless by virtue of the base and actual values sharing

the same units. It is common practice to specify the voltage, VBase, and power, SBase, to be the

primary base values, from which the base current and impedance can be derived. In single-

phase systems:

SBase

IBase ¼ °3:38Þ

A;

VBase

2

VBase

ZBase ¼ : °3:39Þ

SBase

In three-phase systems, the total power, S3f, and the line-to-line voltage, VLL, are readily

available, and the following relations apply:

S3 Base

IBase ¼ p¬¬¬ °3:40Þ

A;

3 VLL Base

2

VLL Base

¼ : °3:41Þ

ZBase

S3 Base

56 MODELLING OF CONVENTIONAL POWER PLANT

Based on Equation (3.37), the per-unit parameters are:

S V

Sp:u: ¼ ; Vp:u: ¼ ;

SBase VBase

°3:42Þ

I Z

¼ ; ¼ :

Ip:u: Zp:u:

IBase ZBase

It should be noted that the transmission line shunt admittances in Equation (3.5) are

converted into per-unit values by using:

Yp:u: ¼ YZBase : °3:43Þ

The transmission-line shunt admittance is sometimes referred to as ˜charging MVAR™,

Qsh. In some application programs, such as power ¬‚ows, is actually supplied in terms of

MVAR as opposed to Siemens (S) or À 1. There is no dif¬culty in transforming Qsh into an

equivalent per-unit shunt admittance if it is assumed that a voltage value of 1 p.u. exists at

both ends of the transmission line:

Qsh p:u:

Yp:u: ¼ ¼ Qsh p:u: : °3:44Þ

2

Vp:u:

Conversely, the information given by a transmission-line parameter program for the shunt

admittance can also be expressed as a charging admittance under the assumption of a 1 p.u.

voltage.

In high-voltage transmission studies, a base power of 100 MVA is normally selected for

the whole system. In contrast, selection of the base voltage is not unique; instead, as many

base voltages are selected as are required to match the number of voltage levels in the

network under study. Having said that, it is important to mention that in some application

studies, such as positive sequence power ¬‚ows, it is not uncommon to have only one base

voltage being selected. This is owing to the fact that the generating plant is modelled as

injections of active and reactive power at the high-voltage bus of the generator transformer.

Also, it is normal to conduct a detailed study for only one voltage level of the network, say

400 kV, with contributions from other parts of the network, operating at different voltage,

treated as either bulk power supply points or loads.

3.2.6 Transmission-line Program: Basic Parameters

A computer program for the calculation of transmission line parameters is given in Program

3.1.(1) The program is general, as far as the number of conductors in the transmission line is

concerned, and caters for up to four bundle conductors per phase and any number of earth wires.

Program written in Matlab1 to calculate transmission-line parameters

PROGRAM 3.1

%***- - - - - Main Program

TransmissionLineData;

Note1: in Matlab1 it is possible to write very long lines, up to 600 characters. Continuation lines are possible

(1)

and are indicated by the use of three periods at the end of the previous line. This convention has been adopted in this

copy.

57

TRANSMISSION LINE MODELLING

[ZSeries,YShunt] = ShortLine(nphc,ngw,nb,bsep,resis,rdext,gmr,...

x,y,f, sigmag,vbase,sbase)

%End main Program

function [ZSeries,YShunt] = ShortLine(nphc,ngw,nb,bsep,...

resis,rdext,gmr,x, y,f,sigmag,vbase,sbase)

[RAD,GMR,RES] = BundleReduction(nphc,ngw,nb,bsep,rdext,gmr,resis);

[YShunt] = PotCoeff(nphc,RAD,x,y,f);

[ZSeries] = Dubanton(nphc,ngw,GMR,RES,x,y,f,sigmag);

[ZSeries] = GroundWireReduction(nphc,ngw,ZSeries);

[ZSeries,YShunt] = PerUnit(nphc,ZSeries,YShunt,vbase,sbase);

function [RAD,GMR,RES] = BundleReduction(nphc,ngw,nb,bsep,rdext,...

gmr,resis);

for ii = 1: nphc + ngw

if nb(ii) == 1

RAD(ii) = rdext(ii);

GMR(ii) = gmr(ii);

elseif nb(ii) == 2

RAD(ii) = sqrt(rdext(ii)*bsep(ii));

GMR(ii) = sqrt(gmr(ii)*bsep(ii));

elseif nb(ii) == 3

RAD(ii) = exp(log(rdext(ii)*bsep(ii)*bsep(ii))/3);

GMR(ii) = exp(log(gmr(ii)*bsep(ii)*bsep(ii))/3);

elseif nb(ii) == 4

RAD(ii) = sqrt(sqrt(rdext(ii)*bsep(ii)*bsep(ii)*bsep(ii)...

*sqrt (2)));

GMR(ii) = sqrt(sqrt(gmr(ii)*bsep(ii)*bsep(ii)*bsep(ii)*sqrt(2)));

end

RES(ii) = resis(ii)/nb(ii);

end

function [YShunt] = PotCoeff(nphc,RAD,x,y,f);

[YShunt] = zeros(nphc,nphc);

omega = 2*pi*f;

eps = 8.854*1e-9;

for ii = 1: nphc

for jj = 1: nphc

if ( ii == jj )

YShunt(ii,ii)=log(2*y(ii)/RAD(ii));

else

YShunt(ii,jj) = log( sqrt ( ( x(ii) - x(jj) )^2 + ...

( y(ii) + y(jj) )^2 ) / sqrt ( ( x(ii) - x(jj) )^2 + ...

58 MODELLING OF CONVENTIONAL POWER PLANT

( y(ii)-y(jj) )^2 ) );

end

end

end

YShunt = i*2*pi*omega*eps*inv(YShunt);

function [ZSeries] = Dubanton(nphc,ngw,GMR,RES,x,y,f,sigmag)

[ZSeries] = zeros(nphc+ngw,nphc+ngw);

mnu = 4*pi*1e-7;

omega = (0+(2*pi*f)*i);

pe = 1/sqrt(omega*mnu*sigmag);

for ii = 1: nphc + ngw

for jj = 1: nphc + ngw

if( ii == jj )

ZSeries(ii,ii) = 1000*( RES(ii) + omega*mnu*...

log((y(ii)+y(jj)+2*pe)/GMR(ii))/(2*pi) );

else

ZSeries(ii,jj) = 1000*omega*mnu*...

log( sqrt((x(ii)-x(jj))^2+(y(ii)+y(jj)+2*pe)^2) /...

sqrt((x(ii)-x(jj))^2+(y(ii)-y(jj))^2 ) )/(2*pi);

end

end

end

function [ZSeries] = GroundWireReduction(nphc,ngw,ZSeries)

for ii = nphc + 1: nphc + ngw

ZSeries(ii,ii) = 1/ZSeries(ii,ii);

for jj = 1: nphc + ngw

if( ii ˜= jj )

ZSeries(jj,ii) = ZSeries(jj,ii)*ZSeries(ii,ii);

for kk = 1: nphc + ngw

if( kk ˜= ii )

ZSeries(jj,kk) = ZSeries(jj,kk) -ZSeries(jj,ii)*...

ZSeries(ii,kk);

if ( jj == nphc + ngw)

ZSeries(ii,kk) = -ZSeries(ii,ii)*ZSeries(ii,kk);

end

end

end

end

end

end

if ngw > 0

for jj = 1: nphc + ngw -1

ZSeries(nphc+ngw,jj) = -ZSeries(nphc+ngw,nphc+ngw)*...

ZSeries(nphc+ngw,jj);

end

59

TRANSMISSION LINE MODELLING

ZSeries = ZSeries(1:nphc,1:nphc);

end

function [ZSeries,YShunt] = PerUnit(nphc,ZSeries,YShunt,vbase,sbase)

zbase = vbase*vbase/sbase;

for ii = 1: nphc

for jj = 1: nphc

ZSeries(ii,jj) = ZSeries(ii,jj)/zbase;

YShunt(ii,jj) = YShunt(ii,jj)*zbase;

end

end

3.2.7 Numerical Example of Transmission Line Parameter Calculation

The basic parameters of a 500 kV, three-phase transmission line of ¬‚at con¬guration are

calculated using the Matlab1 function ShortLine given in Section 3.2.6. There are four

(bundle) conductors per phase and no ground wires in the tower (Arrillaga et al., 1986).

The series impedance and the shunt admittance matrices are calculated in ohms per

kilometres and in per-unit values using a base power of 100 MVA.

Function TransmissionLineData, to read data for the 500 kV, three-phase transmis-

sion line of ¬‚at con¬guration, is as follows:

%transmission line.

%

%nphc = number of phase conductors

%ngw = number of ground wires

%

nphc = 3 ; ngw = 0 ;

%

%Individual Conductors Data

%resis = resistance in ohms per meter

%rdext = external radius in meters

%gmr = geometrical mean radius in meters

%nb = number of bundle conductors per phase -1 to 4

%bsep = separation between conductors in the bundle in meters

%x,y = conductor™s co-ordinates in the tower in meters

%

resis(1) = 0.1379/1000 ; rdext(1) = 1.049/100 ; gmr(1) = 0.817/100 ;

nb(1) = 4 ; bsep(1) = 0.46 ; x(1) = 12.65 ; y(1) = 27.50 ;

resis(2) = 0.1379/1000 ; rdext(2) = 1.049/100 ; gmr(2) = 0.817/100 ;

nb(2) = 4 ; bsep(2) = 0.46 ; x(2) = 0 ; y(2) = 27.50 ;

resis(3) = 0.1379/1000 ; rdext(3) = 1.049/100 ; gmr(3) = 0.817/100 ;

nb(3) = 4 ; bsep(3) = 0.46 ; x(3) = -12.65 ; y(3) = 27.50 ;

%

60 MODELLING OF CONVENTIONAL POWER PLANT

%General Data

%f = frequency

%sigmag = ground™s conductivity

%vbase = base voltage

%sbase = base power

%

f = 50 ; sigmag = 0.01 ; vbase = 500 ; sbase = 100 ;

%

%End of function TransmissionLineData

The series impedance and the shunt admittance matrices in ohms per kilometre are:

2 3

0:0815 þ j0:5435 0:0470 þ j0:2774 0:0470 þ j0:2339

6 7

Zabc ¼ 4 0:0470 þ j0:2774 0:0815 þ j0:5435 0:0470 þ j0:2774 5 kmÀ1 ;

0:0470 þ j0:2339 0:0470 þ j0:2774 0:0815 þ j0:5435

2 3

Àj0:809 Àj0:305

j3:359

6 7

j3:527 Àj0:809 5mS kmÀ1 :

Yabc ¼ 4 Àj0:809

Àj0:305 Àj0:809 j3:359

The geometric imbalances inherent in this transmission line, due to its ¬‚at con¬guration,

are re¬‚ected in the fact that not all mutual reactances have the same value, that is, Xab 6¼ Xac.

Similar effects can be observed in the mutual values of Yabc. Also, The resistive effects

shown in the mutual elements of Zabc are entirely due to the ground return effects. As

expected, this effect is not present in Yabc since capacitive effects are not a function of

ground return.

With reference to a base voltage of 500 kV and 100 MVA, the series impedance and shunt

admittance matrices in per-unit values are:

2 3

0:0326 þ j0:2174 0:0188 þ j0:1110 0:0188 þ j0:0935

6 7

Zabc ¼ 10À3 ‚ 4 0:0188 þ j0:1110 0:0326 þ j0:2174 0:0188 þ j0:1110 5p:u:;

0:0188 þ j0:0935 0:0188 þ j0:1110 0:0326 þ j0:2174

2 3

Àj2:024 Àj0:762

j8:398

6 7

À3

Yabc ¼ 10 ‚ 4 Àj2:024 j8:816 Àj2:024 5p:u:

Àj0:762 Àj2:024 j8:398

3.2.8 Long-line Effects

The transmission line models required for long-distance transmission applications are more

involved than those covered in Section 3.2.4, which are only suitable to represent short to

medium-distance transmission lines (Grainger and Stevenson, 1994). In actual applications,

however, it is not common practice to see transmission lines of more than 300 km without

series compensation, which fall within the category of medium-distance transmission lines

for the purpose of fundamental frequency operation.

However, in some special cases it is desirable to incorporate long-line effects into the

transmission line parameters (Bowman and McNamee, 1964). A case in point would

61

TRANSMISSION LINE MODELLING

be studies relating to placement and sizing of shunt and series compensation of long-

distance transmission. At frequency applications higher than the fundamental frequency, it

is certainly mandatory to incorporate long-line effects since the electrical distance increases

rapidly with frequency (Acha and Madrigal, 2001; Arrillaga et al., 1997). Even transmission

lines of only a few tens of kilometres may be seen as a very long line at 1 kHz.

Calculation of multiconductor transmission line parameters, including long-line effects,

requires the use of formulations derived from the wave propagation equation. This

introduces a degree of extra complexity as these formulations invariably involve square

roots and circular and hyperbolic functions of matrices. Several options are available to

carry out such nonconventional matrix operations, but perhaps the best known method is to

simply apply suitable eigenvector techniques to the relevant transmission line parameter

matrices (Wedephol, 1963). This enables all calculations to be performed in the frame of

reference of the modes and then referred back to the frame of reference of the phases.

Arguably, the best known formulation derived from the wave propagation equation is the

˜ABCD™ parameter formulation:

! ! !

VS AB VR

¼ ; °3:45Þ

C D ÀIR

IS

where

9

A ¼ Tv ‚ Diag°cosh

m lÞ ‚ TÀ1 ; >

>

>

v

À1 >

B ¼ Tv ‚ Diag°zm ‚ sinh

m lÞ ‚ Ti ; =

°3:46Þ

C ¼ Ti ‚ Diag°ym ‚ sinh

m lÞ ‚ TÀ1 ; >

>

>

>

v

;

À1

D ¼ Ti ‚ Diag°cosh

m lÞ ‚ Ti :

In Equations (3.46), Diag is a diagonal matrix; m is the subscript for modes 0, , and ; l is

the length of the line; Tv and Ti are transformation matrices made up of the eigenvectors of

the matrix products ZY and YZ, respectively; and Z and Y are lumped transmission-line

parameters as calculated by the Matlab1 computer program given in Section 3.2.6.

The modal parameters for the propagation constant,

m, and the characteristic impedance

and admittance, zm, and ym, in Equations (3.46) are calculated by ¬rst making Z and Y

diagonal:

)

À1

Zm ¼ Tv Z Ti ;

°3:47Þ

Ym ¼ TÀ1 Y Tv ;

i

and then performing the following operations:

2 3

1=2

°z0 y0 Þ 0 0

6 7

m ¼6 7; °3:48Þ

°z y Þ1=2

4 5

0 0

1=2

°z y Þ

0 0

2 3

°z0 yÀ1 Þ1=2 0 0

0

16 7

¼6 7:

°z yÀ1 Þ1=2

zm ¼ °3:49Þ

0 0

ym 4 5

°z yÀ1 Þ 1=2

0 0

Alternative formulations, derived from the wave propagation equation, are available that

may present advantages in certain applications. The two obvious ones are the impedance

62 MODELLING OF CONVENTIONAL POWER PLANT

and the admittance representations:

! ! !

Z0 Z00

VS IS

¼ ; °3:50Þ

Z00 Z0

VR IR

! ! !

Y0 Y00

IS VS

¼ ; °3:51Þ

Y00 Y0

IR VR

where

9

Z0 ¼ Tv ‚ Diag°zm ‚ coth

m lÞ ‚ TÀ1 ; >

>

>

i

>

Z ¼ Tv ‚ Diag°zm ‚ csch

m lÞ ‚ T ; =

00 À1

i

°3:52Þ

Ti ‚ Diag°ym ‚ coth

m lÞ ‚ TÀ1 ; >

0

Y¼ >

>

>

v

À1 ;

Y00 ¼ ÀTi ‚ Diag°ym ‚ csch

m lÞ ‚ Tv :

3.2.9 Transmission Line Transpositions

High-voltage transmission lines may contain considerable geometric asymmetry, which in

turn causes voltage imbalances at the far end of the line, and transpositions are often used as

a means of balancing the overall impedances of the line (Anderson, 1973; Arrillaga et al.,

1986).

A three-phase transmission line, with a full set of transpositions, consists of three RLC

subsystems, as shown in Figure 3.9, where each section can be viewed as a p-circuit.

Alternatively, if each section is expressed in terms of its ABCD parameters then an

equivalent result can be obtained for the overall transmission line by cascading the

a c b

Section 1 Section 2 Section 3

b a c

c b a

a , b,c

Zshunt S1 Zshunt S2

a ,b,c

Zshunt S3

a , b,c

Yshunt S2 Yshunt S2

a,b,c a, b, c

a,b,c a, b, c a, b, c

Yshunt S1 Yshunt S1 Yshunt S3 Yshunt S3

a, b, c

2 2 2 2 2 2

Transposed transmission line representation and corresponding p sections

Figure 3.9

63

TRANSMISSION LINE MODELLING

individual elements (Anderson, 1973):

!9

! ! ! !

>

VS A1 B1 A2 B2 A3 B3 VR

;>

¼ ‚ ‚ >

=

ÀIR

IS C1 D1 C2 D2 C3 D3

°3:53Þ

! ! !

>

>

VS AB VR

>

¼ : ;

C D ÀIR

IS

Notice that the voltages and currents at the receiving end of section 1 are the voltages and

current at the sending end of section 2, and so on. Notice also in Figure 3.9 that the phase

conductors in sections 1, 2, and 3 occupy different positions in the tower, following the

sense of rotation (a, b, c), (c, a, b), and (b, c, a).

Cascading is also useful for calculating equivalent ABCD parameters of transmission

lines that contain not just transpositions but also shunt and series passive compensation. The

ABCD parameters of the series capacitive compensators are:

! ! !

VS 1Z VR

¼ ; °3:54Þ

0 1 ÀIR

IS

where 1 and 0 are the unit and zero matrices, and Z ¼ Diag°1=j!CÞ: Similarly, the ABCD

parameters of the shunt compensator are:

! ! !

VS 10 VR

¼ ; °3:55Þ

Y 1 ÀIR

IS

where Y ¼ Diag°j!C Þ and Y ¼ Diag°1=j!LÞ for capacitive and inductive shunt compen-

sation, respectively.

3.2.10 Transmission Line Program: Distributed Parameters

In Program 3.2, Program 3.1 is expanded to incorporate long-line effects and discontinuities

along the length of the line such as passive shunt and series compensation. The expansion

is neatly accommodated, leaving the code unchanged, and all added modelling functionality

is coded in two new functions, namely LongLineData and LongLine.

PROGRAM 3.2 Program written in Matlab1 to calculate transmission line parameters,

including long-line effects and passive shunt and series compensation

%***- - - - - Main Program

TransmissionLineData;

LongLineData;

[ZSeries,YShunt,Z012,Y012] = ShortLine(nphc,ngw,nb,bsep,resis,...

rdext,gmr,x,y,f,sigmag,vbase,sbase);

[ZPhase,YPhase] = LongLine(nphc,nsections,length,ZSeries,YShunt,...

ZSe,Ysh);

%End main Program

function [ZPhase,YPhase] = LongLine(nphc,nsect,length,ZSeries,...

YShunt,ZSe,YSh)

64 MODELLING OF CONVENTIONAL POWER PLANT

AUX = eye(nphc*2);

[TV,ZY] = eig(ZSeries*YShunt);

[TI,YZ] = eig(YShunt*ZSeries);

ZModal = inv(TV)*ZSeries*TI;

YModal = inv(TI)*YShunt*TV;

kk = 1;

for ll = 1: nsect

if ( length(ll) > 0 )

[ABCD] = ABCDLine(ll,nphc,length,ZModal,YModal,TV,TI);

else

[ABCD] = ABCDComp(kk,nphc,ZSe,YSh);

kk = kk + 1;

end

AUX = AUX*ABCD;

end

ABCD = AUX;

A = ABCD(1:nphc,1:nphc);

B = ABCD(1:nphc,nphc+1:nphc*2);

C = ABCD(nphc+1:nphc*2,1:nphc);

D = ABCD(nphc+1:nphc*2,nphc+1:nphc*2);

ZPhase(1:nphc,1:nphc) = A*inv(C);

ZPhase(1:nphc,nphc+1:nphc*2) = -B + A*inv(C)*D;

ZPhase(nphc+1:nphc*2,1:nphc) = inv(C);

ZPhase(nphc+1:nphc*2,nphc+1:nphc*2) = inv(C)*D;

YPhase = inv(ZPhase);

%End LongLine function

function [ABCD] = ABCDLine(ll,nphc,length,ZModal,YModal,TV,TI);

Modal = zeros(nphc,nphc);

for ii = 1: nphc

gamma = sqrt(ZModal(ii,ii)*YModal(ii,ii));

gammar = real(gamma*length(ll));

gammai = imag(gamma*length(ll));

fact1 = sinh(gammar);

fact2 = cosh(gammar);

fact3 = sin(gammai);

fact4 = cos(gammai);

Modal(ii,ii) = ((fact2*fact4)+(fact1*fact3)*i);

end

ABCD(1:nphc,1:nphc) = TV*Modal*inv(TV);

for ii = 1: nphc

gamma = sqrt(ZModal(ii,ii)*YModal(ii,ii));

gammar = real(gamma*length(ll));

gammai = imag(gamma*length(ll));

65

TRANSMISSION LINE MODELLING

fact1 = sinh(gammar);

fact2 = cosh(gammar);

fact3 = sin(gammai);

fact4 = cos(gammai);

Modal(ii,ii) = sqrt(ZModal(ii,ii)/YModal(ii,ii))*...

(fact1*fact4+fact2*fact3*i);

end

ABCD(1:nphc,nphc+1:nphc*2) = TV*Modal*inv(TI);

for ii = 1: nphc

gamma = sqrt(ZModal(ii,ii)*YModal(ii,ii));

gammar = real(gamma*length(ll));

gammai = imag(gamma*length(ll));

fact1 = sinh(gammar);

fact2 = cosh(gammar);

fact3 = sin(gammai);

fact4 = cos(gammai);

Modal(ii,ii) = sqrt(YModal(ii,ii)/ZModal(ii,ii))*...

(fact1*fact4+fact2*fact3*i);

end

ABCD(nphc+1:nphc*2,1:nphc) = TI*Modal*inv(TV);

for ii = 1: nphc

gamma = sqrt(ZModal(ii,ii)*YModal(ii,ii));

gammar = real(gamma*length(ll));

gammai = imag(gamma*length(ll));

fact1 = sinh(gammar);

fact2 = cosh(gammar);

fact3 = sin(gammai);

fact4 = cos(gammai);

Modal(ii,ii) = (fact2*fact4+fact1*fact3*i);

end

ABCD(nphc+1:nphc*2,nphc+1:nphc*2) = TI*Modal*inv(TI);

%End ABCDLine function

function [ABCD] = ABCDComp(kk,nphc,ZSe,YSh)

One = eye(nphc) ;

ABCD(1:nphc,1:nphc) = One ;

ABCD(1:nphc,nphc+1:nphc*2) = ZSe(:,:,kk) ;

ABCD(nphc+1:nphc*2,1:nphc) = YSh(:,:,kk) ;

ABCD(nphc+1:nphc*2,nphc+1:nphc*2) = YSh(:,:,kk)*ZSe(:,:,kk) + One ;

%End ABCDComp function

66 MODELLING OF CONVENTIONAL POWER PLANT

3.2.11 Numerical Example of Long Line Parameter Calculation

The parameters of the transmission line in Section 3.2.7 are calculated for the case when the

line is 500 km long. The Matlab1 function LongLine and the data given below, in

LongLineData, are used to such effect.

Function LongLineData, to read data for the 500 kV, three-phase transmission line of ¬‚at

con¬guration is as follows:

%transmission line of Example 1, to include long-line effects and passive

%shunt and series compensation. The line is 500 km long and contains no

%compensation.

%

%Transmission Line Data

%

%nsections = number of sections in the transmission line

%length = total length of transmission line

%

nsections = 1 ;

length(1) = 500 ;

%

%Compensating Plant Data

%

ZSe(:,:,1) = [ 0 0 0 ; 0 0 0 ; 0 0 0 ] ;

YSh(:,:,1) = [ 0 0 0 ; 0 0 0 ; 0 0 0 ] ;

%

%End of function LongLineData

In this example, the self and mutual admittances Y0 and Y00 of the transfer admittance

matrix of Equation (3.51), are:

2 3

1:6428 À j11:4850 À0:6708 þ j4:7380 À0:1371 þ j2:9077

6 7

Y0 ¼ 4 À0:6708 þ j4:7380 1:9417 À j12:7038 À0:6708 þ j4:7380 5p:u:;

À0:1371 þ j2:9077 À0:6708 þ j4:7380 1:6428 À j11:4850

2 3

À1:6336 þ j13:6479 0:6713 À j5:2450 0:1400 À j3:0994

6 7

Y00 ¼ 4 0:6713 À j5:2450 À1:9327 þ j14:9702 0:6713 À j5:2450 5p:u:

0:1400 À j3:0994 0:6713 À j5:2450 À1:6336 þ j13:6479

These parameters were calculated by using accurate expressions derived from the wave

propagation equation. An alternative, approximated, solution, involving the lumped

parameters calculated with Matlab1 function ShortLine and the nodal-based equations

1

Y0 ¼ °Zabc ‚ lengthÞÀ1 þ Yabc ‚ length;

2

and

Y00 ¼ À°Zabc ‚ lengthÞÀ1 ;

67

TRANSMISSION LINE MODELLING

is as follows:

2 3

1:6379 À j10:8189 À0:6711 þ j4:5699 À0:1387 þ j2:8400

6 7

Y0 ¼ 4 À0:6711 þ j4:5699 1:9369 À j12:0023 À0:6711 þ j4:5699 5p:u:;

À0:1387 þ j2:8400 À0:6711 þ j4:5699 1:6379 À j10:8189

2 3

À1:6379 þ j12:9184 0:6711 À j5:0759 0:1387 À j3:0306

Y00 ¼ 4 0:6711 À j5:0759 0:6711 À j5:0759 5p:u:

À1:9369 þ j14:2064

0:1387 À j3:0306 0:6711 À j5:0759 À1:6379 þ j12:9184

It is interesting to notice that even for this relatively long-distance transmission line, little

difference exists between the conductances of the accurate and the approximated solutions.

However, the absolute error in the susceptances is around 5 %.

3.2.12 Symmetrical Components and Sequence Domain Parameters

If Equation (3.1), or its more involved counterparts Equations (3.22) and (3.35), can be

assumed to be perfectly balanced then they can be replaced by the following impedance

matrix equation (Chen and Dillon, 1974):

2 32 32 3

ÁVa ZMM Ia

4 ÁVb 5 ¼ 4 M Z M 54 Ib 5: °3:56Þ

ÁVc MMZ Ic

Such a representation is easily transformed into the sequence domain frame of reference by

using the matrix of symmetrical components and its inverse:

2 3 2 3

111 111

1

Ts ¼ 4 1 h2 h 5 and TÀ1 ¼ 4 1 h h2 5; °3:57Þ

s

3

2 2

1hh 1h h

where h ¼ 1¬120 , and h2 ¼ 1¬240 .

Equation (3.56), written in compact notation, is subjected to the following treatment,

TÀ1 ÁVabc ¼ TÀ1 Zabc Ts TÀ1 Iabc : °3:58Þ

s s s

This yields the sequence domain representation of Equation (3.56),

ÁV012 ¼ Z012 I012 : °3:59Þ

The subscripts 0, 1, and 2 stand for zero, positive, and negative sequence components,

respectively. The following relationships exist between the terms in Equations (3.58) and

(3.59):

9

ÁV012 ¼ TÀ1 ÁVabc ; > =

s

À1

°3:60Þ

I012 ¼ Ts Iabc ;

>

;

Z012 ¼ TÀ1 Zabc Ts :

s

Furthermore, Equation (3.59), in expanded form, is

2 32 32 3

ÁV0 Z0 0 0 I0

4 ÁV1 5 ¼ 4 0 Z1 0 54 I1 5; °3:61Þ

ÁV2 0 0 Z2 I2

68 MODELLING OF CONVENTIONAL POWER PLANT

where

9

Z0 ¼ Z þ 2M; >

=

Z1 ¼ Z À M; °3:62Þ

>

;

Z2 ¼ Z À M:

This is a useful result that enables the calculation of the zero, positive, and negative

sequence impedances from known self and mutual impedances.

The reverse problem “ that where the self and mutual impedances of a perfectly balanced

transmission line are to be determined from known sequence impedances “ is of great

practical interest. Suitable equations can be derived from Equations (3.62):

9

1

Z ¼ °Z0 þ 2Z1 Þ; > =

3 °3:63Þ

>

1 ;

M ¼ °Z0 À Z1 Þ:

3

However, it should be remarked that if Equation (3.1) cannot be assumed to be perfectly

balanced then the use of symmetrical components does not yield a decoupled matrix

equation and the use of symmetrical components is of limited value.

To a limited extent this problem arises when the perfectly balanced counterpart of matrix

Equation (3.36) is represented in the sequence domain. If Equation (3.36) can be assumed to

be perfectly balanced then it is replaced by the following matrix equation:

2 32 32 3

ÁVa ZMMMMM Ia

6 ÁVb 7 6 M Z M M M M 76 Ib 7

6 76 76 7

6 ÁVc 7 6 M M Z M M M 76 Ic 7

6 76 76 7

6 ÁVA 7 ¼ 6 M M M Z M M 76 IA 7: °3:64Þ

6 76 76 7

4 ÁVB 5 4 M M M M Z M 54 IB 5

ÁVC MMMMMZ IC

Using compact notation to represent Equation (3.64), and applying symmetrical

components,

! ! ! ! ! À1 ! !

TÀ1 TÀ1

ÁVabc Z M Ts 0 Iabc

0 0 Ts 0

¼ ;

s s

TÀ1 ÁVABC TÀ1 M Z TÀ1 IABC

0 Ts

0 0 0

s s s

°3:65Þ

we obtain the following result:

! ! !

ÁV012 I012

Z012 M012

¼ : °3:66Þ

0 0

ÁV012 I012

M012 Z012

Equation (3.66) in expanded form is written as

2 32 32 3

ÁV0 Z þ 2M 0 0 3M 0 0 I0

6 ÁV1 7 6 7 6 I1 7

ZÀM

6 76 76 7

0 0 0 0 0

6 ÁV 7 6 7 6 I2 7

6 27 6 76 7

ZÀM

0 0 0 0 0

6 07¼6 76 0 7; °3:67Þ

6 ÁV 7 6 3M 7 6 I0 7

Z þ 2M

6 07 6 76 0 7

0 0 0 0

6 07 6 76 7

4 ÁV1 5 4 5 4 I1 5

ZÀM

0 0 0 0 0

0

0

ZÀM

ÁV2 I2

0 0 0 0 0

69

TRANSMISSION LINE MODELLING

where the sequence domain voltages and currents corresponding to circuit two are primed to

differentiate them from those in circuit one. Also, note the impedance coupling between the

two zero sequence circuits.

3.2.13 Transmission Line Program: Sequence Parameters

Computer Program 3.1 is expanded in Program 3.3 to incorporate sequence parameter

calculations. The new function, SequenceImpedance, is added to the code given in the

program. No additional input data are required.

PROGRAM 3.3 Program written in Matlab1 to calculate transmission-line sequence

parameters

%***- - - - - Main Program

TransmissionLineData;

[ZSeries,YShunt] = ShortLine(nphc,ngw,nb,bsep,resis,rdext,gmr,...

x,y,f, sigmag,vbase,sbase)

[Z012,Y012] = SequenceImpedance(ZSeries,YShunt);

%End main Program

function [Z012,Y012] = SequenceImpedance(ZSeries,YShunt)

TS(1,1) = 1;

TS(1,2) = 1;

TS(1,3) = 1;

TS(2,1) = 1;

TS(2,2) = -0.5-sqrt(3)*0.5*i;

TS(2,3) = -0.5+sqrt(3)*0.5*i;

TS(3,1) = 1;

TS(3,2) = -0.5+sqrt(3)*0.5*i;

TS(3,3) = -0.5-sqrt(3)*0.5*i;

ST = inv(TS);

Z012 = ST*ZSeries*TS;

Y012 = ST*YShunt*TS;

3.2.14 Numerical Example of Sequence Parameter Calculation

The positive, negative, and zero sequence parameters of the transmission line in Section

3.2.7 are:

2 3

0:0702 þ j0:4277 0:0050 À j0:0029 À0:0050 À j0:0029

6 7

¼ 10À3 ‚ 4 À0:0050 À j0:0029 0:0138 þ j0:1122 À0:0101 þ j0:0058 5p:u:;

Z012

0:0050 À j0:0029 0:0101 þ j0:0058 0:0138 þ j0:1122

70 MODELLING OF CONVENTIONAL POWER PLANT

2 3

À0:0002 þ j0:0001 0:0002 þ j0:0001

j0:0053

4 0:0002 þ j0:0001 0:0008 À j0:0005 5p:u:

¼ j0:0101

Y012

À0:0002 þ j0:0001 À0:0008 À j0:0005 j0:0101

As expected, Z012 and Y012 are diagonally dominant and the values on the diagonal

correspond to zero, positive, and negative sequence parameters, respectively. It is important

to notice that no full decoupling of the sequences is possible because of the inherent

geometric imbalances exhibited by the transmission line used in this test case; it is a