PSA LAB EX_1

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EXPERIMENT 1

COMPUTATION OF PARAMETERS AND MODELLING OFTRANSMISSION LINES

1.1 AIM

(i) To determine the positive sequence line parameters L and C per phase per kilometer othree phase single and double circuit transmission lines for different conducarrangements.

(ii) To understand modelling and performance of short, medium and long lines.

1.2 OBJECTIVESi. To become familiar with different arrangements of conductors of a three phase single

double circuit transmission lines and to compute the GMD and GMR for differarrangements.

ii. To compute the series inductance and shunt capacitance per phase, per km of a three phasingle and double circuit overhead transmission lines with solid and bundled conductors.

iii. To become familiar with per phase equivalent of a three phase short and medium lines anto evaluate the performances for different load conditions.

1.3 SOFTWARE REQUIRED

The software required isMatlab 7.5

1.4. THEORETICAL BACK GROUND

1.4.1 Inductance

The inductance is computed from flux linkage per ampere. In the case of the three phase lithe inductance of each phase is not the same if conductors are not spaced equilaterallydifferent inductance in each phase results in unbalanced circuit. Conductors are transposeorder to balance the inductance of the phases and the average inductance per phase is givesimple formulas, which depends on conductor configuration and conductor radius.

General FormulaThe general formula for computing inductance per phase in mH per km of a transmission isgiven by

10133PT605 - POWER SYSTEM & HIGH VOLTAGE LABORATORYANNA UNIVERSITY REGIONAL CENTRE MADURAI Page 1

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L = 0.2 lnD m/D s (1.1)


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whereDm = Geometric Mean Distance (GMD)Ds = Geometric Mean Radius (GMR)

The expression for GMR and GMD for different arrangement of conductors of the transmislines are given in the following section.

I. Single Phase - 2 Wire System

D

GMD = D(1.2)GMR = re-1/4= r (1.3)

r = radius of conductor

II. Three Phase - Symmetrical Spacing:

GMD = D(1.4)GMR = re-1/4 = r

(1.5)r = radius of conductor

III. Three Phase - Asymmetrical Transposed:

A


D

D D

. .

1-2

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DAB

DCA

B

C DBC


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GMD = Geometric mean of the three distances of the asymmetrically placed conductors

= (1.6)

GMR = re-1/4 = r (1.7)

r = radius of conductors

Composite Conductor Lines

Composite conductor is composed of two or more elements or strands electrically in paraThe expression derived for the inductance of composite conductors can be used for the stranand bundled conductors and also for finding GMD and GMR of parallel transmission lines1.4 shows a single phase line with two composite conductors.

Conductor X-with n strands Conductor Y with m

strands

The inductance of composite conductor X is given byLx = 0.2 ln GMD/GMRx (1.8)

where

GMD =(1.9)

GMR x =(1.10)


c b

a n

ba'

m'

c

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ra = r ae-1/4

The distance between elements are represented by D with respective subscripts and ra ,r b and rn have been replaced by Daa, D bb and Dnn respectively for symmetry.

Stranded Conductors:

The GMR for the stranded conductors are generally calculated using equation (1.10). the purpose of GMD calculation, the stranded conductors can be treated as solid conductorthe distance between any two conductors can be taken as equal to as center-to-center dista between the stranded conductors as shown in Fig 1.5, since the distance between the conduis high compared to the distance between the elements in a stranded conductor. This methosufficiently accurate.


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A

Bundle Conductors:

EHV lines are constructed with bundle conductors. Bundle conductors improves power transcapacity and reduces corona loss, radio interference and surge impedance.

The GMR of a bundle conductor is normally calculated using (1.10).GMR for two sub conductor Ds b = (Ds * d)1/2

GMR for three sub conductor Ds b = (Ds * d2)1/3

GMR for four sub conductor Ds b = 1.09 (Ds * d3)1/4

Where Ds is the GMR of each sub conductor and d is the bundle spacingFor the purpose of GMD calculation, the bundled conductor can be treated as a solid conducand the distance between any two conductors can be taken as equal to center-to-center distan


DAB

DBC

DAC B

C

d

d

dd dd

d

d

Fig. Examples of bundles

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between the bundled conductors as shown in Fig 1.7, since the distance between the conductis high compared to bundle spacing.

A B C


DAB DBC

DAC

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Three phase - Double circuit transposed:

A three-phase double circuit line consists of two identical three-phase circuits. The phases

and c are operated with a1-a2, b1-b2 and c1-c2 in parallel respectively. The GMD and GMR arecomputed considering that identical phase forms a composite conductor, For example, phaconductors a1 and a2form a composite conductor and similarly for other phases.

a1 S11c2

H12b1 S22 b2

H23S33 a2

c1

Relative phase position a1 b1c1 c2 b2a2.It can also be a1 b1c1 a2 b2c2.The inductance per phase in milli henries per km isL = 0.2 ln (GMD/GMR L) mH/km.

(1.11)whereGMR L is equivalent geometric mean radius and is given byGMR L = (DSADSB DSC)1/3 (1.12)


1-5

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whereDSA DSB and DSC are GMR of each phase group and given (refer 1.10) byDSA = 4( Ds b Da1a2)2 = [Ds b Da1a2]1/2

DSB = 4 (Ds b D b1b2)2 = [Ds b D b1b2]1/2 (1.13)DSC = 4 (Ds b Dc1c2)2 = [Ds b Dc1c2]1/2

whereDs b = GMR of bundled conductor if conductor a1, a2 . are bundle conductor.Ds b= r a1 = r b1 = r c1 = r a2 = r b2 = r c2 if a1, a2 .. are not bundled conductor.GMD is the equivalent GMD per phase & is given byGMD = [DAB DBC DCA]1/3 (1.14)


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whereDAB,DBC,& DCA are GMD between each phase group A-B, B-C, C-A which are given byDAB = [Da1b1Da1b2Da2b1Da2b2]1/4 (1.15)DBC = [D b1c1D b1c2D b2c1D b2c2]1/4 (1.16)DCA = [Dc1a1Dc2a1 Dc2a1Dc2a2]1/4 (1.17)

1.4.2 Capacitance

A general formula for evaluating capacitance per phase in micro farad per km of atransmission line is given by

C = 0.0556/ln (GMD/GMR) F/km (1.18)

whereGMD is the Geometric Mean Distance which is the same as that defined for inductance uvarious cases.

GMR is the Geometric Mean Radius and is defined case by casebelow:


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(i) Single phase two wires system (for diagram see inductance):

GMD = D

GMR = r (as against r in the case of L)

(ii) Three phase - symmetrical spacing (for diagram see inductance):GMD = DGMR = r in the case of solid conductor

=Ds in the case of stranded conductor to be obtained from manufacturers data.(iii) Three-phase Asymmetrical - transposed (for diagram see Inductance):

GMD = [DAB DBCDCA]1/3 (1.19)GMR = r ; for solid conductor GMR = Ds for stranded conductor

= r bfor bundled conductor wherer b = [r*d]1/2for 2 conductor bundler b = [r*d2]1/3 for 3 conductor bundle

(1.20)


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r b = 1.09 [r*d3]1/4 for 4 conductor bundlewhere

r = radius of each sub conductor d = bundle spacing

(iv) Three phase - Double circuit - transposed (for diagrams see inductance):

C = 0.0556 / ln (GMD/GMRc) F/kmGMD is the same as for inductance as equation (1.14).GMRc is the equivalent GMR, which is given byGMRc = [r Ar B r C ]1/3

(1.21)wherer A,r B and r C are GMR of each phase group obtained asr A = [r b Da1a2]1/2

r B = [r b D b1b2]1/2


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r C = [r b Dc1c2]1/2 (1.22)

where r b GMR of bundle conductor

1.4.3 Line Modeling and Performance Analysis

The following nomenclature is adopted in modelling: z = series impedance per unit length per phase

y = shunt admittance per unit length per phase to neutral.L = inductance per unit length per phaseC = capacitance per unit length per phaser = resistance per unit length per phasel = length of the lineZ = zl = total series impedanceY = yl = total shunt admittance per phase to neutral.

Short line Model and Equations (Lines Less than 80km)

The equivalent circuit of a short transmission line is shown in Fig.1.9

R X IR Is IR

Vs VR Vs VR Fig. Short Line Model Fig. Two port representation of a Line


A B C DConstants

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In this representation, the lumped resistance and inductance are used for modelling and the sadmittance is neglected. A transmission line may be represented by a two port network as shand current and voltage equations can be written in terms of generalised constants known as C D constants.

For the circuit shown above the voltage and currents relationships are given byVs = VR + Z IR

(1.23)Is = IR

(1.24)In terms of A B C D constants

VS A B VR (1.25)

IS = C D IR where


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A = 1, B = Z, C = 0 D= 0.|VR (NL)| - |VR (FL)|

Percentage regulation = --------------------------------------------x 100(1.26) |VR( FL)|

|VR (NL)| = |VS | / ATransmission efficiency of the line = Receiving end power in MW = PR (3) (1.27)Sending end power in MW Ps (3)

Medium Line Model and equations (Lines above 80km):

The shunt admittance is included in this model. The total shunt admittance is divided into twequal parts and placed at the sending and receiving end as in Fig.1.11

Z=R+jX IR Is

Vs Y/2 Y/2VR

Fig.1.11 Nominal Model


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The voltage current relations are given by

Vs = (1+ ZY ) VR + ZIR (1.28)

2Is = Y(1+ZY) VR + (1+ZY) IR (1.29)

4 2In terms of ABCD constants

Vs A B VR = (1.30)

Is C D IR

whereA = (1+ZY ), B=Z ;

2C = Y(1+ ZY) , D=(1+ZY )10133PT605 - POWER SYSTEM & HIGH VOLTAGE LABORATORYANNA UNIVERSITY REGIONAL CENTRE MADURAI Page 17

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4 2The receiving end quantities can be expressed in terms of sending end quantities as

VR D -B Vs (1.31)=

IR -C A IsLong line Model and Equations (lines above 250 km):

In the short and Medium lines, lumped line parameters are used in the model. For accumodelling, the effect of the distributed line parameter must be considered. The voltage current at any specific point along the line in terms of the distance x from the receiving engiven by

V(x) =( VR + ZcIR ) ex ( VR - ZcIR ) e-x2 2 (1.32)

I(x) = ((VR / Zc) + IR ) ex ((VR /ZC) - IR ) e-x(1.33)

2 2In term of Hyperbolic functionsV(x) = VR cosh x + ZcIR sinh x (1.34)I(x) = (1/Zc) VR sinh x + IR cosh x (1.35)where

Zc = z/y is called characteristic impedance = zy is called propagation constant

= + j = zy = ( + jL) (g + jc)


+

-

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is called attenuation constant is called phase constantThe relation between sending and receiving end quantities is given byVs = VR cosh l+ Zc IR sinh lIS = (VR /Zc) sinh l + IR cosh l (1.36)The equivalent model of the long line is given in Fig. 1.12.

Is IR Z = Z sinh l

l Y/2 Y/2 = (1/Zc) tanh (l/2)

Vs


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VR

1.5 RESULT:

(i) The line parameters L and C per phase per kilometer of a three phase single and doubcircuit transmission lines for different conductor arrangements were determined andverified using Mat lab.

(ii) The performance of transmission line was carried out using Mat lab.

1.6 EXERCISES:

1.6.1 A single circuit three phase transposed transmission line is composed of four ACSR 1,272,000 cmil conductor per phase with flat horizontal spacing of 14 m between phases a and b and between phases b and c. The bundle spacing is 45 cm. The

conductor diameter is 3.416 cm.a) Determine the inductance and capacitance per phase per kilometer of the line. b) Verify the results using available program.

1.6.2 A three - phase transposed line composed of one ACSR, 1,43,000 cmil, 4Bobolink conductor per phase with flat horizontal spacing of 11m between phases a and b a between phases b and c. The conductors have a diameter of 3.625 cm and a GMR


Fig. Equivalent model

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1.439 cm. The line is to be replaced by a three conductor bundle of ACSR 477,0cmil, 26/7 Hawk conductors having the same cross sectional area of aluminum as single-conductor line. The conductors have a diameter of 2.1793 cm and a GMR0.8839 cm. The new line will also have a flat horizontal configurations, but it is tooperated at a higher voltage and therefore the phase spacing is increased to 14m

measured from the centre of the bundles. The spacing between the conductors in bundle is 45 cm.

(a) Determine the inductance and capacitance per phase per kilometer of the above twolines.

(b) Verify the results using the available program.(c) Determine the percentage change in the inductance and capacitance in the bundle

conductor system. Which system is better and why?

1.6.3 A 345 kV double circuit three phase transposed line is composed of two ACSR,1,431,000 cmil, 45/7 bobolink conductors per phase with vertical conductor configuraas shown in Fig. 1.13. The conductors have a diameter of 1.427 in and the bundle spacis 18 in.

a) Find the inductance and capacitance per phase per kilometer of the line. b) Verify the results using the available program.

a 16m c

10m 18 b 24m b

9m

c17m a

1.6.4 .Determine the sending end voltage and current using nominal method. Thetransmission line is 100 km long and delivers 10 MW at 66 kV and 0.8 pf. lagging. Afind %regulation and efficiency.

Given that r=0.1 /km , Inductive reactance= 0.2 /km ;Shunt susceptance = 4 x 10-4 siemens.

1.6.5 Determine the receiving end voltage and current of a 3 phase,100 km long,50 Hz linehaving sending end quantities such as 75 MW, 66 kV , 0.99 leading power factor. Alsfind %regulation and efficiency.


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Given that r=0.1 /km , Inductive reactance= 0.2 /km ;Shunt susceptance = 4 x 10-4 siemens.

10133PT605 - POWER SYSTEM & HIGH VOLTAGE LABORATORY

PSA LAB EX_1

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