ee2404 manual

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EE 2404POWER SYSTEM SIMULATION

LABORATORYEEE SSN

LIST OF EXPERIMENTS

1 COMPUTATION OF TRANSMISSION LINE PARAMETERS

2 MODELLING OF TRANSMISSION LINES

3FORMATION OF BUS ADMITTANCE MATRIX BY THE

INSPECTION METHOD

4(A)FORMATION OF BUS IMPEDANCE MATRIX BY THE

INSPECTION METHOD

4(B)FORMATION OF BUS IMPEDANCE MATRIX BY BUS

BUILDING ALGORITHM

5 SOLUTION OF LOAD FLOW AND RELATED PROBLEMSUSING GAUSS - SEIDEL METHOD

6SOLUTION OF LOAD FLOW AND RELATED PROBLEMS

USING NEWTON RAPHSON METHOD

7SOLUTION OF LOAD FLOW AND RELATED PROBLEMS

USING FAST DECOUPLED METHOD

8 FAULT ANALYSIS

9(A)LOAD-FREQUENCY DYNAMICS OF SINGLE-AREA POWER

SYSTEMS

9(B)LOAD-FREQUENCY DYNAMICS OF TWO-AREA POWER

SYSTEMS

10(A)ECONOMIC DISPATCH IN POWER SYSTEMSWITHOUT

LOSS

10(B) ECONOMIC DISPATCH IN POWER SYSTEMSWITH LOSS

11TRANSIENT AND SMALL SIGNAL STABILITY ANALYSIS:

SINGLE- MACHINE INFINITE BUS SYSTEM

12TRANSIENT STABILITY ANALYSIS OF MULTIMACHINE

POWER SYSTEMS13 ELECTROMAGNETIC TRANSIENTS IN POWER SYSTEMS

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COMPUTATION OF TRANSMISSION LINE PARAMETERS

AIM

To determine the positive sequence line parameters inductance (L) and capacitance (C) per phase

per kilometer of a three phase - single and double circuit transmission lines for different conductor

arrangements.

OBJECTIVES

i.

To become familiar with different arrangements of conductors of a three phase single and double

circuit transmission lines and to compute the GMD and GMR for different arrangements.

ii.

To compute the series inductance and shunt capacitance per phase, per km of a three phase single

and double circuit overhead transmission lines with solid and bundled conductors.

SOFTWARE REQUIRED

MATLAB

THEORETICAL BACKGROUND

INDUCTANCE

The general formula for computing inductance per phase in mH per km of a transmission is given

byL = 0.2 ln (Dm/Ds) H/km (1.1)

Where,

Dm= Geometric Mean Distance (GMD)

Ds= Geometric Mean Radius (GMR)The expression for GMR and GMD for different arrangement of conductors of the transmission lines are

given in the following section.

I. Single Phase - 2 Wire System

Fig. 1.1. Conductor arrangement

GMD = D (1.2)

GMR = re-1/4= r (1.3)Where, r = radius of conductor

II. Three Phase - Symmetrical Spacing:

Fig. 1.2. Conductor Arrangement

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GMD = D (1.4)

GMR = re-1/4 = r (1.5)Where, r = radius of conductor

III. Three Phase - Asymmetrical Transposed:

Fig. 1.3. Conductor ArrangementGMD = Geometric mean of the three distances of the asymmetrically placed conductors

3AB BC CAD D D (1.6)

GMR = re-1/4 = r (1.7)Where, r = radius of conductors

Composite Conductor Lines

Fig. 1.4. Single Phase Line With Composite Conductor

The inductance of composite conductor X., is given by

Lx= 0.2 ln (GMD/GMRx) H/km (1.8)

where

' ' ' ' ' '( ...... ).....( ...... )mnaa ab am na nb nm

GMD D D D D D D (1.9)

2

( ...... ).....( ...... )n aa ab an na nb nnGMRx D D D D D D (1.10)

1' 4

a ar r e

The distance between elements are represented by D with respective subscripts and r a , rb and rn havebeen replaced by Daa, Dbb and Dnnrespectively for symmetry.

Stranded Conductors:

Fig. 1.5. Three Phase Line with Stranded Conductors

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Bundle Conductors:

The GMR of a bundle conductor is normally calculated using (1.10).

GMR for two subconductor Dsb

= Dsx d

GMR for three subconductor Dsb

= (Dsxd2)1/3

GMR for four subconductor Dsb

= 1.09 (Dsxd3)1/4

Where Dsis the GMR of each subconductor and d is the bundle spacing

Fig.1.7. Bundled conductor arrangement

IV - Three phase - Double circuit transposed:

Fig. 1.8. Conductor Arrangement

Relative phase position a1b1c1c2b2a2.

It can also be a1b1c1a2b2c2.

The inductance per phase in milli henries per km is

L = 0.2 ln (GMD/GMRL) mH/km. (1.11)

where

GMRLis equivalent geometric mean radius and is given by

GMRL = (DSA DSBDSC)1/3

(1.12)

where

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DSADSBand DSCare GMR of each phase group and given (refer 1.10) by

DSA=4(Ds

bDa1a2)2 = [Ds

bDa1a2]1/2

DSB=4(Ds

bDb1b2)

2 = [Ds

bDb1b2]

1/2 (1.13)

DSC=4(Ds

bDc1c2)2 = [Ds

bDc1c2]1/2

where

Dsb= GMR of bundled conductor if conductor a1, a2. are bundle conductor.

Dsb= ra1 = rb1 = rc1 = ra

2= rb

2 = rc

2if a1, a2.. are not bundled conductor.

GMD is the equivalent GMD per phase & is given by

GMD = [DABDBCDCA]1/3 (1.14)

where

DAB, DBC, & DCAare GMD between each phase group A-B, B-C, C-A which are given by

DAB= [Da1b1 Da1b2 Da2b1 Da2b2]1/4 (1.15)

DBC= [Db1c1 Db1c2 Db2c1 Db2c2]1/4 (1.16)

DCA= [Dc1a1 Dc2a1 Dc2a1 Dc2a2]1/4 (1.17)

CAPACITANCE

A general formula for evaluating capacitance per phase in micro farad per km of a transmission

line is given by

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

Where, GMD is the Geometric Mean Distance which is the same as that defined for inductance undervarious cases.

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

(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 = D

GMR = r in the case of solid conductor

=Dsin the case of stranded conductor to be obtained from manufacturers data.

(iii) Three-phaseAsymmetrical - transposed (for diagram see Inductance):

GMD = [DABDBC DCA]1/3

(1.19)

GMR = r ; for solid conductor

GMR = Dsfor stranded conductor

= rb for bundled conductor

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where

rb= [r*d]1/2 for 2 conductor bundle

rb= [r*d

2]

1/3for 3 conductor bundle (1.20)

rb= 1.09 [r*d3]1/4for 4 conductor bundle

where

r = radius of each subconductor

d = bundle spacing

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

C = 0.0556 / ln (GMD/GMRc) F/km

GMD is the same as for inductance as equation (1.14).

GMRc is the equivalent GMR, which is given by

GMRc = [rA rBrC]1/3

(1.21)

where

rA, rBand rCare GMR of each phase group obtained as

rA= [rbDa1a2]

1/2

rB= [rbDb1b2]

1/2

rC= [rbDc1c2]

1/2 (1.22)

where rbGMR of bundle conductor

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EXERCISES:

1. A three-phase transposed line composed of one ACSR, 1,43,000 cmil, 47/7 Bobolink

conductor per phase with flat horizontal spacing of 11m between phases a and b and

between phases b and c. The conductors have a diameter of 3.625 cm and a GMR of 1.439

cm. The line is to be replaced by a three conductor bundle of ACSR 477,000-cmil, 26/7

Hawk conductors having the same cross sectional area of aluminum as the single-conductor line. The conductors have a diameter of 2.1793 cm and a GMR of 0.8839 cm.

The new line will also have flat horizontal configurations, but it is to be operated at a

higher voltage and therefore the phase spacing is increased to 14m as measured from the

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

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

lines.

(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?

2.

A single circuit three phase transposed transmission line is composed of four ACSR1,272,000 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 3.16 cm.

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

(b)Verify the results using available program.

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 configuration as shown

in Fig. The conductors have a diameter of 1.427 in and the bundle spacing is 18 in.(a)Find the inductance and capacitance per phase per kilometer of the line.

(b)Verify the results using the available program.

(c) If we change the relative phase position to abc - abc, determine the inductance andcapacitance per unit length using available program.

(d)Which relative phase position is better and why?

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MODELLING OF TRANSMISSION LINES

AIM

To understand the modelling and performance of short, medium and long transmission lines.

OBJECTIVES

i.

To become familiar with per phase equivalent of a three phase short and medium lines and to

evaluate the performances for different load conditions.

ii. (a) To become familiar with the theory of long transmission line and study the effect of distributed

parameters on voltage and currents, along the line, (b) calculate the surge impedance and surge

impedance loading.

SOFTWARE REQUIRED

MATLAB


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 phase

C = capacitance per unit length per phase

r = resistance per unit length per phase

l = length of the line

Z = 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.2.1

In this representation, the lumped resistance and inductance are used for modelling and the shunt

admittance is neglected. A transmission line may be represented by a two port network as shown in Fig 2.2

and current and voltage equations can be written in terms of generalised constants known as A B C D

constants. For the circuit in Fig.2.1 the voltage and currents relationships are given by

Vs= VR+ Z IR (2.1)

Is= IR (2.2)

In terms of A B C D constants

S R

S R

V VA B

I IC D

(2.3)

Where, A = 1, B = Z, C = 0 D= 0

( ) ( )

( )

100R NL R FL

R FL

V VPercentageregulation

V

(2.4)

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Transmission efficiency of the line(3 )

(3 )

R

S

P

P

(2.5)

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

The shunt admittance is included in this model. The total shunt admittance is divided into two equal

parts and placed at the sending and receiving end as in Fig.2.3

The voltage current relations are given by

12

S R R

ZYV V ZI

(2.6)

1 14 2

S R R

ZY ZYI Y V I

(2.7)

In terms of ABCD constants

S R

S R

V VA B

I IC D

(2.8)

where 1 ; ; 1 ; 12 4 2

ZY ZY ZYA B Z C Y D

Long line Model and Equations (lines above 250 km):

In the short and medium lines, lumped line parameters are used in the model. For accurate modelling,

the effect of the distributed line parameter must be considered. The voltage and current at any specific

point along the line in terms of the distance x from the receiving end is given by

( )2 2

x xR C R R C RV Z I V Z I

V x e e

(2.10)

( )

2 2

R C R R C Rx xV Z I V Z I

I x e e

(2.11)

In term of Hyperbolic functions

V(x) = VRcosh x + ZcIRsinh x (2.12)I(x) = (1/Zc) VRsinh x + IRcosh x (2.13)where

cZ z / y is called characteristic impedance

zy is called propagation constant

j zy j L g j c

is called attenuation constant is called phase constantThe relation between sending and receiving end quantities is given by

Vs= VRcosh l+ ZcIRsinh lIS = (VR/Zc) sinh l + IRcosh l (2.14)

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The equivalent model of the long line is given in Fig. 2.4.

Lossless Transmission Line

For lossless transmission line, the equations for the rms voltage and currents along the line is given by

V(x) = VRcos x + jZcIRsinx (2.15)

I(x) = j 1/ cZ VRsinx + IRcosx (2.16)

For open circuited line IR= 0 and the no load receiving end voltage is given by

VR(nl) = Vs/ cos l (2.17)For solid short circuit at the receiving end VR= 0, the equation (2.15) and (2.16) reduces to

Vs= jZc IRsinlIs= IRcosl

For a lossless line the surge impedance (SIL) = L C

The load corresponding to the surge impedance at rated voltage is known as surge impedance loading

(SIL) given by

SIL = 3 VRIR* (2.18)

= 3 |VR|2/ Zcfor lossless line Zcis purely resistive (2.19)

SIL =2 (2.20)

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EXERCISES:

1.

A 230 kV, 50 HZ three phase transmissions is 150 km long. The per phase resistance is 0.13 per kmand the reactance is 0.5 per km and the shunt admittance is 3.30 x 10 -690oSiemens per km. Itdelivers 40MW at 220 KV with 0.85 power factor lagging. Use medium line model.(i)

Determine the voltage and current at sending end and also compute voltage regulation and

efficiency.

(ii) Verify the results using the available program

2.

A three phase transmission line has a per phase series impedance of z = 0.03 + j0.4 per km and a perphase shunt admittance of y=j4.0 x 10-6 Siemens per km. The line is 200 km long. Obtain ABCD

parameters of the transmission line. The line is sending 407 MW and 7.833 MVAR at 350 kV. Use

medium model.(i)

Determine the voltage and current at receiving end and also compute voltage regulation and

efficiency.

(ii) Verify the results using the available program

3.

A three phase 50 Hz, 400 kV transmission line is 250 km long. The line parameters per phase per unit

length are found to be

r = 0.032 /kmL = 1.06mH/kmC = 0.011F/kmDetermine the following using the program available use long line model.

(a)

The sending end voltage, current and efficiency when the load at the receiving end is 640 MW at

0.8 power factor logging at 400 kV.

(b)

The receiving end voltage, current, efficiency and losses when 480 MW and 320 MVAR are being

transmitted at 400 kV from the sending end.

(c) The sending end voltage, current and efficiency and losses when the receiving end load impedance

is 230 at 400 kV.(d)

The receiving end voltage when the line is open circuited and is energized with 400 kV at the

sending end. Also, determine the reactance and MVAR of a three phase shunt reactor to be

installed at the receiving end in order to limit the no load receiving end voltage to 400 kV.

(e)

The MVAR and capacitance to be installed at the receiving end for the loading condition in (a) to

keep the receiving end voltage at 400 kV when the line is energized with 400 kV at the sending

end.

(f)

The line voltage profile along the line for the following cases: no load, rated load of 800 MW at

0.8 power factor at sending end at 400 kV, line terminated in the SIL and short circuited line.

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FORMATION OF BUS ADMITTANCE MATRIX BY

THE INSPECTION METHOD

AIM

To understand the formation the bus admittance matrix [Ybus] of a power network by using the

method of inspection, to effect certain required changes on this matrix and to obtain network solution

using this matrix.

OBJECTIVES

i. To write a program to form bus admittance matrix Y,given the impedances of the

elements of a power network and their connectivity (mutual coupling between

elements neglected)

ii. To modify the matrix Y to effect specified changes in the configuration of the

network.

SOFTWARE REQUIRED

MATLAB


The meeting point of various components in a power system is called bus. The bus or bus bar is a

conductor made of Cu or Al having negligible resistance. Hence the bus bar will have zero voltage drop

when it conducts the rated current. Therefore the buses are considered as points of constant voltage in a

power system. When the power system is represented by impedance /reactance diagram it can be

considered as a circuit or network. The buses can be treated as nodes and the voltages of all buses (nodes)can be solved by conventional node analysis technique.

Let Nbe the number of major or principal nodes in the circuit or network. Since the voltages ofnode can be measured only with respect to reference point one of node is considered as reference node, the

network will have (N1) independent voltages. In nodal analysis, the independent voltages are solved bywriting Kirchhoffs current law (KCL) equations. While writing KCL equations for (N1) nodes in thecircuit, the voltage sources in the circuit should be converted to equivalent current sources.

Let V1,V2..Vn be the node voltages of nodes 1, 2, 3,..n respectively. I 11,I22 ,I33 ..Innbe thesum of current sources connected to nodes 1, 2, 3..n respectively.

Let, Yii = Sum of admittances connected to node.

Yij = Negative of the sum of admittance connected between node i and node j.Now the n number of nodal equations for n bus system will be in the form shownbelow

Y11V1+Y12V2+Y13V3+..Y1nVn= I11Y21V1+Y22V2+Y23V3+...Y2nVn=I22

Yn1V1+Yn2V2+Yn3V3+..YnnVn= InnThe above n number of equations can be arranged in the matrix form as shown in below.

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(3.1) = (3.2)Where, Ibus is the vector of injected bus current (i.e., external current sources). The current is

positive when flowing towards the bus and it is negative if flowing away from the bus.

Vbusis the vector of bus voltages measured from the reference node (i.e., node voltages).

Ybusis the bus admittance matrix.

The diagonal element of each node is the sum of admittance connected to it. It is known as the

self-admittance or driving point admittance, i.e.,

= = , i = 1,2,3n (3.3)The off-diagonal element is equal to the negative of all the admittance connected between the

node i and node j. It is known as mutual admittance or transfer admittance, i.e., = = ,i, j = 1,2,3n, ij (3.4)When the bus currents are known, equation (3.2) can be solved for the n bus voltages.

Vbus= Y-1

bus Ibus (3.5)

The admittance matrix obtained with one of the buses as reference is non-singular. Otherwise the

nodal matrix is singular.

Inspection of the bus admittance matrix reveals that the matrix is symmetric along the leading

diagonal, and we need to store the upper triangular nodal admittance matrix only. In a typical power

system network, each bus is connected to only a few nearby buses. Consequently, many off-diagonal

elements are zero. Sucha matrix is called sparse, and efficient numerical techniques can be applied to

compute its inverse. By means of an appropriately ordered triangular decomposition, the inverse of a

sparse matrix can be expressed as a product of sparse matrix factors, thereby giving an advantage in

computational speed, storage and reduction ofround-off errors.

Note: If the transformer is present in the network, get the values for the transformer line

impendence and also off nominal turn ratio.

Addition of line:

Let Yb is the value of admittance added between the node i and node j. This affects the fourelements (i.e., , , , )of the existing Ybus. Thus, the new values for these four elements aredetermined using the following formulas, = + (3.6) = (3.7) = (3.8) = + (3.9)Removal of line:

Let Yb is the value of admittance removed between the node i and node j. This affects the fourelements (i.e., , , , )of the existing Ybus. Thus, the new values for these four elements aredetermined using the following formulas, = (3.10) = + (3.11)

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= + (3.12) = (3.13)Addition of Shunt element:

Let Ysc is the value of admittance added between the node i and node j. This affects only oneelement (i.e.,

) of the existing Ybus . Thus, the new value of this element is determined using the

following formula, = + (3.14)Removal of node (or bus):

Let us consider that kthnode of the existing network can be eliminated. Therefore, the new busadmittance matrix can be determined using the following formula,

= , i, j =1, 2, 3,.,n (3.15)ALGORITHM

Step 1: Get the total number of buses n and number of lines l.Step 2: Get the values of line impedances say from bus i to j for all buses in the network.Step 3: Get the line charging admittance value given in the network.

Step 4: Find the diagonal and off diagonal elements of the bus admittance matrix using

following formulas. = = , i = 1,2,3n = = ,i, j = 1,2,3n, ijWhere, =Diagonal element of the bus admittance matrix.

=Off - diagonal element of the bus admittance matrix.

=Admittance of the line connected between the buses i and jStep 5: Form the bus admittance matrix [Ybus] = (3.16)

Step 6: Print the resultant Ybusmatrix.

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EXERCISES:

1. The parameters of a 4 bus system are as follows. Draw the network and find the Ybus

matrix.

Bus Code Impedance(pu) Line Charging Admittance(pu)1-2 0.2+j0.8 j0.02

2-3 0.3+j0.9 j0.03

2-4 0.25+j1 j0.04

3-4 0.2+j0.8 j0.02

1-3 0.1+j0.4 j0.01

2. Construct the bus admittance matrix for the given network by using inspection method.

3. Determine bus admittance matrix for the network shown in fig.

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FORMATION OF BUS IMPEDANCE MATRIX

BY THE INSPECTION METHOD

AIM

To understand the formation the bus impedance matrix [Zbus] of a power network by usingthe method of inspection, to effect certain required changes on this matrix and to obtain network

solution using this matrix.

OBJECTIVES

i. To write a program to form bus impedance matrix Zbus(by inspection method), given

the impedances of the elements of a power network and their connectivity (mutual

coupling between elements neglected)

SOFTWARE REQUIRED

MATLAB


Zbusmatrix is an important matrix used in different kinds of power system study such as

short circuit study, load flow study etc. We know, from equation (3.2),

= and = = = (4.A.1)Where,Ibus is the vector of injected bus current (i.e., external current sources). The current is

positive when flowing towards the bus and it is negative if flowing away from the bus.

Vbusis the vector of bus voltages measured from the reference node (i.e., node voltages).

Ybusis the bus admittance matrix.is the positive sequence bus impedance matrix.Since Ybusis symmetrical around the principal diagonal, Zbusmust also be symmetrical.

The impedance elements of Zbuson the principal diagonal are called driving-point impedances of

the buses, and the off-diagonal elements are called the transfer impedances of the buses.

ALGORITHM

Step 1:Get the total number of buses n and number of lines l.Step 2: Get the values of line impedances say from bus i to j for all buses in the

network.

Step 3:Get the line charging admittance value given in the network.Step 4:Find the diagonal and off diagonal elements of the bus admittance matrix using

following formulas. = = , i = 1,2,3nand = = ,i, j = 1,2,3n, ijWhere, =Diagonal and off diagonal element of the bus admittancematrix. =Admittance of the line connected between the buses i and j

Step 5:Form the bus admittance matrix [Ybus]

=

Step 6:Find the bus impedance matrix [Zbus] using the relation, = Step 7: Print the resultant Zbusmatrix.

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EXERCISES:

1. The parameters of a 4 bus system are as follows. Draw the network and find the Z bus

matrix.

Bus Code Impedance(pu) Line Charging Admittance(pu)1-2 0.2+j0.8 j0.02

2-3 0.3+j0.9 j0.03

2-4 0.25+j1 j0.04

3-4 0.2+j0.8 j0.02

1-3 0.1+j0.4 j0.01

2. Construct the bus impedance matrix for the given network by using inspection

method.

3. Determine bus impedance matrix for the network shown in fig., by the method of

inspection.

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FORMATION OF BUS IMPEDANCE MATRIX BY

BUS BUILDING ALGORITHM

AIM

To understand the formation the bus impedance matrix Zbusof a power network using bus buildingalgorithm, to effect certain required changes on this matrix and to obtain network solution using this

matrix.

OBJECTIVES

i. To write a program to form bus impedance matrix Zbus(by bus building algorithm),

given the impedances of the elements of a power network and their connectivity

(mutual coupling between elements neglected)

ii. To modify the matrix Zbus to effect specified changes in the configuration of the

network.

SOFTWARE REQUIRED

MATLAB


Zbus building algorithm is a step-by-step procedure which proceeds branch by branch.

Main advantage of this that, any modification of the network elements does not require complete

rebuilding of Zbusmatrix.

Case 1: Adding Zbfrom a new bus p to the reference node

The addition of the new bus p connected to the reference node through Zb without aconnection to any of the buses of the original network cannot alter the original bus voltages when

a current Ipis injected at the new bus.

The voltage Vpat the new bus is equal to IpZb, then ,

(4.B.1)

We note that the column vector of currents multiplied by the new Zbuswill not alter the

voltages of the original network and will result in the correct voltage at the new bus p.

Case 2: Adding Zbfrom a new bus p to an existing bus k

The addition of a new bus p connected through Zbto an existing bus k with Ipinjectedat bus p will cause the current entering the original network at bus k to become the sum of I k

injected at bus k plus the current Ipcoming through Zb, as shown in fig.(4.B.2).

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The current Ipflowing into the network at bus k will increase the original voltage 0bythe voltage IpZkk, i.e., Vk= 0+ IpZkk (4.B.2)and Vpwill be larger than the new Vkby the voltage IpZb . So,

Vp= 0+ IpZkk +IpZb (4.B.3)And substituting for 0, we obtain

Vp= I1Zk1 + I2Zk2 + I3Zk3 +..+INZkN + Ip(Zkk +Zb ) (4.B.4)Since Zbus must be a square matrix around the principal diagonal, we must add a new

column which is the transpose of the new row. The new column accounts for the increase of all

bus voltages due to Ip. The matrix equation is,

(4.B.5)Note that the first N elements of the new row are the elements of row k of Zorigand the first Nelements of the new column are the elements of column k of Zorig.

Case 3: Adding Zb from existing bus k to the reference node

To see how to alter Zorigby connecting an impedance Zb from an existing bus k to thereference node, we add a new bus pconnected through Zbto bus k. Then, we short-circuit bus

pto the reference nodeby letting Vpequal zero to yield the same matrix equation as equation

(4.B.5) except that Vpis zero. So, for the modification we proceed to create a new row and new

column exactly the same as in Case 2, but we then eliminate the ( N + 1 ) row and ( N + 1 )column b y Kron reduction, which is possible because of the zero in the column matrix of

voltages. The each element of Zij(new)

in the new matrix is computed using the formula,

= (4.B.6)Case 4: Adding Zbbetween two existing buses j and k

To add branch impedance Zb between buses j and k already established in Zorig, weexamine figure 4.B.2., which shows these buses extracted from the original network.

Fig.4.B.1. Addition of new bus p connected

through impedance Zb to existing bus k

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The change in voltage at each bus h caused by the injection Ib at bus j and - Ib at buskis given by, Vh= (ZhjZhk) Ib (4.B.7)

Which means that the vector Vof bus voltage changes is found by subtracting columnk from column j of Zorigand by multiplying the result by Ib .Let, V1= 10+V1 (4.B.8)

V1= I1Z11 + I2Z12 + I3Z13 +.++ IjZ1j + IkZ1k ..+ INZ1N + Ib(Z1jZ1k ) (4.B.9)

Similarly, at buses j and k,Vj= I1Zj1 + I2Zj2 + I3Zj3 +.++ IjZjj + IkZjk...+ INZjN + Ib(ZjjZjk ) (4.B.10)Vk= I1Zk1 + I2Zk2 +.++ IjZkj + IkZkk ..+ INZkN + Ib(ZkjZkk ) (4.B.11)and, 0 = 0- 0+ (Zth,jk+ Zb)Ib (4.B.12)

From equations (4.B.10), (4.B.11) and (4.B.12), we get

(4.B.13)

In matrix form,

(4.B.14)In which the coefficient of Ibin the last row is denoted by

Zbb = Zth,jk +Zb = Zjj + Zkk - 2Zjk + Zb (4.B.15)

The new column is column j minus column k of Zorigwith Zbbin the (N+1) row. Thenew row is the transpose of the new column. Eliminating the (N+1) row and (N+1) column of the

square matrix of equation (4.B.16) using the Kron reduction technique using the formula given by

equation (4.B.6). = REMOVING A BRANCH

A single branch of impedance Zbbetween two nodes can be removed from the network by

adding the negative of Zbbetween the same terminating nodes. The reason is, of course, that the

parallel combination of the existing branch (Zb) and the added branch (-Zb) amounts to an

effective open circuit.

Fig.4.B.2. Addition of impedance

Zb

between existing buses j andk

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EXERCISES:

1. Construct the bus impedance matrix for the given network by using bus building

algorithm.

2. Using the method of building algorithm, find the bus impedance matrix for the network

shown in fig.

3. The bus impedance matrix for the network shown in fig is given by

= 0.300 0.200 0.2750.200 0.400 0.2500.275 0.250 0.41875

There is a line outage and the line from bus 1 to

2 is removed. Using the method of building

algorithm, determine the new bus impedance

matrix.

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SOLUTION OF LOAD FLOW AND RELATED PROBLEMS

USING GAUSS-SEIDEL METHOD

AIM

(i)

To understand, the basic aspects of steady state analysis of power systems that is requiredfor effective planning and operation of power systems.

(ii)To understand, in particular, the mathematical formulation of load flow model in complex

form and a simple method of solving load flow problems of small sized system using

Gauss-Seidel iterative algorithm

OBJECTIVES

i. To write a computer program to solve the set of non-linear load flow equations using

Gauss-Seidel Load Flow (GSLF) algorithm and present the results in the format required

for system studies.

ii.

To investigate the convergence characteristics of GSLF algorithm for normally loaded

small system for different acceleration factors.

SOFTWARE REQUIRED

MATLAB


Need For Load Flow Analysis

Load Flow analysis, is the most frequently performed system study by electric utilities.

This analysis is performed on a symmetrical steady-state operating condition of a power systemunder normal mode of operation and aims at obtaining bus voltages and line / transformer flowsfor a given load condition. This information is essential both for long term planning and next day

operational planning. In long term planning, load flow analysis, helps in investigating the

effectiveness of alternative plans and choosing the best plan for system expansion to meet theprojected operating state. In operational planning, it helps in choosing the best unitcommitment plan and generation schedules to run the system efficiently for the next days loadcondition without violating the bus voltage and line flow operating limits.

Description of Load Flow Problem

In the load flow analysis, the system is considered to be operating under steady state

balanced condition and per phase analysis is used.

The network consists of a number of buses (nodes) representing either generating stations

or bulk power substations, switching stations interconnected by means of transmission lines or

power transformers. The bus generation and demand are characterized by complex powers

flowing into and out of the buses respectively. Each transmission line is characterized by its equivalent circuit. The transformer with off-nominal tap ratio is characterized by their equivalent circuit. Shunt compensating capacitors or reactors is represented as shunt susceptance.

Load Flow analysis is essentially concerned with the determination of complex bus

voltages at all buses, given the network configuration and the bus demands. Let the given system

demand (sum of all the bus demands) be met by a specific generation schedule. A generation

schedule is nothing but a combination of MW generation (chosen within their ratings) of the

various spinning generators the total of which should match the given system demand plus the

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transmission losses. It should be noted that there are many generation schedules available to

match the given system demand and one such schedule is chosen for load flow analysis.

The Ideal Load Flow problem is stated as follows:Given: The network configuration (bus admittance matrix), and all the bus power injections (bus

injection refers to bus generation minus bus demand).

To determine: The complex voltages at all the buses.

The steady state of the system is given by the state vector X defined as

X = (12 N V1V2 .VN)T= (

TV

T)T

Once the state of the system is known, all the other quantities of interest in the powernetwork can be computed.

The above statement of Load Flow problem will be modified later after taking into

account certain practical constraints.

Development of Load FlowModel

The Load Flow model in complex form is obtained by writing one complex power

matching equation at each bus.

Referring to Fig 5.1 (b) the complex power injection (generation minus demand) at the kth

bus is equal to the complex power flowing into the network at that bus which is given by

PIk+ jQIk= Pk+ jQk (5.1)

In expanded form

(PGk- PDk) + j (QGk- QDk) = VkIk* (5.2)

The network equation relating bus voltage vector Vwith bus current vector I is

YV = I (5.3)

Taking the kth component of I from (5.3) and substituting for Ik* in (5.2) we get thepower flow model in complex form as

(5.4)

In (5.4) there are N complex variable equations from which the N unknown complex

variables, V1,VNcan be determined.

Classification of Buses

From the Load Flow model in equation (5.4) and from the definition of complex busvoltage, Vkas

Vk= |Vk| k (5.5)one can observe that there are four variables, P I, QI, and |V| associated with each bus. Any twoof these four may be treated as independent variables (that is specified) while the other two may

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be computed by solving power flow equations. The buses are classified based on the variables

specified. Three types of buses classified based on practical requirements are given below:

Slack bus: While specifying a generation schedule for a given system demand, one can

fix up the generation setting of all the generation buses except one bus because of the limitation

of not knowing the transmission loss in advance. This leaves us with the only alternative of

specifying two variables s and |Vs| pertaining to a generator bus (usually a large capacitygeneration bus is chosen and this is called as slack bus) and solving for the remaining (N-1)

complex bus voltages from the respective (N-1) complex load flow equations. Incidentally the

specification of |Vs| helps us to fix the voltage level of the system and the specification of saszero, makes Vsas reference phasor. Thus for the slack bus, both and |V| are specified and PGand QG are to be computed only after the iterative solution of bus voltages is completed.

P-V buses: In order to maintain a good voltage profile over the system, it is customary to

maintain the bus voltage magnitude of each of the generator buses at a desired level. This can be

achieved in practice by proper Automatic Voltage Regulator (AVR) settings. These generatorbuses and other Voltage-controlled buses with controllable reactive power source such as SVC

buses are classified as P-V buses since PG and |V| are specified at these buses. Only one state

variable, is to be computed at this bus. The reactive power generation QG at this bus which is adependent variable is also to be computed to check whether it lies within its operating limits.

P-Q buses: All other buses where both PIand QIare specified are termed as P-Q buses

and at these buses both and |V| are to be computed.

Hence the Practical Load Flow problem maybe stated as:

Given:The network configuration (bus admittance matrix), all the complex bus power demands,

MW generation schedules and voltage magnitudes of all the P-V buses, and voltage magnitude of

the slack bus,

To determine: The bus voltage phase angles of all buses except the slack bus and bus voltage

magnitudes of all the P-Q buses.

Hence the state vector to be solved from the Load Flow model is

X = (12NPV1V2..VNQ)T

(5.6)

Where, NP = N-1

NQ = N-NV1and the NV number of P-V buses and the slack bus are arranged at the end.

Solution to Load FlowProblem

A number of methods are available for solving Load Flow problem. In all these methods,

voltage solution is initially assumed and then improved upon using some iterative process

until convergence is reached. The following three methods will be presented:

(i) Gauss-Seidel Load Flow (GSLF) method

(ii) Newton-Raphson Load Flow (NRLF) method

(iii)

Fast Decoupled Load Flow (FDLF) methodThe first method GSLF is a simple method to program but the voltage solution is updated

only node by node and hence the convergence rate is poor. The NRLF and FDLF methods update

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the voltage solution of all the buses simultaneously in each iteration and hence have faster

convergence rate.

Taking the complex conjugate of equation (5.4) and transferring Vkto the left hand side,

we obtain

(5.7)

Where, k = 1,2, ..(N-1) (Slack bus excluded)Define Ak= (PIkjQIk) / Ykk (5.8)

Bkm= Ykm/ Ykk (5.9)

The voltage equation to be solved during the hth

iteration of G.S method is obtained from (5.7),

(5.8) and (5.9) as

(5.10)

GaussSeidel Load Flow Algorithm

The algorithm for GSLF is given in the flow chart Fig 5.3

Convergence Check:

Referring to Flow chart Fig 5.3, during every iteration h, the maximum change in bus voltage

that has occurred is stored in VMAXas given below

(5.11)

where Vk(h+1)

ek(h+1)

+ j fk(h+1)

= Vk(h+1)

- Vk(h)

The convergence is checked by comparing VMAX with the specified tolerance .

Additional Computation for P-V Bus

The flow chart in Fig.5.3 does not have provision for voltage controlled buses.However, if the link between X and Y in Fig.5.3 is removed and the P-V bus module in Fig.5.4 is

introduced, then P-V buses can be handled.

Referring to Fig.5.3 and Fig.5.4, for each P-V bus during the hth

iteration, before updating

bus voltage, the following computations are made:

Step 1: Adjusting the complex voltage Vk(h) ek

(h)+j fk

(h)to correct the voltage magnitude to the

scheduled value, |Vk|schas follows:

k(h)

= arc tan(fk(h)

/ ek(h)

) (5.12)

Vk(new)(h)

= | Vk|schej

k(h)

(5.13)

Step 2:Compute the reactive power generation using the Vk(new)(h)

as

QG k(h)

= QDk+ Qk(h)

(V(h)

) (5.14)

If the inequality QGkmin QGk

(h) QGk

maxis satisfied, then Vk

(h)is set as V

(h)k(new). Go to

step 3.

If QGk(h)

> QGkmax

, then set QGk(h)

= QGkmax

, go to step 3. (5.15)

If QGk(h)

< QGkmin

, then set QGk(h)

= QGkmin

, go to step 3.

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Step 3:Update QG k(h)

Acceleration Factor

Experience has shown that the number of iterations required for convergence can be

considerably reduced if the correction in bus voltage computed at each iteration is multiplied by a

factor greater than unity (termed as acceleration factor) to bring the voltage closer to the value to

which it is converging.

For example, during the hth iteration the accelerated value of the voltage at kth

bus is

calculated using

Vk, acc(h+1)

= Vk(h)

+ (Vk(h+1)

Vk(h)

) (5.16)

Where, = acceleration factorVk

(h)= accelerated value obtained in the (h-1)

thiteration

Vk(h+1)

= value computed during hth

iteration using equation (5.10)

Then set Vk(h+1)

= Vk,acc(h+1)

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Flow Chart for load-flow solution using Gauss Seidal Method:(When PV buses are absent)

Set convergence criterion |Vmax | =

Set bus count i=1

Compute Vik+1

using the formula

Vik+1

= 1/Yii[(Pi-jQi)/(Vik)*- j=1

i-1YijVj

k+1-

j=i+1nYij Vj

k]

Test for

Slack busC

D

A

B

Start

Read system data,

tolerance level ,acceleration factor

Form bus admittance matrix Ybus

Assume initial bus voltages as Vi=1

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Fig. 5.3 Flow chart for GSLF (P-V busesnot present)

Yes

Is

|Vmax|Set K = K+1

Calculate the line flows, total line

losses, slack bus power

Stop

D

No

Vi+

= Vi+ [Vi

+- Vi

]A

Calculate change in Voltage Vik

= Vik+1

- Vik

Increase bus count i=i+1

Is in

Determine the largest absolute value

of change in voltage using the

formula |Vmax| = | Vik+1

- Vik| for

i=2,3,n

B

C

No

Yes

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Compute Qik+1

using the formula

Qik+1

= -Img{(Vik)*[ j=1

i-1YijVj

k+1 + j=i+1nYij Vj

k]

Test for type

of busB

C

im

PV bus , i>m

Start

Read system data

Form bus admittance matrix

Assume initial values for bus voltages ViFor

i=2,3,m and di0

= m+1,m+2,..n

Set iteration count,k=0

Set bus count i=2

D

PQ bus

Is

Qik+1

imax

Set Qik+1

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Fig. 5.4 P-V Bus Module for GSLF Algorithm.

Compute dik+1

using the

formula di k+1 =< Vi k+1A

Replace Vi by Vi+

Compute Vi+

using the formula

Vik+1

= (1/Yii)[(Pi-jQi)/(Vik)*- j=1

i-1YijVj

k+1

- j=i+1nYij Vj

k]

B

Calculate the line flows,

transmission loss, slack bus power and print the

results

Stop

Yes

Test For

convergence

| Vik+1

- Vik|

Increase the iteration

count k=k+1D

C

Yes

Is in Increase the iterationcount i=i+1

No

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EXERCISES:

1. A simple threebus system is shown in fig. The bus power and voltage specifications aregiven below:

Bus

No.

Voltage

(pu) (deg)

Generator Load Bus

SpecificationP Q P Q

1 1.05 0 - - 0.75 0.225 Slack Bus

2 1.04 - 0.5 - 0 0 P-V Bus

3 - 0 0 2.0 1.25 P-Q Bus

Calculate the slack bus power and line flow using Gauss Seidel iteration technique.

2. Figure shows the one line diagram of a simple three-bus power system with generators at

buses 1 and 3. The magnitude of voltage at bus 1 is adjusted to 1.05 pu. Voltage

magnitude at bus 3 is fixed at 1.04 pu with a real power generation of 200 MW. A load

consisting of 400 MW and 250 Mvar is taken from bus 2. Line impedances are marked in

per unit on a 100 MVA base and the line charging susceptances are neglected. Obtain the

power flow solution by the Gauss-Seidel method including line flows and line losses.

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USING NEWTON-RAPHSON METHOD

AIM

(i) To understand, the basic aspects of steady state analysis of power systems that is required

for effective planning and operation of power systems.


form and a simple method of solving load flow problems of small sized system using

Newton-Raphson iterative algorithm

OBJECTIVES

i. To investigate the convergence characteristics of load flow solutions using NRLF

algorithm for different sized systems and compare the same with that of GSLF algorithm.

SOFTWARE REQUIRED

MATLAB


In general, a nonlinear algebraic equation set may not be able to solve with a direct way. It

may be only possible to be solved some numerical method. That is, it may be possible to arrange

the equations so that the unknowns can be equated to a finite number of functional operations on

known values. Solving the power flow problems require to use iterative techniques. The main

idea in such methods is that it is possible to write a program to compute the next estimates from

the current estimates. The one of the techniques used for the iterative solution of nonlinear

algebraic equations is Newton-Raphson method.

NEWTON-RAPHSON POWER FLOW SOLUTION

Let the current injected at bus i is given by,

Ii= =1 (6.1)Expressing the equation (6.1) in polar form, we have

Ii= =1 < ( + ) (6.2)The complex power at bus i is

PijQi= Vi*Ii (6.3)

Substituting from (6.2) for Iiin (6.3), we get

PijQi= < =1 < ( + ) (6.4)Separating the real and imaginary parts,

Pi= =1 ( + ) (6.5)Qi= - =1 ( + ) (6.6)

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Equations (6.5) and (6.6) constitute a set of non linear algebraic equations in terms of the

independent variables, voltage magnitude in per unit, and phase angle in radians. We have two

equations for each load bus, given by equation (6.5) and (6.6), and one equation for each voltage

controlled bus, given by equation (6.5).

Expanding equation (6.5) and (6.6) in Taylors series about the initial estimate andneglecting all higher order terms results in the following set of linear equations.

(6.7)

In the above equation, bus 1 is assumed to be the slack bus. The Jacobian matrix gives the

linearized relationship between small changes in voltage angle i(k)

and voltage magnitude

() with the small changes in real and reactive power Pi(k) and Qi(k) . Elements of theJacobian matrix are the partial derivatives of (6.5) and (6.6), evaluated at i

(k)and (). In

short form, it can be written as

= 1 23 4 (6.8)For voltage-controlled buses, the voltage magnitudes are known. Therefore, if m buses of

the system arc voltage-controlled, m equations involving Q and V and the correspondingcolumns of the Jacobian matrix are eliminated. Accordingly, there are (n 1) real powerconstraints and (n - 1m) reactive power constraints, and the Jacobian matrix is of order (2n - 2 -m) x (2n - 2 - m). J1is of the order (n - 1) x (n - I). J2is of the order (n - 1) x (n - 1 - m), J3is of

the order (n - 1 - m) x (n - I), and J4is of the order (n -1- m) x (n -1- m).

The diagonal and the off-diagonal elements of J1are

= ( + ) (6.9) = Vi ( + ) , (6.10)The diagonal and the off-diagonal elements of J2are= 2 + ( + ) (6.11) = Vi ( + ) , (6.12)The diagonal and the off-diagonal elements of J3are = ( + ) (6.13)

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= Vi ( + ) , (6.14)The diagonal and the off-diagonal elements of J4are

=

2

(

+

) (6.15)

= Vi ( + ) , (6.16)The terms Pi

(k)and Qi

(k)are the difference between the scheduled and calculated values, known

as the power residuals, given by

Pi(k)

= Pisch

- Pi(k)

(6.17)

Qi(k)

= Qisch

- Qi(k)

(6.18)

The new estimates for bus voltages are

i(k+1)

= i(k)

+ i(k)

(6.19)

(+1)= ()+ () (6.20)The procedure for power flow solution by the Newton - Raphson method is as follows:

1. For load buses, wherePisch

and Qisch

are specified, voltage magnitudes and phase angles

are set equal to the slack bus values, or 1.0 and 0.0, i.e., (0)= 1.0 and i(0)= 0.0. Forvoltage-regulated buses, where

Viand P

i

schare specified, phase angles are set equal to the

slack bus angle, or 0, i.e., i(0)

= 0.

2. For load buses, Pi(k)

and Qi(k)

are calculated from (6.5) and (6.6) and Pi(k)

and Qi(k)

are

calculated from (6.17) and (6.18).

3. For voltage-controlled buses, Pi(k)

and Pi(k)

are calculated from (6.5) and (6.17),

respectively.

4. The elements of the Jacobian matrix (J1, J2,J3, and J4) are calculated from (6.9) - (6.16).

5. The linear simultaneous equation (6.8) is solved directly by optimally ordered triangular

factorization and Gaussian elimination.

6. The new voltage magnitudes and phase angles are computed from (6.19) and (6.20).

7. The process is continued until the residuals Pi(k)

and Qi(k)

are less than the specified

accuracy, i.e.

() (6.21)() (6.22)

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EXERCISES:


Bus

No.

Voltage

(pu) (deg)

Generator Load Bus

SpecificationP Q P Q

1 1.05 0 - - 0.75 0.225 Slack Bus

2 1.04 - 0.5 - 0 0 P-V Bus

3 - 0 0 2.0 1.25 P-Q Bus

Calculate the slack bus power and line flow using Newton-Raphson method.

2. Figure shows the one-line diagram of a simple three-bus power system with generation at

bus 1. The magnitude of voltage at bus 1 is adjusted to 1.05 per unit: The scheduled loads

at buses 2 and 3 are as marked on the diagram. Line impedances are marked in per unit on

a l00-MVA base and the line charging susceptances are neglected. Using the Newton

Raphson method, determine the phasor values of the voltage at the load buses 2 and 3

accurate to four decimal places.

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USING FAST DECOUPLEDMETHOD

AIM

(i) To understand, the basic aspects of steady state analysis of power systems that is required

for effective planning and operation of power systems.


form and a simple method of solving load flow problems of small sized system using Fast

decoupled iterative algorithm

OBJECTIVES

i. To investigate the convergence characteristics of load flow solutions using FDLF

algorithm for different sized systems and compare the same with that of GSLF

algorithm.

SOFTWARE REQUIRED

MATLAB


Power system transmission lines have a very high ratio. For such a system, realpower

changes P are less sensitive to changes in the voltage magnitude and are most sensitive tochanges in phase angle . Similarly, reactivepower is less sensitive to changes in angle and is

mainly dependent on changes in voltage magnitude. Thereforeit is reasonable to set elements J2and J3 of theJacobian matrix tozero. Thus, (6.8)becomes

= 1 00 4 (7.1)Or

P = J1 = (7.2)Q = J4 = (7.3)

Equations (7.2) and (7.3) show that the matrix equation is separated into two decoupled

equations requiring considerably less time to solve compared tothe time required for the solutionor (6.8). Furthermore, considerable simplification can be made to eliminate the need for re-

computing J1 and J4 during each iteration. This procedure results in the decoupled power flow

equations developed by Stott and Alsac.

The diagonal and off diagonal elements of J1are

From equation (6.9), = ( + ) = =1 ( + ) 2 (7.4)

We know form equation (6.6), Qi= - =1 ( + ), the equation(7.4) can be re-written as =Qi 2 (7.5)

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=Qi 2B (7.6)Where B = is the imaginary part of the diagonal elements of the bus

admittance matrix. B is the sum of susceptances of all the elements incident to bus i. In a typicalpower system, the self-susceptance B Qiand we may neglect Qi.

Further simplification is obtainedby assuming,

2 =

, which yields

=B (7.7)From equation (6.10),

= Vi ( + )Under normal operating conditions, (ij) is quite small. By assuming ij - i + j ij ,

the equation (6.10) can be re-written as = Vi B (7.8)Further simplification is obtained by assuming

1,

= ViB (7.9)Similarly, the diagonal elements of J4described by equation (6.15) may be written as= =1 ( + ) (7.10)From equation (6.6) and (7.10), we get,= + (7.11)Since B =

,

may be neglected and the equation (7.11) can becomes

= B (7.12)Similarly from equation (6.16), by assuming ij - i + j ij , yields = B (7.13)With these assumptions, equations (7.2) and (7.3) take the following form

= (7.14) = (7.15)

Here, B and B are the imaginary part of the bus admittance matrix Ybus. Since theelements of this matrix are constant, they need to be triangularized and inverted only once at thebeginning of the iteration. B'is of order of (n - 1). For voltage-controlled buses where andPiare specified and Qiis not specified, the corresponding row and column of Ybusare eliminated.

Thus, B" is of order of (n - 1 - m),where mis the number of voltage-regulated buses.Therefore, in the fast decoupled power flow algorithm, the successive voltage magnitude and

phase angle changes are

= 1 (7.16)

=

"

1

(7.17)

The fast decoupled power flow solution requires more iterations than the Newton-

Raphson method, but requires considerably less time per iteration, and a power flow solution is

obtained very rapidly. This technique is very useful in contingency analysis where numerous

outages are to be simulated or a power flow solution required for on-line control.

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EXERCISES:


BusNo.

Voltage(pu)

(deg) Generator Load BusSpecificationP Q P Q

1 1.05 0 - - 0.75 0.225 Slack Bus

2 1.04 - 0.5 - 0 0 P-V Bus

3 - 0 0 2.0 1.25 P-Q Bus

Calculate the slack bus power and line flow using Fast decoupled Load Flow method.

2. Figure shows the one-line diagram of a simple three-bus power system with generation at

bus 1. The magnitude of voltage at bus 1 is adjusted to 1.05 per unit: The scheduled loads

at buses 2 and 3 are as marked on the diagram. Line impedances are marked in per unit on

a l00-MVA base and the line charging susceptances are neglected. Using the Fast

decoupled Load Flow method, determine the phasor values of the voltage at the load buses

2 and 3 accurate to four decimal places.

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FAULT ANALYSIS

AIM

To become familiar with modelling and analysis of power systems under faulted condition

and to compute the fault level, post-fault voltages and currents for different types of faults, both

symmetric and unsymmetric.

OBJECTIVES

i. To carryout fault analysis for symmetrical and unsymmetrical faults in small systems

using the Thevenins equivalent circuit in the sequence and phase domains at the faultedbus.

ii. To obtain fault analysis report with fault level and current at the faulted point and post-

fault voltages and currents in the network for the following faults

(a)

Three-phase-to- ground(b)Line-to-ground

(c)Line-to-Line

(d)Double-line-to-ground

SOFTWARE REQUIRED

MATLAB


Faults can be defined as the flow of a massive current through an improper path which

could cause enormous equipment damage which will lead to interruption of power, personalinjury, or death. In addition, the voltage level will alternate which can affect the equipment

insulation in case of an increase or could cause a failure of equipment start-up if the voltage is

below a minimum level. As a result, the electrical potential difference of the system neutral will

increase. Hence, People and equipment will be exposed to the danger of electricity which is not

accepted.

In order to prevent such an event, power system fault analysis was introduced. The

process of evaluating the system voltages and currents under various types of short circuits is

called fault analysis which can determine the necessary safety measures & the required protection

system. It is essential to guarantee the safety of public. The analysis of faults leads to appropriateprotection settings which can be computed in order to select suitable fuse, circuit breaker size and

type of relay.

The severity of the fault depends on the short-circuit location, the path taken by fault

current, the system impedance and its voltage level. In order to maintain the continuation of

power supply to all customers which is the core purpose of the power system existence, all faulted

parts must be isolated from the system temporary by the protection schemes. When a fault exists

within the relay protection zone at any transmission line, a signal will trip or open the circuit

breaker isolating the faulted line. To complete this task successfully, fault analysis has to be

conducted in every location assuming several fault conditions. The goal is to determine theoptimum protection scheme by determining the fault currents & voltages.

The main objectives of fault analysis may be stated as follows:

(i) To determine maximum and minimum threephase short circuit currents.

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(ii) To determine the unsymmetrical fault current for single line-to-ground faults, line-to-line

faults, and double line-to-ground faults, and sometimes for open-circuit faults.

(iii) Investigation of the operation of protective relays.

(iv) Determination of rated rupturing capacity of circuit breakers.

(v) To determine fault current distribution and bus-bar voltage levels during fault conditions.

Type of FaultsThere are two types of faults which can occur on any transmission lines; balanced faults

and unbalanced faults also known as symmetrical and asymmetrical faults respectively. Most of

the faults that occur on power systems are not the balanced three-phase faults, but the unbalances

faults. In addition, faults can be categorized as the shunt faults, series faults and simultaneous

faults. In the analysis of power system under fault conditions, it is necessary to make a distinction

between the types of fault to ensure the best results possible in the analysis.

Series Faults

Series faults represent open conductor and take place when unbalanced series impedance

conditions of the lines are present. Two examples of series fault are when the system holds one or

two broken lines, or impedance inserted in one or two lines. In the real world a series faults takes

place, for example, when circuit breakers controls the lines and do not open all three phases, in

this case, one or two phases of the line may be open while the other/s is closed. Series faults are

characterized by increase of voltage and frequency and fall in current in the faulted phases.

Shunt Faults

The shunt faults are the most common type of fault taking place in the field. They involve

power conductors or conductor-to-ground or short circuits between conductors. One of the most

important characteristics of shunt faults is the increment the current suffers and fall in voltage andfrequency. Shunt faults cab be classified into four categories.

1. Line-to-ground fault: This type of fault exists when one phase of any transmission

lines establishes a connection with the ground either by ice, wind, falling tree or any

other incident. 70% of all transmission lines faults are classified under this category.

2. Line-to-line fault: As a result of high winds, one phase could touch anther phase &

line-to-line fault takes place. 15% of all transmission lines faults are considered line-

to-line faults.

3. Double line-to-ground: Falling tree where two phases become in contact with the

ground could lead to this type of fault. In addition, two phases will be involved instead

of one at the line-to-ground faults scenarios. 10% of all transmission lines faults are

under this type of faults.

4. Three phase fault: In this case, falling tower, failure of equipment or even a line

breaking and touching the remaining phases can cause three phase faults. In reality,

this type of fault not often exists which can be seen from its share of 5% of all

transmission lines faults.

The first three of these faults are known as asymmetrical faults.

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ANALYSIS OF FAULTED POWER SYSTEMS

A three-phase balanced fault can be defined as a short circuit with fault impedance called

Zfbetween the ground and each phase. The short circuit will be called a solid fault when Zf is

equal to zero. This type of fault is considered the most sever short circuit which can affect any

electrical system. Fortunately, it is rarely taking place in reality. Fortescue segregated

asymmetrical three-phase voltages and currents into three sets of symmetrical components in1918.

Analyzing any symmetrical fault can be achieved using impedance matrix method or

Thevenins method. Fortescues theorem suggests that any unbalanced fault can be solved intothree independent symmetrical components which differ in the phase sequence. These

components consist of a positive sequence, negative sequence and a zero sequence.

Positive Sequence Components

The positive sequence components are equal in magnitude and displayed from each other

by 120o with the same sequence as the original phases. The positive sequence currents and

voltages follow the same cycle order of the original source. In the case of typical counter

clockwise rotation electrical system, the positive sequence phasor are shown in Fig 8.1. The same

case applies for the positive current phasors. This sequence is also called the abc sequence andusually denoted by the symbol + or 1.

Positive Sequence

Components

Vc1

Vb1

Va1

Negative Sequence Components

This sequence has components that are also equal in magnitude and displayed from each

other by 120o similar to the positive sequence components. However, it has an opposite phase

sequence from the original system. The negative sequence is identified as the acb sequence andusually denoted by the symbol - or 2. The phasors of this sequence are shown in Fig 8.2

where the phasors rotate anti- clockwise. This sequence occurs only in case of an unsymmetricalfault in addition to the positive sequence components,

Negative Sequence

Components

Vb2

Vc2

Va2

Zero Sequence Components

In this sequence, its components consist of three phasors which are equal in magnitude as

before but with a zero displacement. The phasor components are in phase with each other. This is

Fig. 8.1 Positive sequence components

Fig. 8.2: Negative sequence components

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illustrated in Fig 8.3. Under an asymmetrical fault condition, this sequence symbolizes the

residual electricity in the system in terms of voltages and currents where a ground or a fourth wire

exists. It happens when ground currents return to the power system through any grounding point

in the electrical system. In this type of faults, the positive and the negative components are also

present. This sequence is known by the symbol 0.

Zero SequenceComponents

Vb0Vc0 Va0

The following are three sets of components to represent three-phase system voltages as

positive, negative and zero components:Positive: a1 b1 c1V V V

Negative: a2 b2 c2V V V

Zero: a0 b0 c0V V V

The addition of all symmetrical components will present the original system phase

components Va, Vband Vcas seen below:

0 1 2

0 1 2

0 1 2

a a a

b b b b

c c c c

aV V V V

V V V V

V V V V

(8.1)

The a operator is defined below:1 0a (8.2)

The following relations can be driven from equation (8.2):

2

3

1 120

1 0

a

a

From the above definition and using the a operator, it can be translated into a set of

equations to represents each sequence:a) Zero sequence components:

0 00a cbV V V (8.3)

b) Positive sequence components:2

11

1 1

ab

c a

V a V

V aV

(8.4)

c) Negative sequence components:

22

22 2

ab

c a

V aV

V a V

(8.5)

Now, the original system phasors Va, Vband Vccan be expressed in terms of phase acomponents only. Equation (8.1) can be written as follows:

Figure 8.3 Zero sequence components

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0 1 2

20 1 2

20 1 2

a a a a

a a ab

c a a a

V V V V

V V a V aV

V V aV a V

(8.6)

Writing the above equations can be accomplished in a matrix form:

0

2

1

2

2

1 1 11

1

a a

b a

c a

V VV a a V

V a a V

(8.7)

Defining A as:

2

2

1 1 1

1

1

A a a

a a

(8.8)

Equation (8.7) can be written as:

0

1

2

a a

b a

c a

V V

V A V

V V

(8.9)

This equation can be reversed in order to obtain the positive, negative and zero sequences

from the system phasors:

1

0

1

2

a a

a b

a c

V V

V A V

V V

(8.10)

Where1

A is equal to the following:

1 2

2

1 1 11

13

1

A a a

a a

(8.11)

These equations can be applied for the phase voltages and currents. In addition, it can

express the line currents and the line-to-line voltages of any power system under fault conditions.

Fault Analysis in Power Systems

In general, a fault is any event, unbalanced situation or any asymmetrical situation thatinterferes with the normal current flow in a power system and forces voltages and currents to

differ from each other.

It is important to distinguish between series and shunt faults in order to make an accurate

fault analysis of an asymmetrical three-phase system. When the fault is caused by an unbalance in

the line impedance and does not involve a ground, or any type of inter-connection between phase

conductors it is known as a series fault. On the other hand, when the fault occurs and there is an

inter-connection between phase-conductors or between conductor(s) and ground and/or neutral it

is known as a shunt fault.

Statistically, series faults do not occur as often as shunt faults does. Because of this fact

only the shunt faults are explained here in detail since the emphasis in this project is on analysis

of a power system under shunt faults.

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Three-Phase Fault

By definition a three-phase fault is a symmetrical fault. Even though it is the least frequent

fault, it is the most dangerous. Some of the characteristics of a three-phase fault are a very large

fault current and usually a voltage level equals to zero at the site where the fault takes place.

A general representation of a balanced three-phase fault is shown in Figure 8.4 where F is

the fault point with impedances Zf and Zg . Figure 8.5 shows the sequences networks

interconnection diagram.

From Figure 8.5, it can be noticed that the only one that has an internal voltage source is the

positive-sequence network. Therefore, the corresponding currents for each of the sequences can

be expressed as

0

2

1

1

0

0

1.0 0

a

a

a

f

I

I

IZ Z

(8.12)

If the fault impedance Zf is zero,

1

1

1.0 0a

IZ

(8.13)

If equation is substituted into equation

2

1

2

1 1 1 0

1

1 0

af

bf a

cf

I

I a a I

I a a

(8.14)

Solving Equation (8.14), we get

1

1

2

1

1

1

1

1.0 0,

1.0 240,

1.0 120

af a

f

bf a

f

cf a

f

I IZ Z

I a IZ Z

I aIZ Z

(8.15)

Since the sequence networks are short-circuited over their own fault impedance

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0

1 1

2

0

0

a

a f a

a

V

V Z I

V

(8.16)

If Equation is substituted into Equation

2

1

2

1 1 1 0

1

1 0

af

bf a

cf

V

V a a V

V a a

(8.17)

Therefore,

1 1

2

1 1

1 1

240

120

af a f a

bf a f a

cf a f a

V V Z I

V a V Z I

V aV Z I

(8.18)

The line-to-line voltages are

21 1

2

1 1

1 1

1 3 30

3 90

1 3 150

ab af bf a f a

bc bf cf a f a

ca cf af a f a

V V V V a Z I

V V V V a a Z I

V V V V a Z I

(8.19)

If Zf equals to zero,

1

1

1

1.0 0

1.0 240,

1.0 120

af

bf

cf

IZ

I

Z

IZ

(8.20)

The phase voltages becomes,

0

0

0

af

bf

cf

V

V

V

(8.21)

And the line voltages,

0

1

2

00

0

a

a

a

VV

V

(8.22)

Single Line-to-Ground Fault

The single line-to-ground fault is usually referred as short circuit fault and occurs whenone conductor falls to ground or makes contact with the neutral wire. The general representation

of a single line-to-ground fault is shown in Figure 8.6 where F is the fault point with impedances

Zf. Figure 8.7 shows the sequences network diagram. Phase a is usually assumed to be the faulted

phase, this is for simplicity in the fault analysis calculations.

Since the zero-, positive-, and negative-sequence currents are equals as it can be observed

in Figure 8.7. Therefore,

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0 1 2

0 1 2

1.0 0

3a a a

f

I I IZ Z Z Z

(8.23)

Since

0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

I I

I a a I

I a a I

(8.24)

Solving Equation (8.24) the fault current for phase a is

0 1 2af a a aI I I I

(8.25)

0 1 23 3 3

af a a aI I I I (8.26)

From Figure 8.6, it can be observed that, af f af V Z I (8.27)

The voltage at faulted phase a can be obtained by substituting Equation (8.24) into

Equation (8.27). Therefore, 13af f aV Z I (8.28)

but,0 1 2af a a a

V V V V (8.29)

Therefore,

0 1 2 13

a a a f aV V V Z I (8.30)

With the results obtained for sequence currents, the sequence voltages can be obtained

from

0 0

2

1 1

2

22

0 1 1 1

1.0 0 1

0 1

a a

b a

ac

V I

V a a I

a a IV

(8.31)

By solving Equation(8.31), we get,0 0 0

1 1 1

2 2 2

1.0

a a

a a

a a

V Z I

V Z I

V Z I

(8.32)

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If the single line-to-ground fault occurs on phase b or c, the voltages can be found by the

relation that exists to the known phase voltage components,

0

2

1

2

2

1 1 1

1

1

af a

bf a

cf a

V V

V a a V

V a a V

(8.33)

From the above equation (8.33), we get,2

0 1 2

2

0 1 2

bf a a a

cf a a a

V V a V aV

V V aV a V

(8.34)

Line-to-Line Fault

A line-to-line fault may take place either on an overhead and/or underground transmission

system and occurs when two conductors are short-circuited. One of the characteristic of this type

of fault is that its fault impedance magnitude could vary over a wide range making very hard to

predict its upper and lower limits. It is when the fault impedance is zero that the highest

asymmetry at the line-to-line fault occurs.

The general representation of a line-to-line fault is shown in Figure 8.8 where F is the

fault point with impedances Zf. Figure 8.9 shows the sequences network diagram. Phase b and c

are usually assumed to be the faulted phases; this is for simplicity in the fault analysis

calculations.

From Figure 8.9, it can be noticed that

0af

I

bf cf I I (8.35)

bc f bf V Z I

And the sequence currents can be obtained as,

0 0aI (8.36)

1 21 2

1.0 0

a af

I IZ Z Z

(8.37)

If Zf= 0,

1 2

1 2

1.0 0a aI I

Z Z

(8.38)

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The fault currents for phase b and c can be obtained by substituting Equations (8.36) and

(8.37) into Equation (8.24),

13 90bf cf aI I I (8.39)

The sequence voltages can be found similarly by substituting Equations (8.36) and (8.37)

into Equation (8.31)

0

1 1 1

2 2 2 2 1

0

1.0-

a

a a

a a a

V

V Z I

V Z I Z I

(8.40)

Also substituting Equation (8.40) into Equation (8.33)

1 2 11 2

2 2 21 2 11 2

2 21 2 11 2

1.0 ( - )

( - )

( - )

aaf a a

abf a a

acf a a

V V V I Z Z

V a V aV a I aZ a Z

V aV a V a I a Z aZ

(8.41)

Finally, the line-to-line voltages for a line-to-line fault can be expressed as

ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V

(8.42)

Double Line-to-Ground Fault

A double line-to-ground fault represents a serious event that causes a significant

asymmetry in a three-phase symmetrical system and it may spread into a three-phase fault when

not clear in appropriate time. The major problem when analyzing this type of fault is the

assumption of the fault impedance Zf , and the value of the impedance towards the ground Zg.

The general representation of a double line-to-ground fault is shown in Figure 8.10 where

F is the fault point with impedances Zfand the impedance from line to ground Zg . Figure 8.11

shows the sequences network diagram. Phase b and c are assumed to be the faulted phases, this is

for simplicity in the fault analysis calculations.

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From Figure 8.11 it can be observed that

0

( )

( )

af

bf f g bf g cf

cf f g cf g bf

I

V Z Z I Z I

V Z Z I Z I

(8.43)

Based on Figure 8.11, the positive-sequence currents can be found as

12 0

1

2 0

1.0 0

( )( 3 )( )

( ) ( 3 )

af f g

f

f f g

IZ Z Z Z Z

Z ZZ Z Z Z Z

02 1

2 0

( 3 )[ ]( ) ( 3 )

f ga a

f f g

Z Z ZI I

Z Z Z Z Z

(8.44)

20 1

2 0

( )[ ]( ) ( 3 )

fa a

f f g

Z ZI I

Z Z Z Z Z

An alternative method is,

0 1 2

0 1 2

0( )

af a a a

a a a

I I I II I I

(8.45)

If Zfand Zg are both equal to zero, then the positive-, negative-, and zero-sequences can be

obtained from

12 0

1

2 0

1.0 0

( )( )( )

( )

aIZ Z

ZZ Z

02 1

2 0

( )[ ]( )

a aZ

I IZ Z

(8.46)

20 1

2 0

( )[ ]( )

a aZ

I IZ Z

From Figure 8.10, the current for phase a is, 0afI (8.47)

Now, substituting Equations (8.46) into Equation (8.24) to obtain phase b and c fault

currents2

0 1 2

2

0 1 2

bf a a a

cf a a a

I I a I aI

I I aI a I

(8.48)

The total fault current flowing into the neutral is , 03n a bf cf I I I I (8.49)

And the sequences voltages can be obtained by using Equation (8.26)

0 0 0

1 1 1

2 2 2

1.0

a a

a a

a a

V Z I

V Z I

V Z I

(8.50)

The phase voltages are equal to

0 1 2

2

0 1 2

20 1 2

afa a a

bf a a a

cf a a a

V V V V

V V a V aV

V V aV a V

(8.51)

The line-to-line voltages can be obtained from

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ab af bf

bc bf cf

ca cf af

V V V

V V V

V V V

(8.52)

If Zf= 0 and Zg= 0 then the sequence voltages become, and the positive-sequence current

is found by using Equation 8.46

0 1 2 1 11.0a a a aV V V Z I (8.53)

Now the negative- and zero-sequence currents can be obtained from

22

2

00

0

aa

aa

VI

Z

VI

Z

(8.54)

The resultant phase voltages from the relationship given in Equation (8.53) can be

expressed as

0 1 2 130

af a a a a

bf cf

V V V V V V V

(8.56)

And the line-to-line voltages are

0

abf af bf af

bcf bf cf

caf cf af af

V V V V

V V V

V V V V

(8.57)

SHORT CIRCUIT CAPACITY (SCC)

The short-circuit capacity at a bus is a common measure of the strength of a bus. The

short-circuit capacity or the short-circuit MVA at bus k is defined as the product of the

magnitudes of the rated bus voltage and the fault current. The short circuit MVA is used for

determining the dimension of abus bar, and the interruptingcapacity of a circuit breaker.

Based on the above definition. the short - circuit capacity or the short-circuitMVA at bus

k is given by

SCC = 3 103MVA (8.58)Where the line to line voltage is expressed in kilovolts and Ik(F) is expressed inamperes. The symmetrical three-phase fault current in per unit is given by

() . = (0) (8.59)Where (0)is the per unit pre-fault bus voltage, and Xkkis the per unit reactance to the

point of fault. System resistance is neglected and only the inductive reactance of the system is

allowed for. This gives minimum system impedance and maximum fault current and a pessimistic

answer. The base current is given by

= 1033 (8.60)

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Where, SB is the base MVA and VB is the line to line base voltage in kilovolts. Thus the

fault current in amperes is,

= () . =

(0)

103

3 (8.61)

From equations (8.58) and (8.61), we get

SCC =(0)

(8.62)If the base voltage is equal to the rated voltage i.e., = , then,

SCC =(0) (8.63)

The pre-fault bus voltage is usually assumed to be 1.0 p.u., and we therefore obtain fromequation (8.63) the following approximate formula for the short-circuit capacity or the short-

circuit MVA.

SCC = (8.64)

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EXERCISES:

1. The one-line diagram of a simple power system is shown in fig. The neutral of each generator

is grounded through a current-limiting reactor of 0.25/3 per unit on a 100 MVA base. The

system data expressed in per unit on a common 100 MVA base is tabulated below. The

generators are running on no-load at their rated voltage and rated frequency with their emfs in

phase. Determine the fault current and fault MVA when

(a)A balanced three-phase fault at bus 3 through a fault impedance Zf= j0.10 p.u.

(b)A single line-to-ground fault occurs at bus 3 through a fault impedance Zf= j0.10 p.u.

(c)A line-to-line fault occurs at bus 3 through a fault impedance Zf= j0.10 p.u.

(d)A double line-to-ground fault occurs at bus 3 through a fault impedance Zf= j0.10 p.u.

ITEM BASE MVAVOLTAGE

RATINGX1 X2 X0

G1 100 20 kV 0.15 0.15 0.05

G2 100 20 kV 0.15 0.15 0.05

T1 100 20/200 kV 0.10 0.10 0.10

T2 100 20/200 kV 0.10 0.10 0.10

L12 100 220 kV 0.125 0.125 0.30

L13 100 220 kV 0.15 0.15 0.35

L23 100 220 kV 0.25 0.25 0.7125

2. Repeat Q1 by making a fault impedance of Zf= 0.0 p.u.

3.

Repeat Q1 for the fault occurs at bus - 2.

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LOAD-FREQUENCY DYNAMICS OF SINGLE-AREA

POWER SYSTEMS

AIM

To become familiar with the modelling and analysis of load-frequency dynamics of asingle area power system with load-frequency controller (LFC) and to design improved

controllers to obtain the best system response.

OBJECTIVES

i. To study the time response (both steady state and transient) of area frequency deviation

and transient power output change of regulating generator following a small load change

in a single-area power system with the regulating generator under free governor action,for different operating conditions and different system param

ee2404 manual

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