EE 368 Lecture Notes-2011

1

EE 368 POWER SYSTEMS ANALYSIS (3 0 3)

LOAD FLOW ANALYSIS: Formation of a.c. load flow equations. Gauss-Seidel iterative method

of solution. Cartesian and polar forms of load flow equations, formation of the jacobian matrix and

solution using the Newton-Raphson method. Digital computer study of load flow.

FAULT ANALYSIS: Causes of faults, types of faults, 3-phase symmetrical fault calculations,

unsymmetrical shunt and series fault calculations. Simultaneous faults. Applications to digital

computation.

OPERATION AND CONTROL: Characteristics of governors and their operation, speed changer

settings, load-sensitive components of a power station, load-frequency characteristics, Exciter

characteristics, block diagram representation of voltage control systems, voltage and reactive power

control.

STABILITY: Equal area criterion and solution of differential equations.

Mr. E. A. Frimpong

[email protected]

0246665284

1. LOAD FLOW STUDIES

2

Aims When you have completed this chapter, you should be able to

o Explain the important of load flow studies to power utilities,

o Explain the nonlinear nature of load flow equations,

o Formulate the admittance matrix of a power system

o Solve simple load flow problems using the Gauss, Gauss-Seidel and Newton-Raphson

iterative methods

o derive simple equations for load flow and formulate them in a fashion suitable for iterative

analysis,

o perform simple load flow analysis by hand and appreciate why computers perform these

tasks far better than humans and

1.1 Introduction

The purpose on an electric power system is to deliver electric power to consumers in a reliable,

quality, safe and economical manner. The electric power generated are transmitted through power

system components, generally over geographical distances from the generating point to the

consumers. Electric power systems very often have numerous paths over which power can flow,

and the term load flow refers to techniques used to understand how power flows and over what

paths.

In a large integrated power system, it is difficult to assess the voltages and currents associated with

each individual transmission line, although the positions of power generation and consumption are

known. The difficulty lies in the fact that the system equations are nonlinear and cannot be solved

directly. Instead, an iterative approach is used where an initial guess is made to the problem and

then subsequently changed until the solution fits the problem.

Load flow analysis is important to power utilities for the following reasons:

(a) To ensure proper redistribution of power when a line is being removed for maintenance.

(b) For power system planning in other to be able to accommodate future expansion.

(c) To ensure that power system plant is not run above nameplate rating. To keep voltage levels of

certain buses within close tolerances to ensure correct reactive power requirements.

(d) To assess if contingence fault conditions may potentially lead to wide scale system outages.

The objective of any load flow study is to produce the following information:

(a) Voltage magnitude and phase angle at each bus.

(b) Real and reactive power flowing in each element.

(c) Reactive power loading on each generator.

The above objectives are achieved by using the following information:

(a) Branch list of the system connections. That is the impedance of each element, sending-end and

reciving-end nodes. Lines and transformers are represented by their π-equivalent models.

(b) Voltage magnitude and phase angle at one bus, which is the reference point for the rest of the

system.

(c) Real power generated and voltage magnitude at each generator bus.

(d) Real and reactive power demand at each load bus.

The forgoing information is generally available since it either involves readily known data

(impedances, etc.) or quantities which are under the control of power system personnel (active

power output and excitation of generators).

3

Simply stated, the load-flow problem is as follows:

(a) At any bus there are four quantities of interest V , , P and Q .

(b) If any two of these quantities are specified, the other two must not be specified otherwise we

end up with more unknowns than equatioins.

(c) Because records enable the real and reactive power to be accuralty estimated at loads, P and Q

are specified qunatities at loads, which are called PQ buses.

(d) Likewise, the real power output of a generator is controlled by the prime mover and the

magnitude of the voltage is controlled by the exciter, so P and V are specified at generators,

which are called PV buses.

(e) This means that V and θ are unknown at each load bus and θ and Q are unknown at each

generator bus.

(f) Since the system losses are unknown until a solution to the load flow problem has been found,

it is necessary to specify one bus that will supply these losses. This is called the slack (or swing,

or reference) bus and since P and Q are unknown, V and θ must be specified. Ususally, and

angle of o0 is used at the slack bus and all other bus angles are expressed with respect to

slack.

1.2 Bus admittance matrix

The admittance, Y of a transmission line is the inverse of its impedance, Z . For example, the

admittance of a transmission line with resistance pu05.0R and a reactance X of 0.15pu will

have impedance 15.005.0 jjXRZ and admittance is:

pu6j257.71325.615.0j05.0

1Y o

.

3223 YY

3113 YY

1 2

34

2112 YY

4334 YY

4224 YY

10Y 20Y

30Y40Y

11 V,I 22 V,I

33 V,I44 V,I

Fig 1.1 A four-busbar systems showing admittances

Consider the four-bus system of figure 1.1. In terms of the node voltages V1, V2, V3 and V4 and the

given admittances, Kirchoff‟s current law yields,

133112211011 Y)VV(Y)VV(YVI

2442233212122022 Y)VV(Y)VV(Y)VV(YVI

3443232313133033 Y)VV(Y)VV(Y)VV(YVI

343424244044 Y)VV(Y)VV(YVI

4

Rearranging these equations and rewriting them in a matrix for, we obtain

4

3

2

1

3424403424

34342313302313

24232423122012

1312131210

4

3

2

1

V

V

V

V

YYYYY0

YYYYYYY

YYYYYYY

0YYYYY

I

I

I

I

4

3

2

1

44434241

34333231

24232221

14131211

4

3

2

1

V

V

V

V

yyyy

yyyy

yyyy

yyyy

I

I

I

I

.

Thus,

13121011 1,1. YYYcolumnrowinYiey (The sum of all admittances linked to bus 1)

2423122022 YYYYy (The sum of all admittances connected to bus 2)

3423133033 YYYYy (The sum of all admittances connected to bus 3)

34244044 YYYy (The sum of all admittances connected to bus 4)

122112 Yyy (The negative of the admittance between bus 1 and 2)

133113 Yyy (The negative of the admittance between bus 1 and 3)

144114 Yyy (The negative of the admittance between bus 1 and 4)

233223 Yyy (The negative of the admittance between bus 2 and 3)

244224 Yyy (The negative of the admittance between bus 2 and 4)

344334 Yyy (The negative of the admittance between bus 3 and 4)

Example 1.1

Obtain the admittance matrix of the three-busbar system shown in the figure below.

1 2

3

12 2 5 4Y . j

23 1 5 2Y . j 2113 jY 10 0 4 1Y . j 20 0 5 1Y . j

Solution

79.3214.05.21.04.013121011 jjjjYYYy

4.05.24.05.21212 jjYy

Continuing with the rest, the following admittance matrix is obtained:

5

45.225.121

25.175.445.2

2145.279.3

jjj

jjj

jjj

Y

Example 1.2

Given the following line impedances of a three-bus system, obtain its admittance matrix.

Line (bus to bus) puR Xpu

1-2 0.05 0.15

1-3 0.10 0.30

2-3 0.15 0.45

Solution

1 2

1 3

2 3

12 6

0 05 0 15

11 3

0 1 0 3

10 67 2

0 15 0 45

Y j. j .

Y j. j .

Y . j. j .

3 9 2 6 1 3

2 6 2 67 8 0 67 2

1 3 0 67 2 1 67 5

j j j

Y j . j . j

j . j . j

Self Assessment 1.1

Given the following line impedances of a four-bus system, obtain its admittance matrix.

Line (bus to bus) puR Xpu

1-2 0.05 0.15

1-3 0.10 0.30

2-3 0.15 0.45

2-4 0.10 0.30

3-4 0.05 0.15

Answer

9j36j23j10

6j211j666.32j666.03j1

3j12j666.011j666.36j2

03j16j29j3

1.3 The load flow problem

The nodal analysis (refer to any book on electric circuit analysis) is used in the load flow study of

large networks. A balanced three-phase network (a network which has all three phases equally

loaded. Such a network is analysed by studying only one phase) is assumed. The nodal equation for

a system comprising n buses can be expressed in a matrix form as:

VYI Eqn. 1.1

Where V voltages at the nodes or buses,

I currents injected into the nodes or buses, and

Y bus admittance matrix (admittance is the inverse of impedance)

6

The current injected into the ith bus is

n

k

kiki VYI1

n.,..,,i 21 Eqn. 1.2

In practice, active power, reactive power and voltage are specified, but not current. The injected

current and the injected complex power into the ith bus are related by the equation

iiii IVjQP or iiii IVjQP

or i

i

ii IV

jQP

*

Eqn. 1.3

Substituting eqn. 1.3 into eqn.1.2 yields

n.,..,,iVYVjQPn

k

kikiii 211

Eqn. 1.4

Equating real and imaginary parts, we obtain

n

k

kikii VYVReP1

Eqn. 1.5a

n

k

kikii VYVImQ1

Eqn. 1.5b

In power forms, if ii VV and ikikik YY , then

ikikik

n

k

kii cosYVVP 1

Eqn. 1.6a

ikikik

n

k

kii sinYVVQ 1

Eqn. 1.6b

Equations 1.6a and 1.6b give 2n power flow equations at n buses of a power system. Each bus is

characterized by four variables P, Q, V and giving a total of 4n variables. These equations can be

solved for 2n variables if the remaining 2n variables are specified. Practical considerations allow a

power system analyst to fix a priori two variables at each bus. Depending upon which two variables

are fixed a priori, the buses can be classified into three categories: PV bus, PQ bus and Slack bus,

swing bus or reference where V and (or θ) are specified.

Example 1.3

An interconnected cable links generating stations 1 and 2 as shown in the figure below. The desired

voltage profile is flat i.e. .puVV 121 The load demands at the two buses are pujSD 5151

; pujSD 15252 . The station loads are equalized by the flow of power in the cable. Determine

the load flow solution of the system if the cable has impedance pu.j.Z 0500050 . It is given

that generator G1 can generate a maximum of 20.0pu real power.

2DS1DS

1 2

1G 2G

Solution

The power injected into the buses are given by (from eqns. 1.6a and 1.6b):

7

1212211211

2

1111 cosVVYcosVYP

2121122122

2

2222 cosVVYcosVYP

1212211211

2

1111 sinVVYsinVYQ

2121122122

2

2222 sinsin VVYVYQ

We specify the generation at station 1 as .puPG 201 Hence

puPPP DG 51520111

Since the power and voltage of bus 1 are known, it is a PV bus. Bus 2 is thus the slack bus. 2 is

thus set to 0.

o...j.

YY 298490190500050

12211

oooYYY 71.9590.1918029.8490.1929.8490.19211221

From the 1P equation,

o.

.cos..cos.

414

079591929849195

1

1

Substituting 1 into the other equations, we obtain

pu...sin..sin.Q 13041479591929849191

pu...cos..cos.P 86441479591929849192

pu...sin..sin.Q 11141479591929849192

Now

pu..QQQQQQ GGDG 135513055 111111

puPPPP GDG 14.202222

puQQQQ GDG 11.162222

Line loss pu..PPPP DDGG 14025151420202121

Figure 1.2 shows, possibly, the simplest power system configuration. A generator of fixed terminal

voltage VA is connected to a load S via a transmission line which, to keep matters simple, has a

series resistance of R. The problem is to find the amount of real power, PG, generated by the

generator. This seems, at first sight, to be a simple problem, but it will be seen that this is not quite

the case.

?PG

A B

AV BV

SLoad

RcetanresisLine

Figure 1.2 A single busbar power system

It is common in power systems to specify the loads in terms of their power, i.e. so many watts,

kilowatts or megawatts. Thus, the load of figure 1.1 is related to the voltage at busbar B, VB, and the

current through the system, I:

IVS B Eqn. 1.7

8

However, the voltage at busbar B, VB, is related to the voltage at busbar A, VA and the voltage drop

along the transmission line, IR, which in turn is related to the current through the system, I:

IRVV BA Eqn. 1.8

We may solve equations 1.7 and 1.8 for VB. Eliminating I:

B

BA

V

S

R

VV

Eqn. 1.9

which can be written as:

02

RSVVV ABB Eqn. 1.10

And so, to find VB involves a nonlinear equation. Having found VB we may then proceed to find the

power dissipated in the transmission line, and hence find the power produced by the generator at A.

11 jQP 22 jQP

33 jQP 44 jQP

2112 YY

4114 YY 3223 YY 3113 YY

1 2

34 Fig 1.3 A four-busbar power system

This seemingly simple exercise has turned out to involve solving quadratic equations. This is the

fundamental problem in calculating load flows: nonlinear equations. Consider the still relatively

simple network of figure 1.3. Here there are four busbars to consider. If we wished to find the

distribution of power flows along each line, we would need to solve four simultaneous nonlinear

equations. This may no longer be solved by inspection. Instead, we have to use iterative techniques

which are far more easily performed by computer than by hand. The simplest method of iteration is

called the Gauss Method. Equation 1.10 is a simple quadratic equation that may be solved by the

well-known formula:

2

42

RSVVV

AA

B

However, this formula is only applicable to equations with one unknown and hence cannot be used

for solving the equations relating to the system of figure 1.3. To solve a set of nonlinear equations,

we must use an iterative method.

1.4 The Gauss iterative method

1.4.1 Theory

The operation of the Gauss iteration can be appreciated readily by applying it to the solution of the

simple quadratic equation 1.4. Equation 1.4 is rearranged as:

A

B

B VV

RSV Eqn. 1.11

9

However, to make the expression iterative, we will calculate a new value for VB on left hand side of

equation 1.11 by guessing an initial value for VB on the right hand side of equation 1.11. Thus, the

equation becomes:

Ak

B

k

B VV

RSV 1 Eqn. 1.12

where the superscript „k‟ refers to the order in which the values of VB are calculated. Now, by

assuming an initial value for VB and repetitively evaluating equation 1.12, the exact solution of VB

to the equation may be found.

Example 1.4

Calculate the power generated by the generator at busbar A in figure 1.2 given the following

values: Load MW400 , Resistance of transmission line, 5.15R ,

Voltage of busbar kVA 231 . Use the Gauss iterative method and check it by using the quadratic

solution formula.

Solution

To proceed, we use equation 1.12. Taking an initial value of VB as 231kV (i.e. the same as VA), we

get the following results by iteration:

K 1 2 3 4 5 6

VB (kV) 204.16 200.63 200.10 200.02 200.00 200.00

And so we see that to two decimal places, no improvement in the answer occurs for more than 6

iterations. To check the answer, we use the quadratic formula:

2

42

RSVVV

AA

B

,

which gives the answers VB = 200kV or VB = 31kV.

The above example has illustrated another important point regarding iterative equations: the

possibility of multi-valued solutions. When assessing the answers from the quadratic formula, we

intuitively choose the higher value solution since in a power system, we expect busbar voltages to

be approximately the same in magnitude. Hence, VB = 200kV is the correct answer. However, if the

two answers were closer together, this selection process becomes more difficult.

Having found VB, we can calculate the power loss in the transmission as:

MWkV

R

VVP BA

L 625.15

)31()( 22

thus, the generator at A must export

PG = Load + Losses = 400 + 62 = 462MW

Self Assessment 1.2

Try performing the iteration of the previous example using different starting values. You should

find that the result is always 200kV, even for starting values close, but not equal to 31kV.

1.4.2 Application of the Gauss iterative method to multi-terminal power systems

To be of any use to a power system engineer, the Gauss iterative method must be applicable to

more realistic power system configurations. We shall take as our example the system of figure 1.2;

in practice, computer programmes used by power utilities are capable of handling systems with

2000 or more busbars.

10

In our simple system of figure 1.2, we assumed that the transmission line and load are purely

resistive. This allowed us conventionally to ignore any flow of reactive power. However this

quantity is very important in power system analysis and must be considered further. This implies

that loads must be specified in terms of their real and reactive powers, and that the reactance of

transmission lines must also be represented.

The equations are formulated as follows. Taking busbar 3 of figure 1.3 as an example, the total real

power P, and reactive power Q, at busbar 3 are related to the voltage at the bus V3, and the net

current passing through the bus I3, as:

3333 jQPIV

Eqn. 1.13

where the superscript „*‟ denotes the complex conjugate. P is taken to be positive if there is net

generation at a bus, in which case I3 is flowing from the generator into the transmission system as

shown in figure 1.3. I3 may therefore be expressed as:

3

3

33

*

*)(I

V

jQP

or 3

3

33

*I

V

jQP

Eqn. 1.14

However, I3 is composed of the components of current entering or leaving busbar 3 via the lines to

other busbars. Thus, using admittances to represent the transmission lines, equation 1.14 may be re-

expressed as:

443333223113

3

33

*VYVYVYVY

V

jQP

Eqn. 1.15

Note that admittances refer only to the transmission lines and not the busbar loads or generators.

Equation 1.15 may be rearranged into a form suitable for iteration:

443223113

3

33

33

3*

1VYVYVY

V

jQP

YV Eqn. 1.16

Similar to equation 1.16 for the simple system of figure 1.2, iterative use of equation 1.16 allows us

to find the value of V3. In general, for a power system having N buses, the voltage at the kth bus is

given by:

N

1n

nkn

k

kk

kk

k VY*V

jQP

Y

1V kn Eqn. 1.17

where Ykn is the negative of the admittance measured between busbars k and n in the network. Ykk is

the sum of all admittances connected to busbar k. Iterative evaluation of equation 1.17 for all buses

will eventually lead us to the final solution. As described so far, the complex power, jQPS is

specified for all buses. However, no allowance has been made for the power losses, real or reactive,

that occur in the transmission lines; these losses are, of course, load dependent. If we specified S for

all buses, then clearly no solution could be found unless the generation power supply exactly

equalled the load plus the losses. To allow for this, we make one bus within the system independent

of S; this bus is referred to as the slack or swing bus and is usually denoted as being node 1(some

books use 0). Furthermore, the voltage is specified at the slack bus and, hence, equation 1.17 is not

applied to the slack bus.

In load flow analysis, both generation and load are specified. However, the system losses will

depend on the exact solution found by the load flow study. The slack bus is used to ensure that

generation exactly equals load plus losses.

11

Example 1.5

Find the voltages at busbars 2 and 3 of figure 1.4 after one iteration of the Gauss method.

Busbar Voltage Generation(p.u.) Load(p.u.)

1 o099.0 ? 2j1

2 ? 1.0j3.0 1.0j2.0

3 ? 1.0j7.0 0

1 2

3

?V 2

?V 3

09901 j.V 4212 jY

2123 jY 2113 jY

1070 .j.

1020 .j. 21 j

?S 1030 .j.

Fig 1.4 A three-bus power system

Solution

First of all we will construct the admittance matrix from the admittance values given in figure 1.4.

3

2

1

3

2

1

422121

216342

214263

V

V

V

jjj

jjj

jjj

I

I

I

Eqn. 1.18

We will begin by assuming that 2V and 3V are both set to o01 . Hence, applying equation 1.17 to this

problem for 1V and 2V gives us:

323112

2

22

22

2

1VYVY

*V

jQP

YV Eqn. 1.19

and

223113

3

33

33

3

1VYVY

*V

jQP

YV Eqn. 1.20

2.01.01.02.01.03.0222222 jjjjQPjQPjQP DDGG

2.01.022 jjQP

Try to obtain 33 jQP

Substituting in actual values results in the following equations:

0121099.042

01

2.01.0

63

12 jjjj

j

j

jV Eqn. 1.21

12

0121099021

01

1070

42

13 jjj.j

j

.j.

jV Eqn. 1.22

Note: In calculating 3V in eqn. 1.22, we continued to use the initially assumed values of 2V and 3V

even though eqn. 1.21 will give a new 2V . This is a major distinguishing factor between the Gauss

iterative method and other iterative methods. Working out equations 1.21 and 1.22 give:

027097302 .j.V and 130008513 .j.V

Additionally, from equation 1.18,

3211 214263 VjVjVjI Eqn. 1.23

and so

130008512102709730420990631 .j.j.j.jj.jI

062.0429.0 j Eqn. 1.24

Hence, power injected at slack bus 1 is

0610425006204290099011 .j..j.j.*IVS Eqn. 1.25

Note that the solution as shown above is referred to as the Gauss method.

However, another slightly different method can be used in which, after evaluating 2V from

equation 1.21, this new value is then used to calculate 3V from equation 1.22. This approach is

called the Gauss-Seidel method and is a slight improvement on the Gauss method.

Example 1.6

Recalculate busbar voltages 2V and 3V using the Gauss-Seidel iterative method.

Solution

2V is calculated the same way as in the Gauss-Seidel iterative method.

027.0973.021099.021

01

1.07.0

42

13 jjjj

j

j

jV

New V2

1435007151 .j.

13

Example 1.7

(b) Using the Gauss-Seidel iterative method, determine the voltages of busbars 2 and 3 after

one iteration. Assume the following initial voltages 0

2 3 1 0V V .

3 6 2 4 1 2

2 4 3 6 1 2

1 2 1 2 2 4

j j j

Y j j j

j j j

Busbar Voltage Generation(p.u.) Load(p.u.)

1 o099.0 ? 2j1

2 ? 0 25 0 1. j . 1.0j2.0

3 ? 0 4 0 1. j . 0

Solution

2.005.022 jjQP

1.04.033 jjQP

0121099.042

01

2.005.0

63

12 jjjj

j

j

jV

02.097.0 j

02.097.021099.021

01

1.04.0

42

13 jjjj

j

j

jV

08.004.1 j

1.4.3 Limitations of the Gauss iterative method

A drawback to using the Gauss method is the sometimes excessive number of iterations that are

needed before a solution is found. Part of the problem is that this method takes no account of sign

or magnitude of the error existing between iterations. This problem can be met part way by

applying an acceleration factor which multiplies the difference between iterations by a constant in

the range 1 to 2. However, the best solution is to use a technique which takes account of the error

and uses it to modify the next iterative cycle. The Newton-Raphson technique is an example of this

and takes far fewer iterative cycles to reach the solution than Gauss method.

In the Gauss method, the calculation of the values at a new iterative step depends solely on

previous values and the iterative formula. However, in the Newton-Raphson method, the

calculation of a new iterative step also makes use of an estimate of the error from the previous step.

This enables the Newton-Raphson method to converge upon the solution using fewer iterative steps

than the Gauss method.

1.5 The Newton-Raphson method

1.5.1 Theory

The Newton-Raphson method differs from the Gauss-Seidel method in that new iterative updates of

the required busbar voltages are based upon the rate of change of the solution. Initially, we will

simplify the theory in order that the basic principle of this method of iteration is fully understood.

We will begin by considering the d.c. system of figure 1.2. We learned that the relationship

describing voltages, power and line resistance is given by equation 1.10 which is repeated here in a

slightly different form:

14

0VFRSVVV B1BA

2

B Eqn. 1.26

Equation 1.26 shows equation 1.10 as a function, F1, of VB. Of course, when we have found the

correct value of VB to fit our parameters in VA, R and S, then F1 (VB) will be zero. Using, as before,

the index k to refer to successive iterations of VB, we can write:

0VVF k

B

k

B1 Eqn. 1.27

where k

BV is the error between the correct value of VB and its estimate on the „k‟th iteration. In

order to update the value of 1k

BV , we will try to estimate the value of k

BV by using the Taylor

series expansion of F1:

...1111

k

B

k

B

k

B

k

B

k

BBdV

dFVVFVVFVF Eqn. 1.28

Equation 1.28 shows the first two terms in the Taylor series expansion where k

B1 dVdF is the

derivative of F1 with respect to VB evaluated k

BV . Since equation 1.28 equates to zero, we may

evaluate k

BV as:

k

B

1

k

B1k

B

dV

dF

VFV

Eqn. 1.29

Note that equation 1.29 only approximates to k

BV since only the first two terms in the Taylor

series expansion of equation 1.28 were considered.

Example 1.8

Using the same data of example 1.4, calculate the first iterative value of 1k

BV starting with

A

k

B VV initially.

Solution

For kV231VV A

k

B , MW400S and line resistance 5.15 , F1 evaluates to:

RSVVVVF k

BA

k

B

k

B1

62323 104005.151023110231 9102.6 Eqn. 1.30

and

333

A

k

B

k

B

1 1023110231102312VV2dV

dF

Eqn. 1.31

Hence,

kV84.26

10231

102.6

dV

dF

VFV

3

9

k

B

1

k

B1k

B

Eqn. 1.32

Therefore the updated value of BV is:

kV16.20484.26231VVV k

B

k

B

1k

B Eqn. 1.33

Further iterative cycles reveal:

k 1 2 3

15

)kV(VB 204.16 200.10 200.00

These results prove the worth of the Newton-Raphson method. Comparing the above table with the

table for the Gauss method shown in example 1.4, it is clear that the correct solution is reached with

fewer iterative cycles using the Newton-Raphson method (3 iterations compared to 5). Note,

however, that there is a price to be paid for quicker convergence: a more complicated algorithm.

1.5.2 Application to a.c. systems

When applied to a.c. power systems, the equivalent of F1 is a function of two direct variables. In

the Gauss-Seidel case, we saw that the governing equation, equation 1.17, is solved for a real and

imaginary component of Vk, or alternatively, a magnitude and an angle of Vk but in either case there

are two variables to be found.

We will proceed by deriving the equations necessary to solve figure 1.4, the 3 bus power system.

We will take all complex values in the equations to be in polar coordinate form:

kkk VV nnn VV knknk YnY Eqn. 1.34

From equation 1.4 we may write:

N

1n

knknknnkkk YVVjQP Eqn. 1.35

Equation 1.35 may be separated into its real and imaginary components:

N

1n

knknknnkk cosYVVP Eqn. 1.36

N

1n

knknknnkk sinYVVQ Eqn. 1.37

Similar to the Gauss-Seidel case, the slack bus (taken to be bus 1 in figure 1.4) is not considered in

the system equations. Beginning with initial values for the busbar voltages, values for Pk and Qk are

evaluated from equations 1.36 and 1.37. The calculated value for real power, Pk(calc), corresponds to

the term k

B1 VF on the right hand side of equation 1.28 (note, of course, that we will have two

equations of the form of equation 1.28 in the a.c. case, the extra equation being in Qk). Similarly,

the specified value of real power, Pk(spec), corresponds to the left hand side of equation 1.28. Thus,

an equation in P of the form:

k

k

k)calc(k)spec(k V

V

PPP

Eqn. 1.38

may be derived. Notice that the derivative of equation 1.29 becomes a partial derivative in equation

1.38 since Pk is a function of all busbar voltages. An equation similar to equation 1.38 may be

derived for Qk. denoting kP as the difference between the specified and calculated power:

)calc(k)spec(kk PPP Eqn. 1.39

and similarly

)calc(k)spec(kk QQQ Eqn. 1.40

We can write the final expression relating Pk, Qk and Vk for busbars 2 and 3 of figure 1.4:

16

3

2

3

2

3

3

2

3

3

3

2

3

3

2

2

2

3

2

2

2

3

3

2

3

3

3

2

3

3

2

2

2

3

2

2

2

3

2

3

2

V

V

V

Q

V

QQQ

V

P

V

QQQ

V

P

V

PPP

V

P

V

PPP

Q

Q

P

P

Eqn. 1.41

The 44 matrix of partial derivatives is referred to as the Jacobean matrix. By considering the

derivatives of equations 1.36 and 1.37 with respect to the relevant variables (in this case 2 , 3 , 2V

and 3V ), expressions for each element of the Jacobean may be derived. Numerical values of the

Jacobean can then be evaluated using the initial values of the busbar voltages.

Thus far, initial values of busbar voltages have enabled us to calculate P , Q and the Jacobean

matrix. In order to solve for and V , the Jacobean needs to be inverted and then multiplied by

the column vector P and Q . When and V are found, they are added to the initial values for

V to form a more accurate estimate. The iterative process may then be repeated until the correct

busbar voltages are found.

It will be apparent that load flow calculations using the Newton-Raphson method cannot easily be

performed by hand calculation and are better entrusted to a computer.

When written as a computer programme, it is necessary to apply a test which stops the iterative

process when V is less than some preset value which is indicative of the accuracy required. Unlike

the Gauss-Seidel method, the Newton-Raphson method is sensitive to initial values that are far

removed from the correct solution. However, since in power systems busbar voltages are usually

close to 1p.u. is taken as an initial guess.

Extra busbars may be incorporated into equation 1.41 to take account of power systems with

greater than 3 busbars. Each extra busbar will add an extra 2 elements to the column vectors and

add 2 extra columns and 2 extra rows to the Jacobean. Thus, for and N busbar system, the main

burden of the computation will be in evaluating the inverse of a 1N21N2 Jacobean

matrix.

For example, the following steps will be followed when applying Newton-Raphson a three-bus

system such as the one in figure 1.4

Step 1: Calculate )calc(P2 , )calc(P3 , )calc(Q2 and )calc(Q3 from the previous equations

Step 2: Calculate 2P , 3P , 2Q and 3Q

Step 3: For this system, relationship between P , Q and , V is:

17

3

2

3

2

3

3

2

3

3

3

2

3

3

2

2

2

3

2

2

2

3

3

2

3

3

3

2

3

3

2

2

2

3

2

2

2

3

2

3

2

V

V

V

Q

V

QQQ

V

P

V

QQQ

V

P

V

PPP

V

P

V

PPP

Q

Q

P

P

where the 44 matrix is called the Jacobean. Compute the elements of the Jacobean.

Step 4: Invert Jacobean to give 2 , 3 , 2V and 3V

Step 5: Update 2 , 3 , 2V and 3V

Step 6: Go to step 1 and repeat

Assignments

1. Obtain the admittance matrix of the 3-busbar system shown below.

3

13 0 1 0 2Z . j . 23 0 05 0 15Z . j .

12 0 2 0 4Z . j .

0 5cX j . 0 25cX j .

21

2. Obtain the admittance matrix of a power system whose data is given below.

Line (bus to bus) Series admittance Shunt reactance

1-2 2 5j 1j

1-3 1 3j 0 5j .

2-3 1 2j 0

3. Consider the diagram below.

099.0 jV

1 22.01.012 jZ

4.02.023 jZ 4.02.013 jZ

21 j 1.03.0 j

1.06.0 j

1.04.0 j

3

(a) Obtain the admittance matrix.

(b) Find the voltages at busbars 2 and 3 after one iteration, using the Gauss method. Take

the initial values of 1V and 2V to be 001 .

(c) Find the generation at the slack bus.

18

4. (a) Compare Gauss-Seidel and Newton Raphson iterative methods for load flow studies.

(b) What does nnY and nmY signify in an admittance matrix?

(c) Two busbars with generators, interconnected by an overhead line have the following loads

1 20 15S j pu and 2 10 5S j pu . The desired voltages are .puVV 121 The

impedance of the overhead line is 0 002 0 03. j . pu . Determine the load flow solution of

the system if the generator at busbar 1 can generate a maximum real power of 18pu.

19

2. FAULT ANALYSIS

Aims

(a) To introduce the concept of fault analysis, including the usefulness and importance of such

analysis, in electrical power systems.

(b) To convey an understanding, both mathematically and intuitively of how power system

faults are analyzed.

(c) To apply the understanding to some simple power systems.

2.1 Introduction

An essential part of the design of a power supply network is the calculation of the system voltages

and currents which flow in the components when faults of various types occur. The magnitude of

the fault currents gives the engineer the current settings for the protection to be used and the ratings

of the circuit breakers. The idea is to be able to rescue the system from the abnormal conditions

within minimum time.

The main objectives of fault analysis can be summarised as follows:

(a) To determine maximum and minimum three-phase short-circuit currents.

(b) To determine the unsymmetrical fault current for single and double line-to-earth, line-to-

line faults and open-circuit faults.

(c) Investigation of the operation of protective relays.

(d) Determination of rated rupturing capacity of breakers.

(e) To determine fault-current distribution and busbar-voltage levels during faults.

2.2 Types of Faults Faults are undesirable, unpredictable and unavoidable incidents in a power network which have the

capability of changing system values and also destroying power system equipment. If the insulation

of the system should fail at any point, or if a conducting object should come in contact with a bare

power conductor, a will occur. Electrical failure generally implies: insulation failure resulting in a

short-circuit condition and/or conducting-path failure resulting in an open-circuit condition.

Insulation failure is by far the more common type of electrical failure.

The causes of faults are many: they include lightning, wind damage, trees falling across lines,

vehicles colliding with towers or poles, birds shorting out lines, aircraft colliding with lines,

vandalism, small animals entering switchgear, and line breaks due to excessive ice loading. The

principal types of faults are discussed below.

2.2.1 Short-circuited phases

Faults of this type are caused by insulation failure between phase conductors or between phase

conductors and earth, or both, the result being the short-circuiting of one or more phases to earth or

to one another or both. The faults which fall under this category are: three-phase-to-earth fault,

three-phase clear of earth fault, phase-phase fault, double-phase-to-earth fault and Single-phase-to-

earth fault. The most common of these is the single-phase-earth fault (70-80%) followed by phase-

phase and then double-phase-earth. Although a 3-phase fault is the least common, it constitutes the

most severe fault and is also the most amenable to calculations. Figure 2.1 illustrates these faults.

20

Single-phase-earth Double-phase-earth

Phase-phase Three-phase

Three-phase-earth

Fig 2.1 Short-circuited phases

2.2.2 Open-circuited phases

This type of fault is the failure of one or more phases to conduct. The more common cause of this

type of fault are joint failures on overhead lines and cables, and the failure of one or more phases of

a circuit-breaker or isolator to open or close. Single-phase open-circuit, Double-phase open-circuit

and Three-phase open-circuit are the faults which fall under this category. These are illustrated in

figure 2.2. The single-phase and two-phase conditions are of particular interest because they both

tend to produce unbalance of power system currents and voltages with the consequent risk of

damage to rotating plant.

Single-phase open-circuit Double-phase open-circuit

Three-phase open-circuit

Fig 2.2 Open-circuited phases

2.2.3 Simultaneous faults

A simultaneous fault condition, sometimes termed a multiple fault condition, is defined as the

simultaneous presence of two or more faults of similar or dissimilar types at the same or different

points on the power system. Such conditions may result from a common cause, from different but

consequential causes or, extremely rarely, from quite separate and independent causes. The

commonest simultaneous fault condition is undoubtedly the double-circuit overhead-line ( a

double-circuit system is that in which a single tower is made to carry two different three-phase

circuits ) fault in which a common cause (for example lightning or accidental contact) results in a

fault on each of the two circuits (each of the two three-phase circuits) concerned. These two faults,

although possibly geographically coincident, will be electrically separate to an extent determined

by the point of fault and the particular power system configuration.

A simultaneous fault condition of particular interest is that known as the cross-country earth-fault,

in which a single-phase-to-earth fault a one point in the power system occurs coincidentally with a

second single-phase-to-earth fault on another phase and at some other point in the system. This

condition is most commonly experienced on impedance-earthed systems where the second earth-

21

fault may be initiated by the increased healthy-phase voltage resulting from the neutral

displacement produced by the first. As already stated, a simultaneous fault condition may consist of

two different types of fault at the same point, and one example of this is the open-circuit-with-

earth-fault condition in which two faults, namely a single-phase open-circuit and a single-phase-to

earth fault occur coincidentally on the same phase and at the same point in the power system. Such

a condition can occur on an overhead line for example, due to a phase conductor breaking at a point

near to a tower, the conductor on the tower side of the break being held by the suspension insulator

and that on the other side falling to earth.

2.2.4 Winding faults

The types of fault which can occur on machine and transformer windings consist mainly of short-

circuits, from one phase winding to earth, from one phase winding to another or from one point to

another on the same phase winding. The last mentioned condition is known as a short-circuited

turns fault, and is of particular interest from the protection standpoint in that the fault current in the

short-circuited turns may be very large and that in the remainder of the winding very small. The

open-circuited winding condition is quite rare in practice and is usually the result of damage to the

winding as a consequence of a preceding winding short-circuit at or near the point of fault. Open

circuits in transformers may also occur as a result of failure of the tap-change equipment.

2.2.5 Other categories of faults

(a) Changing –fault conditions: The types of faults which have been mentioned above can all

be regarded as fixed fault conditions, in that the type of fault remains unchanged for the duration of

the fault. The great majority of fault conditions are of this type but there are others, known as

changing-fault conditions, in which the type of fault changes during the course of the fault. Such

changing-fault conditions can result from a number of causes, the most common being the

spreading of a fault arc, or of the ionised gases from a fault arc, to other phases and even to other

circuits. A typical example is a single-phase-to-earth fault which develops into a two-phase-earth

fault and possibly, later, into a three-phase fault. The analysis of a changing fault condition presents

no particular difficulty, since the condition can be considered as a succession of fixed fault

conditions, each of which can be analysed individually.

(b) Symmetrical faults: These are faults which affect all phases equally, i.e. three-phase-to-earth

and three-phase clear of earth.

(c) Unsymmetrical faults: These are faults which produce varying effects on some or all phases,

i.e. phase-phase fault, double-phase-to-earth fault and single-phase-to-earth fault

(d) Permanent faults: Faults which do not die out on their own. Such fault occurs when for

example a tree falls on a transmission line.

(e) Transient faults: Faults which die out on their own. For example, a short-circuit caused by

wind.

2.3 Factors affecting fault severity The severity of a power system fault condition may be assessed in terms of the disturbance

produced and the damage caused, the magnitude of the fault current and its duration being of

particular interest, especially in relation to the design and application of power system protection.

The factors which affect fault severity must therefore be given due consideration in all aspects of

power system analysis in order to ensure results which are truly representative of the conditions

which can occur in practice. The factors which normally require to be considered are:

22

(a) Source conditions: These relate to the amount and disposition of all connected generation

(including all other power sources such as interconnections with other systems), the two extremes

of minimum and maximum connected plant being of particular interest. The minimum and

maximum plant conditions are normally those corresponding to the conditions of minimum and

maximum connected load.

(b) Power system configuration: This is determined by the items of plant, namely generators,

transformers, overhead-line and cable circuits etc., assumed to be in service for the particular

condition being investigated and by such other factors as have a bearing on the topology of the

equivalent system network. The system configuration may change during the course of a fault with

consequent changes in the magnitude and distribution of the fault current, typical causes being the

sequential tripping of the circuit-breakers at the two ends of a faulted transmission line and the

sequential clearance of multiple fault conditions

(c) Neutral earthing: Faults which involve the flow of earth current (for example a single-phase

or two-phase fault to earth, a single-phase or two-phase open-circuit) may be influenced

considerably by the system neurtral-earthing arrangements, particularly by the number of neurtral

earthing points and the presence or absence of neutral earthing impedances. Power systems may be

single-point or multiple-point earthed and such earthing may be direct (that is, solid earthing) or via

impedance. Earthing impedance can be used to limit the earth-fault current to a very low and even

negligibly small value, as in the case of a system earthed through a Petersen coil.

(d) Nature and type of fault: From what has already been said, it will be evident that the type of

fault and its position in the power system may have a considerable effect on the magnitude and

distribution of the system fault current this being particularly the case in respect of earth-faults as

compared with phase faults, open-circuits as compared with short-circuits and faults within

machine and transformer windings as compared with similar faults at winding phase-terminals.

Similarly, the effects of a given fault condition may be considerably modified by the simultaneous

presence of one or more other fault conditions as, for example, in the combination of a short-circuit

and an open-circuit phase condition. A further factor which may require consideration is the

possible effect of fault impedance (for example, fault-arc resistance and the ohmic resistance of any

metallic or non-metallic fault path, etc), this being of particular importance in matters relating to

the design and application of distance protection.

The wide range of possible system fault conditions and the many factors which influence them

result in a wide range of possible levels of fault severity, ranging from extremely low levels up to

the maximum levels possible for the system being considered. It is therefore of value, in referring

to fault severity generally, to be able to refer to a standard fault condition, namely the three-phase

short-circuit, and to the level of fault severity produced by this particular fault condition, namely

the three-phase fault level. This level may be expressed in amperes or, as is more usual, in three-

phase MVA (this is obtained from the expression 6

FL 10IV3 , where LV is the nominal line

voltage of the faulted part and FI is the fault current) corresponding to the rated system voltage and

the symmetrical value of the three-phase fault current. The three-phase short-circuit can normally

be regarded as the most severe condition from the point of view of fault severity, and it is

accordingly the maximum possible value of the three-phase fault level which normally determines

the required short-circuit rating of the power system switchgear. A factor which may also have to

be taken into account is the maximum value of the single-phase-to-earth fault current which, in a

solidly-earthed system, may exceed the maximum three-phase fault current.

23

ANALYSIS OF BALANCED THREE-PHASE FAULTS

2-2.1 Balanced three-phase faults

A three-phase fault is a condition of either: (a) all three phases of the system are short

circuited to each other or (b) all three phases of the system are earthed as show in figure 4.

The three-phase short-circuit can normally be regarded as the most severe condition from

the point of view of fault severity, and it is accordingly the maximum possible value of the

three-phase fault level which normally determines the required short-circuit rating of power

system switchgears.

During a three-phase fault, the reactance of a generator is a time varying quantity. It is

dX

in the sub-transient period (one to four cycles),

dX in the transient period (about 30

cycles), and the synchronous reactance dX after that. The purpose of a study determines the

one used. It must be kept in mind that the sub-transient currents can be very large due to the

small size of

dX . Due to symmetry, the three phase currents during a symmetrical fault can

be solved using ordinary circuit theory. If the fault has zero impedance to ground, it is

called a solid fault or bolted fault (all three lines shorted to ground with zero impedance).

2-2.2 Modelling of system components

The main components of power systems which are considered are:

(a) Synchronous machine

It is represented by a constant voltage source behind transient or subtransient reactance.

Normally, both generator and motor subtransient reactance is used to determine the

momentary current flowing on occurrence of a short-circuit. To decide the interrupting

capacity of circuit breakers, subtansient reactance is used for generators and transient

reactance for motors

(b) Transformers

All transformers are considered to be at their normal tap

(c) Lines

Line charging capacitances and other shunt connections to earth are neglected. This is due

to the fact that voltages dip very low and currents drawn by them are small in comparison to

fault current. If the resistances of the lines are smaller than the reactances by a factor of six

or more, the resistances are neglected to obviate the need for complex arithmetic.

(d) Loads

Normally, loads are neglected for the same reason given for case (c).

2-2.3 Analytical tools for three-phase faults

In a balanced three phase circuit, since the information relating to one single phase gives the

information relating to the other two phases as well, it is sufficient to do calculations in a

single phase circuit. There are two common forms used. These are (i) to take any one single

phase of the three phase circuit and (ii) to take an equivalent single phase circuit to

represent the full three phase circuit.

(a) Single phase circuit

Figure 1 shows one single phase „AN‟ of the three phase circuit „ABC N‟. Since the system

is balanced, there is no current in the neutral, and there is no potential drop across the

24

neutral wire. Thus the star point „S‟ of the system would be at the same potential as the

neutral point „N‟.

asp II

asp VV 3

TP

S

A

N

sZ

aE

Z

Figure 1: Single phase circuit

Also, the line current is the same as the phase current, the line voltage is √3 times the phase

voltage, and the total power is 3 times the power in a single phase.

Lp III , 3/Lp VVV and 3/Tp SSS

Working with the single phase circuit would yield single phase quantities, which can then

be converted to three phase quantities using the above conversions.

(b) Equivalent single phase circuit

Of the parameters in the single phase circuit shown in figure 6, the line voltage and the total

power (rather than the phase voltage and one-third the power) are the most important

quantities. It would be useful to have these quantities obtained directly from the circuit

rather than having conversion factors of √3 and 3 respectively. This is achieved in the

equivalent single phase circuit, shown in figure 2, by multiplying the voltage by a factor of

√3 to give Line Voltage directly.

S

A

N

sZ

aL EE 3

asL III 33

asL VV 3

TP

Z

Figure 2: Equivalent single phase circuit

The Impedance remains as the per-phase impedance. However, the Line Current gets

artificially amplified by a factor of√3. This also increases the power by a factor of 23 ,

which is the required correction to get the total power.

Thus, working with the equivalent single phase circuit would yield the required three phase

quantities directly, other than the current which would be LI3 .

2-2.4 Calculation of three-phase short-circuit quantities

Two conditions will be studied.

Case 1: System is unloaded (i.e. generators are not loaded)

Example 2.1

25

In the radial network shown in the figure below, a three-phase fault occurs at point F.

Determine the fault current and the line voltage at the kV11 bus under fault conditions.

G1

G2T1 T2

kV11

Overhead line Cable

kV33 kV6.6

F

Generator 1: MVA10 , 15% reactance

Generator 2: MVA10 , 12.5% reactance

Transformer 1: MVA10 , 10% reactance

Overhead line: kmL 30 , kmjZ 36.027.0 Transformer 2: MVA5 , 8% reactance

Cable: kmL 3 , kmjZ 08.0135.0 Solution

Selecting a system base of MVA100 and base voltages of kV11 in generators, kV33 for

overhead line and kV6.6 for cable, and using the formula: )(

)(

)()(

oldbase

newbase

oldpunewpuS

SXX

Reactance of generator 1 pujj

5.110

10015.0

Reactance of generator 2 pujj

25.110

100125.0

Reactance of transformer 1 pujj

110

1001.0

Reactance of transformer 2 pujj

6.15

10008.0

Overhead line impedance 23

6

21033

10100)36.027.0(30)(

j

V

SZ

base

base

puj 99.0744.0

Cable impedance pujj 55.093.0)106.6(

10100)08.0135.0(3

26

6

The equivalent circuit is shown below

1GX 2GX

1TX LZ

E

2TX CZFI

(a) Total impedance, ZT FCTLTGG IZXZXXX 2121 //

26

08.701.5

82.4674.1

55.093.06.199.0744.0125.1//5.1

j

jjjjjj

pu

Z

EI

T

F

0

0

0

8.70196.08.701.5

01

A

V

SI

base

base

base 8750106.63

10100

3 6

6

AIII basepuactualF 17158750196.0

(b) Total impedance between F and kV11 bus, Z

08.7643.4

14.4674.1

99.0744.06.155.093.01

j

jjjj

Voltage at kV11 bus puZIF

000 688.08.7643.48.70196.0

kVkV 68.91188.0

Example 2.2

In the network shown below, a three-phase fault occurs at point F. Calculate the fault MVA

at F. All given reactance values have been referred to a base of 100MVA. Resistances have

been neglected.

5.0j

3.0j3.0j

5.0j

1.0'' X 1.0'' X

1.0'' X 1.0'' X

1 2

34

F

Solution

The equivalent single-phase network of the generator and line reactances is shown below:

27

12 3 4

F

5.0j

5.0j

3.0j

3.0j

1.0j 1.0j1.0j

1.0j

a

b

c

de

f

12 3 4

F

5.0j

5.0j

3.0j

1.0j

1.0j

a

b

c

e d

f02.0j

06.0j06.0j

12 3 4

F

3.0j

1.0j

1.0j

a

c

d

f

02.0j

56.0j

56.0j

(a) (b)

F

3.0j

1.0j d

56.0j

082.0j

003.0j

016.0j

Total reactance 07.01.0//03.0082.03.0//016.056.0 jjjjjjjX

The fault level at point F MVAZpu

Sbase 143007.0

10100 6

Example 2.3

A generating station is laid down as shown below. The ratings and percentage reactances of

different elements are as indicated. Calculate the kVAs and currents at bus bar D and at the

distance end of the feeder at I.

28

D

H

F

A B C

E

I

X X X

kV6.6

kVMVA 11,5%,5

Generator reactance: 30% at 10MVA

Reactor X: 10% at 10MVA

Feeder has 8% reactance and 6% resistance at 5MVA

Solution

(a) Referring all values to a base of 5MVA

Generator reactance %1510

5%30

MVA

MVA

The equivalent circuit is shown below

%15

%15 %15

%5 %5

%5

Total reactance %5.715//5515//515

Fault level kVAZpu

Sbase 66700075.0

105 6

Fault current AV

levelFault

base

5830106.63

1066700

3 3

3

(b) The equivalent circuit is shown below

29

%15

%15 %15

%5 %5

%5

%5

86 j

Total impedance 86515//5515//515 jj

7.73%4.21%5.206

8655.7

j

jjj

Short-circuit level (i.e. fault level) kVA23400214.0

105 6

Short-circuit current A124010113

10234003

3

Case 2: System is loaded

The voltage behind the sub-transient reactance for the generator is obtained as

dggg XjIVE

00

gE

0

gV

dX

0gI

and for synchronous motors, the voltage behind the subtransient and transient reactances are

given respectively by:

X0

I0

E1

X1

I1 I2

X2

V1V0 V2

mmmm XjIVE

00

dX

mE 0

mV

0mI

Where 0V and

0I are the respective prefault bus voltages and currents.

30

In the above examples, the bus voltages were used to represent the synchronous machines.

These voltages according to the above equations for

gE and

mE or

mE are the voltages

behind the subtansient or transient reatances when the synchronous machines are not

loaded. i.e.

000 mg II .

Example 2.4

A synchronous generator rated 25MVA and a synchronous motor each rate 25MVA, 11kV

having 15% subtransient reactance are connected through transformers and line as shown in

the figure below. The transformers are rated 25MVA, 11/66kV and 66/11kV with leakage

reactance of 10% each. The line has a reactance of 10% on a base of 25MVA, 66kV. The

motor is drawing 15MW at 0.8pf leading and a terminal voltage of 10.6kV when a

symmetrical 3-phase fault occurs at the motor terminals. Find the subtransient current in the

generator, motor and fault path.

M

Generator Motor

T1 T2

Line

Solution

The prefault circuit is shown below

gE

mEM

15.0j 15.0j

1.0j 1.0j 1.0j

kVVm 6.100

pu9636.011

6.10 at a base voltage of kV11

Load pfMW 8.0,15 leading

pfpu 8.0,6.0

25

15 leading at a base of 25MVA

Prefault current, 0110 9.367783.08.0cos8.09636.0

6.0cos

cos

V

PI

28.07536.0

9.367783.01.01.01.015.009636.0 00

00

j

j

IjXVE dmg

0934.00337.1

9.367783.015.009636.0 00

000

j

j

IjXVE mmm

The post fault circuit is shown below

31

gE

mEM

15.0j 15.0j

1.0j 1.0j 1.0j gI

mI

fI

Applying Kirchhoff‟s voltage law,

674.16222.0

45.0

28.07536.0

01.01.01.015.0

j

j

j

jjjj

EI

g

g

8915.66224.015.0

0934.00337.1j

j

j

X

EI

m

m

m

5653.80002.08915.66224.0674.16222.0 jjjIII mgf

Self Assessment 2-2

1. For the system in the figure below, the ratings of the various components are:

Generator: 25MVA, 12.4kV, 10% subtransient reactance

Motor: 20MVA, 3.8kV, 15% subtransient reactance

Transformer T1: 25MVA, 11/33kV, 8% reactance

Transformer T2: 20MVA, 33/3.3kV, 10% reactance

Line: 20 ohms reactance

M

Generator MotorT1 T2

Line

1. The system is loaded so that the motor is drawing 15MWat 0.9 lagging power factor, the

motor terminal voltage being 3.1kV. Find the subtransient current in generator and

motor for a fault at generator bus. ANS: kAkA 93.4,87.8

2. Two synchronous motors are connected to the bus of a large system through a short

transmission line as shown below. The ratings of various components are:

Motors (each): 1MVA, 440V, 0.1 pu transient reactance

Line: 0.05 ohm reactance

Large system: Short circuit MVA at its bus at 440V is 8.

When the motors are the operating at 440V, calculate the short circuit current

(symmetrical) fed into a three-phase fault at motor bus.

M

M

Motor bus

Line

Bus

Large system

ANS: 29.96kA

32

3. A synchronous generator rated 500kVA, 440V, 0.1pu sub-transient reactance is supplying

a passive load 400kW at 0.8 lagging power factor. Calculate the initial symmetrical rms

current for a three-phase fault at generator terminals. ANS: 6.97kA

4. The system shown below is delivering 50MVA at 11kV, 0.8 lagging power factor into a

bus which may be regarded as infinite. Particulars of various system components are:

Generator: 60MVA, 12kV,

dX = 0.35pu

Transformers (each) 80MVA, 12/66kV, X = 0.08pu

Line: 12-ohm Reactance, negligible resistance.

Calculate the symmetrical current that the circuit breaker A and B will be called upon

to interrupt in the event of a three-phase fault occurring at Y near the circuit breaker

B. ANS: 8.319kA

33

2.4 Method of symmetrical components

A power system is normally treated as a balanced three-phase network. In general, when a fault

occurs, the symmetry of a balanced network is upset, resulting in unbalanced currents and voltages

appearing in the network. The exception to this rule is the three-phase fault, which because it

involves all three phases equally at the same time and location, is described as a symmetrical fault.

These unsymmetrical fault conditions can be analysed using the symmetrical component theory.

This method symmetrical components formulates a system of three separate phasor systems which

when superposed give the true unbalance conditions in the circuit. It must be emphasised that the

three phasor systems are essentially artificial and used merely as an aid to calculation. The various

sequence-component voltages and currents do not exist as physical entities in the network.

The method postulates that a three-phase unbalanced system of voltages and currents may be

represented by the following three separate system of phasors:

(a) a balanced 3-phase system (equal in magnitude but 120o degrees apart) in the normal a-b-c

sequence, called the positive phase sequence (pps) system.

(b) a balanced 3-phase system of reversed sequence i.e. a-c-b called the negative phase

sequence (nps) system.

(c) three phasors equal in magnitude and phase revolving in the positive phase rotation called

the zero phase sequence (zps) system.

It is common practice to assign 1, 2, and 0 to the pps, nps and zps components respectively.

Figure 2.3 shows an unbalanced system of voltages (the angular displacement between the three

phases is not 120o) with the corresponding system of symmetrical components

Va

Vb

Vc

Va0

Vc1

Vb1

Va1

Vc2

Vb2

Va2

Vc0

Vb0

systemActual systempps systemnps systemzps

Fig 2.3 Symmetrical components of an unbalanced system

000 cba VVV

If 11 VVa , then 0

11 120VVb (anticlockwise angle is negative) and

0

11 120VVc (clockwise angle is positive).

Similarly, If 22 VVa , then 0

22 120VVb (clockwise angle is positive) and

0

22 120VVc (anticlockwise angle is negative).

It is normal practice to take the a-phase as reference and express the other two phases in terms of

the a-phase quantities. Thus:

2102a1a0aa VVVVVVV

(Phase ‘a’ voltage is the sum of the phase ‘a’ voltage of all the three separate phasors)

21

2

02b1b0bb hVVhVVVVV Eqn. 2.1

34

2

2

102c1c0cc VhhVVVVVV

where 866.05.01201 jh and 866.05.012012 jh

In a matrix form, we can write equation 2.1 as:

2

1

0

2

2

1

1

111

V

V

V

hh

hh

V

V

V

c

b

a

Eqn. 2.2a

c

b

a

V

V

V

hh

hh

V

V

V1

2

2

2

1

0

1

1

111

c

b

a

V

V

V

hh

hh2

2

1

1

111

3

1 Eqn. 2.2b

Similarly,

2

1

0

2

2

1

1

111

I

I

I

hh

hh

I

I

I

c

b

a

or

c

b

a

I

I

I

hh

hh

I

I

I1

2

2

2

1

0

1

1

111

c

b

a

I

I

I

hh

hh2

2

1

1

111

3

1

Eqn. 2.3

Note the new positions of h and2h in the matrices.

Additionally, 1h3 , 1hh 2 , 0hh1 2 , 0hhh 32

.

2.4.1 Representation of plants in the phase-sequence networks

In order to apply symmetrical components to a power system, the various components of the power

system (generators, transformers, lines, etc) must be given impedances which reflect the three

phasor systems. For example, a generator must have a pps impedance, nps impedance and zps

impedance in other for it to be factored into a fault analysis using the method of symmetrical

components.

(a) The synchronous machine

The pps impedance Z1 of the stator winding is the normal transient or sub-transient value, the latter

being about th101 of the synchronous impedance of the machine. The nps impedance Z2 is normally

quite close to the pps impedance under fault conditions. The zps impedance Z0 depends upon the

nature of the connection between the star point of the windings and the earth. Resistors or reactors

are frequently connected between the star point and earth for reasons usually connected with

protective gear and the limitation of over-voltages. Figure 2.4 shows the connection of earthing

impedance.

35

Z0, Z1, Z2

E1

h2E1

hE1

Earthing

impedance, Zn

Figure 2.4 Earthing impedance connection to a synchronous machine

Normally, the only voltage sources appearing in the sequence networks are the pps voltages, as the

generators only generate pps emfs. This is because it is the pps whose sequence is the same as that

of a generator. Whenever earthing impedance is present, three times (3X) its value is added in

series with the zero sequence impedance. The sequence networks representing a synchronous

generator are thus shown in figure 2.5.

3Zn

Z0

I0

E1

Z1

I1

nps

I2

Z2

pps zps

Fig 2.5 Symmetrical components of a synchronous machine

(b) Lines and cables: The pps and nps impedances are the normal balanced values. The zps

impedance depends upon the nature of the return path through the earth. It is also modified by the

presence of an earth wire on the tower which protects the lines against lightning surges. The

following rough guide to the value of Z0 may be used. Typically for an underearth cable pps and

nps impedances are identical but the ratio 53/ 10 ZZ for a 3-cored cable and 1 – 1.25 for a single

core cable.

(c) Transformers: The pps and nps impedance are the normal balanced ones. The zps

connection of transformers is however complicated and depends on the connection of the winding.

Table 2.1 presents the zero-sequence representation of transformers for various winding

arrangements. Zero-sequence currents in the windings on one side of a transformer must produce

the corresponding ampere-turns in the other. The table can be understood by first understanding

figure 2.6.

36

Although in practice, a number of different connections are used, the most common type is the

delta-star connection. In such an arrangement, if there are zps currents present, then they simply

circulate around the delta winding and cannot flow in the lines (or other plant) outside it. This is so

because of the absence of a neutral on the delta side. Zero sequence currents will flow only when

there is a connection to earth.

The following simple rule can be followed to represent zero sequence networks:

Shunt switch

Series switch0Z

Primary side Secondary side

Fig 2.6 Zero sequence representation

Close the shunt switch on a side, if the connection on that side is delta. Close the series switch on a

side if the connection on that side is star earthed. Otherwise, leave the switches open. It must be

noted that on the primary side the series switch is before the shunt but on the secondary side, the

series is after the shunt. Additionally, for generators and motors, a side is closed when it is star

earthed and opened when is star clear of earth.

Table 2.1 Zero-sequence representation of various transformers

imaryPr Secondary

0Z

0Z

0Z

0Z

0Z

Zero-sequence currents free to flow in

both primary and secondary circuits

No path for zero-sequence currents in

primary circuits

Single-phase currents circulate in the

delta but not outside it

No flow of zero-sequence currents

possible

No flow of zero-sequence currents

possible

Tertiary winding provides path for

zero-sequence currents

CONNECTION OF WINDINGS REPRESENTATION PER PHASE COMMENTS

37

Example 2.1

For the power system whose one-line diagram is shown below, sketch the zero sequence network.

m n u v1T 2T1G 2G

nZ

Solution

1ogZ10TZ LZ0 20TZ

20gZ

nZ3

A reference line is initially placed (usually at the top) and starting from the left, the first element

encountered is a generator. Since it is star earthed, its impedance is closed to the reference line. The

impedance value of the generator is 10gZ and since the neutral is earthed through a impedance, 3

times its value ( nZ3 ) is added in series with the 10gZ . The next component encountered is the

transformer, the primary side is delta and so the series which is left open and the shunt switch is

closed, the zeros sequence impedance of the generator 10TZ is then introduced. We then proceed to

the secondary side which is star earthed, so the series switch is closed and the shunt is left open

(there is no need to indicate this opened shunt switch). We then introduce the impedance overhead

line and proceed to the second transformer. The primary side is star earthed so the series switch is

closed and the shunt opened. The impedance of the transformer is introduced and we proceed to the

secondary side where we encounter a delta, so the shunt switch is closed and the series is opened (it

is necessary to indicate an opened series switch). The next component is the second generator

whose impedance is shown but will not be closed to the reference line because it is star clear of

earth.

Example 2.2

Repeat example 1, interchanging the positions of transformer 1 and 2. Generator is replaced with a

star solid-earthed one.

Solution

1ogZ10TZLZ020TZ

20gZ

nZ3

38

Example 2.3

What would be the total zero sequence impedance for the network in example 2 if an earth fault

occurred between 10TZ and 20gZ ?

Solution

The circuit to be used to solve this problem is as indicated below. A line assumed to be carrying

current is dropped on the reference line and is made to exit at the point where the fault occurred.

The total zero sequence impedance is then the impedance encountered by this current as it goes

through the circuit. From the diagram, the impedance encountered is a parallel combination of 20TZ

and 20gZ . The remaining impedances are not included because they have all been open-circuited.

1ogZ10TZLZ020TZ

20gZ

nZ3

i

i

2010

201020100 //

gT

gTgT

ZZ

ZZZZZ

Self assessment 2.1

What is the total zero sequence impedance in example 1 if a fault occurred at the input (primary)

point of transformer 1?

39

2.4.2 Analysis of various faults

In general, we can say that the three sequence networks are as shown in figure 2.7 in which the

sequence impedances are the total impedances which can comprise generators, lines and

transformers. Depending upon the type of fault, the three networks can then be connected in order

to determine the fault currents, voltages, etc. The impedances 0Z , 1Z and 2Z are the total

impedances obtained from the respective sequence networks such as in examples 3 and self

assessment 2.1. You may have realised that you have not been introduced to how to produce the

pps and nps networks as well as finding their impedances. This is because the pps and nps networks

are the normal one-line diagrams (you studied this in EE 365). Their sequence impedances are

found the same way you will find the total zero sequence impedance from a zps network. Do not

worry; the examples below will help you.

Z0

I0

zps

E1

Z1

I1

pps nps

I2

Z2V1V0 V2

Fig 2.7 Unsymmetrical fault components

The next thing we would want to do is develop circuit models for the various fault cases so that

each time such a fault occurs, we will reproduce the appropriate model, substitute the values, and

halleluiah!!!, we are almost through with the analysis.

(a) Single-phase-earth faults: Consider an a-phase-earth fault at point F on a single-end-fed

system as shown in figure 2.8

Ifc

Ifb

Ifa

Vfa

Vfb

Vfc

F

Fig 2.8 A single-phase-earth fault

Note: In the diagram, the three generators and three transformers are just being used to indicate that

each phase is being supplied. In actual fact, it is one three-phase generator, one three-phase

transformer and three separate lines.

40

For this fault, the constraints are:

0Vfa , 0II fcfb

Note: the subscripts „f‟ are just being used to denote the fact that the values being sought for are

faulted values. 0Vfa because the voltage at the fault point is the same as earth voltage which is

zero. 0II fcfb since phase b and c will not be feeding any fault. It must however be noted that

the load currents in these phases will not be zero. It is just that these load currents do not have any

fault component in them.

Using equation 2.3, we then get

fa31

2f1f0f IIII Eqn. 2.4

i.e. from

.3

1

0

0

1

1

111

3

1

2

2

2

1

0

a

a

a

c

b

a

I

I

I

I

I

I

hh

hh

I

I

I

Once again note that the presence or absence of the subscript ‘f’ does not change anything.

This result shows that all the sequence networks are in series since they have the same current (i.e.

fa31

2f1f0f IIII ). The network connection diagram is thus:

Z0

If0

E1

Z1

If1 If2

Z2

If1

Z1

Z2

Z0

If2

If0

E1

Fig 2.9 Sequence network representation of single-phase-earth faults

From figure 2.9, it can be seen that the fault current is given by (by applying Kirchhoff‟s voltage

law):

210

12f1f0ffa

ZZZ

E3I3I3I3I

Eqn. 2.5

The fault point voltages can be likewise obtained. Fig 2.8 is thus the model for single-phase to earth

faults and one of the associated equations is eqn. 2.5.

41

(b) Phase-phase fault: Consider a b-c phase fault as shown in figure 2.10.

Ifc

Ifb

Ifa Vfa

Vfb

Vfc

Fig 2.10 Phase-phase fault

For this fault, the constraints are:

0I fa , fcfb II

Since the fault is not linked to earth, 0VI 0f0f . From the matrix relationship of equation 2.3, we

get

fb

2

31

fc

2

fb31

1f IhhIhhII

fb

2

31

fcfb

2

31

2f IhhhIIhI Eqn. 2.6

i.e. 0II 2f1f

Also 2f1f

2

fb hVVhV

2f

2

1ffc VhhVV

And since for this fault fcfb VV , it thus follows that 2f1f VV . The network connection is thus:

E1

Z1

If1 If2

Z2Vf1 Vf2

Fig 2.11 Sequence network connection of phase-phase fault

It should be noted that for this fault condition, the zps quantities are zero and this is true for any

fault clear of earth.

42

(c) Double-phase-earth fault: Consider a b-c-earth fault as shown in figure 2.12.

Ifc

Ifb

Ifa Vfa

Vfb

Vfc

Fig 2.12 Double-phase-earth fault

Here, the constraints are: 0I fa , 0VV fcfb .

Again, using the matrix relationship of equation 2.3a, we get

3

VVVV fa

2f1f0f

By matrix manipulation, it can also be shown that

0III 2f1f0f

The required sequence network is thus:

Z0

If0

E1

Z1

If1 If2

Z2

Vf1V0 Vf2

Fig 2.13 Sequence network connection of double-phase-earth fault

43

(d) Three-phase-earth fault: Consider the faulted system as shown in figure 2.14.

Ifc

Ifb

Ifa Vfa

Vfb

Vfc

Fig 2.14 Three-phase to earth fault

For this fault, the constraints are:

0VVV fcfbfa and

0III fcfbfa

By using the above constraints in conjunction with the matrix relationships, it can be shown that

0VVV 2f1f0f , 0II 2f0f and fa1f II

It can thus be seen that for this type of fault, only the pps network exist and this is true because as

mentioned before, this is a symmetrical fault. The network is thus:

E1

Z1

If1=Ifa

Vf1=0

Fig 2.14 Sequence network connection of three-phase fault

44

Example 2.4

A three-phase star-connected alternator with its neutral point solidly earthed is excited in such a

way as to give 3.3kV on open circuit. Its positive, negative and zero sequence impedances are j6.0,

j2.0 and j1.2 ohms/phase respectively. A resistive fault occurs between a single line and earth at the

alternator terminals on phase „a‟. The resistance Rf of the fault is .5.0

Show that the fault resistance can be represented by a resistor of value 3Rf in series with the zero

sequence impedances and calculate:

(a) the fault current

(b) the voltage of phase „b‟-earth at the fault point.

Solution

The voltage across the fault resistance 𝑉𝑓 is given by:

fff IRV faf IR But 5.2.3 03

10 eqnfromIIII fafa

0f I3R

0f IR3

Note: 𝐼𝑓 = 𝐼𝑓𝑎 ie., the fault current in the resistance is the phase a fault current.

The sequence network connection is

Z0

I0

E1

Z1

I1 I2

Z2

3Rf

V1 V2V

0

Note: 3𝑅𝑓 is inserted in series with the zero sequence branch. This will be done whenever a fault

resistance or impedance is present.

(a) f021

1021

R3ZZZ

EIII

5.032.1j2j6j

3/3300

o74.804.204

But fault current 021f IIII

4.2043I3 1

45

A2.613

(b) 211

2

0b VhVhVV

111

2

1

2 ZIEhVh

ooo 90674.804.2043/33001201

07.5033.518 j

222 ZIhhV

ooo 90274.804.2041201

53.31671.258 j

000 ZIV

oo 902.174.804.204

47.3908.242 j

47.3908.24253.31671.25807.5033.518 jjjVb

Vj o29.12084.99407.8597.501

Note: 𝑉0 is not the voltage across both 𝑍0 and 3𝑅𝑓 but only 𝑍0 .

Example 2.5

A 100km, 400kV transmission line is subjected to an „a‟ phase to earth fault at 80km. if the fault

resistance is 10 , calculate the fault current.

)61(21

jZZGG

, )8j5.0(Z 0G ,

)31j8.1(ZZ 2L1L and )88j10(Z 0L

Solution

80km

100km

ZG

3

kV400

It has been assumed that there was no load prior to the fault. The impedances given are for the

entire length ie. 100𝑘𝑚 of the line and since the fault occurred at 80𝑘𝑚, then the impedance to be

used in the analysis must be 80

100𝑍.

46

kV3

400E1 and 8.0

100

80

4.78j5.888j108.08j5.0ZZZ 0L0G0

8.30j44.231j8.18.06j1ZZZ 1L1G1

8.30j44.231j8.18.06j1ZZZ 2L2G2

Thus, f210

1210

R3ZZZ

EIII

1038.30j44.28.30j44.244.78j5.83

10400 3

160j38.433

10400 3

71.150411.466 j

Hence, the fault current

0

0 79.7276.472514.451434.1398)71.150411.466(33 jjIIa A

Example 2.6

A synchronous machine A generating 1p.u. voltage is connected through a star-star transformer of

reactance 0.12p.u. to two lines in parallel. The other ends of the lines are connected through a star-

star transformer of reactance 0.1p.u. to a second machine B, also generating 1p.u. voltage. For both

transformers, 021 XXX . Calculate the current fed into the double-line-to-earth fault on the

line-side terminals of the transformer fed from A.

The relevant per unit reactances of all plant, referred to the same base, are as follows: For each line

30.0XX 21 , 70.0X0 . For machine A, 30.0X1 , 20.0X2 , 05.0X0 . For machine B,

25.0X1 , 15.0X2 , 03.0X0

The star points of machine A and of the two transformers are solidly earthed.

A B

F

Solution

47

1F

30.0

12.0

30.0

30.010.0

25.0

AE BE

networksequencePositive

42.0 50.0

1E

1F

2F

20.0

12.0

30.0

30.010.0

15.0

networksequenceNegetive

32.0 40.0

2F

0F12.0

70.0

70.010.0

03.0

networksequenceZero

05.0

.C.O

17.0 48.0

2F

.C.O

From the sequence diagrams, .u.p23.0j50.0j42.0j

50.0j42.0jX1

, .u.p18.0j

40.0j32.0j

40.0j32.0jX2

and

.u.p17.0jX0

The „a‟ phase is taken as reference phasor and the „b‟ and „c‟ phases are assumed to be shorted at

the fault point. From the equivalent circuit for a line-to-line fault,

X0

I0

E1

X1

I1 I2

X2

V1V0 V2

48

u.p

XXXXX

1I

20201

1

u.p18.017.018.017.023.0j

1

.u.p15.3j

20

012

XX

XII

.u.p53.1j

18.017.0

17.015.3j

20

210

XX

XII

u.p62.1j

18.017.0

18.015.3j

21

2

0b hIIhII

u.p72.4.u.p43.2j05.4

53.1j866.0j5.015.3j866.0j5.062.1j

2

2

10c IhhIII

53.1j866.0j5.015.3j866.0j5.062.1j

.u.p72.4u.p43.2j0.4 pujjjIII cbf 86.405.043.20.443.205.4

Example 2.7

An 11kV synchronous generator is connected to a 11/66kV transformer which feeds a 66/11/3.3kV

three-winding transformer through a short feeder of negligible impedance. Calculate the fault

current when a single-phase-to-earth fault occurs on a terminal of the 11kV winding of the three-

winding transformer. The relevant data for the system are as follows:

Generator: .u.p15.0jX1 , .u.p1.0jX2 , .u.p03.0jX0 , all on a 10MVA base; star point of

winding earthed through a 3 resistor.

11/66kV transformer: .u.p1.0jXXX 021 on a 10MVA base; 11kV winding delta connected

and the 66kV winding star connected with the star point solidly earthed.

Three-winding transformer: A 66kV winding, star connected point solidly earthed; 11kV winding,

star connected, star-point earthed through a 3 resistor; 3.3kV winding, delta connected; the three

winding of an equivalent star connection to represent the transformer have sequence impedances,

66kV winding .,u.p04.0jXXX 021 11kV winding .,u.p03.0jXXX 021 3.3kV winding

.,u.p05.0jXXX 021 all on a 10MVA base. Resistance may be neglected throughout.

49

F

3

3

11/66kVp

t

s

o/c

If

Solution

15.0j

05.0j

03.0j

diagramsequencePositive

32.0j

pu1

1F

1.0j 04.0j

1F

pu1

1.0j

05.0j

03.0j

diagramsequenceNegetive

27.0j

2F

1.0j 04.0j

2F

1.0j

05.0j

03.0j

diagramsequenceZero

0 067j .

0F

1.0j 04.0j25.03

25.03

25.03

0F

The positive-, negative-, and zero-sequence networks are shown in the figures above. Much care is

needed with the zero-sequence network owing to the transformer connections. Using the same base

MVA, the 3 earthing resistor will have the following pu value:

25.0

1011

1010323

6

2

base

baseactualpu

V

SRR

For a line-to-earth fault, the equivalent circuit in figure 2.8 is used, but with a slight modification.

From this figure,

f021

1210

R3ZZZ

EIII

0

3 13

0 32 0 27 0 067 0 75

3

0 75 0 66

33 41

1 41

f

o

o

I Ij . j . j . .

. j .

AI f 1575110003

10103

6

i.e.

base

basebasepuactual

V

SIII

33

50

Example 2.8

A synchronous generator with terminal voltage 11.8kV, rating 20MVA and pps transient reactance

.u.p08.0X 1' is subjected to a 3-phase short circuit. Calculate the fault current.

Solution

base

base

S

VupXX

2'

' ).(.)(

557.0

1020

1180008.06

2

Taking E1 to be 3

kV8.11, and from figure 2.14, the fault current

kAX

E

Z

EII f 2.12

557.03

108.11 3

1

1

1

11

Example 2.9

A 25MVA, 11kV generator supplies two motors through two transformers. The one-line diagram of

the system is shown below. If a double-line-earth fault occurs between phases „b‟ and „c‟ at point

G, find the fault current given a fault path reactance of j0.1p.u. The system has the following data:

Generator: ..2.021 upjXX , ..06.00 upjX

11/121kV Transformer: ..0805.0021 upjXXX

121/11kV Transformer: ..0805.0021 upjXXX

Transmission line: 164.021 jXX 494.00 jX

Motor A: ..345.0021 upjXXX

Motor B: ..69.0021 upjXXX

M

M A

B

516.0j

G

Solution

Note: A motor is represented by a constant voltage source behind sub-transient or transient

reactance just like a generator.

2.0j

0805.0j 164.0j 0805.0j

345.0j 69.0j

gE 1mE 2mE

G Positive sequence network

51

2.0j

0805.0j 164.0j 0805.0j

345.0j 69.0j

G Negative sequence network

516.0j3

0805.0j 494.0j 0805.0j

345.0j 69.0j06.0j

G

Zero sequence network

1

2 1

0

0 2 0 0805 0 164 0 0805 0 345 0 69 0 16

0 16

0 69

Z j . j . j . j . // j . // j . j . p.u.

Z Z j . p.u.

Z j . p.u

The equivalent network of a double-line-to-earth with fault impedance is as shown.

Z0

If0

E1

Z1

If1 If2

V1

V0

V2

3Zf

Z2

fZZZZ

EI

3// 021

11

..359.3

1.0369.0//16.016.0

1upj

jjjj

upjjjj

jI

ZZZ

ZZI

f

f.892.2359.3

99.016.0

99.0

3

31

02

0

2

52

467.0359.399.016.0

16.0

31

02

20 jj

jj

jI

ZZZ

ZI

f

21

2

0 hIIhII fb

upupj

jjjjj

.63.172459.5...7005.0.0413.5

892.2866.05.0359.3866.05.0467.0

2

2

10 IhhIII fc

upupj

jjjjj

.73.7459.5...07005.0413.5

892.2866.05.0359.3866.05.0467.0

Fault current pujjjII fcfb 401.17005.0413.57005.0413.5

Assignments

1. For the 400kV system of the figure below, calculate the following:

(a) Va, Vb, Vc, Ia, Ib and Ic for a phase „a‟ to earth fault with no fault resistance.

(b) Vb, Vc and Ib for a „b‟ to „c‟ fault with no fault resistance

(c) Vb, Vc, Ib and Ic for a „b‟ to „c‟ to earth fault with a fault resistance of 5 .

16j0.1ZZ 2s1s , 8.j5.0Z 0s ; 32j8.1ZZ 2L1L , 88j10Z 0L .

[Assume no prefault load flow]

80%

Zs

V

2. Deduce the sequence network connections for the following system under a resistive „a‟ phase

to earth fault.

A B

fZ

3. Two 11kVA, 20MVA three-phase, star-connected generators operate in parallel. The positive,

negative and zero sequence reactances of each being

18.0j ,

15.0.0j and 1.0j respectively.

The star point of one of the generators is isolated and that of the other is earthed through a 2

resistor. A single-line-earth fault occurs at the terminals of one of the generators. Estimate (i)

the fault current, (ii) current in earthing resistor and (iii) the voltage across earthing resistor.

53

3. Operation and Control

Aims

To know the importance of frequency and voltage control

To understand the control of power system frequency and voltage

To perform simple calculations on frequency

3.1 Introduction

The function of an electric power system is to convert energy from one of the naturally available

forms to the electrical form and to transport it to the points of consumption. Energy is seldom

consumed in the electrical form but is rather converted to other forms such as heat, light and

mechanical energy. The advantage of the electrical form of energy is that it can be transported and

controlled with relative ease and with a high degree of efficiency and reliability. A properly

designed and operated power system should therefore meet the following fundamental

requirements:

(a) The system must be able to meet the continually changing load demand for active and

reactive power. Unlike other types of energy, electricity cannot be conveniently stored in

sufficient quantities. Therefore, an adequate “spinning” reserve of active and reactive power

should be maintained and appropriately controlled at all times.

(b) The system should supply energy at minimum cost and with minimum ecological impact.

(c) The “quality” of power supply must meet certain minimum standards with regard to the

following factors:

i. constancy of frequency

ii. constancy of voltage; and

iii. level of reliability

Several levels of control involving a complex array of devices are used to meet the above

requirements.

In power systems, both active and reactive power demands are never steady. Continuous regulation

of the following is therefore required in the operation of the power system:

(a) Steam input to turbogenerators or water input to hydrogenerators to match the active power

demand. Failing this, the machine‟s speed will vary with consequent change in frequency.

The maximum permissible change in power frequency is Hz5.0 .

(b) Excitation of generators to match the reactive power demand with reactive generation.

Failing this, voltages at various system buses may go lower than or beyond the prescribed

limits.

In modern large interconnected systems, automatic generation and voltage regulation equipment are

installed on each generator. The controllers are set for a particular operating condition and they take

care of small changes in load demand without frequency and voltage exceeding the prescribed

limits. With the passage of time, as the change in load demand becomes large, the controllers must

be reset either manually or automatically.

Load frequency and excitation controls can be modelled and analysed independently for the

following reasons.

(a) Small changes in power depend on the change in the internal angle, , of the machine and is

independent of the machine bus voltage whereas the bus voltage depends on machine

excitation but is independent of machine internal angle.

(b) The load frequency controller is slow acting because of the large time constant contributed

by the turbine and generator moment of inertia, and excitation controller is fast acting

because the time constant of the field winding is relatively smaller.

54

3.2 Load-Frequency control

When a consumer connects a new load to a power system, it is initially supplied by the kinetic

energy stored in the rotating masses of the turbine-generators. As the kinetic energy is released, the

rotational speed of the generators falls and the system frequency throughout the interconnected

network begins to drop.

Because there are many generators supplying power into the system, some means must be provided

to allocate the change in demand to the generators. A speed governor on each generating unit

provides the primary speed control function, while supplementary control originating at a central

control centre allocates generation.

In an interconnected system with two or more independently controlled areas, in addition to the

control of frequency, the generation within each area has to be controlled so as to maintain

scheduled power interchange. The control of generation and frequency is commonly referred to as

load-frequency control (LFC).

Load frequency control is important for the following reasons:

(a) Load-frequency control is required so that the numerous power stations-small and large-

connected to the system will run satisfactorily in parallel.

(b) Considerable drop in frequency could result in high magnetizing currents in induction

motors and transformers.

(c) Constancy of speed of motors is important for satisfactory performance.

(d) Constancy of frequency is important for satisfactory performance of electric clocks and

other electric time keeping instruments.

3.2.1 Governor operation and characteristics

The control of frequency and load depends on the turbine speed governor. The traditional governor

control system uses the watt centrifugal governor as a speed sensing device and a hydraulic servo-

system to operate the main supply valve. Thus the control system has high dead band (or zone), i.e.

the speed must change by a certain amount before the main valve begins to open. Additionally,

such systems have a low speed response (0.2-0.3s). Large modern turbogenerators use electro-

hydraulic governing systems. This governing system has high speed response, low dead band, and

accuracy in speed and load control.

An imbalance in mechanical power into, and electrical power out of, the generator is seen as an

acceleration or deceleration of the shaft and hence as a change in rotor speed from its nominal

value. The turbine governor acts upon this variation in speed to regulate the steam or water flow

into the turbine such that when the rotor speed falls below its nominal value, more steam or water is

admitted to the turbine to increase mechanical power, and vice versa on an increase in rotor speed.

The control characteristics of governors are set so that a given change in speed results in a specific

change in turbine power, as described by equation 3.1

fR

1PP refm Eqn 3.1

where mP Change in mechanical power

refP Change to power reference setting, refP

f Change in speed expressed in Hz or per unit

R The „regulation constant‟ or „droop‟

55

With the reference speed constant, so that 0Pref , the change in mechanical power is directly

proportional to the change in speed i.e.

fR

1Pm Eqn 3.2

This characteristic is shown graphically in figure 3.1 for .u.p04.0R (i.e. 4% droop) with refP set

to give .u.p0.1Pm at .u.p0.1f i.e. full load at nominal frequency.

0.2 0.4 0.6 0.8 1.0 1.21.00

O.99

0.98

1.01

1.02

1.03

1.04

Power Output (p.u.)

Sp

eed/f

req

uen

cy (

p.u

.)

mP

f

Fig 3.1 Governor characteristic for pu.R 040

The reference value refP is set via equipment called the speed changer motor, or speeder gear, which

acts upon the governor valves to change the operating point (i.e. valve opening) about which the

governor effects the changes mP in response to frequency deviation f . The effect of changing the

speed gear setting (i.e. refP ) upon the governor characteristic is illustrated in figure 3.2. Effectively,

a family of characteristics is achieved, on any one of which the turbine may be set to operate, by

altering the setting of the speeder gear. In figure 3.2, 1P , 2P and 3P represent different outputs at

various speeder gear settings but for the same speed.

0.2 0.4 0.6 0.8 1.0 1.21.00

O.99

0.98

1.01

1.02

1.03

1.04

Power Output (p.u.)

Sp

eed

/fre

qu

ency

(p

.u.)

1P 2P 3PAlternative speeder

settings

Fig 3.2 Governor characteristic showing effect of speeder motor settings

56

The governor characteristics only completely decide the outputs of the machines when a sudden

change in load occurs or when machines are allowed to vary their outputs according to speed within

a prescribed range in order to keep the frequency constant. This latter mode of operation is known

as free-governor-error action.

The basic concepts of speed governing are best illustrated by considering an isolated generating

unit supplying a local load as shown in figure 3.3. When there is a load change, it is reflected

instantaneously as a change in the electrical torque eT output of the generator. This causes a

mismatch between the mechanical torque mT and the electrical torque eT which in turn results in

speed variations. The governor senses the speed variations and sends actuating signals to the valve

or gate to regulate the steam or water flow into the turbine.

Governor

Turbine G

Load PL

GeneratorValve/gate

Steam or water

Speed

Pe

Pm

Tm

Te

Fig 3.3 Generator supplying isolated load

Tm = mechanical torque Te = electrical torque PL = load power

Pm = mechanical power Pe = electrical power

Example 3.1

A MW500 generator supplies an isolated load of 450MW at a nominal frequency of 50Hz. The

governor operates with a 4% droop. If the load falls to 350MW, what is the new frequency?

Solution

On a base of 500MW, the load of 450MW is .u.p9.0500

450

On the same base, the load of 350MW is .u.p7.0500

350

.u.p2.09.07.0Pm

From Equation 3.2,

mPRf

Hz4.0

Hz50008.0

.u.p008.0

2.004.0

Hence, the new frequency Hzfffnew 4.504.050 Hz4.504.050

57

Example 3.2

Two generators operate in parallel to supply a common load of 750MW. The first has a full load

capacity of 600MW, a governor droop of 5% and is currently generating 450MW. The second has a

full load capacity of 500MW, a governor droop of 4% and is currently generating 300MW. The

initial frequency is 50Hz. What is the new frequency immediately following a load rise of 100MW

and what is the load distribution before supplementary control is applied?

Solution

On a common base of 1100MW

Ignoring losses in the transmission system and assuming that the system load is independent of

frequency, the total load change will be met by the changes in generation of the individual

machines, that is:

.u.p091.01100

100

MW100

PPP 2m1mL

With no change in refP for either machine,

21

2m1mLR

f

R

fPPP

21 R

1

R

1f

Hz204.0

Hz500041.0

.u.p0041.0

088.01

0917.01

091.0

R

1

R

1

Pf

21

L

Hence the new frequency is Hz796.49204.050 .

MW.

MW.

.u.p.

.

.

R

fPm

9548

110004450

04450

09170

00410

1

1

088.01100500

04.0R

0917.01100600

05.0R

2

1

58

MW.

MW.

.u.p.

.

.

R

fPm

2651

110004660

04660

0880

00410

2

2

Note: MW...PP mm 211002651954821 is not exactly 100MW due to approximations

made during calculation.

Example 3.3

Two synchronous generators operate in parallel and supply a total load of 200MW. The ratings of

the machines are 100MW and 200MW and both have governor droop characteristics of 4% from no

load to full load. Calculate the load taken by each machine, assuming free governor action.

Solution

Load supplied by machine A, MWx 10004.0

Load supplied by machine B is MWMWx 20004.0

200

From the above equations,

xx

xx

2002

200

200

10004.0

MWx 67.663

200

Hence load on machine B MW33.13367.66200

3.2.2 Generator-Load model

The increment in power input to a generator-load system is DG PP where tG PP is the

incremental turbine power output, assuming generator incremental loss to be negligible. DP is the

increment in load.

The increment in power input to the system is accounted for in two ways:

(a) Rate of increase of stored KE in the rotating mass of the generator. At scheduled frequency

f 0

, the stored energy is

rke PHW 0 kW-sec.

Where rP is the KVA rating of the generator and H is its inertia constant. Kinetic energy is

proportional to the square of speed (frequency). Therefore, the kinetic energy at frequency

fff 0is given by:

020

20200

2

0

00 2

1Pr)(

)(2)(

f

fH

f

ffffW

f

ffWW kekeke

The rate of change of kinetic energy is

fdt

d

f

HPW

dt

d rke

0

2

(b) As frequency changes, load that is frequency sensitive (motor load) changes. The rate of

change of load with respect to frequency can be regarded as constant for small changes in

frequency. Hence, change in load with frequency can be expressed as

59

fBff

PD

B is positive for a predominantly motor load. Writing a power balance equation, we have

fBfdt

d

f

HPPP r

DG 0

2

Dividing throughout by rP , we obtain

fpuBfdt

d

f

HpuPpuP DG )(

2)()(

0

Taking, the Laplace transform, we obtain

sf

HB

sPsPsF DG

0

2

)()()(

sT

KsPsP

Bs

Bf

H

sPsP

ps

ps

DG

DG

1)()(

1.

21

)()(

0

Where0

2

Bf

HTps i.e. power system time constant.

B

K ps

1 i.e. power system gain.

The figure below shows a complete block diagram of load frequency control of an isolated area.

ST

K

sg

sg

1 ST

K

t

t

1 ST

K

ps

ps

1

R

1

)()( sPsP Gt )(sPD

)(sPc

Speed changer

commandGenerator load

Speed governor system

Turbine

)(sF

Fig 3.4: Block diagram of load frequency control of an isolated area

60

Example 3.4

A 100MVA synchronous generator operates on full load at a frequency of 50Hz. The load is

suddenly reduced by 50MW. Due to time lag in governor system, the main supply valve begins to

close after 0.4s. Determine the change in frequency that occurs in this time. kVAkWH sec/5 of

generator capacity. Neglect change in load due to change in frequency, i.e. .0B

Solution

)(2

0puPf

dt

d

f

HD or

H

fpuPf

dt

d D

2

).( 0

5.0100

50)(

MVA

MWpuPD

5.252

505.0

fdt

d

Hzdtf 10)4.0(5.25.25.24.0

0

4.0

0 and

Hzfff 511500

3.2.3 Control area

It is possible to divide an extended power system into sub-areas in which the generators are tightly

coupled together so as to form a coherent group, i.e. all generators respond in unison to changes in

load or speed changer settings. Such a coherent area is called a control area. In a control area, the

frequency is assumed to be the same throughout, in static as well as dynamic conditions.

In practice, a control area may consist of a single large private company, a government-operated

system such as VRA, or several investor-owned companies bonded together in a power pool. A

control area has a single control centre which operates it.

For purposes of developing a control strategy, a control area can be reduced to a single speed

governor, turbo-generator and load system.

Example 3.5

Determine the drop in the frequency of an unregulated isolated control area because of sudden

increase in load demand. The control area has no inter-ties and no automatic regulation. The

dispatcher is absent. The kinetic energy of the rotating mass before the disturbances is 25,000MWs.

The load demand increases from 5000MW to 5200MW. Assume that ./3.83 HzMWB System

frequency = 60Hz.

Solution

fBfdt

d

f

HPP DG

0

2

ffdt

d

3.83

60

2500022000

024.01.0 ffdt

d

By using one of the methods of solving differential equations:

14.2 1.0 tef

61

The time constant, st 101.0

1 . Therefore in about 5 times the time constant, st 50 .

38.299.04.214.2 5 ef

Hence, the final value of the frequency

Hzfff 62.5738.2600

Self Assessment 3.1

Repeat example 3.5 if by inter-ties, kinetic energy available is MW000,2510 and load before the

disturbance is MW500010 . Note that B was estimated as 1% load reduction per 1% frequency

reduction HzMWHz

MW/33.83

%)1(60

%)1(5000

. Thus, with a load of MW500010 the new

HzMWB /3.833 . Comment on your answer.

Self Assessment 3.2

Obtain the transfer function (figure 3.4)DP

F

for 0)( sPC , 1tsg KK , 100psK , 3R ,

sTsg 4.0 , sTt 5.0 and sTps 20 . DP is in per unit. (a) Use the final value theorem to obtain the

final value of f when puPD 01.0 . (b) Determine the dynamic response of F to a step change

in load puPD 01.0 .

3.2.4 Proportional load frequency control

The use of a regulating factor, or gain, in the feedback branch, here known as R

1is called

proportional-feedback control. It does not completely restore the frequency to 50Hz, but does

reduce the deviation.

The steady state load-frequency characteristic of a speed governor system is obtained from the

transfer function in self assessment 3.2 for the free governor action. The droop of the characteristic

curves

RB 1

1

, i.e.

PKP

Bf D

R

1

1[see figure 3.1]. When

R

1represents the total

regulation factor for a system and B represents the system (control area) load change with

frequency, then R

B1

is known as the area frequency response characteristic(AFRC). AFRC of a

system is determined experimentally.

Example 3.6

Two power systems, A and B, are interconnected through a tie-line and are initially at 60Hz. If there

is a 100MW load change in system A, calculate the change in the steady-state values of frequency

and power transfer. The parameters of the system are as follows:

System A: Stiffness HzMWB /1500

Regulation MWHzR /1500

6

System B: Stiffness HzMWB /1000

62

Regulation MWHzR /1000

6

Solution

BBB PPf

7000

6

1000

1

61000

BBA PPf

100

500,10

6100

1500

1

61500

the changes in frequency in each system must be equal as they are electrically connected. Hence,

BB PP 100500,10

6

7000

6

BP

7000

6

500,10

6

500,10

600

BP

7000

500,101100

MWPB 40500,17

7000100

407000

6 Bf

Hz034.0

3.2.5 Proportional plus integral control

To restore a system completely to its normal frequency, the speed changer setting is adjusted

automatically by feeding a signal from f through an integrator to the speed changer. A block

diagram of integral controller has been shown in figure 3.5.

S

K1 )(sPc)(sF

Fig 3.5 Integral controller

3.3 Reactive power and voltage control

For efficient and reliable operation of power systems, the control of voltage and reactive power

should satisfy the following objectives:

(a) Voltages at the terminals of all equipment in the system are within acceptable limits. Both

utility equipment and consumer equipment are designed to operate certain voltage ratings.

Prolonged operation of an equipment at voltages outside the allowable range could

adversely affect their performance and possibly damage them.

(b) System stability is enhanced to maximize utilization of the transmission system. Voltage

and reactive power control have a significant impact on system stability.

(c) The reactive power flow is minimized so as to reduce RI2 and XI2 losses to a practical

minimum. This ensures that the transmission system operates efficiently, i.e., mainly for

active power transfer.

63

The problem of maintaining voltages within the required limits is complicated by the fact that the

power system supplies power to a vast number of loads and is fed from several generating units. As

loads vary, the reactive power requirements of the transmission system vary. Since reactive power

cannot be transmitted over long distances, voltage control has to be effected by using special

devices dispersed throughout the system. This is in contrast to the control of frequency which

depends on the overall system active power balance. The proper selection and coordination of

equipment for controlling reactive power and voltage are among the major challenges of power

system engineering.

3.3.1 Production and absorption of reactive power

(a) Synchronous generators can generate or absorb reactive power depending on the excitation.

When overexcited, they supply reactive power, and when underexcited, they absorb reactive

power. The capability to continuously supply or absorb reactive power is however limited

by the field current, armature current, and end-region heating limits. Synchronous

generators are normally equipped with automatic voltage regulators which continually

adjust the excitation so as to control the armature voltage.

(b) Overhead lines, depending on the load current, either absorb or supply reactive power. At

loads below the natural load (surge impedance), the lines produce net reactive power; at

loads above the natural load, the lines absorb reactive power.

(c) Underground cables, owing to their high capacitance, have high natural loads. They are

always loaded below their natural loads, and hence generate reactive power under all

operating conditions.

(d) Transformers always absorb reactive power regardless of their loading; at no load, the shunt

magnetizing reactance effects predominate; and at full load, the series leakage inductance

effects predominate.

(e) Loads normally absorb reactive power. A typical load bus supplied by a power system is

composed of a large number of devices. The composition changes depending on the day,

season, and weather conditions. The composite characteristics are normally such that a load

bus absorbs reactive power. Both reactive power and active power of the composite loads

vary as a function of voltage magnitudes. Loads at low-lagging power factors cause

excessive voltage drops in the transmission network uneconomical to supply. Industrial

consumers are normally charged for reactive as well as active power; this gives them an

incentive to improve the load power factor by using shunt capacitors.

(f) Compensating devices are usually added to supply or absorb reactive power and thereby

control the reactive power balance in a desired manner.

3.3.2 Methods of voltage control

The control of voltage levels is accomplished by controlling the production, absorption, and flow of

reactive power at all levels in the system. The generating units provide the basic means of voltage

control; the automatic voltage regulators control field excitation to maintain a scheduled voltage

level at the terminals of the generators. Additional means are usually required to control voltage

throughout the system. The devices used for this purpose may be classified as follows:

(a) Sources or sinks of reactive power, such as shunt capacitors, shunt reactors, synchronous

condensers, and static var compensators(SVCs)

(b) Line reactance compensators, such as series capacitors

(c) Regulating transformers, such as tap-changing transformers and boosters.

3.3.2.1 Shunt capacitors

Shunt capacitors supply reactive power and boost local voltages. They are used throughout the

system and are supplied in a wide range of sizes. The principal advantages of shunt capacitors are

their low cost and their flexibility of installation and operation. The principal disadvantage

64

associated with shunt capacitors is that their reactive power output is proportional to the square of

the voltage. Consequently, the reactive power output is reduced at low voltages when it is likely to

be needed most.

3.3.2.2 Shunt reactors

Shunt reactors are used to compensate for the effects of line capacitance particularly to lime voltage

rise on pen circuit or light load. During heavy loading conditions, some of the reactors may have to

be disconnected. They are usually required for EHV overhead lines longer than 200km.

3.3.2.3 Series capacitors

Series capacitors are connected in series with the line conductors to compensate for the inductive

reactance of the line. This reduces the transfer reactance between buses to which the line is

connected, increases maximum power that can be transmitted, and reduces the effective reactive

power ( XI2 ) loss. Although series capacitors are not usually installed for voltage control as such,

they do contribute to improved voltage control and reactive power balance. The reactive power

produced by a series capacitor increases with increasing power transfer; a series capacitor is self-

regulating in this regard.

3.3.2.4 Synchronous condensers

A synchronous condenser is a synchronous machine running without a prime mover or a

mechanical load. By controlling the field excitation, it can be made to either generate or absorb

reactive power. With a voltage regulator, it can automatically adjust the reactive power output to

maintain constant terminal voltage. It draws a small amount of active power from the power system

to supply losses.

3.3.2.5 Static var compensators

Static var compensators are shunt-connected static generators and/or absorbers whose outputs are

varied so as to control specific parameters of the electric power system. The term “static” is used to

indicate that SVCs, unlike synchronous compensators, have no moving or rotating main

components.

3.3.2.6 Tap-changing transformers

Transformers with tap-changing facilities constitute an important means of controlling voltage

throughout the system at all voltage levels. The taps on these transformers provide a convenient

means of controlling reactive power flow between subsystems. This in turn can be used to control

the voltage profiles, and minimize active and reactive power losses.

The table below summarizes the various voltage control devices and their control action.

Table 3.1 Control action of voltage control devices

Voltage control device Control action

1 Shunt capacitors Generate reactive power

2 Shunt reactors Absorb reactive power

3 Synchronous condensers Generate or absorb reactive power

4 Static var compensators Generate or absorb reactive power

5 Series capacitors Reduce reactive power consumption in lines

6 Tap-changing transformers Raise or lower voltage levels

65

3.4 Excitation systems

The basic function of an excitation system is to provide direct current to the synchronous machine

field winding. In addition, the excitation system performs control and protective functions essential

to the satisfactory performance of the power system by controlling the field voltage and thereby the

field current. Excitation systems have taken many forms over the years of their evolution. They

may be classified into the following three broad categories based on the excitation power source

used: DC excitation systems, AC excitation systems and Static excitation systems.

The control functions include the control of voltage and reactive power flow, and the enhancement

of system stability. The protective functions ensure that the capability limits of the synchronous

machine, excitation system, and other equipment are not exceeded.

3.4.1 Excitation system requirements

The performance requirements of excitation systems are determined by considerations of the

synchronous generator as well as the power system.

The basic requirement is that the excitation system supply and automatically adjust the field current

of the synchronous generator to maintain the terminal voltage as the output varies within the

continuous capability of the generator. In addition, the excitation system must be able to respond to

transient disturbances with field forcing consistent with the generator instantaneous and short-term

capabilities. The generator capabilities in this regard are limited by several factors: rotor insulation

failure due to high field voltage, rotor heating due to high field current, stator heating due to

excessive flux (volts/Hz). The thermal limits have time-dependent characteristics, and the short-

term overload capability of the generators may extend from 15 to 60 seconds. To ensure the best

utilization of the excitation system, it should be capable of meeting the system needs by taking full

advantage of the generator‟s short-term capabilities without exceeding their limits.

From the power system viewpoint, the excitation system should contribute to effective control of

voltage and enhancement of system stability. It should be capable of responding rapidly to a

disturbance so as to enhance transient stability, and of modulating the generator field so as to

enhance small-signal stability (explained in chapter 4).

To fulfil the above roles satisfactorily, the excitation system must meet the following requirements:

Meet specified response criteria.

Provide limiting and protective functions as required to prevent damage to itself, the

generator, and other equipment.

Meet specified requirements for operating flexibility.

Meet the desired reliability and availability, by incorporating the necessary level of

redundancy and internal fault detection and isolation capability.

66

3.4.2 Elements of an excitation system

Figure 3.6 shows the functional block diagram of a typical excitation control system for a large

synchronous generator. The following is a brief description of the various subsystems identified in

the figure.

Limiters and

protective circuits

Generator

Power system

stabilizer

Terminal voltage

transducer and

load compensators

Regulator ExciterRef. To power

system

5

4

3

21

Fig 3.6 Functional block diagram of a synchronous generator excitation control system

(1) Exciter. Provides dc power to the synchronous machine field winding, constituting the

power system stage of the excitation system.

(2) Regulator. Process and amplifies input control signals to a level and form appropriate for

control of the exciter. This includes both regulating and excitation system stabilizing

functions.

(3) Terminal voltage transducer and load compensator. Senses generator terminal voltage,

rectifiers and filters it to dc quantity, and compares it with a reference which represents the

desired terminal voltage. In addition, load (or line-drop or reactive) compensation may be

provided, if it is desired to hold constant voltage at some point electrically remote from the

generator terminal (for example, partway through the step-up transformer).

(4) Power system stabiliser. Provides an additional input signal to the regulator to damp power

system oscillations. Some commonly used input signals are rotor speed deviation,

accelerating power, and frequency deviations.

(5) Limiters and protective circuits. These include a wide array of control and protective

functions which ensure that the capability limits of the exciter and synchronous generator

are not exceeded. Some of the commonly used functions are the field-current limiter,

maximum excitation limiter, terminal voltage limiter, volts-per-Hertz regular and

protection, and underexcitation applied to the excitation system at various locations as a

summing input or a gated input.

67

Assignments

1. Two synchronous generators operate in parallel and supply a total load of 240MW. The ratings

of the machines are 150MW and 200MW and both have governor droop characteristics of 5%

from no load to full load. Calculate the load taken by each machine, assuming free governor

action.

2. A 200MW generator supplies an isolated load of 120MW at a nominal frequency of 50Hz. The

governor operates with a 5% droop. If the load drops by 25%, what is the new frequency?

3. Two generators operate in parallel to supply a common load of 850MW. The first has a full

load capacity of 600MW, a governor droop of 5% and is currently generating 500MW. The

second has a full load capacity of 500MW, a governor droop of 4% and is currently generating

350MW. The initial frequency is 50Hz. What is the new frequency immediately following a

load rise of 100MW and what is the load distribution before supplementary control is applied?

68

4. Stability

Aims

To gain understanding of the types of rotor angle stability

To know the conditions for transient and steady state stability

To perform simple calculations on transient and steady state stability.

4.1 Introduction

All power systems, small or large, have synchronous generators, motors and condensers. A power

system in steady state has frequency 0 and machine rotor angle 0 with respect to a synchronous

rotating axis. After having been subjected to a random disturbance, it is required that it remains in a

state of operating equilibrium. Power system stability may be broadly defined as that property of a

power system that enables it to remain in a state of operating equilibrium under normal operating

conditions and to regain an acceptable state of equilibrium after being subjected to a disturbance.

Instability in a power system may be manifested in many different ways depending on the system

configuration and operating mode. Traditionally, the stability problem has been one of maintaining

synchronous operation. Since power systems rely on synchronous machines for generation of

electrical power, a necessary condition for satisfactory operation is that all synchronous machines

remain in synchronism or, colloquially, “in step”. This aspect of stability is influenced by the

dynamics of generator rotor angles and power-angle relationships.

Instability may also be encountered without loss of synchronism. For example, a system consisting

of synchronous generator feeding an induction motor load through a transmission line can become

unstable because of the collapse of load voltage. Maintenance of synchronism is not an issue in this

instance; instead, the concern is stability and control of voltage.

Power system stability problems may therefore be classified into rotor angle stability and voltage

stability. Only rotor angle stability is studied in this course.

4.2 Rotor angle stability

Rotor angle stability is the ability of interconnected synchronous machines of a power system to

remain in synchronism. Rotor angle stability may be categorized into Steady state(small-

disturbance) stability and Transient stability

4.2.1 Steady state stability

Steady state (small-disturbance) stability is the ability of a power system to maintain synchronism

when subjected to small disturbances as occur continually in normal operation due to small

variations in consumption and generation. Instability that may result can be of two forms: (i) steady

increase in rotor angle due to the lack of sufficient synchronizing torque, or (ii) rotor oscillations of

increasing amplitude due to lack of sufficient damping torque.

In steady state stability problem, we are basically concerned with the determination of upper limits

of machine loadings under condition of gradual changes in load.

4.2.1.1 The swing equation

Assuming that windage (effect of damper windings) and frictional torque is negligible, the motion

of the rotor of a synchronous generator is described by the equation:

ei2

m

2

TTdt

dJ

Where J is rotor moment of inertia

69

iT is shaft torque developed by turbine.

eT is electromagnetic torque developed by synchronous machine.

m is angular displacement of rotor in mechanical radians.

We can derive from the above equation the following equivalent equation called the swing

equation.

ei2

2

0

PPdt

dH2

Where iP is shaft input power in per unit.

eP is electromagnetic power in per unit.

H is inertia constant defined as the stored kinetic energy at synchronous speed in MJ or

MW-sec per unit MVA of machine rating.

MVAMJ

P

JH

r

60 102

radiansinspeedssynchronourotorormechanical

MVAinmechaniceofratingPr

0

is the load angle, power angle, torque angle or internal angle.

4.2.1.2 Power transfer

Consider a single synchronous machine connected to an infinite bus through an external impedance

or reactance. Infinite bus is a system assumed to have large generating capacity so that change in

shaft power of the single machine operating to it does not affect its frequency and change in the

excitation of the single machine does not affect its voltage.

Eeee jXRZ

RI

oV 0

Infinite busSynchronous

machine Fig 4.1 A simple power system

The complex power injected into the infinite bus(or the complex power at its receiving end) if

0R e is given by:

*

RR I0VS

ee

*

ee

*

e

X

90V

X

)90(EV

X

90V

X

)90(EV

90X

VEV

and the active power injected is

e

RX

sinVEP

70

4.2.1.3 Steady state stability limit

The steady state stability limit of the simplest electrical system is defined as the greatest possible

power at its receiving end under a given condition of operation and excitation in the presence of

small disturbances.

We consider the stability limit of the system without excitation control i.e. the excitation is

constant. In this case, the synchronous machine is represented by a constant voltage behind its

synchronous reactance. For a salient machine, it is assumed that dq XX .

EdX eX

oV 0

The electromagnetic power is given as

sinXX

EVP

ed

e

Suppose a small disturbance causes the rotor angle to vary by , i.e. changes from

00 to , the subscript „0‟ denoting steady state condition. Then the change in the

electromagnetic power will be

c

PP

0

e

e

We suppose iP is constant because the governor is slow to act compared to the speed of energy

dynamics. Substituting the above equation into the swing equation we obtain

e0ei2

0

2

0

PPPdt

)(dH2

i0ee2

2

0

PPforPdt

dH2

0

2

2 H2MwhereC

dt

dM

0Cdt

dM

2

2

The characteristic equation is

0CMp2 and its roots areM

Cp

If 0C , the roots are pure imaginary and any small disturbance appearing in the system will result

in continuous oscillations. Line resistance and damper windings of machines ignored in the analysis

cause the system oscillations to decay. The system is therefore stable for a small disturbance so

long as 0P

0

e

.

If 0C both roots are real and one of them is positive. In this case, any small disturbance results

in a periodic rise of the torque angle, and synchronism is soon lost. Thus the system is unstable if

0P

0

e

.

71

At 0cosX

EVC,90 0

o

0 . The angle o

0 90 therefore determines the steady state stability

limit mP , i.e.X

EV90sin

X

EVP o

m .

Example 3.1

For the system where .u.p20.1E.,u.p60.0X.,u.p0.1V.,u.p20.1X ed ,

MVAMWH sec4 and the system frequency Hz50 , calculate the frequency of natural

oscillations if the generator is loaded to (a) 50% and (b) 80% of its maximum power limit.

Solution

(a) For 50% loading,o

0

m

0e

0 305.0P

Psin

radelectrical/pu577.030cos8.1

12.1cos

X

EVC o

0

radelectrical/.u.p0255.0502

42

f2

H2H2M

00

757.40255.0

577.0jj

M

Cp

Natural frequency of oscillations Hz757.02

757.4sec/rad757.4

(b) For 80% loading,o

0

m

0e

0 1.538.0P

Psin

radelectrical/pu4.01.53cos8.1

12.1C o

radelectrical/.u.p0255.0M as before

961.3j0255.0

4.0j

M

Cp

(c) Natural frequency of oscillations Hz637.02

961.3sec/rad961.3

Example 3.2

For the system shown in the figure below, chalculate the limit of steady state power with and

without reactor switch closed.

Mpu.X t 10 puX t 1.0

puX dg 1

1 2gE . pu puEm 0.1

1mgX pu

puX c 1

puX L 25.0

Solution

It can be shown that the system is equivalent to the simplest electrical system.

Case A: Reactor switch is open

Total reactance between generator and motor, dg t L t mgX X X X X X

72

.u.p.... 452110250101

.u.p49.045.2

12.1

X

EEP

mg

m

Case B: Reactor switch is closed. The equivalent circuit is as follows

puEg 2.1 puEm 0.1

puj1 puj 1.0 puj 25.0 puj 1.0 puj1

puj1

a b

c

M

This can be reduced to (by star-to-delta conversion)

puEg 2.1 puEm 0.1

X

M

This gives c

ba

baX

XXXXX

where 35.1j)25.01.01(jXa

1.1j)1.01(jXb

and 1jXc

Therefore

.u.p965.0j485.1j45.2j1j

1.1j35.1j1.1j35.1jX

Steady state power limit, .u.p..

.Pm 2441

9650

121

Self Assessment 3.1

Recalculate the power limit with the capacitive reactor replaced by an inductive reactor of the same

value.

In a generator with controlled field current, the emf, E remains constant for slow increases in the

load. This leads to a fall in the machine terminal voltage. In practice, the field current is controlled

so that the magnitude of the machine terminal voltage remains constant. Under this condition of

excitation, the maximum power is 50-80% higher (depending on the parameters of the generator

and transmission line) than in the case of a constant excitation.

We consider the steady state stability limit of the system where the excitation is controlled by a

regulator with a dead zone (i.e. with a delayed excitation control). Such a regulator acts only after

the voltage drop has exceeded a value called its dead zone. Both andE change under this

condition but since the change in excitation takes place after the change in load, the stability limit is

73

determined from the condition E Constant, i.e.,

0

e

e

PP and the limiting value of the

torque angle is o90 . Note that if the excitation is controlled with a fast-acting regulator then

EE

PPP

0

e

0

e

e

.

Example 3.3

For the system given below, determine the steady state power limit if the terminal voltage of the

generator is held constant at 1.2p.u. by an automatic voltage regulator which does not act fast

enough (i.e. it has sufficient dead zone).

pujX d 5.0

puj 0.1 oV 00.1

Infinite bus

Solution

E 001V

5.0jX d 0.1jX e

02.1 tV

Current injected into the infinite bus1j

0.12.1

jX

VVI

e

t

The machine internal voltage

1j

0.12.15.0j2.1IjXVE dt or

sin8.1j5.0cos8.15.08.1E

Steady state power limit is reached when o90 , i.e., the real part of E is zero, thus o87.7305.0cos8.1 and

o90729.1729.1j87.73sin8.1jsin8.1jE

Steady state power limit .u.p152.15.1

1728.1

XX

EVP

ed

m

4.2.2 Transient stability

Transient stability is the ability the power system to maintain synchronism when subjected to a

severe disturbance such as short-circuit, the tripping of a heavily loaded line, the tripping of a

loaded generator and sudden drop of a large load. A system which is transiently stable may recover

to its original frequency or settle down to a new frequency. Among the problems of transient

stability is the determination of critical clearing time of circuit breakers which isolate the faulty

portion from the system. Knowledge of the critical clearing time helps system planner to coordinate

the relay system so that a given fault is cleared in time. A fault cleared before the critical clearing

time will result in a stable system and a fault cleared after the critical clearing time will result in the

loss of synchronism.

74

The behaviour of the rotor angle in the presence of large disturbance can be determined by solving

the swing equation.

4.2.2.1 Definition of eP in the swing equation

A generator is represented by a constant voltage 'E behind its transient reactance 'Xd . The power

eP in this case for the simplest system is given by

sinX'X

V'EP

ed

e

EdX

eXoV 0

Note that the internal angle of E is not the same as that of E. Thus the angle in the above

equation is an approximation of the torque or load angle.

4.2.2.2 The swing equation

Solution depends on the type of disturbance and also its location in the power system. The practical

approach to the transient stability problem is to simulate a disturbance in the system, obtain a

numerical solution of the swing equation (or equations for a multi-machine case) and then plot

delta against time curve called the swing curve. If delta starts to decrease after reaching a maximum

value, it is normally assumed that the system is stable and the oscillations of delta around the

equilibrium will decay and finally die out. If it is unstable, delta continues to increase.

In the simplest electrical system, large disturbance are generally reflected as the changes in the

transfer reactance. In the study of transient stability, the mechanical input power of the synchronous

generator is assumed to be constant.

4.2.2.3 Equal area criterion

In a system where one machine is swinging with respect to an infinite bus, we can study transient

stability without finding the function )t(f . The general stability criterion is that the system is

stable if at some time t, 0dt

d

and is unstable if 0

dt

d

for a sufficiently long time (more than 1sec

will generally do)

0dt

d

stable

time

0

max

0

Fig 4.2 A stable system

75

This stability criterion can be converted into a simple and easily applicable form for the simplest

electrical system. The swing equation can be restated as:

eiaa

0

2

2

PPPwherePH2dt

d

Multiplying both sides of the swing equation by

dt

d2 , we get

dt

dP

Hdt

d

dt

d2 a

0

2

2

2

0a

d d dP

dt dt H dt

2

0a

dd P d

dt H

Integrating, we obtain

0

2

0a

dd P d

dt H

0

dPHdt

da

0

2

2

1

0

dPHdt

da

0

where 0 is the initial rotor angle before it begins to swing due to disturbance.

From the stability condition, 0dt

d, the condition for stability can be written as:

000

2

1

0

0

dPdP

Hdt

daa . From the figure below,

a

b

c

d e

iP

eIIP

2A

1A

eIP

P

0 c 2

Fig 4.3 Equal area criterion

0000

dPPdP ia This leads to:

76

2121 000

AAAAdPPdPPc

c

eIieIIi

The condition of stability can therefore be stated as: the system is stable if the area under aP

curve reduces to zero at some value of . In other words, a power system will be stable in terms of

transient stability IF AND ONLY IF the accelerating area equals the decelerating area. Hence, the

name equal area criterion of stability.

Consider the figure below.

a

bc

d e

,faultduring

,postfaultandprefault

P

0 c 2

f

PPm

AD

t

. .

.g

eIIP

eIP

Fig 4.4 Pre-fault, during fault and post fault P and the movement of system operating

point when the system is stable

We assume that the fault is cleared with the faulty transmission line back in service. Hence the pre-

fault P also represents the post-fault P curve. Transient stability of the power system can

be analysed by discussing the movement of the system operating point on the P curve, as

follows:

Point a is the initial operating point of the power system where PPm when the generator rotates

at a synchronous speed 0 and the rotor angle is 0 . At the moment when the fault occurs, the

operating point drops from a to b, where eIIm PP . Therefore, the generator starts to accelerate and

the operating point moves along the during-fault P curve. At ctt , the fault is cleared at

operating point c with c . The operating point on the post-fault P curve jumps to point d.

due to the acceleration from point b to c, at this moment, the generator rotates faster than the

reference axis at speed c .

77

At point d, eIm PP . The generator starts to decelerate from speed c . However, because 0 c ,

the rotor angle continues to increase. As a result, the operating point moves along the post-fault

P curve towards point g until it arrives at point e. at point e, the rotor has decelerated to the

synchronous speed 0 from 0 c . However, at point e, eIm PP and the generator continues to

decelerate from the synchronous speed. This results in a decrease in rotor angle and the operating

point moves back along the P curve towards the initial operating point a until it arrives at point

f. finally, the operating point comes back to its initial operating point a. The system is stable in

terms of transient stability.

a

bc

d

e

,faultduring

,postfaultandprefault

P

0 c 2

f

PPm

AD

t

. .eIIP

eIP

.h

bP .

AP

g

Fig 4.5 The case in which the system is unstable in terms of transient stability

The analysis of the unstable condition is similar to that of the stable condition. At point d, (in figure

4.5) the generator starts to decelerate from speed( c corresponding to c ). If c is so great that

generator takes a longer time to reduce its speed back to the synchronous speed, then the operating

point travels along the post-fault P curve and does not stop until it passes point h, as shown in

figure 4.5. At any point under point h – for example, at point e, eIIm PP - the generator starts to

accelerate again from a speed greater than the synchronous speed. Hence, the rotor angle continues to increase and the operating point moves towards point f and g in figure 4.5. The system

loses stability.

From the above analysis, if the system is unstable in terms of transient stability, then it loses

stability in the first swing of rotor angle. Hence, power system transient stability is also called first-

swing stability.

78

The following factors influence system transient stability:

(a) How heavily the generator is loaded

(b) The generator output during the fault. This depends on the fault location and type.

(c) The fault clearance time. The more slowly the fault is cleared, the closer point d is to point h

and the higher the chance that the system will lose transient stability, because the distance for

the operating point to decelerate from point d to h is shorter.

(d) The post-fault transmission reactance.

(e) The generator reactance. A lower reactance increases peak power and reduces initial rotor

angle.

(f) The generator inertia. The higher the inertia, the slower the rate of change in angle. This

reduces the kinetic energy gained during fault; i.e. area accelerating area is reduced.

(g) The generator internal voltage magnitude. This depends on the field excitation.

(h) The infinite bus voltage magnitude.

4.2.2.4 Application of equal area criterion

CASE 1: Sudden loss of one of two parallel lines.

EdX

oV 0

X

XS

Before switch off sinsin mI

ed

eI PXX

VEP

where

2

XX e

Immediately after switching off line,

sinsin mII

d

eII PXX

VEP

mI

i

mIiP

PPP 1

00 sinsin

The P curves are shown below.

79

iP 2A

1A

eIPeP

0 21

eIIP

m

Accelerating area 1

01

dPPA eIIi

Decelerating area 2

12

dPPA ieII

For the system to be stable, it should be possible to find angle 2 such that 21 AA . A limiting

condition is reached when 1

0

2 180 m where

IIm

i

P

P1

1 sin

CASE 2: A 3-phase short-circuit fault occurs at one end of a line, say at point S.

(a) Before occurrence of fault

sinsin mI

ed

eI PXX

EVP

where2

XX e

(b) Upon occurrence of fault: 0eIIP

(c) Circuit breakers at the two ends of the faulted line open at a time 1t (corresponding to angle

1 ), called the clearing time to isolate the faulted line.

iP

eIIP

2A

1A

eIP

0 1 2 m

IIIeP

eP

Fig 4.6 P curves for normal operation and during a 3-phase short-circuit fault at one

end of a line.

dPdPPA ieIIi

1

0

1

01

80

2

1

2

dPPA ieIII sinsin mIII

d

eIII PXX

EVP

Area 1A depends upon the clearing time 1t . This must be less than a certain value called critical

clearing time ct , for the system to be stable. The angle 1 corresponding to ct denoted c is called the

critical clearing angle. It is determined as follows:

m

c

cc

mIII

i

mieIIIieIIiP

PdPPdPdPP

1sin,

00

m

c

c

dPPdP ieIIIi

0

0

m

c

c

dPPdP eIIIii

0

0

m

c

m

c

c

eIIIii PdPdP

0

0

cmmIII

mcmIII

mIII

mIII

mIIIeIII

P

P

P

dP

dPdP

c

m

c

m

c

m

c

m

coscos

coscos

sin

sin

]cos[

00

c

m

m

cmmIIIcmieIIIi coscosPPdPdP

mIII

mmIIIomic

P

cosPPcos

Im

i

P

P1

0 sin and

mIII

i

mP

P1sin

CASE 3: 3-phase short-circuit occurs in the middle of a line.

EdX

oV 0X

2

X

2

X

This is equivalent to the circuit below (using a star-to-delta conversion)

EXXX dII

3

oV 0

2

X

The circuit model of the system during fault is:

(a) Before occurrence of fault: As in case 2

(b) Upon occurrence of fault sinsin mI

II

eII PX

VEP

57

(c) Post-fault eIIIP as in case 2.

iP

eIIP

2A

1A

eIPeP

0 21

eIIIP

m

Fig 4.7 P curves for normal operation, during a 3-phase short-circuit fault at the

middle of a line and after fault clearing

1

01

dPPA eIIi

2

12

dPPA ieIII , for critical clearing angle, we have

m

c

c

dPPdPP ieIIIeIIi

0

00

m

c

c

dPPdPP eIIIieIIi

000 cmIIImmIIImIIcmIImi cosPcosPcosPcosPP

mIImIII

mmIIImIImi

cPP

PPP

coscoscos 00

Where

Im

i

P

P1

0 sin and

mIII

i

mP

P1sin

CASE 4: The circuit breakers of line 2 are reclosed successfully because the fault was

transient and so vanished.

iP

eIIP

aA2

1A

eIVeI PP

eP

0 21

eIIIP

mrc

bA2

58

Fig 4.8 P curves for normal operation, during a 3-phase short-circuit fault at the

middle of a line and after a successful reclosure

We have sinmIeIeIV PPP

mI

i

mP

P1sin

1

0 1

rc m

rci mII mIII i mI iP P sin d P sin P d P sin P d

Example 3.4

A Hz50 generator is delivering 50% of the power that it is capable of delivering

through a transmission line to an infinite bus. A fault occurs that increases the

reactance between the generator and the infinite bus to 500% of the value before the

fault. When the fault is isolated, the maximum power that can be delivered is 75% of

the original maximum value. Determine the critical clearing angel for the condition

described.

Solution

Curves are as shown in figure 4.7

mIImIII

mmIIImIImi

cPP

PPP

coscoscos 00

Dividing the top and bottom by mIP , we obtain

23

30201 coscoscos

rr

rrr mm

c

wheremI

i

P

Pr 1 ,

mI

mII

P

Pr 2 and

mI

mIII

P

Pr 3

X

EVPmI ,

X

EVPmII

5 , mImIII PP 75.0

From the data 75.0,2.05

1,5.0 321 rrr

Also o

IIm

i

P

P305.0sinsin 11

0

o

mIII

mI

mI

i

IIIm

i

P

P

P

P

P

P8.41

75.0

5.0sin1sinsin 11

1

oooo

m 2.1388.41180180 1

Therefore,

o

c

c

34.67

3853.0

2.075.0

2.138cos75.030cos2.0180

302.1385.0

cos

4.2.2.5 Methods of improving system transient stability

The following methods are often employed in practice

(a) Increasing system voltage during disturbance using modern high-speed

excitation systems.

(b) Reducing transfer reactance by reducing conductor spacing, by using series

capacitors(for lines 350km) and increasing the number of parallel lines

between transmission points.

(c) Using high-speed circuit breakers and reclosing breakers.

(d) Fast valving: This measure reduces the mechanical input to the generator after

the occurrence of a fault through fast action from the side of the prime mover.

The result is that the accelerating area is reduced and the decelerating area is

increased.

4.3 Dynamic stability

Dynamic stability has also been widely used as a class of rotor angle stability.

However, it has been used to denote different aspects of the phenomenon in different

literatures. In North American literature, it has been used mostly to denoted small-

signal stability in the presence of automatic control devices(primarily generator

voltage regulators) as distinct from the classical steady-state stability without

automatic controls. In the French and German literature, it has been used to denote

what is termed transient stability. Since much confusion has resulted from use of the

term dynamic stability, both CIGRE and IEEE have recommended that it is not used.

Assignments

1. A synchronous generator represented by a voltage of ..15.1 up in series with a

transient reactance is connected to a large power system with a voltage of ..0.1 up

through a power network. The equivalent transient transfer reactance X between

voltage sources in ..50.0 upj . After the occurrence of a three-phase-to-earth fault

on one of the lines of the power network, two of the line circuit breakers A and B

operate sequentially as follows with corresponding transient transfer reactance

given therein.

(a) Short-circuit occurs at o300 , A operates instantaneously to make

.u.p.X 03

(b) Ato601 , A recloses, ..0.6 upX

(c) Ato752 , A reopens

(d) At o903 , B also opens to clear the fault making ..60.0 upX

Check if the system will operate stably.

2. Calculate the limit of steady state power for the system below

Mpu.X t 10 puX t 1.0

puX dg 1

1 2gE . pu puEm 0.1

1mgX pu

puX c 1

puX L 25.0

puX c 1

3. For the system given below, determine the steady state power limit if the terminal

voltage of the generator is held constant at 1.1p.u. by an automatic voltage

regulator which does not act fast enough(i.e. it has sufficient dead zone).

pujX d 4.0

pujX e 0.1 oV 00.1

Infinite bus

4. A Hz50 generator is delivering 40% of the power that it is capable of delivering

through a double-circuit transmission line to an infinite bus. A three-phase short-

circuit occurs at the generator terminal end of one feeder. Circuit breakers at the

ends of the faulted line trip, and the maximum power that can be transferred is

65% of the original maximum value.

(i) Determine the critical clearing angel for the condition described.

(ii) Find the accelerating and decelerating areas at this angle.