fault calculations

5/19/2018 Fault Calculations

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SUPPLEMENT TO FAULT

CALCULATION

SYMMETRICAL FAULTS

George MatherBA, Dip EE, C Eng, MIEE, AFIMA

1998 George Mather

Published by the Electricity Training Association


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CONTENTS

1 INTRODUCTION

2 FUNDAMENTAL MATHEMATICS

2.1 Complex Numbers

2.2 Linear algebra

2.3 Current in an ac circuit

3 NETWORK ANALYSIS

3.1 Kirchhoffs Laws

3.2 Network reduction

3.2.1 Millmans Theorem

3.2.2 The Principle of Superposition

3.2.3 Thevenins Theorem

3.3 Star - Delta and Delta - Star transformations

4 FAULT CURRENT CALCULATIONS AND PER UNIT METHODS

4.1 Introduction

4.2 Transformers

4.3 Ohmic Methods

4.4 Per Unit Methods

5 WORKED EXAMPLES

6 ANSWERS TO SELF TEST QUESTIONS

APPENDICES:

A MATRIX ALGEBRA

B USE OF SPREADSHEETS

C CURRENT IN AN RL CIRCUIT

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1.0 Introduction

This book is a supplement to Chapter 3 of the Power Systems

Protection correspondence course (PSPC) provided by the Electricity

Training Association. Fault Calculation in general and Chapter 3 of the

Power System Protection Course in particular have gained over the

years an awesome reputation with some PSPC students. The PSPC wasintroduced in the early 1960s and in those days the course was directed

at engineers in the United Kingdom. Nowadays students come from a

wide variety of backgrounds both in academic attainment and

nationality. Sometimes difficulties are exacerbated due to students not

having help available to them locally.

Times have changed since the PSPC course books were written.

Probably all students now have access to a personal computer with

spreadsheet facilities which can make calculations much less tedious.

Used properly, spreadsheets should help students better understand the

basic material. Many early text books in electrical power engineeringconcentrated on network reduction because it was a necessary process

when a slide rule was the most advanced tool available. Knowledge of

network reduction is still advantageous however in order to enable

students understand the behaviour of networks. There is also the

question of how much detail electrical engineers should know about

power system calculations when there are many good power system

analysis systems used in electrical utilities nowadays. It is most

important that engineers understand the output from these systems and

that understanding can only come from good theoretical knowledge

and experience.

Performance of protective gear is a very important aspect of power

system operation but safety is equally important and leads to issues

regarding of the make rating and break rating of switchgear. Protective

gear operating times have decreased on transmission systems as relays

based on digital electronics have been introduced. Faults are

sometimes cleared before the direct current component in the

disturbance has decayed to zero. The subject of initial fault conditions

is dealt with in chapter 2 and in appendix C in more detail. The

introduction of small generation plant connected at medium voltages

requires a good understanding of fault calculation by all designengineers. The first international standard for calculation of fault

current IEC 909 [6] has been issued and this has been implemented in

the United Kingdom as ER G74 [7].

This Supplement has been prepared to give the student a firm basis in

understanding the techniques of network analysis and symmetical fault

calculations. It is a remarkable fact that the course text book contains

only two examples on symmetical faults with one of them being based

upon physical units (ohms and volts). This Supplement recommends

use of per unit quantities but their introduction is delayed until

chapter four after the mathematics and network analysis techniqueshave been covered using physical values in chapters two and three.

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The supplement addresses such fundamental issues as:

why we use complex numbers in ac circuits

systematic ways of solving equations

the dc component and maximum offset in fault current

the best way to express fault rating the need for a common base in per unit calculations

effects of tap change and load

(opionally) an introduction to matrix algebra in fault

calculations

use of spreadsheets

proofs and applications of Thevenins Theorem and

Network transformations

The first four chapters have self test questions ( exercises to the oldfashioned!) against which understanding can be tested.

Use of the supplement is optional but it will not make sense without the

course text book. It covers completely, with one exception - induction

motors , the fundamentals of symmetrical fault calculation and will

provide a sound basis for further study.

Users of the supplement may omit the following sections of the text

book - but will be recommended by the supplement to read parts of

them - 3.1.4, 3.2, 3.3.1, 3.3.2, 3.3.7, 3.3.8, 3.3.9 and 3.3.10.

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2 Fundamental Mathematics

The course text book assumes good knowledge of the mathematics of

complex numbers and linear equations but does little to help the

students overcome difficulties in these subjects. It is not necessary to

be clever at mathematics in order to do fault calculations but a good

working knowledge of complex numbers and linear equations isnecessary. This chapter provides an introduction to and review of these

subjects in the first two sections. The range of material covered is

adequate for the course but in each case these are only the tips of very

interesting icebergs and many students will wish to follow them up

through the bibliography. Those students who feel confident without

this material should at least try to the self test questions.

The third section of the chapter covers why we need to use complex

numbers in electrical circuits. It is worth noting at this stage that

complex numbers are not vectors but the myth that they are vectors was

advanced by some early textbooks. Is easy to see a complex number is

not a vector because a vector does not have a reciprocal whereas a

complex number does; furthermore modern electrical engineering

textbooks draw pseudo vector diagrams of rms quantities and call them

phasor diagrams.

Appendix A introduces matrix algebra and introduces its application to

fault calculation. This material should not be read without a good

understanding of Thevenins Theorem and Nodal Analysis covered in

chapter three.

A Microsoft EXCEL spreadsheet is provided with the supplement to

help with the linear algebra and the more complicated complex number

operations; details of use are provided in appendix B.

Appendix C covers the initial conditions following a fault and

illustrates why making duty of switchgear is very important. This

appendix basically consists of a proof which uses a first order

differential equation. Do not worry if you do not understand the proof

and if you have difficulty do not spend time trying to understand it. Just

believe the result, use it , and return to it at your leisure.

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2.1Complex Numbers

Complex numbers are used in many aspects of electrical power

engineering but they were not discovered by electrical engineers. This

section introduces complex numbers from a fundamental requirement

and then reviews the basic operations used by electrical engineers. More

information on the wide scope of complex numbers may be found in

Speigel[2].

There are no normal, i.e. real, numbers which satisfy the equation z2

+1=0. This is because the equation requires z2 = -1, but z= 1 does

not exist for real numbers. To get round this difficulty we introduce an

operator j which has the property that j2=-1 and we can then solve the

equation as z=j1.

Now consider the quadratic equation z2-2z +10=0. Using the well known

formula for the solution of quadratic equations of the form

az2+bz+c=0

i.e. zb b a

a

c 2 4

2= we find that z thus z=1+j3 and 1-j3

as possible solutions.

= 2 4 40

2

z is a known as complex number. Complex Numbers have a Real

component Re(z) and an Imaginary component Im (z). E.g. in the above

example Re(z)=1 and, for the first case, Im(z)=3. We may plot these

values on an Argand Diagramwhere we plot values of Re(z) along the

x-axis and Im(z) along the y-axis as shown in Fig 2.1.1. Real numbers

may be recognised as a special case of complex numbers having

Im(z)=0.

Fig. 1.1.1 Argand Diargam

x

y

z

We may represent complex numbers in Rectangular form as we have

just shown or we may represent them in Polar Form. Consider first

rectangular form. Addition,and hence subtraction, of complex numbers

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is carried out by summing the real components and the imaginary

components separately.

Thus if z=a+jb and w=c+jd then:

z+w = (a+c)+j(b+d) and z-w = (a-c)+j(b-d)

Multiplicationzw is given by the expansion of (a+jb)(c+jd)

thus zw= ac+jad+jbc+j

2

bd but j

2

=-1zw=(ac-bd)+j(bc+ad).

Example 2.1.1

Calculate z+w, z-w and zw if z=3+j4 and w= 5-j7.

Answer:

z+w = (3+5) +j(4-7) = 8-j3

z-w = (3-5)+j(4 - -7) = -2+j11

zw = (3x5 -4x(-7))+j(3x(-7)+4x5)= 43-j1.

We use the complex conjugateto do division. The complex conjugate is

z*=a-jb i.e. we simply change the sign of Im(z).

It is clear that zz* = a2 +b2.

Consider first the reciprocal of a complex number z=a+jb.

Now 1/z = z*/zz* = (a-jb)/ (a2 +b2 ).

Division is now clearly z/w = z(1/w) hence

z/w=(a+jb)(c-jd)/(c2 +d2 ).

Example 2.1.2

Calculate z/w when z=43-j1 and w=5-j7.

Answer:z

w

j j=

+

+

( )(43 1 5 7

25 49

)

z

w

j

=

+ + ( ) (215 7 301 5

74

)

z

w

j=

+222 296

74

= 3+j4

which agrees with the result in example 2.1.1

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Example 2.1.3

If z=a+jb calculate z2/ 2z.

We should expect the answer to be z/2 just as it would be for real

numbers!.

Answer:

z2 =(a2 -b2) +j 2ab.

1

2 2

2

2

2 2

2 2

2 2 2

2 2

3 2 2 3

2 2

z

a jb

a b

z

z

a jb a b j ab

a b

a ab j a b b

a b

=

+

= +

+=

+ + +

+

( )

( )(( ) )

( )

( )

( )

but since

a (a2

+b2

) = a3

+ab2

and b ( a2

+b2

) = a2

b+b3

then

z2/2z = (a+jb)/2.

Another method:

z

z

z z

zz

z zz

zz

z2

2

2

2 2 2= = =

*

*( *)

*

An obvious result but many students do not take the short cut!

Example 2.1.4

Calculate zw and z/w when z=jb w= jd.

In this case we should be able to ignore complex arithmetic but be aware

of the rules. Hence we should expect zw=-bd and z/w= b/d.

Answer:

Expanded zw=(0+jb)x(0+ jd) hence zw=-bd+j0.

similarly z/w=(0+jb)x(0-jd)/d2=b/d

Example 2.1.5

An electric circuit has two inductive reactances w=j4 and v=j12 in

parallel. Calculate the impedance z (=wv/(w+v)) of the circuit.

Answer z=4x12/(4+12) =j3

Thus we can simplify the calculation by keeping in mind where we are

going.

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Complex numbers in polar form are written as z rej= , Eulers

formula tells us that e jj = +cos sin . Engineers often write z =r

where r is the modulus and is the argument.

To convert from Rectangular form to Polar form r a b= +2 2 and

= tan 1b

a

but note we must always be careful to check the sign of a and b in order

to calculate the angle according to table below

a b degrees

pos pos 0-90

pos neg 270-360

neg pos 90-180

neg neg 180-270

Example 2.1.6

Express z=3+j4 and w=-3+j4 in polar form.

answer:

The modulus of both z and w is 3 42 2 5+ =

For z = =tan .14

3

5313

For w =

= tan .14

312687

Note that the argument for w is (180-53.13)

Once again with z =r and w=s then zw= rs(+) and

z/w=r/s(-). We will now expand z and w using Eulers Formula and

calculate the quotient z/w. The product is left as an exercise for the

student.

z r j= +(cos sin ) w s j= +(cos sin )

z

w

r j

s j

rs j j

s=

+

+=

+

+

(cos sin )

(cos sin )

(cos sin )(cos sin )

(cos sin )

2 2 2

but since cos sin2 2 1+ =

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z

w

r

sj

z

w

r

sj

= + +

= +

((cos cos sin sin ) (sin cos sin cos ))

(cos( ) sin( ))

z

w

r

s= ( )

The reciprocal of a complex number z = r in polar form is simply

1/r

It can be seen easily that multiplication by complex numbers shifts the

resulting vector anticlockwise and division shifts the resulting vector

clockwise.

You always have to consider the best ways of doing multiplication and

division. This of course depends to some extent on how the problem isspecified. If you have rectangular form it is probably worth keeping them

in that form during the calculation.

1.1 Self Test Questions

1 Solve a2+a+1=0. Show that a3=1 and a4=a.

2 Do the indicated calculation in each of the following.

a) (3+j7)(4-j12) and express the result in polar form.

b) (3+j1)2 express the result in rectangular form.

c) +

6 2

1 8

j

j express the result in polar form.

d)( )(

( ) (

3 8 4 10

3 8 4 10

+ )

)

+

+ + +

j j

j jexpress the result in rectangular form.

3 Find the modulus of100

6 8+ j

4 Find the argument of the following complex numbers

a) -5+j0

b) -8-j4

5 Show that z*w = (zw*)* ; why does z*z = zz* ?

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2.2 Linear Equations

Linear equations are used extensively throughout power system

calculations; large systems of equations are solved in modern times by the

digital computer but nevertheless it is useful to be able to solve small

systems using only a pocket calculator or spreadsheet. In this section we

will look at simple methods of solving equations involving at most three

unknowns. The method used will also provide a powerful and systematictool for proving theoretical results. Appendix A will cover matrix algebra

which is used by computer methods of solving power system equations.

The methods introduced in this section will be used in proofs of network

theory to be covered in other sections.

Once understood the process of finding solutions to practical problems of

simultaneous equations is quite tedious. A spreadsheet for solving

equations of up to three variables is included with this supplement.

Description of the spreadshheet is covered in appendix B.

Consider the equations:

3x + 4y = 16

x + 2y = 6

The most elementary method of finding the value of x is to subtract twice

the second equation from the first thus eliminating y:

3x + 4y = 16

2x + 4y = 12

giving x = 4 and, since x+2y = 6, y=1;

This very simple method becomes messy when trying to solve, say,

equations involving complex numbers. Alternatively, to eliminate y we

could multiply the first equation by 2 and the second by 4 and then subract

again in the same way:

thus 6x + 8y = 32

4x + 8y = 24

hence 2x = 8 and x =4, y=1 as before.

This more systematic method leads to a general form.

Consider the system

A11x +A12y = B1 A21x +A22y = B2

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In this system the first suffix denotes a row and the second one denotes a

column. Using the method described above to eliminate y, multiply the

first equation by A22 and the second one by A12.

A11A22x + A12A22y = A22BB1 A21A12x + A22A12y = A12BB2

subtracting the second from the first:

(A11A22- A21A12)x = (A22BB1- A12B2B )

xB A B A

A A A A

B A

B A

A A

A A

=

=

1 22 2 12

11 22 21 12

1 12

2 22

11 12

21 22

The array:

AA A

A A=

11 12

21 22

contains the coefficients of x and y in the original equations.

A11A22- A21A12 is known as the determinant D of an array of dimension

2. It is the difference between the products of the two diagonals of the

array. More details may be found in textbooks on Linear Algebra [1].

The method also leads quickly to Cramers Method for solving linear

equations. In Cramers method the column of coefficients of the unknown

to be found are replaced by those on the right hand side (the Bs).

Hence

x

B A

B A

A A

A A

=

1 12

2 22

11 12

21 22

and y

A B

A B

A A

A A

=

11 1

21 2

11 12

21 22

The method provides an expedient way of solving systems involving two

or three variables but is not recommended when there are more than three

variables.

Example 2.3.1 Solve original problem again using Cramers Method.

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D=

= =

3 4

1 26 4 2

=

= =x

16 4

6 22

32 242

4 and y=

= =

3 16

1 62

18 162

1

We are now in a position to investigate the benefits of the method. First

consider the equations:

3x + 4y - 16z = 0

x + 2y - 6z = 0

There are not enough equations to find the values of the variables which

satisfy them. There are an infinite number of solutions but this does not

prevent a relationship between the variables from being found. From the

example it is clear that by writing the equations as:

3x + 4y = 16z

x + 2y = 6z

it is easily seen that x=4z and y=z.

Cramers Method can be applied to equations involving complex numbers.

Example 2.3.2

Solve: (2+j1)z + (4-j3)w = 12+j16

(4- j7)z + (-3-j1)w = 39-j12

D = (2+j1)(-3-j1)-(4-j7)(4-j3) =(-5-j5)-(-5-j40) = 0+j35

z

j j

j j

D

j j j

D

z j jD

jj

j

w

j j

j j

D

j j j j

D

=j

+

= +

= = ++

= +

=

+ +

= + +

12 16 4 3

39 12 3 1 12 16 3 1 39 12 4 3

20 60 120 165 140 1050 35

3 4

2 1 12 16

4 7 39 12 2 1 39 12 4 7 12 16

( )( ) ( )(

( ) ( )

( )( ) ( )(

)

)

wj j

D

j

jj=

+ =

+

+ = +

( ) ( )90 15 160 20 70 35

0 351 2

Example 2.3.2

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Solve the equations

j10z + j8w = 82

j 4z + j6w = 44

D = j210x6 - j24x8 = -28

z

j

j xj xj jj=

=

=

=

82 8

44 6

28

82 6 44 8

28

140

285

w

j

j j x j x jj=

=

=

=

10 82

4 44

28

10 44 4 82

28

112

284

Cramers rule is useful for proving theoretical results.Read the derivation

of the equations for the two phase to earth fault in paragraph d of section

3.4.4 of the textbook.The example quoted is the most difficult to solve of

the classic fault conditions but this systematic method makes the process

easier to understand.

Example 2.3.2

Fig 3.4.4E and equations 3.4.4.16 and 3.4.4.17 in the text book describe

the conditions at the point of fault for a Two phase to earth fault. We willsolve these equations using Cramers Method.

Answer:

This is one of the standard problems in the application of Symmetrical

component theory to unbalanced faults. It is not necessary to understand

symmetrical components for this answer but the solution to Self Test 2.1.1

is applied.

We are given Ia= 0 (Equation 3.4.4.16) and Vb= Vc = 0 (3.4.4.17)

There are three equations and, as we shall see, three unknowns (I1, I2 andI0) but solving three equations simultaneously leads to quite complicated

manipulation. Instead we will use the voltage equations to obtain

relationships between the variables. We will then substitute these

relationships into the current equation to find direct expressions for the

currents.

Expanding for Vb in symmerical component form

Vb = a2E-a2I1Z1 -aI2Z2-I0Z0=0

(This equation is printed incorrectly in some text books)and Vc= aE-aI1Z1-a2I2Z2-I0Z0=0

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Solve these equations for I1 and I2 in terms of E and I0. First write them in

the form

a Z I aZ I a E I Z

aZ I a Z I aE I Z

2

1 1 2 2

2

0 0

1 1

2

2 2 0 0

+ =

+ =

= D a a Z Z( )2 1 2 (since a4= a)

Using Cramers Method:

I

a E I Z aZ

aE I Z a Z

D

aEZ a I Z Z a EZ aI Z Z

D1

2

0 0 2

0 0

2

2 2

2

0 0 2

2

2 0 0 2

=

=

+

collecting terms:

IZ a a E I Z

Z Z a a

E I Z

Z12

2

0 0

1 2

2

0 0

1

= +

=

+( )( )

( )

Thus I

I Z E

Z01 1

0=

Similarly

I

a Z a E I Z

aZ aE I Z

D

Z E a Z I Z Z E aZ I Z

D2

2

1

2

0 0

1 0 0 1

2

1 0 0 1 1 0 0=

= +

collecting terms IZ Z a a I

Z Z a a

I Z

Z

I Z E

Z2

1 0

2

0

1 2

2

0 0

2

1 1

2

=

= = ( )

( )

From Equation 3.4.4.16

I1 + I2 + I0 = 0

We can now find I1by substituting for I2and I0:

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II Z E

Z

I Z E

Z

giving

I Z Z I Z Z I Z Z EZ EZ

I Z Z Z Z Z Z E Z Z

IE Z Z

Z Z Z Z Z Z

1

1 1

2

1 1

0

1 2 0 1 1 0 1 1 2 0 2

1 2 0 1 0 1 2 0 2

1

0 2

2 0 1 0 1 2

0+

+

=

+ + = +

+ + = +

= +

+ +

( ) (

( )

)

This time substituting to find I2:

E I Z

ZI

I Z

Z

hence

IZ E

Z Z Z Z Z Z

++ + =

=

+ +

2 2

1

2

2 2

0

2

0

2 0 1 0 1 2

0

and ,since I2Z2= I0Z0 , it follows that

IZ E

Z Z Z Z Z Z0

2

2 0 1 0 1 2

=

+ +

2.1 Self test questions

1 Solve for x and y.

7x + y = 26

5x+2y = 25

2 Solve for z and w

4z + 7w = 40-3j

z + 2w = 11-j

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2.3 Current in an a.c circuit

This section investigates the fundamental relationship between voltage andcurrent in an a.c. circuit. The reasons why complex numbers are used forrepresentation of a.c. quantities are included. The section is quite technical andincludes two proofs which involve simple differential equations. You do not

need to read or understand these proofs (which are terminated by a symbol)but they are provided for those students seeking a thorough understanding ofprinciples.

2.3.1 Fundamental Equations

We start with Lenzs Law:

Induced emfs e in a coil are always of such a polarity as to oppose the change

that generated them.

Mathematically we may write this ase= -d/dt

where represents the flux linkages of the coil and t is time

The flux linkages are a function of the current I in the coil and the physicalcharacteristics such the number of turns and diameter etc.

Joseph Henrydemonstrated later that

e= -LdI/dt

where L is the inductanceof the coil. L has units of Voltseconds/amp and isknown as the Henry.

When a sinusoidal voltage V sin t is applied to an ideal coil (with noresistance) then

IV

Lt= sin( )90 .

Proof:

LdI

dtV= sint

hence

dI

dt

V

Lt

IV

Ltdt

=

=

sin

sin

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= IV

Ltcos

but -cos t = sin (t-90)=sin t cos 90-sin 90.cos t

thus IV

Lt= sin( )90

The current lags the voltageby 90.

Note that V and I are peak and NOT rms values.

Resistance R in the circuit adds further complication, in this case

IV

R Xt e

Rt

L=+

+

2 2[sin( ) ] where = tan 1

X

R

cos =+

R

R X2 2

, sin =+

X

R X2 2, and X=L and is known as reactance.

R X2 + 2 is the impedance Z of the circuit.

The proof of this expression is given in appendix C. The ratio R/L is known asthe time constant of the circuit. Figure C.1 of appendix C shows themaximum value the current in an ac circuit can rise to following switch on.This is important for switchgear where making duty is an important safety

design feature. The decay of dc current is also quite important as the faultclearance times of modern transmission equipment reduce.

It is also clear R and X are mutually perpendicular and because of this may berepresented by complex numbers. Thus we may write impedance in polar orrectangular form.

I lags V by an angle This is written as I-. I and V may be either peak orrms values. We will use rms values unless otherwise stated.

There are 3 ways of calculating the powerS in the circuit and we can do thiseither in rectangular or in polar form. Usually S=(P+jQ) where P is real powerand Q is reactive power. Q is positive when the reactive power is inductiveand negative when it is capacitive.

(a) S=I*V. where I* is the complex conjugate of I.In rectangular form:

Let I = I (cos + j sin ) and I* = I ( cos - j sin )

Let V = V ( cos + j sin )

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S= VI ( cos cos +sin sin + j ( sin cos - cos sin )

S = VI ( cos (-) + j sin (-))

note that the sign convention results in lagging vars being treated as positiveie when > .

In Polar Form: S V = VI-x I=

b) S=|I|2 Z

In rectangular form: Let I= a+jb and Z=R+jX, then

S = I*V= I*I Z =(a+jb)(a-jb)(R+jX)=(a2+b2)(R+jX) = |I|2Z

In Polar Form I2 Z where is the impedance angle.

(c) SV

Z=

2

*

In rectangular form: Let V= u+jw and Z=R+jX

S VIVV

Z

u jw u jw

R jX

u w

R jX

V

Z= = =

+

=

+

=*

*

*

( )( )

*

2 2 2

and in polar form SV

Z

= 2

2.3.2 Series and Parallel circuits

For two impedance Z1 and Z2 in series then IV

Z Z=

+1 2 hence Z = Z1 + Z2

in this case of course I flows through both impedances.

Thus V1=IZ1 hence V1 =+

VZ

Z Z

1

1 2

and this represents a potential divider

For two impedances Z1 and Z2 in parallelthe total current I is given by:

IV

Z

V

ZV

Z Z

Z Z= + =

+

1 2

1 2

1 2

( ) hence the impedance of the circuit isZZ Z

Z Z=

+1 2

1 2

Alternatively we may use Admittance conveniently for the parallel circuit.

Admittance Y = 1/Z. Thus I=VY1+VY2= V(Y1+Y2)=VY.

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By comparison of the impedance form and the admittance form it is evidentthat

Y YZ Z

Z Z1 21 2

1 2

+ = +

If we know I the easiest method to calculate the current in each impedancebranch is:

IIZ

Z

IZ

Z Z1 1

2

1 2

= =+

and IIZ

Z

IZ

Z Z2 2

1

1 2

= =+

Example 2.3.2.1

A circuit consists of two impedances Z1 = (1+j8) and Z2 = (3+j4)connected in parallel and series impedance Zs = (1.625+j1.125). The circuitis connected to an ac voltage E = 100v rms.

Calculate (a) the total impedance of the circuit (b) the total current (c) thecurrent flowing in Z1 and Z2 (d) the voltage across Zs and (e) the power andthe reactive power in the circuit.

Answer

Zs

Z1 Z2E

(a) First finding the impedance Zp of the parallel combination.

ZZ Z

Z Z

j j

j j

j

j

j j

jj

Z Z Z j j j

p

s p

=+

= + +

+ + + =

++

= +

= +

= +

= + = + + + = +

1 2

1 2

1 8 3 4

1 8 3 4

29 28

4 12

29 28 4 12

160

220 460

1601375 2 875

1625 1125 1375 2 875 3 4

( )(( )

( ) ( )

( )(

( . . )

( . . ) ( . . ) ( )

)

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(b) Total Current I = V/Z

Ij

jj=

+ =

=

100

3 4

100 3 4

2512 16

( )( A) =20-53.13A

(c) Current in Z1 and Z2

IIZ

Z Z

j j

jj A

I I I j A

1

2

1 2

2 1

12 16 3 4

4 122 5 7 5 7 9 7156

9 5 8 5 12 75 4182

=+

= +

+ = =

= = =

( )( )( . . ) . .

( . . ) . .

(d) Voltage across Zs

Vs = = +

+ = =

VZ

Z

j

jj V V

s 100 1 625 1125

3 437 5 12 5 39 43 18 43

( . . )( . . ) . .

(e) Power

S=|I|2 Z = (122+162 )(3+j4)=(1200+j1600)VA


1 A circuit consists of a combination of two impedances Z1 = (1+j5) andZ2 = (7+j3)connected in series in parallel with an impedance Zp= (4+j8).

The circuit is connected to an ac voltage E = 50v rms.

Calculate (a) the total impedance of the circuit (b) the total current (c) thecurrent flowing in Z1 and Z2 (d) the voltage across Z1 and (e) the power andthe reactive power in the circuit.

Z1

Z2

EZp

2 A 132kV circuit breaker with breaking duty of 25kA rms is designed for useon a 50 hertz system with an X/R ratio of 11.8. Calculate the highest peakkiloampere making duty of the circuit breaker.

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3 Network Analysis

Understanding of fault calculations requires a good understanding of

network analysis; methods of network reduction beyond simple series

and parallel element reductions are based on network theorems. In this

chapter we shall begin with the axioms of Kirchhoffs Laws to analyse

networks and then, using these results as basis, we proceed to cover the

Parallel Generator Theorem; the Principle of Superposition; a directconsequence of this, Thevenins Theorem which is the basis of most

methods of fault calculation, and; finally star-delta and delta-star

transformations. Proofs are provided but may be omitted if you are

short of time. Most of the chapter has been written around two

networks so that the advantages of the different approaches maybe

identified. In order to keep the focus on understanding principles most

calculations are based on resistive ohmic networks. Your spreadsheet

may be used to solve equations but masochists may wish to solve them

longhand!

3.1 Kirchhoffs Laws

We will treat these as axioms for which there is no proof. Kirchhoffs

Current Law states that the currents in branches terminating at a

junction sum to zero. Alternatively we may say the sum of currents

entering the junction equals the sum of currents leaving the junction.

Kirchhoffs Voltage Law states that the voltages around any closed

path sum to zero.

Some authors refer to these as KCL and KVL respectively [5].

There are two basic ways of using these laws directly to solve network

problems. One way, known as mesh current analysis, is to specify the

voltages around meshes and calculate the mesh currents using KVL

and from this deduce the branch currents using KCL. The other way,

known as nodal voltage analysis, is to specify the currents injected at

nodes to calculate the voltages at them using KCL and from this

deduce the branch currents.

3.1.1 Mesh Current Analysis.

Mesh current analysis is illustrated by the following example:

Example 3.1.1.1 Calculate the currents in the following network.

80V

20

130V8

30

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There are two meshes. Let the current round each loop be I1 and I2 .

20 30

880V 130VI1 I2

Using KVL around mesh 1 we can write the equation:

80-20I1 - 8(I1 + I2) = 0 rearranging to become

80 = 28I1 + 8I2

Following a similar procedure around mesh 2 we can write:

130 = 8I1 + 38I2

The solution of these equations using Cramers Method:

Det = (28 x 38) - (8 x 8) =1000

I Det A1

80 8

130 38

2= = and I Det A2

28 80

8 130

3= =

I1= 2A and I2= 3A.

Thus the current in the 8resistor = 5A.

Read section 3.2.1 of your text book up to This simple example has

been solved and noting:

There is a printing error in equations 3.2.1.7 and 3.2.1.9 in sometextbooks. The right hand side in both equations should read 100+j0.

The equation set 3.2.1.6 and 3.2.1.7 are branch current equations but

those in the set 3.2.1.8 and 3.2.1.9 are really mesh current equations.

The solution given is unclear; it is better to use Cramers Method:

Det= (2+j4)(3+j5)-(1+j0)(1+j0)=-15+j22

:

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I

j j

j j

Detj A

I

j j

j j

Detj A

1

2

118 34 1 0

100 0 3 517 707 18164

2 4 118 34

100 0 100 09 932 10 499

=

+ +

+ +

=

=

+ +

+ +

=

. .

. .

Branch current analysis should not be used in general because elimination of

redundant information such as I3in the above textbook example is confusing.

Mesh current analysis provides a more sound basis for creating a

mathematical model but we need to know how many equations are needed

and how to set them up. To determine how many equations are necessary

first short circuit all voltage sources; the number of equations required is the

number of impedance elements e minus the number of nodes v excluding the

reference node. Thus in example 3.1.1 above the number of equations is 3-

1=2! simple. Setting them up just requires including each impedance element

in at least one mesh; each mesh current flows in every element round themesh with currents in adjacent meshes being added or subtracted. Fig

3.2.2A in the textbook shows a good example where the number of equations

required is three i.e. e = 6 and v = 3.

Example 3.1.1.2

Find the current flow in Fig 3.1.1.2.

There are 5 impedance elements and 2 nodes. We therefore need three

equations.

5 10

20100V 100V2

2

I1I2

I3

5 10

20100V 100V

2

2

Fig 3.1.1.2

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The equations for each mesh are:

Mesh 1 25I1- 20I2= 100

Mesh 2 -20I1+ 32I2 + 2I3= 0

Mesh 3 2I2+ 4I3 = 100;

From which:

I1= 5.6A , I2= 2A and I3 = 24A.

The current flow in each impedance element may then be deduced

easily.

3.1.2 Nodal Voltage Analysis

The objective in nodal voltage analysis is to calculate the voltage at

each node and then deduce the branch currents.

Consider the circuit in Fig 3.1.2.1

Y1 Y2

Y4 E1 E2Y5

Y3

V2V

Fig 3.1.2.1

There are two nodes 1 and 2. We know from KCL that the current

flowing into a node sum to zero.

For node 1:

(E1-V1)Y1+ (V2- V1) Y2 + (0 - V1) Y4 = 0.

rearranging:

E1Y1 = (Y1+ Y2+ Y4 ) V1 - Y2 V2

Similarly we can write for node 2:

E2Y3= -Y2 V1 + (Y2+ Y3 +Y5 )V2

Noting that E1 Y1 and E2 Y3 both represent current we can solve the

equations to find V1 and V2.

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These equations may be represented by the figure 3.1.2.2:

Y2

Y3+Y5

Fig 3.1.2.2 has been developed from fig 3.1.2.1 by replacing voltage

sources with constant current source. The constant current source is the

short circuit current at the terminals of the source. i.e.

and; parallel elements Y1 and Y4 have been combined and so have Y3and Y5.

Example 3.1.2

Find the current flowing in each branch of fig 3.1.1.2 using nodal

voltage analysis.

Answer:

.

E1= 100V, E2= 100V, Y1 = 0.2 mho, Y2= 0.1mho, Y3=Y5=0.5 mhoand Y4= 0.05mho.

Thus

20 = (0.2+0.1+0.05) V1 - 0.1V2

20 = 0.35 V1 - 0.1V2

and 50 = -0.1V1 + (0.5+0.5+0.1) V2

50 = -0.1V1+ 1.1V2

Note these equations are written in a systematic form for solution by

Cramers Rule from which V1=72V and V2=52V.

E is equivalent to.. YEY

Y1+Y4 E2Y3E1Y1

Fi 3.1.2.2

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The branch currents are then:

1 (100-72) 0.2 = 5.6A

2 (72-52) 0.1 = 2A

3 (100-52) 0.5 =24A

4 (72-0) 0.05 = 3.6 A

5 (52-0)0.5 = 26A

Thus agreeing with the earlier result!

The rules for setting up nodal current equations are quite simple. An

equation is needed for every node. For each node equation the voltage

coefficient of the corresponding node is the sum of all admittances

connected to the node and for other nodes the voltage coefficient is the

negative of the admittance between them. If two nodes are not connected

directly by a branch the voltage coefficient of the remote node is zero.

Read sections 3.2.3 and 3.2.4 of your text book.

Comments:

The paragraph commencing .. The choice between .. is misleading. In

practice most electrical networks have many more nodes than meshes (e.g.radial distribution networks); the example quoted is not typical. The nodal

voltage method is easier to program than mesh current method.

Modern computer facilities spare you from needing to know advanced

methods of solving equations but references 1 and 4 in the bibliography of

this supplement provide better mathematical background than those quoted

in the textbook.

Self Test Question

Calculate the current in the following circuit:

180V

20

120V5

20

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3.2 Network Reduction Methods

In this section we cover proofs and examples of four network reduction

methods.

3.2.1 The Parallel Generator Theorem

This is sometimes known as Millmans Theorem. It follows as a direct

consequence of nodal voltage analysis and is useful for reducing

voltage sources. Consider the network shown in Fig 3.2.1:

VE Y E Y E Y

Y Y Y=

+ +

+ +

1 1 2 2 3 3

1 2 3

E1 E2 E3

Fig 3.2.1

V

Y1 Y2 Y3

Proof

Applying nodal voltage analysis to the common busbar V:

(E1- V) Y1 + (E2- V) Y2+ (E3- V) Y3 = 0

E1Y1+ E2 Y2 + E3Y3= V( Y1 + Y2 + Y3 )

VE Y E Y E Y

Y Y Y=

+ +

+ +

1 1 2 2 3 3

1 2 3

Example

Consider again example 3.1.1, repeated here for convenience:

20 30

880V 130V

Node 1

We will use the parallel generator theorem to calculate the voltage V.

at node 1

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Now V x=

+ +

+ +

= = =

80

20

130

30

0

81

20

1

30

1

8

250

3025

120

250

30

120

2540V

It is now a simple matter to calculate the current in each element.

Notice that the 8 resistor has been included to calculate V; the zero

voltage element in the numerator would normally be omitted but has

been shown here to clarify the method.

Suppose now the 8resistor was removed. The voltage is now:

V V=

+

+

= =

80

20

130

301

20

1

30

250

305

60

100

The parallel generator theorem can be used to prove a fundamental

principle of fault calculations that:

IV

Zfaultprefault

bus

=

where Zbusis the system impedance of the busbar.

For node 1 in the above example, Zbus= 1/( Y1+Y2+ Y3)

and VE Y E Y

Y Y Ypr efau lt =

+

+ +

1 1 2 2

1 2 3

hence Ifault= E1Y1 + E2 Y2

Calculating the fault current for node 1 in the network of example 3.1.1

is easy. We can see that Ifault is 80/20 + 130/ 30 = 250/30 A. Using the

principle we would expect this result irrespective of the presence of the8 resistor because it is short circuited; but the prefault voltage

conditions are different. We will check both conditions:

First the 8resistor present: Vprefault= 40V ; and

Zbus= 1/(1/8 +1/20+1/30) = 120/25, thus Ifault = 40x25/120=250/30A

Second without the 8resistor: Vprefault = 100V; and

Zbus = 1/ ( 1/20 + 1/30) = 12, thus Ifault = 100/12= 250/30 A

As we should expect!

Page 28 of 85


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This rule applies to any network as we shall see later. In practice fault

level varies marginally with load changes; most networks are operated

in a way to keep the voltage within a small tolerance of a target

voltage and system control devices like generators, transformers and

capacitors are adjusted as the load changes.

3.2.2 The principle of Superposition

This is sometimes quoted in text books as a theorem but as we shall see

it is derived directly from Kirchhoffs laws. Formally it states that:

The voltage and current response of a linear network to a number of

independent sources is the sum of responses obtained by applying each

source once with all other sources set to zero.

Applying this principle to the example 3.1.1.1 using mesh analysis we

have:

20 30

880V 130VI1 I2

First find the current flow with only the 80V source present:

The mesh current equations are :

28I1+ 8I2= 808I1+38I2 = 0

From which I1 = 3.04A and I2= -0.64A

The current flow could be calculated without solving equations; in this

case the impedance Z seen from the 80V source is:

Zx

= ++

=208 30

8 30

500

19

and I1 = 80x19/500=3.04A.

3020

880V I2I1

Page 29 of 85


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I2 may be calculated by current sharing hence:

I I x2 18

30 8304

8

380 64=

+= =. . A

note the reversal of sign when compared with the diagram.

Considering now the 130V source acting alone:

20 30

8 130VI1 I2

With equations

28I1+ 8I2= 0

8I1 + 38I2= 130

from which I1= -1.04A and I2= 3.64A

Adding together the solutions:

I1 = 3.04 - 1.04 = 2AI2= -0.64 + 3.64 = 3A

The current direction convention was kept constant to preserve

consistency. The method when written like this is equivalent to the

mesh current analysis with just one voltage coefficient on the right

hand side at each stage.

It is not necessary to solve equations but network reduction methods

may be applied provided all impedances in the system (including those

of voltage sources for which the voltage has been set to zero) are

included.

The principle of superposition may also be applied to nodal analysis. In

this case all injected currents except one are set to zero and each

current injection is considered separately.

The nodal voltage equations of example 3.1.2 were:

0.35V1- 0.1V2= 20

-0.1V1+ 1.1V = 50

Treating each current source individually:

Page 30 of 85


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0.35V1- 0.1V2= 20

-0.1V1+ 1.1V2= 0

from which V1 = 176/3 V and V2= 16/3 V

and

0.35V1- 0.1V2= 0

-0.1V1+ 1.1V2= 50

from which V1= 40/3 V and V2 = 140/3 V

Summing the two results gives our earlier result:

V1 = (176+40)/ 3 = 72V and V2 = (16+140)/3 = 52V

It is not necessary to solve equations in order to calculate these

voltages. It could be done by injecting current into the reduced network

but this is left as an exercise for the student.

It should be noted that it is not normal practice to solve problems using

the Superposition method in conjunction with the nodal voltage method

but it has been included here to show the general properties of the

superposition method.

3.2.3 Thevenins Theorem

This provides a very powerful method for calculating simple

equivalent networks and load current or fault current in an element of a

network. Thevenins Theorem states:

The current flowing through an impedance ZL connected between any

two points A and B in a network (see fig 3.2.3.1) is given by E/(Z+ZL).

E is the open circuit voltage when ZL is removed and Z is the

impedance of the network as seen from points A and B calculated by

assuming all voltage sources have been short circuited.

B

A

NetworkZL

Fig 3.2.3.1

Thevenins Theorem is a special case of the principle of superposition.

The proof of the Theorem abstract. Instead, we will demonstrate it

using the Principle of Superposition and a familiar example. The

Page 31 of 85


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objective will be to find the current in the 8resistor in the network of

fig 3.1.1 which is reproduced below with a voltage source E in series

with the 8 resistance. First find the value of E which makes the

current in the 8 resistance equal to zero.

20 30

8

80V

130VI1 I2

E

Since E opposes the other two sources we can write equations:

80-E = 28I1+ 8I2130-E = 8I

1+ 38I

2

There are three unknowns. In order to make the current in the 8

resistor equal to zero it is necessary for I2= -I1.

Thus 80 = 20I1+ E

130 = -30I1+ E

from which I1 = -1A , E=100V, and I2= 1A.

Now remove and short the original voltage sources; keeping the same

sign convention we can write equations:

-100 = 28I1+ 8I2-100 = 8I1 + 38I2

Hence

I1= -3A and I2= -2A

Thus the current in the 8resistor is 5A but in the opposite direction to

the normal result. Thus when E acts alone the current in the 8resistor

is -5A but when acting with other voltages the current is zero. This is

because E = 100V opposes exactly the current from the other sources

according to the Principle of Superposition; Thus we conclude that with

E removed a current of 5A will flow in the 8resistor and E =100V is

the open circuit voltage across the 8 resistor (which could be

removed).

We then have from Thevenins Theorem:

IE

Z Z ZL=

+=

+=

100

85 thus Z = 12

Page 32 of 85


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Z must be the equivalent impedance of the network. It is clear from the

situation when E acted alone that Z is the impedance of the network

viewed from node 1. Thus we can represent the circuit as:

The equivalent network is defined by the 100V and the 12 resistor.

The 8resistor is equivalent to a load.

If it is necessary to calculate the current distribution in the network the

easiest method to understand is that by which the voltages are

calculated, for example we know that the voltage across the 8resistor

is 40V; this is the basis of fault calculation used by computer programs

as shown in appendix A. There is a more obscure method by which

the distribution of the 5A in the network is added to the conditions

prevailing during the open circuit condition. It is clear that the 5A

would flow as 3A in the 20 resistor and 2A in the 30 resistor;

adding the results :

Now suppose the load is included in the network and that we want to

calculate the fault current when the 8resistor is short circuited. The

voltage at node 1 before the short circuit is applied was calculated in

section 3.1, using the parallel generator theorem, to be 40V and the

impedance Z is given by:

Z =

+ +

= = = =1

1

8

1

20

1

30

1

25

120

120

25

24

54 8.

20 30

880V

130V

12

100V 8

I1= 3 -1 =2 I2= 2+1 =3

5

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The network with the 8resistor incorporated is now equivalent to fig 3.4.1:

We can now put any new load, including a short circuit, across the

open terminals and obtain the current and voltage conditions. The short

circuit current is the same with or without the load.

Thevenins Theorem is sometimes written based on fig 3.4.2 as:

v = E - iZ

Fault conditions are clearly defined by v=0.

Read Example 2 in section 3.3.10 of the textbook from the paragraph

following Fig 3.3.10I and commencing It should be noted that.. .

Comment: The author makes the explanation of Thevenins Theorem in

the Paragraph commencing Knowing the fault current.. difficult to

understand. The following points will now be clear to you:

Ef is the prefault voltage ( or the open circuit voltage fora fault) and is a property of Estand the loaded network

It is not necessary to reverse the sign of Ef in diagram B

this is a version of the obscure method ,mentioned above;

it is used only to get the current flowing in the correct

direction.

The author fails to mention the real benefit of the method; we could

substitute any value of generator impedance into the network without

needing to recalculate the driving voltage.

Thevenins Theorem can be used to simplify parts of networks;consider the problem of finding the current in the 10resistor in the

following example :

E

Z

i

4.8

40V v

Fig 3.4.2Fig 3.4.1

5 10

20100V

2

2 100V

Fig 3.4.3

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We can break the problem in to three parts; first remove the 10

resistor; then find the equivalent networks for each source and; then

find the current in the 10resistor. The following network stages are

then found:

Replacing the 10resistor we can find the current....

The current flowing is seen easily to be 2A. (i.e. (80-50)/ 15A)

It is now possible to calculate the current in the rest of the network:

Round the left hand loop we can write:

100 = 5I1+ 20 (I1-2) thus I1= 5.6A

For the right hand loop:

100 = 2I2+ 2 ( I2 +2) thus I2= 24A

This result is the same as example 3.1.1.2, as we should expect, but the

method used here is more simple.

4

80V 50V

1

4

80V 50V

110

5

20100V 100V2

2I = 2A

5

20100V 100V

2

2

i.e. 20x5/25= 4

i.e. 100x20/25= 80

I2I1

Page 35 of 85


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Read Example 1 section 3.3.9 in the textbook.

Comment.

There are more efficient methods now available for us to use. Starting

from Fig 3.3.9B, the equivalent of which is reproduced below we

calculate the current flow throughout the network.

As in the previous example first remove the 6.9+j39element....

Next calculate the Thevenin equivalents for the remaining network.

For the left hand side:

Vx j

j

j

Zj x j

jj

=

+

+

=

=

+

+

= +

80800 4 6 26

4 6 79 2

26710 3142

53 2 4 6 26

4 6 79 22 069 17 585

( . )

. .. ( . )

. .. .

For the right hand side:

Vx j

jj

Zj x j

jj

=

+

+

=

=

+

+

= +

80800 2 3 13

2 3 9311340 1718

80 2 3 13

2 3 931701 11225

( . )

.

( . )

.. .

Following the previous example the current in the 6.9+j39element is:

Ij j

j j

I j A

=

j

+ + + + +

=

( ) ( )

( . . ) ( . . ) ( . )

. .

26710 3142 11340 1718

2 069 17 585 1701 11225 6 9 39

14 314 224 421

Note that I flows from left to right in the diagram. The current in theleft and right hand loops can now be calculated.

j 53.2

4.6+j2680.8kV2.3 + j13

j80

80.8kV

j 53.2 6.9+ 39

4.6+j2680.8kV 80.8kV

j80

2.3 + j13

Page 36 of 85


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For the left hand loop let IA be the unknown current.

80800 = j53.2IA+ (4.6+j26)x(IA- (14.314-j224.421))

IA = 55.06-j 1092A

For the right hand loop let IB be the unknown current:

80800 = j80IB + (2.3+j13)x(IB+ (14.314-j224.421))

IB = 24.237-j836.493A

The current in the 4.6+j 26element is 40.746-j 867.088A (Ia-I)

The current in the 2.3+j13element is 38.551-j1061A (IB+ I)

IA and IB are the machine currents at end A and B respectively and the

fault current = IA+ IB= 79.297- 1928A.

Note the small discrepancy from minor errors in the textbook

calculation

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3.3 The Star-Delta and Delta-Star Network Transformations

The Star-Delta and Delta-Star network transformations are very useful

tools for simplifying networks. In this section we will first state and prove

the transformations and then give some examples which will highlight a

common error made by many students.

Za Zb

Zc

Fig 3.3.1

Z2

Z1

Fig 3.3.2

Z3

Star- Delta Transformation

The two networks in Figs 3.3.1 and 3.3.2 are equivalent if:

Z1= Za+ Zc + ZaZc/Zb ;

Z2 = Za+ Zb+ ZaZb/Zc;Z3 = Zb+ Zc+ ZbZc/Za

Delta- Star Transformation

The two networks in Figs 3.3.2 and 3.3.1 are equivalent if:

Za= Z1Z2/(Z1+Z2+Z3);

Zb= Z2Z3/(Z1+Z2+Z3);

Zc= Z1Z3 /(Z1+Z2+Z3)

Proof:

It is a remarkable fact that the solution of the mesh- current problem of the

star network in fig 3.3.1 is equivalent to the coefficients of the nodal

voltage analysis equations of fig 3.3.2. Likewise the solution of the nodalvoltage problem of fig 3.3.2 is equivalent to the coefficients of the mesh

current analysis of fig 3.3.1.

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Za Zb

Zc

Fig 3.3.1.a

I2 , V2I1, V1

Applying mesh current analysis to the network in Fig 3.3.1.a:

V1= (Za+ Zc) I1+ ZcI2 1.1

V2= ZcI1 + ( Zb+ Zc) I2 1.2

Solving these equations by Cramers Rule:

Det = (Za+ Zc) (Zb+ Zc ) - Z2c = ZaZb+ ZaZc+ ZbZc

IZ Z

Z Z Z Z Z ZV

Z

Z Z Z Z Z ZV

IZ

Z Z Z Z Z ZV

Z Z

Z Z Z Z Z ZV

b c

a b a c b c

c

a b a c b c

c

a b a c b c

a c

a b a c b c

1 1

2 1

13

14

= 2

2

+

+ +

+ +

=

+ ++

+

+ +

......... .

......... .

Y2

Y1

Fig 3.3.2a

Y3

I2 , V2I1, V1

Using Fig 3.3.2a, in which admittances are used for convenience, we can

write the nodal voltage equations:

I1= (Y1 + Y2) V1- Y2V2 2.1I2= - Y1 + (Y2 +Y3) V2 2.2

Before solving these equations it is possible to obtain the delta- star

transformation. If the networks are equivalent then the coefficients of V2 in

equations 1.3 and 2.1 must be equal i.e.

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YZ

Z

Z Z Z Z Z Z

Z Z ZZ Z

Z

c

a b a c b c

a b

a b

c

2

2

2

1= =

+ +

= + +

Formulas for Z1 and Z2 may be derived in a similar manner by subtractingthe equivalence of Y2from the coefficients of V1 in equation 1.3 and V2 in

equation 1.4 respectively. This is left as an exercise for the student.

Solving the nodal 2.1 and 2.2 equations by Cramers method we find:

VY Y

Y Y Y Y Y Y I

Y

Y Y Y Y Y Y I

V YY Y Y Y Y Y

I Y YY Y Y Y Y Y

I

1

2 3

1 2 1 3 2 3

1

2

1 2 1 3 2 3

2

22

1 2 1 3 2 3

11 2

1 2 1 3 2 3

2

2 3

2 4

=+

+ ++

+ +

=+ +

+ ++ +

.............. .

............. .

Comparing the coefficients of I2in equations 1.1 and 2.3 it is clear that:

ZY

Y Y Y Y Y Y

Z

Y YY Y

Y

Z

Z Z

Z

Z Z

ZZ Z

Z Z Z

c

c

c

c

=+ +

=

+ +

=

+ +

=+ +

2

1 2 1 3 2 3

1 3

1 3

2

1 3

2

1 3

1 3

1 2 3

1

1

1 1

Formulas for Za and Zbmay be derived from this result and this too is left

as an exercise.

Example 3.3.1

Find the current flow in the following network.

80V

20

130V8

30

First transform the network to a delta.

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130V80V

125

33.3 50

It must be noted that the network inside the box is only an equivalentof

the original. The current in each element is easily calculated:

50element = 130/50 = 2.6A

125element = (130-80) / 125 = 0.4A

33.3element = 80/33.3 = 2.4A

The current from the 130V source = 2.6+0.4 =3A

The current from the 80V source = 2.4-0.4 =2A

These values agree with earlier methods of solving this problem.

Example 3.3.2

A 50km double circuit 132kV overhead line with impedance

0.1814+j0.3920per kilometre has a fault 20km from a termination and in

the network reduction procedure it is necessary to convert the delta to an

equivalent star network.

C A B

Answer:

There are two ways of doing this type of calculation. One is very tedious

and the other very efficient! First the tedious method (followed by 95% of

students and engineers) goes like this:

Impedance AB = 50*(0.1814+j0.3920) = 9.069+j19.602.

Impedance AC = 20*(0.1814+j0.3920) = 3.628+j 7.841

Impedance BC = 30*(0.1814+j0.3920) = 5.442+j11.761

Finding the star element corresponding to AB and AC:

Z Z Z j

Zj j

j

j

j

Z j

AB BC AC

an

an

+ + = +

=+ +

+=

+

+

= +

18138 3 9204

9 069 19 602 3 628 7 841

18138 3 9204

120 795 142 219

18138 3 9204

1814 3 92

. .

( . . )( . . )

. .

. .

. .

. .

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The student completes two other similar calculations to find:

Zcn= 1.088+j2.352 and Zbn= 2.721+j5.881

The efficient methods takes advantage of the fact that all the components

in the calculation have the same X/R ratio. Thus we can convert in terms oflength:

The denominator is twice the line length.

Z jan = + =50 20

100018138 0 39204 1814 3 92

**( . . ) . . j+

With an immense saving in computation!

Other equivalent lengths are LB= 15km and Lc= 6km.

The last example of this section illustrates a common error made by users of

delta-star transformations.

Example 3.3.3

Find the current flow in the 100and 40elements in following network:

C

An error made by many students faced with type of problem is first to

transform the delta ABC to a star equivalent; then calculate the current

flow in the resulting parallel circuits and; then to assume that the current

flow thus calculated represents the solution to the problem. We will do this

calculation. First converting the delta ABC to star:

ZAN= 40x100/200 = 20and likewise ZCN= 30and ZBN= 12

A

B350V

10100

60

1240

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The transformed network is now..

C

I1

The total resistance of the network is easily seen to be 35.

Hence I =10A. Thus I1 = 24x10/64 = 3.75A and I2= 6.25A

Clearly, the network now between terminals ABC is an equivalentand so

we need the current in the original delta. It is easy to see that Vc = 37.5V

and VB= 75V. We can now calculate the current in the original delta:

IAB= (350-75)/40 = 6.875A; IAC = (350-37.5)/100 = 3.125A and;

IBC= (75-37.5)/60 = 0.625A.

There is a better method of finding the current in the elements of the delta

which avoids the very small differences in voltage which occur in practice

.This method find the mesh current in the delta:

100Id+ 60(Id-3.75) + 40(Id- 10) = 0

from which Id = (60x3.75+40x10)/200 = 3.125A

Id is of course the current in the 100element.

A

B350V

20

30 10

12 12

I

I2

3.75A

Id

40

60

100

6.25A

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4.0 FAULT CURRENT CALCULATIONS AND PER UNIT METHODS

4.1 Introduction

We shall investigate the nature of fault conditions and examine best ways of

calculating fault current. All fault calculations would be very simple if there

were no transformers in the power system so basically we need a methodwhich makes them transparent in the calculation. Ohmic methods are

advocated by some authors [3] whilst per-unit methods are used in the PSP

course and are used widely. We will also examine how we should express the

fault current since modern international standards quote them as kiloamperes.

This section covers sections 3.3.2, 3.3.3 and 3.3.7 of the textbook.

4.2 The transformer model

We will first examine the transformer. Assume a ratio of n:1 that is to say the

voltage on the primary side of an ideal transformer winding = Vp and the

voltage on the secondary side = Vs. It is well known that:

Vp Ip= Vs Is but Vp = n Vs and therefore Is = n Ip .

Now consider the effect of impedance.

Let us assume Fig 4.2.1 and Fig 4.2.2 represent equivalent systems in so far

that the voltage at the load terminals V is the same in both cases:

E

Z

n:1

L

O

A

D

II/n II/n

n:1

E

L

O

A

D

V V

z

Fig 4.2.1 Fig 4.2.2

In each case the load current is I amps and the load voltage V volts; thus the

two circuits have equal input and output quantities.

In Fig 4.2.1 E - IZ/n = nV

In Fig 4.2.2 E = n (V + I z) or E - nIz = nV

Thus E - IZ/n = E - nIz

Therefore for the circuits to be equivalent we require Z = n

2

z.

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In general any network element of impedance Z operating at voltage V can be

referred to as an equivalent Zl for a chosen base voltage Vbase by the relation:

Z ZV

V

base'=

2

We are now able to exploit this result which allows us to swap transformers

and impedances around in order to simplify calculations.

4.3 Ohmic Methods

In this method the basic process consists of finding the equivalent impedances

as referred to one reference voltage which in our case will be the voltage level

of the fault.

Consider the following problem of finding I:

E

X

nI/nI

Y

Fig 4.3.1

The result above allows the network to be reduced almost by inspection to:

E/n

X/n2I

Y

Fig 4.3.2

Thus IE

nX

nY

=

+( )2

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Proof :

It is clear from fig 4.3.1 that:

E IXn

nIY

EIX

nnIY

E

nI

X

nY

=

= +

= +

0

2( )

IE

nX

nY

=

+( )2

Example 4.3.1

A 33kV generator of impedance 4.5 feeds an 11kv feeder of impedance0.71via an ideal transformer. Find the 11kV fault current assuming that theimpedances are reactive only.

Answer:

Choosing 11kV as the base voltage the transformation ratio is 3.

The equivalent generator voltage is therefore 33kV/3 and phase voltage is

11000/3V.The generator impedance referred is 4.5/9 = 0.5.

Thus the fault current I = 11000/3/(0.5+0.71) = 5248.64AThe accuracy is preserved for later use!

The result of this example will be explored later but consider first the problem

of finding the current flowing in the following network with two generators of

different voltages.

E1

X

n:1I1 /n

This network has the following equivalent.

I1

Y

1:m I2/m

I2W

I1+ I2

Fig 4.3.3

E2

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I1

E1 /nE2 /m

X/n2

Y

W/m2

I1+ I2

I2

Example 4.3.2

An 11kV circuit having impedance j0.45 is supplied by a nominal 33kVgenerator having an impedance of j18 and a 132kV generator having an

impedance of j76 in a network similar to Fig 4.3.3. Calculate the 11kVfault current for voltages of (a) 33kV and (b) 34kV for the 33kV generator.Assume ideal transformers.

Answer:

Choosing 11kV as base voltage. The transformation ratios are 12 for

132kV and 3 for 33kV

The equivalent impedances of the 132kV generator is 76/144 =0.5278and the equivalent impedance of the 33kV generator is 18/9=2.

(a) Because the equivalent voltages are equal they can be treated as one

behind two impedances in parallel. The parallel combination is thus:

..

..

2 0 5278

2 52780 4176

x

+ =

Ix

A=+

=11000

3 0 4176 0 457320

( . . )

(b) We will do two methods as illustration first Thevenins Theorem and thenthe parallel generator theorem.

Thevenins Theorem

The equivalent voltage of the 33kV machine is now 34/3=11.333kV thus

a current I= 333/3/j2.5278A = -j76.05A flows from it towards the 132kVgenerator. The open circuit voltage at the common terminal is therefore

11.333/3- (j2 x -j76.05) =6391V.

The Thevenin impedance of the network is given by the two generator

impedances in parallel. ie 0.4176(from above). Hence:

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I Af = + =

6391

0 45 0 41767366

. .

Using the Parallel Generator Theorem we can calculate the voltage across

the 11kV feeder directly.

From previous work:

VE Y E Y E Y

Y Y Y=

+ +

+ +1 1 2 2 3 3

1 2 3

in this case E3 = 0 but Y3 = 1/0.45 (the 11kV

impedance)

now V V=

+

+ +

=

11000

3 05278

11333

321

05278

1

2

1

0 45

3314 84. .

. .

.

If =

3314 84

0 45

.

.= 7366A

Returning to reconsider the simple series example No. 4.3.1 there are a few

points worth noting:

The fault MVA supplied by the generator is given by 3VI where Vand I are expressed in kilovalues. ie 3 x 33 x 1.7495 = 100MVA.Of this 100MVA 3I2X was dissipated in the generator windings ie 3

x (1749.52) x 4.5 = 41.322MVA. and 3x (5248.642) x 0.71 =

58.677MVA is dissipated in the 11kV circuit. Thus the fault currentis 5248A the generator output is 58.677MVA but the total

generated magnetic energy = 100MVA of which 41.322MVA is lost

in the machine.

It is clear that Fault Level means the total magnetic energy lost in asystem during a fault. Modern International standards avoid

ambiguity by quoting fault current in amperes. [6]

Note that phase values are always used when calculating faultcurrents. This also applies to the per-unit system which is covered

below.

The percentage voltage dropped in the generator is 41.322% and58.677% is dropped across the 11kv circuit. As we shall see, the

similarity between these values and those above is no coincidence!

Read sections 3.3.2 and 3.3.3 of the textbook.

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4.4 PER UNIT METHODS

Read section section 3.3.7 of your text book up to the end of the paragraph

following equation 3.3.7.11.

Comments:

Vbase is always a phase voltage in three phase systems.Ibase is always a phase (or line ) current.

Taken in stages the proof of equation 3.3.7.3 is, starting from equation

3.3.7.1:

V IZ

V

V

IZ

V

V

V

IZ

V

I

IV

V

I

I

ZI

V

V

V

I

I

Z

Z

V I Z

base base

base base

base

base

base base

base

base

base base base

pu pu pu

=

=

=

=

=

=

.

The values of V and Vbasein equation 3.3.7.9 are not necessarily line to line

values, the equation is equally valid if phase to neutral values are used. Use

of phase to neutral voltage makes the derivation of equation 3.3.7.11 muchclearer; viz:

ZZ

Z

ZI

V

ZS

Vpu

base

base

base

base

base

= = =3 2

This is not very useful because the equation can apply only to impedances

operating at the base voltage. In section 4.2 we have developed a method of

referring impedances Z operating at other voltages V to Z operating at the

voltage base Vbase.

Then:

ZZS V

V V

ZS

Vpu

base base

base

base

L

' = =2

2 2 23

but since V is a phase voltage 3V2 is the square of the operating line voltage

VL.

There are three points worth noting about the expression above:

It is combining two functions of referring to a common voltage base andgiving the Zpuin one simple operation.

It is not quite the same as equation 3.3.7.11 in the textbook which has Zpuon the left hand side. It is normal to use Z instead of Z but we have not

quite finished with Z.

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The expression is evaluated easily if Sbase is quoted in MVA and Vl isquoted in kV. eg for a 33kV circuit of 0.5445 at a base of 100MVA

then Zpu = 0.5445 x 100/ 332= 0.05pu.

Now we are able to calculate the per unit impedance for any operating

voltage but we need another common reference. The textbook states that acommon MVA base should be used without justifying it.

Now we know

ZZS

Vpubase

' =3 2

and Z ZV

V

base'=

2

2

thenZ

Z

S

V

pu base

base

'

' =

3 2

As we are going to manipulate per unit impedances in different combinations

it is necessary to keep the ratio constant for all impedances. The only way todo this of course is to use the same Sbasefor all components. This is because

Vbase will always be determined by the point of fault. The benefit being that

we are no longer limited to a single Vbase ie we may consider faults at

different voltage levels in the same network without considering the base

voltage provided we calculate Zpufor every component using the same Sbaseand the operating voltage. This is fundamentally different from the stong

emphasis placed on base voltage in the textbook.

We can now drop Zpu and from now on we will use only Zpu . It also

follows that every network theorem is just as valid for per unit values as they

are for ohmic values. Having established the outline principle there are just a

few manipulations we need to cover:

4.4.1 Converting from ohmic to per unit impedance

From above ZZxS

kVpu

base

L

= 2

where Sbaseis the Base MVA and kVLis the line operating voltage in

kilovolts.

Proof:

ZZ

ZZx

I

VZx

S

Vx

V

ZxS xV

V V

ZxS

Vpu

base

base

base

base

base

base base

base

base

Lbase

= = = = =3

1

3

2

2 2 2

If S is expressed in units of MVA and VL in kV the ratio is maintained and

the result follows.

Example 4.4.1 Convert 0.1operating at 66kV to per unit at a base of

50MVA.

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Answer:

Zx

pupu = =01 50

660 0011472

..

4.4.2 Changing the base MVA

Plant impedances for transformers and generators are quoted in test data and

on nameplates on the plant rating. It is therefore necessary to refer all plant to

a common base.

Essentially this is covered by equation 3.3.7.13 in the textbook.

Proof:

ZZS

V

Z V ZS

ZV Z S

V S

Z S

S

pu

pu

pu

pu pu

1

1

2

2

1

1

2

2

1 2

2

1

1 2

1

=

=

= =

Thus changing base is a simple ratio calculation.

Example 4.4.2

A 240MVA 400/132kV transformer has a nameplate impedance of 20%.

Calculate the per unit impedance to a 100MVA base.

Answer

Zpu on a base of 100 MVA = 20/100*100/240 = 0.08333pu.

4.4.3 Voltage conversion

It is often necessary to change the voltage base of a quoted per unit

impedance. Typically this is necessary when an overhead line designed foroperation at 132kV is actually operated at, say, 33kV.

In this case ZZ V

Vpu

pu

2

1 1

2

2

2=

Where V1 is the voltage of the quoted per unit impedance Z1pu and V2 is the

operating voltage.

Example 4.4.3

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10km of 132kV overhead line has a per unit impedance to a 100MVA base

of (0.0104 +j 0.0225) pu and is operated at 33kV. Calculate the pu

impedance at 33kV.

Answer:

Z33pu=(0.0104+j0.0225) x 1322/332 = (0.1664+j 0.36)pu

4.4.4 Per Unit Losses

It is often useful to calculate the power losses in a circuit. The per unit

system provides a very convenient way to calculate them.

Lpu= I2pu Zpu

Proof

( ) = = = = =L

L

S

I Z

S

I ZS

Sx

V

V

IV

Sx

ZS

VI Zpu pu pu

3 3 3

3

3

3

2 2

2

2

2

2

2 2

2

Example 4.4.4

An 11kV circuit of resistance 0.726carries 10MVA. Calculate the losses inthe circuit.

Answer: (First by ohmic and then by per unit methods)

1 Ohmic.

I A

Losses x x MVA

= =

= =

10

3110005248

3 524 8 0 726 0 6

7

2

..

. . .

2 Per Unit. Using a base of 100MVA

Zpu= 0.726 x 100/112 = 0.6pu Ipu= 0.1

Lpu= 0.12x 0.6 = 0.006pu = 0.6MVA

4.4.5 Conversion of a specified fault level to Zpu

Fault levels at source busbars are often quoted in MVA. It is therefore

necessary to convert them to impedance in order to compute downstream

fault levels.

We have Vpu = Ipu Zpu

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thus ZV

Ipupu

pu

= or, when the fault level is quoted: ZV

Spu

pu

pu

=

2

Normally the fault level is quoted as MVA without reference to the prefault

voltage. In this event nominal voltage is assumed.

Example 4.4.5

A 132kV infeed busbar operating at 138kV has a short circuit fault current

of 20kA. What is the infeed impedance for a 200 MVA base?

Answer:

Spu= 3 x138x20/200=23.903pu Vbase = 132kV, Vpu = 138/132 = 1.0454pu

Zpu = 1.04542/23.903 = 0.04572pu.

Alternatively:Ibase = 200x10

3/ 3 / 132 = 874.773 A thus Ipu= 20000/874.773 = 22.863pu Zpu= Vpu / Ipu= 1.0454/ 22.863 = 0.04572pu

4.4.6 Load Impedance

It is often necessary to convert load to a per unit impedance.

Let the voltage at the load be V and the power S using the convention that

lagging vars are positive.

Spu = S/Sbaseand Vpu= V/ Vbase

then ZV

Spu

pu

pu

=

2

* .

Proof:

For a three phase system ZV

I

V

S

phase phase= =

3 2

*

since I = V* / S* and VxV*= V2

and ZV

I

V

Sbase

base

base

base

base

= =3 2

Hence ZZ

Z

V

V

S

S

V

Spubase

phase

base

base pu

pu

= =

=

2 2

* *

Example 4.4.6

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Calculate the per unit impedance on a 200MVA base for a load of 150MVA

at 0.85 power factor on a 132kV system operating at 138kV. This example

is taken from example 2 in section 3.3.10 in the textbook.

Answer: (by two methods)

The load angle is arcos (0.85) = 31.788

Vpu = 138/132 = 1.04545.

Spu = 150/200= 0.75= 0.75( 0.85 + j sin 31.788) = 0.6375+j 0.3951

Method I load angle

Zpu= V2pu /S

*pu = 1.045452/ 0.75 = 1.4573

Thus Zpu = 1.4773 (0.85+j sin 31.788) = 1.2387 +j 0.7677pu

Method II Complex load

Zpu = V2pu /S

*pu = 1.045452/ (0.6375 - j 0.3951) = 1.2387 + j0.7677pu

4.4.7 Transformer impedances

Transformers present two types of problem. The first is caused by them

having tap change equipment; the second is caused by non - standard ratios.

First assume a transformer is operating at an off nominal tap ratio a:1.

For example, most 132/33kV transformers in the UK have tap changers ofratio range -20% to 10% fitted to the HV side.

When the tap position is -20%. the per unit terms the ratio a = 0.8

Assuming that 1pu current flows in the secondary and the secondary voltage

is Vputhen the corresponding values on the HV side are 1/a pu and aVpu.

The impedance Zpu viewed from the HV side is then:

v

i

aV

a

a V a Z pu= = =12 2

The transformer test engineer needs to guard against the possibility of this

result distorting the measurement of the true impedance of the transfomer. A

transformer impedance is determined by circulating the exact full load

current in a short circuited secondary winding and measuring the voltage on

the primary winding required to produce the current. The result is known as

impedance voltage and is normally quoted as a percentage of the test

supply voltage; it is easily converted to a per-unit value by dividing by 100

but it must be noted this is based upon the rating of the transformer. However

when the transformer is at the off nominal tap ratio a:1 the secondary

voltage is transformed to aV in the primary winding ; the test engineer avoidsambiguity by expressing the impedance voltage as a percentage of aE (where

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E is the nominal voltage of the primary when a=1); thus keeping the volts per

turn constant. Any variation of transformer impedance from the test is then

a true measure of variation caused by the design of the transformer and the

tap changer.

In accurate power system studies it is necessary to take account of the tap

changer effect (ie a2Z above) and also sometimes necessary to change thetransformer impedance Z. Students are recommended to obtain a full test

certificate for a transformer within their company to see how Z varies with

tap position.

Transformers with non - standard ratios are used sometimes for voltage

control purposes. Typical among these are 33/11.5kV transformers

operating on nominal 33/11kV medium voltage distribution networks.

Methods outlined paragraph 4.4.3 can be used in such a cases.

Example 4.4.7.1

A 33/11.5kV 24MVA transformer having impedance voltage 10% is

operated in a nominal 33/11kV network. Determine the per unit impedance

to be used for doing 11kV fault calculations with a tap changer position of

- 10%.

Answer:

First without the tap changer effect:

Z = (11.5/11)2

x 10 / 100 = 0.1093pu

Taking account of the tap changer .

It should be noted first that when viewed from the 11 kV side the

transformer will have an output voltage of 11.5kV when on the nominal tap.

ie there is an equivalent ratio b = 11/11.5 = 0.9565:1 and this correction

is required throughout the range of the tap changer. The modified a is then

a= (100-10)/100 x b = 0.9x0.9565=0.8609 ; a2= 0.741

Z = a2Z = 0.741x0.1093 = 0.081pu

Clearly we have rebased the impedance by a factor (11.5/11)2 and then

corrected this by modifying the tap changer with the reciprocal! It is

necessary because we need to model the tap changer correctly for voltage

control and the impedance is adjusted to compensate.

The next example illustrates the effect in fault calculation.

Example 4.4.7.2

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Using the transformer in example 4.4.7.1 calculate the fault level in

MVA and the fault current for a fault on the 11kV busbars if the

transformer is operating on its highest tap ( ie raising voltage) and it is

supplied by a circuit of impedance of 0.1pu( on base of 24MVA) from a

33kV source of negligible impedance.

Consider the diagram:

0.1pu 0.081pu

a:1

1pu I/apu

Ipu

Using 11kV as the base voltage a= 0.8609.

Vpu prefault = 1/a .

The total impedance should be referred to the lv side hence Zpu=0.181/a2

Then Ipu= 1/a / (0.181/a2) = 4.756pu The fault MVApu= Ipux Vpu = 5.524pu

The fault MVA is thus 5.524 x 24 = 132.6MVA

Checking that the fault MVA appears as losses we find:

(I/a)2pux Z = (4.756/0.8609)2x 0.181 = 5.524pu

The same fault MVA would have been obtained if 11.5kV had been used

as the reference voltage with a=0.9 and students should check this for

themselves. A choice of voltage base may not always be convenient if there

is 11kV plant on the lower voltage level.

Looking again at example 4.3.1 this time using per unit methods:

A 33kV generator of impedance 4.5 feeds an 11kv feeder of impedance0.71via an ideal transformer. Find the 11kV fault current assuming that theimpedances are reactive only.

Answer Using a 200MVA base.

Noting thatZZxS

kVpu

base

L

= 2 we have

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The generator Zpu = 4.5 x 200/332= 0.826446pu

The 11kv feeder Zpu= 0.71 x 200/112= 1.173553pu

Total Zpu = 2pu. Therefore since Vbase = 11/3 then Vpu = 1pu.

Ipu= 1/Zpu= 1/2 = 0.5pu.

Ibase= 10,497.28A I = 0.5 x 10,497.28 = 5248.64A.

Fault level = 200 x 0.5 = 100MVA.

Losses in the generator are 0.52x 0.82644 = 0.206611pu = 41.322MVA.

Losses in the feeder are 0.52x 1.17355 = 0.293388pu = 58.677MVA.

Voltage dropped in the generator = 0.5 x 0.82644 = 0.41322pu

Voltage dropped in the feeder = 0.5 x 1.17355 = 0.58677pu.

We are now ready to study some examples of fault calculations by drawing

together all the work done so far but first there are some self test questions to

try.


1 Calculate Zpu for an impedance of 0.1 if operating at voltages of 11kV ,

33kV and 132kV for base MVA values of 50, 100 and 200MVA.

2 Find Zbase (in ohms) for a base MVA of 100 at 11kV, 33kV , 132kV and

400kV.

3 A load of 400MW at 0.8pf is connected to a 400kV system. Calculate the

per unit impedance in complex rectangular form for a base of 100MVA if

the voltage at the load terminals is 420kV. Calculate the per unit conductance

and susceptance.

4 A network consists of a generator with a Zpu = 0.01 on a 100MVA base and a

line. The fault level at the remote end of the line is 2625MVA and theprefault voltage is 1.05per unit. What are the losses in the generator and in

the line during the fault. Assume the line has zero resistance.

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5.0 Worked examples of symmetrical fault calculation

5.1 The transformer feeder.

A 33/11kV substation B is equipped with two 12MVA 33/11kV transformers

which are supplied from a 132/33kV substation A by 33kV overhead lines as shown

in fig 5.1.1. The overhead line are each 15km long with impedance per phase of0.091+j0.316. The fault level at the 33kV busbar at substation A is 800MVA. The

transformers at substation B are 12% impedance on rating. Calculate:

The current for fault at the high voltage side of a transformer.

The voltage at the busbar at substation A during the fault.

Use a base MVA of 100.

We will do the calculation two ways, first including resistance and then ignoring it.

First find the per unit impedance of the overhead lines:

Zj x x

jL pu

= +

= +( . . )

. .0 091 0 316 15 100

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