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Fault calculations in a power system
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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
2.3 Self Test Questions
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!
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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
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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:
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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
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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
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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
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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
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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.
4.4 Self Test Questions
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