EE 0308 POWER SYSTEM ANALYSIS
Dr.R.Jegatheesan
Professor, EEE Dept.
POWER SYSTEM OVERVIEW
Power system components, Representation. Single line diagram, per
phase analysis of symmetrical three phase systems, general aspects
relating to power flow, short circuit and stability analysis, per unit
quantities, impedance diagram.
POWER FLOW ANALYSIS
Primitive network, Formation of Bus admittance matrix by inspection
method and singular method. Bus classification – Formulation of
Power Flow problems – Power flow solution using Gauss Seidel
method and Newton Raphson method, Comparison between these
methods. Handling of Voltage controlled buses, Off nominal
transformer ratios and phase shifting transformer.
SYMMETRICAL FAULT ANALYSIS
Formation of Bus Impedance matrix by Z – bus building algorithm. Types of faults
in power systems – Symmetrical fault analysis. Short circuit capacity –
Symmetrical fault analysis through bus impedance matrix.
UNSYMMETRICAL FAULT ANALYSIS
Symmetrical
components
–
sequence
impedance
–
sequence
networks,
Introduction to unsymmetrical faults. Single line to ground fault, line to line and
double line to ground faults – Unsymmetrical fault analysis using bus impedance
matrix.
POWER SYSTEM STABILITY
Introduction to stability studies – classification of stability – Transient stability –
Equal area criterion – critical clearing angle and time – Factors affecting transient
stability. Numerical integration method: Euler method, Modified Euler’s method,
Fourth order Runge Kutta method.
TEXT BOOKS
1. Hadi Sadat, Power System Analysis, Tata Mc Graw Hill Publishing
company, New Delhi, 2002.
2. John.J.Grainger, William D. Stevenson, Power System Analysis, Tata Mc
Graw Hill Publishing company, New Delhi, 2003.
REFERENCE BOOKS
1. Nagarath I.J. and Kothari D.P. Modern Power System Analysis, Tata Mc
Graw Hill Publishing company, New Delhi, 2002.
2. Wadhwa C.L. Electrical Power Systems, New age international (P) Ltd.
Publishers, 1995.
3. Pai M.A. Computer Techniques in Power System Analysis, Tata Mc Graw
Hill Publishing company, New Delhi, 2003.
EE 0308 POWER SYSTEM ANALYSIS
Chapter 1
POWER SYSTEM OVERVIEW
Introduction
A power system consists of a few generating plants, situated close to resources,
supplying electric power to various types of loads spread out over large area,
through large complex transmission and distribution network. Thus a power
system compose of
1. Generation system
2. Transmission system
3. Distribution system
4. Loads
Depending on the fuel used we have Hydro-Electric Power Plants, Thermal Power
Plants and Nuclear Power Plants. Generated supply will be of 11 kV. To have
greater efficiency, transmission is carried out at high voltages of order 230 kV or
400 kV. Power transformers are used to setup the voltage levels. Transmission
system consists of transformers, transmission towers and transmission lines.
Thereafter, voltage levels are reduced in stages. Distribution system supplies
power to different loads. Thus power system network is large, complex and very
expensive. Power system analysis deals with analysis problems associated with
power network. Power Flow Analysis, Short Circuit Analysis and Transient
Stability Study are the main Power System Analysis Problems.
Single line diagram or One-line diagram
Electric power systems are supplied by three-phase generators. Ideally, the generators
are supplying balanced three phase loads. Fig.1.1 shows a star connected generator
supplying star connected balanced load.
IA
Z
EA
EC
0
Z
EB
n
Z
IB
IC
Fig. 1.1 Y- connected generator supplying balanced Y- connected load
A balanced three-phase system is always solved as a single-phase circuit
composed of one of the three lines and the neutral return. Single-phase circuit of
three-phase system considered above is shown in Fig. 1.2.
IA
Z
EA
n
0
Fig. 1.2 Single-phase circuit
Often the diagram is simplified further by omitting the neutral and by indicating
the component parts by standard symbols rather than by their equivalent circuits.
Such a simplified diagram of electric system is called a one-line diagram. It is
also called as single line diagram. The one-line diagram of the simple three-phase
system considered above is shown in Fig. 1.3.
IA
Z
EA
n
0
Fig. 1.2 Single-phase circuit
Load
Fig. 1.3 One-line diagram
Fig. 1.4 shows the one-line diagram of a sample power system.
1
T1
T2
3
2
Load B
Load A
Fig. 1.4 One-line diagram of a sample power system
Fig. 1.4 shows the one-line diagram of a sample power system.
1
T1
T2
3
2
Load B
Load A
Fig. 1.4 One-line diagram of a sample power system
This system has two generators, one solidly grounded and the other grounded
through a resistor, that are connected to a bus and through a step-up transformer
to a transmission line. Another generator, grounded through a reactor, is
connected to a bus and through a transformer to the other end of the
transmission line. A load is connected to each bus.
On the one-line diagram information about the loads, the ratings of the
generators and transformers, and reactances of different components of the
circuit is often given.
Per phase analysis of symmetrical three phase systems
Per phase analysis of symmetrical three phase systems are illustrated through
the following three examples.
EXAMPLE 1.1
A balanced three-phase load connected in star consists of (6+8) Ω impedance in
each phase. It is connected to a three phase supply of 400 V, 50 Hz. Find (i) Phase
current (ii) Line current (iii) Per phase power and (iV) Total power
SOLUTION
Fig. 1.5 Balanced three phase load – 1 phase equivalent
Three phase star connected balanced load and the corresponding single phase
equivalent are shown in Fig. 1.5 (a) and (b)
EXAMPLE 1.2
Repeat the previous problem with the load impedance connected in delta.
SOLUTION
Fig. 1.6 Balanced three phase load – 1 phase equivalent
Three phase delta connected balanced load and the corresponding single phase
equivalent are shown in Fig. 1.6 (a) and (b).
EXAMPLE 1.3
Calculate the line currents in the three wire Y-Y system of Fig. 1.7.
A
(1.5+j6) Ω
+
1100 0 V
(2.5+j4) Ω
-
110 - 2400 V
+
110 - 1200 V
-
B
+
(2.5+j4) Ω
(1.5+j6) Ω
(2.5+j4) Ω
(1.5+j6) Ω
C
Fig. 1.7 Three wire system for Example 1.3
SOLUTION
The single phase equivalent of the given three phase system is shown in Fig. 1.8.
A
(1.5+j6) Ω
+
1100 0 V
(2.5+j4) Ω
-
Fig. 1.8 Single phase equivalent
Total impedance = ( 4 + j10)
Current
1100 0
IA 
 10.2133  68.2 0 A
(4  j10)
Since the source voltages are in positive sequence, the line currents are also in
positive sequence. Therefore
IB  10.2133  188.20 A ;
IC  10.2133  308.20 A  10.213351.80 A
Impedance and reactance diagram
In order to calculate the performance of a power system under load condition or
upon the occurrence of a fault, the one line diagram is used to draw the singlephase or per phase equivalent circuit of the system.
Refer the one-line diagram of a sample power system shown in Fig. 1.4.
1
T1
T2
3
2
Load B
Load A
Fig. 1.4 One-line diagram of a sample power system
Fig.1.9 combines the equivalent circuits for the various components shown in
Fig. 1.4 to form the per-phase impedance diagram of the system.
Fig.1.9 combines the equivalent circuits for the various components shown in
Fig. 1.4 to form the per-phase impedance diagram of the system
1
T1
T2
3
2
Load B
Load A
Fig. 1.4 One-line diagram of a sample power system
The impedance diagram does not include the current limiting impedances shown
in the one-line diagram because no current flows in the ground under balanced
condition.
Fig. 1.9 Per-phase impedance diagram
Fig. 1.9 Per-phase impedance diagram
+
E1
-
+
E2
+
-
Fig. 1.10 Per-phase reactance diagram
E3
-
Per-unit quantities
Absolute values may not give the full significance of quantities. Consider the
marks scored by a student in three subjects as 10, 40 and 95. Many of you may
be tempted to say that he is poor in subject 1, average in subject 2 and good in
subject 3. That is true only when the base for all the marks is 100. If the bases are
10, 50 and 100 for the three subjects respectively then his marks in percentage
are 100,80 and 95 and thus the conclusions are different. Thus, there is a need to
specify base quantity for meaningful interpretation.
Percentage = (actual value / base) x 100
Per-unit quantity = Percentage / 100 = actual value / base
This kind of explanation can be extended to all power system quantities.
In power system we shall deal with voltage, current, impedance and voltampere
or power. When they are large values, we may use kV, ampere, ohm and kVA or
kW as their units. It is to be noted that out of the four quantities voltage, current,
impedance and voltampere if we specify two quantities, other two quantities can
be calculated.
Generally, base voltampere in MVA and base voltage in kV are specified.
For a single-phase system, the following formulas relate the various quantities.
base VA
base MVA x 106
Base current , A 

base voltage, V
base voltage, kV x 1000
=
Base impedance , Ω 
base MVA x 1000
base voltage, kV
base voltage, V base voltage, kV x 1000

base current, A
base current, A
(1.1)
(1.2)
Substituting eq. (1.1) in the above
(base voltage, kV)2 x 1000
Base impedance , Ω 
base MVA x 1000
Thus
(base voltage, kV)2
Base impedance , Ω 
base MVA
(1.3)
Power factor being a dimensionless quantity Base power, MW = base MVA
Per - unit impedance 
actual impedance, Ω
base impedance, Ω
Per-unit impedance = actual impedance x
Base MVA
(Base voltage,kV)2
(1.4)
(1.5)
(base voltage, kV)2
Base impedance , Ω 
base MVA
Per-unit impedance = actual impedance X
(1.3)
Base MVA
(Base voltage,kV)2
(1.5)
For three phase system, when base voltage is specified it is line to line base
voltage and the specified MVA is three phase MVA. Now let us consider a three
phase system. Let Base voltage, kV and Base MVA be specified. Then singlephase
base
voltage,
kV
=
Base
voltage,
kV
/ 3
and
single-phase base MVA=Base MVA / 3. Substituting these in eq. (1.3)
[Base voltage, kV/ 3 ] 2
Base impedance , Ω 
Base MVA / 3
(Base voltage,kV)2

Base MVA
(1.6)
It is to be noted that eq.(1.6) is much similar to eq.(1.3). Thus
Per - unit impedance 
actual impedance
base impedance
= actual impedance X
Base MVA
(Base voltage,kV)2
(1.7)
(base voltage, kV)2
Base impedance , Ω 
base MVA
Per-unit impedance = actual impedance X
(1.6)
Base MVA
(Base voltage,kV)2
(1.7)
EXAMPLE 1.4
A three phase 500 MVA, 22 kV generator has winding reactance of 1.065 Ω. Find
its per-unit reactance.
Solution
222
1.065
Base impedance 
 0.968 Ω ; Per-unit reactance =
 1.1002
500
0.968
Using eq. (1.7), per  unit reactance  1.065 x
500
 1.1002
222
Per-unit impedance = actual impedance X
Base MVA
(Base voltage,kV)2
(1.7)
Per-unit quantities on a different base
Sometimes, knowing the per-unit impedance of a component based on a
particular base values, we need to find the per-unit value of that component
based on some other base values. From eq.(1.7) It is to be noted that the per-unit
impedance is directly proportional to base MVA and inversely proportional to
(base kV)2. Therefore, to change from per-unit impedance on a given base to perunit impedance on a new base, the following equation applies:
Per-unit Znew = per-unit Zgiven
base kVgiv en 2
base MVA new
)
x(
base MVA giv en
base kVnew
(1.8)
Per-unit Znew = per-unit Zgiven
base kVgiv en 2
base MVA new
)
x(
base MVA giv en
base kVnew
(1.8)
EXAMPLE 1.5
The reactance of a generator is given as 0.25 per-unit based on the generator’s of
18 kV, 500 MVA. Find its per-unit reactance on a base of 20 kV, 100 MVA.
Solution
New per-unit reactance = 0.25 x
100
18 2
x(
) = 0.0405
500
20
EXAMPLE 1.6
A single phase 9.6 kVA, 500 V / 1.5 kV transformer has an impedance of 1.302 Ω
with respect to primary side. Find its per-unit impedance with respect to primary
and secondary sides.
Solution
With respect to Primary
Per-unit impedance = 1.302 x
0.0096
 0.05
(0.5)2
With respect to Secondary
Impedance = 1.302 x (
1.5 2
) = 11.718 Ω
0.5
Per-unit impedance = 11.718 x
0.0096
 0.05
(1.5)2
Conclusion
Per-unit impedance of the transformer is same referred to primary as well as
secondary.
Advantages of per-unit calculation
1. Manufacturers usually specify the impedance of a piece of apparatus in
percent or per-unit on the base of the name plate rating.
2. The per-unit impedances of machines of same type and widely different
rating usually lie within narrow range although the ohmic values differ
much.
3. For a transformer, when impedance in ohm is specified, it must be clearly
mentioned whether it is with respect to primary or secondary. The per-unit
impedance of the transformer, once expressed on proper base, is the same
referred to either side.
4. The way in which the three-phase transformers are connected does not
affect the per-unit impedances although the transformer connection does
determine the relation between the voltage bases on the two sides of the
transformer.
EXAMPLE 1.7
A 300 MVA, 20 kV three-phase generator has a subtransient reactance of 20%.
The generator supplies a number of synchronous motors over 64-km
transmission line having transformers at both ends, as shown in Fig. 1.11. The
motors, all rated 13.2 kV, are represented by just two equivalent motors. Rated
inputs to the motors are 200 MVA and 100 MVA for M1 and M2, respectively.
For both motors X” = 20%. The three phase transformer T1 is rated 350 MVA,
230/20 kV with leakage reactance of 10%. Transformer T2 is composed of three
single-phase transformers each rated 127/13.2 kV, 100 MVA with leakage
reactance of 10%. Series reactance of the transmission line is 0.5 Ω/km. Draw
the impedance diagram, with all impedances marked in per-unit. Select the
generator rating as base in the generator circuit.
T2
T1
k
l
m
p
M1
r
M2
n
Fig. 1.11 One-line diagram for Example 1.7
Solution
Base MVA = 300
Base voltage at generator side = 20 kV
Base voltage in transmission line = 230 kV
Line to line voltages of transformer T 2 :
Base voltage at motor side = 230 x
3 x 127 / 13.2  220 / 13.2 kV
13.2
 13.8 kV
220
Base MVA and base voltages at different sections are marked.
300 MVA
T1
20 kV
k
T2
l
230 kV
m
13.8 kV
n
M1
p
r
M2
Per-unit reactance of generator = 0.2
Per-unit reactance of transformer T 1 = 0.1 x
300
 0.0857
350
Per-unit reactance of transmission line = 0.5 x 64 x
Per-unit reactance of transformer T 2 = 0.1 x (
300
 0.1825
2302
220 2
)  0.0915
230
Per-unit reactance of motor M1 = 0.2 x
300
13.2 2
x(
)  0.2745
200
13.8
Per-unit reactance of motor M1 = 0.2 x
300
13.2 2
x(
)  0.549
100
13.8
Per-unit impedance diagram is shown in Fig. 1.12
j 0.0857
k
l
j 0.1815
m
n
j 0.0915
p
r
j 0.2
j 0.2745
+
Eg
-
j 0.549
+
Em1
Fig. 1.12 Per-unit impedance diagram
-
+
-
+ Em2
EXAMPLE 1.8
A transformer rated 200 MVA, 345Y / 20.5 Δ kV connected at the receiving end
of a transmission line feeds a balanced load rated 180 MVA, 22.5 kV, 0.8 power
factor. Determine
(a) The rating of each of three single-phase transformers which when
properly connected will be equivalent to the above three-phase transformer
and
(b) The complex impedance of the load in per-unit, if the base in the
transmission line is 100 MVA, 345 kV.
Solution
(a) Rating of each single-phase transformer: 200 / 3 MVA, (345 / 3 ) / 20.5 kV
i.e 66.7 MVA, 199.2 / 20.5 kV
(b) Load Z =
V
S
2

Base MVA = 100;
22.5 2
 2.81Ω
180
Base voltage at the load side = 20.5 kV
Load in per-unit = 2.81 x
100
36.870  0.669 36.870  0.5352  j4014
2
20.5
EE 0308 Power System Analysis - Problem Set 1
1. A 120 MVA, 19.5 kV generator has Xs = 0.15 per unit and is connected to a
transmission line by a transformer rated 150 MVA, 230 Y/18Δ kV with X =
0.1 per unit. If the base to be used in the calculation is 100 MVA, 230 kV for
the transmission line, find the per unit values to be used for the
transformer and the generator reactances.
2. A transformer’s three phase rating is 5000 kVA, 115/13.2 kV, and its
impedance is 0.007 + j0.075 per unit. The transformer is connected to a
short transmission line whose impedance is 0.02 + j0.1 per unit on a base
of 10 MVA, 13.2 kV. The line supplies a three phase load rated 3200 kW,
13.2 kV with a lagging power factor of 0.8. Draw the per unit impedance
diagram taking the base as 10 MVA, 13.2 kV at the load. If the high tension
voltage is maintained at 115 kV, find the per unit and the actual load
voltage.
3. The one-line diagram of an unloaded power system is shown below.
T1
A
B
C
j 80 Ω
T2
j 100 Ω
E
1
F
2
T3
D
3
Reactances of the two sections of transmission line are shown in the
diagram. The generators and transformers are rated as follows:
Generator 1: 20 MVA, 13.8 kV, X” = 0.2 per unit
Generator 2: 30 MVA, 18 kV, X” = 0.2 per unit
Generator 3: 30 MVA, 20 kV, X” = 0.2 per unit
Transformer T1: 25 MVA, 220Y/13.8Δ kV, X = 10%
Transformer T2: Single-phase units each rated 10 MVA, 127/18 kV, X = 10%
Transformer T3: 35 MVA, 220Y/22Y kV, X = 10%
Draw the reactance diagram with all reactances marked in per unit and with
letters to indicate points corresponding to the one-line diagram. Choose a
base of 50 MVA, 13.8kV in the circuit of generator 1.
4. Draw the impedance diagram of the power system shown below.
j 40 Ω
2
1
j 20 Ω
j 20 Ω
B
A
Δ
Δ
C
3
Mark impedances in per unit. Neglect resistance and use a base of 50 MVA,
138 kV in the 40-Ω line. The ratings of the generator, motors and
transformers are:
Generator 1:
20 MVA,18 kV, X” = 20%
Generator 2:
20 MVA,18 kV, X” = 20%
Synchronous motor 3:
30 MVA, 13.8 kV, X” = 20%
Three phase Y-Y transformers: 20 MVA, 138Y/20Y kV, X = 10%
Three phase Y- Δ transformers: 15 MVA, 138Y/13.8 Δ kV, X = 10%
ANSWERS
1. 0.06667 p.u.; 0.1467 p.u.
2. 0.9317 p.u.; 12.2984 kV
0.014 + j0.15
1.0  00 p.u.
0.02 + j0.1
Vload
2 + j1.5
3.
j0.0826
j0.2
A
C
j0.1033
j0.1667
B
E
j0.5
F
j0.3333
j0.1429
D
j0.2755
4.
A
j0.25
j0.105
j0.0525
j0.25
B
j0.0525
j0.25
j0.25
j0.3333
j0.405
j0.3333
C
j0.3333
j0.405