P UNIT I INTRODUCTION
Need for system planning and operational studies – basic components of a power system.-Introduction to restructuring - Single line diagram – per phase and per unit analysis – Generator - transformer – transmission line and load representation for different power system studies.- Primitive network - construction of Y-bus using inspection and singular transformation methods – z-bus.
UNIT II POWER FLOW ANALYSIS
Importance of power flow analysis in planning and operation of power systems - statement of power flow problem - classification of buses - development of power flow model in complex variables form - iterative solution using Gauss-Seidel method - Q-limit check for voltage controlled buses – power flow model in polar form - iterative solution using Newton-Raphson method .
UNIT III FAULT ANALYSIS – BALANCED FAULTS
Importance of short circuit analysis - assumptions in fault analysis - analysis using Thevenin‟s theorem - Z-bus building algorithm - fault analysis using Z-bus – computations of short circuit capacity, post fault voltage and currents.
UNIT IV FAULT ANALYSIS – UNBALANCED FAULTS
Introduction to symmetrical components – sequence impedances – sequence circuits of synchronous machine, transformer and transmission lines - sequence networks analysis of single line to ground, line to line and double line to ground faults using Thevenin‟s theorem and Z-bus matrix.
UNIT V STABILITY ANALYSIS
Importance of stability analysis in power system planning and operation - classification of power system stability - angle and voltage stability – Single Machine Infinite Bus (SMIB) system: Development of swing equation - equal area criterion - determination of critical clearing angle and time – solution of swing equation by modified Euler method and Runge-Kutta fourth order method.
TEXT BOOKS:
1. Nagrath I.J. and Kothari D.P., „Modern Power System Analysis‟, Tata McGraw-Hill, Fourth Edition, 2011. 2. John J. Grainger and W.D. Stevenson Jr., „Power System Analysis‟, Tata McGraw-Hill, Sixth reprint, 2010. 3. P. Venkatesh, B.V. Manikandan, S. Charles Raja, A. Srinivasan, „ Electrical Power Systems- Analysis, Security and Deregulation‟, PHI Learning Private Limited, New Delhi, 2012.
REFERENCES:
1. Hadi Saadat, „Power System Analysis‟, Tata McGraw Hill Education Pvt. Ltd., New Delhi,
21st
reprint, 2010.
2. Kundur P., „Power System Stability and Control, Tata McGraw Hill Education Pvt. Ltd., New Delhi, 10th reprint, 2010. 3. Pai M A, „Computer Techniques in Power System Analysis‟, Tata Mc Graw-Hill Publishing Company Ltd., New Delhi, Second Edition, 2007. 4. J. Duncan Glover, Mulukutla S. Sarma, Thomas J. Overbye, „ Power System Analysis & Design‟, Cengage Learning, Fifth Edition, 2012. 5. Olle. I. Elgerd, „Electric Energy Systems Theory – An Introduction‟, Tata McGraw Hill Publishing Company Limited, New Delhi, Second Edition, 2012. 6. C.A.Gross, “Power System Analysis,” Wiley India, 2011.
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
7. Generation system
8. Transmission system
9. Distribution system
10. 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 0 n
EC
EB 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
0 n
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.
Load
Fig. 1.3 One-line diagram
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.
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 single-phase or per phase equivalent circuit of the system.
Refer the one-line diagram of a sample power system shown in Fig. 1.4.
Fig.1.5 combines the equivalent circuits for the various components shown in Fig. 1.4 to form the per-phase
impedance diagram of the 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
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 x106
Base current, Abase voltage, V base voltage, kV x 1000
base MVA x 1000
=
(1.1)
base voltage, kV
base voltage, V base voltage, kV x1000
Base impedance , Ω (1.2) base current, A base current, A
Substituting eq. (1.1) in the above
(base voltage, kV)2
x 1000
Base impedance , Ω Thus
base MVA x1000
(base voltage, kV)2
Base impedance , Ω (1.3) base MVA
Power factor being a dimensionless quantity
actual impedance, Ω Per - unit impedance (1.4) base impedance, Ω
Base MVA
Per-unit impedance = actual impedance x (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 single-
phase 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
(Basevoltage,kV)2
Base impedance , Ω (1.6) Base MVA / 3 Base MVA
It is to be noted that eq.(1.6) is much similar to eq.(1.3). Thus
actual impedance Per - unit impedance
baseimpedance
Base MVA
= actual impedance X (Base voltage,kV)2
(1.7)
(base voltage, kV)2
Base impedance , Ω (1.6) base MVA
Base MVA
Per-unit impedance = actual impedance X (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
500
Using eq.(1.7), per unit reactance 1.065 x 1.1002
222
Base MVA
Per-unit impedance = actual impedance X (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 per-unit impedance on a new base, the following equation applies:
base MVA new base kVgiv en 2
Per-unit Znew = per-unit Zgiven base MVA giv x ( base kV ) (1.8) en new
base MVA new base kVgiv en 2
Per-unit Znew = per-unit Zgiven x ( ) (1.8) base MVA giv base kV
en new
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
500 x ( 18
20 )2 = 0.0405
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
0..5
5 )2 = 11.718 Ω
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.
Formation of Y bus ---Primitive network
A power network is essentially an interconnection of several two-terminal
components such as generators, transformers, transmission lines, motors and
loads. Each element has an impedance. The voltage across the element is called
element voltage and the current flowing through the element is called the element
current. A set of components when they are connected form a Primitive network.
A representation of a power system and the corresponding oriented graph are
shown in Fig. 2.1.
1 2 4
3
7
1 2 3
6 4
5 4
2 1
3
0
Fig. 2.1 A power system and corresponding oriented graph
7
1 2 3
6 4
5 4
2 1
3
0
Fig. 2.1 A power system and corresponding oriented graph
Connectivity various elements to form the network can be shown by the
bus incidence matrix A. For above system, this matrix is obtained as
1 2 3 4 5 6 7
1 -1 1
2 -1 1 -1 1 (2.1)
A =
-1 -1
3
4 -1 1 -1
Element voltages are referred as v1, v2, v3, v4, v5, v6 and v7. Element currents
are referred as i1, i2, i3, i4, i5, i6 and i7. In power system network, bus voltages
and bus currents are of more useful. For the above network, the bus voltages are
V1, V2, V3 and V4. The bus voltages are always measured with respect to the
ground bus. The bus currents are designated as I1, I2, I3, and I4. The element
voltages are related to bus voltages as:
7 v1 = - V1
1 2 3 v2 = - V2 6 4
5 4 v3 = - V4
2 v4 = V4 – V3 1
3
v5 = V2 - V3
v6 = V1 – V2 0
v7 = V2 – V4
Expressing the relation in matrix form
-1 V
1 v1
-1
v2
v -1 V2 v
3
=
-1 1
4
v
V3
1
-1
v5
1
-1
6 V4
v 7 1 -1
Thus v = AT V
bus
The element currents are related to bus currents as:
7
I2
6 I3
I4
I1
5 4
2 3 4 1
2
1
3
(2.2)
(2.3)
I1 = - i1 + i6
I2 = - i2 + i5 – i6 + i7
I3 = - i4 – i5
I4 = - i3 + i4 – i7
0
I1 = - i1 + i6
I2 = - i2 + i5 – i6 + i7
I3 = - i4 – i5
I4 = - i3 + i4 – i7
Expressing the relation in matrix form
I1 -1 1
-1 1 -1 1 2
I3 = -1 -1
-1 1 -1 4
Thus Ibus = A i
i1
i2
i3
i4
i5
i6
i7 (2.4)
The element voltages and element impedances are related as:
z11
z z z z z z i1
12 13 14 15 16 17
v1
v2 21 22 23 24 25 26 27 2 v
z31
i3
v3
=
32 33 34 35 36 37
(2.5) 4 41 42 43 44 45 46 47 4 v
v5
z51 52 53 54 55 56 57
i5
6 61 62 63 64 65 66 67 6
z z z z z z
v 7 z
i
71 72 73 74 75 76 77 7
Here zii is the self impedance of element i and zij is the mutual impedance
between elements i and j. In matrix notation the above can be written as
v = z i (2.6)
Here z is known as primitive impedance matrix. The inverse form of above is
i = y v (2.7)
In the above y is called as primitive admittance matrix. Matrices z and y are
inverses of each other.
v = z i (2.6)
i = y v (2.7)
Similar to the above two relations, in terms of bus frame
Vbus = Zbus Ibus (2.8)
Here Vbus is the bus voltage vector, Ibus is the bus current vector and Zbus is
the bus impedance matrix. The inverse form of above is
Ibus = Ybus Vbus (2.9)
Here Ybus is known as bus impedance matrix. Matrices Zbus and Ybus
are inverses of each other.
Derivation of bus admittance matrix It was shown that
v = AT
Vbus Ibus =
A i
i = y v Ibus = Ybus Vbus
Substituting eq. (2.7) in eq. (2.4)
Ibus = A y v Substituting eq. (2.3) in the above
Ibus = A y AT
Vbus Comparing eqs. (2.9) and (2.11)
(2.10) (2.11)
Ybus = A y AT
(2.12)
This is a very general formula for bus admittance matrix and admits mutual
coupling between elements.
(2.3)
(2.4)
(2.7)
(2.9)
In power system problems mutual couplings will have negligible effect and often
omitted. In that case the primitive impedance matrix z and the primitive
admittance matrix y are diagonal and Ybus can be obtained by inspection. This
is illustrated through the seven-elements network considered earlier. When
mutual couplings are neglected
y11 22
y
Ybus = A y AT
y11
= A
33
44
55
66
22
33
44
55
y
77
66
y77
(2.13)
-1
-1
-1
-1 1
1 -1
1 -1
1 -1
-y11
-y22
= -1 1
-y33 -1 1 -1 1
-y44 y44
-1 -1
y55 -y55
-1 1
-1
y66 -y66
y77 -y77
1 2 3 4
1 y11 + y66 - y66 0 0
2 - y66 y22 + y55 + y66+y77 - y55 - y77
Ybus = 3 0 - y55 y44 + y55 - y44
4 0 - y77 - y44 y33 + y44 + y77
1 2 3 4
1 y11 + y66 - y66 0 0
2
- y66 y22 + y55 + y66+y77
- y55
- y77
Ybus = 3 0 - y55 y44 + y55 - y44
4 0
- y77
- y44 y33 + y44 + y77
7
1 2 3 6 4
5 4
2 1
3
0
The rules to form the elements of Ybus are:
The diagonal element Yii equals the sum of the admittances directly
connected to bus i.
The off-diagonal element Yij equals the negative of the admittance
connected between buses i and j. If there is no element between buses
i and j, then Yij equals to zero.
Bus admittance matrix can be constructed by adding the elements one by one.
Separating the entries corresponding to the element 5 that is connected between
buses 2 and 3 the above Ybus can be written as
Ybus =
+
1 2 3 4
1 y11 + y66
- y66 0 0
2 - y66 y22 + y66+y77 0 -y77
3 0 0 y44
- y44
4 0 -y77 - y44 y33 + y44 + y77
1 2 3 4
1 6 2 3
1 0 0 0 0 4
- y55
4
2 0 y55 0 2
1
0 - y55 y55 0
3 3
4 0 0 0 0
0
It can be inferred that the effect of adding element 5 between buses 2 and 3 is to
add admittance y55 to elements Ybus(2,2) and Ybus(3,3) and add – y55 to
elements Ybus(2,3) and Ybus(3,2). To construct the bus admittance matrix
Ybus, initially all the elements are set to zero; then network elements are added
one by one, each time four elements of Ybus are modified.
Example 2.1
Consider the power network shown in Fig. 2.2. The ground bus is marked as 0.
Grounding impedances at buses 1, 2, and 4 are j0.6 Ω, j0.4 Ω and j0.5 Ω
respectively. Impedances of the elements 3-4, 2-3, 1-2 and 2-4 are j0.25 Ω, j0.2 Ω,
j0.2 Ω and j0.5 Ω. The mutual impedance between elements 2-3 and 2-4 is j0.1 Ω.
Obtain the bus admittance matrix of the power network.
1 2
4
3
Fig. 2.2 Power network – Example 2.1
Solution
The oriented graph of the network, with impedances marked is shown in Fig. 2.3.
j 0.5 7
1 2 3
j 0.2 j 0.1 4 4
6 j 0.2
5 j 0.25
j 0.6 2 j 0.4 j 0.5 1
3
0
Fig. 2.3 Data for Example 2.1
Primitive impedance matrix is:
1 2 3 4 5 6 7
1 0.6
2 0.4
z = j 3
0.5
4 0.25
5 0.2 0.1
6 0.2
7 0.1 0.5
Inverting this
1 2 3 4 5 6 7
1 1.6667
2 2.5
3 2.0
y = - j 4 4.0
5 5.5556 -1.1111
6 5.0
7 -1.1111 2.2222
Bus incidence matrix A is:
-1 1
A = -1 1 -1 1
-1 -1
-1 1 -1
Bus admittance matrix Ybus = A y AT
1.6667 -1
2.5
-1
Ybus = - j A
2.0
-1 4.0
-1 1 5.5556 -1.1111
5.0 1 -1
-1.1111
2.2222
1 -1
1 -1
-1.6667
-1 1 - 2.5
- 2.0 -1 1 -1 1
Ybus = - j
- 4.0 4.0 -1 -1
4.4444 - 5.5556 1.1111
-1 1
-1
5.0 - 5.0
1.1111 1.1111 - 2.2222
1 2 3 4
1
2
Ybus = 3
4
- j6.6667 j5.0 0 0
j5.0 - j13.0556 j4.4444 j1.1111
0 j4.4444 - j9.5556 j5.1111
0 j1.1111 j5.1111 - j8.2222