Introduction
A power system can be subdivided into three major parts before supplying electrical
energy to consumers :
• Generation
• Transmission and Subtransmission
• Distribution
A generation of electric power is done by using a generators where one of the
essential components of power systems is the three phase ac generator known as
synchronous generator or alternator
ac generators can generate high power at high voltage. typically 30 kV the size of
generators can vary from 50 MW to 1500 MW.
In a power station several generators are operated in parallel in the power grid to
provide the total power needed. They are connected at a common point called a bus.
After the electric power is generated by a generators a step-up transformers are used
for transmission of power to reduce the power losses . These transformers connecting
to the transmission network is used to transfer electric energy from generating units
at various locations to the distribution system which ultimately supplies the load
.Transmission voltage lines operating at more than 60kV are standardized at 69 kV,
115 kV, 138 kV, 161 kV, 230 kV,345 kV, 500 kV, and 765 kV line-to-line.
Transmission voltages above 230 kV are usually referred to as extra-high voltage
(EHV).
At the receiving end of the transmission lines step-down transformers are used to
reduce the voltage to suitable values for distribution or utilization.
The distribution system is that part which connects the distribution substations to the
consumers ' service entrance equipment.
Loads of power systems are divided into industrial, commercial, and residential.
In addition to generators, transformers, and transmission lines, other devices are
required for the satisfactory operation and protection of a power system. Some of the
protective devices directly connected to the circuits are called switchgear. They
include instrument transformers, circuit breakers, disconnect switches, fuses and
lightning arresters. These devices are necessary to deenergize either for normal
operation or on the occurrence of faults. The associated control equipment and
protective relays are placed on switchboard in control houses.
For a power system to be practical it must be safe, reliable, and economical. Thus
many analyses must be performed to design and operate an electrical system.
However, before going into system analysis we have to model all components of
electrical power systems
Power system representation
The complete circuit diagram for a three phase system is seldom necessary to convey
even the most detailed information about a system. Therefore the interconnection
among the components of the power system may be shown in a simplified diagram is
called a single line diagram.
Single line diagram
A single-line diagram of a power system shows the main connections and
arrangements of components . Power system networks are represented by single-line
diagrams using suitable symbols as shown below :
The impedance diagram on single-phase basis for use under balanced operating
conditions can be easily drawn from the single-line diagram such as shown
The impedance diagram can be abbreviated to reactance diagram when the
resistances and capacitors are neglected.
The per unit quantities (pu)
Because of the large amount of power transmitted , kilowatts or megawatts
and kilovolt-ampere or megavolt-ampere are the common terms. However,
these quantities as well as amperes and ohms are often expressed as a
percent or per unit of a base or reference value specified for each.
In the per-unit system, all quantities are represented as a fraction of the base
value:
actualvalue
Quantityin per  unit
basevalueof quantity
For example
V(pu) =
actual value of voltage (v)
base value of voltage (v)
I(pu) =
actual value of current (A)
base value of current (A)
Z(pu) =
S(pu) =
actual value of impedance(Ω)
base value of impedance(Ω)
actual value of appearant power(VA)
base value of appearant power(VA)
If any two of the four base quantities are specified, the other base values can
be calculated. Usually, base apparent power and base voltage are specified
at a point in the circuit, and the other values are calculated from them. The
base voltage varies by the voltage ratio of each transformer in the circuit but
the base apparent power stays the same through the circuit.
Real power systems are convenient to analyze using their per-phase (since
the system is three-phase) per-unit (since there are many transformers)
equivalent circuits. The per-phase base voltage, current, apparent power, and
impedance are
I base 
S1 ,base
VLN ,base
Zbase 
VLN ,base
I base
V



LN ,base
2
S1 ,base
Where VLN,base is the line-to-neutral base voltage in the three-phase circuit
S1,base is the base apparent power of a single phase in the circuit.
The base current and impedance in a per-unit system can also be expressed
in terms of the three-phase apparent power (which is 3 times the apparent
power of a single phase) and line-to-line voltages (which is √3 times the lineto-neutral voltage):
I base 
Zbase 
S3 ,base
3VLL ,base
VLL ,base
3I base
V


LL ,base

2
S3 ,base
Change base impedance
The per-unit impedance may be transformed from one base to another as:
2
Per  unitZ new
 Vold   Snew 
 per  unitZ old 
 

V
S
 new   old 
Example : a power system consists of one synchronous generator and one
synchronous motor connected by two transformers and a transmission line.
Create a per-phase, per-unit equivalent circuit of this power system using a
base apparent power of 100 MVA and a base line voltage of the generator G1
of 13.8 kV. Given that:
G1 ratings: 100 MVA, 13.8 kV, R = 0.1 pu, Xs = 0.9 pu;
T1 ratings: 100 MVA, 13.8/110 kV, R = 0.01 pu, Xs = 0.05 pu;
T2 ratings: 50 MVA, 120/14.4 kV, R = 0.01 pu, Xs = 0.05 pu;
M ratings: 50 MVA, 13.8 kV, R = 0.1 pu, Xs = 1.1 pu;
L1 impedance: R = 15 , X = 75 .
Solution: To create a per-phase, per-unit equivalent circuit, we need first to
calculate the impedances of each component in the power system in per-unit
to the system base. The system base apparent power is Sbase = 100 MVA
everywhere in the power system. The base voltage in the three regions will
vary as the voltage ratios of the transformers that delineate the regions.
These base voltages are:
Vbase,1  13.8kV Region 1
110
 110kV Region 2
13.8
14.4
 Vbase,2
 13.2kV Region 2
120
Vbase,2  Vbase,1
Vbase,3
The corresponding base impedances in each region are:
Z base,1 
Z base,2 
Z base,3 
VLL ,base 2
S3 ,base
13.8kV 

VLL ,base 2
S3 ,base
VLL ,base 2
S3 ,base
2
100MVA
 1.904Region1
110kV 


 121Region1
100MVA
2
13.2kV 

2
100MVA
 1.743Region1
The impedances of G1 and T1 are specified in per-unit on a base of 13.8 kV
and 100 MVA, which is the same as the system base in Region 1. Therefore,
the per-unit resistances and reactances of these components on the system
base are unchanged:
RG1,pu = 0.1 per unit
XG1,pu = 0.9 per unit
RT1,pu = 0.01 per unit
XT1,pu = 0.05 per unit
There is a transmission line in Region 2 of the power system. The impedance
of the line is specified in ohms, and the base impedance in that region is 121
. Therefore, the per-unit resistance and reactance of the transmission line
are:
15
Rline, system 
 0.124 perunit
121
75
X line, system 
 0.620 perunit
121
The impedance of T2 is specified in per-unit on a base of 14.4 kV and 50
MVA in Region 3. Therefore, the per-unit resistances and reactances of this
component on the system base are:
per  unitZ new  per  unitZ given Vgiven Vnew   Snew S given 
2
RT 2, pu  0.0114.4 13.2  100 50   0.238 perunit
2
X T 2, pu  0.05 14.4 13.2  100 50   0.119 perunit
2
The impedance of M2 is specified in per-unit on a base of 13.8 kV and 50
MVA in Region 3. Therefore, the per-unit resistances and reactances of this
component on the system base are:
per  unitZ new  per  unitZ given Vgiven Vnew   Snew S given 
2
RM 2, pu  0.114.8 13.2  100 50   0.219 perunit
2
X M 2, pu  1.114.8 13.2  100 50   2.405 perunit
2
Therefore, the per-phase, per-unit equivalent circuit of this power system is
shown:
node equation
When the per-unit equivalent circuit of a power system is created, it can be used to
find the voltages, currents, and powers present at various points in a power system.
The most common technique used to solve such circuits is nodal analysis.
In nodal analysis, we use Kirchhoff’s current law equations to determine the voltages
at each node (each bus) in the power system, and then use the resulting voltages to
calculate the currents and power flows at various points in the system.
For example a simple three-phase power system bellow
The per-unit equivalent circuit of this power system:
Note that the per-unit series impedances of the transformers and the transmission
lines between each pair of busses have been added up, and the resulting
impedances were expressed as admittances (Y=1/Z) to simplify nodal analysis.
The voltages between each bus and neutral are represented by single subscripts (V1,
V2) in the equivalent circuit, while the voltages between any two busses are indicated
by double subscripts (V12).
The generators and loads are represented by current sources injecting currents into
the specific nodes. Conventionally, current sources always flow into a node meaning
that the power flow of generators will be positive, while the power flow for motors
will be negative.
According to Kirchhoff’s current flow law (KCL), the sum of all currents entering
any node equals to the sum of all currents leaving the node. KCL can be used to
establish and solve a system of simultaneous equations with the unknown node
voltages.
Assuming that the current from the current sources are entering each node, and that
all other currents are leaving the node, applying the KCL to the node (1) yields:
V1  V2  Ya  V1  V3  Yb  V1Yd  I1
Similarly, for the nodes (2) and (3):
V2  V1  Ya  V2  V3  Yc  V2Ye  I 2
V3  V1  Yb  V3  V2  Yc  V3Y f  I3
Rearranging these equations, we arrive at:
Ya  Yb  Yd V1  YaV2  YbV3  I1
YaV1  Ya  Yc  Ye V2  YcV3  I 2
YbV1  YcV2  Yb  Yc  Y f V3  I 3
In matrix form:
Ya  Yb  Yd

Ya


Yb

 V1   I1 
Ya
Yb
   
Ya  Yc  Ye
Yc
 V2    I 2 
Yc
Yb  Yc  Y f  V3   I 3 
Which is an equation of the form:
YbusV  I
where Ybus is the bus admittance matrix of a system, which has the form:
Ybus
Y11 Y12 Y13 
 Y21 Y22 Y23 
Y31 Y32 Y33 
Ybus has a regular form that is easy to calculate:
1) The diagonal elements Yii equal the sum of all admittances connected to node i.
2) Other elements Yij equal to the negative admittances connected to nodes I and j.
3) The diagonal elements of Ybus are called the self-admittance or driving-point
admittances of the nodes; the off-diagonal elements are called the mutual
admittances or transfer admittances of the nodes.
4) Inverting the bus admittance matrix Ybus yields the bus impedance matrix:
Zbus  Ybus
1
Simple technique for constructing Ybus is only applicable for components that are not
mutually coupled. The technique applicable to mutually coupled components can be
found elsewhere.
Once Ybus is calculated, the solution to (10.15.1) is
1
bus
V Y I
V bus I
A number of techniques can be used to solve systems of simultaneous linear
equations, such as substitution, Gaussian elimination, LU factorization, etc.
A system of n linear equations in n unknowns
Ax  b
where A is an n x n matrix and b is and n-element column vector; the solution will be
1
xA b
where A-1 is the n x n matrix inverse of A.
Example 10.3: a power system consists of four busses interconnected by five
transmission lines. It includes one generator attached to bus 1 and one synchronous
motor connected to bus 3.
The per-phase, per-unit equivalent circuit is shown.
We observe that all impedances are considered as pure reactances to simplify the case
since reactance is much larger than resistance in typical transformers, synchronous
machines, and overhead transmission lines.
Find the per-unit voltage at each bus in the power system and the per-unit current
flow in line 1.
The first step in solving
for bus voltages is to convert the voltage sources into the equivalent current sources
by using the Norton’s theorem. Next, we need to convert all of the impedance
values into admittances and form the admittance matrix Ybus then use it to solve for
the bus voltages, and finally use voltages on buses 1 and 2 to find the current in line
1.
First, we need to find the Norton equivalent circuits for the combination of G1 and T1.
The Thevenin impedance of this combination is ZTH = j1.1, and the short-circuit
current is
Voc 1.110
I sc 

 1.0  80
ZTH
j1.1
The combination of M3 and T2 is shown.
The Thevelin impedance of this combination is ZTH = j1.6, and the short-circuit
current is
I sc 
Voc 0.9  22

 0.563  112
ZTH
j1.6
I sc 
Voc 0.9  22

 0.563  112
ZTH
j1.6
The Norton’s equivalent circuit.
The per-phase, per-unit circuit with the current sources included
The resulting
admittance matrix is:
Ybus
j 5.0
0
j 6.667 
  j12.576
 j 5.0


j
12.5
j
5.0
j
2.5



0
j 5.0 10.625
j 5.0 


j
6.667
j
2.5
j
5.0

j
14.167


The current vector for this circuit is:
 1.0  80 


0

I 
0.563  112


0


The solution to the system of equations will be
 0.989  0.60 
 0.981  1.58 
1
V
V  Ybus
I 
0.974  2.62


 0.982  1.48 
The current in line 1 can be calculated from the equation:
I1  V1  V2  Yline1   0.989  0.60  0.981  1.58     j5.0 
 0.092  25.16