Admittance model and Power flow study.pptx

4/22/2024 EPE 4164 Modelling and analysis of power system 1
Electric networks: Admittance model, Power
flow, Symmetrical faults, Protections

• The purpose of a power system is to deliver the power to the customers require in real
time on demand within acceptable voltage and frequency limits in, a reliable and
economical manner.
• A power flow study (load-flow study) is a steady-state analysis whose target is to
determine the voltages, currents, and real and reactive power flows in a system under
given load conditions. The evaluation of a power system is called power system
analysis
• Power system analysis is an essential part of electrical power system design.
• Calculations and simulations are performed to verify that the electrical system,
including the system components, are correctly specified to perform as intended,
withstand expected stress, and be protected against failures.
• In addition to that, power system analysis helps to understand how the power
system will operate in different configurations and when there are changes to
the system, like capacitor switching (transient), a large motor starting (dynamic), or the
incident energy as a result of an arc flash (static).
• In monitoring power system analysis, we are mainly dealing with power or load flow
analysis, fault analysis, and stability analysis.
Introduction

The functions of power system analysis are:
• To monitor the voltage at various buses, real and reactive power flow
between buses.
• Designing a power system.
• To plan future expansion of of power system.
• Planning a power system for various hypothetical situations. For example,
if a transmission line is being taken offline for maintenance, can the
remaining lines in the system handle the required loads without exceeding
their rated values
• Providing guide lines for optimum operation of power system
• To analyze the behavior of the system under different fault conditions
• To study the ability of the system for larger disturbances (sudden application
of large load).
• To study the ability of the system for small disturbances (routine or small
load changes).

Power Flow Concept
• Consider the three-bus power system.
Generators (G1 and G2 ) are connected to the first
two buses and an electric load is connected to
the third bus.
• The real and reactive power demands are
known for the load bus (3). The generator
voltages are also specified at bus 1 and bus 2.
• The three transmission lines interconnecting the
buses contain both resistance and reactance,
thus currents flow through these lines results in
electrical-losses.

• The two generators (G1 and G2 ) must jointly
supply the total load requirements and the
power losses in the transmission lines.
• The generators are constrained to operate
within their power generation capabilities.
• The generators are also constrained to deliver
the required power at the desired voltage at the
customer loads.
• In addition, there should be no over-loading of
the power system equipments including
transmission lines and transformers.
• Furthermore, there should be no bus voltage
either above or below-specified values of the
bus voltage operating limits.

 Four quantities are associated with each bus
1. Magnitude of voltage(V)
2. Phase angle of voltage(δ)
3. Active power(P)
4. Reactive power(Q)
The load flow problem can be solved with the help of the load
flow equation(Static load flow equation).

 A load flow study is done on a power system to ensure that:
• Generation supplies demand(Load) plus losses.
• Bus voltage magnitude remains close to rated value i.e no bus voltage either
above or below-specified values operating limits.
• Generation operates within specified real and reactive power limits.
• Transmission line and transformer are not overloaded
• It is required for Planning, Operation, Economic Scheduling & Exchange of
power between utilities, expansion of system & also in design stage and
control of an existing system as well as planning its future expansion
• In solving power flow problem, the system is assumed to be operating
under balanced conditions and a single-phase model is used
• The power flow is also required for many other applications such as short-
circuit calculations, transient stability and contingency analysis

• For the network shown, there are some buses
connected with the generators and other buses are
connected to the loads.
• The Real and Reactive power is known at each
Load bus. The Generator Voltages are Also
Specified at the generator buses.
• The Transmission Lines interconnecting the buses
have resistance and inductance. Therefore, the
Electric Current flowing through the lines results in
Electrical Losses.
• The Generators in the System must supply the
Total Electrical Loads plus the Electrical Losses.

 There are some constrains should be considered while running
the system:
1. The Generators Must Operate within their Generation Capabilities.
2. The Generators Must Deliver the required power at the Desired Voltage at
the Loads.
3. There should be no bus voltage either above or below the specified
Voltage operating limits.
4. There Should be no Over-Loading of equipment, including Transmission
Lines and Transformers
 In Case of An Equipment Over-Loaded Or Voltage-Limit Violation.
• The Generation Schedule have to be adjusted and Power Flow in the
transmission lines have to be Re-routed or Capacitor Banks have to be
switched in order to bring the system into its Normal Operating Conditions.
• To Satisfy all the previous requirement for a Reliable Power System Operation,
Power Flow Study is a MUST. The Power flow study is an essential part in
power system Operation, Planning and Design.

Power Flow Analysis Study
• The node voltage method is commonly used for the power system analysis.
The formulation of the network equations results in complex linear equations
in terms of node currents.
• In power systems, powers are known rather than currents. Thus, resulting
equations in terms of power become non-linear and must be solved by
iterative techniques.
• These non-linear equations are known as power flow equations or load flow
equations.
• The power flow programs compute the voltage magnitude and phase
angle at each bus bar in the system under steady-state operation conditions.
• These programs use the bus-voltage data to compute the power flow in the
network and the power losses for all equipment and transmission lines.

 A Load Flow Study Specifically Investigates the:
• Bus bar voltages
• Effect of rearranging circuits and incorporating new circuits on system
loading.
• Effect of injecting in-phase and quadrature boost voltages on system
loading.
• Optimum system running conditions and load distribution.
• Optimum system losses.
• Optimum rating and tap range of transformers.

 Bus Classification
• A bus is a node at which many Transmission lines, Loads,
Generators are connected.
• It is not necessary that all of them be connected to every bus.
• Bus is indicated by vertical line at which no. of components are
connected.
• In load flow study two out of four quantities specified and
other two quantities are to be determined by load flow equation
depending upon that bus are classified.
• Depending on the
quantities that have been
specified, the buses are
classified into 3
categories.

 Load bus or PQ Bus
• A bus at which the Active power and reactive power are specified or
fixed.
• The iteration solves for magnitude(V) and phase angle(δ
• It is pure load bus (no generator at the bus)
• These most common bus comprising almost 80% of all of the busses
in power system.
• It is required to specify only Pd and Qd at such bus as at a load bus
voltage can be allowed to vary within the permissible values

 Generator bus or P-V bus or voltage-controlled buses
• A bus at which the magnitude(V) of the voltage and active power(P) is
defined or fixed.
• The iteration solves for reactive power(Q) and Phase angle(δ) injection
and are determined through load flow equation.
• It is also known as P-V bus.
• This bus is always connected to generator.
• This type of bus is comprises about 10% of all the buses in power system.

 Slack Bus or reference bus or Swing bus (V- 𝛿)
• Voltage magnitude(V) and voltage phase angle(δ) are specified and
iteration solves for real(P) and reactive(Q) power injections.
• Normally there is only one bus of this type is given in power system.
• One generator bus is selected as the reference bus.
• In slack bus voltage angle and magnitude is normally considered 1+j0 p.u.
• In Slack (swing) Bus, generation = load demand + total losses. It is
needed for balancing the power in the network.
• This bus makes up the difference between scheduled loads and generated
power that caused by the losses in the network

 Bus Classification table in Electrical Network

 NETWORK ADMITTANCE AND IMPEDANCE MATRICES
A power system network can be converted into an equivalent
impedance diagram. This diagram forms the basis of power flow (or
load flow) studies and short circuit analysis. The formation of bus
admittance matrix (also known as Ybus matrix) and bus impedance
matrix (also known as Zbus matrix).
These two matrices are related by

 FORMATION OF BUS ADMITTANCE MATRIX
• First step in solving the power flow is to formulate the bus admittance
matrix, often call the Ybus.
• The Ybus gives the relationships between all the bus current injections, I,
and all the bus voltages, V,
I = Ybus V
• The Ybus is developed by applying KCL at each bus in the system to
relate the bus current injections, the bus voltages, and the branch
impedances and admittances

(a) Voltage source with a source impedance
and
(b) its Norton equivalent.

For the time being we shall assume the short line approximation for the
formulation of the bus admittance matrix.
Consider the 4-bus power system shown in
this Fig. This contains two generators G1
and G2 that are connected through
transformers T1 and T2 to buses 1 and 2.
Let us denote the synchronous reactances
of G1 and G2 by XG1 and XG2 respectively
and the leakage reactances of T1 and T2 by
XT1 and XT2 respectively. Let Zij, i = 1, …, 4
and j = 1, …, 4 denote the line impedance
between buses i and j.

Then the system impedance diagram
is as shown in the following Fig. where
Z11 = j(XG1 + XT1) and Z22 = j(XG2 +
XT2). In this figure the nodes with the
node voltages of V1 to V4 indicate the
buses 1 to 4 respectively. Bus 0
indicates the reference node that is
usually the neutral of the Y-connected
system.

The impedance diagram is converted into
an equivalent admittance diagram. In this
diagram Yij = 1/Zij, i = 1, …, 4 and j = 1, …,
The voltage sources EG1 and EG2 are
converted into the equivalent current
sources I1 and I2 respectively using the
Norton’s theorem.

We would like to determine the voltage-current relationships of the network as
shown by Norton equivalent circuit . It is to be noted that this relation can be
written in terms of the node (bus) voltages V1 to V4 and injected currents I1 and I2
as follows:
or
or

COLLEGE OF SCIENCE
AND TECHNOLOGY
• Current injections:
Ii flowing into bus i from generator or load.
Positive if generator; negative if load.
24
I1, I4 will be positive.
I3 will be negative.
I2 will be positive if
gen exceeds load,
otherwise negative.
4/22/2024 EPE 4164 Modelling and analysis of power system

COLLEGE OF SCIENCE
AND TECHNOLOGY
Voltages: Vi is voltage at bus i.
25
 Power system representation Kirchoff’s current law: sum of the
currents at any node must be zero.
and
Ohm’s law:
current flowing through
a conductor is directly proportional
to the potential difference applied
across its ends
14
13
12
1 I
I
I
I 


)
( j
i
ij
ij V
V
y
I 


COLLEGE OF SCIENCE
AND TECHNOLOGY
26
14
13
12
1 I
I
I
I 


)
(
)
(
)
( 4
1
14
3
1
13
2
1
12
1 V
V
y
V
V
y
V
V
y
I 





Power system representation cont …..
Current injections at bus 1

COLLEGE OF SCIENCE
AND TECHNOLOGY
Now collect like terms in the voltages:
27
)
(
)
(
)
( 4
1
14
3
1
13
2
1
12
1 V
V
y
V
V
y
V
V
y
I 





)
(
)
(
)
(
)
( 14
4
13
3
12
2
14
13
12
1
1 y
V
y
V
y
V
y
y
y
V
I 









COLLEGE OF SCIENCE
AND TECHNOLOGY
Repeat for the other four buses:
28
)
(
)
(
)
(
)
( 14
4
13
3
12
2
14
13
12
1
1 y
V
y
V
y
V
y
y
y
V
I 








)
(
)
(
)
(
)
( 24
4
23
3
24
23
21
2
21
1
2 y
V
y
V
y
y
y
V
y
V
I 








)
(
)
(
)
(
)
( 34
4
34
32
31
3
32
2
31
1
3 y
V
y
y
y
V
y
V
y
V
I 








)
(
)
(
)
(
)
( 43
4
43
42
41
3
42
2
41
1
4 y
V
y
y
y
V
y
V
y
V
I 









COLLEGE OF SCIENCE
AND TECHNOLOGY
Repeat for the other four buses:
29
)
(
)
(
)
(
)
( 14
4
13
3
12
2
14
13
12
1
1 y
V
y
V
y
V
y
y
y
V
I 








)
(
)
(
)
(
)
( 24
4
23
3
24
23
21
2
21
1
2 y
V
y
V
y
y
y
V
y
V
I 








)
(
)
(
)
(
)
( 34
4
34
32
31
3
32
2
31
1
3 y
V
y
y
y
V
y
V
y
V
I 








)
(
)
(
)
(
)
( 43
42
41
4
43
3
42
2
41
1
4 y
y
y
V
y
V
y
V
y
V
I 








Notes:
1. yij=yji
2. If branch ij does
not exist, then yij=0.

COLLEGE OF SCIENCE
AND TECHNOLOGY
Write in matrix
form:
30
Define the Y-bus:

























































4
3
2
1
43
42
41
43
42
41
34
34
32
31
32
31
24
23
24
23
21
21
14
13
12
14
13
12
4
3
2
1
V
V
V
V
y
y
y
y
y
y
y
y
y
y
y
y
y
y
y
y
y
y
y
y
y
y
y
y
I
I
I
I

































43
42
41
43
42
41
34
34
32
31
32
31
24
23
24
23
21
21
14
13
12
14
13
12
y
y
y
y
y
y
y
y
y
y
y
y
y
y
y
y
y
y
y
y
y
y
y
y
Y
Define elements of
the Y-bus:













4
43
42
41
34
3
32
31
24
23
2
21
14
13
12
1
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y





































4
3
2
1
4
43
42
41
34
3
32
31
24
23
2
21
14
13
12
1
4
3
2
1
V
V
V
V
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
I
I
I
I

In general the format of the
Ybus matrix for an n-bus
power system is as follows:
where
It is to be noted that Ybus is a symmetric
matrix in which the sum of all the
elements of the kth column is Yk.

Consider the impedance diagram of this
Fig. in which the system parameters are
given in per unit by
The system admittance can then be written in
per unit as
The Ybus is then given
Consequently the bus
impedance matrix is given by
It can be seen that like the Ybus matrix the
Zbus matrix is also symmetric.

We then get the node voltages as:
Solving the equation we get the
node voltages as

Ybus General Form
• The diagonal terms, Yi, are the self admittance terms,
equal to the sum of the admittances of all devices incident
to bus i.
• The off-diagonal terms, Yij, are equal to the negative of the
sum of the admittances joining the two buses.
• With large systems Ybus is a sparse matrix (that is, most
entries are zero)
• Shunt terms, such as with the π-line model, only affect the
diagonal terms.

Two Bus System Example Using the Ybus

Solving for Bus Currents

Solving for Bus Voltages

Electric loads of a power system
• Consumers: industry (22-138 kV), commercial (2.2 – 22kV),
residence (220V – 400V)
• Load Profile: variables (hour-to-hour, daily, weekly, seasonally, etc)
• Load types: inductive loads, resistive loads, and capacitive loads.

END of Lect 8

Admittance model and Power flow study.pptx

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