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Chapter One : Introduction, Single Line Diagram, Per Unit System and Network Matrices.
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INTRODUCTION
Modern Power Systems
Power Producer
generation station
prime mover and generator
step-up transformer
Transmission System
HV transmission lines
switching stations
circuit breakers
transformers
Distribution Utility
distribution substations
step-down transformers
MV distribution feeders
distribution transformers
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System Control
Network Protection
Energy Management
Systems
Energy Control Centre
computer control
SCADA - Supervisory Control And Data Acquisition
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Practical power systems
must be safe
reliable
economical
System Analysis
for system planning
for system operations
requires component modelling
types of analysis
transmission line performance
power flow analysis
economic generation scheduling
fault and stability studies
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System Modelling
Systems are represented on a per-phase basis
A single-phase representation is used for a balanced system
The system is modelled as one phase of a Y-connected network
Symmetrical components are used for unbalanced systems
Unbalance systems may be caused by: generation, network components, loads, or unusual operating conditions such as faults
The per-unit system of measurements is used
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Single Line Diagram
A single line diagram or one-line diagram is a representation of the essentials of a system in a most simplified form. The purpose of one-line diagram is to supply in concise form the significant information about the system. Thus, short line feature can be applied to a large part of the grid network.
Power Component
Symbol =
Generator
Circuit breaker =
Transformer
Transmission line =
Motor
Feeder + load M =
Busbar (substation)
Power components and symbols
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Per-Unit Quantities
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Example
The one-line diagram of a three-phase power system is shown
Use a common base of 100 MVA and 22 kV at the generator draw an impedance diagram with all impedances marked in per-unit the manufacturer’s data for each apparatus is given as follows
G: 90 MVA 22 kV 18%
T1: 50 MVA 22/220 kV 10%
L1: 48.4 ohms
T2: 40 MVA 220/11 kV 6%
T3: 40 MVA 22/110 kV 6.4%
L2: ohms
T4: 40 MVA 110/11 kV 8%
M: 65.5 MVA kV 18.4%
Ld: 57 MVA kV 0.6 pf lag
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Since generator and Transformer voltage bases are the same as their rated values, their per-unit reactances on a 100MVA base are:
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Motor reactance on a 100 MVA, 11 kV base is
Impedance bases for lines 1 and 2 are:
Lines 1 and 2 per-unit reactances are:
The load apparent power at power factor lagging is given by
Hence, the load impedance in ohms is
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Network Matrices
Formation of Bus Admittance Matrix
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Consider node (bus) 1 that is connected to the nodes 2 and 3. Then applying KCL at this node we get
In a similar way application of KCL at nodes 2, 3 and 4 results in the following equations
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Node Elimination by Kron Reduction
Consider an equation of the form Consider an equation of the form
where A is an ( n X n ) real or complex valued matrix, x and b are vectors in either Rn or Cn . assume that the b vector has a zero element in the n th row is given as
We can then eliminate the kth row and kth column to obtain a reduced ( n - 1) number of equations of the form
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The elimination is performed using the following elementary operations
𝑎 𝑘𝑗 𝑛𝑒𝑤 = 𝑎 𝑘𝑗 𝑜𝑙𝑑 − 𝑎 𝑘𝑛 ∗ 𝑎 𝑛𝑗 𝑎 𝑛𝑛
In electrical power systems in which those nodes with zero current injections
are eliminated are said to be KRON reduced. If Ip = 0 in the nodal equations of
the N-bus system, we may directly calculate the elements of the new reduced
bus admittance matrix by choosing Ypp as the pivot and by eliminating bus p
using the formula
where j and k take on all the integer values from 1 to N except p since
row p and column p are to be eliminated.
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Example: The nodal admittance equations of the system in p.u values is
using Y22 as the initial pivot, eliminate node 2 and the corresponding
voltage v2 from 4 * 4 system.
Solution:
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Similar calculations yields the other elements of the KRON-reduced matrix
(1) (3) (4)
(1)
(3)
(4)
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