Power System Analysis lecture course contents

1. Power System
Analysis EPM320
3rd Year Electrical Power
and Machines Dept.
Ahmed A. Daoud, Ph.D.

2. Course Contents
Lecture: 2h/week Tutorial: 2h/week
1- Introduction(Unified Egyptian Network, Power system analysis, Electrical
power systems construction)
2-Electrical Power Transformers(Single phase & three phase
transformers, Equivalent circuit, open and short circuit test, three winding and Auto
transformer, Parallel operation of transformers)
3- Synchronous Generator(synchronous machine, Three phase
generator, Synchronous reactance and equivalent circuit, Real and reactive power
control, Loading capability diagram)
4-Power Systems Economics(Thermal power station, Characteristics of
power generation units, Economic dispatch, Lagrange relaxation method)
5- Load Flow Calculations(Necessity for Power Flow Studies, Conditions
for Successful Operation of a Power System, The Power Flow Equations, Gauss-Seidel
Iterative Method, Newton-Raphson Method, Sparsity of Network Admittance
Matrices)

3. Introduction
The Egyptian Electrical Unified Network (EEUN)in
Egypt is divided into six geographical regions,
namely, Cairo, Canal, Delta, Alexandria/ west delta,
Middle Egypt, and Upper Egypt. The Egyptian
Electricity transmission system is composed of 500
kV, 400 kV, 220 kV, 132 kV, and 66 kV levels.

4. Introduction

5. Introduction

6. Introduction
Power system analysis
An electric power system is made up of electrical components to generate, transmit and use
electric power. This could be the elaborate network that supplies power to a region’s home
and industry through an electrical grid transmission system from generating plants located
faraway. Or, this could even be a captive power plant/micro-grid which generates and
consumes power within the same premises itself. The majority of these systems rely upon
three-phase AC power - the standard for large-scale power transmission and distribution
across the modern world. Specialized power systems are also found in aircraft, electric rail
systems, ocean liners and automobiles that do not always rely upon three-phase AC power.
The planning, design, and operation of these commercial and industrial power systems
requires in-depth engineering studies to evaluate existing and proposed system
performance, reliability, safety, and economics.

7. Introduction
Electrical power systems construction
Generation Transmission Load

8. Per Unit Quantities
There are several reasons for using a per-unit system:
•Similar apparatus (generators, transformers, lines) will have similar per-
unit impedances and losses expressed on their own rating, regardless of
their absolute size.
•Use of the constant is reduced in three-phase calculations.
•Per-unit quantities are the same on either side of a transformer,
independent of voltage level.
•By normalizing quantities to a common base, both hand and
automatic calculations are simplified.
𝐼𝑏𝑎𝑠𝑒 =
𝑆𝑏𝑎𝑠𝑒
𝑉𝑏𝑎𝑠𝑒
For Single phase
𝑍𝑏𝑎𝑠𝑒 =
𝑉𝑏𝑎𝑠𝑒
𝐼𝑏𝑎𝑠𝑒
=
𝑉𝑏𝑎𝑠𝑒
2
𝐼𝑏𝑎𝑠𝑒𝑉𝑏𝑎𝑠𝑒
=
𝑉𝑏𝑎𝑠𝑒
2
𝑆𝑏𝑎𝑠𝑒
For Three phase
𝑆𝑏𝑎𝑠𝑒 = 3𝑉𝑏𝑎𝑠𝑒𝐼𝑏𝑎𝑠𝑒
𝐼𝑏𝑎𝑠𝑒 =
𝑆𝑏𝑎𝑠𝑒
3𝑉𝑏𝑎𝑠𝑒
𝑍𝑏𝑎𝑠𝑒 =
𝑉𝑏𝑎𝑠𝑒
𝐼𝑏𝑎𝑠𝑒
=
3𝑉𝑏𝑎𝑠𝑒
2
𝑆𝑏𝑎𝑠𝑒

9. Per Unit Quantities

10. Per Unit Quantities
G1
G2
M
30 MVA, 15 kV
XG1= j 0.1 pu
25 MVA, 13.8 kV
XG2= j 0.15 pu
25 MVA, 18/138 kV
XT1= j 0.1 pu
25 MVA, 138/13.8 kV
XT2= j 0.1 pu
10 MVA, 12 kV
XM= j 0.1 pu
20 + j60 W
Take 30 MVA, 15 kV as base on generating bus bar draw the equivalent
circuit of the system using per unit quantities.

11. Per Unit Quantities
The terminl voltage of a Y connected load consisting of three equal
impedances of 2030 is 4.4 kV line to line. The impedance of each of
the three lines connecting the load to a bus at a bus station is
1.475W.
(a)Find the line-to-line voltage at the substation bus.
(b)Find the solution by working in per unit on a base of 4.4 kV, 127 A so
that both voltage and current magnitudes will be 1 .0 per unit.
Current rather than KVA is specified here since the latter quantity
does not enter the problem .

12. Bus Admittance Matrix
Nodal Analysis

13. Bus Admittance Matrix
Node (3) 𝑉3𝑌𝑎 + 𝑉3 − 𝑉2 𝑌𝑏 + 𝑉3 − 𝑉1 𝑌𝑐 = 𝐼3
Rearranging
Node (1) 𝑉1 𝑌𝑐 + 𝑌𝑑 + 𝑌
𝑓 − 𝑉2𝑌𝑑 − 𝑉3𝑌𝑐 − 𝑉4𝑌
𝑓 = 0
Node (3) -𝑉1𝑌𝑐 − 𝑉2𝑌𝑏 + 𝑉3(𝑌𝑎 + 𝑌𝑏 + 𝑌𝑐) = 𝐼3
Similarly for nodes (2) and (4)
We obtain:
𝑌11 𝑌12 𝑌13 𝑌14
𝑌21 𝑌22 𝑌23 𝑌24
𝑌31
𝑌41
𝑌32
𝑌42
𝑌33 𝑌34
𝑌43 𝑌44
𝑉1
𝑉2
𝑉3
𝑉4
=
𝐼1
𝐼2
𝐼3
𝐼4

14. Single Line Diagrams

15. Exercises