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WEEK 01: INTRODUCTION TO POWER SYSTEMS


Sections: Intro | Per Unit Quantities

Introduction

Definition. A power system is a network composed of a generating unit, usually made up of generators or a power plant, supplying various loads through a transmission and distribution systems.

Transmission systems include those components that transforms high voltages to low voltages, or vice versa, and the sundries for maintaining these components.

Distribution systems includes substations, whether indoor or outdoor, for parcelling the power generated by the power plant and transmitted by the transmission lines.

Components. The components of a power system are as follows:
1. Generator
2. Motor
3. Transmission Lines
4. Transformers
5. Transducers
6. Protective Devices

Scope. The scope of this module are:
1. The study and analysis of power system faults.
2. The application of symmetrical components and per unit computations to analyze the power system faults.
3. The application of theories and solutions learned from previous studies of basic circuitry, transient analysis and electrical machines.

Objective. At the end of these modules, the student should be able to:
1. Identify, define and distinguish the different components of a power system.
2. Draw the equivalent one-line, reactance, impedance, and sequence diagrams, as well as the symmetrical component phasors of the power system.
3. Analyze and perform mathematical computations on the different faults on unloaded generators, unbalanced power system, and faults through impedance.


Sections: Intro | Per Unit Quantities

Review of Per Unit Computations

Definition. Per Unit is a term applied to computations on a power system using the actual values divided by a pre-selected base.
The basic formulas are:

Base V = KV

Changing the Base. In changing the base from a given base to a new base, two basic formulas are involved:

Example. The reactance of a generator , X" is given at 0.25 per unit based on the generator's nameplate rating of 18kV, 500 MVA. The base for calculation is 20kV, 100 MVA. Find X" on the new base.

Solution:
Using the formula given above:
X" = (0.25)(18/20)²(100/500) = 0.0405 p.u.

NextPer Unit on Transformers


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