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Course OverviewCourse OverviewSynopsis:

The subject deals with topics related to the-

system, power flow analysis, analysis of balanced

and unbalanced faults, power system stability and, ,protection, differential protection and application,distance protection and application. Overall, this

the protection schemes for power systemnetwork.

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Course Outcome

should be able to:

and unbalanced faults techniques. (PLO4-

CTPS-C4)2. Demonstrate power flow analysis using related

software. (PLO3-CS-P4)

. on power system requirement. (PLO11-SD-A3)

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Lecture Plan

WEEKS CONTENTS

1 Per-Unit System

2 Power Flow Analysis

3 Analysis of Balanced Fault

4 Analysis of Unbalanced Fault

6 Application of Power System Stability

7 Load Frequency and Automatic Generation Controls

8 Reactive Power and Voltage Controls

9 Non-Directional Overcurrent and Earth Fault Protection

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ec ure an

WEEK CONTENTS

10 Directional Overcurrent and Earth Fault Relay

11 Differential Protection Scheme

12 Differential Protection Application

13 Distance Protection Scheme

14 Distance Protection Application

Syllabusdetails

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CHAPTER CONTENTS

1 Per-Unit System (3 Hours)

Introduction

Vectors

Operators

Convention Used for Voltage Direction ase uan es an er- n ys em

Transferring Per-Unit Quantities from One Set

of Base Values

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Representation of Electric Power System

One-Line Diagram (OLD)

Definition:A diagram showing the interconnection of

-

power system by standard symbols on a single

phase basis.

Y

B

=

8

- -

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Advantages of OLD

Simplicity

1- represents all 3-s of the balanced system

by their standard symbols

The completion of the circuit through the neutral is

omitted

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Symbols for One-Line Diagram

2-winding power transformerOr

3-winding power transformerOr

LoadOr

(oil/liquid) (OCB)

Air CB (ACB)

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Busbar 3-phase, 3-wire

Transmission line

3-phase wye,

Fuse

3-phase wye,

A

Current

transformer

(CT)

Vor Potential

transformer

PT or VT

Voltmeter

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Impedance (Z) and Reactance (X) Diagram

Impedance (Z = R + jX) diagram is converted fromOLD showing the equivalent circuit of eachcomponen o e sys em. s nee e o ca cu a e e

performance of a system under normal and abnormalconditions i.e. load conditions (Load Flow (LF) studies)or upon e occurrence o a au s or c rcu auanalysis studies).

Reactance (jX) diagram is further simplified from Zdiagram by omitting all static loads, all Rs, themagnetizing I (Im) of each transformer, and thecapacitance (C) of the transmission line. It is applied

only to fault calculations, and not to LF studies.Z and X diagrams sometimes called the Positive-

se uence dia ram.

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Z and X Diagrams

Example: OLD of an EPS

T2T1

Load A

WITH WISDOM WE EXPLORE 13

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Z dia ram corres ondin to the OLD

E1 E2 E3

Gen.3

LoadB

TransformerT2

TransmissionLine

TransformerT1

LoadA

Generators1 and 2

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X diagram corresponding to the OLD

E1 E2 E1

Generators

1 and 2

Transmission

LineTransformer

T2

Gen.

3Transformer

T1

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Per - unit (P.U) Representation

Common quantities used in power system analysis (PSA) are

voltage (V) (in kV), current (I) (in kA), voltamperes (in kVA or

MVA), and impedance (in ). It is very cumbersome to convert

s o eren vo age eve s n a av ng wo or more

levels. P.U. representation is introduced in such a way that the various

p ys ca quan es are expresse as a ec ma rac on or

multiples of base quantities and is defined as:

actual quantityQuantity in per-unit

base value quantity

Example:

For instance, if a base voltage of 275 kV is chosen, actual

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. , , . . ,

1.00, and 1.05 per-unit.

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For 1- systems: The formula relates the various quantities

for 1- system:

(1- )

( )

base kVA (in kVA)Base I (in A)

base V (in kV)LN

LN

1

kVvoltage,base

kVAbaseAcurrent,Base

( )

2

base V (in V)

Base Z (in ohms) base I (in A)

(base V ) (in kV)

LN

LN

LN

Acurrent,base

,impedanceBase

2

LN )kVvoltage,(base

(1- )

(1- ) (1- )

ase n o msbase MVA (in MVA)

Base power (in kW) base kVA (

in kVA)

1MVAbase

Base P, MW1= Base MVA1(3- ) (1- ) ase , = ase

2.)(

base

base

base

upVVAZ

ZactuaZZ

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or - sys ems:The formula relates the various quantities for 3- system:

(3- )base kVABase I (in A)

3 X base V (in kV)LL

LL

3

kVvoltage,baseX3

kVAbaseAcurrent,Base

2

( )

(3- )

(base V ) (in kV)Base Z (in ohms)

base MVA

LL

3

2

LL

MVAbase

)kVvoltage,(baseimpedanceBase

(3- ) (3- )

(3- ) (3- )

Base power (in MW) base MVA

33

33

MVAbaseMWpower,Base

kVAbasekWpower,Base

)( baseVAZactualZ

Z

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.

basebase VZ

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Example:

The base impedance and base voltage for a given

power system are 10 and 400V, respectively.

Solution:

400

rom ms aw,

Base current =10

X40040=

19

1000

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Example:

A 3-, Y-connected system is rated at 100 MVA and 132 kV.

Express 80 MVA of 3- apparent power (S) as a p.u. value

referred to:

(a) the 3- system MVA as base and

(b) the 1- system MVA as base.

(a) For the3- base,

Base MVA = 100 MVA = 1 p.u.

and Base kV = 132 kV = 1 p.u.

so p.u. MVA = 80/100 = 0.8 p.u.

(b) For the1- base,

Base MVA = 100/3 MVA = 33.33 MVA = 1 p.u.

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. . .

so p.u. MVA = (1/3)*(80/33.333) = 0.8 p.u.

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Changing the Base of P.U. Quantities

The Z of individualgeneratorsandtransformersare

generally in terms of%orp.u.quantities based on

.

For PSA, all Zs must be expressed in p.u. on acommon system base. Thus, it is necessary to

convert the p.u. Zs from one base to another

(common base, for example: 100 MVA).

P.U. Z of a circuit

element2

(actual Z in ) * (base MVA)

(base V) in kV

The equation shows that p.u. Z is directly proportional

to the base MVA and inversely proportional to the

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square of the base V.

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Therefore to chan e from old base .u. Z to newbase p.u. Z, the following equation applies:

old newnew old

new old

base kV base MVAP.U. Z P.U. Z base kV base MVA

Example 1:

The reactance X of a generator is given as0.20 p.u. based on the generators nameplate

rating of 13.2 kV, 30 MVA. The base for

new base.

2

23

.x" 0.20 0.306 p.u.

13.8 30

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Analysis of PS problems are greatly simplified by

using single-line Z diagram in which systemparameters are expressed in p.u. The steps to

. .

Step 1: Select a common volt-ampere base for

the entire power system and a voltage base for one.

Step 2:Compute voltage bases for all partsof thePS by correlating the transformation ratios of the

transformer banks.

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nameplate of the equipment) to the common systemvolt-ampere base and the applicable voltage base. In

values, compute base Z for the part of the PS in whichthe equipment is connected and calculate the p.u..

Step 4:Draw a single-line diagramof the PS indicating. .

PS.

.

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Example 4:

Draw the reactance diagram of system shown

below. Assume reactance for the transmission line is

the generator circuit

6.6/66 kV

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A 30 MVA 13.8 kV 3-generator has a sub-transient

reactance (Xd) of 15%. The generator supplies two

.

ends, as shown in OLD below. The motors have rated

inputs of 20 MVA and 10 MVA, both 12.5 kV with x =

20%. The 3-transformer T1 is rated 35 MVA 13.2/116

(/Y) kV with leakage reactance (Xl) of 10%. 3-

transformer T2 is rated at 10 MVA, 116/12.5 (Y/) kV

with Xl of 10%. Series X of the tr. line is 80 . Draw theX diagram with all Xs marked in p.u. Select the

generator rating as base in the generator circuit.

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