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In: Business and Management

Submitted By moontsang
Words 883
Pages 4
Power System Fault and Protection

Page 1

Power System Fault Protection
  

Per Unit Fault Protection

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Per Unit


Problem




Computation for a power system having 2 or more voltage levels become cumbersome Need to convert currents to a different voltage level wherever they flow through a transformer

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Per Unit


Solution




A set of base values is assumed for each voltage class All values express in per unit (i.e. no dimension)

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Per Unit


To completely define a per unit system, minimum 4 base quantities are required
   

Voltage Current Impedance Power

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Per Unit


General


Per Unit= Actual / Base
Vpu=Vactual / Vbase Ipu=Iactual / Ibase Zpu=Zactual / Zbase Spu=Sactual / Sbase Sbase= Vbase Ibase Vbase= Ibase Zbase
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Specific
     

Example

Page 7

Example

Page 8

Example

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Example

Page 10

Example

Page 11

Example

Page 12

Example

Page 13

Example

Page 14

Example

Page 15

Fault
  

Cause of Fault Effect of Fault Fault Level

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Causes of Faults


Transient over voltages


the failure of insulation, resulting in fault current or short-circuit current.

 

Insulation aging External object
 

tree branches bird
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Effect of Fault


Equipment be damaged due to overheating or insulation breakdown
     

Generator Transformer Busbar Cable Circuit Breaker etc.
Page 18

Effect of Fault


Power System becomes unstable




Voltage Reduction for the whole power system. Frequency change


results in cascade tripping of generators.

Page 19

Type of Fault
 

Shunt Fault Series Fault

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Fault Study


 



Determine the maximum fault current for short circuit fault Determine open circuit fault Determine circuit breaker rating Determine scheme of protection

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Fault Level


The fault level is
 

the three-phase fault value expressed in MVA at the system voltage at the fault point in the system

Page 22

Fault Level Limiting


Problem


Large interconnected system




Large number of generators and thus large power rating Large fault current fed into the fault point



Solution


Reduce fault current


Increase reactance of a system
Page 23

Fault Level Limiting


Reactor arrangement


Normal condition


Small voltage drop across Reactor Smaller short circuit current fed into the faulty point Smaller breaking currents for circuit breaker
Page 24



Short circuit condition




Reactance in Generator Circuit

Reactors

Feeders

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Reactance in series with Feeders

Reactors

Feeders

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Busbar reactors in Ring System

Reactors

Feeders

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Busbar reactors in a Tie Bar System

Reactors

Feeders

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Example

Page 29

Example

Page 30

Example

Page 31

Protection


Problem


Fault cause abnormal current flows in the system


many undesirable effects are likely to occur



Solution


remove the fault from the system quickly The equipment that performs the functions of fault clearing from the system



Protection System


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Component for Protection
 

Circuit Breaker Transducer
 

Voltage Transformer Current Transformer



Relay

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Circuit Breaker

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Voltage Transformer (VT)


Voltage transformers are connected across the points at which the voltage is to be measured

Ns Vs  Vp Np

Page 35

Current Transformer (CT)


Primary winding


Consists of a single turn Consist of multiple-turn



Secondary winding


Page 36

Relay


A protective relay is an electrical device


Abnormal condition
 

to initiate isolation of a part of the power system to operate an alarm signal

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Sequence of Protection Operation
 

 

Fault occurs Voltage or current signal transmits to relay Relay operates Circuit Breaker Fault clear

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Type of Protection
  

Over current protection Differential protection Primary and backup protection

Page 39

Over Current Protection


Function
 

monitor the current continuously issue a tripping command to the circuitbreaker when triggered

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Time Over Current Protection


Time over current relay


definite time


ta

ta=constant

In the definite time scheme, the operating ta" time is constant regardless of the current flowing through the relay providing it is higher than the pick-up setting

Iset

(a)

I

Page 41

Time Over Current Protection


Time over current relay


inverse time


The operating time of an inverse time-over current relay is inversely proportional to the level of the fault current ta t a"

Iset

I

Page 42

Inverse Definite Multiple Time Relay (IDMT)


Two possible adjustments :
 

Current setting Time setting

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Discrimination by Time


Objective


To delay between successive relay



Circuit breaker trip time
OCB: 150 ms SF6 CB: 50 ms


Page 44

Discrimination by Time


Relay trip time
 

for EM Relays with OCB: 0.4 – 0.5s for Solid State Relays with vacuum or SF6 switchgear: 0.25s

Page 45

Differential Protection


Compare the current between upstream and downstream of the protected zone
Protection Zone
CT 1

F

F

CT 2

F

IA Trip Coil

IB

Page 46

Differential Protection


Normal condition


Current different = 0



Fault condition
Current different 0  Differential current triggers the trip coil


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Differential Protection


Features








Sensitive to faults only within a particular section of a power system Greatly increase the reliability of the protection of that particular section Insensitive to faults outside that particular section Do not contribute to back-up protection of the remainder of the power system
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Example

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Example

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Solution

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Solution

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Solution

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Primary and Back-up Protection


Main protection



Cover smaller section of the power system Shorter operating time Covering larger section of the power system Longer operating time



Back-up protection
 

Page 54…...

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