New Chapter 2 Phasors and System Arrangements

7/25/2019 New Chapter 2 Phasors and System Arrangements

1/27

CHAPTER 2

PHASORS AND

SYSTEM ARRANGEMENTS

Chapter 2 Page 1 Phasors and System Arrangements


2/27

PHASORS AND SYSTEM ARRANGEMENTS

Relationship between Current And Voltage In Various SystemElements

Resistor Inductor

Capacitor

Complex Impedance

Three Phase Power Loads

Introduction

Problem 1

Relationship Between Currents and Voltages

Characterist ics of Faults

Using Phasor Relationships to Design Directional Relays

Problem 2: V,I Phasor Relationships

Types of Distribut ion Systems

Radial System

Loop System (Ring Main Unit)

Primary Slective System

Secondary Selective System

Substation Bus Arrangements

Single Bus Arrangement

Doube Bus Double Breaker

Main and Transfer Bus

Double Bus Single Breaker

Ring Bus Arrangement

Breaker and a Half Scheme

Reliability Comparisons

IEEE Device Numbers



3/27

RELATIONSHIP BETWEEN CURRENT AND VOLTAGE INVARIOUS SYSTEM COMPONENTS

Resistor

The current and voltage are in phasein a purely resistive circuit, as shown in Figure2-1.

Figure 2-1. Resistor Circuit

Inductor

The current lags the voltage by /2 radians or 90o in a purely inductive circuit, asshown in Figure 2-2.

Figure 2-2. Inductor Circui t



4/27

Capacitor

The current leadsthe voltage by /2 radians or 90oin a purely capacitive circuit, asshown in Figure 2-3.

Figure 2-3. Capacitor Circui t

Complex Impedance

In general, impedance is a complex number of the form Z = R+ jX, where R(resistance) is the real part and X (reactance) is the reactive or imaginary part.

The inductive reactance (XL) in a coil or wire that is measured in ohms () is equal to2 times the frequency (f) times the inductance (L) of the coil that is measured inHenries (H) or XL = 2fL. The capacitive reactance (XC) of a capacitor that ismeasured in ohms () is equal to the reciprocal of 2times the frequency (f) timesthe capacitance that is measured in farads (F) or XC= 1/(2fC).

Because impedance is a complex number it may be represented on the complexnumber plane; however, because resistance is never negative, only the first andfourth quadrants are involved in the analysis. The resistance (R) is located on thepositive real axis, inductive reactance (XL) is located on the positive reactance(imaginary) axis, and capacitive reactance (XC) is located on the negative reactance(imaginary) axis, as shown in Figure 2-4.



5/27

Complex Impedance

Figure 2-4. Complex Impedance

An impedance triangle is often used as a graphical representation of impedance.

The impedance triangle consists of vectors that represent resistance (R) andreactance (jX), as shown in Figure 2-5. The vector Zis the sum of the two vectorsR and jX.

.

Figure 2-5. Impedance Triangle



6/27

THREE-PHASE POWER LOADS

Introduction

A three-phase system has three sources of power with a certain time intervalbetween each source. It is very easy to generate three-phase voltages byconnecting three windings 120o apart on a generator. Figure 2-6 shows a three-phase system in an ABC phase rotation sequence.

Three wires of a three-phase system can provide 173% ( 3) more power than twowires of a single-phase system. When both single-phase and three-phase loads aresupplied from the same power supply, a three-phase, four-wire system (3, 4-wire),which is called a wye (Y) connection, is used to supply power. If there are onlythree-phase loads, a three-phase, three wire system (3, 2-wire), which is called adelta () connection, is used to supply power.

Figure 2-6. Three-Phase Power



7/27

PROBLEM 1:

Given a five breaker bus with power flows as indicated in Figure 2-7, and the systemelements that are indicated below, identify each breaker (insert the letter) with itsrespective system component.

System elements:a. p.f. correction capacitor d. induction motorb. pure resistive heater e. synchronous motorc. generator

Figure 2-7. Power Flow Phasors (Problem 1)



8/27

SOLUTION TO PROBLEM 1:

Figure 2-8. Solution to Problem 1



9/27

Summary

Voltage, Current, and Impedance:

E = IZ I = E/Z Z = E/I = R + j(XL- XC) |Z| = [R2+ (XL- XC)

2]1/2 cos = R/Z = p.f. sin = X/Z tan = X/R

Power (Single-Phase):

S = EI = (P2+ Q2)1/2 P = EI cos = S cos Q = EI sin = S sin

cos = P/S = p.f. sin = Q/S tan = Q/P

Power (Three-Phase):

S = 3 EI =(P2 + Q2)1/2

P = 3 EI cos = S cos

Q = 3 EI sin = S sin

cos = P/S = p.f.

sin = Q/S tan = Q/P

Motors:

kWin = (hpoutx .746 kW/hpout)/() kVAin = kWin/p.f.

I = (hpoutx .746 kW/hpout)/( 3 x kV x x p.f.)

= kVA/( 3 x kV)

Note: If x p.f. .746 then kVAin= hpout

Transformers:

kVA = 3 x kV x I

Ipri = kVA/( 3 x kVpri)

Isec = kVA/( 3 x kVsec)



10/27


RELATIONSHIP BETWEEN CURRENTS AND VOLTAGES

Figure 2-9.

Relationship between Current and Voltage for a System wi th Unity PowerFactor

At unity power factor

Ialeads Vbcby 90 degrees

Ibleads Vcaby 90 degrees

Icleads Vabby 90 degrees

Vbc

Vab

Ic Ib

Ia

Vbc

Van

Vbn

Vcn


11/27

CHARACTERISTICS OF FAULTS

SYSTEM VOLTAGE FAULT ANGLES

7.2 23 kV 20 to 45 lag

23 69 kV 45 to 75 lag

69 230 kV 60 to 80 lag

230 kV and above 75 to 85 lag



12/27

USING PHASOR RELATIONSHIPS TO DESIGNDIRECTIONAL RELAYS

67

67

67

Ia and Van are in Phase

Ia leads Vbcby 90 degrees

Ia

Vbc

Vbc

Ia

Ia lags Ia(before fault)by approx. 60 degrees


Ia lags Ia (before fault)by 180 degrees

Ia lags Vbcby 90 degrees

Vbc

Ia

Back-Feed

Normal Load

Fault on

Transformer

Primary

67

67

67

Ia and Van are in Phase


Ia

Vbc

Vbc

Ia

Ia lags Ia(before fault)by approx. 60 degrees


Ia lags Ia (before fault)by 180 degrees

Ia lags Vbcby 90 degrees

Vbc

Ia

Back-Feed

Normal Load

Fault on

Transformer

Primary

Figure 2-10 Ninety Degrees Connection for Directional Relays



13/27

PROBLEM 2: V, I PHASOR RELATIONSHIPS

Given the circuit diagram that is shown in Figure 2-11a with current flows and V, Iphasor relationships as shown in Figure 2-11b, match the following elements in thecircuit diagram (insert the letter).

a. pure capacitor c. pure resistorb. pure inductor d. generator

A

B

I1

I1

I2

I2

I3

I4

I3

I4

(a) Circuit Diagram

(b) Phasor Relationships

VAB

VAB

Figure 2-11. V, I Phasor Relationships (Problem 3)



14/27

TYPES OF DISTRIBUTION SYSTEMS

RADIAL SYSTEM

(a) (b)

Figure 2-12. Examples of Radial Systems

Advantages

Low initial investment

Reduced Fault level compared to other systems

Simple to operate

Simple Relaying (no need for directional relays)

Disadvantages

No redundancy

Taking out any component for maintenance causes power interruption topart or all of the system

Comments

Used to feed non-critical loads



15/27

LOOP SYSTEM (RING MAIN UNIT)

(a)

(b)

Figure 2-13. Examples of Loop Systems

Advantages

Provides some redundancy

Faults on the high side loop do not cause outages, if the loop is closed, andproper relaying is used

Disadvantages

High initial investment

High Fault level, compared to the radial systems

More complex to operate

Need directional relaying

Comments

Used to feed critical loads



16/27

PRIMARY SELECTIVE SYSTEM

Figure 2-14. Example of a Primary Selective Radial System

Advantages

Relatively low initial investment

Reduced Fault level compared to loop system

Simple to operate

Simple Relaying (no need for directional relays)

Disadvantages

Only provides redundancy for faults on the primary side of the selectorswitch

Comments

Provides some degree of redundancy



17/27

SECONDARY SELECTIVE SYSTEM

(a) (b)

Normally Open Tie Breaker

or

Normally Closed Tie Breaker

Normally Open Tie Breaker

or

Normally Closed Tie Breaker

Figure 2-15. Example of Secondary Selective System

Advantages

Relatively Low initial investment

Same fault level as radial system (if tie breaker is open)

Simple to operate

No need for directional relays (if tie breaker is open)

Disadvantages

High fault level (if tie breaker is closed)

Need directional relaying (if tie breaker is closed)

Comments

Provides good degree of redundancy



18/27

SUBSTATION BUS ARRANGEMENTS

SINGLE BUS ARRANGEMENT

Disc. Switch

Circuit Breaker

Lines

Line

Bus

Disc. Switch

Circuit Breaker

Lines

Line

Bus

Figure 2-16. Example of a Single Bus Arrangement

The single-bus scheme is not normally used for major substations. Dependence onone main bus can cause serious outage in the event of breakers or bus failure. Thestation must be de-energized in order to carry out bus maintenance or add busextensions. Although the protective relaying is relatively simple, the single busscheme is considered inflexible and subject to complete outage.



19/27

DOUBLE BUS - DOUBLE BREAKER

LineLine

Transformer

LineLine

Transformer

Figure 2-17. Examples of Double Bus-Double Breaker Scheme

The double-bus-double-breaker scheme requires two circuit breakers for each feedercircuit. Normally each circuit is connected to both busses. In some cases, half of thecircuits could operate on each bus. For these cases, bus or breaker failure wouldcause the loss of half the circuits. The location of the main bus must be such as toprevent faults spreading to both buses. The use of two breakers per circuit makesthis scheme expensive. However, it represents a high order of reliability when allcircuits are connected to operate on both busses.



20/27

MAIN AND TRANSFER BUS

LineLine

Transformer

Transfer Bus

Main Bus

BusTie

Breaker

LineLine

Transformer

Transfer Bus

Main Bus

BusTie

Breaker

Figure 2-18. Examples of a Main and Transfer Bus Scheme

The main-and-transfer-bus scheme adds a transfer bus to the single-bus scheme.An extra bus-tie circuit breaker is provided to tie the main and transfer busses

together.

When a circuit breaker is removed from service for maintenance, the bus tie circuitbreaker is used to keep that circuit energized. Unless the protective relays are alsotransferred, the bus-ti relaying must be capable of protecting transmission lines orgenerators. This is considered rather unsatisfactory since relaying selectivity is poor.

A satisfactory alternative consists of connecting the line and bus relaying to currenttransformers located on the lines rather than on the breakers. For this arrangementline and bus relaying need not be transferred when a circuit breaker is taken out ofservice for maintenance, with the bus-tie breaker used to keep the circuit energized.

If the main bus is ever taken out of service for maintenance, no circuit breakersremain to protect any of the feeder circuits. Failure of any breaker or failure of themain bus can cause complete loss of service of the station.

Disconnect switch operation with the main-and-transfer-bus scheme can lead to

operator error, injury, and possible shutdown.



21/27

DOUBLE BUS SINGLE BREAKER

Bus tie breaker

Line Line

Bus No. 1

Bus No. 2

Bus tie breaker

Line Line

Bus No. 1

Bus No. 2

Figure 2-19. Example of a Double Bus Arrangement Scheme

This scheme used two main busses, and each circuit includes two bus selectordisconnect switches. A bus tie circuit connects to the two main busses and, whenclosed, allows transfer of a feeder from one bus to the other bus without de-energizing the feeder circuit by operating the bus selector disconnect switches. Thecircuits may all operate from the No. 1 main bus, or half of the circuits may beoperated off either bus. In the first case, the station will be out of service for bus orbreaker failure. In the second case, half of the circuits would be lost for bus orbreaker failure.

In some cases circuits operate from both the No. 1 and No. 2 bus and the bus-tiebreaker is normally operated closed. For this type of operation a very selective busprotective relaying scheme is required in order to prevent complete loss of thestation for a fault on either bus.

Disconnect switch operation becomes quite involved with the possibility of operatorerror, injury and possible shutdown. The double-bus-single-breaker scheme is poorin reliability and is not used for important substations.



22/27

RING BUS ARRANGEMENT

Line

Line

Transformer

Transformer

Line

Line

Transformer

Transformer

Figure 2-20. Example of a Ring Bus Arrangement

In the ring-bus scheme, the breakers are arranged in a ring with circuits connectedbetween breakers. There is the same number of circuits as there are breakers.

During normal operation, all breakers are closed. For a circuit fault, two breakers aretripped, and in the event one of the breakers fails to operate to clear the fault, anadditional circuit will be tripped by operation of breaker-failure backup relays. Duringbreaker maintenance, the ring is broken, but all lines remain in service.

The circuits connecter to the ring are arranged so that sources are alternated withloads. For an extended circuit outage, the line disconnects switch may be openedand the ring can be closed

No changes to protective relays are required for any of the various operatingconditions or during maintenance.

The ring bus scheme is economical in cost, has good reliability, is safe for operation,is flexible, and is normally considered suitable for important substations up to a limitof five circuits. Protective relaying and automatic reclosing are more complex thanfor previous schemes described. It is common practice to build major substationsinitially as a ringbus; for more than five outgoing circuits, the ring bus is usuallydeveloped to the breaker-and-a-half scheme.



23/27

BREAKER AND A HALF SCHEME

TieBreaker

Line

Line

Bus No. 1

Bus No. 2

Line

Line

TieBreaker

Line

Line

Bus No. 1

Bus No. 2

Line

Line

Figure 2-21. Examples of Breaker and a Half Scheme

The breaker-and-a-half scheme, sometimes called the three-switch scheme, hasthree breakers in series between the main busses. Two circuits are connectedbetween the three breakers, hence the term breaker and a half. This pattern isrepeated along the main busses so that one and a half breakers are used for eachcircuit.

Under normal operating conditions, all breakers are closed and both busses areenergized. AA circuit is tripped by opening the two associated circuit breakers. Tiebreaker failure will trip one additional circuit, but no additional circuit is lost if a linetrip involves failure of a bus tie breaker.

Either bus may be taken out of service at any time with no loss of service. Withsources connected opposite loads, it is possible to operate with both busses out ofservice. Breaker maintenance can be done with no loss of service, no relay changes,and simple operation of the breaker disconnects.

The breaker and a half arrangement is more expensive than the other schemes,except the double breaker-double-bus scheme. However, the breaker-and-a-halfscheme is superior in flexibility, reliability, and safety. Protective relaying andautomatic reclosing schemes are more complex than for other schemes.



24/27

RELIABILITY COMPARISONS

The various schemes have been compared to emphasize their advantages anddisadvantages. The basis of comparison to be employed is the economic justificationof a particular degree of reliability. The determination of the degree of reliability

involves an appraisal of anticipated operating conditions and the continuity of servicerequired by the load to be served.



25/27

SUMMARY OF COMPARISON OF SWITCHING SCHEMES

Switching

Scheme

Advantages Disadvantages

Single Bus 1. Lowest Cost 2. Failure of bus or any circuitbreaker results in shutdown ofentire substation

3. Difficult to do any maintenance

4. Bus cannot be extended withoutcompletely de-energizing thesubstation

5. Can be used only where loads canbe interrupted or have other supply

arrangementsDouble bus,

double breaker

1. Each circuit has twodedicated breakers

2. Has flexibility In permittingfeeder circuits to beconnected to either bus.

3. Any breaker can be takenout of service formaintenance

4. High reliability

1. Most expensive

2. Would lose half of the circuits forbreaker failure if circuits are notconnected to both busses

Main and

Transfer

1. Low initial and ultimate cost

2. Any breaker can be takenout of service formaintenance

3. Potential devices may beused on the main bus forrelaying

1. Requires one extra breaker for thebus tie

2. Switching is somewhatcomplicated when maintaining abreaker

3. Failure of bus or any circuitbreaker results in shutdown ofentire substation

Double bus,single breaker

1. Permits some flexibility withtwo operating busses

2. Either main bus may beisolated for maintenance

3. Circuits can be transferredreadily from one bus to theother bus by use of bus-tiebreaker and bus selectordisconnect switches.

1. One extra breaker is required forthe bus tie

2. Four switches are required percircuit

3. Bus protection schemes maycause loss of substation when itoperates if all circuits areconnected to that bus

4. High exposure to bus faults

5. Line breaker failure takes allcircuits connected to that bus outof service

6. Bus-tie breaker failure takes entiresubstation out of service



26/27

SUMMARY OF COMPARISON OF SWITCHING SCHEMES (Cont 'd.)

SwitchingScheme

Advantages Disadvantages

Ring bus 1. Low initial and ultimate cost

2. Flexible operation forbreaker maintenance

3. Any breaker can beremoved for maintenancewithout interrupting load

4. Requires only one breaker

per circuit5. Does not use main bus

6. Each circuit is fed by twobreakers

7. All switching is done withbreakers

1. If a fault occurs during a breakermaintenance period, the ring can beseparated into two sections

2. Automatic reclosing and protectiverelaying circuitry rather complex

3. If a single set of relays are used, thecircuit must be taken out of service

to maintain the relays (Common onall schemes)

4. Requires potential devices on allcircuits since there is no definitepotential reference point. Thesedevices may be required in all casesfor synchronizing, live line, orvoltage indication

5. Breaker failure during a fault on oneof the circuits causes loss of oneadditional circuit owing to operationof breaker-failure relaying

Breaker and a

Half

1. Most flexible operation

2. Highly reliable

3. Breaker failure of bus sidebreakers removes only onecircuit from service

4. All switching is done withbreakers

5. Simple operation; nodisconnect switchingrequired for normal

operation6. Either main bus can be

taken out of service at anytime for maintenance

7. Bus failure does not removeany feeder circuits fromservice

1. 1 breakers per circuit

2. Relaying and automatic reclosingare somewhat involved since themiddle breaker must be responsiveto either of its associate circuits



27/27

New Chapter 2 Phasors and System Arrangements

Documents

Transcript of New Chapter 2 Phasors and System Arrangements