Elements of power system by azlan4477

PREPARED BY: DR. ZULKIFFLI BIN ABDUL HAMID

PART 1 BASIC COMPONENTS

Introduction  Electrical power system:  A network of components designed to transmit and distribute the energy

produced by generators to locations where it is used.  Required to generate and supply electrical energy to consumers.  The three main characteristics of electricity supply are as follows:

 Electricity cannot be stored in bulk form. Thus, power must be generated

continuously so that it meets the demand at a specific voltage and frequency.  Increases in population, industrial expansion, housing areas and etc lead to the continuous increase in demand for power.  Power station is located near to the resources location.

Introduction  The power system today is a complex interconnected network.  The basic components of a power system are:  Generation  Transmission and subtransmission  Distribution  Loads

Introduction Generator bus

Transformer

Load bus

Generator Load Generation

Transmission

Power system representation

Distribution

100 kV to 500 kV

10 kV to 20 kV

10 kV to 30 kV 240 V and 415 V

Generation  Generation is the first part of power system and is responsible for generating electricity through synchronous generators.  The synchronous generators are the essential components of generation and it is located in power plant.  Synchronous generator is driven by prime mover (such as steam turbine, gas turbine, hydraulic turbine, nuclear fuel, diesel engine).  Generation is modeled by an AC voltage source in power system.  Standard generated voltage (TNB) by a power plant is between 10 kV to 25 kV.

Generation Powers supplied by all power stations are based on power demand, which is predicted based on previous demand.

Daily load curve for 24 hours

Method of electricity generation Other energies Method of generation

Process (turbine & generator) Power plant

Electrical energy Source of energy

Thermal energy

Thermal power plant

Coal, oil, gas

Kinetic energy

Hydroeletric power plant, wind power plant

Water, wind

Nuclear energy

Nuclear power plant

Radioactive materials

Solar energy

Solar power plant

Radiation of sunlight

Wave energy

Wave and tidal power plant

Sea wave

Renewable energy

Main sources of energy for electricity generation

Oil

Coal

Gases

Nuclear

Water

Energy source

Advantages

Disadvantages

Coal

Most common source Relatively cheap fuel

The most polluting (CO2) Need antipollution features Non-renewable

Oil

A bit more polluting than However, much more expensive (price natural gas but easy to always fluctuates) transport over long distance Non-renewable

Natural gas

Better and cleaner energy Hard to transport over long distance source than coal Non-renewable

Water

The best source Non-polluting No cost for the ‘fuel’

Nuclear energy

Non-polluting Extremely expensive to build Once built, the cost of the Require elaborate safety system and ‘fuel’ is low expensive training

Renewable energy (wind, solar, wave)

Clean and zero ‘fuel’ cost

Modify & damage ecosystem

High set-up cost Generated output is not very high

Power Plant

Thermal power station (Coal based)

Power Plant

Wind power station

Power Plant

Hydro power station

Roles of power stations  Power stations can be categorized based on their role in power system. There 3 categories:  Base power station  Intermediate power station  Peak power station

Generation Powers supplied by all power stations are based on power demand, which is predicted based on previous demand.

Daily load curve for 24 hours

Roles of power stations  Base power station:  Base load is the minimum amount of power that a utility company must

make available to its customers.  Deliver full power continuously i.e. 24 hours a day and 365 days a year.  Most economical in terms of operating costs i.e. high efficiency output & most reliable.  Nuclear & coal-fired stations are suitable for this purpose.  Intermediate power station:

 Can respond to changes in demand relatively quickly, usually by adding or

removing one or more generating units.  Operate mainly during day time to complement the base stations.  The most economical in terms of production costs, although unreliable in terms of actual energy output.  Hydropower stations are suitable for this purpose.

Roles of power stations  Peak power station:  Generally run only when there is a high demand, known as peak demand  Deliver power for short intervals during the day.  Able to respond very quickly because they can be started up in a few

minutes.  Have high operating cost due to their idling-time and unutilized capital costs but necessary for optimal operation of power system.  Usually gas turbines are used for this purpose.

Transmission  Transmission is the intermediate part of power system and is responsible for transporting the generated power from power plant to substation or from generation to distribution.  High voltage power lines and towers are the essential components of transmission.  Transmission is modeled by an impedance between two buses in power system.  Between the transmission regions, there are many substations for stepping down the voltage from high to low level.  The standard voltage level (TNB) for transmission are 132 kV, 275 kV and 500 kV.

Transmission  Two types of transmission lines:  overhead lines  underground cables

Overhead line

Underground cable

Overhead transmission line  Consist of three conductors suspended by towers or poles.  Generally the lowest-cost method (since most of the insulation is provided by air)  Classified according to the range of voltages:  Low voltage: less than 1 kV (distribution system)  Medium voltage: between 1 kV to ~ 33 kV (distribution system)  High voltage: between 33 kV to ~230 kV (for long distance transmission)  Extra high voltage: over 230 kV ~ 800 kV (for long distance transmission)  Ultra high voltage: higher than 800 kV (long distance transmission)

Underground cables  These lines are designed to be buried underground or under water  Compared to overhead lines, buried cables are:  More expensive (cost of burying cables is higher)  Harder to maintain and repair  Cannot be used for very long distances due to capacitance problems

 Nevertheless, they are increasingly popular in new urban areas where overhead transmission lines are considered to be unsuitable.

Substations  In transmission system, there are many substations for transforming from high to low voltage level (step down) according to consumer requirement.  Inside the substation there are:  Transformer (heart of substation)  Switchgear (on-off switching of the supply)  Bus-bar (copper made node)  Control panels (for control purpose)

 In power system, a substation is represented by transformer :

Substations  TNB’s substation:  Main Intake Substation (PMU) – step up voltage to transmission level (132

kV, 275 kV, 500 kV) from generation level.  Distribution Substation (PPU) – step down voltage from transmission level to distribution level (from 33 kV to 11 kV).  Residential Substation – step down from distribution level to residential level (from 11 kV to 415 V).  Interconnecting substation – connect between different power systems (e.g. HVDC connect Malaysia and Thailand).

Substations

PMU

PPU

Distribution  Distribution is the last part of power system and is responsible for delivering or distributing the electricity from substation to the consumers (load).  Distribution substation and low voltage feeders are the essential components of distribution.  In power system, distribution system can be radial or ring network.  The standard voltage level (TNB) for distribution are 240 V, 415 V, 11 kV and 33 kV.

Loads  Load is the end-component of power system and it receives the electricity from distribution system.  They are the consumers of power system.  Load is modeled by an arrow in power system (to indicate the consumption of electricity).  There are three four categories of loads:  Industrial load  Commercial load  Critical load  Residential load

Loads Type of consumers

Required voltage

Example

Heavy industrial loads

100 kV to 300 kV

Oil and gas platform, automotive manufacturer

Medium industrial loads & commercial buildings

10 kV to 30 kV

Electronic manufacturer, shopping mall, university

Critical loads

10 kV to 30 kV

Hospital, military defense system

Residential loads

240 V to 415 V

Single storey up to three storey building

Transmission grid  Grid system is an overall network that consists of many interconnected power system.  The advantages of grid systems are:  Stability in operation – has sufficient reserved power (reactive power)  Service continuity – can still supply electricity when breakdown happens.  Economy – generators can be scheduled to operate based on demand level.

 In Peninsular Malaysia, the power grid is known as the National Grid.  It is operated and owned by Tenaga Nasional Berhad (TNB).  The system spans the whole of Peninsular Malaysia, connecting electricity generating stations owned by TNB and Independent Power Producers (IPPs).

TNB grid

Transmission grid  IPPs are privately owned power producers. They generate and sell power to TNB.  Examples of IPPs:  YTL Generation Sdn Bhd - Paka, Pasir Gudang  Malakoff Berhad - Lumut Power Station, Segari, Prai Power Station,

Butterworth, Tanjung Bin Power Station, Johor  Genting Sanyen Power Sdn Bhd - Kuala Langat Power Station  There are also two other electrical grids in East Malaysia:  In Sabah: Sabah Electricity Sdn. Bhd. (SESB)  In Sarawak: Sarawak Electricity Supply Corporation (SESCO)

Power system representation  The essential components of power system are:  Generator – it is a synchronous generator and represented by a voltage

source

 Synchronous condenser – it is a synchronous generator which acts as a

capacitor for reactive power supply.

 Transmission line – and modeled by an impedance connected between two

buses.

 Bus – it is like a node or terminal in electric circuit.  Transformer – it is represented by two crossed circles or parallel winding.  Load – it is the consumer of power system and represented by an arrow

connected on a bus.

Power system representation Load Generator

Transformer Motor Transmission line

Power system representation Generator bus

Generator

Generation

Load bus

Transformer

Transmission

Load

Distribution

30-Bus transmission power system Synchronous generator Transmission line Synchronous condenser

Load

13-Bus radial distribution network Substation transformer Substation transformer

PART 2 SYNCHRONOUS GENERATOR

Introduction  One of the essential components in a power system. Also known as alternator.  Designed to operate at synchronous speed, ns. Hence, the name.  A typical generator consists of 2 parts (separated by a small air gap):  An outside stationary stator  An inside rotating rotor

Introduction

Introduction IAR

R IAY

Y IAB

3-phase load

N IF VF

Synchronous generator connected to its load

Winding types  It has 2 types of windings:  Field winding (rotor winding):  Creates magnetic field  Placed on the rotor

 Armature winding (stator winding):  Voltage is induced on it  Placed on the stator  Always connected in Y-connection

Rotor types  There are 2 types of rotor:  Salient pole:  Driven by low-speed hydraulic turbine  Require large no of poles  Posses large diameter to provide space for the poles

 Cylindrical (non-salient/round):  Driven by high-speed steam turbine  No of poles cannot be less than 2  Smaller compared to salient pole

Stator and its winding R Y Stator winding (armature winding)

B N

Salient pole rotor Rotor winding

Slip rings

Slip ring Shaft

Rotor core +

Brush

Shaft Rotor Winding (field winding)

Rotor core

Cylindrical rotor Shaft

Slip rings

Rotor winding

Brush

Rotor core

Slip ring Shaft +

Brush

Rotor slot

Rotor Winding (field winding)

Operation  Principle of Operation:  A dc current is applied to the rotor winding to produce magnetic field.  Rotor is turned by a prime mover, producing a rotating magnetic field

within the air gap.  The rotating magnetic field induces 3Ф voltages within the stator winding  The rotating magnetic field & the rotor rotate at the same speed called synchronous speed, ns and given by:

ns =

120 f P

(in rpm)

f = Freq of induced voltage (Hz) P = No of poles

Operation N

EGR

Y EG

EBG

3-phase load N

I AR

VF → I F → B R → τ app → E G → I A

I AY

I BA S

BR BS

Equivalent circuit IF jXS VF

RF LF

+ _

IA +

Rotor

Vϕ _ Stator

VF = I F RF

Vφ = EG − ( R A + jX S ) I A 48

Equivalent circuit

Xs is 10 to 100 times greater than Ra, and Ra is normally negligible (Ra=0)

Equivalent circuit Synchronous generator

3-phase load IL

IL RA I AR

jXS +

VT EAY

I AB

Vϕ

EAR

EAB jXS

IDR

IDB

jXS

ZΔ

N I AY RA

ZΔ

IDY

Stator Rotor IF

Powers and power factor  Input power supplied to generator:

Pin = τ appω s  The real, reactive and apparent output power delivered by generator is:

P3φ = 3 Vφ I A ⋅ pf = 3 VT I L ⋅ pf = τ out ω s Q3φ = 3 Vφ I A sin θ = 3 VT I L sin θ S 3φ = 3Vφ I a* = 3 Vφ I A ∠θ = 3 VT I L ∠θ  Power factor of generator is:

pf = cos θ

Powers and power factor VT = terminal voltage / line − to − line voltage I L = line current pf = power factor of generator

θ = angle of power factor (negative of angle I a ) τ app = applied torque by prime over / input torque τ out = output torque delivered by generator

Maximum power transfer  From the complex output power supplied by generator:

S 3φ = 3Vφ I a*  From equation of EG, the phase current Ia is:

Vϕ ∠00 = EG ∠δ − ( RA + jX S ) I A∠θ Vϕ ∠0 = EG ∠δ − ( Z S ∠γ ) I A∠θ 0

EG ∠δ − Vϕ ∠0 I A∠θ = Z S ∠γ

Maximum power transfer

So, the real and reactive output power supplied by generator are:

Output powers considering RA

Maximum power transfer

P3ϕ = Q3ϕ =

3 EG Vϕ XS 3 Vϕ XS

sin δ

 EG cos δ − Vϕ   

Output powers ignoring RA Where δ is power angle

Maximum power transfer Pmax(3Ď&#x2022; ) =

3 EG VĎ&#x2022; XS

Maximum output power

Value Pmax is called as the steady-state stability limit or static stability limit (i.e. the maximum power before the machine loses synchronism)

Phasor diagrams EG

Vϕ

EG IA jXS

IA jXS

γ or δ

IA RA

Vϕ Unity PF

Lagging PF

IA γ

IA jXS

IA RA

Vϕ

Leading PF

Voltage regulation  It is defined as the percentage change in terminal voltage from no load to full load.  The voltage regulation for synchronous generator is given by

VR =

EG − Vφ Vφ

× 100%

Infinite bus  Synchronous generators are rarely used to supply individual loads.  Most are connected to large interconnected power networks.  At the point of connection, system voltage is constant in magnitude, phase angle & frequency. Such a point in a power system is referred to as infinite bus.  The infinite bus or grid can also be defined as a power system so large that its voltage & frequency do not vary regardless of how much real & reactive power is drawn or supplied to it.  It can be applied when the power grid is sufficiently large that the action of any one user or generator will not affect the operation of the power grid.

Infinite bus  In an infinite bus:  System frequency is constant, independent of power flow.  System voltage is constant, independent of reactive power consumed or

supplied.

 An infinite bus assumed in many small electrical applications.  As an example:

We take for granted that the voltage supply to a residential outlet will be 240V and 50Hz, but the voltage and frequency are not changed when you turn the TV on.

Infinite bus  In a power plant, the synchronous generators are connected to or disconnected from the infinite bus, depending on the power demand on the grid system.  The operation of connecting a synchronous generator to the infinite bus is known as paralleling with the infinite bus.  Before the alternator (i.e. synchronous generator) can be connected to the infinite bus, however, the incoming alternator and the infinite bus must have the same:  1. Voltage  2. Frequency  3. Phase sequence  4. Phase

[Answer: a) 23.558 kV, 17.1° (b) 807.485 A, -53.43° , 0.596 lagging (c) 136 MW, 3248.85 A, 36.32°

Equivalent circuit

Example 2

Answer: (a) 12.806 kV, 14.47◦ (b) 80.4 MW (c) 3344 A, 36.73◦

PART 3 TRANSFORMERS

Introduction  Transformer is another essential component in a power system.  It converts AC voltage from one level to another through the action of magnetic field.  Convert higher voltage to lower voltage: step-down transformer  Convert lower voltage to higher voltage: step-up transformer

 Enable transmission of electrical energy over great distances (step-up to high voltage so that power is transmitted at low current, so low losses).

Equivalent circuit – ideal  Ideal transformer has no losses.

Ideal transformer

 For an ideal transformer, the voltage & current relationship is given by:

Equivalent circuit - practical ď&#x201A;&#x2014; For a practical transformer, there are losses (i.e. flux leakage, core & copper losses). Taking into account all the losses, the equivalent circuit is given as below:

Per phase equivalent circuit of a practical transformer

Equivalent circuit - approximate ď&#x201A;&#x2014; The equivalent circuit can be simplified by referring to either primary or secondary side. ď&#x201A;&#x2014; Per phase equivalent circuit referred to the primary side (approximate):

Equivalent circuit - approximate ï&#x201A;&#x2014; Where:

V1 = V2' + I 2' ( R1 + jX 1 + R2' + jX 2' )

Equivalent circuit - approximate ď&#x201A;&#x2014; Per phase equivalent circuit referred to the secondary side (approximate)

Equivalent circuit - approximate ï&#x201A;&#x2014; Where:

V1' = V2 + I 2 ( R1' + jX 1' + R2 + jX 2 )

Efficiency & voltage regulation  The efficiency is given by:

Ploss = PCu + Pcore

 Voltage regulation is given by:

VR =

V1 − V V

' 2

V1' − V2 V2

× 100%

Referred to primary Referred to secondary

Pout = V2 I 2 . pf PCu = I 2 RT Pcore

V2 = RC 73

Example 1 A 150-kVA, 2400/240-V single-phase transformer has the parameters as shown in Figure 3.31. (a) Determine the equivalent circuit referred to the high-voltage side. (b) Find the primary voltage and voltage regulation when the transformer is operating at full-load 0.8 power factor lagging and 24o V. (c) Find the primary voltage and voltage regulation when the transformer is operating at full-load 0.8 power factor leading. (d) Determine its efficiency for the condition in (b)

Answer: (a) Zseries = 0.4 + j0.9 (b) 2453.933 V, 0.7â&#x2014;Ś, 2.25% (c) 2387.004 V, 1.44â&#x2014;Ś, -0.54% (d) 94.o6%

Autotransformers  A special purpose transformer or simply called the autotransformer is used when a voltage level need to be changed by only a small amount; for instance, from 110 V to 120 V or from 13.2 kV to 13.8 kV.  This small rise in voltage magnitude is necessary for compensating voltage drop in power system.  It is expensive and wasteful to construct a conventional transformer for such small voltage change. Hence, the autotransformer is needed.  An autotransformer is constructed from a conventional transformer by modifying the connection of primary and secondary winding.

Autotransformers ď&#x201A;&#x2014; A conventional two-winding transformer can be changed into an autotransformer by connecting the primary and secondary windings in series. ď&#x201A;&#x2014; This type of transformer is called as autotransformer.

a) Two-winding transformer

b) reconnected as an autotransformer

Step-up Autotransformer I1

+ +

V1 S 2− w I1 = V1 I2 =

S 2− w V2

I L = I1 + I 2 , I H = I 2 VL = V1 , VH = V1 + V2 77

Step-down Autotransformer I1

V2 _

S 2− w I1 = V1 I2 =

S 2− w V2

_ +

I L = I1 + I 2 , I H = I1

VL = V2 , VH = V1 + V2 78

Power & efficiency of Autotransformer  The kVA rating of autotransformer is:

S auto = VL I L = VH I H  The output power is:

Pout = S auto ⋅ pf  Efficiency:

Pout η= × 100% Pout + Ploss  Note: losses of two-winding transformer and autotransformer are the same.

Autotransformers  Performance of autotransformer is measured in terms of power rating advantage (PRA), which is defined as the ratio of apparent power rating of autotransformer (Sauto) to two-winding transformer (S2-w).  This is given below:

PRA =

S auto N 1 = 1+ 2 = 1+ S 2− w N1 a

Autotransformers ď&#x201A;&#x2014; When a two-winding transformer is connected as an autotransformer, it has smaller impedance compared to the two-winding connection. ď&#x201A;&#x2014; It is common practice to consider an autotransformer as a two-winding transformer with its two winding connected in series, where the impedance is referred to the N1-turn side (primary side).

Autotransformer equivalent circuit with its impedance

Example 2

Answer: (a) 360 kVA (b) 99.31%

Three-phase transformer ď&#x201A;&#x2014; The previous analysis considers single phase transformer. In practice, transformers are connected in three-phase form.

Three-phase transformer  In three-phase form, the primary and secondary sides can have either same or different connection (Y or Δ).

Y-Y connection

Y-Δ connection

Three-phase transformer Δ-Y connection

Δ-Δ connection

Three Winding Transformers  Three-winding transformers are used for:  Supplying two independent loads at different voltages from the same

source  Interconnection of two transmission systems of different voltages

 These windings are called as primary, secondary and tertiary windings.  The tertiary winding is:  Usually used to provide voltage for auxiliary (additional) power purpose in

a substation or to supply a local distribution system  Also connected to the switched capacitor or reactor for the purpose of reactive power compensation

Three Winding Transformers ď&#x201A;&#x2014; Equivalent circuit of three-winding transformer is given as:

Equivalent circuit of three-winding transformer

Three Winding Transformers By solving (1), (2) and (3), the impedances of the three windings referred to primary side are:

Voltage Control of Transformers  Voltage control in transformers is required to:  Compensate for varying voltage drops in the system  Control reactive power flow over transmission lines  Control phase angle and therefore, real power flow.

 Two methods commonly used:  tap-changing transformer  regulating transformers (boosters)

Tap Changing Transformer  Used to control voltage magnitudes at all levels by changing its tap setting. Tap setting is used to change the transformer’s turn ratio, a.  2 types:

Off-load tap changing transformers (OLTC):  Requires disconnection when changing the tap setting  The changes are not frequent (normally due to load growth and seasonal change) Tap-changing under load (TCUL) transformers:  Can change taps when power is connected  The changes are frequent, depends on system condition

Tap Changing Transformer ď&#x201A;&#x2014; Consider diagram below with a step-up transformer at sending end and a stepdown transformer at receiving end of a transmission line, where tS and tR are the tap setting in per-unit. V1â&#x20AC;&#x2122; and V2â&#x20AC;&#x2122; are the supply and load phase voltages referred to high voltage side respectively.

A radial line with tap changing transformers at both ends

Tap Changing Transformer ď&#x201A;&#x2014; If VS and VR are the phase voltages at both ends of the line, then:

ď&#x201A;&#x2014; The phasor diagram for the above equation is shown below:

Voltage phasor diagram

Tap Changing Transformer

Tap Changing Transformer  So, the tap setting at sending end bus is:

V2' ts =

V1' 1−

RPφ + XQφ V1' V2'

 Note: P and Q in the above equation is for single phase load power. V1’ and V2’ are the phase voltages.

Tap Changing Transformer  From the previous equation:  Pφ and Qφ are single-phase real and reactive power of load  Sending-end phase voltage (generator voltage) refer to high voltage side:

V1 V1 = a1 '

VP1 a1 = VS 1

 Receiving-end phase voltage (load voltage) refer to high voltage side:

V = a2V2 ' 2

VP 2 a2 = VS 2 99

23 kV

Answer: tS = 1.08 pu, tR = 0.926 pu

100

Regulating Transformer (Booster)  Used to change voltage magnitude & phase angle, at a certain point in the system by a small amount.  Consists of an exciting transformer and a series transformer  The advantages of regulating transformers are:  The main transformers are free from tapings  It can be used at any intermediate point.  It can be taken out for maintenance without affecting the system.

 Two types of booster:  Voltage magnitude control  Phase angle control

101

Booster - Voltage Magnitude Control ď&#x201A;&#x2014; The connection for phase a is given below ď&#x201A;&#x2014; Other phases have identical arrangement.

Per-phase regulating transformer for voltage magnitude control

102

Booster - Voltage Magnitude Control  The secondary of the exciting transformer is tapped and the voltage obtained from it is applied to the primary of the series transformer.  The corresponding voltage, ΔVan on the secondary of the series transformer is added to the input voltage, Van.  Thus, output voltage is:

 The output voltage magnitude can be adjusted by changing the exciting transformer’s tap.  By changing the switch from position 1 to 2, the polarity of the voltage across the series transformer is reversed, so that the output voltage magnitude can be controlled either to be increased or decreased.

103

Booster - Phase Angle Control  A typical arrangement for phase a is shown below  If the injected voltage, ΔVbc is out of phase with the input voltage, Van then the resultant voltage, Van’ will have a phase shift with respect to the input voltage.

Per-phase regulating transformer for phase angle control

104

Booster - Phase Angle Control  The series transformer of phase a is supplied from the secondary winding of the exciting transformer, which is connected to phase b and c.  The injected voltage ΔVbc is in quadrature with the voltage Van, thus the resultant voltage V’an goes through a phase shift α as shown below

Phasor diagram for phase angle control

105

Phase Angle Control  The output voltage is

 The amount of phase shift can be adjusted by changing the exciting transformer’s tap.  By changing the switch from 1 to 2, the output voltage can be made to lag or lead the input voltage.

106

PREPARED BY: DR. ZULKIFFLI BIN ABDUL HAMID

107

Part 1 Transmission Line Models

108

Introduction ď&#x201A;&#x2014; The purpose of transmission lines is to transfer electric energy from generating stations to the distribution system which then supplies the load.

ď&#x201A;&#x2014; These transmission lines are physically extended over hundreds of kilometres. ď&#x201A;&#x2014; As a result, the associated resistance, inductance & capacitance are distributed along the length of the line.

109

Introduction ď&#x201A;&#x2014; A transmission line is characterized by a series of resistance, inductance & shunt capacitance per unit length. ď&#x201A;&#x2014; It might be modelled as a repeating series of lumped constants, each representing the resistance, reactance & capacitance of a small segment of the entire line.

110

Transmission line model ď&#x201A;&#x2014; In power system, a transmission line is modeled by a single line connected between two buses. ď&#x201A;&#x2014; The current or power of the transmission line flows from sending-end bus to receiving end bus. This is shown below:

VS Ps + jQs

PR + jQR Sending end bus

Receiving end bus

111

Transmission line model  A transmission line is modeled based on its length.  The following are the existing transmission line models according to their length.  Short line model – less than 80 km  Medium line model – between 80 km to 250 km  Long line model – above 250 km

 Two-port network is used when performing analysis on transmission line. In this case, the ABCD parameters are used.

Sending

Receiving

112

Short trans. line model ď&#x201A;&#x2014; The following diagram represents the short transmission line model.

ď&#x201A;&#x2014; In the above figure, VS is the sending end phase voltage, VR is the receiving end phase voltage, IS is the sending phase current, IR is the receiving end phase current, Z is the impedance of transmission line

113

Short trans. line model

z= r + jω L Ω / km series impedance per unit length Z =z ⋅ l Ω total series impedance

114

Solving for receiving end parameters ď&#x201A;&#x2014; Instead of finding the sending end voltage or current, sometimes, we are also required to determine the value of receiving end voltage and current. ď&#x201A;&#x2014; This is done through the following matrix equations:

115

Sending and receiving end parameters S S (3φ )= 3VS I S*= S S (3φ ) ∠θVs − θ Is= PS (3φ ) + jQS (3φ ) S R (3φ = 3V I = S R (3φ ) ∠θVr − θ Ir= PR (3φ ) + jQR (3φ ) ) * R R

Complex power

S S (3φ )

3= VS I S PS (3φ ) S S (3φ ) .cos QS (3φ ) S S (3φ ) .sin θ S θS =

S R (3φ )

3= VR I R PR (3φ ) S R (3φ ) .cos QR (3φ ) S R (3φ ) .sin θ R θR =

Apparent, real and reactive power

116

Sending and receiving end parameters VS = VR =

VS ( L − L ) 3 VR ( L − L ) 3

∠θVs ∠θVr

PFS = cos(θVs − θ Is ) PFR = cos(θVr − θ Ir )

IS Phase voltages (θVr is usually assumed to zero)

Phase current (found from power)

Power factor at sending and receiving end

117

Voltage regulation of transmission line  The voltage regulation of transmission line is calculated as follows:

%VR

VR ( L − L )( NL ) − VR ( L − L )( FL ) VR ( L − L )( FL )

×100%

 At no load, IR = 0, so:

118

Efficiency of transmission line  Transmission efficiency is a parameter which describes how much percentage of input power at sending end is delivered to the receiving end and is expressed in percentage.  The efficiency of transmission line is found as follows:

%η =

PR (3φ ) PS (3φ )

×100%

 Where, the sending end real power PS(3φ) is found as: ∗ S= 3 V = I PS (3φ ) + jQS (3φ ) S (3φ ) S S

 The receiving end real power PR(3φ) is normally given in the question.  The transmission line losses is equal to the difference between PS(3φ) and PR(3φ) .

119

Phasor diagram ď&#x201A;&#x2014; The phasor diagrams for various load power factor are given below:

120

Example 1

Assume the frequency is 60 Hz

Answer: (a) Vs(l-l) = 250 kV, 322.8 MW + j288.6 MVar, 13.6%, 94.4%

121

Medium line model  For medium transmission line, the shunt capacitance cannot be neglected.  It is represented by two lumped capacitors, one before & one after the series impedance, with values equal to half of the total line capacitance.  The nominal pi-model of medium transmission line is given below.

 Where, Y is the shunt admittance and the unit is Siemens (S), VS is the sending end phase voltage, VR is the receiving end phase voltage, IS is the sending phase current, IR is the receiving end phase current, Z is the impedance of transmission line. 122

Medium line model

Note: A and D are in per-unit, B in Ohms and C in Siemens

z= r + jω L Ω / km series impedance per unit length Z =z ⋅ l Ω total series impedance y= G + jωC S / km shunt admittance per unit length Y= y ⋅ l S total shunt admittance l = line length (km)

123

Medium line model

ď&#x201A;&#x2014; Derivation of ABCD parameters are as follows:

124

Medium line model ď&#x201A;&#x2014; So, finally:

ď&#x201A;&#x2014; Where:

125

Example 2

Take the frequency as 60 Hz

126

Example 3

127

Example 4

128

Long line model ď&#x201A;&#x2014; For short & medium lines, reasonably accurate models were obtained by assuming the line parameters to be lumped (combined). ď&#x201A;&#x2014; For lines greater than 250 km, however, the exact effect of the transmission line parameters must be considered. ď&#x201A;&#x2014; These parameters are not lumped, but are uniformly distributed (repetitively) along the length of the line as shown below:

129

Long line model  There are two approaches when performing analysis on long transmission line:  Using exact model  Using equivalent pi-model

Exact model

Equivalent pi-model

130

Long line model â&#x20AC;&#x201C; exact model ď&#x201A;&#x2014; Two important parameters of long transmission line:

(rad/km)

131

Long line model – exact model  Mathematical property:

a + bi = c∠θ = c ∠

θ 2

132

Long line model – exact model  The exact ABCD-parameters for exact long line model are follows:

(In per-unit) (In Ohms) (In Siemens)

133

Long line model – exact model  The following trigonometric identities are used for determination of ABCD parameters

let γl = a + bi cosh(γl ) = (cosh a )(cos b) + j (sinh a)(sin b) sinh(γl ) = (sinh a)(cos b) + j (cosh a)(sin b)  γl  cosh(γl ) − 1 tanh  = sinh(γl ) 2

All angles are in radian

134

Example 5

Answer: γ = 3.209x10-5 + j1.244x10-3 per km Zc = 266.04 – j6.652 Ohm A = D = 0.9312 + j3.51x10-3 B = 4.81 + j96.94 C = –5.724x10-7 + j1.37x10-3

cosh γl = 0.9312 + j3.51x10-3 sinh γl = 8.964x10-3 + j0.3646

135

Long line model – pi model  The transmission line can also be represented by an equivalent pi-circuit.  The circuit is identical to the nominal pi-circuit of medium line except that Z’ and Y’ replace Z and Y.

Equivalent pi-model

136

Long line model – pi model  The ABCD parameters of the equivalent pi-circuit are:

Z sinh(γl ) Z = Z C sinh(γl ) = γl '

(In per-unit) (In Ohms) (In Siemens)

Y' =

2  γl  tanh  ZC 2

 tanh(γl / 2)   = Y   (γl / 2)   cosh(γl ) − 1   = Y   (γl / 2) sinh(γl )  137

Example 6

Answer:

138

Inductance of a Transmission Line ď&#x201A;&#x2014; Inductance of a transmission line is given by:

ď&#x201A;&#x2014; Where:

r = radius of each conductor D = distance between the phases D

139

Inductance of a Transmission Line  From the equation:  If D increases, the ratio D/r increases. Thus the total inductance of the line

increases.  If r increases, the ratio D/r decreases. Thus the total inductance of the line decreases.  Therefore:

 The greater the radius of the conductors in a transmission line, the

lower the inductance of the line.  The greater the spacing between the phases of a transmission line, the greater the inductance of the line.

140

Capacitance of a Transmission Line ď&#x201A;&#x2014; Capacitance of a transmission line is given by:

ď&#x201A;&#x2014; Where:

r = radius of each conductor D = distance between the phases D

141

Capacitance of a Transmission Line  From the equation:  If D increases, the ratio D/r increases. Therefore the total capacitance of

the line decreases.  If r increases, the ratio D/r decreases. Therefore the total capacitance of the line increases.  Therefore:  The greater the spacing between the phases of a transmission line, the

lower the capacitance of the line.  The greater the radius of the conductors in a transmission line, the higher the capacitance of the line.

142

Part 2 Lossless line & compensation techniques

143

Lossless line  The concept of lossless line is used for quick & reasonably accurate calculations  When losses are neglected, simpler expressions for transmission line parameters are obtained.  Transmission lines are generally designed to have low losses.  For a lossless line:

 Hence:

r = 0, g = 0, α = 0 z = j ωL , y = j ωC , γ = j β 144

Lossless line ď&#x201A;&#x2014; The characteristic impedance, ZC (usually called as surge impedance) becomes:

ď&#x201A;&#x2014; Also, the propagation constant, Îł becomes:

145

Lossless line  In general, the velocity of propagation, v (in m/s or km/s) and the wavelength, λ (in m or km) of the voltage & current waves travel along a transmission line are given by:

ω β

λ=

2π

 For a lossless line, they become:

146

Lossless line  The rms voltage and current along the transmission line are given by:

= V ( x) cos ( β x ) ⋅ VR + j ⋅ Z C ⋅ sin ( β x ) ⋅ I R 1 I ( x ) =j ⋅ ⋅ sin ( β x ) ⋅ VR + cos ( β x ) ⋅ I R ZC  At sending end, x = l, therefore:

= VS cos ( β ⋅ l ) ⋅ VR + j ⋅ Z C ⋅ sin ( β ⋅ l ) ⋅ I R 1 IS = j ⋅ ⋅ sin ( β ⋅ l ) ⋅ VR + cos ( β ⋅ l ) ⋅ I R ZC 147

Lossless line – exact model  Therefore, for a lossless transmission line using exact long line model, the matrix equation becomes:

VS   A B  VR  I  =  I  C D  R   S 

 Where:

A= D= cos ( β ⋅ l ) = B jZ C sin ( β ⋅ l ) 1 = C j sin ( β ⋅ l ) ZC

148

Lossless line – pi model  For equivalent pi-model, the ABCD parameters are the same as that of medium line, but the impedance and admittance are found as follows:

VS   A B  VR  I  =  I  C D  R   S 

= Z ' jX =' jZ C sin ( β ⋅ l ) 2  β ⋅l  Y = j tan   ZC 2   '

149

Wave reflection  Basically, when an incident wave (voltage or current wave) travels along a transmission line, some of its portion will be transmitted while the other parts will be reflected back to the source. This is shown below: Incident Wave, e

Transmitted Wave, e’ Reflected Wave, e”

150

Wave reflection  So, there will be two coefficients, namely:  Transmission coefficient:

ct =

2Z L Z L + Zo

cr =

Z L − Zo Z L + Zo

 Reflection coefficient:

 And:

ct = 1 + c r

151

Wave reflection  So, the transmitted wave, e’ is:

e' = α ⋅ e  So, the reflected wave, e’’ is:

e '' = β ⋅ e  Where, e is the incident wave

152

Surge impedance loading (SIL) ď&#x201A;&#x2014; Surge impedance loading (SIL) is the power delivered by a lossless line to a load whose resistance equal to the surge impedance of the line.

Lossless line

153

Surge impedance loading (SIL) ď&#x201A;&#x2014; SIL for a lossless line is given by:

ď&#x201A;&#x2014; At this condition, the receiving end current is given by:-

154

Surge impedance loading (SIL)  In MW, SIL can be found as follows:

SIL =

(kVR (l −l ) ) 2 ZC

in MW

155

Example 7

Answer: (a) β = 0.001259 rad/km, Zc = 290.43 Ohms, v = 2.994 x 105 km/s, λ = 4990 km (b) VS = 356.53 kV, 16.1°, IS = 902.3 A, -17.9°, SS = (800 + j539.672) MVA, VR = 32.87%

156

Line loadability VR

VS Ps + jQs

PR + jQR Sending end bus

Receiving end bus

157

Line loadability ď&#x201A;&#x2014; The sending end complex power is:-

ď&#x201A;&#x2014; The real and reactive powers at the sending end of the line are:-

158

Line loadability ď&#x201A;&#x2014; The receiving end complex power is:

ď&#x201A;&#x2014; The real and reactive powers at the receiving end of the line are:

159

Line loadability ď&#x201A;&#x2014; The real and reactive transmission line losses are:

160

Line loadability  For a lossless line:

X ' = Z C sin( βl )  Therefore for a lossless line, the receiving end powers become:

Lossline line

 As the load increases, the angle δ. The maximum power that can be transmitted to

the load happens at δ = 90.

161

Line loadability  Since a lossless line has X’ = Zc sin βl, the receiving end real power becomes:

 Therefore, the receiving end real power for a lossless line in terms of SIL is written as:

162

Line loadability  Thus, the loadability equations (receiving end power) for a lossless line are as follows:

Lossline line

 Any value for δ greater than 90° will cause overloading of transmission line. Hence, the above equations determine the limit of a transmission line before it becomes unstable.

163

Transmission Capability  The power handling ability of a transmission line is limited by:  Thermal loading limit – depends on line temperature  Stability / loadability limit – depends on line loading (angle δ)

164

Transmission Capability – thermal loading limit  The increase in conductor temperature, due to real power loss, stretches the conductors.  This will increase the conductor sag between transmission towers.  At higher temperatures this may result in irreversible stretching.  The thermal limit is specified by the current carrying capacity of the conductor and is available in the manufacturer’s data.  If the current carrying capacity is denoted by Ithermal, the thermal loading of a line is:

165

Transmission Capability â&#x20AC;&#x201C; thermal loading limit T3(>T2)

T2(>T1)

Safety limit Transmission tower

Transmission tower

As the temperature increases, the conductor sag also increases

166

Transmission Capability – stability limit  Theoretically, maximum power transfer occurs when δ = 90º (for a lossless line).

 However, to assure stability, the practical operating load angle is usually limited to 35º to 45º.  This is because when the reactances of the generators & transformer are added to the line, it will result in a larger δ for a given load.

167

Example 8

Answer: (a) 400 kV (b) 1167 MW

168

Line compensation  On long transmission lines:  Light loads appreciably less than SIL result in a rise of voltage at the

receiving end, and  Heavy loads appreciably greater than SIL will produce a large dip / drop in voltage.

 To increase line loadability and maintain voltages near rated values, inductors & capacitors are used.  Three methods for line compensation:  Shunt reactors  Shunt capacitors  Series capacitor

169

Line compensation – shunt reactors  Shunt reactors (inductors) are widely used to reduce high voltage to a specified value under light load (less than SIL).  The inductors absorb excessive reactive power and reduce overvoltages during light load conditions. Hence, the load voltage is maintained at desired value.  A reactor of reactance XLsh connected at the receiving end of a long transmission line is shown below:

Shunt reactor compensation

170

Line compensation –shunt reactors  From the figure, the receiving end current is:

 From the matrix equation for a lossless line:

 cos( βl ) VS    I  =  j 1 sin( βl )  S  ZC

jZ C sin( βl )  VR  cos( βl )   I R    

171

Line compensation –shunt reactors  Substitute IR into VS :

 Solving for XLsh yields

172

Line compensation – shunt reactors  If VS = VR :

 The reactive power consumed by the shunt reactor in order to maintain the desired receiving end voltage is:

QLsh =

(kVR (l −l ) ) 2 X Lsh

(MVar )

173

Line compensation –shunt reactors  Using the same derivation, the relation between IS & IR is found as follows:

 From the above equation, VS = VR then:

174

Line compensation –shunt reactors  With one reactor only at the receiving end, the voltage profile will not be uniform, and the maximum rise occurs at the midspan.  For VS = VR, the voltage at the midspan is given by:

 Also, the current at the midspan is zero.  Installing reactors at both ends of the line will improve the voltage profile and reduce the tension at midspan.

175

Example 9  Given a lossless line with the following parameters:

ZC = 290.43 Ohms,

βl = 21.641°

a) Calculate the receiving end voltage when line is terminated in an open circuit and is energized with 500 kV at the sending end. b) Determine the reactance and the Mvar of a three-phase shunt reactor to be installed at the receiving end to keep the no load receiving end voltage at the rated value (i.e. at 500 kV).

Answer: a) VR(nl) = 310.57 kV, VR(l-l)(nl) = 537.9 kV b) XLsh = 1519.5 Ohms, QLsh = 164.53 Mvar

176

Line compensation – shunt capacitor  Shunt capacitors are used to supply reactive power & increase voltages during heavy load conditions.  The effect is to maintain the receiving end voltage at a satisfactory level. Shunt capacitors are also used for power factor correction.  Capacitors are connected either directly to a bus bar or to the tertiary winding of a main transformer and are disposed along the route to minimize the losses and voltage drops.  Besides shunt capacitors, static var control & synchronous condensers can also be used for the same purpose.

177

Line compensation – series capacitor  Series capacitors are used to increase line loadability (capability of line to support more loads).  They are connected in series with the line, usually located at the midpoint.  They are used to reduce the series reactance between the load & the supply point.  This results in:  improved transient and steady state stability  more economical loading,  minimum voltage dip (drop) at load buses.

178

Line compensation – series capacitor  Major drawbacks of series capacitor:  Special protective devices are required to protect the capacitors & bypass the

high current produced when a short circuit occurs.  Also, inclusion of series capacitors establishes a resonant circuit that can oscillate at a frequency below the normal synchronous frequency when stimulated by a disturbance. This phenomenon is referred to as subsynchronous resonance (SSR) & may cause considerable damage to the turbine-generator  The subsynchronous resonant frequency can be found using:-

 Where fS is the synchronous frequency, L’ is the lumped line inductance corrected for the effect of distribution & Cser is the capacitance of the series capacitor

179

Line compensation – shunt & series capacitor

SR = PR + jQR SC = –jQC

SL = PL + jQL Load

180

Line compensation – shunt & series capacitor Series capacitor The new receiving end real power including series capacitor is:

Shunt capacitor The reactive power injected by shunt capacitor is:

QC = QR − QL

Where: XCser : reactance of series capacitor X’ : reactance of line [X’ = Zc sin (βl)]

Where: QR : receiving end reactive power QL : load reactive power Receiving end reactive power is:

Compensation percentage:

CP =

X Cser × 100% ' X

The new impedance (exact & pi-model):

Z ' = j ( X ' − X Cser )

Reactance of shunt capacitor:

XC =

VR QC

181

Example 10  A lossless line supplies a load of 1000 MVA, 0.8 power factor lagging at 500 kV. Given the following parameters:

ZC = 290.43 Ohms, a)

βl = 21.641°

Determine the Mvar and the capacitance of the shunt capacitors to be installed at the receiving end to keep the receiving end voltage at 500 kV when the line is energized with 500 kV at the sending end. Only series capacitors are installed at the midpoint of the line providing 40 percent compensation. Find the sending end voltage and voltage regulation. Use equivalent pi-model of lossless line.

Answer: a) δ = 20.044°, QC = 576.85 MVar, C = 6.1 uF b) A = 0.9577, B = j64.26, VS = 326.4 kV, 10.47°, VR = 18%

182

Definitions  A two port network − is an electrical network model with one pair of input terminals and

one pair of output terminals. − It is commonly used to model the voltage and current characteristics of complex electrical networks  Voltage Regulation − is defined as the rise in voltage at the receiving end, expressed as

percentage of full load voltage, when full load at a specified power factor is removed while the sending end voltage is held constant.

− Voltage regulation describes the ability of a system to provide near constant

voltage over a wide range of load conditions.

183