A304SE PPT Slides T111_final

1

Introduction

Lesson 1

Learning objectives

To understand the conventional sources of energy

To understand the alternative sources of energy

2

What is Power System

A system that deals with the business of Generation

Transmission Distribution

of electrical energyIt is the most complex and massive man made

system

The study of electric power systems is concerned with the generation, transmission, distribution, and utilization of electric power .

The first of these – the generation of electric power – involves the conversion of energy from a nonelectrical form (such as thermal, hydraulic, or solar energy) to electric energy.

3

Goal of Power System

High reliability standard Lowest operational cost

Minimum environmental impact

Power System Components

4

The load is not an integral part of the Power System as they are not owned by power generating companies, but it has to be modeled in the PS analysis

Electricity cannot be stored and the supplier has small control over the load at any time. The engineer endeavors to keep the output from the generator equal to the connected load all the time.

There is a continuous demand of power. A large and continuous process of adding to the system exists.

The distribution and nature of fuel. The problem of station siting and transmission distances is an involved exercise in economics.

Prepared by: Mr. John Mateo

5

Power Generation

Power generation takes place in power plants, which may geographically dispersed.

A power plant may contain more than one power generating units.

Energy Sources

Hydrocarbons (oil, gas, natural gas, etc). Water

Solar Nuclear Chemical Wind Tidal

8

Consumption of electricity in Singapore is about 6,151.54 kWh per person (2001 data), placing it as the 29th highest consumer of electricity worldwide, and 4th highest in Asia, according to the nationamaster.comInternet site, using data from the U.S. Department of Energy. The site also ranked Singapore as the 8th highest country worldwide (and 2nd in Asia) in terms of electricity generation ability per person, with 1.45 kW per person. Analysts expect demand to grow at 4-5% annually.

As of end-2003, Singapore’s total installed generation capacity was 8,919 MW, compared with peak demand of 5,139 MW. EMA requires the gencos to maintain 30% reserve capacity, which brings total required capacity to around 6,500 MW. The significant excess capacity is mainly a result of the conversion of oil-powered plants to gas, in an effort to maintain competitiveness in a liberalized market. Some 60% of Singapore’s electricity generation is now fueled by natural gas.

9

Grid Inteconnection

Grid interconnection of is the interconnection of a number of distributed generators owned and operated by different entities

Advantages of Grid Interconnection

Economical as power plants may use less number of generating machines as they can depend on the grids when the need arises.

Companies can take advantage of the most economical cost of power generation.

10

Interconnection increases the amount of current flow when a short circuit occurs in a system and requires the installation of breakers able to interrupt large current flow.

The disturbance caused by a short circuit in one system may spread into the other systems.

Disadvantages of Grid Interconnection

Environmental effects of Grid

Interconnections

Air pollutant emissions. Modest quantities of emissions may be produced during power line construction, but the main influence of grid interconnections on air pollutant emissions will be through the impact of transmission interconnections on which power plants are run where and when in the interconnected nations.

Water pollution impacts, including erosion and water pollutants produced as a result of power line construction and operation, and incremental water pollution from power plant construction, power generation, and fuel extraction/storage.

11

Solid waste impacts, mainly coal ash and high- and low-level nuclear wastes from electricity generation, but also including wastes from fuel extraction and possibly from power line and/or power plant construction.

Land-use impacts, including costs such as the restriction of uses of land through which a power line passes.

Wildlife/biodiversity impacts, including costs such as the potential impacts of power line construction and operation on flora and fauna in the power line area.

Human health impacts, including the impacts of electromagnetic fields (EMFs) from power lines on humans living and working in the power line vicinity (net costs of the interconnection project

12

Distributed power generation

Distributed power generators are small, modular electricity generators typically sited close to customer loads

By siting smaller, more fuel-flexible systems near energy consumers, distributed generation avoids transmission and distribution power losses and provides a choice of energy systems to the utility customer.

Many distributed power systems produce so little noise or emissions that they can be located inside, or immediately adjacent to, the buildings where the power is needed. This greatly simplifies the problems of bringing power to expanding commercial, residential, and industrial areas

Power Transmission Systems

Connects generating plants to consumers Interconnect power grids

High voltage AC transmission High voltage DC transmission

13

Advantage of HV transmission

Lower transmission losses/MW transfer Lower line-voltage drop / km

Higher transmission capacity/ Km Reduced right-of-way requirement /MW

transfer Lower capital and operating costs/ MW

transfer

Power transmission equipments

Step-up transformers Step-down transformers Voltage regulators Phase shifters Transmission lines and cables Circuit breakers Series and shunt reactors and capacitors Lightning arresters Relays Converters/Inverters

14

Power Distribution

Receives electrical energy from the HV/MV levels

Supplies energy to consumers

- at MV/LV levels- at three-phase (industrial)- at single phase (consumers)

Power distribution devices

Distribution transformers Feeder cables

Switches/relays. Etc Lightning arresters Protective relays PT/CT

15

Power Distribution System

Power System Operation and Control

16

Key Operational Goals

Power Balance : Generation must remain constant with demand

Total generation (t) = total demand (t) +b Losses (t)

System security : equipment power flows must not exceed equipment ratings under normal, or single outage condition.

Power Quality Consideration

Frequency Regulation: System frequency must remain within its operational range

fmin ≤ f(t) ≤ fmax

Voltage Regulation : Bus Voltage must remain its operational limits

Vmin ≤ V(t) ≤ Vmax

17

Types of operation and control

Centralized - is based on system-wide data and usually handles slow events

- examples are SCADA and EMS

Decentralized - handles local data and tackles fast events

- protection systems

Examples of Centralized control

Frequency control (regulation) Interchange control

Generation dispatch (control of generating units)

System security assessment and enhancement (both static and dynamic)

Unit commitment (unit’s ON/OFF status)

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Equipment protection against over-voltage and over-current

Generator speed control

Generator terminal voltage control

Examples of DeCentralized control

Power Utility Structure Traditional

Restructured (Deregulated)

Power System Organization

19

Regulated Power utility

20

Traditional Utilities

Operate as monopolies (“have captive customers”)

Government regulated

Have to get regulators approval for rate increases

No incentive to increase generation efficiency “Vertical integrated (VI) business structures

for improved profitability

Deregulated PS Structure

Has been legislated by government that results in

Increased competition (reduced rates)

New technology utilization in generation, transmission and distribution

Increased outside investment in PS structure

21

Lesson 2

Learning objectives

To examine the parameters of a power transmission line

To examine the voltage and current relation in a transmission line

22

Three characteristics, or parameters: its resistance,

inductance, and capacitance.

The dc resistance R of a conductor of length l and cross-sectional area A is

R = pl/A (in ohms)

where p is the resistivity of the material of the conductor in ohm-meters

The temperature dependence of resistance is quantified by the relation

R2 = R1[1 + α(T2 – T1)]

23

Resistivities and Temperature Coefficients of Resistance

The inductance per conductor of a two-wire, single-phase transmission line is given by

(in henrys per meters)

where µ0 = 4p × 10-7 H/m (the permeability of free space), D is the distance between the centers of the conductors, and r is the radius of the conductors. The total, or loop, inductance is then

where r' = re-1/4 is known as the geometric mean radius (GMR) of the conductor.

24

The per-phase (or line-to-neutral) inductance of a three-phase transmission line with equilaterally spaced conductors is

where r is the conductor radius and D is the spacing between conductors

Transposition is the interchanging of the positions of the conductors at regular intervals along the line. Due to unequal inductances in the three phases, leading to unequal voltage dropst he cause by the usual nonsymmetrical spacing

D is equivalent to De, obtained from

De = (DabDbcDca) 1/3

where the distances Dab, Dbc, and Dcaare shown in Figure.

The average per-phase inductance for a transposed line is still given by

25

The shunt capacitance per unit length of a single-phase, two-wire transmission line is given by

(in farads per meter)

is the permittivity of free space

For a three-phase line with equilaterally spaced conductors, the per-phase (or line-to-neutral) capacitance is

26

A transmission-line cable consists of 19 strands of identical copper conductors, each 1.5mm in diameter. The length of the cable is 2km but, because of the twist of the strands, the actual length of each conductor is increased by 5 percent. What is the resistance of the cable? Take the resistivity of copper to be 1.72 ×10-8 Ωm.

Allowing for twist, we find thatl = (1.05)(2000) = 2100m.

The cross-sectional area of all 19 strands is19(/4)(1.5 × 10-3 )2 = 33.576 × 10-6 m2

. R = rl/A = (1.72 × 10-8 × 2100)/(33.576 × 10-6)= 1.076 Ω

27

The per-phase line loss in a 40-km long transmission line is not to exceed 60k Ω while it is delivering 100A per phase. If the resistivity of the conductor material is 1.72 × 10-8 Ω.m, determine the required conductor diameter.

The line loss is to be, at most,I2 R = (100)2 R = 60 × 103

R = 6.substitute A = D2 /4 to R = pl/A yields

D2 /4 = (1.72 × 10-8 )(40 × 103 )/6

D = 1.208cm

28

A sample of copper wire has a resistance of 50 Ω at 10°C. What must be the maximum operating temperature of the wire if the resistance is to increase by at most 10 percent? Take the tenperature coefficient at 10°C to be

α = 0.00409°C-1

Here we have R 1 = 50 Ω and

R 2 = 50 + 0.1 × 50 = 55 Ω. Also, T 1 = 10°C, and we require T 2.

55 = 50[1 + 0.00409(T 2 – 10)] or T 2 = 34.45°C

29

Determine the resistance of a 10km long solid cylindrical aluminum conductor with a diameter of 250 mils, at (a) 20°C and (b) 120°C.


To find the cross-sectional area of the conductor, we note that

250 mils = 0.25 in = 0.35 cmso A = (π /4)(0.635) 2= 0.317cm2

Also, from Table 1, r = 2.83µ Ω.cm and a = 0.0039°C-1 at 20°C

(a) At 20°C, (R20 = rl/A = 2.83 × 10-8 × (10 × 103) / (0.317 × 10-4)

= 8.93 Ω (a) At 120°,

R120 = R20[1 + a(120 – 20)] = 8.93(1 + 0.0039 × 100)= 12.41 Ω


30

A single-circuit, three-phase, 60Hz transmission line consists of three conductors arranged as shown in the Figure. If the conductors are the same as that in Question no 5, find the inductive reactance of the line per kilometer per phase


De = (5 × 5 × 8) 1/3 = 5.848mr = ½ × 0.635 × 10-2m, so that

De/r = (5.848 × 2 × 102)/0.635 = 1841.9and In(De/r) = 7.52.

Hence, we have, for each kilometer of length,L = 2(1/4 + 7.52) × 10-7 × 103= 1.554 mH/km

The inductive reactance per kilometer is then

XL = ω L = 377 × 1.554 × 10-3= 0.5858Ω


31

Discuss the effect of TL parameters on the operation and performance of transmission lines.

Evaluate the losses, efficiency, and voltage regulation of transmission lines

Determine the consequence of such performance characteristics on the operation of a power system.

32

Short line, - considered valid for lines up to 80km long

- the shunt effects (conductance and capacitance) are neglected

- represented by a lumped resistance in series with a lumped inductance

Medium line - generally range from 80 to 240km in length

- the shunt capacitances are lumped at a few predetermined locations along the line;

Long lines - longer than 240km- have uniformly distributed

parameters

33

The short transmission line is represented

by the lumped parameters R and L is the inductance (per phase) of the entire line. The line is shown to have two ends: the sending end (designated by the subscript S) at the generator, and the receiving end (designated R) at the load

34

Quantities of significance here are the voltage regulation and efficiency of transmission. These quantities are defined as follows for lines of all lengths:

Percent voltage regulations == [|VR(no load)| - |VR(load)|]/|VR(load)| × 100

where VR is the receiving-end voltage

Efficiency of transmission = Power at receiving end / power at

sending end= PR/PS

35

In a medium-length transmission line the shunt effect of the transmission line due to the line capacitance is not negligible

Two representations for a medium-length transmission line

a. the nominal-II circuit andb. the nominal-T circuit

36

The parameters of a long line are considered to be distributed over the entire length of the line. One phase (with return through neutral) of a long line, of length L, is shown.

37

The voltage V at any point along this line is given by

where γ = √yz, γ is known as the propagation constanty is the shunt. admittance per unit length of the

line, and z is the series impedance per unit length,

The voltage V at any point along this line

The current I at any point along the line is given by

Zc = √ z/y is called the characteristic impedance of the line

38

A transmission line may be viewed as a four-terminal network.

the terminal voltages and currents are related by

Vs = A VR + B IRIs = C VR + D IR

where the constants A, B, C, and D are called the generalized circuit constants or

ABCD constants and are, in general, complex. By reciprocity, they are related to each other as follows:

AD – BC = 1

39

On a long transmission line, the voltage V and current I everywhere along the line satisfy a relation called the wave equation.

For a lossless transmission line, such that z

and y are purely reactive, the wave equation may be written as

the superscripts + and – denotes, respectively, waves traveling in the +x and –x directions along the transmission line

40

A wave such as V+(t – x/u) that is traveling in the positive x direction is called a forward-

traveling wave,

and one that is moving in the negative x direction is a backward-traveling wave

for a lossy lineZc = √z/y,

For a lossless line Zc = √L/C = Rc

V+ / I+ = √L/C and V- / I- = √L/C

41

A transmission line of total length L that terminates in a resistance RL and is driven by a pulse voltage source having an open-circuit voltage waveform Vs(t) as shown and an internal resistance RS. To determine the terminal voltages V(0,t) and V(L,t) and terminal currents I(0,t) and I(L,t) as functions of time we consider the portion of the line at the load). At x = L, we must have

V(L,t) = RL I(L,t)

If RL = RC; in that case there is no backward-traveling wave and the line is said to be matched at the load. But the

discontinuity in the produced by the load

resistor then results in a wave being reflected in the form of a backward-traveling wave.

42

A voltage reflection coefficient at the load is the ratio of the amplitudes of the backward- and forward-traveling voltage waves at x = L;

a current reflection coefficient at the load may be defined as

The current reflection coefficient is thus the negative of the voltage reflection coefficient

44

A 10km long, single-phase short transmission line has 0.5<60°Ω /km impedance. The line supplies a 316.8kW load at 0.8 power factor lagging. What is the voltage regulation if the receiving-end voltage is 3.3kV?

45

Essential Formula

cos θ = power factor (p.f.)Ohm’s law : V = I x Z

Percent Regulation = (Vno load –Vload )/ Vno

load

x 100

cos -1 0.8 = 36.87°and Z = (0.5∠60°)(10) = 5 ∠ 60°Ω

I = [(316.8 × 103)/(3.3 × 103 × 0.8)]<-36.87°= 120 ∠ -36.87°A

IZ = (5 ∠ 60°)(120 ∠ -36.87°) = (551.77 + j235.69)V

VS = (3300 + j0) + (551.77 + j235.69) = (3851.77 + j235.69)V

|VS| = 3858.97V

Percent voltage regulation = [(3858.97 – 3300)/3300] × 100

= 16.94 percent

46

The per-phase impedance of a short transmission line is (0.3 + j0.4) Ω. The sending-end line-to line voltage is 3300V, and the load at the receiving end is 300 kilowatts per phase at 0.8 power factor lagging.

Calculate (a) the receiving-end voltage and

(b) the line current

Essential Formulas

per-phase voltage : Vs= V/ √ 3Ohm’s law : V = I x R ; P = I x VPercent Regulation = (Vno load –Vload )/ Vno load

x 100

Vs2 = (VRcosΦR + RI)2 + (VR sinΦR + XI)2

47

(a) On a per-phase basis,Vs= 3300/ √ 3 = 1905.25V

andI = (300 × 103)/(0.8) VR = (3.75 × 105)/ VR A

Vs2 = (VRcosΦR + RI)2 + (VR sinΦR + XI)2

1905.252 = [0.8 VR + (0.3 × 3.75 × 105)/ VR] 2 + [0.6 VR× (0.4 × 3.75 × 105)/ VR ] 2

VR = 1805V (b) we have

I = (3.75 × 105)/1905 = 207.75A

What is the maximum power that can be transmitted over a three-phase short transmission line having a per-phase impedance of (0.3 + j0.4) Ω if the receiving-end voltage is 6351 volts per phase and the voltage regulation is not to exceed 5 percent?

48

Essential Formulas:

impedance magnitude: Z 2= R2 + X2

per-phase voltage : Vs= V/ √ 3Ohm’s law : V = I x R ; P = I x V

VS = VR + I(R + jX)I = IR + ICR

On a per-phase basis,VR = 6351V

VS = (1 + 0.05)(6351) = 6668.6VZ = √[(0.3) 2 + (0.4) 2] = 0.5 Ω

Pmax = (6351/0.5) 2(0.5 × 6668.6 / 6351 -0.3)

= 36.3MW/phaseand the maximum total power that can be

transmitted is 3× 36.3 = 108.9MW

49

The per-phase parameters for a 60Hz, 200km long transmission line are R = 2.07Ω, L = 310.8mH, and C = 1.4774µF. the line supplies a 100MW, wye-connected load at 215kV (line-t0-line) and 0.9 power factor lagging. Calculate the sending-end voltage, using the nominal-II circuit representation

Essential Formulas:per-phase voltage : Vs= V/√ 3

Ohm’s law : V = I x R ; P = I x VVS = VR + I(R + jX)I = IR + ICR

50

To use the nominal-II circuit, we first express VR and IR per phase as follows:

VR = (215 × 103)/√3 = 124.13kV

IR = (100 × 106)/(√3 × 215 × 103 × 0.9)

= 298.37<-25.8° I CR = V R /X Rc/2

= (124.12 × 103<0°)/[1/(377 × 0.5 × 1.4774 × 10-6) <90°]

= 34.57<90°A

I = I R + I CR = 298.37<-25.8° + 34.57<90°

= 285 ∠ -29.5°A

R + jXL = 2.07 + j377 × 0.3108 » 117.19 ∠88.98° 10-6

I(R + jXL) = 285<-19.5° × 117.19 ∠ 88.98° = 33.4 ∠ 69.48°kV

VS = VR + I(R + jXL) = 124.13<0° + 33.4<69.48°

= 139.39 ∠ 12.97°kV/phase

51

Determine the ABCD constants for the nominal-T circuit of a transmission line for which R = 10 Ω, X = 20 Ω, and Y = 400µS for each phase.

A = D = (1 + ½YZ)= 1 + j[(4 × 10 -4)/2](10 + j20)

= 0.996<0.115° B = Z(1 + ¼YZ)

= (10 + j20)[1 + ¼ (j4 × 10 -4)(10 + j20)] = 22.25<63.45°W

C = Y = j4 × 10 -4 = 4 × 10 -4<90°S

52

A 138-kV three phase short transmission line has a per-phase impedance of (2 +j4) ohms. If the line supplies a 25 MW load at 0.8 power factor lagging, calculate (a) the efficiency of transmission and (b) the sending end voltage and power factor.

ANSWER: (a) 98.78 percent (b) 139.5 kV, 0.99

53

DIRECT & TRANSFER

SURGE CALCULATION

Overhead transmission lines and cables under transient conditions can be represented by series-connected distributed inductances and shunt-connected distributed capacitances.

They can be considered to have a surge impedance, which at any locations relates the surge voltage and surge current by the equation:

Surge voltage = Surge current x Surge

impedance

54

The surge impedance of an overhead transmission line is of the order of 500 OHMs, and that of a cable of the order of 50 OHMs.

With power transformer, the surge impedance varies from 500 OHM to 800 OHM

BEWLEY LATTICE DIAGRAM

Trace the propagation of a travelling wave in overhead transmission lines or cables.

55

Distance between P1 and P2 (ft) =

Travelling wave velocity (ft per microsecond) x Transit time (microsecond)

where the travelling wave is 550ft per microseconds or 167.6 metre per microsecond

The travelling wave with 1 p.u. magnitude on reaching position P2 is reflected back into the line with a reflection coefficient

where and is refracted into further lines beyond position P2 with a refraction coefficient , where

are the surge impedances of the line P1 P2 and that of the further line beyond P2.

56

Example:

A 132 kV cable is to cross a river, one mile wide through a 132 kV overhead line and is to connect to another 132 kV cable at the other side of the river. Draw the Bewleylattice diagram and trace the surge voltage at the other side of the river crossing.

58

LESSON 3


59

Learning objectives

To simplify complex power system using one-line diagram

To represent power system components with per-unit

The basic components of a power system are generators, transformers, transmission lines, and loads. The interconnections among these components in the power system may be shown in a so-called one-line diagram.

For analysis, the equivalent circuits of the components are shown in a reactance diagram or an impedance diagram


60

The advantage of a one-line representation is its simplicity:

One phase represents all three phases of the balanced system;

the equivalent circuits of the components are replaced by their standard symbols;

and the completion of the circuit through the neutral is omitted.

61

One-line diagram that may serve as the basis for a circuit representation that includes the equivalent circuits of the components of the power system

63

1. A generator can be represented by a voltage source in series with an inductive reactance. The internal resistance of the generator is negligible compared to the reactance.

2. The loads are inductive. 3. The transformer core is ideal, and the

transformer may be represented by a reactance. 4. The transmission line is a medium-length line

and can be denoted by a T circuit. An alternative representation, such as a pi circuit, is equally applicable.

5. The delta-wye-connected transformer T1 may be replaced by an equivalent wye-wye connected transformer (via a delta-to-wye transformation) so that the impedance diagram may be drawn on a per-phase basis.

65

Computations for a power system having two or more voltage levels become very cumbersome when it is necessary to convert currents to a different voltage level whenever they flow through a transformer (the change in current being inversely proportional to the transformer turns ratio).

A set of base values, or base quantities, is assumed for each voltage class, and each parameter is expressed as a decimal fraction of its respective base

Per –Unit : then the ratio of actual to base quantity.

A minimum of four base quantities is required to completely defined a per-unit system; these are voltage, current, power, and impedance (or admittance). If two of them are set arbitrarily, then the other two become fixed.

66

Base current = base voltamperes / base voltage (in amperes)

Base impedance = base voltage / base current (in ohms)

Per-unit voltage = actual voltage / base voltage (per unit, or pu)

Per-unit current = actual current / base current (per unit, or pu)

Per-unit impedance = actual impedance / base impedance (per unit, or pu)

Generators and Per Unit System

A generator rated 1000 VA and 200 V has internal impedance of j10 Ω.

67

The ratings of the generator are chosen as the base values.

Ω===

===

=

=

401000

200

S

VZ

A5200

1000

V

SI

V200V

VA1000S

2

base

2

basebase

base

basebase

base

base

The generator impedance is stamped on the nameplate together with the other ratings.

This generator has impedance of j25%. This means that the per unit impedance is

)(.u.p25.0j100

25j

100

ZZ %

.u.p Ω===

68

The actual impedance in Ω is

Zactual = Zp.u. × Zbase = j0.25 × 40 = j10 Ω

Example

The given generator is short circuited at its terminals. Find the short circuit current and the short circuit power delivered by the generator in p.u., in %, and in the actual units

69

)A.(u.p0.4j25.0j

1

Z

EI

.u.pS

.u.pG

.u.psc −=== r

rr

)A(.u.p0.4j)4j(1IES *

.u.psc.u.pG.u.psc =−×=×= ∗rrr

%400j%100II .u.psc%sc −=×=rr

%400j%100SS .u.psc%sc =×=rr

The actual values calculated from the per unit values are:

VA4000j10000.4jSSS

A0.20j0.50.4jIII

base.u.pscsc

base.u.pscsc

=×=×=

−=×−=×=rr

rr

70

Transformers and Per Unit System

A transformer is rated 2000 VA, 200V/400V, and has an internal impedance of j4.0 Ω as seen from the low voltage side.

The internal impedance of the transformer as seen from the high voltage side is

Ω=

×=

= 0.16j

200

4000.4j

V

VZZ

22

1

2

LVHV

rr

71

Bases for transformer low side and high

side

The transformer per unit impedance is the same, regardless of to which side of the transformer it is referred. The transformer equivalent circuit is reduced to a single impedance:

72

Three Phase System and Per Unit

Calculations

In three phase systems the total apparent power is given by

where VLL is the line-to-line voltage and IL is the line current.

LLL3 IV3S =φ

Similarly,

where

S3φ base is the chosen three phase base power, usually the rated power of the equipment,

or the chosen base power used for all system calculation by the power utility

VLL base is the chosen line-to-line base voltage, usually the rated voltage of the equipment

or the nominal voltage level of a part of the power system

IL base is the line base current, usually calculated from the base power and the base

voltage

baseLLbaseLbase3 IV3S =φ

73

∆ and Y Connection in Per Unit

Calculations

The base values are related through the same relationships as the actual quantities:

baseY

baseL

baseLN

baseL

baseLN

base

baseLL

base Z3I

V3

3

I

V3

I

VZ ====

∆∆

.u.pY

baseY

Y

baseY

Y

base

.u.p ZZ

Z

Z3

Z3

Z

ZZ ====

∆

∆∆

74

The per-unit (pu) impedance of a generator or transformer, as supplied by the manufacturer, is generally based on the rating of the generator or transformer itself. However, such a per-unit impedance can be referred to a new voltampere base with the equation

75

If the old base voltage and new base voltage are the same, the equation simplifies to


The impedance of transmission lines are expressed in ohms, but can be easily converted to pu values on a given volt-ampere base .

76

The base impedance and base voltage for a given power system are 10Ω and 400V, respectively.

Calculate the base kVA and the base current.

77

Base current = base voltage / base impedance

= 400/10

= 40A Base kVA = (base voltage x base

current)/1000= (40 × 400)/1000 = 16kVA

The base current and base voltage of a 345-kV system are chosen to be 3000A and 300kV, respectively.

Determine the per-unit voltage and the base impedance for the system.

78

Base impedance = base voltage / base current (in ohms)

= (300 × 103)/3000

= 100Ω

Per-unit voltage = actual voltage / base voltage

= 345/300 = 1.15 pu

A single-phase, 10kVA, 200V generator has an internal impedance Zg of 2Ω. Using the ratings of the generator as base values, determine the generated per-unit voltage that is required to produce full-load current under short-circuit conditions.

79

In per-unit terms, we haveBase voltage = 200V = 1puBase kVA = 10kVA = 1puBase current = Base kVA / Base voltage

= 10,000/200 = 50A = 1pu

The generated voltage required to produce the rated current under short circuit is IZg = 50 × 2 = 100V; or,

in per-unit terms, 100/200 = 0.5pu

Let a 5kVA, 400/200V transformer be approximately represented by a 2 Ω reactance referred to the low voltage side. Considering the rated values as base quantities, express the transformer reactance as a per-unit quantity.

80

Base voltamperes = 5000VA and base voltage = 200V

Base current = 5000/200 = 25A

Base impedance = 200/25 = 8Ω

Then the per-unit reactance referred to the low-voltage side is

Per-unit reactance = 2/8 = 0.25 pu

A three-phase, wye-connected, 6.25kVA, 220V synchronous generator has a reactance of 8.4 Ω per phase.

Using the rated kVA and voltage as base values, determine the per-unit reactance. Then refer this per-unit value to a 230V, 7.5kVA base

81

Base volt-amperes = 6250 = 1pu and base voltage = 220 = 1puBase current = 6250/(√ 3 × 220) = 16.4 =

1pu

Base reactance = 220/16.4 = 13.4 = 1puPer-unit reactance = 8.4/13.4 = 0.627pu

For the 230V, 7.5kVA basePer-unit reactance = 0.627(220/230)2

(7500/6250) = 0.688pu

A three-phase, 13kV transmission line delivers 8MVA of load. The per-phase impedance of the line is (0.01 + j0.05) pu, referred to a 13kV, 8MVA base. What is the voltage drop across the line?

82

The given base quantities yieldBase kVA = 8000 = 1puBase kV = 13 = 1pu

Then the other base quantities areBase current = 8000/13 √ 3 = 355.3 = 1puBase impedance = 13,000/355.3 = 36.6 = 1pu

From these, we find the actual values asImpedance = 36.6(0.01 + j0.05) = (0.366 + j1.83)ΩVoltage drop = 355.3(0.366 + j1.83) = 130 + j650

= 663.1V

A portion of a power system consists of two generators in parallel, connected to a step-up transformer that links them with a 230kV transmission line. The ratings of these components are

Generator G1 = 10MVA, 12 percent reactance Generator G2 = 5MVA, 8 percent reactance Transformer: 15MVA, 6 percent reactance Transmission line: (4 + j60) Ω, 230kVwhere the reactances are computed on the basis

of the individual component ratings. Express the reactances and the impedance in percent with 15MVA as the base value.

83

For generator G1,

Percent reactance = 12(15/10) = 18 percent

For generator G2, Percent reactance = 8(15/5) = 24 percent

For the transformer, Percent reactance = 6(15/15) = 6 percent

And for the transmission line, from (2.2) and (2.7)

Percent impedance = (4 + j60)[(15 × 10 6)/(230 × 10 3) 2] × 100

= (0.113 + j1.7) percent

Draw an impedance diagram for the system shown in the figure, expressing all values as per-unit values.

84

Choose 50kVA to be the base kVA. For generator G1,Zpu = j0.2(2500) 2(50)/(2500) 2(10) = j1.0pu For generator G2, Zpu = j0.3(2500) 2(50)/(2500) 2(20) = j0.75pu For transformer T1, Zpu = j0.1(2500) 2(50)/(2500) 2(40) =

j0.125pu For the transmission line, Zpu = (50 + j200)50,000/8000 2 =

0.039 + j0.156 pu For transformer T2, Zpu = j0.09(10,000) 2(50)/(8000) 2(80) =

j0.088pu For motor M, kVApu = 25/50 = 0.5pu

Resulting reactance diagram

85

Draw a per-unit reactance diagram for the system shown

Impedance Diagram

86

Choosing 20MVA and 66kV as base values.,

for G1, Xpu = j0.15 pu because its percent rectance is 15 percent with the same kVA base.

Also for G2 and G3,Xpu = (20/10)j0.1 = j0.2pu

For T1 and T2, Xpu = (20/30)j0.15 = j0.1pu For T3, Xpu = (20/2.5)j0.08 = j0.64pu For the line, Xpu = Xline(base kVA)/(base

kV)21000= j60(20,000)/(60) 2(1000) = j0.276pu

87

Per Unit Calculations

The per unit value is the ratio of the actual value and the base value of the same quantity

valuebase

valueactualvalueunitper =

88

Change of Base

When pieces of equipment with various different ratings are connected to a system, it is necessary to convert their impedances to a per unit value expressed on the same base.

The base that we are converting from will be denoted by subscript M, the base we are converting to will be denoted by subscript N. The base impedance for the bases M and N are, respectively

baseM

2

baseM

baseMS

VZ =

baseN

2

baseN

baseNS

VZ =

89

Using the MVA and kV notation,

2

baseN

2

baseM

baseM

baseN

.u.pM.u.pN)kV(

)kV(

MVA

MVAZZ ⋅=

Tutorial Sample

90

Example

Specification

91

After referring from 6.6 kV to the 66

kV side

92

G1 & G2, assume common bases 30

MVA and 6.6 kV

93

On the side of 66 kV transmission line,

assume common bases 30 MVA and 66

kV:

On G3, assume common bases 30

MVA and 6.6 kV:

94

Lesson 4

95

Learning objective

To conduct load flow study in a power system

Power-flow studies, more commonly known as load-flow

studies, are extremely important in evaluating the operation of power systems, controlling them, and planning for future expressions.

A power-flow study yields mainly the real and reactive power and phasor voltage at each bus on the system, although much additional information is available from the computer printouts of typical power-flow studies

96

We assume the short transmission line shown in the Figure has negligible resistance and a series reactance of jXohms per phase. The per-phase sending-end and receiving-end voltages are VS and VR, respectively. We wish to determine the real and reactive power at the sending end and at the receiving end, given that VS leads VR by an angle δ.

Complex power S, in voltamperes, is in general given byS = P + jQ = VI* VA

where I* is the complex conjugate of I. Thus, on a per-phase basis, at the sending end we have

Ss = Ps + jQs = VsI* VA From Fig.(a), I is given by

I = (1/jX)(Vs – VR)so

I* = (1/-jX)(Vs* - VR *)

Ss = (Vs/-jX)(Vs* - VR *)

97

Now, from the phasor diagram of Fig. (b),

VR = | VR |<0°so VR = VR *and Vs = |Vs|< δ

Hence

Ss= (|Vs|2 - | VR ||Vs| ejδ)/(-jX)= (|Vs|| VR |/X)sin δ + j(1/X)(|Vs|2 - |Vs|| VR |cos δ)

Finally, since Ss = Ps + jQs, we may write

Ps = (1/X)(|Vs|| VR |sin δ) W and

QS = (1/X)(|Vs|2- |Vs|| VR |cos δ) var

Similarly, at the receiving end we haveVR = PR + j QR = VR I*

And finally obtain

PR = (1/X)(|Vs|| VR |sin δ) W and

QR = (1/X)(|Vs|| VR |cos δ - | VR |2) var

If the system operates with δ » 0, then the average reactive power flow over the line is given by

Qav = (1/2)(Qs + QR) = (1/2X)(| VS| 2 - | VR | 2) var

This equation shows the strong dependence of the reactive power flow on the voltage difference

98

In an actual power system, explicit analytical solutions are not forthcoming because of load fluctuations on the buses and because the receiving-end voltage may not be known. Then, numerical methods must be used to solve for unknown quantities – generally via an iterative procedure

The figure shows a two-bus system, with the real power represented by solid arrows and the reactive power by dashed arrows. The governing equations for the system are (on a per-phase basis)

S2 = V2I*V 1 = V2+ Z lI

Solving for V2 and eliminating I from these equations, we obtain

V2= V 1 – Z lI = V1- Z lS2*/V2* (A)

99

To solve (A) iteratively, we would assume a value for V and call it V (0). We would substitute this in the right-hand side of (A) and solve for V, calling the new value of V, obtained in this first iteration, V (1). We would then substitute (V(1))* in the right-hand side of (A) and obtain a new value V (2). This procedure would be replaced until convergence to the desired precision was achieved.

Definition of the special buses: 1. A load bus is a bus for which the active and reactive

powers P and Q are known, and |V| and δ are to be found. 2. A generator bus is a bus for which the magnitude of the

generated voltage |V| and the corresponding generated power P are known, and Q and δ are to be

obtained. 3. A swing bus (or slack bus) is a generator bus at which

|V| and d are specified, and P and Q are to be determined. For convenience, we choose V< δ = 1<0° per

unit.

100

The kth (of N) nodal current is

The kth (of N) nodal voltage is

One of the more commonly used iterative numerical procedures

The Gauss and Gauss-Seidel methods are iterative procedures for solving simultaneous (nonlinear) equations.

We illustrate the Gauss method with the following example.

101

y – 3x + 1.9 = 0 y + x2 – 1.8 = 0

To solve with the Gauss method, we rewrite the given equations as

x = y/3 + 0.633 (1) y = 1.8 – x2 (2)

We now make an initial guess of x0 = 1 and y0 = 1, update x with (1), and update y with (2). That is, we compute

x1 = y0/3 + 0.633 = 1/3 + 0.633 = 0.9663 (3)

And

y1 = 1.8 – x02

= 1.8 – 1 = 0.8 (4)

After several iterations, we obtain x = 0.938 and y = 0.917. A few more iterations would bring us very close to the exact results: x = 0.93926 and y = 0.9178. However, it must be pointed out that an “uneducated guess” of the initial values (such as x0 = y0 = 100) would have caused the solution to diverge

102

In certain cases the real power on each of the various buses is also specified. The corresponding reactive power is then determined as needed to maintain each bus voltage. Stated otherwise, we investigate the effect of a particular bus voltage on the reactive power supplied by a given generator to the system.

If we represent the system with its Thevenin equivalent and connect the system to a generator as shown in Fig. 1, then the corresponding phasordiagram resembles those shown in Fig. 2 for leading, lagging, and unitypower factors.

Fig. 1

103

Fig. 2

In the three diagrams of Fig. 2 for a constant power delivered by thegenerator, the component of I in phase with ETh must be constant. It follows from Fig. 2 that with constant power input to the bus, larger magnitudes of bus voltage V1 require larger |Eg|, and the larger |Eg| is obtained by increasing the excitation of the generator. Increasing the bus voltage by increasing |Eg| causes the current to become more lagging. Thus, increasing the voltage specified at a generator bus means that the generator feeding the bus will increase its output of reactive power to the bus. From the standpoint of the operation of the system, we are controlling bus voltage and Q generation by adjustingthe generator excitation.

Another method of controlling bus voltage is by installing shunt capacitor banks at the buses at both the transmission and distribution voltage levels along transmission lines or at substations and loads. Each capacitor bank supplies reactive power at the point at which it is placed. Hence it reduces the line current necessary to supply reactive power to the load and reduces the voltage drop in the line via power-factor improvement. Since the capacitor banks lower the reactive power required from the generators, more realpower output is available.

If a capacitor bank is installed at a particular node, the node (or bus) voltage can be determined from the Theveninequivalent of the system, shown in the figure, and the corresponding phasor diagram. The increase in V1 due to the capacitor bank is approximately equal to |IC|XTh if ETh remains constant.

105

For the system shown, it is desired that |V1| = |V2| = 1pu. The loads, as shown, are S1 = 6 + j10pu and S2 = 14 + j8 pu. The line impedance is j0.05 pu. If the real power input at each bus is 10pu, calculate the power and the power factors at the two ends.

Let V2 = 1<0° and V1 = 1<δ. ThenP1 = P2 = |V1||V2|sinδ/X or

10= (1 × 1)sinδ/0.05from which δ = 30° and V1 = 1<30°.

The reactive power is given byQ1 = |V1|2/X - |V1||V2|cosδ/X

= 1/0.05 – cos30°/0.05 = 2.68pu = -Q2so Qline = Q1 – Q2 = 2Q1 = 5.36pu Thus we have

Load on bus 1 = (6 + j10) + (10 + j5.36) = 16 + j15.36 puPower factor at bus 1 = 0.72 laggingLoad on bus 2 = (14 + j8) + (10 +j5.36) = 24 + j13.36 puPower factor at bus 2 = 0.87 lagging

106

Two buses are interconnected by a transmission line of impedance (0.3 + j1.2) pu. The voltage on one bus is 1<0°, and the load on the other bus is (1 + j0.4) pu. Determine the per-unit voltage on this second bus.

Also calculate the per-unit real and reactive power on the first bus

From the above equation

Iteration V2,pu0 1.0 + j01 0.922 – j0.1082 0.903 – j0.1063 0.9 – j0.1084 0.9 – j0.108

107

Then V2 = 0.9 – j0.108 pu and

I = S2*/V2* = (1 – j0.4)/(0.9 + j0.108) = 1.188<-28.6° pu

so that I* = 1.188<28.6°. This gives us

S1 = P1 + jQ1 = V1I1* = (1<0°)(1.188<28.6°)= 1.188<28.6° = 1.043 + j0.569

Hence, P1 = 1.043 pu and Q1 = 0.569 pu

The voltages on the two buses of Question 2 are to be made equal in magnitude by supplying reactive power at the second bus. How much reactive power must be supplied?

108

We have |V1| = 1, V2 = 1<0° (desired), Zl = 0.03 + j0.12, and S2* = (1 – j0.4).

Then from

V1 = V2 + (Zl/V2*)(S2* + jQ‘2) (1)

Where Q‘2 represents the added reactive power at bus 2.

1 = |1 + (0.03 + j0.12)[1 + j(Q2 – 0.4)]|

Hence, Q2 = 0.259 pu.

The per-unit impedance of a short transmission line is j0.06. The per-unit load on the line is (1 + j0.6) pu at a receiving-end voltage of 1<0° pu. Calculate the average reactive power flow over the line.

109

The sending-end voltage isVS = VR + IZ

= 1<0° + (1 + j0.6)(j0.06) = 0.9658<3.56° pu

Thus, from Qav = (1/2X)(|VS|2 - |VR|2) var ,

= [1/2(0.06)](0.96582 – 12)

= -56 pu

Solve the following equation by the Gauss-Seidel method: x2 – 6x + 2 = 0.

110

We solve the given equation for a, obtainingx = (1/6)x2 = 1/3 = F(x)

and use the initial estimate x(0) = 1. Then, in succeeding iterations we obtain

Iteration 1: x(1) = F(1) = 1/6 + 1/3 = 0.5 Iteration 2: x(2) = F(0.5) = (1/6)(0.5) 2 + 1/3 = 0.375 Iteration 3: x(3) = F(0.375) = (1/6)(0.375) 2 + 1/3 = 0.3568 Iteration 4: x(4) = F(0.3568) = (1/6)(0.3568) 2+ 1/3 = 0.3545 We may now stop since |x(n+1)| - |x(n)| < Î = 0.0023; which seems

to be sufficiently small. The quadraticgives this root as x = 0.35425 to five places

For the two-bus system and with the data as shown and with Y11 = Y22 = 1.6<-80° pu and Y21 = Y12 = 1.9<100° pu, determine the per-unit voltage at bus 2 by the Gauss-Seidel method.

111

The power into the two buses isS1 = (P 1 – 1.1) + j(Q 1 – 0.4) pu

S 2 = -0.5 – j0.3 pu

From

V 2(i+1) = (1/Y22)[(P 2– jQ 2)/(V 2(i))* - Y21V 1] (1)

With the given numerical values, (1) becomesV 2

(i+1) = (1/1.6<-80°)[0.583<149°/(V 2 (i))* -(1.9<100°)(1.1<0°)]= (0.625<80°)[0.583<149°/(V 2 (i))* - 2.09<100°] (2)

To begin the iterations, we let V 2(0) = 1.0<-10° puV 2

(1) = (0.625<80°)[(0.583<139°) – (2.09<100°)] = 1.048<-12.6° pu

The next iteration yieldsV 2

(2)) = (0.625<80°)(0.583<149°/1.048,12.6° - 2.09<100°)= 1.047<-8.6° pu

112

Power System Economics

LESSON 5

Learning objectives

a. To identify the different processes involved in flow and congestion control

113

Of the numerous aspects of power system operation and control, we shall consider only the economical operation of power systems and the control of load frequency, generator voltage, and the turbine governor

Within a power plant, a number of ac generators generally operate in parallel. For the economic operation of the plant, the total load must be appropriately shared by the generating units. Because fuel cost is the major factor in determining economic operation, curves like that of Figure 1 are important to power-plantoperation.

114

Note in the figure that the inverse slope of the curve at any point in the fuel efficiency of the generating unit operating at that point. Maximum fuel efficiency occurs at the point at which the line from the origin is tangent to the curve. Point A in Fig. 1 is such a point for a unit having the input-output characteristic of Fig. 1; there, an output of 250 MW requires an input of approximately 2.1 × 109 Btu/h. Or, we may say that the fuel requirement is 8.4 × 106 Btu/MWh.

115

To obtain the most economic load distribution between two units, we must determine the incremental cost corresponding to a partial shift of load between the units. We first convert the fuel requirement into a dollar cost per megawatthour. Then the incremental cost is determined from the slopes of the input-output curves (Fig.1) for the two units.

From Fig. 1, for each unit,

Incremental fuel cost = dF/dP (in dollars per megawatthour)

where F = input in dollars per hour, P = output in megawatts.

At a given output, this incremental fuel cost is the additional cost of increasing the output by MW.

In a plant having two operating units, generally the incremental fuel cost of one unit will be higher than that of the other. For the most economic operation, load should be transferred from the unit with the higher incremental costs of the two become equal. In a plant with several units, the criterion for load division is that all units must operate at the same incremental fuel cost.

116

To include the effect of transmission-line losses on economical system operation, we must express these losses as a function of plant power output. Fig. 2 shows two plants connected to a three-phase load.

The total transmission loss (for all three phases) is given byPloss = 3(|I1|2R 1 + |I 2|2R 2 + |I 3|2R 3)

where the R’s are the per-phase resistances of the lines, and where

|I 3 | = |I 1 + I 2 | = |I 1 | + |I 2 |) if we assume that I 1 and I 2 are in phase. These

currents may be expressed in terms of P 1 and P 2, the respective plant outputs, as

|I 1 | = P 1 / √ 3|V 1 |cos Φ 1|I 2 | = P 2 / √ 3|V 2 |cos Φ 2

where cos Φ 1 and cos Φ 2 are the power factors at buses 1 and 2, respectively.

117

Ploss = P 1 2 B 11 + 2 P 1 P 2B 12 + P 22

2 B 22

where B 11, B 12, and B 22 are called loss

coefficients or B coefficients and are given by

B 11 = (R 1 + R 3)/|V 1 | 2 cos 2 Φ 1

B 12 = R 13/|V 1 ||V 12|cos Φ 1 cos Φ 2

B 22 = (R 2 + R 3)/|V 12| 2 cos 2 Φ 2For a system of n plants,

For a system of n plants, the total cost of fuel per hour is

and the total power output is

With transmission losses, we must have

118

where P R and P loss are, respectively, the total power received by the load and lost in transmission.

For a given (constant), P R, dP R = 0.

In addition, when the load is allocated among the n plants for minimum fuel cost, dF total = 0. Then

119

The system fuel cost is minimized when the incremental fuel cost for each plant, multiplied by its penalty factor, is the same throughout the system, that, when

To determine the L K

120

A number of automatic controls are used in present-day power systems. These include devices that control the generator voltage, the turbine governor, and the load frequency; there are also computer controls to ensure economic power flow and to control reactive power, among other power-system variables.

Generator voltage control is accomplished by controlling the exciter voltage. The block-diagram representation of a closed-loop automatic voltage regulating system is shown in simplified form

121

The open-loop transfer function G(s) is given by

G(s) = k/(1 + T as)(1 + T es)(1 + T fs)

where T a, T e, and T f are, respectively, the time constants associated with the amplifier, the exciter, and the generator field, and the open-loop gain k is

k = kakekf

Sudden changes in the load cause the turbine speed and, consequently, the generator frequency to change as well. The change in turbine speed occurs when the generator electromagnetic torque no longer equalsthe turbine (or other prime mover) mechanical torque. Thus, the change ∆f in the generator frequency may be used as a control signal for controlling the turbine mechanical output power. The change in the turbine output power as a function of a change in the generator frequency is given by

∆Pm = ∆Pref – (1/R) ∆f

122

where ∆Pm and ∆Pref are, respectively, the changes in the turbine output power and the reference power (as determined by the governor setting), and R is known as the regulation constant.

It is implied that the accelerations and decelerations of the generator rotor are controlled by the turbine governor. However, the frequency deviation ∆f still remains if ∆Pref = 0.

123

Use Fig. 1 to find the fuel requirements for outputs of (a) 100MW and (b) 400MW. Thus verify that point A is probably the maximum fuel-efficiency point.

Hint: We have to show that the fuel requirement is higher than that of point A, that is, 8.4 × 106 Btu/MWh.

124

(a) From Fig. 1, at 100MW output, the fuel input is approximately 1 × 109 Btu/h. Hence,

Fuel requirement = (1 × 109)/100 = 10 × 106 Btu/MWh

(b) Similarly, at 400 MW output, the fuel input is approximately 3.6 × 109 Btu/h. Then

Fuel requirement = (3.6 × 109)/400 = 9.0 × 106 Btu/MWh

Clearly, both values are greater than that for point A.

A certain amount of coal costs $1.20 and produces 106 Btu of energy as fuel for a generating unit. If the input-output characteristic of the unit is that shown in Fig. 1, determine the incremental fuel cost at point A.

Hint: We have to solve for the slope of the line at point A to solve for the incremental cost.

125

Slope at A = (2.2 – 2.0)109/(260 – 234)106

= 7.7 Btu/MWh Thus, Incremental cost = 7.7 × 1.20

= $9.24/MWh

126

For a certain turbine-generator set, R = 0.04pu, based on the generator rating of 100MVA and 60Hz. The generator frequency decreases by 0.02Hz, and the system adjusts to steady-state operation. By how much does the turbine output power increase?

The per-unit frequency change isPer-unit ∆f = ∆f/fbase = -0.02/60

= -3.33 × 10-4 pu Then from ∆Pm = ∆Pref – (1/R) ∆f

yieldsPer-unit ∆Pm = -(1/0.04)(-3.33 × 10-4)

= 8.33 × 10-3 pu

The actual increase in output power is then∆Pm = (8.33 × 10-3)(100) = 0.833MW

127

Graphs of the incremental fuel costs (in dollars per megawatthour) for two generating units in a power plant are shown in Fig. A. These graphs are linear. The plant output ranges from 240MW to 1000MW over a 24-h period. During this period the load on each unit varies from 120MW to 600MW. Plot a curve of incremental fuel cost l versus plant output for minimum-fuel-cost operation.

128

From the figure, we obtaindF1/dP1 = 0.008P1 + 8 (1)dF2/dP2 = 0.009P2 + 6 (2)

At 120MW, dF1/dP1 = 8.96 and dF2/dP2 = 7.08. Therefore, until dF2/dP2 has risen to 8.96, unit 2 should take all the additional load above 120MW.

Using (2), we find that dF2/dP2 is equal to 8.96 when P2 = 328.9MW, at which value

Ptotal = P1 + P2 = 120 + 328.9 = 448.9MW

These values give us the second row of the Table . Similar computations give the remaining rows of the table, whose values are plotted in the figure (dashed lines)

129

For maximum demand from the plant of Question 4, determine how the load should be shared by the two generating units for minimum fuel cost.

Hint: For minimum fuel cost, incremental cost for each plant must be equal, that is,

dF1/dP1 = dF2/dP2

The maximum load isPtotal = 1000 = P1 + P2

For minimum fuel cost, 0.008P1 + 8 = 0.009P2 + 60.008P1 + 8 = 0.009P2 + 6 (2)

Solving ( for P1 and P2) yieldsP1 = 411.76MW and P2 = 588.24MW

130

Determine the incremental fuel cost for each unit for the conditions of Question No. 5

The incremental fuel cost λ is the same for both units.

From dF1/dP1 = 0.008P1 + 8 of Question 4

With P1 = 411.76MW, λ = 0.008 × 411.76 + 8 = $11.29/MWh

Prepared by; Mr. John Mateo

131

Symmetrical Fault Analysis

LESSON 6

Learning objectives

To calculate fault current in a three-phase circuit due to a symmetrical fault

132

The operation of a power system departs from normal after the occurrence of a fault. Faults give rise to abnormal operating conditions – usually excessive currents and voltages at certain points on the system –which are guarded against with various types of protective equipment.



133

1. Determination of the maximum and minimum three-phase short-circuit currents

2. Determination of unsymmetrical faults currents, as in single line-to-ground, double line-toground, line-to-line, and open-circuit faults

3. Determination of the ratings of required circuit breakers

4. Investigation of schemes of protective relaying 5. Determination of voltage levels at strategic points

during a fault


A balanced three-phase short circuit) is an example of a symmetrical fault. Balanced three-phase fault calculations can be carried out on a per-phase basis, so that only single-phase equivalent circuits need be used in the analysis. Invariably, the circuit constants are expressed in per-unit terms, and all calculations are made on a per-unit basis.

134

In short-circuit calculations, we often evaluate the shortcircuit MVA (megavolt-amperes), which is equal to

(√3)VlIf,

where Vl is the nominal line voltage in kilovolts, and

If is the fault current in kiloamperes.

135

The symmetrical trace of a short-circuited stator-current shown, may be divided into three periods or time regimes:

the subtransient period, lasting only for the first few cycles, during which the current decrement is very rapid;

the transient period, covering a relatively

longer time during which the current

decrement is more moderate; and

finally the steady-state period

Suppose now that a generator is loaded when a fault occurs. shows the corresponding equivalent circuit with the fault to occur at point P. The current flowing before the fault occurs is IL, the voltage at the fault is Vf, and the terminal voltage of the generator is Vt.

136

When a three-phase fault occurs at P, the circuit shown becomes the appropriate equivalent circuit (with switch S is closed).

Here a voltage Eg" in series with Xd" supplies the steady-state current IL when switch S is open, and supplies the current to the short circuit through Xd" and Zext when switch S is closed . If we can determine Eg", we can find this current through Xd", which will be i".

With switch S open, we haveEg" = Vt + jILXd" (

which defines Eg", the subtransient internal voltage.

Similarly, for the transient internal voltage we haveEg' = Vt + jILXd'

Clearly Eg" and Eg' are dependent on the value of the load before the fault occurs

137

The figure shows the one-line diagram for a single phase of a system in which a generator supplies a load through a step-up transformer, a transmission line, and a step-down transformer. Calculate the per-unit current. The transformers are ideal.

138

Because the voltage (and current) levels change across the transformers, different base voltages are involved at different parts of the system.

At the generator,Vbase,gen = 480V = 1pukVAbase,gen = 20kVA = 1puIbase,gen = 20,000/480 = 41.67A = 1pu

Zbase,gen = 480/41.67 = 11.52W = 1pu

Along the transmission line,Vbase,line = 480/0.5 = 960V = 1pu

kVAbase,line = 20kVA = 1puIbase,line = 20,000/960 = 20.83A = 1pu

Zbase,line = 960/20.83 = 46.08W = 1puZline = (1 + j3)/46.08 = (0.022 + j0.065)pu

At the loadVbase,load = 960/10 = 96V = 1pukVAbase,load = 20kVA = 1puIbase,load = 20,000/96 = 208.3A = 1pu

Zbase,load = 96/208.3 = 0.4608W = 1puZload = (2 + j5)/0.4608 = (4.34 + j10.85)pu

The total impedance is thenZtotal = Zline + Zload

= (0.022 + j0.065) + (4.34 + j10.85)= 4.362 + j10.915 = 11.75Ð68°

so that I = 1<0°/Ztotal = 1<0°/11.75<68° = 0.085<68

139

An interconnected generator-reactor system is shown in Figure. The base values for the given percent reactances are the ratings of the individual pieces of equipment. A three-phase short-circuit occurs at point F. Determine the fault current and the fault kVA if the busbar line-to-line voltage is 11kV

140

Per phase reactance diagram

Simplified into:

First we arbitrarily choose 50MVA as the base MVA, and find per-unit values for the system reactances, referred to this base. We obtain

XG1 = (50/10)0.10 = 0.5puXG2 = (50/20)0.15 = 0.375puXG3 = (50/20)0.15 = 0.375puX1 = (50/10)0.5 = 0.25puX2 = (50/8)0.4 = 0.25pu

The total reactance from the neutral to the fault at F is, from that diagram,Per-unit reactance

= j[0.5(0.2344 + 0.25)]/[0.5 + (0.2344 + 0.25)] = j0.246Fault MVA = (50 × 103)/0.246 = 203.25MVAFault current = (203.25 × 106)/(√3 × 11 × 103) = 10,668A

141

The phase currents in a wye-connected, unbalanced load are Ia = (44 – j33), Ib = -32 + j24), and Ic = (-40 + j25)A. Determine the sequence currents.

I a0 = 1/3 [(44 – j33) – (32 + j24) + (-40 + j25)]

= -9.33 – j10.67 = 14.17<-131.2°A

I a1 = 1/3 [(44 – j33) – (-0.5 + j0.866)(32 + j24) – (0.5 – j0.866)(-40 + j25)]

= 40.81 – j8.77 = 41.7<-12.1°A I a2 = 1/3 [(44 – j33) + (0.5 – j0.866)(32 +

j24) + (-0.5 + j0.866)(-40 + j25)]= 12.52 – j13.48 = 18.37<-47°A

142

The system shown in figure is initially on no load. Calculate the subtransient fault current that results when a three-phase fault occurs at F, given that the transformer voltage on the high-voltage side is 66kV.

Let the base voltage (on the high side) be 69kV, and the base kVA be 75MVA. Then, in per-unit terms we have,

for generator G 1,X d” = 0.25(75,000/50,000) = 0.375puE G1 = 66/69 = 0.957pu

For generator G2,X d” = 0.25(75,000/25,000) = 0.750puE G2= 66/69 = 0.957pu

For the transformer, X = 0.10pu.

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The reactance diagram for the system before the fault occurs; the fault is simulated by closing switch S. The two parallel subtransient reactances are equivalent to a reactance

X sub = (0.375 × 0.75)/(0.375 + 0.75) = 0.25puTherefore, as a phasor with E G1 as reference, the subtransient current in the short circuit is

i" = 0.957/(j0.25 + j0.10) = -j2.735pu

Example no 5

A three-phase short-circuit fault occurs at point

F in the system shown. Calculate the fault

current.

144

Let the base MVA be 30 MVA and let 33 kV be the

base voltage. The reactances and the impedances in

pu will be:

XG1 = 0.15 ( 30 / 20) = 0.225 pu

XG2 = 0.10 ( 30 / 10) = 0.3 pu

Xtrans = 0.005 ( 30 / 30) = 0.05 pu

Zline = (3 + j15) ( 30 / 332) = (0.086 + j0.4132) pu

With X = (XG1 XG2/ XG1 +XG2) = (j0.225)(j0.3)/j0.2225 +j0.3)

X = j0.079 pu

Per unit Ztotal = 0.0826 + j.079+j0.05+j0.4632 = 0.0826+j.5918

= 0.5975 82°

Then short circuit kVA = 30 x 103 / .5975 = 50.21 MVA

And Short circuit current = (50.21 x 106)/(√3 x 33 x 103)

= 875.5 A

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Symmetrical Components

Lesson 7

Learning objectives

To transform unbalanced voltage and current components into balanced parameters

146

Method of Symmetrical Components

This method is based on the fact that a set of three-phase unbalanced phasors can be resolved into three sets of symmetrical components, which are termed the positive sequence, negative-sequence, and zero-sequence components.

a. positive phase sequence system – a balanced three-phase system in the normal sequence

b. negative phase sequence system - a balanced three-phase system in the reversed sequence

c. zero phase sequence system – three phasorsequal in magnitude and direction revolving in the positive phase

147



In particular, we haveV a= V a1 + V a2 + V a3V b= V b1 + V b2 + V b3V c = V c1 + V c2 + V c3

We now introduce an operator a that causes a counterclcokwiserotation of 120° (just as the j operator produces a 90°rotation), such that

1 + a + a 2 = 0

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V b1 = a 2V a1

V c1 = aV a1

V b2 = aVa2

V c2 = a2V a2

V a0 = V b0 = V co

in terms of components of phase a,V a = V a0 + V a1 + Va2Vb= V a0 + a2 a1 1 + aVa2Vc = V a0 + aV a1 + a2Va2

components of a given sequence in terms of any chosen component.


Solving for the sequence components of aV a0 = 1/3 (V a + V b + Vc) V a1 = 1/3 (V a + aVb + a2Vc) V a 2= 1/3 (V a + a2Vb + aVc)

A quantity (current, voltage, impedance, power) that is given in terms of symmetrical components is sometimes called the sequence quantity, as in “sequence current.”

149

S = V a I a * + VbI b* + V cI c*

= 3(V a0I a0 * V a1 I a1 * V a2 I a2 *)

The sequence power is one-third the power in terms of phase quantities


The phase currents in a wye-connected, unbalanced load are Ia = (44 – j33), Ib = -(32 + j24), and Ic = -(40 + j25)A. Determine the sequence currents.


150

I a0 = 1/3 [(44 – j33) – (32 + j24) + (-40 + j25)]

= -9.33 – j10.67 = 14.17∠-131.2°A I a1 = 1/3 [(44 – j33) – (-0.5 + j0.866)(32 + j24) –

(0.5 – j0.866)(-40 + j25)]= 40.81 – j8.77 = 41.7<-12.1°A

I a2 = 1/3 [(44 – j33) + (0.5 – j0.866)(32 + j24) + (-0.5 + j0.866)(-40 + j25)]

= 12.52 – j13.48 = 18.37 ∠ -47°A


Example

152

Example

153

Example

156

Unsymmetrical Faults

Lesson 8

Learning objectives

To calculate fault current in a three-phase circuit due asymmetrical fault

157

SEQUENCE NETWORKS OF

GENERATORS

(1) Since the generator is designed to supply balanced three phase voltages, the generator voltages are of positive sequence only and hence only appear in the positive sequence network

(2) Since the neutral impedance connected to the generator neutral carries three zero sequence current, the neutral generator impedance is represented by three times its value in the zero sequence network

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Transformer phase representation

Zero-sequence currents free to flow in both primary and secondary circuits

Wye-wye grounded

No path for zero-sequence currents in primary circuit

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Delta-wye

Single phase currents can circulate in the delta but not outside it

Delta - wye

No flow of zero-sequence currents possible

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Delta-delta

No flow of zero-sequence currents possible

Tertiary windings provides path for zero-sequence currents

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Types of Faults

Single line-to-earth fault

Io=1/3(Ia +Ib + Ic)

I1 = 1/3(Ia +a2Ic + aIb) I2 = 1/3(Ia +a2Ib + aIc) Io=I1=I2 = Ia/3

Ib = Ic = 0

Va = E - I1Z1 – I2Z2 -1oZo = 0 E = I1Z1 – I2Z2 -1oZo

If = 3E / (Z1 + Z2 + Zo)

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Sequence inteconnection

Line –to-line fault

164

Io=0 I1 = 1/3 Ib(a - a2)

I2 = 1/3 Ib(a2 - a) I1= E/ (Z1 + Z2)

Sequence inteconnection

165

Line –to-line-to-earth fault

Ia = 0 I1 = E / (Z1 + Z2Zo/(Z2 +Zo)

I2 = - I1 (Zo /(Zo+ Z2 ))

I0 = - I1 (Z2/(Zo+ Z2 ))

If = 3E / (Z1 + 3Zg +Z2 + Zo)

166

Example

A synchronous machine A generating 1 p.u. voltage is connected through star-star transformer, reactance 0.12 p.u. to two lines in parallel. The other ends of the lines are connected through star-star transformer of reactance 0.1 p.u. to a second machine B, also generating 1 p.u. voltage. For both transformers, X1 = X2 = Xo.

167

The relevant per unit reactances of the plant, all referred to the sme base are as follows:

For each line, X1=X2=0.3; X0=0.7

The star points of machine A and B are solidly earthed.

X1 X2 X0

Machine A 0.3 0.2 0.05

Machine B 0.25 0.15 0.03

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Positive- sequence

Negative sequence

169

Zero-sequence

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Example

A 75 MVA 115/33 kV power transformer at the P.T.T. Central Substation in Rayong, Thailand, developed a line-to-ground short-circuit fault at a high voltage winding terminal.

The percentage impedance of the transformer is 12% at 75 MVA, and the transformer connection is delta on the HV windings and star on the LV windings. If the supply impedance of 115 kV powersupply is 3% based on 75 MVA, draw the sequence impedance networks for the ground fault, and calculate the ground fault current. The 115 kV network can be considered as a star connection with centre point solidly earthed.

171

For a single line-to-ground fault on the “A” phase, the positive, negative and zero sequence networks are to be connected in series in order to calculate the sequence short-circuit current,

172

Total sequence impedances = 3 x 3% = 9%If = 1/9% = 11.11 p.u.

Short-circuit current on “A” phase = 3 x 11.11 p.u. = 33.33 x 376.5= 12,549 Ampere

Example 1

Three-phase short-circuit fault at terminals of generator G1 in power system of figure 1.

There are three fault current infeeds: One fault infeed from generator G1, one fault infeed from generator G2 and one fault infeedfrom generator G3.

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Example

Three-phase short-circuit fault at transformer T1 66 kV circuit-breaker terminal in power system of figure 1.

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Example

With reference to the power system of figure 1, the neutral point reactors at the three generators L1, L2 and L3 are all 5 OHMs.

Calculate the fault current for a single line-to-ground short-circuit at transformer T1 66 kV circuit-breaker terminal, that is interrupted by the circuit-breaker.

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LESSON 9

Learning objectives

a. To conduct power system stability study on a power system

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By the stability of a power system we mean

the ability of the system to remain in

operating equilibrium, or synchronism, while disturbances occur on the system. Three types of stability are of concern: steadystate, dynamic, and transient stability

Steady-state stability relates to the response of a synchronous machine to a gradually increasing load.

Dynamic stability involves the response to small disturbances that occur in the system, producing oscillations. If these oscillations are of successively smaller amplitudes, the system is considered dynamically stable. If the oscillations grow in amplitude, the system is dynamically unstable. The source of this type of instability is usually an interaction between control systems. The system’s response to the disturbance may not become apparent for some 10 to 30s.

180

Transient stability involves the response to large disturbances, which may cause rather large changes in rotor speeds, power angles, and power transfers. The system’s response to such a disturbance is usually evident within 1s.

The angular momentum and initial constant play an important role in determining the stability of a synchronous machine. The per-unit inertia constant H is defined as the kinetic energy stored in the rotating parts of the machine at synchronous speed per unit megavoltampere (MVA) rating of the machine. Thus, if G is the MVA rating of the machine, then

GH = ½ Jωs2

181

where J is the polar moment of inertia of all rotating parts in kilogram-(meters-squared), and

ωs is the angular synchronous velocity in electrical radians per second

If M is the corresponding angular momentum, then

M = J ωs

Since ωs = 360f electrical degrees per second,

GH = ½ M ωs = ½ M(360)for M = GH/180f MJ.s/electrical degree

where f is the frequency of rotation.

182

Consider a synchronous generator developing an electromagnetic torque T e(and a corresponding electromagnetic power P e) while operating at the synchronous

speed ωs. If the input torque provided by the prime mover at the generator shaft is T i, then under steady-state conditions (with no disturbance) we have

T e = T IT e ωs = T i ωsT i ωs - T e ωs = P i – Pe = 0

If a departure from steady state occurs, such as a change in load or a fault, the “power in” P i no longer equals the “power out” Pout, and if P i – Pe = 0. Instead, an accelerating torque comes into play. If P a is the corresponding accelerating (or decelerating) power, then

Where Pa is in megawatts, and θ is the angular position of the rotor

The per-unit swing equation is

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An inertia constant H mach based on a machine’s own MVA rating may be converted to a value H systrelative to the system base S syst with the formula

H syst = H mach (S mach /S syst)

A convenient system base value is 100MVA.

The moment of inertia of a synchronous machine is given by WR 2/32.2 slug-(feet-squared), where W is the weight of the rotating part of the machine in pounds, and R is its radius of gyration in feet. Machinery manufacturers generally supply the value of WR 2for their machines.

The swing equation may be written for two machines as

where the subscripts 1 and 2 correspond to machines 1 and 2, respectively. If we denote the relative angle between the two rotor axes by δ, such that δ= δ1 - δ 2 , then

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Algorithm for the Iterations

Critical Clearing Angle

If a disturbance (or fault) occurs in a system, δ begins to increase under the influence of positive accelerating power, and the system will become unstable if δ becomes very large.

There is a critical angle within which the fault must be cleared if he system is to remain stable and the equal-area criterion is to be satisfied. This angle is known as the critical

clearing angle δc.

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For example, consider a system that normally operates along curve A in Fig. If a three-phase short circuit occurs across the line, its curve of power versus power angle will correspond to the horizontal axis. For stability, the critical clearing angle must be such that area A1 = area A2.

187

The inertia constant H for a 60Hz, 100MVA hydroelectric generator is 4.0MJ/MVA. How much kinetic energy is stored in the rotor at synchronous speed? If the input to the generator is suddenly increased by 20MVA, what acceleration is imparted to the rotor?

The energy stored in the rotor at synchronous speed is

GH = 100 × 4 = 400MJ

Using

with Pa = 20MVA of accelerating power and with

M = GH/180f = 400/(180 × 60) = 1/27

d2θ/dt2 = 20 × 27 = 540°/s2

M(d2θ/dt2) = Pi – Pe = Pa

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A 500MVA synchronous machine has H1 = 4.6MJ/MVA, and a 1500MVA machine has H2=3.0MJ/MVA. The two machines operate in parallel in a power station. What is the equivalent H constant for the two, relative to a 100MVA base?

The total kinetic energy of the two machines is

KE = 4.6 × 600 + 3 × 1500 = 6800MJ

Thus, the equivalent H relative to a 100MVA base is

H = 6800/100 = 68MJ/MVA

189

A 100MVA, two-pole, 60Hz generator has a moment of inertia of 50 × 103kgm2. What is the energy stored in the rotor at the rated speed? What is the corresponding angular momentum? Determine the inertia constant H.

The stored energy isKE(stored) = ½ Jωm2

= ½ (50 × 103)[(2 × 3600)/60] 2

= 3553MJ Then

H = KE(stored)/MVA = 3553/100 = 35.53MJ/MVA

M = GH/180f = (100)(35.53)/(180)(60) = 0.329MJ.rad/s

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The kinetic energy stored in the rotor of a 50MVA, six-pole, 60Hz synchronous machine is 200MJ. The input to the machine is 25MW at a developed power of 22.5MW. Calculate the accelerating power and the acceleration

The accelerating power isPa = Pi – Pe = 25 – 22.5 = 2.5MW

Now, also,H = KE(stored)/machine rating in MVA

= 200/50 = 4 M = GH/180f = (50 × 4)/(180 × 60)

= 0.0185MJ.s/degree= 1.06MJ.s/rad

d2δ/dt2 = 2.5/10.06 = 2.356 rad/s2

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If the acceleration of the machine of Question no 4 remains constant for ten cycles, what is the power angle at the end of the ten cycles?

d2δ/dt2 = 2.356. Integration with respect to t yields

dδ/dt = 2.356t + C1

Since dδ/dt = 0 at t = 0, C1 = 0. A second integration now gives

δ= 1.178t2 + C2

At t = 0, let δ =δ0 (the initial power angle). Then

δ = 1.178t2 + δ0 At 60Hz, the time required for ten cycles is t =

1/6 s. For this value of t,δ= 1.178(1/6)2 + δ0 = (0.0327 + δ0) rad

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Question no 6

The generator of Question 4 has an internal voltage of 1.2 pu and is connected to an infinite bus operating at a voltage of 1.0pu through a 0.3pu reactance. A three-phase short circuit occurs on the line. Subsequently, circuit breakers operate and the reactance between the generator and the bus becomes 0.4pu. Calculate the critical clearing angle.

Before the fault, Pmax = (1.2 × 1.0)/0.3 = 4.0 pu

During the fault, Pmax2 = 0 and k1 = 0 for use in . After the fault is cleared, Pmax3 = (1.2 × 1.0)/0.4 = 3.0pu and k2 = 3.0/4.0 for use in

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The initial power angle δ 0 is given by 4sin δ 0 = 1.0,

δ 0 = 0.2527 rad. Define δ 'm = p - δ m (see Fig.). The angle δ

m in is obtained from

sin δ 'm = 1/3.0 and δ m = π - δ 'm

δ m = 2.8 rad.

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Substituting k1, k2, δ 0 and δ m in

cos δ c = (1/0.75)[(2.8 – 0.2527)0.25 – 0 + 0.75cos2.8]

= -0.093

from which δ c = 95.34°

A304SE PPT Slides T111_final

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