power sys. analysis.pdf

EE 0308 POWER SYSTEM ANALYSIS

Dr.R.Jegatheesan Professor, EEE Dept.

POWER SYSTEM OVERVIEW

Power system components, Representation. Single line diagram, per

phase analysis of symmetrical three phase systems, general aspects

relating to power flow, short circuit and stability analysis, per unit

quantities, impedance diagram.

POWER FLOW ANALYSIS

Primitive network, Formation of Bus admittance matrix by inspection

method and singular method. Bus classification – Formulation of

Power Flow problems – Power flow solution using Gauss Seidel

method and Newton Raphson method, Comparison between these

methods. Handling of Voltage controlled buses, Off nominal

transformer ratios and phase shifting transformer.

SYMMETRICAL FAULT ANALYSIS

Formation of Bus Impedance matrix by Z – bus building algorithm. Types of faults

in power systems – Symmetrical fault analysis. Short circuit capacity –

Symmetrical fault analysis through bus impedance matrix.

UNSYMMETRICAL FAULT ANALYSIS

Symmetrical components – sequence impedance – sequence networks,

Introduction to unsymmetrical faults. Single line to ground fault, line to line and

double line to ground faults – Unsymmetrical fault analysis using bus impedance

matrix.

POWER SYSTEM STABILITY

Introduction to stability studies – classification of stability – Transient stability –

Equal area criterion – critical clearing angle and time – Factors affecting transient

stability. Numerical integration method: Euler method, Modified Euler’s method,

Fourth order Runge Kutta method.

TEXT BOOKS

1. Hadi Sadat, Power System Analysis, Tata Mc Graw Hill Publishing

company, New Delhi, 2002.

2. John.J.Grainger, William D. Stevenson, Power System Analysis, Tata Mc

Graw Hill Publishing company, New Delhi, 2003.

REFERENCE BOOKS

1. Nagarath I.J. and Kothari D.P. Modern Power System Analysis, Tata Mc

Graw Hill Publishing company, New Delhi, 2002.

2. Wadhwa C.L. Electrical Power Systems, New age international (P) Ltd.

Publishers, 1995.

3. Pai M.A. Computer Techniques in Power System Analysis, Tata Mc Graw

Hill Publishing company, New Delhi, 2003.

EE 0308 POWER SYSTEM ANALYSIS

Chapter 1

POWER SYSTEM OVERVIEW

Introduction

A power system consists of a few generating plants, situated close to resources,

supplying electric power to various types of loads spread out over large area,

through large complex transmission and distribution network. Thus a power

system compose of

1. Generation system

2. Transmission system

3. Distribution system

4. Loads

Depending on the fuel used we have Hydro-Electric Power Plants, Thermal Power

Plants and Nuclear Power Plants. Generated supply will be of 11 kV. To have

greater efficiency, transmission is carried out at high voltages of order 230 kV or

400 kV. Power transformers are used to setup the voltage levels. Transmission

system consists of transformers, transmission towers and transmission lines.

Thereafter, voltage levels are reduced in stages. Distribution system supplies

power to different loads. Thus power system network is large, complex and very

expensive. Power system analysis deals with analysis problems associated with

power network. Power Flow Analysis, Short Circuit Analysis and Transient

Stability Study are the main Power System Analysis Problems.

CE

CI

AI

BI

BE

Z

n

AE

0 Z Z

Single line diagram or One-line diagram

Electric power systems are supplied by three-phase generators. Ideally, the generators

are supplying balanced three phase loads. Fig.1.1 shows a star connected generator

supplying star connected balanced load.

Fig. 1.1 Y- connected generator supplying balanced Y- connected load

Z

n

AE

0

AI

A balanced three-phase system is always solved as a single-phase circuit

composed of one of the three lines and the neutral return. Single-phase circuit of

three-phase system considered above is shown in Fig. 1.2.

Often the diagram is simplified further by omitting the neutral and by indicating

the component parts by standard symbols rather than by their equivalent circuits.

Such a simplified diagram of electric system is called a one-line diagram. It is

also called as single line diagram. The one-line diagram of the simple three-phase

system considered above is shown in Fig. 1.3.

Fig. 1.2 Single-phase circuit

Z

n

AE

0

AI

Fig. 1.4 shows the one-line diagram of a sample power system.

Fig. 1.3 One-line diagram

Load

Fig. 1.2 Single-phase circuit

Load A

2 3

1

Load B

T2 T1

Fig. 1.4 One-line diagram of a sample power system

Fig. 1.4 shows the one-line diagram of a sample power system.

This system has two generators, one solidly grounded and the other grounded

through a resistor, that are connected to a bus and through a step-up transformer

to a transmission line. Another generator, grounded through a reactor, is

connected to a bus and through a transformer to the other end of the

transmission line. A load is connected to each bus.

On the one-line diagram information about the loads, the ratings of the

generators and transformers, and reactances of different components of the

circuit is often given.

Per phase analysis of symmetrical three phase systems

Per phase analysis of symmetrical three phase systems are illustrated through

the following three examples.

Load A

2 3

1

Load B

T2 T1


EXAMPLE 1.1

A balanced three-phase load connected in star consists of (6+8) Ω impedance in

each phase. It is connected to a three phase supply of 400 V, 50 Hz. Find (i) Phase

current (ii) Line current (iii) Per phase power and (iV) Total power

SOLUTION

Fig. 1.5 Balanced three phase load – 1 phase equivalent

Three phase star connected balanced load and the corresponding single phase

equivalent are shown in Fig. 1.5 (a) and (b)

EXAMPLE 1.2

Repeat the previous problem with the load impedance connected in delta.

SOLUTION

Fig. 1.6 Balanced three phase load – 1 phase equivalent

Three phase delta connected balanced load and the corresponding single phase

equivalent are shown in Fig. 1.6 (a) and (b).

V240-110 0 V120-110 0

C

V0110 0

A

B -

+

+ +

-

-

EXAMPLE 1.3

Calculate the line currents in the three wire Y-Y system of Fig. 1.7.

(2.5+j4) Ω

(2.5+j4) Ω

Fig. 1.7 Three wire system for Example 1.3

(1.5+j6) Ω

(1.5+j6) Ω

(1.5+j6) Ω

(2.5+j4) Ω

V0110 0

A

+

-

SOLUTION

The single phase equivalent of the given three phase system is shown in Fig. 1.8.

Total impedance = ( 4 + j10)

Current A68.210.2133j10)(4

0110I 0

0

A

Since the source voltages are in positive sequence, the line currents are also in

positive sequence. Therefore

A188.210.2133I 0

B ; A51.810.2133A308.210.2133I 00

C

(2.5+j4) Ω

Fig. 1.8 Single phase equivalent

(1.5+j6) Ω

Impedance and reactance diagram

In order to calculate the performance of a power system under load condition or

upon the occurrence of a fault, the one line diagram is used to draw the single-

phase or per phase equivalent circuit of the system.

Refer the one-line diagram of a sample power system shown in Fig. 1.4.

Fig.1.9 combines the equivalent circuits for the various components shown in

Fig. 1.4 to form the per-phase impedance diagram of the system.

Load A

2

3

1

Load B

T2 T1


Fig.1.9 combines the equivalent circuits for the various components shown in

Fig. 1.4 to form the per-phase impedance diagram of the system

The impedance diagram does not include the current limiting impedances shown

in the one-line diagram because no current flows in the ground under balanced

condition.

Fig. 1.9 Per-phase impedance diagram

Load A

2

3

1

Load B

T2 T1


+

-

E3

E1

-

+

E2

-

+

Fig. 1.9 Per-phase impedance diagram

Fig. 1.10 Per-phase reactance diagram

Per-unit quantities

Absolute values may not give the full significance of quantities. Consider the

marks scored by a student in three subjects as 10, 40 and 95. Many of you may

be tempted to say that he is poor in subject 1, average in subject 2 and good in

subject 3. That is true only when the base for all the marks is 100. If the bases are

10, 50 and 100 for the three subjects respectively then his marks in percentage

are 100,80 and 95 and thus the conclusions are different. Thus, there is a need to

specify base quantity for meaningful interpretation.

Percentage = (actual value / base) x 100

Per-unit quantity = Percentage / 100 = actual value / base

This kind of explanation can be extended to all power system quantities.

In power system we shall deal with voltage, current, impedance and voltampere

or power. When they are large values, we may use kV, ampere, ohm and kVA or

kW as their units. It is to be noted that out of the four quantities voltage, current,

impedance and voltampere if we specify two quantities, other two quantities can

be calculated.

Generally, base voltampere in MVA and base voltage in kV are specified.

(1.5)

For a single-phase system, the following formulas relate the various quantities.

1000 xkV voltage,base

10xMVA base

Vvoltage,base

VAbaseA,currentBase

6

= kVvoltage,base

1000xMVA base (1.1)

Acurrent,base

1000xkVvoltage,base

Acurrent,base

Vvoltage,baseΩ,impedanceBase (1.2)

Substituting eq. (1.1) in the above

1000xMVAbase

1000xkV) voltage,(baseΩ,impedanceBase

2

Thus

MVAbase

kV) voltage,(baseΩ,impedanceBase

2

(1.3)

Power factor being a dimensionless quantity Base power, MW = base MVA

Ωimpedance, base

Ω impedance, actualimpedanceunit -Per (1.4)

Per-unit impedance = actual impedance x 2kV)voltage,(Base

MVABase

(1.5)

MVAbase


2

(1.3)

Per-unit impedance = actual impedance X 2kV)voltage,(Base

MVABase

For three phase system, when base voltage is specified it is line to line base

voltage and the specified MVA is three phase MVA. Now let us consider a three

phase system. Let Base voltage, kV and Base MVA be specified. Then single-

phase base voltage, kV = Base voltage, kV / 3 and

single-phase base MVA=Base MVA / 3. Substituting these in eq. (1.3)

MVABase

kV)voltage,(Base

MVABase

kV/ voltage,Base 2

3/

][Ω,impedanceBase

23 (1.6)

It is to be noted that eq.(1.6) is much similar to eq.(1.3). Thus

impedancebase

impedance actualimpedanceunit -Per

= actual impedance X 2kV)voltage,(Base

MVABase (1.7)

MVAbase


2

(1.6)


MVABase (1.7)

EXAMPLE 1.4

A three phase 500 MVA, 22 kV generator has winding reactance of 1.065 Ω. Find

its per-unit reactance.

Solution

Ω0.968500

22impedanceBase

2

; Per-unit reactance = 1.10020.968

1.065

1.100222

500x1.065reactanceunitper(1.7),eq.Using

2


MVABase (1.7)

Per-unit quantities on a different base

Sometimes, knowing the per-unit impedance of a component based on a

particular base values, we need to find the per-unit value of that component

based on some other base values. From eq.(1.7) It is to be noted that the per-unit

impedance is directly proportional to base MVA and inversely proportional to

(base kV)2. Therefore, to change from per-unit impedance on a given base to per-

unit impedance on a new base, the following equation applies:

Per-unit Znew = per-unit Zgiven new

giv en

giv en

new

kVbase

kVbase(x

MVAbase

MVAbase)2 (1.8)

Per-unit Znew = per-unit Zgiven new

giv en

giv en

new

kVbase

kVbase(x

MVAbase

MVAbase)2 (1.8)

EXAMPLE 1.5

The reactance of a generator is given as 0.25 per-unit based on the generator’s of

18 kV, 500 MVA. Find its per-unit reactance on a base of 20 kV, 100 MVA.

Solution

New per-unit reactance = 2)20

18(x

500

100x0.25 = 0.0405

EXAMPLE 1.6

A single phase 9.6 kVA, 500 V / 1.5 kV transformer has an impedance of 1.302 Ω

with respect to primary side. Find its per-unit impedance with respect to primary

and secondary sides.

Solution

With respect to Primary

Per-unit impedance = 1.302 x 0.05(0.5)

0.00962

With respect to Secondary

Impedance = 1.302 x 2)0.5

1.5( = 11.718 Ω

Per-unit impedance = 11.718 x 0.05(1.5)

0.00962

Conclusion

Per-unit impedance of the transformer is same referred to primary as well as

secondary.

Advantages of per-unit calculation

1. Manufacturers usually specify the impedance of a piece of apparatus in

percent or per-unit on the base of the name plate rating.

2. The per-unit impedances of machines of same type and widely different

rating usually lie within narrow range although the ohmic values differ

much.

3. For a transformer, when impedance in ohm is specified, it must be clearly

mentioned whether it is with respect to primary or secondary. The per-unit

impedance of the transformer, once expressed on proper base, is the same

referred to either side.

4. The way in which the three-phase transformers are connected does not

affect the per-unit impedances although the transformer connection does

determine the relation between the voltage bases on the two sides of the

transformer.

EXAMPLE 1.7

A 300 MVA, 20 kV three-phase generator has a subtransient reactance of 20%.

The generator supplies a number of synchronous motors over 64-km

transmission line having transformers at both ends, as shown in Fig. 1.11. The

motors, all rated 13.2 kV, are represented by just two equivalent motors. Rated

inputs to the motors are 200 MVA and 100 MVA for M1 and M2, respectively.

For both motors X” = 20%. The three phase transformer T1 is rated 350 MVA,

230/20 kV with leakage reactance of 10%. Transformer T2 is composed of three

single-phase transformers each rated 127/13.2 kV, 100 MVA with leakage

reactance of 10%. Series reactance of the transmission line is 0.5 Ω/km. Draw

the impedance diagram, with all impedances marked in per-unit. Select the

generator rating as base in the generator circuit.

T2 T1

Fig. 1.11 One-line diagram for Example 1.7

M1

M2 r

p

n m l k

Solution

Base MVA = 300

Base voltage at generator side = 20 kV

Base voltage in transmission line = 230 kV

Line to line voltages of transformer T2 : kV13.2/22013.2/127x3

Base voltage at motor side = 230 x kV13.8220

13.2

Base MVA and base voltages at different sections are marked.

M2

p

n m l k

r

M1

300 MVA

13.8 kV 20 kV

230 kV

T1 T2

Per-unit reactance of generator = 0.2

Per-unit reactance of transformer T1 = 0.1 x 0.0857350

300

Per-unit reactance of transmission line = 0.5 x 64 x 0.1825230

3002

Per-unit reactance of transformer T2 = 0.1 x 0.0915230

2202)(

Per-unit reactance of motor M1 = 0.2 x 0.2745)13.8

13.2(x

200

300 2

Per-unit reactance of motor M1 = 0.2 x 0.549)13.8

13.2(x

100

300 2

j 0.549 j 0.2745

+

Em1

Eg

+

+

-

-

Em2

-

+

Per-unit impedance diagram is shown in Fig. 1.12

Fig. 1.12 Per-unit impedance diagram

j 0.2

j 0.0915 j 0.1815 j 0.0857

k l n m

p r

EXAMPLE 1.8

A transformer rated 200 MVA, 345Y / 20.5Δ kV connected at the receiving end

of a transmission line feeds a balanced load rated 180 MVA, 22.5 kV, 0.8 power

factor. Determine

(a) The rating of each of three single-phase transformers which when

properly connected will be equivalent to the above three-phase transformer

and

(b) The complex impedance of the load in per-unit, if the base in the

transmission line is 100 MVA, 345 kV.

Solution

(a) Rating of each single-phase transformer: 200 / 3 MVA, (345 / 3 ) / 20.5 kV

i.e 66.7 MVA, 199.2 / 20.5 kV

(b) Load Z = 2.81180

22.5

S

V 22

Ω

Base MVA = 100; Base voltage at the load side = 20.5 kV

Load in per-unit = 2.81 x j40140.535236.870.66936.8720.5

100 00

2

EE 0308 Power System Analysis - Problem Set 1

1. A 120 MVA, 19.5 kV generator has Xs = 0.15 per unit and is connected to a

transmission line by a transformer rated 150 MVA, 230 Y/18Δ kV with X =

0.1 per unit. If the base to be used in the calculation is 100 MVA, 230 kV for

the transmission line, find the per unit values to be used for the

transformer and the generator reactances.

2. A transformer’s three phase rating is 5000 kVA, 115/13.2 kV, and its

impedance is 0.007 + j0.075 per unit. The transformer is connected to a

short transmission line whose impedance is 0.02 + j0.1 per unit on a base

of 10 MVA, 13.2 kV. The line supplies a three phase load rated 3200 kW,

13.2 kV with a lagging power factor of 0.8. Draw the per unit impedance

diagram taking the base as 10 MVA, 13.2 kV at the load. If the high tension

voltage is maintained at 115 kV, find the per unit and the actual load

voltage.

3. The one-line diagram of an unloaded power system is shown below.

Reactances of the two sections of transmission line are shown in the

diagram. The generators and transformers are rated as follows:

Generator 1: 20 MVA, 13.8 kV, X” = 0.2 per unit

Generator 2: 30 MVA, 18 kV, X” = 0.2 per unit

Generator 3: 30 MVA, 20 kV, X” = 0.2 per unit

Transformer T1: 25 MVA, 220Y/13.8Δ kV, X = 10%

Transformer T2: Single-phase units each rated 10 MVA, 127/18 kV, X = 10%

Transformer T3: 35 MVA, 220Y/22Y kV, X = 10%

Draw the reactance diagram with all reactances marked in per unit and with

letters to indicate points corresponding to the one-line diagram. Choose a

base of 50 MVA, 13.8kV in the circuit of generator 1.

F E

D

C

B A

1 2

3

T1

T3

T2

j 100 Ω j 80 Ω

4. Draw the impedance diagram of the power system shown below.

Mark impedances in per unit. Neglect resistance and use a base of 50 MVA,

138 kV in the 40-Ω line. The ratings of the generator, motors and

transformers are:

Generator 1: 20 MVA,18 kV, X” = 20%

Generator 2: 20 MVA,18 kV, X” = 20%

Synchronous motor 3: 30 MVA, 13.8 kV, X” = 20%

Three phase Y-Y transformers: 20 MVA, 138Y/20Y kV, X = 10%

Three phase Y- Δ transformers: 15 MVA, 138Y/13.8 Δ kV, X = 10%

C

j 20 Ω j 20 Ω

1

3

2

j 40 Ω

B A

Δ Δ

ANSWERS

1. 0.06667 p.u.; 0.1467 p.u.

2. 0.9317 p.u.; 12.2984 kV

0.014 + j0.15 0.02 + j0.1

2 + j1.5 Vload 1.0 00 p.u.

3.

4.

C

B A

j0.405 j0.405

j0.3333

j0.3333 j0.3333

j0.0525 j0.0525

j0.105

j0.25 j0.25

j0.25 j0.25

F E

D

B A

j0.2755

j0.1429 j0.3333

j0.5

j0.2 j0.0826

j0.1033 j0.1667

C

Documents

power sys. analysis.pdf