Lecture Notes Set No. 1 - Asynchronous Learning Networkaln.fiu.edu/EEL5270/wp-content/uploads/2017/01/lecture-notes1_EEL... · EEL-5270 Electrical Transients in Power Systems Lecture

1-1

Florida International University

Department of Electrical & Computer Engineering

EEL-5270 Electrical Transients in Power Systems

Lecture Notes

Set No. 1

Professor Osama A. Mohammed

Miami, Florida

Spring 2017

EEL-5270 Electrical Transients in Power Systems, Prof. Mohammed, Copyright All rights reserved, 2008-2017

1-2EEL-5270 Electrical Transients in Power Systems, Prof. Mohammed, Copyright All rights reserved, 2008-2017

FLORIDA INTERNATIONAL UNIVERSITY DEPARTMENT OF ELECTRICAL AND COMPUTER ENGINEERING

EEL-5270 Electrical Transients in Power Systems SPRING SEMESTER 2017

INFORMATION SHEET AND COURSE TOPICS

COURSE INSTRUCTOR:

Professor Osama A. Mohammed, Ph.D., Fellow IEEE

Energy Systems Research Laboratory

Department of Electrical & Computer Engineering

Room EC-3983

Florida International University

Miami, Florida 33174 USA

Tel: +1 (305) 348-3040 (Office), Tel: +1 (305) 348-3914 (ESRL Lab)

Fax: +1 (305) 348-3707 (Office)

E-mail: [email protected]

Class Time/Room: F: 2:00-4:30 pm, EC-2410

OFFICE: EC-3983

Office hours F 1:00-2:00 pm

PHONE: (305) 348-3040/2807 (office/staff)

PRE-REQUISITE: EEL-4213 or permission of instructor.

CREDIT HOURS: 3 Hours

TEXT_BOOK AND NOTES:

1. Electrical Transients in Power Systems by Allan Greenwood, Wiley, 1994.

2. Selected lecture notes by Professor Mohammed and other demonstration material and examples will be made

available at the above Web site and/or in class.

Who Should Take This Course?

FIU Electrical and Computer Engineering Students who took EEL4213 (power I) and Graduate Students.

Students at other Universities in Florida or out of State with the course’s prerequisites.

Engineers and technical personnel in Industry preparing for Engineering License

Engineers and technical staff who want to keep current and reach a deep understanding of energy

conversion concepts.

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Objectives:

1. Introducing Electrical Transients in power systems

2. Cover the concepts of traveling waves and propagation

3. Modeling of transmission lines as distributed parameter systems.

4. Discuss issues related to insulation coordination, grounding and limiting of surge effects

5. Develop techniques related to reflections at transition points in lines and cables

6. Multi conductor transients and distributed parameter modeling for components and shielding issues.

7. Involve students in a practical experience through the term project.

Course topics:

Introduction to Electrical Transients in Power Systems (Circuit opening and closing transients, Recovery

transients.

Traveling waves on Transmission Systems (Propagation of Surges)

Modeling of Transmission Lines by Distributed Parameter Concepts (Lossles cases, loss cases,

distrotionless cases, Lines with small losses)

Distortion due to Corona

Energy in Traveling Waves

Characteristics of Traveling Waves (wave shapes, standards, impulse and switching surges, lightning,

Basic Impulse Insulation Level (BIL), Flashover surge generators)

Analytical approximation of Surge Wave Shapes

Transition Points (Lines, Cables, Voltage Buildup, various types of Transition points, terminations)

Lumped Series and Shunt Impedance Transition Points (Junctions at Cables, Substations

Dissimilar Voltage and Current Surges

Surge Arresters (Linear and Nonlinear Characteristics.

Successive Reflections on Transmission Systems

Grounding

Insulation Coordination

Multi-Conductor, Multi-velocity Systems

Transient Performance of Distributed parameter Systems for Transformer, Generator and Motor

Windings

Shielding

Practical Examples, projects

References:

Appropriate lists and copies of technical papers will be distributed or listed for your collection. A list of recent

text-books and other technical record will be suggested to you. However, you are also required to research and

obtained other pertinent materials related to the topics covered.

ASSISTANCE: Please try to see Dr. Mohammed during his listed office hours or through the communication

forum on the web page. If this proves impossible, a personal appointment should arranged by calling my direct

phone number or the ECE department secretary at extension (305-348-2807).

ABSENCE: Class attendance (physical or virtual) is very important and is considered in your overall

performance in the course. Students are responsible for all material covered in that class.


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IMPORTANT RULE: Students are encouraged to discuss the course topics with the professor and with each

other. Any work submitted (Homework, Tests, projects, etc.) should be pledged and signed as the students' own

work, and that there is no any unauthorized help was obtained. Violators will be subject to academic misconduct,

which might lead to dismissal from the university.

GRADING POLICY:

Homework will be assigned regularly, collected and graded. Efforts in homework indicate that you are studying

and caring about the course and therefore can have an impact on your final grade. Time for the mid-term will be

announced one week in advance. Any work submitted must be neat and detailed for partial mark. Your Grade

will be calculated as Follows:

Homework and Class Projects 20%

Mid Term 25%

Term Project/Research Paper 20%

Final Exam 35%

_______

Total 100%

TERM PROJECT:

The project will involve one or more of the topics of this course. Final presentation (oral and written) of the

overall project results will be required. Software available for this class can be utilized for the projects.


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The Electromagnetic Pulse (EMP) Effect


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The Influence of a Charged Cloud


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I. HOW LIGHTNING IS FORMED

AVERAGE THUNDERCLOUD USUALLY BOTTOMS AROUND 2 MILES (10,000 )

AND TOPS OUT AROUND 8 MILES (50,000feet) IN ELEVATION.

TEMPERATURE IS ABOUT 40oF AT BASE

T0 -65 o AT THE TOP.

WARM MOIST AIR ON LEADING EDGE

ASCENDS RAPIDLY FROM 40oF TO -65oF

FORMING ICE CRYSTALS AND WATER

DROPLETS.

IT IS THIS RAPID ASCENT AND

SWIRLING MOTION THAT FORMS THE

ELECTRIC CHARGE. EEL-5270 Electrical Transients in Power Systems, Prof. Mohammed, Copyright All rights reserved, 2008-2017

1-8

THIS CREATES A HIGHLY POSITIVE CHARGE

IN THE TOP OF THE CLOUD (6 MILES AT

-65ºF), A HIGHLY NEGATIVE CHARGE IN THE

LOWER CENTRAL PART (3 MILES AT 0ºF) AND

A SMALL POSITIVE CHARGE AT THE BASE (2

MILES AT 40ºF). THE DIFFERENCES IN

POTENTIAL RESULTS IN THE LIGHTNING

STROKE.


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TYPICAL STROKE TO EARTH IS AROUND 125

MILLION VOLTS, 20,000 AMPS, WITH

TEMPERATURE AROUND 50,000oF.


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POINTS OF INTEREST

YOU CAN TELL HOW NEAR THE

STORM IS BY COUNTING THE

SECONDS BETWEEN LIGHNING AND

THUNDER. DIVIDE THE NUMBER OF

SECONDS BY 5.

THUNDERSTORMS ARE

RESPONSIBLE FOR MAINTAINING

EARTH’S NEGATIVE CHARGE.

LIGHTNING PRODUCES NITROGEN

COMPOUNDS THAT ARE ESSENTIAL

FOR MOST PLANTS.

LIGHTNING KILLS MORE PEOPLE IN

THE U.S. EVERY YEAR THAN

TORNADOES, HURRICANES, OR

FLOODS.


1-11EEL-5270 Electrical Transients in Power Systems, Prof. Mohammed, Copyright All rights reserved, 2008-2017

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AVERAGE THUNDERSTORM DAYS PER

YEAR IN U.S.


EEL-5270 Electrical Transients in Power Systems, Prof. Mohammed, Copyright All rights reserved, 2008-20171-13

ELECTROMAGNETIC COUPLING

ELECTROMAGNETIC COUPLING IS THE MOST

COMMON WAY LIGHNING GETS INTO THE

SYSTEM.

AS A CHARGED CLOUD MOVES INTO THE

AREA (OR BUILDS UP IN THE AREA), THE LINE

BECOMES IMMERSED IN THE

ELECTROSTATIC FIELD.

WHEN LIGHTNING STIKES IN THE VICINITY

(WITHIN A FEW HUNDRED FEET), IT MAY

RESULT IN A COUPLING EFFECT BETWEEN

THE FIELD AROUND THE STROKE AND THE

LINE SIMILAR TO THE WAY THE COILS OF A

TRANSFORMER OPERATE.

THIS RESULTS IN OVERVOLTAGES (SURGES)

ON THE LINE WHICH TRAVEL UNTIL IT FINDS

A PATH TO GROUND. THIS POINTS OUT THE

NEED FOR MAINTAINING A GOOD

GROUNDING SYSTEM.


II. HOW LIGHTNING GETS INTO THE SYSTEM

DIRECT HIT


ELECTROMAGNETIC COUPLING - UNDERGROUND

A DIRECT STRIKE TO EARTH IN THE

VICINITY OF BURIED CABLE OR SPILL OVER

FROM THE OH THROUGH THE RISER CAN

CAUSE AN OVERVOLTAGE ON THE

CONCENTRIC CABLE.

WHEN THESE HIGHER VOLTAGES ARE

PRESENT, THE DIFFERENCE IN POTENTIAL

BETWEEN THE PHASE CONDUCTOR AND

CONCENTRIC NEUTRAL CAN RESULT IN

TREEING AND DAMAGED EQUIPMENT.


PRE-DISCHARGE BUILD UP - B Ø

WHEN SURGE IS PRESENT ON LINE, ALL

3 PHASES ARE AFFECTED.

ARRESTERS, ON ALL PHASES, WHICH

ARE COMMONLY GROUNDED LIMIT THE

SURGE OVER VOLTAGES BETWEEN

PHASES.

IF THE GROUND RESISTANCE IS NOT

LOW WNOUGH THE SURGE REMAINS ON

THE LINE LONGER, POSSIBLY

RESULTING IN DAMAGE TO THE

ARRESTERS AS WELL AS OTHER

EQUIPMENT.


MULTI-GROUNDED WYE CIRCUITS

MULTI-GROUNDED WYE CIRCUITS DEPEND

ON INTERMEDIATE LOW RESISTANCE

GROUND STATIONS FOR THE SYSTEM TO

OPERATE PROPERLY UNDER ABNORMAL

CONDITIONS. BECAUSE OF HIGH SURGE

IMPEDENCE IN THE NEUTRAL CONDUCTOR,

IT SHOULD NOT BE COUNTED ON TO

DISSIPATE THE SURGE & OVERVOLTAGES OR

TO CARRY THE SURGES TO THE SUBSTATION

FOR DISSIPATION.






Comparison of various distribution line protective measures.


B.I.L. - BASIC IMPULSE INSULATION LEVEL

THE ABILITY OF ANY MATERIAL OR DEVICE TO

WITHSTAND A PRE-DETERMINED IMPULSE LEVEL.

EXAMPLES:

PIN INSULATOR - 100 KV

35 KV INSULATOR (POST) - 180 KV

ANY POLE IN SALT SPRAY AREA - 0 KV

CONCRETE POLE - 0 KV

WOOD POLE

WET - 75 KV/FT

DRY - 100 KV/FT

WOOD X-ARM

WET - 75 KV

DRY - 100 KV/FT

AIR - 186 KV/FT

13 KV TRANSFORMER - 95 KV

23 KV TRANSFORMER - 125 KV



BIL OF 35 KV SKIRTED INSULATOR IS 180 KV -

WHEN BOLT EXTENDS OUT TO COVER 2 - 3

SKIRTS, BIL IS REDUCED.

EXAMPLE: BOLTS EXTENDS OVER 3 SKIRTS

180 KV - 90 KV = 90 KV

AIR GAP BETWEEN BOLT & INSULATOR = 1 +

INSULATIONS VALUE OF AIR = 15 KV PER 1

THEREFORE BIL IS APPROXIMATELY 105 KV



CONCRETE POLE

35 KV INSULATOR - 180 KV

CONCRETE POLE - _ 0___

TOTAL BIL 180 KV

WOOD POLE

35 KV INSULATOR - 180 KV

WOOD POLE (WET) - 75 KV/FT

75 KV 180 KV

x_ 3_ 225 KV

225 KV TOTAL BIL 405 KV


INSTALLATION OF GROUND RODS

-----PREFERRED METHOD-----

RODS DRIVEN VERTICALLY IN TANDEM

DOUBLING ROD LENGTH REDUCES

OHM READING BY ABOUT 40%

EXAMPLE:

20 FT DEEP ROD - 30 OHMS

40 FT DEEP ROD - 18 OHMS


INSTALLATION OF GROUND RODS

ALTERNATIVE TO DEEP DRIVEN ROD

CLUSTERED RODS - MULTIPLE RODS DRIVEN

IN PARALLEL AND

CONNECTED TOGETHER

- WHEN -

ONLY WHEN INITIAL ROD(S) ARE DRIVEN TO

REFUSAL BEFORE SPECIFIED RESISTANCE IS

REACHED


ALTERNATIVE TO DEEP DRIVEN ROD

INSTALL GROUND ROD AT ANGLE

RECOMMENDED ONLY FOR AREAS WHERE

HARD ROCK PROHIBITS DRIVING RODS

VERTICALLY DEEP ENOUGH TO ACHIEVE

SPECIFIED OHMS.

EXAMPLE:

- ROD DRIVEN VERTICALLY TO 20 DEPTH

- WHEN DRIVEN AT 45 ANGLE WILL HAVE

28 OF ROD IN CONTACT WITH EARTH

MAY LOWER RESISTANCE BY 20% - 40% OVER

RODS DRIVEN VERTICALLY


TESTING OF GROUND RODS

EQUIPMENT

FUSE TESTER

VIBOGROUNG ( M&S #590-54700-5)

MEGGER EARTH TESTER

(M&S #590-56000-1)


EARTH RESISTANCE VS. INSULATION

RESISTANCE METERS

EARTH RESISTANCE METERS OPERATE ON

FREQUENCIES BETWEEN 97 Hz AND 125 Hz.

INSULATION RESISTANCE METERS

OPERATE ON FREQ. IN 60 Hz RANGE.

INSULATION RESISTANCE METERS ARE

SUSCEPTIBLE TO STRAY GROUND

CURRENTS OR D.C. CURRENTS WHICH

INFLUENCE READINGS.

INSULATION RESISTANCE METERS ARE

NOT TO BE USED FOR EARTH RESISTANCE

TESTING.


FUSE TESTER

3 AMP ------> 25 OHMS

8 AMP ------> 10 OHMS

FUSE IS DESIGN TO BLOW AT 1 - 1/2

TIMES ITS RATED VALUE

3 AMP FUSE WILL BLOW AT 4-1/2 AMPS

8 AMP FUSE WILL BLOW AT 12 AMPS

OHMS LAW E = R X 1 R = E / 1

R = 115 / 4.5 = 25.5 OHMS

R=RESISTANCE I=CURRENT E=VOLTS


VIBROGROUND AND MEGGER EARTH TESTER

NULL BALANCE INSTRUMENTS

USED FOR MEASURING EARTH

RESISTANCE

METHOD FOR USE:

2 POINT METHOD

3 POINT METHOD


2 POINT (DIRECT METHOD)

MEASURE RESISTANCE BETWEEN

GROUND ROD AND SURROUNDING EARTH

USING THE SYSTEM NEUTRAL AS

REFERENCE.


3 POINT (FALL OF POTENTIAL)

MEASURE RESISTANCE OF GROUND ROD

AND SURROUNDING EARTH USING

POTENTIAL AND CURRENT PROBES AS

REFERENCE.


TESTING GROUNDS AT EXISTING LOCATIONS

TEST FOR 25 OHMS OR LESS

IF 25 OHMS OR LESS NOT ACHIEVED, THEN:

OPTIONS:

1. DRIVE ADDITIONAL RODS ON TOP OF

EXITING RODS TO ACHIEVE 25 OHMS.

2. MOVE OVER 10 IF POSSIBLE (6 MINIMUM)

AND CLUSTER RODS 25 OHMS.

3. ABANDON EXISTING INSTALLATION AND

DRIVE NEW RODS TO ACHIEVE 10 OHMS IF POSSIBLE WITH 40 OR LESS (AS

PREVIOUSLY SPECIFIED)


GOOD CONSTRUCTION PRACTICES

RELATING TO

GROUNDING

THERE ARE MANY THINGS WE CAN

DO TO IMPROVE OUR GROUNDING

SYSTEM AND SERVICE

RELIABILITY.


INSTALLATION AND ATTACHMENT

OF GROUND RODS

ALL GROUNDING CONNECTIONS ARE

ELECTRICAL

WIRE BRUSH AND USE INHIBITOR

TIGHTEN COUPLINGS FOR EACH SECTION

OF GROUND ROD

USE SEPARATE CLAMP OR CONNECTOR FOR

EACH CONNECTION


SURGE ARRESTERS AND LEADS

MAKE ALL LEADS AS SHORT AS POSSIBLE

AVOID SHARP BENDS AND KINKS

(EACH KINK INCREASES IMPEDANCE BY 20%)

DO NOT COIL WIRE

AVOID KINKS ON ANY BONDING WIRE


I. Electrical Transients in Power Systems

1. Circuit Closing Transients

What is of importance?

1. The closing of a circuit breaker to

energize a load.

2. The opening of a circuit breaker to

clear a fault.

Consider the following circuit:

Steady state power factor = 222 LR

R

z

R


Assuming an ideal source (small impedance compared

to load)

)sin()( tVtv m

= arbitrary phase angle to allow a general study of the

instant of switching

At closing of the switch, s, we can write:

)sin( tVdt

diLRi m

Rewriting the same equation as:

]sincoscos[sin ttVdt

diLRi m

Laplace transforming both sides gives:

2222

sincos)()()(

s

s

sVoLisLsIsRI m

For initial value of ooi )( i


2222

sincos1)(

s

s

sL

RL

VsI m

Try to put this equation in a simple form. Define.

LR

LV

B

LV

A

m

m

sin

cos

2222)(

ss

Bs

ss

AsI

1L gives after arranging

tm etLR

Vti

sin)(222

The first Term is the Steady State Part

The second Term is the Transient Part


If the breaker closes at time, t, such that

the transient term becomes zero, and the current wave

becomes symmetrical.

If the breaker closes at time, t, when

the transient term is maximum and the peak of i(t)

will approach (2x steady rate)

[Practical consideration in circuit breaker design]

2


2. Circuit Opening (Fault Removal) Transients

Recovery Transient:

The transient accompanies the removal of a fault

from a power system.

Assume

L= inductance of a transmission system (line and a

transformer, etc.)

C= natural capacitance of the system adjacent to the

breaker (capacitance to ground through bushings,

CT’s, main transformer, T-Lines, etc.

Circuit Resistance and other types dissipation elements

are ignored.


• When a fault occurs on a system, a significant (in

fact very large) amount of fault current flows.

• The opening of a circuit breaker contacts by itself

does not interrupt the current. (An arc will be

established between the contacts, the current will

continue to flow. )

• Successful interruption depends upon the control of

the arc.

• In an ac circuit, the current passes through zero

twice per cycle. At one of these instances, current

will be interrupted.

• In the circuit under discussion, the current lags the

voltage by and is limited by the inductance.2


• The voltage at the breaker contacts will be the arc

voltage.

• Ignoring the arc voltage since the source voltage is

sinusoidally varying and is at its peak at the instant

of fault current is interrupted.

Hence:tVtv m cos)(

The interrupted circuit equation is given by:

2

2

)(

dt

vdc

dt

di

dt

dvci

tvvdt

diL

c

c

and

so

Therefore;

tVLc

vLcdt

vdmc

c cos11

2

2

Before the breaker interrupts the fault current at t=0

00)0(

)0(00)0(

00)0(

dt

dvvi

v

ccc

c


Transforming into Laplace

Lc

s

sV

svvsvsvs

m

cccc

1

)()0()0()(

20

22

20

20

2

Lc

10

Applying the initial conditions, gives the following

for )(svc

2220

2

20)(

ss

sVsv m

c

Obtaining the inverse Laplace Transform, yields

ttVtv mc 0220

20 coscos)(

Represents the voltage across the circuit breaker

contacts “s” which is called

“Recovery (Restriking) Transient Voltage”

since then 0

ttVtv omc coscos)(


Recovery Transient (voltage) Across Breaker Contacts For 0

tVtv mc 0cos1)(

It can be easily seen that for, , the value of

would be

t0

)(tvc

mc Vtv 20

• If the natural frequency is high (L or c or

both are small), the voltage will rise very

quickly with time.

• If the rate of application of the recovery voltage

should exceed the rate of build up of the

dielectric strength in the medium between the

contacts of the breaker,

0

cv


the breaker will not be able to hold off and arc from re-

igniting. This results in the breaker and the power

system components carrying the fault current for at least

another 1/2 cycle until the current becomes zero again.

• This phenomenon is of so great importance for

short T-Lines in which both L & C are small.

The effect is called “kilometric fault”


Example: for a particular short transmission line

3101 L Hence, .10400 12 Faradc

.sec10410250

1)(

.1025010104

1

2

6

30

3

2 103

00

periodT

Hzf

If this line is rated 13.8 kv, the voltage could

possibly rise to twice the peak line to neutral value

in a period of

2

0T

23

8.132

20

Tvci.e. K.volts.to neutral

Per Phase

TimePeriod

voltage61023

28.132Rate of

voltage rise =

.sec/103.11 6 kv

Such aralc from practical experience is beyond the

capability of any C.B. (Notice Dielectric strength of

air is 30 kv/cm)

Vel. of Separation of breaker cont..=30

103.11 6

hrmilehrkm

km

cm

/3800/36005.3

.sec/5.3

.sec/1035.0 6

(Mechanically impossible)


Example of Force Calculations (Subsequent to S.C.) or

closing of a breaker

..csi Instantaneous worst case A80003.39522

Ai

upI

cs

cs

3.3952

103

3453

1050034.32(max)

..34.31.02.0

0.1

3

6

..

..

MNewtmF

mF

mF

IBBImeterF

r

IB

r

IH

/8.38086.2978.12

320

3345

8.12/

8.12/

2

80001048000/

/

2,

2

2

7

0

Newton/Meter Cond. Length (H.V. Side


Effect of Current Asymmetry on Recovery Voltage:

• When a circuit breaker is closed at random, it is

likely that the current during the transient period

will lack symmetry as shown below:

• In a three phase system, the voltages are 120 apart.

During the transient period, asymmetry will result in one

or more of the phases.

• The asymmetry will depend on the time in the cycle

at which the fault occurs.


• The circuit breaker interrupt at an instant of the

cycle at which the current is zero.

• The recovery voltage will oscillate around the

instantaneous value of supply voltage.

• With asymmetrical fault current, the supply voltage

(system voltage) will no longer be at its peak. The

transient recovery voltage is not as high as was

shown earlier.

Effect of Current Asymmetry on Recovery Voltage


Effect of Arc Resistance on Recovery Transient

•In cases where the arc voltage drop (across arc

resistance) cannot be ignored, from a quasi-quantitative

stand point, the arc drop is heavily resistive.

•This has the effect of bringing the current to be

interrupted no more in phase with the voltage.

•This has the effect of making the point of zero current

occur at point in the cycle at which the system voltage is

not at its peak. This has the effect of reducing the first

(sharpest) peak of the transient recovery voltage.

Effect of Arc Resistance on Recovery Voltage


3 More Complex Recovery Transients

Consider an elementary power system which may

consist of a source and an unloaded transformer as

shown below:

Such a system can be represented from circuits point of

view as follows:

L1 and C1 represent the inductance and stray

capacitance on the source ride of the breaker. L2 and C2

may represent the inductance and stray capacitance of

the unloaded transformer.

When the circuit breaker opens in such a

power system it completely isolates the load side from

the supply side. After the instant of breaker operation,

the two halves of the system behave independently.


What happens following such a switching operation?

(1) Before the breaker opens, the supply voltage will

divide in proportion to the inductances (capacitors

represent a high reactance path). The voltage across

the capacitors is:

(2) For all practical purposes, the transformer

inductance, L2, will be much larger than the source

inductance, L1. Hence, C1 and C2 have a voltage

vc(t), which is approximately equal to the full

supply voltage v(t).

(3) Following the current interruption, C2 will discharge

through L2 and a natural frequency f2.

Figure (b) on the next page shows that this transient

will be damped by circuit resistance.

)()(21

2 tvLL

Ltvc

222

2

1

CLf


Double Frequency Recovery Voltage

(4) C1, on the source side, charged to a voltage

near source voltage level, will exhibit a

recovery transient voltage wave shape shown

on the next page.


• In this case the natural frequency is smaller than the

case discussed earlier due to the initial charge on the

capacitor C1.

This is shown in figure (a) on the previous page.

(5) The voltage across the breaker is made of the

difference between vc1(t) and vc2(t). The result is

shown in Figure (c) on the previous page, which is

(wave(a)-wave(b))

The above case shows a recovery voltage wave with two

distinct frequencies f1 & f2.

111

2

1

CLf


An even more complex case

Consider the following case which is encountered quite

frequently in the case of a circuit breaker clearing the fault

on the secondary side of a transformer as shown below:

Elementary Power System With Double Frequency

Recovery Voltage.

The equivalent circuit becomes as shown:

Equivalent Circuit of System


In the equivalent above L1 represents the inductance up

to the transformer and L2 is the leakage inductance of the

transformer. C1 and C2 are the system natural

capacitance on either side of the transformer.

Assume that during the period of interest, the

source voltage will remain constant at a value V1 than:

Under 60 Hz, C1 represents a very high

reactance path.

At t=o before fault is cleared

and hence

Also, the fault will be cleared at an instant when

i1(t)=i2(t)=0

So, from the equivalent circuit

Vtv )(

VLL

Lvc

21

2)0(1

00)0(2

cv


dt

diL

dt

diLtvtv

dtic

vtv

dtiic

vdt

diLtvtv

c

tcc

tcc

22

11

202

2101

11

)()(

1)0()(

1)0()()(

2

22

11

0

Since v(t)=V, Laplace transform to the following:

)()(1

)(

)()(

)0()()0()()(

)(1

)(

)0()0(

)(1

)(1

)()(1)0(

)0()(

22

2

2

1

1

222

11

222

22221111

22

1121

11

1

211

1111

sVsVscLs

V

sLsI

ssvccsI

iLssILiLssILs

VsV

sIsc

sV

iLs

v

s

VsI

scsI

scsL

sIsIscs

viLssIL

s

V

cc

c

c

c

c

OR

and

and

and

0

0 0


21211

2211

1

LLLcLB

cLcLA

Let and

,one has

))(())((

1)(

22

221

222

221

22 ss

Bs

sssAVsVc

Using the partial fraction technique one can show that

the inverse Laplace transform is:

tB

AV

tB

AVtvc

222

21

22

22

122

21

21

21

22

21

cos1

cos11

)(2

If , we can obtain a condition under which

the voltage vc2 which will show across the breaker

contacts might reach as high as (3V).

*The algebra gets rather involved quickly as more

circuit elements are added.

The propagation of surges on power T-Lines is

discussed next.

21

Documents

Lecture Notes Set No. 1 - Asynchronous Learning Networkaln.fiu.edu/EEL5270/wp-content/uploads/2017/01/lecture-notes1_EEL... · EEL-5270 Electrical Transients in Power Systems Lecture