EE493 543 Lecture 1 Sym Fault

8/13/2019 EE493 543 Lecture 1 Sym Fault

1/19

EE 493/543 -Power Systems III, Symmetrical Faults,Dr. Sukumar Brahma, NMSU

1

Overview of Fault

Fault means current flowing through a path it isnot intended to flow through.

Temporary faults are caused by lighting, birds,animals.

Permanent faults are caused by insulationfailure, birds, animals.

If a temporary fault occurs, the goal of theprotection scheme is to restore the service afterthe fault clears by itself.

If a permanent fault occurs, the goal of theprotection scheme is to isolate the fault fromthe system and keep the outage area to aminimum.

Sustained fault can result in equipment damageand instability.


2

Overview of Fault

Open Circuit Faults: Broken conductor

Short Circuit Faults: Balanced faults:

Three-phase fault (5%, least frequent)

Unbalanced faults:

Single line to ground faults (SLG-70~80%- most frequent)

Line to line faults (LL)

Double line to ground faults (LLG) We shall study short circuit faults in

this course


2/19


3

Overview of Fault

Depending on the fault location inthe system, either a three-phasefault or a single phase fault canbe the most severe.

There-phase fault is the easiesttype of fault to be analyzed,because it is a balanced fault.

To analyze unbalanced faults, we

need to use symmetricalcomponents


4

Transients in R-L Circuit

The first part of the solution is a sinusoid, the second part is

an exponentially decaying DC component. The amount of DC component depends on the switchinginstant, and the rate of decay depends the ratio R/L.

Next three slides show the solution plotted for VRMS= 230 kV,XL=160 (L = 0.4244 H), R = 32 .

For this data, |Vmax|/|Z| = 2.

)sin(max tV~

R

Li

dt

diLRitV )sin(max

)1()]sin()[sin(||

max

T

t

etZ

Vi

andLRZ 22 )(|| sec),/(tan 1

R

LTRL

Solution

Equation

Where,


3/19


5


0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05-4

-3

-2

-1

0

1

2

Time (sec)

current(kA)

Current profile for switching RL circuit at different instants

alpha=theta

alpha=theta+pi/4

alpha=theta+pi/2

Observe:When =,dc offset iszero (Noasymmetry)

When-=/2,asymmetryis maximum.


6

Amount of dc Offset

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05-4

-3

-2

-1

0

1

2

Time (sec)

current(kA)

Current break-up for switching RL circuit at alpha=theta+pi/2

DC Part

Resultant

AC Part

Observe:When -=/2, at t=0, themagnitude of dc offset equalsthe peak of symmetrical accurrent, making the totalinstantaneous current equal tozero - This is dictated by thephysics of the circuit.

However, the current rises fastand the peak goes to morethan 150% of the peak ofsymmetrical current in the

case shown here.

This peak will depend on the rate of decay of dc current. Slower decaymeans the peak would be higher and vice versa (see next slide).


4/19


7


0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05-4

-3

-2

-1

0

1

2

Time (sec)

current(kA)

Effect of R on the decay of DC component

AC Com ponen t

R=52 Ohm

R=42 Ohm

R=32 Ohm


8

Math shows what figure shows Recall:

If -=/2, then

])2/[sin(||

max LRt

etZ

Vi

At t=0,

Observe: the dc offset is worst - it equals the peak of thesymmetrical current. But the total current at t=0 is zero.

0||||

maxmax Z

V

Z

Vi

If -=0, then

There is no dc offset. Calculate the current (dc, ac, and total) when -=/2, at

t=8 ms.

)][sin(||

max tZ

Vi

)1()]sin()[sin(||

max

T

t

etZ

Vi


5/19


9

Analyzing Asymmetrical Waveform

At any given moment,

Where Iac is the RMS value of the symmetrical part of the current. We saw that in the worst case scenario, maximum value of Idc

occurs at t=0, and it equals the peak of the symmetrical part of accurrent, i.e., 2Iac.

Also, Idc decays according to a time constant T = L/R (see equation-1, slide#4), or T = X/2fR

Therefore, for the worst case scenario, idc(t)= 2Iace-t/T. This

means:

currentassymetrictheofvalueRMStheis)(where

)()(

)()()(

22

tI

tiItI

OR

tititi

RMS

dcacRMS

dcac

cyclesintimeftwhereeItI

eIeIItI

RX

acRMS

T

t

acT

t

acacRMS

,21)(

212)(

)/(

4

22

2


10

Analyzing Asymmetrical Waveform

ac

RX

acRMS IKeItI )(21)( )/(

4

From the previous page,

K() is called the asymmetry factor. The maximum value of K() is at t=0, when the value is 3

Look at Ex# 7.1, page# 360-361 (4th ed.),page#324,325 (3rd ed.)


6/19


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13

Modeling Armature Reaction in Synchronous Generator

2sin

11111)(

'''

''''max

t

Xe

XXe

XXEti

d

T

t

dd

T

t

dd

acdd

The phenomena is mathematically modeled by the followingequation:

Where Td< Td< Td. Let us see how this equation models thephenomena.

At the time of the short circuit, t=0, hence

''''max ")0(,,

2sin)0(

d

RMSRMS

d

acX

EIIort

X

Ei

After a certain time t, when t>>Td, but t


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15

Modeling a Fault

This circuit models the pre-fault and fault conditions by

two sources of value VPF, the pre-fault voltage at P. When the switch is open, the circuit models pre-fault orload condition, and when close, it models the faultcondition.

XdG

G TL M

XdMXTL

IL

P Suppose a 3-ph fault occurs at P inthis system.

+

-

+

-

P

Neutral

jXdG jXdM

jXTL

EGEM

VPF

VPF

+

-

-

+

IF

IL IL


16

Modeling a Fault

fault+

- -

+

P

Neutral

jXdG jXdM

jXTL

EGEM

IMIG

IF

+

-

+

-

P

Neutral

jXdG jXdM

jXTL

EGEM

VPF

VPF

+

-

-

+

IF

Pre-fault

Switchopen

Switchclose

Thus, fault issimulated by thesummation of thesetwo circuits.For small systems,the pre-fault circuitcan be solved withcircuit analysistechniques, but forlarge power systems,it is solved usingload-flow.Fault circuit can besolved usingThevenins theoremfor small systems andusing ZBUS for largesystems.

+

-

+

-

+

-

P

Neutral

VPFjXdG jXdM

EM

ILIL jXTL

EG

IFG=IG+ ILIFM=IM- IL


9/19


17

PFTH

dmdgTL

TH VEjXjXjX

Z

,11

1

dMdGTL

dMFG

jXjXjX

jXII

)(

Analyzing the fault conditions

+

- -

+

P

Neutral

jXdG jXdM

jXTL

EGEM

IMIG

IF

P+

-

ETH=VPF

ZTHjXTL+ jXdG

jXdM

IF

IG IM

FTH

PFF

TH

PFF

ZZ

VI

Z

VI

then,ZisimpedancefaulttheIf F

dMdGTL

dGTLFM

jXjXjX

jXjXII

)(


18

dMdGTL

dMFG

jXjXjX

jXII

)(

The actual fault currents comingfrom generator and motor are the

sum of currents before fault and dueto fault.

Finding the Total Fault Current

+

- -

+

P

Neutral

jXdG jXdM

jXTL

EGEM

IMIG

IFFTH

PFF

ZZ

VI

dMdGTL

dGTLFM

jXjXjX

jXjXII

)(

L

dMdGTL

dMFLGFG I

jXjXjXjXIIII

)(

L

dMdGTL

dGTLFLMFM I

jXjXjX

jXjXIIII

)(

fault

Pre-fault

+

+

-

+

-

+

-

P

Neutral

VPFjXdG jXdM

EM

ILIL jXTL

EG


10/191


19

Example

XdG=j0.2

G TL M

XdM=j0.2X=j0.1

IL

P

Assume EG=1.02/_40 PU, EM=1/_0

0 PU.

3-ph fault occurs at P. Find the totalfault current, fault contribution fromgenerator and motor, and post-faultvoltage at the generator bus.

+

-

+

-

+

-

P

Neutral

VPFJ0.2 J0.2

1/_00

ILIL J0.1

1.02/_40

Solve the pre-fault Circuit

PUj

IL0

00

13.830.14665.0

01402.1

PU

jVPF0

00

1.621.0074

)2.0()13.830.1466(01

VGPF

PU

jVGPF0

00

2.421.0114

)2.0()13.830.1466(402.1


20

Example continued

+

- -

+

P

Neutral

J0.2 J0.2

J0.1

EGEM

IMIG

IF

P+

-ETH

=1.0074/_1.620

ZTHj0.3

J0.2

IF

IG IM

PUVE

PUjjj

Z

PFTH

TH

0

1

62.10074.1

12.02.0

1

3.0

1

PUZ

VI

TH

PFF

038.883951.8

PUj

jII

PUj

jII

FM

FG

0

0

38.885.0375.0

3.0

38.883.3585.0

2.0

FFMFG

LMFM

LGFG

IIICHECK

PUjIII

PUIII

:

50

864.3 0

Solve the fault Circuit

Add currents from both circuitsa l g e b r a i c a l l y

Observe: There is very little differencebetween IFG&IG and IFM&IM. Therefore,pre-fault currents are often neglected.


11/191


21

Example continued

Solve the fault Circuit with Actual currents to get post-fault voltage at the generator bus

+

- -

+

P

Neutral

J0.2 J0.2

J0.1

EGEM

IFMIFG

IF

VFG

PUPU

VVV

PUjjIV

FGGPFG

FGFG

0

00

00

1.620.6716434.042.21.0114

434.0)1.0(864.30)1.0(0


22

Some Observations

As seen in previous slides, load current is negligiblecompared to fault current in most cases. Therefore,the pre-fault voltage VPF (=ETH) is taken as 1/_0

0 PUand IL is neglected. Thus, load flow results are notused in fault calculations.

Sometimes, fault path has arc-resistance. In thatcase, it is included in the Thevenin model.

It is difficult to find Thevenin equivalent impedance ofa large power system in conventional way. As we shallsee, at bus i, the element ZBUS(i,i,) is equal to theThevenin equivalent impedance. Since ZBUS of anysystem is known, it is quite simple to form theThevenin equivalent circuit.

We solved the example using steady state reactances(Xd) for machines. We can find subtransient ortransient currents in the same way, using thecorresponding machine reactances (Xd or Xd) .


12/191


23

Using ZBUS for Fault Analysis

As mentioned before, the fault is simulated by two sources of ValuesVPF and VPF in series. The figure above also shows fault impedanceZF.

The fault can be seen as a superposition of the pre-fault and faultconditions. Pre-fault condition has three sources actingsimultaneously: EG, EM and VPF. VPF is short-circuited.

-VPF is the only source working during fault condition, and EG, EM and

VPF are short-circuited. We find bus-voltages in both cases and add the results algebraically toget the final solution. Use these voltages to find line currents.

This is shown on the next slide.

+

-

+

-

jXdG jXdM

jX12

EGEM

VPF

VPF

+

-

-

+

IF

IL

IL1

23

GZF

jX23


24

Viewing Fault as Superposition of Sources

Circuit with Fault

Fault

+

Pre-fault voltages are V1, V2, V3.

Due to VPF acting alone, thevoltage changes are V1, V2, V3.

Final bus-voltages will be V1+ V1, V2+ V2 and V3+ V3.

jX12IL

+

-

+

-

jXdG jXdM

jX12

EG EM

VPF

+

-

IL

IL

1 2 3

G

V1

V2 V3

Pre-fault

ZF

jX23

jX23

G

jXdG jXdMVPF +

--IF

1 2 31 2 3V1

V2

V3

jX12

IF1-2 IF3-2

ZF

jX23

+

-

+

-

jXdG jXdM

EGEM

VPF

VPF

+

-

-

+

IF

IL

12

3

GZF


13/191


25

Analyzing the Fault Circuit

In this circuit, the only current injection into

a bus is -IF at bus-2. Writing a loop equation at bus-2,

0

0

333231

232221

131211

3

2

1

3

2

1

333231

232221

131211

3

2

1

FI

ZZZ

ZZZ

ZZZ

V

V

V

I

I

I

ZZZ

ZZZ

ZZZ

V

V

V

)2(,, 323222121 FFF IZVIZVIZV

Now using the ZBUS model of the system,

G

jXdG jXdMVPF +

--IF

1 2 31 2 3V1

V2 V3jX12

IF1-2 IF3-2

ZF

jX23

G

jXdG jXdMVPF +

--IF

1 2 31 2 3V1

V2 V3jX12

IF1-2 IF3-2

ZF

jX23

)1(0)( 2

2

VVZI

IZ

VV

PFFF

F

F

PF


26

Analyzing the Fault Circuit (Cont.)

)3(

0)(

22

22

F

PFF

FPFFF

ZZ

VI

IZVZI

Thus, to solve a three-phase fault, use (3) to find IF and use (2) to find

voltage changes at all buses.

In general, in a n-bus system, for a fault at bus k with fault impedance ZF,

.....1,, niIZVandZZ

VI Fiki

Fkk

PFkF

G

jXdG jXdMVPF +

--IF

1 2 31 2 3V1

V2 V3jX12

IF1-2 IF3-2

ZF

jX23

G

jXdG jXdMVPF +

--IF

1 2 31 2 3V1

V2 V3jX12

IF1-2 IF3-2

ZF

jX23 Using equations (1) and (2) from the previousslide:

FTH

PFF

ZZ

VI Compare (3) with this equation we derived on slide#18:

We can see here that Thevenin impedance at a node is the same asthe diagonal element of ZBUS corresponding to that node.


14/19


15/191


29

Example Continued

Bus-2 has a 3-Ph fault. Fault impedance ZF=0

. This systemhas a ZBUS of order-2. We will get YBUS and invert it to get ZBUS.

PUjjj

YBUS0095115

1.0

1

2.0

1)1,1(

+

-

+

-

J0.2 J0.2

1/_00

J0.1

1.02/_40

1 2

G

+

-

+

-

J0.2 J0.2

1/_00

J0.1

1.02/_40

1 2

G

PUjjj

YBUS0095115

1.0

1

2.0

1)2,2(

PUjj

YY BUSBUS0090110

1.0

1)1,2()2,1(

PUjYBUS

1510

1015PUjYZ BUSBUS

12.008.0

08.012.01

PUjZZ

VI

F

PFF

00

22

22 38.88395.8

012.0

62.10074.1

0

0

02

1

38.1780074.1

38.1786716.0

38.88395.8

0

12.008.0

08.012.0j

V

V

I12IM


30

Example Continued

Now add the pre-fault and fault voltages to get final bus voltagesusing (3). Use (4) to calculate line currents.

PUV

V

V

V

V

V

PF

PF

0

0

0

0

0

0

2

1

2

1

2

1

00

434.0

38.1780074.1

38.1786716.0

62.10074.1

42.20114.1

PUjz

VVII FG

00

12

2121 864.3

1.0

0434.0

PUjz

VVI MFM

00

12

2 9052.0

001

+

-

+

-

J0.2 J0.2

1/_00

J0.1

1.02/_40

1 2

G

+

-

+

-

J0.2 J0.2

1/_00

J0.1

1.02/_40

1 2

G

I12IM

Compare these currents with answers on slide#21 and voltages withanswers on slide#22. They tally exactly. Observe that in this method,we use the final voltages to find line currents. Therefore, we do notneed to add the pre-fault load currents separately.


16/19


17/19


18/191


35

Fuses (Continued)

Special Types of Fuses

Fuses for capacitor banks with high frequency inrush current Fuses with very high interruption ratings

Electronic fuses

Characteristics of Fuses

Minimum Melting Characteristics - An average melting time where arcingdoes not occur (link melts). See next slide

Total Clearing Characteristics Time taken to clear the arc - Used incoordinating against minimum melting characteristic of a larger fuse,located toward the current source. See next slide

E-rated fuses are very commonly used in distribution systems.

Fuses are available with rated continuous current ratings of: 0.5, 1, 2, 3,5, 7, 10, 15, 20, 30, 40, 50, 65, 80, 100, 125, 150, 200, 300, and 400,amperes


36

Fuse Characteristics

For the denominator:

Use 300 seconds for links rated upto 100 amperes

Use 600 seconds for links rated 140and 200 amperes

K and T Types of Links:

Speed Ratio of K links are in therange 6.0 to 8.1

Speed Ratio of T links are in therange 10.0 to 13.0

In the given characteristics, thespeed ratio is around 210/31=6.8.

seconds600or300atCurrentMelting

seconds0.1atCurrentMeltingRatioSpeed


19/19


37

Different MM Characteristics


38

Expulsion Power Fuse 7.5kV. to 69kV. 100 to400 Amp

Documents

EE493 543 Lecture 1 Sym Fault