lec5(2)(1) power system

8/18/2019 lec5(2)(1) power system

1/42

POWER SYSTEMS ILecture 5

06-88-590-68

Electrical and Computer Engineering

University of Windsor

Dr. Ali Tahmasebi


2/42

1

Fault Analysis

l The cause of electric power system faults is

insulation breakdown

l This breakdown can be due to a variety of different

factors

– lightning

– wires blowing together in the wind

– animals or plants coming in contact with the wires

– salt spray or pollution on insulators


3/42

2

Fault Types

l There are two main types of faults

– symmetric faults: system remains balanced; these faults

are relatively rare, but are the easiest to analyze so we’ll

consider them first. – unsymmetric faults: system is no longer balanced; very

common, but more difficult to analyze

l The most common type of fault on a three phase

system by far is the single line-to-ground (SLG),followed by the line-to-line faults (LL), double line-

to-ground (DLG) faults, and balanced three phase

faults


4/42

3

Lightning Strike Event Sequence

1. Lighting hits line, setting up an ionized path to

ground

l 30 million lightning strikes per year in US!

l a single typical stroke might have 25,000 amps, with arise time of 10 ms, dissipated in 200 ms.

l multiple strokes can occur in a single flash, causing the

lightning to appear to flicker, with the total event

lasting up to a second.

2. Conduction path is maintained by ionized air after

lightning stroke energy has dissipated, resulting in

high fault currents (often > 25,000 amps!)


5/42

4

Lightning Strike Sequence, cont’d

3. Within one to two cycles (16 ms) relays at both

ends of line detect high currents, signaling circuit

breakers to open the line

l nearby locations see decreased voltages

4. Circuit breakers open to de-energize line in an

additional one to two cycles

l breaking tens of thousands of amps of fault current is no

small feat!

l with line removed voltages usually return to near normal

5. Circuit breakers may reclose after several seconds,

trying to restore faulted line to service


6/42

5

Fault Analysis

l Fault currents cause equipment damage due to both

thermal and mechanical processes

l Goal of fault analysis is to determine the

magnitudes of the currents present during the fault – need to determine the maximum current to insure devices

can survive the fault

– need to determine the maximum current the circuit

breakers (CBs) need to interrupt to correctly size the CBs


7/42

6

RL Circuit Analysis

l To understand fault analysis we need to review the

behavior of an RL circuit

Before the switch is closed obviously i(t) = 0.When the switch is closed at t=0 the current will

have two components: 1) a steady-state value

2) a transient value

= 2 sin ( + )


8/42

7

RL Circuit Analysis, cont’d

1. Steady-state current component (from standard

phasor analysis):

=

2 V sin ( + − )

Where = + = + and = tan

The rms value of this sinusoidal current is:

=


9/42

8


2. Exponentially decaying dc current component (offset

current): =

Where T is the time constant,

=

[sec]

The value of C1 is determined from the initial conditions:

0 = 0 = + =2 sin ( + − )

+

At t = 0 we get = − sin ( − ) which depends on aand it has a maximum at = −


10/42

9



11/42

10


Hence i(t) is a sinusoidal superimposed on a decaying dc

current. The magnitude of idc(0) depends on when the switch

is closed.

For fault analysis we are only concerned with the worst case

which is when the current is at its maximum.


12/42

11


Maximum value of the offset current idc is at = − :

max = =2

= 2

Then the fault current with largest dc offset is:

= 2 sin − + [A]


13/42

12


This current is not a periodic function so we can’t formally

define an rms value. However, as an approximation define:

= + = + 2

=

1 + 2

[A]


14/42

13

If = = , and =

where t is time in cycles, then:

=


Where K (t) is the asymmetry factor and = 1 + 2

Then max = 3 which is at t = 0.


15/42

14

Generator Modeling During Faults

l During a fault the only devices that can contribute

fault current are those with energy storage

l Thus the models of generators (and other rotating

machines) are very important since they contributethe bulk of the fault current.

l General model of one phase of a 3-phase

synchronous machine for fault analysis:

= 2 sin ( + )


16/42

15

Generator Modeling, cont’d

"d

'd

d

The time varying reactance is typically approximated

using three different values, each valid for a different

time period:

X direct-axis subtransient reactance

X direct-axis transient reactance

X dire

=

=

= ct-axis synchronous reactance

(


17/42

16


Instantaneous ac fault current for balanced 3-phase fault on

the generator terminal:

= 2 1 −

1

+ 1 − 1

+ 1 sin ( + − 2)

Where Eg is the rms, line-to-neutral terminal voltage of

unloaded generator before fault.


18/42

17


• At t = 0 (fault moment): 0 = = which is

called subtransient fault current and its duration depends

on (direct-axis subtransient time constant which isusually around 0.035 seconds)

• As time passes (t >> ): decays and the rms faultcurrent is: = which is called transient fault currentand its duration depends on

(direct-axis transient timeconstant which is usually around 1 to 2 seconds)

• As time passes even more (t >> ): steady-state acfault current becomes: ∞ = =


19/42

18


• Maximum dc offset current for each phase (this max.

happens at a = 0):

, =

2

= 2

Where TA is the armature time constant and is usually

around 0.2 seconds.

NOTE: , , ,, and are provided by thegenerator manufacturer.

• Total short circuit current will be = + ()


20/42

19

Generator Short Circuit ac Current

22

2


21/42

20

Generator Short Circuit Example

l A 500 MVA, 20 kV, 3f is operated with an internal

voltage of 1.05 pu. Assume a solid 3f fault occurs

on the generator's terminal and that the circuit

breaker operates after three cycles. Determine thefault current. Assume

" '

" '

A

0.15, 0.24, 1.1 (all per unit)

0.035 seconds, 2.0 seconds

T 0.2 seconds

d d d

d d

X X X

T T

= = =

= =

=


22/42

21

Generator S.C. Example, cont'd

2.0

ac

0.035

ac

6 base ac3

0.2DC

Substituting in the values

1 1 1

1.1 0.24 1.1( ) 1.05

1 10.15 0.24

1.05(0) 7 p.u.0.15

500 10I 14,433 A (0) 101,000 A3 20 10

I (0) 101 kA 2 143 k

t

t

t

e

I t

e

I

I

e

-

-

é ùæ ö+ - +ç ÷ê úè ø

= ê ú

æ öê ú-ç ÷ê úè øë û

= =

´= = =´

= ´ = RMSA I (0) 175 kA=

=

=®

157


23/42

22

Generator S.C. Example, cont'd

0.052.0

ac0.050.035

ac

0.05

0.2DC

RMS

Evaluating at t = 0.05 seconds for breaker opening

1 1 1

1.1 0.24 1.1(0.05) 1.05

1 10.15 0.24

(0.05) 70.8 kA

I (0.05) 143 kA 111 k A

I (0.05

e

I

e

I

e

-

-

-

é ùæ ö+ - +ç ÷ê úè ø

= ê ú

æ öê ú-ç ÷ê úè ø û

=

= ´ =2 2

) 70.8 111 132 kA= + =

= 4.92 p.u.


24/42

23

Network Fault Analysis Simplifications

l To simplify analysis of symmetrical fault currents

in networks we'll make several simplifications:

1. Transmission lines are represented by their series

reactance

2. Transformers are represented by their leakage

reactances

3. Synchronous machines are modeled as a constant

voltage behind direct-axis subtransient reactance

4. Induction motors are ignored if small (


25/42

24

Network Fault Example

For the following network assume a fault on the

terminal of the generator; all data is per unit

except for the transmission line reactance

2

19.5Convert to per unit: 0.1 per unit

138100

line X = =

generator has 1.05

terminal voltage &

supplies 100 MVAwith 0.95 lag pf


26/42

25

Network Fault Example, cont'd

Faulted network per unit diagram

*'a

To determine the fault current we need to first estimate

the internal voltages for the generator and motor For the generator 1.05, 1.0 18.2

1.0 18.20.952 18.2 E 1.103 7.1

1.05

T G

Gen

V S

I

= = Ð °

Ðæ ö= = Ð - ° = Ðç ÷

è ø


27/42

26

Network Fault Example, cont'd

f

The motor's terminal voltage is then

1.05 0 - (0.9044 - 0.2973) 0.3 1.00 15.8

The motor's internal voltage is

1.00 15.8 (0.9044 - 0.2973) 0.2

1.008 26.6

We can then solve as a linear circuit:

1I

j j

j j

Ð ´ = Ð -

Ð - °- ´

= Ð - °

= .103 7.1 1.008 26.60.15 0.5

7.353 82.9 2.016 116.6 9.09

j j

j

Ð ° Ð - °+

= Ð - °+ Ð - °=


28/42

27

Fault Analysis Solution Techniques

l Circuit models used during the fault allow the

network to be represented as a linear circuit

l There are two main methods for solving for fault

currents:1. Direct method: Use prefault conditions to solve for the

internal machine voltages; then apply fault and solve

directly

2. Superposition: Fault is represented by two opposingvoltage sources; solve system by superposition

– first voltage just represents the prefault operating point

– second system only has a single voltage source


29/42

28

Superposition Approach

Faulted Condition

Exact Equivalent to Faulted Condition

Fault is represented

by two equal and opposite voltage

sources, each with

a magnitude equal

to the pre-fault voltage


30/42

29

Superposition Approach, cont’d

Since this is now a linear network, the faulted voltages

and currents are just the sum of the pre-fault conditions

[the (1) component] and the conditions with just a single

voltage source at the fault location [the (2) component].

Pre-fault (1) component equal to the pre-fault

power flow solution, where obviously the pre-fault

“fault current is zero:

I g(1) = I L = pre-

fault generator

current (usually

very small

compared to I f )


31/42

30

Superposition Approach, cont’d

Fault (2) component due to a single voltage source

at the fault location, with a magnitude equal to the

negative of the pre-fault voltage at the fault location.

(1) (2) (1) (2)

(1) (2) (2)

I

0

gg g m m m

f f f f

I I I I I

I I I I

= + = +

= + = +


32/42

31

Two Bus Superposition Solution

f

(1) (1)

(2) f

(2) f

(2)

Before the fault we had E 1.05 0 ,

0.952 18.2 and 0.952 18.2

Solving for the (2) network we get

E 1.05 07

j0.15 j0.15

E 1.05 02.1

j0.5 j0.5

7 2.1 9.1

0.952

g m

g

m

f

g

I I

I j

I j

I j j j

I

= Ð °

= Ð - ° =- Ð -

Ð °= = =-

Ð °= = =-

= - - =-

= Ð 18.2 7 7.35 82.9 j- °- = Ð - °

This matches

what we

calculated

earlier


33/42

32

Extension to Larger Systems

Superposition can be easily extended to larger systems:

Y bus . E = I

WhereY bus : positive-sequence bus admittance matrix

E : vector of bus voltages

I : vector of current sources

If we define: Z bus

= (Y bus

)-1 then:

Z bus . I = E


34/42

33


General form of Y bus :

§ Diagonal elements yii = sum of all admittancesconnected to bus ‘i’ (self-admittance elements)

§ Off-diagonal elements yij = - (sum of admittancesconnected between buses ‘i’ and ‘j’ )

ü In large power systems, Y bus is a sparse matrix (mostelements are zero)


35/42

34


•

Assuming the fault happens at bus n then the element n of the vector I(2) is: In = - If

² (the negative sign indicates

that the current is flowing away from bus n ).

• Other elements of vector I(2) are all zero (since there is

only one source at bus n ).

• We also know that En(2) = -Vf

( Vf = voltage at the fault location before fault happened)

This equation only needs to be solved for the fault circuit (2):

Z bus(2) . I(2) = E(2)


36/42

35

Calculating Fault Current

⋯ ⋯

⋮⋮

⋮⋮

⋮ ⋮… ⋮ ⋮

…

0

0

⋮−⋮0

=

()()

⋮−⋮

()

So for a power system with a total of N buses:

→ − = − → = Z nn is known as the driving point impedance.


37/42

36

Calculating Bus Voltages

• To calculate voltage at any bus k in circuit (2):

() = − = −

Z kn is known as the transfer point impedance.

If pre-faults load currents are assumed to be zero (sincethey are very small compared to I f ) ® () = then:

= − + = 1 −

• To find bus voltages, use superposition:

= () + ()


38/42

37

3-Generator System Fault Example

g1 2 3

For simplicity assume the system is unloaded

before the fault with

E 1.05 0

Hence all the prefault currents are zero.

g g E E = = = Ð °


39/42

38


15 10 0

10 20 5

0 5 9

bus j

-é ùê ú= -ê ú

-ê úû

Y

115 10 0

10 20 5

0 5 9

0.1088 0.0632 0.0351

0.0632 0.0947 0.0526

0.0351 0.0526 0.1409

bus Z j

j

--é ù

ê ú= -ê ú

-ê úë û

é ùê ú=ê úê úû


40/42

39

1

(2)

1.05For a fault at bus 1 we get I 9.6

0.1088

0.1088 0.0632 0.0351 9.6

0.0632 0.0947 0.0526 0

0.0351 0.0526 0.1409 0

1.05 0

0.60 00.337 0

f j I j

j

j

-= = =-

-

é ùé ù

ê úê ú= ê úê úê úê úë ûë û

- Ð °é ùê ú

= - Ð °ê ú- Ð °ê úû

V


¬ Bus voltages at fault circuit (2)


41/42

40


1.05 0 1.05 0 0 0

1.05 0 0.606 0 0.444 0

1.05 0 0.337 0 0.713 0

Ð ° - Ð ° Ð °é ù é ù é ùê ú ê ú ê ú= Ð ° + - Ð ° = Ð °ê ú ê ú ê ú

Ð ° - Ð ° Ð °ê ú ê ú ê úû ë û ë û

V


42/42

41

Circuit Breaker and Fuse Selection

Study section 7.5 of “Glover” (Posted on course website)

Date post:	07-Jul-2018
Category:	Documents
Upload:	parth
View:	218 times
Download:	0 times

lec5(2)(1) power system

Documents