EE 369

POWER SYSTEM ANALYSIS

Lecture 18 Fault Analysis Tom Overbye and Ross Baldick

1

Announcements

 Read Chapter 7.  Homework 12 is 6.43, 6.48, 6.59, 6.61, 12.19, 12.22, 12.20, 12.24, 12.26, 12.28, 12.29; due Tuesday Nov. 25.

 Homework 13 is 12.21, 12.25, 12.27, 7.1, 7.3, 7.4, 7.5, 7.6, 7.9, 7.12, 7.16; due Thursday, December 4.

2

Transmission Fault Analysis

 The cause of electric power system faults is insulation breakdown/compromise.

 This breakdown can be due to a variety of different factors: – Lightning ionizing air, – Wires blowing together in the wind, – Animals or plants coming in contact with the wires, – Salt spray or pollution on insulators. 3

Transmission Fault Types

 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 (considered in EE 368L).

 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

Lightning Strike Event Sequence

1. Lighting hits line, setting up an ionized path to ground  30 million lightning strikes per year in US!

 a single typical stroke might have 25,000 amps, with a rise time of 10  s, dissipated in 200  s.  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

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:  nearby locations see decreased voltages 4. Circuit breakers open to de-energize line in an additional one to two cycles:  breaking tens of thousands of amps of fault current is no small feat!  with line removed voltages usually return to near normal.

5. Circuit breakers may reclose after several

Fault Analysis

 Fault currents cause equipment damage due to both thermal and mechanical processes.

 Goal of fault analysis is to determine the magnitudes of the currents present during the fault: – need to determine the maximum current to ensure devices can survive the fault, – need to determine the maximum current the circuit breakers (CBs) need to interrupt to correctly size the CBs.

7

RL Circuit Analysis

 To understand fault analysis we need to review the behavior of an RL circuit

R L

 ) (Note text uses sinusoidal voltage instead of cos!) Before the switch is closed, 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.

8

RL Circuit Analysis, cont’d

1. Steady-state current component (from standard phasor analysis) Steady-state phasor current magnitude is

I ac



V Z

, 

R

2  ( 

L

) 2 

R

2 

X

2 and current phasor angle is 

Z Z

 Corresponding in stantaneous current is: 

i

ac 

Z Z

) 9

RL Circuit Analysis, cont’d

2. Exponentially decaying dc current component

i dc

 

t T T



L R

The value of

C

1 is determined from the initial conditions:

i

(0)

C

1  

i

ac 

i

dc  2

V

cos(   

Z

) 

Z

2

V

cos(  

Z

) which depends on 

Z

10 

t T

Time varying current

i(t) Superposition of steady-state component and exponentially decaying dc offset.

time 11

RL Circuit Analysis, cont’d

dc current. The magnitude of

i

dc (0) depends on when the switch is closed. For fault analysis we're just concerned with the worst case.

Highest DC c   

i ac



i dc

   2

V Z

2

V

cos( 

t

)  2

V e



t T Z



t

) 

e



t T

)

Z

Z   ,

C

1  2

V Z

12

RMS for Fault Current

The interrupting capability of a circuit breaker is specified in terms of the RMS current it can interrupt.

 2

V



t

) 

e



t T

) is

Z

not periodic, so we can't formally define an RMS value . However, if

T t

then we can approximate the current as a sinusoid plus a time-invarying dc offset. The RMS value of such a current is equal to the square root of the sum of the squares of the indivi dual RMS values of the two current components.

RMS for Fault Current

I RMS 

I

2

ac



I

2

dc

, where

I ac



V Z

,

I dc

 

I

2

ac

 2

ac

 2

t T

2

V Z e



t T

 This function has a maximum value of 3

I ac

.

Therefore the worst case effect of the dc component is included simply by 2 mu ltiplying the ac fault currents by 3.



t T

, 14

Generator Modeling During Faults

 During a fault the only devices that can contribute fault current are those with energy storage.

 Thus the models of generators (and other rotating machines) are very important since they contribute the bulk of the fault current.

 Generators can be approximated as a constant voltage behind a time-varying reactance:

E

'

a

15

Generator Modeling, cont’d

The time varying reactance is typically approximated using three different values, each valid for a different time period: X " d X ' d X d    direct-axis subtransient reactance direct-axis transient reactance dire ct-axis synchronous reactance Can then estimate currents using circuit theory: For example, could calculate steady-state current that would occur after a three-phase short-circuit if no circuit breakers interrupt current. 16

Generator Modeling, cont’d

For a balanced three-phase fault on the generator

i

terminal the ac fault current is (see page 362) ac  2

E a

'        

X

1

d X

1 "

d

   

X

1 1

X

'

d

'

d



X

 

e

 1

d t T d

"  

e



t T d

'        sin(   ) where

T d

"

T d

'   direct-axis su btransient time constant (  0.035sec) direct-axis transient time constant (  1sec) 17

Generator Modeling, cont'd

The phasor current is then

I

ac 

E a

'        

X

1

d X

1 "

d

   

X

1 1

X

'

d

'

d



X

 

e

 1

d t T d

"  

e

 The maximum DC offset is

t T d

'       

I

DC  2

X

"

d E a

'

e



t T A

where

T A

is the armature time constant (  0.2 seconds) 18

Generator Short Circuit Currents

19

Generator Short Circuit Currents

20

Generator Short Circuit Example

 A 500 MVA, 20 kV, 3  is operated with an internal voltage of 1.05 pu. Assume a solid 3  fault occurs on the generator's terminal and that the circuit breaker operates after three cycles. Determine the fault current. Assume

X

"

d T T d

"

A

 0.15,

X

'

d

 0.24,

X

 0.035 seconds,

T d

'  0.2 seconds 

d

 1.1 (all per unit) 2.0 seconds 21

Generator S.C. Example, cont'd

Substituting in the values

I

ac  1.05

    1 1.1

1  0.15

 1  1 0.24 1.1



e

  1 0.24

e



t

2.0

t

0.035

    

I I

ac (0)  base  1.05

0.15

 7 p.u.

 6  3  14, 433 A

I

ac (0)  101,000 A

I

DC (0)  101 kA  2

e t

0.2

 143 k A

I

RMS (0)  175 kA 22

Generator S.C. Example, cont'd

Evaluating at t = 0.05 seconds for breaker opening

I

ac (0.05)  1.05

    1 1.1

1  0.15

 1  1 0.24 1.1



e

 0.05

2.0

 1 0.24

e

 0.05

0.035

I

ac (0.05)  70.8 kA

I

DC (0.05)  143 

e

 0.05

0.2

kA  111 k A     

I

RMS (0.05

)  70.8

2  111 2  132 kA 23

Network Fault Analysis Simplifications

 To simplify analysis of 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 or treated as synchronous machines 5. Other (nonspinning) loads are ignored 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 generator has 1.05

terminal voltage & supplies 100 MVA with 0.95 lag pf Convert to per unit:

X line

 19.5

138 2 100  0.1 per unit 25

Network Fault Example, cont'd

Faulted network per unit diagram To determine the fault current we need to first estimate

I

the internal voltages for the generator and motor For the generator

V T

 1.05,

S G

 1.0 18.2



Gen

 1.05

 *  0.952

  

E

'

a

 1.103 7.1

 26

Network Fault Example, cont'd

The motor's terminal voltage is then 

j

The motor's internal voltage is 1.00

  15.8

   1.008

  26.6

 

j

0.2

  15.8



I

We can then solve as a linear circuit:

f

 1 

j

0.15

  1.008

 

j

0.5

26.6

  7.353

  82.9

  2.016

  116.6

 

j

9.09

27

Fault Analysis Solution Techniques

  Circuit models used during the fault allow the network to be represented as a linear circuit 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 opposing voltage sources; solve system by superposition: – first voltage just represents the prefault operating point – second system only has a single voltage source.

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 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 Obvious the pre-fault “fault current” is zero!

30

Superposition Approach, cont’d

Fault (1) 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.

I g



I f



I

(1)

g



I

(2)

g I

(1)

f



I

(2)

f I m



I

(1)

m



I

(2)

m I

(2)

f

31

Two Bus Superposition Solution

Before the fault we had E f

I

(1)

g

 0.952

   

I

(1)

m

  0.952

  18.2

 Solving for the (2) network we get

I

(2)

g

 E f j0.15

 j0.15

 

j

7

I

(2)

m

 E f j0.5

 j0.5

 

j

2.1

I I

(2)

f g

 

j

7 

j

2.1

 0.952

  18.2

   

j

7

j

9.1

 7.35

  82.9

 This matches what we calculated earlier 32

Extension to Larger Systems

The superposition approach can be easily extended to larger systems. Using the

Y

bus

Y

bus

V



I

we have For the second (2) system there is only one voltage source so is all zeros except at the fault loca tion

I

        0

I

0

f

      However to use this approach we need to first determine I

f

33

Determination of Fault Current

Define the bus impedance matrix

Z

bus as

Z

bus

Y

 1

bus

V



Z

bus

I

Then    

Z Z

11

n

1

Z

1

n Z nn

        0 

I

0

f

     

V

1    

V V

2

n

(2)  1

n

(2) (2)      For a fault a bus i we get -I f

Z ii

 

V f

 

V i

(1) 34

Determination of Fault Current

Hence

I f



V i

(1)

Z ii

Where

Z ii

driving point impedance

ij



j

) transfer point imepdance Voltages during the fault are also found by superposition

V i



V i

(1) 

V i

(2)

V i

(1) are prefault values 35