Lecture 18Markets, Fault Analysis

Professor Tom OverbyeDepartment of Electrical and

Computer Engineering

ECE 476

POWER SYSTEM ANALYSIS

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Announcements

Be reading Chapter 7 HW 7 is 12.26, 12.28, 12.29, 7.1 due October 27 in class.

Correct case for 12.29 was emailed out; demo of OPF during class

US citizens and permanent residents should consider applying for a Grainger Power Engineering Awards. Due Nov 1. See http://energy.ece.illinois.edu/grainger.html for details.

The Design Project, which is worth three regular homeworks, is assigned today; it is due on Nov 17 in class. It is Design Project 2 from Chapter 6 (fifth edition of course). For tower configuration assume a symmetric conductor spacing, with the distance in feet

given by the following formula:

(Last two digits of your EIN+50)/9. Example student A has an UIN of xxx65. Then his/her spacing is (65+50)/9 = 12.78 ft.

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Why not pay as bid?

Two options for paying market participants– Pay as bid– Pay last accepted offer

What would be potential advantages/disadvantages of both?

Talk about supply and demand curves, scarcity, withholding, market power

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In the News: Electricity Price Caps

• Texas (ERCOT) is considering raising the maximum wholesale price cap from $3000/MWh to $6000/MWh to encourage moreelectric supply.

• Average price in 2010 was $40/MWh, down from $86/Mwhin 2008.

• ERCOT is not subject to mostfederal regulations

Source: Wall Street Journal, Oct 3, 2011

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Market Experiments

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Fault Analysis

The cause of electric power system faults is insulation breakdown

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

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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

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 On very high voltage lines faults are practically always single line to

ground due to large conductor spacing

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Worldwide Lightning Strike Density

Source: http://science.nasa.gov/science-news/science-at-nasa/2001/ast05dec_1/

Units are Lightning Flashes per square km per year; Florida istop location in the US; very few on the West Coast, or HI, AK. Thisis an important consideration when talking about electric reliability!

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Lightning Strike Event Sequence

1. Lighting hits line, setting up an ionized path to ground Tens of millions of 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!)

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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 seconds, trying to restore faulted line to service

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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 insure devices

can survive the fault– need to determine the maximum current the circuit

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

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RL Circuit Analysis

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

( )

2 cos( )

v t

V t

Before the switch is closed obviously i(t) = 0.When the switch is closed at t=0 the current willhave two components: 1) a steady-state value2) a transient value

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RL Circuit Analysis, cont’d

ac

2 2 2 2

1. Steady-state current component (from standard

phasor analysis)

2 cos( )i ( )

where ( )

ac

V tt

Z

Z R L R X

VI

Z

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RL Circuit Analysis, cont’d

dc 1

1

ac dc 1

1

2. Exponentially decaying dc current component

i ( )

where T is the time constant,

The value of is determined from the initial

conditions:

2(0) 0 i ( ) i ( ) cos( )

2

tT

tT

Z

t C e

LT RC

Vi t t t C e

Z

VC

Z

cos( ) which depends on Z

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Time varying current

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RL Circuit Analysis, cont’d

dc

1

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

dc current. The magnitude of i (0) depends on when

the switch is closed. For fault analysis we're just

2concerned with the worst case:

( )

VC

Zi t

ac dci ( ) i ( )

2 2( ) cos( )

2(cos( ) )

tT

tT

t t

V Vi t t e

Z Z

Vt e

Z

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RMS for Fault Current

2 2RMS

22 2

2The function i(t) (cos( ) ) is not periodic,

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

as an approximation define

I ( ) ( ) ( )

2

This function has a maximum va

tT

ac dc

tT

ac ac

Vt e

Z

t i t i t

I I e

lue of 3

Therefore the dc component is included simply by

multiplying the ac fault currents by 3

acI

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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

'aE

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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

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Generator Modeling, cont’d

'

"

''

ac

" '

"d

For a balanced three-phase fault on the generator

terminal the ac fault current is (see page 386)

1 1 1

i ( ) 2 sin( )1 1

where

T direct-axis su

d

d

tT

d dda t

T

d d

eX XX

t E t

eX X

'd

btransient time constant ( 0.035sec)

T direct-axis transient time constant ( 1sec)

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Generator Modeling, cont'd

'

"

''

ac

" '

'

DC "

A

The phasor current is then

1 1 1

1 1

The maximum DC offset is

2I ( )

where T is the armature time constant ( 0.2 seconds)

d

d

A

tT

d dda t

T

d d

tTa

d

eX XX

I E

eX X

Et e

X

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Generator Short Circuit Currents

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Generator Short Circuit Currents

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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

" '

" '

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

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Generator S.C. Example, cont'd

2.0

ac0.035

ac

6

base ac3

0.2DC

Substituting in the values

1 1 11.1 0.24 1.1

( ) 1.051 1

0.15 0.24

1.05(0) 7 p.u.0.15

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

3 20 10

I (0) 101 kA 2 143 k

t

t

t

e

I t

e

I

I

e

RMSA I (0) 175 kA

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Generator S.C. Example, cont'd

0.052.0

ac 0.050.035

ac

0.050.2

DC

RMS

Evaluating at t = 0.05 seconds for breaker opening

1 1 11.1 0.24 1.1

(0.05) 1.051 1

0.15 0.24

(0.05) 70.8 kA

I (0.05) 143 kA 111 k A

I (0.05

e

I

e

I

e

2 2) 70.8 111 132 kA