Download Slides

ECE 476
POWER SYSTEM ANALYSIS
Lecture 18
Markets, Fault Analysis
Professor Tom Overbye
Department of Electrical and
Computer Engineering
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.
1
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
2
In the News: Electricity Price Caps
•
•
•
Texas (ERCOT) is considering raising the
maximum wholesale price cap
from $3000/MWh to
$6000/MWh to encourage more
electric supply.
Average price in 2010 was
$40/MWh, down from $86/Mwh
in 2008.
ERCOT is not subject to most
federal regulations
Source: Wall Street Journal, Oct 3, 2011
3
Market Experiments
4
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
5
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
6
Worldwide Lightning Strike Density
Units are Lightning Flashes per square km per year; Florida is
top location in the US; very few on the West Coast, or HI, AK. This
is an important consideration when talking about electric reliability!
Source: http://science.nasa.gov/science-news/science-at-nasa/2001/ast05dec_1/
7
Lightning Strike Event Sequence
1.
Lighting hits line, setting up an ionized path to
ground



2.
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.
Conduction path is maintained by ionized air after
lightning stroke energy has dissipated, resulting in
high fault currents (often > 25,000 amps!)
8
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

4.
Circuit breakers open to de-energize line in an
additional one to two cycles


5.
nearby locations see decreased voltages
breaking tens of thousands of amps of fault current is no
small feat!
with line removed voltages usually return to near normal
Circuit breakers may reclose after several seconds,
trying to restore faulted line to service
9
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
10
RL Circuit Analysis

To understand fault analysis we need to review the
behavior of an RL circuit
v(t ) 
2 V cos( t   )
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
11
RL Circuit Analysis, cont’d
1. Steady-state current component (from standard
phasor analysis)
iac (t ) 
where Z 
I ac
2V cos(t   )
Z
R 2  ( L)2  R 2  X 2
V

Z
12
RL Circuit Analysis, cont’d
2. Exponentially decaying dc current component
idc (t )  C1e
t
T
where T is the time constant, T  L R
The value of C1 is determined from the initial
conditions:
t
2V
i (0)  0  i ac (t )  i dc (t ) 
cos(t     Z )  C1e T
Z
2V
C1  
cos(   Z ) which depends on 
Z
13
Time varying current
14
RL Circuit Analysis, cont’d
Hence i(t) is a sinusoidal superimposed on a decaying
dc current. The magnitude of i dc (0) depends on when
the switch is closed. For fault analysis we're just
2V
concerned with the worst case: C1 
Z
i (t )  i ac (t )  i dc (t )
i (t )


2V
2V t T
cos(t ) 
e
Z
Z
t
2V
(cos(t )  e T )
Z
15
RMS for Fault Current
t
2V
The function i(t) 
(cos(t )  e T ) is not periodic,
Z
so we can't formally define an RMS value. However,
as an approximation define
I RMS (t ) 

2
2
iac
(t )  idc
(t )
2
I ac
2t
2  T
 2 I ac e
This function has a maximum value of 3 I ac
Therefore the dc component is included simply by
multiplying the ac fault currents by 3
16
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
'
Ea
17
Generator Modeling, cont’d
The time varying reactance is typically approximated
using three different values, each valid for a different
time period:
X"d
 direct-axis subtransient reactance
X 'd
 direct-axis transient reactance
Xd
 direct-axis synchronous reactance
18
Generator Modeling, cont’d
For a balanced three-phase fault on the generator
terminal the ac fault current is (see page 386)
iac (t ) 
 1  1
1  
 ' 

e
X
 Xd Xd 
'  d
2Ea 
 t "
1 
 1
Td

e
 X " X ' 
d 
 d
t
Td'



 sin(t   )



where
Td"  direct-axis subtransient time constant (  0.035sec)
Td'  direct-axis transient time constant (  1sec)
19
Generator Modeling, cont'd
The phasor current is then
 1  1
1  
 ' 

e
X
 Xd Xd 
'  d
I ac  Ea 
 t "
1 
 1
Td
 X "  X '  e
d 
 d
t
Td'







The maximum DC offset is
2 Ea' 
I DC (t ) 
e
"
Xd
t
TA
where TA is the armature time constant (  0.2 seconds)
20
Generator Short Circuit Currents
21
Generator Short Circuit Currents
22
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"  0.15, X d'  0.24, X d  1.1 (all per unit)
Td"  0.035 seconds, Td'
 2.0 seconds
TA  0.2 seconds
23
Generator S.C. Example, cont'd
Substituting in the values
1   t 2.0 
1  1


e



1.1  0.24 1.1 

I ac (t )  1.05 

 1  1  e t 0.035

 0.15 0.24 

I ac (0)  1.05
 7 p.u.
0.15
I base

500  106
 14,433 A I ac (0)  101,000 A
3
3 20  10
I DC (0)  101 kA  2 e
t
0.2
 143 k A IRMS (0)  175 kA
24
Generator S.C. Example, cont'd
Evaluating at t = 0.05 seconds for breaker opening
1   0.05 2.0 
1  1


e



1.1  0.24 1.1 

I ac (0.05)  1.05 


0.05
 1  1  e

0.035
 0.15 0.24 

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)  70.82  1112  132 kA
25