ECE 476
Power System Analysis
Lecture 19: Ground, Symmetrical
Components
Prof. Tom Overbye
Dept. of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
overbye@illinois.edu
Announcements
• Read Chapters 8 and 9
• HW 9 is 7.6, 7.13, 7.19, 7.28, 8.4; will be covered by
an in-class quiz on Nov 5
• Second exam is Thursday Nov. 12 during class.
Closed book, closed notes, one new note sheet, first
exam note sheet, and calculators allowed
• Abbott Power Plant tour is Tuesday Nov. 10 during
class (we’ll be meeting at Abbott on the north side,
NW corner of Oak and Armory)
• Design project is assigned today; due on Nov 19;
details are on the website
1
Design Project: Due on Nov 19
• Goal is to optimally add new transmission lines and/or
transformers to fix existing problems and problems
caused by the addition of a new wind farm.
Eagle
Island Electric Company (IEC)
Pheasant
System Losses: 35.22 MW
8 7 MW
1 9 M var
1 .0 1 0 pu
1 .0 1 5 pu
28%
16%
Canary
1 .0 1 8 pu
24%
1 3 0 MW
3 0 M var
Parrot
Turkey
12%
Ostrich
1 5 M var
Sparrow
7 0 M var
NewWind
0 MW
1 6 5 MW
3 0 M var
3%
7%
Finch
18%
Crow
19%
1 3 5 MW
2 0 M var
5 5 MW
1 5 M var
6%
Lark
0 .9 9 6 pu
Heron
Hawk
1 6 1 MW
Finch
1 7 6 MW
1 5 M var
33%
57%
Condor
1 4 0 MW
2 0 M var
1 .0 2 5 pu
47%
Peacock
1 .0 1 0 pu
1 .0 0 4 pu
8 8 MW
1 1 M var
1 .0 1 5 pu
sl a ck
0 .9 9 9 pu
48%
2 1 M var
Piper
35%
0 .9 9 4 pu
2 8 M var
47%
22%
Assignment
will count
as three
regular
homeworks
3 9 6 MW
1 4 0 MW
3 2 M var
1 2 8 MW
0 .9 8 8 pu
22%
53%
1 7 5 MW
4 0 M var
36%
11%
3%
13%
0 .9 8 9 pu
16%
45%
1 .0 0 5 pu
Bluebird
45%
5 1 M var
22%
0 .9 9 7 pu
30%
1 .0 1 4 pu
Oriole
1 .0 0 1 pu
0 .9 9 1 pu
29%
1 .0 1 0 pu
27%
10%
2%
1 1 5 MW
2 5 M var
35%
23%
0 .9 9 3 pu
1 3 0 MW
4 5 M var
0 .0 0 0 pu
1 .0 1 2 pu
0 .9 9 3 pu
1 5 0 MW
3 9 M var
7%
Owl
1 .0 0 2 pu
6 0 MW
1 5 M var
8%
30%
2 6 8 MW
1 2 8 M var
0 .9 9 6 pu
1 3 2 MW
Mallard
28%
Rook
14%
30%
1 .0 1 5 pu
0 .9 9 6 pu
3 5 M var
17%
1 .0 1 4 pu
14%
2%
1 6 0 MW
1 .0 1 5 pu
8%
6%
13%
1 .0 2 3 pu
Cardinal
7%
10%
2 0 0 MW
6 0 M var
1 9 8 MW
3 5 M var
0 .9 9 3 pu
9%
1 .0 1 9 pu
0 .9 9 2 pu
5 0 5 MW
5%
1 1 2 MW
4 0 M var
1 .0 1 5 pu
5 7 M var
8 3 M var
1 1 0 MW
0 .9 9 9 pu 3 0 M var
5 6 M var
Robin
44%
38%
39%
1 .0 0 6 pu
7 8 M var
Woodpecker
17%
1 .0 2 0 pu
24%
46%
19%
3 5 0 MW
Dove
3 0 0 MW
7 5 MW
1 5 M var
1 5 0 MW
7 0 M var
1 .0 1 8 pu
Flamingo
9 5 MW
2 3 M var
1 .0 1 3 pu
13%
1 .0 1 8 pu
Hen
9 0 0 MW
78%
4%
78%
1 1 2 0 MW
1 .0 2 0 pu
2
Symmetrical Faults:
Extension to Larger Systems
The superposition approach can be easily extended
to larger systems. Using the Ybus we have
Ybus V  I
For the second (2) system there is only one voltage
source so I is all zeros except at the fault location


 0 


I   I f 


 0 


However to use this
approach we need to
first determine If
3
Determination of Fault Current
Define the bus impedance matrix Z bus as
Z bus
 Z11
Then 

 Z n1
1
Ybus
V  Z busI
(2) 

V


1
 (2) 


Z1n  0
V

  2 
  I   

 f  

Z nn   0  V (2) 
n 1

  (2) 
Vn 
For a fault a bus i we get -If Zii  V f  Vi(1)
4
Determination of Fault Current
Hence
Vi(1)
If 
Zii
Where
Zii
driving point impedance
Zij (i  j )
transfer point imepdance
Voltages during the fault are also found by superposition
Vi  Vi(1)  Vi(2)
Vi(1) are prefault values
5
Three Gen System Fault Example
For simplicity assume the system is unloaded
before the fault with
E g1  Eg 2  Eg 3  1.050
Hence all the prefault currents are zero.
6
Three Gen Example, cont’d
Ybus
0
 15 10
 j  10 20 5 


5 9 
 0
1
Zbus
0
 15 10
 j  10 20 5 


5 9 
 0
 0.1088 0.0632 0.0351
 j 0.0632 0.0947 0.0526


 0.0351 0.0526 0.1409
7
Three Gen Example, cont’d
1.05
For a fault at bus 1 we get I1 
 j 9.6   I f
 j 0.1088
V (2)
 0.1088 0.0632 0.0351  j 9.6 
 j 0.0632 0.0947 0.0526   0 



 0.0351 0.0526 0.1409   0 
 1.050 
  0.600 


 0.3370
8
Three Gen Example, cont’d
1.050  1.050   00 
V  1.050   0.6060   0.4440 

 
 

1.050  0.3370  0.7130 
9
PowerWorld Example 7.5: Bus 2 Fault
One
Five
Four
Three
7 pu
11 pu
slack
0.724 pu
0.000 deg
0.579 pu
0.000 deg
0.687 pu
0.000 deg
0.798 pu
0.000 deg
0.000 pu
0.000 deg
Two
10
Problem 7.28
SLACK345
0.79 pu
5 pu
RAY345
sla ck
0.78 pu
SLACK138
TIM345
0.70 pu
RAY138
0.83 pu
TIM138
0.61 pu
0.79 pu
0.64 pu
0.52 pu
TIM69
RAY69
PAI69
0.56 pu
0.58 pu
GROSS69
FERNA69
MORO138
0.50 pu
WOLEN69
HISKY69
0.52 pu
0.59 pu
PETE69
0.64 pu
BOB138
DEMAR69
HANNAH69
0.50 pu
UIUC69
BOB69
0.43 pu
0.00 pu
9 pu
LYNN138
0 pu
0.564 pu
0.56 pu
BLT138
0.61 pu
AMANDA69
SHIMKO69
HOMER69
0.24 pu
0.82 pu
BLT69
0.32 pu
HALE69
9 pu
0.62 pu
0.61 pu
0.35 pu
0.62 pu
PATTEN69
ROGER69
0.64 pu
LAUF69
0.68 pu
WEBER69
0 pu
3 pu
0.70 pu
LAUF138
0.77 pu
0.75 pu
SAVOY69
0.77 pu
2 pu
JO138
JO345
BUCKY138
0.77 pu
3 pu
SAVOY138
3 pu
0.84 pu
0.91 pu
11
Analysis of Unsymmetric Systems
• Except for the balanced three-phase fault, faults
result in an unbalanced system.
• The most common types of faults are single lineground (SLG) and line-line (LL). Other types are
double line-ground (DLG), open conductor, and
balanced three phase.
• System is only unbalanced at point of fault!
• The easiest method to analyze unbalanced system
operation due to faults is through the use of
symmetrical components
• First we’ll go brief coverage of ground calculations
12
Grounding
• When studying unbalanced system operation how a
system is grounded can have a major impact on the
fault flows
• Ground current does not come into play during
balanced system analysis (since net current to
ground would be zero).
• Becomes important in the study of unbalanced
systems, such as during most faults.
13
Grounding, cont’d
• Voltages are always defined as a voltage
difference. The ground is used to establish the zero
voltage reference point
–
ground need not be the actual ground (e.g., an airplane)
• During balanced system operation we can ignore
the ground since there is no neutral current
• There are two primary reasons for grounding
electrical systems
1.
2.
safety
protect equipment
14
How good a conductor is dirt?
• There is nothing magical about an earth ground.
All the electrical laws, such as Ohm’s law, still
apply.
• Therefore to determine the resistance of the ground
we can treat it like any other resistive material:
  conductor length
Resistance R 
cross sectional area
  2.65  108 -m for aluminum
  1.68  108 -m for copper
where  is the resistivity
15
How good a conductor is dirt?
  2.65  108 -m for aluminum




 5  1016 -m for quartz (insulator!)
 160 -m for top soil
 900 -m for sand/gravel
 20 -m for salt marsh
What is resistance of a mile long, one inch diameter,
circular wire made out of aluminum ?
2.65  108  1609

R=

0.083
mile
  0.01282
16
How good a conductor is dirt?
What is resistance of a mile long, one inch diameter,
circular wire made out of topsoil?
160  1609
6 
R=
 500  10
2
mile
  0.0128
In order to achieve 0.08 
with our dirt wire
mile
we would need a cross sectional area of
160  1609
6 2
 3.2  10 m
(i.e., a radius of about 1000 m)
0.08
But what the ground lacks in  , it makes up for in A!
17
Calculation of grounding resistance
• Because of its large cross sectional area the earth is
actually a pretty good conductor.
• Devices are physically grounded by having a
conductor in physical contact with the ground;
having a fairly large area of contact is important.
• Most of the resistance associated with establishing
an earth ground comes within a short distance of
the grounding point.
• Typical substation grounding resistance is between
0.1 and 1 ohm; fence is also grounded, usually by
connecting it to the substation ground grid.
18
Calculation of grounding R, cont’d
• Example: Calculate the resistance from a
grounding rod out to a radial distance x from the
rod, assuming the rod has a radius of r:
In general we have
x
R
cross sectional area
but now area changes
 dx
dR 
with length.
2  length  x
x
 dxˆ

x
R  

ln
2  length  xˆ 2  length r
r
19
Calculation of grounding R, cont’d
For example, if r  1.5 inches, length = 10 feet,
and   160   m we get the following values as
a function of x (in meters)
160
x
R 
ln
2  3.05 0.038
The actual values will be
x
R
substantially less since
1m
27.2 
we’ve assumed no current
flowing downward into
10 m
46.4 
the ground
100 m 65.6 
100 km 83.4 
20
Symmetric Components
• The key idea of symmetrical component analysis is to
decompose the system into three sequence networks.
The networks are then coupled only at the point of the
unbalance (i.e., the fault)
• The three sequence networks are known as the
–
–
–
positive sequence (this is the one we’ve been using)
negative sequence
zero sequence
• Presented in paper by Charles .L Fortescue in 1918
(judged as most important power paper of 20th century)
Heydt, G. T.; Venkata, S. S.; Balijepalli, N. (October 24, 2000). "High Impact Papers in Power Engineering,
1900-1999" Proceedings 2000 North American Power Symposium, vol. 1, October 2000. North American Power
Symposium (NAPS). Waterloo, Ontario.
21
Positive Sequence Sets
• The positive sequence sets have three phase
currents/voltages with equal magnitude, with phase
b lagging phase a by 120°, and phase c lagging
phase b by 120°.
• We’ve been studying positive sequence sets
Positive sequence
sets have zero
neutral current
22
Negative Sequence Sets
• The negative sequence sets have three phase
currents/voltages with equal magnitude, with
phase b leading phase a by 120°, and phase c
leading phase b by 120°.
• Negative sequence sets are similar to positive
sequence, except the phase order is reversed
Negative sequence
sets have zero
neutral current
23
Zero Sequence Sets
• Zero sequence sets have three values with equal
magnitude and angle.
• Zero sequence sets have neutral current
24
Sequence Set Representation
• Any arbitrary set of three phasors, say Ia, Ib, Ic can
be represented as a sum of the three sequence sets
I a  I a0  I a  I a
I b  I b0  I b  I b
I c  I c0  I c  I c
where
I a0 , I b0 , I c0 is the zero sequence set
I a , I b , I c is the positive sequence set
I a , I b , I c is the negative sequence set
25