ECE 476 Power System Analysis Lecture 19: Ground, Symmetrical Components Prof. Tom Overbye Dept. of...
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Transcript of ECE 476 Power System Analysis Lecture 19: Ground, Symmetrical Components Prof. Tom Overbye Dept. of...
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
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
2
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.
slack
Island Electric Company (IEC)
350 MW
505 MW
396 MW
268 MW 128 Mvar
115 MW 25 Mvar
150 MW 70 Mvar
110 MW 30 Mvar
150 MW 39 Mvar
60 MW 15 Mvar
55 MW 15 Mvar
130 MW 30 Mvar
175 MW 40 Mvar
140 MW 32 Mvar 176 MW
15 Mvar
165 MW 30 Mvar
132 MW
15 Mvar
160 MW
35 Mvar
112 MW 40 Mvar
200 MW 60 Mvar
95 MW
23 Mvar
75 MW 15 Mvar
198 MW 35 Mvar
87 MW 19 Mvar
161 MW
21 Mvar
135 MW 20 Mvar
140 MW 20 Mvar
88 MW 11 Mvar
130 MW 45 Mvar
128 MW
28 Mvar
45%
22%
33%
35%
4%
13%
23%
14%
13%
28%
35%
8%
17%
13%
57%
28%
6%
3%
6%
16%
30%
22%
10%
27%
12%
30%
44%
9%
5%
46%
78%
45%
53%
47%
19%
900 MW
70 Mvar
51 Mvar
78 Mvar
47%
3%
10%
19%
16%
22%
48%
RobinEagle
Rook
SparrowBluebird
Hawk
Parrot
Turkey
Mallard
Condor
Hen
Dove
Piper
Cardinal
24%
1120 MW
2%
2%
29%
8%
36%
7%
18%
7%
11%
24%
17%
38% 39%
30%
7%
78%
14%
83 Mvar
300 MWSystem Losses: 35.22 MW
0.000 pu
NewWind
0 MW
56 Mvar
0.996 pu
1.018 pu
1.018 pu
1.015 pu
1.019 pu
1.015 pu
1.002 pu
0.997 pu
0.996 pu
1.004 pu
0.988 pu
1.001 pu
1.010 pu
1.020 pu
0.999 pu
0.989 pu
0.993 pu
0.996 pu
1.014 pu0.993 pu
1.010 pu
1.015 pu
1.023 pu
1.015 pu
1.010 pu1.006 pu
0.999 pu
1.018 pu1.020 pu
0.993 pu
1.012 pu
1.013 pu
0.992 pu
0.991 pu
1.005 pu
1.014 pu
1.015 pu0.994 pu
1.025 pu
57 Mvar
Ostrich
Crow
Peacock
Lark Finch
Oriole
Heron
Finch
Owl
Woodpecker
Pheasant
Canary
Flamingo Assignmentwill countas threeregularhomeworks
3
Symmetrical Faults: Extension to Larger Systems
bus
bus
The superposition approach can be easily extended
to larger systems. Using the we have
For the second (2) system there is only one voltage
source so is all zeros except at the fault loca
Y
Y V I
I tion
0
0
fI
I
However to use thisapproach we need tofirst determine If
4
Determination of Fault Current
bus
1bus bus
(2)1
(2)11 1 2
(2)1 1
(2)
(1)f
Define the bus impedance matrix as
0
Then
0
For a fault a bus i we get -I
bus
n
f
n nn n
n
ii f i
V
Z Z V
I
Z Z V
V
Z V V
Z
Z Y V Z I
5
Determination of Fault Current
(1)
f
ii
ij
(1) (2) (1)i
Hence
I
Where
Z driving point impedance
Z ( ) transfer point imepdance
Voltages during the fault are also found by superposition
V are prefault values
i
ii
i i i
VZ
i j
V V V
6
Three Gen 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 gE E
7
Three Gen Example, cont’d
15 10 0
10 20 5
0 5 9bus j
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
busZ j
j
8
Three Gen Example, cont’d
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 0
0.337 0
fj Ij
j
j
V
9
Three Gen Example, cont’d
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
V
10
PowerWorld Example 7.5: Bus 2 Fault
slack
One
Two
ThreeFourFive
0.724 pu 0.579 pu
0.000 pu
0.687 pu
0.798 pu 0.000 deg 0.000 deg
0.000 deg
0.000 deg
0.000 deg
7 pu 11 pu
11
Problem 7.28
slack
SLACK345
SLACK138
RAY345
RAY138
RAY69
FERNA69
DEMAR69
BLT69
BLT138
BOB138
BOB69
WOLEN69
SHIMKO69
ROGER69
UIUC69
PETE69
HISKY69
TIM69
TIM138
TIM345
PAI69
GROSS69
HANNAH69
AMANDA69
HOMER69
LAUF69
MORO138
LAUF138
HALE69
PATTEN69
WEBER69
BUCKY138
SAVOY69
SAVOY138
JO138 JO345
0.79 pu
0.70 pu
0.79 pu
0.83 pu
0.50 pu
0.59 pu0.52 pu
0.43 pu
0.50 pu
0.61 pu
0.61 pu
0.32 pu0.24 pu
0.58 pu
0.56 pu
0.78 pu
0.61 pu
0.64 pu
0.52 pu
0.564 pu
0.56 pu
0.62 pu
0.68 pu
0.70 pu0.75 pu
0.64 pu
0.35 pu 0.62 pu
0.77 pu
0.77 pu 0.77 pu
0.84 pu 0.91 pu
0.64 pu
LYNN138
0.82 pu
0.00 pu
5 pu
9 pu
3 pu
2 pu
0 pu 3 pu
9 pu
3 pu
0 pu
12
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 line-ground (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 calculations13
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.
14
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 systems1. safety
2. protect equipment
15
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:
8
8
onductor lengthResistance
cross sectional area
2.65 10 -m for aluminum
1.68 10 -m for copper
where is the resistivity
cR
16
How good a conductor is dirt?
8
16
2.65 10 -m for aluminum
5 10 -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 wir
8
2
e made out of aluminum ?
2.65 10 1609 R= 0.083 mile0.0128
17
How good a conductor is dirt?
62
What is resistance of a mile long, one inch diameter,
circular wire made out of topsoil?
160 1609 R= 500 10 mile0.0128
In order to achieve 0.08 with our dirt wiremilewe would need a cross sectio
6 2
nal area of
160 16093.2 10 (i.e., a radius of about 1000 m)
0.08But what the ground lacks in , it makes up for in A!
m
18
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.
19
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
xcross sectional area
but now area changes
with length.
R
2
ˆln
2 2ˆ
x
r
dxdR
length x
dx xR
length x length r
20
Calculation of grounding R, cont’d
For example, if r 1.5 inches, length = 10 feet,
and 160 we get the following values as
a function of x (in meters)
160ln
2 3.05 0.038x R
1 m 27.2
10 m 46.4
100 m 65.6
100 km 83.4
m
xR
The actual values will besubstantially less since we’ve assumed no currentflowing downward intothe ground
21
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. 22
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 sequencesets have zeroneutral current
23
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 sequencesets have zeroneutral current
24
Zero Sequence Sets
• Zero sequence sets have three values with equal magnitude and angle.
• Zero sequence sets have neutral current
25
Sequence Set Representation
• Any arbitrary set of three phasors, say Ia, Ib, Ic can be represented as a sum of the three sequence sets
0
0
0
0 0 0
where
, , is the zero sequence set
, , is the positive sequence set
, , is the negative sequence set
a a a a
b b b b
c c c c
a b c
a b c
a b c
I I I I
I I I I
I I I I
I I I
I I I
I I I
26