Lecture 20 - Course Website Directory

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ECE 476
Power System Analysis
Lecture 20: Symmetrical Components,
Unsymmetrical Faults
Prof. Tom Overbye
Dept. of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
[email protected]
Announcements
• Read Chapters 8 and 9
• HW 9 quiz today
• Second exam is Thursday Nov. 12 during class.
Closed book, closed notes, one new note sheet, first
exam note sheet, and calculators allowed
–
–
Exam covers through end of today's lecture
Room 1013 ECEB.
• 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 due on Nov 19; details on website
1
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.
2
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
3
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
4
Zero Sequence Sets
• Zero sequence sets have three values with equal
magnitude and angle.
• Zero sequence sets have neutral current
5
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
6
Conversion from Sequence to Phase
Only three of the sequence values are unique,
I0a , I a , I a ; the others are determined as follows:
 1120
     0
2
3
 1
3
I0a  I0b  I0c (since by definition they are all equal)
I b   2 I a
I c   I a
I b   I a
I c   2 I a
1
 1  1 1
 Ia 
1
 I   I0 1  I +  2   I      1  2
 b a   a   a   
 
 2  1 
 I c 
1
  
 
0

1  Ia
 
  Ia 
2 
   I a 
7
Conversion Sequence to Phase
Define the symmetrical components transformation
matrix
1
1 1


2
A  1 

1   2 


0
0


Ia
I
 Ia 
 
 


Then I  I b  A  I a   A  I   A I s
 






I
 c 
 I a 
 I 
8
Conversion Phase to Sequence
By taking the inverse we can convert from the
phase values to the sequence values
I s  A 1I
1
1 1
1 1 
2
with A  1   
3
1  2  


Sequence sets can be used with voltages as well
as with currents
9
Symmetrical Component Example 1
 I a   100 
Let I   I b   10 
Then
  

 I c   10 
1   100    
1 1
1
1
2
I s  A I  1    10   100 

 

3
1  2    10   0 


 100 
 0 
If I  10   
Is   0 




10  
100
10
Symmetrical Component Example 2
Va   0 
Let V  Vb     
  

Vc     
Then
1   0   0 
1 1
1
1
2
Vs  A V  1         

 

3
1  2        6.12 


11
Symmetrical Component Example 3
 I 0   100 
  
Let I s   I   10


     

 I  
Then
1   100    
1 1


2
I  AI s  1 
  10    

 

1   2        


12
Use of Symmetrical Components
• Consider the following wye-connected load:
I n  I a  Ib  I c
Vag  I a Z y  I n Z n
Vag  ( ZY  Z n ) I a  Z n I b  Z n I c
Vbg  Z n I a  ( ZY  Z n ) I b  Z n I c
Vcg  Z n I a  Z n I b  ( ZY  Z n ) I c
Vag 
Z y  Zn
 

Vbg    Z n
V 
 Z
n
 cg 

Zn
Z y  Zn
Zn
  Ia 
 
Z n  Ib
 
Z y  Z n   I c 
Zn
13
Use of Symmetrical Components
Vag 
Z y  Zn
Zn
 

Z y  Zn
Vbg    Z n
V 
 Z
Zn
n
 cg 

V
 Z I V  A Vs
A Vs  Z A I s

 Z y  3Z n

1
A ZA  
0

0

  Ia 
 
Z n  Ib
 
Z y  Z n   I c 
I  A Is
Zn
Vs  A 1 Z A I s
0
Zy
0
0

0
Z y 
14
Networks are Now Decoupled
V 0 
 Z y  3Z n 0
 

0
Zy
V   
 

0
0
V 

Systems are decoupled
V 0  ( Z y  3Z n ) I 0

V  Zy I
0

I
0
 
0  I 
 

Z y  I 
 
V
 Zy I

15
Sequence diagrams for generators
• Key point: generators only produce positive
sequence voltages; therefore only the positive
sequence has a voltage source
During a fault Z+  Z  Xd”. The zero
sequence impedance is usually substantially
smaller. The value of Zn depends on whether
the generator is grounded
16
Sequence diagrams for Transformers
• The positive and negative sequence diagrams for
transformers are similar to those for transmission
lines.
• The zero sequence network depends upon both
how the transformer is grounded and its type of
connection. The easiest to understand is a double
grounded wye-wye
17
Transformer Sequence Diagrams
18
Unbalanced Fault Analysis
• The first step in the analysis of unbalanced faults is
to assemble the three sequence networks. For
example, for the earlier single generator, single
motor example let’s develop the sequence networks
19
Sequence Diagrams for Example
Positive Sequence Network
Negative Sequence Network
20
Sequence Diagrams for Example
Zero Sequence Network
21
Create Thevenin Equivalents
• To do further analysis we first need to calculate the
thevenin equivalents as seen from the fault location.
In this example the fault is at the terminal of the right
machine so the thevenin equivalents are:
Zth  j 0.2 in parallel with j0.455
Zth  j 0.21 in parallel with j0.475
22
Single Line-to-Ground (SLG) Faults
• Unbalanced faults unbalance the network, but only
at the fault location. This causes a coupling of the
sequence networks. How the sequence networks
are coupled depends upon the fault type. We’ll
derive these relationships for several common
faults.
• With a SLG fault only one phase has non-zero fault
current -- we’ll assume it is phase A.
23
SLG Faults, cont’d
 I af
 f
 Ib
 f
 I c
 ? 
  
  0 
 0 
  
Then since
 I 0f 
1  ? 
1 1
 
1
1 f

2 
0


 I f   1    0  I f  I f  I f  I a
 
3
3
 
2
1 
 0 


 I f 


24
SLG Faults, cont’d
Vaf
Vaf
 f
Vb
 f
Vc
 Z f I af
0

V

1 f
1 1
 



2



1


V


 f 

1   2    

 V f 

This means Vaf  V f0  V f  V f
The only way these two constraints can be satisified
is by coupling the sequence networks in series
25
SLG Faults, cont’d
With the
sequence
networks in
series we can
solve for the
fault currents
(assume Zf=0)
I f
1.050

  j1.964  I f  I 0f
j (0.1389  0.1456  0.25  3Z f )
I  A I s  I af   j 5.8 (of course, Ibf  I cf  0)
26
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