Transcript ECE 310 - University of Illinois at Urbana–Champaign

ECE 476
POWER SYSTEM ANALYSIS
Lecture 19
Balance Fault Analysis, Symmetrical Components
Professor Tom Overbye
Department of Electrical and
Computer Engineering
Announcements



Be reading Chapter 8
HW 8 is 7.6, 7.13, 7.19, 7.28; due Nov 3 in class.
Start working on Design Project. Tentatively due Nov 17 in
class.
1
In the News
•
Earlier this week the Illinois House and Senate
overrode the Governor’s veto of the Smart Grid Bill;
it is now law
•
•
Law authorizes a ten year, $3 billion project to
“modernization” the electric grid in Illinois
•
•
On Tuesday Quinn called it a “smart greed” plan
All ComEd customers and most of Ameren will get
“smart” meters
Effort will be paid for through rate increases over the
period; ComEd estimated the average increase of $36
per year would be offset by electricity savings
2
Network Fault Analysis Simplifications

To simplify analysis of fault currents in networks
we'll make several simplifications:
1.
2.
3.
4.
5.
Transmission lines are represented by their series
reactance
Transformers are represented by their leakage
reactances
Synchronous machines are modeled as a constant
voltage behind direct-axis subtransient reactance
Induction motors are ignored or treated as synchronous
machines
Other (nonspinning) loads are ignored
3
Network Fault Example
For the following network assume a fault on the
terminal of the generator; all data is per unit
except for the transmission line reactance
generator has 1.05
terminal voltage &
supplies 100 MVA
with 0.95 lag pf
Convert to per unit: X line
19.5

 0.1 per unit
2
138
100
4
Network Fault Example, cont'd
Faulted network per unit diagram
To determine the fault current we need to first estimate
the internal voltages for the generator and motor
For the generator VT  1.05, SG  1.018.2
*
I Gen
1.018.2 


  0.952  18.2
 1.05 
'
Ea
 1.1037.1
5
Network Fault Example, cont'd
The motor's terminal voltage is then
1.050 - (0.9044 - j 0.2973)  j 0.3  1.00  15.8
The motor's internal voltage is
1.00  15.8  (0.9044 - j 0.2973)  j 0.2
 1.008  26.6
We can then solve as a linear circuit:
1.1037.1 1.008  26.6
If 

j 0.15
j 0.5
 7.353  82.9  2.016  116.6  j 9.09
6
Fault Analysis Solution Techniques


Circuit models used during the fault allow the
network to be represented as a linear circuit
There are two main methods for solving for fault
currents:
1.
2.
Direct method: Use prefault conditions to solve for the
internal machine voltages; then apply fault and solve
directly
Superposition: Fault is represented by two opposing
voltage sources; solve system by superposition
– first voltage just represents the prefault operating point
– second system only has a single voltage source
7
Superposition Approach
Faulted Condition
Exact Equivalent to Faulted Condition
Fault is represented
by two equal and
opposite voltage
sources, each with
a magnitude equal
to the pre-fault voltage
8
Superposition Approach, cont’d
Since this is now a linear network, the faulted voltages
and currents are just the sum of the pre-fault conditions
[the (1) component] and the conditions with just a single
voltage source at the fault location [the (2) component]
Pre-fault (1) component equal to the pre-fault
power flow solution
Obvious the
pre-fault
“fault current”
is zero!
9
Superposition Approach, cont’d
Fault (1) component due to a single voltage source
at the fault location, with a magnitude equal to the
negative of the pre-fault voltage at the fault location.
I g  I (1)  I g(2)
g
I m  I m(1)  I m(2)
(2)
(2)
I f  I (1)

I

0

I
f
f
f
10
Two Bus Superposition Solution
Before the fault we had E f  1.050,
I (1)  0.952  18.2 and I m(1)  0.952  18.2
g
Solving for the (2) network we get
I g(2)
I m(2)
I (2)
f
Ef
1.050


  j7
j0.15
j0.15
E f 1.050


  j 2.1
j0.5
j0.5
  j 7  j 2.1   j 9.1
I g  0.952  18.2  j 7  7.35  82.9
This matches
what we
calculated
earlier
11
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
12
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
V2 



  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)
13
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
14
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.
15
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 
16
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
17
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 
18
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
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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
20
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.
21
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
22
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
23
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
24
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
mile
  0.01282
In order to achieve 0.08 
with our dirt wire
mile
we would need a cross sectional area of
160  1609
 3.2  106 m 2 (i.e., a radius of about 1000 m)
0.08
But what the ground lacks in  , it makes up for in A!
25
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.
26
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
R
dR 
 dx
2  length  x
x
 dxˆ
x
cross sectional area
but now area changes
with length.

x
R  

ln
2  length  xˆ 2  length r
r
27
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
x
R
The actual values will be
1m
27.2 
substantially less since
10 m
46.4 
we’ve assumed no current
100 m 65.6 
flowing downward into
the ground
100 km 83.4 
28