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EEET2106/EEET2339
Power System Analysis and Control
Lecture – 10
Dr Nuwantha Fernando

Power System Faults

RMIT University© 2014

EEET2106/EEET2339 Power System Analysis & Control - © Dr S Moorthy & Dr N Fernando

2

Faults on Power Systems
Under some abnormal conditions, due to the partial or complete failure of
the insulation at one or more points, abnormally high currents may flow
through the unintended paths in the system
...
When a complete failure of the
insulation occurs we say that there is a solid or bolted short-circuit
...

• Mechanical causes - accidents

RMIT University© 2014

EEET2106/EEET2339 Power System Analysis & Control - © Dr S Moorthy & Dr N Fernando

3

Effects of Faults
When a fault occurs, the current may exceed ten times the normal load current or
more
...


Fault clearing is done within 2-20 cycles
...


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EEET2106/EEET2339 Power System Analysis & Control - © Dr S Moorthy & Dr N Fernando

4

Protection
A protection system detects faulty conditions and initiates circuit
interruption
...
Circuit-breakers have mechanical
arrangements to compress or extend a system of springs and are latched on
when they are closed
...
The
triggering of such devices are done by the “protection system”
...
Various equipment in the system
should be able to withstand the immense destructive forces of the fault
currents - though it may flow only for a few milliseconds
...
To select
appropriate CB and other equipment
we need to know the magnitude and
the characteristics of the current that
would flow when a fault occurs
...


arcing time

Interrupting time

contact parting time

Fault clearing time-event diagram

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EEET2106/EEET2339 Power System Analysis & Control - © Dr S Moorthy & Dr N Fernando

6

Fault Studies
How do we determine the expected fault currents at various points in the
power system? Fault studies or Short-circuit calculations are done to
estimate this important design parameter
...
(Balanced faults)
All three phases are short-circuited
...
Usually, the three phase faults give rise to the maximum
fault current at a point
...


RMIT University© 2014

EEET2106/EEET2339 Power System Analysis & Control - © Dr S Moorthy & Dr N Fernando

7

Transient and sub-transient reactances
X”d= direct axis sub-transient reactance of the machine
X’d = direct axis transient reactance of the machine
Xd = direct axis synchronous reactance of the machine
The machine reactance’s and the time constants are usually given by the
manufacturers
...

The dc component which off-set the ac is given by: i dc  t   I" m e



t
TA

where TA is the machine armature time constant
...

The fault current however, reduces as time progresses and after a few cycles
it is determined by the transient (X’d) reactance
...

Therefore, when we evaluate fault
currents, we should model the
generators according to the time of
interest
...

When load currents are neglected,
= E’g = terminal voltage = 1
...

The sub-transient voltage of the machine is then given by

E"  Vt  jX" IL
g
d

The transient voltage of the machine is given by E'g  Vt  jX'd IL
The above two equations neglect the saliency of the poles of the generator
...
e
...

Example: 1
A 100 MVA, 13
...
15 pu and 0
...
It is connected to bus-A of a large
system and it supplies 75 MW and lagging 27
...
49 kV
...


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EEET2106/EEET2339 Power System Analysis & Control - © Dr S Moorthy & Dr N Fernando

10

Example 1 : solution
Base units are 100 MVA and 13
...

The terminal voltage of the generator is 14
...
05 pu
...
75 + j 0
...

Since S = P +j Q = V I* , I = (P-jQ)/V* = (0
...
278)/1
...

The generator current = (0
...
265) pu
...
15 pu

Bus-k

"
E"  Vt  jX d I L
g

= 1
...
15(0
...
265) =
1
...
61o
...
0955
...
1238
...
23pu Bus-k

Sub-transient model

Transient Model

Transient model

'
'
Eg Vt  jX d I L

= 1
...
23(0
...
265) =
1
...
4o
...

The fault current Ifk and the voltage sag
depends on the impedance seen looking into the
system from bus-k
...


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Bus-k

Bus-k

ETh

EEET2106/EEET2339 Power System Analysis & Control - © Dr S Moorthy & Dr N Fernando

12

Short Circuit Capacity or Fault Levels
The “fault level” of a system at bus-k is defined as the product of the
magnitudes of pre-fault voltage and post-fault current
...

If pre-fault voltage is expressed in line-to-line kV and post fault current in
line kA, then
fault level = 3 VLL I L MVA (three phase)
...

The fault current = 1/ZTh when ZTh is in pu
...
1 pu
...

(a) ) What is the fault level at bus-k?
(b) What is the fault current?
Solution
(a) Fault level = 10 pu or 10,000 MVA
...


Fault current = 10 pu*1
...
43 kA

RMIT University© 2014

EEET2106/EEET2339 Power System Analysis & Control - © Dr S Moorthy & Dr N Fernando

14

Simplifying Assumptions made in Fault Analysis

1
...

Prefault voltages are 1
...
This can be correct only if
there is no current in the circuit before the fault occurs
...
)
3
...
Good approximation in
transmission circuits
...


Synchronous machines are represented by E behind X’ or X”
...
8kV

T2

T1
#1

G2

132kV

132kV

#2

L2

L1
#3

(a) Calculate the fault level at bus-3,
(b) Calculate the post-fault voltages at bus-1 and bus-2
...
8
13
...
2pu*
0
...
15pu*
0
...
0 pu0o
...

Common Base: 1000 MVA
G1

 1000 
0
...
0 pu
 100 

G2

 1000 
015

...
75 pu
 200 

T1

T2

132 kV, 33 kV and 13
...


...

  0
...
424
1000

Lines-1 & 2 reactances are:

15
20
 0
...

17
...
424

Step 2 : Draw the circuit diagram and reduce the circuit
...
0pu

1
...
0
j1
...
861

#3

j0
...
861//j2
...
479
IF = j = 0
...
5

j1
...
861
#1

j2
...
479
#2

1
...
0pu

#3
1
...
57 kA

Current division between G1 & G2
2
...
676
  0
...
398  3
...
676 - 0
...
417 pu
Post-fault bus voltages
V1f = 1 - j3(-j0
...
223pu = 29
...

V2f = 1-j1
...
416) =0
...
2 kV line-to-line
...
The other
two busbar voltages have dropped to a low value as is expected when
there is a fault in the system
...
A more systematic approach which is amenable to computer
programming is necessary
...

In fact, most of the commercial software packages available are based on this
method or a variation of it
...

Example 4:
Consider a 3-bus network shown below
...


Z 1 , Z  2 and Z  3

Find the symmetrical fault currents at all three busbars
...
After converting voltage
sources into current sources, the
network is redrawn below
...
1

ZL1

#1

j0
...
5

#2

j0
...
2

Eg2

ZL2
#3

ZL3
Eg1/j0
...
1

#1

ZL1

j0
...
5

j0
...
2

#2

j0
...

Rule 2: Off-diagonal elements
The element in row k and column l, Ykl= (sum of the admittances of all
the branches between nodes k and ) times (-1)

 IN 

= vector of currents injected into the nodes by the
equivalent current generators
...

Then and the network reduces to:

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EEET2106/EEET2339 Power System Analysis & Control - © Dr S Moorthy & Dr N Fernando

22

Pre-fault conditions
Ig1= -j10

#1

#2

-j5

-j2

-j10

Ig2=-j5

-j4
-j5

17
 Ybus    j  5

 2


5
14
4

2 
4 

6 


#3
Admittances in pu

 I N    Yb u s  V N 
 V10 
 Ig1 


 
0
 Ig2    Ybus   V2 
 0
0 
 
 V3 



 V10 

...

 VN    V20   prefault bus voltages  10 in the simplified case
...



 V2 
 V10 
 Ig1 


 
 V2 0   Z bus Ig2
 
 0
 
 0 
 V3 



Inverting,





V10 , V20 & V30

are the nodal voltages before the fault is applied
...
0782
 Z bus   j 0
...
0552


impedance matrix of the system
...
0437
01126

...
0897

0
...
0897 

0
...

10 10
5   10  pre  fault bus voltage
 Z bus  Ig   Z bus    j    
...

 0  10

 

Thevenin Equivalent Circuit seen from node-k
Z11


...
Z1n

Z 21

...



...


...
Z1n

...


 Z k1

...



...


...
Z kn

...



...

Z n2


...


...


...
Z nn

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The k th diagonal element, Zkk of
the [Zbus] matrix is the Thevenin
impedance of the system
network seen from node-k
...


Zkk=ZTh
Bus-k
Vo

k

=ETh

Eth = Vok

the pre-fault voltage of node k
...

This is a very useful result, since we can obtain Thevenin
equivalent circuit of any circuit in this systematic method
without having to reduce the networks step by step in a
manner peculiar to each network we try to solve
...

Zkk=ZTh
Bus-k

System

Zkk=ZTh
Bus-k

Bus-k
Vok =ETh

o
k

V =ETh

System as seen from node-k

Ifk

Vok
f
Ik 
Z kk

Fault on node-k

Considering three phase faults at all busbars in turn, we have,
when the fault is at bus#1,

f
I1

o
V1
1


 12
...
0782

when the fault is at bus#2,

If
2

o
V2
1


 8
...


when the fault is at bus#3,

If 
3

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o
V3
1

 4
...
2448

EEET2106/EEET2339 Power System Analysis & Control - © Dr S Moorthy & Dr N Fernando

26

Post-fault bus voltages
f
V2

0
f
V1  Z1k Ik
0
f
V2  Z 2k Ik


...


f
V1


...

f
Vn



Post-fault voltages are calculated from the bus
impedance matrix elements, fault current at bus-k and
the pre-fault voltages at the buses
...

0
f
Vk  Z kk Ik

...
0782
 Z bus   j 0
...
0552


Let us continue with the previous example to
calculate the post-fault voltages
...
0437
01126

...
0897

0
...
0897 

0
...


When there is a symmetrical fault on bus #3, the fault current If3 =
1/(j0
...
085
...
0552  j 4
...
77450 o pu
3
f
o
V2  V2  Z 23 I f  1   j 0
...
085  0
...
2448  j 4
...
00000 o pu
3

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EEET2106/EEET2339 Power System Analysis & Control - © Dr S Moorthy & Dr N Fernando

27

Branch currents during fault
Having calculated the post-fault voltages of all the busbars in the
network, it is possible to calculate the branch currents in all the branches
of the network
...

Suppose a branch b having an admittance of Yb is connected between
nodes m and n, then the current in that branch flowing from end m to end
f
f
n is given by: Ib  Yb Vm  Vn





The current in the line between bus #1 and #2 = (0
...
6336)/(j0
...
7045 pu
...
7745-0
...
5)
-j1
...


=

The current in the line between bus #2 and #3 = (0
...
0)/(j0
...
534 pu
...
1 pu and that of generator 2 is j0
...

The current infeed from generator 1 is = (1-0
...
1) = -j2
...

The current infeed from generator 2 is = (1-0
...
2) = -j1
...

The fault current infeeds in all the branches and the nodal voltages when a
solid three phase fault occurs at bus#3 of the system are summarised in the
diagram
...
255

I=-j0
...
832

Vf2=0
...
7745
I=-j1
...
534
#3
If3=-j4
...

Normally a power flow program is used
...
00 0
...
The source reactances may be
subtransient or transient reactances of the machines depending on the time of
interest
...

Step 3: Replace all the loads by the equivalent constant impedances which
take the same active and reactive power as the load at the pre-fault bus
voltages
...

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EEET2106/EEET2339 Power System Analysis & Control - © Dr S Moorthy & Dr N Fernando

31

Summary – Continued 1
Step 4: Assemble the bus admittance matrix  Ybus 
Step 5: Obtain bus impedance matrix by inverting the bus admittance
matrix  Z bus 
Step 6: Calculate the three phase fault current at each busbar
...

0
In the simple approach Vk  1

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EEET2106/EEET2339 Power System Analysis & Control - © Dr S Moorthy & Dr N Fernando

32

Summary – Continued 2
Step 7: Post-fault voltages at all the nodes can be calculated using
the following equation: V f  V 0  Z If
m

m

mk k

The equation applies to node m when there is a fault on busbar k
...

Step 8: Currents flowing in the lines or any plant in the network
during a faulty condition are calculated from the knowledge of the
post-fault voltages of the end busbars and the impedance of the
component

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