Power System Protection Ground faults in isolated neutral Ground

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1
Power System Protection
Ground faults in isolated neutral
systems
2
Briefly about the speaker
• Professor at Norwegian Univ. Science and
Technology – Dept. Electrical Engineering
– Power system transients and protection
– High
Hi h voltage
lt
engineering,
i
i
stress
t
calculations
l l ti
– Recent focus on Power Transformers
• Developer of ATPDraw
• Sabbatical at MTU
– Room 628
628, phone 487
487-2910
2910
– hhoidale@mtu.edu
3
Solid vs. isolated neutral
+ No over-voltages in fault
situations
- High
Hi h ffault
lt currentt
+ Easy fault detection
+ G
Ground
ou d faults
au s pe
persist
ss
• Fast trip and reclosing
- Poor power quality
- Extra stress
- Undefined voltages to
ground. Sqrt(3) rise
during ground faults
faults.
MOV dimensioning.
+ Low fault current
- Difficult fault
detection/location
+ Can continue to operate
during ground fault.
Increased power quality
• Safety issues:
down/broken conductors
4
Why is the fault current low in
i l d neutrall systems?
isolated
?
• Fault current must return through line
capacitances. High return impedance.
Ground faults:
• The most common fault
• Often temporary
– T
Trees/branches,
/b
h
snow/ice/wind,
– Birds
– Lightning
•
If 
3 V f
Z1  Z 2  Z 0  3R f
Simultaneous faults….
5
Usage of isolated neutral
• In Norway
Dy
Dy
– LV system 230 V IT
– MV system 12-24 kV
– Distribution level 66
66-132
132 kV
• The 230 V IT system is gradually replaced by
400 V TN ((solidly
yg
grounded)) due to risk of
undetected ground faults, double ground faults,
and lower loss in a 400 V TN system.
• In
I MV;
MV requirements
i
t to
t detect
d t t a 3k ground
d ffault
lt
• Power quality is very important to the industry
– Fast trip & reclosing not acceptable
6
Zero sequence measurements
• Current I0: Sum Ia, Ib, Ic

– Summing the current numerically
– Residual connection
3I0
3I0
I0
Lower
accuracy
due to CT
differences
– Summation transformer
(Toroidal/Rogowski coil)
• Voltage U0: Sum Ua, Ub, Uc
– Open delta
3U0
– Neutral point (isolated or resonance earth)
U0
7
Isolated neutral network
• Normal to only consider ground capacitance
–
–
–
–
symmetrical
y
system,
y
ignore conductive line charging
no voltage drop along line
imbalance in loads canceled by
y D-coupled
p
transformers
• Fault current mainly returns in the two healthy phases
• Fault resistance important
VN
h = e j120º
hVpa
h2Vpa
Vpa
Rf
If
Cg
8
Consequences
• Fault current is independent on where the fault
occurs
• For calculation of feeder current and voltages
g
capacitances can be concentrated
g to g
ground will rise for the healthy
y
• The voltage
phases
• Easy to detect that there is a ground fault (by
looking at V0), difficult to detect where (by
looking at I0 in various feeders)
9
Phasors and equivalent
• Equivalent circuit during fault (phase A):
N
+
If
Vpa
3Cg
C
A
If 
Rf
VN  V0 
V pa
1  3 j C g R f

V pa  3 jC g
1  3 j C g R f
I f
C
3 j C g
A
B
Before
If
Rf=
Line-voltages
unchanged
B
A
Rf=0
UN
After
10
Zero sequence in single radial
• Same zero-sequence voltage in the entire
system; selectivity is challenging
• The fault is traditionally
y located byy
sectionalizing (often manual)
• Zero sequence current varies along the line:
I0
If
U0  constant
3I0
pos.
Rf
11
From fault to zero sequence current
x
l-x
Vpa
VN
h2·Vpa
h·Vpa
C’g·x
C’g·(l-x)
• Currents:
I a ( x)  (VN  V pa )  jC 'g  (l  x)
I b ( x)  (VN  h 2  V pa )  jC 'g  (l  x)
d
Fault location x=d
x
l
I c ( x)  (VN  h  V pa )  jC 'g  (l  x)  I f  H (d  x)
 VN  j3C 'g  x, x  d
3I 0 ( x)  ( I a  I b  I c )  VN  j3C 'g  (l  x)  I f  H (d  x)  
VN  j3C 'g  (l  x), x  d
ignore voltage drop, no load-effect
12
Multiple radials
• Three radials fed by the same transformer
– Fault current in the faulty feeder is the negative sum
of the current in the two healthy feeders
I 01  VN  jC g1
I 02  VN  jC g 2
I 03  ( I 01  I 02 )
• With two radials the currents are equal in size so
directional protection is required
• With an increasing number of feeders simple over
overcurrent protection becomes possible
13
Directional over
over-current
current relay
• Often the preferred solution
• Measure zero sequence voltage and zero
sequence
q
current in each radial
• Then for each radial:
I01
I02
V0

I03
Forward
direction
Healthy
feeder
V0
I0n
Tripping
T
i i zone
Faulty feeder
14
Relaying logic
• Is zero sequence voltage above threshold?
– V0>Vlim (type 59G over-voltage relay)
• Is zero sequence current above threshold
and in correct quadrant (3rd)?
– I0>Ilim
I0 Ili (type
(t
67N di
directional
ti
l over-currentt relay)
l )
+
59G (V0>Vlim)
-
67N (I0>Ilim)
52 trip
15
Example- earth fault protection
Example
22 kV
28 mi
y
50 miles line in total
Total capacitance is
50 miC’g [F/mi]
14 mi
8 mi
5 nF/mi assumed in
this case.
NB! Much higher in
cable systems.
• Fault current ((independent
p
on location))
– Depends on fault resistance
 I f ,min
min  2.4454.7 A @ 3 k
22 kV / 3
If 


'
R f  1/ ( j3  50mi  C g ) R f  j 4244 
 I f ,max  3.090 A @ 0 
U pa
16
Earth fault line 1 (28 mi)
• P
Partition
titi off the
th earth
th fault
f lt currentt measured
d for
f
each radial
• Healthy lines 2&3 feed 14 + 8 parts into line 1
22/50
14/50
28 mi
50/50
14 mi
8/50
8 mi
17
Earth fault line 2 (14 mi)
• Partition of the earth fault current measured for
each radial
parts into line 2
• Healthyy lines 1&3 feed 28 + 8 p
28/50
28 mi
14 mi
36/50
8/50
50/50
8 mi
18
Earth fault line 3 (8 mi)
• Partition of the earth fault current measured for
each radial
p
into line 3
• Healthyy lines 1&2 feed 28 + 14 parts
28/50
28 mi
14/50
14 mi
8 mi
42/50
50/50
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