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TRANSCRIPT
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SHORT CIRCUIT STUDIES
1
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Short circuit study
Requirements:
1. Determining the fault level at buses
2. Selection of breaker ratings
3. Protective device co-ordination
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Symmetrical components
Unbalanced system of „n‟ related phasors can be resolved to „n‟ system of balanced phasors.
In each balanced phasor, angle between two phasors andmagnitude of each phasor are equal.
Phase Quantities: Ia , Ib & Ic and
Sequence components are :
Ia1 stands for Positive sequence current.
Ia2 stands for Negative sequence current.
Ia0 stands for Zero sequence current.
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Sequence components
Ia1, Ib1 & Ic1 : Same phase sequence as Ia, Ib & Ic
Ia2, Ib2, & Ic2 :Opposite phase sequence as Ia, Ib and Ic
Ia0, Ib0, & Ic0 : All in-phase
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Ia
Ic
Ib
Ia1
Ic1
Ib1
120
120
120 Ia2
Ib2
Ic2
120
120
120
Ia0 Ib0
Ic0
Unbalanced current phasor
Positive sequence current
phasors
Negative sequence current
phasors
Zero sequence current phasor
Sequence components
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Similarly for voltages:
c0c2c1c
b0b2b1b
a0a2a1a
VVVVVVVVVVVV
++=++=++=
=
a0
a2
a1
2
2
c
b
a
VVV
1aa
1aa
111
VVV
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sequencetoPhaseps
aa
aa
T
phasetoSequencespand
aa
aa
aaT
ps
sp
111
1
1
3
1
12012401,1201
1
1
2
2
2
2
2
111
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Therefore no zero sequence exists in line voltagephasor.
Since phase voltage sum is not always zero, in thephase voltage phasor, the zero sequence voltageexists.
alwaysVVV
VVVV
V
V
V
aa
aa
V
V
V
LcLLbLLaL
cbaa
c
b
a
a
a
a
0
3
1
111
1
1
3
1
0
2
2
0
2
1
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I
I
I
a a
a a
I
I
I
I I I I
a
a
a
a
b
c
a a b c
1
2
0
2
2
0
1
3
1
1
1 1 1
1
3
Ia
a
b
Ib
Ic
c
Ia+ Ib + Ic= 0
Therefore no zero sequence current flows
into delta connection.
Sequence Currents:
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Ia a
Ib
Ic
b
c
Star connection Star grounded
Ia
I0 a
3I0
Ib
Ic
I0
I0
b
c
Ia+Ib+Ic = 0, There foreno zero sequence currentflows into star connection
Ia+Ib+Ic may not be zero.Hence path always exists forzero sequence currents.
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Transformer sequence
impedance diagram
Star - Delta +ve and -ve Zero
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Transformer sequence impedance diagram
+ve and -ve ZeroStar grounded-Delta
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Transformer Connection to reduce the 3rd
harmonic content (flow)
• Inrush current (Magnetising current)
• 3rd harmonic content : 40 to 50 % of fundamental.
• No 3rd harmonic current.
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Three winding Transformer
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Sequence Impedance
a
b
c
a’
b’
c’
Zaa
Zbb
Zcc
c
b
a
cccbca
bcbbba
acabaa
'
cc
'
bb
'
aa
I
I
I
ZZZ
ZZZ
ZZZ
V
V
V
Let Zaa = Zbb = Zcc = Zs and Zab = Zac = Zba = Zbc = Zca
= Zcb = ZmTsp Vs = [Z] . Tsp.Is
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V Z Z IV Z Z IV Z Zm I
a s m a
a s m a
a s a
1 1
2 2
0 02
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For Rotating Machines
Za,b,c is not symmetric. Even then, the Z1,2.0 is
diagonalized.
Exercise:
V
V
V
Z Z Z
Z Z Z
Z Z Z
I
I
I
a
b
c
s m m
m s m
m m s
a
b
c
1 2
2 1
1 2
Find the expression for Z1,2,0
and Prove that it is a diagonal matrix:
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Advantages of Sequence Component
De-couples the equation to +ve, -ve and zerosequence and hence easy to solve.
End conditions can be applied easily for fault studiesin sequence components.
Since the equations are de-coupled, same set ofsubroutines can be called for solving the +ve, -ve andzero sequence system.
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Steady State
a
b
c
Ia
Ib
Ic
Ea
Ec Eb
ZnIn
+
+ +
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Positive Sequence Network
c
a
b
Ia1
Ib1
Ic1
Ea
Ec Eb
Z1
Z1
Z1
Z1
EaVa1
Ia1
Reference Bus
Va1 = E a - I a1 Z1
a
+
-
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Negative Sequence Network
c
Ia2
Ib2
Ic2
Z2
Z2
Z2
Z2
Va2
Reference Bus
Va2 = - Ia2 Z2
a
b
a
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Zero Sequence Network
c
a
b
Ia0
Ib0
Ic0
Zg0
3Zn
Va0
Reference Bus
Va3 = - Ia0 Z0
a Zg0
Zg0
Zn 3Ia0 Zg0
Z0
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Generator impedance for fault study :
1.Transient (xd‟) or sub-transient ( xd”) is considered for positivesequence.
2. X2 i.e. Negative sequence which is close to xd”. (Approximately).
3. X0 is small 0.1 to 0.7 times xd”
Typical values on own rating :
Xd 100 to 200 %
Xq 60 to 200 %
Xd‟ 21 to 41 %
Xd” 13 to 30 %
X2 Xd”
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Transmission line :
Positive sequence impedance = Negative sequence impedance
Zero sequence impedance depends upon : Return path, Groundwires and Earth resistively.
Zero sequence reactance is approximately 2 - 2.5 times positivesequence impedance.
R0 is usually large. May be 5 to 10 times also.
B0 is 65 to 80 % of B positive sequence
Transformers :
All are equal i.e. Zpt = Znt = Zzt for transformer.
Zzt 90% of Zpt, if magnetising branch is neglected.
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Fault Representation:
Va1
Vf
Z1
Ia1
Va2
Z2
Ia2
Va1
Z0
Ia0
Va1 = Vf - Z1.Ia1 Va2 = -Z2 . Ia2 Va0 = - Z0 . Ia0
a0
a2
a1
0
2
1f
a0
a2
a1
I
I
I
Z00
0Z0
00Z
0
0
V
V
V
V
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Three phase fault representation:
a
b
c
a
b
c
Vab = 0 Va = 0 Va1 = 0Vbc = 0 Vb = 0 Va2 = 0Vca = 0 Vc = 0 Va0 = 0
a0
a2
a1
0
2
1f
I
I
I
Z00
0Z0
00Z
0
0
V
0
0
0
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Ia1 = Vf Z1 , Ia2 = 0 , Ia0 = 0
a0
a2
a1
2
2
c
b
a
I
I
I
1aa
1aa
111
I
I
I
Ia = Ia1
Ib = a2Ia1
Ic = aIa1
Ia = Ib = Ic
Three phase fault representation:
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Fault Through Impedance :
Zf
a
Zf
a
Zf
a
Z0
Zf
a
Zf
a
Zf
a
0I0IZZ
VI a0a2
f1
fa1
, ,
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Single Line to Ground Fault Representation
a
b
c
Ic
Ib
Ia
Va = 0
Ib = 0
Ic = 0
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2
2
11111
3
1aaaa
III
c
b
a
0
2
1
III
a
a
a
1 10
3II aa 2 ,
3II aa 1 ,
31 II aa 021 III aaa Q
0
2
1
0
2
1
0
2
1
000000
00
III
ZZ
ZV
VVV
a
a
af
a
a
a
111 IZVV afa 222 IZV aa 000 IZV aa
IZIZIZVVVV 002211021 aaafaaa
021 0VVVV aaaa Q
021
021
ZZZ
VIII
faaa
021
3
ZZZ
VI
f
a
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Fault Through Impedance :
a
b
c
Ia
Ic
Ib
Zf
f021
fa0a2a1
3zZZZ
VIII
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Line to Line Fault Representation:
Vb = Vc
Ia = 0
Ib = -Ic
a
b
c
Ic
Ib
Ia
c
b
a
2
2
a0
a2
a1
V
V
V
111
aa1
aa1
3
1
V
V
V
Va1 = 1/3 [Va+aVb+a2Vc] = 1/3 [Va+aVb +a2Vc]
Va2 = 1/3 [Va+a2Vb+aVc] = 1/3 [Va+a2Vb+aVc]
Va1 = Va2
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I
I
I
a a
a a
I
I
I
a
a
a
a
b
c
1
2
0
2
21
3
1
1
1 1 1
Ia0 = 1/3 (Ia +Ib +Ic) = 1/3 (Ib - Ib) = 0
Ia1 = 1/3 (aIb+a2Ic) = 1/3(aIb-a2Ib)
Ia2 = 1/3 (a2Ib + aIc) = 1/3 (a2Ib-aIb)
Ia1 = -Ia2
0
00
00
00
0
0 1
1
0
2
1
0
2
1
a
af
a
a
a
I
I
Z
Z
ZV
V
V
VVa1 = Vf - Z1Ia1
Va2 = Z2Ia2 = Va1 = Vf - Z1Ia1
Z2Ia1 = Vf - Z1Ia1
I
V
Z Zand I I I Va
f
a a a a11 2
2 1 0 00 0, ,
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Fault Through Impedance :
a
b
c
IcIb
Ia
Zf
I I
V
Z Z Za a
f
f
1 2
1 2
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Double Line to Ground Fault Representation:
a
b
cIcIb
Ia
Vb = 0
Vc = 0
Ia = 0
VVV
a a
a aV
a
a
a a
1
2
02
21
3
1
11 1 1
00
1 1 1V V V V V Va a a a a a1 2 03 3 3
, ,
V V Va a a1 2 0
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I
I
I
a a
a a I
I
a
a
a
b
c
1
2
0
2
21
3
1
1
1 1 1
0
Ia1 + Ia2 + Ia0 = Ia =0
Ia1 = 1/3 (aIb + a2Ic)
Ia2 = 1/3 (a2Ib + aIc)
Ia0 = 1/3 (Ib + Ic)
1/3(aIb + a2Ic) + 1/3(a2Ib + aIc) + 1/3(Ib +Ic) = 0
aIb + a2Ic + a2Ib + aIc + Ib + Ic = 0
V
V
V
V Z
Z
Z
I
I
I
a
a
a
f a
a
a
1
2
0
1
2
0
1
2
0
0
0
0 0
0 0
0 0
Va1 = Vf - Z1Ia1
Va2 = - Z2Ia2
Va0 = - Z0Ia0
Va1 = Va2 = Va0
Vf - Z1Ia1 = - Z2Ia2 - Z0Ia0
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V a0
Z1 Z
2Z
0
V f
Iao
V a2V a1
Ia2I a1
Fault Through Impedance:
a
b
c
ZfZf
Zg
IV
Z Z ZZ Z
Z Z Z
Z Z Z
Z Z Z Z
af
f
f
f g
1
12 0
2 0
1 1
2 2
0 0 3
'' '
' '
'
'
'
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Open Faults
Open in phase B & C
b
c
b’
c’
a a’
Ib = 0 and Ic = 0
V V Z I Z Iia
ka
aa a aa f ' '
I
I
I
a a
a a
I
I I I I I
a
a
a
a
a a a a f
1
2
0
2
2
1 2 0
1
3
1
1
1 1 1
0
0
1
3
1
3
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V
V
V
V Z
Z
Z
I
I
I
a
a
a
f a
a
a
1
2
0
1
2
0
1
2
0
0
0
0 0
0 0
0 0
V V Z I
V Z I
V Z I
V V V V V Z I Z I
V V
Z I V Z I Z I Z I
I I IV
Z Z Z Z
a f a
a a
a a
ka
la
kl kl kl zz f aa a
f kl
aa a kl a a a
a a aokl
aa
1 1 1
2 2 2
0 0 0
1 2 01
1 1 1 2 2 0 0
1 2
1 2 0
3
3
3
' '
'
'
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Vkl is the Thevenin equivalent voltage, once the line is removed, between buses k & l.
f
0
f
2
f
1f
'
aa
0
kl
0
lk
0
ll
0
kk0
2
kl
2
lk
2
ll
2
kk2
1
kl
1
lk
1
ll
1
kk1
ZZZZ3Z
ZZZZZ
ZZZZZ
ZZZZZ
Solution Methodology:
1. Do the load flow with the line isolated.
2. Then insert the line with two phase open.
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Open in phase A :
• Ia = 0• Similar type of analysis like double line fault.• Voltage is between two nodes, rather than between
node and ground.• Impedance is the Thevenin‟s equivalent impedance
between nodes.
IV V
ZZ Z
Z Z
fl k l
12 0
2 0
'' '
' '
.
Z Z Z Z Z Z
Z Z Z Z Z Z
Z Z Z Z Z Z
l
kk ll lk kl f
l
kk ll lk kl f
l
kk ll lk kl f
1
1 1 1 1 1
2
2 2 2 2 2
0
0 0 0 0 0
I IZ
Z Zf f
2 1 0
0 2
.
'
' 'I
I Z
Z Zf
f0
1
2
0 2
.'
' '
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Solution Methodology:
1. Form the Y bus for +ve, -ve and zero sequence.2. Do the LU factorisation for +ve, -ve and zero sequence.3. To find the driving point impedance-
In fault study, particular row or column of Z bus is required.
Consider, Yx = z
x=Y-1z
x= Z.z
Let z have 1 in pth location and 0, at all other location.
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x
Z Z Z Z
Z Z Z Z
Z Z Z Z
Z
Z
Z
Yx z
p n
p p pp pn
n n np nn
p
pp
np
11 12 1 1
1 2
1 2
10
1
0
L
L
M
L
M M
Pass in z, 1.0 for the desired bus and 0 else where.Call LU solution - x will be the desired column of Z bus.
4. Find the sequence fault current I1f, I2
f and I0f.
5. Determine the post fault bus voltage
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[Vf] = [Z] [If1]
= [Z] {[I0] - [If]}
= [Z] [I0] + [Z][-If]
= [V0] + [V]
V = [Z] [-If]
Y[V] = [-If]
Solve V using LU solution.
Above can be written as :
[Y]p[V]p = [-If]p , p: positive sequence and similarly for other sequences.
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For open fault, between k & l,
I k I
I
f p f
f
00
0
M
M
M
For faults other than open fault,
I If p f
p
0
0
0
:
:
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Example:
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Grounding Practice in Power System :
Advantages of ungrounded system :
• Ground connection normally doesn‟t carry current. Hence elimination saves the cost.
• Current can be carried in other phases, with fewer interruptions.
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Grounding Practice in Power System :
Limitations of ungrounded system :
• With the increase in voltage and line length current has increased and self clearing nature advantage couldn‟t be seen.
• Arcing ground : Phenomena of alternate clearing and re-striking of the arc, which cause high voltage (surge and transient).
• If grounded, the insulation can be graded in transformer from line to neutral, there by reducing the cost.
• In case of ungrounded system, influence on communication lines is more.
•Ground current can‟t be limited.
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Ungrounded System :a
b
c
3.01.73 1.73
0.0
Perfectly transposed line :Neutral of transformer is at zero potential.
Un-transposed line :Neutral is shifted.
Capacitance grounded :Perfectly transposed line.
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Resistance Grounded System:
The resistance to neutral limits ground current.
Selection of resistance value:
• Amount of ground fault current.
• Power loss in resistor during ground fault.
Power loss: usually expressed as a percentage of system rating.
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R
X
g
= 16 %X
g
= 8%
Z1 = 24% , Z2=24% , Z0=3R+j8
I
Z Z Z j j R j R j
Power Loss I RR j
R
f
f
3 3 100
24 24 3 8
300
3 56
300
3 56
1 2 0
2
2
*
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R
X
g
= 16 %X
g
= 8%
Z1 = 24% , Z2=24% , Z0=3R+j8
I
Z Z Z j j R j R j
Power Loss I RR j
R
f
f
3 3 100
24 24 3 8
300
3 56
300
3 56
1 2 0
2
2
*
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If : in pu, R : in puFor three phase system, power loss
=I R
percentage of phasesystemkVA ratingf2
33
R is in percent p.u.Maximum power loss = ?
Selection : How much be the value of ground fault current ?What should be the percent power loss in the ground resistance ?
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Effectively grounded system:
Solidly grounded : No impedance between neutral and earth
Effectively grounded : As per AIEE standard No.32, Section 32 - 1.05, May 1947 :
X system
X system
R
X system
o0
1 1
30 1 0 . .
X1 =X2 =25%
X1 =7%=X2=X0
X1 =34%X2 =120%
69 kV25 MVA
69 kV
B
A
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For fault atAX
X
For faultatBX
X
Grounded System
X
Xbut no resonance (resonancegrounding)
0
1
0
1
0
1
7
32
127
32 34
127
66
30
=
=+
=
Reactance :
. ;
If Xg = 50 at A , find the nature of grounding at A & B.
Effectively grounded system:
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Resonant - Grounded System :Capacitance current is tuned or neutralised by a neutral reactor or similar device.
0
0
I
I
I
0
I
I 2
Fault current is made zero by selecting the suitable tap on the ground reactor.
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Circuit Breaker Selection
1. Determine the symmetrical current (maximum) for any type of fault.
2. Multiply the current by factor obtained from standard tables depending on the duty cycle.
(2 cycles clearing, 5 cycles clearing etc.)
Use the above current to specify the interrupting current
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References
1. Stagg and A.H. El-Abiad, "Computer Methods in Power System Analysis", Mcgraw-Hill, 13th print, New Delhi,1988.
2. William D. Stevenson "Elements Of Power System Analysis", McGraw-Hill, Fourth edition, New Delhi, 1982.
3. George L. Kusic, "Computer Aided Power System Analysis",Prentice-Hall, International, N.J.,New Delhi, 1989.
4. J. Arrilllaga, C.P. Arnold and B.J.Harker ,"Computer Modeling of Electrical Power Systems", John Wiley and Sons,1983.
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Queries & Discussions
59
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Thank You
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TRANSIENT STABILITY STUDIES
1
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Voltage stabilityRotor angle stability
Study period: 0-10
sec
Mid-term/long-term stability
Study period: seconds to
several minutes
(slow dynamics)
Small signal stability
Non-oscillatory
Sufficient synchronizing torqueOscillatory
Unstable control
action
Transient stability
Large disturbance
(First swing)
Stability
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Power System Operating States
f, v, loading acceptable, load met, n-1 or n-2 contingency acceptable
f, v, loading acceptable,load met n-1 or n-2 contingency not satisfied
f, v, loading not acceptable, load not met
Normal
Alert Restorative
Cascaded
system
syste
m In Extremis Emergency
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Pool control centre
System control centre
To other systemTo other system To other system
Transmission system Power plant
DS DS DS G G G
DS: Distribution System
G : Generator
Control Hierarchy
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Power System Subsystem & Controls
f Ptie Pgtotal
Schedule System generation control Load frequency control with
economic allocation
Other generating units and associated controls
Generating unit’s control
Prime mover & control
Excitation system and control
Generator f/N or
Pg Vt
If
Pm f/N
Transmission Controls Reactive power, Voltage control, HVDC transmission and others
Pg
f Ptie Pg.total
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Stability classification
• Transient stability : Transmission line faults, sudden load change, loss of generation, line switching etc.
• Dynamic stability : Slow or gradual variations. Machine, governor-Turbine, Exciter modeling in detail.
• Steady state stability : Changes in operating condition. Simple model of generator.
66
Power System Stability
Ability of a power system to remain in synchronism. Classification of transients : Electromagnetic and Electromechanical
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Stable Unstable
Stable Unstable
t Sec. t Sec.
t sec.t Sec.
m2
m2
m1 m1
0 0
00
Transient Stability:First swing stability problemMulti-swing stability problem. (normal size)
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Assumptions :
Synchronous speed current and voltage are considered.
DC off set currents, harmonics are neglected.
Symmetrical components approach.
Generated voltage is independent of machine speed.
Circuit parameters are constant at nominal system
frequency.(Frequency variation of parameter neglected).
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Steady State Stability
Stable Unstable
Stable
Unstable
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Mechanical Equation :
J : Moment of inertia of rotor masses (kg-mt2 )
m : Angular displacement of rotor w.r.t. a stationary axis
(mechanical radians)
t : Time (seconds)
Tm : Mechanical or Shaft Torque ( N-m )
Te : Net electrical torque (N-m )
Ta : Net accelerating torque (N-m)
For generator, Tm and Te are +ve.
Jd
d tT T T N mm
a m e*2
2
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Reference Axis
Stationary
m
Reference axis at sm
sm
sm
Rotor axis at rotor
speed
m sm mt
m : Angular displacement of the rotor in mechanical radians.
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)(
.sec
2
2
2
2
2
2
2
2
2
2
ApproxPPdt
dM
wattsPPdt
dJ
perradiansinvelocityAngulardt
d
mNTTdt
dJ
dt
d
dt
d
dt
d
dt
d
emm
emm
m
mm
emm
mm
msm
m
Where M : Inertia constant = at synchronous speed in Joules-sec per mechanical radian.
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Constant is defined as the ratio of stored Kinetic Energyin Mega Joules at synchronous speed and machinerating in MVA
emm
sm
emm
sm
smsm
PPdt
dHS
wattsPPdt
dM
radianmechanicalperMJSH
M
MVAMJS
M
S
JH
2
2
2
2
2
2
2
/2
1
2
1
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em
s
emm
sm
emm
sm
PPdt
dH
puPPdt
dH
S
PP
dt
dH
2
2
2
2
2
2
2
2
2
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s
s
s
em
sem
s
em
em
HtIf
tH
pudt
dH
puPPLet
dt
dPP
dt
dH
puPPtd
d
f
H
reeselectricalinIf
EquationSwingpuPPdt
d
f
H
,2
2,1
2
1
2
180
deg
2
2
2
2
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Pm
Pe = 1 pu
At t = 0, breaker is opened.Initially Pe = 1 pu on machine rating Pm = 1pu and kept unchanged.
In 2H seconds, the speed doubles.
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MVASystem
ratingMachineHH
MVAsystemHRatingMachineH
RatingMachine
KEStoredH
mechsystem
systemmech
Inertia constant (H) is in the range 2 - 9 for varioustypes of machines. Hence H-constant is usuallydefined for machine.
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Relation between H constant and Momentof Inertia is given by:
machineS
NWR
H
Wattslbft
lbftNWR
KE
feetingyrationofRadiusR
poundsinpartrotationalofWeightW
ftlbWR
InertiaofMoment
6
22
22
22
10550
746
60
2
2.322
1
746sec/550
60
2
2.322
1
:
:
2.32
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Example :
Smach = 1333 MVA
WR2 = 5820000 lb - ft2
N = 1800 RPM
1333
60
18002
2.32
5820000
2
110
550
7462
6
H
= 3.2677575 pu (MJ/ MVA)
On 100 MVA base : H = 1333 / 100 = 43.56 (MJ /
MVA)
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G1 ,H1
Pe1 Pm1
G2 ,H2
Pm2 Pe2
112
1
2
1em PP
dt
d
f
H
222
2
2
2em PP
dt
d
f
H
21212
2
21eemm PPPP
dt
d
f
HH
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1 = 2 = ; since the machines swing together
PePtd
d
f
Hm
2
2
H = H1 + H2
Pm = Pm1 + Pm2
Pe = Pe1 + Pe2
G1 and G2 are called coherent machines.
Inertia Constant
HStored energy at rated speed in MWs
MVA rating
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MKS System
ratingMVA
RPMJ
ratingMVA
JH
RPMradianmechanicalinspeedRated
mkgininertiaofMomentJ
MWsJ
wattsJenergyKineticenergyStored
om
om
om
om
2
9
62
2
62
2
1048.510
2
1
60
2sec/:
:
102
1
2
1
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British Units
22
2
2
356.12.32
)(
mkgWR
J
ftlbgyrationofradiusofsquare
partrotatingofWeightWR
Given
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Example:
sec2
/)(
1048.5
77168.27547356.12.32
26
22
H
timestartingMechanical
MWsratingMVAH
energyStored
MVAMWsratingMVA
RPMJH
mtkgWR
J
MVA rating : 555 WR2 : 654158 lb-ft2
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Typical values
Unit Type H Constant
Hydo unit 2 to 4
Thermal unit
2 pole - 3600 RPM 2.5 to 6
4 pole - 1800 RPM 4 to 10
Non coherent machines
112
1
2
1 PePmdt
d
f
H
11
1
2
1
2
em PPH
f
dt
d
222
2
2
2em PP
dt
d
f
H
22
2
2
2
2
em PPH
f
dt
d
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21
2112
12
21
2112
12
21
21
12122
12
2
21
2112
21
2112
2
21
2
21
21
22
21
111
21
2
2
21
2
21
21
21
122112
2
21
2
22
2
11
1
212
2
,
1
1
1
HH
PHPHP
HH
PHPHP
HH
HHH
Where
PPtd
d
f
H
HH
PHPH
HH
PHPH
td
d
fHH
HH
PPHH
HPP
HH
H
td
d
fHH
HH
HH
HPPPPH
dt
d
f
PPH
fPP
H
f
dt
d
ee
e
mm
m
em
eemm
emem
emem
emem
Relative swing (with reference to one machine) is more important, rather than absolute swing.
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Swing Curves
Relative swing (with reference to one machine) is more important, rather than absolute swing.
Absolute Plot
1
o
2
3
T in sec.
4
Relative Plot (i-)
4
3
20
1
T in sec.
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Classical model : (Type 1) Constant voltage behind transient reactance.
E’ = Vt + (0 + jxd’) I
I
E’
Vt
jxd’ I
Ref.Vt
I
E’
jxd’+
-
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Power angle equation jXs
jXs
PE E
Xs
1 2 sin
E1 : Magnitude of voltage at bus1
E2 : Magnitude of voltage at bus2
: 1 - 2
Xs : Reactance
22 E11 E
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Machine Parameters
Synchronous : Steady state, sustained.
Transient : Slowly decaying
Sub-transient : Rapidly decaying
E=?
X=?
"'
"'
""'
0
qoqo
dod
dqqqd
TT
TT
XXXXX
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Parameter Hydro (pu) Thermal (pu)
xd 0.6 - 1.5 1.0 - 2.3
xq 0.4 - 1.0 1.0 - 2.3
xd’
0.2 - 0.5 0.15 - 0.4
xq’
------- 0.3 - 1.0
xd”
0.15 - 0.35 0.12 -0.25
xq”
0.2 - 0.45 0.12 -0.25
Td0’
1.5 - 9.0 s 3.0 -10.0 s
Tq0’
------- 0.5 - 2.0 s
Td0” 0.01 - 0.05 s 0.02 - 0.05 s
Tq0” 0.01 - 0.09 s 0.02 - 0.05 s
Ra 0.002 - 0.02 0.0015 - 0.005
Typical values
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Stability P
Pm
O
Pe=Pmax sin
u 90
0 s 180
0
Stable
• At s ; Pm = Pe ; net accelerating torque = 0.
• Let Pe decrease slightly.
•
increase (acceleration)
comes back to original position.
• Stable region . Hence s is stable operating point.
veisPPdt
d
f
Hem
2
2
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Unstable
· At u; Pm = Pe ; Net accelerating torque = 0
· Let Pe decrease slightly.
·
increases, (acceleration)
· Pe further decreases.
· Chain reaction never comes back to
normal value .
Hence u is unstable operating point.
veisPPdt
d
f
Hem
2
2
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Infinite bus
Generator connected to infinite bus.
· High inertia. H compared to other machines in the system.
· Frequency is constant.
· Low impedance. Xd‟ is very small.
· E‟ is constant and Vt is fixed.
· Infinite fault level symbol.
System
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Example :
200 MW
1.05 pu V
250 MVA 250 MVA Slack bus
1 pu - V
H = 3.2 , Z = 10% on own rating , Xd1 = 25% , tap = 1,
Ra = 0.0 and neglect R.
1. Establish the initial condition.
2. Perform the transient stability without disturbance.
3. Open the transformer as outage & do the study.
4. How long the breaker can be kept open before closing,
without loosing synchronism.
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Load Modeling
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Vary the tap.
Switch on the capacitor.
Determine the response (charge) in load.
Compute the parameters. P = P0 (CP + CI . V + CZ . V2) ( 1+Kf . f)
P varies with time, voltage and frequency.
P0 varies with time - can be constant at a given time of a day.
CP, CI, CZ & Kf are constants.
V & f are known at any time instant.
P is known from measurements.
Solve the non linear problem over a set of measurements.
fo
Po
frequency
Po
we
r
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Let the load be 10,000 MW. i.e. P0 = 10,000
Let for 1 Hz change in frequency, let the load change be 700 MW.
5.3
501100
7
:100
;
%7000,10
700
7001
700)(
)(700
p
f
p
f
p
foo
p
fo
C
baseMVAonthen
frequencyinchangeunitpertheisfpuinisPIf
C
numberpowerCffP
loadindecreaseCfP
What it implies :· Initial load 10,000 MW.· Loss of generation 700 MW· Increase in load 700 MW · Frequency 49 Hz.
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Load Model Parameters
Load model paraters
Measurement based approach
Input: Connected load
Measurement: P,V, f over a period
Out put: Parameters
Component based approach
Industrial
Commercial
residential
Agricultural
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Component pf P/V Q/V P/f Q/f
Air conditioner
3 p central heating0.90 0.088 2.5 0.98 -1.3
Arc furnace 0.7 2.3 1.6 -1.0 -1.0
Industrial Motors 0.88 0.07 0.5 2.5 1.2
Agricultural pumps 0.85 1.4 1.4 5.0 4.0
Residential
Summer 0.9 1.2 2.9 0.8 -2.2
winter 0.99 1.5 3.2 1.0 -1.5
Commercial
summer 0.85 0.99 3.5 1.2 -1.6
winter 0.9 1.3 3.1 1.5 -1.1
Industrial 0.85 0.18 6.0 2.6 1.6
Power plant 0.8 0.1 1.6 2.9 1.8
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Transducer
Regulator Exciter Generator
Limiter + relay
PSS
Excitation System Components
Ref.
Controller
Regulator
Power
amplifier
(Exciter) Plant
Feedback elements
Block Schematic
Et Efd
Vtr Ver
Vref
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Reactive Power Control
Synchronous generators
Overhead lines / Under ground cables
Transformers
Loads
Compensating devices
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Control Devices: Sources /Sinks --- Shunt capacitor, Shunt inductor
(Reactor), Synchronous condenser, and SVC.
Line reactance compensation --- Series capacitor
Transformer -----OLTC, boosters
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Speed Governor Systems :
AGC
Speed
changer
Speed
Governor Valve
/gate
Electrical
System
Energy Supply
steam or water
Turbine Generator
Speed
Tie line Power
Frequencies
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Types of Control: Primary Control : Governor action
Secondary Control : AGC, load frequency control (For selected generators)
1/R 1
1
T
T
ws
ws
1
T Kms D
speed
Turbine Droop(Goveror) Generator +
-
Speed
Ref.
Under Frequency operation:
Vibratory stress on the long low pressure turbine blades Degradation in the performance of plant auxiliaries say,
induction motor.
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Limitations: Only maximum spinning reserve can be achieved
Turbine pickup delay
Boiler slow dynamics
Speed governor delay
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Load Shedding:
Trip signal 49.5
0.4 Hz/s
48
10% load rejection
15% load rejection
50% load rejection
30% load rejection
1 Hz/s
4 Hz/s
2 Hz/s
Other measures:
* Fast valving
* Steam by-passing
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Modules in a Program:1. Data reading
2. Initialization
Steady state load flow
Control block parameter AVR, Gov., Machine, Motor, PSS, VDC, SVC.
3. Disturbance model
4. Control block modeling
5. Machine modeling
6. Load flow solution
7. Protective relay modeling
8. Special functions
Cyclic load
Arc furnace
Re-closure
9. Results output
Report
Graph
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Typical Swing Curve :
Time in seconds
AVR & PSS
Constant Efd
AVR with no PSS
65432
90
60
30
1
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Typical Swing Curve :
0.01
0.090
0.025
5 4 3 2 1
180
120
60
Time in Sec.
Rotor angle
degrees
Integration step size : Typical value : 0.01 seconds, Range : 0.005 to 0.02 seconds
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AVR : Type 1
1
1 1 sT
k
sT
1
21
s=f(Efd)
sk
sT
3
41
1
2 3k sT
PSS
+-
+
Vref
+
- EfdVT
Mapping Range
T1 Tr 0.01 -- 0.10T2 Ta 0.025 -- 0.2
T3 TE 0.5 -- 1.0
T4 TF1 0.5 -- 1.0
K1 KA 25 -- 200
K2 KE -0.5 -- 1.0
K3 KF 0.01 -- 0.1
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AVR : Type 2
Mapping Range
T1 TR 0.01 --- 0.05
T2 TA 0.01 --- 0.05
T3 TE 0.5 --- 1.0
T4 TF1 1.0
T5 TF2 0.1
k1 KA 50--200-400-600
k2 KE 1.0
k3 Kf 0.01 - 0.05
Efd1
1 1 sT
k
sT
1
21
s k
sT sT
3
4 51 1
Vs
+-
+
Vref
SE
1
2 3k sT+
-VT
VRmax
VRmin
-
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AVR TYPE – 5
VT 1
1 0 01 . s
k sT sT
sT sT
1 2 3
4 5
1 1
1 1
( )
+
- +
Vref VRmax
VRmin
Efdmax
Efdmin
1
1 1 sT
Mapping Range
0.01 Tr 0.01 -- 0.025
T1 Ts 0.025
T2 TE 1.0
T3 TF1 0.06
T4 TA 40.0
T5 TF2 0.024
k1 kA 200.0
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Steam Turbine Governing System
ref
+
1/T31/S
+Pref
P5
Pmax
C min
P-up
P-dn
1+sT2
0 k1(1+sT1)
-
+
- Pmin
C max
K1 : 0.05 - 0.04 Pmax : 1.0
T1 : 0.1 Pmin : 0.0
T2 : 0.03 Pup : 0.1
T3 : 0.4 Pdn : -1.0
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P
1/(1+sT1 ) 1/(1+sT2)
k1+k2 k3+k4
(1/1+sT3) (1/1+sT4)
k5+k6 k7+k8
Ps
T1 : 0.26 k1 : 0.2760 k1 + k2 : VHP
T2 : 10.0 k3 : 0.324 k3 + k4 : HP
T3 : 0.5 k5 : 0.4 k5 + k6 : IP
T4 : 99.9 k7 : 0.0 k7 + k8 : LP
Turbine Model
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Hydro Governor
+
1
1 1 sT
-
+
-
1
2T
Pmin
Pmax
Ps
ksT
sT1
3
31.
k2
Transient Droop Compensator
Permanent Droop Compensator
ref
+
+
Cmin P-dn
Cmax
1
s
P-up
k1 : 0.36 Pmax : 1.1 T2 : 0.2
k2 : 0.04 Pmin : 0.0 T3 : 5.0
T1 : 0.05 Pup : 0.1 Pdown : -
1.0
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Hydro Turbine
Ps1-sT1
1+0.5sT1
DM
T1 (T) : 1.0
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Transient Stability Enhancement
Philosophy :
1. Minimize the disturbance influence by minimizing the fault severity and duration.
2. Increase the restoring synchronizing forces.
3. Reduce accelerating torque.
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Transient Stability Enhancement
Methods :
1. High speed fault clearing.
2. Reduction of transmission system reactance.
3. Regulated shunt compensation.
4. Dynamic Braking.
5. Reactor switching.
6. Independent pole operation of circuit breaker.
7. Single pole switching
8. Fast valaving.
9. Generator tripping.
10.Controlled system separation and load shedding.
11.High speed excitation systems.
12.HVDC transmission link control.
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Major references used in the development of Transient Stability Studies Module
1. Dommel, N. Sato “Fast Transient Stability Solutions”, IEEE Transactions on Power Apparatus and Systems, 1972, PP 1643 -1650.
2. W. Dommel, “Digital computer solution of electromagnetic transients in single and multiphase networks”, IEEE Transactions on Power Apparatus and Systems, April 1969, Vol. PAS-88, PP 388 - 399.
3. IEEE Committee Report, “Dynamic Models for Steam and Hydro Turbines in Power System Studies”, IEEE PES Winter Meeting, New York, Jan./Feb. 1973. (Paper T 73 089-0).
4. IEEE Committee Report, “Proposed Excitation System Definitions for Synchronous Machines”, IEEE Transactions on Power Apparatus and Systems, Vol. PAS-88, No. 8, August 1969.
5. IEEE Committee Report, “Computer representation of excitation systems”, IEEE Transactions Power on Apparatus and Systems, June 1968, Vol. PAS-87, PP 1460 - 1464.
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Queries & Discussions
121
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Thank You
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Introduction to Power System Protection
1
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Topics for Discussion Protective relaying
Need for protection
Types of Protection
General Philosophy
Over-Current Relays
Distance Relays
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Introduction
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Protective Relaying Safeguard equipment from damage during
abnormal operating conditions.
Protective Relaying should:
Ensure normal operation
Prevent equipment failure
Limit the effects of disturbances
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Need for Protection Loss of equipment (Permanent or Partial damage)
Loss of production
Revenue loss
Fire hazard, loss of life
Loss of confidence level in using electricity as a commodity
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Functions of Protection Primary function - Prompt removal of faulty
element suffering a short circuit, or when it startsto operates in an abnormal manner
Secondary function, to provide indication oflocation and type of failure
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Types of Protection
Over voltage protection – Lightning/Surgearrestors, Insulation co-ordination
Fault protection (over loading/current) -Application of Relays
Protection for human (safety) – Earthing andEarth mats, clearances, insulation
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Protection Philosophy
Cricket Protection
Position the fielders Designs the system
and sets the relays
Delivers the ball Charges the system
Batsman hits the ball Fault occurs
Mid-off stops/fails Primary relay
operates/ fails
Long-off stops/ fails Backup relay
operates / fails
Match lost ? Loss of
Power/Revenue
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Protection FeaturesSensitive – detection of short circuit or abnormalcondition
Selectivity – ensure that only the unhealthy part ofthe system is disconnected
Speed – to prevent or minimize damage and risk ofinstability of machines
Reliability – to ensure proper action even after longperiods of inactivity and also after repeatedoperations under severe conditions
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Protection System Design
Design of circuit breakers and isolating devices
Transducer (sensing devices)
Relay design
Relay application and co-ordination
Simulation and testing
Battery
Trip Coil
CB
Aux.
Switch
Relay
Relay Contacts
C T
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Protection Statistics Correct and Undesired
Operation as per the settings.
Settings are wrong, not updated
Human Error
Leaving the trip circuit open after test
Open circuited trip coils
Mechanical failure of circuit breaker
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Causes of Protection Failure Current or voltage supply to the relays
DC-tripping voltage
Protective relays
Tripping circuit or breaker mechanism
Circuit breaker
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Equipment Selection Relay selection
Selecting a relay scheme to recognize the existence of a fault within a given protective zone
Initiate circuit breaker operation
Circuit breaker selection
Determine the interruption requirement
Normal current and voltage rating
Selection of breakers and location
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Fault Characteristics
Magnitude
Frequency
Phase angle
Duration
Rate of change
Direction or order of change
Harmonics or the wave shape
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Protection Scheme Design
Divide the system into zones
Design appropriate breakers
Minimize the amount of load being disconnected
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Types of Relays
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Schematic of a Static Relay
COM T NOT AR
TRIPOver-Current :
FCRICT
CT
PP
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Schematic of an NR
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Over-Current Relays
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Basics
Relays operate based on the current sensed.
Can be used as primary or backup protection
May be directional or non-directional
Operation time depends on curve selected
Settings involve selecting the plug setting and the time setting
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Definitions Plug Setting
Plug setting is that value of current abovewhich the relay should operate.
Time dial setting
The time dial setting of the relay provides thediscrimination between the primary and backuprelay.
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Standard Curves
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Curve Equations
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Relay Co-ordination Identification of main and backup relays
Steady state analysis
Determination of plug setting
Determination of definite time setting
Determination of time dial setting
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Distance Relays
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Basics*Distance relays are the most widely used relays fortransmission lines protection.
*The relay measures the distance in terms ofimpedance by measuring the voltage and current.
*Major advantage is that relay acts as primary aswell as remote back up.
*The present digital distance relays offer morefunctionalities in terms of protection, monitoring andcontrol rather than just impedance measurement
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Types of Characteristics Mho
Quadrilateral
Lentical
Quadra-Mho
Elliptical Characteristics
Circular
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Mho Characteristic
X
Roffset
Zone-3
Zone-2
Zone-1(R3X3)(R2X2)
(R1X1)
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Quadrilateral Characteristics
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Impedance Calculation
Relays operate if the impedance seen is withinthe set characteristics known as zones.
Separate zones are present for phase and earthloops.
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PhilosophyZone Direction Protected Reach Time Setting (s)
Zone 1 Forward 80 % of the PL Instantaneous
Zone 2(for 400kV and above)
Forward
For Single Circuit –120 % of the PL
0.35
For Double Circuit –150 % of PL
0.5 to 0.6 – If Z2 reach overreaches 50% of the shortest line;0.35 – otherwise
Zone 2 (for 220kV and below
Forward120% of PL or100% of PL + 50% of ASL
0.35
PL: Protected Line, ASL: Adjacent Shortest Line, ALL: Adjacent Longest Line
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PhilosophyZone Direction Protected Reach Time Setting (s)
Zone 3 Forward 120% 0f (PL + ALL) 0.8 to 1.0
Zone 4 Reverse
10% for the Long line (>100 km)20% for short line (<100 km)
0.5 if Z4 overreaches 50% of reverse shortest line0.35: otherwise
Note : Z2 reach should not encroach the next lower voltage level: If Z3 reach encroaches next voltage level (after considering in feed), Z3 time must
be coordinated: If utility uses carrier blocking scheme, then the Z4 reach may be increased as the
requirement. It should cover the LBB of local bus bar and should be coordinated with Z2 time ofthe all other lines.
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Zone 1 Instrument Transformer errors
Transmission line is not completely balanced
Hence apparent impedance is susceptible to an error of 20%.
Zone 1 is under reached to prevent incorrect operation for fault on next line (Eg. Close to adjacent bus)
Zone 1 reach
Who covers the Remaining
?
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Zone 2 for single circuit line To provide coverage to remaining portion of
protected line
Over reach margin should be min 20% (for reason
discussed in Zone1)
Zone 1 reach
Zone 2 reach What about backup
protection for fault here ?
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Zone 2 for double circuit line
Zone 1 reach
Zone 2 reach What about backup
protection for fault here ?
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Zone 3 Zone 3 protection is backup protection for fault on
adjacent line.
Need to reach the PL + ALL
Zone 1 reach
Zone 2 reach
Considering Errors
Zone 3 reach
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Resistive Reach (Zone 1)
Earth
Should provide maximum coverage considering fault resistance, arc resistance and tower footing resistance.
Should be <4.5 * X1 (X1=Zone 1 reach)
Phase
Reach should be set to provide coverage against all types of anticipated phase to phase faults subject to check of possibility against load point encroachment
Should be <3*X1
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Resistive Reach (Zone 2 and Zone 3)
The philosophy used for Zone 1 is applicable herealso.
Additionally
Due to in-feeds, the apparent fault resistanceseen by relay is several times the actual value,this should be considered before arriving at thesetting.
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Queries and Discussions
161
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Thank You
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Protection Case Studies
1
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Topics for Discussion Overcurrent Relays
Distance Relays
Differential Relays
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Overcurrent relays
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Relay coordination sample
• Generator data
• Generator fault level :
• Maximum : 1000 MVA
• Minimum : 500 MVA
Transmission Line Details
Sl. No From Bus To Bus Impedance(pu)
1 1 2 0.1
2 2 3 0.1
3 3 4 0.1
4 4 5 0.1
~
R1
R2
R3
R4
B-1
B-2
B-3
B-4
B-5
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Relay data
Sl No Relay Name
MaximumLoad Current(A)
CT ratingRelay
Characteristics
1 R1 800 800/1 3s
2 R2 400 400/1 3s
3 R3 200 200/1 3s
4 R4 100 100/1 3s
Relay Pairs
Sl No Primary Relay Backup Relay
1 R4 R3
2 R3 R2
3 R2 R1
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Fault Studies results
Relay Impedance Fault Bus Fault MVAFault Current
in A
R1 0.1 1 1000 52486.388
R2 0.2 2 500 26243.194
R3 0.3 3 333.33 17495.288
R4 0.4 4 250 13121.590
SlNo
PrimaryRelay
Backup Relay
Fault Bus
Backup Relay Current in A
1 R4 R3 413121.590
2 R3 R2 3 17495.288
3 R2 R1 5 26243.194
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Plug setting computationFor Relay R4
% Setting Primary Setting Remarks
50 50 < Load Current
75 75 < Load Current
100 100 = Load Current
RelayLoad
CurrentCT Rating % Setting
Primary Current
R1 800 800/1 100 800
R2 400 400/1 100 400
R3 200 200/1 100 200
R4 100 100/1 100 100
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Time setting computation
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Time setting computation
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Final relay settingsRelay Plug Setting TMS
R1 800 0.6
R2 400 0.47
R3 200 0.29
R4 100 0.05
Operating times for a fault at Bus 5, with a fault current of 10, 500 A
R1 ≈ 1.61 s
R2 ≈ 1.00 s
R3 ≈ 0.51 s
R4 ≈ 0.07 s
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Co-ordinated Curves
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Application
• Backup to distance relays (220/132 kV andbelow)
• HV and LV side of ICT as backup to differentialrelays
• DEF is used in 400 kV line as protection for highimpedance earth faults.
• Non directional OC is also used in 400kV lineunder fuse fail condition
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DOC Setting Philosophy• For 220 kV and 132 kV line with only one main distance
protection, DOC must be set in such a way that for fault at remote bus the DOC (IDMT) is coordinated with Zone 2 time typically 1.1 s
• For DOC at 400 kV side of ICT, it is to be set such that for fault at remote end bus (longest line), DOC (IDMT) is coordinated with Zone 3 time
• Instantaneous DOC can be used at 400 kV side of ICT, set such that
– It does not pickup for fault on 220 kV bus
– It does not pickup for transformer charging current
– Time setting of 0.05 s to 0.1 s can be considered.
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DEF Setting Philosophy• For 220 kV and 132 kV line with only one main distance
protection, DEF (IDMT) must be set in such a way that it is coordinated with Zone 2 time
• For 400 kV line or where two main distance protection is used, DEF is used only for protection against high impedance faults and is coordinated with Zone 3 time
• For DEF at 400 kV side of ICT, it is to be set such that for fault at remote end bus (longest line), DEF (IDMT) is coordinated with Zone 3 time
• Instantaneous DEF can be used at 400 kV side of ICT, set such that
– It does not pickup for fault on 220 kV bus
– It does not pickup for transformer charging current
– Time setting of 0.05 s to 0.1 s can be considered.
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Instantaneous Setting calculation
15000 MVA
315 MVA
Zps=0.12 pu
Zebra
55 km
3φ Fault
Ia = 3225 A
3000 MVA
R2
• Reflected Current = 1.3 * 3225 = 4195 A• Inrush Current = 8*450 = 3600 A• Ip>> = Highest of the above two = 4200 A (Set
value).
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Distance relays
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179
• Sample SystemR1 = 0.0297 Ω/kmX1 = 0.332 Ω/kmR0 = 0.161 Ω/kmX0 = 1.24 Ω/kmZm = 0.528 Ω/km
G
400 kV Station 1 400 kV Station 2 220 kV Station 2
130 km
Twin Moose
80 km
Twin Moose
200 km
Twin Moose
162 km
Twin Moose
4x315 MVA
Z=0.12 pu
21
CCVT input
CT input
Station 5
Station 4
Station 3
Other Lines89 km
Twin Moose
Station 6
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Zone 1• Primary referred
• X1_prim = 0.332 *130*0.8 = 34.5 Ω
• Secondary referred
• X1_sec = 34.5*CTR/PTR, CTR:CT Ratio, PTR: PT Ratio
• Operating Time: Instantaneous
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Zone 2• Protected line is single circuit
• Adjacent Shortest Line: 80 km to Station 5
• X2_Prim = 0.332*130*1.2 = 51.79 Ω (130+26 km)
• X2 covers only 32% of Adjacent Shortest Line.
• Operating Time : 0.35s
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Zone 3• Adjacent Longest Line is 200 km to Station 4
• X3_Prim = 0.332*(130+200)*1.2 = 131.47 Ω
• That is 130 km PL + additional 266 km
• Time Setting
• Check if Zone 3 of PL encroach Zone 3 of adjacentline protection
• Zone 3 of Station 2- Station 5 line is 202.8 km.
• The Two Zone 3 are overlapping and hence must betime coordinated.
• Time = 0.8+ 0.06 (tcb)+0.03 (treset)+0.06 (tsf)= 0.95(set 1s)
• tcb, treset and tsf values are as per CEA report
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Points to be Considered• Obtain the actual line parameters from line impedance
test results. If not available, consider the standard values.
• Check Relay Setting type
• Primary or secondary referred values
• RX or Z
• Computation of Zero sequence compensation factor. (K0, Kn or Kr-Kx
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Points to be Considered• Note the shortest and longest line emanating from
the adjacent substation, along with impedance values
• For double circuit lines, check if it is two single circuit tower or one double circuit tower
• If necessary, carry out system study to study the effect of in feed, mutual coupling, power swing to achieve coordinated setting.
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Load Encroachment Consideration
• Basic Check
• Rated V = 63.5 V and Rated I = 1 A. (Secondary Referred)
• Phase Impedance = (0.85*V)/(1.5*I) = 36 Ω
• R reach must always be less than 36 Ω
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Operating Time coordination
• Zone 1 is primary protection, hence instantaneous operation
• Zone 2 is a overreach zone and hence needs to be delayed.
• If errors involved is less, then Zone 2 can trigger for Zone 1 fault of adjacent line. Hence Zone 2 is to be coordinated with Zone 1.
0
0.35
Tim
e (
s)
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Operating Time Coordination• For fault at 80% of segment B, Segment B zone 1 can
also under reach by 20%
• Hence only Zone 2 of Segment B pick up
• Segment A Zone 2 can also pickup.
• Hence segment A Z2 to be coordinated with segment B Z2
Tim
e (
s)
0
0.35
0.5 to 0.6
Segment A Segment B
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Operating Time coordination
• For 220 kV and below, if Zone 2 of PL can encroach the Zone 2 of ASL, then coordination is achieved by reducing the Zone 2 reach of PL to coordinate with Zone 2 reach of ASL.
Tim
e (
s)
0
0.35
Segment A Segment B
Reduced Zone 2 reach for coordination purpose
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Operating Time coordination• Zone 3 is the backup protection and hence needs to
be operate after Zone 1 or Zone 2 has failed to clear the fault.
• Zone 3 is coordinated with Zone 2 time of adjacent line relay.
Tim
e (
s)
0
0.35
0.5 to 0.6
Segment A Segment B
0.8 to 1.0
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Load Encroachment Consideration• Reach setting is given considering the various “under
reach” effect that can occur. This makes the resistivesetting “high”
• Phase loop measures the phase impedance and hencehas to be set such that it does not trip for abnormal oremergency system loading condition
• Emergency loading condition is to be used to decide theload encroachment point.
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Load Encroachment Consideration• Following criteria may be considered
• 1.5 times the thermal rating of line
OR
• 1.5 times associated bay equipment rating(minimum of all equipment)
• Minimum voltage of 0.85 pu to be considered
Load encroachment angle can be considered as 30(approx. 0.85 pf)
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Power Swing Blocking• During power system disturbances, unbalance
exist between generators which cause theoperating point to oscillate
• R-X trajectory might enter the zones of protection
• Blinder schemes are traditionally used to detectswings and prevent unnecessary tripping
• The rate of change of impedance is used todifferentiate between power swings and faults
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Power Swing Blocking Schemes
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PSB Options
No power swing blocking
Block all zones
Block zone 2 and higher
Block all zones but allow OST
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Stability Studies for PSB Settings
• Identify the generators which are prone to instabilityand the lines where power swings are observed.
• Determine the location where system should beseparated
• Time and reach settings are calculated
• Outermost zone reach should be within the innerboundary
• Outer boundary should be less than load impedance
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Carrier Aided Protection
Transfer
Unblocking
Direct
Blocking
Permissive
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Permissive Under-Reach
A BZ1
Z2
Z1
Z2
Close to A
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Permissive Over-Reach
A BZ1
Z2
Z1
Z2
Close to A
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Queries & Discussions
199
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Thank You
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Principles of Unit Protection
1
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Topics for Discussion
Introduction to Unit Protection
Differential Protection – Simple, Percentage Biased and High Impedance
Applications
Setting Examples
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Meaning
Protection which responds to faults onlyin one zone
No need for time grading
Normally fast acting – not affected byfault severity
Operates based on the comparison ofboundary conditions
Introduction
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Types of Unit Protection
Differential Protection
Utilizes Current
Over-Fluxing Protection
Utilizes Voltage and Frequency
Backup-Impedance Protection
Utilizes Current and Voltage
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Simple Differential Protection
Differential Protection
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Setting Consideration The effect of magnetizing inrush during initial
energization should be considered.
Correction for possible phase shift across thetransformer windings (phase correction)
The effects of different earthing and windingarrangements (filtering of zero sequence currents)
Correction for possible unbalance of signals fromcurrent transformers on either side of the windings(ratio correction)
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Phase Correction and Zero
Sequence Filtering
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Ratio Correction
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Biased Differential Relay
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Requirement of Biased
RelaysErrors which vary proportionally with the operatingconditions such as:
– Errors due to transformer taps
– CT errors
– Ratio Mismatch – Difficulty in obtaining exactlysimilar currents in both secondaries
Transient Errors
– CT Saturation
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Operating Characteristics
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CT
Satu
ration
• During CT saturation magnetizing impedance = zero
• Infinite magnetizing current will flow in the shorted path.
• Secondary of the network acts as open circuit.
• During this situation secondary current Is = zero
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High Impedance Differential Protection
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Comparison of Differential Schemes
Low Impedance High Impedance• Identical CTs not required
• Not stable for CTSaturation for externalfaults
• No protection for over-voltage required
• Stable for CT saturation forexternal faults
• Identical CTs are preferredfor sensitive operation
• May require an MOV toprotect Relay againstvoltage developed duringexternal faults
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Restricted Earth Fault Protection
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Amplification of Neutral
Current
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Currents During External Faults
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Currents During an Internal
Fault
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Transformer Protection
Differential Relays
Mainly percentage biased relays
2nd and 5th harmonic restraints are provided
High Set may be provided on HV winding
Restricted Earth Fault Relay
High Impedance Schemes are popular
Applications
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Modern Day Transformer Protection
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Line/Cable Protection
Differential Protection
– Percentage biased schemes are used
– Inrush currents are considered
– Major challenge was the distance between two ends
Pilot Wire Protection
– Measurements communicated over the pilot wire
– Balanced voltage as well as circulating current schemes are used
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Balanced Voltage Scheme
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Comparison of Pilot Wire
Schemes
Balanced Voltage Circulating Current Works on the tripping
principle
Interruption to the pilotwires causes blocking
Short circuit of pilot wiresleads to tripping
Overcurrent protectionrequired to preventincorrect tripping due toshort circuit of pilot wires
Works on blocking principle
Interruption of the pilotwires causes tripping
Short circuit of pilot wiresleads to blocking
Overcurrent protectionrequired to block operationfor large currents due topilot wire interruption
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Bus-Bar Protection
Differential Schemes
Incoming and outgoing currents are balanced
Multiple CT connections
Both High and Low Impedance Schemes are used
Partial Bus Bar Relay
Retention of supply to healthy portion of the network
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Partial Bus Bar Relay Example
225
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Overall Differential
Protection
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Sample Network
Setting Examples
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General Parameters
Parameter Value Unit
Mid tap Voltage in % 0
Primary side Full load Current at Mid tap Voltage
262.4319405 A
Secondary side Full load current 2624.319405 A
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Secondary Ratio
Compensation
Parameter Value Unit
Secondary side Full load Current on CT secondary 0.874773135 A
Secondary side ratio compensation 1.143153533
Secondary Full load CT secondary current with ratio correction
1 A
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Primary Side Ratio
Compensation
Parameter Value Unit
Primary full Load current At Maximum tap Voltage 249.9351815 A
Primary full Load current At Minimum tap Voltage 276.2441479 A
Primary side Full load Current on CT secondary side at Mid tap Voltage 0.874773135 A
Primary full Load current on CT secondary side At Maximum tap Voltage 0.833117272 A
Primary full Load current on CT secondary side At Minimum tap Voltage 0.920813826 A
Primary side ratio compensation 1.143153533
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Differential Current
Computation
Parameter Value Unit
Primary full Load current on CT secondary with ratio correction At Maximum tap Voltage
0.952380952 A
Primary full Load current on CT secondary with ratio correction At Minimum tap Voltage
1.052631579 A
Differential current at maximum tap 0.047619048 A
Differential current at minimum tap 0.052631579 A
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Restraint Current
Calculation
Parameter Value Unit
Restraint current at maximum tap 0.976190476 A
Restraint current at minimum tap 1.052631579 A
Setting Error at maximum tap 4.87804878 %
Setting Error at minimum tap 5.00 %
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Relay Settings
Setting Value Unit
Min Current 0.15 A
Slope 1 Start 0.75 A
Slope 1 20
Slope 2 Start 1.33 A
Slope 2 75
High Set 12.5 A
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Relay Characteristics
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0 0.5 1 1.5 2 2.5
I dif
fere
nti
al o
n C
T Se
con
dar
y
I restraint on CT Secondary
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REF - CT Selection The voltage across the stabilizing resistor at maximum through fault
current is given by
Vs = k*Ikmax*(RCT + 2. RL)/ n volts
Where:
RCT= current transformer secondary winding resistance
RL = One way maximum lead resistance from the CT to the relaying point
Ikmax = maximum external fault current
k = stabilizing factor used to account for the distorted voltage wave due to faults.
n = transformation ratio of the CT
The knee point voltage of the CTs should be chosen at least 2 times the stability voltage setting
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REF - Voltage Settings
Considerations
Always, relay operating voltage should be less than Vk/2 but more than Vs.
If stability voltage increases beyond Vk, CT enters saturation region for which there is infinite Magnetization (Current), but voltage remains at the same level (no huge change in voltage). This condition is seen as open circuited on the secondary.
Vset >Vs, such that relay will not operate for external fault with CT saturation condition.
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REF - Stabilizing Resistor Design The value of the Iprim at which the relay operates at certain settings
can be calculated as follows:Iprim = n * (Ir + m*(Vs/Vk)*Io) AWhere,n = the transformation ratio of the current transformerIr = the current value representing the relay settingm = the number of current transformers included in the
protectionIo = the magnetizing current of one current transformer at knee
point voltage, Vk. The relay setting current Ir can be computed as
Ir = (Iprim / n) - m*(Vs/Vk)*Io A
The setting of the stabilizing resistor (Rs) must be calculated in the following manner, where the setting is a function of the required stability voltage setting (Vs) and the relay current setting (Ir).
Stabilizing Resistor value, Rs = Vs/Ir
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REF - Metrosil Requirement
The following equations should be used to estimate the peaktransient voltage that could be produced for an internal fault.
Vp = 2 * sqrt(2. Vk(Vf – Vk)) voltsVf = If *(RCT + 2.RL + Rs) voltsWhere:Vp = Peak voltage developed by the CT under internal fault
conditions.Vk = Current transformer knee-point voltage.Vf = Maximum voltage that would be produced if CT saturation
did not occur.If = Maximum internal secondary fault current.RCT = Current transformer secondary winding resistance.RL = One way maximum lead burden from current transformer to
relay.Rs = Stabilizing resistor.
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Queries & Discussions
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Thank You