SHORT CIRCUIT CALCULATIONS REVISITED. pptx

SHORT CIRCUIT
CALCULATIONS
REVISITED
Short Circuit Calculations
IIEE Presentation
FORTUNATO C.
LEYNES
Chairman
Professional Regulatory
Board of Electrical Engineering
Professional Regulation Commission

Short Circuit (Fault) Analysis
FAULT-PROOF SYSTEM
 not practical
 neither economical
 faults or failures occur in any powersystem
In the various parts of the electrical network
under short circuit or unbalanced condition,
the determination of the magnitudes and phase
angles
 Currents
 Voltages
 Impedances
IIEE Presentation

Application of Fault Analysis
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1. The determination of the required mechanical
strength of electrical equipment to withstand
the stresses brought about by the flow of high
short circuit currents
2. The selection of circuit breakers and switch
ratings
3. The selection of protective relay and fuse
ratings

4. The setting and coordination of protective
devices
5. The selection of surge arresters and insulation
ratings of electrical equipment
6. The determination of fault impedances for use
in stability studies
7. The calculation of voltage sags caused resulting
from short circuits
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8. The sizing of series reactors to limit the short
circuit current to a desired value
9. To determine the short circuit capability of
series capacitors used in series compensation of
long transmission lines
10. To determine the size of grounding
transformers, resistances, or reactors
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Per Unit Calculations
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Three-phase Systems
IIEE Presentation
)2
base MVA3
(base voltage, kV
base kVA3
(base voltage, kV )2
1000
ZB LL
ZB LL

Per Unit Quantities
B
IIEE Presentation
pu
B
pu
B
pu
Base impedance (Z )
actual impedance
BaseVoltage (kV )
V
Base Current (I )
actual current
Z 

actual voltage (kV)
I 

Changing the Base of Per Unit
Quantities
pu[new]
B[new]
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[new]
[new]
B[new]
[old]
pu[old]
[old]
pu[old]
Z
Z
base kVA
basekV
Z
base kVA[old]
base kVA[old]
Z

Z()
2
1000

Z base kV 2
1000
Z() 
base kV 2
1000

actual impedance, Z()

Changing the Base of Per Unit
Quantities
B[new]
IIEE Presentation
pu[new]
[new]
[new]
B[new]
[old]
pu[old ]
[old ]
pu[old]
Z
Z
base kVA
basekV
Z
base kVA[old]
base kVA[old]
Z

Z()
2
1000

Z base kV 2
1000
Z() 
base kV 2
1000

actual impedance, Z()

kVA Base for Motors
IIEE Presentation
kVA/hp hp rating
1.00 Induction < 100hp
1.00 Synchronous 0.8 pf
0.95 Induction 100 < 999hp
0.90 Induction > 1000hp
0.80 Synchronous 1.0 pf

SYMMETRICA
L
COMPONENT
S
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Vc1 Va1
Va2
V
Vb2
Unbalanced Phasors
Zero Sequence
Vb1
Positive Sequence
Va0  Vb0 Vc0
c
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V
Vb
Vc0
c1
V
Vc2
Vb0
Vb2
Vb1
Va1
a2
V
Negative Sequence
c2
Va 0
Va

Symmetrical Components of
Unbalanced Three-phase Phasor
a
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b c
a
a
V  
V
Vb Vc 
 a2
V  aV 
b c
a2
2
a1
a0
3
1
3
V 
1
V  aV  a V 
3
V 
1
V
a2
a1
a0
c
a2
a1
a0
b
a2
a1
V V  aV  a2
V
V V  a2
V  aV
V V
Va Va0

In matrix form:
Symmetrical Components of
Unbalanced Three-phase Phasor
 
  2
1
1 1
Vc
Vb   1 a
a

IIEE Presentation
aVa1  
 V 2 
a2 Va2  Va2

a1 
Va  1Va0  Va0
1
1Va 
1 1
3 1 a a Vb 
a2
aVc

 
1 

Sequence Networks
IIEE Presentation
Power System Short Circuit
Calculations

Fault Point
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The fault point of a system is that point to
which the unbalanced connection is
attached to an otherwise balanced system.

Definition of Sequence
Networks
Positive-sequence Network
Ea1 = Thevenin’s equivalent
voltage as seen at the fault
Z1 =
point
Thevenin’s equivalent
impedance as seen from
the fault point
Va1  Ea1 Ia1Z1
Z1
Ea1
Ia1
Va1
+
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-

Negative-sequence Network
Z2 = Thevenin’s equivalent
negative-sequence
impedance as seen at
the fault point
Networks
Va2 Ia 2Z2
Z2
Ia2
Va
2
+
-
IIEE Presentation

Zero-sequence Network
Z0 = Thevenin’s equivalent
zero-sequence impedance
as seen at the fault point
Networks
a0 0
a0
V  I Z Z0
Ia0
Va
0
+
-
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Sequence Network Models of
Power System Components
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Calculations

Synchronous Machines
(Positive Sequence Network)
Z1  jx''d
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(Negative Sequence Network)
2
2
x''d x''q
Z  j Z2
Ia2
Va2
+
-
Where:
x”d = direct-axis sub-transient reactance
x”q = quadrature-axis sub-transient reactance
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(Zero Sequence Network)
Solidly-Grounded Neutral
jX0
n
I0
ground
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I0
Impedance-Grounded Neutral
jX0
n
3Zg
ground
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Ungrounded-Wye or Delta Connected
Generators
jX0
n
I0
ground
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Two-Winding Transformers
Standard
Symbol
P S
ZPS
Equivalent
Positive-seq. Network
P S
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Three-Winding Transformers
t
Zst  Zs Z
Z ps  Zp  Zs
Z pt  Z p  Zt 

pt
st
ps
s
pt st
p ps
Zpt  Zst  Z ps
Zt 
 Z Z
2
1

Z 
1 Z
2
Z 
1
Z  Z  Z 
P
T
Standard
Symbol
Zt
T
Equivalent Positive-Sequence
Network
ZS
Zp S
S P
2
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Transformers
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The negative-sequence network of two-
winding and three-winding transformers
are modeled in the same way as the
positive-sequence network since the
positive-sequence and negative-sequence
impedances of transformers are equal.

Simplified Derivation of Transformer Zero-
Sequence Circuit Modeling
Grounded wye S1 = 1 and S3 = 0
or
S2 = 1 or S4 = 0
Delta S1 = 0 and S3 = 1
or
S2 = 0 and S4 = 1
Ungrounded wye S1 = 0 and S3 = 0
or
S2 = 0 and S4 = 0
cuit Calculations
(Thanks to Engr. Antonio C. Coronel, Retired VP, Meralco, and former member, Board of Electrical Engineering)
P Q
Short Cir
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Simplified Derivation of Transformer
Zero-Sequence Circuit Modeling
P Q
S1
IIEE Presentation
S3 S4
S2
Grounded wye – Grounded wye
S1 = 1
S2 = 1
S3 = 0
S4 = 0
Z

P Q
Grounded wye – Ungrounded wye
S1 = 1
S2 = 0
S3 = 0
S4 = 0
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S1 = 1
S2 = 0
S3 = 0
S4 = 1
P Q
Grounded wye – Delta
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S1 = 0
S2 = 0
S3 = 0
S4 = 0
P Q
Delta – Delta
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Transformers
(Zero-Sequence Circuit Model)
Transformer
Connection
Zero-Sequence
Circuit Equivalent
P Q ZPQ
P Q
P Q ZPQ
P
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Q

Transformer
Connection
Q
P ZPQ
P Q
P Q ZPQ
P
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Q
Transformers
Zero-Sequence
Circuit Equivalent

Transformer
Connection
P Q
R
ZP ZQ
ZR
P
R
Q
P Q ZPQ
Zero-Sequence
Circuit Equivalent
P Q
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Transformers

Transformer
Connection
Zero-Sequence
Circuit Equivalent
ZP ZQ
ZR
P
R
Q
P Q
R
Q
P
R
ZP ZQ
ZR
P
R
Q
Transformers
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Transmission Lines (Positive
Sequence Network)
ra jXL
Z1
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Transmission Lines
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The same model as the positive-sequence
network is used for transmission lines
inasmuch as the positive-sequence and
negative-sequence impedances of
transmission lines are the same

Transmission Lines
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The zero-sequence network model for a
transmission line is the same as that of the
positive- and negative-sequence networks.
The sequence impedance of the model is of
course the zero-sequence impedance of the
line. This is normally higher than the
positive- and negative-sequence impedances
because of the influence of the earth’s
resistivity and the ground wire/s.

Calculations
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Classification of Power System
Short Circuits

Shunt Faults
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Single line-to-ground faults
Double line-to-ground faults
Line-to-line faults
Three-phase faults

Series Faults
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One-line open faults
Two-line open faults

Combination of Shunt and
Series Faults
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Single line-to-ground and one-line open
Double line-to-ground and one-line open faults
Line-to-line and one-line open faults
Three-phase and one-line open faults

Single line-to-ground and two-line open faults
Double line-to-ground and two-line open faults
Line-to-line and two-line open faults
Three-phase and two-line open faults
IIEE Presentation
Combination of Shunt and
Series Faults

Balanced Faults
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Symmetrical or
Three-Phase Faults

Derivation of Sequence Network Interconnections
Vc  Ic Z f
Boundary conditions:
Ia  Ib  Ic 0
VF  Va  Ia Z f  Vb  Ib Z f
Eq’n (1)
Eq’n (2)
Zf Zf
Vf
System A System B
A
B
C
Va Vb Vc
Ia
Ib
Ic
Zf
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IIEE Presentation
Z0
Ia0
Va0
+
-
Z1
Ea1
Ia1
Va1
+
-
Zf
Z2
Ia2
Va2
+
-
Zf
1
1
Z
a1
f
f
Ea1
1 f
Ea1
a1

Ea1
I  I
Z  Z
I f  Ia1 
Zf  0,
I f  Ia 0  Ia1 Ia 2
Z  Z
I 

Unbalanced Faults
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Single Line-to-Ground Faults

System A System B
A
B
C
Zf
Va Vb Vc
Ia
Ib
Ic
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Ib  Ic  0
Va  Ia Z f  0

E
Z0  Z1  Z2  3Zf
Ia0  Ia1  Ia2 a1
Z1
Ea1
Ia1
Va1
+
-
Z2
Ia2
Va2
+
-
Z0
Ia0
Va0
+
-
3Zf
1
IIEE Presentation
0
2
0 1
2
1
2
1
0
I
3Ea1
Ea1
Ea1
a2
a1
a0
Z 2Z
I f  Ia 
If Z f  0 and Z1  Z
I f  Ia  Ia0  Ia1  Ia 2  3Ia1 3Ia0
Z  2Z
Ia 0  Ia1  Ia 2 
Z  Z
Z  Z  Z
 I 
 I
If Z f  0

Unbalanced Faults
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Line-to-Line Faults

Vc  Ib Zf
Zf
System A System B
A
B
C
Va Vb Vc
Ia
Ib
Ic
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Ia  0
Ib  Ic
Vb  IbZ f Vc; or Vb

2
2Z1
E
1 2 f
Ea1
a1
Ia1 a1
If Z f  0 and Z1  Z
Z  Z Z
I 
a1
a1
f
a2
a1
a0
c
The fault current
  j 3I
Ia1  Ia 2
I  a2
 aI
Iao  0;
 I  a2
I  aI
I f  Ib  I
Z1
Ea1 Va1
+
-
Zf
Ia1
Z2
Ia2
Va2
+
-
Z0
Ia0
Va0
+
-
IIEE Presentation

1
IIEE Presentation
1
1
1
2
f [3]
f [3]
f
3
I
I
Z
I
2Z
3Ea1
Ea1


Ea1
3  Ea1
 

  
2 Z
I f   j
if Z f  0

f 



2Z  Z
I   j 3

thus, with Z1  Z2
f [LL]

Unbalanced Faults
IIEE Presentation
Double-to-line Ground Fault

 Ib  Ic Zg
Vb  Ib Zf
Vc  IcZ f  Ib  Ic Zg
Eq’n BC-1
Eq’n BC-2
Eq’n BC-3
Zf
SystemA System B
A
B
C
Va Vb Vc
Ia
Ib
Ic
Zf
Vf
Zg (Ib+Ic)
IIEE Presentation
Ia  0

g
f
Ea1
a1
Z2  Z0  2Z 3Z
 3Zg 
Z  Z Z  Z
2 f 0 f
Z1 Z f 
I 
Z0
Ia0
Va0
+
-
Z2
Ia2
Va2
+
-
Zf Zf Zf+3Zg
Z1
Ea1
Ia1
Va1
+
-
IIEE Presentation

Negative-sequence Component:


 g 
 Z0 2Z f 3Z
2
 Z 0 Z f  3Zg
Ia 2  Ia1

Z
Zero-sequence Component:
The fault current
f a0 a1 a2
a2
a1
a0
c a0 a1 a2
f b
I f  2Ia0 1Ia1  1Ia 2  2Ia0 Ia1  Ia2 
but Ia0  Ia1  Ia2  0; or Ia0  Ia1  Ia2 
thus, I f 3Ia 0
 a2
 aI  a  a2
I
I  2I
 aI  a2
I 
 a2
I  aI I
I  I  I  I

IIEE Presentation






g
Z  Z0  2Z f 3Z
2
Z2  Z f
Ia0  Ia1


1 1 0
IIEE Presentation
0
1
Z
Z1Z0
E
Z
Z Z
Z0 Ea1
a1
a1
a2
Ea1
a1
Z 2
 2Z Z

 




 Z1  Z0 


1 0 


 



Z  Z
Z 
  



0 

 Z1  Z0 
I  I
Z 2
 2Z Z
1 1 0

Z1  Z0 Ea1
Z1 Z0
Z1 1 0
I 
If Z f  Zg  0 and Z1  Z2

0
1
1 0
1
1
1
1
1
1
Z2
Z1Z0
a0
f
Z1Ea1
a1
a 0 a1
3Ea1
Z  2Z
I  3I 
Ea1
Z 2Z

 2Z Z


0 
  Z 
Z

1 0 




E
Z 


Z  Z 
Z 
 
0 


 Z 
Z  Z 
I  I 
1 1 0
1 0
IIEEPresentation
0
0
1
1
0
1 0
1
0
1
1
Z2
Z0 Ea1
a1
Z1Z0
a2
Ea1
Z1Z0
a1
Z 2
 2Z Z


  
 1 0 
Z Z







E

Z 



Z Z
Z 

a1
   
 Z 
 Z1  Z0 
I  I
 2Z Z

Z1  Z0 Ea1
Z  Z
Z 
I 
If Z f  Zg  0 and Z1  Z2

Voltage Rise Phenomenon
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Single-to-line Ground Fault

IIEE Presentation
Unfaulted Phase B Voltage During Single
Line-to-Ground Faults












 









 



 











 
1
1
1 
1
2
0 1
2
1

 1 0 
1

 1 0 
2 
0
 1 0 
 2
1 
neglecting resistances, R1 and R0 ;
 2 0
1 
1 
X1
X0

X

  X0

a2
V  E
Z1
 Z 

Z

  Z0

a2
V  E
 Z 

Z

 Z0
 Z1 
2 
Z0 
 Z1 
 E a 
 2Z1  Z0 
 Z  Z 
Z
2Z  Z
Z a
2Z  Z
 a E
Z
 Z
a1

b
a1


b
a1
Vb  Ea1 a 

Ea1
Ea1
a1
Ea1
Vb  
2Z
a2
V V  a2
V aV
b a0 a1

Unfaulted Phase B Voltage During
Single Line-to-Ground Faults
Neglecting resistances R0 &R1
50
45
40
35
30
25
20
15
10
5
0
-10 -6 -3 -2.8 -2.6 -2.4 -1 0 1 2 5 9 25 45 65 85
-2.2 -2 -1.8 -1.6 -1.4 -1.2
X0/X1
Phase
B
Voltage
(p.u.)
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Fault MVA
1
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3
for singe line - to - ground fault in p.u.:
Z
I
F(SLG)
Z0 2Z1
I 

1
MVAF  IF  MVAbase
where, for E
a1
 1.0p.u.;
for three  phase fault in p.u.:
F(3)

Fault MVA
IIEE Presentation
1
0
0
1
1
I
p.u.
I
p.u.
I
FSLG
FSLG
Z 
3
2Z

3
2Z  Z
Single line to  ground fault MVA :
MVAFSLG IF SLG( p.u.) MVAbase
Z 
1
F 3
Three  phase fault MVA:
MVAF3 IF 3( p.u.) MVAbase

Assumptions Made to Simplify Fault
Calculations
IIEE Presentation
1. Pre-fault load currents are neglected.
2. Pre-fault voltages are assumed equal to 1.0
per unit.
3. Resistances are neglected (only for 115kV &
up).
4. Mutual impedances, when not appreciableare
neglected.
5. Off-nominal transformer taps are equal to1.0
per unit.
6. Positive- and negative-sequence impedances
are equal.

Outline of Procedures for
IIEE Presentation
1 Setup the network impedances expressed in per
unit on a common MVA base in the form of a
single-line diagram
2 Determine the single equivalent (Thevenin’s)
impedance of each sequence network.
3 Determine the distribution factor giving the
current in the individual branches for unittotal
sequence current.

IIEE Presentation
4 Interconnect the three sequence networks for the
type of fault under considerations and calculate
the sequence currents at the faultpoint.
5 Determine the sequence current distribution by
the application of the distribution factors to the
sequence currents at the fault point
6 Synthesize the phase currents from the sequence
currents.

IIEE Presentation
7 Determine the sequence voltages throughout the
networks from the sequence current distribution
and branch impedances
8 Synthesize the phase voltages from the sequence
voltage components
9 Convert the pre unit currents and voltages to
actual physical units

CIRCUIT BREAKING
SIZING
(Asymmetrical Rating Factors)
Momentary Rating
Multiplying Factor = 1.6
Interrupting Rating
Multiplying Factor
IIEE Presentation
8 cycles
5 cycles
3 cycles
1 ½ cycles
= 1.0
= 1.1
= 1.2
= 1.5

EXAMPLE
PROBLEM
In the power system
shown, determine the
momentary and
interrupting ratings for
primary and secondary
circuit breakers of
transformer T2.
IIEE Presentation

Solution:
10
1
0
F(SLG)
x 
3
 20.125  0.05pu
x1 
8
 0.125pu
100 
I  1000 10 pu
100
F(3)
G1:
 
IIEE Presentation
Equivalent 69kVsystem@100MVA: T1:
100
75 
M1, M2 : kVAB  0.95000  4500
or MVAB 4.5
x  0.15 4.5  3.3333pu
x  0.06100  0.80pu
100
100
x" 0.1025 
 0.40pu
x0  0.0625 
 0.24 pu
T2,T3, T3:
30 
I  800  8pu x  0.08100  0.2667pu

Solution:
Positive sequence network:
IIEE Presentation

Solution:
Zero sequence network:
IIEE Presentation

Improper Sequence Network Models
IIEE Presentation

ions
Incorrect Sequence Network Models
Short Circuit Calculat
IIEE Presentation

QUESTIONS
?
IIEE Presentation

IIEE Presentation

SHORT CIRCUIT CALCULATIONS REVISITED. pptx

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