Short Circuit Calculation
Ren Christian M. Santos, REE, MSc
bit.ly/linkedin-rcms
Learning Objectives:
System Representation
Mathematical
Representation
Solution to Short
Circuit Problems
Applications
Fault Types
Shunt Fault
(Unbalance between phases or between phase and ground)
• Single Line-to-Ground Fault (unsymmetrical)
• Line-to-Line Fault (unsymmetrical)
• Double Line-to-Ground Fault (unsymmetrical)
• Three-Phase Fault (symmetrical)
Series Fault
(Unbalance in line impedances which does not involve neutral
or ground, or any inter-connection between phases)
• One Line Open (unsymmetrical)
• Two Lines Open (unsymmetrical)
Shunt Fault Types
Single Line-to-Ground Fault (unsymmetrical)
System
a
b
c
Zf
Boundary Conditions:
• Ib=Ic=0
• Va=ZfIa
Ground
Ia
Shunt Fault Types
Line-to-Line Fault (unsymmetrical)
System
a
b
c
Ib
Zf
Ic
Boundary Conditions:
• Ia=0
• Ib=-Ic
• Vb-Vc =ZfIb
Ground
Shunt Fault Types
Double Line-to-Ground Fault (unsymmetrical)
System
a
b
c
Ib
Zf
Zf
Ic
Boundary Conditions:
• Ia=0
• Vb = (Zf +Zg)Ib+ZgIc
• Vc = (Zf +Zg)Ic+ZgIb
Ground
Zg
Shunt Fault Types
Three-Phase Fault (symmetrical)
System
a
b
c
Ia
Ib
Zf
Ic
Boundary Conditions:
• balanced
• ground current is zero
Ground
Zf
Zf
Shunt Fault Types
Three-Phase Fault (symmetrical)
System
a
b
c
Ia
Ib
Zf
Ic
Boundary Conditions:
• balanced
• ground current is zero
Ground
Zf
Zf
Zg
Series Fault Types
One Line Open (unsymmetrical)
System
a
b
c
Boundary Conditions:
• Ia=0
• Vb-Vb’ =ZfIb
• Vc-Vc’ =ZfIb
Ground
Ia
Ib
Zf
Ic
Zf
a'
b'
c'
Series Fault Types
Two Lines Open (unsymmetrical)
System
a
b
c
Ia
Ib
Ic
Boundary Conditions:
• Ib = Ic=0
• Va-Va’ =ZfIa
Ground
Zf
Learning Objectives:
System Representation
Mathematical
Representation
Solution to Short
Circuit Problems
Applications
Symmetrical Components
In a three-phase system, a given unbalanced network may be
replaced by three sets of balanced networks which are referred
to as the symmetrical components of the original unbalanced
network
• Positive-sequence network
• Negative-sequence network
• Zero-sequence network
Symmetrical Components
𝑰𝒂𝟐
𝑰𝒂𝟏
Zth1
𝑰𝒂𝟎
+
+
Va1
Va2
Zth2
Vth
-
Positive Sequence
Theveninequivalent Network
+
Va0
Zth0
-
-
Negative Sequence
Theveninequivalent Network
Zero Sequence
Theveninequivalent Network
Element Representation
System Representation: Generator
ZG1
ZG2
G
Positive Sequence
Negative Sequence
ZG0
Zero Sequence
System Representation: Line
ZL1
Positive Sequence
ZL2
Negative Sequence
ZL0
Zero Sequence
System Representation: Transformer
ZT1
Positive Sequence
ZT2
Negative Sequence
ZT0
Zero Sequence
(for Yg-Yg)
System Representation: Transformer
ZT1
Positive Sequence
ZT2
Negative Sequence
ZT0
Zero Sequence
(for Yg-Δ)
System Representation: Transformer
ZT1
Positive Sequence
ZT2
Negative Sequence
ZT0
Zero Sequence
(for Δ-Δ)
System Representation: Transformer
ZT1
Positive Sequence
ZT2
Negative Sequence
ZT0
Zero Sequence
(for Y-Y)
Learning Opportunity: Determine the zero sequence connection for
other types of transformers
Learning Objectives:
System Representation
Mathematical
Representation
Solution to Short
Circuit Problems
Applications
Recall
𝑰𝒂𝟐
𝑰𝒂𝟏
Zth1
𝑰𝒂𝟎
+
+
Va1
Va2
Zth2
Vth
-
Positive Sequence
Theveninequivalent Network
+
Va0
Zth0
-
-
Negative Sequence
Theveninequivalent Network
Zero Sequence
Theveninequivalent Network
Symmetrical Components: Mathematical Representation
𝑉𝑎0
1 1
𝑉𝑎1 = 1
3
𝑉𝑎2
1
1
𝑎
𝑎2
1
𝑎2
𝑎
𝑉𝑎
𝑉𝑏
𝑉𝑐
𝑉𝑎
1
𝑉𝑏 = 1
𝑉𝑐
1
1
𝑎2
𝑎
1
𝑎
𝑎2
𝑉𝑎0
𝑉𝑎1
𝑉𝑎2
𝐼𝑎0
1 1
𝐼𝑎1 = 1
3
𝐼𝑎2
1
1
𝑎
𝑎2
1
𝑎2
𝑎
𝐼𝑎
𝐼𝑏
𝐼𝑐
𝐼𝑎
1
𝐼𝑏 = 1
𝐼𝑐
1
1
𝑎2
𝑎
1
𝑎
𝑎2
𝐼𝑎0
𝐼𝑎1
𝐼𝑎2
a=1∠120o
1 + a + a2 = 0
General Steps
Thevenin
SLD with Sequence data
(convert in per unit)
Draw with (+) (-) (0)
Sequence Networks
Compute for Vth and Zth
Symmetrical
Components
(Series and Parallel Circuits)
Connect the Sequence
Networks based on
Boundary Conditions and
Fault Type
Compute for Sequence
currents and voltages
Compute for Actual
currents and voltages
Sample Derivation: SLG
System
a
b
c
Zf
Boundary Conditions:
• Ib=Ic=0
• Va=ZfIa
Ground
Ia
Symmetrical Components
Boundary Conditions:
• Ib=Ic=0
• Va=ZfIa
𝐼𝑎0
1 1
𝐼𝑎1 = 1
3
𝐼𝑎2
1
1
𝑎
𝑎2
1
𝑎2
𝑎
𝐼𝑎
𝐼𝑏
𝐼𝑐
𝐼𝑎0
1 1
𝐼𝑎1 = 1
3
𝐼𝑎2
1
1
𝑎
𝑎2
1
𝑎2
𝑎
𝐼𝑎
0
0
𝐼𝑎0
𝐼
1 𝑎
𝐼𝑎1 = 𝐼𝑎
3 𝐼
𝐼𝑎2
𝑎
𝑰𝒂𝟎 = 𝑰𝒂𝟏 = 𝑰𝒂𝟐
𝟏
= 𝑰𝒂
𝟑
Symmetrical Components
Boundary Conditions:
• Ib=Ic=0
• Va=ZfIa
𝑉𝑎 = 𝑍𝑓 𝐼𝑎
Recall
𝑉𝑎
1 1
𝑉𝑏 = 1 𝑎2
𝑉𝑐
1 𝑎
1 𝑉𝑎0
𝑎 𝑉𝑎1
𝑎2 𝑉𝑎2
𝐼𝑎
1 1
𝐼𝑏 = 1 𝑎2
𝐼𝑐
1 𝑎
1
𝑎
𝑎2
𝐼𝑎0
𝐼𝑎1
𝐼𝑎2
𝑉𝑎0 + 𝑉𝑎1 + 𝑉𝑎2 = 𝑍𝑓 𝐼𝑎0 + 𝐼𝑎1 + 𝐼𝑎2
Recall
𝐼𝑎0 = 𝐼𝑎1 = 𝐼𝑎2
𝑽𝒂𝟎 + 𝑽𝒂𝟏 + 𝑽𝒂𝟐 = 𝟑𝒁𝒇 𝑰𝒂𝟎
Symmetrical Components
3Zf
𝑰𝒂𝟎 = 𝑰𝒂𝟏 = 𝑰𝒂𝟐 =
𝟏
𝑰
𝟑 𝒂
𝑰𝒂𝟎 = 𝑰𝒂𝟏 = 𝑰𝒂𝟐
𝐼𝑎1 = 𝐼𝑎2 = 𝐼𝑎0
𝑉𝑡ℎ
=
𝑍𝑡ℎ1 + 𝑍𝑡ℎ2 + 𝑍𝑡ℎ0 + 3𝑍𝑓
𝑽𝒂𝟎 + 𝑽𝒂𝟏 + 𝑽𝒂𝟐 = 𝟑𝒁𝒇 𝑰𝒂𝟎
𝑽𝒂𝟎 + 𝑽𝒂𝟏 + 𝑽𝒂𝟐 − 𝟑𝒁𝒇 𝑰𝒂𝟎 = 𝟎
𝑰𝒂 = 𝟑𝑰𝒂𝟏
𝟑𝑽𝒕𝒉
=
𝒁𝒕𝒉𝟏 + 𝒁𝒕𝒉𝟐 + 𝒁𝒕𝒉𝟎 + 𝟑𝒁𝒇
Learning Opportunity: Check if currents Ib and Ic are zero
Sequence Network Connection Summary
Fault Type
Connection
Floating
SLG
Series
None
LL
Parallel
Zero Sequence Network
LLG
Parallel
None
3-phase
Series
Negative and Zero Sequence Network
1LO
Parallel
None
2LO
Series
None
Learning Opportunity: Update the table with Fault Impedance (Zf)
Learning Objectives:
System Representation
Mathematical
Representation
Solution to Short
Circuit Problems
Applications
3-bus System
G
G
X1=0.25
X2=0.25
X0=0.10
XH=0.12
XX=-0.04
XY=0.09
X=0.10
X1=0.05
X2=0.05
X0=0.20
2
1
X1=0.10
X2=0.10
X0=0.30
X1=0.10
X2=0.10
X0=0.30
3
Load
X1=0.25
X2=0.25
X0=0.10
Shunt Fault Sequence Network
Positive Sequence
N
G
1∠0o
G
j0.25
1∠0o
j0.25
j0.10
j0.09
j0.12
2
j0.05
-j0.04
1
j0.10
j0.10
3
Negative Sequence
N
j0.25
j0.25
j0.10
j0.09
j0.12
2
j0.05
-j0.04
1
j0.10
j0.10
3
Zero Sequence
N
j0.10
j0.10
j0.10
j0.09
j0.12
2
j0.20
-j0.04
1
j0.30
j0.30
3
Faulted 3-bus system
G
G
X1=0.25
X2=0.25
X0=0.10
XH=0.12
XX=-0.04
XY=0.09
X1=0.25
X2=0.25
X0=0.10
X=0.10
X1=0.05
X2=0.05
X0=0.20
2
F
1
X1=0.10
X2=0.10
X0=0.30
X1=0.10
X2=0.10
X0=0.30
3
Load
Positive Sequence (Shunt Fault)
N
G
1∠0o
G
j0.25
1∠0o
j0.25
j0.10
j0.09
j0.12
2
j0.05
-j0.04
F
1
j0.10
j0.10
3
Positive Sequence
(Shunt Fault)
Thevenin Impedance (F to N)
Zth = {(j0.25+j0.12-j0.04)+[(j0.05)||(j0.10 + j0.10)]}
||
{(j0.25+j0.10)}
Zth = {(j0.33)+[(j0.05)||(j0.20)]} || {(j0.35)}
Zth = {(j0.33)+[j0.04]} || {(j0.35)}
Zth = {(j0.37)} || {(j0.35)}
Ia1
Zth = j0.18
Thevenin Voltage (F to N)
Vth =
1∠0o
Z1
+
j0.18
Va1
Vth
1∠0o
N1
|| - parallel impedance calculation
F1
-
Negative Sequence (Shunt Fault)
N
j0.25
j0.25
j0.10
j0.09
j0.12
2
j0.05
-j0.04
F
1
j0.10
j0.10
3
Negative Sequence
(Shunt Fault)
Thevenin Impedance (F to N)
Zth = {(j0.25+j0.12-j0.04)+[(j0.05)||(j0.10 + j0.10)]}
||
{(j0.25+j0.10)}
Zth = {(j0.33)+[(j0.05)||(j0.20)]} || {(j0.35)}
Zth = {(j0.33)+[j0.04]} || {(j0.35)}
Zth = {(j0.37)} || {(j0.35)}
Ia2
Zth = j0.18
Z2
F2
+
j0.18
Va2
N2
|| - parallel impedance calculation
-
Zero Sequence (Shunt Fault)
N
j0.10
j0.10
j0.10
j0.09
j0.12
2
j0.20
-j0.04
F
1
j0.30
j0.30
3
Zero Sequence
(Shunt Fault)
Thevenin Impedance (F to N)
Zth = {[(j0.10+j0.12)||(j0.09)]-j0.04+
[(j0.20)||(j0.30 + j0.30)]}
||
{(j0.10+j0.10)}
Zth = {[(j0.22)||(j0.09)]-j0.04+
[(j0.20)||(j0.60)]}
||
{(j0.20)}
Zth = {j0.06-j0.04+
[j0.15]}
||
{(j0.20)}
Ia0
Z0
F0
+
j0.09
Va0
Zth = {j0.17} || {(j0.20)}
Zth = j0.09
N0
|| - parallel impedance calculation
-
Single Line to Ground Fault
Ia1
Z1
F1
Ia2
+
j0.18
Z2
Ia0
+
j0.18
Va1
Vth
F2
Z0
F0
+
j0.09
Va2
Va0
1∠0o
N1
N2
-
𝐼𝑎1 = 𝐼𝑎2 = 𝐼𝑎0
-
1∠0
=
𝑗0.18 + 𝑗0.18 + 𝑗0.09 + 3(0)
𝑰𝒂𝟏 = 𝑰𝒂𝟐 = 𝑰𝒂𝟎 = −𝒋𝟐. 𝟐𝟐
N0
-
3ZF
Single Line to Ground Fault
𝐼𝑎1 = 𝐼𝑎2 = 𝐼𝑎0 = −𝑗2.22
𝐼𝑎
1
𝐼𝑏 = 1
𝐼𝑐
1
𝐼𝑎
1
𝐼𝑏 = 1
𝐼𝑐
1
1
𝑎2
𝑎
1
𝑎2
𝑎
𝐼𝑎0
𝐼𝑎1
𝐼𝑎2
1
𝑎
𝑎2
1
𝑎
𝑎2
−𝑗2.22
−𝑗2.22
−𝑗2.22
𝐼𝑎
−𝑗6.66
𝐼𝑏 =
0
𝐼𝑐
0
Single Line to Ground Fault
System
a
b
c
Ia = 6.66 pu
Ib = 0 pu
Ic = 0 pu
Ground
Learning Opportunity: Re-compute when Zf = j0.10
Line to Line Fault
ZF
Ia1
Z1
F1
Ia2
+
j0.18
Z2
F2
j0.18
Va1
Vth
Ia0
+
Z0
F0
+
j0.09
Va2
Va0
1∠0o
N1
N2
-
-
N0
𝐼𝑎0 = 0
𝐼𝑎1
1∠0
= −𝐼𝑎2 =
= −𝑗2.78
𝑗0.18 + 𝑗0.18 + (0)
Learning Opportunity: Why is zero sequence current is zero?
-
Line to Line Fault
𝐼𝑎1
𝐼𝑎0 = 0
1∠0
= −𝐼𝑎2 =
= −𝑗2.78
𝑗0.18 + 𝑗0.18 + (0)
𝐼𝑎
1
𝐼𝑏 = 1
𝐼𝑐
1
1
𝑎2
𝑎
𝐼𝑎
1 1
𝐼𝑏 = 1 𝑎2
𝐼𝑐
1 𝑎
1
𝑎
𝑎2
1
𝑎
𝑎2
𝐼𝑎
0
𝐼𝑏 = −4.82
𝐼𝑐
4.82
𝐼𝑎0
𝐼𝑎1
𝐼𝑎2
0
−𝑗2.78
+𝑗2.78
Line to Line Fault
System
a
b
c
Ia = 0 pu
Ib = -4.78 pu
Ic = 4.78 pu
Ground
Learning Opportunity: Re-compute when Zf = j0.10
Double Line to Ground Fault
Learning Opportunity: Show end to end process for DLG Fault
3-phase Fault
Learning Opportunity: Show end to end process for 3-phase Fault
Series Fault Sequence Network
Positive Sequence (Series Fault)
N
G
1∠0o
G
j0.25
1.1∠0o
j0.25
j0.10
j0.09
j0.12
2
j0.05
-j0.04
F
1
F’
j0.10
j0.10
3
Positive Sequence (Series Fault)
N
G
1∠0o
G
j0.33
1.1∠0o
j0.35
𝐼𝑙𝑜𝑜𝑝
2
j0.05
F
1
j0.20
Thevenin Voltage (F to F’)
𝐼𝑙𝑜𝑜𝑝
1.1∠0 − 1 ∠0
=
= −𝑗0.14
𝑗0.33 + 𝑗0.35 + 𝑗0.05
𝑉𝑡ℎ = −𝑗0.14 ∗ 𝑗0.05 = 0.007
F’
Positive Sequence (Series Fault)
N
G
1∠0o
G
Xd
j0.33
1.1∠0o
j0.35
Xf
Xe
2
j0.05
F
1
j0.20
Thevenin Impedance (F to F’)
𝑗0.33 ∗ 𝑗0.35
𝑋𝑑 =
= 𝑗0.16
𝑗0.33 + 𝑗0.35 + 𝑗0.05
𝑋𝑒 = 𝑋𝑓 = 𝑗0.02
F’
Positive Sequence (Series Fault)
N
j0.16
j0.02
j0.02
2
F
1
j0.20
Thevenin Impedance (F to F’)
𝑍𝑡ℎ = 𝑗0.20 + 𝑗0.02 + 𝑗0.02
𝑍𝑡ℎ = 𝑗0.24
F’
Positive Sequence (Series Fault)
Ia1
Z1
F1
+
j0.24
Va1
Vth
0.007∠0o
F1’
-
Negative Sequence (Series Fault)
Ia2
Z2
F2
+
j0.24
Va2
F2’
-
Zero Sequence (Series Fault)
N
j0.10
j0.10
j0.10
j0.09
j0.12
2
j0.20
-j0.04
F
1
F’
j0.30
j0.30
3
Zero Sequence (Series Fault)
N
Xd
j0.02
j0.20
Xf
Xe
2
j0.20
F
1
j0.60
F’
Zero Sequence (Series Fault)
N
j0.01
j0.01
j0.10
2
F
1
j0.60
Thevenin Impedance (F to F’)
𝑍𝑡ℎ = 𝑗0.60 + 𝑗0.01 + 𝑗0.10
𝑍𝑡ℎ = 𝑗0.71
F’
Zero Sequence (Series Fault)
Ia0
Z0
F0
+
j0.71
Va2
F0’
-
Series Fault Sequence Network
Ia1
F1
j0.24
Z1
Ia2
+
Z2
Ia0
+
j0.24
Va1
Vth
F2
Z0
F0
+
j0.71
Va2
Va2
0.007∠0o
F1’
N
-
F2’
-
F0’
-
One Line Open Sequence Network (1LO)
Ia1
Ia2
+
F1
j0.24
Z1
Z2
F2
+
j0.24
F0
Z0
Va1
Vth
Ia0
+
j0.71
Va2
Va2
0.007∠0o
F1’
F2’
-
-
F0’
N
𝐼𝑎1 =
𝐼𝑎2
0.007∠0
= −𝑗0.017
𝑗0.24 + 𝑗0.24 || 𝑗0.71
𝑗0.71
= −𝐼𝑎1
= 𝑗0.013
𝑗0.24 + 𝑗0.71
𝐼𝑎0
𝑗0.24
= −𝐼𝑎1
= 𝑗0.004
𝑗0.24 + 𝑗0.71
-
One Line Open Sequence Network (1LO)
𝐼𝑎1 =
𝐼𝑎2
0.007∠0
= −𝑗0.017
𝑗0.24 + 𝑗0.24 || 𝑗0.71
𝑗0.71
= −𝐼𝑎1
= 𝑗0.013
𝑗0.24 + 𝑗0.71
𝐼𝑎
1
𝐼𝑏 = 1
𝐼𝑐
1
𝐼𝑎0 = −𝐼𝑎1
1
𝑎2
𝑎
1
𝑎
𝑎2
𝑗0.24
= 𝑗0.004
𝑗0.24 + 𝑗0.71
𝐼𝑎0
𝐼𝑎1
𝐼𝑎2
Learning Opportunity: Compute to Ia, Ib, Ic
Two Line Open Sequence Network (2LO)
Ia1
F1
j0.24
Z1
Ia2
+
Z2
Ia0
+
j0.24
Va1
Vth
F2
Z0
F0
+
j0.71
Va2
Va2
0.007∠0o
F1’
-
F2’
-
N
𝐼𝑎1 = 𝐼𝑎2 = 𝐼𝑎0 =
0.007∠0
𝑗0.24 + 𝑗0.24 + 𝑗0.71
𝑰𝒂𝟏 = 𝑰𝒂𝟐 = 𝑰𝒂𝟎 = −𝒋𝟎. 𝟎𝟎𝟔
F0’
-
Two Line Open Sequence Network (2LO)
𝐼𝑎1 = 𝐼𝑎2 = 𝐼𝑎0 =
0.007∠0
𝑗0.24 + 𝑗0.24 + 𝑗0.71
𝑰𝒂𝟏 = 𝑰𝒂𝟐 = 𝑰𝒂𝟎 = −𝒋𝟎. 𝟎𝟎𝟔
𝐼𝑎
1
𝐼𝑏 = 1
𝐼𝑐
1
1
𝑎2
𝑎
1
𝑎
𝑎2
𝐼𝑎0
𝐼𝑎1
𝐼𝑎2
Learning Opportunity: Compute to Ia, Ib, Ic
Learning Objectives:
System Representation
Mathematical
Representation
Solution to Short
Circuit Problems
Applications
Most common Applications
Protection Coordination
Insulation Coordination (Surge Protection)
Arc-flash Calculations
Stability Analysis
Short Circuit Calculation
Ren Christian M. Santos, REE, MSc
bit.ly/linkedin-rcms