This document discusses line reactance, zero sequence reactance, and mutual zero sequence reactance in power transmission lines. It begins by explaining the basics of inductance and deriving the formula to calculate the inductance of single and three-phase transmission lines. It then shows how to represent the inductance of a three-phase line as a matrix and introduces the concepts of zero sequence impedance and mutual zero sequence impedance. The document uses symmetrical component analysis to derive the zero sequence impedance matrix and provides an example calculation for a specific 132kV transmission line. It explains how to use the zero sequence reactance value to estimate short circuit currents.

1. Line Reactance, Zero
Sequence Reactance and
Mutual Zero Sequence
Reactance
By S. P. Vasekar, Superintending Engineer, MSETCL,
Testing and Communication Circle Aurangabad

2. Line Reactance, Zero Sequence Reactance and
Mutual Zero Sequence Reactance
• Inductance of the transmission line
– Basics of Inductance and inductance of Single and
Three Phase Lines
• Inductance of Three Phase Line as Matrix
– General Theory
• Zero Sequence Impedance of Transmission
Line
– Theory of Sequence Component and It’s
Application
• Zero Sequence Mutual Impedance of
Transmission Line
– Sequence Component Theory as Applied to Double
Circuit Line

3. Inductance of the
Transmission Line
Basics of Inductance and Inductance
of Single and Three Phase Line

4. Inductance – Quick Refresh
V = L * di/dt By definition
V = N* dΦ/dt Faradays law
N*dΦ/dt = L*di/dt
L = N*dΦ / di
Inductance is flux
linkage per ampere
of current
ϕ ϕ
I
Φ
L = NΦ/I

5. What is inductance (Physics)
Whenever current flows there is a loop.
This current loop setups magnetic field.
The magnetic field can be considered
as magnetic field lines and which also
setup loops. These two loops are
always interlinked
Visualization of
magnetic field with
QuickField software

6. Approximate Field Distribution for Circular
Loop
0.0000 mWb0.0173 mWb0.8973 mWb1.7123 mWb2.0173 mWb

7. What is Magnetic Flux Linkage
1000 Amp
Φ = 2 mWb
1000 mm
r = 22 mm
L = ϕ/I = 2*10-6 Henry

8. Magnetic Field By Two Wires
1 m

9. Remember Linkage
• The most important thing is flux
linkage with current loop

10. Inductance of a long straight wire
• Only open straight conductor can not carry any
current
• Hence magnetic flux setup by it also zero
• And it is not possible to derive any expression
for it’s inductance.

11. Magnetic Field Distribution – 1-Ph Line

12. Concept of GMR – Solid Conductor
Area A1
Area A2
A1 = A2
R
R'
R' = R*e (-1/4)
R' = 0.7788*R
ϕ

13. Inductance of Two Wire Line
1 m
L = 4x10-7LogeD/r’ H/m
L = 2x10-7LogeD/r’ H/m

14. Formula for calculation of
inductance of 3-Ph transmission line
1
2
5
7 3
6 4
GMR =
𝟕 𝟕
𝒅𝟏𝟏 ∗ 𝒅𝟏𝟐 ∗ 𝒅𝟏𝟑 ∗ 𝒅𝟏𝟒 ∗ 𝒅𝟏𝟓 ∗ 𝒅𝟏𝟔 ∗ 𝒅𝟏𝟕 ∗
𝟕
𝒅𝟐𝟏 ∗ 𝒅𝟐𝟐 ∗ 𝒅𝟐𝟑 ∗ 𝒅𝟐𝟒 ∗ 𝒅𝟐𝟓 ∗ 𝒅𝟐𝟔 ∗ 𝒅𝟐𝟕 ∗ ⋯ . .
DAB
DBC
DCA
A
B
L = 2*10-7 *Loge(GMD/GMR) H/m
C
GMD =
𝟑
𝑫 𝑨𝑩 ∗ 𝑫𝑩𝑪 ∗ 𝑫𝑪𝑨

15. Calculation of GMR

16. Solution for 132kV Line
with Panther Conductor
• L = 2x10-7*Loge(4.0/0.0086767) H/m
• L = 2x10-7*Loge(461) H/m
• L = 2x10-7*6.1334 H/m
• L = 12.2668x10-7 H/m
• L = 12.2668x10-3 H/Km
• X = 2*π*f*L
• X = 2*3.142*50*12.2668*10-3
• X = 314*12.2668*10-3
• X = 0.3851 Ω / Km

17. Impedance of Line as Matrix
General Equation

18. Effect of Mutual Coupling
N 85
A 100mm2
0.0001m2
μ0 1.257E-05H.m
μr 281
l 160mm
0.16m
L = μ0μrN2A/l
0.0159H
XL = 2*π*f*L
5.0Ohm
N 170
A 100mm2
0.0001m2
μ0 1.257E-05H.m
μr 281
l 160mm
0.16m
L = μ0μrN2A/l
0.0638H
XL = 2*π*f*L
20.0Ohm

19. Mutual Inductance, Mathematical Representation
ɸ
ɸ
N = 170
a = 80 mm
t = 10 mm
N = 170
b = 10 mm
b = 10 mm
a = 80 mm

20. Mutual Impedance – 2 Wire Line

21. 3-Ph Transmission Line – QuickField Problem

22. 3-Ph Transmission Line – Field Simulation

23. Three Phase Line Matrix Notation
Without
Considering
Mutual Coupling
With Mutual
Coupling

24. Generl Equation of the transmission line
parameters (Without Earth Wire)
𝑟 + 𝑅𝐶 𝑅𝐶 𝑅𝐶
𝑅𝐶 𝑟 + 𝑅𝐶 𝑅𝐶
𝑅𝐶 𝑅𝐶 𝑟 + 𝑅𝐶
From last slide we have Vabc – Vabc’ = Zabc*Iabc (All Bold
face represents matrix)
Where Zabc = Rabc + Jxabc and to calculate Zabc from
conductor and line data use
Rabc =
r is resistance (Ω/km) of the
conductor used and RC =
0.0001*π2f for f = 50 Hz RC =
0.04935 Ω/km
𝐿𝑛(
1
𝐺𝑀𝑅
) 𝐿𝑛
1
𝐷𝑎𝑏
+ 𝐺𝐶 𝐿𝑛
1
𝐷𝑎𝑐
+ 𝐺𝐶
𝐿𝑛
1
𝐷𝑏𝑎
+ 𝐺𝐶 𝐿𝑛(
1
𝐺𝑀𝑅
) 𝐿𝑛
1
𝐷𝑏𝑐
+ 𝐺𝐶
𝐿𝑛
1
𝐷𝑐𝑎
+ 𝐺𝐶 𝐿𝑛
1
𝐷𝑐𝑏
+ 𝐺𝐶 𝐿𝑛(
1
𝐺𝑀𝑅
)
Xabc =
M is for converting H/m to Ω/km
and it is 4*π*f*0.0001 for 50Hz
its value is 0.062832.
GC = ρ is earath resistivity
and for ρ = 100 Ω.m and f = 50
Hz GC = 6.8357
M*
658 ∗
𝜌
𝑓
Ln
Ω/km
Ω/km

25. Practical System Calculations
132kV
50 km
0.2 ACSR
50 MW
For 0.2 ACSR
GMR =
8.6767mm
0.008677m
For Above line
Dab = 4m
Dbc = 4m
Dca = 8m
Let ρ = 100Ω.m
f = 50Hz
r = 0.1522Ω/km
0.2015 + j0.7277 0.0493 + j0.3424 0.0493 + j0.2988
Zabc =
0.0493 + j0.3424 0.2015 + j0.7277 0.0493 + j0.3424
0.0493 + j0.2988 0.0493 + j0.3424 0.2015 + j0.7277

26. Zero Sequence Impedance
of Transmission Line
Sequence Component Theory and
Application

27. Symmetrical Components
Resolve Components

28. +Ve Sequence Comp.
-Ve Sequence Comp.
0 Sequence Comp.
Symmetrical Components and Matrix A-1

29. Operator - a
VR
VB
VY
a = -0.5 + j0.866
OR
a = 1.0 ∟1200
= a*VR
= a*a*VR

30. +Ve Sequence Comp.
-Ve Sequence Comp.
0 Sequence Comp.
Symmetrical Components and Matrix A
Va = Va0 + Va1 + Va2
Vb = Vb0 + Vb1 + Vb2
Vc = Vc0 + Vc1 + Vc2
Va = Va0 + Va1 + Va2
Vb = Va0 + a2Va1 + aVa2
Vc = Vc0 + aVa1 + a2Va2
Va
Vb
Vc
Va0
Va1
Va2
1 1 1
1 a2 a
1 a a2
*=
Vabc = A * V012

31. Symmetrical Components and Matrix A
Va
Vb
Vc
Va0
Va1
Va2
1 1 1
1 a2 a
1 a a2
*=
Vabc = A * V012
V012 = A-1 * Vabc V0
V1
V2
Va
Vb
Vc
1 1 1
1 a a2
1 a2 a
*= 1/3

32. U 7.07 ∟ 45.00
V 4.81 ∟-102.74
W 12.44 ∟ 27.25
U0 5.39 ∟21.80
U1 3.61 ∟-33.69
U2 5.83 ∟120.96
+Ve Sequence Comp.
-Ve Sequence Comp.
0 Sequence Comp.
U012 = A-1Uabc
Symmetrical Components and Matrix A-1
Display Matrix FormulaDisplay MatrixDisplay Zero Seq. ComponentsDisplay +Ve Seq. ComponentsDisplay-Ve Seq. ComponentsCheck Solution

33. Sequence Impedances
Vabc = Xabc * Iabc (1f considered Va’ = Vb’ = Vc’ = 0)
A*V012 = Xabc*A*I012
A-1*A*V012 = A-1*Xabc*A*I012
V012 = X012 * I012 Where X012 = A-1*Xabc*A
Or in General Term
Z012 = A-1 * Zabc * A

34. Sequence Impedances of Transmission Line
For 0.2 ACSR
GMR = 8.6767mm
0.00867m
For Above line
Dab = 4m
Dbc = 4m
Dca = 8m
Let ρ = 100Ω.m
f = 50Hz
r = 0.1522Ω/km
0.2015+j0.7277 0.0493+j0.3424 0.0493+j0.2988
Zabc = 0.0493+j0.3424 0.2015+j0.7277 0.0493+j0.3424
0.0493+j0.2988 0.0493+j0.3424 0.2015+j0.7277
0.3002+j1.3835 0.0126-j0.0073 -0.0126-j0.0073
Z012 = -0.0126-j0.0073 0.1522+j0.3999 -0.0251+j0.0145
0.0126-j0.0073 0.0251+j0.0145 0.1522+j0.3999

35. Application of Sequence Impedance Matrix
50 km
I1
r = 0.16/km 5 m
R1 = 8 Ohm 2 m I2
R2 = 8 Ohm
R3 = 8 Ohm I3
V = 80 V

36. Current through un-transposed line
36.3887 16.4190 15.3620
Xabc = 16.4190 36.3887 19.2976
15.3620 19.2976 36.3887
0.03665 -0.01159 -0.00932
Xabc
-1 -0.01159 0.04190 -0.01733
-0.00932 -0.01733 0.04061
1.258575
1.038553
1.116394
Ia
Ib
Ic
0.03665 -0.01159 -0.00932
-0.01159 0.04190 -0.01733
-0.00932 -0.01733 0.04061
80
80
80
Ia
Ib
Ic
X=
=

37. Current through un-transposed line
36.3887 16.4190 15.3620
Xabc = 16.4190 36.3887 19.2976
15.3620 19.2976 36.3887
I0
I1
I2
80
0
0
Ia
Ib
Ic
X=
=
0.0000 + j70.4411 0.3051 - j1.1357 -0.3051 - j1.1357
X012 = -0.3051 - j1.1357 0.0000 + j19.3625 -0.6103 + j2.2714
0.3051 - j1.1357 0.6103 + j2.2714 0.0000 + j19.3625
0.0000 + j0.014223 0.000281 - j0.000755 -0.000281 - j0.000755
-0.000281 - j0.000755 0.0000 + j0.052465 -0.001682 + j0.006115
0.000281 - j0.000755 0.001682 + j0.006115i 0.0000 + j0.052465
From
I012 = X012
-1*V012
And
Iabc = A*I012
1.2586
1.0386
1.1164
I0
I1
I2
=
1.137843
0.0604 - j0.0225
0.0604 + j0.0225
1.138∟00.0
0.064∟-20.4
0.064∟20.4
Or AND

38. What is special about sequence impedance matrix
0.7552 0.3459 0.3029
Zabc = 0.3459 0.7552 0.3459
0.3029 0.3459 0.7552
1.4157 0.0145 0.0145
Z012 = 0.0145 0.4279 0.0290
0.0145 0.0290 0.4279
Off diagonal elements are negligible indicates that there is minimum
mutual coupling if we use symmetrical component analysis
For a 132kV 50Hz, 0.2 ACSR, Vertical
Single Circuit Line Without Earth Wire

39. Short-circuit currents by using X0
0.0000 + j70.4411 0.3051 - j1.1357 -0.3051 - j1.1357
X012 = -0.3051 - j1.1357 0.0000 + j19.3625 -0.6103 + j2.2714
0.3051 - j1.1357 0.6103 + j2.2714 0.0000 + j19.3625
Ia ≈ Ib ≈ Ic ≈ I0 = V0/X0 = 80/70.44 = 1.136
Ia
Ib
Ic
=
1.2586
1.0386
1.1164
And Iavg = 1.137