EE 4006

Power and Electric Distribution

Systems Design Capsule
Lecture #14
22.10.2024

Instructor: Alper Savaşcı, PhD

Inductance of Single-Phase Two-Wire Line

per conductor

per conductor

′
, 𝑟 𝑦 =0.7788 𝑟 𝑦

𝐿=𝐿𝑥 + 𝐿 𝑦
If = , the total circuit inductance:

per conductor
Single-phase two-
wire line

2
Inductance of Three-Phase Three-Wire Line

per conductor

Since the conductor has a simmetry, the

same result is obtained for, and .

3
 Inductance of Composite
Conductors, Unequal Phase
Spacing, Bundled Conductors
Conductor has N Conductor has M
identical identical subconductors,
subconductors, each with radius and
each with radius with return current .
and
with current .

𝜆𝑥 −7 𝐷 𝑥𝑦 𝜆𝑦 −7 𝐷 𝑥𝑦
𝐿𝑥 = =2 ×10 ln 𝐿 𝑦= =2× 10 ln
𝐼 𝐷𝑥𝑥 𝐼 𝐷 𝑦𝑦

𝐿=𝐿𝑥 + 𝐿 𝑦 per circuit

The total inductance L of the single-phase circuit is

4
 √
𝑁 𝑀
𝑁𝑀
𝐷 𝑥𝑦 = ∏
𝑘=1
∏ 𝐷𝑘𝑚
′
Geometric Mean Distance or GMD
between conductors x and y
𝑚=1

√∏ ∏
𝑁 𝑁
𝑁2
Geometric Mean Radius or GMR of conductor
𝐷 𝑥𝑥 = 𝐷𝑘𝑚
𝑘=1 𝑚=1

√∏ ∏
2 𝑀 𝑀
𝑀 Geometric Mean Radius or GMR of conductor
𝐷 𝑦𝑦 = 𝐷𝑘𝑚
′ ′
𝑘=1 𝑚=1

5
6
 2 1 1’ 2’ 3’

√
𝑁𝑀
𝑁 𝑀 𝐷 𝑥𝑦 = √( 𝐷 𝐷 𝐷 )( 𝐷 𝐷 𝐷 )( 𝐷 𝐷 𝐷 )
3 ⋅3
11′ 12′ 13′ 21′ 22 ′ 23 ′ 3 1′ 32′ 33′

𝐷 𝑥𝑦 = ∏
𝑘=1
∏ 𝐷𝑘𝑚
′
¿ √ ( 6⋅(6 .1)(6.2))( (6.1)(6.2)(6.3))( (6.05)(6.15)(6.25))
9

𝑚=1
¿ √ ( 6.1 ) ( 6.2 ) (6.3)( 6)(6.05)(6.15) (6.25)
9 2 2

√(𝐷
3 ⋅3

√
𝑁 𝑁
𝐷 𝑥𝑥 = 11 𝐷1 2 𝐷 1 3 ) ( 𝐷21 𝐷22 𝐷 23 ) ( 𝐷3 1 𝐷3 2 𝐷3 3 )
2
𝑁
𝐷 𝑥𝑥 = ∏ ∏
𝑘=1 𝑚=1
𝐷𝑘𝑚

7
 2 1 1’ 2’ 3’

3
𝐷 𝑦𝑦 = √(𝐷
3 ⋅3
𝐷1 ′ 2 ′ 𝐷1 ′ 3 ′ )( 𝐷2 ′ 1 ′ 𝐷2 ′ 2 ′ 𝐷 2′ 3 ′ )( 𝐷 3 ′1 ′ 𝐷 3 ′ 2′ 𝐷3 ′ 3 ′ )

√
1′ 1′
2 𝑀 𝑀
𝑀
𝐷 𝑦𝑦 = ∏ ∏ 𝐷𝑘𝑚
′ ′
𝑘=1 𝑚=1

The actual configuration can now be replaced by the two equivalent hollow
conductors each with its own geometric mean radius and separated by the
geometric mean distance as shown below:

8
For the line configuration given below, determine , , and in H/m for
the single-phase two-conductor line.

9
 √
𝑁 𝑁
𝑁2
𝐷 𝑥𝑥 = ∏ ∏
𝑘=1 𝑚=1
𝐷𝑘𝑚

√∏ ∏
2 𝑀 𝑀
𝑀
𝐷 𝑦𝑦 = 𝐷𝑘𝑚
′ ′
𝑘=1 𝑚=1

10
 • If the spacings between phases are unequal,
unbalanced flux linkages occur, and the
phase inductances are unequal.
• However, balance can be restored by
exchanging the conductor positions along the
line with a technique called transposition.
Unequal phase
• To calculate inductance of this line, assume
spacings
balanced positive-sequence currents, for
which .

Completely transposed
three-phase line

11
Line transposition
Need for transposition
• Equalizing line inductance
• Reducing radio interference by
canceling out the total flux linkages
over one transposition cycle.

https://electronics.stackexchange.com/
questions/83152/how-do-transposition-
towers-in-transmission-lines-work
12
 Completely transposed
three-phase line

• Linkage flux of phase a in each position

𝜆 𝑎1 =2 ×10
−7
[ 𝐼 𝑎 ln
1
𝐷𝑆
+ 𝐼 𝑏 ln
1
𝐷12
+ 𝐼 𝑐 ln
1
𝐷31 ] Wb-t/m

𝜆 𝑎 2=2 ×10
−7
[ 𝐼 𝑎 ln
1
𝐷𝑆
+ 𝐼 𝑏 ln
1
𝐷 23
+ 𝐼 𝑐 ln
1
𝐷 12 ] Wb-t/m

𝜆 𝑎 3=2 ×10
−7
[ 𝐼 𝑎 ln
1
𝐷𝑆
+ 𝐼 𝑏 ln
1
𝐷31
+ 𝐼 𝑐 ln
1
𝐷 23 ] Wb-t/m

𝜆𝑎 1 + 𝜆 𝑎2 + 𝜆𝑎 2
𝜆 𝑎=
3

¿
2 ×10− 7
3 [
3 𝐼 𝑎 ln
1
𝐷𝑆
+ 𝐼 𝑏 ln
1
𝐷12 𝐷 23 𝐷 31
+ 𝐼 𝑐 ln
1
𝐷 12 𝐷23 𝐷31 ] • Using

[ ]
−7
2 ×10 1 1
¿ 3 𝐼 𝑎 ln − 𝐼 𝑎 ln
3 𝐷𝑆 𝐷 12 𝐷23 𝐷31

[ ]
is the conductor GMR for
¿ 2 ×10
−7
𝐼𝑎 ln
√
3
𝐷12 𝐷23 𝐷31 stranded conductors, or
𝐷𝑆 for solid cylindrical
conductors.
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 √
[ ]
3
𝜆𝑎 −7 𝐷12 𝐷23 𝐷31
𝐿 𝑎= =2 ×10 ln H/m per phase
𝐼𝑎 𝐷𝑆

• The result will be same for other two phases and

• However, only one phase is sufficient to be considered for balanced
three-phase operation of a completely transposed three-phase
line.

Defining

𝐷 𝑒𝑞=√3 𝐷12 𝐷23 𝐷31 the geometric mean distance between phases.

𝐿 𝑎=2 × 10
−7
[ ln
𝐷𝑒𝑞
𝐷𝑆 ] H/m

is the conductor GMR for stranded conductors, or for solid cylindrical

conductors.

14
 𝐷 𝑒𝑞=√ 𝐷12 𝐷23 𝐷31=12.6 m the GMD between
3
,
phases
from table A.4
Flat horzn. spacing
𝐿 𝑎=2 ×10 −7
ln( 12.6
0.0158496 )
× ( 200 ⋅ 103 ) =0.267 𝐻 1 2 3

𝑋 𝑎=2 𝜋 𝑓 𝐿𝑎=( 2 𝜋 ⋅ 60 ) × 0.267 𝐻 =101 Ω 10 m 10 m

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Bundled Conductors

• To calculate inductance, DS is replaced by the GMR of the bundle. Since the

bundle constitutes a composite conductor, calculation of bundle GMR is done by

√
𝑁 𝑁
𝑁2
𝐷 𝑥𝑥 = ∏ ∏
𝑘=1 𝑚=1
𝐷𝑘𝑚

𝐿 𝑎=2 × 10
−7
[ ln
𝐷𝑒𝑞
𝐷𝑆 ] 𝐿 𝑎=2 ×10
−7
[ ln
𝐷𝑒𝑞
𝐺𝑀𝑅 ] H/m

16
• If the conductors are stranded and the bundle spacing d is large
compared to the conductor outside radius, each stranded conductor is
first replaced by an equivalent solid cylindrical conductor with GMR = DS.
• Then the bundle is replaced by one equivalent conductor with GMR = D SL

𝐿 𝑎=2 × 10 −7
[ ln
𝐷𝑒𝑞
𝐷 𝑆𝐿 ] H/m

17
Each of the 806 mm2 conductors in Example 4.4 is replaced by two
403 mm2 (795,000 cmil) ACSR 26/2 conductors, as shown in the
Figure below. Bundle spacing is 0.40 m. Flat horizontal spacing is
retained, with 10 m between adjacent bundle centers. Calculate the
inductive reactance of the line and compare it with that of Example
4.4. Note that line length is 200 km.

18
Table A.4

19
a)

GMR of each phase

GMD of each phase

20
The GMD between phases is given by the cube root of the product of the three-
phase spacings.

The inductance per phase is found as

b) The line reactance for each phase then becomes

21
22
Now consider A 345-kV, three-phase, double-circuit line with phase-conductor’s
GMR of 1.8 cm, and the horizontal conductor configuration shown in Figure 4.35.

(a) Determine the inductance per meter per phase in henries.

m
m
m

23
 m m

H/m

(b) Inductance of one circuit is calculated below:

m m

H/m

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Series Impedances
• In this section, we develop equations suitable for
computer calculation of the series impedances,
including resistances and inductive reactance, for the
three-phase overhead line shown in Figure 4.16.
• This line has three phase conductors a, b, and c,
where bundled conductors, if any, have already been
replaced by equivalent conductors,
• The line has N neutral conductors denoted
by , which all neutral conductors are
connected in parallel and grounded.
• If phase currents are not balanced, there may be a
return current flowing through the grounded neutral
wires and the earth.
• The earth return current will spread out
under the line, seeking the lowest
impedance return path.
• This return current can be represented by a set of
‘‘earth return’’ conductors located directly
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under the overhead conductors as shown
Earth resistivity
• Each earth return conductor carries the negative
of its overhead conductor current which has a
GMR denoted and distance from its overhead
conductor, and resistance .

𝐷 𝑘 𝑘 = 𝐷 𝑘𝑘 m : Earth sensitivity
′ ′

𝐷 𝑘 𝑘 =658.5 √ 𝜌 / 𝑓 m
′ ′
: frequency in hertz.

Ω/m

26
 • We renumber the overhead
conductors from , beginning with
the phase conductors, then
overhead neutral conductors.
• The sum of all the currents are
zero for an operating
transmission line as
(6+ 2 𝑁 )

∑ 𝐼 𝑘=0 : Number of neutral

𝑘=1 conductors.

The flux linking the overhead conductor k

( )
(3 +𝑁 )
𝐷𝑘𝑚
∑
′
−7
𝜆 𝑘=2 ×10 𝐼 𝑚 ln Wb - t/m
𝑚=1 𝐷 𝑘𝑚
In matrix form
𝝀=𝑳 𝑰
𝐿 𝑘𝑚=2 × 10 −7 ln ( 𝐷𝑘 𝑚
𝐷𝑘𝑚
′

)
: (3+N) vector
: (3+N) vector
: (3+N) x (3+N) matrix 27
 Circuit representation of series-phase
impedances Voltage drops across
the conductors.

: Resistance of each overhead conductor.

: Mutual resistance due to the image conductors.

28
The diagonal element of the matrix

And the off-diagonal elements, for , are

29
 Solving the second eqn for

Plug in in the first eqn

30
 For transposed lines

series-phase impedance matrix

31
 Electric Field and Voltage
• The capacitance between conductors in a medium
with constant permittivity can be obtained by
determining the following:
1. Electric field strength, , from Gauss Law
2. Voltage between conductors
3. Capacitance from charge per unit volt ()

• As a step toward computing capacitances of general conductor

configurations,
1. we first compute the electric field of a uniformly charged,
solid cylindrical conductor and
2. the voltage between two points outside the conductor.
3. We also compute the voltage between two conductors in
an array of charged conductors.
32
• Figure 4.18 shows a solid cylindrical conductor
with radius r and with charge q coulombs per
meter (assumed positive in the figure),
uniformly distributed on the conductor surface.
• For simplicity, assume that the conductor is (1)
sufficiently long that end effects are negligible,
and (2) a perfect conductor (that is, zero
resistivity, ρ = 0).
• To determine the electric field outside the
conductor, select the cylinder with radius x > r
and with 1-meter length, shown in Figure 4.18.
• Due to the uniform charge distribution, the
electric field strength Ex is constant on the
cylinder. Also, there is no tangential component
of Ex, so the electric field is radial to the
conductor.

33
• A plot of the electric field lines is also shown in
Figure 4.18. The direction of the field lines,
denoted by the arrows, is from the positive
charges where the field originates, to the
negative charges, which in this case are at
infinity.
• If the charge on the conductor surface were
negative, then the direction of the field lines
would be reversed.
• Concentric cylinders surrounding the conductor
are constant potential surfaces. The potential
difference between two concentric cylinders at
distances D1 and D2 from the conductor center
is

34
• The array of M solid cylindrical
conductors are shown in Figure
4.19. Assume that each conductor m
has an ac charge qm C/m uniformly
distributed along the conductor.
• The voltage Vkim between conductors k
and i due to the charge qm acting alone is

• Using superposition, the voltage Vki between conductors k and i due to all
the changes is

35
 Capacitance: Single-Phase Two-wire Line
• Assume that the conductors are
energized by a voltage source such that
conductor has a uniform charge and,
• Assuming conservation of charge,
conductor has an equal quantity of
negative charge .

Using
volts

36
For a 1-meter line length, the capacitance between conductors is

If

37
Circuit Representations
Circuit representations of the line-to-line and line-to-neutral capacitances are
shown in Figure 4.20.

38
 Capacitance in Three-phase Three-Wire Line
• Next consider the three-phase line with equal phase spacing shown
in Figure
• To determine the positive-sequence capacitance,
assume positive-sequence charges such that

The voltage between conductors and is

Using

39
Using,

The voltage between conductors and is

40
Recall that for balanced positive-
sequence voltages,

The capacitance line-to-neutral per line

The same result can be found for phase -b and -c.

41
Capacitance: Stranded
Conductors Unequal Phase
Spacing, Bundled Conductor
• Practical conductors with resistivities similar to those listed in Table 4.3 have a
small internal electric field. As a result, the external electric field is slightly
altered near the conductor surfaces. Also, the electric field near the surface of a
stranded conductor is not the same as that of a solid cylindrical conductor.
• However, it is normal practice when calculating line capacitance to replace a
stranded conductor by a perfectly con ducting solid cylindrical conductor whose
radius equals the outside radius of the stranded conductor.
• The resulting error in capacitance is small since only the electric field near the
conductor surfaces is affected.
• For three-phase lines with unequal phase spacing, balanced positive sequence
voltages are not obtained with balanced positive-sequence charges. Instead,
unbalanced line-to-neutral voltages occur, and the phase-to-neutral
capacitances are unequal. Balance can be restored by transposing the line

42
Capacitance: Bundled
Conductors

43
Charging current and
Reactive power
• The current supplied to the transmission-line capacitance is called charging
current.
• For a single-phase circuit operating at line-to-line voltage , the charging current
is

• A capacitor delivers reactive power and the reactive power delivered by the
line-to-line capacitance is

44
45
46