Pharos University
EE-272
Electrical Power Engineering 1
“Electrical Engineering Dep”
Prepared By:
Dr. Sahar Abd El Moneim Moussa
Dr. Sahar Abd El Moneim Moussa
1
OVERHEAD TRANSMISSION LINE
OHTL
Dr. Sahar Abd El Moneim Moussa
2
2-2 Inductance of single phase Two Wire line
Ia
ra
Ib
r rb
Dab
The flux linkage in conductor “a” due to current Ia is given by
𝐷𝑎𝑝
−7
∴ 𝑎𝑎 = 2 ∗ 10 𝐼𝑎 (ln
)
𝑊𝑏/𝑚
𝑟′𝑎
where: Dap: is the distance between the conductor “a” and a
point “P” far away from conductor “a”
-
Dr. Sahar Abd El Moneim Moussa
3
-
The flux linkage in conductor “a” due to current Ib is given by
∴ 𝑎𝑏 = 2 ∗ 10
−7
𝐼𝑏 (ln
𝐷𝑏𝑝
)
𝑊𝑏/𝑚
𝐷𝑎𝑏
where: Dab: is the distance between the two conductor.
-
The total flux linkage with conductor “a” due to Ia and Ib is:
𝑎 = 𝑎𝑎 + 𝑎𝑏
𝐷𝑎𝑝
𝐷𝑏𝑝
−7
𝑎 = 2 ∗ 10
𝐼𝑎 𝑙𝑛 ′ + 𝐼𝑏 𝑙𝑛
𝑟𝑎
𝐷𝑎𝑏
Since: Ib=-Ia
𝑎 = 2 ∗
10−7
𝐼𝑎
1
1
𝑙𝑛𝐷𝑎𝑝 + 𝑙𝑛
− 𝑙𝑛𝐷𝑏𝑝 − 𝑙𝑛
𝑟′𝑎
𝐷𝑎𝑏
Dr. Sahar Abd El Moneim Moussa
4
Since: P is a very far point, Dap=Dbp=Dp
1
1
∴ 𝑎 = 2 ∗ 10 𝐼𝑎 𝑙𝑛
− 𝑙𝑛
𝑟′𝑎
𝐷𝑎𝑏
𝐷𝑎𝑏
−7
𝑎 = 2 ∗ 10 𝐼𝑎 ln
𝑟′𝑎
−7
Accordingly:
𝐿𝑎 = 2 ∗
10−7
𝐷𝑎𝑏
𝑙𝑛
𝑟′𝑎
H/m
𝐷𝑎𝑏
𝑟′𝑏
H/m
𝐿𝑏 = 2 ∗ 10−7 𝑙𝑛
Dr. Sahar Abd El Moneim Moussa
5
-
The total inductance of a single phase T.L. is
L =La + Lb
∴𝐿=2 ∗
10−7 (
𝐷𝑎𝑏
𝑙𝑛
𝑟′𝑎
+
𝐷𝑎𝑏
𝑙𝑛
𝑟′𝑏
)
H/m
Since: r’a=r’b= r’
𝐿 = 2 ∗ 10−7 ln
𝐷𝑎𝑏
𝑟′
H/m
Dr. Sahar Abd El Moneim Moussa
6
2-3 Inductance of 3-Phase, 3-Wire with equal spacing
c
La= Lb= Lc
D
D
DD
D
a
b
- The total flux linkage with conductor “a” is due to Ia, Ib, and Ic .
𝑎 = 2 ∗
1
10−7 [𝐼𝑎 𝑙𝑛 ′
𝑟
1
1
1
+ 𝐼𝑏 𝑙𝑛 + 𝐼𝑐 𝑙𝑛 + 𝐼𝑎 + 𝐼𝑏 + 𝐼𝑐 𝑙𝑛 ]
𝐷
𝐷
𝐷𝑃
𝑎
Dr. Sahar Abd El Moneim Moussa
7
Since: Ia + Ib+ Ic=0
𝑎 = 2 ∗
1
10−7 [𝐼𝑎 𝑙𝑛 ′
𝑟
1
+ (𝐼𝑏 + 𝐼𝑐 )𝑙𝑛 ]
𝐷
𝑎
Since: Ib+ Ic= -Ia
𝑎 = 2 ∗
1
10−7 𝐼𝑎 [𝑙𝑛 ′
𝑟
𝑎 = 2 ∗ 10−7 𝐼𝑎 𝑙𝑛
𝐿𝑎(𝑝𝑒𝑟 𝑝ℎ𝑎𝑠𝑒) = 2 ∗ 10
1
− 𝑙𝑛 ]
𝐷
𝑎
𝐷
𝑟′𝑎
−7
wb/m
𝐷
𝑙𝑛 ′
𝑟 𝑎
H/m
Dr. Sahar Abd El Moneim Moussa
8
Example 1 :
Find the inductance of a 3-phase overhead transmission line if the
conductor diameter is 2 cm and the three conductors are placed at
the corners of an equilateral triangle of side 4 m
C
Solution: (Equal Spacing)
D
D
DD
a
D
D= 4m
,d=2 cm
,r= 1 cm=10-2 m
L per phase =2 X 10-7 ln [D / r’]
H/m
L per phase =2 X 10-7 ln [4 / (0.78 x10-2 )] = 1.248 x 10-6
b
H/m
Dr. Sahar Abd El Moneim Moussa
9
2-4 Inductance of 3-Phase, 3-Wire with UNequal
spacing
C
Dac
Dbc
DD
a
La
Lb
b
Dab
Lc
Dr. Sahar Abd El Moneim Moussa
10
-
The total flux linkage with conductor “a” due to Ia, Ib, and Ic is
𝑎 = 2 ∗ 10
−7
1
1
1
1
𝐼𝑎 𝑙𝑛 ′ + 𝐼𝑏 𝑙𝑛
+ 𝐼𝑐 𝑙𝑛
+ 𝐼𝑎 + 𝐼𝑏 + 𝐼𝑐 𝑙𝑛
𝑟𝑎
𝐷𝑎𝑏
𝐷𝑐𝑎
𝐷𝑃
0
𝑎 = 2 ∗
-
10−7
1
1
1
𝐼𝑎 𝑙𝑛 ′ + 𝐼𝑏 𝑙𝑛
+ 𝐼𝑐 𝑙𝑛
𝑟𝑎
𝐷𝑎𝑏
𝐷𝑐𝑎
The total flux linkage with conductor “b” due to Ia, Ib, and Ic is
𝑏 = 2 ∗
10−7
1
1
1
𝐼𝑏 𝑙𝑛 ′ + 𝐼𝑎 𝑙𝑛
+ 𝐼𝑐 𝑙𝑛
𝑟𝑏
𝐷𝑎𝑏
𝐷𝑏𝑐
Dr. Sahar Abd El Moneim Moussa
11
-
The total flux linkage with conductor “c” due to Ia, Ib, and Ic is
𝑐 = 2 ∗ 10
−7
1
1
1
𝐼𝑐 𝑙𝑛 ′ + 𝐼𝑎 𝑙𝑛
+ 𝐼𝑏 𝑙𝑛
𝑟𝑐
𝐷𝑐𝑎
𝐷𝑏𝑐
As we can see: a b  c
Accordingly: La Lb  Lc
Dr. Sahar Abd El Moneim Moussa
12

In case of unequal spacing, the inductances of the three phase
conductors are all different.

This will cause unequal voltage drops across line conductors,

Therefore the voltage phasors at receiving-end substation of
the line will be unbalanced.

To overcome such unbalance , line transposition is made.
Dr. Sahar Abd El Moneim Moussa
13
2-5 Inductance of 3-Phase, 3-Wire with UNequal
spacing Transposed lines
-
B
C
A
C
A
B
A
B
C
The total flux linkage with conductor “a” in the first section
due to Ia, Ib, and Ic is
𝑎𝑰 = 2 ∗
10−7
1
𝐼𝑎 𝑙𝑛 ′
𝑟 𝑎
1
+ 𝐼𝑏 𝑙𝑛
𝐷12
+
1
𝐼𝑐 𝑙𝑛
𝐷13
wb/m
Dr. Sahar Abd El Moneim Moussa
14
-
The total flux linkage with conductor “a” in the second
section due to Ia, Ib, and Ic is
𝑎𝑰𝑰 = 2 ∗
-
10−7
1
𝐼𝑎 𝑙𝑛 ′
𝑟 𝑎
+
1
𝐼𝑏 𝑙𝑛
𝐷23
1
+ 𝐼𝑐 𝑙𝑛
𝐷12
wb/m
The total flux linkage with conductor “a” in the third section
due to Ia, Ib, and Ic is
𝑎𝑰𝑰𝑰 = 2 ∗
10−7
1
𝐼𝑎 𝑙𝑛 ′
𝑟 𝑎
1
+ 𝐼𝑏 𝑙𝑛
𝐷13
+
1
𝐼𝑐 𝑙𝑛
𝐷23
wb/m
Dr. Sahar Abd El Moneim Moussa
15
-
The average value of flux with conductor “a” will be:
𝑎𝑰 + 𝑎𝑰𝑰 + 𝑎𝑰𝑰𝑰
𝑎 =
3
2 ∗ 10−7
1
𝑎 =
3𝐼𝑎 𝑙𝑛 ′ + 𝐼𝑏 + 𝐼𝑐
3
𝑟𝑎
1
1
1
𝑙𝑛
+ 𝑙𝑛
+ 𝑙𝑛
𝐷12
𝐷13
𝐷23
2 ∗ 10−7
1
𝑎 =
[3𝐼𝑎 𝑙𝑛 ′ − 𝐼𝑎
3
𝑟𝑎
2 ∗ 10−7
1
𝑎 =
[3𝐼𝑎 𝑙𝑛 ′ − 3𝐼𝑎
3
𝑟𝑎
𝑎 = 2 ∗ 10
−7
1
𝑙𝑛
]
𝐷12 𝐷13 𝐷23
𝑙𝑛 3
1
𝐷12 𝐷13 𝐷23
]
1
1
𝐼𝑎 [𝑙𝑛 ′ − 𝑙𝑛 3
]
𝑟𝑎
𝐷12 𝐷13 𝐷23
Dr. Sahar Abd El Moneim Moussa
16
𝑎 = 2 ∗
10−7 𝐼𝑎
𝐷𝑒𝑞
𝑙𝑛 ′
𝑟𝑎
Where:
Deq is called the geometric mean distance = GMD =
3
D12 D13 D23
Accordingly;
𝐿𝑎(𝑎𝑣𝑒𝑟𝑎𝑔𝑒) = 2 ∗ 10−7 𝑙𝑛
𝐷𝑒𝑞
𝑟′𝑎
H/m
Dr. Sahar Abd El Moneim Moussa
17
Example 2:
The three conductors of a 3 phase transmission line are arranged in a
horizontal plane and are 4 m apart. The diameter of each conductor is
2 cm. calculate the inductance per phase per m of the line
( This example will be solved in the lecture)
Dr. Sahar Abd El Moneim Moussa
18
2-6 Inductance of Bundled conductors

Bundling of conductors means the use of more than one
conductor per phase. It is a common practice for EHV lines.

Bundling reduces the electric field strength at the conductor
surfaces , which in turn reduces corona and its effects.
Dr. Sahar Abd El Moneim Moussa
19

CORONA EFFECT:
•
Corona is an electrical discharge caused by ionization of the
air surrounding the conductor, which occurs when the
strength of the electric field at the conductor surface exceeds
the dielectric strength of the air.
•
Air at 79 cm pressure and 25 C breaks at 30 kV/cm. If the
electric field strength exceeds this value Corona occurs.
•
Corona is observed by violet-colored light around the
conductor, a hissing noise, and Ozone-smell due to ionization
of air.
Dr. Sahar Abd El Moneim Moussa
20
L per phase=2 X 10-7 ln [Deq / DSL] H/m
Dr. Sahar Abd El Moneim Moussa
21
Where,
-
For two conductors:
a
b
d
C
D
DSL =
𝑟′ 𝑑
For two conductor bundle
Where,
d: is the bundling spacing
Dr. Sahar Abd El Moneim Moussa
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-
For three conductors:
a
b
D
DSL =
3
𝑟′ 𝑑2
C
d
For Three-conductor Bundling
Dr. Sahar Abd El Moneim Moussa
23
-
For four conductors:
a
b
D
DSL = 1.091
4
𝑟′ 𝑑3
C
d
For Four -conductor Bundling
Dr. Sahar Abd El Moneim Moussa
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Example 3:
The 3 phases of a TL are placed at the corners of an equilateral
triangle of side 12.5 m. Each phase consists of 2 bundled
conductors, the conductor diameter is 4.6 cm and the bundle
spacing is 45 cm. Find L.
Solution:
Dab = 12.5 m
D bc = 12.5 m
D ca = 12.5 m
D conductor=4.6 cm
d= 45 cm
C
d
a
b
D
Dr. Sahar Abd El Moneim Moussa
25
Deq = DabDbcDca = 12.5 m
r’ =[ (4.6/2) x 10-2 ] x 0.7788= 1.794 cm
DSL = √r’ d = 8.985 cm
L per phase=2 X 10-7 ln [Deq / DSL] H/m
= 2 X 10-7 ln
1250
8.985
= 9.87 x 10 -7 H/m
Dr. Sahar Abd El Moneim Moussa
26