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Chapter Eight
Transmission Line parameters
Resistance, Inductance , Capacitance
Resistance :The
resistance of a solid round conductor at specified temperature is
given by :(36)
Where
is the conductor resistivity
Is the conductor length
Is the cross-section area
When ac ( alter native current ) flow in an conductor the current
distribution is not uniformly over the conductor cross-sectional area and
the current density is greatest at the surface of the conductor.
This causes the ac resistance to be somewhat high. Than the d.c
resistance. This behavior is known as skin effect. At 60 HZ the a.c
resistance is about 2 percent higher than dc resistance.
*The conductor resistance increase as temper due increase.
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Where
are conductor resistance at
respectively. T is
the temperature constant that depend on the conductor material . for
example T for the aluminum is T=228.
Inductance (L)
Any conductor carrying a current will be surrounded by a magnetic field
change with the change of the current.
Where
is the magnetic linkage flux and this will be change as the
changing of the current (
and this lead to make the permeability
constant for all flux level as in the air.
Henery (H)
(1)
Where
is the flux linkage of the conductor ( web.turn).
I is the current flow in the conductor
*In any conductor carrying a current there is :1- Internal flux
This will be as a concentric circles inside the conductor and caused by
the current that flow in the same conductor.
2- External flux
This will be as a concentric circles outside or around the conductor and
caused by the same current .
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Inductance of conductor to external flux :After the derivative of external linkage flux between
The inductance between the
Two point that location outside
Or external the conductor.
r
X
H/m
Flux outside the conductor
Due to the current I that
Flow in the same conductor
*Inductance between two single- phase conductor :-
=
D
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H/m where
(3)
And in the same manner
H/m where
(4)
And the inductance of the Total system is
And if
Also we can write eq. (3) and (4) by this form
(5)
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*Flux linkage in term of self and Mutual inductance
As in eq. (1)
(6)
Where
is the Total linkage flux in the conductor 1
is the Total linkage flux in the conductor 2
is the self inductance due to
is the Mutual inductance due to
is the mutual inductance due to
is the self inductance due to
Since
(7)
By comparing eq.(7) with eq. (5)
} Self inductance
} Mutual inductance
(8)
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The concept of self and Mutual inductance can be extended to a group of
n conductors. Consider n conductors carrying phase currents
Such that
(9)
Generalizing eq. (6), the linkage of conductor are
(10)
And from eq. (8)
(11)
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*Inductance of composite conductor :The current in each
Su b conductor of
Group A is
b
c
a
And the current in each
Sub conductor of
Group B is
Due to eq. (11) and
Its application
d
m
n
Conductor
A
Has acurrent +I
(12)
Conductor
A
Has acurrent -I
+1n
+1n
The inductance of sub conductor a is
(12)
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As so on
The average conductor in any sub conductor in group A
Since all the sub conductor of conductor A are electrically parallel, so that
the inductance o A will be
(13)
(13)
(14)
(14)
…(
…(
(15)
(15)
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Where
GMD is the Geometric mean Distance
And it is equal for A and B
GMR is the Geometric Mean radius
And it is not equal for A and B in any time
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Ex1) A single phase has two group of conductor A and B where A
consists of 3- wires (sub conductors) a ,b, c each of its have a radius of
0.25 cm ,and the group B consists of two wires d ,e each of its have a
radius of 0.5cm. calculate the inductance of the Total system where the
distance between the sub conductor is as follow :.
So1)
d
6
m
b
6
m
e
c
Group
A
Where
=0.4 m
Group
B
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*Inductance of
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Transmission Line :-
a- Symmetrical spacing :consider one meter length of
line with three conductors, each with
radius
r , symmetrically spacing in a triangle (as example )
configuration as shown in the Fig. below
Assume we have a balanced
D
D
Current
(16)
D
From eq. (11). The Total flux linkage of phase a is
(17)
(17)
Since
Since we have a symmetrical spacing
H/m
(18)
(18)
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b) A symmetrical spacing :with a symmetrical spacing even with balance currents, the voltage drop
due to line inductance with be unbalanced, because the inductance of each
phase will be not equal. Consider one meter length of of
line with
three conductor each of radius r as shown in the fig. below .
The application of eq. (11)
1
2
3
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since we have a balance system
If we have the parameter a
Which is
120
120
120
The equations of inductance a symmetrical spacing as in the
Fig above are :-
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(18)
*Transpose System :One way to solve the problem of a symmetrical spacing and the different
inductance for each phase is :Transpose system which it is :Interchanging the phase configuration every one – third the length so that
each conductor is moved to occupy the next physical in regular sequence.
And by this method the inductance for each phase will be equal.
Such a transposition arrangement is shown in the Fig. below :a
a
c
b
b
a
c
3
c
-1First
Transpose
b
-2Second
Transpose
a
-3third
Transpose
2
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Now let take this example
1
1
2
3
2
3
Since in a transposed line each phase take all three position. The
inductance per phase can be obtained by finding the average value of
eq. (18)
H/m
(19)
(19)
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1
1
2
2
3
3
1
2
3
represent first transpose
represent second transpose
represent third transpose
This is another way to solve the problem of transpose system. From eq.
(18)
( First transpose )
( second transpose )
(third transpose )
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(20)
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(20)
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*Bundle System :Bundle system or bundle conductor consists of a two, three, four sub
conductors symmetrically arranged in configuration as in the fig. below :d
d
d
d
d
d
-1-
-2-
d
-3-
The geometric Mean Radius of each fig. is as below
Two sub
(conductor) =
And so on
(three-sub conductor)
The advantage of the bundle system is
1- Reduce the inductance of the line, which is reduce the line losses and
improves the line capacity and system stability over the use of single
conductor.
2- Reduce the corona losses .
3-Reuce the Radio in deference.
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Ex2) A
double circuit of conductors as shown in the Fig. below.
Calculate the inductance per phase per km if the system is transpose at
equal distance. The diameter of the conductor is 5 cm.
So1)
a
In order to solve the problem
1.5m
We must first know the
b
There are two way to solve the
1.5m
Problem.
c
From the fig. above we see that
3m
So that in order to get a per phase inductance (Balance system) and we
have a transpose system
We must take an equivalent GMR
Where
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In all cases
Where
(first transpose) (second transpose) (third transpose)
OR in other meaning
(general from)
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Ex3)
line, each phase has two parallel conductor and the radius of
the conductor is 0.205 cm. If the distance between the different conductor
is as in the fig. below. Calculate the reactance in ( ohm ) per one km. if the
frequency is 50 HZ and the phases are transpose the location between its
at equal distance.
6.05
0.4 m
6.05
0.4 m
0.4 m
So1)
Transpose states
-1-2-3Not
1) The Distance of sub conductor for each phase is equally
2) The Distance of sub conductor for each phases for the three stages of
complete transpose are equally
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Also
3)
for all cases of tanspose stages ( general form)
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Ex4) what is the bandle system, what is the advantage of it, show that
from the two fig. below by calculating the inductance per phase per km of
the line in each fig. below. Note that the diameter of the conductor is 5 cm
So1)
Fig
-1-
From the Fig. (1)
Fig
-2-
From the fig. (2)
a
0.6m
b
4m
4m
4m
0.6m
c
4m
0.6m
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The bundle system cause a reduction in the inductance
By
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Capacitance :-
Where
C is the capacitance
r
is the charge
V is the voltage
is the radius of the
Conductor.
For one meter length of the conductor.
The electric flux density (D) at a radius X is given
Where
is the charge ( coulomb / one meter ) and
A is the area of the surface of an axial length of
One meter of the conductor.
Along uniformly charged conductor has lines of electric flux extending
radials out from the conductor surface
The electric field intensity E
may be found from the relation
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Where
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E is the electric field intensity V/m
Is the relative permittivity of the vacuum
is the relative permittivity of free space = 8.85
is the relative permittivity of the air = 1
V/M
(21)
The potential difference between cylinder from position
is
define as the work done in moving a unit charge of coulomb from
Through the electric field produced by the charge on the conductor
(22)
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f/m
(23)
* Capacitance of single – phase lines :Assuming conductor q alone to
Have a charge of
The voltage between conductor
1 and 2 is
r
r
D
Now assume only conductor 2 having a charge of
between conductor 2 and 1 is
. The voltage
Since
From the principle of superposition, the potential difference due to
presence of both charge is
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For single – phase line
v/m
(23)
f/m
(24)
Equation (24) give the line – to line capacitance between
the conductor
(between conductor )
( between each
n
Conductor and the neutral )
f/m
(25)
MF/KM
(26)
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*Capacitance of
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lines :-
After take the complete transpose
Of this fig. and make a dative for
The capacitance
F/M
(27)
Mf/km
(28)
* if we have bundle system. The same process as in the inductance
MF/KM
(29)
Note : In the inductance the radius is .
But the capacitance contain radius r .
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*The derive of
Consider one meter length of
line with 3 conductors each of its has
a radius of r as shown in the fig. below
Since we have a balanced
system
(30)
For complete transpose as shown in the fig. above
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Also , similarly we find the average voltage
b
Where
For balance 3
voltage
130
a
130
c
(31)
Mf/km
(32)
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The effect of earth on the capacitance of
transmossion
earth
F/M
(33)
is the capacitance of each phase ( conductor )
To the neutral .( earth)
Note :-
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(34)
*The effect of earth on the capacitance of single-phase
(35)
Also
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Ex5) transmission line of 10km , as shown in the fig. below. The radius
of each conductor is 0.885 cm and the phase are transpose at equal
distance also the resistivity of the line is (
Find the resistance, inductance, capacitance and the charging current of
the line .
5.4
Note that :
The voltage
The line is 33kv
And the frequency
Is 50 Hz
So1)
3m
3m
5.4 m
6.3m
Note that
So that we must take
But in the first let calculate
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In order to get GMR, as we say before there are two way to the solution :1)