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MODULE I
Transmission line
Conductors
Commonly used conductor materials:
The most commonly used conductor materials for over head lines are copper, aluminium, steelcored aluminium, galvanised steel and cadmium copper
...

All conductors used for overhead lines are preferably stranded in order to increase the
flexibility
...
wires
...
In the manufacture of stranded
conductors, the consecutive layers of wires are twisted or spiralled in opposite directions so
that layers are bound together
...
Copper
...
It is always used in the hard drawn form as stranded
conductor
...

Copper has high current density i
...
, the current carrying capacity of copper per unit of Xsectional area is quite large
...
Firstly, smaller X-sectional area of
conductor is required and secondly, the area offered by the conductor to wind loads is
reduced
...
There is
hardly any doubt that copper is an ideal material for transmission and distribution of
electric power
...
Now-a-days the trend is to use aluminium in place of copper
...
Aluminium
...
The relative comparison of the two materials is briefed
below:
(i) The conductivity of aluminium is 60% that of copper
...
For the same resistance, the diameter of aluminium

conductor is about 1·26 times the diameter of copper conductor
...
This often requires the use of higher towers with
consequence of greater sag
...
Therefore, an aluminium conductor has almost one-half the weight of equivalent copper
conductor
...

(iii) Aluminium conductor being light, is liable to greater swings and hence larger cross-arms are
required
...
Considering the combined properties of cost,
conductivity, tensile strength, weight etc
...
Therefore, it is
being widely used as a conductor material
...

3
...
Due to low tensile strength, aluminium conductors produce greater
sag
...
In order to increase the tensile strength, the aluminium conductor is
reinforced with a core of galvanised steel wires
...
C
...
R
...


Fig 1
...
Usually, diameter of both steel and aluminium wires is the same
...
Fig
...
The result of this
composite conductor is that steel core takes greater percentage of mechanical strength
while aluminium strands carry the bulk of current
...
Therefore, steel cored aluminium conductors will produce smaller
sag and hence longer spans can be used
...

TRANSMISSION LINE PARAMETER
An

electric transmission line has four parameters, namely resistance, inductance,

capacitance and shunt conductance
...
Each line element has its own value, and it is not possible to concentrate
or lumped them at discrete points on the line
...
However, the validity of assumption for the analysis on lumped basis may
fail if the line is very long
...
Due to these flux linkages, the conductor possesses inductance
...
e
...


Flux Linkages:
As stated earlier, the inductance of a circuit is defined as the flux linkages per unit
current
...
We shall discuss two important cases of flux linkages
...
Flux linkages due to a single current carrying conductor
...
1
...
This current will set up magnetic field
...
Both these fluxes will contribute to the
inductance of the conductor
...
Refer to Fig
...
2 (ii) where the X-section of the conductor
is shown magnified for clarity
...
2: Internal flux linkage in a cylindrical conductor
If μ (=μ0μr) is the permeability of the conductor, then flux density at the considered point is given
by
Bx   0  r H x



 0 xI
wb/m2
2
2r

(μr=1 for non magnetic material)

Now, flux dφ through a cylindrical shell of radial thickness dx and axial length 1 m is given by

d  B x  1  dx 

 0 xI
dx
2r 2

This flux links with the current Ix only
...
Now let us calculate the flux linkages of the
conductor due to external flux
...
Referring to Fig
...
5, the field intensity at a distance x metres (from
centre) outside the conductor is given by ;

Fig 1
...


d  d 

0I
dx Weber-turns
2x

Total flux linkage of the conductor from surface to infinity


 ext  
r

Over all flux linkage    int   ext 

0I
dx Weber-turns
2x

0 I  0 I

dx
8 r 2x


0 I
2

 1  dx 
    weber-turns/m length
4 r x 

Inductance of Single Phase Two Wire Line
A single phase line consists of two parallel conductors which form a rectangular loop of
one turn
...
The changing flux links the loop and hence the loop possesses inductance
...
But as the X -sectional area of the loop is very
large, even for a small flux density, the total flux linking the loop is quite large and hence the
line has appreciable inductance
...
4: Single phase two wire transmission line
Consider a single phase overhead line consisting of two parallel conductors A and B spaced d
metres apart as shown in Fig
...
7
...
e
...


IA+IB=0
In order to find the inductance of conductor A (or conductor B), we shall have to consider the
flux linkages with it
...
Flux linkages with
conductor A due to its own current

 I
 0 A
2

 1  dx 
  
4 r x 

Flux linkages with conductor A due to current IB


 0 I B  dx 


2 d x 

Total flux linkage with the with conductor A is

A 



 0 I A  1  dx   0 I B  dx
 

2  4 r x  2 d x

 0  1


  ln   ln r  I A  ln   ln d I B 

2  4




0  I A

 I A ln r  I B ln d 

2  4




0  I A

 I A ln d  I A ln r 

2  4



0  I A
d
 I A ln 

2  4
r



Inductance of conductor A,

LA 


0 I A  1
d
 ln 

r
2  4

A

0
2

IA
d
1
 4  ln r  H/m

 14
d
 2  10 ln e  ln 
r

7

 2  10 7 [ln

d
re

 2  10 7 ln

1
4

]

d
H/m
r'

The radius r′ is that of a fictitious conductor assumed to have no internal flux but with the same
inductance as the actual conductor of radius r
...

Loop inductance = 2 LA = 2 × 2 × 10−7 log d/r′ H/m
Note that r′ = 0·7788 r is applicable to only solid round conductor
...
1
...
Let d1, d2 and d 3 be the spacing between the conductors as shown
...
e
...
Consider the flux linkages with conductor
A
...


Fig 1
...
Under such conditions, the flux linkages with
conductor A become:

A 

0
2

A 


 1

 4  ln r  I A  ( I B  I C ) ln d 




0
2

A 
Inductance of conductor A, L A 

A
IA




 1

 4  ln r  I A  I A ln d 




 0 I A  1
d 
  ln 

2  4
r 

0
2

 1
d 
 4  ln r  H/m



putting the value of μ0=4π x 10-7 in the above equation
 d
L A  2  10 7  ln ,
 r


 H/m


(ii)Unsymmetrical spacing
When 3-phase line conductors are not equidistant from each other, the conductor spacing is
said to be unsymmetrical
...
A different inductance in each phase results in unequal voltage drops in
the three phases even if the currents in the conductors are balanced
...
In order that voltage drops are equal in all

conductors, we generally interchange the positions of the conductors at regular intervals along
the line so that each conductor occupies the original position of every other conductor over an
equal distance
...
Fig
...
5 shows
the transposed line
...
The effect of transposition is that each conductor
has the same average inductance
...
5: Transposition of three phase conductor
Above fig
...
5 shows a 3-phase transposed line having unsymmetrical spacing
...
Let us further assume balanced conditions i
...
, IA + IB
+IC = 0
...
The concept of electric potential is extremely
important for the determination of capacitance in a circuit since the latter is defined as the charge
per unit potential
...


Fig 1
...
Consider a long straight cylindrical
conductor A of radius r metres
...
It is desired to find the expression for
VA
...
Therefore, the potential
difference between conductor A and infinity distant neutral plane is given by:
Q
VA  A
2 0




r

dx
x

Capacitance of Single Phase Two Wire Line
Consider a single phase overhead transmission line consisting of two parallel conductors
A and B spaced d metres apart in air
...

Let their respective charge be + Q and − Q coulombs per metre length
...
7: Single phase two wire transmission line
The total p
...
between conductor A and neutral “infinite” plane is
VA 


Q



dx
Q


2 0 r x 2 0
Q
2 0

ln




d

dx
x

d
Volts
r

Similarly, p
...
between conductor B and neutral “infinite” plane is

VB 

Q



dx
Q


2 0 r x 2 0




d

dx
x



Q d
Volts
ln
2 0 r

Both these potentials are w
...
t
...
Since the unlike charges attract each
other, the potential difference between the conductors is

V AB  2V A 

C AB 

2Q
d
ln
2 0 r


Q
 0 F/m
d
V AB
ln
r

Capacitance to neutral: Above equation gives the capacitance between the conductors of a
two wire line
...
Since potential of the mid-point between the conductors is
zero, the potential difference between each conductor and the ground or neutral is half
the potential difference between the conductors
...


C N  C AN  C BN  2C AB 

2 0
d
ln
r

Capacitance of a 3-Phase Overhead Line
In a 3-phase transmission line, the capacitance of each conductor is considered instead of
capacitance from conductor to conductor
...
, symmetrical
spacing and unsymmetrical spacing
...
Fig
...
8 shows the three conductors A, B and C of the 3-phase
overhead transmission line having charges QA, QB and QC per metre length respectively
...
We shall find the capacitance from
line conductor to neutral in this symmetrically spaced line
...
1
...
8 Three phase symmetrically spaced transmission line


VA  
r

QA
2 0 x



dx  
d

QB
2 0 x



dx  
d

Qc
2 0 x

dx

Assuming QA+QB+QC=0

VA 

QA
d
ln
2 0 r

Capacitance of conductor A with respect to neutral

CA 

Q A 2 0

F/m
d
VA
ln
r

Note that this equation is identical to capacitance to neutral for two-wire line
...

(ii) Unsymmetrical spacing
...
1
...
Let us assume balanced conditions i
...
Q A+ QB+ QC = 0
...
9: Unsymmetrically spaced transposed three phase line

VA 

3 d d d
QA
1 2 3
ln
2
r

Capacitance from conductor to neutral is

CA 

QA

VA

2 0
3

ln

d1d 2 d 3
r

Performance of Transmission Line
The transmission lines are categorized as three types1) Short transmission line– the line length is up to 80 km and the operating voltage is < 20 kV
...
Like other electrical system, the transmission network also will have some power
loss and voltage drop during transmitting power from sending end to receiving end
...


Efficiency of transmission line=

Power delivered at receiving end
×100%
Power sent from sending end

Power sent from sending end – line losses = Power delivered at receiving end
...


% regulation=

no load receiving end voltage-full load receiving end voltage
×100%
full load voltage

Every transmission line will have three basic electrical parameters
...
As the transmission line is a set of
conductors being run from one place to another supported by transmission towers, the parameters
are distributed uniformly along the line
...
Frequency of the power is 50 Hz
...
λ = v where f is power frequency, & λ is wave length and v is the speed of light
...


For this reason, the transmission line, with length less than 160 km, the parameters are assumed
to be lumped and not distributed
...

This electrically short transmission lines are again categorized as short transmission line (length
up to 80 km) and medium transmission line(length between 80 and 160 km)
...
Lines with length more than 160
km, the parameters are considered to be distributed over the line
...

TWO PORT NETWORK
A major section of power system engineering deals in the transmission of electrical power from
one particular place (e
...
generating station) to another like substations or distribution units with
maximum efficiency
...
Thus the entire transmission system can be simplified
to a two port network for the sake of easier calculations
...
As the name suggests, a 2 port
network consists of an input port PQ and an output port RS
...
Thus it is essentially a 2 port or a 4 terminal circuit, having

Fig 1
...

Receiving end voltage=VR
Receiving end current=IR
Given to the output port R S
...

Thus the relation between the sending and receiving end specifications are given using ABCD
parameters by the equations below
...

ABCD Parameters (When Receiving End is Open Circuited)
The receiving end is open circuited meaning receiving end current IR = 0
...
Since dimension
wise A is a ratio of voltage to voltage, A is a dimension less parameter
...
e
...
Since dimension wise C is a ratio of current to voltage, its unit is mho
...

ABCD Parameters (When Receiving End is Short Circuited)
Receiving end is short circuited meaning receiving end voltage VR = 0
Applying this condition to equation (1) we get,


= 0+
VS = 0 + BI R
=

(

= 0)

Thus its implies that on applying short circuit condition to ABCD parameters, we get parameter
B as the ratio of sending end voltage to the short circuit receiving end current
...
Thus B is the short circuit resistance and is
given by
B = VS ⁄ IR Ω
...
e
...
Since dimension
wise D is a ratio of current to current, it’s a dimension less parameter
...


For short length, the shunt capacitance of this type of line is neglected and other parameters
like electrical resistance and inductor of these short lines are lumped, hence the equivalent circuit
is represented as given below, Let’s draw the vector diagram for this equivalent circuit, taking
receiving end current Ir as reference
...


Fig
...
11 Representation of a short transmission line

As the shunt capacitance of the line is neglected, hence sending end current and receiving
end current is same, i
...

Is = IR
...
R
...
X
...
R
...
X
...

As per definition of voltage regulation of power transmission line,

%regulation 



Vs  VR
100%
VR

I R R cos  R  I R X sin  R
X 100%
VR

Any electrical network generally has two input terminals and two output terminals
...
This network is called two – port network
...
Mathematically a two port network can be solved by 2
by 2 matrix
...

Hence two port network of transmission line can be represented as 2 by 2 matrixes
...
Voltage and currents of the network can represented as,
Vs  AVR  BI R
I S  CVR  DI R

Where A, B, C and D are different constant of the network
...
It is dimension less
...
This parameter is referred as transfer impedance
...
It has
the dimension of admittance
...
It is
dimensionless
...
As we know that the constant A, B, C and D are related for
passive network as,
AD − BC = 1
...
1 − Z
...


VR' 

VS
A

and as per definition of voltage regulation of power transmission line,

Efficiency of Short Transmission Line
The efficiency of short line as simple as efficiency equation of any other electrical equipment,
that means
%efficienc y 

Power received at receiving end
X100
Power received at receiving end  3I 2R R

R is per phase electrical resistance of the transmission line
...
Due to the line length being considerably
high, admittance Y of the network does play a role in calculating the effective circuit parameters,
unlike in the case of short transmission lines
...

These lumped parameters of a medium length transmission line can be represented using two
different models, namely1) Nominal Π representation
...

Let’s now go into the detailed discussion of these above mentioned models
...
As we can see from the diagram of the Π
network below, the total lumped shunt admittance is divided into 2 equal halves, and each half
with value Y ⁄ 2 is placed at both the sending and the receiving end while the entire circuit
impedance is between the two
...
It
is mainly used for determining the general circuit parameters and performing load flow analysis
...
1
...

IR is the current flowing through the receiving end of the circuit
...
And
I2 is the current through the impedance Z
...


(2)

I2  I3  IR
Now substituting equation (2) to equation (1)
I S  I1  I 3  I R


Y
Y
VS  VR  I R
2
2

(3)

Now by applying KVL to the circuit,
VS  VR  ZI 2
 V R  Z (V R

 (Z

Y
 IR )
2

Y
 1)VR  ZI R
2

(4)

Now substituting equation (4) to equation (3), we get
Is 

Y Y
Y
[( Z  1)V R  ZI R ]  VR  I R
2 2
2

Y
Y
 Y ( Z  1)V R  ( Z  1) I R
4
2

Comparing equation (4) and (5) with the standard ABCD parameter equations we derive the
parameters of a medium transmission line as:

(5)

Y

A   Z  1
2

BZ
Y
C  Y ( Z  1)
4
Y
D  ( Z  1)
2
Nominal T Representation of a Medium Transmission Line
In the nominal T model of a medium transmission line the lumped shunt admittance is placed in
the middle, while the net series impedance is divided into two equal halves and placed on either
side of the shunt admittance
...


Fig
...
13: Nominal T representation of medium transmission line
Here also Vs and VR is the supply and receiving end voltages respectively, and
Is is the current flowing through the supply end
...

Let M be a node at the midpoint of the circuit, and the drop at M, be given by VM
...
Calculations related to circuit parameters (ABCD parameters) of such
a power transmission is not that simple, as was the case for a short transmission line or medium
transmission line
...


Fig
...
14: Long line model
a) Ignoring the shunt admittance of the network, like in a small transmission line model
...

Rather, for all practical reasons we should consider the circuit impedance and admittance to be
distributed over the entire circuit length as shown in the figure below
...
For accurate modeling to determine circuit parameters let us consider the circuit of
the long transmission line as shown in the diagram below
...
1
...
Lets us now consider an element of infinitely small length Δx at a distance x from
the receiving end as shown in the figure 1
...

V = value of voltage just before entering the element Δx
...

V+ΔV = voltage leaving the element Δx
...

ΔV = voltage drop across element Δx
...

Therefore, the voltage drop across the infinitely small element Δx is given by

V  Izx
Now to determine the current ΔI, we apply KCL to node A
...

Therefore, we can write dI ⁄ dx = V y

(2)

Now derivating both sides of eqn (1) with respect to x,
d2 V ⁄ d x2 = z dI ⁄ dx
Now substituting dI ⁄ dx = V y from equation (2)
d2 V ⁄ d x2 = zyV
or d2 V ⁄ d x 2 − zyV = 0

(3)

The solution of the above second order differential equation is given by
...
r
...

dV/dx = √(yz) A1 e x√yz − √(yz)A2e−x√yz

(5)

Now comparing equation (1) with equation (5)

I

dV

dX

y
A1e  X
z

YZ

(6)

Now to go further let us define the characteristic impedance Zc and propagation constant δ of a
long transmission line as

Zc = √(z/y) Ω
δ = √(yz)

Then the voltage and current equation can be expressed in terms of characteristic impedance and
propagation constant as
V = A1 eδx + A2 e−δx
δx

I = A1/ Zc e + A2 / Zc e

(7)
−δx

(8)

Now at x=0, V= VR and I= IR
...

VR = A1 + A2

(9)

IR = A1/ Z c + A2 / Z c

(10)

Solving equation (9) and (10),
We get values of A1 and A2 as,
A1 = (VR + ZCIR) ⁄ 2
And A1 = (VR − ZC IR ) ⁄ 2
Now applying another extreme condition at x=l, we have V = VS and I = IS
...
Such a phenomena does not have much role to play in case of a very short
line, but with increase in the effective length of the conductors, skin effect increases
considerably
...
The
distribution of current over the entire cross section of the conductor is quite uniform in case of a
DC system
...
e
...
In fact there even arises a condition when absolutely
no current flows through the core, and concentrating the entire amount on the surface region,
thus resulting in an increase in the effective electrical resistance of the conductor
...

Why Skin Effect Occurs in Transmission Lines?
Having understood the phenomena of skin effect let us now see why this arises in case of an AC
system
...

Let us initially consider the solid conductor to be split up into a number of annular filaments
spaced infinitely small distance apart, such that each filament carries an infinitely small fraction
of the total current
...

Now during the flow of an alternating current, the current carrying filaments lying on the core
has a flux linkage with the entire conductor cross section including the filaments of the surface as
well as those in the core
...
Thus the flux linkage of the
conductor increases as we move closer towards the core and at the same rate increases
the inductor as it has a direct proportionality relationship with flux linkage
...
The high value of reactance in the inner section results in the current being distributed
in an un-uniform manner and forcing the bulk of the current to flow through the outer surface or
skin giving rise to the phenomena called skin effect in transmission lines
...
1
...

2) Type of material
...

4) Operational frequency
...

When the alternating current is flowing through a conductor, alternating magnetic flux is
generated surrounding the conductor
...
This
circulating current increases the resistance of the conductor and push away the flowing current
through the conductor, which causes the crowding effect
...
The flux due to central conductor links with right side conductor
...
Therefore, the inductance of the elements farther apart is more as
compared to the elements near to each other and hence the current density is less in the
elements farther apart than the current density in the element near to each other
...
This phenomenon is actually referred as proximity effect
...


Series and shunt compensation:

The demand of active power is expressing Kilo watt (kw) or mega watt (mw)
...
All the arrangements in electrical pomes system
are done to meet up this basic requirement
...
This reactive power is expressed in Kilo VAR or Mega VAR
...

These inductive loads are generally electromagnetic circuit of electric motors, electrical
transformers, inductance of transmission and distribution networks, induction furnaces,
fluorescent lightings etc
...
e
...
This ratio is alternatively known as electrical
power factor, and fewer ratios indicates poor power factor of the system
...
And hence reactive power compensation becomes so important
...


Let’s explain in details,
we know that active power is expressed =VIcosθ
where,cosθ is the power factor of the system
...


As the current of the system increases, the ohmic loss of the system increases
...
The crosssection of the conducting parts of the system may also have to be increased for carrying extra
ampere burden, which is also not economical in the commercial point of view
...


The equipments used to compensate reactive power
...

(1) Synchronous condensers
(2) Static capacitors or Capacitor Bank

Synchronous condensers, can produce reactive power and the production of reactive power can
be regulated
...
That is why synchronous condensers, are justified to use only for voltage regulation
of very high voltage transmission system
...
This division enables

the capacitor to run in 1, 2, 1+2=3, 2+2=4, 1+2+2=5 steps
...
These divisions make the static capacitor bank
more expensive but still the cost is much lower them synchronous condensers
...
This is practically and economically possible only by using
small rated capacitors with individual load not by using synchronous condensers
...


Static capacitor can further be subdivided in to two categories,
(a) Shunt capacitors
(b) Series capacitor

Fig
...
17: Series and Shunt Capacitor bank

These categories are mainly based on the methods of connecting capacitor bank with the system
...
There are some specific advantages of using shunt capacitors such as,
a) It reduces line current of the system
...

c) It also reduces system Losses
...

e) It reduces load of the alternator
...


All the above mentioned benefits come from the fact, that the effect of capacitor reduces reactive
current flowing through the whole system
...

series capacitor on the other hand has no control over flow of current
...
Actually, the

capacitive reactance of series capacitor neutralizes the inductive reactance of the line hence,
reduces, effective reactance of the line
...

But series capacitor bank has a major disadvantage
...
Thus series capacitor must
have sophisticated and elaborate protective equipments
...




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