11
12
13
14
15

6
7
8
9
10

1
2
3
4
5

NEED FOR TRANSMISSION LINE

• Generating stations, transmission lines and distribution system are the main components of an electric power system.

• Generating stations and a distribution systems are connected through transmission lines, which also connect from one grid to another.
• Electrical power generated at a voltage of 11 to 25 kv, which is stepped up to the transmission levels in the range 66 to 400 KV or higher.

• The need of distribution systems and transmission lines is to transmit the electric Power from the generating station to the Consumer (Industry or Domestic).
Pic1
TRANSMITTING THE ELECTRICAL POWER

Types of systems:
• D.C (Direct current) system
• A.C (Alternating current) system
TYPES OF D.C SYSTEMS
• D.C Two wire system

• D.C Two wire with mid point earthed system.

• D.C Three wire system
TYPES OF A.C SYSTEMS
1. Single phase A.C system
• Single phase two wire.
• Single phase two wire with mid point earthed.
• Single phase three wire
2. Two phase A.C system
• Two phase three wire
• Two phase four wire

3. Three phase A.C system
• Three phase three wire
• Three phase four wire
D.C SYSTEMS
D.C Two wire system
Pic2
D.C Two wire with mid point earthed system
Pic3
D.C Three wire system with mid point earthed
Pic4
A.C SYSTEMS
Single phase two wire:
Pic5
Single phase two wire with mid point earthed:
Pic6
Single phase three wire with mid point earthed:
Pic7
Two Phase Three wire:
Pic8
Two phase four wire:
Pic9
Three phase three wire:
Pic10
Three phase four wire
Pic12
Advantages of D.C Transmission:
• Requires only two conductors.

• No inductance & capacitance effect.

• No phase displacement & surge problem.
• No skin effect & less corona loss.

• Reduced interference with communicating circuits.

• Potential stress on the insulation is less.
• Loss due to charging current eliminated.

• Used in under ground cables also.

• Voltage regulation is better & No stabilizer required
Disadvantages of D.C Transmission:
• Power can’t be generated at high voltages
• Voltage can’t be stepped up
• Cost of converting and inverting stations are high
• Requires costly switch gear
Advantages of A.C Transmission:
• Voltage can be stepped up and stepped down.

• Electrical power generated at high voltages.

• Maintenance of substation is easier &cheaper

• Cost of switch gear is less
Disadvantages of A.C Transmission:
• More conductor material is required.

• Power loss due to charging current.

• Construction of transmission lines are difficult.

• Synchronizing problems.
• Due to skin effect line resistance will increase

• Due to corona Spacing between conductors should be
kept more.

• In cables alternating current causes sheath loss
The following assumptions are made for
calculating conducting material
• Equal amounts of power transmitted.

• Equal lengths of line

• Equal line loss.

• The max. voltage between two conductors
is same (Vm).
CONDUCTOR MATERIAL REQUIRED FOR OVER HEAD LINE FOR VARIOUS SYSTEMS
D.C system:
1.D.C two wire system with one conductor earthed
Pic12
Vm is the max.potential difference between
conductor and earth in volts.

I is current in amperes, R is the resistance of the
conductor in ohms.

Line voltage = Vm
Power transmitted = P = Vm I watts.
Line loss or power loss = 2 I2 R watts.
2. D.C Two wire system with mid point earthed:
Pic13
Vm is the max.potential difference between
conductor and earth in volts.

I1 is current in amperes, R1 is the resistance of the
conductor in ohms.

Line voltage = 2 Vm

Power transmitted = P = 2 Vm I1 watts.

Line loss or power loss = 2 I12 R1 watts.
Equating for equal power

2 Vm I1= Vm I

I1 = I/2

Equating for equal power loss

2 I12 R1 = 2 I2 R

R / R1 = I12 / I2

= (I / 2)2 / I2

= 1/4 = a1 / a
Amt. of cond. Mat. Req. for D.C 2wire sys. with mid point earthed
Amt. of cond. Mat. Req. for D.C 2wire sys. with one end earthed

=1/4
A.C SYSTEM
Single phase two wire system with one conductor
earthed:
pic14
Vm is the max.potential difference between
conductor and earth in volts.

I2 is current in amperes, R2 is the resistance of the
conductor in ohms and power factor =

Rms voltage =

Power transmitted = P = I2 watts.

Line loss or power loss = 2 I22 R2 watts.
Amt. of cond. Mat.Req.for A.C 2wire sys.with one cond.earthed
Amt. of cond. Mat.Req. for D.C 2wire sys. with one end earthed
. Single phase two wire system with mid point
earthed:
pic15
Vm is the max.potential difference between
conductor and earth in volts.

I3 is current in amperes, R3 is the resistance of the conductor in ohms and power factor =


Rms voltage = 2 =


Power transmitted = P = I3 watts


Line loss or power loss = 2 I32 R3 watts.
Equating for equal power, I3 = Vm I

I3 = I /

Equating for equal power loss, 2 I32 R3 = 2 I2 R

R / R3 = I32 / I2 = 1 / 2 = a3 / a
Amt.of cond. Mat. Req.for A.C 2wire sys.with mid point earthed
Amt.of cond. Mat. Req. for D.C 2wire sys. with one end earthed
3. Three phase three wire system:
Pic16
Vm is the max.potential difference between
conductor and earth in volts.

I5 is current in amperes, R5 is the resistance of the
conductor in ohms and power factor =

Rms voltage =

Power transmitted = P = 3 I5 watts.
Line loss or power loss = 3 I52 R5 watts.
Equating for equal power, 3 I5 = Vm I


I5 = (I / 3)

Equating for equal power loss, 3 I52 R5 = 2 I2 R


R / R5 = 3 I52 / 2 I2 = 1 / 3 = a5 / a
Amt.of cond. Mat.Req.for A.C 3- 3wire system
Amt.of cond Mat. Req.for D.C 2wire sys.with one end earthed
CONDUCTOR MATERIAL REQUIRED FOR UNDERGROUND CABLES FOR VARIOUS SYSTEMS

D.C system:
1.D.C two wire system
Pic17
Vm is the max.potential difference between
conductor and earth in volts.

I is current in amperes, R is the resistance of the conductor in ohms.

Line voltage = Vm
Power transmitted = P = Vm I watts.
Line loss or power loss = 2 I2 R watts.
2.D.C Two wire system with mid point earthed:
Pic18
Vm is the max.potential difference between
conductors.

I1 is current in amperes, R1 is the resistance of the
conductor in ohms.

Line voltage = Vm

Power transmitted = P = Vm I1 watts.

Line loss or power loss = 2 I12 R1 watts.
Equating for equal power

Vm I1= Vm I

I1 = I

Equating for equal power loss

2 I12 R1 = 2 I2 R

R / R1 = I12 / I2

= (I )2 / I2

= 1 = a1 / a
Amt.of cond.Mat.Req.for D.C 2wire sys. with mid point earthed
Amt.of cond.Mat. Req. for D.C 2wire system.
A.C SYSTEM
1. Single phase two wire system
Pic19
Vm is the max.potential difference between
conductors.

I2 is current in amperes, R2 is the resistance of the conductor in ohms and power factor =

Rms voltage =

Power transmitted = P = I2 watts.

Line loss or power loss = 2 I22 R2 watts.
Equating for equal power, I2 = Vm I


I2 = I /

Equating for equal power loss, 2 I22 R2 = 2 I2 R


2 ( I / )2 R2 = 2 I2 R

R / R2 = 2 / = a2 / a

Amt.of cond.Mat.Req.for A.C 2wire system.
Amt.of cond.Mat. Req. for D.C 2wire system.
2. Single phase two wire system with mid point
earthed:
pic20
Vm is the max.potential difference between
conductors.

I3 is current in amperes, R3 is the resistance of the conductor in ohms and power factor =

Rms voltage =

Power transmitted = P = I3 watts.

Line loss or power loss = 2 I32 R3 watts.
Equating for equal power, I3 = Vm I


I3 = I /

Equating for equal power loss, 2 I32 R3 = 2 I2 R


2 ( I / )2 R3 = 2 I2 R

R / R3 = 2 / = a3 /
Amt.of cond.Mat.Req.for 1 phase A.C 2wire system.
Amt.of cond.Mat. Req. for D.C 2wire system.
3. Three phase three wire system:
Pic21
Vm is the max. potential difference between
conductors.

I4 is current in amperes, R4 is the resistance of the conductor in ohms and power factor =


Max voltage between conductor to neutral =
Rms voltage = =


Power transmitted = P = 3 I4 watts.

Line loss or power loss = 3 I42 R4 watts.
Equating for equal power, 3 I4 = Vm I


I4 = I / 3

Equating for equal power loss, 3 I42 R4 = 2 I2 R


3 ( I / 3 )2 R4 = 2 I2 R

R / R4 = 1 / = a4 / a
Amt.of cond.Mat.Req.for A.C 3 phase 3 wire system.
Amt.of cond. Mat. Req. for D.C 2wire system.
COMPARISON OF VARIOUS SYSTEMS OF TRANSMISSION
(The ratio of conductor material required in any system compared with that in the corresponding 2 wire D.C system)
S.no System max.voltage to earth max.voltage bet.cond.
D.C
1. Two wire 1 1
2. Two wire mid point earthed 0.25 1
A.C
1. Single phase two wire
2. Single phase two wire
mid point earthed
3. Three phase three wire
EFFECT OF SUPPLY FREQUENCY AND VOLTAGE ON TRANSMISSION LINES
Supply frequency:
• Frequency must be maintained constant so that A.C
motors do not change in their speeds.

• Almost all large systems the frequency is kept
constant so that synchronous clocks may be operated
directly from the power mains.
• In any country one common supply frequency is adopted to facilitate the interchange of energy between different generating stations and substations.

• Electric lights do not operate satisfactorily at lower frequencies and gives flickering
• At lower frequencies in transformers the no. of flux linkages required is more. Hence more turns an cost. The 25 Hz. Transformers cost is 40% to 50% more than 50 Hz. Transformers
• At Higher frequencies say 50 Hz the reactance of lines and machinery is two times more than at 25 Hz frequency. Due to this poor regulation will occur.
• In case A.C motors, change of frequency causes
change of speed
• From the above effects it is desirable to choose constant frequency of 50 Hz or 60 Hz
Supply Voltage:
• In transmission lines more troubles are due to high currents compared to high voltages

• Some times lower maintenance and operating costs may obtained at higher transmission voltage

• The size of the conductor reduced at high voltages.
• At high voltages spacing between conductors
increases which increases the cost of line supports
EFFECT OF VOLTAGE ON LINE EFFICIENCY

• If the voltage is raised ‘m’ times, the current in the line
required to transmit the same amount of power is reduced ‘m’
times. Line losses I2R are reduced ‘m2’ times and hence line
efficiency is increased.
EFFECT OF VOLTAGE ON LINE DROP

Now Er = Es– IR, where R is the resistance of
the both conductors.
Pic22
is the percentage (%) voltage drop in both the conductors expressed as the %of the sending end voltage.

Therefore the voltage is increased the efficiency is increased for a given line drop.
EFFECT OF VOLTAGE ON VOLUME OF CONDUCTOR MATERIAL
• Increase of voltage by ‘m’ times the current is decreased by
‘m’ times and line loss is decreased by ‘m2 ‘ times.

• To transmit with the same loss as before, the resistance of
the line can be increased ‘m2 ‘ times that is the volume of
the conductor material can be decreased ‘m2 ‘ times.
• For any given line, transmission efficiency and given amount of power to be delivered, the volume of the conductor material is inversely proportional to the square of the voltage and square of the power factor.
volume of conductor material is
A X l ∞ 1/(E2cos2Ф)
Where A = cross section of the conductor
l = length of the line
E = phase voltage
cosФ = power factor (p.f)

Hence there is a great saving of conductor material by employing a high voltage.
EFFECT OF VOLTAGE ON TRANSFORMERS, INSULATORS, SWITCHGEARS AND SUPPORTS
High transmission voltage should
• Increases cost of insulators.

• Increases cost of transformers, switchgears and terminal apparatus.

• Increases cost of supports.
EMPIRICAL FORMULA FOR TRANSMISION VOLTAGE
• The economical transmission voltage between lines in
three phase A.C system is
pic23
Where V = Line voltage in K V
P = Maximum K W per phase to be
delivered to single circuit.
l = Distance of transmission in Km.
• Voltages normally adopted for transmission are
Pic24
IMPORTANCE OF HVDC TRANSMISSION
• With the development of thyristors handling large currents and voltages and associated high voltage D.C transmission (H.V.D.C) technology, it has been possible to use H.V.D.C.
• H.V.D.C Transmission having distinct superiority
over A.C Transmission particularly for
• Long distance bulk power transmission

• System inter-connection between two or more
networks

• Submarine cables
Features of H.V.D.C Transmission
• HVDC power transfer can be controlled
quickly and accurately
• No problems of reactive power flow
• Voltage fluctuations are less
• Transmission losses are less
• It requires costly and complex substations and controls
• From above facts, HVDC Transmission are
considered for bulk power transmission over long
distances, Submarine cable and system
inter-connections only
REQUIREMENTS OF CONDUCTOR MATERIAL
Main requirements of conducting material are
• High electrical conductivity.

• High tensile strength.

• Low specific gravity

• Economical cost and long life

• Low specific resistance
• Should not be brittle.

• Non corrosive.

• Should be flexible

• Free from oxides

• Should withstand high wind pressure

• Should have smooth surface to reduce corona
TYPES OF TRANSMISSION LINE CONDUCTOR
1. Solid conductors.

2. Stranded conductors.

3. Hallow conductors.

4. Bundled conductors.
SOLID CONDUCTORS
• Solid conductors are mainly available in the form of
solid copper and copper clad steel.
Pic25
• Over weight as compared to other conductors

• Solid wires except of smaller sizes are difficult to handle and transport.

• Skin effect is more

• They are only used for smaller cross sections.

• Solid aluminum conductors are not used
because of mechanical reasons.
STRANDED CONDUCTORS
• Standard conductors have greater flexibility and
mechanical strength as compared to single
conductor of same cross section.
Pic26
• Stranded conductors usually have a central wire around
which there are successive layers of 6, 12, 18, 24… wires.

• The number of strands is given by the formula
N = 3n(n+1)+1
where ‘n’ is the number of layers.
• Diameter of the stranded conductor can be determined by
D = (2n+1)d
where ‘d’ is the diameter of the each stranded.
‘n’ is the no. of layers

• In manufacture process, adjacent layers are
spiraled in opposite directions, so that the
layers are bound together.

• Stranded conductors are easier to handle
and transport.
• Standard conductors are mainly available in the
following forms
• ACSR ( Aluminium Conductors Steel Reinforced )
• AAC ( All Aluminium Conductors )
• AAAC (All Aluminium Alloy Conductors )
• ACAR (Aluminium Conductors Alloy Reinforced )
• Out of the above ACSR conductors are used for
transmission lines
HALLOW CONDUCTORS
• Hallow conductors gives large diameter.
Pic27
• Tensile strength and flexibility are less.

• Corona loss is reduced.

• Skin effect is less.

• Major portion of the conductor surface is exposed to
atmosphere causes more wind pressure and too much
ice formation on surface
BUNDLED CONDUCTORS
• Generally there is no difference in construction between
stranded and bundled conductors.


• Only difference is Bundled conductors have very large
number of strands than stranded conductors.
Relative merits between solid, stranded
and hallow conductors
pic28
CURRENT DISTORTION EFFECTS

• When a conductor is carrying a steady current (D.C) it will be distribute uniformly over the whole cross section.

• In practice the alternating current does not distribute
uniformly over the whole cross section of the conductor, but
it is distorted due to,
a) Skin, b) Proximity and c) spirality effects.
SKIN EFFECT


• The distribution of alternating currents through a conductor is not uniform, because the current density increases from the middle towards the outer layers.

• Due to A.C flux, the reactance of the central core of the conductor increase, therefore the impedance increases.

• Hence current confines to the upper layers of the conductor.
• This crowding of current towards the surface of the
conductor is known as skin effect.
Pic29
EFFECTS OF SKIN EFFECT
• At lower frequencies the effect is very small.

• For higher frequencies or solid conductors of large
diameter, the effect is considerable.

• Due skin effect the effective area of conductor through
which current flow is reduced, which increases resistance.

• This effect is much smaller for stranded conductor.

• This effect facilitate to use hallow conductors.
PROXIMITY EFFECT
• Consider two conductors placed side by side one is
going and other one is for return of current.

• The fluxes are not distribute uniformly, the currents are
also not distribute uniformly, so that the currents are
crowded to one side of the conductors.
• This effect is known as proximity effect as shown.
Pic30
Concentration of current due to conductor B on A
EFFECTS OF PROXIMITY EFFECT
Like skin effect, the proximity effect increases the resistance and decreases the reactance of the conductor.

The effect is more for a)Large diameter conductors,
b) high frequencies and c) closeness of the conductors.

At normal frequency and for wide spacing
of conductors, this effect is negligible.

4. This effect is eliminated for stranded conductors.
SPIRALITY EFFECT
• This effect is associated only for stranded conductors,
where the effecting length of the conductor is increased,
which is known as spirality effect.

• The magnitude of this effect depends on the size and
method of construction of the conductor.
EFFECTS OF SPIRALITY EFFECT
• Due to this effect the effective length of the
conductor increases.

• Spirality effect increases both resistance and internal
reactance of the standard conductors

• This effect is very small at normal frequency such as at
50 Hz frequency and can be ignored in the case of non-
magnetic conductors.
PROPERTIES OF THE CONDUCTOR MATERIAL OF VARIOUS TYPES
Properties Copper Aluminum Steel Alu.(1.3)&Steel(1.4)

Sp.gravity 8,900 2,700 7,860 3,450 3,700
Ult.tensile 40 18 40 – 320 120 120
str.(kg/mm2)
Sp.res.at 200C 0.01786 0.0287 0.178
Ω-m/mm2(ρ)
Res.temp.(α) 0.0038 0.004 0.00496
Coe.at 200C
PROPERTIES OF SOLID COPPER (HARD DRAWN)CONDUCTOR

SWG no. Dia.of (d) Nominal (mm2) Ct.rat.for temp.rise. (A)
conductor. Conductor. X - Sect.area(a) 27.80C 55.50C

10 3.251 8.30 37 50

9 3.658 10.51 44 60

8 4.064 12.97 52 70

7 4.470 15.70 58 81
SWG no. Dia.of (d) Nominal (mm2) Ct.rat.for temp.rise. (A)
conductor. Conductor. X - Sect.area(a) 27.80C 55.50C

6 4.877 18.68 68 92

5 5.385 22.77 78 107

4 5.893 27.27 91 122

3 6.401 32.18 103 139

2 7.010 38.60 117 158
SWG no. Dia. of (d) Nominal (mm2) Ct.rat.for temp.rise. (A)
conductor. Conductor. X - Sect.area(a) 27.80C 55.50C
1 7.620 45.60 133 181

0 8.230 53.19 150 203

00 8.839 61.36 167 226

000 9.449 70.72 184 250

0000 10.160 81.07 205 278
The resistance of conductor/m can be calculated both at 27.80C and 55.50C is

• ρ27.8 = ρ20 (1+ α20 (27.8 – 20)
• R27.8 = ρ27.8 (l /a) or R55.5 = ρ55.5 (l /a)

• ρ55.5 = ρ55.5 (1+ α20 (55.5 – 20)

‘a’ is given in the table and take ‘l’ = 1 metre
A.C.S.R CONDUCTOR

Code No. and dia. Dia. of cond. App.Cary.cap.
Word of wire in mm in amp.at
400C 450C

Squirrel 6/2.11 6.33 115 107

Gopher 6/2.36 7.08 133 123

Weasel 6/2.59 7.77 150 139
Code No. and dia. Dia.of cond. App.Cary.cap.
Word of wire in mm in amp.at
40C 45C

Ferret 6/3.00 9.00 181 168

Rabbit 6/3.55 10.05 208 193

Mink 6/3.66 10.98 234 217

Beaver 6/3.99 11.97 261 242

Raccon 6/4.09 12.27 270 250

Otter 6/4.22 12.66 281 260

Cat 6/4.50 13.50 305 283

Hare 6/4.72 14.15 324 300

Leopard 6/5.28 15.84 375 348

Tiger 30/7x2.36 16.52 382 354

Wolf 30/7x2.59 18.13 430 398

Panther 30/7x3.00 21.00 520 482

Lion 30/7x3.18 22.26 555 515

Bear 30/7x3.35 23.45 595 552

Goat 30/7x3.71 25.97 680 630
The resistance of conductor/m can be
calculated both at 40C and 45C.

Here both aluminum and steel wires are there, so
resistance of both are calculated in the same way
given at hard drawn solid copper conductors some
times manufacturer supplied the resistance along
with the conductors.
ALL ALUMINUM CONDUCTORS
(A.A.C)STRANDED.

Code No. and dia. Dia.of cond. App.Cary.cap.
Word of wire in mm in amp.at
40C 45C
Mantis 3/3.00 6.468 116 108

Aphis 3/3.55 7.223 133 123

Weevil 3/3.66 7.894 147 136

Lady bird 7/2.79 8.37 178 165

Ant 7/3.10 9.30 204 189

Fly 7/3.40 10.20 229 212

Blue battle 7/3.65 10.95 252 234

Earwig 7/3.78 11.34 264 245

Grass hopper 7/3.91 11.73 275 255

Clegg 7/4.17 12.51 298 276

Wasp 7/4.39 13.17 318 295

Catter piller 19/3.53 10.59 460 386

Chafer 19/3.78 18.90 504 468

Spider 19/3.99 19.95 540 504

Cockroach 19/4.22 21.10 575 534

Butter fly 19/4.65 23.25 655 608

Moth 19/5.00 25.00 720 660

Locust 19/5.36 26.80 790 734


Maybug 37/4.09 28.63 850 790

Scorpion 37/4.20 29.89 895 830

The resistance of conductor/m can be calculated both at 400C and 450C, in the same way as hard drawn solid copper conductor.


INDUCTANCE OF SINGLE PHASE OVERHEAD LINE
Definition of inductance:
• It is defined as flux linkages per ampere and
Measured in Henry.
Inductance, L= N j / I Henry.
• A single phase line consists of two parallel conductors which form a rectangular loop of one turn.

• A changing magnetic flux is set up when an alternating current (A.C) flows through the conductors.

• This flux links the loop and hence loop
possesses inductance.
• Consider a single phase overhead line consisting of two parallel conductors of radius ‘r’ m spaced ‘D’ m apart as shown in figure below.
Pic31
• In order to find the inductance of conductor we shall
have to consider flux linkages with it.

• Let I be the current in the conductor and Ix be the
current through the ring of thickness ‘dx’ at a
distance ‘x’ from the centre of the conductor.
Current density
In the ring
Pic32
Assuming uniform current density, we have
Pic33
Field strength at a distance x from the center of the conductor
Pic34
Flux through a cylindrical shell of radial thickness dx and
axial length 1 m long
pic35
The linkage of the shell
Pic36
The linkage inside the conductor
Pic37
Consider the flux between the conductors ( Flux external to the conductor)
The field strength at a distance x from the centre
Pic38
Flux linkage out side the conductor
Pic39
Total flux linkage / conductor
= Flux linkage inside the conductor+
Flux linkage outside the conductor
Pic40
Inductance / conductor = Flux linkage / ampere
Pic41
• If the conductor is non-magnetic material, then
Pic42
PROBLEMS:
A single phase line has two parallel conductors 2 meters apart. The diameter of each conductor1.2 cm. Calculate the loop inductance per km of the line.

Solution:
Given data:
spacing of conductor, d = 2 m = 200 cm
Radius of the conductor, r = 1.2/2 = 0.6 cm
Loop inductance per meter length of line
= 10-7 [µr + 4 loge (d/r)] H
= 10-7 [1 + 4 loge (200/0.6)] H
= 24.23 x 10-7 H
Loop inductance per km. length of line
= 24.23 x 10-7 x 1000 H
= 24.23 x 10-4 H
= 2.423 m H
. A single phase line has two parallel conductors 3 meters apart. The radius of each conductor being 1 cm. Calculate the loop inductance per km of the line if the material of the conductor is (i) copper (ii) steel with permeability of 100.

Solution:
Given data:
spacing of conductor, d = 3 m = 300 cm
Radius of the conductor, r = 1 cm
Loop inductance per meter length of line
= 10-7 [µr + 4 loge (d/r)] H
(i) With copper conductor, µr = 1
Loop inductance per meter length of line
= 10-7 [1 + 4 loge (300/1)] H
= 23.8 x 10-7 H
Loop inductance per k.meter length of line
= 23.8 x 10-7 x 1000 H
= 23.8 x 10-4 H
= 2.38 m H
(ii) With steel conductor, µr = 100
Loop inductance per meter length of line
= 10-7 [100 + 4 loge (300/1)] H
= 122.8 x 10-7 H
Loop inductance per km length of line
= 122.8 x 10-7 x 1000 H
= 122.8 x 10-4 H
= 12.28 m H
Inductance of three phase transmission line
Three phase symmetrical spaced conductors: (Equally spaced)
• A,B,C are three conductors equally spaced at distance of ‘D’ meters and radius of the conductors is ‘r’ meters.
Pic43
• Inductance /conductor (phase)/meter

= 10-7[0.5 + 2 loge (D/r] H/m.

• Inductance /conductor (phase)/ Km

= 10-4 [0.5 + 2 loge (D/r] H/ Km.
Three phase asymmetrical spaced conductors: (unequal spacing)
• A,B,C are three conductors unequally spaced at distance of ‘DAB’, ‘DBC’, ‘DCA’, meters and radius of the conductors is ‘r’ meters.
Pic44
Problem:
1) Find the inductance of three phase transmission line using 1.24 cm diameter conductors when these are placed at the corners of an equilateral triangle of each side 2meters. (i) inductance/conductor (phase)/meter and (ii) inductance/conductor (phase)/Km.
• Inductance /conductor (phase)/meter

= 10-7[0.5 + 2 loge [( 3 DABDBCDCA ) / r ]] H/m.

• Inductance /conductor (phase)/ k.meter

=10-4[0.5 + 2 loge [( 3 DABDBCDCA ) / r ]] H/ Km.
Solution:
Given data:
Conductor spacing D = 2 m
= 200 cm pic45
Conductor radius r =1.24/2
= 0.62 cm
• Inductance /conductor (phase)/m

= 10-7[0.5 + 2 loge (200/0.62] H/m.

= 12 x 10-7 H.
• Inductance /conductor (phase)/ Km

= 12 x 10-7 x 1000 H.

= 12 x 10-4 H

= 1.2 x 10-3 H

= 1.2 m H
2) Find the inductance of three phase transmission line using 1.24 cm diameter conductors when these are placed at the corners of triangle of sides 2m, 2.5m, 4.5m.
(i) inductance/conductor (phase)/meter and (ii) inductance/conductor (phase)/k.m.
Solution:
Given data:
Fig. Shows the three conductors A, B, C
of three phase line placed at the corners of triangle of sides 2m, 2.5m, 4.5m.
pic46
DAB = 2 m, DBC = 2.5 m, DCA = 4.5 m.

Radius of conductor, r = 1.24/2 = 0.62 cm

Equivalent equilateral spacing,
Deq = 3 DAB DBC DCA

= 3 2 x 2.5 x 4.5
= 2.82 m
= 282 cm
(i) Inductance /conductor (phase)/meter = 10-7[0.5 + 2 loge [( 3 DABDBCDCA ) / r ]] H/m. = 10-7[0.5 + 2 loge (Deq/r)] =10-7[0.5 + 2 loge(282/0.62) = 12.74 x 10-7 H = 12.74 x 10-7 H (ii) Inductance /conductor (phase)/ Km
=10-4[0.5+2 loge [( 3 DABDBCDCA ) / r ]]H/Km.
=10-4[0.5+2 loge (Deq/r)]
=10-4[0.5+2 loge (282/0.62)] = 12.74 x 10-4 H
= 1.274 x 10-3 H = 1.274 m H.
Need for transposition of over head lines
• Generally the transmission line conductors of three phase three wire system are irregularly spaced.
• Because it is difficult to maintain an exact height of the conductor from the ground on the towers or poles.
• Due to the above the transmission line conductor constants ( R, L, C ) are not the same for all phases
• In order to equalize these line constants, the three lines must be transposed (change the position) in such a way that the three lines occupies a position relatively other wires for l/3 of the total length.
• Fig. Shown below a transposed lines.
Pic47
Effects of transposition
• The transposition in transmission lines will balance the capacitance of the line, so that electro statically induced voltage is balanced in the complete length of the line.

• It is also balance the inductance of the line and hence the electromagnetically induced voltage is balanced in the complete length of the line.

• By proper co ordination of transposition of power and near by communication lines, the induced voltages can be reduced to very small and it also reduces the noise in the near by communication lines.
Inductance in a transposed line
In a three phase transposed line the
(i) inductance/conductor (phase)/m
=10-7[0.5 + 2 loge [(D= 3 DABDBCDCA ) / r ]] H/m.
(ii) inductance/conductor (phase)/Km
=10-4[0.5 +2 loge [(D= 3 DABDBCDCA ) / r ]]H/km.
Here D is known as Geometrical mean distance (GMD).
Problem :
• Find the inductance of three phase transmission line
using 2 cm diameter conductors when these are placed
at the corners of triangle of sides 3.6m, 5.4m, 8.1m. The conductors are transposed at regular intervals.
(i) inductance/conductor (phase)/m and
(ii) inductance/conductor (phase)/Km.
Solution:
Given data:
Fig. Shows the three conductors A, B, C of three phase
line placed at the corners of triangle of sides 3.6m,
5.4m, and 8.1m.
Pic48
DAB = 3.6 m, DBC = 5.4 m, DCA = 8.1 m.

Radius of conductor, r = 2 / 2 = 1 cm

Geometric mean distance,

D = 3 DAB DBC DCA

= 3 3.6 x 5.4 x 8.1

= 5.4 m

= 540 cm
Inductance /conductor (phase)/meter

= 10-7[0.5 + 2 loge (GMD ) / r ] H.

= 10-7[0.5 + 2 loge (540 / 1)] H.

= 13.09 x 10-7 H.

= 13.09 x 10-7 H.
Inductance /conductor (phase)/ Km

= 10-4[0.5 + 2 loge (GMD ) / r ] H.

= 10-4[0.5 + 2 loge (540) / 1) H.

= 13.09 x 10-4 H. = 1.309 x 10-3 H.

= 1.309 m H.
Capacitance
• It is defined as the ratio of the charge in coulombs on a capacitor to he potential difference (P.D) applied across the capacitor is known as capacitance of the capacitor.
• It is measured in Farads (F) and is usually expressed in (micro) µF or pF.
Pic49
Capacitance of single phase over head line:
• Consider a single phase transmission line consisting of
two parallel conductors A and B spaced ‘d’ meters
apart in air.
Pic50
Let their respective charge be + Q and – Q
coulombs per meter length.

Consider at point ‘p’

The electrical intensity due to conductor A

= Q / (2πЄ0Єr x )

The electrical intensity due to conductor B

= Q / (2πЄ0Єr (d - x))
• The total intensity at ‘p’ due to A and B is

Ex = Q / (2πЄ0Єr x ) + Q / (2πЄ0Єr (d - x))

• The p.d between the conductors is equal to the work done in moving one coulomb from one conductor to the other against the electrostatic force.
• If one coulomb is moved a distance dx then the work done is Ex dx joules.

• There fore the P.D between the conductors



V = Q / (2πЄ0Єr) [ ]
V=Q/(2πЄ0Єr)[logex–loge (d - x) ]rd-r

=Q/(2πЄ0Єr){[loge(d-r)-loge(d-d+r)]-[loger–loge (d –r)]}

=Q/(2πЄ0Єr) [loge (d - r) – loge r – loger + loge (d –r)]

=Q/(2πЄ0Єr)[2 loge(d-r)-2 loger]

=2 Q/(2πЄ0Єr)[ loge(d-r) - loger]
V = Q/(πЄ0Єr)[ loge(d-r) - loger]

Neglect r compared to d, then we have

V = Q/(πЄ0Єr)[ loge( d / r)]

Capacitance/loop/meter = charge/P.D = Q/V

= Q/(Q (loge( d / r)) /(πЄ0Єr) = πЄ0Єr/ loge( d / r)
Capacitance/loop/meter = charge/P.D = Q/V

= πЄ0Єr/ loge( d / r) F/m

Capacitance/loop/Km = πЄ0Єr/ loge( d / r) x 103

= [(22/7)x1x8.854x10-12x103]/ loge( d / r)

{ Here, Є0 = 8.854 x 10-12, Єr = 1}
= (2781 x 10-9)/ loge( d / r) F/km

= ( 0.02781) / loge( d / r) μF/km

Capacitance/loop/km = ( 0.02781) / loge( d / r) μF/km

Similarly, Capacitance/conductor/km

= ( 0.0556) / loge( d / r) μF/km
Problems :
1) Calculate the capacitance/km of a pair of parallel conductors,
5 mm in diameter and spaced uniformly 20 cm apart in air.
Solution :
Given data :

Diameter of the conductor = 5 mm
Radius of the conductor, r = 5/2 = 2.5 mm = 0.25 cm
Spacing between conductor, d = 20 cm
Capacitance/loop/km = 0.02781/loge(d/r)

= 0.02781/loge(20/0.25)

= 0.02781/loge(80)

= 0.006342 µ F.
2) calculate the capacitance/km of a pair of parallel
conductors, 10 mm in diameter and spaced uniformly
2 m apart in air.
Solution :
Given data :

Diameter of the conductor = 10 mm
Radius of the conductor, r = 10/2 = 5 mm = 0.5 cm
Spacing between conductor, d = 2 m = 200 cm
Capacitance/loop/km = 0.02781/loge(d/r)

= 0.02781/loge(200/0.5)

= 0.02781/loge(400)

= 0.00464 µ F.
Inductance of three phase transmission line
Three phase symmetrical spaced conductors: (Equally spaced).
• A,B,C are three conductors equally spaced
at distance of ‘D’ meters and radius of the
conductors is ‘r’ meters.
Pic51
Capacitance/conductor/km
=0.0556/loge(D/r) µF
Three phase unsymmetrical spaced conductors: (un equal spacing)
• A,B,C are three conductors unequally spaced at distance of ‘DAB’, ‘DBC’, ‘DCA’, meters and radius of the conductors is ‘r’ meters.
Pic52
Geometrical mean distance

D = 3 DAB DBC DCA

Capacitance / conductor / km

= 0.0556 / loge [( 3 DAB DBC DCA )/ r] µ F
Problem :
1) Calculate the capacitance of the each conductor of a 400 km, three phase line with an equilateral configuration. The spacing between conductors is 3 m and radius of the conductor is 1 cm.

Solution :
Given data :
Length of the line L = 400 km
Spacing between conductor, D = 3m = 300 cm
Radius of the conductor, r = 1cm
Capacitance / conductor / km = 0.0556 / loge (D/r)

= 0.0556 / lobe (300/1)

= 9.747 x 10-3 µ F / km

Total Capacitance / conductor = 9.747 x 10-3 x 400

= 3.898 µ F
Calculate the capacitance / km of a three phase line completely transposed with a triangle configuration. The spacing between conductors are 1 m, 1.25m and 2.125 m. Diameter of the conductor is 1.25 cm.

Solution :
Given data :
DAB = 1m, DBC = 1.25 m, DCA= 2.125m

Geometric mean distance D = 3 DAB DBC DCA

D = 3 1 x 1.25 x 2.125 = 1.385 m =138.5 cm
Diameter of the conductor = 1 cm

Radius of the conductor r = ½ = 0.5 cm

Capacitance / conductor / km

= 0.0556/ loge[( 3 DABDBCDCA ) / r ]

=0.0556/loge(138.5/0.5)

=0.0556/5.624=0.0098µF/km
3) Calculate the capacitance / phase or line to neutral capacitance of a three phase line completely transposed with horizontal configuration. The spacing between conductors are 2 m, 2.5m and 4.5 m. Diameter of the conductor is 1.25 cm.

Solution :
Given data :

DAB = 2m, DBC = 2.5 m, DCA= 4.5m
Geometric mean distance D = 3 DAB DBC DCA

= 3 2 x 2.5 x 4.5

= 2.82 m = 282 cm
Radius of the conductor, r = 1.25 /2 = 0.625 cm
Line to neutral capacitance or
Capacitance / conductor / km

= 0.0556/ loge[( 3 DABDBCDCA ) / r ]

= 0.0556/loge(282/0.625)

= 0.0556/6.112=0.00909µF/km
Resistance, inductance and capacitance of the three phase over head line conductors at different voltages.
(Values apply to double circuit lines, A.C.S.R conductors).
1.Rated voltage kv 20 30 60
2.Nominal x section 35 150 35 150 95 240
Al ( mm2)
3.Diameter of the 8.1 17.3 8.1 17.3 13.4 21.7
conductor (mm)
4.Resistance at 200C 0.87 0.2 .87 0.2 0.32 0.13
(ohm/km)
5.Reactance (2 fl) at 0.39 0.34 0.4 0.35 0.4 0.37
50 Hz (ohm/km)
6.Mutual cap. (10-9 F/km) 9.7 11.2 9.5 10.9 9.5 10.3
7.cap.To earth(10-9 F/km) 3.4 3.6 3.5 3.7 3.8 4.0
Values apply to double circuit lines, A.C.S.R conductors).
1.Rated voltage kv 110 220
2.Nominal x section 120 300 340 2x240
Al ( mm2)
3.Diameter of the 15.7 24.2 28.1 2x21.7
conductor (mm)
4.Resistance at 200C 0.25 0.10 0.09 0.062
(ohm/km)
5.Reactance (2 fl) at 0.41 0.38 0.40 0.32
50 Hz (ohm/km)
6.Mutual cap. (10-9 F/km) 9.3 10 9.5 11.5
7.cap.To earth(10-9 F/km) 4.0 4.2 4.8 6.3
Over head transmission lines
• The transmission line constants Resistance, Inductance
and Capacitance are uniformly distributed along the line
• The transmission lines are mainly classified in to
three types, namely Short, Medium and Long
transmission line
SHORT TRANSMISSION LINE
• The length of the lines less than 80 kms and line voltage is less than 20 kv are termed as short transmission lines.

• Due to smaller length and lower voltage the capacitance effects are very small and hence it can be neglected.
MEDIUM TRANSMISSION LINE
• The length of the lines lies between 80 to 240 kms
and line voltage is lies between 20 to 100 kv are
termed as medium transmission lines.


• Due to sufficient length and line voltage the
capacitance effects are taken into account.
LONG TRANSMISSION LINE
• The length of the lines are more than 240 km and
line voltage is greater than 100 kv are termed as
long transmission lines.

• Here also the capacitance effects are taken into account.
REASONS FOR LUMPED CONSTANTS IN SHORT AND MEDIUM TRANSMISSION LINES
• The transmission line constants resistance (R), inductance (L) and capacitance (C) are proportional to the length of the line and hence they are uniformly distributed along the line as shown in fig.
Pic53
In the above fig.1

Where, r = resistance / unit length

L = inductance / unit length

c = capacitance / unit length

g = conductance / unit length.
• For short transmission lines the effect of capacitance is
very small, hence it may be neglected and other constants are lumped instead of distributed, because the current flowing through the line is same the distributed drops are equal to lumped drops, so they are lumped.


• For medium transmission lines the capacitance is assumed to be lumped at one or more points in the line.
VOLTAGE REGULATION
• When a transmission line carrying current, there is voltage drop in line due to resistance and inductance of the line.

• Due to this, the receiving end voltage (VR) is less than the sending end voltage (VS).

• The voltage drop in the line = VS – VR.

• The ratio of voltage drop (VS – VR) to the receiving end (VR)
is known as voltage regulation.

i e voltage regulation = (VS – VR) / VR.
PERCENTAGE VOLTAGE REGULATION
• The voltage regulation is expressed in percentage (%) then it is known as percentage (%) regulation.


• i. e percentage voltage regulation = pic54
APPROXIMATE FORMULA FOR PERCENTAGE
VOLTAG REGULATION
Pic55
From the vector diagram.

OC is nearly equal to OF

OC = OF
= OA + AF
= OA + AG + GF
= OA + AG + BH

There fore, VS = VR + IR cosΦR+IXL sinΦR ----(i)
But voltage regulation = (VS – VR) / VR

There fore, (VS – VR) = IR cosΦR+IXL sinΦR

Voltage regulation = (VS – VR) / VR

= (IR cosΦR+IXL sinΦR) / VR
• Percentage (%) voltage regulation for lagging p. f.

= pic56



• Percentage (%) voltage regulation for leading p. f.

= pic57
CLASSIFICATION OF TRANSMISSION LINES
• Short transmission lines (STL).
• Medium transmission lines (MTL).
• Long transmission lines (LTL).
SHORT TRANSMISSION LINES
● OH line up to 80km
● Voltage level > 20 kv
● Capacitance effect negligible
● Only Resistance & Inductance are taken into account
MEDIUM TRANSMISSION LINE
● Line Voltage is high >20kv and <>100kv
● Line constants are considered uniformly distributed
● OH line is more than 160 km
SHORT TRANSMISSION LINE
• Line Diagram:
Pic58
PHASOR DIAGRAM
I = Load Current in amps
R = Loop Resistance
XL=Loop Reactance
VR=Receiving end voltage pic59
Vs=Sending end voltage
SENDING END VOLTAGE(VS)
From the phasor diagram …The right angled triangle ODC
Pic60
VOLTAGE REGULATION:
Pic61
EFFICIENCY :
Receiving end power = VR IR CosFR
Sending end power = VR IR CosFR + I2R
Line losses = I2R
Transmission efficiency = pic62

TUTORIAL PROBLEMS
1) A 3 Phase line delivers 3600 KW at a power factor 0.8 lagging to a load when the sending end voltage is 33KV if Resistance & Reactance are 5.31 ohm and 5.5 ohms respectively.
Determine
a) The receiving end voltage
b) Line current
c) Transmission efficiency
Given data:
Power/phase = 3600/3 = 1200KW
Resistance… R = 5.31ohms
Reactance… X = 5.54ohms
Power factor. CosøR = 0.8
Sending end voltage Vs=33000/√3=19052
To be found:
Receiving end voltage VR=?
Line current, IL=?
Transmission Efficiency htr =?
Power delivered/phase P = VR IR cosøR

I = P / VR cosøR

I=(1200*1000)/VR*0.8

I=(150*105)/VR

Sending end voltage Vs = VR+IRcosøR +IXSinøR

19052=VR + (15*105 *5.31 * 0.8)/VR+ (15*105 *5.54 * 0.6)/ VR

VR2-19052 VR+11358000=0

By solving above equation

VR=18435 volts
Power delivered/phase P = VR IR cosøR

I = P / VR cosøR

I=(1200*1000)/VR*0.8

I=(150*105)/VR

Sending end voltage Vs = VR+IRcosøR +IXSinøR

19052=VR + (15*105 *5.31 * 0.8)/VR+ (15*105 *5.54 * 0.6)/ VR

VR2-19052 VR+11358000=0

By solving above equation

VR=18435 volts
a) Line voltage at receiving end

VR = √3* x18435=31930.volts

b) Line current…,I=15x105 / 18435 = 81.36Amps

Line losses = 3 I2R=3x81.362 x 5.31

=105447 watts

c) Transmission efficiency
=[Receiving end power/(Receiving
end power +line losses)]*100

=[3600/(3600+105.447)]*100
2) A load of 4000kw at 1100 volts is being received from
a single phase transmission line at a p.f of 0.9 lag if
R=0.18 ohms & X= 0.02ohms respectively. Neglect
capacitance of the line. Calculate
a) Sending end voltage
b) Current
c) Power factor
Solution:
Given data:
Load P = 4000 Kw
VR= 1100volts
R = 0.018 ohms
XL = 0.02 ohms
IR = P / VR cos F R
IR= [(4000*1000)/(1100*0.9)]
=4040Amps
Cos øR=0.9, Sin øR=0.4352

Resistance for Go and Return conductor
R=2x0.018 = 0.036ohms

Reactance X = 2x0.02 = 0.04ohms
VS2 =(VR cos F R + IR)2 + (VR sin FR +IX )2
= (1100X0.9+4040X0.036)2+(1100X0.4352+4040X0.4)
= 5677075.8
VS = 2382.66
• Sending end p.f cos Fs = pic63
= 0.47 lag
Medium Transmission Line(Nominal Π(pie) method)
• Line Diagram:
Pic64
As shown in the figure the following points can be observed
• In the above method capacitance of each conductor
(L to N) is divided in two halves.
• Half of the capacitance is lumped at the sending end
and other half at the receiving end
• Capacitance at sending end has no effect on the line drop
• Charging current must be added to line current to obtain
sending end current
As per line diagram

I= Load Current in amps
R= Loop Resistance
XL=Loop Reactance
VR=Receiving end voltage
Vs=Sending end voltage
C= Capacitance/Phase
COS Fr = receiving end p.f
COS Fs = sending end p.f
PHASOR DIAGRAM (Π method)
Pic65
Sending End Voltage
Pic66
Sending End Current
Pic67
Medium Transmission Line (Nominal T Method)
Line diagram
Pic68
Nominal T Method
• Whole line capacitance is assumed to be concentrated at the middle point of the line
• Half the line resistance and reactance are lumped on its either side
• Full charging current flows over half the line
• One phase of 3 phase transmission line is shown in the line diagram
PHASOR DIAGRAM (Nominal T method)
Pic69
Sending End Current :
Pic70
Sending End Voltage:
Pic71
EXERCISE PROBLEMS
• A 3-phase, 50-Hz Overhead transmission line 100km long has the following constants:
Resistance/Km/Phase = 0.1 ohm
Inductive Reactance/km/phase = 0.2 ohms
Capacitive susceptance /km/phase = 0.04X10-4 siemen
Determine a) the sending end current b) sending end voltage c) sending end power factor d) transmission efficiency, when supplying a balanced load of 10,000 KW at 66kv, pf =0.8 lagging Solve in T method
SOLUTION
Given Data:
Total Resistance / Phase R = 0.1X100=10 ohms
Total Reactance / Phase XL= 0.2 X 100 = 20 ohms
Capacitive Susceptance Y = 0.04 X 10-4 X 100
= 4 X 10-4
Receiving end Voltage / Phase VR= 66000 / √3 = 38105 V
Pic72
MEDIUM TRANSMISSION LINE (T METHOD)
Line Diagram:
Pic73
PHASOR DIAGRAM (T method)
Pic74
Receiving end voltage, VR = V R + j0 = 38,105 V
Load current IR = I R ( cosΦR – j sin Φ R )
= 109 ( 0.8 – j 0.6 )
= 87.2 – j 65.4
Voltage across C, V1 = VR + IR Z/2
= [38,105 + ( 87.2 – j65.4) (5+j10)]
= 39,125 + j545
Charging current, IC = jYV1
= j4 X 10-4(39,195 + j 545)
= 0.218+j15.6
Sending end current, IS = IR +IC
= (87.2-j65.4)+(-0.218+j15.6)
= 87.0-j 49.8= 100 -29047/
Sending end current = 100A
Sending end voltage = VS = V1+ IS Z/2
= (39,195 + j545)+(87.0-j 49.8)(5+j10)
=40128 + j1170 = 40145 1040/
Line value of sending end voltage = √3x40145
= 69533 V
= 69.533 kv Referring to phasor diagram in Fig.

θ1= angle between VR and VS = 1040/
θ2= angle between VR and IS = 29047/
ΦS= angle between VS and IS = θ1+ θ2 = 31027/

Sending end power factor
Cos ΦS = Cos 31027/
= 0.853 lag
Sending end power = 3VSIS cos ΦS
= 3X40,145X100X0.853
= 10273105 W
=10273.105 kW

Power delivered =10,000 kW
Pic75
• A 3-phase, 50-Hz Overhead transmission line 150km long has the following constants:
Resistance/Km/Phase = 0.1 ohm
Inductive Reactance/km/phase = 0.5 ohms
Capacitive shunt admittance /km/phase=3X10-6 siemen
Determine a) the sending end current b) sending end voltage c) sending end power factor when supplying a balanced load of 50 MW at 110kv, pf =0.8 lagging Solve in Π method
SOLUTION
Given Data:
Total Resistance / Phase R = 0.1X150=15 ohms
Total Reactance / Phase XL= 0.5 X 150 = 75 ohms
Capacitive Admittance/ Phase Y = 3 X 10-6X 150 = 45 X 10-5
Receiving end Voltage / Phase VR = 110X1000 / √3 = 38105 V
Pic76
Taking receiving end voltage as the reference phasor , we have
Pic77
Pic78
Pic79
CHARGING CURRENT (IC)IN TRANSMISSION LINES
• The capacitance effect is considered in over head
lines over 100km and above.
• Due to capacitance effect current produces is usually called charging current
• Charging current flows in quadrature with the voltage
• Charging current flows through the line even though
receiving end is open circuited
• It has maximum value at sending end and decreases
continuously as the receiving end is approached
IC=VRY
• Due to the charging current there will be power loss in
the line
POWER LOSS DUE TO CHARGING CURRENT
• The Value of charging current at a point P “x” km from the
sending end. Total length of line “l” km as shown below fig.
pic80
pic81
pic82
FERRANTI EFFECT
• In long transmission line when the line is open circuited or very lightly loaded , a rise of voltage occurs at the receiving end is known as ferranti effect.
• The above pressure rise is due to the emf of self inductance of the charging current.
• To determine the magnitude of pressure rise one half of the total line capacitance will be assumed to be concentrated at the receiving end
DERIVATION OF RISE IN VOLTAGE AT RECEIVING END
Line Diagram
Pic83
VECTOR DIAGRAM:
Pic84
With reference to the above vector diagram the sending end voltage is less than the receiving end voltage
Neglecting Resistance drop
Rise in voltage=OP-OQ=ICX
IC=VRY/2
Pic85
Pic86
CORONA IN TRANSMISSION LINES:

• When the voltage between the conductors of a overhead line exceeds a certain value, The air surrounding it is broken & ionization starts.

• The voltage at which ionization starts is called disruptive critical voltage ( Vd)
• If the voltage exceeds the disruptive critical voltage, a hissing noise is heard and the conductors are surrounded by a luminous envelope.

• The voltage at which this occurs is called visual critical voltage (Vv ).

• The phenomena of the hissing noise,the violet glow and production of ozone is called corona
DEFINITION OF CORONA:

• Corona is due to the bombardment of air molecules with subsequent dislodging of electrons, by ionised particles.
OR
• The phenomenon of violet glow,hissing noise & production of ozone gas in an overhead transmission line is known as corona.
• If the conductors are uniform and smooth, same conditions will hold good through out the length of the conductor.

• If the conductors are not uniform & smooth rough points will appear brighter

• In case of DC system the positive conductor will have a uniform glow & will brighter than negative one.
• The luminous blue envelope is nothing but breakdown of air into ions under high electrostatic stress.

• The breakdown starts first near the surface of the conductor, because the electrostatic stress is maximum at the surface.

• The thickness of the conducting layer of air increases with the increase of supply voltage.
FACTORS AFFECTING CORONA
Corona is affected by the following factors, namely

1. Atmosphere:
• In stormy weather conditions, the no. of ions in
the atmosphere is more as compared to fair
weather conditions.
• Hence corona will occur in stormy weather at
much less voltage compared with to fair weather.
2. Conductor:
• Corona decreases with increase in diameter
of the conductor
• Corona is more in standard conductors as
compared to solid conductors.
• Corona is more in rough and dirty surface of the
conductors as compared to smooth conductors.
3. Spacing between the conductor:
• If the spacing between the conductors is large
as compared with their diameter, there is no
formation of corona.
4. Line voltage:
• At low voltage, there is no Corona, but at
higher line voltages corona effect will appear
Disruptive critical voltage

The minimum phase to neutral voltage at which corona occurs is known as disruptive critical voltage

It is calculated by the following equation :

In case of two parallel conductors the max potential gradient is given by

g max = Vc / r loge (d/r)
where Vc = The r.m.s value of voltage of
the conductor to neutral

r = The radius of the conductor in cm.

d = The spacing of conductor in cm.

• The breakdown strength of air at 760mm pressure
and temperature of 25oC is 30kv/cm (max) or
21.1 kV/cm (r. m. s) and is denoted by go
• The value of go depends on the density of air
• The dielectric strength of air is proportional to its density over a wide range

• The dielectric strength of air is directly proportional to the barometric pressure ‘ b ’ & inversely proportional to the absolute temperature ‘ t ’.
The breakdown strength at a barometric pressure of ‘b’ cm mercury and temperature of toc becomes δ g0 where

δ =3.92b / ( 273+t )

If Vdo is the voltage to the neutral at which breakdown
occurs, then

VC =go δ r loge (d/r) kv(rms)to neutral
In practice it is necessary to allow for the condition of
surface of the wire so that

VC =go δ mo r loge(d/r)

Where
mo = roughness factor or irregularity factor
= 1 for smooth and clean wires
= 0.98 to 0.93 for rough wires
= 0.87 to 0.8 for stranded wires
g = Max potential gradient or breakdown value of air
= 30kV(max)/cm or 21.2kV(r.m.s)/cm.
δ = density of the air surrounding the conductor
= 3.92b/273+t 0 c b is the pressure in cm of mercury
r = radius of the conductor in cm.
d = distance between the conductor in cm.
Therefore
V c =21.2 mc δ loge(D/r) kV (r.m.s) to neutral.
POWER LOSS DUE TO CORONA
• Formation of corona is always accompanied by energy loss which is dissipated in the form of
Light, Heat, Sound and Chemical action.

• The formation of corona is associated with loss of power, which will have some effect on the efficiency of the line.
• Corona will not have any appreciable effect on the line regulation.
• The surface voltage gradient at line conductor exceeds the critical breakdown stress,corona appears and energy is dissipated in the form of light and heat.

• The energy loss dissipated in the form of light and heat is known as corona power loss and this is derived from the AC network reducing the transmission energy.
The power loss due to corona is given by
Pic87
Where f = Supply frequency in Hz

V = Phase neutral voltage in r.m.s

VC = Disruptive critical voltage r.m.s per phase
EFFECTS OF CORONA
The effects of corona are

1.There is a definite dissipation of power , although it is
not so important except under abnormal weather
conditions like storm etc.

2. Corrosion due to ozone formation.

3. The current drawn by the line due to corona loss is
4. This may cause interference with neighboring
communication circuits.

5. If corona is present the effective capacitance of the
conductor increases.

6. Corona works as a safety valve for surges.
7. Corona discharge reduces the dielectric strength .
8. The reduction in dielectric strength makes easier for
the flash over between insulators and conductors.
9. Transmission efficiency is reduced due to corona
METHODS OF REDUCING CORONA
• Corona effects are observed at a working voltage of 33kV or above.Therefore , careful design should be made to avoid corona

• Corona effects are observed mainly at the substations or on busbars rated for 33kV and higher voltages

• Otherwise highly ionised air due to corona may cause flashover in the insulators or between the phases, causing considerable damage to the equipment
• The effects of corona can be decreased either by increasing the size of the conductor or the space between the conductors.
1. By increasing conductor size:
• By increasing conductor size, the voltage at
which corona occurs is raised and hence
corona effects are considerably reduced

• This is one of the reasons that ACSR conductors
which have a large cross sectional area are used
in the transmission lines .For E.H.V. Lines,
Bundled conductors are used
2. By increasing conductor spacing:
• By increasing spacing between the conductors,
the voltage at which corona occurs is raised .
• If the distance between the conductors is increased
the tower costs and reactance drop increases.
• It is not advisable to increase the spacing.
Hence size of the conductor is increased .
• In a special conductor construction , one or more layers of copper strands are spiralled round a core of twisted copper I beam which is shown in the fig.1
Pic88
• Thus by means of I beam core,strength is added to the hollow conductor without the addition of deadweight.
• Another design which in use consists of number of tongued and grooved rectangular copper sections. Which spiralled along the length of the conductor to form a hollow tube as shown in the fig.2
Pic89
APPLICATIONS OF HOT LINE TECHNIQUE
• The following are main applications of hot line technique

1. Changing of broken and flashed out discs
2. Connecting the temporary jumpers to the line
3. Testing of transformers, bushings and other
equipments for leakage.
4. Polarity test on 11KV and 33Kv feeders for
parallel operation
5. For removing and replacement of cross arms
6. Maintenance of contaminated insulators.
HOT LINE TECHNIQUE
• In hot line technique, the line men work on the
energized transmission line with safety for repair of
fitting and equipments on the pole
In hot line technique, the tools must be lighter and stranger
Wooden species are mainly used as a hot line tools.
Wooden species are mainly used as a hot line tools.
Generation of Electrical Energy
• Energy available in the nature in different forms.
• Conversion of available energy into Electrical energy is
known as generation of electrical energy.
ENERGY FROM SOURCE
Pic90
NUCLEAR POWER PLANT
Pic91
SOLAR PANELS
Pic92
Hydro Power Plant
Pic93
WIND MILLS
Pic94
WIND POWER
Pic95
WAVE POWER
Pic96
HYDRO POWER PLANT
Pic97
NUCLEAR POWER PLANT.
Pic98
WIND POWER
Pic99
ELECTRIC POWER TRANSMISSION
Pic100
AC POWER TRANSMISSION
• AC power transmission is the transmission of electric power by alternating current.
• Usually transmission lines use three phase AC current. Single phase AC current is sometimes used in a railway electrification system.
• In urban areas, trains may be powered by DC at 600 volts.
• Electricity is usually transmitted over long distance through overhead power transmission lines.
• Underground power transmission is used only in densely populated areas due to its high cost of installation, maintenance, because the high reactive power produces large charging currents and difficulties in voltage management.
• Today, transmission-level voltages are usually considered
to be 110 kV and above.
• Lower voltages such as 66 kV and 33 kV are usually
considered sub-transmission voltages but are occasionally
used on long lines with light loads.
• Voltages less than 33 kV are usually used for distribution.
• Voltages above 220 kV are considered extra high voltage
and require different designs compared to equipment used at lower voltages.
BULK POWER TRANSMISSION
• Transmission efficiency is improved by increasing the voltage using a step-up transformer, which reduces the current in the conductors.
• The reduced current flowing through the conductor reduces the losses in the conductor.
• since, according to Joule's Law, the losses are proportional to the square of the current, halving the current makes the transmission loss ¼th the original value.
LOSSES
• Transmitting electricity at high voltage reduces the fraction of energy lost to Joule heating.
• For a given amount of power, a higher voltage reduces the current and thus the resistive losses in the conductor.
• Long distance transmission is typically done with overhead lines at voltages of 115 to 1,200 kV. However, at extremely high voltages, more than 2,000 kV between conductor and ground, corona discharge losses are so large that they can offset the lower resistance loss.































































































COMPARISON OF VARIOUS SYSTEMS OF TRANSMISSION
(The ratio of conductor material required in any system compared with that in the corresponding 2 wire D.C system)
S.no System max.voltage to earth max.voltage bet.cond.
D.C
1. Two wire 1 1
2. Two wire mid point earthed 0.25 1
A.C
1. Single phase two wire
2. Single phase two wire
mid point earthed
3. Three phase three wire

NEED FOR TRANSMISSION LINE

• Generating stations, transmission lines and distribution system are the main components of an electric power system.

• Generating stations and a distribution systems are connected through transmission lines, which also connect from one grid to another.
• Electrical power generated at a voltage of 11 to 25 kv, which is stepped up to the transmission levels in the range 66 to 400 KV or higher.

• The need of distribution systems and transmission lines is to transmit the electric Power from the generating station to the Consumer (Industry or Domestic).
Pic1
TRANSMITTING THE ELECTRICAL POWER

Types of systems:
• D.C (Direct current) system
• A.C (Alternating current) system
TYPES OF D.C SYSTEMS
• D.C Two wire system

• D.C Two wire with mid point earthed system.

• D.C Three wire system
TYPES OF A.C SYSTEMS
1. Single phase A.C system
• Single phase two wire.
• Single phase two wire with mid point earthed.
• Single phase three wire
2. Two phase A.C system
• Two phase three wire
• Two phase four wire

3. Three phase A.C system
• Three phase three wire
• Three phase four wire
D.C SYSTEMS
D.C Two wire system
Pic2
D.C Two wire with mid point earthed system
Pic3
D.C Three wire system with mid point earthed
Pic4
A.C SYSTEMS
Single phase two wire:
Pic5
Single phase two wire with mid point earthed:
Pic6
Single phase three wire with mid point earthed:
Pic7
Two Phase Three wire:
Pic8
Two phase four wire:
Pic9
Three phase three wire:
Pic10
Three phase four wire
Pic12
Advantages of D.C Transmission:
• Requires only two conductors.

• No inductance & capacitance effect.

• No phase displacement & surge problem.
• No skin effect & less corona loss.

• Reduced interference with communicating circuits.

• Potential stress on the insulation is less.
• Loss due to charging current eliminated.

• Used in under ground cables also.

• Voltage regulation is better & No stabilizer required
Disadvantages of D.C Transmission:
• Power can’t be generated at high voltages
• Voltage can’t be stepped up
• Cost of converting and inverting stations are high
• Requires costly switch gear
Advantages of A.C Transmission:
• Voltage can be stepped up and stepped down.

• Electrical power generated at high voltages.

• Maintenance of substation is easier &cheaper

• Cost of switch gear is less
Disadvantages of A.C Transmission:
• More conductor material is required.

• Power loss due to charging current.

• Construction of transmission lines are difficult.

• Synchronizing problems.
• Due to skin effect line resistance will increase

• Due to corona Spacing between conductors should be
kept more.

• In cables alternating current causes sheath loss
The following assumptions are made for
calculating conducting material
• Equal amounts of power transmitted.

• Equal lengths of line

• Equal line loss.

• The max. voltage between two conductors
is same (Vm).
CONDUCTOR MATERIAL REQUIRED FOR OVER HEAD LINE FOR VARIOUS SYSTEMS
D.C system:
1.D.C two wire system with one conductor earthed
Pic12
Vm is the max.potential difference between
conductor and earth in volts.

I is current in amperes, R is the resistance of the
conductor in ohms.

Line voltage = Vm
Power transmitted = P = Vm I watts.
Line loss or power loss = 2 I2 R watts.
2. D.C Two wire system with mid point earthed:
Pic13
Vm is the max.potential difference between
conductor and earth in volts.

I1 is current in amperes, R1 is the resistance of the
conductor in ohms.

Line voltage = 2 Vm

Power transmitted = P = 2 Vm I1 watts.

Line loss or power loss = 2 I12 R1 watts.
Equating for equal power

2 Vm I1= Vm I

I1 = I/2

Equating for equal power loss

2 I12 R1 = 2 I2 R

R / R1 = I12 / I2

= (I / 2)2 / I2

= 1/4 = a1 / a
Amt. of cond. Mat. Req. for D.C 2wire sys. with mid point earthed
Amt. of cond. Mat. Req. for D.C 2wire sys. with one end earthed

=1/4
A.C SYSTEM
Single phase two wire system with one conductor
earthed:
pic14
Vm is the max.potential difference between
conductor and earth in volts.

I2 is current in amperes, R2 is the resistance of the
conductor in ohms and power factor =

Rms voltage =

Power transmitted = P = I2 watts.

Line loss or power loss = 2 I22 R2 watts.
Amt. of cond. Mat.Req.for A.C 2wire sys.with one cond.earthed
Amt. of cond. Mat.Req. for D.C 2wire sys. with one end earthed
. Single phase two wire system with mid point
earthed:
pic15
Vm is the max.potential difference between
conductor and earth in volts.

I3 is current in amperes, R3 is the resistance of the conductor in ohms and power factor =


Rms voltage = 2 =


Power transmitted = P = I3 watts


Line loss or power loss = 2 I32 R3 watts.
Equating for equal power, I3 = Vm I

I3 = I /

Equating for equal power loss, 2 I32 R3 = 2 I2 R

R / R3 = I32 / I2 = 1 / 2 = a3 / a
Amt.of cond. Mat. Req.for A.C 2wire sys.with mid point earthed
Amt.of cond. Mat. Req. for D.C 2wire sys. with one end earthed
3. Three phase three wire system:
Pic16
Vm is the max.potential difference between
conductor and earth in volts.

I5 is current in amperes, R5 is the resistance of the
conductor in ohms and power factor =

Rms voltage =

Power transmitted = P = 3 I5 watts.
Line loss or power loss = 3 I52 R5 watts.
Equating for equal power, 3 I5 = Vm I


I5 = (I / 3)

Equating for equal power loss, 3 I52 R5 = 2 I2 R


R / R5 = 3 I52 / 2 I2 = 1 / 3 = a5 / a
Amt.of cond. Mat.Req.for A.C 3- 3wire system
Amt.of cond Mat. Req.for D.C 2wire sys.with one end earthed
CONDUCTOR MATERIAL REQUIRED FOR UNDERGROUND CABLES FOR VARIOUS SYSTEMS

D.C system:
1.D.C two wire system
Pic17
Vm is the max.potential difference between
conductor and earth in volts.

I is current in amperes, R is the resistance of the conductor in ohms.

Line voltage = Vm
Power transmitted = P = Vm I watts.
Line loss or power loss = 2 I2 R watts.
2.D.C Two wire system with mid point earthed:
Pic18
Vm is the max.potential difference between
conductors.

I1 is current in amperes, R1 is the resistance of the
conductor in ohms.

Line voltage = Vm

Power transmitted = P = Vm I1 watts.

Line loss or power loss = 2 I12 R1 watts.
Equating for equal power

Vm I1= Vm I

I1 = I

Equating for equal power loss

2 I12 R1 = 2 I2 R

R / R1 = I12 / I2

= (I )2 / I2

= 1 = a1 / a
Amt.of cond.Mat.Req.for D.C 2wire sys. with mid point earthed
Amt.of cond.Mat. Req. for D.C 2wire system.
A.C SYSTEM
1. Single phase two wire system
Pic19
Vm is the max.potential difference between
conductors.

I2 is current in amperes, R2 is the resistance of the conductor in ohms and power factor =

Rms voltage =

Power transmitted = P = I2 watts.

Line loss or power loss = 2 I22 R2 watts.
Equating for equal power, I2 = Vm I


I2 = I /

Equating for equal power loss, 2 I22 R2 = 2 I2 R


2 ( I / )2 R2 = 2 I2 R

R / R2 = 2 / = a2 / a

Amt.of cond.Mat.Req.for A.C 2wire system.
Amt.of cond.Mat. Req. for D.C 2wire system.
2. Single phase two wire system with mid point
earthed:
pic20
Vm is the max.potential difference between
conductors.

I3 is current in amperes, R3 is the resistance of the conductor in ohms and power factor =

Rms voltage =

Power transmitted = P = I3 watts.

Line loss or power loss = 2 I32 R3 watts.
Equating for equal power, I3 = Vm I


I3 = I /

Equating for equal power loss, 2 I32 R3 = 2 I2 R


2 ( I / )2 R3 = 2 I2 R

R / R3 = 2 / = a3 /
Amt.of cond.Mat.Req.for 1 phase A.C 2wire system.
Amt.of cond.Mat. Req. for D.C 2wire system.
3. Three phase three wire system:
Pic21
Vm is the max. potential difference between
conductors.

I4 is current in amperes, R4 is the resistance of the conductor in ohms and power factor =


Max voltage between conductor to neutral =
Rms voltage = =


Power transmitted = P = 3 I4 watts.

Line loss or power loss = 3 I42 R4 watts.
Equating for equal power, 3 I4 = Vm I


I4 = I / 3

Equating for equal power loss, 3 I42 R4 = 2 I2 R


3 ( I / 3 )2 R4 = 2 I2 R

R / R4 = 1 / = a4 / a
Amt.of cond.Mat.Req.for A.C 3 phase 3 wire system.
Amt.of cond. Mat. Req. for D.C 2wire system.
COMPARISON OF VARIOUS SYSTEMS OF TRANSMISSION
(The ratio of conductor material required in any system compared with that in the corresponding 2 wire D.C system)
S.no System max.voltage to earth max.voltage bet.cond.
D.C
1. Two wire 1 1
2. Two wire mid point earthed 0.25 1
A.C
1. Single phase two wire
2. Single phase two wire
mid point earthed
3. Three phase three wire
EFFECT OF SUPPLY FREQUENCY AND VOLTAGE ON TRANSMISSION LINES
Supply frequency:
• Frequency must be maintained constant so that A.C
motors do not change in their speeds.

• Almost all large systems the frequency is kept
constant so that synchronous clocks may be operated
directly from the power mains.
• In any country one common supply frequency is adopted to facilitate the interchange of energy between different generating stations and substations.

• Electric lights do not operate satisfactorily at lower frequencies and gives flickering
• At lower frequencies in transformers the no. of flux linkages required is more. Hence more turns an cost. The 25 Hz. Transformers cost is 40% to 50% more than 50 Hz. Transformers
• At Higher frequencies say 50 Hz the reactance of lines and machinery is two times more than at 25 Hz frequency. Due to this poor regulation will occur.
• In case A.C motors, change of frequency causes
change of speed
• From the above effects it is desirable to choose constant frequency of 50 Hz or 60 Hz
Supply Voltage:
• In transmission lines more troubles are due to high currents compared to high voltages

• Some times lower maintenance and operating costs may obtained at higher transmission voltage

• The size of the conductor reduced at high voltages.
• At high voltages spacing between conductors
increases which increases the cost of line supports
EFFECT OF VOLTAGE ON LINE EFFICIENCY

• If the voltage is raised ‘m’ times, the current in the line
required to transmit the same amount of power is reduced ‘m’
times. Line losses I2R are reduced ‘m2’ times and hence line
efficiency is increased.
EFFECT OF VOLTAGE ON LINE DROP

Now Er = Es– IR, where R is the resistance of
the both conductors.
Pic22
is the percentage (%) voltage drop in both the conductors expressed as the %of the sending end voltage.

Therefore the voltage is increased the efficiency is increased for a given line drop.
EFFECT OF VOLTAGE ON VOLUME OF CONDUCTOR MATERIAL
• Increase of voltage by ‘m’ times the current is decreased by
‘m’ times and line loss is decreased by ‘m2 ‘ times.

• To transmit with the same loss as before, the resistance of
the line can be increased ‘m2 ‘ times that is the volume of
the conductor material can be decreased ‘m2 ‘ times.
• For any given line, transmission efficiency and given amount of power to be delivered, the volume of the conductor material is inversely proportional to the square of the voltage and square of the power factor.
volume of conductor material is
A X l ∞ 1/(E2cos2Ф)
Where A = cross section of the conductor
l = length of the line
E = phase voltage
cosФ = power factor (p.f)

Hence there is a great saving of conductor material by employing a high voltage.
EFFECT OF VOLTAGE ON TRANSFORMERS, INSULATORS, SWITCHGEARS AND SUPPORTS
High transmission voltage should
• Increases cost of insulators.

• Increases cost of transformers, switchgears and terminal apparatus.

• Increases cost of supports.
EMPIRICAL FORMULA FOR TRANSMISION VOLTAGE
• The economical transmission voltage between lines in
three phase A.C system is
pic23
Where V = Line voltage in K V
P = Maximum K W per phase to be
delivered to single circuit.
l = Distance of transmission in Km.
• Voltages normally adopted for transmission are
Pic24
IMPORTANCE OF HVDC TRANSMISSION
• With the development of thyristors handling large currents and voltages and associated high voltage D.C transmission (H.V.D.C) technology, it has been possible to use H.V.D.C.
• H.V.D.C Transmission having distinct superiority
over A.C Transmission particularly for
• Long distance bulk power transmission

• System inter-connection between two or more
networks

• Submarine cables
Features of H.V.D.C Transmission
• HVDC power transfer can be controlled
quickly and accurately
• No problems of reactive power flow
• Voltage fluctuations are less
• Transmission losses are less
• It requires costly and complex substations and controls
• From above facts, HVDC Transmission are
considered for bulk power transmission over long
distances, Submarine cable and system
inter-connections only
REQUIREMENTS OF CONDUCTOR MATERIAL
Main requirements of conducting material are
• High electrical conductivity.

• High tensile strength.

• Low specific gravity

• Economical cost and long life

• Low specific resistance
• Should not be brittle.

• Non corrosive.

• Should be flexible

• Free from oxides

• Should withstand high wind pressure

• Should have smooth surface to reduce corona
TYPES OF TRANSMISSION LINE CONDUCTOR
1. Solid conductors.

2. Stranded conductors.

3. Hallow conductors.

4. Bundled conductors.
SOLID CONDUCTORS
• Solid conductors are mainly available in the form of
solid copper and copper clad steel.
Pic25
• Over weight as compared to other conductors

• Solid wires except of smaller sizes are difficult to handle and transport.

• Skin effect is more

• They are only used for smaller cross sections.

• Solid aluminum conductors are not used
because of mechanical reasons.
STRANDED CONDUCTORS
• Standard conductors have greater flexibility and
mechanical strength as compared to single
conductor of same cross section.
Pic26
• Stranded conductors usually have a central wire around
which there are successive layers of 6, 12, 18, 24… wires.

• The number of strands is given by the formula
N = 3n(n+1)+1
where ‘n’ is the number of layers.
• Diameter of the stranded conductor can be determined by
D = (2n+1)d
where ‘d’ is the diameter of the each stranded.
‘n’ is the no. of layers

• In manufacture process, adjacent layers are
spiraled in opposite directions, so that the
layers are bound together.

• Stranded conductors are easier to handle
and transport.
• Standard conductors are mainly available in the
following forms
• ACSR ( Aluminium Conductors Steel Reinforced )
• AAC ( All Aluminium Conductors )
• AAAC (All Aluminium Alloy Conductors )
• ACAR (Aluminium Conductors Alloy Reinforced )
• Out of the above ACSR conductors are used for
transmission lines
HALLOW CONDUCTORS
• Hallow conductors gives large diameter.
Pic27
• Tensile strength and flexibility are less.

• Corona loss is reduced.

• Skin effect is less.

• Major portion of the conductor surface is exposed to
atmosphere causes more wind pressure and too much
ice formation on surface
BUNDLED CONDUCTORS
• Generally there is no difference in construction between
stranded and bundled conductors.


• Only difference is Bundled conductors have very large
number of strands than stranded conductors.
Relative merits between solid, stranded
and hallow conductors
pic28
CURRENT DISTORTION EFFECTS

• When a conductor is carrying a steady current (D.C) it will be distribute uniformly over the whole cross section.

• In practice the alternating current does not distribute
uniformly over the whole cross section of the conductor, but
it is distorted due to,
a) Skin, b) Proximity and c) spirality effects.
SKIN EFFECT


• The distribution of alternating currents through a conductor is not uniform, because the current density increases from the middle towards the outer layers.

• Due to A.C flux, the reactance of the central core of the conductor increase, therefore the impedance increases.

• Hence current confines to the upper layers of the conductor.
• This crowding of current towards the surface of the
conductor is known as skin effect.
Pic29
EFFECTS OF SKIN EFFECT
• At lower frequencies the effect is very small.

• For higher frequencies or solid conductors of large
diameter, the effect is considerable.

• Due skin effect the effective area of conductor through
which current flow is reduced, which increases resistance.

• This effect is much smaller for stranded conductor.

• This effect facilitate to use hallow conductors.
PROXIMITY EFFECT
• Consider two conductors placed side by side one is
going and other one is for return of current.

• The fluxes are not distribute uniformly, the currents are
also not distribute uniformly, so that the currents are
crowded to one side of the conductors.
• This effect is known as proximity effect as shown.
Pic30
Concentration of current due to conductor B on A
EFFECTS OF PROXIMITY EFFECT
Like skin effect, the proximity effect increases the resistance and decreases the reactance of the conductor.

The effect is more for a)Large diameter conductors,
b) high frequencies and c) closeness of the conductors.

At normal frequency and for wide spacing
of conductors, this effect is negligible.

4. This effect is eliminated for stranded conductors.
SPIRALITY EFFECT
• This effect is associated only for stranded conductors,
where the effecting length of the conductor is increased,
which is known as spirality effect.

• The magnitude of this effect depends on the size and
method of construction of the conductor.
EFFECTS OF SPIRALITY EFFECT
• Due to this effect the effective length of the
conductor increases.

• Spirality effect increases both resistance and internal
reactance of the standard conductors

• This effect is very small at normal frequency such as at
50 Hz frequency and can be ignored in the case of non-
magnetic conductors.
PROPERTIES OF THE CONDUCTOR MATERIAL OF VARIOUS TYPES
Properties Copper Aluminum Steel Alu.(1.3)&Steel(1.4)

Sp.gravity 8,900 2,700 7,860 3,450 3,700
Ult.tensile 40 18 40 – 320 120 120
str.(kg/mm2)
Sp.res.at 200C 0.01786 0.0287 0.178
Ω-m/mm2(ρ)
Res.temp.(α) 0.0038 0.004 0.00496
Coe.at 200C
PROPERTIES OF SOLID COPPER (HARD DRAWN)CONDUCTOR

SWG no. Dia.of (d) Nominal (mm2) Ct.rat.for temp.rise. (A)
conductor. Conductor. X - Sect.area(a) 27.80C 55.50C

10 3.251 8.30 37 50

9 3.658 10.51 44 60

8 4.064 12.97 52 70

7 4.470 15.70 58 81
SWG no. Dia.of (d) Nominal (mm2) Ct.rat.for temp.rise. (A)
conductor. Conductor. X - Sect.area(a) 27.80C 55.50C

6 4.877 18.68 68 92

5 5.385 22.77 78 107

4 5.893 27.27 91 122

3 6.401 32.18 103 139

2 7.010 38.60 117 158
SWG no. Dia. of (d) Nominal (mm2) Ct.rat.for temp.rise. (A)
conductor. Conductor. X - Sect.area(a) 27.80C 55.50C
1 7.620 45.60 133 181

0 8.230 53.19 150 203

00 8.839 61.36 167 226

000 9.449 70.72 184 250

0000 10.160 81.07 205 278
The resistance of conductor/m can be calculated both at 27.80C and 55.50C is

• ρ27.8 = ρ20 (1+ α20 (27.8 – 20)
• R27.8 = ρ27.8 (l /a) or R55.5 = ρ55.5 (l /a)

• ρ55.5 = ρ55.5 (1+ α20 (55.5 – 20)

‘a’ is given in the table and take ‘l’ = 1 metre
A.C.S.R CONDUCTOR

Code No. and dia. Dia. of cond. App.Cary.cap.
Word of wire in mm in amp.at
400C 450C

Squirrel 6/2.11 6.33 115 107

Gopher 6/2.36 7.08 133 123

Weasel 6/2.59 7.77 150 139
Code No. and dia. Dia.of cond. App.Cary.cap.
Word of wire in mm in amp.at
40C 45C

Ferret 6/3.00 9.00 181 168

Rabbit 6/3.55 10.05 208 193

Mink 6/3.66 10.98 234 217

Beaver 6/3.99 11.97 261 242

Raccon 6/4.09 12.27 270 250

Otter 6/4.22 12.66 281 260

Cat 6/4.50 13.50 305 283

Hare 6/4.72 14.15 324 300

Leopard 6/5.28 15.84 375 348

Tiger 30/7x2.36 16.52 382 354

Wolf 30/7x2.59 18.13 430 398

Panther 30/7x3.00 21.00 520 482

Lion 30/7x3.18 22.26 555 515

Bear 30/7x3.35 23.45 595 552

Goat 30/7x3.71 25.97 680 630
The resistance of conductor/m can be
calculated both at 40C and 45C.

Here both aluminum and steel wires are there, so
resistance of both are calculated in the same way
given at hard drawn solid copper conductors some
times manufacturer supplied the resistance along
with the conductors.
ALL ALUMINUM CONDUCTORS
(A.A.C)STRANDED.

Code No. and dia. Dia.of cond. App.Cary.cap.
Word of wire in mm in amp.at
40C 45C
Mantis 3/3.00 6.468 116 108

Aphis 3/3.55 7.223 133 123

Weevil 3/3.66 7.894 147 136

Lady bird 7/2.79 8.37 178 165

Ant 7/3.10 9.30 204 189

Fly 7/3.40 10.20 229 212

Blue battle 7/3.65 10.95 252 234

Earwig 7/3.78 11.34 264 245

Grass hopper 7/3.91 11.73 275 255

Clegg 7/4.17 12.51 298 276

Wasp 7/4.39 13.17 318 295

Catter piller 19/3.53 10.59 460 386

Chafer 19/3.78 18.90 504 468

Spider 19/3.99 19.95 540 504

Cockroach 19/4.22 21.10 575 534

Butter fly 19/4.65 23.25 655 608

Moth 19/5.00 25.00 720 660

Locust 19/5.36 26.80 790 734


Maybug 37/4.09 28.63 850 790

Scorpion 37/4.20 29.89 895 830

The resistance of conductor/m can be calculated both at 400C and 450C, in the same way as hard drawn solid copper conductor.


INDUCTANCE OF SINGLE PHASE OVERHEAD LINE
Definition of inductance:
• It is defined as flux linkages per ampere and
Measured in Henry.
Inductance, L= N j / I Henry.
• A single phase line consists of two parallel conductors which form a rectangular loop of one turn.

• A changing magnetic flux is set up when an alternating current (A.C) flows through the conductors.

• This flux links the loop and hence loop
possesses inductance.
• Consider a single phase overhead line consisting of two parallel conductors of radius ‘r’ m spaced ‘D’ m apart as shown in figure below.
Pic31
• In order to find the inductance of conductor we shall
have to consider flux linkages with it.

• Let I be the current in the conductor and Ix be the
current through the ring of thickness ‘dx’ at a
distance ‘x’ from the centre of the conductor.
Current density
In the ring
Pic32
Assuming uniform current density, we have
Pic33
Field strength at a distance x from the center of the conductor
Pic34
Flux through a cylindrical shell of radial thickness dx and
axial length 1 m long
pic35
The linkage of the shell
Pic36
The linkage inside the conductor
Pic37
Consider the flux between the conductors ( Flux external to the conductor)
The field strength at a distance x from the centre
Pic38
Flux linkage out side the conductor
Pic39
Total flux linkage / conductor
= Flux linkage inside the conductor+
Flux linkage outside the conductor
Pic40
Inductance / conductor = Flux linkage / ampere
Pic41
• If the conductor is non-magnetic material, then
Pic42
PROBLEMS:
A single phase line has two parallel conductors 2 meters apart. The diameter of each conductor1.2 cm. Calculate the loop inductance per km of the line.

Solution:
Given data:
spacing of conductor, d = 2 m = 200 cm
Radius of the conductor, r = 1.2/2 = 0.6 cm
Loop inductance per meter length of line
= 10-7 [µr + 4 loge (d/r)] H
= 10-7 [1 + 4 loge (200/0.6)] H
= 24.23 x 10-7 H
Loop inductance per km. length of line
= 24.23 x 10-7 x 1000 H
= 24.23 x 10-4 H
= 2.423 m H
. A single phase line has two parallel conductors 3 meters apart. The radius of each conductor being 1 cm. Calculate the loop inductance per km of the line if the material of the conductor is (i) copper (ii) steel with permeability of 100.

Solution:
Given data:
spacing of conductor, d = 3 m = 300 cm
Radius of the conductor, r = 1 cm
Loop inductance per meter length of line
= 10-7 [µr + 4 loge (d/r)] H
(i) With copper conductor, µr = 1
Loop inductance per meter length of line
= 10-7 [1 + 4 loge (300/1)] H
= 23.8 x 10-7 H
Loop inductance per k.meter length of line
= 23.8 x 10-7 x 1000 H
= 23.8 x 10-4 H
= 2.38 m H
(ii) With steel conductor, µr = 100
Loop inductance per meter length of line
= 10-7 [100 + 4 loge (300/1)] H
= 122.8 x 10-7 H
Loop inductance per km length of line
= 122.8 x 10-7 x 1000 H
= 122.8 x 10-4 H
= 12.28 m H
Inductance of three phase transmission line
Three phase symmetrical spaced conductors: (Equally spaced)
• A,B,C are three conductors equally spaced at distance of ‘D’ meters and radius of the conductors is ‘r’ meters.
Pic43
• Inductance /conductor (phase)/meter

= 10-7[0.5 + 2 loge (D/r] H/m.

• Inductance /conductor (phase)/ Km

= 10-4 [0.5 + 2 loge (D/r] H/ Km.
Three phase asymmetrical spaced conductors: (unequal spacing)
• A,B,C are three conductors unequally spaced at distance of ‘DAB’, ‘DBC’, ‘DCA’, meters and radius of the conductors is ‘r’ meters.
Pic44
Problem:
1) Find the inductance of three phase transmission line using 1.24 cm diameter conductors when these are placed at the corners of an equilateral triangle of each side 2meters. (i) inductance/conductor (phase)/meter and (ii) inductance/conductor (phase)/Km.
• Inductance /conductor (phase)/meter

= 10-7[0.5 + 2 loge [( 3 DABDBCDCA ) / r ]] H/m.

• Inductance /conductor (phase)/ k.meter

=10-4[0.5 + 2 loge [( 3 DABDBCDCA ) / r ]] H/ Km.
Solution:
Given data:
Conductor spacing D = 2 m
= 200 cm pic45
Conductor radius r =1.24/2
= 0.62 cm
• Inductance /conductor (phase)/m

= 10-7[0.5 + 2 loge (200/0.62] H/m.

= 12 x 10-7 H.
• Inductance /conductor (phase)/ Km

= 12 x 10-7 x 1000 H.

= 12 x 10-4 H

= 1.2 x 10-3 H

= 1.2 m H
2) Find the inductance of three phase transmission line using 1.24 cm diameter conductors when these are placed at the corners of triangle of sides 2m, 2.5m, 4.5m.
(i) inductance/conductor (phase)/meter and (ii) inductance/conductor (phase)/k.m.
Solution:
Given data:
Fig. Shows the three conductors A, B, C
of three phase line placed at the corners of triangle of sides 2m, 2.5m, 4.5m.
pic46
DAB = 2 m, DBC = 2.5 m, DCA = 4.5 m.

Radius of conductor, r = 1.24/2 = 0.62 cm

Equivalent equilateral spacing,
Deq = 3 DAB DBC DCA

= 3 2 x 2.5 x 4.5
= 2.82 m
= 282 cm
(i) Inductance /conductor (phase)/meter = 10-7[0.5 + 2 loge [( 3 DABDBCDCA ) / r ]] H/m. = 10-7[0.5 + 2 loge (Deq/r)] =10-7[0.5 + 2 loge(282/0.62) = 12.74 x 10-7 H = 12.74 x 10-7 H (ii) Inductance /conductor (phase)/ Km
=10-4[0.5+2 loge [( 3 DABDBCDCA ) / r ]]H/Km.
=10-4[0.5+2 loge (Deq/r)]
=10-4[0.5+2 loge (282/0.62)] = 12.74 x 10-4 H
= 1.274 x 10-3 H = 1.274 m H.
Need for transposition of over head lines
• Generally the transmission line conductors of three phase three wire system are irregularly spaced.
• Because it is difficult to maintain an exact height of the conductor from the ground on the towers or poles.
• Due to the above the transmission line conductor constants ( R, L, C ) are not the same for all phases
• In order to equalize these line constants, the three lines must be transposed (change the position) in such a way that the three lines occupies a position relatively other wires for l/3 of the total length.
• Fig. Shown below a transposed lines.
Pic47
Effects of transposition
• The transposition in transmission lines will balance the capacitance of the line, so that electro statically induced voltage is balanced in the complete length of the line.

• It is also balance the inductance of the line and hence the electromagnetically induced voltage is balanced in the complete length of the line.

• By proper co ordination of transposition of power and near by communication lines, the induced voltages can be reduced to very small and it also reduces the noise in the near by communication lines.
Inductance in a transposed line
In a three phase transposed line the
(i) inductance/conductor (phase)/m
=10-7[0.5 + 2 loge [(D= 3 DABDBCDCA ) / r ]] H/m.
(ii) inductance/conductor (phase)/Km
=10-4[0.5 +2 loge [(D= 3 DABDBCDCA ) / r ]]H/km.
Here D is known as Geometrical mean distance (GMD).
Problem :
• Find the inductance of three phase transmission line
using 2 cm diameter conductors when these are placed
at the corners of triangle of sides 3.6m, 5.4m, 8.1m. The conductors are transposed at regular intervals.
(i) inductance/conductor (phase)/m and
(ii) inductance/conductor (phase)/Km.
Solution:
Given data:
Fig. Shows the three conductors A, B, C of three phase
line placed at the corners of triangle of sides 3.6m,
5.4m, and 8.1m.
Pic48
DAB = 3.6 m, DBC = 5.4 m, DCA = 8.1 m.

Radius of conductor, r = 2 / 2 = 1 cm

Geometric mean distance,

D = 3 DAB DBC DCA

= 3 3.6 x 5.4 x 8.1

= 5.4 m

= 540 cm
Inductance /conductor (phase)/meter

= 10-7[0.5 + 2 loge (GMD ) / r ] H.

= 10-7[0.5 + 2 loge (540 / 1)] H.

= 13.09 x 10-7 H.

= 13.09 x 10-7 H.
Inductance /conductor (phase)/ Km

= 10-4[0.5 + 2 loge (GMD ) / r ] H.

= 10-4[0.5 + 2 loge (540) / 1) H.

= 13.09 x 10-4 H. = 1.309 x 10-3 H.

= 1.309 m H.
Capacitance
• It is defined as the ratio of the charge in coulombs on a capacitor to he potential difference (P.D) applied across the capacitor is known as capacitance of the capacitor.
• It is measured in Farads (F) and is usually expressed in (micro) µF or pF.
Pic49
Capacitance of single phase over head line:
• Consider a single phase transmission line consisting of
two parallel conductors A and B spaced ‘d’ meters
apart in air.
Pic50
Let their respective charge be + Q and – Q
coulombs per meter length.

Consider at point ‘p’

The electrical intensity due to conductor A

= Q / (2πЄ0Єr x )

The electrical intensity due to conductor B

= Q / (2πЄ0Єr (d - x))
• The total intensity at ‘p’ due to A and B is

Ex = Q / (2πЄ0Єr x ) + Q / (2πЄ0Єr (d - x))

• The p.d between the conductors is equal to the work done in moving one coulomb from one conductor to the other against the electrostatic force.
• If one coulomb is moved a distance dx then the work done is Ex dx joules.

• There fore the P.D between the conductors



V = Q / (2πЄ0Єr) [ ]
V=Q/(2πЄ0Єr)[logex–loge (d - x) ]rd-r

=Q/(2πЄ0Єr){[loge(d-r)-loge(d-d+r)]-[loger–loge (d –r)]}

=Q/(2πЄ0Єr) [loge (d - r) – loge r – loger + loge (d –r)]

=Q/(2πЄ0Єr)[2 loge(d-r)-2 loger]

=2 Q/(2πЄ0Єr)[ loge(d-r) - loger]
V = Q/(πЄ0Єr)[ loge(d-r) - loger]

Neglect r compared to d, then we have

V = Q/(πЄ0Єr)[ loge( d / r)]

Capacitance/loop/meter = charge/P.D = Q/V

= Q/(Q (loge( d / r)) /(πЄ0Єr) = πЄ0Єr/ loge( d / r)
Capacitance/loop/meter = charge/P.D = Q/V

= πЄ0Єr/ loge( d / r) F/m

Capacitance/loop/Km = πЄ0Єr/ loge( d / r) x 103

= [(22/7)x1x8.854x10-12x103]/ loge( d / r)

{ Here, Є0 = 8.854 x 10-12, Єr = 1}
= (2781 x 10-9)/ loge( d / r) F/km

= ( 0.02781) / loge( d / r) μF/km

Capacitance/loop/km = ( 0.02781) / loge( d / r) μF/km

Similarly, Capacitance/conductor/km

= ( 0.0556) / loge( d / r) μF/km
Problems :
1) Calculate the capacitance/km of a pair of parallel conductors,
5 mm in diameter and spaced uniformly 20 cm apart in air.
Solution :
Given data :

Diameter of the conductor = 5 mm
Radius of the conductor, r = 5/2 = 2.5 mm = 0.25 cm
Spacing between conductor, d = 20 cm
Capacitance/loop/km = 0.02781/loge(d/r)

= 0.02781/loge(20/0.25)

= 0.02781/loge(80)

= 0.006342 µ F.
2) calculate the capacitance/km of a pair of parallel
conductors, 10 mm in diameter and spaced uniformly
2 m apart in air.
Solution :
Given data :

Diameter of the conductor = 10 mm
Radius of the conductor, r = 10/2 = 5 mm = 0.5 cm
Spacing between conductor, d = 2 m = 200 cm
Capacitance/loop/km = 0.02781/loge(d/r)

= 0.02781/loge(200/0.5)

= 0.02781/loge(400)

= 0.00464 µ F.
Inductance of three phase transmission line
Three phase symmetrical spaced conductors: (Equally spaced).
• A,B,C are three conductors equally spaced
at distance of ‘D’ meters and radius of the
conductors is ‘r’ meters.
Pic51
Capacitance/conductor/km
=0.0556/loge(D/r) µF
Three phase unsymmetrical spaced conductors: (un equal spacing)
• A,B,C are three conductors unequally spaced at distance of ‘DAB’, ‘DBC’, ‘DCA’, meters and radius of the conductors is ‘r’ meters.
Pic52
Geometrical mean distance

D = 3 DAB DBC DCA

Capacitance / conductor / km

= 0.0556 / loge [( 3 DAB DBC DCA )/ r] µ F
Problem :
1) Calculate the capacitance of the each conductor of a 400 km, three phase line with an equilateral configuration. The spacing between conductors is 3 m and radius of the conductor is 1 cm.

Solution :
Given data :
Length of the line L = 400 km
Spacing between conductor, D = 3m = 300 cm
Radius of the conductor, r = 1cm
Capacitance / conductor / km = 0.0556 / loge (D/r)

= 0.0556 / lobe (300/1)

= 9.747 x 10-3 µ F / km

Total Capacitance / conductor = 9.747 x 10-3 x 400

= 3.898 µ F
Calculate the capacitance / km of a three phase line completely transposed with a triangle configuration. The spacing between conductors are 1 m, 1.25m and 2.125 m. Diameter of the conductor is 1.25 cm.

Solution :
Given data :
DAB = 1m, DBC = 1.25 m, DCA= 2.125m

Geometric mean distance D = 3 DAB DBC DCA

D = 3 1 x 1.25 x 2.125 = 1.385 m =138.5 cm
Diameter of the conductor = 1 cm

Radius of the conductor r = ½ = 0.5 cm

Capacitance / conductor / km

= 0.0556/ loge[( 3 DABDBCDCA ) / r ]

=0.0556/loge(138.5/0.5)

=0.0556/5.624=0.0098µF/km
3) Calculate the capacitance / phase or line to neutral capacitance of a three phase line completely transposed with horizontal configuration. The spacing between conductors are 2 m, 2.5m and 4.5 m. Diameter of the conductor is 1.25 cm.

Solution :
Given data :

DAB = 2m, DBC = 2.5 m, DCA= 4.5m
Geometric mean distance D = 3 DAB DBC DCA

= 3 2 x 2.5 x 4.5

= 2.82 m = 282 cm
Radius of the conductor, r = 1.25 /2 = 0.625 cm
Line to neutral capacitance or
Capacitance / conductor / km

= 0.0556/ loge[( 3 DABDBCDCA ) / r ]

= 0.0556/loge(282/0.625)

= 0.0556/6.112=0.00909µF/km
Resistance, inductance and capacitance of the three phase over head line conductors at different voltages.
(Values apply to double circuit lines, A.C.S.R conductors).
1.Rated voltage kv 20 30 60
2.Nominal x section 35 150 35 150 95 240
Al ( mm2)
3.Diameter of the 8.1 17.3 8.1 17.3 13.4 21.7
conductor (mm)
4.Resistance at 200C 0.87 0.2 .87 0.2 0.32 0.13
(ohm/km)
5.Reactance (2 fl) at 0.39 0.34 0.4 0.35 0.4 0.37
50 Hz (ohm/km)
6.Mutual cap. (10-9 F/km) 9.7 11.2 9.5 10.9 9.5 10.3
7.cap.To earth(10-9 F/km) 3.4 3.6 3.5 3.7 3.8 4.0
Values apply to double circuit lines, A.C.S.R conductors).
1.Rated voltage kv 110 220
2.Nominal x section 120 300 340 2x240
Al ( mm2)
3.Diameter of the 15.7 24.2 28.1 2x21.7
conductor (mm)
4.Resistance at 200C 0.25 0.10 0.09 0.062
(ohm/km)
5.Reactance (2 fl) at 0.41 0.38 0.40 0.32
50 Hz (ohm/km)
6.Mutual cap. (10-9 F/km) 9.3 10 9.5 11.5
7.cap.To earth(10-9 F/km) 4.0 4.2 4.8 6.3
Over head transmission lines
• The transmission line constants Resistance, Inductance
and Capacitance are uniformly distributed along the line
• The transmission lines are mainly classified in to
three types, namely Short, Medium and Long
transmission line
SHORT TRANSMISSION LINE
• The length of the lines less than 80 kms and line voltage is less than 20 kv are termed as short transmission lines.

• Due to smaller length and lower voltage the capacitance effects are very small and hence it can be neglected.
MEDIUM TRANSMISSION LINE
• The length of the lines lies between 80 to 240 kms
and line voltage is lies between 20 to 100 kv are
termed as medium transmission lines.


• Due to sufficient length and line voltage the
capacitance effects are taken into account.
LONG TRANSMISSION LINE
• The length of the lines are more than 240 km and
line voltage is greater than 100 kv are termed as
long transmission lines.

• Here also the capacitance effects are taken into account.
REASONS FOR LUMPED CONSTANTS IN SHORT AND MEDIUM TRANSMISSION LINES
• The transmission line constants resistance (R), inductance (L) and capacitance (C) are proportional to the length of the line and hence they are uniformly distributed along the line as shown in fig.
Pic53
In the above fig.1

Where, r = resistance / unit length

L = inductance / unit length

c = capacitance / unit length

g = conductance / unit length.
• For short transmission lines the effect of capacitance is
very small, hence it may be neglected and other constants are lumped instead of distributed, because the current flowing through the line is same the distributed drops are equal to lumped drops, so they are lumped.


• For medium transmission lines the capacitance is assumed to be lumped at one or more points in the line.
VOLTAGE REGULATION
• When a transmission line carrying current, there is voltage drop in line due to resistance and inductance of the line.

• Due to this, the receiving end voltage (VR) is less than the sending end voltage (VS).

• The voltage drop in the line = VS – VR.

• The ratio of voltage drop (VS – VR) to the receiving end (VR)
is known as voltage regulation.

i e voltage regulation = (VS – VR) / VR.
PERCENTAGE VOLTAGE REGULATION
• The voltage regulation is expressed in percentage (%) then it is known as percentage (%) regulation.


• i. e percentage voltage regulation = pic54
APPROXIMATE FORMULA FOR PERCENTAGE
VOLTAG REGULATION
Pic55
From the vector diagram.

OC is nearly equal to OF

OC = OF
= OA + AF
= OA + AG + GF
= OA + AG + BH

There fore, VS = VR + IR cosΦR+IXL sinΦR ----(i)
But voltage regulation = (VS – VR) / VR

There fore, (VS – VR) = IR cosΦR+IXL sinΦR

Voltage regulation = (VS – VR) / VR

= (IR cosΦR+IXL sinΦR) / VR
• Percentage (%) voltage regulation for lagging p. f.

= pic56



• Percentage (%) voltage regulation for leading p. f.

= pic57
CLASSIFICATION OF TRANSMISSION LINES
• Short transmission lines (STL).
• Medium transmission lines (MTL).
• Long transmission lines (LTL).
SHORT TRANSMISSION LINES
● OH line up to 80km
● Voltage level > 20 kv
● Capacitance effect negligible
● Only Resistance & Inductance are taken into account
MEDIUM TRANSMISSION LINE
● Line Voltage is high >20kv and <>100kv
● Line constants are considered uniformly distributed
● OH line is more than 160 km
SHORT TRANSMISSION LINE
• Line Diagram:
Pic58
PHASOR DIAGRAM
I = Load Current in amps
R = Loop Resistance
XL=Loop Reactance
VR=Receiving end voltage pic59
Vs=Sending end voltage
SENDING END VOLTAGE(VS)
From the phasor diagram …The right angled triangle ODC
Pic60
VOLTAGE REGULATION:
Pic61
EFFICIENCY :
Receiving end power = VR IR CosFR
Sending end power = VR IR CosFR + I2R
Line losses = I2R
Transmission efficiency = pic62

TUTORIAL PROBLEMS
1) A 3 Phase line delivers 3600 KW at a power factor 0.8 lagging to a load when the sending end voltage is 33KV if Resistance & Reactance are 5.31 ohm and 5.5 ohms respectively.
Determine
a) The receiving end voltage
b) Line current
c) Transmission efficiency
Given data:
Power/phase = 3600/3 = 1200KW
Resistance… R = 5.31ohms
Reactance… X = 5.54ohms
Power factor. CosøR = 0.8
Sending end voltage Vs=33000/√3=19052
To be found:
Receiving end voltage VR=?
Line current, IL=?
Transmission Efficiency htr =?
Power delivered/phase P = VR IR cosøR

I = P / VR cosøR

I=(1200*1000)/VR*0.8

I=(150*105)/VR

Sending end voltage Vs = VR+IRcosøR +IXSinøR

19052=VR + (15*105 *5.31 * 0.8)/VR+ (15*105 *5.54 * 0.6)/ VR

VR2-19052 VR+11358000=0

By solving above equation

VR=18435 volts
Power delivered/phase P = VR IR cosøR

I = P / VR cosøR

I=(1200*1000)/VR*0.8

I=(150*105)/VR

Sending end voltage Vs = VR+IRcosøR +IXSinøR

19052=VR + (15*105 *5.31 * 0.8)/VR+ (15*105 *5.54 * 0.6)/ VR

VR2-19052 VR+11358000=0

By solving above equation

VR=18435 volts
a) Line voltage at receiving end

VR = √3* x18435=31930.volts

b) Line current…,I=15x105 / 18435 = 81.36Amps

Line losses = 3 I2R=3x81.362 x 5.31

=105447 watts

c) Transmission efficiency
=[Receiving end power/(Receiving
end power +line losses)]*100

=[3600/(3600+105.447)]*100
2) A load of 4000kw at 1100 volts is being received from
a single phase transmission line at a p.f of 0.9 lag if
R=0.18 ohms & X= 0.02ohms respectively. Neglect
capacitance of the line. Calculate
a) Sending end voltage
b) Current
c) Power factor
Solution:
Given data:
Load P = 4000 Kw
VR= 1100volts
R = 0.018 ohms
XL = 0.02 ohms
IR = P / VR cos F R
IR= [(4000*1000)/(1100*0.9)]
=4040Amps
Cos øR=0.9, Sin øR=0.4352

Resistance for Go and Return conductor
R=2x0.018 = 0.036ohms

Reactance X = 2x0.02 = 0.04ohms
VS2 =(VR cos F R + IR)2 + (VR sin FR +IX )2
= (1100X0.9+4040X0.036)2+(1100X0.4352+4040X0.4)
= 5677075.8
VS = 2382.66
• Sending end p.f cos Fs = pic63
= 0.47 lag
Medium Transmission Line(Nominal Π(pie) method)
• Line Diagram:
Pic64
As shown in the figure the following points can be observed
• In the above method capacitance of each conductor
(L to N) is divided in two halves.
• Half of the capacitance is lumped at the sending end
and other half at the receiving end
• Capacitance at sending end has no effect on the line drop
• Charging current must be added to line current to obtain
sending end current
As per line diagram

I= Load Current in amps
R= Loop Resistance
XL=Loop Reactance
VR=Receiving end voltage
Vs=Sending end voltage
C= Capacitance/Phase
COS Fr = receiving end p.f
COS Fs = sending end p.f
PHASOR DIAGRAM (Π method)
Pic65
Sending End Voltage
Pic66
Sending End Current
Pic67
Medium Transmission Line (Nominal T Method)
Line diagram
Pic68
Nominal T Method
• Whole line capacitance is assumed to be concentrated at the middle point of the line
• Half the line resistance and reactance are lumped on its either side
• Full charging current flows over half the line
• One phase of 3 phase transmission line is shown in the line diagram
PHASOR DIAGRAM (Nominal T method)
Pic69
Sending End Current :
Pic70
Sending End Voltage:
Pic71
EXERCISE PROBLEMS
• A 3-phase, 50-Hz Overhead transmission line 100km long has the following constants:
Resistance/Km/Phase = 0.1 ohm
Inductive Reactance/km/phase = 0.2 ohms
Capacitive susceptance /km/phase = 0.04X10-4 siemen
Determine a) the sending end current b) sending end voltage c) sending end power factor d) transmission efficiency, when supplying a balanced load of 10,000 KW at 66kv, pf =0.8 lagging Solve in T method
SOLUTION
Given Data:
Total Resistance / Phase R = 0.1X100=10 ohms
Total Reactance / Phase XL= 0.2 X 100 = 20 ohms
Capacitive Susceptance Y = 0.04 X 10-4 X 100
= 4 X 10-4
Receiving end Voltage / Phase VR= 66000 / √3 = 38105 V
Pic72
MEDIUM TRANSMISSION LINE (T METHOD)
Line Diagram:
Pic73
PHASOR DIAGRAM (T method)
Pic74
Receiving end voltage, VR = V R + j0 = 38,105 V
Load current IR = I R ( cosΦR – j sin Φ R )
= 109 ( 0.8 – j 0.6 )
= 87.2 – j 65.4
Voltage across C, V1 = VR + IR Z/2
= [38,105 + ( 87.2 – j65.4) (5+j10)]
= 39,125 + j545
Charging current, IC = jYV1
= j4 X 10-4(39,195 + j 545)
= 0.218+j15.6
Sending end current, IS = IR +IC
= (87.2-j65.4)+(-0.218+j15.6)
= 87.0-j 49.8= 100 -29047/
Sending end current = 100A
Sending end voltage = VS = V1+ IS Z/2
= (39,195 + j545)+(87.0-j 49.8)(5+j10)
=40128 + j1170 = 40145 1040/
Line value of sending end voltage = √3x40145
= 69533 V
= 69.533 kv Referring to phasor diagram in Fig.

θ1= angle between VR and VS = 1040/
θ2= angle between VR and IS = 29047/
ΦS= angle between VS and IS = θ1+ θ2 = 31027/

Sending end power factor
Cos ΦS = Cos 31027/
= 0.853 lag
Sending end power = 3VSIS cos ΦS
= 3X40,145X100X0.853
= 10273105 W
=10273.105 kW

Power delivered =10,000 kW
Pic75
• A 3-phase, 50-Hz Overhead transmission line 150km long has the following constants:
Resistance/Km/Phase = 0.1 ohm
Inductive Reactance/km/phase = 0.5 ohms
Capacitive shunt admittance /km/phase=3X10-6 siemen
Determine a) the sending end current b) sending end voltage c) sending end power factor when supplying a balanced load of 50 MW at 110kv, pf =0.8 lagging Solve in Π method
SOLUTION
Given Data:
Total Resistance / Phase R = 0.1X150=15 ohms
Total Reactance / Phase XL= 0.5 X 150 = 75 ohms
Capacitive Admittance/ Phase Y = 3 X 10-6X 150 = 45 X 10-5
Receiving end Voltage / Phase VR = 110X1000 / √3 = 38105 V
Pic76
Taking receiving end voltage as the reference phasor , we have
Pic77
Pic78
Pic79
CHARGING CURRENT (IC)IN TRANSMISSION LINES
• The capacitance effect is considered in over head
lines over 100km and above.
• Due to capacitance effect current produces is usually called charging current
• Charging current flows in quadrature with the voltage
• Charging current flows through the line even though
receiving end is open circuited
• It has maximum value at sending end and decreases
continuously as the receiving end is approached
IC=VRY
• Due to the charging current there will be power loss in
the line
POWER LOSS DUE TO CHARGING CURRENT
• The Value of charging current at a point P “x” km from the
sending end. Total length of line “l” km as shown below fig.
pic80
pic81
pic82
FERRANTI EFFECT
• In long transmission line when the line is open circuited or very lightly loaded , a rise of voltage occurs at the receiving end is known as ferranti effect.
• The above pressure rise is due to the emf of self inductance of the charging current.
• To determine the magnitude of pressure rise one half of the total line capacitance will be assumed to be concentrated at the receiving end
DERIVATION OF RISE IN VOLTAGE AT RECEIVING END
Line Diagram
Pic83
VECTOR DIAGRAM:
Pic84
With reference to the above vector diagram the sending end voltage is less than the receiving end voltage
Neglecting Resistance drop
Rise in voltage=OP-OQ=ICX
IC=VRY/2
Pic85
Pic86
CORONA IN TRANSMISSION LINES:

• When the voltage between the conductors of a overhead line exceeds a certain value, The air surrounding it is broken & ionization starts.

• The voltage at which ionization starts is called disruptive critical voltage ( Vd)
• If the voltage exceeds the disruptive critical voltage, a hissing noise is heard and the conductors are surrounded by a luminous envelope.

• The voltage at which this occurs is called visual critical voltage (Vv ).

• The phenomena of the hissing noise,the violet glow and production of ozone is called corona
DEFINITION OF CORONA:

• Corona is due to the bombardment of air molecules with subsequent dislodging of electrons, by ionised particles.
OR
• The phenomenon of violet glow,hissing noise & production of ozone gas in an overhead transmission line is known as corona.
• If the conductors are uniform and smooth, same conditions will hold good through out the length of the conductor.

• If the conductors are not uniform & smooth rough points will appear brighter

• In case of DC system the positive conductor will have a uniform glow & will brighter than negative one.
• The luminous blue envelope is nothing but breakdown of air into ions under high electrostatic stress.

• The breakdown starts first near the surface of the conductor, because the electrostatic stress is maximum at the surface.

• The thickness of the conducting layer of air increases with the increase of supply voltage.
FACTORS AFFECTING CORONA
Corona is affected by the following factors, namely

1. Atmosphere:
• In stormy weather conditions, the no. of ions in
the atmosphere is more as compared to fair
weather conditions.
• Hence corona will occur in stormy weather at
much less voltage compared with to fair weather.
2. Conductor:
• Corona decreases with increase in diameter
of the conductor
• Corona is more in standard conductors as
compared to solid conductors.
• Corona is more in rough and dirty surface of the
conductors as compared to smooth conductors.
3. Spacing between the conductor:
• If the spacing between the conductors is large
as compared with their diameter, there is no
formation of corona.
4. Line voltage:
• At low voltage, there is no Corona, but at
higher line voltages corona effect will appear
Disruptive critical voltage

The minimum phase to neutral voltage at which corona occurs is known as disruptive critical voltage

It is calculated by the following equation :

In case of two parallel conductors the max potential gradient is given by

g max = Vc / r loge (d/r)
where Vc = The r.m.s value of voltage of
the conductor to neutral

r = The radius of the conductor in cm.

d = The spacing of conductor in cm.

• The breakdown strength of air at 760mm pressure
and temperature of 25oC is 30kv/cm (max) or
21.1 kV/cm (r. m. s) and is denoted by go
• The value of go depends on the density of air
• The dielectric strength of air is proportional to its density over a wide range

• The dielectric strength of air is directly proportional to the barometric pressure ‘ b ’ & inversely proportional to the absolute temperature ‘ t ’.
The breakdown strength at a barometric pressure of ‘b’ cm mercury and temperature of toc becomes δ g0 where

δ =3.92b / ( 273+t )

If Vdo is the voltage to the neutral at which breakdown
occurs, then

VC =go δ r loge (d/r) kv(rms)to neutral
In practice it is necessary to allow for the condition of
surface of the wire so that

VC =go δ mo r loge(d/r)

Where
mo = roughness factor or irregularity factor
= 1 for smooth and clean wires
= 0.98 to 0.93 for rough wires
= 0.87 to 0.8 for stranded wires
g = Max potential gradient or breakdown value of air
= 30kV(max)/cm or 21.2kV(r.m.s)/cm.
δ = density of the air surrounding the conductor
= 3.92b/273+t 0 c b is the pressure in cm of mercury
r = radius of the conductor in cm.
d = distance between the conductor in cm.
Therefore
V c =21.2 mc δ loge(D/r) kV (r.m.s) to neutral.
POWER LOSS DUE TO CORONA
• Formation of corona is always accompanied by energy loss which is dissipated in the form of
Light, Heat, Sound and Chemical action.

• The formation of corona is associated with loss of power, which will have some effect on the efficiency of the line.
• Corona will not have any appreciable effect on the line regulation.
• The surface voltage gradient at line conductor exceeds the critical breakdown stress,corona appears and energy is dissipated in the form of light and heat.

• The energy loss dissipated in the form of light and heat is known as corona power loss and this is derived from the AC network reducing the transmission energy.
The power loss due to corona is given by
Pic87
Where f = Supply frequency in Hz

V = Phase neutral voltage in r.m.s

VC = Disruptive critical voltage r.m.s per phase
EFFECTS OF CORONA
The effects of corona are

1.There is a definite dissipation of power , although it is
not so important except under abnormal weather
conditions like storm etc.

2. Corrosion due to ozone formation.

3. The current drawn by the line due to corona loss is
4. This may cause interference with neighboring
communication circuits.

5. If corona is present the effective capacitance of the
conductor increases.

6. Corona works as a safety valve for surges.
7. Corona discharge reduces the dielectric strength .
8. The reduction in dielectric strength makes easier for
the flash over between insulators and conductors.
9. Transmission efficiency is reduced due to corona
METHODS OF REDUCING CORONA
• Corona effects are observed at a working voltage of 33kV or above.Therefore , careful design should be made to avoid corona

• Corona effects are observed mainly at the substations or on busbars rated for 33kV and higher voltages

• Otherwise highly ionised air due to corona may cause flashover in the insulators or between the phases, causing considerable damage to the equipment
• The effects of corona can be decreased either by increasing the size of the conductor or the space between the conductors.
1. By increasing conductor size:
• By increasing conductor size, the voltage at
which corona occurs is raised and hence
corona effects are considerably reduced

• This is one of the reasons that ACSR conductors
which have a large cross sectional area are used
in the transmission lines .For E.H.V. Lines,
Bundled conductors are used
2. By increasing conductor spacing:
• By increasing spacing between the conductors,
the voltage at which corona occurs is raised .
• If the distance between the conductors is increased
the tower costs and reactance drop increases.
• It is not advisable to increase the spacing.
Hence size of the conductor is increased .
• In a special conductor construction , one or more layers of copper strands are spiralled round a core of twisted copper I beam which is shown in the fig.1
Pic88
• Thus by means of I beam core,strength is added to the hollow conductor without the addition of deadweight.
• Another design which in use consists of number of tongued and grooved rectangular copper sections. Which spiralled along the length of the conductor to form a hollow tube as shown in the fig.2
Pic89
APPLICATIONS OF HOT LINE TECHNIQUE
• The following are main applications of hot line technique

1. Changing of broken and flashed out discs
2. Connecting the temporary jumpers to the line
3. Testing of transformers, bushings and other
equipments for leakage.
4. Polarity test on 11KV and 33Kv feeders for
parallel operation
5. For removing and replacement of cross arms
6. Maintenance of contaminated insulators.
HOT LINE TECHNIQUE
• In hot line technique, the line men work on the
energized transmission line with safety for repair of
fitting and equipments on the pole
In hot line technique, the tools must be lighter and stranger
Wooden species are mainly used as a hot line tools.
Wooden species are mainly used as a hot line tools.
Generation of Electrical Energy
• Energy available in the nature in different forms.
• Conversion of available energy into Electrical energy is
known as generation of electrical energy.
ENERGY FROM SOURCE
Pic90
NUCLEAR POWER PLANT
Pic91
SOLAR PANELS
Pic92
Hydro Power Plant
Pic93
WIND MILLS
Pic94
WIND POWER
Pic95
WAVE POWER
Pic96
HYDRO POWER PLANT
Pic97
NUCLEAR POWER PLANT.
Pic98
WIND POWER
Pic99
ELECTRIC POWER TRANSMISSION
Pic100
AC POWER TRANSMISSION
• AC power transmission is the transmission of electric power by alternating current.
• Usually transmission lines use three phase AC current. Single phase AC current is sometimes used in a railway electrification system.
• In urban areas, trains may be powered by DC at 600 volts.
• Electricity is usually transmitted over long distance through overhead power transmission lines.
• Underground power transmission is used only in densely populated areas due to its high cost of installation, maintenance, because the high reactive power produces large charging currents and difficulties in voltage management.
• Today, transmission-level voltages are usually considered
to be 110 kV and above.
• Lower voltages such as 66 kV and 33 kV are usually
considered sub-transmission voltages but are occasionally
used on long lines with light loads.
• Voltages less than 33 kV are usually used for distribution.
• Voltages above 220 kV are considered extra high voltage
and require different designs compared to equipment used at lower voltages.
BULK POWER TRANSMISSION
• Transmission efficiency is improved by increasing the voltage using a step-up transformer, which reduces the current in the conductors.
• The reduced current flowing through the conductor reduces the losses in the conductor.
• since, according to Joule's Law, the losses are proportional to the square of the current, halving the current makes the transmission loss ¼th the original value.
LOSSES
• Transmitting electricity at high voltage reduces the fraction of energy lost to Joule heating.
• For a given amount of power, a higher voltage reduces the current and thus the resistive losses in the conductor.
• Long distance transmission is typically done with overhead lines at voltages of 115 to 1,200 kV. However, at extremely high voltages, more than 2,000 kV between conductor and ground, corona discharge losses are so large that they can offset the lower resistance loss.