EE2303 Transmission & Distribution Syllabus

EE2303
TRANSMISSION & DISTRIBUTION
V Sem, BE (EEE)
Regulation 2008
Dr.P.Valsalal
Associate Professor
DEEE, Anna University
Chennai 600 025
1
Transmission & Distribution

Syllabus
UNIT I: Introduction (9 Hrs)
2

- DC Transmission (1880s)
- AC Transmission
- HVAC
- HVDC –world’s first HVDC –Gotland
scheme in Sweden in1954(20MW,
96km cable sea return), -100kV
Historical Background
3

Structure of electric
Power System
-Generation
-Transmission
-Distribution
4

Generation
-Conventional
-Non-conventional
5

Conventional
-Thermal power station
-Nuclear power station
-Hydro-electric power station
6

Power Plants - Locations
Thermal - Neyveli, Ennore, Tuticorin
Hydro - Kundha, Pykara,
Kadamparai
Nuclear - Kalpakkam
7

8
Environmental Problems

Non-conventional
- Wind
- Solar
- Tidal
- Geothermal
- Bio-mass
9

Power System Network
10

Transmission / Distribution
Systems
• Overhead lines (OHL)
• Underground cables (UGC)
11

Overhead lines
-Less costly
-Power loss is due to only
loss
R
I2
12

- Tappings can be made at desired
points, so cost of providing service
mains is also reduced
- OHL has considerable inductance
which causes voltage drop.
13

Overhead Conductors
• Aluminum
• Copper
• ACSR
14

Span
Sag
Insulator
Tension Tower
Supporting
Tower Tension Tower
Definition of Parameters
15

Transmission Lines -Types
• 400kV :Moose (Twin, Quad)
• 230kV : Zebra, Kundha
• 110kV : Leopard, Panther
• LT : AAC 7/3.35, 7/4.09
[ACSR 30/7/3.35
No.of strands/steel str./dia]
16

• Grid: The transmission system of an
area (or state)
• Regional Grid: Different grids are
interconnected through tie lines
• National Grid: Different regional grids
are further inter-connected
• Each grid operates independently
17

Regional grids
-Eastern
-Western
-Northern
-Southern
-North Eastern
18

Transmission Voltage Level
Tamil Nadu
110kV, 230kV, 400kV
Other states
132kV, 220kV, 400kV, and
765kV
19

Typical Construction – O/H
• Towers / Poles – to increase
phase to ground distance and
limit exposure to the public
• Cross arms – separate phases
from each other and from ground
potential
20

• Insulators – to separate phase
voltage from ground potential
• Ground/Static Wire - for
lightning protection
21

Distribution Voltage Level
11kV, 400 V and 230V
22

Standard voltages
Classification Range Standard levels
LV <1kV 400V (3-ph)
230V (1-ph)
MV 1kV-100kV
11kV, 22kV, 33kV
and 66kV
HV 100-300kV 110kV and 230kV
EHV 300-765kV 400kV and 765kV
UHV >765kV 1200 kV
23

330kV Transmission Line
24

Dual Circuit Transmission
Tower
25

Transmission Lines with Sag
26

Underground cable
-Less prone to natural hazards
-More costly and so limited to
only densely populated places
-Dielectric loss and sheath loss
are also significant
27

-Lower series impedance and
higher shunt capacitance
-Cables have low inductance ,
so lower voltage drop
-Fault detection and rectification
is very difficult
-Limited distance due to Ferranti
effect
28

-Beautification of the city
-Land : people will object
29

Cross Sections of UGC
XPLE cable
30

3 Transmission Cable
3 ½ cable up to 33kV
31

UG Cable
Single core above
33kV
Horizontal config.
32

400kV Cable
33

400kV Cable Installation
34

Cable Laying operation
Sand, tape, sand and compaction
Pilot wire - protection
35

400kV Cable Joints–Joint Bay
36

Transformer
Surge
arrester
Transformer
Cooling
fan
37

Transformer
38

Potential Transformer
Step down the
Voltage to the
required Level
Used for both
metering and
Protection
Purposes
39

A 69 kV Substation
Circuit
breaker
Disconnect Disconnect
Current CT
Bus bar
Circuit
breaker
Current CT
Bus bar
Circuit
breaker
Current CT
Bus bar
40

Disconnect Switch
Open
Open
41

Surge Arrester
42

Earthing Materials
43

Earthing Rods
44

High Voltage DC
Transmission
• High-voltage dc lines are used to
transport large amounts of energy
over a long distance.
• A representative application is
the Pacific DC Intertie, which
interconnects the Los Angeles
area with Oregon.
Cont…
45

• The voltage of the DC Intertie is
±500 kV and the maximum
energy transport is 3100 MW.
• More than one hundred dc
transmission systems operate
around the world, one of the
oldest and most famous is the
cable interconnection between
England and France.
Cont…
46

HVDC –world’s first HVDC –
Gotland scheme in Sweden
in1954(20MW, 96km cable
sea return), -100kV
In 1965 – Japan – 50 & 60 Hz
system – zero length (Back to
Back), 300MW, 250kV in
either direction
Cont…
47

First converter station using
exclusively Thyristors – Eel River
Scheme in Canada-60Hz, 1972,
320MW,80kV, zero length
48

Few in - India
- 1500MW +/-500kV 814km long
bipole between Rihand-dadri in
Northern region
- 2X250MW,70kV Vindhyachal
back- to-back connecting
Northern and Western regions
Cont…
49

- 2X500MW, 140kV Chandrapur
back-to-back link connecting
Southern and Western regions
50

AC
filter
AC
filter
DC
filter
DC
filter
DC Line (+)
DC Line (-)
Converter
Transformer Inductance Inductance
Converter
Transformer
Concept of HVDC
Transmission
51

12-Pulse Bridge Converter
52

A Schematic of a Bipolar HVDC
System identifying main
Components
53

Merits
- More power can be
transmitted per conductor per
circuit.
- Use of Ground Return Possible
- Smaller Tower Size
Cont…
54

- Higher Capacity available for
cables
- No skin effect
- Less corona and radio
interference
- No Stability Problem
- Asynchronous interconnection
possible
Cont…
55

- Lower short circuit fault levels
- Tie line power is easily
controlled
56

Demerits
- Difficulty of breaking DC
currents (high cost of DC
breakers)
- Inability to use transformers
- High cost of conversion
equipment
Cont…
57

- Generation of harmonics,
requires AC & DC filters / costly
- Complexity of control
- Difficulty of high power
generation
- Reactive power requirement
- Absence of overload capacity
58

Applications
1.Long Distance bulk transmission
system
2. Underground or Underwater
cables
Cont…
59

3. Asynchronous interconnection
of AC systems
4. Control and stabilization of
power flow
60

In South America and Japan
(50 and 60 Hz )networks
It would be impossible to
exchange power between them
with an AC line or cable. HVDC
is then the only solution.
61

Comparison of costs of
AC and DC Transmission
62

Break- even Distance
OHL – 800km
Submarine cables – 25km
UGC – 50km
Cont…
63

-HVAC needs only transformers
-HVDC needs very expensive
converters and inverters (capital
cost is very high)
-Below 800km, HVAC is
cheaper
-Above 800km, greater the
distance, the more money
saved.
64

HVDC Links
1. Monopolar
2. Bipolar
3. Homopolar
65

Monopolar Link
66

Bipolar Link
67

Homopolar Link
68

Comparison
Monopolar – cost-wise good
(cable tr)
Bipolar – If one pole is isolated
due to fault,pole can operate
with ground withhalf the rated
load
Cont…
69

- Equal to double circuit
transmission line
- Less harmonics interference
than monopolar
- Power reversal is possible
Homopolar – reconnection is
done
70

FACTS devices
Flexible AC transmission systems (FACTS)
controllers have been mainly used for solving
various power system steady state control
problems. However, recent studies reveal that
FACTS controllers could be employed to
enhance power system stability in addition to
their main function of power flow control.
71

MECHANICAL DESIGN
OF TRANSMISSION
LINES
72

Performance- Electrical
Design
 Power Flow Analysis
 Short Circuit Analysis
 Power System Stability
 Overvoltage Protection
73

Performance- Mechanical
Design
Line conductors /supports
Cross arms
74

Transmission Line
Grounding
rod
Foundation
Tower
Shield conductors
Insulator Bundled
conductor
(four conductors)
75

Line Supports
Loads due to
Conductors and insulators
(including ice and wind loads
on the conductors) together
with wind load on the support
itself.
76

Types of Line supports
-Wooden poles
-RCC poles
-Steel tubular poles
-Steel towers
77

Wooden Poles
-Made up of chemically treated
wood
-Used for distribution purpose
-Very economical
-Susceptible to decay
78

Wooden Poles
79

Reinforced Cement
Concrete Poles
-Stronger and costlier than
wooden poles
-They have long life and little
maintenance
-Used up to 33kV lines
80

RCC Poles
81

Steel Tubular Poles
-Light weight
-Long life
-High strength to weight ratio
-Used up to 33kV
82

Steel Tubular Poles
83

Contd…
84

Sag and Tension
Calculations
-Parabolic form
-Catenary (curve similar to
hyperbolic cosine curve)
85

Supports at same level
(Parabolic form)
86

Let
l = length of span, i.e., horizontal
distance between supports, m
S = sag at mid span, m
T = conductor tension (assumed
constant over the whole span)N
w = Conductor weight, N/m
87

Parabolic form
when
S
y  2
l
x 
2
)
2
(l
a
S 
2
ax
y 
2
4 l
S
a  (5.2a)
2
4 






l
x
S
y (5.2b)
88

Load is uniformly distributed : N/m
Total weight =
Total weight is taken by two supports at two
ends
Total Force in A :
Total Force in B :
Bending Moment (BM) = Force X distance
BM at O = 0
2
wl
2
wl
wl
w
89

Consider the equilibrium of half
line OB. Assuming that the
conductor is almost horizontal and
taking moments about B.
90

A
O
B
Showing bending moment
91

BM due to reaction=
BM due to self weight =
BM due to horizontal force =
2
2
l
wl
2
1
2
2

l
wl
S
T 
92

0
4
2
2
2




 TS
l
wl
l
wl
8
4
2
2
wl
wl
TS 

;
8
2
wl
TS 
T
wl
S
8
2

93

T
w
a
2
 (5.4a)
a
l
S
T
w

 2
4
2
T
wx
y
2
2
 (5.4b)
94

Effect of wind and ice
95

96

Ice covered conductor
97

Consider 1m length of conductor
d = diameter of conductor, m
t= radial thickness of ice, m
Overall diameter of ice covered
conductor is D=d + 2t
Volume of ice per metre length of
conductor 3
2
2
)
(
4
m
d
D 

98

Density of Ice = 915
Area of Ice = C/S area of conductor along
with Ice - C/S of conductor
3
/ m
kg
 
2
2
2
2
4
4
4
d
D
d
D
Ai 






 
   
d
t
t
d
t
d
Ai 



 
 2
2
2
4
99

Weight of Ice = kg/m
N/m
N/m
N/m
 
 d
t
t
Ai
i 


 915
 
d
t
t 

 81
.
9
915
 
d
t
t 

 14
.
3
15
.
8976
 
d
t
t 

 4
10
82
.
2
100

-The wind pressure is assumed to
act horizontally on the projected
area of the ice covered conductor.
-The projected area is D sq.m/
metre length of conductor
- For a wind pressure of p N/sq.m
of projected area, the wind load
is
w
F
m
N
pD
Fw /
 (5.6)
101

m
N
F
w
w
F w
i
t /
]
)
[( 5
.
0
2
2



102

i
w
w
w
F



tan
T
l
F
S t
8
2

(5.7)
The vertical sag is 
cos
S
103

Total length of conductor
104

Consider a small right angled triangle
on the parabola
2
ax
y 
105

2
2
2
)
(
)
(
)
( dy
dx
ds 

2
2
2
2
4
1
1 x
a
dx
dy
dx
ds
















)
)(
2
1
(
)
(
)
4
1
( 2
2
5
.
0
2
2
dx
x
a
dx
x
a
ds 



106

k
x
a
x
s 


3
2 3
2
.
0
,
.
0
,
0 

 k
Therefore
s
x
3
2 3
2
x
a
x
s 

(5.8)
107

s is curve distance from o to Point
P(x,y). Let the length of line be z.
Substitute and
We get length of half line z/2
2
l
x  2
4
l
S
a 
l
S
l
l
l
S
l
z 2
3
4
2
3
4
2
3
8
16
2
2
2







l
S
l
z
2
3
8


108

Substituting the value of S:
T
l
F
S t
8
2

2
2
4
2
8
3
8
T
l
F
l
l
z t











 2
2
2
24
1
T
l
F
l
z t
109

Support at different level
110

Substituting the co-ordinates of point C in
eqn
2
1
c
l
l
x 
 (5.14)
T
wl
S c
8
2
 (5.15)
2
2
4
l
x
S
y 
h
S
y 

1
2
2
1
1
4
c
l
Sx
y 
111

Substituting the value of S and
2
2
1
4
c
l
Sx
h
S 

S
l
h
S
x c
4
2
2
1


1
x
T
wl
l
h
T
wl
l
l
c
c
c
c
8
4
8
2
2
2
2
2









112

Simplifying,
lw
Th
l
lc
2


113

Problem 1
An overhead line, over a river
crossing, is supported by two
towers 50m and 80m above
water level. The horizontal span
is 300m.
114

If the weight of the conductor
is 8.28N/m and the tension in
the conductor is 19620 N, find
the height of the mid-point of
the line above water level.
115

Given Data
Height of towers = 50m and 80m
Horizontal span, l= 300m
Weight of conductor = 8.28N/m
Tension = 19620N
Height of mid point of line above
water level??
116

Length of conductor
Since both the supports are
on the same side of point O
wl
Th
l
lc
2


 
300
28
.
8
50
80
19620
2
300






c
l
m
lc 9
.
773

l
lc 2

117

B
C
O
300m
50m
80m
773.9m
lc
P
Diagram
118

Horizontal distance between O and C
Height of point C above O
Horizontal distance between O and P
m
96
.
86
300
2
9
.
773



m
T
wx
595
.
1
19620
2
96
.
86
28
.
8
2
2
2





m
96
.
236
150
96
.
86 


119

Height of P above O
Height of mid point P above point C
Height of mid point P above water level
m
847
.
11
19620
2
96
.
236
28
.
8 2




m
252
.
10
595
.
1
847
.
11 


m
252
.
60
50
252
.
10 


120

With all values
121

Problem 2
An overhead line has ACSR conductor of
1.95cm diameter and a span of 244m.
The allowable tension is . Find
sag in still air condition with no ice
covering, vertical sag when there is an ice
covering of 0.96cm thickness and a
horizontal wind pressure of 382 N/sq.m of
projected area. Ice weighs 8920 .
N
4
10
56
.
3 
3
/ m
N
122

Cond….
The line is carried by insulator
strings 1.43m long. What should
be the height of lowest cross-
arm to give a minimum ground
clearance of 7.62m under bad
weather conditions? The
conductor weight is 0.847 kg/m.
123

Given Data
Diameter of conductor = 1.95cm
Span = 244m
Allowable tension =
Thickness of ice covering = 0.96cm
Horizontal wind pressure, p = 382 N/sq.m
Weight of Ice = 8920
Weight of conductor = 0.847 kg/m
= 0.847×9.81=8.31N/m
N
4
10
56
.
3 
3
/ m
N
124

To be calculated
-Sag in still air condition??
-Vertical sag with ice??
-Height of lowest arm??
125

Sag,
m
T
wl
S 74
.
1
10
56
.
3
8
244
31
.
8
8 4
2
2






m
d 2
10
95
.
1 


m
t 2
10
96
.
0 


  m
t
d
D 2
2
10
87
.
3
10
96
.
0
2
95
.
1
2 








m
N
pD
Fw /
78
.
14
10
87
.
3
382 2




 
126

    m
N
d
D
wi /
82
.
7
10
95
.
1
87
.
3
2230
4
8920 4
2
2
2
2





 


 
   
  m
N
F
w
w
F w
i
t /
88
.
21
78
.
14
82
.
7
31
.
8
5
.
0
2
2
5
.
0
2
2







m
T
l
F
S t
57
.
4
10
56
.
3
8
244
88
.
21
8 4
2
2






127

o
i
w
w
w
F
y 5
.
42
82
.
7
31
.
8
78
.
14
tan
tan 1
1










 

m
S
g
Verticalsa 37
.
3
5
.
42
cos
57
.
4
cos 

 
128

Height of lowest cross arm
= minimum ground clearance +
vertical sag + length of insulator
string
= 7.62+3.37+1.43= 12.42m
129

Weight of conductor
Sag is directly proportional to
weight per unit length of the
conductor. Ice and wind loads
also increase the sag.
Factors affecting sag
130

Span
A longer span causes more sag.
Sag is proportional to square of
span.
T
l
F
S t
8
2

2
l
S 
Cond….
131

Conductor Tension
Sag is inversely proportional to
conductor tension
An increase in conductor tension
causes more stresses in the
conductor and more load on
insulators and towers.
Cond….
132

Temperature
A decrease in temperature reduces
the sag. If accompanied by snow
and wind, the sag may increase.
Heavy snow and high wind
pressure create worst conditions
for the line.
Cond….
133

Ground clearance
Electricity rules specify
minimum clearance between
ground and conductor.
Cond….
134

Clearanc
e to
ground
400V 11kV 33kV 66kV 132kV 220kV 400kV
Across
Street
(m)
5.8 5.8 6.1 6.1 6.1 7.0 8.4
Along
Street
(m)
5.5 5.5 5.8 6.1 6.1 7.0 8.4
Other
areas
(m)
4.6 4.6 5.2 5.5 6.1 7.0 8.4
135

Stringing Chart
-The curves of sag and tension
with temperature variation are
called stringing charts.
-These are useful in erecting the
transmission line conductors at
specified temperatures and
loading conditions.
136

Stringing Chart
137

Sag Template
It is a plot of curves on
transparent paper which is used
for locating the tower positions.
138

Sag template
139

Discussions
140
1.The power is not generated at very high
voltage level. Why?
2.Why is power transmitted at very high
voltage?
3.Why is negative polarity conductor used
in HVDC applications?
4.HVDC is preferred over HVAC, other than
economic point of view - discuss?

Selected References
1. B.R.Gupta, ‘Power system analysis and
design’, S. Chand & Company Ltd, New
Delhi, 3rd Edition.
2. R.K.Rajput, ‘Power system Engineering’,
Laxmi Publications (P) Ltd. New Delhi, 1st
Edition.
3. S.M.Singh, ‘Electric Power, Generation,
Transmission and Distribution’, PHI
Learning Private Limited, New Delhi, 2nd
Ed. 141

EE2303 Transmission & Distribution Syllabus

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