UNESCO-NIGERIA TECHNICAL &
VOCATIONAL EDUCATION
REVITALISATION PROJECT-PHASE II
NATIONAL DIPLOMA IN
ELECTRICAL ENGINEERING TECHNOLOGY
ELECTRICAL POWER SYSTEM (II)
COURSE CODE: EEC 232
YEAR II- SEMESTER III
THEORY
Version 1: December 2008
1
TABLE OF CONTENTS
Week 1
1.0
Methods of Electricity Generation ------------------------ 1
1.1
1.2
1.3
1.4
1.5
1.6
Hydro Electrical power station
Advantages
Disadvantage
Choke of site hydro electrical power station
Formation of hydro electrical power station
Calculation of hydro electrical power station
Week 2
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1.7 Nuclear power station
1.8 Advantages
1.9 Disadvantage
1.10 Choke of site Nuclear power
Week 3
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1.11 Maqneto hydro Dynamic (MHD) power station
1.12 Basic principle of MHD Generators
Week 4
Week 5
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1.13 Typical power system
1.14 Mayor components in Typical power system
1.15 National Grid system in Nigeria
1.16 Different Voltage levels
-------------------------------------------------------------------------20
1.17 Load curves
1.18 Importance of load curves
1.19 Factors influencing cost of electrical supply
1.20 Variable loads
1.21 Units Generated per annum
Week 6
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2.1 Line supports
Week 7
------------------------------------------------------------------------35
2.2 Contraction of undergrad cables
2.3 Contraction of cables
2.4 Insulating materials for cables
Week 8
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2.5 Constrictions of conductors
2.6 Common conductor materials
Week 9
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2.7 Transmission and its parameters
2.8 Capacitances
2.9 Single phase line calculation
Week 10
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2.10 Transmission line parameters
2.11 Resistance
2.12 Short single phase calculation
2.13 Short 3- phase
Week 11
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2.14 Performance of transmission lines
2.15 Introduction
2.16 Classification of overhead transmission line
2.17 Per unit Quantities
2.18 Advantage of per unit quantities
Week 12
Performance of short and medium transmission line --------- 60
2.19 Per unit system
2.20 Per unit in single phase circuit
2.21 Unit quantities for 3 phase system
Week13
Performance of transmission line----------------------------------64
3.1 Introduction
3.1 Equivalent cot for transmission line
Wee 14
--------------------------------------------------------------------------66
3.2 Nominal T method
Week 15
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3.3 Nominal  method
3.4 Long transmission line
1. Methods of Electricity Generation
Week 1
1.1 HYDRO-ELECTRICAL POWER STATION
A generating station which utilities the potential energy of at a high level for the generation of
electrical (energy is known as a hydro-electric power station.)
Hydro-electric power station are generally located in hilly areas were dams can be built
conveniently and large waters reservoirs can be obtained. In a hydro-electric power station, water
head is created by constructing a dam across a river of lake. From the dam, water is led to a water
turbine. The water turbine captures the energy in the falling water and changes the hydraulic energy
(i.e product of head and flow of waters) into mechanical energy ay the turbine shaft. The turbine
drives the alternator which converts mechanical energy into electrical energy.
1
1. Methods of Electricity Generation
Week 1
Hydro-electric power station is becoming very popular because the reserves of fuels (i.e coal and
oil) are depleting day by day. These system have added importance for flood control, storage water
for irrigation and water for drinking purposes.
2
1. Methods of Electricity Generation
Week 1
1.2 ADVANTAGES

It requires no fuel as water is used for the generation of electrical energy.

It is quite neat and clean as no smoke or ash is produced

It requires very small running charges because water is the source of energy which is
available free of cost.

It is comparatively simple in construction and requires less maintenance.

It does not require a long starting time like a steam, power station. In fact such plants can
be put into service instantly.

It is robust and has a longer life.

Such plants serve many purposes. Like irrigation & flooed control

Although such plants require the attention of highly skilled persons at the time of
construction. Yet for operation, a few experienced persons may do the job well.
1.3 DISADVANTAGES

It involves high capital cost due to construction of dam

There is uncertainty about the availability of huge amount of water due to
dependence on weather conditions

Skilled and experienced hands are required to build the plant

It requires high cost of transmission lines as the plant is located in highly areas which
are quite away from the consumers.
1.4 CHOICE OF SITE FOR HYDRO-ELECTRIC POWER
The following point should be taken into account while selecting the site for a hydroelectric power station.
3
1. Methods of Electricity Generation

Week 1
Availability of water: since the primary requirement of a hydro-electric power
station is the availability of huge quantity of water, such plants should be built at a
place (e.g rivers, canal) where adequate water is available at a good head.

Storage of water: There are wide variations in water supply from river or canal
during the year. This makes it necessary to store water by constructing a dam, in
order to ensure the generating of power continuously throughout the year. The
storage helps equalizing the flow of water any excess quantity of waters at a certain
period of the year can be made available during times of very low flow in the river.

Cost and types of land: The land for the construction of the plant should be
available at a reasonable price. Also, the bearing capacity of the ground should be
adequate to withstand the weight of heavy equipment to be installed.

Transportation facilities: The site selected for a hydroelectric plant
should be accessible by rail and road so that necessary equipment and machinery
could easily transport.
1.5 FORMATION OF HYDRO-ELECTRIC POWER STATION
The constituents of a hydro-electric plant are:
Hydraulic structures

Water turbines

Electrical equipment
4
1. Methods of Electricity Generation
Week 1
1.6 CALCULATION IN HYDRO-ELECTRIC POWER STATION
Examples 1 A hydro-electric generating station is supplied from a reservoir of capacity 5 x
116 cubic meters at a head of 200 meters. Find the total energy available in kwh if the overall
efficiency is 75%.
Solution:
Weight of water; W= volume of water x density = (5 x106) x 100
(Since mass of 1m3 of water is 1000kg), also 5 x 109 kg x 9.81N
Electrical energy available = W x H x Y x 7 overall
= (5 x 199 x 9.81) x (200) x (0.75) watt sec.
= (5 x 109 x 9.81) x (200) x 0.75) in kwh = 2.044 x 106kwh
Example 2
It has been estimated that a minimum run off of approximately 94m3/sec will be
available at a hydraulic project with a head of 39m. determine
(i)
Firm capacity
(ii)
Yearly gross output. Assume the efficiency of the plant to be 80%.
Solution:
Weight of water available; W = 94 x 1000 = 94000kg/sec
Water head, H = 39m
Work done/sec = W x H = (94000 x 9.81) x (39) watts sec
= 35963000W
= 35963 kw (gross plant capacity)
(i)
Firm capacity; = plant efficiency x gross plant capacity
= 0.80 x 35963 = 28770kw
5
1. Methods of Electricity Generation
(ii)
Week 1
Yearly gross output;
= firm capacity x hours in a year
= 28770 x 8760 = 252 x 106 kwh.
6
Methods of Electricity Generation
Week 2
1.7 NUCLEAR POWER STATION
This is a generating station which nuclear energy is being converted into electrical
energy.
In nuclear power station, heavy elements such as uranium (U235) or thorium (Th232)
are subjected to nuclear fission in a special apparatus known as a reactor. The heat energy
thus, released is utilized in raising steam at high temperature and pressure. The steam runs
the steam turbine which converts steam energy into mechanical energy into electrical
energy.
The most important feature of a nuclear power station is that huge amount of
electrical energy can be produced from a relatively small amount of nuclear fuel as
compared to other conventional types of power stations. It has been found that complete
fission of 1kg of uranium (U235) can produce as much energy as can be produced by the
burning of 4,500 tons of high grade coal. Although the recovery principal nuclear fuels (i.e
uranium and thorium) is difficult and expensive, yet the total fuels are considerably higher
than those of conventional fuels, such as coal, oil and gas. At present, energy crisis gripping
7
Methods of Electricity Generation
Week 2
us and, therefore, nuclear energy can be successfully employed for producing low cost
electrical energy on a large scale to meet the growing commercial and industrial demands.
1.8 ADVANTAGES

The amount of fuel required is small. Therefore, there is a considerable saving in the
cost of fuel transportation

A nuclear power plant requires less space as compared to any other types of same
size.

It has low running charges as a small amount of fuel is used for producing bulk
electrical energy.

This type of plant is very economical for producing bulk electric power.

It can be located near the load centres because it does not require large quantities of
water.

There are large deposits of nuclear fuels available all over the world. Therefore, such
plants can ensure continued supply of electrical energy for thousand of years.

It ensures reliability of operations.
1.9 DISADVANTAGES

The fuel used is expensive and is difficult to recover

The capital cost on a nuclear plant is very high as compared to other types of
plants.

The erection and commissioning of plant requires greater technical knowhow.

The fission by products are generally radioactive and may cause a dangerous
amount of radioactive pollution.
8
Methods of Electricity Generation

Week 2
Maintenance charges are high due to lack of standard station. And also attract
high salaries of the trained personnel employed to handle the plant.

Nuclear power plants are not well suited for varying loads as the reactor does
not respond to the load fluctuations efficiently.

The disposal of the by-products, which are radioactive, is a
problem. They are either to be disposed off in a deep trench or in a sea away
from sea-shore.
1.10 CHOICE OF SITE FOR NUCLEAR POWER
The following point should be kept in view while selecting the site for a nuclear
power station;
9
Methods of Electricity Generation

Week 2
Availability of water: As sufficient water is required for cooling purposes,
therefore, the plant site should be located were sample quantity of water is available
e.g across a river or by-sea-side.

Disposal of waste: The waste produced by fission in a nuclear power station is
generally radioactive which must be disposed off properly to avoid health hazards.
The waste should either be buried in a deep trench or disposed off is sea quite for
away from the sea shore. Therefore, the site selection for such plant should have
adequate arrangement for the disposal of radioactive waste.

Distance from populated areas: The site for a nuclear power station should far
away from the populated areas as there is a danger of presence of radioactivity in the
atmosphere near the plant. However, as a precautionary measure, a dome is used in
the plant which does not allow the radioactivity to spread by wind or underground
water ways.

Transportation facilities: In this situation the site should have adequate
facilities in order to transport the heavy equipment during erection and facilitate the
movement of the workers employed in the plant.
10
Generation of Electric Energy
Week 3
1.11 MAQNETOHYDRO POWER STATION
The generation of power by (MHD) generator is based on the fundamental laws of
electro-magnetic induction. That is to say, when a current carrying conductor moves
across a magnetic field (cut the lines of force), an electromotive force (e mf) is
induced in the conductor, which produces an electric current that flows if the external
circuit is complete.
Similarly in MHD when an ionized gas flows across the lines of field a voltage gas
flows across the lines of magnetic field a voltage is induced
.the ionized gas acts like an electric conductor.
An MHD power generation system
11
Generation of Electric Energy
1.12
Week 3
BASIC PRINCIPLES OF MHD GENERATOR
MHD generator is an efficient heat engine, which directly converts thermal energy
into electricity. MHD is one of the current techniques employed in power generation
where the efficiency is high as 60% compared to about 35% efficiency of
conventional thermal power stations.
It requires a working fluid and one of such fluid is partially ionized gas. The principle
is shown in figure above.
The MHD generator consists of a pipe whose diameter increases from one end
to another. The diameter is couple to magnetic poles and electrodes at the end .the
ionized fluid are perpendicular to the direction of motion of a fluid conductor. At very
high temperature, the gas becomes conductive. Conductivity is increased when the
gas is seeded. That means, when small amount of vaporized metal such as potassium
is added to the gas. The chamber is also supplied with compressed air. The gas
expands after passing through the nozzle to almost atmospheric pressure .energy is
induced when ionized gas passes through the magnetic field. The energy generated
subsequently causes the flow of current in the electric circuit.
An inverter is used to
convert the direct current (D.C) into alternative current (A.C) furthermore the
remaining current from the MHD, which is about 2310C, is used to a steam turbine.
Nigeria currently does not generate current from MHD generation even though the
potentials are enormous reserves of high grade coal estimated at about 1357 million
12
Generation of Electric Energy
Week 3
tones. There is the need to set up a research and development center in the field of
MHD. For instance, a 5mw experiment plant can be set up which is expandable to
accommodate more units as expertise is improved.
The advantages accruable from using MHD generation are as follows;- High efficiency of energy conversion of about 60% compared to conventional steam
power plants.
- High reliability, as is no moving part.
- The MHD is suitable for use as base load, an especially peak loading as it can be
used as in capacity just after starting.
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Methods of Electricity Generation
Week 4
1.13 SCHEMATIC DIAGRAM OF A TYPICAL POWER SYSTEM
Step-up
Transformer
Generator
Bus (1)
Step-down
Transformer
Distribution
Transformer
3.3KV
Consumers
Terminal
Generation
Transmission
Distribution
Figure 1.1
Transmission and distribution of electrical power is summed up as
follows:
“Electrical power is usually generated and transmitted in-3phase. As for the
distribution, a 3-phase or 1-phase could be used depending on the need of the
customer. A could be seen from Fig 1.1 power is supplied from the generator, say
11KV which is stepped- up by the step-up transformer to about 132KV. This high
voltage is used to transmit electricity over long distance so as to minimize power
losses at a far end of the line. For the purpose of security, continuity, repairs and
maintenance, transmission lines are designed and constructed in duplicate. The 314
Methods of Electricity Generation
Week 4
phase wire overhead high-voltage transmission line terminates in step-down
transformers in substation, usually situated outside the city, “The voltage is stepped
down to 33KV. At the substation the voltage is reduced from 33KV to 3.3KV, 3-wire
for primary distribution. Secondary distribution is at 400/240V, where voltage is
further stepped down from 3.3KV to 400V at the distribution sub-stations. Feeders
radiating from distribution sub-station supply power to distribution network in their
respective areas. If the distribution networks happen to be at a great distance from the
substation, then they are supplied from secondaries of distribution transformers which
are either pole-mounted or else housed in kiosks at suitable points of the distribution
networks. The most common system for secondary distribution is 400/240V, 3-, 4
wire system. The single phase residential highlighting load is connected between any
one line and the neutral whereas 3-phase, 400V motor load is connected across 3phase line” [9].
Generating voltages: 6.6KV, 11KV, 13.2KV or 33KV
High voltage transmission: 330KV, 132KV, 66KV, 6.6KV 3.3KV
Low voltage distribution: A.C 415/240V, 3 -, 4 wires
:
50Hz + 1% and – 1%
UK
:
50Hz
USA
:
60Hz
Standard frequency: Nigeria
15
Methods of Electricity Generation
Week 4
1.14 MAJOR COMPONENTS IN A TYPICAL POWER SYSTEM LAYOUT
 Generating Stations (Alternators)
 Step- Up Transformers
 Step-Down Transformers
 Bus bars
 Transmission Lines
 Distribution Lines
1.15 NATIONAL GRID SYSTEM IN NIGERIA
Electricity generated at various locations in the country is often remotely
located from major load centers; this is so because the power stations must be
cited as close as possible to their fuel sources. This calls for centralized or gird
connected to operations of the power sources. In Nigeria, the power generation
stations are linked to the National Control Centre (NCC) located at Oshogbo,
Osun State, where supervision, control of system and load allocation to the
various parts of the country is connected.
A new Supervisory National Control Centre (SNCC) at Shiroro, Niger
State has been completed by POWER HOLDING COMPANY OF NIGERIA
(PHCN) and is expected to offer computerized control of all grid operations,
which the old and National Control Centre could not provide. The installation
and commissioning of the Supervisory Control and Data Acquisition (SCADA)
16
Methods of Electricity Generation
Week 4
project at the SNCC will enable the operators to view various control their
switching. The operators will also have at their disposal the grid data with
which to supervise and control the system.
The power generated from the stations is transported to the load centers
via grid lines, which are high voltage levels of 330kv and 132kv. As most of
the generators produce power at lower voltage, step-up power transformers are
utilized to raise it to high levels. To maintain the stability in grid operations and
system stability, the system frequency and voltage have to be continuously
adjusted in response to emergent system events. Switching of generation is
considered with the following criteria enforced:

System frequency of 50Hz 1%

Voltage at 330kv level + 10% and -15%

Voltage at 132kv level +5% and -15%

Spinning reserve capacity to be 200MW.
Power is transmitted at a very high voltage levels so as to minimize losses in
view of far distance of the transmission lines. Transmission losses ion 1998 has
been estimated to be 9.77% of the total energy production. As at 199, PHCN
had a total route length of 5,000km for 330KV transmission lines and 6000km
for 132KV lines. As at 1972, PHCN had only 1,262km and 1,012km for 330kv
and 132 lines respectively.
17
Methods of Electricity Generation
Week 4
1.16 DIFFERENT VOLTAGE LEVEL ACCORDING TO COMSUMPTION
When power is generated and eventually and transmitted to the various
load canters, distribution lines are utilized to convey electricity to the numerous
consumers. Since very few industrial consumers tap power at 132kv levels
different ratings of power and distribution transformers are required to step
down the voltages to cover levels with majority of the consumers taking power
at 415V/240V. In the distribution sub-sector, injection substations are required
to step down the voltage from 132KV to 330KV and 11KV levels. The 330KV
lines are then used to convey power to urban and rural areas and also feed some
industrial customers. 11KV lines are usually use as the major distribution
network in both urban and rural set-up. The voltage is finally stepped down to
415/240V using distribution transformer for majority of commercial and
domestic consumers. As the demand exceed generations, the difference in
power has to be shed systematically across the country, so that electric power is
rotated among various load centre, which is referred to as load shedding. Loadshedding is done so as to prevent the power system from total collapse. Load
allocation is centrally done through the National Control Centre and it is the
duty of the distribution sub-sector to ensure that the allocated power is shared
judiciously. Load shedding is mostly carried out on 11KV feeders and in
18
Methods of Electricity Generation
Week 4
certain cases at 33KV levels depending on the amount of load to be shed as
instructed by the NCC.
19
Methods of Electricity Generation
Week 5
1.17 LOAD CURVES
Load curve us a graphical representation of load in (kw) with respect to time in hours.
The load curve shows the variation of load on the power station.
The load on a power station is never constant; it varies from time to time. These load
variation during the whole day (24hours) are recorded half-hourly or hourly and are plotted
against time on the graph, on which it could be daily, weakly, monthly and annually.
Load in Kw
12
2
4
6
8
10
12
2
4
6
8
10
12
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Methods of Electricity Generation
Week 5
30
25
20
15
10
5
6am
8am 1pm
0
3
3pm
6pm
10
12
6am (time(hours)
30
25
20
15
10
5
8
24
21
Methods of Electricity Generation
1.18
Week 5
THE IMPORTANCE OF LOAD CURVES
Load curves and load duration curves help in the following:

The estimation of generating cost

The selecting the proper sequence in generating units should be run

The selection of economical sizes of generating units

As an aid to forecasting power generating needs for future expansion

the daily load curve different hours of the day

the area under the daily load curve gives the number of units generated in the day. i.e
units generated/day = Area (in kwh) under daily load C

The highest point on the daily load curve represents the maximum demand on the
station on that day.

The area under the daily load curve divided by the total number of hours gives the
average load on the station in the day.
 Average load =

Area in (kwh) under daily load curve
24 hours
The ration of the area under the load curve to the total area of rectangle in which it is
contained gives the load factor.
 Load factor =

Load factor =
Area in (kwh) under daily load curve = Average load x 24
Max. demand
Max. demand x 24
Area in (kwh) under daily load curve
Total area in rectangular in which the load curve is contained
22
Methods of Electricity Generation
Week 5
1.19 FACTORS INFLUENCING COST OF ELECTRICITY SUPPLY
These factors determine the size and the necessary capital investment needed to set-up a plant
and also the various factors are used in computing parameters of power plant loads, these factors
are: Demand, Demand factors, average demand, connected load, plant use factor, maximum
demand, diversity factor, load factor
Plant use factor: It is the ratio of kwh generated to the product of plant capacity and the number
of hours for which the plant was in operation. i.e plant use factor
Plant capacity x hours of use
Demand: This is defined as load requirement over a small period of time.
Demand factor factor: It is the ratio of maximum demand on the power station to its
connecting load.
Demand factor factor = maximum demand
connected load
Connected load: it is the sum of rating in kw of equipment installed in the consumers premises.
Maximum demand: It is the greatest demand of load on the power station during a given
period.
Average demand load: average loads occurring on the power station in a given period (day,
month or year)
Daily average load = No of units (kwh) generated in a day
24 hours
Monthly average load = No of units (kwh) generated in a month
Number of hours in a month
Yearly average load = No of units (kwh) generated in a year
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Methods of Electricity Generation
Week 5
24 x 365 = 8760 hrs
Diversity factor =
Sum of individual max. demands
Max. demand of power station
Plant capacity factor: It is the ratio of actual energy produced to the maximum possible energy that
could have been produced during a given period.
Plant capacity factor: it is the ratio of actual energy produced to the maximum possible energy
that could have been produced during a given period.
Plant Cap. Factor =
Actual energy produced
Max. energy that could have been produced
1.20 VARIABLE LOADS (BASE, INTERMEDIATE AND PEAK LOAD)
24
Methods of Electricity Generation
Week 5
Base Load: Base load is the load below, which the demand never falls and must be supplied
100% of all the time. The power plants use to supply the load at the base portion of the load curve
are called as base load power plants. Base load plants run continuously throughout the year, have
large capacities and are expected to be very efficient. Base load plants are also expected to have low
maintenance costs. Hydropower stations and nuclear power stations are often used to provide base
loading requirement.
Intermediate Load: AN intermediate load is the load that satisfied the average load demand.
Thermal power stations are usually employed to satisfy this load.
Peak Load: peak load is the load that will satisfy the maximum load demand in a given period.
Peak load plants should be capable of quick starting, should be of a smaller capacity and should be
for a short period in the whole year. Gas turbine units, diesel generating units and pumped hydrosystem of medium sizes are often used and pumped hydro-system of medium sizes are often used to
satisfy peak loading requirements. Gas turbine generating units are popular because they can be
started few minutes.
1.21 UNITS GENERATED PER ANNUM
It is often required to find the kwh generated per annum from maximum demand and load factor.
The procedures are:Load factor = Average load
Max. demand

average load = M.D. x L.F
 Units generated/annum = average load (kw) x hours in a year
= max. (kw) x LF x 8760
Example: The maximum demand on a power station is 100MW if the annual load factor is 40%,
calculate the total energy generated in a year.
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Methods of Electricity Generation
Week 5
Solution: Energy generated year = M.D. L.F x Hrs in a year
= (100 x 10) x (0.4) x (24 x 365) kwh
= 3504 x 105kwh
Example:
A generating station has a connected load of 43mw and maximum demand of 20mw, the units
generated being 61x106 per annum. Calculated (i) the demand factor and (ii) load factor.
Solution:
(i)
Demand factor = Max. demand = 20
Connected load
43
(ii)
Average demand = units generated/annum
Hours in a year
61.5 x106
8759
=
= 0.4565
7020kw
 Load factor = Average demand =
Max. demand
7020
20 x 103
Example: An individual consumer having a maximum demand of 100Kw consumed 500,000 units
of electricity/annum. Determine his load factor.
Solution:
Load factor = A.D
M.D
= 500,000
100 x 365 x 24
=
57.1%, or 0.571
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Methods of Electricity Generation
Week 5
Example: A power station has to supply load an fallows
Time (hours) 0 – 6
30
6 – 12
12 – 14
14 – 18
18 – 24
90
60
100
50
a.
Draw the load curve
b.
Draw the load duration curve
c.
Choose a suitable generating unit to supply the load
d.
Calculate load factor
27
Methods of Electricity Generation
Week 5
Solution
(a) load curve
100
90
80
70
60
50
40
30
20
10
(b) Load duration curve
(a) load curve
100
90
80
70
60
50
40
30
20
10
(b) Load duration curve
28
Methods of Electricity Generation
Week 5
Example: A power station has to meet the following demand
A – 200Kw between 8am and 6pm
B – 100 Kw between 6am and 10am
C – 50 kw – between 6 am and 10am
D – 100 kw between 10 am & 6pm & then between 6pm & 6am
Plot the daily load curve and determine (i) diversity factor
(ii) unit generated per day (iii) land factor
Solution: The table for the load cycle:Time
0 -6
6–8
(hours)
Loads (kw)
A
B
100kw
C
50kw
D
100 kw
Total Load
100kw
150kw
8 – 10
10 – 18
18 – 24
200kw
200kw
-
100kw
50kw
35kw
100kw
300kw
100 kw
100kw
From the maximum demand = 350kw
Sum of individual max demand of groups
= 200 + 100 + 150 + 100 = 450 kw
(i) D.f = 450 = 1.286
350
(ii) Unit generated/day = area (kwh) under load curve
= (100 x 6) + (150 x2) + 350 x 2) + (300 x 8) + (100 x 6) = 4600kwh
(iii) Average load = units generated/day = 4600 191.7kw
No of hours in a day 24
Load factor = average load = 191.7 = 0.548 or 54.8%
A max demand 350
Example:
The daily demands of three consumers are:Time Consumer1 Consumer2 Consumer3
29
2.0 Principle of Transmission, Distribution Week 6
and Protection
2.1 LINE SUPPORTS
The supporting structure for over head line conductors are various type of poles
and towers called lines supports.
In general, the line support should have the following properties:
High mechanical strength to withstand the weight of conductors and wind
loads etc.
Light in weight without the lost of mechanical strength.
Cheap in cost and economical to maintain.
Longer life.
Easy accessibility of conductors for maintenance.
30
2.0 Principle of Transmission, Distribution Week 6
and Protection
The line support used for transmission and distribution of electric power are of
various types including wooden poles, steel poles R.C.C and lattice steel towers. The
choice of supporting structure for a particular case depend on the line span, Xsectional area, line voltage, cost and local conditions.
Wooden poles. These are made of seasoned wood (Sal or Chir) and are suitable
for lines of moderate X – sectional area of relatively shorter span, say up to 50
meters. Such support is cheap and, easily available, provide insulating properties
and therefore, are widely used for distribution purposes in rural areas as an
economical proposition. The wooden poles generally tend to rot below the ground
level, causing foundation failure. In order to prevent this, the portion of the pole
below the ground level is impregnated with preservative compounds like creosote
oil. Double pole structure of the ‘A’ or ‘H’ type are often used (see fig.8.2) to
obtain a higher transverse strength than could be economically provided by means
of single poles.
The main objections to wooden supports are:

Tendency to rot below the ground level.

Comparatively smaller life (20-25 years).

Cannot be used for voltages higher than 20kv.

Less mechanical strength and.
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
Require periodical inspection.

2. Steel poles; The steel poles are often used as a substitute for wooden poles. The
posses’ greater mechanical strength, longer life and permit longer spasm to be
used .such poles are generally used for distribution purposes in the cities. This
type of supports needs to be galvanized or pointed in other to prolong its life.
The steel poles are of three types viz,.

Rail poles

tubular poles and

Rolled steel joints.
3. RCC poles. The reinforced concrete poles have become very popular as line
support in recent year. The have greater mechanical strength ,longer life and
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permit longer spans than steel poles .moreover, they give good out look require
little maintenance and have good insulating properties.Fig.8.3 shows
R.C.C.poles for single and double circuit .The holes in the poles facilitate the
climbing of poles and at the same reduce the weight of line support .The main
difficulty with the use of these poles is the high cost of transport owing to
their heavy weight .therefore, each poles are often manufactured at the site in
order to avoid heavy cost of transportation.
4. Steel towers: In practice, wooden, steel and reinforced concrete poles are used
for distribution purpose at low voltages, say up to 11kv. However, for long
distance transportation at higher voltage, steel towers are invariably employed.
Steel have greater mechanical strength, longer life, can withstand most severe
climate conditions and permit the use of longer spans. The risk of interrupted
service due to broken or punctured insulation is considerably reduced owing to
longer spans. Tower footings are usually grounded by driving rods into the
earth. This minimizes the lightning conductor.
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2.2 CONSTRUCTION OF UNDERGROUND CABLES
An underground cable essentially consists of one or more conductors covered with
suitable insulation and surrounded by a protecting cover.
Although several types of cables are available, the type of cable to be will depend
upon the working voltage and services requirements. In general, a cable must fulfill
the following necessary requirements:
(i)
The conductor used in cables should be tinned stranded copper or aluminum
of high conductivity. Stranding is done so that conductor may become
flexible carry more current.
(ii)
The conductor size should be such that the cable carries the desired load
current without overheating and causes voltage drop within permissible
limits.
(iii)
The cable must have proper thickness of insulation in order to give high
degree of safety and reliability at the voltage for which it is designed.
(iv)
The cable must be provided with suitable mechanical protection so that it
may withstand the rough use in laying it.
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2.3 CONSTRUCTION OF CABLES
Show the general construction of a 3-conductor cable. The various parts are:

Cores or Conductors. A cable may one have one or more than one core
(conductor) depending upon the type of service for which it is intended. For
instance, the 3 phase conductor cable shown in figure is used for 3-phase
service. The conductors are made of tinned copper or aluminum and are
usually stranded in order to provide flexible to the cable.
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
Insulation. Each core or conductor is provided with a suitable thickness of
layer depending upon the voltage to be withstood by the cable. The
commonly used materials for insulation are impregnated paper, varnished or
mineral compound.

Metallic sheath: In order to protect the cable from moisture, gases or other
damaging liquids (acid or alkalis) in the soil and atmosphere, a metallic
sheath of lead or aluminum is provided over the insulation as shown in
figure.

Bedding: Over the metallic sheath is applied a layer of bedding which
consists of a fibrous material like jute or Hessian tape. The purpose of
bedding is to protect the metallic sheath against corrosion and from
mechanical injury due to armouring.

Armouring: over the bedding, armouring is provided which consists of one
or two layers of galvanized steel wire or steel material (like jute) similar to
bedding is provided over the armouring.
It may not be out of place to mentioned here that bedding, armouring and
are only applied to the cables for the protection of conductor insulation
protect the metallic sheath from mechanical
Injury.
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2.4 Insulating materials for cables.
Satisfactory operation of a cable depends to a great extent upon the
characteristic of insulation used. Therefore, the proper choice of insulating
material is of considerable importance. In general, the insulating materials
cables should have the following properties.
i.
High insulation resistance to avoid leakage current
ii.
High dielectric strength to avoid electrical breakout of cables.
iii.
High mechanical strength to withstand the mechanical handling
of
cables.
Non-hydroscopic i.e. it should not absorb moisture from air or soil.
The moisture tends to decrease the insulating resistance and hastens the breakdown of
the cable. In case of the insulation material is hygroscopic; it must be enclosed in a
waterproof covering like lead sheath.
Non - inflammable.
Low cost so as to make the underground system a viable proposition. Unaffected by
acids and alkalis to avoid any chemical action. One insulating material to be uses
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2.5 CONSTRUCTION OF CONDUCTORS
The conductor is one of the important items as most of the capital outlay is invested
for it. Therefore, proper choice of material and size of conductor is of considerable
importance.
The conductor material used for transmission and distribution of electric power
should have the following properties:
 High electrical conductivity.
 High tensile strength in order to withstand mechanical stresses.
 Low cost so that it can be used for long distances.
 Cross arm which provide support to the insulators.
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 Miscellaneous items such as phase plates, danger plate, lightning arrestors,
anti-climbing wires etc.
All above requirements are not found in a single material. Therefore while selecting a
conductor material for a particular case, a compromises made between the cost and
the required electrical and mechanical properties.
2.6 Commonly used of conductor materials.
The most commonly used conductor materials for over head lines are copper,
aluminum, steel-cored aluminum, galvanized steel and cadmium copper. The choice
of a particular material will depend upon the cost, the required electrical and
mechanical properties and the local conditions.
All conductors used for over head lines are preferably stranded in order to increase
flexibility. In stranded conductors, there is generally one central wire and round this,
successive layers of wires containing 6, 12, 18, 24…… wires. Thus, if there are n
layers, the total number of individual wires is 3n (n+1) +1. In the manufacture of
stranded conductors, the consecutive layers of wires are twisted or spiraled in
opposite direction so that layers are bound together.
1.
Copper. Copper is an ideal material for over head lines owing to its high
electrical conductivity and greater tensile strength. It is always in the hard drawn form
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as stranded conductors. Although hard drawing decrease the electrical conductivity
slightly yet it increases the tensile strength considerably.
Copper has high current density i.e., the current carrying capacity of copper per
unit of X- sectional area id quite large. This leads to two advantages. Firstly, smaller
X-sectional area of conductor is required and secondly, the area offered by the
conductor to wind load is reduced. Moreover, this metal is quite homogeneous,
durable and high scrap value.
There is hardly any doubt that copper is an ideal material for transmission and
distribution of electric power. However, due to its higher cost and non- availability, it
is rarely used for these purpose. Now– a – days the trend is to use aluminium in place
of copper.
2.
Aluminum. Aluminum is cheap and light as compared to copper but it has
much smaller conductivity and tensile strength. The relative comparison of
the two materials is briefed below:

The conductivity of aluminum is 60% that of copper. The smaller
conductivity of aluminum means that for any particular transmission
efficiency, the X - sectional area of conductor must be lager in aluminum
than in copper. For the same resistance, the diameter of aluminum conductor
is about 1.26 times the diameter of the copper conductor.
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The increase X- section of Aluminum exposes a greater surface to wind
pressure and, therefore, supporting towers must be design for greater transverse
strength. This often requires the used of higher towers with consequence of greater
sag.

The specific gravity of aluminum (2.71gm/cc) is lower than that of
copper (8.9gm/cc). Therefore, an aluminum conductor has a most one-half
the weigh of equivalent copper conductor. For this reason, the supporting
strictures for aluminium need not be made so strong as that of copper
conductors.
 Aluminum conductor being light, is liable to greater swings and hence
larger cross- arms are required.

Due to lower tensile strength and higher co - efficient of linear expansion
of aluminum, the sag is greater in aluminum conductors.
Considering the combined properties of cost, conductivity, tensile strength, weight
etc., aluminum has an edge over copper. Therefore, it is being used as a conductor
material. It is particularly profitable to use aluminium for heavy-current transmission
where the conductors’ size is large and its cost forms a major proportion of the total
cost of complete installation.
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3.
Steel cored aluminium. Due to low tensile strength, aluminium conductors
produce greater sag. This prohibit their used for larger span and makes them
unsuitable for distance transmission. in order to increase the tensile strength , the
aluminium conductor is reinforced with a core of galvanized steel wires. The
composite conductor thus obtained is known
as steel core aluminium and is
abbreviated as A.C.S.R.(aluminium conductor reinforced).
Steel – cored aluminium conductors consists of central core of galvanized steel
wires surrounded by a number of aluminium strands. Usually diameter of both steel
and aluminium wires is the same. The X- section of the two metal is generally in the
ratio of 1:6but can be modified to 1:4in order to get more tensile strength for the
conductor. Fig 8.1 shows steel cored aluminium conductor having
one steel wire
surrounded by six wires of aluminium. The result of this composite conductors is that
steel cored takes greater percentage of mechanical strength while the aluminium
strand carry the bulk of current. The steel cored aluminum conductors have the
following advantages:
 The reinforcement with steel increases the tensile strength but at the same
time keeps the composite conductors light. Therefore steel cored aluminium
conductors produce smaller sag and hence longer span can be used.
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 Due to smaller sag with steel cored aluminium conductors. Towers of
smaller height can be used.
4. Galvanized steel. Steel have high tensile strength. Therefore, galvanized steel
conductors can be used for extremely long span or short line section exposed to
abnormally high stresses due to climatic conditions. They have been found very
suitable in rural areas where cheapness is the main consideration. Due to poor
conductivity and high resistance of steel, such conductors are not suitable for
transmitting high large power over a long distance. However, they can be used to
advantage for transmitting a small power over a small distance were the size of the
copper conductor desirable from economic considerations would be too small and
thus unsuitable for used because of poor mechanical strength.
5.
Cadmium copper. The conductor material being employed in certain cases is
copper alloyed with cadmium. An addition of 1%or2% cadmium to copper increases
the tensile strength by about 50% and the conductivity is only reduced by 15% below
that of pure copper. Therefore, cadmium copper conductor can be used for
exceptionally long spans. However, due to high cost of cadmium, such conductors
will be economical only for lines of small X- section i.e., where the cost of conductor
material is comparatively small compared with the cost of supports.
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2.7 TRANSMISSION LINE PARATMETERS
(Resistance, Inductance and Capacitances)
Consist of its ohmic resistance, inductance capacitance between its conductors.
These parameters are distributed along its entire length. Reactance of single – phase
transmission line;
Inductance/conductor = 1/2(1+logh%) x 10-2 H/M
Reactance/conductor X = 2 – f x 1/2(1-log %) x 10-3 ΩM
Reactance of 3 – phase transmission line
D2
D
D
D1
D3
D
(a) Symmetrical Spacing
(b) Unsymmetrical Spacing
The inductance per conductor of 3-phase line with conductor symmetrical spaced is
same as inductance per KM of single-phase with equivalent spacing. If they are
unsystematically spaced the D is calculated
3 D1D2D3
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2.8 CAPACITANCE OF TRANSMISSION LINES
For single phase transmission lines
Cn
2
Loge D/V
F/M r = radius of each conductor
Taking ground effect into consideration
h – height of conductors
Cn =
F/M d – spacing between conductors
2
d
logh 1 + d2/4h2
For 3- transmission line: (1)
= Cn
Symmetrical spaced conductors
2
Loge D/r
(2)
F/M
Asymmetrical spaced conductors
The D is calculated as;
D = 3 D1D2D3
SHORT SINGLE PHASE LINE CALCULATIONS
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2.10 TRANSMISSION LINE PARATMETERS
(Resistance, Inductance and Capacitances)
Transmission line parameters also known as constants of transmission line, consist of
its ohmic resistance, inductance capacitance between its conductors. These parameters
are distributed along its entire length. Reactance of single – phase transmission line;
Inductance/conductor = 1/2(1+logh%) x 10-2 H/M
Reactance/conductor X = 2 – f x 1/2(1-log%) x 10-3 ΩM
Reactance of 3 – phase transmission line
D2
D
D
D1
D3
D
(a) Symmetrical Spacing
(b) Unsymmetrical Spacing
The inductance per conductor of 3-phase line with conductor symmetrical spaced is
same as inductance per KM of single-phase with equivalent spacing. If they are
unsystematically spaced the D is calculated
3 D1D2D3
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2.11 CAPACITANCE OF TRANSMISSION LINES
For single phase transmission lines
Cn
2
Loge D/V
F/M r = radius of each conductor
Taking ground effect into consideration
h – height of conductors
Cn =
F/M d – spacing between conductors
2
d
logh 1 + d2/4h2
For 3- transmission line: (1)
= Cn
Symmetrical spaced conductors
2
Loge D/r
(2)
F/M
Asymmetrical spaced conductors
The D is calculated as;
D = 3 D1D2D3
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2.12 SHORT SINGLE PHASE LINE CALCULATIONS
In this circuit, capacitance is neglected and line constants are lumped together, since they work at
relatively low voltage values
Es
- Sending voltage
ER is voltage references vector
ER-
- Voltage at receiving end
From  ABD
R
- Resistance of both conductors
X
-Reactance of both conductor = wL I=I <R
CosR -Power factor
Es = ER + IZ
= (CosR – j sinR)
Z = Z <
Z = R + jx
Es = ER + I (Cos R J SinR) (R +jX)
Es = ER + IR(CosR + I x sinR) + (I x cosR – IRsinR)2
If P.F is leading I = I <R = IcosR + j sinR)
Sending end P.F = s = (R +)
Example:
A single phase has a reactance of 6 < 500 and supplies a load of 100A, 400V at 0.9
lagging. Calculate the sending-end voltage.
Solutions:
E = 4000 <0 Z = 6< 50
R = cos-1 0.9 = 250501
I = 100 < 250500
Voltage drop = IZ = 100 x 6< (500 – 250 501)
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= 600<2410
=600[0.9123+jo.409)
=547.38+j245.6
Es = (4,000+ jo)+(547.38+j245.6)
Es= 4554V
2.13 SHORT THREE-PHASE TRANSMISSION LINE CONSTANTS
Calculations are done following same procedure for single-phase calculations. It is
regarded as 3 x single-phase units. Line voltage at sending end is multiple by 3. Active
power at receiving end = 3 x VL x IL x cos. Reactive power at receiving end = 3 x VL x IL
x sin.
Example: A-3 50Hz, transmission line has a resistance of 10 per phase and an inductance of 40mH
per phase supplies a load of 2000KW at 0.8 lagging and 22KV at the receiving end calculate the:
(a) Sending end voltage and power factor
(b) Transmission efficiency
(c) Regulation
Solution:
VR/Phase =
22.00
√3
Line current =
2000 x 103
(√3 x 22,000 x0.8)
= 65.64
XL= 2 x 50 x 40 x10-3 = 12.56Ω
R = 10 Ω Zph = (10 + j 12.56)
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= 1019.2 + j 265.5
a) Vs = 12,701.7 + (1019.2 + 265.5)
Vs = 13721+j 265.5 = 13724 <1.12
Sending end line voltage = 13724 x √3 =23.77KV
 = 1.10, CosR = 0.8, R = 36.90
s = 1.10 + 3690 = 38.00, cosR =0.788
Total power loss = 312R
= 3 x 65.62 x 10 = 129.1KW
Input power = 2000 + 1291.1 = 2129.1KW
Transmission efficiency r = 2000 = 0.9393 = 93.93%
2129.1
Line regulation
Regulation = Vs – VR x 1000
VR
Regulation = 23.77 – 22 x 100 = 8%
22
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2.14 PERFORMANCE OF TRANSMISSION LINES
2.15 INTTRODUCTION
The important consideration in the design and operation of a transmission line are the determination
of voltage drop, line losses and efficiency of transmission. These values are greatly influenced by
the lines constants R1 L and C of transmission line. For instance, the voltage drop in the line
depends upon the values of above three line constants. Similarly, the resistance of transmission line
conductors is the most important cause of power loss in the line and determines the transmission
efficiency. In this chapter, we shall develop formular by which we can calculate voltage regulation,
line losses and efficiency of transmission lines. These formular are important for two principal
reasons. Firstly, they provide an opportunity to understand the effects of the line on bus voltages and
the flow of power. Secondly, they help in developing an overall understanding of what is occurring
on electric power system.
2.16 CLASSIFICATION OF OVERHEAD TRANSMISSION LINES
A transmission line has three constant R, L and C distributed uniformly along the whole length of
the line. The resistance and inductance from the series impedance, the capacitance existing between
conductors for 1-phase line or from a conductor to neutral for a 3-phase line forms a shunt path
throughout the length of the line. Therefore, capacitance effects introduce complications in
transmission line calculations. Depending upon the manner in which capacitance is taken into
account, the overhead transmission lines are classified as:

Short transmission lines: when the length of an overhead transmission line is up to
about 50km and the line voltage is corporately low (<20kV), it is usually considered as a
short transmission line. Due to smaller length and lower voltage, the capacitance effects are
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small and hence can be neglected. Therefore, while studying the performance of a short
transmission line, only resistance and inductance of the line are taken into account.

Medium transmission lines: When the length of an overhead transmission line is about
50-150km and the line voltage is moderately high (>20KV< 100kV), it is considered as a
medium transmission line. Due to sufficient length and voltage of the line, the capacitance
effects are taken into account. For purposes of calculations, the distribution capacitance of
the line is divided and lumped in the form of condensers shunted across the line and at one or
more points.

Long transmission lines: When the length of an overhead transmission line is more
than 150 km and line voltage is very high (>100kV), it is considered as a long transmission
line. For the treatment of such a line, the line constants are considered uniformly distributed
over the whole length of the line and rigorous methods are employed for solution.
It may be emphasized here that exact solution of any transmission line must consider
the fact that the constants of the line are not lumped but are distributed uniformly throughout
the length of the line. However, reasonable accuracy can be obtained by considering these
constants as lumped for short
2.17 PER UNIT QUANTITIES
Dimension less numbers that are used to represent powers, voltage, currents impedance etc. referred
to a common base are called per – unit quantities. Power-system quantities such as voltage, current,
power and impedance are often in per unit or percent of specification base values. If 300V is taken
as base voltage, then the voltage of 60V is (60/300) = 0.2 per unit or 20%.
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Calculations are thereby made with the per unit quantities instead of the real quantities. Per quantity
is therefore calculated as follows:
Per unit quantity =
actual quantity
Base value of quantity
The base value has the same units with the actual value, thereby making the per unit quantity
dimensionless.
Example: for a single phase circuit, the following wing base value are given:
a.
Base current: 5 amperes
b.
Base voltage: 100volts
Calculate the base impedance and express 20A, 0.2A 50V and 2 in their per unit forms.
Solution:
Base impedance =
20A:
20 = 4p.u
5
0.2A:
0.2 = 0.04p
5
50V:
50 = 0.2 p.u
100
2 :
2 = 0.1 p.u
20
Base Voltage =
Base Current
100
5
= 20
Per Unit quantities for single-phase circuit
Select base KV and base KVA
KVpu = Actual KV
Base KV
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Base impedance (ohms) = I = Base KVA
Base KV
Base impedance (ohms) = Z = Base KV x 1000
Base Current I
= Base KV x Base KV x 1000
Base KVA
= (Base KV)2 x 100……………….. (i)
Base KVA
Base power KW = Base KVA
P.U impedance = Zpu =
=
Actual impedance
Base impedance
Actual impedance x Base KVA…………………….(ii)
(Base KV)2 x 1000
For change of base refer to (i) above
Zpu referred to new base = Zpu referred to old base
x Base KV old x Base KVA new
Base KV new
Base KVA old
Unit quantities for the three phase system
In 3 circuit the base voltage represents line to neutral voltage and currents represent line
currents
Base KVA = 3 -  KVA
3
Base Current = Base KVA
Base KV
Base impedance (ohms) = Base KV2 x 1000
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Base KVA
Consider 3 -  base KVA and base KV is taken as line KV,
Base current =
Base KVA
√3 x Base KV
Base impedance = Base KV x 100 2
√3
Base KVA/3
=
Base KV2 x 1000
Base KVA
Example: Given a base of 11KV and base KVA 1000, current 5 ohms into per unit.
Base KV2 x 1000
Base KVA
Answer: Base impedance =
112
100
=
x 100
121
12Ω = 1 pu as base impedance
5 Ω = 5 0.0413 p.u
121
Example: A 11 KV, 20,000 KVA generator has a reactance of 0.18 p.u referred to its rating
as bases. The new bases for calculation are 160KV and 40,000 KVA. Calculate the new p.u?
Solution: New Zp.u = Z.p.u old x
Old Base KV2 x New Base KVA
New Base KV
Old Base KVA
 Xnew = 0.05 x 112 x 40,000
160
20,000
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= 0.15 x 0.00472 x 2 =
Xnew = 0.0014 p.u
Example: A 3- circuit X.Y.Z are connected as shown
X
Y
Transformer
Z
Ω400
Load
Transformer
A
B
Transformer A is rated 15,000KVA and has ratio 11/22KV, leakage reactance of 6%.
Transformer B is rated 15,000KVA and has ration 22/4.4 KV and leakage reactance of 10%.
Load resistance is 400Ω. Calculate base impedance of circuit X, Y and Z as well as load
impedance referred to X and Y.
Solution: KVs’ are chosen to tally with transformer ratio so as to satisfy calculations
Base impedance =
Base KV2 x 1000
Base KVA
For circuit X, base impedance = 112 x 1000 = 8.06Ω
15,000
For circuit Y, base impedance = 222 x 1000 = 32.3Ω
15,000
For circuit Z, base impedance = 442 x 1000 = 1.29Ω
15,000
Loading impedance = 400Ω
Impedance p.u =
400 = 310 p.u
1.29
Load impedance referred to circuit Y = 400 x 22
In p.u =
= 10,000Ω
4.4
10,000 = 309.6 p.u
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32.3
Load impedance referred to circuit X = 10,000x 11 = 2500Ω
22
In p.u = 2500 = 310p.u
8.06
2.18 ADVANTAGE OF PER UNIT SYSTEM
1. Per-unit system has greatly simplified calculation in electrical power system.
2. Transformer equated circuit are greatly simplified by specifying base quantities. The
possibility of making same errors is eliminated.
3. For easy identification and analysis manufacturers usually specify transformer and
machines in per unit.
4. Since micro-machine are built to represent actual machines for the purpose of research,
the per unit system provides a good basis for comparison.
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2.19 PER UNIT QUANTITIES SYSTEM
Dimension less numbers that are used to represent powers, voltage, currents
impedance etc. referred to a common base are called per – unit quantities. Powersystem quantities such as voltage, current, power and impedance are often in per unit
or percent of specification base values. If 300V is taken as base voltage, then the
voltage of 60V is (60/300) = 0.2 per unit or 20%.
Calculations are thereby made with the per unit quantities instead of the real
quantities. Per quantity is therefore calculated as follows:
Per unit quantity =
actual quantity
Base value of quantity
The base value has the same units with the actual value, thereby making the per unit
quantity dimensionless.
Example: for a single phase circuit, the following wing base values are given:
a.
Base current: 5 amperes
b.
Base voltage: 100volts
Calculate the base impedance and express 20A, 0.2A 50V and 2 in their per unit
forms.
Solution:
Base impedance =
Base Voltage
Base Current
=
100
5
= 20
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20A:
20 = 4p.u
5
0.2A:
0.2 = 0.04p
5
50V:
50 = 0.2 p.u
100
2:
2 = 0.1 p.u
20
2.2O PER UNIT QUANTITIES FOR SINGLE-PHASE CIRCUIT
Select base KV and base KVA
KVpu = Actual KV
Base KV
Base impedance (ohms) = I = Base KVA
Base KV
Base impedance (ohms) = Z = Base KV x 1000
Base Current I
= Base KV x Base KV x 1000
Base KVA
= (Base KV)2 x 100……………….. (i)
Base KVA
Base power KW = Base KVA
P.U impedance = Zpu =
Actual impedance
Base impedance
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2.0 Principle of transmission, Distribution and Week
protections
12
=
Actual impedance x Base KVA…………………….(ii)
(Base KV)2 x 1000
For change of base refer to (i) above
Zpu referred to new base = Zpu referred to old base
x Base KV old x Base KVA new
Base KV new
Base KVA old
2.21 UNIT QUANTITIES FOR THE THREE PHASE SYSTEM
In 3 circuit the base voltage represents line to neutral voltage and currents
represent line currents
Base KVA = 3 -  KVA
3
Base Current = Base KVA
Base KV
Base impedance (ohms) = Base KV2 x 1000
Base KVA
Consider 3 -  base KVA and base KV is taken as line KV,
Base current =
Base KVA
√3 x Base KV
Base impedance = Base KV x 100
√3
Base KVA/3
2
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2.0 Principle of transmission, Distribution and Week
protections
12
=
Base KV2 x 1000
Base KVA
Example: Given a base of 11KV and base KVA 1000, current 5 ohms into per
unit.
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3.0 Performance of Short and Medium Week
Transmission Line
13
3.1 INTRODUCTION
Performance of transmission lines under normal operating conditions. The analytical expressions for
currents and voltages and the equivalent circuits for transmission lines are first developed for
“short” lines and for “long” lines (where the effects of distributed line capacitance must be taken
into account). A simplification is presented in the treatment of long lines that greatly clarifies their
analysis and reduces the amount of work necessary for calculations. Problems relating to the
regulation and losses of lines and their operation under conditions of fixed terminal voltages are then
considered. The circle diagrams are developed for short lines, long lines, the general equivalent
circuit, and for the general circuit using ABCD constants. The circle diagrams are revised from the
previous editions of the book to conform to the convention for reactive power which is now
accepted by the American Institute of Electrical Engineers, so that lagging reactive power is positive
and leading reactive power is negative.
When determining the relations between voltages and currents on a three-phase system it is
customary to treat them on a („per phase” basis. The voltages are given from line
to neutral, the currents for one phase, the impedances for one conductor, and the equations written
for one phase. The three-phase system is thus reduced to an equivalent Single-phase system.
However, vector relationships between voltages and currents developed on this basis are applicable
to line-to-line voltages and line currents if the impedance drops are multiplied by d/3 for three-phase
systems and by 2 for single-phase two-wire systems.
Most equations developed will relate the terminal conditions at the two ends of the line since they
are of primary importance. These terminals will be called the sending end and receiving end with
reference to the direction of normal flow of power, and the corresponding quantities designated by
the subscripts 5‟ and R.
3.2 EQUIVALENT CIRCUITS FOR TRANSMISSION LINES
1. Short Transmission Lines
For all types of problems it is usually safe to apply the short transmission line analysis to lines up to
30 miles in length or all lines of voltages less than about 40 kv. The importance of distributed
capacitance and its charging current varies not only with the characteristics of the line
hut also with the different types of problems. For this reason no definite length can be stipulated as
the dividing Point between long and short lines of transmission lines
Recisrd by:
R. F. Lawrence
Neglecting the capacitance a transmission line can be trcatcd as simple, lumped, constant
impedance,
Z = R +jX = zs = rs + jxs
Where
z = series impedance of one conductor in ohms per mile
T* = resistance of one conductor in ohms per mile
x* = inductive reactance of one conductor in ohms per
mile
s = length of line in miles
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3.0 Performance of Short and Medium Week
Transmission Line
13
The corresponding “per phase” or equivalent single-phase
circuit is shown in Fig. 1 together with the vector diagram
Z=ts=(r+jx)s
EQUIVALENT TRANSMISSION CIRCUIT TO NEUTRAL
Fig. l-Equivalent circuit and vector diagram for short transmission lines. Relating the line current
and the line-to-neutral voltages at the two ends of the line. The analytical expression for this
relationship is given by the equation: Es=ER+ZI (1)
Throughout this chapter, the following symbols are used: J$--is a vector quantity
,!is the absolute magnitude of the quantity E-is the conjugate of the vector quantity
2. Long Transmission Lines
The relative importance of the charging current of the line for all types of problems varies directly
with the voltage of the line and inversely with the load current. To appreciate this fully it is
necessary to consider the analysis of “long” lines.
A “long” transmission line can be considered as an infinite number of series impedances and shunt
capacitances connected as shown in Fig. The current IR is unequal to Is in both magnitude and
phase position because some current is shunted through the capacitance between phase
65
3.0 Performance of Short and Medium Week
Transmission Lines
14
3.2 NOMINAL T METHOD
In this method, the whole line capacitance is assumed to be concentrated at the middle
point of the line and half the line resistance and reactance are lumped on its side as
shown in fig 10.11. Therefore, in this arrangement, full charging current flow over
half the line in fig 10.11, one phase of 3 phase transmission line is shown as it is
advantageous to work in phase instead of line to line values.
DIAGRAM
Let IR – load current per phase;
R = resistance per phase
XL = inductive reactance per phase
C= capacitance per phase
Cos R = receiving end power factor (lagging)
Vs = sending end voltage/phase
V1 = voltage across capacitor C
The phasor diagram for the circuit is shown in fig 10.12 Taking the receiving end
voltage VR as the reference phasor, we have,
Receiving end voltage
Load current
VR = VR + j0
IR = IR (cosR – jsinR)
Voltage across C, V1 = VR + IR Z12
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3.0 Performance of Short and Medium Week
Transmission Lines
14
=VR + IR (cosR–jsinR) R + j XL
2
2
Capacitance current
IC = j w CV1 = j 2πf C V1
Sending end current
Is = IR + IC
Sending end voltage,
Vs = V1 + Is Z = V1 + Is R + j XL
2
2
Example: A 3-phase, 50-Hz overhead transmission line 100 km long has the
following constants:
Resistance/km/phase
=
0.1Ω
Inductive reactance susceptance/km/phase
=
0.2 Ω
Determine (i) The sending end current (ii) sending end voltage (iii) sending end
power factor and (iv) transmission efficiency when supplying a balanced load of
10,000 kw at 66 kv, p.f 0.8 lagging, use nominal T method.
Solution figs (i) and (ii) show the circuit diagram and phasor and diagram of the line
respectively.
DIAGRAM
Total resistance/phase,
Total reactance/phase
R = 0.1 x 100 = 10 Ω
XL = 0.2 x 100 = 20 Ω
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3.0 Performance of Short and Medium Week
Transmission Lines
14
Capacitive susceptance, Y = 0.04 x 10-4 x 100 = 4 x 10-4S
Receiving end voltage/phase
VR = 66,000/ √3 = 38105V
Load current,
IR =
10,000 x 103
√3 x 66x103 x 0.8
= 109A
cosR = 0.8; sinR = 0.6
Impedance per phase,
Z = R + j XL = 10 + j 20
(i) Taking receiving end voltage as the reference phasor
Receiving end voltage,
VR = VR + j 0 = 38, 105 V
IR = IR (cos - sinR) = 109 (0.8-j 0.6) =87.2–j
Load current,
65.4
V1 = VR+IR Z/2 = 38, 105+ (87.2 – j 65.4) (5+j 10)
Voltage across C,
= 38,105 + 436+J 872+J327 + 654 = 39,195+ J 545
Charging current, Ic= j Y V1 = j 4 x 10-4 (39, 195 + j545) 0 – 0.218 j 15.6
Sending end current
Is = IR Ic – (87.2 – 65.4) + (-0.218 + j 15.6)
= 87.0 – 49.8 = 100< - 29047’A
 Sending end current = 100A
(ii) Sending end voltage Vs = V1 + Is Z/2 = (39.195 + 545) +(87.0-49.8) (5+j 10)
= 39,195 + j 545 + 434.9 + j 870 – j 249 + 498
= 40128 + j 1170 = 40145 < 10 40’V
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3.0 Performance of Short and Medium Week
Transmission Lines
14
 Line value of sending end voltage = 40145 x √3 = 69,533 V = 69.533 kV
(iii)
Referring to phasor diagram in fig 10.14
1 = angle between VR and VS = 10 40’
2 = angle between VR and IS = 290 47’

S = angle between Vs and Is
= 1+ 2 = 1040’ + 29 47 = 31027’
 Sending end power factor, cosS = cos 31027’ = 0.853 lag
(iv)
Sending end power = 3 VsIs cos s = 3 x 40,145 x 100 x 0.853
= 10273105 w = 10273.105 kw
Power delivered
= 10,000kw
Transmission efficiency =
10,000 x 100 = 97.34%
10273.105
Example: A 3-phase, 50 Hz transmission line 100 km long delivers 20 MW at
0.9 p.f lagging and at 110kv. The resistance and reactance of the line phase km are
0.2Ω and 0.4 Ω respectively. while capacitance admittance is 2.5 x 10-6
Siemens/km/phase. Calculate (i) the current and voltage at the sending end (ii)
efficiency of transmission. Use nominal T method.
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3.0 Performance of Short and Medium Week
Transmission Lines
14
Solution.
Total resistance/phase, R = 0.2 x 100 = 20 Ω
Total reactance/phase, X = 0.4 x 100 = 40 Ω
Total capacitance admittance/phase Y = 2.5 x 10-6x100 = 2.5 x 10-4S
Phase impedance, Z = 20 + j 40
DIAGRAM
Receiving end voltage, VR = 110 x 103 √3 = 63508 V
Load current, IR =
20 x 106
√3 x 110 x 103 x 0.9
(i)
= 116.6A
Taking receiving end voltage as the reference phasor diagram 10.15 (ii)
we have,
IR = VR + j 0 = 63508 V
Load current,
IR = IR (cos – j sin) =116.6 (0.9 – j 0.435) = 105-
j50.7
Voltage across C, V1 = VR + IR Z/2 = 63508 + (105- j50.7) (10 +j 20)
= 63508 + (2064 + j 1593) = 65572 + j 1593
Charging current, IC = j Y V1 = j 2.5 x 10-4(65572 + j 1593) = - 0.4 j 16.4
= (104.6 – j 34.3) = 110 < - 180 9’ A
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3.0 Performance of Short and Medium Week
Transmission Lines
14
Sending end current = 110 A
Sending end voltage
Vs = V1 + Is Z/2
= (65572 + j1593) + (104.6 j34.3) (10 + j 20)
= 67304 + j 3342
Magnitude of Vs= (67304) 2 + (3342)2 = 67387 V

Line value of sending end voltage
= Line value of sending end voltage
= 67387 x 3 = 116717 kV
(ii)
Total line losses for the three phases
= 3 I2s R/2 + 3IR2 R/2
= 3 x (110)2 x 10 + 3 (116.6)2 x 10
= 0.770 x 106 W = 0.770 MW
Transmission efficiency =
20
20 + 0.770
x 100 = 96.29%
71
3.0 Performance of Short and Medium Week
Transmission Line
15
3.3 NOMINAL Π METHOD
In this method capacitance of each conductor (i.e, line ti neutral) is divided into
two halves; one half being lumped at sending end and the other half at the
receiving end as shown in fig. it is obvious that capacitance at the sending end
has on the line drop. However, its charging current must be added to line
current in order to obtain the total sensing end current.
DIAGRAM
Let
IR = load current per phase
R = resistance per phase
XL = inductive reactance per phase
C = capacitance per phase
CosR = receiving end power factor (lagging)
Vs = sending end voltage per phase
The * phasor diagram for the circuit is shown in fig 10.17 taking the receiving
voltage as the reference phasor, we have,
VR = VR + j 0
Load current, I R = I R (cosR R j sinR)
Charging current at load end is
72
3.0 Performance of Short and Medium Week
Transmission Line
15
IC1 = j w (C/2) VR = j π f.C V R
Line current,
IL = IR + IC1
Sending end voltage, Vs = VR + IL Z = VR + IL (R + j XL)
Charging current at the sending ends is
IC2 = IC1 = j w (C/2) VS = j π f.C VS
Sending end current,
IS = IL + IC2
Example: A 3-phase, 50Hz, 150 km line has a resistance, inductive reactance
and capacitive shunt admittance of 0.1Ω, 0.5 Ω and 3 x 10 -6 per km per phase.
If the line delivers 50 MW at kV and 0.8 p.f lagging, determine the sending end
voltage and current. Assume a nominal π circuit for the line.
Solution: Fig 10.18 shows the circuit diagram for the line.
Total resistance/phase,
R = 0.1 x 150 = 150 Ω
Total reactance/phase
XL = 0.5 x 150 = 75 Ω
Capacitance admittance/phase
Y = 3 x 10-6 x 150 = 45 x 10-5S
Receiving end voltage/phase, 110 x 103/ 3 = 63.508V
Load current,
IR =
50 x 106
√3 x 110 x 103 x 0.8
= 328A
cosR = 0.8; sinR = 0.6
DIAGRAM
73
3.0 Performance of Short and Medium Week
Transmission Line
15
Taking receiving voltage as the reference phasor, we have
VR= VR + j 0 = 63,508 V
Load current
IR = (cosR – j sinR) = 328 (0.8 – j 0.6) = 262.4 – j196.8
Charging current at the load end is
IC1 = VR j Y = 63,508 x j 45 x 10-5 = j 14.3
2
2
Line current
IL = IR + IC1 = (262.4 – j 196.8) + j14.3 = 262.4 – j 182.5
Sending end voltage,
Vs = VR + IL Z = VR + IL (R + j XL)
= 63,508 + (262.4 – j 182.5) (15+j 75)
= 63,508 + 3936 + j 19,680 – j 2727.5 + 13,687
= 81,131 + k 16.942.5 = 82,881 < 110 47’ V
 Line to line sending end voltage = 82,881 x √3 =1,43,550 V = 143.55kV
Charging current at the sending ends is
IC2 = j Vs Y/2 = (81,131 + j 16,942.5) j 45 x 10-5
2
= - 3.81 + j 18.25
Sending end current
Is = IL + IC2 = (262.4 – j 182.5) + (-3.81 + j
18.25)
= 258.6 – J 164.25 = 306.4 < - 32.40A
74
3.0 Performance of Short and Medium Week
Transmission Line
15
 Sending end current
= 306.4 A
Example 10.14. A 100 km long 3-phase, 50-Hz transmission line has the
following line constants
Resistance/phase = 0.1Ω
Reactance/phase = 0.5Ω
Susceptance/phase/km = 10 x 10-6 S
If the line supplies load of 20 MW at 0.9 p.f lagging at 66 kv at the receiving
end, calculate by nominal π method:
(i)
Sending end power factor
(ii)
Regulation
(iii)
Transmission efficiency
Solution. Fig 10.19 shows the circuit diagram for line
Total resistance/phase,
R = 0.1 x 100 = 10 Ω
Total reactance/phase
XL = 0.5 x 100 = 50 Ω
Capacitance admittance/phase
Y = 3 x 10-6 x 100 = 10 x 10-4 S
Receiving end voltage/phase, VR=66 x 103/ √3 = 38105V
Load current,
IR =
20 x 106
√3 x 66 x 103 x 0.9
= 195A
cosR = 0.9; sinR = 0.435
DIAGRAM
75
3.0 Performance of Short and Medium Week
Transmission Line
15
Taking receiving voltage as the reference phasor, we have
VR= VR + j 0 = 38105V
Load current
IR = (cosR – j sinR) = 195 (0.9 – j 0.435)=176 – j 85
Charging current at the load end is
IC1 = VR j Y = 38105 x j 10 x 10-4 = j 19
2
2
Line current
IL = IR + IC1 = (176 – j .85) + j 19 = 176 – j 66
Sending end voltage,
Vs = VR + IL Z = VR + IL (R + j XL)
= 38,105 + (176 – j 66) (10 + j50)
= 38,105 + (5060 + j 8140)
= 43,925 + j 8140 = 43,925 < 10.650V
Sending end line to line voltage = 43,925x √3 x = 76 x 103 V = 76kV
Charging current at the sending ends is
IC2 = j Vs Y/2 = (43,165 + j 8140) 10x 10-4
2
= - 4.0 + j 21.6
Sending end current
Is = IL + IC2 = (176 – j 66) + (-4.0 + j 21.6)
= 172 – j 44.4 = 177.6 < - 14.50A
(i) Referring to phasor diagram in fig 10.20
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3.0 Performance of Short and Medium Week
Transmission Line
15
1 = angle between VR and Vs = 10.650
2 = angle between VR and Vs = 10.650
S = angle between Vs and Is = 2 +1
=14.5 + 10.650 = 25.150
 Sending end p.f cos S = cos 25.15 = 0.905 lag
(ii) % voltage regulation = Vs–VR x 100 = 43925–38105 x100 = 15.27%
VR
38105
(iii)
Sending end power = 3 Vs Is cosS = 3 x 43925 x 177.6 x 0.905
= 21.18 x 10-6 W = 21.18 MW
Transmission efficiency = (20/21.18) x 100 = 94%
3.4 LONG TRANSMISSION LINES
It is well known that line constants of the transmission line are uniformly
distributed over the entire length of the line .However; reasonable accuracy can
be obtained in line calculation for short and medium lines by considering these
constants as lumped. If such an assumption of lumped constants is applied to
long transmission lines having length excess of about 150km, it is found that
serious errors are introduced in the performance calculation. Therefore, in order
to obtain fair degree of accuracy in the performance calculations of lone line,
the line constants are considered as uniformly distributed throughout the length
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3.0 Performance of Short and Medium Week
Transmission Line
15
of the line. Rigorous mathematical treatment is required for the solution of such
line.
DIAGRAM
Fig shows the equivalent circuit of a 3-phase long transmission line on a
phase-neutral basis. The whole line length is divided into n sections, each
section having line constants Ith/n of those for the whole line. The following
points may be noted:
(i)
The line constants are uniformly distributed over the entire length of line
as is actually the case.
(ii)
The resistance and inductive reactance are the series elements
(iii)
The leakages susceptance (B) and leakage conductance (G) are shunt
elements. The leakage susceptance is due to the fact that capacitance
exists between line and neutral. The leakage conductance takes into
account the energy losses occurring through leakage over the insulators
or due to corona effect between conductors.
Admittance = √G2 + B2
78