Power System Eng

1

Elrctrical Power Engineering

Syllabus

Electrical power generation …………………………..10 hoursPower transmission …………………………………...20 hours Insulation and Corona ………………………………….8 hoursPower cable …………………………………………...10 hoursSwitchgear ……………………………………………. 6 hoursElectrical part of a power station ………………………6 hours

References

1- Electrical power systems. { A.E. Guile , W. Paterson } Volume one

2- Elements of power system analysis . { William D. Stevenson , SR. }

3- A course in electrical power . { M.L. Soni and P.V. Gupta }

4- Electrical power systems. { Weedy , B.M. }

5- Fundamentals of power system . { S.K. Agarwala }

6- Principles of power system . { V.K. Mehta }

7- Power system analysis . { Charles A. Gross }

8- The Transmission and distribution of electrical energy. { Cotton }

9- Power system analysis and design . { J. Duncan Glover }

2

Structure of electrical power systems

The structure of electric power systems consists of the following main parts :

1- Generating stations ( power station ) : Bulk electric power is produced by special plants known as generating stations or power plants . This part divided in two section :a- Mechanical section – It is sources of mechanical energy as

( Boiler, Turbine….) .

b- Electrical section – as ( Alternators, transformers, protection apparatus, controls system, and measuring instruments….. ) .

2- Transmission lines: are the connecting links between the power stations and the distribution system and lead to other power systems over interconnections .There are two types:

a- Overhead transmission line .

b-Under ground cables .

3- Distribution systems: It connects all the individual loads to the transmission line.

ELECTRICAL POWER GENERATION

Energy Resources

The resources of electrical energy ( primary fuels ) may be divided into :

a - Thermal : as Coal, Oil, Natural gas, Nuclear, and Solar

b- Non Thermal : as Hydropower , Wind , and direct solar conversion .

3

While considering primary fuels in meeting the world’s energy requirements we may therefore include the following :

i- Solid fuels ( Coal )ii- Liquid fuels ( Oil )iii- Natural gasiv- Hydropowerv- Nuclear

The above Fig. shows the possible changes in percentage shares of primary fuels in meeting the world’s energy requirement over period of 50 years ( 1950-2000 )

4

Location of power station

The best geographical location of the generating plants would be close to the electrical load centers , i.e, the regions where the major energy demand exists . However, the locations of the primary fuel conventional energy sources do not necessarily coincide with load centers.

We are therefore faced with following choice ;

1- Build the power plants close to the energy sources and then transport the electrical energy to the load centers

2- Build the power plants close to load centers and transport the fuel from the source locality.

The electrical power grid makes the first alternative technically feasible . However, in reality , the actual choice will be based upon a combination of technical , economical and environmental factors.

Type of power station

There are four type of power station :

1- Thermal power station ( Steam power station ), the primary fuels for this type are (oil or coal or natural gas )

2- Hydro-Electric power station , this type installed where resources of water available at sufficient head.

3- Nuclear power station , the primary fuels for this type is ( Nuclear fusion of Uranium )

4- Diesel power station , in this type , diesel engine is used as the prime mover .

5

Thermal power station ( Steam power station )

A generating station which converts heat energy of coal or oil into electrical energy is known as a steam power station.

The schematic arrangement of steam power station is shown in below fig.

6

Main parts of power station are :

Boilers : It is a device meant for production steam under pressure.

There are two type of boiler:

a) Fire tube boiler ; b ) Water tube boiler

For electrical power stations , generally water tube boilerare used.

Turbine : It is the main part of the mechanical system of the power station from which the mechanical energy can be obtained.

Alternator ( Synchronous generator and exciter) : It is the main part of electrical system from which the electrical energy can be obtained.

2-pole ; 3000 r.p.m ( 50 HZ )

; 3600 r.p.m ( 60 HZ )

Power 1000 MW ; 11-33 KV ; 3-phase ; stare connection

Air cooling up to 40 MW , above 40 MW using hydrogen for cooling

Main exciters are compound d.c generators of 115 V or 230 V .

Generally , power factor is 0.8 lagging .

7

Selection of site :

The water supply should be available .

The site should be easily accessible.

Stability of foundation.

Land should be available at a reasonable price and further extension should be possible

Restrictions of surrounding.

Hydro-Electric power station

A generating station which utilizes the potential energy of water at a high level for the generation of electrical energy is known as hydro-electric power station.

The schematic arrangement of a hydro-electric plant is shown in below fig.

8

* Kinetic energy of water is its energy in motion ( flow of water ) and is a function of mass and velocity .

* Potential energy is a function of the difference in level of water between two points ( head ) .

Choice of site :

1- The station should be situated where a sufficient flow and head of water are available .

2- The site should be easily a accessible

There are three different methods of classification hydro-electric plants :

According to quality of water available , according to this classification the plants may be divided into :

a) Run-off river plant without pond age.

b) Run-off river plant with pond age.

c) Reservoir plants.

According to available head , according to this classification the plants may be divided into :

a) High head H > 300 m

b) medium head 30 < H < 300 m

c) Low head H < 30 m

According to nature of load. , according to this classification the plants may be divided into :

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a) Base load.

b) Peak load.

c) Pumped storage plants for peak load.

Main parts of hydro-electric plants:

1- Reservoir : Its purpose is to store water .

2- Dam : The function of dam is to provide a head of water to be utilized in the water turbine.

3- Penstocks : These are pipes of large diameter, taking water from the intake works to the power house ( turbine )

4- Prime mover : It is converted kinetic energy of ( turbine ) water into mechanical energy .

5- Power house : It consists of two main parts : * Turbine

* Electric equipment ( generator ).

Nuclear power station

A generating station in which nuclear energy is converted into electrical energy is known as a nuclear power station.

The schematic arrangement of nuclear power plant and nuclear reactor is shown in below fig.

10

11

Choice of site :

1- As in the case of steam power station the water should be available .

2- It should be situated a way from the populated area from the safety point of view.

3- The site should be easily accessible.

4- There should be sufficient space to get rid of radio active waste.

5- It should be situated as near to the load center as possible to avoid losses in transmission

Fuels:

Fuels generally used in reactor are natural Uranium ( U235

92 ) ,

Plutonium ( Pu239

94) and Uranium ( U

233

92 ) . Only (U

235

92 ) is naturally

available .

Natural uranium occurs in three isotopes (U235

92 , 0.7%) , (U

238

92,

99.3%) and (U234

92 . minute traces ) . Of these isotopes U

235

92 is very

easily and readily fissionable.

Pu239

94 and U

233

92are not found in nature , but can formed in the

nuclear reactors during fission process from U238

92 and Thorium Th

232

90

due to the absorption of neutrons with out fission .

Elements of a nuclear power station :

A nuclear power station differs from a conventional steam (thermal)

power station only in the steam generating part (i.e boiler) , the boiler

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of steam power station is replaced by the nuclear reactor and heat

exchanger.

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 .

Moderator : It is used to moderate or reduce the neutron speeds to value that increase the probability of fission occurring . made from Graphite heavy water or beryllium.

Shielding : the purpose of shielding is to give protection from the deadly radiation.

Control rod : by this rod can be controlled chain reaction . The materials of rod are Cadmium , Boron .

Coolant : A coolant transfer heat production in side the reactor to a heat exchanger . The coolants commonly used are gas ( air ,

Hydrogen , helium ) , water , heavy water , liquid metals ( sodium ) .

Diesel Power Station

A generating station in which diesel engine is used as the prime mover for the generating of electrical energy is known as diesel power station .

The fig. below shows the schematic arrangement of a typical diesel power station .

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Advantages

1- The design and layout of the plant are quite simple.

2- It occupies less space , and it can be located at any place .

3- It can be started quickly and can pick up load in a short time .

4- There are no standby losses , and it requires less quantity of water for cooling and less operating staff .

5- The overall cost is much less than that of steam power station of the same capacity .

6- The thermal efficiency of the plant is higher than that of asteam power station ( 35% )

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Disadvantages

1- The plan has high running charges as the fuel ( i.e diesel ) used is costly .

2- The plant does not work satisfactory under overload conditions for a longer period .

3- The plant can only generate small power .

4- The cost of lubrication and maintenance charges are generally high .

GAS Turbine Power Plant

A generating station which employs gas turbine as the prime mover for the generation of electrical energy is known as a gas turbine power plant .

In gas turbine , air is used as the working fluid . The air is compressed by compressor and is led to the combustion chamber where heat is added to air, thus raising its temperature . The hot and high pressure air from the combustion chamber is then passed to the gas turbine where it expands and does the mechanical energy .

The fig. below shows the schematic arrangement of a gas turbine power plant .

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Advantages

1- It is simple in design , and much smaller in size as compared to steam power station of the same capacity since no boiler and their auxiliaries such as feed water arrangement are required .

2- The initial and operation costs are much lower than that of equivalent steam power station .

3- It requires comparatively less water as no condenser is used .

4- The maintenance charge are quite small because it is simple in construction and operation as compared to steam turbine .

5- It can be started quickly from cold conditions .

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6- There are no standby losses .These losses occur in a steam power station because boiler is kept in operation even when the steam turbine is supplying no load .

Disadvantages

1- There is problem for starting the unit. It is because before thestarting the turbine , the compressor required power from some external source .

2- The overall efficiency of such plants is low ( about 20 % ) because the exhaust gases from the turbine contain sufficient heat , and a greater part of power developed by the turbine is used in driving the compressor .

3- The temperature of combustion chamber is quite high (3000˚ F) so that its life is comparatively reduced .

Load Curves and Factors

In choosing the type of generation ( thermal , hydro-electric and nuclear) ,and to select the size and number of generating units , a number of points have to be considered . Some of these are :

The kind of fuel available and its cost, availability of suitable sites for a hydro station , and the nature of load to be supplied.

The load which a power system has to supply is never constant because of variable demands at different time of the day .The variations can be see from the predication load curve.

load curve :

It is a graphic record showing the demand of the power for every

instant during the hour , the day , the month or the year .

The figure below , represent the daily load curve

17

Important terms and factors

The variable load problem has introduction the following terms and factors in power plant generating :

1. Connected load : It is the sum of continuous rating of all the equipments connected in supply system.

2. Demand : Demand of an installation or system is the load that is drawn from the supply at a specified interval of time , it is expressed in KWs , KVA or Amperes.

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3. Maximum demand : It is the greatest demand of load on the power station during a given period .

Max. demand is very important to determine the installed capacity of the station . The station must be capable meeting the max. demand .

4. Demand factor = loadConnected

demandMax.

Demand factor < 1 , it is used to determine the capacity of the plant equipment. and may indicate the degree to which the total connectedload is operated simultaneously .

5. Average load : The average of load occurring on the power station in a given period ( day or month or year ) is known as average load or average demand.

Daily average load = hours

dayingeneratedKWhunitsofNo

24

)(

= hours

KWhdayaduringgeneratedenergyTotal

24

= hours

curvedailytheunderArea

24

19

Also :

Monthly average load = monthainhoursofNo

KWhmonthaduringgeneratedenergyTotal

Yearly average load = )8760( hyearainhoursofNo

KWhyearaduringgeneratedenergyTotal

In general average load = periodTinhoursofNo

periodTduringgeneratedenergyTotal

6- Load factor (L.F.) = )(loaddemandMax

demandAverage

during a certain

period

= demandMax

periodTinhoursofNo

periodTaduringgeneratedenergyTotal

.

.

= periodTinhoursofNodemandMax

periodTaduringgeneratedenrgyTotal

..

Note : If T = 24 hours , the L.F is called daily load factor

20

L.F < 1 ; L.F demandMax.

1

Cost Plant Capacity of Station Max. Demand

Cost of plant stationpowerofFL.

1

i.e load factor is very important to determine the overall cost

of plant.

And it indicates the degree to which the peak load is sustained during the period .

7 – Diversity factor ( D.F ) = stationpowerondemandMax

demandindividualofSum

.

.max

Fig ( a ) 1. 2-6 100 KW = y1

2. 2-6 100 KW = y2

3. 2-6 100 KW = y3

Y

yyyFD

321.

= Y

y3 =

300

300 = 1 ( Very bad D.F )

Fig ( b ) 3100

300. FD ( Very good D.F )

21

KW KW

y1

y2 Y

y1 y2 y3 Y y3

0 8 16 24 hours 0 14 16 hours

Fig. b Fig. a

D.F 1 ; D.F demandMax.

1

Cost of plant FD.

1

8 – Capacity factor = producebeenhavecouldthatenergyMax

producedenergyActual

.

22

= TdemandMax

TdemandloadAverage

.

)(

= CapacityPlant

demandAverage

If the period is one year ,

Annual Capacity factor = 8760capacityPlant

outputKWhAnnual

9- Reserve Capacity = Plant Capacity – Max. Demand

10 - Plant use factor = useofhourscapacityPlant

KWhinoutputStation

EX. 20 MW power station , produce annual output of 6.5 610 KWh and remains in operation for 2100hours in year , find the plant use factor.

plant use factor = 2100)1020(

10)105.6(6

36

= 0.154

= 15.4 %

23

Load Duration Curve

When the load elements of a load curve are arranged in the order of descending magnitudes , the curve thus obtained is called a load duration curve . It gives the data in more presentable form.

Figures below represent i) daily load curve , ii) daily load duration curve

Area under daily load curve = Area under daily load duration curve .

= Total energy generated ( KWh ) on the day

24

Load curves and selection of the number and sizes of the generation units

The number and size of generating units are selected in such a way that they correctly fit the station load curve as shown :

Time Units in operation

0-----7 1

7-----12 1 + 2

12-----14 1

14-----17 1 + 2

17-----22.30 1 + 2 + 3

22.30-----24 1 + 2

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The important points in the selection of generator units are:

The selection of units should be approximately fit the annual( yearly) load curve of the station.

The capacity of the plant should be made 15 to 20% more than the maximum demand.

One unit should be kept as a spare generating unit ( stand by unit ).

By using identical units ( having the same capacity ) ensure saving in cost of station .

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The load curve can be fit very accurately if large number and small capacity of units are selected , this is one side , in other side , the investment cost per KW of capacity increases as the size of the units decreases .

Inter Connected Grid System

The connected of several generating station in parallel is known as inter connected system

Example: A power station is to supply three consumers . The daily demand of three consumers are given below :

Time (hours) Consumer (1) Consumer 2) Consumer (3)

0 – 6 200 KW 100 KW No - load

6 – 14 600 KW 1000 KW 400 KW

14 – 18 No - load 600 KW 400 KW

18 – 24 800 KW No - load 600 KW

Plot the load curve of power station and , find :

1- Load factor of individual consumer.

2- Diversity factor of power station .

3- Load factor of power station .

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Solution :

(KW)

2400

2000 2000

1600

1200 1400

800 1000

400 300

6 14 18 24 hours

1- load factor of consumer = 100.

/

dayinhoursdemandMax

dayconsumedEnergy

load factor of consumer(1) =

%25.5610024800

68004086006200

load factor of consumer(2) =

%8.45100241000

604600810006100

load factor of consumer(3) =

%3.5810024600

66004400840060

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2- Diversity factor = stationpowerondemand

demandsindiviualofSum

.max

.max

From load curve , the Max. demand on power station is 2000 KW

Diversity factor = 2.12000

6001000800

3- Load factor of power station =

%9.62100242000

6140041000820006300

Example: A power station is to supply two loads . The daily load curve of the station is as shown in figure below . If the load factor of the two loads are 0.416 and 0.458 respectively , and the diversity factor of the power station is 1.142 , find :

1- The load factor of power station .

2- The energy consumed per day and the maximum demand of each load .

KW

1400

1200

1000

800

600

400

200

6 12 14 20 24 hours

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Solution :

1- load factor ( L.F ) = 100.

/)(

dayinhoursdemandMax

daygeneratedconsumedEnergy

Load factor of power station =

241400

)6006(2

1)4006(

2

140010400640018

= 49.0241400

16600

2- Let : E1 – energy consumed by load 1

E2 – energy consumed by load 2

M1 – Max. demand of load 1

M2 – Max. demand of load 2

L.F1 , L.F2 - load factor of loads 1 and 2

124

11.

M

EFL

; 124

1416.0

M

E

10 M1 = E1 ……….. (1)

30

224

2458.0

M

E

11 M2 = E2 …………………(2)

Diversity factor = stationpowerondemand

demandsindiviualofSum

.max

.max

1400

21142.1

MM

1600 = M1 + M2 ………….(3)

Energy generated from power station = E1 + E1

16600 = E1 + E2 …………….(4)

Put (3) and (4) in (2)

11 ( 1600 – M1 ) = 16600 – E1

17600 – 16600 – 11 M1 + E1 = 0 ……….(5)

Put (1) in (5)

1000 – 11 M1 + 10 M1 = 0

M1 = 1000 KW

From (3) M2 = 1600 – 1000 = 600 KW

From (1) E1 = 10 x 1000 = 10000 KWh

From (2) E2 = 11x 600 = 6600 KWh

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Discussion Questions

1- Draw the important components of a steam power station . 2- What factors are taken into account to select the site of steam power station . 3- State the main parts of a thermal power station , and discuss the function of each part .

4- Draw the important components of a Hydro-Electric power station .

5- What factors are taken into account to select the site of Hydro-Electric power station .

6- Draw the important components of a nuclear power station .

7- What factors are taken into account to select the site of nuclear power station . 8- What do you understand from the load curve and what information are

converged by a load curve. 9- Discuss the important points to be taken into consideration while selecting the

size and number of units generation . 10- Explain the terms , connected load , load factor , plant use factor , diversity

factor .

Tutorial Problems

Q1 A power station is to supply three loads. The daily cycle of loads are given below:

Time in hours 0 - 6 6 - 10 10 - 12 12 - 18 18 - 22 22 - 24Loads A in MW 10 15 20 30 15 5

Time in hours 0 - 10 10 - 20 20 - 24Loads B in MW 0 15 0

Time in hours 0 - 4 4 - 10 10 - 14 14 - 16 16 - 20 20 - 24Loads C in MW 5 15 50 10 20 10

1. Draw the load and duration curves of power station.2. Calculate: i- load factor of each load.

ii- load factor of the power station. iii- Diversity factor of the power station.

iv- The capacity factor when the station is having reserve set 20% of the max. Demand.

32

Q2 A power station is to supply two loads . The daily load curves of two loads are shown in figures below :

MW MW

25 25

20 20

15 15

10 10

5

4 12 16 20 24 hours 12 16 24 hours

Load curve A Load curve B

1- Plot the load and duration curves of power station .2 - Find :

i- Load factor of each load.ii- Load factor and Diversity factor of power station.iii- The reserve capacity and Capacity factor of the power station .If the

power station consists of three units , the installed capacity of each unit is 20 MW.

Q3 A Power station is to supply two loads . The daily of loads are given below.

22-2416-228-166-80-6Time in hours353018124Load A in W

20-2416-2010-162-100-2Time in hours

102520158Load B in W

Draw the load and duration curves of power station, and find:1- Load factor of each load.2- Load factor of the power station.3- Diversity factor of the power station.

4 - The capacity factor when the station is having reserve set 20% of the max. Demand.

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Prediction of load and Energy Requirements

The nature of load to be supplied affects the choice of plant to a considerable extent. The load which a power system has to supply is never constant because of variable demands at different time of the day . The variations can be seen from the predicated LOAD CURVE.

The minimum capacity of generating station must be such as to meet the maximum demand At the same time it is essential for the power system to maintain reliability and continuity of power supply at all time.

Therefore to determine the prediction of load and energy requirement must be draw the load curve of demand. There are some methods for this purpose ,are:

1- Load survey

2- Methods of extrapolation

3- Mathematical methods

4- Mathematical methods using economic parameters.

Choice of Type, Size and Number of Generator Units

In choosing the type of generation ( thermal, hydro-electric and nuclear ) , and to select the size and number of generating units, a number of points have to be considered. Some of these are: The kind of fuel available and its cost, abundant quantity of cooling water should be available ( for thermal station ), availability of suitable sites ( for hydro station ), and the nature of load to be supplied.

The important points in the selection of generator units are:

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The selection of units should be approximately fit the annual ( yearly) load curve of the station.

The capacity of the plant should be made 15 to 20% more than the maximum demand.

One unit should be kept as a spare generating unit ( stand by unit ).

By using identical units ( having the same capacity ) ensure saving in cost of station .

٣٣

Power Transmission

There are two types of transmission system:

a- Overhead transmission line .

b- Under ground cables .

Overhead transmission line

An overhead transmission line consists of conductors , insulators , support structures , and in most cases , shield wire ( ground wire ) .

Economic choice of transmission voltage

……………( 1 )cos1 VIP phase

IF cos constant

,

,,,

).(..1 { loss

gearsswitchandinsulatorstowersrstransformeofcostRVregulationVoltagecostconductorofASCphase IVP

The transmission losses may be computed from the approximate formulas :

٣٤

var2

22

2

22

tr

trtrlloss

tr

trtrlloss

V

QPXQ

wattV

QPRP

….( 2 )

Where : lR - resistance of line

lX - reactance of line

trtr QandP are active and reactive transmission power ( power of load )

Above discussion and formulas informs that the cost for lost energy decreases with increased voltage level . However , the fixed costs of (towers , insulators transformers and switchgears ) increase with voltage .

Therefore , for every transmission line , there isoptimum transmission voltage , beyond which there is nothing to be gained in the matter of economy.

The fig. blow shown that the total transmission costs will therefore be minimized at a certain voltage level. It is called economical transmission voltage .

٣٥

transmissioncosts

total costs

fixed costs

cost for lost energy economical voltage

transmission voltage level

The empirical formula to find the economical transmission voltage between lines in a 3-phase ac system is :

150

362.05.5

PlV …..( 3 )

Where V – line voltage in KV . P – maximum KW per phase to be

delivered to single circuit . l – length of transmission line .

The larger power to be transmitted and greater the distance over which they must be wheeled , the higher must be the operation voltage level chosen .

٣٦

Standard voltage levels

Low voltage transmission (KV) : 11 , 22 , 33 , 66

High voltage transmission (KV) : 110 , 132 , 220 , 275 , 330

Extra high voltage transmission (KV) : 380 , 400 , 500 ,750 , 1000

1100-1500 under research

Conductor material

Characteristics of material ;1- High electrical conductivity .2- High tensile strength .3- Low cost .

Better materials : 1- Copper (Cu): Conductivity and tensile strength of copper

are high , but cost of material also high .

2- Aluminum (AL) :a- conductivity of AL is 60% of that of Cu .b- Coefficient of expansion is high .c- Weak (low) tensile strength .

٣٧

d- Lower cost and lighter weight of an AL conductor compared with a Cu .

e- For the same resistance , AL conductor has a large diameter than an Cu conductor , i.e low effect of corona .

Therefore aluminum has replaced copper as the most common conductor metal for overhead transmission .

Symbols identifying different types of AL conductors are as follows : AAC : all- aluminum conductor .AAAC : all- aluminum -alloy conductor .ACAR : aluminum conductor alloy –reinforced .ACSR : aluminum conductor , steel – reinforced .

The most common conductor types is ACSR , which consists of layers aluminum strands surrounding a central core of steel strands as shown in fig. below .

Steel strands

Aluminum strands

٣٨

Stranded conductors are easier to manufacture , since larger conductor size can be obtained by simply adding successive layers of strands , and also easier to handle and more flexible than solid conductors .

The number of strands depends on the number of layers and on whether all the strands are the same diameter . (The strands are uniform diameter) :

Total No. of strands ( sN ) = )1(31 nn …( 4 )

Where n – is the number of layers around the central strand .

No.of layers ( n )

1 2 3 4 5

Ns, including the Single center strand

6+1=7 18+1=19 36+1=37 60+1=61 90+1=91

First layer

Second layer

٣٩

Stranded conductors usually have central wire (core) around which are successive layers of ( 6 , 12 , 18 ,24 ) wires as shown in above fig.

The equivalent diameter of stranded conductor is given by :

dnD )21( ……………..( 5 ) Where d - is the diameter of the strand.

EX. Stranded conductor 19/2.9 mm . calculate the equivalent diameter of conductor .

Solution : The diameter of one strand (d) = 2.9 mm No. of stranded in conductor (Ns) = 19 Therefore for No. of layer = 2 , distributed

as follow : = [ 1 + 6 + 12 ] core 1st layer 2nd layer

or can be calculated by eq. ( 4 ) as follow :

)1(3119 nn

nn 3318 2 062 nn

0)2)(3( nn 2 n

dnD )21( mm5.149.2)221(

٤٠

Parameters of overhead transmission lines

Transmission lines (T.L) are basically circuits with distributed constants ( parameters ) : series resistance , series inductance , shunt capacitance , and shunt conductance are distributed a long the whole length of line ,as shown in below fig.

Series resistance accounts for ohmic ( RI 2 ) line losses.Series impedance , including resistance and inductive reactance , gives rise to series – voltage drops along the line .Shunt capacitance gives rise to line – chargingcurrents . Shunt conductance accounts for ( GV 2 ) line losses due to leakage currents between conductors or between conductors and ground over insulators , addition to corona losses . Shunt conductance of overhead lines is usually neglected.

٤١

The same current flows through the upper and lower part of each section the resistance and inductance of both can be combined and the equivalent circuit of below fig. is obtained . Thus :

., 111111 etcllLrrR

Then if each section of the line is of equal length corresponding to unit length ( say one meter ) of the line we will have :

RRRR .......321 , resistance per unit length of the line i.e. ohms/loop meter . Similarly ,

LLLL .....321 henrys/loop meter , GGGG .....321 mhos/loop meter , andCCCC ......321 farads/loop meter

Section of unit length

٤٢

Reaistance

The d.c resistance of various type of conductors at specified temperature T is found by :

A

lR T

Tdc

. …………….( 6 )

Where T - conductor receptivity at temperature T ( m. ) .

l – conductor length (m) . A – conductor cross- sectional area ( 2m ) .

The resistance of T.L conductor is effected by the following factors :

1- Spiraling of a stranded conductor :

For stranded conductors , alternate layers of strands are spiraled in opposite directions to hold the strands together . Spiraling makes the strands length 2% longer than the actual conductor length . As a result , the dc resistance of a stranded is 1 or2% larger than that calculated from eq. ( 6 ) for a specified conductor length .

٤٣

2- Temperature :

Temperature correction are determined by :

1

2

1

2

tT

tT

R

R

……………..( 7 )

Where 21 RandR are the resistances of conductor at temperatures 21 tandt respectively .

T = 234.5 for annealed copper of 100% conductivity = 241.5 for hard-drawn copper of 97.3%

conductivity .= 228.1 for hard-drawn aluminum of 61%

conductivity .

3- Frequency ( skin effect ) :

For dc , the current distribution ( current density ) is uniform throughout the conductor cross section , and eq. (6) is valid . However , for ac , the current distribution is nonuniform . As frequency increases , the current in a solid cylindrical conductor tends to crowd toward the conductor surface , with smaller current density at the conductor center. This phenomenon is called skin effect .The increase in resistance due to skin effect can be determined from tables or curves .

٤٤

Inductance :

It is defined as the flux linkages per unit current . i.e :

henrysI

L

…………………( 8 )

Where - is flux linkage ( Weber-turns ) I - is current ( amperes )

For instantaneous value of current and flux linkage,

iL

[ and I are in phase , L is real . ]

This is true provided flux linkage of circuit vary linearly with current , i.e , the magnetic circuit has constant permeability ( ) . If permeability is not constant we have to make use of the second fundamental equation :

dt

diL

dt

dLi

dt

de

…………………( 9 )

Where e – is induced voltage ( volt ) . L – is constant of proportionality , inductance

( henrys ) .

٤٥

dt

di- is rate of changes of current

( amps/second ) .

Mutual inductance between two circuits is defined as flux linkages of one circuit per unit current in the other circuit .

henrysI

M2

1212

……………….(10)

Where 12 - is flux linkages with circuit 1 (Weber-turns)

2I - is current in circuit 2 (amps)

Inductance of a solid cylindrical conductor due to internal flux :

dx

1.dxA

INmmf lHmmf

)(m

AT

l

IN

l

mmfH

٤٦

Where H – field intensity . l - length of path of flux .

xl 2N – No. of turns of conductor { for our analysis N=1

(one conductor )}

IlH

Ampere's law : [ The mmf around any closed path

( xl 2 ) equals the current enclosed by the path ( xI ) ]

Or xx IHx 2

Where xH - is tangent of contour , it is constant.

xI - is the portion of total current enclosed by the contour .If current density in the conductor is uniform ;

Ir

xI x 2

2

I - total current in the conductor .

mATIr

xI

xr

x

x

IH x

x /22.2 22

2

٤٧

We know : HB Where B – flux density . - permeability of conductor .

And , HB r 0

Where r - relative permeability

0 - permeability of free space

meterhenrys /104 70

2

2/

2mWbI

r

xBx

Also , we know : AB

Where - flux ( Weber )

A – area (2m )

The differential flux d per – unit length of conductor in the cross - hatched rectangle of width dx shown in fig. above , is :

٤٨

dABdAB x .;

mWbdxr

IxdA

r

Ixd /1.

22 22

Where d - flux enclosed in element of thickness dx per meter ( axial ) length of conductor .

So that flux linkage per meter length of conductor :

mTWbdxr

Ixd

r

xd /.

2 4

3

2

2

Integrating above eq. from x = 0 to x = r determines

the total flux linkages .int in side the conductor :

mTWbIx

r

I

dxxr

IdxI

r

x

r

rr

/.842

22

0

4

4

0

34

04

3

.int

For nonmagnetic conductor , r = 1 ;

therefore , 7

0 104

٤٩

mTWbII

/.102

1

8

104

87

70

.int

The internal inductance .intL per – unit length of conductor due to this flux linkage is then :

meterhenrysI

L /2

10 7.int

.int

…(11)

Flux linkages between two points external to an isolated Conductor ( Inductance of a solid cylindrical conductor due to external flux ) :

conductor carried 1D 1P current = I x dx

2D

2P

lHINmmf 1;2 Nxl

N – No. of turn

٥٠

IIHx xx 2

mAT

x

IH x /

2

AHB

2/

2mWb

x

IBx

dABdAB x .;

The flux links with current I in the thickness dx :

mWbdxx

Id /

2

The flux linkage per meter :

dd ( flux external to the conductor links all the current in the conductor . )

The flux linkages between 1P and 2P :

2

1

/.ln22 1

212

D

D

mTWbD

DIdx

x

I

For nonmagnetic conductor , r = 1 ;

٥١

therefore , 7

0 104

mTWbD

DI /.ln102

1

2712

mhenrys

D

D

IL /ln102

1

271212

…..(12)

Inductance 12L , this only for flux linkage of an isolated conductor which lie between the points

1P and 2P distant 1D and 2D respectively from the center of the conductor as shown in fig.

Inductance of single – phase Two – wire line :

conductor

radius 2r conductor

radius 1r

For external flux only ( using eq. 12 )

٥٢

mhenrysr

DL ext /ln102

1

71

For internal flux only ( eq. 11 )

meterhenrysL /2

10 7

.int1

Therefore , total inductance of the circuit due to current in conductor 1 only is :

extLLL int1

1

771 ln10210

2

1

r

DL

1

4/17

1

7 lnln102ln4

1102

r

De

r

D

14/1

74/1

1

7 ln102ln102re

De

r

D

But , 7788.04/1 e

1

71 7788.0

ln102r

DL

mhenrysr

DL /ln102

1

71

……( 13 )

Where , 7788.01 r

1r is called Geometric mean radius ( G.M.R )

٥٣

1r = 0.7788 times the radius of conductor .

Also, the inductance due to current in conductor No.2 :

mhenrysr

DL /ln102

2

72

Total inductance for the complete circuit :

21

721 lnln102

r

D

r

DLLL

21

27 ln102

rr

D

21

27 ln

2

1104

rr

D2/1

21

27 ln104

rr

D

21

7 ln104rr

D

If the radius of the two conductor is same , i.e :

1r = 2r = r , therefore rrr 21

٥٤

)/(/ln104 7 mHmhenrysr

DL

And ,

KmmHr

DL /ln4.0

...........( 14 )

Sometimes this inductance is called inductance per loop length , it is double the inductance per conductor in a single phase line .

Flux linkages of one conductor in a group :

0.......321 nIIII

٥٥

If 11 p denotes all the flux linkages of conductor 1

due to its own current 1I , internal and external , upto point P . By using eq. (11) and (12) :

mTWbr

PDI

r

PDI

Ip

/.ln102

10ln22

1

11

7

7

1

11

111

where 1r = 0.7788 1r

Also , 21 p - is flux linkages with conductor 1 due to current in conductor 2 ( 2I ) , but excluding flux beyond point (P) is equal to the flux produced by 2I

between the point P and conductor 1 (i.e. the flux linkages due to 2I with in limiting distance pD2 and

12D from conductor 2 ) .

12

22

721 ln102

D

DI p

p

Similarly for pnp 131 ........,

The flux linkages p1 with conductor 1 due to

nIIII ,...,, 321 , but excluding flux beyond point P is:

n

npn

pppp D

DI

D

DI

D

DI

r

DI

113

33

12

22

1

11

71 ln....lnlnln102

٥٦

npnppp

nn

DIDIDIDI

DI

DI

DI

rI

ln.....lnlnln

1ln....

1ln

1ln

1ln

102

332211

1133

122

117

But , 1321 ....... nn IIIII

npn

ppp

nn

p

DIII

IDIDIDI

DI

DI

DI

rI

ln)......

(.....lnlnln

1ln....

1ln

1ln

1ln

102

132

1332211

1133

122

11

71

np

pnn

np

p

np

p

nn

p

D

DI

D

DI

D

DI

DI

DI

DI

rI

)1(1

22

11

1133

122

11

71

ln......lnln

1ln....

1ln

1ln

1ln

102

Now if point (P) moves infinity, terms such as :

np

pn

np

p

np

p

D

D

D

D

D

D )1(21 .....;

approach the value 1 and ,

ln(1) = 0

٥٧

nn D

ID

ID

Ir

I113

312

21

17

1

1ln....

1ln

1ln

1ln102

Wb.T / m

Also denoting 1r as 11D we have :

nn D

ID

ID

ID

I113

312

211

17

1

1ln....

1ln

1ln

1ln102

Wb.T / m ….(15)

Inductance of composite conductor lines :

conductor A conductor B carried current I carried current - I

٥٨

Applying eq. (15) to segment (filament ) (a) of conductor A , we obtain for flux linkage of filament (a)

qabaaa

apabaaa

Dq

I

Dq

I

Dq

I

DP

I

DP

I

DP

I

1ln.....

1ln

1ln102

1ln.....

1ln

1ln102

7

7

qqa

qba

qaa

p

ap

p

ab

p

aa

DDD

DDDI/1/1/1

/1/1/1

7

ln.....lnln

1ln.....

1ln

1ln

102

Where q

aaqaa

q

aa

DDD

/1

/1

/1

ln1

ln1

ln

mTWbDDDD

DDDDI

papacabaa

qqacabaaa

a /......

.....ln102 7

mHDDDD

DDDDp

pIL

papacabaa

qqacabaaaa

a /.....

.....ln102

/7

……(16)aL - Inductance of filament (a) .

In similar manner we may write :

mHDDDD

DDDDp

pIL

pbpbcbbba

qqbcbbbabb

b /.....

.....ln102

/7

٥٩

Where :

xyx

xyxy

zyxzyx

y ln)(ln

lnln)/(ln

lnlnln)(ln

The average inductance of the filaments of conductor A is :

p

LLLLL pcba

average

.....

Conductor A is composed of (p) filaments electrically in parallel ;

1- If , averagepcba LLLLL

p

LL a

A

Where AL , inductance of conductor A .

2 - If pcba LLLL

2

......

P

LLLL

P

LL pcbaaverage

A

mH

DDDD

DDDD

DDDDDDDD

DDDDDDDD

L

pppcpbpa

qpcpbpap

pbpbcbbbaapacabaa

qbcbbbabpq

qacabaaa

A /

).........(

).........(

)........)(.....(

)........().....(

ln1022

7

…………(17)

٦٠

The numerator is called Geometric mean distance ( G.M.D ) between conductors A and B , and denoted

mD , but denominator is called Geometric mean radius ( G.M.R ) and denoted SD .

meterhenrys

D

DL

SA

mA /ln102 7 …..(18)

meterhenrys

D

DL

SB

mB /ln102 7

If conductors A and B are identical i.e. SSBSA DDD

KmmHD

D

mHD

DLLL

S

m

S

mBA

/ln104.0

/ln104

7

7

.…(19)

Where L – inductance of line (or loop)

٦١

Inductance of three-phase lines :

A) Equilateral spacing :

0 cba IIIBy using eq.(15)

mTWbD

ID

Ir

I cbaa /.1

ln1

ln1

ln102 7

)( cba III

mTWbr

DIaa /.ln102 7

mHr

DLa /ln102 7

……….(20)

rr 7788.0aL - Inductance per phase of 3-phase line .

IF conductor is stranded , SDr , therefore :

mHD

DL

Sa /ln102 7

.........(21)

٦٢

B) Unsymmetrical spacing :

In this case , the flux linkages and inductances of the phases are different , since the circuit becomes unbalance . Balance of the three phases can be restored by exchanging the positions of the conductors at regular intervals along the line so that , the conductor of each phase occupies each of the three positions 1,2 and 3 for about one-third of its length , as illustrated in below fig. Such an exchange of conductor position is called transpositions

٦٣

1- Transposed conductor with unsymmetrical spacing:

Assume 321 ,, aaa - are the flux linkages of conductor (a) in positions 1 , 2 , and 3 respectively and by using eq. (15) :

3112

71

1ln

1ln

1ln102

DI

DI

rI cbaa

1223

72

1ln

1ln

1ln102

DI

DI

rI cbaa

2331

73

1ln

1ln

1ln102

DI

DI

rI cbaa

Then average flux linkages of conductor (a) :

3321 aaa

a

312312312312

7 1ln

3

11ln

3

11ln102

DDDI

DDDI

rI cbaa

3312312

3312312

7 1ln

1ln

1ln102

DDDI

DDDI

rI cba

But , )( cba III

mTWbr

DDDIaa /.ln102

33123127

٦٤

mH

r

DDD

IL

a

aa /ln102

33123127

mH

D

DL

s

eqa /ln102 7

………(22)

Where , rDDDDD seq ;3312312

cba LLL ( because of the circuit is balance )

Average phase inductance = cba LLL 3

1

aa L

L

3

3

mH

D

D

s

eq /ln102 7

2- Untransposed conductors with unsymmetrical spacing :

By using eq. (15) , flux linkage of conductor (a)

3112

7 1ln

1ln

1ln102

DI

DI

rI cbaa

٦٥

And inductance of phase (a) , a

aa I

L

3112

7 1ln

1ln

1ln102

DI

I

DI

I

rL

a

c

a

ba

0 cba III

If aI is taken as a reference phasor :

120;240;0 acabaa IIIIII

And ,

2

3

2

1

240sin240cos2401

j

jI

I

a

b

2

3

2

1

120sin120cos1201

j

jI

I

a

c

Thus summarizing :

mHD

Dj

r

DDLa /ln3ln102

31

1231127

mHD

Dj

r

DDL b /ln3ln102

12

2312237

٦٦

mHD

Dj

r

DDLc /ln3ln102

23

3123317

...........(23)

IF conductor is stranded , SDr , in above equations.Inductance of bundled conductors :

It is common practice for Extra high voltage (EHV) lines to use more than one conductor per phase , a practice called bundling . Bundling reduces the electric field strength at conductor surfaces , which in turn reduces or eliminates corona , and also reduces the series reactance ( Z ) of the line by increasing the GMR of the bundle . Fig. below shows common EHV bundles consisting of two , three , or four conductors .

٦٧

d - bundle spacing .To calculate inductance of bundle conductor , SD in eq.( 21 , 22 , 23 ) replaced by the GMR of the bundle ,

which is denoted by bSLD .

1- For two – conductor bundle :

)()()( 4 22 22dDdDdDD SSS

bSL

2- For three – conductor bundle :

3 29 323 3 )()(2

dDdDddDD SSSbSL

3- For four – conductor bundle :

4 3

4 316 434 4

09.1

2)2()2(2

dD

dDdDdddDD

S

SSSbSL

IF conductor is solid , rDS in above three condition

٦٨

Inductance of parallel – circuit three – phase lines with symmetrical spacing :

a a a C

b b b b

c C c a

incorrect correct Tower arrangement of conductors

Flux linkage of phase (a) conductors :

D

D D D3 2D D3

D 30° 30° D

D

٦٩

DDI

DDI

DrI

c

ba

a1

ln3

1ln

3

1ln

1ln

2

1ln

1ln

102 7

22

7

3

1ln

3

1ln

2

1ln102

DI

DI

rDI cbaa

2

7

3

1ln

2

1ln102

DII

rDI cbaa

But , 0 cba III

2

7

3

1ln

2

1ln102

DI

rDI aaa

r

DI

rD

DI aaa 2

3ln102

2

3ln102 7

27

Inductance of phase (a) :

mHr

D

IL

a

aa /

2

3ln102 7

……(24)

Inductance of each conductor aLaora 2)()(

mHr

D

r

D/

2

3ln104

2

3ln1022 77

٧٠

EX.1 A 3-phase , 50Hz , 132 KV overhead transmission lines has conductor diameter of 4 cm each , are arranged in a horizontal plane as shown in fig . supplies a balanced load , assume the line is completely transposed . Find the inductance per Km per phase .

Phase A Phase B Phase C

4m 4m

Solution :

3CABCABeqm DDDDD

m04.58443 rrDS 7788.0

cm5576.1

2

47788.0

277

105576.1

04.5ln102ln102

s

m

D

DL

mH /1053.11 7KmmH /153.1

٧١

EX.2 A 3-phase , 50Hz , 400 KV overhead transmission lines are arranged in a horizontal plane , each phase has two – strand bundle conductors , the diameter of each strand is 25mm , as shown in the fig. below. Find the inductance per Km per phase .

0.3m 25mm

6m 6m Phase A Phase B Phase C

Solution :

3CABCABeqm DDDDD

m56.712663

rrDS 7788.0

mm735.9

2

257788.0

dDD SbSL

m054.03.010735.9 3

054.0

56.7ln102ln102 77

s

m

D

DL

KmmHmH /988.0/1088.9 7

٧٢

Discussion Questions

1 – The resistance of a transmission line is effected by some factors . Explain these factors .

2 – What is the effect of skin effect in the resistance of transmission line conductors .

3 – Show that the inductance of a conductor due to internal flux is given by :

mHL /102

1 7

4 –What is the effect of temperature in the resistance of transmission line conductors .

5 – Derive in expression for the inductance per phase for a 3- phase overhead transmission line when conductors are of equilateral spacing .

6 – What is the effect of unsymmetrical spacing of conductors in a 3-phase transmission line .

7 – State the advantage of bundle conductor lines over single conductor lines .

Tutorial Problems

Q1. A 3-phase , 50 Hz , 110 KV , overhead transmission line consists of three solid conductors of 3 cm diameter positioned on the corners of triangle with sides of 2 m , 2.5 m , 3.125 m .

If the conductor of each phase of this line is replaced by three-strand bundle conductor has the same equivalent area of one solid conductor , and the space between the strands of bundle is 0.2 m .

Find the inductance for two conditions . Assume the internal magnetic flux of conductor is neglected and the line is transposed.

٧٣

Q2. A 3-phase , 132 KV , 50 Hz , 100 Km , single circuit bundle conductor transmission line as shown in the figure below . If thediameter of each strand is 1 cm , and the conductors are regularly transposed . Determine , the inductance per phase per Km .

6 m

5 m

0.18 m