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Electrical Power Systems

References

- Electrical power
- ( Dr\ S.L. Uppa

) - Elements of power system analysis
- ( William

Stevenson ) - Modern power system analysis
- ( I.J.Nagrath ,

D.R. Kothari )

Elements of power system

- 1. Power stations .
- 2. Substations .
- 3. Busbars .
- 4. Primary T.L and secondary T.L .

Step Up Transformer

Step down Transformer

Power Station

Transmission Line 1

T L 2

S

11/500 kV

500/220 kV

220/66 kV

Busbars

Loads

66/11 kV

11kV/380V

Elements of Power System

Standard voltages

- Generation voltages .
- 3.3 , 6.6 , 11 , 33 KV
- 2. Transmission line voltage .
- 11 , 33 , 66 , 110 , 132 , 165 ,

220 , 380 , 400 , 500 , 750 KV - 3. Distribution high voltages .
- 11 , 6.6 KV
- 4. Distribution low voltages .
- 380 , 220 V

Standard voltages In Egypt

- T.L voltages 66 , 220 KV and 500 KV from

High Dam to Cairo

Power stations

- Types of power stations
- Thermal power stations .
- Hydro power stations .
- Nuclear power stations .
- Gas power stations .

Thermal power stations

- It dependents on coal and petrol to heat the

water in big boilers under high pressure to

transfer the water to steam.

Turbine

Generator

BB

Condenser

Cooler

Coal Burner

Boiler

Pump

Thermal Power Station

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Advantages of this stations .

- It has low construction ( primary ) cost .
- It uses small area to construct .
- No trembles ( vibrations ) .
- It can be constructed in minimum time compared to

Hydro and Nuclear power stations . - It can be constructed near to the load .

Disadvantages

- Running costs are high because it uses coal and

solar . - The response is very low to supply the increasing

in load . - It is not clean and causes more pollution .
- It has low efficiency ( 25 40 ) .

Some of consideration must be taken

- It must near to source of water .
- It must near to transmitted tools .
- It needs to strong land which has low price .
- We can extend the station.

Hydro power stations

- It depends on two deferent levels in the river.

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Advantages of this stations .

- Running cost is very low because it depends on

water . - The response is very high to supply the sudden

increasing in load. - It is clean .
- Efficiency is equal ( 90 95 ) .

Disadvantages

- It has high construction cost .
- It has high T.L costs .
- Constructed far from the loads .
- Takes more time to construct .

Nuclear power stations

- When Uranium 235 is crashed with neutrons,

releasing neutrons and heat energy . These

neutrons then participate in the chain reaction

of fashioning more atoms.

Control rots

Heat exchanger

Steam

Turbine

Generator

CO2

Reactor

Water

Condenser

Fuel rots (Uranium 235)

Nuclear Power Station

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- Advantages and disadvantages are similar to

hydro power station added to that it has higher

protection cost and it is constructed in the

desert.

Gas power stations

- It depends on the outage of gases from the

refine petrol factory which produces high

pressure gas, (in the Max and Tebein in Egypt) .

Series impedance of T.L .

- It has four parameters resistance, inductance,

capacitance and conductance. - Conductance is between the conductors and between

conductors and the ground through the insulators

as a leakage current. - The resistance and inductance are uniformly

distributed along the line.

Types of conductors .

- Copper conductor .
- Aluminum conductor .
- Aluminum conductor, steel reinforced
- Aluminum conductors have replaced copper because

of the much lower cost and lighter weight.

Resistance

- Resistance causes power loss in the T.L.
- For uniform resistance ,

Relation between resistance and temperature rise .

t

t2

t1

R

R1

R2

T

The influence of skin effect on resistance

- Uniform distribution of current throughout the

cross-section of a conductor exists only for D.C.

- In A.C, as increased of frequency, the

non-uniformly of distribution of current becomes

more appearance. This case is called skin effect.

- The alternating flux induces higher voltages

acting on the interior filaments than are induced

on filaments near the surface of the conductor .

J

r

r

r

Bundled conductors

- The trend toward ever higher voltages for T.L has

stimulated interest in the use of two or more

conductors per phase. - Such a line said to be composed of " bundled "

conductors.

- Usually the spacing of conductors of a phase is

about ( 10 ) times the diameter of one conductor

, that is about ( 8 to 20 in ). - The advantages of bundling are reduced reactance

because of increased self SGM and reduce voltage

drop and voltage gradient which result in reduced

radio interference

Transmission Circuit Calculations

Short Transmission line

- In the case of a short transmission line the

capacitance and conductance to earth may be

neglected. - Leaving only the series resistance and inductance

to be taken into consideration. - The current entering the line at the sending-end

termination is equal to the current leaving at

the receiving-end, and this same current flows

through all the line sections. - The R and L parameters may therefore be regarded

as ' lumped ' .

- The equivalent circuit diagram and the vector

diagram for a short line are shown in fig.( 6.1 )

in which

Fig.( 6.1 a ) Equivalent circuit for a short

transmission line

Vs

Ir ZL

Ir XL

Vr

Øs

Ir R

Ør

Is Ir

Fig.( 6.1 b ) Vector diagram for a short

transmission line .

- The currents IS and IR will be equal in magnitude

but not in phase. - R is obtained from a knowledge of the line length

,the size of conductor and the specifics

resistance of the conductor material , - while XL is calculated from the conductor

spacing and radius .

- Referring to the equivalent circuit
- Hence, if the receiving-end conditions are known

the necessary sending-end voltage may be

calculated .

- It will be noted that ( 6.1a ) and ( 6.1b ) are

phasor equations , a more approximate method

involving scalar quantities is as follows

Referring to the vector diagram,

- However (IR XL) and (IR R) are very much less

than VR and the small voltage is in quadrate with

the much larger VSX , - The voltage regulation of the line is given by

the rise in voltage when full loads is removed ,

or

Example

A three-phase line delivers 3 MW at 11 KV for a

distance of 15 Km . Line loss is 10 of power

delivered , load power factor is 0.8 lagging .

frequency is 50 Hz , 1.7 m equilateral spacing of

conductors . Calculate the sending-end voltage

and regulation .

Solution

- Assuming that the conductors are manufactured

from copper having a resistance of 0.0137 ohms

per meter for a cross-sectional area of 1 mm2 ,

the conductor cross-section is 80 mm2

corresponding to a radius of 5 mm .

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Medium Transmission line

- It has been mentioned in section 6.2 that the

capacitance of medium length lines is

significant. - When the effect of capacitance is not negligible

, it may be assumed to be concentrated at one or

more definite points along the line. - A number of localized capacitance methods have

been used to make approximate line performance

calculations.

- The following methods are more commonly used
- These methods of calculation give reasonably

accurate results for the solution of most

transmission-line problems .

Nominal T method .

- In a nominal T method the total line capacitance

is assumed to be concentrated at the middle point

of the line . The T representation of a line is

shown in fig.( 6.12 ).

- With the usual meanings of the quantities given

in fig.( 2 ) , - Voltage at the mid-point of the line .
- Current in the capacitor ,
- Sending-end current ,

- Sending-end voltage ,
- Equations ( 6.13.1 ) and ( 6.13.2 ) give the

sending-end current and sending-end voltage

respectively. - Other quantities , such as phase shift, power

input, efficiency, regulation, etc, can be

determined in the usual manner .

Phasor diagram

- The phasor diagram of the nominal T circuit can

be drawn for a lagging power factor as - Phasor diagram of a nominal T network

Vs

Iab

Vab

IsXl

Øs

IsR

Vr

Ør

IrR/2

IrXl/2

Is

Ir

Iab

- In the phasor diagram

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Nominal p method .

- This method assumed that one-half of the total

line capacitance is concentrated at each end of

the line and the total resistance and inductive

reactance are concentrated at the center . - Fig.( 6.14 ) shows the nominal p representation

of the line.

- From fig.( 6.14),
- Voltage at the sending-end ,

- Sending-end current,
- Equations ( 6.14.1 ) and ( 6.14.2 ) give the

sending-end voltage and current respectively .

The other calculations can be made in the usual

manner.

Phasor diagram

- The phasor diagram of a nominal p circuit is can

be drawn for a lagging power factor of the load

as

Vs

Iab

Icd

Øs

I Z

IXl

Vr

Is

Ør

I

IR

Ir

Iab

Phasor diagram of a nominal II network

Example

A three-phase, 50 Hz, transmission line, 40 km

long delivers 36 Mw at 0.8 power factor lagging

at 60 kv (phase). The line constants per

conductor are , Shunt leakage may be neglected.

Find the sending-end voltage , current , phase

angle, and the efficiency . Use (a) nominal T

method, (b) nominal p method.

Solution

Phase voltage at the receiving-end

Power per phase

- Therefore, the receiving-end current ,
- Taking as the reference phasor,
- Reference per phase,

- Inductive reactance per phase ,
- Series impedance per phase,
- Shunt admittance per phase

- Calculation by nominal T method
- The nominal T circuit for the line is shown in

Fig.(6.1). - The current in the capacitor ,

- The current at the sending-end ,
- Voltage drop in the left-hand half of the line ,

- Voltage at the sending-end ,
- Fig.(6.16)Phasor diagram

- Sending-end line voltage ,
- Phase difference between and ,
- Sending-end power factor ,

- Power loss in the line ,
- Transmission efficiency ,

- Alternatively , transmission efficiency may be

calculated as follows

- Calculation by nominal p method
- The nominal circuit for the line is shown in

Fig.( 6.14)

- Voltage drop per phase ,
- Voltage at the sending-end per phase ,

- Sending-end line voltage ,

- Sending-end current ,

- Sending-end power factor ,
- Power loss in the line ,
- Transmission efficiency ,

Example

- A three-phase , 50 Hz , 150 km line operates at

110 Kv between the lines at the sending-end. The

total inductance and capacitance per phase are

(0.2 H) and (1.5 µF) . Neglecting losses

calculate the value of receiving-end load having

a power factor of unity for which the voltage at

the receiving-end will be the same as that at the

sending-end . Assume one-half of the total

capacitance of the line to be concentrated at

each end . - The circuit for the given line is shown in

fig.(6.17) . It is a nominal representation .

Solution

- Inductive reactance per phase ,
- Series impedance per phase ,
- Shunt admittance per phase ,

- Fig.(6.17) Illustrating example (6.9).

- Current in the load-end capacitor ,
- Let the load current be Ir . Since the load power

factor is unity , - Current through the inductive reactance ,

- Sending-end voltage ,

General Network Constants

Introduction

- A network having two input and two output

terminals is known as a two-port network . It may

also be called a two-terminal-pair network or

quadriple network . In fig.(1 . a,b) represent

the input pair terminals and ( c,d ) the output

pair terminals . The two pairs of terminals are

usually shown to be enclosed in a box . - Fig.( 1 ) Two-port network .

- A circuit consisting of any arrangement of its

components is connected to these terminals . - Where A , B , C , D are called the general

network constants of the system . These constants

are known by other names like transmission

parameters , chain parameters and auxiliary

network constants .

- Equation (1) can be put in the matrix form as

- The matrix is called the transfer

matrix or transmission matrix of the network

Cascaded network

- The overall A , B , C , D constants for several

2-port networks connected in cascade ( or chain

arrangement ) can be found out easily . Fig.( 2

) shows two cascaded networks , and one that is

the equivalent of both . The constants of the two

component networks are A1 , B1 , C1 , D1 and A2 ,

B2 , C2 , D2 . Let the constants for the

equivalent network be A0 , B0 , C0 , D0 . - Fig.( 2 ) Two cascaded networks and their

equivalents.

- Let Va and Ia be the voltage and current

respectively at the junction (a) of the two

networks. - For the network ( 1 ) ,

- Substituting the values of Va and Ia from the

first set of equations in the second set , we

have

- The sending-end voltage and current for the

equivalent network with constants A0 , B0 , C0 ,

D0 are given by - Equating the constants of Vr and Ir , the

overall constants for the two networks in cascade

are

- Matrix method . For the first network ,
- But Va and Ia are the input voltage and current

respectively of the second network , so that

- Combining these equations ,
- For the equivalent network ,
- Comparing equations (2.9) and (2.10) we get ,

Relations between A,B ,C ,D constants

- The relations between A, B , C , D constants of

a passive , linear and bilateral network can be

found with the help of reciprocity theorem .

First a voltage V is applied to the input

terminals keeping the output terminals short

circuited fig.( 3 ,a ) . Since under short

circuit Vr 0 , equations ( 1.1 ) give

- Now , the voltage V is applied to the output

terminals and the input terminals are short

circuited fig.( 3 ,b ) . The directions of flow

of currents at the input and output terminals are

reversed and the sending-end voltage Vs becomes

zero. Equation ( 1.1 ) become

- Since the network is passive , by the reciprocity

theorem , - Fig.( 3 )

- Combining equations ( 3.1 ), ( 3.3 ) , ( 3.4 )

and ( 3.5 ) we get , - Dividing both the sides of the above equation by

-V/B we get , - Equation ( 3.6 ) is of one of the required

relations between the network constants. This

relation may also be put in the determinant form

as

Series impedance circuit

- A circuit having a series impedance Z is shown

in fig.(4) . Such a case is found in a short

transmission line where the line capacitance is

negligible and the shunt admittance Y is zero .

A transformer with magnetizing current neglected

can also be represented by such a circuit . - Fig.( 4 ) Series impedance circuit .

- For the network shown in fig.( 4 ) we may write

- By comparing these equations with the general

equations (1.1) and (1.2) the general constants

for the series impedance network can be written

as - The transfer matrix for the network is

Shunt admittance circuit

- Fig,( 5 ) , shows a transmission network with a

shunt admittance Y . Such a network may represent

the magnetizing current circuit of a transformer

or a shunt capacitor . - Fig.( 5 ) Shunt admittance circuit

- For the network shown in fig.( 5 ) we may write
- Hence,

Half T network

- A half T network is shown in fig.( 6 ) .
- Fig.( 6 ) Half-T network .

- Hence,

- Matrix method , The half T network can be

considered as the cascade connection of two

sections . One section is a series impedance Z

and the other a shunt admittance Y . The overall

constants are obtained from the matrix product of

the transfer matrices of each section in the

correct order.

Overhead lines and its mechanical characteristics

Introduction

- An overhead line comprises mainly of i )

conductor, ii ) supports , iii ) insulators and

pole fittings . The function of overhead lines is

to transmit electrical energy , and the important

characteristics which the line conductors must

have are - a) High electrical conductivity .
- b) High tensile strength .
- c) Low density .
- d) Low cost .
- The metals which posses the above properties are

copper , aluminum and steel , which are used

either alone or in combination .

Types of conductors

- Copper
- The most common conductor used for transmission

is copper , because it is twice as strong as

soft drawn copper - The merits of this metal as a line conductor are

- i ) It has a best conductivity in comparison to

other metals .

- ii ) It has higher current density , so for

the given current rating , lesser cross-sectional

area of conductor is required and hence it

provides lesser cross-sectional area to wind

loads . - iii ) The metal is quite homogeneous .
- iv ) It has low specific resistance .
- v ) It is durable and has a higher scrap value .

- Aluminum
- Next to copper aluminum is the conductor used in

order of performance as far as the conductivity

is concerned .Its merits and demerits are - i ) It is cheaper than copper .
- ii ) It is lighter in weight .
- iii ) It is second in conductivity ( among the

metals used for transmission ) . Commercial

hard-down aluminum wire at standard temperature

has approximately 60.6 per cent conductivity in

comparison to standard annealed copper wire . - iv ) For same ohmic resistance , its diameter

is about 1.27 times that of copper .

- v ) At higher voltages it causes less corona loss

. - vi ) Since the diameter of the conductor is more

, so it is subject to greater wind pressure due

to which greater is the swing of the conductor

and greater is the sag . - vii ) Since the conductors are liable to swing,

so it requires larger cross arms . - viii ) As the melting point of the conductor is

low , so the short circuit etc. will damage it . - ix ) Joining of aluminum is much more difficult

than that of any other material .

- Steel
- No doubt it has got the greatest tensile strength

, but it is least used for transmission of

electrical energy as it has got high resistance.

It has the following properties - i ) It is lowest in conductivity .
- ii ) It has high internal reactance .
- iii ) It is much subjected to eddy current and

hysterisis loss. - iv ) In a damp atmosphere it is rusted .
- Hence its use is limited .

- Aluminum conductor with steel
- reinforced ( A.C.S.R )
- An aluminum conductor having a central core of

galvanized steel wires is used for high voltage

transmission purposes. - This is done to increase the tensile strength of

aluminum conductor. The galvanized steel core is

covered by one or more strands of aluminum wires.

- The steel conductors used are galvanized in order

to prevent rusting.

- Thus the steel reinforced aluminum conductor has

less sag and longer span than the copper

conductor line since it has high tensile strength

. - The aluminum steel conductor has a larger

diameter than any other type of conductor of same

resistance . - For all calculation purposes , it is assumed that

the current is passing only in the aluminum

section .

Line supports

- The line supports are poles and the chief

requirements for such supports are - i ) They must be mechanically strong with factor

of safety of 2.5 to 3 . - ii ) They must be light in weight without the

loss of strength . - iii ) They must have least number of parts .
- iv ) They must be cheap .
- v ) Their maintenance cost should be minimum .
- vii ) They must have longer life .
- viii ) They must be of pleasing shape .

- The different types of poles which can be used as

line supports are - a. Wooden poles .
- b. Steel tubular poles
- c. Reinforced concrete poles .
- d. Steel towers .
- Fig.(1)Single phase single-circuit

Spacing between the conductors

- The most suitable spacing between the conductors

can be arrived at by mathematical calculations.

- It can only be obtained by empirical formulae

which have been obtained from practical

considerations.

Fig.(7)Three-phase single circuit horizontal

disposition of conductor and steel towers

- Generally the following formulae are used for

obtaining spacing between the conductors - Where,
- Vkv is voltage in kilovolts .
- d is diameter of conductor
- in inches .
- w is weight of conductor
- in lb. per foot run .
- D is sag in feet .

Sag-tension calculations for the overhead lines

- The theory of sag tension calculation is based on

the fact that when a wire of uniform

cross-section is suspended between two points at

the same level , the wire sags down - The line between the two points must be so

tensioned.

Fig.(9)Representation of sag in the conductor

suspended between two points

- Let , ( L ) be the length of the conductor POQ ,

suspended between the supports P and Q at the

same level and having a distance L between them . - Let , O be the lowest point of the centenary so

formed, D be the maximum sag , and let - w be the weight of the conductor per unit

length . - T be the tension at any point A of the

conductor . - To be the tension at point O of the conductor

, which is taken as origin . - S be the length of the conductor OA .

Effect of ice covering and wind over the line

- Under the severest conditions of ice covering and

wind , the stress over the line is increased to

the maximum . The ice covering over the conductor

increase the weight of the conductor per unit

length . Let , ( d cm ) be the diameter of the

conductor and ( r cm ) be the radial thickness of

ice.

Fig.(10) Representation of conductor covered

with ice

- Cross-sectional area of the conductor
- Overall cross-sectional area when covered with

ice - Sectional area of the ice

- Density of ice
- Weight of ice per meter length
- The effect of wind is allowed for by assuming

that the wind is blowing with a velocity of (

80.45 km ) per hour across the line . It is

equivalent to a pressure of ( 33.7 km ) per

square meter of the projected surface to the line

to ice .

- The projected surface per meter length of the

conductor - So, wind pressure Pw per meter run of the line

in a horizontal direction ,

- Fig.(11)Representation of resultant force acting

on the conductor . - So , the resultant force Wi acting on the

conductor from figure , is given as

Example

An overhead line has a span of 220 meters , the

lines conductor weights 684 km . per 1,000

meters . Calculate the max. sag in the line , if

the maximum allowable tension in the line is

1,450 kg

Solution

Maximum sag

Weight per unit length

Max. sag