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Electrical Power Systems

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Title: Electrical Power Systems


1
Electrical Power Systems
2
References
  • Electrical power
  • ( Dr\ S.L. Uppa
    )
  • Elements of power system analysis
  • ( William
    Stevenson )
  • Modern power system analysis
  • ( I.J.Nagrath ,
    D.R. Kothari )

3
Elements of power system
  • 1. Power stations .
  • 2. Substations .
  • 3. Busbars .
  • 4. Primary T.L and secondary T.L .

4
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
5
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

6
Standard voltages In Egypt
  • T.L voltages 66 , 220 KV and 500 KV from
    High Dam to Cairo

7
Power stations
  • Types of power stations
  • Thermal power stations .
  • Hydro power stations .
  • Nuclear power stations .
  • Gas power stations .

8
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.

9
Turbine
Generator
BB
Condenser
Cooler
Coal Burner
Boiler
Pump
Thermal Power Station
10
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11
Advantages of this stations .
  1. It has low construction ( primary ) cost .
  2. It uses small area to construct .
  3. No trembles ( vibrations ) .
  4. It can be constructed in minimum time compared to
    Hydro and Nuclear power stations .
  5. It can be constructed near to the load .

12
Disadvantages
  1. Running costs are high because it uses coal and
    solar .
  2. The response is very low to supply the increasing
    in load .
  3. It is not clean and causes more pollution .
  4. It has low efficiency ( 25 40 ) .

13
Some of consideration must be taken
  1. It must near to source of water .
  2. It must near to transmitted tools .
  3. It needs to strong land which has low price .
  4. We can extend the station.

14
Hydro power stations
  • It depends on two deferent levels in the river.

15
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16
Advantages of this stations .
  1. Running cost is very low because it depends on
    water .
  2. The response is very high to supply the sudden
    increasing in load.
  3. It is clean .
  4. Efficiency is equal ( 90 95 ) .

17
Disadvantages
  1. It has high construction cost .
  2. It has high T.L costs .
  3. Constructed far from the loads .
  4. Takes more time to construct .

18
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.

19
Control rots
Heat exchanger
Steam
Turbine
Generator
CO2
Reactor
Water
Condenser
Fuel rots (Uranium 235)
Nuclear Power Station
20
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21
  • Advantages and disadvantages are similar to
    hydro power station added to that it has higher
    protection cost and it is constructed in the
    desert.

22
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) .

23
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.

24
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.

25
Resistance
  • Resistance causes power loss in the T.L.
  • For uniform resistance ,

26
Relation between resistance and temperature rise .
t
t2
t1
R
R1
R2
T
27
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.

28
  • 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
29
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.

30
  • 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

31
Transmission Circuit Calculations
32
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 ' .

33
  • 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
34
Vs
Ir ZL
Ir XL
Vr
Øs
Ir R
Ør
Is Ir
Fig.( 6.1 b ) Vector diagram for a short
transmission line .
35
  • 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 .

36
  • Referring to the equivalent circuit
  • Hence, if the receiving-end conditions are known
    the necessary sending-end voltage may be
    calculated .

37
  • 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,

38
  • 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

39
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
40
  • 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 .

41
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42
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43
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.

44
  • The following methods are more commonly used
  • These methods of calculation give reasonably
    accurate results for the solution of most
    transmission-line problems .

45
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 ).

46
  • 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 ,

47
  • 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 .

48
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
49
  • In the phasor diagram

50
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51
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.

52
  • From fig.( 6.14),
  • Voltage at the sending-end ,

53
  • 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.

54
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
55
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
56
  • Therefore, the receiving-end current ,
  • Taking as the reference phasor,
  • Reference per phase,

57
  • Inductive reactance per phase ,
  • Series impedance per phase,
  • Shunt admittance per phase

58
  • Calculation by nominal T method
  • The nominal T circuit for the line is shown in
    Fig.(6.1).
  • The current in the capacitor ,

59
  • The current at the sending-end ,
  • Voltage drop in the left-hand half of the line ,

60
  • Voltage at the sending-end ,
  • Fig.(6.16)Phasor diagram

61
  • Sending-end line voltage ,
  • Phase difference between and ,
  • Sending-end power factor ,

62
  • Power loss in the line ,
  • Transmission efficiency ,

63
  • Alternatively , transmission efficiency may be
    calculated as follows

64
  • Calculation by nominal p method
  • The nominal circuit for the line is shown in
    Fig.( 6.14)

65
  • Voltage drop per phase ,
  • Voltage at the sending-end per phase ,

66
  • Sending-end line voltage ,

67
  • Sending-end current ,

68
  • Sending-end power factor ,
  • Power loss in the line ,
  • Transmission efficiency ,

69
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
70
  • Inductive reactance per phase ,
  • Series impedance per phase ,
  • Shunt admittance per phase ,

71
  • Fig.(6.17) Illustrating example (6.9).

72
  • Current in the load-end capacitor ,
  • Let the load current be Ir . Since the load power
    factor is unity ,
  • Current through the inductive reactance ,

73
  • Sending-end voltage ,

74
General Network Constants
75
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 .

76
  • 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 .

77
  • Equation (1) can be put in the matrix form as
  • The matrix is called the transfer
    matrix or transmission matrix of the network

78
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.

79
  • Let Va and Ia be the voltage and current
    respectively at the junction (a) of the two
    networks.
  • For the network ( 1 ) ,

80
  • Substituting the values of Va and Ia from the
    first set of equations in the second set , we
    have

81
  • 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

82
  • Matrix method . For the first network ,
  • But Va and Ia are the input voltage and current
    respectively of the second network , so that

83
  • Combining these equations ,
  • For the equivalent network ,
  • Comparing equations (2.9) and (2.10) we get ,

84
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

85
  • 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

86
  • Since the network is passive , by the reciprocity
    theorem ,
  • Fig.( 3 )

87
  • 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

88
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 .

89
  • 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

90
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

91
  • For the network shown in fig.( 5 ) we may write
  • Hence,

92
Half T network
  • A half T network is shown in fig.( 6 ) .
  • Fig.( 6 ) Half-T network .

93
  • Hence,

94
  • 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.

95
Overhead lines and its mechanical characteristics
96
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 .

97
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 .

98
  • 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 .

99
  • 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 .

100
  • 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 .

101
  • 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 .

102
  • 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.

103
  • 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 .

104
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 .

105
  • 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

106
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
107
  • 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 .

108
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.

109
Fig.(9)Representation of sag in the conductor
suspended between two points
110
  • 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 .

111
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
112
  • Cross-sectional area of the conductor
  • Overall cross-sectional area when covered with
    ice
  • Sectional area of the ice

113
  • 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 .

114
  • The projected surface per meter length of the
    conductor
  • So, wind pressure Pw per meter run of the line
    in a horizontal direction ,

115
  • Fig.(11)Representation of resultant force acting
    on the conductor .
  • So , the resultant force Wi acting on the
    conductor from figure , is given as

116
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
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