Power Flow/ Circle
Diagrams/Corona Issues
UNIT-2
• The flow of active and reactive power over a transmission line can be handled computationally.
• The Iocus of complex sending-and receiving-end power is a circle.
• Since circles are convenient to draw, the circle diagrams area useful aid to visualize the load
flow problem over a single transmission.
The expressions for complex number receiving-and sending-end powers are
A
B
The units for Sp and Ss are MVA (three-phase) with voltages in KV line.
• SR and SS are each composed of two phasor components
• (one a constant phasor and the other a phasor of fixed magnitude but variable angle.
• The loci for SR and SS would, therefore, be circles drawn from the tip of constant phasors as
centres.
In equation (A) the centre of receiving-end circle is located at the tip of the phasor.
(Either in polar coordinates or in terms of rectangular
coordinate)
Horizontal coordinate of the centre
The radius of the receiving-end circle is
Vertical coordinate of the centre
• The centre is located by drawing
OCR at an angle (β-δ) in the positive
direction from the negative MWaxis
• From the centre CR , the receivingend circle is drawn with the radius
• The operating point M is located on
the circle by means of the received
real power PR.
• The corresponding QR or ƟR can be
immediately read from the circle
diagram.
• The torque angle δ can be read in
accordance with the positive
direction indicated from the
reference line.
Similarly, in equation (B) the centre of Sending-end circle is located at the tip of the phasor.
(Either in polar coordinates or in terms of rectangular
coordinate)
Horizontal coordinate of the centre
The radius of the sending-end circle is
Vertical coordinate of the centre
• The centre is located by drawing
OCS at an angle (β-δ) from the
positive MW-axis
• From the centre CS , the sendingend circle is drawn with the radius
• The operating point N is located on
the circle and the sending end real
power PS is shown in the circle
diagram.
• The corresponding QS or ƟS can be
immediately read from the circle
diagram.
• The torque angle δ can be read in
accordance with the direction
indicated from the reference line.
Corona
The phenomenon of violet glow, hissing noise and production of ozone gas in an
overhead transmission line is known as corona
When an alternating potential difference is applied across two conductors whose
spacing is large as compared to their diameters, there is no apparent change in the
condition of atmospheric air surrounding the wires if the applied voltage is low.
However, when the applied voltage exceeds a certain value, called critical disruptive
voltage, the conductors are surrounded by a faint violet glow called corona.
*If the conductors are polished and smooth, the corona glow will be uniform
throughout the length of the conductors, otherwise the rough points will appear
brighter
*With d.c. voltage
The positive wire has uniform glow about it, while the negative conductor has spotty
glow
Theory of corona formation
❖ Some ionisation is always present in air due to cosmic rays, ultraviolet radiations and
radioactivity.
❖ Therefore, under normal conditions, the air around the conductors contains some
ionised particles (i.e., free electrons and +ve ions) and neutral molecules.
❖ When p.d. is applied between the conductors, potential gradient is set up in the air
which will have maximum value at the conductor surfaces.
❖ Under the influence of potential gradient, the existing free electrons acquire greater
velocities. The greater the applied voltage, the greater the potential gradient and more
is the velocity of free electrons.
❖ When the potential gradient at the conductor surface reaches about 30 kV per cm
(max. value), the velocity acquired by the free electrons is sufficient to strike a neutral
molecule with enough force to dislodge one or more electrons from it.
❖ This produces another ion and one or more free electrons, which is turn are accelerated
until they collide with other neutral molecules, thus producing other ions.
❖ Thus, the process of ionisation is cumulative. The result of this ionisation is that either
corona is formed or spark takes place between the conductors.
Factors Affecting Corona
It is the minimum phase-neutral voltage at which corona occurs.
Consider two conductors of radii r cm and spaced d cm apart.
If V is the phase-neutral potential, then potential gradient at the conductor surface
is given by:
The breakdown strength of air at 76 cm pressure and temperature of 250 C (Standard
conditions) is 30 kV/cm (max) or 21·2 kV/cm (r.m.s.) and is denoted by go . If Vc is the
phase-neutral potential required under these conditions, then,
It is the minimum phase-neutral voltage at which corona glow appears all along the line
conductors.
Formation of corona is always accompanied by energy loss which is dissipated in the
form of light, heat, sound and chemical action. When disruptive voltage is exceeded, the
power loss due to corona is given by
Advantages and Disadvantages of Corona
Methods of Reducing Corona Effect
Problem 1
A 3-phase line has conductors 2 cm in diameter spaced equilaterally 1 m apart. If the
dielectric strength of air is 30 kV (max) per cm, find the disruptive critical voltage for the
line. Take air density factor δ= 0·952 and irregularity factor mo= 0·9.
Problem 2
A 132 kV line with 1·956 cm dia. conductors is built so that corona takes place if the line
voltage exceeds 210 kV (r.m.s.). If the value of potential gradient at which ionisation occurs
can be taken as 30 kV per cm, find the spacing between the conductors. (3 Phase line and
smooth conductors)
Mechanical Design of
Overhead Lines
UNIT-3
Commonly used conductor materials. The most commonly used conductor materials for
overhead lines are copper, aluminium, steel-cored aluminium, galvanised steel and
cadmium copper.
The supporting structures for overhead line conductors are various types of poles and
towers called line supports.
wooden poles, steel poles, R.C.C. poles and lattice steel towers
Note: The choice of supporting structure for a particular case depends upon the
line span, X-sectional area, line voltage, cost and local conditions.
Wooden poles.
• These are made of seasoned wood (sal or chir) and are suitable for lines of moderate Xsectional area and of relatively shorter spans, say upto 50 metres.
• Such supports are cheap, easily available, provide insulating properties and, therefore,
are widely used for distribution purposes in rural areas as an economical proposition.
The main objections to wooden supports are :
(i)
tendency to rot below the ground level
(ii) comparatively smaller life (20-25 years)
(iii) cannot be used for voltages higher than 20 kV
(iv) less mechanical strength and
(v) require periodical inspection.
Wooden poles.
Steel poles
The steel poles are often used as a substitute for wooden poles.
They possess greater mechanical strength, longer life and permit longer spans to be used.
Such poles are generally used for distribution purposes in the cities.
This type of supports need to be galvanised or painted in order to prolong its life.
The steel poles are of three types viz., (i) rail poles (ii) tubular poles and (iii) rolled steel
joints.
RCC poles.
They have greater mechanical
strength, longer life and permit
longer spans than steel poles.
Moreover, they give good outlook,
require little maintenance and have
good insulating properties.
The holes in the poles facilitate the
climbing of poles and at the same
time reduce the weight of line
supports.
The main difficulty with the use of
these poles is the high cost of
transport owing to their heavy
weight.
Therefore, such poles are often
manufactured at the site in order to
avoid heavy cost of transportation.
Single circuit
Double circuit
RCC poles.
Steel towers.
•
For
long
distance
transmission
at
higher
voltage, steel towers are
invariably employed. Steel
towers
have
greater
mechanical strength, longer
life, can withstand most
severe climatic conditions
and permit the use of
longer spans.
•
The risk of interrupted
service due to broken or
punctured insulation is
considerably
reduced
owing to longer spans.
•
Tower footings are usually
grounded by driving rods
into the earth. This
minimizes the lightning
troubles as each tower acts
as a lightning conductor.
Single circuit
Double circuit
Types of
Towers
Suspension
Tower
Tension
Tower
Angle
Tower
End
Tower
This type
of towers exists
in the beginning and at
the end of the line which
exposed to tension in one
side.
End Tower
Insulators
The overhead line conductors should be
supported on the poles or towers in such a way
that currents from conductors do not flow to
earth through supports i.e., line conductors
must be properly insulated from supports.
Desirable properties
(i)
(i)
High mechanical strength in order to withstand conductor
load, wind load etc.
High electrical resistance of insulator material in order to
avoid leakage currents to earth.
(iii) High relative permittivity of insulator material in order that
dielectric strength is high.
(iv) The insulator material should be non-porous, free from
impurities and cracks otherwise thepermittivity will be lowered.
(v)
High ratio of puncture strength to flashover.
The most commonly used
material for insulators of
overhead line is porcelain
but glass, steatite and
special
composition
materials are also used to a
limited extent.
Porcelain is produced by
firing
at
a
high
temperature a mixture of
kaolin, feldspar and
quartz. It is stronger
mechanically than glass,
gives less trouble from
leakage and is less effected
by changes of temperature
Types of Insulators
Pin type insulators
As the name suggests, the pin type insulator is
secured to the cross-arm on the pole.
There is a groove on the upper end of the
insulator for housing the conductor.
The conductor passes through this groove and
is bound by the annealed wire of the same
material as the conductor
Pin type insulators are used for
transmission and distribution
of electric power at voltages up
to 33 kV. Beyond operating
voltage of 33 kV, the pin type
insulators become too bulky
and hence uneconomical.
flash-over
Causes of insulator failure.
Insulators are required to withstand both
mechanical and electrical stresses.
The latter type is pirmarily due to line voltage and
may cause the breakdown of the insulator.
The electrical breakdown of the insulator can occur
either by flash-over or puncture. In flashover, an arc
occurs between the line conductor and insulator pin
(i.e., earth) and the discharge jumps across the *air
gaps, following shortest distance.
The insulator is generally dry and its surfaces have
proper insulating properties. Therefore, arc can only
occur through air gap between conductor and
insulator pin.
Fig. shows the arcing distance (i.e. a + b + c) for the
insulator.
In case of flash-over, the insulator will continue to
act in its proper capacity unless extreme heat
produced by the arc destroys the insulator
For pin type
insulators, the
value of safety
factor is about
10.
In case of puncture, the discharge occurs
from conductor to pin through the body of
the insulator. When such breakdown is
involved,
the insulator is permanently destroyed due to
excessive heat.
In practice, sufficient thickness of porcelain
is provided in the insulator to avoid puncture
by the line voltage.
Suspension type insulators.
The cost of pin type insulator increases rapidly
as the working voltage is increased. Therefore,
this type of insulator is not economical beyond
33 kV.
For high voltages (>33 kV), it is a usual practice
to use suspension type insulators
It consist of a number of porcelain
discs connected in series by metal
links in the form of a string.
The conductor is suspended at the
bottom end of this string while the
other end of the string is secured to
the cross-arm of the tower.
Each unit or disc is designed for
low voltage, say 11 kV.
The number of discs in series would
obviously depend upon the working
voltage.
• E.g.. If the working voltage is 66
kV, then six discs in series will
be provided on the string
Advantages (Suspension type insulators)
(i)
Suspension type insulators are cheaper than pin type insulators for voltages beyond 33
kV.
(ii) Each unit or disc of suspension type insulator is designed for low voltage usually 11 kV.
Depending upon the working voltage, the desired number of discs can be connected in series.
(iii) If any one disc is damaged, the whole string does not become useless because the
damaged disc can be replaced by the sound one.
(iv) The suspension arrangement provides greater flexibility to the line. The connection at the
cross arm is such that insulator string is free to swing in any direction and can take up the
position where mechanical stresses are minimum.
(v) In case of increased demand on the transmission line, it is found more satisfactory to
supply the greater demand by raising the line voltage than to provide another set of
conductors.
The additional insulation required for the raised voltage can be easily obtained in the
suspension arrangement by adding the desired number of discs.
(vi) The suspension type insulators are generally used with steel towers. As the conductors
run below the earthed cross-arm of the tower, therefore, this arrangement provides partial
protection from lightning.
Strain insulators.
When there is a dead end of the line or
there is corner or sharp curve, the line
is subjected to greater tension. In
order to relieve the line of excessive
tension, strain insulators are used.
For low voltage lines (< 11 kV),
shackle insulators are used as strain
insulators
The discs of strain insulators are used in
the vertical plane. When the tension in
lines is exceedingly high, as at long river
spans, two or more strings are used in
parallel.
Shackle insulators.
Used for low voltage distribution lines. Such
insulators can be used either in a horizontal
position or in a vertical position.
They can be directly fixed to the pole with a
bolt or to the cross arm.
Testing of Insulators
Insulating Material
The main cause of failure of overhead line insulator, is flash over, occurs in between line and
earth during abnormal over voltage in the system.
During this flash over, the huge heat produced by arcing, causes puncher in insulator body.
Viewing this phenomenon the materials used for electrical insulator, has to posses some
specific properties
Properties of Insulating Material
1.It must be mechanically strong enough to carry tension and weight of conductors.
1.It must have very high dielectric strength to withstand the voltage stresses in High Voltage
system.
1.It must possesses high Insulation Resistance to prevent leakage current to the earth.
1.The insulating material must be free from unwanted impurities.
1.It should not be porous.
1.There must not be any entrance on the surface of electrical insulator so that the moisture or
gases can enter in it.
1.There physical as well as electrical properties must be less effected by changing temperature.
Causes of Insulator Failure
Testing of Insulators
Cracking of Insulator
Due to changing climate conditions, these different materials in the insulator expand and
contract in different rate. These unequal expansion and contraction of porcelain, steel and
cement are the chief cause of cracking of insulator.
Defective Insulation Material
If the insulation material used for insulator is defective anywhere, the insulator may have a
high chance of being puncher from that place.
Porosity in The Insulation Materials
If the porcelain insulator is manufactured at low temperatures, it will make it porous,
and due to this reason it will absorb moisture from air thus its insulation will decrease
and leakage current will start to flow through the insulator which will lead to insulator
failure
Improper Glazing on Insulator Surface
If the surface of porcelain insulator is not properly glazed, moisture can stick over it. This
moisture along with deposited dust on the insulator surface, produces a conducting path. As
a result the flash over distance of the insulator is reduced. As the flash over distance is
reduced, the chance of failure of insulator due to flash over becomes more.
Causes of Insulator Failure
Testing of Insulators
Flash Over Across Insulator
If flash over occurs, the insulator may be over heated which may ultimately results into
shuttering of it.
Mechanical Stresses on Insulator
If an insulator has any weak portion due to manufacturing defect, it may break from that
weak portion when mechanical stress is applied on it by its conductor. These are the main
causes of insulator failure.
Insulator Testing
According to the British Standard, the electrical insulator must undergo the following
tests:
1. Flashover tests of insulator
2. Performance tests
3. Routine tests
Testing of Insulators
Flashover Test
There are mainly three types of flashover test performed on an insulator
Power Frequency Dry Flashover Test of Insulator
1.First the insulator to be tested is mounted in the manner in which it would be used
practically.
1.Then terminals of variable power frequency voltage source are connected to the both
electrodes of the insulator.
1.Now the power frequency voltage is applied and gradually increased up to the specified
value. This specified value is below the minimum flashover voltage.
1.This voltage is maintained for one minute and observe that there should not be any
flash-over or puncher occurred.
The insulator must be capable of sustaining the specified minimum voltage for one minute
without flash over.
Testing of Insulators
Power Frequency Wet Flashover Test or Rain Test of Insulator
1.In this test also the insulator to be tested is mounted in the manner in which it would be
used practically.
1.Then terminals of variable power frequency voltage source are connected to the both
electrodes of the insulator.
1.After that the insulator is sprayed with water at an angle of 45o in such a manner that its
precipitation should not be more 5.08 mm per minute. The resistance of the water used for
spraying must be between 9 kΩ 10 11 kΩ per cm3 at normal atmospheric pressure and
temperature. In this way we create artificial raining condition.
1.Now the power frequency voltage is applied and gradually increased up to the specified
value.
1.This voltage is maintained for either one minute or 30 second as specified and observe that
there should not be any flash-over or puncher occurred. The insulator must be capable of
sustaining the specified minimum power frequency voltage for specified period without flash
over in the said wet condition.
Testing of Insulators
Power Frequency Flashover Voltage test of Insulator
1.The insulator is kept in similar manner of previous test.
1.In this test the applied voltage is gradually increased in similar to that of previous tests.
1.But in that case the voltage when the surroundings air breaks down, is noted.
Impulse Frequency Flashover Voltage Test of Insulator
The overhead outdoor insulator must be capable of sustaining high voltage surges caused by
lightning etc. So this must be tested against the high voltage surges.
1.The insulator is kept in similar manner of previous test.
1.Then several hundred thousands Hz very high impulse voltage generator is connected to the insulator.
1.Such a voltage is applied to the insulator and the spark over voltage is noted.
1.The ratio of this noted voltage to the voltage reading collected from power frequency flash over
voltage test is known as impulse ratio of insulator.
This ratio should be approximately 1.4 for pin type insulator and 1.3 for suspension type insulators.
Testing of Insulators
Performance Test of Insulator
Temperature Cycle Test of Insulator
1.The insulator is first heated in water at 70oC for one hour.
1.Then this insulator immediately cooled in water at 7oC for another one hour.
1.This cycle is repeated for three times.
1.After completion of these three temperature cycles, the insulator is dried and the glazing
of insulator is thoroughly observed. After this test there should not be any damaged or
deterioration in the glaze of the insulator surface.
Puncture Voltage Test of Insulator
1.The insulator is first suspended in an insulating oil.
1.Then voltage of 1.3 times of flash over voltage, is applied to the insulator.
A good insulator should not puncture under this condition
Testing of Insulators
Porosity Test of Insulator
1.The insulator is first broken into pieces.
1.Then These broken pieces of insulator are immersed in a 0.5 % alcohol solution of fuchsine
dye under pressure of about 140.7 kg ⁄ cm2 for 24 hours.
1.After that the sample are removed and examine.
The presence of a slight porosity in the material is indicated by a deep penetration of the dye
into it
Mechanical Strength Test of Insulator
The insulator is applied by 2½ times the maximum working strength for about one minute.
The insulator must be capable of sustaining this much mechanical stress for one minute
without any damage in it.
Testing of Insulators
Routine Test of Insulator
Each of the insulator must undergo the following routine test before they are
recommended for using at site.
Proof Load Test of Insulator
In proof load test of insulator, a load of 20% in excess of specified maximum working
load is applied for about one minute to each of the insulator.
Corrosion Test of Insulator
1.In corrosion test of insulator, The insulator with its galvanized or steel fittings is suspended into a
copper sulfate solution for one minute.
1.Then the insulator is removed from the solution and wiped, cleaned.
1.Again it is suspended into the copper sulfate solution for one minute.
1.The process is repeated for four times.
Then it should be examined and there should not be any disposition of metal on it.
Potential Distribution over Suspension Insulator String
• mutual capacitance
or self-capacitance
• shunt capacitance
(i) The voltage impressed on a string of suspension insulators does not distribute itself
uniformly across the individual discs due to the presence of shunt capacitance.
(ii) The disc nearest to the conductor has maximum voltage across it. As we move towards the
cross-arm, the voltage across each disc goes on decreasing.
(iii) The unit nearest to the conductor is under maximum electrical stress and is likely to be
punctured. Therefore, means must be provided to equalise the potential across each unit.
(iv) If the voltage impressed across the string were d.c., then voltage across each unit would be
the same. It is because insulator capacitances are ineffective for d.c.
String Efficiency
Note: String efficiency is an important consideration since it decides the potential
distribution along the string. The greater the string efficiency, the more uniform is the
voltage distribution.
Methods of Improving String Efficiency
By using longer cross-arms.
The value of string efficiency depends upon the value of K i.e., ratio of shunt
capacitance to mutual capacitance. The lesser the value of K, the greater is the string
efficiency and more uniform is the voltage distribution. In practice, K = 0·1 is the
limit that can be achieved by this method.
By using longer cross-arms.
By using a guard ring.
By grading the insulators
In this method, insulators of different dimensions are so chosen that each has a
different capacitance. The insulators are capacitance graded i.e. they are assembled in
the string in such a way that the top unit has the minimum capacitance, increasing
progressively as the bottom unit (i.e., nearest to conductor) is reached. Since voltage
is inversely proportional to capacitance, this method tends to equalize the potential
distribution across the units in the string.
By using a guard ring.
The potential across each unit in a string can be equalised by using a guard ring which
is a metal ring electrically connected to the conductor and surrounding the bottom
insulator.
The guard ring is contoured in such a way that shunt capacitance currents i1, i2 etc. are
equal to metal fitting line capacitance currents i’1', i'2 etc. The result is that same charging
current I flows through each unit of string. Consequently, there will be uniform potential
distribution across the units.
In a 33 kV overhead line, there are three units in the string of insulators. If the
capacitance between each insulator pin and earth is 11% of self-capacitance of
each insulator, find (i) the distribution of voltage over 3 insulators and (ii) string
efficiency.
Corona
The phenomenon of violet glow, hissing noise and production of ozone gas in an
overhead transmission line is known as corona
When an alternating potential difference is applied across two conductors whose
spacing is large as compared to their diameters, there is no apparent change in the
condition of atmospheric air surrounding the wires if the applied voltage is low.
However, when the applied voltage exceeds a certain value, called critical disruptive
voltage, the conductors are surrounded by a faint violet glow called corona.
*If the conductors are polished and smooth, the corona glow will be uniform
throughout the length of the conductors, otherwise the rough points will appear
brighter
*With d.c. voltage
The positive wire has uniform glow about it, while the negative conductor has spotty
glow
Theory of corona formation
❖ Some ionisation is always present in air due to cosmic rays, ultraviolet radiations and
radioactivity.
❖ Therefore, under normal conditions, the air around the conductors contains some
ionised particles (i.e., free electrons and +ve ions) and neutral molecules.
❖ When p.d. is applied between the conductors, potential gradient is set up in the air
which will have maximum value at the conductor surfaces.
❖ Under the influence of potential gradient, the existing free electrons acquire greater
velocities. The greater the applied voltage, the greater the potential gradient and more
is the velocity of free electrons.
❖ When the potential gradient at the conductor surface reaches about 30 kV per cm
(max. value), the velocity acquired by the free electrons is sufficient to strike a neutral
molecule with enough force to dislodge one or more electrons from it.
❖ This produces another ion and one or more free electrons, which is turn are accelerated
until they collide with other neutral molecules, thus producing other ions.
❖ Thus, the process of ionisation is cumulative. The result of this ionisation is that either
corona is formed or spark takes place between the conductors.
Factors Affecting Corona
It is the minimum phase-neutral voltage at which corona occurs.
Consider two conductors of radii r cm and spaced d cm apart.
If V is the phase-neutral potential, then potential gradient at the conductor surface
is given by:
The breakdown strength of air at 76 cm pressure and temperature of 250 C (Standard
conditions) is 30 kV/cm (max) or 21·2 kV/cm (r.m.s.) and is denoted by go . If Vc is the
phase-neutral potential required under these conditions, then,
It is the minimum phase-neutral voltage at which corona glow appears all along the line
conductors.
Formation of corona is always accompanied by energy loss which is dissipated in the
form of light, heat, sound and chemical action. When disruptive voltage is exceeded, the
power loss due to corona is given by
Advantages and Disadvantages of Corona
Methods of Reducing Corona Effect
Problem 1
A 3-phase line has conductors 2 cm in diameter spaced equilaterally 1 m apart. If the
dielectric strength of air is 30 kV (max) per cm, find the disruptive critical voltage for the
line. Take air density factor δ= 0·952 and irregularity factor mo= 0·9.
Problem 2
A 132 kV line with 1·956 cm dia. conductors is built so that corona takes place if the line
voltage exceeds 210 kV (r.m.s.). If the value of potential gradient at which ionisation occurs
can be taken as 30 kV per cm, find the spacing between the conductors. (3 Phase line and
smooth conductors)
Why Sag? - Safe tension- save conductor material?
The difference in level between points of supports and the lowest point on the
conductor is called sag.
A conductor suspended between two equilevel supports Aand B.
Note: The conductor sag should be kept to a minimum in order to reduce the conductor
material required and to avoid extra pole height for sufficient clearance above ground
level.
It is also desirable that tension in the conductor should be low to avoid the mechanical
failure of conductor and to permit the use of less strong supports.
low conductor tension and minimum sag ??
Calculation of Sag
Tension-conductor weight, effects of wind, ice loading and temperature variations.
Note: standard practice to keep conductor tension less than 50% of its ultimate tensile
strength i.e., minimum factor of safety in respect of conductor tension should be 2
Case A: When supports are at equal levels
(Assuming that the curvature is so
small that curved length is equal to
its horizontal projection (i.e., OP= x))
Two forces acting on OP
Case B :When supports are at unequal levels.
In hilly areas, we generally come across conductors suspended between supports at
unequal levels
Case B :When supports are at unequal
levels.
Effect of wind and ice loading
A conductor may have ice
coating and simultaneously
subjected to wind pressure.
Effect of wind and ice loading
Problem 1
Problem 2
A transmission line has a span of 150 m between level supports. The conductor has a
cross-sectional area of 2 cm2. The tension in the conductor is 2000 kg. If the specific
gravity of the conductor material is 9·9 gm/cm3 and wind pressure is 1·5 kg/m length,
calculate the sag. What is the vertical sag?
This is the value of slant sag in a direction
making an angle ɵ with the vertical.
Problem 3
A transmission line has a span of 200 metres between level supports. The conductor
has a cross-sectional area of 1·29 cm2, weighs 1170 kg/km and has a breaking stress
of
4218 kg/cm2. Calculate the sag for a safety factor of 5, allowing a wind pressure of
122 kg per square metre of projected area. What is the vertical sag?
Problem 4
The towers of height 30 m and 90 m respectively support a transmission line conductor
at water crossing. The horizontal distance between the towers is 500 m. If the tension in
the conductor is 1600 kg, find the minimum clearance of the conductor and water and
clearance mid-way between the supports. Weight of conductor is 1·5 kg/m. Bases of
the towers can be considered to be at water level.
Under Ground Cables
UNIT4
The underground cables have several advantages such as less liable to damage through storms or lightning, low
maintenance cost, less chances of faults, smaller voltage drop and better general appearance.
However, their major drawback is that they have greater installation cost and introduce insulation problems at
high voltages compared with the equivalent overhead system. For this reason, underground cables are employed where
it is impracticable to use overhead lines.
The chief use of underground cables for many years has been for distribution of electric power in congested urban
areas at comparatively low or moderate voltages.
An underground cable essentially consists of one or more conductors covered with suitable insulation and
surrounded by a protecting cover.
NECESSARY REQUIREMENTS
(i)
The conductor used in cables should be tinned stranded copper or aluminium of high conductivity. Stranding
is done so that conductor may become flexible and carry more current.
(ii)
The conductor size should be such that the cable carries the desired load current without overheating and
causes voltage drop within permissible limits.
(iii)
The cable must have proper thickness of insulation in order to give high degree of safety and reliability at
the voltage for which it is designed.
(iv)
The cable must be provided with suitable mechanical protection so that it may withstand the rough use in
laying it.
(v)
The materials used in the manufacture of cables should be such that there is complete chemical and physical
stability throughout
Construction of Cables
(i) Cores or Conductors. A cable may have one or more than one core (conductor) depending upon the type of
service for which it is intended. For instance, the 3-conductor cable shown in Fig is used for 3-phase service. The
conductors are made of tinned copper or aluminium and are usually stranded in order to provide flexibility to the
cable.
(ii) Insulation. Each core or conductor is provided with a suitable thickness of insulation, the thickness of layer
depending upon the voltage to be withstood by the cable. The commonly used materials for insulation are
impregnated paper, varnished cambric or rubber mineral compound.
(iii) Metallic sheath. In order to protect the cable from moisture, gases or other damaging liquids (acids or alkalies)
in the soil and atmosphere, a metallic sheath of lead or aluminium is provided over the insulation as shown
(iv) Bedding. Over the metallic sheath is applied a layer of bedding which consists of a fibrous material like jute
or hessian tape. The purpose of bedding is to protect the metallic sheath against corrosion and from mechanical
injury due to armouring.
(v) Armouring. Over the bedding, armouring is provided which consists of one or two layers of galvanised steel
wire or steel tape. Its purpose is to protect the cable from mechanical injury while laying it and during the course
of handling. Armouring may not be done in the case of some cables.
(vi) Serving. In order to protect armouring from atmospheric conditions, a layer of fibrous material (like jute)
similar to bedding is provided over the armouring. This is known as serving.
Insulating Materials for Cables
(i) High insulation resistance to avoid leakage current.
(ii) High dielectric strength to avoid electrical breakdown of the cable.
(iii) High mechanical strength to withstand the mechanical handling of cables.
(iv) Non-hygroscopic i.e., it should not absorb moisture from air or soil. The moisture tends to decrease the insulation
resistance and hastens the breakdown of the cable. In case the insulating material is hygroscopic, it must be enclosed
in a waterproof covering like lead sheath.
(v) Non-inflammable.
(vi) Low cost so as to make the underground system a viable proposition.
(vii) Unaffected by acids and alkalies to avoid any chemical action.
The principal insulating materials used in cables are rubber, vulcanised India rubber, impregnated paper, varnished
cambric and polyvinyl chloride.
Rubber - milky sap of tropical trees - relative permittivity varying between 2 and 3, - dielectric strength is about 30
kV/mm and resistivity of insulation is 1017 Ω m - major drawbacks - readily absorbs moisture, maximum safe
temperature is low (about 38ºC), soft and liable to damage due to rough handling and ages when exposed to light - pure
rubber cannot be used as an insulating material.
Vulcanised India Rubber (V.I.R.)- prepared by mixing pure rubber with mineral matter
such as zine oxide, red lead etc., and 3 to 5% of sulphur. The compound so formed is rolled into thin sheets and cut into
strips. The rubber compound is then applied to the conductor and is heated to a temperature of about 150ºC. The whole
process is called vulcanisation and the product obtained is known as vulcanised India rubber.
Vulcanised India rubber has greater mechanical strength, durability and wear resistant property than pure rubber.
Its main drawback is that sulphur reacts very quickly with copper and for this reason, cables using VIR insulation
have tinned copper conductor. The VIR insulation is generally used for low and moderate voltage cables.
Impregnated paper It consists of chemically pulped paper made from wood chippings and impregnated with
some compound such as paraffinic or napthenic material. This type of insulation has almost superseded the
rubber insulation. It is because it has the advantages of low cost, low capacitance, high dielectric strength and high
insulation resistance.
The only disadvantage is that paper is hygroscopic and even if it is impregnated with suitable compound, it absorbs
moisture and thus lowers the insulation resistance of the cable. For this reason, paper insulated cables are always
provided with some protective covering and are never left unsealed.
Polypropylen
e Laminated
Paper
Varnished cambric It is a cotton cloth impregnated and coated with varnish. This type of insulation is also known as
empire tape.
The cambric is lapped on to the conductor in the form of a tape and its surfaces are coated with petroleum jelly
compound to allow for the sliding of one turn over another as the cable is bent.
As the varnished cambric is hygroscopic, therefore, such cables are always provided with metallic sheath. Its dielectric
strength is about 4 kV/mm and permittivity is 2.5 to 3.8.
Polyvinyl chloride (PVC) This insulating material is a synthetic compound. It is obtained from the polymerisation of
acetylene and is in the form of white powder. For obtaining this material as a cable insulation, it is compounded with
certain materials known as plasticizers which are liquids with high boiling point. The plasticizer forms a gell and
renders the material plastic over the desired range of temperature.
Polyvinyl chloride has high insulation resistance, good dielectric strength and mechanical toughness over a wide range
of temperatures. It is inert to oxygen and almost inert to many alkalies and acids.
Therefore, this type of insulation is preferred over VIR in extreme enviormental conditions such as in cement
factory or chemical factory. As the mechanical properties (i.e., elasticity etc.) of PVC are not so good as those of
rubber, therefore, PVC insulated cables are generally used for low and medium domestic lights and power installations.
Classification of Cables
Cables for underground service may be classified in two ways according to
(i) the type of insulating material used in their manufacture
(ii) the voltage for which they are manufactured.
(i) Low-tension (L.T.) cables — upto 1000 V
(ii) High-tension (H.T.) cables — upto 11,000 V
(iii) Super-tension (S.T.) cables — from 22 kV to 33 kV
(iv) Extra high-tension (E.H.T.) cables — from 33 kV to 66 kV
(v) Extra super voltage cables — beyond 132 kV
Single-core low tension
cable.
A cable may have one or more than one core depending upon the
type of service for which it is intended.
It may be
(i)
single-core
(ii)
two-core
(iii)
three-core
(iv)
four-core etc.
For a 3-phase service, either 3-single-core cables or three-core cable
can be used depending upon the operating voltage and load demand.
- Low voltages (up to
6600 V)
Cables for 3-Phase Service
For three phase service, either three-core cable or three single core cables may be used.
For voltages up to 66 kV, 3-core cable (i.e., multi-core construction) is preferred due to economic reasons.
However, for voltages beyond 66 kV, 3-core-cables become too large and unwieldy and, therefore, single-core
cables are used.
1.
Belted cables — upto 11 kV
2. Screened cables — from 22 kV to 66 kV
3. Pressure cables — beyond 66 kV.
Belted cables
These cables are used for voltages upto 11kV but in
extraordinary cases, their use may be extended upto 22kV
- leakage current causes local heating
Screened cables
(copper woven
fabric tape)
Firstly, the perforations in the metallic
screens
assist
in
the
complete
impregnation of the cable with the
compound and thus the possibility of air
pockets or voids (vacuous spaces) in the
dielectric is eliminated. The voids if
present tend to reduce the breakdown
strength of the cable and may cause
considerable damage to the paper
insulation.
(perforated
aluminium foil)
H-type cables.
Secondly, the metallic screens increase
the heat dissipating power of the cable.
(Designed by H. Hochstadter)
S.L. (separate lead) type cables
There is no overall lead sheath but only armouring and
serving are provided
Firstly, the separate sheaths minimise the possibility of core-tocore breakdown.
Secondly, bending of cables becomes easy due to the elimination
of overall lead sheath.
Limitations of solid type cables
(voltage limit for solid type cables is 66 kV)
As a solid cable carries the load, its conductor temperature increases and the cable compound (i.e., insulating
compound over paper) expands. This action stretches the lead sheath whichmay be damaged.
When the load on the cable decreases, the conductor cools and a partial vacuum is formed within the cable sheath. If
the pinholes are present in the lead sheath, moist air may be drawn into the cable. The moisture reduces the dielectric
strength of insulation and may eventually cause the breakdown of the cable.
Voids are always present in the insulation of a cable, the voids are formed as a result of the differential expansion and
contraction of the sheath and impregnated compound. The voids nearest to the conductor are the first to break
down, the chemical and thermal effects of ionisation causing permanent damage to the paper insulation.
Pressure cables
Oil-filled cables.
Gas pressure cables.
(for voltages beyond 66 kV)
Oil-filled cables.
The oil under pressure (it is the same oil used for
impregnation) is kept constantly supplied to the channel by
means of external reservoirs placed at suitable distances (say
500 m) along
the route of the cable.
Oil under pressure compresses the layers of paper insulation
and is forced into any voids that may have formed between
the layers.
Due to the elimination of voids, oil-filled cables can be used
for higher voltages, the range being from 66 kV upto 230 kV.
Oil-filled cables are of three types viz., single-core conductor
channel, single-core sheath channel and three-core fillerspace channels.
The oil under pressure is supplied to the channel by
means of external reservoir. As the channel is made of
spiral steel tape, it allows the oil to percolate between
copper strands to the wrapped insulation. The oil
pressure compresses the layers of paper insulation and
prevents the possibility of void formation. The system
is so designed that when the oil gets expanded due to
increase in cable temperature, the extra oil collects in
the reservoir
- channels are composed of perforated metal-ribbon tubing and are at earth potential
The oil-filled cables have three principal advantages.
Firstly, formation of voids and ionisation are avoided.
Secondly, allowable temperature range and dielectric strength are increased.
Thirdly, if there is leakage, the defect in the lead sheath is at once indicated and the possibility of earth faults is
decreased.
(However, their major disadvantages are the high initial cost and complicated system of
Laying)
Gas pressure cables.
The voltage required to set up ionisation inside a void increases as the pressure is increased.
Therefore, if ordinary cable is subjected to a sufficiently high pressure, the ionisation can be altogether eliminated.
At the same time, the increased pressure produces radial compression which tends to close any voids. This is the
underlying principle of gas pressure cables.
The construction of the cable is similar to that of an ordinary solid type except
that it is of triangular shape and thickness of lead sheath is 75% that of solid
cable.
The triangular section reduces the weight and gives low thermal resistance but
the main reason for triangular shape is that the lead sheath acts as a pressure
membrane.
The sheath is protected by a thin metal tape. The cable is laid in a gas-tight
steel pipe. The pipe is filled with dry nitrogen gas at 12 to 15 atmospheres.
The gas pressure produces radial compression and closes the voids that may
have formed between the layers of paper insulation. Moreover, maintenance
cost is small and the nitrogen gas helps in quenching any flame. However, it
has the disadvantage that the overall cost is very high.
(designed by Hochstadter, Vogal and
Bowden)
Laying of Underground Cables
There are three main methods of laying underground cables viz., direct laying, draw-in system and the solid system
Direct laying.
- trench of about 1·5 metres deep and 45 cm wide is dug
- trench is covered with a layer of fine sand (of about
10 cm thickness) and the cable is laid over this sand
bed
The sand prevents the entry of moisture from the ground and
thus protects the cable from decay
(i) It is a simple and less costly method.
(ii) It gives the best conditions for dissipating the heat generated in the cables.
(iii) It is a clean and safe method as the cable is invisible and free from external disturbances.
Direct laying.
Disadvantages
(i) The extension of load is possible only by a completely new excavation which may cost as much as the original
work.
(ii) The alterations in the cable network cannot be made easily.
(iii) The maintenance cost is very high.
(iv) Localisation of fault is difficult.
(v) It cannot be used in congested areas where excavation is expensive and inconvenient.
Draw-in system.
Solid system.
In this method of laying, the cable is laid in open
pipes or troughs dug out in earth along the cable
route.
The troughing is of cast iron, stoneware, asphalt or
treated wood.
After the cable is laid in position, the troughing is
filled with a bituminous or asphaltic compound
and covered over.
Cables laid in this manner are usually plain lead
covered because troughing affords good
mechanical protection.
The cable conductor is provided with a suitable thickness
of insulating material in order to prevent leakage current.
The path for leakage current is radial through the
insulation.
The opposition offered by insulation to leakage current is
known as insulation resistance of the cable.
For satisfactory operation, the insulation resistance of the
cable should be very high
Problems
A single-core cable has a conductor diameter of 1cm and insulation thickness of 0·4
cm. If the specific resistance of insulation is 5 × 1014 Ω-cm, calculate the insulation
resistance for a 2 km length of the cable.
The insulation resistance of a single-core cable is 495 MΩ per km. If the core diameter is 2·5 cm and resistivity of
insulation is 4·5 × 1014 Ω-cm, find the insulation thickness.
Capacitance of a Single-Core Cable
Let the charge per metre
axial length of the cable
be Q coulombs and ε be
the permittivity of the
insulation
material
between core and lead
sheath.
The work done in moving a unit positive charge from point P through a distance dx in the direction of electric field is
Ex dx.
Therefor the potential difference V between conductor and sheath
Problems
A single core cable has a conductor diameter of 1 cm and internal sheath diameter of 1·8 cm. If impregnated paper
of relative permittivity 4 is used as the insulation, calculate the capacitance for 1 km length of the cable.
Calculate the capacitance and charging current of a single core cable used on a 3-phase, 66 kV system. The cable is
1 km long having a core diameter of 10 cm and an impregnated paper insulation of thickness 7 cm. The relative
permittivity of the insulation may be taken as 4 and the supply at 50 Hz.
Under operating conditions, the insulation of a cable is subjected to electrostatic forces. This is known as
dielectric stress.
The dielectric stress at any point in a cable is the potential gradient (or electric intensity) at that point.
The electric intensity at a point x metres from the centre of
the cable is
By definition, electric intensity is equal to potential gradient.
Therefore, potential gradient g at a point x metres from the centre of cable
is
We know potential difference V between conductor and sheath is
Potential gradient varies inversely as the distance x.
Therefore, potential gradient will be maximum when x is minimum i.e., when x = d/2 or at the surface of the
conductor.
On the other hand, potential gradient will be minimum at x = D/2 or at sheath surface.
Problems
A 33 kV single core cable has a conductor diameter of 1 cm and a sheath of inside diameter 4 cm. Find the
maximum and minimum stress in the insulation
For safe working of the cable, dielectric strength of the insulation should be more than the maximum stress
The values of working voltage V and internal sheath diameter D have to be kept fixed at certain values due to
design considerations. This leaves conductor diameter d to be the only variable in exp.(i)
For low and medium voltage cables, the value of conductor diameter arrived at by this method (i.e., d = 2V/gmax) is
often too small from the point of view of current density.
Therefore, the conductor diameter of such cables is determined from the consideration of safe current density.
For high voltage cables, designs based on this theory give a very high value of d, much too large from the point of
view of current carrying capacity and it is, therefore, advantageous to increase the conductor diameter to this value.
There are three ways of doing this without using excessive copper :
Find the most economical value of diameter of a single-core cable to be used on 50 kV, single-phase system. The
maximum permissible stress in the dielectric is not to exceed 40 kV/cm.
Grading of Cables
The process of achieving uniform electrostatic stress in the dielectric of cables is known as grading of
cables.
•
Electrostatic stress in a single core cable has a maximum value (gmax) at the conductor surface and goes
on decreasing as we move towards the sheath.
•
The maximum voltage that can be safely applied to a cable depends upon gmax i.e., electrostatic stress at
the conductor surface.
•
If a dielectric of high strength is used for a cable, it is useful only near the conductor where stress is
maximum. But as we move away from the conductor, the electrostatic stress decreases, so the dielectric
will be unnecessarily over strong
Unequal stress distribution in a cable is undesirable for two reasons
•
Firstly, insulation of greater thickness is required which increases the cable size.
•
Secondly, it may lead to the breakdown of insulation.
Capacitive Grading
The process of achieving uniformity in the dielectric stress by using layers of different dielectrics is known as
capacitance grading.
Intersheath grading
In this method of cable grading, a homogeneous dielectric is used, but it is divided into various layers by placing
metallic intersheaths between the core and lead sheath. The intersheaths are held at suitable potentials which are
in-between the core potential and earth potential.
This arrangement improves voltage distribution in the dielectric of the cable and consequently more uniform
potential gradient is obtained.
Capacitance Grading
In capacitance grading, the homogeneous dielectric is replaced by a composite dielectric.
The composite dielectric consists of various layers of different dielectrics in such a manner that relative permittivity εr of
any layer is inversely proportional to its distance from the centre.
In practice, two or three dielectrics are used in the decreasing order of permittivity ; the dielectric of highest
permittivity being used near the core.
ε1 > ε2 > ε3
✓ Obviously, V > V′ i.e., for given dimensions of the cable, a graded cable can be worked at a greater
potential than non-graded cable.
✓ For the same safe potential, the size of graded cable will be less than that of non-graded cable.
Note:
As the permissible values of gmax are peak values, therefore, all the voltages in above expressions should be
taken as peak values and not the r.m.s. values.
Intersheath Grading
In this method of cable grading, a homogeneous
dielectric is used, but it is divided into various
layers by placing metallic intersheaths between
the core and lead sheath.
The intersheaths are held at suitable potentials
which are in between the core potential and earth
potential.
This arrangement improves voltage distribution
in the dielectric of the cable and consequently
more uniform potential gradient is obtained.
Maximum stress between core and intersheath 1 is
Since the dielectric is homogeneous, the maximum stress in
each layer is the same
As the cable behaves like three capacitors in series, therefore, all the potentials are in phase i.e. Voltage between
conductor and earthed lead sheath is
Disadvantages
Firstly, there are complications in fixing the sheath potentials.
Secondly, the intersheaths are likely to be damaged during transportation and installation which might result in local
concentrations of potential gradient.
Thirdly, there are considerable losses in the intersheaths due to charging currents. For these reasons, intersheath
grading is rarely used.
Problem 1
A single-core lead sheathed cable is graded by using three dielectrics of relative permittivity 5, 4 and 3 respectively.
The conductor diameter is 2 cm and overall diameter is 8cm. If the three dielectrics are worked at the same maximum
stress of 40 kV/cm, find the safe working voltage of the cable.
What will be the value of safe working voltage for an ungraded cable, assuming the same conductor and overall
diameter and the maximum dielectric stress ?
As the maximum stress in the three dielectrics is the same,
Permissible peak voltage for the cable
Thus for the same conductor diameter (d) and the same overall dimension (D), the graded cable can be operated
at a voltage (57·84 − 39·20) = 18·64 kV (r.m.s.) higher than the homogeneous cable — an increase of about
47%.
Problem 2
A single core cable of conductor diameter 2 cm and lead sheath of diameter 5.3 cm is to be used on a 66 kV, 3-phase
system. Two intersheaths of diameter 3·1 cm and 4·2 cm are introduced between the core and lead sheath. If the
maximum stress in the layers is the same, find the voltages on the intersheaths
Vph (between core and sheath)
Capacitance of 3-Core Cables
The capacitance of a cable system is much more important than that of overhead line because in cables
(i)
conductors are nearer to each other and to the earthed sheath
(ii)
they are separated by a dielectric of permittivity much greater than that of air
Although core-core capacitance Cc and core-earth capacitance Ce can be obtained from the empirical formulas for
belted cables, their values can also be determined by measurements
Measurement 1
The three cores are bunched together (i.e. commoned) and the capacitance is measured between the bunched cores and
the sheath. The bunching eliminates all the three capacitors Cc, leaving the three capacitors Ce in parallel.
If C1 is the measured capacitance,
Knowing the value of C1, the value of Ce can be determined.
Measurement 2
Two cores are bunched with the sheath and capacitance is measured between them and the third core.
This test yields 2Cc + Ce.
As the value of Ce is known from first test and C2 is found experimentally, therefore, value of Cc can be determined.
Measurement 3
If we desire CN = Ce + 3Cc
The capacitance between two cores or lines is measured with the third core free or
connected to the sheath
This eliminates one of the capacitors Ce so that if C3 is the measured capacitance,
then,
Problems 1
The capacitance per kilometre of a 3-phase belted cable is 0·3 μ F between the two cores with the third core
connected to the lead sheath. Calculate the charging current taken by five kilometres of this cable when
connected to a 3-phase, 50 Hz, 11 kV supply.
Problem 2
The capacitances of a 3-phase belted cable are 12·6 μF between the three cores bunched together and the lead
sheath and 7·4 μF between one core and the other two connected to sheath. Find the charging current drawn by the
cable when connected to 66 kV, 50 Hz supply.
UNIT-5
Distribution systems – General aspects. Kelvin's
A.C distribution – single phase and three
Techniques of voltage control and power
improvement . Introduction to Distribution loss –
trends in transmission and distribution systems
Law –
phasefactor
Recent
Distribution Systems – General Aspects
That part of power system which distributes electric power for local use is known as distribution system.
Feeders. A feeder is a conductor which connects the substation (or localised generating station) to the area where
power is to be distributed. Generally, no tappings are taken
from the feeder so that current in it remains the same
throughout. The main consideration in the design of a feeder
is the current carrying capacity
Distributor. A distributor is a conductor from which tappings
are taken for supply to the consumers. In Fig., AB, BC, CD
and DA are the distributors. The current through a distributor
is not constant because tappings are taken at various places
along its length. While designing a
distributor, voltage drop along its length is the main
consideration since the statutory limit of voltage variations is
± 6% of rated value at the consumers’ terminals
Service mains. A service mains is generally a small cable
which connects the distributor to the consumers’ terminals.
Radial feeder
Loop feeder
Parallel feeder
Classification of Distribution Systems
Nature of current: According to nature of current, distribution system may be classified as (a) d.c. distribution
system (b) a.c. distribution system
Type of construction: According to type of construction, distribution system may be classified as (a) overhead
system (b) underground system.
The overhead system is generally employed for distribution as it is 5 to 10 times cheaper
than the equivalent underground system. In general, the underground system is used at places where overhead
construction is impracticable or prohibited by the local laws
Scheme of connection. According to scheme of connection, the distribution system may be classified as (a) radial
system (b) ring main system (c) inter-connected system.
Each scheme has its own advantages and disadvantages
A.C. Distribution
1) Primary distribution system
The most commonly used primary distribution voltages are
11 kV, 6·6 kV and 3·3 kV. Due to economic considerations,
primary distribution is carried out by 3-phase, 3-wire system.
Primary Network
2) Secondary distribution system
The secondary distribution employs
400/230 V, 3-phase, 4-wire system.
The voltage between any two phases
is 400 V and between any phase and
neutral is 230 V.
The single phase domestic loads are
connected between any one phase
and the neutral, whereas 3-phase 400
V motor loads are connected across
3- phase lines directly.
Secondary Distribution
D.C. Distribution
2-wire d.c. system
3-wire d.c. system
This system of distribution consists of two wires. One
is the outgoing or positive wire and the other is the
return or negative wire. The loads such as lamps,
motors etc. are connected in parallel between the two
wires as shown in Fig. This system is never used for
transmission purposes due to low efficiency but may be
employed for distribution of d.c. power.
It consists of two outers and a middle or neutral
wire which is earthed at the substation. The voltage
between the outers is twice the voltage between
either outer and neutral wire as shown in Fig. The
principal advantage of this system is that it makes
available two voltages at the consumer terminals
viz., V between any outer and the neutral and 2V
between the outers. Loads requiring high voltage
(e.g., motors) are connected across the outers,
whereas lamps and heating circuits requiring less
voltage are connected between either outer and the
neutral
Overhead Versus Underground System
Overhead Versus Underground System
Connection Schemes of Distribution System
Radial System
The radial system
is employed only
when power is
generated at low
voltage and the
substation
is
located at the
centre of the load.
Limitations
(a) The end of the distributor nearest to the feeding point will be heavily loaded.
(b) The consumers are dependent on a single feeder and single distributor. Therefore, any fault on the
feeder or distributor cuts off supply to the consumers who are on the side of the fault away from the
substation.
(c) The consumers at the distant end of the distributor would be subjected to serious voltage
fluctuations when the load on the distributor changes. (Due to these limitations, this system is used for
short distances only)
Ring main system.
In this system, the primaries of
distribution transformers form a loop.
The loop circuit starts from the
substation bus-bars, makes a loop
through the area to be served, and
returns to the substation. The
distributors are tapped from different
points M, O and Q of the feeder through
distribution
transformers
Advantages
(a) There are less voltage fluctuations at consumer’s terminals.
(b) The system is very reliable as each distributor is fed via two feeders (distributor from point
M is supplied by the feeders SLM and SRQPONM).. In the event of fault on any section of
the feeder, the continuity of supply is maintained. For example, suppose that fault occurs at
any point F of section SLM of the feeder. Then section SLM of the feeder can be isolated for
repairs and at the same time continuity of supply is maintained to all the consumers via the
feeder SRQPONM.
Interconnected system
When the feeder ring is
energised by two or
more
than
two
generating stations or
substations, it is called
inter-connected system
Advantages
(a) It increases the service reliability.
(b) Any area fed from one generating station during peak load hours can be fed from the other generating
station. This reduces reserve power capacity and increases efficiency of the system.
Requirements of a Distribution System
Proper voltage
•
•
•
•
Voltage variations at consumer’s terminals should be as low as possible.
Low voltage causes loss of revenue, inefficient lighting and possible burning out of motors.
High voltage causes lamps to burn out permanently and may cause failure of other appliances.
The statutory limit of voltage variations is ± 6% of the rated value at the consumer’s terminals.
(Thus, if the declared voltage is 230 V, then the highest voltage of the consumer should not exceed
244 V while the lowest voltage of the consumer should not be less than 216 V.)
Availability of power on demand
Power must be available to the consumers in any amount that they may require from
time to time.
Reliability
The reliability can be improved to a considerable extent by (a) interconnected system
(b) reliable automatic control system (c) providing additional reserve facilities.
Design Considerations in Distribution System
❖ Good voltage regulation of a distribution network is probably the most important factor
responsible for delivering good service to the consumers.
Feeders. A feeder is designed from the point of
view of its current carrying capacity while the
voltage drop consideration is relatively
unimportant. It is because voltage drop in a
feeder can be compensated by means of voltage
regulating equipment at the substation
Distributors. A distributor is designed from the
point of view of the voltage drop in it. It is
because a distributor supplies power to the
consumers and there is a statutory limit of
voltage variations at the consumer’s terminals
(± 6% of rated value). The size and length of
the distributor should be such that voltage at
the consumer’s terminals is within the
permissible limits.
HT RING MAIN UNIT
Economic Choice of Conductor Size – Kelvin’s
Law (Economics of Power Transmission)
The cost of conductor material required for designing a transmission line is a very
considerable part of the total cost of a transmission line. Therefore, the determination of
proper size of the conductor for the transmission line is very important. The proper size of
the conductor for transmission line is determined by the Kelvin’s law (given by Lord Kelvin
in 1881).
The Kelvin’s law states that the most economical area of conductor is that for which the
total annual cost of transmission line is minimum.
The total annual cost of the transmission line can be divided into two parts viz. −
•Annual charges on capital cost
•Annual cost of energy wasted in conductor
Annual Charges on Capital Cost
These annual charges are on the account of interest and depreciation on the capital
cost of complete installation of the transmission line.
In case of overhead transmission system, the complete installation cost will be the annual
interest and depreciation on the capital cost of conductors, pole supports and insulators
and the cost of their erection.
Now, for the overhead transmission system, the cost of conductor is proportional to the
area of cross-section, the cost of insulator is constant, and the cost of pole (or tower)
supports and their erection is partly constant and partly proportional to cross-sectional
area of the conductor.
Therefore, the annual charges on the capital cost of an overhead transmission line can be
expressed as −
Where, P1 and P2 are the constants and a is the cross-sectional area of the conductor.
Annual Cost of Energy Wasted in Conductor
This cost is on the account of energy wasted in the conductor due to I2R losses. Suppose a
constant current in the conductor throughout the year, thus the power loss in the conductor is
directly proportional to the resistance.
Since, the resistance of the conductor is inversely proportional to the cross-sectional area of
the conductor. Hence, the energy lost in the conductor is inversely proportional to the crosssectional area.
Therefore, the annual cost of energy wasted in the conductor of overhead transmission line
can be expressed as −
Where, P3 is a constant.
Now, from the equations (1) and (2), the total annual cost of the transmission line is given
by,
Hence, the Kelvin’s law can also be stated as the most economical area of conductor is
that for which the variable part of annual charges on capital cost is equal to the annual
cost of energy wasted in the conductor.
The figure shows the graphical representation of the Kelvin’s law. Here, m is the lowest
point on the curve, which represents the most economical area of cross-section, i.e., lowest
annual cost of transmission line.
Limitations of Kelvin’s Law
In practice, the following are the limitations of the Kelvin’s law −
•The Kelvin’s law does not involve several physical factors such as safe current density,
mechanical strength, corona loss, etc.
•The conductor size determined by the Kelvin’s law may not always be practicable one
since it may be too small for the safe carrying of necessary current.
•It is not easy to estimate the energy loss in the transmission line without actual load
curves, which are not available at the time of estimation.
•The assumption that the annual charges on the account of interest and depreciation on the
capital cost is in the form of 𝑃1 + 𝑎𝑃2 is strictly speaking not true. Therefore, the interest
and depreciation on the capital cost cannot be determined accurately.
D.C. Distribution
Types of D.C. Distributors
Distributor fed at one end
❖
❖
❖
❖
Distributor fed at one end
Distributor fed at both ends
Distributor fed at the centre
Ring distributor.
The following points are worth noting in a
singly fed distributor :
(a) The current in the various sections of
the distributor away from feeding point
goes on decreasing. Thus current in
section AC is more than the current in
section CD and current in section CD
is more than the current in section DE.
Single line diagram of a d.c. distributor AB fed at the
end A (also known as singly fed distributor) and
loads I1, I2 and I3 tapped off at points C, D and E
respectively.
(b) The voltage across the loads away from
the feeding point goes on decreasing.
Thus in Fig., the minimum voltage
occurs at the load point E.
(c) In case a fault occurs on any section of
the distributor, the whole distributor
will have to be disconnected from the
supply mains. Therefore, continuity of
supply is interrupted.
Distributor fed at both ends
•
In this type of feeding, the distributor is
connected to the supply mains at both ends
and loads are tapped off at different points
along the length of the distributor.
•
The voltage at the feeding points may or
may not be equal.
•
Fig. shows a distributor AB fed at the ends
A and B and loads of I1, I2 and I3 tapped off
at points C, D and E respectively.
•
•
Here, the load voltage goes on decreasing
as we move away from one feeding point
say A, reaches minimum value and
then again starts rising and reaches
maximum value when we reach the other
feeding point B.
If a fault occurs on any feeding point of
the distributor, the continuity of supply
is maintained from the other feeding
point.
•
In case of fault on any section of the
distributor, the continuity of supply is
maintained from the other feeding
point.
•
The area of X-section required for a
doubly fed distributor is much less than
that of a singly fed distributor.
•
•
The minimum voltage occurs at some load
point and is never fixed. It is shifted with
the variation of load on different sections
of the distributor.
Distributor fed at the centre.
In this type of feeding, the centre of the
distributor is connected to the supply
mains as in Fig. It is equivalent to two
singly fed distributors, each distributor
having a common feeding point and
length equal to half of the total length.
Ring mains
In this type, the distributor is in the form of a closed
ring as in Fig. It is equivalent to a straight distributor
fed at both ends with equal voltages, the two ends
being brought together to form a closed ring. The
distributor ring may be fed at one or more than one
point.
Methods of feeding a distributor
(i)
Concentrated loading
(ii)
Uniform loading
(iii)
Both concentrated and uniform loading
The concentrated loads are those which act on particular points of the distributor.
E.g.. A common example of such loads is that tapped off for domestic use.
Distributed loads are those which act uniformly on all points of the distributor.
(Ideally, there are no distributed loads).
E.g.. Large number of loads of same wattage connected to the distributor at equal
distances.
In d.c. distribution calculations, one important point of interest is the determination of point
of minimum potential on the distributor.
The distributor is so designed that the minimum potential on it is not less than 6% of rated
voltage at the consumer’s terminals
D.C. Distributor Fed at one End—Concentrated Loading
Single line diagram of a 2-wire
d.c. distributor AB fed at one end
A and having concentrated loads
I1, I2, I3 and I4 tapped off at points
C, D, E and F respectively
A 2-wire d.c. distributor cable AB is 2 km long and supplies loads of 100A, 150A,200A and 50A
situated 500 m, 1000 m, 1600 m and 2000 m from the feeding point A. Each conductor has a resistance
of 0·01 Ω per 1000 m. Calculate the p.d. at each load point if a p.d. of 300 V is maintained at point A.
The currents in the various sections of the distributor
are :
Uniformly Loaded Distributor Fed at One End
Single line diagram of a 2-wire d.c. distributor AB fed at one
end A and loaded uniformly with i amperes per metre length.
It means that at every 1 m length of the distributor, the load
tapped is i amperes. Let l metres be the length of the
distributor and r ohm be the resistance per metre run.
Consider a point C on the distributor at a distance x metres
from the feeding point A as shown in Fig.
Then current at point C is
?
Thus, in a uniformly loaded distributor fed at one end, the total voltage drop is equal to that produced by the
whole of the load assumed to be concentrated at the middle point.
A 2-wire d.c. distributor 200 metres long is uniformly loaded with 2A/metre. Resistance of single
wire is 0·3 Ω/km. If the distributor is fed at one end, calculate :
(i)
the voltage drop up to a distance of 150 m from the feeding point
(ii) the maximum voltage drop
Distributor Fed at Both Ends — Concentrated Loading
((i) equal voltages (ii) unequal voltages)
Two ends fed with equal voltages
As we move away from one of
the feeding points, say A, p.d.
goes on decreasing till it reaches
the minimum value at some load
point, say E, and then again
starts rising and becomes V volts
as we reach the other feeding
point B.
All the currents tapped off between points A and E
(minimum p.d. point) will be supplied from the
feeding point A while those tapped off between B and
E will be supplied from the feeding point B.
Point of minimum potential
Note: The point of minimum
potential, current comes from
both ends of the distributor
The load point where the
currents are coming from both
sides of the distributor is the
point of minimum potential i.e.
point E in this case.
Two ends fed with unequal voltages.
A 2-wire d.c. street mains AB, 600 m long is fed from both ends at 220 V. Loads of 20 A,
40 A, 50 A and 30 A are tapped at distances of 100m, 250m, 400m and 500 m from the
end A respectively. If the area of X-section of distributor conductor is 1cm2, find the
minimum consumer voltage. Take ρ = 1·7 × 10−6 Ω cm
It is clear that currents are coming to
load point E from both sides i.e. from
point D and point F. Hence, E is the
point of minimum potential
Uniformly Loaded Distributor Fed at Both Ends
There can be two cases viz. the distributor fed at both ends
with (i) equal voltages (ii) unequal voltages.
Distributor fed at both ends with equal voltages
Consider a distributor AB of length l metres,
having resistance r ohms per metre run and
with uniform loading of i amperes per metre
run as shown in Fig. Let the distributor be fed
at the feeding points A and B at equal voltages,
say V volts. The total current supplied to the
distributor is il. As the two end voltages are
equal, therefore, current supplied from each
feeding point is i l/2 i.e.
Distributor fed at both ends with unequal voltages
Consider a distributor AB of length l metres having resistance r ohms per metre run and with a uniform loading of
i amperes per metre run as shown in Fig. Let the distributor be fed from feeding points A and B at voltages VA
and VB respectively.
Let the point of minimum potential C is situated at a distance x metres from the feeding point A. Then current
supplied by the feeding point A will be i x.
A 2-wire d.c. distributor AB 500 metres long is fed from both ends and is loaded uniformly at the rate of 1·0 A/metre. At feeding point A,
the voltage is maintained at 255 V and at B at 250 V. If the resistance of each conductor is 0·1 Ω per kilometre, determine :
(i) the minimum voltage and the point where it occurs
(ii) the currents supplied from feeding points A and B
Distributor with Both Concentrated and Uniform Loading
There are several problems where a distributor has both concentrated and uniform loadings.
In such situations, the total drop over any section of the distributor is equal to the sum of
drops due to concentrated and uniform loading in that section.
A 2-wire d.c. distributor AB, 900 metres long is fed at A at 400 V and loads of 50 A, 100 A and 150 A
are tapped off from C, D and E which are at a distance of 200 m, 500 m and 800 m from point A
respectively. The distributor is also loaded uniformly at the rate of 0.5 A/m. If the resistance of
distributor per metre (go and return) is 0.0001 Ω, calculate voltage (i) at point B and (ii) at point D.
Note: This problem can be solved in two stages. First, the drop at any point due to concentrated loading is found. To this is added the
voltage drop due to uniform loading.
Drops due to concentrated loads
Drops due to uniform loading
Ring Distributor
For the purpose of calculating voltage distribution, the distributor can be considered as consisting of a
series of open distributors fed at both ends. The principal advantage of ring distributor is that by proper
choice in the number of feeding points, great economy in copper can be affected
A 2-wire d.c. ring distributor is 300 m long and is fed at 240 V at point A. At point B, 150 m from A, a load of 120 A is
taken and at C, 100 m in the opposite direction, a load of 80 A is taken. If the resistance per 100 m of single conductor is
0·03 Ω, find :
(i) current in each section of distributor
(ii) voltage at points B and C
A.C. Distribution
A.C. Distribution Calculations
(i)
In case of d.c. system, the voltage drop is due to resistance alone. However, in a.c. system, the voltage drops are
due to the combined effects of resistance, inductance and capacitance.
(ii)
In a d.c. system, additions and subtractions of currents or voltages are done arithmetically but in case of a.c.
system, these operations are done vectorially.
(iii) In an a.c. system, power factor (p.f.) has to be taken into account. Loads tapped off form the distributor are
generally at different power factors. There are two ways of referring power factor viz
(a) It may be referred to supply or receiving end voltage which is regarded as the reference
vector.
(b) It may be referred to the voltage at the load point itself.
Note: Voltages, currents and impedances are expressed in complex notation and the
calculations are made exactly as in d.c. distribution.
Methods of Solving A.C. Distribution Problems
In a.c. distribution calculations, power factors of various load currents have to be
considered since currents in different sections of the distributor will be the vector sum of
load currents and not the arithmetic sum.
The power factors of load currents may be given
(i) w.r.t. receiving or sending end voltage or
(ii) w.r.t. to load voltage itself.
(i) Power factors referred to receiving end voltage.
Consider an a.c. distributor AB
with concentrated loads of I1
and I2 tapped off at points C
and B as shown in Fig. Taking
the receiving end voltage VB as
the reference vector, let
lagging power factors at C and
B be cos φ1 and cos φ2 w.r.t. VB.
Let R1, X1 and R2, X2 be the
resistance and reactance of
sections AC and CB of the
distributor.
Here, the receiving end voltage VB is taken as the reference vector. As power
factors of loads are given w.r.t. VB, therefore, I1 and I2 lag behind VB by φ1 and φ2
respectively.
(ii) Power factors referred to respective load voltages
The power factors of loads
in the previous Fig. are
referred to their respective
load voltages. Then φ1 is
the phase angle between
VC and I1 and φ2 is the
phase angle between VB
and I2. The vector diagram
under these conditions is
shown in Fig.
A single phase a.c. distributor AB 300 metres long is fed from end A and is loaded as under :
(i)
100 A at 0·707 p.f. lagging 200 m from point A
(ii) 200 A at 0·8 p.f. lagging 300 m from point A
The load resistance and reactance of the distributor is 0·2 Ω and 0·1 Ω per kilometre. Calculate the total voltage drop
in the distributor. The load power factors refer to the voltage at the far end.
A single phase distributor 2 kilometres long supplies a load of 120 A at 0·8 p.f. lagging at its far end
and a load of 80 A at 0·9 p.f. lagging at its mid-point. Both power factors are referred to the voltage at
the far end. The resistance and reactance per km (go and return) are 0·05 Ω and 0·1 Ω respectively. If
the voltage at the far end is maintained at 230 V, calculate :
(i) voltage at the sending end
(ii) phase angle between voltages at the two ends.
3-Phase AC Distribution Calculations
A 3-phase ring main ABCD fed at A at 11 kV supplies balanced loads of 50 A at 0.8 p.f. lagging at B, 120 A at
unity p.f. at C and 70 A at 0·866 lagging at D, the load currents being referred to the supply voltage at A. The
impedances of the various sections are :
Section AB = (1 + j 0·6) Ω ; Section BC = (1·2 + j 0·9) Ω
Section CD = (0·8 + j 0·5) Ω ; Section DA = (3 + j 2) Ω
Calculate the currents in various sections and station bus-bar voltages at B, C and D.
Let current in section AB be (x + j y).
∴ Current in section BC,
IBC
=(x + j y) − 50 (0·8 − j 0·6) = (x − 40) + j (y + 30)
Current in section CD,
=(x − 160) + j (y + 30)
ICD
=[(x − 40) + j (y + 30)] − [120 + j 0]
Current in section DA,
=(x − 220·6) + j (y + 65)
IDA
=[(x − 160) + j (y + 30)] − [70 (0·866 − j 0·5)]
The 3-phase, 4-wire system is widely used for distribution of electric power in commercial and industrial buildings.
The single phase load is connected between any line and neutral wire while a 3-phase load is connected across the
three lines. The 3-phase, 4-wire system invariably carries unbalanced loads
The following points may be noted carefully :
(i) Since the neutral wire has negligible resistance, supply neutral N and load neutral N′ will be at the same
potential. It means that voltage across each impedance is equal to the phase voltage of the supply.
However, current in each phase (or line) will be different due to unequal impedances.
(ii) The amount of current flowing in the neutral wire will depend upon the magnitudes of line currents and their
phasor relations. In most circuits encountered in practice, the neutral current is equal to or smaller than
one of the line currents. The exceptions are those circuits having severe unbalance.
Note: In actual practice, we never have an unbalanced 3-phase, 4-wire load. Most of the 3-phase loads (e.g.
3-phase motors) are 3-phase, 3-wire and are balanced loads. In fact, these are the single phase loads on
the 3-phase, 4-wire supply which constitute unbalanced, 4-wire Y-connected load.
Eg.1 Non-reactive loads of 10 kW, 8 kW and 5 kW are connected between the neutral and the red, yellow
and blue phases respectively of a 3-phase, 4-wire system. The line voltage is 400V. Calculate (i) the
current in each line and (ii) the current in the neutral wire
Resolving the three currents along x-axis and y-axis, we have,
Resultant horizontal component
Note: The three line currents are
of
different
magnitude
but
o
displaced 120 from one another.
The current in the neutral wire will
be the phasor sum of the three
line currents.
Resultant vertical component
As shown in Fig., current in neutral wire is
Eg.2 A 3-phase, 4-wire system supplies power at 400 V and lighting at 230 V. If the lamps is use require
70, 84 and 33 amperes in each of the three lines, what should be the current in the neutral wire ? If a 3phase motor is now started, taking 200 A from the lines at a p.f. of 0·2 lagging, what should be the total
current in each line and the neutral wire ? Find also the total power supplied to the lamps and the motor.
Lamp load alone If there is lamp load alone,
the line currents in phases R,Y and B are 70
A, 84A and 33 A respectively. These currents
will be 120o apart (assuming phase sequence
RYB)
Both lamp load and motor load
The current in each line is the phasor sum of the line currents due to lamp load and
motor load.
Active component of motor current
= 200 × cos φm = 200 × 0·2
= 40 A
Reactive component of motor current = 200 × sin φm = 200 × 0·98 = 196 A
Eg.3 The three line leads of a 400/230 V, 3-phase, 4-wire supply are designated as R, Y and B respectively. The
fourth wire or neutral wire is designated as N. The phase sequence is RYB.
Compute the currents in the four wires when the following loads are connected to this supply :
From R to N : 20 kW, unity power factor
From Y to N : 28·75 kVA, 0·866 lag
From B to N : 28·75 kVA, 0·866 lead
If the load from B to N is removed, what will be the value of currents in the four wires ?
The current IR is in phase with VRN, current IY lags behind its phase voltage VYN by cos−1
(0·866) = 30o
and the current IB leads its phase voltage VBN by cos−1 (0·866) = 30o
The current in the neutral wire will be equal to the phasor sum of the three line currents IR, IY
and IB. Resolving these currents along x-axis and y-axis,
When load from B to N removed.
The current in the neutral wire will be equal to the phasor sum of the three line currents IR, IY
and IB. Resolving these currents along x-axis and y-axis,
Methods of Voltage Control
(i) By excitation control (Generating Station)
(ii) By using tap changing transformers
(iii) Auto-transformer tap changing
(iv) Booster transformers
(v) Induction regulators
(vi) By synchronous condenser (Transmission Line)
Tirril
Regulator
Brown-Boveri Regulator
In this type of regulator, exciter field rheostat is varied continuously or in small steps instead of being first
completely cut in and then completely cut out as in Tirril regulator. For this purpose, a regulating
resistance is connected in series with the field circuit of the exciter. Fluctuations in the alternator voltage
are detected by a control device which actuates a motor. The motor drives the regulating rheostat and
cuts out or cuts in some resistance from the rheostat, thus changing the exciter and hence the alternator
voltage
Tap-Changing
Transformers
Off load tap-changing
transformer
On-load tap-changing transformer
Limitation
s
• During switching, the
impedance of transformer is
increased and there will be a
voltagesurge.
• There are twice as many
tappings as the voltage steps.
Booster
Transformer
Auto-Transformer Tapchanging
Induction
Regulators
Voltage Control by Synchronous
Condenser
The voltage at the receiving end of a transmission line can be controlled by installing specially designed
synchronous motors called synchronous condensers at the receiving end of the line.
By changing the excitation of a synchronous motor, it can be made to take a leading power factor. A
synchronous motor at no load and taking a leading power factor is known as a synchronous condenser. It is so
called because the characteristics of the motor then resemble with that of a condenser
The synchronous condenser supplies wattless leading kVA to the line depending upon the excitation of the motor.
This wattless leading kVA partly or fully cancels the wattless lagging kVA of the line, thus controlling the voltage drop
in the line. In this way, voltage at the receiving end of a transmission line can be kept constant as the load on the
system changes
Without synchronous
condenser
With synchronous
condenser
From the vector diagram, the relation between V1 and V2 is given by ;
POWER FACTOR
IMPROVEMENT
Phasor Diagram
Power Triangle
Power Factor
Meter
CAUSES OF LOW POWER
FACTOR
i. The induction motors work at a low lagging power factor at light loads and improved power factor with increased
loads.
ii. The transformers have a lagging power factor because they draw magnetizing current.
iii. Miscellaneous equipment like arc lamps, electric discharge lamps, welding equipment etc., operatesat a low
power factor.
iv. The industrial heating furnaces operate at a low lagging power factor.
v. The variation of load on the power system also causes low power factor.
EFFECTS OR DISADVANTAGES OF LOW
POWER FACTOR
•
•
•
•
•
•
Effect on transmission lines
Effect on transformers
Effect on switchgear and bus bar
Effects on generators
Effect on prime movers
Effect on existing power systems:
ADVANTAGES OF POWER FACTOR
IMPROVEMENT
• The kW capacity of the prime movers is better utilized due to decreased reactive power.
• This increases the kilowatt capacity of the alternators, transformers, and the lines.
• The efficiency of the system is increased.
• The cost per unit decreases.
• The voltage regulation of the lines is improved.
• Power losses in the system are reduced due to reduction in load current.
• Investment on generators, transformers, transmission lines and distributors per kilowatt of the load supplied is
reduced.
• kVA demand charges for large consumers undergoes reduction.
METHODS OF IMPROVING POWER
FACTOR
i. Static capacitors
ii. Synchronous condensers, and
iii. Phase advancers
Static capacitors
Calculation of Shunt Capacitor Rating
Leading kVAr (Qc) supplied by
power
factor
correction
equipment as
Advantages of shunt capacitor:
• Losses are low.
• It requires less maintenance, because there are no rotating parts.
• Easy installation.
Disadvantages of shunt capacitor:
• Less service life.
• Easily damaged due to excess of voltage.
• Its repair is uneconomical.
• Difficult to control switching position of capacitors because of removing or adding the capacitors in the
circuit for different power factors.
SYNCHRONOUS
CONDENSER
The rated kVA of
synchronous
condenser,
Advantages
• A synchronous condenser has an inherently sinusoidal wave form and the harmonic voltage does not exist.
• It can supply as well as absorb kVAr.
• The power factor can be varied smoothly.
• It allows the over-loading for short periods.
• The high inertia of the synchronous condenser reduces the effect of sudden changes in the system load and
improves the stability of the system.
• It reduces the switching surges due to sudden connection or disconnection of lines in the system
Disadvantages
• Power loss increases.
• Uneconomical for small ratings.
• It is not possible to add or take away the units and to alter the rating of the synchronous
condenser.
• The cost of maintenance is high.
• It produces noise.
Phase Advancers
There are special commutator machines, which are used to improve the power factor of the induction motor.
When the supply is given to the stator of an induction motor, it takes a lagging current. Therefore, the induction
motor has low lagging power factor. For compensating this lagging current, a phase advancer (mounted on
same shaft) is used. It supplies m.m.f. to the rotor circuit at slip frequency
Advantages
1. The lagging kVAr drawn by the motor is reduced by compensating the stator lagging current at slip frequency.
2. Wherever the use of synchronous motors is not suitable, a phase advancer can be used.
The major drawback of this kind of a compensator is that it is not economical for motors with low ratings (less
than 200 hp).