1. Why the concept of self GMD is not applicable for capacitance calculation?
The self geometric mean distance (GMD) method is not applicable for capacitance calculation
because the inner radius of the conductor is not considered in capacitance calculations. The
self GMD method is only used to evaluate inductance.
2. What is skin effect?
Skin effect is the tendency of alternating electric current (AC) to flow near the surface of a
conductor, rather than through its core. This happens when the thickness of the conductor is
comparable to or greater than the skin depth.
3. Point out any two reasons for line loss in transmission line.
In AC transmission, the primary causes are the line's resistance, reactance due to its
inductance and capacitance, and corona discharge.
4. Mention the significance of surge impedance loading.
Surge impedance loading, commonly called SIL, is a quantity used
by system operators as a benchmark to determine whether a
transmission line is acting as a capacitance that injects reactive
power (VARs) into the system or as an inductance that consumes
VARs, thus contributing to reactive power losses in the system.
Surge impedance loading (SIL) is a parameter used to evaluate the performance
of a transmission line and to minimize power losses. SIL is significant because it
helps to:
Determine if a line is consuming or producing reactive power: By
comparing SIL to the actual real power flow, system operators can determine if a
line is consuming or producing reactive power.
Optimize transmission line performance: SIL helps engineers to optimize
transmission line performance.
Minimize power losses: SIL helps engineers to minimize power losses.
Ensure uniform current and voltage: SIL is the ideal load because the
current and voltage are uniform along the line.
Ensure in-phase current and voltage waves: SIL ensures that the wave
of current and voltage is in phase.
SIL is calculated by dividing the square of transmission voltage by the line's surge
impedance.
5. Define sag.
Sag is defined as the difference in level between points of supports and the lowest point on the
conductor.
6. What is tower spotting?
Tower spotting is the process of selecting the type, height, and location of transmission line
towers.
7. Name the different types of of cables.
Low-tension (L.T.) cables — upto 1000 V
High-tension (H.T.) cables — upto 11,000 V
Super-tension (S.T.) cables — from 22 kV to 33 kV
Extra high-tension (E.H.T.) cables — from 33 kV to 66 kV
Extra super voltage cables — beyond 132 kV
8. Compare the overhead lines and underground cables.
Overhead Lines Underground Cables
An underground cable is an insulated
An overhead line is the one that uses bare
cable which is buried under the earth
conductors supported on a pole or tower
surface for power transmission and
for power transmission and distribution.
distribution.
Undergrounds cables use insulated
Overhead lines use bare conductor.
conductor.
The current carrying capacity of overhead Underground cables have lesser current
lines is comparatively more. carrying capacity.
Overhead lines are relatively less Underground cables are comparatively
expensive. expensive.
9. State kelvin’s law.
The most economical conductor size is one for which annual cost of energy loss is equal to
annual interest and depreciation on the capital investment of the condcutor material. This is
known as Kelvin's law.
10. List the objectives of FACTS.
The main objectives behind the FACTS based controllers are
1. The power transfer capability of transmission systems is to be increased.
2. The power flow is to be kept over the designated routes.
PART- B
11.(A) Derive the expression for inductance of three phase line with unsymmetrical spacing.
Inductance of Three Phase Line with Unsymmetrical Spacing but Transposed
Now consider a three phase line having three conductors but not spaced equilaterally. The problem of
finding the inductance in this case is difficult. The flux linkage and the corresponding inductance will
not be same in each phase. Due to this different inductance per phase there is imbalance in the circuit
though the currents in each phases are balanced.
The drops in the three phases due to these inductances are observed to be different. Thus at the
receiving end we will not get the same voltage.
In order to achieve balance under this case, transposition of transmission line is preferred after a
certain fixed distance. This is shown in the Fig.
The positions of the conductors are exchanged at regular interval along the line so that each conductor
occupies the original position of every other conductor over an equal distance. This exchange of
conductor positions is called transposition. Thus balance in the three phase is restored.
The average inductance of one conductor is obtained by finding the flux linkages of a conductor for
each position that is occupied during a complete cycle of transposition. Then the average flux
linkages are obtained.
Now let us find the flux linkages of conductor x which is in position 1 whereas conductor y and z are
in positions 2 and 3 respectively.
Conductor x is in position 2 whereas conductors y and z are in positions 3 and 1 respectively.
Conductor x is in position 3 whereas conductor y and z are in positions 1 and 2 respectively.
(OR)
11. (B) Determine the capacitance per phase of the doubled circuit line as shown in figure, the
diameter is 2.193cm
12.(A) draw the phasor diagram of a short and medium transmission lines and derive an
expression for voltage regulation and transmission efficiency.
Three Phase Short Transmission Line
Usually transmission of electrical power is done using 3 phase system. It can be treated as three single
phase units with each conducting unit transmitting one third of the total power. The analysis of 3
phase system is done by considering only one phase as similar conditions can be obtained in three
phase system.
The expressions obtained for regulation, efficiency in case of single phase system are valid in case of
3 phase system. Here we have to take the phase voltages for VS and VR. Similarly R and XL are
resistance and inductive reactances per phase respectively.
Consider a star connected generator supplying a balanced star connected load through a transmission
line. This is shown in the Fig.
The load p.f. has considerable effect on the regulation and efficiency of transmission line.
Consider the approximate expression for VS
This expression is valid for lagging power factor. In case of leading power factor the expression for
voltage regulation is given by,
When the load power factor is lagging or unity such that IR cos ϕR > IXL sin ϕR then the voltage
regulation is positive. It indicates that the receiving end voltage V R is less than sending end voltage
VS.
With decrease in the power factor in case of lagging loads, the voltage regulation of the line increases
for given values of VR and I.
If the load power factor is such that IXL sin ϕR > IR sin ϕR, then voltage regulation is negative which
indicates that receiving end voltage VR is more than the sending end voltage VS.
For leading power factor loads, the voltage regulation of the line decreases with decrease in p.f.
The phasor diagram for lagging load, unity p.f. load and leading p.f. load is shown in the Fig.
The angle between VR and VS is called power angle and denoted as δ.
The power delivered to the load depends upon the power factor. For single phase circuit power is
given by,
It can be seen that for a given power transmitted (P) at the voltage (VR) at the receiving end, the load
current I is inversely proportional to the load power factor, cos ϕR. If the load p.f. is low then current
and hence losses will be more which will decrease transmission efficiency.
Medium Transmission line
The short transmission lines have smaller lengths and the power transmitted by these lines are
comparatively at lower voltages. Due to this, while analysing short transmission lines, the effect of
capacitance is neglected. With increase in length and voltage of transmission line, the capacitive
effects are dominant and they can not be neglected.
The medium transmission lines are having length lying between 50 to 150 km and they operate at
voltage greater than 20 kV. Hence in making the analysis of medium transmission line, we have to
take into account the effect of capacitance for better accuracy.
The capacitance is uniformly distributed along the length of transmission line. For the simplicity in
the calculations, the line capacitance is assumed to be limped at one or more points.
1. End Condenser Method
In this method the line capacitance is lumped or concentrated near the load or at the receiving end.
This method overestimates the effect of capacitance.
In case of 3 phase systems, it is always convinient to represent a single phase instead of line to line
values. Fig. shows one phase of three phase transmission line.
The corresponding phasor diagram shown in the Fig.
Let VS = Sending end voltage
VR = Receiving end voltage
IR = Load current per phase
XL = Inductive reactance per phase
C = Capacitance per phase
cos ϕR = Receiving end power factor
As VR is the reference vector,
The load current IR is lagging behind VR by an angle of ϕR
(OR)
12. (B) A three phase , 50 HZ power transmission line has line resistance of 30 ohm and
inductive reactance of 70 ohm per phase. The capacitive susceptance is 4x10-4 mho per phase. If the
load at the receiving end is 50 MVA at 0.8 pf lagging with 132kV line voltage. Calculate (i) Voltage
and current at sending end. (ii)Regulation and (iii) Efficiency of the line for this load. USe Nominal
π method.
13.(A)What are the different types of towers? Explain any two methods in detail?
Types of Towers
The structures which are used to support the transmission lines are called towers or line supports. The
performance of transmission line depends also on the design of the towers. The towers must be
designed so that they can carry the load of the conductors with extreme loading conditions along with
the insulators.
The basic requirements of a tower are,
i) Must be mechanically strong, capable of carrying load of conductors and insulators with extreme
loading conditions.
ii) Maintenance cost must be low.
iii) Must be long lasting having longer life.
iv) Must be light in weight.
v) Must have minimum number of members.
vi) Easily accessible for erection of conductors.
vii) Must be economical.
viii) Should not affect appearance of the locality.
The various types of towers are,
1. Wooden poles : The towers use wooden structure hence simplest and cheapest. These structures
are used for low voltages and for distribution purposes only. The Fig. 4.11.1 shows a wooden pole
structure. The main limitation of such a tower is that it is elastic and tend to decay hence its life is
short.
A metallic cap is used at the top to protect it from decay. The underground portion of the tower must
be properly treated with a preservative.
2. Reinforced concrete poles : In recent times, the use of reinforced concrete poles for the towers is
popular. These towers are mechanically very strong and their maintenance cost is low. Their life is
very long. These towers are attractive in appearance. But these towers are very heavy and hence cost
of transportation is very high. These are used for systems upto 33 kV voltage level.
3. Steel tubular poles : These towers are designed for high and extra high voltage transmission lines.
Due to steel poles, these towers can be used for longer spans. These towers have low base area hence
preferred in localities where base area is an important consideration. For longer life, such towers must
be galvanized or painted regularly. The Fig. 4.11.2 shows the structure of a tower using steel tubular
poles. These poles can have different diameters throughout its length. These are used for the systems
upto 33 kV voltage level.
4. Latticed steel towers : The steel towers are very commonly used as these are suitable for the
systems with voltages higher than 33 kV and useful for long spans. Generally self supporting broad
base lattice galvanized steel towers are used. These are mechanically very strong and have longer life.
The lattice structures are light in weight and economical. The structures can be easily transported and
can be erected in very inaccessible locations. These are useful for crossing valleys, railway lines,
rivers etc. These are fabricated from galvanized angle iron sections and then transported seperately to
the site where towers are assembled and erected. The towers are finally painted to provide protection
from adverse weather conditions. The steel towers are further classified as,
i) Self supporting ii) Guyed towers and iii) Chainnet type towers.
The main disadvantage of such towers is more right of way (ROW) which is the area of land required
along the lengh of the line keeping the towers in the middle. For higher voltage lines, more right of
way is required. The Fig. 4.11.3 shows the various types of steel towers.
(OR)
13.(B) Derive an expression for sag of a line supported between two supports of the same height.
Definition of Sag
Sag : The difference in levels between point of support and the lowest point on the conductor is called
sag.
Consider an overhead transmission line suspended between the two supports A and B as shown in the
Fig. 4.2.1.
1. supports at Equal Level
Consider a conductor supported by the supports A and B which are located at same level. This is
shown in the Fig. 4.3.1.
The point O is the lowest point on the trajectory. Mathematically it can be proved that point O is at
the midspan.
Let
L = Length of span in metres
W = Weight per unit length of the conductor in kg/m
T = Tension in the conductor in kg
Consider a point P on the conductor and let point O is origin. Hence the co-ordinates of point P are
(x, y).
The length of span L is large compared to sag S hence the shape of conductor takes the form of
parabola.
Let l = Half span length = L/2
As the curve is very small due to small sag, it can be assumed that the length OP of the conductor is
same as the × co-ordinate of point P.
l (OP) = x
Now there are two external forces acting on the portion OP of the conductor,
1. The tension T.
2. The weight wx which acts at a distance of x/2 from the point O or P, as OP = x.
The tension T acts in horizontal direction at point O.
Taking moments of these two forces about point P and equating them we get,
The equation shows that the trajectory is parabolic in nature. At the support A and B, the vertical
distance y from the origin O indicates the sag S.
. At A or B, x = l = L/2 and y = S
Substituting in equation (4.3.1),
where
L = Total span length and T = Tension in conductor
14.(A) Explain the methods of grading of cables with neat diagrams and equations.
Grading of Cables
. The process of obtaining uniform distribution of stress in the insulation of cables is called grading of
cables.
The unequal distribution of stress has two effects,
1. Greater insulation thickness is required, which increases the cost and size.
2. It may lead to the breakdown of insulation.
Hence the grading of cables is done.
There are two methods of grading the cables which are,
1. Use of intersheaths for grading
2. Capacitance grading
1. Use of Intersheaths for Grading
In this method of grading, in between the core and the lead sheath number of metallic sheaths are
placed which are called intersheaths. All these intersheaths are maintained at different potentials by
connecting them to the tappings of the transformer secondary. These potentials are between the core
potential and earth potential. Generally lead is used for these sheaths as it is flexible and corrosion
resistance but as its mechanical strength is less, aluminium also can be used. Aluminium is low
weight and mechanically strong but it is much costlier than lead.
Using the intersheaths, maintaining at different potential, uniform distribution of stress is obtained in
the cables.
Consider a cable with core diameter d and overall diameter with lead sheath as D. Let two
intersheaths are used having diameter d 1 and d2 which are kept at the potentials V 1 and
V2 respectively.
The intersheaths and stress distribution is shown in the Fig. 6.8.1.
Let V1 = Voltage of intersheath 1 with respect to earth
V2 = Voltage of intersheath 2 with respect to earth
It has been proved that stress at a point which is at a distance x is inversely proportional to distance x
and given by,
gx = Q / 2π εx = k / x ... (6.8.1)
where k is constant.
So electric stress g1 between the conductor and intersheath 1 is,
g1 = k1 / x where k1 = Constant
Now potential difference between core and the first intersheath is V – V1.
The potential difference between intersheath 2 and outermost sheath is V 2 only as potential of
intersheath is maintained at V2 with respect to earth.
Choosing proper values of V1 and V2, glmax, g2max etc. can made equal and hence uniform distribution
of stress can be obtained.
2. Capacitance Grading
The grading done by using the layers of dielectrics having different permittivities between the core
and the sheath is called capacitance grading.
In intersheath grading, the permittivity of dielectric is same everywhere and the dielectric is said to be
homogeneous. But in case of capacitance grading, a composite dielectric is used.
Let d1= Diameter of the dielectric with permittivity ε1
and D = Diameter of the dielectric with permittivity ε2 This is shown in the Fig. 6.8.3.
The stress at a point which is at a distance x is inversely proportional to the distance x and given by,
gx = Q / 2π εx
Hence the stress at any point in the inner dielectric is,
g1 = Q / 2π ε1x
Similarly the dielectric stress in the outer dielectric is,
g2 = Q / 2π ε2x
Hence the total voltage V can be expressed as,
The stress is maximum at surface of conductor i.e. x = d/2.
(OR)
14.(B) An 11 kV, three phase, underground feeder of 2 km long uses three single core cables. The diameter of
each conductor is 28mm and an insulation thickness of 4.4mm and the relative permittivity of 4. Determine
(i) Capacitance of the cable per phase, (ii) Charging current per phase, (iii) Total charging KVAR and
(iv)Determine loss per phase if the power factor of Unloaded cable is 0.04
15.(A) What is Ring main Distributor? Find the current supplied at points A and B of the ring main distributor
shown in fig. the loads are at unity power factor.
In this system, the feeder covers the whole area of the supply in the ring fashion and finally
terminates at the substation from where it is started. The feeder is in closed loop form and looks like a
ring hence the name given to the system is ring main distribution system.
(OR)
15.(B)Describe the different types of FACTS controllers with necessary diagram.
FACTS Devices
The various FACTS devices are as given below.
1. Static Synchronous Compensator (STATCOM)
2. Static Synchronous Generator (SSG)
3. Static VAR Compensator (SVC)
4. Thyristorized switched or controlled reactor (TSR/TCR)
5. Thyristor switched capacitor
6. Static VAR Generator or absorber (SVG)
7. Static VAR System (SVS)
8. Thyristor Controlled Braking Resistor (TCBR)
9. Static Synchronous Series Compensator (SSSC)
10. Interline Power Flow Controller (IPFC)
11. Thyristor Controlled or switched series capacitor or series reactor (TCSC/TSSC/TCSR/TSSR)
12. Unified Power Flow Controller (UPFC)
13. Thyristor Controlled Phase Shifting Transformer (TCPST)
14. Interphase Power Controller (IPC)
15. Thyristor Controlled Voltage Limiter (TCVL)
16. Thyristor Controlled Voltage Regulator (TCVR)
1. Static Shychronous Compensator (STATCOM)
It is shunt connected static VAR compensator whose capacitive or inductive output current can be
controlled independently of the ac system voltage. It is shown in the Fig. 7.29.1.
In case of voltage source converter, the ac output voltage is controlled in such a way as the proper
reactive current will flow for any bus voltage. The dc capacitor voltage is automatically adjusted as
per the requirement so that it acts as voltage source for the converter. The harmonics in the system
can be absorbed by designing STATCOM as an active filter.
It is a three phase inverter driven by voltage across capacitor and the three phase output voltages are
in phase with ac system voltages. The difference in the amplitudes of the voltages gives how much
current flows. The reactive power and its polarity can be changed by controlling the voltage. The
performance of STATCOM is better than SVC.
With depression in voltage, STATCOM will still supply high reactive power by using its over current
capability. The large capacitor present acts as storage device and can continue to deliver some energy
for short duration just like synchronous condenser.
The use of STATCOM needs Gate Turn Off (GTO) thyristors which are costly as compared to
normal thyristors.
2. Static VAR Compensator (SVC)
In STATCOM, converters are used while in SVC thyristors without gate turn off capability are used.
It is shunt connected static VAR generator or absorber. The output of SVC is adjusted to control
capacitive or inductive current in order to control or maintain certain parameters, normally magnitude
of bus voltage of the power systems. A basic model of SVC is shown in the Fig. 7.29.2.
The separate equipments are present in SVC for lagging and leading VARs. It is a low cost substitute
for STATCOM. In STATCOM, the most reactive power that is delivered is product of voltage and
current whereas in case of SVC, it is the square of voltage divided by the impedance. The reactive
power capability steeply falls off as a function of square of voltage in this case.
3. Thyristor Controlled Series Capacitor (TSCS)
It is a capacitive reactance type of compensator consisting of a series capacitor bank connected in
parallel with a thyristor controlled reactor so as to provide smooth variable capacitive reactance.
It is important type of FACTS controller based on thyristors without the gate turn off capability. It is
shown in the Fig. 7.29.3.
The Thyristor Controlled Reactor (TCR) is connected across a series capacitor. When the firing angle
of TCR is 180°, the reactor is nonconducting and the series capacitor has its normal impedance. If the
firing angle is decreased from 180°, the capacitive impedance increases. The reactor is fully
conducting when the firing angle is 90°. In this case, the total impedance is inductive as the designed
value of reactor impedance is less than the impedance of series capacitor. The TCSC helps in limiting
fault current for a firing angle of 90°.
For geting best performance from TCSC, it has different sized smaller capacitors or several equal
capacitors instead of a single large unit.
4. Unified Power Flow Controller (UPFC)
It is a combination of Static Synchronous Compensator (STATCOM) and Static Synchronous Series
Compensator (SSSC). These two are coupled through a dc link and allows bidirectional flow of real
power between series output terminals of SSSC and shunt output terminals of the STATCOM. These
can be controlled to provide real and reactive series line compensation without an external electrical
energy source. It is shown in the Fig. 7.29.4.
The meaning of UPFC is angular, unconstrained injection of series voltage to control selectively the
line voltage, impedance and angle. Alternatively, real and reactive power flow in the line is
controlled. The independent controllable shunt compensation is also provided by UPFC.
UPFC can be made more effective by connecting additional stroage shunt as a super conducting
magnet connected to the dc link through the electronic interface. The controlled exchange of real
power is possible in case of UPFC.
PART- C
16.(A) A three phase circuit line consists of 7/4-5 mm hard drawn copper conductors. The arrangements of
the conductors is shown in fig. the line is completely transposed. Calculate inductive reactance per phase per
km of the system.
(OR)
16.(B) A 2 wire DC Street mains AB, 600 m long if fed from both ends at 220V. Loads of 20A, 40A,
50 A and 30 A are tapped at distances of 100m, 250m, 400m and 500m from the end A respectively.
If the area of cross section of distributor conductor is 1 square centimeter, find the minimum
consumer voltage, take ρ = 1.7x 10-6 Ω- cm.
